Simplicial Complexes in Finite Groups. (Cortona. July 2000)

Transcription

1 Simplicial Complexes in Finite Groups (Cortona. July 2000)

2 Firenze, September These notes contain the material presented at a Scuola Matematica Interuniversitaria summer course held in Cortona in July For a large part they are based on earlier unpublished seminar notes by Luis Ezquerro and myself on the same subject. Thus, except for the errors, Luis should be regarded as a co-author. I am graetly indebted to Jurgen Pulkus, whose sensible suggestions and comments led to many improvements both in presentaton and contents. I also thank the students of the course who pointed out several misprints and inaccuracies. Carlo Casolo Dipartimento di Matematica U. Dini Viale Morgagni 67/A I Firenze casolo@math.unifi.it 1

3 Chapter 1 Simplicial Complexes Definition. Let V be a finite non empty set. An (abstract) simplicial complex with vertex set V is a collection Γ of non empty subsets of V satisfying the following conditions: 1. {v} Γ for each v V ; 2. if τ σ Γ then τ Γ. Let Γ be a simplicial complex with vertex set V. We then write V = vert(γ), and call simplices the elements of Γ. More specifically, if σ Γ, and σ = n + 1, we say that σ is a n simplex (or also that σ is a simplex of dimension n). We define the dimension of Γ as the largest dimension of its simplices. For a n-simplex σ, we usually write σ = {v 0, v 1,..., v n }, where v i V. If σ is a simplex, any = τ σ is called a face of σ (by definition, τ Γ). Vertices (i.e. the elements of V ) are identified with the 0-simplices. A subcomplex of a complex Γ is a subset of Γ which is closed by taking all faces of its elements (clearly a subcomplex is in turn a complex). If n is the dimension of Γ and 0 q n, then the set of all simplices of Γ of dimension at most q is a complex, and it is called the q-skeleton of Γ. Examples. 1) Let A be a finite graph with vertex set V. The clique complex K(A) associated to A is the simplicial complex on Vert(K(A)) = V whose simplices are the non-empty subsets of V that determine a complete subgraph (a clique ) of A. 2) Let V be a finite dimensional vector space over a finite field F. Then the set of all non-empty linearly independent subsets of V is a simplicial complex, whose dimension is dim F (V ) 1. 3) Let G be a finite group and F = {H 0,..., H n } a family of subgroups of G. The coset complex Γ F associated to F is defined in the following way. The vertices 2

4 of Γ are all the right cosets H i g with g G and i = 0,..., n. For 0 q n, a q-simplex is any set of q + 1 distinct cosets {H i0 g 0,..., H iq g q } such that H i0 g 0... H iq g q. Since any element in the intersection can be taken as a representative for the same cosets, we have that a q-simplex is any family {H i0 g,..., H iq g} with g G and 0 i 0 < i 1 <... < i q n. Polyhedra. An affine combination of points P 0,..., P m in an euclidean space R n is m λ i P i i=0 with m λ i = 1. An affine combination is convex if furthermore λ i 0 for all i = 0,..., m. The convex hull of a set of points is the set of all convex combinations of them. The points P 0,..., P m in R n are affinely independent if none of them is an affine combination of the others (or, in other words, if the vectors P 1 P 0,..., P m P 0 are linearly independent). If P 0,..., P m are affinely independent then their convex hull is denoted by σ = P 0,..., P m and it is called an affine m-simplex. The points P 0,..., P m are the vertices of σ, a face of σ is a simplex of the form P i0,..., P iq with {P i0,..., P iq } {P 0,..., P m }. We leave it as an exercise to prove that a simplex determines its vertices. A geometric simplicial complex (see [54]) is a collection K of affine simplices in an euclidean space such that if σ, τ K then any face of σ is in K and σ τ is either empty or a face of both σ and τ. A topological space X is called a polyhedron if it is homeomorphic to a space K = σ K σ for some geometric complex K. Now, to any geometric complex K there is a naturally associated abstract complex K a whose simplices are the set of the vertices of the simplices of K. Conversely, it is easy to associate to any abstract complex Γ a geometric complex U(Γ) and its underlying topological space. In R n the points E 0 = 0, E 1 = (1, 0,..., 0), E 2 = (0, 1,..., 0),..., E n = (0, 0,..., 1) are affinely independend; the simplex E 0, E 1,..., E n is called the standard n- simplex and denoted by n. Let now Γ be an abstract complex with V = V ert(γ) = {v 0, v 1,..., v n }. Then define a map u : Γ n which assignes to each simplex σ = {v i0, v i1,..., v iq } of Γ the face u(σ) = E i0, E i1,..., E iq of n. Finally put U(Γ) = σ Γ u(σ). Clearly U(Γ) is a geometric complex. A topological space which is homeomorphic to U(Γ) is called a geometrical realization of Γ. Up to homeomorphisms, we denote by Γ a geometric realization of Γ. 3 i=0

5 The geometrical realization allows to introduce concepts and language from topology in the study of abstract complexes. In these notes, I will refer from time to time to the geometric realization of a complex, as it might provide a more intuitive point of view, but I will generally try to avoid when possible topological subtlety, by keeping a more combinatorial approach. This will have the cost of losing some information on the topological features of the complexes we will study. Notation. Given an n-dimensional complex Γ, for each 0 q n we denote by S q (Γ) the set of all q-simplices of Γ, and by α q the cardinality of S q (Γ). Definition. Let Γ be a complex of dimension n. The Euler-Poincarè characteristic of Γ is the integer n χ(γ) = ( 1) q α q. The reduced Euler-Poincarè characteristic is defined as: χ(γ) = χ(γ) 1. q=0 Definition. Let Γ and Γ be simplicial complexes with vertex set V and V respectively. A simplicial map f : Γ Γ is an application f : V V which maps all simplices of Γ into simplices of Γ (possibly of smaller dimension), i.e. such that {x 0, x 1,..., x q } Γ {fx 0, fx 1,..., fx q } Γ. f : Γ Γ is an isomorphism of simplicial complex if it is bijective and both f and f 1 are simplicial maps. A simplicial map f : Γ Γ of abstract simplicial complexes induces in a standard way a continuous map f : U(Γ) U(Γ) of the geometric realizations as we have described above. In fact, for every simplex σ of Γ the vertices of u(σ) are by construction affinely independent. Thus the restriction f vert(σ) determines a unique affine map f σ : u(σ) U(Γ) which is continuous. This assignement is clearly consistent on faces, and allows, by piecing together the f σ, to define f. It is then an immediate consequence that geometric realizations of isomorphic simplicial complexes are homeomorphic. Also, it is clear that the converse is not true. Barycentric subdivision Let σ = P 0,..., P m be an affine simplex in R n ; the barycenter of σ is the point B σ = 1 m + 1 (P P m ). Given a geometric complex K, one constructs a new complex Sd(K) whose vertex set is the set of all barycenters of the simplices of K, and for q 1 the q-simplices 4

