Peter Weyl Theory for General Compact Groups

On $V_1\oplus V_2$ with the inner product $⟨(v_1,v_2),(w_1,w_2)⟩=⟨v_1,w_1⟩+⟨v_2,w_2⟩$ define $({\pi }_1\oplus {\pi }_2)(k)(v_1,v_2)=({\pi }_1(k)v_1,{\pi }_2(k)v_2)$; the group law holds componentwise, and $\mathrm{\parallel }({\pi }_1\oplus {\pi }_2)(k)(v_1,v_2){\mathrm{\parallel }}^2=\mathrm{\parallel }{\pi }_1(k)v_1{\mathrm{\parallel }}^2+\mathrm{\parallel }{\pi }_2(k)v_2{\mathrm{\parallel }}^2=\mathrm{\parallel }(v_1,v_2){\mathrm{\parallel }}^2$ shows that each operator is an isometry, bijective with inverse $({\pi }_1\oplus {\pi }_2)(k^{-1})$, hence unitary by [F7], while strong continuity follows from $\mathrm{\parallel }({\pi }_1\oplus {\pi }_2)(k)(v_1,v_2)-({\pi }_1\oplus {\pi }_2)(k_0)(v_1,v_2){\mathrm{\parallel }}^2=\mathrm{\parallel }{\pi }_1(k)v_1-{\pi }_1(k_0)v_1{\mathrm{\parallel }}^2+\mathrm{\parallel }{\pi }_2(k)v_2-{\pi }_2(k_0)v_2{\mathrm{\parallel }}^2\to 0$. The disjoint union of bases of $V_1$ and $V_2$ is a basis of $V_1\oplus V_2$, so the dimension is $d_1+d_2$ by [F5], and expanding the inner product gives $c_{(v_1,v_2),(w_1,w_2)}^{{\pi }_1\oplus {\pi }_2}(k)=⟨{\pi }_1(k)v_1,w_1⟩+⟨{\pi }_2(k)v_2,w_2⟩=c_{v_1,w_1}^{{\pi }_1}(k)+c_{v_2,w_2}^{{\pi }_2}(k)$ for all vectors and all $k$, which is (1).

Fix orthonormal bases $(e_i{)}_{i\le d_1}$ of $V_1$ and $(f_j{)}_{j\le d_2}$ of $V_2$ [F2]; the elementary tensors $e_i\otimes f_j$ span $V_1\otimes V_2$ because $v\otimes w={\sum }_{i,j}⟨v,e_i⟩⟨w,f_j⟩e_i\otimes f_j$ by the expansion of $v$ and $w$, and they are linearly independent because a relation ${\sum }_{i,j}{\lambda }_{ij}{e}_i\otimes f_j=0$ returns ${\lambda }_{i_0{j}_0}=0$ when one applies the linear functional induced by the bilinear form $(v,w)\mapsto ⟨v,e_{i_0}⟩⟨w,f_{j_0}⟩$ through [F4]; hence they form a basis and $\dim (V_1\otimes V_2)=d_1{d}_2$ by [F5]. Declaring this basis orthonormal makes $V_1\otimes V_2$ a finite-dimensional complex Hilbert space, and the same expansions give $⟨v\otimes w,v^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }}⟩={\sum }_{i,j}⟨v,e_i⟩⟨w,f_j⟩\overline{⟨v^{\mathrm{\prime }},e_i⟩⟨w^{\mathrm{\prime }},f_j⟩}=⟨v,v^{\mathrm{\prime }}⟩⟨w,w^{\mathrm{\prime }}⟩$ for all elementary tensors, so such an inner product exists and is unique with this property because the elementary tensors span.

