数理解析研究所講究録
1072 巻 1998 年 50-61

50Complex Ruelle Operator
andHyperbolic Complex Dynamical Systems
Shigehiro Ushiki
Graduate School of Human and Environmental Studies
Kyoto University
京都大学大学院人間環境学研究科宇敷重広1.

Decomposition of Complex Ruelle operator

be a hyperbolic rational mapping. We assume that allLet
the attractive periodic points of are fixed points, all the critical points, is inculed in C.
of are non-degenerate, and that the Julia set of $R,$,
Let $N$ denote the number of attractive fixed points and let
denote the attractive basin ofdenote the attractive fixed points. Let. Let
denote the set of critical points of .
, such that
, $N$ , let denote an oriented multicurve in
For $k=1,$,, whereis an open set satisfying
. Let
.
and
and
, let $O_{0}(o)$ denote the space of functions
For open set
holomorphic inand has an analytic extention to a neighbourhood of
belongs to the closure of
the closure of , and satisfies $g(\infty)=0$ if. We have the following decomposition of holomorphic functions. The
direct sum in the theorem means the uniqueness of the decomposition.
$R:\overline{\mathbb{C}}arrow\overline{\mathbb{C}}$

$R$

$R$

$J_{R}$$a_{1},$$\cdots$$a_{N}$$A_{k}$$a_{k}$$R$$C_{R}$$\cdots$$A_{k}$$\gamma_{k}$$\gamma_{k}=\partial\Omega_{k}$$R^{-1}(\Omega_{k})\subset\Omega_{k},$$\Omega_{k}$$C_{R}\cap\Omega_{k}\cap A_{k}=\phi$$\Omega=\mathrm{n}_{k=1}^{N}\Omega_{k}$$\gamma=\bigcup_{k=1}^{N}\gamma k$$g:Oarrow \mathbb{C}$$O\subset\overline{\mathbb{C}}$

$O$

$O$

$\infty$$O$

THEOREM 1.1$O_{0}( \Omega)=\bigoplus_{1k=}o0N(\Omega_{k})$

PROOF

Let

$\Omega_{k}\cup A_{k}=\overline{\mathbb{C}}$$g\in O_{0}(\Omega)$. Then$g$.

can be expressed as$g(x)= \frac{1}{2\pi i}\int\gamma\frac{g(\tau)}{\tau-x}d\mathcal{T}$,$x\in\Omega$.

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For $k=1,$$\cdots$, $N$ , let$\Gamma_{k}$:$\mathcal{O}_{0}(\Omega)arrow \mathcal{O}_{0}(\Omega_{k})$be defined by,$( \mathrm{r}_{kg})(_{X})=\frac{1}{2\pi i}\int\gamma k\frac{g(\tau)}{\tau-x}d\tau$Asis bouded on
infinity. Hence$g(\tau)$$\gamma_{k},$$x\in\Omega_{k}$is holomorphic inand vanishes at the. As, we have the decomposition$\Gamma_{k}g$$\Gamma_{k}g\in \mathcal{O}_{0}(\Omega_{k}\rangle$.

$\Omega_{k}$$\gamma=\bigcup_{k=1}^{N}\gamma_{k}$$g= \sum_{k=1}^{N}\Gamma_{k}g$.

To prove the uniqueness of the decomposition, assume$k=1,$, $N$ , and$g_{k}\in \mathcal{O}_{0}(\Omega_{k})$for$\cdots$$\sum_{k=1}^{N}g_{k}=0$.

Then is holomorphic inand at the same time it can be analytically, sinceextendedis holomorphic in. This shows
that
, N.
is constant for $k=1,$
takes value zero at theinfinity if the infinity belongs to the domain of its definition. Therefore,$g_{k}=0$ for all $k=1,$, $N$ , except one. But the exceptional one must be
zero since.
$\Omega_{k}$$g_{k}$$\dot{\mathrm{t}}\mathrm{o}A_{k}$$A_{k}$$-g_{k}=\Sigma_{j\neq kg_{j}}$$\cdots$$g_{k}$$\mathrm{H}_{\mathrm{o}\mathrm{W}\mathrm{e}}\mathrm{V}\mathrm{e}\mathrm{r}.’g_{k}$$\cdots$$\Sigma_{k=1}^{N}g_{k}=0$

THEOREM 1.2$\Gamma_{k}$:,$\mathcal{O}_{0}(\Omega)arrow o_{0}(\Omega_{k})$$k=1,$$\cdots,$$N$

are projections.
PROOF

For all
is holomorphic in. If $j\neq k$ , then
. Therefore,. As we saw in the previous theorem,$g\in \mathcal{O}_{0}(\Omega),$$\Gamma_{k}^{2}g=\Gamma_{k}g$$g\in \mathcal{O}_{0}(\Omega)$$\gamma_{j}\subset A_{j}\subset\Omega_{k}$$\Omega_{k}$, hence we havefor all$\Gamma_{j}\Gamma_{k}g=0$$\Sigma_{k=1}^{N}\Gamma_{k}=id$

DEFINITION 1.3$\mathcal{O}_{0}(\Omega)$$\Gamma_{k}g$

We define complex Ruelle operator.