6 are the affine simplices B σ0, B σ1,..., B σq where σ 0,... σ q are simplices of K such that σ 0 σ 1... σ q. The complex Sd(K) just defined is called the barycentric subdivision of K. It is easy to verify that the underlying space Sd(K) coincides with K. Now, let P be a finite partially ordered set. The complex of chains of P is the simplicial complex (P ) whose vertices are the elements of P, and simplices are all the chains (i.e. linearly ordered subsets) of P. Thus, for q 0, the q-simplices of (P ) are all the chains of length q of P, and the dimension of (P ) is the maximal length of a chain in P. This construction for complexes is quite general. In fact, let Γ be a given abstract complex. Then we can view Γ as a partially ordered set, where the order relation among simplices is just the set inclusion. Then the complex of chains (Γ) is called the barycentric subdivision of Γ. It is not difficult to show that if Γ is a geometric realization of Γ then its barycentric subdivision (in the geometric sense) Sd( Γ ) is a geometric realization of (Γ). Thus Γ and (Γ) have homeomorphic geometric realizations. Let P, P be partially ordedered sets. An application f : P P is an order preserving map if, for all x, y P x y f(x) f(y). An isomorphism of posets is a bijection f : P P such that f and its inverse f 1 are order preserving. It is clear that an order preserving map on partially ordered sets induces in a natural way a simplicial map on the corresponding complexes of chains. Examples. 1) Let V be a finite dimensional vector space over a finite field. The set of all subspaces of V is finite, and ordered by inclusion. It then gives rise to a complex whose elements are all the chains (usually called flags) of subspaces V 0 < V 1 <... < V q. The dimension of this complex is the dimension n of V as a vector space. Observe that, in this complex, all maximal simplices have the same dimension n, and that the group of linear applications of V acts naturally as a group of simplicial maps on the complex. 2) Let G be a finite group. Then the set L(G) of all subgroups of G ordered by inclusion, yields a complex of chains on which the group Aut(G) acts as a group of simplicial automorphisms. Observe that, apart from rather restricted cases, in these examples the maximal simplices have different dimensions. 5

7 1.1 Homology Definition. A Chain Complex is a family C = { C n, n n Z } where, for each n, C n is an abelian group and n : C n C n 1 are group homomorphisms such that n n 1 = 0 (that is Im n ker n 1 ). If C is a chain complex, then for all n Z the n-th homology group of C is defined as H n (C) = ker n /Im n+1. We refer to the homology of C to mean the family of groups H (C) = {H n (C) n Z}. Let Γ be a simplicial complex. An orientation of Γ is a partial ordering on Vert(Γ) which induces a total ordering on each simplex of Γ. A complex with a fixed orientation is said to be oriented. Observe that a total ordering on Vert(Γ) is always an orientation of Γ; thus every complex can be oriented. Given an oriented complex Γ, we recall the fundamental construction of the associate integral chain complex { C q (Γ), q q Z }. We fix n to denote the dimension of the complex Γ. For 0 q n, and for each element σ of S q (Γ), we now introduce the symbol [σ] = [v 0, v 1,..., v q ] where v 0, v 1,..., v q are the vertices of σ written in the order given by the orientation. Then, for 0 q n, we denote by C q (Γ) the free abelian group with free generating set whose elements are all symbols [σ] with σ S q (Γ). For q n + 1 we set C q (Γ) = 0. We now define the maps q, called the differentiations (of degree q) - For q = 0 we define 0 : C 0 (Γ) 0 to be the zero map. - For 1 q n, q : C q (Γ) C q 1 is the homomorphism obtained by extending by Z-linearity the mapping q q ([x 0, x 1,..., x q ]) = ( 1) i [x 0,..., x i,..., x q ] where the notation x i means that the element x i has been removed. i=0 Theorem 1.1 Let Γ be a simplicial complex of dimension n. Then, with the definitions given, the sequence C (Γ), : 0 C n (Γ) C n 1 (Γ) C 1 (Γ) C 0 (Γ) 0 is a chain complex. Proof. We have only to show that, for each q, Im q+1 ker q, that is q q+1 = 0. We have ( q+1 ) q q+1 ([x 0, x 1,..., x q+1 ]) = q ( 1) i [x 0,..., x i,..., x q+1 ] = i=0 6

8 ( 1) i q ([x 0,..., x i,..., x q+1 ]) = q+1 = i=0 q+1 i 1 = ( 1) i ( 1) j [x 0,.., x j,.., x i,.., x q+1 ] + i=0 j=0 q+1 j=i+1 ( 1) j 1 [x 0,.., x i,.., x j,.., x q+1 ] = 0 Now, For each q we put Z q (Γ) = ker q B q (Γ) = Im q+1 the group of q-cycles of Γ the group of q-boundaries of Γ The q-th homology group of Γ is the factor group H q (Γ) = Z q(γ) B q (Γ) We denote by H (Γ) the sequence of groups H q (Γ). It is a fundamental fact in algebraic topology that the homology of a simplicial complex coincides with the homology of its geometrical realization (see [54], [51]). In particular it does not depend on the chosen orientation. In the following we will always assume that a complex is already oriented. Let Γ be a complex of dimension n. Then the differentiation n+1 is the zero map, whence H n (Γ) = Z n (Γ) C n (Γ) is a subgroup of a free abelian group, and so it is free (possibly trivial). We make this observation a proposition. Proposition 1.2 Let Γ be a complex of dimension n, then H n (Γ) is a free abelian group. In several arguments, it is more convenient to work with the reduced homology H (Γ). It is defined via the augmented chain complex C (Γ), : 0 C n (Γ) C n 1 (Γ) C 0 (Γ) Z 0 where q = q for q 1, 1 = 0, and 0 is defined by 0 λ v v = λ v. v S 0 (Γ) v S 0 (Γ) It is easy to verify that H q (Γ) = H q (Γ) for q 1, and H 0 (Γ) = H 0 (Γ) Z. 7

9 Definition. Let A be a finitely generated abelian group. Then A is the direct product of cyclic subgroups. The number of infinite cyclic group that appear in such a decomposition is an invariant of A, that we call the rank of A, and denote by r 0 (A). The following two facts are not difficult to prove. In the first statement, the tensor product is as Z-modules, and the dimension is that of a Q-vector space. Proposition 1.3 Let A be a finitely generated abelian group. Then r 0 (A) = dim(a Q) Lemma 1.4 Let A be a finitely generated abelian group, let A 1 A, and write A 2 = A/A 1. Then r 0 (A) = r 0 (A 1 ) + r 0 (A 2 ). This fact can be restated by saying that given an exact sequence 0 A 1 A A 2 0 of finitely generated abelian groups, then r 0 (A) = r 0 (A 1 ) + r 0 (A 2 ). Theorem 1.5 Let Γ be an oriented simplicial complex of dimension n. Then χ(γ) = n ( 1) q r 0 (H q (Γ)). q=0 Proof. Consider the chain complex associated to Γ: 0 C n (Γ) C n 1 (Γ)... C 1 (Γ) C 0 (Γ) 0 with differentials i. Then, for each 0 q n, we have α q = r 0 (C q (Γ)) (by definition of C q (Γ)) and an exact sequence: 0 Z q (Γ) C q (Γ) B q 1 (Γ) 0 where we put B 1 = 0. Thus, by Lemma 1.4 r 0 (C q (Γ)) = r 0 (Z q (Γ)) + r 0 (B q 1 (Γ)). On the other hand, by the very definition of homology groups, for each 0 q n, we have the exact sequence and so, again by Lemma B q (Γ) Z q (Γ) H q (Γ) 0 r 0 (Z q (Γ)) = r 0 (B q (Γ)) + r 0 (H q (Γ)) 8

