Let \(G\) be a locally compact topological group, this is a survey on some elementary results on the representation theory when \(G\) is compact. From now on are going to assume all groups compact and all the Hilbert spaces separable.

Since \(G\) is compact by averaging any Radon measure on \(G\) we can construct a left invariant Haar measure \(dy\); i.e. one which satisfies

\[\int_G f(xy) dy = \int_G f(y)\,dy,\qquad f \in L^1(G,dy)\]

for all \(x\in G\). Without loss of generality one chooses \(dy\) to satisfy \(\int_G dy = 1\).

Given a strongly-continuous unitary representation \(\pi: G \to \mathcal{B}(\mathcal{H})\), we denote by \(Hom_G (\mathcal{H})\) the space of intertwining operators, that is, operators \(T: \mathcal{H} \to \mathcal{H}\) such that \(\pi(x)T = T \pi(x)\) for all the elements of our group. Analogously for two representations \(\pi_1,\pi_2\) we denote the space of intertwining operators by

\[Hom_{G}( \mathcal{H}_{\pi_1}; {\mathcal{H}}_{\pi_2}).\]

If \(\pi\) is a representation, we denote by \(\mathcal{H}_\pi\) the space on which \(G\) acts.

Definition: A representation of \(G\) on a Hilbert space \(\mathcal{H}_\pi\) is called irreducible if \(\mathcal{H}_\pi\) does not have any non-trivial closed invariant subspace.

Definition: Two unitary representations are called equivalent if there exist an invertible intertwining operator between them, we denote this “equivalence relation” by \(\cong\).

Since the group has a finite measure there is no harm in assuming all representations unitary. This is due to Weyl’s unitary trick; it consists of taking any inner product on the Space on which \(G\) acts and averaging it, i.e. defining a new inner product by formula

\[\langle u,v \rangle_G =\int_G\langle \pi(x)u,\pi(x)v \rangle\,dx.\]

What is obtained is a inner product for which \(\pi(x)\) is a unitary operator for every element \(x \in G\).

We recall one of the fundamental result in representation theory.

Theorem: (Schur’s Lemma) Let \(G\) be a locally compact group. A unitary representation of \(G\) is irreducible if and only if \(Hom_G(\mathcal{H},\mathcal{H})\) contains only multiples of the identity.

Suppose \(\pi_1,\pi_2\) are irreducible representations. Then either \(Hom_G(\mathcal{H}_{\pi_1}; \mathcal{H}_{\pi_2})\) is one dimensional and \(\pi_1 \cong \pi_2\), or, otherwise \(Hom_G(\mathcal{H}_{\pi_1}; \mathcal{H}_{\pi_2})\) is trivial.

Proof: Notice that both the kernel and the image of a non-zero intertwining operator \(U\) are invariant subspaces of \(\mathcal{H}\), consequently \(U\) is bijective. By the open mapping theorem \(U\) must be invertible in \(Hom_G(\mathcal{H},\mathcal{H})\), so the later is a Banach division algebra over \(\mathcal C\), by the Gelfand-Mazur theorem the only Banach algebra that satisfies this is \(\mathcal{C}\). To prove the second part of the statement, by the same arguments one checks that any nonzero intertwining operator must be an isomorphism, plus the observation that \(U^*U \in Hom_G(\mathcal{H}_{\pi_1})\) hence \(U = \lambda id_{\mathcal{H}_{\pi_1}}\) for some non zero constant \(\lambda \geq 0\), normalizing \(U\) one gets the desired unitary equivalence. If \(U,S \in Hom_G(\mathcal{H}_{\pi_1},\mathcal{H}_{\pi_2})\), then \(S^{-1}U \in Hom_G(\mathcal{H}_{\pi_1})\), therefore \(U\) and \(S\) must be multiples of each other. \(\square\)

Let \(\pi\) be a representation of \(G\), given \(u,v \in \mathcal{H} = \mathcal{H}_\pi\) we define

\[Tu = \int_G \langle u,\pi(x) v \rangle\pi(x)v\,dx.\]

Since \(\langle Tu,u \rangle = \int_G \lvert \langle u,\pi(y)u \rangle\rvert^2 dy \geq 0\), \(T\) is a positive bounded operator in \(\mathcal{H}\), it is constructed in such a way that it intertwines \(\pi\) with itself, and it is non zero since \(Tv \not= 0\). If \(\pi\) is irreducible, then \(T\) must be a multiple of the identity, this is a key point since

Proposition: \(T:\mathcal{H} \to \mathcal{H}\) is a compact self-adjoint intertwining operator.

