Peter Weyl

8/10/2019 Peter Weyl

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A compact groupGhas a unique normalized both left and right invariant Haar

measure; and ifGacts continuously on a finite dimensional vector space V,Valso

has an inner product which is invariant under the action ofG. Because, starting

from an arbitrary inner product, say P(, ), we define

v, u:=

G

P(gu, gv)dg,

(integrating with respect tothe Haar measure) which is automatically invariant.

1 Matrix coefficient

Let G

GL(V) be a representation, where V is n-dimensional. Now, choose

a vector v V and any linear functional L : V C [note that it is given by

L(x) =< x, w >, once you fix an inner product, for some w V.] Now, observe

that we have two cases: ifv, w are linearly dependent, i.e. sayw = v: vcan be

extended to a basis, with respect to that basis,has a matrix form, and L((g)v) =

11(g) < v, v>. Again, ifv and w are not dependent, thenv, wcan be a part of a

basis, with respect to which if we consider the matrix from of, then

L((g)v) =< 11(g)v +12(g)w + , w>=12(g)< w, w> .

So, functions of the form L((g)v)are scalar multiples of coefficients of the matrix

form of with respect to particular bases.

Such functions will be called matrix coefficients. If for a continuous function

f on G, {(g)f : g G} spans a finite dimensional vector space V over C, then

fis a matrix coefficient, for (, V) is a finite dimensional representation ofG and

f(g) = L((g)f), where L : V C : (1) is a linear functional. Thus, itis immediate that the sum of matrix coefficients is also a matrix coefficient. And

also, for a matrix coefficientg L((g)v), all its right translates are in the linear

span of the coefficient of the matrix form of, which is finite dimensional. So,

one can invariantly define a matrix coefficients to be a continuous function whose

right translates form a finite dimensional subspace ofC(G).

2 Peter-Weyl Theorem

T1 (Peter-Weyl) Given a compact group G, a continuous function f on

it, and any positive real number, there is an elementin the representative ring,i.e., in the C-algebra generated by matrix coefficients, such that||f || < .

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Proof. Let G be a compact group. Let f :G C be continuous.

1. Since fis conts, and G is compact, for given there is an open nbd. Uof

identity such that (x)(y)(x1y U |f(x) f(y)|< /2).

2. There is continuous : G C supported onUsuch that

G(g)dg =1 and

(g1

) =(g).3. Have compact self adjoint operatorT : L2(G) L2(G) : i.e.

T(x) =

G(g)(g1x)dg. This means that Tcommutes with projections (g),

where(g)(x) =(g1x).

Note:

< T, >=

G

G

(g1h)(g)dg(h)dh;

and

< , T >=

G

(g)

G

(h1g)(h)dhdg.

But the two right side terms are equal by the assumptions on . So, T is self-

adjoint.

4. It means that in the spectral decomposition

L2(G) =ker T (E),

(here bar means the closure of the direct sum) where each 0, and each Ebeing eigenspaces, are finite dimensional; and also, as T commutes with (g),

these E are G-modules.

5. So,T L2(G) = T E = E.

6. As is evident, ||T f f|| /2, andT f E. The inequality is by this:

T f(x) f(x) =

G

{f(g)(g1x) f(x)}dg

=

G

{f(g)(x1g) f(x)}dg

=

G

{f(xg)(g) f(x)(g)}dg,

whence

|T f(x) f(x)| supgU

|f(xg) f(x)| /2.

7. Each E is inside the representative ring: let 1, , n be an orthogonal basisof it; since it is G-stable, one has, for any E, i(g

1x) =

jci j(g)j(x),

whence, in matrix notation,

(g1x) =(ci j(g))(x),

which means that as (x) = (ci j(g1))(ci j(g))(x), the matrix (ci j(g)) is always

invertible. So, i(g) =

jci j(g1)j(e), which means that the elements of the

orthonormal basis are linear combinations of the coefficients of a representation.

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More so, by the remerk in the previous section, as E isG-stable finite dimen-

sional, functions in it are matrix coefficients. So,T f is approximated in L2 norm

by elements of the representative ring.

Precisely, there is a coefficient1 of a linear representation such that for a

scalar multiple of that, we have||T f ||2

< /2.

8. So, one has such an element in the representative ring that ||f ||2 < ,

since the L2-norm is less than or equal to the sup norm. However, for any there

is such a; so, we can choose such a that||f ||2 <

2C, whereC=||||2. Write

=(f ).

So,

|(x)|2 = |

G

(g1x)(g)dg|2

= |

G

(g)(x1g)dg|2

= |

G

(xg)(g)dg|2 =| < Lx, > |2

|| Lx||22 ||||

22

=

G

|(xg)|2dg C2

= C2

G

|(g)|2dg = C2||||22.

So, ||T f T|| C ||f ||2 < /2. Again, as was chosen to be linear

combination of elements of Es, it follows that T is also in the representative

ring. Lastly, one sees that

||f T|| ||f T f|| + ||T f T|| < /2 + /2 = .

3 Applications

T2 Every compact Lie group has a faithful representation.

Proof. LetUbe an open neighbourhood of the identity which doesnt contain any

non trivial Lie subgroup. There is by normality of the group a continuous functionwhich is two outside Uand 0 at the identity. Now, by Peter-Weyl theorem there

is a representative function which vanishes at the identity and is greater than 1

outside U. Let that be given by g L((g)v). So, the kernel of is contained

in U; but the kernel is a closed subgroup i.e. a Lie subgroup, contradicting the

choice ofU, unless it is trivial.

