Contents

Chapter 1. Basic Representation Theory 3 1. Unitary Representations 3 2. Cyclic representations 10 3. The left regular representation 12 4. The ax + b-group, wavelets and square integrable representations 14 5. Generalizations of the Wavelet transform 18

2 CONTENTS

CHAPTER 1 Basic Representation Theory

1. Unitary Representations

In this section G will be a locally compact Hausdorff topological group with left Haar measure λ.

Definition 1. A representation of G in a topological vector space V ≠ {0} is a homorphism ρ : G → GL(V), the space of continuous invertible linear maps V → V, such that

G ∋ x ↦ ρ(x)u ∈ V

is continuous for all u ∈ V. If V is a Hilbert space let d(ρ) = dim(V). ρ is called finite dimensional if d(ρ) < ∞. If dim(V) = 1 then ρ is called a character.

Remark 1. If ρ is a character, then ρ is nothing but a continuous homomorphism ρ : G → ℂ^{∗}.

Let ρ be a representation of G. If nothing else is said then V_{ρ} will denote the corresponding topological vector space. Define a map G × V_{ρ} → V_{ρ}, (a, u) ↦ a · u := ρ(a)u. This defines a separately continuous action of G on V_{ρ} such that each of the maps u ↦ a · u is linear. On the other hand if (a, u) ↦ a · u is a separately continuous action of G on a topological vector space V by linear transformation, then we can define a representation of G on V by ρ(a)u = a · u. Thus a representation is just an separately continuous action of G on a topological vector space given by linear transformations. We will sometimes write (ρ, V) for a representation ρ in V.

Definition 2. Let ρ, τ be representations of G in the topological vector space V_{ρ} respectively V_{τ}. 1. A subspace W ⊂ V_{ρ} is invariant if ρ(a)W ⊂ W.

2. A continuous linear map T : V_{ρ} → V_{τ} is called an intertwining operator if T ρ(a) = τ(a)T for all a ∈ G. Denote by Hom_{G}(ρ, τ) the set of intertwining operators between ρ and τ.

3. We say that ρ and τ are equivalent, denoted by ρ ≃ τ if Hom_{G}(ρ, τ) contains an isomorphism.

4. irreducible if the only closed invariant subspaces of V are {0} and V.

5. τ is a sub-representation of ρ if there exists an injective T ∈ Hom_{G}(ρ, τ).

6. τ is a quotient of ρ if there exists a surjective T ∈ Hom_{G}(ρ, τ).

Lemma 1.1. Let ρ, τ be representations of G and let T ∈ Hom_{G}(ρ, τ). Then \overline{Im(T)} ⊂ V_{τ} is τ-invariant and Ker(T) ⊂ V_{ρ} is ρ-invariant. In particular the following holds:

1. If ρ is irreducible, then either T is zero or injective.

2. If τ is irreducible then either T is zero or T has a dense image.

Proof. Let u = T (v) ∈ Im(T) and a ∈ G. Then τ(a)u = τ(a)T (v) = T (τ(a)v) ∈ Im(T). Hence τ(G)Im(T) ⊂ Im(T) ⊂ \overline{Im(T)}. As τ(a) is continuous it follows that τ(a) \overline{Im(T)} ⊂ \overline{Im(T)} for all a ∈ G and hence \overline{Im(T)} is invariant. Assume now that u ∈ Ker(T) and a ∈ G. Then T (ρ(a)u) = τ(a)T (u) = 0 as T (u) = 0. If ρ is irreducible then, as Ker(T) is closed, it follows by the definition that either Ker(T) is {0} or Ker(T) = V_{ρ}. In the first case T is injective and in the second case T is the zero map. (2) follows in the same way.

Definition 3. Let ρ be a representations of G. Then ρ is called:

1. unitary representation if V_{ρ} is a Hilbert space and ρ(a) is unitary for each a ∈ G,

2. two unitary representations ρ and π are unitary equivalent if there exists a unitary isomorphism U ∈ Hom_{G}(ρ, π)

Notice that if ρ is a unitary representation of G then

ρ(a)^{∗} = ρ(a)^{−1} = ρ(a^{−1}) .

Example 1.1. Let G = γ_{n} the permutation group of n-elements. Let V = ℝ^{n}. Define a representation ρ : G → O(n) by

ρ(g)(x_{1}, …, x_{n}) = (x_{g^{−1}(1)}, …, x_{g^{−1}(n)}) .

This representation is not irreducible if n > 1 because the space ℂv with v = (1, …, 1) is invariant.

Example 1.2. Let G = SO(n) and let ℂ^{k}[ℝ^{n}] be the space of polynomials of degree ≤ k. Let ℂ_{k}[ℝ^{n}] be the subspace of polynomials of degree k. This is exactly the space of polynomials that are homogeneous of degree k i.e.,

p(λx) = λ^{k}p(x) .

Define a representation ρ of G on ℂ^{k}[ℝ^{n}] by

[ρ_{k}(g)p](x_{1}, …, x_{n}) := p(g^{−1}x_{1}, …, g^{−1}x_{n}) .

This representation is not irreducible because ℂ_{k}[ℝ^{n}] is invariant. Thus the restriction of ρ_{k} to the space of polynomial that are homogeneous of degree n is a representation of G. But even this representation is not irreducible.

Lemma 1.2. The multiplication operator M : ℂ^{k−2}[ℝ^{n}] → ℂ^{k}[ℝ^{n}], p(x) ↦ (x_{1}^{2}+…+x_{n}^{2})p(x) = ||x||^{2} p(x) is an intertwining operator. In particular it follows that Im(M) is an invariant subspace. If n ≥ 2 then Im(M) ≠ ℂ^{k}[ℝ^{n}].

Proof. We have

(Mρ_{k−2}(g)p)(x) = ||x||^{2}ρ_{k−2}(g)p(x)
= ||x||^{2}p(g^{−1}x)
= ||g^{−1}x||^{2}p(g^{−1}x)
= ρ_{k}(g)(Mp)(x)

where we have used that ||g^{−1}x|| = ||x|| because g, and hence also g^{−1}, is orthogonal. That Im(M) is invariant follows now from lemma 1.1. If n ≥ 2 then x_{1}^{k} ∉ Im(M) and hence Im(M) ≠ ℂ^{k}[ℝ^{n}].

Example 1.3. Let G = ℝ^{n}. Let e_{y}(x) = e^{−ix·y}, y ∈ ℝ^{n}. Then each of the maps y ↦ e_{y} is a one dimensional representation of ℝ^{n}. We can define two representations on the Hilbert space L^{2}(ℝ^{n}). The first of them is the regular representation

L_{x}(f)(y) = f(y − x), f ∈ L^{2}(ℝ^{n}) .

The other is

U_{x}F (y) = e_{x}(y)F (y) .

As FL_{x}f(y) = e_{x}(y)Ff(y) it follows that the Fourier transform is an unitary isomorphism of representations.

Example 1.4. Consider now the canonical action of G on the spaces C(G), C_{c}(G), and C_{∞}(G). Let f ∈ C(G) and ϵ > 0 and let K be a compact set. For each x ∈ K let U(x) ∈ N (e) be symmetric and such that

|f(y^{−1}x) − f(x)| < ϵ/2 ∀y ∈ U (x)

Let V (x) be symmetric and such that V (x)^{2} ⊂ U(x). As K is compact there are finitely many x_{1}, …, x_{n} ∈ K such that with V_{j} = V (x_{j}) we have K ⊂ ∪_{j}V_{j}x_{j}. Let V = ∩V_{j}. Let z ∈ K. Then there is a j such that z = vx_{j} ∈ V_{j}x_{j}. It y ∈ V then y^{−1}z = (y^{−1}v)x_{j} ∈ V_{j}^{2}x_{j} ⊂ U(x_{j}). Hence

|f(y^{−1}z) − f(z)| ≤ |f(y^{−1}vx_{j}) − f(x_{j})| + |f(vx_{j}) − f(x_{j})| < ϵ

Hence f is uniformly continuous on K. Define a representation of G on C_{c}(G) by

L_{x}(f)(y) = f(x^{−1}y) .

It is clear that L_{xy} = L_{x}L_{y} and the above consideration shows that x ↦ L_{x}f is a continuous. For f ∈ C_{∞}(G) and x ∈ G let L_{x}(f)(y) = f(x^{−1}y). Let ϵ > 0 be given. Then there exists a compact set K ⊂ G such that |f(x)| < ϵ/2 for x ∉ K. The above shows also that there exists a V ∈ N (e), symmetric, such that |f(y^{−1}x) − f(x)| < ϵ/3 for all x ∈ K and all y ∈ V. We claim that

||L_{y}f − f||_{∞} < ϵ ∀y ∈ V .

If x ∈ K then |f(y^{−1}x) − f(x)| < ϵ/2. If both x and y^{−1}x are not in K then |f(y^{−1}x) − f(x)| < ϵ. Assume now that y^{−1}x ∈ K and x ∉ K. As x = y(y^{−1}x), y^{−1}x ∈ K, and V is symmetric it follows that

|f(y^{−1}x) − f(x)| = |f(y^{−1}x) − f(y(y^{−1}x))| < ϵ/2 .

It follows that L_{y} is a continuous representation of G on C_{∞}(G).

Lemma 1.3. Let ρ, τ, π be representations T ∈ Hom_{G}(ρ, τ) and S ∈ Hom_{G}(τ, π). Then ST ∈ Hom_{G}(ρ, π).

Proof. We have for all a ∈ G: ST ρ(a) = Sτ(a)T = π(a)ST.

Lemma 1.4. Let ρ and τ be unitary representations and T ∈ Hom_{G}(ρ, τ). Then T^{∗} ∈ Hom_{G}(τ, ρ).

Proof. Let a ∈ G, u ∈ V_{ρ} and w ∈ V_{τ}. Then T^{∗}τ(a) = (τ(a)^{∗}T)^{∗} = (τ(a^{−1})T)^{∗} = (T ρ(a^{−1}))^{∗} = ρ(a)T^{∗} because both ρ and τ are unitary.

Lemma 1.5. Unitary equivalence is an equivalence relation on the set of all unitary representations of G.

Proof. The follows by the last two lemma noticing that id ∈ Hom_{G}(ρ, ρ) and that T^{∗} is an isomorphism if and only if T is an isomorphism with T^{∗−1} = (T^{−1})^{∗}.