10 where B n (Γ) = 0. Therefore χ(γ) = n q=0 ( 1)q r 0 (C q (Γ)) = n q=0 ( 1)q r 0 (Z q (Γ)) + n q=0 ( 1)q r 0 (B q 1 (Γ)) = n q=0 ( 1)q r 0 (Z q (Γ)) + n q=0 ( 1)q+1 r 0 (B q (Γ)) = n q=0 ( 1)q (r 0 (Z q (Γ)) r 0 (B q (Γ))) = n q=0 ( 1)q r 0 (H q (Γ)) Definition. A complex Γ is acyclic if H 0 (Γ) = Z and H q (Γ) = 0 for all q 1. Equivalently, if its reduced homology is zero (i.e. Hq (Γ) = 0 for all q 0). By Theorem 1.5, if Γ is an acyclic complex then χ(γ) = 1. Let X, Y be oriented simplicial complexes, and let f : X Y be a simplicial map. Then, for each simplex σ = {x 0,..., x q } S q (X), its image fσ = {fx 0,..., fx q } is a simplex of Y. Quite possibly, the dimension of fσ could be less than that of σ. Also, if fσ has the same dimension q, the ordering fx 0,..., fx q might not be the one compatible with the given orientation in Y ; thus we denote by π σ the permutation on the indices {0,..., q} such that fx πσ0 <... < fx πσq in the ordering of Y. Then, for q 0, the map f induces a group homomorphism f q : C q (X) C q (Y ) by setting, for all σ S q (X), f q (σ) = and extending by Z-linearity. { sign(πσ )[fσ] if fσ S q (Y ) 0 if fσ S q (Y ) One also defines f 1 to be the zero map in the ordinary case, and the identity 1 Z in reduced homology. With the given definition, it is not difficult to check the following fundamental property. Proposition 1.6 Let X, Y be oriented complexes. For all q 0 let q, q denote the q-differentiation on X and on Y respectively. Let f : X Y be a simplicial map. Then for all q 0: q f q = f q 1 q. 9

11 This propositions shows that, for each q 0, one has q f q (Z q (X)) = f q 1 q (Z q (Y )) = 0 that is f q (Z q (X)) Z q (X). Also f q (B q (X)) = f q q+1 (C q+1 (X)) = q+1 f q+1 (C q+1 (X)) B q (Y ). These observations allow to define, for each q 0 an application f q : H q (X) H q (Y ) as the homomorphism induced by f q, that is, for all a Z q (X), f q (a + B q (X)) = f q (a) + B q (Y ). A morphism h : C D, of two chain complexes of abelian groups C = {C n, n } and D = {D n, δ n } is a family {h n n Z} of group homomorphisms h n : C n D n, such that h n 1 n = δ n h n for all n Z. A sequence 0 A B C 0 of morphisms of chain complexes is called exact if it arises from short exact sequences of abelian groups 0 A n B n C n 0 for all n 0. Thus, Proposition 1.6 states that any simplicial map induces a morphism of chain complexes. Observe that if X and Y are simplicial complexes, and f : C (X) C (Y ) is a morphism of chain complexes, then the images f 0 (x) of the vertices x V ert(x) need not be vertices of Y. This is the case if f is induced by a simplicial map. Then, we say that a morphism f : C (X) C (Y ) is a chain and vertex morphism if f 0 maps V ert(x) (as a distinguished set of free generators of C 0 (X)) in V ert(y ). The following result is useful in many cases. We omit the proof which is obtained by chasing along map diagrams, and it is not difficult. Proposition 1.7 Given a short exact sequence 0 A B C 0 of chain complexes, there is a long exact sequence... H q+1 (C) H q (A) H q (B) H q (C) H q 1 (A)... Proof. See Rotman [51], pages 93,94. A particular case occurs in the direct sum of chain complexes: Proposition 1.8 Let C and D be chain complexes. Then H (C D) = H (C) H (D). 10

12 We now consider the case in which a complex X is covered by two (or more) subcomplexes X 1 and X 2. This simply means that X = X 1 X 2. In order to make the statements more complete, it is convenient to deal also with the empty set (which might occur as X 1 X 2 ). So we adopt the convention that, for all q 0, C q ( ) = H q ( ) = 0 (and that χ( ) = 0). Lemma 1.9 Let X 1 and X 2 be subcomplexes of the complex X such that X 1 X 2 = X. Then for each q 0 there is an exact sequence 0 C q (X 1 X 2 ) C q (X 1 ) C q (X 2 ) C q (X) 0. Proof. Consider the four simplicial inclusions i 1 : X 1 X 2 X 1, i 2 : X 1 X 2 X 2, j 1 : X 1 X, j 2 : X 2 X. Observe that the definition of the groups C q implies that C q (X 1 X 2 ) = C q (X 1 ) C q (X 2 ) and C q (X) = C q (X 1 ) + C q (X 2 ). Then we have natural definitions for homomorphisms i : C q (X 1 X 2 ) C q (X 1 ) C q (X 2 ) j : C q (X 1 ) C q (X 2 ) C q (X) given by, for any s S q (X 1 X 2 ) and any s 1 S q (X 1 ), s 2 S q (X 2 ) i(s) = (i 1 (s), i 2 (s)) j(s 1, s 2 ) = j 1 (s 1 ) + j 2 (s 2 ). By the remark made above, it is immediate that j is surjective, and i is injective. Also, if s S q (X 1 X 2 ), then j(i(s)) = j(s, s) = s s = 0, that is Im(i) ker(j). Conversely, given a = λ s s C q (X 1 ) and b = µ r r C q (X 2 ), write a = a + c a, b = b +c b, where c a, c b are the contribution given by generators in S q (X 1 ) S q (X 2 ). If j(a, b) = 0, we then have 0 = a + b = a + b + (c a + c b ) and so, by linear independence, a = b = 0 = c a + c b. Thus a = b and a, b belong to the group generated by S q (X 1 ) S q (X 2 ). In other words (a, b) = (a, a) = i(a), proving that ker(j) Im(i). In conclusion the sequence is exact. 0 C q (X 1 X 2 ) C q (X 1 ) C q (X 2 ) C q (X) 0 Theorem 1.10 (Mayer-Vietoris) Let X 1, X 2 be subcomplexes of the complex X such that X = X 1 X 2. Then there is the long exact sequence:... H q+1 (X) H q (X 1 X 2 ) H q (X 1 ) H q (X 2 ) H q (X)... 11