$L:\mathcal{O}_{0}(\Omega)arrow$by$(Lg)(x)=y \in R^{-1}()\sum_{x}\frac{g(y)}{(R(y))^{2}},$,$g\in$$O_{0}(\Omega)$,$x\in\Omega$.

.

Note thatand $R’(y)\neq 0$ as we assumed is hyperbolicand contains no critical points. As indicated by [1], the complex Ruelle$R^{-1}(x)\subset\Omega$$\Omega$$R$

52
operator can be expressed as an integral operator of the form :.

$(Lg)(X)= \frac{1}{2\pi i}\int\gamma\frac{g(\tau)}{R’(\mathcal{T})(R(\mathcal{T})-X)}d\mathcal{T}$This formula is easily verified by applying the Cauchy’s theorem about
residues and it shows that $Lg\in O_{0}(\Omega)$ . Comparing with the PerronFrobenius operator, we see that the spectral radius of is smaller than
1.
$L$

$L$

DEFINITION 1.4$L_{ij}$:$O_{0}(\Omega_{j})arrow \mathcal{O}_{0}(\Omega_{i})$is defined by$L_{ij}=\Gamma_{i}\mathrm{o}L|_{\mathcal{O}_{\mathrm{o}(\Omega)}}j$

The Ruelle operator can be expressed as an $N\cross N$ matrice of operators:$L=(L_{ij})$

The components$L_{ij}$are computed as follows.

PROPOSITION 1.5

If $i\neq j$ , then for$g_{j}\in \mathcal{O}_{0}(\Omega_{j})$and$(L_{ijg_{j})}(_{X})=- \sum_{ic\in c_{R}\mathrm{n}A}\frac{g_{j}(_{C)}}{R’’(_{C})(R(C)-X)}$

PROOF

As

$g_{j}\in o_{0}(\Omega_{j})$and$L_{ijg_{j}}$.

,we can apply the residue theorem to the complement of
PROPOSITION 1.6$A_{i}$,

is defined by$(L_{ijg_{j})}(X)= \frac{1}{2\pi i}\int\gamma i\frac{g_{j}(_{\mathcal{T})}}{R’(\mathcal{T})(R(\mathcal{T})-X)}d\tau$at the critical points in$x\in\Omega_{i}$$\Omega_{i}$. The residues

give the formula.

For$g_{j}\in o_{0}(\Omega_{j})$and for$x\in\Omega_{j}$,.

$(L_{jj})gj(X)= \sum\frac{g_{j}(y)}{(R(y))^{2}}y\in R-1(x)’+C\in C_{R}\sum_{\mathrm{n}\Omega_{j}}\frac{g_{j}(_{C)}}{R^{;;}(_{C})(R(C)-X)}$

PROOF
2.

In this case, we can apply the residue theorem to$\Omega_{j}$.

M\"obius transformation and complex Ruelle operator

53

In this section, we observe the behavior of t.he complex Ruelle operator under a coordinate change of the Riemenn sphere by a M\"obiustransformation.
be a M\"obius transformation of the Riemann sphere.
Let $M$ :Let $\alpha=M^{-1}(\infty),$ $\beta=M(\infty)$ , and $\tilde{R}=M\circ R\circ M^{-1}$ . We set, and assume. In order to avoid confusion, we
denote the complex Ruelle operator defined in the previous section by
associated to the rational mapping . Now, we define a “complex Ruelle
operator” associated to the M\"obius transformation $M$ .
$\overline{\mathbb{C}}arrow\overline{\mathbb{C}}$$\tilde{\Omega}_{k}=$$M(\Omega_{k}),\tilde{\Omega}=M(\Omega)$$\alpha\not\in\Omega$$L_{R}$$R$

DEFINITION 2. 1
$L_{M}$:

is defined by$o_{0}(\Omega)arrow \mathcal{O}_{0}(\tilde{\Omega})$for$g\in \mathcal{O}_{0}(\Omega)$$(L_{Mg})( \tilde{X})=,\frac{g\mathrm{o}M^{-1}(\tilde{x})}{(M(M^{-1}(_{\tilde{X}))})^{2}}.$and$\tilde{x}\in\tilde{\Omega}$’.

