# Pentagram Rigidity for Centrally Symmetric Octagons

Richard Evan Schwartz Supported by N.S.F. Grant DMS-2102802

###### Abstract

In this paper I will establish a special case of a conjecture that intertwines
the deep diagonal pentagram maps and Poncelet polygons. The special case is
that of the $3$-diagonal map acting on affine equivalence classes of centrally
symmetric octagons. This is the simplest case that goes beyond an analysis of
elliptic curves. The proof involves establishing that the map is Arnold-
Liouville integrable in this case, and then exploring the Lagrangian surface
foliation in detail.

##  1 Introduction

###  1.1 Conjecture and Results

A Poncelet polygon in the projective plane is a polygon whose vertices lie in
one conic section and whose edges lie in lines tangent to another conic
section. A polygon in the projective plane is convex if it is convex in the
ordinary sense when viewed in an affine patch. In an affine patch, a convex
Poncelet polygon is inscribed in one ellipse and circumscribed about another.

Let $(n,k)$ be relatively prime integers with $n\geq 7$ and $k\in(2,n/2)$.
Given an $n$-gon $P_{0}$, we let $P_{1}=T_{k}(P_{0})$ be the $n$-gon obtained
by intersecting the successive $k$-diagonals of $P_{0}$. Figure 1 shows this
for $(n,k)=(8,3)$. The map $T_{k}$ is generically defined and invertible. The
maps $T_{k}$ and $T_{k}^{-1}$ are always defined on convex $n$-gons. Figure 1
(right) gives an example where $P_{0}$ is convex but $P_{1}=T_{3}(P_{0})$ is
not.

Figure 1: $P_{0}$ and $P_{1}=T_{3}(P_{0})$.

Starting with $P_{0}$ we define the $k$-diagonal orbit to be the bi-infinite
sequence $\\{P_{j}\\}$ where $P_{j}=T_{k}^{j}(P_{0})$. In [16] I proved that
if $P$ is Poncelet then $T_{k}(P)$ and $P$ are projectively equivalent. This
means in particular that if $P_{0}$ is convex Poncelet then $P_{j}$ is convex
for all $j\in\mbox{\boldmath{$Z$}}$. The Pentagram Rigidity Conjecture says
that the converse is also true.

###### Conjecture 1.1

Let $P_{0}$ be a convex $n$-gon and let $\\{P_{j}\\}$ be its $k$-diagonal
orbit. Then $P_{0}$ is convex Poncelet if and only if $\\{P_{j}\\}$ is convex
for all $j\in\mbox{\boldmath{$Z$}}$.

The restriction that $n$ and $k$ are relatively prime is perhaps not strictly
necessary, but the result can fail when $n$ and $k$ are both even. We take
$k\geq 3$ because the map $T_{2}$, the extensively studied pentagram map,
preserves convexity. So does $T_{2}^{-1}$.

In [17] I proved Conjecture 1.1 for the case of $8$-gons with $4$-fold
rotational symmetry. In this toy case, the relevant moduli space is foliated
by $T_{3}$ invariant elliptic curves. The case of $7$-gons with bilateral
symmetry would have a similar story. In this paper I will prove the first
really nontrivial case of the conjecture, a case that goes beyond the analysis
of elliptic curves. An even-sided polygon $P$ is centrally symmetric if it is
invariant with respect to the map $p\to-p$. The Poncelet polygons with an even
number of sides are always centrally symmetric. See [4].

######  Theorem 1.2 (Main)

Let $P_{0}$ be a convex centrally symmetric $8$-gon and $\\{P_{j}\\}$ its
$3$-diagonal orbit. Then

  * •

$P_{0}$ is inscribed in an ellipse iff $P_{j}$ is convex for all $j\leq 0$.

  * •

$P_{0}$ is circumscribed about an ellipse iff $P_{j}$ is convex for all $j\geq
0$.

In particular, $P_{0}$ is Poncelet if and only if $P_{j}$ is convex for all
$j\in\mbox{\boldmath{$Z$}}$.

Theorem 1.2 implies, in particular, that being inscribed or circumscribed in
an ellipse is a $T_{3}$ invariant condition. This invariance, for general
$8$-gons, is already a theorem in my paper with Serge Tabachnikov [18]. In
§8.2, when I treat the case of inscribed and circumscribed $8$-gons, I will
obtain further structural information about the inscribed and circumscribed
cases.

###### Theorem 1.3

Let $P_{0}$ be a convex centrally symmetric $8$-gon which is not Poncelet but
which is still inscribed in an ellipse. Let $\\{P_{k}\\}$ be its $3$-diagonal
orbit. Then $P_{k}$ converges to a convex Poncelet $8$-gon as $k\to-\infty$
and (if the iterates are all defined) to a star-convex Poncelet $8$-gon as
$k\to+\infty$. Up to affine transformation, the two limits have the same
vertex set and the vertex orders are related by the star-reordering:
$12345678\to 14725836$.

Let ${\cal P\/}_{8,2}$ denote the space of affine equivalence classes of
centrally symmetric $8$-gons. We choose coordinates so that ${\cal P\/}_{8,2}$
is an open dense subset of $\mbox{\boldmath{$R$}}^{4}$. One step in the proof
of Theorem 1.2 is proving the following integrability result. (See §2.1 for
definitions.)

###### Theorem 1.4

The action of $T_{3}$ on ${\cal P\/}_{8,2}$ has an invariant (singular)
symplectic form and $2$ algebraically independent rational invariants which
Poisson commute with respect to it.

