Julia sets for polynomial diffeomorphisms of ${\Bbb C}^2$ are not semianalytic

TL;DR

This paper proves that for polynomial diffeomorphisms of complex two-space with positive entropy, their Julia sets and those of their inverses are not semi-analytic, highlighting complex geometric properties.

Contribution

It establishes that Julia sets for such polynomial diffeomorphisms are not semi-analytic, revealing new geometric constraints in complex dynamics.

Findings

Julia sets are not semi-analytic for positive entropy diffeomorphisms

Julia sets of the inverse map share the same non-semi-analytic property

The result applies to all polynomial diffeomorphisms with positive entropy

Abstract

For any polynomial diffeomorphism $f$ of ${\Bbb C}^2$ with positive entropy, neither the Julia set of $f$ nor of its inverse $f^{-1}$ is semi-analytic.

Equations

$$\rho (z,\bar{z})=\sum _{I,J}{c}_{I,J}{z}^I{\bar{z}}^J$$

$$\rho (z,\bar{w})=\sum _{I,J}{c}_{I,J}{z}^I{\bar{w}}^J$$

$$Q_w:=\{z\in U:\rho (z,\bar{w})=0\},$$

$$W^{ss}(p)=\{p\}\cup \{q\in C^2:\underset{n\to \infty }{\lim }\frac{1}{n}\log (\mathrm{dist}(f^n(p),f^n(q)))=\log |\alpha |\}$$

$$W_{\mathrm{loc}}^{ss}(p):=\{q\in W^{ss}(p):f^n(q)\in U\mathrm{for}\mathrm{all}n\ge 0\}.$$

$$\rho (x,y,\bar{x},\bar{y})=x^{k}u(x,y)+{\bar{x}}^k\overline{u(x,y)}+\Psi (x,y,\bar{x},\bar{y})$$

$$\eqalign{\rho(f(x,&y))=\rho(x_{1},y_{1},\bar{x}_{1},\bar{y}_{1})=\cr&=\rho(\beta x+\cdots,\alpha y+\cdots,\bar{\beta}\bar{x}+\cdots,\bar{\alpha}\bar{y}+\cdots)=\beta^{k}x^{k}+\bar{\beta}^{k}\bar{x}^{k}+\Psi_{1}}$$

Full text

Fornæss and Sibony [FS2] showed that, for a generic Hénon map, the Julia set is neither smooth nor semianalytic. In [BK] we showed that $J^{+}$ can never be $C^{1}$ smooth. However, the Julia sets in [N] and [Uch1–3] have singular points and thus are not $C^{1}$, and this non-smoothness was our motivation for the present Theorem.

**§1. Levi flat hypersurfaces. ** Let $U\subset{{C}}^{k}$ be an open subset. A function $\rho$ on $U$ is said to be real analytic if for every $q\in U$, $\rho$ can be written as a real power series which converges in a neighborhood of $q$. Let us suppose that $q=0$ and write

[TABLE]

where $I=(i_{1},\dots,i_{k})$ is a $k$-tuple of nonnegative integers, and $z^{I}=z_{1}^{i_{1}}\cdots z_{k}^{i_{k}}$, and similarly for $J$ and $\bar{z}^{J}$. We may treat $z$ and $\bar{z}$ as independent variables and write

[TABLE]

The reality condition on $\rho$ is that $c_{I,J}=\overline{c_{J,I}}$, which means that $\rho(z,\bar{w})=\overline{\rho(w,\bar{z})}$. A set $X$ is real analytic if it can be written locally as $X\cap U=\{\rho=0\}$. A point $x_{0}\in X$ is said to be regular if $X$ is a smooth manifold in a neighborhood of $x_{0}$. We write $Reg(X)$ for the set of regular points, and $Reg(X)$ is dense in $X$ (see [BM]), although the dimension may be different at different points.

