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Here's a question out of idle curiosity. Let $G$ be a topological group. Is it possible for both $G$ and (a model of) $BG$ to be finite CW complexes? (Apart from the obvious example of $G$ being [up to homotopy] the trivial group.)

A comment is that the fibration sequence $G \to EG \to BG$ shows that $\chi(G)\chi(BG) = 1$ in this case, so $\chi(G) = \pm 1$. This rules out plenty of examples, e.g. Lie groups, finite groups...

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    $\begingroup$ My guess is that this not possible. In fact I think if you look at the Serre spectral sequence for the fibration you mention you can see that if the cohomology of G is bounded, then the cohomology of BG must be unbounded (hence it can't be finite complex). $\endgroup$ – Chris Schommer-Pries Oct 2 '13 at 21:34
  • $\begingroup$ Can a topological group be a finite CW complex and not a Lie group? $\endgroup$ – David E Speyer Oct 2 '13 at 22:47
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    $\begingroup$ Yes, Chris. First reduce to the case when $G$ is connected, replacing $BG$ by its (finite) universal covering space. Now if $BG$ is not contractible then its top homology (with coefficients in a suitable field) is in some positive dimension $d$. The top homology of $G$ is in some positive dimension $e$. The spectral sequence shows that the contractible space $EG$ has nontrivial $H_{d+e}$, contradiction. $\endgroup$ – Tom Goodwillie Oct 3 '13 at 0:20
  • $\begingroup$ @DavidSpeyer it depends on what you mean by "be" ;) See answers below. $\endgroup$ – Jesper Grodal Oct 11 '13 at 5:59
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To expand my comment: No it is not possible.

First suppose that $G$ is connected and that both $G$ and $BG$ have the homotopy types of finite complexes. If $G$ is not contractible, then let $k>0$ be minimal such that $\pi_k(G)$ is nontrivial. We have $H_k(G)=\pi_k(G)=\pi_{k+1}(BG)=H_{k+1}(BG)$. Choose a prime $p$ such that the latter finitely generated abelian group is nontrivial mod $p$. From now on let homology be with mod $p$ coefficients. Let $d$ be maximal such that $H_d(BG)$ is nontrivial. Let $e$ be maximal such that $H_e(G)$ is nontrivial. The spectral sequence of the fibration has a nontrivial group at $E^2_{d,e}$ that will not go away. Contradiction. So $G$ connected implies $G$ contractible.

If $G$ is not connected then $\pi_0G$ is finite. Let $G_0$ be the component of the identity in $G$. Then $BG_0$, the universal cover of $BG$, is also a finite complex, so by the previous paragraph $G_0$ is contractible. That is, $G$ is (up to homotopy) discrete and $BG$ is the classifying space of a finite group. For any cyclic subgroup $C$ of the finite group $G$, $BC$ is a finite cover of $BG$ and therefore finite. The cohomology of $BC$ is such that it cannot be equivalent to a finite complex unless $C$ is trivial. So $G$ is trivial (contractible).

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    $\begingroup$ Incidentally, there is a very interesting theory of homotopy types that both contain groups and compact manifolds, but not Lie groups. The smallest one that's not even rationally a Lie group has dimension >1100. $\endgroup$ – Allen Knutson Oct 3 '13 at 4:51
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A topological group is homogenous. In a finite CW complex, the interior of every top dimensional cell is a topological manifold. Therefore, a finite CW complex which is a toplogical group is a topological manifold. By Hilbert's fifth problem,a topological group which is a topological manifold is a Lie group (not necessarily connected).

Finite CW complexes are also compact, so your group is a compact Lie group.

Positive dimensional Lie groups have torus subgroups, hence have $\chi(G)=0$, so $BG$ is not a finite CW complex as you say.

Zero dimensional Lie groups have $\chi(G) = |G|$, so $BG$ is only a finite CW complex if $|G|=1$.

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    $\begingroup$ WHat is useful is that the interior of every top dimensional cell is a topological manifold, no? To conclude that $%\chi(G)=0$ it is simplest to use the fact that $G$ is paralellizable, maybe. $\endgroup$ – Mariano Suárez-Álvarez Oct 2 '13 at 23:03
  • $\begingroup$ @MarianoSuárez-Alvarez Thanks, that's what I meant to write. $\endgroup$ – David E Speyer Oct 2 '13 at 23:05
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    $\begingroup$ The limitation of this answer (contrary to Tom's) is that it takes a strict definition of "is" in "is a finite CW complex". Namely you take it to mean "homeomorphic to" rather than "homotopy equivalent to". A finite loop space is a space homotopy equivalent to both a topological group and a finite CW complex. These need not in general be homotopy equivalent to compact Lie groups -- see e.g., section 3 of my survey paper math.ku.dk/~jg/papers/icm.pdf for a summary. $\endgroup$ – Jesper Grodal Oct 9 '13 at 18:59
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As stated in previous answers, the answer to the question is no. Here is another viewpoint, which gives additional information about the relationship between $G$ and $BG$:

A pair $(G,BG)$ where $G$ is a topological group homotopy equivalent to a finite CW complex, and $BG$ its classifying space, is called a finite loop space in the literature. So the question is if a finite loop space $G$ can have classifying space $BG$ homotopy equivalent to a finite CW complex, and the answer to this is no.

