Every unitary representation of a compact group is a direct sum of irreducible representations.

Question

I've read nice proofs of a few different variants of the Peter-Weyl theorem and its corollaries. For instance I know that for $G$ a compact group, $L^2(G)$ is a Hilbert space direct sum of the matrix coefficients of the irreducible representations of $G$, all of which are finite dimensional.

However, there's one fact I'm not sure how to prove: If $G$ is a compact group and $\pi : G \rightarrow \mathcal{U}(H)$ is a unitary representation of $G$ on a Hilbert space $H$, then $(\pi, H)$ is a Hilbert space direct sum of irreducible representations.

Answer 1

So I really only know the proof of Follands "A Course in Abstract Harmonic Analysis", but I really like it. Here's a sketch:

Define an operator on your Hilbert space $ H $ as $ T(v) = \int_{G} \pi(g)(u) \langle v, \pi(g)u \rangle dg $ where $ u \in H $ is an arbitrary unit vector. We claim now, that $ T $ is a compact, positive, non-zero operator and also an intertwiner, i.e. $ T \circ \pi(g) = \pi(g) \circ T $. Positivity, non-zero and intertwiner are quite straightforward. To show compactness of $ T $, you approximate $ T $ (in norm) by compact operators. (This happens analogue to the approximation of Riemann integrals by Riemann sums; this is one way to define vector valued integrals as used in the defintion of $ T $.)

Since $ T $ is compact and positive, it has an eigenvalue with finite dimensional eigenspace $ V \subset H $. Since $ T $ preserves $ V $ and $ T $ is an intertwiner, we have that $ \pi(G) $ preserves $ V $. Now as $ V $ is finite dimensional, we can decompose $ V $ into irreducible subrepresentations (just by usual representation theory arguments). This shows that $ H $ really has an irreducible subrepresentation.

Now the rest follows just by using Zorns lemma: If we consider the family of mutually orthogonal irreducible subrepresentations, ordered by inclusions, we get that there is a maximal element $ \oplus_{i \in I} V_{i} $. If $ \oplus_{i \in I} V_{i} $ was not maximal, there would be a non-zero element in the orthogonal complement, which then would contradict the maximality of $ \oplus_{i \in I} V_{i} $.