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

Note, according to McClure - Restriction maps in equivariant K-theory a similar result holds in equivariant K-theory. Perhaps there is a way to derive the above from McClure's theorem?

I asked this question on stackexchange and didn't manage to get a (complete) answer nor solve it myself. However, Qiaochu Yuan offered a proof for the non-torsion elements.

  • 5
    $\begingroup$ As Qiaochu suggests, $H^*(BG)$ injects into $H^*(NT)$ (normalizer of torus). A splitting is given by the Becker-Gottlieb transfer. The composite of a map and the transfer is the Euler characteristic of the fibers and $\chi(G/NT)=1$, so the transfer yields a splitting. $\endgroup$ Sep 21, 2020 at 14:54
  • 2
    $\begingroup$ The Becker–Gottlieb observation can be found in the appendix to this paper—link.springer.com/chapter/10.1007/BFb0070640—which unfortunately points out on its second page, in Theorem 2, that the sequence $1 \to T \to N_G(T) \to W_G \to 1$ does not typically split. The last page suggests that it may still be an open question whether the Serre spectral sequence of $BT \to BN \to BW$ collapses. $\endgroup$
    – jdc
    Sep 22, 2020 at 12:33
  • 1
    $\begingroup$ There is something maddening about this, actually: there's a string of results from Borel (and Serre, in one paper) that for compact, connected Lie groups $G$, it is equivalent for $G$ to have $p$-torsion, for $BG$ to have $p$-torsion, for the order of $W$ to be divisible by $p$, and for $G$ to contain an elementary $p$-group (which can be taken of order $\leq p^3$ not contained in any maximal torus. But his proof, which he apologizes for, doesn't use the existence of some such subgroup $P$ to detect the $p$-torsion in $H^*BG \to H^*BP$. Rather, he reduces to the simply-connected case... $\endgroup$
    – jdc
    Sep 22, 2020 at 12:41
  • 2
    $\begingroup$ "Les résultats de cet article mettent ainsi en évidence quelques liens entre la structure de groupe d'un groupe de Lie et sa topologie. Cependant, l'auteur regrette vivement d'en avoir été réduit à établir plusieurs propositions d'énoncé général à l'aide de vérifications utilisant la classification. 'You have a low shopkeeping mind. You think of things that would never come into a gentleman's head.' 'That's the Swiss national character, dear lady.' (B. Shaw)." $\endgroup$
    – jdc
    Sep 22, 2020 at 12:46
  • 1
    $\begingroup$ @jdc's reference: Curtis, Wiederhold, and Williams - Normalizers of maximal tori. $\endgroup$
    – LSpice
    Oct 4, 2020 at 15:34

2 Answers 2


After the heavy lifting done by people on MSE and in the comments, I think it's not too bad to finish off the proof that the answer is yes.

As argued by Ben Wieland in the comments, we reduce to showing that for any short exact sequence of topological groups $$U(1)^n \to G \to W $$ where $W$ is finite, we have that $H^\ast(BG;\mathbb Z)$ injects into the product of $H^\ast(BF;\mathbb Z)$ over all finite subgroups $F \subseteq G$. The jist of the argument is going to be replacing $U(1)^n$ with $(Q/\mathbb Z)^n$, and then arguing that every finitely-generated subgroup of the resulting extension $G_\ast$ is finite.

The first thing to note is that for some $m \in \mathbb Z$, there exist maps of exact sequences $$\require{AMScd} \begin{CD} (\mathbb Z/m)^n @>>> G_m @>>> W \\ @VVV @VVV @VV{=}V\\(\mathbb Q /\mathbb Z)^n @>>> G_\ast @>>> W \\ @VVV @VVV @VV{=}V\\ U(1)^n @>>> G @>>> W \end{CD}$$

The middle line exists and maps to the bottom line because (1) $\mathbb Q/\mathbb Z$ is the torsion subgroup of $U(1)$, and so must be preserved by the action of $W$, and (2) the discrete group quotient $U(1)^\delta / (\mathbb Q/\mathbb Z)$ is a rational vector space, so no matter the action of $W$, the cohomology of $W$ with values in this quotient vanishes by Maschke's theorem. Thus $H^\ast(BW; \underline{(\mathbb Q/\mathbb Z)^n}) \to H^\ast(BW; \underline{U(1)^{\delta,n}})$ is an isomorphism and in particular the class classifying the extension is hit.

