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Let $G$ be a nonabelian group, with classifying space $BG$.
Motivation: We can compute its homology, $H_\ast(BG)=H_\ast(G)$. It would be nice to see some equivariant computations, like $H_\ast^G(BG)$ where $G$ acts by conjugation, but I need a particular model of $BG$ to work with. In any case, if $G$ acted freely it would reduce to computing the homology of $BG/G$. Although this action won't be free, I still wonder what $BG/G$ looks like.

Preferred Explicit model: Let $EG$ be the weakly contractible space (constructed simplicially using the elements of $G$), and consider the action of $G\times G$ on $G$ by $(g_1,g_2)\cdot g=g_1gg_2^{-1}$. Then we recover the classifying space $BG$ as $EG/G$ (with $G=G\times\lbrace 1\rbrace$), and we get the space $BG/G$ as $EG/(G\times G)$.
[[Edit]: As pointed out, the outcome will depend on the model. If this isn't a "good" model, then I'll settle for a better one! (Although I would want to understand this model).

So I have this space $BG/G$, dividing out the classifying space by the conjugation $G$-action. Here is where I get some big help: By the Kan-Thurston theorem, there exists a group $K$ such that $BG/G$ and $BK$ have the same homologies.

What can $K$ be? (Note that if $G$ were abelian then we'd trivially have $K=G$).
Is there a deeper connection between these two spaces?

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I think your claim about the equivariant homology is wrong. Even a trivial group action does not preserve equivariant homology. Instead, $EG/G\times G$ is equivariantly homotopic to $BG\times G$, simply because it is a quotient of a contractible space by a $G\times G$ action, so $K=G\times G$. –  Will Sawin Jan 25 '13 at 3:40
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I'm not sure this a very natural thing to consider since $BG/G$ can depend on which model you are using for $EG$. The problem is that $G$ doesn't act freely on $BG$. In other words, the construction is not homotopy invariant. Furthermore, your action is homotopically trivial in the sense that there is a space $Z$, which we consider as a trivial $G$-space and a finite chain of $G$-equivariant weak homotopy equivalences to $BG$. –  John Klein Jan 25 '13 at 3:41
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I don't think the Kan-Thurston theorem will help in understanding $BG/G$. Note that the Kan-Thurston theorem applies to any connected space: any such $X$ is a $BK$ from a homology point-of-view. Furthermore, your $K$ is not unique, since we can taking the cartesian product of it with any homologically trivial $BH$. –  John Klein Jan 25 '13 at 3:51
    
@Will: I didn't make any claim about equivariant homology -- it was just a motivator, appearing maybe ambiguous: if the action is free then $H_\ast^G(X)=H_\ast(X/G)$, but here our action isn't free and just makes me wonder what $BG/G$ is anyway. For your latter statement, I am not familiar with equivariant-homotopy and so am probably missing something, but $G\times G$ doesn't act freely on $EG$, and moreover if $G$ were abelian then you would be saying $BG/G=BG$ and $B(G\times G)$ have the same homologies? –  Chris Gerig Jan 25 '13 at 4:00
    
@John: But I specified the exact model to use here. And although $K$ isn't unique I would at least want some example of it (I would guess that it is closely related to $G$). I did not understand the statement involving this space Z though. –  Chris Gerig Jan 25 '13 at 6:51
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3 Answers

up vote 12 down vote accepted

What you should take as a model is the homotopy quotient $EG \times_G BG$. From the homotopy sequence of the fibration $EG \times_G BG \to BG$ (projection on first factor), you get that $EG \times_G BG$ is aspherical and a short exact sequence

$$ 1 \to \pi_1 (BG) \to \pi_1 (EG \times_G BG) \to \pi_1 (BG) \to 1. $$

Since the action of $G$ on $BG$ has a fixed point, this sequence (which is completely natural in $G$) is split. The induced action of the base $G$ on the fibre $G$ is by conjugation. So: $EG \times_G BG \cong BK$; $K = G \ltimes_{ad} G$, with the conjugation action.

EDIT: Tom Goodwillie pointed out that $G \ltimes_{ad} G \cong G \times G$.

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@Johannes: I think Chris wants to use a specific model for the quotient $BG/G$; your model is not the one he seems to be interested in. –  John Klein Jan 25 '13 at 12:37
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The semidirect product for the conjugation action is isomorphic to $G\times G$. –  Tom Goodwillie Jan 25 '13 at 13:40
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@Johannes: It seems to me the action of $G$ on $BG$, induced by conjugation, is trivial up to homotopy, so $EG \times_G BG$ is weak equivalent to $BG \times BG$. –  John Klein Jan 25 '13 at 14:25
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The space $BG/G$, with the model you describe, has been studied, but is not completely understood. It's $\pi_1$ identifies with the abelianization, but there may be higher homotopy.

It identifies with a hocolim over the Quillen category with objects all subgroups and morphisms given with conjugation and inclusion.

I'm not sure the group $K$ will have a very meaningful description (but I hope I'm wrong).

The question you ask actually came up earlier on the Don Davis algtop list, see

http://www.lehigh.edu/~dmd1/post04.html

and look under "Question about G and BG", and the responses. To my regret I can't find the notes by Bill Dwyer referred to here, though they have been in my possession.

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Dumb question: Why does $\pi_1(BG/G)=G_{ab}$? With this being true, and $H_1(BK)=H_1(BG/G)$, we have $K_{ab}=G_{ab}$, so $K$ does look connected to $G$ (perhaps $K=G_{ab}$). –  Chris Gerig Jan 27 '13 at 18:56
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I thought about the case when $G$ is a free group. I believe that in this case $H_2(BG/G)$ is the second exterior power of $G^{ab}$ (which is also $H_2(BG^{ab})$), and $H_3(BG/G)$ is the kernel of $$ \oplus (C\otimes C)\to G^{ab}\otimes G^{ab},$$

where $C$ ranges over representatives of conjugacy classes of maximal cyclic subgroups of $G$ (the cokernel of this same map is yet another description of $H_2$), and $H_n(BG/G)$ is trivial for all $n>3$.

EDIT A useful observation is that if $H$ is a subgroup of $G$ then the fixed point set $(BG)^H$ is $B(Z_GH)$, where $Z_GH$ is the centralizer of $H$ in $G$.

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