PROPOSITION 2.2$L_{M^{-1}}.=L_{M}^{-1}$,$L_{\tilde{R}}=L_{M^{\mathrm{O}}RM^{-1}}L\circ L$.

First equality is easily verified by computing
PROOF
directly. Second equality is easily verified similarly by theanddefinition of the complex Ruelle operator. However, we would like to givea proof for the operator defined as an integral operator. Let.
,
Then we have, forand$L_{M^{-1}}\mathrm{o}L_{M}$$L_{M^{\circ}}L_{M^{-1}}$$\tilde{g}\in \mathcal{O}_{0}(\tilde{\Omega})$$x\in\Omega$$\tilde{x}\in\tilde{\Omega}$$(L_{M^{-1}} \tilde{g})(x)=\frac{\tilde{g}\circ M(x)}{((M^{-1})/(M(_{X))})^{2}}$,$(L_{R}L_{M}-1 \tilde{g})(X)=\frac{1}{2\pi i}\int_{\gamma}\frac{(L_{M^{-1}}\tilde{g})(\tau)}{R’(\tau)(R(\mathcal{T})-X)}d_{\mathcal{T}}$,$= \frac{1}{2\pi i}\int_{\gamma}\frac{\tilde{g}\mathrm{o}M(\tau)}{R’(\tau)(R(\mathcal{T})-X)((M-1)\prime\circ M(\tau))^{2}}d\tau$and$(L_{M}L_{RM^{-}}L1\tilde{g})(\tilde{x})$$= \frac{1}{(M’\circ M^{-}1(\tilde{X}))2}\frac{1}{2_{7\ulcorner}i}\int_{\gamma}\frac{\tilde{g}\circ M(\tau)d_{\mathcal{T}}}{R’(\tau)(R(\mathcal{T})-M-1(\tilde{X}))((M^{-}1)/M\circ(\mathcal{T}))2}$

54.

$= \frac{1}{2\pi i}\int_{\gamma}\frac{\tilde{g}\circ M(\mathcal{T})(M’(\tau))^{2}d\mathcal{T}}{(M’\circ M^{-}1(_{\tilde{X}}))^{2}R(\tau)(R(\tau)-M^{-}1(\tilde{x}))},$On the other hand, by a change of variables $\sigma=M(\tau)$ , we have$(L_{\tilde{R}} \tilde{g})(\tilde{x})=\frac{1}{2\pi i}\int_{\gamma}\frac{\tilde{g}(\sigma)d\sigma}{\tilde{R}’(\sigma)(\tilde{R}(\sigma)-\tilde{X})}$$= \frac{1}{2\pi i}\int_{\gamma}\frac{\tilde{g}\mathrm{o}M(\mathcal{T})M’(\mathcal{T})d_{\mathcal{T}}}{M’(R(\mathcal{T}))R/(\mathcal{T})(M^{-}1)\prime \mathrm{o}M(\mathcal{T})(M\mathrm{o}R(\mathcal{T})-\tilde{X})}$

Hence we obtain$(L_{M}L_{R}L_{M}-1\tilde{g})(\tilde{x})-(L_{\overline{R}}\tilde{g})(\tilde{x})$$= \frac{1}{2\pi i}\int_{\gamma}\frac{\tilde{g}\circ M(\mathcal{T})(M’(\mathcal{T}))2}{R(\tau)},\cross$$\cross(\frac{1}{(M’\circ M^{-}1(_{\tilde{X})})^{2}(R(_{\mathcal{T}})-M^{-}1(\tilde{x}))}-\frac{1}{M’\circ R(_{\mathcal{T})}(M\circ R(\tau)-\tilde{X})}\mathrm{I}^{d\tau}\cdot$

As $R’(\tau)\neq 0$ and $M’\circ R(\tau)\neq 0$ for, the integrand can have poles
only at. The residues at such points are, by setting$\tau\in\Omega$$\tau\in R^{-1}\circ M^{-}1(\tilde{x})\cap\Omega$$x=M^{-1}(\tilde{x})$and $y\in R^{-1}(x)$ , computed as$\frac{\tilde{g}\circ M(y)(M’(y))^{2}}{R(y)},(\frac{1}{(M’(x))2R’(y)}-\frac{1}{M’\mathrm{o}R(y)(M\circ R(y)R/(y))},)=0$.

Hence the proposition follows.

DEFINITION 2.3

Components:, wherefor$L_{M,ij}$bythe projection.

