Compact groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 10 April 2010

Compact groups

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

Suppose that $V$ is a representation of $G$ in a complete locally convex vector space. Let $(,) : V \otimes V \to ℂ$ be an inner product on $V$ and define a new inner product $⟨,⟩ : V \otimes V \to ℂ$ by $⟨v_{1}, v_{2}⟩ = \int_{G} (g v_{1}, g v_{2}) d μ (g), v_{1}, v_{2} \in V .$

Under the innder product $⟨,⟩$ the representation $V$ is unitary. If $V$ is a finite dimensional representation of $G,$ $\begin{array}{rcl} V : & G & \to & M_{n} (ℂ) \\ g & \mapsto & V (g), \end{array} then \begin{array}{rcl} \end{array}$ V : G → M n ℂ 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 $C (G)$ 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 $C (G)$ on $G$ by $f v = \int_{G} f (g) g v d μ (g), f \in C (G), v \in V .$ The complete locally convex assumption on $V$ is necessary to define the integral in (???)

If $V$ is a representation of $G$ define $V^{fin} = \{v \in V | the G -module generated by v is finite dimensional\} .$

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

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

		Proof.
	Let $f \in {C (G)}^{rep} .$ Let $v, w$ be vectors in a finite dimensional representation $V$ such that $f (g) = ⟨v, g w⟩$ for all $g \in G .$ Let $\{v_{1} \dots v_{k}\}$ be an orthonormal basis of $V$ and let $W$ be the vector space of linear combinations of the functions $f_{j} = ⟨v_{j}, g w⟩, 1 \leq j \leq k .$ 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 $1 \leq i \leq k,$ $(x f_{i}) (g) = \tilde{f_{i}} (x^{- 1} g) = ⟨v_{i}, x^{- 1} g w⟩ = ⟨x v_{i}, g w⟩ = ⟨\sum_{j = 1}^{k} c_{j} v_{j} g w⟩ = \sum_{j = 1}^{k} c_{j} f_{j} (g)$ for some constants $c_{j} \in ℂ .$ So the $G$ -module generated by $f$ is contained in the finite dimensional representation $W .$ So $f \in {C (G)}^{fin} .$ So ${C (G)}^{rep} \subseteq {C (G)}^{fin} .$ Let $f \in {C (G)}^{fin}$ and let $f_{1} = f, f_{2}, \dots, f_{k}$ be orthonormal basis of the finite dimensional representation $W$ generated by $f .$ Then $f (g) = (g^{- 1} f_{1}) (1) = \sum_{j = 1}^{k} ⟨f_{j}, g^{- 1} f_{1}⟩ f_{j} (1), where c_{j} = ⟨f_{j}, g^{- 1}, f_{1}⟩ .$ Define a new finite dimensional representation W of $G$ which has orthonormal basis $\{\}$ 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 $g_{1} (g_{2})$ w = g1 g2 w , for all $g_{1}, g_{2} \in G .$ Since $⟨⟩$ w j g w i = fj g -1 fi , $f (g) = ⟨\sum_{j = 1}^{k} c_{j}⟩$ w j g w 1 where cj = fj 1 and so $f \in {C (G)}^{rep} .$ So ${C (G)}^{fin} \subseteq {C (G)}^{rep} .$ $□$

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

${C (G)}^{rep}$ is dense in $C (G),$ under the topology defined by the $\sup$ norm.
$V^{fin}$ is dense in $V$ for all representations $V$ of $G .$
$G$ is linear, ie there is an injective map $i : G \to {GL}_{n} (ℂ)$ for some $n .$
Let $\hat{G}$ be an index set for the finite dimensional representations of $G .$ For each finite dimensional irreducible representation $G^{λ}, λ \in \hat{G},$ fix an orthonormal basis $\{v_{i}^{λ} | 1 \leq i \leq d_{λ}\}$ of $G^{λ} .$ Define $M_{i j}^{λ} \in {C (G)}^{rep}$ by $M_{i j}^{λ} = ⟨v_{i}^{λ}, g v_{j}^{λ}⟩, g \in G .$ Then $\begin{array}{rcl} \oplus_{λ \in \hat{G}} G^{λ} \otimes G^{λ} & \to & {C (G)}^{rep} \\ v_{i}^{λ} \otimes v_{j}^{λ} & \mapsto & M_{i j}^{λ} \end{array}$ is an isomorphism of $G \times G$ -modules.
The map $\begin{array}{rcl} \oplus_{λ \in \hat{G}} M_{d_{λ}} (ℂ) & \to & {C (G)}^{rep} \\ E_{i j}^{λ} & \mapsto & M_{i j}^{λ} \end{array}$ is an isomorphism of algebras.