Let us recall the spectral theorem for normal operators as this will be needed. Let H be a Hilbert space and T ∈ B(H). Then T is normal if T^{∗}T = T T^{∗}. T is self-adjoint it T = T^{∗} and positive if (T u, u) ≥ 0 for all u ∈ H. Let σ(T) be the set of complex numbers such that T − λid is not invertible. Then σ(T) is bounded. If T is self-adjoint then σ(T) ⊂ ℝ and if T is positive, then σ(T) ⊂ ℝ_{0}^{+} = {s ∈ ℝ | s ≥ 0}.

Definition 4. Let A be a σ-algebra on a set Ω, and let H be a Hilbert space. A resolution of the identity is a mapping

E : A → B(H)

such that the following holds for all A, B ∈ A:

1. E(∅) = 0 and E(Ω) = id

2. E(A) is a self-adjoint projection

3. E(A ∩ B) = E(A)E(B)

4. If A ∩ B = ∅ then E(A ∪ B) = E(A) + E(B)

5. For every u, v ∈ H the set-function

A ∋ A ↦ E_{u,v}(U) = (E(A)u, v) ∈ ℂ

is a complex measure on Ω.

Theorem 1.6 (The Spectral Theorem). Let T ∈ B(H) be normal. Then there is a unique resolution of the identity E on the Borel subset of σ(T) such that for all u, v ∈ H

(T u, v) = ∫ λdE_{u,v}

Furthermore, every projection E(A) commutes with every S ∈ B(H) that commutes with T.

If E is a resolution of the identity as above, then we write

T = ∫ λdE .

If T ∈ B(H) is normal and f : σ(T) → ℂ is bounded and measurable, then

β(u, v) := ∫_{σ(T)} f(λ)dE_{u,v}

is well defined, linear in u and conjugate-linear in v. Furthermore |β(u, v)| ≤ ||f||_{∞}||u|| ||v||. Hence β is bounded and it exists a bounded linear operator, denoted by f(T) such that

(f(T)u, v) = ∫_{σ(T)} f(λ) dE_{u,v} .

We write

f(T) = ∫_{σ(T)} f(λ) dE .

Theorem 1.7 (Symbolic Calculus for T). Let T ∈ B(H) be normal. The mapping f ↦ f(T) is a homomorphism of the algebra of all bounded Borel functions on σ(T) into B(H). Furthermore the following holds:

1. The constant function λ ↦ 1 is carried to id

2. the function λ ↦ λ maps to T

3. f̄(T) = f(T)^{∗}

4. ||f(T)|| ≤ ||f||_{∞} .If f is continuous then ||f(T)|| = ||f||_{∞}.

5. If f_{n} → f uniformly then ||f_{n}(T) − f(T)|| → 0 as n → ∞

6. If S ∈ B(H) commutes with T then Sf(T) = f(T)S

Our first application of the symbolic calculus is the following lemma.

Lemma 1.8. Let ρ be a unitary representation of G and T ∈ Hom_{G}(ρ, ρ) normal. If F : σ(T) → ℂ is continuous then F (T) ∈ Hom_{G}(ρ, ρ).

Proof. This follows immediately from (6).

Lemma 1.9. Let ρ and τ be unitary representations of G. Let T ∈ Hom_{G}(ρ, τ) be an isomorphism. Let T = U|T | be the polar decomposition of T, then |T | = \sqrt{T^{∗}T} ∈ Hom_{G}(ρ, ρ) and U ∈ Hom_{G}(ρ, τ).

Proof. We have T^{∗}T ∈ Hom_{G}(ρ, ρ) is self adjoint positive and injective. By Lemma 1.8 it follows that T = \sqrt{T^{∗}T} ∈ Hom_{G}(ρ, ρ). Define V := \sqrt{T^{∗}T} ∘ T^{−1} : V_{τ} → V_{ρ} which is possible as T is an isomorphism. We have V ∈ Hom_{G}(τ, ρ) and, using that \sqrt{T^{∗}T} is self-adjoint:,

V^{∗}V = T^{∗−1}\sqrt{T^{∗}T} ∘ \sqrt{T^{∗}T} ∘ T^{−1} = id_{V_{τ}}

and similarly V V^{∗} = id_{V_{ρ}}. Hence V is unitary. Let U = V^{∗} ∈ Hom_{G}(ρ, τ). Then T = U|T | and the claim follows.

Lemma 1.10. Let ρ and τ be unitary representations of G. Then ρ and τ are equivalent if and only if they are unitary equivalent.

Proof. If ρ and τ are unitary equivalent then they are equivalent. Let now T ∈ Hom_{G}(ρ, τ) be an isomorphism. Write T = U|T |. Then according to Lemma 1.9 we have U ∈ Hom_{G}(ρ, τ) an unitary isomorphism.

Let ρ be an unitary representation of G. Let [ρ] be the equivalence class of ρ, i.e., the set of all unitary representations of G that are equivalent to ρ. We let Ĝ be the set of all equivalence classes of irreducible unitary representations of G. We will quite often identify representations with their equivalence class just as in the case of L^{p}-spaces.

Theorem 1.11 (Schur’s Lemma). Let ρ be an irreducible unitary representation of G and T : V_{ρ} → V_{ρ} an intertwining operator. Then there exists a λ ∈ ℂ such that T = λid

Proof. Let T ∈ Hom_{G}(ρ, ρ). If T is non-zero then Ker(T) is invariant. Hence T is injective. Similarly T has dense image. Furthermore T^{∗} ∈ Hom_{G}(ρ, ρ) and hence

T_{1} := \frac{1}{2} (T + T^{∗}) , T_{2} = \frac{1}{i}(T − T^{∗}) ∈ Hom_{G}(ρ, ρ) .

As T = T_{1}+iT_{2} and both T_{1} and T_{2} are self adjoint we can assume that T is self adjoint. Let T = ∫ λdE(λ) be the spectral decomposition of T. Let λ_{0} be such that E(λ_{0}) ≠ 0. Then E(λ_{0}) is a non-zero projection commuting with ρ(G). Hence Im(E(λ_{0})) = V. It follows that T = λ_{0}id.

Corollary 1.12. Let Z(G) denote the center of G. If ρ is an irreducible unitary representation of G then there exists a continuous homomorphism χ_{π} : Z(G) → 𝕋 such that

π(z) = χ_{π}(z)id

for all z ∈ Z(G). The character χ_{π} is called the central character of ρ.

Corollary 1.13. Assume that G is abelian. Then every irreducible unitary representation is onedimensional.

Corollary 1.14. Let G = ℝ^{n} then every irreducible unitary representation of G is of the form x ↦ e^{ix·y} for some y ∈ ℝ^{n}. In particular Ĝ ≃ ℝ^{n}.

Proof. Assume first that n = 1. Schur’s Lemma implies that every irreducible representation of ℝ is one dimensional. Let χ : ℝ → ℂ^{∗} be a one dimensional representation. Assume that χ is differentiable. Then

χ′(t) = lim_{h→0} \frac{χ(t + h) − χ(t)}{h}
= ( lim_{h→0} \frac{χ(h) − χ(0)}{h}) χ(t)
= χ′(0)χ(t) .

Hence

χ(t) = e^{χ′(0)t}

as χ(0) = 1. As χ(0) = 1 there exists a t_{0} > 0 such that

∫_{0}^{t_{0}} χ(s) ds ≠ 0 .

Then

χ(t) ∫_{0}^{t_{0}} χ(s) ds = ∫_{0}^{t_{0}} χ(t + s) ds = ∫_{s}^{t_{0}+s} χ(u) du

But the last integral is differentiable as a function of s. Hence

χ(t) = (∫_{0}^{t_{0}} χ(s) ds)^{−1} ∫_{s}^{t_{0}+s} χ(u) du

is also differentiable. If χ is unitary, then |χ(s)| = 1 for all s ∈ ℝ. But this is only possible if χ′(0) ∈ iℝ. Let now n > 0. Then χ(t_{1}, …, t_{n}) = χ(t_{1}) · . . . · χ(t_{n}) and each t_{j} ↦ χ(t_{j}) is a character of ℝ and hence of the form χ(t_{j}) = e^{it_{j}y_{j}}. Let y = (y_{1}, …, y_{n}). Then the claim follows.

Corollary 1.15. Let G = 𝕋 then every irreducible representation of G is of the form z ↦ z^{n} for some n ∈ ℤ. In particular Ĝ ≃ ℤ.

Proof. Let χ be a character on 𝕋. Define χ̃ : ℝ → ℂ^{∗} by χ̂(t) = χ(e^{it}). Then χ̂ is a character on ℝ and hence of the form

χ̂(t) = e^{ixt}

for some x ∈ ℝ. But χ(2π) = 1 = e^{2πix} and hence x ∈ ℤ.

Lemma 1.16. Let ρ be a unitary representation of G. Let W ⊂ V_{ρ} be invariant. Then

W^{⊥} := {u ∈ V | ∀w ∈ W : (u, w) = 0}

is invariant and W ⊕ W^{⊥} = V.

Proof. Let u ∈ W^{⊥} and w ∈ W. Then
(ρ(a)u, w) = (u, ρ(a)^{∗}w) = (u, ρ(a^{−1})w) = 0 .

Hence ρ(a)u ∈ W^{⊥}.

Definition 5. Let ρ be a unitary representation of G. The functions ρ_{u,v}(x) := (u, ρ(x)v), u, v ∈ V_{ρ}, are called coefficient functions or matrix coefficients of ρ.

The coefficient functions ρ_{u,v} are continuous and bounded. Hence if μ ∈ M_{b}(G) it follows that ρ_{u,v} ∈ L^{1}(G). Let β(u, v) := ∫ (π(x)u, v) dμ(x) = ∫ \overline{ρ_{u,v}(x)} dμ. Then β is linear in the first factor and conjugate linear in the second factor. Furthermore

|β(u, v)| ≤ ∫ |(π(x)u, v)| d|μ|(x)
≤ ||u|| ||v|| ||μ|| .

Thus β is bounded. Hence there is a bounded linear map T : V → V, ||T || ≤ ||μ|| such that

β(u, v) = (T u, v) ∀u, v ∈ V .

Define π(μ) := T. We will write π(μ) = ∫ π(x) dμ(x) .

Definition 6. Let (A, || · ||,^{∗}) be an involutive Banach algebra. A start representation of A is a homomorphism of ∗-algebra ρ : A → B(H) such that ||ρ(a)|| ≤ ||a|| for all a ∈ A.

Theorem 1.17. Let π be a unitary representation of G. Define π : M_{b}(G) → B(V_{ρ}) by π(μ) = ∫ π(x) dμ(x). Then π is a ∗-representation of M_{b}(G).

