Fixed-point index, the incompatibility theorem, and torus parametrization
Andrey M.\ Mishchenko
The fixed-point index of a homeomorphism of Jordan curves measures the number of fixed-points, with multiplicity, of the extension of the homeomorphism to the full Jordan domains in question. The now-classical Circle Index Lemma says that the fixed-point index of a positive-orientation-preserving homeomorphism of round circles is always non-negative. We begin by proving a generalization of this lemma, to accommodate Jordan curves bounding domains which do not disconnect each other. We then apply this generalization to give a new proof of Schramm's Incompatibility Theorem, which was used by Schramm to give the first proof of the rigidity of circle packings filling the complex and hyperbolic planes. As an example application, we include outlines of proofs of these circle packing theorems. We then introduce a new tool, the so-called torus parametrization, for working with fixed-point index, which allows some problems concerning this quantity to be approached combinatorially. We apply torus parametrization to give the first purely topological proof of the following lemma: given two positively oriented Jordan curves, one may essentially prescribe the images of three points of one of the curves in the other, and obtain an orientation-preserving homeomorphism between the curves, having non-negative fixed-point index, which respects this prescription. This lemma is essential to our proof of the Incompatibility Theorem.
fpindex.tex
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\begin{document}

\title{Fixed-point index, the incompatibility theorem, and torus parametrization}
\author{Andrey M.\ Mishchenko}
\thanks{The author was partially supported by NSF grants DMS-0456940, DMS-0555750, DMS-0801029, DMS-1101373.  This article is partially adapted from the Ph.D.\ thesis \cite{mishchenko-thesis} of the author.  MSC2010 subject classification: primary: 54H25, secondary: 52C26.}
\email{misHchea@umich.edu}
\date{\today}

\begin{abstract}
The fixed-point index of a homeomorphism of Jordan curves measures the number of fixed-points, with multiplicity, of the extension of the homeomorphism to the full Jordan domains in question.  The now-classical Circle Index Lemma says that the fixed-point index of a positive-orientation-preserving homeomorphism of round circles is always non-negative. We begin by proving a generalization of this lemma, to accommodate Jordan curves bounding domains which do not disconnect each other. We then apply this generalization to give a new proof of Schramm's Incompatibility Theorem, which was used by Schramm to give the first proof of the rigidity of circle packings filling the complex and hyperbolic planes. As an example application, we include outlines of proofs of these circle packing theorems.

We then introduce a new tool, the so-called torus parametrization, for working with fixed-point index, which allows some problems concerning this quantity to be approached combinatorially. We apply torus parametrization to give the first purely topological proof of the following lemma: given two positively oriented Jordan curves, one may essentially prescribe the images of three points of one of the curves in the other, and obtain an orientation-preserving homeomorphism between the curves, having non-negative fixed-point index, which respects this prescription. This lemma is essential to our proof of the Incompatibility Theorem.
\end{abstract}

\maketitle

\tableofcontents



\section{Introduction\threestars}
\label{chap:intro fpi}

This article is concerned with a topological quantity, the so-called \emph{fixed-point index} of a homeomorphism of Jordan curves, which has proven useful in the study of various areas of complex analysis.  We begin with its definition:

\begin{definition}
A \emph{Jordan curve} is a homeomorphic image of a topological circle $\bbS^1$ in the complex plane $\bbC$.  A \emph{Jordan domain} is a bounded open set in $\bbC$ with Jordan curve boundary.  We use the term \emph{closed Jordan domain} or \emph{compact Jordan domain} to refer to the closure of a Jordan domain.  We define the \emph{positive orientation} on a Jordan curve as usual.  That is, if $K$ is a closed Jordan domain, then as we traverse $\partial K$ in what we call the \emph{positive} direction, the interior of $K$ stays to the left.

%\begin{jct}
%\label{jct}
%A Jordan curve $\gamma$ separates the plane into two connected components, one bounded and one unbounded.  The bounded one is a topological open disk and the unbounded one is a punctured topological open disk.  The boundary of either is exactly $\gamma$.
%\end{jct}

%\noindent See \cite{MR1160354}*{Theorem 10.2}.

Let $K$ and $\tilde K$ be closed Jordan domains.  Let $\phi:\partial K \to \partial \tilde K$ be a homeomorphism of Jordan curves which is fixed-point-free and orientation-preserving.  We call such a homeomorphism \emph{indexable}.  Then $\{\phi(z) - z\}_{z\in \partial K}$ is a closed curve in the plane which misses the origin.  It has a natural orientation induced by traversing $\partial K$ positively.  Then we define the \emph{fixed-point index} of $\phi$, denoted $\eta(\phi)$, to be the winding number of $\{\phi(z) - z\}_{z\in \partial K}$ around the origin.
\end{definition}

Two examples are shown in Figures \ref{fig:ex disjoint then winding 0} and \ref{fig:ex neg winding forced}.  We remark that the fixed-point index depends crucially on the choice of homeomorphism, and also on the way that the sets $K$ and $\tilde K$ sit on top of each other.  It is a worthwhile exercise to construct an indexable homeomorphism $\partial K\to \partial \tilde K$, for $K$ and $\tilde K$ as in Figure \ref{fig:ex neg winding forced}, having fixed-point index unequal to $-1$.\medskip





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\caption[Two closed Jordan domains $K$ and $\tilde K$ so that any indexable homeomorphism $\phi:\partial K \to \partial \tilde K$ satisfies $\eta(f) = 0$]{
{\bf Two closed Jordan domains $K$ and $\tilde K$ so that any indexable homeomorphism $\phi:\partial K \to \partial \tilde K$ satisfies $\eta(\phi) = 0$.}  The arrows on $\partial K$ and $\partial \tilde K$ indicate the positive orientations on these Jordan curves.  In this case $\phi$ is indexable so long as it is orientation-preserving; the fixed-point-free condition is automatic because $\partial K$ and $\partial \tilde K$ do not meet.  The dashed arrow represents a vector of the form $\phi(z) - z$.  The vector $\phi(z) - z$ must always point ``to the right,'' so the curve $\{\phi(z) - z\}_{z\in \partial K}$ has winding number $0$ around the origin, thus $\eta(\phi) = 0$.
}
\label{fig:ex disjoint then winding 0}
\end{figure}



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\label{fig:ex neg winding forced}
{\bf An indexable homeomorphism $\phi:\partial K \to \partial \tilde K$ so that $\eta(\phi) = -1$.}  Suppose we insist that $\phi$ identifies the respective corners as shown.  Then tracing the path of the dashed vectors $\phi(z) - z$ as $z$ traverses $\partial K$ positively, we see that $\phi(z) - z$ winds once clockwise around the origin, thus $\eta(\phi)=-1$.
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\end{figure}







Fixed-point index has found applications for example in the theories of circle packing \cite{MR1207210}, Koebe uniformization \cite{MR1207210}, and Sierpinski carpets \cite{MR2900233}*{Section 12}.  In all of these settings, it has been applied to prove powerful existence, rigidity, and uniformization statements.  Most recently, the current author has used fixed-point index, including torus parametrization, to prove rigidity statements for collections of possibly-overlapping round disks, see \citelist{\cite{mishchenko-thesis} \cite{mishchenko-rigidity-2012}}.\medskip

The fixed-point index measures the following topological quantity: suppose that $K$ and $\tilde K$ are closed Jordan domains, and $\Phi : K \to \tilde K$ is a homeomorphism, having finitely many fixed points, which restricts to an indexable homeomorphism $\partial\Phi : \partial K \to \partial \tilde K$.  There is a well-understood notion of the \emph{multiplicity} of a fixed point of $\Phi$.  Then the fixed-point index $\eta(\partial\Phi)$ counts the number of fixed-points of $\Phi$, with multiplicity.  For more discussion along these lines, see \cite{MR1207210}*{Section 2}.\medskip

In this article, we describe a new technique for working with fixed-point index, which we call the \emph{torus parametrization} of a pair of Jordan curves, defined in Section \ref{chap:torus}.  We apply torus parametrization to give a new, elementary proof of the following fundamental lemma:

\begin{threep}
\label{threep}
\label{3p}
\label{lem:3p}
\label{lem:three points}
\label{tppl}
Let $K$ and $\tilde K$ be compact Jordan domains in transverse position, with boundaries oriented positively.  Let $z_1,z_2,z_3\in \partial K\setminus \partial \tilde K$ appear in counterclockwise order, similarly $\tilde z_1,\tilde z_2, \tilde z_3\in \partial \tilde K\setminus \partial K$.  Then there is an indexable homeomorphism $\phi:\partial K \to \partial \tilde K$ sending $z_i \mapsto \tilde z_i$ for $i=1,2,3$, so that $\eta(\phi) \ge 0$.
\end{threep}

\noindent Two Jordan domains are in \emph{transverse position} if their boundary Jordan curves cross wherever they meet, c.f.\ Definition \ref{transverse position} in Section \ref{sec background}.  The example given in Figure \ref{fig:ex neg winding forced} shows that if we prescribe the images of four points, then a negative fixed-point index may be forced.

A version of the Three Point Prescription Lemma \ref{threep} is stated in \cite{MR2131318}*{Lemma 8.14}, but we have not been able to fill in the details of the argument.  The idea of the approach is as follows: first, any Riemann mapping $\Phi:\Omega \to \tilde\Omega$ between open Jordan domains extends to a homeomorphism $\partial \Phi :  \partial \Omega \to \partial \tilde \Omega$ of their boundaries, and we may prescribe the images of three points of $\partial \Omega$ in $\partial \tilde \Omega$ by post-composing with self-biholomorphisms of $\tilde\Omega$.  Next, it is known that any isolated fixed point of a holomorphic map has non-negative multiplicity, see \cite{MR1207210}*{Section 2}.  Thus the map $\partial \Phi$, if it is indexable, has non-negative fixed-point index, because the fixed-point index of $\partial \Phi$ counts the fixed points of $\Phi$ with multiplicity, completing the argument.  However, it is not clear how to deal with possible fixed points in the induced boundary map $\partial \Phi$.  Our proof of Lemma \ref{3p} uses only induction and plane topology arguments and is given in Section \ref{sec:3p}.\medskip

We also state and prove a new fundamental lemma on fixed-point index, generalizing the well-known Circle Index Lemma \ref{cil} which states that the fixed-point index of an indexable homeomorphism between circles is always non-negative.  The Circle Index Lemma was a crucial ingredient in all of the applications of fixed-point index described above.  In our generalization, round disks are replaced by closed Jordan domains which do not disconnect each other.  In particular, the closed Jordan domains $K$ and $\tilde K$ are said to \emph{cut each other} if $K\setminus \tilde K$ or $\tilde K \setminus K$ is disconnected.  Then:

\begin{lemma}
\label{no cut index}
Let $K$ and $\tilde K$ be closed Jordan domains in transverse position, which do not cut each other, having boundaries oriented positively.  Let $\phi : \partial K \to \partial \tilde K$ be an indexable homeomorphism.  Then $\eta(\phi) \ge 0$.
\end{lemma}

\noindent The proof appears at the end of Section \ref{incompat section}.\medskip

As an example of the power of fixed-point index, we apply the Three Point Prescription Lemma \ref{threep} and Lemma \ref{no cut index} to prove a version of the Incompatibility Theorem of Schramm \cite{MR1076089}*{Theorem 3.1}, as described in Section \ref{incompat section}.  The Incompatibility Theorem is then easily applied in Section \ref{cp proofs} to prove some well-known rigidity theorems for circle packings.  The ideas for these proofs are borrowed from \citelist{\cite{MR1207210} \cite{MR2131318}*{Chapter 8}}.\medskip

\noindent {\bf Acknowledgments.}  Thanks to Jordan Watkins for many fruitful discussions, especially for pointing us strongly in the direction of what we call \emph{torus parametrization}.  Thanks also to Mario Bonk for many fruitful discussions, especially for helpful suggestions that greatly simplified the proof of Lemma \ref{no cut index}.


\section{Background lemmas and definitions\threestars}
\label{sec background}

In the upcoming discussion it will be useful to have access to two well-known lemmas on fixed-point index.  This section is devoted to introducing these two lemmas.\medskip

Our first background lemma says essentially that ``the fixed-point index between two circles is always non-negative'':

\begin{cil}
\label{cil}
\label{lem:circle index lemma}
Let $K$ and $\tilde K$ be closed Jordan domains in $\bbC$, with boundaries oriented positively, and let $\phi:\partial K \to \partial \tilde K$ be an indexable homeomorphism.  Then the following hold.
\begin{enumerate}
\item We have $\eta(\phi)=\eta(\phi^{-1})$.
\item If $K\subseteq \tilde K$ or $\tilde K\subseteq K$, then $\eta(\phi)=1$.
\item If $K$ and $\tilde K$ have disjoint interiors, then $\eta(\phi) = 0$.
\item If $\partial K$ and $\partial \tilde K$ intersect in exactly two points, then $\eta(\phi)\ge 0$.
\end{enumerate}
As a consequence of the above, if $K$ and $\tilde K$ are closed disks in the plane, then $\eta(\phi) \ge 0$.
\end{cil}

\noindent Lemma \ref{cil} can be found in \cite{MR1207210}*{Lemma 2.2}, with a clear and complete proof.  There it is indicated that a version of the lemma appeared earlier in \cite{MR0051934}.\medskip



The moral of our second background lemma is that fixed-point indices ``add nicely'':


\begin{additivity}
\label{lem:indices add}
\label{additivity}
\label{ial}
Suppose that $K$ and $L$ are interiorwise disjoint closed Jordan domains which meet along a single positive-length Jordan arc $\partial K \cap \partial L$, similarly for $\tilde K$ and $\tilde L$.  Then $K\cup L$ and $\tilde K \cup \tilde L$ are closed Jordan domains.

