Timeline for Tracking spectral sequence differentials
Current License: CC BY-SA 3.0
4 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 19, 2012 at 12:32 | comment | added | Ralph | $H^1(G;M)$ is just the quotient of the derivations $Der(G,M)$ by the subgroup $Ider(G,M)$ of inner derivations. Also note that the seven-term-exact sequence yields the exact sequence $0 \to H^1(G/K;M) \to H^1(G;M) \to H^1(K;M)^{G/K} \to H^2(G/K;M^K)$ that could be of help. | |
| Apr 19, 2012 at 12:10 | comment | added | Earthliŋ | That has it somewhat backwards. I want to calculate the dimension of $H^1(G,M)$, with twisted coefficients from $K$ and $G/K$, which happen to be abelian. I thought the best way of getting at the cohomology of $G$ would be via the s.s. Indeed, the terms are easy to calculate, but the result is obscured by the differential. Maybe it would have been wiser to bite the bullet and calculate $H^1(G,M)$ directly via the bar resolution...? | |
| Apr 19, 2012 at 11:02 | history | answered | Mark Grant | CC BY-SA 3.0 |