数理解析研究所講究録 1269 巻 2002 年 124-136

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Dynamics of Polynomial Automorphisms ofHerman ring$\mathbb{C}^{2}$:

神貞介 (Teisuke Jin)

東京大学大学院数理科学研究科(University of Tokyo, Graduate School of Mathematical Sciences)
Abstract
Herman ring is aperiodic set which is biholomorphic to an annulus and rotates irrationally by iteration.
Though the structure is known well its existence is unknown. We wiu show that there are no Herman rings undersome conditions in the dynamics of the title.

1Introduction
In this paper we denote $z=(x, y)\in \mathbb{C}^{2}$ . Take an appropriate. Let $Pj(y)$ be polynomials such that degree
for $j=1$ ,, $m$ . We$f_{j}(x, y)=(y,pj(y)-\delta jx)$ generalized Henon mappings, where.
Moreover we define$m\in \mathrm{N}$$d_{j}>1$$\mathrm{c}\mathrm{a}\mathbb{I}$$\ldots$$\delta_{\mathrm{j}}\neq 0$$f=f_{m}\mathrm{o}\cdots \mathrm{o}f_{1}$,$\delta$$=\delta_{1}\cdots\delta_{m}$,$d=d_{1}\cdots d_{m}$Note that is the Jacobian determinant of $f$, and thatme also generalized
In [FM] Friedland and Milnor classified the polynomial automorphisms of$\delta$$f_{j}^{-1}$.

\bullet an

.

.

maps.
into three types:$\mathrm{H}6\mathrm{n}\mathrm{o}\mathrm{n}$$\mathbb{C}^{2}$affine mapping: (x,$y)\mapsto(\mu_{11}x+\mu_{12}y+\lambda_{1},\mu_{21}x+\mu_{22}y+\lambda_{2})$ ,an elementary mapping: (x,$y)\mapsto(\mu_{1}x+\lambda, \mu_{2}y+p(x))$ ,composite of generalized H6non mappings: (x, $y)\mapsto f(x,$y).

Since the dynamical structures of the former two mappings are simple, they were investigated sufficiently in [FM].
So we study the last one.
We define $K^{\pm}=$ { $z\in \mathbb{C}^{2}|\{f^{\pm n}(z)|n$
is boundd},, $K=K^{+}\cap K^{-}$ , $J=J^{+}\cap J^{-}$ . They
are closed invariant sets and are important objects in dynamical systems. Moreover we define.
For $R$ $>0$, we define $V^{+}= \{z\in \mathbb{C}^{2}||x|>\max\{|y|, R\}\}$ , $V^{-}= \{z\in \mathbb{C}^{2}||y|>\max\{|x|, R\}\}$ and$V=\{z\in \mathbb{C}^{2}||x|, |y|\leq R\}$ . It is known thatfor sufficiently large $R$ $>0$ .
We define the Green functionsas (cf. [BS1, Section 3])$J^{\pm}=\partial K^{\pm}$$\in \mathrm{N}\}$$I^{\pm}=\mathbb{C}^{2}\backslash K^{\pm}$$K^{\pm}\subset V\cup V^{\pm}$$G^{\pm}$$G^{\pm}(z)= \lim_{narrow\infty}\frac{1}{d^{n}}\log^{+}||f^{\pm n}(z)||$.

are non-negative continuous plurisubharmonic functions such that $G^{\pm}(z)>0$ if and only if,
are pluriharmonic, and.
Before we define Herman ring, we state aclassification of Fatou components. We proceed with the following$G^{\pm}$$z\in I^{\pm}$$G^{\pm}\mathrm{o}f=d^{\pm 1}\cdot G^{\pm}$volume property.

.
.
.

Proposition 1.1. (FM, Lemma 3.7]) Denote by

or 0, Vol

$if|\delta|<1$ ,then$\mathrm{V}\mathrm{o}\mathrm{l}(K^{+})=\infty$$if|\delta|=1$ ,then$\mathrm{V}\mathrm{o}\mathrm{l}(K^{+})=\mathrm{V}\mathrm{o}\mathrm{l}($$if|\delta|>1$ ,then$\mathrm{V}\mathrm{o}\mathrm{l}(K^{+})=0$, Vol$\mathrm{V}\mathrm{o}\mathrm{l}()$$($the usual Lebesgue volume in$)=0$,$)<\infty$,$($$)=\infty$or 0.

$\mathbb{C}^{2}$. We have$G^{\pm}|_{I^{\pm}}$

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In this paper we assume $|\delta|<1$ , i.e. dissipative. Then only
can have non-empty internals by the above
proposition. We call each component of int
Fatou component. Its classification theorem is as follows.
$K^{+}$$K^{+}$

Theorem 1.2. ([BS2, Section 5]) Each connected component of int $K^{+}is$ classified as follows.
wandering domain$\{$

periodic domain$\{recurrentdomain\{\begin{array}{l}basinofasinkSiegelcylinderHemancylinder\end{array}non- recurrentdomain$

Before we define the names, we mention about existence and non-existence of the above domains.
As far as the author knows, it is unknown whether wandering domains exist or not. Non-recurrent domains). Of course there are basins of sinks. Fornaess
exist and have been investigated only alittle ([Hak, Ul, U2,[FS,
2]
It is unknown whether Herman cylinders
cylinders.
and Sibony investigated in
Section that there are Siegel
exist or not.
The only known fact with respect to non-existence of Herman cylinder is as follows: if $f$ is uniformly hyPer, Theorem 5.6]).
bolic on , then Fatou components consist of basins of finite sinks (cf.
,Let us return to the definition of the names. Fatou component is wandering if $f^{n}(U)\cap U=\emptyset$ for any$f^{p}(U)=U$
say
.
We
is
if
recurrent there arefor some. We call period for the minimum
periodic if.
compact $C\subset U$ and $z\in U$ such that $f^{n}(z)\in C$ for infinitely many
, we defineFor$\mathrm{W}]$$J$$[\mathrm{B}\mathrm{S}1$$U$$p\in \mathrm{N}$$n\in \mathrm{N}$

$p$

$U$

$\mathrm{P}$$n\in \mathrm{N}$$E\subset \mathbb{C}^{2}$$W^{s}(E)=\{z\in \mathbb{C}^{2}|d(f^{n}(z), f(E))arrow \mathrm{O}(narrow\infty)\}$,$W^{u}(E)=\{z\in \mathbb{C}^{2}|d(f^{n}(z), f(E))arrow \mathrm{O}(narrow-\infty)\}$$W_{0}^{s}(E)=.\cup W^{s}(C)c\subset E\cdot \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{t}$’$W_{0}^{u}(E)=.\cup W^{u}(C)C\mathrm{C}E\cdot \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{t}$.

