Chain Complexes over Principal Ideal Domains

Transcription

1 Chan Compexes over Prncpa Idea Domans Dem Fachberech 3 (Mathematk und Informatk der Unverstät Bremen zur Erangung des akademschen Grades Doktor der Naturwssenschaften (Dr. rer. nat. vorgeegte Dssertaton von Gerrt Grenzebach Gutachter: Prof. Dr. Dmtry N. Fechtner-Kozov (Unverstät Bremen Prof. Dr. Martn Henk (Otto-von-Guercke-Unverstät Magdeburg Engerecht am: 26. November 23 Tag des Promotonskooquums: 23. Januar 24

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3 Phantase st wchtger as Wssen, denn Wssen st begrenzt. (Abert Ensten

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5 Introducton Orgnay, chan compexes arse n agebrac topoogy to characterse topoogca spaces by speca agebrac nvarants caed homoogy groups. Forgettng about underyng spaces, chan compexes are studed abstracty n homoogca agebra as an agebrac structure. Gven a topoogca space X, one aways obtans a chan compex and so caed snguar homoogy groups H n (X as descrbed n Hatcher (28. A speca case are geometrca smpca compexes (or more genera -compexes for whch a more prmtve homoogy theory exsts caed smpca homoogy whch yeds the same homoogy groups. Furthermore, a partcuar chan compex caed ceuar exsts for any CW-compex, and ts ceuar homoogy groups correspond to the snguar homoogy groups (cf. Hatcher, 28. A geometrca smpca compex conssts of smpces and can be but up out of these n a gueng process. If a smpces are gued together n a speca way, a smpca compex s sad to be sheabe. It s a partcuar property of sheabe smpca compexes that ther compete homoogca nformaton s we-known (cf. Kozov, 28. The noton of sheabty s more generay defned for ce compexes and seems to arse at frst n artces by Sanderson (957 and Rudn (958 athough smar deas seem to be even oder (cf. Bng, 95, for exampe. In the begnnng, sheabty was defned ony for pure compexes, but ths concept has been generased to nonpure compexes by Björner and Wachs (996. Sheabty has deveoped to be an mportant too n combnatorcs, topoogy and geometry. For exampe, the Upper Bound Conjecture for convex poytopes By McMuen and Staney, the Upper Bound Conjecture was formuated by Motzkn at the November meetng of the Amercan Mathematca Socety n Evanston n 956. Motzkn... conjectured (mpcty that f P s any d-dmensona convex poytope wth n vertces and f -dmensona faces, then f c (n, d.... Ths s the content of Motzkn s conjecture, known as the upper bound conjecture (UBC for convex poytopes. (cf. Staney, 975, p. 36f., T. S. Motzkn... formuated what has come to be known as The Upper Bound Conjecture. (cf. McMuen, 97, p. 79, V

6 by Motzkn has been proven 2 by McMuen (97 usng a ater pubshed resut by Bruggesser and Man (97 that the boundary compex of a convex poytope s sheabe. Snce smpca compexes and CW-compexes yed dfferent chan compexes, one may ask whether t s possbe to defne sheabty for abstract chan compexes. Because the homoogy of sheabe smpca compexes s we-known, one may aso ask what speca propertes a chan compex mght have f t comes from a sheabe smpca compex, or, more generay, whch propertes must be fufed by a chan compex to get compete nformaton about ts homoogy. We are concerned wth these questons n the second part of ths thess. It turns out that our noton of sheabty for chan compexes does not suffce to determne homoogy competey, so we specase sheabe chan compexes by further condtons nspred by smpca compexes. The frst part of ths thess deas wth mappng cones of chan compexes and s aso motvated by smpca compexes. Gven some smpca compex, a cone over s obtaned by addng a new vertex and extra smpces contanng ths new vertex. Ths yeds a new smpca compex whch corresponds to a mappng cone over the chan compex C beongng to. We generase the noton of a cone to chan compexes abandonng the geometrca dea of an apex and compare t wth mappng cones. In partcuar, we use mappng cones to construct a cone over a gven chan compex. The text comprses fve chapters. In Chapter the basc concepts ke smpca compexes, chan compexes and homoogy are ntroduced. In partcuar, homoogy of sheabe smpca compexes s dscussed. At the end of ths chapter, we determne the homoogy modue H d (C of a fnte chan compex C of order d. The Chapters 2 and 3 contan most of my own resuts. Chapter 2 deas wth acycc chan compexes, cones and mappng cones. At frst, an abstract defnton of a cone s gven, but mosty, we are concerned wth mappng cones whch w be used to construct a cone over a gven chan compex. In 957, n an abstract pubshed n the Buetn of the Amercan Mathematca Socety, T. S. Motzkn made the foowng conjecture.... The Upper Bound Conjecture (U.B.C.. (cf. McMuen and Shephard, 97, p. 52. A abstracts of ths November meetng are coected by Youngs ( The proof s aso pubshed n a book about convex poytopes by McMuen and Shephard (97. The Upper Bound Conjecture for spheres has been proven by Staney (975. VI

7 Fnay, t turns out that not every mappng cone s a cone matchng our defnton, but conversey there aso exst cones whch cannot be obtaned as a mappng cone. But t s possbe to name certan condtons on whch a mappng cone s a cone, see Theorem 2.7. In Chapter 3 we are concerned wth sheabe chan compexes. In partcuar, the exstence of a speca sheng and sheabty of -skeetons are shown. But n genera, there s no nformaton about the homoogy of sheabe chan compexes, therefore we cam addtona condtons on them whch mtate other propertes of smpca compexes. Ths eads to the noton of reguar and totay reguar chan compexes. We obtan compete homoogca nformaton for totay reguar chan compexes whch have a specfc augmentaton map ɛ, ths resut s noted n Theorem 3.3. In the end, we consder mappng cones over sheabe or reguar chan compexes. It turns out that the mappng cone over a sheabe (or reguar chan compex s sheabe (or reguar, respectvey, too, see Theorem 3.34 and Theorem A smar resut for totay reguar chan compexes foows wthout further work as a coroary. A short concuson of a our resuts s gven n Chapter 4. The foowng Appendx (Chapter A contans a short overvew about CW-compexes. Some of the resuts are jont work wth Björn Waker. Prmary, these are the noton of crtca bass eements and the proof of Theorem.6 as we as the noton of an abstract cone for chan compexes and the proof of Lemma 2.6. Together wth most of the content of Chapter 3 these resuts have been pubshed n Grenzebach and Waker (24. VII

