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Let $\mathcal{C}$ be the category of bounded cochain complexes of $R$-modules for a commutative ring $R$. I am trying to prove the following formula involving determinant $\text{Det}(F)$ of a map of chain complexes $F^* : C^* \rightarrow C^*$:

$$\text{Det}(F) = \prod_{i = 0}^n \text{det}(F^i : C^i \rightarrow C^i)^{-1^i}$$

I am hoping there is a notion of determinant of chain complexes such that this formula will hold, but I also want to construct it instead of proving merely that it exists. There should be a wedge product $\Lambda^i C^*$. We should be able to define $\Lambda^n C^*$ for some $n$, maybe.

By the way, this shows up in the Lefchetz fixed point theorem.

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    $\begingroup$ Related: mathoverflow.net/q/7124 $\endgroup$
    – Z. M
    Commented Dec 3, 2021 at 20:18
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    $\begingroup$ Wedge products of chain complexes are probably not what you hope they are. At least with the usual tensor product, the swap $C^* \otimes D^* \stackrel\sim\to D^* \otimes C^*$ acts as $c \otimes d \mapsto (-1)^{\deg c \deg d}d \otimes c$. A consequence is that $\bigwedge^i(M[1]) = (S^iM)[i]$ for any $R$-module $M$, as the antisymmetriser becomes the symmetriser in odd degree. In particular, $\bigwedge^i C^*$ will not be $0$ for $i \gg 0$ in general. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Dec 3, 2021 at 21:21
  • $\begingroup$ Actually I've been told that you get the divider powers instead of the symmetric power if $C^*$ is concentrated either in positive or negative degree. I don't really understand why, but this distinction disappears if you assume $R$ contains $\mathbf Q$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Dec 3, 2021 at 21:25

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