webhost.bridgew.eduwebhost.bridgew.edu/voussa/images/talk at slu.pdf · Measure and integration G - 2nd countable locally compact topological group. Existence of µ left Haar measure

Contents

Some technicalities in Abstract Harmonic Analysis.

Continuous wavelets theory.

Previous work.

My class of group (semi-direct product exponential solvable liegroups).

Examples.

Vignon S. Oussa (Saint Louis University) Admissibility of quasiregular representation of a class of semidirect exponential lie groups.January 2011 1 / 44

Contents



Previous work.


Examples.


Contents



Previous work.


Examples.


Contents



Previous work.


Examples.


Contents



Previous work.


Examples.


Measure and integration

G - 2nd countable locally compact topological group.

Existence of µ left Haar measure on G . Invariant under lefttranslations µ (xE ) = µ (E ) .

Example: when G is an open subset of Rn , left Haar measure isdx

jdet(JLx )j, where dx is the Lebesgue measure.

Modular function is a continuous homomorphism: ∆ : G ! (0,∞) .

1 µ (Ex) = ∆ (x) µ (E ) andRG f(xa)dx = ∆

a1

RG f(x)dx .

2 If G is a connected Lie group, ∆ (x) = jdetAd (x)j1 .

G is unimodular if ∆ (G ) = 1 and Haar measure is both left and rightinvariant.









a1

RG f(x)dx .











a1

RG f(x)dx .











a1

RG f(x)dx .









Modular function is a continuous homomorphism: ∆ : G ! (0,∞) .1 µ (Ex) = ∆ (x) µ (E ) and

RG f(xa)dx = ∆

a1

RG f(x)dx .










RG f(xa)dx = ∆

a1

RG f(x)dx .










RG f(xa)dx = ∆

a1

RG f(x)dx .




Examples

Examples1 G = (Rnf0g,) Haar measure is dxjx j with ∆G (R) = 1) G isunimodular.

2 "ax+b" group: groups of transformations is nonunimodular.∆G (x , a) = ea and Haar measure is 1

a2 da dx .

G =

a x0 1

: a > 0, x 2 R

.

3 H is Heisenberg group is unimodular.

H =

8<:24 1 y z0 1 x0 0 1

35 : x , y , z 2 R

9=; .Haar measure Lebesgue measure on R3.


Representation theory

A unitary representation π : G ! U (Hπ) is a strongly continuousgroup homomorphism into groups of unitary operators on Hπ

Left regular representation of R in L2 (R): (Lx f) (y) = f(y x).Quasiregular representation of ax+b group:

(τ (b, a) f) (x) = 1pa fxba

.

Schrodinger representation of H in L2 (R):(π (z , y , x) f) (t) = e i (zxt+xy )f (t x) .

A character is a unitary representation into the circle group T.

A representation π is irreducible , the only π-stable closedsubspaces are trivial i.e. 0 and Hπ.

π1 ' π2 () 9 U : Hπ1 ! Hπ2 , U unitary linear operator such thatU π1(x) = π2(x) U.bG : the dual of G is the set of all unitary equivalence classes ofirreducibles unitary representations of G .




Left regular representation of R in L2 (R): (Lx f) (y) = f(y x).

Quasiregular representation of ax+b group:


.










.










.










.










.










.




π1 ' π2 () 9 U : Hπ1 ! Hπ2 , U unitary linear operator such thatU π1(x) = π2(x) U.

bG : the dual of G is the set of all unitary equivalence classes ofirreducibles unitary representations of G .






.






Examples

1-dimensional Abelian group: bR ' R.

Heisenberg group: H =R3,

with operation

(z , y , x) (r , s, t) = (z + r + xs, y + s, x + t) .

bH ' fx 2 R : x 6= 0g[R2.

"ax+b" group: G = Ro (0,∞) where (x , a) (y , b) = (x + ay , ab) .

bG ' f1g [R.

Euclidean Motion group: G = CoT with operationz , e iθ

x , e iα

=zxe iθ, e iθe iα

. bG ' (0,∞) [Z.


Examples



with operation

(z , y , x) (r , s, t) = (z + r + xs, y + s, x + t) .

bH ' fx 2 R : x 6= 0g[R2.


bG ' f1g [R.


x , e iα


. bG ' (0,∞) [Z.


Examples



with operation

(z , y , x) (r , s, t) = (z + r + xs, y + s, x + t) .

bH ' fx 2 R : x 6= 0g[R2.


bG ' f1g [R.


x , e iα


. bG ' (0,∞) [Z.


Examples



with operation

(z , y , x) (r , s, t) = (z + r + xs, y + s, x + t) .

bH ' fx 2 R : x 6= 0g[R2.


bG ' f1g [R.


x , e iα


. bG ' (0,∞) [Z.


