\documentclass{uicthesi} \usepackage{graphicx} \usepackage{amssymb} \usepackage{epstopdf} \DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png} \newtheorem{thm}{Theorem} \newtheorem{prop}{Proposition} \newtheorem{claim}{Claim} \newtheorem{lemma}{Lemma} \newtheorem{ex}{Example} \newtheorem{proof}{Proof} \newtheorem{cor}{Corollary} \newtheorem{obs} {Observation} %\theoremstyle{definition} %\newtheorem{def}[thm]{Definition} \newtheorem{defn}{Definition} \newtheorem{prob}{Problem} % MATH ----------------------------------------------------------- \newcommand{\norm}[1]{\left\Vert#1\right\Vert} \newcommand{\abs}[1]{\left\vert#1\right\vert} \newcommand{\set}[1]{\left\{#1\right\}} \newcommand{\Real}{\Bbb R} \newcommand{\To}{\longrightarrow} \newcommand{\union}{\ensuremath{\displaystyle\bigcup}} \newcommand{\orbit}{\ensuremath {\mathcal O}} % \newcommand{\done}{\hfill \ensuremath{\clubsuit}\medskip} \newcommand{\done}{\hfill \ensuremath{\Box}} \newcommand{\cl}[1]{{\overline #1 }} \font\namefont=cmr10 scaled\magstep2 \usepackage{graphicx} \begin{document} \title{Conjugation Problems for Hirsch Foliations} \author{Joseph H. Shive} \pdegrees{B.S. (University of Illinois at Chicago 1991)\\M.S. (University of Illinois at Chicago 2001)} \degree{Doctor of Philosophy in Mathematics} \maketitle \copyrightpage \dedication {\null\vfil {\large \begin{center} Dedicated to the Memory of Donna J. Shive (1938-2001) \end{center}} \vfil\null} \acknowledgements { Any list of acknowledgements will, of course be incomplete. Ther are professors who took extra time to explain concepts, friends and family who have provided emotional and material support, and fellow students with whom I have discussed mathematics. I will try to thank the most important few: \noindent My advisor Steve Hurder, for his patience and guidance over the years. \noindent My dad Jim, and his wife Charlotte, and my brothers and sisters Geoff and Lois, Peg and Peter, Jon and Shirley, who have all supported me in every way possible. \noindent John, Tony, Gaspar and Ken, for their good friendship. \vspace{60pt} {\hfill JHS} } \tableofcontents \listoffigures \summary In this thesis, we study the problem of when two $C^{r}$-foliations of codimension one on compact manifolds which are topologically conjugate must be $C^{r}$-conjugate, or at least $C^{r}$-conjugate on exceptional minimal sets. The transverse geometry of an exceptional minimal set in codimension one is that of a geometric Cantor set, and for a Markov minimal set, there is a finite set of linearly contracting generators for the induced holonomy pseudogroup. Our main result gives a solution of the conjugacy problem for Markov minimal sets in terms of the asymptotic ratio function defined on the endset of the typical leaf in the minimal set. The solution is obtained by studying the conjugacy problem first on Cantor sets in the line, and then extending and interpreting this solution in the context of maps between foliations. The second part of this thesis is the investigation of the conjugacy problem for a class of codimension one foliations which generalize a construction by M. Hirsch. \newcounter{ctr} % \parindent=0pt \chapter{Introduction} \section{Introduction} In this thesis, we study the problem of when two $C^{r}$-foliations of codimension one on compact manifolds which are topologically conjugate must be $C^{r}$-conjugate, or at least $C^{r}$-conjugate on exceptional minimal sets. The transverse geometry of an exceptional minimal set in codimension one is that of a geometric Cantor set, and for a Markov minimal set, there is a finite set of linearly contracting generators for the induced holonomy pseudogroup. Our main result gives a solution of the conjugacy problem for Markov minimal sets in terms of the asymptotic ratio function defined on the endset of the typical leaf in the minimal set. The solution is obtained by studying the conjugacy problem first on Cantor sets in the line, and then extending and interpreting this solution in the context of maps between foliations. The second part of this thesis is the investigation of the conjugacy problem for a class of codimension one foliations which generalize a construction by M. Hirsch. \section{Some background for the casual reader} \subsection{Cookie Cutters} A casual reader of this thesis, if there is one, might be familiar with geometric objects called fractals. Fractals are described in the popular literature as self-similar sets, i.e. a small part of the set will be a scaled down replica of the whole set. Perhaps the simplest example of a self-similar fractal is the Cantor middle-third set. To construct the middle third set, we begin with the interval $I=[0,1]$. We subdivide $I$ into thirds, and remove the middle third $G = (\frac{1}{3},\frac{2}{3})$. After this removal, we're left with $I_0 = [0,\frac{1}{3}]$ and $I_1 = [\frac{2}{3},1]$. We repeat this process on both $I_0$ and $I_1$, removing the middle thirds, which we call $G_0$ and $G_1$, respectively from both of them. The second approximation consists of the intervals $I_{00} = [0,\frac{1}{9}]$, $I_{01}= [\frac{2}{9}, \frac{1}{3}]$, $I_{10}= [\frac{2}{3}, \frac{7}{9}]$, and $I_{01}= [\frac{8}{9},1]$. We remove the middle thirds from each of these four intervals, $G_{00}$, $G_{01}$, $G_{10}$, and $G_{01}$ respectively, obtaining eight disjoint closed intervals, $I_{000}$, $I_{001}$, $I_{010}$, $I_{011}$, $I_{100}$, $I_{101}$, $I_{110}$, and $I_{111}$,each with length $\frac{1}{27}$. If $w$ is any finite string of 0s and 1s, we then $I_w$ will eventually be defined in this process. We continue this an infinite number of times to obtain $C$. The intervals $I_w$ shrink down to points in $C$. \begin{figure}[htbp] \centering \includegraphics[width=0.5\textwidth]{./images/middlethird.png} \caption{The Middle Third Set} \end{figure} We say $G_w$ is the {\em gap} of $I_w$. Note that each of the subsets $C_w = C\bigcap I_w$ are similar to the whole set $C$, since we simply repeated the same construction on each of the sub-intervals as we did on the whole set, so we call $C_w$ a {\em clone} and $I_w$ a {\em clone interval}. The self-similarity of $C$ shows us that if we take either of the clones, $C_0$ or $C_1$, and stretch them out by a factor of 3, we get $C$ back. That is, $C$ is preserved by the map \[ F:x \mapsto \left\{ \begin{array}{lr} 3x & x\in [0,\frac{1}{3}]\\ 3x-2& x\in [\frac{2}{3} ,1] \end{array}\right.\] \begin{figure}[htbp] \centering \includegraphics[width=0.4\textwidth]{./images/3xmod1.png} \caption{ {The function $F$ which preserves the middle third set}} \end{figure} $F$ has two right inverses, $$F_0^{-1}:x\mapsto \frac{x}{3}$$ $$F_1^{-1}:x\mapsto \frac{x+2}{3}$$ We can use $F$ to define the intervals: \begin{eqnarray*} I_0=F_0^{-1}[I] & ~ & I_1=F_1^{-1}[I]\\ I_{00}=F_0^{-1}[I_0] & ~ & I_{01}=f_0^{-1}[I_1] \end{eqnarray*} and so forth, where $F_0^{-1}[I]=\set{F_0^{-1}(x)|x\in I}$. In general for $w$ a finite word, $I_{iw}=F_i^{-1}[I_w]$, and conversely $F[I_{iw}] = I_w$. As the sub-intervals shrink down to points in $C$, we see that for $F(x_{\epsilon_0\epsilon_1...})=x_{\epsilon_1...}$. We note that $C$ is totally disconnected, perfect, compact, and has the cardinality of the reals. This classifies $C$ up to homeomorphism; any topological space with these properties is homeomorphic to $C$. We call any set which is homeomorphic to the middle-third set Cantor set. A cookie cutter function set (see page~\pageref{'cookcut'}) is a generalization of the function $F$ defined above. We let $I_0=[0,a]$ and $I_1=[b,1]$ be disjoint intervals, and let $h_0:I_0\To I$ and $h_1:I_1\To I$ be functions whose derivatives are bounded above 1. Then define \begin{eqnarray*} I_0=h_0^{-1}[I] & ~ & I_1=h_1^{-1}[I]\\ I_{00}=h_0^{-1}[I_0] & ~ & I_{01}=h_0^{-1}[I_1] \end{eqnarray*} and for $w$ any finite string of 0's and 1's, $I_{iw}=h_i^{-1}[I_w]$. Then, as with the middle third set, the intervals will shrink down to points in a Cantor set, which we'll call $C$ again. \begin{figure}[htbp] \centering \includegraphics[width=0.45\textwidth]{./images/cookcut.png} \caption{ {A cookie cutter map with two generators}} \end{figure} So we have a Cantor set, $C$, defined in a similar manner to the middle third set. The map, \[ h:x \mapsto \left\{ \begin{array}{lr} h_0(x) & x\in [0,a]\\ h_1(x) & x\in [b ,1] \end{array}\right.\] acts on $C$ by stretching both $C_0$ and $C_1$ out and laying them back down on top of all of $C$. \medskip If we know the lengths of all the gaps and clone intervals, of course, we then know how to construct $C$. If we know the relative sizes of the intervals, we can construct $C$ up to affine rescaling. For all finite words $w$, we let $l(w)$, $g(w)$, and $r(w)$ denote $\frac{|I_{w0}|}{|I_w|}$, $\frac{|G_w|}{|I_w|}$, and $\frac{|I_{w1|}}{|I_w|}$ respectively, where $|I_w|$ is the length of $I_w$. Then, all we need, besides $l(w)$, $g(w)$, and $r(w)$ for all $w$, to construct $C$ is the interval $I$ (the convex hull of $C$). Thus we have the {\em ratio geometry function}, $ w\mapsto (l(w), g(w), r(w) ) = (\frac{|I_{w0}|}{|I_w|}, \frac{|G_w|}{|I_w|}, \frac{I_{|w1|}}{|I_w|})$ whose domain is $\{0,1\}^{\mathbb N}$, and whose range is contained in the unit two-simplex $\{(x,y,z)|x+y+z < 1\}$. The middle third set has ratio geometry $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ for every word $w$. \subsection{Foliations and the Hirsch Example} In this thesis, we'll look at Cantor sets which arise in the context of dynamical systems, and ultimately at Cantor sets which arise in the context of foliations. We think of a foliation as a manifold with a local product structure. The examples we consider will all be three dimensional manifolds with two dimensional ( codimension 1) foliations. Informally, we say that the neighborhood of any point looks like a stack of papers, or the pages in a book. These stacks overlap, so that papers in one stack continue on into the neighboring stack, but as you move from one stack to another the corresponding pages might get closer together or farther apart. The ``stacks of papers'' are formally called {\em foliation charts} and each individual page is called a {\em plaque}. If we start with one page in one stack, and follow it in to a neighboring stack, and from there to another neighboring stack, we build up the {\em leaf}, which consists of all the plaques which you can ever reach this way. A leaf will be a two-dimensional manifold embedded in the ambient three-manifold. For a more formal definition, see Section~\ref{sec-foliations}. The examples of foliations we consider will all be generalizations of the {\em Hirsch foliation} \cite{Hirsch1975}, which we describe in detail in Section~\ref{sec-hirsch}. We obtain the Hirsch foliation by starting with a solid torus and, from the interior, removing another solid torus which wraps around twice. This gives us a manifold, foliated by two-holed disks, with two transverse toruses as boundary components. The exterior component is a torus which wraps around once, the interior component is a torus which wraps around twice. We then glue the exterior boundary component to the interior component to obtain a foliated manifold without boundary. In order to preserve the foliation, the gluing map must identify latitudinal circles from the interior boundary component to latitudinal circles on the exterior component. In the longitudinal direction, the gluing map reduces to a 2 to 1 local homeomorphism of the circle. \begin{figure}[htbp] \centering \includegraphics[width=0.25\textwidth]{./images/hirsch.png} \caption{ {The Hirsch Foliation: We glue the top of the cylinder to the bottom to get a solid torus with another solid torus removed. Then we glue the inside component to the outside component to get the Hirsch foliation. }} \end{figure} We think of the two-holed disks $D_i= D_2\times \{t\}$ as (non-standard) plaques of $M$. We think of these two-holed plaques as pairs of pants. The waist corresponds to an exterior circle, the cuffs correspond to an interior circle. As we glue the interior component to the exterior component, we form ${\mathcal F}$ by successively sewing these waist of one pair of pants, corresponding to the point $h(x)$, to the cuff of another pair corresponding to the point $x$. We see that a typical leaf looks like a tree made of tubing. If the leaf corresponds to a periodic orbit of $h$, then at some point we'll sew a cuff to a waist that has already been sewn in, so the leaf will be a tree made of tubing. If the leaf doesn't correspond to a periodic point of $h$, the leaf is an infinite tree with no handles. \begin{figure}[htbp] \centering \includegraphics[width=0.7\textwidth]{./images/leaf.png} \caption{ {As we sew pairs of pants together to get the leaves, we see that the leaves will be coarsely equivalent to a tree. }} \end{figure} \subsection{The conjugation problem} One of the basic questions about foliations asked by H.B.~Lawson in his survey on foliations (see sections 5 and 8 of \cite{Lawson1974}) is: if two foliations are homeomorphic, are they necessarily diffeomorphic? Let $(M_1, {\mathcal F}_1)$ and $(M_2, {\mathcal F}_2)$ be foliations of transverse differentiability class $C^{k + \lambda}$ and $\Phi \colon M_1 \to M_2$ a $C^r$-homeomorphism, for $r \leq k$, which maps the leaves of ${\mathcal F}_1$ to the leaves of ${\mathcal F}_2$. The conjugacy problem for foliations is to find conditions on the foliations and the map $\Phi$ which are sufficient to imply that the map $\Phi$ is also $C^{k + \lambda}$. The study of conjugacy problems has a long tradition in dynamical systems. One of the most influential results is due to D.~Anosov, who showed in his Thesis \cite{Anosov1969} that there are $C^{\infty}$-flows on 3-manifolds which are $C^1$ conjugate but cannot be $C^2$ conjugate. The work of S.~Hurder and A.~Katok \cite{HK1990} showed that the crucial question is the regularity of the weak-stable foliations associated to these Anosov flows. The foliations themselves are transversally $C^{1 + \lambda}$ for any real number $0 < \lambda < 1$, but if the weak-stable foliations are $C^2$ then the flow is smoothly conjugate to an algebraic model. One of the key tools for the study of the conjugacy problem is the so-called ``bootstrapping process'' for proving the regularity of a map conjugating two smooth hyperbolic dynamical systems on compact manifolds, introduced in a series of foundational papers by R.~de~la~Llave, J.~Marco, and R.~Moriyon \cite{deLlave1987,deLlave1992,LMM1986,LM1988,MarcoMoriyon1987a,MarcoMoriyon1987b}. This was further developed for the study of the conjugations between stable foliations by B.~Hasselblatt in the papers \cite{Hasselblatt1992,Hasselblatt1994,Hasselblatt1997} who gave obstacles to regularity in the form of ``bunching data'' for the eigenvalues at periodic points of the hyperbolic maps. The foundational papers of D.~Sullivan \cite{Sullivan1988,Sullivan1992} on the conjugacy problem for geometrically defined Cantor sets and ``cookie-cutter'' dynamical systems used the ``ratio geometry'' for an exceptional minimal set to define a scaling function. The scaling function introduced by Feigenbaum to study $k$ to 1 maps of the circle with dense minimal sets. The scaling function is defined on the dual Cantor set associated to the dynamical system, and classifies the local differentiable structure of the Cantor set. These methods were subsequently applied by T.~Bedford and A.M.~Fisher \cite{BF1997}, to study the ``scenery process'' during which they flushed out the proofs from Sullivan's original work. The study of the geometry and dynamics of a foliation $\mathcal F$ on a compact manifold encompasses all of the issues with the dynamical systems defined by a flow (as a flow defines a foliation with 1-dimensional leaves) plus much more. When the leaves have higher dimension, the geometry of the leaves can make the dynamics of the foliation far more complicated than that encountered in the study of flows. This complexity forces the study of foliations to focus on ``model problems'', which are typical examples where a solution provides a model for more general cases. In this thesis we study the conjugacy problem for the ``Hirsch foliations''. The original Hirsch foliation was a construction of a codimension one foliation on a compact 3-manifold such that the foliation was real analytic and had an exceptional minimal set.\cite{Hirsch1975} One of the points of this thesis is to present a generalization of the Hirsch construction which results in a very broad class of dynamical behavior for the resulting foliations. These will be our model problems which will motivate our discussion of the conjugacy problem. The theory we develop will mostly solve problem 1, while only laying a possible background for problems 2 and 3. \begin{prob}\label{prob1} Let $(M_1, {\mathcal F}_1)$ and $(M_2, {\mathcal F}_2)$ be generalized Hirsch foliations of transverse differentiability class $C^{k + \lambda}$ and $\Phi \colon M_1 \to M_2$ a $C^r$-homeomorphism, for $r \leq k$, which maps the leaves of ${\mathcal F}_1$ to the leaves of ${\mathcal F}_2$. Find conditions which imply that $\Phi$ is $C^{k + \lambda}$. \end{prob} Ghys and Tusboi considered the conjugacy problems for $C^2$ foliations on compact manifolds in \cite{GhysTsuboi1988}. Their basic technique was similar to the classical method used in Shub and Sullivan \cite{ShubSullivan1985}, and is based on the observation that a $C^2$-map which commutes with a linear contraction is itself linear. Another purpose of this thesis is to develop and apply the more general techniques of bootstrapping and ratio geometry, which were developed for the study of standard dynamical systems, to the study of the conjugacy problems for foliations, so that it applies to foliations whose differentiability is at least $C^1$. Note that Cantwell and Conlon exhibited foliations of all degree of differentiability which cannot be smoothed to a higher degree in \cite{CantwellConlon1988a,CantwellConlon1994}. The boot strapping techniques used in this thesis applies to the holonomy of Hirsch foliation, not just in a neighborhood of the minimal set, but everywhere holonomy element which doesn't have a non-hyperbolic fixed point. \begin{prob}\label{prob2} What can be said about the conjugacy problem for generalized Hirsch foliations whose holonomy has special points, and is not completely hyperbolic? \end{prob} Finally, the work in this thesis suggests a very general problem, which we only begin to solve. \begin{prob}\label{prob3} Classify the holonomy pseudogroups which arise from the generalized Hirsch foliations. \end{prob} \subsection{The conjugation problem for cookie cutters.