# First homology group of the general linear group

The abelianization of the general linear group $$GL(n,\mathbb{R})$$, defined by $$GL(n,\mathbb{R})^{ab} := GL(n,\mathbb{R})/[GL(n,\mathbb{R}), GL(n,\mathbb{R})],$$ is isomorphic to $$\mathbb{R}^{\times}$$. This follows from the fact that $$[GL(n,\mathbb{R}),GL(n,\mathbb{R})] \cong SL(n,\mathbb{R})$$, so that $$GL(n,\mathbb{R})^{ab} \cong \mathbb{R}^{\times}$$ by the first isomorphism theorem.

Since for any group $$G$$, the homology group $$H_1(G;\mathbb{Z}) \cong G^{ab}$$, it turns out that $$H_1(GL(n,\mathbb{R});\mathbb{Z}) \cong \mathbb{R}^{\times}$$.

I wonder if it possible to obtain this result by directly computing $$H_1(GL(n,\mathbb{R});\mathbb{Z})$$ as the quotient $$Z_1(GL(n,\mathbb{R}))/B_1(GL(n,\mathbb{R}))$$ of 1-cycles by 1-boundaries.

What do $$C_1$$, $$Z_1$$ and $$B_1$$ look like for $$GL(n,\mathbb{R})$$? Is a direct computation of $$H_1(GL(n,\mathbb{R}))$$ possible?

In case this question turns out to be too elementary for MO, I apologize in advance. In that case I will quickly move my post to MSE.

• try to copy the proof of the fact that $H_1$ is the Abelianization in this particular case. Commented May 31, 2019 at 0:33

This more of a comment, but here it goes anyway.

Following Brown's book Cohomology of Groups on page 36, we see that with respect to the standard resolution and any group $$G$$: $$C_2(G)\xrightarrow{\partial}C_1(G)\xrightarrow{0}\mathbb{Z}\to 0,$$ where $$\partial[g|h]=[h]-[gh]+[g]$$.

Consequently, $$Z_1(G)=C_1(G)=\{[g]\ |\ g\in G\}$$ and $$B_1(G)=\{[h]-[gh]+[g]\ |\ h,g\in G\}$$.

So $$H_1(G)\cong G/[G,G]$$. I am not sure what more you could expect in any special case of this.

In the case of a complex reductive (affine algebraic) group $$G$$, the quotient $$G/[G,G]$$ will always be isomorphic to an algebraic torus (isomorphic to a product of $$\mathrm{GL}_1$$'s), but that has nothing to do with group cohomology (it's the central isogeny theorem; see Milne's book Algebraic Groups for example). A special case of this general fact is $$G=\mathrm{GL}_n(\mathbb{C})$$ where $$G/[G,G]\cong \mathbb{C}^*$$. The fact that you get $$\mathbb{R}^*$$ for $$G=\mathrm{GL}_n(\mathbb{R})$$ is analogous to this case (in fact it is true for any field).

• It's true that, algebraically, $G/[G, G]$ is a torus; but does it follow that $G(\mathbb R)/[G(\mathbb R), G(\mathbb R)]$ is always the group of $\mathbb R$-points of a torus? Commented Jun 1, 2019 at 0:06
• @LSpice With respect to $\mathbb{R}$-loci, this is true at least for $GL_n$, which is all I was saying since that was the case of interest of the OP. The more general statement I mentioned, $G/DG\cong T/(T\cap DG),$ can be found in Milne's book on Algebraic Groups (Chapter 22 in the version on his website). Commented Jun 1, 2019 at 0:53
• Indeed, I agree that it's true for $\operatorname{GL}_n$. At the risk of belabouring the point, what I meant to say was that the algebraic-group equality $G/DG \cong T/(T \cap DG)$ that you mention doesn't obviously (or not obviously to me) imply that $G(\mathbb R)/D(G(\mathbb R)) \cong (T/(T \cap DG))(\mathbb R)$, which seemed to be your claim. (Note particularly the denominator $D(G(\mathbb R)) = [G(\mathbb R), G(\mathbb R)]$, as opposed to $(DG)(\mathbb R)$.) Commented Jun 1, 2019 at 0:55
• @LSpice As I mentioned, I do not think I made that claim (I apologize if my use of language is causing confusion). I made the general claim in the cases where the central isogeny theorem holds, for example $\mathbb{C}$, and then said that the reason one gets $\mathbb{R}^*$ in the case of interest of the OP "is a real version of this". Commented Jun 1, 2019 at 1:03
• Anyway, I will edit my answer to make this more clear. Commented Jun 1, 2019 at 1:19