Proof. We have to show that π(μ ⋆ ν) = π(μ)π(ν) and that π(μ^{∗}) = π(μ)^{∗}. First we have
(π(μ ⋆ ν)u, v) = ∫ (π(x)u, v)dμ ⋆ ν
= ∫∫ (π(xy)u, v) dμ(x)dν(y)
= ∫∫ (π(x)π(y)u, v) dμ(x)dν(y)
= ∫ (π(μ)π(y)u, v) dν(y)
= ∫ (π(y)u, π(μ)^{∗}v) dν(y)
= (π(ν)u, π(μ)^{∗}v)
= (π(μ)π(ν)u, v) .

As this holds for all u, v ∈ V_{ρ} it follows that π(μ ⋆ ν) = π(μ)π(ν). Similarly we get

(π(μ)u, v) = ∫ (π(x)u, v) dμ(x)
= ∫ (u, π(x^{−1})v) dμ(x)
= ∫ \overline{(π(x^{−1})v, u)} dμ(x)
= \overline{∫ (π(x)v, u) dμ^{∗}(x)}
= \overline{(π(μ^{∗})v, u)}
= (u, π(μ^{∗})v) .

Hence π(μ)^{∗} = π(μ^{∗}).

If μ = λ_{f} is a L^{1}(G)-function we write π(f) := π(λ_{f}). Then

(π(f)u, v) = ∫_{G} f(x)(π(x)u, v) dλ(x) .

We therefore get

Theorem 1.18. Let ρ be a unitary representation of G. Then the map ρ : L^{1}(G) → B(V_{ρ}) given by

(π(f)u, v) := ∫_{G} f(x)(π(x)u, v) dλ(x), u, v ∈ V_{ρ}

is a ∗-representation of the Banach algebra L^{1}(G)

Example 1.5. Let G = ℝ^{n} and take as a representation the left regular representation. Let f ∈ L^{1}(ℝ^{n}) and g, h ∈ L^{2}(ℝ^{n}). Assume first that g, h ∈ C_{c}(ℝ^{n}). Then

(L(f)g, h) = ∫ f(u)(L_{u}g, h) dλ_{n}(u)
= ∫ f(u) ∫ g(x − u)\overline{h(x)} dλ_{n}(x) dλ_{n}(u)
= ∫ (∫ f(u) g(x − u) dλ_{n}(u)) \overline{h(x)} dλ_{n}(x)
= (f ⋆ g, h) .

As the space of compactly supported functions is dense in L^{2}(G) and |(f ⋆ g, h) | ≤ ||f||_{1}||g||_{2}||h||_{2} it follows that e_{y}(f)g = f ∗ g for all f ∈ L^{1}(ℝ^{n}) and all g ∈ L^{2}(ℝ^{n}). If instead of the left regular representation we take the representation U we get

(U(f)g, h) = ∫ f(u)(U(u)g, h) dλ_{n}(u)
= ∫ f(u) ∫ e^{−iu·x}g(x)\overline{h(x)} dλ_{n}(x) dλ_{n}(u)
= ∫ (∫ f(u)e^{−iu·x} dλ_{n}(u)) g(x)\overline{h(x)} dλ_{n}(x)
= ∫ f̂(x)g(x)\overline{h(x)} dλ_{n}(x)
= (f̂g, h̄) .

Hence

U(f) = f̂.

Finally take e_{y}. Then e_{y}(f) is a complex number – as the inner product in ℂ is just multiplication of complex numbers – we get

e_{y}(f) = ∫ f(u) e_{y}(u) dλ_{n}(u)
= ∫ f(u) e^{−iy·u} dλ_{n}(u)
= f̂(y) .

Lemma 1.19. Let A ⊂ ℝ^{n} be measurable such that λ_{n}(A) > 0. Let

L^{2}(A) = {F ∈ L^{2}(ℝ^{n}) | F (x) = 0 for almost all x ∈ ℝ^{n} \ A}

and let

L_{A}^{2}(ℝ^{n}) := F^{−1}L^{2}(A) .

Then L_{A}^{2}(ℝ^{n}) is invariant under the left regular representation. The projection pr_{A} : L^{2}(ℝ^{n}) → L_{A}^{2}(ℝ^{n}) is given by

pr_{A}(f) = F^{−1}([Ff]χ_{A})

where χ_{A} is the indicator function of A. If λ_{n}(A) ≠ 0 and λ_{n}(ℝ^{n} \ A) ≠ 0, then L^{2}(ℝ^{n}) ≃_{ℝ^{n}} L_{A}^{2}(ℝ^{n}) ⊕ L_{ℝ^{n}\A}^{2}(ℝ^{n}).

Proof. We have f̂(ω) = 0 if and only if e^{ix·ω}f̂(ω) = 0. Hence { \atop L_{A}^{2}(ℝ^{n})} is invariant under the representation U. As F is an intertwining operator it follows that L_{A}^{2}(ℝ^{n}) is invariant. Let f ∈ L^{2}(ℝ^{n}). Then

f̂ = f̂χ_{A} + f̂χ_{ℝ^{n}\A} .

Hence L^{2}(ℝ^{n}) ≃ L_{A}^{2}(ℝ^{n}) ⊕ L_{ℝ^{n}\A}^{2}(ℝ^{n}) as a Hilbert space. But [U(x)F]χ_{B} = U(x)[χ_{B}] for every B ⊂ ℝ^{n} and F ∈ L^{2}(ℝ^{n}). Hence F ↦ F χ_{A} is an intertwining operator. Similarly F ↦ F χ_{ℝ^{n}\A} is an intertwining operator. It follows that L^{2}(ℝ^{n}) ≃ L_{A}^{2}(ℝ^{n}) ⊕ L_{ℝ^{n}\A}^{2}(ℝ^{n}) as a ℝ^{n}-representation. Let pr_{A}(f) = F^{−1}([Ff]χ_{A}). Then pr_{A} : L^{2}(ℝ^{n}) → L^{2}(ℝ^{n}) as the Fourier transform is an unitary isomorphism L^{2}(ℝ^{n}) ≃ L^{2}(ℝ^{n}). Furthermore as χ_{A}^{2} = χ_{A} it follows that p̂r_{A} : L^{2}(ℝ^{n}) → L^{2}(ℝ^{n}), F ↦ F χ_{A} is the orthogonal projection onto L^{2}(A). As the Fourier transform is an unitary isomorphism and

pr = F^{∗} ∘ p̂r ∘ F

It follows that pr is the orthogonal projection onto L_{A}^{2}(ℝ^{n}).

We cite the following theorem that we will not prove but use later on:

Theorem 1.20. The invariant subspaces in L^{2}(ℝ^{n}) are exactly the spaces L_{A}^{2}(ℝ^{n}) with A ⊂ ℝ^{n} measurable and λ_{n}(A) > 0.

Exercise

2. Cyclic representations

Let J be an index set. If J is finite and (ρ_{j}, H_{j}) is a representation of G in a topological vector space H_{j} then we can always define a representation ⊕ρ_{j} : a ↦ ⊕ρ_{j}(a) of G in ⊕H_{j}. In case J is infinite more care is needed. Assume that for each j ∈ J we have given a Hilbert space H_{j}. Let

⨁_{j∈J} H_{j} := {(u_{j})_{j∈J} | ∀j ∈ J : u_{j} ∈ H_{j} and ∑_{j∈J} |u_{j}|_{j}^{2} < ∞}

where index j indicates the norm in the Hilbert space H_{j}. Then ⨁ H_{j} is a Hilbert space with inner product ((u_{j}), (v_{j})) = ∑(u_{j}, v_{j})_{j} and corresponding norm

|| ⨁ u_{j}|| = \sqrt{∑_{j∈J} ||u_{j}||^{2}} .

Notice that if (u_{j})_{j} ∈ ⊕H_{j} then u_{j} ≠ 0 for at most countably many j ∈ J. If for each j ∈ J we have T_{j} ∈ B(H_{j}) define ⊕T_{j} : ⊕H_{j} → ⊕H_{j} by

(⨁_{j} T_{j}) ((u_{j})_{j∈J}) := (T_{j}u_{j})_{j∈J} .

We say that the family (T_{j})_{j} is uniformly bounded if there exists a constant M such that ||T_{j}|| ≤ M for all j.

Lemma 1.21. Assume that (T_{j}) is uniformly bounded. Then ⊕_{j}T_{j} ∈ B(⊕H_{j}). If ||T_{j}|| ≤ M for all j ∈ J, then || ⊕ T_{j}|| ≤ M. We have T^{∗} = ⊕T_{j}^{∗}, so in particular if T_{j} is unitary for each j then ⊕T_{j} is unitary. If (S_{j}) is another family of uniformly bounded operators S_{j} : H_{j} → H_{j} then (⊕S_{j})(⊕T_{j}) = ⊕S_{j}T_{j}.

Proof. Let T = ⊕T_{j} and let u = (u_{j})_{j} ∈ ⊕H_{j}. Then
||T u||^{2} = ∑_{j} ||T_{j}u_{j}||^{2}
≤ ∑_{j} ||T_{j}||^{2}||u_{j}||^{2}
≤ M^{2} ∑_{j} ||u_{j}||^{2}
= M^{2}||u||^{2} .

That (⊕S_{j})(⊕T_{j}) = ⊕S_{j}T_{j} follows directly from the definition. Assume now that each T_{j} is unitary. Then T = ⊕T_{j} is continuous as ||T_{j}|| = 1 for all j. Let u = (u_{j}), v = (v_{j}) ∈ ⊕H_{j}. Then

(T u, v) = ∑(T_{j}u_{j}, v_{j})
= ∑(u_{j}, T_{j}^{∗}v_{j})
= (u, T^{∗}v)

Hence T^{∗} = ⊕T_{j}^{∗}.

Let now (ρ_{j})_{j} be a unitary representation of G in H_{j}. Define ⊕_{j}ρ_{j} : G → U(⊕H_{j}) by (⊕ρ_{j})(a) := ⊕ρ_{j}(a). Then ⊕ρ_{j}(a) is unitary.

Lemma 1.22. ρ = ⨁_{j} ρ_{j} is an unitary representation of G.

Proof. By the last lemma it follows that ρ(a) is unitary and that ρ(ab) = ρ(a)ρ(b). To show that a ↦ ρ(a)u is continuous it is enough to show that at e. Let (u_{j})_{j} ∈ ⊕H_{j}. Then there exists at most countable infinite subset L ⊂ J such that u_{j} = 0 for j ∉ L. Let ϵ > 0 be given. Then there exists a finite subset K ⊂ L such that

∑_{j∉L} ||u_{j}||^{2} < ϵ^{2}/4 .