A real hypersurface is said to be Levi flat if it is (locally) pseudoconvex from both sides. Recall that $G^{+}$ is pluriharmonic on the set $\{G^{+}>0\}$, so: If the set $J^{+}=\partial\{G^{+}>0\}$ is $C^{1}$ smooth at some point, then it is Levi flat there. A real analytic set is said to be Levi flat if it is Levi flat at each regular point. If $X$ is real analytic, Levi flat hypersurface, then at each regular point, there is a local holomorphic coordinate system such that $X$ is locally given as $\{z_{1}+\bar{z}_{1}=0\}$. At singular points, the situation is more complicated.

A semianalytic set is given locally as a finite, disjoint union of sets of the form $\{r_{j}=0,s_{j}>0\}$, where $r_{j}$ and $s_{j}$ are real analytic. (See Bierstone and Milman [BM] for further information on semianalyticity.) The following Lemma allows us to replace a semianalytic set $J^{+}$ by an analytic set $X$.

(i) $X\cap V=J^{+}\cap V\neq\emptyset$ for some open set $V\subset U$,

(ii) $X$ is locally invariant in the sense that $f(X)\cap U\subset X$.

*Proof. * If $J^{+}$ is semianalytic in a neighborhood of $p$, then there is an open set $U$ containing $p$ such that $J^{+}\cap U$ is a finite union of sets $X_{j}=\{r_{j}=0,s_{j}>0\}$, where $r_{j},s_{j}$ are real analytic on $U$. We may assume that $p$ belongs to the closure of each $X_{j}$. Passing to an iterate of $f$, we may assume that each $X_{j}$ is invariant, in the sense that $f(X_{j}\cap U)\subset X_{j}$. Since the set of regular points is dense, we may suppose that at least one of the $X_{j}$ intersects $Reg(J^{+})$. Let us set $\rho:=r_{j}$ and $X:=\{\rho=0\}$. We may assume that $X$ is an irreducible subvariety of $U$; otherwise, pass to a component. We have local invariance (ii). Since $X$ intersects $Reg(J^{+})$, we may choose an open set $V$ such that condition (i) holds, and such that $V\cap J^{+}$ consists of regular points. Since $J^{+}\cap V$ is regular, it is Levi flat, and since $X$ is irreducible, it is Levi flat, too.

Let us discuss the hypersurface $X=\{\rho=0\}$, where $\rho(z,\bar{w})$ converges for $z,w\in U$. If for fixed $w\in U$, $\rho(z,\bar{w})=0$ for all $z$, we say that $X$ is Segrè degenerate at $w$. If $X$ is not degenerate at $w\in U$, then the Segrè variety, which is defined as

[TABLE]

is a proper subvariety of $U$. (In other words, the condition that $w$ is Segrè degenerate means that the Segrè variety is the whole open set $U$.) We may choose the defining function $\rho$ to be minimal, which means that if $\rho^{\prime}$ is any other defining function, then $\rho^{\prime}=h\rho$. The family of Segrè varieties is independent of the choice of minimal defining function.

At this stage, we can conclude that $J^{\pm}$ cannot be algebraic.

*Proof. * Let us suppose that $J^{+}=\{\rho(z,\bar{z})=0\}$ is defined by a real polynomial. A regular point $w\in J^{+}$ is Segrè nondegenerate, so $Q_{w}$ is a proper subvariety of ${{C}}^{2}$, which is contained in $J^{+}$. On the other hand, this is not possible, since by [BS] there is no subvariety of ${{C}}^{2}$ which is contained in $K^{+}$.

The set of Segrè degenerate points is a complex subvariety of codimension at least 2 (see [PSS, §3]). Thus in ${{C}}^{2}$, the Segrè degenerate points are isolated, so we may assume that $U$ is sufficiently small that all points of $X\cap U-\{p\}$ are Segrè nondegenerate.

