\documentstyle[12pt,amssymb]{amsart} \makeatletter % Bring back \eqalign from plain TeX (for less than full-width alignments) % \def\eqalign#1{\null\,\vcenter{\openup\jot\m@th \ialign{\strut\hfil$\displaystyle{##}$&$\displaystyle{{}##}$\hfil \crcr#1\crcr}}\,} \makeatother \def\g{\gamma} \def\om{\omega} \def\s{\sigma} \def\S{\Sigma} \def\cl{{\mathrm {cl}}} \def\HD{{\mathrm{HD}}} \def\Arg{{\mathrm {Arg}}} \def\HDess {\HD_{{\mathrm{ess}}}} \def\HDhorseshoe{\HD_{{\mathrm{horseshoe}}}} \def\mod{{\mathrm {mod}}} \def\Di{{\mathrm{Distor}}} \def\Cap{{\mathrm{Cap}}} \def\a{\alpha} \def\La{\Lambda} \def\z{\zeta} \def\k{\kappa} \def\e{\varepsilon} \def\b{\beta} \def\f{\varphi} \def\d{\delta} \def\la{\lambda} \def\th{\vartheta} \def\C{{\Bbb C}} \def\D{{\Bbb D}} \def\Z{{\Bbb Z}} \def\R{{\Bbb R}} \def\T{{\cal T}} \def\B{{\cal B}} \def\Crit{{\mathrm {Crit}}} \def\U{{\cal U}} \def\M{{\cal M}} \def\N{{\cal N}} \def\L{{\cal L}} \def\1{{\boldsymbol 1}} \def\and{{\mathrm and}} \def\ov{\overline} \def\for{{\mathrm for}} \def\dist{{\mathrm {dist}}} \def\length{{\mathrm {length}}} \def\h{{\mathrm h}} \def\supp{{\mathrm {supp}}} \def\Vol{{\mathrm {Vol}}} \def\diam{{\mathrm {diam}}} \def\Const{{\mathrm {Const}}} \def\deg{{\mathrm {deg}}} \def\Comp{{\mathrm {Comp}}} \def\ess{{{\mathrm {ess}}}} \def\diaml{\diam_{\Re\log}} \def\Tel{{\mathrm {Tel}}} \def\distl{\dist_{\Re\log}} \def\Jac{{\mathrm Jac}} \textheight=574pt \textwidth=432pt \oddsidemargin=18.88pt \evensidemargin=18.88pt \topmargin=14.21pt \begin{document} \title[Julia sets of holomorphic Collet--Eckmann maps] {On measure and Hausdorff dimension of Julia sets for holomorphic Collet--Eckmann maps} \author{Feliks Przytycki} \address{Institute of Mathematics\\ Polish Academy of Sciences\\ ul. \'Sniadeckich 8\\ 00 950 Warszawa, Poland} \email{feliksp@@impan.impan.gov.pl} \thanks{The author acknowledges support by Polish KBN Grant 2 P301 01307 ``Iteracje i Fraktale, II''. He expresses also his gratitude to the MSRI at Berkeley (partial support by NSF grant DMS-9022140) and ICTP at Trieste, where parts of this paper were written.} \keywords{Math. Rev. 58F23} \date{(version of July 26, 1995)} \maketitle \begin{abstract} Let $f:\ov\C\to\ov\C$ be a rational map on the Riemann sphere , such that for every $f$-critical point $c\in J$ which forward trajectory does not contain any other critical point, $|(f^n)'(f(c))|$ grows exponentially fast (Collet--Eckmann condition), there are no parabolic periodic points, and else such that Julia set is not the whole sphere. Then smooth (Riemann) measure of the Julia set is 0. For $f$ satisfying additionally Masato Tsujii's condition that the average distance of $f^n(c)$ from the set of critical points is not too small, we prove that Hausdorff dimension of Julia set is less than 2. This is the case for $f(z)=z^2+c$ with $c$ real, $0\in J$, for a positive line measure set of parameters $c$. \end{abstract} \section*{Introduction} It is well-known that if $f:\ov\C\to\ov\C$ a rational map of the Riemann sphere is