(2.43,-1.4567188)(2.585,-1.5717187)(2.66,-1.6217188)
(2.735,-1.6717187)(2.875,-1.7467188)(2.94,-1.7717187)
(3.005,-1.7967187)(3.175,-1.8567188)(3.28,-1.8917187)
(3.385,-1.9267187)(3.62,-1.9867188)(3.75,-2.0117188)
(3.88,-2.0367188)(4.155,-2.0767188)(4.3,-2.0917187)
(4.445,-2.1067188)(4.735,-2.1267188)(4.88,-2.1317186)
(5.025,-2.1367188)(5.255,-2.1417189)(5.34,-2.1417189)
(5.425,-2.1417189)(5.66,-2.1117187)(5.81,-2.0817187)
(5.96,-2.0517187)(6.165,-2.0017188)(6.22,-1.9817188)
(6.275,-1.9617188)(6.425,-1.8917187)(6.52,-1.8417188)
(6.615,-1.7917187)(6.75,-1.6967187)(6.79,-1.6517187)
(6.83,-1.6067188)(6.91,-1.4767188)(6.95,-1.3917187)
(6.99,-1.3067187)(7.055,-1.0917188)(7.08,-0.96171874)
(7.105,-0.83171874)(7.105,-0.50171876)(7.08,-0.30171874)
(7.055,-0.10171875)(7.015,0.17328125)(7.0,0.24828126)
(6.985,0.32328126)(6.95,0.47328126)(6.93,0.54828125)
(6.91,0.62328124)(6.865,0.80328125)(6.84,0.90828127)
(6.815,1.0132812)(6.77,1.1732812)(6.75,1.2282813)
(6.73,1.2832812)(6.69,1.3832812)(6.67,1.4282813)
(6.65,1.4732813)(6.615,1.5482812)(6.6,1.5782813)
(6.585,1.6082813)(6.555,1.6632812)(6.54,1.6882813)
(6.525,1.7132813)(6.5,1.7532812)(6.47,1.7682812)
$R_{\theta}$$K$$\tilde{K}$$\theta$$w$$z_{\theta}$$\tilde{z}_{\theta}$

Figure 13.  Two convex closed Jordan domains $K$ and $\tilde{K}$ in transverse
position, with boundaries meeting at six points. As $\theta$ varies
positively, the ray $R_{\theta}$ scans around the boundaries of both $K$ and
$\tilde{K}$ positively.

###### Proof.

For the following construction, see Figure 13. Let $w$ be a common interior
point of $K$ and $\tilde{K}$. Let $R_{\theta}$ be the ray emanating from the
point $w$ at an angle of $\theta$ from the positive real direction. Let
$P_{i}$ be the points of $\partial K\cap\partial\tilde{K}$ where $\partial K$
is entering $\tilde{K}$, and let $\tilde{P}_{i}$ be those where
$\partial\tilde{K}$ is entering $K$. Define
$w^{\prime},R^{\prime}_{\theta},P^{\prime}_{i},\tilde{P}^{\prime}_{i}$
analogously for $K^{\prime}$ and $\tilde{K}^{\prime}$. Identify
$\mathbb{S}^{1}$ with the interval $[0,2\pi]$ with its endpoints identified,
and define a homeomorphism $\mathbb{S}^{1}\to\mathbb{S}^{1}$, denoting the
image of $\theta\in[0,2\pi]$ by $\theta^{\prime}$, so that $R_{\theta}$ hits a
point $P_{i}$ if and only if $R^{\prime}_{\theta^{\prime}}$ hits a point
$P^{\prime}_{i}$, similarly for $\tilde{P}_{i}$ and $\tilde{P}^{\prime}_{i}$.
Define homeomorphisms $R_{\theta}\to R^{\prime}_{\theta^{\prime}}$ piecewise
linearly on the components $R_{\theta}\setminus(\partial
K\cup\partial\tilde{K})\to R^{\prime}_{\theta^{\prime}}\setminus(\partial
K^{\prime}\cup\partial\tilde{K}^{\prime})$. Then these homeomorphisms glue to
an orientation-preserving homeomorphism $\mathbb{C}\to\mathbb{C}$ sending
$\\{K,\tilde{K}\\}$ to $\\{K^{\prime},\tilde{K}^{\prime}\\}$. ∎

Lemma 9.2 is very much false if we omit the condition that the Jordan domains
are convex. Which _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 _meanders_ 666Thanks to Thomas Lam for informing us of
the topic of meander theory. We are fortunate that our setting is nice enough
that a statement like Lemma 9.2 is possible. The clean construction we use in
our proof is due to Nic Ford and Jordan Watkins.

We now take a moment to introduce some notation we use throughout the rest of
the article. Let $\gamma$ be an oriented Jordan curve. Let $a,b\in\gamma$ be
distinct. Then $[a\to b]_{\gamma}$ is the oriented closed sub-arc of $\gamma$
starting at $a$ and ending at $b$. Then for example $[a\to
b]_{\gamma}\cap[b\to a]_{\gamma}=\\{a,b\\}$ and $[a\to b]_{\gamma}\cup[b\to
a]_{\gamma}=\gamma$.

Throughout the rest of this section, let $\\{A,B\\}$ and
$\\{\tilde{A},\tilde{B}\\}$ be pairs of overlapping closed disks in general
position. We label $\\{u,v\\}=\partial A\cap\partial B$ as in the preceding
section, see Figure 11 on p. 11 for a reminder. Label $\tilde{u}$ and
$\tilde{v}$ analogously. We denote $E=A\cap B$ and
$\tilde{E}=\tilde{A}\cap\tilde{B}$, and loosely refer to these as _eyes_. The
rest of this section consists of the proofs of an assortment of lemmas about
these disks, which we give without further comment.

######  Lemma 9.3.

The Jordan curves $\partial E$ and $\partial\tilde{E}$ meet exactly 0, 2, 4,
or 6 times.

###### Proof.

That they meet an even number of times is a consequence of the general
position hypothesis. There is an immediate upper bound of 8 meeting points
because each of $\partial E$ and $\partial\tilde{E}$ is the union of two
circular arcs. Suppose for contradiction that $\partial E$ and
$\partial\tilde{E}$ meet 8 times. Thus every meeting point of one of the
circles $\partial A$ and $\partial B$ with one of $\partial\tilde{A}$ and
$\partial\tilde{B}$ lies in $\partial E\cap\partial\tilde{E}$. It follows that
$\partial(A\cup B)$ does not meet $\partial(\tilde{A}\cup\tilde{B})$. But
these are Jordan curves, so then we have either that one of $A\cup B$ and
$\tilde{A}\cup\tilde{B}$ contains the other, or that they are disjoint. They
cannot be disjoint because $\partial E$ and $\partial\tilde{E}$ meet (8 times)
by hypothesis, so suppose without loss of generality that
$\tilde{A}\cup\tilde{B}$ is contained in $A\cup B$, in particular in its
interior by the general position hypothesis. Then the sub-arc $\partial
E\cap\partial A$ must enter the region $\tilde{A}\cup\tilde{B}$ somewhere, so
that it may intersect $\partial\tilde{E}$, a contradiction. ∎

$\tilde{E}$$v$$\tilde{u}$$E$$\tilde{v}$$u$

(a)

$\tilde{E}$$v$$\tilde{v}$$E$$\tilde{u}$$u$

(b)

Figure 14.  The possible relevant topological configurations for two generally
positioned eyes whose boundaries meet at six points. There are two remaining
possibilities, not depicted, obtained by simultaneously swapping $u$ with $v$
and $\tilde{u}$ with $\tilde{v}$, which are irrelevant by symmetry .

######  Lemma 9.4.

Suppose that $\partial E$ and $\partial\tilde{E}$ meet 6 times, that
$A\setminus B$ and $\tilde{A}\setminus\tilde{B}$ meet, and that $B\setminus A$
and $\tilde{B}\setminus\tilde{A}$ meet. Then
$\\{E,u,v,\tilde{E},\tilde{u},\tilde{v}\\}$ are in one of the two topological
configurations represented in Figure 14, up to possibly simultaneously
swapping $u$ with $v$ and $\tilde{u}$ with $\tilde{v}$.

###### Proof.

By Lemma 9.2, if $\partial E$ and $\partial\tilde{E}$ meet 6 times then they
are in the topological configuration shown in Figure 15. We denote by
$\epsilon_{i}$ the connected components of $\partial
E\setminus\partial\tilde{E}$, and by $\tilde{\epsilon}_{i}$ the connected
components of $\partial\tilde{E}\setminus\partial E$, labeled as in Figure 15.
We consider the indices of the $\epsilon_{i}$ and $\tilde{\epsilon}_{i}$ only
modulo $6$. For example, we write $\epsilon_{2+5}=\epsilon_{1}$.

$\tilde{\epsilon}_{2}$$\epsilon_{3}$$\tilde{\epsilon}_{4}$$\epsilon_{5}$$\tilde{\epsilon}_{6}$$\epsilon_{1}$$\epsilon_{2}$$\tilde{\epsilon}_{5}$$\epsilon_{6}$$\tilde{\epsilon}_{1}$$\epsilon_{4}$$\tilde{\epsilon}_{3}$

Figure 15.  The components of $\partial E\setminus\partial\tilde{E}$ and
$\partial\tilde{E}\setminus\partial E$ for two transversely positioned convex
closed Jordan domains $E$ and $\tilde{E}$ meeting at six points. The solid
curve represents $\partial E$, and the dashed curve represents
$\partial\tilde{E}$.

Proposition 8.2 allows us to make the following observation:

######  Observation 9.5.

Neither $\tilde{u}$ nor $\tilde{v}$ may lie in $E$, and neither $u$ nor $v$
may lie in $\tilde{E}$.

To see why, note that if $\partial E$ and $\partial\tilde{E}$ meet six times,
then by the pigeonhole principle at least one of $\partial A$ and $\partial B$
must meet $\partial\tilde{E}$ at least three times. Thus at least one of
$\spadesuit$g and $\clubsuit$g must occur. Thus $\tilde{u}$ and $\tilde{v}$
lie outside of at least one of $A$ and $B$, but $E=A\cap B$, thus neither
$\tilde{u}$ nor $\tilde{v}$ lies in $E$. The other part follows identically.

Thus we may assume that $u\in\epsilon_{1}$. Then $v$ lies along $\epsilon_{3}$
or $\epsilon_{5}$. By relabeling the $\epsilon_{i}$ and switching the roles of
$u$ and $v$ as necessary, we may assume that $v\in\epsilon_{3}$. Our proof
will be done once we show that neither $\tilde{u}$ nor $\tilde{v}$ may lie
along $\tilde{\epsilon}_{2}$. Suppose for contradiction that $\tilde{u}$ lies
along $\tilde{\epsilon}_{2}$. Then $\tilde{v}$ lies along either
$\tilde{\epsilon}_{4}$ or $\tilde{\epsilon}_{6}$. If
$\tilde{v}\in\tilde{\epsilon}_{4}$, then the circular arc $[v\to u]_{\partial
E}$ meets the circular arc $[\tilde{v}\to\tilde{u}]_{\partial\tilde{E}}$ three
times, a contradiction. Similarly, if $\tilde{v}\in\tilde{\epsilon}_{6}$, then
the circular arc $[v\to u]_{\partial E}$ meets the circular arc
$[\tilde{u}\to\tilde{v}]_{\partial\tilde{E}}$ three times, also a
contradiction. Thus $\tilde{u}\not\in\tilde{\epsilon}_{2}$. The argument is
the same if we had initially let $\tilde{v}\in\tilde{\epsilon}_{2}$. ∎

######  Lemma 9.6.

The following four statements hold.

  1. (1)

If $[\tilde{u}\to\tilde{v}]_{\partial\tilde{E}}\subset A$ and
$[\tilde{v}\to\tilde{u}]_{\partial\tilde{E}}$ meets $\partial A$, then
$B\setminus A$ and $\tilde{B}\setminus\tilde{A}$ are disjoint.

  2. (2)

If $[\tilde{v}\to\tilde{u}]_{\partial\tilde{E}}\subset B$ and
$[\tilde{u}\to\tilde{v}]_{\partial\tilde{E}}$ meets $\partial B$, then
$A\setminus B$ and $\tilde{A}\setminus\tilde{B}$ are disjoint.

