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I'm interested in a proof (or counter-example) of the following:

Let $G$ be a topological group. If $G$ contains torsion then $H^n(BG)\neq 0$ for infinitely many $n$.

I know this is true for discrete groups, but I'm wondering if an analogous argument can be made for groups with a topology.

In the simplest case $G=C_k$ is cyclic of order $k<\infty$, and one computes $H^*(BC_k)\cong \mathbb{Z}[u]/(ku)$ for a class $u$ of degree $2$. The approach I know is to compute the Serre spectral sequence for the universal bundle $S^1 \to S^\infty \to \mathbb{C}P^\infty$, and notice that if we include $C_k \to S^1$ as the $k$-th roots of unity then $BC_k \simeq S^\infty/C_k$ and we get a fibre bundle $S^1/C_k \to BC_k \to \mathbb{C}P^\infty$ along with a map of fibrations from the universal bundle which is a map of degree $k$ in the fibres; now study the induced map on spectral sequences.

Now if $G$ is an arbitrary group with $k$-torsion, it has a subgroup isomorphic to $C_k$. If $G$ is discrete we can complete the argument using group cohomology $H^*_{Grp}(G)\cong H^*(BG)$, since in this context it's easy to see that $cd(G) \geq cd(C_k)=\infty$. However if $G$ is not discrete then group cohomology is more subtle (see this question), because the usual definition computes $H^*(K(G,1))$ instead of $H^*(BG)$. Is there an analogous argument using an appropriate definition of $H^*_{TopGrp}(G)$?

Alternatively, is there an argument that avoids group cohomology altogether, and instead uses geometry (for Lie groups) or algebraic topology? I tried thinking about how the inclusion $C_k \to G$ induces a map on spectral sequences (like in the case $G=S^1$), but if $G$ is not connected then the coefficients are potentially twisted, and in either case there are too many unknowns for me to see what to do.

William
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1 Answers1

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There's an argument involving the Euler characteristic that may or may not apply in your situation. Namely, there's a fibration $$G \to EG \to BG$$ with $EG$ contractible. So if $H^* BG$ is zero in all but finitely many degrees and thus $\chi(BG)$ is a finite number, we would have $$\chi(BG) = \frac{1}{\chi(G)}.$$ This would only make sense if $\chi(G) = \pm 1$. However, for example if $G$ is a connected compact Lie group, then $\chi(G) = 0$ and $\chi(BG)$ wouldn't make sense.

JHF
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  • I think this argument also applies to connected compact topological groups, since you can prove $\chi(G)=0$ using Lefschetz fixed-point theorem (as in this question), and by extension any topological group with compact components since $\chi$ is additive for disjoint unions. Moreover I think it also applies to an arbitrary Lie group because the identity component is homotopy equivalent to its maximal compact subgroup. Does that sound right? – William Apr 11 '19 at 00:26
  • Sounds right to me. – JHF Apr 11 '19 at 14:47