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 Peter-Weyl 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 $C(G)$ denote the set of continuous complex-valued functions on $G$.
Let $(V, \phi)$ be a representation of $G$. The matrix coefficient map is the map \[M_V:\End(V) \to C(G)\] where \[M_V(T)(g) := \tr(\phi(g) \circ T)\]
Recall that $\End(V) \cong V \otimes V^*$. Let $\{e_i\}$ be a basis for $V$ with dual basis $\{e^j\}$ of $V^*$. Then \[M_V(e_i \otimes e^j)(g) = \tr (\phi(g) e_i \otimes e^j) = \tr (e^j \phi(g) e_i) = e^j \phi(g) e_i\] This is just the $(i,j)$th entry in the matrix $\phi(g)$ in this basis. So the matrix coefficient map $M_V$ generalizes literal matrix coefficients.
Let $G,H$ be compact groups. Let $U$ be an irrep of $G$ and $W$ an irrep of $H$. Then $U \otimes W$ is an irreducible $G\times H$ representation (where the action is given by $(g,h) \cdot u \otimes w = (gu) \otimes (hw)$).

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.

$M_V:\End(V) \to C(G)$ is $(G \times G)$-linear

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.

If $V$ is irreducible, then $M_V:\End(V) \to C(G)$ is injective.

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.

  1. $M_{V^*}(T^*) = M_V(T)^*$
  2. $M_{\overline V}(\overline T) = \overline{M_V(T)}$
  3. $M_{V \oplus W}(T \oplus S) = M_V(T) + M_W(S)$
  4. $M_{V \otimes W}(T \otimes S) = M_V(T) \cdot M_W(S)$
  5. $M_{\Hom(V,W)}(S \circ \bullet \circ T^*) = M_V(T)^* \cdot M_W(S)$
  6. $M_{V^G}(Av_G \circ T) = av(M_V(T))$
  1. $M_{V^*}(T^*)(g) = \tr(\phi(g^{-1})^T \circ T^T) = \tr(\phi(g^{-1}) \circ T) = M_V(T)^*(g)$
  2. $M_{\overline V}(\overline T)(g) = \tr(\overline{\phi(g)} \circ \overline T) = \overline{M_V(T)(g)}$
  3. The computation here looks a bit longer, but it's pretty straightforward \[\begin{aligned} M_{V \oplus W}(T \oplus S)(g) &= \tr((\phi(g) \oplus \psi(g)) \circ (T \oplus S))\\ & = \tr((\phi(g) \circ T) \oplus (\psi(g) \circ S))\\ &= \tr(\phi(g) \circ T) + \tr (\psi(g) \circ S)\\ &= M_V(T)(g) + M_W(S)(g) \end{aligned}\]
  4. This one also looks a bit long, but it's also straightforward \[\begin{aligned} M_{V \otimes W}(T \otimes S)(g) &= \tr((\phi(g) \otimes \psi(g)) \circ (T \otimes S))\\ & = \tr((\phi(g) \circ T) \otimes (\psi(g) \circ S))\\ &= \tr(\phi(g) \circ T) \cdot \tr (\psi(g) \circ S)\\ &= M_V(T)(g) \cdot M_W(S)(g) \end{aligned}\]
  5. This follows from the first and fourth identities, using the fact that $\Hom(V,W) \cong W \otimes V^*$.
  6. 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 left-invariant, this is just equal to \[\begin{aligned} M_{V^G}(Av_G \circ T)(g) &= \int_G \tr_V(T \circ \phi(g))dg\\ &= av(\tr_V(T \circ \phi(g)))\\ &= av(M_V(T)) \end{aligned}\]
(Orthogonality of matrix coefficients)

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 well-defined adjoint map $S \mapsto S^\dagger$ for $S \in \End(V)$. Note that because our inner product is $G$-invariant, $\phi(g)^\dagger = \phi(g^{-1})$.