13 Proof. By Proposition 1.8, the homology of C (X 1 ) C (X 2 ) is H (X 1 ) H (X 2 ) Consider then the short exact sequence of Lemma 1.9, and apply Proposition 1.7 to get the long sequence. There is a similar statement for reduced homology, which is in fact more suitable for calculations. Here, the convention for the empty set is H q ( ) = 0 for q 0, and H 1 ( ) = Z (while, if X, H 1 (X) = 0). Theorem 1.11 Let X 1, X 2 be subcomplexes of the complex X such that X = X 1 X 2. Then there is the long exact sequence:... H q+1 (X) H q (X 1 X 2 ) H q (X 1 ) H q (X 2 ) H q (X)... where the final maps are H 1 (X) H 0 (X 1 X 2 ) H 0 (X 1 ) H 0 (X 2 ) H 0 (X) H 1 (X 1 X 2 ) 0. Definition. Let X be a complex. A path from a vertex v to a vertex w of X is just a path in the 1-skeleton of X viewed as a (undirected) graph; i.e. a sequence of 1-simplices σ 0,..., σ k such that s 0 = {v = v 0, v 1 }, s 1 = {v 1, v 2 },..., s k 1 = {v k 1, v k }, s k = {v k, v k+1 = w}. X is connected if every two vertices of X are joined by a path. A subcomplex Y of X is a connected component if Y is connected and there is no path joining any vertex v V ert(y ) to any w V ert(x) \ V ert(y ). Observe that in our definition of path we make no reference to an orientation. It is not hard to verify that a complex is connected if and only if its geometrical realization is path-connected in the topological sense. Also it is easy to see that the connected components of a complex are uniquely determined. Exercises. 1) A circuit in a complex is a non empty path joining a vertex v to itself. Prove that if s 0,..., s k is a path in X from the vertex v to w, then (given a suitable orientation) 1 (s s k ) = v w. Thus prove that every circuit is associated to a 1-cycle. 2) Let F = {H 0,..., H n } be a family of proper subgroups of the group G, and let Γ F be the associated coset complex. Prove that Γ F is connected if and only if H 0,..., H n = G. Proposition 1.12 Assume the complex X has d connected components X 1,..., X d. Then, for each q 0; H q (X) = H q (X 1 ) H q (X 2 ) H q (X d ). 12

14 Proof. This is obtained by an easy application of the Mayer-Vietoris sequence and induction on d. Proposition 1.13 Let Γ be a connected complex, and let V =Vert(Γ). Then i) B 0 (Γ) = { v V λ vv v V λ v = 0 }. ii) H 0 (Γ) = Z. Proof. i) Clearly, B = { v V λ vv v V λ v = 0 } is a subgroup of C 0 (Γ), and it contains B 0 (Γ), since for all [v, w] S 1 (Γ), ([v, w]) = v w B. Fix a vertex a in Γ. Then for all v V ert(γ) there is a sequence s o, s 1,..., s k of 1-simplices of Γ connecting v to a. Then, by choosing a suitable orientation, (s o + s s k ) = v a B 0 (Γ). Now, if v V λ v = 0 then λ v v = λ v v λ v a = λ v (v a) B 0 (Γ) v V v V v V v V proving that B 0 (Γ) = B. ii) By i) the map au : C 0 (Γ) Z defined by au ( v V λ vv ) = v V λ v is a surjective homomorphism with kernel B 0 (Γ). Then H 0 (Γ) = C 0 (Γ)/B 0 (Γ) = Z. Corollary 1.14 For any complex Γ, H 0 (Γ) = Z d where d is the number of connected components of Γ. If (X, x 0 ) and (Y, y 0 ) are pointed topological spaces, then their wedge X Y is the quotient space of their disjoint union, where the basepoints are identified. A similar construction can be given for simplicial complexes: a wedge X Y of connected complexes X and Y is a complex whose vertex set is the disjoint union of V ert(x) and V ert(y ) with a single vertex identified, and whose simplices are just those of X and those of Y. Exercise. Let Y, Z be connected complexes. Prove that, for n 1: H n (Y Z) = H n (Y ) H n (Z), and H 0 (Y Z) = Z. More generally, let Y, Z be connected subcomplexes of the complex X = Y Z, and assume that Y Z is the set of faces of a single simplex. Prove that H (X) = H (Y ) H (Z). What we have described in this section is the integral homology of a complex, which is the most common in topology. Fixed an abelian group A it is possible to similarly define homology with coefficients in A, by considering for each q 0, instead of a free abelian group C q (Γ) the Z-tensor product C q (Γ; A) = C q (Γ) A, and differential maps q 1. The homology groups of the resulting chain complex 13

15 are denoted by H q (Γ; A). In topology it is often useful the case in which one takes as A some (integral) homology group of the complex. In group theory one is interested in the case in which A is a field; we have seen that if a group G acts as a group of simplicial maps on a complex Γ, then there are natural induced actions of G on the homology groups. If coefficients are taken in a field this gives raise to linear representations of G. For those who are familiar with the theory of Z-modules, we recall that the relation between integral and arbitrary homology is described by the so-called Universal Coefficients Theorem: H q (Γ; A) = (H q (Γ) A) T or(h q 1 (Γ), A) In particular if A is torsion-free, then H q (Γ; A) = H q (Γ) A. Thus H q (Γ; Q) = 0 if and only if H q (Γ) is a finite group; we then say that Γ is Q-acyclic if, H 0 (Γ) = Z and H q (Γ) is finite for all q 1. Another case that we like to mention is when A = Z p is the additive group of the field of p elements. We say that Γ is Z p -acyclic if H 0 (Γ; Z p ) = Z p and H q (Γ; Z p ) = 0 for q 1. By the UCT Theorem this happens if and only if H 0 (Γ) = Z and, for all q 1, H q (Γ) is a finite group of order not divided by p. In fact, if G is an abelian group, then G Z p = G/pG and T or(g, Zp ) = {x G px = 0}. Thus, since H q (Γ) is a finitely generated abelian group, H q (Γ; Z p ) = (H q (Γ) Z p ) T or(h q 1 (Γ), Z p ) is trivial if and only if H q (Γ) is a finite group with no elements of order p. 1.2 Homotopy equivalence In this section, following [51], we describe a way of establishing that two complexes have the same homology. Thus, let X and Y be simplicial complexes, and denote by q and δ q the differentiation in X and in Y respectively. Definiton. Two simplicial maps f, g : X Y are (homotopy) equivalent (written f g) if for each q 0 there exists a homomorphism π q : C q (X) C q+1 (Y ) such that δ q+1 π q + π q 1 q = f q g q where π 1 = 0. Homotopy equivalence of simplicial maps is readily seen to be an equivalence relation (exercise). Theorem 1.15 If f, g : X Y are equivalent simplicial maps then f q = g q for all q 0. 14

