In this sectionGalways denotes a locally compact group.
Definition 2.4.1. (a) A positive Radon measureµonGis calledleft invariant
We likewise define right invariance by Z
(b) A positive left invariant Radon measure µon Gis called a (left) Haar integral, resp., a(left) Haar measure, if 06=f ≥0 forf ∈Cc(G) implies
Z
G
f(x)dµ(x)>0.
In the following we shall denote Haar measures onGbyµG.
Remark 2.4.2. One can show that every locally compact group G possesses a Haar measure and that for two Haar measuresµandµ0 there exists aλ >0 withµ0=λµ([Neu90], [HiNe91]).
If G is compact and µ a Haar measure on G, then µ(G) is finite positive, so that we obtain a unique Haar probability measure onG. We call this Haar measurenormalized.
Example 2.4.3. (a) IfGis a discrete group, thenCc(G) is the space of finitely supported functions onG, and the counting measure
Z
G
f dµ:=X
g∈G
f(g) is a Haar measure onG. If, in addition, Gis finite, then
Z is a normalized Haar measure.
(b) ForG=Zwe obtain in particular a Haar measure by Z
Z
f dµZ:=X
n∈Z
f(n).
(c) On G=Rn, the Riemann, resp., Lebesgue integral defines a Haar mea-sure by
is a Haar measure.
is a Haar measure onG. Note that a continuous function with compact support onR× vanishes in a neighborhood of 0, so that the integral is defined.
Lemma 2.4.4. IfµGis a Haar measure onGandh∈C(G)withR
Proposition 2.4.5. Let µG be a Haar measure on G. Then there exists a continuous homomorphism
∆G:G→(R×+,·) with (ρg)∗µG= ∆G(g)−1µG for g∈G.
Proof. Since left and right multiplications onGcommutes, the Radon measure (ρg)∗µG is also left invariant and satisfies
depends continuously on g, we note that for a fixed g ∈ G, we actually in-tegrate only over supp(f)g−1. For any compact neighborhood K of g0, the subset supp(f)K−1ofGis compact (it is the image of the compact product set supp(f)×K under the continuous map (x, y)7→xy−1), and for anyg∈K we
so that the continuity ing0follows from Lemma 2.3.7. That ∆G is a homomor-phism is an immediate consequence of the definition:
(ρgh)∗µG = (ρhρg)∗µG= (ρh)∗(ρg)∗µG
= ∆G(g)−1(ρh)∗µG= ∆G(g)−1∆G(h)−1µG
(cf. Exercise 2.2.7).
Definition 2.4.6. The function ∆G is called themodular factor ofG. Clearly, it does not depend on the choice of the Haar measure µG. A locally compact group Gis calledunimodularif ∆G = 1, i.e., each left invariant Haar measure is also right invariant, hence biinvariant.
Proposition 2.4.7. A locally compact groupGis unimodular i f it satisfies one of the following conditions:
(a) Gis compact.
(b) Gis abelian.
(c) Its commutator group(G, G)is dense.
Proof. (a) In this case ∆G(G) is a compact subgroup ofR×+, hence equal to{1}.
(b) Follows from the fact thatρg=λg for anyg∈G.
(c) Since R×+ is abelian, (G, G)⊆ker ∆G. If (G, G) is dense, the continuity of ∆G implies that ∆G= 1.
Lemma 2.4.8. Let Gbe a locally compact group,µG a Haar measure and ∆G be the modular factor. Then we have forf ∈L1(G, µG)the following formulas:
Proof. (a) is the definition of the modular factor.
(b) Using (a), we obtain (ρg)∗µG= ∆G(g)−1µG. Since we also have (ρg)∗∆G= I is a Haar integral. In view of the Uniqueness of Haar measure, there exists a C >0 with
It remains to show thatC= 1. We apply the preceding relation to the compactly supported functionfe(x) :=f(x−1)∆G(x)−1 to find
Proposition 2.4.9. Let G be a locally compact group and µG a (left) Haar measure on G. OnL2(G) :=L2(G, µG)we have two continuous unitary repre-sentations of G. Theleft regular representation
πl(g)f :=f◦λ−1g and the right regular representation
πr(g)f :=p
∆G(g)f◦ρg.
Proof. The continuity of the left regular representation follows from Corol-lary 2.3.9. For the right regular representation we apply Proposition 2.3.8 to the left action ofGonGdefined byσg(x) :=xg−1=ρg−1x. Then
eδ(g, x) := d (σg)∗µG
dµG (x) = d (ρ−1g )∗µG
dµG (x) = ∆G(g) is a continuous function onG×G, which implies the continuity ofπr.