###  1.2 Context

The pentagram map, $T_{2}$, is one of the best known discrete completely
integrable system. See for instance, [1], [2], [3], [5], [6], [7], [9], [10],
[11], [12], [13], [14], [15], [20], [21]. The Pentagram Rigidity Conjecture
would be a statement about the global topology of an abelian foliation
associated to a completely integrable system if the map $T_{3}$ were known to
be completely integrable.

In a certain sense, though not in a sense useful for our purposes, certain
integrability results are known for $T_{k}$ for all $k\geq 2$. It is proved in
[3] that $T_{k}$ is completely integrable when defined on the space of
twisted, corrugated polygons. These are certain maps from the integers into
$\mbox{\boldmath{$R$}}\mbox{\boldmath{$P$}}^{n}$ for various $n$. A general
integrability result, Theorem 6.2 in [7], covers the action of $T_{3}$ on the
space of projective classes of ordinary polygons, in the sense that this is a
subspace of a larger space where the map is completely integrable. However, it
seems difficult to extract from the work in [3] and [7] an explicit
integrability result like Theorem 1.4.

Concerning Theoren 1.4, I found the invariant symplectic form on ${\cal
P\/}_{8,2}$, as well as the two invariant quantities, by inspired guesswork. I
did not try to use the various machines that are now known to work with
$T_{2}$ – e.g., cluster algebras, refactoring, Lax pairs, the Kenyon-Goncharov
dimer systems, and spin networks.

Theorem 1.2 involves an analysis of the geometry and topology of the level
sets of the invariants. The essential point is that the level sets which
intersect the subspace of ${\cal P\/}_{8,2}$ containing the convex polygons
also contain points outside this space. The dynamics on these level sets turns
out to be essentially torus motion, and the dynamics mixes up the subset of
convex points with the subset of non-convex points unless the starting point
represents a Poncelet $8$-gon.

It is surprising to me that this integrable structure coexists with the
hyperbolic nature of the action of $T_{3}$ on the set of inscribed and
circumscribed $8$-gons, but there is no contradiction here. This sort of
coexistence also appears e.g. in the phase space of elliptical billiards. We
have a higher dimensional analogue of the same phenomenon.

###  1.3 Pictures of Orbits

Figure 2 (left) shows a planar projection of the first $2^{15}$ points of an
orbit that starts with a point representing a convex $8$-gon.

Figure 2: A torus orbit and a beautiful orbit.

We prove that (after blowing up some points in a suitable sense) the metric
completion of this orbit is a torus. The darkly shaded part of the orbit is
the set corresponding to convex $8$-gons.

Figure 2 (right) shows a planar projection of the first $2^{19}$ points of an
orbit that does not contain points representing convex $8$-gons. This orbit
appears to lie on a higher genus surface with a singular flat structure. This
kind of orbit is not relevant for our analysis but it makes a beautiful
picture and hints at some additional structure we do not explore in this
paper.

###  1.4 Organization

In §2 I will give some background about integrable systems, and also recall
the definition of the resultant of two polynomials. In §3 I will prove Theorem
1.4. In §3.6 I give a $6$-step outline of the proof of Theorem 1.2. In §4-8 I
carry out the $6$ steps of the outline. In §8.2 I prove Theorem 1.3.

Why is this paper so long? Theorem 1.4 has a short proof, but then the
challenges begin. One difficulty is that we need to understand every single
orbit that contains points representing convex $8$-gons, so the usual appeal
to Sard’s Theorem to show that most level sets are smooth will not help us. We
need to use some computer algebra, and this boils down to computing a bunch of
resultants. Even after we know that every relevant orbit is smooth, we have to
deal with the fact that the level sets have many components to them because
they are divided up by the coordinate hyperplanes, where the invariants and
the symplectic form blow up. We need to compute the topology of every
component and then assemble the components. The final difficulty is that
$T_{3}$ sometimes blows up or blows down various ends of our level set
components.

The interested reader can download the computer program I wrote, which does
experiments with the $3$-diagonal map on ${\cal P\/}_{8,2}$. See
http://www.math.brown.edu/$\sim$res/Java/OCTAGON.tar

###  1.5 Acknowledgements

I would like to thank Misha Bialy, Dan Cristofaro-Gardiner, Misha Gehktman,
Anton Izosimov, Boris Khesin, Curtis McMullen, Valentin Ovsienko, Dan Reznik,
Joe Silverman, Sergei Tabachnikov, and Max Weinreich for helpful
conversations.

##  2 Preliminaries

###  2.1 Complete Integrability

Here we discuss the very basics of integrable systems.

Invariant Functions and Level Sets: Let $U\subset
V\subset\mbox{\boldmath{$R$}}^{4}$ be two open sets and suppose we have a
smooth map $T:U\to V$. Our map $T$ will be a suitable restriction of maps
closely related to $T_{3}$, and $U,V$ will be subsets of ${\cal P\/}_{8,2}$, a
space we identify with an open dense subset of $\mbox{\boldmath{$R$}}^{4}$.

Suppose that $F_{1},F_{2}:V\to\mbox{\boldmath{$R$}}$ are two smooth
$T$-invariant functions. The map $\Psi=(F_{1},F_{2})$ gives us a smooth and
$T$-invariant map from $V$ into $\mbox{\boldmath{$R$}}^{2}$. We call a pair
$(x,y)\in\mbox{\boldmath{$R$}}^{2}$ a regular value if the gradients $\nabla
F_{1}$ and $\nabla F_{2}$ are nonzero and linearly independent for all
$p\in\Psi^{-1}(x,y)$. Otherwise we call $(x,y)$ a singular value. When $(x,y)$
is a regular value the level set $\Sigma=\Psi^{-1}(x,y)\subset V$ is a smooth
surface. Furthermore $T(\Sigma\cap U)\subset\Sigma$.