16 Proof. Let f g and, for each q 0 let π q : C q (X) C q+1 (Y ) be a homomorphism such that δ q+1 π q + π q 1 q = f q g q, and π 1 = 0. Let z Z q (X), then (f q g q )(z) = (δ q+1 π q + π q 1 q )(z) = δ q+1 π q (z) Imδ q+1 = B q (Y ). Hence f q (z + B q (X)) = f q (z) + B q (Y ) = g q (z) + B q (Y ) = g q (z + B q (X)), and so f q = g q. Definition. Two simplicial complexes X and Y are equivalent (written X Y ) if there exist simplicial maps f : X Y and g : Y X such that g f 1 X and f g 1 Y. (Of course it is possible to define similarly the more general notion of equivalence of chain complexes - see [51] - but we will not insist on that.) Complexes having homeomorphic geometrical realizations are easily seen to be equivalent. In particular, any complex is equivalent to its barycentric subdivision. Theorem 1.16 If X and Y are equivalent complexes then for all q 0 H q (X) = H q (Y ). Proof. Let f : X Y and g : Y X be morphisms such that g f 1 X and f g 1 Y. Then, for q 0, we have from Theorem 1.15, g q f q = (g f) q = (1 X ) q = 1 Hq(X). Similarly, f q g q = 1 Hq(Y ). Hence H q (X) = H q (Y ). A simplicial map is called a constant if all vertices have the same image. We say that a complex is (chain) contractible if it is equivalent to the complex of a single point. Proposition 1.17 Every contractible complex is acyclic; in particular its Euler characteristic is 1. For a complex X the following conditions are equivalent (i) X is contractible; (ii) the identity map 1 X is equivalent to a constant; (iii) for each q 0 there exists a homomorphism c q : C q (X) C q+1 (X) and a vertex a of X such that q+1 c q +c q 1 q = 1 Cq(X) for q 1, and 1 c 0 (v) = v a for all v Vert(X). 15

17 Proof. The first claim follows immediately from Theorem Suppose that X is contractible. Then it is equivalent to a point {v}, and there exist simplicial maps f : X {v} and g : {v} X such that g f 1 X and f g 1 {v}. But then g f is just the constant application which maps every x V ert(x) in g(v), and this proves (i) (ii). Suppose now that the identity map 1 X is equivalent to a constant a : X X, where for all x V ert(x), a(x) = a for a fixed vertex a. Then by definition, for each q 0 there exists a homomorphism π q : C q (X) C q+1 (X) such that q+1 π q + π q 1 q = (1 X ) q a q. Now, (1 X ) q = 1 Cq(X), a q = 0 for q 1, and a 0 (x) = a for all x V ert(x). We get (iii) by simply renaming π q by c q. Hence (ii) (iii). Finally, assume that (iii) holds. Put π q = c q for all q 0, π 1 = 0, and let Y be the complex with the single vertex a. Define f : X Y and g : Y X by setting f(v) = a for all v V ert(x) and g(a) = a. Then f g = 1 Y and g f = a. For q 1 we have q+1 π q + π q 1 q = 1 Cq(X) = (1 X ) q a q. Also, for all v V ert(x) we have ( 1 π 0 + π 1 0 )(v) = ( 1 c 0 )(v) = v a, whence 1 π 0 + π 1 0 = (1 X ) 0 a 0. Thus a = g f 1 X, proving (iii) (i). Remark. Our definition of contractible simplicial complex is indeed equivalent to acyclicity. This is because the chain groups C q (X) are finitely generated free abelian groups. In the literature on complexes and finite groups, a contractible complex is usually one whose geometric realisation is homotopy equivalent to a point in the topological sense. We have, however, given a homological definition because it seems closer to the combinatorial point of view of these notes. Moreover, all the results that we will present, proving the contractibility of a certain complex, unless explicitely stated, hold in the topological sense also. 1.3 Examples Let n 0 and I n = {v 0,..., v n } be a set of cardinality n + 1. We denote by n the complex of all non-empty subsets of I n (a geometric realization of it is just the standard n-simplex with all its faces). As an exercise show that n is acyclic. This is in fact a particular case of what is called a cone. Definiton. A complex X is a cone if there exists a vertex x of X such that σ {x} X for all σ X. Observe that the complexes n defined above are cones. Proposition 1.18 Let the complex X be a cone. Then X is contractible. In particular χ(x) = 1. Proof. Let v be a vertex of the cone X such that σ {v} X for all σ X. Fix an orientation on X such that v is the least element. 16

18 For each q 0 define a homomorphism π q : C q (X) C q+1 (X) by setting, for all q-simplex σ = [x 0,..., x q ] S q (X): π q (σ) = { [v, x0,..., x q ] if x 0 v 0 otherwise and extending by Z-linearity. Let q 1. Then, if x 0 v: = [x 0,..., x q ] + If x 0 = v: ( q+1 π q + π q 1 q )(σ) = q+1 π q (σ) + π q 1 q (σ) = q = q+1 ([v, x 0,..., x q ]) + π q 1 ( 1) i [x 0,..., x i,..., x q ] = i=0 q ( 1) i+1 [v, x 0,..., x i,..., x q ] + i=0 = [x 0,..., x q ] = σ. q ( 1) i [v, x 0,..., x i,..., x q ] = q ( q+1 π q + π q 1 q (σ) = π q 1 q )(σ) = π q 1 ( 1) i [x 0,..., x i,..., x q ] = [x 0,..., x q ]. Thus q+1 π q + π q 1 q = 1 if q 1. For q = 0 we have 1 π 0 (x) = x v for all x Vert(x)\{v}, and 1 π 0 (v) = 0 = v v. By Proposition 1.17, X is contractible. i=0 i=0 Suppose, for example, that P is a partially ordered set with a greatest (or a least) element. Then the complex of chains (P ) is a cone, and so by Propositon 1.18 it is contractible. We will obatin significant generalizations of this fact in chapter 3. Let n 0. The complex of all proper faces of n+1 is called the n-dimensional sphere, and is denoted by S n. As an abstract complex it is the set of all proper non empty subsets of a set of cardinality n + 2. Proposition 1.19 Let S n be the n-dimensional sphere. Then 1) if n 1, then H 0 (S n ) = H n (S n ) = Z and H q (S n ) = 0 for 0 q n. 2) H 0 (S 0 ) = Z Z and H q (S 0 ) = 0 for q 1. Proof. S 0 is the set of two disjoint vertices, while S 1 is the set of vertices and edges of a triangle. In these cases the statement is verified by a straightforward calculation. 17