Induced representation

Let H < G , and (π,Hπ) representation of H, µ a quasi-invariantmeasure of G/H.

Consider Hilbert completion F of F0,(f : G ! Hπ : f(xh) = π (h)1 [f(x)] , andR

G/H jf (g)j2 dµ (gH) < ∞

).

Inner product on F = L2 (G/H,Hπ) is

hf, gi def=ZG/Hhf(x), g(x)iHπ

dµ (x) .

Let y 2 G such that dµ (yx) = q (y) dµ (x) and τ = IndGH (π)

(τ (y) f) (x) = q (y)1/2 fy1x

.



Let H < G , and (π,Hπ) representation of H, µ a quasi-invariantmeasure of G/H.Consider Hilbert completion F of F0,(

f : G ! Hπ : f(xh) = π (h)1 [f(x)] , andRG/H jf (g)j

2 dµ (gH) < ∞

).



dµ (x) .


(τ (y) f) (x) = q (y)1/2 fy1x

.





2 dµ (gH) < ∞

).



dµ (x) .


(τ (y) f) (x) = q (y)1/2 fy1x

.





2 dµ (gH) < ∞

).



dµ (x) .


(τ (y) f) (x) = q (y)1/2 fy1x

.


Example

G = Ro (0,∞) with (x , a) (y , b) = (x + ay , ab) . Considerτ = IndG(0,∞)1

F ' L2 (R)φ : Ro (0,∞)! C : φ ((x , 1) (1, a)) = φ (x , 1) , andR

Rjφ (x , 1)j2 d (x , 1) < ∞

.

d (x , 1) is not invariant under the action of G but is quasi-invariant.

(τ (x , 1)φ) (y , 1) = φ (y x , 1)(τ (1, h)φ) (y , 1) = h1/2φ

h1y , 1

.


Example



Rjφ (x , 1)j2 d (x , 1) < ∞

.


(τ (x , 1)φ) (y , 1) = φ (y x , 1)(τ (1, h)φ) (y , 1) = h1/2φ

h1y , 1

.


Example



Rjφ (x , 1)j2 d (x , 1) < ∞

.


(τ (x , 1)φ) (y , 1) = φ (y x , 1)(τ (1, h)φ) (y , 1) = h1/2φ

h1y , 1

.


Tensor products

Given H,K Hilbert spaces, HK =nT : K ! H : T is bounded linear, ∑k kT (bk )k2H < ∞

owhere

fbkgk ONB in K.

hS ,T i = Trace (T S) , rank one operator: u v : K ! H suchthat

(u v)w = hw, vi u.Rank one operators are dense in HK.Given representations π, σ of G , dene π σ representation ofG G acting in Hπ Hσ

(π σ) (x , y) (u v) = π (x) σ (y) (u v)= π (x) u σ (y) v.


Tensor products


owhere

fbkgk ONB in K.hS ,T i = Trace (T S) , rank one operator: u v : K ! H suchthat

(u v)w = hw, vi u.

Rank one operators are dense in HK.Given representations π, σ of G , dene π σ representation ofG G acting in Hπ Hσ



Tensor products


owhere


(u v)w = hw, vi u.Rank one operators are dense in HK.

Given representations π, σ of G , dene π σ representation ofG G acting in Hπ Hσ



Tensor products


owhere


(u v)w = hw, vi u.Rank one operators are dense in HK.Given representations π, σ of G , dene π σ representation ofG G acting in Hπ Hσ



Direct integral of Hibert spaces

Let (X , µ) be a measure space and fHxgx2X a eld of Hilbert spaces.Dene H the vector space of all measurable sections s such that

ksk2 def=ZXks(x)k2 dµ (x) < ∞

H =R X Hxdµ (x) is a Hilbert space with inner product

hs, ti =ZXhs (x) , t (x)iHx dµ (x) .

Notice if Hx = H for all x ,R X Hxdµ (x) ' L2 (X ,H) .

ExampleVia Fourier transforms, it is known that

L2 (R) 'Z

RCdx

L2 (H) 'Z

RL2 (R) L2 (R) jλj dλ.









L2 (R) 'Z

RCdx

L2 (H) 'Z










L2 (R) 'Z

RCdx

L2 (H) 'Z










L2 (R) 'Z

RCdx

L2 (H) 'Z



Plancherel and Fourier transforms

Fourier transform: (operator-valued transform)π (f) =

RG f(x)π(x)dx , and f 2 L1 (G ) ,π 2 bG .

Existence ofνG , fDπgπ2bG , νG is a measure on bG and fDπgπ2bG

is measurable eld of operators such that the Fourier transformextends from L1 (G ) \ L2 (G ) to a unitary operator on L2 (G ) .Plancherel transform: P : L2 (G )!

