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YANPEI LIU INTRODUCTORY MAP THEORY Kapa & Omega, Glendale, AZ USA 2010 Yanpei LIU Institute of Mathematics Beijing Jiaotong University Beijing 100044, P.R.China Email: ypliu@bjtu.edu.cn Introductory Map Theory Kapa & Omega, Glendale, AZ USA 2010 This book can be ordered in a paper bound reprint from: Books on Demand ProQuest Information & Learning (University of Microfilm International) 300 N.Zeeb Road P.O.Box 1346, Ann Arbor MI 48106-1346, USA Tel:1-800-521-0600(Customer Service) http://wwwlib.umi.com/bod Peer Reviewers: L.F.Mao, Chinese Academy of Mathematics and System Science, P.R.China. J.L.Cai, Beijing Normal University, P.R.China. H.Ren, East China Normal University, P.R.China. R.X.Hao, Beijing Jiaotong University, P.R.China. Copyright 2010 by Kapa & Omega, Glendale, AZ and Yanpei Liu Many books can be downloaded from the following Digital Library of Science: http: //www.gallup.unm.edu/~smarandache/eBooks-otherformats.htm ISBN: 978-1-59973-134-6 Printed in America Preface Maps as a mathematical topic arose probably from the four color problem|Bir1, Orel] and the more general map coloring problem|HiC1, Rinl, Liu11] in the mid of nineteenth century although maps as poly- hedra which go back to the Platonic age. I could not list references in detail on them because it is well known for a large range of readers and beyond the scope of this book. Here, I only intend to present a comprehensive theory of maps as a rigorous mathematical concept which has been developed mostly in the last half a century. However, as described in the book[Liu15| maps can be seen as graphs in development from partition to permutation and as a basis extended to Smarandache geometry shown in [Mao3-4]. This is why maps are much concerned with abstraction in the present stage. In the beginning, maps as polyhedra were as a topological, or ge- ometric object even with geographical consideration|Kem1]. The first formal definition of a map was done by Heffter from [Hef1] in the 19th century. However, it was not paid an attention by mathematicians until 1960 when Edmonds published a note in the AMS Notices with the dual form of Heffter’s in |Edm1,Liu3]. Although this concept was widely used in literature as [Liul—2, Liu4—6, Rin1-3, Stal-2, et all, its disadvantage for the nonorientable case involved does not bring with some convenience for clarifying some related mathematical thinking. Since Tutte described the nonorientability in a new way [Tut1- 3], a number of authors begin to develop it in combinatorization of continuous objects as in [Lit1, Liu7-10, Vin1-2, et al]. The above representations are all with complication in construct- ing an embedding, or all distinct embeddings of a graph on a surface. lv Preface However, the joint tree model of an embedding completed in recent years and initiated from the early articles at the end of seventies in the last century by the present author as shown in [Liul—2] enables us to make the complication much simpler. Because of the generality that an asymmetric object can always be seen with some local symmetry in certain extent, the concepts of graphs and maps are just put in such a rule. In fact, the former is corresponding to that a group of two elements sticks on an edge and the later is that a group of four elements sticks on an edge such that a graph without symmetry at all is in company with local symmetry. This treatment will bring more advantages for observing the structure of a graph. Of course, the later is with restriction of the former because of the later as a permutation and the former as a partition. The joint tree representation of an embedding of a graph on two dimensional manifolds, particularly surfaces(compact 2-manifolds without boundary in our case), is described in Chapter I for simplifying a number of results old and new. This book contains the following chapters in company with re- lated subjects. In Chapter I, the embedding of a graph on surfaces are much concerned because they are motivated to building up the theory of abstract maps related with Smarandache geometry. The second chapter is for the formal definition of abstract maps. One can see that this matter is a natural generalization of graph em- bedding on surfaces. The third chapter is on the duality not only for maps themselves but also for operations on maps from one surface to another. One can see how the duality is naturally deduced from the abstract maps described in the second chapter. The fourth chapter is on the orientability. One can see how the orientability is formally designed as a combinatorial invariant. The fifth chapter concentrates on the classification of orientable maps. The sixth chapter is for the classification of nonorientable maps. From the two chapters: Chapter V and Chapter VI, one can see Preface v how the procedure is simplified for these classifications. The seventh chapter is on the isomorphisms of maps and pro- vides an efficient algorithm for the justification and recognition of the isomorphism of two maps, which has been shown to be useful for de- termining the automorphism group of a map in the eighth chapter. Moreover, it enables us to access an automorphism of a graph. The ninth and the tenth chapters observe the number of distinct asymmetric maps with the size as a parameter. In the former, only one vertex maps are counted by favorite formulas and in the latter, general maps are counted from differential equations. More progresses about this kind of counting are referred to read the recent book|Liu7] and many further articles|«Bax1, BeG1, CaL1-2, ReL1-3, etc]. The next chapter, Chapter XI, only presents some ideas for ac- cessing the symmetric census of maps and further, of graphs. This topic is being developed in some other directions|KwL1-2] and left as a subject written in the near future. From Chapter XII through Chapter XV, extensions from basic theory are much concerned with further applications. Chapter XII discusses in brief on genus polynomial of a graph and all its super maps rooted and unrooted on the basis of the joint tree model. Recent progresses on this aspect are referred to read the articles [Liu13-15, LiP1, WaL1-2, ZhL1-2, ZuL1, etc]. Chapter XIII is on the census of maps with vertex or face par- titions. Although such census involves with much complication and difficulty, because of the recent progress on a basic topic about trees via an elementary method firstly used by the author himself we are able to do a number of types of such census in very simple way. This chapter reflects on such aspects around. Chapter XIV is on graphs that their super maps are particularly considered for asymmetrical and symmetrical census via their semi- automorphism and automorphism groups or via embeddings of graphs given [Liu19, MaL1, MaT1, MaW1, etc]. Chapter XV, is on functional equations discovered in the census of a variety of maps on sphere and general surfaces. Although their vi Preface well definedness has been done, almost all of them have not yet been solved up to now. Three appendices are compliment to the context. One provides the clarification of the concepts of polyhedra, surfaces, embeddings, and maps and their relationship. The other two are for exhaustively calculating numerical results and listing all rooted and unrooted maps for small graphs with more calculating results compared with those appearing in [Liu14], |Liu17] and |Liu19]. From a large amount of materials, more than hundred observa- tions for beginners probably senior undergraduates, more than hun- dred exercises for mainly graduates of master degree and more than hundred research problems for mainly graduates of doctoral degree are carefully designed at the end of each chapter in adapting the needs of such a wide range of readers for mastering, extending and investigat- ing a number of ways to get further development on the basic theory of abstract maps. Although I have been trying to design this book self contained as much as possible, some books such as [DiM1], [Mss1] and [GaJ1] might be helpful to those not familiar with basic knowledge of permutation groups, topology and computing complexity as background. Since early nineties of the last century, a number of my former and present graduates were or are engaged with topics related to this book. Among them, I have to mention Dr. Ying Liu[LpL1], Dr. Yuan- qiu Huang[|HuL1], Dr. Junliang Cai[CaL1-2], Dr. Deming Li|LiL1], Dr. Han Ren[ReL1-3], Dr. Rongxia Hao[HaC1, HaL1], Dr. Zhaox- iang Li|LiQ1-2], Dr. Linfan Mao|MaL1, MaT1, MaW1], Dr. Er- ling Wei[WiL1-2], Dr. Weili He[HeL1], Dr. Liangxia Wan[WaL1-2], Dr. Yichao Chen|CnL1, CnR1], Dr. Yan Xu|XuL1-2], Dr. Wen- zhong Liu[LwL1-2]|, Dr. Zeling Shao[ShL1], Dr. Yan Yang[YaL1-2], Dr. Guanghua Dong{DoL1], Ms. Ximei Zhao[ZhL1-2], Mr. Lifeng Li[LiP1], Ms. Huiyan Wang[WgL1], Ms. Zhao Chai[CiL1], Mr. Zi- long Zhu[ZuL1], et al for their successful work related to this book. On this occasion, I should express my heartiest appreciation of the financial support by KOSEF of Korea from the Com?MaC (Com- Preface vu binatorial and Computational Mathematics Research Center) of the Pohang University of Science and Technology in the summer of 2001. In that period, the intention of this book was established. Moreover, I should be also appreciated to the Natural Science Foundation of China for the research development reflected in this book under its Grants(60373030, 10571013, 10871021). Y.P. Liu Beijing, China Jan., 2010 Contents POAC. ca scc eames ENVIO o aM UpR NEUEN EDU a RS iii Chapter I Abstract Embeddings ............................ 1 [I.1 Graphs and networks use a a ag ace xr aC cb doin atrae dea ce a os 1 L2 AG MERMTTT-—-——————— 9 iin Cds 25 s68 eer ien t duri dod bb 16 1.4 Abstract representation ............ 0.00 ccc cece e nee 22 L5 Smarandache 2-manifolds with map geometry ......... 27 Activities on Chapter I .............................sess 32 D5 Observations used Cao ater ido c aac eae aca edd ce c a 32 Er D e E E E E E 34 IS Reséearchês:srirssecrsisriisirrid Cada bed OR pound 37 Chapter II Abstract Maps ..................... 00 cee eee eee Al [ll Ground S016. eras e rdeda ive ane an iaiia euet dx 41 I2 Basi permutations essesi qaa Y asit ede d abe daa dara 43 [I.3 Conjugate BIO. oo a9 RO au Que ROPA ag RR 46 I4 TEGBSILTVEOBXIOITIS 5 uec dk Ier RC er Ip E E E 40 IL5 Included angles.. n... qu schemi ca ide ro ata? enrich 55 Activities on Chapter II ................................ 57 [I.6 Observations. tet atias.ose eae eni at dar datio eed ad eee taree 57 WD ee hee atu et ee went TTE E I TOI TT TT 58 ILS ResearchESeereeisreredenionit ernaia na eta oy Eia 59 Chapter III Duality .s.cucidese eases lotti hr o XCR e CR eoe d 65 DELI Dual DS senesini anene tbi ru ade arsit ad padre RA 65 X Contents II.2 Deletion of an edge us uomexexa xu erae ed d a x eb a redes 72 TILES Addition of an CACC. 16.20. deGeeuktewdetetideauatadadd 85 HIA Basie transformati oss siniesigivastacegaesatxnadeeaw’ 96 Activities on Chapter III ............................se. 98 [II.5 Observations. uc daa s tin prt n a qitod Sr da Oa ora 98 ILU EXErCigES s sodatn Exe E RM vd sees Red Padi vd 99 [HT TRGSOREC DO e erscinas quac tanya dee dp ac opui a vay det 101 Chapter IV — Orientability .......................Lsssessss 103 IN d Ornéntation ua sica aca don at Fo a deb S fec rae t har 103 1352 Basic GuivaleliCeus s quem aua t ri o Redon he 30 9 ae cay 107 IS 9 Euler characteristic ssa eene Sa a epe RX eadavassdea cs 113 IV.4 Pattern examples ssa inane aan eave SO OR RR) 116 Activities on Chapter IV ...........................s. 119 Es QC NEST T ET OE OT 119 T5. Exercises ote ries dui ice diesen d n o oni depu m bo aa dic 120 IV rc Researches: eo cem ekle EO RC RC E CP ec e db oes 122 Chapter V Orientable Maps ....................... suse. 124 V.1 Butterflies usus tase dard i b aee RR LR eg oar rat 124 352 Simplified DubbertliQg. viua dasiooben ER ERA I eee ao does 126 Veo Reduced rules iuoaaueus d e? Vie waa nir ara q apto edad 130 V.4 Ornentable principled) order hore EC PpCOO e a 134 Veo Qrentable PONS: ziehe pda por dex Edu eoe 137 Activities on Chapter V isse ek xa rarae das 139 V.6 Observations Lia auis cte dot eoo herd eet tul 139 Quid 169 1: er araa raS a N OO OUT LOTO Dll S 140 A S Researches 5.4500 pupa ibt doce dcl opui dc ape Rice EXE LUCR 142 Chapter VI Nonorientable Maps ........................ 145 BAN IZ IU OTETTOTTEEOQTE TORTE 145 Contents xi VI.2 Simplified DAIIOS. :scwsaeyvsdsn esaecue i aue ades xad 149 VI.3 Nonorientable rules iua ua qe iratiag ea eda ow oer as 151 VLA Nonorientable principles «usos cese mr Rs 156 VI.5 Nonorientable perius, iussi eerte e 157 Activities on Chapter VI ........................ssess 159 VI:S Observations ks aa od Race Pda ade e d rada qr ibas d 159 VLO ExerciSES PT 160 DANN. 01: Mm 162 Chapter VII Isomorphisms of Maps .................... 164 MIL EACOmiti AUIID uoces aw bolo DE Ea d acl A ODE A oto 164 VIIL.2 Isomorphism OHODPOEL, sa dai o S rea ed ea dura 168 VILS Hee ni MTM" 172 MESS Justification ecd param nha a dci ru dd RE RE ELE ew ak 177 VIL Pattern examples... ci reeacea nde rk OR P S e nian 180 Activities on Chapter VII ..........................se. 185 VIO ODSOPVOUOPIB o uada ex tod ERO bar acatecta Pars qnos 185 VIDT Ss cece usb piedad mb dez vua tite Ru acr prp ard; 186 VILS: Researches ceo zd vxo arceat XR Re xo Pici c aao 188 Chapter VIII Asymmetrization ......................... 190 VLE AG OHIOTISDESHEIS eo ee eta pha etam irt bereit tnb 190 VIIL2 Upper bound of group order............LsLssuuueuue. 193 VIIL3 Determination of the group...................000- 196 VULA ROODE S mesire minaaa usec natn ee etui dud subi 201 Activities on Chapter VIII .........................s. 206 MILL QOS VA HONS s us a in iq e Perna c epa re opa eg s 206 MILIA BISOBCISEBS s cued t dp pe VC RRPER C rc Pe Re Pee E Rs 207 ATI. F6ear hes yak esee S Roe ee acd Ren eem Eo s 209 xli Contents Chapter IX Rooted Petal Bundles ....................... 212 IX.1 Orientable petal bundles... n.n.. Lor re 212 [IX.2 Planar pedal bundles «544» 9223 e re heard 2IT IX.3 Nonorientable pedal bundles....................sse. 220 IX.4 The number of pedal bundles....................... 226 Activities on Chapter IX .............................. 230 [X.5 Observations. ........ ununun nananana nananana 230 [X.6 Exercises 410g Da mesi vases be dat ESE KESER ESNA NEE ERDRE Ea 231 IX. 0 Resear Chics. cessie era PX VOCE HOER VICES E SER RUP ex ra Ro 202 Chapter X Asymmetrized Maps ......................... 235 X.1 Orientable equation........... 0... cece eee nes 235 X.2 Planar rooted I DES e io pst duuybur dehet dri rrara 243 A.9 Nonorientable equation :2. 2 voor ter RE RR RR RA RR 250 X.4 Gross GuabtlOfb. v au Sog serasa saunu abes don debo OU 255 Ao The number of rooted mapa-ssese sat ex bns 258 Activities on Chapter X ............................... 261 X.6 Observations TEE 261 bm cut RRRREPR 262 X8 Researches s conc agar eta esr acl Papi Sept edic abd RR i 265 Chapter XI Maps with Symmetry ....................... 268 AL] Symmetric relation |. eovescs ones hoa E hee oO hk 268 XI An annHlcablOleessensequad rest Ibsiea ora RE E d. 270 XI.3 Symmetric principle >... cu antowy aire, dane aca acne DR 212 XI.4 General examples........... 0... cc ccc eee ee ee 274 Activities on Chapter XI .........................Lsss. 278 XI.5 CDS GB aeuo ert leger nis Ee enr Pb Ehe a d 278 RAD PRMNNCE T PITT 279 AXI T ReséarchéS. Lese wie chee deck arate a weet e Aten Po POR weld 280 Contents xiii Chapter XII Genus Polynomials ......................... 282 AIL1 Associate surfaces 1i. cessere ert eres 282 AIL Layer division of a surface Lec o pet E v ERR 285 AL Handle polymomials 22a pide ded debo er eode 289 XII.4 Crosscap polynomials ua seien eame coun Fee ec es n 290 Activities on Chapter XII ..........................ss 292 XIL.5 Observations. bore nd ducentas pb ROS Roe RU Id dod 292 PU Gores PRINS actu anes OTI IT LITT 293 X.T CSC 001. ss sec crnctndcteenaakiensdk tated earners 294 Chapter XIII Census with Partitions .................... 297 PTL Planted trees aue dedere dace asm dida Soit cine: 297 Al 2 Hamiltonian cubic Mapes discs ee bo Le rh RR ARORAA 305 ATTE S Halin maps aereo es esie qe REA Era tog adie ele POR 307 XIIL4 Biboundary inner rooted maps ................sue 310 II General maps ce eectesce qu tad biciaces VAL CARA E aces 315 XIHI.6 Pane ean: acus acr ai ended he raton Rc p Iu weg 917 Activities on Chapter XIII ............................ 323 AIL. E TODSEPUAUIONSS osa arra ta da wires Strap e CR pro pos 323 RAULS BXGPCISOS sad queas P Eh POS na ian dator mda E E 324 AULD Researches Lu doses paca is aa xe x ORC IE OR re ecl 325 Chapter XIV Super Maps of a Graph ................... 326 XIV.1 Semi-automorphisms on a graph .................. 326 XIV.2 Automorphisms on a graph scs ovesoerriecet Ene dn 329 XIV.3 Relationships oio qd eic doe P Desc RR Ra m ORAL 332 XIV.4 Nonisomorphic super maps................00000ee- 334 XIV.5 Via rooted super MANS .2iiaacdccesnerganeascae dares 336 Activities on Chapter XIV .........................s 341 XIV .G Observallofis.l.2 26r red e Ee x esent 341 xlv Contents PAN AC ECGs S s casses Ped y ebd e me dd Edd SE Se ds 342 XIV:8 Researches Pm 334 Chapter XV Equations with Partitions ................. 345 XV.1 The meson functional xxv esee xe ees 345 XV.2 General maps on the sphere.............Luuuuusuuus 350 XV.3 Nonseparable maps on the sphere.................. 303 XV.4 Maps without cut-edge on furfaces................. 307 XV.5 Eulerian maps on the sphere.................00000- 361 XV.6 Eulerian maps on Surfaces... acceso etre 365 Activities on Chapter XV ........................ssss. 370 XV.7 CbservatiOfigesscie ee d va e Deren Od Re ia e 370 DoW oy BOT o MMC orl beg X1 040 state ened cian ener ai he ahaoe 372 Appendix I Concepts of Polyhedra, Surfaces, Embeddings and Maps 52 eme 374 DT Pol Waite care eae bol eae paren aE 374 Pa ees aver ono Oe owe oo eee ee 377 Ax.I.3 To DOOOLHE Sos w id dade aar Quiet go M e co ool eed 381 CY Ri PH" 384 Appendix II Table of Genus Polynomials for Embeddings and Maps of Small Size............ 389 Ax.IL1 Triconnected cubic graphs...................00005 389 A IL2- Dot Sie xoc iocos Suc sex dea tances hoes nd eu es 398 AILES WHOIS Ss wdc cre veces pear kese EREE E RES oe 401 oe TEE Link Dultidlóges cs exercere tan caet bce ER prati e 403 Ax.IL5 Complete bipartite graphs....................0005 405 Ax.ILO Complete graphs. «ce ehe h ge teens ntn 407 Contents XV Appendix III Atlas of Rooted and Unrooted Maps for Small €Papligosad esses x AR aad rA ARRA 409 Ax.IIL1 Bouquets Bm of size 4> M> 1................. 409 Ax.IIL2 Link bundles Lm of size 7 > m > 3....... ees 415 Ax.IIL3 Complete bipartite graphs Km,n, 4 > m,n > 3...432 Ax. ILA Wheels W, of order 6 > n > 4... severe 437 Ax 111.5 Complete graphs Kn, 5 > m > 4... esee 447 Ax.III.6 Triconnected cubic graphs of size in [6,15]....... 452 Biblogřaphy ssuresansde s heeatenoa sis du adbn Sa du EDSR sa ad uar dtd 472 Terminology cr kiant boa OnE vp" 480 Chapter I Abstract Embeddings e A graph is considered as a partition on the union of sets obtained from each element of a given set the binary group B = {0,1} sticks on. e A surface, i.e., a compact 2-manifold without boundary in topol- Ogy, ls seen as a polygon of even edges pairwise identified. e An embedding of a graph on a surface is represented by a joint tree of the graph. A joint of a graph consists of a plane extended tree with labelled cotree semi-edges. Two semi-edges of a cotree edge has the same label as the cotree edge with a binary index. An extended tree is compounded of a spanning tree with cotree semi-edges. e Combinatorial properties of an embedding in abstraction are par- ticularly discussed for the formal definition of a map. L1 Graphs and networks Let X bea finite set. For any x € X, the binary group B = {0,1} sticks on x to obtain Bx = {x(0), z(1)). x(0) and z(1) are called the ends of x, or Bx. If Bx is seen as an ordered set (x(0), z(1)), then 2 Chapter I Abstract Embeddings x(0) and z(1) are, respectively, initial and terminal ends of x. Let Ade vo Hn (1.1) rcx i.e., the disjoint union of all Bx, x € X. X is called the ground set. A (directed) pregraph is a partition Par= (Pi, Po, ---} of the ground set AX ,1.e., A= ye, (1.2) i>1 Bx (or (x(0), x(1))), or simply denoted by z itself, x € X, is called an (arc) edge and P;, i > 1, a node or vertex. A (directed) pregraph is written as G — (V, E) where V —Par and E = B(X) ={Bal|x € X] (= {(x(0), x(1))] € Xj). If X is a finite set, the (directed) pregraph is called finite; otherwise, infinite. In this book, (directed) pregraphs are all finite. If X = Ø, then the (directed) pregraph is said to be empty as well. An edge (arc) is considered to have two semiedges each of them is incident with only one end (semiarcs with directions of one from the end and the other to the end). An edge (arc) is with two ends identified is called a selfloop (di-selfloop); otherwise, a link (di-link). If t edges (arcs) have same ends (same direction) are called a multiedge (multiarc), or t-edge (t-arc). Example 1.1 There are two directed pregraphs on X = (x], 1 B. Par; = ((z(0)), {x(1) }}; Par; = {{x(0), x(1)}}. They are all distinct pregraphs as well as shown in Fig.1.1. L1 Graphs and networks 3 Xx T Parı Pars Fig.1.1 Directed pregraphs of 1 edge Further, pregraphs of size 2 are observed. Example 1.2 On X = {21,22}, the 15 directed pregraphs are as follows: Par, = ((1(0)). 00100). {2(0)}, {z2(1)}}; Par, = {{21(0), x1(1)}, {x2(0)}, 122(1) 1: Pars = {{21(0), x2(0)}, {z1(1)}, 102(1) 1: Par, = {{21(0), xo(1)}, {z1(1)}, 122(0) 1; Pars = {{21(0)}, {21(1), £2(0)}, {za(1) 1: Pars = {{21(0)}, {21(1), 29(1) }, 122(0) 1; Par; = {{21(0)}, {21(1)}, {z2(1), z2(0)}}; Pars = {{1(0), x1(1), z2(0)}, 122(1) } }: Parg = {{1(0), 21(1), z2(1)}, 122(0) fF: Pario = {{21(0), 22(0); z2(1)}, {£1(1)}}; Pari = {{21(0)}, {21(1), %2(0), z2(1)}}; Paria = {{x1(0), 21(1), va(0), vo(1) }}; Paris = {{21(0), x1(1)}, {22(0), vo(1) 3: Paria = {{21(0), x2(0)}, {21(1), a(1) fF: Paris = {{21(0), x2(1)}, {1(1), v2(0) fF. Among the 15 directed pregraphs, Pars, Par4, Pars; and Pare are 1 pregraph; Pars and Parg are 1 pregraph; Parjo and Par; are 1 pregraph; Par;4 and Parı5 are 1 pregraph; and others are 1 pregraph each. Thus, there are 9 pregraphs in all(as shown in Fig.1.2). 4 Chapter I Abstract Embeddings T T y y { ] x 1 2 Par, Paro Para Tı T2 "o we I1 T2 Par4 Pars Parę X1 V V Parz Parg Parg x L] M? I Pario Pari Parj2 x zl 7 2i qi Paria Pari4 Paris Fig.1.2 Directed pregraphs of 2 edges Now, Par= (P4, Po,---} and B are, respectively, seen as a map- ping z => P, z € P, i > 1 anda mapping z +> Z, Z Æ z, {z,z} € B(X). The composition of two mappings a and (3 on a set Z is defined L1 Graphs and networks 5 to be the mapping (aß)z = U ay, z € Z. (1.3) yeBz Let Vypargy be the semigroup generated by Par=Par(X) and B = B(X). Since the mappings a =Par and B have the property that y € az & z € ay, it can be checked that for any z, y € B(X), what is determined by dy € Vipargy, 2 € vy is an equivalence. If B( X) itself is a equivalent class, then the semi- group V (pa; g} is called transitive on X' = B(X). A (directed)pregraph with V pag transitive on X is called a (directed ) graph . A (directed)pregraph G = (V, E) that for any two vertices u,v € V, there exists a sequence of edges e1, €2,---,e, for the two ends of e;, i = 2,3,---,s—1, are in common with those of respective e;_; and e;44 where u and v are, respectively, the other ends of e; and eg, is called connected . Such a sequence of edges is called a trail between u and v. A trail without edge repetition is a walk. A walk without vertex repetition is a path. A trail, walk, or path with u = v is, respectively, a travel, tour, or circuit. Theorem 1.1 A (directed)pregraph is a (directed)graph if, and only if, it is connected. Proof Necessity. Since Par’ = Par, k > 1, and B* = B, k > 1, by the transitivity, for any two elements y, z € X, there exists y such that z € yy and there exists an integer n > 0 such that y = (BPar)" B = (BPar) --- (BPar) B, (1.4) TL where BPar appears for n times. Therefore, the (directed)pregraph is connected. Sufficiency. If a (directed)pregraph is connected, i.e., for any two elements x,y € X, their incident vertices u,v € V, have edges e1, €2,°°*,@s, Such that e;, 2 = 2,83,---,s5— 1, is in common with e;. 4 6 Chapter I Abstract Embeddings and e;4,1. Of course, u and v are, respectively, the ends of e1 and es. Thus, y € yz wherey — (ParB)?B. This implies that the semigroup V,parg; is transitive on A. Therefore, the (directed)pregraph is a (directed)graph. O It is seen from the theorem that (directed) graphs here are, in fact, connected (directed) graphs in most textbooks. Because discon- nectedness is rarely necessary to consider, for convenience all graphs, embeddings and then maps in what follows are defined within con- nectedness in this book. A network N is such a graph G = (V, E) with a real function w(e) € R,e € E on E, and hence write N = (G;w). Usually, a network N is denoted by the graph G itself if no confusion occurs. Finite recursion principle On a finite set A, choose ag € A as the initial element at the Oth step. Assume a; is chosen at the ith, i > 0, step with a given rule. If not all elements available from a; are not yet chosen, choose one of them as a;,4, at the i+ 1st step by the rule, then a chosen element will be encountered in finite steps unless all elements of A are chosen. Finite restrict recursion principle Ona finite set A, choose ag € A as the initial element at the Oth step. Assume a; is chosen at the ith, 2 > 0, step with a given rule. If a; is not available from aj, choose one of elements available from a; as a;,4 at the i + Ist step by the rule, then ap will be encountered in finite steps unless all elements of A are chosen. The two principles above are very useful in finite sets, graphs and networks, even in a wide range of combinatorial optimizations. A G = (V, E) with V = Vi + V2 of both V; and Vz independent, i.e., its vertex set is partitioned into two parts with each part having no pair of vertices adjacent, is called bipartite. Theorem 1.2 A graph G = (V, E) is bipartite if, and only if, G has no circuit with odd number of edges. Proof Necessity. Since G is bipartite, start from vy € V ini- L1 Graphs and networks 7 tially chosen and then by the rule from the vertex just chosen to one of its adjacent vertices via an edge unused and then marked by used, according to the finite recursion principle, an even circuit (from bipar- tite), or no circuit at all, can be found. From the arbitrariness of vo and the way going on, no circuit of G is with odd number of edges. Sufficiency. Since all circuits are even, start from marking an arbitrary vertex by 0 and then by the rule from a vertex marked by b € B = {0,1} to mark all its adjacent vertices by b = 1 — b, according to the finite recursion principle the vertex set is partitioned into Vo = (v € V| marked by 0} and V; = (v € V| marked by 1}. By the rule, Vy and Vj are both independent and hence G is bipartite. L From this theorem, a graph without circuit is bipartite. In fact, from the transitivity, any graph without circuit is a tree. On a pregraph, the number of elements incident to a vertex is called the degree of the vertex. A pregraph of all vertices with even degree is said to be even . If an even pregraph is a graph, then it is called a Euler graph. Theorem 1.3 A pregraph G = (V, E) is even if, and only if, there exist circuits C1, C5, --- , C4, on G such that E-Ci tC» C, (1.5) where n is a nonnegative integer. Proof Necessity. Since all the degrees of vertices on G are even, any pregraph obtained by deleting the edges of a circuit from G is still even. From the finite recursion principle, there exist a nonnegative integer n and circuits Ci, C5, --- , Cn, on G such that (1.5) is satisfied. Sufficiency. Because a circuit contributes 2 to the degree of each of its incident vertices, (1.5) guarantees each of vertices on G has even degree. Hence, G is even. [] The set of circuits (C;|1l € i € n} of G in (1.5) is called a circuit partition, or written as Cir=Cir(G). Two direct conclusions of Theorem 1.3 are very useful . One is the case that G is a graph. The 8 Chapter I Abstract Embeddings other is for G is a directed pregraph. Their forms and proofs are left for the reader. Let N — (G;w) be a network where G — (V, E) and w(e) — —w(e) € Zn = (0,1,---,n — 1}, ie., mod n, n > 1, integer group. For examples, Zı = {0}, Zə = B = {0,1} etc. Suppose x, = —z, € Zn, v € V, are variables. Let us discuss the system of equations Eu + £y = w(e) (mod n), e = (uv) € E (1.6) On Zn. Theorem 1.4 System of equations(1.6) has a solution on Z, if, and only if, there is no circuit C such that X wle) ) £0 (mod n) (1.7) ecC on N. Proof Necessity. Assume C is a circuit satisfying (1.7) on N. Because the restricted part of (1.6) on C has no solution, the whole system of equations (1.6) has to be no solution either. Therefore, N has no such circuit. This is a contradiction to the assumption Sufficiency. Let £o = a € Zn, start from vg € V reached. Assume vi € V reached and x; = a; at step i. Choose one of e; = (vi, vi41) € E without used(otherwise, backward 1 step as the step i). Choose v;+1 reached and e; used with a;,, = a; + w(e;) at step i + 1. If a circuit as {e€9,€1,---, e, ej = (vj, Uj41),0 € j € lvi = vo, occurs within a permutation of indices, then from (1.7) Qj41 = aj + w(ei) = a1 + wle) + wle) I.2 Surfaces 9 Because the system of equations obtained by deleting all the equations for all the edges on the circuit from (1.6) is equivalent to the original system of equations (1.6), in virtue of the finite recursion principle a solution of (1.6) can always be extracted . O When n = 2, this theorem has a variety of applications. In [Liu5], some applications can be seen. Further, its extension on a nonAbelian group can also be done while the system of equations are not yet linear but quadratic. 1.2 Surfaces In topology, a surface is a compact 2-dimensional manifold with- out boundary. In fact, it can be seen as what is obtained by identifying each pair of edges on a polygon of even edges pairwise. For example, in Fig.1.3, two ends of each dot line are the same point. The left is a sphere, or the plane when the infinity is seen as a point. The right is the projective plane. From the symmetry of the sphere, a surface can also seen as a sphere cutting off a polygon with pairwise edges identified. The two surfaces in Fig.1.3 are formed by a polygon of two edges pairwise as a. Sphere(Plane) Projective plane Fig.1.3 Sphere and projective plane 10 Chapter I Abstract Embeddings Surface closed curve axiom A closed curve on a surface has one of the two possibilities: one side and two sides. A curve with two sides is called a double side curve ; otherwise, a single side curve . As shown in Fig.1.3, any closed curve on a sphere is a double side curve(In fact, this is the Jordan curve axiom). However, it is different from the sphere for the projective plane. there are both a single(shown by a dot line) and a double side curve. How do we justify whether a closed curve on a surface is of single side, or not? In order to answer this question, the concept of contractibility of a curve has to be clarified. If a closed curve on a surface can be continuously contracted along one side into a point, then it is said to be contractible, or homotopic to 0. Torus Klein bottle Fig.1.4 Torus and Klein bottle It is seen that a single side curve is never homotopic to 0 and a double side curve is not always homotopic to 0. For example, in Fig.1.4, the left, i.e., the torus, each of the dot lines is of double side but not contractible. The right, i.e., the Klein bottle, all the dot lines are of single side , and hence, none of them is contractible. A surface with all closed curves of double side is called orientable; otherwise, nonorientable . For example, in Fig.1.3, the sphere is orientable and the projec- I.2 Surfaces 11 tive plane is nonorientable. In Fig.1.4, the torus is orientable and the Klein bottle is nonorientable. The maximum number of closed curves cutting along without destroying the continuity on a surface is called the pregenus of the surface. In view of Jordan curve axiom, there is no such closed curve on the sphere. Thus, the pregenus of sphere is 0. On the projective plane, only one such curve is available (each of dot lines is such a closed curve in Fig.1.3) and hence the pregenus of projective plane is 1. Similarly, the pregenera of torus and Klein bottle are both 2 as shown in Fig.1.4. Theorem 1.5 The pregenus of an orientable surface is a non- negative even number. A formal proof can not be done until Chapter 5. Based on this theorem, the genus of an orientable surface can be defined to be half its pregenus, called the orientable genus. The genus of a nonorientable surface, called nonorientable genus, is its pregenus itself. The sphere is written as aa^! where a^! is with the opposite direction of a on the boundary of the polygon. Thus, the projective plane, torus and Klein bottle are, respectively, aa, aba~!b~! and aabb. In general, p Op = ajbja; b7 | l T = abia bi lasboa; b3 - . aybya;, b; ! and q Qe = li Aili = 41010202 * ` AgAg (1.9) i=l denote, respectively, a surface of orientable genus p and a surface of nonorientable genus q. Of course, Oo, Q1, O1 and Qə are, respectively, the sphere, projective plane, torus and Klein bottle. It is easily checked that whenever an even polygon is with a pair of its edges in the same direction, the polygon represents a nonori- 12 Chapter I Abstract Embeddings entable surface. Thus, all Op, p > 0, orientable and all Q,, q = 1, are nonorientable. Forms (1.8) and (1.9) are said to be standard. If the form of a surface is defined by its orientability and its genus, then the operations 1-3 and their inverses shown as in Fig.1.5-7, do not change the surface form. Fig.1.5 Operation 1: Aaa | < A Fig.1.6 Operation 2: AabBab < AcBc Fig.1.7 Operation 3: AB 4 ( (Aa)( a 1B) In fact, what is determined under these operations is just a topo- logical equivalence, denoted by ~top. Notice that A and B are all linear order of letters and permitted I.2 Surfaces 13 to be empty as degenerate case in these operations. 'The parentheses stand for cyclic order when more than one cyclic orders occur for distinguishing from one to another. Relation 0 On a surface (A, B), if A is a surface itself then (A, B) = ((A)z(B)z 7) = ((A)(B)). Relation 1 (AxByCaz !Dy )) ~top ((ADCB)(zxyr |y !)). Relation 2 (AzBz) ^y ((AB™')(xz)). Relation 3 (Axzyzy !z |) ~top ((A)(zx)(yy)(zz)). In the four relations, A, B, C, and D are permitted to be empty. B^ = b7! . -b3 'b7'b7' is also called the inverse of B = bibob3 - - bs, s > 1. Parentheses are always omitted when unnecessary to distin- guish cyclic or linear order. On a surface S, the operation of cutting off a quadrangle aba! 57! and then identifying each pair of edges with the same letter is called a handle as shown in the left of Fig.1.8. If the quadrangle aba~!b~! is replaced by aa, then such an oper- ation is called a crosscap as shown in the right of Fig.1.8. Handle Crosscap Fig.1.8 Handle and crosscap The following theorem shows the result of doing a handle on an orientable surface. Theorem 1.6 What is obtained by doing a handle on an ori- entable surface is still orientable with its genus 1 added. 14 Chapter I Abstract Embeddings Proof Suppose S is the surface obtained, then p B ovs | [ a0; 5; cap ibat! ( Relation 0) i=1 p pad. si aed ; ~top ] [aio b; xx 0ga55185,115,5, (Relation 1) i=1 p eis EET ^" top ] [ aita; b; Qp410p 4105 10544 (Operation 1) i1 p+1 = li ajbja; ^b; |. i=l This is the theorem. [] In the above proof, x and x! are a line connecting the two boundaries to represent the surface as a polygon shown in Fig.1.8. 'This procedure can be seen as the degenerate case of operation 3. In what follows, observe the result by doing a crosscap on an orientable surface. Theorem 1.7 Onan orientable surface of genus p, p > 0, what is obtained by doing a crosscap is nonorientable with its genus 2p + 1. Proof Suppose N is the surface obtained, then p N Step li a;bia; |b; xaax | (Relation 0) i=1 p ~top li ajbja; b; xx cc (Relation 2) i=1 p ^"top li ajbja; 1b; xx" cc, cc36363 (Relation 3) i=2 2p+1 ~top II cici. (Relation 3 by p — 1 times). i=1 I.2 Surfaces 15 This is the theorem. O By doing a handle on a nonorientable surface, 2 more genus should be added with the same nonorientability. Theorem 1.8 On a nonorientable surface, what is obtained by doing a handle is nonorientable with its genus 2 added. Proof Suppose N is the obtained surface, then q N op li aja;zaba tb tx! (Relation 0) i-l q ™top II ajajzz laba |b | (Relation 1) i=1 q ~top li aja;aba !b ! (Operation 1) i=l q+2 ~top li cici. (Relation 3). i—1 This is the theorem. O By doing a crosscap on a nonorientable surface, 1 more genus produced with the same nonorientability Theorem 1.9 On a nonorientable surface, what is obtained by doing a crosscap is nonorientable with its genus 1 added. 16 Chapter I Abstract Embeddings Proof Suppose N is the obtained surface, then q N “top II aja;zaarx | (Relation 0) i=l q ^" top li ajajzx ‘aa (Relation 2) i=l q ™top II a,;ajaa (Operation 1) i=1 qi ^" top li aiai. (Relation 3). i=l This is the theorem. E I.3 Embedding Let G = (V, E, V = (vy vo, --- v4], be a graph. A point in the 3-dimensional space is represented by a real number t as the parame- ter, e.g., (x,y, z) = (t, t2, t3). Write the vertices as Ui = (£i, Yi, zi) = CER such that t; Æ tj, i # j, 1 <i,j <n, and an edge as (u,v) =ut Av, 1€ A € I, i.e., the straight line segment between u and v. Because for any four vertices vj, vj, vj and vy, Li Xj Li — XQ Li — Tk det | yi—Vj Wi—W Yi Yk £j Zj £i — 2| £j — Zk tj—tj tj—t ti— tr =det | 2-0 -P P-E j fa PP 9-5 L3 Embedding 17 m (ti E DOR = ti) (ti — tr) (tk = ti) (tk = bb = tı) * 0, 1.8: the four points are not coplanar, any two edges in G has no intersection inner point. A representation of a graph on a space with vertices as points and edges as curves pairwise no intersection inner point is called an embedding of the graph in the space. If all edges are straight line segments in an embedding, then it is called a straight line embedding. Thus, any graph has a straight line embedding in the 3-dimensional space. Similarly, A surface embedding of graph G is a continuous injection 4G of an embedding of G on the 3-dimensional space to a surface S such that each connected component of S — uG is homotopic to 0. The connected component is called a face of the embedding. In early books, a surface embedding is also called a cellular embedding. Because only a surface embedding is concerned with in what follows, an embedding is always meant a surface embedding if not necessary to specify. A graph without circuit is called a tree. A spanning tree of a graph is such a subgraph that is a tree with the same order as the graph. Usually, a spanning tree of a graph is in short called a tree on the graph. For a tree on a graph, the numbers of edges on the tree and not on the tree are only dependent on the order of the graph. They are, respectively, called the rank and the corank of the graph. The corank is also called the Betti number, or cyclic number by some authors. The following procedure can be used for finding an embedding on a surface. First, given a cyclic order of all semiedges at each vertex of G, called a rotation. Find a tree(spanning, of course) T' on G and distin- guish all the edges not on T by letters. Then, replace each edge not of T' by two articulate edges with the same letter. From this procedure, G is transformed into G without changing the rotation at each vertex except for new vertices that are all ar- ticulate. Because G is a tree, according to the rotation, all lettered articulate edges of G form a polygon with pairs of edges, and hence 18 Chapter I Abstract Embeddings a surface in correspondence with a choice of indices on each pair of the same letter. For convenience, G with a choice of indices of pair in the same letter is called a joint tree of G. Theorem 1.10 A graph G = (V, E) can always embedded into a surface of orientable genus at most |/9/2]|, or of nonorientable genus at most 0, where 8 is the Betti number of G. Proof It is seen that any joint tree of G is an embedding of G on the surface determined by its associate polygon. From (1.8) for the orientable case, the surface has its genus at most |20/4| = |8/2 |]. From (1.9) for the nonorientable case, the surface has its genus at most 28/2 = B. o In Fig.1.9, graph G and one of its joint tree are shown. Here, the spanning tree T is represented by edges without letter. a, b and c are edges not on T. Because the polygon is abcacb ~top c lb tcbaa (Relation 2) ~top aabbcc (Theorem 1.7), the joint tree is, in fact, an embedding of G on a nonorientable surface of genus 3. b c G G Fig.1.9 Graph and its joint tree Because any graph with given rotation can always immersed in the plane in agreement with the rotation, each edge has two sides. As known, embeddings of a graph on surfaces are distinguished by the rotation of semiedges at each vertex and the choice of indices of the two semiedges on each edge of the graph whenever edges are labelled L3 Embedding 19 by letters. Different indices of the two semiedges of an edge stand for from one side of the edge to the other on a face boundary in an embedding. Theorem 1.11 A tree can only be embedded on the sphere. Any graph G except tree can be embedded on a nonorientable surface. Any graph G can always be embedded on an orientable surface. Let no(G) be the number of distinct embeddings on orientable surfaces, then the number of embeddings on all surfaces is 2" 9no(G), no(G) = T=)", (1.10) i22 where O(G) is the Betti number and n; is the number of vertices of degree 7 in G. Proof On a surface of genus not 0, only a graph with at least a circuit is possible to have an embedding. Because a tree has no circuit, it can only embedded on the sphere. Because a graph not a tree has at least one circuit, from Theorem 1.10 the second and the third statements are true. Since distinct planar embeddings of a joint tree of G with the indices of each letter different correspond to distinct embeddings of G on orientable surfaces and the number of distinct planar embeddings of joint trees is no(G) = [TC - 0» i22 Further, since the indices of letters on the B(G) edges has 2°) of choices for a given orientable embedding and among them only one choice corresponds to an orientable embedding, the fourth statement is true. [] For an embedding u(G) of G on a surface, let v(uG), e(uG) and @(wG) are, respectively, its vertex number, or order, edge number, or size and face number, or coorder. Theorem 1.12 Fora surface S, all embeddings 4(G) of a graph G have Eul(uG) = v(uG) — e(uG) + d(uG) the same, only dependent 20 Chapter I Abstract Embeddings on S and independent of G. Further, 2— 2p, p 2 0, when © has orientable genus p; Eul(uG) — jc wd (1.11) when S has nonorientable genus q. Proof For an embedding u(G) on S, if it has at least 2 faces, then by connectedness it has 2 faces with a common edge. From the finite recursion principle, by the inverse of Operation 3 an embedding u(G,) of Gi on S with only 1 face on S is found. It is easy to check that Eul(uG) —Eul(uG). Similarly, by the inverse of Operation 2 an embedding u(Go) of Go on S with only 1 vertex is found. It is also easy to check that Eul(wG) =Eul(wG’). Further, by Operation 1 and Relations 1-3, it is seen that Eul(uGo) =Eul(O,), p > 0; or Eul(Q,), q > 1 according as S is an orientable surface in (1.8); or not in (1.9). From the arbitrariness of G, the first statement is proved. By calculating the order, size and coorder of Op, p > 0; or Qq, q > 1, (1.11) is soon obtained. So, the second statement is proved. O According to this theorem, for an embedding u(G) of graph G, Eul(uG) is called its Euler characteristic, or of the surface it is on. Further, g(wG) is the genus of the surface u(G) is on. If a graph G is with the minimum length of circuits c, then from Theorem 1.12 the genus y(G) of an orientable surface G can be embedded on satisfies the inequality v(G) - «1 2) etie (1.12) ce 2 2 — 2 and the genus 4(G) of a nonorientable surface G can be embedded on satisfies the inequality 2— (V(G) - « - 5) €3(0) <8. (1.13) If a graph has an embedding with its genus attaining the lower(upper) bound in (1.12) and (1.13), then it is called down(up)-embeddable. In L3 Embedding 21 fact, a graph is up-embeddable on nonorientable, or orientable sur- faces according as it has an embedding with only 1 face, or at most 2 faces. Theorem 1.13 All graphs but trees are up-embeddable on nonorientable surfaces. Further, if a graph has an embedding of nonorientable genus / and an embedding of nonorientable genus k, | < k, then for any 2, | « i « k, it has an embedding of nonorientable genus t. Proof For an arbitrary embedding of a graph G on a nonori- entable surface, let T' be its corresponding joint tree. From the nonori- entability, the associate 20 (G)-gon P has at least 1 letter with different indices(or same power of its two occurrences!). If P = Q4, q = B(G), then the embedding is an up-embedding in its own right. Otherwise, by Relation 2, or Relation 3 if necessary, whenever s^!s or stst oc- curs, it is, respectively, replaced by ss or sts- t. In virtue of no letter missed in the procedure, from the finite recursion principle, P' = Qg, q = B(G), is obtained. This is the first statement. From the arbitrariness of starting embedding in the procedure of proving the first statement by only using Relation 2 instead of Relation 3 (AststB eo, Ass ! Btt by Relation 2), because the genus of the surface is increased 1 by 1, the the second statement is true. O The second statement of this theorem is also called the inter- polation theorem. The orientable form of interpolation theorem is firstly given by Duke[Duk1]. The maximum(minimum) of the genus of surfaces (orientable or nonorientable) a graph can be embedded on is call the maximum genus(minimum genus) of the graph. The- orem 1.13 shows that graphs but trees are all have their maximum genus on nonorientable surfaces the Betti number with the interpola- tion theorem. The proof would be the simplest one. However, for the orientable case, it is far from simple. many results have been obtained since 1978(see |Liu1-2], [LiuL1], [HuanL1] and [LidL1]) in this aspect. On the determination of minimum genus of a graph, only a few of graphs with certain symmetry are done(see Chapter 12 in |Liu5-6]). 22 Chapter I Abstract Embeddings L4 Abstract representation Let G = (V, X), V = (vi, v3, , Un}, X = (z,,25,:-, Lm} CV{X}V = {{u, v} Vu, v € V), be a graph. For an embedding u(G) of G on a surface, each edge has not only two ends as in G but also two sides. Let o be the operation from one side to the other and 8 be the operation from one end to the other. From the symmetry between the two ends and between the two sides, a^ ee um] (1.14) where 1 is the identity. By considering that the result from one side to the other and then to the other end and the result from one end to the other and then to the other side are the same, t.e., Bo = a. (1.15) Further, it can be seen that K = (1,0, 8, y}, y = a, is a group, called the Klein group where (ab)? = (a8)(a8) = (aBa) 6 — (aab) = (aa) (8) =, (1.16) Thus, an edge x € X of G in an embedding pu(G) of G becomes Ka = {x,ax, Bx, yz}, as shown in Fig.1.10. x Ba —— - ——— y m ax apa Fig.1.10 An edge sticking on K In fact, let X=) Kx; (1.17) i=1 L4 Abstract representation 23 where summation stands for the disjoint union, then a and (@ can both be seen as a permutation on 4, i.e., m a =| [aa a 8 = aa a i=1 iA The vertex x is on deals with the rotation as {(x, Px, Dp es -), (ax, aP Ix, oP is Jh (1.18) as shown in Fig.1.11(when its degree is 4). pes BPx Px | aPx ax ox E Px BP x Ba <= ij) ePi ABP 2a aP3x Pea o] BP3x Fig.1.11 The rotation at a vertex It is seen that P is also a permutation on X. The set of elements in each cycle of this permutation is called an orbit of an element in the cycle. For example, the orbit of element x under permutation P is denoted by (x)p. From (1.18), (x)p ( \(ax)p =, LEX. (1.19) The two cycles at a vertex in an embedding have a relation as (az, Pax, Paz,- -) =(axz,aP !z, o/P ?z, ..-) (1.20) —a(z,P lx, P ?z,...). For convenience, one of the two cycles is chosen to represent the vertex, l; Cs (x, Px, P?y... 27 24 Chapter I Abstract Embeddings Or (ox, P tr, P 7, - --). Theorem 1.14 aP = Pta. Proof By multiplying the two sides of (1.20) by o from the left and then comparing the second terms on the two sides, aPa =P. By multiplying its two sides by a from the right, the theorem is soon obtained. [] Since a and f are both permutations on ¥, y = aß and P* = Py are permutations on A as well. Let (e P aP ige) be the cycle of P* involving x. From the symmetry between x and x, the cycle of P* involving (x is (B, P* x, P** Bx, 3e -) which has the same number of elements as that involving x does. Because P*(Gx) = PaG(Gx) = Pax and from Theorem 1.14 Paz = aP 1z = ay(yP x = ayP* Iz = BP's, we have P* (G2) = BP* "a. (1.21) Furthermore, because P**(Gx) = P*(P*(Gx)) and from (1.21) P*(P*(Ba)) = P'(8P* 2), by (1.21) for P*~'x instead of x, we have P* (Bx) = (P (P*^2)) = B(P* 72). L4 Abstract representation 25 On the basis of the finite restrict recursion principle, a cycle is found. Therefore, (Bx, P* Bv, P Bx, -.-) = B(a,P* x, P* "z,---). (1.22) This implies c)» (fo = 0, (1.23) for z € X. Theorem 1.15 (P* = P* !g. Proof A direct result of (1.22). O Based on (1.22), it is seen that the face involving x of the em- bedding represented by P is ((z, P*z, P*^z, .--), (Gx,P* "Bx, P* -- ]. (1.24) Similarly to vertices, based on (1.22) and (1.23), the face can be represented by one of the two cycles in (1.24). Example 1.3 Let G = K4, i.e. , the complete set of order 4. Given its rotation Ua, y, z), (Bz, l, yw), (4l, u, By), (Bx, Ww, au) y, as shown in Fig 1.12. Its two faces are (x, Gu, Gl, yz) and (y, Qu, AW, al, VY, e; Bus, "yas. N Fig.1.12 A rotation of Ky 26 Chapter I Abstract Embeddings Thus, it is an embedding of K4 on the torus O, = (ABA !B ^J) as shown in Fig.1.13. ys V1 of Xx UA v — Á EN v3 |B Fig.1.13 Embedding determined by rotation Further, another rotation of K4 is chosen for getting another embedding of K4. Example 1.4(Continuous to Example 1.3) Another embedding of K4 is shown as in Fig.1.14. Its rotation is {(x, y, 2), (8z, L, yw), (u, vy, 71), (Bz, w, yu) }. Its two faces are (x, Bu, pl yz) and (az, w, Bz, ay, au, aw, al, By). This is an embedding of Ky on the Klein bottle Nz = (ABA B) ~top= (AABB) L5 Smarandache 2-manifolds with map geometry 27 as shown in Fig.1.14. ys V1 of NX UA v2 —— A A K EN U3 B |) ey Fig.1.14 Embedding distinguished by rotation Such an idea is preferable to deal with combinatorial maps via algebraic but neither geometric nor topological approaches. I5 Smarandache 2-manifolds with map geometry Smarandache system The embedding of a graph on surface enables one to construct finitely Smarandache 2-manifolds, i.e., map geometries on surfaces. A rule in a mathematical system (X; R) is said to be Smaran- dachely denied if it behaves in at least two different ways within the same set X, i.e., validated and invalided, or only invalided but in mul- tiple distinct ways. A Smarandache system (X; R) is a mathematical system which has at least one Smarandachely denied rule in R (see [Mao4] for de- tails). Particularly, if (3; ) is nothing but a metric space (M; p), then such a Smarandache system is called a Smarandache geometry, seeing references [Maol|-[Mao4] and [Smal]—|Sma2). Example 1.5(Smarandache geometry) Let R? be a Euclidean plane, points A,B € R? and l a straight line, where each straight 28 Chapter I Abstract Embeddings line passes through A will turn 30° degree to the upper and passes through 5 will turn 30? degree to the down such as those shown in Fig.1.15. Then each line passing through A in Fı will intersect with l, lines passing through B in F5 will not intersect with l and there is only one line passing through other points does not intersect with /. Fi 4 4996-77 Fig.1.15 Then such a geometry space R? with queer points A and B is a Smarandache geometry since the axiom given a line and a point exterior this line, there is one line parallel to this line is now replaced by none line, one line and infinite lines. A more general way for constructing Smarandache geometries is by Smarandache multi-spaces ({Mao3]). For an integer m > 2, let (Xy; R4), (Eo; R3), +++; (Em; Rm) be m mathematical systems different two by two. A Smarandache multi-space is a pair (35 R) with i=1 i=l Such a multi-space naturally induce a graph structure with V(G) = iX de, ey Xl Example 1.16(|Mao5]) Let n bean integer, Z; = ({0,1,2,---,n— 1, +) an additive group (modn) and P = (0,1,2,---,n — 1) a permu- tation. For any integer 7,0 < i < n — 1, define Zin = P(Z) such that P'(k) +; P'(I) = P'(m) in Zj41 if k+l =m in Zi, where +; L5 Smarandache 2-manifolds with map geometry 29 denotes the binary operation +; : (P'(Kk), P'(I)) —^ P'(m). Then we with O = {+;, 0 € i € n-1] isa Smarandache multi-space underlying a graph Kn, where Z; = Z4 for integers 1 <i € m. Map geometry A nice model on Smarandache geometries, namely s-manifolds on the plane was found by Iseri in [Ise1l] defined as follows, which is in fact a case of map geometry. An s-manifold is any collection C(T', n) of these equilateral trian- gular disks T;,1 < i < n satisfying the following conditions: (i) each edge e is the identification of at most two edges ej, e; in two distinct triangular disks T;, Tj, 1 < i,j <n andi Æj; (ii) each vertex v is the identification of one vertex in each of five, six or seven distinct triangular disks. These vertices are classified by the number of the disks around them. A vertex around five, six or seven triangular disks is called an elliptic vertex, an Euclidean vertex or a hyperbolic vertex, respectively. Fig.1.16 In a plane, an elliptic vertex O, a Euclidean vertex P and a hyperbolic vertex Q and an s-line L1, Lə or L3 passes through points O, P or Q are shown in Fig.1.16(a), (b), (c), respectively. The map geometry is gotten by endowing an angular function u: V(M) — |0, 47) on an embedding M for generalizing Iseri’s model 30 Chapter I Abstract Embeddings on surfaces following, which was first introduced in |Mao2]. Map geometry without boundary Let M be a combinato- rial map on a surface S with each vertex valency> 3 and u : V(M) > [0, 4), i.e., endow each vertex u, u € V(M) with a real number u(u), 0 < ulu) < DOR The pair (M, u) is called a map geometry without bound- ary, ulu) an angle factor on u and orientable or non-orientable if M is orientable or not. Map geometry with boundary Let (M,u) be a map geom- etry without boundary, faces fi, fo,--:, f; € F(M), 1 < L< e(M) — 1. If S(M) \ (fu fo, fi) is connected, then (M, u)™ = (S(M) \ { fi, foi fit, p) is called a map geometry with boundary fi, fo, fi, and orientable or not if (M, u) is orientable or not, where S(M) de- notes the underlying surface of M. pu (u)u(u) = 2n pu (u)u(u) > 2n Fig.1.17 Certainly, a vertex u € V(M) with py(u)u(u) < 27, = 27 or > 2r can be also realized in a Euclidean space R?, such as those shown in Fig.1.17. A point u in a map geometry (M, u) is said to be elliptic, Eu- clidean or hyperbolic if py (u)u(u) < 27, py (u)u(u) = 2x or py(u)u(u) > 2r. If u(u) = 60°, we find these elliptic, Euclidean or hyperbolic ver- tices are just the same in Iseri’s model, which means that these s- manifolds are a special map geometry. If a line passes through a point £u E W with the entering ray and equal to 180° only when u is Euclidean. For convenience, we always assume that a line passing through an elliptic point turn to the left and a hy- perbolic point to the right on the plane. Then we know the following u, it must has an angle L5 Smarandache 2-manifolds with map geometry 31 results. Theorem 1.16 Let M be an embedding on a locally orientable surface with |M| > 3 and py(u) > 3 for Vu € V(M). Then there exists an angle factor u : V(M) — [0,47) such that (M, u) is a Smarandache geometry by denial the axiom (E5) with axioms (E5) for Euclidean, (L5) for hyperbolic and (R5) for elliptic. Theorem 1.17 Let M be an embedding on a locally orientable surface with order> 3, vertex valency> 3 and a face f € F(M). Then there is an angle factor u : V(M) — [0, 47) such that (M, u)! is a Smarandache geometry by denial the axiom (E5) with axioms (E5) for Euclidean, (L5) for hyperbolic and (R5) for elliptic. A complete proof of Theorems 1.16-1.17 can be found in refer- ences in [Mao2-4]. It should be noted that the map geometry with boundary is in fact a generalization of Klein model for hyperbolic geometry, which uses a boundary circle and lines are straight line seg- ment in this circle, such as those shown in Fig.1.18. uu s A p di. Ea. Fig.1.18 Activities on Chapter I 1.6 Observations O1.1 Let X be a finite set and B be a binary set. Is (Bx|v € X} a pregraph or a graph? If unnecessary, what condition does a pregraph, or a graph satisfy? O1.2 Let X be a finite set, B be a binary set and Me Ba LEX For a permutation w on 4 such that V? = 1, is ((z,yz]|x € X} a graph? If unnecessary, when is it a graph? O1.3 How many orientable, or nonorientable surfaces can a hexagon represent? List all of them. O1.4 How many orientable, or nonorientable surfaces can a 2k- gon represent? How to List them all. O1.5 In [Liul], an embedding of a graph G on the nonori- entable surface of genus 3(G) is constructed in a way from a specific tree on G. Now, how to get such an embedding from any tree on G. O1.6 Suppose G has an embedding on an orientable surface of genus k. If k is not the maximum genus of G, how to find an embedding of G on an orientable surface of genus k + 1. O1.7 Any embedding of a graph G — (V, X) is a permutation on {1,a, 8, o 8X = X as shown by (1.17). However, a, 8 and y = a are each a permutation on ¥, but not an embedding of G in general. When does each of them determine an embedding of G. I.6 Observations 33 O1.8 If permutation P on X is an embedding of a graph, to show that for any x € X, there does not exist an integer m such that ax = pra. O1.9 If permutation P on X is an embedding of a graph, to show PaP =a. O1.10 Observe that permutation 8 as P satisfies both O1.8 and 01.9. However, permutation a as P satisfies 01.9 but not O1.8. O1.11 Let S be a set of permutations on ¥ and Wg be the group generated by S. If for any x,y € X, there exists a Y € Vg such that x = wy, then the group Ys is said to be transitive on X. Observe that if permutation P on X is an embedding of a graph, then group Vr,I-—1P,o,0), is transitive on X. An embedding of a graph G = (V, E) can be combinatorially represented by a rotation system o(V) on the vertex set V of G and a function on the edge set as A: E — B, B = {0,1}, denoted by GLA). In order to determine the faces of an embedding, an immersion of the graph and a rule should be established. An immersion of a graph is such a representation of the graph in the plane that vertices injects into the plane on an imaged circle and edges are straight line segments between their two ends. Travel and traverse rule From a point on one side of an edge, travel as long as on the same side until at the middle of a edge with A = 1, then traverse to the other side. O1.12 By the T' T-rule(i.e., the travel and traverse rule) on an immersion, the initial side the starting point is on can always be encountered to get a travel as a set of edges met on the way. O1.13 By the TT-rule on an immersion of a graph, a set of travels can always be found for any edge occurs exactly twice. 34 Activities on Chapter I On a graph G — (V, E), a subset of edges C C E that V has a 2-partition V = Vi + V2 with the property: C = {(u,v) € Flue Vive Vo} (1.25) is called a cocycle of G. O1.14 For an immersion G,(A) of graph G = (V, E), the em- bedding of G determined by G, (A) is orientable if, and only if, the set telVe € E, € (e) = 1} is a cocycle of G. 1.7 Exercises E1.1 For a graph G, prove that G has no odd circuit(a circuit with odd number of edges) if, and only if, for a tree on G, G has no odd fundamental circuit. E1.2 Prove that a graph G — (V, E) has no odd fundamental circuit if, and only if, E itself is a cocycle. Let [ be a nonAbelian group. The identity is denoted by 1. Write To = (£|£? = 1,€ € T}, ie., the set of all elements of order 2. For a pregraph G = (V, E), let z, € I'o,v € V, be variables on the vertex set V and w(e) € To, e € E be a weight function on the edge set E. On the network N = (G; w), its incidence equation is Tuly = w(e) ec E. (1.26) E1.3 Prove that if the incidence equation has a solution, then it has at least |l'o|, i.e. , the number of elements in To, solutions. E1.4 Prove that the incidence equation has a solution if, and only if, G has no circuit C such that w(C) = [[ «o #1. e€C E1.5 Let o(G) be the number of connected components on pre- graph G. Prove that if the incidence equation has a solution, then it I7 Exercises 35 has Pol" solutions. E1.6 For an orientable surface, provide a procedure for deter- mining its orientable genus and then estimate an upper bound of the operations necessary. E1.7 For a nonorientable surface, provide a procedure for de- termining its nonorientable genus and then estimate an upper bound of the number of operations necessarily used. E1.8 Let G bea graph of order 2 with all its edges selfloops but only one. Prove that G is not up-embeddable on orientable surfaces if, and only if, each vertex is incident with odd number of selfloops. E1.9 According to the orientable and nonorientable genera, list all embeddings of K4, the complete graph of order 4. A graph is called i-separable if it has i, i > 1, vertices such that it is not connected anymore when the 7 vertices with their incident edges are deleted. A set of i vertices separable without a proper subset separable for a graph is called an i-cut of the graph. A graph which has an i-cut without (i — 1)-cut is said to be i- connected. E1.10 Prove that if a 3-connected graph G is planar (embed- dable on the sphere), then it has exact 2 embeddings on the sphere. For a planar graph G — (V, E), and u,v € V, if G has the form Q= Mo. X (u, v) E E; (1.27) Gil JG» ~~ {(u,v)}, M (u, v) g B and Gi( 6G» = (u,v) (1.28) such that G and Gə are both with at least 2 edges, then {u,v} is called a splitting pair of G, and Gi and Gg, its splitting block. If a splitting block at a splitting pair has no proper subgraph is still a splitting block, then it is said to be standard|Macl]. 