Invariant Symplectic Form: We fix some level set $\Sigma$ and assume it is a
regular level set. A symplectic form on $V$ is a smooth, closed and non-
degenerate $2$-form. Let $\omega$ be such a form. This form is called
$T$-invariant if $T^{*}(\omega)=\omega$. Here $T^{*}$ is the usual pullback
operation. Given $\omega$ and $F_{j}$, there is a unique vector field $X_{j}$
such that

$\omega(X_{j},V)=D_{V}F_{j}.$

Here $D_{V}F_{j}$ is the directional derivative of $F_{j}$ in the direction of
$V$. The vector field $X_{j}$ is called the Hamiltonian of $F_{j}$. Because
$\omega(X_{j},X_{j})=0$, the vector field $X_{j}$ is tangent to the level set
of $F_{j}$. Also, the flow generated by $F_{j}$ is symplectic; it preserves
$\omega$.

The functions $F_{1}$ and $F_{2}$ Poisson commute if

$\omega(X_{1},X_{2})=0$ (1)

everywhere. The vectors $X_{1},X_{2}$ are linearly independent at some point
iff the gradients $\nabla F_{1},\nabla F_{2}$ are linearly independent at this
point. When this happens, the restriction of $\omega$ to $\Sigma$ is $0$. That
is, $\Sigma$ is Lagrangian. The vector fields $X_{1}$ and $X_{2}$ define
commuting flows that preserve both $\Sigma$ and $\omega$.

Translation Surface Structure: We can use these commuting flows to define
coordinate charts from $\Sigma$ into $\mbox{\boldmath{$R$}}^{2}$ in a
canonical way. We start with some point $p\in\Sigma$, which we map to the
origin. Each point $q\in\Sigma$ sufficiently near to $p$ defines two numbers
$a_{1}(p,q)$ and $a_{2}(p,q)$ such that one can reach $q$ by starting at $p$
and flowing for time $a_{1}(p,q)$ along $X_{1}$ and then for time $a_{2}(p,q)$
along $X_{2}$. The coordinate chart is given by $q\to(a_{1}(p,q),a_{2}(p,q)).$
The commuting nature of the flows combines with the linear independence to
show that this map really is a local coordinate chart about $p$.

By construction, the overlap functions for our coordinate chart are
translations. Thus $\Sigma$ has the structure of translation surface (without
singular points). Since everything is $T$-invariant, the map $T:\Sigma\cap
U\to\Sigma$ is a translation in these coordinates. In this case we would call
$T$ completely integrable with respect to the pair $(U,V)$.

In our case we have a rational map $T$ on $\mbox{\boldmath{$R$}}^{4}$ and we
will construct two invariant rational functions $F_{1}$ and $F_{2}$. We also
will find a $T$-invariant and rationally defined symplectic form. All these
objects have singularities; they are only defined on open dense subsets of
$\mbox{\boldmath{$R$}}^{4}$. However, we will always restrict our attention to
suitable pairs $(U,V)$ of open sets where everything is everywhere defined.

###  2.2 Resultants

For ease of typesetting, I will describe the resultant for two polynomials of
degree $2$ and $3$. The construction is meant to work for general polynomials
of degree $m$ and $n$. See §2 of [19] for more details. Given $2$ polynomials
$P=a_{2}x^{2}+a_{1}x+a_{0}$ and $Q=b_{3}x^{3}+b_{2}x^{2}+b_{1}x+b_{0},$ the
resultant is defined to be

${\rm res\/}(P,Q)={\rm det\/}\left[\begin{matrix}a_{2}&a_{1}&a_{0}&0&0\cr
0&a_{2}&a_{1}&a_{0}&0\cr 0&0&a_{2}&a_{1}&a_{0}\cr b_{3}&b_{2}&b_{1}&b_{0}&0\cr
0&b_{3}&b_{2}&b_{1}&b_{0}\end{matrix}\right]$ (2)

This number vanishes if and only if $P$ and $Q$ have a common (complex) root.

In the multivariable case, one can treat two polynomials $P(x_{1},...,x_{n})$
and $Q(x_{1},...,x_{n})$ as elements of the ring $R[x_{n}]$ where
$R=\mbox{\boldmath{$C$}}[x_{1},...,x_{n-1}]$. The resultant ${\rm
res\/}(P,Q,x_{n})$ computes the resultant in $R$ and thus gives a polynomial
in $\mbox{\boldmath{$C$}}[x_{1},...,x_{n-1}]$. The polynomials $P$ and $Q$
simultaneously vanish at $(x_{1},...,x_{n})$ only if the resultant ${\rm
res\/}(P,Q,x_{n})$ vanishes at $(x_{1},...,x_{n-1})$.

##  3 The Main Ideas

###  3.1 The Map in Coordinates

Every member $P\in{\cal P\/}_{8,2}$ has a canonical representative with
vertices

$(1,0),\hskip 4.0pt(a,b),\hskip 4.0pt(0,1),\hskip 4.0pt(-d,c),\hskip
4.0pt(-1,0),\hskip 4.0pt(-a,-b),\hskip 4.0pt(0,-1),\hskip 4.0pt(d,-c).$ (3)

We label these vertices $v_{0},....,v_{7}$. We call $p=(a,b,c,d)$ the
coordinates if $P$. When $(a,b)=(c,d)$ our point lies in the space ${\cal
P\/}_{8,4}$ of affine equivalence classes of $8$-gons with $4$-fold rotational
symmetry.

