Harmonic Analysis EPvan den Ban Master Class Course MRI Fall Intro duction For an intro duction to the following material we refer to the rst lecture given by the author One may also confer Sug which will b e the basic reference for the rest of this course Haar measure Let X b e a lo cally compact Hausdor space Denition A p ositive Radon integral on X is a linear functional I C X C such that c f I f Remark By the Riesz representation theorem I is actually the integral asso ciated with a regular Borel measure We shall refer to this measure as the Radon measure asso ciated with I and we shall use the notation Z I f f x dx G Lemma Let I b e a p ositive Radon measure on X Then a f g I f I g f g C X c b f C X jI f j I jf j c c The map f I f is continuous on C G c Proof For a use that g f and the denition of I b First assume that f is realvalued and dene continuous functions by f maxf and f maxf Then f f f and jf j f f Hence jI f j jI f I f jI f I f I jf j Now assume that f is complex valued and that I f R Then I Im f Hence jI f j jI Re f j I jRe f j I jf j Sug M Sugiura Unitary Representations and Harmonic Analysis nd ed NorthHollandKo ddans ha Amsterdam by what we proved ab ove Let f now b e arbitrary Then there exists a z C with jz j and zI f Then I zf R hence jI f j jI zf j I jzf j I jf j It remains to establish c Let K X b e a compact subset Then wemust show that there exists a C such that for all f C K wehave jI f j C kf k There exists a nonnegative c function g C X such that g on K Hence for all f C K wehave jf j jf jg kf k g c c and it follows that jI f j I jf j kf k I g which establishes the desired estimate Note that by taking the inmum over g one obtains the estimate jI f j kf k volume K By a Gspace we shall from now on mean a lo cally compact Hausdor space X together with a continuous action of G on X We agree to write L for the induced action of G on C X Thus if C X then L x g g x The Gspace X is called homogeneous if there is only one Gorbit ie if x X is a p oint then G x X One readily veries that in this setting the stabilizer G fg G gx x g x is a closed subgroup of G Moreover the map G X g gx induces a homeomorphism GG X Thus any homogeneous Gspace is of the form GH with H a closed subgroup of x G ConverselyifH is a closed subgroup then GH is a homogeneous space in a natural way Example The unit sphere S in R is a homogeneous space for the sp ecial orthogonal group SO Let e Then the stabilizer of e in SO is the group of rotations with axis Re which is isomorphic to SO The sphere is thus identied with the quotientSOSO This is the starting p oint for the spherical harmonics to b e discussed later Lemma Let X be a Gspace and let C X Then g L is a continuous map c g G C X c Proof Let g G Then wemust show that L L is close to in C X for g close to g g c where L and we see that without loss of generality w L L L g No g g g gg wemayaswell assume that g e Let C b e a compact neighb ourho o d of e in G Then the set K C supp is compact in X Moreover for g C the function L has supp ort contained g in K Fix Then wemust show that there exists an op en neighb ourho o d U of e in C such that jL kL k sup x xj g g K xK for all x U The map g x g xiscontinuous hence for every x K there exists an op en neighb ourho o d N x in X and an op en neighb ourho o d U e in C G such that x x jg y y j for all y N g U By compactness of K there exists a nite collection x x x x of p oints in K such that the sets N cover K Let U be the intersection of the sets n x j U Then U is an op en neighb ourho o d of e in G and one readily checks that holds for g U j Corollary Let X be a Gspace and let I b e a p ositive Radon integral on G Then for C X wehavethatg I L denes a continuous function G C c g A p ositive Radon integral I on a Gspace X is called invariant if for all C X and c g G wehave I L I g or equivalentlyif denotes the measure asso ciated with I Z Z g xdx x dx X X Lemma Let I be a Ginvariant p ositive nontrivial Radon measure on a Ghomogeneous space X Let f C X and supp ose that f everywhere Then I f f c Proof Supp ose that f is as ab ove and I f but f Then there exists a p oint x X such that f x Hence there exists a compact neighb ourho o d U x and a p ositve constant Let C suppg b e the compact supp ort such that f on U Let now g C X g c of g Then every p ointofC is contained in Gx and by compactness there exists a nite subset S G such that the sets sU s S cover C For every s S there exists a such that on s P sU wehave g L f Hence g L f and it follows that s s s s s sS X I g I L f s s sS tegral It follows from this the