Let $\phi:\partial K\to \partial \tilde K$ and $\psi:\partial L\to \partial\tilde L$ be indexable homeomorphisms.  Suppose that $\phi$ and $\psi$ agree on $\partial K \cap \partial L$.  Let $\theta:\partial (K\cup L) \to \partial (\tilde K \cup \tilde L)$ be induced via restriction to $\phi$ or $\psi$ as necessary.  Then $\theta$ is an indexable homeomorphism and $\eta(\theta) = \eta(\phi) + \eta(\psi)$.
\end{additivity}






\begin{figure}
\centering

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\caption[An illustration of fixed-point index additivity]
{
\label{fig:lem indices add 1}
{\bf An illustration of fixed-point index additivity.}
}

\end{figure}






\begin{proof}
\label{ex:indices add}
The situation is as depicted in Figure \ref{fig:lem indices add 1}.  We may consider $\eta(\phi)$ to be $1/2\pi$ times the change in argument of the vector $\phi(z) - z$, as $z$ traverses $\partial K$ once in the positive direction.  Then as $z$ varies positively in $\partial K$ and in $\partial L$ the contributions to the sum $\eta(\phi) + \eta(\psi)$ along $\partial K \cap \partial L$ cancel.
\end{proof}\medskip

Next, we wish to make precise our notion of \emph{transverse position}:

\begin{definition}
\label{transverse position}
Two Jordan curves $\gamma$ and $\tilde\gamma$ are in \emph{transverse position} if for any point $z\in \gamma \cap \tilde\gamma$ where they meet, they \emph{cross transversely}, which here means that there is an open neighborhood $U\subset \bbC$ of $z$ and a homeomorphism $\phi : U \to \bbD$ between $U$ and the open unit disk $\bbD$, so that $\phi(\gamma \cap U) = \bbR \cap \bbD$ and $\phi(\tilde\gamma \cap U) = i \bbR \cap \bbD$.  Two (open or closed) Jordan domains are said to be in \emph{transverse position} if their boundary Jordan curves are in transverse position.
\end{definition}


\noindent Note that if two Jordan curves are in transverse position, then they meet finitely many times, by a compactness argument.\medskip


Finally, it will be helpful to have available to us the terminology of the following definition:

\begin{definition}
Suppose that $X_1,\ldots,X_n$ and $ X'_1,\ldots, X'_n$ are all subsets of $\bbC$.  Then we say that the collections $\{X_1,\ldots,X_n\}$ and $\{X'_1,\ldots, X'_n\}$ are in the same \emph{topological configuration} if there is an orientation-preserving homeomorphism $\varphi:\bbC \to \bbC$ so that $\varphi(X_i) =  X'_i$ for all $1\le i\le n$.  In practice the collections of objects under consideration will not be labeled $X_i$ and $ X'_i$, but there will be some natural bijection between them.  Then our requirement is that $\varphi$ respects this natural bijection.  We will say that certain conditions on some objects \emph{uniquely determine} their topological configuration if any two collections of objects satisfying the given conditions are in the same topological configuration.
\end{definition}

\noindent For example, considering a point $z\in \bbC$ and an open Jordan domain $\Omega$, the condition that $z\in \Omega$ uniquely determines the topological configuration of $\{z, \Omega\}$, but the condition that $z\not \in \Omega$ does not uniquely determine the topological configuration of $\{z, \Omega\}$.  (We may have $z\in \partial \Omega$, or $z\in \bbC \setminus (\Omega \cup \partial \Omega)$, and these situations are topologically distinct.)



The following lemma says that when working with fixed-point index, we need to consider our Jordan domains only ``up to topological configuration.''

\begin{lemma}
\label{prop:index invariance}
Suppose $K$ and $\tilde K$ are closed Jordan domains.  Let $f:\partial K \to \partial \tilde K$ be an indexable homeomorphism.  Suppose that $K'$ and $\tilde K'$ are also closed Jordan domains, so that $\{K,\tilde K\}$ and $\{K',\tilde K'\}$ are in the same topological configuration, via the homeomorphism $\varphi: \bbC\to \bbC$.  Let $f':\partial K'\to \partial \tilde K'$ be induced in the natural way, explicitly as $f' = \varphi|_{\partial \tilde K} \circ f \circ \varphi\inv|_{\partial K'}$.  Then $f'$ is indexable with respect to the usual orientation on $\partial K'$ and $\partial \tilde K'$, and $\eta(f) = \eta(f')$.
\end{lemma}

\begin{proof}
The following is well-known.  For a reference, see Chapters 1 and 2 of \cite{MR2850125}.
\begin{fact}
Every orientation-preserving homeomorphism $\bbC\to \bbC$ is homotopic to the identity map via homeomorphisms.
\end{fact}
\noindent Thus let $H_t:\bbC\times [0,1]\to \bbC$ be such a homotopy from the identity to $\varphi$.  Explicitly, for fixed $t$ we have that $H_t$ is an orientation-preserving homeomorphism $\bbC\to \bbC$, with $H_0$ equal to the identity on $\bbC$ and $H_1 = \varphi$.

Let $K_t = H_t(K)$ and $\tilde K_t = H_t(\tilde K)$.  Then $K_t$ and $\tilde K_t$ are closed Jordan domains, because $H_t$ is a homeomorphism.  Let $f_t:\partial K_t\to \partial \tilde K_t$ be induced in the natural way, explicitly as $H_t|_{\partial \tilde K} \circ f \circ H_t\inv|_{\partial K_t}$.  Let $\gamma_t = \{f_t(z) - z\}_{z\in \partial K_t}$.  Then tautologically $\eta(f)$ is the winding number of $\gamma_0$ around the origin, and $\eta(f')$ is the winding number of $\gamma_1$ around the origin.

Every $\gamma_t$ is a closed curve because $\partial K_t$ is a closed curve and $f_t$ is continuous.  Once we establish that no $\gamma_t$ passes through the origin Lemma \ref{prop:index invariance} will be proved because we have an induced homotopy from $\gamma_0$ to $\gamma_1$, and two curves homotopic in $\bbC\setminus \{0\}$ have the same winding number around the origin.  Suppose for contradiction that $0\in \gamma_t$.  Then there is a $z\in \partial K_t$ so that $f_t(z) = z$.  Thus $H_t\circ f\circ H_t\inv(z)= z$, and so $f(H_t\inv (z)) = H_t\inv (z)$, contradicting the fixed-point-free condition on $f$.
\end{proof}








\section{The Incompatibility Theorem\threestars}
\label{incompat section}

In this section, we state and prove the Incompatibility Theorem of Oded Schramm, appearing in \cite{MR1076089}*{Theorem 3.1}.  Before doing so, we need some preliminary definitions:\medskip

First, a \emph{topological rectangle} is a closed Jordan domain $R$ with four marked points on its boundary $\partial R$, which we naturally call its \emph{corners}.  A \emph{side} of $R$ is a closed sub-arc of $\partial R$ having two corners of $R$ as its endpoints, and containing no other corner of $R$.  We abuse notation slightly and use the same symbol, in this case $R$, to refer both to a topological rectangle and to its constituent closed Jordan domain.  We define \emph{topological triangles}, their \emph{corners}, and their \emph{sides} analogously, and will employ the same abuse of notation.

A \emph{packing} of a topological rectangle $R$ consists of a finite collection $\{K_1,\ldots,K_n\}$ of closed Jordan domains so that the following hold:

\begin{itemize}
\item Every $K_i$ is contained in $R$.
\item The $K_i$ are pairwise interiorwise disjoint, and any two of them meet at at most one point.
\item Each of the $K_i$ meets $\partial R$ at at most one point, and no $K_i$ meets a corner of $R$.
\item For every connected component $U$ of $R\setminus \cup_{i=1}^n K_i$, we have that the closure of $U$ is a topological triangle $T$ each of whose sides is contained in one of the $\partial K_i$, or in a side of $R$.  Then every corner of $T$ is either a corner of $R$, or an intersection point of some pair of the $K_i$.
\end{itemize}

\noindent See Figure \ref{clown1} for two examples of packings of topological rectangles.  Let $S_a,S_b,S_c,S_d$ denote the sides of $R$.  The \emph{contact graph} of the packing of $R$ by $K_1,\ldots,K_n$ is the graph having vertices $v_1,\ldots,v_n$ corresponding to the $K_1,\ldots,K_n$ and $v_a,v_b,v_c,v_d$ corresponding to the sides of $R$, both in the natural way, so that two distinct vertices share an edge if and only if the corresponding sets meet.  Note that for example, the contact graph of a packing of a rectangle is always a triangulation of a square, that is, a triangulation of a topological closed disk having four boundary edges.

%\noindent The \emph{contact graph} of such a collection $\{K_1,\ldots,K_n\}$ is the graph having a vertex $v_i$ for each $K_i$, so that distinct $v_i$ and $v_j$ share an edge if and only if $K_i$ and $K_j$ meet.

%Suppose that the collection $\{K_1,\ldots,K_n\}$ packs $R$.  Then for every corner $z$ of $R$, there is exactly one of the $K_1,\ldots,K_n$, which we will denote $K_{(z)}$, which meets the closure of that connected component of $R\setminus \cup_{i=1}^n K_i$ which contains $z$.  See Figure \ref{clown1} for two examples of packings of topological rectangles by closed Jordan domains.  Note that in general, even if $z_1$ and $z_2$ are distinct corners of $R$, it may be that $K_{(z_1)} = K_{(z_2)}$.


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\caption[Two topological rectangles packed with shapes]
{
\label{clown1}
{\bf Two topological rectangles packed with shapes.}
}
\end{figure}

Next, suppose that we have packings of topological rectangles $R$ and $\tilde R$ by collections of closed Jordan domains $K_1,\ldots,K_n$ and $\tilde K_1,\ldots,\tilde K_n$, respectively.  The two packings are said to be in \emph{transverse position} if:

\begin{itemize}
\item For every pair of integers $i$ and $\tilde i$, both between $1$ and $n$, with possibly $i=\tilde i$, we have that $K_i$ and $\tilde K_{\tilde i}$ are in transverse position as closed Jordan domains.
\item For every $1\le i\le n$, we have that $K_i$ and $\tilde R$ are in transverse position as closed Jordan domains, as are $\tilde K_i$ and $R$.
\item No intersection point of a pair of the sets $\partial K_1,\ldots,\partial K_n,\partial R$ lies on any $\partial \tilde K_i$ nor on $\partial \tilde R$.  Similarly, no intersection point of a pair of the sets $\partial \tilde K_1,\ldots,\partial \tilde K_n,\partial \tilde R$ lies on any $\partial K_i$ nor on $\partial R$.
\end{itemize}

\noindent For two examples of packings in transverse position, see Figure \ref{clown2}.\medskip

Finally, two closed Jordan domains $K$ and $\tilde K$ which are in transverse position are said to \emph{cut} one another if at least one of the sets $K\setminus \tilde K$ and $\tilde K\setminus K$ is disconnected.  Though it is not important for us, it holds for such $K$ and $\tilde K$ that $K\setminus \tilde K$ is connected if and only if $\tilde K \setminus K$ is.\medskip

We are now ready to state our version of the Incompatibility Theorem originally due to Schramm.  The original appears in \cite{MR1076089}*{Theorem 3.1}, where it is used to give the first proof of the rigidity of circle packings filling the complex plane.

When processing our statement of the Incompatibility Theorem \ref{incompat}, it may be helpful to keep the following in mind: in Schramm's original formulation, our notion of shapes \emph{cutting one another} is referred to as \emph{incompatibility}.  Then the Incompatibility Theorem \ref{incompat} may be remembered as, ``if two incompatible rectangles are packed in the same combinatorial way, then at least one pair of corresponding shapes of the two packings will be incompatible.''  The precise statement given in \cite{MR1076089} is somewhat different from the one we give here.  Our statement communicates the main idea of the theorem, and suffices for our applications.  Our proof uses fixed-point index, with the main new tool being Lemma \ref{no cut index}.  The original proof by Schramm is via different methods, but is also elementary.


