Representations of Compact Groups (Part 2)
I wrote about some basic results concerning representaions of compact lie groups earlier. Today, I'll be exploring this topic more. I'll prove the PeterWeyl Theorem, which helps us understand irreducible representations of a compact group using the regular representation of $G$ on $L^2(G)$. First, I'll start by generalizing characters to matrix coefficients.
Matrix Coefficients
Let $n = \dim U$ and $m = \dim W$. Let $V \subseteq U \otimes W$ be a nonzero subrepresentation. Since $V$ is nonzero, it must contain some nonzero vector $u \otimes w$. Since $U$ is an irrep of $G$, the smallest $G$invariant subspace containing $\C u$ is all of $U$. So we can find $g_1, \ldots, g_n \in G$ such that $g_1u, \ldots, g_nu$ is a basis of $U$. Similarly, we can find $h_1, \ldots, h_m \in H$ such that $h_1w, \ldots, h_mw$ is a basis of $w$. Since $V$ is a subrepresentation, we know that $(g,h) (u \otimes w) \in V$ for any $(g,h) \in G \times H$. Thus, $(g_iu) \otimes (h_jw) \in V$ for all $i,j$. This means that $V$ contains a basis for $U \otimes W$, so $V$ is all of $U \otimes W$. Thus, the only nonzero $(G \times H)$subrepresentation of $U \otimes W$ is the entire space $U \otimes W$. So $U \otimes W$ is irreducible.
Earlier, we gave $\End(V) = \Hom(V,V)$ the structure of a $G$representation, whenever $(V, \phi)$ is a $G$representation. We essentially conjugated the matrix by $\phi(g)$. Instead of multiplying by the same $\phi(g)$ on both sides, we can actually define an action of $(G \times G)$ on this space, and this action will turn out to be useful. We define $(G \times G$)actions on $\End(V)$ and $C(G)$ by setting $(g,h) \cdot T = \phi(h) \circ T \circ \phi(g)^{1}$ and $((g,h) \cdot f)(x) = f(g^{1}xh)$ respectively.
Let $T \in \End(V)$. Then $M_V(T)(g) = \tr(\phi(g) \circ T)$. Let $(h_1, h_2) \in G \times G$. \[\begin{aligned} M_V((h_1,h_2)\cdot T)g &= \tr(\phi(g) \circ ((h_1,h_2) \cdot T))\\ &= \tr(\phi(g) \circ \phi(h_2) \circ T \circ \phi(h_1)^{1})\\ &= \tr(\phi(h_1^{1}gh_2) \circ T)\\ &= M_V(T)(h_1^{1}gh_2)\\ &= ((h_1,h_2)\cdot (M_V(T)))(g) \end{aligned}\] So $M_V$ is $(G \times G)$linear.
Clearly if $V$ is irreducible, then $V^*$ is also an irreducible $G$representation. By our lemma, this means that $V \otimes V^* \cong \End(V)$ is an irreducible $(G \times G)$representation. Since $M_V$ is $(G \times G)$linear, $\ker M_V$ must be a subrepresentation. Thus, $M_V$ is either zero or injective. $M_V$ is clearly nonzero since it sends the identity matrix to $\chi_V$, which is nonzero.
By observing that $M_V(\Id) = \chi_V$, we see that matrix coefficient maps generalize characters. Matrix coefficients share a lot of nice properties with characters. Just as our operations on representations gave us operations on characters, we can also find corresponding operations on matrix coefficient maps. And we will prove an orthogonality relationship between matrix coefficients generalizing the orthogonality of characters.
