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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Oct 17, 2011 at 8:05 comment added Chris Gerig (flip the M and M' entries around actually; but I guess the typo is evident)
Oct 16, 2011 at 3:42 history edited Chris Gerig CC BY-SA 3.0
added 275 characters in body
Oct 16, 2011 at 3:26 vote accept Xiao-Gang Wen
Oct 16, 2011 at 1:24 comment added Chris Gerig .... No, it relates $H^*(G_1\times G_2,M\otimes M')$ to $H^*(G_1,M)$ and $H^*(G_2,M')$. Take $G_2=0$ and $M=\mathbb{Z}_2$ and $M'=\mathbb{Z}$.
Oct 16, 2011 at 0:34 comment added Xiao-Gang Wen Dear Chris: Kunneth formula relates $H^*[G_1\times G_2, M]$ to $H^*[G_1, M]$ and $H^*[G_2, M]$. I do not know how Kunneth formula relates $H^*[G, Z_2]$ to $H^*[G, Z]$.
Oct 16, 2011 at 0:20 history answered Chris Gerig CC BY-SA 3.0