36 Activities on Chapter I E1.11 Prove the following three statements. (i) Let A and B be two standard splitting blocks of a splitting pair, then they are no edge in common; (ii) For a splitting pair of a 2-connected planar graph G — (V, E), the standard splitting block decomposition of the edge set E, i.e., E-ELJEU--Uz such that E; N E; = 0, 1 < i # j € s, and the induced subgraphs of Ej;, lxi € s, are all standard splitting blocks of the splitting pair, is unique. (iii) Let b; be the number of standard splitting blocks of the ith splitting pair, 2 = 1,2,---,m, m be the number of splitting pairs, then the number of embeddings of a 2-connected planar graph on the sphere is obi o [[@ u D. j=l For a graph G, let II(G) be the set of rotation systems of inner vertices on a joint tree and w,(i), m € II, be the boundary polygon with its 9(G) pairs of letters whose indices are determined by a binary number i with G(G) digits by the rule: the two indices of the lth letter are same or different according as the lth digit of is 0 or 1. Define EM £u,()) = M. aa (1.29) k=-6(G) where xz is an undeterminate and l; if T ) Nto O , —k; a een Un er 84 (1.30) 0, otherwise. Now, let Qr = {w,(i)| 0 € i < 299 — 1) (1.31) and £(0,) = 2. E(w,(7)). (1.32) I.8 Researches 37 E1.12 Prove that the coefficient of z^ in &G)— M; &(0;) n€llI(G) is the number of embeddings of G on the surface of relative genus k, -6(G) < k < |). A graph of order 2 without selfloop is called a link bundle. E1.13 Let Lm be the link bundle of size m > 1. Determine (Lm). A graph of order 1 is also called a bouquet, or a loop bundle. E1.14 Let Bm be a bouquet of size m, m > 1. Determine E(Bm). A graph of order 2 is called a bipole. Of course, a link bundle is a bipole which has no selfloop. E1.15 Let P, be a bipole of size m, m > 1. Determine (Pm). 1.8 Researches The set(repetition at most twice of elements permitted) of edges appearing on a travel can be shown to have a partition of each subset forming still a subtravel except probably the travel itself. Such a partition is called a decomposition of a travel into subtravels. However, it is not yet known if any travel can be decomposed into tours except only the case that its induced graph has a cut-edge. R1.1 Prove, or disprove, the conjecture that a travel with at most twice occurrences of an edge in a graph has a decomposition into tours if, and only if, the induced subgraph of the travel is without cut-edge. Because a circuit is restricted from a tour by no repetition of a vertex, the following conjecture would look stronger the last one. R1.2 Prove, or disprove, the conjecture that a travel with at 38 Activities on Chapter I most twice occurrences of an edge in a graph has a decomposition into circuits if, and only if, the induced subgraph of the travel is without cut-edge. However, it can be shown from Theorem 1.3 that any tour has a decomposition into circuits. The above two conjectures are, in fact, equivalent. Because a cut-edge is never on a circuit, the necessity is always true. A travel with three occurrences of an edge permitted does not have a decomposition into circuits in general. For example, on the graph determined by Par= {{x(0), y(0)}, (z(1), y(1)}}, the travel zx ry | where x = (x(0), z(1)) and y = (y(0), y(1)) has no circuit decomposition. Furthermore, the two conjectures are apparently right when the graph is planar because each face boundary of its planar embedding is generally a tour whenever without cut-edge. R1.3 For a given graph G and an integer p, p > 0, find the number n,(G) of embeddings of G on the orientable surface of genus p. The aim is at the genus distribution of embeddings of G on ori- entable surfaces, i.e., the polynomial [c/2] Po(G) = 5 | m(G)2”, p=0 where o is the Betti number of G. For p = 0, no(G) can be done based on |Liu6]. If G is planar, O1.11 provides the result for 2-connected case. Others can also be derived. As to justify if a graph is planar, a theory can be seen from Chapters 3,5 and 7 in [Liu5]. Generally speaking, not easy to get the complete answer in a short period of time. However, the following approach would be avail- able to access this problem. Choose a special type of graphs, for instance, a wheel(a circuit Cn all of whose vertices are adjacent to an extra vertex), a generalized Halin graph(a circuit with a disjoint tree I.8 Researches 39 except for all articulate vertices forming the vertex set of the circuit) and so forth. Of course, the technique and theoretical results in 1.3 can be employed to calculate the number of distinct embeddings of a graph by hand and by computer. R1.4 Orientable single peak conjecture. The coefficients of the polynomial in R1.3 are of single peak , i.e., they are from increase to decrease as p runs from 0 to |a/2]|(> 3), n, (G). 'The purpose here is to prove, or disprove the conjecture not nec- essary to get all n,(G), 0 € p € [c /2](2 3). R1.5 Determine the number of distinct embeddings, which have one, or two faces, of a graph on orientable surfaces. R1.6 Fora given graph G and an integer q, q > 1, find the num- ber (G) of distinct embeddings on nonorientable surfaces of genus q. The aim is at the genus polynomial of embeddings of G on nonori- entable surfaces: Py(G) = 5 (G), where o is the Betti number of G. Some pre-investigations for G is that a wheel, or a generalized Halin graph can firstly be done. R1.7 Fora graph G, justify if it is embeddable on the projective plane, and then determine 744(G) according to the connectivity of G. R1.8 For a graph embeddable on the projective plane, determine how many sets of circuits such that for each, all of its circuits are essential if, and only if, one of them is essential in an embedding of G on the projective plane. R1.9 Nonorientable single peak conjecture. The coefficients of the polynomial in R1.6 are of single peak in the interval [o, 0] where c is the Betti number of G. 40 Activities on Chapter I R1.10 For a given type of graphs G and an integer p, find the number of distinct embeddings of graphs in G on the orientable surface of genus p. Further, determine the polynomial lo(G)/2| Po(G) = $, mp (G)a” p=0 where c(G) = max{a(G)|G € G}. R1.11 For a given type of graphs G and an integer q, q > 1, find the number of embeddings of graphs in G on the nonorientable surface of genus g. Further, determine the polynomial o(G) Py(G) = 5 fig(G)x4 where o(G) = max{a(G)|G € G}. R1.12 For a set of graphs with some fixed invariants, extract sharp bounds(lower or upper) of the orientable minimum genus and sharp bounds(lower or upper) of orientable maximum genus. Here, invariants are chosen from the order (vertex number), size(edge number), chromatic number (the minimum number of colors by which vertices of a graph can be colored such that adjacent vertices have distinct colors), crossing number(the minimum number of crossing in- ner points among all planar immersions of a graph), thickness (the minimum number of subsets among all partitions of the edge set such that each of the subsets induces a planar graph), and so forth. R1.13 For a set of graphs and a set of invariants fixed, pro- vide sharp bounds(lower or upper) of minimum nonorientable genus of embeddings of graphs in the set. Chapter II Abstract Maps e A ground set is formed by the Klein group K = {1, a, 8, a3} stick- ing on a finite set X. e A basic permutation is such a permutation on the ground set that no element x is in the same cycle with az. e The conjugate axiom on a map is determined by each vertex con- sisting of two conjugate cycles for o, as well as by each face con- sisting of two conjugate cycles for 9. e The transitive axiom on a map is from the connectedness of its underlying graph. e An included angle is determined by either a vertex with one of its incident faces, or a face with one of its incident vertices. II.1 Ground sets Given a finite set X = (z1,25,:--, £m}, called the basic set, its elements are distinct. Two operations œ and @ on X are defined as for any z € X, az # Bx, ax, Gx d X and o?z = afar) = z, B?x = B(8x) = x. Further, define a3 = fa = y such that for any x € X, yx ~#ax, px and yx d X. 42 Chapter II Abstract Maps Let aX = [ox|Vvzr € X}, 8X = iüx|Vx € X} and yX = {ya|\Vx € X}, then o, B and y determine a bijection between X and, respectively, aX, GX and yX. By a bijection is meant a one to one correspondence between two sets of the same cardinality. In other words, X()ex-x(|ex-x[(]|yx-8 aX ()BX = BX()\yX 24X[(]ox - (2.1) JaX| = [BX| = |yX| = |X]. The set X UaX UGX UyX, or briefly A = V(X), is called the ground set. Now, observe set K = 4{1,a,8,y}. Its elements are seen as permutations on the ground set X. Here, 1 is the identity. From o? =p =1 and af = Ba, 34? = (aB)(aB) = a(88)a = o? = 1, and hence a= (Bx, Vay); =: =: Il H = Il ra (£i, azi) (zi, Bai) | | (ovi, voi); (2.2) ll =: = II H m jai Si (zi, Yi) lio. aTi). i=1 It is easily seen that K is a group, called Klein group because it is isomorphic to the group of four elements discovered by Klein in geom- sc =: Il H etry. For any z € X, let Kx = {x, ax, Bx, yx}, called a quadricell. Theorem 2.1 For any basic set X, its ground set is Ty in (2.3) LEX where the summation represents the disjoint union of sets. Proof From (2.1), for any z,y € X and z Z y, Ke( Ky = (x, arf, 32) ( Ku, ay, Bv, yy} = 9. IL2 Basic permutations 43 From (2.1) again, |) Kc2 X c oX - 8X e 4X =X. rcx Therefore, (2.3) is true. O Furthermore, since for any x € X, Ring eA eR ow) = he, (2.4) we have Xy Hye Koc Ki ycoa X zEßX teyX This implies that the four elements in a quadricell are with symmetry. II.2 Basic permutations Let Per(X), or briefly Per, be a permutation on the ground set ¥. Because of bijection, according to the finite strict recursion principle, for any x € X, there exists a minimum positive integer k(x) such that Per (95 = Ds i.e., Per contains the cyclic permutation , or in short cycle, C= Per?z, +--+ , Per* (9715). Write {x£ }per as the set of all elements in the cycle (z)pa. Such a set is called the orbit of x under permutation Per. The integer k(x) is called the order of x under permutation Per. If for any x € X, az É {x}Per, (2.5) then the permutation Per is said to be basic to a. Example 2.1 From (2.2), for permutations a and D on the ground set, o is not basic, but 8 is basic. 44 Chapter II Abstract Maps Let Par= (Xi, Xo,---,X,} be a partition on the ground set X, then Per = | [(x) (2.6) i=1 determines a permutation on 4, called induced from the partition Par. Here, (X;), 1 € i € s, stands for a cyclic order arranged on X;. This shows that a partition (.X;|1 <i € s} has Ixi- 212 Ie - 0^ (2.7) s-l i23 induced permutations. In (2.7), n;, à > 3, is the number of subsets in Par with 7 elements. Theorem 2.2 Let Par= (X;|]1 € i € s) be a partition on the ground set V(X). If Par has an induced permutation basic, then all of its induced permutations are basic. Further, a partition Par has its induced permutation basic if, and only if, for x € X, there does not exist Y € Par such that {x ax} or {8x, yz} C YAN Kr: (2.8) Proof Because the basicness of a permutation is independent of the order on cycles, the first statement is proved. Assume an induced permutation Per of a partition Par is basic. From (2.5), for any x € X, in virtue of Ke ={z,axr}+ 18r; yr}, no Y €Par exists such that (2.8) is satisfied. This is the necessity of the second statement. Conversely, because for any x € X, no Y € Par exists such that (2.8) is satisfied, it is only possible that x and az are in distinct subsets of partition Par. Therefore, ax Z (x]pa. Based on (2.5), this is the sufficiency of the second statement. [] On the basis of this theorem, induced basic permutations can be easily extracted from a partition of the ground set. IL2 Basic permutations 45 Example 2.2 Let ¥ = {z,az,(@x,yx} = Kx. There are 15 partitions on X as Par, = (iz); {az}, {8r}, Uy]; Pato =4 {e097}, {Orhi Pars = {{x, Bx}, {az}, {yr}}; Par, = (iz, yx}, {ax}, (82) 5 Pars = (1682, ax}, {x}, {yz}}: Parę = ((x (ox, yx}, {Gx}}; Par; = {{z}, {ax}, (82, yx}}; Pare ed wr ore) Parg = (iz, Bx, yx}, (6x) ); Pari = {{z, ax, yx}, {ax}}; Part = liz. 102,09 0mFE Pario = {{x£, ax, Bx, yz E Pars = {1r arh (eae) | Parya = {{z, Bx}, fax, yz] ; Paris = {{z, 7x}, fax, Bx}}. From Theorem 2.2, induced basic permutations can only be extracted from Par,, Pars, Pary, Pars, Parę, Parj, and Par;; among them. Since each of these partitions has no subset with at least 3 elements, from (2.7) it only induces 1 basic permutation. Hence, 7 basic permutations are induced in all. Based on Theorem 2.2, a partition that induces a basic permu- tation is called basic as well. For a partition Par on X, if every pair of x and oz, x € X, deals with the element x in Par, then this partition determines a pregraph if any. For example, in Example 2, there are only Pon = Pars = Pary = Pate ={{7},4 0211: Partie = Paria = Paris = {{x, 8£}} form 2 premaps of size 1 and others meaningless among the 15 par- 46 Chapter II Abstract Maps titions. Further, each of the 2 premaps is a graph. The result is the same as in Example 1.1. II.3 Conjugate axiom Let Per; and Per» be two permutations on the ground set A. If for z € X, (Peroz)per, = Per(z)pg-: = Pero(®) pop, (2.9) then the two orbits (x)per, and (Perox)pa, of Per, are said to be con- jugate. For a permutation 7 on the ground set X, if aP =P ‘a, (2.10) then (P, a)(or for the sake of brevity, P) is called satisfying the con- jugate axiom. Theorem 2.3 For a basic permutation P on the ground set X, the two orbits (x)p and (ax)p for any x € X are conjugate if, and only if, (P, a) satisfies the conjugate axiom. Proof Necessity. Because of orbits (x)p and (ax)p conjugate, from (2.9), (ax)p = a(x)5'. Hence, Pax = aP 12, i.e., Pa = «P1. This implies (2.10). Sufficiency. Since P satisfies (2.10), P(oz) = P(o'P)P-!z = P(P-la)P-!z = (PP aP lax = aP™tr. By induction, assume that P!(ax) = aP~'x, 1 > 1, then we have Plax) = PP'(ox) = PPr = P(aPP')P s ( then by (2.10)) (PP ap Vz sap ity, II.3 Conjugate axiom 4T Hence, (ax)p = a(x)5'. From (2.9), orbits (z)p and (ax)p are conju- gate. This is the sufficiency. [] Unlike Theorem 2.2, for a partition on the ground set X, from one of its induced permutations satisfying the conjugate axiom, it can not be deduced to others. A premap, denoted by (A5, P), is such a basic permutation P on the ground set X that the conjugate axiom is satisfied for (P, a). Example 2.3 By no means any basic partition is in companion with a basic permutation. Among the 7 basic partitions as shown in Example 2.2, only the induced permutations of Par,, Parj4 and Paris are premaps. Because (P, 3) is not necessary to satisfy the conjugate axiom on a premap (4X, P), a is called the first operation and (3, the second . Thus, X should be precisely written as Vg if necessary. Based on the basicness and Theorem 2.3, any premap P has the form as II o}. (2.11) LEX p where Xp is the set of distinct representatives for the conjugate pairs {{x}p, {x}5} of cycles in P. And further, x= Y {xhp{a}p. (2.12) For convenience, one of two conjugate orbits in {{a}p, {x }p} is chosen to stand for the pair itself as a vertex of the premap. Example 2.4 Let X = {21,22}, then X = (x1am 21,21, 25, AX2, BL, YL}. Choose Pi = (21, 8x1) (o3, 21) (22) (o2) (822) (2) and Pa = (z1, 821, T2) (o1, AX2, yz1) (x2) (^yza). 48 Chapter II Abstract Maps The former has 3 vertices (x1, 971), (v2) and (8x3). The latter has 2 vertices (2, 02, £2) and (zs). Lemma 2.1 If permutation P on Xag is a premap, then P'8—g8p* (2.13) where P* = Py, y = af. Proof Because P*3 = Pa = Pa, from the conjugate axiom, p*g —op = BBoP^' (Al 6? = 1) = B((PaB) ^) = gp *, Therefore, the lemma holds. C This lemma tells us that although 8 does not satisfy the conju- gate axiom for permutation P in general, 9 does satisfy the conjugate axiom for permutation 7*. Lemma 2.2 If permutation P on A, is a premap, then per- mutation P* = Py is basic for 6. Proof Because the 4 elements in a quadricell are distinct, r Æ Bats Case 1 P*x # Gx. Otherwise, from P*r = Gx, Pa(Gxr) = P*x = Bx. A contradiction to that P is basic for a. Case 2 (P*)*x Z Bx. Otherwise, P*z = P* !8z. From Lemma 2.1, P*x = BP*x. A contradiction to that P*x and GP*z are in the same quadricell. In general, assume by induction that Case l: (P*)'y Z Bx, 1 < | € k, k 2 2, is proved. To prove Case k+1 (P*)'*!z 4 Bx. Otherwise, P*"z = P* 18x. From Lemma 2.1, P*^ !(P*z) = 8(P*z). A contradiction to the induction hypothesis. II.4 Transitive axiom 40 In all, the lemma is proved. [] Theorem 2.4 Permutation P on X is a premap (Xap, P) if, and only if, permutation P* = Py is a premap (A54, P*). Proof Necessity. Since permutation P is a premap (4,5, P), P is basic for a and satisfies the conjugate axiom. From Lemma 2.2 and Lemma 2.1, P* is basic for @ and satisfies the conjugate axiom. Hence, P* is a premap (A4, P"). Sufficiency. Because P** = P*y = P, the sufficiency is right. O On the basis of Theorem 2.4, the vertices of premap P* are de- fined to be the faces of premap P. The former is called the dual of the latter. Since P** = P, the latter is also the dual of the former. Example 2.5 For the two premaps Pı and P> as shown in Example 2.4, we have Pt = (z1,021)(821, yz1) (22, Yr) (822, 022) and P5 = (71, 023,95, 943) (Da, (35, Dto, ^21). Because T has 2 vertices (r1, axı) and (x2, yx2) and P3 has 1 vertex (24, a4, £2, yX2), we seen that premaps Pı and P» have, respectively, 2 faces and 1 face. II.4 ‘Transitive axiom For a set of permutations T = (r;|1 € i € k}, k > 1 on X, let s k Wr = {yv = III. im) EZ meEn,s >21}, (214) l=1 j=1 where Z is the set of integers and II is the set of all permutations on {1,2,---, k}, k>1. 50 Chapter II Abstract Maps Since all elements in V7 are permutations on X, they are closed for composition(or in other word, multiplication) with the associative law but without the commutative law. Further, it is seen that a permutation in V7 if, and only if, its inverse is in V. The identity is the element in V7 when all i;(7) = 0, 7 € ILin (2.14). Therefore, V7 is a group in its own right, called the generated group of T. Let Y be a permutation group on X. If for any z,y € X, there exists an element Y € V such that x = wy, then the group V is said to be transitive , or in other words, the group W satisfies the transitive axiom. Now, consider a binary relation on X, denoted by ~y, that for any z,y € X, zr yyy eJ eY. xax. (2.15) Because the relation ~y determined by (2.15) for a permutation group V on & is a equivalence, & is classified into classes as V/~yw. Theorem 2.5 A permutation group V on X is transitive if, and only if, for the equivalence ~y determined by (2.15), |¥/ ~y | = 1. Proof Necessity. From the transitivity, for any x,y € X, there exists Y € V such that x = wy. In view of (2.15), for any z, y € X, x ~y y. Hence, for ^, |X/ ey | = 1. Sufficiency. Because |X/ ~y | = 1, for any z,y € X, exists i» € V such that x = wy. Therefore, the permutation group V on X is transitive. L For a premap, the pregraph with the same vertices and edges as the premap is is called its under pregraph. Conversely, the premap is a super premap of its under pregraph. Let (Xag, P) be a premap. If permutation group V;, J = (o, B, P), on the ground set X, g is transitive, i.e., with the transitive axiom, then the premap is called a map . Lemma 2.3 Let M = (Xag, P) be a premap. For any x,y € II.4 Transitive axiom 51 Xap, exists Y € Uz, J = (o, 0, P), such that x = wy if, and only if, there is a path from the vertex v, x is with to the vertex v, y is with in the under pregraph of M. Proof Necessity. In view of (2.14) with the conjugate axiom, write V = a Pra Ba Pa” GB MY Ba Pha, where 6;,0; € {0,1}, 1 < i < s, andl; € Z, 1 € i € s. Because as Phas and a? Pha?! are, respectively, acting on vertices vy and vy, V determines a trail from v, to v, of s — 1 edges. Since there is a trail between two vertices if, and only if, there is a path between them, the necessity is done. Sufficiency. let (us, Us_1, +++, V1),Us = Vy, V1 = Vz, be a path from vy to v; in the under pregraph of M. Then, there exist ô; o; € {0,1}, 1<i<s, andl; E€ Z, 1€ 4 € s, such that y — a Pig. -" Bad P's a and z = wy. From (2.14), the sufficiency is done . [] Theorem 2.6 A premap is a map if, and only if, its under pregraph is a graph. Proof From the transitive axiom and Lemma 2.3, its under pre- graph is a graph. This is the necessity. Conversely, from the connectedness and Lemma 2,3 , the premap satisfies the transitive axiom and hence its under pregraph is a graph. This is the sufficiency. [] Example 2.6 In Fig.1.2, the pregraph determined by Par; has 2 super premaps: P, = (z1)(oa1) (y21) (821) (2, x2) (0/2, 822) and P, = (z1) (azı) (y21) (821) (x2, 822) (o2; *y2) as shown in Fig.2.1. 52 Chapter II Abstract Maps (22, 22) (z2, Bx) Pi P» Fig.2.1 Two super premaps From Theorem 2.6, none of the two super premaps is a map in Fig.2.1. However, the pregraph determined by Parı2 is a graph in Fig.1.2. From Theorem 2.6, each of its super premaps is a map as shown in the following example. Example 2.7 In Fig.2.2, there are 273! = 24 distinct embed- dings of the graph determined by Par» in Fig.1.2. On the associate(or boundary) polygon of the joint tree, the pair of a letter is defined to be of distinct powers when x and yx appear; the same power otherwise. yi yr T2 vU vU vU f NE Ya NE fe YVL2Q T2 YyT2 (a) Oo (b) Oo (c) O1 II.4 Transitive axiom x A Tj U f£ o NOn T2 (e) O1 Ba TIN 7 vU $ Nyx2 X2 (h) Ni x2 my Z vU 7 b T2 (k) No yTı TIN 7 vU 25 b T2 (n) Ni wy TIN vU f Nu "yw (f) Oo T2 U N X1 x2 (i) No A . Nro Via X 54 ON OZ ANZ N a2 4 Noam Jor AN (p) Ni (q) No hd M r1\ Z7 aN Z7 v U 4 Nu» f£ No Bx» T2 (s) No (t) No We ANZ ANS AX (v ) No (w) Ni Fig.2.2 All embeddings of a graph Chapter II Abstract Maps NY (r) Mi WE d gn Ly Joan © (u) Ni Bx EX J A . Nro Bx2 (x) No In Fig.2.2, the graph determined by Parjo(Fig.1.2) has 6 ori- entable embeddings (a—f). Here, (a), (b), (d) and (f) are the same map on the sphere Op ^o (%12_). And, (c) and (e) are the same map on the torus O4 ~top (tirar. are, in fact, 2 maps. 151). Hence, such 6 distinct embeddings Among the 18 nonorientable embeddings, 10 are on the projective plane and 8 are on the Klein bottle. On the projective plane, (g), (h), (i), (D, (m), (n), (p) and (r) are the same map (Ny ep (x12, /xo23)). II.5 Included angles 55 And, (u) and (w) are another map (Nj = (21%2%12%2)). On the Klein bottle, (i), (k), (o) and (q) are the same map (N2 = (xizsz1 'a)). And, (s), (t), (v) and (x) are another map (No ~top (%1%1%2%2)). Therefore, there are only 4 maps among the 18 embeddings. II.5 Included angles Let M = (Xa B(X), P) be a premap. Write k = |X5,5(.X)/ ov, |, J = 1oa,B,P],and k k Xa p(X) = 3 Aa), X= ^. xi i=l i=1 where A4,5,(X) € Xa8/ ~w,, o; and fj; are, respectively, a and 8 restricted on Æw; s, (Xi), à = 1,2,---, k. Further, k M — 3 Mi, Mi = (Xs (X), Pi), (2.16) i=l where M; is a map and P; is P restricted on X4, a (Xi), i — 1,2,---, k. This enables us only to discuss maps without loss of generality. Lemma 2.4 Any map (X, P) has that (Pa)? = 1. Proof From the conjugate axiom, (Pa)? = (Pa)(Pa) = P(aP)a = (PP (ao) = 1. This is the conclusion of the lemma. O Lemma 2.5 Any map (¥,P) has that (P*8)? = 1. Proof Because P*3 = Py3 = Pa( 66) = Po, from Lemma 2.4, the conclusion is obtained. O On the basis of the above two lemmas, on a map M = (Xag, P), any z € Xa g has that (r, Pos = (F Drm) (2.17) 56 Chapter II Abstract Maps Thus, (x, Paz), or (P*Gx,x), is called an included angle of the map. For an edge Ka = {x, ax, Bx, yx} of M, (x, ax] and {8x, yx} are its two ends , or semiedges. And, (x, 8x} and {az, yx} are its two sides, or cosemiedges. Theorem 2.7 Fora (4(X),P), let V = (v|v = (x) pU (ox jp, Vr € X} and F = (flf = {x}p-U {Bx}p-, Vr € X). If z, and v; are, respectively, in v and f as representatives, then A= p» aP Ix, 4 {Pry,aty}+--- veV + (Plz, aP~*2,}) = (zn BP ay} + {Pray Bas} +. feF + {Pt lay, BP* rg). (2.18) Proof From X = „eyv and the conjugate axiom, v = (zy, Pazo} + {Px,,PaPx,}+--- HOP zu PaP zu) = (z,, P^ Iz.) + {Ptn aty} +>: HIP aP tuk This is the first equality. The second equality can similarly be derived from X = } jep f and Lemma 2.1(the conjugate axiom for P* with 8). O It is seen from the theorem that the numbers of included angles, semiedges and co-semiedges are, each, equal to the sum of degrees of vertices. Since every edge has exactly 2 semiedges, this number is 2 times the size of the map. Activities on Chapter II II.6 Observations O2.1 For a set of 4k elements, observe how many ground sets can be produced. O2.2 By a set of 12 elements as an example, observe how many basic permutations satisfy the conjugate axiom. O2.3 In Example 2, observe which nonbasic partition has a permutation with the conjugate axiom. O2.4 Provide two embeddings with the same under graph. O2.5 For an embedding, observe if its mirror image is the same as itself. How about for a map? O2.6 For a map (A,,5, P), observe the orbits of permutations Pa and aP. Whether, or not, they are a map. O2.7 For a map (A5, P), observe the orbits of permutations PG and BP. Whether, or not, they are a map. O2.8 On a map (4,5, P), whether, or not, the permutation Pa is a map on the same ground set. A map with each of its face a quadrangle is called a quadrangu- lation. A map with only triangular faces is a triangulation. In general, a map with all vertices of the same degree is said to be vertex regular, or primal regular. If all faces are of the same degree, the the map is said to be face regular , or dual regular. A primal regular map with its vertex degree i, i > 1, is called an i-map. A 58 Activities on Chapter II dual regular map with its face degree 7, 7 > 1, is called a 7*-map. A triangulation and a quadrangulation are, respectively, a 3*-map and a 4*-map in their own right. O2.9 Whether, or not, the under graph of a 4*-map is always bipartite. Furthermore, whether, or not, the under graph of a (2k)*- map, k > 3, is always bipartite. O2.10 Observe that for any integer i, ? > 1, whether, or not, there always exists an 7-map and an i*-map. If the degree of any vertex(or face) of a map is only an integer i, or j, j Zz i, i,j > 1, then the map is called a (i, j)-map (or (i*, 7*)- map). Similarly, the meanings of a (i, 7*)-map and a (i*,7)-map are known. O2.11 For any i,j, i Æ (i, j)-map, a (i*, j*)-map, or a (i, j = 1, whether, or not, there is a j')map. II.7 Exercises E2.1 For two permutations Per; and Per» on a set of 4 elements with Per,;? = Per? = 1, list all the generated groups V (Per, Pero} Show if each of them is isomorphic to the Klein group. E2.2 Let A and u be permutations on the set A and M = u? = 1. Prove that if the generated group V5, is isomorphic to the Klein group, then A is a ground set if, and only if, A = Ay + A» +--+ Ag, k > 1, such that Aly, and uļ|4, are both have two orbits, and A|4, Æ Bx L9. —1 E2.3 For a map (45,5, P), prove that a(x),p = (ax)p., E2.4 For a map (Xag, P), prove that (z),p and (Gx),p are conjugate. E2.5 To prove that all planar embeddings of K4, i.e., the com- plete graph of order 4, are the same map. A map is said to be nonseparable if its under graph is nonsepa- II.8 Researches 59 rable, i.e. , no cut-vertez(its deletion with incident edges destroys the connectedness in a graph). E2.6 List all nonseparable maps of size 4. A map with all vertices of even degree is called a Euler map. If the face set of a map can be partitioned into two parts each of which no two faces have an edge in common, then the face partition is called a edge independent 2-partition . E2.7 Provide and prove a condition for the face set of a Euler map having an edge independent 2-partition. E2.8 Prove the following statements: (i) The permutation 8 on the ground set ¥ is a map if, and only it = Kr: (ii) The permutation a on the ground set A is a map if, and only if, X = Ke = {x, pz, az, yz}, i.e., B is the first operation. E2.9 Fora map M = (a6, P), (i) Provide a map M and an integer i, i > 2, such that P’ is not a map; (ii) For i, i > 2, provide the condition such that P’ is still a map. E2.10 Let C and D be, respectively, the sets of 3-maps and 3*-maps. For the size given, provide a 1—to-1 correspondence between them. IL.8 Researches From Theorem 1.10 in Chapter I, any graph has an embedding on a surface (orientable or nonorientable). However, if an embedding is restricted to a particular property, then the existence is still necessary to investigate. If a map has each of its faces partitionable into circuits, 60 Activities on Chapter II then it is called a favorable map. If a graph has an embedding which is a favorable map, then the embedding is also said to be favorable|Liu12]. R2.1 Conjecture. Any graph without cut-edge has a favorable embedding. It is easily checked and proved that a graph with a cut-edge does not have a favorable embedding. However, no graph without cut- edge is exploded to have no favorable embedding yet. Some types of graphs have be shown to satisfy this conjecture such as Kn, n > 3; Kmn, m,n > 2, Qn, n 7 2, planar graphs without cut-edge etc. A map which has no face itself with a common edge is said to be preproper . It can be shown that all preproper maps are favorable. However, the converse case is unnecessary to be true. R2.2 Conjecture. Any graph without cut-edge has a preproper embedding. Similarly, it is also known that any graph with a cut-edge does not have a preproper embedding. And, Kn, n > 3; Kmn, m,n > 2, Qn, n > 2, planar graphs without cut-edge etc are shown to satisfy the conjecture as well. Furthermore, if a map has each of its faces a circuit itself, then it is called a proper map, or strong map. Likewise, proper embedding, or strong embedding . It can be shown that all proper maps are preproper. However, the converse case is unnecessary to be true. R2.3 Conjecture. Any graph without cut-edge has a proper embedding. For proper embeddings as well, it is known that any graph with a cut-edge does not have a proper embedding. And, Kn, n > 3; Kmn, m,n > 2, Qn, n 2 2, planar graphs without cut-edge etc are all shown to satisfy this conjecture. Although conjectures R2.1—R2.3 are stronger to stronger, be- cause R2.3 has not yet shown to be true, or not, the the two formers are still meaningful. If a favorable(proper) embedding of a graph of order n has at II.8 Researches 61 most n — 1 faces, then it said to be of small face . Now, it is known that triangulations on the sphere have a small face proper embedding. Because triangulations of order n have exactly 3n — 6 edges and 2n — 4 faces, all the small face embeddings are not yet on the sphere for n > 4. R2.4 Conjecture. Any graph of order at least six without cut- edge has a small face proper embedding. Because it is proved that K; has a proper embedding only on the surfaces of orientable genus 1(torus) and nonorientable genus at most 2(Klein bottle) |WeL1], they have at least 5 faces and hence are not of small face. R2.5 Conjecture. Any nonseparable graph of order n has a proper embedding with at most n faces. If a map only has i-faces and i + 1-faces, 3 < i < n — 1, then it is said to be semi-regular. R2.6 Conjecture. Any nonseparable graph of order n, n > 7, has a semi-regular proper embedding. In fact, if a graph without cut-edge has a cut-vertex, then it can be decomposed into nonseparable blocks none of which is a link itself. If this conjecture is proved, then it is also right for a graph without cut-edge. Some relationships among these conjectures and more with new developments can be seen in [Liu12]. R2.7 For an integer i > 3, provide a necessary and sufficient condition for a graph having an 7-embedding, or 2*-embedding. Par- ticularly, when 7 = 3,4 and 5. First, start from 2 = 3 with a given type of graphs. For instance, choose G — K,, the complete graph of order n. For 3*-embedding, on the basis of Theorem 1.12( called Euler formula), a necessary condition for Ky, n > 3, having an 3*-embedding is —1 n- 20 D 5 =2— 2p and 36 — n(n - 1), where $ and p are, respectively, the face number of an 3*-embedding 62 Activities on Chapter II and the genus of the orientable surface the embedding is on. It is known that the condition is still sufficient. If nonorientable surfaces are considered, the necessary condition n- 20 D 5g = 2~q and 36 = n(n - 1), u (n — 3)(n — 4) n —3)(n—4 a — — 3 where q is the nonorientable genus of the surface an 3*-embedding is on is not sufficient anymore for g > 1. because when n = 7, K; would have an 3*-embedding on the surface of nonorientable genus q = 2. However, it is shown that Ky is not embeddable on the Klein bottle([Lemma 4.1 in |Liu11]). It has been proved that except only for this case, the necessary condition is also sufficient. More other types of graphs, such as the complete bipartite graph Kmmn, n-cube Qn and so forth can also be seen in [Liu11]. R2.8 For 3 < i,j < 6, recognize if a graph has an (i, j)- embedding ( or (i*, j*)-embedding). More generally, investigate the upper or/and the lower bounds of i and j such that for a given type of graphs having an (i*, j*)- embedding. R2.9 Given two integers 7,7 not less than 7, justify if a graph has an (i, j)-embedding(or (7*, j*)-embedding). If a proper map has any pair of its faces with at most 1 edge in common, then it is called a polygonal map . R2.10 Conjecture. Any 3-connected graph has an embedding which is a polygonal map. From (a) and (b) in Fig.2.3, this conjecture is not valid for non- separable graphs. The two graphs are nonseparable. The graph in (a) has a multi-edge, but that in (b) does not. It can be checked that, none of them has a polygonal embedding. II.8 Researches 63 (a) (b) Fig.2.3 Two graphs without polygonal embedding If an embedding of a graph has its genus(orientable or nonori- entable) minimum among all the embeddings of the graph, then it is called a minimum (orientable or nonorientable) genus embedding. Based on the Euler formula, a minimum genus embedding has its face number maximum. So, a minimum (orientable or nonorientable) genus embedding is also called a maximum (orientable or nonorientable) face number embedding. Maximum face number implies that the average length of faces is smaller, and hence the possibility of faces being circuits is greater. This once caused to guess that minimum genus embeddings were all proper. However, a nonproper minimum genus embedding of a specific graph can be constructed by making 1 face as greater as possible with all other faces as less as possible. In fact, for torus and projective plane, all maximum face number embeddings are shown to be proper. For surfaces of big genus, a specific type of graphs were provided for all of their maximum face number embed- dings nonproper|Zhal]. R2.11 Conjecture. Any nonseparable regular graph has a max- imum face number embedding which is a proper map. A further suggestion is to find an embedding the lengths of all faces are nearly equal. The difference between the maximum length 64 Activities on Chapter II and the minimum length of faces in a map is call the equilibrium of the map. An embedding of a graph with its equilibrium minimum is called an equilibrious embedding of the graph. R2.12 Conjecture. Any 3-connected graph has an equilibrious embedding which is proper. An approach to access the conjecture is still for some types of graphs, e.g., planar graphs, Halin graphs, Hamiltonian graphs, further graphs embeddable to a surface with given genus etc. Chapter III Duality e The dual of a map (A45, P) is the map (A4, Pap) and vice versa. e The deletion of an edge in a map is the contraction of the corre- sponding edge in the dual map and vice versa. e The addition of an edge to, the inverse of deleting an edge in, a map is splitting off its corresponding edge on, the inverse of contracting an edge in, the dual map and vice versa. e The deletion of an edge with its inverse, the addition, and the dual of deletion, the contraction of an edge with its inverse, split- ting off an edge are restricted on the same surface to form basic transformations. II.1 Dual maps On the basis of IL2, for a basic permutation 7 on the ground set Xag, (Xag, P) is a premap if, and only if, (X44, P*) is a premap where P* = Py, y = af( Theorem 2.4). The latter is called the dual of the former. Since P” = P* Ba = (PagB)Ba = P(aBBa) = P, the former is also the dual of the latter. 66 Chapter III Duality Because the transitivity of two elements in the ground set X45 on a premap (Xap, P) under the group Yy, J = {P, a, 8), determine an equivalence, denoted by ~y,, the restriction of P on a class Xa pl Yyy is called a transitive block. Theorem 3.1 Premap M» = (X, Per;) is the dual of premap Mı = (4, Peri) if, and only if, K TNI — x / Wy (3.1) where Jo = (Pers, a, 8), J = (Peri, a, 8), and Per? = Peryy = Peri“, y = af. Proof Necessity. Since M» is the dual of Mı, Pers = Perjo/f. From Per, = PerjaG € V;, V; = Yz. Hence, (3.1) holds. This is the necessity. Sufficiency. Since Mj is a premap and Pers = Peria, M» is also a premap, and then the dual of M; by considering Theorem 2.4. This is the sufficiency. [] From this theorem, the duality between Mı and M» induces a 1-to-1l correspondence between their transitive blocks in dual pair. Because each transitive block is a map, it leads what the dual map of a map is. The representation of a premap by its transitive blocks is called its transitive decomposition. Example 3.1 Map Li = ({x, ax, zx, YTY, DESDE y= ap, and its dual . Lp = ({x, 0x, ax, yz}, (x, ax)). Or, in the form as II.1 Dual maps 67 Lı = (e,v) and L1 = (e, f), where v = (x, Gx), f = (x, ox), e = (z,oz, Bx, yz), y= af, and e* = [z,0r,ozx,yr), as shown in Fig.3.1. In the following figure, two As and two Bs are, respectively, iden- tified on the surface. Fig.3.1 Map and its dual This figure shows what a dual pair of maps looks like. It is a generalization of a dual pair of maps on the plane. A map with its under pregraph a selfloop is called a loop map. It is seen that L4 and its dual Li in Fig.3.1 are both loop maps. In a premap M = (X,P), if a vertex v = {(x)p, (ax)p} is transformed into two vertices v, = ((z, Px, ---, Pix), (az, aPiz, ---,oPz)) = ((x)p, (ax)p} 68 Chapter III Duality and U9 = Pity, pity. LN xD), (aPi*ty, aP zx, pues oJ *?3)) = ((P^*!z)p, (o/P?*1g)p), j = 0, with other vertices unchanged for permutation 7 becoming permuta- tion P’. It is seen that permutation 7' is basic and with the conjugate axiom as well. Hence, M’ = (AX, P") is also a premap. Such an op- eration is called cutting a vertex. If elements at vı are not transitive with elements at v in M’, then elements at v; and elements at və are said to be cuttable in M. ‘The vertex v is called a cutting vertex in M; otherwise, noncuttable . If there are two elements cuttable in a map, the the map is said to be cuttable in its own right. In virtue of Theorem 3.1, cuttability and noncuttability are con- cerned with only maps without loss of the generality of premaps. Lemma 3.1 A map M is cuttable if, and only if, its dual M* is cuttable. Proof Necessity. From M = (X, P) cuttable, vertex is assumed to cut into two vertices as v = (x, Px, ---, Pix) and . . vg = (Pi **n, Pia, ... Pa) for obtaining premap M' = (X, P") = Mi + M» where vj is on M; = (Xi, Pi), A = X + A5, Pj is the restriction of P’ on Xj, = 1,2. It can be checked that Mı and M» are both maps. Thus, on M* = (X,P*), vertex v* = (z, A, yP!z, PH, B, yP7 1a) can be cut into two vertices vt = (z, A, YP!) IIL.1 Dual maps 69 and us (Pitta, B, yP lr) where A and B are, respectively, linear orders of elements in X; and AX». This attains the premap M*' = (X,P") = Mi + M3 where A = X 4X, P*; is the restriction of P*' on %,i = 1,2. The necessity is obtained. Sufficiency. From the duality, it is deduced from the necessity. If two elements in the ground set of a map M are not transitive in M' obtained by cutting a vertex on M, they are said to be cuttable; otherwise, noncuttable. It can be checked that the noncuttability de- termines an equivalence, denoted by ~ne on the ground set of M. The restriction of M on each Xap “ne is called a noncuttable block. If all noncuttable blocks and all cutting vertices of a map deal with vertices such that two vertices are adjacent if, and only if, one is a noncuttable block and the other is a cutting vertex incident to the block, then the graph obtained in this way is called a cutting graph of the map. It is easily shown that the cutting graph of a map is always a tree. For a face f = (x)p, of a premap M = (4’,P), if there has, and only has, an integer l > 0 for transforming f into f= (z, Trta (P4)'x) and f- (PA E rg (Py) x) such that x and (P)'*!z are not transitive at all, then f is called a cutting face of M. From the procedure in the proof of Lemma 3.1, For a cutting vertex of a premap, there has, and only has, a corresponding cutting face in the dual of the premap. Theorem 3.2 Two maps M and N are mutually dual if, and only if, their cutting graph are the same and the corresponding non- 70 Chapter III Duality cuttable blocks are mutually dual such that a cutting vertex of one corresponds to a cutting face of the other. Proof Necessity. Because maps M and N are mutually dual, from the procedure in the proof of Lemma 3.1, there is a 1-to-1 corre- spondence between their noncuttable blocks such that two correspond- ing blocks are mutually dual. There is also a 1-to-1 correspondence between their cutting vertices such that the cyclic orders of their blocks at two corresponding cutting vertices are in correspondence. There- fore, their cutting graphs are the same. This is the necessity. Sufficiency. Because maps M and N have the same cutting graph, a tree of course, in virtue of the correspondence between cut- ting vertices and cutting faces, the sufficiency is deduced from Lemma 3.1. [] Note 3.1 Tow trees are said to be the same in the theorem when trees as maps(planar of course) are the same but not the iso- morphism of trees as graphs(the latter can be deduced from the former but unnecessary to be true from the latter to the former). On a premap M = (4,P), if an edge Kx is incident with two faces, i.e., yx € (x )p4 U (Ox) ps, then it is said to be single ; otherwise, i.e., yv € ijo, U (O1) p, (with only one face), double. An edge with distinct ends is called a link; otherwise, a loop . Clearly, single link, single loop, double link and double loop. Further, a double link is called a harmonic link, or singular link according as yx € (x)p,, or not. Similarly, a single loop is called a harmonic loop, or singular loop according as yx € (x)py, or not. Theorem 3.3 For an edge e, = {z, oz, Gx,yx} of premap M = (Xa B, P) and its corresponding edge ež = (x, Bx, ax, yx} of the dual M* = (Koa P*), P* = P4, IIL1 Dual maps 71 (i) e; is a single link if, and only if, e* is a single link; (ii) e is a harmonic link if, and only if, e* is a harmonic loop; (iii) e, is a singular link if, and only if, e* is a singular loop; (iv) e; is a double loop if, and only if, e* is a double loop. Proof Necessity. (i) Because e, is a link, (x)p and (yx)p belong to distinct vertices. And because e; is a single edge, (x)p. and (yx)p, belong to distinct face. By the duality, ež is a single link as well. (ii) Because e, is a double link, in spite of (x)p and (yx)p belong- ing to distinct vertices, yx, or ax € (z)p.. And because of harmonic k link, the only opportunity is yz € (r)p. From the duality, ež is a harmonic loop. (iii) Because e, is a singular link, in spite of (x)p and (yx)p belonging to distinct vertices, ax € (x)p«. In virtue of a as the second operation of M*, e% is a singular loop. (iv) Because e, is a double loop, Gx € (x)p and ax € (x)p.. From the symmetry between o and 0, M and M*, ež is a double loop as well. Sufficiency. From the symmetry in duality, i.e., M = (M*)*, it is obtained from the necessity. [] On the basis of this theorem, the classification and the dual re- lationship among edges are shown in Table 3.1. 12 Chapter II Duality L Single à Single L i i D D n o Harmonic Harmonic o n u u b b k l Singular Singular il k e e S S i Harmonic Harmonic i n n L g g L ő 7 Singular Singular e. ô o o Double ———— Double P P " * M Dual relation M Table 3.1 Duality between edges In the table above, harmonic links will be classified into segmen- tation edges and terminal links and harmonic loops into shearing loops and terminal loops in III.2 to have additional two dual pairs of edges: segmentation edges and shearing loops, terminal links and terminal loops. II.2 Deletion of an edge Let M = (Xa p(X), P)be a premap and e; = Kx = (x, az, pr, yx}, € X, an edge. What is obtained by deleting the edge e, from M is denoted by M-a (290) Er) (3.2) where P_, is the permutation restricted from P on Xa g(X) — Ka. Lemma 3.2 Permutation P_, is determined in the following II.2 Deletion of an edge 73 way as when e, is not a selfloop, Px(and aP~'z), if y = P !r(and aPx); Psy = 4 Pyz(fl aP yz), (3.3) if y = P yz(and aPyz); Py, otherwise, and when e, is a selfloop with yx € (x)p, Px(and aP1z), if y =P !z(and aPzx); P uy = < Pyz(and o/P- az), (3.4) if y =P !rz(and aPyz); Py, otherwise, otherwise, i.e., yx € (x)p, yx is replaced by (x in (3.4). Proof When e, is not a selfloop. Because only vertices (r)p and (yx)p are, respectively, changed in M — e, from M as (P'2)p_, = (Poe, Pa, Pa) and (Pyr), m (P^ yz, Pyz, a) Pz) (Fig.3.2(a)=(b)). This implies (3.3). When e; is a selfloop with yx € (r)p. Because only vertex (x)p is changed in M — e, from M as (P^!z)p | = (P Iz, Px,- , P yz, Pya,---,P 2) (Fig.3.2(c)(d)), or (P !z)p | — (P Ig, Pte P 18x, Pha, +++, Px) (Fig.3.2(c)—(d) in parentheses) according as yx € (x)p, or not. The former is (3.4). The latter is what is obtained from (3.4) with yx is replaced by Oz. [] 74 Chapter II Duality In Fig.3.2, the left two figures are parts of the original map and the right two figures, the results by deleting the edge Kz. Further, Fig.3.3-9 are all like this without specification. Pes Pyx NS LZ m d poi x xcd (P Z182) DK CU. (c) Plyg Pys "S LZ =. p | Ply (P 18m) TP wr Px Y Z Tur ET lo Pig (d) Fig.3.2 Deletion of an edge Lemma 3.3 For a premap M = (4,P), M — e, = (8 — II.2 Deletion of an edge T5 Kz,P ,)is also a premap. And, the number of transitive blocks in M — e, is not less than that in M. Proof Because P is basic for a, from Lemma 3.2 P. , is also basic for a. Because P satisfies the conjugate axiom for a, from Lemma 3.2 and Theorem 2.3 P_, is also satisfies the conjugate axiom for a. The first statement is done. Because any nontransitive pair of elements in M is never transi- tive in M — e,, the second statement is done. E If e, is an edge of a premap M such that M — e, has more transitive blocks than M does, then e; is called a segmentation edge. If an edge has its one end formed by only one semiedge of the edge itself, then it is called an terminal link. From the symmetry of elements in a quadricell, (x) can be assumed as the 1-vertex incident with a terminal link without loss of generality. Since (Py)yz = Px = x, yx € (x)p4. Hence, a terminal link is always a harmonic link. However, a harmonic link is unnecessary to be a terminal link. This point can be seen in the following theorem. Theorem 3.4 Fora map M = (X,P), M—e, = (X—Kz,P.,) is a map if, and only if, e, is not a harmonic link of M except for terminal link. Proof When (z,*x) € (x)p4, i.e., e; is a terminal link, Because no isolated vertex in any premap, from Lemma 3.3, M — e, = (X — Kz,P.,)isa map. In what follows, this case is not considered again. Necessity. Suppose M — e, is a map, but e, is a harmonic link of M. Because P4^!z and Py Iz + x((x, yx) € (x)py) for group Vy, J' = (P..,,0, 3}, are not transitive on the set ¥ — Kx, M — e, is not a map. This is a contradiction to the assumption. Sufficiency. Because M is a map, P_, is basic. From Theorem 2.3, P_, satisfies the conjugate axiom. Then, based on Table 3.1, two cases should be discussed for the transitivity. (i) When e; is a single edge(including single loops!) or singular link of M. Because e, is not a cut-edge of its under graph G(M), 16 Chapter III Duality G(M — ez) is connected. From Theorem 2.6, P_, for group V is transitive on the set A — Ka. Thus, M — e, is a map. (ii) When e, is a double loop of M. Because (Pa) !a)p , = ((Py)~*z, Tad, PES BPBz, P*yz, ---, (Py) az), we have Px = B(Py)x with (Py) tax and P8x = B(8Pxz) are transitive in M — e,. Hence, M — e, is a map as well. [] From the proof of the theorem, a much fundamental conclusion is soon deduced. Corollary 3.1 In à map M, an edge e, is a segmentation edge if, and only if, e, is a harmonic link except for terminal links. And, e, is a harmonic link if, and only if, it is a cut-edge of graph G(M ). Proof To prove the first statement. Necessity. Because e, is a segmentation edge, G(M — e,) is not connected. e, is only a link. On the basis of Table 3.1, e, is also a double edge. And, because e, is not singular, e, is only harmonic. Clearly, a terminal link is not a segmentation edge in its own right. Sufficiency. Because e, is not a terminal link, Px is distinct form x and yx is distinct from Pyx. And, because e, is a harmonic link, Px and P^z are not transitive in M — ez. Thus, e, is a segmentation edge. To prove the second statement. Because it can be shown that ex is a terminal link of M if, and only if, e, is an articulate edge of graph G(M), a cut-edge as well. This statement is deduced from the first statement. L] Let M = (Xa p(X), P) be a pregraph and e; = Kx = (x, az, Gx, yx}, € X, be an edge. The contraction of ey from M, denoted by M ee, = (Xa p(X) = Ka, Pex); is defined to be the dual of M* — ež where ež = (x, Bx, ax, yx}, the corresponding edge of e, in the dual M* of M. In other words, Per = FI III.2 Deletion of an edge 77 Lemma 3.4 Per is determined by the following (i-iii): (i) When e, is a link. For y € X,9(X) — Ka, Pyz(and P^ !z), if y= P^!z(and aPyz); Percy = < Pz(and oP-wz), if y= P^ yz(and oPz); (3.5) Py, otherwise, shown as in Fig.3.3(a)—9(b). (ii) When e; is a harmonic loop. For y € %,3(X) — Ka, Pyx(and o/P^!z), if y= P^ !z(and aP yz); P..y = 4 Pr(and aP az), if y= P^ yz(and aPz); (3.6) Py, otherwise, shown as in Fig.3.3(c)=—> (d). (iii) When e, is a singular , or double loop. For y € Xa p(X) — A. o/P- 8x(and o/P^!z), if y = P Iz(and P^ !6z); Pag =< Prud. (3.7) if y = aPx(and aP pz); Py, otherwise, shown as in Fig.3.3(e)=—>(f). Proof (i) When e, is a link. In the dual M* of M, from the duality, (a) p» = (a, Pyx, (PyY^, +++, (Py) 2) and (yx) p+ = (yx, Pa, (PyY a, - --, (Py) ya), Or (n)pees Un Pau oes UP) ab qm Pu PP and hence P*,, is only different from P* = Py at vertices (Pyx)p:, = (Pyr, (PyYz, ---, (Py) 12) 78 Chapter III Duality and (Pz)p., = (Px, (Py)*y2,--+, (Py) ya) or at vertex (Px) px, = (PIE 32999 (Pa) ya, Pa, as (Py) 1a) with their conjugations according as e, is single, or double. By con- sidering Per = P* y. Perly) = Pal P yx) P* Y(P ya) = P'(Py yz) = Px for y = P^ yz and Perly) = TP Ix) = P* A4(P Ix) =P (Psy) ix = Pyx for y = P ‘x. From the conjugate axiom, the cases for y = aP ‘x and y = a(P yz) in the parentheses of (3.5) are also obtained. Then, for other y, Pay) = P'.yy = (Py)yy = Py in the both cases. Therefore, (3.5) is true. (ii) When e; is a harmonic loop. In a similar way to (3.5) for e; single, (3.6) is also obtained. ii) When e, is a singular, or double loop. In the dual M* of M, (x) p» = (x, Pyz, (PyYz, +++, (Py) 1x) and (ax) p: = (ax, PIAL, (Py az, my (Py) lox), Or (x) p» = Ge Pyz, ttt3 (Py) lax, QT, Pad; E (Py) x), and hence (z)p*, = (Pyz, (Pyz, ---, (Py) 12) II.2 Deletion of an edge 79 and (ax)p», = (Pyaz, (Py)’az, +++, (Py) lox) or (x)p», = (P»yz,---, (Py) lox, Pyaz,---,(Py)7'z) with their conjugations according as e, is singular, or double. By considering Pes = P*.7, Pat) =P el a Pr) =P nmtapa = P* (Pax) = P* ((Py) lax) =Pron = Pox for y = aPx and Posl y) = Pa = P* (P's) = PŁ (P3) te) = Paz = Paix = oP 18x for y = P-!x. From the conjugate axiom, the cases for y = P x and a@P Gx are also obtained. Then, for all other y, Palu) = P' yy = (Py)yy = Py. This is (3.7). O NZ 4 a pd Pte nee 80 R Ptr Ppa L NI m c P a == IN 5 Px PlBr S (e) Chapter II Duality Ptr | Pyx Px | Playa Ply | Tz | Px * . NY TP Bx Sg-1 (f) Fig.3.3 Contraction of an edge IIL2 Deletion of an edge 81 From Lemma 3.4, it is seen that in the constriction of edge e, on a premap only if e, is not a selfloop, two vertices (z)p and (yx)p are composed of one vertex (P7lz)p,, = (Px, P»yz, ---, P ys, Pax,---,P 72) (Fig.3.3(a)— (b)); if e; is a harmonic loop, vertex (x)p is divided into two vertices (P^ ix)p,. = (P Iz, Pya P72) and (P^ yz)p,, = (P ya, Pa, «P ^na) (Fig.3.2(c)=(d)); and if e, is a singular, or double loop, vertex (x)p becomes vertex (Px)p,, = (Pzp P 18x, P t -- , o/P Bx) (Fig.3.3(e)>(f)). Lemma 3.5 For a premap M, M ee, is always a premap. And, the number of transitive blocks in M e e, is not less than that in M. Proof From Lemma 3.3 and the duality, the first statement is true. Because any nontransitive pair of elements in M is never tran- sitive in M ee, from Lemma 3.4, the second statement is true. L] If a harmonic loop e; has (x) py = (x), or (yx)py = (yx), then it is called a terminal loop. If the two elements of a co-semiedge appear in a vertex in succession, then the edge is called a twist loop. Lemma 3.6 For an edge e, of a map M, ex is a terminal loop if, and only if, ež is an terminal link in M*. And, e, is a twist loop if, and only if, ež is a twist loop. Proof A direct result deduced from the duality. [] Theorem 3.5 For an edge e, of a map M = (X4 X), P), M ee, is a map if, and only if, e; is not a harmonic loop but terminal loop. 82 Chapter II Duality Proof Because for a terminal loop ez, M ee, is always a map. In what follows, this case is excluded. Necessity. Suppose M ee, is a map but e, is a harmonic loop. Since ež is a harmonic link in M*(Table 3.1), from Theorem 3.4 and Lemma 3.1, P~'x and Pz are, respectively, belonging to two distinct transitive blocks of M. From Lemma 3.4(ii), M ee; has two transitive blocks. This contradicts to that M ee, is a map. Sufficiency. Since e, is not a harmonic loop, only two cases should be considered as e; is not a loop or e; is a singular loop. For the former, in spite of a single or double edge, from Lemma 3.4(i), M ee, is a map. For the latter, from Lemma 3.4(iii), M ee, = M — e, is also a map. Therefore, the theorem is done. [] If a loop e, has that P^ !v and Px are in distinct noncuttable blocks, then it is called a shearing loop . From Theorem 3.5, all shear- ing loops are harmonic. However, the converse case is unnecessarily true. Corollary 3.2 In a map M, an edge e; is a shearing loop if, and only if, ež is a harmonic, but not terminal loop in M*. Proof A direct result of Theorem 3.5. E Theorem 3.6 The dual of premap M —e, is the premap M*ee*, where M* is the dual of M and ež in M* is the corresponding edge of e, in M. Proof(1) Because M* ee? is the dual of (M*)* —e'2—-M — ex, by the symmetry of the duality the theorem holds. [] However, if the contraction of e; on M is defined by (3.5-7), then the theorem can also be proved. Proof(2) Based on Table 3.1, four cases should be discussed. (i) In M, e, is a single link, and hence ež is a single link in M*. Since (Pyx)p 4, = (P72) pz, (Pr)p = (PX) pz, II.2 Deletion of an edge 83 and (Pyx)p_, = (P*z)p: y, (M — e,)* = M* © æ. (ii) In M, e, is a harmonic link, and hence ež is a harmonic loop in M*(Dually, in M, ex is a harmonic loop, and hence e; is a harmonic link in M*). Now, (z)p, = (x)p». According as e; is a terminal link or not, a transitive block of M becomes one or two transitive blocks in M — e,. Meanwhile, According as e? is a terminal loop or not, a transitive block of M* becomes one or two transitive blocks of M* e ež. By considering the changes in vertices and faces, (M — er)“ = M* e ež is found. (iii) In M, e, is a singular link, and hence ež is a singular loop in M*(Dually, In M, e, is a singular loop, and hence ež is a singular link in M*). Since (P~'x)p_, = (P^'z)p;, and (P^ yz)p , = (P^ yx)p; ,, in view of (Pyx)p_,y = (Pyx)p:, we have (M —e;)' = M*eej. (iv) In M, e, is a double loop, and hence e? is a double loop in M*. Since (Px)p_, = (Pz)p;, and (Papa = (Px)p:., we have (M — ez)“ = M*ee;. O Corollary 3.3 In a map, an edge is a harmonic link if, and only if, the corresponding edge in its dual is a harmonic loop. And, an edge is a segmentation edge if, and only if, the corresponding edge in its dual is a shearing loop. Proof A direct result of Theorem 3.6. [] 84 Chapter III Duality Example 3.2 Map M = (Kz + Ky + Kz,P) where P = (a, By, yz) (y, 2,72) and its dual M* = (K*z + K*y + K*z, P*) where D* = Py = CA AT, AZ, VY, z); are, respectively, shown in Fig.3.4(a) and (b). Here, K = {1, a, 86, y} and K* = {1, 3,a,7} are used to distinguish o and f. Map M — e; = (Ky + Kz, P.) where P_z = (By,yz)(y, 2) and its dual (M — e,)* = (K*y + K*z,(P_z)*), where (P_z)* = Pry = (y,Gz,ay, yz), are, respectively, shown in Fig.3.4(c) and (d). It is easily seen that (M — e,)* = M* è ež. II.3 Addition of an edge 85 Fig.3.4 Duality between deletion and contraction IIIL.3 Addition of an edge Let M = (Xap, P) be a premap, e; = Kx = {z, az, Bx, yx}, and x € Xag. Write as M + ez = (Xag Kx, P), where P,, is determined from P in the following manner. For any y € X and two given angles (l, Pal) and (t, Pat), if l and t are not at the same vertex, or at the same vertex and e, as a harmonic loop(assume t € (l)p without loss of generality), then t(and ax), if y = x(and at); Pat(and x), if y = ax(and aPat); Pizy = 4 l(and Bx), if y = yx(and al); (3.8) Pal(and yx), if y = Gx(and aPal); Py, otherwise, 86 Chapter III Duality /UÀ Xx > \ JAN \~ f AN Pal (a) For (3.8) (b) For (3.9) Fig.3.5 Appending an edge shown in Fig.3.5(a), otherwise, i.e., e, is a double, or singular loop (assume t € (l)p without loss of generality), t(and ax), if y = x(and at); Pat(and x), if y = ox(ando'Pat); Pizy = $ Wanda), if y = yz(and al); (3.9) Pal(and Gx), if y = yz(and aPal); Py, otherwise, as shown in Fig.3.5(b). Such a transformation from M into M + e, is called appending an edge eyr. Lemma 3.7 For à premap M = (4,P), M+e, = (¥ + Kz,P,,)is also a premap. And, the number of transitive blocks in M 4- e, is not greater than that in M. Proof From (3.8) and (3.9), P,, is basic. In virtue of Theorem 2.3, it suffices to show that the orbits of ,, are partitioned into conjugate pairs for the conjugate axiom. In fact, if | and t are at distinct vertices, then 7?,, is obtained from P in replacing two vertices II.3 Addition of an edge 87 (t)p and (l)p by respective (t)p,, = (t, Pt,---, Pat, x) and (U)p,, = (, Pl,---,aPal, yx), or Oz is substituted for yx. If | and t are at the same vertex, then P, is obtained from P in replacing the vertex (l)p by (lp,, — (L Pl,---,aPat, x,t, Pt,---,aPal, yz) or (p, = (U, Pl, ---,aPat, r,t, Pt,---,aPal, Bx) according as e, is a harmonic loop or not. This shows that the orbits of P,, are partitioned into conjugate pairs for a. E Note 3.2 On the degenerate case t = l, if e, is a harmonic loop, then (Dg, = (L, PL oPal yx, £); otherwise, i.e., e; is a twist loop, OG, =U, F eo Pal. bre): Theorem 3.7 For a premap, not a map, M = (X5, P), the number of transitive blocks of M + e, is less than that of M if, and only if, e; ia a segmentation edge. If M is a map, then M + e, is also a map. Proof Since the number of components of graph G(M + e;) is less than that of G(M) if, and only if, e, is a cut-edge which is not articulate, the first statement is deduced from Corollary 3.1. Because the transitivity between two elements in the ground set of M under appending an edge is unchanged, the second statement is valid. [] Note 3.3 Let M' = (X + Kx, P) = M + e; for M = (€,P). Because P' ,, = P, M is obtained by the deletion of the edge e, from 88 Chapter III Duality M', i.e., M = M'—e,. Therefore, the operation of appending an edge on a premap is the inverse of the corresponding edge deletion. Now, another operation for increasing by an edge on a premap is considered. This is the splitting an edge seen as the inverse of edge contraction. Let M = (Xa B, P) be a premap. Suppose (l, Pal) and (t, Pat) are two angles. For x £ Xag, let M oe, = (Xag + Kx, Por), where Pox is determined by P in the following manner. The transformation from M into M oe, is called splitting an edge e, and ez, the splitting edge of M. Lemma 3.8 Let!c {t}pU {at}p. Ifl ¢ (t)p, U (Bt)p., then Pat(or ax), if y = x(or aPat); l(or x), if y = az(or al), Pory = 4 Pol(or Gx), if y = yx(or aPal); (3.10) t(or yx), if y = Gx(or at); Py, otherwise. The edge e, is a single link as shown in Fig.3.6; Otherwise, t.e., | € (t) py U (Gt) p,, then Pat(or ax), if y = x(or aPat); l(or x), if y= az(or al); Pory = $ t(or Bx), if y = Oz (or at); (3.11) Pol(or Bx), if y = yz(or aPal); Py, otherwise, or yx replaced by Gz to attain, respectively, e, as a singular or har- monic link shown in Fig.3.7. III.3 Addition of an edge 89 y X ALA 7N: Da zu G E (b) Moe; Fig.3.6 1 Z (t)py U (Ot)p, uv d EN ^ N Ey ü x Q 4 \ x ^N A 3 E (a) M (b) Moe; Fig3.7 | € (t)p, U (Bt)p-, 90 Chapter III Duality Proof Since | € {t}p U {at}p, l and t are at the same vertex. Thus, e; is a link. If | ¢ (t)p, U (Bt)p4, ie., | is in a face different from that t is in, or in other words, e, is single, then by the reason as (Pox)ex is different from only the vertex (Pont) Paje = (Pat, ol, Pal,---, at) where Pozz = Pat shown in Fig.3.6, from Lemma 3.4(i), (Por)exr = P. Otherwise, according as e, is singular or harmonic, (Pox)ex is different from only the vertex (Pozx)tp,.)