  3. (3)

If $[u\to v]_{\partial E}\subset\tilde{A}$ and $[v\to u]_{\partial E}$ meets
$\partial\tilde{A}$, then $\tilde{B}\setminus\tilde{A}$ and $B\setminus A$ are
disjoint.

  4. (4)

If $[v\to u]_{\partial E}\subset\tilde{B}$ and $[u\to v]_{\partial E}$ meets
$\partial\tilde{B}$, then $\tilde{A}\setminus\tilde{B}$ and $A\setminus B$ are
disjoint.

###### Proof.

We prove only (1), as (2), (3), (4) are symmetric restatements of it. Suppose
the hypotheses of (1) hold. Then both $\tilde{u}$ and $\tilde{v}$ lie in $A$.
Thus the circular arc $[\tilde{v}\to\tilde{u}]_{\partial\tilde{E}}$ meets
$\partial A$ either exactly twice or not at all, in fact exactly twice because
of the hypotheses. But
$[\tilde{v}\to\tilde{u}]_{\partial\tilde{E}}=[\tilde{v}\to\tilde{u}]_{\partial\tilde{B}}$.
Thus $[\tilde{u}\to\tilde{v}]_{\partial\tilde{B}}$ does not meet $\partial A$,
and has its endpoints lying in $A$, so
$[\tilde{u}\to\tilde{v}]_{\partial\tilde{B}}\subset A$.

From our definitions of $\tilde{u}$ and $\tilde{v}$, it is easy to check that
$\partial(\tilde{B}\setminus\tilde{A})$ is the union of the arcs
$[\tilde{u}\to\tilde{v}]_{\partial\tilde{B}}$ and
$[\tilde{u}\to\tilde{v}]_{\partial\tilde{E}}$. It follows that
$\partial(\tilde{B}\setminus\tilde{A})$ is contained in $A$. Thus
$\tilde{B}\setminus\tilde{A}$ is contained in $A$, and so is disjoint from
$B\setminus A$. ∎

$\tilde{E}$$E$$u$$v$$\tilde{v}$$\tilde{u}$

Figure 16.  A topological configuration of two eyes which guarantees that
$A\setminus B$ and $\tilde{A}\setminus\tilde{B}$ do not meet, and that
$B\setminus A$ and $\tilde{B}\setminus\tilde{A}$ do not meet.

######  Lemma 9.7.

Suppose $\\{E,u,v,\tilde{E},\tilde{u},\tilde{v}\\}$ are in the topological
configuration depicted in Figure 16. Then $A\setminus B$ and
$\tilde{A}\setminus\tilde{B}$ do not meet, and $B\setminus A$ and
$\tilde{B}\setminus\tilde{A}$ do not meet.

###### Proof.

The curves $\partial\tilde{A}\setminus\partial\tilde{E}$ and
$\partial\tilde{B}\setminus\partial\tilde{E}$ both have $\tilde{u}$ and
$\tilde{v}$ as their endpoints and otherwise avoid $\tilde{E}$. Thus each must
cross $\partial E$ twice. These four crossings together with the points
$\partial E\cap\partial\tilde{E}$ accounts for all eight possible intersection
points between $\partial A\cup\partial B$ and
$\partial\tilde{A}\cup\partial\tilde{B}$. Thus the arc
$[\tilde{v}\to\tilde{u}]_{\partial\tilde{E}}$ does not meet $\partial B$.
Because this arc meets $B\supset E$, we conclude that
$[\tilde{v}\to\tilde{u}]_{\partial\tilde{E}}$ is contained in $B$. Note that
$[\tilde{u}\to\tilde{v}]_{\partial\tilde{E}}$ meets $\partial B$. Thus by part
(1) of Lemma 9.7 we get that $A\setminus B$ and $\tilde{A}\setminus\tilde{B}$
are disjoint. That $B\setminus A$ and $\tilde{B}\setminus\tilde{A}$ are
disjoint follows by symmetry. ∎

######  Lemma 9.8.

Suppose $u\in\tilde{E}$ and $\tilde{u}\in E$. Then $A\setminus B$ and
$\tilde{A}\setminus\tilde{B}$ do not meet, or $B\setminus A$ and
$\tilde{B}\setminus\tilde{A}$ do not meet.

###### Proof.

Suppose for contradiction that $u\in\tilde{E}$ and $\tilde{u}\in E$, but that
$A\setminus B$ and $\tilde{A}\setminus\tilde{B}$ meet, and that $B\setminus A$
and $\tilde{B}\setminus\tilde{A}$ meet.

######  Observation 9.9.

Neither of $B\setminus A$ and $\tilde{B}\setminus\tilde{A}$ contains the
other.

To see why this is true, note that $u$ is in fact an interior point of
$\tilde{E}$ by the general position hypothesis, and that
$u\in\partial(B\setminus A)$. Thus $B\setminus A$ meets the exterior of
$\tilde{B}\setminus\tilde{A}$. A similar argument gives that
$\tilde{B}\setminus\tilde{A}$ meets the exterior of $B\setminus A$.

We are supposing for contradiction that $B\setminus A$ and
$\tilde{B}\setminus\tilde{A}$ meet, and by Observation 9.9 neither of them
contains the other. Thus if we can show that $\partial(B\setminus A)$ and
$\partial(\tilde{B}\setminus\tilde{A})$ do not meet we will have derived a
contradiction, as desired.

Note that Proposition 8.2 applies. This allows us to make the following
observation.

######  Observation 9.10.

Either $\diamondsuit$a or $\diamondsuit$e occurs, either $\heartsuit$a or
$\heartsuit$d occurs, either $\spadesuit$a or $\spadesuit$e occurs, and either
$\clubsuit$a or $\clubsuit$d occurs.

We prove that either $\diamondsuit$a or $\diamondsuit$e occurs, and the other
parts of the observation follow similarly. Because
$\tilde{E}\subset\tilde{A}$, we may eliminate any candidate topological
configurations where $u\not\in\tilde{A}$. This eliminates $\diamondsuit$d,
$\diamondsuit$f, $\diamondsuit$g, and $\diamondsuit$h. Next, because
$\tilde{u}\in\partial\tilde{A}$, we may eliminate any candidate topological
configurations where $\partial\tilde{A}$ does not meet $E$, as this would
preclude $\tilde{u}\in E$. This eliminates $\diamondsuit$b and
$\diamondsuit$c, leaving us with only the two claimed possibilities. Thus the
remainder of our proof breaks into cases as follows.

######  Case 1.

Suppose that both $\diamondsuit$a and $\heartsuit$a occur.

Then $\partial(A\setminus B)$ is contained in $\tilde{A}$, and
$\partial(B\setminus A)$ is contained in $\tilde{B}$. Thus
$\partial(A\setminus B)\cup\partial(B\setminus A)$ is contained in
$\tilde{A}\cup\tilde{B}$. But $\partial(A\cup B)$ is contained in
$\partial(A\setminus B)\cup\partial(B\setminus A)$, thus in
$\tilde{A}\cup\tilde{B}$. We conclude that $A\cup
B\subset\tilde{A}\cup\tilde{B}$. Now
$\tilde{u}\in\partial(\tilde{A}\cup\tilde{B})$ and $E\subset A\cup B$, so by
the general position hypothesis we get a contradiction to $\tilde{u}\in E$.

######  Case 2.

Suppose that both $\diamondsuit$a and $\heartsuit$d occur.

Then $u\in\tilde{E}$ and $v\in\tilde{A}\setminus\tilde{B}$. One of the
following two sub-cases occurs.

######  Sub-case 2.1.

Suppose that $\spadesuit$a occurs.

Then $\partial A$ does not meet $\tilde{A}\setminus\tilde{B}$. But $v$ lies on
$\partial A$, contradicting $v\in\tilde{A}\setminus\tilde{B}$.

######  Sub-case 2.2.

Suppose that $\spadesuit$e occurs. Then one of $\clubsuit$a and $\clubsuit$d
occurs.

From $\spadesuit$e and that $u\in\tilde{E}$ and
$v\in\tilde{A}\setminus\tilde{B}$, it follows that $\partial(B\setminus
A)\cap\partial A=[u\to v]_{\partial E}$ does not meet
$\partial(\tilde{B}\setminus\tilde{A})$. If $\clubsuit$a occurs, then
$\partial B\supset\partial(B\setminus A)\cap\partial B$ does not meet
$\partial(\tilde{B}\setminus\tilde{A})$. If $\clubsuit$d occurs, then via
$u\in\tilde{E}$ and $v\in\tilde{A}\setminus\tilde{B}$ we get that
$\partial(B\setminus A)\cap\partial B=[u\to v]_{\partial B}$ does not meet
$\partial(\tilde{B}\setminus\tilde{A})$. In either case $\partial(B\setminus
A)$ and $\partial(\tilde{B}\setminus\tilde{A})$ do not meet, giving us a
contradiction.

Cases (1) and (2) together rule out $\heartsuit$a, $\spadesuit$a, and
$\clubsuit$a by symmetry, so the only remaining case is the following.

######  Case 3.

Suppose that $\diamondsuit$e, $\heartsuit$d, $\spadesuit$e, and $\clubsuit$d
occur.

By $\diamondsuit$e and $\heartsuit$d we have that $u\in\tilde{E}$ and
$v\in\mathbb{C}\setminus(\tilde{A}\cup\tilde{B})$. Then from $\spadesuit$e and
$\clubsuit$d we get that neither $\partial(B\setminus A)\cap\partial A=[u\to
v]_{\partial A}$ nor $\partial(B\setminus A)\cap\partial B=[u\to v]_{\partial
B}$ meets $\partial(\tilde{B}\setminus\tilde{A})$, again giving us the desired
contradiction. ∎

##  10\. Torus parametrization

In this section we introduce a tool, called _torus parametrization_ that
allows us to work with fixed-point index combinatorially. This will allow us
to systematically and relatively painlessly handle the remaining case
analysis.

Let $K$ and $\tilde{K}$ be closed Jordan domains in transverse position, so
that $\partial K$ and $\partial\tilde{K}$ meet at $2M\geq 0$ points, with
boundaries oriented as usual. Let $\partial
K\allowbreak\cap\allowbreak\partial\tilde{K}=\allowbreak\\{P_{1},\allowbreak\ldots,\allowbreak
P_{M},\allowbreak\tilde{P}_{1},\ldots,\allowbreak\tilde{P}_{M}\\}$, 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 $\mathbb{S}^{1}$ with an
orientation and let $\kappa:\partial K\to\mathbb{S}^{1}$ and
$\tilde{\kappa}:\partial\tilde{K}\to\mathbb{S}^{1}$ be orientation-preserving
homeomorphisms. We refer to this as fixing a _torus parametrization_ for $K$
and $\tilde{K}$.

We consider a point $(x,y)$ on the 2-torus
$\mathbb{T}=\mathbb{S}^{1}\times\mathbb{S}^{1}$ to be parametrizing
simultaneously a point $\kappa^{-1}(x)\in\partial K$ and a point
$\tilde{\kappa}^{-1}(y)\in\partial\tilde{K}$. We denote by
$p_{i}\in\mathbb{T}$ be the unique point $(x,y)\in\mathbb{T}$ satisfying
$\kappa^{-1}(x)=\tilde{\kappa}^{-1}(y)=P_{i}$, similarly
$\tilde{p}_{i}\in\mathbb{T}$. 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 $(x_{0},y_{0})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$. Then we
may draw an image of $\mathbb{T}=\mathbb{S}^{1}\times\mathbb{S}^{1}$ by
letting $\\{x_{0}\\}\times\mathbb{S}^{1}$ be the vertical axis and letting
$\mathbb{S}^{1}\times\\{y_{0}\\}$ be the horizontal axis. Then we call
$(x_{0},y_{0})$ a _base point_ for the drawing. See Figure 17 for an example.