} Sullivan's solution of the conjugation problem for cookie cutters uses the {\em ratio geometry} to define the {\em scaling function}, which categorizes the differentiable structure. The ratio geometry is of an interval $I_\omega$ is given by the relative lengths of $I_{\omega 0}$, $G_{\omega}$, and $I_{\omega 1}$. So we get the ordered triplet $(\frac{|I_{\omega 0}|}{|I_\omega|},\frac{|G_{\omega }|}{|I_\omega|},\frac{|I_{\omega 1}|}{|I_\omega|})$. We normally just think of this as a function of $\omega$. These ratios look a little bit like difference quotients for $f^{-1}$, but, for instance, if we write $\omega = w_n w_{n-1}\dots w_2 w_1$, then $I_{\omega 0}=f_{w_n}^{-1}\circ f_{w_{n-1}}^{-1}\circ\dots\circ f_{w_2}^{-1}\circ f_{w_1}^{-1}\circ f_0^{-1}\circ f^n[I_\omega]$. As it turns out, the conjugacy class of $f$ determines the ratio geometry exponentially closely, which is our first main theorem for cookie cutters, which is proved on page~\pageref{lin->exp:pf}. \begin{thm} \label{lin->exp}(Sullivan) If two cookie cutters are $C^{1+\lambda}$ conjugate, then their ratio geometries are exponentially equivalent. \end{thm} A corollary of this is that as we add symbols to the left of $\omega$, the ratio geometry converges exponentially fast to a limiting geometry which we call the scaling function. This gives us our second main theorem for cookie cutters which is proved on page~\pageref{lin->scl:pf}. \begin{thm} \label{lin->scl}(Sullivan) $C$ and $C'$ are exponentially equivalent cookie cutters if and only if they have the same scaling function. \end{thm} Taken together, these two theorems imply that $C^{1+\lambda}$-conjugate cookie cutters have the same scaling function. As we add symbols to the left of $\omega$, the intervals $I_{w_{n+k},\dots w_{n+2} w_{n+1} \omega}$ gets smaller and smaller, so in as much as the ratio geometry looks like difference quotients, the scaling function looks like derivatives. But besides the fact that we aren't really forming a difference quotient, $I_{w_{n+k},\dots w_{n+2} w_{n+1} \omega}$ will bounce around depending on whether $I_{w_{n+k}}$ is 0 or 1. Even though, in a neighborhood of the minimal set, the converse of the above two theorems are true (page~\pageref{scl->lin:pf}), hence the scaling function is just what is needed to classify the $C^{1+\lambda}$ structure: \begin{thm} \label{scl->lin}(Sullivan) Let $C$ and $C'$ be two exponentially equivalent labeled Cantor sets. Then they're $C^{1+\lambda}$ conjugate in some neighborhood of $C$. \end{thm} Theorems \ref{lin->scl} and \ref{scl->lin} taken together imply that two Cantor sets with the same scaling function are $C^{1+\lambda}$-conjugate on some open neighborhood. Furthermore, using the bootstrapping method, the scaling function $\phi:\Sigma^{-} _{2} \To{int \Delta^{2}}$ classifies the $C^{k+\lambda}$ structure (page~\pageref{cookieboot:pf}), where $\Sigma^{-} _{2} $ is the set of left-infinite strings on two symbols, and $\Delta^{2}$ is the 2-simplex $\set{(x,y,z)\in \Real^{3}|x^{2}+y^{2}+z^{2}\le 1}$. \begin{thm} (Sullivan) \label{cookieboot}Let $\phi$ be a $C^{1+\lambda}$ conjugation between $C^{k+\lambda}$ cookie cutters. Then $\phi$ is itself $C^{k+\lambda}$ \end{thm} \subsection{Dynamically defined Cantor sets on the circle} Let $f$ be a cookie cutter function on any interval, say $I=[0,1]$. After identifying $-A$ with $1$, we can think of $I$ as a subset of the circle $[-A,1]/_{(A\sim1)}$. Then if we extend $f$ to a function $F$ as pictured in Figure~\ref{fig:cookiecircle}, we get a 2 to 1 cover of the circle with the same minimal set as $f$. We see that the fixed interval $H=(-A,0)$ acts as a trap for forward iterations of $F$. The gaps of $C$ are all backwards iterates of the fixed interval. This was the function Hirsch described in his construction. \begin{figure}[htbp] \centering \includegraphics[width=0.6\textwidth]{./images/flat1.png} \caption{ {A cookie cutter on the circle}}\label{fig:cookiecircle} \end{figure} On $I_0$ and $I_1$, $F=f$ so the dynamics of $F$ are the same, and the discussion about ratio geometry and the scaling function still applies. We've extended $f$ to the gaps $G$ and $H$. But the backwards iterates of $G$ still consists of all the gaps in $I_0\cup I_1$, and the only forward iterate of the interval $G$ is the interval $H$. On $H$, as we've drawn $F$, there's an attracting fixed point, $p$, but the backwards orbit of $p$ approaches the Cantor set $C$. \begin{figure}[htbp] \centering \includegraphics[width=0.6\textwidth]{./images/exceptional1.png} \caption{ {A map with a period two interval. $f[G_1]=G_2$ and $f[G_2]=G_1$, so $G_1 \cup G_2$ acts as a trap for the dynamics of $f$. If $f'>1$ off of $G_1 \cup G_2$, then the backwards images of $G_1 \cup G_2$ form the gaps of an invariant Cantor set.}}\label{fig:per2int} \end{figure} In general, the way we construct a $C^1$ function, $f$ on a one dimensional manifold which generates a Cantor set, $C$ is we have an interval, $G$, that traps the dynamics of $f$. Once we land in $G$, we can't get back out. The backwards iterates of $G$ are the gaps of $C$. For a cookie cutter on the interval, $f$ isn't even defined on $G$, so once we land in $G$, we can't iterate $f$ at all. When we extended $f$ to the circle, the trap is the interval $H$. Once the orbit of a point lands in $H$, it stays there, and so is asymptotic to a fixed point. We could also trap the dynamics of $f$ by having a periodic interval. In Figure~\ref{fig:per2int}, we graph a 2 to 1 local diffeomorphism on $S^1$ with a period 2 interval, $G_1$. $f[G_1]=G_2$ and $f[G_2]=G_1$. Once an orbit of a point lands in $G_1\cup G_2$, it stays there. If $f$ is hyperbolic off of $G_1\cup G_2$, then the inverse orbit of $G_1\cup G_2$ is a dense set of open intervals, and hence forms the gaps of a closed totally disconnected space, and as it turns out are the gaps of a Cantor set, $C$. $C$ is the minimal set for $f$. Just as we did with the cookie cutter, we can use the dynamics of $f$ to define a structure of nested sub-intervals and gaps. We start with the three closed intervals $I_0$, $I_1$, and $I_2$. But instead of labeling the gaps according to the interval they're inside of, as we did for the cookie cutter, we label them according to the interval they're next to. We arbitrarily choose to label them according to the interval to their right. Remember that we've identified the endpoints to get a circle, so that $f$ is continuous on $I_1$. Also $I_0$ and $I_2$ share an endpoint. \begin{figure}[htbp] \centering \includegraphics[width=.6\textwidth]{./images/cantorset_3gap1.png} \caption{ {The nested interval structure of a function with a period two interval. As we iterate the process, the intervals $I_{w}$ will nest down to an invariant Cantor set.}} \end{figure} We restrict $f$ to three domains to get the three diffeomorphisms acting on intervals: $$f_0=f|_{I_0} \hspace{.3in}f_1=f|_{G_1\cup I_1} \hspace{.3in}f_2=f|_{G_2\cup I_2}$$ We'll label the sub-intervals in a similar manner as we did for cookie cutters. Starting with $I_{00}$ and moving to the right, we get $$ I_{00}=f_0^{-1}(I_0) \hspace{.3in}G_{00}=f_0^{-1}(G_0) \hspace{.3in}I_{01}=f_0^{-1}(I_1)$$ $$ I_{12}=f_1^{-1}(I_2) \hspace{.3in}I_{10}=f_1^{-1}(I_0)$$ $$ I_{21}=f_2^{-1}(I_1) \hspace{.3in}G_{22}=f_2^{-1}(G_2) \hspace{.3in}I_{22}=f_2^{-1}(I_2)$$ Continuing in this manner, for a $w$ finite string of 0s 1s and 2s, if we've defined the interval $I_w$ or the gap $G_w$, then we define $I_{iw}=f_i^{-1}(I_w)$ whenever $I_w$ is in the domain of $f_i^{-1}$. We define $G_{iw}$ similarly. This is an example of a {\em hyperbolic Markov exceptional minimal set}, which we'll shorten to Markov exceptional sets even though one could talk about non-hyperbolic Markov exceptional minimal sets. While the notation is trickier, all of the discussion about the ratio geometry and scaling function applies to any Markov exceptional set. The intervals shrink down to points in the Cantor set, and so define a labeling on points of $C$. But since $I_2$ and $I_0$ overlap at their endpoint, the coding is not unique, and because not all words are realized, the labeling on $C$ is a semiconjugacy to a subshift. We can still define the ratio geometry, and scaling function which classify the differentiable structure, giving us the same picture as for the cookie cutters. See page~\pageref{Mark-disc} for a discussion of the following theorems. The proofs are almost identical to theorems \ref{lin->exp}, \ref{lin->scl}, and \ref{cookieboot} requiring only changing to the notation outlined on page~\pageref{sclMark}. We don't need to state an analogue to theorem~\ref{scl->lin} since we already stated it in a more general setting. \begin{thm}\label{M-lin->exp} Let $C$ and $C'$ be $C^{1+\lambda}$ conjugate Markov exceptional sets. Then the ratio geometry of $C$ is exponentially equivalent to the ratio geometry of $C'$. \end{thm} \begin{thm}\label{M-lin->scl} Let $C$ and $C'$ be Markov exceptional sets. Then they are exponentially equivalent if and only if they have the same scaling function. \end{thm} \begin{thm}\label{M-boot} Let $\phi$ be a $C^{1+\lambda}$ conjugation between $C^{k+\lambda}$ Markov exceptional sets. Then $\phi$ is itself $C^{k+\lambda}$. \end{thm} \subsection{The Conjugation Problem for Hirsch Foliations} The transverse dynamics of the Hirsch foliation is given by an orientation preserving two to one local diffeomorphism, of the circle $h$. A leaf corresponds to a total orbit of $h$, and a point of period $k$ corresponds to a handle composed of $k$ plaques. We will choose $M$ to be a Hirsch foliation whose holonomy function $h$ has a Markov exceptional set. Then the earlier discussion must apply to $h$. For instance, we can use a Markov basis for $h$ to define a ratio geometry on the transversal of $M$. In this situation, the theory of Markov exceptional sets applies to the transversals of $M$, so we have the following theorems. (See page~\pageref{foliatdisc}.) \begin{thm}\label{foliatexp} Let $F:(M_1,{\mathcal F}_1)\to F:(M_2,{\mathcal F}_2)$ be a $C^{1+\lambda}$ diffeomorphism. Then the transverse ratio geometry of ${\mathcal F}_1$ is exponentially equivalent to the transverse ratio geometry of ${\mathcal F}_2$. \end{thm} This theorem is trivially true as a special case of theorem~\ref{M-lin->exp}. But the context of the foliation gives us another geometrical interpretation of the ratio geometry. The two branches of $h^{-1}$ are both holonomy elements. That means there exists a path $\gamma_0$ so that we apply $h_0^{-1}$ to a sub-interval, $J$, of the transverse circle by flowing $J$ along the path $\gamma_0$, and likewise for $h_1^{-1}$. The holonomy along the catenation the paths $\gamma_i$ is given by iterating $h_0^{-1}$ and $h_1^{-1}$. So instead of defining the scaling function on an abstract dual Cantor set, we can define it on a set of infinitely long holonomy paths. \begin{thm}For a foliation with a Markov exceptional set, as we flow along an infinitely long path with contracting holonomy, the transverse ratio geometry will converge to the scaling function of the transverse minimal set. \end{thm} \begin{obs} For a Hirsch foliation, this gives us a nice geometric interpretation for the dual Cantor set on which the scaling function is defined. The paths we flow along, in general, will go to an end of the leaf. (Though it might also go around a handle.) In particular, if we choose a leaf $L$ with no handles, then the domain of the scaling function is a subset of the endset of $L$. \end{obs} \begin{thm} Let ${\mathcal F}_1$ and ${\mathcal F}_2$ be $C^{1+\lambda}$ conjugate foliations with Markov exceptional sets. Then they have the same scaling function. \end{thm} \begin{thm} Let $\phi$ be a $C^{1+\lambda}$ diffeomorphism between $C^{k+\lambda}$ foliations ${(M_1,\mathcal F}_1)$ and $ (M_2,{\mathcal F}_2)$ with Markov minimal sets. Then, $\phi$ is itself transversally $C^{k+\lambda}$ in a neighborhood of the exceptional minimal set. \end{thm} By applying a smoothing lemma, we have the application: \begin{thm} Let $\phi$ be a $C^{1+\lambda}$ diffeomorphism between $C^{k+\lambda}$ foliations ${(M_1,\mathcal F}_1)$ and $ (M_2,{\mathcal F}_2)$ with Markov minimal sets. Then $\phi$ is $C^0$ close to a $C^{k+\lambda}$ function in a neighborhood of the minimal set. \end{thm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \chapter{Definitions and Concepts} \section{Lipschitz and H\"older Continuity} \begin{defn} A map $\phi: X \longrightarrow Y$ between two metric spaces is said to be {\em Lipschitz} if it only scales distances by a bounded amount. That is there exists a constant $K$ such that for all $x$ and $y \in X$, \[ d(\phi(x),\phi(y)) \le Kd(x,y)\] \end{defn} For maps of the real line, we can reformulate this definition to say that the difference quotient is bounded: \[ \left| \frac{\phi(x)-\phi(y)}{x-y} \right| \le K \] $K$ is called the {\em Lipschitz constant} for $\phi$. We say that $\phi$ is less than $K$ in the Lipschitz norm. So if $K$ is minimal, then $K$ is the Lipschitz norm of $\phi$. A Lipschitz map can easily be non-differentiable $-$ for instance a piecewise linear map is Lipschitz $-$ but since the difference quotient is bounded, a Lipschitz map is better behaved than an arbitrary continuous map. If we let the domain of $\phi$ be the compact interval $I = [a,b]$, then by the Mean Value Theorem, if $\phi$ is $C^1$, then $\phi$ is Lipschitz. Furthermore, a Lipschitz map is $C^0$, so if we restrict our domains to compact intervals, the Lipschitz condition provides an intermediate level of smoothness between $C^0$ and $C^1$. \begin{ex} A differentiable map on a compact interval $I= [a,b]$ is Lipschitz. By the mean value theorem, for $x, y \in [a,b]$, there exists $z \in [x,y]$ such that $f(x) - f(y) = f'(z)(x-y)$ so $C={\displaystyle \max_{z\in[a,b]} }f'(z)$ works as the Lipschitz constant. \end{ex} \begin{ex} Let $\phi: I \longrightarrow {\Real}$ be piecewise Lipschitz. That is let $I = [a,b]$ and $ a=a_0 < a_1 < a_2 < \dots < a_N = b$. On each interval $I_i= [a_i,a_{i+1}]$, let $|\phi(x) - \phi(y)| \le C_i |x-y|$. Then $\phi$ is Lipschitz on $I$. \end{ex} \noindent \textbf{Proof:} Let $i < j$, $x \in I_i$ and $y \in I_j$, then \begin{eqnarray*} |\phi(y) - \phi(x)| & \le & |\phi(y) - \phi(a_j)| + |\phi(a_j) - \phi(a_{j-1})| + \dots + |\phi(a_{i+1})- \phi(x)|\\ & \le& C_i|y-a_j| + C_{i-1}|a_j-a_{j-1}| + \dots + C_i|a_{i+1}-x|\\ & \le& \max C_i \left[ (y-a_j) +(a_j- a_{j-1}) + \dots (a_{i+1}-x)\right]\\ &= & C|y-x| \end{eqnarray*} Note that this example can be generalized. We break the interval $I$ up into an infinite number of sub-intervals. We require that $\phi$ be Lipschitz on each interval $I_i$ and that the Lipschitz constants $C_i$ are all bounded. Then $\phi$ is Lipschitz on all of $I$. \begin{prop} Let the series $\sum f_i(x)$ be absolutely convergent on $[0,1]$. Let $\phi$ be a Lipschitz map on $[0,1]$. Set $\phi(0) = f_i(0) = 0$. Then the series $\sum\phi \circ f_i$ is also absolutely convergent. \end{prop} \noindent \textbf{Proof:} \begin{eqnarray*} \sum |\phi \circ f_i(x)| & \le & \sum \left|\phi(f_i(x)) - \phi(f_i(0))\right|\\ & \le & \sum K|f_i(x) - f_i(0)|\\ & \le & \sum K|f_i(x)| \end{eqnarray*} which converges since $f_i$ is absolutely convergent. \done \begin{prop} Let $f$ and $g$ be Lipschitz maps on a compact interval $I$. Then $f+g$, $f\cdot g$, and $f \circ g$ are all Lipschitz. \end{prop} \noindent \textbf{Proof:} \begin{enumerate} \item $ f + g$ is Lipschitz by the triangle inequality.\\ \item $|f(x) g(x) - f(z) g(z)| \le |f(x)|\cdot |g(x)- g(z)| + {|g(z)| \cdot |f(x) -f(z)|}$\\ \item $ |f\circ g(x) - f\circ g(z)| \le C_f|g(x)-g(z)| \le C_fC_g|x-z|$. \end{enumerate} \done %====================================================================== \begin{defn} Let $0 < \lambda \le 1$. Then $\phi$ is {\em H\"older continuous} of degree $\lambda$ if there exists numbers $0<\delta<1$ and $K>0$ such that when $|x-y|<\delta$, $|\phi(x) - \phi(y)| \le K |x-y|^\lambda$. $K$ is called the {\em H\"older constant} for $\phi$. \end{defn} The H\"older condition is a generalization of the Lipschitz condition which allows us to define more intermediate degrees of smoothness between $C^0$ and Lipschitz. Note that H\"older of degree $\lambda = 1$ is exactly the same as Lipschitz. The following two propositions justify the statement that H\"older continuity provides fractional degrees of smoothness between $C^0$ and Lipschitz. \begin{prop} For $\lambda \le \kappa$, $C^\kappa$ implies $C^\lambda$ I.e. if $\phi: I \longrightarrow J$ be a map of compact intervals which is H\"older of degree $\kappa$, then $\phi$ is also H\"older of degree $\lambda$. \end{prop} \noindent \textbf{Proof:} Let $x= a_0 < a_1< \dots ~1$, but we will restrict our attention to functions with compact domains, in which case H\"older of degree $>1$ implies that the function is locally constant. The following example shows why we define H\"older continuity as a local condition. \begin{ex} The identity$:{\Real} \longrightarrow {\Real}$, doesn't satisfy a global H\"older condition. If $\lambda < 1$, and $\lim _{n\rightarrow \infty}x_n = \infty$, then $\lim _{n\rightarrow \infty}\frac{|x_n-0|}{|x_n-0|^\lambda} = \lim_{n\rightarrow \infty}x_n^{1-\lambda}= \infty$. \done \end{ex} \begin{ex} Since the tangent line for $f(x) =\sqrt x$ is vertical at $x=0$, (and hence the difference quotient is unbounded,) $f$ is neither differentiable nor Lipschitz on the interval $[0,1]$. $f$ is H\"older of degree $\frac{1}{2}$ though. \end{ex} \noindent \textbf{Proof:} Without loss of generality, take $x\ge y$. Then we need to show that $\sqrt x - \sqrt y \le \sqrt{x-y}$. \begin{eqnarray*} x & \le & x + 2\sqrt{x-y}\sqrt y \\ & \le & (x-y) + 2\sqrt{x-y}\sqrt y + y\\ &=& (\sqrt{x-y} +\sqrt y)^2\\ \sqrt x & \le &\sqrt{x-y} + \sqrt y\\ \sqrt x - \sqrt y &\le & \sqrt{x-y} \end{eqnarray*} \done \begin{prop} If $\phi$ is Lipschitz, then $\sqrt\phi$ is H\"older of degree $\frac{1}{2}$. \end{prop} \begin{ex} Unlike Lipschitz maps, composition with H\"older maps need not preserve absolute convergence. Let $f_n(x) = \frac{x}{n^2}$ and $ \phi(x)= \sqrt x$. Then $\sum f_n(x) = x \sum \frac{1}{n^2}$ which converges absolutely, but $\sum \phi \circ f_n(x) = \sqrt x \sum \frac{1}{n}$ which doesn't converge. \done \end{ex} While H\"older continuous functions don't necessarily preserve absolute convergence, they do preserve geometric sums. This is exactly what we will need to prove the regularity theorems. \begin{prop} Let $r<1$, $\lambda <1$, and $|f_n(x)| \le r^n$. Let $\phi$ be H\"older of degree $\lambda$. Further let $\phi(0) = f_n(0) = 0$. Then $\phi \circ f_n(x)$ is also bounded by a geometric series, and hence is summable. \end{prop} \noindent \textbf{Proof:} \begin{eqnarray*} \sum|\phi \circ f_n(x)| & \le &\sum C|f_n(x)|^\lambda\\ & \le & \sum Cr^{\lambda n}\\ & \le & \frac{C}{1-r^\lambda} \end{eqnarray*} \done %************************************************************************** The H\"older condition gives us intermediate degrees of smoothness between $C^0$ and $C^1$. We extend this to get intermediate degrees between $C^k$ and $C^{k+1}$ simply by requiring the $k^{th}$ derivative to satisfy the H\"older condition: $f$ is $C^{k+\lambda}$ if $|f^{(k)}(x) - f^{(k)}(y)| \le C_f|x-y|^\lambda$. Again we note that this notation is insufficient when $\lambda = 1$, since $C^{k+1}$ doesn't just mean that the $k^{th}$ derivative is Lipschitz, but that the $ k+1^{st}$ derivative is continuous. When it's clear from context we'll use this notation for the Lipschitz case too. For instance when we say a map is $C^{0+1}$ we'll mean Lipschitz, as opposed to $C^1$ which means once differentiable. Recall that $f$ is less than $K$ in the $C^k$ norm if $f^{(j)}(x) \le K$ for all $j \le k$ and all $x \in I$. We say that $K$ bounds $f$ in the $C^{k+ \lambda}$ norm if in addition K is a H\"older constant for $f^{(k)}$. \begin{prop} Let $f$ and $g$ be two $C^{k+\lambda}$ maps. Then $f + g$, $f \cdot g$, and $f\circ g$ are all $C^{k+\lambda}$ too. \end{prop} \noindent \textbf {Proof:} \noindent i) f+g is $C^{k+\lambda}$: Let $|x-z| < \min(\delta_f,\delta_g)$. Then \begin{eqnarray*} \left| (f+g)^{(k)} (x) - (f+g)^{(k)}(z) \right| & \le &|f^{(k)}(x) -f^{(k)}(z)| + g^{(k)}(x) - g^{(k)}(z)|\\ & \le& C_f|x-z|^\lambda + C_g |x-z|^\lambda\\ &\le& K|x-z|^\lambda \end{eqnarray*} \noindent ii) $ (f\cdot g)$ is $ C^{k+\lambda}$: \begin{eqnarray*} \left| (f\cdot g)^{(k)} (x) - (f\cdot g)^{(k)}(y) \right| & \le &\sum_{j=0}^{k} {k \choose j}\left| f^{(j)}(x)g^{(k-j)}(x) - f^{(j)}(y)g^{(k-j)}(y) \right| \\ & \le & \sum_{j=0}^{k} {k \choose j} \left[\left| f^{(j)}(x)(g^{(k-j)}(x)-g^{(k-j)}(y))\right|\right.\\ & & \hspace*{1cm}+\left.\left| g^{(k-j)}(y)(f^{(j)}(x)-f^{(j)}(y))\right| \right]\\ &\le& M|x-y|^{\lambda} \end{eqnarray*} \noindent iii) $f \circ g$ is $C^{k+\lambda}$: By induction on $k$. For $k=1$, \begin{eqnarray*} |f\circ g'(x)- f\circ g' (z)| & =& |f'g(x)g'(x)- f'(g(z))g'(z)|\\ & \le& |f'g(x)| |g'(x)- g'(z)| + |g'(z)| |f'g(x)-f'g(z)|\\ & \le & K|x-z |^\lambda \end{eqnarray*} So the theorem is true for $k=1$. Now suppose it's true for $k 0$ so that for any $\alpha \in {\mathcal A}$ the set $\{\beta \in {\mathcal A} \;| \; U_{\alpha} \cap U_{\beta} \not= \emptyset \}$ has cardinality at most $m({\mathcal A})$ \par \indent {\textbf {(2)}} There are local coordinate charts $\phi_{\alpha} : U_{\alpha} \rightarrow (-1,1)^m $, so that each map $\phi_{\alpha}$ admits an extension to a homeomorphism $\tilde{\phi}_{\alpha} : \tilde{U}_{\alpha} \rightarrow (-2,2)^m$ where $\tilde{U}_{\alpha}$ contains the closure of the open set $U_{\alpha}$ \indent {\textbf {(3)}} For each $z \in (-2,2)^q$, the pre-image $ \tilde{\phi}_{\alpha}^{-1} ((-2,2)^p\times\{z\}) \subset \tilde{U}_{\alpha}$ is the connected component containing $\tilde{\phi}_{\alpha}^{-1}(\{0\} \times \{z\})$ of the intersection of the leaf of $\mathcal F$ through $\phi_{\alpha}^{-1}(\{0\} \times \{z\})$ with the set $\tilde{U}_{\alpha}$. The extensibility condition in~(2) is made to guarantee that the topological structure on the leaves remains tame out to the boundary of the chart $\phi_{\alpha}$. The collection $\{(U_{\alpha},\phi_{\alpha})\;|\;\alpha \in {\mathcal A}\}$ is called a {\em regular foliation atlas} for $\mathcal F$. The inverse images $$ {\mathcal P}_{\alpha}(z) = {\phi}_{\alpha}^{-1} ( (-1,1)^p \times \{z\}) \subset {U}_{\alpha}$$ are topological discs contained in the leaves of $\mathcal F$, called the {\em plaques} associated to this atlas. {\em We will assume that the covering is chosen so that all plaques have diameter at most 1.