R bG Hπ Hπ dνG (π)extending map f 7! fπ (f) Dπgπ2bG

kfk2L2(G ) =ZbG Trace

π (f) Dπ (π (f) Dπ)

dνG (π)

Inverse plancherel transform: f 2 L2(G ),

f(x) =ZbG Trace (π (f) Dπ π(x)) dνG (π) .






is measurable eld of operators such that the Fourier transformextends from L1 (G ) \ L2 (G ) to a unitary operator on L2 (G ) .

Plancherel transform: P : L2 (G )!R bG Hπ Hπ dνG (π)

extending map f 7! fπ (f) Dπgπ2bGkfk2L2(G ) =

ZbG Trace


dνG (π)












dνG (π)












dνG (π)




Examples

1 1-dimensional abelian: G = R, νG equivalent with Lebesguemeasure on R, Dπ = Id.

2 Heisenberg group H, νG is concentrated on

fx 2 R : x > 0g[ fx 2 R : x < 0g , dνG (x) = jx j dx , and

Dπ = Id.3 "ax+b" group, νG point mass measure concentrated on f1,1g .bG = nIndGRχ1

oacting in L2 (0,∞) and (D1f) (t) = et f(t).

4 Exponential solvable group: G = CoR, such that R acts on C

by spiralst (x + iy) = (x + iy) et(1+i ).

The dual of G is parametrized by a circle T.bG = nIndGC χz : z = e iθoacting in L2 (0,∞)

(De iθ f) (t) = e2t f(t) and νG is equivalent to Lebesgue measure on

T.


Examples




Dπ = Id.

3 "ax+b" group, νG point mass measure concentrated on f1,1g .bG = nIndGRχ1






T.


Examples










T.


Examples










T.Vignon S. Oussa (Saint Louis University) Admissibility of quasiregular representation of a class of semidirect exponential lie groups.January 2011 11 / 44

Continuous Wavelets theory

DenitionLet (π,Hπ) be a unitary representation of G , Wη : Hπ ! L2 (G ) withWηφ

(g) = hφ,π(g)ηi . If Wη denes an isometric embedding, η is an

admissible vector for π, or η is a mother wavelet and Wη is acontinuous wavelet transform.

TheoremA representation (π,Hπ) is admissible if and only if it is equivalent witha subrepresentation of the left regular representation λG .

Proof.

Wη (π(x)f) (y) = hπ(x)f,π(y)ηi =f,π(x1y)η

=Wηf

x1y

=

λG (x)Wηf(y) .

Corollaryπ is admissible if and only if π < λG or π ' λG .


Admissibility of regular representation on the real line

1 (λR (x) f) (y) = f (y x) and Wη : L2 (R)! L2 (R) .

2Wηf

(x) = hf,λR (x) ηi = (f η) (x) .

3 Assume Wη is an isometry

Wηf 2L2(R) =

ZR

\(f η) (ξ)2 dξ =

ZR

bf (ξ)2equal 1 a.ez | bη (ξ)2dξ

=Z

R

bf (ξ)2 Contradiction bη /2 L2bR .

4 λR is not an admissible !



1 (λR (x) f) (y) = f (y x) and Wη : L2 (R)! L2 (R) .2Wηf

(x) = hf,λR (x) ηi = (f η) (x) .


Wηf 2L2(R) =

ZR

\(f η) (ξ)2 dξ =

ZR


=Z

R






(x) = hf,λR (x) ηi = (f η) (x) .


Wηf 2L2(R) =

ZR

\(f η) (ξ)2 dξ =

ZR


=Z

R






(x) = hf,λR (x) ηi = (f η) (x) .


Wηf 2L2(R) =

ZR

\(f η) (ξ)2 dξ =

ZR


=Z

R




Admissibility of regular representation on integers group.

1 (λZ (k) f) (n) = f (n k) and Wη : l2 (Z)! l2 (Z) with

k 7! hf,λZ (k) ηi = (f η) (k) .

2 Assume Wη is an isometry Wηf 2L2(Z) =

12π

Z π

π

bf (z)2 bη (z)2 dz = bf 2T

) bη (z)2 = 1 a.e) jbη (z)j2 = 1.

3 η = F11[π,π]

) η (k) = 1

2π

R ππ e

ikzdz =0 if k 6= 01 if k = 0

η = ( , 0, 1, 0, ) Schauder basis element!




k 7! hf,λZ (k) ηi = (f η) (k) .


12π

Z π

π


) bη (z)2 = 1 a.e) jbη (z)j2 = 1.

