Homology of Chain Complexes

Homology of Chain Complexes Skripte zur Vorlesung Prof. Dr. Klaus Johannson I. Properties of Chain Complexes. 1. Abelian Groups. 2. Chain Complexes. II. Properties of Homology. 3. Homology of Quotients. 4. Mapping Cylinders and Chain Homotopies 5. Cohomology of Duals. 6. Tensor Product for Chain Complexes. 7. Homology of Tensor Products. III. Applications. 8. Weak Lefschetz Fixpoint Theorem. 9. Strong Lefschetz Fixpoint Theorem. 10. Homology for Group Actions

1. Abelian Groups. In this section we discuss some basic results of abelian groups and their dual groups. We introduce chain complexes built from them and we define homology and cohomology groups. 1. Classification. We begin with some basic results concerning abelian groups. So let G be a finitely generated abelian group. In many respects G behaves like a vector space. But there is one important difference, namely the possibility of finite elements. Here we say an element g G is a finite element (or: an element of finite order or: a torsion element) if ng = 0, for some n Z with n 0. Otherwise g is called an infinite element (or: an element of infinite order). Given this notation define Tor(G) := {g G g is finite } {0} Free(G) := G/Tor(G). It turns out that Tor(G) is a subgroup and that Free(G) is isomorphic to a subgroup Free(G) in G with Free(G) Tor(G) = {0} and Free(G) Tor(G) = G. The shorthand expression for this fact is G = Free(G) Tor(G) If Tor(G) = {0}, i.e., if Free(G) = G, we say G is a free abelian group. Note further that the torsion subgroup Tor(G) is a finite subgroup. Remark. Analyzing the groups Free(G) and Tor(G) further one finds out that Free(G) = Z... Z, Tor(G) }{{} = Z m1... Z mn. m times This decomposition is not unique. For instance, Z m Z n is ismorphic to Z mn if gcd(m,n) = 1 (it is not cyclic if gcd(m,n) 1). We get a unique decomposition with the additional requirement m 1 m 2... m n ( *). For historical reasons, the number m is called the Betti number and the numbers m 1,...,m n are called the torsion coefficients of G. See below for a calculation of those numbers. * see [Seifert-Threlfall, Lehrbuch der Topologie. 86,87]

2. Abelian Groups 2. Generating Abelian Groups. Let M be a set. Then define F(M) := { a m m a m Z and a m = 0, for almost all m M } where a m m denotes the formal sum such as 2a + 3b 4c. Define a composition + : F(M) F(M) F(M) by setting ( a m m) + ( b n n) = c p p where { am, if p = m and m n c p = b n, if p = n and m n a m + a n, if p = m = n. Definition. F(M) as constructed above is called the free abelian group generated by the set M. Example. It is very easy to create a finitely generated abelian group and to find its Betti and torsion numbers. One starts with a collection of letters such as a,b,c and a collection of sums such as 3a + 5b 2c,2a + b + 5c. Given these data the abelian group G as the quotient group of the free abelian generated by a,b,c modulo the (normal) subgroup generated by the sums 3a + 5b 2c, 2a + b + 5c. Note that the group G is completely represented by the matrix: [ ] 3 5 2 2 1 5 Here the rows are labelled by the sums r 1 = 3a + 5b 2c, r 2 = 2a + b + 5c and the columns are labelled by the generators a, b, c. Recall that every integer matrix can be put in normal form, using row- and column-operations alone. For instance, we have the following operations for the matrix above [ ] [ ] [ ] 3 5 2 1 4 7 1 0 7 2 1 5 2 1 5 2 7 5 [ ] [ ] [ ] 1 0 0 1 0 0 1 0 0 2 7 19 0 7 5 0 2 5 [ ] [ ] 1 0 0 1 0 0 0 2 1 0 0 1

