Issues in understanding proof of Peter-Weyl theorem for compact groups

Question

I'm doing a course in representation theory, and I'm having trouble with two things. First, we make the following definition.

Let $(\pi, V)$ be a representation of $G$, then we define $$M_{\pi} = \text{Span} \{ \ell(\pi(g)v) \ \vert \ v \in V, \ell \in V^{\ast} \}.$$ The elements of $M_{\pi}$ are matrix coefficients of $\pi$. My question is where does $M_{\pi}$ live, what is it a subspace of?

Secondly, I'm struggling to understand the proof of the Peter Weyl theorem, that is the matrix coefficients are dense in $L^2(G)$, and therefore $$L^2(G) \cong \widehat{\bigoplus}_{(\pi,V)} V^{\ast} \otimes V.$$

The proof is as follows. They prove first that for a $\mu$-measurable function $f$, where $\mu$ is taken to be the Haar measure, that since $\| f \|_1 \leq \| f \|_2$ it follows that $L^2(G) \subseteq L^1(G)$. Is this immediate? Then applying a result proved earlier, they assert that since $C(G)$ is dense in $L^1(G)$ it follows that the matrix coefficients of a compact group $G$ are dense in $C(G)$, with sup norm. This seems reasonable. The main source of confusion is that they then assert the conclusion since we proved previously that if $M$ is a vector space spanned by the union of $M_{\pi}$ over all irreducible representations ($\pi,V)$ of $G$, then $$M \cong \bigoplus_{(\pi,V)} V^{\ast} \otimes V.$$ I don't see how this follows and what is the qualitative significance of the fact that if $M$ is a vector space spanned by the union of $M_{\pi}$ over all irreducible representations ($\pi,V)$ of $G$, then $$M \cong \bigoplus_{(\pi,V)} V^{\ast} \otimes V?$$

Answer 1

My question is where does $M_{\pi}$ live, what is it a subspace of?

It is a subspace of $L^2(G)$: each element of $M_\pi$ is a function $G\to\mathbb{C}$. Namely, the function that sends $g$ to $\ell(\pi(g)v)$. (I can see how you were confused by this; the notation in your definition is $M_\pi$ is really unclear that we are talking about the function $g\mapsto\ell(\pi(g)v)$ rather than just a single number $\ell(\pi(g)v)$.)

They prove first that for a $\mu$-measurable function $f$, where $\mu$ is taken to be the Haar measure, that since $\| f \|_1 \leq \| f \|_2$ it follows that $L^2(G) \subseteq L^1(G)$. Is this immediate?

Yes. If $f\in L^2(G)$, that means $\|f\|_2<\infty$, and hence $\|f\|_1<\infty$ as well so $f\in L^1(G)$.

I don't see how this follows and what is the qualitative significance of the fact that if $M$ is a vector space spanned by the union of $M_{\pi}$ over all irreducible representations ($\pi,V)$ of $G$, then $$M \cong \bigoplus_{(\pi,V)} V^{\ast} \otimes V?$$

At this point, you have already shown the matrix coefficients are dense in $L^2(G)$. The vector space $M$ is defined as the subspace of $L^2(G)$ spanned by all matrix coefficients of all irreducible representations. So since $M$ contains all the matrix coefficients, $M$ is also dense in $L^2(G)$. Since $M$ is a dense subspace of $L^2(G)$ and $L^2(G)$ is complete, $L^2(G)$ is the completion of $M$. Now if we know that $$M \cong \bigoplus_{(\pi,V)} V^{\ast} \otimes V,$$ we can conclude that $L^2(G)$, the completion of $M$, is isomorphic to the completed direct sum $$\widehat{\bigoplus}_{(\pi,V)} V^{\ast} \otimes V$$ which is exactly what we wanted to prove.