3 η = F11[π,π]

) η (k) = 1

2π

R ππ e

ikzdz =0 if k 6= 01 if k = 0





k 7! hf,λZ (k) ηi = (f η) (k) .


12π

Z π

π


) bη (z)2 = 1 a.e) jbη (z)j2 = 1.

3 η = F11[π,π]

) η (k) = 1

2π

R ππ e

ikzdz =0 if k 6= 01 if k = 0



Admissibility of the regular representation

Theorem

(Hartmut Fuhr 2002) Wη (f) = f η where η(x) = η(x1) and\Wη (f) (π) = bf (π) bη (π) D1π .

Wη (f) 2 = \Wη (f)

2 = ZbG bf (π) bη (π) D1π

2HSdµ (π)

η is admissible , \Wη (f)

2 = RbG bf (π) 2HS dµ (π), D1π bη (π) is anisometry for µ-a.e π on Hπ.

Theorem(Hartmut Fuhr 2002)

1 Let G be nonunimodular with type I regular representation G. ThenG has an admissible vector.

2 Let G be a unimodular group, such that λG has an admissible vector.Then G is discrete.


Admissibility of the regular representation

Theorem

(Hartmut Fuhr 2002) Wη (f) = f η where η(x) = η(x1) and\Wη (f) (π) = bf (π) bη (π) D1π .

Wη (f) 2 = \Wη (f)

2 = ZbG bf (π) bη (π) D1π

2HSdµ (π)

η is admissible , \Wη (f)

2 = RbG bf (π) 2HS dµ (π), D1π bη (π) is anisometry for µ-a.e π on Hπ.

Theorem(Hartmut Fuhr 2002)

1 Let G be nonunimodular with type I regular representation G. ThenG has an admissible vector.

2 Let G be a unimodular group, such that λG has an admissible vector.Then G is discrete.Vignon S. Oussa (Saint Louis University) Admissibility of quasiregular representation of a class of semidirect exponential lie groups.January 2011 15 / 44

The "ax+b" group (classic example)

1 G = Ro (0,∞) with ∆G (x , a) = a1 (nonunimodular group).τ = IndGR+ (1) acting in L2 (R)

(τ (b, a) f) (x) = jaj1/2 fx ba

.

2 Wη : L2 (R) ! L2 (RoR+)

Wηf

(b, a) = jaj1/2

ZRf(x)η (a1 (x b))dx .

3 τ ' π+ π contained in λG '

π+ 1L2(R)

π 1L2(R)

thus admissible.

Calderon Condition: η is admissible ifR

R+ jbη(w)j2 dwjw j = 1.




(τ (b, a) f) (x) = jaj1/2 fx ba

.

2 Wη : L2 (R) ! L2 (RoR+)

Wηf

(b, a) = jaj1/2



π+ 1L2(R)

π 1L2(R)

thus admissible.






(τ (b, a) f) (x) = jaj1/2 fx ba

.

2 Wη : L2 (R) ! L2 (RoR+)

Wηf

(b, a) = jaj1/2



π+ 1L2(R)

π 1L2(R)

thus admissible.






(τ (b, a) f) (x) = jaj1/2 fx ba

.

2 Wη : L2 (R) ! L2 (RoR+)

Wηf

(b, a) = jaj1/2



π+ 1L2(R)

π 1L2(R)

thus admissible.


R+ jbη(w)j2 dwjw j = 1.Vignon S. Oussa (Saint Louis University) Admissibility of quasiregular representation of a class of semidirect exponential lie groups.January 2011 16 / 44

Generalization of "ax+b" group

1 (Laugasen, Weaver, Weiss and Wilson 2002, Fuhr, 2005)G = Rn oH, H < GL(n,Rn).

1 If G is nonunimodular, H-orbits in Rn are locally closed and for a.e.x 2 Rn , the stabilizer Hx in H is compact, then τ = IndGH1 acting inL2 (Rn , dx) has an admissible vector.

2 If τ has an admissible vector, then G is nonunimodular and for a.e.x 2 Rn , the stabilizer Hx in H is compact.

2 N not abelian but nilpotent simply connected

1 (Liu and Peng, 1997) N Heisenberg group, H = R.2 (Fuhr, 2005) N homogeneous, H = R.3 (Currey 2007) N arbitrary, H abelian, Ad (H) completely reducible over

R.








R.








R.








R.







1 (Liu and Peng, 1997) N Heisenberg group, H = R.

2 (Fuhr, 2005) N homogeneous, H = R.3 (Currey 2007) N arbitrary, H abelian, Ad (H) completely reducible over

R.







1 (Liu and Peng, 1997) N Heisenberg group, H = R.2 (Fuhr, 2005) N homogeneous, H = R.