Let $E$ be an irreducible represetation of $G$. The Hilbert-Schmidt inner product on $\End(E)$ is given by $\inrp TS_{HS} := \tr_E (T \circ S^\dagger)$.
  1. The Hilbert-Schmidt inner product is $(G \times G)$-invariant
  2. $M_E: \End(E) \to C(G)$ is unitary (up to a scalar factor) with respect to the Hilbert-Schmidt inner product on $\End(E)$ and the inner product we have been using on $C(G)$
  1. 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, \[\begin{aligned} \inrp{(g,h)T}{(g,h)S} &= \tr_E(\phi(g) \circ T \circ \phi(h)^{-1} \circ (\phi(g) \circ S \circ \phi(h)^{-1})^\dagger)\\ &= \tr_E(\phi(g) \circ T \circ \phi(h)^{-1} \circ \phi(h^{-1})^\dagger \circ S^\dagger \circ \phi(g)^\dagger)\\ &= \tr_E(\phi(g) \circ T \circ \phi(h)^{-1} \circ \phi(h) \circ S^\dagger \circ \phi(g^{-1}))\\ &= \tr_E(\phi(g) \circ T \circ S^\dagger \circ \phi(g^{-1}))\\ &= \tr_E(T \circ S^\dagger)\\ &= \inrp T S \end{aligned}\]
  2. 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 Peter-Weyl 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 Peter-Weyl theorem is a generalization of this result to compact groups. Before proving the theorem, we need some preliminaries.

Let $(V, \phi)$ be a (possibly infinite-dimensional) representation of $G$. We say that $v \in V$ is $G$-finite if $Gv = \{gv\;:\;g \in G\}$ lies in a finite-dimensional subspace of $G$.
Let $f \in C(G)$. The following are equivalent:
  1. $f$ is left-$G$-finite
  2. $f$ is right-$G$-finite
  3. $f$ is $(G \times G)$-finite
  4. $f$ 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 finite-dimensional 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 finite-dimensional 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 $M_V$
Let $C(G)^{fin}$ denote the subset of $C(G)$ consisting of functions which satisfy these equivalent conditions.
Note that $C(G)^{fin}$ is a subalgebra of $C(G)$. Suppose $f, g \in C(G)^{fin}$. By the fourth condition, we must have representations $V,W$ with $S \in \End(V), T \in \End(W)$ such that $f = M_V(S)$, $g = M_W(T)$. Then $f+g = M_{V \oplus W}(S \oplus T)$, $f \cdot g = M_{V \otimes W}(S \otimes T)$ and $\overline f = M_{\overline V}(\overline S)$.
Let $L^2(G)^{fin}$ denote the subset of $L^2(G)$ consisting of left-$G$-finite vectors.
Eventually we will prove that $L^2(G)^{fin} = C(G)^{fin}$. But that is not obvious (at least to me) yet.
The following are equivalent
  1. $C(G)^{fin}$ is dense in $C(G)$
  2. $C(G)^{fin}$ is dense in $L^2(G)$
  3. $L^2(G)^{fin}$ is dense in $L^2(G)$
  4. For every $e \neq g \in G$, there exists an irreducible representation $(V, \phi)$ of $G$ such that $\phi(g) \neq \Id$
  5. $C(G)^{fin}$ 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 finite-dimensional subspace of $C(G)$ by definition of $L^2(G)^{fin}$. So $\tilde f$ is a vector in a finite-dimensional 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 Peter-Weyl theorem will tell us that these equivalent conditions are true. But we need to prove a few more technical lemmas before we can prove it.
Let $X$ be a compact spaces, with a nowhere-vanishing measure $\mu$. Assume without loss of generality that $\mu(1) = 1$. Let $K \in C(X \times X)$ and define \[\begin{aligned} T_K: L^2(X) &\to L^2(X)\\ T_K(f)(x) &:= \int_X K(x,y) f(y) \; dy \end{aligned}\] $T_K$ defines a continuous map $L^2(X) \to L^2(X)$ of operator norm at most $\|K\|_{L^2(X \times X)}$. $T_K$ is compact. If $K(x,y) = \overline{K(y,x)}$, then the operator is self-adjoint as well.

