PSI* 2016/17 - DL 4 - algèbre symplectique Première partie

p≥1Mpp p
IpMpM∈ MpMRpM
tM
(|)Rpkk
u v U V (u|v) = tUV =tV U
MtMM = I
n= 2m J ∈ M2m
J=0−Im
Im0
n= 2m M ∈ M2m
tMJM =J
J
M∈ M2mM=A B
C DA, B, C, D ∈ Mm
M A, B, C, D
(tAC tBD
tAD −tCB =Im
D Q ∈ MmM=ImQ
0ImA−QC 0 C D
M D det M= 1
B, D ∈ MmtBD s1, s2∈Rs16=s2v1, v2∈Rm
(D−s1B)v1= 0 (D−s2B)v2= 0 (Dv1|Dv2)
M v ∈RmDv = 0 Bv = 0
s∈RD−sB
det M= 1 Im0
−sImIm
M P
N N N−1
∀λ∈C, λ 6= 0, P (λ) = λ2mP(1/λ)
λ0∈CM d 1
λ0λ0
1
λ0M
d
−1

m= 2 ∈ M4
C
6= 1
C
φRnx∈Rnkφp(x)kn∈Nφp
φ p
M∈ MnC
M
Ω∈ Mm0−Ω
Ω 0 
M
M∈ MnR
M
MC
M