A basic result (see Pinchuk, Shafikov and Sukhov [PSS]) is that if $X$ is Levi flat, then for $w\in X$, then each proper Segrè variety $Q_{w}$ is contained in $X$. We say that $p$ is *dicritical * if there are infinitely many distinct varieties $Q_{q}$ passing through $p$. Since $X$ is irreducible, it follows that if infinitely many varieties $Q_{q}$ contain $p$, then all varieties $Q_{q}$ contain $p$. We will make use of the following result:

Proof. If $r_{0}$ is a saddle point, then by [BS2], the stable manifold $W^{s}(r_{0})$ is dense in $J^{+}$. Since there are infinitely many saddle points, we may suppose that $r_{0}\neq p$. Let $q\in W^{s}(r_{0})\cap X-\{p\}$ be a regular point of $X$. We may assume that $q$ is Segrè nondegenerate, so that $Q_{q}$ is a complex subvariety of $X$. Further, since the leaves of the complex foliation of a Levi flat hypersurface are unique, it follows that $W^{s}(r_{0})$ intersects $Q_{q}$ in an open set. If $p$ is dicritical, then $p\in Q_{q}$. On the other hand, since $p$ is fixed, it cannot belong to $W^{s}(r_{0})$. Thus $\hat{W}^{s}(r_{0}):=W^{s}(r_{0})\cup Q_{q}$ is a complex manifold which is strictly larger than $W^{s}(q_{0})$. (Note that we may desingularize $\hat{W}^{s}(r_{0})$ if $p$ is a singular point of $Q_{q}$.) Now recall that $W^{s}(r_{0})$ is uniformized by ${{C}}$, and the only Riemann surface which strictly contains ${{C}}$ is the Riemann sphere, which is compact. Since ${{C}}^{2}$ can contain no compact, complex manifolds, we have a contradiction, by which we conclude that $Q_{q}$ cannot contain $p$. Thus $p$ is not dicritical.

Let us suppose that the multipliers of $Df$ at $p$ are $|\alpha|<|\beta|$, with $|\alpha|<1$. Then the strong stable set of $p$ corresponding to the multiplier $\alpha$ is defined as

[TABLE]

By the Strong Stable Manifold Theorem, $W^{ss}(p)$ is a complex submanifold of ${{C}}^{2}$ which is uniformized by ${{C}}$. The local strong stable manifold is defined as

[TABLE]

Let us choose coordinates $(x,y)$ near $p=(0,0)$ so that the coordinate axes are the eigenspaces for $Df(p)$. Then if we take $U=\{|x|<r_{1},|y|<r_{2}\}$ to be a small bidisk, then $W^{ss}_{\rm loc}(p)$ is the connected component of $W^{ss}(p)\cap U$ which contains $p$.

We conclude this section by showing that the local strong stable manifold coincides with the Segrè variety through $p$.

*Proof. * The Jacobian determinant of $f$ is a nonzero constant $\delta$. As was shown in [BK], if $J^{+}$ is semianalytic, we must have $|\delta|\leq 1$. Let $\alpha$, $\beta$ be the multipliers of $Df$ at $p$. Since $|\alpha\beta|=|\delta|<1$, we may suppose that $|\alpha|<1$. If $|\beta|<1$, then $p$ is an attracting fixed point, which means that $p$ belongs to the interior of $K^{+}$. Since $p\in J^{+}=\partial K^{+}$, we must have $|\beta|\geq 1$. Thus the eigenvalues are distinct, and we may diagonalize $Df(p)$. We may suppose that $p=(0,0)$, and $f(x,y)=(x_{1},y_{1})=(\beta x+\cdots,\alpha y+\cdots)$. Further, we may choose local coordinates such that $W^{ss}_{\rm loc}(p)=\{x=0\}$.

