Consider part II of the Peter-Weyl Theorem (see this wikipedia page for more information):

Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.

What does the conclusion have to with $\rho$? The conclusion seems to have nothing to do with the given unitary representation $\rho$. I always thought Peter-Weyl said that $\rho$ splits into a orthogonal direct sum of finite dimensional unitary representations, not the Hilbert space.

  • $\begingroup$ Look at $Gv = \overline{span}(\{ \rho(g)v, g \in G\})$. If $Gv $ is a non-trivial subspace, since $\rho(g)$ is unitary $\rho(g)Gv^\perp$ is orthogonal to $\rho(g)Gv$ and $\rho = (\rho|_{Gv},\rho|_{Gv^\perp})$ $\endgroup$ – reuns Mar 21 '19 at 2:32

When it says "$H$ splits" it means $H$ as a Hilbert space with a unitary action of $G$. In other words, the direct sum decomposition is indeed a decomposition of the representation $\rho$, not just of the Hilbert space $H$.


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