hyperbolic, i.e. expanding on its Julia set $J=J(f)$ namely $|(f^n)'|>1$ for an integer $n>0$, then Hausdorff dimension $\HD(J)<2$. The same holds for a more general class of subexpanding maps, namely such maps that all critical points in $J(f)$ are non-reccurrent, supposed $J(f)\not=\ov\C$, see [U]. On the other hand there is an abundance of rational maps with $J\not=\ov\C$ and $\HD(J)=2$, [Shi]. Recently Chris Bishop and Peter Jones proved that for every finitely generated not geometrically finite Kleinian groups for the Poincar\'e limit set $\Lambda$ one has $\HD(\Lambda)=2$. As geometrically finite exhibits some analogy to subexpanding in the Kleinian Groups -- Rational Maps dictionary, the question arised, expressed by Ch. Bishop and M. Lyubich at MSRI Berkeley conference in January 1995, isn't it true for every non-subexpanding rational map with connected Julia set, that $HD(J)=2$ ? Here we give a negative answer. For a large class of ``non-uniformly" hyperbolic so called Collet--Eckmann maps, studied in [P1], satisfying an additional Tsujii condition, $ \HD(J)<2$. \medskip\noindent {\bf Notation.} For a rational map $f:\ov\C\to\ov\C$ denote by $\Crit(f)$ the set of all critical points of $f$, i.e.points where $f'=0$. Let $\nu:=\sup\{\hbox{multiplicity of}f^n \hbox{at} c: c\in\Crit(f)\cap J\}$. Finally denote by $\Crit'(f)$ the set of all critical points of $f$ in $J(f)$ which forward trajectories do not contain other critical points. We prove in this paper the following results: \medskip\noindent {\bf Theorem A.} Let $f$ be a rational map on the Riemann sphere $f:\ov\C\to\ov\C$, and there exist $\lambda>1, C>0$ such that for every $f$-critical point $c\in \Crit'(f)$ $$ |(f^n)'(f(c))|\ge C\lambda^n, \eqno (0.1) $$ there are no parabolic periodic points, and $J(f)\not=\ov\C$. Then $\Vol(J(f)=0$, where Vol denotes Riemann measure on $\ov\C$. \medskip\noindent {\bf Theorem B.} In the conditions of Theorem A assume additionally that $$ \lim_{t\to\infty}\limsup_{n\to\infty}{1\over n}\sum_{j=1}^n \max(0,-\log\bigl(\dist(f^j(c),\Crit(f))\bigr)-t) = 0. \eqno (0.2) $$ Then $\HD(J(f))<2$. \medskip\noindent For $f(z)=z^2+c, c\in [-2,0]$ real, it is proved in [T] that (0.1) and (0.2) are satisfied for a positive measure set of parameters $c$ for which there is no sink in the interval $[c,c^2+c]$. Tsujii's condition in [T] called there {\it weak regularity} is in fact apparently stronger than (0.2). The set of subexpanding maps satisfying (0.1) and weak regularity has measure 0, [T]. Thus Theorem B answers Bishop--Lyubich's question. \medskip\noindent {\bf Remark.