The top line exists and maps to the middle line because when we choose a cocycle $W \times W \to (\mathbb Q/\mathbb Z)^n$, we see that because $W$ is finite, the cocyle has finite image, and every finitely-generated subgroup of $(\mathbb Q/\mathbb Z)^n$ is finite — thus the cocycle lives in some finite, $W$-invariant subgroup $(\mathbb Z/m)^n \subseteq (\mathbb Q / \mathbb Z)^n$ (using that $(\mathbb Z/m)^n \subseteq (\mathbb Q/\mathbb Z)^n$ is the $m$-torsion subgroup and so must be $W$-invariant).

By similar reasoning, we see that every finitely-generated subgroup of $G_\ast$ is finite. Therefore, because homology commutes with filtered colimits, we have $H_\ast(BG_\ast) = \varinjlim_{G' \subseteq G_\ast} H_\ast(BG')$, where the colimit is over finite subgroups (or even just those finite subgroups $G' = G_m$ of the form given above), and we use any constant coefficients.

Now, because $B(\mathbb Q/\mathbb Z)^n \to BU(1)^n$ is a homology isomorphism with any finite coefficients (this can be checked in several ways), we see that $BG_\ast \to BG$ is likewise a homology isomorphism with finite coefficients by the Serre spectral sequence. So the composite map $\varinjlim H_\ast(BG_m) \to H_\ast(BG_\ast) \to H_\ast(BG)$ is an isomorphism with finite coefficients. By the universal coefficient theorem, the map $H^\ast(BG) \to \prod H^\ast(BG_m)$ is injective with finite coefficients. Since Qiaochu has already shown that it is an injection on nontorsion elements, it follows that this map is an injection on integral cohomology (noting that these things are sufficiently finite to be safe).

Note that most of the above boiled down to facts about the extension $U(1)^n \to G \to W$, not really facts about (co)homology — the only thing we really needed was the isomorphism $H_\ast(B(\mathbb Q/\mathbb Z)) \cong H_\ast(BU(1))$ with finite coefficients.

In fact, Theorem 5.7 of Becker and Gottlieb's original paper The transfer map and fiber bundles (DOI) is actually stated for general cohomology theory, and implies by Ben Wieland's argument that $\Sigma^\infty BG$ splits off of $\Sigma^\infty BN(T)$, so the reduction to extensions $T \to N \to W$ with $T$ a torus and $W$ finite holds for an arbitrary homology or cohomology theory $E$.

It's not hard to show that if $E$ is a spectrum with trivial rationalization, then $B(\mathbb Q/\mathbb Z)^n \to BU(1)^n$ is and $E$-homology or cohomology equivalence. So the above argument shows that in this case, we have that $N_\ast \to N$ is an $E$-homology and $E$-cohomology equivalence, where $N_\ast$ fits in the extension $(\mathbb Q/\mathbb Z)^n \to N_\ast \to W$ as above. Moreover, since $BN_\ast = \varinjlim BN'$ where the colimit is over finite subgroups, and since this is a homotopy colimit, we have $E_\ast(BN) = \varinjlim E_\ast(BN')$, so that $\bigoplus E_\ast(BN') \to E_\ast(BN)$ is surjective. For cohomology, there are potential $\varprojlim^1$ issues.

Thus the statement we get is:

Theorem: Let $E$ be a spectrum with trivial rationalization, and let $G$ be a compact Lie group. Then $\bigoplus_{F \subseteq G} E_\ast(BF) \to E_\ast(BG)$ is surjective, where the sum is over finite subgroups $F \subseteq G$.

It would be nice if this could be upgraded to a statement about all spectra by considering also the rationalization, but that seems unpromising because of Maschke's theorem — Qiaochu's argument for nontorsion classes is more subtle, it seems.