 $M_{V^*}(T^*) = M_V(T)^*$
 $M_{\overline V}(\overline T) = \overline{M_V(T)}$
 $M_{V \oplus W}(T \oplus S) = M_V(T) + M_W(S)$
 $M_{V \otimes W}(T \otimes S) = M_V(T) \cdot M_W(S)$
 $M_{\Hom(V,W)}(S \circ \bullet \circ T^*) = M_V(T)^* \cdot M_W(S)$
 $M_{V^G}(Av_G \circ T) = av(M_V(T))$
 $M_{V^*}(T^*)(g) = \tr(\phi(g^{1})^T \circ T^T) = \tr(\phi(g^{1}) \circ T) = M_V(T)^*(g)$
 $M_{\overline V}(\overline T)(g) = \tr(\overline{\phi(g)} \circ \overline T) = \overline{M_V(T)(g)}$
 The computation here looks a bit longer, but it's pretty straightforward \[\begin{aligned} M_{V \oplus W}(T \oplus S)(g) &= \tr((\phi(g) \oplus \psi(g)) \circ (T \oplus S))\\ & = \tr((\phi(g) \circ T) \oplus (\psi(g) \circ S))\\ &= \tr(\phi(g) \circ T) + \tr (\psi(g) \circ S)\\ &= M_V(T)(g) + M_W(S)(g) \end{aligned}\]
 This one also looks a bit long, but it's also straightforward \[\begin{aligned} M_{V \otimes W}(T \otimes S)(g) &= \tr((\phi(g) \otimes \psi(g)) \circ (T \otimes S))\\ & = \tr((\phi(g) \circ T) \otimes (\psi(g) \circ S))\\ &= \tr(\phi(g) \circ T) \cdot \tr (\psi(g) \circ S)\\ &= M_V(T)(g) \cdot M_W(S)(g) \end{aligned}\]
 This follows from the first and fourth identities, using the fact that $\Hom(V,W) \cong W \otimes V^*$.
 This computation is the trickiest, but it's still not too bad. It's pretty much all tricks we've used before \[\begin{aligned} M_{V^G}(Av_G \circ T)(g) &= \tr_{V^G}(\phi(g) \circ Av_G \circ T)\\ &= \tr_{V^G}(T \circ \phi(g) \circ Av_G) \end{aligned}\] Since $T \circ \phi(g) \circ Av_G$ acts as $T \circ \phi(g) \circ Av_G$ on $V^G$ and acts as $0$ on the orthogonal complement of $V^G$, we can trace over $V$ instead of $V^G$. So we see that \[\begin{aligned} M_{V^G}(Av_G \circ T)(g) &= \tr_V(T \circ \phi(g) \circ Av_G)\\ &= \tr_V\left( T \phi(g) \int_G \phi(h)dh\right)\\ &= \int_G \tr_V(T \circ \phi(gh))dh \end{aligned}\] Since our measure is leftinvariant, this is just equal to \[\begin{aligned} M_{V^G}(Av_G \circ T)(g) &= \int_G \tr_V(T \circ \phi(g))dg\\ &= av(\tr_V(T \circ \phi(g)))\\ &= av(M_V(T)) \end{aligned}\]
Let $E, F$ be nonisomorphic irreducible representations of $G$. Let $T \in \End(E)$ and $S \in \End(F)$. Then \[\inrp{M_F(S)}{M_E(T)} = 0\]
Just like the proof for characters, this proof is pretty simple given the constructions we defined above. \[\begin{aligned} \inrp{M_F(S)}{M_E(T)} &= av(M_F(S) \cdot \overline{M_E(T)})\\ &= av(M_F(S) \cdot M_{\overline E}(\overline T))\\ &= av(M_{F \otimes \overline E})(S \otimes \overline T)\\ &= M_{(F \otimes \overline E)^G}(Av_G \circ (S \otimes \overline T)) \end{aligned}\] Recall that $\overline E$ is isomorphic to $E^*$. Thus, $F \otimes \overline E \cong F \otimes E^* \cong \Hom(E,F)$. This isomorphism shows us that $(F \otimes \overline E)^G = \Hom(E,F)^G = \Hom_G(E,F)$, which is $0$ by Schur's lemma. Thus, the inner product must be $0$.
