Peter Weyl Theory for General Compact Groups — Examples

If $n=0$, $U(0)$ is the trivial group, $\ker \rho =K$ and the representation factors through the trivial finite quotient; also $\ker {\pi }_i\subseteq K$ for any index $i$. Hence assume $n\ge 1$. First $\rho$ is continuous in operator norm: for an orthonormal basis $e_1,\dots ,e_n$ of ${\mathbb{C}}^n$ and $x={\sum }_i{x}_i{e}_i$ with $\mathrm{\parallel }x\mathrm{\parallel }\le 1$, $\mathrm{\parallel }(\rho (k)-\rho (k_0))x\mathrm{\parallel }\le {\sum }_i\mathrm{\mid }{x}_i\mathrm{\mid }\mathrm{\parallel }(\rho (k)-\rho (k_0))e_i\mathrm{\parallel }\le ({\sum }_i\mathrm{\parallel }(\rho (k)-\rho (k_0))e_i{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}$ by Cauchy–Schwarz, and the right side tends to $0$ as $k\to k_0$ because the finitely many orbit maps are continuous [F3]; by [F2] the group $U(n)$ is a finite-dimensional real Lie group, so its no-small-subgroups lemma provides an open identity neighbourhood $V\subseteq U(n)$ containing no subgroup other than $\{e\}$, and $W:={\rho }^{-1}(V)$ is an open neighbourhood of the identity of $K$.

By [F1] the subgroups $\ker {\pi }_i$ form an open normal neighbourhood basis at the identity, so $\ker {\pi }_i\subseteq W$ for some $i$; then $\rho (\ker {\pi }_i)$ is a subgroup of $U(n)$ by [F4] and lies in $V$, hence $\rho (\ker {\pi }_i)=\{e\}$ by the choice of $V$, that is, $\ker {\pi }_i\subseteq \ker \rho$. Since $\ker {\pi }_i$ is open, its cosets are open, and $\ker \rho$ is a union of cosets of $\ker {\pi }_i$, so $\ker \rho$ is open as well.

The inclusion $\ker {\pi }_i\subseteq \ker \rho$ implies that $\rho$ factors as $\rho =\bar{\rho }\circ q$, where $q:K\to F:=K\mathrm{/}\ker {\pi }_i$ is the quotient homomorphism and $\bar{\rho }(q(k)):=\rho (k)$; here $q$ is a surjective continuous homomorphism and $F$ is finite because the first isomorphism theorem gives $F\cong {\pi }_i(K)\subseteq G_i$ with $G_i$ finite [F4]. Every matrix coefficient $c_{v,w}^{\rho }$ is constant on each coset of $\ker {\pi }_i$, since $\rho$ is, hence is locally constant and factors through the finite group $F$; this proves the lemma. The Axiom of Choice is consumed through the countable choice assumed by the Lie-group example [F2] and the profinite presentation; the factorisation argument itself is choice-free apart from those inputs.

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.

The map $x\mapsto (p_F(x){)}_F$ from $K$ to the set of compatible tuples is a topological group isomorphism onto , with inverse given by the compatible tuple's coordinates: it is a group homomorphism because each $p_F$ is, it is injective because the coordinates determine $x$, it is surjective because the coordinates of a compatible tuple define an element of $K$ whose projections are the given ones, and both directions are continuous for the product and inverse-limit topologies by [F1]; hence $K$ is profinite by [F2], and the kernels $\ker p_F$ form an open normal neighbourhood basis at the identity. By [F4] each $\ker p_F$ is the closed subgroup of tuples trivial in the coordinates of $F$ and is canonically isomorphic to ${\prod }_{i\mathrm{\notin }F}{G}_i$; moreover $p_F$ is surjective with $K\mathrm{/}\ker p_F\cong K_F$ by [F6].