For each j ∈ K choose U_{j} ∈ N (e) such that

||ρ_{j}(a)u_{j} − u_{j}||^{2} < ϵ^{2}/2|K| .

Let U := ∩_{j∈K}U_{j} ∈ N (e). If a ∈ U then

||ρ(a)u − u||^{2} = ∑ ||ρ_{j}(a)u_{j} − u_{j}||^{2}
= ∑_{j∈K} ||ρ_{j}(a)u_{j} − u_{j}||^{2} + ∑_{j∉K} ||ρ_{j}(a)u_{j} − u_{j}||^{2}
< ϵ^{2}/2 + 2 ∑_{j∉K} ||u_{j}||^{2}

< \frac{ϵ^{2}}{2} + 2\frac{ϵ^{2}}{4} = ϵ^{2}

Thus a ↦ ρ(a)u is continuous.

The main problems in the representation theory are:

1. Classify all the irreducible unitary representations up to equivalence, i.e., determine Ĝ

2. Decompose a given unitary representation into irreducible representations. In many cases it is not enough to consider direct sums of representations. One need what is called a direct integral of representations where one replaces the index set J by a Borel space X, but we will not go into that here.

At this point we can only prove the following theorem. But first we need some preparations:

Definition 7. A representation π is called cyclic if there exits a vector u ≠ 0 in V_{π} such that the smallest closed subspace of V_{π} containing u is V_{π} it self.

Let [π(G)u] be the closed subspace generated by π(G)u. Then u is a cyclic vector if and only if [π(G)u] = V_{π}. Notice that if π is irreducible, then every non-zero vector is cyclic.

Theorem 1.23. Let ρ be a unitary representation of G. Then ρ is a direct sum of cyclic subrepresentions.

Proof. Let u ∈ V_{o} be non-zero. Let V (u) = [ρ(G)u]. Then ρ|V (u) is a cyclic sub-representation of ρ. If V (u) = V_{ρ} then we are done. If V (u) ≠ V_{ρ} let S be the collection of all sets U = {V_{i}}_{i∈I} where V_{i} is a cyclic sub-representation of V_{ρ} and V_{i} ⊥ V_{j} if i ≠ j. Then {V (u)} ∈ S so S ≠ ∅. For U, W ∈ S let U ≤ W if U ⊂ W. Let C be a chain in S. Then

V := ⋃_{U∈C} U ∈ S

and U ≤ V for all U ∈ C. It follows that S has a maximal element. Let U ∈ S be maximal. Let

W := ⨁_{U∈U} U .

Then each U is cyclic. We claim that W = V_{ρ}. If W ≠ V_{ρ}. Then W^{⊥} ≠ {0} is G-invariant. Let u ∈ W^{⊥}, u ≠ 0, and let V (u) = [ρ(G)u] as before. Then V (u) is cyclic and perpendicular to all U ∈ U. Hence U ∪ {V (u)} ∈ S and U ≨ U ∪ {V (u)} contradicting the maximality of U. Hence W = V_{ρ}.

As an example of cyclic decomposition we show that a L^{2}-function on ℝ^{n} is cyclic if and only if the Fourier transform does not vanish on a set of positive measure,

Lemma 1.24. A function f ∈ L^{2}(ℝ^{n}) is cyclic if and only if f̂(y) ≠ 0 for almost all y ∈ ℝ^{n}.

Proof. Assume that f̂(y) ≠ 0 for almost all y ∈ ℝ^{n}. Assume that g ⊥ [L(G)f]. Then 0 = (L_{z}f, g) for all x ∈ ℝ^{n}. This implies

0 = (L_{x}f, g)
= (L̂_{x}f, ĝ)
= ∫ f̂(ω)\overline{ĝ(ω)} e^{−ix·ω} dλ_{n}(ω)
= F(f̂\overline{ĝ})(x)

But the Fourier transform is injective on L^{2}(ℝ^{n}). Thus f̂(ω)\overline{ĝ(ω)} = 0 for almost all ω. As f̂(ω) ≠ 0 for almost all ω it follows that ĝ(ω) = 0 for almost all ω. But then g = 0.

Assume that there is a set A of positive measure such that f̂(ω) = 0 for all ω ∈ A. Let g ∈ L_{A}^{2}(ℝ^{n}) and let x ∈ ℝ^{n}. Then

(L_{x}f, g) = (L̂_{x}f, ĝ)
= ∫ e^{−iω·x}f̂(ω)\overline{ĝ(ω)} dλ_{n}(ω)
= 0

because f̂(ω)\overline{g(ω)} = 0 for all ω ∈ ℝ^{n}. Hence L_{A}^{2}(ℝ^{n}) is perpendicular to [L(ℝ^{n})f] so in particular f can not be cyclic.

Exercise

1. Let (H_{j})_{j∈J} be a family of Hilbert spaces with norm || · ||_{j} and inner product (·, ·)_{j}. Show that ⨁_{j∈J} H_{j} := {(u_{j})_{j∈J} | ∀j ∈ J : u_{j} ∈ H_{j} and ∑_{j∈J} |u_{j}|_{j}^{2} < ∞} is a Hilbert space.

2. A function φ : G → ℂ is called positive definite if for all x_{1}, …, x_{n} ∈ G the matrix (φ(x_{i}^{−1}x_{j}))

is positive semidefinite. Show the following:

(a) The function φ is positive definite if and only if ∑_{i,j=1}^{n} c_{i}\overline{c}_{j}φ(x_{i}^{−1}x_{j}) ≥ 0 for all x_{1}, …, x_{n} ∈ G and all c_{1}, …, c_{n} ∈ ℂ.

(b) Let E be the space of functions f : G → ℂ of the form f = ∑_{j=1}^{n} c_{j}L(x_{j})φ. Define a bilinear form (·, ·) on E by

(∑_{i=1}^{n} c_{i}L(x_{i})φ, ∑_{j=1}^{k} d_{j}L(y_{j})φ) := ∑_{i,j} c_{i}d̄_{j}φ(y_{j}^{−1}x_{i}) .

Show that (f, f) ≥ 0 for all f ∈ E.

(c) Let N = {f ∈ E | (f, f) = 0}. Then (·, ·) defines an inner product on E/N. Let || · || be the corresponding norm on E/N.

(d) For f ∈ E and x ∈ G let π(x)f(y) = f(x^{−1}y). Then π(G)N ⊂ N.

(e) Define π_{φ}(x) : E/N → E/N by π_{ϖ}(x)[f] := [π(x)f] where [f] denotes the equivalence class of f. Then ||π_{φ}(x)f|| = ||f||. In particular π_{φ} extends to a continuous map on V, the completion of E/N in the norm || · ||, to V. Show that π_{φ} is a continuous representation of G.

3. The left regular representation

From now on G stands for a locally compact Hausdorff topological group and X is a locally compact Hausdorff Lindeloff space. We assume that G acts separately continuous on X. Let μ_{X} be a positive quasi-invariant Radon measure on X. Define Δ_{X} : G → ℝ^{+} such that L_{a}μ_{X} = Δ_{X}(a)μ_{X}. Thus

∫_{X} f(a · x) dμ_{X}(x) = Δ_{X}(a) ∫ f(x) dμ_{X}(x)

for all f ∈ L^{1}(X). Then Δ_{X} is a continuous homomorphism. Define

L_{a}f(x) := Δ_{X}(a)^{1/2}f(a^{−1} · x) , f ∈ L^{2}(X) = L^{2}(X, μ_{X}) .

Lemma 1.25. Let X be a locally compact Hausdorff T_{3a} space. Let K ⊂ X be compact, K^{o} ≠ ∅, and x ∉ K. Then there exists a f ∈ C_{c}(X), such that f(k) = 1, k ∈ K, and f(x) = 0.

Theorem 1.26. L is a (continuous) unitary representation of G in L^{2}(X).

Proof. It is clear that L_{ab} = L_{a}L_{b} and a ↦ L_{a}f is continuous for each f ∈ L^{2}(X) (see problem ??). If f, g ∈ L^{2}(X) and a ∈ G then

(L_{a}f, g) = Δ_{X}(a)^{1/2} ∫ f(a^{−1} · x)\overline{g(x)} dμ_{X}
= ∫ f(a^{−1} · x)\overline{g(a · (a^{−1} · x))} dμ_{X}
= Δ_{X}(a)^{1/2}Δ_{X}(a)^{−1} ∫ f(x)\overline{g(a · x)} dμ_{X}
= ∫ f(x)\overline{Δ_{X}(a)^{−1/2}g(a · x)} dμ_{X}(x)
(f, L_{a^{−1}}g) .

Thus L_{a} is unitary. Let f ∈ C_{c}(X). Let ϵ > 0 be given. Let K = Supp(f). Choose V ∈ N (e) compact and symmetric. Then V · K = V^{−1} · K = L ⊂ X is compact. Let C = ∫ χ_{L}(x) dμ_{X}(x) = μ_{X}(L) ≥ 0. We can assume that C > 0 otherwise there is nothing to prove. As K is compact we can find a symmetric neighborhood U ∈ N (e), U ⊂ V such that for all a ∈ U and x ∈ K:

|Δ_{X}(a)^{1/2} − 1| < \frac{ϵ}{2(||f||_{∞} + 1)\sqrt{C}}

|f(a^{−1}x) − f(x)| ≤ \frac{ϵ}{2\sqrt{C}} .

Hence

|Δ_{X}(a)^{1/2}f(a^{−1}x) − f(x)| ≤ |Δ_{X}(a)^{1/2} − 1||f(a^{−1}x)| + |f(a^{−1}x) − f(x)|
< \frac{ϵ}{\sqrt{C}} .

Thus

∫ |L_{a}f(x) − f(x)|^{2} dμ_{X} = ∫ |L_{a}f(x) − f(x)|^{2}χ_{L}(x) dμ_{x}(x)
< \frac{ϵ^{2}}{C} ∫ χ_{L}(x) dμ_{X}(x)
= ϵ^{2} .

Thus

||L_{a}f − f||^{2} < ϵ ∀a ∈ U .

Let now g ∈ L^{2}(X) be arbitrary. Choose f ∈ C_{c}(X) such that ||g − f||_{2} < ϵ/3. Choose U ∈ N (e) such that ||L_{a}f − f||_{2} < ϵ/3 for all a ∈ U. Then

||L_{a}g − g||_{2} ≤ ||L_{a}g − L_{a}f||_{2} + ||L_{a}f − f||_{2} + ||f − g||_{2} < ϵ .

