representations of locally compact groupsjultika.oulu.fi/files/nbnfioulu-201311201890.pdfgroups. the...

Representations of Locally CompactGroups

Master’s ThesisAntti Rautio

Department of Mathematical SciencesUniversity of Oulu

2013

Contents

Introduction ii

1 Banach Algebras 11.1 Banach and C*-algebras . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Gelfand Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 The Spectral Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 15

2 Locally Compact Groups 182.1 Haar measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.2 Convolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3 Representation Theory 293.1 Hilbert Space Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 293.2 Unitary Representations . . . . . . . . . . . . . . . . . . . . . . . . 313.3 The Gelfand-Raikov Theorem . . . . . . . . . . . . . . . . . . . . . 39

4 Compact Groups 514.1 Representations of Compact Groups . . . . . . . . . . . . . . . . . . 514.2 The Peter-Weyl Theorem . . . . . . . . . . . . . . . . . . . . . . . . 54

i

Introduction

The topic of this thesis is representation theory. The idea of representation theoryis to represent an algebraic object, such as a locally compact group or an algebra, asa more concrete group or algebra consisting of matrices or operators. In this way wecan study an algebraic object as collection of symmetries of a vector space. Hencewe can apply the methods of linear algebra and functional analysis to the studyof groups and algebras. Representation theory also provides a generalization ofFourier analysis to groups. The applications of representation theory are diverse,both within pure mathematics and outside of it. For example in the book [17]abstract harmonic analysis is applied to number theory. Outside of mathematicsrepresentation theory has been used in physics, chemistry and even engineering,for the latter see for instance [2].

The theory of representations of finite groups was initiated in the 1890’s by peo-ple like Frobenius, Schur and Burnside. In the 1920’s representations of arbitrarycompact groups, and finite-dimensional (possibly nonunitary) representations ofthe classical matrix groups were investigated by Weyl and others. In the 1940’smathematicians such as Gelfand started to study (possibly infinite-dimensional)unitary representations of locally compact groups. Other important figures in rep-resentation theory include Harish-Chandra, Kirillov and Mackey. More on thehistory of representation theory can be found in [11].

Chapter 1 covers the results of Banach algebra and C*-algebra theory that weneed for representation theory. The main theorem of this chapter is the spectraltheorem for normal operators. In Chapter 2 we study locally compact groups andpresent basic results of Haar measures. Using the Haar measure we can defineconvolution of functions. The properties of this convolution are then investigated.We conclude the chapter with the construction of approximate identities.

In Chapter 3 we get to the main theme of the thesis, that is representation the-ory. We present the basic concepts of unitary representations of locally compactgroups. The first important result is Schur’s lemma, which describes irreducibil-ity of a representation in terms of commuting operators. Then we describe theconnection between unitary representations of a locally compact group and non-degenerate *-representations of the group algebra. In the last part of the chapter

ii

INTRODUCTION iii

we study functions of positive type. We establish a correspondence between thesefunctions and cyclic unitary representations. Then we can prove the last majorresult of the chapter, the Gelfand-Raikov theorem, which guarantees that locallycompact groups have enough irreducible representations to separate points.

The representations of compact groups are particularly well behaved, whichwe shall show in Chapter 4. We summarize the results of this chapter in thePeter-Weyl theorem.

The main references used were [8] for Banach algebra theory, [17] for the spec-tral theorem and its application to Schur’s lemma, and [5] for locally compactgroups and representation theory.

Chapter 1

Banach Algebras

Banach and C*-algebras have an important role in the representation theory oflocally compact groups. In this chapter we cover some of the basic theory of Banachalgebras. Then we shall focus on commutative Banach algebras and the Gelfandtheory of these algebras. We prove the Gelfand-Naimark theorem for commutativeunital C*-algebras. We conclude the chapter by using the aforementioned theoremto prove the spectral theorem for normal operators, which will play a crucial rolein representation theory.

1.1 Banach and C*-algebrasIn this text the scalar field will be C.

Definition 1.1.1. An algebra A is a vector space over C that is also a ring, withaddition being the vector addition, and for every x and y in A and λ, µ ∈ C theidentity

(λx)(µy) = λµ(xy)

holds. A subalgebra is a linear subspace of A that is also a subring of A.We shall denote the Banach dual of a normed space A by A∗.A normed linear space (A, ‖·‖) that is also an algebra is called a normed algebra

if‖xy‖ ≤ ‖x‖‖y‖

for every x and y in A. A normed algebra is a Banach algebra if it is also a Banachspace.

An algebra is commutative if xy = yx for all x and y in A. An algebra is unitalif there exists an element e ∈ A such that ex = xe = x for all x ∈ A. We shalldenote the identity element of a unital algebra by e. An element x of an unital

1

CHAPTER 1. BANACH ALGEBRAS 2

algebra is invertible if there exists an element y such that xy = yx = e. Denotey = x−1 and A× = {x ∈ A : x invertible in A}.

Definition 1.1.2. Let A be an algebra over C. An involution is a mapping∗ : x 7→ x∗ from A to A such that

(a) (x+ y)∗ = x∗ + y∗ and (λx)∗ = λx∗,

(b) (xy)∗ = y∗x∗ and (x∗)∗ = x

for all x, y ∈ A and λ ∈ C. This makes A into a *-algebra. A normed algebra (Ba-nach algebra) with an involution is called a normed *-algebra (Banach *-algebra)if the involution is isometric, that is if ‖x∗‖ = ‖x‖ for all x ∈ A.

Some algebras do not have an identity. However an algebra A can always beembedded into an algebra with identity. Let Ae = A ⊕ C. With multiplicationdefined by (x, λ)(y, µ) = (xy + µx + λy, λµ), norm ‖(x, λ)‖ = ‖x‖ + |λ|, andinvolution (x, λ)∗ = (x∗, λ), the space Ae becomes a unital Banach *-algebra withidentity (0, 1). This is called adjoining an identity to A.

A Banach algebra A with involution x 7→ x∗ is called a C*-algebra, if its normsatisfies the equation ‖x∗x‖ = ‖x‖2 for all x ∈ A. A closed subalgebra B of aC*-algebra A is called C*-subalgebra if x∗ ∈ B whenever x ∈ B. A C*-algebra isa Banach *-algebra since ‖x‖2 = ‖x∗x‖ ≤ ‖x∗‖‖x‖ implies ‖x‖ ≤ ‖x∗‖ and hence‖x‖ = ‖x∗‖ for every x ∈ A.

Example 1.1.3. Let X be locally compact Hausdorff space. We denote by Cb(X)the set of bounded continuous complex valued functions on X. A continuouscomplex valued function vanishes at infinity if for every ε > 0 there exists acompact subset K ⊂ X such that |f(x)| < ε whenever x ∈ X \ K. Denotethe set of all continuous functions that vanish at infinity by C0(X). The setsuppf = {x ∈ X : f(x) 6= 0} is the support of a function f . Denote by Cc(X) theset of all continuous functions that have compact support. All of the sets Cb(X),C0(X) and Cc(X) are algebras with pointwise addition, multiplication and scalarmultiplication. The norm is the supremum norm given by

‖f‖∞ = supx∈X|f(x)|.

The involution for these spaces is the complex conjugation f 7→ f given byf(x) = f(x). With this norm Cb(X) and C0(X) become commutative C*-algebras,whereas Cc(X) is complete only when X is compact. If X is not compact, thenonly Cb(X) is unital.


Example 1.1.4. LetH be a Hilbert space and denote the set of bounded operatorson H by L(H). Now L(H) is a C*-algebra since if T ∈ LH), then

‖T ∗‖ = sup‖u‖=1

‖T ∗u‖ = sup‖u‖=1

sup‖v‖=1

|〈T ∗u, v〉| = sup‖u‖=1

sup‖v‖=1

|〈u, Tv〉| ≤ ‖T‖,

so ‖T ∗T‖ ≤ ‖T‖2, and

‖T ∗T‖ = sup‖u‖=1

sup‖v‖=1

|〈Tu, Tv〉| ≥ sup‖u‖=1

‖Tu‖2 = ‖T‖2.

Therefore ‖T ∗T‖ = ‖T‖2.

The modest looking equality that ties the multiplication, involution and normof a C*-algebra turns out to have massive implications. Namely the (commutative)Gelfand-Naimark theorem, which we shall prove, states that every commutativeC*-algebra is of the form C0(X) for some locally compact Hausdorff space X, andmore generally every C*-algebra is a C*-subalgebra of L(H) for some Hilbert spaceH by the (noncommutative) Gelfand-Naimark theorem.

The algebras we study in this chapter are all normed algebras.When trying to understand an algebra, one natural question we may ask is,

assuming the algebra is unital, what can we say about the invertible elements? Ifthe algebra is a Banach algebra then the first nontrivial invertible element couldbe e − x for some ‖x‖ < 1 since then the series e +

∑∞n=1 x

n is convergent and isthe inverse of e−x, which we will prove in a slightly more general form in Lemma1.1.7. Modifying this example using scalar multiplication we obtain

λ−1(e− λ−1x)−1 = (λe− x)−1

if ‖x‖ < λ. This motivates our next definition.

Definition 1.1.5. For an element x ∈ A, the spectrum of x in A is

σA(x) = {λ ∈ C : λe− x 6∈ A×}.

The complement ρA(x) = C \σA(x) is called the resolvent set of x. For x ∈ A, thenumber

r(x) = inf{‖xn‖1/n : n ∈ N}

is called the spectral radius of x.

Clearly r(x) ≤ ‖x‖. In the definition of spectral radius the infimum can in factbe replaced by a limit.

Lemma 1.1.6. For every x ∈ A, r(x) = limn→∞ ‖xn‖1/n.


Proof. It is sufficient to show that for every ε > 0 there exists N(ε) ∈ N suchthat ‖xn‖1/n < r(x) + ε for every n ≥ N(ε). Let ε > 0. Pick k ∈ N such that‖xk‖1/k < r(x)+ε/2. Any n can be expressed in the form n = p(n)k+q(n), wherep(n) ∈ N, 0 ≤ q(n) ≤ k − 1. Therefore

p(n)

n=

1

k

(1− q(n)

n

)→ 1

k,

as n→∞. Hence ‖xk‖p(n)/n‖x‖q(n)/n → ‖xk‖1/k as n→∞. Therefore there existsnk ∈ N such that ‖xk‖p(n)/n‖x‖q(n)/n < ‖xk‖1/k +ε/2 for all n ≥ nk. It follows that

‖xn‖1/n ≤ ‖xk‖p(n)/n‖x‖q(n)/n < ‖xk‖1/k + ε/2 < r(x) + ε

for all n ≥ nk.

We already alluded to the following generalization of the geometric power series.

Lemma 1.1.7. Let A be a Banach algebra and let x ∈ A with r(x) < 1. Thene− x is invertible in A and

(e− x)−1 = e+∞∑n=1

xn.

Proof. Fix any η such that r(x) < η < 1. Then ‖xn‖1/n ≤ η for all n ≥ Nfor some N ∈ N. Then ‖xn‖ ≤ ηn for all n ≥ N , and since η < 1 the series∑∞

n=1 ‖xn‖ converges. Since A is complete, the sequence of partial sums ym =e+∑m

n=1 xn, m ∈ N converges in A with limit y = e+

∑∞n=1 x

n. Indeed, ‖y−ym‖ ≤∑∞n=m+1 ‖xn‖. Now

(e− x)ym = ym(e− x) = e− xm+1

for all m. Because ym → y and xm → 0 as m→∞, we conclude that (e− x)y =y(e− x) = e.

Note that if ‖x‖ < 1, then r(x) < 1 and the results of the above lemma hold.As a corollary to the above construction we gain some insight to the topology

of the set of invertible elements.

Lemma 1.1.8. Let A be a normed unital algebra.

(i) If x, y ∈ A× are such that ‖y − x‖ ≤ 12‖x−1‖−1, then

‖y−1 − x−1‖ ≤ 2‖x−1‖2‖y − x‖.

Moreover x 7→ x−1 is a homeomorphism of A×.


(ii) If A is complete, then A× is open, and if x ∈ A such that ‖x− e‖ < 1, thenx ∈ A×.

Proof. (i) If x and y are such that the inequality holds, then

‖y−1‖ − ‖x−1‖ ≤ ‖y−1 − x−1‖ ≤ ‖y−1‖‖x− y‖‖x−1‖ ≤ 1

2‖y−1‖,

so ‖y−1‖ ≤ 2‖x−1‖, and therefore

‖y−1 − x−1‖ ≤ ‖y−1‖‖x− y‖‖x−1‖ ≤ 2‖x−1‖2‖y − x‖.

Hence the bijection x 7→ x−1 of A× is continuous, and since it is its own inverse,it is a homeomorphism.

(ii) If ‖x− e‖ < 1, then by Lemma 1.1.7 we have e− (e− x) = x ∈ A×. Nowlet x be any element of A×, and let ‖y − x‖ < ‖x−1‖−1. Then

‖e− x−1y‖ ≤ ‖x−1‖‖x− y‖ < 1.

By what we have shown x−1y ∈ A×, and hence y ∈ A×. Therefore A× is open inA.

The following theorem justifies the the name spectral radius for r(x), and it isalso one of the most fundamental results in the theory of Banach algebras.

Theorem 1.1.9. Let A be a Banach algebra and x ∈ A. Then the spectrum σA(x)is a non-empty compact subset of C and

max{|λ| : λ ∈ σA(x)} = r(x).

Proof. First note that σA(x) is closed. This is true since A× is open, and ρA(x)is the inverse image of A× with respect to the continuous function λ 7→ λe − x.Moreover σA(x) is bounded, since if |λ| > r(x), then r((1/λ)x) < 1 and hence byLemma 1.1.7 λ(e− (1/λ)x) = λe−x ∈ A×, so σA(x) ⊂ {λ ∈ C : |λ| ≤ r(x)}. ThusσA(x) is compact.

Let us show next that σA(x) 6= ∅. Take any l ∈ A∗. We shall consider thefunction on ρA(x) defined by

f(λ) = l((λe− x)−1).

If λ, µ ∈ ρA(x), then

(λe−x)−1 = (λe−x)−1(µe−x)(µe−x)−1 = (λe−x)−1((µ−λ)e+λe−x)(µe−x)−1

= ((µ− λ)(λe− x)−1 + e)(µe− x)−1 = (µ− λ)(λe− x)−1(µe− x)−1 + (µe− x)−1.


Now if λ 6= µ, we have

f(λ)− f(µ)

λ− µ= −l((λe− x)−1(µe− x)−1).

Since l is continuous and y 7→ y−1 is continuous on A×,

limλ→µ

f(λ)− f(µ)

λ− µ= −l((µe− x)−2),

so in particular the function f is analytic on ρA(x). If |λ| > ‖x‖, then

(λe− x)−1 =

(λ

(e− 1

λx

))−1

=1

λ

(e− 1

λx

)−1

=1

λ

∞∑n=0

λ−nxn,

so

‖(λe− x)−1‖ ≤ 1

|λ|

∞∑n=0

(‖x‖|λ|

)n=

1

|λ|1

1− |λ|−1‖x‖,

which tends to zero as |λ| → ∞. Thus |f(λ)| ≤ ‖l‖‖(λe− x)−1‖, so f vanishes atinfinity.

Now assume σA(x) = ∅. Then clearly f is bounded on the closed disk |λ| ≤ ‖x‖since it is continuous. It follows that f is bounded on the whole complex plane, andso by Liouville’s theorem it is constant. Since f vanishes at infinity, we have f = 0.Because l ∈ A∗ was arbitrary, we get l((λe − x)−1) = 0 for each λ ∈ ρA(x) andall l ∈ A∗, so by Hahn-Banach theorem (λe − x)−1 = 0, which is a contradiction.Therefore σA(x) is nonempty.

Let s(x) = sup{|λ| : λ ∈ σ(x)}. Now s(x) ≤ r(x), since σA(x) ⊂ {λ ∈ C :|λ| ≤ r(x)}. Assume that s(x) < r(x). Then pick µ such that s(x) < µ < r(x).By what we have shown above, for l ∈ A∗ the function f(λ) = l((λe − x)−1) isanalytic on ρ(x), and in particular on the domain U = {λ : |λ| > s(x)}. Now for|λ| > ‖x‖, we have

f(λ) =∞∑n=0

λ−(n+1)l(xn).

This series is the Laurent series of f on the domain |λ| > ‖x‖. Since f is analyticon U , the uniqueness of the Laurent series implies that

∞∑n=0

l(xn)µ−(n+1)

converges. Therefore l(xn)µ−(n+1) → 0 as n → ∞. So for each l ∈ A∗ the set ofcomplex numbers

{l(xn)µ−(n+1) : n ∈ N}


is bounded. Denote by y ∈ A∗∗ the functional y(l) = l(y), where y ∈ A and l ∈ A∗.Letting yn = µ−(n+1)xn we see that supn∈N yn(l) < ∞ for each l ∈ A∗, so by theBanach-Steinhaus theorem there exists C > 0 such that ‖µ−(n+1)xn‖ ≤ C for alln ∈ N. Hence ‖xn‖ ≤ Cµn+1, so

r(x) = limn→∞

‖xn‖1/n ≤ limn→∞

(Cµn+1)1/n = µ.

This is a contradiction, so s(x) = r(x).