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\caption
{
\label{topo conf incompat rect}
}
\end{figure}


\begin{incompat}
\label{incompat}
Let $R$ and $\tilde R$ be topological rectangles having sides $S_a,\allowbreak S_b,\allowbreak S_c,\allowbreak S_d$ and $\tilde S_a,\allowbreak \tilde S_b,\allowbreak \tilde S_c,\allowbreak \tilde S_d$ respectively, in the topological configuration depicted in Figure \ref{topo conf incompat rect}.  Suppose that we are given packings of $R$ and $\tilde R$ by collections of closed Jordan domains $K_1,\ldots,K_n$ and $\tilde K_1,\ldots,\tilde K_n$ respectively, in transverse position, so that the packings are combinatorially equivalent in the following precise sense: denoting the contact graphs of the packings by $G$ and $\tilde G$ respectively, we insist that the following holds: letting $v_a,v_b,v_c,v_d,v_1,\ldots,v_n$ and $\tilde v_a,\tilde v_b, \tilde v_c, \tilde v_d, \tilde v_1, \ldots, \tilde v_n$ denote the vertex sets of $G$ and $\tilde G$ in the natural way, we have that $G$ and $\tilde G$ are isomorphic via the identification $v_i \mapsto \tilde v_i$ for $i=a,b,c,d,1,\ldots,n$.  Then, there is an $1\le i\le n$ so that $K_i$ and $\tilde K_i$ cut each other.
\end{incompat}

\noindent For example, the packings shown in Figure \ref{clown1} share a contact graph.  In Figure \ref{clown2} we have overlaid them in two different ways, in both cases ensuring that the hypotheses of Theorem \ref{incompat} are satisfied.  We see that in each case, a pair of corresponding closed Jordan domains cut one another.\medskip




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\caption[Shapes cutting each other]
{
\label{clown2}
{\bf Shapes cutting each other.}  We have drawn $R$ and $\tilde R$ on top of each other in two different ``incompatible'' ways, and in both cases some pair of corresponding shapes are ``incompatible.''
}
\end{figure}

The rest of this section consists of a proof of the Incompatibility Theorem \ref{incompat}.  Our proof relies on the following simple lemma, which appeared in the introduction but is restated here for the convenience of the reader:

\renewcommand{\thelemmafree}{\ref{no cut index}}
\begin{lemmafree}
Let $K$ and $\tilde K$ be closed Jordan domains in transverse position, which do not cut each other, having boundaries oriented positively.  Let $\phi : \partial K \to \partial \tilde K$ be an indexable homeomorphism.  Then $\eta(\phi) \ge 0$.
\end{lemmafree}


\noindent We now give the proof of Theorem \ref{incompat} assuming Lemma \ref{no cut index}:

First, note that there is a natural bijection between the $U_f$ and the $\tilde U_f$, where we write $\{U_f\}_{f\in F}$ to denote the connected components of $R \setminus \partial R \cup \bigcup_{i=1}^n K_i$, similarly $\{\tilde U_f\}_{f\in F}$.  Moreover, for fixed $f$, we have that $U_f$ and $\tilde U_f$ are topological triangles, and that the corners of $U_f$ correspond in a natural way to those of its partner $\tilde U_f$.  (To see this, one may consider the graphs $G$ and $\tilde G$.  Each is the 1-skeleton of a triangulation of a topological closed disk, and because the graphs are isomorphic, we get that the triangulations are combinatorially equivalent.  Then the $U_f$ are in natural bijection with the faces $F$ of this combinatorial triangulation, as are the $\tilde U_f$.)

Given such a pair $U_f$ and $\tilde U_f$, we have that $U_f$ and $\tilde U_f$ are in transverse position as closed Jordan domains, because the packings we began with are in transverse position.  For every such pair $U_f$ and $\tilde U_f$, orient $\partial U_f$ and $\tilde\partial U_f$ positively, and let $\phi_f : \partial U_f \to \partial \tilde U_f$ be an indexable homeomorphism identifying corresponding corners, satisfying $\eta(\phi_f) \ge 0$.  We may do so by the Three Point Prescription Lemma \ref{tppl}.

Next, for every $1\le i\le n$, orienting $\partial K_i$ and $\partial \tilde K_i$ positively, we obtain an indexable homeomorphism $\phi_i : \partial K_i \to \partial \tilde K_i$ by restriction to the $\phi_f$ as necessary.  Also, via the same procedure, orienting $\partial R$ and $\partial \tilde R$ positively, we obtain an indexable homeomorphism $\phi_R : \partial R \to \partial \tilde R$.  Then, by the Index Additivity Lemma \ref{ial}, we have:

\[
\eta(\phi_R) = \sum_{i=1}^n \eta(\phi_i) + \sum_{f\in F} \eta(\phi_f)
\]

\noindent Now, as we saw in Figure \ref{fig:ex neg winding forced}, we have that $\eta(\phi_R) = -1$.  On the other hand, every $\eta(\phi_f)$ is non-negative by construction, and the $\eta(\phi_i)$ are non-negative by Lemma \ref{no cut index}, which gives us a contradiction.\medskip

To complete the proof of the Incompatibility Theorem \ref{incompat}, we require the proofs of several lemmas.  The proof of the Three Point Prescription Lemma \ref{tppl} is given in Section \ref{sec:3p} using torus parametrization.  Our proof of Lemma \ref{no cut index} is inspired\footnote{Thanks to Mario Bonk for suggesting this line of proof, greatly simplifying the required arguments.} by an argument given by Schramm in \cite{MR1207210}*{Proof of Lemma 2.2} in support of the Circle Index Lemma \ref{cil}.  We first need to prove a topological lemma on Jordan domains which do not cut each other:

\begin{lemma}
\label{uniqa}
Suppose $K$ and $\tilde K$ are closed Jordan domains in transverse position which do not cut each other, whose boundaries meet at least twice.  Then the topological configuration of $\{K,\tilde K\}$ is determined by how many times $\partial K$ and $\partial \tilde K$ meet.
\end{lemma}

\begin{proof}
First, suppose without loss of generality that $K$ is the closed unit disk $\bar\bbD$.  Let $2m\ge 2$ be the number of intersection points of $\partial K$ with $\partial \tilde K$.  Note also that we may without loss of generality pick these arbitrarily along $\partial K = \partial \bbD$.  Label these $z_1,\ldots,z_{2m}$ in clockwise order around $\partial K = \partial \bbD$.  This brings us to the situation of Figure \ref{jon1}.

Orient $\partial K$ and $\partial \tilde K$ as usual.  We now follow what happens as we traverse $\partial \tilde K$.  By relabeling we may suppose that $\partial \tilde K$ enters $K$ at $z_1$.  Because the interior of $\tilde K$ stays to the left of $\partial \tilde K$, and because $\partial \tilde K$ crosses $\partial K$ at every point where the two curves meet, it follows that $\partial \tilde K$ exits $K$ at $z_2$.  The same reasoning allows us to conclude that $\partial \tilde K$ enters $K$ at $z_3$, etc., so we get that $\partial \tilde K$ enters $K$ at $z_i$ for all odd $1\le i\le 2m$, and exits $K$ at $z_i$ for all even $1\le i\le 2m$.  This brings us to the situation of Figure \ref{jon2}.

We now consider where $\partial \tilde K$ goes after it crosses $z_1$.  Denote by $z_i$ the point of $\partial K \cap \partial \tilde K$ at which it arrives immediately after crossing $z_1$, noting that then $i$ is even.  We wish to establish that then $i=2$, so suppose for contradiction that $i\ne 2$.  This brings us to the situation of Figure \ref{jon3}.  Then $[z_1\to z_i]_{\partial \tilde K}$ disconnects $K = \bar\bbD$ into two components, call them $A_1$ and $A_2$.  We have that every connected component of $K \setminus \tilde K$ must then be completely contained in one of $A_1$ and $A_2$.  But $K\setminus \tilde K$ is connected by hypothesis, so one of $A_1$ and $A_2$ must be disjoint from $K\setminus \tilde K$.  We then get a contradiction, because, keeping careful track of the orientation of $\partial \tilde K$, we see that there are points of $K \setminus \tilde K$ immediately counterclockwise from $z_1$ along $\partial K$, and points of $K \setminus \tilde K$ immediately clockwise from $z_2$ along $\partial K$, and these lie in different components of $K \setminus [z_1\to z_i]_{\partial \tilde K}$ unless $i=2$.  The same reasoning allows us to conclude that, for any odd $1\le j\le 2m$, after entering $K$ at $z_j$, the curve $\partial \tilde K$ exits $K$ at $z_{j+1}$, bringing us to the situation of Figure \ref{jon4}.

By the same reasoning as in the previous paragraph, we get that, for any even $1\le j\le 2m$, after exiting $K$ at $z_j$, the curve $\partial \tilde K$ enters $K$ at $z_{j+1}$, adopting the convention that $z_{2m + 1} = z_1$.  However, in this case, there are two ways to connect $z_j$ to $z_{j+1}$: roughly speaking, we may either travel clockwise around the complement of $K=\bar\bbD$ from $z_j$ to $z_{j+1}$, or counterclockwise.  If we connect every such pair $z_j$ and $z_{j+1}$ with ``clockwise'' arcs, then the resulting orientation on $\partial \tilde K$ is not positive with respect to the Jordan domain it bounds, see Figure \ref{jon5}.  Thus suppose without loss of generality, by relabeling if necessary, that $z_m$ and $z_1$ are connected with a ``counterclockwise'' arc.  Now for the remaining pairs $z_j,z_{j+1}$, with $j$ even, there are no choices (up to topological equivalence) about how to draw the connecting sub-arcs of $\partial \tilde K$ between them, and we arrive at the situation of Figure \ref{jon6}.
\end{proof}


\begin{remark}
Which \emph{a priori} topological configurations can occur for two Jordan curves in transverse position is a poorly understood question, and is known as the study of \emph{meanders}.  We are fortunate that our setting is nice enough that a statement like that of Lemma \ref{uniqa} is possible.  Thanks to Thomas Lam for informing us of the topic of meander theory.
\end{remark}



\begin{figure}
\centering
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}
\caption
{
\label{fig:jon}{\bf The construction of the topologically unique pair $K$ and $\tilde K$ of transversely positioned closed Jordan domains, not cutting each other, having boundaries meeting at 8 points.}  The orientation on $\partial K = \partial \bbD$ is the usual, positive one, that is, the counterclockwise one.  In all cases, dashed arcs are sub-arcs of $\partial \tilde K$.
}
\end{figure}




\begin{proof}[Proof of Lemma \ref{no cut index}]
\phantomsection \label{no cut index proof}
In light of Lemmas \ref{prop:index invariance} and \ref{uniqa}, we may suppose that $K$ and $\tilde K$ are as in Figure \ref{no cut index figure}, of course drawn with the correct number of meeting points between $\partial K$ and $\partial \tilde K$.  Recalling that $\eta(f)$ counts the winding number of the vector $f(z) - z$ around the origin, we consider when it is possible for $f(z) - z$ to be a positive real number.  If $z$ does not lie on the intersection of the left side of the rectangle $\partial K$ with the interior of $\tilde K$, then $f(z) - z$ certainly has either a negative real component, or a non-zero imaginary component.  Thus pick a $z$ lying in this intersection.  Then only way for $f(z) - z$ to be real and positive is for $f(z)$ to lie on the semicircular sub-arc of $\partial \tilde K$ to the right of $z$, as in the figure.  But then, considering the orientations on $\partial K$ and $\partial \tilde K$, we get that the vector $f(z) - z$ must always be turning counterclockwise at such a $z$, as we traverse $\partial K$ in the positive direction.  We conclude that whenever the curve $\{f(z) - z\}_{\partial K}$ crosses the positive real axis, it does so in the positive direction, thus the winding number of this curve around the origin is non-negative.
\end{proof}

\begin{figure}
\centering
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{
\label{no cut index figure} We see that whenever $f(z) - z$ is positive and real, as we traverse $\partial K$ and $\partial \tilde K$ positively, the point $f(z)$ is moving upward, and the point $z$ is moving downward, so the vector $f(z) - z$ is winding counterclockwise, thus in the positive direction.
}
\end{figure}

\begin{remark}
The same Figure \ref{no cut index figure} can be used to show that under the hypotheses of Lemma \ref{no cut index}, we get that $\eta(f) \le 2$.  Thus the only fixed-point indices which can be achieved in this setting are $0, 1, 2$, and all three of these occur.  We do not use these facts, so working out the details is left as an exercise for the interested reader.
\end{remark}










\section{Proof of rigidity and uniformization of circle packings\threestars}
\label{sec:rigid in plane}
\label{cp proofs}

As an example of the power of fixed-point index, in this section we prove three rigidity and uniformization theorems on circle packings.  Theorem \ref{rigid in sphere} is usually credited to Koebe \cite{koebe-1936}, Andreev \cite{MR0273510}, and Thurston\footnote{Originally at his talk at the International Congress of Mathematicians, Helsinki, 1978, according to \cite{MR1303402}*{p.\ 135}.  See also \cite{thurston-gt3m-notes}*{Chapter 13}.}.  All of their proofs were via methods different from ours.  Theorems \ref{rigid in plane} and \ref{thm:no luv} are originally due to Schramm \cite{MR1076089}*{Rigidity Theorems 1.1, 5.1}, who proved them essentially via the Incompatibility Theorem \ref{incompat}, although his proof of the Incompatibility Theorem is not via fixed-point index techniques.  The first ones to study circle packings by fixed-point index techniques were He and Schramm in \cite{MR1207210}, although they did not proceed via the Incompatibility Theorem, instead applying other normalizations to the packings in question, to similarly argue by contradiction.  For further references on circle packing, see for example the articles \citelist{\cite{MR1303402} \cite{MR2884870}} and their bibliographies.\medskip

We begin with some basic definitions.  A \emph{circle packing} is a collection $\calP = \{D_i\}$ of pairwise interiorwise disjoint round closed disks in the Riemann sphere $\hat\bbC$.  The \emph{contact graph} of a circle packing is the graph having a vertex for every disk of the packing, so that two vertices share an edge if and only if the corresponding disks meet.  Then the following hold:

\begin{theorem}
\label{rigid in sphere}
Suppose that $\calP$ and $\tilde\calP$ are circle packings in $\hat\bbC$, sharing a contact graph that triangulates the 2-sphere $\bbS^2$.  Then $\calP$ and $\tilde\calP$ differ by a M\"obius or anti-M\"obius transformation.
\end{theorem}

\begin{theorem}
\label{rigid in plane}
Suppose that $\calP$ and $\tilde\calP$ are circle packings which are locally finite in $\bbC$, sharing a contact graph that triangulates a topological open disk.  Then $\calP$ and $\tilde\calP$ differ by a Euclidean similarity.
\end{theorem}

\begin{theorem}
\label{thm:no luv}
There cannot be two circle packings $\calP$ and $\tilde \calP$ sharing a contact graph triangulating a topological open disk, so that one is locally finite in $\bbC$ and the other is locally finite in the hyperbolic plane $\bbH^2$, equivalently the open unit disk $\bbD$.
\end{theorem}