$L_{M,ijg_{j}}=\tilde{\Gamma}_{i}L_{Mg_{j}}$PROPOSITION 2.4$g_{j}\in o_{0}(\Omega_{j})$If$\infty\in\Omega_{j}$$O_{0}(\Omega_{j})arrow O_{0}(\tilde{\Omega}_{i})$$\tilde{\Gamma}_{i}$:, then$(L_{M,jjg_{j})}( \tilde{X})=(LMg_{j})(\tilde{x})+{\rm Res}\tau=\infty\frac{g_{j}(_{\mathcal{T})}}{M’(\tau)(M(\mathcal{T})-\tilde{X})}$

If$\infty\not\in\Omega_{j}$, then$(L_{M,jjg_{j}})(_{\tilde{X}})=(L_{Mg_{j}})(\tilde{X})$

If

$i,$

$\neq j$and$\infty\in\Omega_{i}$$\infty\not\in\Omega_{i}$.

, then$L_{M,ij}=0$

If $i\neq j$ andis defined
denote$O_{0}(\tilde{\Omega})arrow o_{0}(\tilde{\Omega}_{i})$.

, then$(L_{M,ijg_{j}})( \tilde{X})=-{\rm Res} \mathcal{T}=\infty\frac{g_{j}(_{\mathcal{T})}}{M’(\tau)(M(\mathcal{T})-\tilde{X})}$

.

.

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These formulas are easily verified by a direct computation
PROOF
by applying the residue theorem.

3.

Partial complex Ruelle operator

In this section, we examine a diagonal component of the complex Ruelle operator. The Fredholm determinant and the resolvent of the adjoint:of the complex Ruelle operadiagonal component
tor can be computed in a similar manner as is given by [1] and [2].
For the sake of simplicity, we assume $a_{1}=\infty$ , andis an attactive
fixed point with eigenvalue a satisfying $0<|\sigma|<1$ . We define the partial
Ruelle operator as follows.
$L_{ii}$$\mathcal{O}_{0}(\Omega_{i})arrow O_{0}(\Omega_{i})$$a_{1}$

DEFINITION 3.1
defined byPartial Ruelle operator$L_{11}$:$O_{0}(\Omega_{1})arrow O_{0}(\Omega_{1})$$(L_{11g})(X)= \frac{1}{2\pi i}\int\gamma_{1}\frac{g(\tau)d_{\mathcal{T}}}{R’(\mathcal{T})(R(\mathcal{T})-X)}$

is.

An expicit formula for the partial Ruelle operator is given by proposition 1.6. We shall consider the dual operator.
DEFINITION 3.2
The dual spaceofis the space ofcontinuous, complex linear, and holomorphic functional :.
The topology ofis understood as the uniform convergence in aneighborhood of the closure of . A functional is said to be holomorphic
if the value
is holomophic with respect to the parameter for aholomorphic family of functions .
$O_{0}^{*}(\Omega_{1})$$o_{0}(\Omega_{1})$$F$$O_{0}(\Omega_{1})arrow \mathbb{C}$$\mathcal{O}_{0}(\Omega_{1})$$\Omega_{1}$$F[g_{\mu}]$$\mu$$g_{\mu}$

PROPOSITION 3.3, such that

For any$F\in O_{0}^{*}(\Omega_{1})$, there exists an$f\in \mathcal{O}_{0}(\overline{\mathbb{C}}\backslash$$\Omega_{1})$$F[g]= \frac{1}{2\pi i}\int_{\gamma_{1}}f(\mathcal{T})g(\mathcal{T})d\tau$

PROOF

,

for

$g\in O_{0}(\Omega_{1})$.

In fact, the so called Cauchy transform$f(z)=F[ \frac{1}{z-\zeta}]$gives such a function. As
parametrized by$\mathcal{O}_{0}(\Omega_{1})$$\frac{1}{z-\zeta}$

is a holomorphic family of functions in. $f(z)$ is holomorphic in
and$z\in\overline{\mathbb{C}}\backslash \Omega_{1}$$\overline{\mathbb{C}}\backslash \Omega_{1}$56$f(\infty)=0$ ,hence$f\in O_{0}(\overline{\mathbb{C}}\backslash \Omega_{1})$. For$g\in O_{0}(\Omega_{1})$,$F[g]=F[ \frac{1}{2\pi i}, \int_{\gamma 1}\frac{g(z)}{z-\zeta}d_{Z}]$$= \frac{1}{2\pi i}\int_{\gamma_{1}}g(Z)F[\frac{1}{z-(}]dz=\frac{1}{2\pi i}\int_{\gamma}1g(_{Z})f(z)dz$

Note that such function$f(z)\in \mathcal{O}_{0}(\overline{\mathbb{C}}\backslash \Omega_{1})$is unique since$\frac{1}{2\pi i}\int_{\gamma_{1}}f(_{\mathcal{T}})\frac{1}{z-\tau}d\mathcal{T}=f(z)$The dual operator
PROPOSITION 3.4
represented by integral operator
:
, by
for
and
$\mathcal{L}_{11}^{*}$$f\in O_{0}(\overline{\mathbb{C}}\backslash \Omega_{1})$.