It turns out that the Banach algebras (over the complex field) that do not havenon-invertible elements other than zero are rather simple to describe.

Theorem 1.1.10 (Gelfand-Mazur theorem). Let A be a Banach algebra, and sup-pose each nonzero element is invertible. Then A is isomorphic to C.

Proof. Let a ∈ A. Since σA(a) is nonempty, pick λ ∈ σA(a). Now λe− a 6∈ A×, soby assumption λe− a = 0. Hence a = λe, so every element is a scalar multiple ofthe identity, so A is isomorphic to C.

By the above we should have a look at the non-invertible elements of an algebra.It follows from the defintion that if x is not invertible, then e 6∈ xA or e 6∈ Ax,so one of the sets xA or Ax is a proper subset of A. In fact such a set has somealgebraic structure.

Definition 1.1.11. A subset I of an algebra A is an ideal if I is a subspace of Aand aI ⊂ I and Ia ⊂ I for all a ∈ A. An ideal I is called proper if I 6= A. Anideal M is called maximal if it is proper and if I is an ideal of A such that M ⊂ Iand M 6= I, then I = A.

Every proper ideal of a unital algebra is in fact contained in a maximal ideal.

Lemma 1.1.12. Let I be a proper ideal of a unital algebra A. Then I is containedin some maximal ideal M .

Proof. Let I be a proper ideal of A. Let L be the set of all ideals L of A such thatI ⊂ L and e 6∈ L. Now L is nonempty since I ∈ L. The set L is an ordered setwith the inclusion order. We shall show that L satisfies the hypothesis of Zorn’slemma. Let K be a totally ordered subset of L and put L =

⋃{K : K ∈ K}.

Then e 6∈ L and L is an ideal since K is totally ordered. So L ∈ L and L is anupper bound for K. Hence, by Zorn’s lemma L has a maximal element M . If Jis a proper ideal containing M , then by maximality J = M , so M is a maximalideal.


Remark 1.1.13. Suppose A is a commutative. Then an element x ∈ A is invert-ible if and only if x 6∈ M for every maximal ideal M . Indeed, x 6∈ A× if and onlyif xA is a proper ideal, which is equivalent to xA ⊂M for some maximal ideal Mby the previous lemma.

Recall that if I is a closed subspace of A, then the quotient norm on A/I isdefined by

‖x+ I‖ = infa∈I‖x+ a‖.

The quotient of a normed (Banach) algebra by a closed ideal is again a normed(Banach) algebra.

Lemma 1.1.14. Assume I is a closed ideal of a normed algebra A. Then A/I,equipped with the quotient norm, is a normed algebra. If A is a Banach algebra,then so is A/I.

Proof. Since I is a closed subspace, we have that A/I is a Banach space if A iscomplete, so all we need to check is that the inequality ‖ab‖ ≤ ‖a‖‖b‖ holds forthe quotient norm. Now for any x, y ∈ A,

‖(x+ I)(y + I)‖ = ‖xy + I‖ = infz∈I‖xy + z‖ ≤ inf

a,b∈I‖(x+ a)(y + b)‖

≤ infa,b∈I‖x+ a‖‖y + b‖ = ‖x+ I‖‖y + I‖,

so the proof is complete.

Since our algebras have topological structure, closed ideals are of particularinterest.

Lemma 1.1.15. Let A be a Banach algebra and I ⊂ A be a proper ideal. Then

I ∩ {x ∈ A : ‖x− e‖ < 1} = ∅.

In particular I is also a proper ideal and every maximal ideal is closed in A.

Proof. If x ∈ A is such that ‖x− e‖ < 1, then e− (e− x) = x ∈ A×, so x 6∈ I.Now clearly I is a subspace, and if x ∈ I and a ∈ A, then

ax = a(limnxn) = lim

naxn ∈ I,

where xn ∈ I for every n ∈ N and limn xn = x, so I is an ideal. By the first partof the lemma, I does not contain e, so I is a proper ideal.

If M is a maximal ideal, then M ⊂M ⊂ A, so M = M .


1.2 Gelfand TheoryIn this subsection we study commutative Banach algebras. The main tool is theset of nonzero multiplicative functionals.

Definition 1.2.1. A linear functional ϕ on an algebra A is multiplicative ifϕ(xy) = ϕ(x)ϕ(y) for all x, y ∈ A. In other words ϕ is an algebra homomorphismbetween A and C. We denote the set of all non-zero multiplicative functionals ona Banach algebra A by ∆(A).

Let us prove some basic properties of multiplicative functionals.

Lemma 1.2.2. Let A be a Banach algebra with identity e. Suppose ϕ ∈ ∆(A).Then ϕ(e) = 1 and ϕ(x) 6= 0 for every invertible element x ∈ A.

Proof. Pick x ∈ A such that ϕ(x) 6= 0. Then ϕ(x)ϕ(e) = ϕ(xe) = ϕ(x) so dividingby ϕ(x) yields ϕ(e) = 1. If x ∈ A is invertible, then ϕ(x−1)ϕ(x) = ϕ(x−1x) =ϕ(e) = 1. Hence ϕ(x) 6= 0.

Remark 1.2.3. Because ψ(e) = 1 for every ψ ∈ ∆(Ae), each ϕ ∈ ∆(A) has aunique extension ϕ ∈ ∆(Ae) given by

ϕ(x+ λe) = ϕ(x) + λ, x ∈ A, λ ∈ C.

Let ∆(A) = {ϕ : ϕ ∈ ∆(A)}. Morever, let ϕ∞ denote the homomorphism from Aeto C with kernel A, that is, ϕ∞(x+ λe) = λ. Then

∆(Ae) = ∆(A) ∪ {ϕ∞}.

To see this, let ψ ∈ ∆(Ae) and ψ 6= ϕ∞. Then ψ|A ∈ ∆(A) since ψ is nonzero.Hence ψ = ψ|A. Identifying ∆(A) with ∆(A) ⊂ ∆(Ae) we always regard ∆(A) asa subset of ∆(Ae). In this sense, ∆(Ae) = ∆(A) ∪ {ϕ∞}.

It is worth noting that we do not need to assume that a multiplicative functionalis continuous. These mappings are in fact automatically continuous.

Lemma 1.2.4. Let A be a Banach algebra. Every ϕ ∈ ∆(A) is a bounded linearfunctional on A. In particular, ‖ϕ‖ ≤ 1 and ‖ϕ‖ = 1 if A is unital.

Proof. If |λ| > ‖x‖ then λe−x is invertible, so λ−ϕ(x) = ϕ(λe−x) 6= 0. In otherwords ϕ(x) 6= λ so |ϕ(x)| ≤ ‖x‖. Therefore ‖ϕ‖ ≤ 1 and ‖ϕ‖ = 1 if A is unital,since ϕ(e) = 1.


We will prove another automatic continuity result in Lemma 3.2.14. Automaticcontinuity of various maps, such as homomorphisms and derivations, is a field ofresearch in its own right, see for instance [4] and [19].

Observe that the above theorem says that ∆(A) is a subset of the closed unitball of A∗ or the unit sphere if A is unital.

It should be noted that an algebra does not necessarily have any nonzero multi-plicative functionals. For instance if H is a Hilbert space of dimension greater thanone, then ∆(L(H)) = ∅. We give a sketch of a proof. Note first that if ϕ ∈ ∆(A)and x is a nilpotent element, that is xn = 0 for some n, then ϕ(x)n = ϕ(xn) = 0,so ϕ(x) = 0. Now let {eλ}λ∈Λ be an orthonormal basis for H. Assume first thatdimH is even or infinite. Then there exists a partition of Λ to disjoint sets Λ1 andΛ2 with the same cardinality. Let β : Λ1 → Λ2 be a bijection. Now define oper-ators A(

∑λ∈C αλeλ) =

∑λ∈C∩Λ1

αλeβ(λ) and B(∑

λ∈C αλeλ) =∑

λ∈C∩Λ2αλeβ−1(λ).

It is easy to verify that A2 = 0 = B2, and (A + B)2 = I. Hence ϕ(I) =ϕ((A + B)2) = (ϕ(A) + ϕ(B))2 = 0, so ϕ = 0. If dimH = n is odd (or moregenerally finite) and greater than one, then let A(x1, . . . , xn−1, xn) = (e2, . . . , en, 0)and B(e1, . . . , en) = (0, . . . , 0, e1). Then An = 0 and B2 = 0, but (A+B)n = I, soϕ(I) = 0 for every multiplicative functional ϕ.

However a commutative unital Banach algebra has maximal ideals, and thoseideals correspond to multiplicative functionals, which we prove in Theorem 1.2.8.Because of this the results of Gelfand theory are often stated for commutativealgebras. The proofs of some of the following results do not seem to depend oncommutativity, however these statements are quite meaningless if the spectrum isempty. Hence we shall assume that the spectrum is nonempty, which is true whenthe algebra is commutative.

The set ∆(A) becomes a topological space when we give it the relative weak*topology from A∗. The space ∆(A) is often called the spectrum of A. Reader mayfind it confusing to use the term spectrum in two different contexts. However thetwo notions are in fact related, as we shall see.

It is important to know what kind of space the spectrum ∆(A) is.

Theorem 1.2.5. Let A be a Banach algebra. Then

(i) ∆(A) is compact Hausdorff if A has an identity;

(ii) ∆(A) is a locally compact Hausdorff space;

(iii) ∆(Ae) = ∆(A) ∪ {ϕ∞} is the one-point compactification of ∆(A).

Proof. (i) Since ∆(A) is subset of the unit ball, which is compact in the weak*-topology, it is sufficient to show that ∆(A) is closed in A∗. To see this, let (ϕλ)be a net in ∆(A) such that ϕλ → ϕ ∈ A∗ in the weak*-topology, so in other


words ϕλ(x) → ϕ(x) for every x ∈ A. Let x, y ∈ A. Now ϕ(xy) = limλ ϕλ(xy) =limλ ϕλ(x)ϕλ(y) = limλ ϕλ(x) limλ ϕλ(y) = ϕ(x)ϕ(y). Therefore ϕ is multiplica-tive. On the other hand ϕ(e) = limλ ϕλ(e) = limλ 1 = 1 so ϕ is not zero andϕ ∈ ∆(A). Hence ∆(A) is closed and is compact.

(ii) Now we assume A does not have an identity. We denote the basic neigh-borhoods of ∆(A) and ∆(Ae) by U and Ue, respectively. Then, for ϕ ∈ ∆(A),ε > 0 and a finite subset F ⊂ A,

Ue(ϕ, F, ε) =

{U(ϕ, F, ε) ∪ {ϕ∞} if |ϕ(x)| < ε for all x ∈ F ,U(ϕ, F, ε) otherwise.

Therefore the topology on ∆(A) coincides with the relative topology of ∆(Ae).Now if ϕ ∈ ∆(A) ⊂ ∆(Ae), we may find open disjoint neighborhoods U and Vsuch that ϕ ∈ U and ϕ∞ ∈ V . Now X \ V is compact in ∆(Ae), so it is alsocompact in ∆(A) and ϕ ∈ U ⊂ X \ V . Hence ∆(A) is locally compact.

(iii) Let x ∈ A and ε > 0. Now

Ue(ϕ∞, x, ε) = {ϕ∞} ∪ {ϕ ∈ ∆(A) : |ϕ(x)| < ε}= ∆(Ae) \ {ψ ∈ ∆(Ae) : |ψ(x)| ≥ ε}.

Now the sets {ψ ∈ ∆(Ae) : |ψ(x)| ≥ ε}, x ∈ A are closed in ∆(Ae) and hence com-pact. Finite union of such sets is compact too. Therefore the complement of a basicneighborhood Ue(ϕ∞, F, ε) is compact, so ∆(Ae) is the one-point compactificationof ∆(A).

Using the spectrum we get a rather natural representation for A.

Definition 1.2.6. For x ∈ A, we define x : ∆(A)→ C by x(ϕ) = ϕ(x). Then x isa continuous, since if ϕλ → ϕ, then x(ϕλ) = ϕλ(x)→ ϕ(x) = x(ϕ). The functionx is called the Gelfand transform of x. The mapping

ΓA : A→ C(∆(A)), x 7→ x

is called Gelfand representation of A.

It is easy to verify that the Gelfand representation is an algebra homomorphism.We will prove some important results for the Gelfand representation.

Theorem 1.2.7. Let A be a Banach algebra and Γ be the Gelfand representationof A.

(i) Γ maps A into C0(∆(A)) and is norm decreasing;

(ii) Γ(A) separates the points of ∆(A).


Proof. (i) If A is unital, then ∆(A) is compact, so C0(∆(A)) = C(∆(A)). Assumenow that A does not have an identity. Then ∆(Ae) is the one-point compactifica-tion of ∆(A) and x(ϕ∞) = 0 for x ∈ A, so x ∈ C0(∆(A)). Also

‖Γ(x)‖∞ = ‖x‖∞ = supϕ∈∆(A)

|x(ϕ)| = supϕ∈∆(A)

|ϕ(x)| ≤ supϕ∈∆(A)

‖ϕ‖‖x‖ ≤ ‖x‖.

(ii) If ϕ1 6= ϕ2, then necessarily there exists x ∈ A such that ϕ1(x) 6= ϕ2(x).Therefore x(ϕ1) 6= x(ϕ2) so Γ(A) separates points of ∆(A).

The spectrum of a commutative Banach algebra is sometimes called the max-imal ideal space of the algebra, which is an appropriate name by the followingtheorem.

Theorem 1.2.8. For a commutative unital Banach algebra A, the map

ϕ 7→ kerϕ = {x ∈ A : ϕ(x) = 0}

is a bijection between ∆(A) and the set of maximal ideals of A.

Proof. If ϕ ∈ ∆(A), then kerϕ is a maximal ideal. Let ϕ1, ϕ2 ∈ ∆(A) and assumenow that kerϕ1 = kerϕ2, and denote this ideal by I. Since e /∈ I and I is maximal,we can express any x ∈ A uniquely as

x = λe+ y, y ∈ I, λ ∈ C.

Now since ϕ(e) = 1 for any ϕ ∈ ∆(A), we get

ϕ1(x) = λϕ1(e) + ϕ1(y) = λ = λϕ2(e) + ϕ2(y) = ϕ2(x)

for every x ∈ A, so ϕ1 = ϕ2 and ϕ 7→ kerϕ is injective.Let M be a maximal ideal of A. Now M is closed in A, so A/M is a Banach

algebra. We shall show that if x+M 6= M for some x ∈ A, then x+M ∈ (A/M)×.First if x+M 6= M for some x ∈ A, then x ∈ A \M .

Let K = {m+ ax : m ∈M,a ∈ A} ⊂ A. Now K is in fact an ideal in A, sinceif m1,m2 ∈M , a1, a2 ∈ A and λ ∈ C, then

m1 + a1x+m2 + a2x = m1 +m2 + (a1 + a2)x ∈ K

and by commutativity

(m1 + a1x)a2 = m1a2 + (a1a2)x ∈ K.

Also K 6= M since x = 0 + ex ∈ K.


Since M ⊂ K and M 6= K, we have K = A due to maximality. Thereforethere exists m0 ∈ M and a0 ∈ A such that e = m0 + a0x, so e − a0x ∈ M .Therefore e + M = e + (a0x − e) + M = a0x + M = (a0 + M)(x + M), so(x+M)−1 = a0 +M ∈ A/M . By the Gelfand-Mazur theorem A/M is isomorphicto C. If we denote the quotient map from A to A/M by q and the isomorphismfrom A/M to C by i, then M = ker i ◦ q.

As we promised earlier, the spectrum of an element and the spectrum of analgebra are indeed related.

Theorem 1.2.9. Let A be a commutative unital Banach algebra. For each x ∈ Ax(∆(A)) = σA(x).

Proof. If λ ∈ ρA(x), then 0 6= ϕ(x− λe) = ϕ(x)− λ, so ϕ(x) ∈ C \ ρA(x) = σA(x).Therefore ϕ(x) ∈ σA(x) for every ϕ ∈ ∆(A), so x(∆(A)) ⊂ σ(x).

Conversely if λ ∈ σA(x), then I = (λe− x)A is a proper ideal in A and henceit is contained in some kerϕ for some ϕ ∈ ∆(A). It follows that λ ∈ x(∆(A)).

A stronger relation between the spectrum of an element and an algebra will becontained in the proof of the spectral theorem.

Now we turn our attention to C*-algebras.

Lemma 1.2.10. Let A be a C*-algebra. Then the Gelfand homomorphism is a*-homomorphism; that is, x∗ = x.

Proof. We have to show that ϕ(x∗) = ϕ(x) for ϕ ∈ ∆(A) and x ∈ A. We mayassume that A has identity A. Let

ϕ(x) = α + iβ and ϕ(x∗) = γ + iδ,

α, β, γ, δ ∈ R. Towards a contradiction, assume that β + δ 6= 0 and let

y = (β + δ)−1(x+ x∗ − (α + γ)e) ∈ A.

Now since e = e∗∗ = (e∗e)∗ = e∗e = e∗, we have y∗ = y and

ϕ(y) = (β + δ)−1(α + iβ + γ + iδ − (α + γ)) = i.