,are lower than 1Letsink if both eigenvalues ofbe aperiodic point with period . We callis a basin of a sink if
in modulus, source if greater, saddle point if one lower and another greater. We say$U=W^{s}(z_{1})$ for some sink.
We call $D$
Siegel disk if $D$ satisfies the following properties: 7) is aperiodic set with period and therefor (6 , where is aunit disk on ais abijective holomorphic map : A $arrow D$ such that
. Here, we have to take the maximum
for some
complex plane and is an irrational rotation, i.e.
$V$ with respect to inclusion. We call $U$ Siegel cylinder if $U=W_{0^{S}}(D)$ for some Siegel disk V.
We call it
Herman ring if $H$ satisfies the following properties: $H$ is aperiodic set with period andfor ( $\in A$ , where $A=\{\zeta\in \mathbb{C}|$there is abijective holomorphic map : $Aarrow H$ such that$r_{1}<|\zeta|<r2\}$ is an annulus and. Here, we have to takefor some
is an irrational rotation, i.e.
the maximum $H$ with respect to inclusion. We call $U$ Heman cylinder if $U=W_{0}^{s}(H)$ for some Herman ring ??.
We have arrived at agood position to describe our question.
$p$$z_{1}$$Df^{\mathrm{p}}(z_{1})$$z_{1}$$U$$z_{1}$$\subset \mathbb{C}^{2}$$p$$f^{p}(\varphi(\zeta))=\varphi(b\zeta)$$\varphi$$b=e^{:\pi\theta}$$b$$\Delta$$\Delta$$\theta\in \mathrm{R}\backslash \mathrm{Q}$$\subset \mathbb{C}^{2}$$p$$\varphi$$b$$f^{p}(\varphi(\zeta))=\varphi(b\zeta)$$b=e^{:\pi\theta}$$\theta\in \mathbb{R}\backslash \mathrm{Q}$

Problem. ([BFGK, Problem 10.2.2(i)]) In case of dissipative, does aH\’enon map admit aHerman ring (Herman
cylinder) ?

In section 2, we will investigate several properties of aHerman cylinder. In particular, Proposition 2.9 will giveaclassification. In Theorem 3.1 of section 3, we will show that one of the types in the classification is impossible.
Perhaps it might be aclue either to prove there are no Herman rings or to construct aHerman ring.

2Fundamental properties of Herman cylinder
2.1 Functional properties
Note that we assume $|\delta|<1$ in this paper. The following is aknown fact

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Proposition 2.1. ([BS2, section 5]) Definesuch that $\Phi(A$ x$A\cross \mathbb{C}arrow W_{0}^{s}(H)$. Then there exists a biholomorphic rrgapoL.

$L( \zeta, \eta)=(b\zeta, \frac{\delta^{\mathrm{p}}}{b}\eta)$$\{0\})=\mathrm{i}\mathrm{t}$or d$f^{p}\circ\Phi=\Phi$$\Phi$:

Proposition 2.2. Let be an arbitrary Fatou component argd $M$ simply connected one dimensional complex
mnifold in . Then $M\cap U$ is simply connected.
$U$$\mathbb{C}^{2}$

Proof. Assume that $M\cap U$ is not simply connected. Then there is apoint

which is surrounded by
on $M$ . By perturbing $M$ to $M’$ , we can takewhich is surrounded by $M’\cap U$ on $M’$ . We
recall the Green function, which vanishes on $M’\cap U$ and is positive on. It contradicts the maximum$M\cap U$$z_{1}\in M\backslash U$$z_{2}\in M’\backslash K^{+}$$G^{+}$$M’\backslash K^{+}$

principle.

Cl

We define $u_{a}(\eta)=u(a, \eta)=G^{-}\circ\Phi(a, \eta)$ for $a\in A$,. Then is asubharmonic function on C.
In general, let be anon-negative bharmonic function. We define the oderby$\eta\in \mathbb{C}$$v$$u_{a}$$\mathrm{o}\mathrm{f}v$$\mathrm{s}.\mathrm{u}$$\mathrm{o}\mathrm{r}\mathrm{d}v=\lim_{rarrow}\sup_{\infty}\frac{\log\max_{|\eta|=r}v(\eta)}{1\mathrm{o}\mathrm{g}r}$

Let be the order of , then we say$v$$\rho$$v$

ismean type of order if$\rho$$0< \lim_{rarrow}\sup_{\infty}\frac{\max_{|\eta|=r}v(\eta)}{r^{\rho}}<\infty$

Proposition 23. For $a\in A$$u_{a}$

mean typeis$u_{a}$.

order:$\rho=\mathrm{o}\mathrm{r}\mathrm{d}u_{a}=\frac{1\mathrm{o}\mathrm{g}d}{\log(\mathrm{l}/|\delta|)}$Proof. It is sufficient to show that,.

is of mean type under the assumption that$\rho=*_{\log 1\prod\delta \mathrm{T}}^{1d}\cdot$.

$| \mathrm{I}=\sup_{\zeta|a|}\lim_{rarrow}\sup_{\infty}\frac{\max_{|\eta|=r}u_{\zeta}(\eta)}{r^{\rho}}\leq 1\dot{\mathrm{m}}\sup_{rarrow\infty}\max\frac{\max_{|\eta|=r}u_{\zeta}(\eta)}{r^{\rho}}|\zeta|=|a|$

For $r>1$ , we take$n\in \mathrm{Z}$such that$1/|\delta|^{\mathrm{p}n}\leq r$$<1/|\delta|^{p\langle n+1)}$.

$| \zeta|=|a|||=|a|\max\max_{\eta}\frac{u_{\zeta}(\eta)}{r^{\rho}}\leq\max\max_{\mathrm{P}(\mathrm{n}+1)}\frac{G^{-}\mathrm{o}\Phi(\zeta,\eta)}{(1/|\delta|^{\mathrm{p}n})^{\rho}}|\zeta|=|a||\eta|=1/|\delta|$$= \max\max_{1^{\mathrm{p}(\mathrm{n}+1)}}\frac{G^{-}\circ f^{-p(n+1)}\mathrm{o}\Phi \mathrm{o}L^{n+1}(\zeta,\eta)}{d^{pn}}|\zeta|=|a||\eta|=1/|\delta$$= \max \mathrm{m}\mathrm{m}\frac{d^{\mathrm{p}(n+1)}\cdot G^{-}\circ\Phi(b^{n+1}\zeta,\eta)}{d^{pn}}|\zeta|=|a||\eta|=1$$=d^{\mathrm{p}} \max\max u_{\zeta}(\eta)|\zeta|=|a||\eta|=1^{\cdot}$

Therefore we have
$1 \dot{\mathrm{m}}\sup_{rarrow\infty}\frac{\max_{|\eta|=r}u_{a}(\eta)}{r^{\rho}}<\infty$.

Similarly we can compute as follows.
$| \zeta|=a|\inf \mathrm{J}\mathrm{i}\mathrm{m}\sup_{farrow\infty}\frac{\max_{|\eta|=r}u_{\zeta}(\eta)}{r^{\rho}}\geq \mathrm{J}\mathrm{i}\mathrm{m}\sup_{farrow\infty}\min\frac{\max_{|\eta|=r}u_{\zeta}(\eta)}{r^{\rho}}|\zeta \mathrm{I}=|a|$

For $r>1$ , we take$n\in \mathrm{Z}$.

such that $1/|\delta|^{pn}\leq r<1/|\delta|^{p(n+1)}$ .