When we apply the map $T_{3}$ we initially get a polygon $P^{\prime}$ with
vertices $v_{0}^{\prime},...,v_{7}^{\prime}$, where

$v_{k}^{\prime}=\overline{v_{k+1}v_{k+4}}\cap\overline{v_{k+2}v_{k+5}},$

where the indices are taken mod $8$. One could use other labeling conventions,
but this one leads to a nice formula for $T_{3}$.

$T_{3}=A\Delta A\Delta,$ (4)

where

$A(a,b,c,d)=(-b,-a,-d,-c),$ (5)

and after setting $e=ac+bd$,

$\Delta=\bigg{(}\frac{b(c+d+1)}{c(e+a+c+1)},\frac{d(e+b+c)}{c(e+a+c+1)},\frac{d(a+b+1)}{a(e+a+c+1)},\frac{b(e+a+d)}{a(e+a+c+1)}\bigg{)}$
(6)

The map $\Delta$ is especially nice because all its component functions are
positive. Here are some symmetries of our map:

$\displaystyle I(A\Delta)I=A\Delta,\hskip 30.0ptI(a,b,c,d)=(c,d,a,b).$ (7)
$\displaystyle JT_{3}J=IT_{3},\hskip 30.0ptJ(a,b,c,d)=(b,a,d,c).$

Since $I$ commutes with $A\Delta$, we see that $IT_{3}I=T_{3}$. Also, $J$
commutes with $T_{3}^{2}$ because

$JT_{3}^{2}J=(JT_{3}J)(JT_{3}J)=(IT_{3})(IT_{3})=(IT_{3}I)T_{3}=T_{3}^{2}.$

When it comes time to prove Theorem 1.2 we will use the action of $J$ to
reduce the number of cases we have to consider.

###  3.2 Some Important Functions

Let ${\cal I\/}\subset{\cal P\/}_{8,2}$ denote the subset consisting of
equivalence classes of $8$-gons which are inscribed in a conic section. Let
${\cal I\/}^{*}\subset{\cal P\/}_{8,2}$ denote the subset consisting of
equivalence classes of $8$-gos circumscribed about a conic section. Define
functions

$g^{*}_{ab}=a-b,\hskip 20.0ptg^{*}_{cd}=c-d,\hskip
20.0ptg_{ab}=\frac{1-a^{2}-b^{2}}{ab},\hskip
20.0ptg_{cd}=\frac{1-c^{2}-d^{2}}{cd}.$ (8)

A computation shows that

$(a,b,c,d)\in{\cal I\/}\hskip 3.0pt\Longleftrightarrow\hskip
3.0ptg_{ab}+g_{cd}=0,\hskip 10.0pt(a,b,c,d)\in{\cal I\/}^{*}\hskip
3.0pt\Longleftrightarrow\hskip 3.0ptg^{*}_{ab}+g^{*}_{cd}=0.$ (9)

The first of these equations has a simple geometric interpretation. From the
way we have normalized our $8$-gons, the point $(a,b)$ lies on the ellipse
$x^{2}+y^{2}+g_{ab}xy=0$. The point $(-d,c)$ lies on the ellipse
$x^{2}+y^{2}-g_{cd}xy=0$. If these points lie on the same ellipse then the
constants are the same, meaning that $g_{ab}+g_{cd}=0$. I don’t know a nice
geometric interpretation for the second equation, but it is an easy
calculation.

As we will see in §8.2, the functions $g_{ab}$, etc., play a special role in
the analysis of inscribed and circumscribed $8$-gons. They also serve as the
building block for a $T_{3}$-invariant function we call $G$. I discovered that
the function

$G(a,b,c,d)=2(g_{ab}+g_{cd})(g_{ab}^{*}+g_{cd}^{*})$ (10)

is an invariant for $T_{3}$.

After finding $G$, I looked for a second invariant. Referring to the
literature on the pentagram map, say [15], I guessed that the polynomial
expression $(O_{8}/E_{8})^{1/6}$ is an invariant for $T_{3}$. The functions
$O_{8}$ and $E_{8}$ are (now) called the odd and even Casimirs for the
$T_{2}$-invariant Poisson structure.

After finding these invariants, I played around with them until I came up with
algebraically nicer ones, $F_{1}$ and $F_{2}$, described in the next section.
The invariants $F_{1}$ and $F_{2}$ satisfy the relations:

$F_{2}-F_{1}=G,\hskip 30.0pt\frac{F_{1}}{F_{2}}=(O_{8}/E_{8})^{1/6}.$ (11)

Remark: In spite of this discussion about the origins of $F_{1}$ and $F_{2}$,
the reader need not know anything about $O_{8}$ and $E_{8}$ to understand
$F_{1}$ and $F_{2}$.

###  3.3 Proof of Theorem 1.4

Let $e=ac+bd$. Our two basic invariants for $T_{3}$ are $F_{1}$ and $F_{2}$
where

$F_{1}=\frac{(1+a-b)(1+c-d)(e+b-c)(e+d-a)}{abcd},$
$F_{2}=\frac{(1-a+b)(1-c+d)(e-b+c)(e-d+a)}{abcd}.$ (12)

Referring to the maps in Equation 7, these functions obey the symmetries:
$F_{j}\circ I=F_{j}$ and $F_{3-j}=F_{j}\circ J$. One can calculate in
Mathematica directly that $F_{1}$ and $F_{2}$ are invariants for $A$ and for
$\Delta$, and hence for $T_{3}$. One evaluation suffices to check that $\nabla
F_{1}$ and $\nabla F_{2}$ are linearly independent at some point, Hence
$F_{1}$ and $F_{2}$ are algebraically independent.

The invariant symplectic form is.