Corollary 2.4.10. For a locally compact groupG, the left regular representation is injective. In particular, Ghas a faithful continuous unitary representation.
Proof. For g 6= 1, pick disjoint open neighborhoods U of 1 and V of g with gU ⊆V. Let 0≤f ∈Cc(G) be non-zero with supp(f)⊆U. Then
hπl(g)f, fi= Z
G
f(g−1x)f(x)dµG(x) = 0
because supp(πl(g)f) =gsupp(f)⊆gU ⊆V intersectsUtrivially. On the other hand the definition of Haar measure implieskfk2>0, so that πl(g)6=1.
Exercises for Section 2.4
Exercise 2.4.1. Letλ=dX denote Lebesgue measure on the spaceMn(R)∼= Rn
2 of real (n×n)-matrices. Show that a Haar measure on GLn(R) is given by dµGLn(R)(g) = 1
|det(g)|ndλ(g).
Hint: Calculate the determinant of the linear mapsλg: Mn(R)→Mn(R), x7→
gx.
Exercise 2.4.2. LetG= Aff1(R)∼=RnR×denote the affine group ofR, where (b, a) corresponds to the affine mapϕb,a(x) :=ax+b. This group is sometimes called theax+b-group. Show that a Haar measure on this group is obtained by
Z
G
f(b, a)dµG(b, a) :=
Z
R
Z
R×
f(b, a)db da
|a|2.
Show further that ∆G(b, a) =|a|−1, which implies thatGis not unimodular.
Exercise 2.4.3. We consider the group G:= GL2(R) and the real projective line
P1(R) ={[v] :=Rv: 06=v∈R2}
of 1-dimensional linear subspaces of R2. We write [x : y] for the line R x
y
. Show that:
(a) We endow P1(R) with the quotient topology with respect to the map q: R2\ {0} →P1(R), v 7→[v]. Show that P1(R) is homeomorphic toS1. Hint: Consider the squaring map on T⊆C.
(b) The map R→ P1(R), x7→ [x: 1] is injective and its complement consists of the single point ∞ := [1 : 0] (the horizontal line). We thus identify P1(R) with the one-point compactification ofR. These are the so-called homogeneous coordinatesonP1(R).
(c) The natural action of SL2(R) on P1(R) by g.[v] := [gv] is given in the coordinates of (b) by
g.x=σg(x) := ax+b
cx+d for g= a b
c d
.
(d) There exists a unique Radon measureµwith total massπonP1(R) which is invariant under the group O2(R). Hint: IdentifyP1(R) with the compact group SO2(R)/{±1} ∼=T.
(e) Show that, in homogeneous coordinates, we have dµ(x) = 1+xdx2. Hint:
cosx −sinx sinx cosx
.0 = −tanx, and the image of Lebesgue measure on ]−π/2, π/2[ under tan is 1+xdx2.
(f) Show that the action of SL2(R) onP1(R) preserves the measure class ofµ.
Hint: Show that σg(x) := ax+bcx+d satisfies σg0(x) = (cx+d)1 2 and derive the formula
δ(σg)(x) =d((σg)∗µ)
dµ = 1 +x2
(a−cx)2+ (b−dx)2, δ(σg)(∞) = 1 c2+d2. (g) The density function also has the following metric interpretation with
re-spect to the euclidean norm onR2:
δ(σg)([v]) = kg−1vk2 kvk2 .
The corresponding unitary representations of SL2(R) on L2(P1(R), µ) defined by
πs(g)f :=δ(σg)12+is(σg)∗f
(cf. Example 2.2.7) form the so-calledspherical principal series.
Exercise 2.4.4. LetX be a locally compact space,µa positive Radon measure onX,Ha Hilbert space andf ∈Cc(X,H) be a compactly supported continuous function.
(a) Prove the existence of theH-valued integral I:=
Z
X
f(x)dµ(x), i.e., the existence of an elementI∈ Hwith
hv, Ii= Z
X
hv, f(x)idµ(x) for v∈ H.
Hint: Verify that the right hand side of the above expression is defined and show that it defines a continuous linear functional onH.
(b) Show that, ifµis a probability measure, then I∈conv(f(X)).
Hint: Use the Hahn–Banach Separation Theorem.