Because ${\pi }_1(k)$ and ${\pi }_2(k)$ are unitary, the elementary-tensor formula of step 1.2 and the action [F4] give $⟨({\pi }_1\otimes {\pi }_2)(k)(v\otimes w),({\pi }_1\otimes {\pi }_2)(k)(v^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }})⟩=⟨{\pi }_1(k)v,{\pi }_1(k)v^{\mathrm{\prime }}⟩⟨{\pi }_2(k)w,{\pi }_2(k)w^{\mathrm{\prime }}⟩=⟨v,v^{\mathrm{\prime }}⟩⟨w,w^{\mathrm{\prime }}⟩=⟨v\otimes w,v^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }}⟩$ for all elementary tensors; both sides are sesquilinear and the elementary tensors span, so the identity holds on all of $V_1\otimes V_2$, each $({\pi }_1\otimes {\pi }_2)(k)$ is a bijective linear isometry, hence unitary by [F7], and $c_{v\otimes w,v^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }}}^{{\pi }_1\otimes {\pi }_2}(k)=c_{v,v^{\mathrm{\prime }}}^{{\pi }_1}(k)c_{w,w^{\mathrm{\prime }}}^{{\pi }_2}(k)$ on elementary tensors. For finite expansions $u={\sum }_a{\alpha }_a{v}_a\otimes w_a$ and $u\mathrm{\prime }={\sum }_b{\beta }_b{v}_b\mathrm{\prime }\otimes w_b\mathrm{\prime }$, sesquilinearity instead gives $c_{u,u\mathrm{\prime }}^{{\pi }_1\otimes {\pi }_2}(k)={\sum }_{a,b}{\alpha }_a\overline{{\beta }_b}{c}_{v_a,v_b\mathrm{\prime }}^{{\pi }_1}(k)c_{w_a,w_b\mathrm{\prime }}^{{\pi }_2}(k)$, a finite sum of products. For strong continuity write $u={\sum }_a{c}_a{v}_a\otimes w_a$ as a finite sum; then [F3] gives $\mathrm{\parallel }({\pi }_1\otimes {\pi }_2)(k)u-({\pi }_1\otimes {\pi }_2)(k_0)u\mathrm{\parallel }\le {\sum }_a\mathrm{\mid }{c}_a\mathrm{\mid }(\mathrm{\parallel }{\pi }_1(k)v_a\mathrm{\parallel }\mathrm{\parallel }{\pi }_2(k)w_a-{\pi }_2(k_0)w_a\mathrm{\parallel }+\mathrm{\parallel }{\pi }_1(k)v_a-{\pi }_1(k_0)v_a\mathrm{\parallel }\mathrm{\parallel }{\pi }_2(k_0)w_a\mathrm{\parallel })\to 0$ as $k\to k_0$, using $\mathrm{\parallel }x\otimes y\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ from step 1.2 and the strong continuity of ${\pi }_1$ and ${\pi }_2$; hence ${\pi }_1\otimes {\pi }_2$ is strongly continuous, which completes (2).

The trivial representation $1_K(k)z=z$ on $\mathbb{C}$ is a continuous unitary representation whose matrix coefficient at the unit vector $1$ is the constant function $1$. Now let $\pi$ be finite-dimensional continuous unitary on $V$ with orthonormal basis $e_1,\dots ,e_d$, and let $\sigma (k)$ be the linear operator whose matrix in this basis is the entrywise conjugate of that of $\pi (k)$, that is $⟨\sigma (k)e_i,e_j⟩=\overline{⟨\pi (k)e_i,e_j⟩}$ for all $i,j$; then $\sigma (k)=J\pi (k)J$, where $J({\sum }_i{a}_i{e}_i)={\sum }_i\overline{a_i}{e}_i$ and $J^2=I$, so $\sigma (kh)=J\pi (k)\pi (h)J=(J\pi (k)J)(J\pi (h)J)=\sigma (k)\sigma (h)$, so $\sigma$ is a homomorphism, and $\sigma (k)$ is unitary because its matrix is the conjugate of the unitary matrix of $\pi (k)$; each matrix entry $k\mapsto ⟨\sigma (k)e_i,e_j⟩$ is continuous as the conjugate of a continuous function, so $\sigma$ is strongly continuous, since $\mathrm{\parallel }(\sigma (k)-\sigma (k_0))x\mathrm{\parallel }\le {\sum }_i\mathrm{\mid }{x}_i\mathrm{\mid }\mathrm{\parallel }(\sigma (k)-\sigma (k_0))e_i\mathrm{\parallel }\to 0$ for $x={\sum }_i{x}_i{e}_i$ by [F3]; expanding coefficients in the basis gives $\overline{c_{v,w}^{\pi }(k)}={\sum }_{i,j}\overline{⟨v,e_i⟩\stackrel{‾}{⟨w,e_j⟩}⟨\pi (k)e_i,e_j⟩}={\sum }_{i,j}\overline{⟨v,e_i⟩}⟨w,e_j⟩\overline{⟨\pi (k)e_i,e_j⟩}=⟨\sigma (k)Jv,Jw⟩=c_{Jv,Jw}^{\sigma }(k)$ for all $v,w\in V$ and all $k$; this proves (3).