$||= \min_{\zeta}\max_{\eta}\frac{u_{\zeta}(\eta)}{r^{\rho}}\geq\min_{\zeta|a|11=|a||}\max_{=1=|a||\eta|1/1^{\delta}1^{\mathrm{p}\mathrm{n}}}\frac{G^{-}\circ\Phi(\zeta,\eta)}{(1/|\delta|^{p(n+1)})^{\rho}}$$=| \zeta|=|a|\min|\eta|1/|\delta \mathrm{I}^{\mathrm{p}\mathrm{n}}\max_{=}\frac{G^{-}\mathrm{o}f^{-\mathrm{p}n}\mathrm{o}\Phi \mathrm{o}L^{n}(\zeta,\eta)}{d^{p(n+1)}}$$=| \zeta|=|a|\min||=1\max_{\pi}\frac{d^{pn}\cdot G^{-}\mathrm{o}\Phi(b^{n}\zeta,\eta)}{d^{\mathrm{p}(n+1)}}$$=d^{-p} \mathrm{m}\dot{\mathrm{m}}\max u_{\zeta}(\eta)|\zeta|=|a||\eta|=1^{\cdot}$

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Let us show that the last side is positive. Assume for some$a’$with $|a’|=|a|$ ,$u_{a’}||\eta|\leq 1\equiv 0$. Then$u(\{b^{-n}a’\}\cross\{\eta\in \mathbb{C}||\eta|\leq 1/|\delta|^{pn}\})$$=u(L^{-n}(\{a’\}\mathrm{x}\{\eta\in \mathbb{C}||\eta|\leq 1\}))$$=d^{n}u(\{a’\}\mathrm{x}\{\eta\in \mathbb{C}||\eta|\leq 1\})=0$,
is dense inSince
On the other hand, ($;-|_{V}+>0$ and the range of the non-constant holomorphic map. It is acontradiction.
So
Therefore we have$\{\zeta\in \mathbb{C}||\zeta|=|a|\}\cross \mathbb{C}$$\bigcup_{n=0}^{\infty}\{b^{-n}a’\}\mathrm{x}\{\eta\in \mathbb{C}||\eta|\leq 1/|\delta|^{pn}\}$$u_{a}\equiv 0$$\Phi_{a}$.

is contained in$V\cup V^{+}$.

$u_{a}\not\equiv 0$$\lim_{rarrow}\sup_{\infty}\frac{\max_{|\eta|=r}u_{a}(\eta)}{r^{\rho}}>0$.
$\square$

2.2 Formal classification
Let $C=\{c_{1}, \ldots, c_{n}\}$ be afinite ordered subset of ametric space with ametric$d(c_{j}, c_{j+1})<\epsilon$ for any $1\leq j<n$ .

$d$. We call$C$g-chain if

be a positive decreasing sequence converging to 0. Take $\epsilon j^{-}chainCj=\{cj1, \ldots, cjn_{j}\}$.
Lemma 2.4. Let
is compact. Then the -limit set:converges
Assume$\{\epsilon_{j}\}_{j\in \mathrm{N}}$$and\overline{\bigcup_{j=1}^{\infty}C_{j}}$$\{c_{j1}\}_{j\in \mathrm{N}}$$\omega$$k=1j=k\cap\cup^{c_{j}}\infty\overline{\infty}$

is a connected compact set.

and E2 such that $E=$
is not connected. Then there exist compact sets
converges in . Then there. Observe that $d(E_{1}, E_{2})>0$ . We may assume
and.
which accumulates on
is asequencewhich accumulates on
decreases to 0, there is asequence
On the other hand, because. It is acontradiction.

Proof. Assume$E_{1}\cup E_{2}$$E_{1}$$E= \bigcap_{k=1}^{\infty}\overline{\bigcup_{j=k}^{\infty}C_{j}}$$E_{1}$$\{c_{j1}\}_{j\in \mathrm{N}}$$E_{1}\cap E_{2}=\emptyset$$E_{2}$$\{c_{jk_{j}}\}_{j\in \mathrm{N}}$$\{\mathrm{z}\mathrm{r}*$$\{c_{j}\iota_{j}\}_{j\in \mathrm{N}}$$\{\epsilon_{j}\}$$\in$$\square$$\overline{\cup C_{j}}|\min\{d(w, E_{1}), d(w, E_{2})\}\geq \mathrm{d}\{\mathrm{E}\mathrm{x}, E_{2})/3\}$

Lemma 2.5. Let

curve

be a closed subset andwhich winds once.

$X\subset \mathbb{R}^{2}$$\Gamma\subset \mathbb{R}^{2}\backslash X$$\mathrm{Y}$a compact component of X. Then there is a simple closed$\mathrm{Y}$Proof. At first we show that there is$\epsilon$$>0$ such that the subset$\mathrm{o}\mathrm{f}X$

which can be joined to$\mathrm{Y}$by g-chain on $X$ iscompact.
and $r>0$ withconverging to 0and
Assume the contrary. Take apositive decreasing sequence$C_{j}\subset
X$
point of
start
the
that
such
-chainwe can take
. By the assumption, for any
thattheconnected
lemma,
concludewecan$C_{j}\subset
B(w_{1},2r)$. By the previous
and
and
is
$B(w_{1},r)$. It is acontradiction.
exceeds
component of $X$ containing
is at least far
be the compact set which can be joined to by -chain. Each point on
Let
once.
winds
which
from . It is not difficult to find asimple closed curve$w_{1}\in \mathrm{Y}$$\{\epsilon_{j}\}$$C_{j}$$j\in \mathrm{N}$$\mathrm{Y}\subset B(w_{1}, r)$$\epsilon_{j}$$C_{j}\backslash B(w_{1}, r)\neq\emptyset$$w_{1}$$w_{1}$$\mathrm{Y}$$\mathrm{Y}’\subset X$$X\backslash \mathrm{Y}’$$\epsilon$

We proceedjvith investigatingthe structure of aHerman cylinder.
We define $K=\Phi^{-1}(K^{-})$ , $K_{a}=\{\eta\in \mathbb{C}|\Phi(a, \eta)\in K^{-}\}$ for $a\in A$ . Note that$u(\zeta, \eta)=0\},\overline{K}_{a}=\{\eta\in \mathbb{C}|u_{a}(\eta)=0\}$

Definition 2.6. We say$\overline{K}_{a}$$\overline{K}_{a}$is bridged if the component ofis bridged.

(2) The component of$\overline{K}_{a}$$\tilde{K}=\{(\zeta, \eta)\in A\mathrm{x}\mathbb{C}|$.

Lemma 2.7. The following are equivalent.
(1)$\epsilon$$\square$$\mathrm{Y}’$$\Gamma\subset \mathbb{C}\backslash X$$\mathrm{Y}’$containing 0is not a point$\overline{K}_{a}$containing 0is unbounded.

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(3)

$\overline{K}_{a}$has an unbounded component.

Proof. (2)$\Rightarrow(1)$. Assume the component ofcontaining 0is bounded, i.e. the component is contained infor some $r>0$ . Take an increasing sequencesuch that$\overline{K}_{a}$$B(0,r)=\{\eta\in \mathbb{C}||\eta|<r\}$$b^{-n_{\mathrm{j}}}a$$\{n_{j}\}_{j\in \mathrm{N}}\subset \mathrm{N}$converges to . Then$a$$\{b^{-n_{j}}a\}\mathrm{x}\tilde{K}_{b^{-\mathrm{n}_{\mathrm{j}_{0}}}}=L^{-n_{j}}(\{a\}\mathrm{x}\tilde{K}_{a})=\{b^{-n_{\mathrm{j}}}a\}\mathrm{x}(b/\delta^{p})^{n_{\dot{f}}}\tilde{K}_{a}\subset\tilde{K}$.