Exercise 2.4.5. Letσ:K× H → H,(k, v)7→k.vbe a continuous action of the compact group K by affine maps on the Hilbert space H. Show thatσ has a fixed point. Hint: For any orbit K-orbitK.v, define thecenter of massby
c(v) :=
Z
K
k.v dµK(k),
where µK is a normalized Haar measure onK (cf. Exercise 2.4.4).
Conclude that for any continuous unitary representation (π,H) of K each continuous 1-cocyclef:K→ His a coboundary, i.e., of the formf(k) =π(k)v−
v for somev∈ H. Hint: Use Exercise 2.2.2.
Chapter 3
Reproducing Kernel Spaces
In Chapter 2 we have seen how Hilbert spaces and continuous unitary represen-tations can be constructed on L2-spaces. An L2-spaceL2(X, µ) of a measure space (X,S) has the serious disadvantage that its elements are not functions onX, they are only equivalence classes of functions modulo those vanishing on µ-zero sets. However, many important classes of unitary representations can be realized in spaces of continuous functions. In particular for infinite dimensional Lie groups, this is the preferred point of view because measure theory on infi-nite dimensional spaces has serious defects that one can avoid by using other methods.
The main concept introduced in this chapter is that of a reproducing kernel Hilbert space. These are Hilbert spaces H of functions on a setX for which all point evaluationsH →C, f 7→f(x), are continuous linear functions. Repre-senting these functions according to the Fr´echet–Riesz Theorem by an element Kx∈ H, we obtain a function
K: X×X →C, K(x, y) :=Ky(x)
called the reproducing kernel of H. Typical questions arising in this context are: Which functions onX×X are reproducing kernels (such kernels are called positive definite) and, if we have a group action on X, how can we construct unitary representations on reproducing kernel spaces.
Throughout this chapterKdenotes either Ror C.
3.1 Hilbert Spaces with Continuous Point Eval-uations
Definition 3.1.1. LetX be a set.
(a) Consider a Hilbert spaceH which is contained in the space KX of K -valued functions on X. We say thatHhascontinuous point evaluations if for
53
eachx∈X the linear functional
evx:H →K, f 7→f(x)
is continuous. In view of the Fr´echet–Riesz Theorem, this implies the existence of someKx∈ Hwith
f(x) =hf, Kxi for f ∈ H, x∈X.
The corresponding function
K:X×X →C, K(x, y) :=Ky(x)
is called the reproducing kernel of H. As we shall see below, H is uniquely determined byK, so that we shall denote it byHK to emphasize this fact.
(b) A function K: X ×X → K is called a positive definite kernel if for each finite subset{x1, . . . , xn} ⊆X the matrix K(xi, xj)
i,j=1,...,n is positive semidefinite. For a functionK:X×X →Cwe write K∗(x, y) :=K(y, x) and say thatK ishermitian(orsymmetric forK=R) ifK∗=K.
We writeP(X,K) for the set of positive definite kernels on the setX.
Remark 3.1.2. (a) OverK=C, the positive definiteness of a kernelKalready follows from the requirement that for all choicesx1, . . . , xn ∈Xandc1, . . . , cn ∈ Cwe have
n
X
j,k=1
cjckK(xj, xk)≥0 because this implies thatKis hermitian (Exercise 3.1.1).
ForK=R, the requirement of the kernel to be hermitian is not redundant.
Indeed, the matrix
(Kij)i,j=1,2=
0 1
−1 0
,
considered as a kernel on the two element setX ={1,2}, satisfies
2
X
j,k=1
cjckK(xj, xk) = 0
forx1, . . . , xn∈X andc1, . . . , cn∈R, butK is not hermitian.
(b) For any positive definite kernelK∈ P(X) andx, y∈ P(X), the positive definiteness of the hermitian matrix
K(x, x) K(x, y) K(y, x) K(y, y)
implies in particular that
|K(x, y)|2≤K(x, x)K(y, y) (3.1)
In the following we call a subsetS of a Hilbert space H total if it spans a dense subspace.
Theorem 3.1.3. (Characterization Theorem)The following assertions hold for a function K:X×X →K:
(a) If K is the reproducing kernel of H ⊆ KX, then the following assertions hold:
(1) K is positive definite.
(2) {Kx:x∈X} is total in H.
(3) For any orthonormal basis(ej)j∈J, we haveK(x, y) =P
j∈Jej(x)ej(y).
(b) If K is positive definite, then H0K := span{Kx: x ∈ X} ⊆ KX carries a unique positive definite hermitian form satisfying
hKy, Kxi=K(x, y) for x, y∈X. (3.2) The completionHK of H0K permits an injection
ι:HK→KX, ι(v)(x) :=hv, Kxi
whose image is a Hilbert space with reproducing kernelK that we identify withHK.