The trivial representation $1_K$ is a finite-dimensional continuous unitary representation whose matrix coefficient at a unit vector is the constant function $1$, so every constant function lies in $R(K)$ by [F1]; $R(K)$ is closed under pointwise addition and scalar multiplication because it is by definition a linear span of coefficient functions, and pointwise addition of representatives computed at each $k$ agrees with the sum in the function space of [F3].

If $f=c_{v,w}^{\pi }$ and $f^{\mathrm{\prime }}=c_{v^{\mathrm{\prime }},w^{\mathrm{\prime }}}^{\sigma }$ are matrix coefficients of finite-dimensional continuous unitary representations, then [F2] gives $f(k)f^{\mathrm{\prime }}(k)=c_{v,w}^{\pi }(k)c_{v^{\mathrm{\prime }},w^{\mathrm{\prime }}}^{\sigma }(k)=c_{v\otimes v^{\mathrm{\prime }},w\otimes w^{\mathrm{\prime }}}^{\pi \otimes \sigma }(k)$ for every $k$, a matrix coefficient of the finite-dimensional continuous unitary representation $\pi \otimes \sigma$, hence an element of $R(K)$; general products in $R(K)$ follow by bilinear expansion of finite linear combinations of coefficients, so $R(K)$ is closed under pointwise multiplication.

If $f=c_{v,w}^{\pi }$, then [F2] exhibits $\bar{f}$ as a matrix coefficient of a finite-dimensional continuous unitary representation of $K$, hence $\bar{f}\in R(K)$; since conjugation is additive and conjugate-linear, $\bar{f}\in R(K)$ for every finite linear combination $f$ of matrix coefficients, so $R(K)$ is closed under complex conjugation.

If $f=c_{v,w}^{\pi }$ and $g\in K$, then for every $k\in K$ the unitarity [F4] and the homomorphism property give $f(g^{-1}k)=⟨\pi (g{)}^{-1}\pi (k)v,w⟩=⟨\pi (k)v,\pi (g)w⟩=c_{v,\pi (g)w}^{\pi }(k)$ and $f(kg)=⟨\pi (k)\pi (g)v,w⟩=c_{\pi (g)v,w}^{\pi }(k)$, so the left translate $k\mapsto f(g^{-1}k)$ and the right translate $k\mapsto f(kg)$ are again matrix coefficients of $\pi$; by linearity of translation on functions the same holds for every $f\in R(K)$, so $R(K)$ is closed under left and right translation.

Collecting steps 1.1, 1.2, 1.3 and 1.4: $R(K)$ is a complex linear subspace of the continuous complex functions on $K$ containing the constants and closed under pointwise multiplication, complex conjugation and translation, so it is a self-adjoint unital complex function algebra on $K$ by [F3].

Fix $g\in K$ and $h\in L^2(K)$; for every $x$ the function $y\mapsto \phi (x{y}^{-1})h(yg)$ is integrable by Cauchy–Schwarz and Haar invariance, and the substitution $y^{\mathrm{\prime }}=yg$ together with right invariance of $\mu$ and the identity $x(y^{\mathrm{\prime }}{g}^{-1}{)}^{-1}=xg{y}^{\mathrm{\prime }-1}$ gives $(C_{\phi }\rho (g)h)(x)={\int }_K\phi (x{y}^{-1})h(yg)d\mu (y)={\int }_K\phi (xg{y}^{\mathrm{\prime }-1})h(y^{\mathrm{\prime }})d\mu (y^{\mathrm{\prime }})=(\rho (g)C_{\phi }h)(x)$ for almost every $x\in K$, whence $C_{\phi }\rho (g)=\rho (g)C_{\phi }$ in $L^2(K)$; this is (1).

