Let $V$ be a finite-dimensional real linear space, and let $K$ be a compact subgroup of $GL(V)$ (with the usual topology); then is there a basis of $V$ such that every $f\in K$ is an orthogonal matrix under this basis?
Compact subgroups of the general linear group
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linear-algebra
topological-groups
1 Answers
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Let $\mu$ the normalized Haar measure on $G$, and $\langle\cdot\mid\cdot\rangle$ the usual inner product on $\mathbb R^n$ (we assume that $V=\mathbb R^n$). Define an other inner product $\langle\cdot\mid \cdot\rangle_G$ by $$\langle x\mid y\rangle_G:=\int_G \langle gx\mid gy\rangle\mu(dg).$$ Since the Haar measure is invariant by translation, we have for all $g\in G$: $\langle gx\mid gy\rangle_G=\langle x\mid y\rangle$. Let $M'$ such that $\langle x\mid y\rangle_G=^tyM'x$. We have $^tgM'g=M'$ for all $g\in G$, and since $M'$ is positive definite, we can find an invertible matrix $M$ such that $M'=^tMM$. We have $^tg^tMMg=^tMM$ so $^t(MgM^{-1})\cdot (MgM^{-1})=I_n$, therefore $MGM^{-1}\subset O(n)$.
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1Nice -- how did you come up with this? – 2012-01-07
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1Note that this trick works quite a bit more generally: if $\rho: G \to L(H)$ is an uniformly bounded and strongly continuous representation of an [amenable group](http://en.wikipedia.org/wiki/Amenable_group) then there's an unitary operator such that $U\rho U^{-1}$ is an unitary operator. Dixmier's unitarizability conjecture is the question about the converse: if every strongly continuous and uniformly bounded representation on a Hilbert space is unitarily equivalent to an unitary operator, is then $G$ amenable? – 2012-01-07
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2@joriki: it's a standard trick in the representation theory of compact groups. A more sophisticated version of it goes by the name of [Weyl's unitarian trick](http://en.wikipedia.org/wiki/Unitarian_trick). – 2012-01-07
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0@t.b.: Thanks!$ $ – 2012-01-07
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1+1 A nice trick. Very much similar in spirit to a proof of Maschke's theorem, wherein you can use an orthogonal complement after you have made sure that the inner product is $G$-equivariant. In that case we can simply take the average, because the group is finite. I'm sure @joriki has seen that. – 2012-01-07