Let \(G\) be a compact Lie group and let \(\rho:G\to GL(V)\) be a finite-dimensional continuous complex representation. Show that \(\rho\) is equivalent to a unitary representation.
Solution
Choose any Hermitian inner product \(\langle-,-\rangle_0\) on \(V\). Let \(\mu\) be normalized Haar probability measure on the compact group \(G\). Define \[ \langle v,w\rangle_G =\int_G\langle \rho(g)v,\rho(g)w\rangle_0\,d\mu(g). \] This is again a Hermitian inner product: positivity follows because the integrand is continuous and nonnegative, and for \(v\ne0\) its value at the identity is positive.
For \(h\in G\), left invariance of Haar measure gives \[ \begin{aligned} \langle\rho(h)v,\rho(h)w\rangle_G &=\int_G\langle\rho(g)\rho(h)v,\rho(g)\rho(h)w\rangle_0\,d\mu(g)\\ &=\int_G\langle\rho(gh)v,\rho(gh)w\rangle_0\,d\mu(g)\\ &=\langle v,w\rangle_G. \end{aligned} \] Thus every \(\rho(h)\) is unitary with respect to \(\langle-,-\rangle_G\).
Choose an orthonormal basis for this invariant Hermitian form. In that basis, all matrices \(\rho(g)\) are unitary. Hence the original representation is equivalent to a unitary representation.