Let $K\leq GL_n(\mathbb{R})$ be a compact Lie subgroup. I need to prove that $K$ is a conjugate of a subgroup of $O(n)$. The hint is to use the Haar measure, but I really don't see how to do this.
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Let us prove that there exists a positive definite quadratic form on $\mathbf R^n$ which is $K$-invariant. Let $q(x)=\int _K q_0(k.x)d\mu$, with $\mu$ the Haar measure on $K$, and $q_0$ the euclidian quadratic form (it is obviously $q$ invariant). Let us check that $q$ is a quadratic form. Indeed $q(x)=<x,x>$, with $<x,y>=\int _K <k.x, k.y>_0d\mu$, where $<x,y>_0$ is the euclidian scalar product, so $q$ is a quadratic form. To prove that $q$ is positive definite note that on the orbit $Kx$, $q_0$ achieve its minimum, by compacity. Furthermore, this minimum is strictly positive if $x$ is not $0$ as $q_0(kx)=0\implies x=0$. So $K$ is contained in the orthogonal group of $q$, which is conjugate to the orthogonal group, as there exists a base on which $q$ is the standard euclidian form, $q(x)=<P(x),P(x)>$ and $O(q)=PO(n)P^{-1}$.
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1 The Cartan-Iwasawa-Malcev theorem 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. Since we know that $O(n)$ is a maximal compact subgroup of $GL(n,\mathbb{R})$, see here, it follows that $K$ is conjugate to a subgroup of $O(n)$.
Reference for the proof (including Haar measure): Theorem $3$ in the notes of Pete L. Clark.
- $\begingroup$ Link to lecture notes is broken... $\endgroup$dohmatob– dohmatob2022-02-14 12:25:41 +00:00Commented Feb 14, 2022 at 12:25