This orthogonality is a very nice result, but it only applies to distinct irreducible representations. You might wonder: what can we say about $\inrp{M_E(S)}{M_E(T)}$ for $S,T \in \End(E)$? It turns out that we can put an inner product on $\End(E)$ with respect to which $\inrp{M_E(S)}{M_E(T)} = \inrp ST$! This is about as nice a result as you could hope for. First, we'll construct the inner product, and then we'll show that the matrix coefficient map is unitary with respect to this inner product.
Let $E$ be an irreducible representation of $G$. A $G$invariant inner product on $E$ defines an isomorphism $E^* \xrightarrow{\sim} \overline E$. If $E$ is irreducible, these are both irreducible. So Schur's lemma tells us that there is only one such isomorphism, up to scalar multiplication. So $E$ has only one $G$invariant inner product, up to scalar multiplication. Thus, we have a welldefined adjoint map $S \mapsto S^\dagger$ for $S \in \End(V)$. Note that because our inner product is $G$invariant, $\phi(g)^\dagger = \phi(g^{1})$.
 The HilbertSchmidt inner product is $(G \times G)$invariant
 $M_E: \End(E) \to C(G)$ is unitary (up to a scalar factor) with respect to the HilbertSchmidt inner product on $\End(E)$ and the inner product we have been using on $C(G)$
 Recall that the $(G \times G)$ action on $\End(E)$ is given by $(g,h) \cdot T = \phi(g) \circ T \circ \phi(h)^{1}$. Thus, \[\begin{aligned} \inrp{(g,h)T}{(g,h)S} &= \tr_E(\phi(g) \circ T \circ \phi(h)^{1} \circ (\phi(g) \circ S \circ \phi(h)^{1})^\dagger)\\ &= \tr_E(\phi(g) \circ T \circ \phi(h)^{1} \circ \phi(h^{1})^\dagger \circ S^\dagger \circ \phi(g)^\dagger)\\ &= \tr_E(\phi(g) \circ T \circ \phi(h)^{1} \circ \phi(h) \circ S^\dagger \circ \phi(g^{1}))\\ &= \tr_E(\phi(g) \circ T \circ S^\dagger \circ \phi(g^{1}))\\ &= \tr_E(T \circ S^\dagger)\\ &= \inrp T S \end{aligned}\]
 We saw earlier that $\End(E)$ is an irreducible $(G \times G)$representation. Since $M_E$ is $(G\times G)$linear, this means it must be injective.
The PeterWeyl Theorem
One of the nice features of the representation theory of finite groups is that we have the regular representation of the group on itself, and the regular representation has a nice decomposition as a sum of all irreducible representations (up to isomorphism, counted with multiplicity). The PeterWeyl theorem is a generalization of this result to compact groups. Before proving the theorem, we need some preliminaries.
 $f$ is left$G$finite
 $f$ is right$G$finite
 $f$ is $(G \times G)$finite
 $f$ is in the image of the matrix coefficient map $M_V$ for some $V$

We will show that $4 \implies 3 \implies 2 \implies 4$. The proof that $4 \implies 3 \implies 1 \implies 4$ is essentially the same.
 $4 \implies 3:$ Since $M_V$ is a $(G \times G)$linear map and has a finitedimensional domain, every function in its image must be $(G \times G)$finite
 $3 \implies 2:$ The right$G$action is the action of $(1 \times G) \subseteq (G \times G)$, so $(G \times G)$finiteness implies right$G$finiteness.
 $2 \implies 4:$ Let $f$ be right$G$finite. Then $f$ is contained in a finitedimensional representation $V \subsetneq C(G)$ of $G$ (with the right$G$action). Let $\alpha \in V^*$ be the linear functional $\alpha(h) := h(e)$. Now, we see that \[M_V(f \otimes \alpha)(g) = \tr_V(\phi(g) \circ f \otimes \alpha) = \alpha(g \cdot f) = (g \cdot f)(e) = f(g)\] Thus, $f$ is in the image of the matrix coefficient map $M_V$
 $C(G)^{fin}$ is dense in $C(G)$
 $C(G)^{fin}$ is dense in $L^2(G)$
 $L^2(G)^{fin}$ is dense in $L^2(G)$
 For every $e \neq g \in G$, there exists an irreducible representation $(V, \phi)$ of $G$ such that $\phi(g) \neq \Id$
 $C(G)^{fin}$ separates points of $G$. That is to say, for every pair $g \neq h \in G$, there is a function $f \in C(G)^{fin}$ such that $f(g) \neq f(h)$
$1 \implies 2$: Since $C(G)$ is dense in $L^2(G)$, if $C(G)^{fin}$ is dense in $C(G)$, it must also be dense in $L^2(G)$.