Probably also some statement about cohomology is possible ….

In fact, it's not hard to extend the statement to arbitrary $G$-spaces. That is:

Theorem: Let $E$ be a spectrum with trivial rationalization, let $G$ be a compact Lie group, and let $X$ be a $G$-space. Then $\bigoplus_{F \subseteq G} E_\ast(X_{hF}) \to E_\ast(X_{hG})$ is surjective, where the sum is over finite subgroups $F \subseteq G$.

  • $\begingroup$ Thanks Tim! So I'm curious, can you derive McClure's theorem from this argument? $\endgroup$ Sep 29, 2020 at 18:21
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    $\begingroup$ @overcaffeinated That's an interesting question. Certainly some modification would be needed -- I think that if you could show $(K/m)^\ast(BG) \to \prod_F (K/m)^\ast(BF)$ to be injective for each $0 \neq m \in \mathbb Z$, this would imply that $K^\ast(BG) \to \prod_F K^\ast(BF)$ is injective too. This reduces to the case where $E$ has trivial rationalization as above. But I'm not sure how to translate from the filtered-colimit statement to the cohomology-injectivity statement because I don't know how to control the $\varprojlim^1$ term. $\endgroup$
    – Tim Campion
    Sep 29, 2020 at 18:38
  • $\begingroup$ I tried to add links to the comments, but couldn't find Qiaochu's comment (which @BenWieland also references). $\endgroup$
    – LSpice
    Oct 4, 2020 at 15:40
  • $\begingroup$ Thanks! I added a link to Qiaochu's MSE answer. $\endgroup$
    – Tim Campion
    Oct 4, 2020 at 16:06
  • $\begingroup$ Just a note that partly in response to overcaffeinated's comment here, I worked out a second answer which, as noted in a comment on that answer, suffices to recover McClure's result, as well as answering the question independently, without using Qiaochu's MSE answer. $\endgroup$
    – Tim Campion
    Feb 10, 2021 at 17:34

$\DeclareMathOperator\Image{Image}$Here's the most general result I think I can muster. I've split it out into a second answer in order to keep the answer to the original question more self-contained.

Theorem 1: Let $G$ be a compact Lie group, let $X$ be a $G$-space, and let $E$ be a spectrum. Then the following hold, where $F$ ranges over finite subgroups of $G$:

  1. The image of $\bigoplus_F E_\ast(X_{hF}) \to E_\ast(X_{hG})$ contains all of the torsion;

  2. The kernel of $E^\ast(X_{hG}) \to \prod_F E^\ast(X_{hF})$ is contained in the subgroup of divisible elements.

This follows from the following two more precise theorems:

Theorem 2: Let $G$ be a compact Lie group, and let $X$ be a $G$-space. Let $N \subseteq G$ be the normalizer of a maximal torus $T \subseteq G$, and let $W = N / T$ be the Weyl group. Then $\Sigma^\infty_+ X_{hG}$ splits off of $\Sigma^\infty_+ X_{hN}$.

Proof: The splitting is given by the Becker-Gottlieb transfer: the fiber of $X_{hN} \to X_{hG}$ is $G/N$, the same as the fiber of $BN \to BG$, which has Euler characteristic 1.

Theorem 3: Let $N$ be an extension of a finite group $W$ by a torus $T$, and let $E$ be a spectrum and $m \in \mathbb N_{\geq 2}$. Then the following hold, where $F$ ranges over finite subgroups of $N$:

  1. $\varinjlim_F (E/m)_\ast(X_{hF}) \to (E/m)_\ast(X_{hN})$ is an isomorphism;

  2. $(E/m)^\ast(X_{hN}) \to \varprojlim_F (E/m)^\ast(X_{hF})$ is an isomorphism.