$\omega=\frac{1}{ab}{\bf d\/}a\wedge{\bf d\/}b+\frac{1}{cd}{\bf
d\/}c\wedge{\bf d\/}d.$ (13)

One can see directly that $A^{*}(\omega)=-\omega$. We will describe the
calculation that shows $\Delta^{*}(\omega)=-\omega$. The two facts together
imply that $\omega$ is $T_{3}$ invariant. Let $e_{1},e_{2},e_{3},e_{4}$ be the
standard basis vectors on $\mbox{\boldmath{$R$}}^{4}$. Let $\Psi$ denote the
Jacobian of $\Delta$, namely the $4\times 4$ matrix of partial derivatives.
Let

$(a^{\prime},b^{\prime},c^{\prime},d^{\prime})=\Delta(a,b,c,d).$

We compute

$\Delta^{*}(\omega)(e_{i},e_{j})=\frac{1}{a^{\prime}b^{\prime}}(\Psi_{1i}\Psi_{2j}-\Psi_{2i}\Psi_{1j})+\frac{1}{c^{\prime}d^{\prime}}(\Psi_{3i}\Psi_{4j}-\Psi_{4i}\Psi_{3j})=^{(!)}-\omega(e_{i},e_{j}).$

The equality with an exclaimation point is, for each $(i,j)$, a big
Mathematica calculation that miraculously works out.

Given the simple nature of $\omega$ we can write down the Hamiltonian vector
field $X_{\phi}$ for a function
$\phi:\mbox{\boldmath{$R$}}^{4}\to\mbox{\boldmath{$R$}}$ in an explicit way.
We have

$X_{\phi}=(-ab\phi_{b},ab\phi_{a},-cd\phi_{d},cd\phi_{c}).$ (14)

Here $\phi_{a}=\partial\phi/\partial a$, etc. $X_{\phi}$ is defined wherever
$\phi$ is.

Let $X_{j}$ be the Hamiltonian of $F_{j}$. Another Mathematica calculation
shows that $\omega(X_{1},X_{2})=0$. Now we have established all the points of
Theorem 1.4.

###  3.4 Positivity

We would like to know something about how our invariants behave on convex
points – i.e. those points $(a,b,c,d)$ which represent convex $8$-gons. Let
${\cal C\/}$ be the set of convex points.

######  Lemma 3.1 (Positivity)

Every factor of $F_{1}$ and every factor of $F_{2}$ is positive on $\cal C$.
In particular, $F_{1}$ and $F_{2}$ are positive on $\cal C$.

Proof: Let $(a,b,c,d)\in\cal C$. The convexity gives the constraints

$a,b,c,d>0,\hskip 10.0pt|a-b|<1,\hskip 10.0pt|c-d|<1,\hskip 10.0pta+b>1,\hskip
10.0ptc+d>1.$

Let $e=ac+bd$, as in the definition of $F_{1}$ and $F_{2}$. To finish the
proof we just need to show that the $4$ quantities $e+b-c$ and $e-b+c$ and
$e+a-d$ and $e+d-a$ are positive. By symmetry it suffices to show that
$\phi:=e+b-c>0$ when the constraints above are satisfied.

If we replace $(a,b,c,d)$ by $(ta,tb,tc,td)$ with $t<1$ then we strictly
decrease $\phi$ and we retain the constraints as long as
$\min(ta+tb,tc+tc)>1$. Thus, it suffices to prove that when $\min(a+b,c+d)=1$
(and the other constraints are satisfied) we have $\phi>0$. When $a+b=1$ we
compute that $\phi=(a-1)(1-c+d)>0$. When $c+d=1$ we have $\phi=2b-c+ac-bc$.
Observe that $\partial\phi/\partial c=-1+a-b<0$. Since $c<1$ we have
$\phi>a+b-1>0$. The last expression is what we get by setting $c=1$.
$\spadesuit$

###### Lemma 3.2

$(a,b,c,d)\in{\cal C\/}-{\cal I\/}-{\cal I\/}^{*}$ lies on one of four
connected components, depending only on the signs of $g_{ab}+g_{cd}$ and
$g_{ab}^{*}+g_{cd}^{*}$.

Proof: The map

$\psi(a,b,c,d)=(g_{ab},g_{cd},g^{*}_{ab},g^{*}_{cd})=(a^{\prime},b^{\prime},c^{\prime},d^{\prime})$

is a diffeomorphism from $\cal C$ to $(-2,2)^{4}$. The geometric reason $\psi$
is a diffeomorphism is that $\lambda\cap\mu$ is a single point where $\lambda$
is a line of slope $1$ and $y$-intercept in $(-1,1)$ and $\mu$ is an ellipse
through the points $(\pm 1,0)$ and $(0,\pm 1)$. Our diffeomorphism $\psi$ maps
the points of ${\cal I\/}$ and ${\cal I\/}^{*}$ respectively into the
hyperplanes $a^{\prime}+b^{\prime}=0$ and $c^{\prime}+d^{\prime}=0$.
$\spadesuit$

###  3.5 Integral Curves

Let $X_{G}=X_{2}-X_{1}$ be the Hamiltonian of $G=F_{2}-F_{1}$. We let $\cal X$
denote be the set of $(a,b,c,d)$ such that all the factors of $F_{1},F_{2},G$
are nonzero.

###### Lemma 3.3

The Hamiltonian $X_{G}$ never vanishes on $\cal X$.