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9 Acknowedgements Whe wrtng ths thess, I got a ot of support by severa peope whom I woud ke to thank a. Frst of a, I want to thank Dmtry Fechtner-Kozov and Eva-Mara Fechtner for brngng chan compexes to my attenton. I aso woud ke to thank Martn Henk for actng as the second referee. Furthermore, I am ndebted to Emanuee Deucch and to Ingof Schäfer for many hepfu dscussons and advces whch heped me to understand some parts of my work better. Speca thanks go to Björn Waker wth whom I dd the frst steps nto chan compexes whch became the bass for my work. I woud aso ke to thank Roman Bruckner, Tm Haga, Phpp Nemann, Johannes Nüße, Tanja Schnder, Sören Schuze and my brothers Arne and Caas Grenzebach. A of them have read bg parts of my manuscrpt and dscovered severa weak ponts. In partcuar, the greatest thanks go to Arne and Caas for ther wonderfu support n desgnng the ayout of ths manuscrpt whch probaby woud ook much worse wthout them. Furthermore, I woud ke to thank Martn Dugosch, Raf Donau, Steffen Hahn, Juane Lehmann and Franzska Pott who heped me n varous ways. I am gratefu to Brgt Feddersen and Nadja Bäker for ther support n a admnstratve matters. And I am thankfu for the technca support by Torsten Radeke and Maro Drewes. Furthermore, I am ndebted to the Graduate Center ProUB at the Unversty of Bremen, n partcuar to Jua Geseer. Fnay, I want to thank a peope who have encouraged and supported me, especay my parents, my brothers Arne and Caas wth Cauda, Chrstna and Phpp Nemann as we as Jasmn and Whem Kandt. IX

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11 Contents. Bascs: Smpca and Chan Compexes.. Smpca Compexes Smpces Geometrc Smpca Compexes Chan Compexes Premnares about Rngs and Free Modues Man Concepts Pure Chan Compexes Subcompexes Chan Compexes obtaned from Smpca Compexes..3. Homoogy Reduced Homoogy An Augmentaton Map for Smpca Compexes About Augmentaton Maps for Chan Compexes Homoogy of Sheabe Smpca Compexes Crtca Bass Eements n Free Chan Compexes Acycc Chan Compexes, Cones and Mappng Cones Acycc Chan Compexes and Cones Mappng Cones A Frst Sght Mappng Cones over Free Chan Compexes Trva Mappng Cones Constructng a Cone The Smpca Case The Genera Case for Chan Compexes Not Every Cone s a Mappng Cone! Mappng Cones over Eementary Chan Compexes A Fna Comment on Cones XI

12 Contents 3. Sheabe and Reguar Chan Compexes Sheabe Chan Compexes Defnton and Frst Exampes Monotoncay Descendng Shengs Skeetons of Sheabe Chan Compexes We-ordered Bases Reguar Chan Compexes Defnton and Exampes Skeetons of Reguar Chan Compexes About Reduced Homoogy Homoogy of Totay Reguar Chan Compexes Mappng Cones over Sheabe Chan Compexes Concuson 93 A. CW-Compexes 95 A.. Terms and Notons A.2. Sheabty of Reguar Ce Compexes A.3. Orented CW-Compexes A.4. Ceuar Chan Compexes Bbography Index 7 XII

13 Lst of Fgures.. Smpces of dmenson,, 2 and Standard -smpex and standard 2-smpex Exampes for smpca compexes Budng up a smpca compex by gueng smpces Exampes of cones wth apex v Constructng a cone over a smpca compex by addng a new vertex v A tetrahedron s a cone over a trange A geometrca cone over a smpex S equas a smpex (S, v n the cone over A.. The crce S as ce compex XIII

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15 Symbos and Abbrevatons We use the foowng terms: N: natura numbers,, 2,... Z: nteger numbers R: rea numbers F q : fnte fed wth q eements Z n : the nteger numbers mod n : empty set d: the dentty map from a set to tsef =: somorphc : homotopy equvaent # A: number of eements n a fnte set A A B or A B: set-theoretca contanment, not necessary proper A \ B: reatve compement of a set B n a set A A B: ntersecton of sets A and B A B: unon of sets A and B A B: dsjont unon of sets A and B A B: drect sum of modues A and B A B: wedge sum of topoogca spaces A and B e,..., e m : modue generated by the eements e,..., e m A : modue generated by a eements n a set A XV

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17 . Bascs: Smpca and Chan Compexes In ths chapter we w ntroduce chan compexes. However, we w expan smpca compexes frst snce our approach to chan compexes s motvated by them... Smpca Compexes We ntroduce some background knowedge about geometrc smpca compexes. For more detaed nformatons, we refer to the books by Maunder (97, Secton 2.3, Hatcher (28, Secton 2., Kozov (28, Secton 2.2, Spaner (966, Secton 3. and Mac Lane (975, Chapter II Smpces Reca that (n + ponts p, p,..., p n n R m for some m n are sad to be affney ndependent f the vectors p p, p 2 p,..., p n p are neary ndependent. Defnton.. A geometrc n-smpex s the convex hu of (n + affney ndependent ponts p, p,..., p n n R m for some m n. The ponts p, p,..., p n are caed vertces. The convex hus of the subsets of {p, p,..., p n } are caed subsmpces or faces of the n-smpex. The number n s the dmenson of the n-smpex. If n, subsmpces of dmenson (n are caed facets. Remark.2. Because {p, p,..., p n }, any smpex contans the empty smpex as a subsmpex. We denote t by and set dm( := for ts dmenson. Basc exampes for smpces are ponts, nes, tranges and tetrahedra (smpces of dmenson,, 2 and 3, see Fgure..

18 . Bascs: Smpca and Chan Compexes p 3 p 2 p p p p p p p p 2 Fgure..: Smpces of dmenson,, 2 and 3 (pont, ne, trange and tetrahedron If we take the convex hu of the standard unt vectors n R n+, we get a unque smpex of dmenson n N, the standard n-smpex: n := { (t, t,..., t n R n+ n } t k = and t k for a k. k= The standard -smpex and the standard 2-smpex are shown n Fgure.2. Defnton.3. Let A := {p, p,..., p n } be the set of vertces of a geometrc n-smpex. If A s totay ordered, we ca the n-smpex ordered. If p < p < < p n wth respect to a tota orderng of A, we denote the n-smpex by [p,..., p n ]. The facet of A whch does not contan the vertex p s denoted by [p,..., ˆp,..., p n ]. Remark.4. For exampe, the ndces of the vertces p yed a tota orderng on a set A := {p, p,..., p n }. e 2 e 3 e 2 e e Fgure.2.: Standard -smpex and standard 2-smpex 2

19 .. Smpca Compexes..2. Geometrc Smpca Compexes Defnton.5. A geometrc smpca compex s a fnte set of smpces n R m for some m such that. every subsmpex of a smpex n s tsef a smpex n, 2. the ntersecton of any two smpces of s a subsmpex of each of them. For brevty, we w sometmes skp the noton geometrc and just say smpca compex nstead. Exampes for smpca compexes are shown n Fgure.3. Defnton.6. Let be a smpca compex. A subcompex s a subset Λ of such that a subsmpces of any smpex n Λ are aso contaned n Λ. Remark.7. By Aeksandrov (965, p. 52, a smpca compex s connected f t s not the unon of two nonempty dsjont subcompexes. An exampe for a not connected subcompex s gven n Fgure.3(a. Defnton.8. The dmenson of a smpca compex s the maxmum of the dmensons of ts smpces. Defnton.9. A smpex n a smpca compex s caed maxma f t s not a subsmpex of any other smpex n. A smpca compex s caed pure f a of ts maxma smpces have the same dmenson. An exampe for a pure smpca compex s shown n Fgure.3(b. To determne the dmenson of any smpca compex, t suffces to consder a ts maxma smpces. (a not connected (b pure (c sheabe Fgure.3.: Exampes for smpca compexes 3