19 Thus let n 2, and observe that S n is connected, whence H 0 (S n ) = Z. We then proceed by induction on n. Fix a vertex v 0 of S n, and let X = S n \ {{v 1,..., v n+1 }} be the complex obtained by removing from S n the face opposite to v 0 (this subcomplex is called the star of v 0 and is the set of all simplices that are faces of some simplex that has v 0 as a vertex). By its definiton, X is a cone, and thus it is contractible. Let Y be the complex of all simplices of S n that do not contain v 0. Y is just the set of all nonempty subsets of {v 1,..., v n+1 }, so it is of type n and it is also contractible. Thus H q (X) = H q (Y ) = 0 for q 1, and H 0 (X) = H 0 (Y ) = Z. Clearly S n = X Y, and Z = X Y is just S n 1. By inductive hypothesis we then have H 0 (Z) = H n 1 (Z) = Z and H q (Z) = 0 for 0 q n 1. Mayer-Vietoris theorem now gives the exact sequence: 0 H n (S n ) H n 1 (Z) H n 1 (X) H n 1 (Y ) H n 1 (S n ) H 1 (S n ) H 0 (Z) H 0 (X) H 0 (Y ) H 0 (S n ) 0 which reduces to 0 H n (S n ) Z 0 0 H 1 (S n ) Z Z Z Z 0 yielding easily the desired conclusion. Corollary 1.20 If n is even then χ(s n ) = 2; if n is odd χ(s n ) = 0. The statement of Proposition 1.19 can be more concisely expressed by using reduced homology: for n 1 H n (S n ) = Z and H q (S n ) = 0 for q n 1 Definiton. Given n 0, a complex Γ is n-spherical if Hq (Γ) = 0 for q n. A complex is spherical if it is n-spherical for some n 0. Exercise. Let 0 n m. Study the homology of the set X of all non-empty subsets of the set {0, 1,..., m} that have cardinality at most n. Our final example is considering complexes of dimension 1. These are just (undirected and loop-free) graphs. We leave as an exercise to prove the following. Proposition 1.21 Let X be a complex of dimension 1. Then the following conditions are equivalent. i) X is connected and χ(x) = 1; ii) X is acyclic; 18

20 iii) X is contractible; iv) X is connected and does not have any circuit; v) for each pair of vertices x, y of X there exists one and only one path connecting x to y. A 1-dimensional complex satisfying the properties of Proposition 1.21 is called a tree. 1.4 Carriers We now describe briefly an important tool, that has wide applications and which we will use in Chapter 3. In what follows, X and Y are given abstract simplicial complexes. Definitions. 1) A carrier of X in Y is a map C that assigns to each simplex σ of X a subcomplex C(σ) of Y such that if τ is a face of σ then C(τ) C(σ). A carrier C is contractible (acyclic) if C(σ) is contractible (acyclic) for all σ X. A morphism of chain complexes g : C (X) C (Y ) is carried by C if, for all q 0, g q (σ) C q (C(σ)) for any σ S q (X). Lemma 1.22 Let C be an acyclic carrier from X to Y. Then (i) There exists a morphism g : C (X) C (Y ), carried by C. (ii) If C is contractible, then any two morphism g, f : C (X) C (Y ) carried by C are equivalent. Proof. (sketch). (i) For each x V ert(x) choose a vertex y C(x) (which is not empty); let g 0 (x) = y and extend by linearity to obtain a homomorphism g 0 : C 0 (X) C 0 (Y ). Then, proceed by induction to construct homomorphisms g n : C n (X) C n (Y ), such that g n 1 n = n g n. Let g n 1 be defined. For all σ S n (X), the complex C(σ) is acyclic, thus n 1 g n 1 n (σ) = g n 2 n 1 n (σ) = 0, whence g n 1 n (σ) ker n 1 = Im n, as H n 1 (C(σ)) = 0. (the differentials here and in many following occurences are the restrictions to C(σ)). Then choose s C n (C(σ)) such that n (s) = g n 1 n (σ), and put g n (σ) = s. Do the same for all σ S n (X) and then extend by linearity to define g n. The homomorphisms g n thus constructed define a morphism g : C (X) C (Y ), which by construction satisfyies g = g, and is carried by C. (Observe that, for the proof, it is enough to assume that for all σ X, Hi (C(σ)) = 0 for 1 i < dim(σ).) (ii) One constructs homomorphisms π n : C n (X) C n+1 (Y ) satisfying n+1 π n + π n 1 n = f n g n in a similar way, proceeding by induction. For each x V ert(x), as C(x) is contractible, there exists a homomorphisms π0 x : C 0(C(x)) C 1 (C(x)) 19

21 such that for all y V ert(c(x)), 1 π0 x (y) = y a for a fixed vertex a of C(x). Since f 0 (x) g 0 (x) C 0 (C(x)), we define π 0 (x) = π0 x(f 0(x) g 0 (x)), and extend by linearity. If π k is defined for all k < n, let us define π n. Let σ S n (X). Then n (π n 1 n (σ) (f n (σ) g n (σ))) = 0. Now, the simplices that appear in n (σ) are faces of σ, and so π n 1 n (σ) C n (C(σ)). Also, f n (σ) g n (σ) C n (C(σ)). Therefore π n 1 n (σ) (f n (σ) g n (σ)) belongs to C n (C(σ)) ker n. Again, since C(σ) is acyclic, there exists s C n+1 (C(σ)) C n+1 (Y ) such that n+1 (s) = π n 1 n (σ) (f n (σ) g n (σ)). We then set π n (σ) = s and extend by linearity to define a homomorphism π n : C n (X) C n+1 (Y ). The π n thus constructed satisfy the required conditions to establish that f and g are equivalent. 20

22 Chapter 2 Group Actions Let a group G act by permutations on a set Ω (not necessarily faithfully). For each x Ω and each g G we denote by x g the image of x under the permutation induced by g. We also fix the following notations: - x G = { x g g G } (the orbit of x under G); - G x = { g G x g = x } (the stabilizer of x in G); - Ω g = { x Ω x g = x } (the set of fixed points of g); - Ω G = { x Ω x g = x for all g G} (the set of fixed points of G). Recall that the distinct orbits of G on Ω form a partition of Ω, and that for each x Ω, x G = G : G x. We remind now a well known formula that we will use later on. Lemma 2.1 Let the group G act by permutations on the (finite) set Ω. Then Ω g = t G g G where t is the number of distinct orbits of G on Ω. Proof. We count the order of the set S = { (x, g) Ω G x g = x }. Let x 1, x 2,..., x t be a set of representatives of the distinct orbits. We have Ω g = S = G x = g G x Ω t x G i G xi = i=1 t G : G xi G xi = t G. i=1 21