.

$L_{11}^{*}$:

isdefined,$O_{0}^{*}(\Omega_{1})arrow \mathcal{O}_{0}^{*}(\Omega_{1})$$\mathcal{O}_{0}(\overline{\mathbb{C}}\backslash \Omega_{1})arrow O_{0}(\overline{\mathbb{C}}\backslash \Omega_{1})$$z\in\overline{\mathbb{C}}\backslash \Omega_{1}$$( \mathcal{L}_{11}^{*}f)(_{Z})=\frac{1}{2\pi i}\int_{\gamma_{1}},\frac{f(R(\tau))d_{\mathcal{T}}}{R(\tau)(z-\tau)}$$= \frac{f(R(Z))}{R(z)},-\sum_{1c\in C_{R}\mathrm{n}A}\frac{f(R(C))}{R’’(c)(_{Z}-c)}$

PROOF
we have.

The proof is almost same as in [1]. By a direct computation,$( \mathcal{L}_{11}^{*}f)(Z)=(L_{11}^{*}F)[\frac{1}{z-\zeta}]=F[L11[\frac{1}{z-\zeta}]]$$=F[ \frac{1}{2\pi i}f_{\gamma_{1}}\frac{d\tau}{R’(\tau)(R(\mathcal{T})-\zeta)(z-\mathcal{T})}\exists$$= \frac{1}{2\pi i}\int_{\gamma_{1}}f(\zeta)d\zeta\frac{1}{2\pi i}f_{\gamma_{1}}\frac{d\tau}{R’(\tau)(R(\tau)-\zeta)(z-\mathcal{T})}$$== \frac{1}{\frac{2\pi i1}{2\pi i}}\int_{\int_{\gamma_{1}}}\gamma 1\frac{}{R’(\tau)(z-\tau)}\frac{d\tau}{R’(\mathcal{T}),f(R(^{(\tau_{\mathcal{T}})}\mathcal{T}z))-d}\frac{1}{2\pi i}f_{\gamma}1\frac{f(\zeta)d\zeta}{R(\tau)-\zeta}$$= \frac{f(R(z))}{\frac{f(R(R’(z)z))}{R’(z)}}-\Sigma_{C\in C_{R}}=-\Sigma_{c}\in c_{R}\cap A(\mathrm{n}A_{1}{\rm Res}_{\infty)^{\frac{\tau=cf(R}{R(c)}}},,\frac{f(R(\mathcal{T}))}{(z-(c)^{()}R^{J})\tau c)(z-\tau)}$.

and the poles at the critical ponts in theNote thatlast line of the above calculation cancel out.
$\mathcal{L}_{11}^{*}f\in O_{0}(\overline{\mathbb{C}}\backslash \Omega_{1})$4. Fredholm determinant of the adjoint Ruelle operator
In this section, we compute the Fredholm determinant and the resol. The calculation is almost same as invent of the adjoint operator[1].
$\mathcal{L}_{11}^{*}$

57

Let
denote the number of critical points in. Let
be the critical points in$n_{1}$$A_{1}$. And let$\{c_{1}, \cdots , c_{n_{1}}\}$$A_{1}$$H(x, z; \lambda)=\sum_{n=0}^{\infty}\frac{\lambda^{n}}{(R^{\mathrm{O}}n)’(Z)(R\circ n(Z)-X)}$,and let$M( \lambda)=(\delta_{ij}+\frac{\lambda}{R^{\prime/}(C_{i})}H(Cj, R(ci);\lambda))_{i,j=1}n_{1}$be an$n_{1}\cross n_{1}$matrice.

THEOREM 4.1

The Fredholm determinant$D_{11}(\lambda)$$D_{11}( \lambda)=n=1\prod\infty(1-\sigma n+1\lambda)\det M(\lambda)$

It is meromorphic in
essential spectrum.

$\mathbb{C}$and holomorphic forof$\mathcal{L}_{11}^{*}$

is given by.
.