20 . Bascs: Smpca and Chan Compexes F F 2 F 3 F 4 F 5 F F F F 3 F 2 F 2 F F 3 F F 3 F 2 F 5 F 4 Fgure.4.: Budng up a smpca compex by gueng smpces. The chosen smpces F, F 2, F 3, F 4 and F 5 are shown at the top. Beneath the gueng process s ustrated. For 2 k 5, any ntersecton ( k = F Fk s cooured orange. F 2 F 4 One can magne that a connected geometrc smpca compex can be but up from ts maxma smpces by gueng them together. Mathematcay, gueng smpces means that equa dmensona proper subsmpces of them are dentfed. To bud up a geometrc smpca compex n ths way, we take t geometrc smpces F, F 2,..., F t of arbtrary dmensons whch w be the maxma smpces n the constructed compex. We start wth F, then we add F 2 by dentfyng some proper subsmpex of F 2 wth a proper subsmpex of F of the same dmenson. We proceed n ths way, n whch any ntersecton ( k = F Fk must be a smpca compex. An exampe for ths procedure s shown n Fgure.4. If n ths process every smpex F k s gued ony aong facets,.e. subsmpces of dmenson (dm F k, we get a speca smpca compex (cf. Kozov, 28, p. 2. Defnton.. Let be a smpca compex of dmenson d. An order of ts maxma smpces F, F 2,..., F t s caed a sheng (or a sheng order f d = or f each subcompex ( k = F Fk s pure and (dm F k -dmensona for 4

21 .2. Chan Compexes a k {2,..., t}. If such a sheng exsts, the smpca compex s sad to be sheabe. Remark.. Every sheabe smpca compex wthout maxma smpces of dmenson s connected. Fgure.3(c on page 3 shows an exampe. Exampe.2.. Every smpex s a sheabe smpca compex wth exacty one maxma smpex. 2. Let [p,..., p n ] be an ordered n-smpex. The smpca compex whose maxma smpces are a ts facets [p,..., ˆp,..., p n ] =: s sheabe because the ntersecton of any two facets and j s the subsmpex [p,..., ˆp,..., ˆp j,..., p n ] of dmenson (n 2. In fact, every orderng of a smpex facets s a sheng. For ater use we ntroduce the foowng notons. Defnton.3. A geometrc smpca compex s ordered f ts set of vertces,.e. ts set of -smpces, s totay ordered. Defnton.4. Let be a sheabe smpca compex whose maxma smpces F, F 2,..., F t are ordered n a n sheng. A maxma smpex F s caed a spannng smpex (wth respect to the chosen sheng f dm(f = or f the smpex F s gued aong a ts facets. Remark.5. The orderng of the maxma bass eements n a sheng can aways be rearranged n such a way that a spannng smpces come ast (cf. Björner and Wachs, 996, Second Rearrangement Lemma 2.7. Remark.6. We obtan an ordered smpca compex from any geometrc smpca compex by numberng ts vertces..2. Chan Compexes.2.. Premnares about Rngs and Free Modues In ths secton, we ntroduce some basc terms whch are essenta for our further work. We refer to the books by Bosch (24, Lang (22 and especay by Oejekaus and Remmert (974. 5

22 . Bascs: Smpca and Chan Compexes Any rng R s mpcty supposed to be commutatve and unta, the unt w be denoted by. A prncpa dea doman s a commutatve and unta rng whch s an ntegra doman and whose deas are a prncpa. Reca that a modue M over R s free f there exsts a famy (x I n M whch s neary ndependent and generates the modue M,.e. M = I Rx. Such a famy (x I s caed a bass of M. Free modues over prncpa dea domans have the foowng property (cf. Lang, 22, p. 46. Theorem.7. If M s a free modue over a prncpa dea doman R, then every submodue of M s free. For fntey generated modues,.e. modues of the form M = n = Re for some n N, we reca the foowng terms by Oejekaus and Remmert (974, pp. 9 and : the generatng number gen R M s the mnma number of eements whch generate M, the degree of freedom dgf R M s the maxma number of neary ndependent eements n M. The next two theorems can be found n Oejekaus and Remmert (974, pp. 2 and 5 and w be hepfu for our cacuatons n Secton.6. Theorem.8. Let R be an ntegra doman. Let M and N be fntey generated R-modues. For any R-near map ϕ : M N hods: dgf R (M = dgf R (ker ϕ + dgf R (m ϕ. Theorem.9. Let R be an ntegra doman. If M s a fntey generated free R- modue, then gen R M = dgf R M Man Concepts We start by defnng chan compexes whch are our centra term. The defnton bases on work by Webe (994, p. 2, Hton and Stammbach (97, p. 7, Kozov (28, p. 5, Cartan and Eenberg (956, p. 58 and Mac Lane (975, p

23 .2. Chan Compexes Defnton.2. For every ν Z et C ν be a modue over a rng R. Let ν : C ν C ν, ν Z, be R-near maps such that ν ν+ =,.e. m ν+ ker ν. The sequence C + C... 2 C C C... s a chan compex whch we denote by C. The R-modues C ν are caed chan modues, and the maps ν are boundary maps. Smar defntons usng abean groups nstead of R-modues can be found n the books by Dod (972, p. 6, Gefand and Mann (996, p. 25, Hatcher (28, p. 6, Massey (99, p. 254, Maunder (97, p. 7 and Spaner (966, p. 57. Remark.2. A chan compex C over R can be seen as a Z-graded R-modue wth an R-near map d : C C such that d(c ν C ν and d d = (cf. Cartan and Eenberg, 956, p. 58. Remark.22. The eements of a chan modue C ν are often caed ν-chans, see Hton and Stammbach (97, p. 8, for exampe. Now we characterse some speca chan compexes, n parts we foow Spaner (966, p. 57. Defnton.23. Let C be a chan compex over R wth chan modues C ν, ν Z. If every chan modue C ν s free over R wth bass Ω ν, then C s a free chan compex. We ca Ω := ν Z Ω ν ts bass and denote such a chan compex by (C, Ω. If addtonay each chan modue C ν s zero or fntey generated, we ca (C, Ω fntey free. Remark.24. Sometmes, we w denote a free chan compex (C, Ω shorty by C, keepng ts bass Ω n mnd. Remark.25. In the foowng, we w mosty consder free chan compexes (C, Ω over a prncpa dea doman R wth a fxed bass Ω := ν Z Ω ν. Then the submodues ker( ν and m( ν are free for a ν Z by Theorem.7. Defnton.26. Let C be a chan compex wth chan modues C ν, ν Z. If C ν = for a ν <, the chan compex s sad to be nonnegatve. A nonnegatve chan compex s caed postve by Hton and Stammbach (97, p. 26, Massey (99, p. 288 and Mac Lane (975, p. 4; Cartan and Eenberg (956, p. 75 ca t eft postve. 7