., = (Pat, ol, Pal, ---, at) or (Pat,---,al,t,---,aPal) shown in Fig.3.7. From Lemma 3.4(i) again, (Por)ex = P. O Lemma 3.9 Let! ¢ (t)p U {at}p. If t and l are not transitive on M, then Pol(or ax), if y = x(or o/Pal); t(or x), if y= az(or at); Pat =< Pul(or pr); iu yrtor Pot; (3.12) l(or yx), if y = Gz(or al); Py, otherwise as shown in Fig.3.8 where e, is a harmonic loop. Otherwise, i.e., t and / are transitive on M, then Pol(or ax), if y = x(or aPal); Pat(or x), if y= ax(or aPat); Pay = 4 Wor Bx), if y = yz(or al); (3.13) t(or yx), if y = Bor at); Py, otherwise, as shown in Fig.3.9 where e; is a singular, or harmonic loop according as it is incident with two faces, or one face. II.3 Addition of an edge 91 mal DNV Aa ee a s (a) M (b) M o ex Fig.3.8 t and l nontransitive RA dox, D | Les a oe | Mo oe (a) M (b) Moe; Fig.3.9 t and l transitive Proof Since | € {t}p U {at}p, l and t are at the same vertex. Thus, e; is a link. If | ¢ (t)p, U (Gt)p,, i.e., lis in a face different from that t is in, or in other words, e, is single, then by the reason as 92 Chapter II Duality (Poxr)ex is different from only the vertex (Pork) (Poser = (Pat,-++, al, Pad, ---, at) where Pozz = Pat shown in Fig.3.6, from Lemma 3.4(i), (Por)ex = P. Otherwise, according as e, is singular or harmonic, (Pox)ex is different from only the vertex (Port Pda = (Pat,:-+,al, Pal, ---, at) or (Pat,---,al,t,---,aPal) shown in Fig.3.7. From Lemma 3.4(i) again, (Por)ex = P. O Lemma 3.10 If M — (X,7P) is a premap, then for any x € X, Moe, = (X + Ka, Pox) is also a premap. And, Moe, has the number of its transitive blocks not greater than M does. Proof From Lemmas 3.8-9, permutation P., is basic and parti- tioned into conjugate pairs for a. Then by Theorem 2.3, M oz is also a premap. This is the first statement. Because splitting an edge does not changing the transitivity of any pair of elements in X, the second statement holds. L Lemma 3.11 Edge e, is a harmonic loop on Moe; if, and only if, Port and P,yx are not transitive on M. Proof Necessity. Because the splitting edge e, is a harmonic loop on M, it is in the case (3.12). As shown in Fig. 3.8, P,,x(— L) and P.,ax(= t) are not transitive on M. Hence, by the symmetry among elements in Kx, Porz and Porya are not transitive on M either. Sufficiency. Because P,,0r(— l) and P,,ax(= t) are not tran- sitive on M, only l ¢ {t}p U {at}p is possible. This is the case for (3.12). Thus, e; is a harmonic loop. O Theorem 3.8 For a premap M = (Xap, P) not a map, the number of transitive blocks in M o e, is less than that in M if, and only if, e, is a harmonic loop. If M is a map, then M oe, is also a map. IIL3 Addition of an edge 93 Proof From Lemma 3.11, the number of transitive blocks in M o e, is less than that in M if, and only if, e, is a segmentation edge in M -—- e,. This is the first statement. Because splitting an edge in a map does not changing the transitivity, the second statement is obtained. [] Lemma 3.12 Edgee, = {z, az, Gx, yx} is appended in premap M if, and only if, edge ež = {x, Gx, ax, yx} is split to in premap M*. Proof Necessity. (1) If e, is a single edge of M + ez, i.e., yx ¢ (z)p,,4 U (0x) p,,7, then (x)p;, and (yz)p;, are two vertices on (M + €r)“. Hence, from (3.10-11), ež is the splitting edge from the vertex (Pi) Pp = US NR UIS Pu OMS (Pin) ya) on M*. (2) Otherwise, i.e., e; is a double edge on M + ex. Thus, yx € (z)p, 4 U (Bz)p,,,. Two cases should be considered. (i) If yz € (x)p,,4, then e, is a segmentation edge on M + €x. From Corollary 3.1, P?,v and P!,yz are not transitive on M. Furthermore, from (3.12), ež is a splitting edge(a harmonic loop) on M*. * x (ii) If yx € (Gx)p,,,, then from (3.13), e? is a splitting edge on M*. Sufficiency. (1) If ež is not a loop on M*oe?, then from (3.10-11), M has a face (Pottery = Phat (P5) m Phys em OP) YE): From (3.6), e, is an appending edge between angle (aP, ‘a, P*,yx) and angle (aP*, "yz, P?) in a face of M, as shown in (3.8-9). (2) Otherwise, i.e., ež is a loop. From (3.12-13), there are two faces on M with an angle each. Such two angles determine the ap- pending edge e, on M. [] 94 Chapter III Duality Theorem 3.9 The dual of a premap M + e, is the premap M* oe? where M* is the dual of M and ej is the dual edge in M* oe; corresponding to e; in M + e;. Proof A directed result of Lemma 3.4. [] From what has been discussed above, both the following diagrams (Tab: P) c (Tap tas Pa q q (3.14) Con ar We dq XP and (Mog P) => (Mus Ka, Pe) ü ü (3.15) EP eo. ASR Peer) are commutative. Example 3.3 Map M = (Kx--Ky,P), P = (x,y)(yy, yx) and its dual M* = (K*z + K*y,P*), P* = Py = (x,y, az, yy) (y, yz), are, respectively, shown in (a) and (b) of Fig.3.10. Notice that œ and B have distinguished roles in K = {1,a, 8, y} and K* = {1, 8, a, 7}. Map M +e; = (Kz + Ky + Kz, P42), Pig = (a y, a), Ba, yy) and its dual (M + e,)* = (K*z + K*y + K*z, (P42)®), (Pi = Pirs (2, Bz, vy, yz) yo. By), are, respectively, shown in (c) and (d) of Fig.3.10. According to (3.13), e? is the splitting edge at the pair of angles (x, ay) and (yy, 8x) in M*. Therefore, M* oe, = (K*x + K*y + K*z, (z, zx, az,vyy)(y,yz)). II.3 Addition of an edge If By is seen as y, then M* oet = K AUS (c) M+e, 95 (M +e)". (d) (M + ez)* Fig.3.10 Duality between appending and splitting an edge Based on Theorem 3.9, in a premap M = (X, P), splitting an edge e, attains M o e; which is just M* + e? obtained by appending the edge e; in its dual M* = (X*, P"). 96 Chapter III Duality IIL4 Basic transformation In a premap M = (X,5,P), the deletion of a single edge e; is called basic deleting an edge and its result is denoted by M — e,. The contraction of a link e, is called basic contracting an edge, and its result is denoted by M ey ez. The two operations are, in all, called basic subtracting an edge. Similarly, appending a single edge is called basic appending an edge, and splitting a link is called basic splitting an edge. Such two operations are, in all, called basic adding an edge. Apparently, M +, ez and M oy e, are the results of basic adding an edge e, on M in their own right. Basic subtracting and basic adding an edge are in all called basic transformation. From what we have known above, A premap becomes another premap under basic transformation. Theorem 3.10 Suppose M' is a premap obtained by basic transformation from premap M, then M’ is a map if, and only if, M is à map. Proof Because a single edge is never a harmonic link, from The- orem 3.4 the theorem holds for basic deleting an edge. Because a link is never a harmonic loop, from Theorem 3.5, the theorem holds for basic contracting an edge. Then, from Theorem 3.7-8, the theorem holds for basic adding an edge. [] Furthermore, for basic transformation, the following conclusion can also be done. Theorem 3.11 Let M = (X,P) bea map and M* = (X*,7P*), its dual. Then, for any single edge e; in M, (M —» e,)* = M*ey e? and for a single edge e, not in M, (M +, e,)* = M* op ež. Conversely, for any link e; in M, M ey e; = M* —, ež and for a link e, not in M, M op ey = M" Hye. Proof Based on the duality between edges as shown in Table 3.1, the statements are meaningful. From Theorem 3.6 and Theorem 3.9, the fist statement is true. In virtue of the duality, the second IIL.4 Basic transformation statement is true. 97 [] From this theorem, the following two diagrams are seen to be commutative: (Tap: P) EP] and (Tap P) GP (Aap 8, Pu) * us (X* Iu e ex) ———— Mag- Kr, Pres) +pex * * 9567. @ber (3.16) (3.17) On the basis of basic transformation, an equivalence can be es- tablished for classifying maps in agreement with the classification of surfaces. Activities on Chapter III III.5 Observations O3.1 Observe the condition for two permutations Per; and Per» satisfying (3.1) on the same ground set. O3.2 Given a tree(e.g., the star of 5 edges), observe how many maps are there for their dual maps all having the tree as under graph. O3.3 Write the dual of map M = (Kz, (x, Bx)). If a map has its dual with the same under graph, then it is said to be self-dual for the graph. Observe if M is self-dual for its under graph. O3.4 Provide a map which is cuttable and its under graph with 8, cut-vertex. 03.5 How to distinguish the cuttability of a map and the sep- arability of its under graph? O3.6 Is the under graph of the dual of a Eulerian map bipar- tite? If yes, explain the why; otherwise, by an example. O3.7 Is the dual of a preproper map(no double edge) always a preproper map? If yes, explain the why; otherwise, by an example. O3.8 Is the dual of a proper map(each face formed by a circuit in its under graph) always a proper map? If yes, explain the why; otherwise, by an example. 03.9 Is the dual of a polygonal map(two face with at most one edge in common) always a polygonal map? If yes, explain the why; III.6 Exercises 99 otherwise, by an example. 03.10 Isthe under graph of the dual of a map with 3-connected under graph still 3-connected? If yes, explain the why; otherwise, by an example. III.6 Exercises E3.1 Prove that the under graph of a map M = (X,P) isa tree if, and only if, its dual M* has the following three properties: (i) M* has only one vertex; (ii) For any x € X, x and yz are in the same orbit of P*; (iii) For any y € (x)p., there is no subsequence z, y, yx, yy or x, Yy, yx, y in its cycle. For a map M = (X (X), P) and $ C X, let M| Ks] = (Ks, P[Ks]) where [Ks] is the restriction of P on Ks = KS, and M[Ks] is said to be induced on Ks from M. Generally speaking, M[Ks] is not a map, but always a premap. A cocircuit of a graph with all of its edges incident to the same vertex is called a proper cocircuit . E3.2 Let M — Eg = M|X — Ks], Es = le;|Vxz € S). Prove that M — Es, S C X, is a map if, and only if, there is no proper cocircuit of G|M] on graph G(M |Ks]). A proper circuit of a map M is such a set C of edges that C* = {e*|\Ve € C) is a proper cocircuit of G|M*]. E3.3 Let H = G/M] and My be the set of all maps whose under graphs are H. Prove that if C is a proper circuit of a map M, then it is a proper circuit of all N € My. E3.4 Let M e Es = (M* — E%)* where M* is the dual of M and Ez = [ei|Vr € S). Prove that M e Es is a map if, and only if, Eg is a proper circuit of M. E3.5 Prove that a map M is on the sphere if, and only if, each face of its dual M* corresponds to a proper cocircuit of M. 100 Activities on Chapter III If a map has its dual Eulerian, then it is called a dual Eulerian map. E3.6 Prove that a map is a dual Eulerian map if, and only if, each of its faces is incident with even number of edges. If a preproper dual Eulerian map has each of its faces partition- able into circuits, then it is called a even assigned map . A map is called bipartite when its under graph is bipartite. E3.7 Prove that a dual Eulerian map is bipartite if, and only if, it is even assigned. E3.8 Prove that a quadrangulation is bipartite if, and only if, it is without loop. For a loopless quadrangulation Q = (44,5, Q), from E3.8, its vertex set V can be partitioned into two subsets Vi and V5 such that the two ends of each edge are never in the same subset. Such a subset of vertices is called an independent set. X it and d stand for the sets of elements in Xa g incident to, respectively, Vj and V;|Dehl, Gaul, MuS1]. E3.9 Let o = 8Qw, y — aß. Prove that (cr, Qox), x € Xap, is an angle. Angles (oz,0 Qox) and (x, Qox) as shown in E3.9 are called an independent pair . Thus, each face in a quadrangulation has exactly two independent pairs of angles. E3.10 Let Kı = {1, a1, 61,71}, 1 = o11. And, a; = Q and Bı = GQy7(= c as shown in E3.9). Prove (i) Kı is the Klein group of four elements; Gi) Xi, = 95 55 Kw v; €Vi ye(vj o (ii) Qi = (s Qi) is a map, where Q; is the restriction of O on xo cu Similarly to E3.10, from V5, another map Q» can be deduced. Q1 and Q» are called the incident pair of the quadrangulation. III.7 Researches 101 E3.11 Prove that the two maps in the incident pair of a quad- rangulation are mutually dual. E3.12 Prove that any planar quadrangulation is loopless. E3.13 Let A and B be the sets of, respective all planar quad- rangulations and all dual pairs of planar maps. Establish a 1-to-1 correspondence between A and B(i.e., bijection). IIL7 Researches For a map, if the basic deletion of an edge can not be done anymore, then the map is said to be basic deleting edge irreducible. similarly, if the basic contraction of an edge can not be done on a map anymore, then the map is said to be basic contracting irreducible. R3.1 Given the size, determine the number of self-dual maps as an integral function of the size, or provide a way to list all the self- dual maps of the same size and deduce a relation among the numbers of different sizes. R3.2 Given the size, determine the number of maps all basic deleting irreducible as an integral function of the size, or provide a way to list all such maps with the same size and deduce a relation among the numbers of different sizes. R3.3 Given the size, determine the number of maps all basic contracting irreducible as an integral function of the size, or provide a way to list all such maps with the same size and deduce a relation among the numbers of different sizes. R3.4 For any given graph, determine the number of maps all basic deleting irreducible with the same under graph, or provide a way to list all such maps with the same size and deduce a relation among the numbers of different sizes. R3.5 For any given map, determine the number of all basic deleting irreducible maps obtained from the map by basic deletion, or 102 Activities on Chapter III provide a way to list all such maps with the same size and deduce a relation among the numbers of different sizes. R3.6 Fora given graph, determine the number of maps all basic contracting irreducible with the same under graph, or provide a way to list such maps and deduce a relation among the numbers of different sizes. R3.7 For a given map, determine the number of all basic con- tracting irreducible maps obtained from the map by basic contraction, or provide a way to list such maps and deduce a relation among the numbers of different sizes. If a map is basic both deleting and contracting irreducible, then it is said to be basic subtracting irreducible. R3.8 Given the size, determine the number of basic subtracting irreducible maps as an integral function of the size, or provide a way to list all such maps with the same size and deduce a relation among the numbers of different sizes. R3.9 Fora given graph, determine the number of maps all basic subtracting irreducible with the same under graph, or provide a way to list such maps and deduce a relation among the numbers of different sizes. R3.10 Find a relation between triangulations and quadrangu- lations. If a map has each of its faces pentagon, then it is called a quin- quangulation. Similarly, the meaning of a hexagonalization. R3.11 Justify whether or not a triangulation has a spanning quinquangulation or hexagonalization. If do, determine its number. R3.12 Even assigned conjecture: A bipartite graph without cut-edge has a super map even assigned. Chapter IV Orientability e The orientability is determined by the orientation for each edge with two sides; otherwise, nonorientability. e The basic equivalence is defined via basic transformations to show that the orientability is an invariant in an equivalent class. This equivalence is, in fact, the elementary equivalence on surfaces. e The Euler characteristic is also shown to be an invariant in an equivalent class. e Two examples show that none of the orientability(nonorientability as well) and the Euler characteristic can determine the equivalent class. IV.1 Orientation Let M = (X44, P) be a map(from Theorem 2.6, without loss of generality for a premap), and Y7, I = (y, P}, y = aß = Ga, be the group generated by the set of permutations /. Now, it is known that the number of orbits of P on X,,5 is double the number of vertices on M and the number of orbits of P^ on X4, is double the number of faces on M. Because P, Py € Vr, the number of orbits of the group Vr; on Xag is not greater than any of their both. 104 Chapter IV Orientability Lemma 4.1 The number of orbits of the group V; on Xap is not greater than 2. Proof Because Py € Wy, for any x € Xa, {r}p, € {r)en Here, {x}p, and {x}, are the orbits of, respectively, the permutation Py and the group V; on Xap. For any chosen element z € Xag, from P € V, for any y € {x}p,, (y) e4 € (x]w,, and from y € v, Uv», € ovde, € {rhor In view of Theorem 2.6, at least half of elements at each vertex belong to (x)w,. Therefore, (x), contains at least half of elements in X, 5. Similarly, {ax}, contains at least half of elements in 4X, 5. In consequence, based on the basicness of P for o, V; has at most 2 orbits on A a. O According to this lemma, a map M = (ag, P) has only two possibilities: group Vr is with two, or one, orbits on Xag. The former is called orientable, and the later, nonorientable . From the proof of the lemma, an efficient algorithm can be es- tablished for determining all the orbits of group V; on the ground set. Actually, in an orientable map, because V; has two orbits for o, the ground set is partitioned into two parts of equal size. It is seen from Lemma 4.1 that each quadricell (7.e., edge) is distinguished by two elements in each of the two orbits. And, the two elements of an edge in the same orbit have to be with different ends of the edge. Thus, each of the two orbits determines the under graph of the map. Example 4.1 Consider map M = (X,P) where X = Kr + Ky+ Kz+Ku+ Kv+ Kw and P = (a, y, z)(yz, u,v) (Yu, yy, w)(yw, yu, yz) as shown in Fig.4.1(a). Its two faces are (2, ^99, yv, yz) IV.1 Orientation 105 and (yx, Y, W, YU, V, VYY, Z, u). In fact, for this map, group V; has two orbits. One is (x, yw, yv, yz, YL, y, w, yu, v, yy, z, u]. The other is what is obtained from it by multiplying o to each of all its elements. Thus, M is orientable. Fig.4.1(b) shows that M is an embedding of the complete graph of order 4 on the torus (yuy lu). Fig.4.1 An embedding of Ky Corollary 4.1 If V;, ] = (P, af], has two orbits on A5, then they are conjugate for both o and f. Proof It is known from Lemma 4.1 that the two obits have the same number of elements, i.e., half of Xa g. Because y € {x}w, if, and only if, ay € (ax]«, and for any Kx, ox and Bx are always in the same orbit of V, this implies that {ar}y, = (Ox)w, different from [x y, and hence the conclusion. O Example 4.2 Consider map N = (X, Q) where X = Kr + Ky+ Kz+ Ku+ Kv+ Kw and Q = (x,y, z)(yz, u, v) (yu, By, w) (vw, yu, ya) 106 Chapter IV Orientability as shown in Fig.4.2(a). That is obtained from the map M in Fig.4.1(a) in the replacement of cycle (^v, yy, w) by cycle (zv, Gy, w). Here, N has also two faces (z, yw, yv, yz) and (yz, y, Bv, Qu, Bu, YY, Z, u). Because By € {y}ay € (v) w,, ,,; from the corollary group V, 9} has only one orbit, t.e., hpa = &. Therefore, N is nonorientable. It is seen from Fig.4.2(b) that N is an embedding of the complete graph of order 4 on the surface (yuyu') ep (yyuu), i.e., Klein bottle . Fig.4.2 An embedding of Ky on the Klein bottle Theorem 4.1 A map M = (X, P) is nonorientable if, and only if, there exists an element x € X such that Gx € {x}w,, or ax € {x}u, where I = (y, P}. Proof Necessity. Suppose ax ¢ {x}w,, then V; has at least two orbits. However from Lemma 4.1, it has exactly two orbits . Thus, M is never nonorientable. The necessity holds IV.2 Basic equivalence 107 Sufficiency. Because Gx € [x]w,, from Corollary 4.1 it is only possible to have {x} = X, i.e., V; has only one orbit. Hence, M is nonorientable. This is the sufficiency. [] This theorem enables us to justify the nonorientability and hence the orientability of a map much simple. If there exists a face (x)p,, denoted by S,, such that ox € S,, or there exists a vertex (r)p, denoted by S,, such that Gx € S, on M, then M is nonorientable(as shown in Example 2). Otherwise, From y € $, via acting P, or y, for getting z Z Sz, Sx is extended into S,U{z}p,U {z}p which is seen as a new S to see if y,ay € S, or y, Gy € S. If it does, then M is nonorientable; otherwise, do the extension until |S| = |X|/2, or S= X. Theorem 4.2 A map M = (X,7) is orientable if, and only if, its dual M* = (X*, P*) is orientable. Proof Because P* = Py € Woo (y = ap = Ba), Vip = Yi py. So, for any z € A = X*, (zjwe,,,, = (]w,, ,.,. This is the conclusion of the theorem. [] IV.2 Basic equivalence First, observe the effect for the orientability of a map via basic transformation. For a map M = (X,7P) and its edge ez, let M —p e, and M eye, be, respectively, obtained by basic deleting and basic contracting the edge e, on M. From Theorem 3.10, M —» e; and M ej e, are both a map . 108 Chapter IV Orientability SS : Pe — ls 8] (a) Single link e; (b) Single loop e; Fig.4.3 Basic deleting an edge Lemma 4.2 If M' is the map obtained by basic subtracting an edge from M, then M' is orientable if, and only if, M is orientable. Proof First, to prove the theorem for M' = M —, eg. Necessity. From M' = M — e, = (X', P) orientable, group V' = Wy, py has two orbits on X' = X—Kz, i.e., {Px}w and {Pax}w. Because e; is single, Pyx € {Px}w and PGx € (PoxWw. So, group V = V.» has two orbits [je = {Pxr}w U (x, yr} and (axis = {Par}w U (oz, Br} on ¥, i.e., M is orientable. Sufficiency. Because e, is a single link (Fig.4.3(a)), or single loop (Fig.4.3(b)), in virtue of that group V has two orbits {x} and (ox]w on X, group WV’ has two orbits {Pr}w = (z]e — (x, yr} and [Por jw = {ar} — lov, Bx} on X’, i.e., M' is orientable. Then, to prove the theorem for M' = M e, e,. On the basis of Theorem 3.11, the result is directly deduced from that for M' = M —p Er. E 109 IV.2 Basic equivalence Whenever that two new angles occur in the deletion of an edge with 4 angles lost is noticed, the edge appending as the inverse of deletion is done between the two angles. And then the same case comes for basic deleting and basic appending an edge. In this sense, Lemma 4.3 in what follows is seen as a direct result of Lemma 4.2. However, it is still proved in an independent way. For basic appending an edge, since the edge is only permitted to be a single link or a single loop, this operation is, in fact, done by putting the edge in the same face. Let map M = (X, P) have a face (y)Py = (Yo Vi; * +, Ys) = M +p ez where yo = y, yı = (Py)y, +++, Ys = (Py) !y. Denote M +; ey when appending the edge e, in between angles (y, Pay) and (yi, Pay;), 0i xs. From (3.10), M +i e; = M ty e, 0 < i < s, are all maps(Fig.4.4). Eme QE Eae A Vi-i SS Yl-1 m z= ZEN P= (b) M +i ex 110 Chapter IV Orientability -— = ; A. ] A i (d) M +141 ex (c) M +i ex Fig.4.4 Basic appending an edge Lemma 4.3 Maps M+ ;e, = M+ per, 0 <i € s, are orientable if, and only if, M is orientable. sr Peg Proof necessity. Since M' = M +; ex are all orientable, group V' = Wy. py has two orbits (zw and (ow on Xt’ = X + Kx. Because e, is a single link (Fig.4.4(a) and (c)), or single loop (Fig.4.4(b) and (d)), P'r € (x)w and P'az € (ox. Hence, group V = Y; p} has two orbits {P't}y = {x}w — ix yx} and {P'ax}y = far}w — (oz, Bx} on X. This implies that M is orientable. Sufficiency. Since e, is a single link (Fig.4.4(a) and (c)), or single loop (Fig.4.4(b) and (d)), the two orbits {r}w = {y}v + {2,72} and lox jw = {ay}y + (oc, Gx} of group V' on X" are deduced from the two orbits {y}w and {ay}w [] of group V on X. Therefore, M' is orientable. IV.2 Basic equivalence 111 As for basic splitting an edge, whenever that two new angles occur in the contraction of an edge with 4 angles lost is noticed, the edge splitting seen as the inverse of contraction is done between the two angles. Next, consider how to list all possibilities for basic splitting from a given angle. For a map M = (#,P), let (y)p = (Yo, Yi; tts Yl-1,Ul,°°°; Üs) yo = y, s > 0, be a vertex. Denote by M o;e, the result obtained from M by basic splitting an edge between angles (y, Pay) and (yi, ay;_1) where y = yo and Pay = ays. From Theorem 3.10, M o;e, = M+ pez, 0 € i € s, are all maps (Fig.4.5). AA ANN á AAN (c) M o e (d) M os e Fig.4.5 Basic splitting an edge Lemma 4.4 Fora map M = (%,P) and x ¢ X, map M oiez = M +p ez, 0 € i € s, is orientable if, and only if, M is orientable. Proof Necessity. Because M' = M o; ez = (X',P),0x i s, is orientable, group Y = V, py has two obits {x}w and {ar} on 112 Chapter IV Orientability AX" = X + Kr. Since e, is a single link (Fig.4.5(b), (c) and (d)), or a double link (Fig.4.5(a) and (c)), P'r € {x}w and P'az € {ar}w. Therefore, group V = V, p; has two orbits UP'zY, = (xw — {x, yx} and (P'ox)y = {ar}w — (oc, Bx} on X. This implies that M is orientable. Sufficiency. Because e, is a single link (Fig.4.5(b), (c) and (d)), or a double link (Fig.4.5(a) and (c)), the two orbits (xw = {y}w+ {x, yx} and {ar}w = {ay}wt+{az, Bx} of group V' on X' are deduced from the two orbits {y}w and {ay}w of group V on X. Therefore, M' is orientable . O Corollary 4.2 If M' is the map obtained by basic adding an edge from map M, then M' is orientable if, and only if, M is orientable. Proof A direct result of Lemma 4.3 and Lemma 4.4. E The operation of basic appending an edge between two successive angles of a face in a map is also called increasing duplition (Fig.4.4(b) and (d)), and its inverse operation, decreasing duplition. And du- ally, the operation of basic splitting an edge is also called increasing subdivision(Fig.4.5(b) and (d)), and its inverse operation, decreasing subdivision. Corollary 4.3 A premap M' obtained by increasing duplition, increasing subdivision, decreasing duplition, or decreasing subdivision from a map M is still a map with the same orientability of M. Proof The results for decreasing duplition and decreasing sub- division are derived from Lemma 4.2. Those for increasing duplition and increasing subdivision are from Corollary 4.2. [] If map Mı can be obtained from map M^» via a series of basic adding and/or basic subtracting an edge, then they are called mutually basic equivalence, denoted by Mı ~pe Mo. IV.3 Euler characteristic 113 Theorem 4.3 If maps Mı ~pe M», then Mj is orientable if, and only if, M» is orientable. Proof A direct result of Lemma 4.2 and Corollary 4.2. [] Since ~pe is an equivalent relation, maps are partitioned into classes of basic equivalence, in short equivalent class. Theorem 4.3 shows that the orientability of maps is an invariant in the same equiv- alent class. IV.3 Euler characteristic For a map M = (X,), let v = v(M), € = e(M) and d= d(M) are, respectively, the order(vertex number), size(edge number) and co-order(face number) of M, then x(M) =v—e+¢ (4.1) is called the Euler characteristic of M. Theorem 4.4 Let M* be the dual of a map M, then x(M7) = x(M). (4.2) Proof Because v(M*) = $(M), e(M*) = e(M) and ¢(M*) = v(M), (4.2) is obtained from (4.1). O Lemma 4.5 For a map M = (X,P) and an edge ez, £ € X, let M — e, and M ee, be, respectively, obtained from M by deleting and contracting the edge e,, then X(M — ez), if e, is single; «M) =4 (4.3) X(M e e;), if e, is a link. Proof From Theorem 3.11 and Theorem 4.4, only necessary to consider for one of M — e, and M e e}. Here, the former is chosen. To prove x(M — ez) = x(M) for e, single. 114 Chapter IV Orientability Because e; is single, v(M — ez) = v(M), (M — ez) = e( M) — 1 and ó(M — er) = ¢(M) — 1. From (4.1), X(M — ex) = v(M) — (e(M) — 1) + (6(M) — 1) = v(M) — (M) + &(M) - x(M). This is just what is wanted to get. [] Corollary 4.4 For any map M, x(M) < 2. Proof By induction on the co-order ¢(M). If M has only one face, i.e., (M) = 1, then In view of the connectedness, e(M) > v(M) — 1. In consequence, x(M) € v(M) — (v(M)—1) 1 — 2. Thus, the conclusion is true for 9(M) = 1. In general, i.e., 96(M) > 2. Because of the transitivity on a map, there exists a single edge e, on M. From Lemma 4.5, M' = M — e, has X( M^) = x(M). Since 6(M") = 6(M) —1, by the induction hypothesis x(M") € 2. That is x(M) < 2, the conclusion. O For an indifferent reception, because the order, size and co-order of a map can be much greater as the map is much enlarged. The con- clusion would be unimaginable. In fact, since the deletion od a single edge does not change the connectivity with the Euler characteristic unchange and the size of a connected graph is never less than its order minus one, this conclusion becomes reasonable. Corollary 4.5 For basic subtracting an edge e; on a map M, x(M —» ex) = x(M) and x(M e, ex) = x(M). IV.3 Euler characteristic 115 Proof A direct result of Lemma 4.5. [] Lemma 4.6 Fora map M = (4,P) and an edge ez, x ¢ X, let M + e and M o e, be obtained from M via, respectively, appending and splitting the edge e,, then «un - 1 X(M + e), if e, is single; 4.4 X(M o ez), if e, is a link. us Proof From Theorem 3.11 and Theorem 4.4, only necessary to consider for one of M +e, and M oez. The former is chosen. To prove x(M + ex) = x(M). Because e, is single, then v(M + e;) = v(M), e(M + er) = e(M) +1 and ó(M + e,) = ¢(M) +1. From (4.1), x(M + ex) ae (e(M) + 1) + (6(M) +1) moe ) - (M) + (M) x(M). Therefore, the lemma is true. [] Corollary 4.6 For basic adding an edge e; on a map M, x( M+» er) = x(M) and x(M op e;) = x(M). Proof A direct result of Lemma 4.6. [] For a map M = (4,7) and an edge ez, x € X, let Mj.) and Mp4; be obtained from M by, respectively, increasing subdivision and increasing duplition for edge ez, and Miesz] and Mj 4, by, respectively, decreasing subdivision and decreasing duplition fox edge e,. From Corollary 4.3, they are all maps. Corollary 4.7 For increasing subdivision and increasing dupli- tion, X(Mj) = x(M); x(Mpaj) = x(M) (4.5) and for decreasing subdivision and decreasing duplition, X(Miex}) = x(M); x(Mpia) = x(M). (4.6) 116 Chapter IV Orientability Proof Because increasing subdivision and increasing duplition are a special type of basic adding an edge, from Corollary 4.5, (4.5) holds. Because decreasing subdivision and decreasing duplition are a special type of basic subtracting an edge, from Corollary 4.6, (4.6) holds. The corollary is obtained. [] The following theorem shows that the Euler characteristic is an invariant in the basic equivalent classes of maps. Theorem 4.5 If maps M; ~pe Mo, then x(M1) = x(M3). (4.7) Proof Because the basic transformation consists of basic sub- tracting and basic adding an edge, from Corollary 4,5 and Corollary 4.6, (4.7) is obtained. [] IV.4 Pattern examples Pattern 4.1 Consider the map M = (X, P) where XY = Ka + Ky 4 Kz 4 Ku 4 Kw + Kl and P = (x,y, z)(al, yz, Qw)(Bl, yy, au)(w, yu, 8x), shown in Fig.1.13. By deleting the single edge e, on M, let M; = (X1, 1) = M—be;, then Xj = Ky + Kz + Ku+ Kw + Kl and P, = (y, z)(al, yz, Bw) (BL, yy, oru) (w, yu). By contracting the double link e; on Mi, let Mo = (Xz, P2) = M; e, ez, then Xo = Ky + Ku + Kw + Kl and Po = (y, Bw, al)(8l, yy, oru) (w, yu). By contracting the double link e; on Mə, let M3 = (A3, P3) = Mp €y ej, then 43 = Ky + Ku + Kw and P; = (y, Bw, yy, ou) (w, yu). IV.4 Pattern examples 117 By contracting the double link e, on M3, let M4 = (A4, P4) = Ms € eu, then X4, = Ky + Kw and Py = (y, Bw, yy, aw). Now, M, has only one vertex and only one face and hence any basic transformation for subtracting an edge can not be done. It is a map on the torus (Fig.4.6). Fig.4.6 A map basic equivalent to M Pattern 4.2 Again, consider the map N — (X, Q) where X — Ket+Ky+ Kz- Ku+ Kw+ KI and Q = (x,y, z)(al, yz, Bw) (Gl, By, au)(w, yu, Bx), as shown in Fig.1.14. By deleting the single edge e; on M, let Ny = (41, Q1) = N—pez, then 4% = Ky + Kz + Ku + Kw + Kl and Q; = (y, z) (al, yz, Bw) (BL, By, au) (w, yu). By contracting the double link e; on Nj, let No = (A5, Q2) = N, @, e,, then X = Ky + Ku + Kw + KI and Qə = (y, Bw, al) (Bl, By, au) (w, yu). By contracting the double link e; on No, let N3 = (A35, Q3) = No €y ej, then A3 = Ky + Ku + Kw and Qa = (y, pw, By, ou) (w, yu). 118 Chapter IV Orientability Finally, By contracting the double link e, on Ns, let N4 (A4, Q4) = N3 ey ey, then X4, = Ky + Kw and Q4- (y, fw, By, aw). Now, the basic transformation can not be done anymore on N4. N4 is a map on the Klein bottle, as shown in Fig.4.7. B 1 w XL Vv A A By B Fig.4.7 A map basic equivalent to N From the two patterns, it is seen that M4 ^^y. N4, and hence M ^y. N. Although their Euler characteristic are the same, i.e., x(M) = x(M4) =1 -2+1 = x(N1) = x(N), their orientability are different. Activities on Chapter IV IV.5 Observations 04.1 Fora map M = (X,,4, P), observe how many orbits does the group Yia p} have on the ground set Yag? What condition is it for its transitivity. O4.2 Fora map M = (A,,4, P), observe how many orbits does the group Vygp.;, ^ = ap, have on the ground set Xag? What condition is it for its transitivity. 04.3 Fora map M = (X,,4, P), observe how many orbits does the group V5», have on the ground set Xag? What condition is it for its transitivity. O4.4 Fora map M = (A,,4, P), observe how many orbits does the group Vyapy, y = o, have on the ground set Xag? What condition is it for its transitivity. 04.5 If map M = (4,P) is orientable, is M — e, always ori- entable for any ez, r € X? If yes, explain the reason; otherwise, by an example. 04.6 If map M = (X,7P) is nonorientable, is M — e, always nonorientable for any ez, x € X? If yes, explain the reason; otherwise, by an example. 04.7 If map M = (X,7) is orientable, is M e e, always ori- entable for any ez, x € X? If yes, explain the reason; otherwise, by an example. 04.8 If map M = (X,P) is nonorientable, is M e e, always 120 Activities on Chapter IV nonorientable for any ez, x € X? If yes, explain the reason; otherwise, by an example. 04.9 If map M = (4,P) is orientable, is M + e, always ori- entable for any ez, x ¢ X? If yes, explain the reason; otherwise, by an example. 04.10 If map M = (4,P) is nonorientable, is M + ey always nonorientable for any ez, x ¢ X? If yes, explain the reason; otherwise, by an example. 04.11 If map M = (X,7P) is orientable, is M o e, always ori- entable for any ez, x ¢ X? If yes, explain the reason; otherwise, by an example. 04.12 If map M = (X, P) is nonorientable, is M o e, always nonorientable for any ez, x ¢ X? If yes, explain the reason; otherwise, by an example. 04.13 Show by example that a face of an orientable map M does not correspond to a cocycle on its dualM*. IV.6 Exercises E4.1 ‘Try to prove Lemma4.1 in three different manners. E4.2 For a map M, prove that there exists a nonnegative inte- ger n and basic transformations 71, 7,::-,75, such that M* = | [rM (4.8) i=1 where M* is the dual of M. E4.3 Foramap M = (Xap, P), the group Yip}, y = aß, with two orbits on X45 is known. Prove that x(M) = 0 (mod 2). (4.9) E4.4 Prove that a mapM = (AX,,5, P) is nonorientable if, and IV.6 Exercises 121 only if, there exist x,y € X such that Ky N TE iual > 2. E4.5 Try to prove Corollary 4.4 in two different manners. E4.6 Ifa map M = (X45,P) has only one face, prove that M is nonorientable if, and only if, there exists an x € X such that ax € [z)p, where y = af. For a map M = (Xag, P), let A be the set of all orbits of P on Xag. Graph Gu = (V, E) where V = A and E = ((A, B)3x € YX,x € A and yz € B). And, Gy is called the subsidiary graph of M. E4.7 Fora map M = (Xa p, P), prove that the group V(45pj has two orbits on Xag if, and only if, the subsidiary graph Gm of M has two connected components. For a map M = (Aap P), y = a, let fi = ((zi)p4 (Cri) Py}, i = 1,2,---,¢, be all the faces of M. If a set S = {s;|1 € i € $) of orbits of permutation Py on X», satisfies |S N f;| = 1,i = 1,2,---,¢, then S ia called a face representative of M. Let graph Gs = (V, E) be with V = S as a face representative of M and e — (s,t) € E as a pair of faces s and t with an edge in common. On E, define a weight function 0, there exists x € s such that yz € t; w(e) — | 1, otherwise for e = (s,t) € E. And, (Gs, w) is called an associate net of M. On an associate net (Gs, w), Gs = (V, E), of a map M, if there exists a label (v) € {0,1} for vertex v € V such that for any e = (u,v) € E, l(u) + l(v) = w(e) (mod 2), then the associate net (Gs, w) is said to be balanced. E4.8 For a map M, prove that if one of its associate nets is balanced, then all of its associate net are balanced. 122 Activities on Chapter IV E4.9 Prove that a map is orientable if, and only if, the map has an associate net balanced. For a graph G = (V, E), A C E, if V has a 2-partition, i.e., V «WU V3 and Vj NV = Q, such that A= ((u,v) € E|u € Vi, v € Va}, then A is called a cocycle of G. E4.10 Prove that a map M is orientable if, and only if, for a face representative S of M, its associate net (Gs, w) has (e € E|w(e) = 1} as a cocycle. EA4.11 Prove that a map M has its Euler characteristic 2 4 if, and only if, each of its faces corresponds to a cocycle of the under graph of its dual M*. IV.7 Researches R4.1 Characterize that the under graph of a map has an super map with its Euler characteristic 1. R4.2 Characterize that the under graph of a map has an super map with its Euler characteristic 0. RA4.3 For any orientable map, characterize that the under graph of the map has an super orientable map with its Euler characteristic 0. R4.4 For a vertex regular map and a given integer g < 1, char- acterize that the under graph of the map has a super map with its Euler characteristic g. R4.5 For a vertex regular orientable map and a given integer g < 0, characterize that the under graph of the map has a super map with its Euler characteristic 2g. A graph which has a spanning circuit ia called a Hamiltonian graph. Such a spanning circuit is called a Hamiltonian circuit of the IV.7 Researches 123 graph. If a map has its under graph Hamiltonian, then it is called a Hamiltonian map. R4.6 For a Hamiltonian map and a given integer g < 1, char- acterize that the under graph of the map has a super nonorientable map with its Euler characteristic g. R4.7 For a Hamiltonian map and a given integer g < 0, char- acterize that the under graph of the map has a super orientable map with its Euler characteristic 2g. For a vertex 3-map(or cubic map), if it has only i-face and j-face, iz j, i,j > 3, then it is called an (i, j) ;-map. R4.8 For a given integer g < 1, determine the number of (3, 4) j-map of order n(n > 1) with Euler characteristic g. R4.9 For a given integer g € 1, determine the number of (4, 5) j-map of order n(n > 1) with Euler characteristic g. R4.10 For a given integer g < 1, determine the number of (5,6) j-map of order n(n > 1) with Euler characteristic g. R4.11 Given a graph G of order n(n > 4), determine the con- dition for G have a super (n — 1, n) s-map. On a (n — 1, n) j-map of order n(n > 4), let 61 be the number of (n — 1)-faces. If its Euler characteristic is g € 1, then n and $; should satisfy the following condition: (n — 1)l(n(n — g) + $1), (4.10) i.e., n— 1 is a facer of n(n — g) + 1. R4.12 Given an integer g < 1, for any positive numbers n and ġı satisfying (4.10), determine if there exists a (n — 1, n);-map with its Euler characteristic g. Chapter V Orientable Maps e Any irreducible orientable map under basic subtracting edges is defined to be a butterfly. However, an equivalent class may have more than 1 butterflies. e The simplified butterflies are for the standard orientable maps to show that each equivalent class has at most 1 simplified butterfly. Reduced rules are for transforming a map(unnecessary to be ori- entable) into another butterfly, if orientable, in the same equivalent class. A basic rule is extracted for deriving all other rules. e Principles only for orientable maps are clarified to transform any map to a simplified butterfly in the same equivalent class. Hence, each equivalent class has at least 1 simplified butterfly. e Orientable genus instead of the Euler characteristic is an invariant in an equivalent class to show that orientable genus itself determine the equivalent class. V.1 Butterflies On the basis of Chapter IV, this chapter discusses orientable maps with a standard form in each of basic equivalent classes. If an V.1 Butterflies 125 orientable map has only one vertex and only one face, then it is called a butterfly. Lemma 5.1 In each of basic equivalent classes, there exists a map with only one vertex. Proof For a map M = (A55, P), if M has at least two vertices, from the transitive axiom, there exists an x € Xa g such that (x)p and (yx)p, y = af, determine two distinct vertices. Because e, is a link, by basic contracting ez, M' = M ey e, ~pe M. Then, M' has one vertex less than M does. In view of Theorem 3.10, M' is also a map. If M' does not have only one vertex, the procedure is permitted to go on with M" instead of M. By the finite recursion principle a map M" with only one vertex can be found such that M’ ~pe M. This is the lemma. L A map with only one vertex is also called a single vertex map , or in brief, a petal bundle . Lemma 5.2 In a basic equivalent class of maps, there exists a map with only one face. Proof For a map M = (Xap, P), if M has at least two faces, from the transitive axiom, there exists an x € Xa 8 such that (x). and (yx)py, y = af, determine two distinct faces. Because e, is single, by basic deleting ez, M' = M —p e, ^y. M. Now, M’ has one face less than M does. From Theorem 3.10, M' is still à map . Thus, if M' is not with only one face, this procedure is allowed to go on with M' instead of M. By the finite recursion principle, a map M" with only one face can be finally found such that M’ ~pe M. The lemma is proved. [] In fact, on the basis of Theorem 3.6, Lemma 5.1 and Lemma 5.2 are mutually dual. Furthermore, what should be noticed is the independence of the orientability for the two lemmas. 126 Chapter V Orientable Maps Theorem 5.1 For any orientable map M, there exists a but- terfly H such that H ~pe M. Proof If M has at least two vertices, from Lemma 5.1, there exists a single vertex map L ~pe M. In virtue of Theorem 4.3, L is still orientable. if L has at least two faces, from Lemma 5.2, there exists a single face map H ~pe L. In virtue of Theorem 4.3, H is still orientable. Since H has, finally, both one vertex and one face, H is a butterfly. Therefore, H ~pe L ~pe M. This is the theorem. [] This theorem enables us only to discuss butterflies for the Basic equivalence classes of maps without loss of generality. V.2 Simplified butterflies Let O; = (Xk, Tr), k > 0, where 4a u k iiec 3 C Ky), 35 k 1 61) i=l and (0), = k = 0; — k E (I [Go wo vos vu), 24 & > 1. p i=l It is east to check that all Oi, k > 0, are maps. And, they are called O-standard maps. When k = 1 and 2, O; and O» are, respectively given in (a) and (b) of Fig.5.1. V.2 Simplified butterflies 127 T1 yi yy2 | Vi ^w yai Ur E — < vy. yv2 DW Á | i 71 y2 r2 (a) Oi (b) O2 Fig.5.1 Two O-standard maps Note 5.1 When k = 0, Oo = (00,0) is seen as the degenerate case of a map with no edge. For example, what is obtained by basic deleting an edge on £y = (Ka, (x, yx)) is just Oo. Usually, it is seen as the map with only one vertex without edge, or called the trivial map. Lemma 5.3 For any k > 0, O-standard map QO, is orientable. Proof When k = 0, from Oo = (Kz, (x, yx)) —p ex, Oo ™be (Kr; (2, yx). Because (zjw(,.,,,, = (v, yr} and [o], = lom, Bx) are two orbits, (Kx, (x, yx)) is orientable. In view of Theorem 4.3, Oy is ori- entable. For k > 1, assume, by induction, that Oy 4 = (A4, Jk-1) is orientable. From (5.1) and (5.2), group V5, ,.,j has two orbits as Unibw, ug = {x yrl <i <k-1} and [ozije;, 4} = {02i Brill <i E k — 1}. For Oy = (Ax, Jr), from (5.2), Jy = (Ji); tis Yk, YLk, YYk). Group Vry} has only two orbits as {Ci} v5.4} = {T1} es, 1 U Uis Yk Ys Yu) = {rn yal <i < kj 128 Chapter V Orientable Maps and {axri}u,, 4} = lomijeu, ,,) U loti; our, Bxx, Fyn} = {ax;, bxi < i < k — 1} losa Dx. DUI = for, Gali < k}. Therefore, Og, k > 1, are all orientable. [] Lemma 5.4 For any k > 0, O-standard map Ox has only one face. Proof When k = 0, since Op = (Ka, (x,7yx)) —y e;, Oo should have one face which is composed from the two faces (x) and (ax) of (Kz,(x,vyx)). Therefore, Op has only one face(seen as a degenerate case because of no edge). For any Os = C sz 1; trom (5.1) and (5:2), (Xy = ED oYUi 315 Uis ooo Tk, Y Uk Y ks Yk) is a face of O}. However, since T {rial = gl; O; has only this face. [] From (5.2), each O-standard map has only one vertex (Oy is the degenerate case of no incident edge). Based on the above two lemmas, any O-standard map is a butterfly. Because of the simplicity in form for them, they are called simplified butterflies . Since for any k > 0, simplified butterfly O; has 2k edges, one vertex and one face, its Euler characteristic is (Ox) = 2 — 2k. (5.3) Lemma 5.5 For any two simplified butterflies O; and O;, 7, j > 0, Oi ~pe O; if, and only if, i = j. Proof Because the sufficiency, t.e., the former O; ~a O; is de- rived from the latter i = j, is natural, only necessary to prove the necessity. V.2 Simplified butterflies 129 By contradiction. Suppose i Æ j, but O; ~pe O;. Because of the basic equivalence, from Theorem 4.5, x(Oi) = x(O;). However, from (5.3) and the condition i Z j, x(Oi) = 2 — 2i 2 2 — 2j = x(O;). This is a contradiction. O Theorem 5.2 In each of the basic equivalent classes of ori- entable maps, there is at most one map which is a simplified butterfly. Proof By contradiction. Suppose simplified butterflies O; and O;, i Æ j, i,j = 0, are in the same class. However, this is a contradic- tion to Lemma 5.5. [] In the next two sections of this chapter, it will be seen that in each basic equivalent class of orientable maps, there is at least one map which is a simplified butterfly. On the basis of Theorem 4.5, two butterflies of the same size have the same Euler characteristic. Do they all simplified butterflies? However, the answer is negative! Example 5.1 Observe map M = (4, J) where X = Kzı + Kyi + Krz + Kye and J= (zr yi; La, Yo, YL1, VY1, YL2, "Yy2)- Because the face (11) 74 = Uis "YU, 92, ^YU2, t1, Y1, "2, y») has 8 elements, half the elements of ground set, M has only one face. Hence, M is a butterfly, but not a simplified butterfly. Actually, the simplified butterfly with the sane Euler characteristic of M is O». 130 Chapter V Orientable Maps V.3 Reduced rules Although butterflies are necessary to find a representative for each basic equivalent class of orientable maps, single vertex maps are restricted in this section for such a classification based on Lemma 5.1. For convenience, the basic equivalence between two maps are not distinguished from that between their basic permutations. In other words, (A, P1) ~pe (A5, P5) stands for P, ~pe P5. Lemma 5.6 For a single vertex map M = (X, J), if I = (R,x, yz, 8S) where R and S are two linear orders on X, then J pe CF); (5.4) as shown from (a) to (b) in Fig.5.2. Proof Because J = (R,2,72,8), (Jyye = Je = ym, i.e., (yx) 7, = (ya) is a face. Because e; is a single edge, by basic deleting e; on M, M'—M —ye, — (X — Krt, J'), J’ =(R,S). From M ex M UP tige d RB O This lemma enables us to transform a single vertex map into another single vertex map with one face less in a basic equivalent class. Fig.5.2 Reduced rule (5.4) V.3 Reduced rules yri — — n E WN YY | -——— Gi —— C E D yz Fig.5.3 Reduced rule (5.5) Lemma 5.7 BHor(O0,5,2),1 — Ux, B, C, 7% D. yy, E) where A, B, C, D and E are all linear orders on X, then d Nbe [A D CD. B, asus qe n, as shown from (a) to (b) in Fig.5.3. (5.5) Proof Four steps are considered for each step as a claim. Claim 1 J ~re (E, A,2,2, D, C, yz, yz, B). Proof For the angle pair (az, Jx) and (Gz, Jyx) of J = (A, m, B,y, C, yz, D, yy, E), by basic splitting e;(a link), get I ewe (D, yy, E, A, x, 2) Uye B, y, ©, yx). Then, since e, is a link, by basic contracting e, on J, get di “be Jo = KON E 2D Come. This is the conclusion of Claim 1. Here, Jı and Jz are, respectively, shown in (a) and (b) of Fig.5.4. Claim 2 Jorn (y, A, x, B, E, yy, D,C,y2). Proof For the pair of angle (az, J2z) and angle between E and A on Rh (E, A, x, z, D,C, yx, yz, B), by basic splitting e,(a link), get Jo c» B = (A, z,z,y)(vy, D,C, yz, yz, B, E). Then, since e; is a link, by basic contracting e; on J3, get J3 ^v bc JA E (y; A m, B, E, qu D, C, vx). 131 132 Chapter V Orientable Maps This is the conclusion of Claim 2. Here, 73 and J4 are, respectively, shown in (a) and (b) of Fig.5.5. Claim 3 Ja ~ve (B, E, z, A, y, yz, yy, D,C) Fig.5.6 For Claim 3 Proof For the angle pair (ay, Jay) and (o ‘yy, yy) on Ja V.3 Reduced rules 133 (y, A,x, B, E,yy, D,C, yx), by basic splitting e,(a link), get JA epe J5 = (A, x, B, E, z\(yz, yy, D, C, y£, y). Then, since ez is a link, by basic contracting e, on J3, get Js ~ve Jo = (B, E, z, A, y, yz, yy, D, C). This is the conclusion of Claim 3. Here, Js and Js are, respectively, shown in (a) and (b) of Fig.5.6. Claim 4 Js; ~pe (A, D, C, B, E, z, 2, yz, yz). Proof For the angle pair (az, Jez) and (ay, yz) of Js = (B, E, z, A, y, yz, vy, D, C), by basic splitting e, (a link), get Ie ~ve Iz = (A, y, x) (ym, yz, yy, D, C, B, E, z). Then, since e, is a link, by basic contracting e, on J7, get Ur ~ve Jg = (A, DC, B, E, s £, yz, ye). This is the conclusion of Claim 4. Here, Jz and Jg are, respectively, shown in (a) and (b) of Fig.5.7. Fig.5.7 Claim 4 On the basis of the four claims above, T ~pe So ~pe Ja ~be Jo ~be Jg = (A, D,C, B, E, x,y, yz, yy). 134 Chapter V Orientable Maps This is (5.5). O An attention should be paid to that Lemma 5.6 and Lemma 5.7 are both valid for orientable and nonorientable maps. They are called reduced rules for maps. More precisely, They are explained as in the following. Reduced rule 1 A map with its basic permutation J is basic equivalent to what is obtained by leaving off such a successive elements (xz, ysr), rE X, on J. Reduced rule 2 A map with its basic permutation J in the form as (A, z.B, y, C, yx, D, yy, E) is basic equivalent to what is ob- tained by interchanging the linear order B between x and y and the linear order D between yx and yy, and then leaving off x,y, yx and yy and putting (z, y, yz, yy) behind E on J. V.4 Orientable principles 'This section is centralized on discussing the basic equivalent classes of orientable maps. The main purpose is to extract that there is at least one simplified butterfly in each class. From the first sec- tion of this chapter, it is known that each class is considered for only butterflies without loss of generality. Theorem 5.3 For a butterfly M = (Xag, ), y = af, if no x,y € X exist such that J = (A, x, B,y, C, yz, D, yy, E) where A, B, C, D and E are all linear orders on Xag, then M ~re Op, (5.6) i.e., the trivial map(the degenerate simplified butterfly without edge). Proof For convenience, in a cyclic permutation J on Xag, if two elements x,y € Xa g arein the formas J = (A, x, B,y,C, yx, D, yy, E), then they are said to be interlaced; otherwise, parallel. V.4 Orientable principles 135 Claim If any two elements are parallel on J, then there is an element z € Xag such that (z, yx) C J, ie., (z,"yz) is a segment of J itself. Proof By contradiction. If no such an elements exists on AX, then for any zı € X, there is a nonempty linear order Bı on X such that J = (Aj, z, Bi, yz, C1) where A; and C are some linear orders on Xa g. Because Bı Z 0, for any x2 € By, on the basis of orientability and r9 and 2, parallel, the only possibility is yr. € Bı. From the known condition, there is also a linear order B» Z Ø on Xag such that Bı = (Ao, £2, Bo, 422, C3) where A> and C3 are segments on Bj, i.e., some linear orders on X. Such a procedure can only go on to the infinity. This is a contradiction to the finiteness of Xag. Hence, the claim is true. If J Æ Ú, then from the claim, there exists an element x in J such that J = (A,z,yz, B). However, because (yz), = (yz) is a face in its own right, J has to be with at least two faces. This is a contradiction to that M = (4, J) is a butterfly. Hence, the only possibility is J = 0, i.e., (5.6) holds. O Actually, this theorem including the claim in its proof is valid for any orientable single vertex map. Therefore, it can be seen that the reduced rule(Lemma 5.6) and the following corollary are valid for any map (orientable or nonorientable) as well. Corollary 5.1 Let S = (A, x, yx, B) be a segment on a vertex of a map M. And let M’ be obtained from M by substituting (A, B) for S and afterward deleting Kx from the ground set. Then, M' is a map. And, M’ e. M. Proof Because it is easy to check that M’ = M —y M, from Theorem 3.10, M' is a map. This is the first statement. In view of basic deletion of an edge as a basic transformation, M’ ~pe M. [] Corollary 5.2 Let S be a segment at a vertex of a map M. If for each element x in S, yx is also in S and any two elements in 136 Chapter V Orientable Maps S are not interlaced, then there exists an element y in S such that S = (A, y, yy, B). Proof In the same way of proving Theorem 5.3, the conclusion is soon obtained. [] Theorem 5.4 Ina butterfly M = (2,5, J), if there are x,y € Xa g such that J = (A, x, B, y, C, yz, D, yy, E), then there is an inte- ger k > 1 such that Mac Or: (5.7) i.e., the simplified butterfly with 2k edges. Proof Based on Reduced rule 2(Lemma 5.7), J Nig (ADO BE m uT y). Let H = (A, D,C, B, E). From Corollary 5.1, assume H is not in the form as S without loss of generality. From Corollary 5.2 and Theorem 5.3, H has two possibilities: H = Ø, or there exist two elements zi and y; interlaced in A. If the former, then J ~pe (2, y, yz, yy), ie. , M ~a Oi. Oth- erwise, i.e., the latter, then KA = (A, X1; By, Yis C, ^Y 31; IA, YY; Fi). An attention should be paid to that E, = (Fi, x,y, yx, yy). In this case, from Lemma 5.7, vi ~be (Ai, Dy C^ Bis E, X1; Y1, Yv1, yyı) = (A1, IA, C, Bı, fu LLY, YT, VY, Ti, Y1, "1, yi). Let Hı = (Aj, Di, Ci, By, F3), then for Hı instead of H, go on the procedure. According to the principle of finite recursion, it is only possible to exists an integer k > 1 such that (5.7) holds. E This theorem shows that each basic equivalent class of orientable maps has at least one map which is a simplified butterfly. By considering Theorem 5.2, each basic equivalent class of ori- entable maps has, and only has, an integer k > 0 such that the sim- plified butterfly of size k is in the class V.5 Orientable genus 137 V.5 Orientable genus Although Euler characteristic of a map is an invariant for basic transformation, a basic equivalent class of maps can not be determined by itself. This is shown from the map M in Example 4.1 of section IV.4 and the map N in Example 4.2 of section IV.4. They both have the same Euler characteristic. However, they are not in the same basic equivalent class of maps because M is orientable and N is nonorientable. Now, a further invariant should be considered for a class of ori- entable maps under the basic equivalence. Theorem 5.5 For any orientable map M = (4, P), there has, and only has, an integer k > 0, such that its Euler characteristic is x(M) = 2 — 2k. (5.8) Proof From Theorem 5.1 and Theorem 4.5, only necessary to discuss butterflies. From Theorem 5.2 and Theorem 5.4, M has, and only has, an integer k > 0, such that M ~pe Og. Therefore, from (5.3), X(M) = 2 — 2k. This is (5.8). O Corollary 5.3 The Euler characteristic of an orientable map M = (X, P) is always an even number, i.e., X(M) — 0 (mod 2). (5.9) Proof A direct result of Theorem 5.5. [] Since Euler characteristic is an invariant of a basic equivalent class, the integer k in (5.8) is an invariant as well. From Theorem 5.5, k determines a basic equivalent class for orientable maps. Since each orientable map in the basic equivalent class determined by k can be seen as an embedding of its under graph on the orientable surface of 138 Chapter V Orientable Maps genus k, k is also called the genus, or more precisely, orientable genus of the map. Of course, only an orientable map has the orientable genus. From what has been discussed above, it is seen that although Euler characteristic can not determine the basic equivalent class for all maps, the Euler characteristic can certainly determine the basic equivalent class for orientable maps. Activities on Chapter V V.6 Observations O5.1 Think, is there a butterfly which has 3 edges? Further, is there a butterfly with some odd number of edges? If yes, provide an example. Otherwise, explain the reason. O5.2 Observe that any butterfly with 2 edges is a simplified butterfly. Explain the reason. O5.3 Provide a butterfly of at most 5 edges , which is not a simplified butterfly. O5.4 Observe that is there a superfluous operation among the four operations: basic deleting, basic appending, basic contracting and basic splitting an edge in the basic transformation for the basic equivalence? If no, indicate the role of each of them. If yes, indicate the superfluous operation with the why. O5.5 Think, do some three of the four operations: basic delet- ing, basic appending, basic contracting and basic splitting an edge in the basic transformation determine an equivalence? If do, provide an example. Otherwise, explain the reason. O5.6 Among the eight operations of the above four with in- creasing duplition, decreasing duplition, increasing subdivision and decreasing subdivision, how many groups of these operations are there for determining the basic equivalence. O5.7 Can an equivalence be determined by basic deleting and basic appending an edge? If yes, observe some of invariants under the 140 Activities on Chapter V equivalence. Otherwise, explain the reason. O5.8 Can an equivalence be determined by basic contracting and basic splitting an edge? If yes, observe some of invariants under the equivalence. Otherwise, explain the reason. O5.9 Can an equivalence be determined by basic increasing duplition and decreasing duplition of an edge? If yes, observe some of invariants under the equivalence. Otherwise, explain the reason. O5.10 Can an equivalence be determined by basic increasing subdivision and decreasing subdivision of an edge? If yes, observe some of invariants under the equivalence. Otherwise, explain the rea- son. O5.11 Can the procedure of proving Lemma 5.7 by four steps be improved to that by three steps? If yes, provide a proof of three steps. Otherwise, explain the why. V.7 Exercises E5.1 By basic deleting and basic appending an edge only, prove (A, z, B, y, C, yz, D, yy, E) ^w (A, D, C, B, E, z, y, yz, yy). E5.2 Prove that a map has its orientable genus 1 if, and only if, M ~he (x, Y, 4, yy) (Bz, bt, bz, at). E5.3 Provide two maps of order 3 with orientable genus 1. And, explain they are distinct. For a set of operations A and a set of maps B not necessary to be closed under A, if for a map M € B there is no such a map N € B of size less than the size of M that N can be obtained from M by operations in A, then M is called irreducible under A. E5.4 For an integer k > 0, determine all the irreducible sin- gle vertex maps of orientable genus k under basic deleting and basic appending an edge. V.7 Exercises 141 E5.5 For an integer n > 1, determine all the irreducible ori- entable single vertex maps of size k under basic deleting and basic appending an edge. E5.6 For an integer n > 1, determine all the irreducible ori- entable single vertex maps of size k under basic contracting and basic splitting an edge. E5.7 For an integer k > 0, determine all the irreducible single vertex maps of orientable genus k under basic contracting and basic splitting an edge. E5.8 Prove that a complete graph K, of order n > 3 has an orientable single face embedding if and only if, ("5') is even. E5.9 Prove that a complete graph K, of order n > 3 has an orientable two face embedding if and only if, a is odd. E5.10 Prove that a complete bipartite graph Km,n of order n + m > 3 has an orientable single face embedding if, and only if, (m — 1)(n — 1) is even. E5.11 Prove that a complete bipartite graph Km,n of order n + m > 3 has an orientable two face embedding if, and only if, (m — 1)(n — 1) is odd. An n-cubeis the graph formed by the skeleton of the n-dimensional cuboid. The order of an n-cube is 2” and the size, n2"-. E5.12 Prove that any n-cube, n > 2, has an two face embed- ding. A map of orientable genus 0 is also said to be planar. E5.13 Prove that a map M is planar if, and only if, each face of M corresponds to a cocycle(See 1.6) in the under graph G(M*) of its dual M*. E5.14 Prove that a map M = (Xap, P) is orientable if, and 142 Activities on Chapter V only if, P can be transformed into cyclic permutation J such that k k I = (TL [Gov [Loz vm) (5.10) i=l i—1 V.8 Researches If the travel formed by a face in a map can be partitioned into tours (travel without edge repetition), then the face is said to be pan- tour. À map with all of its faces pan-tour is called a pan-tour map . Because any tour can be partitioned into circuits, a pan-tour map is, in fact, a favorable map as mentioned in 2.8. A pre-proper embedding corresponds to what is called a tour map because each face forms a tour in its under graph. R5.1 Characterize and recognize that a graph has a super map which is an orientable pan-tour map. R5.2 Characterize and recognize that a graph has a super map which is an orientable tour map. R5.3 Orientable pan-tour conjecture. Prove, or improve, that any nonseparable graph has a super map which is an orientable pan- tour map. R5.4 Orientable tour conjecture. Prove, or improve, that any nonseparable graph has a super map which is an orientable tour map. R5.5 Characterize and recognize that a graph has a super map which is an orientable pre-proper map. R5.6 Orientable proper map conjecture. Prove, or improve, that any nonseparable graph has a super map which is an orientable proper map. The orientable minimum pan-tour genus , usually called ori- entable pan-tour genus, of a graph is the minimum among all orientable genera of its super pan-tour maps. Similarly, the orientable pan-tour V.8 Researches 143 marimum genus of a graph is the maximum among all orientable gen- era of its super pan-tour maps. R5.7 Determine the orientable maximum pan-tour genus of a graph. R5.8 Determine the orientable maximum tour genus of a graph. R5.9 Determine the orientable pan-tour genus of a graph. R5.10 Determine the orientable tour genus of a graph. Although many progresses has been made on determining the orientable maximum genus of a graph, the study on determining ori- entable (maximum) pan-tour genus, or orientable (maximum) tour genus of a graph does not lead to any notable result yet. This sug- gests to investigate their bounds(upper or lower) for some class of graphs. R5.11 Characterize the class of graphs in which each graph has its orientable maximum pan-tour genus equal to its orientable maximum genus. Find the least upper bound of the absolute difference between the orientable maximum pan-tour genus and the orientable maximum genus for a class of graphs with the two genera not equal. R5.12 Characterize the class of graphs in which each graph has its orientable maximum tour genus equal to its orientable maximum genus. Find the least upper bound of the absolute difference between the orientable maximum pan-tour genus and the orientable maximum genus for a class of graphs with the two genera not equal. R5.13 Characterize the class of graphs in which each graph has its orientable pan-tour genus equal to its orientable genus. Find the least upper bound of the absolute difference between the orientable pan-tour genus and the orientable genus for a class of graphs with the two genera not equal. R5.14 Characterize the class of graphs in which each graph has its orientable tour genus equal to its orientable genus. Find the least upper bound of the absolute difference between the orientable 144 Activities on Chapter V tour genus and the orientable genus for a class of graphs with the two genera not equal. Chapter VI Nonorientable Maps e Any irreducible nonorientable map under basic subtraction of an edge is defined to be a barfly. However, an equivalent class may have more than 1 barflies. e The simplified barflies are for standard nonorientable maps to show that each equivalent class has at most 1 simplified barfly. e Nonorientable rules are for transforming a barfly into another barfly in the same equivalent class. A basic rule is extracted for deriving from one to all others. e Principles only for nonorientable maps are clarified to transform any nonorientable map to a simplified barfly in the same equivalent class. Hence, each equivalent class has at least 1 simplified barfly. e Nonorientable genus instead of the Euler characteristic is an invari- ant in an equivalent class to show that nonorientable genus itself determines the equivalent class. VI.1 Barflies This chapter concentrate on discussing the basic equivalent classes of nonorientable maps by extracting a representative for each class. On the basis of Lemma 5.1 and Lemma 5.2, Only maps with a single 146 Chapter VI Nonorientable Maps vertex and a single face are considered for this purpose without loss of generality. A nonorientable map with both a single vertex and a single face is called a barfly. The barfly with only one edge is the map consisted of a single twist loop, i.e., NU = (Ka, (x, Bxr)). Fig.6.1 Barflies with two edges Example 6.1 ‘Two barflies of size two. Let Nn” = (Kr + Ky, Z4) and NP = (Kx + Ky, T2)(shown, respectively, in (a) and (b) of Fig.6.1) where Ti = (£, Y, Y£, BY), To = G5 Da ug. Du). Because of (tiny = (x, By, ax, ay) and (2) ty = Gs OQ, y, ay), each of Nn? and NP has exactly one face. And, since (tomm = (z)v,,, = Kat Ky, they are both nonorientable. As mentioned in the last section, for convenience, the scope of maps considered here for the specific purpose should be enlarged to all nonorientable one vertex maps from barflies. VI.1 Barflies 147 Lemma 6.1 For a single vertex map M = (Xa»e, P), M is nonorientable if, and only if, there exists an x € Xa p such that Bx € {x}p. Proof Necessity. By contradiction. Assume that for any x € Xap, there always has yx € {x}p, y = af, then az ¢ {r}u,p,,- From 4.1, M is not nonorientable. Sufficiency. Since x € Xag and Bx € {x}p, from Corollary 4.1 and only one vertex, Yip} has only one orbit on X. Hence, M is nonorientable. L This lemma can easily be employed for checking the nonori- entability of a one vertex map. Example 6.2 Six barflies of three edges. Let N89) —UNgq Ky + Kz,Z;), i = 1,2,---,6(shown, respectively, in (a,b,---,f) of Fig.6.2) where x, Bx, y, By, 2, Bz), 7,y, 2, By, Bz, Bz), 2, y, 2, Bz, yz, By), T, Bx, Y, 2, Bz, By), T, Y, Bx, 2, By, Bz), £, Y, Z, Px, yy, yz). Because Gx € {x}z7, € Ew d ^, — ag, i= 1,2,---,6, from Lemma 6.1, they are all nonorientable. Since = dy Si T= ( ga mST T= dg eT Dt opo ag. ao); L,Y, 2, "Y, y, oz), z, By, ax, yz, Bz, ay), 3, 0x Y, YZ, 02, da. £, ay, BZ,, YX, y, az), 225 = e7 OZ, By, QTL, YY; rar =| = = = the maps (Kx + Ky + Kz, Ti), i = 1,2,---,6, are all with only one face. Therefore, they are all barflies. 