Compact Groups

Unitary Representations

In case $E$ is an infinite dimensional (separable) Hilbert-space all operators in $\Hom(E),\Hom(E,F),\Gl(E)$, etc. will be assumed to be bounded unless otherwise stated.
Suppose $\Psi:G\rar\Gl(E)$ is a representation of the compact group $G$ in some Hilbert-space $E$. We will also presuppose that for any $x\in E$ the mapping $g\mapsto\Psi(g)x$ is continuous - this implies that $g\mapsto\Psi(g)x$ is integrabel with respect to any bounded Borel measure on $G$. Suppose $\mu$ is a Haar probability measure on $G$, then $$ (x,y)\mapsto\int_G\la\Psi(g)x,\Psi(g)y\ra\,\mu(dg) $$ is another Euclidean product on $E$ and all mappings $\Psi(g)$, $g\in G$, are unitary with respect to this product. Thus it's not to much a restriction to stick to unitary representations. Given such a representation $\Psi$ we say that it's irreducible if every invariant subspace $F$ of $E$ is either $\{0\}$ or dense. Of course, if $E$ is finite dimensional we retain the definition given in the chapter on finite groups. But in general this makes some difference: for example, for an intertwining operator $A\in\Hom_G(\Psi,\Phi)$ of two irreducible representations we have: $\ker A$ is either $\{0\}$, in which case $A$ is one-one, or dense, implying that $A=0$; also $\im A$ is either $\{0\}$, i.e. $A=0$ or dense. Thus if $A\neq0$, then $A$ is one-one with dense image but this doesn`t necessarily mean that $A$ is an isomorphism. For $A$ to be an isomorphism, we have to have another condition: $\im A$ must be closed, which by the open-mapping theorem and the homomorphism theorem holds if $A^*$ is onto.

If for all $x\in E$ the mapping $g\mapsto\Psi(g)x$ is continuous, then by the uniform boundedness theorem the mapping $G\times E\rar E$, $(g,x)\mapsto\Psi(g)x$ is continuous.

First we are going to prove Schur's lemma in the general case of a unitary representation $\Psi$ of a group $G$ in an arbitrary Hilbert space $E$

Let $E$ be Hilbert-space and $\Psi:G\rar\OO(E)$ an irreducible orthogonal representation.

A bounded self-adjoint operator $A\in\Hom_G(\Psi)$ is a multiple of the identity.
If $E$ is complex and $\Psi:G\rar\UU(E)$ unitary and irreducible, then $\Hom_G(\Psi)=\C1_E$.

$\proof$ 1. We have to prove that a bounded self-adjoint operator $A$ on $E$ which commutes with all unitary operators $\Psi(g)$, $g\in G$, is a multiple of the identity: Suppose not, then the spectrum of $A$ contains at least two points $a\neq b$. Choose bounded continuous functions $f_a,f_b:\R\rar\R$ such that $f_af_b=0$, $f_a(a)=1$ and $f_b(b)=1$; it follows that $f_a(A)f_b(A)=0$ and $f_a(A),f_b(A)\neq0$. Put $F\colon=\im f_a(A)$, then for all $g\in G$: $$ \Psi(g)(F)=\Psi(g)(f_a(A)(E))=f_a(A)\Psi(g)(E)\sbe F \quad\mbox{hence}\quad \Psi(g)(\cl{F})\sbe\cl{F}~. $$ Hence either $F=\{0\}$ or $\cl F=E$. Since $f(A)\neq0$, $F$ must be different from $\{0\}$ and since $f_b(A)\neq0$ and $F\sbe\ker f_b(A)$, $F$ cannot be dense.
2. We decompose $A\in\Hom(E)$ as follows: $A=B+iC$ where $B\colon=(A+A^*)/2$ and $C\colon=(A-A^*)/2i$. $B$ and $C$ are self-adjoint and for all $g\in G$ we have: $\Psi(g)(B+iC)\Psi(g)^{-1}=B+iC$. Since $\Psi$ is unitary both $B_1\colon=\Psi(g)B\Psi(g)^{-1}$ and $C_1\colon=\Psi(g)C\Psi(g)^{-1}$ are self-adjoint and for all $x\in E$: $\la(B-B_1)x,x\ra+i\la(C-C_1)x,x\ra=0$, it follows from the facts that $\la(B-B_1)x,x\ra,\la(C-C_1)x,x\ra\in\R$, that $\la(B-B_1)x,x\ra=\la(C-C_1)x,x\ra=0$ and thus by lemma $B_1=B$ and $C_1=C$. $\eofproof$

This gives us also an immediate generalization of corollary and corollary:

Every irreducible unitary representation of a commutative group $G$ is of dimension $1$ and thus every character is a homomorphism $\chi:G\rar S^1$. A unitary representation $\Psi:G\rar\UU(E)$ is irreducible if and only if every bounded self-adjoint operator $A:E\rar E$ commuting with all $\Psi(g)$, $g\in G$, is a multiple of the identity, i.e. $\{A\in\Hom_G(\Psi):\,A^*=A\}=\R1_E$.

Next we prove the generalization of proposition for compact groups. In order to achieve this we have to make a little digression

A vector $x_0\in E$ is called a cyclic vector of the unitary representation $\Psi:G\rar\UU(E)$ if its orbit $\{\Psi(g)x_0:\,g\in G\}$ is a dense subset of $E$. If $\Psi$ admits a cyclic vector, then we say that $\Psi$ is cyclic.

Since the space $F$ generated by the orbit is clearly invariant, every non zero vector of an irreducible unitary representation is a cyclic vector and thus $\Psi$ is cyclic.

Every unitary representation of a group $G$ in a Hilbert-space $E$ is the Hilbert-space sum of a pairwise orthogonal cyclic family of unitary sub-representations $(\Psi_\a)$.

$\proof$ Let ${\cal H}$ be the set of all families $(F_\a)$ of closed, pairwise orthogonal, invariant subspaces such that for all $\a$: $\Psi|F_\a$ is cyclic. Since $\{0\}\in{\cal H}$ we have: ${\cal H}\neq\emptyset$. Moreover, put $$ (F_\a)_{\a\in I}\leq(G_\b)_{\b\in J}:\Lrar\forall\a\in I\ \exists\b\in J:\ G_\b\supseteq F_\a~. $$ then $({\cal H},\leq)$ is ordered inductively and thus by Zorn's lemma there is a maximal family $E_\g$. Suppose $\cl{\bigoplus E_\g}\neq E$, then there is some $x_0\in(\bigoplus E_\g)^\perp$ such that $\norm{x_0}=1$. Define $E_0$ to be the closure of $\lhull{\Psi(g)x_0:\ g\in G}$, then we have $E_0\perp E_\g$, because for any $x\in E_\g$: $$ \la\Psi(g)x_0,x\ra=\la x_0,\Psi(g^{-1})x\ra=0~. $$ Thus the family $E_0\cup(E_\g)$ is strictly greater than the family $(E_\g)$, which contradicts the maximality of $(E_\g)$. $\eofproof$

Every unitary representation $\Psi$ of a compact group $G$ in a Hilbert space $(E,\la.,.\ra)$ is the Hilbert-space sum of pairwise orthogonal, finite dimensional and irreducible unitary sub-representations.

$\proof$ By proposition we may assume that $\Psi$ is cyclic with cyclic vector $x_0$, $\norm{x_0}=1$. Let $\mu$ be the normalized Haar measure on $G$ and define another Euclidean product by $$ [x,y]:=\int_G\la\Psi(g)x,x_0\ra\la x_0,\Psi(g)y\ra\,\mu(dg)~. $$ Since $x_0$ is a cyclic vector, this is indeed an Euclidean product; moreover $$ |[x,y]|\leq\norm{\Psi(g)x}\,\norm{\Psi(g)y}=\Vert x\Vert\,\norm y $$ and thus there exists a strictly positive self-adjoint operator $K:E\to E$, such that $\la Kx,y\ra=[x,y]$. We will show that $K$ is compact: So let $x_n$ be any sequence in the unit ball of $E$ which converges weakly to $x$, then \begin{eqnarray*} \norm{Kx_n}^2 &=&\la Kx_n,Kx_n\ra =[x_n,Kx_n]\\ &=&\int_G\la\Psi(g)x_n,x_0\ra\la x_0,\Psi(g)Kx_n\ra\,\mu(dg)\\ &=&\int_G\la\Psi(g)x_n,x_0\ra\la K\Psi(g^{-1})x_0,x_n\ra\,\mu(dg)\\ &=&\int_G\la\Psi(g)x_n,x_0\ra \int_G\la\Psi(h)\Psi(g^{-1})x_0,x_0\ra\la x_0,\Psi(h)x_n\ra\,\mu(dh)\,\mu(dg)~. \end{eqnarray*} By bounded convergence, this converges to $$ \int_G\la\Psi(g)x,x_0\ra \int_G\la\Psi(h)\Psi(g^{-1})x_0,x_0\ra\la x_0,\Psi(h)x\ra\,\mu(dh)\,\mu(dg) =\norm{Kx}^2 $$ Thus $$ \norm{Kx_n-Kx}^2 =-\la Kx_n,Kx\ra-\la Kx,Kx_n\ra+\norm{Kx_n}^2+\norm{Kx}^2 \to0~. $$ By the spectral theorem for compact self-adjoint operators we can find a sequence of pairwise orthogonal subspaces $E_n$ of finite dimension and strictly positive numbers $\l_n$, such that $$ E=\cl{\bigoplus E_n} \quad\mbox{und}\quad K|_{E_n}=\l_n id_{E_n} $$ The eigen-spaces $E_n$ of $K$ are $\Psi$-invariant: we will prove that $K$ commutes with all operators $\Psi(g)$. \begin{eqnarray*} \la K\Psi(g)x,y\ra &=&[\Psi(g)x,y] =\int_G\la\Psi(hg)x,x_0\ra\la x_0,\Psi(h)y\ra\,\mu(dh)\\ &=&\int_G\la\Psi(h)x,x_0\ra\la x_0,\Psi(hg^{-1})y\ra\,\mu(dh)\\ &=&[x,\Psi(g^{-1})y] =\la Kx,\Psi(g^{-1})y\ra =\la\Psi(g)Kx,y\ra~. \end{eqnarray*} Finally by proposition we may decompose the finite dimensional representations $\Psi(.)|E_n$ into the direct sum of pairwise orthogonal irreducible sub-representations. $\eofproof$

