Endomomorphisms of Chain Complexes of vector spaces and determinants - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T13:36:52Z http://mathoverflow.net/feeds/question/117658 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/117658/endomomorphisms-of-chain-complexes-of-vector-spaces-and-determinants Endomomorphisms of Chain Complexes of vector spaces and determinants Hiro 2012-12-30T17:51:38Z 2012-12-30T20:13:57Z <p>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 $K$.</p> <p>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 $n$.</p> <p>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 $g_{\ast}$).</p> <p>Moreover, assume that $\det (f_{n}) = \det (g_{n})$ for any $n$.</p> <p>My question is:</p> <hr> <p>QUESTION</p> <p>Under the above conditions, does the next equation hold up to sign?</p> <p>$\prod_{n} \det (H_{n}(f_{\ast}))^{(-1)^{n}} = \prod_{n} \det (H_{n}(g_{\ast}))^{(-1)^{n}}$.</p> <hr> <p>Note that if the chain complex $C_{\ast}$ is bounded above, the statement can be proved as follows: </p> <p>$\prod_{n} \det (H_{n}(f_{\ast}))^{(-1)^{n}} = \prod_{n} \det (f_{n})^{(-1)^{n}} = \prod_{n} \det (g_{n})^{(-1)^{n}} = \prod_{n} \det (H_{n}(g_{\ast}))^{(-1)^{n}}$</p> <p>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.</p> <p>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.</p> <h1>By Sawin's answer</h1> <p>There is a counterexample if one does not admit the difference of sign.</p> <p>Please give me any advice.</p> http://mathoverflow.net/questions/117658/endomomorphisms-of-chain-complexes-of-vector-spaces-and-determinants/117663#117663 Answer by Will Sawin for Endomomorphisms of Chain Complexes of vector spaces and determinants Will Sawin 2012-12-30T20:02:36Z 2012-12-30T20:02:36Z <p>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.</p>