Let
be apositive sequence decreasing to 0. We take -chain
inso that thestarting point
isand $C_{j}\subset\{b^{-n_{j}}a\}\mathrm{x}B(0,2r)$ . Moreover we can assume
for any sufficiently large because of the hypothesis (2). By Lemma 2.4 we can conclude that the component ofcontaining 0exceeds $B(0, r)$ . It is acontradiction.
(3). Take an increasing sequencesuch thatconverges to . Then$\{\epsilon_{j}\}_{j\in \mathrm{N}}$$\{b^{-n_{\mathrm{j}}}a\}\mathrm{x}(b/\delta^{p})^{n_{\dot{f}}}\tilde{K}_{a}$$C_{j}$$\epsilon_{j}$$(b^{-n_{\mathrm{j}}}a, 0)$$\mathrm{o}\mathrm{f}C_{j}$$C_{j}\not\subset\{b^{-n_{f}}a\}\mathrm{x}B(0, r)$$j$$\overline{K}_{a}$$\Rightarrow(2)$$b^{n_{j}}a$$\{n_{j}\}_{j\in \mathrm{N}}$$a$$\{b^{n_{j}}a\}\mathrm{x}\tilde{K}_{b^{\mathrm{n}_{j}}a}=\{bnja\}\mathrm{x}(\delta^{p}/b)^{n_{\dot{f}}}\tilde{K}_{a}$Let $E$ be an unbounded component of. We can takepositive sequence decreasing to 0. We take -chain
inconverges $(a, 0)$ andandthe component ofcontaining 0is not apoint.
$\overline{K}_{a}$$r$$C_{j}$$\epsilon_{\mathrm{j}}$$C_{j}\subset\{b^{n_{\dot{f}}}a\}\mathrm{x}B(0,2r)$

.
. Let

be a
so that the starting point of. By Lemma 2.4, we can conclude that$>0$ such that $B(0, r)$ $\cap E\neq\emptyset$$\{b^{n_{\dot{f}}}a\}\mathrm{x}$$\{\epsilon \mathrm{j}\}_{\mathrm{j}\in \mathrm{N}}$$(\delta^{p}/b)^{n_{\mathrm{j}}}E$$C_{j}\not\subset\{b^{n_{j}}a\}\mathrm{x}B(0,r)$$C_{j}$$\overline{K}_{a}$$\square$Lemma 2S. For a $\in A$ the following hold.
(1) If(2) If$\overline{K}_{a}$$\tilde{K}_{a}$has no compact components, then so isis bridged, then so is$\overline{K}_{\zeta}$

for any$\overline{K}_{\zeta}$$|\zeta|=|a|$for any$|\zeta|=|a|$.

.

Pmof. The proof of (2) is similar to the previous lemma. So we give only the proof of (1).

To prove (1), we show thathas compact component then so is, for $|a’|=|a|$ . By Lemma 2.5, thereis acurve which surrounds the component and never intersects. Takesurrounded by . Then there
existssuch thatnever intersects .
Consider . Note that $u\zeta(\eta)=0$ if and only if$u_{\zeta}(\eta)>0$ . When{. We define
is sufficiently small,for any with. Recall that is harmonic inand continuous
on C. So has zero points inside of , i.e. for withhas at least one compact component inside
of .
Take
such that $|b^{n}a’-a|\leq\epsilon$. Then$\mathrm{i}\mathrm{f}\overline{K}_{a}$$\tilde{K}_{a’}$$\Gamma$$\tilde{K}_{a}$$\epsilon>0$$\eta_{1}\in\overline{K}_{a}$$\overline{K}$$\Gamma_{\epsilon}=\{\zeta\in A||\zeta-a|\leq\epsilon\}\mathrm{x}\Gamma$$\eta\in\tilde{K}$$\mathrm{u}\mathrm{c}$$\Gamma$.

$\epsilon>0$$c=\dot{\mathrm{m}}\mathrm{n}_{(\zeta,\eta)\in\Gamma}$$u_{\zeta}(\eta_{1})<c$$|\zeta-a|\leq\epsilon$$\zeta$$\Gamma$$u\zeta$$\mathbb{C}\backslash \tilde{K}_{\zeta}$$u\zeta$$|\zeta-a|\leq\epsilon,\tilde{K}_{\zeta}$$\langle$$\Gamma$$n\in \mathrm{N}$$\overline{K}_{a’}=(b/\delta^{p})^{n}\overline{K}_{b^{\mathrm{n}}a’}$

Because$\tilde{K}_{b^{\mathrm{n}}a’}$has compact$\mathrm{c}\mathrm{o}\mathrm{m}\mu \mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}$so is$\tilde{K}_{a’}$

.

.

Cl

We obtain the following classification.

Proposition 2.9. For
(1)

(2)

$\overline{K}_{a}$$\overline{K}_{a}$$a\in A\overline{K}_{a}$is classified into the following three types:

has no compact components,

is bridged and has compact components,(3) each component ofMoreover, for any$\langle$with$\tilde{K}_{a}$is compact.

$|\zeta|=|a|,\tilde{K}_{a}$and$\tilde{K}_{\zeta}$

are classified into the same category.

2.3 Continuity about irrational rotation
In the following, we show several kinds of continuities of sets along irrational rotation.

Lemma 2.10. Take , $a’\in A$ with $|a|=|a’|$ . Let
Then
$a$

$\{n_{j}\}_{j\in \mathrm{N}}\subset \mathrm{N}$be a sequence such that $ab^{n_{g}}arrow a’$ as $jarrow\infty$ .

$k=1j=k \cap\cup\infty\infty(\frac{\delta^{p}}{b})^{n_{\mathrm{j}}}\tilde{K}_{a}\subset\tilde{K}_{a’}$.

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Proof. Since$\{ab^{n_{\mathrm{j}}}\}\mathrm{x}(\frac{\delta^{p}}{b})^{n_{j}}\overline{K}_{a}=L^{n_{j}}(\{a\}\mathrm{x}\overline{K}_{a})\subset\overline{K}$,

we have$j=k \cup\{ab^{n_{j}}\}\mathrm{x}\infty(\frac{\delta^{p}}{b})^{n_{j}}\overline{K}_{a}\subset\overline{K}$,.

$k=1j=k \cap\cup\{ab^{n_{\mathrm{j}}}\}\cross\infty\infty(\frac{\delta^{p}}{b})^{n_{j}}\overline{K}_{a}\subset\{a’\}\mathrm{x}\overline{K}_{a’}$On the other hand, we take$\eta\in k=1j=k\cap\cup\infty\infty(\frac{\delta^{p}}{b})^{n_{j}}\overline{K}_{a}$This means that for any$>0$ and$\epsilon$$k\in \mathrm{N}$.

, there is $l\geq k$ such that$d$(,$( \frac{\delta^{p}}{b})^{n\iota}\overline{K}_{a})<\frac{\epsilon}{2}$$\eta$.

Then, there exists $j\geq k$ such that$d$($(a’, \eta)$,$\{ab^{n_{\mathrm{j}}}\}\mathrm{x}(\frac{\delta^{p}}{b})^{n_{j}}\overline{K}_{a}$)$<\epsilon$.

This implies$(a’, \eta)\in\cap\cup\{ab^{n_{\mathrm{j}}}\}\mathrm{x}k=1j=k\infty\infty(\frac{\delta^{p}}{b})^{n_{j}}\overline{K}_{a}$.