Proof: Our proof is something of an overkill but it shows some beautiful and
useful formulas. Recall that $G=2(g^{*}_{ab}+g^{*}_{cd})(g_{ab}+g_{ab})$. We
also define $h=\log(ac/bd)$. We compute

$X_{G}\cdot\nabla
g^{*}_{ab}=2(g_{ab}^{*}+g_{cd}^{*})(1-a+b)(1+a-b)(a+b)/(ab).$ (15)
$X_{G}\cdot\nabla
g^{*}_{cd}=2(g_{ab}^{*}+g_{cd}^{*})(1-c+d)(1+c-d)(c+d)/(cd).$ (16)
$X_{G}\cdot\nabla g_{ab}=-2(g_{ab}+g_{cd})(1-a+b)(1+a-b)(a+b)/(ab).$ (17)
$X_{G}\cdot\nabla g_{cd}=-2(g_{ab}+g_{cd})(1-c+d)(1+c-d)(c+d)/(cd).$ (18)
$X_{G}\cdot\nabla
h=2(g_{ab}^{*}+g_{cd}^{*})\big{(}(1+a^{2}+b^{2})/(ab)+(1+c^{2}+d^{2})/(cd)\big{)}.$
(19)

Equations 15 – 18 all vanish on $\cal X$ only if $a+b=c+d=0$. But then both
$ab$ and $cd$ are negative. This makes Equation 19 nonzero. $\spadesuit$

Say that a $G$-curve is a curve tangent to $X_{G}$ everywhere. The above
result says that $\cal X$ is foliated by $G$-curves. Since one of the
functions considered in Lemma 3.3 is strictly monotone along a given
$G$-curve, we see that these curves exit every compact subset of $\cal X$ at
both ends.

There is a $2$-dimensional $T_{3}$-invariant set $\cal Y$ consisting entirely
of points where $X_{1}$ and $X_{2}$ are linearly dependent. The set $\cal Y$
is given by:

$ac+bd+1=ac^{2}+ca^{2}+bd^{2}+db^{2}=0.$ (20)

After a lot of trial and error I found that on $\cal Y$ we have

$0=Y(F_{1},F_{2})=512+216F_{1}F_{2}+192(F_{1}+F_{2})-30(F_{1}+F_{2})^{2}+(F_{1}+F_{2})^{3}.$
(21)

This equation was hard to guess but it is easy to verify algebraically.

An easy exercise in calculus shows that $Y>0$ on the region where
$F_{1},F_{2}>0$. Hence $\cal Y$ is disjoint from any level set which
intersects $\cal C$.

###  3.6 Proof of the Main Theorem: Outline

Step 1, Linear Independence: In §4 we prove that $X_{1}$ and $X_{2}$ are
linearly independent througout ${\cal X\/}-{\cal Y\/}$. Hence ${\cal
X\/}-{\cal Y\/}$ is foliated by smooth surfaces.

Step 2, The Cylinder and the Nice Loop: Let ${\cal X\/}_{+}\subset{\cal X\/}$
denote the subset where all factors of $F_{1},F_{2},G$ are positive. Let
$L_{+}$ be a level set in ${\cal X\/}_{+}$. Let ${\cal U\/}$ denote the set
$(a,b,c,d)$ where $a,b,c,d>0$ and $\max(a+b,c+d)=1$. In §5, we prove that
every $G$-curve in $L_{+}$ intersects $\cal U$ exactly once, and that
$L_{+}\cap\cal U$ is a single loop which we call the nice loop. The nice loop
is smooth away from the $2$ points on it satisfying $a+b=c+d=1$. From all this
we deduce that $L_{+}$ is a cylinder. For points in ${\cal P\/}_{8,4}$,
treated in [17], the set $L_{+}$ is an arc: a single $G$-curve.

Step 3, Intrinsic Boundedness and Concavity: In §6 we prove that each level
set $L_{+}$ in ${\cal X\/}_{+}$ is bounded with respect to its intrinsic flat
structure coming from its integrability. We also prove that $L_{+}$ is locally
concave near its intrinsic boundary. We prove this by showing that
$\iota_{5}=A\Delta A\Delta A$ is an isometry of the sub-cylinder
$L_{+}^{\prime}\subset L_{+}$ bounded by the nice loop and the end of $L_{+}$
which abuts $(0,0,0,0)$. The map $\iota_{5}$ swaps the ends of
$L_{+}^{\prime}$ and thus allows us to convert info about the nice loop into
info about one end of $L_{+}$. We prove that $\Delta$ preserves $L_{+}$ and
swaps its ends. This gives us info about the other end of $L_{+}$.

Figure 3: The instrinsic structure of $L(3,4)$.

The bottom cylinder in Figure 3 is a numerical plot of $L_{+}$, in its
intrinsic flat structure, corresponding to the parameters
$(F_{1},F_{2})=(3,4)$. I drew this picture by numerically integrating linear
combinations of $X_{1}$ and $X_{2}$. The vertical foliation corresponds to
$X_{G}=X_{2}-X_{1}$. We call the points of non-concavity cusps. They
correspond to the non-smooth points on the nice loop.

Step 4, Analytic Continuation: In §7 we use the structure of $L_{+}$ and
symmetry to see that $L_{+}$ is one component of an invariant torus
$\widehat{L}$. Figure 3 shows $\widehat{L}$ for the parameters
$(F_{1},F_{2})=(3,4)$. The opposite sides of the parallogram are glued
together in this picture.

Here is the construction. Let $L_{-}=A(L_{+})$. This cylinder is disjoint from
$L_{+}$. We show the existence of $2$ intrinsically convex disks – white in
Figure 3 – which serve as a mutual analytic continuation which glues $L_{+}$
and $L_{-}$ together. The middle strip in Figure 3 shows $L_{-}$. Then there
is one more cylinder $M$ which touches $L_{+}$ and $L_{-}$ at the cusps.
Finally, there are $2$ more convex disks which glue $M$ to $L_{+}$ and $2$
which glue $M$ to $L_{-}$.