$|\lambda|<|\sigma|^{-2}$$\mathcal{L}_{11}^{*}$

has noThis theorem follows immediately from proposition 4.3 below. We
assume that the backward orbits of critical points do not intersect withthe curve . Let$\gamma_{1}$$\Omega_{R}=\overline{\mathbb{C}}\backslash (\Omega_{1}\cup\bigcup_{n=0}\{_{Z}\infty\in\Omega_{1}|R’(R^{\mathrm{o}n}(Z))=0\})$Then$\mathcal{O}_{0}(\Omega_{1})\subset o_{0}(\Omega_{R})$. Define$\mathcal{L}_{R}^{*}$

:

.

by

$\mathcal{O}_{0}(\Omega_{R})arrow o_{0}(\Omega_{R})$$( \mathcal{L}_{R}^{*}f)(_{Z})=\frac{1}{2\pi i}\int\gamma_{1}’\frac{f(R(\tau))d_{\mathcal{T}}}{R(\tau)(z-\tau)}$. We see immediately that the image of, and. Therefore,
andcoincide on
the same spectrum. Define a operator :for$f\in \mathcal{O}_{0}(\Omega_{R})$$o_{0}(\Omega_{1})$$\mathcal{L}_{R}^{*}$$\mathcal{L}_{R}^{*}$$\mathcal{O}_{0}(\Omega_{1})$$\mathcal{L}_{11}^{*}$$\mathcal{K}$$( \mathcal{K}f)(_{Z})=\frac{f(R(Z))}{R(z)},$,$\mathcal{K}$$\mathcal{L}_{R}^{*}$$\mathcal{L}_{11}^{*}$$O_{0}(\Omega_{R})arrow \mathcal{O}_{0}(\Omega_{R})$$f\in O_{0}(\Omega R),$ $z\in\Omega_{R}$

PROPOSITION 4.2 The spectrum of
determinant is given by

is included in
and
has
by

is

$\{\sigma^{n+1}\}_{n1}^{\infty}=$$\det(I-\lambda \mathcal{K})=\prod_{n=}\infty 1(1-\sigma^{n+}\lambda 1)$

’

.

.

and the Fredholm

58

The eigenfunction$f_{n}(z)$for eigenvalue$\lambda^{-1}=\sigma^{n+1}$$f_{n}(Z)= \frac{1}{\varphi’(z)(\varphi(Z))n}$is given by,

where holomorphic function :is the Schr\"oder’s function of theformnear the infinity and satisfying
the Schr\"oder’s equation $\varphi(R(z))=\sigma^{-1}\varphi(z)$ .
$\Omega_{R}arrow\overline{\mathbb{C}}$$\varphi$$\varphi(z)=z+a_{0}+\frac{a_{2}}{z}+\cdots+\frac{a_{n}}{z^{n}}+\cdots$

PROOF
The fact thatis an eigenfunction for eigenvalue
is immediately verified by using the Shr\"oder’s equation. Eigenfunction
by using the function equations
can be extended to$\sigma^{n+1}$$f_{n}(z)$$f_{n}(z)$$\Omega_{R}$,and$\mathcal{K}f_{n}=\sigma^{n+1}f_{n}$$( \mathcal{K}f_{n})(z)=\frac{f_{n}(R(_{Z}))}{R(z)}$$\cdot$As the eigenfunctions, whereform a complete basis ofdenotes the connected component ofcontaining the infinity, andas the eigenfunctions are determined from a germ at the infinity of theeigenfunction by the function equation above, the Fredholm determinantis given by the formula in the proposition.
$\{f_{n}\}_{n=1}^{\infty}$$O_{0}(\Omega_{R}^{\infty})$$\Omega_{R}$$\Omega_{R}^{\infty}$

Define linear maps$\mathcal{G}$:$o_{0}(\Omega_{R})arrow \mathbb{C}^{n_{1}}$$\mathcal{G}f=(\frac{f(R(c_{j}))}{R’(_{C_{j}})},)_{j=1}^{n}1$$\mathcal{F}\alpha=\sum_{=j1}^{n_{1}}\frac{\alpha_{j}}{z-c_{j}}$

,

,

and

$\mathcal{F}$:$\mathbb{C}^{n_{1}}arrow o_{0}(\Omega_{R})$$f\in O_{0}(\Omega R)$$\alpha=(\alpha_{j})\in \mathbb{C}^{n_{1}}$

by

,

.

We have$\mathcal{L}_{R}^{*}=\mathcal{K}-\mathcal{F}\mathcal{G}$.

The Fredholm determinant of the adjoint Ruelle operator
as follows.

is computed$\mathcal{L}_{R}^{*}$PROPOSITION 4.3$D_{11}(\lambda)=\det(I-\lambda \mathcal{L}_{R}^{*})=\det(I-\lambda \mathcal{K})\det M(\lambda)$

where$M(\lambda)=I_{n_{1}}+\lambda \mathcal{G}(I-\lambda \mathcal{K})^{-1}\mathcal{F}$

.