Therefore$\{a’\}\mathrm{x}\cap\cup k=1j=k\infty\infty(\frac{\delta^{p}}{b})^{n_{\mathrm{j}}\infty\infty}\overline{K}_{a}\subset\cap\cup\{ab^{n_{j}}\}\mathrm{x}k=1j=k(\frac{\delta^{p}}{b})^{n_{j}}\overline{K}_{a}$.

We obtain the assertion.
We define$\square$$\overline{I}=A\mathrm{x}\mathbb{C}\backslash \overline{K}$and$\overline{I_{a}}=\mathbb{C}\backslash \overline{K}_{a}$for $a\in A$ . Under the hypothesis of the above lemma, we have

$k= \mathrm{l}j=k\cup \mathrm{i}\mathrm{n}\mathrm{t}\cap\infty\infty(\frac{\delta^{p}}{b})^{n_{j}}\overline{I}_{a}\supset\overline{I}_{a’}$.

More precisely we obtain the following.

Proposition 2.11. Take , $a’\in A$ with $|a|=|a’|$ . Let$jarrow\infty$ . Then each componentof$a$$k=\mathrm{l}\cup \mathrm{i}\mathrm{n}\mathrm{t}\infty$

is either in agreement with a component of$\overline{I}_{a’}$$\{n_{j}\}_{j\in \mathrm{N}}\subset \mathrm{N}$$j=k \cap\infty(\frac{\delta^{p}}{b})^{n_{\mathrm{j}}}\overline{I_{a}}$or contained in$\overline{K}_{a’}$.

be a sequence such that $ab^{n_{j}}arrow a’$ as130Proof. Take an arbitrary component$I_{1}$

ffom.

$k=1j=k \cup \mathrm{i}\mathrm{n}\mathrm{t}\cap\infty\infty(\frac{\delta^{p}}{b})^{n_{j}}\tilde{I_{a}}$We may assume contains some component
such that for any $j\geq k$ ,$I_{1}$$\mathrm{o}\mathrm{f}\tilde{I_{a’}}$. Take acompactly contained open set$V\subset\tilde{I_{a’}}$. Then there is$k\in \mathrm{N}$, i.e.

$( \frac{\delta^{p}}{b})^{n_{j}}\tilde{I_{a}}\supset V$

Since

$u$

is continuous on A$\mathrm{x}\mathbb{C}$.

$\tilde{I}_{ab^{\mathrm{n}_{j}}}\supset V$,$u_{ab^{\mathrm{n}_{j}}}|_{V}arrow u_{a’}|_{V}$uniformly as $jarrow\infty$ . On the other hand, since$u_{ab^{\mathrm{n}_{\mathrm{j}}}}$is harmonic on$\tilde{I_{ab^{\mathrm{n}_{\mathrm{j}}}}}$,$u_{ab^{\mathrm{n}_{\dot{f}}}}\in H(V)$for any $j\geq k$ . Because$V\subset\subset I_{1}$is arbitrary,$u_{a’}\in \mathcal{H}(I_{1})$contains some component of , on which
is apositiv\^e harmonic function, and. Because is connected, coincides with some component of .

Recall that$\mathbb{C}\backslash \overline{I_{a’}}$.

$I_{1}$$\tilde{I_{a’}}$$u_{a’}$$I_{1}$$I_{1}$$u_{a’}$

vanishes onCl$\tilde{I_{a’}}$For $c>0$ , we define,$\tilde{I_{a}’}=\{\eta\in \mathbb{C}|u_{a}(\eta)>c\}$which is asubset of$\overline{I}_{a}$. The next lemma tells that$\tilde{I_{a}’}$Lemma 2.12. Take , $a’\in A$ with $|a|=|a’|$ . lLetas
. Then$a$plays role similar to$\{n_{j}\}_{n\in \mathrm{N}}\subset \mathrm{N}$$\overline{I_{a}}$.

be an increasing sequence such that$ab^{n_{\mathrm{j}}}arrow a’$$jarrow\infty$

intk=1$$\cap$\tilde{I_{a’}}\subset\cup$$j=k \infty\infty(\frac{\delta^{p}}{b})^{n_{\dot{f}}}\tilde{I_{a}’}\subset\cup \mathrm{i}\mathrm{n}\mathrm{t}\cap(k=\mathrm{l}\mathrm{j}=k\frac{\delta^{p}}{b})^{n_{\dot{f}}}\tilde{I_{a}}\infty\infty$.

Moreover, each component of the middle side is either in agreement with some component$K_{a’}$.

Proof It is sufficient to show the left inclusion.
$\eta\in\overline{I_{a}’}$if and only if $u_{a}(\eta)>c$ . Therefore$\eta\in(\frac{\delta^{p}}{b})^{n_{f}}\tilde{I_{a}’}\Leftrightarrow(ab^{n_{\mathrm{j}}}, \eta)\in\{ab^{n_{\dot{f}}}\}\mathrm{x}(\frac{\delta^{p}}{b})^{n_{j}}\tilde{I_{a}’}$$\Leftrightarrow(ab^{n_{\mathrm{j}}}, \eta)\in L^{n_{f}}(\{a\}\mathrm{x}\tilde{I_{a}’})$$\Leftrightarrow L^{-n_{\dot{f}}}(ab^{n_{\mathrm{j}}}, \eta)\in\{a\}\mathrm{x}\tilde{I_{a}’}$$\Leftrightarrow u(L^{-n_{\dot{f}}}(ab^{n_{\dot{f}}}, \eta))>c$.

$\Leftrightarrow ff^{\iota_{\dot{f}}}u(ab^{n_{\mathrm{j}}}, \eta)>c$

Take$\eta_{1}\in\overline{I}_{a’}$. Then there is$\epsilon_{1}>0$

with$B(\eta_{1},3\epsilon_{1})\subset\tilde{I_{a’}}$. Note that$u_{a’}|_{B(\eta_{1},3\epsilon_{1})}>0$

Since

$u$

is continuous, there is$k_{1}\in \mathrm{N}$.

such that for any $j\geq k_{1}$ ,$u_{ab^{\mathrm{n}_{\mathrm{j}}}}|_{B(\eta_{1},2\epsilon_{1})}>0$.

of$\tilde{I_{a’}}$or contained in

131
Moreover there exists$k_{2}\in \mathrm{N}$for arbitrary $j\geq k_{2}$ ,$u_{ab^{n_{j}}}|_{B(\eta_{1\prime}\epsilon_{1})}> \frac{c}{d^{n_{j}}}$,$\cdot.\mathrm{e}$.

$B( \eta_{1}, \epsilon_{1})\subset(\frac{\delta^{p}}{b})^{n_{j}}\overline{I_{a}’}$.

Therefore we have$\eta_{1}\in \mathrm{i}\mathrm{n}\mathrm{t}\cap j=k_{2}\infty(\frac{\delta^{p}}{b})^{n_{j}}\overline{I’}_{a}$.
ClThis implies the left inclusion in the assertion.

3In case thathas no compact components$\overline{K}_{a}$We have the following non-existence of Herman rings.