(c) Kis positive definite if and only if there exists a Hilbert spaceH ⊆KX with reproducing kernelK.
Proof. (a)(1) ThatKis hermitian follows from
K(y, x) =Kx(y) =hKx, Kyi=hKy, Kxi=K(x, y).
Forc∈Kn we further have X
j,k
cjckK(xj, xk) =X
j,k
cjckhKxk, Kxji=kX
k
ckKxkk2≥0.
This proves (1).
(2) Iff ∈ His orthogonal to eachKx, thenf(x) = 0 for eachx∈X implies f = 0. Therefore{Kx:x∈X}spans a dense subspace.
(3) If (ej)j∈J is an ONB ofH, then we have for eachy∈X the relation Ky=X
j∈J
hKy, ejiej=X
j∈J
ej(y)ej, and therefore
K(x, y) =Ky(x) =X
j∈J
ej(y)ej(x).
(b) We want to put D X
j
cjKxj,X
k
dkKxkE :=X
j,k
cjdkK(xk, xj), (3.3) so that we have to show that this is well-defined.
So let f = P
jcjKxj and h = P
kdkKxk ∈ H0K. Then we obtain for the right hand side
X
j,k
cjdkK(xk, xj) =X
j,k
cjdkKxj(xk) =X
k
dkf(xk). (3.4) This expression does not depend on the representation of f as a linear combi-nation of the Kxj. Similarly, we see that the right hand side does not depend on the representation ofhas a linear combination of theKxk. Therefore
hf, hi:=X
j,k
cjdkK(xk, xj)
is well-defined. SinceK is positive definite, we thus obtain a positive semidefi-nite hermitian form onH0K. From (3.4) we obtain forh=Kxthe relation
hf, Kxi=f(x) for x∈X, f ∈ H0K. Ifhf, fi= 0, then the Cauchy–Schwarz inequality yields
|f(x)|2=|hf, Kxi|2≤K(x, x)hf, fi= 0, so thatf = 0. ThereforeH0K is a pre-Hilbert space.
Now letHK be the completion ofH0K. Then
ι:HK →KX, ι(v)(x) :=hv, Kxi
is an injective linear map because the set {Kx: x∈X} is total inHK0, hence also inHK. NowHK ∼=ι(HK)⊆KX is a Hilbert space with continuous point evaluations and reproducing kernelK.
(c) follows (a) and (b).
Lemma 3.1.4. (Uniqueness Lemma for Reproducing Kernel Spaces) If H ⊆ KX is a Hilbert space with continuous point evaluations and reproducing kernel K, thenH=HK.
Proof. SinceKis the reproducing kernel ofH, it contains the subspace H0K :=
span{Kx:x∈X}ofHK, and the inclusionη:H0K → His isometric because the scalar products on the pairs (Kx, Ky) coincide. Nowη extends to an isometric embedding ηb: HK → H, and since H0K is also dense in H, we see that ηb is surjective. Forf ∈ HK we now have
η(fb )(x) =hbη(f), KxiH=hf, KxiHK=f(x), so thatbη(f) =f, and we conclude thatHK=H.
Definition 3.1.5. The preceding lemma justifies the notation HK for the unique Hilbert subspace of KX with continuous point evaluations and repro-ducing kernelK. We call it the reproducing kernel Hilbert space defined byK.
Lemma 3.1.6. If HK ⊆KX is a reproducing kernel space andS⊆X a subset with
K(x, x)≤C for x∈S, then
|f(x)| ≤√
Ckfk for x∈S, f ∈ HK.
In particular, convergence in HK implies uniform convergence on S.
Proof. Forf ∈ HK andx∈S, we have
|f(x)|=|hf, Kxi| ≤ kfk · kKxk=kfkp
hKx, Kxi=kfkp
K(x, x)≤√ Ckfk.
Exercises for Section 3.1
Exercise 3.1.1. Show that, ifA∈M2(C) satisfies c∗Ac≥0 for c∈C2, thenA∗=A.
Exercise 3.1.2. Let X be a non-empty set and T ⊆ X ×X be a subset containing the diagonal. Then the characteristic function χT ofT is a positive definite kernel if and only if T is an equivalence relation.
Exercise 3.1.3. Show that if K is a positive definite kernel and c >0, then HcK =HK as subspaces ofKX. Explain how their scalar products are related.