Complete Reducibility for Compact Groups

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a compact Hausdorff group and let $\mu$ be its normalized Haar probability measure (Normalized Haar probability on a compact group); this is the only place the Axiom of Choice is consumed by the definition.

Let $V$ be a finite-dimensional complex vector space and let $\rho :K\to \mathrm{GL}(V)$ be a continuous finite-dimensional complex representation: a homomorphism of groups such that $\rho$ is continuous when $\mathrm{GL}(V)\subseteq \mathrm{End}(V)$ carries the topology induced by a norm on $\mathrm{End}(V)$. In finite dimension any two norms on $\mathrm{End}(V)$ induce the same topology (A chosen algebraic basis identifies a finite-dimensional normed space with a coordinate space), so the continuity requirement does not depend on the norm chosen.

Let $h_0$ be a Hermitian inner product on $V$ that is linear in the first variable and conjugate-linear in the second (Real and complex inner-product spaces and their induced length). The averaged Hermitian form of the pair $(\rho ,h_0)$ is the map $h:V\times V\to \mathbb{C}$ defined by

$h(v,w):={\int }_K{h}_0(\rho (k)v,\rho (k)w)d\mu (k)\mathrm{.}$

Well-definedness and conventions. The form $h_0$ is linear in the first and conjugate-linear in the second variable, and so is $h$: for each fixed $k$ the integrand is linear in $v$ and conjugate-linear in $w$, and these properties pass through the integral. Hermitian symmetry likewise passes to the limit because the integrand of $h(w,v)$ is the complex conjugate of the integrand of $h(v,w)$ for every $k$. For fixed $v,w$ the integrand $k\mapsto h_0(\rho (k)v,\rho (k)w)$ is continuous: $\rho$ is continuous, evaluation $g\mapsto \rho (g)v$ is therefore continuous, and $h_0$ is continuous on the finite-dimensional space $V\times V$ (The inner product is jointly continuous). A continuous complex function on the compact space $K$ is bounded and integrable against the Borel probability measure $\mu$, so the displayed integral is a finite complex number and $h$ is a sesquilinear form, linear in its first variable and conjugate-linear in its second. Whether $h$ is positive definite and $\rho (K)$-invariant is a theorem, not a convention: those two properties are proved in Averaging a Hermitian form unitarizes a finite-dimensional compact-group representation.

Let $G$ be a topological group, let $\pi :G\to U(H)$ be a strongly continuous unitary representation of $G$ on a complex Hilbert space $H$ (Strongly continuous unitary representations, invariant linear subspaces and intertwiners, Hilbert space), and let $M\subseteq H$ be a closed invariant subspace. Then the orthogonal complement $M^{\perp }$ (Orthogonality and the orthogonal complement) is a closed invariant subspace of $H$. This assertion is choice free.

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a compact Hausdorff topological group (Topological group: multiplication and inversion are continuous, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) with normalized Haar probability measure $\mu$ (Normalized Haar probability on a compact group). Let $V$ be a finite-dimensional complex vector space, let $\rho :K\to \mathrm{GL}(V)$ be a continuous finite-dimensional complex representation of $K$ (A finite-dimensional representation $\rho :G\to \mathrm{GL}(V)$ over a field, and its degree), let $h_0$ be a Hermitian inner product on $V$ that is linear in the first variable and conjugate-linear in the second (Real and complex inner-product spaces and their induced length, Real and complex inner product spaces, with the inner product linear in the first argument), and let $h(v,w):={\int }_K{h}_0(\rho (k)v,\rho (k)w)d\mu (k)$ be the averaged form of the pair $(\rho ,h_0)$ (Averaged Hermitian form for a compact group). Then

1. $h(v,v)>0$ for every $v\in V$ with $v\mathrm{\ne }0$; that is, $h$ is positive definite, and
2. $h(\rho (g)v,\rho (g)w)=h(v,w)$ for every $g\in K$ and all $v,w\in V$; that is, $h$ is $K$-invariant.

Consequently $h$ is an inner product on $V$, and every $\rho (g)$ is a unitary operator of the finite-dimensional inner product space $(V,h)$ (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces), so the representation $\rho$ is unitary for the averaged form $h$.

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a compact Hausdorff topological group (Topological group: multiplication and inversion are continuous, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not), let $V$ be a finite-dimensional complex vector space, and let $\rho :K\to \mathrm{GL}(V)$ be a continuous finite-dimensional complex representation (A finite-dimensional representation $\rho :G\to \mathrm{GL}(V)$ over a field, and its degree, Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis). Then $\rho$ is completely reducible (A completely reducible representation as a finite direct sum of irreducible subrepresentations): there are finitely many irreducible subrepresentations $V_1,\dots ,V_r\subseteq V$ (Subrepresentations, direct sums of representations, and irreducibility) with $V=V_1\oplus \cdots \oplus V_r,$ the empty direct sum being allowed, so the zero representation is completely reducible.