(0.11,0.03578125) (0.26,-0.31421876)
(0.335,-0.48921874)(0.46,-0.7892187)(0.51,-0.9142187)
(0.56,-1.0392188)(0.735,-1.2892188)(0.86,-1.4142188)
(0.985,-1.5392188)(1.235,-1.7142187)(1.36,-1.7642188)
(1.485,-1.8142188)(1.71,-1.8892188)(1.81,-1.9142188)
(1.91,-1.9392188)(2.135,-1.9642187)(2.26,-1.9642187)
(2.385,-1.9642187)(2.585,-1.9142188)(2.66,-1.8642187)
(2.735,-1.8142188)(2.835,-1.6642188)(2.86,-1.5642188)
(2.885,-1.4642187)(2.91,-1.2392187)(2.91,-1.1142187)
(2.91,-0.9892188)(2.785,-0.7392188)(2.66,-0.6142188)
(2.535,-0.48921874)(2.26,-0.21421875)(2.11,-0.06421875)
(1.96,0.08578125)(1.86,0.31078124)(1.91,0.38578126)
(1.96,0.46078125)(2.135,0.56078124)(2.26,0.5857813)
(2.385,0.61078125)(2.61,0.7107813)(2.71,0.78578126)
(2.81,0.86078125)(2.985,1.0857812)(3.06,1.2357812)
(3.135,1.3857813)(3.16,1.6107812)(3.11,1.6857812)
(3.06,1.7607813)(2.91,1.8607812)(2.81,1.8857813)
(2.71,1.9107813)(2.485,1.9607812)(2.36,1.9857812)
(2.235,2.0107813)(1.985,2.0357811)(1.86,2.0357811)
(1.735,2.0357811)(1.51,2.0107813)(1.41,1.9857812)
(1.31,1.9607812)(1.11,1.9107813)(1.01,1.8857813)
(0.91,1.8607812)(0.71,1.7357812)(0.61,1.6357813)
(0.51,1.5357813)(0.36,1.3607812)(0.31,1.2857813)
(0.26,1.2107812)(0.16,1.0357813)(0.11,0.93578124)
(0.06,0.8357813)(0.01,0.61078125)(0.01,0.48578125)
(0.01,0.36078125)(0.035,0.18578126)(0.11,0.03578125) (4.11,0.23578125)
(4.11,0.48578125) (4.11,0.61078125)(4.06,0.86078125)(4.01,0.98578125)
(3.96,1.1107812)(3.86,1.3107812)(3.81,1.3857813)
(3.76,1.4607812)(3.66,1.6107812)(3.61,1.6857812)
(3.56,1.7607813)(3.385,1.8607812)(3.26,1.8857813)
(3.135,1.9107813)(2.91,1.9357812)(2.81,1.9357812)
(2.71,1.9357812)(2.51,1.7857813)(2.41,1.6357813)
(2.31,1.4857812)(2.235,1.2107812)(2.26,1.0857812)
(2.285,0.9607813)(2.335,0.73578125)(2.36,0.6357812)
(2.385,0.53578126)(2.435,0.33578125)(2.46,0.23578125)
(2.485,0.13578124)(2.51,-0.08921875)(2.51,-0.21421875)
(2.51,-0.33921874)(2.435,-0.63921875)(2.36,-0.81421876)
(2.285,-0.9892188)(2.235,-1.2642188)(2.26,-1.3642187)
(2.285,-1.4642187)(2.385,-1.6142187)(2.46,-1.6642188)
(2.535,-1.7142187)(2.685,-1.8142188)(2.76,-1.8642187)
(2.835,-1.9142188)(3.035,-1.9642187)(3.16,-1.9642187)
(3.285,-1.9642187)(3.51,-1.9142188)(3.61,-1.8642187)
(3.71,-1.8142188)(3.86,-1.6892188)(3.91,-1.6142187)
(3.96,-1.5392188)(4.01,-1.3392187)(4.01,-1.2142187)
(4.01,-1.0892187)(4.035,-0.83921874)(4.06,-0.71421874)
(4.085,-0.58921874)(4.11,-0.33921874)(4.11,-0.21421875)
(4.11,-0.08921875)(4.11,0.08578125)(4.11,0.23578125)
$K$$u$$\tilde{K}$$\tilde{u}$$P_{1}$$P_{2}$$\tilde{P}_{1}$$\tilde{P}_{2}$

(a)

$p_{1}$$p_{2}$$\tilde{p}_{1}$$\tilde{p}_{2}$$(\kappa(u),\tilde{\kappa}(\tilde{u}))$

(b)

Figure 17.  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 $\mathbb{T}$ 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
$\mathbb{T}$ 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.

Suppose that $\phi:\partial K\to\partial\tilde{K}$ is an orientation-
preserving homeomorphism. Then $\phi$ determines an oriented curve $\gamma$ in
$\mathbb{T}$ for us, namely its graph
$\gamma=\\{(\kappa(z),\tilde{\kappa}(f(z))\\}_{z\in\partial K}$, 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 $\tilde{p}_{i}$.

The nicest thing about torus parametrization is that it allows us to compute
$\eta(\phi)$ easily by looking at the curve $\gamma$ associated to $\phi$. In
particular, 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
$\\{\tilde{\kappa}(\tilde{u})\\}\times\mathbb{S}^{1}$ and
$\mathbb{S}^{1}\times\\{\kappa(u)\\}$ divide $\mathbb{T}$ into two simply
connected open sets $\Delta_{\uparrow}(u,\gamma)$ and
$\Delta_{\downarrow}(u,\gamma)$ as shown in Figure 18. 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 $\\#p_{\downarrow}(u,\gamma)$ 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
$\\#p_{\uparrow}(u,\gamma)$, $\\#\tilde{p}_{\downarrow}(u,\gamma)$, and
$\\#\tilde{p}_{\uparrow}(u,\gamma)$ in the analogous way. Denote by
$\omega(\alpha,z)$ the winding number of the closed curve
$\alpha\subset\mathbb{C}$ around the point $z\not\in\alpha$. Then:

$(\kappa(u),\tilde{\kappa}(\tilde{u}))$(0.0,-2.6325) (0.08,-2.4725)
(0.11,-2.4225)(0.165,-2.3375)(0.19,-2.3025)
(0.215,-2.2675)(0.28,-2.1925)(0.32,-2.1525)
(0.36,-2.1125)(0.445,-2.0425)(0.49,-2.0125)
(0.535,-1.9825)(0.65,-1.8825)(0.72,-1.8125)
(0.79,-1.7425)(0.895,-1.6275)(0.93,-1.5825)
(0.965,-1.5375)(1.035,-1.4325)(1.07,-1.3725)
(1.105,-1.3125)(1.19,-1.1925)(1.24,-1.1325)
(1.29,-1.0725)(1.38,-0.9675)(1.42,-0.9225)
(1.46,-0.8775)(1.565,-0.7775)(1.63,-0.7225)
(1.695,-0.6675)(1.825,-0.5675)(1.89,-0.5225)
(1.955,-0.4775)(2.075,-0.4025)(2.13,-0.3725)
(2.185,-0.3425)(2.305,-0.2875)(2.37,-0.2625)
(2.435,-0.2375)(2.56,-0.1925)(2.62,-0.1725)
(2.68,-0.1525)(2.8,-0.0975)(2.86,-0.0625)
(2.92,-0.0275)(3.035,0.0525)(3.09,0.0975)
(3.145,0.1425)(3.235,0.2325)(3.27,0.2775)
(3.305,0.3225)(3.37,0.4125)(3.4,0.4575) (3.43,0.5025)(3.48,0.5875)(3.5,0.6275)
(3.52,0.6675)(3.555,0.7625)(3.57,0.8175)
(3.585,0.8725)(3.62,0.9775)(3.64,1.0275)
(3.66,1.0775)(3.7,1.1625)(3.72,1.1975)
(3.74,1.2325)(3.795,1.3075)(3.83,1.3475)
(3.865,1.3875)(3.945,1.4675)(3.99,1.5075)
(4.035,1.5475)(4.125,1.6075)(4.17,1.6275)
(4.215,1.6475)(4.325,1.6925)(4.39,1.7175)
(4.455,1.7425)(4.59,1.7825)(4.66,1.7975)
(4.73,1.8125)(4.84,1.8425)(4.88,1.8575)
(4.92,1.8725)(5.015,1.9175)(5.07,1.9475)
(5.125,1.9775)(5.215,2.0425)(5.25,2.0775)
(5.285,2.1125)(5.35,2.1925)(5.38,2.2375)
(5.41,2.2825)(5.47,2.3875)(5.5,2.4475) (5.53,2.5075)(5.57,2.6025)(5.58,2.6375)
(5.59,2.6725)(5.61,2.7375)(5.62,2.7675)
(5.63,2.7975)(5.655,2.8475)(5.67,2.8675)
(5.685,2.8875)(5.705,2.9275)(5.71,2.9475)
(5.715,2.9675)(5.73,3.0025)(5.74,3.0175)
(5.75,3.0325)(5.765,3.0625)(5.82,3.1875)
$\Delta_{\uparrow}(u,\gamma)$$\Delta_{\downarrow}(u,\gamma)$$\gamma$$\mathbb{S}^{1}\times\\{\tilde{\kappa}(\tilde{u})\\}$$\\{\kappa(u)\\}\times\mathbb{S}^{1}$$p_{1}$$\tilde{p}_{1}$$p_{2}$$\tilde{p}_{2}$

Figure 18.  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.

######  Lemma 10.1.

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
$\mathbb{T}$. Suppose that $\phi(u)=\tilde{u}$, where
$u\not\in\partial\tilde{K}$ and $\tilde{u}\not\in\partial K$. Then:

(3) $\displaystyle\eta(\phi)=w(\gamma)$ $\displaystyle=\omega(\partial
K,\tilde{u})+\omega(\partial\tilde{K},u)-\\#p_{\downarrow}(u,\gamma)+\\#\tilde{p}_{\downarrow}(u,\gamma)$
(4) $\displaystyle=\omega(\partial
K,\tilde{u})+\omega(\partial\tilde{K},u)+\\#p_{\uparrow}(u,\gamma)-\\#\tilde{p}_{\uparrow}(u,\gamma)$

###### Proof.

Suppose $\gamma_{0}$ is any oriented closed curve in
$\mathbb{T}\setminus\\{p_{1},\ldots,p_{M},\tilde{p}_{1},\ldots,\tilde{p}_{M}\\}$.
Then the closed curve
$\\{\tilde{\kappa}^{-1}(y)-\kappa^{-1}(x)\\}_{(x,y)\in\gamma_{0}}$ 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 $\\{\tilde{\kappa}^{-1}(y)-\kappa^{-1}(x)\\}_{(x,y)\in\gamma_{0}}$. First:

######  Observation 10.2.

If $\gamma_{1}$ and $\gamma_{2}$ are homotopic in
$\mathbb{T}\setminus\\{p_{1},\ldots,p_{M},\tilde{p}_{1},\ldots,\tilde{p}_{M}\\}$
then $w(\gamma_{1})=w(\gamma_{2})$.

This is because the homotopy between $\gamma_{1}$ and $\gamma_{2}$ in
$\mathbb{T}\setminus\\{p_{2},\ldots,p_{M},\tilde{p}_{1},\ldots,\tilde{p}_{M}\\}$
induces a homotopy between the closed curves
$\\{\tilde{\kappa}^{-1}(y)-\kappa^{-1}(x)\\}_{(x,y)\in\gamma_{1}}$ and
$\\{\tilde{\kappa}^{-1}(y)-\kappa^{-1}(x)\\}_{(x,y)\in\gamma_{2}}$ in the
punctured plane $\mathbb{C}\setminus\\{0\\}$.

If $\gamma$ has orientation induced by traversing $\partial K$ and
$\partial\tilde{K}$ positively, then the following is a tautology.

######  Observation 10.3.

$\eta(\phi)=w(\gamma)$

Orient $\partial\Delta_{\downarrow}(u,\gamma)$ as shown in Figure 18. Then
$\partial\Delta_{\downarrow}(u,\gamma)$ is the concatenation of the curve
$\gamma$ traversed backwards with
$\mathbb{S}^{1}\times\\{\tilde{\kappa}(\tilde{u})\\}$ and
$\\{\kappa(u)\\}\times\mathbb{S}^{1}$, where the two latter curves are
oriented according to the positive orientation on $\mathbb{S}^{1}$.