,

59

PROOF
$\det(I-\lambda \mathcal{L}_{R}^{*})=\det(I-\lambda \mathcal{K}+\lambda \mathcal{F}\mathcal{G})$$=\det(I-\lambda \mathcal{K})\det(.I+\lambda(I-\lambda \mathcal{K})^{-1}\mathcal{F}\mathcal{G})$$=\det(I-\lambda \mathcal{K})\det(I_{n_{1}}+\lambda \mathcal{G}(I-\lambda \mathcal{K})-1F)$$=\det(I-\lambda \mathcal{K})\det M(\lambda)$The$n_{1}\cross n_{1}$matrice$M(\lambda)$.

is computed as follows.

$M(\lambda)=I_{n_{1}}+\dot{\lambda}\mathcal{G}(I-\lambda \mathcal{K})^{-}1F$$=I_{n_{1}}+\lambda \mathcal{G}(\Sigma_{n=0}\infty\lambda^{n_{\mathcal{K}^{n}}})\mathcal{F}$$=I_{n_{1^{+}}}\lambda \mathcal{G}(\Sigma_{n=}^{\infty}0^{\frac{\lambda^{n}}{(R^{\circ n})’(z)(R^{\circ n}(z)-c_{i})}})^{n}j=11$$=I_{n_{1}}+ \lambda(\frac{1}{R’’(c_{i})}\Sigma_{n=0}^{\infty}\frac{\lambda^{n}}{(R^{\circ n})’(C_{i})(R^{\circ}n(C_{i})-c_{j})})i,j=n11$$=( \delta_{ij}+\frac{\lambda}{R’’(c_{i})}H(C_{j}, R(c_{i});\lambda))^{n_{1}}i,j=1$

As the spectrum of isis meromorphic in and holomorphic in. This completes the proof of the proposition
4.3 and the Theorem 4.1.
$\mathcal{K}$$\{\sigma^{n+1}\}_{n=1}^{\infty},$$M(\lambda)$$\mathbb{C}$$\{\lambda||\lambda|<\sigma^{-2}\}$

5. The resolvent of the partial adjoint Ruelle operator
The resolvent of the parttial adjoint Ruelle operator
can be computed in an analougous manner as in [1] and [2]. First, we compute theresolvent function of operator .
$\mathcal{L}_{11}^{*}$$\mathcal{K}$The function $H(x, z;\lambda)$ defined in the previous
PROPOSITION 5.1
section is the resolvent function of , i.e.,$\mathcal{K}$$H(x, z; \lambda)=(I-\lambda \mathcal{K})^{-1}\frac{1}{z-x}$$= \sum_{n=0}^{\infty}\lambda n\mathcal{K}^{n_{\frac{1}{z-x}}}=\sum n=0\infty\frac{\lambda^{n}}{(R^{\mathrm{o}n})’(z)(R^{\circ(Z)-}nX)}$

where.
$x,z$ and$x\in\Omega_{1},$$z\in\Omega_{R}$and$\lambda\in \mathbb{C}\backslash \{\sigma^{-k}\}_{k=2}^{\infty}$. $H(x, z;\lambda)$ is holomorphic in$\lambda$

PROOF

,

As is easily observed, we have$\mathcal{K}^{n}f(z)=\frac{f(R^{\mathrm{o}n}(z))}{(R^{\mathrm{o}n})/(Z)}$

.

60
Let
:
, be the complete system of eigenfunctions$n=1,2,$ofgiven by propositon 4.2. For each
, we can expand thefunction $(z-x)-1\in O_{0}(\Omega_{R})$ in the form
$f_{n}$$\Omega_{R}arrow \mathbb{C},$$\cdots$$\mathcal{K}$$x\in\Omega_{1}$$\frac{1}{z-x}=\sum_{n=1}^{\infty}b_{n}(_{X})f_{n}(z)$Observe that$b_{n}(x)$is holomorpic in$\Omega_{1}$.

. With this expression, we have$(I- \lambda \mathcal{K})-1\frac{1}{z-x}=\sum_{n=0}\infty\lambda n\mathcal{K}^{n}\frac{1}{z-x}$$= \sum_{n=0}^{\infty}\lambda^{n}\mathcal{K}n_{\sum ,k}\infty=1b_{k}(X)f_{k}(z)=k=1\sum\infty b_{k}(x)n\sum_{=0}^{\infty}\lambda^{n_{\mathcal{K}^{n}}}fk(z)$$= \sum_{k=1}^{\infty}b_{k}(_{X})\sum_{=n0}\infty(\lambda\sigma)^{n}k+1fk(z)=\sum_{k1}\infty=b_{k}(X)\frac{1}{1-\lambda\sigma^{k+1}}f_{k()}z$.

