What is $H_1(\operatorname{GL}_2(\mathbb{R}))$?

Question

I am trying to compute the cohomology groups of $\operatorname{GL}_2(\mathbb{R})$. I know that $\operatorname{GL}_2(\mathbb{R})$ has two connected components, so $H_0(\operatorname{GL}_2(\mathbb{R})) = \mathbb{R}^2$. Also, $\operatorname{GL}_2(\mathbb{R})$ is homotopically equivalent $O(2)$, the group of orthogonal matrices. Since $O(2)$ is 1 dimensional, this tells me that $H_k(\operatorname{GL}_2(\mathbb{R})) = 0$ for $k > 1$. But how do I compute $H_1(\operatorname{GL}_2(\mathbb{R})) = H_1(O(2))$?

Answer 1

First use the fact that the connected components of a Lie group are all diffeomorphic: if $G_0$ is the connected component of the identity and $G_1$ is another connected component, then for any $g \in G_1$, $L_g|_{G_0} : G_0 \to G_1$ is a diffeomorphism. In particular, the connected component of $O(2)$ which doesn't contain the identity is diffeomorphic to the one that does, namely $SO(2)$. Now recall that $SO(2)$ is diffeomorphic to $S^1$, so $O(2)$ is diffeomorphic to $S^1\sqcup S^1$ and hence

$$H_1(O(2)) \cong H_1(S^1\sqcup S^1) \cong H_1(S^1)\oplus H_1(S^1) \cong \mathbb{R}^2.$$

In fact, once you know that $O(2)$ is a closed one-dimensional manifold with two connected components, it must be diffeomorphic to $S^1\sqcup S^1$ by the classification of one-dimensional manifolds.