(1.69,2.5796876) (1.89,2.4196875)
(1.99,2.3396876)(2.28,2.0546875)(2.47,1.8496875)
(2.66,1.6446875)(2.89,1.3196875)(2.93,1.1996875)
(2.97,1.0796875)(3.09,0.8246875)(3.17,0.6896875)
(3.25,0.5546875)(3.44,0.3296875)(3.55,0.2396875)
(3.66,0.1496875)(3.845,-0.0403125)(3.92,-0.1403125)
(3.995,-0.2403125)(4.145,-0.4453125)(4.22,-0.5503125)
(4.295,-0.6553125)(4.465,-0.8553125)(4.56,-0.9503125)
(4.655,-1.0453125)(4.82,-1.1903125)(4.89,-1.2403125)
(4.96,-1.2903125)(5.13,-1.4053125)(5.23,-1.4703125)
(5.33,-1.5353125)(5.48,-1.6753125)(5.53,-1.7503124)
(5.58,-1.8253125)(5.665,-1.9453125)(5.7,-1.9903125)
(5.735,-2.0353124)(5.81,-2.1303124)(5.85,-2.1803124)
(5.89,-2.2303126)(5.97,-2.3103125)(6.01,-2.3403125)
(6.05,-2.3703125)(6.13,-2.4153125)(6.17,-2.4303124)
(6.21,-2.4453125)(6.295,-2.4603126)(6.34,-2.4603126)
(6.385,-2.4603126)(6.5,-2.4353125)(6.57,-2.4103124)
(6.64,-2.3853126)(6.82,-2.2553124)(6.93,-2.1503124)
(7.04,-2.0453124)(7.235,-1.8803124)(7.32,-1.8203125)
(7.405,-1.7603126)(7.635,-1.5903125)(7.78,-1.4803125)
(7.925,-1.3703125)(8.155,-1.1853125)(8.24,-1.1103125)
(8.325,-1.0353125)(8.49,-0.8753125)(8.57,-0.7903125)
(8.65,-0.7053125)(8.765,-0.5353125)(8.8,-0.4503125)
(8.835,-0.3653125)(8.88,-0.2203125)(8.89,-0.1603125)
(8.9,-0.1003125)(8.93,0.0446875)(8.95,0.1296875)
(8.97,0.2146875)(9.005,0.3746875)(9.02,0.4496875)
(9.035,0.5246875)(9.045,0.7096875)(9.04,0.8196875)
(9.035,0.9296875)(8.98,1.1296875)(8.93,1.2196875)
(8.88,1.3096875)(8.725,1.5196875)(8.62,1.6396875)
(8.515,1.7596875)(8.305,2.0246875)(8.2,2.1696875)
(8.095,2.3146875)(7.9,2.5396874)(7.81,2.6196876)
(7.72,2.6996875)(7.445,2.8696876)(7.26,2.9596875)
(7.075,3.0496874)(6.76,3.1896875)(6.63,3.2396874)
(6.5,3.2896874)(6.09,3.3846874)(5.81,3.4296875)
(5.53,3.4746876)(5.115,3.5346875)(4.98,3.5496874)
(4.845,3.5646875)(4.53,3.5846875)(4.35,3.5896876)
(4.17,3.5946875)(3.845,3.5596876)(3.7,3.5196874)
(3.555,3.4796875)(3.275,3.3746874)(3.14,3.3096876)
(3.005,3.2446876)(2.79,3.1446874)(2.71,3.1096876)
(2.63,3.0746875)(2.485,3.0146875)(2.42,2.9896874)
(2.355,2.9646876)(2.195,2.8796875)(2.1,2.8196876)
(2.005,2.7596874)(1.89,2.6896875)(1.87,2.6796875)
(1.85,2.6696875)(1.825,2.6396875)(1.82,2.6196876)
(1.815,2.5996876)(1.8,2.5646875)(1.79,2.5496874)
(1.78,2.5346875)(1.77,2.5146875)(1.77,2.4996874) (1.47,-1.4403125)
(1.61,-1.4003125) (1.68,-1.3803124)(1.92,-1.2903125)(2.09,-1.2203125)
(2.26,-1.1503125)(2.49,-1.0453125)(2.55,-1.0103126)
(2.61,-0.9753125)(2.74,-0.8903125)(2.81,-0.8403125)
(2.88,-0.7903125)(3.02,-0.6653125)(3.09,-0.5903125)
(3.16,-0.5153125)(3.295,-0.3603125)(3.36,-0.2803125)
(3.425,-0.2003125)(3.58,-0.0753125)(3.67,-0.0303125)
(3.76,0.0146875)(3.965,0.0946875)(4.08,0.1296875)
(4.195,0.1646875)(4.455,0.2946875)(4.6,0.3896875)
(4.745,0.4846875)(4.96,0.6396875)(5.03,0.6996875)
(5.1,0.7596875)(5.24,0.8746875)(5.31,0.9296875)
(5.38,0.9846875)(5.52,1.1046875)(5.59,1.1696875)
(5.66,1.2346874)(5.79,1.3496875)(5.85,1.3996875)
(5.91,1.4496875)(6.055,1.5446875)(6.14,1.5896875)
(6.225,1.6346875)(6.37,1.7096875)(6.43,1.7396874)
(6.49,1.7696875)(6.615,1.8246875)(6.68,1.8496875)
(6.745,1.8746876)(6.86,1.9096875)(6.91,1.9196875)
(6.96,1.9296875)(7.07,1.9396875)(7.13,1.9396875)
(7.19,1.9396875)(7.3,1.9396875)(7.35,1.9396875)
(7.4,1.9396875)(7.5,1.9246875)(7.55,1.9096875)
(7.6,1.8946875)(7.68,1.8596874)(7.71,1.8396875)
(7.74,1.8196875)(7.835,1.6996875)(7.9,1.5996875)
(7.965,1.4996876)(8.115,1.1346875)(8.2,0.8696875)
(8.285,0.6046875)(8.45,0.2146875)(8.53,0.0896875)
(8.61,-0.0353125)(8.705,-0.2703125)(8.72,-0.3803125)
(8.735,-0.4903125)(8.695,-0.7453125)(8.64,-0.8903125)
(8.585,-1.0353125)(8.41,-1.4003125)(8.29,-1.6203125)
(8.17,-1.8403125)(7.975,-2.1903124)(7.9,-2.3203125)
(7.825,-2.4503126)(7.7,-2.6303124)(7.65,-2.6803124)
(7.6,-2.7303126)(7.43,-2.8553126)(7.31,-2.9303124)
(7.19,-3.0053124)(6.87,-3.1503124)(6.67,-3.2203126)
(6.47,-3.2903125)(6.115,-3.3803124)(5.96,-3.4003124)
(5.805,-3.4203124)(5.51,-3.4453125)(5.37,-3.4503126)
(5.23,-3.4553125)(4.995,-3.4903126)(4.9,-3.5203125)
(4.805,-3.5503125)(4.575,-3.5853126)(4.44,-3.5903125)
(4.305,-3.5953126)(4.07,-3.6003125)(3.97,-3.6003125)
(3.87,-3.6003125)(3.57,-3.5553124)(3.37,-3.5103126)
(3.17,-3.4653125)(2.845,-3.3803124)(2.72,-3.3403125)
(2.595,-3.3003125)(2.34,-3.1453125)(2.21,-3.0303125)
(2.08,-2.9153125)(1.92,-2.7403126)(1.89,-2.6803124)
(1.86,-2.6203125)(1.815,-2.5003126)(1.8,-2.4403124)
(1.785,-2.3803124)(1.745,-2.2553124)(1.72,-2.1903124)
(1.695,-2.1253126)(1.65,-2.0153124)(1.63,-1.9703125)
(1.61,-1.9253125)(1.57,-1.8353125)(1.55,-1.7903125)
(1.53,-1.7453125)(1.505,-1.6703125)(1.5,-1.6403126)
(1.495,-1.6103125)(1.485,-1.5553125)(1.48,-1.5303125)
(1.475,-1.5053124)(1.48,-1.4603125)(1.49,-1.4403125) (1.47,-1.4403125)
(1.61,-1.4003125) (1.68,-1.3803124)(1.92,-1.2903125)(2.09,-1.2203125)
(2.26,-1.1503125)(2.49,-1.0453125)(2.55,-1.0103126)
(2.61,-0.9753125)(2.74,-0.8903125)(2.81,-0.8403125)
(2.88,-0.7903125)(3.02,-0.6653125)(3.09,-0.5903125)
(3.16,-0.5153125)(3.295,-0.3603125)(3.36,-0.2803125)
(3.425,-0.2003125)(3.58,-0.0753125)(3.67,-0.0303125)
(3.76,0.0146875)(3.965,0.0946875)(4.08,0.1296875)
(4.195,0.1646875)(4.455,0.2946875)(4.6,0.3896875)
(4.745,0.4846875)(4.96,0.6396875)(5.03,0.6996875)
(5.1,0.7596875)(5.24,0.8746875)(5.31,0.9296875)
(5.38,0.9846875)(5.52,1.1046875)(5.59,1.1696875)
(5.66,1.2346874)(5.79,1.3496875)(5.85,1.3996875)
(5.91,1.4496875)(6.055,1.5446875)(6.14,1.5896875)
(6.225,1.6346875)(6.37,1.7096875)(6.43,1.7396874)
(6.49,1.7696875)(6.615,1.8246875)(6.68,1.8496875)
(6.745,1.8746876)(6.86,1.9096875)(6.91,1.9196875)
(6.96,1.9296875)(7.07,1.9396875)(7.13,1.9396875)
(7.19,1.9396875)(7.3,1.9396875)(7.35,1.9396875)
(7.4,1.9396875)(7.5,1.9246875)(7.55,1.9096875) (1.69,2.5796876)
(1.89,2.4196875) (1.99,2.3396876)(2.28,2.0546875)(2.47,1.8496875)
(2.66,1.6446875)(2.89,1.3196875)(2.93,1.1996875)
(2.97,1.0796875)(3.09,0.8246875)(3.17,0.6896875)
(3.25,0.5546875)(3.44,0.3296875)(3.55,0.2396875)
(3.66,0.1496875)(3.845,-0.0403125)(3.92,-0.1403125)
(3.995,-0.2403125)(4.145,-0.4453125)(4.22,-0.5503125)
(4.295,-0.6553125)(4.465,-0.8553125)(4.56,-0.9503125)
(4.655,-1.0453125)(4.82,-1.1903125)(4.89,-1.2403125)
(4.96,-1.2903125)(5.13,-1.4053125)(5.23,-1.4703125)
(5.33,-1.5353125)(5.48,-1.6753125)(5.53,-1.7503124)
(5.58,-1.8253125)(5.665,-1.9453125)(5.7,-1.9903125)
(5.735,-2.0353124)(5.81,-2.1303124)(5.85,-2.1803124)
(5.89,-2.2303126)(5.97,-2.3103125)(6.01,-2.3403125)
(6.05,-2.3703125)(6.13,-2.4153125)(6.17,-2.4303124)
(6.21,-2.4453125)(6.295,-2.4603126)(6.34,-2.4603126)
$K$$\tilde{K}$$P_{i}$$\kappa^{-1}(x_{0})$$\kappa^{-1}(x_{1})$$\tilde{\kappa}^{-1}(y_{0})$$\tilde{\kappa}^{-1}(y_{1})$(5.41,-1.3603125)
(5.58,-1.2303125) (5.665,-1.1653125)(5.835,-1.0853125)(5.92,-1.0703125)
(6.005,-1.0553125)(6.185,-1.0403125)(6.28,-1.0403125)
(6.375,-1.0403125)(6.61,-1.0603125)(6.75,-1.0803125)
(6.89,-1.1003125)(7.21,-1.1853125)(7.39,-1.2503124)
(7.57,-1.3153125)(7.945,-1.4953125)(8.14,-1.6103125)
(8.335,-1.7253125)(8.67,-1.8903126)(8.81,-1.9403125)
(8.95,-1.9903125)(9.33,-2.0553124)(9.57,-2.0703125)
(9.81,-2.0853126)(10.225,-2.0503125)(10.4,-2.0003126)
(10.575,-1.9503125)(10.785,-1.8703125)(10.82,-1.8403125)
(10.855,-1.8103125)(10.915,-1.7553124)(10.94,-1.7303125)
(10.965,-1.7053125)(11.005,-1.6553125)(11.02,-1.6303124)
(11.035,-1.6053125)(11.055,-1.5553125)(11.07,-1.4803125)

Figure 19.  The local picture near $P_{i}$. This allows us to compute the
“local fixed-point index” $w(\zeta(p_{i}))$ of $f$ near $P_{i}$.

