It is a well known fact that the matrix groups $GL_n(\Bbb R), SL_n(\Bbb R), \dots$ can be considered as submanifolds of $\Bbb R^{n^2}$.
I did not yet attend a lecture on Lie groups, so I don’t know much more about this viewpoint other than this fact. Still I am wondering whether one can distinguish them by singular (co-)homology.
I could not find a result on the internet computing the simplicial homology of some matrix groups, partly because of noise of the form „computing simplicial homology via matrices“. Of course, if this was asked before I will happily delete it. Moreover please note that this question is not meant in the way of asking you to do the computations for me. I would like to know a reference to a place, where the homology of these groups are computed or see a short argument (need not be accessible to me) why they are equal.
Thank you for your time!
Best Answer
There are actually a few things that can mean the "cohomology of a topological group $G$" in different contexts. It could mean
consider $G$ as a space (forgetting group structure), and take its singular cohomology $H^*(G)$.
consider $G$ as a group (forgetting the topology) and take its group cohomology $H^*_{grp}(G)$
take the singular cohomology of its classifying space $H^*(BG)$.
If $G$ is discrete it's a theorem that the 2) and 3) agree, but if $G$ has a non-trivial topology then usually all three are different. I'm assuming you really mean case 1).
By polar decomposition there are homotopy equivalences $GL_n(\mathbb{R}) \simeq O(n)$ and $SL_n(\mathbb{R})\simeq SO(n)$, so if you're interested in homotopy invariants you can consider these spaces instead. One thing to notice is that, topologically, $O(n)$ is just a disjoint union of two copies of $SO(n)$, so $H_*(O(n)) \cong H_*(SO(n)) \oplus H_*(SO(n))$ (of course as a group $O(n)$ is a non-trivial extension of $SO(n)$ by $\mathbb{Z}/2$). The interesting question is then whether $H_*(SO(n))\cong H_*(SO(m))$ for $n\neq m$, and this is answered in section 7 of "The integral homology and cohomology rings of $SO(n)$ and $Spin(n)$" by Pittie, though I haven't deciphered the result far enough to tell whether they really are all different but it seems so.
If you want an answer to 1) or 3) in terms of cohomology then I recommend Mimura and Toda's "Topology of Lie Groups" as a reference, as they compute the cohomology of $O(n)$ and $SO(n)$ as well as $BO(n)$ and $BSO(n)$, and I forget but they might actually compute the homology as well (most sources I know only compute the cohomology of the classifying spaces).
The cohomology of the classifying spaces is a very different story than just the groups because now the functorial map $B\iota\colon BSO(n) \to BO(n)$ is not an inclusion of a component but a double-cover. The $\mathbb{Z}/2$ cohomologies of $BO(n)$ and $BSO(n)$ are distinguished by the fact that the first universal Steifel-Whitney class vanishes in $H^1(BSO(n);\mathbb{Z}/2)$, and the integral cohomolgies can be distinguished by the fact that there is an integral Steifel-Whitney class $W_2 = \beta(w_1)\in H^2(BO(n);\mathbb{Z})$ but not in the cohomology of $SO(n)$, as well as the existence of an Euler class $e \in H^{n}(BSO(n);\mathbb{Z})$. Moreover $H^*(BO(n)) \ncong H^*(BO(m))$ and $H^*(BSO(n)) \ncong H^*(BSO(m))$ when $n\neq m$. I believe the (singular) cohomologies of $O(n)$ and $SO(n)$ have similar distinctions but I don't have access to Mimura and Toda at the moment to make sure.
As for option 2), I don't know the computations of the group cohomologies $H^*_{grp}(GL_n(\mathbb{R}))$ and $H^*_{grp}(SL_n(\mathbb{R})$, it may be similar to the link that E. KOW posted in the comments about $GL_n(\mathbb{Z})$.