The flat torus $\widehat{L}$ is the union of the cylinders $L_{+}$, $L_{-}$,
$M$, and the $6$ connecting disks. The geodesics labeled $h$ and $g_{+}$ and
$g_{-}$ separate the open cylinders, because the connecting disks are convex.
In particular, $L_{+}$ is contained in the cylinder bounded by $g_{+}$ and
$h$, and this cylinder has less than half the height of the whole torus. We
show that $A\Delta(h)=g_{+}$ and (hence) $T_{3}(h)=(A\Delta)^{2}(h)=g_{-}$.
Since $g_{-}$ and $h$ are distinct, the vertical component of the translation
induced by $T_{3}$ is nontrivial.

For points in ${\cal P\/}_{8,4}$, the set $\widehat{L}$ is an analytic loop
comprised of $L_{+},L_{-},M$. It is one component of the real points of an
elliptic curve. Here $L_{+}$ and $L_{-}$ have the same length, so $L_{+}$ has
less than half the length of $\widehat{L}$.

Step 5, Vertical Displacement: In §8.1 we show that a counterexample to
Theorem 1.2 lying in ${\cal C\/}-{\cal P\/}_{8,4}-{\cal I\/}-{\cal I\/}^{*}$
leads to a $p\in{\cal X\/}_{+}$ whose $T_{3}$ orbit, either in the forwards
direction or in the backwards direction, remains entirely in $L_{+}$. Since
$L_{+}$ is contained in a strip of less than half the total height of
$\widehat{L}$, and since the vertical component of the translation induced by
$T_{3}^{\pm 1}$ is nontrivial, we see that both the forwards and the backwards
$T_{3}$-orbit of any point in $L_{+}$ eventually exit $L_{+}$. This is a
contradiction. One gets the same contradiction for points of ${\cal
P\/}_{8,4}-{\cal I\/}-{\cal I\/}^{*}$.

Step 6, Inscribed and Circumscribed Cases: In §8.2 we show that ${\cal
I\/}^{*}$ is foliated by invariant sets which (when completed) are
holomorphically equivalent to the Riemann sphere. Under this equivalence, the
map $T_{3}$ acts as a hyperbolic linear fractional transformation. The
attracting fixed point in each level set is convex Poncelet and the repelling
fixed point is the star-reordering of the attracting fixed point. This
establishes Theorem 1.3 and all the statements of Theorem 1.2 pertaining to
$8$-gons in ${\cal I\/}\cup{\cal I\/}^{*}$.

##  4 Linear Independence

###  4.1 Reduction to 9 Cases

We carry out Step 1 of the outline in §3.6. Let $\cal X$ and $\cal Y$ be the
sets from the previous chapter. (See §3.5 for definitions.) In this chapter we
prove that the Hamiltonians $X_{1}$ and $X_{2}$ are linearly independent at
each point of ${\cal X\/}-{\cal Y\/}$. Throughout the proof, $N(\cdot)$
denotes the numerator function.

Using Equations 15 and 16 we see $X_{G}\cdot\mu=0$, where

$\mu=(\alpha,-\alpha,-\beta,\beta),\hskip
15.0pt\alpha=\frac{4ab(c+d)}{(a-b+1)(-a+b+1)},\hskip
10.0pt\beta=\frac{4cd(a+b)}{(c-d+1)(-c+d+1)}$

When the vectors $X_{1}$ and $X_{2}$ are linearly dependent we have
$X_{j}\cdot\mu=0$. To prove the linear independence result, we consider
polynomials

$f_{uv}=N({\bf d\/}u\wedge{\bf d\/}v(X_{1},X_{2})),\hskip
30.0ptg=N(X_{1}\cdot\mu).$ (22)

Here $u,v\in\\{a,b,c,d\\}$. The denominators are monomials in $a,b,c,d$ and so
they do not vanish in $\cal X$. These polynomials all vanish when $X_{1}$ and
$X_{2}$ are linearly dependent. For each $e\in\\{a,b,c,d\\}$ we define

$h_{e}={\rm res\/}(f_{ab},g,e).$ (23)

Up to signs, we have

$h_{b}(a,b,c,d)=h_{a}(b,a,d,c),\hskip
15.0pth_{c}(a,b,c,d)=h_{a}(c,d,a,b),\hskip
15.0pth_{d}(a,b,c,d)=h_{a}(d,c,b,a),$
$h_{a}=(-1+b)(1+b)(b-c)(b-d)(c+d)^{2}\times\phi_{1}\phi_{2}^{2}\phi_{3}^{2}\phi_{4},$
$\phi_{1}(a,b,c,d)=bc-c^{2}+bd+cd$ $\phi_{2}(a,b,c,d)=b-c+bc-c^{2}-bd-cd$
$\phi_{3}(a,b,c,d)=\phi_{2}(-a,-b,-c,-d).$
$\phi_{4}(a,b,c,d)=1-c^{2}+2bd+b^{2}cd-bc^{2}d+b^{2}d^{2}+bcd^{2}.$

If $X_{1}$ and $X_{2}$ are linearly dependent, then one factor from each
expression $h_{a},h_{b},h_{b},h_{c}$ vanishes. We will go through the $9$
factors for $h_{a}$ and eliminate all of them from consideration. If we have
shown that the $k$th factor of $h_{a}$ cannot vanish then by symmetry we have
also shown this for each of $h_{b},h_{c},h_{d}$.