######  Observation 10.4.

If $\mathbb{S}^{1}\times\\{\tilde{\kappa}(\tilde{u})\\}$ and
$\\{\kappa(u)\\}\times\mathbb{S}^{1}$ are oriented according to the positive
orientation on $\mathbb{S}^{1}$, then
$w(\mathbb{S}^{1}\times\\{\tilde{\kappa}(\tilde{u})\\})=\omega(\partial
K,\tilde{u})$ and
$w(\\{\kappa(u)\\}\times\mathbb{S}^{1})=\omega(\partial\tilde{K},u)$.

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:

$\displaystyle w(\partial\Delta_{\downarrow}(u,\gamma))$
$\displaystyle=w(\mathbb{S}^{1}\times\\{\tilde{\kappa}(\tilde{u})\\})+w(\\{\kappa(u)\\}\times\mathbb{S}^{1})-w(\gamma)$
$\displaystyle=\omega(\partial
K,\tilde{u})+\omega(\partial\tilde{K},u)-\eta(\phi)$

For every $i$ let $\zeta(p_{i})$ and $\zeta(\tilde{p}_{i})$ be small squares
around $p_{i}$ and $\tilde{p}_{i}$ respectively in $\mathbb{T}$, oriented as
shown in Figure 18. By _square_ we mean a simple closed curve which decomposes
into four “sides,” so that on a given side one of the two coordinates of
$\mathbb{S}^{1}\times\mathbb{S}^{1}=\mathbb{T}$ 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 $\Delta_{\downarrow}(u,\gamma)$.

######  Observation 10.5.

The curves $\Gamma$ and $\partial\Delta_{\downarrow}(u,\gamma)$ are homotopic
in $\mathbb{T}\setminus\\{p_{1},\ldots,\allowbreak
p_{M},\allowbreak\tilde{p}_{1},\ldots,\allowbreak\tilde{p}_{M}\\}$. Also
$w(\delta)=0$. It follows that:

$w(\partial\Delta_{\downarrow}(u,\gamma))=w(\Gamma)=\sum_{p_{i}\in\Delta_{\downarrow}(u,\gamma)}w(\zeta(p_{i}))+\sum_{\tilde{p}_{i}\in\Delta_{\downarrow}(u,\gamma)}w(\zeta(\tilde{p}_{i}))$

See Figure 18 for an example. On the other hand, the following holds.

######  Observation 10.6.

$w(\zeta(p_{i}))=1$, $w(\zeta(\tilde{p}_{i}))=-1$

To see why, suppose that $\zeta(p_{i})=\partial([x_{0}\to
x_{1}]_{\mathbb{S}^{1}}\times[y_{0}\to y_{1}]_{\mathbb{S}^{1}})$. Then up to
orientation-preserving homeomorphism the picture near $P_{i}$ is as in Figure
19. We let $(x,y)$ traverse $\zeta(p_{i})$ positively starting at
$(x_{0},y_{0})$, keeping track of the vector
$\tilde{\kappa}^{-1}(y)-\tilde{\kappa}^{-1}(x)$ as we do so. The vector
$\tilde{\kappa}^{-1}(y_{0})-\tilde{\kappa}^{-1}(x_{0})$ points to the right.
As $x$ varies from $x_{0}$ to $x_{1}$, the vector
$\tilde{\kappa}^{-1}(y)-\tilde{\kappa}^{-1}(x)$ rotates in the positive
direction, that is, counter-clockwise, until it arrives at
$\tilde{\kappa}^{-1}(y_{0})-\kappa^{-1}(x_{1})$, which points upward.
Continuing in this fashion, we see that
$\tilde{\kappa}^{-1}(y)-\tilde{\kappa}^{-1}(x)$ 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 3. The proof that equation 4 holds is similar. ∎

$p_{1}$$\tilde{p}_{1}$$\tilde{s}_{1}$$\tilde{s}_{2}$$s_{1}$$s_{2}$

(a)

$p_{1}$$\tilde{p}_{1}$$(\kappa(u),\tilde{\kappa}(\tilde{u}))$$\tilde{t}_{1}$$\tilde{t}_{2}$$\tilde{t}_{3}$$t_{1}$$t_{2}$$t_{3}$

(b) $(\kappa(u),\tilde{\kappa}(u))=(s_{1},\tilde{s}_{2})$

Figure 20.  Two drawings of a torus parametrization for two eyes whose
boundaries meet exactly twice. There is some choice of base point giving the
drawing on the left. The drawing on the right is the same torus
parametrization drawn using the base point
$(\kappa(u),\tilde{\kappa}(u))=(s_{1},\tilde{s}_{2})$.

##  11\. Proof of Propositions 11.1 and 11.5

In this section, we prove the two remaining outstanding propositions to
complete the proof of our Main Index Theorem 5.3 and thus our main rigidity
results.

######  Proposition 11.1.

Let $\\{A,B\\}$ and $\\{\tilde{A},\tilde{B}\\}$ be pairs of overlapping closed
disks in the plane $\mathbb{C}$ in general position. Suppose that neither of
$E=A\cap B$ and $\tilde{E}=\tilde{A}\cap\tilde{B}$ contains the other. Suppose
further that $A\setminus B$ and $\tilde{A}\setminus\tilde{B}$ meet, and that
$B\setminus A$ and $\tilde{B}\setminus\tilde{A}$ meet. Then there is a
faithful indexable homeomorphism $\epsilon:\partial E\to\partial\tilde{E}$
satisfying $\eta(e)=0$.

###### Proof.

If $\partial E$ and $\partial\tilde{E}$ do not meet, we get that $E$ and
$\tilde{E}$ are disjoint. Then any indexable homeomorphism $\epsilon:\partial
E\to\partial\tilde{E}$ satisfies $\eta(\epsilon)=0$. Thus suppose that
$\partial E$ and $\partial\tilde{E}$ meet. Fix a torus parametrization for $E$
and $\tilde{E}$ via $\kappa:\partial E\to\mathbb{S}^{1}$ and
$\tilde{\kappa}:\partial\tilde{E}\to\mathbb{S}^{1}$. As before denote by
$p_{i}$ the points of $\partial E\cap\partial\tilde{E}$ where $\partial E$ is
entering $\tilde{E}$, and by $\tilde{p}_{i}$ those where $\partial\tilde{E}$
is entering $E$. Note that $\partial E$ and $\partial\tilde{E}$ meet at
exactly 2, 4, or 6 points by Lemma 9.3. The proof breaks into these three
cases.

######  Case 1.

Suppose that $\partial E$ and $\partial\tilde{E}$ meet at exactly two points.

Then with an appropriate choice of base point, the torus parametrization for
$E$ and $\tilde{E}$ is as shown in Figure 20a. The points
$s_{1},s_{2}\in\mathbb{S}^{1}$ in Figure 20a are exactly the topologically
distinct places where $\kappa(u)$ may be, similarly
$\tilde{s}_{1},\tilde{s}_{2}\in\mathbb{S}^{1}$ for
$\tilde{\kappa}(\tilde{u})$. A choice of
$(s_{j},\tilde{s}_{\tilde{j}})=(\kappa(u),\tilde{\kappa}(\tilde{u}))$
completely determines the topological configuration of
$\\{E,\tilde{E},u,\tilde{u}\\}$, and conversely every possible topological
configuration of those sets is achieved via this procedure. By Lemma 9.8 we
may suppose without loss of generality that $u\not\in\tilde{E}$, thus that
$\kappa(u)=s_{1}$.

$(\kappa(u),\tilde{\kappa}(\tilde{u}))$$t_{1}$$t_{2}$$t_{3}$$\tilde{t}_{1}$$\tilde{t}_{2}$$\tilde{t}_{3}$$p_{1}$$\tilde{p}_{1}$

(a) $(\kappa(u),\tilde{\kappa}(\tilde{u}))=(s_{1},\tilde{s}_{1})$

$(\kappa(u),\tilde{\kappa}(\tilde{u}))$$t_{1}$$t_{2}$$t_{3}$$\tilde{t}_{1}$$\tilde{t}_{2}$$\tilde{t}_{3}$$p_{1}$$\tilde{p}_{1}$

(b) $(\kappa(u),\tilde{\kappa}(\tilde{u}))=(s_{1},\tilde{s}_{2})$

Figure 21.  Graphs of homeomorphisms $\epsilon$ giving $\eta(\epsilon)=0$ for
a pair of eyes whose boundaries meet twice. The torus parametrizations are
drawn using the indicated choice of base point.

Suppose first that $\tilde{\kappa}(\tilde{u})=\tilde{s}_{2}$. In Figure 20b we
redraw the torus parametrization for $E$ and $\tilde{E}$ using the base point
$(s_{1},\tilde{s}_{2})=(\kappa(u),\tilde{\kappa}(\tilde{u}))$. Then the points
$t_{1},t_{2},t_{3}\in\mathbb{S}^{1}$ are exactly the topologically distinct
places where $\kappa(v)$ may be, similarly
$\tilde{t}_{1},\tilde{t}_{2},\tilde{t}_{3}\in\mathbb{S}^{1}$ for
$\tilde{\kappa}(\tilde{v})$.

######  Observation 11.2.

A choice of
$(s_{j},\tilde{s}_{\tilde{j}})=(\kappa(u),\tilde{\kappa}(\tilde{u}))$ and a
subsequent choice of
$(t_{k},\tilde{t}_{\tilde{k}})=(\kappa(v),\tilde{\kappa}(\tilde{v}))$ together
completely determine the topological configuration of
$\\{E,\tilde{E},u,\tilde{u},v,\tilde{v}\\}$. Conversely every possible
topological configuration of
$\\{E,\tilde{E},u,\tilde{u},v,\allowbreak\tilde{v}\\}$ is achieved by some
choice of $(s_{j},\tilde{s}_{\tilde{j}})$, and then a subsequent choice of
$(t_{k},\tilde{t}_{\tilde{k}})$, for $(\kappa(u),\tilde{\kappa}(\tilde{u}))$
and $(\kappa(v),\tilde{\kappa}(\tilde{v}))$ respectively.

$p_{1}$$\tilde{p}_{1}$$p_{2}$$\tilde{p}_{2}$$\tilde{s}_{1}$$\tilde{s}_{2}$$\tilde{s}_{3}$$\tilde{s}_{4}$$s_{1}$$s_{2}$$s_{3}$$s_{4}$

(a)

$p_{1}$$\tilde{p}_{1}$$p_{2}$$\tilde{p}_{2}$$t_{1}$$t_{2}$$t_{3}$$t_{4}$$t_{5}$$\tilde{t}_{1}$$\tilde{t}_{2}$$\tilde{t}_{3}$$\tilde{t}_{4}$$\tilde{t}_{5}$

(b) $(\kappa(u),\tilde{\kappa}(\tilde{u}))=(s_{1},\tilde{s}_{1})$

$p_{1}$$\tilde{p}_{1}$$p_{2}$$\tilde{p}_{2}$$t_{1}$$t_{2}$$t_{3}$$t_{4}$$t_{5}$$\tilde{t}_{1}$$\tilde{t}_{2}$$\tilde{t}_{3}$$\tilde{t}_{4}$$\tilde{t}_{5}$

(c) $(\kappa(u),\tilde{\kappa}(\tilde{u}))=(s_{1},\tilde{s}_{2})$

$p_{1}$$\tilde{p}_{1}$$p_{2}$$\tilde{p}_{2}$$t_{1}$$t_{2}$$t_{3}$$t_{4}$$t_{5}$$\tilde{t}_{1}$$\tilde{t}_{2}$$\tilde{t}_{3}$$\tilde{t}_{4}$$\tilde{t}_{5}$

(d) $(\kappa(u),\tilde{\kappa}(\tilde{u}))=(s_{1},\tilde{s}_{3})$

Figure 22. The situation if two eyes’ boundaries meet four times. Figure 22a
shows the torus parametrization for $E$ and $\tilde{E}$ with some suitable
choice of base point. Figures 22b–22d give graphs of homeomorphisms $\epsilon$
giving $\eta(\epsilon)=0$, with torus parametrizations drawn using base point
$(\kappa(u),\tilde{\kappa}(\tilde{u}))=(s_{j},\tilde{s}_{\tilde{j}})$ as
indicated.