$2 \implies 3$: Since $C(G)^{fin} \subseteq L^2(G)^{fin}$, then $C(G)^{fin}$ being dense implies that $L^2(G)^{fin}$ is dense as well.
$3 \implies 4$: Consider $e \neq g \in G$. We will start by finding a function in $L^2(G)^{fin}$ on which $g$ acts nontrivially. First, note that there must be some function $f \in C(G)$ such that $g \cdot f \neq f$. Now, suppose for contradictin that $g$ acts trivially on all of $L^2(G)^{fin}$. Since the action is continuous and $L^2(G)^{fin}$ is dense, $g$ must act trivially on all of $L^2(G)$, which is a contradiction. Thus, there is some $\tilde f \in L^2(G)^{fin}$ upon which $g$ acts nontrivially. Then $G \tilde f$ is contained in a finitedimensional subspace of $C(G)$ by definition of $L^2(G)^{fin}$. So $\tilde f$ is a vector in a finitedimensional representation of $G$ upon which $g$ acts nontrivially.
$4 \implies 5$: Let $g \neq h \in G$. By (4), there is some irreducible representation $(V, \phi)$ such that $\phi(gh^{1}) \neq \Id$. Consider $M_V(\phi(h^{1}))$. \[\begin{aligned} M_V(\phi(h^{1}))(g) &= \tr_V(\phi(g) \circ \phi(h^{1}))\\ &= \chi_V(gh^{1})\\ M_V(\phi(h^{1}))(h) &= \tr_V(\phi(h) \circ \phi(h^{1}))\\ &= \chi_V(\Id) \end{aligned}\] Since we can pick a metric on $V$ with respect to which $\phi(gh^{1})$ is unitary, it must have complex eigenvalues, and those eigenvalues must be roots of unity. Thus, the only way for $\tr_V(\phi(gh^{1}))$ to equal $\tr_V(\Id)$ is if all of the eigenvalues are 1, which means that $\phi(gh^{1}) = \Id$. But this is impossible. Thus, $\chi_V(gh^{1}) \neq \chi_V(\Id)$. So $M_V(\phi(h^{1}))$ separates $g$ and $h$. And by definition, $M_V(\phi(h^{1}))$ is in $C(G)^{fin}$.
$5 \implies 1$: We observed earlier that $C(G)^{fin}$ is a subalgebra of $C(G)$. Clearly $C(G)^{fin}$ contains a nonzero constant function. Thus, the Stone Weierstrass theorem tells us that $C(G)^{fin}$ is dense in $C(G)$ if and only if it separates points.
The operator norm of $T_K$ is \[\begin{aligned} \T_K\_{op} &:= \sup_{f \in L^2(X), \f\_{L^2(X)}=1}\T_K(f)\_{L^2(X)}\\ &= \sup_{\f\_{L^2(X)}=1} \left\x \mapsto \int_X K(x,y) f(y)\;dy\right\_{L^2(X)}\\ &= \sup_{\f\_{L^2(X)}=1} \left(\int_{X}\left(\int_X K(x,y) f(y) \;dy\right)^2 \;dx \right)^{1/2} \end{aligned}\] By Holder's inequality, \[\int_X K(x,y) f(y)\;dy \leq \left(\int_X K(x,y)^2\;dy\right)^{1/2} \f\_{L^2(X)}\] Therefore, \[\T_K\_{op} \leq \left(\int_X \int_X K(x,y)^2\;dydx\right)^2 = \K\_{L^2(X \times X)}\] Note that since $T_K$ is a bounded linear operator, it is continuous.