} One thinks of the plaques as ``tiling stones'' which cover the leaves in a regular fashion. The plaques are indexed by the {\em complete transversal} $\displaystyle \mathcal T = \bigcup_{\alpha \in {\mathcal A}} {\mathcal T}_{\alpha} $ associated to the given covering, where ${\mathcal T}_{\alpha} = (-1,1)^q$. The charts $\phi_{\alpha}$ define tame embeddings $$ t_{\alpha} = \phi_{\alpha}^{-1}(\{0\} \times \cdot) : {\mathcal T}_{\alpha} \rightarrow U_{\alpha} \subset V $$ We will implicitly identify the set $\mathcal T$ with its image in $V$ under the maps $\; t_{\alpha}$, though the union of these maps may not be not injective, but is at most finite-to-one. Finally, the fourth condition ensures that the leaves are $C^2$-manifolds: \indent {\textbf {(1.4)}} For each $z \in (-2,2)^q$, and $\beta$ so that $\phi_{\alpha}((-1,1)^p \times \{z\}) \cap U_{\beta} \not= \emptyset$ the transition function $\phi_{\alpha \beta, z}$ is $C^2$ uniformly in the parameter $z$, where $$ \phi_{\alpha \beta, z} = \phi_{\beta} \circ \phi_{\alpha}^{-1} \colon (-2,2)^p \times \{z\} \cap \phi^{-1}(U_{\alpha} \cap U_{\beta}) \longrightarrow (-2,2)^p $$ The foliation $\mathcal F$ is said to be $C^r$ if the foliation charts $\{\phi_{\alpha}\;|\;\alpha \in {\mathcal A}\}$ can be chosen to be $C^r$-diffeomorphisms. \subsection{The holonomy groupoid} A pair of indices $\alpha$ and $\beta$ is {\em admissible} if $U_{\alpha} \cap U_{\beta} \not= \emptyset$. For each admissible pair $\alpha, \beta $ define $${\mathcal T}_{\alpha\beta} = \{z \in {\mathcal T}_{\alpha} = (-1,1)^m \mbox{ such that } {\mathcal P}_\alpha (\phi_\alpha^{-1}(z) ) \cap U_{\beta} \not= \emptyset\}.$$ Then there is a well-defined transition function $\gamma_{\alpha\beta} \colon {\mathcal T}_{\alpha\beta} \longrightarrow {\mathcal T}_{\beta \alpha}$, which for $x \in {\mathcal T}_{\alpha\beta} $ is given by taking the plaque through $\phi_\alpha^{-1}(x)$ and projecting it onto the transversal of $U_\beta$. The formula for $\gamma_{\alpha\beta}(x)$ is \[ \gamma_{\alpha\beta}(x) = \pi_\beta \circ \phi_{\beta}[{\mathcal P}_\alpha (\phi^{-1}(x) ) \cap U_{\beta}] \in {\mathcal T}_{\beta \alpha}\] \noindent where $\pi_\beta$ is the projection of $\phi_\beta(U)$ onto ${\mathcal T}_{\beta}$. The continuity of the charts $\phi_{\alpha}$ implies that each $\gamma_{\alpha\beta}$ is continuous; in fact, one can see that $\gamma_{\alpha\beta}$ is a local homeomorphism from ${\mathcal T}_{\alpha\beta}$ onto ${\mathcal T}_{\beta \alpha}$ and that $\gamma_{\alpha\beta}^{-1}=\gamma^{}_{\beta\alpha}$. A {\em leaf-wise path} $\gamma$ is a continuous map $\gamma : [0,1] \rightarrow M$ whose image is contained in a single leaf of $\mathcal F$. Suppose that a leafwise path $\gamma$ has initial point $\gamma(0) = t_{\alpha}(z_0)$ and final point $\gamma(1) = t_{\beta}(z_1)$, then $\gamma$ determines a local holonomy map $h_{\gamma}$ by composing the local holonomy maps $\gamma_{\alpha\beta}$ along the plaques which $\gamma$ intersects. $h_{\gamma}$ is a local homeomorphism from a neighborhood of $z_0$ to a neighborhood of $z_1$. More generally, if the initial point $\gamma(0) $ lies in the plaque ${\mathcal P}_{\alpha}(z_0)$ and $\gamma(1)$ lies in the plaque ${\mathcal P}_{\beta}(z_1)$, then $\gamma$ again defines a local homeomorphism $h_{\gamma}$. Note that the holonomy of a concatenation of two paths is the composition of their holonomy maps. We say that two leafwise paths $\gamma_1$ and $\gamma_2$ with $\gamma_1(0) = \gamma_2(0)$ and $\gamma_1(1) = \gamma_2(1)$ have the same holonomy if $h_{\gamma_1}$ and $h_{\gamma_2}$ agree on a common open set about $z_0$. Define an equivalence relation on pointed leafwise paths by specifying that $\gamma_1 \sim_h \gamma_2$ if $\gamma_1$ and $\gamma_2$ have the same holonomy. This definition is independent of choice of foliation charts. If $U_\alpha$, $U_\beta$, and $U_\delta$ are three foliation charts with $U_\alpha \cap U_\beta \cap U_\delta \ne \emptyset$, then to evaluate $\gamma_{\beta\delta} \circ \gamma_{\alpha\beta} (z)$, take the plaque through ${\mathcal P}_\alpha(\phi_\alpha^{-1}(z))$ and project it onto the transversal ${\mathcal T}_\beta$ to get $\gamma_{\alpha\beta}(z)$. Now to apply $\gamma_{\beta\delta}$ to $\gamma_{\alpha\beta}(z)$ we take the plaque ${\mathcal P}_\beta( \phi^{-1}_\beta (\gamma_{\alpha\beta}(z))$, which by definition intersects ${\mathcal P}_\alpha(\phi_\alpha^{-1}(z))$, and project it to the transversal ${\mathcal T}_\delta$. To evaluate $\gamma_{\alpha\delta}(z)$ we project the plaque ${\mathcal P}_\alpha(\phi_\alpha^{-1}(z))$ to the transversal ${\mathcal T}_\delta$. But since ${\mathcal P}_\beta( \phi^{-1}_\beta (\gamma_{\alpha\beta}(z))$ intersects ${\mathcal P}_\alpha(\phi_\alpha^{-1}(z))$, they both project to the same point on ${\mathcal T}_\delta$. So $\gamma_{\beta\delta} \circ \gamma_{\alpha\beta} (z)$ and $\gamma_{\alpha\delta}(z)$ agree on their common domain. Now let $\gamma:[0,1]\longrightarrow M$ be a leafwise path in $M$. Choose two coverings of its image by plaques, ${\mathcal U}_\alpha = \{ {\mathcal P}_{\alpha_1} \dots {\mathcal P}_{\alpha_k}\}$ and ${\mathcal U}_\beta = \{ {\mathcal P}_{\beta_1} \dots {\mathcal P}_{\beta_l}\}$. Define the holonomy function $\gamma_{\alpha_1\dots\alpha_k} = \gamma_{\alpha_{k-1}\alpha_k} \circ \dots \circ \gamma_{\alpha_2\alpha_3} \circ \gamma_{\alpha_1\alpha_2}$ and $\gamma'_{\alpha_1\dots\alpha_k} = \gamma_{\alpha_{k-1}\alpha_k} \circ \dots \circ \gamma_{\alpha_r\alpha_{r+1}} \circ \gamma_{\beta\alpha_r} \circ \gamma_{\alpha_{r-1}\beta}\circ\dots \circ \gamma_{\alpha_1\alpha_2}$. Now $\gamma_{\beta\alpha_r} \circ\gamma_{\alpha_{r-1}\beta} =\gamma_{\alpha_{r-1}\alpha_r}$ on their common domain so $\gamma{\alpha_1\alpha_k}=\gamma'{\alpha_1\alpha_k}$ on their common domain. By induction the holonomy functions defined by ${\mathcal U}_\alpha$ and ${\mathcal U}_\beta$ are both equivalent to the one defined by ${\mathcal U}_\alpha \cup {\mathcal U}_\beta$ and hence are both equivalent to each other. The {\em holonomy groupoid} ${\mathcal G}_{\mathcal F}$ is the set of $\sim_h$ equivalence classes of pointed leafwise paths for $\mathcal F$, equipped with the topology whose basic sets are generated by ``neighborhoods of leafwise paths'' (cf. section~2, \cite{Winkelnkemper}). The manifold $M$ embeds into ${\mathcal G}_{\mathcal F}$ by associating to $x \in M$ the constant path $*x$ at $x$. \begin{obs} If $f:M_1\longrightarrow M_2$ is a leaf preserving $C^k$ diffeomorphism, and let $U_1,\dots,U_k$ be a cover of $M_1$ by foliation charts. Then $f[U_1], \dots f[U_k]$ are foliation charts on $M_2$ with coordinate functions $\psi_k=\phi_k\circ f^{-1}$. Let $V_\alpha$ be a foliation chart on $M_2$ with coordinate function $\psi_\alpha$. Then, since $f$ is a diffeomorphism, $\psi_\alpha\circ\psi_j^{-1}=\psi_\alpha\circ f\phi_j^{-1}$ is $C^k$. And since $f$ is leaf preserving, $\psi_j^{-1}[(-2,2)^p\times \{z\}]= f\circ\phi_j^{-1}[(-2,2)^p\times \{z\}]$ is the connected component of $L_{\phi^{-1}[(0,z)]}\cap U_j$. $f$ is a leaf-preserving diffeomorphism, so $\psi^{-1}_j [ (-2,2)^p\times \{z\}]$ is a connected component of $L_{\phi^{-1}[(0,z)]}\cap f[U_j]$. \end{obs} \section{Limit Sets And Invariant Sets} \subsection{Orbits} For any dynamical system we will have concepts of limit sets and invariant sets. In the simplest case we iterate a function $f:X\To X$. If $f(x)=x$ then $x$ is a {\em fixed point} of $f$. If there is some natural number $k$ so that $f^{k+1}(x)=f^k(x)$, then $c$ is an {\em eventually fixed point}. So $x$ is an eventually fixed point precisely when some iterate of f is a fixed point. If there is some natural number $r$ so that $f^r(x)=x$, then x is a {\em periodic point}. If $r$ is the smallest number for which $f^r(x)=x$, then $x$ is a periodic point with period $r$. If there exists numbers $k$ and $r$ so that $f^{k+r}(x)=f^k(x)$ with $r$ minimal, then $x$ is an {\em eventually periodic} point of period $r$. So a periodic point of period $r$ is a fixed point for $f^r$, and an eventually periodic point is an eventually fixed point for $f^r$. ${\mathcal O}^+ (x)$, the {\em forward orbit} of $x$, consists of $x$ and all of its iterates. So ${\mathcal O}^+ (x)= \set{x,f(x),\dots,f^n(x)\dots}$. While we define the forwards orbit as a set, sometimes we'll abuse the language and mean the {\em sequence} $x,f(x),\dots,f^n(x)\dots$. If $f$ is not defined globally, it's possible that $f^k(x)$ is not defined, so the orbit of $x$ could be a finite sequence. If $ x$ is a periodic point, we call ${\mathcal O}(x)$ a periodic orbit. The {\em backwards orbit} of $x$, ${\mathcal O}^-(x)$, consists of all points which have $x$ in their forwards orbits. So ${\mathcal O}^-(x) = \union_{n=1}^{\infty} f^{-n}(x)$. Unless $f$ is invertible, it does not generally make sense to think of the backwards orbit as a sequence, but if $f^-1$ has $k$ branches, for instance, we will think of the sequence $\dots f_{i_3}^{-1}, f_{i_2}^{-1}, f_{i_1}^{-1}$ as determining one of many possible sequences which we'll call {\em a} backwards orbit. The {\em total orbit} of $x$, ${\mathcal O}_{tot} (x)$ consists of all points which we can arrive at by successively applying $f$ or any branch of $f^{-1}$. ${\mathcal O}_{tot} (x)$ is the set of all the points $y$ so that there exists $m$ and $n$ with $f^m(x) = f^n(y)$ which is equal to the union of the backwards orbits of all the iterates of $x$. For a pseudogroup, $\Gamma$, acting on $X$, we define ${\mathcal O}(x) = \set{\gamma(x)|\gamma\in\Gamma}$. This is analogous to the total orbit of a point for a function, but in general we don't have analogues for the forwards orbit or the backwards orbit. However given a basis ${\mathcal B}$ which generates $\Gamma$, we adopt the convention that applying an element of ${\mathcal B}$ takes us forwards, so $\orbit^+(x) = \set{\gamma_m\dots\gamma_2\gamma_1| \gamma_i\in {\mathcal B} }$, and $\orbit^-(x) = \set{\gamma_m^{-1}\dots\gamma_2^{-1}\gamma^{-1}_1| \gamma_i\in {\mathcal B} }$. We can think of a foliation as a dynamical system as well, but instead of having discrete time periods in which we iterate functions, we have a continuum of time during which we can flow along a leaf. As a matter of fact, we can think of the holonomy pseudogroup acting on a total transversal as a discretization of the continuous dynamics along the leaves. So a leaf of a foliation is analogous to the orbit of a point in a pseudogroup. Again, we don't normally have a way to define forwards and backwards directions, but in most of our examples, we'll have a basis for the holonomy pseudogroup, which will then define forwards and backwards directions. Furthermore all of the basis elements in these examples will have expanding dynamics, so for these examples, independently of the basis, we can define forward directions to have expanding holonomy, and backwards directions have contracting holonomy. \subsection{Minimal Sets} An {\em invariant set} of a dynamical system is a set which is fixed by that system. For a single function $f$, the invariant set is a set $K$ so that $f(K)=K$. For a pseudogroup, $\Gamma$, an invariant set is a set $K$ such that $\Gamma(K)=K$, i.e. $\union_{\gamma\in\Gamma}\gamma(K)=K$. Motivated by the observation that $K $ is an invariant set for a pseudogroup $\Gamma$ if and only if $K$ is a union of orbits. The corresponding term for a foliation is an {\em ${\mathcal F}$-saturated } set. A ${\mathcal F}$-saturated set is a union of leaves. \begin{prop} Let $K$ be an invariant set for a homeomorphism $f:X\To X$. Then $\cl K$ is an invariant set as well. \end{prop} \noindent \textbf{Proof:} Let $x\in \cl{K} $. Then there exists a sequence $x_n\longrightarrow x$ with $x_n\in K$. By continuity of $f$, $f(x_n)\longrightarrow f(x)$. Since $f(x_n)\in K$, $f(x)$ must be an element of $\cl K$. Conversely, by a similar argument, $f^{-1}(x)$ is an element of $\cl K$ as well, so $\cl K$ is truly an invariant set. \done For an arbitrary pseudogroup, $\Gamma$, there's no reason that the closure of an invariant set must be invariant. The closure of an invariant set might not even be contained in the domain of $\Gamma$. And even if it is, if $x_n\longrightarrow x$, we can't be sure that every $x_n$ and $x$ are all in the domain of the same element of $\Gamma$. For instance, if $X=[-1,1]$, $\gamma_1:x\mapsto \frac{1}{2}x+1$ on $[-1,0)$ and $\gamma_2:x\mapsto x$ on $(0,1]$. Then $(0,1]$ is an invariant set, but it's closure, $[0,1]$ isn't. A {\em minimal set} is a closed invariant set which contains no proper closed invariant subsets. \begin{prop} Let $\Gamma$ be a symmetric pseudogroup generated by a finite basis whose domains are all closed, then any minimal set $K$ for $\Gamma$ is the closure of an orbit. \end{prop} \noindent \textbf{Proof:} We've seen that any invariant set must be a union of orbits. So for any $x\in K$, we need to show that $\cl {\orbit (x)}$ is invariant. Then $\cl {\orbit (x)} $ is a subset of $ K$, but it can't be a proper subset, so $K$ must equal $\cl {\orbit (x)} $. So let $y\in \cl {\orbit (x)}$. Then there must a sequence $\zeta_n(x) \longrightarrow y$ Then since $\Gamma$ is generated by a finite basis, there must be a $\gamma$ in the basis so that an infinite subsequence $\zeta_{n_j}(x)$ which are all in the domain of $\gamma$ and since the domain of $\gamma$ is closed, $y$ must be in the domain of $\gamma$ as well. Hence $\gamma\zeta_{n_j}(x) \longrightarrow \gamma y$, and so $\Gamma(\cl{\orbit(x)})\subset \cl{\orbit(x)}$. Similarly, $\zeta_{n_{j+1}}(x)$ must be in the range of $\gamma$, so $\gamma^{-1}\zeta_{n_{j+1}}(x) \longrightarrow \gamma^{-1} y$, and hence $ \cl{\orbit(x)} \subset \Gamma(\cl{\orbit(x)})$. \done While the minimal set must be the closure of an orbit, the converse is not true. For any point $x$ not in a minimal set, $\cl {\orbit (x)}$ will not be minimal. An {\em exhaustion sequence} for a leaf $L$ is an increasing sequence of connected compact sets $$K_1 \subset K_2 \subset \cdots K_n \subset \cdots \subset L$$ whose union is all of $L$. The $\omega$-limit set of a leaf $L$ is the intersection $\displaystyle \omega(L) = \bigcap_{n=1}^{\infty} \;\; \overline{L-K_n} \;\;$ where the closures are formed with respect to the topology on $V$. \begin{prop} $\omega(L)$ is a compact, saturated set, independent of the choice of exhaustion sequence. Moreover, if $L - K_n$ is connected for all $n$ then $\omega(L)$ is also connected. \end{prop} A leaf $L$ is {\em proper} if the inclusion $L \hookrightarrow V$ induces from $V$ the metric topology on $L$. It is an easy exercise that a leaf is proper exactly when $L \cap \omega(L) = \emptyset$. An {\em end} $\epsilon$ of a non-compact manifold $L$ is determined by a choice of an open neighborhood system of $\epsilon$, which is a collection $\{U_\alpha\}_{\alpha \in A}$ such that \indent $\bullet$ each $U_\alpha$ is an open subset of $L$ with non-compact closure, \indent $\bullet$ each finite intersection $U_{\alpha_1} \cap \ldots \cap U_{\alpha_q}$ is connected and nonempty, \indent $\bullet$ the infinite intersection $\cap_1^{\infty} U_{\alpha_i} = \emptyset$. \noindent Two open neighborhood systems $\{U_\alpha\}_{\alpha \in A}$ and $\{U_\beta\}_{\beta \in B}$ define the same end if for every $\beta \in B$ there exists $\alpha \in A$ such that $U_\beta \subset U_\alpha$. Given an open neighborhood system $\{U_\alpha\}_{\alpha \in A}$ of $\epsilon$, the $\epsilon$-limit set $\displaystyle {\lim}_{\epsilon}(L) = \bigcap_{\alpha \in A} \overline U_\alpha $ Clearly, for each end $\epsilon$, we have $\displaystyle {\lim}_{\epsilon}(L) \subset \omega(L)$. But $\omega(L)$ may include more points than just the union of the $\epsilon$-limit sets of $L$. An end $\epsilon$ of $L$ is {\em proper} if $L$ is not contained in $\displaystyle {\lim}_{\epsilon}(L)$, and $\epsilon$ is {\em totally proper} if $\displaystyle {\lim}_{\epsilon}(L)$ is a union of proper leaves. A leaf $L'$ is said to be the {\em asymptote} of a leaf $L$ if $\omega(L) = L'$. Note this implies that $\omega(L') = \emptyset$ and hence $L'$ must be compact. A compact, non-empty, $\mathcal F$-saturated set $X$ is {\em minimal} for $\mathcal F$ if each leaf of $X$ is dense in $X$. Equivalently, $X$ is minimal with respect to the properties that it be closed, non-empty and $\mathcal F$-saturated. Zorn's Lemma implies that for each end $\epsilon$ of $L$, there is a minimal set contained in ${\lim}_{\epsilon}(L)$. \subsection{Paths To Infinity.} Let $\gamma:[0,\infty) \To L$ be a leaf-wise path. We say that $\gamma$ is a {\em path to infinity} if for any compact set $K\subset L$, there exists $m$ so that if $t>m$ then $\gamma(t)\not\in K$. There is a natural map from the set of paths to infinity to the set of ends. If $(K_n)$ is an exhaustion sequence, then for large enough $m$, the path $\gamma$ restricted to the interval $[m,\infty)$ remains outside of $K_n$, and hence must stay in a connected component of $L-K_n$. Therefore, if $\epsilon$ is the end which is associated to $\gamma$, it also makes sense to call $\gamma$ a {\em path to the end $\epsilon$}. Note that if $\gamma$ is a path to the end $\epsilon$, we could define the omega limit set $\omega(\gamma)$ in the obvious way, but it need only be true that $\omega(\gamma)\subset {\lim}_{\epsilon}(L)$, not that $\omega(\gamma)= {\lim}_{\epsilon}(L)$ \subsection{Cantor Sets} We recall the construction of the Cantor middle-third set, $C$. Begin with the interval $I=[0,1]$. We subdivide $I$ into thirds, and remove $G = (\frac{1}{3},\frac{2}{3})$. After this removal, we're left with $I_0 = [0,\frac{1}{3}]$ and $I_1 = [\frac{2}{3},1]$. We continue in this way, removing the middle thirds from both $I_0$ and $I_1$. The second approximation consists of the intervals $I_{00} = [0,\frac{1}{9}]$, $I_{01}= [\frac{2}{9}, \frac{1}{3}]$, $I_{10}= [\frac{2}{3}, \frac{7}{9}]$, and $I_{01}= [\frac{8}{9},1]$. We continue this an infinite number of times to obtain $C$. \begin{figure}[htbp] \centering \includegraphics[width=0.5\textwidth]{./images/middlethird.png} \caption{the middle third set} \end{figure} To define $C$ more rigorously, we let $w$ be a string of $k$ zeroes and ones. Assume $I_w$ has been defined. Remove $G_w$, the interior of the middle third of $I_w$ being left with $I_{w0}$ and $I_{w1}$. This defines $I_w$ inductively for any finite string $w$. Then the $n^{th}$ approximation, $C_n=\bigcup_{|w|=n}I_w$. This gives us a decreasing sequence of compact sets, $C_n \subset C_{n-1}$. We define \[C= \bigcap_{n = 1}^\infty C_n = \bigcap_{n = 1}^\infty \bigcup_{|w|=n} I_w\] \noindent We say $G_w$ is the {\em gap} of $I_w$. Note that each of the subsets $C_w = C\bigcap I_w$ are similar to the whole set $C$. So we call $C_w$ a {\em clone} or {\em cylinder set} and $I_w$ a {\em clone interval}. We let $x_{\epsilon_0\epsilon_1...} = \bigcap_{n=1}^\infty I_{\epsilon_0\epsilon_1...\epsilon_n}$. With the dictionary order on $\{ 0, 1 \} ^{\mathbb N}$, this equips the middle third set with an order preserving homeomorphism \[\phi: \{0,1\}^{\mathbb N} \longrightarrow C\] \[\epsilon_0\epsilon_1...\epsilon_n \mapsto x_{\epsilon_0\epsilon_1...