3 (Currey 2007) N arbitrary, H abelian, Ad (H) completely reducible overR.








R.


How to settle admissibility of an arbitrary representation

ProblemUnder which conditions is τ equivalent with a subrepresentation ofλG ?

SolutionAssume G is non discrete and nonunimodular.

1 Describe parameters of bG, compute Plancherel transform, Plancherelmeasure and eld of operators Dπ.

2 Decompose λG ' P λG P1 =R bG m (π)π dµG (π) , describe

multiplicity m and measure µG .

3 Decompose τ ' T τ T 1 =R bG m0 (π)π dνG (π) , and describe

multiplicity m0 and measure νG4 Is νG << µG ? If answer is no stop τ is not admissible.5 If answer to 4 yes then τ is admissible.


















multiplicity m0 and measure νG

4 Is νG << µG ? If answer is no stop τ is not admissible.5 If answer to 4 yes then τ is admissible.









multiplicity m0 and measure νG4 Is νG << µG ? If answer is no stop τ is not admissible.

5 If answer to 4 yes then τ is admissible.











My class of group

N oHN connected, simply connected non commutative Nilpotent Liegroup. H a connected abelian Lie group with Lie algebra h =RA1 RAk .

H < Aut (N), all Aj , ad (Aj ) is diagonalisable. Eigenvalues ofad (h) /2 iR. Existence of xed ordered basis for nC : fZigni=1 =set of eigenvectors for ad (Aj ) .

ad (Ak ) (Zj ) = [Ak ,Zj ] = νj (1+ iαj )Zjexp tAk Zj = exp (tνj (1+ iαj ))Zj .

G = N oH is an exponential solvable Lie group.(n, h)(n0, h0) = (n (h n0) , hh0). Left Haar measurejdet (Ad (n, h))j1 dndh.τ = IndGH (1) acting in L2(N); Left regular representation λG actingin L2 (N oH) .Problem: Under which conditions τ admissible?


My class of group

N oHN connected, simply connected non commutative Nilpotent Liegroup. H a connected abelian Lie group with Lie algebra h =RA1 RAk .H < Aut (N), all Aj , ad (Aj ) is diagonalisable. Eigenvalues ofad (h) /2 iR. Existence of xed ordered basis for nC : fZigni=1 =set of eigenvectors for ad (Aj ) .




My class of group



G = N oH is an exponential solvable Lie group.(n, h)(n0, h0) = (n (h n0) , hh0). Left Haar measurejdet (Ad (n, h))j1 dndh.

τ = IndGH (1) acting in L2(N); Left regular representation λG actingin L2 (N oH) .Problem: Under which conditions τ admissible?


My class of group



G = N oH is an exponential solvable Lie group.(n, h)(n0, h0) = (n (h n0) , hh0). Left Haar measurejdet (Ad (n, h))j1 dndh.τ = IndGH (1) acting in L2(N); Left regular representation λG actingin L2 (N oH) .

Problem: Under which conditions τ admissible?


My class of group





Examples and counter-examples

Examples: n = R span fZ ,Y ,Xg with [X ,Y ] = Z , h = Rk ,

H = (0,∞)k and N = H.A

X 1Y 1Z 0

or

AX iY 1 iX + iY 1+ iZ 2

or

A BX iY 1 1X + iY 1 1Z 2 0

Counter-examples: solvable but not exponential:A

X iY iX + iY iZ 2

or

A BX iY 1 i 1X + iY 1+ i 1Z 2 2

.


Examples and counter-examples

Examples: n = R span fZ ,Y ,Xg with [X ,Y ] = Z , h = Rk ,

H = (0,∞)k and N = H.A

X 1Y 1Z 0

or

AX iY 1 iX + iY 1+ iZ 2

or

A BX iY 1 1X + iY 1 1Z 2 0

Counter-examples: solvable but not exponential:

AX iY iX + iY iZ 2

or

A BX iY 1 i 1X + iY 1+ i 1Z 2 2

.


Construction of dual of G

Coadjoint action on g: g l (X ) = lg1Xg

. g 2 G , l 2 g.bG ' fOrbG (λ) : λ 2 gg . (Kirilov, Konstant)

Parametrize bG using cross-section for coadjoint orbits.(Currey 1992) Construct jump index e f1, 2, , dimNg, ϕidenties directions in g in which G acts exponentially".

1 Cross-section for coadjoint orbits is smooth manifold for xed"adaptable basis"

Σ = fl 2 Ω : l (Zj (l)) = 0, j 2 enϕ and jbj (l)j = 18j 2 ϕ g .

2 H-invariant cross-section for the N-orbits,

Λ ' n/N = ff 2 Ωe : f (Zj (f )) = 0 8j 2 e g .