This shows that $H(x, z;\lambda)$ has an analytic extension to the domain.

$\Omega_{1}\cross$$\Omega_{R}\cross(\mathbb{C}\backslash \{\sigma^{-k}\}_{k2}^{\infty}=)$The resolvent function $E(x, z;\lambda)$ is defined by$E(_{X,z;} \lambda)=(I-\lambda \mathcal{L}_{11}*)^{-1}\frac{1}{z-x}$$E(x, z;\lambda)$

is holomorphic in$x$,$x\in\Omega_{1},$$z\in\overline{\mathbb{C}}\backslash \Omega_{1},$$\lambda\in \mathbb{C},$$E(x, \infty;\lambda)=0$.

and , and meromorphic in .
$z$$\lambda$

PROPOSITION 5.2,$(I-\lambda \mathcal{L}_{1}^{*})1=-1(I-\lambda \mathcal{K})-1-\lambda(I-\lambda \mathcal{K})-1\mathcal{F}(M(\lambda))^{-}1\mathcal{G}(I-\lambda \mathcal{K})^{-}1$

where$M(\lambda)=I_{n_{1}}+\lambda \mathcal{G}(I-\lambda \mathcal{K})^{-}1\mathcal{F}$

PROOF

is an

$n_{1}\cross n_{1}$matrice.

By a direct computation.

$(I-\lambda \mathcal{L}_{11}*)^{-1}=(I-\lambda \mathcal{K}+\lambda \mathcal{F}\mathcal{G})^{-1}$$=(I-\lambda \mathcal{K})^{-1}(I+\lambda \mathcal{F}\mathcal{G}(I-\lambda \mathcal{K})-1)^{-}1$$=(I-\lambda \mathcal{K})^{-}1(I+\Sigma_{k=}\infty(1-\lambda)k(\mathcal{F}\mathcal{G}(I-\lambda \mathcal{K})-1)^{k})$$=(I-\lambda \mathcal{K})^{-1}(I-\lambda \mathcal{F}\Sigma_{k}\infty(=0-\lambda \mathcal{G}(I-\lambda \mathcal{K})-1)^{k}\mathcal{G}(I-\lambda \mathcal{K})^{-}1)$$=(I-\lambda \mathcal{K})^{-}1-\lambda(I-\lambda \mathcal{K})^{-}1\mathcal{F}(M(\lambda))-1\mathcal{G}(I-\lambda \mathcal{K})-1$

As is easily verified,in this proposition is same as
the beginning of the previous section.
$M(\lambda)$$M(\lambda)$Let$H_{1}(z; \lambda)=((I-\lambda \mathcal{K})^{-}1\frac{1}{z-c_{i}})i=1=(n_{1}H(_{C}i, Z;\lambda))^{n_{1}}i=1$.

given in61

be a row vector and let$H_{2}(x; \lambda)=\mathcal{G}(I-\lambda \mathcal{K})-1\frac{1}{z-x}=\mathcal{G}H(x, z;\lambda)=(\frac{H(x,R(_{C}i),\lambda)}{R’(C_{i})},\cdot)_{i=1}^{n}1$

be a column vector. With all these things together, we find the explicit
expression of the resolvent function $E(x, z;\lambda)$ .

THEOREM 5.3$E(X, z;\lambda)=H(x, z;\lambda)+\lambda H_{1}(_{Z\lambda)((\wedge))H_{2}(};M-1X;\lambda)$.

The resolvent function is holomorphic in, and in, andmeromorphic in
C. The poles are the zeros of the Fredholm determinant.
$x\in\Omega_{1}$$z\in\Omega_{R}$$\lambda\in$$D_{11}(\lambda)$

References[1] G.M.Levin, M.L.Sodin, and P.M.Yuditski: A Ruelle Operator for a
Real Julia Set, Communications in Mathematical Physics, 141, 119132 (1991).
[2] G.Levin, M.Sodin, and P.Yuditski: Ruelle operators with rational weights for Julia sets, Journal d’analyse math\’ematiques, Vol.
63(1994),303-331.
[3] M.Tsujii: A transversality condition for quadratic family at ColletEckmann parameter, Problems in Complex Dynamical Systems,
RIMS Kokyuroku,1998.
$1042_{\mathrm{P}\mathrm{p}},99-101$