148 Chapter VI Nonorientable Maps Theorem 6.1 For any nonorientable map M, there exists a barfly N such that M ^y N. (6.1) (e) NO) (f) NO Fig.6.2 Barflies of three edges Proof From Lemma 5.1, by basic transformation M can be transformed into a single vertex map. From Theorem 4.3, in view of VI.2 Simplified barflies 149 the nonorientability of M, the single vertex map is also nonorientable. Then from Lemma 5.2, by basic transformation the single vertex can be transformed into a single face map. From Theorem 4.3, this map is also nonorientable and hence a barfly N which satisfies(6.1). O This theorem enables us to restrict ourselves only to transform a barfly into another barfly under the basic equivalence. VI.2 Simplified barflies Let Qi = (X, Zi), l > 1, where l Ape Kay, (6.2) i=1 and l Tı = (J [ (xi, Be), (6.3) i=] they are called N-standard map . When k = 1, 2, 3 and 4, Q; = N™, (s = NO. Qs = NO and Q4 are, respectively, shown in (a), (b), (c) and (d) of Fig.6.3. Lemma 6.2 For any | > 1, N-standard maps Q; are all nonori- entable. Proof Because all Qj, | > 1, are single vertex map and G2, € {x1}, L> 1, from Lemma 6.1, they are all nonorientable. [] Lemma 6.3 For any | > 1, N-standard maps Q are all with only one face. Proof Because Qj = NO, Q4 = NO? and Q3 = N98. from the two examples above, they are all with only one face. Their faces are (21), = (21, 021), (€1) try = ((11)1,, 22, 022) = (1, 021, 12, 02) and (215, = ( (51) T9523, 095) = (91,021, Ta, 025, 23; QU). 150 Chapter VI Nonorientable Maps Assume, by induction, that [ilie quies (kūri ted Oba, seres AE qoe for | > 4. Since Z; = ((z1)3, ,, 21, xi), (z1)z4 = (Gina (Ziy)azi-1,-+:). And since (Zry)oxiLi = bxi = zi, (iy) = Zjo(fxi) = axı and (Lyon = 16x = xy, (Xr — Cyan TI, ox) — (245 OX1,°°°, 2-1, (11-1, Tl, aa). Therefore, all N-standard maps are with only one face. [] | | Bay (a) Ni (b) No T1 T1 Buy ENS "N l EDS 7A Ne Bag Bug (c) Ns (d) Na Fig.6.3 Simplified barfly VI.3 Nonorientable rules 151 Because each N-standard map has only one face, from the two lemmas above, it is known that for any N-standard map is a barfly. And because each of such barflies has a simpler form, it is called a simplified barfly. Since for any | > 1, the simplified barfly Q; is with l edges, 1 vertex and 1 face, its Euler characteristic is x(Q) 22-1. (6.4) Theorem 6.2 For any basic equivalent class of nonorientable maps, there exists at most one map which is a simplified barfly. Proof By contradiction. Assume that there are two simplified barflies Q; and Q;, i # j, i,j > 1, in a basic equivalent class of nonorientable maps. From Theorem 4.5 and (6.4), x(Qi) =2-i=2- j = x(Q;). This implies i = 7. A contradiction to the assumption. El In the following sections, it will be shown that there exists at least one map which is a simplified barfly in each basic equivalent class of nonorientable maps. VI.3 Nonorientable rules As mentioned above, this section is for establishing two basic rules of transforming a nonorientable single vertex map into another nonorientable single vertex map within basic equivalence. Lemma 6.4 Fora nonorientable single vertex map M = (Xap, T), if Z = (A, x, B, Bx, C) where A, B and C are segments of linear order in the cycle Z on Xag, then T ^. (A,aB", C, 2, Br). (6.5) Note 6.1 For a segment B = (Zz,T?z,- --, Z?z), Bx = Tir, 152 Chapter VI Nonorientable Maps of linear order in the cycle of Z on Z, 5, from Theorem 2.3, (aX "p. Toga s T (ad^ = (o?z, aD? 1x, -- , o x) (6.6) sab Proof Two steps expressed by claims are considered for trans- forming a nonorientable single vertex map into another nonorientable single vertex map under the basic equivalence. Claim 1 (A,z, B, 8z,C) ~pe (A, 9B 1, By, aC" 1, y). Proof By basic splitting an edge e, between the two angles (az, Zx) and (C, A)(i.e., the angle between C and A) on Z = (A,x, B, Bx, C), L ™be (A, T, y) (YY; B, pz, C) Uum ac yack) = (y, A, x) (yz, o B1, By, aC"), as shown in (a) and (b) of Fig.6.4. Because e, is a link in Z} = (y, A, z)( yz, o.B !, 8y, aC 1), by basic contracting e,, Tı ^vbc (y, A, oB^l, By, aC!) = (A, oB^!, By, aC", y), as shown in (c) of Fig.6.4. Claim 2 7» ~pe (A, a B-t, C, Bx, x) where T, = (A,aB !, By, aC 1,3). Proof By basic splitting e, between the two angles (yy, T28y) and (ay, Izy) on (A,aB™, By, aC", y), To XYbe (A, aB, By, v) (ya, aC, y) = (A, aBn S, By, g) (ay, C, Bx) = Gs A, aB}, By) (ay, C, Bur); VL3 Nonorientable rules 153 as shown in (d) and (e) of Fig.6.4. Because e, is a link in Z3 = (x, A, aB^!, By) (ay, C, Bx), by basic contracting €y, T3 ^w (2, A,B, C, Bx) = (A,aB",C, Bx, x), as shown in (f) of Fig.6.4. From Claim 1 and Claim 2, 1 ^vbc Ti XY be To ~Y be Ts ^^c (A, B^ 1, C, Bax, x). This is (6.5). O On the basis of the procedure in the proof of the lemma, the two claims show the following rules as basic equivalent transformations. Nonorientable rule 1 On a nonorientable map M = (A,,5, P) unnecessary to have a single vertex, if Gx € (x)p, then the map M' obtained by translating r and (x in a direction via, respectively, seg- ments C and D, and then by substituting aC~! and aD! for, respec- tively, C(without Gx) and D(without x) on (z)p is basic equivalent to M, ie., M! ^ M. aB! 154 Chapter VI Nonorientable Maps =A > BE Fig.6.4 Claim 1 and Claim 2 This is, in fact, the Claim 1 above. However, from the proof of Claim 2 a much simpler rule can be extracted. Nonorientable rule 2 On a nonorientable map M = (X,7) unnecessary to have a single vertex, if Gx € (x)p, then the map M’ obtained by translating x(or Gx) via a segment C, and then by sub- stituting aC"! for C without 6x(or x) on (x)p is basic equivalent to M, ie., M' ~pe M. It is seen that Nonorientable rule 1 can be done by employing Nonorientable rule 2 twice. Therefore, Nonorientable rule 2 is funda- mental. From this point of view, the proof of Lemma 6.4 can be done only by Nonorientable rule 2. Lemma 6.5 For a nonorientable single vertex map (4,5, Z), VL3 Nonorientable rules 155 y = a6, it L= (A, z, B£,Y, zZ, YY, yz) where A is a segment of linear order on Xag, then Zoos (Am, Dau. Dy 2, De. (6.7) Proof By basic splitting e; between the two angles (ax, 8x) and (az, yy) on (A, c, Bx, y, 2, "yy. yz), T = (A, £, Br, y, 2, YY, 72) ~be (YY, 72, A, m, t) (t, B, y, 2) = (yy, yz, A, x, t) (yv, Bt, az, ay) = (t, yy, yz, A, c) ys, Bt, az, oy). Because e, is a link in 7, = (t, yy, yz, A, x) yz, Gt, az, ay), by con- tracting ezr, T, ~be (t, "yy, yz, A, Bt, az, ay) = (A, Bt, az, ay, t, yy, yz). By substituting (A, 8t), (ay, t, yy) and (Ø) for , respectively, A, B and C in (6.5), T4 ~pe (A, Bt, By, at, y, z, Bz). Further, by substituting (A, Gt), (at) and (z, 8z) for, respectively, A, B and C in (6.5), d ^vbc (A, t, Bt, y, Bus 2,02) = (At Dora Dues Oa. This is (6.7). O This lemma shows that in a map M = (a8, P), Y = aß, if (x)p = (x, Bx, Y, Z, VY, VZ, A), then the map obtained by substituting (y, Gy, z, Bz) for (y, z, yy, yz) on (xz)p is basic equivalent to M, i.e., M' ~be M. This is usually called nonorientable rule 3. 156 Chapter VI Nonorientable Maps Actually, nonorientable rule 3 can be also deduced from Nonori- entable rule 2. Although Nonorientable rule 2 is fundamental, Nonori- entable rule 1 and nonorientable rule 3 are more convenient for recur- sion. VI.4 Nonorientable principles In this section, barflies are only considered for this classification because it has been known that there is no loss of generality for general nonorientable maps. Lemma 6.6 Ina barfly N = (Xag, T), there exists an element x € X, such that T = (A, x, B, Bx, C), (6.8) where A, B and C are segments of Z on X, 5. Proof By contradiction. Since A, B and C are permitted to be empty, if no x € X such that Z satisfies (6.8), then from only one vertex, for any x € X, it is only possible that yx € (x)z and Bx £ (x)z. Therefore, (z)y,,., and (Gx)w,,., are the two orbits of Vz 4 on X. 'Thus, M is orientable. This is a contradiction to the nonorientability of a barfly. o Theorem 6.3 For any barfly N = (Xa g,T), there exists an integer | > 1 such that T ~pe Qı- (6.9) Proof From Lemma 6.6 and Lemma 6.4, it can assumed that L ~be (A, | [tv;, 62;)), j=l where 2 is as great as possible in this form. Naturally, i > 1. From the maximality of 7 and only one vertex, x € A if, and only if, yx € A. VI.5 Nonorientable genus 157 Two cases have to be discussed. Case 1 If no element in A is interlaced, then from Corollary 5.2 and Corollary 5.1, (6.9) holds. Here, l = i. Case 2 Otherwise, by Lemma 5.7(the reduced rule). it can be assumed that i t Z ^» (B, | | (23, 625 | [( 25 vus v3); j=1 j=1 where t is as great as possible in this form. Naturally, t > 1. From the maximality of t, no element in B is interlaced. By Corollary 5.2 and Corollary 5.1, i t T ~pe ( qK Eu li Ux Ae j=l j=l By nonorientable rule 3, 2t--i I ~pe (| | (27, 82) =D jal From (6.2) and (6.3), this is (6.9) where l = 2t + i. O On the basis of Theorem 6.1 and Theorem 6.3, it is know that there is at least one simplified barfly in each of basic equivalent classes for nonorientable maps. VI.5 Nonorientable genus Now, let us go back to general nonorientable maps for the in- variants of determining the basic equivalent classes for nonorientable maps. Theorem 6.4 For any nonorientable map N = (X,P) ina basic equivalent class, there has, and only has, an integer l > 1 such that the Euler characteristic id AD mde d (6.10) 158 Chapter VI Nonorientable Maps Proof From Theorem 6.3, there is a simplified barfly in a basic equivalent class of barflies. From Theorem 6.1, in each basic equivalent class of nonorientable maps, there has an integer l > 1 such that Q; is in this class. On the other hand, from Theorem 6.2, only Q; is in this class. Therefore, from (6.4) and Theorem 4.5, (6.10) is obtained. O This integer | = 2 — y(N) > 1 is called the nonorientable genus of the class N is in, or of N. Now, it is seen from Chapters IV, V and VI that if the orientabil- ity of a map is defined to be 1, when the map is orientable; —1, when the map is nonorientable, then the relative genus of a map is the prod- uct of its orientability and its absolute genus (orientable genus, if the map is orientable; nonorientable genus, if the map is nonorientable). Thus, a basic equivalent class of maps(orientable and nonorientable) is determined by only its relative genus. Activities on Chapter VI V.6 Observations O6.1 Think, is the map obtained by deleting an edge on a nonorientable map still nonorientable? If yes, explain the reason. Oth- erwise, provide an example. O6.2 Think, the absolute genus of a map obtained via deleting an edge on a map(orientable and nonorientable) is at most 1 less than that of the original map. Explain the why. O6.3 Think, is the map obtained by contracting an edge on a nonorientable map still nonorientable? If yes, explain the reason. Otherwise, provide an example. O6.4 Think, the absolute genus of the map obtained via con- tracting an edge on a map(orientable and nonorientable) is at most 1 less than that of the original map. Explain the why. O6.5 Observe if a nonorientable map can always be trans- formed into a barfly via only basic deleting and basic appending an edge. If it can, explain the reason. Otherwise, discuss what type of nonorientable maps can be done. O6.6 Observe if a nonorientable map can always be trans- formed into a barfly via only basic contracting and basic splitting an edge. If it can, explain the reason. Otherwise, discuss what type of nonorientable maps can be done. O6.7 Observe if there are other standard maps than simpli- fied barflies for the classification of nonorientable maps under basic 160 Activities on Chapter VI equivalence. O6.8 Consider how to derive the nonorientable rule 3 only from the nonorientable rule 2. O6.9 For a map M = (X45, P), if the linear order (A, T; B; Gz, C) c (x)p is replaced by (A, aB^ 1, C, x, Bx) to transform the permutation P on Xa g into permutation P’ on Xag, then M’ = (X, P") is also a map. Is M' basic equivalent to M? If yes, explain the reason. Otherwise, provide an example. O6.10 Observe that all the nonorientable rules 1-3 are valid for any nonorientable map not necessary to be of single vertex. VILT Exercises E6.1 By basic deleting and basic appending edge, prove that (A, z, B, oz, C) ^ (A, x, B, C, o). = (Xab, P) and an element x € Xap, the linear order (x,y, yY, yy} € (x)p is replaced by (x, Bx, y, By) C (x)p for obtaining M' = (X, P’). Prove that M' ~pe M. E6.3 By basic deleting and basic appending an edge, prove E6.2 For a given nonorientable map M (A, £, Y, YT, VY, 2, az) ™be (A, T, AL, Y, QY, Z, az). E6.4 List all barflies of three edges rather than those in Exam- ple 2. E6.5 Prove that for any two barflies, one can always be trans- formed into another only by the nonorientable rule 2. The irreducibility appearing in what follows is in agreement with that in section V.7. VI.7 Exercises 161 E6.6 Determine all irreducible barflies under the basic deleting and basic appending an edge. E6.7 For a given integer / > 1, determine all the irreducible maps of absolute genus / under the basic deleting and basic appending an edge. E6.8 Determine all irreducible barflies under the basic con- tracting and basic splitting an edge. E6.9 For a given integer / > 1, determine all the irreducible maps of absolute genus / under the basic contracting and basic split- ting an edge. E6.10 Prove that for any nonorientable map M = (4, P), there has, and only has, an integer l > 1 such that S (I [(ai, Yi, bti, "YYyi), X s--15 Bts31); i=1 P be M 1=2s+1,s>0; (6.11) CI [Gs va 8m 9). W l = 25,5 > 1. i=1 Here, when s = 0, E i=l E6.11 Prove that for any nonorientable map M = (4, P), there has, and only has, an integer / > 1 such that P ™be (24; 22,777, Xk, Dus dE ga, £21) k 1 = qI Ti, li Ba) i=1 i=k by the nonorientable rules instead of Lemma 6.4. E6.12 Prove that for any graph G, but a tree, G has its maxi- mum nonorientable genus Iu(G) = (G) — v(G) +1 (6.12) 162 Activities on Chapter VI where e(G) and v(G) are, respectively, the size(edge number) and the order(vertex number) of G. VI.8 Researches Similarly to Chapter V, among all nonorientable embeddings of a graph, the one with minimum (maximum) of absolute genus is called a minimum (maximum) genus embedding. The genus of a minimum(maximum) genus embedding on nonori- entable surfaces for a graph is called the minimum (maximum) nonori- entable genus of the graph. The minimum nonorientable genus of a graph is also called the nonorientable genus of the graph. If the minimum genus embedding is a nonorientable pan-tour(favorable) map, the the genus is called the nonorientable pan-tour(favorable) genus. And the likes, nonorientable pan-tour maximum genus, nonori- entable tour genus (or nonorientable preproper genus), nonorientable tour maximum genus, etc. R6.1 Justify and recognize if a graph has a nonorientable em- bedding which is a pan-tour map. R6.2 Justify and recognize if a graph has a nonorientable em- bedding which is a tour map. R6.3 Determine the least upper bound and the greatest lower bound of the nonorientable pan-tour genus(or genera) for a graph(or a set of graphs). R6.4 Determine the least upper bound and the greatest lower bound of the nonorientable tour genus(or genera) for a graph(or a set of graphs). R6.5 Determine the least upper bound and the greatest lower bound of the nonorientable proper genus(or genera) for a graph(or a set of graphs). Because it looks no much possibility to get a result simple as VLS Researches 163 shown in (6.12) for determining the nonorientable pan-tour maximum genus, nonorientable tour maximum genus and nonorientable proper maximum genus of a graph in general, only some types of graphs are available to be considered for such kind of result. R6.6 Determine the least upper bound and the greatest lower bound of the nonorientable pan-tour maximum genus(or genera) for a graph(or a set of graphs). R6.7 Determine the least upper bound and the greatest lower bound of the nonorientable tour maximum genus(or genera) for a graph(or a set of graphs). R6.8 Determine the least upper bound and the greatest lower bound of the nonorientable proper maximum genus(or genera) for a graph(or a set of graphs). R6.9 Nonorientable pan-tour conjecture (prove, or disprove). Any nonseparable graph has a nonorientable embedding which is a pan-tour map. R6.10 Nonorientable tour map conjecture (prove, or disprove). Any nonseparable graph has a nonorientable embedding which is a tour map. R6.11 Nonorientable proper map conjecture (prove,or disprove). Any nonseparable graph has a nonorientable embedding which is a proper map. R6.12 Nonorientable Small face proper map conjecture(prove, or disprove). A nonseparable graph of order n has a nonorientable embedding which is a proper map with n — 1 faces. Chapter VII Isomorphisms of Maps e An isomorphism is defined for the classification of maps. A map is dealt with an isomorphic class of embeddings of the under graph of the map. e Two maps are isomorphic if, and only if, their dual maps are iso- morphic with the same isomorphism. e T'wo types of efficient algorithms are designed for recognizing if two maps are isomorphic. e Primal trail codes, or dual trail codes are used for justifying the isomorphism of two maps. e Two pattern examples show how to to recognize and justify if two maps are isomorphic. VIL1 Commutativity In view of topology, the basic equivalent classes of maps are, in fact, a type of topological equivalent classes of 2-dimensional closed compact manifolds without boundary, or in brief surfaces. Two embeddings of a graph explained in Chapter I are distinct if they are treated as 1-dimensional complexes to be non-equivalent under a topological equivalence. VIL1 Commutativity 165 If a map is dealt with an embedding of a graph on a surface, then two distinct maps are, of course, distinct embeddings of their under graph. However, the conversed case is not necessary to be true. This Chapter is intended to introduce a type of combinatorial equivalence which is still seen as a type of topological equivalence but different from that for embeddings of a graph. In general, the equivalence between two maps can be deduced from that between two embeddings of their under graph. However, the coversed case is not necessary to be true. For two maps Mi = (A4,4(X1), Pi) and Mz = (As,5( X2), P2), if there exists a 1—to-1 correspondence (i.e., bijection) T : Xap Xi) — Xa p(X) between Xa 5(.X1) and Xa (X2) such that for any z € Xag X1), Tog) =ar z); (Bx) era c Pim) = Prin. (7.1) then 7 is called an isomorphism from M to M». Lemma 7.1 If is an isomorphism from Mı to M», then its inverse T~! exists, and 7^! is an isomorphism from M» to Mj. Proof Sincer is a bijection, 7^! exists. And, 7^! is also a 1-to-1 correspondence from M» to Mi. For any y € Xa p(X2), let x = 7 ly € Xq,3(X1). Because y = Tx and 7 is an isomorphism for M; to M», from (T.1), Thar) = Qy, T(Bx) — By, T(Pi2) E Poy. Further, because 7^! exists, then T (By) = Bx = B(r y), T (Py) —Tiz-—T(rl). is an isomorphism from M» to Mj. = r (ay)-—arca(r y), 1 This implies 7^! 1 Based on this lemma, 7, or 7T ^ can be called an isomorphism between M, and Mo. 166 Chapter VII Isomorphisms of Maps Examle 7.1 Let Mı = (X1, P1) where (= Kzit Kyt Ka+ Ku and Pi = (tutii) (U1, ya) (Cuy: and M» = (A5, P2) where Xz = Kt + Ky + Kzı + Kuz and P» = (yo, 22, £2) (YU2, Bx2) (auz, Bz2)(By2) as shown in Fig.7.1. First, let r(z1) = x9. from the first two relations in (7.1) and the property of Klein group, if 7 is an isomorphism between M; and M», then T(oui)e arm) = 5, T(Bz1) = B(ra1) = B», T(yzi) = v(21) = 712, i.e., TUR 44) = Kus. Then, from the third relation of (7.1), T(y1) = r(Piz1) = Pyr(z1) = Pore = ys. Thus, T(A yi) = Ky». Similarly, from r(z1) = T(Piy1) = Pot (yi) = Pays = 22, T(K 2) = Kz, and from T(ui) = T(Pryzi) = Pot (y21) = Pyyzo = us, T(Kuj) = Kus. Finally, check that if the 1—0-1 correspondence 7 from 4X, to Xə satisfies TP; = P3. In fact, from the conjugate axiom, it is only necessary to have TPy = (TH TY1, TA) Ts TVA) (TO, ryz) TI) = (Xa, Y2, 22) (u2, 22) (Gua, yz) (Vy2) = (Yo, 22, 3) (YUz, Bx2)(AU2, 322) (Bye) = P3: VIL1 Commutativity 167 Therefore, 7 is an isomorphism between M; and Mj. Bur | T" —_— Í j ü i (a) Mi (b) M5 Fig.7.1 Two isomorphic maps Note 7.1 If the two maps Mı and M» in Example 1 are, re- spectively, seen as embeddings of their under graphs G, and Go, then they are distinct. If Ka is represented by x = (ætt, x7!) where at! = {x azr} and x^! = {8x, yx}, then the vertices 1G, have their rotation as E n e, a uj), (zi E ul D, (yi P, and hence uG; is on the projective plane (u1, u1). And the vertices of [G5 have their rotation as (235, ES Uy (z1 E uj p’ (zi E ul p? (yi 2 and hence G2 is on the projective plane (u1, u1) as well. However, the induced 1-to-1 correspondence r|,(r|,(s1) = s», s = r,y, z, u) from u(Gi) to u(G2) has T]. rh yt at) = (ae ye, 2a") v (ad, a ya). This implies that uG and 1G» are distinct. Theorem 7.1 Let M; = (A&,,5(.X1), P1) and M» = (Xa ol X2), P2) be two maps. For a bijection 7 : Vo,g(X1) —9 | A,5(.X2), T is an iso- 168 Chapter VII Automorphisms of Maps morphism if, and only if, the diagrams Xapi) —— Kap(Xa) | "| (7.3) Xq,a(X1) ———— Xa,(X2) for q = 9» = a, 11 = N = D, and for n = Pı and no = Po, are all commutative , i.e., all paths with the same initial object and the same terminal object have the same effect. Proof Necessity. From the first relation in (7.1), for any x € Xa pl Xı), T(ax) = a(rz). That is to say the result of composing the mappings on the direct path Xa p(X) ^ Xa p(X) ^ Xa p(X) is the same as the result of composing the mappings on the direct path Xa pl X1) 4 a,a(X2) 5 o,8 (X2). Therefore, (7.2) is commutative for 7; = 7» = a. Similarly, from the second and the third relations in (7.1), the commutativity for rj = no = D, and for n = Pı and y» = P; are obtained. Sufficiency. On the basis of (7.2), the three relations in (7.1) can be induced from the commutativity for m = r» = a, for m = m = f, and for m = Pı and nz = P». This is the sufficiency E VIL2 Isomorphism theorem Because the isomorphism between two maps determines an equiv- alent relation, what has to be considered for the equivalence is the equivalent classes, called isomorphic classes of maps. Two maps are said to be different if they are in different isomorphic classes. In order to clarify the isomorphic classes of maps, invariants should be inves- tigated. In this and the next sections, a sequence of elements with VIL2 Isomorphism theorem 169 its length half the cardinality of the ground set. In fact, this implies that the isomorphic class can be determined by a polynomial of degree as a linear function of half the cardinality of the ground set for both orientable and nonorientable maps. Lemma 7.2 If two maps Mı and M» are isomorphic, then Mj is orientable if, and only if, M» is orientable. Proof Let M; = (Xi, Pi), i = 1,2. Assume 7 is an isomorphism from Mı to Mə. From (7.2), ra = ar,TB = Br and TP, = Por, ie., rar =a, 707 0drPg = Ps. Necessity. Since Mj is orientable, from Theorem 4.1, permutation group V; = Yip, a} has two orbits (v1), and (o1)y,, 21 € X, on A. And, since rat~! = a and TOTH = fj, Tyr =la) = r(or Tor = (rat!) (7677) = a By considering 7P,7~! = Po, for any v, € V, mr = We € Wo. Therefore, V» also has two orbits on %2, i.e., (v2)w, and (@x2)y,, where £2 = TX, € A». This implies that Mə is orientable as well. Sufficiency. Because of the symmetry of 7 between Mı and Mo, the sufficiency is deduced from the necessity. [] For a map M = (X,7) where v(M), e(M) and ¢(M) stand for, respectively, the order(vertex number), the size(edge number) and the coorder(face number) of M. Lemma 7.3 If two maps Mı and M» are isomorphic, then v(Mi) = v(Ms), €(Mi) = (M3), (Mi) = 6(M3). — (7.3) Proof Let M; = (Xi, Pi), i = 1,2. Assume 7 is an isomorphism from Mj to M». From the commutativity for r4 = Pı and nz = P» in 170 Chapter VII Automorphisms of Maps (7.2), TP47-! = P». Then, for any integer n > 1 by induction, FUP ur =r Therefore, for any zı € A4, TX] = 2», T(z1)p, = (TT1)rpir- = (£2) pp. Because a 1-to-1 correspondence on vertices between M, and M» is induced from this, v( M) = v(M). Similarly, from rr ! = * and (Pi r = (P T(Piy)r t = Poy. Further, for any integer n > 1, r(P1y)"r! = (Poy)". This provides T(X1) Pry = (TT1) Piy- = (22)0 as a 1-to-1 correspondence on faces between M; and M». Therefore, (Mı) = (Mp). Finally, from ror ^! =a and rfr! = 8 and hence ryr ^! = 4, for any z; € Aj, x» = Tx, implies TKx, = Kay. This provides a 1-to-1 correspondence on edges between M; and Mə. therefore, e(Mi) = e(M3). o For a map M = (X, P), the Euler characteristic given by (4.1) is X(M) = v(M) — «(M) + 6(M) where v(M), e(M) and ¢(M) are, respectively, the order, the size and the co-order of M. Corollary 7.1 If two maps Mı and Mə are isomorphic, then x(Mi) = x(M3). (7.4) VIL2 Isomorphism theorem EE Proof A direct result of Lemma 7.3. E For a map M = (X5, P), let M* = (X7 5, P*) be the dual of M. It is, from Chapter III, known that M* = (Xa, PY). Theorem 7.2 Maps M; and M» are isomorphic if, and only if, their duals Mf and M5 are isomorphic. Proof Let M; = A i = 1,2, then M7 = (CP i = 1,2, where AU = X and Pt = Pq, i = 1,2. Necessity. Suppose 7 is an isomorphic between M; and Mo, then from Theorem 7.1, tat =o; TET "=p, Par | = P». On the basis of this, for any xı € a = a and %2 = T71 € AD. = Xa TK x, = T{£1, 821, 021, y$i] = rx, TB2 4, Tax, TYL} = (22, Üz2, 05, 13) = K* T9, and TP -—T(PWaui-rTYyrOj = (rPir--) (rr) = Pay = Pp. This implies that the diagram x0 T y a m (7.5) B (1)* T (2)* 2^ ————————À5 A a 7] are all commutative for 71 = 1j = 0, for m = 1j = a, and for m = TX and 72 = P3. therefore, from Theorem 7.1, 7 is an isomorphism between Mf and M5 in its own right. Sufficiency. from the symmetry of duality, the sufficiency is de- duced form the necessity. [] ES. Chapter VII Automorphisms of Maps — (x Let M; = (4555 and P = Piy, i = LR , Pi), and M? = uu F where a = A l Corollary 7.2 A bijection 7 : x ) — x9 is an isomorphism between maps Mı and Mg if, and only iE is an isomorphism between maps M; and M5. Proof A direct result in the proof of Theorem 7.2. [] VIL3 Recognition Although some invariants are provided, they are still far from determining an isomorphism between two maps in the last section. In fact, it will be shown in this section that an isomorphism between two maps can be determined by the number of invariants dependent on their size, i.e., a sequence of invariants in a number as a function of their size. In order to do this, algorithms are established for justifying and recognizing if two maps are isomorphic. In other words, an isomor- phism can be found between two maps if any; or no isomorphism exits at all otherwise. Generally speaking, since the ground set of a map is finite, i.e., its cardinality is 4e, € is the size of the map, in a theoretical point of view, there exists a permutation which corresponds to an isomorphism among all the (4c)! permutations if any, or no isomorphism at all between two maps otherwise. However, this is a impractical way even on a modern computer. Our purpose is to establish an algorithm directly with the amount of computation as small as possible without counting all the permu- tations. Here, two types of algorithms are presented. One is called vertez- algorithm based on (7.2). Another is called face-algorithm based on (7.5). Their clue is as follows. For two maps M; = (43, P1) and Mz = VIL3 Recognition 173 (A45, P2), from Lemma 7.3, only necessary to consider |À3| = |%| because the cardinality is an invariant under an isomorphism. First, choose x, € AX, and y; € Xə (a trick should be noticed here!). Then, start, respectively, from zı and yı on Mı and M» by a certain rule (algorithms are distinguished by rules ). Arrange the orbits Milta and IA ia as cycles. If 721) Vera, = wo (7.6) can be induced from y; = 7(z1), then stop. Otherwise, choose another yıla trick!). Go no the procedure on M» until every possible y; has be chosen. Finally, if stops at the latter, then it is shown that M; and M» are not isomorphic, and denoted by M; Z Mə; otherwise, an isomorphism between M; and M^» is done from (7.6), denoted by M, = M». Algorithm 7.1 Based on vertices, determine if two maps are isomorphic. Given two maps P = (X,7) and Q = (Y, Q), and their order, size and co-order are all equal(otherwise, not isomorphic!). In conve- nience, for any x € X, let |x| = |[x)p|, ie. , the valency of vertex (x)p. Initiation Given x € X, choose y € y. Let r(x) = y and TKzr = Ky. Label both x and y by 1. Naturally, Kz = Ky = K1 = (1, o1, 81, 41) (Here, the number 1 deals with a symbol!). Label (x)p by 0, then x = 1 is the first element coming to vertex 0. By (v, ty) denote that t, is the first element coming to vertex v. Let S be a sequence of symbols storing numbers and symbols and l, the maximum of labels on all the edges with a label. Here, S = Ó, | — 1 and the minimum of labels among all labelled but not passed vertices n = 0. If vertex (y1)p = (1)p, the maximum vertex label m = 0; otherwise, label vertex (y1)p by 1, m = 1. Proceeding When all vertices are labelled as used, then go to Halt (1). 174 Chapter VII Automorphisms of Maps For n, let sp and sg be, respectively, the number of edges without label on (ytn)p and (4t,)o. If sp Z: Sg, when no y can be chosen, then goto Halt (2); other- wise, choose another y and then goto Initiation. In the direction starting from ytn, label those edges by /+1,---,/+ S,8 = Sp = Sg È 0 in order. Thus, two linear orders of elements with numbers labelled (ts, PYtn, ttg P-MWt,) and Fir Qytn, ar) Q wo are obtained. If the two are not equal, when no y is available to choose, then goto Halt (2); otherwise, choose another y and then goto Initiation. Put this linear order into S as last part and then substitute the extended sequence for S. In the meantime, label K(l-4- 1), K(l 4- 2), s K(l4- s) on P and Q. Substitute l + s for l. Mark vertex n as used. Substitute n 4- 1 for n. Let r be the number of vertices without label in (o (E L))ps s (E 489, and label them as m + 1, - -,m +r in order. Substitute m + r for m. Go on the Proceeding. Halt (1) Output S. (2) P and Q are not isomorphic. About Algorithm 7.1, from the way of choosing y, each element in the ground set is passed through at most once. So there exists a constant c such that the amount of computation is at most c|AX |. Since the worst case is for y chooses all over the ground set ,the total amount of computation is at most c|.X|?. Because of |X| = 4e where e is the size of the map, this amount is with its order as the size squared, i.e., O(€?). As described above, if checking all possibilities of |V|!, by Stirling VIL3 Recognition 175 formula, I| ~ Vre P y|- >> O(e!) >> O(e) soe) when |V| = |X| = 4e is large enough. Thus, this algorithm is much efficient. Algorithm 7.2 Based on faces, determine if two maps are iso- morphic. Given two maps P = (Xap, P) and Q = (Vag, Q), and their order, size and co-order are all equal(otherwise, not isomorphic!). For convenience, let ¥ = Xag, Y = Y, and for any x € X, let |x| = l{z}p,|, i.e., the valency of face (x)p, where y = af. Initiation Given z € X, choose y € Y. Let r(x) = y and TKr = Ky. Label both x and y by 1. Naturally, Kx = Ky = K1 = (1,01, 81, 41] (Here, the number 1 deals with a symbol!). Label (z)piga by 0, then x = 1 is the first element coming to face 0. By (t) denote that t£; is the first element coming to face f. Let T' be a sequence of symbols storing numbers and symbols and /, the maximum of labels over all the edges with a label. Here, T = 0, l = 1 and the minimum of labels among all labelled but not passed faces n = 0. If face (y1)p, = (1)p,, the maximum face label m = 0; otherwise, label face (y1)p, by 1, m — 1. Proceeding When all faces are labelled as used, then go to Halt (1). For n, let sp and sg be, respectively, the number of edges without label on (7tn)py and (415) o4. If sp # Sg, when no y can be chosen, then goto Halt (2); other- wise, choose another y and then goto Initiation. In the direction starting from ytn, label those edges by /+1,---,/+ S,8 = Sp = Sg È 0 in order. Thus, two linear orders of elements with numbers labelled has Pyryts, UU Py ytn) 176 Chapter VII Automorphisms of Maps and (yin, Dyin, ++, Qu Ita) are obtained. If the two are not equal, when no y is available to choose, then goto Halt (2); otherwise, choose another y and then goto Initiation. Put this linear order into S as last part and then substitute the extended sequence for S. In the meantime, label A(/ +1), K(l 4 1), ---, K(l+ s) on P and Q. Substitute l + s for l. Mark n as used. Substitute n + 1 for n. Let r be the number of vertices without label in (y(L-4- 1))p,---, (y+ 8), and label them as m + 1, - -,m +r in order. Substitute m + r for m. Go on the Proceeding. Put this linear order into T' as last part and then substitute the extended sequence for T. In the meantime, label K(I +1), K(l 4- 2), 5s K(L+ s) on P and Q. Substitute | + s for l. Mark face n as used. Substitute n 4- 1 for n. Let r be the number of faces without label in (*(L4- L)p, (E 8)», and label them as m + 1,-- m +r in order. Substitute m + r for m. Go on the Proceeding. Halt (1) Output T. (2) P and Q are not isomorphic. About Algorithm 7.2, it can be seen as the dual of Algorithm 7.1. The amount of its computation is also estimated as O(c?). Note 7.2 This two algorithms suggest us that whenever a cyclic order of edges at each vertex is given, an efficient algorithm for justi- fying and recognizing if two graphs are isomorphic within the cyclic order at each vertex can be established. By saying an algorithm effi- cient, it is meant that there exists an constant c such that the amount of its computation is about O(c^), € is the size of the graphs. If without considering the limitation of a cyclic order at each ver- tex, no efficient algorithm for an isomorphism of two graphs has been VILA Justification 177 found yet up to now. However, a new approach is, from what has been discussed here, provided for further investigation of an isomorphism between two graphs. VILA Justification In this section, it is shown that the two algorithms described in the last section can be used for justifying and recognizing whether, or not, two maps are isomorphic. Lemma 7.4 Let S and T are, respectively, the outputs of Al- gorithm 7.1 and Algorithm 7.2 at Halt (1), then (i) Elements in S and T are all in the same orbit of group Yip} on X. (ii) S forms an orbit of group Usp, on X if, and only if, T forms an orbit of group Yip} on «; (ii) S forms an orbit of group V». on X if, and only if, for any z E S, yz ES. Proof (i) From the proceedings of the two algorithms, it is seen that from an element only passes through y and P( Algorithm 7.1), or y and P^(Algorithm 7.2) for getting an element in S, or T. Because y, P, Py € VU rp and ~? = 1 , elements in S and T are all in the same orbit of group Vp. on «X. (ii) Necessity. Because S forms an orbit of group V». on X, and from Algorithm 7.1, S contains half the elements of X, by Lemma 4.1, group Yip, has two orbits on A. This implies in the orientable case. Thus, from (i), T forms an orbit of group ipy on X as well. Sufficiency. On the basis of duality, it is deduced from the neces- sity. (ii) Necessity. Since S forms an orbit of group Yip} on ¥ and 5 contains only half the elements of X, by Lemma 4.1, group Vp. has two orbits on X. From the orientability, for any z € S, yx € S. Sufficiency. Since for any x € 5, yx € S, and S only contains 178 Chapter VII Automorphisms of Maps half the elements of X, by Corollary 4.1, it is only possible that S itself forms an orbit of group V». on X. [] For nonorientable maps, such two algorithms have their outputs S and T also containing half the elements of ¥ but not forming an orbit of group V (p... Lemma 7.5 Let S and T are, respectively, the outputs of Al- gorithm 7.1 and Algorithm 7.2 at Halt (1). And , let Gs and Gr be, respectively, the graphs induced by elements in S and T, then Gs = Gr = G(P). Proof From Lemma 7.4(i), by the procedures of the two algo- rithms, because the intersection of each of S and T with any quadricell consists of two elements incident the two ends of the edge, S, T' as well, is incident to all edges with two ends of each edge in map P. Therefore, Gs = Gr = G(P). O Theorem 7.3 The output S of Algorithm 7.1 at Halt (1) in- duces an isomorphism between maps P and Q. Halt (2) shows that maps P and Q are not isomorphic. Proof Let r be a mapping from X to Y such that the image and the co-image are with the same label. From the transitivity of a map, 7 is a bijection. Because 7T Kr = Krz, x € X, then rar | =a and rÜr ! = B. And in the Proceeding, for labelling a vertex (x)p, T(r)p = (Trx)g. From Lemma 7.5, this implies that 7PT ! = Q. Based on Theorem 7.1, 7 is an isomorphism between P and Q. This is the first statement. By contradiction to prove the second statement. Assume that there is an isomorphism 7 between P and Q. If r(x) = y, then by Algorithm 7.1 the procedure should terminate at Halt (1). However, a termination at Halt (2) shows that for any x € 4X, there is no elements in Y corresponding to x in an isomorphism between maps P and Q, and hence it is impossible to terminate at Halt (1). This is a contradiction. VILA Justification 179 Therefore, the theorem is true. [] Although the theorem below has its proof with a similar reason- ing, in order to understand the precise differences the proof is still in a detailed explanation. Theorem 7.4 The output T of Algorithm 7.2 at Halt (1) in- duces an isomorphism between maps P and Q. Halt (2) shows that maps P and Q are not isomorphic. Proof Let r be a mapping from X to Y such that the image and the co-image are with the same label. From the transitivity of a map, T is a bijection. Because T Kx = Krz, x € X, then rar ! = o and rÜT ! = 8. And in the Proceeding, for labelling a face (x)p,, T(x)p, = (rz)g4. From Lemma 7.5, this implies that TP»yr ! = Qy. Based on Theorem 7.2, 7 is an isomorphism between P and Q. This is the first statement. By contradiction to prove the second statement. Assume that there is an isomorphism 7 between P and Q. If r(x) = y, then by Algorithm 7.2 the procedure should terminate at Halt (1). However, a termination at Halt (2) shows that for any x € Æ, there is no elements in Y corresponding to x in an isomorphism between maps P and Q, and hence it is impossible to terminate at Halt (1). This is a contradiction. Therefore, the theorem is true. H If missing what is related to y in Algorithm 7.1 and Algorithm 7.2, then for any map M = (X, P), the procedures will always termi- nate at Halt (1). Thus, their outputs S and T are, respectively, called a primal trail code and a dual trail code of M. When an element x and a map P should be indicated, they are denoted by respective S,(P) and T}(P). Theorem 7.5 Let P = (X,P) and Q = (V, Q) be two given maps. Then, they are isomorphic if, and only if, for any x € X chosen, there exists an element y € Y such that S (P) = S,(Q), or 180 Chapter VII Automorphisms of Maps T,(P) = T,(Q). Proof Necessity. Suppose 7 is an isomorphism between maps P = (#,P) and Q = (¥,Q). For the given element r € X, let y = iud From Theorem 7.3, or Theorem 7.4, $,(P) = S,(Q), or T,(P) = T,(Q). Sufficiency. From Theorem 7.3, or Theorem 7.4, it is known that by Algorithm 7.1, or Algorithm 7.2, their outputs induces an isomorphism between P = (X, P) and Q = (Y, Q). O Note 7.3 In justifying whether, or not, two maps are isomor- phic, the initial element xr can be chosen arbitrarily in one of the two maps to see if there is an element y in the other such that Se P) = 540) or TIL P) = TAQ): This enables us to do for some convenience. In addition, based on Theorem 7.5, all isomorphisms between two maps can be found if any. VIL5 Pattern examples Here, two pattern examples are provided for further understand- ing the procedures of the two algorithms described in the last section. Pattern 7.1 Justify whether, or not, two maps M; = (A3, P1) and M» = (A5, P5) are isomorphic where X= Krit Ky, Pi = (zy yy By)(yoi) and X = Kra + Kyo, P» = (yo, £2, By2) (yv2). First, for Mi, choose x = z,. By Algorithm 7.1, find S,(Mj). Let P, = (zy, yi, By) (y21) = wv. VIL5 Pattern examples 181 Initiation xı = 1, Kay = {1,a1, 61,41), u=0,v= 1, Seg lL=0, m=1. Proceeding Step 1 Pı = (1, y1, 8y1) (71). yı = 2, Ky = 12 a2 02h u = 0, v = L S02 802), l=2, se mL Step2 P, = (1,2,82)(31). ues og SL, S = (1,2, 82, y1), [52,2 1, w= L Halt (1) Output: S,(Mi) = S = (1,2, 82, y1). Then, for Mə, because a link should correspond to a link and a vertex should correspond to a vertex with the same valency, y has only two possibilities for choice, i.e., zo and ox». Choose y = x». By Algorithm 7.1, find S,(M»). Let P» = (ya, £2, By2)(yx2) = uv. Initiation z2 = 1, Kz = {1,a1, 61, y1}, u=0,v= 1, S=0, l=0, m=k Proceeding Step 1 Pa= (yi, 1, 8y2) (1). By» = 2, K By. = 12,02,02,521. u = 0, U= L S = (1,2, 62), 122, n2 1, m— 1. Step 2 'P = (2,1, 82)(71). u=0, v=1, S = (1,2, 82, y1}, L=2, n=1, mL 182 Chapter VII Automorphisms of Maps Halt (1) Output: 5,0M5) 9 5 = 11,2,02, 1X. Since S,(M1) = S,(M»3) and y = x2, an isomorphism from Mi to M^» is found as 7}: Ti Kai = Kx, T Ky = K yp. Then, choose y = ox». By Algorithm 7.1, find S,(M»). Let P5 = (axo, AY2, YY2) (Üx2) = uv. Initiation az = 1, Kaz = {1,a1, 01,1}, u= 0,v = 1, S20. l=0, m=1. Proceeding Step 1 Py = (1, ayz, yyz) (31). ayz = 2, Kay: = 12,02,82, 21, u= 0, v= 1, 91,960), ET m e, Step 2 Pi = (1,2, 82)(71). u=0, v= 1, S = (1,2, 82, y1}, L=2, n=1, w= 1, Halt (1) Output: S,(M2) = S = (1,3,82,71): Since S,(M1) = S,(M») and y = oz», an isomorphism from Mı to Mo is found as 73: ToK a4 = Karte, ToKyy = Kaye. In consequence, there are two isomorphisms between M; and Mo above in all. Since 2 € S,(Mi) but y2 ¢ S,(M;), by Lemma 7.4(iii), Mı, Mə as well, is nonorientable. Pattern 7.2 Justify whether, or not, Mı = (43, P1) and M5 = (Xə, P5) are isomorphic where A= hay t Kyn Tp ovo) mn Va) VIL5 Pattern examples 183 and A» = Kus Kyo, Po = (yo, 22, YY2) (722). First, for Mi, choose x = gı. By Algorithm 7.2, find T;(Mj). Let Pry = (21,741, ) (V1) = f9. Initiation Xj = L, Kx, = {1,a1, 81, y1}, f = 0, g = 1, Tel1e0 m=0. Proceeding Step 1 Piy = (1,71, y1) (791). y= 2; Ky = {2, a2, 82,72), f = 0, g= 1, Tei(Lo1,2$, (22, 21, mel. Step 2 Piy = (1,71, 2)(72). 509,921 PSA i527 q els m= 1. Halt (1) Output: TaM) & T = 1,41, 2,42). Then, for M», because a link should be corresponding to a link and a vertex should be corresponding to a vertex with the same va- lency, y only has two possibilities for choosing, i.e. , r9 and az». Choose y = x». By Algorithm 7.2, find T,(M»). Let Pay = (x2, 7X2, YY2) (yo) = fg. Initiation LQ = 1, Kx = {1,a1, DL y1}, f = 0, g = 1, T=), l=0, m=0. Proceeding Step 1. Poy = (1,71, yy) (a). yy = 2, Kyy = 12,02, 82,42), f 20, g— 1, T=(1,7L2) l=2 n=l; m=] 184 Chapter VII Automorphisms of Maps Step 2 Poy = (1,71, 2)(72). f=0; g= 1, Pai). 82 n=1, m= 1. Halt (1) Output: T,(M2) = T = (1,71, 2, 72). Since T;(M1) = T,(M2) and y = x, an isomorphism from M; to M^» is found as 7}: Ty Kg, Kay, MK yy = Ky. Then, choose y = ax. By Algorithm 7.2, find T,(M2). Let Pay = (a, Bx», aye) (Gy2) = f9. Initiation ax = 1, Kax = {1,a1, 61,71}, f=0,g=1, Te 120, 20. Proceeding Step 1 Poy = (1,71, aye) (Byz). ays = 2, Kyy = {2, a2, 82,72}, f=0, g=1, T'ecLet2Lgje wel quel Step 2 Poy = (1,71, 2)(72). f=0, 9=1, T = Lo d A doma, n=1, mem Halt (1) Output: T,(M5) = T = (1,31, 2,72). Since 7,(Mi) = T,(M») and y = ox», an isomorphism from Mı to Me» is found as m: To)Kx| = Kart, T» Ky1 = Kaye. In consequence, there are two isomorphisms between M; and Mo in all. By Lemma 7.4(iii), Mi, M» as well, is orientable. Activities on Chapter VII VII.6 Observations O7.1 Observe that whether, or not, an isomorphism between two maps is always mapping a link to a link and a loop to a loop. If it is, describe the reason. Otherwise, by an example. O7.2 Observe that whether, or not, an isomorphism between two maps is always mapping an element incident with a vertex of valency 7 to an element incident with a vertex of valency i. If it is, describe the reason. Otherwise, by an example. O7.3 If missing rar ! = a or TT! = 8 in (7.2), whether, or not, T is still an isomorphism. If it is, describe the reason. Otherwise, by an example. O7.4 Provide two distinct embeddings of a graph which are two isomorphic maps. O7.5 Observe that how many non-isomorphic maps among all embeddings of the complete graph of order 4. O7.6 List all non-isomorphic maps of size 3 and find the dis- tribution by relative genus. O7.7 Explain the differences between non-isomorphic maps with the same under graph and distinct embeddings of the graph by exam- ples. O7.8 Explain the differences between non-isomorphic graphs and distinct embeddings of these graphs by examples. 186 Activities on Chapter VII O7.9 Let 7 and 7» are two isomorphisms between two maps. Observe whether, or not, their composition 7175 l7 is also an isomor- phism. If it is, describe the reason. Otherwise, by an example. O7.10 Observe some algebraic properties on the set of all iso- morphisms between two maps. O7.11 Observe that for two maps Mı = (43,1) and Mə = (A, P2), is the composition of two permutations Pı and P» a map? If it is, describe the reason. Otherwise, by an example. O7.12 Observe some algebraic properties of the sets of all maps and premaps on the same ground set under the composition of per- mutations. VII.7 Exercises If an edge is with its two ends of valencies ? and j, then it is called a (i, j)-edge, 0 < i, j € 2e. If its two incident faces are of valencies | and s, then it is called a (l, s)'-edge, 0 € s,t < 2e. Here, e is the size of a map. E7.1 Let m;;(M) and n;;(M) are, respectively, the numbers of (i, j)-edge and (i, 7)*-edge in map M. Prove that if maps M; and M» are isomorphic, then for any i and j, 0 € i, j € 2e, mi(Mi) = mjj(M3) and ni; (Mi) = nij( Mə). E7.2 Prove that a bijection between the basic sets of two maps is an isomorphism of the two maps if, and only if, it induces both the correspondences between their vertices and between their faces. E7.3 Design an algorithm which is different from Algorithm 7.1 and Algorithm 7.2 for justifying an isomorphism between two maps such that its computation amount is in the same order as their’s. E7.4 Prove that Algorithm 7.1 and Algorithm 7.2 are with the computation order O(e) in justifying an isomorphism of two maps which have a triangular face, or a vertex of valency 3 where e€ is the VIL.7 Exercises 187 size of the maps. E7.5 Prove that Algorithm 7.1 and Algorithm 7.2 are with the computation order O(e) in justifying an isomorphism of two maps which have only one —articulate vertea(a vertex of valency 1), or only one loop where e is the size of the maps. E7.6 Determine the number of non-isomorphic butterflies of size m > 1, or establish a method to list them. E7.7 Determine the number of non-isomorphic barflies of size m > 1, or establish a method to list them. E7.8 On the basis of Algorithm 7.1, design an algorithm for justifying an isomorphism between two planar graphs(not maps!), and estimate its computation order. E7.9 On the basis of Algorithm 7.2, design an algorithm for justifying an isomorphism between two planar graphs(not maps!), and estimate its computation order. For any map M, let T be a spanning tree of its under graph G(M). Each co-tree edge is partitioned into two semi-edges seen as edges. Because what is obtained is just a tree when each of such semi-edges is seen with a new articulate vertex, it is a joint tree cor- responding to an embedding of its under graph G(M), also called an joint tree of M. If x and (Gz are in the same direction for an edge X = Kx partitioned along the joint tree, then it is said to be with the same sign, denoted by X and X; otherwise, different signs, denoted by X and X-!, or X^! and X. The cyclic order of letters with signs of such semi-edges partitioned into is called an joint sequence of the map. E7.10 Prove that a graph G is planar if, and only if, there exists an joint sequence of maps whose under graph is G such that each letter is with different signs and no two letters are interlaced. E7.11 Prove that for any complete graph Kn, n > 1, there is no joint tree for all maps whose under graph are K, such that it corresponds to a simplified butterfly. 188 Activities on Chapter VII VILS Researches R7.1 Discuss whether, or not, there exits a number, indepen- dent on the size of a map considered, of invariants within isomorphism of maps for justifying and recognizing an isomorphism between two maps. R7.2 For a given graph G and an integer g, determine the number of distinct embeddings of G on the surface of relative genus g, and the number of non-isomorphic maps among them. R7.3 For a given type of graphs G and an integer g, find the number of distinct embeddings of graphs in G on the surface of relative genus g, and the number of non-isomorphic maps among them. R7.4 Determine the number of non-isomorphic triangulations of size m > 3. R7.5 Determine the number of non-isomorphic quadrangula- tions of size m > 4. R7.6 For an integral vector (no, n4, --- , Noi- -), find the num- ber of non-isomorphic Euler planar maps each of which has n»; vertices of valency 2i, à > 1. Because it can be shown that two graphs G4 and Gs are isomor- phic if, and only if, for a surface they can be embedded into, there exist embeddings p1(G 1) and u3(G3) isomorphic, this enables us to investigate the isomorphism between two graphs. The aim is at an efficient algorithm if any. R7.7 Suppose map Mj is an embedding of Gi on an orientable surface of genus g, justify whether, or not, there is an embedding M» of graph Gə such that Mə and M; are isomorphic. R7.8 Suppose map Mj is an embedding of G, on a non-orientable surface of genus g, justify whether, or not, there is an embedding M» of graph Gə such that Mə and M; are isomorphic. R7.9 According to [Liul], any graph with at least a circuit has a non-orientable embedding with only one face. Justify whether, or VII.8 Researches 189 not, two graphs G4 and Gs have two respective single face embeddings which are isomorphic. R7.10 Justify whether, or not, a graph has two distinct single face embeddings which are isomorphic maps. A graph is called up-embeddable if it has an orientable embedding of genus which is the integral part of half the Betti number of the graph. Because of the result in [Liul], unnecessary to consider the up-embeddability for non-orientable case. R7.11 Determine the up-embeddability and the maximum ori- entable genus of a graph via its joint sequences. R7.12 For a given graph G and an integer g, justify whether, or not, the graph G has an embedding of relative genus g. Chapter VIII Asymmetrization e An automorphism of a map is an isomorphism from the map to itself. All automorphisms of a map form a group called its au- tomorphism group. of a map is, in fact, the trivialization of its automorphism group. e A number of sharp upper bounds of automorphism group orders for a variety of maps are provided. e The automorphism groups of simplified butterflies and those of simplified barflies are determined. e The realization of of a map is from rooting an element of the ground set. VIIL1 Automorphisms An isomorphism of a map to itself is called an automorphism . Let 7 be an automorphism of map M = (X, P). If for x € X, T(x) =y and x Æ y, then two elements x and y play the same role on M, or say, they are symmetric. Hence, an automorphism of a map reflects the symmetry among elements in the ground set of the map. Lemma 8.1 Suppose 7; and 72 are two automorphisms of map M, then their composition 7,7» is also an automorphism of map M. VIIL1 Automorphisms 191 Proof Because 7, is an automorphism of M = (Xag, P), from [62) qr So mm ed ups eem. Similarly, for 75, ed E = =] T20T, =Q, ToDT, =p, PTa =P. Therefore, for 7175, (n)a(ri2) = (TaT2)a(73 ^n) = (mar; ri | = IQT] = a, (7172) 8 (7172) = (7172) 8 (Ty Ti ") = n (Trz) = nmn =P, and (ris)P(ri2) = (T172)P (ra^) = n(r4Pr; ri! =P =P. This implies that for 7,72, (7.2) is commutative. From Theorem 7.1, T|T2 is an automorphism of M as well. [] On the basis of the property on permutation composition, auto- morphisms satisfy the associate law for composition. Because an automorphism 7 is a bijection, it has a unique inverse denoted by 7~!. Because T 'ar=r (ror!) —(r rar ir) «a, and similarly, y peg PD from Theorem 7.1, 7^! is also an automorphism. If an element x € ¥ has r(x) = x for a mapping(particularly, an automorphism) 7, then z is called a fixed point of r. If every element 192 Chapter VIII Asymmetrization is a fixed point of 7, then 7 is called an identity . Easy to see that an identity on ¥ is, of course, an automorphism of M, usually said to be trivial . By the property of a permutation, an identity is the unity of automorphisms, always denoted by 1. In summary, the set of all automorphisms of a map M forms a group, called the automorphism group of M, denoted by Aut( M). Its order is the cardinality of the set aut( M) = |Aut(M)]|, i.e., the number of elements in Aut(M) because of the finiteness. Theorem 8.1 Let 7 be an automorphism of map M = (&,P). If 7 has a fixed point, the 7 = 1, i.e., the identity. Proof Suppose x is the fixed point, ie. , r(x) = x. Because 7 is an isomorphism, From (7.1), Tau) ewr(w) eu, (8x) = Br(2) = Ba T(Px) = P(T(£)) = Pz, i.e., ax, Dx and x are all fixed points. Then for any V € Via gp}, T(v(z)) = v(r(z)) = v(z). Therefore, from transitive axiom, every element on ¥ is a fixed point of 7. This means that 7 is the identity. [] In virtue of this theorem, the automorphism induced from T(x) = y can be represented by 7 = (x — y). Example 8.1 Let us go back to the automorphisms of the maps described in Pattern 7.1 and Pattern 7.2. If Mı and M» in Pattern 7.1 are taken to be M = (Kz + Ky, (x,y, By)(yz)) = Mı, then it is seen that only one nontrivial automorphism 7 = (x — az) exists. Thus, its automorphism group is Aut(M) = (1, (x ^ ax)}, VIII.2 Upper bound of group order 193 i.e., a group of order 2. Then, maps M; and M» in Pattern 7.2 are taken to be M = (Kz 4 Ky, (xz, y, yy)(yz)) = Ms, it has also only one nontrivial automorphism 7 = (x — ox). So, its automorphism group is Aut(M) = (1, (x ^ ax)}, a group of order 2, as well. However, maps M; and M» here are not isomorphic. In fact, it is seen that M, is nonorientable with relative genus —1. and M^» is orientable of relative genus 1. VIIL2 Upper bounds of group order Because the automorphism group of a combinatorial structure with finite elements is an finite permutation group in its own right, its order must be bounded by an finite number. And, because there are n! permutations on a combinatorial structure of n elements, the order of its automorphism group is bounded by n!. However, n! is an exponential function of n according to the Stirling approximate formula, it is too large for determining the auto- morphism group in general. Now, it is asked that is there an constant c such that the order of automorphism group is bounded by n°, or denoted by O(n°), if there is, then such a result would be much hopeful for the determination of the group efficiently. In matter of fact, if the order of automorphism group is O(n*), c is independent of n for a structure with n elements, then an efficient algorithm can be designed for justifying and recognizing if two of them are isomorphic in a theoretical point of view. Lemma 8.2 For any map M = (4,P), the order of its auto- morphism group is aut(M) < |X| = 4e(M) (8.1) 194 Chapter VIII Asymmetrization where e(M) = 1| | is the size of M. Proof From Theorem 8.1, M has at most |X| = 4e(M) auto- morphism, i.e., (8.1). O The bound presented by this lemma is sharp, i.e., it can not be reduced any more. For an example, the link map L = (Ka, (x)(»yx)). The order of its automorphism group is 4 = |Kz| = e(L). Lemma 8.3 For an integer i > 1, let v;( M) be the number of i-vertices(vertex incident with i semi-edges) in map M = (X, P), then aut(M) | 2ivi(M), (8.2) i.e., aut( M) is a factor of 2iv;( M). Proof Let r € Aut(M) be an automorphism of M. For x € X, (x)p is an i-vertex, assume T(x) = y. From the third relation of (7.1), (y)p is also an i-vertex. Then, the elements of X; = [x|Vx € X,|{x}p| = i) can be classified by the equivalent relation £ Aut Y 4> dr € Aut(M)x = Ty induced from the group Aut(M ). From Theorem 8.1, Aut(G) has a bijection with every equivalent class. This implies that each class has aut(M) elements. Therefore, aut(M) | ||. Because |X;| = 2iv;( M), (8.2) is soon obtained. O This lemma allows to improve, even apparently improve the bound presented by Lemma 8.1 for a map not vertex-regular(each vertex has the same valency). Lemma 8.4 For an integer j > 1, let ġ;(M) be the number of j-faces of map M = (X, P), then aut(M) | 2j6;(M), (8.3) VIII.2 Upper bound of group order 195 i.e., aut(M) is a factor of 27¢;(M). Proof Let r € Aut(M) be an automorphism of M. For x € X, (x)p, is a j-face, assume T(x) = y. From the first two relations of (7.1), (yz) = yy. Then from this and the third relations, r((Py)x) = a Thus, (y)p4 is also a j-face. And, the elements of X; = {x|Vx € X,|{x}p,| = j} can be classified by the equivalence £ ~aut €— dr € Aut(M)x = Ty induced from the group Aut(M ). Further, from Theorem 8.1, Aut(G) has a bijection with every equivalent class. This leads that each class has aut(M) elements. Therefore, aut(M) | |: Because |X;| = 2j¢;(/), (8.3) is soon obtained. O This lemma allows also to improve, even apparently improve the bound presented by Lemma 8.1 for a map not face-regular(each face has the same valency). Theorem 8.2 Let v;(M) and ġ;(M) be, respectively, the num- bers of i-vertices and j-faces in map M = (X, P), i,j > 1, then aut(M) | (2iv;, 26; | Vi, i» 1, Vj, j > 1), (8.4) where (2iv; 276; | Vi, i > 1, Vj, j > 1) represents the greatest common divisor of all the numbers in the parentheses. Proof From Lemma 8.3, for any integer i > 1, aut( M) | 2iv;(M). From Lemma 8.4, for any integer j > 1, aut(M) | 2i$;(M). By combining the two relations above, (8.4) is soon found. O 196 Chapter VIII Asymmetrization Based on this theorem, the following corollary is naturally de- duced. Corollary 8.1 Let v;(M) and ¢;(M) be, respectively, the num- bers of i-vertices and j-faces in map M = (X, P), i, j > 1, then aut(M) < (2iv;, 2jó; | Vi, i> 1, Vj, j > 1). (8.5) Proof A direct result of (8.4). O Corollary 8.2 For map M = (4,P), e(M) is its size, then aut( M) | 4e(M). (8.6) Proof Because 4e(M) 2 25 ini(M) =2X_jġ;(M) i>1 jel = 5 2ivi(M) = 5 2jġ;(M), i21 j21 we have (2iv;, 26; | Vi, i> 1, Vj, j > 1) | 4e(M). Hence, from Theorem 8.2, (8.6) is soon derived. El VIII.3 Determination of the group In this section, the automorphism groups of standard maps, t.e., simplified butterflies and simplified barflies, are discussed. First, observe the orientable case. For Oj= (fi) = (Aart AM, (1, Yi TY) VIIL3 Determination of the group 197 by Algorithm 7.1, Sa (01) = Lys YL yw = 1, 2, 71, 32; Sas, (O1) = 1, Byi, 71, ayı = 1,2, 1, 72; Say, (O1) = L oy 71, 8j = 1,2, 31, 72; S45,(01) = 1, 71,71, y1 = 1,2, y1, 2; Bul CO) = Ly a= T9, eye Dag 1) e, gy. T a m 1,2 1,52: Ban tO) em T o Loa = T, 2 eu Bay (Oi) ec, Lue 12 01, 59. Uea Sx, (O1) = Sas (O1) 2s Sex, (O1) = S45, (01) = SO) = Say, (O1) = S54, (O1) — S44,(01) —L2,y1,32. Thus, O; has its automorphism group of order 8, t.e., aut(Qi) m4x(2x1l)2zs A map with a non-trivial automorphism group is said to be sym- metrical. If a map with its automorphism group of order 4 times its size, the it is said to be completely symmetrical. It can be seen that O4 is completely symmetrical. However, none of Og, k > 2, is completely symmetrical although they are all symmetrical. Theorem 8.3 Forsimplified butterflies(orientable standard maps) Ox = (A, Je), k = 1, where k Ay = X Ci - Kyi) i=1 and k Jk = (LIGA Vi, yi, yi), 1=1 198 Chapter VIII Asymmetrization we have 2k, if k > 2; 8.7 8, ifk=1. o0 aut(O,) = i Proof From the symmetry between (zi, yi, Y£i, Yyi), 1 > 2, and (z1, Uy, Y21, "Yy1) in Jk, k > 2, only necessary to calculate Sz, (Ox), Sa (Ok), Sy(Ox), S44 (0x), Sarı (Or), San(Ox) Sayı (Ok), and Say (Ox) by Algorithm 7.1. From Algorithm 7.1, k S3, (OX) = Lys Y1, vu | [ea vo veo vyd i=l k = 1,2, 71,72, [ [Gi — 1), 2i, 52i — 1), 72i) i=2 k -H« ((2i — 1), 2i, (2i — 1), 21), k Saxil (Or) = = L; Q (T [I (xi, Yis YLi VY) BY, Y1, ayı i—1 + Sr (Ok), k Dn (Ox) = 1, YY, [i Vi; Yi, Vs); yl, yı =l * S5, (Ox), k Sa, (Ox) = 1, QUA, yl, at] [(z: Yi, VX; vyi)), 1 i=l # S, (Ox), k S5 (Ox) = 1.92471; | [ener toyga i=1 * S3, (Ox), VIIL3 Determination of the group 199 k Say (Ox) =1 , QT], Q (ften yi yen 099) 5 01,6 * Dei (Ox), k Bul (Ox) = um Sog b y. £1, V1, yz j=l Se (0x); k Say (Ox) = L Bun al, QT], a(] [ (ai. Vi; "Yi, qu)) | =l k -:1,251,92, [ [ (i= 1), 24,922 =1),;72) 22 = utt Because two automorphisms are from S%,,(Ox) = Sr (Or), Oy have 2 x k = 2k automorphisms altogether. Hence, when k > 2, aut(O;) = 2k. When k = 1, aut(O1) = 8 is known. O Then, observe the nonorientable case. for Qi = (45,23) = (IK, (21, 821)), by Algorithm 7.1, Sx, (Q1) = LG Don (1) = LoL S85, (Q1) x LOR Pais (Qi) = L 8l i.e., aut(Q1) = Theorem 8.4 For simplified barflies Qı = (45,71), | > 1, where l i=1 200 Chapter VIII Asymmetrization and Tı = Tic: zi, i=l we have ato s a (8.8) Proof From the symmetry of (xj, Oxi), à > 2, and (x1, 92) in Tı, l > 2, only necessary to calculate S5 (Qi); Sazı (Qi); Spa (Qi) and S54, (Qi) by employing Algorithm 7.1. From Algorithm 7.1, l Sali) = 1, 81, | [ (4 Baa) i=2 l ES 1, 61,2, 82, | | (wi, Bai) i=3 l = | [G6 i=1 l Dads (Qi) = = 1 Q qK Tpu) me 1—2 * Dai G1); l Out Qi) — um da. i1—2 Sz (Q1), VIILA4 Rootings 201 l S291) = 1,81, a( e 1—2 2 = 1, 61, L, BI, li (yi, o) = ] [G0 i=1 = Oy, (Qi). Because two automorphisms are from Sys, (Qi) = Sa (Q1), Qi has 2 x | = 2l automorphisms altogether. Hence, when / > 2, aut(Qi) e. When l = 1, aut(Q1) = 4 is known. O Similarly, the two theorems can also be proved by employing Algorithm 7.2. VIIL.4 Rootings For a given map M = (X, P), if a subset R C X is chosen such that an automorphism of M with R fixed, i.e., an element of R does only correspond to an element of R, then M is called a set rooted map. The subset R is called the rooted set of M, and an element of R is called a rooted element . Theorem 8.5 For a set rooted map MÈ = (X,7), R is the rooted set, aut(M®) | |R]. (8.9) Proof Assume that all elements in R are partitioned into equiv- alent classes under the group Aut( MP). From Theorem 8.1, each class has aut( MÈ) elements. Therefore, (8.9) is satisfied. O 202 Chapter VIII Asymmetrization Corollary 8.3 For a set rooted map MÈ? = (X,7), R is the rooted set, aut(MP) < |R]. (8.10) Proof A direct result of (8.9). E For a given map M = (X, P), if a vertex v,, x € X, is chosen such that an automorphism of M has to be with v, fixed, t.e., an element incident with v, has to correspond to an element incident with vz, then M is called a vertex rooted map. The vertex v, is called the rooted vertex of M, and an element incident with v,, rooted element . Corollary 8.4 For a vertex rooted map M" = (X, P), v, is the rooted vertex, aut( M") | 2|{a}p]. (8.11) Proof This is (8.9) when R = (xp U {ax}p. O For a given map M = (&, P), if face fr, x € X, is chosen such that an automorphism of M has f, fixed, i.e. , an element incident with f, should be corresponding to an element incident with f,, then M is said to be a face rooted map. The face f, is called the rooted face of M. An element in rooted face is called an rooted element. Corollary 8.5 Fora face rooted map M" = (X, P) with rooted face fr, aut(M^) | 2] (x1, ]. (8.12) Proof This is (8.9) when R = {x}p,U (ox) p,. O For given map M = (X, P), if edge e,, x € X, is chosen such that an automorphism of M is with e, fixed, t.e., an element in e, is always corresponding to an element in e;, then M is called an edge rooted map. Edge e; is the rooted edge of M. An element in the rooted edge is also called a rooted element. VIILA4 Rootings 203 Corollary 8.6 For an edge rooted map M* = (X, P) with the rooted edge ez, aut( M*^) | |Kz|. (8.13) Proof The case of (8.9) when R= Ka. O For a given map M = (4,P), an element x € X is chosen such that an automorphism of M is with x as a fixed point, then M is called a rooted map . The element x is the root of M. The vertex, the edge and the face incident to the root are, respectively, called the root vertex, the root edge and the root face. Corollary 8.7 For a rooted map M" = (X, P) with its root z, aut( M") = 1. (8.14) Proof The case of (8.9) when R = {x}. E This tells us that a rooted map does not have the symmetry at all. The way mentioned above shows such a general clue for transforming a problem with symmetry to a problem without symmetry and then doing the reversion. Example 8.2 Map Mı = (Kz + Ky, (2)(y2, y, yy)) has 4 distinct ways for choosing the root. Because M; has the following 4 primal trail codes Ox = lo, y1, 2,72) = Dar Du = 1, 2, 32, 114 = S Bx Sy = 1,71, 2,92, = Say Soy = 1, 2, 1,32, = Soy, the 4 ways of rooting are shown in Fig.8.1(a-d) where the root is marked at its tail. Example 8.3 Map M» = (Kx + Ky, (x)(y2, y, Gy)) has 4 dis- tinct ways for choosing the root. Because Mə has the following 4 204 Chapter VIII Asymmetrization primal trail codes Sz = 19, Y1, 2, 82, = Sar, Sys = 1,2, 82,, 1, = Sg; S ex LL 52. = og Hil elus the 4 ways of rooting are shown in Fig.8.2(a-d) where the root is marked at its tail. | | (c) (d) Fig.8.1 Rootings in Example 1 VIILA Rootings NG Aan NNO 14 (c) (d) Fig.8.2 Rootings in Example 2 Activities on Chapter VIII VIIL5 Observations O8.1 Observe that how many embeddings the complete graph K4 of order 4 has. How many non-isomorphic maps are there among them?. O8.2 Consider the automorphism of the cube. Observe that what happens to the automorphism of those obtained by deleting, contracting, splitting and appending an edge on the cube. O8.3 Consider the automorphism of the octahedron. Observe that what happens to the automorphism of those obtained by deleting, contracting, splitting and appending an edge on the octahedron. O8.4 Consider the automorphism of the dodecahedron. Ob- serve that what happens to the automorphism of those obtained by deleting, contracting, splitting and appending an edge on the dodeca- hedron. O8.5 Consider the automorphism of the icosahedron. Observe that what happens to the automorphism of those obtained by deleting, contracting, splitting and appending an edge on the icosahedron. O8.6 Observe the duality between O8.2 and O8.3 and between O8.4 and O8.5. O8.7 Consider how to justify is there an edge in a map so that the order of the automorphism group of what is obtained by deleting the edge on the map does not change. O8.8 Consider how to justify is there an edge in a map so VIII.6 Exercises 207 that the order of the automorphism group of what is obtained by contracting the edge on the map does not change. O8.9 Consider how to justify is there an edge in a map so that the order of the automorphism group of what is obtained by splitting the edge on the map does not change. OS8.10 Consider how to justify is there an edge in a map so that the order of the automorphism group of what is obtained by appending the edge on the map does not change. O8.11 Provide a map whose automorphism group is the cyclic group. O8.12 List all maps of size 3 with their automorphisms of order VIII.6 Exercises From the last two chapters, it is seen that the automorphism of a map provides a foundation for justifying is the map isomorphic to another. This is an pattern example for the automorphism of a general combinatorial structure. For instance, a graph, a network, a combinatorial design, a lattice, a group, a ring, a field etc. E8.1 Observe whether, or not, an automorphism of a map in- duces an automorphism of its under graph. E8.2 Determine the automorphism group of the map (X, P) where k X=) (Koi Ky) i=1 and k k P= (LI Ti; Yi SIK is VYi)) i=l i=l for k > 2. 