1 Abelian Groups 3 This means that the generators a,b,c can be changed, using additions and substractions of generators alone, to the generators a,b,c with a = 0,c = 0 and no condition on b. Thus G = Z =< b > is the free cyclic group Z generated by b. If the 1 in the first column were a 3 instead, then 3a = 0 and G would be G =< b > < a 3a = 0 > = Z Z/3Z = Z Z 3. Remark. If we wanted to we could iterate the above construction. Indeed, given a group (G, ), we can repeat the formal definition and formally define a sum by setting F(G) := { a m m a m Z and a m = 0, for almost all m G } + : F(G) F(G) F(G) ( a m m) + ( b n n) = c p p as before. So far we have not done anythng different. But notice that G comes with a product : G G G which has not been used yet. We can use it to define a multiplication by setting as required by the distributative law. : F(G) F(G) ( a m m) ( b n n) := c p p Definition. Let G be a group. Then F(G),+, ) constructed above is a ring. It is called the group ring generated by the group (G, ).

4. Abelian Groups 3. Constructing Abelian Groups. There are three important constructions for abelian groups that we single out, the direct sum, the quotient, and the dual groups. Definition. Let G,H be two abelian groups. Then G H denotes the abelian group (G H,+) whose composition is given by (g 1,h 1 ) + (g 2,h 2 ) = (g 1 + g 2,h 1 + h 2 ). Notation: G H is called the direct sum of G and H. Definition. Let H be a subgroup of the abelian group G. Then G/H denotes the abelian group (G/H, +) whose composition is given by [g] + [h] := [g + h]. Notation: G/H is called the quotient group of G modulo H. Definition. Let G be an abelian group. Then Hom(G, Z) := { ϕ : G Z ϕ is a homomorphism } is the abelian group whose composition is given by (ϕ + ψ)(g) = ϕ(g) + ψ(g). Notation. Hom(G, Z) is called the dual group of G. It is also denoted by G. Remark. It is easily verified that G H, G/H and Hom(G, Z), together with the compositions given above, are abelian groups again. Remark. Every homomorphism ϕ : G G between abelian group induces a dual homomorphism ϕ : (G ) G, f f ϕ, between the dual groups, because (f 1 + f 2 ) ϕ(x) = f 1 ϕ(x) + f 2 ϕ(x), for all x G. Definition. Let C,D be two abelian groups. Then define: (1) F(C D) := the free abelian group generated by the set C D.

1 Abelian Groups 5 (2) := the equivalence relation on F(C D) generated by: (3) C D := F(C D)/. (a,b) + (c,b) (a + c,b) (a,b) + (a,c) (a,b + c) r(a,b) (ra,b) (a,rb) Notation. C D is called the tensor product of C and D. Remark. C D, together with the addition defined above, is an abelian group. Its elements are equivalence classes represended by α ij c i d j, where c i d j := [(c i,d j )]. Lemma. Let C, D be abelian groups and let C D be the tensor product. Then the map π : C D C D, (c,d) c d is a bilinear map with the property: for all bilinear maps ϕ, there is a linear map ψ that makes the following diagram commutative: ϕ C D E π ր ψ C D Proof. Follows directly from the construction. Example. Z 3 Z 5 = 0 and Z m Z n = Zgcd(m,n). (by definition of Z n, it follows that Z 1 = 0).