Therefore for all t ∈ R,

ϕ(y + tie) = ϕ(y) + ti = (t+ 1)i,

so |t + 1| = |ϕ(y + tie)| ≤ sup{|l(y + tie)| : l ∈ A∗, ‖l‖ ≤ 1} = ‖y + tie‖. Sincey = y∗, the C*-norm property gives

(t+ 1)2 ≤ ‖y + tie‖2 = ‖(y + tie)(y + tie)∗‖ = ‖(y + tie)(y − tie)‖= ‖y2 + t2e‖ ≤ ‖y2‖+ t2.


It follows that 2t+ 1 ≤ ‖y2‖ for every t ∈ R, which is impossible (take for instancet = ‖y2‖). This shows that β + δ = 0, so δ = −β. Hence

ϕ((ix)∗) = ϕ(−ix∗) = −iϕ(x∗) = −i(γ + iδ) = −β − iγ.

On the other hand ϕ(ix) = i(α+ iβ) = −β + iα. Going through exactly the samearguments with ix in place of x we obtain α + (−γ) = 0, so γ = α. This showsthat ϕ(x∗) = ϕ(x).

For commutative C*-algebras the spectral radius coincides with the norm ofthe element.

Lemma 1.2.11. Let A be a commutative C*-algebra and x ∈ A. Then r(x) = ‖x‖.

Proof. First note that if y∗ = y, then ‖y2‖ = ‖y‖2, so

‖y2n‖ = ‖y‖2n

for all n ≥ 0.Now (xx∗)m = xm(x∗)m for all nonnegative integers m. Hence

‖x‖2n = ‖xx∗‖2n−1

= ‖x2n(x∗)2n‖1/2 = ‖x2n(x2n)∗‖1/2 = ‖x2n‖.

Therefore‖x2n‖2−n = ‖x‖

for all n, so r(x) = ‖x‖.

The above argument also works for a normal operator T ∈ L(H). That is ifT ∈ L(H) such that T ∗T = TT ∗, then r(T ) = ‖T‖.

Now we can present the main theorem of this subsection.

Theorem 1.2.12 (Gelfand-Naimark Theorem). For a commutative unital C*-algebra A the Gelfand homomorphism is an isometric *-isomorphism from A ontoC(∆(A)).

Proof. It is sufficient to show that Γ is an isometry and surjective. If x ∈ A, then

‖x‖∞ = supϕ∈∆(A)

|x(ϕ)| = supϕ∈∆(A)

|ϕ(x)| = maxλ∈σA(x)

|λ| = r(x) = ‖x‖.

Now since Γ is an isometry and A is complete, we have that Γ(A) is closed inC(∆(A)). On the other hand Γ(A) separates points and contains constants, so byStone-Weierstrass we have Γ(A) = C(∆(A)).

The above theorem can also be stated and proved for commutative C*-algebraswithout an identity element. Then the Gelfand homomorphism is an isometric *-isomorphism from A onto C0(∆(A)). However we only need the unital case toprove the spectral theorem.


1.3 The Spectral TheoremIn linear algebra there are many useful results that are called spectral theoremsthat describe how a matrix can be diagonalized. For example a square matrix Tis normal if and only if there exists a unitary matrix U such that T = UDU∗

where D is a diagonal matrix. In this chapter we shall give a generalization of thistheorem for (possibly infinite-dimensional) operators.

Assume that T is normal. Then we denote the smallest, closed, self-adjoint,unital subalgebra containing T by AT . This is the closure of the algebra generatedby T , T ∗ and I, and is commutative because T is normal. It is not difficult tosee that if T = UDU∗ is a normal matrix, then AT is isomorphic to Cn withpointwise operations, where n is the number of distinct eigenvalues of T . HenceAT ”diagonalizes” to Cn. This is the form of the spectral theorem which we shallgeneralize.

In the next theorem we denote σ(T ) to be the spectrum with respect to L(H)and σA(T ) to be the spectrum with respect to the subalgebra AT .

Theorem 1.3.1 (Spectral Theorem). Let T ∈ L(H) be a normal operator. Thenthere exists an isometric *-isomorphism Φ : C(σ(T ))→ AT such that Φ(iσ(T )) = T ,where iσ(T ) : σ(T )→ C is the inclusion mapping iσ(T )(z) = z.

Proof. Consider the Gelfand transform of T , that is,

T : ∆(AT )→ Cγ 7→ γ(T ).

Now T is continuous. Moreover, if T (γ1) = T (γ2), then we have

γ1(T ∗) = γ1(T ) = γ2(T ) = γ2(T ∗).

Thus γ1 and γ2 agree on a unital subalgebra of L(H) generated by T and T ∗,and by continuity they agree on AT , so γ1 = γ2. Therefore T is injective. Nowsince ∆(AT ) is compact and T is continuous and injective, we have that ∆(AT ) ishomeomorphic to its image, which in fact is σA(T ). That is

T : ∆(AT )→ σA(T )

is a homeomorphism.Next consider the map

Ψ : C(σA(T ))→ C(∆(AT ))

f 7→ f ◦ T .


The map Ψ is an isometric *-isomorphism. Now

f ◦ T (γ) = f(T (γ)) = f(T (γ)) = f ◦ T (γ),

so Ψ is an isometric *-isomorphism. We now define Φ = Γ−1Ψ, so that the diagram

C(σA(T )) C(∆(AT ))

AT

Ψ

ΦΓ

commutes. Being a composition of isometric *-isomorphisms, Φ is also an isometric*-isomorphism. We consider the effect of Φ on a function f ∈ C(σA(T )). Firstnote that Ψ(f)(γ) = f(T (γ)) = f(γ(T )). Now since the Gelfand transform Γ isan isomorphism, for every f ∈ C(σ(T )) there exists unique P ∈ AT such thatΨ(f) = Γ(P ), so P = Γ−1Ψ(f) = Φ(f). So for every γ ∈ ∆(AT )

f(γ(T )) = Ψ(f)(γ) = Γ(P )(γ) = γ(P ) = γ(Φ(f)).

In particular γ(Φ(iσ(T ))) = iσ(T )(γ(T )) = γ(T ) and γ(Φ(1)) = 1 = γ(I) for everyγ ∈ ∆(AT ), so Φ(iσ(T )) = T and Φ(1) = I, which implies Φ(iσ(T )) = T andΦ(1) = I. It remains to show that σA(T ) = σ(T ). Clearly σ(T ) ⊂ σA(T ), since ifλI − T is invertible in AT then it is clearly invertible in L(H) as well.

Now let λ ∈ σA(T ), ε > 0 be arbitrary and choose f ∈ C(σA(T )) such that‖f‖∞ ≤ 1, f(λ) = 1 and f(µ) = 0 whenever |λ− µ| ≥ ε. Let P = Φ(f). Since Φis an isometry and f is zero outside a ball centered at λ, we have

‖(T − λI)P‖ = ‖Φ−1((T − λI)P )‖∞ = ‖(iσA(T ) − λ)f‖∞ ≤ ε.

Thus if T − λI is invertible, it would follow that

1 = ‖f‖∞ = ‖P‖ = ‖(T − λI)−1(T − λI)P‖≤ ‖(T − λI)−1‖‖(T − λI)P‖ ≤ ‖(T − λI)−1‖ε.

Since ε was arbitrary, this forces ‖(T − λI)−1‖ to infinity. Hence T − λI is notinvertible, so indeed λ ∈ σ(T ).

From the spectral theorem we get a useful corollary.

Lemma 1.3.2. Let T be a normal operator on a complex Hilbert space. Thefollowing are equivalent:

(i) σ(T ) is a point.


(ii) T is a scalar multiple of the identity operator.

(iii) AT = C.

Proof. (i) implies (ii): Now

T = Φ(iσ(T )) = Φ(i{λ}) = Φ(λχ{λ}) = λΦ(χ{λ}) = λI.

(ii) implies (iii): Since T = λI, we have AT = CI = C.(iii) implies (i): If σ(T ) has more than one point, then there exists f, g ∈

C(σ(T )) \ {0}, such that f vanishes outside of an open ball centered at someλ1 ∈ σ(T ) and g vanishes outside of an open ball centered at some differentλ2 ∈ σ(T ), and furthermore fg = 0. So Φ(f)Φ(g) = Φ(fg) = 0, which is acontradiction, since in AT = C all nonzero elements are invertible.

Chapter 2

Locally Compact Groups

The principal objects on which abstract harmonic analysis takes place are thelocally compact groups. The fundamental feature of a locally compact group,without which we could do little, is the existence and uniqueness of a translationinvariant measure λ. Such a measure also gives the space L1(λ) the structure ofa Banach *-algebra. These are the key ideas of this chapter. We conclude thechapter with the construction of approximate identities.

2.1 Haar measureDefinition 2.1.1. A topological group is a group G equipped with a topology suchthat the group operations are continuous, that is (x, y) 7→ xy is continuous fromG×G to G and x 7→ x−1 is continuous from G to G.

We shall denote the unit of a topological group by e. If A ⊂ G and x ∈ G, wedefine

Ax = {yx : y ∈ A}, xA = {xy : y ∈ A}, A−1 = {y−1 : y ∈ A},

and if B ⊂ G then we define

AB = {xy : x ∈ A, y ∈ B}.

We say that A is symmetric if A−1 = A.

Theorem 2.1.2. Let G be a topological group.

(i) For every neighborhood U of e there is a symmetric neighborhood V of e suchthat V V ⊂ U .

(ii) If A and B are compact sets in G, so is AB.

18

CHAPTER 2. LOCALLY COMPACT GROUPS 19

Proof. (i) Since (x, y) 7→ xy is continuous at e it follows that for every neighbor-hood U of e there exists neighborhoods V1 and V2 of e with V1V2 ⊂ U . We canchoose the desired set to be V = V1 ∩ V2 ∩ V −1

1 ∩ V −12 , which is clearly symmetric

and V V ⊂ V1V2 ⊂ U .(ii) The set AB is the image of a compact set A×B under the continuous map

(x, y) 7→ xy, hence AB is compact.

If f is a function on a topological group G and y ∈ G, we define the left andright translates of f through y by

Lyf(x) = f(y−1x), Ryf(x) = f(xy).

Here we use y−1 in Ly and y in Ry so that the maps y 7→ Ly and y 7→ Ry are grouphomomorphisms:

Lyx = LyLx, Ryz = RyRz.

We say that a function f on G is left (respectively right) uniformly continuousif ‖Lyf − f‖∞ → 0 (respectively ‖Ryf − f‖∞ → 0) as y → e. We shall denotethe set of bounded left (right) uniformly continuous functions on G by LUC(G)(RUC(G)).

Theorem 2.1.3. If G is a topological group, then Cc(G) ⊂ LUC(G) ∩RUC(G).

Proof. We shall prove f ∈ RUC(G). The argument for f ∈ LUC(G) is similar.Let f ∈ Cc(G) and ε > 0 and denote K = suppf . For every x ∈ K there exists aneighborhood Ux of e such that |f(xy)−f(x)| < 1

2ε for all y ∈ Ux, and there exists

a symmetric neighborhood Vx of e such that VxVx ⊂ Ux. The family {xVx}x∈Kis an open cover of K, so there x1, . . . , xn ∈ K such that K ⊂

⋃nk=1 xkVxk . Let

V =⋂nk=1 Vxk . We claim that ‖Ryf − f‖∞ < ε for y ∈ V .

If x ∈ K then there exists k such that x ∈ xkVxk , so xy ∈ xkVxkVxk ⊂ xkUxk .But then

|f(xy)− f(x)| ≤ |f(xy)− f(xk)|+ |f(xk)− f(x)| < 1

2ε+

1

2ε = ε.

Similarly, if xy ∈ K, then xy ∈ xkVxk for some k and x = xyy−1 ∈ xkVxkVxk ⊂xkUxk , so

|f(xy)− f(x)| ≤ |f(xy)− f(xk)|+ |f(xk)− f(x)| < ε.

If x, xy 6∈ K, then f(x) = f(xy) = 0.

Definition 2.1.4. By a locally compact group we shall mean a topological groupwhose topology is locally compact and Hausdorff.


In this text the Borel sets of a topological space are generated by open sets.

Definition 2.1.5. A left (respectively right) Haar measure on G is a nonzerocountably additive measure µ on G that satisfies the following properties:

(i) µ(xE) = µ(E) (µ(Ex) = µ(E)) for every x ∈ G and for every Borel setE ⊂ G;

(ii) µ(K) <∞ for every compact K;

(iii) µ(E) = inf{µ(U) : E ⊂ U,U open} for every Borel E ⊂ G;

(iv) µ(E) = sup{µ(K) : K ⊂ E,K compact} for every open E ⊂ G.

Theorem 2.1.6. Let G be a locally compact group.

(i) There exists a left Haar measure λ on G.

(ii) If λ is a left Haar measure on G, then λ(U) > 0 for every nonempty openset U , and

∫fdλ > 0 for every f ∈ C+

c (G) = {f ∈ Cc(G) : f ≥ 0} \ {0}.

(iii) If λ and µ are left Haar measures on G, then there exists c ∈ (0,∞) suchthat µ = cλ.

The proof can be found for instance in [5, p. 37, Theorem 2.10.]. In this bookan invariant nonzero positive functional on Cc(G) is constructed, and then by theRiesz representation theorem it is given by an appropriate measure. From now onwe always assume that G is locally compact.

Example 2.1.7.

(1) dx/|x| is a Haar measure on the multiplicative group R \ {0}.

(2) The ax+ b group is the group of affine transformations x 7→ ax+ b of R witha > 0 and b ∈ R. On G dadb/a2 is a left Haar measure and dadb/a is a rightHaar measure.

(3) Lebesgue measure∏

i<j dαij is a left and right Haar measure on the groupof n × n real matrices (αij) such that αij = 0 for i > j and αii = 1 for1 ≤ i ≤ n. This is the group of upper triangular matrices of with diagonalentries all equal to 1. When n = 3 the group is often called the Heisenberggroup.

(4) On the group GL(n,R) = {T ∈ L(Rn) : detT 6= 0} | detT |−ndT is a leftand right Haar measure, where dT is Lebesgue measure on Rn2 , where weinterpret Rn2 as the vector space of all real n× n matrices.


(5) It can be proved that the special linear group SL(2,R) = {T ∈ GL(2,R) :detT = 1} has the Iwasawa decomposition, that is

SL(2,R) =

{(1 x0 1

)(√a 0

0 1/√a

)(cos θ − sin θsin θ cos θ

): x, θ ∈ R, a > 0

}.

Using this decomposition a left and right Haar measure is given by dθ2π

dxdaa2

.

For the proof of (5) see [6, p. 255, 17.].

Definition 2.1.8. Let (X,µ) be a measure space and let 0 < p < ∞. Let Lp(µ)denote the set of all µ-measurable functions f (or rather their equivalence classes)such that |f |p is µ-integrable. In particular L1(µ) is the set µ-integrable functions.For 1 ≤ p <∞ we set

‖f‖p =

(∫|f |pdµ

)1/p

, f ∈ Lp(µ).

Let L∞(µ) denote the set of all essentially bounded functions (or rather theirequivalence classes), that is functions f that coincide with a bounded functionalmost everywhere with respect to µ. For f ∈ L∞(µ) we set

‖f‖∞ = inf{a ∈ R : µ({x : f(x) > a}) = 0}.

When G is not σ-compact, the Haar measure is not σ-finite. This results insome technical complications in the measure theory. We will mention some ofthese problems and explain why they are not serious.

Firstly here is a useful lemma.

Lemma 2.1.9. If G is a locally compact group, then G has an open, closed andσ-compact subgroup.

Proof. Let U be a symmetric compact neighborhood of e. Then H =⋃∞n=1 U

n isan open subgroup. Hence it is also closed since the cosets yH are also open andX \H =

⋃y 6∈H yH.

Now let G be a non-σ-compact locally compact group, with left Haar measureλ. By the previous lemma there is a subgroup H that is open, closed and σ-compact. Let Y be a subset of G that contains exactly one element from each leftcoset of H, so that G is a disjoint union of the sets yH, y ∈ Y . It is not difficultto see that the restriction of λ to the Borel subsets of H is a left Haar measureon H. Moreover, this restriction determines λ entirely. First of all, it determinesλ on the Borel subsets of each coset yH, since λ(yE) = λ(E). One might thenthink that for every Borel E ⊂ G one would have λ(E) =

∑y∈Y λ(E ∩ yH). In

fact what happens is the following.


Lemma 2.1.10. Suppose E ⊂ G is a Borel set. If E ⊂⋃∞j=1 yjH for some

countable set {yj} ⊂ Y , then λ(E) =∑∞

j=1 λ(E ∩ yjH). If E ∩ yH 6= 0 foruncountably many y, then λ(E) =∞.

Proof. The first claim follows from the countable additivity of the Haar measure.By outer regularity it suffices to assume that E is open. In this case λ(E∩yH) > 0whenever E∩yH 6= 0 since E∩yH is open. If this happens for uncountably manyy, then if we write

{y : λ(E ∩ yH) > 0} =∞⋃n=1

{y : λ(E ∩ yH) >

1

n

}we see that for some n there are uncountably many y for which λ(E ∩ yH) > 1

n,

and it follows that λ(E) =∞.

The above lemmas allow some theorems valid for σ-finite spaces to be general-ized to general locally compact groups.