} In [DPU] it is proved that for every rational map $f\ov\C\to\ov\C, c\in\Crit'$ $$ \limsup_{n\to\infty} {1\over n}\sum_{j=1}^n-\log\dist(f^j(c),\Crit(f))\le C_f $$ where $C_f$ depends only on $f$. Here in the condition (0.2) it is sufficient, for Theorem B to hold, to have a positive constant instead of 0 on the right hand side, unfortunately apparently much smaller than $C_f$. Crucial in proving Theorems A and B is the following intermediate result: \medskip\noindent {\bf Theorem 0.1 (on the existence of pacim)}, see [P1]. Let $f:\ov\C\to\ov\C$ satisfies the assumptions of Theorem A. Let $\mu$ be an $\a$-conformal measure on the Julia set $J=J(f)$ for an arbitrary $\a>0$. Assume there exists $0<\lambda'<\lambda$ such that for every $n\ge 1$ and every $c\in \Crit'(f)$ $$ \int{d\mu \over \dist (x,f^n(c))^{(1-1/\nu)\a} }< C^{-1}(\lambda')^{\a n/\nu}. \eqno (0.3) $$ Then there exists an $f$-invariant probability measure $m$ on $J$ absolutely continuous with respect to $\mu$ (pacim). \medskip\noindent Recall that a probability measure $\mu$ on $J$ is called $\a$-conformal if for every Borel $B\subset J$ on which $f$ is injective $\mu(f(B)=\int_B |f'|^\a d\mu$. In particular $|f'|^\a$ is Jacobian for $f$ and $\mu$. The number $\a$ is called the exponent of the conformal measure. If $\Vol(J)>0$ then the restriction of Vol to $J$, normalized, is 2-conformal and obviously satisfies (0.3). If $\HD(J)=2$ then by [P1] we know there exists a 2-conformal measure $\mu$ on $J$ but we do not know whether it is not too singular, namely whether it satisfies (0.3). Under the additional assumption (0.2) we shall prove that it is so for every $\a$-conformal measure. \medskip\noindent {\bf Notation.} Const will denote various positive constants which may change from one formula to another, even in one string of estimates. \section{More on pacim. Proof of Theorem A.} {\bf Proposition 1.1.} In the situation of Theorem 0.1 there exists $K>0$ such that $\mu$-a.e. ${dm \over d\mu} \ge K$. \medskip\noindent {\bf Proof.} In Proof of Theorem 0.1 [P1] one obtains $m$ as a weak* limit of a subsequence of the sequence of measures ${1\over n}\sum_{j=0}^{n-1}f_*^j(\mu)$. It is sufficient to prove that there exists $K>0$ and $n_0>0$ such that for $\mu$-a.e. $y\in J(f)$ $$ {df^n_*(\mu) \over d\mu}(y)={\cal L}^n(\1) \ge K. \eqno (1.1) $$ Here $\cal L$ denotes the transfer operator, which can be defined for example by ${\cal L}(\f)(y)=\sum_{f(z)=y}|f'(z)|^{-\a}\f(z)$. $\1$ is the constant function of value 1. We can assume $y\notin\bigcup_{n>0} f^n(\Crit(f))$ because $$ \mu(\bigcup_{n>0} f^n(\Crit(f)))=0 \eqno (1.2) $$ If a critical value for $f^n$ were an atom then a critical point would have $\mu$ measure equal to $\infty$. It is sufficient to prove (1.1) for $y\in B(x,\d)\cap J(f)$ for an {\it a priori} chosen $x$ and an arbitrarily small $\d$ and next to use the fact that there exists $m\ge 0$ such that $f^m(B(x,\d))\supset J(f)$ (called {\it topological exactness}). Indeed $$ \L^n(\1)(w)=\sum_{f^m(y)=w}\L^{n-m}(\1)(y)|(f^m)'|^{-\a}\ge\ (\sup|(f^m)'|)^{-\a}\L^{n-m}(y_0) $$ where $y_0\in f^{-m}(\{w\})\cap B(x,\d)$. Recall the estimate from [P1]. For an arbitrary $\g>1$ there exists $C>0$ such that for every $x\in J(f)$ $$ {\cal L}^n(\1)(x)\le C+C \sum_{c\in\Crit(f)\cap J} \sum_{j=0}^{\infty} {\g^j\la^{-j\a/\nu}\over \dist (x,f^j(f(c)))^{(1-1/\nu)\a} }. \eqno (1.3) $$ By the assumptions (0.1) and (0.3) the above function is $\mu$-integrable if $\g$ is small enough. Pay attention to the assumption (0.3). It concerns only $c\in\Crit'$. Fortunately there is only a finite number of summands in (1.3) for which $f^{j_0}(c)\in\Crit, j_0\ge j$ . Each summand is integrable because up to a constant it is bounded by $\L^j(\1)$. So $$\sum_{c\in\Crit(f)\cap J} \sum_{j=s}^{\infty} {\g^j\la^{-j\a/\nu}\over \dist (x,f^j(f(c)))^{(1-1/\nu)\a} }\to 0 \ \ \mu-{\rm a.e.