First suppose $\phi \in C(K)$. Its kernel $(x,y)\mapsto \phi (x{y}^{-1})$ is product-measurable by the finite-rectangle product-measurability argument in L² convolution on a compact group is Hilbert–Schmidt. For $h_1,h_2\in L^2(K)$ the function $(x,y)\mapsto \phi (x{y}^{-1})h_1(y)\overline{h_2(x)}$ lies in $L^1(K\times K)$ because its absolute integral is at most $\mathrm{\parallel }\phi {\mathrm{\parallel }}_2\mathrm{\parallel }{h}_1{\mathrm{\parallel }}_2\mathrm{\parallel }{h}_2{\mathrm{\parallel }}_2$ and $\mu$ is a probability, so Fubini and the definition of $C_{\phi }$ give $⟨C_{\phi }{h}_1,h_2⟩={\int }_K{\int }_K\phi (x{y}^{-1})h_1(y)\overline{h_2(x)}d\mu (y)d\mu (x)$.

By definition of ${\phi }^{\ast }$ one has $\overline{{\phi }^{\ast }(y{x}^{-1})}=\phi (x{y}^{-1})$ for all $x,y\in K$, so expanding the definition of $C_{{\phi }^{\ast }}$ in the second slot and applying Fubini to the same $L^1$ function as in step 1.2 gives $⟨h_1,C_{{\phi }^{\ast }}{h}_2⟩={\int }_K{\int }_K{h}_1(y)\overline{{\phi }^{\ast }(y{x}^{-1})}\overline{h_2(x)}d\mu (x)d\mu (y)={\int }_K{\int }_K\phi (x{y}^{-1})h_1(y)\overline{h_2(x)}d\mu (y)d\mu (x)=⟨C_{\phi }{h}_1,h_2⟩$ for all $h_1,h_2\in L^2(K)$.

For continuous $\phi$, the identity of step 2.1 exhibits $C_{{\phi }^{\ast }}$ as an adjoint of the bounded operator $C_{\phi }$, so $C_{\phi }^{\ast }=C_{{\phi }^{\ast }}$ by uniqueness of the Hilbert adjoint. For general $\phi \in L^2(K)$ choose ${\phi }_n\in C(K)$ converging in $L^2$ to $\phi$, as in the convolution supplier [F4]. Haar inversion gives $\mathrm{\parallel }{\phi }_n^{\ast }-{\phi }^{\ast }{\mathrm{\parallel }}_2=\mathrm{\parallel }{\phi }_n-\phi {\mathrm{\parallel }}_2$, and Cauchy–Schwarz gives $\mathrm{\parallel }{C}_a\mathrm{\parallel }\le \mathrm{\parallel }a{\mathrm{\parallel }}_2$. Hence $C_{{\phi }_n}\to C_{\phi }$ and $C_{{\phi }_n^{\ast }}\to C_{{\phi }^{\ast }}$ in operator norm; the adjoint norm identity in [F8] passes $C_{{\phi }_n}^{\ast }=C_{{\phi }_n^{\ast }}$ to the limit. Thus $C_{\phi }^{\ast }=C_{{\phi }^{\ast }}$ for every $L^2$ kernel. If ${\phi }^{\ast }=\phi$ almost everywhere, independence of the representative gives $C_{{\phi }^{\ast }}=C_{\phi }$, and $C_{\phi }$ is self-adjoint.

Let $z\in \ker (C_{\phi }-\lambda I)$ for some $\lambda \in \mathbb{C}$ and let $g\in K$; by (1) the operators $C_{\phi }$ and $\rho (g)$ commute, hence $C_{\phi }(\rho (g)z)=\rho (g)C_{\phi }z=\lambda \rho (g)z$, so $\rho (g)z$ lies in the same eigenspace, and applying this to $g^{-1}$ gives $\rho (g)\ker (C_{\phi }-\lambda I)=\ker (C_{\phi }-\lambda I)$; the case $\lambda =0$ is the kernel $\ker C_{\phi }$. The Axiom of Choice is consumed through the normalized Haar probability and the cited Hilbert-space and adjoint suppliers; the computations above are choice-free apart from those inputs.

