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Consider the restriction of the ordinary group cohomology $H^*(BG,\mathbb{Z})$, where $G$ is a compact Lie group and $BG$ is its classifying space, to finite subgroups $F < G$. If we consider the product of all such restrictions $$H^*(BG,\mathbb{Z}) \to \prod_F H^*(BF,\mathbb{Z}),$$ is this map injective?

EDIT: I asked this question at mathoverflow, and Tim Campion provided an argument for the torsion elements, which together with Qiaochu's answer below one has a complete solution to the question, so yes the map is injective.

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  • $\begingroup$ Does a derived-functor characterization of cohomology make this clearer? $\endgroup$ Sep 10, 2020 at 18:05
  • $\begingroup$ If it helps, I managed to find what appears to be an analogous result for equivariant K-theory. See theorem A here core.ac.uk/download/pdf/82743963.pdf $\endgroup$
    – wzzx
    Sep 10, 2020 at 18:10

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If the cohomology of $BG$ is torsion-free (which happens e.g. when $G = U(n)$ but not when $G = O(n)$) then it injects into the cohomology of $BT$ where $T \to G$ is any maximal torus (and in fact rationally $H^{\bullet}(BG, \mathbb{Q})$ is precisely the Weyl group invariants $H^{\bullet}(BT, \mathbb{Q})^W$). So in this case it suffices to know that this is true for a torus. The question for tori further reduces to the case of a single circle $S^1$, and now we want to know whether the map

$$H^{\bullet}(BS^1, \mathbb{Z}) \to \prod_{n \ge 1} H^{\bullet}(B \mathbb{Z}/n, \mathbb{Z})$$

is injective. Now, I believe but don't know how to prove that in positive degree the restriction

$$\mathbb{Z} \cong H^{2k}(BS^1, \mathbb{Z}) \to H^{2k}(B\mathbb{Z}/n, \mathbb{Z}) \cong \mathbb{Z}/n$$

is just reduction $\bmod n$, which if true would mean the answer is yes in the torsion-free case. In the general case we at least get that the kernel of the map consists at worst of torsion classes. Hopefully someone who actually knows group cohomology can say more from here.

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    $\begingroup$ Yes, I agree. To prove that for $S^1$ the map is restriction, consider that the Chern class of the $S^1$ bundle maps to the generator of $H^2(B\mathbb{Z}/n,\mathbb{Z})$. I am quite interested in the torsion case as well though... $\endgroup$
    – wzzx
    Sep 10, 2020 at 18:23
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    $\begingroup$ @wzzx: here's a thought. I wonder if it suffices to consider finite subgroups of the normalizer $N(T)$ of a maximal torus (which I guess are extensions of subgroups of the Weyl group $W$ by an abelian group or something like that)... if it was known whether $H^{\bullet}(BG) \to H^{\bullet}(BN(T))$ was injective this would at least not be doomed from the outset. Probably the torsion primes are known for both? $\endgroup$ Sep 10, 2020 at 18:31
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    $\begingroup$ @QiaochuYuan: Another way to prove the restriction map is reduction $\pmod{n}$, is to look at the spectral sequence for the fibration $S^1\rightarrow L_n^\infty\rightarrow \mathbb{C}P^\infty$, where $\mathbb{C}P^\infty = BS^1 = S^\infty/S^1$ and $L_n^\infty = B\mathbb{Z}/n = S^\infty/(\mathbb{Z}/n)$. $\endgroup$ Sep 10, 2020 at 18:56
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    $\begingroup$ According to iumj.indiana.edu/docs/32036/32036.asp , all the torsion in the $O(n)$ and $SO(n)$ cases is $2$-torsion, and can be described as the image of a Bockstein homomorphism. I think someone who read this paper carefully could probably figure out whether this injected into $N(T)$, as you describe. $\endgroup$ Sep 14, 2020 at 13:43

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