Here is an example that is useful to consider. Let G = R × Rd, where Rd

is R with discrete topology. We can take H = R × {0} to be the subgroupas in the Lemma 2.1.9 and Y = {0} × Rd. To obtain Haar measure λ on Gjust take the Lebesgue measure on each horizontal line R × {y} and add themtogether as in Lemma 2.1.10. In particular Y is closed and λ(Y ) = ∞, but theintersection of Y with any coset of H, or with any compact set, has measure 0.Hence λ is not inner regular on Y . It also shows that λ is not quite the productof the Haar measures on R and Rd. Indeed the Haar measure of R is the familiarLebesgue measure µ and the Haar measure of Rd is the counting measure ν, so(µ× ν)(Y ) = µ({0})ν(R) = 0 · ∞ = 0 if we go by the convention 0 · ∞ = 0.

We will need three fundamental theorems in measure theory that do not holdfor general non-σ-compact spaces. These theorems are the Fubini’s theorem, theRadon-Nikodym theorem, and the duality of L1(µ) and L∞(µ). We will not givedetailed explanations why we can use these. These matters are discussed in [5,p. 43-46]. Also the third chapter of [7] covers the measure theory necessary forintegration on locally compact spaces.

We shall need Fubini’s theorem to reverse the order of integration in double in-tegrals

∫G

∫Gf(x, y)dλ(x)dλ(y). If the function f vanishes outside some σ-compact

set E ⊂ G×G, then there is no problem in doing this. Indeed the projections E1

and E2 of E onto the first and second factors are also σ-compact, and E ⊂ E1×E2.Therefore we can replace G × G by the σ-compact space E1 × E2, and then wemay apply Fubini’s theorem. This hypothesis usually holds when f is constructedfrom functions on G that belong to Lp(G) for some p < ∞, for such functionsvanish outside some σ-compact set

⋃∞j=1 yjH by Lemma 2.1.10. For instance when

dealing with convolution we consider functions of the form f(x, y) = g(x)h(x−1y).


If g vanishes outside some σ-compact A and h vanish outside some σ-compact B,then f vanishes outside A× AB, where AB is σ-compact.

Some kind of Radon-Nikodym theorem is necessary if we wish to obtain theaforementioned duality. A proof can be found in [7, Theorem 12.17.].

When µ is not σ-finite it is generally false that L∞(µ) = L1(µ)∗ with the usualdefinition of L∞(µ). However we can modify the definition of L∞(µ) to make theduality hold in the case of Haar measure on a locally compact group. A set E ⊂ Xis locally Borel is E ∩ F is Borel whenever F is Borel and µ(F ) < ∞. A locallyBorel set is locally null if µ(E ∩ F ) = 0 whenever F is Borel and µ(F ) < ∞. Anassertion about points of X is true locally almost everywhere if it is true except ona locally null set. A function f : X → C is locally measurable if f−1(A) is locallyBorel for every Borel set A ⊂ C. We now (re-)define L∞(µ) to be the set of locallymeasurable functions that are bounded locally almost everywhere. Functions thatagree locally almost everywhere are considered equivalent. The norm

‖f‖∞ = inf{c : |f(x)| ≤ c locally almost everywhere}makes L∞(µ) a Banach space. Now L∞(µ) = L1(µ)∗. In the case of Haar measureλ on a locally compact group the lemmas 2.1.9 and 2.1.10 can be used. The proofcan also be found in [7, Theorem 12.18.].

Henceforth L∞(µ) will always denote the space defined above. When µ isσ-finite the definition coincides with the usual definition of L∞(µ).

The following approximation result will be needed.

Theorem 2.1.11. Let 1 ≤ p <∞. Then Cc(G) is a dense subspace in Lp(µ).

For the proof see [7, Theorem 12.10.].Let G be a locally compact group with left Haar measure λ. If for every x ∈ G

we define λx(E) = λ(Ex), then λx is again a left Haar measure. By the uniquenessof Haar measure, there exists a number ∆(x) > 0 such that λx = ∆(x)λ, and∆(x) is independent of the original choice of λ. To see this, let µ and ν areleft Haar measures, c > 0 such that ν = cµ, and ∆1(x),∆2(x) > 0 such thatµ(Ex) = ∆1(x)µ(E) and ν(Ex) = ∆2(x)ν(E). Then for measurable set E ⊂ Gwith 0 < µ(E), ν(E) <∞ we have

∆1(x)cµ(E) = cµ(Ex) = ν(Ex) = ∆2(x)ν(E) = ∆2(x)cµ(E)

so dividing by cµ(E) we get ∆1(x) = ∆2(x). The function ∆ : G→ (0,∞) is calledthe modular function of G. We shall denote the multiplicative group of positivereal numbers by R×.Theorem 2.1.12. The modular function ∆ is a continuous homomorphism fromG to R×. Moreover, for any f ∈ L1(λ),∫

Ryfdλ = ∆(y−1)

∫fdλ.


For the proof see [5, Proposition 2.24.].A group is called unimodular if ∆ = 1. Compact groups are unimodular, since

the only compact subgroup of R× is {1}.The formula

dλ(x−1) = ∆(x−1)dλ(x)

is useful when making substitutions in integrals.

2.2 ConvolutionsFrom now on we shall assume that each locally compact group G is equipped witha fixed left Haar measure λ. We shall usually write dx for dλ(x), |E| for λ(E),and Lp(G) for Lp(λ).

The group operation on G together with the Haar measure can be used todefine another operation on L1(G).

Definition 2.2.1. If f, g ∈ L1(G), then the convolution of f and g is the functiondefined by

f ∗ g(x) =

∫f(y)g(y−1x)dy.

Sometimes the functions can be also taken from spaces other than L1(G).

By applying Fubini’s theorem we see that f ∗ g is integrable for almost everyx and that ‖f ∗ g‖1 ≤ ‖f‖1‖g‖1, for∫ ∫

|f(y)g(y−1x)|dxdy =

∫ ∫|f(y)g(x)|dxdy = ‖f‖1‖g‖1

by the left invariance of the Haar measure.The integral f ∗ g can be expressed in several different forms.

f ∗ g(x) =

∫f(y)g(y−1x)dy

=

∫f(xy)g(y−1)dy

=

∫f(y−1)g(yx)∆(y−1)dy

=

∫f(xy−1)g(y)∆(y−1)dy.


The convolution product is associative, since if f, g, h ∈ L1(G), then

(f ∗ g) ∗ h(x) =

∫(f ∗ g)(y)h(y−1x)dy =

∫ ∫f(z)g(z−1)dzh(y−1x)dy

=

∫f(z)

∫g(z−1y)h(y−1x)dydz =

∫f(z)

∫g(y)h(y−1z−1x)dydz

=

∫f(z)(g ∗ h)(z−1x)dz = f ∗ (g ∗ h)(x).

Remark 2.2.2. Convolution is commutative if and only if the underlying groupG is abelian. Indeed if G is abelian, then

f ∗ g(x) =

∫f(y)g(y−1x)dy =

∫g(xy)f(y−1)dy = g ∗ f(x).

Before showing the converse, observe that supp(f ∗ g) ⊂ (suppf)(suppg). Indeedif (f ∗ g)(x) 6= 0 for some x ∈ G, then there must be some y ∈ G such thatf(xy)g(y−1) 6= 0, so f(xy) 6= 0 and g(y−1) 6= 0. Therefore we have x = xyy−1 ∈(suppf)(suppg). Hence supp(f ∗ g) ⊂ (suppf)(suppg).

Now if G is nonabelian, then there exists x, y ∈ G such that xy 6= yx. SinceG is Hausdorff, there exists open disjoint neighborhoods W and W ′ of xy andyx respectively. Now by the joint continuity of the group operation there existsrelative compact neighborhoods U1 and U2 of x and V1 and V2 of y such thatU1V1 ⊂ W and V2U2 ⊂ W ′. Denoting U = U1∩U2 and V = V1∩V2 we get UV ⊂ Wand V U ⊂ W ′. Hence supp(χU ∗ χV ) ⊂ UV ⊂ W and supp(χV ∗ χU) ⊂ W ′, soχU ∗ χV 6= χV ∗ χU .

We will need the following lemma for integration.

Lemma 2.2.3 (Minkowski’s inequality for integrals). Let 1 ≤ p < ∞ and let(X,A, µ) and (Y,B, ν) be σ-finite measure spaces. Let φ be a complex valued A×Bmeasurable function on the product X × Y . Then(∫ ∣∣∣∣∫ φ(x, y)dν(y)

∣∣∣∣p dµ(x)

)1/p

≤∫ (∫

|φ(x, y)|pdµ(x)

)1/p

dν(y)

in the sense that if the right side is finite, then the left side exists, and the inequalityholds. The inequality can also be written as∥∥∥∥∫ φ(·, y)dν(y)

∥∥∥∥p

≤∫‖φ(·, y)‖pdν(y).


Proof. When p = 1 the claim follows from Fubini’s theorem. So assume p > 1 andlet

C =

∫ (∫|φ(x, y)|pdµ(x)

)1/p

dν(y) <∞.

It follows that∫|φ(x, y)|pdµ(x) < ∞ for almost all y. If q = p/(p − 1) and

g ∈ Lq(µ), then ∫|g(x)φ(x, y)|dx ≤ ‖g‖q

(∫|φ(x, y)|pdµ(x)

)1/p

by Hölder’s inequality. Thus∫ ∫|g(x)φ(x, y)|dµ(x)dν(y) ≤ C‖g‖q.

By Fubini’s theorem it follows that∫|g(x)φ(x, y)|dν(y) <∞

for almost all x. Since g ∈ Lq(µ) was arbitrary we see that∫|φ(x, y)|dν(y) <∞

for almost all x and so h(x) =∫φ(x, y)dν(y) exists for almost all x. By Fubini’s

theorem ∣∣∣∣∫ g(x)h(x)µ(x)

∣∣∣∣ ≤ ∫ ∫ |g(x)φ(x, y)|dν(y)dµ(x) ≤ C‖g‖q

Therefore there exists h′ ∈ Lp(µ) with ‖h′‖p ≤ C such that∫g(x)h(x)dµ(x) =

∫g(x)h′(x)dµ(x)

for each g ∈ Lq(µ).

Although we stated the previous theorem for σ-finite spaces, by what we havediscussed we can also use it for locally compact groups that are not necessarilyσ-compact.

We also need to know how convolution behaves when performed for functionsother than L1(G).


Theorem 2.2.4. Suppose 1 ≤ p ≤ ∞, f ∈ L1(G) and g ∈ Lp(G). Then we havef ∗ g ∈ Lp(G) and ‖f ∗ g‖p ≤ ‖f‖1‖g‖p.

Also if f, g ∈ Cc(G), then f ∗ g ∈ Cc(G).

Proof. By Minkowski’s inequality and left-invariance of the Lp norm for integralswe have

‖f ∗ g‖p =

∥∥∥∥∫ f(y)Lyg(·)dy∥∥∥∥p

≤∫|f(y)|‖Lyg(·)‖pdy = ‖f‖1‖g‖p

whenever 1 ≤ p < ∞. When p = ∞ we have |f ∗ g(x)| ≤∫|f(y)g(y−1x)|dy ≤

‖f‖1‖g‖∞.Let f, g ∈ Cc(G), x ∈ G and ε > 0. Denote

∫g(z−1)dz = C. Now f is

uniformly left continuous, so there exists a neighborhood U of e such that ‖Lyf −f‖∞ < ε/C whenever y ∈ U . Therefore

|f ∗ g(x)− f ∗ g(y)| =∣∣∣∣∫ [f(xz)− f(yz)]g(z−1)dz

∣∣∣∣≤

∫‖Lx−1f − Ly−1f‖∞|g(z−1)|dz = ‖Lyx−1f − f‖∞C < ε

whenever y ∈ Ux, proving that f ∗g is continuous. On the other hand supp(f ∗g) ⊂(suppf)(suppg) by Remark 2.2.2, so f ∗ g has compact support.

When G is discrete, the function δ defined by δ(e) = 1 and δ(x) = 0 wheneverx 6= e satisfies δ ∗ f = f ∗ δ = f for any f . Such function does not exist if G isnot discrete. However there is a net of functions with this kind of property. Butbefore we can prove that, let us show that the translation on Lp(G) is continuous.

Theorem 2.2.5. If 1 ≤ p < ∞ and f ∈ Lp(G), then ‖Lyf − f‖p → 0 and‖Ryf − f‖p → 0 as y → e.

Proof. First assume g ∈ Cc(G) and that V is a fixed compact neighborhood of e.Now K = (suppg)V ∪ V (suppg) is a compact set, and Lyg and Ryg are supportedin K whenever y ∈ V . Now

‖Lyg − g‖p ≤ µ(K)‖Lyg − g‖∞ → 0

as y → e, and similarly we get ‖Ryg − g‖p → 0.Now suppose f ∈ Lp(G). Then if ε > 0 is arbitrary, then there exists g ∈ Cc(G)

such that ‖f − g‖p < ε. We have ‖Lyf‖p = ‖f‖p and ‖Ryf‖p = ∆(y)−1/p‖f‖p ≤C‖f‖p for y ∈ V since V is compact. Hence

‖Ryf − f‖p ≤ ‖Ry(f − g)‖p + ‖Ryg − g‖p + ‖g − f‖p ≤ (C + 1)ε+ ‖Ryg − g‖pwhere the term ‖Ryg − g‖p → 0 when y → e. The case for Ly goes the sameway.


Theorem 2.2.6. Let U be a neighborhood base at e in G. For each U ∈ U let ψUbe a function such that suppψU ⊂ U , ψU ≥ 0, ψU(x) = ψU(x−1) and

∫ψU = 1.

Then ‖f ∗ ψU − f‖p → 0 as U → e if 1 ≤ p <∞ and f ∈ Lp(G) or if p =∞ andf ∈ RUC(G). Also ‖ψU ∗ f − f‖p → 0 as U → e if 1 ≤ p <∞ and f ∈ Lp(G) orif p =∞ and f ∈ LUC(G).

Proof. Since ψU(x−1) = ψU(x) and∫ψU = 1, we have

f ∗ ψU(y)− f(y) =

∫f(yx)ψU(x−1)dx− f(y)

∫ψU(x)dx

=

∫[Rxf(y)− f(y)]ψU(x)dx.

Then by Minkowski’s inequality for integrals we have

‖f ∗ ψU − f‖p ≤∫‖Rxf − f‖pψU(x)dx ≤ sup

x∈U‖Rxf − f‖p.

Hence ‖f ∗ ψU − f‖p → 0 by the previous theorem or by right uniform continuityof f if p =∞. The second claim follows in the same way, since

ψU ∗ f(y)− f(y) =

∫ψU(x)f(x−1y)dx−

∫ψU(x)f(y)dx

=

∫[Lxf(y)− f(y)]ψU(x)dx.

Remark 2.2.7. We do not need the symmetry of ψU when we prove that ψU ∗f →f . This will be relevant later.

A family {ψU} of functions as in the previous theorem is called an approximateidentity. There are plenty of approximate identities. For instance, if we take thesets U to be compact and symmetric and then take ψU = |U |−1χU , or we couldtake the ψU ’s to be continuous, or even smooth in some circumstances.

Chapter 3

Representation Theory

In this chapter we present the basic concepts of the theory of unitary represen-tations of locally compact groups. The main results that we prove are Schur’slemma and the Gelfand-Raikov theorem concerning the existence of irreduciblerepresentations. The key idea in the proof of the latter claim is the correspon-dence between cyclic representations and functions of positive type. We will alsotouch on the correspondence between unitary representations of groups and non-degenerate *-representations of the group algebra.

3.1 Hilbert Space TheoryIn this section we recall some of results concerning Hilbert spaces necessary forrepresentation theory.

Let V and X be complex vector spaces. A map T : V → X is antilinear ifT (αu + βv) = αTu + βTv for all α, β ∈ C and u, v ∈ V . A map B : V × V → Xis sesquilinear if B(·, v) is linear for every v ∈ V and B(u, ·) is antilinear for everyu ∈ V . A sesquilinear map from V × V to C is called a sesquilinear form on V .Sesquilinear maps are completely determined by their values on the diagonal.

Lemma 3.1.1 (The Polarization Identity). Suppose B : V×V → X is sesquilinear,and let Q(v) = B(v, v). Then for all u, v ∈ V,

B(u, v) =1

4[Q(u+ v)−Q(u− v) + iQ(u+ iv)− iQ(u− iv)].

Proof. Simply expand the expression on the right and collect the terms.

A sesquilinear form B on V is called Hermitian if B(v, u) = B(u, v) for allu, v ∈ V and positive (semi-definite) if B(u, u) ≥ 0 for all u ∈ V . A sesquilinearform B on a normed space V is called bounded if there exists M ≥ 0 such that|B(u, v)| ≤M‖u‖‖v‖ for every u, v ∈ V .

29

CHAPTER 3. REPRESENTATION THEORY 30

Lemma 3.1.2. A sesquilinear form B is Hermitian if and only if B(u, u) ∈ R forall u ∈ V. Every positive form is Hermitian.