Apply the profinite factorisation lemma [F3] to the profinite presentation of $K$ as : there is a finite $F\subseteq I$ with $\ker p_F\subseteq \ker \rho$, and $\rho$ factors as $\rho ={\rho }_F\circ p_F$ for a representation ${\rho }_F$ of the finite group $K_F$; since the $\ker p_F$ form an identity neighbourhood basis by step 1.1, the finite set $F$ may be chosen so that $\ker p_F$ lies inside any prescribed identity neighbourhood of $K$. Every matrix coefficient $c_{v,w}^{\rho }(k)=⟨{\rho }_F(p_F(k))v,w⟩$ by [F5] depends only on $p_F(k)$, hence only on the coordinates in $F$, and is constant on the cosets of $\ker p_F$; this proves the lemma.

By [F1] the product $K$ is a profinite group, hence compact Hausdorff, with normalized Haar probability, and the kernels of the coordinate projections form an open normal neighbourhood basis. By [F2] every continuous finite-dimensional unitary representation of $K$ factors through some $p_F$, and every matrix coefficient then depends on the coordinates in $F$ only; by [F3] such a pullback is irreducible exactly when the representation of the finite group $K_F$ is, so the finite-coordinate irreducibles of $K$ are exactly the pullbacks of the irreducibles of the groups $K_F$. Consequently $R(K)$ is the union over finite $F$ of the pullbacks $p_F^{\ast }R(K_F)$: one inclusion is [F2], and conversely a pullback of an element of $R(K_F)$ is a finite linear combination of pullbacks of matrix coefficients, hence a representative function of $K$. Since $R(K_F)=C(K_F)$ for the finite group $K_F$ by [F4], $R(K)$ is exactly the algebra of continuous functions depending on finitely many coordinates.

The general theory applies to the compact Hausdorff group $K$ by [F5]: $R(K)$ is uniformly dense in $C(K)$ and the normalized coefficient family is an orthonormal basis of $L^2(K)$, while the regular representation decomposes into its isotypic blocks. The density also follows directly from the description of $R(K)$ in step 1.1: it is a unital self-adjoint algebra of continuous functions separating points by [F6], so the unital Stone-Weierstrass theorem gives uniform density; point separation uses only the finitely many coordinates in which two points differ, and no countability of $I$ is needed or claimed. This completes the verification. The Axiom of Choice enters exactly through the suppliers and coefficient-family choices of [A1].

By [F1] the group $\mathbb{T}$ is a compact abelian topological group, so it carries normalized Haar probability and every irreducible continuous unitary representation of it is subject to the compact theory; let $\pi$ be such a representation on a nonzero Hilbert space $H$; for all $s,t\in \mathbb{T}$ the operators commute, $\pi (t)\pi (s)=\pi (ts)=\pi (st)=\pi (s)\pi (t)$, so every $\pi (t)$ is a bounded self-intertwiner and [F3] makes it a scalar $\chi (t)$ times the identity; then every one-dimensional subspace of $H$ is $\pi (\mathbb{T})$-invariant, so irreducibility forces $\dim H=1$, and $\chi :\mathbb{T}\to \mathbb{T}$ is a continuous character because $\pi$ is strongly continuous and unitary [F5], with $\mathrm{\mid }\chi (t)\mathrm{\mid }=1$. The normalized matrix coefficient of the one-dimensional representation $\pi =\chi$ at the unit vector is $\sqrt{1}⟨\chi (t)e_1,e_1⟩=\chi (t)$; this proves (1).

By [F2] the family $\{z^n:n\in \mathbb{Z}\}$ is orthonormal with closed linear span $L^2(\mathbb{T})$, that is, it is a complete orthonormal family; this is (2).