23 Recall that an automorphism of a simplicial complex Γ is a bijective simplicial map Γ Γ (in this case it is easy to prove that the inverse map is again simplicial). Clearly, the set Aut(Γ) of all automorphisms of a complex Γ is a group under composition. A group G acts simplicially on Γ if there is a group homomorphism G Aut(Γ). In this case we say that Γ is a G-complex. Observe that if G acts on Γ, then, for all q 0 and all g G, the map g q is an automorphism of H q (Γ), and the correspondence g g q defines a linear representation (over the integers in ordinary homology) G Aut(H q (Γ)). If f is a simplicial map on a complex Γ of dimension n, then for each 0 q n we denote by S q (Γ) f the set of all q-simplices of Γ fixed by f. (Observe that it is not always the case that f actually permutes the elements of S q (Γ), as it might map some of those to simplices of smaller dimension; however it certainly induces permutations if it is an isomorphism). Now, denote by Γ f the union of all S q (Γ) f (i.e. the set of all simplices of Γ fixed by f). In general, Γ f might well not be a complex (think for instance to a rotation of a triangle, it is clearly simplicial and fixes the triangle itself but none of its proper faces). This remark leads to the following definition. Definition. Let Γ be a simplicial complex. A simplicial map f : Γ Γ is admissible if for every simplex σ = {x 0, x 1,..., x n } in Γ {fx 0, fx 1,..., fx n } = σ fx i = x i, i = 0, 1,..., n. The following facts are easily verified Proposition 2.2 Let f : Γ Γ be an admissible simplicial map on the complex Γ. Then Γ f is a subcomplex of Γ and, for any positive q, S q (Γ) f = S q (Γ f ). Proposition 2.3 Let P be a poset and let f : P P be an order preserving map. Then f induces an admissible simplicial map on (P ). In particular, every simplicial map f : Γ Γ on the complex Γ induces an admissible map on the barycentric subdivision of Γ. It is also clear that the inverse of an admissible automorphism is admissible, and that the composition of admissible automorphisms is admissible. Thus, the set of admissible automorphisms of a complex Γ is a subgroup of Aut(Γ) (prove that it is a normal subgroup). Proposition 2.4 Let G be a group of admissible automorphisms of the complex Γ. Let n be the dimension of Γ, and for 0 q n, let t q be the number of orbits of G on S q (Γ). Then n χ(γ g ) = G g G q=0( 1) q t q = ( 1) dim(σ) G σ σ Γ 22

24 Proof. By Lemma 2.1 we have g G S q(γ) g = G t q for all 0 q n. Hence g G χ(γ g ) = g G = n ( 1) q S q (Γ g ) = q=0 n ( 1) q G t q = G q=0 n ( 1) q S q (Γ) g = g G q=0 n ( 1) q t q. On the other hand, if for a 0 q n, σ 1,..., σ tq are representatives of the orbits of G on S q (Γ), we have t q G = t q i=1 G : G σ i G σi = σ S G q(γ) σ. Thus q=0 n χ(γ g ) = G g G q=0( 1) q t q = ( 1) dim(σ) G σ. σ Γ Corollary 2.5 Let G be a group of admissible automorphisms of the complex Γ. Assume that χ(γ g ) = 1 for all 1 g G. Then χ(γ) 1 (mod G ). Proof. By the previous Proposition (and with the same notations) n G q=0( 1) q t q = χ(γ) + χ(γ g ) = χ(γ) + G 1. g Lefschetz number Let A be a free abelian group of finite rank, and {x 1,..., x n } be a free system of generators of A. For any f End(A) are defined coefficients α ij Z such that for every i = 1,..., n n f(x i ) = α ij x j. The trace of f is then the integer j=1 T r(f) = n α ii. i=1 23

25 As in the familiar case of vector spaces over a field, one verifies that the trace of f does not depend on the free system of generators. In general, the trace of an endomorphism f of a finitely generated abelian group G is defined as the trace of the endomorphism f induced by f on the factor group G/T (G), where T (G) denotes the torsion subgroup of G (i.e. the set of all the elements of finite order of G; it is a well known and easy fact that G/T (G) is a finitely generated free abelian group). Lemma 2.6 Let G be a finitely generated abelian group, f End(G), G 1 a subgroup of G such that f(g 1 ) = G 1, and G 2 = G/G 1. Then f induces by restriction a homomorphism f 1 of G 1, and a homomorphism f 2 of G 2 by the rule f 2 (g + G 1 ) = f(g) + G 1, and we have T r(f) = T r(f 1 ) + T r(f 2 ). Proof. Exercise. Now, let Γ be a simplicial complex of dimension n, and f : Γ Γ a simplicial map. For each 0 q n, f induces a homomorphism f q on C q (Γ). We consider the free base S q (Γ) of C q (Γ). Since f q permutes the elements of S q (Γ), or sends some of them to 0, the rows of the matrix representing f q with respect to S q (Γ) all have α q 1 zero entries and the remaining entry is either 1 or 0. Thus, the trace of f q is the number of 1-entries in the diagonal, i.e. the number of q-simplices fixed by f q. Hence, for all 0 q n, we have T r(f q ) = S q (Γ) f. The Lefschetz number Λ(f) of f is now defined as follows Λ(f) = n ( 1) q T r(f q ). q=0 Similarly if, for 0 q n, f q is the endomorphism induced by f on H q (Γ), and f = (f 0,..., f n ), we set Λ(f ) = n ( 1) q T r(f q ). q=0 Theorem 2.7 Let f : X X be a simplicial map on the complex X. Then Λ(f) = Λ(f ). 24

26 Proof. Let n be the dimension of X, and consider the chain complex with differentiations q 0 C n (X) C n 1 (X)... C 1 (X) C 0 (X) 0. Since, for all q, q f q = f q 1 q, we have that f q (Z q (X)) Z q (X) and f q (B q (X)) B q (X); so the restriction fq Z of f q to Z q (X) is a homomorphism of Z q (X), and the restriction fq B to B q (X) is a homomorphism of B q (X). Thus, by applying Lemma 2.6 to the short exact sequence 0 Z q (X) C q (X) B q 1 (X) 0 we get (setting f 1 B = 0) T r(f q ) = T r(fq Z ) + T r(fq 1). B On the other hand, the definition of homology groups is the exact sequence 0 B q (X) Z q (X) H q (X) 0 and, again applying Lemma 2.6 (here: fn B = 0), T r(fq Z ) = T r(fq B ) + T r(f q ). Putting things together Λ(f) = = = = n n ( 1) q T r(f q ) = ( 1) q (T r(fq Z ) + T r(fq 1)) B = q=0 q=0 q=0 n n ( 1) q T r(fq Z ) + ( 1) q T r(fq 1) B = q=0 q=0 n n ( 1) q T r(fq Z ) + ( 1) q+1 T r(fq B ) = q=0 n n ( 1) q (T r(fq Z ) T r(fq B )) = ( 1) q T r(f q ) = Λ(f ). q=0 q=0 In the statement of the next Theorem, and in what follows, we adopt the convention: χ( ) = 0. 25

27 Theorem 2.8 Let f : Γ Γ be an admissible simplicial map on the complex Γ. Then Λ(f) = χ(γ f ). Proof. Λ(f) = n ( 1) q T r(f q ) = q=0 n ( 1) q S q (Γ) f = q=0 n ( 1) q S q (Γ f ) = χ(γ f ). q=0 Corollary 2.9 Let f : Γ Γ be a simplicial map such that Λ(f) 0. Then 1. If f is admissible, it admits a fixed point (on V ert(γ)) 2. f fixes at least one simplex of Γ. Proof. 1) Follows immediately from the previous Theorem. 2) Consider the simplicial map f induced by f on the barycentric subdivision (Γ) of Γ. Then f is admissible. Also, for 0 q n = dim(γ), we have H q (Γ) = H q ( (Γ)), and f q = f q. Hence Λ(f ) = Λ(f ) = Λ(f ) = Λ(f) 0. By point 1), f has a fixed point on (Γ), i.e. f fixes some simplex of Γ. Corollary 2.10 Let f : Γ Γ be an admissible simplicial map. If Γ is acyclic then χ(γ f ) = 1 (in particular, f fixes some vertices of Γ). Proof. By hypothesis we have H q (Γ) = 0 for all q 1, and H 0 (Γ) = Z. Therefore, T r(f q ) = 0 for all q 1, and T r(f 0 ) = 1 (in fact, if v V ert(γ) then v + B 0 (Γ) is a generator of H 0 (γ) = Z, also f(v) is a vertex and so 0 (f(v) v) = 0, and f 0 (v + B 0 (Γ)) = f(v) + B 0 (Γ) = v + B 0 (Γ)). By Theorem 2.7 and Theorem 2.8 we get χ(γ f ) = Λ(f) = Λ(f ) = T r(f 0 ) = 1. Observe that Corollary 2.10 remains true if one replaces the hypothesis of acyclicity of Γ by just assuming that Γ is connected and H q (Γ) is a finite group for all q 1 (or, in other words, Γ is Q-acyclic). 26