24 . Bascs: Smpca and Chan Compexes Defnton.27. Let (C, Ω be a fntey free and nonnegatve chan compex. If C d = for some d N and C ν = for a ν > d, we say that (C, Ω s fnte of order d. We wrte ord(c, Ω = d or shorty ord(c = d. The zero compex (Z, s sad to be fnte of order =: ord(z,. There exst speca maps between chan compexes over R whch respect ther structure. Ths term s standard n terature and can, for exampe, be found n the books by Cartan and Eenberg (956, p. 59, Dod (972, p. 6, Eenberg and Steenrod (98, p. 24, Gefand and Mann (996, p. 4, Hatcher (28, p., Hton and Stammbach (97, p. 7, Kozov (28, p. 52, Massey (99, p. 255, Maunder (97, p. 9, Spaner (966, p. 58 or Webe (994, p. 2. Defnton.28. Let B, C be chan compexes over R wth boundary maps B ν and C ν, respectvey. A chan map f : B C s a coecton of R-near maps f ν : B ν C ν such that f ν B ν = C ν f ν for a ν Z. A chan map f : B C s caed a chan somorphsm f each near map f ν : B ν C ν s an somorphsm. Then the chan compexes B and C are sad to be somorphc. For exampe, et (C, Ω be a free chan compex. If we permute the bass eements n some bass Ω ν, we obtan a chan somorphsm (C, Ω (C, Ω. In fact, we w never change the bass of any free chan compex (C, Ω except for such permutatons. Remark.29. Takng a chan compexes over some rng R as objects and a chan maps between them as morphsms gves a category of chan compexes, cf. Hton and Stammbach (97, p. 8 or Spaner (966, p Pure Chan Compexes Defnton.3. Let C ν be a free chan modue wth bass Ω ν = {e ν I ν }, ν Z, n a free chan compex (C, Ω over some rng R. For any eement x = Iν a e ν C ν, the support of x s the set of a bass eements e ν wth coeffcent a =,.e. supp(x := {e ν a = } Ω ν. The boundary of x s the support of ν (x: bd(x := supp ( ν (x Ω ν. 8

25 .2. Chan Compexes Remark.3. Snce the modue C ν s free, we have C ν = I ν Re ν. Therefore, for any eement x = Iν a e ν C ν, amost every coeffcent a vanshes. Hence, for any eement n a free chan compex, support and boundary are fnte sets. Remark.32. For any x C ν the foowng equvaences hod: supp(x = x =, bd(x = x ker( ν. Defnton.33. Let (C, Ω be a fnte chan compex of order d whose chan modues C ν have the bases Ω ν = {e ν,..., eν k ν } for k ν or Ω ν =. The chan compex (C, Ω s caed pure f each bass eement e ν Ω for ν (d and k ν s contaned n the boundary of some bass eement e ν+, k ν+. Defnton.34. A bass eement e Ω of a free chan compex (C, Ω s caed maxma f t s not contaned n the boundary of any other bass eement. Remark.35. If (C, Ω s a fnte chan compex of order d, a bass eements of Ω d are maxma. Furthermore, f (C, Ω s even a pure chan compex, the maxma bass eements are exacty those n Ω d Subcompexes For the defnton of subcompexes, we refer to Webe (994, p. 6. Defnton.36. Let C be a chan compex. A chan compex B s a subcompex of C f each chan modue B ν s a submodue of C ν and each boundary map δ ν : B ν B ν s the restrcton of the boundary map ν : C ν C ν to B ν. Remark.37. Every subcompex B of a free chan compex (C, Ω over a prncpa dea doman s aso free accordng to Theorem.7. If U, W are subcompexes of a chan compex C, then ther ntersecton U W, whose chan modues are (U W ν := U ν W ν, s aso a subcompex of C. Another subcompex of C, obtaned from U and W as we, s ther sum U + W, havng the chan modues (U + W ν := U ν + W ν. 9

26 . Bascs: Smpca and Chan Compexes Every chan compex C contans the foowng subcompexes: the zero compex Z wth a chan modues Z ν =, the kerne compex ker( wth chan modues ( ker( ν = ker( ν, the mage compex m( wth chan modues ( m( ν = m( ν+. In a cases, the restrcton of the boundary map to the subcompex s zero. Furthermore, we get a sequence of subcompexes Z m( ker( C. We are nterested n some partcuar subcompexes of nonnegatve chan compexes. Defnton.38. Let C be a nonnegatve chan compex. For N, the -skeeton sk (C s a subcompex of C whose chan modues are ( sk (C ν = C ν for ν and ( sk (C = otherwse. For a free and nonnegatve chan ν compex (C, Ω, the -skeeton s denoted by sk (C, Ω. Remark.39. If (C, Ω s a free and nonnegatve chan compex, each -skeeton sk (C, Ω s free wth bass ν= Ω ν. If (C, Ω s even fntey free and nonnegatve wth C =, then sk (C, Ω s a fnte chan compex of order. Partcuary, f (C, Ω s a fnte chan compex of order d, then the d-skeeton sk d (C, Ω s equa to (C, Ω. Moreover, for any pure fnte chan compex (C, Ω of order d wth C =, ts -skeeton sk (C, Ω s pure, too. Now we ntroduce subcompexes whch are anaogues to smpces n a smpca compex. Defnton.4. Let (C, Ω be a free and nonnegatve chan compex wth bass Ω. For any µ N and a bass eement e µ Ω µ, et (C µ e, Ω e µ denote the free, nonnegatve subcompex of (C, Ω whose chan modues ( C µ e have the ν foowng bases ( Ω µ e ν : ( Ωe µ = for ν µ + and ν <, ν ( Ωe µ µ = {eµ }, ν = bd(e for ν µ. ( Ωe µ e (Ω e µ ν+ Such a subcompex (C µ e, Ω e µ s caed eementary.

27 .3. Homoogy Remark.4. Each eementary subcompex (C µ e, Ω e µ s a pure and fnte chan compex of order µ. In partcuar, ( Ω µ e µ = bd(eµ. We defne the order of e µ as the order of the fnte subcompex (C µ e, Ω e µ,.e. ord(e µ := ord(c e µ, Ω e µ = µ. If there s some bass eement ek λ Ω λ for whch ek λ Ω e µ Ω e κ hods for some λ < mn{µ, κ}, ths mpes Ω e λ k Ω µ e Ω e κ,.e. the eementary subcompex (C e λ k, Ω e λ k s contaned n the chan compex generated by Ω µ e Ω e κ Chan Compexes obtaned from Smpca Compexes There s a standard way to construct a nonnegatve chan compex over some rng from an ordered smpca compex (cf. Hatcher, 28, pp Let be a smpca compex of dmenson n whose set of vertces {v, v,..., v k } s ordered such that v < v + for a (k. Ths smpca compex generates a fnte chan compex (C, Ω of order n as foows. For every ν-smpex [v, v,..., v ν ] wth < < < ν k, there s exacty one bass eement e... ν n the bass Ω ν of the chan modue C ν. Hence, there are as many bass eements n Ω ν as ν-smpces n. In partcuar, every bass eement n Ω n corresponds to a smpex of dmenson n n, and the chan modue C s generated by (k + eements e,..., e k. Furthermore, we have C ν = for a ν > n and a ν <. A boundary map ν : C ν C ν for ν n s gven by e... ν ν = ( e ν whereas (e = for a {,,..., k}. For ν > n and ν, each boundary map ν s the zero map. It s easy to show that ν ν = for a ν Z (cf. Hatcher, 28, p Homoogy The noton of homoogy s centra n homoogca agebra. We foow Gefand and Mann (996, p. 25 and Hton and Stammbach (97, p. 8.