The rest of this section consists of the proofs of these three theorems.  Before moving on, we make one note:

\begin{remark}
The statement for locally finite packings in the hyperbolic plane $\bbH^2 \cong \bbD$ analogous to Theorem \ref{rigid in plane} also holds.  However, to apply our techniques to prove it, one would need to show that two combinatorially equivalent packings, both locally finite in $\bbH^2$, having contact graphs triangulating $\bbH^2$, induce a homeomorphism, or at least some appropriately behaved identification, on the boundary $\partial_\infty \bbH^2 \cong \partial \bbD$.  This turns out to be true (in fact, it follows from the rigidity theorem we discuss in this remark), but non-trivial.
\end{remark}

\begin{proof}[Proof of Theorem \ref{rigid in sphere}]
The argument is by contradiction.  The main idea is to superimpose the packings $\calP$ and $\tilde\calP$ on the Riemann sphere in a convenient way.  In particular, we will isolate two topological quadrilaterals $Q$ and $\tilde Q$ so that they are packed in combinatorially equivalent ways by disks of $\calP$ and $\tilde \calP$, but so that $Q$ and $\tilde Q$ cut each other, as in Figure \ref{clown2}.  Then the simple observation that two round disks cannot cut each other will give us our desired contradiction, via the Incompatibility Theorem \ref{incompat}.\medskip


Let $X = (V,E,F)$ be the common triangulation of $\bbS^2$ which $\calP$ and $\tilde\calP$ realize.  Suppose for contradiction that $\calP$ and $\tilde \calP$ are not equivalent under any M\"obius or anti-M\"obius transformation.  For the first part of the proof, we apply a sequence of normalizations to $\calP$ and to $\tilde\calP$.  Let $f_0 = \left<v_1,v_2,v_3\right> \in F$ be a face of $X$.  We first normalize via a M\"obius transformation so that $D_i = \tilde D_i$ for $i=1,2,3$.  Here $D_i\in \calP$ is the disk corresponding to the vertex $v_i\in V$ of $X$, similarly $\tilde D_i\in \tilde\calP$.  In particular, the correct M\"obius transformation is the one that sends the intersection points of the $\partial D_i$ to the corresponding ones of the $\partial \tilde D_i$.

Our next normalization is in our initial choice of $f_0$ and our labeling of the $v_i$, as per the following observation:

\begin{observation}
\label{obs:old prf}
Let $v_4$ denote the vertex of $X$ other than $v_1$ so that $\left<v_2,v_3,v_4\right>$ is a face of $X$.  Then there is some choice of $f_0 = \left<v_1,v_2,v_3\right>$ so that the disks $D_4$ and $\tilde D_4$ are not equal after our normalization identifying $D_i = \tilde D_i$ for $i=1,2,3$.
\end{observation}

\noindent If there were no such choice of $f_0$ then in fact every pair of corresponding disks $D_i$ and $\tilde D_i$ would coincide after our first normalization, and so $\calP$ and $\tilde \calP$ coincide.

An \emph{interstice} of the packing $\calP$ is a connected component of $\hat\bbC \setminus \calP$.  Every interstice is necessarily a curvilinear triangle, because the packing's contact graph triangulates $\hat\bbC$.  For our next normalization, we insist that $\infty$ lies in the interstice formed by $D_1=\tilde D_1,D_2= \tilde D_2,D_3=\tilde D_3$ which contains no other disks of either packing.  Finally, we insist that $D_1=\tilde D_1,D_2=\tilde D_2,D_3=\tilde D_3$ all have Euclidean radius 1, and that $D_2$ and $D_3$ are tangent at a point lying on the horizontal axis, so that $D_1$ lies to their left.  The situation for $\calP$ is depicted in Figure \ref{fig:calp}.\medskip

\begin{figure}
\centering
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\caption[The packing $\calP$ after some normalizations]
{
\label{fig:calp}
{\bf The packing $\calP$ after some normalizations.}  The disks of $\calP$ all lie between $D_1,D_2,D_3$.  Note that the interstice formed by $D_1, D_2, D_3$ on $\hat\bbC$ is the outside region in these figures.  The disk $D_4$ is ``the first disk of $\calP\setminus \{D_1,D_2,D_3\}$ we get to if we start scanning from the right.''  In \subref{fig:qq} the topological quadrilateral $Q$ is outlined in bold.
}
\end{figure}


%We summarize the picture as it stands after these normalizations.  First, we return to the plane $\bbC$.  We insist that the Euclidean center of $D_1$ and the tangency point between $D_2$ and $D_3$ both lie on the horizontal axis, with the center of $D_1$ to the left of the tangency point between $D_2$ and $D_3$.  The collections $\calP$ and $\tilde\calP$ are still made up of closed disks, since the point we sent to infinity does not lie in any disk of either packing.  Corresponding pairs of disks $D_i$ and $\tilde D_i$ are equal for $i = 1,2,3$, and these disks are the ``boundary disks'' of both packings.  The disks $\calP$ accumulate around some point $z_\infty$ in the plane, as do the disks of $\tilde\calP$ around $\tilde z_\infty$.  The points $z_\infty$ and $\tilde z_\infty$ may be the same or may be different.  Finally, every face of $X$ corresponds to an interstice in each packing, except for $f= \left<v_1,v_2,v_3\right>$: the disks $D_i$, equivalently $\tilde D_i$, for $i=1,2,3$, form an interstice in $\hat\bbC$, but not in $\bbC$.  Recall that for us, an interstice is a topological open disk that meets no disks of the packing.




From now on we work in the plane $\bbC$, in the sense that $\infty \in \hat\bbC$ will not move again for the remainder of the proof.  Note that every face $f$ of $F$ corresponds to some interstices $U_f\subset \bbC$ and $\tilde U_f\subset \bbC$ of $\calP$ and $\tilde \calP$ respectively, except for $f_0$, for which the interstices $U_{f_0} = \tilde U_{f_0}$ contain $\infty$.  Let $Q$ be the topological quadrilateral shown in Figure \ref{fig:qq}.  More precisely, let $V_Q = V\setminus \{v_1,v_2,v_3\}$, and let $F_Q = F\setminus \{f_0 = \left<v_1,v_2,v_3\right>, \left<v_2,v_3,v_4\right>\}$.  Then we define $Q = \bigcup_{v\in V_Q} D_v \cup \bigcup_{f\in F_Q} U_f$.  Define the analogous objects for $\tilde\calP$ in the obvious way.\medskip

We now apply one final transformation to $\calP$.  First, suppose without loss of generality that the Euclidean radius of $D_4$ is larger than that of $\tilde D_4$.  Then translate every disk of $\calP$ to the right by a small amount $\varepsilon>0$, leaving the disks of $\tilde\calP$ unchanged.  Denote this transformation by $T_\varepsilon$.  We will discuss more precise requirements on $\varepsilon$ later.  The situation is depicted in Figure \ref{fig:qqmain}.  The essential point is that there is an open interval of values that $\varepsilon>0$ may take so that after all of our transformations, the topological quadrilaterals $Q$ and $\tilde Q$ are arranged qualitatively as in Figure \ref{fig:qq3}.  In particular, we may choose $\varepsilon$ so that the packings of $Q$ and $\tilde Q$ by the remaining disks of $\calP$ and $\tilde\calP$ are in transverse position, because there are only finitely many values of $\varepsilon>0$ for which this fails.  Then our desired contradiction follows immediately from the Incompatibility Theorem \ref{incompat}, because round closed disks cannot cut one another.
\end{proof}



\begin{figure}
\centering
\subfloat[]
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\label{fig:qq3}
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\caption[The interaction between $\calP$ and $\tilde{\calP}$ before and after applying $T_\varepsilon$]
{
\label{fig:qqmain}
{\bf The interaction between $\calP$ and $\tilde\calP$ before and after applying $T_\varepsilon$.}  In \subref{fig:qq2} we see the superimposition of the $D_i$ with the $\tilde D_i$ before applying $T_\varepsilon$ to $\calP$.  The disks $D_i$ are drawn solid, and the disks $\tilde D_i$ are drawn dashed.  In \subref{fig:qq3} we see the relative positions of $Q$ and $\tilde Q$ after applying $T_\varepsilon$ to $\calP$.
}
\end{figure}


\begin{proof}[Proof of Theorem \ref{rigid in plane}]
The proof of Theorem \ref{rigid in plane} proceeds along the same lines, except that after our first round of normalizations identifying $D_i$ and $\tilde D_i$ for $i=1,2,3$, and sending a point of their common interstice to $\infty$, the remaining disks of $\calP$ accumulate around a point $z_\infty\in \bbC$, as do those of $\tilde \calP$ around a point $\tilde z_\infty\in \bbC$.  The points $z_\infty$ and $\tilde z_\infty$ may coincide or may be different.  We define and apply $T_\varepsilon$ as before, this time making sure that $z_\infty$ and $\tilde z_\infty$ differ after applying $T_\varepsilon$.

Next, pick small disjoint neighborhoods $W$ and $\tilde W$ of $z_\infty$ and $\tilde z_\infty$ respectively, and contained in $Q$ and $\tilde Q$ respectively.  Then, let $V_L$ be the set of vertices $v\in V$ so that both $D_v\subset W$ and $\tilde D_v\subset \tilde W$.  Remove vertices from $V_L$ until the sub-triangulation of $X$ having vertices $V\setminus V_L$ is a triangulation of a topological closed disk.  Let $F_L$ be the set of faces of $X$ corresponding to interstices formed by disks whose vertices are in $V_L$.  Let $L$ (which stands for leftovers) be the union $\bigcup_{v\in V_L} D_v \cup \bigcup_{f\in F_L} U_f\cup z_\infty$, and define $\tilde L$ similarly.  Then $\{D_v\}_{v\in V\setminus V_L}$ together with $L$ form a packing of the topological quadrilateral $Q$ by closed Jordan domains, as do $\{\tilde D_v\}_{v\in V\setminus V_L}, \tilde L$ in $\tilde Q$.  Furthermore, because $L$ and $\tilde L$ are disjoint by construction, these two domains do not cut each other.  Then we get our desired contradiction by the Incompatibility Theorem \ref{incompat} as before.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:no luv}]
The adaptation here is along similar lines as the adaptation to prove Theorem \ref{rigid in plane}.  Suppose for contradiction that $\calP$ is locally finite in $\bbC$, and $\tilde\calP$ is locally finite in $\bbH^2 \cong \bbD$.  This time, after our normalizations, the disks of $\calP$ accumulate around a single point $z_\infty$, and the disks of $\tilde\calP$ accumulate around some round circle $C$ contained in the bounded region in the plane formed between $D_1=\tilde D_1,D_2=\tilde D_1,D_3=\tilde D_3$.  This time, ensure that we chose $\varepsilon$ so that $z_\infty$ does not lie on the circle $C$.  We define $L$ and $\tilde L$ by throwing away disks of our circle packings, as before, but this time, either $L$ and $\tilde L$ are disjoint, or one contains the other.  In either case, the two do not cut each other, and the conclusion of the proof proceeds as before.
\end{proof}















































































\section{Torus parametrization\threestars}
\label{chap:torus}



Before defining torus parametrization, it will be helpful to have access to the following simple lemma:





\begin{lemma}
\label{lem:gen pos lemma}
Suppose $K$ and $\tilde K$ are closed Jordan domains in transverse position.  Suppose that $z\in \partial K \cap \partial \tilde K$.  Orient $\partial K$ and $\partial \tilde K$ positively as usual.  Then one of the following two mutually exclusive possibilities holds at the point $z$.
\begin{enumerate}
\item \label{case:lem:orientations:1} The curve $\partial \tilde K$ is entering $K$, and the curve $\partial K$ is exiting $\tilde K$.
\item \label{case:lem:orientations:2} The curve $\partial K$ is entering $\tilde K$, and the curve $\partial \tilde K$ is exiting $K$.
\end{enumerate}
Thus as we traverse $\partial K$, we alternate arriving at points of $\partial K \cap \partial \tilde K$ where (\ref{case:lem:orientations:1}) occurs and those where (\ref{case:lem:orientations:2}) occurs, and the same holds as we traverse $\partial \tilde K$.
\end{lemma}



\begin{proof}
Let $z\in \partial K \cap \partial \tilde K$.  We may assume, by applying a homeomorphism, that locally near $z$ the picture looks like Figure \ref{fig:in out in out}, with $\partial K$ oriented down-to-up as shown.  Then $K$ lies to the left of $\partial K$.  Now, certainly $\partial \tilde K$ is either entering or exiting $K$ at $z$.  Suppose $\partial \tilde K$ is entering $K$ at $z$.  Then $\partial \tilde K$ is oriented right-to-left, and so $\tilde K$ is below $\partial \tilde K$.  Thus $\partial K$ is exiting $\tilde K$, and case (1) occurs.  Similarly, if $\partial \tilde K$ is exiting $K$ at $z$ then $\partial K$ is entering $\tilde K$ at $z$, so case (2) occurs.
\end{proof}




\begin{figure}
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\caption[A meeting point between two Jordan curves in transverse position]
{
\label{fig:in out in out}
{\bf A meeting point between two Jordan curves in transverse position.}  The orientation shown on $\partial K$ implies that $K$ lies to the left.  Depending on the orientation chosen for $\partial \tilde K$ we will get that $\tilde K$ lies above $\partial K$ or below it.
}
\end{figure}




































We are now ready to define \emph{torus parametrization}.  Throughout the definition, refer to Figure \ref{tpex1} for an example.


\begin{definition}
Let $K$ and $\tilde K$ be closed Jordan domains in transverse position, so that $\partial K$ and $\partial \tilde K$ meet at $2M\ge 0$ points, with boundaries oriented as usual.  Let $\partial K \cap \partial \tilde K = \{P_1,\ldots,P_M,\tilde P_1,\ldots,\tilde P_M\}$, where $P_i$ and $\tilde P_i$ are labeled so that at every $P_i$ we have that $\partial K$ is entering $\tilde K$, and at every $\tilde P_i$ we have that $\partial \tilde K$ is entering $K$.  Imbue $\bbS^1$ with an orientation and let $\kappa:\partial K\to \bbS^1$ and $\tilde \kappa : \partial\tilde K \to \bbS^1$ be orientation-preserving homeomorphisms.  We refer to this as fixing a \emph{torus parametrization} for $K$ and $\tilde K$.