latter equality b eing a consequnece of the invariance of the in that I contradiction Corollary Let the hyp othesis of the ab ove lemma b e fullled Then Z hf jg i I f g f xg x dx X denes a p ositive denite Hermitian inner pro duct on C X c We denote the completion of C X with resp ect to the ab ove inner pro duct by L X Let c L denote the left represenation of G on X Then byinvariance of the measure the map L is x unitary for the inner pro duct on C x and therefore has a unique extension to a unitary map c L L X L X x The group G may b e viewed as a homogeneous space for its left right action A left right invariantpositive nontrivial Radon integral on X is called a left right Haar integral on G Theorem Let G b e a lo cally compact group Then there exists a left right Haar integral I on G If I is a second left right Haar integral on G then I cI for a p ositive constant c Proof For the existence part of this theorem we refer the reader to BI We will prove uniqueness If f C X then we dene the function f C X by f xf x If J is a c c BI Bourbaki Integration Chapter right Haar integral then we dene the left Haar integral J by J f J f Thus wehavea bijection b etween left and rightHaarintegrals Let now I and J b e a left and a right Haar integral resp ectively Then it suces to show that J cI for some c Let and denote the measures asso ciated with I and J resp ectively and let f C Gbesuch that I f Dene c Z D x I f f y x d y f G Then D is a continuous function on Gby Lemma f Let g C G Then c Z J g g xy d x G for every y G Multiplying the ab ove identity with f y andintegrating with resp ect to dy we obtain Z Z Z Z I f J g f y g xy d xdy f y g xy dy d x G G G G where the change of order of integration is allowed byFubinis therorem Note that the integrand is compactly supp orted as a function on G G Using left invariance of and after that again changing the order of integration wemay rewrite the integral as Z Z f x y d x g y dy f D g f G G Hence J g I D g for every g C GLet f f C Gbetwo functions with I f f c c j ws from the ab ove that I D D g for all g hence D D Thus it then it follo F f f f follows that D D is a continuous function which is actually indep endent of the particular f choice of f Substituting x e in the denition of D wenow infer that f D eI f J f for all f in the set S of functions C GwithI The set S is readily seen to b e dense c in C G so that is actually valid for all f C G Hence J D eI c c Remark The existence and the uniqueness of Haar measure is much easier to establish when G is a Lie group ie G p ossesses the structure of C manifold compatible with its top ologyandsuch that the map G G G x y xy is smo oth ie C Examples of Lie groups are SUn SOn SLn C GLn C SLn R GLn R For simplicitywe assume n denotes that G is connected Let n dim G and x a nontrivial nform T G Here T G e e x the cotangentspace of G at an element x G Let l G G g xg and dene for x G x n the nform T G by x x T l x e x e Then x is a nowhere vanishing smo oth dierential nform on G hence denes an x orientation on G We dene I C G C by the oriented integral c Z I f f f C G c G One readily veries that I is a p ositiveRadonintegral which is nontrivial Moreover one also readily checks that l for all g G from which it follows that I is a left Haar integral g For further details we refer to BD or Wa Let I b e a left Haar integral on G Then for every x G we dene the Radon integral R I x by R I I R Then one readily veres that R I isaleftHaarintegral use that R x x x x and L commute Hence R I xI for a uniquely determined x The function y x G which is readily seen to b e indep endentoftyhe particular choice of I is called the mo dular function of G Exercise Show that is a continuous group homomorphism A group G with is called unimo dular Thus a group is unimo dular if its left Haar measure is a right Haar measure For obvious reasons the Haar measure is said to b e biinvariant in this case Exercise Show that a compact group G is unimo dular Hint use that the image G is a compact group Representations From nowonG will always b e a lo cally compact top ological group Exercise Let b e a representation of G in a Banach space V Show that the following conditions are equivalent a G V V is continuous b For every x G the map x is coninuous and for every v V the map G V x x v is continuous Hint to show that a follows from b use the BanachSteinhaus theorem Lemma Let X be a Ghomogeneous space and let I b e a nontrivial invariant p ositive Radon integral on X Then the representation L of G in L X dened in the previous section is continuous Proof In view of