28 . Bascs: Smpca and Chan Compexes Defnton.42. Let C be a chan compex over some rng R wth boundary maps ν : C ν C ν. The homoogy modues of C are H ν (C := ker( ν m( ν+ for a ν Z, and ther eements are caed homoogy casses. Remark.43. In terature, the homoogy modues are often caed homoogy groups. Ths name usuay occurs when chan compexes are defned wth abean groups nstead of R-modues. Remark.44. If (C, Ω s a free chan compex over a prncpa dea doman R, then ker( ν and m( ν are free R-modues for a ν Z. However, the homoogy modues H ν (C are not necessary free and may contan torson eements. By Hton and Stammbach (97, p.8, eements of ker( ν are caed ν- cyces and eements of m( ν+ are caed ν-boundares. Usng ths notons, we can say that each homoogy cass of H ν (C s represented by a ν-cyce. Two ν-cyces n the same homoogy cass are homoogous, and ther dfference s a ν-boundary. Remark.45. Let B, C be chan compexes over a rng R and f : B C a chan somorphsm. Then both chan compexes have the same homoogy,.e. H ν (B = H ν (C for a ν Z. In partcuar, the homoogy of any free chan compex (C, Ω w not change f we permute bass eements n some bass Ω ν..4. Reduced Homoogy In ths secton, we defne reduced homoogy for chan compexes over prncpa dea domans. In ths case, the reduced homoogy modue of ndex s wthout any further condtons reated to the usua homoogy modue. Defnton.46. Let R be a prncpa dea doman and (C, Ω a free and nonnegatve chan compex over R. If we repace the boundary map : C by an R-near map ɛ : C R wth ɛ =, we get an augmented chan 2

29 .4. Reduced Homoogy compex over R (cf. Hatcher, 28, p. :... n+ C n n n Cn C2 C C ɛ R η. The homoogy modues of ths chan compex are the reduced homoogy modues H ν (C for a ν Z. For ν, we obtan H ν (C = H ν (C. In partcuar, we have H (C = ker(ɛ/m( and H (C = ker(η/m(ɛ. Furthermore, H ν (C = for a ν 2. We ca ɛ an augmentaton map. Remark.47. If the augmentaton map ɛ s surjectve, we obtan H (C = snce then ɛ(x s a unt n R for some x C. If ɛ =, then H (C = R. If ɛ s nether the zero map nor surjectve, the homoogy modue H (C s not free and contans ony torson eements. If ɛ =, reduced homoogy s the same as the usua homoogy apart from H (C = R. But f ɛ =, whch s ony possbe for C =, the reduced homoogy modue H (C = ker(ɛ/ m( dffers from H (C. Snce m(ɛ s free by Theorem.7, we obtan m(ɛ = R. Hence, we get C = ker( = ker(ɛ Re for some e C wth ɛ(e generatng m(ɛ (cf. Oejekaus and Remmert, 974, p. 8. As m( ker(ɛ, we have H (C = H (C R. Lemma.48. Let (C, Ω be a free and nonnegatve chan compex over a prncpa dea doman R. Then ts reduced homoogy modue H (C s ndependent of the choce of an augmentaton map ɛ =. ɛ Proof. Let C ɛ R and 2 C R be nonzero augmentaton maps, ɛ = ɛ 2. For the moment, we denote the zeroth reduced homoogy modue by H (C, ɛ for {, 2}. Hence, we get and therefore H (C, ɛ = H (C, ɛ 2. H (C, ɛ R = H (C = H (C, ɛ 2 R However, the homoogy modue H (C depends on the choce of ɛ. For exampe, we consder the fnte chan compex (C, Ω over Z whose bass s Ω = Ω = {e}. Let ɛ and ɛ 2 be augmentaton maps wth ɛ (e = and ɛ 2 (e = 2. Then we get H (C, ɛ = and H (C, ɛ 2 = Z 2. 3

30 . Bascs: Smpca and Chan Compexes.4.. An Augmentaton Map for Smpca Compexes Let be an ordered smpca compex of dmenson n. From, we obtan a fnte chan compex (C, Ω of order n over some prncpa dea doman R as descrbed n Secton.2.5. Reca that we have a one-to-one correspondence between the ν-smpces n and the bass eements n Ω ν Ω for every ν Z. In partcuar, the bass eements n Ω = {e, e,..., e k } correspond to the vertces of. To defne an augmentaton map, we reca that (e = e e for any bass eement e Ω. Snce we need ɛ =, we defne an R-near map for any eement x = k = α e C as foows (cf. Hatcher, 28, p. : ɛ : C R, x = k k α e α. = = In partcuar, ɛ(e =, so ɛ s surjectve. Ths s a canonca choce of ɛ. Of course, one can aso choose ɛ (x = k = α or ɛ r (x = r k = α for any fxed r R \ {} nstead. Snce r s not a zero dvsor and thus ker(ɛ r = ker(ɛ, ths choce does not change the reduced homoogy modue H (C, but we get torson n H (C f r s not a unt n Z About Augmentaton Maps for Chan Compexes For any free and nonnegatve chan compex over a prncpa dea doman R, the augmentaton map ɛ does not need to be unque. But n contrast to smpca compexes there exst chan compexes for whch ɛ must be. Indeed, t s aways possbe to take ɛ =, but we are nterested n ɛ = because n that case the reduced homoogy modue H (C s dfferent from H (C. We consder some exampes.. Let (C, Ω be a fnte chan compex of order over Z whose chan modues have the bases Ω = {e } and Ω = {e }. Let (e = e. We must defne ɛ(e =, hence H (C = H (C and H (C = Z. 2. Consder the fnte chan compex (C, Ω of order over Z wth chan modue bases Ω = {e, e 2 } and Ω = {e, e 2 }. Let (e = e + e 2 and (e2 = e + 2e 2. Because e 2 e maps to e 2, we ony get ɛ =. As above, H (C = H (C and H (C = Z. 4

31 .4. Reduced Homoogy 3. Agan, et (C, Ω be a fnte chan compex of order over Z whose chan modues have the bases Ω = {e, e 2 } and Ω = {e, e 2, e 3, e 4 }. Let (e = e + e 2 + 2e 3 and (e2 = e 3 + e 4. Then, H (C = Z 2. Now, there are at east two dfferent ways to defne an augmentaton map C Z. a ɛ s gven by: ɛ (e =, ɛ (e 3 =, ɛ (e 2 =, ɛ (e 4 =. b ɛ 2 s defned n the foowng way: ɛ 2 (e =, ɛ 2(e 3 =, ɛ 2 (e 2 =, ɛ 2 (e 4 =. In both cases H (C = Z and H (C =, as both augmentaton maps are surjectve. If there s a bass eement e Ω whch s not contaned n the boundary of any bass eement of Ω, then an augmentaton map ɛ = aways exsts and can be obtaned by defnng ɛ(e =. In partcuar, ɛ can be defned ths way for every fnte chan compex of order. For chan compexes of order d we treat two speca cases. Theorem.49. Let (C, Ω be a fnte chan compex of order d over a prncpa dea doman R wth char R =. Let Ω = {e,..., e k } be the bass of ts chan modue C wth k. Then the foowng statements are equvaent: ( # bd(x 2 for every x C \ ker. (2 An augmentaton map ɛ : C R exsts such that ɛ(e µ = for a e µ Ω. Proof. ( (2: If C =, we can set ɛ(e µ = for a e µ Ω, and we are done. So et Ω = {e,..., e k } wth k. For each e µ, whch s not contaned n the boundary of any e λ, we defne ɛ(e µ =. Hence, we assume wthout oss of generaty that the -skeeton of (C, Ω s pure. Let (e k λ = a λ e for λ k. = 5