We consider a point $(x,y)$ on the 2-torus $\bbT = \bbS^1\times \bbS^1$ to be parametrizing simultaneously a point $\kappa\inv(x) \in \partial K$ and a point $\tilde\kappa\inv(y) \in \partial \tilde K$.  We denote by $p_i\in \bbT$ be the unique point $(x,y)\in \bbT$ satisfying $\kappa\inv(x) = \tilde\kappa\inv(y) = P_i$, similarly $\tilde p_i\in \bbT$.  Note that by the transverse position hypothesis no pair of points in $\{p_1,\ldots,p_M, \tilde p_1,\ldots,\tilde p_M\}$ share a first coordinate, nor a second coordinate.

Suppose we pick $(x_0,y_0)\in \bbS^1\times \bbS^1$.  Then we may draw an image of $\bbT = \bbS^1\times \bbS^1$ by letting $\{x_0\}\times \bbS^1$ be the vertical axis and letting $\bbS^1 \times \{y_0\}$ be the horizontal axis.  Then we call $(x_0,y_0)$ a \emph{base point} for the drawing.
\end{definition}

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\caption[A pair of closed Jordan domains $K$ and $\tilde K$ and a torus parametrization for them, drawn with base point $(\kappa(u),\tilde\kappa(\tilde u))$]
{
\label{tpex1}
{\bf A pair of closed Jordan domains $K$ and $\tilde K$ and a torus parametrization for them, drawn with base point $(\kappa(u),\tilde\kappa(\tilde u))$.}  The key points to check are that as we vary the first coordinate of $\bbT$ positively starting at $u$, we arrive at $\kappa(P_1)$, $\kappa(\tilde P_1)$, $\kappa(P_2)$, and $\kappa(\tilde P_2)$ in that order, and as we vary the second coordinate of $\bbT$ positively starting at $\tilde\kappa(\tilde u)$, we arrive at $\tilde\kappa(P_1)$, $\tilde \kappa(\tilde P_2)$, $\tilde\kappa(P_2)$, and $\tilde\kappa(P_1)$ in that order.
}
\end{figure}





Suppose that $\phi:\partial K \to \partial \tilde K$ is an orientation-preserving homeomorphism.  Then $\phi$ determines an oriented curve $\gamma$ in $\bbT$ for us, namely its graph $\gamma = \{(\kappa(z),\tilde\kappa(\phi(z))\}_{z\in \partial K}$, with orientation obtained by traversing $\partial K$ positively.  Note that $\phi$ is fixed-point-free if and only if its associated curve $\gamma$ misses all of the $p_i$ and $\tilde p_i$.  Pick $u\in \partial K$ and denote $\tilde u = \phi(u)$.  Then if we draw the torus parametrization for $K$ and $\tilde K$ using the base point $(\kappa(u),\tilde\kappa(\tilde u))$, the curve $\gamma$ associated to $\phi$ ``looks like the graph of a strictly increasing function.''  The converse is also true: given any such $\gamma$, it determines for us an orientation-preserving homeomorphism $\partial K\to \partial \tilde K$ sending $u$ to $\tilde u$, which is fixed-point-free if and only if $\gamma$ misses all of the $p_i$ and $\tilde p_i$.







\newcommand{\Num}{\operatorname{Num}}


\newcommand{\nump}{\#p}
\newcommand{\numq}{\#\tilde p}



Suppose that $\phi(u) = \tilde u$, equivalently that $(\kappa(u),\tilde\kappa(\tilde u)) \in \gamma$.  The curve $\gamma$ and the horizontal and vertical axes $\{\tilde\kappa(\tilde u)\}\times \bbS^1$ and $\bbS^1\times \{\kappa(u)\}$ divide $\bbT$ into two simply connected open sets $\Delta_\uparrow(u,\gamma)$ and $\Delta_\downarrow(u,\gamma)$ as shown in Figure \ref{fig:proving amazing index}.  We suppress the dependence on $\tilde u$ in the notation because $\tilde u = \phi(u)$.  If neither $u\in \partial \tilde K$ nor $\tilde u\in \partial K$ then every $p_i$ and every $\tilde p_i$ lies in either $\Delta_\downarrow(u,\gamma)$ or $\Delta_\uparrow(u,\gamma)$.  In this case we write $\nump_\downarrow(u,\gamma)$ to denote $|\{p_1,\ldots,p_M\} \cap \Delta_\downarrow(u,\gamma)|$ the number of points $p_i$ which lie in $\Delta_\downarrow(u,\gamma)$, and we define $\nump_\uparrow(u,\gamma)$, $\numq_\downarrow(u,\gamma)$, and $\numq_\uparrow(u,\gamma)$ in the analogous way.  Denote by $\omega(\alpha,z)$ the winding number of the closed curve $\alpha\subset \bbC$ around the point $z\not\in\alpha$.\medskip

The following central lemma says that given an indexable homeomorphism $\phi: \partial K \to \partial \tilde K$, we may read off its fixed-point index $\eta(\phi)$ simply by examining the curve $\gamma$ associated to $\phi$ in the way we just described:


\begin{lemma}
\label{prop:computing index from torus}
Let $K$ and $\tilde K$ be closed Jordan domains.  Fix a torus parametrization of $K$ and $\tilde K$ via $\kappa$ and $\tilde\kappa$.  Let $\phi:\partial K \to \partial \tilde K$ be an indexable homeomorphism, with graph $\gamma$ in $\bbT$.  Suppose that $\phi(u) = \tilde u$, where $u\not\in \partial \tilde K$ and $\tilde u\not\in \partial K$.  Then:
\begin{align}
\label{eq 1:prop:computing index from torus eq}\eta(\phi)  = w(\gamma)& = \omega(\partial K, \tilde u) + \omega(\partial \tilde K, u ) - \nump_\downarrow(u,\gamma) + \numq_\downarrow(u,\gamma)\\
\label{eq 2:prop:computing index from torus eq}& = \omega(\partial K, \tilde u) + \omega(\partial \tilde K, u) + \nump_\uparrow(u,\gamma) - \numq_\uparrow(u,\gamma)
\end{align}
\end{lemma}


\noindent The remainder of the section is spent proving Lemma \ref{prop:computing index from torus}.\medskip

Suppose $\gamma_0$ is any oriented closed curve in $\bbT \setminus \{p_1,\ldots,p_M,\tilde p_1,\ldots,\tilde p_M\}$.  Then the closed curve $\{\tilde\kappa\inv(y) - \kappa\inv(x)\}_{(x,y)\in \gamma_0}$ misses the origin, and has a natural orientation obtained by traversing $\gamma_0$ positively.  We denote by $w(\gamma_0)$ the winding number around the origin of $\{\tilde\kappa\inv(y) - \kappa\inv(x)\}_{(x,y)\in \gamma_0}$.

\begin{observation}
\label{lem:homotopy classes}
If $\gamma_1$ and $\gamma_2$ are homotopic in $\bbT \setminus \{p_1,\ldots,p_M,\tilde p_1,\ldots,\tilde p_M\}$ then $w(\gamma_1) = w(\gamma_2)$.
\end{observation}

\noindent This is because the homotopy between $\gamma_1$ and $\gamma_2$ in $\bbT\setminus \{p_2,\ldots,p_M,\tilde p_1,\ldots,\tilde p_M\}$ induces a homotopy between the closed curves $\{\tilde\kappa\inv(y) - \kappa\inv(x)\}_{(x,y)\in \gamma_1}$ and $\{\tilde\kappa\inv(y) - \kappa\inv(x)\}_{(x,y)\in \gamma_2}$ in the punctured plane $\bbC\setminus \{0\}$.

Suppose that $\phi:\partial K\to \partial \tilde K$ is a fixed-point-free orientation-preserving homeomorphism.  Let $\gamma$ be its graph in $\bbT$.  If $\gamma$ has orientation induced by traversing $\partial K$ and $\partial \tilde K$ positively, then the following is a tautology.

\begin{observation}
\label{ex:never gonna reference this lol}
$\eta(\phi) = w(\gamma)$
\end{observation}








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\caption[A homotopy from $\partial\Delta_\downarrow(u,\gamma)$ to $\Gamma$]
{
\label{fig:proving amazing index}
{\bf A homotopy from $\partial\Delta_\downarrow(u,\gamma)$ to $\Gamma$.}  Here the orientation shown on $\gamma$ is the opposite of the orientation induced by traversing $\partial K$ positively.
}
\end{figure}


Orient $\partial \Delta_\downarrow(u,\gamma)$ as shown in Figure \ref{fig:proving amazing index}.  Then $\partial \Delta_\downarrow(u,\gamma)$ is the concatenation of the curve $\gamma$ traversed backwards with $\bbS^1\times \{\tilde\kappa(\tilde u)\}$ and $\{\kappa(u)\}\times \bbS^1$, where the two latter curves are oriented according to the positive orientation on $\bbS^1$.

\begin{observation}
If $\bbS^1\times \{\tilde\kappa(\tilde u)\}$ and $\{\kappa(u)\}\times \bbS^1$ are oriented according to the positive orientation on $\bbS^1$, then $w(\bbS^1\times \{\tilde\kappa(\tilde u)\}) = \omega(\partial K, \tilde u)$ and $w(\{\kappa(u)\}\times \bbS^1) = \omega(\partial \tilde K, u)$.
\end{observation}
\noindent It is also easy to see that if we concatenate two closed curves $\gamma_1$ and $\gamma_2$ that meet at a point, we get $w(\gamma_1\circ \gamma_2) = w(\gamma_1) + w(\gamma_2)$.  Thus in light of the orientations on $\partial \Delta_\downarrow(u,\gamma)$ and all other curves concerned we get:
\begin{align*}
\label{prop:computing index from torus eq}
w(\partial \Delta_\downarrow(u,\gamma))& = w(\bbS^1\times \{\tilde\kappa(\tilde u)\}) + w(\{\kappa(u)\}\times \bbS^1) - w(\gamma) \\
& = \omega(\partial K,\tilde u) +  \omega(\partial \tilde K, u) - \eta(\phi)
\end{align*}

For every $i$ let $\zeta(p_i)$ and $\zeta(\tilde p_i)$ be small squares around $p_i$ and $\tilde p_i$ respectively in $\bbT$, oriented as shown in Figure \ref{fig:proving amazing index}.  By \emph{square} we mean a simple closed curve which decomposes into four ``sides,'' so that on a given side one of the two coordinates of $\bbS^1\times \bbS^1 = \bbT$ is constant.  Pick the rectangles small enough so that the closed boxes they bound are pairwise disjoint and do not meet $\partial \Delta_\downarrow(u,\gamma)$.

Let $\Gamma$ be the closed curve in $\Delta_\downarrow(u,\gamma)$ obtained in the following way.  First, start with every loop $\zeta(p_i)$ and $\zeta(\tilde p_i)$ for those $p_i$ and $\tilde p_i$ lying in $\Delta_\downarrow(u,\gamma)$.  Let $\delta_0$ be an arc contained in the interior of $\Delta_\downarrow(u,\gamma)$ which meets each $\zeta(p_i)$ and $\zeta(\tilde p_i)$ contained in $\Delta_\downarrow(u,\gamma)$ at exactly one point.  It is easy to prove inductively that such an arc exists.  Let $\delta$ be the closed curve obtained by traversing $\delta_0$ first in one direction, then in the other.  Then let $\Gamma$ be obtained by concatenating $\delta$ with every $\zeta(p_i)$ and $\zeta(\tilde p_i)$ contained in $\Delta_\downarrow(u,\gamma)$.
\begin{observation}
The curves $\Gamma$ and $\partial \Delta_\downarrow(u,\gamma)$ are homotopic in $\bbT\setminus \{p_1,\ldots,\allowbreak p_M,\allowbreak\tilde p_1,\ldots,\allowbreak\tilde p_M\}$.  Also $w(\delta) = 0$.  It follows that:
\[
w(\partial \Delta_\downarrow(u,\gamma)) = w(\Gamma) = \sum_{p_i\in \Delta_\downarrow(u,\gamma)} w(\zeta(p_i)) + \sum_{\tilde p_i\in \Delta_\downarrow(u,\gamma)} w(\zeta(\tilde p_i))
\]
\end{observation}

\noindent See Figure \ref{fig:proving amazing index} for an example.  On the other hand, the following holds.





