Timeline for Determinant of chain complexes
Current License: CC BY-SA 4.0
7 events
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| Dec 3, 2021 at 21:25 | comment | added | R. van Dobben de Bruyn | 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 $\mathbf Q$. | |
| Dec 3, 2021 at 21:21 | comment | added | R. van Dobben de Bruyn | 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. | |
| Dec 3, 2021 at 20:28 | history | edited | user30211 | CC BY-SA 4.0 |
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| Dec 3, 2021 at 20:21 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Dec 3, 2021 at 20:20 | history | edited | user30211 | CC BY-SA 4.0 |
added 157 characters in body
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| Dec 3, 2021 at 20:18 | comment | added | Z. M | Related: mathoverflow.net/q/7124 | |
| Dec 3, 2021 at 20:14 | history | asked | user30211 | CC BY-SA 4.0 |