We are currently working under the assumption that
$(\kappa(u),\tilde{\kappa}(\tilde{u}))=(s_{1},\tilde{s}_{2})$. For every
choice of
$(t_{k},\tilde{t}_{\tilde{k}})=(\kappa(v),\tilde{\kappa}(\tilde{v}))$ we hope
to find a faithful indexable homeomorphism $\epsilon:\partial
E\to\partial\tilde{E}$ so that $\eta(\epsilon)=0$.

######  Observation 11.3.

Suppose we have drawn the parametrization for $E$ and $\tilde{E}$ using
$(s_{j},\tilde{s}_{\tilde{j}})=(\kappa(u),\tilde{\kappa}(\tilde{u}))$ as the
base point. Then finding a faithful indexable homeomorphism $\epsilon:\partial
E\to\partial\tilde{E}$ amounts to finding a curve $\gamma$ in
$\mathbb{T}\setminus\\{p_{1},\ldots,\tilde{p}_{M},\tilde{p}_{1},\ldots,\tilde{p}_{M}\\}$
which “looks like the graph of a strictly increasing function,” from the
lower-left-hand corner $(\kappa(u),\tilde{\kappa}(u))$ to the upper-right-hand
corner, passing through
$(\kappa(v),\tilde{\kappa}(\tilde{v}))=(t_{k},\tilde{t}_{\tilde{k}})$. Having
fixed such a curve $\gamma$, we may compute $\eta(\epsilon)$, where $\epsilon$
is the homeomorphism associated to $\gamma$, using Lemma 10.1.

In our current situation $\kappa(u)=s_{1}$ implies that $u\not\in\tilde{K}$,
and $\tilde{\kappa}(\tilde{u})=\tilde{s}_{2}$ implies that $\tilde{u}\not\in
K$. Thus by Lemma 10.1 we wish to find curves $\gamma$ so that both $p_{1}$
and $\tilde{p}_{1}$ lie in the upper diagonal $\Delta_{\uparrow}(u,\gamma)$,
or both lie in the lower diagonal $\Delta_{\downarrow}(u,\gamma)$. Figure 21b
depicts such a $\gamma$ for every $(t_{k},\tilde{t}_{\tilde{k}})$ except for
$(t_{2},\tilde{t}_{2})$. Suppose
$(t_{2},\tilde{t}_{2})=(\kappa(v),\tilde{\kappa}(\tilde{v}))$. Then
$v\in\tilde{K}$ and $\tilde{v}\in K$, so we get a contradiction by Lemma 9.8.
From now on points $(t_{k},\tilde{t}_{\tilde{k}})$ which are handled via Lemma
9.8 will be labeled with an asterisk, as in Figure 21b.

Next suppose that
$(\kappa(u),\tilde{\kappa}(\tilde{u}))=(s_{1},\tilde{s}_{1})$. The situation
is depicted in Figure 21a. Then $u\not\in\tilde{K}$ and $\tilde{u}\in K$, so
to achieve $\eta(\epsilon)=0$ we wish to find curves $\gamma$ so that
$p_{1}\in\Delta_{\downarrow}(u,\gamma)$ and
$\tilde{p}_{1}\in\Delta_{\uparrow}(u,\gamma)$. This time there are four
$(t_{k},\tilde{t}_{\tilde{k}})$ for which this is not possible. For
$(\kappa(v),\tilde{\kappa}(\tilde{v})=(t_{2},\tilde{t}_{1}),(t_{2},\tilde{t}_{3})$
we again get contradictions via Lemma 9.8. The following observation will be
helpful for
$(\kappa(v),\tilde{\kappa}(\tilde{v})=(t_{1},\tilde{t}_{3}),(t_{3},\tilde{t}_{1})$.

######  Observation 11.4.

Choose $(s_{j},\tilde{s}_{\tilde{j}})=(\kappa(u),\tilde{\kappa}(\tilde{u}))$
and draw our torus parametrization for $E$ and $\tilde{E}$ using
$(\kappa(u),\tilde{\kappa}(\tilde{u}))$ as the base point. Then a choice of
$(t_{k},\tilde{t}_{\tilde{k}})=(\kappa(v),\tilde{\kappa}(\tilde{v}))$ defines
for us four “quadrants,” namely
$[\kappa(u)\to\kappa(v)]_{\mathbb{S}^{1}}\times[\tilde{\kappa}(\tilde{u})\to\tilde{\kappa}(\tilde{v})]_{\mathbb{S}^{1}}$
the points “below and to the left of” $(t_{k},\tilde{t}_{\tilde{k}})$, etc.
Then which of the two arcs $\partial A\cap\partial E$ and $\partial
B\cap\partial E$, and which of $\partial\tilde{A}\cap\partial\tilde{E}$ and
$\partial\tilde{B}\cap\partial\tilde{E}$, a point $P_{i}$ or $\tilde{P}_{i}$
lies on is determined by which quadrant $p_{i}$ or $\tilde{p}_{i}$ lies in.

For example, suppose
$(\kappa(v),\tilde{\kappa}(\tilde{v})=(t_{1},\tilde{t}_{3})$. Then $p_{1}$ and
$\tilde{p}_{1}$ lie in the lower-right-hand quadrant
$[\kappa(v)\to\kappa(u)]_{\mathbb{S}^{1}}\times[\tilde{\kappa}(\tilde{u})\to\tilde{\kappa}(\tilde{v})]_{\mathbb{S}^{1}}$,
so both $P_{1}$ and $\tilde{P}_{1}$ lie on $\partial E\cap\partial B=[v\to
u]_{\partial E}$ and on
$\partial\tilde{E}\cap\partial\tilde{A}=[\tilde{u}\to\tilde{v}]_{\partial\tilde{E}}$.
Also $[\tilde{v}\to\tilde{u}]_{\partial E}$ is contained in $E$, because both
$\tilde{v}$ and $\tilde{u}$ are, and no $p_{i}$ nor any $\tilde{p}_{i}$ lies
the two upper quadrants
$[\tilde{v}\to\tilde{u}]_{\mathbb{S}^{1}}\times\mathbb{S}^{1}$. Then we get a
contradiction via Lemma 9.6. A similar argument gives us a contradiction via
Lemma 9.6 for $(\kappa(v),\tilde{\kappa}(\tilde{v})=(t_{3},\tilde{t}_{1})$.
From now on points $(t_{k},\tilde{t}_{\tilde{k}})$ which are handled via Lemma
9.6 in this way will be labeled with a diamond, as in Figure 21a. This
completes the proof of Proposition 11.1 when $\partial E$ and
$\partial\tilde{E}$ meet at exactly two points.

######  Case 2.

Suppose that $\partial E$ and $\partial\tilde{E}$ meet at exactly four points.

Lemma 9.2 guarantees that with a correct choice of base point, the torus
parametrization for $E$ and $\tilde{E}$ is as in Figure 22a. As before, we may
suppose without loss of generality that $u\not\in\tilde{K}$, thus
$\kappa(u)=s_{1}$, by Lemma 9.8 and relabeling the $s_{i}$ if necessary. Thus
we have the possibilities
$\tilde{\kappa}(\tilde{u})=\tilde{s}_{1},\tilde{s}_{2},\tilde{s}_{3},\tilde{s}_{4}$
to consider. The cases
$(\kappa(u),\tilde{\kappa}(\tilde{u}))=(s_{1},\tilde{s}_{2})$ and
$(\kappa(u),\tilde{\kappa}(\tilde{u}))=(s_{1},\tilde{s}_{4})$ are symmetric by
Figure 23. Figures 22b–22d give the solutions for
$\tilde{\kappa}(u)=\tilde{s}_{1},\tilde{s}_{2},\tilde{s}_{3}$, modulo some
remaining special cases.

$u=\kappa^{-1}(s_{1})$$\tilde{u}=\tilde{\kappa}^{-1}(\tilde{s}_{2})$

(a)

$u=\kappa^{-1}(s_{1})$$\tilde{u}=\tilde{\kappa}^{-1}(\tilde{s}_{4})$

(b)

Figure 23.  The topological configurations of $\\{E,u,\tilde{u}\\}$ leading to
the cases
$(\kappa(u),\tilde{\kappa}(u)=(s_{1},\tilde{s}_{2}),(s_{1},\tilde{s}_{4})$. We
see that these are equivalent via a rotation, because
$\eta(\epsilon)=\eta(\epsilon^{-1})$.

Points $(t_{k},\tilde{t}_{\tilde{k}})$ labeled with an asterisk or a diamond
are handled via Lemma 9.8 or 9.6 respectively as before. Suppose
$(\kappa(u),\tilde{\kappa}(\tilde{u}))=(s_{1},\tilde{s}_{1})$, and
$(\kappa(v),\tilde{\kappa}(\tilde{v}))=(t_{1},\tilde{t}_{1})$ in Figure 22b.
Then the upper-right-hand quadrant defined for us by $(t_{1},\tilde{t}_{1})$
contains all four points $p_{1},p_{2},\tilde{p}_{1},\tilde{p}_{2}$, thus the
circular arcs $[v\to u]_{\partial E}$ and
$[\tilde{v}\to\tilde{u}]_{\partial\tilde{E}}$ meet four times, a
contradiction. All points that are handled in this way are labeled with a
small circle. Finally, if
$(\kappa(u),\tilde{\kappa}(\tilde{u}))=(s_{1},\tilde{s}_{3})$ and
$(\kappa(v),\tilde{\kappa}(\tilde{v}))=(t_{3},\tilde{t}_{3})$ in Figure 22d,
we get a contradiction via Lemma 9.7.

######  Case 3.

Suppose that $\partial E$ and $\partial\tilde{E}$ meet at exactly six points.

Then Lemma 9.4 restricts us to two cases to consider. These are handled in
Figure 24. This completes the proof of Proposition 11.1. ∎

$(\kappa(u),\tilde{\kappa}(\tilde{u}))$$p_{1}$$p_{2}$$p_{3}$$\tilde{p}_{1}$$\tilde{p}_{2}$$\tilde{p}_{3}$$(v,\tilde{v})$

(a)

$(\kappa(u),\tilde{\kappa}(\tilde{u}))$$p_{1}$$p_{2}$$p_{3}$$\tilde{p}_{1}$$\tilde{p}_{2}$$\tilde{p}_{3}$$(v,\tilde{v})$

(b)

Figure 24.  Torus parametrizations for the eyes depicted in Figure 14. Both
are drawn with base point $(\kappa(u),\tilde{\kappa}(\tilde{u}))$. Each of the
two curves is the graph of a faithful indexable homeomorphism
$\epsilon:\partial E\to\partial\tilde{E}$ satisfying $\eta(\epsilon)=0$.

######  Proposition 11.5.

Let $\mathcal{D}=\\{D_{1},\ldots,D_{n}\\}$ and
$\tilde{\mathcal{D}}=\\{\tilde{D}_{1},\ldots,\tilde{D}_{n}\\}$ be as in the
statement of our Main Index Theorem 5.3. That is, they are thin disk
configurations in the plane $\mathbb{C}$ in general position, realizing the
same pair $(G,\Theta)$ where $G=(V,E)$ is a graph and $\Theta:E\to[0,\pi)$. In
addition, suppose that for all $i,j$ the sets $D_{i}\setminus D_{j}$ and
$\tilde{D}_{i}\setminus\tilde{D}_{j}$ meet. Suppose that there is no $i$ so
that one of $D_{i}$ and $\tilde{D}_{i}$ contains the other. Suppose that for
every disjoint non-empty $I,J\subset\\{1,\ldots,n\\}$ so that $I\sqcup
J=\\{1,\ldots,n\\}$, there exists an eye $E_{ij}$ with $i\in I$ and $j\in J$
so that one of $E_{ij}$ and $\tilde{E}_{ij}$ contains the other. Then for
every $i$ we have that any faithful indexable homeomorphism
$\delta_{i}:\partial D_{i}\to\partial\tilde{D}_{i}$ satisfies
$\eta(\delta_{i})\geq 1$. Furthermore there is a $k$ so that $D_{i}$ and
$D_{k}$ overlap for all $i$, and so that one of $E_{ij}$ and $\tilde{E}_{ij}$
contains the other if and only if either $i=k$ or $j=k$.