} \ \hbox{as} s\to\infty . \eqno (1.4) $$ Fix from now on an arbitrary $x\in J(f)$ for which (1.4) holds, $(dm/d\mu) (x)\ge 1$ and $x\notin\bigcup_{n>0}\f^n(\Crit(f))$ (possible by (1.2) and by $\int(dm/d\mu)d\mu=1$). We need now to repeat from [P1] a part of Proof of Theorem 0.1: For every $y\in B(x,\d)$ and $n>0$ $$ \eqalign{ \L^n(\1)(y)&=\sum_{y'\in f^{-n}(y), {\rm regular}}|(f^n)'(y')|^{-\a} + \sum_{(y',s) {\rm singular}}\L^{n-s}(\1)(y')|(f^s)'(y')|^{-\a}\cr &=\sum_{{\rm reg},y}+\sum_{{\rm sing},y} .} \eqno (1.5) $$ We shall recall the definitions of {\it regular} and {\it singular}: Take an arbitrary subexponentially decreasing sequence of positive numbers $b_j, j=1,2\dots$ with $\sum b_j=1/100$. Denote by $B_{[k}$ the disc $B(x,(\prod_{j=1}^k (1-b_j))2\d)$. We call $s$ the {\it essentially critical time} for a sequence of compatible components $W_j=\Comp f^{-j}(B_{[j})$, compatible means $f(W_j)\subset W_{j-1}$, if there exists a critical point $c\in W_s$ such that $f^s(c)\in B_{[s}$. We call $y'$ {\it regular} in (1.5) if for the sequence of compatible components $W_s, s=0,1,\dots,n, W_n\ni y'$ no $s0 $$ The name {\it regular} concerned formally $y'\in f^{-n}(y)$ but in fact it concerns the branch of $f^{-n}$ mapping $y$ to $y'$ not depending on $y\in B(x,\d)$. By distortion of any branch $g$ of $f^{-n}$ on a set $U$ we mean $$\sup_{z\in B}|g'(z)|/\inf_{z\in B}|g'(z)|.$$ This proves Proposition 1.1.\hfill$\square$ \medskip\noindent {\bf Corollary 1.2} In the situation of Theorem 0.1 for measure-theoretic entropy $\h_m(f)>0$. \medskip\noindent {\bf Proof.} Denote $dm/d\mu$ by $u$. Consider an open set $U\subset \ov\C$ intersecting $J(f)$ such that there exist two branches $g_1$ and $g_2$ of $f^{-1}$ on it. Then by the $f$-invariance of $m$ we have $\Jac_m(g_1)+\Jac_m(g_2) \le 1$ ($=1$ if we considered all branches of $f^{-1}$). $\Jac_m(g_i)$ means Jacobian with respect to $m$ for $g_i$. We have $m(U)>0$ because $\mu$ does not vanish on open sets in $J$ (by the topological exactness of $f$ on $J$) and by Proposition 1.2. At $m$-a.e. $x\in U$ $$ \Jac_m(g_i)(x)=u(g_i(x))|g_i'(x)|u(x)^{-1}>0, $$ (we used here also (1.4)). Hence $\Jac_m(g_i)<1$, so $\Jac_m(f)>1$ on the set $g_i(U), i=1,2$ of positive measure $m$. Now we use Rokchlin's formula and obtain $$ \h_m(f)=\int\log \Jac_m(f) dm >0 .\eqno\square $$ Let $\chi_m=\int\log|f'|dm$ denote characteristic Lyapunov exponent. \medskip\noindent {\bf Corollary 1.3} In the situation of Theorem 0.1, $\chi_m>0$. \medskip\noindent {\bf Proof.} This Corollary follows from Ruelle's inequality $\h_m(f)<2\chi_m$, see [R]. \medskip\noindent {\bf Proof of Theorem A.