\epsilon_n}\] \noindent We note that $C$ is totally disconnected, perfect, compact, and has the cardinality of the reals. This classifies $C$ up to homeomorphism; any topological space with these properties is homeomorphic to $C$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{Bernoulli Shift Spaces} We have seen that points in the middle third set, $C$, can be labeled with infinite strings of 0s and 1s, that is elements of the set $\Sigma_2=\set{0,1}^{\mathbf N}$. We can also model the dynamics of $f$, the function which generated $C$. Define the {\em Bernoulli shift space} on $k \ge 2$ symbols, $\Sigma_k= \{ 0,1\dots,k-1 \}^{\mathbb N}$ be the set of all right infinite words from the alphabet $\{0,1, \dots,k-1 \}$. If we equip $\Sigma_k$ with the product topology, so the basic open sets are strings which begin with the same prefix, then $\Sigma_k$ is homeomorphic to $C$. A basic open set is of the form \[ U_{a_1\dots a_m} = \left\{ \omega \in X_k | \omega = a_1a_2\dots a_m w_{m+1} w_{m+2}\dots \right\} \] We call $U_{a_1\dots a_m}$ a {\em cylinder set} of $\Sigma_k$. The {\em (right) shift map}, $\sigma$ is a $k$ to 1 local homeomorphism which acts on $\Sigma_k$ by forgetting the first component. \[ \sigma : w_1 w_2 w_3 \dots \mapsto w_2 w_3 \dots \] $\sigma$ is a $k$ to 1 covering map $\Sigma_k \longrightarrow \Sigma_k$. A {\em subshift} is a closed subset $W \subset \Sigma_k$ which is invariant under $\sigma$. In the sections about dynamically defined Cantor sets, all of our examples will be conjugate to a subshift. Subshifts which come up in the context of geometric examples tend to be Markov, and hence the periodic points are dense. Again we define a clone of a subshift to be the set of all words which begin with a given finite word. \begin{ex} Let $X= \{0,1\}^{\mathbb N}$. Let $W = \left\{ w_1w_2\dots \in X_2 | \right.$ if $w_i=0$, then $\left. w_{i+1}=1 \right\} $. $W$ is a subshift of $X$. \end{ex} \begin{ex} Via the inclusion $\{0,1\}^{\mathbb N} \longrightarrow \{0,1,2\}^{\mathbb N}$, we can view $\Sigma_2$ as a subshift of $\Sigma_3$. The dynamics of $\Sigma_2$ are also closely related to the dynamics of $\Sigma_4$. Let $\phi: \Sigma_4\To\Sigma_2$ be defined by \begin{eqnarray*} 0 & \mapsto &00\\ 1 & \mapsto &01\\ 2 & \mapsto&10 \\ 3 & \mapsto&11 \end{eqnarray*} Then $\phi$ conjugates $\sigma$ on $\Sigma_4$ to $\sigma^2$ on $\Sigma_2$ \end{ex} \begin{ex}[A subshift with no periodic orbits.] Let $X$ be a subshift without periodic orbits, and let $x\in X$. Then for any finite string of zeroes and ones, $w$, there must be a bound on the number of times in a row that $w$ appears in $x$. If there is no bound on the number of times $w$ appears in a row, then for any $m\in {\mathbb N}$, there exists some $k$ so that $\sigma^k(x) = (w)^mx_ix_{i+1}\dots$. Then since $X$ is closed, the periodic point $www...$ is in $X$. On the other hand, if we start with such an element $x$, we find that $\{\sigma^n(x)\}$ is a countable set which is invariant under $\sigma$, but which has no periodic points . Since $\sigma$ is continuous we take the closure of this set, which produces a subshift which, since no block repeats in an unbounded way, still has no periodic points. By alternating digits from this (possibly countable) set with arbitrary digits, we get an uncountable subshift which still has no periodic points. Now we construct an infinite word $x$ which has no finite strings occurring three times in a row: Let $x_1 = 0$, $x_2=01$, $x_4=0110$, $x_8=0110$ $1001$, inductively define $x_{2^{m+1}} = x_{2^m} \overline{x_{2^m}}$, where $\overline 0 = 1$ and $\overline 1 = 0$ and define $\omega= \lim_{k\rightarrow\infty} x_k$. We see that $\omega$ is composed of two types of blocks of four digits, $0110$ and $1001$. So there are at most eight ($2^3$) possible blocks of length 12 beginning in the $4k$th position. But we also see that $0110$ does not repeat 3 times in a row, and neither does $1001$. So there are only six possible blocks of length twelve beginning in the $4k$th position: $$0110 0110 1001$$ $$0110 1001 0110$$ $$0110 1001 1001$$ and three more which you get by swapping the 0s and 1s in the above three. By inspecting these six blocks we see that $\omega$ contains no blocks of size 1, 2 or 3 which repeats more than twice in a row. Now if we remove every other digit from $\omega$, we obtain $\omega$ back again, so this shows that no block of size $2^k$ or $3 \cdot 2^k$ repeats more than twice in a row. For instance, if a block of size six repeated 3 times in a row, we could remove every other digit from $\omega$ and get a block of size 3 which repeats three times in a row. Now we look at blocks of length $4k+1$. Let $B$ be a block of length $4k+1$ which begins in $4m+1$th spot. Then $B$ begins with 0110 or 1001, and the next three digits after $B$ are either 110 or 001. So $B$ doesn't repeat twice in a row. Similar arguments work for all blocks of length $4k+1$ starting in the $4k+i$th position, as well as blocks of length $4k+3$ starting in the $4k+i$th position and so no block of odd length greater than one repeats twice in a row. And once again, since removing every other digit from $\omega$ gives $\omega$ back again, no even block repeats three times in a row either, so $\omega$ has the desired property. \done \end{ex} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \subsection{Labeled Cantor sets.} We could try to construct Cantor sets similarly to the way we constructed the middle third set: Start with an interval $I$ in the line, remove an open interval from the interior of $I$ leaving $I_0$ and $I_1$, and so forth. This process won't guarantee the creation of a Cantor set though; for the resulting set to be totally disconnected, we need to make the set of gaps dense in $I$. Instead, we will define a Cantor set to be any set which is homeomorphic to $\{ 0,1 \}^{\mathbb N}$ (and hence to the middle third set). These are all the totally disconnected perfect sets with the cardinality of the continuum. We will consider Cantor sets which are embedded in a closed interval. We say the Cantor set $C\subset \Real$ is {\em labeled} by the shift space $\Sigma_k$ if there is an order preserving homeomorphism $\phi: \Sigma_k \To C$. The point $x$ in $C$ is labeled by the infinite word $\omega\in\Sigma_k$. We write $x_\omega $ for the point $\phi(\omega)$. A labeled Cantor sets inherits a nested structure of gaps and clones from its labelings. Let $(C, \phi)$ be a labeled Cantor set conjugate to $X_2$. We write $x_{w_1w_2\dots} = \phi(w_1w_2\dots)$. Let $w$ be a finite string of 0s and 1s. Then we define $I_w$ as the closed interval \[I_w = [x_{w00\dots},x_{w(k-1)(k-1)\dots}] \] and $G_w$ as the open interval \[G_w = (x_{w0(k-1)(k-1)\dots},x_{w(k-1)000\dots}) \] We write $|I_w|$ and $|G_w|$ for the respective lengths of $I_w$ and $G_w$. If $w$ contains $m$ symbols, then we say $I_w$ (or $G_w$) is a level $m$ clone (or gap). So in general, a lower level clone (gap) will be longer than a higher level clone (gap). A Cantor set embedded in the line can have many different labelings on it. The set of clone intervals will be different for two different labelings, while the set of gaps is dependent only on the underlying set. The middle third set is an example of a labeled Cantor set. We note that the underlying Cantor set $C$ can have many different labelings on it. Since a Cantor set is closed, its complement is open, so the set of gaps depends only on $C$ and not on the labeling of $C$. The set of clone intervals depends on the choice of labeling (as does the labeling of gaps and clone intervals by finite words.) We can also define a Cantor set which is labeled by a subshift, but for an arbitrary subshift it would be a little more difficult to extend the labels to the gaps. We can still define the clone $C_w = \phi(w \epsilon_k \epsilon_{k+1}\dots)$ and then we can define $I_w$ as the convex hull of $C_w$. The gaps might be a little more difficult to label, but we could describe the gaps by the clones which they're between. The set $\{ 0,1 \}^{\mathbb N} $ comes equipped with the (left) shift map \[\sigma:\{ 0,1 \}^{\mathbb N} \longrightarrow \{0,1\}^{\mathbb N}\] \[\epsilon_1\epsilon_2\epsilon_3\epsilon_4\dots\mapsto \epsilon_2\epsilon_3\epsilon_4\dots. \] The map $\sigma$ is a two-to-one local homeomorphism. Let $C$ be a labeled Cantor set. The shift map on $\{ 0,1 \}^{\mathbb N} $ induces a map $\sigma : C\rightarrow C$, with $\sigma(x_{\epsilon_1\epsilon_2\epsilon_3\epsilon_4\dots})= x_{\epsilon_2\epsilon_3\epsilon_4\dots} $. We also call this map on $C$ the shift map. Let $C$ and $C'$ be labeled Cantor sets in ${\Real}$. Then the map which preserves the labeling is the unique order-preserving homeomorphism $\phi: C\rightarrow C'$ which commutes with the shift map. We write $\phi(x_{\epsilon_1\epsilon_2\epsilon_3\epsilon_4\dots}) = x'_{\epsilon_1\epsilon_2\epsilon_3\epsilon_4\dots}$. Note that we use $\sigma$ to denote the shift map on $ \{ 0, 1\}^{\mathbb N}$ as well as the map which it induces on $C_1$. To review our situation, so far we have a Cantor set, $C$, defined in a similar manner to the middle third set. The shift map, $\sigma$, acts on $C$ by stretching both $C_0$ and $C_1$ out and laying them back down on top of all of $C$. If we know the lengths of all the gaps and clone intervals, of course, we then know how to construct $C$. If we know the relative sizes of the intervals, we can construct $C$ up to affine rescaling. For all finite words $w$, we let $l(w)$, $g(w)$, and $r(w)$ denote $\frac{|I_{w0}|}{|I_w|}$, $\frac{|G_w|}{|I_w|}$, and $ \frac{|I_{w1|}}{|I_w|}$ respectively. Then, all we need, besides $l(w)$, $g(w)$, and $r(w)$ for all $w$, to construct $C$ is the interval $I$ (the convex hull of $C$). Thus we have the {\em ratio geometry function}, $ w \mapsto (l(w), g(w), r(w) ) = (\frac{|I_{w0}|}{|I_w|}, \frac{|G_w|}{|I_w|}, \frac{I_{|w1|}}{|I_w|})$ whose domain is $\{0,1\}^{\mathbb N}$, and whose range is contained in the unit two-simplex $\{(x,y,z)|x+y+z < 1\}$. \section{Bounded Geometry.} We say that $C$ has bounded geometry if there exists $\alpha$ and $\beta$ with $ 0 < \alpha <\beta <1$ such that for all $w$, $l(w)$, $g(w)$, and $r(w)$ are all between $\alpha$ and $\beta$. Bounded geometry is analogous to hyperbolicity of $\sigma$, or equivalently to hyperbolicity of both branches of $\sigma^{-1}$. Suppose we require $\sigma $ to be hyperbolic, say $\sigma_0^{-1}$ and $\sigma_1^{-1}$ both extend to hyperbolic functions, $f_0$ and $f_1$. We take $f_i \le \beta \le 1$. Then the difference quotient, say for $\sigma_0^{-1}$, over the interval $I_w$, we get $\frac{|I_{0w}|}{|I_w|}\le \beta$, whereas bounded geometry requires that $l(w) = \frac{|I_{w0}|}{|I_{w}|}\le \beta$. We note that if $C$ has bounded geometry, then $\alpha^n \le |I_w| \le \beta^n$. The middle third set, of course, has bounded geometry with $\alpha = \beta = \frac{1}{3}$. \begin{prop} \label{bdd-holder} Let $C$ and $C'$ be labeled Cantor sets with bounded geometry. Let $\phi:C \rightarrow C'$ be the homeomorphism induced by the shift map. \begin{displaymath} \begin{array}{rcl} I & \stackrel{\sigma}{\longrightarrow} & I\\ \phi\downarrow& &\downarrow \phi\\ I' & \stackrel{\sigma'}{\longrightarrow} & I'\\ \end{array} \end{displaymath} Then $\phi$ is H\"older continuous. \end{prop} \noindent \textbf{Proof:} Let $ x,y \in C$. Then there exists a word $w \in \{ 0, 1 \}^k$, for some $k$, such that $G_w \subset [x,y] \subset I_w$. We denote $\phi (G_w)= H_w$ and $\phi (I_w)= J_w$. Then $|\phi(x) - \phi(y)| \le |J_w| \le \beta^n \le\beta \alpha^{\lambda n} \le \beta|G_w|^\lambda \le \beta|x-y|^\lambda$, where $\lambda = \log_\alpha \beta$. \done \section{Cookie Cutters II} \label{'cookcut'} The middle third set can be generated dynamically, as the minimal closed invariant set of the affine map \[ h:x \mapsto \left\{ \begin{array}{lr} 3x & x\in [0,\frac{1}{3}]\\ 3x-2& x\in [\frac{2}{3} ,1] \end{array}\right.\] \begin{figure}[htbp] \centering \includegraphics[width=0.4\textwidth]{./images/3xmod1.png} \caption{ {The function $x\mapsto 3x \pmod{1}$ which has the middle third set as an invariant set.}} \end{figure} Alternately, we consider the two branches of $h^{-1}$: \[ h_0^{-1}= \frac{x}{3}\] \[ h_1^{-1}= \frac{x+2}{3}\] \noindent Then $C$ is the omega limit set of $ \{h_0^{-1},h_1^{-1}\}$. We note that $h$ affinely scales $I_i$ onto $I$ for $i= 0,1$. Let $\omega = \omega_1 \omega_2 \dots \in \{0,1\}^{\mathbb N}$. Then \begin{eqnarray*} f(I_{\omega_1\omega_2 \dots \omega_k}) & = &I_{\omega_2 \dots \omega_k}\\ f(G_{\omega_1\omega_2 \dots \omega_k}) & = &G_{\omega_2 \dots\omega_k}\\ f(x_{\omega_1\omega_2 \dots}) & = &x_{\omega_2 \omega_3\dots}\\ f_i^{-1}(I_{\omega_1\omega_2 \dots \omega_k}) & = &I_{i\omega_1\omega_2 \dots \omega_k}\\ f_i^{-1}(G_{\omega_1\omega_2 \dots \omega_k}) & = &G_{i\omega_1\omega_2 \dots \omega_k}\\ f_i^{-1}(x_{\omega_1\omega_2 \dots}) & = &x_{i\omega_1\omega_2 \dots} \end{eqnarray*} So $h|_C$ commutes with the {shift map} $\sigma$ on $\{0,1\}^{\mathbb N}$. \[\sigma: \epsilon_1 \epsilon_2 \dots \mapsto \epsilon_2 \epsilon_3\dots\] A cookie cutter Cantor set is a generalization of this process with generating maps defined on disjoint intervals. The generating maps are required to be hyperbolic local diffeomorphisms. \begin{figure}[htbp] \centering \includegraphics[width=0.5\textwidth]{./images/cookcut.png} \caption{ {A cookie cutter map with two generators}} \end{figure} We let $I_k = [a_k,b_k]$, for $0\le k \le m-1$ be $m$ disjoint intervals. Say that $b_k < a_{k+1}$. A {\em cookie cutter} is a map $F: \cup I_k \longrightarrow I$ which restricts to bijections on each $I_k$. We also require that there exist $\lambda$ and $\gamma$ with $0<\lambda <\gamma <1$ such that $\frac{1}{\gamma} \le |F'| \le \frac{1}{\lambda}$. Equivalently, we require that $0< \lambda \le |{F_i^{-1}}'| \le \gamma <1$, where the $F_k^{-1}$'s are the $m$ branches of $F^{-1}$. Then we let $C$ be the minimal closed set for $F$. By construction, $I_k = F_k^{-1}(I)$. For $w = \epsilon_0 \epsilon_1 \dots \epsilon_n$ a finite string from the alphabet $\{0, \dots, m-1\}$, we define $I_w = F_{\epsilon_0}^{-1}F_{\epsilon_1}^{-1}\dots F_{\epsilon_n}^{-1}(I)$. Then $C= \bigcup_n \bigcap_{|w|=n} I_w$. Again we have the same situation as the middle third set,with $F:C\longrightarrow C$ being conjugate to $\sigma :\{0,\dots , m-1\}^{\mathbb N} \longrightarrow \{0,\dots , m-1\}^{\mathbb N}$. If $F$ is $C^{k+\lambda}$, we call $F$ a $C^{k+\lambda}$ {\em cookie cutter Cantor set} . We say that $C$ is generated by $F_0^{-1} \dots F_{m-1}^{-1}$ and that $C$ has $m$ generators. We use hyperbolicity to guarantee that the resulting set is totally disconnected. $F$ induces a labeling on $C$ so that $F$ is conjugate to the shift map. We say that $C$ is an affine Cantor set if $F$ is affine on each subinterval. For the sake of simplicity, we will mostly consider cookie cutters with two generators. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{prop} \label{cookie-bdd} A cookie cutter Cantor set has bounded geometry. \end{prop} \noindent \textbf{Proof:} Let $C$ be a cookie cutter set with $k$ generators. We will use the Mean Value Theorem to show that $\frac{|I_w|}{|G^j_w|}$ is bounded above, and hence $\frac{|G^j_w|}{|I_w|}$ is bounded below. The proof will apply to $\frac{|I_{wj}|}{|I_w|}$ as well. When replacing $G^j_w$ by $I_{wj}$. Then since all of these values add up to one, and they are all bounded below, they're all bounded above too. $F^n(I_{w}) = I$ and $F^n(G^j_w) = G^j$ so there exists $z_0$ and $z_1 \in I_w$ so that $|I|= {F^n}'(z_0) |I_w|$ and $|G^j| ={F^n}'(z_1)|G^j_w|$ so \begin{eqnarray*} \frac{|I|}{|G^j|} &=& \frac{F^n(I_w)}{F^n(G^j_w)}\\ &=& \frac{{F^n}'(z_0)|I_w|}{{F^n}'(z_1)|G^j_w|}\\ \frac{|I_w|}{|G^j_w|} &=& \frac{{F^n}'(z_1)|I|}{{F^n}'(z_0)|G^j|}\\ \log \frac{|I_w|}{|G^j_w|}& = & \log {{F^n}'(z_1)}- \log {{F^n}'(z_0)} + \log{\frac{|I|}{|G^j|}}\\ & = & \sum_{i=0}^{n-1}\left| \log F'(F^i(z_1)) - \log F'(F^i(z_0))\right| + \log \frac{|I|}{|G^j|}\\ &\le &\sum_{i=0}^{n-1} K|F^i(z_1) - F^i(z_0)|^\lambda + \log \frac{|I|}{|G^j|}\\ &\le & \sum_{i=0}^{n-1} K \gamma^{i\lambda} + \log \frac{|I|}{|G^j|}\\ & \le & K' \end{eqnarray*}. \done \section{Exceptional Minimal Sets} For a pseudogroup acting on either the circle or the line, an {\em exceptional minimal set} is a minimal set which is homeomorphic to the Cantor set. For a foliation, an exceptional minimal set is a minimal set whose intersection with the transversal is a Cantor set. Of particular interest are Markov exceptional minimal sets. All of the known examples of $C^k$ foliations with exceptional minimal sets are Markov if $k\ge2$. Definition: Let $\Gamma$ be a pseudogroup of local $C^{1+\lambda}$ diffeomorphisms of a compact 1-dimensional manifold $T$. Let $C$ be an exceptional minimal set for $\Gamma$. % which is disjoint from the boundary of $T$. $C$ is {\em Markov} if there exist elements $\gamma_1, \dots, \gamma_m$ of $\Gamma$ and closed intervals $I_1\dots, I_m$ of $T$ such that \begin{enumerate} \item Int $I_k$s are pairwise disjoint, \item $C \subset \cup_{k=1}^m I_k$, \item $C \cap$ Int $I_k \ne \emptyset$ for each $k$, \item the domain of $\gamma_k$ contains $I_k$ for each $k$, \item $\gamma_k|I_k\cap C$'s generate $\Gamma| C$, and \item if $\gamma_k(I_k) \cap$ Int $I_j \ne\emptyset$, then $\gamma_k(I_k) \supset I_j$ \end{enumerate} We call $(I_0,\dots I_m;\gamma_0, \dots \gamma_m)$ a Markov basis. We call a Markov basis hyperbolic if $1<\frac{1}{\beta}\le\gamma_k'\le\frac{1}{\alpha}$. A Markov exceptional minimal set is hyperbolic if it can be generated by a hyperbolic Markov basis. Since we don't require the $I_k$'s to be pairwise disjoint, (they might share endpoints,) there might be a countable number of points which have two different labelings on them. % \vspace{2 in} To apply the theory of cookie cutters to Markov exceptional sets, we need to modify the notation a little bit. We let $C$ be a Markov exceptional set with basis ${\mathcal B}=\set{(\gamma_k,I_k)}$, $k=0\dots M$. Without loss of generality, we assume that $I_k$ is no bigger than it needs to be. That is, for each $k$ an end points of $\gamma_k(I_k)$ is an endpoint of $\gamma_j$, for some $j$, with $I_j\subset\gamma_k(I_k)$. If not we shrink the domain of $\gamma_k$ so that it If we only need to repeat this process a finite number of times, in which case we're left with ${\mathcal B}$, with smaller domains, but which still is a basis for $C$. If we need to repeat this an infinite number of times, then the intervals in question shrink down to a point, which means they weren't necessary in ${\mathcal B}$ in the first place. So we assume that the domains of ${\mathcal B}$ are no bigger than they need to be, which means that $\gamma_j(I_j)$ doesn't have gaps on its ends. Then if $I_{j_1}\subset\gamma_{j_2}(I_{j_2})$, we define $I_{j_2j_1}=\gamma_{j_2}^{-1}(I_{j_1})$. And if $I_{j_2}\subset\gamma_{j_3}(I_{j_3})$ as well, we define $I_{j_3j_2j_1}=\gamma_{j_3}^{-1}(I_{j_2j_1})$. We see that $I_w$ shrinks down to a point as the length of $w$ increases to infinity. And that \[C=\bigcap_n \hspace{.1in} \union_{|w|=n}I_w\] where $w$ is a word $j_{n}j_{n-1}\dots j_{1}$ with $j_{k}\subset\gamma_{k-1}(I_{j_{k-1}})$ Thus $C$ is labeled by the subshift $\Sigma_{\mathcal B}=\set{j_{n}j_{n-1}\dots j_{1}\subset\Sigma_M:j_{k}\subset\gamma_{k-1}(I_{j_{k-1}})}$. Now unlike a cookie cutter, the nested interval structure of a Markov exceptional set need not be homogeneous. There could be a different number of level-1 gaps in each cylinder set. If there are $M$ elements in the basis, there could be as many as $M$ different local versions of the nested interval structure. We'll label the gaps according to the interval to their right. If $I_{jk}$ has a gap to its left which is also in $I_j$, we label that gap $G_{jk}$. Then $G_{ijk}=\gamma_i^{-1}(G_{jk})$ which will be directly to the left of $I_{ijk}$, and so on. In this way we label all the gaps in Dom$({\mathcal B})$. We could just as easily named the gaps according to the intervals on their right. We just arbitrarily chose the interval on their left. We also note that since the domains of $({\mathcal B})$ might share a common end point, we might have a countable number of points with two labelings on them. So the map $\phi:\Sigma_{\mathcal B}\To C$ is continuous, but not a homeomorphism. Hence $\phi$ is only a semiconjugacy between $\sigma$ and $\Gamma$. \label{sclMark} \section{The scaling function for a Markov exceptional set.