Compact groups

## Compact groups

Let $G$ be a compact Lie group and let $\mu$ be a Haar measure on $G.$ Assume that $\mu$ is normalised so that $\mu \left(G\right)=1.$ The algebra ${C}_{c}\left(G\right)$ (under convolution) of continuous complex valued functions on $G$ with compact support is the same as the algebra $C\left(G\right)$ of continuous functions on $G.$ The vector space $C\left(G\right)$ is a $G$-module with $G$-action given by $xf g =f x -1 g , for x∈ G ,f∈ C G .$ The group $G$ acts on $C\left(G\right)$ in two ways, $L g f x =f g -1 x , and R g f x =f xg ,$ and these two actions commute with each other.

Suppose that $V$ is a representation of $G$ in a complete locally convex vector space. Let $\left(,\right):V\otimes V\to ℂ$ be an inner product on $V$ and define a new inner product $⟨,⟩:V\otimes V\to ℂ$ by $v 1 v 2 = ∫G g v 1 g v 2 dμ g , v 1 , v 2 ∈ V.$

Under the innder product $⟨,⟩$ the representation $V$ is unitary. If $V$ is a finite dimensional representation of $G,$ $V : G → M n ℂ g ↦ V g , then$V : G M n g V g =V g -1 t , is another finite dimensional representation of $G.$

Every finite dimensional representation of a compact group is unitary and completely decomposable.

The representation $C\left(G\right)$ is an example of an infinite dimensional representation of $G$ which is not unitary.

If $V$ is a representation of $G$ in a complete locally convex normed vector space $V$ then the representation $V$ can be extended to be a representation of the algebra (under convolution) of continuous functions $C\left(G\right)$ on $G$ by $fv= ∫G f g gvdμ g , f∈ C G ,v∈ V.$ The complete locally convex assumption on $V$ is necessary to define the integral in (???)

If $V$ is a representation of $G$ define

The vector space ${C\left(G\right)}^{\mathrm{rep}}$ of representative functions consists of all functions $f:G\to ℂ$ given by $f g = v gw ,$ for some vectors $v,w$ in a finite dimensional representation of $G.$

Let $G$ be a compact group. Then ${C\left(G\right)}^{\mathrm{fin}}={C\left(G\right)}^{\mathrm{rep}}.$

 Proof. Let $f\in {C\left(G\right)}^{\mathrm{rep}}.$ Let $v,w$ be vectors in a finite dimensional representation $V$ such that $f\left(g\right)=⟨v,gw⟩$ for all $g\in G.$ Let $\left\{{v}_{1},\dots ,{v}_{k}\right\}$ be an orthonormal basis of $V$ and let $W$ be the vector space of linear combinations of the functions ${f}_{j}=⟨{v}_{j},gw⟩,1\le j\le k.$ Since $v$ can be written as a linear combination of the ${v}_{j},$ the function $f$ can be written as a linear combination of the ${f}_{j}$ and so $f\in W.$ For each $1\le i\le k,$ $x fi g = fi ~ x -1 g = vi x -1 gw = x vi gw = ∑ j=1 k cj vj gw =∑ j=1 k cj fj g$ for some constants ${c}_{j}\in ℂ.$ So the $G$ -module generated by $f$ is contained in the finite dimensional representation $W.$ So $f\in {C\left(G\right)}^{\mathrm{fin}}.$ So ${C\left(G\right)}^{\mathrm{rep}}\subseteq {C\left(G\right)}^{\mathrm{fin}}.$ Let $f\in {C\left(G\right)}^{\mathrm{fin}}$ and let ${f}_{1}=f,{f}_{2},\dots ,{f}_{k}$ be orthonormal basis of the finite dimensional representation $W$ generated by $f.$ Then Define a new finite dimensional representation W of $G$ which has orthonormal basis $\left\{{}_{}\right\}$w 1 … w k and $G$ action given by $g$w i =∑ j=1 k fj g -1 fi w j , 1≤i≤k. It is straightforward to check that ${g}_{1}\left({g}_{2}\right)$w = g1 g2 w , for all ${g}_{1},{g}_{2}\in G.$ Since $⟨{}_{}⟩$w j g w i = fj g -1 fi , $f g = ∑ j=1 k cj$w j g w 1 where   cj = fj 1 and so $f\in {C\left(G\right)}^{\mathrm{rep}}.$ So ${C\left(G\right)}^{\mathrm{fin}}\subseteq {C\left(G\right)}^{\mathrm{rep}}.$ $\square$