Proof of Theorem 1 from Theorems 2 and 3: By Theorem 2, it will suffice to consider the case where $G = N$ is an extension of a finite group by a torus. Theorem 3 establishes Theorem 1 for $E/m$. Then (1) follows by considering the natural short exact sequence $0 \to E_{\ast}(-)/m \to (E/m)_\ast(-) \to E_{\ast-1}(-)^{\text{$m$-tor}} \to 0$, and the argument for (2) uses a similar exact sequence.

The proof of Theorem 3 will follow from a series of lemmas. For the remainder, we let $U(1)^n \to N \to W$ be an extension of a finite group by a torus, and we let $(\mathbb Q/\mathbb Z)^n \to N_\ast \to W$ and $(C_q)^n \to N_q \to W$ be the subextensions which exist by the analysis in my other answer. We fix a spectrum $E$, $m \in \mathbb N_{\geq 2}$, and an $N$-space $X$.

Lemma 4: The fiber of $X_{hN_\ast} \to X_{hN}$ is $B\mathbb Q^n$, and in particular this map is an $(E/m)_\ast$ and $(E/m)^\ast$ equivalence.

Proof: This comes via a diagram chase from the fiber sequence $B\mathbb Q^n \to B(\mathbb Q/\mathbb Z)^n \to BU(1)^n$.

Lemma 5: We have $(E/m)_\ast(X_{hN_\ast}) \cong \varinjlim_q (E/m)_\ast(X_{hN_q})$ canonically, and a canonical short exact sequence $0 \to \varprojlim^1 (E/m)^{\ast+1}(X_{hN_q}) \to (E/m)^\ast(X_{hN_\ast}) \to \varprojlim (E/m)^\ast(X_{hN_q}) \to 0$.

Proof: By the analysis in my other answer, we have $N_\ast = \varinjlim N_q$. Therefore $BN_\ast = \varinjlim BN_q$, and it follows that $X_{hN_\ast} = \varinjlim X_{hN_q}$. The lemma follows by the usual formulas for homology and cohomology of a filtered colimit.

Lemma 6: If $E$ is $m$-torsion, then $\varprojlim^1 E^\ast(X_{hN_q}) = 0$.

Proof of Theorem 3: This follows from Lemmas 4, 5, and 6, once we note that $E/m$ is $m^2$-torsion.

It remains to prove Lemma 6.

Lemma 7: Let $q,r \in \mathbb Z$, and consider the inclusion $C_{qr}^n \xrightarrow i (\mathbb Q/\mathbb Z)^n$. Consider also the inclusion $C_q^n \xrightarrow j C_{qr}^n$ with quotient $C_r^n$. Let $A$ be an $r$-torsion and $q$-torsion abelian group. Then $H^\ast(ij;A)$ is injective and $\Image(H^\ast(ij;A)) = \Image(H^\ast(j;A))$.

Proof: Direct calculation. More precisely, $H^\ast(BC_q;A)$ and $H^\ast(BC_{qr};A)$ both have $A$ in all degrees; the inclusion of $H^\ast(B(\mathbb Q/Z);A)$ is an isomorphism onto the even degrees, and $H^\ast(j;A)$ kills the odd classes while being an isomorphism on even classes. Then one extends this analysis to $n > 1$.

Lemma 8: Let $A \xrightarrow i B \xrightarrow j C$ be maps of chain complexes. Suppose that $ji$ is injective and $\Image(ji) = \Image(i)$. Then the sequence of homologies $H_\ast(A) \xrightarrow{i_\ast} H_\ast(B) \xrightarrow{j_\ast} H_\ast(C)$ has $i_\ast$ injective and $Image(j_\ast i_\ast) = Image(i_\ast)$.

Proof: Diagram chase.

Corollary 9: Fix $s \in \mathbb Z$, and consider the maps $H^s(BN_\ast;A) \xrightarrow i H^s(BN_{qr};A) \xrightarrow j H^s(BN_q;A)$. For $q$, $r$ sufficiently divisible by $m$ and $A$ $m$-torsion, we have that $ji$ is injective and $\Image(ji) = \Image(j)$.