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\caption[The local picture near $P_i$]
{
\label{fig:proving local winding pi}
{\bf The local picture near $P_i$.}  This allows us to compute the ``local fixed-point index'' $w(\zeta(p_i))$ of $f$ near $P_i$.
}
\end{figure}


\begin{observation}
\label{lem:local windings for pi qi}
$w(\zeta(p_i)) = 1$, $w(\zeta(\tilde p_i)) = -1$
\end{observation}

\noindent To see why, suppose that $\zeta(p_i) = \partial ( [x_0 \to x_1]_{\bbS^1} \times [y_0\to y_1]_{\bbS^1})$.  Then up to orientation-preserving homeomorphism the picture near $P_i$ is as in Figure \ref{fig:proving local winding pi}.  We let $(x,y)$ traverse $\zeta(p_i)$ positively starting at $(x_0,y_0)$, keeping track of the vector $\tilde\kappa\inv(y) - \tilde\kappa\inv(x)$ as we do so.  The vector $\tilde\kappa\inv(y_0) - \tilde\kappa\inv(x_0)$ points to the right.  As $x$ varies from $x_0$ to $x_1$, the vector $\tilde\kappa\inv(y) - \tilde\kappa\inv(x)$ rotates in the positive direction, that is, counter-clockwise, until it arrives at $\tilde \kappa\inv (y_0) - \kappa\inv (x_1)$, which points upward.  Continuing in this fashion, we see that $\tilde\kappa\inv(y) - \tilde\kappa\inv(x)$ makes one full counter-clockwise rotation as we traverse $\zeta(p_i)$.  The proof that $w(\zeta(\tilde p_i)) = -1$ is similar.  Combining all of our observations establishes equation \ref{eq 1:prop:computing index from torus eq}.  The proof that equation \ref{eq 2:prop:computing index from torus eq} holds is similar.\qed























































\section{Proof of the Three Point Prescription Lemma \ref{lem:3p}\threestars}
\label{sec:3p}

We restate the lemma for the convenience of the reader.  Let $K$ and $\tilde K$ be compact Jordan domains in transverse position.  Let $z_1,z_2,z_3\in \partial K\setminus \partial \tilde K$ appear in counterclockwise order, similarly $\tilde z_1,\tilde z_2, \tilde z_3\in \partial \tilde K\setminus \partial K$.  An indexable homeomorphism $\phi:\partial K\to \partial \tilde K$ is called \emph{faithful} if it sends $z_i\mapsto \tilde z_i$ for $i=1,2,3$.  We wish to find a faithful $\phi$ with $\eta(\phi)\ge 0$.  We refer to the $z_i$ and the $\tilde z_i$ as our \emph{constraint points}.

%Two arcs $\gamma \subset \partial K$ and $\tilde \gamma \subset \partial \tilde K$ are called \emph{separable} if there exists a faithful $f$ which does not identify any points of $\gamma$ with any points of $\tilde \gamma$.

We proceed by induction on the number of intersection points $\partial K \cap \partial \tilde K$, noting that this number is always even.  The Circle Index Lemma \ref{cil} takes care of the cases where $\partial K$ and $\partial \tilde K$ meet $0$ or $2$ times.  Thus suppose that $\partial K$ and $\partial \tilde K$ meet at least $4$ times.


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\caption[``Pulling apart'' two sub-curves of $\partial K$ and $\partial \tilde K$ ``crossing minimally'']
{
\label{fig:pull me}
{\bf ``Pulling apart'' two sub-curves of $\partial K$ and $\partial \tilde K$ ``crossing minimally.''}  Note that \emph{a priori} the orientations on $\partial K$ and $\partial \tilde K$ may be arbitrary, so the open Jordan domain formed between the two curves on the left may be contained in both $K$ and $\tilde K$, in just one of them, or in neither.
}
\end{figure}

The main idea of the proof is to find sub-arcs of $\partial K$ and $\partial \tilde K$ which ``cross minimally,'' and ``pull them apart,'' see Figure \ref{fig:pull me}.  When ``pulling these sub-arcs apart,'' we leave the rest of $\partial K$ and $\partial \tilde K$ fixed, as in Figure \ref{fig:pull me 2}.  We then apply the induction hypothesis to the resulting Jordan domains $K'$ and $\tilde K'$, obtaining some indexable homeomorphism $\phi':\partial K'\to \partial \tilde K'$ with $\eta(\phi')\ge 0$.  Finally, we use $\phi'$ to construct $\phi$, arguing that the fixed-point index is preserved or increased in this last step.  The details of the proof will be worked out in a torus parametrization.  This is because the torus parametrization allows us to work systematically and somewhat combinatorially through many cases, and unfortunately much of the proof is case analysis.  Also unfortunately, it is often difficult to discuss the plane topological pictures under consideration using prose, so much of the proof involves chasing the definitions of various terms and mathematical expressions around.



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{
\label{fig:pull me 2}
{\bf The compact Jordan domains under consideration, before and after ``pulling the sub-arcs apart.''}  Note that in this case, we had three choices of which arcs to pull apart, and each choice results in a different pair $K',\tilde K'$.
}
\end{figure}





%\subsection*{Part 1: Notation and terminology}

\subsection{Initial set-up}

Two intersection points $p_0,\tilde p_0\in \partial K \cap \partial \tilde K$ are called \emph{adjacent in $\partial K$} if one of the two arcs $[p_0\to \tilde p_0]_{\partial K}$ and $[\tilde p_0\to p_0]_{\partial K}$ does not contain any other intersection points $\partial K\cap \partial \tilde K$.  Note that two intersection points of $\partial K \cap \partial \tilde K$ may be adjacent only if $\partial K$ enters $\tilde K$ at one and leaves $\tilde K$ at the other, so our notation is in keeping with our conventions from Section \ref{chap:torus}.  We define \emph{adjacency in $\partial \tilde K$} similarly.

Two intersection points $p_0,\tilde p_0\in \partial K\cap \partial \tilde K$ are called \emph{doubly adjacent} if they are adjacent both in $\partial K$ and in $\partial \tilde K$.  Then the notion of sub-arcs of $\partial K$ and $\partial \tilde K$ ``crossing minimally'' which we informally described before is exactly captured by the property of double adjacency.  If $\partial K \cap \partial \tilde K$ meet, then there is always at least one doubly adjacent pair of points in $\partial K \cap \partial \tilde K$.


From now on, we fix a torus parametrization of $\partial K$ and $\partial \tilde K$ in $\bbT = \bbS^1\times \bbS^1$ via $\kappa:\partial K \to \bbS^1$ and $\tilde\kappa:\partial \tilde K \to \bbS^1$.  Let $p_0,\tilde p_0$ be doubly adjacent.  Then we denote $P_0 = (\kappa(p_0),\tilde\kappa(p_0))$ and $\tilde P_0 = (\kappa(\tilde p_0), \tilde\kappa(\tilde p_0))$ as usual, so $P_0\in \bbT$ parametrizes the point $p_0 \in \partial K\cap \partial \tilde K$, similarly $\tilde P_0$ for $\tilde p_0$.  We can now see how we plan to apply induction:

\begin{observation}
\label{obs:construct induction}
Let the torus parametrization $\bbT'$ be obtained from $\bbT$ by deleting $P_0$ and $\tilde P_0$.  Then $\bbT'$ parametrizes a pair $\partial K'$ and $\partial \tilde K'$, where $K'$ and $\tilde K'$ are compact Jordan domains in transverse position, meeting two times fewer than do $K$ and $\tilde K$.
\end{observation}

\noindent Thus our strategy will be as follows: first, apply the induction hypothesis to $\bbT'$, to get a faithful indexable $\gamma'$ so that $w(\gamma') = \eta(\phi') \ge 0$; then, ``reinsert'' the points $P_0$ and $\tilde P_0$, in such a way that the fixed-point index is preserved or increased.


\subsection{Partitioning $\bbT$}

We will refer to the collection of the sets $\bbS^1 \times \{\tilde\kappa(\tilde z_i)\}$ and $\{\kappa(z_i)\}\times \bbS^1$ for $i=1,2,3$ as our \emph{grid lines}, and to their pairwise intersection points as our \emph{lattice points}.  Furthermore we refer to the points $(\kappa(z_i),\tilde\kappa(\tilde z_i))$, for $i=1,2,3$, as our \emph{constraint lattice points}.  If we draw our torus parametrization with a lattice point chosen to be the base point, then the grid lines actually divide the drawing into a grid, with nine disjoint rectangular cells in total.  We will refer to these as our \emph{lattice cells}.  Within any lattice cell there are well-defined left, right, up, and down directions.

From now on, whenever we draw the torus parametrization $\bbT$, we will always pick one of the lattice points for the base point.  Suppose for the rest of this paragraph that we have fixed such a drawing.  Then we may refer to the lattice cells according to their position in this drawing in the natural way, for example we may refer to the bottom-left lattice cell.  We denote the lattice cells by $C_\uparrow, C_\nearrow, C_\to, \ldots, C_\gets, C_\nwarrow$ in the natural way, and write $C_\bullet$ to denote the central lattice cell.  Then the union of the closed upper-left, middle-left, and lower-left lattice cells is called the \emph{left lattice column}.  We define the \emph{central} and \emph{right lattice columns} analogously, as well as the \emph{bottom}, \emph{central}, and \emph{top lattice rows}.  We again emphasize that these descriptors depend crucially on the choice of base point for the drawing of the torus parametrization.

As usual, we denote $P_i = (\kappa(p_i),\tilde\kappa(p_i))$ and $\tilde P_i = (\kappa(\tilde p_i),\tilde \kappa(\tilde p_i))$, where $\{p_i,\tilde p_i\} = \partial K \cap \partial \tilde K$, with the same notational conventions as in Section \ref{chap:torus}.  We will apply Lemma \ref{prop:computing index from torus} frequently.  Before moving on to the next part of the proof, we make an observation that will simplify things later:

\begin{observation}
\label{obs:wrap}
Suppose that all of the $P_i$ and $\tilde P_i$ lie in a single lattice row, or in a single lattice column.  Then Lemma \ref{lem:3p} holds, without the need for induction.
\end{observation}


\begin{figure}
\centering
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\caption[The situation if all $P_i$ and $\tilde P_i$ lie in a single lattice row]
{
\label{fig:first steps}
{\bf The situation if all $P_i$ and $\tilde P_i$ lie in a single lattice row.}  Here we have supposed that they lie in the bottom lattice row, in which case they in fact lie in some open rectangle $R_0$ contained in the bottom lattice row.
}
\end{figure}


\noindent To see why, we walk through Figure \ref{fig:first steps}.  We suppose that all of the $P_i$ and $\tilde P_i$ lie in a single lattice row, call it $R$.  The same argument will work if they all lie in a single lattice column.  Without loss of generality, we have drawn the torus parametrization so that $R$ is the bottom lattice row, and so that the base point is a constraint lattice point.  We may do so by picking the correct constraint lattice point to serve as our base point.  Then the other two constraint lattice points are the ones along the diagonal, drawn in the figure as dots.  As usual, a faithful indexable homeomorphism $\phi:\partial K \to \partial \tilde K$ can be parametrized by a ``strictly increasing'' curve
\begin{itemize}
\item from the bottom-left corner of the bottom-left cell,
\item to the top-right corner of the top-right cell,
\item passing through the two other constraint lattice points, and
\item missing all of the $P_i$ and $\tilde P_i$.
\end{itemize}
Thus the $\gamma$ we have drawn in Figure \ref{fig:first steps} parametrizes such a homeomorphism.  Furthermore, because there are finitely many $P_i$ and $\tilde P_i$, and all of them lie in the bottom lattice row $R$, in fact they lie in an open rectangle $R_0$ as depicted in Figure \ref{fig:first steps}.  Thus $\eta(\phi) \ge 0$ by Lemma \ref{prop:computing index from torus}.





\subsection{The adjacency box}

Let $p_0$ and $\tilde p_0$ be adjacent in $\partial K$, and let $a_0$ denote the arc connecting $p_0$ and $\tilde p_0$ along $\partial K$ which contains no other intersection points $\partial K \cap \partial \tilde K$.  Thus $a_0$ is equal to one of $[p_0\to \tilde p_0]_{\partial K}$ and $[\tilde p_0\to p_0]_{\partial K}$.  There is a unique such $a_0$ so long as $\partial K$ and $\partial \tilde K$ meet at least four times, as is our running assumption.  We call this $a_0$ the \emph{short adjacency arc of $p_0,\tilde p_0$ in $\partial K$}, and define the \emph{short adjacency arc $\tilde a_0$ of $p_0,\tilde p_0$ in $\partial \tilde K$} similarly.

Continuing with the notation of the last paragraph, let $a$ be obtained from $a_0$ by extending it slightly on both ends, so that $a \subset \partial K$ is a closed Jordan arc containing $p_0$ and $\tilde p_0$, but no other points of $\partial K\cap \partial \tilde K$, and no constraint point $z_i$ which was not already contained in $a_0$.  There is a topologically unique such $a$.  Then we say that $a$ is the \emph{adjacency arc of $p_0,\tilde p_0$ in $\partial K$} of $p_0,\tilde p_0$.  We define the \emph{adjacency arc $\tilde a$ of $p_0,\tilde p_0$ in $\partial \tilde K$} similarly.

Again continuing with the notation of before, let $A = \kappa(a)\times \tilde\kappa(\tilde a) \subset \bbT = \bbS^1\times \bbS^1$ be the closed rectangle so that $(x,\tilde x)\in A$ if and only if $\kappa\inv(x)\in a $ and $\tilde\kappa\inv(\tilde x)\in \tilde a $.  Because $a$ and $\tilde a$ are essentially unique, we get also that there is a topologically unique such $A$.  Then $A$ is called the \emph{adjacency box} of $p_0,\tilde p_0$.