The spectral theorem [F1] applied to $T=C_{\phi }$ gives that $\mathrm{\Sigma }$ is finite or countably infinite, that every $E_{\lambda }$ is finite dimensional, and that the closed linear span $M$ of ${\bigcup }_{\lambda \in \mathrm{\Sigma }}{E}_{\lambda }$ equals $(\ker C_{\phi }{)}^{\perp }$; distinct eigenspaces are orthogonal by [F2]; [F5] applied to the closed subspace $M$ gives $L^2(K)=M\oplus M^{\perp }$ with $M=(\ker C_{\phi }{)}^{\perp }$, so $M^{\perp }=\ker C_{\phi }$ and hence $L^2(K)=\ker C_{\phi }\oplus M$ as an orthogonal decomposition; since the $E_{\lambda }$ are pairwise orthogonal closed subspaces with closed linear span $M$, the canonical map ${\widehat{⨁}}_{\lambda \in \mathrm{\Sigma }}{E}_{\lambda }\to M$ is a unitary isomorphism by [F4], so $L^2(K)=\ker C_{\phi }\oplus {\widehat{⨁}}_{\lambda \in \mathrm{\Sigma }}{E}_{\lambda }$ is an orthogonal Hilbert-space direct sum; this is (1).

By the choice ${\phi }^{\ast }=\phi$ and [F3], every $E_{\lambda }$ and $\ker C_{\phi }$ is invariant under the right regular representation $\rho$; each $E_{\lambda }$ is closed and finite dimensional by step 1.1, and by [F6] the restriction of $\rho$ to $E_{\lambda }$ is a strongly continuous unitary representation of $K$, hence a finite-dimensional continuous unitary representation; this is (2), and it also exhibits $\rho$ as the Hilbert direct sum of these subrepresentations and the kernel by [F4]. The Axiom of Choice is consumed through the normalized Haar probability, the compact self-adjoint spectral theorem and the cited Hilbert-space suppliers; the argument above is choice-free apart from those inputs.

By [F1] choose an open symmetric neighbourhood $U$ of $e$ with $g\mathrm{\notin }U\cdot U$ and by [F2] a continuous ${\phi }_0:K\to [0,1]$ with ${\phi }_0(e)=1$ and vanishing outside $U$; put $\phi (k):=\frac{1}{2}({\phi }_0(k)+{\phi }_0(k^{-1}))$, so $\phi \ge 0$ is continuous with $\phi (e)>0$, vanishing outside $U$ and $\phi (k^{-1})=\phi (k)$, hence ${\phi }^{\ast }=\phi$; put $\psi :=\phi \ast \phi \in C(K)$. Then $\psi$ is real and satisfies ${\psi }^{\ast }={\phi }^{\ast }\text{\hspace{0.17em}⁣}\ast {\phi }^{\ast }=\psi$ by [F3], and $\psi =C_{\phi }\phi$ because $C_{\phi }\phi (x)={\int }_K\phi (x{v}^{-1})\phi (v)d\mu (v)={\int }_K\phi (w)\phi (w^{-1}x)d\mu (w)=\psi (x)$ under the substitution $w=x{v}^{-1}$, which preserves $\mu$ by [F4]; moreover $\psi (e)={\int }_K\phi (v)\phi (v^{-1})d\mu (v)={\int }_K{\phi }^{2}d\mu >0$ because $\phi$ is continuous, positive at $e$ and $\mu$ is positive on the nonempty open set where $\phi >0$, while $\psi (g)={\int }_K\phi (v)\phi (v^{-1}g)d\mu (v)=0$ since $\phi (v)\phi (v^{-1}g)\mathrm{\ne }0$ would give $v\in U$ and $v^{-1}g\in U$, hence $g\in U\cdot U$ by symmetry of $U$; finally $C_{\phi }$ and $C_{\psi }$ are compact self-adjoint with the spectral decomposition $L^2(K)=\ker C_{\phi }\oplus {\widehat{⨁}}_{\lambda \mathrm{\ne }0}{E}_{\lambda }$ into finite-dimensional $\rho$-invariant pieces by [F5].