208 Activities on Chapter VIII E8.3 Determine the automorphism group of the map (X, P) where : Ay KS ge] and Peru E xa, Dai) lor k > 2. A primal matching of a map is defined to be such a set of edges that any pair of its edges have no common end. If a primal matching is incident to all the vertices on the map, then it is said to be perfect. A dual matching of a map is such a set of edges that any pair of its edges have no incident face in common. If a dual matching is incident to all the faces on the map, then it is said to be perfect as well. By no means any map has a perfect primal matching. One having a prefect primal matching is called a primal matching map. By no means any map has a perfect dual matching either. One having a prefect dual matching is called a dual matching map. A map which is both of primal matching and of dual matching is said to be of bi-matching. E8.4 Suppose M is a primal matching map and P, a perfect primal matching of M. Let n;;(P) be the number of (i, j)-edges in P. Prove that for any i, j, ni;(P) #0, E8.5 Suppose M is a dual matching map and P*, a perfect dual matching. Let n; ;(P*) be the number of (i, 7)*-edges in P*. Prove that for any i, j, ni ;(P*) * 0, aut(M) | 2(i + j)n; ;(P"). E8.6 Design an algorithm for justifying does a primal matching map, a dual matching map, or a bi-matching map have an non-trivial automorphism. Explain their efficiency. Provide a condition for the VIII.7 Researches 209 automorphism group of a primal matching map, a dual matching map, or a bi-matching map with, respectively, a primal matching, a dual matching, or a bi-matching rooted as the same as without rooting. For a map M, if a set of its faces is pairwise without common edge, then it is said to be independent. If an independent face set is pairwise without common vertex, it is called a cavity . If a cavity is spanning, i.e., all vertices of M are incident with the cavity, then it is called a full cavity . E8.7 Recognize if a vertex 3-regular(i.e., cubic) map has a full cavity. E8.8 For a vertex 3-regular map M with a full cavity H, let h;(H) be the number of i*-faces, à > 1, in this cavity. Prove that for any integer i > 1, A;(H) > 0, aut(M) | 6ih;(H). E8.9 Recognize whether, or not, a map has a full cavity. For a face f in a cavity H, its H-valency is the number of its incident primal semi-edges except for those in the boundary. E8.10 For a map M, let h;(H;j) be the number of j-faces of H-valency i. Prove that aut(M) | 2(¢ + 27)h;(H; j). E8.11 For a given integer k, find the number of non-isomorphic butterflies of relative genus k(k > 0), or of barflies of relative genus k(k « 0). VIII.7 Researches R8.1 Given 3 integers m > 1, g and s > 1, determine the 210 Activities on Chapter VIII number of primal matching maps of relative genus g with size 7 and the order s of their automorphism groups. R8.2 Given 3 integers m > 1, g and s > 1, determine the number of dual matching maps of relative genus g with size m and the order s of their automorphism groups. R8.3 Given 3 integers m > 1, g and s > 1, determine the number of bi-matching maps of relative genus g with size m and the order s of their automorphism groups. The first three problems should be considered for starting from g = 0, 1, and then —1. Particularly, the three problems for self-dual maps should be firstly studied before the general cases. RS8.4 Find the cubic maps of size m > 7 with a given relative genus and the maximum order of their automorphism groups R8.5 Find the maps of size m > 1 with a given relative genus and the order 1 of their automorphism groups. R8.6 Prove, or disprove, the conjecture that for a given relative genus, almost all maps have their automorphism groups of order 1. R8.7 Given three integers m > 1, g and s > 1, determine the full cavity maps of size m with relative genus g and the order of their automorphism groups s. If a map has a set of edges inducing a Hamiltonian circuit on its under graph, then it is called a primal H-map . If a map has e set of edges inducing a Hamiltonian circuit on the under graph of its dual, then it is called a dual H-map . If a map is both a primal H-map and a dual H-map, then it is called a double H-map. R8.8 Given three integers m > 1, g and s > 1, determine the primal H-maps of size m with relative genus g and their automorphism group of order s. R8.9 Given three integers m > 1, g and s > 1, determine the VIII.7 Researches 211 dual H-maps of size m with relative genus g and their automorphism group of order s. R8.10 Given three integers m > 1, g and s > 1, determine the double H-maps of size m with relative genus g and their automorphism group of order s. Chapter IX Rooted Petal Bundles e A petal bundle is a map which has only one vertex, or in other words, each edge of self-loop. e From decomposing the set of rooted orientable petal bundles, a lin- ear differential equation satisfied by the enumerating function with size as the parameter is discovered and then an explicit expression of the function is extracted. A quadratic equation of the enumerating function for rooted petal bundles on the surface of orientable genus 0 is discovered and then an explicit expression is extracted. e From decomposing the set of rooted nonorientable petal bundles, a linear differential equation satisfied by the enumerating function with size as the parameter is discovered in company with the ori- entable case and then a favorable explicit expression of the function is also extracted. e The numbers of orientable, nonorientable and total petal bundles with given size are, separately, obtained and then calculated for size not greater than 10. IX.1 Orientable petal bundles IX.1 Orientable petal bundles 213 A single vertex map is also called a petal bundle, its under graph is a bouquet. In this section, the orientable rooted petal bundles are investigated for determining their enumerating function with size as a parameter by a simple form. Let D be the set of all non-isomorphic orientable rooted petal bundles. For convenience, the trivial map 0 is assumed to be in D. Now, D is divided into two classes: Dj and Dy, i.e., D = Di + Dy (9.1) where (D); = (9) and Dy, of course, consists of all petal bundles in D other than 9. Lemma 9.1 Let Day = (D — a|VD € Dy}. Then, Din = D. (9.2) Proof For any D = (X, P) € Dim, there exists a D' = (4",P’) € Dy such that D = D'—a'. Because D' is orientable, group Y = Yi pn has two orbits {r'}p and {ar }p on X". Because yr’ € {r'}p, D has also two orbits {r}p = {r'}p — {r yr} and far}p = {ar’}p — far’, 8r), and hence D is orientable as well. From Theorem 3.4, petal bundle D' leads thatD is a petal bundle. This implies that Diy) C D. Conversely, for any D = (X,P) € D, P = (r,Pr,--.,P r), D' = (X + Kr', P^) where T' =(r' r'r,Pr,- Pr ; 20) Because D is orientable, group V = Y; p} has two orbits {r}w and {ar}y on X. Thus, group V' = V, has two orbits (r'hw = {r}y + {r yr'd and {ar tw = (or) + (ar, 8r) 214 Chapter IX Rooted Petal Bundles on X’. This means that D’ is also orientable. Because D’ has only one vertex, D' € D. And, from D' Æ 0, it is only possible that D' € Dg. Therefore, in view of D = D' — a', D € Dap. [] From the last part in the proof of this lemma, for any D = (r)7 € D, D' has 2m(D) 4-1 distinct choices such that D' = D; = D+e; € Dy, 0 € i € 2m(D), and hence D = D' — a! where e; = Kr’ and Do — (r^, P. T, JT, MEE Jm DEM i= 0; Dj — Um Ttt, gm qr’, Tr A oe, 1<i<2m(D)-1; Dom(D) m (r; T, Ir, my ge. yr’), i = 2m(D), for ^y = aß. Lemma 9.2 Let H(D) = (Dj;|i = 0,1,2,---,2m(D)} for D € D. Then, Dy — 5 H(D). (9.3) DED Proof Because of Lemma 9.1, it is easily seen that the set on the left hand side of (9.3) is a subset of the set on the right. Conversely, from H(D) C Dy, for any D € D, the set on the right hand side of (9.3) is also a subset on the left. O The importance of Lemma 9.2 is that (9.3) provides a 1—to-1 correspondence between the sets on its two sides. This is seen from the fact that for any two non-isomorphic petal bundles Dı and D», (D) NH(D2) = 0. On the basis of Lemmas 9.1-2, the enumerating functions of sets Dı and Dy can be evaluated as a function of D's as fol) y 7 (9.4) where m(D) is the size of D. Because Di only consists of the trivial map ? and v has no edge, folz) 7 1. (9.5) IX.1 Orientable petal bundles 215 Lemma 9.3 For Dy, 2dfp foal) = £ fp + 2x T (9.6) Proof From Lemma 9.2, Tola) = ` a) DED = ť (Zemo + gen) DED =g ` Po 4 On ` m(D)aztP) DED DED = £t fp + 2,287» dx where fp = fp(x). This is (9.6). [] Theorem 9.1 The differential equation about h dA gu cs E , l 2a? = l + (1—2)h; (9.7) ho = h|z=0 = 1 is well defined in the ring of infinite series with integral coefficients and finite terms of negative exponents. And, the solution is h = fp(x). Proof Suppose h = Ho + Hix + Hox? + -+ Hz" +--+, for H; € Z,, i 2 0. According to the first relation of (9.7), via equating the coefficients of the terms with the same power of z on its two sides, the recursion —1 + Ho = 0; Hı — Ho = 0; (9.8) Hm = (2m — 1) Aya, M22 is soon found. From this, Hp = 1(the initial condition!), Hı = 1, ---, and hence all the coefficients of h can be determined. Because only addition and multiplication are used in the evaluation, all Hm, m > 1, are integers from integrality of Ho. This is the first statement. 216 Chapter IX Rooted Petal Bundles As for the last statement, from (9.1) and (9.5-6), it is seen that h = fp(x) satisfies the first relation of (9.7). And from the initial condition hy = fp(0) = 1, we only have that h = fp(x) by the first statement. L In fact, from (9.8), i (2m)! eet) = eur E: mm! where m > 0. Further, from Theorem 9.2, -14 xo TEN Pimp (9.9) Example 9.1 From (9.9), there are 3 orientable rooted petal bundles of size 2. However, there are 2 orientable non-rooted petal bundles as shown in (a) and (b) of Fig.9.1. In (a), based on primal trail code(or dual trail code), only 1 rooted(r; as the root) element. In (b), 2 rooted(ro and r3 as the roots) elements. Fig.9.1 Petal bundles of size 2 IX.2 Planar pedal bundles 217 IX.2 Planar pedal bundles Petal bundles are here restricted to those of genus 0, i.e., planar pedal bundle. Rooted is still considered. Because orientable petal bundles can be partitioned into classes by genus as Dew (9.10) k>0 where D; is the set of rooted petal bundles with orientable genus k. What is discussed in this section is Dp. For convenience, the trivial map ¥V is included in Dy. For this, Dp should be partitioned by the valency of root-face into classes as Dy > (9.11) s>0 where F,, s > 0, is planar rooted petal bundles with the root-face of valency s. Lemma 9.4 Let S(the trivial map 0 is included) and T(V is excluded) be two set of rooted maps. If for any S = (X,P) € S — 0, there exist an integer k > 1 and maps S; = (X, P) € T, 1 €i k, such that k w=) x, (9.12) i=l and P is different from P;, 1 < i < k, only at vertex (r)p = ((r1) Py (72) Past tts (Tk) Px) (9.13) where r = ri. Then, 1 fs(x) = 1- f) (9.14) where fs(x) and fr(x) are the enumerating functions of, respectively, S and 7. Proof First, S is classified based on k mentioned above, k > 0, SUY Ss k>0 1.8: 218 Chapter IX Rooted Petal Bundles Naturally, So = (0). Then, because any M; = (Vk, Qk) E Sk, k > 1, has the form as shown in (9.12) and (9.13) (4 and P are, respectively, replaced by Vy and Q;), we have fs, (x) = qe) My €Sy 5 q m1) m(S2) + +m(Sk) ($1,583.94) S,€T, 1<i<k = (fr(2))*. Therefore, by considering fs,(x) = 1, fs(x) = X fs,(2) k>0 =14+ Y ro) k>1 E 1 1— fr(x) Notice that since x is an undeterminate, it can be considered for the values satisfying | fr(x)| < 1. This lemma is proved. O If S and 7 are, respectively, seen as Dp and F}, it can be checked that the condition of Lemma 9.4 is satisfied, then 1 = 1f) Further, another relation between fp,(r) and fz,(x) has to be found. (9.15) Lemma 9.5 Let Fay = {D — a|VD € 7j. Then, Fa = Do. (9.16) Proof Because 0 Z F,, for any D € 7, from the planarity of D, D' = D—ais planar and from D with a single vertex, D' = D—a is with a single vertex. Hence, D' € Do. This implies that Fa; C Do. IX.2 Planar pedal bundles 219 Conversely, for any D' = (X', P") € Do, in view of a single vertex, P= (r)p. Let D= D'az-(X'- Kr,P) where P = (r,(rp,^r). Naturally, D is of single vertex. Because D is obtained from D' by appending an edge, from Corollary 4.2 and Lemma 4.6, the planarity of D' leads that D is planar. And, from (r)p, = (r), D € Fy. Since D' = D — a, D' € Fy). This implies that Dy ta [] Because this lemma provides a 1-to-1 correspondence between Fı and Dy, it is soon obtained that faz) = 3e pou DEF, =r »» pO) (9.17) Dc€Dg eeu In virtue of (9.17) and (9.15), = ER (9.18) Theorem 9.2 Let h® = fp,(x) be the enumerating function of planar rooted petal bundles with the size as the parameter, then A?) = Sn (9.19) m (m 4- 1)! Proof From (9.18), it is seen that h) satisfies the quadratic equation about h as zh? — h 4-1 — 0. It can be checked that only one of its two solutions is in a power series with all coefficients non-negative integers. That is (). 1— vl-—4v h 2x 220 Chapter IX Rooted Petal Bundles By expanding \/1 — 4x into a power series, (9.19) is soon found via rearrangement. El From the quadratic equation, a non-linear recursion can be de- rived for determining the coefficients of h. However, a linear recursion can be extracted for getting a simple result. This is far from an uni- versal way. Example 9.2 From known by (9.19), the number of planar rooted petal bundles of size 3 is 5. However, there are 2 planar non- rooted petal bundles altogether, shown in (a) and (b) of Fig.9.2. In (a), by primal trail codes(or dual trail codes), 3. Their roots are r1, r and r3. In (b), 2. Their roots are r4 and rs. TQ] =2 TA-— c wd 7 xL ze NEP 2n z (a) (b) Fig.9.2 Planar petal bundles of size 3 IX.3 Nonorientable pedal bundles The central task of this section is to determine the enumerat- ing function of nonorientable rooted petal bundles with size as the parameter. Let U be the set of all nonorientable rooted petal bundles. Be- cause the trivial map is orientable, 0 is never in U. In other words, any map in U does have at least one edge. Now, & is partitioned into IX.3 Nonorientable pedal bundles 221 two classes: Uy = (M|VM € U, M — a orientable} and Uy = (M|VM € U, M — a nonorientable], lEs U = ug. (9.20) First, the decomposition of the two sets Ur and My should be investigated. Lemma 9.6 Let Um = (M — a|VM € U}. Then, Hay =D (9.21) where D is the set of all orientable rooted petal bundles given by (9.1). Proof For M = (X,P) € Uy, if M 7 92, then M' = (XP^) where 4’ = X + Kr’ and P’ is different from P only at the vertex (r')p = (r', Br’, r, Pr, P?r, ..., Pr) such that M = M' — a’. From M' € Mı, M € M. If M = 9, then M = (Kr',(r', Br’) € ut such that M = M' — a'. Meanwhile, M € D. Hence, U € D. Conversely, for M = (4,P) € D, let M' = (X', P') such that X' = X + Kr’. Because M has a single vertex, Pi = (r)m — (r^, Br. r, Pr, P’r, Eds po. Therefore, M’ has a single vertex as well. And, since r^, Gr’ € {r’}w, M' € U. By reminding that M = M' — a’ is orientable, M’ € (f. Thus, M € Uy. This implies that D C U1. E Lemma 9.7 For any D = (X, F) € D, r = r(D), let B(D) = { B;|0 € i € 2m(D)) where m(D) is the size of D, and Br vim. der (r'r, Br, F'r,---), 1€ ix 2m(D) — 1; (r', v, , Fr,- , 8r), i = 2m(D). B;(D) = (9.22) 222 Chapter IX Rooted Petal Bundles 'Then, Uu — 5 B(D). (9.23) DED Proof For any M = (Z, P) € Uh, because Deu M is only some B;, 1 € i € 2m(D) in (9.22) such that Pr’ = r, or P?r' = r(Here, r' and r are, respectively, the roots of M and D). Therefore, M is also an element of the set on the right hand side of (9.23). Conversely, for an element M in the set on the right of (9.23), from Lemma 9.6, M € Uy. This is to say that M is also an element of the set on the left hand side of (9.23). [] Example 9.3 Let D = (Kz,(z,yx)) € D. Then, D is of size l, be mi) 1, Three rooted petal bundles Bo(D), B4(D), and B»(D) € ttj are produced from D and shown as, respectively in (a), (b), and (c) of Fig.9.4 where r' = y and r = z. Fig.9.3 Nonorientable petal bundles from orientable ones Lemma 9.8 Let Unn = (M — a|VM € Up}. Then, Un) =U. (9.24) Proof For M = (X,P) € Um, let M’ € Uy such that M = M' — a. Because M' is a nonorientable petal bundle and M’ € Uy, M is a nonorientable petal bundle as well, i.e., M € U. IX.3 Nonorientable pedal bundles 223 Conversely, for any M = (X, P) € U, there exists M' = (X',P^) € Uy such that M = M' — d', e.g., X! = X + Kr’, P! = (r' yr, (rp). Therefore, M € Ui. [] Further, observe that for a map M € U, how many non-isomorphic maps M’ € Uyare there such that M = M’ — a’. Two cases should be considered: (1) r', yr’ and r are in the same orbit of P’; (2) r', Br’ and r are in the same orbit of P’. (1) Based on the rule of rooting, because yr’ only has 2m( M) 4-1 possible positions, 7.e., yr! = Pir! P'r, P'(Pr), T: Ppp, then Ce Ae (r)p), i = 0; TT, r! P'r,---), HM) = E (9.25) 1<i<2m(M) —-1; (r', (r)p, yr), i = 2m(M). (2) Based on the rule of rooting, because 8r’ also has 2m(M) + lpossible positions, t.e., Br! 2 lp! P'r, P'(Pr), I Pipe, then —- (r OF , (r)p), 1 = 0; m: POTE BF. P'r, -Js 1<i<2m(M) -1; (r', (r)p, 8r), i 2 2m(M). Ji(M) = (9.26) Example 9.4 Let M = (Kz,(x,0x)) € U. The map M has only one edge, i.e., m( M) = 1. Six nonorientable petal bundles Jo(M), (M), and I2(M), with Jo(M), (M), and Jo(M) € Uy are produced for M and shown as, respectively, in (a), (b), and (c), with (d), (e), and (f) of Fig.9.4 where p andres. 224 Chapter IX Rooted Petal Bundles Fig.9.4 Nonorientable petal bundles from nonorientable ones Lemma 9.9 For any M €U, let Er = U(M)fo < i € 2m(M)}; (9.27) J(M) = {Jj(M)|0 < j € 2m(M)}. Then, Un = X_(T(M) + J(M)). (9.28) Meu Proof For any M' = (4',P’) € Un, because M = M'—d' EU, M' is only some J;, 0 < i € 2m(M) — 1 in (9.25), or some J;, 0 < j € 2m(M) — 1 in (9.26) such that Pr’ = r, or P?r' = r(Here, r' and r are, respectively, the roots of M’ and M). Therefore, M is also an element of the set on the right hand side of (9.28). Conversely, for an element M in the set on the right of (9.28), from Lemma 9.8, M € Uy. This is to say that M is also an element of the set on the left hand side of (9.28). E Lemma 9.10 Let S and 7 be two sets of maps. If for any T € T, there exists a set C(T) C S such that IX.3 Nonorientable pedal bundles 225 (i) for any T € 7, |£C(T)| = am(T) + b and for S € S, m(T) = m(S) — c, where a b and c are constants and m(T) is an isomorphic invariant, e.g., the size; and (y S= S UD TET then fs(x) = z°(bfr + art) (9.29) Proof Because £L(T) provides a mapping from a map in 7 toa subset of S and the cardinality of the subset is only dependent on the parameter of the enumerating function(by (i)), and (ii) means that the mapping provides a partition on S, then x spe 3 (am(T) + b)a™) TET =b >», LU) -- ax ` mT TET TET d = bfr + ur dg This is (9.29) by multiplying x^ to the two sides. E Theorem 9.3 The enumerating function g = fulx) of nonori- entable rooted petal bundles in the set U with size as the parameter satisfies the equation as dh 4,238 = (1 — 2x)g — z(h + 22 —); o da (9.30) dg Jz 2=0 E 1, k where h = fp(x) is the enumerating function of orientable rooted petal bundles given by (9.9). Proof Because (9.23) and (9.28) provides the mappings from maps D € D and U € YU to, respectively, a subset of Ut with 2m(U)+1 elements and a subset of Uy with 2(2m(U) +1) = 4m(U) +4 elements, 226 Chapter IX Rooted Petal Bundles where D and U are 1 edge less than their images, from Lemma 9.10, dA dh = 2p—)- 2g? — fu(x) 9 x(h + emt th + 2x qm and — dg dg fu, (x) = «(2g + Ar) = 2xg 1,778 By (9.20) again, dh d g = zh 4- 227 — 4 2zg 4+ Ag? dx dx Via rearrangement, (9.30) is soon obtained. [] IX.4 The number of pedal bundles Because (9.9) provides the number of orientable rooted petal bun- dles with size m, m > 0, t.e., (2m — 1)! Hg ee a7 a 2m-l(m, — 1)! (9.31) for m > 1. Of curse, Hp = 1. Here, the number of nonorientable rooted petal bundles with size m is evaluated only by the equation shown in (9.30). Let Gm be the number of nonorientable rooted petal bundles with size m, m > 1. In fact, Gm, m > 1, are determined by the recursion as Gm = (4m — 2)Gm-1 + Hm, mM > 2; i (4m — 2) 88-1 m (9.32) Gil. The solution of the recursion (9.32) is obtained, i.e., Muriel Z Cm = Saim I] mt He-» " = (9.33) Mu 25 Ie =2 Ju IX.4 The number of pedal bundles 227 Example 9.5 When m = 1, there are one orientable rooted petal bundle of 1 edge, i.e., M = (Kx, (x, yx)) and one nonorientable rooted petal bundle of 1 edge, i.e., N = (Ka, (x, Gx)) € U. By appending an edge Ky on M, 3 non-isomorphic nonorientable rooted petal bundles of 2 edges are produced and shown in (a-c) of Fig.9.3. By appending an edge Ky on N, 6 non-isomorphic nonorientable rooted petal bundles of 2 edges are produced and shown in (a-f) of Fig.9.4. Then, Gə = 9 which is in agreement with that provided by (9.32), or (9.33). Now, Hm, Gm, and HO for m < 10 are listed in Table 9.1. 1 1 1 1 2 3 9 2 3 15 105 5 4 105 1575 14 5 945 29295 42 6 10395 654885 132 7 135135 17162145 A29 8 2027025 516891375 | 1430 9| 34459425 17608766175 | 4862 10 | 654729075 669787843725 | 16796 Table 9.1 Numbers of rooted petal bundles in 10 edges Lemma 9.11 For an integer m > 1, the number of non-isomorphic nonorientable rooted petal bundles with size m is Gm = (2? —1) Hm (9.34) where Hm is given by (9.31). Proof By induction. When m = 1, from H; = 1, Gi = 1. (9.34) is true. Assume Gi satisfies (9.34) for any 1 € k < m — 1, m > 2. Then, 228 Chapter IX Rooted Petal Bundles from (9.32), Gm = (4m ES 2) Gi Suas = (4m - 2)((27 7 = 1) Hm) + Hm. Since it can, from (9.31), be seen that Hz = (2m = 1)Hm-1, we have H Ca = lám =a] = H,, (4m - 2)( > + Am — 2 — 90-1 DET |) Hn mam) = (2(27^ — 1) + 1)Hm = (27 — 1)Hm This is (9.34). O Theorem 9.4 For an integer m > 1, the number of non- isomorphic rooted petal bundles with size m is 2" (2m — 1)!! (9.35) where " (2m — 1)!! = | [Q: - 1). (9.36) Proof Because of (9.34), the number of non-isomorphic petal bundles with m edges is Hm + Gm = 2™ Hn: (9.37) By substituting (9.31) into (9.37), we have (2m — 1)! 2m H. — 9m NO eye O * 9m-T(m — 1)! zc Ll (2m — 1)! (m — 1)! = 2" (2m — 1)!!. IX.4 The number of pedal bundles 229 This is (9.35). O The theorem above reminds the number of embeddings of the bouquet of size m 2™(2m — 1)! derived from (1.10) as a special case. This is (m— 1)! times the number of rooted petal bundles with m edges. Activities on Chapter IX IX.5 Observations A map with only one face is called a unisheet. O9.1 Observe that there is a 1—to-1 correspondence between the set of petal bundles and the set of unisheets. O9.2 Think that what types of graphs can be as the under graph of a unisheet and what type of graphs are not as the under graph of a unisheet. O9.3 Is any planar graph a under graph of a planar unisheet? O9.4 Think about three ways to justify a map which is a planar petal bundle. O9.5 Discuss how to determine that a petal bundle is on the projective plane. O9.6 Discuss how to determine that a petal bundle is on the torus. O9.7 Discuss how to determine that a petal bundle is on the Klein bottle. O9.8 Discuss how to determine that a unisheet is on the plane. O9.9 Discuss how to determine that a unisheet is on the pro- jective plane. O9.10 Discuss how to determine that a unisheet is on the torus. O9.11 Discuss how to determine that a unisheet is on the Klein IX.6 Exercises 231 bottle. IX.6 Exercises E9.1 Show that for any graph G, there exists a unisheet U such that G is the under graph of U. E9.2 Prove that the number of rooted unisheet with size m is TIL (2m - 1)! = [ [Gi - 1). i=l E9.3 Prove that a unisheet is planar if, and only if, its under graph is a tree. E9.4 For an integer m > 1, determine the number of rooted petal bundles of size m on the torus. E9.5 For an integer m > 1, determine the number of rooted petal bundles of size m on the projective plane. A graph is said to be unicyclic if it has only one cycle. E9.6 Prove that a unisheet is on the projective plane if, and only if, its under graph is unicyclic. A graph is said to be eves-cyclic if it has two fundamental circuits incident. E9.7 Prove that a unisheet is on the torus if, and only if, its under graph is eves-cyclic. E9.8 For an integer m > 1, determine the number of non- isomor- phic rooted petal bundles with size m on the projective plane. E9.9 For an integer m > 1, determine the number of non- isomor- phic rooted petal bundles with size m on the torus. E9.10 For an integer m > 1, determine the number of non- isomorphic rooted petal bundles with size m on the Klein bottle. 232 Activities on Chapter IX E9.11 For an integer m > 1, determine the number of non- isomorphic rooted unisheets with size m on the projective plane. E9.12 For an integer m > 1, determine the number of non- isomorphic rooted unisheets with size m on the Klein bottle. A map of order 3 is also called tri-pole map. E9.13 For a given integer m > 1, determine the number of all non-isomorphic rooted tri-pole maps with size m in the plane. IX.7 Researches R9.1 For two integers m > 1 and p > 2, determine the number of rooted petal bundles with size m on the surface of orientable genus p. R9.2 For two integers m > 1 and q > 3, determine the number of rooted petal bundles with size m on the surface of nonorientable genus q. R9.3 For two integers m > 1 and p > 2, determine the number of rooted unisheets with size m on the surface of orientable genus p. R9.4 For two integers m > 1 and q > 3, determine the number of rooted unisheets with size m on the surface of nonorientable genus q. A map of order 2 is also called bi-pole map. R9.5 For a given integer m > 1, determine the number of all non-isomorphic orientable rooted bi-pole maps with size m R9.6 For a given integer m > 1, determine the number of all non-isomorphic nonorientable rooted bi-pole maps with size m R9.7 For two integers m > 1 and p > 0, determine the number of all non-isomorphic orientable rooted bi-pole maps with size m of IX.7 Researches 233 the surface of genus p. R9.8 For two integers m > 1 and q > 1, determine the number of all non-isomorphic nonorientable rooted bi-pole maps with size m on the surface of genus q. R9.9 For a given integer m > 1, determine the number of all non-isomorphic orientable rooted tri-pole maps with size m R9.10 For a given integer m > 1, determine the number of all non-isomorphic nonorientable rooted tri-pole maps with size m R9.11 For two integers m > 1 and p > 1, determine the num- ber of all non-isomorphic orientable rooted tri-pole maps with size m of the surface of genus p. R9.12 For two integers m > 1 and q > 1, determine the num- ber of all non-isomorphic nonorientable rooted tri-pole maps with size m on the surface of genus q. R9.13 For two integers n > 5 and p > s(n) where (n — 3)(n — 4) s(n) = E i.e., the least integer not less than the fractional (n — 3)(n — 4)/12(or called the up-integer of the fractional), determine the number of rooted maps whose under graph is the complete graph of order n with ori- entable genus p. R9.14 For two integers n > 5 and p > t(n) where t(n) = [EAE determine the number of rooted maps whose under graph is the com- plete graph of order n with nonorientable genus q. R9.15 For two integers n > 3 and p > c(n) where c(n) = (n — 4)2 5 +1, determine the number of rooted maps whose under graph is the n-cube with orientable genus p. 234 Activities on Chapter IX R9.16 For two integers n > 3 and q > d(n) where dn) (n-—4)29-*4 determine the number of rooted maps whose under graph is the n-cube with nonorientable genus q. R9.17 For three integers m, n > 3 and p > r(n) where (m — 2)(n — 2) r(n) — p[—— 92 determine the number of rooted maps whose under graph is the com- plete bipartite graph of order m + n with orientable genus p. R9.18 For three integers m, n > 3 and q > l(n) where i(n) = [E th. determine the number of rooted maps whose under graph is the com- plete bipartite graph of order m + n with nonorientable genus q. Chapter X Asymmetrized Maps e From decomposing the set of rooted orientable maps, a quadratic differential equation satisfied by the enumerating function with size as the parameter is discovered and then a recursion formula is extracted for determining the function. A quadratic equation of the enumerating function in company with its partial values for rooted maps on the surface of orientable genus 0 is discovered with an extra parameter and then an explicit ex- pression of the function with only size as a parameter is via char- acteristic parameters extracted for each term summation free. e From decomposing the set of rooted nonorientable maps, a nonlin- ear differential equation satisfied by the enumerating function with size as the parameter is discovered in company with the orientable case and then a recursion formula is extracted for determining the function. e The numbers of orientable, nonorientable and total maps with given size are, in all, obtained and then calculated for size not greater than 10. X.1 Orientable equation It is from Corollary 8.7 shown that a map with symmetry be- 236 Chapter X Asymmetrized Maps comes a map without symmetry whenever an element is chosen as the root. Such a map with a root is called a rooted map. Rooting is, in fact, a kind of simplification in mathematics, par- ticularly in recognizing distinct combinatorial configurations for re- ducing the complexity. As soon as the rooted case is done, the general case can be re- covered by considering the symmetry in a suitable way. For maps, the estimation of the order of the automorphism group of a map in the last chapter and the efficient algorithm for justifying and recognizing if two maps are isomorphic in Chapter VII provide a theoretical foundation for transforming rooted maps into non-rooted maps. This will be seen in the next chapter. The main purpose of this chapter is to present some methods for investigating non-planar rooted maps as appendix to the monograph Enumerative Theory of Maps|Liu7] in which most pages are for planar maps, particularly rooted. Let M be the set of all orientable rooted maps. For M = (&,P) € M, let Uz = (x)p = (2, Pr; Pla) (10.1) be the vertex incident with x € X. The root is always denoted by r. The rooted edge which is incident with r is denoted by a = Kr. The rooting of M — a is taking P?r as its root where ô = min(i|P'r g Kr,i > 1). (10.2) In fact, 1, if P ; "a A (10.3) 2, othewise. In virtue of Theorem 3.4, M — a is a map if, and only if, a is not a harmonic loop except terminal link (or segmentation edge) of M. Now, let us partition M into three parts: Mr, My and Mp, oa M = Mı + Mu + Mir (10.4) X.1 Orientable equation 237 where Mr = {V}, i.e., consisted of the trivial map, My and Mr are, respectively, consisted of those with a as a segmentation edge and not. Lemma 10.1 Let May — (M — a|VM € My}. Then, May — M x M (10.5) where x stands for the Cartesian product of two sets. Proof For any M = (X,P) € M x M, let M = Mi + Ms, Mj = (4, Pi), i = 1,2. Assume M' = (X', P^) such that X' = X - Kr' and P’ is different from P; or Pı only at, respectively, Up — (r)p = a, T2, Pore, ubi , Px ro) or UBr! = (Gr’)p = (a Br, T1, Pira, DI Piin). Since M’ € M and its rooted edge a’ = Kr’ is a segmentation edge, M' € My. It is checked that M = M' — a’. Therefore, M € Map. Conversely, for any M € May, we have M' € My such that M = M' — qd! where a’ = Kr’. From M' € My, M = M, + Mə where Mi, M» € M. This implies that M € M x M. [] It is seen from this lemma that there is a 1—to-1 correspondence between M(€ Muy, or M x M) and M'(€ My). Hence, For M = (X, P) € My, because M — a is a map (Theorem 3.4), from (10.3), the root r((M — a) of M — a has two possibilities: when Pr(M) Z yr(M), r(M — a) = Pr(M); otherwise, r(M — a) = P?r(M). Let M = (¥,P) = M —a where X = X — Kr and P are different from P only at (Fis = (Pr, DE ive Por), if a is not a loop; (P4r)s = (Pyr, T?wyr, e sal II à, 18 not a loop 238 Chapter X Asymmetrized Maps otherwise, 7.e., when a is a loop, (P?r, P3r, ..., P lr), 34 yr = Pr; (T)s = (Pr, E pe pg TS P), if yr = Pu ss Since M(X, P) is orientable, group V = Vy, p; has two orbits {r}w and {ar}y on 4. For M = (X,P) = M —a, group V = V5, also has two orbits Ug = {r}w — tr, yr} and {ar}; = (lore — (or, Gr}. So, M is also orientable. Furthermore, for every element y € {f}y, is there exactly one position of a = Kr, i.e., yr is in the angle (ay, P), for M € vm such that M = M — a. This means that in Mj, there are {Fal = 514] = 2m) where m(M ) is the size of M, non-isomorphic maps for producing M. By considering for the case Pr = yr, in My, there are 2m(M) +1 non-isomorphic maps for M altogether. Example 10.1 Let M = (Kz + Ky + Kz, (x,y, yz)(z, yx, yy)) where f = x is the root shown in Fig.10.1(a). Since it is orientable, the orbit of group V which 7 is in can be written as a cyclic permutation as (f)g = (x, y, Y2, Z, YT, VY). Then, Fig.10.1(b-h) presents all the 2m(M) +1 = 2x 3+1=7 maps in Mm, obtained by appending a = Kr on map M where (b) is for Pr = yr and (c-h) are for those obtained by appending a = Kr in the order of (#)g from M. 239 X.1 Orientable equation (b) (a) 2 E] S 2 SN (d) (c) (f) (e) 240 Chapter X Asymmetrized Maps Ne AE LIN 4 (g) (h) EM Fig.10.1 New maps obtained by appending an edge Lemma 10.2 Let Mam = (M — a|VM € Mm}. Then, Many = M. (10.7) Proof Because for any M € Myr, M —a is also a map (Theorem 3.4), then Mam € M. Conversely, for any M € M, any one, e.g., M' of the 2m(M) +1 maps obtained by appending a’ from M in the above way is with M' € Mym. Because M = M' — d', then M € Many. [] For convenience, let H(M) be the set of all the 2m(M) + 1 maps in Mın, obtained from M by appending an edge in the above way. From Theorem 8.1, they are all mutually nonisomorphic in the sense of rooting. Lemma 10.3 For Mın, Mu = » H(M). (10.8) MEM Proof For any M = (X,P) € Mm, let M = (X,P) = M — a. Because a is not a segmentation edge, from Theorem 3.4 and Corollary 3.1, M € .M. By orientability, because r € {7}% where v= Vy py there exists y € {7} such that Py = yr, or Pr = yr. Because l{7}¢| = 2mM, the former has 2m(M) possibilities and the latter, only one. This is the 2m(M) + 1 possibilities in H(M). Further, X.1 Orientable equation 241 because M € M, M is an element of the set on the right hand side of (10.8). Conversely, for any M € H(M), M € M, Since M = M —a € M, by Theorem 3.4 and Corollary 3.1, a is not a segmentation edge. Therefore, M € Myr. [] Furthermore, (10.8) provides a 1-to-1 correspondence between the sets on its two sides. This enables us to construct all orientable maps with the rooted edge not a segmentation edge from general ori- entable maps with smaller size. In order to determine the number of non-isomorphic orientable rooted maps in M with size m > 0, the enumerating function of set M jui)e y qu (10.9) MEM has to be investigated for a simpler form in infinite power series where m(M) is the size of M. In the series form of (10.9), the coefficient of the term with 7”, m > 0, is just the number of non-isomorphic orientable rooted maps with size m. From (10.4), f(x) = faux) + f(x) + fux). (10.10) Lemmas above enable us to evaluate f(x), fq, (x) and fy (x) as functions of f = f(z). First, because M, contains only one map V and m(V) = 0, fm, (£) contributes the constant term 1 of f, i.e., fon) = 1 EUH Lemma 10.4 For My, Proof According to the 1-to-1 correspondence between My and M and that the former is with its size 1 greater than the latter 242 Chapter X Asymmetrized Maps in the correspondence, by (10.6), didus =g » x a MEM (11) =i ` iM) MEMxM MEM = of’. This is (10.12). o Lemma 10.5 For Mm, 2 E Satale )= Bf +22 (10.13) Proof From the 1—to-1 correspondence between My and Mm and that the former is with its size 1 greater than the latter in the correspondence, and then by Lemma 10.3 and Lemma 9.10, m(M) Emin (x = a MEM yy) df = Iy o(f + 2L) =axf+ og dx This is (10.13). O Theorem 10.1 The differential equation about f df Di S - l " dz Goana (10.14) fo = flz-o = 1 is well defined in the ring of infinite power series with all coefficients nonnegative integers and the terms of negative powers finite. And, the solution is f = f(x). X.2 Planar rooted maps 243 Proof Suppose f = Fo + Fix + Fave + Eme" +--+, Be Z,, i 2 0. Based on the first relation of (10.14), by equating the coefficients on the two sits with the same power of x, the recursion —1 + fo = 0; P HER 0 ma (10.15) Fm = (2m — 1)Fm-1 + ` FiFm-i-i i—0 qmi 2 is soon extracted. Then, Fp = 1(the initial condition), Fi = 2, ---, all the coefficients of f can uniquely found from this recursion. Because only addition and multiplication are used for evaluating all the coef- ficients from the initial condition that Fo is an integer, Fm, m > 1, must all be integers. This is the first statement. For the last statement, from (10.10) and (10.11-13), it is seen that f = f(a) satisfies the first relation of (10.14). And, fo = fm(0) = 1 is just the initial condition. By the uniqueness in the first statement, the only possibility is f = f(x). E Although the form of the equation in Theorem 10.1 is rather simple, because of the occurrence of f?, it is far from getting the solution directly. In fact, it is an equation in the Riccati's type. It has no analytic solution in general. X.2 Planar rooted maps Let 7 bethe set of all planar rooted maps. Because it looks hard to decompose 7 into some classes so that each class can be produced by 7 with only size as the parameter. Now, another parameter for a map M, i.e., the valency of the rooted vertex n( M), is introduced. The enumerating function of 7 is gaa) ay (10.16) MeT 244 Chapter X Asymmetrized Maps where m(M ) is still the size of M. Assume that 7 is partitioned into three classes: 79, 7; and 75, i.e., T=HtTh+h (10.17) where Jp = (0), Ti and J, are the sets of planar rooted maps with the rooted edge, respective a loop and a link(not loop). For M € T, let a= Kr be the rooted edge of M with the root r — r(M). For maps M; € 7 (1-1 and 2), let a; = Kr; be the rooted edge of M; with the root r; = r(M;j). The 1-addition of two maps Mı = (A, Pı) and M» = (A5, Po) is to produce the map Mı +- My = Mı U Ms with the root r = ri provided M; N Mə = (v, ] where v, = ((r1)p,, (r2)»,). Lemma 10.6 Let Ta) = {M — a|VM € Ti}, then des sT (10.18) where 7 x 7 = (M4 + M3|VM;, Mə € T Y is called the 1-product of T with itself. Proof For any M = (4,P) € Ti), let MW = (X', P) € 1; such that M' — a/ = M. Because a’ = Kr’ is a loop, (r)p — Us pue. Dentes yr, E Ptr), From the planarity, M’—a’ = M,+- Mo, Mj = (Xi, Pi), i = 1,2, where X = X +% = X' — Kr’, Pi and P» are different from P only at (r)p becoming, respectively, (rijp, = (P^r', P(P'r’), «+, PY yr!) where y = af and (rap, = (P'yr', P(P'yr’), ++, P' y. This implies M € 7 x: 7. Conversely, for M € T x- T, because M = Mı + Ms, let M' = M +a’, a’ = Kr’, such that (r)p = (r5 (rp Yr, (2) Pa) X.2 Planar rooted maps 245 then M' € T and M = M' — q'. Since a’ is a loop, M' € Ti and hence MET). O Because this lemma presents a 1—to-1 correspondence M = Mı + M» between M € Ta) and Mj, Mo € T with m(M) = m(Mij) +m(M2) and n(M) = n( Mı) + n(M»), the enumerating function of Tay fnm y)- ` g M) mM) MeTx T gta) mE), ei Mp a pos (10.19) Tn 1VÀ n( M 2 ( > qM ROM) MeT = rae y). Then, from the 1-to-1 correspondence between 7; and Za) with the former of size 1 greater than the latter and the former of the rooted vertex valency 2 greater than the latter, the enumerating function of T; is fa sg) eu Pa bay) ec wy T Gs) (10.20) However, for 72, the correspondence between 7; and 7» = (M e a|lVM € To} with the former of size 1 greater than the latter is not of 1-to-1 where M e a is the contraction of the rooted edge a on map M. The root on M ea is defined to be Pyr when Pyr 4 yr; or (P4)?r otherwise. Because a is a link, this is a basic transformation. According to Chapter V, M — a is planar if, and only if M is. Hence, Toy = T. (10.21) Further, observe what a correspondence between 75 and 7 is. For M = (X,P) € T, let (r)p = (r, Pr, --, P" )-17) where n(M) is the valency of the rooted vertex v, on M. By splitting a link at v, all those obtained are still planar because this operation is a basic transformation. For doing this, there are n(M) 4- 1 possibilities altogether. 246 Chapter X Asymmetrized Maps Let M; = (Xi, Pi), 0 € i € n(M) be all the n(M) + 1 maps obtained from M by splitting an edge at the rooted vertex v, = (r)p as (ro, (r)p), while vari = (770); (ri, Pr,- , p OD715). while vgn, = (yr1, r); (ra, P?r, . .., p 719). while Un ceu T IPI CD IE (10.22) (Tr(M)-15 pat). while var 45, = (YTs(my-5 ^7 Jal M) (Tr(a), while Ubrany = (Yrn(m), (T)P). Lemma 10.7 For a map M € 7, let K(M) = (Mili 20,1,2,---, n(M)] where M;, 0 € i € n(M), are given by (10.22). Then, T; = M KM). (10.23) MET Proof For any M € T5, because a = Kr is a link, from (10.21), Mea c T and from (10.22, M € K(M ea). Therefore M is an element of the set on the right hand side of (10.23). Conversely, for M is an element of the set on the right of (10.23), there exists a map M’ € T such that M € K(M"). Because all maps in K(M) are planar if, and only if, M' is planar, M € 7 as well. Moreover, from (10.22), the rooted edge a is a link, M € 75. This meas that M an element of the set on the left hand side of (10.23).0 Because (10.23) presents a 1—to-1 correspondence between maps with the same size on its two sides and the valency of rooted vertex of M;(0 € à € n(M)) is n(M) — i for any M € T, the enumerating X.2 Planar rooted maps 247 function of 75 is IuG)e yey MEJ n(M) =a DD re MeT i-0 (10.24) ff yn M) — MeT zy = to — yt {j= " o — yt) where t = t(x, y) and to = t(x, 1). Theorem 10.2 The enumerating function t = t(x, y) of planar rooted maps satisfies the equation as zy^(1— yt? — (1— y - zy))t +rytot (1— y) =0 (10.25) where ty = t(x, 1). Proof From (10.17), t= fax.) + fÍn(z.vy) + fIa(*,y). Because 79 = {J} and V has no edge, fz,(z,y) = 1. From (10.20) an (10.24), zy t=1+ry t + lto — yt). Via rearrangement of terms, (10.25) is soon found. O Although (10.25) is a quadratic equation, because the occurrence of tọ which is also unknown and the equation becomes an identity when y = 1, complication occurs in solving the equation directly. The discriminant of the equation (10.25), denoted by D(z, y), is D(z,y) = (zy^ — y + 1} — 4(y — 1) (y — 1 — xyto) —1-2gy 4 (1— 2z)y? + (a? + 2x 248 Chapter X Asymmetrized Maps —H(x))y® + H(z)y* (10.26) where H (x) = Ax?tg + x? — 4a. (10.27) Assume that D(z, y) has the form as D(x,y) = (1 — 0y) (1 ay + by’) = 1— (20 — a)y + (6? — 2a0 + by? J-0(a0 — 2b)y? + 0?by. (10.28) By comparing with (10.26), Qe 2: (10.29) and ] — 2x = 0(4 — 30) + b; a” + 2x — H (x) = 0(2(0 — 1)0 — 2b); (10.30) H(x) = 0?b. Then, an equation about b with 0 as the parameter is found as 1 7 (1 — 46 c 36 b) + 1— 40 4- 36? — b — 6b = 2(0 — 1)0? — 20b. By rearrangement, it becomes b? — (100? — 160 + 6)b + (96% — 320? -420? — 240 4- 5) — 0. (10.31) The discriminant of (10.31) is (100? — 160 + 6)? — 4(90* — 320? + 420? — 240 + 5) = 640* — 1920? + 2080? — 960 + 16 = (80? = 120 + 4?. Therefore, b= (@—1)’,or 96° —140 +5 = (90 —5)(0 — 1). X.2 Planar rooted maps 249 The latter has to be chosen in our case. By the last relation of (10.30), H(x)«6^(00 —5)(0 —1). From the first relation of (10.30) and (10.27), the expressions for x and bọ with parameter 0 are extracted as x = (30 — 2)(1— 0); T A0 — 3 (10.32) 0^ 130 = 2)? This enables us to get bọ as a power series of x as 2x9" 2m)! b = — r". 10.33 (e) 2. m!(m + 2)! E ( ) by eliminating the parameter 0 via Lagrangian inversion. More about this method can be seen in the monograph[Liu8}. Example 10.2 For m = 2, it is known from (10.33) that the number of non-isomorphic planar rooted maps is the coefficient of x. That is 9. Because there are 4 non-isomorphic planar maps of size 2 as shown in Fig.10.2. The arrows on the same map represent the roots of non-isomorphic rooted ones. Such as there are, respectively, 2, 2, 1 and 4 non-isomorphic rooted maps in (a), (b), (c) and (d) of Fig.10.2 to get 9 altogether. V 615 (a) (c) (d) Fig.10.2 Planar rooted maps of two edges 250 Chapter X Asymmetrized Maps X.9 Nonorientable equation Let Nm be the set of nonorientable rooted maps of size m. Of course, m > 1. And, let Nm be partitioned into ND and AD. i.e., Nm = NO + ND (10.34) where Ni) = {NIN € Nm, N — a orientable} and NO = Nm — N® = (N[N € Nn, N — a nonorientable}, a = e,(N) is still the rooted edge. Lemma 10.8 Let Mẹ’ = {N — a|VN € NP), then ALD e Ma (10.35) where M,,,_1 is the set of orientable rooted maps of size m — 1, m > 1. Proof Because of the nonorientability of N € ND and the ori- entability of N — a, from Corollary 3.1, a is not a segmentation edge. Based on Theorem 3.4, N — a is always an orientable map. So, for any N € VT, N € Mai, m > 1. This implies MP C Mmi. Conversely, for any N = (X, P) € M41, by appending an edge a’ on N, N' = (X',P") is obtained where X' = X + Kr' and P’ is different from P only at the vertex (r)p ES ge B (rip): Since Kr’ is not a segmentation edge, from Theorem 3.7, N' is a map. And since Gr’ € {r'}w, V' = Vipa}, from Theorem 4.1, N’ is nonorientable. Further, because N = N’ — a’ is orientable and m(N) +1=m(N’) =m, N € NR. This implies Mm- CMP. O For any M = (X, P) € Mm, since M is orientable, assume [re = (rar, oscar] V; € V = Vp, i = 1,2,---,2m(M) — 1 = 2m — 1. By appending the edge r’, A(M) LAo( M), A1( M), UO -, Aom(M)} X.3 Nonorientable equation 251 is obtained where A;(M) = M+e,, = (£i, Pi) such that A; = X + Kr; and P; is determined in the following manner: Bro in the angle (o/P !r, r) brili = 1,2,---, 2m(M) — 1) in the angle (o/P- ir, vir), Dro, (w in the angle (r, aP'r). Because Br; € (ri) w, where V; = Vy, 53, à = 0,1,---,2m(M), from Theorem 4.1, A; are all nonorientable. Because A;(M) — e,, € Mm, Ai(M) € NO, 0 € i € 2m(M). From Lemma 10.8, NO:- M^ AM) (10.36) MEM, m > 0. Of course, Mo consists of only the trivial map. For MP, two cases should be considered: M® and NW), ie., ND — NOY 4 VD) (10.37) where N® = {NWN € N™® a =e, is a terminal, or segmentation edge] and AD = {NWN € N a is neither terminal nor segmentation edge]. Of course, MIP = MP — NO, Lemma 10.9 Let MY = {N - aNN € AG. Then, NO = M Max Nat XY Nux Mu nytng=m—-1 nytng=m—-1 n1,n920 n1,n920 + > Nig X Na nytng=m-1 n1,n920 (10.38) where x represents the Cartesian product of sets. Proof Easy to see except for noticing that N — a has a transitive block which is the trivial map when a is a terminal edge. [] 252 Chapter X Asymmetrized Maps Lemma 10.10 Let NS) ={N-alVNeE ND}. Then, NO = Ne (10.39) where m > 2. Proof Because a = e, is neither terminal nor segmentation edge, from Theorem 3.4, N € ND. By the nonorientability and the size m—1,NeE Mni; dB N e A ad On the other hand, for any N = (X, P) € N41, we have N' = (X', P^) such that 4’ = A + Kr' and P’ is different from P only at the vertex (rp. = (r’, yr’, (r)p). Since N is nonorientable and a’ = Kr’ is neither terminal nor segmentation edge, N' € ND. Thus, N = N'—a' € NAP. This implies Nm-1 C ACD. In consequence, the lemma is proved. L One attention should be paid to is that when m = 1, there is only one nonorientable map (Kr, (r, 8r)), and (Kr, (r, 8r)) € ND. Thus, (10.39) is meaningful only for m > 2. On the basis of this lemma, it is necessary to see how many N' E€ NO can be produced from one N € Nm such that N = N'— d. Because N is nonorientable, let I = {r, Vir, Vor, - -- , Vo 1r] be consists of half the elements in {r}y = A, V = V, p}, such that for any x € I, Kx 1I = (z,»yz). Two cases are now considered. Case 1 For any N = (X, P) € Nn, let B(N) — LBo(N), PY(N), Ba(N),- -+ , Ba (N)] where Bj(N) = (Xi, Pi) =N + e, j — 0,1,2,---, 2m, have Gro in the angle (o/P^ ir, r), Brj(j — 1,2,---,2m — 1) in the angle (o/P ^ y;r, jr), Brom in the angle (r, aP tr). X.3 Nonorientable equation 253 Case 2 For any N = (X, P) € Nn, let C(N) = {Co(N), C (IN), C(N), ++, Com(N) } where C;(N) = (5, Qj) = N + er, j =0,1,2,---, 2m, have yro in the angle (o/P^!r, r}, yrj(j = 1,2,---,2m — 1) in the angle (o/P! y;r, jr), rom in the angle (r, o/P^!r). On the basis of Lemma 10.10, from the conjugate axiom, Ni = V (G(N) + C(N)) (10.40) NENm for m > 1. Because N = M +M 4- ---, the enumerating function —- 5.6, De” = fuo (2) + faz) m>1 NeN (10.41) = fox) + fym (z) + fyo (x). Lemma 10.11 For NMO = NO LAG adf dx where fm = f(x) is the enumerating function of orientable rooted maps determined by equation (10.14). fyo(x) = afm + 2x (10.42) Proof On the basis of (10.35-36), from Lemma 9.10, the lemma is obtained . O Lemma 10.12 For A) = NI +N) fon (x) = 2xfuty + eff (10.43) where fm = f(x) as in (10.10) and fy = f(x) as in (10.41). Proof A direct result of Lemma 10.9. [] 254 Chapter X Asymmetrized Maps Lemma 10.13 For NV“) = NO +N? SEE fno (a) = 2z fu + ad (10.44) Proof On the basis of Lemma 10.10 with its extension (10.40), from Lemma 10.11, (10.44) is soon obtained. O Theorem 10.3 The following equation about f art — a(x) f — xf? — 2xb(x); df dx (10.45) =1 y=0 where hz) = f - 20M is well defined in the ring of power series with all coefficients non- negative integers and negative powers finite. And, the solution is f = fw). Proof Let f = Nyx + Nox? + Naz? 4- ---, then from (10.45) all the coefficients can be determined by the recursion on = 1-27 — 2z fm; Ny = (4m = 2)Nm-1 T (2m — LP m-—1 m—2 2 Nos —i NiNm- —i E 2 dd D ! (10.46) m > 2; N=, where Fm, m > 0, are known in (10.15). Because all Nm, m > 1, determined by (10.46) are positive integers, the former statement is true. The latter is directly deduced from (10.41-44). [] X.4 Gross equation 255 X.4 Gross equation Let Rm be the set of general(orientable and nonorientable) rooted maps with size m, m > 0. Of course, Rp consists of only the trivial map. For m > 1, Rm is partitioned into two subsets RIN ) and «RU ) 1. €., Rm = RY + RDP (10.47) where RN) = {RIVR € Rm, e, (R) is a terminal link or segmentation edge} and RID) = (R|VR € Rm, e,(R) is neither terminal nor segmentation edge]. Of course, qi = Rm- RI Lemma 10.14 Let RẸ = {R — aVR € RO, then R= CN Rakha (10.48) ny+tng=m—1 n1,n920 m. 1. Proof Forany R € RW, because a = e,(R) is a terminal link or a segmentation edge, R — a has two transitive block (when a is a terminal link, the trivial map is seen as a transitive block in its own right), R— a = R4 + Rə and R4 € Rm, Ro € Rm. In other words, the set on the left hand side of (10.48) is a subset of the set of its right. Conversely, for any Ry = (44,71) € Rn, and Ry = (X2, P2) € Rn, by appending a = e,, R = (X,P) is obtained where ¥ = X, + A5 -- Kr and P is different from Pı and P2 only at the vertices (r)p = (r, (r1)p,) and (yr)p = (yr, (r2)p,). It is easily checked that R € Rm, m = nı +n2+ 1. In other words, the set on the right hand side of (10.48) is a subset of the set on the left. O 256 Chapter X Asymmetrized Maps Since R = Ro + 4 + Ra» - ---, the enumerating function fra) = 9 C, Da" m»0 RERm (10.49) = fr (x) + from (at) + fgo(x), where RY = RMN 9... and RY = RID 4 RD +- First, because Ro consists of only the trivial map, fus) m 1. (10.50) Then, from Lemma 10.14, fg (x) = z fR, (10.51) where fr = fr(£). In order to evaluate fpg (x), R® has to be decomposed. Lemma 10.15 Let Ri = LR — avVR € RD}, then RIO = Rina, (10.52) where m > 1. Proof For any R' € qr ) because a’ = e, is neither terminal link nor segmentation edge, from Theorem 3.4, R = R' — a! € Ry_1. This implies RoC Ree: Conversely, for any R = (X, P) € Rm-1, by appending the edge a’ = Kr’, R' = (X', P’) is obtained where X' = X + Kr' and P’ is dif- ferent from P only at the vertex (r’)p: = (r', yr’, (r)p). From Theorem 3.7, R! € Rm. Because a’ is neither terminal link nor segmentation edge and R = R' — a^, R € RIP. This implies Rast de 'The lemma is proved. [] Based on this, what should be further considered for is how many Rc RID, can be produced from one R € Rm such that R = R — q'. X.4 Gross equation 257 Because R is a map(orientable, or nonorientable), let I = {r, yir, bar, ++, amar) be the set of elements in correspondence with a primal trail code, or dual trail code. For any x € I, Kx NI = {x, yx} has two possibilities as cases. Case 1 For any R= (X, P) € Rm, let D(R) = {Do(R), D,(R), D»(R), uS Do, (R)] where D,(R) = (4), Pj) = RF e,, j =0,1,2,---,2m, have Gro in the angle aP r), Br;(j — 1,2,---,2m — 1) in the angle (aP7ty;r, jr), Brom in the angle (r, aP tr). Cases 2 For any R= (X,P) € Rm, let E(R) = {Eo(R), A(R), EX(R), +++, E (R)] where Ej(R) = (Yj, Qj) = RF erns J =0,1,2,---,2m, have yro in the angle (aP~'r,r), Yr; = 1,2,--.,2m — 1) in the angle (oP- sr, jr), Vom in the angle (r, o/P !r). Based on Lemma 10.15, from the conjugate axiom, T Riu = >, (D(R) + E(R) (10.53) RERm for m > 1. Because R = RP + RY +--+, from Lemma 10.15 with its extension (10.53) and Lemma 9.10, the enumerating function fram (x) = 2z fg + s lR (10.54) 258 Chapter X Asymmetrized Maps Theorem 10.4 The equation about f d [o = 14 (1-22)f - xf’, (10.55) fo = f(0-1 is well defined in the ring of power series with coefficients all nonneg- ative integers and terms of negative power finite. And, the solution is f = fr(z). Proof In virtue of the initial condition of equation (10.55), as- sume f = Ro + Rix + Rox? +--+. Of course, Ro = fo = 1. Further, from equation (10.55), the recursion m-1 Rm = (4m — 2 Fd + Ss ee eee i=0 (10.56) mi: Hy 1 is soon found for determine all the coefficients Rm, m > 0. It is easily checked that all of them are positive integers and hence the former statement is true. The latter is a direct result of (10.50—51) and (10.53). O X.5 The number of rooted maps First, let Om = (Fo, Fi, -++ , Fm), m => 0, be the m+1 dimensional vector where Fm, m > 0, are the number of non-isomorphic orientable rooted maps with size m. And, oÈ, = (Fm, Fm-1, +, Fo) called the reversed vector of the vector om. Easy to check that Om = ((0w) )^ = (em) ) = om (10.57) TIL m where T the transposition of a matrix. The recursion (10.15) for determining Fm, m > 0, becomes Fy, = (2m — 1) Fi + Onit B! (10.58) Fy=1. X.5 The number of rooted maps 259 By (10.58), the number of non-isomorphic orientable rooted maps with size m, m > 1, can be calculated. In the first column of Table 10.1, Fin, m < 10, are listed. Then, let ôm = (N1, No, --- , N1) where Npn is the number of non- isomorphic nonorientable rooted maps with size m for m > 1. The recursion (10.46) for determining Nm, m > 1, becomes Nm = (4m — 2) Ng.1 + (2m — 1) Fs iina atoma os m. 2: N,=1 (10.59) where 2,4,» is given in (10.58). By (10.59), the number Nn can be calculated for m > 1. In the second column of Table 10.1, Nm, m < 10, are listed. Finally, let pm = (Ro, Ri,*--, Rm) where Rm is the number of non-isomorphic general maps with size m for m > 0. The recursion (10.56) becomes Rm = (4m — 2)Rm-1 + Pm—1P met m > 1; (10.60) m=i By (10.60), the number Rm can be calculated for m > 0. In the third column of Table 10.1, Rm, m < 10, are listed. From those numbers in Table 10.1, it is also checked that the enumerating functions f(x), f(z) and fr(x) of, respectively, non- isomorphic orientable, nonorientable and general(orientable and nonori- entable) rooted maps with size as the parameter satisfy the relation as fr(x) = f(x) + fw(x). (10.61) 260 mL 0 1 2 3 4 5 6 7 8 9 0 Chapter X Asymmetrized Maps 1 0 1 2 1 3 10 14 24 74 223 297 706 4190 4896 8162 92116 100278 110410 2339894 2450304 1708394 67825003 69533397 29752066 2217740030 2247492096 576037442 80952028936 | 81528066378 12277827850 3268104785654 | 3280382613504 Table 10.1 Numbers of rooted maps with size less than 11 Activities on Chapter X X.6 Observations O10.1 Because a surface can be seen as a polygon with even edges pairwise identified in the plane, think that whether, or not, a map on a surface rather than plane can always represented by a planar one. If it can, explain the reason, or provide an example otherwise. O10.2 For a rooted map M, let m(M) and l(M) be, respec- tively, the size and the valency of root-face in M, observe the numbers of non-isomorphic planar rooted maps for m( M), I(M) < 3. Similarly, do the same with the valency of root-vertex s(M) in- stead of (M). A map with all its faces of 3-valent except for the root-face is called a near triangulation. O10.3 Fora near triangulation T, let m(T) and I(T) be, respec- tively, the size and the root-face valency of T. Observe the numbers of non-isomorphic planar rooted near triangulations for m, l < 4. Similarly, do the same with the valency of root-vertex s instead of l. A map with all its faces of valency 4 except for the root-face is called a near quadrangulation. O10.4 . For a near triangulation Q, let m(Q) and l(Q) be, respectively, the size and the root-face valency of T'Q. Observe the numbers of non-isomorphic planar rooted near quadrangulation for m, | 262 Activities on Chapter X Similarly, do the same with the valency of root-vertex s instead of l. A planar map from which the result of deleting all the edges on a face is a tree is called a Halin map. The face is said to be specific. O10.5 Evaluate the number of non-isomorphic rooted Halin maps with the root on the specific face by the parameters: the size m and the valency l of the specific face for m > 6 and l > 3. 