28 2.2 Orbits Let G be a group of admissible automorphisms of the complex X. Then each S q (X) partitions into its G-orbits. This suggests the possibility of thinking of a complex whose elements are the orbits of G on X. However, what appears to be a natural definition for it does not always work. Let us look at a simple example. Suppose {x, y} is an edge of X; then we would like {x G, y G } to be an edge in a complex of orbits. The simplest way is to ask whether {x G, y G } = {x, y} G. To see that this is not always the case, think of a 1-sphere X = {x, y, z, [x, y], [y, z], [z, x]} and to the automorphism g of X that permutes cyclically x, y, z. Then both S 0 (X) and S 1 (X) have a unique orbit under G = g, whence the set of orbits cannot be given a structure of a complex (if we want that orbits of 1-simplices be again 1-simplices). We are then lead to consider the following properties for a group of automorphisms G of a complex X. (A) For all x V ert(x) and all g G, if {x, x g } X then x = x g. (B) for every simplex {x 0, x 1,..., x q } of X (vertices not necessarily distinct) and g 0, g 1,..., g q G, if {x g 0 0, xg 1 1,..., xgq q } X then there exists g G such that {x g 0 0, xg 1 1,..., xgq q } = {x g 0, xg 1,..., xg q}. Lemma 2.11 Property (B) implies property (A). Proof. Suppose the group G satisfies (B) in its action on the complex X, and let x V ert(x) and g G be such that {x, x g } X. Then there exists h G such that {x, x g } = {x h, x h }, and this forces x = x g. Note that property (A) is stronger than admissibility, but it is still verified by automorphisms of posets. (Observe that the 1-sphere example considered before shows that admissibility does not imply property (A)). Lemma 2.12 Let G be a finite group of automorphisms of a poset P. satisfies (A) on the complex of chains (P ). Then G Proof. Let x P and g G be such that {x, x g } is a chain in P, say x x g. Since g has finite order m and is order preserving we get x x g... x gm = x whence x g = x. Exercise. Assume that G is a group of automorphisms of the complex X. Prove that G satisfies property (A) if and only if for all σ X and all g G, every vertex in σ σ g is fixed by g. Definition. A group of automorphisms G of a complex X is regular if all subgroups of G satisfy property (B) in the induced action. 27

29 Theorem 2.13 Let G be a group of automorphisms of the complex X satisfying (A). Then G is regular on the barycentric subdivision (X). Proof. It will be enough to prove that G satisfies property (B), since (A) is a property which is obviously hinerited by all subgroups of G. Let Σ = {σ 0, σ 1,..., σ q } be a simplex of (X). Then σ 0,..., σ q are simplices of X and we have chosen indices so that σ 0 σ 1... σ q. Let g 0, g 1,..., g q be elements of G such that Σ = {σ g 0 0, σg 1 1,..., σgq q } (X). Then, since the elements of G are automorphisms, σ g 0 0 σg σgq q. We prove by induction on q that there exists g G such that Σ = Σ g. If q = 0 we have nothing to prove. Thus, let q 1. By inductive hypothesis we have a g G such that {σ g 1 1,..., σgq q } = {σ g 1,..., σg q }. Now σ g 1 1 is the least element of the chain, so σ g 0 0 σg 1 1 = σg 1. Let h = g 0g 1. Then σ0 h σ 1. If x is a vertex of σ 0 then {x, x h } σ 1 is a simplex in X. By property (A) this forces x = x h. Hence σ 0 = σ0 h, that is σg 0 0 = σg 0, thus completing the proof. Corollary 2.14 Let G be a finite group of automorphisms of a poset P. Then G induces a regular group of automorphisms on the barycentric subdivision of (P ). Corollary 2.15 Let G be a group of automorphisms of the complex X. Then G induces a regular group of automorphisms on the second barycentric subdivision of X. For groups satifying (B) we may now define the orbit complex. Proposition 2.16 Let G be a regular group of automorphisms of the complex X, and let X/G be the set of all the G-orbits on X (we mean the set of the orbits on all subsets S q (X)). Then X/G has a natural structure of simplicial complex. Proof. Exercise. Definition. The complex X/G defined in Proposition 2.16 is called the orbit complex of G over X. Observe that, if G is regular on the complex X and σ X, then the G-orbit σ G is a simplex of X/G whose faces are the orbits of the faces of σ. Thus, writing τ G σ G it means that there exist g, h G such that τ g σ h. 28

30 Assume G is a regular group of automorphisms of a complex X. We are now interested in describing relations among the homology of X and of the orbit complex X/G. Observe first that the projection π : X X/G defined by mapping each simplex s of X in its G-orbit, π(s) = s G, is a surjective simplicial map. Extending by linearity we have surjective homomorphisms π : C (X) C (X/G) and in turn homomorphisms on homology π : H (X) H (X/G) such that π (s + B (X)) = π(s) + B (X/G), for all s Z (X). Now consider the application σ defined by setting σ(s) = g G sg for all s C q (X); it is clearly an endomorphism for each q 0. Lemma 2.17 With the above notations, ker σ = ker π. Proof. Observe that the Lemma will follow if for each simplex s X, an integral combination c = g G n gs g belongs to ker σ if and only if it belongs to ker π. Now, for all simplices s, σ(s) = σ(s g ). Thus σ(c) = ( g G n g)σ(s), and c ker σ if and only if g G n g = 0. On the other hand π(c) = ( g G n g)s G and again c ker π if and only if g G n g = 0. Theorem 2.18 There exists a homomorphism µ : H (X/G) H (X) such that, for all s Z (X) µ π (s + B (X)) = σ(s) + B (X) π µ (s G + B (X/G)) = G (s G + B (X/G)) Proof. For every q 0 we have C q (X/G) = C q (X)/ ker π = C q (X)/ ker σ = Imσ C q (X) explicitely, we have the monomorphism µ : C q (X/G) C q (X) given by µ(s G ) = µ(π(s)) = σ(s), which is well-defined. The induced homomorphism µ : H q (X/G) H q (X) given by, for all s Z q (X), µ (s G + B q (X/G)) = σ(s) + B q (X) 29