For the proof of Proposition 11.5, we need to establish three geometric
lemmas:

$\pi-\measuredangle(d_{-1},D)$$\pi-\measuredangle(d_{+1},D)$$\measuredangle(d_{-1},d_{+1})$$m(d_{-1})$$m(d_{+1})$$m(D)$(4.07,0.94328123)
(4.02,1.1432812) (3.995,1.2432812)(3.895,1.3932812)(3.82,1.4432813)
(3.745,1.4932812)(3.57,1.5682813)(3.47,1.5932813)
(3.37,1.6182812)(3.22,1.6932813)(3.07,1.8432813) (4.97,0.94328123)
(4.97,1.1432812) (4.97,1.2432812)(5.02,1.4182812)(5.07,1.4932812)
(5.12,1.5682813)(5.295,1.6682812)(5.42,1.6932813)
(5.545,1.7182813)(5.72,1.7932812)(5.87,1.9432813) (4.57,0.44328126)
(4.72,0.54328126) (4.795,0.59328127)(4.945,0.59328127)(5.02,0.54328126)
(5.095,0.49328125)(5.22,0.31828126)(5.27,0.19328125)
(5.32,0.06828125)(5.37,-0.18171875)(5.37,-0.30671874)
(5.37,-0.43171874)(5.395,-0.6067188)(5.47,-0.75671875)

Figure 25.  The image of the Möbius transformation described in the proof of
Lemma 11.6.

######  Lemma 11.6.

Suppose that $D,d_{-1},d_{+1}$ are closed disks in the plane $\mathbb{C}$ in
the topological configuration depicted in Figure 27a. Then
$\pi+\measuredangle(d_{-1},d_{+1})<\measuredangle(d_{-1},D)+\measuredangle(d_{+1},D)$.

###### Proof.

Let $m$ be a Möbius transformation sending a point on the bottom arc of
$\partial D\setminus d_{-1}\cup d_{+1}$ to $\infty$, so that $m(D)$ is the
lower half plane. Then the images of the disks under $m$ are as depicted in
Figure 25. We see that
$(\pi-\measuredangle(d_{-1},D))+(\pi-\measuredangle(d_{+1},D))+\measuredangle(d_{-1},d_{+1})<\pi$
and the desired inequality follows. ∎

######  Lemma 11.7.

Suppose that $D,d_{-1},d_{+1}$ are closed disks in the plane $\mathbb{C}$ in
the topological configuration depicted in Figure 26. Then
$\measuredangle(d_{-1},D)+\measuredangle(d_{+1},D)<\pi+\measuredangle(d_{-1},d_{+1})$.

###### Proof.

This is proved similarly to Lemma 11.6, see Figure 26. ∎

$\pi-\measuredangle(d_{-1},D)$$\pi-\measuredangle(d_{+1},D)$$\measuredangle(d_{-1},d_{+1})$$m(d_{-1})$$m(d_{+1})$$m(D)$$d_{-1}$$d_{+1}$$D$$m$(10.39,-0.2203125)
(10.49,-0.0703125) (10.54,0.0046875)(10.69,0.1046875)(10.79,0.1296875)
(10.89,0.1546875)(11.115,0.1796875)(11.24,0.1796875)
(11.365,0.1796875)(11.64,0.2046875)(11.79,0.2296875)
(11.94,0.2546875)(12.165,0.3296875)(12.24,0.3796875)
(12.315,0.4296875)(12.465,0.5546875)(12.54,0.6296875)
(12.615,0.7046875)(12.715,0.8546875)(12.79,1.0796875) (9.49,-0.2203125)
(9.39,-0.0703125) (9.34,0.0046875)(9.115,0.1046875)(8.94,0.1296875)
(8.765,0.1546875)(8.465,0.2046875)(8.34,0.2296875)
(8.215,0.2546875)(8.015,0.3546875)(7.94,0.4296875)
(7.865,0.5046875)(7.74,0.7046875)(7.69,0.8296875)
(7.64,0.9546875)(7.59,1.1296875)(7.59,1.2796875) (9.99,0.5796875)
(10.19,0.6796875) (10.29,0.7296875)(10.365,0.9046875)(10.34,1.0296875)
(10.315,1.1546875)(10.24,1.4046875)(10.19,1.5296875)
(10.14,1.6546875)(10.09,1.8796875)(10.09,2.1796875) (2.69,-1.8203125)
(2.79,-2.0203125) (2.84,-2.1203125)(2.99,-2.2453125)(3.09,-2.2703125)
(3.19,-2.2953124)(3.315,-2.2953124)(3.39,-2.2203126) $\infty$$m$

Figure 26.  A Möbius transformation chosen to prove Lemma 11.7. Here $\partial
m(D)=\mathbb{R}$, and $m(D)$ is the lower half-plane.

$d_{-1}$$d_{+1}$$D$

(a)

$d_{-1}$$d_{+1}$$D$

(b)

Figure 27.  The topological configurations for which we prove Lemma 11.8.

######  Lemma 11.8.

Suppose that $D,d_{-1},d_{+1}$ are closed disks in the plane $\mathbb{C}$ in
one of the two topological configurations depicted in Figure 27. In either
case, we get that both $\measuredangle(d_{-1},D)$ and
$\measuredangle(d_{+1},D)$ are strictly greater than
$\measuredangle(d_{-1},d_{+1})$.

###### Proof.

Suppose that the disks are in the configuration depicted in Figure 27a. Let
$m$ be a Möbius transformation sending a point on $\partial d_{+1}\setminus D$
to $\infty$. We may suppose without loss of generality that $m(d_{+1})$ is the
lower half-plane. Then the image of our disks under $m$ is as in Figure 28,
where $\theta_{1}=\measuredangle(d_{+1},D)$ and
$\theta_{2}=\measuredangle(d_{-1},d_{+1})$. It is then an easy exercise to
show that $\theta_{2}<\theta_{1}$ because the two circles $\partial m(d_{-1})$
and $\partial m(D)$ meet in the upper half-plane. The other inequality follows
by symmetry. The case where the disks are in the configuration depicted in
Figure 27b follows from the first case after applying a Möbius transformation
sending a point in the interior of $D\cap d_{-1}\cap d_{+1}$ to $\infty$. ∎

###### Proof of Proposition 11.5.

Recalling notation from before, if $D_{i}$ and $D_{j}$ overlap then
$E_{ij}=D_{i}\cap D_{j}$, similarly $\tilde{E}_{ij}$, and a homeomorphism
$\delta_{i}:\partial D_{i}\to\partial\tilde{D}_{i}$ is called _faithful_ if it
restricts to homeomorphisms $D_{j}\cap\partial
D_{i}\to\tilde{D}_{j}\cap\partial D_{i}$ for all $j$.

######  Claim 11.9.

Let $i,j$ be so that $\tilde{E}_{ij}\subset E_{ij}$. Denote $A=D_{i}$,
$B=D_{j}$, $\tilde{A}=\tilde{D}_{i}$, $\tilde{B}=\tilde{D}_{j}$. Then both
$\spadesuit$c and $\clubsuit$c occur. Also one of $\diamondsuit$d,
$\diamondsuit$e, $\diamondsuit$g occurs, and one of $\heartsuit$d,
$\heartsuit$e, $\heartsuit$g occurs. Furthermore at least one of
$\diamondsuit$g and $\heartsuit$g occurs.

To see why, note first that both $\spadesuit$c and $\clubsuit$c occur, because
these are the only candidates in Figure 10 where $\tilde{A}\cap\tilde{B}$ is
contained in the respective one of $A$ and $B$. Note the following by Lemmas
11.8

(5)
$\measuredangle(A,B)=\measuredangle(\tilde{A},\tilde{B})<\measuredangle(\tilde{A},B)$

and the following by Lemma 11.6.

(6)
$\pi+\measuredangle(\tilde{A},\tilde{B})<\measuredangle(A,\tilde{A})+\measuredangle(\tilde{B},A),\qquad\pi+\measuredangle(\tilde{A},\tilde{B})<\measuredangle(\tilde{A},B)+\measuredangle(\tilde{B},B)$

Next, because $\tilde{A}\cap\tilde{B}$ contains part of $\partial\tilde{A}$
and part of $\partial\tilde{B}$, both of these circles must pass through
$A\cap B$. Noting that $\diamondsuit$f cannot occur because
$\tilde{A}\not\subset A$, we conclude that one of $\diamondsuit$a,
$\diamondsuit$d, $\diamondsuit$e, $\diamondsuit$g, and $\diamondsuit$h occurs.
If either of $\diamondsuit$a and $\diamondsuit$h occurs, then Lemma 11.8
implies that $\measuredangle(\tilde{A},B)<\measuredangle(A,B)$, contradicting
5. This leaves us with only the claimed possibilities $\diamondsuit$d,
$\diamondsuit$e, and $\diamondsuit$g. By symmetry we also get that one of
$\heartsuit$d, $\heartsuit$e, and $\heartsuit$g occurs.

Finally, note by Lemma 11.7 that if $\diamondsuit$d or $\diamondsuit$e occurs
then we get
$\measuredangle(\tilde{A},A)+\measuredangle(\tilde{A},B)<\pi+\measuredangle(A,B)$,
and if $\diamondsuit$d or $\diamondsuit$e occurs then we get
$\measuredangle(\tilde{B},A)+\measuredangle(\tilde{B},B)<\pi+\measuredangle(A,B)$.
We get that if neither of $\diamondsuit$g and $\heartsuit$g occurs, then we
may combine these two inequalities with 6 to arrive at a contradiction,
establishing Claim 11.9.

$\theta_{1}$$\theta_{2}$$m(d_{+1})$$m(d_{-1})$$m(D)$

Figure 28.  The image of the Möbius transformation described in the proof of
Lemma 11.8.

Moving on, pick $1\leq i\leq n$. By the hypotheses of Proposition 11.5 there
is a $j$ so that one of $E_{ij}$ and $\tilde{E}_{ij}$ contains the other,
without loss of generality so that $\tilde{E}_{ij}\subset E_{ij}$. Let
$\delta_{i}:\partial D_{i}\to\partial\tilde{D}_{i}$ be a faithful indexable
homeomorphism. Continuing with the notation of Claim 11.9, regardless of which
of $\diamondsuit$d, $\diamondsuit$e, and $\diamondsuit$g occurs, there is a
point $z\in\partial A\cap\partial E$ so that $z$ lies in the interior of
$\tilde{A}$. Furthermore note that $\delta_{i}(z)\in\partial\tilde{E}$ by the
faithfulness condition, and that $\tilde{E}\subset A$ by our hypotheses, so
$\delta_{i}(z)$ lies in the interior of $A$. Thus if we draw a torus
parametrization for $A$ and $\tilde{A}$ using
$(\kappa(z),\tilde{\kappa}(\delta_{i}(z)))$ as the base point, Lemma 10.1
implies that $\eta(\delta_{i})\geq 1$, because $\partial A$ and
$\partial\tilde{A}$ meet exactly twice. This establishes the first part of
Proposition 11.5.

Next, let $H_{\mathrm{u}}$ be the undirected simple graph having
$\\{1,\ldots,n\\}$ as its vertex set, so that $\left<i,j\right>$ is an edge in
$H_{\mathrm{u}}$ if and only if $D_{i}$ and $D_{j}$ overlap and one of
$E_{ij}$ and $\tilde{E}_{ij}$ contains the other. Note that $H_{\mathrm{u}}$
is connected, otherwise we could pick $I$ to be the vertex set of one
connected component of $H_{\mathrm{u}}$ and $J$ to be
$\\{1,\ldots,n\\}\setminus I$ to contradict the hypotheses of Proposition
11.5.