The Great Orthogonality Theorem

Now the proof of the great orthogonality theorem for compact groups just mimics the proof of the great orthogonality theorem for finite groups in finite dimensional spaces; instead of normalized sums we simply use integration with respect to the normalized Haar measure!

Suppose $\Psi$ and $\Phi$ are two irreducible unitary representations of the compact group $G$ in the Hilbert spaces $E$ and $F$ respectively. 1. If $\Psi$ and $\Phi$ are not equivalent, then for all $g\in G$, all $x,u\in E$ and all $y,v\in F$: \begin{eqnarray*} &&\int_G\la\Psi(h)x,u\ra\la\Phi(h^{-1}g)y,v\ra\,\mu(dh)=0 \quad\mbox{i.e.}\\ &&\int_{G}\Psi(h)\otimes\Phi(h^{-1}g)\,\mu(dh)=0~. \end{eqnarray*} where $\otimes$ is the tensor product, cf. section, and the Euclidean product on $E\otimes F$ satisfies $\la x\otimes y,u\otimes v\ra=\la x,u\ra\la y,v\ra$.
2. If $\Psi=\Phi$ and $\dim(E)=l$, then for all $g\in G$, all $x,u\in E$ and all $y,v\in F$: \begin{eqnarray*} &&\int_G\la\Psi(h)x,u\ra\la\Psi(h^{-1}g)y,v\ra\,\mu(dh) =\frac1l\la\Psi(g)y,u\ra\la x,v\ra~. \quad\mbox{i.e.}\\ &&\Big(\int_G\Psi(h)\otimes\Psi(h^{-1}g\Big)x\otimes y\,\mu(dh) =\Big(\frac1l\Psi(g)\otimes1\Big)y\otimes x~. \end{eqnarray*}

$\proof$ This is just a copy of the proof of theorem: Fixing $g\in G$, $u\in E$ and $y\in F$ we define a linear operator $J:E\to F$ by $$ Jx\colon=\int_G\la\Psi(h)x,u\ra\Phi(h^{-1}g)y\,\mu(dh)~. $$ Then for all $g^\prime\in G$ we have \begin{eqnarray*} J\Psi(g^\prime)x &=&\int_G\la\Psi(h)\Psi(g^\prime)x,u\ra\Phi(h^{-1}g)y\,\mu(dh)\\ &=&\int_G\la\Psi(hg^\prime)x,u\ra\Phi(h^{-1}g)y\,\mu(dh)\\ &=&\int_G\la\Psi(h)x,u\ra\Phi(g^\prime h^{-1}g)y\,\mu(dh) =\Phi(g^\prime)Jx~. \end{eqnarray*} Thus for all $g^\prime\in G$: $J\Psi(g^\prime)=\Phi(g^\prime)J$. Since $\Psi(g)^*=\Psi(g^{-1})$, it follows that for all $g^\prime\in G$: $\Psi(g^\prime)^{-1}J^*=J^*\Phi(g^\prime)^{-1}$, thus: $\Psi(g^\prime)J^*=J^*\Phi(g^\prime)$ and therefore we get for the operator $R=J^*J:E\to E$: $$ R\Psi(g^\prime) =J^*\Phi(g^\prime)J =\Psi(g^\prime)J^*J =\Psi(g^\prime)R~. $$ By theorem we know that $E$ and $F$ are finite dimensional and by Schur's Lemma: $R=\l id_{E}$, where $\l\geq0$. If there exist $g\in G$, $u\in E$ and $y\in F$ such that $\l > 0$, then $R$ must be an isomorhism, thus $J$ is one-one and as $J(E)$ is a non-trivial $G$-invariant sub-space of $F$: $J(E)=F$, which means that $\Psi$ and $\Phi$ are equivalent.
2. If $\Psi=\Phi$, then $J=\l id_E$ and thus $\l=l^{-1}\tr J$. The trace of $J$ is \begin{eqnarray*} \tr J &=&\int_G\la\Psi(h^{-1}g)y,\Psi(h)^*u\ra\,\mu(dh)\\ &=&\int_G\la\Psi(h^{-1})\Psi(g)y,\Psi(h^{-1})u\ra\,\mu(dh)\\ &=&\int_G\la\Psi(g)y,u\ra\,\mu(dh) =\la\Psi(g)y,u\ra~. \end{eqnarray*} Hence for all $x,v\in E$: $$ \int_G\la\Psi(h)x,u\ra\la\Psi(h^{-1}g)y,v\ra\,\mu(dh) =\la Jx,v\ra =\frac1l\la\Psi(g)y,u\ra\la x,v\ra~. $$ $\eofproof$

If $\Phi$ is equivalent to $\Psi$, i.e. $\Phi=U\Psi U^*$ for some isometry $U$, then $Jx=\tfrac1n\la\Psi(g)U^*y,u\ra Ux$, i.e. $J$ is a multiple of $U$.

Let $\psi\colon=\tr\Psi$ and $\vp\colon=\tr\Phi$ be the characters of $\Psi$ and $\Phi$ respectively - cf. section, then both are continuous and we obtain from the above relations and $\tr(\Psi(h)\otimes\Phi(h^{-1}g))=\psi(h)\vp(h^{-1}g)$: \begin{equation} \int_G\psi(h)\vp(h^{-1}g)\,\mu(dh)=0 \quad\mbox{and}\quad \int_G\psi(h)\psi(h^{-1}g)\,\mu(dh)=\frac1l\psi(g)~. \end{equation} In particular for $g=e$ we infer by observing $\psi(h^{-1})=\cl{\psi(h)}$ and $\psi(e)=l$: \begin{equation} \int_G\psi(h)\cl{\vp(h)}\,\mu(dh)=0 \quad\mbox{and}\quad \int_G|\psi(h)|^2\,\mu(dh)=1~. \end{equation} Which can be stated equivalently \begin{equation}\label{goteq1}\tag{GOT1} \psi*\vp=0,\quad \psi*\psi=\tfrac1l\psi, \quad\psi(e)=l \quad\mbox{and}\quad \la\psi,\vp\ra=0,\quad \norm{\psi}=1~. \end{equation} where $l$ is the dimension of the irreducible representation $\Psi$, which features the character $\psi$, i.e. $\psi(g)=\tr\Psi(g)$.

Two irreducible unitary representations $\Psi$ and $\Phi$ of a compact group $G$ are equivalent, if and only iff their characters coincide.

$\proof$ Suppose $\Psi$ and $\Phi$ are not equivalent but their characters coincide, then we infer from the relation \eqref{goteq1}: $0=\psi*\psi(e)=l^{-1}\psi(e)=1$. $\eofproof$

A finite dimensional representation $\Psi:G\rar\UU(E)$ is irreducible iff $\tr\Psi$ is a character of $G$, which exactly holds iff $\norm{\tr\Psi}=1$. Cf. corollary.

The results of subsection on isotypic components and SALCs also hold for compact groups!

If $\Psi:G\rar\UU(E)$ and $\Phi:G\rar\UU(F)$ are representations, then $$ \la\tr\Psi,\tr\Phi\ra=\dim\Hom_G(\Psi,\Phi), $$ in particular $\dim\Hom_G(\Psi)=\sum n_j^2$, where $n_j$ is the multiplicity of the irreducible representation featuring the character $\chi_j$. Cf. corollary.

Suppose $\Psi,\Phi$ are unitary representations of the compact group $G$ in the Hilbert-spaces $E$ and $F$ respectively. If $A:E\rar F$ denotes any bounded linear operator, then putting $$ J_g(A)x\colon=\int_G\Phi(h^{-1}g)A\Psi(h)x\,\mu(dh) $$ we get a bounded linear operator $J_g(A)\in\Hom_G(\Psi,\Phi)$ 2. If $A\in\Hom_G(\Psi,\Phi)$, then $A^\sharp\colon=J_e(A)=A$. 3. The mapping $J_e:\Hom(E,F)\rar\Hom_G(\Psi,\Phi)$ is a projection onto $\Hom_G(\Psi,\Phi)$. 4. If both $\Psi$ and $\Phi$ are irreducible, then $$ A^\sharp=\left\{\begin{array}{cl} 0&\mbox{if $\Psi$ and $\Phi$ are not equivalent}\\ \frac{\tr A}{\dim E}\,1_E&\mbox{if $\Psi=\Phi$} \end{array}\right. $$

$\proof$ Cf. proposition. $\eofproof$

Peter-Weyl Theorem

Cf. F. Peter and H. Weyl. The Peter-Weyl Theorem generalizes the Character Theorem for finite groups to arbitrary compact groups.