} The ratio geometry is defined the same for a Markov exceptional set as it is for a cookie cutter. If $I_{j_k j_{k-1}\dots j_1 w}$, (respectively $G_{j_k j_{k-1}\dots j_1 w}$) is a subinterval of $I_{j_k j_{k-1}\dots j_1 }$, then $\frac{\abs{ I_{j_k j_{k-1}\dots j_1 w}}}{\abs{I_{j_k j_{k-1}\dots j_1 }}}$, (resp. $\frac{ \abs{G_{j_k j_{k-1}\dots j_1 w}}}{\abs{I_{j_k j_{k-1}\dots j_1 }}}$) is a component of the ratio geometry. In particular, $I_{j_1}$ has a finite number of sub-intervals and a finite number of sub-gaps. We write the intervals as $I_{j_1 k_1},I_{j_1 k_2}\dots I_{j_1 k_m}$ with $k_11$ and $f_1' (x)>1$ for $x\not\in (a,b)\bigcup(c,d)$ Then the system $\{f_0,f_1\}$ Generates a Markov exceptional set. For ease of discussion, we'll write $m=\frac{1}{2}$ and \[ f(x) = \left \{ \begin{array}{lcr} f_0(x)& & x \in [0, \frac{1}{2}]\\ f_1(x)& & x\in [\frac{1}{2},1] \end{array} \right. \] and ignore that $f$ is not well-defined at $\frac{1}{2}$. Then $(a,b)\bigcup(c,d)$ acts as a trap for $f$. As soon as $x\in(a,b)$, the forward orbit $f(x),f^2(x),f^3(x),\dots $ will ping pong between $(a,b)$ and $(c,d)$, and the backwards orbit of $(a,b)\bigcup(c,d)$ will consist of a countable number of disjoint intervals. % \vspace{4 in} If we let $f$ be $C^{1+\lambda}$ with $\frac{1}{\alpha}>f'(x)>\frac{1}{\beta}>1$ for $x\not \in(a,b)\cup(c,d)$ then $\overline{\orbit(1)}$ will be a Cantor set. We would have two ``level zero" gaps, $(a,b)$ and $(c,d)$. $f_1^{-1}(a,b)=(c,d)$ and $f_0^{-1}(c,d)=(a,b)$, so there are only 2 ``level one" gaps as well, $f_0^{-1}(a,b)$ and $f_1^{-1}(c,d)$. But there are 4 level-two gaps, $f_0^{-2}(a,b)$, $f_1^{-1}f_0^{-1}(a,b)$, $f_0^{-2}(c,d)$ and $f_0^{-1}f_1^{-1}(c,d)$. There will be eight level-3 gaps, and so forth. Since we bounded the derivatives of $f'$, we have $\alpha^2(b-a)\le f_0^{-2}(a,b)\le\beta^2(b-a)$ and so on. Furthermore, we have $\begin{array}{lcr} f_0^{-1}[0,a]\subset[0,a] & &f_1^{-1}[0,a]=[\frac{1}{2},c]\\ f_0^{-1}[b,\frac{1}{2}]\subset[0,a]& &f_1^{-1}[b,\frac{1}{2}]\subset[d,1]\\ f_0^{-1}[\frac{1}{2},c]\subset[0,a]& &f_1^{-1}[\frac{1}{2},c]\subset[d,1]\\ f_0^{-1}[d,1]=[0,\frac{1}{2}] & &f_1^{-1}[d,1]\subset[d,1] \end{array}$ So these four intervals form a Markov basis for $f$. Hence f has a hyperbolic exceptional minimal set. \subsection{A period two interval} If we define the circle, $S^1$, to be the interval $[0,1]$ with its end points identified, then since $f_0(\frac{1}{2})=1$ and $f_1(\frac{1}{2})=0$, $f$ defines a function on $S^1$ which is no longer ill-defined at $\frac{1}{2}$. So we can think of $f$ as a function on the circle with the discussion above still holding, but since $f_0(\frac{1}{2})=1$, and $f_1(\frac{1}{2})=0$, and we've identified the points 1 and 0 in order to form the circle, we only need 3 subintervals, as shown. We could define this example a little more generally. The fixed point need not be at 0, and the point $m$ whose image is the fixed point need not be $\frac{1}{2}$, but for ease of discussion, we'll keep these values. \begin{figure}[htbp] \centering \includegraphics[width=0.5\textwidth]{./images/exceptional1.png} \caption{ {A map with a period two interval}} \end{figure} \eject We now let $f(0)=0$, $I_0=[o,a]$, $I_1=[b,c]$, and $I_2=[c,0]$ We define $f_i$ to be $f$ restricted to the domain $I_i$. So now $f_0$ and $f_1$ are defined differently than they were on $I$. We have \begin{figure}[htbp] \centering \includegraphics[width=0.8\textwidth]{./images/cantorset_3gap1.png} \caption{ {The minimal set of a period two interval}} \end{figure} \begin{eqnarray*} f_0[I_0] & = & I_0\cup G_0 \cup I_1 \\ f_1[I_1] & = &I_2\cup I_0 \\ f_2[I_2] & = &I_1\cup G_1 \cup I_2 \end{eqnarray*} Notice that $I_0$ and $I_1$ are adjacent sub-intervals. The right endpoint of $I_2$ is the same as the left endpoint of $I_1$. $f$ stretches $I_0$ over $I_0\cup G_0 \cup I_1$. $f$ translates and stretches $I_1$ so that the left ``half" of $I_1$ maps to $I_2$, and the right ``half" of $I_1$ maps to $I_0$. And $f$ stretches $I_2$ to cover $I_1\cup G_1 \cup I_2$. Since $f_0[I_0]=I_0\cup G_0 \cup I_1$, we define \begin{eqnarray*} I_{00} & = & f_0^{-1}[I_0]\\ G_{00} & = & f_0^{-1}[G_0]\\ I_{01} & = & f_0^{-1}[I_1] \end{eqnarray*} We define sub-intervals of $I_1$ and $I_2$ similarly, so that \begin{eqnarray*} I_0 & = & I_{00}\cup G_{00}\cup I_{01}\\ I_1 & = & I_{12}\cup I_{10}\\ I_2 & = & I_{21}\cup G_{21}\cup I_{22} \end{eqnarray*} Now we form the ratios $$\frac{|I_{00}|}{|I_0|},\frac{|G_{00}|}{|I_0|},\frac{|I_{01}|}{|I_0|}$$ $$\frac{|I_{12}|}{|I_1|},\frac{|I_{10}|}{|I_1|}$$ $$\frac{|I_{21}|}{|I_2|},\frac{|G_{21}|}{|I_2|},\frac{|I_{22}|}{|I_2|}$$ and note that $$\frac{|I_{00}|}{|I_0|}+\frac{|G_{00}|}{|I_0|}+\frac{|I_{01}|}{|I_0|}=1$$ $$\frac{|I_{12}|}{|I_1|}+\frac{|I_{10}|}{|I_1|}=1$$ $$\frac{|I_{21}|}{|I_2|}+\frac{|G_{21}|}{|I_2|}+\frac{|I_{22}|}{|I_2|}=1$$ We also get \begin{eqnarray*} I_{00} & = & I_{000}\cup G_{000}\cup I_{001}\\ I_{01} & = & I_{012}\cup I_{010}\\ I_{12} & = & I_{121}\cup G_{121}\cup I_{122} \end{eqnarray*} and so forth, where $I_{i\omega} = f_{i}^{-1}[I_\omega]$ when defined. Then the ratio geometry $$\frac{|I_{000}|}{|I_{00}|}+\frac{|G_{000}|}{|I_{00}|}+\frac{|I_{001}|}{|I_{00}|}=1$$ $$\frac{|I_{012}|}{|I_{01}|}+\frac{|I_{010}|}{|I_{01}|}=1$$ $$\frac{|I_{121}|}{|I_{12}|}+\frac{|G_{121}|}{|I_{12}|}+\frac{|I_{122}|}{|I_{12}|}=1$$ and so forth. Continuing this process, we see will encode the intervals and gaps with finite words $\omega\in\bigcup\{0,1,2\}^n$. But not all words in $\bigcup\{0,1,2\}^n$ are allowed. If $\omega$ indexes a gap, then $\omega$ must end with a 0 or a 2. More specifically, if $\omega$ encodes one of these intervals and there's a 0 in the $n$th spot, then there must be either a 0 or 1 in the $n+1$st spot. Similarly, a 1 must be followed by a 0 or a 2, and a 2 must be followed by a 1 or a 2. Now since $F$ is hyperbolic, we see that the lengths of level-$n$ intervals and gaps must be bounded as they were for cookie cutters. If $\omega$ is a word of length $n$, then $\alpha<\frac{|I_{\sigma i}|}{|I_\sigma|}<\beta$, and $\alpha^n<|I_\sigma|<\beta^n$. The same bounds also work for gaps. Now the lengths of the clone intervals go to 0 exponentially fast, and a co-gap consists of at most two clone intervals, so the lengths of the co-gaps go to zero exponentially fast as well. Therefore the invariant set we get will be totally disconnected. The end points are in the invariant set, and any point in the invariant set can be approximated by end points, so the set is perfect. We also see that the backwards orbit of zero is dense, so that this set is indeed a Markov exceptional set. If $x\in C$, we can encode $x$ by the subshift of $\{0,1,2\}^{\mathbb N}$ which is given by the constraints that a 0 or 1 must follow a 0, a 0 or 2 must follow a 1, and a 1 or 2 must follow a 2. Notice that this encoding is not entirely unique, since $x_{\sigma0000000000\dots}$ and $x_{\sigma2222222222\dots}$ are the same point. We can use variations of this example to get a whole class of examples of Markov exceptional set defined on the circle. We could let $f$ have a period three interval so that $f(a_1,b_1)=(a_2,b_2)$, $f(a_2,b_2)=(a_3,b_3)$, and $f(a_3,b_3)=(a_1,b_1)$. If $f$ is hyperbolic off of these three intervals, then $f$ will again generate an Markov exceptional set. $f$, as constructed, has 3 level zero gaps, 3 level-1 gaps, 6 level-2 gaps, 12 level-3 gaps, etc. We could also let $f$ be 3 to 1. If we play this ping pong game with two intervals, so $f(a,b)=(c,d)$ and $f(c,d)=(a,b)$, bounding the derivatives as we did before, then the forward iterates of $f$ are still trapped in $(a,b) \cup(c,d)$. And since we make $f$ hyperbolic off of these intervals, f still generates a Markov exceptional set. Constructed this way, $f$ has two level-0 gaps. But it now has four level-1 gaps. (In general, the inverse image of two intervals would contain 6 intervals, since f is 3 to 1. This is true of our two intervals, $(a,b)$ and $(c,d)$, but these six intervals include $(a,b)$ and $(c,d)$, so there are only four new intervals, which are level-1 gaps. There are 12 level-2 gaps, 36 level-3 gaps and so on. We could keep constructing such examples by using $n$ to 1 functions, with as many periodic intervals as we want. In fact we can derive the topological types of these examples by starting with the function $x\mapsto nx \pmod{1}$ and ``thickening up" one (or more) periodic orbit to get a periodic interval. \subsection{A free group acting on the circle} Let $\phi$ and $\psi$ be diffeomorphisms of the circle. Let $\phi$ have exactly two fixed points: $u_-$, which is repelling, and $u_+$, which is attracting. Similarly let $\psi$ have two fixed points $v_-$, which is repelling, and $v_+$, which is attracting. Further let $U_-$, $U_+$, $V_-$, and $V_+$ be disjoint neighborhoods of $u_-$, $u_+$, $v_-$, and $v_+$ respectively so that $\phi(v_-) \in U_+$, $\phi(v_+)\in U_+$, $\phi^{-1}(v_-) \in U_-$, $\phi^{-1}(v_+)\in U_-$, $\psi(u_-) \in V_+$, $\psi(u_+)\in V_+$, $\psi^{-1}(u_-) \in V_-$, $\psi^{-1}(u_+)\in V_-$ Then $\Gamma$, the free group generated by $\phi$ and $\psi$, acts on the circle with a Cantor set as the minimal set. \noindent \textbf{Proof:} Let $x$ be any point other than $u_-$. Then $\phi^n(x) \longrightarrow u_+$. So $u_+$ is in the closure of the orbit of $x$. But $\psi(u_-) \ne u_-$, so $\phi^n\circ\psi(u_-)\longrightarrow u_+)$ as well, hence $u_+$ is in the closure of any orbit. So we only need to show that $\overline{{O}(u_+)}$ is totally disconnected and perfect. So let $x = \gamma_j\circ\dots\circ\gamma_1(u_+)$ and $y=\xi_k\circ\dots\circ\xi_1(u_+)$ be two different points in $\orbit(u_+)$, where $\gamma_i$ and $\xi_i \in \{\phi, \phi^{-1}, \psi, \psi^{-1}\}$. Since $u_+$ is a fixed point for $\phi$, we take $\gamma_1$, $\xi_1 \in \{\psi, \psi^{-1}\}$. Let $r$ be the smallest number for which $\gamma_{j-r} \ne \xi_{k-r}$. Then $\gamma_{j-r}\circ\dots\circ\gamma_1(u_+)$ and $\xi_{k-r}\circ\dots\circ\xi_1(u_+)$ are in different neighborhoods $U_+, U_-, V_+, V_-$. E.g. if $\gamma_{j-r}=\phi$ and $\xi_{k-r}=\psi^{-1}$, then $\gamma_{j-r}\circ\dots\circ\gamma_1(u_+)\in U_+$ and $\xi_{k-r}\circ\dots\circ\xi_1(u_+)\in V_-$. We note that $\orbit(u)\in U_+\cup U_- \cup V_+ \cup V_-$, so $\gamma_{j-r}\circ\dots\circ\gamma_1(u_+)$ and $\xi_{k-r}\circ\dots\circ\xi_1(u_+)$ have a neighborhood, $N$ between them which contains no elements of $\orbit(u_+)$. Therefore $\gamma_j\circ\dots\circ\gamma_1(u_+)$ and $\xi_k\circ\dots\circ\xi_1(u_+)$ have a neighborhood, $\gamma_j\circ\dots\circ\gamma_{j-r}(N)$ between them which also contains no points of $\orbit(u_+)$. That is $\orbit(u_+)$ is nowhere dense, and therefore $\overline{\orbit(u_+)}$ is also nowhere dense, and hence it's totally disconnected. Now to show that $\orbit(u_+)$ is perfect, we let $x\ne u_+, x\in \orbit(u_+)$ then since $u_+$ is the attracting fixed point for $\phi$, $\phi^n(x)\longrightarrow u_+$. Then if $\gamma\in\Gamma$, $\gamma\phi^n(x)\longrightarrow\gamma(u_+)$. Therefore $\orbit(u_+)$ is perfect, and hence $\overline{\orbit(u_+)}$ is an exceptional minimal set. \section{The Hirsch Foliation} \label{sec-hirsch} Hirsch's construction gives us a rich source of examples. We obtain the Hirsch foliation \cite{Hirsch1975} by starting with a solid torus and removing from the interior another solid torus which wraps around twice. This gives us a manifold, foliated by two-holed disks, with two transverse toruses as boundary components. We then glue the exterior boundary component to the interior component to obtain a foliated manifold without boundary. To describe this construction more rigorously, we'll start with $D_2$, the closed disk with two holes. $D_2$ has three circular boundary components which we'll call the two interior components $S_0$ and $S_1$ and the exterior component $S_2$. We'll also view $D_2$ as a pair of pants, in which case we'll call $S_0$ and $S_1$ the cuffs and we'll call $S_2$ the waist. We form the cartesian product of $D_2$ with a closed interval. $D_2\times [a,b]$ is a solid cylinder with two open solid cylinders removed from the interior. We decompose the boundary of $D_2\times [a,b]$ into five pieces: one exterior vertical cylinder, $S_0\times[a,b]$, two interior vertical cylinders $S_1\times[a,b]$ and $S_2\times[a,b]$, the bottom slice $D_2\times \set{a}$, and the top slice $D_2\times \set{b}$. We think of $D_2\times[a,b]$ as a foliation whose leaves are the level sets $D_2\times\set{x}$ where $x \in [a,b]$. The top and bottom slices are boundary leaves, and the vertical boundary cylinders are transverse. We form the foliated manifold $N$ by gluing the bottom slice to the top slice with a half twist. (That is we glue $S_1 \times \set{a}$ to $S_2\times \set{b}$ and we glue $S_1 \times \set{b}$ to $S_2\times \set{a}$). Since we glued with a half twist, we connected the two interior boundary cylinders together into one torus. $N$ is a solid torus with an open solid torus which wraps around twice removed from the inside. Since we glued the boundary leaves of $D_2\times [a,b]$ together, we still have a foliation whose leaves are pairs of pants. $N$ has two boundary components: the exterior boundary torus, $T_e$ and the interior boundary torus, $T_i$, which wraps around twice. We will identify the circle $[a,b]/_{a\sim b}$ with a longitudinal circle $S_e \in T_e$ which wraps around once, and hence hits each leaf once. We use $S_e$ to index the leaves of $N$. For $x\in S_e$, we let $P_x$ be the leaf of $N$ whose waist intersects $S_e$ at the point $x$. We could actually view $N$ as a fiber bundle over $D_2$, whose fibers are circles. Then we could choose $S_e$ to be the fiber over a point of $S_2$. But it suits our needs just to think of $N$ as a foliation with two transverse boundary toruses. We'll call the circles of intersection between leaves of $N$ and $T_e$ {\em latitudinal} circles, and we'll call circles which are parallel to $S_e$ {\em longitudinal} circles. We'll name longitudinal and latitudinal circles for $T_i$ similarly. So longitudes are transverse circles and latitudes are leaf-wise circles. % \vspace*{4in} We now form the Hirsch foliation by gluing $T_e \in N$ to $T_i$, using a diffeomorphism $H: T_e \longrightarrow T_i$. We require $H$ to take latitudinal circles of $T_e$ to latitudinal circles of $T_i$. That is $H$ preserves the the foliation on the transverse boundary components. So the quotient manifold $(M,{\mathcal F})$ remains a foliated manifold. We use the same two-holed discs, or pairs of pants, which were leaves of $N$, as non-standard plaques of $M$. These plaques are non-standard not only because they are not homeomorphic to open balls, but also because they need not intersect in regular position. We will see that there must be at least one plaque which intersects itself, and at least one pair of plaques which intersect each other in two components. % \vspace{3.6 in} $T_e$ and $T_i$ have been glued together to produce a single transverse torus, $T$, in the quotient manifold $M$. We define $S_i\subset T_i$ in $N$ by $S_i = H^{-1}(S_e)$. Then $S_e$ and $S_i$ get glued together in $M$ to form the transverse circle $S\in T$. $S$ still hits the waist of each plaque exactly once, so we can still index these plaques by $S$. For $x\in S$, we let ${\mathcal P}_x$ be the plaque whose waist hits $S$ at the point $x$. $S$ hits either cuff of each plaque exactly once as well. In general, for $t\in S$, the plaque ${\mathcal P}_t$ has three boundary components, $S_0 \times \{t\}$, $S_1 \times\set{t}$, and $S_2 \times \{t\}$; each of which intersects exactly one other plaque. (Recall that $S_0 \times \{a\} = S_1 \times \{b\}$.) As $H$ preserves the leaves on the boundary torus, $\pi\circ H$ restricted to $S$ is a two to one local homeomorphism $h:S\longrightarrow S$, where $\pi$ is leaf-wise projection onto $S$ in $N$. Depending on our choice of gluing map $H$, $h$ could be any two to one local diffeomorphism. As it happens, $h$ completely determines both the transverse and the leaf-wise dynamics of ${\mathcal F}$. For any $t\in S$, $S_2 \times \{t\}$ intersects one of the cuffs of. So ${\mathcal P}_{t}$ intersects ${\mathcal P}_{h(t)}$. Similarly if $h_i^{-1}$ either of the two branches of $h^{-1}$, then ${\mathcal P}_{t}$ intersects ${\mathcal P}_{h_i^{-1}(t)}$ \subsection{The Shape of Leaves} For $s\in S$, what does the leaf, $L_s$, containing $s$ look like? We think of the plaques ${\mathcal P}_s$ as pairs of pants whose waist is the exterior circle $S_2 \times \{s\}$, and the cuffs are the interior circles $S_0 \times \{s\}$ and $S_1 \times \{s\}$. In General, ${\mathcal P}_s$ intersects three other plaques, ${\mathcal P}_{h(s)}$, ${\mathcal P}_{h_0^{-1}(s)}$ and ${\mathcal P}_{h_1^{-1}(s)}$. So we build up the leaf by sewing the cuffs of ${\mathcal P}_s$ to the waists of ${\mathcal P}_{h_0^{-1}(s)}$ and ${\mathcal P}_{h_1^{-1}(s)}$ and by sewing the waist of ${\mathcal P}_s$ to ${\mathcal P}_{h(s)}$. We continue in this fashion. In the $r$th step, we obtain $L_s^r$, the union of all plaques within a distance $r$ from ${\mathcal P}_s$. $L_s^r$ consists of all plaques parametrized by ${\mathcal O}_r(s) = \{t\in S \mid $ there exists $j$, $k > 0$ with $j + k = r $ such that $ h^j(s) = h^k(t) \}$. Thus $L_s$ consists of all the plaques parametrized by the total orbit of $s$, ${\mathcal O}(s) = \{t\in S |$ there exists $j$, $k > 0$ such that $ h^j(s) = h^k(t) \}$. That is $L_s = \bigcup_{t\in{\mathcal O}_s} {\mathcal P}_t$. We see that $L_s^r$ is a tubular neighborhood of a finite tree, the root of which is the waist of ${\mathcal P}_{h^r(s)}$, and the ends of which consist of $2^{r+1} + 2^r - 1$ cuffs. If $s$ is an eventually periodic point of $h$, then one of these cuffs will eventually be sewn to the root of $L_s^r$. % \vspace{4 in} So $L_s$ consists either of a tubular neighborhood of an infinite tree, possibly with one handle depending on whether $s$ is an eventually periodic point of $h$. In either case, $L_s$ is quasi-isometric to a tree. In the first case, $L_s$ is homeomorphic to a sphere minus a Cantor set. In the second case, $L_s$ is homeomorphic to a torus minus a Cantor set. The Cantor set we remove corresponds to the set of ends of $L_s$. \subsection {The holonomy of the Hirsch foliation} The non-standard plaques we use come up naturally in the construction, and they made it easy to describe the shape of leaves, but we also use them because they are parametrized by $S$, making $S$ a totally transverse circle. The holonomy on $S$ with respect to these plaques is easy to describe since the holonomy on $N$ is trivial, and $H$ preserves the foliation. We define a map $h:S \longrightarrow S$, with $h(t)=r$. $h$ is a two to one map of the circle. The map $h$ is the holonomy function of ${\mathcal F}$ on $S$. The self-intersecting plaques mentioned above correspond to fixed points of $h$, and the pairs of plaques which intersect in two components correspond to period two points. % \vspace{4 in} We note that $h$ is the holonomy function on $S$, so all of the dynamics of ${\mathcal F}$ are encoded by $h$. We saw that a leaf corresponds to a total orbit of $h$, and a point of period $k$ corresponds to a handle of length $k$ in the plaque metric. $L_s^r$ is an exhaustion sequence for $L_s$, and the plaques of $L_s^r$ are parametrized by ${\mathcal O}_r(s)$, which itself is an exhaustion sequence for ${\mathcal O}(s)$. So $\omega(L_s) = \bigcup_{t \in \omega(s)}L_t$. If we fix the point $s\in S $ as a base-point of $L_s$. An end of $L_s$ corresponds either to the sequence $$s, h(s), h^2(s) \dots$$ or a sequence $$s, h(s), h^2(s) \dots h^k(s), h_{i_{k+1}}^{-1}(h^k(s)), h_{i_{k+2}}^{-1}(h^k(s)) \dots$$ The end which corresponds to the first sequence will be called the {\em expanding end}. An end which corresponds to a sequence of the second type will be called an {\em eventually contracting end}. An end which only corresponds to a backwards orbit of $s$ will be called a {\em contracting end}. Note that the designation of an end as eventually contracting is not natural as it depends on the arbitrary choice of a base-point. An invariant set of ${\mathcal F}$ corresponds to an invariant set of $h$. If $X\subset M$ is an invariant set for ${\mathcal F}$, then $X\cup S$ will be an invariant set for $h$, and if $W \subset S$ is an invariant set for $h$, then $X=\set{ {\mathcal P_x} | x\in W} = \set{ {L_x} | x\in W} $ is an invariant set of ${\mathcal F}$. Before gluing the boundary toruses together, the holonomy of $N$ is trivial. This means that for paths that stay away from the torus $T$, the holonomy of $M$ is trivial as well. The only non-trivial holonomy we get