(Peter-Weyl) Let $G$ be a compact Lie group. Then

1. ${C\left(G\right)}^{\mathrm{rep}}$ is dense in $C\left(G\right),$ under the topology defined by the $\mathrm{sup}$ norm.
2. ${V}^{\mathrm{fin}}$ is dense in $V$ for all representations $V$ of $G.$
3. $G$ is linear, ie there is an injective map $i:G\to {\mathrm{GL}}_{n}\left(ℂ\right)$ for some $n.$
4. Let $\stackrel{^}{G}$ be an index set for the finite dimensional representations of $G.$ For each finite dimensional irreducible representation ${G}^{\lambda },\lambda \in \stackrel{^}{G},$ fix an orthonormal basis $\left\{{v}_{i}^{\lambda }\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}1\le i\le {d}_{\lambda }\right\}$ of ${G}^{\lambda }.$ Define ${M}_{ij}^{\lambda }\in {C\left(G\right)}^{\mathrm{rep}}$ by $M ij λ = viλ g vjλ , g∈ G.$ Then $⊕ λ∈ G ^ Gλ ⊗ Gλ → C G rep viλ ⊗ vjλ ↦ M ij λ$ is an isomorphism of $G×G$-modules.
5. The map $⊕ λ∈ G^ M dλ ℂ → C G rep E ij λ ↦ M ij λ$ is an isomorphism of algebras.
and (a), (b), (c), (d) and (e) are all equivalent.

 Proof. (b)$⇒$ (a) is immediate. (a)$⇒$ (b): Note that ${C\left(G\right)}^{\mathrm{fin}}V\subseteq {V}^{\mathrm{fin}}.$ Since ${C\left(G\right)}^{\mathrm{fin}}$ is dense in $C\left(G\right),$ the closure of ${C\left(G\right)}^{\mathrm{fin}}V$ contains $C\left(G\right)V.$ Let ${f}_{1},\dots ,{f}_{2}$ be a sequence of functions in $C\left(G\right)$ such that $\mu \left({f}_{i}\right)=1$ and the sequence approaches the $\delta$ function at 1, ie the function ${\delta }_{1}$ which has $\mathrm{supp}\left({\delta }_{1}\right)=\left\{1\right\}.$ If $v\in V$ then the sequence ${f}_{1}v,{f}_{2}v,\dots$ approaches $1v=v$ and so $v$ is in the closure of $C\left(G\right)V.$ So the closure of $C\left(G\right)V$ is $V.$ So ${V}^{\mathrm{fin}}$ is dense in $V.$ The following method of making this more precise is given by Brocker and tom Dieck. An operator $K:C\left(G\right)\to C\left(G\right)$ is compact if for every bounded $B\subseteq C\left(G\right),$ every sequence $\left({f}_{n}\right)\subseteq K\left(B\right)$ converges in $K\left(B\right).$ An operator $K:C\left(G\right)\to C\left(G\right)$ is a symmetric operator if $⟨K{f}_{1}{f}_{2}⟩=⟨{f}_{1},K{f}_{2}⟩$ for all ${f}_{1},{f}_{2}\in C\left(G\right).$ $\square$

See Brocker-tom Dieck Theorem 2.6. If $K:C\left(G\right)\to C\left(G\right)$ is a compact symmetric operator then

1. $∥K∥=\mathrm{sup}\left\{∥Kf∥\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}∥f∥\le 1\right\}$ or $-∥K∥$ is an eigenvalue of $K,$
2. All eigenspaces of $K$ are finite dimensional,
3. ${\oplus }_{\lambda }C{\left(G\right)}_{\lambda }$ is dense in $C\left(G\right).$