A sub-algebra of $L_2(G)$ and its $\g$-invariant sub-spaces

Suppose $\Psi:G\rar\UU(E)$ is an irreducible representation of the compact group $G$ in the $l$-dimensional Hilbert-space $E$. Then, as in the case of a finite group we may define the coordinate-functions $$ \psi_{jk}(g)\colon=\la\Psi(g)e_k,e_j\ra~. $$ i.e. $(\psi_{jk}(g))_{j,k=1}^l$ is the matrix of $\Psi(g)$ with respect to the orthonormal basis $e_1,\ldots,e_l$. Since $\Psi$ is unitary, we have: $$ \psi_{jk}(g) =\la\Psi(g)e_k,e_j\ra =\la e_k,\Psi(g)^*e_j\ra =\la e_k,\Psi(g)^{-1}e_j\ra =\cl{\la\Psi(g^{-1})e_j,e_k\ra} =\cl{\psi_{kj}(g^{-1})}~. $$ If $\Phi$ is another inequivalent irreducible representation with coordinate-function $\vp_{rs}$, then the great orthogonality theorem simply states that: \begin{equation}\label{pwteq1}\tag{PWT1} \psi_{jk}*\vp_{rs}=0,\quad \psi_{jk}*\psi_{rs}=\tfrac1l\psi_{js}\d_{kr}~. \end{equation} Putting $g=e$, we infer \begin{equation}\label{pwteq2}\tag{PWT2} \psi_{jk}(e)=\d_{jk},\quad \la\psi_{jk},\vp_{rs}\ra=0,\quad \la\psi_{jk},\psi_{rs}\ra=\tfrac1l\d_{jr}\d_{ks}~. \end{equation} Thus the $l^2$ coordinate-functions of $\Psi$ form an orthogonal set and coordinate-functions for inequivalent irreducible representation are mutually orthogonal.
Now let $\r$ be any element in the dual $\wh G$ of $G$ and let $\Psi_\r$ be an irreducible representation of class $\r$ of the compact group $G$ in some Hilbert-space $E_\r$ of dimension $n(\r)$. We choose some orthonormal basis $e(\r)_j$, $1\leq j\leq n(\r)$, of $E_\r$ and define the continuous coordinate-functions $\psi(\r)_{jk}:G\to\C$ by $\psi(\r)_{jk}\colon=\la\Psi_\r(g)e(\r)_k,e(\r)_j\ra$. For the dual representation we therefore get: $$ \psi(\bar\r)_{jk}=\bar\psi(\r)_{jk}~. $$ In order to simplify notation we drop the argument $\r$ whenever there is not to much a chance of confusion. For any $\r\in\wh G$ put \begin{equation}\label{pwteq3}\tag{PWT3} \ell(\r)_k\colon=\lhull{\psi(\r)_{jk}:\,j=1,\ldots,n(\r)} \quad\mbox{and}\quad {\cal A}(\r)\colon=\lhull{\psi(\r)_{jk}:j,k\leq n(\r)} \end{equation}

The following holds:

The space ${\cal A}(\r)$ is a sub-space of $C(G)$ and it only depends on $\r\in\wh G$.
$\bar\psi(\r)_{kj}(g)=\psi(\r)_{jk}(g^{-1})$.
$\psi(\r)_{jk}(e)=\d_{jk}$.
For all $\r\neq\r^\prime$: $\psi(\r)_{jk}*\psi(\r^\prime)_{rs}=0$.
$\psi(\r)_{jk}*\psi(\r)_{rs}=\frac1{n(\r)}\d_{kr}\psi(\r)_{js}$.
For all $\r\neq\r^\prime$: $\la\psi(\r)_{jk},\psi(\r^\prime)_{rs}\ra=0$.
$\la\psi(\r)_{jk},\psi(\r)_{rs}\ra=\frac1{n(\r)}\d_{jr}\d_{ks}$.

$\proof$ Compare corollary. We only need to prove the first assertion; all the other assertions are direct consequences of either \eqref{pwteq1} or \eqref{pwteq2}: So let $\Phi$ be any irreducible representation equivalent to $\Psi$, i.e.: $\Phi(g)=U\Psi(g)U^*$, for some isometry $U$. Then $$ \la\Phi(g)e_k,e_j\ra =\la U\Psi(g)U^*e_j,e_k\ra =\la\Psi(g)U^*e_j,U^*e_j\ra\in{\cal A}(\r)~. $$ $\eofproof$

Now we look at the left-regular representation $$ \g:G\rar\Hom(L_2(G))\quad\g(g)f(x)=f(g^{-1}x), $$ restricted to the sub-spaces ${\cal A}(\r)$ and $\ell(\r)_k$: For $g\in G$ and $\r\in\wh G$ let $\Psi^{\r}(g)$ denote the matrix $$ \Psi^{\r}(g)=(\psi(\r)_{jk})_{j,k=1}^{n(\r)}\in\UU(n(\r)) $$ That`s simply the matrix of $\Psi_\r(g)$ with respect to a given basis $e(\r)_1,\ldots.e(\r)_{n(\r)}$ of $E_\r$. Hence $\Psi^{\r}:G\to U(n(\r))$ is a unitary representation, i.e. $\Psi^{\r}(gh)=\Psi^{\r}(g)\Psi^{\r}(h)$ or \begin{equation}\label{pwteq4}\tag{PWT4} \psi(\r)_{jk}(gh)=\sum_{m=1}^{n(\r)}\psi(\r)_{jm}(g)\psi(\r)_{mk}(h)~. \end{equation} Analogously to corollary we get

${\cal A}(\r)$ is a two sided ideal in $L_2(G)$, i.e. for all $f\in L_2(G)$ and all $\psi\in{\cal A}(\r)$: $f*\psi,\psi*f\in{\cal A}(\r)$. 2. For all $k=1,\ldots,n(\r)$ the spaces $\ell(\r)_k$ are invariant under the left-regular representation $\g$ and $\g(.)|\ell(\r)_k$ is equivalent to the contra-gredient representation $\bar\Psi^\r$, i.e. it is of class $\bar\r$. Thus the restriction of the left-regular representation to ${\cal A}(\r)$ is the sum of $n(\r)$ irreducible representations all equivalent to the contra-gredient representation $\bar\Psi^\r$.

$\proof$ We will drop the $\r$! The first statements follow from proposition and it remains to verify that $\ell(\r)_k$ is $\g$-invariant and equivalent to the contra-gredient representation. By \eqref{pwteq4} we have: $$ \g(g)\psi_{jk}(x) =\psi_{jk}(g^{-1}x) =\sum_m\psi_{jm}(g^{-1})\psi_{mk}(x) \quad\mbox{i.e.}\quad \g(g)\psi_{jk}\in\ell_k~. $$ By proposition the functions $\sqrt{n}\,\psi_{jk}$, $1\leq j\leq n$ form an orthonormal basis fo $\ell_k$. Let us denote by $(\g_{st})$ the matrix of the linear mapping $\g(g):\ell_k\rar\ell_k$ with respect to this basis, i.e. by proposition: \begin{eqnarray*} \g_{st}&=&n\la\g(g)\psi_{tk},\psi_{sk}\ra\\ &=&n\sum_m\psi_{tm}(g^{-1})\la\psi_{mk},\psi_{sk}\ra =\psi_{ts}(g^{-1}) =\bar\psi_{st}(g)~. \end{eqnarray*} Hence $\g(.)|\ell_k$ is of class $\bar\r$. $\eofproof$

For any compact group $G$ the following statements hold:

$(\ell(\r)_k)$, $1\leq k\leq n(\r)$, is a family of minimal left ideals in $L_2(G)$.
For $\r\neq\r^\prime$ we have: $\ell(\r)_k*\ell(\r^\prime)_k=\{0\}$.
$\ell(\r)_k*\ell(\r)_j=\ell(\r)_j$.
The spaces $\ell(\r)_k$ are pairwise orthogonal in $L_2(G)$ and $$ {\cal A}(\r)=\bigoplus_{k=1}^{n(\r)}\ell(\r)_k~. $$
$\{\sqrt{n(\r)}\psi(\r)_{jk}:\,j,k=1,\ldots,n(\r)\}$ is an orthonormal basis of ${\cal A}(\r)$.
$({\cal A}(\r),+,*)$ is isomorphic to $(\Ma(n(\r),\C),+,\cdot)$.
$({\cal A}(\r))$, $\r\in\wh G$, is a family of minimal two sided ideals in the convolution algebra $L_2(G)$.

