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The problem: Let $X$ be a product of equidimensional spheres of arbitrary dimension, say $k$, and $G$ a finite group acting freely on $X$. Assume that the induced $G$-action on the $Z_2$-cohomology ring of $X$ is trivial. We then have a Serre spectral sequence with the $E_2$-page given by the expression $H^*(G; H^*(X; Z_2))$ and converging to $H^*(X/G; Z_2)$.

The first non-zero differential is certainly $d^{k+1}$. Let $M$ be the image of the transgression (hence $M$ is a subgroup of $H^{k+1}(G; Z_2))$. Now suppose that $M = 0$.

How do I deduce that $E_{\infty} = E_{k+2} = E_2$?

Background: I've been trying to learn some spectral sequences these past few days. After a while I had decided I kind of have a grasp of what is going on and have been trying to understand some research-level examples. The one described above comes from the paper Free compact group actions on products of spheres by R. Oliver. In fact, he takes G to be the alternating group on four letters, but I don't think this has anything to do with my question. (My guess is that it somehow follows from the multiplicative structure of the spectral sequence.)

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To see that $E_{\infty}=E_{k+2}$: your $E_2$ page is supported in the 0th and kth rows, so any differentials beyond $d^{k+1}$ must be zero because they start and/or end at entries that are 0 (and have been since the $E_2$ page). Remember that the differential $d^j$ goes $j$ to the right and $j-1$ downwards.

You're right that $E_{k+2}=E_2$ follows from the fact that the transgression is 0. In fact, since the coefficients are trivial (and the cohomology rings of $G$ and $X$ are both finitely generated in each degree), then $E_2 = H^\ast(G;H^\ast(X)) \cong H^\ast(G) \otimes H^\ast(X)$, and under this identification all the differentials must satisfy a signed Leibniz rule.


I'm trying to make sense of your "Serre SS", since it's not the one I'm familiar with. The one I'm used to has $E_2^{p,q} = H^p(B;H^q(F;\mathbb{Z})) \Rightarrow H^{p+q}(E;\mathbb{Z})$; here, $F \hookrightarrow E \rightarrow B$ is a fibration, and we assume either that $\pi_1(B)=0$ or at least that $\pi_1(B)$ acts trivially on $H^q(F)$. (Otherwise you get twisted coefficients, which are awesome but harder to deal with.)

In the setup you've got, it sounds like you have a fiber bundle with discrete fiber (i.e., a covering space) $G \hookrightarrow X \rightarrow X/G$, where by $G$ I actually mean the underlying set of the group $G$. So $H^0(G;\mathbb{Z}) = \mathbb{Z}^{\oplus |G|}$ and $H^q(G;\mathbb{Z})=0$ for $q>0$. I'd imagine that the assumption of triviality of the induced $G$-action on $H^*(X)$ would guarantee that the coefficients are untwisted, although I'm not sure this is right. If it is, the $E_2$ term of my Serre SS for this would only have nonzero entries along the bottom row; this reflects the close connection between the (co)homology of a base space and a covering space. So the SS has no nonzero differentials, again because all possible differentials $d^i$ ($i\geq 2$) either start at a trivial group or end at a trivial group or both. So $E_2 = E_\infty$.

I guess probably that paper is using some fancier SS for group actions that I've never heard of. Since usually the condition for untwisted coefficients is that $\pi_1(B)$ acts trivially on $H^q(F)$, I'm inclined to think that perhaps your cohomology is actually group cohomology or something. I don't know much about this, but I'm pretty sure that $H^q_{group}(G)=H^q_{sing}(BG)$, where $BG=K(G,1)$ is the classifying space of $G$. This would mesh with what I've said so far, because $\pi_1(K(G,1))=G$.

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Note that since G is finite and the action is free, it is also properly discontinuous. This is what is necessary for the projection to be a covering space. – Aaron Mazel-Gee Sep 28 '10 at 15:30

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