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Discrete Holomorphic Local Dynamical Systems

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72 Eric Bedford<br />

for some constant C. On the other hand, the Jacobian δ is constant, so we have<br />

Vol( f n j W0)=|δ| 2n j Vol(W0).<br />

Since f n jW0 ⊂ V ∩{|y| < e δdn j } the first estimate needs to dominate the second<br />

estimate. This is not possible when n j is large, so we conclude that we must have<br />

v = 0onint(K + ). ⊓⊔<br />

1.6 Böttcher Coordinate<br />

We would like to define a function ϕ + = limn→∞ y 1/dn<br />

n , because if we have such a<br />

function, then we will have ϕ + ◦ f =(ϕ + ) d . In order to make this work, we consider<br />

the product<br />

y 1/dn<br />

n<br />

= y0<br />

� y1<br />

y d 0<br />

� 1/d � y2<br />

By the estimate (2), yn/yd n−1 = 1 + q(x j,y j)<br />

yd j<br />

the following:<br />

Theorem 1.29. The limit<br />

ϕ + (x,y) := y lim<br />

y d 1<br />

n−1<br />

∏ n→∞<br />

j=0<br />

� 1/d 2<br />

···<br />

�<br />

yn<br />

y d n−1<br />

� 1/d n<br />

is close to 1 for (x,y) ∈ V + ,sowehave<br />

�<br />

1 + q(x j,yj)<br />

y d j<br />

� 1<br />

d j<br />

exists uniformly on V + and defines a holomorphic function there. Further, we have<br />

ϕ + ◦ f =(ϕ + ) d .<br />

Theorem 1.30.<br />

(1) If γ is a path in U + which starts at a point of V + ,thenϕ + has an analytic<br />

continuation along γ.<br />

(2) There exists no function ψ which is holomorphic on f −1 V + and which is equal<br />

to ϕ + on V + .<br />

Proof. It is clear from the definitions that G + = log|ϕ + | on V + .NowletG ∗ be a<br />

pluriharmonic conjugate for G + in a neighborhood of the starting point of γ, such<br />

that ϕ + = exp(G + + iG ∗ ). Now we may continue G ∗ pluri-harmonically along γ,<br />

which gives us the desired analytic continuation of ϕ + .<br />

The function ψ ◦ f −1 is defined on V + , and it is equal to ϕ + ◦ f −1 on fV + . Thus<br />

(ψ ◦ f −1 ) d =(ϕ ◦ f −1 ) d = ϕ + on fV + . Thus ϕ ◦ f is a dth root of ϕ + on V + .On<br />

the other hand, by the formula, we see that ϕ + ∼ y for (x,y) ∈ V + for |y| large. Since<br />

y does not have a dth root on {|y| > M}, this is a contradiction. ⊓⊔<br />

.

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