$\proof$ Again we drop the $\r$! 1. For each finite Borel measure $\nu$ we have by \eqref{pwteq4}: \begin{eqnarray*} \nu*\psi_{jk}(x) &=&\int_G\psi_{jk}(y^{-1}x)\,\nu(dy)\\ &=&\sum_m\Big(\int_G\psi_{jm}(y^{-1})\,\nu(dy)\Big)\psi_{mk}(x)\in\ell_k~. \end{eqnarray*} Hence $\ell_k$ is a left ideal in the convolution algebra $L_2(G)$. Now assume $\ell$ is another left ideal different from $\{0\}$ and $\ell\sbe\ell_k$. Then $\ell$ is a finite dimensional sub-space of $L_2(G)$, in particular it`s closed and therefore we have for any $x\in G$: $\g(x)f=\d_x*f\in\ell$. Hence $\ell$ is invariant under the left-regular representation. As $\g:G\to U(\ell(\r)_k)\in\bar\r$ is irreducible we must have: $\ell=\ell_k$.
2. and 3. immediately follow from proposition 4. and 5. respectively.
4. These assertions follow from proposition 5., 2., 3., 7. and 5.
5. and 6. follow from proposition 7.
7. Let $E^{jk}$ denote the matrix with just one entry different from zero and that`s the entry in its $j$-th row and $k$-th column, which is $1$. Then we have $E^{jk}\cdot E^{rs}=\d_{kr}E^{js}$ and by proposition 5.: $\psi_{jk}*\psi_{rs}=(1/n)\d_{kr}\psi_{js}$. Hence an isomorphism is given by $$ \psi(\r)_{jk}\mapsto\frac1{n(\r)}E^{jk} $$ 8. ${\cal A}(\r)$ is a two sided ideal by corollary. Thus by 7. the assertion follows from the fact that the algebra $(\Ma(n(\r),\C),+,\cdot)$ is simple. $\eofproof$

For $\psi(\r)_k\colon=n(\r)\psi(\r)_{kk}\in\ell(\r)_k$ prove that:

$\psi(\r)_j*\psi(\r)_k=\psi(\r)_j\d_{jk}$.
$\psi(\bar\r)_k=\bar\psi(\r)_k$.
$\psi(\r)_k(g^{-1})=\bar\psi(\r)_k(g)$.
$\psi(\r)_k(e)=n(\r)$.
$\norm{\psi(\r)_k}_2=\sqrt{n(\r)}$.
$\ell(\r)_k=L_2(G)*\psi(\r)_k=\ell(\r)_k*\psi(\r)_k$.

As the neutral elmement in $(\Ma(n(\r),\C),+,\cdot)$ with respect to multiplication is the identity, the neutral element in ${\cal A}(\r)$ is \begin{equation}\label{pwteq5}\tag{PWT5} u(\r) \colon=\sum_{k=1}^{n(\r)}n(\r)\psi(\r)_{kk} =n(\r)\tr\Psi^\r =n(\r)\tr\Psi_\r~. \end{equation}

Prove that $\la f*g,h\ra=\la g,\check f*h\ra$, where $\check f(x)\colon=\cl{f(x^{-1})}$. Conclude that the convolution operator $A_f:g\mapsto f*g$ is normal iff $\check f*f=f*\check f$ - which is always true for e.g. class functions $f$ - and self-adjoint iff for all $x\in G$: $\bar f(x)=f(x^{-1})$.

As $G$ is unimodular, this follows from proposition. However here is a copy of the argument: \begin{eqnarray*} \int_G f*g(y)\bar h(y)\,\mu(dy) &=&\int_G\int_G f(y)g(y^{-1}x)\bar h(x)\,\mu(dx)\,\mu(dy) =\int_G\int_G f(y)g(x)\bar h(yx)\,\mu(dx)\,\mu(dy)\\ &=&\int_G g(x)\int_G \cl{\check f(y^{-1})h(yx)}\,\mu(dy)\,\mu(dx) =\int_G g(x)\int_G \cl{\check f(y)h(y^{-1}x)}\,\mu(dy)\,\mu(dx)\\ &=&\int_G g(x)\cl{\check f*h}(x)\,\mu(dx) =\la g,\check f*h\ra~. \end{eqnarray*}

For the neutral element $u(\r)\in{\cal A}(\r)$ the following holds:

$u(\bar\r)=\bar u(\r)$ and $u(\r)(e)=n(\r)^2$.
$u(\r)(g^{-1})=\bar u(\r)(g)$.
The linear map $P_{\bar\r}:L_2(G)\to{\cal A}(\r)$, $P_{\bar\r}f:=f*u(\r)=u(\r)*f$ is the orthogonal projection onto ${\cal A}(\r)$.

$\proof$ 1. 2. All the assertions follow immediately from the definition and exam.
3. First we verify that $u$ is a class function, i.e. $u(x)=u(gxg^{-1})$ - cf. e.g. proposition: $$ u(gxg^{-1}) =n\tr\Psi(gxg^{-1}) =n\tr(\Psi(g)\Psi(x)\Psi(g^{-1})) =n\tr(\Psi(g)\Psi(x)\Psi(g)^{-1}) =n\tr(\Psi(x)) =u(x)~. $$ This implies $u(gx)=u(xg)$ and thus \begin{eqnarray*} f*u(x) &=&\int_G f(xg^{-1})u(g)\,\mu(dg) =\int_G f(g^{-1})u(gx)\,\mu(dg)\\ &=&\int_G f(g^{-1})u(xg)\,\mu(dg) =\int_G f(g)u(xg^{-1})\,\mu(dg) =u*f(x)~. \end{eqnarray*} By exam we have: $P_{\bar\r}(L_2(G))={\cal A}(\r)$. On the other hand the adjoint of the convolution operator $f\mapsto f*u$ is the convolution operator $f\mapsto f*\skew1\check{\bar u}$. Hence by 2. $P_{\bar\r}$ is self-adjoint and $P_{\bar\r}^2f=(f*u)*u=f*(u*u)=f*u=P_{\bar\r} f$. Therefore $P_{\bar\r}$ is the orthogonal projection onto ${\cal A}(\r)$. $\eofproof$

Isotypic components

Let $\Psi:G\to U(E)$ a unitary representation of $G$ and $(\Psi_j)_{j\in J}$ the family of its irreducible sub-representations in sub-spaces $E_j$ of $E$. Then is the orthogonal direct sum of Hilbert-spaces $$ E=\bigoplus_{j\in J}E_j~. $$ For any $\r\in\wh G$ we denote by $E(\r)$ the isotypic component $$ E(\r):=\bigoplus\{E_j:\,\Psi_j\in\r\}~. $$ We obviously have $E=\bigoplus\{E(\r):\,\r\in\wh G\}$. The following is the counterpart for compact group to lemma

The space $E(\r)$ is the smallest closed $\Psi$-invariant sub-space of $E$ containing all $\Psi$-invariant closed sub-spaces $F$ such that $\Psi(.)|F$ is irreducible and of class $\r$. Moreover, any irreducible sub-representation of $\Psi(.)|E(\r)$ is of class $\r$.

$\proof$ $E(\r)$ is by definition closed and $\Psi$-invariant. We proceed as in lemma: Let $J:F\rar E$ be the canonical inclusion and $P_j:E\rar E_j$ the orthogonal projection. Then both $J\Psi(g)=\Psi(g)J$ and $P_j\Psi(g)=\Psi(g)P_j$ and thus $P_jJ:F\rar E_j$ is intertwining. Since $F$ and $E_j$ are irreducible and finite dimensional we infer that $P_jJ$ is either an isomorphism or the zero map. If it is zero, then $F\sbe E_j^\perp$ and if $P_jJ$ is an isomorphism, then $\Psi(.)|F$ and $\Psi(.)|E_j$ are equivalent. Hence $F$ is orthogonal to all sub-spaces $E_j$ whose class is different from $\r$, which means that $F$ must be a sub-space of $E(\r)$. This also shows that any irreducible sub-representation $\Psi(.)|F$ of $\Psi(.)|E(\r)$ is of class $\r$. $\eofproof$

As in the finite group case it follows that the decomposition into isotypic components does not depend on the decomposition in irreducible sub-spaces and this in turn implies that the decomposition in isotypic components is unique. However the decomposition of an isotypic component in its irreducible sub-spaces is not unique - cf. e.g. exam.

Let $G$ be a compact group with dual $\wh G$. Then there is a family ${\cal A}(\r)$, ${\r\in\wh G}$, of minimal two sided finite dimensional ideals in the convolution algebra $L_2(G)$ such that $L_2(G)$ is the orthogonal Hilbert-space direct sum of algebras, i.e. for all $\r^\prime\neq\r$: ${\cal A}(\r)*{\cal A}(\r^\prime)=\{0\}$ and $$ L_2(G)=\bigoplus_{\r\in\wh G}{\cal A}(\r)~. $$ Each algebra ${\cal A}(\r)$ is a sub-space of $C(G)$ and it is isomorphic to $\Ma(n(\r),\C)$. If $u(\r)$ denotes the neutral element in ${\cal A}(\r)$, i.e. $u(\r)=n(\r)\chi_\r$, then the orthogonal projection $P_{\bar\r}:L_2(G)\to{\cal A}(\r)$ is given by $$ P_{\bar\r} f =f*u(\r)=u(\r)*f =n(\r)\g(\chi_\r)f, $$ where $\g(\chi_\r)$ denotes the extension of the left-regular representation to $L_1(G)$ - cf. \eqref{chaeq2}. ${\cal A}(\r)$ are also the isotypic components $E(\bar\r)$ of the left-regular representation $\g:G\rar\Hom(L_2(G))$, $\g(g)f(x)\colon=f(g^{-1}x)$.