010.6 ‘Try, by edge contraction, to determine the enumerating function of general rooted maps with the size as the parameter. O10.7 ‘Try, by edge contraction, to determine the enumerating function of rooted petal bundles with the size an the root-face valency as the two parameters. 010.8 Try, by directly solving the equation (9.7), to determine the enumerating function of orientable rooted petal bundles h. 010.9 ‘Try, by directly solving the equation (9.30), to deter- mine the enumerating function of nonorientable rooted petal bundles g. O10.10 Observe the numbers of general Eulerian rooted maps with the size smaller. X.7 Exercises For a map, if the result of deleting all inner vertices(not articulate vertex, i.e., a vertex of valency 1, or a terminal) of a spanning tree is, itself, a travel with only one vertex of valency probably greater than 2, then it is called a pan-Halin map. Because this travel is still a map and becomes a petal bundle via decreasing subdivision, such a petal bundle is called the base map of the pan-Halin map. A pan-Halin map of which the base map is of size 2p and ori- entable genus p > 0, or of size q and nonorientable genus q > 1, is said to be pre-standard . If a pre-standard pan-Halin map has its base X.7 Exercises 263 map with each edge incident with at least one terminal of the tree, then it is said to be standard. Let pg be the set of all pre-standard pan-Halin rooted maps. The root rg for H € Hy is chosen be an element incident with the vertex and the face of the base map of H. For any H = (X,P) € f, the tree T on H is seen as a planted tree(a plane tree with the root-vertex of valency 1) with its root rr = P(Py)'rx where t = min{s|(Py)*ry incident with a terminal of T}. E10.1 Given the partition of vertices according to their va- lencies on a planted tree j = (j1,js,-:::,), Le, ji, i > 1, is the number of unrooted vertices of valency i, prove that the number of non-isomorphic planted trees with the partition is (n — 1)! g! n=14+)S j; i>1 where i.e., the order, and j! = [[;., j! E10.2 Given the vertex partition s = (55,53,::-), prove that the number of non-isomorphic pre-standard pan-Halin rooted maps with the partition and their base maps of size m on a surface of ori- entable genus p is pm ( " + 2p — 1 \ /s3\ nlm 2p— 1 mj s! n+2=% s; and s! = [ [ si. i>2 s>2 where E10.3 Given the vertex partition s = (s2, 53,::-), Prove that the number of non-isomorphic pre-standard pan-Halin rooted maps 264 Activities on Chapter X with the partition and their base maps of size m on a surface of nonori- 9m m+@q—1)\ (83\ nln q—1 mj s! n+2=%_ s; and s! = | [ si. i2 S22 entable genus q is where E10.4 Given the vertex partition s = (s2,53,---), prove that the number of non-isomorphic standard pan-Halin rooted maps with the partition and their base maps of size m on a surface of orientable genus p is pm (M — 1\ /s3\n!m 2p—1/\m/ s! n+2=%_ s; i22 where and s > 0,s #0, m > 2p 7 1. E10.5 Given the vertex partition s = (55, 53,::-), prove that the number of non-isomorphic standard pan-Halin rooted maps with the partition and their base maps of size m on a surface of nonori- 9m m — 1N (s3\n!m q—1/NmJ/ s! n+2=%_ s; i22 entable genus q is where ands >0,s40,m>q2>1. E10.6 Evaluate the number of near triangulations of size m on the projective plane. E10.7 Evaluate the number of near triangulations of size m on the Klein bottle. X.8 Researches 265 E10.8 Evaluate the number of rooted quadrangulations of size m on the projective plane. E10.9 Evaluate the number of near quadrangulations of size m on the Klein bottle. E10.10 Determine the enumerating function of rooted petal bundles with the size as the parameter on the torus. E10.11 Determine the enumerating function of rooted petal bundles with the size as the parameter on the projective plane. E10.12 Determine the enumerating function of orientable two vertex rooted maps with size as the parameter. E10.13 Determine the enumerating function of nonorientable two vertex rooted maps with size as the parameter. E10.14 Establish an equation satisfied by the enumerating func- tion of general non-separable rooted maps. E10.15 Establish an equation satisfied by the enumerating func- tion of general Eulerian rooted maps. A map is said to be loopless if its under graph has no self-loop. E10.16 Establish an equation satisfied by the enumerating func- tion of general loopless rooted maps. A map is said to be simple if its under graph has neither self-loop nor multi-edge. E10.17 Establish an equation satisfied by the enumerating func- tion of general simple rooted maps. X.8 Researches R10.1 Given a relative genus g Z O(the case g = 0 is solved), determine the number of rooted near triangulations of size m > |g| on 266 Activities on Chapter X a surface of genus g. A rooted map with all vertices of the same valency except for probably one vertex is said to be near regular. Among them, near 3-regular and near 4-regular are often encountered in literature. Although near triangulations or near quadrangulations are, re- spectively, the dual maps of rear 3-regular, or near 4- regular maps, they are still considered for most convenience from a different point of view. R10.2 Given a relative genus g # O(the case g = 0 is solved), determine the number of rooted near 3-regular maps of size m > |g| on a surface of genus g. R10.3 Given a relative genus g 4 O(the case g = 0 is solved), determine the number of rooted near quadrangulations of size m > |g| on a surface of genus g. R10.4 Given a relative genus g 4 0(the case g = 0 is solved), determine the number of rooted near 4-regular maps of size m > |g| on a surface of genus g. R10.5 Given a relative genus g Z O(the case g = 0 is solved), determine the number of non-separable rooted maps of size m > |g| on a surface of genus g. R10.6 Given a relative genus g Z O(the case g = 0 is solved), determine the number of Eulerian rooted maps of size m > |g| on a surface of genus g. R10.7 Given a relative genus g Z O(the case g = 0 ia solved), determine the number of non-separable Eulerian rooted maps of size m 2 |g| on a surface of genus g. For the problems above, another parameter / > 1 is absolutely necessary in almost all cases. it is the valency of the extra vertex, or face according as the regularity is for vertices, or faces. R10.8 Given a relative genus g Z O(the case g = 0 is known), X.8 Researches 267 find a relation between general maps and quadrangulations on a sur- face of genus g. R10.9 Given a relative genus g Z 0(the case g = 0 is known), find a relation between general maps and triangulations on a surface of genus g. R10.10 Given a relative genus g 4 O(the case g = 0, a 1-to- 1 correspondence between loopless planar rooted maps of size m — 1 and 2-connected planar rooted triangulations of 2m — 1 unrooted faces should be found, but now unknown yet), find a relation between loopless rooted maps and triangulations on a surface of genus g. R10.11 Present an expression of the solution for equation (10.14) by special functions, particularly the hyperbolic geometric function. Chapter XI Maps with Symmetry e A relation between the number of rooted maps and the order of the automorphism group of a map is established. e A general procedure is shown for determining the group order dis- tribution of maps with given size via an example as an application of the relation. e A principle for counting unrooted maps from rooted ones is pro- vided. e Dased on the principle, a general procedure is shown for determin- ing the genus distribution of unrooted maps with given size via two examples. e Conversely, rooted maps can be also determined via unrooted maps. XI.1 Symmetric relation First, observe how to derive the number of non-isomorphic un- rooted maps from that of non-isomorphic rooted maps when the auto- morphism group is known, or in other words, how to transform results without symmetry to those with symmetry. XL1 Symmetric relation 269 Theorem 11.1 Let no(U; I) be the number of non-isomorphic rooted maps with a given set of invariants including the size in the set of maps U considered. If the order of automorphism group of each map M in U is independent of the map M itself, but only dependent on U and I, denoted by aut(U; I), then the number of non-isomorphic unrooted maps with J in M is aut(U; D)no(Ut; I) m(U; I) = Ae (11.1) where e € Í is the size. Proof Let map M = (X,7P) € U. From Theorem 8.1, for any rca, |X.| = Hyl dr € Aut(M), y = 7x)| = aut(M). (11.3) In view of Corollary 8.2, M itself produce |.|]X| | 4e — |X,| aut(M) no(M) (11.3) non-isomorphic rooted maps. Therefore, there are 4e u; I) = — m1) = à, M Meu 4e = — —n,(U;I aut(U; I ji non-isomorphic rooted maps in U. Via rearrangement, (11.1) is soon obtained. [] In chapter VIII, efficient algorithms are established for finding the automorphism group of a map, this enables us to get how many non-isomorphic rooted maps from a unrooted map by (11.1). However, from Chapter IX and Chapter X, it is unnecessary to know the automorphism group for counting rooted maps. This en- ables us to enumerate unrooted maps via automorphism groups by employing (11.1). 270 Chapter XI Maps with Symmetry Problem of type 1 For a set of maps M known the number of non-isomorphic rooted maps with a given size, determine the num- ber of non-isomorphic unrooted maps with the given size according to the orders of their automorphism groups, or in other words, the distribution of unrooted maps on the orders of their automorphism groups. Although this problem does not yet have general progress in present, a great amount of results for rooted case have already pro- vided reachable conditions for the problem. XL2 An application In what follows, provide a general procedure for solving the prob- lem of type 1 via the determination of the distribution of rooted petal bundles on the orders of the automorphism groups of corresponding unrooted maps on the basis of Chapter IX. From Table 9.1 at the end of Chapter IX, the number of non- isomorphic planar rooted petal bundles with size 4 is H ©) = 14, shown n (a-n) of Fig.11.1. In virtue of Corollary 8.2, the orders of their automorphism groups are possibly 1, 2, 4, 8 and 16 only 5 cases. Case 1 aut(M) 21, M = (Kz 4- Ky + Kz + Kt, J). None. Case 2 aut(M) = 2. 8 planar rooted petal bundles: ida = (By IR 2,20, yt, "yy; ; Sa = (2,9, yy, 2, "92, yt, o yt; | Se = (2592, 9s 2 "2," ys o yos ig en us do ^y d, 6 et ey DW. YY, YL, Z, YZ, t, yt T, Y, Z, YZ, t, Yt, YY, YT); T Y; Z, YZ; YY, t, yt, YT); T, Y, YY, Z, t, YZ, Yt, YT); Tou M omm MP om P c Si Iz Js = Jr Tee eed eei eri shown in (a-h) of Fig.11.1. XL2 An application 271 Case 3 aut(M) = 4. 4 planar rooted petal bundles: Jo = (t, Y, VY, YT, Z, t, yt, yz); Jio = ras YT, y, 2, t, yt, YF, yy); Ju -(yzt,ytyz,yy,yx); Ju = (2, Y2 Va TE ts t), shown in (i-l) of Fig.11.1. Case 4 aut(M) = 8. 2 planar rooted petal bundles: Jig (55935 09. 2599, 05 y8)5 lace (5,35 09 2,38, 0; 8,9). shown in (m, n) of Fig.11.1. Case 5 aut(M) = 16. None. This procedure can be done for determining the automorphism groups of a unrooted maps via their primal trail codes, or dual trail codes, by computers and then via the collection of the same class of them according to the orders of the groups. NIZ. NIZ NZ Lx AN AN (a) (b) (c) NZ NIS NZ. AIX AIN VN (d) (e) (f) 262 Chapter XI Maps with Symmetry Fig.11.1 Planar petal bundles of size 4 XL3 Symmetric Principle Whenever the distribution of rooted maps on the orders of au- tomorphism groups is given for a set of maps, the number of non- isomorphic unrooted maps can be soon extracted Theorem 11.2 Let no;(.Mt; I) be the number of non-isomorphic rooted maps with the set of invariants / and the order of their auto- morphism groups 7 in a set of maps M for i|4e, 1 < i € 4e, where e is XL3 Symmetric Principle 273 the size, then the number of non-isomorphic unrooted maps in M is m(M;1)= M: TRU (11.4) 4e i|4e 1<i<4e Proof Let ny;(M; I) be the number of non-isomorphic unrooted maps with the set of invariants J and the order of their automorphism groups i in the set of maps M for i|4e, 1 < i < 4e, where e is the size, then nil MET) ` ny; (Mt; I). (11.5) i|4e 1<i<4e From Theorem 11.1, each unrooted map M € M, Aut(M) = i, produces 4e d non-isomorphic rooted maps. Therefore, ingil M; I) nai (M; I) = = (11.6) By substituting (11.6) into (11.5), (11.4) is soon obtained. O On the choice of the set of invariants J, two types should be mentioned. One is that the set J consists of only the size and the genus for determining the genus distribution of non-isomorphic maps in a set of maps M. The other is that the set J consists of only the size and the orders of automorphisms for determining the symmetric distribution of non-isomorphic maps in a set of maps M. Problem of type 2 For a set of maps M with the number of non-isomorphic maps given, determine the number of non-isomorphic under graphs of maps in M. Although the justification of whether, or not, two graphs are isomorphic is much far from easy, a feasible approach to it is presented from the above discussion. Because the under graphs are isomorphic 274 Chapter XI Maps with Symmetry if the two maps are isomorphic, the only thing we have to do is to classify non-isomorphic maps by their isomorphic under graphs. On the other hand, for a graph, it is also possible to discuss how many non-isomorphic rooted maps are with the graph as their under graph, and then to discuss how many non-isomorphic unrooted maps are with the graph as their under graphs, and finally to classify maps according to the isomorphism of their under graphs. XL4 General examples On the basis of the 15 orientable rooted petal bundles of size 3 and the 9 nonorientable rooted petal bundles of size 2(in Table 9.1 at the end of Chapter IX), a general procedure is established for deter- mining the genus distribution of them. Orientable case Let M = (Ka+Ky4+ Kz, J), 1<i< 15. genus 0 5 orientable rooted petal bundles shown in (a-e) of Fig.11.2. Here, d = (£, YT, Y, YY, Z, YZ); Jr = [35:05 99) b mor es) with the order of its automorphism group aut(M) = 6 are one un- rooted map; Js = (zx, yx, yz, yz, vu) Ja = (2, y, WY, VE, 2, V2); Is = (CY eye VY.) with the order of its automorphism group aut(M) = 4 are one un- rooted map. Genus 1 10 orientable petal bundles shown in (f-o) of Fig.11.2. Here, Jes GG eoa a = E qr ave yarn da ec ndo mou y ce pop d ses Jio = Dish x m t var s Jı = (£, Y, YT, YY, £;^y«) XL4 General examples 275 with the order of its automorphism group aut(M) = 2 are one un- rooted map; Sia = (2,5, 2) Vs VE) 72); Sia = (2, Y; YE, 2, Vs V2); Fis = (2, Y, 2, YE, 72, VY) with the order of its automorphism group aut(M) = 4 are one un- rooted map; and Jis = (£, Y, 2, YL, VY, YZ) with the order of its automorphism group aut(M) = 12 is one unrooted map itself. All are listed in Table 11.1 shown the genus distribution, group order distribution as well, of orientable unrooted maps. 276 Chapter XI Maps with Symmetry Fig.11.2 Orientable petal bundles of size 3 Nonorientable case Let N = (Kx + Ky, Ji), 1<i< 9. Genusl 5 nonorientable rooted petal bundles shown in Fig.9.4(e, a,c) and in Fig.9.3(a,c). Here, eh = Ut By, pu. y) with the order of its automorphism group aut( N) = 8 is one unrooted map itself; Jo = (x, Bz,y, vy); Ja = (a, Bx, yy, y); Ja = (2,2, y, By); Js = (2, yz, By, y) XL4 General examples 277 with the order of its automorphism group aut( N) = 2 are one unrooted map. Genus? 4 nonorientable rooted petal bundles shown in Fig.9.3(b) and Fig.9.4(b,d,f). Here, Je = (x, By, yz, y); Fr = (2, 7y, Bz, y); Js = 0, By); Jo m (2, D. D, y) with the order of its automorphism group aut(N) = 4 are 2 unrooted maps. All are listed in Table 11.2 shown the genus distribution, group order distribution as well, of nonorientable unrooted maps. Table 11.2 Distributions of nonorientable petal bundles Activities on Chapter XI XI.5 Observations O11.1 Given all the 54 planar rooted maps of size 3, find their distribution according to the orders of automorphism groups. O11.2 Given all the 40 outerplanar rooted maps(the root inci- dent with the outer face) of size 3, find their distribution according to the orders of automorphism groups. O11.3 Observe how many non-isomorphic rooted maps whose under graph is 4, i.e., the complete graph of order 4. O11.4 Observe the distribution of all the rooted maps whose under graph is Ky according to the orders of their automorphism groups. O11.5 Observe the genus distribution of all orientable maps whose under graph is K4. O11.6 Observe the genus distribution of all nonorientable maps whose under graph is K4. O11.7 Given all the 56 planar Euler rooted maps of size 4, find how many non-isomorphic unrooted maps among them. O11.8 Given all the 27 planar 4-regular rooted maps of co- order 4, find how many non-isomorphic unrooted maps among them. O11.9 For m > 5, try to determine the number of planar un- rooted petal bundles of size m. O11.10 For m 2 4, try to determine the number of orientable XI.G Exercises 279 unrooted petal bundles of size m. O11.11 For m > 3, try to determine the number of nonori- entable unrooted petal bundles of size m. XI.6 Exercises E11.1 Prove that the number of non-isomorphic outerplanar rooted maps(the root is on the outer face) of size m is 2m Jin)! (m+ 1)!m! for m > 1. E11.2 Prove that the number of planar 4-regular rooted maps of co-order n -- 1 is 2(2n — 2)! (n+ (n — 1)! n—1 forn > 1. E11.3 Prove that the number of non-isomorphic planar rooted maps of size m is gg 6) (n+1)(n+2)\m E11.4 Prove that the number of planar loopless rooted trian- culations of size 3m is for m > 0. 27H (35m m!(2m + 2)! for m > 1. E11.5 Prove that the number of non-isomorphic planar Euler rooted maps of size m is 8 x2?-l0m)l mI(m + 2)! for m > 1. 280 Activities on Chapter XI E11.6 Prove that the number of non-isomorphic planar non- separable rooted maps of order p and co-order q is (2p 4- q — 5)!(2q 4- p — 5)! (p — 1)!(g — 1)! (2p — 3)! (2g — 3)! where p, q > 2. E11.7 Prove that the number of non-isomorphic planar simple rooted maps of size m is Y A(2m + 1) (2m — i — 4)! es CL ea 2)! (2m — i + L)!m! where m > 2. E11.8 Prove that the number of non-isomorphic planar 3-connected rooted maps of size m > 6 is (—1)2 + fa where Rm, m > 2, are determined by the recursion (7m — 22) Rra + 200m. — 1) Rms 2m with the initial conditions R; = —1 and R» = 2. R= m>3 2 E11.9 For m > 4, determine the genus distribution of ori- entable rooted petal bundles with size m. E11.10 For m > 5, determine the genus distribution of nonori- entable rooted petal bundles with size m. E11.11 For m > 4, determine the number of non-isomorphic outerplanar unrooted maps with size m. XI.7 Researches R11.1 Determine the number of non-isomorphic planar 4-regular unrooted maps of co-order n 4- 1 for n > 1. XI.7 Researches 281 R11.2 Determine the number of non-isomorphic planar loop- less unrooted triangulations of size 3m, m > 2. R11.3 Determine the number of non-isomorphic planar Euler unrooted maps of size m > 2. R11.4 Determine the number of non-isomorphic planar non- separable unrooted maps of size m > 2. R11.5 Prove, or disprove, the conjecture that almost all trees have the order of their automorphism group 1 when the size is greater enough. R11.6 Prove, or disprove, the conjecture that almost all maps with a given relative genus have the order of their automorphism group 1 when the size is large enough. R11.7 Prove , or disprove, the conjecture that for a positive integer gle, g > 2, almost no orientable map is with the order of automorphism group g when e is large enough. R11.8 Prove , or disprove, the conjecture that for a positive integer gle, g > 2, almost no nonorientable map is with the order of automorphism group g when e is large enough. R11.9 Determine the genus distribution of 4-regular rooted maps of co-order n+ 1, n > 1. R11.10 Determine the genus distribution of loopless rooted tri- angulations of size 3m, m > 2. R11.11 Determine the genus distribution of Euler rooted map with size m > 2. R11.12 Determine the genus distribution of nonseparable rooted map with size m > 2. Although corresponding problems about genus distribution can also posed for unrooted case, they would be only suitable after the solution of rooted case in general. Moreover, the genus distributions of maps with under graphs in some chosen classes can also be investigated. Chapter XII Genus Polynomials e The set of associate surfaces of a graph are constructed to deter- mine all of its distinct embeddings, or its super maps as well. e A layer division of an associate surface of a graph is defined for establishing an operation to transform this surface into another associate surface. A procedure can be constructed for listing all other associate surfaces from an associate surface by this operation without repetition. e A principle of determining the genus polynomial, called handle polynomial, of a graph is provided for the orientable case. e The genus polynomial of a graph for nonorientable case, also called crosscap polynomial, is derived from the handle polynomial of the graph. XII.1 Associate surfaces Given a graph G — (V, E) and a spanning tree T', the edge set E is partitioned into Er(tree edge) and Er(cotree edge), i.e., E = Er + Er. Let Er = [ili = 1,2,-..,0), 8 = B(G) be the Betti number(or cyclic number) of G. If i = (u[i], v[i]), then iu and 2, are, respectively, meant the semi-edges of i incident with u[i] and vļi]. XL1 Symmetric relation 283 Write @ = (V + Vj, Er + Ej), where V, = (v, v;1 € i € 8) and FE, = {(uli], vi), (vli], vj) 1 < i € 8). Because G” is a tree itself, G” is called an expanded tree of T' on G, and denoted by To, or T in general case[Liu13-14]. Let ô = (01,605, ---, 0g) be a binary vector, or as a binary number of 2 digits. Denoted by T? that T, edges (u[i],v;) and (v[i], v;) are labelled by i with indices: +(always omitted) or —, 1 < i € B, where 0; = 0 means that the two indices are the same; otherwise, different. Then, ô is called an assignment of indices on T For v € V, let o, be a rotation at v and og = {0,|Wv € V}, the rotation of G, then Î, determine an embedding of T on the plane. Theorem 12.1 For any o asa rotation and ô as an assignment of indices, T? determines a joint tree. Proof By the definition of a joint tree, it is soon seen. [] According to the theory described in Chapter 1, the orientabil- ity and genus are naturally defined to be that of its corresponding embedding. Lemma 12.1 Joint tree T5 is orientable if, and only if, ô = 0. Proof Because 6 = 0 implies each label with its two occurrences of different indices, the lemma is true. [] On a joint tree T$ , the surface determined by the boundary of the infinite face on the planar embedding of T, with ô on label indices is said to be an associate. Lemma 12.2 The genus of a joint tree Tu is that of its associate surface. Proof Only from the definition of orientability of a joint tree.L] Two associate surfaces are the same is meant that they have the same assignment with the same cyclic order. Otherwise, distinct. Let 284 Chapter XII Genus Polynomials F(@) be the set of distinct surfaces on Ig = {1,2,---, 8}. For a surface F € F((3) and a tree T on a graph G, if there exists an joint tree 7? such that F is its associate surface, then F is said to be admissible . Let Fr(8) be the set of all distinct associate surfaces. Given two integers p, p > 0, and q, q > 1, let Fr(8;p)(or Fr(0;q),q = 1), p = 0, be all distinct admissible surfaces of ori- entable genus p(or nonorientable genus q). Theorem 12.2 For any integer p > 0( or q > 1), the cardinal- ity |Fr(G; p)|(or |-Fr(8; q)|) is independent of the choice of tree T on G. Further, it is the number of distinct embeddings of G on a surfaces of orientable genus p(or nonorientable genus q). Proof According to O1.14, a 1-to-1 correspondence between two sets of embeddings generated by two distinct spanning trees can be found such that same embeddings are in correspondence. This implies the theorem. [] Because of \Fr(8)| = X Er p) +X |Fr(6; a)l, p20 q>1 the following conclusion is found from the theorem. Corollary 12.1 The cardinality |Fr(8)| is independent of the choice of tree T on G. Further, it is the number of distinct embeddings of G. From Lemma 12.1, the nonorientability of an associate surface can be easily justified by only checking if it has a label 7 with the same index, i.e., ó(i) = 1. Theorem 12.3 There is a 1-to-1 correspondence between as- sociate surfaces and embeddings of a graph. Proof First, we can easily seen that each embedding determines an associate surface. Then, we show that each associate surface is XIL2 Layer division of a surface 285 determined by an embedding. Because of Theorem 12, this statement is derived. go From what is mentioned above, it is soon seen that the problem of determining the genus distribution of all embeddings for a graph is transformed into that of finding the number of all distinct admissible associate surfaces in each elementary equivalent class and the prob- lem on minimum and maximum genus of a graph is that among all admissible associate surfaces of the graph. All of them are done on a polygon. XIL2 Layer division of a surface Given a surface S — (A). it is divided into segments layer by layer as in the following. The Oth layer contains only one segment, i.e., A(— Ay). The 1st layer is obtained by dividing the segment Ag into l segments, i.e., S = (A1, Ao,---, Ai), where Aj, As, ---, Aj, are called the 1st layer segments. Suppose that on k — 1st layer, the k — 1st layer segments are where m,_1) is an integral k — 1-vector satisfied by 1 Ana- Tipi) < (nı, N2, M ny) < Aag with Ig = (1, 1, "ied. 1 asit. IN k-i) = (Ni, No, 1ta Ny), Ni = l4 = Nay); No = lana N3 = lan eux m Ny. a4 = lAng 9? then the kth layer segments are obtained by dividing each k— 1st layer segment as Ay An ity Ån (12.1) 1p Rk) (1) Ang, jj where lg = (1g 1,1) € Qs i) € Nay = QN o ay Ne) 286 Chapter XII Genus Polynomials and N; = lA, ay 1 €i < Ny. Segments in (L1) are called sons of Ap, .," Conversely, Aj, is the father of any one in (12.1). A layer segment which has only one element is called an end segment and others, principle segments. For an example, let S = (1, —7,2, —5,8, —1,4, —6,5, —2,6,7, —3, —4). Fig.12.1 shows a layer division of S and Tab.12.1, the principle seg- ments in each layer. Layers Principle segments Oth layer 1st layer Asi 7959 146 567 4) 2nd layer ; H = (—2;6), I = (-3; 4) Tab.12.1 Layers and principle segments (1, —7,2 — 5;3, 1,4, —6, 5; —2, 6, 7, —3 — 4) (;—7,2;-5) | (3, —1;4, —6; 5) (—2, 6; 7; —3, —4) Fig.12.1 A layer division of S For a layer division of a surface, if principle segments are dealt with vertices and edges are with the relationship between father and son, then what is obtained is a tree denoted by T'. On T, by adding cotree edges as end segments, a graph G = (V, E) is induced. For example, the graph induced from the layer division shown in Fig.12.1 is as Y= 14, B,C, D, EJ GA (12.2) XIL2 Layer division of a surface 287 and E =1{6, 6,c,0,¢, 7,9, h;1,2,9,4, 5,0, T]; (12.3) where a= (A,B), b= (A, C),c = (A, D),d = (B, E), e — (C,F), = (0,G),g = (ID, H),h — (D,I), and 1 = (B, F),2 = (E, H),3 = (F,D,4 = (G,I), 5 = (B,C),6 = (G, H),7 = (D, E). By considering Er = {a,b,c,d,e, f, g, h, Er = (1,2,3,4,5,6,7),0; = 0,2 = 1,2,---,7, and the rotation o implied in the layer division, a joint tree a is produced. Theorem 12.4 A layer division of a surface determines a joint tree. Conversely, a joint tree determines a layer division of its associate surface. Proof From the procedure of constructing a layer division, a joint tree is determined. Conversely, it is natural. [] Then, an operation on a layer division is discussed for trans- forming an associate surface into another in order to visit all associate surfaces without repetition. A layer segment with all its successors is called a branch in the layer division. The operation of interchanging the positions of two layer segments with the same father in a layer division is called an exchanger. Lemma 12.3 A layer division of an associate surface of a graph under an exchanger is still a layer division of another associate surface. Conversely, the later under the same exchanger becomes the former. Proof From the correspondence between layer divisions and as- sociate surfaces, the lemma can be obtained. [] On the basis of this lemma, exchanger can be seen as an operation on the set of all associate surfaces of a graph. 288 Chapter XII Genus Polynomials Lemma 12.4 The exchanger is closed in the set of all associate surfaces of a graph. Proof From the correspondence between joint trees and layer divisions, the conclusion of the lemma is seen. [] Lemma 12.5 Let .A(G) be the set of all associate surfaces of a graph G, then for any S1, S9 € A(G), there exist a sequence of exchangers on the set such that Sı can be transformed into So. Proof By considering the joint trees and layer divisions, the lemma is right. E] If .A(G) is dealt as the vertex set and an edge as an exchanger, then what is obtained is called the associate surface graph of G, and denoted by H(G). From Theorem 12.3, it is also called the surface embedding graph of G. Theorem 12.5 In H(G), there is a Hamilton path. Further, for any two vertices, H(G) has a Hamilton path with the two vertices as ends. Proof By arranging an order, an Hamiltonian path can be ex- tracted based on the procedure of the layer division. [] First, starting from a surface in A(G), by doing exchangers at each principle in one layer to another, a Hamilton path can always be found in considering Theorem 12.3. This implies the first statement. Further, for chosen $1, $5 € A(G) = V(H(G)) adjective, starting from from S1, by doing exchangers avoid Ss except the final step, on the basis of the strongly finite recursion principle, a Hamilton path between Sı and $5 can be obtained. This implies that H(G) has a Hamilton circuit and hence the last statement. This theorem tells us that the problem of determining the mini- mum, or maximum genus of graph G has an algorithm in time linear on H(G). XIL3 Handle polymomials 289 XIL3 Handle polynomials Let S(G) be the set of associate surfaces of a graph G and S,(G), the subset of S(G) with genus g. The the enumerating function Jmax (Giz) = So |S (Ale (12.4) 9=Jmin is called the genus polynomial of G where gmin and gmax are, respec- tively, the minimum and maximum genus of G for orientable, or nonorientable case. In orientable case, u(G;x) = y(G; x) is called the handle polynomial. In nonorientable case, v(G; y) = 7(G; y) is the crosscap polynomaal. On the basis of the theory described in 12.1 and 12.2, (12.4) is in fact the genus distribution of embeddings of G. Because the enumerating function of super rooted maps of G is a constant times the genus polynomial y(G; z), for the enumeration of naps by genus it is enough only to discuss y(G; z). Lemma 12.6 An orientable associate surface of a graph with- out two letters interlaced has a letter x such that zx! is a segment of the surface. Proof Let (x,x~') be a segment of the surface with minimum of letters. If it does not contain only the letter x, then there is another letter y in it. Because of x and y noninterlaced, the segment (y, y^ 1), or (y !,y), is a subsegment of (x,x~'). However, it has at least one letter less than the minimum. [] Lemma 12.7 An orientable associate surface of a graph is with genus 0 if, and only if, no two letters are interlaced. Proof On the basis of Lemma 12.6, by the finite recursion prin- ciple the lemma can soon be found. [] Theorem 12.6 If an orientable associate surface of a graph has two letters interlaced, i.e., in form as ArByCxz- ! Dy ! E, then its 290 Chapter XII Genus Polynomials genus is k, k > 1, if, and only if, the orientable genus of ADC'BE is k — 1. Proof On the basis of Relation 1 in L2, the theorem is soon found. [] According this theorem, a linear time algorithm can be designed for classifying the orientable associate surfaces of a graph G by their genus. Let N;(G) be the number of orientable associate surfaces of G with genus 7, 2 > 0 Theorem 12.7 The handle polynomial of G is u(G; x) = ` Ni(G)z* (12.5) 0xix 2| where 0 is the Betti number of G. Proof From (12.4), the theorem follows. O XII.4 Crosscap polynomials Let £5, = (S55;]1 € j < si} where s; is the number of orientable 20-gons (surfaces) with genus i, then Fog = X Fig. (12.6) 0s, Given a surface 5 € Fog, S induces 228 1 nonorientable surfaces. Let Ng be the set of all nonorientable surfaces induced by S. then the polynomial ós(y) = X. INS) (12.7) 1<j<6 is called the nonorientable form of S where N;(S) is the subset of Ns, se For a graph G with Betti number 5, the set of all associate ori- entable surfaces of determined by joint trees of G is denoted by S(G). XIL4 Crosscap polymomials 291 Let S5;(G) for 6 € As be the subset of S(G) with nonorientable form ô where As is the set of all nonorientable forms of surfaces in S(G). Theorem 12.8 The crosscap polynomial of a graph G is v(G;y) = X. |Ss(G)|d(y). (12.8) EAs Proof From (12.4) and (12.7), the theorem is deduced. O Theorem 12.9 Ifa nonorientable associate surface of a graph is in form as ArBzxC, then its genus is k, k > 1, if, and only if, the genus of AB-1C is l k — 1, if AB-!C is nonorientable; k—1 12.9 , otherwise. 2 Proof From Relation 2 in I.2, the theorem is soon found. [] According to Theorem 12.9, a linear time algorithm can also be designed for determining the genus of a surface and then classify nonorientable associate surfaces of a graph by genus. Hence, the cross- cap polynomial expressed by (12.8) can soon be found. Activities on Chapter XII XII.5 Observations O12.1 For the bouquet of size 3, list all its associate surfaces and then classify them by genus, orientable and nonorientable. O12.2 For K,, list all of its associate surfaces and then classify them by genus, orientable and nonorientable. O12.3 Compare the two sets of associate surfaces obtained in 012.1 and 012.2. O12.4 Observe how many layer divisions of the surface (aa bb lec). O12.5 List all orientable surfaces of 4-gons. O12.6 For each orientable surface obtained in O12.5, find its nonorientable form. O12.7 List all orientable surfaces of 6-gons. O12.8 For each orientable surface obtained in O12.7, find its nonorientable form. 012.9 Find the nonorientable form of surface (aa !bb cc). O12.10 Find the nonorientable form of surface (abca 1b 1c 5). O12.11 Find the nonorientable form of surface (abcc b ta~t). XII.6 Exercises 293 XILO Exercises A graph is called a necklace if it is Hamiltonian and the result of deleting all 2-edges is a perfect matching. E12.1 Find the handle polynomial of a necklace with order 2n, no. E12.2 Find the crosscap polynomial of a necklace with order 2n, n > 2. E12.3 Show that the nonorientable form of surface (a,b, +++ dnbna b an n) for n > 1 is a((1+ z)?^ — z?"), E12.4 Show that the nonorientable form of surface (a1a9 os aiu ts a,') for n > 1 is z((1-4 zr)" — x"). E12.5 Show that the nonorientable form of surface (aya; asaz eH ana) for n > 4 is (1-F x)" — 1. E12.6 For all surfaces of 20-gons, 6 > 4, with genus 0, Show that their nonorientable forms are (1-4 x)? — 1. A graph is called a ladder if it has a Hamiltonian circuit and all edges not on the circuit are geometrically parallel. E12.7 Find the handle polynomial of a ladder with m edges not on the Hamiltonian circuit. 294 Activities on Chapter XII E12.8 Find the crosscap polynomial of a ladder with m edges not on the Hamiltonian circuit. A graph is called a Ringel ladder if it is cubic without multi- edge and consists of a ladder with the two 2-edges each of which is subdivided into 3 edges by two vertices and the other two edges are interlaced on the hamiltonian circuit E12.9 Find the handle polynomial of a ladder with m edges not on the Hamiltonian circuit. E12.10 Find the crosscap polynomial of a ladder with m edges not on the Hamiltonian circuit. XII.7 Researches R12.1 Find the handle polynomial of the bouquet B, of size m, m > 1 by joint tree model. R12.2 Find the crosscap polynomial of the bouquet of size m, m > 1 by joint tree model. R12.3 Find the handle polynomial of the wheel W,, of order n, n 2 4 by joint tree model. R12.4 Find the crosscap polynomial of the wheel W,, of order n, n 2 4 by joint tree model. R12.5 Find the handle polynomial of the complete graph Kn of order n, n > 4 by joint tree model. R12.6 Find the crosscap polynomial of the complete graph Kn of order n, n > 4 by joint tree model. R12.7 Find the handle polynomial of the complete bipartite graph Km,n of order m+n, m,n > 3 by joint tree model. R12.8 Find the crosscap polynomial of the complete bipartite graph Km,n of order m+n, m,n > 3 by joint tree model. R12.9 Find the handle polynomial of the n-cube of order n, XII.7 Researches 295 n > 3 by joint tree model. R12.10 Find the crosscap polynomial of the n-cube of order n, n > 3 by joint tree model. R12.11 For the n-cube Q,, n > 3, prove that the minimum genus Yn = Jmin(Qn) with 7,_1 satisfies the relation Grint Qn) — 2" (n — 3) T Jmin(Qn-1) from an associate surface of Qn-ı with genus 7y,_; to get an associate surface of Qn with genus Yn. R12.12 For the complete bipartite graph Am», M > n > 4, prove that the minimum genus Ym,n = Jmin(Km,n) with Ymn—1 satisfies the relation ool Pas) = 2 T T TE, = where EE, m= 00 fn), 1(2 Mn: 2n, 2117/2), : 3(2 /n,2 f[n/2]; 2/n, 2] |n/2 |); [um = 2 m = 2(mod 4); 4 [== |, m - 02n), 12h. 2 Mn/21), 3(2 Jn, 2|Lm/2];2In,2 Jln/2]) from an associate surface of Km n-1 with genus Ym,n-1 to get an asso- ciate surface of Km, n with genus Ym,n- R12.13 For the complete graph K,, n > 5, prove that the minimum genus ^; = Jmin( Kn) with *, 4 satisfies the relation n— 4 Gal Ka) E ( 6 ) ale Grint A usd) 296 Activities on Chapter XII where P5 n= 230 fLn/6)),3(2ILn/6]), 5Q f1n/6]) pes Ed n = 4(mod 6); |^ e zn n = 0,1(2|[n/6]), 3(2 J|n/6]), 5(2]|n/6]) from an associate surface of KK, , with genus *,.., to get an associate surface of K, with genus Yn. Chapter XIII Census with Partitions e The planted trees are enumerated with vertex partition vector in an elementary way instead as those methods used before. e A summation free form of the number of outerplanar rooted maps is derived from the result on planted trees. e On the basis of the result for planted outerplanar maps, the num- bers of Hamilton cubic rooted maps is determined. e The number of Halin rooted maps with vertex partition is gotten as a form without summation. e Biboundary inner rooted maps on the sphere are counted by, an explicit formula with vertex partitions. e On the basis of joint tree model, the number of general rooted maps with vertex partition can also expressed via planted trees in an indirected way. e The pan-flowers which have pan-Halin maps as a special case are classified according to vertex partition and genus given. XIIL.1 Planted trees A plane tree is such a super planar rooted map of a tree. A planted tree is a plane tree of root-vertex valency 1. In Fig.13.1, (a) shows a plane tree and (b), a planted tree. 298 Chapter XIII Census with Partitions Let T be a planted tree of order n with vertices vo, v1, U2, --, Un, Nn > 1, where vo is the rooted vertex. The segment recorded as travelling along the face boundary of T from vp bach to itself and then vg left off is called a V-code of T when v; is replaced by i for i—1,2,---,n as shown in Fig.13.2 and Fig.13.3. (a) A plane tree (b) A planted tree Fig.13.1 Plane tree and planted tree A sequence of numbers is said to be polyhedral if each adjacent pair of numbers occurs twice. It is easily seen that a V-code of a planted tree is a polyhedral segment. The vector n = (n4,n5,::-,ni,:--), where n;(? > 1) is the number of unrooted vertices of valency 2, is called the vertex partition of a planted tree. For a sequence of nonnegative integers n4, no, * -:, ni, --- denoted by a vector n = (n1, n2,- , Ni, +), if 3. 2- ini = 1, (13.1) i>1 then n is said to be feasible. Let n = (nj, n5, ma) XIII.1 Planted trees 299 where nı — 1, when: = l; ng—-ı + 1, when; = k — 1; yg um | (13.2) ny —1, when t = k; nj, otherwise for a k > 2 and ni, ng > 0, then n' is called a reduction of n. Lemma 13.1 A reduction n’ of a sequence of nonnegative in- tegers n is feasible if, and only if n is feasible. Proof By considering (13.2), we have the eqaulity as 3. Q-)ni- M (2-)n = -1 — (2 — k) + (2- k1) 20. i>1 i21 ' his leads to the lemma. [] The sequence n = (1, 1) is feasible but no reduction can be done. So, it is called irreducible. Lemma 13.2 Any feasible sequence n has nı > 0. Proof By contradiction. Suppose n is feasible but n; — 0. Be- cause of 3. 2-ini2 (2- ini x 0. i21 i22 This contradicts to (13.1), the feasibility. O Lemma 13.3 Any feasible sequence n # ngo can always be transformed into n, only by reductions. Proof Because of n Z ny, Lemma 13.2 enables us to get a reduc- tion. Whenever the reduction is not ng, another reduction can also be done from Lemma 13.1. By the finite recursion principle, the lemma is done. [] Theorem 13.1 For a nonnegative integer sequence n = (n4, n, s Ni, +++), there exists such a planted tree that n; unrooted vertices are of valency i(i > 1) if, and only if, n is feasible. 300 Chapter XIII Census with Partitions Proof Necessity. Suppose T' is such a planted tree. Because n; is the number of unrooted vertices with valency i, i > 1 in T, the size of T' is Y i21 1c ini 22V 7n. i21 i21 and hence This means that n satisfies (13.1), i.e., n is feasible. Sufficiency. First, it is seen that the irreducible sequence is the vertex partition of the planted tree whose under graph is a path of two edges. Then, by following the inversion of the procedure in the proof of Lemma 13.3, a planted tree with a given feasible sequence can be found. [] For a polyhedral segment L with 1 as both starting and ending numbers on the set N = {1,2,3,---,n}$, n > 1, let the vector be the point partition of L where n; be the number of occurrences of i in L, i >l. In a polyhedral segment L, if vuv is a subsegment of L, then u is said to be contractible. The operation of deleting u and then identify- ing v, or in other words vuv is replaced by v, is called contraction. If L can be transformed into a single point, then L is called a celluliform. If the point partition of L satisfies (13.1), then L is said to be feasible as well. It can be seen that any celluliform is a feasible segment but con- versely not necessary to be true. In what follows, the notation bellow is adopted as O- Cm) HC) o enm where s > 2, n; > 0 are all integers and 8 i-1 n= ) Ni, Oj-1 = X nj. i=1 j=1 XIII.1 Planted trees 301 Notice that when s — 2, it becomes the combination of choosing ni from m. Example 13.1 In Fig.13.2, two distinct planted trees of order 5 are with vertex partition n = (3,0,2) satisfying (13.1). (a) is with sequence 123242151 and (b), 121343531. Here, 1 5 1 5! 5 \3,0,2 ~ 531021 ang (a) 123242151 ) 121343531 Fig.13.2 Trees with n = (3,0, 2) In Fig.13.3, six distinct planted trees of order 5 shown by (a-f) are with vertex partition n = (2,2, 1) satisfying (13.1). Here, lf 5 1 5! 5221) 522m1 For a feasible segment of numbers on N, the occurrences of i € N divides the segment into sections in number equal to that of times of its occurrences. Each of the sections is called an i-section. If a feasible segment on N is with the property that all numbers less than 7 have occurred before the first occurrence of i, 1 <i € n, then it is called favorable. Denote by 11 a 1-to-1 correspondence between two sets. Lemma 13.4 Let 7, be the set of all planted trees of order n+1 with vertex partition n and £n, the set of all favorable celluliforms on N with point partition n, then J, 1&1 Lp. 302 Chapter XIII Census with Partitions Proof Necessity. For T € 7,, it is easy to check that its V-code A(T) is uniquely a favorable celluliform, i.e., u(T) € Ln. Sufficiency. Let u € £n. Because of the uniqueness of the great- est point which is contractible, a point can be done by successfully contracting the greatest points. By reversing the procedure, a tree T(u) € 1, is done. O Theorem 13.2 The number of nonisomorphic planted trees of order n + 1 with vertex partition n is where n! = | |m! (13.5) (a) 123432521 (b) 1232345421 (c) 123435321 (d) 123214541 (e) 121345431 (f) 123432151 Fig.13.3 Trees with n = (2, 2, 1) XIII.1 Planted trees 303 Proof On the basis of Lemma 13.4, it suffices to discuss the set of all favorable celluliforms £,. Since each favorable celluliform has n possibilities to choose the minimum point and different possibilities correspond to different ways of choosing n from n elements, the set of all ways is partitioned into classes. A way is represented by number sequence of length n with repetition as occurrence in the natural order. Two ways A and B are equivalent if, and only if, there exists a number i € N such that A + i(mod n) is B in cyclic order. A way starting from 1 is said to be standard. Because of each class with n ways in which only the standard way enables us to form the V-code of a planted tree, the theorem is soon obtained. [] In Example 13.1, Fig.13.2 and Fig.13.3 show two cases of (13.4). Only take n = (3,0, 2) as an example. There are 10 ways of combina- tions of choosing 2 points with 3 occurrences each and 3 points with 1 occurrence each from 5 points numbered by 1,2,3,4 and 5 as 1) 111222345; (2) 111233345; (3) 111234445; 4) 111234555; (5) 122233345; (6) 122234445; 7) 122234555; (8) 123334445; (9) 123334555; 10) 123444555 "m LOS LS OS in which 2 classes are divided as C = {(1), (5), (8), (10), (4)} and C5 = {(2), (6), (9), (3), (7)} because of (5) = 222333451, (8) = 333444512, (10) = 444555123 and (4) = 555111234 as (1) = 111222345 for Ci, and the like for C5. For a general outerplanar rooted map M = (Xa B(X), P) with (r)p, on the specific circuit where r is the root and y = ag and its dual M* = (Xa4(.X), Py) with root r as well without loss of generality, let Hy be the map obtained from M* by transforming the vertex (r)p, of M* into vertices (r), ((Py)r), +++, (Py) !r). Such an operation is called articulation. The root rg of Hm is taken r as shown in 304 Chapter XIII Census with Partitions Fig.13.4 in which bold lines are on M and dashed lines, on Hm. Here, multiedges are permitted. Fig134 M ana Hy Further, it is easily checked that Hy), is a planted tree of size which is equal to the size of M. Lemma 13.5 An outerplanar rooted map M of order n 4- 1 with face partition s is 1-to-1 corresponding to a planted tree Hm with vertex partition t = s + nl,. Proof By the procedure of getting Hm from M, it is seen that the number of 7-faces in M is the same as that in Hy for i > 1. For i = 1, Hy has n — 1 articulate vertices greater than sı. In virtue of the nonseparability of M, sı = 0. Conversely, it is still true and hence the lemma. [] An attention which should be paid to is that all articulate edges in Hymy are 1—to-1 corresponding to all edges on the root-face boundary of M. Theorem 13.3 The number of nonisomorphic outerplanar rooted maps of order n with face partition s is ee en aso where s, +n — n, i.e., 5; = 0, because of no articulate vertex and s is the number of unrooted faces. Proof On the basis of Lemma 13.5, the theorem is obtained from Theorem 13.2. [] XIIL2 Hamiltonian cubic maps 305 Since a bipartite map has all of its faces of even valency, its face partition s is of all s; = 0 when 7 is even. Corollary 13.1 The number of nonisomorphic outerplanar rooted bipartite maps of order 2m with face partition s is ji —1 ————— iui (13.7) 2m 4- s 1 Ns 4- (2m — 1)1, where s, +n — n, i.e., 5; = 0, because of no articulate vertex and s is the number of unrooted faces. [] A map is said to be simple if it has neither selfloop nor multiedge. Corollary 13.2 The number of nonisomorphic outerplanar rooted simple maps of order n with face partition s is 1 ire — 1 —_ 13.8 m Me nm Es where s, +n — n, i.e., 5; = 0, because of no articulate vertex and s is the number of unrooted faces. [] Corollary 13.3 The number of nonisomorphic outerplanar rooted bipartite maps of order 2m with face partition s is 1 —1 ————— E (13.9) 2m 4- s — 1 Xs 4- (2m — 1)1, where s, +n — n, i.e., 5; = 0, because of no articulate vertex and s is the number of unrooted faces. [] XIII.2 Hamiltonian cubic map For saving the space occupied, this section concentrate to discuss on Hamiltonian planar rooted quadregular maps as super maps of a Hamiltonian planar graph, and then provide a main idea for general 306 Chapter XIII Census with Partitions such maps. A map is said to be quadregular if each of its vertices is of valency 4. A Hamiltonian planar rooted quadregular map with the two edges not on the Hamiltonian circuit not success in the rotation at each vertex is called a quaternity. Let Mi = (Xa vel X1), J1) and Mz = (Xa vel X2), J2) be two rooted maps with their roots, respectively, rı and ra. Assume (rı)zy and (72) my are with the same length. The map obtained by identifying rı and arg with Kr; = Karo as well as (J1y)'ri and (aJey)'r2 for i > 1 is called the boundary identification of Mj and M», denoted by I(Mi, Mz). The operation from Mı and M» to I(Mi, Mə) is called boundary identifier. A boundary identification of two outerplanar cubic rooted maps is a quaternity because of Mı and Mə both outerplanar and cubic with its root rı = rə. Lemma 13.6 Let O, and Z, be the sets of all, respectively, quaternities and boundary identifiers with face partition s, then there is a 1-to-1 correspondence between Q, and Z;. Proof By considering the inverse of a boundary identifier, a qua- ternity becomes two cubic outerplanar maps whose boundary identi- fication is just the quaternity with the same face partition s. This is the lemma. [] From the proof of this lemma, it is seen the identity Lemma 13.7 The number of nonisomorphic outerplanar cubic rooted maps of order n with face partition s is crei (non) sa for s € Scu», the set of all the vectors available as the face partition of an outerplanar cubic map. XIIL3 Halin maps 307 Proof From Theorem 13.3, the conclusion is true. [] Theorem 13.4 The number of nonisomorphic quaternities of order n with face partition s is p OMGE) v. $1:52€ cub 8—8]-82 where a; — |s;|, called the absolute norm of s;, i.e., the sum of all the absolute values of entries in s; for i = 1,2. Proof Sincethe set of all quaternities of order n is the Cartesian product of the set of all cubic outerplanar rooted maps and itself, the formula (13.12) is soon obtained. [] 'This method can be also employed for the case when the bound- ary is cubic and further for others with observing boundary combina- torics. XIII.3 Halin maps If a graph can be partitioned into a tree and a circuit whose vertex set consists of all articulate vertices of the tree, then it is called a Halin graph. A planar Halin map is a super map of a Halin graph on the surface of genus 0 such that the circuit forms a face boundary. Let H = (X,,5(X), J) be a planar Halin rooted map with (r) gy, y = af, as the face formed by the specific circuit where r is the root. The associate planted tree denoted by Ty is obtained by deleting all the edges Kr, K(Jy)r, +++, K(Jy) 1r on the circuit. Lemma 13.8 A planar Halin rooted map with vertex partition u of the specific circuit with length n is 1-to-1 corresponding to a planted tree with vertex partition v = u + (n — 1)(1, — 1). Proof By considering the procedure from a Halin map H to a tree 308 Chapter XIII Census with Partitions Ty, carefully counting the numbers of vertices with the same valency and comparing them of H with those of Ty, the lemma is found. OU Theorem 13.5 The number of nonisomorphic planar Halin rooted map with vertex partition u of the specific circuit with length "C 1 lu 4-n—1 AH) = Tena " (n — 1 a) Hey where |u| is the absolute norm of u. n is Proof On the basis of Lemma 13.8, by Theorem 13.2, the con- clusion of the theorem is done. [] Let Hy = (%o,9(X1), A) and Hə = (Xo,3(X2), J2) be two planar Halin rooted maps with |{ri}7,4| = {r2} nyl, the boundary identifica- tion of Hı and H^» is called a double leaf. A graph with a specific circuit of all vertices of valency 4 is called a quadcircularity. A super map of a quadcircularity is a quadcircula- tion. Lemma 13.9 A planar rooted quadcirculation M is a double leaf if, and only if, the map obtained from M by deleting all edges on the specific circuit can be partitioned into two trees such that each of vertices on the circuit is articulate of both the trees. Proof Since a double leaf is obtained by boundary identifier from two Halin maps, the conclusion is directly deduced. [] Lemma 13.10 A planar rooted quadcirculation with vertex partition u of the specific circuit of length n is 1-to-1 corresponding to a pair of planar Halin rooted maps H, and Hə with vertex partitions, respectively, s and £ such that u =s +t-— (n — 1)(21, — 14). (13.14) where 1; is the vector of all entries 0 but the i-th 1 for i = 3,4. Proof By considering that u does not involve n — 1 unrooted XIIL3 Halin maps 309 3-vertices in s and £ each and involves n — 1 unrooted 4-vertices, the formula (13.14) holds. O Theorem 13.6 The number of nonisomorphic double leafs with vertex partition u of the specific circuit of valency n is ` A(HS) ACHL) (13.15) S-t—u-(n—1)(213—14) 8,t€Sq] where Sq is the set of vectors available as vertex partitions of planar Halin maps. Proof On account of Lemma 13.10, the theorem is soon derived from Theorem 13.2 for the Cartesian product of two sets. [] Given a nonseparable graph G with a cocircuit C* of an orien- tation defined, if G is planar in companion with such a orientation then G is said to have the C*-oriented planarity, or cocircuit oriented planarity. A planar super map of such a graph is called a cocircuit oriented map. If each edge on the cocircuit is bisectioned and then snip off each new 2-valent vertex as two articulate vertices in a cocir- cuit oriented map M so that what obtained is two disjoint plane trees, then M is called a cocircular map. The root is always chosen to be an element in an edge on the cocircuit in a cocircular map. Lemma 13.11 A cocircular map with the oriented cocircuit of n+ 1 edges and the vertex partition u is 1—to-1 corresponding to a pair of planted trees (T1, T5) with vertex partitions u and us such that uj; = uU», = n. Proof By considering the uniqueness of a cocircular map com- posed from two planted trees, the coclusion is directly deduced. UO Let U, be the set of all integer vectors feasible to a planted tree with n unrooted articulate vertices. Theorem 13.8 The number of nonisomorphic cocircular maps 310 Chapter XIII Census with Partitions with the oriented cocircuit of n edges and given vertex partition u is iua IY /lul (a) m Uy 42 €Un uy | [us] uj tug=u Proof Based on Lemma 13.11, the formula (13.16) is derived from Theorem 13.2. L A cocirculation is such a planar rooted map which has a cocircuit oriented. For this type of planar maps, the number of nonsepsrable ones can be determined from maps with cubic boundary of root-face. More interestingly, maps with cubic boundary of root-face can be transformed into maps with root-vertex valency as a parameter. In view of this, many types of planar maps with cubic boundary can be known from what have been done for counting maps with size and root-vertex valency as two parameters. XIIL4 Biboundary inner rooted maps A map is said to be biboundary if it has a circuit C that two trees are obtained by deleting all the edges on C. In view of this, a Hamilton cubic map is a uniboundary map because it is not necessary to have two connected components as all the edges on the Hamiltonian circuit are deleted. Here, only planar case is considered. LetM = (X, 7)be a biboundary map, r = r(M) is its root. The length of boundary is m and the vertex partition vector of nonbound- ary vertices is n = (ni,n5,-:--), nj, i > 1, the number of i-vertices not on the boundary. Assume M; = (4, 41) and My = (A5, J) are two submaps of M. Denote byC = (Kr, Kgr,-:-,Ky™ 'r) the boundary circuit of M where i J^yx;, if yx; not incident with an inner edge of Mo; pti = aT Dxz;, otherwise. XIII.4 Biboundary inner rooted maps 311 To= TIm =T and 2z;=y'r,i=1,2,---,m—1. Let rı = r(Mi) =r and ra = r(M2) = aq" tr. This means the root-vertex of Mı adjacent to that of M» in M. Such a root is said to be inner rooted. First, denoted by Bm, m > 6, the set of all biboundary rooted maps with the boundary lengthm. Lemma 13.12 Let Wm, and Wm, are, respectively, unibound- ary maps of boundary lengths mı > 3 and mə > 3, then a pair of (WA, Wo}, Wi € Wm, and Ws € Wma, m = m; + Mo, composes of mı Sm (Mı) = M (13.17) biboundary maps in B,,. And, this combinatorial number is deter- mined by the recursion as Tn» os Mə — 1 [ims [=|] mz d (13.18) 1 Ma =1, m 21; FA =1, m = 0. Proof By induction on m», m» > 2, for any m; > 1. First, check the case of my = 2, for m, > 1, that [2] -[] [o] [2 Jem From the fact that if o vertices in the first segment, then the second segment has to have mj, vertices; if 1 vertex in the first segment, then the second segment has to have m, — 1 vertices; ---; if mı vertices in the first segment, then the second segment has to have 0 verticesm;. They are all together m; 4- 1. Thus, (13.18) is true for m» — 2. Then, assume £4, 1(mij), m2 > 3, have been determined by (13.18). To prove tm,(mi) is determined by (13.18). Because of mı + 1 occurrences for putting mı vertices in to m» segments as 312 Chapter XIII Census with Partitions when m vertices in the first segment, then no vertex in all other m» — 1 segments and hence 5,,, 1(0) ways; when mı — 1 vertices in the first segment, then i vertex in all other m; — 1 segments and hence Sm,-1(1) ways; ---; when 0 vertices in the first segment, then m; ver- tices in all other m» — 1 segments and hence s,,, 1(mm1i) ways. They are Sm (M1) = 357 9 Sm—1 (i) ways all together. That is My mə | _ Mə — 1 [m SE r | mza i=0 By the induction hypothesis, Sm (mı) is determined. The lemma is true. g Then, denote by Dm, m > 6, the set of all biboundary rooted maps with the boundary length m. Lemma 13.13 Let Mm, and Mm, be the uniboundary rooted maps of boundary lengths, respectively, mı > 3 and mə > 3, then a pair {M), Mo}, Mı € Mm, and M» € Mm, composes Ls (mi) = ( ri ) (13.19) biboundary rooted maps in Dm, Mm = m44-m». And, this combinatorial number is determined by m \ _ = my —1 (m)-X| ; i; (13.20) i=0 where the terms on the right hand side of (13.19) are given in Lemma 19.13. Proof By induction on mg for any mı > 1. First, when m» = 2, By considering for assigning rm vertices in M» edges on the boundary in the order determined that when 1, 2, -+, my, vertices in the first edge(incident with the root), the second edge has to have, respectively, mı — 1, m4 — 2, ---, 0 vertices, we have to(m) = s1(0) + si(1) Tec si(mi — 1) = mı. XIII.4 Biboundary inner rooted maps 313 Then, assume £4, 1(mij), m2 > 3, have been determined by (13.20). To prove that tma (m1) = $m4-1(0) + Sma—1(1) = Pee $m;-1(mà 2 b is determined by (13.20) as well. Because of the first edge in M» edges allowed to have mi, mı — 1, ---, 1 vertex, the other mə — 1 edges only allowed to have, respectively, 0, 1, ---, mı — 1 vertex. This implies mM — 1 mM — 1 mM — 1 m epe pem mi] Hence, (13.20) is right. [] Denote by Qi, 1 € i € tm (Mı) the t4,(mi) biboundary inner rooted maps mentioned in this lemma with M (Mi, M5) = (Qi, Qs, +>, Qi, (m)]- Lemma 13.14 Let Hm = ( M(Mi, M3)|IV(Mi, M») € Mm, x M miz m + M2 = my}, then Des wee (13.21) HEH i.e., Hm is a partition of Dm. Proof For any D € Dm, it is known from biboundary maps that there exist mı and m2, Mı + Mə = m, such that Mı € Mm, and M» € Mm, compose of D. Thus, D = (Mi, M3) € Hin. Conversely, for any Q € H, H € Hm, because of Q composed from two uniboundary maps M; € Mm, and M» € Mma, m, + m2 = m, there exist D € Dm such that D = Q. In summary, the lemma is obtained. [] In what follows, observe how many nonisomorphic uniboundary maps of boundary length m with vertex partition vector n. Lemma 13.15 The number of uniboundary rooted maps of boundary length m, m > 3, and nonboundary vertex partition vector 314 Chapter XIII Census with Partitions (m t n — 1)! "im — Tin (13.22) n(m, n) = where n = |n| 2 n1 +n +--+.. Proof Let M = (X, J) bea uniboundary rooted maps of bound- ary length m, m > 3, and nonboundary vertex partition vector n. Its root is r. Because of cubicness on the boundary, Jr is incident with an articulate vertex of the tree. Let Jr be the root to make the tree planted. Because of all 1-vertices of the planted tree on the boundary, its vertex partition vector is n+ (m — 1)1,, where 1, is the vector of all entries 0 but the first entry 1. Since a planted tree with vertex parti- tion is 1-to-1 corresponding to a uniboundary rooted map of boundary length m, m > 3, and vertex partition vector n, from Theorem 13.2, the number of nonisomorphic uniboundary rooted maps f boundary length m, m > 3, and vertex partition vector n is (om (m t n — 1)! nh, bh) = 2 aaa. S XT ý (m +n — 1)!