If $V$ be an irreducible component of $Q_{p}$, and $V$ is not the same as $W^{ss}_{\rm loc}(p)$, then we may choose $U$ sufficiently small that $Q_{p}\cap W^{ss}_{\rm loc}(p)=Q_{p}\cap\{x=0\}=\{(0,0)\}=\{p\}$. Thus for some positive integer $\mu$ we may choose a root $x^{1/\mu}$ and represent $V$ locally as a Puiseux expansion $V=\{y=\sum_{j=1}^{\infty}a_{j}x^{j/\mu}\}$. The local invariance of $V$ at $p=(0,0)$ means that we will have $y_{1}=\sum_{j=1}^{\infty}a_{j}x_{1}^{j/\mu}$. If $a_{j_{0}}$ is the first nonvanishing coefficient, we must have $\alpha=\beta^{j_{0}/\mu}$. But this is impossible since $j_{0}/\mu>0$, and $|\alpha|<1\leq|\beta|$. It follows, then that the only irreducible component of $Q_{p}$ is $\{x=0\}$.

**§2. Multipliers at a fixed point. **

Proof. By Proposition 1.5, we know that the multipliers of $Df$ at $p$ are $|\alpha|<1$ and $|\beta|\geq 1$. We must show that $|\beta|>1$. If not, then $|\beta|=1$. First, we observe that $\beta$ cannot be a root of unity. For in that case, $p$ is a semi-attracting, semi-parabolic fixed point. By Ueda [U1] and Hakim [H], there is a semi-parabolic basin ${\cal B}$ with $\partial{\cal B}=J^{+}$. However, the boundary of ${\cal B}$ has a fractal “cusp” at $p$ (reminiscent of the cauliflower Julia set) and is not semianalytic. We conclude that $\beta^{k}\neq 1$ for all nonzero integers $k$.

Now let us use coordinates from the proof of Lemma 1.5. Since $Q_{(0,0)}=\{x=0\}$, we may write $\rho(x,y,0,0)=x^{k}u(x,y)$, where $u(x,y)$ is a holomorphic function with $u(0,0)=1$. This means that

[TABLE]

where in the expansion of $\rho$, all of the purely holomorphic terms are contained in $x^{k}u(x,y)$, and $x^{k}$ is the purely holomorphic part of lowest order. Now there is a real analytic unit $h(x,y,\bar{x},\bar{y})$ such that $\rho\circ f=h\,\rho$, and $h(0,0)=c\neq 0$ is real. Thus the purely holomorphic part of lowest order are $cx^{k}$. On the other hand, as in the proof of Lemma 1.5, we have

[TABLE]

Thus we see that the purely holomorphic terms of lowest order are $\beta^{k}x^{k}$, from which we conclude that $\beta^{k}$ is real, which is a contradiction.

If $p$ is a saddle point, we let $W^{u}(p)$ be the unstable manifold at $p$. If the multipliers are $|\alpha|<1<|\beta|$, then there is a holomorphic uniformization $\psi_{p}:{{C}}\to W^{u}(p)\subset{{C}}^{2}$ such that $\psi_{p}(0)=0$, and $\psi_{p}(\beta\zeta)=f(\psi_{p}(\zeta))$. We set $J_{p}:=\psi_{p}^{-1}(J^{+}\cap W^{u}(p))$. By the invariance of $J^{+}$ it follows that $J_{p}$ is invariant under $\zeta\mapsto\beta\zeta$.

Proof. $J_{p}$ is a semianalytic set of dimension one in ${{C}}$. Thus it has a tangent cone at the origin. Since it is invariant under $\zeta\mapsto\beta\zeta$, we conclude that $\beta\in{{R}}$, and $J_{p}$ consists of a finite number of (infinite) rays emanating from the origin. We must show that there are exactly two rays whose union forms a line passing through the origin. We consider the points $r\in J_{p}$ which correspond to transverse intersections between $W^{s}(p)$ and $W^{u}(p)$. By [BLS4] this set is dense in $J_{p}$. Let $\Delta_{0}\subset{{C}}$ denote a small disk about the origin, and let $\Delta\subset{{C}}$ denote a disk about $r$, small enough that $J_{p}\cap\Delta$ is a single segment $I$ which divides $\Delta$ into halves $\Delta^{\prime}$ and $\Delta^{\prime\prime}$.