By [F2] each $z^n$ is a continuous character, hence a one-dimensional continuous unitary representation of $\mathbb{T}$, and step 1.1 shows that its normalized matrix coefficient is $z^n$ itself; therefore $\{z^n\}\subseteq \mathcal{B}$. Since $\mathcal{B}$ is an orthonormal basis by [F4] and $\{z^n\}$ is a complete orthonormal subfamily by step 1.2, [F6] gives $\mathcal{B}=\{z^n:n\in \mathbb{Z}\}$; consequently the unitary dual of $\mathbb{T}$ is exactly $\{z^n:n\in \mathbb{Z}\}$, which is in bijection with $\mathbb{Z}$ because $z^n=z^m$ fails at $[t]=1\mathrm{/}(2(n-m))$ when $n\mathrm{\ne }m$. The Pontryagin dual of the compact abelian group $\mathbb{T}$ is discrete by [F7], consistent with this dual being the discrete family of integer characters; the example does not recompute ${\mathrm{Hom}}_{cts}(\mathbb{T},\mathbb{T})$. The Peter-Weyl decomposition of $L^2(\mathbb{T})$ is therefore the Hilbert direct sum of the one-dimensional blocks $\mathbb{C}{z}^n$, $n\in \mathbb{Z}$, the classical Fourier series decomposition, and the Parseval identity of the compact theory specializes to the classical Parseval identity for Fourier series and the expansion to Fourier inversion in $L^2$ (The Parseval identity for Fourier series); this completes the verification. The Axiom of Choice is inherited through the cited suppliers.

First record the representative formula for $f\in L^1(\mathbb{R})$ and $h_0\in C_c(\mathbb{R})$: the function $x\mapsto F(x):={\int }_{\mathbb{R}}f(y)h_0(x-y)d{\lambda }_1(y)$ represents the class $f\ast h_0$, because for $u_n\in C_c$ with $\mathrm{\parallel }{u}_n-f{\mathrm{\parallel }}_1\to 0$ one has $u_n\ast h_0\to f\ast h_0$ by [F3], while $\mathrm{\mid }F(x)-(u_n\ast h_0)(x)\mathrm{\mid }\le {\int }_{\mathbb{R}}\mathrm{\mid }f-u_n\mathrm{\mid }(y)\mathrm{\mid }{h}_0(x-y)\mathrm{\mid }d{\lambda }_1(y)$ and hence $\mathrm{\parallel }F-u_n\ast h_0{\mathrm{\parallel }}_1\le \mathrm{\parallel }f-u_n{\mathrm{\parallel }}_1\mathrm{\parallel }{h}_0{\mathrm{\parallel }}_1\to 0$ by [F4] applied to the nonnegative product-measurable function $\mathrm{\mid }f-u_n\mathrm{\mid }(y)\mathrm{\mid }{h}_0(x-y)\mathrm{\mid }$ and translation invariance [F1], so $F$ and $f\ast h_0$ differ only on a null set.

Let $U_n:=(-1\mathrm{/}(n+1),1\mathrm{/}(n+1))$ and choose $e_n:=e_{U_n}$ from the net of [F2]; for every $n$ and every $x$ the representative of $g\ast e_n$ from step 1.1 gives $(g\ast e_n)(x)={\int }_{\mathbb{R}}g(y)e_n(x-y)d{\lambda }_1(y)={\int }_{\mathbb{R}}g(x-z)e_n(z)d{\lambda }_1(z)$ by the substitution $z=x-y$ and translation invariance, while the hypothesis with shift $t=x$ gives $g(x-z)=g(-z)$ for almost every $z$, so $(g\ast e_n)(x)={\int }_{\mathbb{R}}g(-z)e_n(z)d{\lambda }_1(z)=:c_n$ is a constant independent of $x$; thus $g\ast e_n$ equals the constant $c_n$ almost everywhere.

By [F3] each $g\ast e_n$ is an $L^1$ class. Step 2.1 identifies it with the constant $c_n$; since ${\lambda }_1(\mathbb{R})=\mathrm{\infty }$, integrability forces $c_n=0$. The neighbourhoods $U_n=(-1\mathrm{/}(n+1),1\mathrm{/}(n+1))$ are cofinal in the identity neighbourhoods: every such neighbourhood contains some $U_m$, and $U_n\subseteq U_m$ for $n\ge m$. The right approximate-identity convergence in [F2] therefore gives $\mathrm{\parallel }g\ast e_n-g{\mathrm{\parallel }}_1\to 0$. Since $g\ast e_n=0$ for every $n$, $\mathrm{\parallel }g{\mathrm{\parallel }}_1=0$ and $g=0$ almost everywhere.