Proof: Using Lemma 7 as a base case, use Lemma 8 to induct through the pages of the Serre spectral sequences for the fibrations over $BW$. This is a first-quadrant spectral sequence, so for fixed $s$ it stabilizes at a finite page. The statement can be tested on associated gradeds, so there are no extension problems.

Corollary 10: Assume that $E$ is bounded below and $m$-torsion, and fix $s \in \mathbb Z$. Consider the maps $E^s(X_{hN_\ast}) \xrightarrow i E^s(X_{hN_{qr}}) \xrightarrow j E^s(X_{hN_q})$. For $q,r$ sufficiently divisible by $m$, we have that $ji$ is injective and $\Image(ji) = \Image(j)$.

Proof: Using Corollary 9 as a base case, use Lemma 8 inductively to walk through the Atiyah–Hirzebruch spectral sequences for the fibrations over $BN_\ast$, $BN_{qr}$, and $BN_q$ respectively (which all have fiber $X$). Since we are assuming that $E$ is bounded below, this is essentially a first-quadrant spectral sequence so the argument goes in the same way as before.

Proof of Lemma 6: That this follows from Corollary 10 in the case where $E$ is bounded below can be seen in two ways — either from the eventual injectivity of $E^\ast(X_{hN_\ast}) \to E^\ast(X_{hN_q})$, or from the fact that sequence is Mittag–Leffler. When $E$ is not bounded below, we simply pass to a suitable connective cover of $E$, since we are always taking the cohomology of a suspension spectrum, which is bounded below.

  • 1
    $\begingroup$ $H^\ast(BG;\mathbb Z)$ is finitely-generated in each degree. In particular, it contains no nonzero divisible elements. So we recover the result of the my first answer, without relying on Qiaochu's answer. Similarly, $ku^\ast(BG)$ is finitely-generated in each degree by the Atiyah-Hirzebruch spectral sequence. For a space $X$, $ku^\ast(X) = KU^\ast(X)$ for $\ast = 0,1$, so $KU^\ast(BG)$ is finitely-generated in each degree. In particular, it contains no nonzero divisible elements, and we recover McClure's result. $\endgroup$
    – Tim Campion
    Oct 4, 2020 at 2:57
  • $\begingroup$ Here's a criterion which covers most of the cases I'd care to think about: Suppose that $E$-cohomology refines to a graded $R$-module-valued functor for some PID $R$ such that $E_\ast$ is finitely-generated over $R$ in each degree. Then $E^\ast(X)$ is finitely-generated over $R$ in each degree if $X$ has finitely many cells in each dimension. If each residue field of $R$ has finite characteristic, then any finitely-generated $R$-module is free of nonzero divisible elements. So in this case, $E^\ast(X_{hG}) \to \prod_F E^\ast(X_{hF})$ is an injection. $\endgroup$
    – Tim Campion
    Oct 7, 2020 at 17:43
  • $\begingroup$ I might be missing something obvious, but can you expand on the second part of lemma 4 ? The identification of the fiber is clear, and the fact that it has trivial $E/m$-co/homology is too; but the point about $E/m$-co/homology equivalence is slightly less so (to me). In case $E/m$ is bounded below I think I can make it work with some Serre-Atiyah-Hirzebruch spectral sequence game, but without that assumption I (think I) lose my nice convergence properties so I'm not sure how that works $\endgroup$ Feb 8, 2021 at 10:13
  • $\begingroup$ I guess for cohomology you can actually get away with it by the same trick as in your last paragraph, but it seems trickier for homology $\endgroup$ Feb 8, 2021 at 10:42
  • $\begingroup$ @MaximeRamzi I don't seem to recall thinking carefully about convergence of any spectral sequences here beyond "first-quadrant" arguments, so I probably overlooked this. As you say, for cohomology it's ultimately safe for us to assume that $E$ is bounded below, so it ends up not being an issue. It's nice to know somebody's reading at least parts of this argument carefully! I fear it gets more and more telegraphic. I should probably do a proper literature search to see if this is worth publishing / figure out some collaboration to do so... $\endgroup$
    – Tim Campion
    Feb 10, 2021 at 5:16

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