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 finite-rank operators between Hilbert spaces is a compact operator. So it is sufficient to show that $T_K$ is the limit of a sequence of finite-rank 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 Stone-Weierstrass 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 finite-rank. 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 self-adjoint. \[\begin{aligned} \inrp{T_K(f)}{g} &= \int_X T_Kf(x) \overline{g(x)}\;dx\\ &= \int_X \int_X K(x,y) f(y)\;dy \;\overline {g(x)}\;dx\\ &= \int_X f(y) \int_X K(x,y) \overline {g(x)}\;dx\;dy\\ &= \int_X f(y) \overline{\int_X K(y,x) g(x)\;dx}\;dy\\ &= \int_X f(y) \overline {T_Kg(y)}\;dy\\ &= \inrp f {T_K(g)} \end{aligned}\]

Recall that for a compact self-adjoint 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 $L^2(G)$.

Let $k \in C(G)$, and let $K(x,y) = k(y^{-1}x)$. Then the operator $T_K$ is known as convolution, and is denoted $f * k := T_K(f)$.

Note that if $k^* = \overline k$ (i.e. $k(g^{-1}) = \overline{k(g)}$), then our convolution operator is self-adjoint.

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 left-invariant

Let $f \in C(G)$ and $\epsilon > 0$. Let $e \in U \subseteq G$ be open with $U = U^{-1}$ and $|f(x) - f(xy)| \leq \epsilon$ for all $x \in G, y \in U$. Then there exists a real-valued function $u_U \in C(G)$ such that $u_U \geq 0, u_U^* = u_U$, $\int_G u_U \; d\mu = 1$, and $u_U$ is zero outside of $U$. For this function, we have \[\|f * u_U - f\|_\infty \leq \epsilon\]

Constructing such a function $u_U$ is fairly straightforward. Pick any nonzero continuous real-valued 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 left-invariant, 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 \[\begin{aligned} |f*u_U(x) - f(x)| &\leq \int_U u_U(t) \Big|f(xt) - f(x)\Big|\;dt \leq \epsilon \end{aligned}\] Therefore, $\|f*u_U - f\|_\infty \leq \epsilon$.

Now, we can finally prove the Peter-Weyl theorem

  1. $C(G)^{fin}$ is dense in $C(G)$
  2. $C(G)^{fin}$ is dense in $L^2(G)$
  3. $L^2(G)^{fin}$ is dense in $L^2(G)$
  4. For every $e \neq g \in G$, there exists an irreducible representation $(V, \phi)$ of $G$ such that $\phi(g) \neq \Id$
  5. $C(G)^{fin}$ 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 real-valued and $u = u^*$, we also have $\overline u = u^*$. So convolution with $u$ is a compact, self-adjoint operator. Thus, $f*u$ is in the image of a compact, self-adjoint 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 $L^2(G)$.

(Peter-Weyl Decomposition)

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 finite-dimensional representation $V$. Since every finite-dimensional 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 Peter-Weyl decomposition.

Note that $\End(E_i) \cong E_i \otimes E_i^* \cong E_i^{\oplus \dim E_i}$. So the Peter-Weyl decomposition is also written \[\widehat{\bigoplus_i} \; E_i^{\oplus \dim E_i} \cong L^2(G)\]

Application: $L^2(S^1)$

As a simple example of the Peter-Weyl 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 one-dimensional. 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 Peter-Weyl theorem tells us that these give us a basis of $L^2(S^1)$. So the Peter-Weyl theorem generalizes Fourier series to functions on arbitrary compact groups.

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    Hello, Have you been looking for financing options for your new business plans, Are you seeking for a loan to expand your existing business, Do you find yourself in a bit of trouble with unpaid bills and you don’t know which way to go or where to turn to? Have you been turned down by your banks? MRS. DOROTHY JEAN INVESTMENTS says YES when your banks say NO. Contact us as we offer financial services at a low and affordable interest rate of 2% for long and short term loans. Interested applicants should contact us for further loan acquisition procedures via

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    categories of investment

    Loan Offer
    Mining Plan
    Business Finance Plan
    Binary option Trade Plan
    Forex trade Plan
    Stocks market Trade Plan
    Return on investment (ROI) Plan
    Gold and Silver Trade Plan
    Oil and Gas Trade Plan
    Diamond Trade Plan
    Agriculture Trade Plan
    Real Estate Trade Plan

    Mrs. Dorothy Pilkenton Jean
    Financial Advisor on Bank Instruments,
    Private Banking and Client Services
    Email Address:
    Operation: We provide Financial Service Such As Bank Instrument


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