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a compact Hausdorff group with normalized Haar probability measure $\mu$ (Normalized Haar probability on a compact group). Under the assumed Axiom of Choice the Axiom of Countable Choice holds as well (AC supplies the countable and dependent choices used in Banach integration).

Let $\pi :K\to U(H)$ and $\sigma :K\to U(J)$ be strongly continuous unitary representations of $K$ on complex Hilbert spaces $H$ and $J$ (Strongly continuous unitary representations, invariant linear subspaces and intertwiners, Hilbert space), and let $T\in \mathcal{B}(H,J)$ be a bounded linear operator (A bounded linear operator between normed spaces). The pairing below is linear in the first variable and conjugate-linear in the second (Real and complex inner-product spaces and their induced length).

The scalar integrals. Fix $v\in H$ and $w\in J$ and put

$b_v(w):={\int }_K⟨\sigma (k)T\pi (k{)}^{-1}v,w⟩d\mu (k)\mathrm{.}$

The integrand is continuous on $K$. Indeed, $k\mapsto T\pi (k{)}^{-1}v$ is continuous because $\pi$ is strongly continuous and $T$ is bounded, and for a continuous map $\psi$ into $J$ the map $k\mapsto \sigma (k)\psi (k)$ is continuous because every $\sigma (k)$ is an isometry: for fixed $k_0$,

$\mathrm{\parallel }\sigma (k)\psi (k)-\sigma (k_0)\psi (k_0)\mathrm{\parallel }\le \mathrm{\parallel }\psi (k)-\psi (k_0)\mathrm{\parallel }+\mathrm{\parallel }(\sigma (k)-\sigma (k_0))\psi (k_0)\mathrm{\parallel },$

and both terms tend to zero as $k\to k_0$. A continuous complex function on the compact space $K$ is bounded and Borel, so the displayed integral is a finite complex number; and $\mid ⟨\sigma (k)T\pi (k{)}^{-1}v,w⟩\mid \le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }w\mathrm{\parallel }$ for every $k$ by the Cauchy–Schwarz inequality (Cauchy–Schwarz: $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, with equality exactly for dependent pairs), so $\mathrm{\mid }{b}_v(w)\mathrm{\mid }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }w\mathrm{\parallel }$. For fixed $v$ the assignment $w\mapsto b_v(w)$ is conjugate-linear and $\mathrm{\parallel }w\mathrm{\parallel }\mapsto \mathrm{\parallel }{b}_v(w)\mathrm{\parallel }$ is bounded by $\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$; hence

${\phi }_v(w):=\overline{b_v(w)}={\int }_K⟨w,\sigma (k)T\pi (k{)}^{-1}v⟩d\mu (k)$

is a bounded linear functional on $J$ of norm at most $\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$.

The definition. Under Countable Choice the Hilbert space $J$ has the Riesz representation property (Riesz representation for Hilbert spaces): there is a unique vector $A(T)v\in J$ with ${\phi }_v(w)=⟨w,A(T)v⟩$ for every $w\in J$, equivalently

$⟨A(T)v,w⟩=b_v(w)={\int }_K⟨\sigma (k)T\pi (k{)}^{-1}v,w⟩d\mu (k)(v\in H,w\in J)\mathrm{.}$

This determines $A(T)$ as a map $H\to J$, the Haar average of $T$ with respect to the pair $(\sigma ,\pi )$.

Well-definedness. For each $v$ the vector $A(T)v$ is unique, so $A(T)$ is a well-defined function. It is linear: if $v=a{v}_1+b{v}_2$ then, by the conjugate-linearity in $f$ of the representing vector recorded in Riesz representation for Hilbert spaces, the vector representing ${\phi }_{a{v}_1+b{v}_2}=\bar{a}{\phi }_{v_1}+\bar{b}{\phi }_{v_2}$ is $aA(T)v_1+bA(T)v_2$; and $\mathrm{\parallel }A(T)v\mathrm{\parallel }=\mathrm{\parallel }{\phi }_v\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$ because the Riesz representation is isometric, so $A(T)$ is a bounded linear operator with $\mathrm{\parallel }A(T)\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }$. The assignment $A$ is itself linear in $T$: for fixed $v,w$ the integrand is linear in $T$, so the scalar integral is, and equality of the weak matrix elements together with uniqueness of the Riesz vector gives $A(aT+b{T}^{\mathrm{\prime }})=aA(T)+bA(T^{\mathrm{\prime }})$. Thus $A:\mathcal{B}(H,J)\to \mathcal{B}(H,J)$ is a well-defined map.