Proof. If B is Hermitian, then B(u, u) = B(u, u) ∈ R. For any sesquilinear form wehave Q(au) = |a|2Q(u) whenever a ∈ C, so if B(u, u) ∈ R, then by the polarizationidentity

B(v, u) =1

4[Q(v + u)−Q(v − u)− iQ(v + iu) + iQ(v − iu)]

=1

4[Q(u+ v)−Q(u− v)− iQ(u− iv) + iQ(u+ iv)] = B(u, v).

The second assertion follows from the first one.

Lemma 3.1.3 (The Schwarz and Minkowski Inequalities). Let B be a positivesesquilinear form on V, and let Q(u) = B(u, u). Then

|B(u, v)|2 ≤ Q(u)Q(v), Q(u+ v)1/2 ≤ Q(u)1/2 +Q(v)1/2.

Proof. The usual proofs of these inequalities do not depend on the definiteness, sothey apply for positive forms.

An operator on a Hilbert space is called unitary if it is surjective and 〈Tu, Tv〉 =〈u, v〉. An operator is called self-adjoint if T ∗ = T . If 〈Tu, u〉 ≥ 0 then theoperator T is positive. Every positive operator is self-adjoint since 〈·, ·〉T = 〈T ·, ·〉is a positive form.

Theorem 3.1.4. If H is a Hilbert space and B : H × H → C is a boundedHermitian sesquilinear form, then there exists a bounded, self-adjoint operator T ∈L(H) such that B(u, v) = 〈Tu, v〉.

Proof. The map u 7→ B(u, v) defines a bounded functional for every v ∈ H, soby the Frechet-Riesz representation theorem for each v ∈ H there exists a vectorvB ∈ H such that B(u, v) = 〈u, vB〉. Now the map T (v) = vB is linear andbounded, since

〈u, T (αv + v′)〉 = B(u, αv + v′) = αB(u, v) +B(u, v′)

= α〈u, T (v)〉+ 〈u, T (v′)〉 = 〈u, αT (v) + T (v′)〉,

so T is indeed linear and ‖Tv‖ = sup‖u‖=1 |〈Tv, u〉| = sup‖u‖=1 |B(u, v)| ≤ M‖v‖.Also

〈u, Tv〉 = B(u, v) = B(v, u) = 〈v, Tu〉 = 〈Tu, v〉

so T is self-adjoint.


Recall that an operator T ∈ L(H) is compact if the image of any bounded sub-set of H is relatively compact, that is, its closure is compact. It will be importantto know the properties of compact operators when developing the representationtheory of compact groups. The proofs of the following theorems concerning com-pact operators can be found in [10, Chapter 8].

Theorem 3.1.5. Let H be a Hilbert space.

(i) Every finite rank operator is compact.

(ii) The set of compact operators in L(H) is closed in the operator topology.

Theorem 3.1.6. If T ∈ L(H) is self-adjoint and compact, there exists an or-thonormal basis for H consisting of eigenvectors of T . Each eigenspace is finitedimensional.

Let {Hα}α∈A be a family of Hilbert spaces. The direct sum⊕

α∈AHα is theset of all v = (vα)α∈A in the Cartesian product

∏α∈AHα such that

∑‖vα‖2 <∞.

This condition implies that vα = 0 for all but a countably many α. The space⊕α∈AHα is a Hilbert space with the inner product

〈u, v〉 =∑α∈A

〈uα, vα〉,

and the summands Hα are embedded in it as mutually orthogonal closed sub-spaces.

3.2 Unitary RepresentationsDefinition 3.2.1. Let G be a locally compact group. A unitary representationof G is a homomorphism π from G into the group U(Hπ) of unitary operatorson some Hilbert space Hπ that is continuous with respect to the strong operatortopology. In other words a map π : G → U(Hπ) such that π(xy) = π(x)π(y)and π(x−1) = π(x)−1 = π(x)∗, and for which x 7→ π(x)u is continuous from G toHπ for any u ∈ Hπ. The space Hπ is called the representation space of π and itsdimension is called the dimension or degree of π.

In this thesis we are concerned almost exclusively with unitary representations.It is worth noting that strong continuity is implied by the seemingly less re-

strictive condition of weak continuity, namely, that x 7→ 〈π(x)u, v〉 should becontinuous from G to C for every u, v ∈ Hπ. This is true since the strong andweak operator topologies coincide on U(Hπ). Indeed, if (Tα) is a net of unitary


operators converging to T in the weak operator topology of U(Hπ), then for anyu ∈ Hπ

‖(Tα − T )u‖2 = ‖Tαu‖2 − 2Re〈Tαu, Tu〉+ ‖Tu‖2 = 2‖u‖2 − 2Re〈Tαu, Tu〉.

The last term converges to 2‖Tu‖2 = 2‖u‖2, so ‖(Tα − T )u‖ → 0.

Example 3.2.2. Left translations yield the left regular representation πL of G onL2(G), which is defined by

[πL(x)f ](y) = Lxf(y) = f(x−1y).

Similarly one can define the right regular representation πR on L2(G) (with leftHaar measure)

[πR(x)f ](y) = ∆(x)1/2Rxf(y) = ∆(x)1/2f(yx).

In this text the regular representation that we treat is the left one. In factthese two are equivalent in a sense that will be explained below.

Any unitary representation π of G on Hπ defines another representation π onthe dual space H∗π of Hπ, namely π(x) = π(x−1)′ where the prime denotes thetranspose. This representation π is called the contragedient of π. Let v′ = 〈·, v〉and denote the inner product on the dualH∗π by 〈u′, v′〉′ = 〈v, u〉. Now if u, v ∈ Hπ,then by the formula T ′v′ = (T ∗v)′ we get

〈π(x)u′, v′〉′ = 〈π(x−1)′u′, v′〉′ = 〈(π(x)u)′, v′〉′ = 〈v, π(x)u〉 = 〈π(x)u, v〉.

Hence the contragedient of π is something like the ”complex conjugate” of π.

Definition 3.2.3. If π1 and π2 are unitary representations of G, an intertwiningoperator for π1 and π2 is a bounded linear map T : Hπ1 → Hπ2 such that Tπ1(x) =π2(x)T for all x ∈ G. The set of all intertwining operators is denoted by C(π1, π2).Two representations π1 and π2 are (unitarily) equivalent if C(π1, π2) contains aunitary transformation U : Hπ1 → Hπ2 , so that π2(x) = Uπ1(x)U−1. By unitarytransformation we simply mean a linear surjective isometry.

Example 3.2.4. The left and right regular representations are unitarily equiva-lent, and the intertwining operator T ∈ C(πL, πR) is given by

Tf(y) = ∆(y−1)1/2f(y−1).

We shall write C(π) for C(π, π). This is the space of bounded operators on Hπ

that commute with π(x) for every x ∈ G. It is called the commutator or centralizerof π. The commutator is in fact a *-algebra that is closed in the weak operatortopology, that is, a von Neumann algebra.


If a unitary representation π of G is of the form

π(x) =

(π1(x) 0

0 π2(x)

)(3.1)

where π1 and π2 are unitary representations of G, then sometimes it is better toanalyze π1 and π2 if we wish to understand π.

Definition 3.2.5. A closed subspaceM of Hπ is called an invariant subspace forπ if π(x)M ⊂ M for all x ∈ G. If M 6= {0} is invariant, the restriction of π toM,

πM(x) = π(x)|M,

defines a representation of G onM, called a subrepresentation of π. If π admitsa closed invariant subspaceM that is nontrivial, that isM 6= {0} andM 6= Hπ,then π is called reducible, otherwise π is irreducible.

If {πα}α∈A is a family of unitary representations, their direct sum⊕

πα isthe representation π on H =

⊕Hπα defined by π(x)(

∑vα) =

∑πα(x)vα, where

vα ∈ Hπα . In this case the spaces Hπα , as subspaces of H, are invariant under π,and each πα is a subrepresentation of π.

Theorem 3.2.6. IfM is invariant under π, then so isM⊥.

Proof. If u ∈M and v ∈M⊥, then

〈u, π(x)v〉 = 〈π(x)∗u, v〉 = 〈π(x−1)u, v〉 = 0,

so π(x)v ∈M⊥.

As a corollary if π has a nontrivial invariant subspaceM, then π is the directsum of πM and πM⊥ . This result is false for non-unitary representations. For ex-

ample, π(t) =

(1 t0 1

)defines a representation of R on C2, and the only nontrivial

invariant subspace is the one spanned by (1, 0).

Definition 3.2.7. If π is a representation of G and u ∈ Hπ, the closed linear spanMu of {π(x)u : x ∈ G} in Hπ is called the cyclic subspace generated by u. ClearlyMu is invariant under π. IfMu = Hπ, then u is called the cyclic vector for π. Arepresentation π is called a cyclic representation if it has a cyclic vector.

Remark 3.2.8. Every irreducible representation is a cyclic representation. Fur-thermore every nonzero vector in the representation space is cyclic. To see this,pick any u 6= 0. Now Mu 6= {0}, so by irreducibility it is the whole space.


However a cyclic representation is not necessarily irreducible. Consider the therepresentation ρ of R on C2 given by

ρ(x) =

(cosx − sinxsinx cosx

).

This is a unitary cyclic representation with cyclic vector (1, 0), since ρ(π/2)(1, 0) =(0, 1). It is not irreducible, since

ρ(x) =

(cosx − sinxsinx cosx

)=

(−1/√

2 i/√

2

i/√

2 −1/√

2

)(eix 00 e−ix

)(−1/√

2 −i/√

2

−i/√

2 −1/√

2

).

Here the invariant subspaces are

U{(x, 0) : x ∈ C} = {(−x, ix) : x ∈ C}

andU{(0, y) : y ∈ C} = {(iy,−y) : y ∈ C}.

Theorem 3.2.9. Every unitary representation is a direct sum of cyclic represen-tations.

Proof. Let π be a representation on Hπ. By Zorn’s lemma, there is a maximalcollection {Mα}α∈A of mutually orthogonal cyclic subspaces of Hπ. If there isa nonzero u ∈ Hπ orthogonal to all the subspaces Mα, the cyclic subspace gen-erated by u would also be orthogonal to the subspaces Mα, since 〈π(x)u,m〉 =〈u, π(x−1)m〉 = 0 whenever x ∈ G and m ∈ Mα. This contradicts maximality.Hence Hπ =

⊕Mα, and π =

⊕πMα .

One may observe that if π is a unitary representation as in (3.1), then every

π(x) commutes with nontrivial elements(λI 00 µI

), where λ, µ ∈ C may differ.

This suggests a relationship between the reducibility of a representation and theintertwining operators.

Theorem 3.2.10. LetM be a closed subspace of Hπ and let P be the orthogonalprojection ontoM. ThenM is invariant under π if and only if P ∈ C(π).

Proof. If P ∈ C(π) and v ∈ M, then π(x)v = π(x)Pv = Pπ(x)v ∈ M, so M isinvariant under π. Conversely, ifM is invariant, then π(x)Pv = π(x)v = Pπ(x)vfor v ∈ M and π(x)Pv = 0 = Pπ(x)v for v ∈ M⊥, since by Theorem 3.2.6π(x)v ∈M⊥. Hence π(x)P = Pπ(x).

The picture is completed by Schur’s lemma, which is one of the fundamentaltheorems in the subject.


Theorem 3.2.11 (Schur’s Lemma).

(a) A unitary representation π of G is irreducible if and only if C(π) containsonly scalar multiples of the identity.

(b) Suppose π1 and π2 are irreducible unitary representations of G. If π1 and π2

are equivalent then C(π1, π2) is one-dimensional. Otherwise C(π1, π2) = {0}.

Proof. (a) If π is reducible, then C(π) contains nontrivial projections.Now let π be irreducible and T ∈ C(π). First we assume that T is normal.

Assume H is nontrivial. Let λ ∈ σ(T ). Then we can find f ∈ C(σ(T )) such thatf 6= 0 and f vanishes outside an open neighborhood of λ. Let Φ : C(σ(T ))→ ATbe the isometry of the spectral theorem. Then W = Φ(f)H is invariant underπ. This is so since Φ(f) is a limit of polynomials in T , T ∗ and I, each of whichcommutes with π(g) for every g ∈ G. In other words if Φ(f) = limn→∞ pn(T ),where pn(T ) is a polynomial in T , T ∗ and I, we have

π(g)Φ(f)H = π(g) limn→∞

pn(T )H = limn→∞

pn(T )π(g)H = Φ(f)H,

so W = Φ(f)H is also invariant. Since π is irreducible and f is nontrivial, we haveW = H (otherwise we would have W = {0}, and so Φ(f) = 0).

Now suppose that σ(T ) is not a singleton. Then there exists some µ ∈ σ(T )distinct from λ, so we can pick two nonzero functions f, h ∈ C(σ(T )) such thattheir supports are disjoint. But then

{0} = Φ(h)Φ(f)H

and W 6= H, since otherwise we would have Φ(h)H = Φ(h)W = {0} so Φ(h) = 0which is a contradiction. Hence σ(T ) contains at most one point, so T = λI.

Now let T ∈ C(π). Then T = A+iB, where A = 12(T +T ∗) and B = 1

2i(T−T ∗)

are self-adjoint and therefore normal. Furthermore A,B ∈ C(π) so A = cI andB = dI. Therefore T = (c+ id)I proving that C(π) ∈ CI when π is irreducible.

(b) If T ∈ C(π1, π2) then T ∗ ∈ C(π2, π1) because

T ∗π2(x) = [π2(x−1)T ]∗ = [Tπ1(x−1)]∗ = π1(x)T ∗.

It follows that T ∗T ∈ C(π1) and TT ∗ ∈ C(π2), so T ∗T = cI and TT ∗ = dI forsome c, d ∈ R. In fact c = d, since if c 6= 0 then c2I = T ∗TT ∗T = dT ∗T = cdIso cI = dI. Similarly if d 6= 0 then d2I = TT ∗TT ∗ = cTT ∗ = cdI so cI = dI. IfT ∗T = 0, then ‖Tu‖2 = 〈T ∗Tu, u〉 = 0 for all u ∈ Hπ1 . Hence, either T = 0 orc−1/2T is unitary. This shows precisely that C(π1, π2) = {0} when π1 and π2 areinequivalent, and that C(π1, π2) consists of scalar multiples of unitary operators.If T1, T2 ∈ C(π1, π2) are unitary then T−1

2 T1 = T ∗2 T1 ∈ C(π1), so T−12 T1 = cI and

T1 = cT2, so dim C(π1, π2) = 1.


When studying reducibility it is often more convenient to work with operators,as we shall see.

As an immediate corollary we get a description of the irreducible representa-tions of abelian groups.

Corollary 3.2.12. If G is abelian, then every irreducible representation of G isone-dimensional.

Proof. If π is a representation of G, the operators π(x) all commute with oneanother and so belong to C(π). If π is irreducible, we have π(x) = cxI for eachx ∈ G. But then every one-dimensional subspace of Hπ is invariant, so dimHπ =1.

The representation theory of locally compact abelian groups is well understood.The irreducible representations form a group G called the dual group of G. Formore on this topic see [5, Chapter 4].

Irreducible unitary representations of a locally compact group are the basicbuilding blocks of harmonic analysis associated to that group, just like primenumbers are the building blocks of integers. However the relationship between anarbitrary unitary representation of a group and the irreducible unitary represen-tations of that group is not quite as straightforward as the one between an integerand its factorization to prime numbers. For starters it is not obvious that a givengroup has any irreducible representations other than the trivial one-dimensionalrepresentation π0(x) = 1. But in fact there are enough irreducible unitary repre-sentations to separate points of G. This is the Gelfand-Raikov theorem, which weshall prove at the end of this chapter. Hence the basic questions of representationtheory of G are the following.

(i) Describe all the irreducible unitary representations of G, up to equivalence.

(ii) Determine how arbitrary unitary representations of G can be built fromirreducible ones.

(iii) Given a specific unitary representation of G such as the regular representa-tion, show concretely how to build it out of irreducible ones.

The answer to (i) naturally depends strongly on the particular group. The irre-ducible representations have been determined for many groups, however we do notdiscuss any of these examples in this text. See for instance [5] or [9].

As to question (ii), one might wish that every unitary representation wouldbe a direct sum of irreducible subrepresentations. When the group is compactthis is true as we shall see, but it is not true generally. Consider the left regularrepresentation of R on L2(R), [πL(x)f ](t) = f(t− x). This representation has no


irreducible subrepresentations. If there was one it would be one-dimensional byCorollary 3.2.12, so the invariant subspace would be of the form {cf : c ∈ C} forsome f ∈ L2(R)\{0}. But then we would have f(t−x) = [πL(x)f ](t) = cxf(t) forsome cx ∈ T = {z ∈ C : |z| = 1}, so |f | is a constant function. This is impossiblefor f ∈ L2(G) unless f = 0. For cases such as these one needs a direct integral ofirreducible representations. Direct sum is a special case of this. We will not gointo this direct integrals in this text, see for instance [5, Section 7.4.].

The Peter-Weyl theorem answers question (iii) for the left regular representa-tion of compact groups.

If G is a locally compact group, then L1(G) is a Banach *-algebra under theconvolution product and the involution f ∗(x) = ∆(x−1)f(x−1).

Definition 3.2.13. Let A be a Banach *-algebra and H a Hilbert space. Amapping φ : A → L(H) is a nondegenerate *-representation of A on H if it is*-homomorphism and φ(A)H = {φ(a)v : a ∈ A, v ∈ H} is dense in H.

