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Dec
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comment Behaviour of cohomology groups under extension of scalars
If we are willing to pass to homology, and $R$ is flat, then it works. There is a Kunneth spectral sequence that will express the left hand side in terms of various Tor terms of $R$ over $\hat R$, which vanish. I think the cohomology question probably works if $R$ is flat and we make some finiteness assumptions...
Dec
25
answered Behaviour of cohomology groups under extension of scalars
Nov
16
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Nov
16
comment Transgression in terms of k-invariant for chain complexes
I'm also confused, especially by what "multiplication by" means, as this is a map from H^n(X) to H^(n+1) (BG). But I'm very interested if there is a general description of this transgression map. I suggest that M is meant to be the top cohomology group of X. Also perhaps multiplication means composition product, but then shouldn't it be Ext(M,k) rather than (k,M)?
Sep
17
comment Generation of cohomology of graded algebras
I may just be confused with the gradings/notation, but doesn't it seem like this should follow directly from the cobar complex being of finite type (since A is)? EDIT: never mind, I was indeed just confused.
Sep
15
comment Computations in modular cohomology of finite groups
You're welcome! By the way, I just posted the thesis to the Arxiv (arxiv.org/abs/1509.03910) to make it less of a pain to track down.
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Sep
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answered Computations in modular cohomology of finite groups
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Mar
3
revised On the cohomology of a finite covering map
added 13 characters in body
Mar
3
comment On the cohomology of a finite covering map
Yes, that's a much quicker way to see it than the chain-level argument I had in mind. Of course spectral sequence for $X\to X/G\to BG$ will not collapse without some assumptions... if you want a totally general answer (i.e. with integer coefficients) there will be no way to avoid doing a spectral sequence. I pointed out in my answer (because I thought the questioner was thinking of it) that the spectral sequence for $G\to X\to X/G$ does collapse, but without giving any useful information -- sorry that wasn't clear.
Mar
2
comment On the cohomology of a finite covering map
Chris: the confusion here is that some people are interpreting $H^*(G)$ to be the singular cohomology of $G$, and others the group cohomology. I don't know which one the questioner intended (but group cohomology would make more sense).
Mar
2
comment On the cohomology of a finite covering map
Right, perfect! Of course, in this case (free action) the Borel construction is homotopy-equivalent to the orbit space $X/G$. A small quibble: it's not a principle bundle (the fiber is $X$, not $G$) but merely a fiber bundle with structure group $G$. Of course, it's still a fibration so we get a Serre spectral sequence as desired.
Mar
2
comment On the cohomology of a finite covering map
(added) The thing that goes wrong in that example is that 2, the order of $G$, is not a unit in the coefficient ring.
Mar
2
comment On the cohomology of a finite covering map
Careful, hypotheses are needed here (to show that $H^*(X/G)$ is the $G$ invariants of $H^*(X)$). Example: take the covering space $S^2\to\mathbb{R}P^2$ with integer coefficients. Then $H^2(X)=0$ but $H^2(X/G)\neq0$.