As pointed out in the previous answers, to get that $BG$ does not have a finite dimensional model, it is easy to reduce to the case where $G$ is connected (using that $BG$ is not finite for $G$ a finite group), so let's assume this.

Now, by the structure of Hopf algebras (Milnor-Moore) $H^*(G;{\mathbb Q})$ is an exterior algebra on a number $r$ of odd dimensional generators. The number $r$ is called the (rational) rank, and agrees with the usual notion of rank of compact Lie groups. Hence by a spectral sequence argument

$H^*(BG;{\mathbb Q})$ is a polynomial algebra on $r$ generators.

So the non-finiteness of $BG$ is a corollary of the following well-known fact:

Fact: A non-contractible connected finite loop space (or even H-space) has positive rank $r$.

Proof of Fact: The claim e.g., follows from a more general statement saying the rank is also equal to the number of odd degree generators for the mod $p$ cohomology (see Kane: Homology of Hopf Spaces Section 13-3, which uses the Bockstein spectral sequence to deduce this). To get the more limited statement of the rank being positive one can give a more pedestrian argument: Suppose that $G$ is non-contractible. Then (since its a simple space and a CW complex) $\tilde H^*(G;{\mathbb Z}) \neq 0$. If this cohomology is torsion free, then the rational cohomology is non-trivial, and we are done. So suppose there is torsion. If there is non-trivial $p$-torsion in $\tilde H^*(G;{\mathbb Z})$ then, by the universal coefficient theorem, $H^*(G;{\mathbb F_p})$ has cohomology in two consecutive degrees. So $H^*(G;{\mathbb F_p})$ has to have generators as a ring in odd degrees. But then the structure of Hopf algebras algebras shows that the Euler characteristic of $G$ has to be zero (since $H^*(G;{\mathbb F_p})$ contains an exterior tensor summand on a odd degree class, if $p$ is odd, and a truncated polynomial algebra on an odd degree class, truncated at $x^{2^k}$ for some $k$, if $p=2$). But $\chi(G) =0$ implies $\tilde H^*(G;{\mathbb Q}) \neq 0$ as wanted.

Additional info: There has been a lot of work on finite loop spaces, starting with the work of Hopf and Serre in the 1940's and 1950's, since they are generalizations of compact Lie groups. (Hopf looked at the weaker notion of H-spaces.) There is by now in fact a classification of (connected) finite loop spaces; see Section 3 of my survey paper http://www.math.ku.dk/~jg/papers/icm.pdf

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The answer is always no unless $G$ is trivial. In fact, I can generalize your statement slightly: I only need to assume that $G$ and $BG$ have merely the homotopy type of finite complexes.

I will outline the argument below for the case when $G$ is connected. (I know how to extend the argument I give below to disconnected $G$, although it is not published). My argument is purely homotopy theoretic.

In my paper, "The dualizing spectrum of a topological group" (here's a link: http://www.math.wayne.edu/~klein/quinn.pdf) I introduced the dualizing spectrum $D_G$ of a topological group. This is given by the homotopy fixed points of the left translation action of $G$ on the suspension spectrum of $G_+ = G \amalg \text{pt}$. That is, $$ D_G := \text{maps}_G(EG,\Sigma^\infty G_+) \, . $$ I then showed that if $$ F \to E \to B $$ is a fibration with $F,E,B$ homotopy finite, connected and based, then $$ D_{\Omega E } \simeq D_{\Omega B } \wedge D_{\Omega F } $$ where $\Omega E$ is a suitable topological group model for the loop space of $E$.

Let's consider the case of a connected topological group $G$ such that both $G$ and $BG$ are homotopy finite (finitely dominated is actually good enough here).

Then we have the universal bundle $$ G \to EG \to BG $$ and we get $$ S^0 \simeq D_{\Omega G} \wedge D_G $$ (here I'm using the fact that $D_{\Omega EG}$ is the sphere spectrum up to homotopy--this is not hard to check).

From the last equation, it's relatively straightforward to check that both $D_G$ and $D_{\Omega G}$ are sphere spectra up to homotopy. Furthermore, if $D_G \simeq S^{-d}$ it must be the case that $D_{\Omega G} \simeq S^d$.

But another theorem in my paper shows that whenever $G$ is a topological group with $BG$ finitely dominated, then $BG$ is a Poincaré space of dimension $d$ if and only if $D_G$ is weak equivalent to $S^{-d}$.

It follows from this that $G$ is a Poincaré space of dimension $d$ and $BG$ is a Poincar\'e space of dimension $-d$.

The only possibility then is that $G$ and $BG$ are Poincaré spaces of dimension $0$. But since we are assuming that $G$ is connected, it follows that $G$ has the homotopy type of a point.

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