Σ ' Λ/H cross-section of H orbits in Λ.


Cross-sections

Theorem(Oussa) Let K = StabH (Λ) < H with its Lie algebra k.The smoothmanifold

Σ = f(f , k, 0) : f 2 Σ and k 2 kgis a cross-section for almost all G-orbits, and if π : g ! n thenΣ = π (Σ) ' Λ/H is cross-section in n for the N-orbits.

Proof.(sketch) Dene cross section mapping λ : Ωe ! Λ,ν (f ) = f1 j n : f (Zj ) 6= 0g , and Λν = ff 2 Λ : ν (f ) = ν g .Existence of ν f1, , ng such that Λν is dense and zariski open.Dene Ων = π1 λ1 (Λν) which is dense and zariski open in g. ApplyCurrey jump index method.


The irreducible representations of G

1 Construct real polarizing subalgebra p(f ) for each f 2 Λ. DeneDf = exp (p(f ) \ n) . For each f 2 Λ, πf = IndNDf χf is irreduciblerep of N. bN ' fπf : f 2 Λg .

2 Let K be the stabilizer of Λ in H. Dene extension eπf of πf whereeπf : NK ! U (Hf ) . eπf (n, h) = πf (n)C (h, f ), andδ = det (Ad jN/Df )

C (h, f )φ(x) = δ(h)1/2φ(h1 x).

3 Given (f , σ) 2 Σ and ρf ,σ = IndGNK (eπf χσ) with χσ 2 bK . ρf ,σ isirreducible representation of G

bG ' ρf ,σ : (f , σ) 2 Σ





C (h, f )φ(x) = δ(h)1/2φ(h1 x).


bG ' ρf ,σ : (f , σ) 2 Σ





C (h, f )φ(x) = δ(h)1/2φ(h1 x).


bG ' ρf ,σ : (f , σ) 2 Σ


Plancherel measure and decomposition of left regularrepresentation

Theorem(Currey 2005)

1 A measure dvG (λ, σ) on Σ

dµ (λ, σ) =jPfe (λ, σ)j

(2π)dim g+d ∏j2ϕ

j1+ iαj jdλdσ

2 A measurable eldρλ,σ,Hλ,σ

of irreducible representations and a

measurable eld fDλ,σg of positive, self-adjoint, and semi-invariantoperators acting in Hf ,σ such that

(Dλ,σf) (h) = ∆ (h) f (h) .



Theorem(Currey 2005)

1 A measure dvG (λ, σ) on Σ

dµ (λ, σ) =jPfe (λ, σ)j

(2π)dim g+d ∏j2ϕ

j1+ iαj jdλdσ

2 A measurable eldρλ,σ,Hλ,σ

of irreducible representations and a

measurable eld fDλ,σg of positive, self-adjoint, and semi-invariantoperators acting in Hf ,σ such that

(Dλ,σf) (h) = ∆ (h) f (h) .



TheoremLet λG be the left regular representation of G acting in L2(G ). LetHλ,σ = L2 (H/K ,Hλ) and ρλ,σ = IndGNK (eπλ χσ)

λG 'Z

Σkρλ,σ 1Hλ,σ

dµ (λ, σ)

acting in

PL2 (G )

=Z

ΣkHλ,σ Hλ,σ dµ (λ, σ) .

.


Decomposition of the quasiregular representation

τ = IndGH (1) , f 2 L2 (N)

τ(n, 1)f (m) = f(n1m),

τ(1, h)f (m) = δ(h)12 f(h1 m).

Fact

Dening bτ(n) = Fτ(n)F , bτ(n) acts in F L2 (N) such thatbτ = R Λ bτλ jPf(λ)j dλ acting inR

Λ Hλ Hλ jPf(λ)j dλ

bτλ(n)bf(λ) = πλ(n) bf (λ)bτλ(h)bf(λ) = δ(h)12Ch1 λ(h, h

1 λ)bf(h1 λ)Ch1 λ(h1, h1 λ).


Decomposition of the quasiregular representation

LemmaK is non trivial if dim (cent (g) \ h) = 0 then

C (,λ) 'Zk

mMk=1

σ dµK (σ) acting in

Zk

mMk=1

C dµK (σ)

τ 'Z

Σ

Zk

ρλ,σ 1Cm

dµK (σ) jPf(λ)j dλ.

When m is nite, m is equal to a power of 2.

Lemma

If K is trivial then τ 'R

Σk

ρλ,σ 1l2(N)dµK (σ) jPf(λ)j dλ.


