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Given an $n$-dimensional real vector space $V$, is a choice of faithful representation of the orthogonal group $O(n)$ on $V$ equivalent to a choice of inner product on $V$?

I think this should be true. Certainly a choice of inner product on $V$ determines a group $O(V)$ which is isomorphic to $O(n)$ via a choice of orthonormal basis on $V$. However, I am not sure how to define an inner product on $V$ using a given faithful representation of $O(n)$.

ಠ_ಠ
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    Weyl's unitary trick? Start with any inner product $\langle-,-\rangle_1$, then define $\langle u,v\rangle_2=\int_{{\rm O}(n)}\langle gu,gv\rangle_1,{\rm d}g$. – anon Oct 09 '16 at 22:17
  • Interesting; I don't know much representation theory so I haven't seen this before. It would be nice if there was a purely linear-algebraic argument though, or one that didn't involve integration. – ಠ_ಠ Oct 09 '16 at 22:43
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    Perhaps you could define $r_i$ to be the diagonal matrix with $1$ in all entries except $-1$ in the $i$th, then define $s_i=\sum_{j\ne i}\rho(r_j)$, and use ${s_1v,s_2,\cdots,s_nv}$ as a putative orthogonal basis to begin with. – anon Oct 09 '16 at 22:56
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    Any faithful rep defines an inner product (by Weyl's unitary trick), actually unique up to scalar multiplication. But this inner product does not determine the faithful rep... for instance any automorphism of $O(V)$ determines a representation which leaves invariant the scalar product. So no, this is not true, be careful with the use of "equivalent". – YCor Oct 10 '16 at 06:52
  • I doubt there is an easy argument without using integration. Maybe we can cheat and use representation theory, but this will be far from elementary. – YCor Oct 10 '16 at 06:54
  • Ok thanks! In that case, if you or arctic tern would like to post this as an answer I would be happy to accept it. – ಠ_ಠ Oct 10 '16 at 21:38

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Fix a basis of $V$. Now you can represent each element of $O(n)$ as a matrix, and you can read the orthonormal bases from their columns.

More concretely, fix any $U\in O(n)$ (represented in $V$). Starting from a basis $v_1,\ldots,v_n$ of $V$, define $$ w_j=\sum_{k=1}^nU_{kj}v_k. $$ Then $w_1,\ldots,w_n$ will be an orthonormal basis. Now, given $x,y\in V$, we define $$ \langle x,y\rangle_U^{\phantom U}=\sum_{j=1}^nx_jy_j, $$ where $x=\sum_jx_jw_j$, $y=\sum_jy_jw_j$.

Martin Argerami
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    I have no idea what you claim to prove, but this certainly does not prove that there is a scalar product invariant under the whole group. – YCor Oct 10 '16 at 06:48