6. Abelian Groups 4. Calculating Dual Groups. Proposition. The following holds: (1) (G 1 G 2 ) = G 1 G 2, if G 1,G 2 are abelian groups. (2) Z = Z. (3) Z m = {0}. Proof. ad (1) Let f (G 1 G 2 ). Then f is a homomorphism f : G 1 G 2 Z. Set f i := f G i, i = 1,2. Then f 1 G 1,f 2 G 2 and so f 1 f 2 G 1 G 2. It is easily verified that the assignment f f 1 f 2 defines an isomorphism (G 1 G 2 ) G 1 G 2. ad (2) Define a map ϕ : Z Z, f f(1). Then ϕ(f 1 + f 2 ) = (f 1 + f 2 )(1) = f 1 (1) + f 2 (1) = ϕ(f 1 ) + ϕ(f 2 ) Thus ϕ is a homomorphism. Moreover, f ker(ϕ) f(1) = 0 f(1 +... + 1) = 0. This means f(x) = 0, for all x Z. Thus ϕ is injective. Finally, for every a Z, let f a : G Z be the extension of the map f a (1) = a, Then f a G with ϕ(f a ) = f a (1) = a. Hence ϕ is surjective. Thus, altogether, ϕ is an isomorphism. ad (3) Let f Z m. Then f is a homomorphism f : Z m Z. Set a = f(1). Then 0 = f[0] = f[m] = mf(1) = ma and so a = 0 since m 0. It follows that f(x) = 0, for all x Z. Hence Z m = {0}. Corollary. Let G be a finitely generated abelian group. Then G = Free(G). Proof. By the classification of finitely generated abelian groups we have G = Z... Z Z m1... Z mn and so G = Z... Z (Z m1 )... (Z mn ) = Z... Z = Z... Z = Free(G). We collect the following results without proof. Proposition. The dual of a homomorphism has the following properties. (1) id = id (2) (ψ ϕ) = ϕ ψ (3) if ϕ is an isomorphism so is ϕ (4) if ϕ is the zero homomorphism so is ϕ

(5) if ϕ is surjective, then the dual ϕ is injective. 1 Abelian Groups 7 (6) if ϕ is an injective homomorphism, then the dual ϕ need not be surjective. Warning. The dual of an injective homomorphism need not be surjective. 5. Exact Sequences of Abelian Groups. Definition. A sequence ϕ m+1 ϕ m 0... A m+1 Am Am 1... 0 of abelian groups A m and homomorphisms ϕ m : A m A m 1 is called an exact sequence if im(ϕ m+1 ) = ker(ϕ m ) Theorem. Let be a short exact sequence of abelian group. Suppose D is a free abelian group. Then the induced sequences are also short exact sequences. Proof. 0 A B C 0 0 A D B D C D 0 0 Hom(A,D) Hom(B,D) Hom(C,D) 0 Theorem. Let 0 A B ϕ C 0 be a short exact sequence of finitely generated, free abelian group. Then there is a homomorphism ψ : C B with ϕψ = id: 0 A B ψ ϕ C 0 Notation. ψ is called a splitting homomorphism since Proof. B = ker(ϕ) im(ψ).

8. Abelian Groups Five Term Theorem. In the following commutative diagram of abelian groups and homomorphisms, suppose that (1) the two horizontal rows are exact sequences and that (2) the maps α,β and δ,ǫ are isomorphisms: A i B j C k D l E Then γ is an isomorphism. α β γ δ ǫ A i B j C k D l E Proof. The theorem follows from the following two claims. Claim 1. Suppose β,δ are surjective and ǫ is injective. Then γ is surjective. Let c C be an arbitrary element. Then we have k (c) = δ(d), for some d D (since δ is surjective) l(d) = 0 (since ǫ is injective and since ǫl(d) = l δ(d) = l k (c) = 0) d = k(c), for some c C (since the top-row is exact) k (c γ(c)) = 0, (since k (c ) k γ(c) = k (c) δk(c) = k (c ) = 0). c γ(c) = j (b ), for some b B (by exactness) b = β(b), for some b B (since β is surjective) γ(c + j(b)) = c (since γ(c + j(b)) = γ(c) + γj(b)γ(c) + j β(b) = γ(c) + j (b ) = c ) and so γ is surjective since c was arbitrary.

1 Abelian Groups 9 Claim 2. Suppose β,δ are injective and α is surjective. Then γ is injective. Let c ker(γ). Then we have δk(c) = 0 (since δk(c) = k γ(c)) k(c) = 0 (since δ is injective) c = j(b), for some b B β(b) = i (a ), for some a A (since j β(b) = γj(b) = γ(c) = 0) a = α(a) (since α is surjective) i(a) b = 0 (since β is injective and since β(i(a) b) = βi(a) β(b) = i α(a) β(b) = i (a ) β(b) = 0) b = i(a) c = 0 (since c = j(b) = ji(a) and since ji = 0) It follows that γ is injective. This finishes the proof of the theorem. The material of this section is taken from [Hatcher, Algebraic Topology].