# Why is BG infinite dimensional for G finite ?

If $G \neq \lbrace 1 \rbrace$ is a finite group with classifying space $BG$ then there are infinitely many i such that $H^i(BG,\mathbb{Z}) \neq 0$. This can be found, for example, there:

Non-vanishing of group cohomology in sufficiently high degree

As a consequence, the CW-complex $BG$ (unique up to homotopy) can not be of finite dimension.

Question: Are there alternative proofs for this observation. In particular, I would be interested in knowing if there is a purely topological proof without homological algebra.

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Well, there's the more general statement that any space with bounded homotopy groups is homologically infinite-dimensional. –  Aaron Mazel-Gee Nov 1 '11 at 18:23
I rather like Johannes Ebert's answer to the linked question. It is purely topological (using the theory of characteristic classes). –  Mark Grant Nov 1 '11 at 18:47
The homological proof seems to be the easiest... But maybe this would work: First show this for BA, where A is cyclic (you could use an explicit model for BA to do this...). Then show that the inclusion of any subgroup H in G gives rise to a map of rings H^*G--->H^*H that displays the latter as a finitely generated module over the former. (To prove this, look hard at the fibration U(n)/G ---> BG--->BU(n) arising from a faithful unitary representation of G, say via G---> S_n ---> U(n)). Then combine these two facts! –  Dylan Wilson Nov 1 '11 at 18:59
(BTW the above is from notes for a course being taught by Steve Mitchell, you should check this out for lots more fun group cohomology stuff: math.washington.edu/~mitchell/Quillen/quillen.html) –  Dylan Wilson Nov 1 '11 at 19:01
(Also the reason I didn't post this as an answer is because, while "topological", it certainly does not avoid homological algebra!! Spectral sequences make an appearance...) –  Dylan Wilson Nov 1 '11 at 19:03

## 3 Answers

I believe there's an argument using Euler characteristic. Let $G$ be a finite group, $BG=K(G,1)$ the classifying space, and $EG=\widetilde{BG}$ the universal cover, which is contractible. Then $\chi(EG)=1$. Now, if $H_{\ast}(G)=H_{\ast}(BG)$ were finite, then $\chi(BG)$ would be an integer (use whatever field coefficients you prefer). But by the multiplicativity of Euler characteristic, then $\chi(BG)|G|=\chi(EG)=1$, so $\chi(BG)=1/|G|$, a contradiction. I forget who this argument is attributed to. Also, one may see geometrically that any finite group has a classifying space with finitely many cells in each dimension, so if $H_{\ast}(BG)$ is infinite, it must be non-vanishing in infinitely many dimensions (i.e. not infinite rank in a single dimension).

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This is also the core of Akita's lovely argument that the Torelli group has infinite-dimensional cohomology groups. –  JSE Nov 1 '11 at 19:20
This simple idea is even nicer to me if we change the question: that is, it gives a proof that "a finite-dimensional contractible space can't have a free action of G". In other words, a fixed-point theorem. –  Elizabeth S. Q. Goodman Nov 3 '11 at 3:57
Nice, but the first argument only shows that $BG$ cannot be finite. The question was, if it can be finite-dimensional. I think one cannot prove that it cannot be finite-dimensional without actually computing the cohomology of cyclic groups. –  Andreas Thom Jan 10 '12 at 6:42
What result for multiplicativity of the Euler characteristic are you using? I know of two neither of which seems to apply here. Both require homology of the base, fibre and total space to be finite dimensional (which is the case here); one additionally assumes the base is a finite CW-complex, the other result assumes the action of the fundamental group of the base on the homology of the fibre is trivial. Both of those results can be found in Serre's Homologie Singulière Des Espaces Fibrés. –  Omar Antolín-Camarena Sep 5 '13 at 22:12
Omar, yes, this result only applies when the complex is the base is a finite CW-complex, as pointed out by Andreas Thom. –  Ian Agol Sep 5 '13 at 22:49

For every subgroup $H\subset G$, $BH$ occurs as a covering space of $BG$. If $BG$ were finite-dimensional then every covering space would be finite-dimensional. But for $C_p$ cyclic of prime order $p$ the space $BC_p$ has nontrivial mod $p$ cohomology in infinitely many (in fact all) dimensions. This can be made pretty geometric: there is a nice cell structure with one cell in every dimension and manifolds (lens spaces) as the odd-dimensional skeleta ...

EDIT By the way, this also yields the more general statement that $BG$ cannot be finite-dimensional unless $G$ is torsion-free.

EDIT In response to a comment here are some details: Make $C_p$ act on $S^{2n-1}$, the unit sphere in $\mathbb C^n$, freely by $p$th roots of $1$. This sphere is $(2n-2)$-connected and the union as $n\to\infty$ is contractible, so the orbit space is a model for $BC_p$. One can describe a cell structure in $S^{2n-1}$ with $p$ cells in every dimension up to $2n-1$ yielding a cell structure on the orbit space with one cell in every dimension up to $2n-1$, so that $BC_p$ gets one cell in every dimension. You can work out the boundary maps and see that the mod $p$ cohomology is nontrivial in all dimensions. Or you can save some trouble by using Poincare duality, since these odd-numbered skeleta are manifolds.

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Hi, Tom, I did not see why a space have cells in every dimension than it's homology or cohomology has to be nontrivial in infinite many dimensions. Sort of why geometric dimension are the same(?) as cohomology or homology dimension. –  Xiaolei Wu Nov 17 '11 at 5:04

A proof based on fixed point theory: If $BG=EG/G$ is finite-dimensional, then $EG$ is as well and has the homology of a point. Choose a non-trivial Sylow subgroup $P$ of $G$. By a well-known theorem of P.A. Smith, the fixed point set $EG^P$ is non-empty, contradicting the free action of $G$ (and hence $P$) on $EG$. Consequently $BG$ must be of infinite dimension.

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