Let $f(z)$ be a rational function of degree $d \geq 2$, with complex coefficients. I am interested in fully invariant measures for the dynamical system $(\mathbb C_\infty,f)$, where $\mathbb C_\infty$ is the Riemann sphere. By a fully invariant measure, I mean a probability measure $\mu$ such that $f^\ast \mu = d \mu$. (Such a measure also satisfies $f_\ast \mu =\mu$). We can limit ourselves to ergodic ones (those which are not barycenter of two other fully invariant measures).

Those measures which have finite support are easy to describe: from the classical theory of Montel-Fatou-Julia, there exists a larger finite subset if $E$ such that $f^{-1}(E)=E$ and it has at most two elements; every fully invariant measure with finite support has support in $E$, so there are $0$, $1$ or $2$ finite support fully invariant ergodic measures.

From results of Ljubisch and Freyre-Lopez-Mane, there is also the so-called "natural measure" $\mu$, which is defined as the limit of $(\frac{1}{d}f^\ast)^n \nu$ where $\nu$ is any smooth measure on $\mathbb C_\infty$, or any Dirac measure $\delta_x$ for $x \not \in E$. It is ergodic, has the Julia set for support, and many other nice properties.

My question is

Are there always other ergodic fully invariant measures? If so, how to prove it and construct one?

  • $\begingroup$ Does the full invariance property $f^*\mu=d\mu$ have any ramification about the metric entropy of $\mu$? There is another paper (link.springer.com/content/pdf/10.1007/BF02584743.pdf) by Mañé which proves that the measure of maximal entropy is unique. So if $h_\mu=\log(d)$, the only answer is the construction of Lyubich and Freire-Lopes-Mañé which yields an MME by taking iterated preimages. $\endgroup$
    – KhashF
    Mar 23, 2020 at 22:49
  • $\begingroup$ What exactly do you mean by the pullback measure $f^*(\mu)$? Measures are usually pushed forward. $\endgroup$ Mar 24, 2020 at 1:07
  • $\begingroup$ Since you mention the measure of maximal entropy as an example, I assume that you mean that, for a set where the function is injective, you take the pushforward for the corresponding inverse branch, scaled by 1/d to get another probability measure. Then I believe that the measure of maximal entropy is the only such "balanced" measure on the Julia set. $\endgroup$ Mar 24, 2020 at 1:16
  • $\begingroup$ @Lasse You are right that measures are usually pushed forward. But in this case, where $f$ is a finite map of degree $d$, that is the fibers of $f$ has always $d$ points (for a finite number of fibers we need to count the points with multiplicity for this to be true, and we do), the we $f^\ast \mu = \mu (f_\ast)$ where $f_\ast(z)=\sum_{y \in f^{-1}(z)} f(y)$, the points $y$ being counted with multiplicity. $\endgroup$
    – Joël
    Mar 24, 2020 at 11:02

3 Answers 3


The unique measure of maximal entropy $\mu_f$ supported on the Julia set of a rational map $f$ of degree $d \geq 2$ is indeed the unique balanced measure for $f$, i.e., the only probability measure $\mu$ not charging the exceptional set and satisfying $f^*\mu =d \cdot \mu$. As you already noticed in the comments, uniqueness of a measure with this property is explicitly stated in the mentioned paper by Freire, Lopes and Ma\~ne in their Theorem, part (d) (page 46). The proof of this statement is on p. 55 and the argument goes as follows: for any balanced measure $\mu$ it is shown that $\mu$ is absolutely continuous with respect to $\mu_f$ and the ergodicity of $\mu_f$ implies that $\mu=\mu_f$ (existence and ergodicity of $\mu_f$ are proved earlier in the paper). No assumption of non-atomicity, no reference to critical points or classification of Fatou components is employed in the proof of this uniqueness statement.

Another way to prove uniqueness of balanced measure is to use potentials of measures on the Riemann sphere $\mathbb{C}_\infty$ introduced as in

F. Berteloot, V. Mayer, Rudiments de dynamique holomorphe, Vol. 7 of Cours Sp\'ecialis\'es, Soci\'et\'e Math\'ematique de France, Paris (2001)

They give a streamlined treatment based on prior results by Fornaess and Sibony, Hubbard and Papadopol, Ueda and others. Consider the cone $\mathcal{P}$ of functions $U$ plurisubharmonic on $\mathbb{C}^2$ and satisfying $U(tz)=c\log|t|+U(z)$ with a constant $c=c(U) >0$. Each such function defines a positive measure $\mu_U$ on $\mathbb{C}_\infty$ by $\langle \mu_U, \Phi \rangle =\int_{\mathbb{C}_\infty}(U \circ \sigma)\frac{i}{\pi}\partial\bar{\partial}\Phi$ for every smooth test function $\Phi$ with support in the domain of definition of the section $\sigma$ of the natural projection $\Pi: \mathbb{C}^2\setminus \{0\} \to \mathbb{C}_\infty$. Furthermore, every positive measure $\nu$ on $\mathbb{C}_\infty$ is defined by a function $U \in \mathcal{P}$ (unique if required to satisfy $\sup_{\|z\|\leq 1}U(z)=0$), specifically by $U(z)=\int_{\mathbb{C}_\infty}\log\frac{|z_1w_2-z_2w_1|}{\|w\|}d\nu([w])$ (Th\'eor`eme VIII.9 in this reference). This is called the potential of $\nu$.

