Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am trying to find a relatively elementary proof of the following result: If $(K,d^K)$ and $(L,d^L)$ are acyclic chain complexes of free abelian groups, then their tensor product $(K \otimes L,d^{K \otimes L})$ is acyclic. I am familiar with the usual proof of this result in terms of the Kunneth tensor product formula, but since that result is considerably more general than this one, I was wondering if there is a simpler proof that works in this case. Anyone have ideas?

share|improve this question
    
A naive comment: is it true that acyclic complexes of free abelian groups are contractible? If so, the tensor product ought to respect homotopy equivalences... –  Qiaochu Yuan Jun 1 '12 at 3:45
    
Solving the problem via homotopy equivalences was actually my first idea for a solution, but I could not figure out an obvious equivalence. –  Joshua Jun 1 '12 at 4:03
    
@Joshua: You should not forget to upvote useful answers to your questions. You haven't upvoted any answer so far. –  Marc Palm Jun 1 '12 at 12:46

2 Answers 2

It suffices to suppose that one of your complexes, say $K$, is an acyclic complex of free abelian groups. Then, if $L$ is any complex of abelian groups, $K \otimes L$ is acyclic.

This can be easily seen as follows: Assume for the moment we know that $K$ is contractible, i.e. there is a chain equivalence $f: K \to 0$. Then $f \otimes id_L:K \otimes L \to 0 \otimes L=0$ is a also a chain equivalence (see MacLane, Homotopy, Cor. V.9.2). Hence $H_\ast(K \otimes L) \cong H_\ast(0)=0$ shows that $K \otimes L$ is acyclic.

The fact that acyclic complexes of free abelian groups are contractible is Lemma VI.3.2 of Brown: Cohomology of Groups. But the proof is simple enough to be posted here. First note that the short exact sequence $$0 \to \operatorname{im} d_{i+1} \to K_i \to \operatorname{im} d_i \to 0$$ splits since as a subgroup of the free abelian group $K_{i-1}$, $\operatorname{im} d_i$ is a free abelian group and hence a projective $\mathbb Z$-module. Write $$K_i = \operatorname{im} d_{i+1} \oplus A_i$$ such that $d_i$ maps $A_i$ isomorphicaly onto $\operatorname{im} d_i$. Let $e_i: \operatorname{im} d_i \to A_i$ be the inverse map. Now define $h_i:K_i \to K_{i+1}$ by $$h_i|\operatorname{im} d_{i+1} := e_{i+1},\;\;\;h_i|A_i := 0$$ Let $x=d+a \in K_i$ with $d \in \operatorname{im} d_{i+1}, a \in A_i$. Since $d_i(x)=d_i(a)$ we obtain
$$(h_{i-1}d_i + d_{i+1}h_i)(x) = e_i(d_i(a)) + d_{i+1}(e_i(d))= a + d=x.$$ Hence $h: id_K \simeq 0$ is the searched homotopy.

Added: The argument holds verbatim for complexes over a PID.

share|improve this answer

The homotopy approach works if one of your chain complexes, say $K$, is bounded below (assuming homological indexing). In this case $K$ is chain homotopy equivalent to the $0$ complex (see Munkres, Elements of Algebraic Topology, Theorem 46.2) In fact, let $f: 0\to K$ and $g:K\to 0$ be the unique maps, which must then be the homotopy inverses. In particular then, there must be a degree 1 $D:K\to K$ such that $\text{id}_K=\text{id}_K-gf=\partial D+D\partial$ (so $D$ is the chain contraction). Now consider $\bar D=D\otimes \text{id}_L:K\otimes L\to K\otimes L$. Then for any $x\otimes y\in K\otimes L$, we have \begin{align*} (\partial\bar D+\bar D \partial)(x\otimes y) & =\partial (Dx\otimes y)+\bar D(\partial x\otimes y+(-1)^{|x|}x\otimes \partial y)\newline & =\partial Dx\otimes y+(-1)^{|x|+1}Dx\otimes\partial y+D\partial x\otimes y+(-1)^{|x|}Dx\otimes \partial y\newline & =\partial Dx\otimes y+D\partial x\otimes y\newline & =((\partial D+D\partial)(x))\otimes y\newline & =\text{id}_K(x)\otimes y=x\otimes y \end{align*} So in other words $\bar D$ is a chain contraction of $K\otimes L$

Sorry - I don't know how to do aligned equations on here. I'm sure someone will come spruce that up (please?).

share|improve this answer
    
@Greg Friedman: The align environment works - the only trick is to replace \\ with \newline. Probably eqnarray works too (but I didn't try that yet). –  Mark Grant Jun 1 '12 at 5:48
    
Why do you require one of the complexes to be bounded below ? –  Ralph Jun 1 '12 at 8:48
    
@Mark. Thanks - that explains why it looked like it was almost working. @Ralph - the boundedness is usually needed because the homotopy operators are (in the most hands-on proofs) constructed inductively and you need some place to start the induction. Have a look at the proof of the theorem in Munkres that I cited. Conceivably this is a simple enough case that there's some way around this, but I'll have to think about it/hunt around. –  Greg Friedman Jun 1 '12 at 18:32
    
Oops, I should have read Ralph's answer first before writing the last comment! I guess boundedness isn't necessary here :-) –  Greg Friedman Jun 1 '12 at 18:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.