Finally let $v\in L^2(\mathbb{R},{\lambda }_1)$ satisfy $\mathrm{\mid }v(x-t)\mathrm{\mid }=\mathrm{\mid }v(x)\mathrm{\mid }$ for every $t$ and almost every $x$; then $g:=\mathrm{\mid }v{\mathrm{\mid }}^2$ lies in $L^1(\mathbb{R},{\lambda }_1)$ by [F6] and satisfies $g(x-t)=g(x)$ for every $t$ and almost every $x$, so step 3.1 applied to this $g$ gives $g=0$ almost everywhere, that is, $v=0$ almost everywhere, which is the stated consequence; this proves the lemma. The Axiom of Choice is consumed through the approximate identity net of [F2] and through those measure-theoretic suppliers of [F1] that need it, the complete-measure, dilation-reflection and Radon-measure theorems being proved under the Axiom of Countable Choice; the translation, convolution and subsequence arguments are choice-free apart from those inputs.

Suppose $\lambda$ has a nonzero irreducible unitary subrepresentation on a closed invariant subspace $M\subseteq L^2(\mathbb{R})$; for $t\in \mathbb{R}$ the operator $\lambda (t){\mathrm{\mid }}_M$ is a bounded self-intertwiner of the restriction because $\lambda (t)\lambda (s)=\lambda (s)\lambda (t)$ for all $s,t$ (the group is abelian), so [F2] makes it a scalar $\chi (t)$ times the identity, and then every one-dimensional subspace of $M$ is invariant, so irreducibility [F3] forces $M=\mathbb{C}v$ with $\mathrm{\parallel }v\mathrm{\parallel }=1$ and $\lambda (t)v=\chi (t)v$ for all $t$; by [F6] this gives $\mathrm{\mid }v(x-t)\mathrm{\mid }=\mathrm{\mid }v(x)\mathrm{\mid }$ for every $t$ and almost every $x$, so $v=0$ almost everywhere by [F5], contradicting $\mathrm{\parallel }v\mathrm{\parallel }=1$; hence $\lambda$ has no nonzero irreducible unitary subrepresentation.

If $\lambda$ were a Hilbert direct sum of finite-dimensional irreducible unitary subrepresentations, then by [F4] at least one summand would be nonzero because $L^2(\mathbb{R})\mathrm{\ne }\{0\}$ by [F1], and that summand would be a nonzero irreducible unitary subrepresentation of $\lambda$, contradicting step 1.1; therefore no such decomposition exists, and the refuted statement fails. The Axiom of Choice enters through the cited Schur lemma, the Hilbert-direct-sum and regular-representation suppliers and the Lebesgue-measure facts of [F1] (the complete-measure and Radon-measure theorems are proved under the Axiom of Countable Choice); the vanishing argument itself is choice-free apart from those inputs.

In this scalar case the direct integral ${\int }_{\mathbb{R}}^{\oplus }\mathbb{C}d\xi$ means the Hilbert space of measurable scalar sections $h(\xi )$ with $\int \mathrm{\mid }h(\xi ){\mathrm{\mid }}^{2}d\xi <\mathrm{\infty }$, modulo null equality, with its integral inner product; this is exactly $L^2(\mathbb{R},d\xi )$. Let its fiberwise action be $(D(t)h)(\xi )=e^{-2\pi it\xi }h(\xi )$. Unit modulus makes each $D(t)$ unitary, the exponential addition law makes it a representation, and dominated convergence with majorant $4\mathrm{\mid }h{\mathrm{\mid }}^2$ proves strong continuity. On Schwartz inputs [F7] gives ${\mathcal{F}}_2\lambda (t)=D(t){\mathcal{F}}_2$; both sides are bounded operators, so density extends this identity to every $L^2$ class. The surjective unitary ${\mathcal{F}}_2$ therefore realizes $\lambda$ as the asserted direct integral of one-dimensional characters. For a fixed frequency the character has spatial modulus1, whose squared integral over $\mathbb{R}$ is infinite, so it is no nonzero vector in the original $L^2$ space. This explicit scalar integral supplies the motivating contrast without assuming general direct-integral decomposition or uniqueness theory.