Theorem 3.1. The case (1) in Proposition 2.9 is impossible.
Corollary 3.2. The case that$\tilde{K}_{a}$is connected is impossible.

is holomorphic on(cf. [MNTU, Section 7.3]).
To prove the theorem we use aBottcher function$\log|\varphi^{-}(z)|=G^{-}(z)$for $z\in V^{+}$ . Thereandfor sufficiently large $R>0$ , and satisfies$|w|(w\in\mathbb{C})$
$(x,
y)\in
V^{+}$
is sufficiently large,. Whenis $M>1$ such that $1/M\leq|\varphi^{-}(x, y)|/|x|\underline{<}M$ for.
$\{z\in V^{+}|\varphi^{-}(z)=w\}$ is asimply connected one dimensional complex manifold in. Note that $\log|\psi_{a}|=u_{a}$ .
We use anotation$V^{+}$$\varphi^{-}$$\varphi^{-}$$\varphi^{-}\circ f^{-1}(z)=(\varphi^{-}(z))^{d}$$V^{+}$$\psi\zeta(\eta)=\psi(\zeta, \eta)=\varphi^{-}\circ\Phi(\zeta, \eta)$

Lemma 33. If$\overline{K}_{a}$has no compact components, then$\nabla u_{a}(\eta)\neq(0,$.
for any
Proof. Assume the contrary i.e. there is0)$\eta\in\overline{I}_{a}$.
such that$\{\eta\in\mathbb{C}|u_{a}(\eta)>c\}$ is at most $\max\{1,2\rho\}$By Proposition 2.3, for any $c>0$ the number of components of(cf. [Hay, Theorem 8.9]). Since the number is monotone increasing along $c>0$ , we can take $c>0$ so that the
number attains its maximum.
. In fact, let us assume the contrary, i.e. there ishas no zero points in
Thensuch thatwith $\nabla u_{a}(\eta_{1})=(0,0)$ . Define $c’=u_{a}(\eta_{0})$ . There are $n\geq 2,0\leq\theta<2\pi$ and$\eta_{0}\in\overline{I}_{a}$$\frac{\partial u}{\partial\eta}(\eta_{0})=0$$\overline{I’}_{a}=\{\eta\in \mathbb{C}|u_{a}(\eta)>c\}$$\frac{\partial u}{\partial\eta}$$\epsilon_{1}>0$$\eta_{1}\in\overline{I’}_{a}$$u_{a}(\eta_{0}+t\exp(i(\theta+2\pi j/n)))>c’$,$u_{a}(\eta_{0}+t\exp(i(\theta+2\pi(j+1/2)/n)))<c’$ ,for any $0\leq j<n$ and $0<t<2\epsilon_{1}$ . Define$I_{1}=${$\eta\in \mathbb{C}|u_{a}(\eta)>c’$,$\eta$

is in the component ofis aconnected open set. Moreover,Because of the definition of ,which joins
Therefore there is an arc$c$$I_{1}$$\overline{I’}_{a}$$I_{1}$

containing$\eta_{0}$}.

is simply connected because so is$\overline{I}_{a}$.

$\Gamma\subset I_{1}$$\eta_{0}+\epsilon_{1}\exp(i\theta)$

and $\eta_{0}+\epsilon_{1}\exp(i(\theta+2\pi/n))$ .

so that $u_{a}\geq c’$ on
We can extend in aneighborhood of 770 and obtain aclosed curve
inside of on which $u_{a}<c’$ . It is acontradiction.
as $jarrow\infty$ . Then forbe asequence such thatLet$\Gamma’$$\Gamma$$\Gamma’$. But there is apoint$\Gamma’$$\{n_{j}\}_{j\in \mathrm{N}}\subset \mathrm{N}$$\eta\in(\delta^{p}/b)^{n_{j}}I_{a}^{\overline{\prime}}$$ab^{n_{\mathrm{j}}}arrow a$$u_{ab^{n_{j}}}(\eta)=u(ab^{n_{j}}, \eta)$$=u(L^{n_{j}}(a, (b/\delta^{p})^{n_{\mathrm{j}}}\eta))$$=d^{-n_{j}}u(a, (b/\delta^{p})^{n_{j}}\eta)$$=d^{-n_{\mathrm{j}}}u_{a}((b/\delta^{p})^{n_{\mathrm{j}}}\eta)$

.

,

132

$\frac{\partial u_{ab^{\mathrm{n}_{j}}}}{\partial\eta}(\eta)=d^{-n_{\mathrm{j}}}(\frac{b}{\delta^{p}})^{n_{j}}\frac{\partial u_{a}}{\partial\eta}((b/\delta^{p})^{n_{j}}\eta)\neq 0$

On the other hand, by Lemma 2.12, there are$\epsilon_{0}>0$

and$k\in \mathrm{N}$.

such that for any $j\geq k$ ,$B( \eta_{0}, \in 0)\subset(\frac{\delta^{p}}{b})^{n_{j}}\tilde{I_{a}’}$.

Because is continuous,converge to uniformly inas $jarrow\infty$ .
When harmonic functions
converge to anon-constant harmonic function,
uniformly on.
then antiholomorphic functions
converge touniformly. By Hurwitz theorem, each zero point of.
.
is an accumulation point of zero points of. But
has no zero points infor any $j\geq k$ . It is
acontradiction.
$u$$B(\eta_{0},\epsilon_{0})$$u_{a}$$u_{ab^{\mathrm{n}_{\mathrm{j}}}}$$u_{ab^{\mathrm{n}_{\mathrm{j}}}}$$B(\eta_{0},\epsilon_{0})$$u_{a}$$\partial u\mathrm{n}$$\frac{\partial u}{\partial\eta}$$\overline{\partial}^{\frac{f}{\eta}}$$\frac{\partial u}{\partial\eta}$$\partial u\mathrm{n}$$\partial u\mathrm{n}$$\overline{\partial}^{\frac{f}{\eta}}$$\overline{\partial}^{\frac{f}{\eta}}$$B(\eta_{0},\epsilon 0)$$\square$We use anotation H$=\{\xi\in \mathbb{C}$|${\rm Re}\xi>0\}$.

Lemma 3.4. Assume
has no compact components. Then for arry component
number of components of$\tilde{K}_{a}$$I_{0}$

of$\tilde{I_{a}}$argdfor any $c>0$ the$\{\eta\in I_{0}|u_{a}(\eta)>c\}$is exactly one.

Proof. Since I0 is simply connected, there is g$\in O(I_{0})$such that ${\rm Re} g=u_{a}$ . Then

$g:I_{0}arrow \mathrm{E}$.

By the previous lemma, for each c $>0$ its level set$\{\eta\in I_{0}|u_{a}(\eta)=c\}$

is aset of smooth simple arcs, whose all ends go to infinity. Therefore :is locally biholomorphic andproper. This implies is bijective. Hence the above each level set consists of single arc. We obtain the required$g$$I_{0}arrow \mathrm{f}\mathrm{f}\mathrm{i}$$g$result.

Cl

Lemma $3S$. Assume$\tilde{K}_{a}$to definehas no compact components. &t$\log\psi_{a}$:$I_{0}$

be an arbitrary component of$\tilde{I_{a}}$. Then it is possible$I_{0}arrow \mathrm{E}$by analytic continuation. Moreover it is biholomorphic.