$\proof$ By corollary it suffices to show that the isotypic components $E(\bar\r)$ of the left-regular representation is a subset of ${\cal A}(\r)$. So let $F$ be a sub-space of $L_2(G)$ such that $\g(.)|F$ is of class $\bar\r$ - in particular $F$ is finite dimensional and thus closed. By corollary $\g(.)|F$ is equivalent to $\g(.)|\ell(\r)_k$. Hence there is an isomorphism $T:\ell(\r)_k\rar F$ such that $\g(g)|F\circ T=T\circ \g(g)|\ell(\r)_k$. Putting $f_j\colon=T(\psi_{jk})$ we get by \eqref{pwteq4}: $$ \g(g)f_j =\g(g)T\psi(\r)_{jk} =T\g(g)\psi(\r)_{jk} =T\Big(\sum_{l=1}^{n(\r)}\psi(\r)_{jl}(g^{-1})\psi(\r)_{lk}\Big) =\sum_{l=1}^{n(\r)}\psi(\r)_{jl}(g^{-1})f_l $$ Since by proposition: $\psi(\r)_{jk}(g^{-1})=\bar\psi(\r)_{kj}(g)$ and $\la\psi(\r)_{mm},\psi(\r)_{jl}\ra=\d_{ml}\d_{mj}/n(\r)$ we conclude that \begin{eqnarray*} u(\r)*f_j(x) &=&\int_G u(\r)(g)f_j(g^{-1}x)\,\mu(dg) =\int_G u(\r)(g)\g(g)f_j(x)\,\mu(dg)\\ &=&\int_G\Big(n(\r)\sum_m\psi(\r)_{mm}(g)\Big)\Big(\sum_l\psi(\r)_{jl}(g^{-1})f_l(x)\Big)\,\mu(dg)\\ &=&n(\r)\sum_{l,m}\Big(\int_G\psi(\r)_{mm}(g)\psi(\r)_{jl}(g^{-1})\mu(dg)\Big)f_l(x)\\ &=&n(\r)\sum_{l,m}\la \psi(\r)_{mm},\psi(\r)_{lj}\ra f_l(x) =\sum_{l,m}\d_{ml}\d_{mj} f_l(x) =f_j(x)~. \end{eqnarray*} This shows that $F\sbe{\cal A}(\r)$ and therefore $E(\bar\r)\sbe{\cal A}(\r)$. $\eofproof$

Characters

For each $\r\in\wh G$ we denoted by $u(\r)$ the neutral element in the minimal two sided ideal ${\cal A}(\r)$. The continuous function \begin{equation}\label{chaeq1}\tag{CHA1} \chi_\r:G\to\C,\quad\chi_\r\colon=n(\r)^{-1}u(\r)=\tr\Psi_\r~. \end{equation} is said to be the character of the class $\r$ or the character of ${\cal A}(\r)$ or a character of $G$.

As $\tr(\Psi(gxg^{-1}))=\tr(\Psi(g)\Psi(x)\Psi(g)^{-1})=\tr\Psi(x)$ characters are class functions, i.e. $\chi\in Z(G)$.

If $f:G\rar\C$ is any function then $$ f_c(x)\colon=\int_G f(yxy^{-1})\,\mu(dy) $$ is a class function. 2. Prove that $P:f\mapsto f_c$ is the orthogonal projection $L_2(G)\rar Z(G)$.

We summarize the properties of characters:

Let $\chi_\r$ be a character of the class $\r$. Then the following holds:

$\chi_{\bar\r}=\bar\chi_\r$.
$\chi_\r(e)=n(\r)$.
For $\r\neq\r^\prime$ we have: $\chi_\r*\chi_{\r^\prime}=0$.
$\chi_\r*\chi_\r=\frac1{n(\r)}\chi_\r$.
For $\r\neq\r^\prime$ we have: $\la\chi_\r,\chi_{\r^\prime}\ra=0$.
$\Vert\chi_\r\Vert_2=1$ and $\Vert\chi_\r\Vert_\infty=n(\r)$.
$\chi_\r$, $\r\in\wh G$ is an orthonormal basis of the center $Z(G)$ of $L_2(G)$.

$\proof$ 1. to 6. follow from exam. As for the last assertion we notice that the center of $\Ma(n(\r),\C)$ ist $\C 1$ and thus the center of $L_2(G)$ is by theorem just the orthogonal Hilbert-space direct sum $$ \bigoplus_{\r\in\wh G}\C\chi_\r, $$ in other words: the characters form an orthonormal basis of the center $Z(G)$ of the convolution algebra $L_2(G)$. $\eofproof$

If $\Psi:G\rar\UU(E)$ is any representation in a finite dimensional Hilbert-space $E$, then the multiplicity of the irreducible representation of class $\r$ is $$ m(\r)=\la\tr\Psi(g),\chi_\r\ra~. $$

Characters of some commutative groups

For a commutative group $G$ the dual $\wh G$ is the set of characters, i.e. the set of all homomorphisms $\chi:G\rar S^1$, which by itself is a group, the group operation being pointwise multiplication.

Every character $\chi:\R^n\rar S^1$ of the additive group $(\R^n,+)$ is given by $$ \chi(x)=e^{i\la x,y\ra}, $$ where $y\in\R^n$. Thus the dual of $(\R^n,+)$ is isomorphic to $(\R^n,+)$.
Every character $\chi:\TT^n\rar S^1$ is given by $$ \chi(z)=\prod z_j^{y_j}, $$ where $y\in\Z^n$. Thus the dual of $\TT^n$ is isomorphic to $(\Z^n,+)$.
Every character $\chi:\Z^n\rar S^1$ is given by $$ \chi(x)=\prod z_j^{x_j}, $$ where $z\in\TT^n$. Thus the dual of $\Z^n$ is isomorphic to $\TT^n$.
Suppose $n,p\in\N$, $p\geq2$. Every character $\chi:\Z_p^n\rar S^1$ is given by $$ \chi(x)=e^{i\la x,y\ra}, $$ where $y\in\Z_p^n$. Thus the dual of $\Z_p^n$ is isomorphic to $\Z_p^n$.

An algebra homomorphism on $L_1(G)$

We `extend` a unitary representation $\Psi:G\rar\UU(E)$ to an algebra homomorphism $\Psi:(L_1(G),+,*)\rar\Hom(E)$: For any $\nu\in M(G)$ we define a bounded linear operator $\Psi(\nu)\in\Hom(E)$ by \begin{equation}\label{chaeq2}\tag{CHA2} \forall x\in E:\quad \Psi(\nu)x\colon=\int_G\Psi(g)x\,\nu(dg)~. \end{equation} In particular: $\Psi(\d_g)x=\Psi(g)x$. In case $d\nu=f\,d\mu$ for some $f\in L_1(G)$ we write $\Psi(f)$ instead of $\Psi(\nu)$. If $f_n\geq0$ is a sequence in $L_1(G)$ such that $\int f_n\,d\mu=1$ and the probability measures with density $f_n$ converge weakly to $\d_g$ for some $g\in G$, then $\Psi(f_n)x$ converges to $\Psi(g)x$. In analogy to finite groups we get the following generalization of lemma:

For all $f,g\in L_1(G)$: $\Psi(f)\Psi(g)=\Psi(f*g)$, i.e. $f\mapsto\Psi(f)$ is an algebra homomorphism from $(L_1(G),+,*)$ into $\Hom(E)$.
If $f\in Z(G)$, then $\Psi(f)\in\Hom_G(\Psi)$.
If $f\in Z(G)$ and if $\Psi$ is irreducible with character $\chi$ and dimension $l$, then: $$ \Psi(f)=\frac1l\la f,\bar\chi\ra id_E \quad\mbox{and in particular}\quad \Psi(l\bar\chi)=id_E~. $$

$\proof$ 1. Putting $z=xy$ i.e. $x=zy^{-1}$ we have by definition and Fubini: \begin{eqnarray*} \Psi(f)\Psi(g) &=&\int_G\int_G f(x)g(y)\Psi(x)\Psi(y)\,\mu(dx)\,\mu(dy)\\ &=&\int_G\int_G f(zy^{-1})g(y)\Psi(z)\,\mu(dy)\,\mu(dz) =\int_G f*g(z)\Psi(z)\,\mu(dz) =\Psi(f*g)~. \end{eqnarray*} 2. If $f$ is in the center $Z(G)$ of $L_2(G)$, then for all $F\in L_1(G)$: $f*F=g*F$ and thus by 1.: $\Psi(f)\Psi(F)=\Psi(F)\Psi(f)$ and thus by the remarks above: for all $g\in G$: $\Psi(f)\Psi(g)=\Psi(g)\Psi(f)$.
3. If in addition $\Psi$ is irreducible, then by Schur’s Lemma: $\Psi(f)=\a id_E$, i.e.: $\tr\Psi(f)=\a l$; on the other hand: $$ \tr\Psi(f) =\int_G f(g)\chi(g)\,\mu(dg) =\la f,\bar\chi\ra~. $$ Hence $\a=\la f,\bar\chi\ra/l$. $\eofproof$

The following extends exam:

Suppose $\Psi:G\rar\UU(E)$ is irreducible and $\dim E=l$. 1. Show that for all $g\in G$ and all $x,y\in E$: $$ \int_G\Psi(g)\otimes\Psi(g)^*x\otimes y\,\mu(dg)=\frac1ly\otimes x~. $$ 2. Let $e_1,\ldots,e_n$ be an orthonormal basis for $E$ and put $E^{jk}x\colon=\la x,e_k\ra e_j$ and define the exchange operator or swap operator $P_{ex}\in\Hom(E\otimes E)$ by $$ P_{ex} \colon=\sum_{j,k}E^{kj}\otimes E^{jk} =\sum_{j,k}E^{kj}\otimes E^{kj*}~. $$ Verify that $P_{ex}(x\otimes y)=y\otimes x$, that the eigen-values of the unitary operator $P_{ex}$ are $\pm1$ and that for all $A\in\Hom(E)$: $[A\otimes A,P_{ex}]=0$. 3. Cf. exam and prove that for all $A\in\Hom(E)$: $\tr_1((A\otimes1)P_{ex})=A$.