32 . Bascs: Smpca and Chan Compexes Because ɛ =, we have to sove the foowng system of near equatons to defne ɛ. a a 2 a k ɛ(e a 2 a 22 a 2k ɛ(e2 =. (..... a k a k 2 a k k ɛ(e k By transformaton of the rows we get a bock matrx ã ã,j+ ã,k ã 22 ã 2,j+ ã 2,k ã jj ã j,j+ ã j,k..... for some j k and ã = for a j. Note that j < k because otherwse we woud get a row ( ãk k havng ony one entry ã k k =. Gettng such a row means that there exsts an eement x C wth # bd(x = whch s a contradcton! We defne an augmentaton map ɛ by choosng vaues ɛ(e µ for a bass eements e µ Ω teratvey n the foowng way. At frst, we defne ã := j = ã =, choose some r k ɛ(e k := ã r k. R \ {} and set We proceed by teraton. Assume that for some (j + < k k eements r R \ {} are chosen for a k k such that ɛ(e = ã r. We choose some r k R \ {} such that ã,k r k + k =k ã, r = for a ndces j, for whch there s at east one coeffcent ã,t = for some (k t k. Ths s possbe because each equaton 6

33 .4. Reduced Homoogy ã,k r k + k =k ã,r = has at most one souton r k f there s some ã,t =. Hence, there are at most j nonzero eements n R whch cannot be chosen. We set ɛ(e k := ã r k then. For each ν j, we defne ( ɛ(e ν := j, =ν Then we have ɛ(e µ = for a µ k. ã k =j+ ã ν, r =. (2 (: We assume that some x C \ ker exsts such that # bd(x =. Then (x = ρe for some e µ Ω and ρ R \ {}. Hence, ɛ(e µ = hods for every augmentaton map ɛ whch s a contradcton to (2. Remark.5. The above theorem s not true for chan compexes defned over some fnte fed F q. We show exampes for F 2, F 3 and F 4, but n the same manner one can create exampes for any fnte fed. Let (C, Ω be a fnte chan compex of order over F 2 whose chan modues have the bases Ω := {e } and Ω := {e, e 2, e 3 }. Let (e = e + e 2 + e 3, then we must set ɛ(e = for some {, 2, 3} and any augmentaton map ɛ. Let (C, Ω be a fnte chan compex of order over F 3 whose chan modues have the bases Ω := {e, e 2 } and Ω := {e, e 2, e 3, e 4 }. Let (e = e + e 3 + e 4 and (e2 = e 2 + e 3 + 2e 4. Then we must set ɛ(e = or ɛ(e 2 = for any augmentaton map wth ɛ(e 3 = and ɛ(e4 =. Let (C, Ω be a fnte chan compex of order over F 4, whch may be generated by and σ over F 2. Let ts chan modues have the bases Ω := {e, e 2, e 3 } and Ω := {e, e 2, e 3, e 4, e 5 }, and et (e = e + e 4 + e 5, (e2 = e 2 + e 4 + σe 5 and (e3 = e 3 + e 4 + (σ + e 5. Then agan t s mpossbe to get an augmentaton map ɛ wth ɛ(e = for a {, 2, 3, 4, 5}. Hence, f R s a prncpa dea doman wth char R =, we can ony prove a weaker statement. 7

34 . Bascs: Smpca and Chan Compexes Theorem.5. Let (C, Ω be a fnte chan compex of order d over a prncpa dea doman R. Let Ω = {e,..., e k } be the bass of ts chan modue C wth k.. If # bd(x 2 for every x C \ ker, then an augmentaton map ɛ = exsts. 2. If an augmentaton map ɛ : C R exsts such that ɛ(e = for a e Ω, then # bd(x 2 for every x C \ ker. Proof. The proof of the second statement s the same as the proof for (2 ( n the precedng Theorem.49. To prove the frst statement we have to show that, for any augmentaton map ɛ, there s some e Ω such that ɛ(e =. If the -skeeton of (C, Ω s not a pure fnte chan compex of order, nothng remans to be done as we can defne ɛ(e = for any maxma bass eement e Ω. Hence, we assume that the -skeeton of (C, Ω s pure and et Ω := {e,..., e k } be the bass of C. As above et (e k = a e for k. = To defne an augmentaton map ɛ we have to sove a system of near equatons as n Equaton (. on page 6. The souton of ths system cannot be unque because otherwse there must be an eement x C wth # bd(x = whch s mpossbe by assumpton. Hence, there s some e k Ω such that ɛ(e k =, and therefore ɛ =..5. Homoogy of Sheabe Smpca Compexes The homoogy of sheabe geometrc smpca compexes s we-known, see Björner and Wachs (996, Chapter 4 or Björner (992, Theorem wth a drect proof for pure sheabe smpca compexes, for exampe. In ths secton, we consder ony homoogy over Z because ths s the usua case n terature. We w recover the statement of the next theorem n a more genera context n Secton

35 .5. Homoogy of Sheabe Smpca Compexes Theorem.52. Let be a sheabe geometrc smpca compex of dmenson d. Let ts maxma smpces be ordered n a sheng. For d et there be n spannng smpces of dmenson. Then the homoogy modues of are H ( = Z n for d, H ( = Z n +, H ( = f < or > d. Remark.53. Every smpca compex of dmenson s sheabe. If conssts of n vertces, we get H ( = Z n accordng to Hatcher (28, Proposton 2.7. Proof of Theorem.52. Let Σ be the set of spannng smpces except the ones of dmenson. Let be the subcompex of whch contans a ts smpces except the maxma smpces of dmenson. By Kozov (28, Theorem 2.3, the subcompex s homotopy equvaent 2 to a wedge of spheres: S dm σ. σ Σ Because topoogca spaces whch are homotopy equvaent have somorphc homoogy modues (cf. Hatcher, 28, p., we get H ( ( = dm H S σ σ Σ for a d. By Hatcher (28, Coroary 2.25, we obtan H ( ( = H S dm σ. σ Σ The homoogy of a sphere s we-known, cf. Hatcher (28, p. 4, Massey (99, p. 86 or Spaner (966, p. 9, for exampe. For, we have H (S = Z and H (S n = f = n. Hence, we get for d: H ( = H ( = Z n. Because the subcompex s path-connected, we obtan H ( = Z and therefore H ( = Z n +. Remark.54. An anaogous statement for pure sheabe smpca compexes has been proven by Björner (992, p The noton of a homotopy equvaence s expaned by Hatcher (28, p. 3, for exampe. 9

36 . Bascs: Smpca and Chan Compexes.6. Crtca Bass Eements n Free Chan Compexes In ths secton et a chan compexes be defned over some prncpa dea doman R. Defnton.55. Let (C, Ω be a fntey free chan compex over R whose chan modues C ν have bases Ω ν = or Ω ν = {e ν,..., eν k ν } for some k ν. Let Γ be the set of a ts maxma bass eements: Γ := { e Ω e bd( f for a f Ω }. Let the bass eements n each bass Ω ν be ordered n such a way that the eements of Ω ν \ Γ come frst. A maxma bass eement e ν k (Ω ν Γ s caed crtca f there exst eements a R for (k such that ν (ek ν k = a ν (e ν, = precrtca f there exst eements a R for k wth a k = such that a k ν (ek ν k = a ν (e ν, = noncrtca f e ν k s nether crtca nor precrtca. Remark.56.. A crtca bass eement s aways precrtca, and for a bass eement e ν k wth k = both notons concde snce the empty sum s zero. For k 2, a precrtca eement e ν k s crtca f the coeffcent a k can be chosen to be a unt n R. Hence, the two notons are equvaent f the prncpa dea doman R s a fed. 2. supp ( a k ν (e ν k = supp ( ν (e ν k f a k =. 3. It s possbe to change the orderng of a bass Ω ν n such a way that a noncrtca bass eements come frst. 2