Decomposition of the quasi regular representation

LemmaAssuming K is non trivial K1 = cent (g) \ h. If dim (K1) > 0 thenC (,λ) '

R k?1

CmdµK/K1 (σ)

τ 'Z

Σ

Zk?1

ρλ,σ 1Cm

dµK/K1 (σ) jPf(λ)j dλ.

k?1 is the subspace of linear functionals in h vanishing on k1.


Decomposition of the quasi regular representation

Proof.

(Sketch) Decompose Λdi¤eo' Σ H/K . L2 (N) is unitarily equivalent

with

Z ΛHf Hf jPf(f )j df '

Z Σ

Hλz | Z H/KHhλ Hhλdωhλ(h)

jPf(λ)j dλ

'Z

ΣHλ jPf(λ)j dλ.

Identify Hλ withR k L

2H/K ,Hλ Cm

dσ via explicit isomorphism

isometry using elementary tensors intertwining

Z H/K

bτλ dm (λ) with

Zk

ρλ,σ 1Cm

dµK (σ) .


Summary

Let k1= cent (g) \ h

Left regular rep : λG 'Z

Σkρλ,σ 1Hλ,σ

dµ (λ, σ)

dim (k1) = 0) τ 'Z

Σ

Z k

ρλ,σ 1Cm dµK (σ) jPf(λ)j dλ.

dim (k1) > 0) τ 'Z

Σ

Z k?1

ρλ,σ 1Cm dµK/K1 (σ) jPf(λ)j dλ


Results

TheoremIf dim (k1) = 0, the quasiregular representation of G is contained insidethe regular representation of G. In other words, τ < λG or τ ' λG .

TheoremIf dim (k1) > 0, the quasiregular representation of G is notquasi-equivalent with the left regular representation of G .

Corollary

If dim (k1) > 0, the quasiregular representation of G is never admissible.However, assuming dim (k1) = 0, the quasiregular representation isadmissible if and only if G is nonunimodular.


Admissibility conditions

Assuming G is nonunimodular and dim (k1) = 0.

DenitionsLet Hλ,σ = L2 (H/K ,Hλ) , m 2N [ f∞g anddµ (λ, σ) = dµK (σ) jPf(λ)j dλ.

D =Z

ΣkHλ,σ Cm dµ (λ, σ) ,

G =Z

ΣkHλ,σ Hλ,σ dµ (λ, σ)

P(L2 (G )) =Z

ΣkHλ,σ Hλ,σ dµ (λ, σ) .

Observation: µ µ.


DenitionDene isometric embeddings: S : D! G and W : G!P(L2 (G )),

Sλ,σ

v

λ,σ

α1λ,σ, α2

λ,σ, , αm

λ,σ

= v

λ,η

m

∑j=1

αj

λ,σ

bk

Wλ,σ (uλ,σ vλ,σ) =

dµ (λ, σ)

dµ (λ, σ)

1/2

uλ,σ vλ,σ.

FactIf m is innite(

S =R

Σk Sλ,σdµ (λ, σ)

W =R

ΣkWλ,σdµ (λ, σ)are unitary operators.


Observation: P1 W S eT intertwines τ with λG .

P L2 G

L2 N

F

T S W

T

L2 G

P

P 1 W S T

F L2 N


Admissible conditions(cont)

Put Kk = Hλ,σ vk ' Hλ,σ ' L2 (Rs) , andfvkg is ONB. Have chain ofisometries intertwining rep.

Dk =Z

ΣKk dµ (λ, σ)) D '

mMk=1

Dk

Dk ' L2 (Σ, Hλ,σ) ' L2 (ΣRs)

L2 (N) 'mMk=1

eT1 L2 (ΣRs)=

mMk=1

L2 (N)

k.

D 'm timesz |

L2 (ΣRs) L2 (ΣRs) L2 (ΣRs)


Results on admissible conditions on mulitiplicity-freesubspaces

Theorem

Given η 2 L2 (N)k such that Tλ,σbη (λ, σ) = q (λ, σ) vk , η is admissiblefor τjL2 (N)k if and only if D1λ,σWλ,σSλ,σTλ,σbη (λ, σ) 2

HS= 1 µ-a.e.

Theorem

(Reformulation) Given η 2 L2 (N)k such that Tλ,σ (bη (λ, σ)) = f2 L2 (ΣRs) is admissible for τjL2 (N)k if and only if D1λ,σWλ,σf

(λ, σ, )

L2(Rs)

= 1 µ-a.e.


Proof.

(sketch)Wηϕ

(n, h) =

P1WSTF (ϕ)

P1WSTF (η)

(n, h) ,

letting (TωF )ϕ (ω) = f (ω) vk , Wηϕ

2L2(G ) =

[Wηϕ 2L2(Σ)

=ZΣ

WωSωTωFϕ (ω) (WωSωTωFη (ω))D1ω

2HS dµ (ω)

=ZΣ

kWωSωf (ω) vkk2HS D1ω (WωSωq (ω) vk )

2HSdµ (ω)

=ZΣ

kWωSωf (ω)k2equal 1 µ-a.ez | D1ω Wωq (ω)

2dµ (ω)

=) [Wηϕ

2L2(Σ)

= kϕk2 .