Suppose $\Psi:G\rar\UU(E)$ is irreducible and $\dim E=l$. For any $A\in\Hom(E)$ put $f(x)\colon=l\tr(A\Psi(x))$. Prove by means of exam that $A=\Psi(f)$ and conclude that $\Psi:L_2(G)\rar\Hom(E)$ is onto, i.e. any operator $A$ is a linear combination of the unitary operators $\Psi(x)$, $x\in G$.

Density of characters in $Z(G)\cap C(G)$

Finally we will prove that the characters span a dense sub-space of $C(G)\cap Z(G)$ of $C(G)$. By lemma $Z(G)\cap C(G)$ is a closed sub-space of $C(G)$.

For $f\in C(G)$ and $g\in L_2(G)$ we have $$ f*g=\sum_{\r\in\wh G}\left(\sum_{j,k=1}^{n(\r)}n(\r)\la g,\psi(\r)_{jk}\ra f*\psi(\r)_{jk}\right) $$ and the series converges uniformly on $G$. In particular if $g\in Z(G)$: $$ f*g=\sum_{\r\in\wh G}\la g,\chi_\r\ra f*\chi_\r $$ and the series converges uniformly on $G$.

$\proof$ By theorem the set $$ \Big\{\sqrt{n(\r)}\psi(\r)_{jk}:\,\r\in\wh G,j,k\leq n(\r)\Big\} $$ is an orthonormal basis for $L_2(G)$ and for continuous $f$ the mapping $L_2(G)\to C(G)$, $g\mapsto f*g$ is a bounded linear map. $\eofproof$

1. $f\in Z(G)$ if and only if for all $x\in G$: $\mu(\{y:f(xy)=f(yx)\})=1$. 2. A continuous function $f:G\to\C$ lies in the center $Z(G)$ of $L_2(G)$ if and only if for all $x,y\in G$: $f(xy)=f(yx)$.

$\proof$ If one of the functions $f,g$ is continuous, then $f*g,g*f\in C(G)$. Now $$ f*g(x)=\int_Gf(xy)g(y^{-1})\,\mu(dy)\\ \quad\mbox{and}\quad g*f(x)=\int_Gf(yx)g(y^{-1})\,\mu(dy)~. $$ If $f\in Z(G)$ and $g\in C(G)$, then for all $x\in G$: $f*g(x)=g*f(x)$ and therefore $\mu(\{y:f(xy)=f(yx)\})=1$. $\eofproof$

For any neighborhood $U$ of $e$ there is a continuous function $f_c:G\rar[0,1]$ satisfying: $f_c\in Z(G)$, $f_c(e)=1$ and $\supp(f_c)\sbe U^3$.

$\proof$ First we will show that for each neighborhood $U$ of $e$ there is a neighborhood $U_0$ of $e$ such that for all $x\in G$. $xU_0x^{-1}\sbe U^3$. We choose a neighborhood $U$ of $e$ such that $x_0\notin U^3$. For each $x\in G$ we can find a neighborhood $U_x$ of $e$ satisfying $xU_xx^{-1}\sbe U$. Consequently there are neighborhoods $V_x$ of $x$ such that $V_xU_xV_x^{-1}\sbe U^3$. As $\{V_x:\,x\in G\}$ is an open covering of $G$, there are $x_1,\ldots,x_n\in G$ such that $$ G=\bigcup_{j=1}^n V_{x_j}. $$ Put $U_0=\bigcap_jU_{x_j}$, then for all $x\in G$: $V_{x_j}U_0V_{x_j}^{-1}\sbe U^3$, i.e. all $x\in G$: $xU_0x^{-1}\sbe U^3$.
Now choose a continuous function $f:G\rar[0,1]$ such that $f(e)=1$ and $\supp(f)\sbe U_0$ and define (cf. exam $$ f_c(x):=\int_f(yxy^{-1})\,\mu(dy). $$ Then $f_c(e)=1$, $f_c(x)\in[0,1]$, $\supp(f_c)\sbe\bigcup_{y\in G}y^{-1}U_0y\sbe U^3$ and $$ f_c(xy) =\int_Gf_0(zxyz^{-1})\,\mu(dz) =\int_G f_0(zyxyy^{-1}z^{-1})\,\mu(dz) =f_c(yx) $$ $\eofproof$

Suppose $G$ is a compact group with dual $\wh G$. Then the following holds

$\{\chi_\r:\,\r\in\wh G\}$ separates the points of $G$.
$\lhull{\chi_\r:\,\r\in\wh G}$ is a dense sub-space of the sub-space $C(G)\cap Z(G)$ of $C(G)$.

$\proof$ 1. It suffices to prove that $\{\chi_\r:\,\r\in\wh G\}$ separates any $x_0\neq e$ and $e$: by lemma we can find for any neighborhood $U$ of $e$ a non negative function $f_c\in C(G)\cap C(G)$ such that $f_c(e)=1$ and $\supp(f_c)\sbe U$. Hence $f_c*f_c(e) > 0$ and $f_c*f_c(x_0)=0$ if $x_0\notin UU^{-1}$. Suppose that for all $\r$: $\chi_\r(e)=\chi_\r(x_0)$, then by corollary: $$ 0 =f_c*f_c(x_0) =\sum_{\r\in\wh G}\la f_c,\chi_\r\ra f_c*\chi_\r(x_0) =\sum_{\r\in\wh G}\la f_c,\chi_\r\ra f_c*\chi_\r(e) =f_c*f_c(e) > 0~. $$ 2. We verify that the closure $C\colon=\lhull{\chi_\r:\,\r\in\wh G}$ is closed under point wise multiplication of functions, i.e. $\chi_\r\cdot\chi_{\r^\prime}\in C$. Now there is a finite subset $\L\sbe\wh G$ and $n_\l\in\N$ such that $$ \chi_\r\cdot\chi_{\r^\prime} =\tr(\Psi_\r\otimes\Psi_{\r^\prime}) =\sum_{\l\in\L}n_\l\chi_\l\in C $$ The constant function $1$ is the character of the trivial representation and $\bar\chi_\r=\chi_{\bar\r}$. By the proof of the Stone-Weierstraß Theorem $C$ is dense in the sub-space $C(G)\cap Z(G)$ of $C(G)$. $\eofproof$

We don`t have to invoke Stone-Weierstraß, because we can construct an approximate central unit: Suppose $g\in Z(G)$ is continuous and $\e > 0$. Since $g$ is uniformly continuous there is a symmetric neighborhood $U$ of $e$ such that for all $xy^{-1}\in U$: $|g(y)-g(x)| < \e$. By lemma there is a continuous function $f_U:G\rar\R_0^+$ such that $\supp(f_U)\sbe U$, $f_U\in Z(G)$ and $\int f_u\,d\mu=1$. Thus for all $x\in G$: \begin{eqnarray*} |f_U*g(x)-g(x)| &=&\Big|\int f_U(xy^{-1})g(y)-g(x)\,\mu(dy)\Big|\\ &\leq&\int f_U(xy^{-1})|g(y)-g(x)|\,\mu(dy) < \e \end{eqnarray*} Now $f_U*g\in Z(G)$ and by corollary $f_U*g\in\cl{C}$ and thus $g\in\cl{C}$.

Projection Theorem

In this section we prove a formula for the orthogonal projection on the isotypic components of a given representation $\Psi:G\rar\UU(E)$ in a finite dimensional Hilbert-space $E$. In case of the left-regular representation the Peter-Weyl Theorem gave us both the isotypic components: ${\cal A}(\bar\r)$ and the orthogonal projections: $$ P_{\r}f =n(\r)\chi_{\bar\r}*f =n(\r)\g(\chi_{\bar\r})f =n(\r)\g(\bar\chi_{\r})f $$ where $\g(\chi)$ is the extension of the left-regular representation to the convolution algebra $L_1(G)$ - cf. subsection.