37 .6. Crtca Bass Eements n Free Chan Compexes Crtca bass eements are an anaogue to spannng smpces n sheabe smpca compexes, as the foowng emma shows. Lemma.57. Let be an ordered smpca compex of dmenson d and (C, Ω the correspondng fnte chan compex of order d over R obtaned from as descrbed n Secton.2.5. Then any bass eement ek ν Ω whch corresponds to a spannng smpex of s crtca. Proof. For ν d, et each bass Ω ν be ordered n such a way that the maxma bass eements correspondng to spannng smpces come ast,.e. Ω ν = {e ν,..., eν m ν, em ν ν +,..., eν k }{{} ν }. spannng smpces Then, for each bass eement ek ν Ω ν wth (m ν + k k ν, we have bd(ek ν m ν = bd(eν. We consder the smpca subcompex whch conssts of a maxma smpces of except ts spannng smpces. Accordng to Kozov (28, p. 23, ths subcompex s coapsbe, so there exsts a homotopy equvaence to a snge pont (cf. Kozov, 28, p. 94. Hence, the correspondng subcompex (Ĉ, Ω of (C, Ω whose chan modues have the bases Ω ν := {e ν,..., eν m ν } has the same reduced homoogy modues,.e. H (Ĉ = for a Z. For a bass eements e µ k whch correspond to a spannng smpex, we consder the subcompex Ĉ(e µ k whose bases are Ω(e µ k ν = Ω ν for ν = µ and Ω(e µ k µ = Ω µ {e µ k }. If µ =, there s nothng to do. For µ, we get µ (e µ k m Ĉ(e µ µ k ker Ĉ(e µ µ µ k = ker µ = m µ, µ Ĉµ Ĉµ wrtng nstead of ɛ for the augmentaton map here f µ =. Hence, there s some x µ k Ĉ µ such that µ (x µ k = µ(e µ k. Therefore, the bass eement eµ k s crtca. In a pure fnte chan compex (C, Ω of order d, a the precrtca eements n the chan modue bass Ω d can be seen as the generators of homoogy n dmenson d. If a bass eements of Ω d are ether noncrtca or crtca, we can name a bass of H d (C. At frst, we treat a speca case whch does not arse at smpca compexes of dmenson or hgher. 2

38 . Bascs: Smpca and Chan Compexes Lemma.58. Let (C, Ω be a pure fnte chan compex of order d over R and et a bass eements n Ω d = {e d,..., ed k d } be crtca. Then Ω ν = for a ν = d, and H d (C = R k d s generated by e d,..., ed k d. Proof. By nducton we get d (e d = for a k d. As (C, Ω s pure, there are no bass eements whch are not contaned n any eementary subcompex (C e d, Ω e d. Remark.59. A assumptons of Lemma.58 are fufed by every fnte chan compex of order. To name a bass of H d (C for pure fnte chan compexes of order d whch have at east one noncrtca bass eement n Ω d, we foow Björner (992, p. 254, who has performed ths task for the speca case of sheabe smpca compexes, and generase hs proof to chan compexes. To formuate the theorem we ntroduce a new notaton. Let C ν be a chan modue generated by Ω ν := {e ν,..., eν k ν } and consder some ρ = k ν = a e ν C ν. Then we denote the coeffcent of e ν by ρ(e ν := a. Theorem.6. Let (C, Ω be a pure fnte chan compex of order d and Ω d be a bass of C d wth k d eements. Let there be n < k d crtca eements g,..., g n n Ω d and a other bass eements be noncrtca. Let Ω d be ordered n such a way that the noncrtca eements come frst: Then the foowng hods:. H d (C = R n. Ω d = {e,..., e m, g,..., g n }, m + n = k d. 2. For n, there exst unque d-cyces ρ,..., ρ n n H d (C = ker( d such that ρ (g j = δ j. 3. For n, {ρ,..., ρ n } s a bass of H d (C. Proof. At frst, we consder the case n =,.e. the bass Ω d has ony noncrtca eements, so Ω d = {e,..., e m }. In partcuar, we have d (e = for a m then. We assume that there exsts an eement x C d \ {} such that d (x =. Then # supp(x 2 hods. Let x = m = a e and defne := max{ m a = } 2. So we get a d (e = < ( a d (e. Hence, 22

39 .6. Crtca Bass Eements n Free Chan Compexes e s not noncrtca whch s a contradcton. Therefore, ker( d =,.e. H d (C =. For n, the frst statement s a consequence of the second and thrd, so we start provng the second statement usng nducton. Havng n crtca eements, we get Ω d = {e,..., e m, g,..., g n }. We consder the subcompex (Ĉ, Ω of (C, Ω whose bass s Ω := m = Ω e. Its chan modues are Ĉ d = e,..., e m and Ĉ ν = C ν for (d ν. As a chan compex, (Ĉ, Ω s pure and fnte of order d havng ony noncrtca eements, because the ony crtca eements n Ω d are g,..., g n. Therefore, H d (Ĉ =. For every n there s some ρ Ĉ d such that d ( ρ = d (g. Hence, ρ := g ρ ker( d = H d (C. Snce ρ Ĉ d = e,..., e m, the ony crtca eement contaned n supp(ρ s g. In partcuar, we have { f = j, ρ (g j = δ j = f = j. Hence, the second statement s proven up to unqueness. Let σ ker( d = H d (C wth σ (g j = δ j. So we get σ ρ = m = c e wth coeffcents c R. We concude σ ρ ker ( d Ĉd = Hd (Ĉ =. Therefore σ = ρ, so we have shown unqueness. It remans to prove that {ρ,..., ρ n } generates H d (C and s neary ndependent. Let n = a ρ = wth a R. Because ρ = g ρ, we get = n = a g n = a ρ. }{{} Ĉ d = e,...,e m As {e,..., e m, g,..., g n } s a bass of C d, we concude a = for a, so the eements ρ,..., ρ n are ndependent. Let σ H d (C = ker( d C d. We set τ := σ n = σ(g ρ ker( d. For j n, the coeffcent τ(g j of g j n τ s τ(g j = σ(g j n = σ(g ρ (g j }{{} =δ j = σ(g j σ(g j =. Therefore, τ e,..., e m = Ĉ d. So we get τ ker ( d = Hd (Ĉ =. Ĉd Ths yeds σ = = n σ(g ρ,.e. {ρ,..., ρ n } generates H d (C. Because of ndependence, {ρ,..., ρ n } s a bass of H d (C and hence, H d (C = R n. 23

