# Classifying Spaces of Matrix Lie Groups

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 $$K(SL(n,\mathbb{Z}),1)$$?

• $SL(n, \mathbb{R})$ has $SO(n)$ as a maximal compact subgroup, so any model for $BSO(n)$ is a model for $BSL(n, \mathbb{R})$. Commented Apr 10, 2023 at 14:26
• Great, thank you! Is $BSL(n,\mathbb{Z})$ less understood? Commented Apr 10, 2023 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. Commented Apr 10, 2023 at 15:11
• I see. You’re saying that we could look for the Eilenberg-MacLane space $K(SL(n,\mathbb{Z}),1)$. Commented Apr 10, 2023 at 15:46
• Yes, exactly. If $G$ is a discrete topological group, then $BG = K(G, 1)$. Commented Apr 10, 2023 at 16:22

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.
• Thank you for the response! I suppose that this shifts the question to understanding the classifying space of $SL(n,\mathbb{Z}/3)$. Commented Apr 13, 2023 at 1:43
• 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! Commented Apr 13, 2023 at 14:29