For all $f,g\in L_1(G)$ and all $F\in L_\infty(G)$ we have by definition of the convolution: $$ \int_G F.f*g\,d\mu=\int_G\int_G F(xy)f(x)g(y)\,\mu(dx)\,\mu(dy)~. $$

Let $\Psi:G\rar\UU(E)$ be a representation of $G$ in the finite dimensional Hilbert-space $E$. Then the orthogonal projection $P_\r$ onto the isotypic component $E(\r)$ is given by: $$ P_\r\colon =n(\r)\Psi(\bar\chi_\r) =n(\r)\int_G\cl{\chi_m(h)}\Psi(h)\,\mu(dh)~. $$ where $n(\r)$ is the dimension of the irreducible representation of class $\r$.

$\proof$ 1. We verify that $P_\r$ is self-adjoint: Since $\chi_\r(h^{-1})=\bar\chi_\r(h)$ we get $$ P_\r^* =n(\r)\int_G\chi_\r(h)\Psi(h)^*\,\mu(dh) =n(\r)\int_G \bar\chi_j(h^{-1})\Psi(h^{-1})\,\mu(dh) =P_\r $$ 2. $P_\r$ is a projection, the family of these projections is mutually orthogonal and $\Psi(.)|\im P_\r$ is a representation: From lemma and proposition we infer that: $$ P_\r P_{\r^\prime} =n(\r)n(\r^\prime)\Psi(\bar\chi_\r*\bar\chi_{\r^\prime}) =n(\r)\d_{jk}\Psi(\bar\chi_\r) =\d_{\r \r^\prime}P_\r~. $$ Thus the operators $P_\r$, $\r\in\wh G$, are pairwise orthogonal projections onto sub-spaces. Put $E_\r\colon=\im P_\r$; since $\bar\chi_\r$ is a class function, we have by lemma: $\Psi(g)P_\r=P_\r\Psi(g)$. Hence $\Psi_\r(.)\colon=\Psi(.)|E_\r$ is a representation of $G$ in $E_\r$.
3. All sub-representations of $\Psi_\r$ are of class $\r$: for all $\r^\prime,\r$ we have by exam: \begin{eqnarray*} \la\tr\Psi_\r(.),\chi_{\r^\prime}\ra &=&\la\tr\Psi(.)P_\r,\chi_{\r^\prime}\ra =\la\tr(P_\r\Psi(.)),\chi_{\r^\prime}\ra\\ &=&\int_G\tr(P_\r\Psi(g))\cl{\chi_{\r^\prime}(g)}\,\mu(dg)\\ &=&n(\r)\int_G\int_G\cl{\chi_\r(h)} \cl{\chi_{\r^\prime}(g)}\tr\Psi(hg)\,\mu(dh)\,\mu(dg)\\ &=&n(\r)\int_G\cl{\chi_\r*\chi_{\r^\prime}}(g)\tr\Psi(g)\,\mu(dg)~. \end{eqnarray*} By proposition this is $0$ for $\r^\prime\neq\r$ and $\la\Psi(.),\chi_\r\ra$ for $\r^\prime=\r$. But the latter is just the multiplicity $m(\r)$ of $\r$ in $E(\r)$. Hence any irreducible sub-representation of $\Psi_\r$ must have character $\chi_\r$ and its multiplicity is $m(\r)$. By lemma the means that the isotypic component $E(\r)$ equals $E_\r$. $\eofproof$

Compute the trace of $P_\r$ from it`s definition and show that it is given by: $\tr P_\r=n(\r)m(\r)$, where $m(\r)$ denotes the multiplicity of the irreducible representation of class $\r$ in $\Psi$.

Fourier Transform

This is more or less just a copy of section for finite groups! Again we start with an arbitrary unitary representation $\Psi:G\rar\UU(E)$ of the compact group $G$ in a Hilbert-space $E$. For every $\nu\in M(G)$ remember the definition \eqref{chaeq2} of $\Psi(\nu)$: $$ \Psi(\nu)\in\Hom(E)\quad x\mapsto\int_G\Psi(g)x\,\mu(dg)~. $$ and for $f\in L_1(G)$: $d\nu\colon=f\,d\mu$ and $$ \Psi(f)\colon=\int_Gf(g)\Psi(g)x\,\mu(dg)~. $$ We sum up the properties of this extension of the representation $\Psi:G\rar\UU(E)$ to $M(G)$:

The mapping $\Psi:M(G)\to\Hom(E)$ is linear and we have:

It`s a contraction, i.e.: $\norm{\Psi(\nu)}\leq\Vert\nu\Vert$ - on the left hand side we mean the operator-norm!
It`s an extension, i.e.: for all $g\in G$: $\Psi(\d_g)=\Psi(g)$.
It`s an algebra homomorphism, i.e. for all $\nu,\l\in M(G)$: $\Psi(\nu*\l)=\Psi(\nu)\Psi(\l)$.
If for all $g\in G$ and all Borel sets $A$: $\nu(A)=\nu(gAg^{-1})$, then $\Psi(f)\in\Hom_G(\Psi)$. Moreover, if $\Psi$ is irreducible with character $\chi$ and dimension $l$, then: $$ \Psi(\nu)=\frac1l\la f,\bar\chi\ra id_E \quad\mbox{and in particular}\quad \Psi(l\bar\chi)=id_E~. $$
$\Psi(\skew1\check{\bar\mu})=\Psi(\mu)^*$, where $\check\nu(A)\colon=\nu(A^{-1})$.

$\proof$ By lemma we only need to check the properties 1., 4. and 5.: As for the first we get by the definition of the norm on $M(G)$: $$ \norm{\Psi(\nu)x} \leq\int\norm{\Psi(g)x}\,|\nu|(dg) =\Vert x\Vert\,\Vert\nu\Vert~. $$ 4. If $\nu(A)=\nu(gAg^{-1})$, then $\nu(gA)=\nu(Ag)$ and thus for all $\l\in M(G)$ and all $F\in C(G)$: $\int F(xy)\,\nu(dx)=\int F(yx)\,\nu(dx)$, from which we infer that $$ \int F(xy)\,\nu(dx)\,\l(dy) =\int F(yx)\,\nu(dx)\,\l(dy) =\int F(xy)\,\nu(dy)\,\l(dx) =\int F(xy)\,\l(dx)\,\nu(dy), $$ i.e. $\nu*\l=\l*\nu$. By 3.: $\Psi(\nu)\Psi(\l)=\Psi(\l)\Psi(\nu)$ and by 2. we get for all $g\in G$: $\Psi(\nu)\Psi(g)=\Psi(g)\Psi(\nu)$. If in addition $\Psi$ is irreducible, then $\Psi(\nu)=\a.id_E$ and as in lemma: $\tr\Psi(\nu)=\a l$; on the other hand: $$ \tr\Psi(\nu) =\int_G \chi(g)\,\nu(dg) =\la f,\bar\chi\ra \quad\mbox{hence}\quad \a=\frac1l\la f,\bar\chi\ra~. $$ 5. The last property is checked by straight insertion \begin{eqnarray*} \la\Psi(\nu)^*x,y\ra &=&\la x,\Psi(\nu)y\ra =\cl{\la\Psi(\nu)y,x\ra}\\ &=&\int\cl{\la\Psi(g)y,x\ra}\,\bar\nu(dg) =\int\la x,\Psi(g)y\ra\,\bar\nu(dg)\\ &=&\int\la\Psi(g^{-1})x,y\ra\,\bar\nu(dg) =\int\la\Psi(g)x,y\ra\,\skew1\check{\bar\nu}(dg) =\la\Psi(\skew1\check{\bar\nu})x,y\ra~. \end{eqnarray*} $\eofproof$