and (a), (b), (c), (d) and (e) are all equivalent.

Proof.

(b) $\Rightarrow$ (a) is immediate.

(a) $\Rightarrow$ (b): Note that ${C (G)}^{fin} V \subseteq V^{fin} .$ Since ${C (G)}^{fin}$ is dense in $C (G),$ the closure of ${C (G)}^{fin} V$ contains $C (G) V .$ Let $f_{1}, \dots, f_{2}$ be a sequence of functions in $C (G)$ such that $μ (f_{i}) = 1$ and the sequence approaches the $δ$ function at 1, ie the function $δ_{1}$ which has $supp (δ_{1}) = \{1\} .$ If $v \in V$ then the sequence $f_{1} v, f_{2} v, \dots$ approaches $1 v = v$ and so $v$ is in the closure of $C (G) V .$ So the closure of $C (G) V$ is $V .$ So $V^{fin}$ is dense in $V .$

The following method of making this more precise is given by Brocker and tom Dieck.

An operator $K : C (G) \to C (G)$ is compact if for every bounded $B \subseteq C (G),$ every sequence $(f_{n}) \subseteq K (B)$ converges in $K (B) .$ An operator $K : C (G) \to C (G)$ is a symmetric operator if $⟨K f_{1} f_{2}⟩ = ⟨f_{1}, K f_{2}⟩$ for all $f_{1}, f_{2} \in C (G) .$

$□$

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

$∥K∥ = \sup \{∥K f∥ | ∥f∥ \leq 1\}$ or $- ∥K∥$ is an eigenvalue of $K,$
All eigenspaces of $K$ are finite dimensional,
$\oplus_{λ} C {(G)}_{λ}$ is dense in $C (G) .$

Proof.

(b) The reason eigenspaces are finite dimensional: let $x_{1}, x_{2}, \dots$ be an orthonormal basis. Then $K x_{i} = λ x_{i} .$ So ${∥K x_{i} - K x_{j}∥}^{2} = |λ^{2}| {∥x_{1} - x_{j}∥}^{2} = 2 {∥λ∥}^{2}$ and this never goes to zero.

(c) If not then $U^{⊥} = ()$ ⊕ λ C G λ ⊥ is nonzero. Then $K : U^{⊥} \to U^{⊥}$ is a compact symmetric operator. So this operator has a finite dimensional eigenspace. This is a contradiction. So $U^{⊥} = 0 .$ So $\oplus_{λ} C {(G)}_{λ}$ is dense in $C (G) .$

Take $K$ to be the operator given by convolution by an approximation $φ$ to the $δ$ function. Then $K f$ is close to $f,$ ${∥K f - f∥}_{\infty} = |\int_{G} (δ (g) f (x g) - f (g)) d μ (g)| \leq \int_{G} ε δ (g) d μ (g) = ε$ $= {∥δ (1) - 1∥}_{\infty} \leq ε$ and $K f$ can be approximated by the action of $φ$ on finite dimensional subspaces.