Hence a ↦ L_{a}g is continuous for all g ∈ L^{2}(X).

The representation L_{a} is called the left regular representations of G in L^{2}(X). If we take X = G then we speak of the left regular representation of G. One of the main problems of harmonic analysis on G is to understand this representation. In particular to decompose it into irreducible representations. Let now μ ∈ M_{b}(G) and f, g ∈ C_{c}(X). Then

(L(μ)f, g) = ∫_{G}(L_{a}f, g) dμ(a)
= ∫_{G} ∫_{X} f(a^{−1}x) \overline{g(x)} dμ_{X}(x)dμ(a)
= ∫_{X} (∫_{G} f(a^{−1}x) dμ(a)) \overline{g(x)} dμ_{X}(x)
= (μ ⋆ f, g) .

Hence L(μ)f = μ ⋆ f.

Let G = γ_{n} the group of permutations of n objects. Let X = {1, …, n}. We can view ℂ^{n} as the set of maps X → ℂ, z(n) = z_{n}. The counting measure μ({k}) = 1 is G invariant and the corresponding norm is

||z||^{2} = ∑_{j=1}^{n} |z(n)|^{2} = ∑_{j=1}^{n} |z_{n}|^{2} .

Thus L^{2}(X) = ℂ^{n}. Writing the left regular representation as an action on ℂ^{n} we have

σ · (x_{1}, …, x_{n})^{t} := (x_{σ^{−1}(1)}, …, x_{σ^{−1}(n)})^{t} .

This representation is not irreducible if n > 1. Let W = ℂ ∑_{j=1}^{n} e_{j}. Then W is G-invariant but W ≠ L^{2}(X).

4. The ax + b-group, wavelets and square integrable representations

In this subsection let G be the ax + b group. Recall that the multiplication and inverse in G is given by

(a, b)(α, β) = (aα, aβ + b)

and

(a, b)^{−1} = (1/a, −b/a) .

The ax + b-group acts on ℝ by (a, b) · t = at + b. Let (a, b) ∈ G and f ∈ C_{c}(ℝ). Then

∫ f((a, b) · t) dt = ∫ f(at + b) dt = a^{−1} ∫ f(u) du

where we have used u = at + b, and du = adt. According to Theorem 1.26 we can define a unitary representation of G on L^{2}(ℝ) by

ρ(x)f(t) = a^{−1/2}f(x^{−1} · t) = a^{−1/2}f (\frac{t − b}{a}) , x = (a, b) ∈ G .

Let H_{+}^{2} = L_{ℝ^{+}}^{2}(ℝ) be the classical Hardy space of functions with f̂(t) = 0 for t < 0 and let H_{−}^{2} be the space of L^{2}-functions such that f̂(t) = 0 for t > 0. Then L^{2}(ℝ) = H_{+}^{2} ⊕ H_{−}^{2}.

Theorem 1.27. Let the notation be as above. The spaces H_{+}^{2} and H_{−}^{2} are invariant and irreducible. Finally L^{2}(ℝ) ≃_{G} H_{+}^{2} ⊕ H_{−}^{2}.

Proof. As a > 0 we get:

{ \atop ρ(a, b)f}(ω) = a^{−1/2} ∫ f (\frac{t − b}{a}) e^{itω} dt
= a^{1/2} ∫ f(u) e^{i(au+b)ω} du (let u = \frac{t − b}{a}, du = \frac{dt}{a})
= e^{ibω}f̂(aω) .

Let V ⊂ L^{2}(ℝ) be an invariant subspace, V ≠ 0. As G contains ℝ as a subgroup and ρ|ℝ is just the regular representation of ℝ it follows that there exists a subset A ⊂ ℝ, not of measure zero with respect to the Lebesgue measure, such that V = L_{A}^{2}(ℝ). Let t ∈ A. t ≠ 0. Assume that t > 0. Then it follows by the above that ℝ^{+} = ℝ^{+}t ⊂ A. If we can find t < 0, t ∈ A, then it follows in the same way, that −ℝ^{+} ⊂ A. Hence we have only the three possibilities: A = ℝ^{+}, A = −ℝ^{+}, and A = ℝ (up to measure zero). The first case corresponds to H_{+}^{2}, the second case corresponds to H_{−}^{2} and the third case corresponds to L^{2}(ℝ).

The following discussion on wavelets can be extended to all of L^{2}(ℝ) but we will only concentrate on the Hardy space and leave the natural extension as an exercise. We normalize the Haar measure on G to be

dx = \frac{dadb}{\sqrt{2π}a^{2}}

Definition 8. Let ψ ∈ H_{+}^{2}. Then ψ is called a wavelet if ψ̂ ∈ L^{2}(ℝ, \frac{dt}{|t|}).

4. THE ax + b-GROUP, WAVELETS AND SQUARE INTEGRABLE REPRESENTATIONS 15

Let ψ ∈ L^{2}(ℝ) be a wavelet function and assume that C_{π} = ∫_{0}^{∞} \frac{|ψ̂(ω)|^{2}}{ω} \frac{dω}{\sqrt{2π}} > 0. Then

W_{ψ}f(a, b) := (f, ρ(a, b)ψ) = a^{−1/2} ∫_{−∞}^{∞} f(t)\overline{ψ (\frac{t − b}{a})} \frac{dt}{\sqrt{2π}}

is called the wavelet transform of f. Then W f := W_{ψ}f : G → ℂ is continuous. We will now show that W f is square integrable.

Lemma 1.28. Let the notation be as above. Then

W_{ψ}f(a, b) = a^{1/2} ∫ f̂(ω)\overline{ψ̂(aω)} e^{iωb} \frac{dω}{\sqrt{2π}} .

Proof. This follows from the general theory of the Fourier transform (Add exact reference).

Theorem 1.29. Let f ∈ L^{2}(ℝ) and let ψ ∈ L^{2}(ℝ). Then W_{ψ}f ∈ L^{2}(G) if and only if ψ is a wavelet functions. In that case

(W_{ψ}f, W_{ψ}g) = C_{ψ}(f, g) .

for all f, g ∈ L^{2}(ℝ).

Proof. Notice first that W_{ψ}f(a, b) ≥ 0 hence we can use Tonelli’s theorem to integrate first over b. We get with F (ω) = F_{a}(ω) = \overline{f̂(x)}ψ̂(aω):

∫_{−∞}^{∞} W_{ψ}f(a, b)\overline{W_{ψ}f(a, b)} \frac{db}{\sqrt{2π}} = a ∫_{−∞}^{∞} (∫ f̂(ω)\overline{ψ̂(aω)} e^{iωb} \frac{dω}{\sqrt{2π}} ) \overline{(∫ f̂(ω′)\overline{ψ̂(aω′)} e^{iω′b} \frac{dω}{\sqrt{2π}} )}\frac{db}{\sqrt{2π}}
= a ∫_{−∞}^{∞} \overline{F̂(b)}F̂(b) \frac{db}{\sqrt{2π}}
= a ∫_{−∞}^{∞} \overline{F (b)}F (b) \frac{db}{\sqrt{2π}}
= a ∫_{−∞}^{∞} f̂(b)\overline{f(b)} \overline{ψ̂(ab)}ψ̂(ab) \frac{db}{\sqrt{2π}}

Using Tonelli’s Theorem again we get

∫_{0}^{∞} ∫_{−∞}^{∞} W_{ψ}f(a, b)\overline{W_{ψ}f(a, b)} \frac{db}{\sqrt{2π}} \frac{da}{a^{2}} = ∫ ∫ f̂(b)\overline{f̂(b)} |ψ̂(ab)|^{2} \frac{da}{a} \frac{db}{\sqrt{2π}}
= C_{ψ} ∫ f̂(b)\overline{f̂(b)} \frac{db}{\sqrt{2π}}
= C_{ψ}(f, f)

Hence W_{ψ}f ∈ L^{2}(G) if and only if C_{ψ} < ∞ and in that case ||W_{ψ}f|| = \sqrt{C_{ψ}}||f||. The second claim follows by the same arguments by now using that W_{ψ}f\overline{W_{ψ}g} ∈ L^{1}(G) and hence Fubini’s Theorem is applicable.

As a special consequence of this result we get:

Lemma 1.30. Let ψ ∈ H_{+}^{2} be a wavelet. Then

H_{+}^{2} ∋ f ↦ W_{ψ}f ∈ L^{2}(G)

is continuous with norm C_{ψ} > 0. Furthermore W_{ψ} intertwines the representation ρ and the left regular representation on G, i.e. W_{ψ}ρ(a) = ρ(a)W_{ψ}.

Proof. The only thing that we have to show is the intertwining property. Let x ∈ G. Then

W_{ψ}(ρ(x)f)(y) = (ρ(x)f, ρ(y)ψ) = (f, ρ(x^{−1}y)ψ) = L_{x}W_{ψ}f(y) .

We use this as a motivation for the following definition:

Definition 9. Let G be a Hausdorff locally compact topological group and let ρ be an irreducible unitary representation of G. Then ρ is called square integrable if there exists a ψ ∈ V, ψ ≠ 0, such ρ_{u,ψ} ∈ L^{2}(G) for all u ∈ V_{ρ} and the map

V_{ρ} ∋ u ↦ ρ_{u,ψ} ∈ L^{2}(G)

is continuous. In this case ψ is called a wavelet vector.

Lemma 1.31. Assume that ρ is square integrable, and let ψ be a wavelet vector. Then the map W_{ψ} : V_{ρ} → L^{2}(G) is a continuous intertwining operator.

Proof. We only have to show that W_{ψ} is intertwining. But

W_{ψ}(ρ(x)u)(y) = (ρ(x)u, ρ(y)ψ) = (u, ρ(x^{−1}y)ψ) = L_{x}W_{ψ}u(x) .

Definition 10. Let X be a set and let H be a Hilbert space of functions f : X → V where V is a Hilbert space. The space H is called a reproducing Hilbert space if the evaluation maps ev_{x} : H → V, f ↦ f(x), are continuous for all x ∈ X.

Lemma 1.32. Let H be a reproducing Hilbert space. Define K(x, y) := ev_{x}ev_{y}^{∗} : V → V. Let K_{y}(x) := K(x, y). Then the following holds:

1. K(x, y) : V → V is continuous and linear, i.e. K(x, y) ∈ Hom(V, V)

2. (f, K_{y}v) = (f(x), v) for all f ∈ H and all v ∈ V.