(n — m1)! where n = |n| = ni + n3 4- ---. By considering that nı = 0 in n, the lemma is done. C On the basis of the above tree lemmas, the main result of this section can be gotten. Theorem 13.9 The number of biboundary inner rooted maps of boundary length m > 6 and nonboundary vertex partition is Y Mə (n! s TI — 1)! plz y p Mao 13.23 m (n? + m3 — 1)! E (mg = 1)!n?! where L = ((r4, m», nl, n?)| m; + ma = m, n! + n? = n, m4, m; > 3). Proof For any given m, and m», mı +M = m, with n! and n?, XIIL5 General maps 315 nl +n? =n, from Lemma 13.14-15, Dm can be classified into 5 (n! 4- m4 — 1)! (n? 4- m5 — 1)! (m; — l)mn!! (me — 1)!n?! (m; ,ma,n!,n?)e£ classes. From Lemma 13.13, each class has Cm) nonisomorphic biboundary inner rooted maps of boundary length m and nonboundary vertex partition vector n. Thus, the theorem is proved. L XIII.5 General maps Based on the joint tree model shown in Chapter XII, it looks general maps on surfaces in a closed relation with joint trees. In this section, only orientable case is considered as an instance. Because of the independence with a tree chosen, general maps with a cotree marked are particularly investigated. For the convenience for description, all maps are assumed to have no articulate edge. Let M = (X, J) be a map with cotree edges aj = Ka, a2 = Kx, ---, aj = Ka; marked where | = (M) is the Betti number of M. The root of M is chosen on a cotree edge, assume r = r(M) = x4. Another map Hy = (Xy, Jz) is constructed as l Xa =X +Y (Ksi + Kt; — ai) (13.24) i=1 where Ks; = zi, oui, 95i, ysjsand Kt; = {yz;, 82i, Bt, yu}, 1 <1 < l, ; Jg is defined as T { (x)z, when x € X; (x), when z ¢ X. (13.25) 316 Chapter XIII Census with Partitions Lemma 13.16 For any rooted map M with a cotree marked, the map Hy is a planted tree with the number of articulate vertices two times the Betti number of M. Proof Because of connected without circuit on Hy, Hy is a tree in its own right. Since the number of cotree edges is the Betti number of M, from the construction of Hm and no articulate edge on M, the number of articulate vertices on Hm is two times the Betti number of M. go Let M (l; n) be the set of all general rooted maps with a cotree of size | marked and vertex partition n including the root-vertex. And, let H(n) be the set of all planted trees with articulate vertices two times the number of cotree edges and vertex partition n excluding the root-vertex. Lemma 13.17 There is a 1-to-1 correspondence between M (l; n) and H(n + (21 — 1)1,) as the set of joint trees. Proof For M € M(l; n), it is easily seen that the corresponding Hy is just a joint tree of M and hence Hy € H(n + (21 — 1)1,). Conversely, for H € H(n+(2l—1)1,), in virtue of a joint tree with its articulate vertices are pairwise marked as cotree edges of the cor- responding map M as H — Hy, by counting the valencies of vertices, it is checked that M € M(l; n). Therefore, the lemma is true. [] This lemma enables us to determine the number of general rooted maps with a cotree marked with vertex partition given. Theorem 13.10 The number of rooted general maps with a cotree marked for a vertex(root-vertex included) partition n given is (n + 21 — 2)! ME n)| = “Ol iym ^ (13.26) where | is the Betti number(the size of cotree) and n = |n]. Proof A direct result of Lemma 13.17 and Theorem 13.2. OU XIII.G Pan-flowers 317 XIIL.G Pan-flowers A map is called a pan-flower if it can be seen as a standard petal bundle added a tree such that only all vertices are in the inner parts of edges on the petal bundle. The petal bundle is seen as the boundary of a pan-flower. Because of not a circuit for the base graph of a petal bundle in general, a pan-flower is reasonably seen as a generalization of a map with a boundary, or a boundary map. A pan-Halin map is only a special case when the petal bundle is asymmetric. For convenience, the petal bundle in a pan-flower is called the base map. If a edge of a fundamental circuit on the underlying graph of base graph is allowed to have no articulate vertex of the tree, then the pan-flower is said to be pre-standard. If the edge has at least one articulate vertex of the tree, then the pan-flower is standard. This section is concerned with pan-flowers in the two classes foe a vertex partition vector given. Let Hpsu be the set of all rooted pre-standard pan-flowers, where the root is chosen to be an element incident to the vertex of base map. For any H = (X, J) € Hy, the tree Ty is always seen as a planted tree whose root is first encountered on the rooted face of H starting from the root of H. Otherwise, the first encountered at the root- vertex of H from the root. Lemma 13.18 Let Hpsu(p;s) be the set of all rooted pre- standard pan-flowers with vertex partition vector s = (s2, 53,-:-) ona surfaces of orientable genus p, then jı —Ópa jit 2p . |Hpsu(p; 8)| = 2/179» ie E P) A (13.27) where 71(j) is the set of planted trees with vertex partition vector j = (ju J2,°++), such that s; = ji, i A 3; s3 = Ja ja +1, pz 1,82 0, j > 0, but s # 0 and j z 0. Further, 6,1 is the Kronecker symbol, i.e., dpi = 1, when p = 1; 0p = 0, otherwise. Proof On the basis of pre-standardness, it is from the defini- tion of pan-flowers seen that an element in the set on the left hand 318 Chapter XIII Census with Partitions side of (13.27) has an element in the set on the right hand side in correspondence. In what follows, to prove that each T = (Vv, J) € 71(j), produces | +2 51-2 f d ) maps in Hpsu(p; s). 951— Denied by (r); (v1), (2), +++, (2j) all the articulate vertices of T, where 0 < h < ly < --- « l4, such that x; = (Jar, i = 1,,2,---, 91, r = r(T) is the root of T. First, by considering that the underlying graph of base map has 2p loops, only one vertex, its embedding on the orientable surface of genus 2p has exactly one face. Because of the order of its automor- phism group 8 when p = 1, only one possible way; 2p when p > 2, two possible ways. Then, the assignment of the jı + 1 articulate vertices, (r), (x;), i = 1,2,--.,7]1 of T on the base map has the number of ways as choosing 2p — 1 from 7, + 2 intervals with repetition allowable. That 1S A-2-t(2p-1)—-1Y _ fjit+2p 2p—1 2p—1/- Finally, since each of elements r, x;, i = 1,2,---, jı, has 2 ways: one side (ax, ax), or the other (x, Gx}, they have gjı +1 ways altogether. In summary of the three cases, The aim reaches at. [] On the basis of this lemma, by employing a result in 813.1, the following theorem can be deduced. Theorem 13.11 The number of pre-standard rooted pan-flowers with vertex partition vector s = (s2, 53,:--) and its base involving m missing vertices on an orientable surface of genus p, p > 1, is 2p —1 | pete d à 0) E (13.28) where s! = [[,.5 5;! = 59!53! --- and n+ 2 = $5.5 si. XIII.G Pan-flowers 319 Proof From Lemma 13.18, this number is - m=+2p-— 1 , 9 a pi )n@ where j = (jı, j2, ja, -) such that 7; +1 = m, j3 = 53 — m and ji = si, i 3, 22. Then from Theorem 13.2, we have (a —1)! — n! OR T a Dlss mia S3 \ n!m Am) s^ where n = n' — 1 = 905,91 — 1 = $5255; — 2. By substituting this into the last, (13.28) is obtained. E Let Tii: s) be the set of all pre-standard rooted pan-flowers with vertex partition vector s = (s2, 53, : --) on a nonorientable surface of genus q, s > 0, but s #0. Lemma 13.19 For Hps(q; 8), q > 0, s > 0, but s Z 0, we have mt+q-1 (Hpsu(q; s)| = z-ten[ ed Jm (13.29) where j = (m— 1, 52, 83—m, S4, 85, -) and m is the number of trivalent vertices(i.e., missing vertices on its base). Proof Lj Similarly to the proof of Lemma 13.18. However, an attention should be paid to that the size of the base is q, q > 1, instead of 2p, p > 1 and the order of automorphism group of the base is 2q when q > 2; 4 when q= 1. [] similarly to Theorem 13.11, we have Theorem 13.12 The number of pre-standard rooted pan-flowers with vertex partition vector s = (55,53,---) and its base involving m missing vertices on a nonorientable surface of genus q, g > 1, is jm-&i 1 (" Tue ) (5) nim. (13.30) q—-1 mj s! 320 Chapter XIII Census with Partitions where «F2 = » sys and 8| e [pga Proof Similarly to the proof of Theorem 13.11, from Lemma 13.19 and Theorem 13.2, the theorem is done. L For standard pan-flowers, let Hs be the set of all such maps. From the definition, if the base map has m missing vertices, then it is not possible on an orientable surface of genus greater then m/2, or on a nonorientable surface of genus greater than m. Lemma 13.20 Let Hsy(p;.s) be the set of all standard rooted pan-flowers with vertex partition vector s = (89, 83,---) on an ori- entable surface of genus p. If maps in H,y(p;s) have its base with m 2 2p missing vertices, then we have : — 9m—ôp 1+1 m= : Pax) zr img — asap where 7,(5), as above, is the set of panted trees with vertex partition vector j = (ji, Jo, ja.) for jj = m — 1, ja = 83 m, ji = si, i 3, $23 Proof For any H € Hsy(p;s), from the definitions of standard- ness and pan-flowers, it is seen that there exists a planted tree in 71(7) corresponding to H. Thus, it suffices to prove that any planted tree in T = (X, J) € T(J) produces Qm— pitt m — 1 2p—1 maps in Hsy(p; s). First, an attention should be paid to that maps in Hsn(p; s) are with their base of size 2p on an orientable surface of genus p. Since the order of automorphism group of the base is 4p when p > 2; 8 when p = 1, the base has, respectively, 2 ways when p > 2; 1 way to choose its root. Second, since the number of missing vertices on the base is m, T' must have m — 1 unrooted articulate vertices. Let them be incident with (21), +++, (@m_1). From the standardness again, there are m — XIIL6 Pan-flowers 321 1 intervals for choice in the linear order ((r), (x1), (£2), +++, (zi 1)). Thus, any 2p — 1 points insertion divides the linear order into 2p nonempty segments. This has m -—1 2p— 1 distinct ways. Third, notice that each of the m articulate edges including the root-edge in T' has 2 choices, and hence pL distinct choices altogether. In summary rom the three cases, the lemma is soon found. UO Based on this, we have Theorem 13.13 The number of standard rooted pan-flowers with vertex partition vector s = (s2, 53,:--) and their base map of m, m > 2p, unrooted vertices on an orientable surface Sp of genus p > 1, us m — 1X (/s3\nlm oF aa »" E ) (5) Fi (13.32) where n + 2 = 55.55; Proof Similarly to the proof of Theorem 13.11. However, by Lemma 13.20 instead of Lemma 13.8. [] At a look again for the nonorientable case. Lemma 13.21 Let H.y(q; s) be the set of standard rooted pan- flowers with vertex partition vector s = (s2, 53, --) on a nonorientable surface of genus q > 1. If each map in Hsy(q; s) has its base map of m unrooted vertices, then we have Han(q;s)| = 27-9 CNO (13.33) where n(j) zx A )L Jj = 63m Jaen A = m— L J3 = 63 — M, Jli = s 13, i22. 322 Chapter XIII Census with Partitions Proof Similarly to the proof of Lemma XII.2.3. However, what an attention should be paid to is the base of size q instead of 2p for a surface of genus q. [] Thus, we can also have Theorem 13.14 The number of standard rooted pan-flowers with vertex partition vector s = (s2, 53,---) and their bases of m un- rooted vertices on a nonorientable surface of genus q, q > 1, is E ! ee (s P) (5) ma (13.34) where s 20, $20, m2q21andn-c2-—95/.5si. Proof Similarly to the proof of Theorem 13.13. However, by Lemma 13.21 instead of by Lemma 13.20. [] Activities on Chapter XIII XIIL.7 Observations O13.1 Observe the number of plane rooted trees of size n > 0 given. O13.2 Observe the number of outerplanar rooted maps of size n 2 0 given. O13.3 Observe the number of wintersweets of size n > 0. O13.4 To show a relationship between outer planar maps and trees. O13.5 Observe how to evaluate the number of plane rooted trees with the number of articulate vertices m > 2 and size n > 1 given. O13.6 Consider what relation have the enufunction with the number of nonrooted vertices, nonrooted faces and the enufunction of vertex partition and genus as parameters. O13.7 Observe the number of rooted plane tree with root- vertex valency and vertex partition. O13.8 Observe the number of rooted plane trees with the num- ber of articulate vertices and vertex partition of other vertices. O13.9 Observe the difference between boundary maps and non- boundary maps. O13.10 observe the automorphism groups of a map and one of its boundary map by . 324 Activities on Chapter XIII XIII.S Exercises E13.1 Determine the enufunction of planted trees with vertex partition as parameters by establishing and solving a equation. A wintersweet is a rooted map with the property that it becomes a tree if missing circuits only at nonrooted terminal vertices of the tree. E13.2 Establish an equation satisfied by the vertex partition function of Wintersweets and then try to solve the equation. A rooted map is called unicyclic if it has only one circuit. E13.3 Establish an equation satisfied by the vertex partition function of unicyclic maps when the root is on a circuit and then try to solve the equation. E13.4 Establish an equation satisfied by the vertex partition function of Halin rooted maps and then try to solve the equation. E13.5 Establish an equation satisfied by the vertex partition function of outerplanar rooted maps and then try to solve the equation. E13.6 Establish an equation satisfied by the face partition func- tion of outerplanar rooted maps when the root is not on the circuit and then try to solve the equation. E13.7 Establish an equation satisfied by the face partition func- tion of planar rooted petal bundles and then try to solve the equation. E13.8 Establish an equation satisfied by the face partition func- tion of outerplanar rooted maps when the root is on the circuit and then try to solve the equation. E13.9 Establish an equation satisfied by the vertex partition function of unicyclic maps when the root is not on a circuit and then try to solve the equation. E13.10 Establish an equation satisfied by the face partition function of planar rooted supermaps of bouquets and then try to solve XIII.7 Researches 325 the equation. E13.11 Establish an equation satisfied by the vertex partition function of planar rooted maps with two vertex disjoint circuits and then try to solve the equation. XIII.9 Researches R13.1 Determine the vertex partition function of general rooted maps on the sphere. R13.2 Determine the vertex partition function of general rooted maps on all surfaces. R13.3 Determine the vertex partition function of general Eu- lerian rooted maps on all surfaces. R13.4 Determine the vertex partition function of 2-edge con- nected rooted maps on the sphere. R13.5 Determine the vertex partition function of 2-edge con- nected rooted maps on all surfaces. R13.6 Determine the vertex partition function of nonseparable rooted maps on the sphere. R13.7 Determine the vertex partition function of nonseparable rooted maps on all surfaces. R13.8 Determine the vertex partition function of loopless rooted maps on the sphere. R13.9 Determine the vertex partition function of loopless rooted maps on all surfaces. R13.10 Determine the vertex partition function of rooted tri- angulations on all surfaces. Chapter XIV Super Maps of a Graph e A semi-automorphism of a graph is a bijection from its semiedge set to itself generated by the binary group sticking on all edges such that the partitions in correspondence. e An automorphism of a graph is a bijection from the edge set to itself such that the adjacency on edges in correspondence. e The semi-automorphism group of a graph is different from its au- tomorphism group if, and only if, a loop occurs. e Nonisomorphic super rooted and unrooted maps of a graph can be done from the embeddings of the graph via its automorphism group or semi-automorphism group of the graph. XIV.1 Semi-automorphisms on a graph A pregraph is considered as a partition on the set of all semiedges as shown in Chapter I. Let G = (4¥,6;7) be a pregraph where X, ô and 7 are, respectively, the set of all semiedges, the permutation determined by edges and the partition on X. Two regraphs Gi = (4, ĉi; m1) and G5 = (X2, ô2; 75) are said to XIV.1 Semiedge automorphisms on a graph 327 be seme-isomorphic if there is a bijection 7 : YA; — A» such that X —————À A» "| |» (14.1) p) ———9 QA) are commutative for y = ô and m where m, is induced from 7 on ^1(41). The bijection 7 is called a semi-automorphism between Gi and Go. Example 14.1 Given two pregraphs G = (4, 61; 7) where 8 = S210, (D) à = [ [(@:(0), 2:1); m= {X)1<i<8} i with Xi = (1(0), 26(0), ze(1)}, X2 = {x1(1), v2 (0), z3(0)}, Xs = (z2(1), 25(0), 77(0)}, X4 = {x4(0), v5 (1); v7 (1), va(0) Xs = {x4(1), v5(1), x8(1)} and H = (YV, 69; 72) where y= Suo (1)}, d = | [(v0) (2); micis) i with Yı = {y4(0), ye(0), y7(O)}, Yo = {y5(0), yo(1), y7(1), ys(0)}, Ys = {y3(0), ys(1), ys(1)}, Ya = {ye(0), ys(1), ya) J, Ys = {y1(0), (1), y2(1)}- Let T: A — Y be a bijection with (14.1) commutative for y; = ôi, i = ] and 2, as X1 X2. X3 X4 UH X6 XT Tg doy2 Ó2U4 Y3 Ys Yo Yı Y7 Ys 328 Chapter XIV Super Maps of a Graph Because of TX; = {7x1(0), TX6(0), rzo(1)) = {y2(1), (0), y1(1)} = YS, ), ya(1), y3(1)} = Y ,7£5(0), 7£7(0)} = (y4(0), ye(0), v; (0)) = Yi, 9 8 N Qut, © n>" 4 8 w Qm. =) io od ww —— | mma m ea bo aet mss c 4 A | 4 —— 8 A c 8 A m nex” 8 p LL 8 a oO — —— = {yo(1), yr (1), ys (0), ys(O)} = Y2, TX5 = (7za(1), rz4(1), Teg(1)} = {y3(0), ys(1), us(1)) = Ys; we have 74,71 = TT, i.e., (14.1) is commutative for yi = 7;, i = 1 and 2. Therefore, 7 is a semi-isomorphism between G and H. Lemma 14.1 If two pregraphs G and H are semi-isomorphic, then they have the same number of connected components provided omission of isolated vertex. Proof By contradiction. Suppose G and H are semi-isomorphic with a semi-isomorphism 7 : G — H but G = Gi + Ga with two components: G4, and Gj and H, a component itself. From the the commutativity of (14.1), H has two components as well. This contra- dicts to the assumption that H is a component itself. [] If G — H, then a semi-isomorphism between G and H is called a semi-automorphism of G. Lemma 14.1 enables us to discuss semi- automorphism of only a graph instead of a pregraph without loss gen- erality. Lemma 14.1 allows us to consider only graphs instead of pre- graphs for semi-automorphisms. Moreover for the sake of brevity, only graphs of order greater than 4 are considered as the general case in what follows. Theorem 14.1 The set of all semi-automorphisms of a graph forms a group. Proof Because of all semi-automorphisms as permutations act- ing on the set of semi-edges, the commutativity leads to the closedness in the set of all semi-automorphisms under composition with the as- XIV.2 Automorphisms on a graph 329 sociate law. Moreover, easy to check that the identity permutation is a semi-automorphisms and the inverse of a semi-automorphism is still a semi-automorphism. This theorem holds. [] This group in Theorem 14.1 is called the semi-automorphism group of the graph. Example 14.2 In Example 14.1, the pregraph G = (X, 61; 77) is a graph. It is easily checked that _ ( T1 T2 T3 X4 X5 Te T7 Tg | _ = E fo U3 L4 Ly Ó]Xg T7 2» = (xe(0), xe(1)) is a semi-automorphism on G. It can also be checked that the semi- automorphism group of G is Autyi(G) = {7;|0 € i < 11} where To = 1, the identity; Ti = (£5, £7); que (£4, £8); T3 = (£5, £7) (24, £8); Ta = (£2, £3) (£4, 125) (zv, 01238); T5 = (£2, £3) (25, $118) (£7, 0174); XIV.2 Automorphisms on a graph Now, let us be back to the usual form of a graph G = (V, E) where V and E are, respectively, the vertex and edge sets. In fact, if X; as described in 8XIV.1 is denoted by v;, then V = (vo|i = 0.1.2. --) and E = (zj|j = 0,1,2,---}. An edge-isomorphism of two pregraphs G; = (Vj, Ei), i = 1 and 330 Chapter XIV Super Maps of a Graph 2, is defined as a bijection 7: E, —> E» with diagram FA —— , E» "| |” (14.2) V —“— VW commutative where 7;, i = 1 and 2, are seen a mapping 2^ — V;. When G = G4 = G^», an edge-isomorphism between G and G2 becomes an edge-automorphism on G. Lemma 14.2 If two pregraphs G and H are edge-isomorphic, then they have the same number of components provided omission of isolated vertex. Proof Similar to the proof of Lemma 14.1. [] This lemma enables us to discuss only graphs instead of pre- graphs for edge-isomorphisms or edge-automorphisms. Theorem 14.2 All edge-automorphisms of a graph G forms a group, denoted by Autee(G). Proof Similar to the proof of Theorem 14.1. [] Example 14.3 The graph G In Example 14.2 has its Aut..(G) = {7;|0 € i < 5) where To = 1, the identity; An isomorphism, or in the sense above vertez-isomorphism, be- tween two pregraphs G; = (Vj, Ei), i = 1 and 2, is defined as a bijection XIV.3 Relationships 331 T: Vj — V which satisfies that the diagram VW ———3À V2 «| e (14.3) Téi BiG Wx Vi ——— eV Ve is commutative where é;(v;) = Ein, for vj € Vi, i = 1 and 2. When G = Gi = Go, a isomorphism between G4 and G^ is called an automorphism of G. Lemma 14.3 If two pregraphs G and H are isomorphic, then they have the same number of components. Proof Similar to the proof of Lemma 14.2. [] This lemma enables us to discuss only graphs instead of pre- graphs for isomorphisms or automorphisms. Theorem 14.3 all automorphisms of a graph G form a group, denoted by Aut(G). Proof Similar to the proof of Theorem 14.2. [] The group mentioned in this theorem is called the automorphism group of G. Example 14.4 The graph G In Example 14.2 has its Aut(G) = {7;|0 € i < 1) where i To = 1, the identity; Tı = (X3, X5). Because of no influence on the automorphism group of a graph when deleting loops, or replacing multiedge by a single edge, 7;, ? — 1,2 and 3 in Example 14.3 are to the identity 7) and 74 and 75 in Example 14.3 to 7, here. 332 Chapter XIV Super Maps of a Graph XIV.3 Relationships Fundamental relationships among those groups mentioned in the last section are then explained for the coming usages. Theorem 14.4 Auty:(G) ~ Aute(G) if, and only if, G is loop- less. Proof Necessity. By contradiction. Suppose Autyr(G) ~ Autee(G) with an edge-automorphism 7 but G has a loop denoted by z — (z(0), z(1)). Assume 7(I) = l without loss of generality. However, both the semi-automorphisms: 7; and 7» corresponding to 7 are found as T(x), when x Z z; T(z) = 4 z(0), when x = z(0); 1), when x = z(1) ( and T(x), when x Z z; ( 0), when x = z(1); ), when x = z(0). This implies Autyr(G) 2^ Autee(G), a contradiction. Sufficiency. Because of no loop in G, the symmetry between two ends of a link leads to Autyr(G) ~ Autec(G). E mlr)= 4 z From the proof of Theorem 14.4, the following corollary can be done. Corollary 14.1 Let l be the number of loops in G, then Autne(G) ~ S. x Autee(G) where Sə is the symmetric group of order 2. Proof Because of exact two semi-automorphisms deduced from an edge-automorphism and a loop, the conclusion is done. L From this corollary, we can soon find autne(G) = 2! x autelG). (14.4) XIV.3 Relationships 333 Because of no contribution of a loop to the automorphism group of G, the graph G has its automorphism group Aut(G) always for that obtained by deleting all loops on G. Theorem 14.5 Aute.(G) ~ Aut(G) if, and only if, G is simple. Proof Because of no contribution of either loops or multiedges to the automorphism group Aut(G), the theorem holds. [] In virtue of the proof of Theorem 12.5, a graph with multi-edges G has its automorphism group Aut(G) always for its underlying simple graph, i.e., one obtained by substituting a link for each multi-edge on G. Lemma 14.4 Let G be a graph with i-edges of number nj, i > 2. Then, its edge-automorphism group Autee(G) = 5 m;S; x Aut(G) i»2 where S; is the symmetric group of order i, i > 2. Proof In virtue of S,, as the edge-automorphism group of link bundle Pm of size m, m > 2, the lemma is done. [] On the basis of Lemma 14.4, we can obtain Corollary 14.2 Let l and m; be, respectively, the number of loops and 7-edges, | > 1,2 > 2 in G, then autne(G) = 2'nmeaut(G) (14.5) where Nme = Mne(G) = Y ilm; i>1 which is called the multiplicity of G. Proof By considering Corollary 14.1, the conclusion is done. O 334 Chapter XIV Super Maps of a Graph XIV.4 Nonisomorphic super maps For map M = (Xas, P), its automorphisms are discussed with asymmetrization in Chapter VIII. Let M(G) be the set of all noniso- morphic maps with underlying graph G. Lemma 14.5 For an automorphism Ç on map M = (X45(X), P), we have exhaustively lax) € Autne(G) and Caļsx) € Autur(G) where G = G(M), the under graph of M, and 6(X) = X + 6X. Proof Because Xal X) = (X + 6X) + (aX + aó X) = 6(X) + aó(X), by Conjugate Axiom each Ç € Aut(M) has exhaustively two possibilities: C|5.x) € Autye(G) and Ga|;(x) € Autyr(G). O On the basis of Lemma 14.5, we can find Theorem 14.6 Let £,(G) be the set of all embeddings of a graph G on a surface of genus g(orientable or nonorientable), then the number of nonisomorphic maps in £,(G) is 1 = — X 14. where ®(7) = (M € €,(G)|7(M) = M or ra(M) = M}. Proof Suppose X1, X», ---, Xm are all the equivalent classes of X = €,(G) under the group Autye(G) x (a), then m = m,(G). Let S(x) = {T € Autye(G) x (a)| T(x) = x} be the stabilizer at z, a subgroup of Autyr(G) x (a). Because |Autar(G) x (o)] = ISG)I [Xi], x; E X;,1=1,2,---,m, we have m|Autur(G) x (a)] = 25 ISI Xi]. (1) XIV.4 Nonisomorphic super maps 330 By observing |S(x;)| independent of the choice of x; in the class X;, the right hand side of (1) is 2/1525; 2 1 rcx LEX TES (a E Y 2,1 (2) 7€Autne(G)x (a) z—r(x) = P5, Ie». T€ Autyrc(G)x (o) From (1) and (2), the theorem can be soon derived. O The theorem above shows how to find nonisomorphic super maps of a graph when the semi-automorphism group of the graph is known. Theorem 14.7 Let G be a graph with / loops and m; multi- edges of multiplier i and €,(G), the set of all embeddings of G on a surface of genus g(orientable or nonorientable), then the number of nonisomorphic maps in €,(G) is 1 m,(G) = Uns) P |®(r)| (14.7) qn Uthf where ®(7) = (M € £,(G)|T(M) = M or To(M) = M} and nme is the multiplicity of G. Proof On the basis of Theorem 14.6, the conclusion is soon de- rived from Corollary 14.2. [] Corollary 14.3 Let G be a simple graph. Then, the number of nonisomorphic maps in £,(G) is my(G) = zr PN le) (14.8 where ®(7) = (M € €,(G)|7(M) = M or rTa(M) = M} and nme is the multiplicity of G. 336 Chapter XIV Super Maps of a Graph Proof A direct result of Theorem 14.7 via considering G with neither loop nor multi-edge. [] XIV.5 Via rooted super maps Another approach for determining nonisomorphic super maps of a graph is via rooted ones whenever its distinct embeddings are known. Theorem 14.8 Fora graph G, let R,(G) and €,(G) be, respec- tively, the sets of all nonisomorphic rooted super maps and all distinct embeddings of G with size e(G) on a surface of genus g(orientable or nonorientable). Then, 2e(G) mRA(G)| ES TOL (14.9) Proof Let M,(G) be the set of all nonisomorphic super maps of G. By (11.3), we have RO=- D A MEM, (G) i.e., 4e(G) 2x autne(G) 2 x autne(G) MEM, (GC) aut(M) By considering that 2 x autn(G) = |Autr(G) x (a)| is |(Autne(G) x (0))|u] x [Autur(G) x (a) | and (Autne(G) x (0))|u = Aut(M) from Lemma 14.5, we have |R,(G)| is - 4O — : s |Autgr(G) x mi, 2 ^ tar(G) x (a) (M)| |. 2e(G) m autyr(G) |£4(G)]. XIV.5 Via rooted super maps 337 This is (14.9). O This theorem enables us to determine all the super rooted maps of a graph when the semi-automorphism group of the graph is known. Theorem 14.9 For a graph G with l, | > 1, loops and m; multi-edges of multiplier 7, i > 2, let R,(G) and £,(G) be, respec- tively, the sets of all nonisomorphic rooted super maps and all distinct embeddings of G with size e(G) on a surface of genus g(orientable or nonorientable). Then, «(G) 2! - 1n neaut(G) where Nme is the multiplicity of G. mRA(G)| — £5 (G)] (14.10) Proof A direct result of Theorem 14.8 from Lemma 12.2. O Corollary 14.4 For a simple graph G, let R,(G) and €,(G) be, respectively, the sets of all nonisomorphic rooted super maps and all distinct embeddings of G with size e(G) on a surface of genus g(orientable or nonorientable). Then, 2e(G) RCI = ESI. (14.11) Proof The case of | = 0 and m; = 0, i > 2, of Theorem 14.9. O Corollary 14.5 The number of rooted super maps of a simple graph G with n; vertices of valency 2, i > 1, on orientable surfaces is m [ke-n (14.12) i22 where e is the size of G. Proof Because of the number of distinct embeddings on ori- entable surfaces S > é(G) = [[(G - 09" (14.13) g20 i22 338 Chapter XIV Super Maps of a Graph known, Corollary 14.4 leads to the conclusion. [] Corollary 14.6 The number of rooted super maps of bouquet Bm, m > 1, on orientable surfaces is (2m)! 2m! (14.14) Proof Because of the number of all distinct embeddings of B,, on orientable surfaces (2m ,)! and the order of its semi-automorphism group 2"! known, the conclusion is deduced from Theorem 14.8. CO In virtue of petal bundles all super maps of bouquets, (14.14) is in coincidence with (9.9). The number of nonisomorphic super maps of a graph can also be derived from rooted ones. Theorem 14.10 Fora given graph G, let £&(G) be the set of all its nonequivalent embeddings with automorphism group order k. Then we have the number of all nonisomorphic unrooted supper maps of G is 1 nu(G) = TORTEK > ie (14.15) i|4e 1<i<de where € = €(G) is the size of G and I(G) is the number of loops in G. Proof On the basis of Theorem 14.8, we have 2e(G) [Ri(G)| = TIA) |E:(G)| where R;(G) is rooted super maps of G with automorphism group order i. By Theorem 14.4 and Corollary 14.1, «(G) I;(G)| = AC laut(G) iei (G)]. Because of 4€(G')/i rooted maps produced by an unrooted map XIV.5 Via rooted super maps 339 in R;(G) as known in the proof of Theorem 11.1, we have ] a e(G) acc; iO = iG) MO iau (aj O eG). ~ 2/6 aut(G) Overall possible ;|4e(G) is the conclusion of the theorem. O Further, this theorem can be generalized for any types a set of graphs. Theorem 14.11 For a set of graphs G, the number of noniso- morphic unrooted super maps of all graphs in G is nu(9) = Y^ sert *. d&(G)l. (14.16) GEG i|4e(G) 1<i<4e(G) Proof From Theorem 14.10 overall G € G, the theorem is soon done. Oo For a given genus g of an orientable or nonorientable surface, let E.(G;g) be the set of all nonequivalent embeddings of a graph G on the surface with automorphism group order k. Theorem 14.12 For a given genus g of an orientable or nonori- entable surface, the number of all nonisomorphic unrooted supper maps of a G on the surface is w(G: 9) = sperma È EG) (14.17) i|4e 1<i<4e where € = €(G) is the size of G. Proof By classification of maps and embeddings as well with genus, from Theorem 14.11 the theorem is done. [] Furthermore, this theorem can also generalized for any types a set of graphs. 340 Chapter XIV Super Maps of a Graph Theorem 14.13 For a given genus g of an orientable or nonori- entable surface, the number of nonisomorphic unrooted super maps of all graphs in a set of graphs Gp with a given property P is 1 mn mel) = gy 2; il£;(G; g)]. (14.18) 1<i<4e(G) Proof A particular case of Theorem 14.11. [] Activities on Chapter XIV XIV.6 Observations Let B, be the bouquet of n loops for n > 1. O14.1 Find the semi-automorphism group of B, for n > 1. O14.2 Find the edge-automorphism group of B, for n > 1. O14.3 Find the automorphism group of D, for n > 1. Let D,, be the dipole which is of order two and size m for m > 1 without loop. O14.4 Show Autne(Dm) ~ Autee(Dm) ~ Sm where Sm is the symmetric group of order m for m > 1. O14.5 Find a condition for Aut(D,,) ~ Autyr(D,,), m > 1. O14.6 Determine the number of super maps of K4. O14.7 Determine the number of rooted super maps of K5. 014.8 Determine the number of super maps of K3 3 rooted and unrooted. 014.9 Determine the number of super maps of K4, rooted and unrooted. O14.10 Suppose A;,, be the number of distinct embeddings of Bı, | > 1, on the orientable surface of genus p > 0, determine the number of rooted and unrooted super maps of Bı on the orientable 342 Activities on Chapter XIV surface of genus p > 0. XIV.7 Exercises E14.1 Show that the number of rooted super maps of Kn, the complete graph of order n, n > 4, is (n — 2)*-l. E14.2 Show that the number of rooted super maps of Km,n, the complete bipartite graph of order m+n, m,n > 3, is o= apes. E14.3 Let 7, be the set of non-isomorphic trees of order m, n 2 2. Show that I6-D" — Qn-1) y^ I (2n = 1) ed |Aut(T)| | n!(n4- 1)" the number of rooted plane trees of order n — 21 n;, Where n; is the number of vertices of valency i, i > 1. E14.4 Determine the distribution of rooted super maps of the complete graph Kn, n > 4, by automorphism group orders. E14.5 Determine the distribution of rooted super maps of the complete bipartite graphs K(m, n), m,n > 3, by automorphism group orders. E14.6 Determine the distribution of rooted super maps of su- per cube Q,, n > 4, by automorphism group orders. E14.7 Determine the distribution of rooted super maps of the complete tripartite graph Kọ, m,n, l,m,n > 2, by automorphism group orders. E14.8 Determine the distribution of rooted super maps of the complete equi-bipartite Ky», n > 3, by automorphism group orders. XIV.8 Reseaches 343 E14.9 Determine the distribution of rooted super maps of the complete equi-tripartite /€(n,n,n), n > 2, by automorphism group orders. E14.10 Determine the distribution of rooted super maps of the complete quadpartite graph Kk lmn, k,l, m,n > 2, by automorphism group orders. E14.11 Determine the distribution of rooted super maps of the complete equi-quadpartite graph Knnnn, n > 2, by automorphism group orders. E14.12 Determine the distribution of rooted super maps of su- per wheel Wn, n > 4, by automorphism group orders. XIV.8 Researches R14.1 For given integer n > 1, determine the distribution of outer planar graphs of order n by the order of their semi-automorphism groups. R14.2 For given integer n > 1, determine the distribution of Eulerian planar graphs of order n by the order of their semi-automorphism groups. R14.3 For given integer n > 1, determine the distribution of general planar graphs of order n by the order of their semi-automorphism groups. R14.4 For given integer n > 1, determine the distribution of nonseparable planar graphs of order n by the order of their semi- automorphism groups. R14.5 For given integer n > 1, determine the distribution of cubic planar graphs of order n by the order of their semi-automorphism groups. R14.6 For given integer n > 1, determine the distribution 344 Activities on Chapter XIV of 4-regular planar graphs of order n by the order of their semi- automorphism groups. R14.7 For given integer n > 1, determine the distribution of nonseparable graphs of order n by the order of their semi-automorphism groups. R14.8 For given integer n > 1, determine the distribution of general graphs of order n by the order of their semi-automorphism groups. R14.9 For given integer n > 1, determine the distribution of Elerian graphs of order n by the order of their semi-automorphism groups. R14.10 For given integer n > 1, determine the distribution of general graphs of order n by the order of their semi-automorphism groups. R14.11 For given integer n > 1, determine the distribution of cubic graphs of order n by the order of their semi-automorphism groups. R14.12 For given integer n > 1, determine the distribution of 4-regular graphs of order n by the order of their semi-automorphism groups. R14.13 For given integer n > 1, determine the distribution of 5-regular graphs of order n by the order of their semi-automorphism groups. Chapter XV Equations with Partitions e The meson functional is used for describing equations discovered from census of maps via vertex, or face, partition as parameters. e Functional equations are extracted form the census of general maps and nonseparable maps with the root-vertex valency and the vertex partition vector on the sphere. e Dy observing maps without cut-edge on general surfaces, a func- tional equation has also be found with vertex partition. e Functional equations satisfied by the vertex partition functions of Eulerian maps on the sphere and general surfaces are derived from suitable decompositions of related sets of maps. e All these equations can be shown to be well definedness. However, they are not yet solved in any way. XV.1 The meson functional Let fy) E Ry}, where y= (yi, TAT J be a function, and V(f,y) > 0, i—1,2,--. A transformation is established as 1 y! "TP y convinced y? = 1 > yo. 346 Chapter XV Equations with Partitions Since J is a function from the function space F with basis y {1,y,y?, =+} to the vector space V with basis {yo, y1, ys, +-}, it is called the meson functional. i.e., the Blissard operator. For any Vi = > ay’, L= 1,2, j20 it is easy to check that Je + v2) = $ (ay + az) E j20 = ) Q1jYj + 1 02,8; j20 j20 = fut fu y y Hence, the meson functional is linear. zii The inverse of the meson functional J is denoted by J Dye y y E y), i = 1,2,--., convinced J yo = 1, or simply yọ = 1. However, 1 y is seen as a vector in V. Two linear operators called /eft an dright projection, denoted by, respectively, S, and Ry, are defined in the space V as: let v = 2? 50 474; € V, then Syv = SoU ES Lig qus j20 1 Ryu = > j 4-197: j21 (15.1) In other words, if y; is considered as the vector with all entries 0 but only the i-th 1, then the matrices corresponding to S, and R, are, respectively, as XV.1 The meson functional 347 where te when j = 1; Le (j —1)1)1, when j > 2 for 1; being the infinite vector of all entries 0 but only the (j — 1)-st 1 and HL deos) (15.2b) where 0, when 7 = 1; T. = . z jor when j > 2 for the super index ‘T’ as the transpose. Easy to check that I 0° = n= (1), m~( where / is the identity. je o À 2), (15.3) -1 Theorem 15.1 For v = v(yos yi,:::) € V, let f(y) = J v, y i [ss [roo f me so Proof M equating the coefficients of terms in same type on the two sides, the theorem is done. [] If f(x,y) is a function with two types of unknowns, and assume f(x, y) € V(z, y), a bilinear space, then it is easily checked that [ [ 1e» -| [| fe» (15.5) Denoted by F(z,y) the function in (15.5). Conversely, for F(z, y) € 'R(x, y), we have fay =f | Fe) (15.6) DU dy because of interchangeable between J and J l z y 348 Chapter XV Equations with Partitions Let f(z) € R{z}. The following two operators on f as r—y and any f = ——— (15.8) are, respectively, called the (x, y)- difference and (x, y)-difference of f with respect to z. Lemma 15.1 For any function f(z) € R{z}, let f = f(z), then Oba) eeu. (15.9) Proof Because of the linearity of the two operators O; , and 6; ,, this enables us only to discuss f(z) = z", n » 0. Then, it is seen On y2f = Ox eee v 7 yx” n+l _ cyt! = mmm p" y” => LY t—y = zyó,.yz^ = zyó,,f. This is what we want to prove. [] Theorem 15.2 For any f € R{z}, we have ^82. ylz f) — 0b pl2f) = xa y epl’). (15.10) Proof From (15.7) and (15.8), the left hand side of (15.10) is wy" (^ f (x7) — v fü) - ay PG) - Ea) zi y? PREPO -PP = From (15.7), this is the right hand side of (15.10). O XV.1 The meson functional 349 For a set of maps .A, let falz, y) = b» gm UD yn) (15.11) AEA where m(A) and n(A) are, respectively, the invariant parameter and vector on A. Let F4(x,y) be such a function of two unknowns that falz, y) = ] Pe» (15.12) The powers of x and y in FA4(x,y) are, respectively, called the first parameter and the second parameter. Theorem 15.3 Let S and 7 be two sets of maps. If there is a mapping A(T) = (51, Sn --, Smry4if such that 5; and {i m(T) +2— i} are with a 1-1 correspondence from 7 to S for any T € 7, where i and m(T) + 2 — i are the contributions to, respectively, the first and the second parameters, i = 1,2, ---,m(T) + 1, with the condition as S — S(T), TET then Fs(x,y) = £yôsy(z fr) (15.13) where fr = fr(z) = fr(z,y). Proof From the definition of A, we have (T)+1 Fs(z, y) = ys > gy) He a n) TeT i=l m(T)+1 ,m(I)41 x yY n = 2y 2: — TET y = usu fr). This is (15.13). [] Theorem 15.4 Let Sand 7 be two sets of maps. If there exists a mapping A(T) = (51,55, --, S4, 71) such that S; and (i, m(T) - i) 350 Chapter XV Equations with Partitions are in a 1-1 correspondence for T € T, where i and m(T) + 2 — i are the contributions to, respectively, the first and the second parameters, i —1,2,---,m(T) — 1, with the condition S=) (T) TET then Fs(z, y) = Os y (fr) (15.14) where fr = fr(z) = fr(z,y). Proof From the definition of A, we have m(T)—1 12-607) E > >, o TeT i=1 op me) yx" ry" n -ayy = TET This is (15.14). D XV.2 General maps on the sphere A map is said to be general if both loops and multi-edges are allowed. Of course, the vertex map ? is also treated as degenerate. Let M p be the set of all rooted general planar maps. For any M € Msep; let a = e,(M) be the root-edge. Then, Mp can be divided into three classes: JM ass JV au and MW us 0:8 IVA ao em dV us T VE T Mi (15.15) such that Mp, consists of a single map Ù, Ma, = (M|VM € Msep, a is a loop]. Of course, Ma, = UMIVM € M, ais a link) XV.2 General maps on the sphere 351 in its own right. Lemma 15.2 Let M; have = (M — a|VM € Mep}. Then, we (gep)1 geP1 Nb), = Meer OM ge (15.16) where © is the 1-production as defined in $2.1. Proof For a map M € Msp); because there is a map Me Mis) such that M = M — à,à = e,(M), the root-edge of M, by considering the root-edge à as a loop we see that M = M,+Mb, pro- vided Mı N Mə = o, the common root-vertex of M and M. Since Mi and M» are allowed to be any maps in M 0, this implies that M € M,.. © M. Conversely, for any M € Mj © M, since M = Mj- M», Mi, M» € Mp, we may always construct a map M by adding a loop à at the common vertex of M; and M» as the root-edge of M such that Mı and M^» are in different domains of the loop. Of course, M isa general map. Because the root-edge of M isa loop added, M €. M However, it is easily seen that M — M — à. Therefore, M € Migs, In consequence, the lemma is proved. D sp Including the vertex map 8gep1 * For Msp, because the root-edges are all links we consider the set M (gp) = {M ea|VM € Mas, a = e (M), the root-edge as usual. The smallest map in Mep, is the link map L = (Kr, (r)(agr)) and it is seen that Lea = 2. Thus, 0 € Miep, For any M € My, because the root-edge of M is not a loop we know that M ea € Nas. Conversely, for any M € Mep we may always construct a map Me Msp by splitting the root-vertex of M into two vertices with a new edge a as the root-edge connecting them. This implies that Mea=Me M gep),- Therefore we have M eep)y — IM: (15.17) 352 Chapter XV Equations with Partitions Lemma 15.3 For Mep, we have Men = >, {ViM|0<i<m(M)} (15.18) MeMgep where m(M) is the valency of the root-vertex of M and V; is the operator defined in 87.1. Proof For any M € Msep, because the root-edge a is a link, we may assume a = (01, 02) such that 01 = (r, S) and o» = (apr, T). Let M be the map obtained by contracting the root-edge a into a vertex 6 = (T, S) as the root-vertex of M. It is easily checked that M € Ms from (15.17) and that M = Vis M, 0 « |S| « m(M) where m(M) = |S| + |T|, and |Z|, Z = S or T, stands for the cardi- nality of Z. That implies M is a member of the set on the right hand side of (15.18). Conversely, for any M in the set on the right, because there exist a map Me M, and an integer 1,0 < i < m(M), such that M = V,M, we may soon find that M € Mep, by considering that the root-edge of M is always a link and that M € Mep as well. Thus, M EM as: Therefore the lemma follows. [] Bep From the two Lemmas above we are now allowed to determine the contributions of M,.,,,7 = 0, 1, 2, to the enufunction mana = >, anp (15.19) MEM gep where n(M) = (nı( M), na( M), --- , nj(M),---), ni(m) is the number of vertices of valency i in M and m(M), the valency of the root-vertex of M. XV.3 Nonseparable maps on the sphere 353 First, since 9 has neither non-rooted vertex nor edge we soon see that IMgep, = L (15.20) Then, by Lemma 15.3, IMgep, = Tg. (15.21) where g = gu, (x, y) defined by (15.19). Further, from Lemma 15.3 m(M)-1 OM ep, =| ` pya Ma ya Y MEMgep i=1 By Theorem 15.3, ÜMgep, — s | (5.0). (15.22) y Theorem 15.5 The enufunction g defined by (15..5) satisfies the following functional equation: g—-1-4zgLz J (uos (2g) ): (15.23) y Proof According to (15.15), from (15.20-22) the theorem is soon obtained. [] XV.3 Nonseparable maps on the sphere Let M,, be the set of all rooted nonseparable planar maps with the convention that the loop map L4 = (Kr, (r, o fir)) is included but the link map L = (Kr, (r)(ar)) is not for convenience. Then, M,, is divided into two parts Ms and M,,,,2.€., Mas = Masy + Mas; (15.24) such that M s consists of only the loop map Lı. 354 Chapter XV Equations with Partitions Lemma 15.4 A map M € M,,, M z L4 if, and only if, its dual M* € Ms- Proof By contradiction. Assume M = (Xa B, P) € Mus, M z L; and its dual M* = (X54, PaB) Z Mns. Let v = (x, Paßz,---, Pal") be a cut-vertex of M*. Then we have a face f* = (x, Px,- --,P”x) on M* such that there exists an integer 7,1 < j € n, on f* satisfying Pix = (Paz for some i,1 € i € m, ie, v; = v5, = v*. However, f* is a vertex of M which has the face v* having the symmetry and hence f* is a cut-vertex of M. A contradiction to the assumption appears. The necessity is true. Conversely, from the duality the sufficiency is true as well. O For any M € M,,, let m(M) be the valency of the root-vertex and n(M) = (mi( M), na( M), ---), ni((:M) be the number of nonrooted vertices of valency 2,2 > 1. From the nonimputability, the root-edge a = (v1, vg) of any map M in M,,, is always a link. The map M e a obtained by contracting the root-edge a in M has the same number of faces as M does. Lemma 15.5 For any M € M,,, there is an integer k > 1 with k M ea — 3 Mi (15.25) i=1 such that all M; are allowed to be any map in M,, and that M;, i = 1,2,---,k, does not have the form (15.25) for k > 1. Proof In fact, from what were mentioned in 96.2, we see that k is the root-index of M and that all M;,1 < i € k, do not have the form (15.2) for k > 1. From the nonseparability of M, by considering that all vertices of M; except for the root-vertex are the same as those of M for i = 1,2,---,k, since M; does not have the form (15.2) for XV.3 Nonseparable maps on the sphere 355 k > 1, the root-vertex is not a cut-vertex for i = 1,2,---,k. That implies all M;, 1 < i < k, are allowed to be any map in M,, including the loop map. The lemma follows. [] Now let us write k Mx = { X :Mi| VM; € Mus, 1 <i € k}, (15.26) i=1 and M ns), = {M e a\VM € Masi}, (15.27) where a = e,, the root-edge of M. Lemma 15.6 For M,,,, we have Mors), = >> Mz; k>1 My, = ME (15.28) where is the inner 1v-production. Proof By the definition of inner 1v-product, the last form of (15.28) is easily seen. From Lemma 15.5 we can find that Mins, = U M x. k>1 Moreover, for any i, 7,2 Æ j, we always have Mi () Me= d. Therefore, the first form of (15.28) is true. O Based on the two lemmas above, we are allowed to evaluate the contributions of Mas and M,,, to the enufunction fm, of Mas with vertex partition, t.e., Telam >, ee. (15.29) M €.Mns 356 Chapter XV Equations with Partitions where m(M) is the valency of root-vertex and n(M) = (m(M), ni(M), ---) with n;(M) being the number of nonroot-vertices of valency 7,7 > 1. Since M, consists of only the loop map, which has the root- vertex of valency 2 without nonrooted vertex, we have NEL (15.30) For M,,,, we have to evaluate the function f(z,2) = ` g 04) a aM) (15.31) MEM ys, where s(M) is the valency of the nonrooted vertex vg, incident with the root-edge e, of M. By considering that for M € M,,,, m(M ea) = (m(M) —1) + (s(M) — 1), we soon find M€ Miss, where s(M) is the contribution of the valency of the nonrooted end of the root-edge of M to the valency of the root-vertex of M — Mea, M € M. Because s(M) is allowed to be any number between 1 and m(M) — 1, from Lemma 15.2 we have m(M)-1 5 od m(M Zi n( M jeje qp ox «UB Cy (2 yee k>1 \MeMns = By Theorem 15.4, TUB) = 2z > Ti k>1 £20 p,2f i L- Ongar where f = f(u) = f, (u, y), and hence Ma, = [fen XV.4 Maps without cut-edge on surfaces 391 YOryf =, /—— | (15.32) Theorem 15.6 The enufunction of M,, defined by (15.29) sat- isfies the following functional equation: nay matte | IL. 15.33 f p ( ) Proof Since fuu, = fts, + fMns,» from (15.30) and (15.32) the theorem is obtained. [] XV.4 Maps without cut-edge on surfaces In this section, only maps without cut-edge(or, 2-connected)are considered. Let M be the set of all(including both orientable and nonorientable) rooted maps without cut-edge. Classify M into three classes as M = Mo 4 Mt Mo (15.34) where Mọ consists of only the vertex map 0, My, is of all with the root-edge self-loop and, of course, Mg is of all with the root-edge not self-loop. Lemma 15.7 The contribution of the set Mo to f = fm(z.y) J5 eL, (15.35) where fo = f(x, y). Proof Because of V neither cut-edge nor nonrooted vertex, m(0) = 0 and n(2) = 0. Thus, the lemma is obtained. O In order to determine the enufunction of Mı, how to decompose Mı should first be considered. Lemma 15.8 For Mı, we have May =M, (15.36) 358 Chapter XV Equations with Partitions where M = (M — a|VM € Mı}, a= Kr(M). Proof Because of Lı = (r, yr) € Mı, y = ag, we have Lı — a = 0 € M. For any 5 € My), since there exists M € M such that 5 = M — a, by considering the root-edge of M not cut-edge, it is seen 5 € M. Thus, M) € M. Conversely, for any M = (¥, 7) € M, a new edge a’ = Kr' is added to the root-vertex (r)7 for getting 5;, whose root-vertex is (r'r, Utt Le, am. ne MEE gos), where 0 <i < m(M) — 1. Because of S; — a’ = M, we have S; € Mı. Hence, M € Mq. [] From this lemma, it is seen that each map M = (X, J) in M not only produces S; € M,,0 € i € m(M) — 1 but also Sm € Mı nonisomorphic to them. Its root-vertex is (r^, (r).7, yr’). For M € M, let Lemma 15.9 The set Mı has a decomposition as M,= p» SM, (15.38) MEM where Sy is given from (15.37). Proof First, for M € Mı, because of M' = M — a € My), Lemma 15.8 enables us to have M' € M. Via (15.37), M € Sw is obtained. Thus, what on the left hand side of (7.4.5) is a subset of that on the right hand side. Conversely, for a map M on the left hand side of (15.38), because of the root-edge a self-loop,we have M € Mı. Thus, the set on the left hand side of (15.38) is a subset of that on the right hand side. O On the basis of this lemma, we have Lemma 15.10 For gı = gm: (x.y) = fn (22, y), we have Of ee (15.39) Rh sc e XV.4 Maps without cut-edge on surfaces 359 where f = f, y). Proof From Lemma 15.9, = p» (m(M) + Yay. MEM By Lemma 9.10, we get Of — m2 ad This is the conclusion of the lemma. [] In what follows, Mə is considered. Lemma 15.11 For Mz, let Mi) = (M ea|VM € Mo}, then M (2) — M —23, (15.40) where Ù is the vertex map. Proof For any M € Mv), there is a map M’ € Mp such that M = M' ed'. Because of a’ neither cut-edge nor self-loop, M € M. And, sice the link map Lo = (Kr, (r)(yr)) € M», May CM — 9. Conversely, for any M = (X, J) € M —9, let U;,1 be obtained by splitting the root-vertex (r)5 of M with an additional edge a’ = Kr’ whose two ends are (7, r, --- , Jr) and (yr', girl e Vim 1 < i < m(M). Because of a’ not cut-edge, U; € Mə, 1 < i € m(M). And, because of M = U; e a Me Mo) Thus, M C Ü m M (o. L] For any M = (4,7) € M — 9, let Uy = (Uil x à € m(M)), (15.41) where U; is appeared in the proof of Lemma 15.11. Lemma 15.12 The set M? has the following decomposition: M:= M Um, (15.42) MEM-v 360 Chapter XV Equations with Partitions where Um is given from (15.41). Proof First, for any M € Mo, from Lemma 15.5, M' = M èa € M — 2 and further M € Um. This implies that My = U Us. MEeEM-v Then, for any bMi, M» € M — 0, because of M, not isomorphic to M», Um, (um - 0. Thus, (15.42) is right. The lemma is obtained. O This lemma enables us to determine the contribution of Mə to Lemma 15.13 For f» = fm,(x, y), we have y where f = f(2) = fíu(z. y). Proof From Lemma 15.12, m(M) fz -| 2. ( `. qim 2i) yM), Y MeM-9 i=l By employing Theorem 15.4, (15.43) is obtained. E On the basis of those having been done, the main result can be deduced in what follows. Theorem 15.7 The functional equation about f o NET +=- s | ydes (15.44) ox y is well defined on the field (9t; z, y}. And, its solution is f = f(x) = XV.5 Eulerian maps on the sphere 361 Proof The first statement can be proved in a usual way except for involving a certain complication. The second statement is derived from (15.34) in companion with (15.35), (15.39), and (15.43). [] XV.5 Eulerian maps on the sphere A map is called Eulerian if all the valencies of its vertices are even (or say, all vertices are even). Let U be the set of all the rooted planar Eulerian maps with the convention that the vertex map Ó is in U for convenience. Further, U is divided into 3 classes: Up, U and Wh, i.e., U = Ue o Us o Ub (15.45) such that Uy = (0), or simply write {J} = 0, and U, = {U|VU € U with a = e,(U) being a loop}. Lemma 15.14 Any Eulerian map (not necessarily planar) has no cut-edge. Proof By contradiction. Assume that a Eulerian map M has a cut-edge e = (u,v) such that M = Mi U e U M», Mı n M» = Ø, where Mı and M» are submaps of M with the property that Mi is incident tou and Ms, to v. From the Eulerianity of M, u and v are the unique odd vertex in Mı and Mo, respectively. This contradicts to that both Mı and M» are a submap of M because the number of odd vertices in a map is even. E Lemma 15.15 Let Uy) = {U — a|VU € Ui} where a = e, (U) is the root-edge. Then, we have Un; = Uu (15.46) where © is the 1v-production. 362 Chapter XV Equations with Partitions Proof Because for U € M, the root-edge a is a loop, we see that U — a = U414-Us where U; and U; are in the inner and outer domain of a respectively. Of course, it can be checked that both U and U» are maps in U. Thus, the set on the left hand side of (15.46) is a subset of that on the right. On the other hand, for any U = U44-Uo, U1, Us € U, we may uniquely construct a map U’ by adding a loop at the common vertex of U; and Uy. The root-edge of U’ is chosen to be the loop such that U and Us are respectively in its inner and outer domains. It is easily checked that U’ is a Eulerian map and hence U’ € U. However, U = U'—a € Un. That implies the set on the right hand side of (15.46) is a subset of that on the left as well. O For any map U € 0f», we see that the root-edge a of U has to be a link. From Lemma 15.14, if U ea = U;+U> such that the root-vertex is the common vertex, then the valencies of the vertices in both U; and U» are odd. Further for any U € U, if U = U44-Us, then the valencies of the common vertex between U; and U»5 are both even as well. Lemma 15.16 Let Uo) = {U ea|VU € Lj where a is the root-edge of U. Then, we have Uo, =U — ù (15.47) where J? = U for simplicity. Proof From what has just been discussed, the set on the left hand side of (15.47) is a subset of that on the right. Conversely for any U € U — 9, we may always construct a map U' by splitting the root-vertex into o; and o9 with the new edge a’ = (01, 02) as the root-edge of U’ such that the valencies of o1 and os in U' are both even. However, U = U' ea' € Uy). That implies the set on the right hand side of (15.47) is a subset of that on the left. O For a map U € U — 0, assume the valency of the root-vertex ois 2k, k > l,without loss of generality. The map U’ obtained by XV.5 Eulerian maps on the sphere 363 splitting the root-vertex into o; and o» with the new edge a’ = (01, 02) such that the valency p(o1; U') = 2i and hence p(o»; U") = 2k — 2i + 2 is denoted by Upi, i = 1,2,---,k. From Lemma 15.14, we see that the procedure works and that all the resultant maps Uj, are also Eulerian maps. Lemma 15.17 For Lf», we have [U| = ` Ua] i=1,2,---,m(U)}| (15.48) Ucu-) where 2m(U) is the valency of the root-vertex of U. Proof First, we show that for any U € Uy, it appears in the set on the right hand side of (15.48) only once. Assume that a — (o, v) is the root-edge of U and that p(o) — 2s and p(v) — 2t. Let U' be the map obtained by contracting the root-edge a, i.e., U' — U ea. Then, there is the only possibility that U = U [2s] in the set on the right hand side of (15.48). Then, we show that for any map U in the set on the right hand side of (15.48), it appears also only once in U2. This is obvious from Lemma 15.16 because all elements are distinguished and they are all maps in U by considering the Eulerianity with the root-edges being links. [] In what follows, we see what kind of equation should be satis- fied by the enufunction u of rooted planar Eulerian maps with vertex partition. Write gy g ga (15.49) Ucu where 2m(U) is the valency of the root-vertex as mentioned above and n(U) = (na(U),---, na(U),---), na; (U) is the number of nonrooted vertices of valency 22, i > 1. Theorem 15.8 The function u defined in (15.5) satisfies the 364 Chapter XV Equations with Partitions following functional equation: u — 1-4 z^) a J (ul Qu(/2))) (15.50) y where u(z) = u|;—; = u(z, y). Proof The contribution of Uo to u is ugo =1 (15.51) since 2m(0) = 0 and n(v) = 0. From Lemma 15.15, the contribution of Mı to u is uj = n b» gm yal) UU) = a^. (15.52) The contribution of U to u is denoted by us. Let ii(z) = ` AITE UEU where 2j(U) is the valency of the nonrooted end of the root-edge and n(U) = (fi3(U), R4(U), «fii (U),-- -), alU) is the number of vertices of valency 2i except for the two ends of the root-edge. It is easily seen that R(U) = n(U) — ea) where e»;(yj is the vector with all the components 0 except only for the j(U)-th which is 1. In addition, it can be verified that u» = fao (15.53) From Lemma 15.17, we have m(U) ü(z) E ` »» gi ,2m(U)-2i*2 ya. Ucu—o i=l XV.6 Eulerian maps on the surfaces 365 By Theorem 15.3, ü(z) = a?220,» s (u( V/t)). Then by (15.53), we find that ug = x" J y ôy alul Vt). (15.54) From (15.51), (15.52) and (15.54), the theorem is obtained. UO XV.6 Eulerian maps on the surfaces Let Megu be the set of all orientable Euler rooted maps on sur- faces. Because of no cut-edge for any Eulerian map, Eulerian maps are classified into three classes as Mẹ ,, ME; and M, such that M, consists of only the vertex map J, Mi has all its maps with the root-edge self-loop and Ma = Meu Mia m Miny- (15.55) Naturally, MŽ; has all maps with the root-edge a link. The enufunction g = fmz,(2,y) is of the powers 2m and n = (n2, n4, -+ -) of, respectively, x and y as the valency of root-vertex and the the vertex partition operator. i Lemma 15.18 For MÌ „ we have go = 1, (15.56) where go = fmo (2^, y). Proof Because of V with neither root-edge nor nonrooted vertex, m(0) = 0 and n(0) = 0. The lemma is done. O In order to determine the enufunction of ML, a suitable decom- position of ML. should be first considered. Lemma 15.19 For ML, we have MJ = Meu; (15.57) 366 Chapter XV Equations with Partitions where M, = (M — a|VM € Mi), a = Kr(M), the root-edge. Proof Because of Ly = (r, yr) € ML yo, L1—a = 9 € Meu is seen. For any S € Ma). since there is a map M € Megu such that S = M — a, from M as a Eulerian map, S € Megu is known. Thus, MS C Meu. Conversely, for any M = (X,J) € Meu, By adding a new root-edge a’ = Kr’ at the vertex (r)5 to get S;, whose root-vertex is (ries, ir, oyr!, fr, cs, g20007M), 0 € i € 2m(M) — 1. From S; — à! = M, S; € Ml, Thus, Mga € MË. Oo In the proof of this lemma, it is seen that each map M = (X, J) in Mg produces not only S; € Mt p0 < i € 2m(M) — 1, but also Som € ME, nonisomorphic to them. Its root-vertex is (r^, (r) z, yr’). For M € Megu, let Lemma 15.20 Set Mi, has the following decomposition: Meu = ` Sm, (15.59) M € Mngu where Sm is given from (15.58). Proof First, for any M € ML, from Lemma 15.19, M' = M — a € Megu, and hence M € Sw. Thus, Mia = U Sw. M € Mmgu Then, for any Mı, M» € Mey, because of nonisomorphic be- tween them, Sm ( Su = 9. Therefore, the conclusion of the lemma is true. [] On the basis of this lemma, the following lemma can be seen. XV.6 Eulerian maps on the surfaces 367 Lemma 15.21 For gı = gMbu(z.y) = fm, (2°, y), we have ð gı = xg 2255) (15.60) where g — 9 Ms UU V) = anes y). Proof From (15.59), g= >) (2m(M) + 1)s" "Dy. M €.M gu By employing Lemma 9.10, (15.60) is obtained. [] In what follows, M3. is investigated. Lemma 15.22 For Mj,,, let MB = (M ea|VM € Mju}, then M, = Mea — 9, (15.61) where Ù is the vertex map. Proof Because of L4 € M 0 ¢ MC. (Q)). Then, M2, C M Eu E Ü. Conversely, for any M = (X,P) € Mru — 9, since Mo; = (A + Kra;,P3;) € M, where the two ends of as; = Kroj is obtained by splitting the root-vertex (r)p of M, i.e., (2j) Pa; — (raj; r, Pr, TT (p)? and | (9725) Po; — ("yro;, Ptr, ey (Pj umo, 1<%t< m-—1. Because of M = Mo; è ao;, Mo; € MI s Thus, For any M = (X, J) € Mru — 2, let Muy = {Mp,;|1 € j € m(M)), (15.62) where Mo;, 1 € j € m(M) — 1, have appeared in the proof of Lemma 15.22. 