Timeline for Antisymmetric product of complexes
Current License: CC BY-SA 4.0
8 events
when toggle format | what | by | license | comment | |
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Jun 24, 2021 at 8:29 | vote | accept | MKR | ||
Jun 24, 2021 at 4:22 | answer | added | user1092847 | timeline score: 2 | |
Jun 24, 2021 at 3:09 | history | edited | MKR | CC BY-SA 4.0 |
added 350 characters in body
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Jun 24, 2021 at 0:11 | comment | added | MKR | That's right. I'm just writing the $n=2$ case, and it can be checked that $S^2A$ is indeed in the kernel of $A\otimes B \rightarrow \Lambda^2 B$. | |
Jun 23, 2021 at 13:31 | comment | added | Z. M | The sequence you wrote does not seem to be quite correct, if I am not mistaken: you should replace the first term $S^2A$ by the divided power $\Gamma^2A$ (there is no obvious natural map $S^2A\to A\otimes B$ if $2$ is not invertible). In fact, the Koszul exact sequence [Illusie, Complexe Cotangent et Déformations I, V.4.3.2] gives rise to the long exact sequence \[0\to\Gamma^n A\to\Gamma^{n-1}A\otimes B\to\Gamma^{n-2}A\otimes\bigwedge\nolimits^2B\to\dots\to\bigwedge\nolimits^nB\to\bigwedge\nolimits^nC\to0\]and yours is the special case when $n=2$. | |
Jun 23, 2021 at 6:53 | history | edited | MKR | CC BY-SA 4.0 |
Edited a little to make it clear...
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Jun 23, 2021 at 6:46 | history | edited | MKR | CC BY-SA 4.0 |
edited body
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Jun 23, 2021 at 6:10 | history | asked | MKR | CC BY-SA 4.0 |