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Let $\phi:\mathbb{P}^1\to\mathbb{P}^1$ be a rational function of degree $d\geq2$. How can one prove, using the normalized spherical measure, that $$\int_{\mathbb{P}^1(\mathbb{C})}|(\phi^n)'(z)|\ d\mu (z) \sim d^n$$ as $n\to\infty$?

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  • $\begingroup$ Could you give some motivation? $\endgroup$
    – j.c.
    Commented Mar 9, 2010 at 16:47
  • $\begingroup$ I don't know what generalized spherical measure is, but isn't your integrand just $(\phi^n)^* \omega$? Since $\phi$ acts by $d$ on $H^2(\mathbb{P}^1)$ , and any $2$-form on a surface is closed, this should exactly be $d^n \int \omega$, with no asymptotic needed. I'm probably missing something basic, though. $\endgroup$ Commented Mar 9, 2010 at 20:14
  • $\begingroup$ @ David Speyer For a complex operator $A$ $|det A|^2 =det A_{\mathbf R}$, so integrand differs from the preimage under $n$-th iteration of $\phi$. $\endgroup$
    – Petya
    Commented Mar 10, 2010 at 1:01

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If I understand your problem correctly, i.e. that $\mu$ is the usual metric on the Riemann sphere, then you're asking if essentially all orbits are expansive, at least with respect to that metric.

This will almost always be the case. For example, it will be so whenever the Julia set is holomorphically removable. The only case which is doubtful is when the Julia set is the whole of the Riemann sphere and the forward orbit of the critical point(s) are dense in the sphere -- which can happen [there are nice non-Lattes examples of this by M. Rees, I think the paper dates from 1984 or so]. It could still be the case that that is not enough to dampen the integral, if return times to a given neighborhood are too long.

I have a sneaky feeling that this might happen only in the case of an invariant line field -- and a little Googling brings up a recent paper by Xiaoguang Wang showing that this (essentially) doesn't happen.

[Take the conjectures with a grain of salt, last time I did complex dynamics was 13 years ago!]

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