40 . Bascs: Smpca and Chan Compexes In genera, there are aso precrtca eements whch are not crtca. In ths case we ony know that H d (C = R n, but we cannot name a bass. To prove the foowng theorem we w need the generatng number gen R M and the degree of freedom dgf R M whch we ntroduced n Secton.2. for any fntey generated R-modue M. Theorem.6. Let (C, Ω be a pure fnte chan compex of order d and Ω d be a bass of C d wth k d eements. Let there be n < k d precrtca eements g,..., g n n Ω d and et Ω d be ordered n such a way that the noncrtca eements come frst: Then H d (C = R n. Ω d = {e,..., e m, g,..., g n }, m + n = k d. Proof. The case n = s aready proven. So et there be n precrtca eements,.e. Ω d = {e,..., e m, g,..., g n }. As above we consder the subcompex (Ĉ, Ω of (C, Ω wth bass Ω := m = Ω e. It s pure and fnte of order d wthout precrtca eements, hence H d (Ĉ = ker ( d Ĉd =. For every n, there s an eement ρ Ĉ d and some a R \ {} such that d ( ρ = d (a g because a g are precrtca. Therefore we get ρ := a g ρ ker( d = H d (C. We show that the eements ρ,..., ρ n are neary ndependent. Let n = c ρ = wth c R. Snce ρ = a g ρ, we get = n = c a g n = c ρ. }{{} Ĉ d = e,...,e m Because {e,..., e m, g,..., g n } s a bass of C d, we get c a = for a, so c =. Hence, the eements ρ,..., ρ n are ndependent,.e. ker( d contans at east n ndependent eements. So dgf R (ker d n. Theorem.8 s vad for the boundary map d : C d C d, hence dgf R (C d = dgf R (ker d + dgf R (m d. (.2 Snce C d s a free R-modue generated by (m + n eements, we get by Theorem.9 that dgf R (C d = gen R (C d = m + n. Ths yeds dgf R (m d = dgf R (C d dgf R (ker d m. (.3 We know that m d m d snce Ĉ d s a submodue of C d. Therefore, we ( Ĉd have dgf R m d dgfr (m d. Appyng Theorem.8 to the boundary Ĉd 24

41 .6. Crtca Bass Eements n Free Chan Compexes map d Ĉd : Ĉ d Ĉ d yeds ( ( m = dgf R (Ĉ d = dgf R ker d + dgf }{{ Ĉd R m d dgfr (m d, (.4 } Ĉd = usng dgf R (Ĉ d = gen R (Ĉ d = m by Theorem.9. By the Equatons (.3 and (.4 we get dgf R (m d = m. Thus, dgf R (ker d = n by Equaton (.2. Snce ker d s a free R-modue, we get gen R (ker d = n due to Theorem.9. Fnay, ths mpes H d (C = R n. 25

42

43 2. Acycc Chan Compexes, Cones and Mappng Cones In ths chapter, we consder the reatonshp between cones and mappng cones. Most of ths s motvated by geometrc smpca compexes. At frst, we ntroduce a speca type of chan compexes named acycc whose homoogy s very smpe: The ony nonzero homoogy modue s H (C whch s free and generated by a snge eement. Snce a speca knd of acycc smpca compexes s gven by smpca cones, we generase the noton of a cone to genera chan compexes, forgettng the geometrca descrpton from the smpca case. For any smpca cone there exsts a descrpton as a mappng cone of an dentty chan map. Most of ths chapter deas wth mappng cones for chan compexes, and t turns out that not every cone can be regarded as a mappng cone as we as not every mappng cone s a cone tsef. So ths correspondence n the smpca case s qute speca. 2.. Acycc Chan Compexes and Cones Our defnton of acycc chan compexes refers to Björner (992, p. 253 or Eenberg and Steenrod (98, p. 7. Defnton 2.. A nonnegatve chan compex C over some rng R s acycc f the foowng hods for ts homoogy groups: H (C = R, H ν (C = for ν. Smar defntons can be found n Cartan and Eenberg (956, p. 75, Massey (99, p. 288 and Munkres (984, p. 45. A more genera, but unusua defnton s gven by Hton and Stammbach (97, p

44 2. Acycc Chan Compexes, Cones and Mappng Cones v v v Fgure 2..: Exampes of cones wth apex v Remark 2.2. Let (C, Ω be a free and acycc chan compex over a prncpa dea doman R. If there exsts an augmentaton map ɛ =, then H (C = accordng to Secton.4. If ɛ s even surjectve, then H (C =, too. Some speca smpca compexes beng acycc are cones. A smpca cone has a dstngushed vertex v beng the apex of the cone, and each maxma smpex S n ths smpca compex has exacty one facet whch does not contan the vertex v. For exampe, smpces themseves are cones (cf. Munkres, 984, p. 44. Further exampes are shown n Fgure 2.. To defne the concept of cones for chan compexes, we want to abandon the geometrca dea of an apex. Ths eads to the foowng defnton. Defnton 2.3. Let (C, Ω be a fnte chan compex of order d over a prncpa dea doman R. For a ν d, et Ω ν := {e ν,..., eν k ν } = be a bass of the chan modue C ν. The chan compex (C, Ω s a cone f the foowng condtons are fufed:. For every ν {,..., d}, there s a nonempty subset S ν Ω ν such that a supp( ν e ν j e ν S ν\{e ν j } supp( ν e ν for every eν j S ν, b for every e ν k Ω ν \ S ν, there s an eement τ k C ν+ such that ν+ τ k = c k e ν k + r k wth c k unt n R and r k S ν. 2. # supp( x 2 for a x C \ ker. 3. There s a subset {e} = S Ω wth # S = such that the foowng hods: For every e k Ω \ S there s an eement τ k C such that τ k = c k e k + d ke wth c k unt n R and d k =. 28

45 2.. Acycc Chan Compexes and Cones Remark d k = n the cone condton 3 foows from the cone condton Let Γ be the set of a maxma bass eements of (C, Ω as n Defnton.55. Then Γ Ω ν S ν for a ν. In partcuar, Ω d = S d. 3. ker( ν S ν = {} for a ν d because of the cone condton a. Remark 2.5. For any cone (C, Ω, we can aways defne a nonzero augmentaton map ɛ n the foowng way, usng the cone condton 3: We set ɛ(e := for the ony bass eement e S Ω. For every bass eement e k Ω \ S, we have an eement τ k C such that τ k = c k e k + d ke wth some unt c k R and d k =. Therefore, we defne ɛ(e k := c k d k. Snce ɛ(e =, ths augmentaton map s surjectve, hence H (C =. Furthermore, we have ɛ(e = for a bass eements e Ω. Indeed, the above Defnton 2.3 mpes that cones are acycc, as the foowng emma shows. Lemma 2.6. A cone (C, Ω s acycc. Proof. By defnton, a cone s aways a fnte chan compex of order d for some d N. If d =, then Ω = S. Hence, we get H ν (C = for ν and H (C = R snce # S =. Let d. At frst, we show that ker( ν = m( ν+ for ν. For ν > d, there s nothng to do. For ν = d, we get S d = Ω d and concude ker( d = {} due to Remark 2.4. Therefore, H d (C =. For ν (d, we take an arbtrary eement σ ker ν : σ = k ν a e ν = a e ν + a e ν. = e ν S ν e ν S ν By defnton, there s some τ C ν+ for every e ν S ν such that ν+ τ = c e ν + r wth some unt c n R and r S ν. Snce ν ν+ =, we have 29