Continuing the previous post, there are still much things to be said about the decomposition of the regular representation on $$L^2(G)$$, by the Peter-Weyl theorem, there exist unique constants $$m_\pi \in \mathbb N$$ such that

$L^2(G) \cong \bigoplus_{\pi \in \widehat G} m_\pi \cdot \mathcal{H}_\pi.$

The goal is to show that the constants appearing in the decomposition actually correspond to the dimension of $$\mathcal{H}_\pi$$, from now on we denote the later by $$d_\pi = \text{dim} (\mathcal{H}_\pi)$$.

We begin by noting some properties of the matrix coefficients of representations.

Let $$\pi$$ be a representation, given $$u,v \in \mathcal{H}_\pi$$ define $$\pi_{v,u}(x) = \langle v, \pi(x)u \rangle$$ this is the matrix coefficient of $$\pi$$ at $$u,v$$. Since the representation is strongly continuous this is a (uniformly) continuous function on $$G$$. If $$\{ e_i\}_i$$ is a orthonormal basis of $$\mathcal{H}_\pi$$, we denote $$\pi_{e_j,e_i}= \pi_{i,j}$$.

The following point of view is very fruitful, fix a normal vector $$v \in \mathcal{H}_\pi$$ and consider $$C_\pi u(x) = \pi_{v,u}(x)$$, as an operator $$C_\pi: \mathcal{H} \to L^2(G)$$.

Remark: for all $$x,y \in G$$,

$C_\pi ( \pi(y)u )(x) = \langle u,\pi(y^{-1}x)v \rangle = C_\pi u (y^{-1}x),$

this shows that $$C_\pi$$ intertwines $$\pi$$ and the left regular representation.

We show for further use the adjoint of $$C_\pi: L^2(G) \to \mathcal{H}$$. Given $$f \in L^2(G)$$, the adjoint of $$C_\pi$$ is, in the weak sense,

${C_\pi}^* (f) = \int f(x)\pi(x)v\, dx.$

First note that if two representations $$\pi,\sigma$$ aren’t equivalent, then by Schur’s lemma $$\langle \pi_{u,v},\sigma_{u',v'} \rangle = \langle {C_\sigma}^*C_\pi u,u' \rangle = 0$$, so matrix coefficients corresponding to nonequivalent representations are mutually orthogonal.

Theorem: for all $$u,u',v,v' \in \mathcal{H}_\pi$$ one has

$\langle \pi_{u,v},\pi_{u',v'} \rangle_{L^2(G)} = \frac 1 d_\pi \langle u,u'\rangle\langle v',v \rangle.$

In particular $$\{ \frac 1 {\sqrt d_\pi} \pi_{i,j} \mid \pi \in \widehat G , u,v \in \mathcal{H}_\pi\}$$ is an orthonormal set.

Proof: Define $$A_{v,v'}u = \int_G \langle u,\pi(x)v \rangle \pi(x)v' dx$$. One checks that $$A_{v,v'}$$ is a intertwining operator, hence by Schur’s Lemma $$A_{v,v'} = d_\pi^{-1} \text{Tr}(A_{v,v'}) id_{\mathcal{H}_\pi}$$, note that the trace of $$A_{v,v'}$$ only depends on $$v,v'$$. We first prove the assertion with $$u = u'$$, let

$B*u(v,v') = \langle \pi*{u,v},\pi*{u,v'} \rangle*{L^2(G)}.$

Note that $$B_u(v,v') = \langle A_{u,u}v',v \rangle = \langle A_{v,v'}u,u \rangle$$, so in particular

$\text{Tr}(A_{u,u}) \langle v',v \rangle = \text{Tr}(A_{u,u}) \langle u,u \rangle.$

Taking $$u$$ of unitary norm, one checks that $$\text{Tr}(A_{v,v'}) = \langle v',v \rangle$$, so $$B_u(v,v') = \frac 1 {d_\pi} \langle v',v \rangle \langle u,u \rangle$$ using the polarization identity one gets the desired result.$$\square$$

Let $$L^2_\pi(G)$$ be the subspace generated by the matrix coefficients of $$\pi$$, by the previous theorem $$\{\pi_{i,j}\}$$ is a basis of $$L^2_\pi(G)$$ so the later has dimension $$d_\pi$$. As we saw before $$L^2_\pi(G) \perp L^2_\sigma(G)$$ for in-equivalent representations. If $$U \pi(x) = \sigma(x) U$$ are equivalent, by a simple calculation $$\pi_{v,u} = \sigma_{Uv,Uu}$$ so $$L^2_\pi(G)$$ only depends on the unitary class of $$\pi$$.

Theorem: For a compact group $$G$$,

$L^2(G) = \bigoplus_{\pi \in \widehat G} L^2_\pi(G),$

were each summand is invariant under the action of the left regular representation. Moreover for each $$\pi \in \widehat G$$ we have $$L^2_\pi(G) \cong_G d_\pi \cdot \mathcal{H}_\pi$$, here $$d_\pi = \text{dim}(\mathcal{H}_\pi)$$.

Before prving this theorem, we prove a lemma which has interest on its own.

Lemma: The matrix coefficients of irreducible representations of $$G$$ form a unital $$\mathbb C$$-algebra, closed under complex conjugation.

Proof: Denote by $$\mathcal{A}$$ the closure of the linear span of $$\{ \pi_{u,v} \mid \pi \in \widehat G , u,v \in \mathcal{H}_\pi \}$$. If $$\pi, \sigma$$ are two representations, then $$(\pi\oplus\sigma)_{u\oplus u', v\oplus v'} = \pi_{u,v} + \sigma_{u',v'}$$, so by the Peter-Weyl theorem any matrix coefficient belongs to the closed span of the coefficient of irreducible representations.

Denote by $$\pi\otimes \sigma$$ the tensor product of $$\pi$$ and $$\sigma$$ acting on $$\mathcal{H}_\pi \otimes \mathcal{H}_\sigma$$. then one checks that $$\pi\otimes\sigma_{u\otimes u', v \otimes v'} = \pi_{u,v} \cdot \sigma_{u',v'}$$, by the previous remark the product of matrix coefficients is in $$\mathcal{A}$$.

$$\mathcal{A}$$ is closed under complex conjugation since $$\overline {\pi_{u,v}}=\pi^\dagger_{u,v}$$

where $$\pi^\dagger$$ is the contra-gradient representation of $$\pi$$. This representation is realized as follows; let $$\mathcal{H}^\dagger$$ be equal to $$\mathcal{H}$$ as a set, with complex multiplication defined by $$\lambda \cdot u = \overline \lambda u$$ and inner product $$\langle u,v \rangle_ {\mathcal{H}^\dagger} = \langle v,u \rangle$$, then $$\pi^\dagger(x) u = \pi(x) u$$ is a representation of $$G$$ acting on $$\mathcal{H}^\dagger$$. Let $$\pi : G \to \mathbb C$$ be the trivial representation on $$\mathbb C$$, then $$\pi_{1,1}$$ is a unit for $$\mathcal{A}$$. $$\square$$

Remark: The Gelfand-Raikov theorem implies that $$\mathcal{A}$$ separates points, by the Stone-Von Neumann theorem $$\mathcal{A}$$ is dense in the space of continuous functions on $$G$$.

Combining all these facts, we have proven the theorem.

1. Folland, G. B. A course in abstract harmonic analysis. CRC Press. 2016.