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Apr 26, 2021 at 7:02 history edited David Roberts CC BY-SA 4.0
Fixed/added links
Apr 17, 2011 at 14:40 comment added Allen Hatcher @ Dave Anderson: The isomorphism $ H^\ast(X/\Gamma;{\Bbb Q}) = H^\ast(X;{\Bbb Q})^\Gamma $ is a simple application of the transfer homomorphism. For a textbook proof see my algebraic topology book, Proposition 3G.1, page 321.
Apr 17, 2011 at 5:47 comment added Dave Anderson The reference I know for the general fact $H^\ast(X/\Gamma;{\Bbb Q}) = H^\ast(X;{\Bbb Q})^\Gamma$ (for $\Gamma$ a finite group acting freely) is Grothendieck's Tohoku paper, which of course uses a lot of spectral sequences. Grothendieck remarks that this can be proved much more easily, presumably along the lines Allen K. indicates. Does anyone know an earlier/easier reference? (Cf. the answer here: mathoverflow.net/questions/18898/… )
Apr 15, 2011 at 9:38 comment added José Figueroa-O'Farrill I added the missing asterisks: it is safer to use \ast than *.
Apr 15, 2011 at 9:38 history edited José Figueroa-O'Farrill CC BY-SA 3.0
Added the missing asterisks: safer to use \ast than *
Apr 15, 2011 at 9:26 history answered Ralph CC BY-SA 3.0