in $M$ is created by the gluing map $H$. If $P_x\subset M$ is any of the two-holed plaques, and $\gamma$ is a path which stays in the interior of $P_x$, then the holonomy along $\gamma $ is trivial. Likewise, if $\gamma$ stays in $P_x$ and begins and ends at the same boundary component of $P_x$, the holonomy is trivial as well. The only $\gamma$ can be a path in $P_x$ whose holonomy is non-trivial is if it begins at one boundary component of $P_x$, and ends in another component. Since a path between the cuffs of $P_x$ can be homotoped to a path from one cuff to the waist to the other cuff, we need only consider paths from the waist to one cuff. We may as well consider paths that begin and end in the transverse circle $S$. We fix a point $x\in S$. Then ${\mathcal P}_x$ is the plaque whose waist hits $S$ at $x$. We think of $S$ as being the circle $[a,b]|_{a\sim b}$ embedded into the manifold $M$. If $\gamma$ begins at $x\in S$ and travels from the waist of the plaque $P_x$ to one of its cuffs, then the endpoint of $\gamma$ is, in general, a different point of $S$. Indeed, thinking of $\gamma$ as a curve in $N$, instead of a curve in $M$, and thinking of $N$ as $D_2\times [a,b]$ with the top and bottom identified, then $\gamma$ begins at the point $(w_0,x)$ with $w_0\in S_e\subset D_2$ and $x\in[a,b]$, and $\gamma$ ends at either $(w_1,x)$ with $w_1\in S_i\subset D_2$ or $(w_2,x)$ with $w_2\in S_i$. We'll let $\gamma_0$ be the path that ends at $(w_0,x)$ and $\gamma_1$ the path that ends at $(w_1,x)$ . This labeling really only works for $x\ne a$, since we could label that plaque either $P_a$ or $P_b$. But the point is that $\gamma$ is a path from the exterior component of $D_2\times\set{x}$ to one of its interior components. So returning to the manifold $M$, there are two paths from the waist of $P_x$ to one of its cuffs. We call these two paths $\gamma_0$ and $\gamma_1$. The holonomy along these two paths are the two right inverses of the map $h$, which we get by projecting the gluing map $H$ to a longitudinal circle of $N$. \subsection{The Hirsch Example With Linear Generators} Given any $k$ to 1 local homeomorphism of the circle $h$, the Hirsch construction produces a foliation whose transverse dynamics are given by iterating $h$ and the $k$ branches of $h^{-1}$. We examine the dynamics of some possible two to one functions $h:S^1 \longrightarrow S^1$ in the context of the Hirsch foliation, where $S^1=[a,b]/_{a\sim b}$. In the linear case, $h(x)=2x \bmod 1 $, each leaf is dense, so the $\omega(L) = M$. (A leaf parametrized by an rational number will be a tree made of tubing, with one handle added. A leaf parametrized by a irrational number will be a tree with no handles.) To show that the backwards orbits of $h(x) = 2x \bmod 1\ $ are dense, we note that $h$ has two inverse maps: \[h_0^{-1} : x \longrightarrow \frac{x}{2}\] \[h_1^{-1}: x \longrightarrow \frac{x+1}{2}\] Since $h_0$ contracts to $0$ it suffices to show that the backwards orbit of $0$ is dense. But the backwards orbit of 0 consists of all the dyadic numbers, and hence is dense. The limit set of an end is also dense. \begin{ex} Take $[a,b] = [0,1]$, and use a gluing map $H$ which projects to the map $x\mapsto 2x\pmod1$. Then, for any $x\in S$, the waist of the plaque $P_x$ is sewn to one of the cuffs of $P_{2x}$. The path which starts at $x$, then travels up the leg of $P_{2x}$, and ends at $2x\in S$ has $h:x\mapsto 2x\pmod1$ as a holonomy function. A path in the opposite direction will have one of the branches of $h^{-1}=\frac{1}{2}x\pmod1$ as its holonomy. If we take $x\in(0,1)\subset S$, then there are two paths, $\gamma_0^{-1}$ and $\gamma_1^{-1}$, from the waist of $P_x$ to one of its cuffs. The holonomy along $\gamma_0^{-1}$ will be the function $\frac{x}{2}$ and the holonomy along $\gamma_1^{-1}$ will be the function $\frac{x+1}{2}$. This means that the leaf $L_x$ has one direction with expanding holonomy. Locally $L_x$ has two directions with contracting holonomy, Globally it has an infinitely branching number of directions with contracting holonomy. If you take any two sub-intervals of $S$, and flow them along a path from the waist to one of the cuffs of a plaque $P_x$, then either of the two subintervals will be contracted by a factor of $\frac{1}{2}$, so their relative lengths don't change. If we do this with the appropriate intervals to define the ratio geometry of $h$, we get a constant value of $\frac{1}{2}$. \end{ex} % \vspace*{6 in} \subsection{Limit Sets of Ends and Paths} An end $\epsilon$ corresponds to an infinite branch of the tree, the limit set of $\epsilon$ is determined by a decreasing neighborhood of that branch. Let $\gamma$ be a path to the end $\epsilon$. Then depending on $\epsilon$, the limit set of $\gamma$ could be dense a countable set or a Cantor set. To generate a Cantor set, we need only specify the set of gaps. This will consist of the interval $(\frac{1}{4} , \frac{3}{8})$ along with all of its iterates under the maps \begin{eqnarray*} h_0^{-1} : x & \longrightarrow & \frac{x}{2}\\ h_1^{-1}: x & \longrightarrow & \frac{x+1}{2} \end{eqnarray*} Since the image of any point is dense under these maps, the specified set of gaps must be dense, and hence its compliment is totally disconnected. We see that some of the iterates are contained in earlier iterates. Taking this into account, we see that the endpoints consist of elements of $[0,1]$ whose base two expansion terminate either with 011 or 01, and which contain no instance of 010. If an instance of 010 occurs, that means the gap got translated into the interior of an earlier iterate. This shows that our desired set is perfect, and hence is a Cantor set. To find a path whose limit set actually generates this set, we simply concatenate all finite sequences which contain no instance of 010, interspersing an extra 1 when necessary. One example is the sequence $\dots 100$ $011$ $001$ $1$ 000 11 10 01 1 00 1 1 0. \subsection{The Hirsch Example With A Cookie Cutter} If the gluing map, $H$, projects to a cookie cutter, $h$, on the transverse circle, $S$, then we have much the same situation. Globally a leaf, in general, has one direction with expanding holonomy, as you travel up the tree; and an infinitely branching number of directions with contracting holonomy, as you travel down the tree. Locally a leaf, in general, has two directions with contracting holonomy. Actually if the leaf is part of the minimal set, then it's holonomy is strictly expanding as you travel up the tree, and strictly contracting as you travel down the tree, but for leaves off of the minimal set, this is only a general picture. Either way, as you travel up the tree starting from the plaque $P_x$, you successively hit the plaques $P_{h(x)}$, $P_{h^2(x)}$, $P_{h^3(x)}$, etc. As you travel down the tree starting from the plaque $P_x$, depending on which branches you choose, you might hit the plaques $P_{h_0^{-1}(x)}$, $P_{h_1^{-1}h_0^{-1}(x)}$, $P_{h_1^{-1}h_1^{-1}h_0^{-1}(x)}$, etc. Since $h$ is a cookie cutter on $S$, it doesn't matter that $S$ and $h$ arose in the context of the Hirsch foliation. $h$ is a cookie cutter on the circle $S$, therefore all of the theory of cookie cutters applies to $h$. $h$ has a Cantor set $C$ as a minimal set. We can write $S= H\cup I$ where $I=I_0\cup G\cup I_1$. Then for $w$ any finite string of 0s and 1s, we can define $I_{0w}=h_0^{-1}[I_w]$, and so forth. This gives us the structure of nested intervals $I_{w}=I_{w0}\cup G_w \cup I_{w1}$, which we use to label $C$. $C$ will have bounded geometry, and the scaling function will classify the differentiable structure of $C$. But now we can also view $h$ as the holonomy of $M$ in the expanding direction, and $h_0^{-1}$ and $h_1^{-1}$ as the holonomy of $M$ in the contracting directions. At least, off of the interval $H$, these are the expanding and contracting directions respectively. So let $x_\omega\in S $ and take any two subintervals $J$ and $K$ of $I_0 \cup G\cup I_1$. Choose a branch of the leaf to travel in the contracting direction, say we use a path from $P_{x_\omega} $ to $P_{x_{0\omega}} $ to $P_{x_{0\omega}}$ to $P_{x_{10\omega}}$ to $P_{x_{110\omega}}$, etc. As we flow $J$ and $K$ down this path, say we get the intervals $J_0$ and $K_0$, $J_{10}$ and $K_{10}$, $J_{110}$ and $K_{110}$, etc. Each time we add a 0 or a 1 to the left, that's the same as flowing from the waist to one of the cuffs of whichever plaque we've arrived at. So we are traveling to an end of the leaf $L_x$ in the contracting direction. Since the derivatives of $h_1^{-1}$ are above by $\beta< 1$, the intervals $J_w$ and $K_w$ are exponentially small. That means that the holonomy functions $h_i^{-1}$ are getting closer and closer to linear, and the ratio of $J_w$ and $K_w$ converges exponentially fast. But the intervals we're interested in are $I_0$, $G$, and $I_1$ and $I$. As we flow $I_0$ and $I$, for instance, in a contracting direction on the leaf, their ratios converge exponentially fast to the first component of the scaling function. So, having fixed the base point $x$, the scaling function can be defined on the set of infinitely long paths beginning from $x$ with contracting holonomy instead of some abstract dual Cantor set. These paths correspond to contracting ends of the leaf $L_x$, possibly with the handle thrown in as well if $L_x$ has a handle. \section{The Double Suspension} The double suspension can be used to construct a foliated manifold with a holonomy group, acting on a transverse circle, generated by two diffeomorphisms $\phi$ and $\psi$ of the circle. We describe the double suspension informally by starting with $\Sigma_2 \times S^1$. Where $\Sigma_2$ denotes the two-holed torus. We then cut $\Sigma_2$ twice, once along each handle, and glue the cuts back together using $\phi$ and $\psi$. % \vspace{4 in} More formally we note that when we cut $\Sigma_2$, we're left with $D_3$, a disk with three holes. So we begin with $D_3 \times S^1$. This is a foliated manifold with four boundary components, each of which is a transverse torus. We glue these together pairwise, using a gluing map which is $\phi$ in the transverse direction for the first pair, and a gluing map which is $\psi$ in the transverse direction for the second pair. We see that we get a foliation whose holonomy is the free group generated by $\phi$ and $\psi$ acting on the transverse circle. % \vspace{4 in} \section{Generalizing Hirsch's Construction} Let $f_1, \dots ,f_n$ be local diffeomorphisms of the circle. Then we can modify the Hirsch foliation to produce a foliation whose dynamics are given by $f_1$ through $f_n$. Since $f_k:S^1 \longrightarrow S^1$ is a local diffeomorphism, it must be an $r_k $ to $1$ map. As before we begin with a solid torus, and remove $2n+1 $ worm holes from the center, so that we have a manifold with one exterior boundary component, and $2n+1$ interior boundary components, all of which are tori. We label the interior boundary components $T_0, \dots , T_{2n}$. We remove the worm holes in such a way that $T_0$ through $T_n$ each wraps around once longitudinally, and $T_{n+k}$ wraps around $r_k$ times. We glue $T_0$ to the exterior torus with a map that is the identity in the longitudinal direction. We then glue $T_k$ to $T_{n+k}$ using $f_k$ in the longitudinal direction. \begin{ex} We defined the double suspension of two circle maps in this way. \end{ex} \begin{ex} Let $X$ be a solid torus with two solid toruses removed from the inside, one which wraps around twice, and one which wraps around three times. We then have a foliated manifold with three boundary components, whose leaves are 5-holed latitudinal discs. If we identify the two interior boundary components, we get a foliated manifold with a transverse torus as the boundary. We think of the five-holed discs as non-standard plaques which are parametrized by a longitudinal circle on the boundary. The transverse dynamics will be given by a 3 to 2 local homeomorphism of the circle. The dynamics of such a map are difficult to analyze. In general they will not be Markov, because there are two ways of going forwards and three ways of going backwards.\end{ex} We can further generalize Hirsch's construction. Let $(M,{\mathcal F})$ be any codimension one foliation of a compact manifold and let $T$ be a solid torus with a transverse boundary. We can then hollow out solid toruses from the middle of $T$ and glue their boundary components together. We then get a new codimension one foliation whose holonomy is generated by the holonomy of $\mathcal F$ along with the gluing maps used in the construction. Note that whenever we have a transverse circle, we can take a neighborhood of the transverse circle to get a transverse torus. \begin{ex} Let $R$ be a Reeb component. We can hollow out two solid toruses from $R$, one which wraps around once, and one which wraps around twice. If we glue their boundaries together with a cookie cutter function, we get a foliation whose minimal set is the boundary torus, but which has a local exceptional set which is Markov and hyperbolic. \end{ex} \begin{ex} Let $N_1$ be a solid torus with another solid torus removed from the inside. Let the interior torus wrap around twice in the longitudinal direction, but only once in the latitudinal direction. This interior torus is unknotted as a subset of $\Real^3$. We form the Hirsch foliation on the manifold $M_1$ by identifying the boundary components using the diffeomorphism $H_1$, which projects to the 2 to 1 map $h$ on the transverse circle. Getting a holonomy function $h$ acting on the transverse circle $S$. Let $N_2$ also be a solid torus with another solid torus removed from the inside. But this time, let the interior torus be knotted as a trefoil which wraps around twice in the longitudinal direction and three times in the latitudinal direction. Again, we form a Hirsch foliation on the manifold $M_2$ by gluing the interior component to the exterior component, using the diffeomorphism $H_2$. If $H_2$ projects to the same map $h$, on the transverse circle, then the leaves of $M_1$ are identical to the leaves of $M_2$, and the transverse dynamics of $M_1$ is identical to the transverse dynamics of $M_2$, but the ambient manifolds are not diffeomorphic. \end{ex} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \chapter{Results on $\mathbf S^1$ and $\mathbf I$} \section{The Conjugation Problem for Cookie Cutters} \subsection{Automorphisms of Cookie Cutter Sets} We give a partial classification of the outer automorphisms of cookie cutter Cantor sets. For cookie cutter Cantor sets with two affine generators, such diffeomorphisms have to be given by a finite number of clone maps. This also holds true for labeled Cantor sets with the proper pinching on the ratio geometry, and hence for cookie cutter set with the proper pinching on the derivative. \begin{thm} Let $C$ be an affine cookie cutter set with two generators. Let $\phi: I\longrightarrow I$ be an order preserving $C^1$ diffeomorphism with $\phi(C) = C$. Then there is a finite cover of $C$ by clone intervals $I_{w_0}, \dots I_{w_m}$ so that for each $w_i$, $\phi$ restricts to a clone map on $C_{w_i}$. \end{thm} \noindent \textbf{Proof:} We'll look at a sufficiently small gap, $G$, of $C$. Then by the smoothness of $\phi$, we have a good approximation for the sizes of images of nearby gaps. This will show that we have only one choice for the image of each nearby gap, and that will define a clone map. % \vspace*{4 in} We let $I_0 = [0,\alpha]$ and $I_1=[1-\beta,1]$. Then $f_0= \frac{1}{\alpha}x$, $f_1= \frac{1}{\beta}x$ and $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ are the two rates of expansion. Without loss of generality, we let $\alpha \le \beta$. We let $\gamma$ be the relative length of all the gaps (and so $\alpha +\beta +\gamma = 1$). Let $a\in C$. We show that there is a clone containing $a$ on which $\phi$ restricts to a clone map. Then since $a$ is an arbitrary element, we can cover $C$ by such clones, and since $C$ is compact, we can take a finite sub-cover. So we let $a\in C$, and choose $\epsilon$ sufficiently small. I.e. $\epsilon <\min \left( \frac{\gamma \phi'(a)}{2\alpha + \gamma}, \frac{(1-\beta)\phi'(a)}{1+\beta}\right)$. Now we choose $U$ to be a small enough neighborhood of $a$, that for $u\in U$, $|\phi'(a)-\phi'(u)| \le \epsilon$. Then we take $I_\omega \subset U$ to be a clone interval in $U$ with $a \in I_\omega$. Since $\phi$ preserves $C$, it must take gaps to gaps. So we set $G_{\omega'}= \phi(G_\omega)$, and claim that $\phi(G_{w0})= G_{w'0}$ and $\phi(G_{w1})= G_{w'1}$. Then, by induction on $m$, we show that $\phi(G_{\omega \epsilon_0 \dots \epsilon_m})=\phi(G_{\omega'\epsilon_0\dots\epsilon_m})$. And since the gaps are dense, this will show that $\phi$ is a clone map on $C_\omega$. Since $G_w$ and $G_{w'}$ are gaps in an affine Cantor set, there are integers $j$, $k$, $j'$, and $k'$ so that \begin{eqnarray*} |G_\omega| & = & \alpha^j\beta^k\gamma \\ |G_{\omega'}| & = & \alpha^{j'}\beta^{k'}\gamma \end{eqnarray*} and $$ \frac{|G_{\omega'}|}{|G_\omega|} = \alpha^{j'-j}\beta^{k'-k}= \phi'(z) $$ and hence $$ |\alpha^{j'-j} \beta^{k'-k}-\phi'(a)| < \epsilon. $$ We first show that $\phi(G_{\omega 0}) \subset I_{\omega' 0}$. We suppose that this is not so. Then since $\phi$ is order preserving, and $\phi(G_\omega) = G_{\omega'}$, $\phi(G_{\omega 0})$ is to the left of $I_{\omega'}$, and hence contains the gap, $G_\gamma$, directly to the left of $I_{\omega'}$. Now, since $G_\gamma \not\subset I_{w'0}$, $G_\gamma$ has to be a lower level (and hence bigger) gap than $G_{\omega'}$, and so \[ |\phi(I_{\omega 0})| > (\alpha +\gamma)\alpha^{j'}\beta^{k'}.\] Therefore \begin{eqnarray*} \frac{\alpha + \gamma}{\alpha} [\phi'(a)-\epsilon]& < &\frac{\alpha + \gamma}\alpha^{j'-j}\beta^{k'-k}\\ & <& \frac {|\phi(I_{\omega 0 })|}{|I_{\omega 0}|}\\ & =& \phi'(z)\\ & \le& \phi'(a) + \epsilon \end{eqnarray*} This inequality implies that $\epsilon > \frac{\gamma \phi'(a)}{2\alpha + \gamma}$, which, of course, contradicts $\epsilon$ being sufficiently small. Now we show that $\phi(G_{\omega 0})= G_{\omega' 0}$. Again we suppose otherwise. Then since $\phi(G_{\omega 0 }) \subset I_{\omega 0}$, \[|\phi(G_{\omega 0})| \le \alpha^{j+1}\beta^{k+1}\gamma.