This is a weak operator integral. Only scalar functions are integrated in the definition: for fixed $v$ and $w$, the continuous function $k\mapsto ⟨\sigma (k)T\pi (k{)}^{-1}v,w⟩$ is integrated against the scalar measure $\mu$, and the vector $A(T)v$ is then produced by the Riesz representation theorem. No operator-norm continuity of the orbit $k\mapsto \sigma (k)T\pi (k{)}^{-1}$ of an arbitrary bounded operator $T$ is assumed or asserted. For finite-rank $T:H\to J$, it follows from Finite-rank conjugation orbits are operator-norm continuous as follows. Equip $H\oplus J$ with the sum inner product; it is a Hilbert space because Cauchy sequences converge in each coordinate. The representation $\tau (k)(v,w)=(\pi (k)v,\sigma (k)w)$ is unitary and strongly continuous. The bounded finite-rank operator $S(v,w)=(0,Tv)$ satisfies $\tau (k)S\tau (k{)}^{-1}(v,w)=(0,\sigma (k)T\pi (k{)}^{-1}v)$. The cited lemma makes this conjugation orbit norm continuous, and restriction to $H\oplus \{0\}$ followed by projection onto $J$ does not increase operator norm, proving the claimed continuity for $T$. Only the Axiom of Choice is used, through normalized Haar measure and Countable Choice. That $A$ is an idempotent contraction onto the bounded intertwiners, with operator norm one exactly when ${\mathrm{Hom}}_K(H,J)\mathrm{\ne }\{0\}$, is the content of Haar averaging projects contractively onto the bounded intertwiners ↗, which justifies the present definition.

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a compact Hausdorff group with normalized Haar probability measure $\mu$ (Normalized Haar probability on a compact group), and let $\pi :K\to U(H)$ and $\sigma :K\to U(J)$ be strongly continuous unitary representations of $K$ on complex Hilbert spaces $H$ and $J$ (Strongly continuous unitary representations, invariant linear subspaces and intertwiners, Hilbert space). Write ${\mathrm{Hom}}_K(H,J):=\{T\in \mathcal{B}(H,J):T\pi (g)=\sigma (g)T\text{ for every }g\in K\}$ for the space of bounded intertwiners and let $A:\mathcal{B}(H,J)\to \mathcal{B}(H,J)$ be the Haar averaging operator of Haar averaging of bounded operators as a weak operator integral, so that $⟨A(T)v,w⟩={\int }_K⟨\sigma (k)T\pi (k{)}^{-1}v,w⟩d\mu (k)$ for all $T\in \mathcal{B}(H,J)$, $v\in H$, $w\in J$. Then:

1. $A$ is idempotent, $A\circ A=A$, and its range is exactly ${\mathrm{Hom}}_K(H,J)$: $A(T)$ intertwines for every bounded $T$, and $A(T)=T$ for every bounded intertwiner $T$;
2. $A$ is a contraction, $\mathrm{\parallel }A(T)\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }$ for every $T\in \mathcal{B}(H,J)$, hence $\mathrm{\parallel }A\mathrm{\parallel }\le 1$;
3. $\mathrm{\parallel }A\mathrm{\parallel }=1$ if ${\mathrm{Hom}}_K(H,J)\mathrm{\ne }\{0\}$, and $\mathrm{\parallel }A\mathrm{\parallel }=0$ if ${\mathrm{Hom}}_K(H,J)=\{0\}$.

Assume the Axiom of Choice. Let $K$ be a compact Hausdorff group with normalized Haar probability measure $\mu$ (Normalized Haar probability on a compact group) and let $f\in L^2(K,\mu ;\mathbb{C})$ be a class (Complex Haar L^p spaces and compactly supported functions). Then the left convolution $C_f$ on $L^2(K,\mu ;\mathbb{C})$,

$(C_{f}h)(x)={\int }_{K}f(x{y}^{-1})h(y)d\mu (y),$

is a well-defined bounded linear operator whose definition does not depend on the chosen measurable representative of $f$ and which is Hilbert–Schmidt with $\mathrm{\parallel }{C}_f{\mathrm{\parallel }}_{HS}=\mathrm{\parallel }f{\mathrm{\parallel }}_2$; in particular $C_f$ is compact. The operator acts on the regular representation only; no assertion is made about integrating a convolution kernel on an arbitrary irreducible representation.