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Let $G$ be a finite group. Let $X$ and $Y$ be two $G$-CW complexes with known integer graded $G$-equivariant Bredon cohomology with constant coefficient systems. Is there any Künneth formula for this cohomology theory to calculate the cohomology of the $G$-complex $X \times Y$ with constant coefficient systems? In my case, $X$ is a free $G$-space.

Any reference will be appreciated. Thank you.

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    $\begingroup$ Repeating my comment made in another question, before I saw this one: I don't think I've seen it written out (or done so myself), and it's been a while since I thought about it, but I think it goes like this: If you view Bredon cohomology as Mackey-functor valued, then there's a Kunneth spectral sequence whose $E_2$ term involves Tor groups taken in the category of Mackey functors, i.e., derived functors of the box product. It converges to the cohomology of the product, with appropriate finiteness conditions to get convergence. $\endgroup$
    – Steve Costenoble
    Commented Jul 10, 2015 at 19:39
  • $\begingroup$ Also, this should work when the coefficient system is the Burnside ring Mackey functor; I'm not sure how it needs to be modified for other coefficient systems. $\endgroup$
    – Steve Costenoble
    Commented Jul 10, 2015 at 19:40
  • $\begingroup$ @SteveCostenoble : Thank you for your valuable comments. But it'll be great if you can produce some formula of Kunneth type for constant coefficient sysmtem. $\endgroup$
    – Surojit Ghosh
    Commented Jul 11, 2015 at 9:54
  • $\begingroup$ @SteveCostenoble: Can you please give a proof or reference for RO(G)-graded and Z-graded Kunneth formula with the Burnside ring Mackey functor? $\endgroup$
    – Surojit Ghosh
    Commented Jul 12, 2015 at 7:19

2 Answers 2

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In Martin Fluch thesis, Theorem 3.67, there is a Künneth formula for Bredon homology.

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Since $X$ is free, the Bredon cohomology of $X\times Y$ agrees with the usual cohomology of the orbit space. There is a homotopy pullback square $$ \begin{array}{ccc} (X\times Y)/G & \rightarrow & Y_{hG} \\ \downarrow & & \downarrow \\ X/G & \rightarrow & BG \end{array} $$ where $Y_{hG}$ denotes the Borel construction, or homotopy orbits. The Eilenberg-Moore spectral sequence of this homotopy pullback square is a Kunneth spectral sequence abutting to what you want, but it starts from the Bredon cohomology of $X$ (which is the same as the usual cohomology of $X/G$) and the cohomology of the Borel construction on $Y$, regarded as modules over $H^*(BG)$. As far as I understand, convergence is not guaranteed either... So this is probably not what you want, but I thought it might be worth mentioning.

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  • $\begingroup$ :What can you say about the convergence when $X$-is $S(\xi)$? Where $\xi$ is the irreducible representation of $Z/n$ given by multiplication by $e^{2i \pi /n}$ $\endgroup$
    – Surojit Ghosh
    Commented Jul 13, 2015 at 2:54
  • $\begingroup$ Assuming I understand correctly that $G=Z/n$ is acting on the circle $X$ in the standard way, the EMSS seems overkill. In that case $(X\times Y)/G$ is just the mapping torus (see Hatcher Ex. 1.2.11) of the map given by the action of a generator of $G$ on $Y$. The Mayer-Vietoris sequence expresses the cohomology of $(X\times Y)/G$ in terms of the cohomology of $Y$ and the action of a generator of $Z/n$ on $H^*Y$. $\endgroup$
    – Gustavo Granja
    Commented Jul 13, 2015 at 20:55
  • $\begingroup$ There is a natural map $f : S(\xi)_+ \wedge S^V \rightarrow S(\xi^j)_+ \wedge S^V$. Then for what $n$, $\tilde{H}^n_{Z/n}(f)$ is nonzero?Where $V$ is a representation of $Z/n$ and $f(z,v) = (z^j ,v) , z \in S(\xi) , v \in S^V$. $\endgroup$
    – Surojit Ghosh
    Commented Jul 13, 2015 at 23:56
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    $\begingroup$ Your question in the comment above is not addressed by my answer as the domain and range of $f$ are not free $G$-spaces. If this is the question you are interested in, I suggest you rephrase your original question. $\endgroup$
    – Gustavo Granja
    Commented Jul 14, 2015 at 0:06

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