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Recall the following localization theorem, as stated in Hsiang's Cohomology Theory of Compact Transformation Groups:

Theorem. Let $G=(S^1)^k$ be a torus, $X$ a paracompact $G$-space with finite cohomological dimension, and $F$ the fixed point set of $X$. Then the following localized restriction homomorphism $$ S^{-1} H^*_G(X;\mathbb{Q}) \to S^{-1}H^*_G(F;\mathbb{Q}), $$ where $S = H^*(BG;\mathbb{Q})-\{0\} = \mathbb{Q}[t_1, \ldots, t_k] - \{0\}$, is an isomorphism.

Something similar happens when $G$ is an elementary abelian $p$-group.

On the other hand, Hsiang points out in a couple of places in his book (e.g. top of p. 44, Remark (i) on p. 68) that those mentioned above the only classes of groups for which one can expect such a theorem to work. He presents an example that it is indeed an unreasonable expectation if we consider compact connected non-abelian Lie groups. A key feature of those is that they admit irreducible representations of degree greater than $1$.

This has me wondering/confused: what about groups which do not fall into either of those two families (i.e. neither ($p$-)tori nor compact connected non-abelian), e.g. products $G = (S^1)^k \times (\mathbb{Z}_p)^l$? Does this sort of groups admit a localization theorem as above, say for cohomology with $\mathbb{Z}$ coefficients?

I would think that this last assertion is true (with the proof going through essentially without changes), but Hsiang really seems to emphasise that this cannot happen. On top of that, other sources I looked at (e.g. Allday-Puppe's Cohomological Methods in Transformation Groups) also have localization theorems formulated for ($p$-)tori only. What am I missing?

EDIT: Thanks to Oscar Randall-William's comments I realized why Hsiang's proof doesn't work without changes even for $G=\mathbb{Z}_p$ if we consider cohomology with integral coefficients. Namely, in this case $H^*(BG;\mathbb{Z})=\mathbb{Z}[x]/(px)$ and $S=H^*(BG;\mathbb{Z})-\{0\}$ is not a multiplicative system in that. I still don't know why (whether?) a different choice of $S$ would also be bad, though.

P.S. I posted a version of this question on math.stackexchange a couple of days ago, but it gained almost no traction at all (15 views, and most of those by me I guess). Hopefully it is more suitable here.

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  • $\begingroup$ Do you see why the localisation theorem fails for $p$-torus actions and $\mathbb{F}_\ell$-cohomology, with $p \neq \ell$? $\endgroup$
    – Oscar Randal-Williams
    Commented May 9, 2016 at 8:33
  • $\begingroup$ @Oscar Randal-Williams: I understand why the proof Hsiang gives doesn't work in this case. It boils down to understanding the intersection of $S$ with the kernel of the map $H^*(BG;k) \to H^*(BH;k)$ induced by the inclusion $H \to G$, where $H \subsetneq G$. If this is non-empty for any such $H$, then the conlusion follows from a "general" localization theorem "$S^{-1}H^*_G(X) \cong S^{-1}H^*_G(X^S)$" (which is true for any compact Lie group $G$!)... $\endgroup$
    – user3158840
    Commented May 9, 2016 at 19:25
  • $\begingroup$ ...Now, if $G$ is a $p$-torus and $k=\mathbb{F}_{\ell}$ for $\ell \neq p$, then the said kernel is zero, because $H^*(BG;k)$ vanishes above dimension $0$ and the said map is the identity on $H^0$. And $0$ is most certainly not in $S$. Having said that, I don't think I understand the reason why it fails. $\endgroup$
    – user3158840
    Commented May 9, 2016 at 19:27
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    $\begingroup$ Let $G=\mathbb{Z}/p$ act on $X=\mathbb{Z}/p$ by translation. Then $H^*_G(X;\mathbb{F}_\ell) = H^*(*;\mathbb{F}_\ell) = \mathbb{F}_\ell$. On the other hand $X^G = \emptyset$. But $H^*(G;\mathbb{F}_\ell) = \mathbb{F}_\ell$ is already a field, so contains no multiplicative subset which can be inverted to make $H^*_G(X;\mathbb{F}_\ell)=\mathbb{F}_\ell$ and $H^*_G(X^G;\mathbb{F}_\ell)=0$ agree. $\endgroup$
    – Oscar Randal-Williams
    Commented May 10, 2016 at 8:13
  • $\begingroup$ @Oscar Randal-Williams: Oh, neat! I was so fixated on understanding why this should work, I wasn't looking for counterexamples. I also now understand why Hsiang's proof doesn't work without changes for $G=\mathbb{Z}_p$ and cohomology with $\mathbb{Z}$ coefficients: $S=H^*(BG;\mathbb{Z})-\{0\}$ is not a multiplicative system, since there are elements in the $0$-th gradation that will be zero divisors. But I still don't see what breaks down if I get rid of those and take $S=H^*(BG;\mathbb{Z})-H^0(BG;\mathbb{Z})\cup \{1\}$. $\endgroup$
    – user3158840
    Commented May 10, 2016 at 8:38

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