368 Chapter XV Equations with Partitions Lemma 15.23 Set Mpy” has the following decomposition: Mau? = b Mm (15.63) MEMgu— Ü where M m is given from (15.62). Proof First, for any M € MS as because of M = Mea c Mru — v, Lemma 15.22 tells us that M € M(M). Thus, 2 Mew = |] Mu. M €.M gu Ü Then, for any Mi, Ma € Mguu — 9, because of nonisomorphic between M; and Mo, Mm (| Mas, # 9. This implies (15.63). O On the basis of this lemma, the following conclusion can be seen. Lemma 15.24 For go = fm (v, y), we have g2 = 2 f Pip (15.64) y where g = g(z) = fa@ou(2?, y)- Proof From Lemma 15.23, 92 = ` Ji McMmgu-? j= m(M) x 2j, FM) (M)+2— ye, 1 By employing Theorem 15.3, E 2 gs =T f» 042 2g( Vz). y This is the lemma. O Now, the main result of this section can be described. XV.6 Eulerian maps on the surfaces 369 Theorem 15.9 The functional equation about g ont 98 _ 1 4 (1— z?)g pco (15.65) ðr? — / is well defined on the field L{R; x,y}. Further, its solution is g = 9 Ms UU Y) E fT Mga (5 9): Proof The last conclusion is deduced from (15.55), in compan- ion with (15.56), (15.60), and (15.64). 'The former conclusion is a result of the well definedness for the equation system obtained by equating the coefficients on the two sides of (15.65). [] Activities on Chapter XV XV.7 Observations O15.1 Let f = fuļx.y) be the vertex partition function of a set U of maps. Suppose U, = {U + a'| VU € U} where a’ = Kr’ is the root-edge of U' = U +a’. To evaluate f+ = fu, (x, y) from f. O15.2 Let A be the set of supermaps of K4, the complete graph of order 4. To evaluate f(z, y). O15.3 Let B; be the set of all bipartite rooted maps with the root-edge a cut-edge and B(1) = (B — a|VB € Bı}. Suppose the vertex partition function f of all bipartite rooted maps on surfaces is known. To evaluate fs, (x, y) from f. O15.4 Fora set of maps M, observe the relationship between vertex partition function f(x,y) and the enufunction fm(x, y) of two parameters: the valency of root-vertex and the size. O15.5 Fora set of maps M, observe the relationship between vertex partition function f(x,y) and the enufunction fm(x, y) of two parameters: the valency of root-vertex and the order. O15.6 Fora set of maps M, observe the relationship between vertex partition function f(x,y) and the enufunction fm(x, y) of two parameters: the valency of root-vertex and the coorder. O15.7 Consider the conditions satisfied by the face partition of petal bundle with size n > 1. O15.8 Consider the conditions satisfied by the face partition of a supermap of the complete graph K,, with order n > 4. XV.8 Exercises 371 O15.9 Show a 1-to-1 correspondence between the two sets U’ and M where U’ = {U —alW = (4,7) € U,a = Kr(U),yr € (r)7,r =r(U)}, U consists of all rooted maps on the projective plane and M is of all rooted planar maps. O15.10 Observe the relationship between the vertex partition function of plane rooted trees and the face partition function of out- erplanar rooted maps. O15.11 Observe the existence of a tree for a given vector n as its vertex partition. XV.8 Exercises E15.1 Prove that the vertex partition function of planted trees satisfies the functional equation about f as n=0- f pr. (15.66) E15.2 Solve the functional equation (15.66) in a direct way. E15.3 Ifthe vertex partition function of Halin rooted maps is taken to have the root-face valency instead of root-vertex valency, then show that the function satisfies the functional equation about f as NC" T y^ f? featy to | TR (15.67) where f = f(x) = f(x, y) and f, = f(y). E15.4 Provide à method for solving the functional equation shown in (15.67). E15.5 Prove that the vertex partition function of Wintersweets with root not on a circuit satisfies the following functional equation about f as LY3 (1-41 N =1 +a | viuo) (15.68) 372 Activities on Chapter XV E15.6 Find a way for solving the functional equation shown in (15.68). E15.7 Prove that the vertex partition function of unicyclic maps with root not on the circuit satisfies the functional equation about f as f E LT + s | vdes, (15.69) y where 7, is the vertex partition function of planted trees, which is known in El15.1. E15.8 Solve the functional equation shown in (15.69). E15.9 Show that the following functional equation about f is satisfied by the vertex partition function of outerplanar rooted maps as acero ies | ybeo(2f) (15.70) y where 1 p= z:- V1 — 4z?). x E15.10 Solve the functional equation shown in (15.70). E15.11 Find a functional equation satisfied by the vertex par- tition function of general planar rooted maps. XV.9 Researches R15.1 For given orientable genus p Z 0, determine a functional equation satisfied by the vertex function of a set of maps on the surface of genus p. R15.2 For given nonorientable genus q > 1, determine a func- tional equation satisfied by the vertex function of a set of maps on the surface of genus q. R15.3 Determine a functional equation satisfied by the vertex partition function of nonseparable rooted maps on the Klein bottle. XV.9 Researches 313 R15.4 Determine a functional equation satisfied by the vertex partition function of nonseparable rooted maps on the torus. R15.5 Determine a functional equation satisfied by the vertex partition function of bipartite rooted maps on the torus. R15.6 Solve the functional equation about f as sf Lo | viue (15.71) y R15.7 Solve the functional equation about f as Gt [1o | ierat. (15.72) y y R15.8 Solve the functional equation about f as (faeit -PA= | (fy +avbrslef)). (15.73) y y R15.9 Solve the functional equation about f as 1 f= [= 15.74 y Le D ote R15.10 Solve the functional equation about f as 1 — 0,55 z fe J l eei) O (15.75) y 1— 205 a (Ef. n) = vy Oo fo) R15.11 Solve the functional equation about f as face ; | Mel (15.76) y 1— Ony te R15.12 Solve the functional equation about f as fax E (15.77) y (1 ~~ 0,2. o fuz)? = (1052 yf yz)? R15.13 Solve the functional equation about f as (lg fel aè | ifo (15.78) y Appendix I Concepts of Polyhedra, Surfaces, Embeddings and Maps This appendix provides a fundamental of basic concepts of poly- hedra, surfaces, embeddings and maps from original to developed as a compensation for Chapters I-II. Only those available in the usage from combinatorization to algebraication are particularly concentrated on. Ax. Polyhedra A polyhedron P is a set (C;|1 < i € k}, k > 1, of cycles of letters such that each letter occurs exactly twice with the same power (or index) or different powers: 1(always omitted) and —1 and denoted by P = ((Ci1 € i € k}). It is seen as a set of all the cycles in any cyclic order. This is a general statement of Heffter's|Hefl| (and more than half a century later Edmonds’|Edm1] as dual case) which has the minimal- ity of no proper subset as a polyhedron for the convenience usages. A polyhedron is orientable if there is an orientation of each cycle, clockwise or anticlockwise, such that the two occurrences of each letter with different powers; nonorientable, otherwise. The support of polyhedron P = ({C;|1 < i € k]) is the graph G = (Vp, Ep) with a weight w on Ep where Vp = {Cill € i € k}, (Ci, C;) € Ep if, and only if, C; and Cj, 1 € i, j € k, have a common letter, and (L1) (e) l 0, when two powers are different; wW — 1, otherwise Ax.I.1 Polyhedra 375 for e € Ep. The set of all the edges with weight 1 is called the 1-set of the polyhedron. Theorem I.1 A polyhedron P = ((C;|1 € i € k}) is orientable if, and only if, one of the following statements is satisfied: (1) What obtained by contracting all edges of weight 0 on the support is a bipartite graph; (2) No odd weight fundamental circuit is on the support; (3) No odd weight circuit is on the support; (4) The 1-set forms a cocycle; (5) The equation system about x; = zc;, C; € Vp, on GF(2) for (C;, Cj) € Ep has a solution. Proof Because P is orientable, the two occurrences of each let- ter are with different powers. Since the weights of all edges are the constant 0, the equation system (I.2) has a solution of x; = 0 for all C; € Vp, 1 € i € k. Further, by considering that the consistency of equation system (1.2) is not changed from switching the orientation of a cycle between ce clockwise and anticlockwise while interchanging the weights between 0 and 1 of all the edges incident with the cycle on the support, statement (5) is satisfied for any orientable polyhedron. On the basis of statement (5), from a solution of equation system (1.2) the vertices of Gp are classified into two classes by x; = 1 or 0: 1-class or 0-class respectively. According to equation (1.2), each edge with weight 1 has its two ends in different classes and hence the 1-set is a cocycle. This is statement (4). On the basis of statement (4), since any circuit meets even num- ber of edges with a cocycle, all circuits are with even weight. This means no odd weight circuit. Therefore, statement (3) is satisfied. On the basis of statement (3), the statement (2) is naturally deduced because a fundamental circuit is a circuit in its own right. 376 Appendix I: Concepts of Polyhedra and Maps On the basis of statement (2), by contracting all edges of weight 0 in each fundamental circuit on the support, (1) is satisfied. On the basis of statement (1),the vertices are partitioned into two classes by the equivalence that two vertices are joined by even weight path. By switching the orientation of all vertices in one of the two classes and those in the other class unchanged, a polyhedron without weight 1 edge is found. This implies that P is orientable. In summary, the theorem is proved. [] On the support Gp — (Vp, Ep) of a polyhedron P, the operation of switching the orientations of all vertices in a subset of Vp between clockwise and anticlockwise and the weights of all edges incident with just one end in the subset interchanged between 0 and 1 is called a switch on P. Theorem I.2 The orientability of a polyhedron does not change under switches. Proof From the definition of orientability, the conclusion of this theorem is true. [] Let T' be a minimal set of edges having an edge in common with all cocycles in the support of a polyhedron. In fact, it can seen that T is a spanning tree. All the polyhedra obtained by switching on a polyhedron P are seen to be the same as P; different, otherwise. From Theorem I.2, in order to discuss all different polyhedra it enables us only to consider all such polyhedra of the support with weight 0 on all tree edge for a spanning tree chosen independently in any convenient way. Such a polyhedron is said to be classic. Theorem I.3 A classic polyhedron is orientable if, and only if, all edges as letters have their two occurrences with different powers. A classic polyhedron is nonorientable if, and only if, the set of letters each of which has its two occurrences with same power does not contain a cocycle. Ax.l2 Surfaces 377 Proof The first statement is deduced from Theorem I.1(3). The second statement is by contradiction derived from Theorem I.1 and Theorem I.2. [] Now, a polyhedron(always summed to be classic below) P is con- sidered as a permutation formed by its cycles. Let ó be the permuta- tion with each cycle only consists of the two occurrences of each letter in P. Then, the dual, denoted by P*, of P is defined to be P* — Pó such that their supports are with the same weight. The cycles in P are called faces and those in P* are vertices. Cycles in 6 are edges. Let v(P), e(P) and ¢(P) be, respectively, the number of vertices, edges and faces on P, then v(P) — e(P) - G(P) is the Eulerian characteristic of P. The graph which is formed by vertices and edges of P is called a skeleton of P. Of course, the skeleton of P is the support of P*. Theorem I.4 P* isa polyhedron and P** = P. P* is orientable if, and only if, so is P with the same Eulerian characteristic. Proof It is easily checked that P* is a polyhedron from P as a polyhedron. Since 5? = 1, the identity, we have pete puqpiéeP umm This is the first statement. From Theorem I.3, the second statement is obtained. [] Ax.L2 Surfaces Surfaces seen as polyhedral polygons can be topologically classi- fied by a type of equivalence. Let P be the set of all such polygons. For P = (((Aj))à > 1}) € P, the following three operations including their inverses are called elementary transformation: Operation 0: For (A;) = (Xaa !Y), (Ai) & (XY) where at least one of X and Y is not empty; Operation 1: For (A;) = (XabY ab) (or (XabYb~'a~')), (Ai) & (XaYa)(or XaY a); 378 Appendix I: Concepts of Polyhedra and Maps Operation 2: For (A;) = (Xa) and (A;) = (a !Y), i z j, ({(A;), (4;)}) & (XY) where at least one of X and Y is not empty. Particularly, ({(A;), (A;)}) = (XaYa !) when both (X) and (Y) are polyhedra. If a polyhedron P can be obtained by elementary transformation into another polyhedron Q, then they are called elementary equiva- lence, denoted by P ~a Q. In topology, the elementary equivalence is topological in 2-dimensional sense. Lemma I.1 For P € P, there exists a polyhedron Q — (X) € P where X is a linear order such that P ~a Q. Proof Let P = ({(A,)|1 € i € k]). If k = 1, P is in the form as Q itself. If k > 2, by employing Operation 2 step by step to reduce the number of cycles 1 by 1 if any, the form Q can be found. [] Lemma I.2 For P € P, if P = ((A)(B)) with both (A) and (B) as polyhedra, then for any x ¢ AU B, P ~a ((A)z(B)c 1). Proof It is seen that P = (AB) ~a (Azz ^! B) (by Operation 0) ^a (CAx) (z^! B)) (by Operation 2) = ((A)e(B)a7)). 7 From Lemmas I.1-2, for classifying P it suffices to only discuss polygons as Q. Lemma L3 Let Q = (ArByCz !Dy |), then Q ~a (ADzyBzx !Cy |). (L3) Proof It is seen that Q ^a ((Azz)(z !ByCz ! Dy !)) (by Operation 2) ~e (zADy !z !ByC) (by Operation 2) = (ADzyBzx Cy ). o Ax.L2 Surfaces 319 Lemma L4 Let Q = (AxByCz !Dy |), then Q ~a (BAzxyzr |DCy )). (1.4) Proof It is seen that Q ^a ((x Dy Axz)(ByCz")) (by Operation 2) m~e (BAzzz |DCz !) (by Operation 2) = (BAzyz DOY"): n Lemma L5 Let Q = (AzByCz Dy b) then Q ~a (ADCBzyz y). (1.5) Proof From Lemma I.4 and then Lemma I.3, the lemma is soon done. gO According to Lemma I.5, if A is replaced by EA in polyhedron (ADC B), then the relation is soon derived as Relation 1: (ArByCx !Dy !E) ~a (ADCBEzyx y |). Lemma I.6 Let Q = (AzBz) € P, then Q ~a (AB ' rz). Proof It is seen that Q ~a ((Azz)(z ^ Bz)) = ((zAx) (x B™z)) (by Operation 2) ~el (zAB 1z) = (AB xx) (by Operation 2). gO According to Lemma I.6, if A is replaced by C A in polyhedron (AB), then the relation is soon derived as Relation 2: (ArBzC) ~a (AB !Czz). Lemma I.7 LetQ = (Axyz M zz) € P, thenQ ~a (Aryzyzz). 380 Appendix I: | Concepts of Polyhedra and Maps Proof It is seen that Q ~a ((zAzyt)(t 1x y !z)) (by Operation 2) ~e CAxytyrt) (by Operation 2) = (Axyzyzz). go According to Lemma 1.7, then by Relation 2 twice for x and y, the relation is soon derived as Relation 3: (Aryx !y !zz) ea (Avxyyzz). Lemma L8 If Q € P orientable not as (AxByCx Dy 1E), then Q ~a (xxz~')(= Op). Proof Because Q is not in the above form, Q has to be in form as (Azz 1B). If both A and B are empty, them Q ~a (zx 3); otherwise, Q ~a (AB). Because (AB) still satisfies the given condition, by the finite recursion principle, (zz !) can be found. O Theorem I.5 For any Q € P orientable, if Q 4. (zx 1), then there exists an integer p > 1 such that Q ~el qI viyi; y; )(— Op). (1.5) i=l Proof Because Q ~a (AziByiCz, Dy, E), by Relation 1 we have Q ~a (ADCBExzqyizi yi |). If (ADOBE) ca (zr), the Q a (tiyr Yr): That is the case p= 1. Otherwise (ADCBE) = (A1z5B1yo C325 D1y3 E1). Because (ADCBEzyyizi yi!) = (AirsBiyyCizg Dy; Ej)mwii yi is still in the given condition. By the finite recursion principle, (1.5) is found. E Ax.L3 Embeddings 381 Theorem I.6 For any Q € P nonorientable, there exists an integer q > 1 such that Q ~al qJI £;£;)(= Qq). (1.6) Proof Because Q is nonorientable, there is a letter x; in Q such that Q = (Az,Bz,C). By Relation 2, D ~a (AB !Czqz). If (AB^1C) ~a (xxl), then by Operation 0 we have Q ~a (2121). This is the case of q = 1. Otherwise, there exits an integer k > 1 such that k Q ~el aJI ze Qu and (A) %q (xz!) is orientable. By Theorem L5, there exists an integer s > 1 such that (A) ~a (T nud -w E= O,). Thus, by Relation 3 for s times, we have 2s+k Q ~a (|| zx) Q2. i—1 This is g=2s+k > 1. [] On the basis of Lemma I.8 and Theorems I.5-6, surfaces in topol- ogy are in fact the classes of polyhedra under the elementary equiv- alence. Surfaces Oo, Op, p > 1, are, respectively, orientable standard surfaces of genus 0, p, p > 1. Surfaces Q,, q È 1, are nonorientable standard surfaces of genus q. Ax.l3 Embeddings An embedding(i.e., cellular embedding in early references particu- larly in topology and geometry) of a graph is such a polyhedron whose skeleton is the graph. 382 Appendix I: | Concepts of Polyhedra and Maps The distinction of embeddings are the same as polyhedra. Pre- cisely speaking, two distinct embeddings on a 2-dimensional manifold are not equivalent topologically in 1-dimensional sense. According to Ax.I.1, all embeddings always imply to be classic. For a graph G = (V, E), Heffter-Edmonds' model of an embed- ding of G by rotation system at vertices, in fact, only for orientable case [Hef1] and [Edm1]. Let c = {o,|v € V) be the rotation system on G where o, is the cyclic order of semi-edges at v € V. Then, by the following procedure to find an embedding of G: Procedure I.1 First, put different vertices in different position marked by a hole circle or a bold point on the plane. Draw lines for edges such that no interior point passes through a vertex and o; is in clockwise when v is a hole circle; in anticlockwise, otherwise. Then, by travelling along an edge in the rule: passing through on the same side when the two ends of the edges are in same type; crossing to the other side, otherwise. Find all cycles such that each edge occurs just twice. The set of cycles is denoted by Pg. Lemma I.9 Pg is a polyhedron. Proof Because it is easily checked from the definition of a poly- hedron. [] Lemma I.10 Fg is orientable. Proof Because the dual is orientable, from Theorem L4, the lemma is true. L Theorem I.7 The dual of Pg is an orientable embedding of G. Proof Because the support of the dual of Pg is G itself, the theorem is deduced. [] However, Pg in general is not classic except for all vertices are of same type. Ax.L3 Embeddings 383 Theorem I.8 For a given rotation system c of a graph G, let Pa(a;0) be the polyhedron obtained by the procedure above for all vertices of same type, then P¢(a;0) is unique. Proof From the uniqueness of classic polyhedron in this case, the theorem is done. Ll On the basis of Theorem L8, it suffices only to make all vertices with the same type, e.g., in clockwise. Further, in order to extend to nonorientable case, on account of Theorem 1.3, edges in a set not containing cocycle are marked for crossing one side to the other in the Heffter-Edmonds' model. The marked edges are called twist. This model as well as the Procedure I.1 here is called an expansion. Theorem I.9 The dual of what is obtained in an expansion is a unique nonorientable embedding of G for twist edges fixed. Proof Because one obtained in an expansion is a classic poly- hedron, from the uniqueness of the dual of a polyhedron, the theorem deduced. [] Theorem I.10 All embeddings of a graph G obtained by ex- pansions for all possible rotation system and twist edges in a subset of the cotree T' of a given spanning tree T' on G are distinct. Proof Asa result of Theorem I.9. [] This theorem enables us to choose a spanning tree T' on a graph G for discussing all embeddings of G on surfaces. Let Tj and T5 be two spanning trees of a graph G. The sets of all embeddings of G as shown in Theorem 1.10 for Tj and T» are, respectively, denoted by €; and £». Theorem I.11 Let £f and £j be, respectively, the subsets of £ and Ez on surfaces of genus g(orientable g = p > 0, or nonorientable g=q2 1). Then €; =. 384 Appendix I: | Concepts of Polyhedra and Maps Proof Because of Theorem L.10, it suffices only to discuss ex- pansions for T, and T». Since |T;| = |T5|, Theorems I.8-9 implies the theorem. [] For an embedding P € £7, if P ¢ £3, then there exits a twist edge e in T». By doing a switch with the fundamental cocircuit con- taining e for 7», an embedding P’ in the same distinct class with P is found. If no twit edge is in T», then P’ is the classic embedding in £7 corresponding to P. Otherwise, by the finite recursion, a classic em- bedding Q € £3 in the same distinct class with P is finally found. In this way, the 1-to-1 correspondence between Ef and £3 is established. The last two theorems form the foundation of the joint tree model shown in |Liu13-14]. Related topics are referred to |Sta1-2]. Ax.l.4 Maps Maps as polyhedra or embeddings of its underlying graph had been being no specific meaning until 1979 when Tutte(William T., 1917-2002) clarified that a map is a particular type of permutation on a set formed as a union of quadricells[Tut1—3]. All quadricells are with similar construction that four elements have the symmetry as a straight line segment with two ends and two sides. This idea would go back to Klein(Felix, 1849-1925) who consid- ered a triangulation of an embedding on a surface by inserting a vertex in the interior of each face and each edge and then connecting all line segments from a vertex in the interior of a face to all vertices on the boundary of the face. It is seen that each edge is adjacent to four triangles called flags as a quadricell. So, such a pattern of map used in this course can be named as Klein-Tutte's model. Related topics are referred to [Vin1-2]. Now, we have seen that a surfaces is determined by an elementary class of polyhedra, an embedding is by a distinct class of polyhedra and a map is by an isomorphic class of embeddings. The distinction of em- beddings is based on edges labelled by letters, or numbers. This is also Ax.L4 Maps 385 a kind of asymmetrization. But edges on a map are without labelling. Isomorphic maps are combinatorially considered with symmetry. So, a map is an isomorphic class of embeddings of its underlying graph. Let G = (V, E) be a graph. As shown in Sect. 1.1, V = Par(X) and E = (Bx|r € X) where Par(X) is a partition on B(X) = UxzexBxz, Bx = {x(0), x(1)} for a set X. Two graphs G4 = (Vi, E) and G» = (Vo, E>) are isomorphic if, and only if, there exists a bijection t: X1 — Xə such that the diagrams . a A | ba (L7) X, ———— Xo for c; = Bj;Par; i = 1,2, are commutative. Let Aut(G) be the automorphism group of G. On the other hand, a semi-arc isomorphism between two graphs G, = (Vi, E4) and G2 = (Vo, E2) is defined to be such a bijection 7: By(X4) — B3( X3) that B(X) ————— Bs(X3) hy, it as) T for c; = B; Pary i = 1,2, are commutative. Let Aut;/5(G) be the semi-arc automorphism group of G. Theorem I.12 If Aut(G) and Aut;/5(G) are, resp., the auto- morphism and semi-arc automorphism groups of graph G, then Auty/2(G) = Aut(G) x Sj (1.9) where / is the number of self-loops on G and 55 is the symmetric group of degree 2. Proof Because each automorphism of G just induces two semi- arc isomorphisms of G for a self-loop, the theorem is true. [] 386 Appendix I: | Concepts of Polyhedra and Maps For map M = (Xa, P), its automorphisms are discussed with asymmetrization in Chapter VIII. Let M(G) be the set of all noniso- morphic maps with underlying graph G. Lemmal.11 Foran automorphism Ç on map M = (A,5(X), P), we have exhaustively G|p(x; € Auty2(G) and Go|g(x; € Auty/2(G) where G = G(M), the underlying graph of M, and B(X) = X + BX. Proof Because Xa 5(X) = (X + BX) 4 a B(X), by Conjugate Axiom each ¢ € Aut( possibilities: G[p(x) € Autj/5(G) and Go|np(x) (aX +a3X) = B(X)+ M) has exhaustively two c Auty/2(G). [] Theorem 1.13 Let €,(G) be the set of all embeddings of a graph G on a surface of genus g(orientable or nonorientable), then the number of nonisomorphic maps in £,(G) is ™(O)= se o MO (1.10) TEAuty /2(G) where ®(7) = (M € &(G)|r(M) = M or rTa(M) = M}. Proof Suppose X;, X», --:, Xm are all the equivalent classes of X = €,(G) under the group Auty/2(G) x (o), then m = m,(G). Let S(z) = {7 € Autij(G) x (a)| T(z) = x) be the stabilizer at x, a subgroup of Aut; /;(G) x (a). Because |Aut4/(G) x (a)| 2 [S(2;)l Xi], gee X, t= 1,2, -*-, m, we have m|Auty/2(G) x (a)| = De IS (xi) || Xi]. (1) By observing |S(x;)| independent of the choice of x; in the class X;, Ax.L4 Maps 387 the right hand side of (1) is 2,5)25, 5,1 rcx LEX TES (x) E 2 (2) TEAuty /2(G)x (a) z—T(x) », Ie. T€Aut4/2(G) x (a) From (1) and (2), the theorem can be soon derived. O The theorem above shows how to find nonisomorphic super maps of a graph when the automorphism group of the graph is known. Theorem 1.14 Fora graph G, let R,(G) and €,(G) be, respec- tively, the sets of all nonisomorphic rooted super maps and all distinct embeddings of G with size e(G) on a surface of genus g(orientable or nonorientable). Then, 2e(G) RG)| = auty /2(G) Exa (1.11) Proof Let M,(G) be the set of all nonisomorphic super maps of G. By (11.3), IR (G)| ES 4e(G) 2 x auti/5(G) 7 2 x autı/2(G') aut( M) MeM,(G) By considering that 2 x auty/2(G) = |Auty/2(G) x (a)l = [(Autijs(G) x (a))|u| x |Auti2(G) x (a) (M)| and (Aut;/5(G) x (o))|u = Aut(M), we have 4e(G) [R (G)| = ERES » |Autij2(G) x (o)(M)| n2 MeM,(G) A). le (a) H auty /2 (G) 388 Appendix I: | Concepts of Polyhedra and Maps This is (1.11). O This theorem enables us to determine all the super rooted maps of a graph when the automorphism group of the graph is known. How- ever, the problem of finding an automorphism of a graph is much more difficult than that of finding an automorphism of a map on the basis of Chapter VIII in general. For asymmetric graphs the two theo- rems above provide results much simpler. More results are referred to [MLW1]. Appendix II Table of Genus Polynomials for Embeddings and Maps of Small Size For a graph G, let pe(x), a(x) and utc (x) be, respectively, the orientable genus distributions of embeddings, super maps and rooted super maps of G, or called orientable genus polynomials. Similarly, let qc(zx .), vc(x !) and v5(x !) be, respectively, the nonorientable genus distributions of embeddings, super maps and rooted super maps of G, or called nonorientable genus polynomials. Ax.IL1 Triconnected cubic graphs First, list all nonisomorphic 3-connected cubic graphs from size 6 through 15. Size 6 390 Appendix Il: Table of Genus Polynomials for Embeddings and Maps Size 9 C931 C92 Size 12 C121 C12,2 C12.3 C124 Size 15 C353 C15,3 C154 C15,5 C15,6 Ax.IL1 Triconnected cubic graphs 391 Q9 S V C45,7 C15,8 C15,9 C35,10 Cis C152 C1543 C1514 In what follows, the orientable and nonorientable genus poly- nomials of embeddings, super maps and rooted super maps of 3- connected cubic graphs shown above are provided. Case of size 6: Orientable Po, (x) = 2+ 14z, Bos (z) = 1+ 2a, Mo, (@) = 1+ 72. Nonorientable qos, (z 1) = 14a + 422? + 562°, Ve. (eo) = 2a Ba" 432", VO, (2 *) = "x + 21a? + 282°. 392 Appendix Il: Table of Genus Polynomials for Embeddings and Maps Case of size 9: Orientable Doy, () = 2 + 38x + 242”, Ho, (x) = 1-4 5x + 22”, Mo, (x) = 3+ 57a + 3627; Do, (x) = 40x + 2427, pos Qc) = 20 ae Ho, (x) = 10x + 6x”. Nonorientable Gcy,(a~*) = 22x + 1222? + 4242? + 39227, vc, (£t) = 3x + 12x? + 282? + 23x, Vo, (X 1) = 33x + 1832? + 6362? + 588^; qo, (x!) = 12x + 1082? + 4322? + 40824, ve, (x 1) = x + 2x? + 62? + 62^; Vo, (2°) = 3x + 272^ + 1082? + 102. Case of size 12: Orientable PC (@) = 2+ 70x + 1842”, Boc) = 1+ 15z- 2827, Ho, (2) = 12 + 420% + 110427; Ax.IL1 Triconnected cubic graphs 393 Dc, (x) = 64x + 19227, I C13,5 (x) = 4z + 122, Hp, (x) = 128x + 3842”; Posle) = 56x + 2002”, ics (m) = 5x + 1327, Ho, (x) = 84x + 30027; Dc, (X) = 2 + 54x + 2002”, Jic, (m) = 1+ 52x + 82°, MO, (X) = 14+ 27x + 1002. Nonorientable Cia (X 7) = 30x + 2422? + 14482? + 32722 + 29442", Vc, (m 1) = Tx + Ada? + 217a? + 4522 + 382°, V... (£71) = 180x + 14522? + 8688x? + 1963224 + 176647; qc, (X 7) = 12x + 1802? + 13602? + 331227 + 30722”, Vc, (2 1) = £ + 92? + 642? + 14924 + 13725, Vp, (X 1) = 24a + 36027 + 2720x? + 6624x + 61447"; qc, (X 7) = 10x + 1582? + 12722? + 32962 + 32002”, Voss (X 1) = 2a + Ma? + 572? + 1332 + 11825; Vp, |) = 15a + 23727 + 19087? + 2944" + 48002”; qc, (7 .) = 24x + 192z* + 12882? + 326427 + 31682, Vc, (x 1) = £ + Tz? + 24r? + 58x + 405; Vo, (X 1) = 12z + 962? + 644r” + 16322 + 15842°. Appendix II: Table of Genus Polynomials for Embeddings and Maps Case of size 15: Orientable Dc, (x) = 2 + 102z + 664? + 25622, Jc, (m) = 1+ 27a + 17627 + 682°, Lc, (x) = 30 + 15302 + 996027 + 384025; ( PCs, r) = 720 + 6642? + 288r, Hess (x) = 20x + 1802" + 67227, Hes, (m) = 1080x + 99602? + 43202; Dc, (t) = 56x + 64827 + 32027, HCis,3 (x) 12x + 96x? d 4425, Hc, (x) = 420x + 48602? + 240027; ( Dc, (x) = 80x + 688z? + 25627, Kesal T] = liz + 9327 + 32x3, Ho, (£) = 600x + 51602? + 19202; Dc, (£) = 2 + 118x + 6482? + 25622, 14 27z + 8827 + 362°, 15 + 885x + 4860x? + 19202 T Ho s (2) = (2) = Dc, (0) = 2+ 110x + 688x? + 22427, (£) = 1 + 142 + 602" + 202°, CL) = ( ( ( Uc, (£) = 10 + 550x + 344027 + 11205. Dc, (5) = 2 + 78x + 65677 + 28827, ,(v) 9 1-4 142 + 812? + 2425, r | \= Dc, (x) = 96x + 672x? + 25627, Hass) = 9r + 492? + 182°, His (m) = 3602 + 25202? + 9602°: Ax.IL1 Triconnected cubic graphs 395 PoigaUt) = 48r + 656x? 4 3202, HC15,9 (x) = 8rd 59x? 4r 2523, Ho, (£) = 180x + 2460x? + 1200z°; DC, 4 (1) = 88x + 64827 + 2882°, HC45,10 (x) = 5x + 31r’ + 162°, Hr, (£) = 220x + 16202 + 1202; Pois (£) = 2 + 70x + 63227 + 3202°, HCis,11 (x) = = 1] + 5x + 28x" + 102°, MCs (£) = 3 + 1052 + 9482? + 48027; PCis, a= 72% + 6324? T 3202, UCs S (X) = 6x + 24a? + 142°, Ho, (©) = 108x + 948x? + 48055. PCs (£) = 48x + 72027 + 25627, HC, (x) = 2x + 152” SF 6x’ ; Hc, G0) = 302 + 45027 + 16025; PCs (£) = 40x + 6642? + 32027, jig) =L + Tx? e 229, He, uu) = 10x + 1662? + 802. Nonorientable qc, (x 7) = 38x + 39427 + 33362? + 1274424 + 270082? + 209922°, Vc (x 1) = 10x + 1042? + 8382? + 322027 + 67682? + 53002; Vos 1) = 570a + 59102? + 500402? + 1911602* + 4051202? + 314880z°: 396 Appendix Il: Table of Genus Polynomials for Embeddings and Maps qc, (£t) = 10x + 214x? + 25762? + 11664z* + 274242? + 226247°, VCs (x 1) = 4v + 602? + 6761? + 29882 + 69524? + 56882; Vp, (x |) = 150a + 32102? + 386402? + 1749602* + 4113602? + 3393602°; Ciz3(@') = 6x + 1582? + 21882? + 1091224 + 275042? + 237442, Vc, (x 1) = 2x + 27a? + 3132? + 146624 + 35722? + 30442; Vox 1) = 45a + 11852? + 164102? + 818402* + 2062802? + 17808025; qc, (07 .) = 12x + 244z* + 28162? + 12224z* + 274562? + 217602, Vc, (x 1) = 2x + 33a? + 36827? + 15652 + 34802? + 27362; Voy, 4(@ 1) = 90x + 18302? + 211202? + 916802* + 2059202? + 1632002°; qc, (0 |) = 38x + 410? + 34962? + 129522 + 268802? + 207362°, Voy, (x1) = 8x + T6a? + 5242? + 17682 + 34602? + 26522; VO s (27) = 3857 + 30752? + 262202? + 971402* + 2016002? + 15552025, Cizg(@ |) = 38x + 402? + 34482? + 1304024 + 270722? + 205122°, Ve, (x 1) = 6r + 44a? + 3192? + 11572? + 23542? + 174425; Vox 1) = 190x + 20102? + 172402? + 652002* + 1353602? + 102560; Ax.IL1 Triconnected cubic graphs 397 qc, (0 |) = 32x + 312x? + 28002? + 1180024 + 272002? + 223682°, VCs (x 1) = 5x + 35a? + 2672? + 10772 + 23582? + 18662; VO,,,(@ +) = 160x + 15602? + 140002? + 5900074 + 1360002? + 1118402°; qc, (£t) = 12x + 260x? + 297621? + 124322 + 273282? + 215042°, Vc, (x 1) = 2x + 212? + 2072? + 82824 + 1772x° + 13822°; vL, (a!) = 45x + 9752? + 111602? + 2662024 + 1924802° + 806402°; QC, (x .) = 4a + 1322? + 20492? + 1072024 + 276162? + 240002°, Vc, (m |) = £ + 16x? + 1522? + 753x* + 18112? + 155925; Voy, (X |) = 15x + 4952? + 16502? + 402002* + 1035602? + 90000z°: QC, ao (0 7) = 12x + 2522? + 28642? + 121362 + 272642? + 219846, Voss) = £ + Ma? + 1242? + 5172 + 11542? + 9412; Vern (0 1) = 30x + 63027 + T1602? + 30340x* + 6815602? + 54960; QC, (£71) = 30a + 2822? + 25602? + 112402 + 271682? + 232322°, Vos u (27t) = 2x + 17x? + 922? + 351z* + 1542? + 62425; VO (©) = 45a + 4232 + 38402? + 168602* + 407522? + 348482°, 398 Appendix Il: Table of Genus Polynomials for Embeddings and Maps QC,s ao (0 7) = 12x + 2202? + 24802? + 1124024 + 272642? + 232969, Vo, s (t 1) = 2x + Ma? + 902? + 34324 + T5625 + 63825; VoU 1) = 18r + 3302? + 37202? + 1686074 + 408962? + 34944x; dC, (2 1) = 120x? + 22322? + 11568x + 27936x" + 226562°, Vo, (t 1) = 4a? + 28x? + 14474 + 3072? + 25925; Vos 1) = 1927 + 13952? + 72302 + 174602? + 14160; dC, (xL) = 4x + 120x? + 19002? + 104402 + 276642? + 243842, Vo, (X 1) = a + 22? + 16x? + 62 + 14225 + 11125; Ver, (#1) = a + 302? + 474r? + 261024 + 69162? + 60392°. Ax.IL2 Bouquets Let Bm be the bouquet of size m, m > 1. Case of m — 1: Orientable Ax.IL2 Bouquets 399 Nonorientable Case of m — 2: Orientable pp,(x) = 4-F 2x, HB» (x) =1+2, Hp, (x) = 2+ x. Nonorientable dp, (x t) = 10x + 827, vele) = 2g + 2z^, Vp, (£~!) = 5g + 42”. Case of m — 3: Orientable pp,(x) = 40 + 80x, lp (t) = 2+ 3x, Lp, (x) =5+10z. Nonorientable qp, (xz |) = 176x + 336z? + 3282°, 400 Appendix Il: Table of Genus Polynomials for Embeddings and Maps vg,(x^l) = 5x + 8x? + 82%, Vg, (x 1) = 22x + 422? + 412°. Case of m — 4: Orientable pp,(x) = 672 + 3360x + 10082”, bop, (2) = 3 +102 + 42”, Uf, (m) = 14 + 70x + 212°. Nonorientable qp,(x |) = 44642 + 145922? + 331202? + 23424a**, vg,(z +) = 129 + 33z? + 642? + 4724, vp, (x 1) = 93x + 304z? + 6902? + 48827. Case of m — 5: Orientable pp,(x) = 16128 + 161280 + 18547227, Lp, (x) = 6 + 35x + 382”, UJ, (x) = 42 + 420x + 48327. Nonorientable qp,(x |) = 1482242 + 718080z? + 27456002? + 447744024 + 31599362°, vp,(z |) = 33x + 131z? + 4422? + 686z^ + 4732°, vg, (x 1) = 386x + 187077 + 71502? + 116602* + 82292”. Ax.II.3 Wheels 401 Case of m — 6: Orientable PB (XL) = 506880 + 8870400x + 24837120x? + 570240027, La, (xz) = 124+ 1327 + 3282? + 822°, Lp, (x) = 132 + 2310x + 6468z? + 14852°. Ax.IL3 Wheels Let W, be the wheel of order n, n > 4, i.e., all vertices are of valency (or degree) 3 but one and all 3-valent vertices form a circuit. Case of n — 4: Orientable pw,(x) = 2 + 142, pw,(z) = 1+ 2z, by, (2) = 14+ Te. Nonorientable qw,(x 5) = 14x + 42x? + 562°, uma J= 2x + 8x? + 325, Viy,(@ |) = Tx + 212? + 282°. Case of n= 5: Orientable pw, (x) = 2 + 58x + 362°, pw, (2) = 1+ 8a + 42’, Uy, (x) = 4 + 1162 + 722°. 402 Appendix Il: Table of Genus Polynomials for Embeddings and Maps Nonorientable qw, (£7}) = 28x + 1762? + 6402? + 59624, vy, (xt) = 4r + 18x? + 522? + 482$, viy, (x 1) = 56x + 352z? + 12802? + 1192z*. Case of n — 6: Orientable pw, (x) = 2 + 190x + 57627, ima = 1-- 14x + 412’, py, (0) = 4+ 380x 4-115227. Nonorientable qw,(x 1) = 52x + 5802? + 40802? + 9880z* + 92162, vw,(z +) = 6x + 38x? + 2272? + 539x + 4942, Vi, (x 7) = 104x + 11602? + 81602? + 197602* + 1843227. Case of n = T: Orientable pw, (x) = 2 + 550x + 4968z? + 216027, pw; (x) = 14 34x + 240z? + 1062, Uy (x) = 4 + 1100x + 99362? + 432027. Nonorientable qw,(x 1) = 94x + 1680z? + 194822? + 87536z* + 2054962? + 1695526. Ax.II.4 Link bundles 403 vw, (£71) = 8x + 89a? + 8782? + 3829x* + 8788x? + 72412°, Vi (x 7) = 188x + 33602? + 389642? + 175072x* + 4109922? + 3391042. Case of order n — 8: Orientable pw, (x) = 2 + 1484x + 311782? + 5949627, uw, (£) = 1 + 63x + 11762? + 224627, Ly, (x) = 4 + 2968x + 623562? + 1189927. Ax.ILA4 Link bundles Let L,, be the link bundle of size m, m > 3. A link bundle is a graph of order 2 without loop. Case of size m = 3: Orientable Duls)e2-p2u, pn(x)-1-z, Bp, (£) = l- lm. Nonorientable ina) = 6x + 627, victa t = laos, v; (£71) = 32 + 32°. 404 Appendix Il: Table of Genus Polynomials for Embeddings and Maps Case of size m = 4: Orientable pr,(x) = 6 + 30z, Hr, (z) = 1-4 2z, ui, (£) = 1+ 5z. Nonorientable qr, (x .) = 36x + 96z? + 12027, vns ess 35", v; (x |) = 6x + 162? + 202°. Case of size m = 5: Orientable pr,(x) = 24 + 360x + 1922”, ir.(2) = 14-32 + 32", ur, (£) = 1+ 15x + 82". Nonorientable qr, (x |) = 240x + 12002? + 38402? + 3360274, vp, (zx 1) = 2x + Tx? + Mz? + 142%, v; (x!) = 10x + 50x? + 1602? + 14014. Case of size m = 6: Orientable pr, (x) = 120 + 4200x + 100802, Dr, (x) = 1 + 6x + 102°, us, (x) = 1+ 35a + 842”. Ax.IL5 Complete bipartite graphs 405 Nonorientable qn, (x ) = 1800x + 144002? + 841202? + 184320z* + 1617602, vp,(z |) = 3x + 142? + 482? + 9627 + 722°, vi (x |) = 15x + 1207? + 7012? + 1536x* + 13482”. Case of size m = T: Orientable pr,(x) = 720 + 50400 + 3376802? + 12960025, pz, (x) = 14 8x + 31x? + 162°, ur, (£) = 1 + 70x + 46927 + 18027. Ax.IL5 Complete bipartite graphs Let Km,n be the complete bipartite graph of order m+n, m,n > Case of order m+n — 6: Orientable DK,s(x) = 40x + 24x". Ux (t) = 2x + z?, Hk, (£) = 10x + 6x”. Nonorientable qK,,(') = 122 + 108z? + 4322? + 4082+, Vigna) — vd 227 | 6x? | 62, Vi. (1-1) = 3x + 27x? + 1082? + 1022. 406 Appendix Il: Table of Genus Polynomials for Embeddings and Maps Case of order m -4- n — T: Orientable Dk, (x) = 156x + 22442? + 105622, pcs (0) = 3x + 162? + 102°, Heale) = 26x + 314a? + 17627. Nonorientable qi, (x |) = 12x + 432? + 68522? + 3628827 + 933602? + 807842°, Vk, (X 3) Y) = g + Az? + 3325 + 15624 + 35825 + 31726, Vic, (@ 1) = 2x + 72x? + 11422? + 6048x* + 155602? + +134642°. Case of order m+n = 8: Orientable PK,,(@) = 108x + 249842? + 5650202? + 108950414, HK, (T) = 2x + 252? + 3182? + 53027, Hic, (£) = 3x + 6942? + 156952? + 302642. Dk,,(x) = 240x + 315842? + 2908802? + 1136642", pcs (x) = x + 3327 + 2252? + 10527, Hg, (x) = 10x + 156627 + 121202? + 473627. Ax.IL6 Complete graphs 407 Ax.IL6 Complete graphs Let , be the complete graph of order n, n > 4. Case of order n = 4: Because of K4 = Wi, seen from Case of order n = 4 in Ax.II.3. Case of order n = 5: Orientable PK,(@) = 462x + 49742? + 324027, ju, (£) = 6x + 312? + 132, Up (m) = 77x + 8292? + 3902°. Nonorientable qx, (x .) = 54x + 1320x? + 174902? + 8466024 + +208776x° + 17758829, v,(x 1) = Qe + 112? + 992? + AVI + 9552? + 7962, Vy. (xL) = 9x + 220x7 + 29152? + 14110z* + 347962? + 4-295982. Case of order n = 6: Orientable Dk, (x) = 1800x + 6545762? + 24613800.X? +41242502087 + 415825922°. 408 Appendix Il: Table of Genus Polynomials for Embeddings and Maps Nonorientable qx,(x 1) = 24a + 45602? + 3709202? + 1082844027 + 1922645762° + 1927543032x5 + 119055609602" + 42386101920? + 798313881604? + 592442818562”, Appendix III Atlas of Rooted and Unrooted Maps for Small Graphs In the symbol X : a,b,c for a map appearing under a figure below, X is the under graph of the map, a — oy or qy are, respectively, orientable or nonorientable genus y, b is the series number with two digits and c is the number of ways to assign a root. And, r on a surface is for zl, or —z, x = 1,2,--+. Ax.III.1 Bouquets Bm of size 4> m > 1 Case m = 1: Orientable genus 0 1 CE» 1 B1:00—01—01 Nonorientable genus 1 1 B; : q1 — 01 — 01 410 Appendix III: | Atlas of Rooted and Unrooted Maps Case m — 2: Orientable genus 0 Bə : o0 — 01 — 02 Orientable genus 1 Bə : ol — 01 — 01 Nonorientable genus 1 Ax.III.1 Bouquets Bm of size4 2 m 2-1 Nonorientable genus 2 H By: q2 — 01 — 02 Case m = 3: Orientable genus 0 2 Bz : 00 — 01 — 02 Orientable genus 1 All EM Bə : q2 — 02 — 02 B3 : o0 — 02 — 03 1 B; : ol — 01 — 04 B; : ol — 02 — 03 B3 : ol — 03 — 03 412 Appendix III; | Atlas of Rooted and Unrooted Maps Nonorientable genus 1 B; : q1 — 01 — 06 Bs : q1 — 02 — 06 B3 : q1 — 03 — 06 3 B3:q1—04—-03 B3:q1—05— 01 Nonorientable genus 2 Bz: q2 — 02 — 12 Bz : q2 — 03 — 06 1 1 Ax.III.1 Bouquets Bm of size 4>m> 1 413 3 B; : q2 — 07 — 03 Bs : q2 — 08 — 03 Nonorientable genus 3 3 B; : q3 — 04 — 03 B; : q3 — 05 — 06 B3 : q3 — 06 — 06 1 B3 : q3 — 08 — 03 414 Appendix III: Atlas of Rooted and Unrooted Maps Case m — 4: Orientable genus 0 B; : 00 — 01 — 02 Orientable genus 1 Bg: o1 — 02 — 08 Bag: 01 — 03 — 08 1 1 B; : o1 — 04 — 08 B4:01— 05 — 16 B4: 01 — 06 — 08 1 1 1 B4:01 — 07 — 04 B4:01— 08 —04 B; : ol — 09 — 04 Ax.IIL2 Link bundles L,,,7 > m > 3 415 B4:01 — 10 — 02 Orientable genus 2 eI B4:02—01— 04 Bg : 02 — 02 — 08 B4: 02 — 03 — 08 wi By: 02 — 04 — 01 Ax.IIL2 Link bundles Lm, 6 > m > 3 Case m = 3: Orientable genus 0 416 Appendix III: Atlas of Rooted and Unrooted Maps hol Ls : o0 — 01 — 01 Orientable genus 1 L3 : ol — 01 — 01 Nonorientable genus 1 Ls : q1 — 01 — 03 Nonorientable genus 2 Ax.IIL2 Link bundles L,,,7 > m > 3 417 Ls : q2 — 01 — 03 Case m = 4: Orientable genus 0 L4:00—01— 01 Orientable genus 1 L4:01—01—01 L4:01—02—04 418 Appendix III: Atlas of Rooted and Unrooted Maps Nonorientable genus 1 La :q1—01-02 L4 :q1-— 02 = 04 Nonorientable genus 2 Lg: q2 — 01 — 02 Lg: q2 — 02 — 08 L4: q2 — 03 — 02 L, : q2 — 04 — 04 Nonorientable genus 3 Ax.IIL2 Link bundles L,,,7 > m > 3 419 La : q3 — 01 — 08 La : q3 — 02 — 08 La : q3 — 03 — 04 Case m = 5: Orientable genus 0 wi L5:01 — 01 — 05 Ls; : ol — 02 — 05 Ls; : ol — 03 — 05 Orientable genus 2 420 Appendix III: Atlas of Rooted and Unrooted Maps px] L5:02—01 — 02 L5:02— 02 — 05 L5:02—03—01 Nonorientable genus 1 Ls : q1 — 01 — 05 Ls : q1 — 02 — 05 Nonorientable genus 2 Ls; : q2 — 01 — 05 Ls : q2 — 02 — 10 Ls : q2 — 03 — 10 Ls : q2 — 04 — 05 Ls : q2 — 05 — 10 Ls : q2 — 06 — 05 Ax.IIL2 Link bundles L,,,7 > m > 3 421 Ls : q2 — 07 — 05 Nonorientable genus 3 bol Ls : q3 — 01 — 10 Ls : q3 — 02 — 20 Ls : q3 — 03 — 20 L5:43 — 07—10 L5:q3 — 08 — 10 Ls : q3 — 09 — 20 422 Appendix III: | Atlas of Rooted and Unrooted Maps wi Ls :q3— 10 — 10 L5:q03—11— 10 Ls : q3 — 12 — 05 Ls :q3—13—-05 Lg:g8—14— 05 Nonorientable genus 4 Ls :q4— 01 — 10 Ls : q4 — 02 — 10 Ls :q4— 03 — 10 L5:q4 — 04 — 10 L5:q4 —05 — 10 L5:q4 — 06 — 10 Ax.IIL2 Link bundles L,,,7 > m > 3 423 L5 :94 — 07 — 05 Ls : q4 — 08 — 20 Ls :q4— 09 — 10 Ls :g4 — 10 — 05 Ls :q4— 11—05 Ls : q4 — 12 — 20 Case m = 6: Orientable genus 0 424 Appendix III: Atlas of Rooted and Unrooted Maps Lg:01—04—06 Lg:01—05—06 . Lg:01 — 06 — 06 Orientable genus 2 Ax.IIL2 Link bundles L,,,7 > m > 3 425 Le : 02 — 03 — 03 E 5 3 2 2M : 02 — 09 — 06 Íargd-07-19- Le:02—08—12 Le 3 2 4 Case m = T: Orientable genus 0 5 2 6 i 6 1 5 2 Ly : o0 — 01 — 01 Orientable genus 1 Orientable genus 2 SEE — 28 L; : 02 — 15 — 28 = 4 3 1 3 1 2 6 1 6 1 NEA ; $ L7 : 02 — 14 1x \ SEDAN A = oye CAN 5 m" i'd AM HUE 0 S: AR: 1 6 = Al Ms it AL Mn y V px [ine] gent A xb AA N o 9 Rd S © i : Af NE OW : Orientable genus 3 a 5 3 2 5 4 o3 i A Ne k - ut S| | CN z j=) | eo IN 10 = bs I Re) -] KO o H ho, N = Oo 3 - 2 I 6 5 L7 : 03 — 01 — AX. WM | V Im ho vet sy Ly Im Ist iml Xf M : QU : r- J <H - | Aum e | ine) o r- = dy. oe. gh. QAP PAN PUAJY | Ax :dx:a: gx ! AP QP PAP AP: Ui ap: wa: pO 5 2 1 2 XN € 1 6 5 L7:03—10— 2 6 5 Lz: 432 Appendix III: | Atlas of Rooted and Unrooted Maps Ax.III.3 Complete bipartite graphs Ky, 7,4 > m,n > 3 Case m+n — 6: Orientable genus 1 K35:0l— 1-109 K33:01 —02—01 Orientable genus 2 K33:02— 01 — 06 Nonorientable genus 1 K33:g1—01-—03 Nonorientable genus 2 Ax.IIL3 Complete bipartite graphs Km n,4 > m,n > 3 433 K33:q02— 01 — 18 K33:q2— 02 — 09 Nonorientable genus 3 K35:q3— 01— 18 Kost 09 — 18 K33:q3— 03 — 36 K35 : q3 — 04— 09 K3 : q3 — 05 — 18 K3 3 : q3 — 06 — 09 Nonorientable genus 4 434 Appendix III: | Atlas of Rooted and Unrooted Maps K33:g4 — 01 — 03 K3 3 :q4— 02—18 K3 3 : q4 — 03 — 36 K3 3: q4— 04— 18 K3 3 : q4 — 05 — 09 K3 3 :q4— 06 — 18 Case m -- n — t: Orientable genus 1 K43:01-— 01 — 06 K43:01— 02 — 12 K43:01— 03 — 08 Orientable genus 2 435 Ax.III.3 Complete bipartite graphs Km n,4 > m,n > 3 — 03 — 12 Ka3 : 02 K45:00—02—94 K43:02—01— 02 K43:02— 05 — 24 K4,3 : 02 — 06 — 24 K43 : 02 — 04 — 24 K43:02— 08 — 24 K4,3 : 02 — 09 — 04 K43:02— 07 — 12 48 K4,3 : 02 — 12 — K43:02— 11 — 24 — 48 K4,3 : 02 — 10 436 Appendix III: | Atlas of Rooted and Unrooted Maps K43:02— 13 — 24 K43:02— 14 — 48 K43:02— 15 — 08 K43:02—16—24 Orientable genus 3 K4,3 : 03 —01 — 24 K43:03 — 02 — 24 K43:03 — 03 — 08 Ax.III.4 Wheels W, of order 6 > n > 4 437 K43:08 — 04 — 24 K4,3 : 03 — 05 — 48 K4,3 : 03 — 06 — 08 K43 : 03 — 07 — 08 Ka, : 03 — 08 — 24 K43 : 03 — 09 — 06 K4,3 : 03 — 10 — 02 Ax.IIL.A4 Wheels W,,5 >n > 4 Case n = 4(i.e., the complete graph K, of order 4): Orientable genus 0 438 Appendix III: | Atlas of Rooted and Unrooted Maps W4:00—01-— 01 Orientable genus 1 W3 : ol — 01 — 03 W3 : ol — 02 — 04 Nonorientable genus 1 W, : q1 — 01 — 06 W, : q1 — 02 — 01 Nonorientable genus 2 Ax.III.4 Wheels W, of order 6 > n > 4 439 W3 : q2 — 01 — 06 W3 : q2 — 02 — 12 W3 : q2 — 03 — 03 Nonorientable genus 3 Wz : q3 — 01 — 12 W3 : q3 — 02 — 12 W3 : q3 — 03 — 04 Case n = 5: Orientable genus 0 wi Ws : 00 — 01 — 04 440 Appendix III; | Atlas of Rooted and Unrooted Maps Orientable genus 1 W; : ol — 02 — 16 W; : ol — 03 — 04 W; : ol — 04 — 32 Ws : ol — 05 — 16 Ws : ol — 06 — 16 W; : o1 — 07 — 08 Ws : ol — 08 — 08 Orientable genus 2 Ax.III.4. Wheels W, of order 6 > n > 4 441 Ws : 02 — 01 — 16 Ws : 02 — 02 — 08 Ws : 02 — 03 — 16 1 eI Ws : 02 — 04 — 32 Case n = 6: Orientable genus 0 We : 00 — 01 — 04 Orientable genus 1 442 Appendix III: Atlas of Rooted and Unrooted Maps We : ol — 03 — 40 We : ol — 06 — 40 We : ol — 07 — 20 We : ol — 08 — 20 Wa : o1 — 10 — 40 We : o1 — 11 — 40 We : ol — 12 — 20 Ax.III.4 Wheels W, of order 6 > n > 4 443 Ws : o1 — 13 — 20 Ws : 01 — 14 — 20 Orientable genus 2 We : 02 — 01 — 20 We : 02 — 02 — 20 We : 03 — 03 — 20 We : 02 — 04 — 20 We : 02 — 05 — 04 We : 02 — 06 — 40 444 Appendix III: Atlas of Rooted and Unrooted Maps We : 02 — 07 — 40 We : 02 — 08 — 40 We : 02 — 09 — 20 Wa : 02 — 10 — 20 We : 02 — 11 — 40 We : 02 — 12 — 40 Ws : 02 — 13 — 40 Ws : 02 — 14 — 40 Ws : 02 — 15 — 40 Ws : 02 — 16 — 20 We : 02 — 17 — 40 We : 02 — 18 — 20 Ax.III.4 Wheels W, of order 6 > n > 4 445 Ws : 02 — 19 — 20 Ws : 02 — 20 — 20 Ws : 02 — 21 — 40 We : 02 — 22 — 40 We : 02 — 23 — 40 Ws : 02 — 24 — 40 We : 02 — 25 — 40 We : 02 — 26 — 40 We : 02 — 27 — 20 We : 02 — 28 — 20 We : 02 — 29 — 40 We : 02 — 30 — 40 446 Appendix III: Atlas of Rooted and Unrooted Maps We : 02 — 31 — 40 We : 02 — 32 — 40 We : 02 — 33 — 20 We : 02 — 34 — 20 We : 02 — 35 — 20 We : 02 — 36 — 20 Ws : 02 — 37 — 20 We : 02 — 38 — 20 We : 02 — 39 — 20 We : 02 — 40 — 04 We : 02 — 41 — 04 Ax.IIL5 Complete graphs K„ of order 5 > n > 4 44T Ax.III.5 Complete graphs of order 5 > n > 4 Size 4: (K4 = Wa, W4 is known in Ax.IIL4). Size 5: Orientable genus 1 K; : o1 — 02 — 20 Ks : ol — 03 — 10 Ks : ol — 04 — 02 We : ol — 05 — 20 We : ol — 06 — 05 Orientable genus 2 K; : 02 — 02 — 20 Ks : 02 — 03 — 20 448 Appendix III: Atlas of Rooted and Unrooted Maps K; : 02 — 04 — 20 K; : 02 — 05 — 40 Ks : 02 — 06 — 20 K; : 02 — 07 — 05 K; : 02 — 08 — 40 K; : 02 — 09 — 20 K; : 02 — 10 — 20 Ks : 02 — 11 — 40 Ks : 02 — 12 — 20 K; : 02 — 13 — 20 K; : 02 — 14 — 40 K; : 02 — 15 — 40 Ax.III.5 Complete graphs Kn of order 5 > n > 4 449 Ks : 02 — 16 — 40 K; : 02 — 17 — 40 Ks : 02 — 18 — 20 3 1 ou wi 6 2 Ks : 02 — 19 — 40 K; : 02 — 20 — 40 Ks : 02 — 21 — 20 K; : 02 — 22 — 40 K; : 02 — 23 — 40 K; : 02 — 24 — 20 K; : 02 — 27 — 20 Ks : 02 — 25 — 10 K; : 02 — 26 — 40 450 Appendix III: Atlas of Rooted and Unrooted Maps Ks : 02 — 28 — 20 Ks : 02 — 29 — 40 Ks : 02 — 30 — 10 K; : 02 — 31 — 20 Orientable genus 3 K; : 03 — 01 — 40 K; : 03 — 02 — 20 K; : 03 — 03 — 40 Ax.III.5 Complete graphs Kn of order 5 > n > 4 451 K; : 03 — 04 — 40 K; : 03 — 05 — 20 K; : 03 — 06 — 40 K; : 03 — 07 — 40 K; : 03 — 08 — 20 K; : 03 — 09 — 10 K; : 03 — 10 — 40 Ks : 03 — 12 — 20 452 Appendix III: Atlas of Rooted and Unrooted Maps Ax.III.6 Triconnected cubic graphs of size in [6, 15] Size 6: (C51 = K4 = W4, W4 is known above). Size 9: (Co 2 = K33) C91: Orientable genus 0 E C9, : 00 — 01 — 03 Orientable genus 1 Co, : o1 — 01 — 09 C9, : ol — 02 — 18 C94 :01— 03 — 18 Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 453 EI Co, : ol — 04 — 09 C9, : 01 — 05 — 18 Orientable genus 2 Ü53 :02— d — 18 C51 : 02 — 02 — 18 C2: Orientable genus 1 Co,2 : 01 — 01 — 01 Co,2 : o1 — 02 — 09 Orientable genus 2 454 Appendix III: Atlas of Rooted and Unrooted Maps Co,9 : 02 — 01 — 06 Size 12: (C124 is the cube) C121: Orientable genus 0 Ci24 : 00 — 01 — 12 Orientable genus 1 Ci24 : ol — 01 — 24 Ci24 : ol — 02 — 24 Ci24 : o1 — 03 — 12 Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 455 Ci24 of ol = 04 — 24 Ci24 :ol— 05 = 48 Ci24 H ol = 06 — 24 Ci24 : ol — 07 — 48 Ci24 : ol — 08 — 24 Ci24 : ol — 09 — 48 Ci24 : ol — 10 — 24 Ci24 : ol — 11 — 24 Ci24 : o1 — 12 — 48 456 Appendix III; | Atlas of Rooted and Unrooted Maps Ci24 : ol — 13 — 12 Ci24 : ol — 14 — 24 Ci24 : o1 — 15 — 12 Orientable genus 2 Ci24 :02 — 01 — 28 Ci24 : 02 — 02 — 48 Ci24 : 02 — 03 — 48 Ci24 : 02 — 04 — 48 Ci24 : 02 — 05 — 48 Ci24 : 02 — 06 — 24 Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 457 Ci24 : 02 — 0T — 24 Ci24 : 02 — 08 — 24 Ci24 : 02 — 09 — 48 Ci24 :02— 10 — 48 Ci24 : 02 — 11 — 48 Ci24 : 02 — 12 — 48 Ci24 : 02 — 15 — 48 Ci24 B 02 — 16 — 48 Ci24 : 02 — 17 — 48 Ci24 : 02 — 18 — 48 458 Appendix III: Atlas of Rooted and Unrooted Maps 1 Ci24 : 02 — 25 — 48 Ci24 : 02 — 27 — 24 Ci24 : 02 — 28 — 24 Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 459 C123: Orientable genus 1 2 C2 :01 — 01 — 08 C2 : ol — 02 — 24 C122 : ol — 03 — 48 C123 :ol— 04 = 48 Orientable genus 2 C2 :02 — 01 — 24 C2 : 02 — 02 — 48 C122 :02 — 03 — 24 460 Appendix III; | Atlas of Rooted and Unrooted Maps 1 C23 $ 02 = 04 — 24 C2 :02— 05 = 48 C123 H 02 = 06 — 24 C123 : 02 — 07 — 24 C2 : 02 — 08 — 08 C122 : 02 — 09 — 48 C12 $ 02 — 10 — 48 C2 :02— 11— 16 C12,2 : 02 —12— 48 Orientable genus 1 Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 461 Ci2,3 :01 — 01 — 12 C12,3 : ol — 02 — 24 Ci2,3 : ol — 03 — 24 Ci2,3 : o1 — 04 — 12 C123 :01 — 05 — 12 Orientable genus 2 Ci2,3 :02 — 01 — 24 C12,3 : 02 — 02 — 03 Ci2.3 :02 — 03 — 24 462 Appendix III; | Atlas of Rooted and Unrooted Maps Ci2,3 $ 02 = 04 — 24 C12,3 : 02 — 05 = 48 C123 ` 02 =i 06 — 24 Ci2,3 : 02 — 07 — 48 C12,3 :02 — 08 — 24 C123 : 02 — 09 — 24 C2,3 : 02 — 10 — 06 Ci2,3 : 02 — 11 — 24 Ci»,3 :02 — 12 — 24 Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 463 3 C12,3 :02— 13 = 03 C124: Orientable genus 0 Ci24 :00— 01— 01 Orientable genus 1 Ci24 :01 — 01 — 03 Ci24 : ol — 02 — 12 Ci» : ol — 03 — 03 464 Appendix III; | Atlas of Rooted and Unrooted Maps Ci24 : o1 — 04 — 08 Ci24 :01 — 05 — 01 Orientable genus 2 Ciz x 02 = 01 = 04 Ci24 :02 — 02 — 24 Ci24 : o2 az 03 —12 C24 : 02 — 04 — 04 Cua : 02—05 — 12 054 : 02 — 06 — 24 Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 465 Ci24 : 02 — 07 — 08 Ci24 : 02 — 08 — 12 Size 15: Two chosen where C4514 is the Petersen graph. Cis qi: Orientable genus 0 wi 6 5 Cii : 00 — 01 — 03 Orientable genus 1 C1511 :01 — 01 — 30 C1511 : ol — 02 — 30 Cikin :02 — 03 — 15 466 Appendix III; | Atlas of Rooted and Unrooted Maps C15,11 :01 — 04 — 15 Cisai :01 — 05 — 15 Orientable genus 2 BI 6 6 5 : 02 — 03 — 60 Cis; : 02 — 04 — 30 Cisa1i :02— 05 — 15 Cisai : 02 — 06 — 30 1 5 bol ou I C1511 : 02 — 07 — 60 C1511 : 02 — 08 — 30 Cikin : 02 — 09 — 30 Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 467 C1511 : 02 — 10 — 30 Cisai :02 — 11 — 30 Cisi : 02 — 13 — 60 Cis. : 02 — 14 — 30 Cisai :02 — 15 — 30 1 6 Cisi : 02 — 16 — 03 C15,11 :02 — 17 — 60 Cisai : 02 — 18 — 30 Cis, : 02 — 19 — 30 Cis ii : 02 — 20 — 30 Cikin : 02 — 21 — 30 468 Appendix III: Atlas of Rooted and Unrooted Maps Orientable genus 3 bol NI Ol I Ol ou I 5 I C1511 : 03 — 01 — 60 Cis ii : 03 — 02 — 60 Cikin : 03 — 03 — 60 Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 469 [x] 1 6 C1511 : 03 — 05 — 60 Cisai : 03 — 06 — 60 Cisai : 03 — 07 — 60 Ci5,11 : 03 — 08 — 30 Cisai : 03 — 09 — 30 C15,11 : 03 — 10 — 30 C1514: Orientable genus 1 470 Appendix III; Atlas of Rooted and Unrooted Maps C15,14 :ol — 01 = 10 Orientable genus 2 C15,14 : 02 — 02 — 30 Cis,4 : 02 — 03 — 30 C15,14 : 02 — 04 — 20 C15,14 : 02 — 05 — 30 Cis,4 : 02 — 06 — 10 Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] AT] C15,14 : 02 — 07 — 06 Orientable genus 3 I ou 6 2 Cis,14 : 03 — 01 — 60 Cis,4 : 03 — 02 — 20 Bibliography Baxter, R.J. 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Terminology (i, j)-edge, 186 (i, j)-map, 58 (i, j) s-map, 123 (i*, 7*)-map, 58 (l, s)*-edge, 186 (x, y)-difference, 348 (x, y)-difference, 348 1-addition, 244 1-product, 244 C*-oriented planarity, 309 H-valency, 209 i-connected, 35 i-cut, 35 i-map, 57 i-section, 301 i-vertex, 194 j-face, 194 j'-map, 58 n-cube, 141 N-standard map, 149 O-standard map, 126 s-manifold, 29 V-code, 298 1-set, 375 1st level segment, 285 2-partition, 34, 122 3-map, 123 absolute genus, 158 absolute norm, 307 admissible, 284 appending an edge, 86 arc, 2 articulate vertex, 187, 262 articulation, 303 assignment, 283 associate net, 121 associate surface, 283 associate surface graph, 288 automorphism, 190, 331 automorphism group, 192, 331 balanced, 121 barfly, 146 Base map, 317 base map, 262 basic adding, 96 basic appending, 96 basic contracting, 96 basic contracting irreducible, 101 basic deleting, 96 basic deleting edge irreducible, 101 basic equivalence, 112 basic partition, 45 basic permutation, 43 basic set, 41 basic splitting, 96 basic subtracting, 96 basic subtracting irreducible, 102 basic transformation, 96 Betti number, 17 bi-matching map, 208 bi-pole map, 232 biboundary, 310 480 Terminology bijection, 42, 101 bipartite, 100 bipartite graph, 6 bipole, 37 Blissard operator, 346 boundary identification, 306 boundary identifier, 306 bouquet, 37, 213 branch, 287 butterfly, 125 cavity, 209 cellular embedding, 381 celluliform, 300 chromatic number, 40 circuit, 5 circuit partition, 7 class of basic equivalence, 113 co-order, 113 cocircuit oriented map, 309 cocircuit oriented planarity, 309 cocircular map, 309 cocirculation, 310 cocycle, 34, 122 commutative, 168 completely symmetrical, 197 composition, 4 conjugate axiom, 46 conjugate), 46 connected, 5 contractible, 10 contractible point, 300 contracting, 76 contraction, 300 coorder, 19, 169 481 corank, 17 cosemiedge, 56 crosscap polynomial, 289 crosscap, 13 crossing number, 40 cubic, 209 cut-vertex, 59 cuttable, 68, 69 cutting, 68 cutting face, 69 cutting graph, 69 cutting vertex, 68 cycle, 43 cyclic number, 17 cyclic permutation, 43 decomposition, 37 decreasing duplition, 112 decreasing subdivision, 112 degree, 7 deleting, 72 different indices, 19 different signs, 187 digraph, 5 dipole, 341 directed pregraph, 2 distinct, 283 distinct, 376 double edge, 70 double H-map, 210 double leaf, 308 double link, 70 double loop, 70 double side curve), 10 down-embeddable, 20 482 dual, 49, 65, 377 dual Eulerian map), 100 dual H-map, 210 dual map, 66 dual matching, 208 dual matching map, 208 dual regular, 57 dual trail code, 179 edge, 2 edge independent partition), 59 edge rooted, 202 edge-automorphism, 330 edge-isomorphism, 329 efficient, 176 elementary equivalence, 378 elementary transformation, 377 embedding, 381 embedding), 17 empty pregraph, 2 end segment, 286 end, 1, 56 enumerating function, 241 equilibrious embedding, 64 equilibrium, 64 equivalent class, 113 Euler characteristic, 20 Euler characteristic(Euler charac- teristic), 113 Euler graph, 7 Eulerian, 361 Eulerian characteristic, 377 even, 361 even assigned conjecture, 102 even assigned map, 100 Terminology even pregraph, 7 eves-cyclic, 231 exchanger, 287 expanded tree, 283 expansion, 383 face, 17, 49, 377 face regular, 57 face representative, 121 face rooted, 202 face-algorithm, 172 face-regular, 195 fan-flower, 317 father, 286 favorable embedding, 60 favorable map, 60 favorable segment, 301 feasible segment, 300 feasible sequence, 298 finite pregraph), 2 finite recursion principle), 6 finite restrict recursion principle, 6 first operation, 47 first parameter, 349 fixed point, 191 flag, 384 full cavity, 209 general, 350 generalized Halin graph, 38 generated group, 50 genus polynomial, 289 genus, 138 graph, 5 Terminology ground set, 2, 42 Halin graph, 307 Halin map, 262 Hamilton map, 123 Hamiltonian circuit, 122 Hamiltonian graph, 122 handle polynomial, 289 handle, 13 harmonic link, 70 harmonic loop, 70 hexagonalization), 102 homotopic), 10 identity, 192 immersion, 33 incidence equation, 34 incident pair, 100 included angle, 56 increasing duplition, 112 increasing subdivision, 112 independent face set, 209 independent pair, 100 independent set, 100 induced, 44, 99 infinite pregraph, 2 initial end, 2 inner rooted, 311 inner vertex, 262 interlaced, 134 interpolation theorem, 21 irreducible, 140, 299 isomorphic, 385 isomorphic class, 168 isomorphism, 165, 330 483 joint sequence, 187 joint tree, 18, 187 Klein group, 22 Klein model, 31 ladder, 293 left projection, 346 link, 70 link bundle, 37, 403 link map, 194 loop, 70 loop bundle, 37 loop map, 67 loopless map, 265 map, 50 map geometry, 29 maximum genus, 21 maximum genus embedding, 162 maximum nonorientable face num- ber embedding, 63 maximum nonorientable genus, 162 maximum orientable face number embedding, 63 meson functional, 346 minimum genus, 21 minimum genus embedding, 162 minimum nonorientable genus, 162 minimum nonorientable genus em- bedding, 63 minimum orientable genus embed- ding, 63 multiplicity, 333 near quadrangulation, 261 484 near regular, 266 near triangulation, 261 necklace, 293 network, 6 node, 2 noncuttable, 68, 69 noncuttable block, 69 nonorientable form, 290 nonorientable, 10, 104, 374 nonorientable favorable genus, 162 nonorientable genus, 11, 158, 162 nonorientable pan-tour conjecture, 163 nonorientable pan-tour genus, 162 nonorientable pan-tour maximum genus, 162 nonorientable preproper genus, 162 nonorientable proper map conjec- ture, 163 nonorientable rule, 153 nonorientable rule 3, 155 nonorientable single peak conjec- ture, 39 nonorientable small face proper map conjecture, 163 nonorientable tour genus, 162 nonorientable tour map conjecture, 163 nonorientable tour maximum genus, 162 nonseparable, 58 odd circuit, 34 orbit, 23, 43 order, 19, 40, 43, 169, 192 Terminology orientable genus polynomial, 389 orientable, 10, 104, 374 orientable genus, 11, 138 orientable minimum pan-tour genus, 142 orientable pan-tour conjecture, 142 orientable pan-tour maximum genus, 143 orientable proper map conjecture, 142 orientable single peak conjecture, 39 orientable tour conjecture, 142 pan-Halin map, 262 pan-tour face), 142 pan-tour map, 142 parallel, 134 partition, 2 path, 5 petal bundle, 125, 213 planar graph, 35 planar map, 141 planar pedal bundle, 217 plane tree, 297 planted tree, 263, 297 point partition, 300 polygonal map, 62 polyhedral sequence, 298 polyhedron, 374 pre-standard, 317 pre-standard pan-Halin map), 262 prefect dual matching, 208 prefect primal matching, 208 pregenus, 11 Terminology pregraph, 2 premap, 47 preproper map, 60 primal H-map, 210 primal matching, 208 primal matching map, 208 primal regular, 57 primal trail code, 179 principle segment, 286 problem of type 1, 270 problem of type 2, 273 proper circuit, 99 proper cocircuit, 99 proper embedding, 60 proper map, 60 quadcircularity, 308 quadcirculation, 308 quadrangulation, 57 quadregular map, 306 quadricell, 42 quaternity, 306 quinquangulation, 102 rank, 17 reduced rule, 134 reduction, 299 relative genus, 158 reversed vector, 258 right projection, 346 Ringel ladder, 294 root, 203 root edge, 203 root face, 203 root vertex, 203 485 rooted edge, 202 rooted element, 201, 202 rooted face, 202 rooted map, 203, 236 rooted set, 201 rooted vertex, 202 rotation, 17 same, 283, 376 same sign, 187 second operation, 47 second parameter, 349 segmentation edge, 75 self-dual, 98 semi-arc isomorphism, 385 semi-automorphism, 328 semi-automorphism group, 329 semi-isomorphic, 327 semi-isomorphism, 327 semi-regular map, 61 semiedge, 56 separable, 35 set rooted), 201 sharp, 194 shearing loop, 82 side, 56 simple map, 265, 305 simplified barfly, 151 simplified butterfly, 128 single edge, 70 single link, 70 single loop, 70 single peak, 39 single side curve, 10 single vertex map, 125 486 singular link, 70 singular loop, 70 size, 19, 40, 169 skeleton, 377 small face favorable embedding, 61 small face proper embedding, 61 Smarandache geometry, 29 Smarandache multi-space, 28 son, 286 spanning cavity, 209 spanning tree, 17 specific face, 262 splitting, 88 splitting block, 35 splitting edge, 88 splitting pair, 35 standard, 317 standard, 303 standard form, 12 standard pan-Halin map), 263 standard splitting block, 35 standard splitting block decompo- sition, 36 standard surface, 381 straight line embedding, 17 strong embedding, 60 strong map, 60 subsidiary graph, 121 super premap, 50 support, 374 surface closed curve axiom, 10 surface embedding graph, 288 surface embedding, 17 switch, 376 Terminology symmetric, 190 symmetrical map, 197 terminal, 262 terminal end, 2 terminal link, 75 terminal loop, 81 thickness, 40 tour, 5, 142 tour map, 142 trail, 5 transitive, 5, 33, 50 transitive axiom, 50 transitive block, 66 transitive decomposition, 66 travel and traverse rule, 33 travel, 5 tree, 17 tri-pole map, 232 triangulation, 57 trivial, 192 trivial map, 127 TT-rule, 33 twist, 383 twist loop, 81 under pregraph, 50 uniboundary, 310 unicyclic, 324 unisheet, 230 up-embeddable, 20, 189 up-integer, 233 vertex, 2, 47, 377 vertex partition, 298 vertex regular, 57 Terminology vertex rooted, 202 vertex-algorithm, 172 vertex-isomorphism, 330 vertex-regular, 194 walk, 5 wheel, 38 wintersweet, 324 487 Abstract: A Smarandache system (3; R) is such a mathematical system with at least one Smarandachely denied rule F in R such that it behaves in at least two different ways within the same set X, i.e., validated and invalided, or only invalided but in multiple dis- tinct ways. A map is a 2-cell decomposition of surface, which can be seen as a connected graphs in development from partition to permutation, also a basis for constructing Smaran- dache systems, particularly, Smarandache 2-manifolds for Smarandache geometry. As an introductory book, this book contains the elementary materials in map theory, including embeddings of a graph, abstract maps, duality, orientable and non-orientable maps, iso- morphisms of maps and the enumeration of rooted or unrooted maps, particularly, the joint tree representation of an embedding of a graph on two dimensional manifolds, which enables one to make the complication much simpler on map enumeration. All of these are valuable for researchers and students in combinatorics, graphs and low dimensional topology. In 9 1 QU Leu ss) | 46