3. K_{y}(·)v ∈ H for all v ∈ V and the set of finite linear combinations ∑ c_{j}K_{y_{j}}(·)v_{j} is dense in H 4. K(x, y)^{∗} = K(y, x)

Proof. As ev_{x} : V → V is continuous and linear for all x ∈ X it follows that ev_{x}^{∗} : V → H is continuous and linear. Hence K(x, y) : V → V is continuous and linear. We have K(·, y)v = ev_{y}^{∗}v ∈ H for all y ∈ V. Let u ∈ V and y ∈ X, then

(f(y), v) = (ev_{y}f, v)
= (f, ev_{y}^{∗}v)
= (f, K(·, y)v)

Hence (2) follows. Assume that f ∈ H is perpendicular to all the linear combinations ∑ c_{j}K_{y_{j}}(·)v_{j}. Then in particular

(f(y), v) = (f, K(·, y)v) = 0

for all y ∈ X and all v ∈ V. But then (by taking v = f(y)) it follows that f(y) = 0 for all y ∈ X. Hence f = 0. Now (3) follows. (4) follows directly from the definition as

K(x, y)^{∗} = (ev_{x}ev_{y}^{∗})^{∗} = ev_{y}ev_{x}^{∗} = K(y, x) .

Definition 11. The map K : X × X → Hom(V, V) is called the reproducing kernel of V.

Theorem 1.33. Let ρ and τ be square integrable. Let ψ ∈ V_{ρ}, and η ∈ V_{τ} be wavelet vectors Then the following holds:

1. If ρ and τ are not equivalent, then (W_{ψ}u, W_{η}v) = 0 for all u ∈ V_{ρ} and all v ∈ V_{τ}.

2. Let f ∈ L^{2}(G). Then ρ(f) is defined and W_{ψ}^{∗}f = ρ(f)ψ.

3. If ρ = τ. Then there exists a constant C_{ψ,η} such that

(W_{ψ}u, W_{η}v) = C_{ψ,η}(u, v)

for all u, v ∈ V_{ρ}.

4. Im(W_{ψ}) is a closed reproducing Hilbert subspace of L^{2}(G) contained in L^{2}(G) ∩ C(G).

5. Let c_{ψ} = 1/\sqrt{C_{ψ,ψ}} and U_{ψ} : V_{ρ} → L^{2}(G), u ↦ c_{ψ}W_{ψ}u = W_{c_{ψ}ψ}u. Then U_{ρ} is an unitary isomorphism V_{ρ} ≃ Im(W_{ψ}).

6. Assume that ρ = τ and that C_{ψ,η} ≠ 0. Then

u = \frac{1}{C_{ψ,η}} ρ(W_{ψ}u)η = \frac{1}{C_{ψ,η}} ∫ (u, ρ(a)ψ)ρ(a)η da,

(weakly) for all u ∈ V_{ρ}. In particular

u = \frac{1}{C_{ψ,ψ}} ρ(W_{ψ}u)ψ = \frac{1}{C_{ψ,ψ}} ∫_{G}(u, ρ(a)ψ)ρ(a)ψ da .

7. The reproducing kernel of kernel Im(W_{ψ}) is given by K(x, y) = W_{ψ}ψ(y^{−1}x)
Proof. Define β : V_{ρ} × V_{τ} → ℂ by

β(u, v) := (W_{ψ}u, W_{η}v) = ∫ W_{ψ}u(x) \overline{W_{η}v(x)} dx .

As the wavelet transform is continuous it follows that

|β(u, v)| = |(W_{ψ}u, W_{η}v)| ≤ ||W_{ψ}u|| ||W_{η}v|| ≤ ||W_{ψ}||||W_{η}|| ||u|| ||v|| .

There is therefore a continuous linear map T : V_{ρ} → V_{τ} such that

β(u, v) = (T u, v) .

Let a ∈ G then, using that the wavelet transform is intertwining,

(T ρ(a)u, v) = β(ρ(a)u, v)
= ∫ [W_{ψ}ρ(a)u](x)\overline{W_{η}v(x)} dx
= ∫ W_{ψ}u(a^{−1}x)\overline{W_{η}v(x)} dx
= ∫ W_{ψ}u(x)\overline{W_{η}v(ax)} dx
= ∫ W_{ψ}u(x)\overline{W_{η}τ(a^{−1})v(x)} dx
= β(u, τ(a^{−1})v)
= (T u, τ(a)^{∗}v)
= (τ(a)T u, v) .

As this holds for all u and v it follows that T ρ(a) = τ(a)T. Thus T is an intertwining operator. As ρ is assumed irreducible it follows that either T = 0 or T is injective. Furthermore T^{∗} : V_{τ} → V_{ρ} is intertwining and by Schur’s Lemma T^{∗}T = λid and T T^{∗} = μid for some λ, μ ∈ ℂ. If T ≠ 0 then λ and μ are both non-zero. Let v ∈ V_{τ}. Define u ∈ V_{ρ} by u = μ^{−1}T^{∗}v then T u = v and hence T is surjective. Thus T ≠ 0 implies that ρ and τ are equivalent. Now (1) follows. Let f ∈ Im(W_{ψ}) and u ∈ V_{ρ}. Then

(W_{ψ}^{∗}f, u) = (f, W_{ψ}u)
= ∫ f(a) \overline{(u, ρ(a)ψ)} da
= ∫ f(a)(ρ(a)ψ, u) da
= (ρ(f)ψ, u)

Hence the claim.

(3) If ρ = τ it follows by Schur’s Lemma that T = C_{ψ,η} ∈ ℂ and hence (2) follows. Notice that C_{ψ.,ψ} > 0. Define c_{ψ} := 1/\sqrt{C_{ψ,ψ}} > 0. Then

||W_{ψ}u − W_{ψ}v|| = c_{ψ}||u − v|| . (4.1)

Let (f_{n}) be a Cauchy sequence in Im(W_{ψ}). Let f_{n} ∈ V_{ρ} be such that u_{n} = W_{ψ}f_{n}. Then (f_{n}) it follows by (4.1) that (f_{n}) is a Cauchy sequence in V_{ρ}. Let f = lim f_{n}. Then, again by (4.1) , it follows that lim u_{n} = W_{ψ}f ∈ Im(W_{ψ}). Hence Im(W_{ψ}) is closed in L^{2}(G) and Hence a Hilbert subspace.

(4) Let f = W_{ψ}u ∈ Im(W_{ψ}). Then f(a) = (u, ρ(a)ψ) is continuous. In particular the point evaluations f ↦ f(a) are well defined on Im(W_{ψ}) and Im(W_{ψ}) ⊂ C(G). We have

|f(a)| = |(u, ρ(a)ψ)| ≤ ||u||||ψ|| = (c_{ψ}^{−1}||ψ||)||f||

by (4.1). Hence the point evaluations are continuous linear maps.

(5) By definition we have

(U_{ψ}u.U_{ψ}v) = (u, v)

for all u, v ∈ V_{ρ}. Thus U_{ψ} is unitary. Hence (6) follows from (2).

(7) Let f = U_{ψ}u ∈ Im(W_{ψ}). Then

f(y) = U_{ψ}u(y)
= c_{ψ}^{−1}(u, ρ(y)ψ)
= c_{ψ}^{−1}(U_{ψ}^{∗}f, ρ(y)ψ)
= c_{ψ}^{−1}(f, U_{ψ}(ρ(y)ψ))
= (f, W_{ψ}(ρ(y)ψ))
= ∫ f(x)\overline{(ρ(y)ψ, ρ(x)ψ)} dx
= ∫ f(x)\overline{(ψ, ρ(y^{−1}x)ψ)} dx
= ∫ f(x)\overline{W_{ψ}ψ(y^{−1}x)} dx .

5. Generalizations of the Wavelet transform

Let H be a locally compact Hausdorff topological group and let π : H → GL(n, ℝ) be a continuous homomorphism. Then we can define the semi-direct product G = H ×_{π} ℝ^{n} where the product is given by

(a, x)(b, y) = (ab, π(a)y + x) .

Lemma 1.34. Define Δ_{π} : H → ℝ^{+} by Δ_{π}(h) := | det(π(h))|^{−1}. Then Δ_{π} is a continuous homomorphism. Let d_{H}h denote a left Haar measure on H and d_{n}y the normalized Lebesgue measure f ↦ ∫ f(y) \frac{dy}{(2π)^{n/2}} on ℝ^{n}. Then

C_{c}(G) ∋ f ↦ μ(f) := ∫_{H} ∫_{ℝ^{n}} f(h, y)Δ_{π}(h) d_{n}yd_{H}h

defines a left invariant positive measure on G.

Proof. It is clear that Δ_{ρ} is a continuous homomorphism. Let f ∈ C_{c}(G), b ∈ H and t ∈ ℝ^{n}. Then

(b, t) · μ(f) = ∫_{H} ∫_{ℝ^{n}} f((b, t)(h, y))Δ_{π}(h) d_{n}yd_{H}h
= ∫_{H} ∫_{ℝ^{n}} f(bh, π(b)y + t) Δ_{π}(bh)Δ_{π}(b)^{−1} d_{n}yd_{H}h
= ∫_{H} ∫_{ℝ^{n}} f(bh, u)Δ_{π}(bh) d_{n}yd_{H}h

where we have used the change of variables u = π(b)y + t with d_{n}u = |det π(b)| d_{n}y = Δ_{π}(b)^{−1} d_{n}y. The claim follows now by using Fubini’s Theorem and the fact that d_{H}h is left invariant.

Lemma 1.35. Define π^{t} : H → GL(n, ℝ) by π^{t}(h) = π(h^{−1})^{t}. Then π^{t} : H → GL(n, ℝ) is a continuous homomorphism.

Proof. Let h, k ∈ H. Then π^{t}(hk) = π(k^{−1}h^{−1})^{t} = (π(k^{−1})π(h^{−1}))^{t} = π^{t}(h)π^{t}(k). As transposition is a continuous map on GL(n, ℝ) it follows that π^{t} : H → GL(n, ℝ) is continuous.