Proof For sufficiently large c $>0$ ,is defined on$\psi_{a}$$I_{0}’=\{\eta\in I_{0}|u_{a}(\eta)>c\}$,because the set is contained in $\Phi^{-1}(V^{+})$ . Since is simply connected, 10g
is well-defined on .
We take 9used in the previous proof. Because ${\rm Re}\log\psi_{a}=u_{a}$ , $\log\psi_{a}-g$ is apurely imaginary constant.
Then
can be analytic continued to because is connected. Since :is biholomorphic, so is$I_{0}’$$\log\psi_{a}$$\log\psi_{a}$$I_{0}$$\psi_{a}$$I_{0}’$

$g$

.

.

$\overline{I}$Proposition 3.6. For any $a\in A$ there are$N\in \mathrm{N}$and a closed curve$\gamma:[0, N]arrow\overline{I}$such that$\pi_{A}0\gamma(t)=a\exp(2\pi\dot{|}t)$$\pi_{A}$$I_{\mathit{0}}arrow \mathrm{H}$$\square$Let us investigate the structure ofwhere$I_{0}’$:A$\mathrm{x}\mathbb{C}$$arrow A$

is a mural projection,

133
Note that we take the minimum $N\geq 1$ when we use the proposition.

has . We know $q’ \leq\max\{1,2\rho\}<\infty$ . Takesuch thatto distinct components of . There is small

Proof. Assume$q’$$\overline{I}_{a}$$\eta_{1}$,$\ldots$,$\epsilon_{0}>0$$\overline{I}_{a}$$T=\cup\{(ae^{it}, \eta_{j})j=1q’|0\leq t\leq 2\epsilon_{0}\}\subset\overline{I}$We define asequence$\{n_{j}\}_{j}\subset \mathbb{Z}$as follows. We take$n_{1}\in \mathbb{Z}$$n_{2}\in \mathbb{Z}$.

such that$\epsilon_{0}\leq\arg ab^{n_{1}}-\arg a\leq 2\epsilon_{0}$Then we takeso that any two of them belong$\eta_{q’}\in\overline{I}_{a}$$(\mathrm{m}\mathrm{o}\mathrm{d} 2\pi)$.

such that$\epsilon_{0}\leq\arg ab^{n_{2}}-\arg ab^{n_{1}}\leq 2\epsilon_{0}$$(\mathrm{m}\mathrm{o}\mathrm{d} 2\pi)$

By repeating the procedure, we return to the starting point, i.e. there is$0<\arg a-\arg ab^{n_{k}}\leq 2\epsilon_{0}$such that$k\in \mathrm{N}$$(\mathrm{m}\mathrm{o}\mathrm{d} 2\pi)$.

.

along $T$ . Chooseto, draw an arc ffom
Then we can draw arcs in as follows. For, then draw arcs joining the two points
are in the same component ofandso that. By repeating thealongtoin the component. In the sequel draw an arc from,.
,toprocedure, we can draw an arc ffom eachcontaining the same , we can draw aarc i thethe arc returns to the component ofIf for some
$N$ times thecomponent joining the start point and the end point, and obtain aclosed curve. Otherwise repeat
can draw a
weSo
point.
above procedure and at last some end point arrives at the same component of its start
closed curve.
Finally by perturbing the closed curve we obtain as required.
$(ab^{n_{1}}, \eta_{j\mathrm{o}})$$(a, \eta_{j\mathrm{o}})$$\overline{I}$$\eta_{j\mathrm{o}}$$\overline{I}_{ab^{\mathrm{n}_{1}}}$$(\delta^{p}/b)^{n_{1}}\eta_{j_{1}}$$(\delta^{p}/b)^{n_{1}}\eta_{j_{1}}$$\eta_{j\mathrm{o}}$$L^{n_{1}}(T)$$(ab^{n_{2}}, \eta_{j_{1}})$$(ab^{n_{1}}, \eta_{j_{1}})$$\overline{I_{a}}$$\eta_{q’}$$\ldots$$\eta_{1}$$\overline{I}_{a}$$\mathrm{n}$$\eta_{j}$$\eta_{j}$$\square$$\gamma$:
has no compact components, 7in Proposition 3.6 is unique in the following sense. &t
Lemma 3.7. Ifcomponent
in
the
same
are
$[0, N’]arrow\overline{I}be$andanother closed curve satisfying the same condition, and.
component
same
the
in
$t\in[0,
N]$
are
,
and
of
any
$N=N’$
Then
andforof .
$’\sqrt$$\overline{K}_{a}$$\gamma’(0)$$\gamma(0)$$\overline{I_{a\exp(2\pi it)}}$$\gamma’(t)$$\gamma(t)$$\overline{I_{a}}$Proof. We may suppose $N\leq N’$ . Let us assume for some $t\in[0, N]$ ,, and derive acontradiction.
By iteration of $L^{-1}$ , we may assume$\gamma(t)$and$\gamma’(t)$are in distinct components of$\tilde{I_{a\exp(2\pi\dot{l}t)}}$$\gamma$,$\gamma’\subset\Phi^{-1}(V^{+})$. Then$\psi$

is defined in aneighborhood of and$\gamma$$\gamma’$.

The set{t
is open. In fact, take$t_{1}$$\in[0,$N]|$7(\mathrm{t})$and$\gamma’(t)$are in the same component of

ffom the set. Then there is an arc C$\subset\overline{I}_{a\exp(2\pi\dot{|}t_{1})}$$\overline{I}_{a\exp(2\pi}:t)$joining$\gamma(t_{1})$}and$\gamma’(t_{1})$. Because$|\zeta|=|a|\cup\overline{I_{\zeta}}$is open in $\{\zeta\in A||\zeta|=|a|\}\mathrm{x}\mathbb{C}$, aneighborhood of $C$ is also contained in the above open set. So in a.
are joined by an arc inandneighborhood of ,define$t_{1}$$t_{2}= \min$(a,$\eta_{2})=7(\mathrm{t}$) and (a,$\overline{I_{a\exp(2\pi it)}}$$\gamma’(t)$$\gamma(t)${t$\in[0,$N]|$\eta_{2}’)=7;(\mathrm{t}\mathrm{i})$$7(\mathrm{t})$and. Take$\epsilon_{1}$,$\gamma’(t)$are in distinct components of$\epsilon_{2}>0$$\{a_{2} \exp(2\pi it)|-\epsilon_{1}<t\leq 0\}$$\{a_{2} \exp(2\pi it)|-\epsilon_{1}<t\leq 0\}$$\overline{I}_{a\exp(2\pi\dot{\iota}t)}$such that$\mathrm{x}$,$B(\eta_{2}, \epsilon_{2})\subset\Phi^{-1}(V^{+})\subset\overline{I}$,$\mathrm{x}B(\eta_{2}’, \epsilon_{2})\subset\Phi^{-1}(V^{+})\subset\overline{I}$$(\{a_{2}\exp(2\pi it)|-\epsilon_{1}<t\leq 0\}\cross\partial B(\eta_{2},\epsilon_{2}))\cap\gamma=\emptyset$,$(\{a_{2}\exp(2\pi it)|-\epsilon_{1}<t\leq 0\}\mathrm{x}\partial B(\eta_{2}’, \epsilon_{2}))\cap\gamma’=\emptyset$

.

