7. Homology of Tensor Products.

Transcription

1 7. Homology of Tensor Products. What is the homology H n (C D) of the tensor product of two chain complexes C and D? The answer is that this question is not quite well posed since there are various homology groups that can be associated to a tensor product, namely the total homology, the vertical and the horizontal homology. We concentrate on the total homology In this section we show H n Tot(C D). Weak Künneth Theorem. H n Tot(C D) = (Tot (H p (C) H q (D))) n, provided H p C and H q D are free groups, for all indices. Notation. Given two sequences A p, B q of abelian groups we denote by Tot(A p B q ) the sequence given by (Tot(A p B q )) n := (A p B q ). p+q=n If C, D are chain complexes,then Tot(C D) is the chain complex from the previous section 6.

2 2. Homology of Tensorproducts Let C, C be two chain complexes and let Z p, Z p denote the cycles of C p resp. C p. Lemma 0. The assignment [z p ] [z q] [z p z q], z p Z p, z p Z p defines a homomorphism Θ : H p C H q C H p+q (C C ). It is called the Künneth map. Proof. Use the formula for the boundary map to see that z p z q is a indeed a cycle of C C. Compute (z p + d p+1 ) z q = z p z q + (d p+1 z q), z p (z q + d q+1) = z p z q + ( 1) p (zp d q+1). This proves lemma 0.

3 7 Homology of Tensor Products 3 Let B p, Z p, C p denote the boundaries, cycles and chains, respectively. The same with B p, Z p, C p. Denote by Z, D the chain complexes given by Z n := Z n d n := n Z n = 0 D n := B n 1 d n := n 1 B n 1 = 0 The proof is based on the long exact homology sequence. We begin as follows: 0 Z p j C p B p 1 0. is a short exact sequence of abelian groups. It splits since B p 1 is free. 0 Z C C C D C 0 is a short exact sequence of chain complexes. H m+1 (D C ) β m+1 H m (Z C ) H m (C C ) H m (D C ) β m H m 1 (Z C )

4 4. Homology of Tensorproducts is exact (since it is a portion of the induced long exact homology sequence). Here β m denotes the connecting homomorphism. 0 coker β m+1 α H m (C C ) ker β m 0 is short exact. This proves the following Lemma 1. There is a short exact sequence 0 coker β m+1 α H m (C C ) α ker β m 0. (where β m is the connecting homomorphism specified above). We need to determine ker and coker and the map α.

5 Lemma 2. 7 Homology of Tensor Products 5 H m (Z C ) = (Tot (Z p H q C )) m and H m (D C ) = (Tot (D p H q C )) m. with isomorphisms induced by inclusions. Proof. is short exact. 0 Z q C q B q Z p Z q Z p C q 0 id Z p B q 1 0 ց Z p C q 1 has horizontal and vertical exactness (Z p is free). (1) ker(id ) = Z p Z q C p C q and (2) im(id ) = Z p B q 1 Z p C q 1.

6 6. Homology of Tensorproducts Z m (Z C ) = (Tot Z p Z q) m B m (Z C ) = (Tot Z p B q) m. and Now, 0 B q Z q H q C 0 is short exact 0 (Tot(Z p B q)) m (Tot(Z p Z q)) m (Tot(Z p H q C )) m 0 is short exact. 0 B m (Z C ) Z m (Z C ) (Tot(Z p H q C )) m 0 is a short exact sequence. H m (Z C ) = B m (Z C )/Z m (Z C ) The same with H m (D C ). This proves the lemma 2. = (Tot(Z p H q C )) m

7 7 Homology of Tensor Products 7 Lemma 3. The following diagram commutes: H m+1 (D C ) β m+1 H m (Z C ) = = (Tot (D p+1 H q C )) m j id (Tot (Z p H q C )) m (Recall B p = D p+1 and vertical maps are isomorphisms by lemma 2). Proof. The map β m+1 is the connecting map from lemma 1. It is defined via the following zig-zag diagram (Tot(C p+1 C q)) 0 id m (Tot (D p+1 C q)) m ց (Tot (Z p C q)) j id m (Tot (C p C q)) m as follows: (1) Choose a cycle from D C. This cycle has the form (recall D p+1 = B p ) b p z q, b p B p, z q Z q, p + q = m.

8 8. Homology of Tensorproducts (2) Pull this cycle back to C C, to obtain an element c p+1 z q, where c p+1 = b p. (3) Push this element down to obtain the element (c p+1 z q) = c p+1 z q ± c p+1 z q = b p z q. (4) Pull this element back to Z C, to obtain b p z q. This is the same element we started with. This proves the lemma 3.

9 Lemma 4. 7 Homology of Tensor Products 9 coker β m = (Tot (H p C H q C )) m, ker β m+1 = 0. Proof. 0 B p Z p H p C 0 is a short exact sequence of free groups. 0 B p H q C Z p H q C H p C H q C 0 is short exact (since H q C is supposed to be free). 0 (Tot(B p H q C )) m Tot(Z p H q C )) m (Tot(H p C H q C )) m 0 is short exact. 0 H m+1 (D C ) β m+1 H m (Z C ) (Tot(H p C H q C )) m 0 is short exact (apply lemma 3). This proves both equations from lemma 4.

10 10. Homology of Tensorproducts Now, we finish the proof of the theorem as follows: (1) We put lemma 4 and lemma 2 together. (2) We note that the map coker β m+1 H m (C C ) in sequence (*) from lemma 2 is induced by inclusion Z C C C. Therefore, the first map of the Künneth sequence is just the Künneth map Θ. This proves the Weak Künneth Theorem. The material of this section has been taken from [Munkres, Elements of Algebraic Topology] and [Spanier, Algebraic Topology (1966) p. 227pp.