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Continuous Wavelet Transform on the Hyperboloid - Université de ...

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get <strong>the</strong> more elaborate expressi<strong>on</strong><br />

λ(a, χ) = dcoshχ 1/a<br />

dcoshχ = cosh χ<br />

. (46)<br />

a<br />

√1+a 2 −2 sinh 2 χ<br />

Thus, <strong>the</strong> acti<strong>on</strong> of <strong>the</strong> dilati<strong>on</strong> operator <strong>on</strong> <strong>the</strong> functi<strong>on</strong> is<br />

D a ψ(x) ≡ ψ a (x) =λ 1 2 (a, χ)ψ(d<br />

−1<br />

a x) =λ 1 2 (a, χ)ψ(x 1<br />

a<br />

) (47)<br />

with x a ≡ (χ a ,ϕ) ∈ H 2 + and it reads<br />

ψ a (x) = √ 1 a<br />

sinh χ 1/a<br />

sinh χ<br />

cosh pχ<br />

ψ(x 1 ).<br />

cosh pχ 1/a<br />

a<br />

One easily checks using (45) that D a is unitary in L 2 (H 2 +).<br />

Finally, <strong>the</strong> hyperbolic wavelet functi<strong>on</strong> can be written as<br />

ψ a,g (x) =U g D a ψ(x) =U g ψ a (x).<br />

Accordingly, <strong>the</strong> hyperbolic c<strong>on</strong>tinuous wavelet transform of a signal (functi<strong>on</strong>)<br />

f ∈ L 2 (H+ 2 ) is <strong>de</strong>fined as:<br />

W f (a, g)=〈ψ a,g |f〉 (48)<br />

∫<br />

= [U g D a ψ](x)f(x)dµ(x) (49)<br />

H+<br />

2<br />

∫<br />

= ψ a (g −1 x)f(x)dµ(x) (50)<br />

H 2 +<br />

where x ≡ (χ, ϕ) ∈ H 2 + and g ∈ SO 0 (1, 2).<br />

In <strong>the</strong> next secti<strong>on</strong>, we show how this expressi<strong>on</strong> can be c<strong>on</strong>veniently interpreted<br />

and studied as a hyperbolic c<strong>on</strong>voluti<strong>on</strong>.<br />

6.1 C<strong>on</strong>voluti<strong>on</strong>s <strong>on</strong> H 2<br />

Since H+ 2 is a homogeneous space of SO 0 (1, 2), <strong>on</strong>e can easily <strong>de</strong>fine a c<strong>on</strong>voluti<strong>on</strong>.<br />

In<strong>de</strong>ed, let f ∈ L 2 (H+ 2 )ands ∈ L1 (H+ 2 ), <strong>the</strong>ir hyperbolic c<strong>on</strong>voluti<strong>on</strong><br />

is <strong>the</strong> functi<strong>on</strong> of g ∈ SO 0 (1, 2) <strong>de</strong>fined as<br />

∫<br />

(f ∗ s)(g) =<br />

H 2 +<br />

f(g −1 x)s(x)dµ(x). (51)<br />

15

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