Analyse harmonique sur les groupes et les espaces symétriques

R
L2(R)
f C∞ˆ f
ˆ
f(x) = 1
√2πZR
f(y)eixydy f(x) = 1
√2πZR
e−ixy ˆ f(y)dy
x→eixy
d
dx
R R
C∗
1
SL(2,R)
G
(x, y)→xy−1ge G
G
G ϕg(x) = gxg−1
Ad(g)∈End(g)ϕge G gad(X)
Ad e gEnd(g)g[X, Y ] =
ad(X)(Y) [ ,]
[X, [Y, Z]] + [Y, [Z, X]] + [Z, [Y, X]] = 0
Sl(n, R)sl(n, R) = {X∈
M(n, R); trace(X) = 0}g∈Sl(n, R)X, Y gAd(g)X=gXg−1
[X, Y ] = XY −Y X
g×gκ(X, Y ) = trace(ad(X)ad(Y)) G
gκg
X=G/H G
σ H G
σ H
g h G H σ σ
g=h+q q ={X∈g;σ(X) = −X}

gH ∈Xqgσ(g)−1
Xreg
XxΩ = H.x H
νΩ
νΩ
(Ξ, m)ξ∈Ξ Θξ
HD(X)
GXFξC∞HXreg
D(X)
νH.x =ZΞ Fξ(x)Θξdm(ξ) (∗)
x=eX
f(eX) = RΞcξΘξ(f)dm(ξ)cξ
G=RΞ = R<Θξ, f > f
X=H=H×H/diagonale(H×H)
X=G/H G H G
D(X)
HheXq=ih
H=Sl(2,R)
Sl(2,R)
H=Sl(2,R)h=X=x1x2+x3
x2−x3−x1;xj∈R
h
t=Y(θ) = 0θ
−θ0;θ∈Ra=X(t) = t0 0−t;t∈R
HhH
hreg h
T=y(θ) = cos θ sin θ
−sin θ cos θ ;θ∈RA=a(t) = ε et0 0ε e−t;t∈Ret ε =±1
hreg =Ad(H)areg ∪Ad(H)treg Hreg =∪h∈H(hAregh−1∪hTregh−1)
exp H
h
a t
Hhreg X
hHa t
X∈b b =a t H.X
h/b= [h, X]σH.X h/bσH.X ([Y, X],[Z, X]) = [X, [Y, Z]]
h/bβH.X =σH.X
2π
H.X

H.X(t) = {x1x2+x3
x2−x3−x1;−x2
1+x2 2+x2 3=−t2}βH.X(t)=dx2dx3
|x1|
H.Y (θ) = {x1x2+x3
x2−x3−x1;−x2
1+x2 2+x2 3=θ2}βH.Y (θ)=dx1dx2
|x3|
f∈ D(h)C∞
fM({)(X) = ∈πβH.X
H C∞hreg
breg b=t a
areg C∞a
n∈Nlim
θ→0+(id
d θ )n(M(f))(Y(θ))+ lim
θ→0−(id
d θ )n(M(f))(Y(θ)) = lim
t→0(d
d t)n(M(f))(X(t))
lim
θ→0
d
d θ (sign(θ)M(f))(Y(θ)) = −2f(0)
I(h)
M D(h)I(h)
I(h)0I(h)Hh
r > 0
ZH.X
(1+ kξk2)−rdβH.X (ξ)<∞
ˆ
βH.X
S(h)HHh∗
H X →∂(X)∂(X)ϕ(Y) =
d
dt (ϕ(X+tY ))t=0
∂(p)ˆ βH.X =p(iX)ˆ βH.X
ˆ
βH.X
hreg
ˆ
βH.Y (λ)(Y(θ)) = e−iλθ
2iθ sign(λ)ˆ
βH.Y (λ)(X(t)) = e−|tλ|
|2t|sign(λ)
ˆ
βH.X(s)(Y(θ)) = 0 ˆ βH.X(s)(X(t)) = eist +e−ist
|2t|
fh hreg =
Ad(H)areg ∪Ad(H)treg
Zh
f(X)dX =ZR|2θ| M(f)(Y(θ))dθ +1
2ZR|2t| M(f)(X(t))dt
M(f)(X) = 2πβH.X (f) = 2πˆ βH.X (ˆ
ˇ
f)
=ZR|2θ|ˆ βH.X (Y(θ)) ˆ βH.Y (θ)(f)dθ +1
2ZR|2t|ˆ
βH.X (X(t)) ˆ βH.X(t)(f)dt
h

ˆ
βH.X (f)
|2t|ˆ βH.X (X(t)) j(X)
X j(X(t))1/2=sinh t
tj(Y(θ))1/2=sin θ θ
Θn(ε exp X) = ε1+nˆ βH.Y (n)(X)j(X)1/2
Θ+
s(ε exp X) = ˆ
βH.X(s)(X)j(X)1/2Θ−
s(ε exp X) = εˆ
βH.X(s)(X)j(X)1/2
ΘnΘ±
s
HD(H)
Θ±
0= lim
n→O±Θn
f∈ D(H)Hreg
MH(f)(ε exp X) = j(X)1/2M(f◦exp)(X)H I1−I5
X∈breg
b=a t n∈N∗
Fn(εexp X) = εˆ βH.X (Y(n)|2n|F±
0= lim
n→0±Fn
s∈Rε=±1Fε,s(εexp X)) = X
Y∈b;exp Y =1
ˆ
βH.Y (X(s)) |2s|
F±,s =−(F1,s ±F−1,s)
FnF±
0F±H
H
D(H)
f∈ D(H)
MH(f)(x) = X
n∈Z;n6=0
Fn(x)Θn(f)−i(Θ+
O(f)−Θ− O(f))+ 1
2Zs>0
(F+,s(x)Θ+
s(f)+F−,s(x)Θ− s(f))ds
ϕ∈ D(H)
2πϕ(e) =
X
n∈Z;n6=0 |n|Θn(ϕ) + 1
2Zs>0
stanh(π s
2) Θ+
s(ϕ)ds +1
2Zs>0
scoth(π s
2) Θ−
s(ϕ)ds
ΘnΘ±
sH
Hπ H
H(h, v)→π(h)v
HH H
H π
π(ei)i∈IHϕ∈ D(H)
T r(π(ϕ)) = Pi∈I< π(ϕ)ei, ei> H H
D(H)
ΘnΘ±
s
H

2×2
GC/GR
GC/GR