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I’m currently studying the classifying spaces of some of the matrix Lie groups. I’ve come across a post here that describes the classifying spaces for $SO(n)$, $SU(n)$, $GL(n)$, and $Sp(n)$. I’ve also seen computations for these in other places.

I'm interested in the classifying space for $SL(n,\mathbb{Z})$. As discussed in the comments, this is the same as the Eilenberg-MacLane space $K(SL(n,\mathbb{Z}),1)$. There is a general construction of these spaces given in Hatcher's algebraic topology. However, I can't determine from this if $K(SL(n,\mathbb{Z}),1)$ is some common space. Is there a different method that determines $K(SL(n,\mathbb{Z}),1)$?

slowspider
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    $SL(n, \mathbb{R})$ has $SO(n)$ as a maximal compact subgroup, so any model for $BSO(n)$ is a model for $BSL(n, \mathbb{R})$. – Michael Albanese Apr 10 '23 at 14:26
  • Great, thank you! Is $BSL(n,\mathbb{Z})$ less understood? – slowspider Apr 10 '23 at 14:42
  • $SL(n, \mathbb{Z})$ is a discrete group, so a connected topological space $X$ is a model for $BSL(n, \mathbb{Z})$ if $\pi_1(X) = SL(n, \mathbb{Z})$ and all higher homotopy groups vanish. I don't know of any explicit examples of such spaces. – Michael Albanese Apr 10 '23 at 15:11
  • I see. You’re saying that we could look for the Eilenberg-MacLane space $K(SL(n,\mathbb{Z}),1)$. – slowspider Apr 10 '23 at 15:46
  • Yes, exactly. If $G$ is a discrete topological group, then $BG = K(G, 1)$. – Michael Albanese Apr 10 '23 at 16:22
  • It may be possible, but I don't see how. – Michael Albanese Apr 11 '23 at 11:08
  • It is "almost" the quotient $SL(n,{\mathbb Z})\backslash SL(n,{\mathbb R})/SO(n)$, except the action of $SL(n,{\mathbb Z})$ on the symmetric space $SL(n,{\mathbb R})/SO(n)$ is not free. But there is an easy way to fix this using, for instance, the fact that $SL(n,{\mathbb Z})$ contains a finite index torsion-free subgroup. – Moishe Kohan Apr 12 '23 at 14:45
  • @MoisheKohan Could you expand on how we can use that $SL(n,\mathbb{Z})$ has a finite index torsion-free subgroup to resolve the issue that the action is not free? – slowspider Apr 12 '23 at 15:37
  • Do you know what Borel construction is? – Moishe Kohan Apr 12 '23 at 15:56
  • I know a little bit about it. In this case, it seems like you are wanting to use the fact that if X is a topological space that G acts freely upon, then $(X\times EG)/G$ is homotopic to $X/G$, but I can’t see how to use that here. – slowspider Apr 12 '23 at 16:15
  • You mention that you cannot determine "if $K(SL(n,\mathbb Z),1)$ is some common space". That all depends on what is "common". In homotopy theory there is a general, very well understood, "common" construction of the Eilenberg-Maclane space $K(G,1)$ associated to any discrete group. In a few very special cases, $K(G,1)$ is homotopy equivalent to a space that was "common" before the emergence of the work of Eilenberg-and Maclane: $K(\mathbb Z,1)$ is homotopy equivalent to the circle; $K(\mathbb Z^n,1)$ is homotopy equivalent to the $n$-dimensional torus $(S^1)^n$; and a few others. – Lee Mosher Apr 12 '23 at 19:30
  • @LeeMosher I was hoping that $K(SL(n,\mathbb{Z}),1)$ would be homotopic to a space that was "common" before Eilenberg and MacLane, just like $K(\mathbb{Z}^n,1)$ is homotopic to $(S^1)^n$. What you mentioned is a good point though, not having such a homotopy is what makes the Eilenberg-MacLane construction useful. – slowspider Apr 12 '23 at 20:26

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Here is an answer of sorts. The quotient $X=SL(n,{\mathbb R})/SO(n)$ is a symmetric space of noncompact type, hence, a contractible manifold. The group $\Gamma=SL(n,{\mathbb Z})$ acts on $X$ properly discontinuously but not freely (via the left multiplication of cosets): Stabilizers of points are finite. The group $\Gamma$ contains a torsion-free subgroup $\Gamma_0$ of finite index, the kernel of the natural projection $$ \Gamma\to SL(n,{\mathbb Z}/3). $$ (I think, this is due to I.Schur.) Hence, $\Gamma_0\backslash X$ is a classifying space for $\Gamma_0$. To get one for $\Gamma$ itself one can do the following. Take the classifying space for the finite group $SL(n,{\mathbb Z}/3)$, it is the quotient $SL(n,{\mathbb Z}/3)\backslash Y$, where $Y$ is a contractible CW complex. (This should be treated as a black box.) You get a free proper action of $\Gamma$ on $X\times Y$ where the group $\Gamma$ acts on $X$ as above and acts on $Y$ via the homomorphism $\Gamma\to SL(n,{\mathbb Z}/3)$. Then the quotient $$ \Gamma\backslash (X\times Y) $$ is a classifying space for $\Gamma$. What is it good for, I do not know.

Moishe Kohan
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  • Thank you for the response! I suppose that this shifts the question to understanding the classifying space of $SL(n,\mathbb{Z}/3)$. – slowspider Apr 13 '23 at 01:43
  • @slowspider: The real question is: what would you like to know about this classifying space? There are explicit constructions (especially if the group is finite), what else do you need? Compute cohomology groups? What else? – Moishe Kohan Apr 13 '23 at 03:07
  • Like my response to Lee said, I was wondering if $K(SL(n,\mathbb{Z}),1)$ is homotopic to a space that was "common" before Eilenberg and MacLane, just like $K(\mathbb{Z}_n,1)$ is homotopic to $(S^1)^n$. That doesn't appear to be the case, so what you have is great! – slowspider Apr 13 '23 at 14:29