Next, we will show that $T_K$ is compact. Recall that the limit of a sequence of finiterank operators between Hilbert spaces is a compact operator. So it is sufficient to show that $T_K$ is the limit of a sequence of finiterank operators.
Consider the set of functions \[\left\{x,y \mapsto \sum_i f_i(x) g_i(y)\;\; f_i, g_i \in C(X)\right\} \subseteq C(X \times Y)\] This is clearly a subalgebra of $C(X \times Y)$ containing a nonzero constant function, so the StoneWeierstrass theorem tells us that these functions are dense in $C(X \times Y)$. Since $C(X \times Y)$ is dense in $L^2(X \times Y)$, these functions are also dense in $L^2(X \times Y)$. Thus, any $T_K$ is a limit of operators $T_{K_i}$ for $K_i$ in this subalgebra. Now, we just have to show that $T_{K_i}$ is finiterank. Since a linear combination of operators of finite rank must also have finite rank, it is sufficient to show that $T_{f_1f_2}$ has finite rank (in fact, it has rank 1). \[T_{f_1f_2}(f)(x) = \int_X f_1(x)f_2(y)f(y)\;dy \propto f_1(x)\]
Finally, we note that if $K(x,y) = \overline {K(y,x)}$, then $T_K$ is selfadjoint. \[\begin{aligned} \inrp{T_K(f)}{g} &= \int_X T_Kf(x) \overline{g(x)}\;dx\\ &= \int_X \int_X K(x,y) f(y)\;dy \;\overline {g(x)}\;dx\\ &= \int_X f(y) \int_X K(x,y) \overline {g(x)}\;dx\;dy\\ &= \int_X f(y) \overline{\int_X K(y,x) g(x)\;dx}\;dy\\ &= \int_X f(y) \overline {T_Kg(y)}\;dy\\ &= \inrp f {T_K(g)} \end{aligned}\]
Recall that for a compact selfadjoint operator $T: \mathcal{H} \to \mathcal{H}$ on a Hilbert space $\mathcal H$, the spectral theorem tells us that we have a decomposition of $\mathcal{H}$ into countably many orthogonal eigenspaces with real eigenvalues. In the case we care about, our Hilbert space is $L^2(G)$.
Note that if $k^* = \overline k$ (i.e. $k(g^{1}) = \overline{k(g)}$), then our convolution operator is selfadjoint.
Another nice property of convolution is that it is $G$linear with respect to the left$G$action on $C(G)$. (i.e. $g \cdot (f * k) = (g \cdot f)*k$) \[\begin{aligned} g \cdot (f * k)(x) &= (f * k) (g^{1}x)\\ &= \int_G k(y^{1}g^{1}x)f(y)\;dy\\ &= \int_G k(t^{1}x)f(g^{1}t)\;dt\\ &= (g \cdot f * k)(x) \end{aligned}\] Above, we used the fact that our measure on $G$ is leftinvariant
Constructing such a function $u_U$ is fairly straightforward. Pick any nonzero continuous realvalued function $w \in C(G)$ with support inside $U$. Consider $\tilde w =(w + w^*)^2$. This is clearly continuous, nonnegative, and $\tilde w = \tilde w^*$. Since it is continuous, $\int_G \tilde w\;d\mu$ is finite. So we can set \[u_U = \frac 1 {\int_G \tilde w\;d\mu} \tilde w\]
Now, we'll show that $u_U$'s convolution with $f$ satisfies the desired bound. \[ f*u_U(x)  f(x) = \int_G u_U(y^{1}x) f(x)\;dy  f(x) \] Since $u_U = u_U^*$, we can take the inverse of $u_U$'s argument. \[ f*u_U(x)  f(x) = \int_G u_U(x^{1}y) f(x)\;dy  f(x) \] Since our measure is leftinvariant, we can do a change of variables $t = x^{1}y$. \[ f*u_U(x)  f(x) = \int_G u_U(t) f(xt)\;dt  f(x) \] Since $u_U$ integrates, to 1, $f(x) = \int_G u_U(t) f(x)\;dt$. Thus, \[\begin{aligned} f*u_U(x)  f(x) &= \int_G u_U(t) f(xt)  u_U(t)f(x)\;dt\\ &= \int_G u_U(t)\left[f(xt)  f(x)\right]\;dt \end{aligned}\] Since $u_U$ is zero outside of $U$, we can restrict this integral to $U$ without changing its value. But then $f(xt)  f(x) \leq \epsilon$. So \[\begin{aligned} f*u_U(x)  f(x) &\leq \int_U u_U(t) \Bigf(xt)  f(x)\Big\;dt \leq \epsilon \end{aligned}\] Therefore, $\f*u_U  f\_\infty \leq \epsilon$.