\subsection{Enumerating the possible locations of the adjacency box}


Pick doubly adjacent $p_0,\tilde p_0$, having adjacency arcs $a_0,\tilde a_0$ and let $A = \kappa(a_0)\times \tilde\kappa(\tilde a_0)\subset \bbT$ be their adjacency box.  Let $b_0 = \partial K \setminus a_0$ and $\tilde b_0 = \partial \tilde K \setminus b_0$, and set $B = \kappa(b_0)\times \tilde\kappa(\tilde b_0) \subset \bbT$.  Then the parametrization in $\bbT$ of every point in $\partial K \cap \partial \tilde K \setminus \{p_0,\tilde p_0\}$ lies $B$.  However, we have basically no \emph{a priori} information about how these points are arranged in $B$.  However:

\begin{observation}
Suppose that $z_1,z_2,z_3 \in a_0$.  Then $b_0\times \bbS^1\supset B$ is contained in a lattice column, regardless of which lattice point is chosen for the base point of our drawing of the torus parametrization.  Thus Lemma \ref{lem:3p} holds by Observation \ref{obs:wrap}.
\end{observation}

\noindent Thus we may assume without loss of generality that some $z_i$ does not lie in $a$.  Similarly, one of the $\tilde z_i$ lies outside of $\tilde a$.  This allows us to conclude the following:

\begin{observation}
\label{obs:left col}
We may suppose without loss of generality that $A$ meets the left lattice column.
\end{observation}

\noindent To see why, let $z_i\in z_1,z_2,z_3$ be so that $z_i \not \in a_0$.  Then traverse $\partial K$ starting from $z_i$, positively, until arriving at the initial endpoint of $a_0$.  Let $z_j$ be the last one of $z_1,z_2,z_3$ crossed during this traversal.  Then we may draw the torus parametrization using $(\kappa(z_j),\tilde\kappa(\tilde z_j))$ as the base point.

After relabeling if necessary, we suppose from now on that our torus parametrization is drawn with $(\kappa(z_1),\tilde\kappa(\tilde z_1))$ as our base point.  In particular, this means that $z_1\not \in a$, so our adjacency box $A$ does not meet the left, nor the right grid line $\{\kappa(z_1)\}\times \bbS^1$ of our torus parametrization.

Because $A$ is a topological rectangle in the natural way, it has well-defined left, right, up, and down directions.  More precisely, we suppose that the positive orientation on the first coordinate of $\bbT = \bbS^1\times \bbS^1$ goes left-to-right, and on the second coordinate goes down-to-up.  Then:

\begin{observation}
Under our running assumptions, the topological location of $A$ in the torus parametrization is completely determined by two pieces of information:
\begin{itemize}
\item which of the cells of the left lattice column contains the lower-left corner $A_\swarrow$ of $A$, and
\item which of the nine lattice cells contains the upper-right corner $A_\nearrow$ of $A$.
\end{itemize}
\end{observation}

\noindent We will denote the cases for the topological location of $A$ in the following way: let $r$ equal one of $\swarrow, \gets, \nwarrow$, and let $s$ equal one of $\uparrow, \nearrow, \to, \ldots, \gets, \nwarrow, \bullet$.  Then we say that case $(r,s)$ occurs if and only if $A_\swarrow$ lies in lattice cell $C_r$, and $A_\nearrow$ lies in lattice cell $C_s$.  \emph{A priori} this gives us $3\times 9 = 27$ possibilities for $A$.  Fortunately we will see that many of the cases are handled in more or less the same way.  For reference we have depicted the 27 cases in Figure \ref{alcohol}.

\begin{figure}
\centering
\subfloat[\label{partwine}]
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}\quad
\subfloat[\label{partbeer}]
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\caption[The 27 possible topologically distinct locations for the adjacency box $A$]
{\label{alcohol}
{\bf The 27 possible topologically distinct locations for the adjacency box $A$.}  Both torus parametrizations are drawn with $(\kappa(z_1),\tilde\kappa(\tilde z_1))$ as the base point.  In \subref{partwine} we have cases $(r,s)$ where $t\ne \nearrow, \to, \searrow$.  The adjacency box $A$ is shown with a solid boundary for the cases $s = \nwarrow, \gets, \swarrow$, and with a dashed boundary for the cases $r = \uparrow, \bullet, \downarrow$, except for the case $(\swarrow, \uparrow)$, which is shown with a solid boundary.  In \subref{partbeer}, we have cases $(r,s)$ where $s = \nearrow, \to, \searrow$.  Here the adjacency box $A$ is drawn with a solid boundary, except for the case $(\swarrow, \nearrow)$, which is shown with a dashed boundary.  When a sample location for the adjacency box is drawn hanging off of the bottom of $\bbT$, or over the top, we think of it as ``wrapping around'' to the cell vertically opposite, in accordance with the identification of the top and bottom grid lines of $\bbT$.
}
\end{figure}





\subsection{First application of induction, when $\gamma$ misses $A$}

For the remainder of the proof, let $\bbT'$ be the torus parametrization obtained from $\bbT$ by simply deleting $P_0$ and $\tilde P_0$, and let $\gamma'$ be the curve in $\bbT'$ obtained inductively as described in Observation \ref{obs:construct induction}, so that in particular $w(\gamma') \ge 0$.  We wish to construct $\gamma$ in $\bbT$ from $\gamma'$, so that $w(\gamma) \ge w(\gamma') \ge 0$.  In comparing $w(\gamma)$ with $w(\gamma')$, the germane issue is the effect of the presence of $P_0,\tilde P_0$ in $\bbT$ compared to their absence from $\bbT'$.  In light of Lemma \ref{prop:computing index from torus}, there are precisely two ways that reinserting these points can change the fixed-point indices under consideration:
\begin{itemize}
\item First, they may affect the winding numbers $\omega(\partial K', \tilde z'_1)$ and $\omega(\partial \tilde K', z'_1)$ as compared to $\omega(\partial K, \tilde z_1)$ and $\omega(\partial \tilde K, z_1)$.  Here $K', \tilde K'$ are the closed Jordan domains parametrized by the torus parametrization $\bbT'$.
\item Second, they may affect the sum $ - \nump_\downarrow(z_1,\gamma) + \numq_\downarrow(z_1,\gamma)$, equivalently the sum $\nump_\uparrow(z_1,\gamma) - \numq_\uparrow(z_1,\gamma)$, as compared to the respective sums in $\bbT'$.
\end{itemize}

\noindent We have the following nice observation:

\begin{observation}
\label{obs:no left}
The winding numbers $\omega(\partial K, \tilde z_1)$ and $\omega(\partial K', \tilde z'_1)$ are equal.
\end{observation}

\noindent This is because $\omega(\partial K,\tilde z_1)$ is determined completely by whether we arrive first at a $\kappa(p_i)$ or a $\kappa(\tilde p_i)$ as we traverse the first coordinate $\bbS^1$ of $\bbT$ positively starting at $\kappa(z_1)$, and this is invariant under reinserting $P_0,\tilde P_0$, as the points $p_0$ and $\tilde p_0$ occur immediately after one another, in some order, as we traverse $\partial K$ starting at $z_1$.\noindent

The following claim will prove useful:

Next:

\begin{claim}
\label{obs:no meetingz}
Suppose that $\gamma = \gamma'$ does not meet $A$.  Then $w(\gamma) = w(\gamma') \ge 0$.
\end{claim}


\begin{proof}
There are two cases, handled differently:

\begin{mermaidcase}
\label{mermaid1}
The box $A$ does not meet the bottom/top grid line $\bbS^1 \times \{\tilde\kappa(\tilde z_1)\}$.
\end{mermaidcase}

\noindent In this case, similarly to Observation \ref{obs:no left}, we get that $\omega(\partial \tilde K, z_1) = \omega(\partial \tilde K', z'_1)$.  Also, because $\gamma = \gamma'$ does not meet $A$, we have that $P_0$ and $\tilde P_0$ either both lie above $\gamma$ in $\bbT$, or both lie below it.  Thus $ - \nump_\downarrow(z_1,\gamma) + \numq_\downarrow(z_1,\gamma) = - \nump_\downarrow(z_1,\gamma') + \numq_\downarrow(z_1,\gamma')$, completing the proof of the claim by Lemma \ref{prop:computing index from torus}.

\begin{mermaidcase}
\label{mermaid2}
The box $A$ crosses the bottom/top grid line $\bbS^1 \times \{\tilde\kappa(\tilde z_1)\}$.
\end{mermaidcase}

\noindent In this case, the bottom/top grid line $\bbS^1\times \{\tilde \kappa(\tilde z_1)\}$ cuts $A$ into two connected components $A_\downarrow$ and $A_\uparrow$, one of which contains $P_0$ and the other of which contains $\tilde P_0$.  The ``upper half'' of $A$, which we have denoted $A_\uparrow$, actually lies in the bottom lattice row of $\bbT$, and furthermore lies beneath $\gamma$ because $\gamma$ does not meet $A$.  Similarly $A_\downarrow$ lies in the upper lattice row of $\bbT$ and above $\gamma$.  There are two sub-cases:

\begin{mermaidsubcase}
\label{oceans}
$P_0\in A_\downarrow$, $\tilde P_0\in A_\uparrow$
\end{mermaidsubcase}

\noindent In this case $- \nump_\downarrow(z_1,\gamma) + \numq_\downarrow(z_1,\gamma)$ in $\bbT$ is $1$ less than $-\nump_\downarrow(z_1,\gamma') + \numq_\downarrow(z_1,\gamma')$ in $\bbT'$.  On the other hand   $\omega(\partial \tilde K, z_1) = 1$ and $\omega(\partial \tilde K',z'_1) = 0$.  Thus the total change between indices in the two settings cancels out and we get $w(\gamma) = w(\gamma')$ by Lemma \ref{prop:computing index from torus}.

\begin{mermaidsubcase}
$P_0\in A_\uparrow$, $\tilde P_0\in A_\downarrow$
\end{mermaidsubcase}

\noindent This sub-case is handled in the same way as Sub-case \ref{oceans}, except that this time $\omega(\partial \tilde K, z_1) + 1 = \omega(\partial \tilde K',z'_1)$ and $- \nump_\downarrow(z_1,\gamma) = - \nump_\downarrow(z_1,\gamma') + 1$.  This completes the proof of Claim \ref{obs:no meetingz}.
\end{proof}

\noindent Much of the rest of the proof of Lemma \ref{lem:3p} will be spent trying to reduce to the situation handled by Claim \ref{obs:no meetingz}.




\subsection{Moving $P_0,\tilde P_0$, or adjusting $A$}



We continue with the notation from before.  \emph{A priori} there are exactly four topologically distinct ways that $P_0$ and $\tilde P_0$ may be arranged in the adjacency box $A$ of $p_0,\tilde p_0$, as depicted in Figure \ref{fig:four brothers}.



\begin{figure}
\centering
\subfloat[\label{broa}]
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\subfloat[\label{brob}]
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\subfloat[\label{broc}]
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\caption[The \emph{a priori} possible arrangements of $P_0,\tilde P_0$ in the adjacency box $B$]
{
\label{fig:four brothers}
{\bf The \emph{a priori} possible arrangements of $P_0,\tilde P_0$ in the adjacency box $B$.}
}
\end{figure}



The following observation will be helpful later:

\begin{observation}
\label{obs:move p}
We may move $P_0$ and $\tilde P_0$ as we please within $A$, without changing the topological configuration of $K$, $\tilde K$, and the constraint points, so long as the following two requirements are satisfied:
\begin{itemize}
\item We must not change which of the cases of Figure \ref{fig:four brothers} occurs.
\item Each of $P_0$ and $\tilde P_0$ must remain in the same respective connected component of $A\setminus \cup\{\mathrm{grid\ lines}\}$.
\end{itemize}
\end{observation}

\noindent During the eventual induction step, it will often be desirable to move the points $P_0$ and $\tilde P_0$.  By Observation \ref{obs:move p} we may do so, within limits, without affecting any relevant aspects of the situation.  It will also be useful to note the following:

\begin{observation}
\label{obs:diff comp}
The points $P_0$ and $\tilde P_0$ lie in different connected components of $A\setminus \cup\{\mathrm{grid\ lines}\}$, unless $A = A\setminus \cup\{\mathrm{grid\ lines}\}$.
\end{observation}

\noindent This is by our construction of $A$: if Observation \ref{obs:diff comp} fails, then necessarily $A$ meets all three vertical grid lines, thus $z_1 \in a$, contradicting our running assumptions.

