Why is BG infinite dimensional for G finite ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T02:39:41Z http://mathoverflow.net/feeds/question/79741 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79741/why-is-bg-infinite-dimensional-for-g-finite Why is BG infinite dimensional for G finite ? TJ 2011-11-01T18:18:23Z 2011-11-17T05:54:16Z <p>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: </p> <p><a href="http://mathoverflow.net/questions/64688/non-vanishing-of-group-cohomology-in-sufficiently-high-degree/64702#64702" rel="nofollow">http://mathoverflow.net/questions/64688/non-vanishing-of-group-cohomology-in-sufficiently-high-degree/64702#64702</a></p> <p>As a consequence, the CW-complex $BG$ (unique up to homotopy) can not be of finite dimension. </p> <p>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.</p> http://mathoverflow.net/questions/79741/why-is-bg-infinite-dimensional-for-g-finite/79749#79749 Answer by Agol for Why is BG infinite dimensional for G finite ? Agol 2011-11-01T19:10:46Z 2011-11-01T19:10:46Z <p>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). </p> http://mathoverflow.net/questions/79741/why-is-bg-infinite-dimensional-for-g-finite/79755#79755 Answer by Tom Goodwillie for Why is BG infinite dimensional for G finite ? Tom Goodwillie 2011-11-01T19:35:23Z 2011-11-17T05:54:16Z <p>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 ...</p> <p>EDIT By the way, this also yields the more general statement that $BG$ cannot be finite-dimensional unless $G$ is torsion-free.</p> <p>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.</p> http://mathoverflow.net/questions/79741/why-is-bg-infinite-dimensional-for-g-finite/79780#79780 Answer by unknown (google) for Why is BG infinite dimensional for G finite ? unknown (google) 2011-11-01T22:42:34Z 2011-11-01T22:42:34Z <p>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.</p>