} Suppose $\Vol(J(f))>0$. After normalization we obtain a 2-conformal measure $\mu$ on $J(f)$ and by Theorem 0.1 and Corollary 1.3 a pacim $m$ with $\chi_m>0$. By Pesin's Theory [Pesin] in the iteration in dimension 1 case [Le] ([Le] is on the real case, but the complex one is similar), for $m$-a.e. $x$, there exists a sequence of integers $n_j\to\infty$ and $r>0$ such that for every $j$ there exists a univalent branch $g_j$ of $f^{-n_j}$ on $B_j:=B(f^{n_j}(x),r)$ mapping $f^{n_j}(x)$ to $x$ and $g_j$ has distortion bounded by a constant. By $\chi_m>0 \ \diam g_j(B(f^{n_j}(x),r)\to 0$. (This follows also automatically from the previous assertions by the definition of Julia set [GPS].) Now we can forget about the invariant measure $m$ and go back to $\Vol$. Because $J(f)$ is nowhere dense in $\ov\C$, there exists $\e>0$ such that for every $z\in J(f)$ $$ {\Vol (B(z,r)\setminus J(f)) \over \Vol (B(z,r))}>\e. $$ Bounded distortion for $g_j$ on $B(z,r)$, $z=f^{n_j}(x)$ allows to deduce that the same part of each small disc$\approx g_j(B_j)$ around $x$ is outside $J(f)$ , up to multiplication by a constant. This is so because we can write for every $X\subset B(z,r), y\in B(z,r)$ $$ \Vol g_j(X)\approx |g_j'(y)|^2 \Vol(X) \eqno (1.6) $$ where $\approx$ means up to the multiplication by a uniformly bounded factor. So $x$ is not a density point of $J(f)$. On the other hand a.e. point is a density point. So $\Vol J(f)=0$ and we arrived at a contradiction.\hfill$\square$ \section{Proof of Theorem B.} As mentioned in the Introduction, the following result is crucial: \medskip\noindent {\bf Lemma 2.1.} Under the conditions of Theorem B, i.e. in the situation of Theorem A and assuming (0.2), the condition (0.3) holds for every $\a$-conformal measure, $\a>0$. \medskip\noindent {\bf Proof.\ Step 1.} Denote the expression from (0.2) $$ \max\bigl(0,-\log\inf_{c\in\Crit'(f)}\dist(f^n(c),\Crit(f))-t\bigr) $$ by $\f_t(n)$. Consider the following union of open-closed intervals $$ A_t':=\bigcup_n (n,n+\f_t(n)\cdot K_f] \ \hbox{and write} \ A_t:=\Z_+\setminus A_t', $$ for an arbitrary constant $K_f> \nu/\log\la$. By (0.2) for every $a>0$ there exist $t>0$ and $n(a,t)$ such that for every $n\ge n(a,t)$ $$ A_t\cap [n,n(1+a)]\not=\emptyset \eqno (2.1) $$ Moreover, fixed an arbitrary integer $M>0$, we can guarantee for every $n'\ge n(1+a),n\ge n(a,t)$ $$ \sharp (A_t \cap \{j\in [n,n']: j \hbox{divisable by}M\}) \ge {1\over 2M} (n'-n). \eqno (2.2) $$ Observe that for every $n_0,n$, (2.1) transforms into $$ [n_0+n,n_0+n+a(n_0+n)]=[n_0+n,n_0+n+a({n_0 \over n}+1)n]. $$ The result is that if $n \ge bn_0$ for an arbitrary $b>0$ then $$ A_t\cap [n_0+n,n_0+n+a(b^{-1}+1)n]\not=\emptyset. \eqno (2.3) $$ Denote in the sequel $a(b^{-1}+1)$ by $a'$. \medskip\noindent {\bf Step 2.