Consider the complex disks in ${{C}}^{2}$ given by ${\cal D}_{0}:=\psi_{p}(\Delta_{0})$ and ${\cal D}:=\psi_{p}(\Delta)$. Since ${\cal D}$ is transverse to $W^{s}(p)$ at $\psi_{p}(r)$, we may apply the Lambda Lemma to conclude that there are disks ${\cal D}_{j}\subset f^{j}({\cal D})$ containing $f^{j}(\psi_{p}(r))$ which may be written as graphs over ${\cal D}_{0}$, and ${\cal D}_{j}\to{\cal D}_{0}$ in the $C^{1}$ topology. Let $\gamma_{j}:=f^{j}(\psi_{p}(I))\cap{\cal D}_{j}$. This is a smooth curve which divides ${\cal D}_{j}$ into halves ${\cal D}_{j}^{\prime}$ and ${\cal D}_{j}^{\prime\prime}$, corresponding to the partition $\Delta=\Delta^{\prime}\cup I\cup\Delta^{\prime\prime}$. It follows that the $\gamma_{j}$ converge to a smooth curve $\gamma_{0}\subset{\cal D}_{0}$. Now the Green function $G^{+}$ cannot vanish on all of ${\cal D}_{j}$, so it must be strictly positive everywhere on, say, ${\cal D}_{j}^{\prime}$. Since $G^{+}$ is continuous, we have $G^{+}|_{{\cal D}_{j}^{\prime}}\to G^{+}|_{{\cal D}_{0}^{\prime}}$. If this limit of positive harmonic functions vanishes at a point of ${\cal D}_{0}^{\prime}$, then it must vanish everywhere on ${\cal D}_{0}^{\prime}$. We know that $G^{+}$ cannot vanish everywhere on ${\cal D}_{0}$, so we conclude that it must be strictly positive everythere on either ${\cal D}_{0}^{\prime}$ or ${\cal D}_{0}^{\prime\prime}$. We conclude that $\gamma_{0}\subset J^{+}\cap{\cal D}_{0}$. Thus we conclude that $J_{p}$ contains a line, and any additional rays of $J_{p}$ must lie on the other side, $\Delta_{0}^{\prime\prime}$. However, if there is a ray inside $\Delta^{\prime\prime}_{0}$, there are points of $\Delta^{\prime\prime}_{0}$ where $G^{+}>0$, and we may apply the same argument to ${\cal D}^{\prime\prime}_{j}$ to conclude that all of $J_{p}$ must lie inside the line.

Proof. If $L\subset{{C}}^{2}$ is a complex line, then by [FM], $L\cap J^{+}$ is compact. Since $J^{+}$ is semianalytic of pure dimension 3, it follows that for generic $L$, $J^{+}\cap L$ has real dimension $\leq 1$. In fact, it has pure dimension 1 since there can be no component of dimension zero (which would be an isolated point) because $J^{+}$ is the boundary of $\{G^{+}>0\}$. Now let us fix a complex line $L_{0}$ such that $J^{+}\cap L_{0}$ consists of a finite number of segments and closed curves. We will show that there exists a line $L$ arbitrarily close to $L_{0}$ such that $K^{+}\cap L$ contains interior.