The nondegeneracy condition can be easily seen to be equivalent with thecondition that for every v ∈ H \ {0} there exists a ∈ A such that φ(a)v 6= 0.Indeed if there exists v ∈ H \ {0} such that φ(a)v = 0 for every a ∈ A, then〈v, φ(a)u〉 = 〈φ(a∗)v, u〉 = 0 for every a ∈ A and u ∈ H, so v ∈ (φ(A)H)⊥.Hence φ(A)H is not dense in H. On the other hand if v ∈ (φ(A)H)⊥, then0 = 〈v, φ(a)u〉 = 〈φ(a∗)v, u〉 for all a ∈ A and u ∈ H. Hence φ(a)v = 0 for everya ∈ A, so by assumption v = 0. Therefore φ(A)H is dense in H.

Note that *-representations of Banach *-algebras are not assumed to be contin-uous in any topology. They are in fact automatically continuous by the followinglemma.

Lemma 3.2.14. Let A be a Banach *-algebra and B a C*-algebra. If φ : A→ Bis *-homomorphism, then ‖φ‖ ≤ 1.

Proof. If A is not unital we can adjoin an identity to it. Now φ(eA) is an identityof φ(A), so we may assume that B is unital and φ(eA) = eB.

For every x ∈ A we have σ(φ(x)) ⊂ σ(x), so

‖φ(x)‖2 = ‖φ(x∗x)‖ = r(φ(x∗x)) ≤ r(x∗x) ≤ ‖x∗x‖ ≤ ‖x‖2.

Here the first two equalities hold for C*-algebras (recall Lemma 1.2.11).

Any unitary representation π of G determines a representation of L1(G), stilldenoted by π, in the following way. If f ∈ L1(G), we define a bounded operatorπ(f) on Hπ by

π(f) =

∫f(x)π(x)dx.


We will explain how this integral should be interpreted. For any u ∈ Hπ, we defineπ(f)u by specifying its inner product with an arbitrary v ∈ Hπ, which is given by

〈π(f)u, v〉 =

∫f(x)〈π(x)u, v〉dx.

Since 〈π(·)u, v〉 is a bounded continuous function on G, the integral on the rightis an ordinary integral of a function in L1(G). It is clear from the above formulathat 〈π(f)u, v〉 depends linearly on u and antilinearly on v and that |〈π(f)u, v〉| ≤‖f‖1‖u‖‖v‖, so π(f) really is a bounded operator onHπ with norm ‖π(f)‖ ≤ ‖f‖1.

Example 3.2.15. Let πL be the left regular representation of G. Then πL(f) isgiven by convolution with f on the left, since if f ∈ L1(G) and g, h ∈ L2(G), then

〈πL(f)g, h〉 =

∫f(x)

(∫[πL(x)g](y)h(y)dy

)dx

=

∫ (∫f(x)g(x−1y)dx

)h(y)dy =

∫(f ∗ g)(y)h(y)dy = 〈f ∗ g, h〉.

Theorem 3.2.16. Let π be a unitary representation of G. The map f 7→ π(f)is nondegenerate *-representation of L1(G) on Hπ. Moreover, for x ∈ G andf ∈ L1(G),

π(x)π(f) = π(Lxf), π(f)π(x) = ∆(x−1)π(Rx−1f).

Proof. The correspondence f 7→ π(f) is clearly linear. Now for u, v ∈ Hπ we have

〈π(f ∗ g)u, v〉 =

∫(f ∗ g)(x)〈π(x)u, v〉dx

=

∫ ∫f(y)g(y−1x)〈π(x)u, v〉dxdy =

∫ ∫f(y)g(x)〈π(yx)u, v〉dxdy

=

∫ ∫f(y)g(x)〈π(y)π(x)u, v〉dxdy =

∫f(y)

∫g(x)〈π(x)u, π(y)∗v〉dxdy

=

∫f(y)〈π(g)u, π(y)∗v〉dy =

∫f(y)〈π(y)π(g)u, v〉dy = 〈π(f)π(g)u, v〉,

〈π(f ∗)u, v〉 =

∫∆(x−1)f(x−1)〈π(x)u, v〉dx =

∫f(x)〈π(x−1)u, v〉dx

=

∫〈u, f(x)π(x)v〉dx = 〈u, π(f)v〉 = 〈π(f)∗u, v〉,


〈π(x)π(f)u, v〉 = 〈π(f)u, π(x)∗v〉 =

∫f(y)〈π(xy)u, v〉dy

=

∫f(x−1y)〈π(y)u, v〉dy = 〈π(Lxf)u, v〉,

〈π(f)π(x)u, v〉 =

∫f(y)〈π(y)π(x)u, v〉dy =

∫f(y)〈π(yx)u, v〉dy

= ∆(x−1)

∫f(yx−1)〈π(y)u, v〉dy = ∆(x−1)〈π(Rx−1f)u, v〉.

This shows that π is a *-homomorphism. To see that π is nondegenerate, supposeu ∈ Hπ \ {0}. Pick a compact neighborhood V of e such that ‖π(x)u− u‖ < ‖u‖for x ∈ V , and set f = |V |−1χV . Then

‖π(f)u− u‖ =1

|V |sup‖v‖=1

∣∣∣∣∫V

〈π(x)u− u, v〉dx∣∣∣∣ < ‖u‖

and in particular π(f)u 6= 0.

Conversely nondegenerate *-representation of L1(G) defines a unitary repre-sentation of G.

Theorem 3.2.17. Suppose π is a nondegenerate *-representation of L1(G) on aHilbert space H. Then π arises from a unique unitary representation of G on Hin the way we described above.

We will not prove this claim, as we don’t need this theorem. The proof can befound in [5, Theorem 3.11.]. The idea is that if {ψU} is an approximate identityin G, then π(x) should be the limit of π(LxψU).

3.3 The Gelfand-Raikov TheoremIt is not obvious where one should look for nontrivial irreducible unitary represen-tations for a group G. In this section we shall describe a method of turning thegroup algebra L1(G) into Hilbert spaces on which the group G acts unitarily. Infact every cyclic unitary representation, and in particular every irreducible unitaryrepresentation, arises in this way up to unitary equivalence.

Definition 3.3.1. A function of positive type on a locally compact group G isa function φ ∈ L∞(G) that defines a positive linear functional on the Banach*-algebra L1(G), that is∫

(f ∗ ∗ f)φ ≥ 0 for all f ∈ L1(G).


We have∫(f ∗ ∗ f)φ =

∫ ∫∆(y−1)f(y−1)f(y−1x)φ(x)dydx

=

∫ ∫f(y)f(yx)φ(x)dydx =

∫ ∫f(x)f(y)φ(y−1x)dxdy.

It turns out that every function of positive type agrees locally almost every-where with a continuous function, as we shall see. Let

P = P(G) = the set of all continuous functions of positive type on G.

Theorem 3.3.2. If φ is of positive type, then so is φ.

Proof. For any f ∈ L1(G), we have∫(f ∗ ∗ f)φ =

∫ ∫f(y)f(yx)φ(x)dydx =

∫[(f)∗ ∗ f ]φ ≥ 0.

There is a beautiful connection between functions of positive type and unitaryrepresentations. The following theorem provides the first clue of this.

Theorem 3.3.3. If π is a unitary representation of G and u ∈ Hπ, let φ(x) =〈π(x)u, u〉. Then φ ∈ P.

Proof. Since representations are assumed to be strongly continuous, we have |〈π(x)u, u〉−〈π(y)u, u〉| ≤ ‖π(x)u − π(y)u‖‖u‖ → 0 as y → x, so φ is continuous. Alsoφ(y−1x) = 〈π(y−1)π(x)u, u〉 = 〈π(x)u, π(y)u〉, so if f ∈ L1(G),

∫ ∫f(x)f(y)φ(y−1x)dxdy =

∫ ∫〈f(x)π(x)u, f(y)π(y)u〉dxdy

=

∫(f ∗ ∗ f)(x)〈π(x)u, u〉dx = 〈π(f ∗ ∗ f)u, u〉

= 〈π(f)u, π(f)u〉 = ‖π(f)u‖2 ≥ 0.

Corollary 3.3.4. If f ∈ L2(G), let f(x) = f(x−1). Then f ∗ f ∈ P .

Proof. Let πL be the left regular representation. Then

〈πL(x)f, f〉 =

∫f(x−1y)f(y)dy =

∫f(y−1x)f(y)dy = f ∗ f(x).

Hence f ∗ f ∈ P .


We shall show that every nonzero function of positive type arises from a unitaryrepresentation. If φ 6= 0 is of positive type, it defines a positive semi-definiteHermitian form on L1(G) by

〈f, g〉φ =

∫(g∗ ∗ f)φ =

∫ ∫f(x)g(y)φ(y−1x)dxdy,

which clearly satisfies|〈f, g〉φ| ≤ ‖φ‖∞‖f‖1‖g‖1. (3.2)

Let N = {f ∈ L1(G) : 〈f, f〉φ = 0}. By the Schwarz inequality f ∈ N if and onlyif 〈f, g〉φ = 0 for all g ∈ L1(G). The form 〈·, ·〉φ therefore induces an inner producton the quotient space L1(G)/N , still denoted by 〈·, ·〉φ. We denote the Hilbertspace completion of L1(G)/N by Hφ, and we denote the image of f ∈ L1(G) inL1(G)/N ⊂ Hφ by f . By the inequality (3.2)

‖f‖Hφ ≤ ‖φ‖1/2∞ ‖f‖1.

Now, if f, g ∈ L1(G) and x ∈ G,

〈Lxf, Lxg〉φ =

∫ ∫f(x−1y)g(x−1y)φ(z−1y)dydz

=

∫ ∫f(y)g(y)φ((xz)−1(xy))dydz = 〈f, g〉φ.

In particular, Lx(N ) ⊂ N , so the operators Lx yield a unitary representation πφof G on Hφ that is determined by

πφ(x)f = (Lxf)∼ (f ∈ L1(G)).

It is easy to verify that the corresponding representation of L1(G) on Hφ is givenby πφ(f)g∼ = (f ∗ g)∼.

Theorem 3.3.5. Given a function φ of positive type on G, let Hφ the Hilbert spacedetermined as above by the Hermitian form and let πφ be the unitary representationof G on Hφ. There is a cyclic vector ε for πφ such that πφ(f)ε = f for all f ∈ L1(G)and φ(x) = 〈πφ(x)ε, ε〉 locally almost everywhere.

Proof. Let {ψU} be an approximate identity. Now {ψ∗U} is a left approximateidentity, that is ψ∗U ∗ f → f for all f ∈ L1(G). Therefore for any f ∈ L1(G),〈f , ψU〉φ =

∫(ψ∗U ∗ f)φ→

∫fφ. Also ‖ψU‖Hφ ≤ ‖φ‖

1/2∞ ‖ψU‖1 = ‖φ‖1/2

∞ . It followsthat the functional f 7→ limU〈f , ψU〉φ is bounded on L1(G)/N , so it extends toa bounded functional on the completion Hφ. Therefore lim〈v, ψU〉φ exists for all


v ∈ Hφ, and hence that ψU converges weakly in Hφ to an element ε such that〈f , ε〉 =

∫fφ for all f ∈ L1(G).

If f, g ∈ L1(G) and y ∈ G, we have

〈g, πφ(y)ε〉φ = 〈πφ(y−1)g, ε〉φ = 〈(Ly−1g)∼, ε〉φ

=

∫g(yx)φ(x)dx =

∫g(x)φ(y−1x)dx,

and hence〈g, f〉φ =

∫〈g, πφ(y)ε〉φf(y)dy = 〈g, πφ(f)ε〉φ.

It follows that f = πφ(f)ε for all f ∈ L1(G). It also follows that if 〈g, πφ(y)ε〉 =

0 for all y ∈ G, then by the above 〈g, f〉φ = 0 for all f ∈ L1(G), so the linear span{πφ(y)ε : y ∈ G} is dense in Hφ and ε is a cyclic vector. Moreover

∫〈ε, πφ(y)ε〉φf(y)dy = lim

∫〈ψU , πφ(y)ε〉φf(y)dy = lim〈ψU , πφ(f)ε〉φ

= lim〈ψU , f〉φ = 〈ε, f〉φ = 〈f , ε〉φ =

∫φ(y)f(y)dy

for every f ∈ L1(G), and hence

〈πφ(y)ε, ε〉φ = 〈ε, πφ(y)ε〉φ = φ(y) locally almost everywhere.

Corollary 3.3.6. Every function of positive type agrees locally almost everywherewith a continuous function.

Corollary 3.3.7. If φ ∈ P then ‖φ‖∞ = φ(e) and φ(x−1) = φ(x).

Proof. We have φ(x) = 〈π(x)u, u〉 for some π and u, so |φ(x)| = |〈π(x)u, u〉| ≤‖u‖2 = φ(e) and φ(x−1) = 〈π(x−1)u, u〉 = 〈u, π(x)u〉 = φ(x).

Theorems 3.3.3 and 3.3.5 establish a correspondence between cyclic representa-tions and functions of positive type. Note that in Theorem 3.3.3 we didn’t assumeπ was cyclic, however the expression 〈π(·)u, u〉 only depends on the subrepresen-tation of π on the cyclic subspace generated by u. Moreover representations withthe same associated function of positive type are equivalent.

Theorem 3.3.8. Suppose π and ρ are cyclic representations of G with cyclicvectors u and v. If 〈π(x)u, u〉 = 〈ρ(x)v, v〉 for all x ∈ G, then π and ρ areunitarily equivalent. More precisely there exists a unitary T ∈ C(π, ρ) such thatTu = v.


Proof. Define T [π(x)u] = ρ(x)v. This extends to a well-defined isometry from thespan of {π(x)u : x ∈ G} to the span of {ρ(x)v : x ∈ G}. To check that T iswell-defined, let

∑ni=1 αiπ(xi)u = 0. Now∥∥∥∥∥

n∑i=1

αiρ(xi)v

∥∥∥∥∥2

=n∑

i,j=1

αiαj〈ρ(x−1j xi)v, v〉 =

n∑i,j=1

αiαj〈π(x−1j xi)u, u〉

=

∥∥∥∥∥n∑i=1

αiπ(xi)u

∥∥∥∥∥2

= 0,

so∑n

i=1 αiρ(xi)v = T [∑n

i=1 αiπ(xi)u] = 0 so T is well-defined. By the above it isalso an isometry. By continuity it extends to a unitary map from Hπ to Hρ. Sinceρ(y)T [π(x)u] = ρ(yx)v = T [π(y)π(x)u] we have ρ(y)T = Tπ(y), so T ∈ C(π, ρ).Also Tu = T [π(e)u] = ρ(e)v = v.

Corollary 3.3.9. If π is a cyclic representation of G with cyclic vector u andφ(x) = 〈π(x)u, u〉, then π is unitarily equivalent to the representation πφ.

Proof. If u is a cyclic vector of π, then by Theorem 3.3.5 φ(x) = 〈π(x)u, u〉 =〈πφ(x)ε, ε〉, so we can apply the above theorem.

The set P of continuous functions of positive type is a convex cone. We singleout two subsets of P for special attention. Let

P1 = {φ ∈ P : ‖φ‖∞ = 1} = {φ ∈ P : φ(e) = 1},

P0 = {φ ∈ P : ‖φ‖∞ ≤ 1} = {φ ∈ P : 0 ≤ φ(e) ≤ 1}.

These are bounded convex sets, and denote

E(Pj) = the set of extreme points of Pj, (j = 0, 1).

The extreme points of P1 are of particular interest because of the following theo-rem.

Theorem 3.3.10. If φ ∈ P1, then φ ∈ E(P1) if and only if the representation πφis irreducible.

Proof. Suppose πφ is reducible, say Hφ = M⊕M⊥ where M is nontrivial andinvariant under πφ. Let ε ∈ Hφ be a cyclic vector for πφ. Since ε is cyclic, it cannotbelong toM orM⊥, so ε = u+ v with u ∈M, v ∈M⊥ and u 6= 0 6= v. But then

φ(x) = 〈πφ(x)ε, ε〉φ = 〈πφ(x)u, u〉φ + 〈πφ(x)v, v〉φ = c1ψ1(x) + c2ψ2(x)


where ψ1, ψ2 ∈ P1, c1 = ‖u‖2, c2 = ‖v‖2, and c1 + c2 = φ(e) = 1. It remains toshow that ψ1 6= ψ2.

Suppose towards contradiction that 〈πφ(·)v, v〉φ = c〈πφ(·)u, u〉φ for some con-stant c, which must necessarily be positive. Choose δ > 0 such that

δ <c‖u‖2

‖v‖+ c‖u‖.

It follows that δ‖v‖ < c‖u‖2 − δc‖u‖.Since ε is cyclic in Hφ, there exists α1, . . . , αn ∈ C and x1, . . . xn ∈ G such that∥∥∥∥∥

n∑k=1

αkπ(xk)ε− u

∥∥∥∥∥ < δ.