} Observe now that if $n\in A_t$ then for every $c\in\Crit'(f)$ there exist branches $g_s, s=1,2,\dots,n-1$ of $f^{-s}$ on $B_n:=B(f^n(c),\d)$ with uniformly bounded distortions, where $\d=\e\exp-t\nu$ for a constant $\e$ small enough. Sometimes to exhibit the dependence on $n$ we shall write $g_{s,n}$. Indeed, define $g_s$ on $B_{[s}=B(f^n(c),\prod_{j=1}^s (1-b_j)2\d)$ for $s=1,2,\dots,n-1$ according to the procedure described in Proof of Theorem A. If there is an obstruction, namely $s$ an essential critical time, then for every $z\in B_{[s}$ $$ |g_{s-1}'(z)| \le \la^{-s}\th^s \le \exp (-s\nu/K_f) \eqno (2.4) $$ for $\th>1$ arbitrarily close to 1 (in particular such that $K_f>{\nu\over \log\la-\log\th}$) and for $s$ large enough. The constant $\th$ takes care of distortion. (2.4) holds for $z=f^s(q)$, where $q$ is the critical point making $s$ critical time, without $\th$ by (0.1) (with the constant $C$ instead). The small number $\e$ takes care of $s$ small, which cannot be then essential critical. The inequality (2.4) and rooting ($1/\nu$ to pass from $s-1$ to $s$) imply $\f_t(f^{n-s}(c))\ge s/K_f$, so $n\notin A_t$, a contradiction. \medskip\noindent {\bf Step 3.} By uniformly bounded distortion for the maps $g_{j,n}$, $n\in A_t$ we obtain (compare (1.6)) for every $n_0>0$ large enough, $ c\in\Crit'$ $$ \mu B(f^{n_0}(c), r_j)\approx r_j^\a \eqno (2.5) $$ for a sequence $r_j, j=1,2,\dots$ such that $$ r_1>\exp -Lbn_0 \eqno (2.6) $$ $$ r_{j+1} > r_{j}^{1+ \s} \eqno (2.7) $$ $$ \and \ r_{j+1}0$ is an arbitrary constant close enough to 0. This gives $$ r_{j+1}/r_j \ge \exp (-2(\log L)a'n_j) \eqno (2.9) $$ To conclude we need to know that $r_j$ shrink exponentially fast with $n_j\to\infty$. For that we need the following fact (see for example [GPS], find the analogous fact in Proof of Theorem A): \medskip\noindent {\bf (*)} For every $r>0$ small enough and $\xi, C>0$ there exists $m_0$ such that for every $m\ge m_0, x\in J(f)$ and branch $g$ of $f^{-m}$ on $B(x,r)$ having distortion less than $C$ we have $\diam g(B(x,r))<\xi r$. \medskip\noindent Apply now (2.2) to $n=n_0, n'=n_j+n_0$. We obtain a ``telescope": For all consecutive $\tau_1,\tau_2,\dots,\tau_{k(j)}\in A_t\cap [n_0, n_j+n_0]$ divisible by $M$ $$ g_{\tau_{i+1}-\tau_i, \tau_{i+1}} (B(f^{\tau_{i+1}}(c),\d))\subset B(f^{\tau_i}(c),\d/2) $$ for $M\ge m_0$ from (*). Hence using (2.2) $$ r_j\le 2^{-n_j/2M} . \eqno (2.10) $$ Denote $2a'\log L$ by $\g$ and $(\log 2)/2M$ by $\g'$. (2.9) and (2.10) give $$ r_{j+1} \ge r_j\exp -\g n_j \ge r_j (\exp -\g'n_j)^{\g/\g'} \ge r_j^{1+\g/\g'}. $$ As $\g'$ is a constant and $\g$ can be done arbitrarily small if $a$ is small enough, we obtain (2.7). The condition (2.8) follows from the fact that for $n_0$ large enough all $n_{j+1}-n_j$ are large enough to apply (*). \medskip\noindent {\bf Conclusion.} We obtain $$ \int{d\mu\over\dist(x,f^{n_0}(c))^{(1-1/\nu)\a}} \le $$ $$ \mu(\ov\C \setminus B(f^{n_0}(c), r_1)) {1\over r_1^{(1-1/\nu)\a}} + \sum_{j\ge 2} \mu(B(f^{n_0}(c), r_{j-1})\setminus B(f^{n_0}(c), r_j)) {1\over r_j^{(1-1/\nu)\a}} \le $$ $$ \exp (Lbn_0(1-1/\nu)\a) +\Const\sum_{j\ge 2} {r_{j-1}^\a \over r_j^{(1-1/\nu)\a}} \le $$ $$ (\exp (Lb(1-1/\nu)\a))^{n_0} + \Const\sum_{j\ge 2} r_{j-1}^\a r_{j-1}^{-(1-1/\nu)\a (1+\s)}. $$ The latter series has summands decreasing exponentially fast for $\s$ small enough so it sums up to a constant, hence the first summand dominates. We obtain the bound by $(\la')^{n_0}$ with $\la'>1$ arbitrarily close to 1. Thus (0.3) has been proved.