Proof: \(G\) is compact so for any \(v \in \mathcal{H}\) the function \(\pi(x)v\) is uniformly continuous in \(x\). Let \(\varepsilon \ge 0\), we choose open sets \(V_1,\dots,V_n\) and \(x_i \in V_i\) such that for all \(i=1,\dots,n\):

\[\lVert \pi(x)v - \pi(x_i) \rVert\ \ge \varepsilon \qquad \forall x \in V_i .\]

Note that

\[\begin{aligned} \lVert \langle u,\pi(x)v \rangle\pi(x)v - \langle u,\pi(x_i)v \rangle\pi(x_i)v \rVert &\\ \leq& \lVert \langle u, [\pi(x)v-\pi(x_i)v] \rangle\pi(x)v \rVert\\ \qquad +& \lVert \langle u,\pi(x_i)v \rangle [\pi(x)v-\pi(x_i)v]\rVert\\ \leq& \varepsilon \lVert u\rVert \end{aligned}\]

Define \(T_\varepsilon = \sum_{i=1}^n \int_{V_i} \langle u,\pi(x_i)v \rangle\pi(x_i)v \,dx\), the range of this operator is spanned by \(\{ \pi(x_i)v \}_{i=1}^n\). The previous calculation shows that \(\lVert T - T_\varepsilon \rVert \le \varepsilon\) hence \(T\) can be approximated by finite range operators, in particular \(T\) is compact. \(\square\)

Corollary: Every irreducible representation of a compact group is finite-dimensional.

Now we have the main ingredients for the structure theorems of representations of compact groups.

One comment before the main theorem. If \(\mathcal{H}\) is a representation of \(G\), then any invariant subspace \(E\) admits an orthogonal complement; i.e. since the representation is unitary \(E^\perp\) is also an invariant subspace and \(\mathcal{H} = E \oplus E^\perp\). If \(\mathcal{H}\) is finite-dimensional, one could continue to divide \(\mathcal{H}\) into smaller invariant subspaces, and eventually one ends with a decomposition of \(\mathcal{H}\) consisting only of irreducible subrepresentations. In the infinite-dimensional case this process can go on indefinitely, but the case of compact groups is very similar to the case when \(G\) is finite, more precisely.

Theorem (Peter-Weyl Theorem): Let \(\pi:G\to \mathcal{H}\) be a representation of a compact group. Then there exist irreducible invariant subspaces \(\{\mathcal{H}_i \}_{ i\in I}\) such that \(\mathcal{H} = \bigoplus_i \mathcal{H}_i.\)

Proof: As usual we are in need of the axiom of choice. We consider the partially ordered set (by inclusion)

\[\mathcal{C } = \{ \bigoplus_{i \in I} \mathcal{H}_i \mid \mathcal{H}_i \text{ is a closed irreducible invariant subspace of } \mathcal{H} \}.\]

Recall that \(T\) is a (non-zero) compact self-adjoint operator, by the spectral theorem, \(\mathcal{H}\) decomposes as the direct sum of its eigenspaces, with \(\ker(T-\lambda)\) being finite dimensional for \(\lambda \not= 0\), we fix one of those eigenspaces. Since the later is finite dimensional it can be decomposed as a sum of irreducible subspaces, so \(\mathcal{C}\) is not vacuous. It is clear that all chains in \(\mathcal{C}\) have an upper bound; So \(\mathcal{C}\) has a maximal element \(E = \oplus_i \mathcal{H}_i \in \mathcal{C}\). If we take \(v\in E^\perp\) and define \(Tu = \int \langle u,\pi(x)v \rangle\pi(x)v\,dx\) as before, repeating the previous argument one gets a new finite dimensional subspace of \(E^\perp\), which contradicts the maximality of \(E\); so \(E = \mathcal{H}\). \(\square\)

Now given an arbitrary representation \(\mathcal{H}\) of \(G\), and a decomposition into irreducible representations \(\mathcal{H} = \bigoplus_i \mathcal{H}_i\); we can group equivalent representations in the decomposition, say

\[\mathcal{H} = \bigoplus_{\pi \in \widehat G} \left( \bigoplus_{\{ i:\mathcal{H}_i \cong \mathcal{H}_\pi \}} \mathcal{H}_i \right).\]

To put it another way we get an equivalent formulation.

Theorem: Let \(\mathcal{H}\) be a representation of a compact group, then there exist constants \(m_\pi \in \mathbb N \cup \{ \infty\}\) such that \(\mathcal{H}\) decomposes as a direct sum of finite dimensional irreducible representations

\[\mathcal{H} \cong \bigoplus_{\pi \in \widehat G}\ m_\pi \cdot \mathcal{H}_\pi.\]