Now, we can finally prove the PeterWeyl theorem
 $C(G)^{fin}$ is dense in $C(G)$
 $C(G)^{fin}$ is dense in $L^2(G)$
 $L^2(G)^{fin}$ is dense in $L^2(G)$
 For every $e \neq g \in G$, there exists an irreducible representation $(V, \phi)$ of $G$ such that $\phi(g) \neq \Id$
 $C(G)^{fin}$ separates points of $G$. That is to say, for every pair $g \neq h \in G$, there is a function $f \in C(G)^{fin}$ such that $f(g) \neq f(h)$
Earlier, we proved that these statements are all equivalent. So we only have to prove one of them. We will prove 3. Let $f \in L^2(G)$. We want to show that we can approximate $f$ by elements of $L^2(G)^{fin}$. Since $CG)$ is dense in $L^2(G)$, it is enough to show that we can approximate continuous functions. So let $f$ be continuous.
We saw that we can approximate $f$ with convolutions $f * u$. Since $u$ is realvalued and $u = u^*$, we also have $\overline u = u^*$. So convolution with $u$ is a compact, selfadjoint operator. Thus, $f*u$ is in the image of a compact, selfadjoint operator, so we can approximate it by sums of elements in the nonzero eigenspaces of $\cdot *u$. Since convolution is left$G$linear, the eigenspaces are $G$finite. Thus, we have shown that we can approximate any $f \in L^2(G)$ by $G$finite $L^2(G)$ functions, so $L^2(G)^{fin}$ is dense in $L^2(G)$.
Let $\{(E_i, \phi_i)\}_i$ be a set of representatives of the isomorphism classes of irreducible representations of $G$. The unitary embeddings $M_{E_i}:\End(E_i) \to L^2(G)$ induce an isomorphism of $(G \times G)$representations \[\widehat {\bigoplus_i} \;\End(E_i) \xrightarrow{\sim} L^2(G)\] (the hat over the direct sum sign denotes the completion of the direct sum into a Hilbert space)
Our orthogonality results show us that this map is injective. Furthermore, we note that the image of this map is precisely $C(G)^{fin}$. Suppose $f \in C(G)^{fin}$. Then $f \in \im M_V$ for some finitedimensional representation $V$. Since every finitedimensional representation can be written as a direct sum of irreducible representations, $\im M_V$ is contained in $\bigoplus_i \im M_{E_i}$. Therefore, the image of $\bigoplus_i\End(E_i)$ in $L^2(G)$ is $C(G)^{fin}$, which is dense. This implies the PeterWeyl decomposition.
Application: $L^2(S^1)$
As a simple example of the PeterWeyl theorem, we can consider the circle group $S^1 = U(1)$. Because $S^1$ is abelian, the problem simplifies a lot. Recall that all irreducible representations of abelian groups are onedimensional. Irreducible representations equal their characters and matrix coefficients.
Recall that the irreducible representations of $S^1$ are given by $\phi_k:e^{i\theta} \mapsto e^{ik\theta}$ for $k \in \Z$. The PeterWeyl theorem tells us that these give us a basis of $L^2(S^1)$. So the PeterWeyl theorem generalizes Fourier series to functions on arbitrary compact groups.
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