\hfill$\square$ \medskip\noindent {\bf Remark 2.2.} The only result in our disposal on the abundance of non-subexpanding maps satisfying (0.1) and (0.2) is Tsujii's one concerning $z^2+c, c$ real (see Introduction). For this class however the exponential convergence of $$\diam\Comp f^{-n_j}(B(f^{n_j+n_0}(0),\d)$$ to 0 follows from [N] (the component containing $f^{n_0}(c)$). So restricting our interests to this class we could skip (2.2) and considerations leading to (2.10) above. By [N] $\diam\bigl(\Comp( f^{-n}(B(x,\d)))\cap \R\bigr) < C\tilde\la^{-n}$ for some constants $C>0, \tilde\la>1, \d$ small enough and every component Comp. Just the uniform convergence of the diameters to 0 as $n\to\infty$ follows from [P1], but I do not know how fast is it. \medskip\noindent {\bf Proof of Theorem B.} Suppose that $\HD(J)=2$. Then there exist a 2-conformal measure $\mu$ on $J$. This follows from the existence of an $\a$-conformal measure for $\a=\HDess(J)$, where $\HDess$ is the essential Hausdorff dimension which can be defined for example as supremum of Hausdorff dimensions of expanding Cantor sets in $J$, see [DU] [P2] and [PUbook], and from $\HDess(J)=\HD(J)$, see [P1]. The former holds for every rational map, the latter was proved in [P1] only for Collet--Eckmann maps. By Lemma 2.1 the condition (0.3) holds, hence there exists a pacim $m\ll\mu$. Moreover $\chi_m>0$ by Corollary 1.3. As in Proof of Theorem A by Pesin Theory there exists $X\subset J$, $m(X)=\mu(X)=1$, such that for every $x\in X$ there exists a sequence of integers $n_j(x)\to\infty$, $r>0$ and univalent branches $g_j$ of $f^{-n_j}$ on $B(f^{n_j}(x),r)$ mapping $f^{n_j}$ to $x$ with uniformly bounded distortion. Write $B_{x,j}:=g_j (B(f^{n_j}(x),r))$. We obtain for every $x\in X$ applying (1.6) to Vol and $\mu$ (similarly as in Proof of Theorem A) $$ \mu(B_{x,j})\le\Const\Vol(B_{x,j})\le\Const(\Vol B(x,\diam B_{x,j})) . $$ If $\Vol X=0$ then there exists a covering of $X$ by discs $B(x_t,\diam B_{x_t,j_t}), t=1,2,\dots$ which union has $\Vol<\e$ for $\e$ arbitrarily close to 0, of multiplicity less than a universal constant (Besicovitch's theorem). Hence $$ \e\ge\Const\sum_t\Vol B(x_t,\diam B_{x_t,j_t})\ge\Const \mu \sum_t B_{x_t,j_t}\ge 1, $$ a contradiction. Hence $\Vol J\ge \Vol X>0$. This contradicts Theorem A that $\Vol J=0$ and the proof is over. Remark that we could end the proof directly: As in Proof of Theorem A we show that no point of $X$ is a point of density of the $\Vol$ measure. Hence $\Vol X=0$. (I owe this remark to M. Urba\'nski.)\hfill$\square$ \begin{thebibliography}{}\frenchspacing \bibitem[BJ]{BJ} Ch. Bishop, P. Jones: Hausdorff dimension and Kleinian groups. 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