Define now representations ρ and ρ̂ of G in L^{2}(ℝ^{n}) in a following way

ρ((h, y))f(x) = Δ_{π}(h)^{1/2}f(π(h^{−1})(x − y))
ρ̂((h, y))F (ω) = Δ_{π}^{t}(h)^{1/2}e^{−i(y,ω)}F (π^{t}(h)^{−1}ω)
= Δ_{π}(h)^{−1/2}e^{−i(y,ω)}F (π(h)^{t}ω) .
Lemma 1.36.  1.  Lemma 1.37. ρ_{π} and ρ̂_{π} are unitary representations of H and

{ \atop (ρ_{π}((h, y))f)}(ω) = ρ̂_{π}(h, y)f̂(ω) . (5.1)

Thus F ∈ Hom_{H}(ρ_{π}, ρ_{π}_{t}) is a unitary isomorphism.

Proof. That ρ_{π} is a unitary representation follows from Lemma 1.26. As the Fourier transform is a unitary isomorphism the statement about ρ̂_{π} follows if we can prove (5.1). For the Fourier transform we get:

{ \atop (ρ_{π}((h, y))f)}(ω) = ∫ [ρ_{π}((h, y))f](x)e^{−i(x,ω)} d_{n}x
= Δ_{π}(h)^{1/2} ∫ f(π(h)^{−1}(x − y)) e^{−i(x,ω)} d_{n}x
= Δ_{π}(h)^{−1/2} ∫ f(u)e^{−i(π(h)u+y,ω)} d_{n}u (u = π(h)^{−1}(x − y))
= Δ_{π}(h)^{−1/2}e^{−i(y,ω)} ∫ f(u)e^{−i(u,π(h)^{t}ω)} d_{n}u
= Δ_{π}^{t}(h)e^{−i(y,ω)}f̂(π^{t}(h)^{−1}ω)
= ρ̂_{π}((h, y))f̂(ω) .

Definition 12. Let G be a topological group and X a topological space such that G acts on X. Let μ be a quasi-invariant measure on X. We say that the action is ergodic if for all G-invariant sets A ⊂ X we have μ(X \ A) = 0 or μ(A) = 0.

Example 1.6. Let G = ℝ and X = 𝕋^{2} = {(z, w) ∈ ℂ^{2} | |z| = |w| = 1}. Let a, b ∈ ℝ be such that a/b is irrational. Define an action of G on X by

t · (z, w) = (e^{iat}z, e^{ib}w) .

Then it can be shown that ℝ · x is dense in X for all x. In particular the measure of the complement of any orbit is zero.

Theorem 1.38. Let the notation be as before. Then the following holds:

1. If A ⊂ ℝ^{n} is measurable, invariant under π^{t}(H), and with non-zero measure, then L_{A}^{2}(ℝ^{n}) is G-invariant and non-zero.

2. If V ⊂ L^{2}(ℝ^{n}) is non-zero and invariant, then there exists a A as in (2) such that V = L_{A}^{2}(ℝ^{n}). V is irreducible if and only if the action of π^{t}(H) on A is ergodic.

3. Assume that A is homogeneous under π^{t}(H) and of positive measure. Let ψ ∈ L^{2}(ℝ^{n}). Then ψ is a wavelet vector if and only if H ∋ h ↦ ψ(π(a)^{t}ω) ∈ ℂ is in L^{2}(H) for all ω ∈ A. In particular if A is open, H^{ω} is compact for one (and hence all) ω ∈ ℝ^{n}, and ψ̂ ∈ C_{c}(A), then ψ is a wavelet vector. In particular the set of wavelet vectors in dense in L_{A}^{2}(ℝ^{n}).

Proof. (1) By abuse of notation we let L^{2}(A) = {f ∈ L^{2}(ℝ^{n}) | f|ℝ^{n} \ A = 0 a.e.}. Then L^{2}(A) is invariant under ρ̂_{π}. But then L_{A}^{2}(ℝ^{n}) = F^{∗}(L^{2}(A)) is invariant under ρ_{π} as F is an intertwining operator.

(2) Let V ⊂ L^{2}(ℝ^{n}) be invariant. Then as ℝ^{n} ⊂ G there exists a measurable set B ⊂ ℝ^{n} such that L_{B}^{2}(ℝ^{n}) = V. Let

A = ⋃_{h∈H} π^{t}(h)B .

If A \ B has positive measure that V can not be invariant. Hence A is measurable and V = L_{A}^{2}(ℝ^{n}). Assume that L_{A}^{2}(ℝ^{n}) is irreducible. If there exists a invariant subset B such that λ(A \ B) > 0 and λ(B) > 0, then V = L_{A\B}^{2}(ℝ^{n}) ⊕ L_{B}^{2}(ℝ^{n}) and both L_{A\B}^{2}(ℝ^{n}) and L_{B}^{2}(ℝ^{n}) non-zero and invariant, contradicting the irreducibility of V. Hence A is ergodic. Assume that A is ergodic. If V is not irreducible, then we can decompose V = W_{1} ⊕ W_{2} with W_{i} closed, non-zero, and invariant. As before it follows that W_{i} = L_{B_{i}}^{2}(ℝ^{n}). Let A_{i} = B_{i} ∩ A. Then A_{i} is invariant and (as W_{i} ⊂ V) W_{i} = L_{A_{i}}^{2}(ℝ^{n}). As W_{i} ≠ {0} it follows that λ(A_{i}) > 0. Hence A is not ergodic. The proof of (3) is exactly the same as the proof of Theorem 1.29: Let W_{ψ}f(h, x) = ∫ f(y)Δ_{π}(h)ψ(h^{−1}(y− x)) d_{n}y. Let C_{ψ} = ∫ |ψ̂(π(a)^{t}ω)|^{2} da. Then C_{π} is independent of ω: Let ω′ ∈ A. Then there exists a b ∈ H such that ω′ = π(b)^{t}ω. Hence

∫ |ψ̂(π(a)^{t}ω′)|^{2} da = ∫ |ψ̂(π(ba)^{t}ω)|^{2} da = ∫ |ψ̂(π(a)^{t}ω)|^{2} da = C_{ψ}

Let F (ω) = F_{h}(ω) = \overline{f̂(ω)}ψ̂(π(h)^{t}ω). As W_{ψ}f(a, b)\overline{W_{ψ}f(a, b)} ≥ 0 we can use Tonelli’s Theorem. Hence

∫ W_{ψ}f(h, x)\overline{W_{ψ}f(h, x)} d_{n}x = Δ_{π}(h)^{−1} ∫ (∫ f̂(ω)\overline{ψ̂(π(h)^{t}ω)} e^{ix·ω} d_{n}ω) \overline{(∫ f̂(ω′)\overline{ψ̂(π(h)^{t}ω′)} e^{ix·ω′} d_{n}ω′)}Δ_{π}(h)d_{n}x
= ∫ \overline{F̂(x)}F̂(x) d_{n}x
= ∫ \overline{F (ω)}F (ω) d_{n}ω
= ∫ f̂(ω)\overline{f̂(ω)} \overline{ψ̂(π(h)^{t}ω)}ψ̂(π(h)^{t}ω) d_{n}ω

Using Fubini’s Theorem again we get

∫_{H} ∫ W_{ψ}f(h, x)\overline{W_{ψ}f(h, x)} Δ_{π}(h) d_{n}x dh = ∫ ∫ f̂(ω)\overline{f̂(ω)} |ψ̂(π(h)^{t}ω)|^{2} dhd_{n}ω
= C_{ψ} ∫ f̂(ω)\overline{f̂(ω)} d_{n}ω
= C_{ψ}(f, f)

Hence ||W_{ψ}f|| < ∞ if and only if C_{ψ} < ∞ and in that case

||W_{ψ}f|| = \sqrt{C_{ψ}}||f|| .

Let now g ∈ L^{2}(ℝ^{n}) be arbitrary. By the first part we know that both W_{ψ}f and W_{ψ}g are in L^{2}(G). Hence W_{ψ}f\overline{W_{ψ}g} ∈ L^{1}(G) and we can use Fubini’s theorem to integrate first over ℝ^{n}. Then same calculation as above now shows that

(W_{ψ}f, W_{ψ}g) = C_{ψ}(f, g) .

Example 1.7. Let n ≥ 2 and let H_{n+1} = {g ∈ GL(n + 1, ℝ) | g = λa, λ ∈ ℝ^{+}, a ∈ SO_{o}(1, n)}. We can then describe H as

H_{n+1} = {g ∈ GL(n + 1, ℝ) | ∃λ > 0 : g^{t}I_{1,n}g = λI_{1,n}, det g > 0, g_{11} > 0} .

Write elements in ℝ^{1+n} as (t, x) with x ∈ ℝ^{n} and t ∈ ℝ. Let β((t, x), (s, y)) = st−x·y. Then SO(1, n) is the subgroup of GL(n+1, ℝ) that leaves β invariant and SO_{o}(1, n) is the connected component containing the identity element. Let

C = {(t, x) | t > 0, t^{2} − ||x||^{2} > 0} = {u ∈ ℝ^{n+1} | β(u, u) > 0, u_{1} > 0}

the solid forward light cone. Then H_{n+1} · C = C. If we replace SO(1, n) by SO_{o}(1, n) then we would also have to consider the backward light cone and use C ∪ −C.

Definition 13. Let V be a real vector space. A subset C ⊂ V is convex if λu + (1 − λ)v ∈ C for all u, v ∈ C and λ ∈ [0, 1]. C is a convex cone if C is convex and ℝ^{+}C ⊂ C.

Lemma 1.39. Let C ⊂ ℝ^{n} be a open convex cone invariant under a subgroup H ⊂ GL(n, ℝ). Let C^{∗} = \{u ∈ ℝ^{n} | ∀v ∈ \overline{C} \ \{0\} : (u, v) > 0\} be the dual cone. Then C^{∗} is a invariant convex open cone in ℝ^{n} invariant under the group H^{t} = {g^{t} ∈ H | g ∈ H}.

Proof. Embed SO(n) into SO(1, n) (and hence also H_{n+1}) by

SO(n) ∋ A ↦ ( \begin{matrix}1 & 0 \\ 0 & A \end{matrix} ) ∈ SO(1, n) .

Denote the image also by SO(n).

Lemma 1.40. Let the notation be as above. Then the following holds:

1. Let u, v ∈ C then β(u, v) > 0.

2. C is a open convex cone in ℝ^{1+n}

3. H_{n+1} acts transitively on C and H_{n+1}^{e_{1}} = SO(n) is compact.

4. C^{∗} = C.

Proof. (1) Let u = (t, x), v = (s, y) ∈ C. As t^{2} − ||x||^{2} > 0 and t > 0 it follows that t > ||x||. Similarly s > ||y||. From this it follows that

ts − (x, y) > ||x|| ||y|| − (x, y) ≥ 0 (5.2)

by the Cauchy-Schwart’s inequality.