Now, if a measure $\nu$ is balanced, then its potential $U$ satisfies $F^*U=d\cdot U$ Lemme VIII.12), hence $\frac{1}{d^n}F^{*n}U=U$ for every $n$. Here $F$ denotes a lift of $F$ to $\mathbb{C}^2$. Taking limits in $L^1_{loc}$ as $n \to \infty$ we get $U=G_f$ (Th\'eor`eme VIII.15), the potential of the Lyubich-Freire-Lopes-Ma\~ne measure $\mu_f$. Lifts are not unique, but this does not cause a problem.

If you relax the assumption on a measure supported on Julia set to $f_*\mu = \mu$, then there can be more measures satisfying it, even ergodic ones, besides the measure $\mu_f$. Of course the entropy will be less than $\log d$, sometimes even $0$. For more details on this see

S. P. Lalley, Brownian motion and the equilibrium measure on the Julia set of a rational mapping, Ann. Probab. 20, 4 (1992), 1932--1967.

  • 1
    $\begingroup$ I mentioned critical points and non-atomicity because there was a complaint that (depending on the definition of the pullback) the definition of "fully invariant" could be subtly different from "balanced", since some points have different numbers of preimages from $d$. My point was that this is irrelevant due to the absence of exceptional points in the Julia set (of course, once you go into the proof, it is likely irrelevant in any case). $\endgroup$ Mar 26, 2020 at 10:06

The measure of maximal entropy is the unique measure that is "fully invariant" in your sense. I believe that this already follows from the original proofs - indeed, it is well-known that if you take a point mass at some non-exceptional point, and keep pulling back, you will converge to the measure of maximal entropy. This should be enough to deduce the claim.

In the paper "Conformal and harmonic measures on laminations associated with rational maps" by Lyubich and Kaimanovich, the theorem on the existence of the measure of maximal entropy is stated as follows.

THEOREM. Any rational map f has a unique balanced measure $\kappa$. Moreover, $\operatorname{supp}(\kappa) = J(f)$, and the preimages of any point $z\in J(f)$ (excluding, possibly, two exceptional points) are equidistributed with respect to $\kappa$:

$$ \lim_{n\to\infty} \frac{1}{d^n} \sum_{\zeta\colon f^n(\zeta)=z} \delta_{\zeta} = \kappa,$$

where the limit is taken with respect to the weak topology on the space of probability measures on $J(f)$.

(Here a "balanced" measure is a fully-invariant measure, in your terminology, supported on the Julia set.)

  • $\begingroup$ Thank you and +1. The Theorem you quote from Lyubich and Kaimanovich is Theorem 4.2 page (68), but attributed to Lyubish 93 (and Brolin 65 but that's just for polynomial maps). Surely Lyubish 93 is in fact Lyubish 83: I have a paper by Lyubish with the same tithe and in the same journal published with the same page numbers, but published in 1983. But I can't find where the uniqueness of a balanced measure is asserted in Lyubish. Only is stated the uniqueness of a measure of maximal entropy. $\endgroup$
    – Joël
    Mar 24, 2020 at 15:27
  • $\begingroup$ So the situation is now that I am sure that there are no other fully invariant measure, but I still do not have a proof of it. You suggested that it may follows from the fact that I recalled that for all non-exceptional $x$, $\lim \frac{1}{d^n} (f^\ast)^n \delta_x = \kappa$. Would you have a proof of that? $\endgroup$
    – Joël
    Mar 24, 2020 at 15:30
  • $\begingroup$ @Joël I believe that the existence of MME is established independently by Lyubich and Freire-Lopes-Mañé. If you take a look at the latter, (link.springer.com/content/pdf/10.1007/BF02584744.pdf) on the first page you immediately see the construction of $\kappa$ as $\lim\frac{1}{d^n}(f^n)^*\delta_x$. The uniqueness is proven in another paper of Mañé (link.springer.com/content/pdf/10.1007/BF02584743.pdf). $\endgroup$
    – KhashF
    Mar 24, 2020 at 15:53
  • $\begingroup$ Thank you KhashF. Unfortunately, I do not have access on the paper on Mañé, and it is not reviewed on mathscinet. But from the title, and the world "maximal" in it, I deduce that he proves the uniqueness of a measure of maximal entropy $\log d$. That's not enough to answer my question, since we do not know that a measure of maximal entropy is necessarily fully invariant. $\endgroup$
    – Joël
    Mar 24, 2020 at 16:45
  • $\begingroup$ Continuation of the preceding comment. In Freire-Lopes-Mañé, there is a uniqueness statement. The natural measure $\kappa$ is the unique measure satisfying that for every Borel set $A \subset \mathbb C_\infty$ such that $f_{|A}$ is injective, $\kappa(f(A))=d\kappa(A)$. If we knew that every fully invariant measure satis fies this property, we would be done. But unfortunately, this is false. If $x$ is a fixed exceptional point (for example $\infty$ when $f$ is a polynomial), then $\delta_x$ is fully invariant, and $f$ is injective on the Borel set $x$, but we have $\delta_x(f(x))=\delta_x(x)$. $\endgroup$
    – Joël
    Mar 24, 2020 at 16:52

The way that the paper by Freire-Lopes-Mañé makes sense of $f^*\mu=d\mu$ is the following: ''For any Borel subset $A$ of $\Bbb{C}_\infty$ with $f\restriction_A$ injective, one has $\mu(f(A))=d.\mu(A)$.'' (See p. 46 of this paper.)

