Recently, I read a theorem of existence of conformal measure for the rational map.

I did not understand two places in the proof. The author claims that there exists an open set $V\subset \hat{C}\setminus J(R)$, such that each inverse branch $R_{j}^{-n}$ of $R^{n}$ is a single valued function. And also the inverse orbits $$R_{j_1}^{-1}(V), R_{j_2}^{-1}R_{j_1}^{-1}(V),\dots R_{j_k}^{-1}R_{j_(k-1)}^{-1}(V)\dots R_{j_1}^{-1}(V)$$ are disjoint and the orbits will uniformly converge to $J(R)$, if $V$ is a subset in Siegel disk and Herman ring. there is at most one exceptional inverse orbit.

I was confused with this argument for a very long time, I did know how to give a complete proof. any reference and comments will be appreaciated.

EDIT(Thanks for professor Eremenko's advice.): this arguement is form theorem 3 (page 740) in Sullivan's paper. http://download.springer.com/static/pdf/268/chp%253A10.1007%252FBFb0061443.pdf?auth66=1398441124_3dd347091ebaae9a952145910da7220b&ext=.pdf

  • $\begingroup$ If you are asking MO to help you with reading the paper, then tell us what paper are you reading. $\endgroup$ Apr 23, 2014 at 15:26
  • $\begingroup$ Thank you, professor. This argument is from Theorem3 (page 740)in Dennis Sullivan's paper,and here is download.springer.com/static/pdf/268/… a link $\endgroup$ Apr 23, 2014 at 15:57

1 Answer 1


Usually this argument is justified like this.

First. The critical values of $f^n$ are the forward orbits of critical points. This follows from the chain rule. On the set of normality, forward orbits of critical points either tend to cycles, or lie on some closed curves (in singular domains). Therefore there is always a simply connected open set $V$ on the set of normality that is disjoint of these forward orbits.

Second. This implies that all inverse branches of all iterates are well defined and holomorphic on $V$.

Third. These branches, all together make a normal family because they avoid a large set (Julia set). The limits of sequences of these branches must be constant if there are no singular domains, because the inverse orbits tend to $J$ and $J$ has no interior. There is an exceptional orbit if $V$ is in the singular domain.

This argument was already used by Fatou.

EDIT. To address your questions: that preimages of $V$ under different branches of the same $f^{-n}$ are disjoint is clear. For all preimages under $f^{-m}$ and $f^{-n}$ to be disjoint $V$ should not be in an invariant component. You can always find such $V$, unless all components are comletely invariant. In this last case the components are non-singular, the dynamics inside such component is clear, and again you can choose such $V$.

At the time of Fatou, the existence of singular components was doubtful, but of course Fatou knew that there is such possibility, and always took them into account in his proofs. The term "singular domain" belongs to Fatou. That they can be actually be present, Siegel showed in 1944. In other words, Fatou knew the classification of possible periodic components but he did not know whether all possibilities of this classification can be actually present. He also did not know whether wandering domains can exist or not, but this does not affect the argument we discuss.

  • $\begingroup$ Thank you, professor. Whether the singular domain (Siegel disk, Herman ring) was already known at Fatou's time. The argument of $V$ is always disjoint with $f^{-n}(V)$ for all $n$ seems very obvious, however I can only prove it through the Classification of periodic Fatou component, I do not know how to prove it shortly. $\endgroup$ Apr 24, 2014 at 1:29

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