The symmetric condition on $K$ translates to $φ (g) = φ (g^{- 1})$ and the compactness condition translates to $\int_{G} φ (g) d μ (g) = 1 .$

Note that ${∥f∥}_{2}^{2} = \int f (g)$ f g dμ g ≤∫ f g f g dμ g ≤ f ∞ 2 . So the $L^{2}$ and $\sup$ norms compare. For norms of operators ${∥δ * f∥}_{\infty} \leq {∥δ∥}_{\infty} {∥f∥}_{\infty} .$

(c) $\Rightarrow$ (a): If $i : G \to {GL}_{n} (ℂ)$ is an injection then the algebra ${C (G)}^{alg}$ generated (under pointwise multiplication) by the functions $i_{i j}$ and i ij , where $i_{i j} (g) = {i (g)}_{i j} and i_{i j} (g) =$ i ij g for g∈ G, is contained in ${C (G)}^{fin} .$ This subalgebra separates points of $G$ and is closed under pointwise multiplication and conjugation and so, by the Stone-Weierstrass theorem, is dense in $C (G) .$ So ${C (G)}^{fin}$ is dense in $C (G) .$

(a) $\Rightarrow$ (c): The elements of $C (G)$ distinguish the points of $G$ and so the functions in ${C (G)}^{rep}$ distinguish the points of $G .$ For each $g \in G$ fix a function $f_{g}$ such that $(g f_{g}) (1) = f_{g} (g^{- 1}) \neq f_{g} (1)$ and let $V g$ 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 $K_{1} \supseteq K_{2} \supseteq \dots, where K_{j} = \ker (V_{g_{1}} \oplus \dots \oplus V_{g_{j}}),$ and $K_{i} \neq K_{i + 1 .}$ Since each $K_{i}$ is a closed subgroup of $G,$ and $G$ is compact there is a finite $n$ such that $K_{n} = \{1\} .$ 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 $GL (W) .$

(d) By construction this is an algebra isomorphism. After all the algebra multiplication is designed to extend the $G \times G$ module structure, and this is a $G \times G$ module homomorphism since $\begin{matrix} ((x \otimes y) (v_{i}^{λ} \otimes v_{j}^{λ})) (g) & = & (Φ (x v_{i}^{λ} \otimes y v_{j}^{λ})) (g) \\ = & ⟨x v_{i}^{λ} \otimes g y v_{j}^{λ}⟩ \\ = & ⟨v_{i}^{λ} \otimes x^{- 1} g y v_{j}^{λ}⟩ \\ = & M_{i j}^{λ} (x^{- 1} g y) \\ = & (L_{x} R_{y} M_{i j}^{λ}) (g) . \end{matrix}$

$□$

Note that $Tr (E_{i j}^{λ}) = ⟨v_{i}^{λ}, v_{j}^{λ}⟩ = δ_{i j} .$ Consider the $L^{2}$ norm on ${C (G)}^{rep} .$ $\begin{matrix} {∥f∥}_{2}^{2} & = & \int_{G} f (g) \end{matrix}$ f g dμ g = ∫G f g f* g -1 dμ g where f* g =f g -1 = f* f* 1 . More generally, ${⟨f_{1}, f_{2}⟩}_{2} = (f_{1} * f_{2}) (1) .$ Now $\begin{array}{rcl} τ : & {C (G)}^{rep} & \to & ℂ \\ f & \mapsto & f (1) \end{array}$ is a trace on ${C (G)}^{rep},$ ie $τ (f_{2} * f_{1}) = τ (f_{1} * f_{2})$ for all $f_{1}, f_{2} \in {C (G)}^{rep} .$ In fact this is the trace of the action of ${C (G)}^{rep}$ on itself: $\begin{matrix} τ & = & \int_{G} f (g) g h |_{h} d μ (g) \\ = & \int_{G} f (g) δ_{g 1} d μ (g) \\ = & \int_{g} f (1) d μ (g) = f (1) μ (G) = f (1) . \end{matrix}$ Now consider the action of $\oplus_{λ} M_{d_{λ}} (ℂ)$ on itself. Then, if $f = ({\hat{f}}^{λ})$ then $τ (f) = \sum_{λ \in \hat{G}} d_{λ} Tr (f^{λ}) .$ So ${∥f∥}_{2}^{2} = (f * f^{*}) (1) = τ (f * f^{*}) = τ ({\hat{f}}^{λ} {({\hat{f}}^{λ})}^{t}) = \sum_{λ \in \hat{G}} d_{λ} Tr ({\hat{f}}^{λ} {({\hat{f}}^{λ})}^{t}) .$ Note that $Tr ({Id}_{λ}) = d_{λ}$ and $τ ({Id}_{λ}) =$ ????.

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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