By the above we have∣∣∣∣∣n∑k=1

αk〈π(xk)u, u〉 − 〈u, u〉

∣∣∣∣∣ =

∣∣∣∣∣〈n∑k=1

αkπ(xk)ε− u, u〉

∣∣∣∣∣≤

∥∥∥∥∥n∑k=1

αkπ(xk)ε− u

∥∥∥∥∥ ‖u‖ < δ‖u‖,

so ‖u‖2 − δ‖u‖ < |∑n

k=1 αk〈π(xk)u, u〉|. On the other hand we have∣∣∣∣∣n∑k=1

αk〈π(xk)v, v〉

∣∣∣∣∣ =

∣∣∣∣∣n∑k=1

αk〈π(xk)ε, v〉 − 〈u, v〉

∣∣∣∣∣≤

∥∥∥∥∥n∑k=1

αkπ(xk)ε− u

∥∥∥∥∥ ‖v‖ < δ‖v‖.

Combining the above inequalities we have∣∣∣∣∣n∑k=1

αk〈π(xk)v, v〉

∣∣∣∣∣ < δ‖v‖ < c‖u‖2 − δc‖u‖ < c

∣∣∣∣∣n∑k=1

αk〈π(xk)u, u〉

∣∣∣∣∣ ,so∑n

k=1 αk〈π(xk)v, v〉 6= c∑n

k=1 αk〈π(xk)u, u〉, which implies that for some kαk〈π(xk)v, v〉 6= cαk〈π(xk)u, u〉 and hence ψ1 6= ψ2. This shows that φ is notextreme.

Conversely, suppose πφ is irreducible, but that φ = ψ + ψ′ with ψ, ψ′ ∈ P .Then for any f, g ∈ L1(G), we have

〈f, f〉ψ = 〈f, f〉φ − 〈f, f〉ψ′ ≤ 〈f, f〉φ


and hence

|〈f, g〉ψ| ≤ 〈f, f〉1/2ψ 〈g, g〉1/2ψ ≤ 〈f, f〉1/2φ 〈g, g〉

1/2φ .

Thus the map (f, g) 7→ 〈f, g〉ψ induces a bounded Hermitian form on Hφ, soby Theorem 3.1.4 there is a bounded self-adjoint operator T on Hφ such that〈f, f〉ψ = 〈T f , g〉φ for all f, g ∈ L1(G). Now if x ∈ G and f, g ∈ L1(G) we have

〈Tπφ(x)f , g〉φ = 〈T (Lxf)∼, g〉φ = 〈Lxf, g〉ψ = 〈f, Lx−1g〉ψ= 〈T f , (Lx−1g)∼〉φ = 〈T f , πφ(x−1)g〉φ = 〈πφ(x)T f , g〉φ.

Therefore, T ∈ C(πφ), so by Schur’s lemma, T = cI and 〈f, g〉ψ = c〈f, g〉φfor all f, g. Letting g be an approximate identity we get

∫fψ = c

∫fφ for every

f ∈ L1(G). This implies ψ = cφ and hence ψ′ = (1− c)φ, so φ is extreme.

Recall the following theorems from functional analysis.

Theorem 3.3.11 (Alaoglu’s Theorem). The norm closed unit ball of the dual ofa normed space is weak* compact.

Theorem 3.3.12 (The Krein-Milman Theorem). If C is a compact convex subsetof a real or complex locally convex Hausdorff space X, then C is the closed convexhull of its extreme points.

The proof of Alaouglu’s theorem can be found in [12, p. 229, Theorem 2.6.18.]and the proof of the Krein-Milman theorem in [12, p. 265, Theorem 2.10.6.].

The condition∫

(f ∗ ∗ f)φ ≥ 0 is clearly preserved under weak* limits, so P0

is a weak* closed subset of the closed unit ball in L∞(G). By Alaoglu’s theoremP0 is compact, and then by Krein-Milman theorem it is the closed convex hull ofits extreme points. However P1 is in general not weak* closed, although if G isdiscrete, then the point mass δe at e is in L1(G), and

∫δeφ = φ(e) implies that P1

is weak* closed. In spite of this the conclusion of the Krein-Milman holds for P1

too.

Lemma 3.3.13. E(P0) = E(P1) ∪ {0}.

Proof. Suppose φ1, φ2 ∈ P0, c1, c2 > 0 and c1 + c2 = 1. If c1φ1 + c2φ2 = 0, thenc1φ1(e)+c2φ2(e) = 0, which implies that φ1(e) = φ2(e) = 0 and hence φ1 = φ2 = 0.Thus 0 ∈ E(P0).

Now suppose φ ∈ E(P1) and c1φ1+c2φ2 = φ. Then c1φ1(e)+c2φ2(e) = φ(e) = 1.This implies that φ1(e) = φ2(e) = 1, since otherwise c1φ1 + c2φ2 < c1 + c2 = 1which is a contradiction. Therefore φ1, φ2 ∈ P1. Since φ ∈ E(P1) we have φ1 = φ2,so φ ∈ E(P0).


Finally, no element of E(P0) \ (E(P1) ∪ {0}) is extreme in E(P1), since if φ ∈E(P0)\ (E(P1)∪{0}) then 0 < φ(e) < 1, so φ is interior to the line segment joining0 and φ/φ(e).

Theorem 3.3.14. The convex hull of E(P1) is weak* dense in P1.

Proof. Suppose φ ∈ P1. Now φ ∈ P1 ⊂ P0 = coE(P0) = co(E(P1) ∪ {0}), so φ isthe weak* limit of a net of functions φα of the form c1ψ1 + · · · + cnψn + cn+10 =∑n

j=1 cjψj, where ψ1, . . . , ψn ∈ E(P1), c1, . . . , cn+1 ≥ 0 and∑n+1

j=1 cj = 1.Now ‖ limφα‖∞ = 1 and ‖φα‖∞ ≤ 1, so lim ‖φα‖∞ ≤ 1. In fact we have

lim ‖φα‖ = 1. Towards contradiction assume that c = lim ‖φα‖ < 1. Now for someα0 we have |‖φα‖ − c| < 1−c

2whenever α ≥ α0. It follows that ‖φα‖∞ < 1+c

2< 1

whenever α ≥ α0. Now since the set {f ∈ L∞(G) : ‖f‖∞ ≤ 1+c2} is weak* closed,

we have ‖ limφα‖∞ ≤ 1+c2< 1, which is a contradiction.

Now if we set φ′α = φα/φα(e), we have

φ′α =1

φα(e)

n∑j=1

cjψj,1

φα(e)

n∑j=1

cj =φα(e)

φα(e)= 1.

Thus φ′α is in the convex hull of E(P1) and φ = limφ′α.

Next we show that the weak* topology that P1 inherits from L∞(G) coincideswith the topology of uniform convergence on compact subsets of G.

Definition 3.3.15. On Cb(G) the topology of compact convergence on G is thetopology of uniform convergence on compact subsets of G. A neighborhood baseat the function φ0 consists of sets of the form

N(φ0; ε,K) = {φ : |φ(x)− φ0(x)| < ε for x ∈ K},

where ε > 0 and K ⊂ G is compact.

The proof of the aforementioned remarkable claim relies on the following lemma.

Lemma 3.3.16. Suppose X is a Banach space and B is a norm-bounded subset ofX∗. On B, the weak* topology coincides with the topology of compact convergenceon X.

Proof. The weak* topology is the topology of pointwise convergence on X, soit is weaker than the topology of compact convergence. On the other hand, ifλ0 ∈ B, ε > 0 and K ⊂ X is compact, let C = sup{‖λ‖ : λ ∈ B} and δ = ε/3C.


Then there exists ξ1, . . . , ξn ∈ K such that the balls B(ξj, δ) cover K. If λ ∈ Band ξ ∈ K then ‖ξ − ξj‖ < δ for some j, so that

|λ(ξ)− λ0(ξ)| < |λ(ξ − ξj)|+ |(λ− λ0)(ξj)|+ |λ0(ξj − ξ)|≤ ‖λ‖‖ξ − ξj‖+ |(λ− λ0)(ξj)|+ ‖λ0‖‖ξj − ξ‖

<2ε

3+ |(λ− λ0)(ξj)|

so the weak* neighborhood⋂nj=1{λ : |(λ − λ0)(ξj)| < ε/3} of λ0 is contained in

the neighborhood N(λ0; ε,K) for the topology of compact convergence.

As a corollary we obtain the following lemma.

Lemma 3.3.17. Suppose φ0 ∈ P1 and f ∈ L1(G). For every ε > 0 and everycompact K ⊂ G there is a weak* neighborhood Φ of φ0 in P1 such that |f ∗ φ(x)−f ∗ φ0(x)| < ε for all φ ∈ Φ and x ∈ K.

Proof. By Corollary 3.3.7 we have f ∗φ(x) =∫f(xy)φ(y−1)dy =

∫(Lx−1f)φ. Since

x 7→ Lx−1f is continuous from G to L1(G), {Lx−1f : x ∈ K} is compact in L1(G),and we can apply Lemma 3.3.16.

Lemma 3.3.18. If φ ∈ P1, |φ(x)− φ(y)|2 ≤ 2− 2Reφ(yx−1).

Proof. We have 〈π(x)u, u〉 for some unitary representation π and some unit vectoru ∈ Hπ, so

|φ(x)− φ(y)|2 = |〈[π(x)− π(y)]u, u〉|2 = |〈u, [π(x−1)− π(y−1)]u〉|2

≤ ‖π(x−1)u− π(y−1)u‖2 = 2− 2Re〈π(x−1)u, π(y−1)u〉= 2− 2Re〈π(yx−1)u, u〉 = 2− 2Reφ(yx−1).

Theorem 3.3.19. On P1, the weak* topology coincides with the topology of com-pact convergence on G.

Proof. If f ∈ L1(G) and ε > 0, there is a compact K ⊂ G such that∫G\K |f | <

14ε.

If φ, φ0 ∈ P1 and |φ− φ0| < ε/2‖f‖1 on K then∣∣∣∣∫ (fφ− fφ0)

∣∣∣∣ ≤ ∫K

|f ||φ− φ0|+∫G\K|f ||φ− φ0|

<1

2ε+

∫G\K|f |(‖φ‖+ ‖φ0‖) <

1

2ε+ 2

1

4ε = ε,

so compact convergence on G implies weak* convergence.


Conversely suppose φ0 ∈ P1, ε > 0 and K ⊂ G is compact. We wish to finda weak* neighborhood Φ of φ0 in P1 such that |φ − φ0| < ε on K when φ ∈ Φ.First, if η > 0 there is a compact neighborhood V of e such that |φ0(x)− φ0(e)| =|φ0(e)− 1| < η for all x ∈ V . Let

Φ1 =

{φ ∈ P :

∣∣∣∣∫V

(φ− φ0)

∣∣∣∣ < η|V |}.

Now Φ1 is a weak* neighborhood of φ0 since χV ∈ L1. If φ ∈ Φ1, then∣∣∣∣∫V

(1− φ)

∣∣∣∣ ≤ ∣∣∣∣∫V

(1− φ0)

∣∣∣∣+

∣∣∣∣∫V

(φ0 − φ)

∣∣∣∣ < 2η|V |. (3.3)

Also, if φ ∈ Φ1 and x ∈ G, we have

|χV ∗ φ(x)− |V |φ(x)| =∣∣∣∣∫V

[φ(y−1x)− φ(x)]dy

∣∣∣∣ ≤ ∫V

|φ(y−1x)− φ(x)|dy.

By Lemma 3.3.18, Schwarz inequality and inequality (3.3), the right side of theabove inequality is bounded by∫

V

[2− 2Reφ(y)]1/2dy ≤(∫

V

[2− 2Reφ(y)]dy

)1/2

|V |1/2

≤ 21/2

∣∣∣∣∫V

(1− φ)

∣∣∣∣1/2 |V |1/2 < 2|V |√η.

By Lemma 3.3.17, there exists a weak* neighborhood Φ2 of φ0 in P1 such that|χV ∗ φ(x)− χV ∗ φ0(x)| < η|V | for φ ∈ Φ2 and x ∈ K. Hence, if φ ∈ Φ1 ∩ Φ2 andx ∈ K, |φ(x)− φ0(x)| is bounded by

1|V |

[||V |φ(x)− χV ∗ φ(x)|+ |χV ∗ (φ− φ0)(x)|+|χV ∗ φ0(x)− |V |φ0(x)|

]≤ 1

|V |(2|V |√η + |V |η + 2|V |√η) = η + 4

√η

Therefore, if we choose η so that η + 4√η < ε and take Φ = Φ1 ∩ Φ2, we are

done.

One more simple approximation result is needed for the proof of the Gelfand-Raikov theorem.

Theorem 3.3.20. The linear span of Cc(G)∩P includes all functions of the formf ∗ g with f, g ∈ Cc(G). It is dense in Cc(G) in the uniform norm, and dense inLp(G) (1 ≤ p <∞) in the Lp(G) norm.


Proof. By the Corollary 3.3.4 the set Cc(G)∩P includes all functions of the formf ∗ f with f ∈ Cc(G), where f(x) = f(x−1). By polarization, its linear span alsoincludes all functions of the form f ∗ h and hence all functions of the form f ∗ gwith f, g ∈ Cc(G). Now {f ∗ g : f, g ∈ Cc(G)} is dense in Cc(G) in the uniformnorm and Lp norm because g can be taken to be the approximate identity ψU , andCc(G) is itself dense in Lp(G).

Now we can prove the main result of this section.

Theorem 3.3.21 (The Gelfand-Raikov Theorem). If G is any locally compactgroup, the irreducible unitary representations of G separate points on G. Thatis, if x, y ∈ G with x 6= y, there is an irreducible representation π such thatπ(x) 6= π(y).

Proof. If x 6= y there exists f ∈ Cc(G) such that f(x) 6= f(y), and by Theorem3.3.20 f can be taken to be a linear combination of functions of positive type. ByTheorems 3.3.14 and 3.3.19, there is a linear combination g of extreme points of P1

that approximates f on the compact set {x, y} closely enough so that g(x) 6= g(y).Hence there must be an extreme point φ of P1 such that φ(x) 6= φ(y). Theassociated representation πφ is irreducible by Theorem 3.3.10 and it satisfies

〈πφ(x)ε, ε〉 = φ(x) 6= φ(y) = 〈πφ(y)ε, ε〉

so πφ(x) 6= πφ(y).

When the group G is neither abelian nor compact the irreducible representa-tions may be infinite-dimensional, and often the finite-dimensional representationsdo not separate points of G. An example of such group a group is SL(2,R), forthe proof see for instance [18, p. 113, Corollary 3.].

The construction of unitary representations of a group from functions of posi-tive type is in fact very similiar to the Gelfand-Naimark-Segal construction in thetheory of C*-algebras. Indeed in the language of C*-algebras a state is a positivelinear functional of norm 1, and the GNS construction states that for every state φof a C*-algebra A, there exists a cyclic *-representation πφ of A with cyclic vectorξ such that ρ = 〈πφ(·)ξ, ξ〉. Moreover irreducible *-representations of A correspondto pure states, which are the extreme points in the state space. The proof can befound for instance in [3, p. 31, Theorem 7.7.]. The GNS construction is at theheart of the proof of the noncommutative Gelfand-Naimark theorem.

An examination of the ideas of this chapter reveals the importance of locallycompactness in representation theory. Haar measure allows us to construct thefirst good unitary representations, the regular representations, and perhaps evenmore importantly to consider the group algebra, which had an essential role in theconstruction of cyclic and irreducible representations.


Outside of locally compact groups no comparable representation theory isknown. Every topological Hausdorff group does have isometric Banach repre-sentations, namely the action given by left (right) translation by group elementson the left (right) uniformly continuous functions LUC(G) (RUC(G)). This re-sult is known as Teleman’s theorem. However Banach representations are far lessgeometric and useful than representations on Hilbert spaces. Almost none of theideas and results of this chapter work for Banach representations. For more onTeleman’s theorem see [14].

Chapter 4

Compact Groups

In classical harmonic analysis a function on a compact interval [−π, π] (or theunit circle T) is represented or approximated by sums of trigonometric functions(or complex exponentials). These simpler functions provide an orthonormal basisfor the space L2([−π, π]) (or L2(T)). In this chapter we present some of thebasic theory of representations of compact groups. The results of this chapterare summarized in the Peter-Weyl theorem, which among other things gives anorthonormal basis of functions for L2(G) for a compact group G.

Throughout this chapter G is a compact group with a normalized Haar measure|G| = 1, which is both left and right invariant.

4.1 Representations of Compact GroupsWe begin by proving some basic results about unitary representations on compactgroups. The following lemma is important.

Lemma 4.1.1. Suppose π is a unitary representation of the compact group G. Fixa unit vector u ∈ Hπ, and define the operator T on Hπ by

Tv =

∫〈v, π(x)u〉π(x)udx.

Then T is positive, nonzero and compact, and T ∈ C(π).

Proof. For any v ∈ Hπ we have

〈Tv, v〉 =

∫〈v, π(x)u〉〈π(x)u, v〉dx =

∫|〈v, π(x)u〉|2dx ≥ 0,

so T is positive. Moreover, if we take u = v, |〈u, π(x)u〉|2 is strictly positive on aneighborhood of e, so 〈Tu, u〉 > 0 and hence T 6= 0.