},

134
By Lemma 3.5, for each
{a2 $\exp(2\pi it)$ | $-\epsilon_{1}<t<0$ },containing and . We choose the branches of the logarithms so that$\zeta\in$is well-defined in the component of$\log\psi_{\zeta}$$\eta_{2}’$$\eta_{2}$$\log\psi\zeta(\eta_{2})$to \langle . Then$\overline{I_{\zeta}}$varies continuously with respect$\log\psi_{\zeta}(\eta_{2}’)$varies continuously on $\zeta\in\{a_{2}\exp(2\pi:t)|-\epsilon_{1}<t<0\}$. Moreover since
uniformly in aneighborhood of , there is such that$\eta_{2}’$$\psi_{\zeta}$

converges to$\psi_{a_{2}}$as \langle$arrow a_{2}$$\xi_{2}$$\log\psi_{\zeta}(\eta_{2}’)arrow\xi_{2}$as$\zetaarrow a_{2}$.

On the other hand, there is$\eta_{3}$in the component ofcontaining 772 such that$\overline{I_{a_{2}}}$$\log\psi_{a_{2}}(\eta_{3})=\xi_{2}$where$\log\psi_{a_{2}}$is defined so thatas$\log\psi_{\zeta}(\eta_{2})arrow\log\psi_{a_{2}}(\eta_{2})$,$\zetaarrow a_{2}$. Observe that$\log\psi_{\zeta}(\eta_{3})arrow\xi_{2}$as$\zetaarrow a_{2}$, because$\psi$

is continuous in aneighborhood of$(a_{2}, \eta_{3})$Therefore, bothandconverge toHence
and $\sqrt(N)$ are in the same component of
to $\sqrt(0)$ , i.e. $N=N’$ .
$\log\psi_{\zeta}(\eta_{2}’)$$\log\psi_{\zeta}(\eta_{3})$$\gamma(N)$Lemma $3S$. AssumeThen$\Phi(\gamma)$of$\tilde{I}_{a}$$\log$

has no compact components. Ta&an arbitrary closed curve

has finite components, there issuch that bothe 7and
. We know by Lemma 3.7 that for each $t\in[0, N]$$\tilde{I_{a}}$$q\in \mathrm{N}$$\gamma(t)$and$\psi_{\zeta}$$\square$is trivial in $\pi_{1}(I^{-})$ .

Proof. Since$\overline{I_{a}}$$K\sim a$.

. It contradicts with the injectivity of.
. We can draw acurve in the component fiom $\sqrt(N)$$\xi_{2}$$L^{q}(\gamma)$$\gamma$as in Proposition 3.6.

intersect acommon component(mod N)$L^{q}(\gamma(t+\mathrm{t}\mathrm{o}))$are in the same component offor some.
We can draw acurve in.betweenand. We can extend the curve along to asnip, i.e.
and $L^{q}(\gamma(t+t_{0}))$ are loeally homotopic. since eachof is simply connected, we canjoin the homotopies and have that 7andare homotopic in .
On the other hand, there is an isomorphism :such that$\tilde{I}_{\exp(2\pi\dot{|}t)}$$t_{0}\in \mathrm{R}$$\overline{I_{a\exp(2\pi\cdot t)}}$$\gamma(t)$$L^{q}(\sqrt(t+t\mathrm{o}))$$t$$\gamma(t)$$\tilde{I_{\zeta}}$$\mathrm{c}\mathrm{o}\mathrm{m}\mu \mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}$$\overline{I}$$L^{q}(\gamma)$$\alpha$$\pi_{1}(I^{-})arrow \mathrm{Z}[_{\mathrm{a}}^{1}]$$\alpha(f(C))=\frac{1}{d}\alpha(C)$for any $C\in\pi_{1}(I^{-})$ (cf. [MNTU, Section 7.3]).

Since 7and$L^{q}(\gamma)$

are homotopic in$\tilde{I}=\Phi^{-1}(I^{-})$,$\Phi(\gamma)$

and$\Phi(L^{q}(\gamma))$are homotopic in$\frac{1}{d^{\mathrm{p}q}}\alpha(\Phi(\gamma))=\alpha(f^{m}(\Phi(\gamma)))=\alpha(\Phi(L^{q}(\gamma)))=\alpha(\Phi(\gamma))$$I^{-}$. We obtain.

Therefore $\alpha(\Phi(\gamma))=0$ .

$\square$

Proof of Theorem 3.1. Take 7as in Proposition 3.6. By iterationLet$\pi_{\mathbb{C}}$:$\mathrm{o}\mathrm{f}L^{-1}$be a natural projection. For each $t\in[0, N]$ , let. By Lemma 3.5, we have

$A\cross \mathbb{C}arrow \mathbb{C}$$\pi \mathrm{c}\circ\gamma(t)$.

$\log\psi_{a\exp(2\pi \mathrm{u})}$We choose the branch of the logarithm so that$\psi a\exp(2\pi u)$ and $\psi a\exp(2\pi:(t+1))$ as
different functions:$I_{t}$$I_{t}arrow \mathrm{E}$, we may assume$\psi$

is defined on the curve.
containing

be the component of$\tilde{I_{a\exp(2\pi u)}}$.

$\log\psi a\exp(2\pi\dot{l}t)(\pi \mathrm{c}\mathrm{o}\gamma(t))$

varies continuously. Here, we regard

135
and
Then in general,fact,
a1-formis
thereIn, they coincide.

do not have to be coincide. But sincesuch that$\log\psi a\exp(2\pi i\cdot N)$$\log\psi_{a\exp(2\pi i\cdot 0)}$$\pi_{1}(I^{-})$$\omega$in$I^{-}$$\pi_{1}(I^{-})\ni C\mapsto\int_{C}\omega$and$\int\omega=\log\varphi^{-}$$\Phi(\gamma)$

is trivial in$= \alpha(C)\in \mathbb{Z}[\frac{1}{d}]$(indefinite integral) (cf. [MNTU, Section 7.3]). Therefore$\log\psi_{a\exp(2\pi i\cdot N\rangle}(\pi_{\mathrm{C}}\circ\gamma(0))$$=\log\psi a\exp(2\pi:\cdot N)(\pi \mathrm{c}\circ\gamma(N))$$= \log\psi_{a\exp(2\pi i\cdot 0)}(\pi_{\mathbb{C}}\circ\gamma(0))+\int_{\gamma}\Phi^{*}\omega$$=\log\psi a\exp(2\pi:\cdot 0)(\pi_{\mathrm{C}}0\gamma(0))$

Take an appropriate
such that$\xi$$\in \mathbb{H}$.

so that Re4 is sufficiently large. Then for each $t\in[0, N]$ , there is aunique point$\eta_{t}\in I_{t}$$\log\psi_{a\exp(2\pi it)}(\eta_{t})=\xi$.

Then$[0, N]\ni t\mapsto(a\exp(2\pi it), \eta_{t})\in\overline{I}$is aclosed curve and satisfies$\psi(a\exp(2\pi it), \eta_{t})=e^{\xi}$.

for any $t\in[0, N]$ .
Therefore$[0, N]\ni t\mapsto\Phi(a\exp\langle 2\pi it)$,$\eta_{t})\in W_{0}^{s}(H)$is anon-contractible closed curve, and satisfies.

$\{\Phi(a\exp(2\pi it), \eta_{t})|t\in[0, N]\}\subset(\varphi^{-})^{-1}(e^{\xi})$

This contradicts with Proposition 2.2.

$\square$4Acknowledgments
The author would like to thank to Prof. Shishikura for his advises.

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