Let $H$ be the directed graph obtained from $H_{\mathrm{u}}$ in the following
way. Suppose $\left<i,j\right>$ is an edge in $H_{\mathrm{u}}$. Denote
$A=D_{i}$, $B=D_{j}$, $\tilde{A}=\tilde{D}_{i}$, $\tilde{B}=\tilde{D}_{j}$.
Then $\left<i\to j\right>$ is an edge in $H$ if and only if one of
$\diamondsuit$g and $\spadesuit$g occurs. In particular Claim 11.9 implies
that if $\left<i,j\right>$ is an edge in $H_{\mathrm{u}}$ then at least one of
$\left<i\to j\right>$ and $\left<j\to i\right>$ is an edge in $H$, and
possibly both are.

######  Claim 11.10.

Suppose that $\left<i\to j\right>$ is an edge in $H$. Then $\left<i,j\right>$
is the only edge in $H_{\mathrm{u}}$ having $i$ as a vertex.

To see why, observe first that if $\diamondsuit$d or $\diamondsuit$e occurs
then one intersection point $\partial A\cap\partial\tilde{A}$ lie in the
interior of $B$, and if $\diamondsuit$g occurs then both do. Suppose without
loss of generality that $\tilde{D}_{i}\cap\tilde{D}_{j}\subset D_{i}\cap
D_{j}$. Then both intersection points $\partial
D_{i}\cap\partial\tilde{D}_{i}$ lie in the interior of $D_{j}$. For
contradiction let $k\neq j$ so that $\left<i,k\right>$ is an edge in
$H_{\mathrm{u}}$. There are two cases.

######  Case 1.

Suppose that $\tilde{D}_{i}\cap\tilde{D}_{k}\subset D_{i}\cap D_{k}$.

Then one or both points $\partial D_{i}\cap\partial\tilde{D}_{i}$ lie in the
interior of $D_{k}$. Then there is a point in the interior of $D_{i}$ which
lies in the interiors of both $D_{j}$ and $D_{k}$, a contradiction.

######  Case 2.

Suppose that $D_{i}\cap D_{k}\subset\tilde{D}_{i}\cap\tilde{D}_{k}$.

Then by a symmetric restatement of Claim 11.9 we get that both points
$\partial D_{i}\cap\partial D_{k}$ lie in $\tilde{D}_{i}$. On the other hand
$\tilde{D}_{i}\cap\partial D_{i}$ is contained in the interior of $D_{j}$ by
$\diamondsuit$a. Thus there are points interior to all of $D_{i},D_{j},D_{k}$,
a contradiction.

This establishes Caim 11.10.

Thus $H$ is either the graph on two vertices $\\{i,j\\}$ having one or both of
$\left<i\to j\right>$ and $\left<j\to i\right>$ as edges, or is a graph having
$\\{k,i_{1},\ldots,i_{n-1}\\}$ as vertices and exactly the edges
$\left<i_{\ell}\to k\right>$ for $1\leq\ell<n$. The last part of Proposition
11.5 follows. ∎

This completes the proofs of the main results of this article.

##  12\. Generalizations, open problems, and conjectures

We conclude the article with some general conjectures which are directly
related to the new results of this article.

First, we discuss eliminating the thinness condition from the hypotheses of
our theorem statements. Most simply, it seems likely that our Main Theorems
1.5 and 1.4 should continue to hold with the thinness condition completely
omitted. In this direction, we conjecture the following fixed-point index
statement for non-thin configurations of disks:

######  Conjecture 12.1.

Suppose that $\mathcal{D}$ and $\tilde{\mathcal{D}}$ are disk configurations
in $\mathbb{C}$ realizing the same incidence data. Then any faithful indexable
homeomorphism $\phi:\partial\mathcal{D}\to\partial\tilde{\mathcal{D}}$
satisfies $\eta(\phi)\geq 0$.

Note that something stronger than Conjecture 12.1 would be required to prove
the corresponding generalizations of our main results on disk configurations
using the methods of this article: in particular, we would probably need to
generalize the notion of an isolated subsumptive subset of the common index
set of $\mathcal{D}$ and $\tilde{\mathcal{D}}$. However, it is plausible that
this is a workable approach.

We remark at this point that in the present author’s thesis, see [mishchenko-
thesis], the definition of _thin_ used in the statements of our Main Theorems
1.5 and 1.4 is slightly weaker than the one we have given here: there we call
a disk configuration _thin_ if given three disks from it, the intersection of
their interiors is empty. The proofs there are essentially the same, without
any interesting new ideas, but there are technically annoying degenerate
situations to deal with, so we do not work at this level of generality here.

More strongly, we make two conjectures which together would subsume all other
currently known rigidity and uniformization statements on disk configurations.
First:

######  Conjecture 12.2.

Suppose that $\mathcal{C}$ and $\tilde{\mathcal{C}}$ are disk configurations,
locally finite in $\mathbb{G}$ and $\tilde{\mathbb{G}}$ respectively, where
each of $\mathbb{G}$ and $\tilde{\mathbb{G}}$ is equal to one of $\mathbb{C}$
and $\mathbb{H}^{2}$, with the _a priori_ possibility that
$\mathbb{G}\neq\tilde{\mathbb{G}}$. Suppose that $\mathcal{C}$ and
$\tilde{\mathcal{C}}$ share a contact graph $G=(V,E)$. Suppose further that
$calC$ and $\tilde{\mathcal{C}}$ fill their respective spaces, in sense that
every connected component of $\mathbb{G}\setminus\cup_{D\in\mathcal{C}}D$ or
of
$\tilde{\mathbb{G}}\setminus\cup_{\tilde{D}\in\tilde{\mathcal{C}}}\tilde{D}$
is bounded. Then $\mathbb{G}=\tilde{\mathbb{G}}$.

Second:

######  Conjecture 12.3.

Suppose that $\mathcal{C}$ and $\tilde{\mathcal{C}}$ are disk configurations,
both locally finite in $\mathbb{G}$, where $\mathbb{G}$ is equal to one of
$\mathbb{C}$ and $\mathbb{H}^{2}$. Suppose that $\mathcal{C}$ and
$\tilde{\mathcal{C}}$ realize the same incidence data $(G,\Theta)$. Suppose
further that some maximal planar subgraph of $G$ is the 1-skeleton of a
triangulation of a topological open disk. Then $\mathcal{P}$ and
$\tilde{\mathcal{P}}$ differ by a Euclidean similarity if
$\mathbb{G}=\mathbb{C}$ or by a hyperbolic isometry if
$\mathbb{G}=\mathbb{H}^{2}$.

We also make the natural conjecture analogous to Conjecture 12.3 for disk
configurations on the Riemann sphere. It seems plausible that a fixed-point
index approach could work to prove Conjectures 12.2 and 12.3. An alternative
approach to try to prove Conjecture 12.2 is via vertex extremal length
arguments, along the lines of [MR1680531]*Uniformization Theorem 1.3 and
[MR1331923].

Finally, we conjecture that Conjecture 12.3 is the best possible uniqueness
statement of its type, in the following precise sense:

######  Conjecture 12.4.

Let $\mathcal{C}$ be a disk configuration which is locally finite in
$\mathbb{G}$, where $\mathbb{G}$ is one of $\hat{\mathbb{C}}$, $\mathbb{C}$,
or $\mathbb{H}^{2}$. Let $(G,\Theta)$ be the incidence data of $\mathcal{C}$.
Suppose that no maximal planar subgraph of $G$ is the 1-skeleton of a
triangulation of a topological open disk. Then there are other locally finite
disk configurations in $\mathbb{G}$ realizing $(G,\Theta)$ which are not
images of $\mathcal{C}$ under any conformal or anti-conformal automorphism of
$\mathbb{G}$.

The most promising tool to prove Conjecture 12.4 would be a good existence
statement taking incidence data $(G,\Theta)$ as input.

$D_{1}$$D_{2}$$\tilde{D}_{1}$$\tilde{D}_{2}$(4.37,1.52) (4.33,1.65)
(4.31,1.715)(4.32,1.835)(4.35,1.89) (4.38,1.945)(4.48,2.03)(4.55,2.06)
(4.62,2.09)(4.845,2.13)(5.0,2.14) (5.155,2.15)(5.465,2.15)(5.62,2.14)
(5.775,2.13)(6.095,2.075)(6.26,2.03) (6.425,1.985)(6.69,1.88)(6.79,1.82)
(6.89,1.76)(7.045,1.64)(7.1,1.58) (7.155,1.52)(7.21,1.425)(7.21,1.39)
(7.21,1.355)(7.195,1.3)(7.18,1.28) (7.165,1.26)(7.125,1.22)(7.1,1.2)
(7.075,1.18)(7.03,1.145)(7.01,1.13) (6.99,1.115)(6.955,1.09)(6.94,1.08)
(6.925,1.07)(6.905,1.06)(6.89,1.06) (3.93,-1.48) (3.96,-1.68)
(3.975,-1.78)(3.885,-1.975)(3.78,-2.07) (3.675,-2.165)(3.3,-2.2)(3.03,-2.14)
(2.76,-2.08)(2.285,-1.93)(2.08,-1.84) (1.875,-1.75)(1.555,-1.595)(1.44,-1.53)
(1.325,-1.465)(1.25,-1.325)(1.29,-1.25) (1.33,-1.175)(1.405,-1.07)(1.44,-1.04)
(1.475,-1.01)(1.525,-0.975)(1.57,-0.96) (4.45,-1.38) (4.42,-1.53)
(4.405,-1.605)(4.57,-1.745)(4.75,-1.81) (4.93,-1.875)(5.34,-1.95)(5.57,-1.96)
(5.8,-1.97)(6.17,-1.95)(6.31,-1.92) (6.45,-1.89)(6.65,-1.805)(6.71,-1.75)
(6.77,-1.695)(6.83,-1.58)(6.83,-1.52) (6.83,-1.46)(6.825,-1.36)(6.82,-1.32)
(6.815,-1.28)(6.805,-1.21)(6.8,-1.18) (6.795,-1.15)(6.79,-1.105)(6.79,-1.06)
(3.77,1.46) (3.78,1.56) (3.785,1.61)(3.695,1.715)(3.6,1.77)
(3.505,1.825)(3.3,1.895)(3.19,1.91) (3.08,1.925)(2.805,1.94)(2.64,1.94)
(2.475,1.94)(2.145,1.895)(1.98,1.85) (1.815,1.805)(1.595,1.71)(1.54,1.66)
(1.485,1.61)(1.415,1.52)(1.4,1.48) (1.385,1.44)(1.385,1.345)(1.4,1.29)
(1.415,1.235)(1.455,1.145)(1.48,1.11) (1.505,1.075)(1.545,1.015)(1.56,0.99)
(1.575,0.965)(1.605,0.935)(1.65,0.92)

Figure 29.  A counterexample to Theorem 5.3 if we allow
$\measuredangle(D_{1},D_{2})\neq\measuredangle(\tilde{D}_{1},\tilde{D}_{2})$.
Any indexable $\phi:\partial(D_{1}\cup
D_{2})\to\partial(\tilde{D}_{1}\cup\tilde{D}_{2})$ making the shown
identifications gives $\eta(f)=-1$.

Finally, we consider other directions in which our Main Index Theorem 5.3
could be generalized. First, one may hope to weaken the condition that
$\mathcal{D}$ and $\tilde{\mathcal{D}}$ realize the same incidence data,
insisting only that they share a contact graph. Figure 29 provides an explicit
small-scale counterexample. Alternatively, we may hope to prove a theorem
analogous to Theorem 5.3 for collections of shapes other than metric closed
disks. For example, if $K$ and $\tilde{K}$ are compact convex sets in
$\mathbb{C}$ having smooth boundaries, one of which is the image of the other
by translation and scaling, then $\partial K$ and $\partial\tilde{K}$ meet at
most twice, so the Circle Index Lemma 3.2 applies. This gives hope for a
generalization of Theorem 5.3 in this direction. Schramm has proved rigidity
theorems for packings by shapes other than circles using related ideas, for
example in [MR1076089].

## References