Peter-Weyl for a profinite group

Let $\pi$ be a continuous finite-dimensional unitary representation of $K$; by [F2] it factors as $\pi =\bar{\pi }\circ q$ through $F=K\mathrm{/}N$ finite with $N$ open normal. Then $\bar{\pi }$ is a finite-dimensional continuous unitary representation of the finite group $F$, and every coefficient of $\pi$ is $c_{v,w}^{\pi }=⟨\bar{\pi }(q(\cdot ))v,w⟩=c_{v,w}^{\bar{\pi }}\circ q$, so the coefficient space of $\pi$ is the pullback under $q$ of the finite-dimensional coefficient space of $\bar{\pi }$, of dimension at most $(\dim \pi {)}^2$, since basis expansion gives at most $(\dim \pi {)}^2$ spanning coefficients; equality holds for irreducible $\pi$ by Schur orthogonality [F8]. This coefficient space consists of functions constant on the cosets of $N$, hence locally constant. If $K$ were also a finite-dimensional Lie group, [F3] would give an open identity neighbourhood $U$ containing no subgroup except $\{e\}$, and by the neighbourhood basis of [F1] an open normal subgroup $N\subseteq U$; then $N=\{e\}$ is open and $K$ is discrete, hence finite because it is compact; so an infinite profinite group is not a Lie group.

The irreducible continuous finite-dimensional unitary representations of $K$ are exactly the pullbacks $\bar{\pi }\circ q$ of the irreducible representations $\bar{\pi }$ of the finite quotients $K\mathrm{/}N$: an irreducible $\pi$ is finite dimensional [F4], hence factors through such a quotient by step 1.1, and $\bar{\pi }$ is irreducible because a proper nonzero $\bar{\pi }$-invariant subspace would pull back to a proper nonzero $\pi$-invariant subspace; conversely, if $\bar{\pi }$ is irreducible then the invariant subspaces of $\pi =\bar{\pi }\circ q$ and of $\bar{\pi }$ correspond bijectively, because $q$ is surjective so $\pi (K)=\bar{\pi }(F)$ [F6]. Likewise $R(K)$ is the union of the pullbacks $q^{\ast }R(F)$: every representative function factors through a finite quotient by step 1.1, and conversely a pullback of a representative function of $F$ is a representative function of $K$ because composition with the continuous homomorphism $q$ turns matrix coefficients of representations of $F$ into matrix coefficients of their pullbacks. Since every function on a finite group is a representative function [F5], a function that factors through a finite quotient is automatically in $R(K)$; combined with [F7], $R(K)$ consists exactly of the locally constant functions on $K$.

By [F4] the algebra $R(K)$ is uniformly dense in $C(K)$ and the normalized coefficient family of $K$ is an orthonormal basis of $L^2(K)$; by step 2.1 the classes in the dual are exactly the pullbacks of the irreducible representations of the finite quotients, so the Peter-Weyl theorem holds for $K$ verbatim. No countability of $K$ or of $\widehat{K}$ is asserted, and the finite quotients of a profinite group may have arbitrarily complicated finite representation theory. The Axiom of Choice is consumed through the normalized Haar measure and the cited suppliers.