Now let $\Psi^\r$, $r\in\wh G$, denote a complete set of pairwise in-equivalent irreducible representations of $G$ of dimension $n(\r)$. Denote by $\psi(\r)_{jk}$ the coordinate functions of the representation $\Psi^\r$ and by $\Psi_\r(g)$ the matrix $(\psi(\r)_{jk}(g))_{j,k=1}^{n(\r)}$, i.e. the homomorphism $\Psi_\r:G\rar\UU(n(\r))$ is just a matrix version of $\Psi^\r$. Then the function $\wh f:\wh G\rar\C$ defined by \begin{equation}\label{ftreq1}\tag{FTR1} \F_G(f)(\r)\colon=\wh f(\r)\colon=\int_G f(h)\Psi_\r(h)^*\,\mu(dh) =\int_G f(h)\Psi_\r(h^{-1})\,\mu(dh)\in\Ma(n(\r),\C) \end{equation} is said to be the Fourier transform of $f$ at $\r\in\wh G$. $\wh f(\r)$ is more or less the same as $\Psi_\r(f)$. Of course, we may as well define the Fourier transfrom of any measure $\nu\in M(G)$ $$ \F_G(\nu)(\r)\colon=\wh\nu(\r) \colon=\int_G \Psi_\r(g)^*\,\nu(dg)\in\Ma(n(\r),\C), $$ which, for example gives us: $\wh\d_x(\r)=\Psi_\r(x)^*$. Since $\cl{\psi(\r)_{kj}(h)}=\psi(\r)_{jk}(h^{-1})$, the entries of this matrix are given by $$ \wh\nu(\r)_{jk} =\int_G \psi(\r)_{jk}(h^{-1})\,\nu(dh) =\la\psi(\r)_{kj},\nu\ra~. $$ We know that the functions $\sqrt{n(\r)}\psi(\r)_{jk}$, $\r\in\wh G$, $j,k\in\{1,\ldots,n(\r)\}$, form an orthonormal basis for $L_2(G)$, and therefore: \begin{eqnarray*} f&=&\sum_{\r\in\wh G}\sum_{j,k}\Big\langle f,\sqrt{n(\r)}\psi(\r)_{jk}\Big\rangle\sqrt{n(\r)}\psi(\r)_{jk}\\ &=&\sum_{\r\in\wh G}\sum_{j,k}n(\r)\la f,\psi(\r)_{jk}\ra\psi(\r)_{jk} =\sum_{\r\in\wh G}\sum_{j,k}n(\r)(\wh f(\r))_{kj}\psi(\r)_{jk}~. \end{eqnarray*} Thus we have the Fourier inversion formula: \begin{equation}\label{ftreq2}\tag{FTR2} \forall x\in G:\quad f(x)=\sum_{\r\in\wh G}n(\r)\tr\Big(\wh f(\r)\Psi_\r(x)\Big), \end{equation} and for the norm we obtain: \begin{equation}\label{ftreq3}\tag{FTR3} \Vert f\Vert^2 =\sum_{\r\in\wh G}\sum_{j,k}n(\r)\tr\Big(\wh f(\r)\wh f(\r)^*\Big) =\sum_{\r\in\wh G}\sum_{j,k}n(\r)\norm{\wh f(\r)}_{HS}^2~. \end{equation} We will also identify $\wh G$ with the set of all in-equivalent irreducible representations $\Psi$ and we will therefore write for the Fourier transform of $f\in L_2(G)$ and its inverse: \begin{eqnarray*} \forall\Psi\in\wh G&\quad& \wh f(\Psi)=\int_G f(x)\Psi(x)^*\,\mu(dx)\quad\mbox{and}\\ \forall x\in G&\quad& f(x)=\sum_{\Psi\in\wh G}\dim(\Psi)\tr\Big(\wh f(\Psi)\Psi(x)\Big)~. \end{eqnarray*} On the discrete set $\wh G$ we define the function space ${\cal M}(\wh G)\colon=\prod_{\r\in\wh G}\Ma(n(\r),\C)$ and the Hilbert-space $$ L_2(\wh G) \colon=\{A\in{\cal M}(\wh G):\,\sum_{\r\in\wh G}n(\r)\norm{A(\r)}_{HS}^2 < \infty\} $$ with Euclidean product $$ \la A,B\ra\colon=\sum_{\r\in\wh G}n(\r)\tr(A(\r)B(\r)^*)~. $$

Verify that the Fourier transform of $\check f(x)\colon=\cl{f(x^{-1})}$ is given by the adjoint $\wh f(\Psi)^*$ of $\wh f(\Psi)$.

If $f$ is central, i.e. $f\in Z(G)$ and $\chi\colon=\tr\Psi$ the character of $\Psi$, then by proposition: $$ \wh f(\Psi)=\frac1{\dim(\Psi)}\la f,\chi\ra $$ and thus the Fourier inversion formula yields: $$ f(x) =\sum_{\Psi\in\wh G}\dim(\Psi)\tr\Big(\frac1{\dim(\Psi)}\la f,\chi\ra\Psi(x)\Big) =\sum_{\Psi\in\wh G}\la f,\chi\ra\chi(x)~. $$ which restates that the characters form an orthonormal basis of $Z(G)$ - of course, the convergence of the right hand side series is in $L_2(G)$ and in general not pointwise!

Convolution and Fourier transform

Suppose $G$ is a commutative compact group and $f,g\in L_2(G)$, then the Fourier coefficient of the convolution of these functions is given by \begin{eqnarray*} \wh{f*g}(\chi) &=&\la f*g,\chi\ra =\int_G\int_G f(xy^{-1})g(y)\cl{\chi(x)}\,\mu(dy)\,\mu(dx)\\ &=&\int_G\int_G f(xy^{-1})\cl{\chi(xy^{-1})}g(y)\cl{\chi(y)}\,\mu(dx)\,\mu(dy)\\ &=&\int_G \la f,\chi\ra g(y)\cl{\chi(y)}\,\mu(dy) =\la f,\chi\ra\la g,\chi\ra =\wh f(\chi)\wh g(\chi)~. \end{eqnarray*} Thus in the non-commutative case we conjecture that the Fourier transform of the convolution of two functions $f,g\in L_2(G)$ is the product of the matrices $\wh f(\Psi)$ and $\wh g(\Psi)$, i.e. \begin{equation}\label{ftreq4}\tag{FTR4} \forall\Psi\in\wh G:\quad \wh{f*g}(\Psi)=\wh f(\Psi)\wh g(\Psi)~. \end{equation} Indeed, we just need to copy the proof above replacing the homomorphism $\chi:G\rar S^1$ with the irreducible representations $\Psi:G\rar\UU(\dim(\Psi))$: \begin{eqnarray*} \wh{f*g}(\Psi) &=&\int_G f*g(x)\Psi(x^{-1})\,\mu(dx)\\ &=&\int_G\int_Gf(xy^{-1})g(y)\Psi(x^{-1}y)\Psi(y^{-1})\,\mu(dy)\,\mu(dx)\\ &=&\int_G \Big(\int_Gf(xy^{-1})\Psi(x^{-1}y)\,\mu(dx)\Big)g(y)\Psi(y^{-1})\,\mu(dy)\\ &=&\wh f(\Psi)\int_G g(y)\Psi(y^{-1})\,\mu(dy) =\wh f(\Psi)\wh g(\Psi)~. \end{eqnarray*}

The Fourier transform of $L_xf(y)\colon=f(x^{-1}y)$ is given by $$ \wh d_x(\Psi)\wh f(\r)=\Psi(x)^*\wh f(\Psi)~. $$

The Fourier transform $f\mapsto\wh f$ is an isometry of $L_2(G)$ onto the Hilbert-space $L_2(\wh G)$.

$\proof$ We know from \eqref{ftreq3}, that $f\mapsto\wh f$ is an isometry into. It remains to verify that this isometry is onto. To this end put for given $A\in L_2(\wh G)$: $$ f(x)=\sum_\Psi\dim(\Psi)\tr\Big(A(\Psi)\Psi(x)\Big)~. $$ Then we claim that for each $\Phi\in\wh G$: $\wh f(\Phi)=A(\Phi)$. We verify this by straight calculation: $$ \wh f(\Phi)=\int_G\sum_\Psi\dim(\Psi)\tr\Big(A(\Psi)\Psi(x)\Big)\Phi(x)^*\,\mu(dx) $$ and therefore we need to calculate the matrix entries of the right hand side. By \eqref{caleq2} we get \begin{eqnarray*} \dim(\Psi)\Big(\sum_{x}\tr\Big(A(\Psi)\Phi(x)\Big)\Phi(x)^*\Big)_{rs}\,\mu(dx) &=&\dim(\Psi)\int_G\sum_{j,k}a_{jk}(\Psi)\psi_{kj}(x)\vp_{rs}(x^{-1})\,\mu(dx)\\ &=&\dim(\Psi)\sum_{j,k}a_{jk}(n)\psi_{kj}*\vp_{rs}(e) =\d_{\Psi,\Phi}a_{rs}(\Phi) \end{eqnarray*} i.e. $\wh f(\Phi)=A(\Phi)$. $\eofproof$

The Fourier transform $\F_G:L_1(G)\to{\cal M}(\wh G)$ is injective.

$\proof$ Suppose $\F_G(f)=0$, then for all $g\in L_2(G)$: $f*g\in L_2(G)$ and $\F_G(f*g)=0$ and therefore $f*g=0$. As this holds for all $g\in L_2(G)$ we must have $f=0$. $\eofproof$

For each $f\in L_1(G)$ and each $\e > 0$ the set $$ \Big\{\r\in\wh G:\,\norm{\wh f(\r)} > \e\Big\} \quad\mbox{is finite.} $$

$\proof$ 1. Assume $f\in L_2(G)$, then we conclude from theorem: \begin{eqnarray*} |\{\r\in\wh G:\,\norm{\wh f(\r)} > \e\}| &\leq&|\{\r\in\wh G:\,\norm{\wh f(\r)}_{HS}^2 > \e^2\}|\\ &\leq&\e^{-2}\sum_\r\norm{\wh f(\r)}_{HS}^2 \leq\e^{-2}\norm f_2^2 \end{eqnarray*} 2. For $f\in L_1(G)$ we have by proposition: $\norm{\wh f(\r)}\leq\norm f_1$ and thus \begin{eqnarray*} &&|\{\r\in\wh G:\,\norm{\wh f(\r)} > \e\}|\\ &&\leq|\{\r\in\wh G:\,\norm{(\wh f-\wh g)(\r)} > \e\}|+ |\{\r\in\wh G:\,\norm{\F_G(g)(\r)} > \e\}|\\ &&\leq|\{\r\in\wh G:\,\norm{f-g}_1 > \e\}|+ |\{\r\in\wh G:\,\norm{\wh g(\r)} > \e\}|~. \end{eqnarray*} Now choose $g\in L_2(G)$ such that $\norm{f-g}_1 < \e$. $\eofproof$

Representation of Compact Groups