51

CHAPTER 4. COMPACT GROUPS 52

Since G is compact, x 7→ π(x)u is uniformly continuous. Hence, given ε > 0,we can find a partition of G into disjoint sets E1, . . . , En and points xj ∈ Ej suchthat ‖π(x)u−π(xj)u‖ < 1

2ε for x ∈ Ej. Indeed there is a neighborhood V of e such

that ‖π(x)u−u‖ < 12ε for every x ∈ V . Now let U be a symmetric neighborhood of

e such that UU ⊂ V . The translates of U cover G, so by compactness there existsa finite subcover {gjU}nj=1. Then let E1 = g1U and Ej = gjU \

⋃j−1k=1 gkU whenever

1 < k ≤ n. The finite subcover can be chosen so that every Ej is nonempty, so wemay pick xj ∈ Ej for every j. Now if x ∈ Ej, then x−1

j x ∈ Ug−1j gjU ⊂ V . Hence

‖π(x)u− π(xj)u‖ = ‖π(x−1j x)u− u‖ < 1

2ε for every x ∈ Ej. Now we have

‖〈v, π(x)u〉π(x)u− 〈v, π(xj)u〉π(xj)u‖≤ ‖〈v, [π(x)− π(xj)]u〉π(x)u‖+ ‖〈v, π(xj)u〉[π(x)− π(xj)]u‖< ε‖v‖

for x ∈ Ej, so if we set

Tεv =n∑j=1

|Ej|〈v, π(xj)u〉π(xj)u

we have

‖Tv − Tεv‖ ≤∑j

∫Ej

‖〈v, π(x)u〉π(x)u− 〈v, π(xj)u〉π(xj)u‖dx < ε‖v‖

for all v. But the range of Tε is the linear span of {π(xj)u}n1 , so Tε has finite rank.Therefore T is compact, being the norm limit of operators of finite rank.

Finally T ∈ C(π) because

π(y)Tv =

∫〈v, π(x)u〉π(yx)udx =

∫〈v, π(y−1x)u〉π(x)udx

=

∫〈π(y)v, π(x)u〉π(x)udx = Tπ(y)v.

Theorem 4.1.2. If G is compact, then every irreducible representation of G isfinite-dimensional, and every unitary representation of G is a direct sum of irre-ducible representations.

Proof. Suppose π is irreducible, and let T be as in Lemma 4.1.1. By Schur’slemma, T = cI with c 6= 0. So the identity operator Hπ is compact, and hencedimHπ <∞.


Now let π be an arbitrary unitary representation of G. Since T is compact,nonzero, and self-adjoint, it has a nonzero eigenvalue λ whose eigenspace Eλ isnecessarily finite-dimensional, since the restriction of T to Eλ is λI and com-pact. Since T ∈ C(π), Eλ is invariant because for any x ∈ G and u ∈ Eλ wehave Tπ(x)u = π(x)Tu = λπ(x)u, so indeed π(x)u ∈ Eλ. Hence π has a finite-dimensional subrepresentation. But every finite-dimensional representation is adirect sum of irreducible representations. To see this, if a finite-dimensional rep-resentation (π,H) has an invariant subspaceM with 0 < dimM < dimH, thenπ is the direct sum of two subrepresentations with dimension smaller than dimH.Since dimH is finite, this process of decomposing eventually ends.

Now by Zorn’s lemma there is a maximal family {Mα} of mutually orthogonalirreducible invariant subspaces for π. If N is the orthogonal complement

⊕Mα,

then N is invariant, and by the above argument πN has an irreducible invariantsubspace, contradicting maximality unless N . Thus Hπ =

⊕Mα.

We denote G the set of unitary equivalence classes of irreducible unitary rep-resentations of G. This is indeed a valid set since by the above theorem therepresentations are all finite-dimensional. We denote the equivalence class of π by[π]. Writing ” [π] ∈ G” will be a convenient shorthand for the statement ”π is anirreducible unitary representation of G”.

It is worth noting that if ρ is a possibly nonunitary representation of the com-pact group G on a Hilbert space H, then there is an inner product on H withrespect to which ρ is unitary. To see this, if 〈·, ·〉 is the inner product on H thendefine a new inner product by

〈u, v〉ρ =

∫〈ρ(x)u, ρ(x)v〉dx.

Then 〈·, ·〉ρ is a ρ-invariant inner product, for

〈ρ(y)u, ρ(y)v〉ρ =

∫〈ρ(xy)u, ρ(xy)v〉dx =

∫〈ρ(x)u, ρ(x)v〉dx = 〈u, v〉ρ.

Moreover by Theorem 3.1.4 there exists a positive P ∈ L(H) such that 〈u, v〉ρ =〈Pu, v〉. By the spectral theorem P has the unique square root

√P = S. Now

x 7→ Sρ(x)S−1 defines a unitary representation of G on H since if u, v ∈ H andx ∈ G, then

〈Sρ(x)S−1u, Sρ(x)S−1v〉 = 〈Pρ(x)S−1u, ρ(x)S−1v〉= 〈ρ(x)S−1u, ρ(x)S−1v〉ρ = 〈S−1u, S−1v〉ρ = 〈PS−1u, S−1v〉 = 〈u, v〉.

A close study of the above argument shows that the claim can be generalized.A locally compact group is called amenable if the space of bounded functions


L∞(G) admits an invariant mean, that is there exists a functional m ∈ L∞(G)∗

with m(1) = ‖m‖ = 1 and m(Lxf) = m(f) for every x ∈ G and f ∈ L∞(G). Allcompact groups and abelian groups are amenable. More on amenable groups canbe found on [13], [15] and [1, Chapter G].

A representation ρ : G → L(H) is a uniformly bounded representation ifsupx∈G ‖ρ(x)‖ < ∞. In the example above a strongly continuous representationof a compact group G is uniformly bounded since ρ(G)u is compact and hencebounded for every u ∈ H, so by the Banach-Steinhaus theorem supx∈G ‖ρ(x)‖ <∞. Now 〈u, v〉ρ = m(ϕu,v) defines a ρ-invariant inner product, where ϕu,v(x) =〈ρ(x)u, ρ(x)v〉 is a bounded continuous function. If we denote |ρ| = supx∈G ‖ρ(x)‖,then the inequalities |ρ|−1‖u‖ ≤ ‖u‖ρ ≤ |ρ|‖u‖ hold, so again we find a positiveinvertible S such that Sρ(x)S−1 is unitary for every x ∈ G. A group is called unita-rizable if for every uniformly bounded representation (π,H) we can find S ∈ L(H)such that x 7→ Sπ(x)S−1 is a unitary representation. By what we just showedamenable groups are unitarizable.

Naturally one may ask if the converse holds, that is is every unitarizable groupamenable. This was conjectured by Dixmier in 1950, and it is still open. Somepartial results have been obtained, see for instance [16].

4.2 The Peter-Weyl TheoremWe shall define a non-abelian analog of the trigonometric functions and complexexponentials of classical harmonic analysis.

Definition 4.2.1. If π is any unitary representation of G, the functions

φu,v(x) = 〈π(x)u, v〉 (u, v ∈ Hπ)

are called matrix elements or matrix coefficients of π. If u and v are members ofan orthonormal basis {ej} for Hπ, φu,v is one of the entries of the matrix for π(x)with respect to that basis, namely

πij(x) = φej ,ei(x) = 〈π(x)ej, ei〉.

We denote the linear span of the matrix coefficients of π by Eπ.

The space Eπ is a subspace of C(G) and hence also of Lp(G) for all p.

Theorem 4.2.2. The space Eπ depends only on the unitary equivalence class ofπ. It is invariant under left and right translations. If dimHπ = n < ∞ thendim Eπ ≤ n2.


Proof. If T is a unitary equivalence of π and π′, so that π′(x) = Tπ(x)T−1, then〈π(x)u, v〉 = 〈π′(x)Tu, Tv〉. Now

φu,v(y−1x) = 〈π(y−1x)u, v〉 = 〈π(x)u, π(y)v〉 = φu,π(y)v(x),

and likewise φu,v(xy) = φπ(y)u,v(x). Finally if dimHπ = n, then Eπ is clearlyspanned by the n2 functions πij.

Theorem 4.2.3. If π = π1 ⊕ · · · ⊕ πn then Eπ =∑n

j=1 Eπj .

Proof. Clearly Eπj ⊂ Eπ for all j. On the other hand if u =∑uj and v =

∑vj

with uj, vj ∈ Hπj , then 〈π(x)uj, vi〉 = 0 for every i 6= j and hence φu,v =∑φuj ,vj ∈∑

Eπj .

Note that in the above theorem the sum∑n

j=1 Eπj need not be direct.The matrix coefficients of irreducible representations can be used to make an

orthonormal basis for L2(G). Let dπ = dimHπ, and denote the trace of a matrixA by TrA.

Theorem 4.2.4 (The Schur Orthogonality Relations). Let π and π′ be irreducibleunitary representations of G, and consider Eπ and Eπ′ as subspaces of L2(G).

(a) If [π] 6= [π′] then Eπ⊥Eπ′.

(b) If {ej} is any orthonormal basis for Hπ then {√dππij : i, j = 1, . . . , dπ} is

an orthonormal basis for Eπ.

Proof. If A is any linear map from Hπ to Hπ′ , let

A =

∫π′(x−1)Aπ(x)dx.

Then

Aπ(y) =

∫π′(x−1)Aπ(xy)dx =

∫π′(yx−1)Aπ(x)dx = π′(y)A,

so A ∈ C(π, π′). Given v ∈ Hπ and v′ ∈ Hπ′ , let us define A by Au = 〈u, v〉v′.Then for any u ∈ Hπ and u′ ∈ Hπ′ ,

〈Au, u′〉 =

∫〈Aπ(x)u, π′(x)u′〉dx

=

∫〈π(x)u, v〉〈v′, π′(x)u′〉dx

=

∫φu,v(x)φu′,v′(x)dx.


We now apply Schur’s lemma. If [π] 6= [π′] then A = 0, so Eπ⊥Eπ′ . This proves(a). If π = π′ then A = cI, so if we take U = ei, u

′ = ei′ , v = ej and v′ = ej′ we get∫πij(x)πi′j′(x) = c〈ei, ei′〉 = cδii′ .

Butcdπ = TrA =

∫Tr[π(x−1)Aπ(x)]dx = TrA,

and since Au = 〈u, ej〉ej′ we have TrA =∑〈Aek, ek〉 =

∑〈ek, ej〉〈ej′ , ek〉 = δjj′ .

Hence ∫πij(x)πi′j′(x) =

1

dπδii′δjj′

so {√dππij} is an orthonormal set. Since dim Eπ ≤ d2

π, it is a basis.

We observed in Theorem 4.2.2 that Eπ is invariant under the left and righttranslations L and R. Note that since G is compact we have πR(x) = Rx. For thenext theorem we shall simplify our notation by letting L = πL be the left regularrepresentation and R = πR be the right regular representation. One may then askwhat are the irreducible subrepresentations of L and R on Eπ.

Theorem 4.2.5. Suppose π is irreducible. For i = 1, . . . , dπ let Ri be the linearspan of πi1, . . . , πidπ (the ith row of the matrix (πij)) and let Ci be the linear span ofπ1i, . . . , πdπi (the ith column). Then Ri (respectively Ci) is invariant under the right(left) regular representation, and RRi (LCi) is equivalent to π (π). The equivalenceis given by ∑

cjej 7→∑

cjπij (∑

cjej 7→∑

cjπij).

Proof. In terms of the basis {ej} for Hπ, π is given by

π(x)

(dπ∑j=1

cjej

)=

dπ∑k,j=1

πkj(x)cjek.

Moreover π(yx) = π(y)π(x), so πij(yx) =∑

k πik(y)πkj(x). In other words,Rxπij =

∑k πkj(x)πik, so

Rx

(dπ∑j=1

cjπij

)=

dπ∑j,k=1

πkj(x)cjπik.

Comparing the two above lines we see that π is equivalent to RRi . In the sameway, for left translations we see that

Lx

(dπ∑j=1

cjπji

)=

dπ∑j,k=1

πjk(x)cjπki,


and since π is unitary, we have πjk(x−1) = πkj(x).

Now letE = the linear span of

⋃[π]∈G

Eπ.

So E consists of finite linear combinations of matrix coefficients of irreducible rep-resentations. By Theorem 4.2.3, E is also the the linear span of matrix coefficientsof finite-dimensional representations of G. The space E could be considered as thespace of ”trigonometric functions” on G.

Theorem 4.2.6. E is an algebra.

Proof. It is sufficient to show that if [π], [ρ] ∈ G and πij, ρkl are matrix coefficients,then πijρkl is a matrix coefficients of some finite-dimensional representation of G.We shall construct another representation π⊗ ρ of G using the Kronecker productfor matrices. That is if A = [aij] is an n×m matrix and B is a p× q matrix, thenthe Kronecker product A⊗B is the np×mq block matrix

A⊗B =

a11B · · · a1mB... . . . ...

an1B · · · anmB

.

It is easy to verify that (A⊗B)(C ⊗D) = AC ⊗BD if one can form the matricesAC and BD, and hence (A ⊗ B)−1 = A−1 ⊗ B−1 if A−1 and B−1 exist, and(A⊗B)∗ = A∗⊗B∗. Now define the new representation by (π⊗ρ)(x) = π(x)⊗ρ(x)on Cnm, where n = dimπ and m = dim ρ. By the above mentioned properties ofthe Kronecker product this is a unitary representation of G. Moreover it is quiteclear from the resulting matrix

(π ⊗ ρ)(x) =

π11(x)ρ(x) · · · π1n(x)ρ(x)... . . . ...

πn1(x)ρ(x) · · · πnn(x)ρ(x)

that πij(x)ρkl(x) appears as a matrix coefficient. Indeed the desired coefficient is〈(π ⊗ ρ)(x)e(j−1)m+l, e(i−1)m+k〉. Now by Theorem 4.1.2, π ⊗ ρ is a direct sum ofirreducible representations, so by Theorem 4.2.3, we have πijρkl ∈ E .

Remark 4.2.7. Equivalently we could have defined π ⊗ ρ in the above proof bythe action (π ⊗ ρ)(x)T = π(x)Tρ(x−1), where T is a n×m matrix.

We are almost done with proving the Peter-Weyl theorem.


Theorem 4.2.8. E is dense in C(G) in the uniform norm, and dense in Lp(G)in the Lp norm for p <∞.

Proof. It is enough to show that E is dense in C(G) since C(G) is dense in Lp(G).But E is an algebra that separates points by the Gelfand-Raikov theorem, is closedunder conjugation since every representation has a contragredient and containsconstant functions because of the trivial representation of G on C. Therefore byStone-Weierstrass E is dense in C(G).

The original proof by Hermann Weyl and Fritz Peter (1927) is in fact older thaneither Gelfand-Raikov theorem (1943) or Stone-Weierstrass (1937). The first proofcan be found in [5, Theorem 5.11.]. The proof is based on studying convolutionoperators Tψf = ψ ∗ f on L2(G), where ψ is a continuous symmetric function.The operator is proven to be compact using Arzela-Ascoli theorem, so by thespectral theorem for compact operators L2(G) can be seen as direct sum of finite-dimensional eigenspaces of Tψ. Moreover each eigenspace is invariant under righttranslations, and it will follow that the eigenspaces are contained in E . HenceE ∩ Range(Tψ) will be uniformly dense in Range(Tψ), and taking the union ofranges of Tψ as ψ runs through an approximate identity is dense in C(G), provingthe theorem.

Combining Theorem 4.2.8 with the Schur orthogonality relations, we see thatL2(G) is the orthogonal direct sum of the spaces Eπ as [π] ranges over G, andthat we obtain an orthonormal basis for L2(G) by fixing an element π of eachirreducible equivalence class [π] and taking the matrix coefficients correspondingto an orthonormal basis of Hπ. In the statement of the Peter-Weyl theorem weassume that one representation has been picked from each equivalence class.

The main theorem is a summary of the results of this chapter.

Theorem 4.2.9 (The Peter-Weyl Theorem). Let G be a compact group. Then Eis uniformly dense in C(G), L2(G) =

⊕[π]∈G Eπ, and

{√dππij : i, j = 1, . . . , dπ, [π] ∈ G}

is an orthonormal basis for L2(G). Each [π] ∈ G occurs in the right and left regularrepresentations of G with multiplicity dπ. More precisely, for each i = 1, . . . , dπthe subspace of Eπ (respectively Eπ) spanned by the ith row (ith column) of thematrix (πij) ((πij)) is invariant under the right (left) regular representation, andthe latter representation is equivalent to π.

As an application of the ideas of this thesis we obtain a characterization ofcompact groups.

Corollary 4.2.10. Every compact group is a product of closed subgroups of uni-tary matrices.


Proof. Consider the mapping

F : G→∏

[π]∈G

π(G) ⊂∏

[π]∈G

U(Hπ), g 7→ (π(g))π∈G,

where the image of F has the product topology. By Gelfand-Raikov theorem thisis injective. Note that strong operator topology on U(Hπ) coincides with thenorm topology since Hπ is finite-dimensional. So the function F is continuousand a homomorphism since π(xy) = π(x)π(y) for every [π] ∈ G. Hence G istopologically isomorphic to its image F (G).

Using the Peter-Weyl theorem a Fourier transform can be defined for functionson a compact group. Moreover the transform has some of the same propertiesas the classical Fourier transform, such as taking convolutions of functions topointwise products. More on Fourier analysis on compact groups can be found in[5, p. 133-138].

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