 Proof. (b) The reason eigenspaces are finite dimensional: let ${x}_{1},{x}_{2},\dots$ be an orthonormal basis. Then $K{x}_{i}=\lambda {x}_{i}.$ So $K xi -K x j 2 = λ2 x1 - xj 2 =2 λ 2$ and this never goes to zero. (c) If not then ${U}^{\perp }={\left(\right)}^{}$ ⊕ λ C G λ ⊥ is nonzero. Then $K:{U}^{\perp }\to {U}^{\perp }$ is a compact symmetric operator. So this operator has a finite dimensional eigenspace. This is a contradiction. So ${U}^{\perp }=0.$ So ${\oplus }_{\lambda }C{\left(G\right)}_{\lambda }$ is dense in $C\left(G\right).$ Take $K$ to be the operator given by convolution by an approximation $\phi$ to the $\delta$ function. Then $Kf$ is close to $f,$ $Kf-f ∞ = ∫G δ g f xg -f g dμ g ≤ ∫G εδ g dμ g =ε$ $= δ 1 -1 ∞ ≤ε$ and $Kf$ can be approximated by the action of $\phi$ on finite dimensional subspaces. The symmetric condition on $K$ translates to $φ g =φ g -1$ and the compactness condition translates to $∫G φ g dμ g =1 .$ Note that $f 22 =∫f g$f g dμ g ≤∫ f g f g dμ g ≤ f ∞ 2 . So the ${L}^{2}$ and $\mathrm{sup}$ norms compare. For norms of operators ${∥\delta *f∥}_{\infty }\le {∥\delta ∥}_{\infty }{∥f∥}_{\infty }.$ (c)$⇒$ (a): If $i:G\to {\mathrm{GL}}_{n}\left(ℂ\right)$ is an injection then the algebra ${C\left(G\right)}^{\mathrm{alg}}$ generated (under pointwise multiplication) by the functions ${i}_{ij}$ and ${}_{}$i ij , where $i ij g = i g ij and i ij g =$ i ij g for g∈ G, is contained in ${C\left(G\right)}^{\mathrm{fin}}.$ This subalgebra separates points of $G$ and is closed under pointwise multiplication and conjugation and so, by the Stone-Weierstrass theorem, is dense in $C\left(G\right).$ So ${C\left(G\right)}^{\mathrm{fin}}$ is dense in $C\left(G\right).$ (a) $⇒$(c): The elements of $C\left(G\right)$ distinguish the points of $G$ and so the functions in ${C\left(G\right)}^{\mathrm{rep}}$ distinguish the points of $G.$ For each $g\in G$ fix a function ${f}_{g}$ such that $\left(g{f}_{g}\right)\left(1\right)={f}_{g}\left({g}^{-1}\right)\ne {f}_{g}\left(1\right)$ and let $Vg$ be the finite dimensional representative of $G$ generated by ${f}_{g}.$ By choosing ${g}_{i}\notin {K}_{i-1}$ we can find a sequence ${g}_{1},{g}_{2},\dots$ of elements of $G$ such that $K1 ⊇ K2 ⊇… , where Kj =ker V g1 ⊕ … ⊕ V gj ,$ and ${K}_{i}\ne {K}_{i+1.}$ Since each ${K}_{i}$ is a closed subgroup of $G,$ and $G$ is compact there is a finite $n$ such that ${K}_{n}=\left\{1\right\}.$ Then $W={V}_{{g}_{1}}\oplus \dots \oplus {V}_{{g}_{n}}$ is a finite dimensional representation of $G$ with trivial kernel. So there is an injective map from $G$ into $\mathrm{GL}\left(W\right).$ (d) By construction this is an algebra isomorphism. After all the algebra multiplication is designed to extend the $G×G$ module structure, and this is a $G×G$ module homomorphism since $x⊗y viλ ⊗ vjλ g = Φ x viλ ⊗y vjλ g = x viλ ⊗gy vjλ = viλ ⊗ x -1 gy vjλ = M ij λ x -1 gy = Lx Ry M ij λ g .$ $\square$

Note that $Tr E ij λ = v i λ v j λ = δ ij .$ Consider the ${L}^{2}$ norm on ${C\left(G\right)}^{\mathrm{rep}}.$ $f 22 = ∫G f g$f g dμ g = G f g f* g -1 dμ g where f* g =f g -1 = f* f* 1 . More generally, ${⟨{f}_{1},{f}_{2}⟩}_{2}=\left({f}_{1}*{f}_{2}\right)\left(1\right).$ Now $τ : C G rep → ℂ f ↦ f 1$ is a trace on ${C\left(G\right)}^{\mathrm{rep}},$ ie $\tau \left({f}_{2}*{f}_{1}\right)=\tau \left({f}_{1}*{f}_{2}\right)$ for all ${f}_{1},{f}_{2}\in {C\left(G\right)}^{\mathrm{rep}}.$ In fact this is the trace of the action of ${C\left(G\right)}^{\mathrm{rep}}$ on itself: $τ = ∫G f g gh | h dμ g = ∫G f g δ g1 dμ g = ∫g f 1 dμ g =f 1 μ G =f 1 .$ Now consider the action of ${\oplus }_{\lambda }{M}_{{d}_{\lambda }}\left(ℂ\right)$ on itself. Then, if $f=\left({\stackrel{^}{f}}^{\lambda }\right)$ then $τ f = ∑ λ∈ G ^ dλ Tr fλ .$ So $f 22 = f* f* 1 =τ f* f* =τ f^ λ f^ λ t = ∑ λ∈ G^ dλ Tr f^ λ f^ λ t .$ Note that $\mathrm{Tr}\left({\mathrm{Id}}_{\lambda }\right)={d}_{\lambda }$ and $\tau \left({\mathrm{Id}}_{\lambda }\right)=$ ????.