(2) Let u = (t, x), v = (s, y) ∈ C, and λ ∈ [0, 1]. Then

(λt + (1 − λ)s)^{2} − ||λx + (1 − λ)y||^{2} = λ^{2}t^{2} + (1 − λ)^{2}s^{2} + 2λ(1 − λ)ts
−λ^{2}||x||^{2} − (1 − λ)^{2}||y||^{2} − 2λ(1 − λ)(x, y)
= λ^{2}(t^{2} − ||x||^{2}) + (1 − λ)^{2}(s^{2} − ||y||^{2})
+2λ(1 − λ)(ts − (x, y))
> 0

by (5.2), 0 ≤ λ ≤ 1, and the fact that (t, x), (s, y) ∈ C. Hence λu + (1 − λ)v ∈ C. Hence C is convex. As β(λu, λu) = λ^{2}β(u, v) it follows that ℝ^{+}C ⊂ C. Thus C is a convex cone. C is open by definition and the continuouty of β.

(3) It is clear that SO(n) ⊂ SO(1, n)^{e_{1}}. Assume that g ∈ SO(1, n) fixes e_{1}. Let g = (g_{1}, g_{2}, …, g_{n+1}) where g_{j} are the column vectors. Then

ge_{1} = g_{1} = 1 .

Hence

g = ( \begin{matrix}1 & ∗ \\ 0 & A \end{matrix} )

with A ∈ SL(n, ℝ). But

gI_{1,n}g^{t} = I_{1,n}

and hence

g^{t} = I_{1,n}g^{−1}I_{1.n} .

Hence it follows that also g_{1,j} = 0 for j ≥ 2. Hence

g = ( \begin{matrix}1 & 0 \\ 0 & A \end{matrix} ) .

But then ||Ay|| = ||y|| for all y ∈ ℝ^{n}. As det A = 1 it follows that A ∈ SO(n). We now show that SO(1, n) acts transitively on C. For s ∈ ℝ let

a_{s} = ( \begin{matrix}cosh(s) & 0 & sinh(s) \\ 0 & I_{n−1} & 0 \\ sinh(s) & 0 & cosh(s) \end{matrix} )

Then a_{s} ∈ SO(1, n). and

a_{s}e_{1} = cosh(s)e_{1} + sinh(s)e_{n+1}

Let u = (t, x) ∈ C. Then – after multiplying by β(u, u)^{−1/2} > 0 if necessary – we can assume that β(u, u) = 1. Then t^{2} − ||x||^{2} = 1. Hence there exists a s > 0 such that cosh(s) = t and sinh(s) = ||x||. Then there exists a k ∈ SO(n) such that k(sinh(t)e_{n+1}) = x as SO(n) acts transitively on any sphere in ℝ^{n} centered around zero. It follows that

ka_{t}e_{1} · e_{1} = u.

(4) Let u = (t, x) ∈ C and v = (s, y) ∈ C. Then v′ = (s, −y) ∈ C and by (1)

ts ± (x, y) > 0 .

It follows that (u, v) > 0 and hence u ∈ C^{∗}. On the other hand let u = (t, x) ∈ C^{∗}. Then u ≠ 0. Notice that e_{1} = (1, 0, …, 0) ∈ C. Hence

(e_{1}, u) = t > 0.

If x = 0 then u = (t, 0) ∈ C and we are done. If x ≠ 0 then (||x||, x) ∈ \overline{C} \ \{0\} and hence

t||x|| − ||x||^{2} = ||x||(t − ||x||) > 0 .

It follows that t > ||x||. Thus (t, x) ∈ C..

Corollary 1.41. Let A = {a_{s} | s ∈ ℝ}. Then SO(1, n) = SO(n)ASO(n).

Corollary 1.42. The spaces L_{C}^{2}(ℝ^{n+1}) and L_{−C}^{2}(ℝ^{n+1}) possesses wavelet vectors.

Let us now look at the full space L^{2}(ℝ^{n+1}). Let

N = {v ∈ ℝ^{n+1} | β(v, v) = 0}

and

X = {v ∈ ℝ^{n+1} | β(v, v) < 0}

Then both N and X are invariant under H_{n+1}. Furthermore N has measure zero. Obviously

ℝ^{n+1} = C ˙∪ − C ˙∪X ˙∪N .

The set N has measure zero and therefore does not contribute to the decomposition of L^{2}(ℝ^{n+1}). We embed H_{n} into H_{n+1} by

A ↦ ( \begin{matrix}A & 0 \\ 0 & 1 \end{matrix} ) .

Then obviously Ae_{n+1} = e_{n+1}.

Lemma 1.43. The set X is homogeneous under H_{n+1} and the stabilizer of e_{n+1} ∈ X is H_{n}, i.e. X = H_{n+1}/H_{n}.

Proof. Notice first that

a_{s}e_{n+1} = sinh(s)e_{1} + cosh(s)e_{n+1} .

Let u = (t, x) ∈ {v ∈ ℝ^{n+1} | β(v, v) < 0}. Then after multiplying by (−β(u, u))^{−1/2} if necessary we can assume that ||x||^{2} − t^{2} = −1. Choose s ∈ ℝ such that cosh(s) = ||x|| and sinh(s) = t. Then we can find k ∈ SO(n) such that k(cosh(s)e_{n+1}) = x. Hence

ka_{s}e_{n+1} = u .

If ge_{n+1} = e_{n+1} then as above it follows first of all that g ∈ SO_{o}(1, n)

g = ( \begin{matrix}A & 0 \\ 0 & 1 \end{matrix} ) .

As

gI_{1,n}g^{t} = ( \begin{matrix}AI_{1,n−1}A^{t} & 0 \\ 0 & 1 \end{matrix} ) = I_{1,n}

it follows that A ∈ O(1, n − 1) but as g ∈ SO_{o}(1, n − 1) it follows that det A > 0 and A_{11} > 0. Hence A ∈ SO_{o}(1, n − 1).

Corollary 1.44. Let A = {a_{s} | s ∈ ℝ}. Then SO(1, n) = SO(n)ASO(1, n − 1).

Theorem 1.45. As a representation of ℝ^{+}SO(1, n) we have

L^{2}(ℝ^{n+1}) ≃_{H_{n+1}} L_{C}^{2}(ℝ^{n+1}) ⊕ L_{−C}^{2}(ℝ^{n+1}) ⊕ L_{X}^{2}(ℝ^{n+1}).

Each of the spaces on the right is irreducible. The spaces L_{C}^{2}(ℝ^{n+1}) and L_{−C}^{2}(ℝ^{n+1}) have wavelet vectors but not so the space L_{X}^{2}(ℝ^{n+1}). If we replace H_{n+1} by the group H̃_{n+1} = ℝ^{∗}O(1, n) then L^{2}(ℝ^{n+1}) decomposes into two irreducible parts

L^{2}(ℝ^{n+1}) ≃ L_{C∪−C}^{2}(ℝ^{n+1}) ⊕ L_{X}^{2}(ℝ^{n+1}) .

Proof. Everything except the last statement has been proved now. That part follows by noticing that −I_{n+1}C = −C.

Exercise

In those exercises Ω will stand for a open convex cone in ℝ^{n}.

1. Let G = γ_{3} and X = {1, 2, 3}. Show that L^{2}(X) decomposes into two irreducible representation, one of them, the trivial representation being one dimensional, and the other being two dimensional

2. Let Δ be the modular function of G. Define for a ∈ G and f ∈ L^{2}(G), R(a)f(x) := Δ(a)^{1/2}f(xa). Show that R is a unitary representation of G. It is called the right regular representation of G. Furthermore show that the map L^{2}(G) → L^{2}(G), f ↦ Δ^{−1/2}f^{∨}, is an unitary isomorphism intertwining the left regular representation L and the right regular representation R.

3. Show that Ω ⊂ ℝ^{n} is a convex cone if and only if λu + μv ∈ Ω for all λ, μ ∈ ℝ^{+} and u, v ∈ Ω.

4. Let Ω be a open convex cone invariant under the closed subgroup H ⊂ GL(n, ℝ). Show the

following:

(a) Ω^{∗} ≠ {0} if and only if Ω does not contain any straight lines and in that case Ω^{∗∗} = Ω.

(b) Let u ∈ Ω, then H^{u} is compact

5. Let V be the space Sym_{n} of symmetric n × n-matrices. Let Sym_{p,q}, p + q, be the subset of symmetric matrices with signature (p, q). Recall that GL(n, ℝ) acts on Sym_{n} by

g · A = gAg^{t}

and each Sym_{p,q} is an open orbit. Show the following:

(a) C = Sym_{n,0} and −C = Sym_{0,n} are symmetric cones and GL(n, ℝ) acts transitively on Sym_{n,0}and Sym_{0,p}.

(b) The stabilizer of every point in C and −C is compact and hence each of the spaces L_{C}^{2}(V) and L_{−C}^{2}(V) possesses a wavelet vector.

(c) The set

V \ ⋃_{0≤p,q≤n, p+q=n} Sym_{p,q} = {A ∈ V | det A = 0} .

In particular ⋃_{0≤p,q≤n, p+q=n} Sym_{p,q} is open and dense in V and

L^{2}(V) = ⨁_{0≤p,q≤n, p+q=n} L_{Sym_{p,q}}^{2}(V)

is the decomposition of L^{2}(V) into irreducible subspaces. Only the spaces L_{C}^{2}(V) and L_{V}^{2} (V) possess wavelet vectors.

6. Let V be as in the last problem. Define a bilinear form (·, ·) on V by

(X, Y) = T r(XY) .

Show the following:

(a) (·, ·) is an inner product on V.

(b) C^{∗} = C, so C is self dual.

7. Let V be the space of self adjoint complex n × n matrixes considered as a real vector spaces. Let

H = {A ∈ GL(n, ℂ) | A = λB, λ > 0, B ∈ SL(n, ℂ), B^{∗} = B} .

Define an action of H on V by

g · A = gAg^{∗} .

(a) Find the open orbits in V and show that the union is dense.

(b) Let C be the set of positive definite elements in V. Thus (Au, u) > 0 for all u ≠ 0. (Question: Do we use positive definite for those elements and for ≥ 0 positive semi-definite?). Show that C and −C and open convex cones.

(c) If u ∈ C ∪ −C then H^{u} is compact.

(d) If O is an orbit distinct from C and −C then the stabilizer of a point is not compact.

(e) Describe the decomposition of L^{2}(V) into irreducible representations of H ×_{semidirect} ℝ^{n} and decide which on those parts have wavelet vectors.