Otherwise, for all $L$ in a neighborhood of $L_{0}$, $J^{+}\cap L$ is simply connected, and thus it must be a tree. We let ${\cal E}_{L}\neq\emptyset$ denote the set of endpoints of this tree. Now recall that in a neighborhood of $L_{0}\cap J^{+}$, $J^{+}$ is a finite union of disjoint strata $M_{j}=\{r_{j}=0,s_{j}>0\}$. If $M_{j}$ has dimension 3, then for generic $L_{0}$, $M_{j}\cap L$ will have dimension 1 for all $L$ near $L_{0}$. Thus the endpoints ${\cal E}_{L}$ can come only from $M_{j}$ of dimension 2. Since $J^{+}$ is Levi flat, $M_{j}$ must be complex analytic. Further, since $J^{+}$ is Levi flat, we may follow it globally to conclude that $M$ is contained inside $\tilde{M}$, which is a local variety inside the boundary of $J^{+}$. Thus $\tilde{M}\subset J^{+}$ is a subvariety of ${{C}}^{2}$. However, there is no complex subvariety contained in $K^{+}$ (see [BS] or [FS1]), which is a contradiction.

We conclude from the contradiction that for some $L$ close to $L_{0}$, $J^{+}\cap L$ is not simply connected and thus divides $L$ into (at least) two connected components. Only one of these components can be unbounded, so we let $\omega\subset L$ denote a bounded component of the complement of $J^{+}\cap L$. On the other hand, $G^{+}\geq 0$ vanishes on $J^{+}$, so by the maximum principle, $G^{+}=0$ on $\omega$, so $\omega\subset K^{+}=\{G^{+}=0\}$.

*Proof * We continue to let $\psi_{p}:{{C}}\to W^{u}(p)$ denote the uniformization of $W^{u}(p)$, and we define $g(\zeta):=G^{+}(\psi_{p}(\zeta))$, which is subharmonic on ${{C}}$ and satisfies the functional equation $g(\beta\zeta)=d\cdot g(\zeta)$. By Lemma 2.2, we may assume that $J_{p}$ is the real axis. Thus on the upper/lower half plane, $g(\zeta)=c^{\pm}\Im(\zeta)$ for some constants $c^{+}\geq 0$ and $c^{-}\leq 0$, which are not both zero. By the functional equation, we have $c^{+}\Im(\beta\zeta)=d\cdot c^{+}\Im(\zeta)$ if $\beta>0$, so $\beta=d$ in this case. If $\beta<0$, then we have $c^{+}=-c^{-}$, and $\beta=-d$.

Now we will show that one of the $c^{\pm}$ is zero, so we must have $\beta=d$. By Lemma 2.3, we may choose a $L\subset{{C}}^{2}$ such that $K^{+}\cap L$ contains an interior component $\omega$. We may choose a point $r\in W^{s}(p)\cap\partial\omega$ which is a regular point of $\partial\omega$. Further, we may suppose that $L$ is transverse to $W^{s}(p)$ at $r$. Now we let $\Delta\subset L$ denote a small disk containing $r$, so that $\Delta\cap\partial\omega$ is a smooth arc which divides $\Delta$ into two open components. We have $G^{+}=0$ on $\omega\cap\Delta$ and $G^{+}>0$ on the complementary component. Now we apply the Lambda Lemma as we did in Lemma 2.2, and we conclude that $G^{+}=0$ on one of the components of the complement of ${\cal D}_{0}\cap J^{+}\subset W^{u}(p)$. Thus we have $c^{+}=0$ or $c^{-}=0$.

Our Theorem is a consequence of Lemma 2.4: *Proof of the Theorem. * If $J^{+}$ is semianalytic, then by [BK], $f$ must be dissipative (volume decreasing). Then by [BS2], there can be at most one fixed point $p\in{\rm int}(K^{+})$. Thus every fixed point, with at most one exception, is contained in $J^{+}$. By Lemma 2.4, $d$ is a multiplier for $Df$ at each fixed point, except possibly one. However, by Proposition 5.1 of [BK], this is not possible, so $J^{+}$ cannot be semianalytic.

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Eric Bedford

Stony Brook University

Stony Brook, NY 11794

ebedford@math.stonybrook.edu

Kyounghee Kim

Florida State University

Tallahassee, FL 32306

kim@math.fsu.edu