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3.1 characters

The character of a representation (V, ), which we will denote by V, is sim-

ply the trace of. It is easy to check that for irreducible representations V, W,

HomG(V, W) = 0 ifW and V are inequivalent and Cif they are equivalent. This

is calledSchurs lemma. Now, take the inequivalent case, dimV = n, dimW = m,and take anyn mmatrix (Tkl). Then define the map A : V Wby

Ai j =

G

(

kl

ik(g)Tkll j(g)dg.

This isG-invariant, so by Schurs lemma, is zero. Choose Tkl to be 1 at st-th place

and zero elsewhere, so that

Gis(g)tl(g)dg =0.

IfVis a representation then the operatorp given by v

Ggvdg is a projection

operator onto the fixed pointsVG, so that its trace isdimVG. G acts onHom(V, W)

by (g f)(v) =g(f(g1v)), so that the homomorphisms that are fixed are precisely

theG-morphism i.e. Hom(V, W)G =H omG(V, W). Thus,

dimVG =T r(p) =T r(

G

(g)dg) =

G

T r((g))dg =

G

V(g)dg.

Thus,

dimHom(V, W)G =

G

Hom(V,W)(g)dg =

G

VW(g)dg =

G

WVdg.

Keeping in mind the above discussion, two irreducible representations V and W

are equivalent or not according as

GWVdg is 1 or 0.

Let us now apply this to representation of compact groups. Representationsof compact groups have invariant inner product i.e can thus be broken up into

direct sum of irreducible representations: any finite dimensional representation V

is represented aski=1

niViwhereViare the inequivalent irreducible components of

V and ni are the multiplicities. Thus,V =k

i=1niVi , whence V, V =

in2i

.

This also means that a representation is irreducible if and only ifV, V = 1. It

follows, thus, that ifVis irreducible, then so isV. AndV VasG Gmodule

is irreducible, since GG

VVVV =

GG

VVVV

=

G

VV

2 =1.

Let me denote by R(G) the ring of representative functions, called the rep-

resentation ring. Now, given a representation V ofG , one has the function sV :

V V C(G), where the tensor is taken over complex numbers and C(G) de-

notes the complex valued continuous functions on G: The function sVis defined

by sV( v) =d,vwhere

d,v :G C: g (gv).

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Henceforth we will denote by S(V) the image ofsV inC(G). If (V, ) and (W, )

are inequivalent irreducible representation, then for V, W, and v

V, u Wand with respect to fixed orthonormal bases with respect to G-invariant

inner products, (g)v = (

ji j(g)vj)i and (gv) =

i jaii j(g)vj; and similarly,

(gu) = kl

bk

kl(g)u

l. So, theL

2inner product ofs

V(v) and s

W(u) is

(gv)(gu)dg =

ijkl

aibkvjul

i j(g)kl(g)dg =0.

Now, the image ofsVis spanned bydei

,ej = sV(ei ej). But

dei

,ej (g) =ei ((g)ej) =i j.

Hence the image is the span ofi j. Now, this map is G G-module map, and

for Virreducible, since V V is an irreducible G G-module, this map is an

isomorphism. Hence, we get a map

s:V G

(V V) R(G).

This map is obviously one-one, since the image of one of the summands is or-

thogonal to that of another. We will show that this map is onto. To show that, it

is enough to show that any representative function is in the direct sum of finitely

many S(V). But, for a representative function f, f(g) = g.f(1), and if f gen-

erates the finite dimensional subspaceV, then letei denote an orthonormal basis

of V with respect to invariant inner product, and ei the dual basis, if we write

g.f(x) = iai(g)ei(x), then sV(ej f)(g) =aj(g) and thus,f(g) =

i

sV(ei f)(g)ei(1),

meaning that f is in S(V), which proves the above contention. So, we have the

isomorphism

s:V G

(V V) R(G).

This also means that as a G-module,R(G) is semisimple, and an isotypical com-

ponent is the direct sum of an irreducible representation its dimension times. By

Peter-Weyl theorem, the Hilbert space L2

(G) is the closure ofR(G) i.e.,

L2(G) =

V G(V V),

the sum inside bracket being taken dimension ofVtimes.

At this stage we need to make a lemma for future reference:

L.1 If B is a G-submodule of R(G) then B is closed in both sup and inner

product topology.

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Proof. We have already proved that R(G) is semisimple, which shows that B is

the direct sum ofB S(V) for irreducible V. Now, let fbe in the closure (inner

product topology) ofBinR(G). Then ifVdenotes the projection ontoS(V), then

for anyb B, note that

V(f b), V(f b) f b, f b.

This means thatV(f) is in the closure ofV(B) = B S(V); but the latter being

finite dimensional is closed, whence V(f) B. Now, only finitely many V(f)

are non zero; so, f B. Now, if something is closed in the inner product topology,

it is automatically closed in the sup norm topology.

By Theorem2any compact Lie group G has a faithful representation, say ,

and with respect to a basis if the coefficients of are denoted by i jthen the alge-

bra over C generated by {i j, i j}i j is a unitalC algebra separating points, and by

Stone Weierstrass theorem is dense inC(G). And as this algebra isG-submodule

ofR(G), we conclude by Lemma1that the representative ring is generated as analgebra over the complex numbers by {i j,i j}i j. Thus,

T3 The representative ring of a compact Lie group (or a compact group

admitting a faithful representation) is finitely generated over the field of complex

numbers.

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