Discrete Holomorphic Local Dynamical Systems

248 Tien-Cuong Dinh and Nessim Sibony
Proof. Assume there is θ0 > 0 such that {τ ≥ θ0} is not contained in the orbit of
EZ. We claim that there is a maximal value θ0 satisfying the above property. Indeed,
by definition, τ(a) is smaller than or equal to the average of τ on the fiber of a. So,
we only have to consider the components of {τ ≥ θ0} which intersect U and there
are only finitely many of such components, hence the maximal value exists.
Let E be the union of irreducible components of {τ ≥ θ0} which are not contained
in the orbit of EZ. Sinceθ0 > 0, we know that E ⊂ Z. Wewanttoprove
that E is empty. If a is a generic point in E, it does not belong to the orbit of EZ
and we have τ(a) =θ0. Ifb is a point in f −1 (a), thenb is not in the orbit of EZ.
Therefore, τ(b) ≤ θ0. Sinceτ(a) is smaller than or equal to the average of τ on
f −1 (a), we deduce that τ(b)=θ0, and hence f −1 (a) ⊂ E. By induction, we obtain
that f −n (a) ⊂ E ⊂ Z for every n ≥ 1. Hence, a ∈ EZ. This is a contradiction. ⊓⊔
The following example shows that in general the orbit of E is not an analytic
set. We deduce that in general polynomial-like maps are not homeomorphically
conjugated to restrictions on open sets of endomorphisms of P k (or polynomial
maps of C k such that the infinity is attractive) with the same topological degree.
Example 2.25. Denote by D(a,R) the disc of radius R and of center a in C.Observe
that the polynomial P(z) := 6z 2 + 1 defines a ramified covering of degree 2 from
D := P −1 (D(0,4)) to D(0,4). The domain D is simply connected and is contained
in D(0,1). Letψ be a bi-holomorphic map between D(1,2) and D(0,1) such that
ψ(0)=0. Define h(z,w) :=(P(z),4ψ m (w)) with m large enough. This application
is holomorphic and proper from W := D × D(1,2) to V := D(0,4) × D(0,4). Its
critical set is given by zw = 0.
Define also
g(z,w) := 10 −2� exp(z)cos(πw/2),exp(z)sin(πw/2) � .
One easily check that g defines a bi-holomorphic map between W and U := g(W).
Consider now the polynomial-like map f : U → V defined by f = h ◦ g−1 . Its topological
degree is equal to 2m; its critical set C is equal to g{zw = 0}. Theimageof
C1 := g{z = 0} by f is equal to {z = 1} which is outside U.Theimageofg{w = 0}
by f is {w = 0}∩V.
The intersection {w = 0}∩U contains two components C2 := g{w = 0} and
C ′ 2 := g{w = 2}. They are disjoint because g is bi-holomorphic. We also have
f (C2) ={w = 0}∩V and f −1 {w = 0} = C2. Therefore, E = {w = 0}∩V,
f −1 (E ) ⊂ E and f −1 (E ) �=E ∩ U since f −1 (E ) does not contain C ′ 2 . The orbit
of E is the union of C2 and of the orbit of C ′ 2 .Sincemis large, the image of C′ 2 by
f is a horizontal curve very close to {w = 0}. It follows that the orbit of C ′ 2 is a
countable union of horizontal curves close to {w = 0} and it is not analytic.
It follows that f is not holomorphically conjugate to an endomorphism (or a
polynomial map such that the hyperplane at infinity is attractive) with the same
topological degree. If it were, the exceptional set would not have infinitely many
components in a neighbourhood of w = 0.