\] Since $\beta > \alpha$, we have that \begin{eqnarray*} \phi'(a) - \epsilon& \le& \phi'(z)\\ & =& \frac{|\phi(G_{\omega 0})|}{|G_{\omega 0} |}\\ &\le& \beta \alpha^{j'-j}\beta^{k'-k}\\ &<& \beta(\phi'(a) + \epsilon) \end{eqnarray*} This inequality implies that $\epsilon > \frac{\phi'(a)(1-\beta)}{1+\beta}$ which, of course, again contradicts $\epsilon$ being small. This shows that $\phi(G_{\omega 0}) = G_{\omega' 0}$. The same argument shows that $\phi(G_{\omega 1}) = G_{\omega' 1}$ and hence $\phi$ is a clone map on $C_\omega$. \done \begin{ex} For cookie cutters with more than two generators this theorem becomes false. We let $C$ be the middle third Cantor set, as generated by the following function \[ f(x) = \left \{ \begin{array}{lcr} 9x& & x \in [0, \frac{1}{9}]\\ 9x-2& & x\in [\frac{2}{9},\frac{1}{3}]\\ 3x & & x\in [\frac{2}{3},1] \end{array} \right. \] We define the function $\phi : C \longrightarrow C$. \end{ex} \[ \phi(x) = \left \{ \begin{array}{lcr} 3x& & x \in [0, \frac{1}{9}]\\ x+\frac{4}{9}& & x\in [\frac{2}{9},\frac{1}{3}]\\ \frac{1}{3}x+\frac{2}{3} & & x\in [\frac{2}{3},1] \end{array} \right. \] We see that any clone interval with left end point at 0 is of the form $I = [0, \frac{1}{3^{2n}}]$, and $\phi(I)= [0, \frac{1}{3^{2n-1}}]$ which, with our chosen function, is not a clone interval. We have a similar situation at the point $x= \frac{2}{3}$. Note that while locally $\phi$ isn't an element of the pseudogroup, the germ of $\phi $ at a point will be the germ of a pseudogroup element. \begin{thm} Let $C$ be a labeled Cantor set generated by two generators, with bounded geometry $\alpha \le r \le \beta$ with $\beta^3 < \alpha^2$. If $\phi$ is a $C^{1+\lambda}$ outer automorphism of $C$, (i.e $\phi$ extends to a $C^{1+\lambda}$ map of $I$) then $\phi$ restricts to a piecewise clone map. \end{thm} \noindent \textbf{Proof:} We proceed as before, by pegging down one gap, and showing that nearby gaps must fall into place. Let $\phi$ be such a map. Take $a \in C$. Let $\epsilon$ be sufficiently small. ($\epsilon \le \frac {\alpha^2 -\beta^3} {\alpha^2+\beta^3} \phi'(a)$.) For any clone interval $I_w$ which contains $a$, and is small enough that $|\phi'(u)-\phi'(a)| \le \epsilon$, we claim that $\phi$ restricts to a clone map on $I_w$. Let $G_{w'}= \phi(G_w)$ and show that $\phi(I_{w0}) = I_{w'0}$ and that $\phi(I_{w1}) = I_{w'1}$. Suppose $\phi(I_{w0}) \not \subset I_{w'0}$. Then $I_{w'0} \cup G_v \subset \phi(I_{w0})$ where $G_v$ is the gap directly to the left of $I_{w'}$. In which case \begin{eqnarray*} \phi'(a)+\epsilon &>& \frac{|\phi(I_{w0})|}{|I_{w0}|}\\ &>& \frac{|I_{w'0}|+|G_v|}{|I_{w0}|}\\ &=& \frac{|I_{w'0}|}{|I_{w0}|}+\frac{|G_v|}{|I_{w0}|}\\ &=& \frac{|I_{w'0}|}{|I_{w'}|}\cdot\frac{|I_{w'}|}{|G_{w'}|}\cdot \frac{|G_{w'}|}{|G_{w}|}\cdot \frac{|G_w|}{|I_w|}\cdot\frac{|I_w|}{|I_{w0}|}\\ & & +\frac{|G_{w_1'\dots w_k'}|}{|I_{w_1'\dots w_k}|} \cdot\frac{|I_{w_1'\dots w_k'}|}{|I_{w_1'\dots w_{k+1}}|}\cdot\dots\cdot \frac{|I_{w_1'\dots w_{l-i}'}|}{|I_{w_1'\dots w_l}|}\cdot \frac{|I_{w'}|}{|G_{w'}|}\cdot \frac{|G_{w'}|}{|G_{w}|}\cdot \frac{|G_w|}{|I_w|}\cdot\frac{|I_w|}{|I_{w0}|}\\ &\ge& \frac{\alpha^2}{\beta^2} \left( \phi'(a)-\epsilon \right) + \frac{\alpha}{\beta^{l-k}} \cdot \frac{1}{\beta}\cdot \left(\phi'(a)-\epsilon \right)\cdot\frac{\alpha}{\beta}\\ &\ge& \left( \frac{\alpha^2}{\beta^2} + \frac{\alpha^2}{\beta^3} \right) \left( \phi'(a)-\epsilon \right) \end{eqnarray*} and so \begin{eqnarray*} \phi'(a) + \epsilon & \ge&\left( \frac{\alpha^2}{\beta^2} + \frac{\alpha^2}{\beta^3} \right) \left( \phi'(a)-\epsilon \right)\\ \left( 1+\frac{\alpha^2}{\beta^2}+\frac{\alpha^2}{\beta^3}\right) \epsilon& \ge &\left( \frac{\alpha^2}{\beta^2} + \frac{\alpha^2}{\beta^3}-1 \right)\phi'(a)\\ \epsilon & > & \frac{\alpha^2\beta +\alpha^2-\beta^3}{\alpha^2\beta +\alpha^2+\beta^3} \cdot \phi'(a)\\ & > &\frac{\alpha^2-\beta^3}{\alpha^2+\beta^3} \cdot \phi'(a) \end{eqnarray*} which contradicts $\epsilon$ being small which goes to show that $\phi(I_{w0}) \subset I_{w'0}$. We now show that $\phi(G_{w0})= G_{w'0}$. Suppose not. Then since $\phi(G_{w0}) \subset I_{w'0}$, $\phi(G_{w0}) = G_{w'0\epsilon_1\dots\epsilon_k}$ and hence \begin{eqnarray*} \phi'(a) - \epsilon& < & \frac{|\phi(G_{w0})|}{|G_{w0}|}\\ &=& \frac{|G_{w'0\epsilon\dots\epsilon_k}|}{|G_{w0}|}\\ &=& \frac{|G_{w'0\epsilon\dots\epsilon_k}|}{|I_{w'0\epsilon_1\dots\epsilon_k}|}\cdot \frac{|I_{w'0\epsilon_1\dots\epsilon_k}|}{|I_{w'0\epsilon_1\dots\epsilon_{k-1}}|}\dots \frac{|I_{w'0\epsilon_1}|}{|I_{w'0}|}\cdot \frac{|I_{w'0}|}{|I_{w'}|}\cdot \frac{|I_{w'}|}{|G_{w'}|}\cdot \frac{|G_{w'}|}{|G_w|}\cdot \frac{|G_w|}{|I_w|}\cdot \frac{|I_w|}{|I_{w0}|}\\ &<& \frac{\beta^{k+1}}{\alpha}(\phi'(a)+\epsilon)\cdot \frac{\beta}{\alpha}\\ &<& \frac{\beta^3}{\alpha^2} (\phi'(a)+\epsilon) \end{eqnarray*} This gives us that \[\epsilon >\frac{\alpha^2-\beta^3}{\alpha^2+\beta^3}\phi'(a)\] \noindent This is a contradiction hence $\phi(G_{w0})=G_{w'0}$. Similarly $\phi(G_{w1})=G_{w'1}$, and by induction $\phi(G_{\omega \alpha}) =G_{\omega' \alpha}$. Therefore $\phi$ restricts to a clone map on $C_w$. \done Like the theorem for affine Cantor sets, the proof given here relies on the fact that $C$ is generated by two generators. The counter example given, however, doesn't provide a counter example for this version of the theorem though, since in the example $\alpha = \frac{1}{9}$ and $\beta = \frac{1}{3}$, so $\beta^3>\alpha^2$. \subsection{$C^{1+\lambda}$ Equivalence} In \cite{Sullivan1988,Sullivan1992}, Sullivan used the scaling function, as defined by Feigenbaum in a different setting, to categorize the $C^{1+\lambda}$ structure of cookie cutter Cantor sets. $C^{1+\lambda}$ cookie cutters have bounded geometry. The ratio geometry converges along backwards orbits of $F$ to the scaling function, which is a H\"older continuous function from left-infinite words. If two cookie cutters are $C^{1=\alpha1+\lambda }$ conjugate, then their ratio geometries are exponentially close, which means that they have the same scaling function. Using the boot-strapping method, if $\phi$ is any $C^{k+\lambda} $ conjugation between $C^{k+\lambda}$ cookie cutters, then $\phi$ must itself be $C^{k+\lambda}$ as well. There's a partial converse. Given any H\"older scaling function, we can construct a cookie cutter which has that scaling function, and any two cookie cutters with the same scaling function are $C^{1+\lambda}$ conjugate in a neighborhood of the minimal set. Proposition~\ref{bdd-holder} shows that if two Cantor sets have bounded geometry, the label-preserving map, $\phi$, is H\"older. But the ratio geometry determines the Cantor set, up to affine rescaling. And since $C$ is a perfect set, this is enough to determine the smooth structure in a neighborhood of $C$. By Taylor's theorem, since the length of $I_w$ is exponentially small, the non-linear part of $\phi$ is also exponentially small, which means that the ratio geometry is determined exponentially closely by the smooth structure of $C$. If the conditions of the following proposition are met, we say that $C$ and $C'$ are {\em exponentially close}. \begin{prop} Let $\phi:C\longrightarrow C'$ be the label preserving map between Cantor sets with bounded geometry. We write $J_w= \phi (I_w)$. Then the following conditions are equivalent: There exists $\xi <1$ and $K>0$ such that for every word $w$ of length n, 1) $\left| \frac{|I_{wi|}}{|I_w|}-\frac{|J_{wi}|}{|J_w|}\right| \le K\xi^n$ for $i\in\{0,1\}$ 2) $(1- K)\xi^n\frac{|I_{wi}|}{|I_w|}\le \frac{|J_{wi}|}{|J_w|}\le (1+ K\xi^n)\frac{|I_{wi}|}{|I_w|}$ 3) $e^{- K \xi^n}\frac{|I_{wi}|}{|I_w|}\le\frac{|J_{wi}|}{|J_w|} \le e^{ K \xi^n}\frac{|I_{wi}|}{|I_w|}$ 4) for any finite word $\alpha$, $e^{- k\xi^n} \frac{|I_{w\alpha}|}{|I_w|}\le\frac{|J_{w\alpha}|}{|J_w|} \le e^{ k\xi^n} \frac{|I_{w\alpha}|}{|I_w|}$ 5) for any $u, v, x, y \in I_w$, $e^{- K\xi^n}\cdot \frac{u-v}{x-y}\le\frac{\phi(u)- \phi(v)}{\phi(x)-\phi(y)} \le e^{ K\xi^n} \cdot \frac{u-v}{x-y}$ \end{prop} \noindent \textbf{Proof:} We prove the implications for the upper bounds, and the lower bounds will be the same.\\ 1) $\Rightarrow$ 2): \begin{eqnarray*} \frac{|J_{wi}|}{|J_w|}& \le & \frac {|I_{wi}|}{|I_w|} + K\xi^n\\ & = & \frac {|I_{wi}|}{|I_w|}\left|1+ \frac{|I_w|}{|I_{wi}|}K\xi^n\right|\\ & \le & \frac {|I_{wi}|}{|I_w|}\cdot \left| 1 +\frac{K}{\alpha}\xi^n \right| \end{eqnarray*} 2) $\Rightarrow$ 1): \begin{eqnarray*} \frac{|J_{wi}|}{|J_w|}& \le & \frac {|I_{wi}|}{|I_w|}(1 + K\xi^n)\\ &=&\frac {|I_{wi}|}{|I_w|} + K\xi^n \frac {|I_{wi}|}{|I_w|}\\ &\le&\frac {|I_{wi}|}{|I_w|} +K\xi^n \gamma \end{eqnarray*} 3) $\Rightarrow$ 2): \begin{eqnarray*} \frac {|J_{wi}|}{|J_w|}& \le& e^{ K\xi^n}\frac {|I_{wi}|}{|I_w|} \\ & = &[1+O(\xi^n)]\frac {|I_{wi}|}{|I_w|} \end{eqnarray*} 2) $\Rightarrow$ 3) \begin{eqnarray*} \frac{|I_{wi}|}{|I_w|} \frac{|J_w|}{|J_{wi}|}& \le & 1+ K\xi^n\\ & \le & 1 + K \xi^n +\frac{(K\xi^n)^2}{2} + \dots \\ & = & e^{K\xi^n} \end{eqnarray*} Now we need to show that there exists $K_2$ such that $e^{-K_2\xi^n} \le \frac{|I_{wi}|}{|I_w|} \frac{|J_w|}{|J_{wi}|}$, for which it suffices to show that $e^{-K_2\xi^n} \le 1-K\xi^n$. So choose $K_1 > K$. Then $e^{-K_1\xi^n} \le 1-K_1 + C{K_1}^2\xi^{2n}$. Now choose $N$ big enough so that for $n>N$, $K_1 \xi^n - C{K-1}^2 \xi^{2n} > K\xi^n$. Then $e^{-K_1\xi^n} < 1-K\xi^n$. Now replacing $K_1$ with $K_2 = \max_{n \le N}(K_n, K)$, we have $e^{-{K'}'\xi^n} \le 1-K \xi^n$ for all $n > 0$. This completes the proof that 1), 2), and 3) are equivalent. 5) $\Rightarrow$ 4) $\Rightarrow$ 3) is immediate, so now we need to show that 3) $\Rightarrow$ 4) $\Rightarrow$ 5). \noindent 3) $\Rightarrow$ 4) Write $\alpha = v_0v_1 \dots v_m$. Then \begin{eqnarray*} \frac{|J_{w\alpha}|}{|J_w|}& = & \frac{|J_{w v_0}|}{|J_w|} \frac{|J_{w v_0v_1}|}{|J_{w v_0}|} \dots \frac{|J_{w v_0 \dots v_m}|}{|J_{w v_0 \dots v_{m-1}}|}\\ &\le & e^{K \xi^n} \frac{|I_{w v_0}|}{|I_w|} e^{K \xi^{n+1}} \frac{|I_{w v_0v_1}|}{|I_{w v_0}|} \dots e^{K \xi^{n+m}} \frac{|I_{w v_0 \dots v_m}|}{|I_{w v_0 \dots v_{m-1}}|}\\ & = & e^{K \xi^n(1+\xi + \xi^2 + \dots +\xi^m)} \frac{|I_{w v_0}|}{|I_w|} \frac{|I_{w v_0v_1}|}{|I_{w v_0}|} \dots \frac{|I_{w v_0 \dots v_m}|}{|I_{w v_0 \dots v_{m-1}}|}\\ & \le & e^{\frac{K}{1-\xi} \xi^n} \frac{|I_{w\alpha}|}{|I_w|} \end{eqnarray*} \noindent 4) $\Rightarrow$ 5): We have $x$, $y$, $u$, and $v \in I_w$. If we show that $\frac{|\phi(x) - \phi(y)|}{|J_w|} \le e^{ K \xi^n} \frac{ |x-y|}{|I_w|}$, then since $x$ and $y$ are arbitrary elements of $I_w$, this applies to $u$ and $v$ also, and hence \begin{eqnarray*} \frac{|\phi(x) - \phi(y)|}{|\phi(u) - \phi(v)|} & = & \frac{|\phi(x) - \phi(y)|}{|J_w|} \frac{|J_w|}{|\phi(u) - \phi(v)|}\\ & \le& e^{ K\xi^n} \frac{|x-y|}{|I_w|} e^{ k \xi^n} \frac{|I_w|}{|u-v|}\\ &=& e^{ 2K\xi^n} \frac{|x-y|}{|u-v|} \end{eqnarray*} Now write $[x,y] = \bigcup_{t=1}^\infty \overline{G_t}$ where each $ G_t$ is a gap contained in $[x,y]$. Write $H_t = \phi(G_t)$. Then \begin{eqnarray*} \frac{|\phi(x)- \phi(y)|}{|J_w|}& = & \frac{\sum |H_t|}{|J_w|}\\ & = & \sum \frac{|H_t|}{|J_w|}\\ & \le & \sum e^{ K \xi^n} \frac{|G_t|}{|I_w|}\\ & = & e^{K \xi^n} \frac{|x-y|}{|I_w|} \end{eqnarray*} \done \setcounter{ctr}{\value{thm}} \setcounter{thm}{\ref{lin->exp}} \addtocounter{thm}{-1} \begin{thm}Let $C_1$ and $C_2$ be $C^{1+\lambda}$ conjugate cookie cutters. Then they are exponentially equivalent. \end{thm} \setcounter{thm}{\value{ctr}} \label{lin->exp:pf} \noindent \textbf{Proof:} For any finite word $w$, Taylor's theorem implies that $ |\phi (I_w)| = \phi'(a) |I_w| + O(|I_w|^{1+\lambda})$. So \begin{eqnarray*} \frac{|\phi(I_{w0})|}{|\phi(I_w)|}& \le& \frac{\phi'(a)|I_{w0}| + C|I_{w0}|^{1+\lambda}}{\phi'(a)|I_{w}| - C|I_{w}|^{1+\lambda}}\\ & =& \frac{|I_{w0}| + D_1|I_{w0}|^{1+\lambda}}{|I_w| \left| \left. {1-D_1 |I_w|^\lambda}\right. \right| }\\ & =& \frac{|I_{w0}| + D_1|I_{w0}|^{1+\lambda}}{|I_w|}(1+ O(|I_w|^\lambda)).\\ &\le& \frac{|I_{w0}|}{|I_w|} + D_1 \frac{|I_{w0}|}{|I_w|}|I_w|^\lambda +D_2 \frac{|I_{w0}|}{|I_w|}|I_w|^\lambda + D_1D_2|I_{w0}|^{1+\lambda}|I_w|^\lambda \\ &\le &\frac{|I_{w0}|}{|I_w|} + D_1\gamma|I_{w0}|^\lambda + D_2\gamma|I_w|^\lambda+ D_1D_2|I_{w0}|^{1+\lambda}|I_w|^\lambda\\ &\le& \frac{|I_{w0}|}{|I_w|}+ D_1\gamma\gamma^{(n+1)\lambda} + D_2\gamma\gamma^{n\lambda} + D_1D_2\gamma^{n(1+\lambda)}\gamma^{n\lambda}\\ &\le& \frac{|I_{w0}|}{|I_w|} + D_3\gamma^{\lambda n}. \end{eqnarray*} \noindent Where $D_1 = \frac{C}{\min \phi'(a)}$, $D_2$ is the $O$-constant for $\frac{1}{1-D_1x}$ and $D_3$ is the maximum of $D_1$, $D_2$ and $D_1\cdot D_2$. \done \setcounter{ctr}{\value{thm}} \setcounter{thm}{\ref{lin->scl}} \addtocounter{thm}{-1} \begin{thm} The scaling function converges for a cookie cutter set, and depends only on the exponential class of the ratio geometry.\end{thm} \setcounter{thm}{\value{ctr}} \label{lin->scl:pf} \noindent \textbf{Proof:} We let $\omega = \dots\epsilon_3\epsilon_2\epsilon_1$ be a left infinite word of zeroes and ones. We write $R_k(\omega) = \left( \frac{|I_{\epsilon_k\dots\epsilon_3\epsilon_2\epsilon_10}|} {|I_{\epsilon_k\dots\epsilon_3\epsilon_2\epsilon_1}|},\frac{|G_{\epsilon_k\dots\epsilon_3\epsilon_2\epsilon_1}|} {|I_{\epsilon_k\dots\epsilon_3\epsilon_2\epsilon_1}|},\frac{|I_{\epsilon_k\dots\epsilon_3\epsilon_2\epsilon_11}|} {|I_{\epsilon_k\dots\epsilon_3\epsilon_2\epsilon_1}|} \right)$. Then since $F$ is itself $C^{1+\lambda}$, $C_0$ and $C_1$ are both conjugate to $C$ via $F$. Therefore, by Theorem~\ref{lin->exp}, the ratios are only changed by an exponentially small amount, i.e. \[|R_k(\omega)| -C\gamma^k \le |R_{k+1}(\omega)| \le |R_k(\omega)| + C\gamma^k \] \noindent which implies that \[|R_{k}(\omega)| -\frac{C}{1-\gamma}\gamma^k \le |R_{k+m}(\omega)| \le |R_k(\omega)| -\frac{C}{1-\gamma}\gamma^k \] Therefore $R_k(\omega) $ is a Cauchy sequence and hence converges. Furthermore we see that convergence is exponentially fast. \done The converse of Theorem~\ref{lin->exp} is true as well, but as an intermediate step, we show the weaker condition that $\phi$ is Lipschitz. \begin{lemma} Suppose that for all $w$ of length $k$, $e^{- a_k} \frac{|J_{w0}|}{|J_w|} \le \frac{|I_{w0}|}{|I_w|} \le e^{a_k} \frac{|J_{w0}|}{|J_w|}$. Where $ (a_k)$ is a summable sequence; say $\sum a_k = A$. Then $\phi$ is Lipschitz. \end{lemma} \noindent \textbf {Proof:} Since the set of gaps, $C_1^c$, is dense, we only need to prove that $\phi $ is Lipschitz on $C_1^c$. That is $\phi$ is Lipschitz on each gap with a uniform Lipschitz constant. Write $H_w = \phi(G_w)$. Then \begin{eqnarray*} |H_{ \epsilon_1 \epsilon_2\dots \epsilon_n}| &=& \frac{|H_{ \epsilon_1 \epsilon_2\dots \epsilon_n}|}{|J_{ \epsilon_1 \epsilon_2\dots \epsilon_n}|} \frac{|J_{ \epsilon_1 \epsilon_2\dots \epsilon_n|}}{|J_{ \epsilon_1 \epsilon_2\dots \epsilon_{n-1}}|} \frac{|J_{ \epsilon_1 \epsilon_2\dots \epsilon_{n-1}}|}{|J_{ \epsilon_1 \epsilon_2\dots \epsilon_{n-2}}|} \dots \frac{|J_{\epsilon_1}|}{|J|}|J| \\ % &\le & e^{- a_n}\frac{|G_{ \epsilon_1 \epsilon_2 \dots \epsilon_n}|}{|I_{\epsilon_1 \epsilon_2 \dots \epsilon_n}|} e^{- a_{n-1}} \frac{|I_{\epsilon_1 \epsilon_2 \dots \epsilon_n}|}{|I_{ \epsilon_1 \epsilon_2 \dots \epsilon_ {n-1}}|} e^{- a_{n-2}} \frac{|I_{\epsilon_1 \epsilon_2 \dots \epsilon_{n-1}}|}{|I_{ \epsilon_1 \epsilon_2\dots \epsilon_{n-2}}|} \dots e^{- a_0}\frac {|I_{\epsilon_1}|}{|I|} |J|\\ % &\le & e^A \left[ \frac{|G_{ \epsilon_1 \epsilon_2 \dots \epsilon_n}|}{|I_{ \epsilon_1 \epsilon_2 \dots \epsilon_n}|} \frac{|I_{ \epsilon_1 \epsilon_2 \dots \epsilon_n}|}{|I_{ \epsilon_1 \epsilon_2\dots \epsilon_{n-1}}|} \frac{|I_{ \epsilon_1 \epsilon_2\dots \epsilon_{n-1}}|}{|I_{ \epsilon_1 \epsilon_2\dots \epsilon_{n-2}}|} \dots \frac{|I_{\epsilon_1}|}{|I|}\right] |J|\\ % & = & e^A\frac{|J|}{|I|}|G_w| \end{eqnarray*} \done We'll also need the following two extension lemmas: \begin{lemma} Let $h: X \rightarrow Y$ be a H\"older continuous map from a metric space $X$ to a complete metric space $Y$. Then, since $h$ preserves Cauchy sequences, $h$ extends to the closure of $X$. \end{lemma} \begin{prop}\label{extlemma}Let $C_1$ and $C_2$ be Cantor sets with Lebesgue measure 0 embedded in $I=[0,1]$. Let $\phi: C_1 \rightarrow C_2$ be an order preserving homeomorphism which is $C^{1+\lambda}$, that is $g(x) = \lim_{x\rightarrow y, x\in C_1}\frac{\phi(x)-\phi(y)}{x-y}$ exists and is H\"older. Then $\phi $ extends to a $C^{1+\lambda}$ map $I \rightarrow I$ \end{prop} Let $(a,b)$ be a gap in $C_1$. Since $\phi$ is order preserving $(\phi(a),\phi(b))$ is a gap in $C_2$. To extend $\phi$ to $(a,b)$, we extend $g$ in such a way that $\int_a^b g(x)dx = \phi(b) - \phi(a)$. Then $\hat{\phi} (x) = \int_a^x g(t)dt$ extends $\phi$. So let $m=\frac{a+b}{2}$. Set $g(m)$ to $2D(b,a)- .5g(a)-.5g(b) = 2D(b,a)- .5D(a,a)-.5D(b,b)$. And set $g$ to be linear on the intervals $(a,m)$ and $(m,b)$. Then \[\int_a^b g(x)dx = \phi(b) - \phi(a)\] Furthermore \begin{eqnarray*} |g(m)-g(a)|& =& |2D(b,a)-.5D(a,a)-.5D(b,b)-D(a,a)|\\ & \le& 1.5|D(a,b)-D(a,a)| + .5|D(a,b)-D(b,b)| \\ &\le& 2^\lambda C|m-a|^\lambda. \end{eqnarray*} % \vspace*{4 in} Since $g$ is linear on $[a,m]$, this shows that $g$ is $C^\lambda$ on $[a,m]$. Similarly, $g$ is $C^\lambda$ on $[m,b]$. Extend $g$ to each gap of $C_1$ in this way, and then g is $C^\lambda$ on all of $I$: Let $xlin}} \addtocounter{thm}{-1} \begin{thm} Let $C_1$ and $C_2$ be labeled Cantor sets with bounded geometry. Let $\phi:C_1\longrightarrow C_2$ be the map which preserves labels. If $\phi$ only changes the ratio geometry by exponentially small amounts, then $\phi$ extends to a $C^{1+\lambda}$ map in a neighborhood of $C_1$. \end{thm} \setcounter{thm}{\value{ctr}} \label{scl->lin:pf} \noindent \textbf {Proof:} We proceed in three steps: a) $\phi'$ is defined on $C_1$. That is $lim_{x \rightarrow y, w \in C_1} \frac {\phi(x) - \phi(y)}{x - y} $ exists. b) $\phi'$ is H\"older on $C_1$. c) By Proposition~\ref{extlemma}, $\phi$ extends to a $C^{1+\lambda}$ function of the interval $I$. \noindent \textbf {Proof of a)} We first show that $\phi'$ is defined on $C_1$. Let $x$, $x_n\in C_1$ and $lim_{n\rightarrow \infty}x_n= x$. We show that $\frac {\phi(x) - \phi(x_m)}{x -x_m}$ is a Cauchy sequence, and hence converges. Given $r \in {\mathbb N}$, Choose $N$ so that for $m > N$, $|x -x_m| \le \gamma^r$. In particular, if $I_w$ is the level-$r$ clone interval containing $x$, then $I_w$ contains $x_m$ for all $m > N$. So if $m, n>N$, then \begin{eqnarray*} \left|\frac {\phi(x) - \phi(x_m)}{x - x_m} - \frac {\phi(x) - \phi(x_n)}{x- x_n}\right| & =& \frac {|\phi(x) - \phi(x_m)|}{|x - x_m|}\cdot \left|1-\frac{x - x_m} {x - x_n} \cdot\frac {\phi(x) - \phi(x_n)}{\phi(x) - \phi(x_m)}\right|\\ &\le& C|1-e^{k \xi^n}y|\\ &=& O(\xi^n) \end{eqnarray*} and hence $\frac {\phi(x) - \phi(x_m)}{x - x_m}$ converges. \noindent \textbf{Proof of b)} $\phi'$ is H\"older. Let $x,y \in C_1$. Let $w \in \{0,1\}^{\mathbb N}$ be the word such that $G_w \subset [x,y] \subset I_w$. Choose sequences $x_n \longrightarrow x$ and $y_n \longrightarrow y$. We write $D(a,b)$ for the difference quotient $\frac{\phi(a)- \phi(b)}{a-b}$. Then $|\phi'(x)-\phi'(y)| \le |\phi'(x)-D(x,x_n)| + |D(x,x_n)- D(y,y_n)| +|D(y,y_n)-\phi'(y)|$. The first and last term on the right hand side tend to zero, and by the previous paragraph, the middle term is $O(\xi^r)$. So $|\phi'(x)-\phi'(y)| \le K\xi^r = K\alpha^{\mu r} \le K|G_w|^\mu \le K|x-y|^\mu$. \done While this theorem works to extend the label preserving map to an open neighborhood of the Cantor set, it cannot be used to build conjugations of cookie cutters. If $C_1$ and $C_2$ are exponentially close, then the label preserving map, $\phi$ extends to a $C^{1+\lambda}$ map, $\hat{\phi}$ on $I$. But while $\hat{\phi}$ conjugates $F_1$ to $F_2$ on $C_1$, in general it won't conjugate them off of $C_1$. To do so we would have extend $\phi$ over a single gap and use the dynamics of $F_1$ and $F_2$ to push it around to the other gaps. But then there's no way to ensure that the derivative on one gap will match the nearby gaps, so the function wouldn't necessarily be differentiable on the Cantor set. In fact theorem~\ref{bootstrapping} will show that if $F_1$ and $F_2$ are $C^{k+\lambda}$, then any $C^{1+\lambda}$ of extension $\phi$ to the whole interval is automatically $C^{k+\lambda}$. The particular extension we described was only $C^2+1$, so if $\phi$ is $ C^3$ and we used the dynamics of $F_1$ and $F_2$ to define $\phi$ on the whole interval. We could extend $\phi$ to be $C^{k+\lambda}$ on any one gap, and use the dynamics of $F_1$ and $F_2$ to define $\hat{\phi}$ on the other gaps, then $\hat{\phi}$ wouldn't even be $C^1$ on the Cantor set. \subsection{$C^{k+\lambda}$ Equivalence} \begin{thm}\label{bootstrapping} Let $F$ and $G$ be $C^{k+\lambda}$ maps of 1 dimensional manifolds. Let $F(a) = a$, ${F}'(a) \ne 1$ and let $h$ be a $C^{1+\lambda}$ diffeomorphism conjugating $F$ to $G$. Then $h$ is $C^{k+\lambda}$ on a neighborhood, $U$ of $a$. In fact any $U$ such that $F'$ is bounded away from 1 on $U$, and $G'$ is bounded away from 1 on $h(U)$ will work. \begin{displaymath} \begin{array}{rcl} X & \stackrel{F}{\longrightarrow} & X\\ h\downarrow& &\downarrow h\\ Y & \stackrel{G}{\longrightarrow} & Y\\ \end{array} \end{displaymath} \end{thm} We will prove the result for ${F}'(a) > 1$. Then if $F'(a) <1$, we can apply the result to $F^{-1}$ and $G^{-1}$. Following de la Llave[], we write $h$ as $G \circ h \circ F^{-1}$, which we can then write as $G^n \circ h \circ F^{-n}$. We then linearize $h$ and show that $G^n(h'(0))\cdot F^{-n}$ is a good approximation of $h$. We use this approximation, along with the dynamics of $F$ and $G$ to show that $h$ is $C^{k+\lambda}$. We begin with a couple of lemmas which show that $\sum_{n=1}^\infty F^{-n}(x)$ is $C^{k+\lambda}$. \begin{lemma}Let $\eta$ be a $C^{k+\lambda}$ function, and let $f_n :I\rightarrow I$ be a sequence of positive functions which are uniformly bounded in the $C^{k+\lambda}$ norm and let {$f_n'(x) \le \gamma < 1$} for all n. Write $F_0 = f_0$. Set $F_k = f_k \circ f_{k-1}\circ \dots \circ f_1\circ f_0$. Then there exists a $D_k$ such that for all n, the $k^{th}$ derivative $F_n^{(k)}(x) \le D_k\gamma^n$. Moreover, if we write $H_n = \eta\circ F_n$ then there exists $D_k$ such that for all n, $H_n^{(k)}(x) \le D_k\gamma^n$. \end{lemma} \noindent \textbf {Proof:} First we prove the result for $F_n$. We proceed by