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Let $C_{\ast} : \cdots \to A_{2} \to A_{1} \to A_{0} \to 0$ be a chain complex of finite dimensional vector spaces over a field $K$.

And let $f_{\ast} : C_{\ast} \to C_{\ast}$ and $g_{\ast} : C_{\ast} \to C_{\ast}$ be two chain endomorphisms of $C_{\ast}$ satisfying $\det (f_{n}) \neq 0$ and $\det (g_{n}) \neq 0$ for any $n$.

Assume that the homology group $H_{n}(C_{\ast})$ is zero for all but finitely many $n$. Then, the induced homomorphism $H_{n}(f_{\ast}) : H_{n}(C_{\ast}) \to H_{n}(C_{\ast})$ is zero for all but finitely many $n$, and so the alternating product $\prod_{n}\det (H_{n}(f_{\ast}))^{(-1)^{n}}$ is well-defined (of course, the same holds for $g_{\ast}$).

Moreover, assume that $\det (f_{n}) = \det (g_{n})$ for any $n$.

My question is:


QUESTION

Under the above conditions, does the next equation hold up to sign?

$\prod_{n} \det (H_{n}(f_{\ast}))^{(-1)^{n}} = \prod_{n} \det (H_{n}(g_{\ast}))^{(-1)^{n}}$.


Note that if the chain complex $C_{\ast}$ is bounded above, the statement can be proved as follows:

$\prod_{n} \det (H_{n}(f_{\ast}))^{(-1)^{n}} = \prod_{n} \det (f_{n})^{(-1)^{n}} = \prod_{n} \det (g_{n})^{(-1)^{n}} = \prod_{n} \det (H_{n}(g_{\ast}))^{(-1)^{n}}$

Here, the first and the last equation can be shown by using induction on the length of $C_{\ast}$, snake lemma and the multiplicativity of $\det$ for short exact sequences. The middle equation is the result of the assumption.

The problem is that, in general, $\prod_{n} \det (f_{n})^{(-1)^{n}}$ and $\prod_{n} \det (g_{n})^{(-1)^{n}}$ are not well-defined.

By Sawin's answer

There is a counterexample if one does not admit the difference of sign.

Please give me any advice.

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1 Answer 1

up vote 2 down vote accepted

This is false. Let $C_*$ be the complex where each $C_i$ is two-dimensional and all the maps have rank one, so that there is homology only in degree $0$. Let $f$ be the identity, and let $g$ be multiplication by $-1$. Then since they are acting on two-dimensional vector spaces, their determinants are the same everywhere. But the homology group is one-dimensional, so the determinants of their action on homology are $+1$ and $-1$ respectively, which are different.

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Tanks for your answer! How about if one admits the difference of sign? –  Hiro Dec 30 '12 at 20:10
    
There's nothing special about $-1$ in this example. For instance, by taking the $C_i$ to be $n$-dimensional with differentials alternating between rank 1 and rank $n-1$, you can get any root of unity. With a little more work you should be able to get any invertible scalar as the quotient of the two determinants. –  Eric Wofsey Dec 30 '12 at 20:28
1  
Actually this same chain complex gets you any invertible scalar. Let $g$ act on each two-dimensional vector space with eigenvalues $\lambda$ and $\lambda^{-1}$ with the eigenspaces being the image of the chain map and a complement, alternatingly. Let $f$ be the indentity again. –  Will Sawin Dec 30 '12 at 20:51
    
Thank you for your kind! I did not think one can make any difference! –  Hiro Dec 30 '12 at 20:55

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