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.

  • $\begingroup$ I do not understand: What is the difference between your question and the statement of the PW theorem? $\endgroup$ – Moishe Kohan Jul 2 '14 at 13:10
  • $\begingroup$ @studiosus Well I guess it depends on what exact formulation of the PW theorem you have in mind. The sources I've been reading give the label "PW theorem" to something like the statement in the first paragraph of my post. $\endgroup$ – dessin d'enfant terrible Jul 2 '14 at 13:13

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} $.

  • $\begingroup$ Great! So the real issue is just the non-emptiness of the Zorn's lemma chain. $\endgroup$ – dessin d'enfant terrible Jul 2 '14 at 13:16
  • $\begingroup$ T is what you call a hilbert schmidt operator right? $\endgroup$ – dessin d'enfant terrible Jul 2 '14 at 13:31

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