Next, recall that we had latitude in choosing the adjacency box $A$ of $p_0,\tilde p_0$.  With this in mind, we make the following observation:

\begin{observation}
\label{obs:replace A}
We may replace $A$ by any topological rectangle contained in $A$, having sides ``parallel to'' those of $A$ in the natural sense, so long as the interior of this new rectangle continues to contain $P_0$ and $\tilde P_0$.  This replacement preserves the germane features of $A$ in that every argument we have made in this proof so far continues to work under this replacement.
\end{observation}


We now apply these observations to describe how to complete the proof in a crucial case:


\begin{claim}
\label{claim:monster1}
Suppose that $\gamma$ meets only a single connected component of the set $A\setminus \cup\{\mathrm{grid\ lines}\}$, and furthermore that this component contains a corner of $A$.  Then we may reduce to Claim \ref{obs:no meetingz}, completing the proof of Lemma \ref{lem:3p}.
\end{claim}


\begin{proof}
First note that $\gamma\cap A$ has only a single connected component, because $\gamma$ is ``strictly increasing'' in $\bbT$, thus also in the single connected component of $A\setminus \cup\{\mathrm{grid\ lines}\}$ which $\gamma$ meets.  This also implies that if the component of $A\setminus \cup\{\mathrm{grid\ lines}\}$ which meets $\gamma$ contains a corner of $A$, then it must contain either the upper-left corner $A_{\nwarrow}$ of $A$, or the lower-right corner $A_\searrow$ of $A$, or both of these.  Suppose it contains the upper-left corner.  The argument will be the same if it contains the lower-right corner.  A potential drawing of the situation is depicted in Figure \ref{fig:shells}.


\begin{figure}
\centering
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\caption[Moving $P_0$ and $\tilde P_0$ to find a replacement for $A$ not meeting $\gamma$]
{
\label{fig:shells}
{\bf Moving $P_0$ and $\tilde P_0$ to find a replacement for $A$ not meeting $\gamma$.}  Here $(x_1,\tilde x_1)$ is the point in $\bbT$ where $\gamma$ enters $A$, and $(x_2,\tilde x_2)$ is the point where $\gamma$ exits $A$.
}
\end{figure}

Then it is clear that within the constraints described in Observation \ref{obs:move p} it is possible to move $P_0$ and $\tilde P_0$ if necessary so that both lie to the right of $(x_2,\tilde x_2)$ and below $(x_1,\tilde x_1)$, and then it is clear how to choose $A^*$ our replacement rectangle for $A$.  This completes the proof of Claim \ref{claim:monster1} by Observation \ref{obs:replace A}.
\end{proof}






\subsection{Restricting the possibilities further}


Let $p_0,\tilde p_0$ be doubly adjacent, with short adjacency arcs $a_0$ and $\tilde a_0$.  The unoriented union $a_0\cup \tilde a_0$ is an unoriented Jordan curve.  The orientations on $a_0$ and $\tilde a_0$ may or may not agree, and may or may not induce the positive orientation on $a_0\cup \tilde a_0$.  Let $U$ be the open Jordan domain bounded by $a_0\cup \tilde a_0$.  Then $U$ is called the \emph{adjacency domain} of $p_0,\tilde p_0$.  We make the following observation:

\begin{observation}
\label{obs:nice da}
There exist at least two distinct pairs of doubly adjacent points in $\partial K \cap \partial \tilde K$ whose adjacency domains are contained in $\tilde K$.
\end{observation}

\noindent This is easy to show by arguments similar to those given in the proof of Lemma \ref{uniqa}.  Then we get the following as an immediate corollary:

\begin{observation}
\label{obs:nice direction}
There exist at least two distinct pairs of doubly adjacent points in $\partial K\cap \partial \tilde K$, so that if $p_0,\tilde p_0$ is such a pair, then their short adjacency arc in $\partial K$ is $a_0 = [p_0\to \tilde p_0]_{\partial K}$.
\end{observation}

\noindent We may restate Observation \ref{obs:nice direction} in the following way:

\begin{observation}
\label{obs:nice sides}
There exist at least two distinct pairs of doubly adjacent points in $\partial K \cap \partial \tilde K$, so that if $P_0,\tilde P_0\in \bbT$ parametrize one of these pairs, then $P_0$ lies to the left of $\tilde P_0$ in their adjacency box $A$.
\end{observation}

\noindent Thus without loss of generality, for the remainder of the proof, we restrict our attention to the arrangements in Figures \ref{broa}, \ref{brob}, ignoring the possibilities of Figures \ref{broc}, \ref{brod}.


\subsection{Completing the proof if a constraint lattice point lies in the adjacency box}


\begin{claim}
\label{claim:monster2}
Suppose that $A$ contains a constraint lattice point.  Then we may move $P_0$ and $\tilde P_0$ around in such a way that we get $w(\gamma)\ge 0$, completing the proof of Lemma \ref{lem:3p}.
\end{claim}

\begin{proof}
The following will be useful to keep in mind:

\begin{observation}
The curve $\gamma$ cannot cross grid lines, except at constraint lattice points.
\end{observation}

\noindent We prove Claim \ref{claim:monster2} in four cases:

\begin{latecase}
\label{latecase1}
The arrangement in Figure \ref{broa} occurs, and $A$ does not meet the bottom/top grid line of $\bbT$.
\end{latecase}

\noindent It is not hard to see that then $P_0$ and $\tilde P_0$ are on different sides of $\gamma$ in our drawing of $\bbT$, and furthermore that $P_0$ lies above $\gamma$ and $\tilde P_0$ below it.  Then, arguing straightforwardly as in the proof of Claim \ref{obs:no meetingz}, Case \ref{mermaid1}, we get $w(\gamma) = w(\gamma') + 1$.\vspace{10pt}

For the remaining cases, the following will be the key observation:

\begin{observation}
\label{obs:why am i doing this}
Suppose that $P_0$ lies in a connected component of $A \setminus \cup \{ \mathrm{grid\ lines} \}$ through which $\gamma$ passes.  Then we may decide whether $P_0$ lies above or below $\gamma$ in our drawing of $\bbT$, in the sense that we may move $P_0$ within the constraints given in Observation \ref{obs:move p} to arrange either situation.  The same is true for $\tilde P_0$.
\end{observation}

\begin{latecase}
\label{latecase2}
The arrangement in Figure \ref{broa} occurs, and $A$ crosses the bottom/top grid line of $\bbT$.
\end{latecase}



\noindent We argue similarly to the proof of Claim \ref{obs:no meetingz}, Case \ref{mermaid2}.  In this case it can always be arranged that $P_0$ lies below $\gamma$ and that $\tilde P_0$ lies above it.  There are three cases to check, and in each it is clear how to proceed.  The cases are shown in Figure \ref{fig:keyboardme0}.  Again we get $w(\gamma) = w(\gamma') + 1$.

\begin{figure}
\centering
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\caption[The three possibilities for $A$ when $A$ contains a constraint lattice point and meets the bottom/top grid line, and Figure \ref{broa} occurs]
{
\label{fig:keyboardme0}
{\bf The three possibilities for $A$ when $A$ contains a constraint lattice point and meets the bottom/top grid line, and Figure \ref{broa} occurs.}
}
\end{figure}

\begin{latecase}
\label{latecase3}
The arrangement in Figure \ref{brob} occurs, and $A$ does not meet the bottom/top grid line of $\bbT$.
\end{latecase}

\noindent As per our usual arguments, we have $\omega(\partial \tilde K, z_1) = \omega(\partial \tilde K', z'_1)$ and $\omega(\partial K, \tilde z_1) = \omega(\partial K', \tilde z'_1)$.  Thus we will be done if we can argue that we may arrange so that $P_0$ and $\tilde P_0$ lie on the same side of $\gamma$ in our drawing of $\bbT$.  There are five cases to check, and in each it is clear how to proceed.  Figure \ref{fig:keyboardme1} shows each case.



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\subfloat[\label{fig:keyboardme1c}$(\swarrow,\nearrow,\brob)$]
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\subfloat[\label{fig:keyboardme1e}$(\gets,\nearrow,\brob)$]
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\caption[The five possibilities for $A$ when $A$ contains a constraint lattice point and does not meet the bottom/top grid line, and Figure \ref{brob} occurs]
{
\label{fig:keyboardme1}
{\bf The five possibilities for $A$ when $A$ contains a constraint lattice point and does not meet the bottom/top grid line, and Figure \ref{brob} occurs.}
}
\end{figure}


\begin{latecase}
\label{latecase4}
The arrangement in Figure \ref{brob} occurs, and $A$ crosses the bottom/top grid line of $\bbT$.
\end{latecase}

\noindent In this case we have that $\omega(\partial K, \tilde z_1) = \omega(\partial K', \tilde z'_1)$ as per Observation \ref{obs:no left}, but unfortunately $\omega(\partial \tilde K, z_1) = \omega(\partial \tilde K', z'_1) - 1$.  Thus we will be done if we can argue that we may arrange so that $P_0$ lies above $\gamma$, and $\tilde P_0$ below it, in our drawing of $\bbT$.  There are only three cases to check.  Figure \ref{fig:keyboardme2} shows each.  This completes the proof of Claim \ref{claim:monster2}.
\end{proof}


\begin{figure}
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\subfloat[\label{fig:keyboardme2b}$(\nwarrow,\to,\brob)$]
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}\qquad
\subfloat[\label{fig:keyboardme2c}$(\gets,\searrow,\brob)$]
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\caption[The three possibilities for $A$ when $A$ contains a constraint lattice point and does not meet the bottom/top grid line, and Figure \ref{brob} occurs]
{
\label{fig:keyboardme2}
{\bf The three possibilities for $A$ when $A$ contains a constraint lattice point and does not meet the bottom/top grid line, and Figure \ref{brob} occurs.}
}
\end{figure}





\subsection{Completing the proof of Lemma \ref{lem:3p}}

We summarize the situation so far.  There are $3\times 9 = 27$ possibilities for the location of $A$ in $\bbT$.  Most of the cases fall into one of the following three categories.  For a given type $(s,t)$ of $A$, it is quick and easy to check the conditions that define membership in the categories.  We also refer the reader back to Figure \ref{alcohol}, where each of the 27 cases is drawn.

\begin{category}
\label{tech1}
The intersection $A \cap (C_\swarrow \cup C_\bullet \cup C_\nearrow)$ is empty.
\end{category}

\noindent Then we are done by Claim \ref{obs:no meetingz}, because $\gamma$ is restricted to lie in $C_\swarrow \cup C_\bullet \cup C_\nearrow$.  Next:

\begin{category}
\label{tech3}
The set $A\setminus \cup\{ \mathrm{grid\ lines} \}$ has exactly one connected component lying in $C_\swarrow \cup C_\bullet \cup C_\nearrow$, and this connected component contains a corner of $A$.
\end{category}

\noindent Then we are done by Claim \ref{claim:monster1}, again because $\gamma$ is restricted to lie in $C_\swarrow \cup C_\bullet \cup C_\nearrow$.  Finally:

\begin{category}
\label{tech2}
The adjacency box $A$ contains a constraint lattice point.
\end{category}

\noindent This is exactly handled by Claim \ref{claim:monster2}.\medskip

In Figure \ref{table:chair2} we list the $27$ possible locations of $A$, and for each falling into one of these three categories, we indicate which.



\begin{figure}
\centering
\subfloat{
\scalebox{.8}
{%   | r | s | t || technique |
\begin{tabular}{|c|c||c|}
\hline
\multicolumn{1}{|c}{$r$} & \multicolumn{1}{c||}{$s$} & \multicolumn{1}{|c|}{Category}\\
\hline\hline
$\nwarrow$ & $\uparrow$ & \ref{tech1} \\
$\nwarrow$ & $\nearrow$ & \ref{tech2} \\
$\nwarrow$ & $\to$ & \ref{tech3} \\
$\nwarrow$ & $\searrow$ & \ref{tech2} \\
$\nwarrow$ & $\downarrow$ & \ref{tech2} \\
$\nwarrow$ & $\swarrow$ & \ref{tech2} \\
$\nwarrow$ & $\gets$ & \\
$\nwarrow$ & $\nwarrow$ & \ref{tech1} \\
$\nwarrow$ & $\bullet$ & \ref{tech3} \\
\hline
\end{tabular}
}
}\quad
\subfloat{
\scalebox{.8}
{%   | r | s | t || technique |
\begin{tabular}{|c|c||c|}
\hline
\multicolumn{1}{|c}{$r$} & \multicolumn{1}{c||}{$s$} & \multicolumn{1}{|c|}{Category}\\
\hline\hline
$\gets$ & $\uparrow$ & \ref{tech2} \\
$\gets$ & $\nearrow$ & \ref{tech3} \\
$\gets$ & $\to$ &  \\
$\gets$ & $\searrow$ & \ref{tech3} \\
$\gets$ & $\downarrow$ & \\
$\gets$ & $\swarrow$ & \ref{tech2} \\
$\gets$ & $\gets$ & \ref{tech1} \\
$\gets$ & $\nwarrow$ & \ref{tech1} \\
$\gets$ & $\bullet$ & \ref{tech2} \\
\hline
\end{tabular}
}
}\quad
\subfloat{
\scalebox{.8}
{%   | r | s | t || technique |
\begin{tabular}{|c|c||c|}
\hline
\multicolumn{1}{|c}{$r$} & \multicolumn{1}{c||}{$s$} & \multicolumn{1}{|c|}{Category}\\
\hline\hline
$\swarrow$ & $\uparrow$ & \ref{tech3} \\
$\swarrow$ & $\nearrow$ & \ref{tech3} \\
$\swarrow$ & $\to$ & \ref{tech3} \\
$\swarrow$ & $\searrow$ & \ref{tech2} \\
$\swarrow$ & $\downarrow$ & \ref{tech2} \\
$\swarrow$ & $\swarrow$ & \ref{tech2} \\
$\swarrow$ & $\gets$ & \ref{tech2} \\
$\swarrow$ & $\nwarrow$ & \ref{tech2} \\
$\swarrow$ & $\bullet$ & \ref{tech3} \\
\hline
\end{tabular}
}
}
\caption[The cases to check for Lemma \ref{lem:3p}]
{
\label{table:chair2}
{\bf The cases to check for Lemma \ref{lem:3p}.}
}
\end{figure}


Case $(\gets,\downarrow)$ is easily handled in a manner similar to the proof of Claim \ref{claim:monster1}.  If case $(\gets, \to)$ occurs, then our running assumption that $P_0$ lies to the left of $\tilde P_0$ in $A$ implies that $w(\gamma) = w(\gamma') + 1$ by Lemma \ref{prop:computing index from torus} as usual.

We have no way to deal with case $(\nwarrow, \gets)$ directly.  However, recall from Observation \ref{obs:nice sides} that we had two choices for $P_0,\tilde P_0$ for which every step of the proof until this point go through.  Because the adjacency arc $\tilde a_0$ contains two constraint points $\tilde z_i$ if case $(\nwarrow,\gets)$ occurs, it can occur for only one of these two choices of $P_0,\tilde P_0$, so we do not need to handle this case directly.  This completes the proof of Lemma \ref{lem:3p}.\qed




































































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\end{document}