induction on $k$. For $k=1$, $F_n'(x) = f_{n-1}'[F_{n-1}(x)] \cdot \dots \cdot f_1'[F_0(x)]\cdot f_0'(x) \le \gamma^{n+1}$ so the lemma applies with $D_1 = \gamma$. So suppose that for all $r < k$, there exists a $D_r$ such that for all n, $F_n^{(r)} \le D_r\gamma^n$. Then for $n \ge 2$, \eject \begin{eqnarray*} |F_n^{(k)}(x)|&=&(f_n \circ F_{n-1})^{(k)}(x)\\ &=&\left|\sum_{j=1}^k \sum_{a_1+\dots+a_j=k} \sigma_{a_1...a_j}f_n^{(j)}[F_{n-1}(x)] \cdot F_{n-1}^{(a_1)}(x)\cdot \dots \cdot F_{n-1}^{(a_j)}(x)\right|\\ % &\le & \left|f_n'\circ F_{n-1}(x)\cdot F_{n-1}^{(k)}(x)\right| \\ % & & + \sum_{j=2}^k \sum_{a_1+\dots+a_j=k} \sigma_{a_1...a_j}\left| f_n^{(j)}[F_{n-1}(x)] \cdot F_{n-1}^{(a_1)}(x) \cdot \dots \cdot F_{n-1}^{(a_j)}(x)\right|\\ % & \le&\gamma| F_{n-1}^{(k)}(x)| + \sum_{j=2}^k \sum_{a_1+\dots+a_j=k} \sigma_{a_1...a_j} CD_{a_1}\gamma^{n-1}\dots D_{a_j} \gamma^{(n-1)}\\ % &\le &\gamma|F_{n-1}^{(k)}(x)| + \sum_{j=2}^k \sum_{a_1+\dots+a_j=k} \sigma_{a_1...a_j} CD_{a_1}\dots D_{a_j} \gamma^{j(n-1)}\\ % &\le&\gamma|F_{n-1}^{(k)}(x)| + \sum_{j=2}^k \sum_{a_1+\dots+a_j=k} K_1 \gamma^{2n}\\ % &\le &K_2\gamma^{2n} + \gamma|F_{n-1}^{(k)}(x)|\\ % &\le & K_2\gamma^{2n} + \gamma[K \gamma^{2(n-1)}+ \gamma|F_{n-2}^{(k)}(x)|]\\ % & = & K_2[\gamma^{2n} + \gamma^{2n-1}] + \gamma^2|F_{n-2}^{(k)}(x)|\\ % & = & K_2\left[ \gamma^{2n} + \gamma^{2n-1} + \gamma^{2n-2} + \dots +\gamma^{2n-n}\right] + \gamma^{n+1}|F_0^{(k)}(x)|\\ % &\le &D_k \gamma^n \end{eqnarray*} And so the result holds true for $F_n$. Now we prove the result for $H_n = \eta\circ F_n$. Again we proceed by induction on $k$. For $k=1$ we have $H_n'(x) = \eta'(x) f_n'(x_{n-1})f_{n-1}'(x_{n-2})\dots f_1'(x) \le C_{\eta} \gamma^n$. So we suppose that $|H^{(k)}(x)| \le D_{k}\gamma^n$. Then \begin{eqnarray*} |H^{(k+1)}(x)|&= & |(\eta_n\circ F_n)^{(k+1)}(x)|\\ &= & \left(\eta'(x_n)F_n'(x)\right)^{(k)}\\ &= &\sum_{j=0}^{k}{ k\choose j}(\eta'F_n(x))^{(j)}F_n'(x)^{(k-j)}\\ &= &\sum_{j=0}^{k} { k\choose j} (\eta'F_n(x))^{(j)}F_n(x)^{(k-j+1)}\\ &\le&\sum_{j=0}^{k}M_1D_{k-j}\gamma^n \\ &\le&M_2\gamma^n \end{eqnarray*} \begin{lemma} With the same setup as above, there exists an $M$ such that for all $n$, $|H_n^{(k)}(x) -H_n^{(k)}(z)| \le M\gamma^{n \alpha} |x-z|^\lambda$. \end{lemma} \noindent \textbf {Proof:} First we note that it is sufficient to show the result for $F_n$. By the above lemma, there exists an integer $m$ so that ${H_{n-m}^n}' = \left(\eta\circ f_n \circ f_{n-1} \circ \dots \circ f_{n-m} \right)'(x) \le \gamma$. So if the lemma is true for $F_n$, then by replacing $f_{n-m}$ with $H_{n-m}$, we get $ |H_n^{(k)}(x) -H_n^{(k)}(z)| \le M\gamma^{(n-m) \lambda} |x-z|^\lambda = \frac{M}{\gamma^{m\lambda}} \gamma^{n\lambda}|x-z|^\lambda$. We show the result for $F_n$ by induction on $ k$. For $k=0$, we have ${|F_n(x)-F_n(z)|} = F_n'(w)|x-z| \le \gamma^n|x-z|$. Now suppose that for $ r < k$, there exists a $D_r$ such that for all n, $|F_n^{(r)}(x) -F_n^{(r)}(z)| \le D_{r}\gamma^n|x-z|$. Then $ \left| F_n^{(k)}(x) -F_n^{(k)}(z)\right| $ is estimated by \begin{eqnarray*} &\le& \left|f_n'\circ F_{n-1}(x)F_{n-1}^{(k)}(x) - f_n' \circ F_{n-1}(z)F_{n-1}^{(k)}(z)\right|\\ % & & + \left| f_n^{(k)}\circ F_{n-1}(x)[F_{n-1}'(x)]^k - f_n^{(k)}\circ F_{n-1}(z)[F_{n-1}'(z)]^k\right| \\ % & & + \sum_{j=2}^{k-1} \sum_{a_1+\dots+a_j=k}\sigma_{a_1...a_j} \left| f_n^{(j)}F_{n-1}(x)F_{n-1}^{(a_1)}(x) \dots F_{n-1}^{(a_j)}(x) -f_n^{(j)}F_{n-1}(z)F_{n-1}^{(a_1)}(z) \dots F_{n-1}^{(a_j)}(z)\right| \\ % &\le &|f_n'(F_{n-1}(x))| \left|F_{n-1}^{(k)}(x)-F_{n-1}^{(k)}(z)\right| % + |F_{n-1}^{(k)}(z)| \left|f_n'(F_{n-1}(x)-f_n'(F_{n-1}(z)\right| \\ % & &+ |f_n^{(k)}F_{n-1} (x)| \left| [F_{n-1}'(x)]^k-[F_{n-1}'(z)]^k\right| + |[F_{n-1}'(z)]^k|\left|f_n^{(k)}F_{n-1} (x)-f_n^{(k)}F_{n-1} (z)\right|\\ % & &+ \sum_{j=2}^{k-1} \sum_{a_1+\dots+a_j=k}\sigma_{a_1...a_j}\left|f_n^{(j)}(F_{n-1}(x) -f_n^{(j)}(F_{n-1}(z)\right|\cdot \\ % & &\left[|F_{n-1}^{(r_1)}(*) \dots F_{n-1}^{(r_j)}(*)|\right. \\ % & & +|F_{n-1}^{(r_1)}(x) - F_{n-1}^{(r_1)}(z)| |f_n^{(j)}F_{n-1}(*)F_{n-1}^{(r_2)}(*) \dots F_{n-1}^{(r_j)}(*)|\\ % & & + \dots +\\ % & &\left. +|F_{n-1}^{(r_j)}(x) - F_{n-1}^{(r_j)}(z)| |f_n^{(j)}F_{n-1}(*)F_{n-1}^{(r_1)}(*) \dots F_{n-1}^{(r_{j-1})}(*)|\right] \\ % &\le &\gamma |F_{n-1}^{(k)}(x) - F_{n-1}^{(k)}(z)|\\ % & & +D_k \gamma^{n-1}C\left|F_{n-1}(x)-F_{n-1}(z)\right| + C\left|F_{n-1}'(x) - F_{n-1}'(z) \right|[F_{n-1}'(x)]^{k-1}\\ % & & + [F_{n-1}'(x)]^{k-2}[F_{n-1}'(z)]\\ % & &+ \dots +\\ % & &+\left| [F_{n-1}(z)]^{k-1}\right| +\gamma^{k(n-1)}C\left|F_{n-1}(x)-F_{n-1}(z)\right|^\lambda\\ % & &+\sum_{j=2}^{k-1} \sum_{a_1+\dots+a_j=k} C\left|F_{n-1}(x) - F_{n-1}(z) \right| D_{r_1} \dots D_{r_j}\gamma^{2(n-1)}\\\ % & &+F_{n-1}^{(r_1+1)}(w)|x-z|C D_{r_2} \dots D_{r_j} \gamma^{2(n-1)}\\ % & & + \dots +\\ % & &+ F_{n-1}^{(r_j+1)}(w)|x-z|C D_{r_1} \dots D_{r_{j-1}} \gamma^{2(n-1)}\\ % &\le& \gamma |F_{n-1}^{(k)}(x) - F_{n-1}^{(k)}(z)|\\ % & & + M\gamma^{2n} + M\gamma^{n-1}|x-z| + \gamma^{(n-1)k} C \gamma^{\lambda(n-1)}|x-z|^\lambda + M\gamma^n|x-z|\\ % &\le& \gamma |F_{n-1}^{(k)}(x) - F_{n-1}^{(k)}(z)| + M\gamma^{2n\lambda}|x-z|^\lambda\\ % &\le& M \gamma^{(n-1)\lambda}|x-z|^\lambda \end{eqnarray*} \done The second inequality is obtained by multiple applications of the inequality $|ab-cd|\le |a||b-d| + |d||a-c|$. Each $*$ stands for either $x$ or $z$. Note that we have no control over the size of $M$. This is of no matter, since we're dealing with a fixed $k$. Now we recall a theorem from calculus: \begin{thm} Suppose that $\{ f_n \}$ is a sequence of $C^1$ functions on $[a,b]$ and that $\{ f_n \}$ converges to $f$. Suppose, moreover, that $\{ f'_n \}$ converges uniformly to some function $g$. Then $f$is $C^1$ and $f'(x) = g$. \end{thm} \begin{cor} Let $\eta$ be a $C^{k+\lambda}$ function, and let $f_n :\rightarrow I$ be a sequence of functions which is uniformly bounded in the $C^{k+\lambda}$ norm. Let $H^n(x) = \eta \circ f^n(x)$. Then $H^n$ converges to a $C^{k+\lambda}$ function $H$. \end{cor} \begin{thm}[Livsic $C^{k+\lambda}$ Theorem] Let I be a subinterval of ${\Real}$. Let $\eta :I\longrightarrow I$ and ${F:I\longrightarrow I}$ be $C^{k+ \lambda}$ maps. Let $F$ have a fixed point $a$ and let $F'(y) > 1$. Then if $\psi$ satisfies the equation \[ \psi(F(x))- \psi(x) = \eta(x) \] then $\psi$ is itself $C^{k+\lambda}$ on a neighborhood of $a$. \end{thm} \noindent \textbf {Proof:} Since $$\psi(x) = \psi(F^{-1}(x))+\eta(F^{-1}(x))$$ we get $$\psi(x) = \psi(F^{-n}(x) )+\sum_{k=1}^{n-1}\eta(F^{-i}(x)) = \psi(a) + \sum_{k=1}^{\infty} \eta(F^{-i}(x))$$ We then differentiate this sum term by term to get a candidate for the $k^{th}$ derivative. By the previous two lemmas, $ \sum_{i=1}^{\infty} (\eta \circ F^{-i}(x))$ exists and is $C^{k+\lambda}$. Since this sum is $C^\lambda$, it must be uniformly convergent, and hence is the $k^{th}$ derivative. \done Now we return to theorem~\ref{bootstrapping}. Differentiating the formula \begin{eqnarray*} h(f(x))&=&g(h(x)) \\ h'(f(x))f'(x)&=&g'(h(x))h'(x)\\ \Rightarrow \log (h' \circ f(x))- \log (h'(x))& = &\log g'(h(x))- \log f'(x)\\ \end{eqnarray*} We apply the Livsic $C^{k+ \lambda}$ theorem with $\log h'$ as $\psi$ and $ \log g' h- \log f'$ as $\eta$. Since $ g'$ and $f'$ are $C^{k-1+\lambda}$, and $h$ is $C^{1+\lambda}$, $\eta$ is $C^{1+\lambda}$. So by the Livsic theorem, $\psi = \log h'$ is $C^{1+\lambda}$. Hence $h$ is $C^{2+\lambda}$. But since $h$ is $C^{2+\lambda}$, we repeat the above argument to show that $h$ is $C^{3+\lambda}$. We repeat this process until we have shown that $\psi$ is $C^{k-1+\lambda}$, and hence $h$ is $C^{k+\lambda}$. We note that this whole proof is valid in any neighborhood $U$ of the fixed point $a$, as long as $f' > 1$ on $U$ and $g' >1$ on $h(U)$. As a corollary, we obtain \setcounter{ctr}{\value{thm}} \setcounter{thm}{\ref{cookieboot}} \addtocounter{thm}{-1} \begin{thm} Let $F: I_0 \bigcup I_1 \longrightarrow I$ and $G: J_0 \bigcup J_1 \longrightarrow J$ be two $C^{k+\lambda}$ cookie cutters. Let $\phi: I \longrightarrow J$, conjugating $F$ to $G$ be $C^{1+\lambda}$. For convenience' sake, we take $I = [0,1]$. Since $F'>1$ and $G'>1$, the theorem applies with 0 as the fixed point, and $I_0$ as $U$. So $\phi$ is $C^{k+\lambda}$ on $I_0$. Similarly, using 1 as the fixed point, $\phi$ is $C^{k+\lambda}$ on $I_1$. \end{thm} \setcounter{thm}{\value{ctr}} \label{cookieboot:pf} While we're concerned primarily with the minimal set of a cookie cutter, if $f$ and $g$ are $C^{k+\lambda} $ cookie cutters on the circle and $\phi$ is a $C^{1+\lambda}$ diffeomorphism, the bootstrapping method allows us to say that $\phi $ is in fact $C^{k+\lambda}$ in some neighborhood of any hyperbolic fixed point. But since $\phi= g\circ\phi f^{-1}$, if $\phi$ is $C^{k+\lambda}$ on the neighborhood $U$, then $\phi$ is $C^{k+\lambda} $ on the neighborhood $f[U]$ as well. So by iterating $f$ on $U$, we can extend the differentiability of $phi$, not just on the minimal set, but actually on any interval which contains a hyperbolic fixed point, but no non-hyperbolic fixed points. For instance, if $f$ has only one fixed point, $x_0$, in the interior of the fixed interval $H$, then $\phi$ will be $C^{k+\lambda}$ everywhere except possibly at $x_0$. And if $x_0$ is hyperbolic as well, then $\phi$ is $C^{k+\alpha}$ everywhere. \begin{cor} Let $M$ and $N$ be two $C^{k+\lambda}$ manifolds. Let $M$ have a resilient leaf, $L$. Let $\phi:M\longrightarrow N$, transversally $C^{1+ \lambda}$, induce conjugations of the holonomy maps. (This forces $\phi(L)$ to be a resilient leaf of $N$.) Then $\phi$ is transversally $C^{k+\lambda}$. \end{cor} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The Conjugation Problem for Markov Exceptional Sets} Hyperbolic Markov exceptional minimal sets are generalizations which retain all of the key properties of cookie cutter Cantor sets. The Markov condition ensures that we have a geometry of nested gaps and co-gaps, as we did with a cookie cutter. Hyperbolicity ensures that the estimates we made for cookie cutters on the ratio geometry and the derivatives also work for exceptional Markov sets. So the general picture we have for cookie cutters still applies to Markov exceptional sets. We can still define a ratio geometry which converges to a scaling function. The scaling function will classify the $C^{1+\lambda}$ structure. And we may boot strap to show that two $C^{k+\lambda}$ Markov exceptional sets with the same $C^{1+\lambda}$ structure necessarily have the same $C^{k1+\lambda}$ structure. Because the nested interval structure of a Markov exceptional minimal set is not uniform, the definition of the ratio geometry function is just a little trickier than for the special case of Cookie Cutter sets. But the Markov condition ensures that the structure of gaps and clones gets preserved by the pseudogroup $\Gamma$. Since there may be more than one level-one subgap of the interval, we don't label the gaps according to the interval of which they are a sub-gap, but instead we (arbitrarily) choose to label the gaps with the same label as the interval to their right. So if there is a gap $H\subset I_0$ which is directly to the left of $I_{01}$, then we'll label $H=G_{01}$. \label{Mark-disc}In our discussion of the $C^{1+\lambda}$ and $C^{k+\lambda}$ structure of a cookie cutter set, we used the nested structure of gaps and clones, and the hyperbolicity of the defining function. We have all of these tools in the slightly more general case of a Markov exceptional set, so we can repeat those theorems with minor modifications. There are only two issues in adapting the proofs which we gave for cookie cutters with two generators to the corresponding theorems on hyperbolic Markov exceptional minimal sets. One is purely notational. Instead of having the gap $G_w$ as a subgap of $I_w$, we write $G_{wi_k}$ as a subgap of $I_w$. We need to replace $f$ and $f^n$ with the appropriate functions $\gamma_i$ and $\gamma_{i_1}\circ\dots\circ\gamma_{i_n}$, depending on the domain. The other issue is that for a given word $w$, $I_w$ (or $G_w$) might not be defined. If it is not defined, we don't have to consider it, if it is defined, all of the statements we made about cookie cutters will still be true when adapted to the new notation. In particular, the following theorems, which are generalizations of the theorems we proved for cookie cutters are still true, and with the changes noted above, their proofs are still valid. Propositions \ref{bdd-holder} and \ref{cookie-bdd} hold as well for subshifts as it does for full shifts. \begin{prop} Let $C$ and $C'$ be two Cantor sets, both labeled by the same subshift $X$ and let $C$ and $C'$ both have bounded geometry. Then the label preserving map $\phi: C \longrightarrow C'$ is H\"older continuous. \end{prop} \noindent \textbf{Proof:} The proof is identical to that of Proposition~\ref{bdd-holder}. \begin{prop} A hyperbolic Markov exceptional minimal set has bounded geometry. \end{prop} \noindent \textbf{Proof:} As in Proposition~\ref{cookie-bdd} \, the gap $G_w^j = \gamma_{w_{n-1}}\circ \gamma_{w_{n-2}}\circ \dots \circ \gamma_0[G^j]$ and $I_w = \gamma_{w_{n-1}}\circ \gamma_{w_{n-2}}\circ \dots \circ \gamma_0[I_{w_0}]$ and since each $\gamma_i$ is hyperbolic, we can apply the mean value theorem to get the result. \setcounter{ctr}{\value{thm}} \setcounter{thm}{\ref{M-lin->exp}} \addtocounter{thm}{-1} \begin{thm} Let $C$ and $C'$ be $C^{1+\lambda}$ conjugate Markov exceptional sets. Then the ratio geometry of $C$ is exponentially equivalent to the ratio geometry of $C'$. \end{thm} The proof is identical to that of theorem~\ref{lin->exp} \begin{thm} Let $C$ and $C'$ be $C^{1+\lambda}$-conjugate Markov exceptional sets. Then they have the same scaling function. \end{thm} \begin{thm} Let $\phi$ be a $C^{1+\lambda}$ conjugation between $C^{k+\lambda}$ Markov exceptional sets. Then $\phi$ is itself $C^{k+\lambda}$ \end{thm} \setcounter{thm}{\value{ctr}} \begin{thm} Let $C_1$ and $C_2$ be two Markov exceptional sets defined by the pseudogroups $\Gamma_1 = <\gamma_1^1, \dots , \gamma_k^1>$ and $\Gamma_2 = <\gamma_1^2 \dots , \gamma_k^2>$, and let $\Gamma_1$ and $\Gamma_2$ induce the same subshift on $\{1,\dots k \}^{\mathbb N}$. Further let the label preserving map $\phi:C_1\longrightarrow C_2$ change the ratio geometry by an exponentially small amount. Then $\phi$ extends to a $C^{1+\lambda}$ map on a neighborhood of $C_1$. \end{thm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% In the case where C is a Markov exceptional set with basis $(I_0, \dots I_k ; \gamma_0, \dots \gamma_k)$ and let $\Gamma$ be the pseudogroup generated by $ (\gamma_0, \dots \gamma_k)$. Then, just like the case of a cookie cutter, since the $I_j$s are disjoint and $\gamma_j:I_j \longrightarrow \bigcup_{i=0}^k$, we can combine the $\gamma_j$s into one function $F:\bigcup_{i=0}^k I_1\longrightarrow T$. Then since $\Gamma$ is generated by $F$, $\Gamma$ has one forward end and an uncountable number of backwards ends. \section{The Conjugation Problem For Hirsch Foliations} We will let $(M_1,{\mathcal F_1})$ and $(M_2,{\mathcal F_2})$ be $C^{r+\lambda}$ codimension 1 manifolds with hyperbolic minimal sets $X_1$ and $X_2$. \begin{thm} Let $f:M_1\To M_2$ be a leaf preserving $C^{m}$ map for $m\le r+\lambda$. Then there exists total transversals ${\mathcal T_1 }\subset M_1$ and ${\mathcal T_2 }\subset M_2$ such that $f$ induces a $C^{1+\lambda}$ map $\hat{f}: {\mathcal T_1 } \To {\mathcal T_2 }$. \end{thm} \noindent \textbf{Proof:} Let $(V_1, \dots V_n)$ be a cover of $M_2$ by foliation charts $\psi_j:V_j\To \Real^{n-1}\times\Real^1$. Then $(f^{-1}(V_1), \dots f^{-1} (V_k))$ is an open cover of $M_1$. Let $(U_1,\dots, U_d)$ be a cover of $M$ by foliation charts $\phi_i:U_i\To \Real^{n-1}\times\Real^1$ so that for all $i$, there exists a $j$ with $f[U_i]\subset V_j$. Then in local coordinates, we have $$ \psi_j f \phi_i^{-1}(x,y)= (F_1(x,y),F_2(y))$$ where $F_1:\Real^{n-1}\times\Real^1\To \Real^{n-1}$ and $F_2:\Real^1 \To \Real^1 $ are $C^m$ functions. So we define $\hat{f}(y)$ to be $F_2(y)$ in local coordinates. Then $\hat{f}$ is $C^m$ map on ${\mathcal T^1}$. \done We now return to the theorems stated in the introduction. We let $(M_1,{\mathcal F}_1)$ be a $C^{r+\lambda}$ foliation with a Markov exceptional set, and let ${\mathcal B_1}= \set {\gamma_1,\dots,\gamma_k}$ be a Markov basis for the holonomy pseudogroup $\Gamma_1$. If $\phi: (M_1,{\mathcal F}_1)\To (M_2,{\mathcal F}_2)$ is a $C^{r+\lambda}$ diffeomorphism, then $\hat{\phi}$ conjugates the $\Gamma_1$ to $\Gamma_2$, the holonomy pseudogroup of $ {\mathcal F}_2$. In particular $(M_2,{\mathcal F}_2)$ has a Markov exceptional set as well, and ${\mathcal B_2}= \set {\phi\gamma_1\phi^{-1},\dots,\phi\gamma_k\phi^{-1} }$ is a Markov basis for it's holonomy. $\Gamma_2$ needs not be hyperbolic, but it is eventually hyperbolic, and the fixed points of the minimal set of $\Gamma_2$ are hyperbolic fixed points. We can then use ${\mathcal B_1}$ and ${\mathcal B_2}$ to define the respective ratio geometries. Since $\Gamma_2$ is eventually hyperbolic, that's sufficient to show that the ratio geometry of $C_1=X_1\cup{\mathcal T^1}$ and $C_2=X_2\cup{\mathcal T^2}$ are exponentially equivalent. \setcounter{ctr}{\value{thm}} \setcounter{thm}{\ref{foliatexp}} \addtocounter{thm}{-1} \label{foliatdisc} \begin{thm} Let $F:(M_1,{\mathcal F}_1)\to F:(M_2,{\mathcal F}_2)$ be a $C^{1+\lambda}$ diffeomorphism. Then the transverse ratio geometry of ${\mathcal F}_1$ is exponentially equivalent to the transverse ratio geometry of ${\mathcal F}_2$. \end{thm} \begin{thm} As we flow along an infinitely long path with contracting holonomy, the transverse ratio geometry will converge to a scaling function. \end{thm} \begin{obs} For a Hirsch foliation, this gives us a nice geometric interpretation for the dual Cantor set on which the scaling function is defined. The paths we flow along, in general, will go to an end of the leaf. (Though it might also go around a handle.) In particular, if we choose a leaf $L$ with no handles, then the domain of the scaling function is a subset of the endset of $L$. \end{obs} \begin{thm} Let ${\mathcal F}_1$ and ${\mathcal F}_2$ be $C^{1+\lambda}$ conjugate foliations with Markov exceptional sets. Then they have the same scaling function. \end{thm} And because the fixed points of $\Gamma_2$ in $C_2$ are hyperbolic, the boot strapping process works, and so the scaling function classifies the transverse $C^{r+\lambda}$ structure of the foliation. \begin{thm} Let $\phi$ be a $C^{1+\lambda}$ diffeomorphism between $C^{r+\lambda}$ foliations $({M_1,\mathcal F}_1)$ and $ (M_2,{\mathcal F}_2)$ Then by the corresponding theorem on Markov exceptional sets, $\phi$ is itself transversally $C^{r+\lambda}$ in a neighborhood of the exceptional minimal set. We can then apply a smoothing lemma along the leaves to get that $\phi$ is $C^{k+\lambda}$ in a neighborhood of the minimal set. \end{thm} \setcounter{thm}{\value{ctr}} It would be an interesting result if we could say that $\phi$ is $C^{k+\lambda}$ in the leaf-wise direction, but it's not clear that there's any reason this should be true. However we can approximate $\phi$ as closely as we like with a smooth map, ${\overline{\phi_\epsilon}}$ \vfill \eject \nocite{*} \bibformb \bibliography{thesis} \newpage \vita \parindent=0pt \makebox[3.5in][l]{\textbf {BORN: } August 10, 1968, Chicago, Illinois.} {\textbf {CITIZENSHIP: } USA.} \textbf {EDUCATION: } { \leftskip=.6in \parindent=-.3in \parskip=.05in \textbf {B.S. in Mathematics}, University of Illinois at Chicago, June, 1991. \textbf {M.S. in Mathematics}, University of Illinois at Chicago, December, 2001. \textbf {Ph.D. in Mathematics}, University of Illinois at Chicago, August, 2005. } \textbf {EMPLOYMENT:} {\leftskip=.6in \parindent=-.3in \parskip=.05in \textbf {Adjunct Professor}, Daley College, Chicago, Illinois (2001-2005) \textbf {Teaching Assistant}, UIC Mathematics Department (1991-1998; 2004-2005) \textbf {Tutor}, East Village Youth Program (1991-1997) \vspace{.04 in} \mbox{} } \vspace{-.2in} \textbf {TALKS:} { \leftskip=.6in \parindent=-.3in \parskip=.05in \textit {Foliations 2000, Banach Center, Warsaw, Poland}, June 2000. \textit {Graduate Student Seminar, UIC}, November, 1998. \textit {Graduate Student Seminar, UIC}, October, 1997. \end{document}