Results on admissible conditions

Theorem

Letting ϕ, η 2 L2 (N) ,ϕ= ∑k2I

ϕk and η = ∑k2I

ηk such that for each k, ηk

is admissible for τjL2 (N)k thenDWηk

f,Wηjf 0E= 0 if j 6= k.

Corollary

let η 2 L2 (N) , η = ∑k2I

ηk , Tλ,σ bηk (λ, σ) fk 2 L2 (Rs) , η is admissible

for τ if and only if for each k , D1λ,σWλ,σfk (λ, σ, ) L2(Rs)

= 1 µ a.e.


Example1: real eigenvalues

n = span fZ ,Y ,Xg , h = span fA,Bg , n = Lie (N) , h = Lie (H) andG = N oH.

(z , y , x) , ea, eb((r , s, t) , eu , ev )

= (z + e2ar + xseab , y + eabs, x + ea+bt

, ea+u , eb+v )

[X ,Y ] = Z , [A,X ] = X , [A,Y ] = Y , [A,Z ] = 2Z ,

[B,X ] = X , [B,Y ] = Y , and [B,Z ] = 0.

Λ = f(l1, 0, 0) : l1 6= 0g , and Σ = f(1, 0, 0, 0, σ) : σ 2 RBg .

∆ (z , y , x , a, b) = e4a ) G is nonunimodular.

τ 'R f1gR

ρ1,σ 1C2dσ acting inR f1gR

L2R2C2dσ.


Example1: real eigenvalues (cont)

λG 'R f1gR

ρ1,σ 1L2(R2)dσ(2π)7

acting inR f1gR

L2R2 L2 (R2) dσ

(2π)7.

τ < λG and dim (cent (g) \ h) = 0 thus, τ is admissible.

Field of operators: (D1,σφ) (x) = ∆ (x)φ (x) , φ 2 L2R2.

L2 (N) 'eT1 L2 f1g RR2 eT1 L2 f1g RR2

.

η = η1 + η2 2 L2 (N) such that eT (ηk ) = fk , η is admissible , D11,σW1,σfk (1, σ, ) 2L2(R2)= 1 a.e.


Example 2 complex eigenvalues

n = span fZ1,Z2,Y1,Y2,X1,X2g h = fA,Bg , n = Lie (N) , h = Lie (H)and G = N oH.

[X1 + iX2,Y1 + iY2] = Z1 + iZ2,

[X1 iX2,Y1 iY2] = Z1 iZ2,

[A,X1 + iX2] = (1+ i) (X1 + iX2) ,

[A,Y1 + iY2] = (1 i) (Y1 iY2) ,[A,Z1 + iZ2] = (2+ 2i) (Z1 + iZ2) .

Λ = f(l1, l2, 0, 0, 0, 0) : l1 6= 0, l2 6= 0g , and Σ = T.

Modular function: ∆ (z1, z2, y1, y2, x1, x2, a) = e8a.


Example 2 complex eigenvalues (cont)

Plancherel measure: dµ (z) = 2dz(2π)10

τ 'Z

Tρz 1l2(N)dµ (z) on

Z TL2R3 l2 (N)dµ (z)

λG 'Z

Tρz 1L2(R3)dµ (z) on

Z TL2R3 L2 (R3)dµ (z) .

τ ' λG and dim (cent (g) \ h) = 0 thus, τ is admissible.

Field of operators: (Dzφ) (x) = ∆ (x)φ (x) , for φ 2 L2R3.


Example 2 complex eigenvalues (cont)

Let l2 (N) 'Mk2N

C.

IdentifyR

TL2R3 l2 (N)dµ (z) with

Mk2N

L2TR3, dµ

.

Let L2 (N) =Mk2N

eT1 L2 TR3. Given η = ∑

k2N

ηk , such thateT (ηk ) = fk then η is admissible if and only if for every k 2N , D1z Wz fk (z , ) 2L2(R3)

= 1 for almost every z 2 T.


What is next?

Develop an algorithm for construction of continuous wavelets(admissible vectors) on multiplicity free spaces.

Develop an algorithm for construction of continuous wavelets(admissible vectors) on L2 (N) .


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webhost.bridgew.eduwebhost.bridgew.edu/voussa/images/talk at slu.pdf · Measure and integration G - 2nd countable locally compact topological group. Existence of µ left Haar measure