Sommes d`automorphismes 1. Cas particulier, f est un

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EK K f∈ L(E)f
E
f E
a0 1 f=af + (1 −a)f
f
x∈E k ∈Nfk(x)=0 f= + (f−)f−
E1f x ∈E k
fk(x)=0 x= (f−)(−x−f(x)− · · · − fk−1(x))
a∈E(fk(a))k∈NE
(an)E a0=a a1=f(a)−a an=f(an−1)−an−2n≥2an=Pn(f)(a)
Pn∈Z[X]n(an)E g, h
E g a0(a2k−1, a2k)
h(a2k, a2k+1)n f (an) = g(an) + h(an)f=g+h
BE f B ∪ {0}
f
CxBk∈Nfk(x) = 0 D
xBfk(x)∈ B k∈NB(f)
CfD D E =D \ f(D)
F=∩k∈Nfk(D) = D \ ∪k∈Nfk(E)fF D F
Ea={fk(a), k ∈N}aE B =F ∪ C ∪ (∪a∈E Ea)
E f
f E
B0(f)B=B0∪B00 E f(B00) (f)
CEB C ϕ E
ϕ(C) = Bf0=ϕ◦f f B∪{0}f0=g0+h0g0, h0
E f =ϕ−1◦f0
E
EF2F2f∈ L(E)f
Edim(E)=1 f=
dim(E)6= 1
F
f=

E={0}f= + E n > 1g E
gn=g+g Xn−X−1
E Xn−X−1F2g g −g
E E BE
C D B
BϕB C D g, h E
Bg(x) = ϕ(x)h(x) = x+ϕ(x)x∈ C g(x) = x+ϕ(x)h(x) = ϕ(x)x∈ D
x∈ B g◦h(x) = h◦g(x) = x g(x) + h(x) = x g h
g+h=
F {e}
f(e) = e a Be ϕ E a e
Bf0=f◦ϕB ∪ {0} B (f0)
f0B F0f0f0
E f =f0◦ϕ