One observation is that such an ergodic measure $\mu$ is either supported on the Julia set or is one of those measures with finite support that you mentioned in your question. To see this, notice that if ${\rm{supp}}(\mu)\not\subseteq\mathcal{J}$, then $\mu$, being ergodic, must assign zero mass to the backward invariant closed subset $\mathcal{J}$. So there are two cases:

  1. Suppose ${\rm{supp}}(\mu)\subseteq\mathcal{J}$. It is not hard to see that in this case the support must indeed coincide with the Julia set $\mathcal{J}$: If the open subset $\mathcal{J}-{\rm{supp}}(\mu)$ of the Julia set is non-empty, by Montel's Theorem the union $\bigcup_nf^{-n}\left(\mathcal{J}-{\rm{supp}}(\mu)\right)\cap\mathcal{J}$ coincides with $\mathcal{J}$; but it is of measure zero, a contradiction.

  2. Suppose $\mu(\mathcal{J})=0$, so $\mu$ assigns $1$ to the Fatou set $\mathcal{F}$. Recall two major theorems: The Absence of Wandering Fatou Components and The Classification of Periodic Fatou Components. The measure $\mu$ must assign a positive measure to one of the finitely many periodic Fatou components $U$; such a component must be a member of a cycle -- say of period $p$ -- of either immediate attracting (or super-attracting) basins, immediate parabolic basins, or finally, a cycle of rotation domains (Siegel disks or Herman rings). In the latter case, $f$ is injective on $U$ and hence $f^*\mu=d\mu$ implies $\mu(U)=\mu(f^p(U))=d^p.\mu(U)$ contradicting $\mu(U)>0$. The same idea could be employed to show $\mu(U)=0$ if $U$ is the immediate basin of a parabolic periodic point: The dynamics is injective near such a point (is in the form of $z\mapsto z+1$ in a suitable local coordinate). Finally, let us consider the case where $U$ is the immediate basin of attraction for a periodic point $z_0$ of period $p$. If $\mu(U)>0$, one can consider the system $\left(U,f^p\restriction_U,\frac{1}{\mu(U)}.\mu\restriction_U\right)$. As all orbits converge to $z_0$, the non-wandering set of this system is $\{z_0\}$. Hence the support of $\frac{1}{\mu(U)}.\mu\restriction_U$ is $\{z_0\}$; that is, $\mu(U-\{z_0\})=0$. We conclude that if $\mu(\mathcal{F})=1$, the support of $\mu$ consists of finitely many attracting cycles. So ${\rm{supp}}(\mu)$ is a completely invariant finite subset of the sphere and hence lies in the exceptional set of the complex systems which is of cardinality at most two. Examples are $\mu=\delta_\infty$ when $f$ is a polynomial or a measure of the form $\mu=\frac{1}{2}\left(\delta_0+\delta_\infty\right)$ when $f(z)=\frac{1}{z^d}$.

    Definitely, the first case where $\mu$ is a fully invariant measure with the Julia set $\mathcal{J}$ as its support is more interesting. I agree with @Lasse Rempe-Gillen that the only such a measure in this situation is the measure of maximal entropy. This could be verified directly in some of the well known cases of the Julia dynamics. For instance, suppose $f$ is in the shift locus: $(\mathcal{J},f\restriction_{\mathcal{J}})$ is topologically conjugate to the one-sided shift $\left(\{0,\dots,d-1\}^{\Bbb{N}_0},\sigma\right)$ on $d$ symbols. The only fully invariant measure of the shift system is the $\left(\frac{1}{d},\dots,\frac{1}{d}\right)$-Bernoulli measure (whose pullback is the measure of maximal entropy on $\mathcal{J}$). That is because if $\mu$ is fully invariant, for any choice of symbols $x_0,\dots,x_{k-1}\in\{0,\dots,d-1\}$, the iterates $\sigma,\dots,\sigma^k$ of the left shift $\sigma$ are all injective on the cylinder set $[x_0,\dots,x_{k-1}]$. Hence $1=\mu\left(\{0,\dots,d-1\}^{\Bbb{N}_0}=\sigma^k([x_0,\dots,x_{k-1}])\right)=d^k.\mu\left([x_0,\dots,x_{k-1}]\right)$; so $\mu$ assigns $\frac{1}{d^k}$ to a cylinder set $[x_0,\dots,x_{k-1}]$.


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