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

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