Every compact Lie subgroup of $GL(n, \mathbb{R})$ is conjugate into $O(n)$

Question

I've seen the theorem that every compact Lie subgroup of $GL(n, \mathbb{R})$ is conjugate into $O(n)$, however I'm struggling with a part of the proof. The idea of the proof is to use the Haar measure, compare for example the answer of this question: Compact Lie subgroup of $GL_n(\mathbb{R})$.

So, let $K < GL(n, \mathbb{R})$ be a compact Lie subgroup with Haar measure $\mu$, and denote $\left< \cdot, \cdot\right>$ any scalar product on $\mathbb{R}^n$. Then define for $v, w \in \mathbb{R}^n$: $$B(v, w) = \int_{K}\left< gv,gw \right> d\mu(g)$$ What I'm struggling with is to see how $B$ is $K$-invariant.

For any $k \in K$ we have $$B(kv, kw) = \int_{K}\left< gkv,gkw \right> d\mu(g).$$ I know that the Haar measure is left invariant. Now, if $\mu$ were also right invariant, I could simply replace $d\mu(g)$ with $d\mu(gk)$ and we'd be done. But how do I use left-invariance only?

Answer 1

Things get easier if you use that natural way to act on bilinear forms. This means that in defining $B(v,w)$ you should integrate $\langle g^{-1}\cdot v,g^{-1}\cdot w\rangle d\mu(g)$. Then you should compute $(k\cdot B)(v,w)=B(k^{-1}\cdot v,k^{-1}\cdot w)$, which works out nicely using left invariance.

Alternatively, compactness of $K$ implies that the Haar measure is also right invariant. (The failure of invariance is measured by a homomorphism $K\to\mathbb R_{>0}$ which has to be trivial by compactness of $K$.)

Answer 2

The $K$-invariavce is proved in Proposition $2$ of Pete Clark's exposition, see here, starting with a product which is obviously $K$-invariant, and then using the idea of "averaging". He also proves the Cartan-Iwasawa-Malcev theorem which asserts that every connected Lie group (and indeed every connected locally compact group) admits maximal compact subgroups and that they are all conjugate to one another.