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.