###  4.2 A Case by Case Analysis

The flow trick: If $X_{1}$ and $X_{2}$ are linearly dependent at a point, then
they are linearly dependent along the $G$-curve through this point, and
multiples of $X_{G}$, because flow along $X_{G}$ is a symplectomorphism. If
some function $\zeta$ vanishes at a point of linear dependence, we can assume
that $\zeta$ vanishes along the whole flowline through the point. Otherwise we
perturb along the flowline and produce a new point of linear dependence where
$\zeta\not=0$. In particular, $X_{j}\cdot\nabla\zeta=0$ for $j=1,2$ and that
$X_{G}\cdot\nabla\zeta=0$.

Cases 1 and 2: Suppose $b=1$. By the flow trick, $X_{j}\cdot(0,1,0,0)=0$ for
$j=1,2$. The nonzero solutions are:

  1. 1.

$(a,c)=(1+d,1)$ and $(a,c)=(-1,1+d)$. In both cases $F_{2}=0$.

  2. 2.

$a=1-(d^{2}/2)$ and $\psi:=1/(1+d)-c=0$.

Just the second case gives points in $\cal X$. By the flow trick,
$\zeta:=X_{G}\cdot\nabla\psi=0$. But $\zeta$ is a nontrivial rational function
in $d$ and hence cannot vanish along the flowline. Hence $b\not=1$. The
symmetry $p\to-p$ swaps Cases 1 and 2. Hence $b\not=\pm 1$. By symmetry,
$a,b,c,d\not=\pm 1$.

Case 3: Suppose $\psi:=b-c=0$. We have $X_{G}\cdot\nabla\psi=-4(a-d)(a+d)=0$
by the flow trick. This forces $d=\pm a$. But $G(a,b,b,\pm a)=0$, so these
points are not in $\cal X$. By symmetry we have also ruled out the possibility
that $a-d=0$.

Case 4: Suppose $b-d=0$. Note that $a\pm c\not=0$ because
$(a,b,a,b)\not\in{\cal X\/}$ and $G(a,b,-a,b)=0$. Setting $d=b$ and using the
flow trick, we have

$0=\frac{X_{G}\cdot(0,1,0,-1)}{g_{ab}+g_{cd}}=\frac{4b(a-c)}{a+c}\times(a-b+c).$
(24)

Hence $a-b+c=0$. By the flow trick

$0=u_{j}(a,b):=ab(a-b)X_{j}\cdot(1,-1,1,0).$ (25)

Here $u_{j}$ is a polynomial for $j=1,2$. By the flow trick both $u_{1}$ and
$u_{2}$ vanish along the flowline. But ${\rm res\/}(u_{1},u_{2},a)$ and ${\rm
res\/}(u_{1},u_{2},b)$ are both nontrivial expressions, which contradicts
this. Hence $b-d\not=0$. By symmetry $a-c\not=0$.

Case 5: Suppose $c+d=0$. We first compute

$0=g(a,b,c,-c)=c^{4}(1-a+b)(1+a-b)(a+b).$ (26)

Since $|a-b|\not=1$ we must have $a+b=0$. We set $b=-a$ and $d=-c$. By the
flow trick we have $X_{G}\cdot(1,1,0,0)=X_{G}\cdot(0,0,1,1)=0$. Hence

$0=ac^{2}X_{G}\cdot(1,1,0,0)+a^{2}cX_{G}\cdot(0,0,1,1)=-4(a+c)(a^{2}+c^{2}+4a^{2}c^{2}).$
(27)

Hence $a=-c$. But $G(a,b,c,d)=G(a,-a,-a,a)=0$. This is not in $\cal X$.

Case 6: Suppose $\phi_{1}=0$. This gives

$b=\frac{c^{2}-cd}{c+d}.$ (28)

Note that $c+d\not=0$ by Case 5. By the flow trick,

$0=\frac{(a+b)X_{G}\cdot\nabla\phi_{1}}{-4(a-b+c-d)}=\zeta,\hskip
30.0pt2c+c^{2}d-d^{3}+a(c+d)^{2}.$

Hence $\zeta=0$. Solving for $a$, we get

$a=\frac{-2c-c^{2}d+d^{3}}{(c+d)^{2}}.$

The functions $f_{ab}$ and $f_{ac}$ are rational functions in $c,d$. But both
resultants ${\rm res\/}(N(f_{ab}),N(f_{ac}),c)$ and ${\rm
res\/}(N(f_{ab}),N(f_{ac}),d)$ are nontrivial, so we have only isolated
solutions to $\phi_{1}=0$, contradicting the flow trick.

Cases 7 and 8: Suppose $\phi_{2}=0$. Solving for $b$ we have

$b=\frac{c(1+c+d)}{1+c-d}.$ (29)

$|c-d|\not=1$, and also $c+d+1\not=0$ because $b\not=0$. By the flow trick,

$0=X_{2}\cdot\nabla\phi_{2}=\frac{(1+c)\zeta_{1}^{2}\zeta_{2}}{a(1+c-d)(1+c+d)},\hskip
12.0pt\zeta_{1}=a+ac-d-ad+cd+d^{2}=0.$

Here $\zeta_{2}$ is a large polynomial. We note that $\zeta_{1}$ is one of the
factors of $F_{2}(a,b,c,d)$ when $b$ is as in Equation 29. Hence
$\zeta_{1}\not=0$. Hence $\zeta_{2}=0$ along the flowline. Solving
$\zeta_{2}=0$ we get

$a=\frac{d+5cd+7c^{2}d+3c^{3}d+4c^{2}d^{2}-d^{3}+cd^{3}}{-1-c+c^{2}+c^{3}-2d-6cd-4c^{2}d+d^{2}+cd^{2}+2d^{3}}.$
(30)

(The reader can infer the formula for $\zeta_{2}$ from this expression.) We
write

$F_{ab}=(1+c-d)^{2}\zeta_{3}\zeta_{4},\hskip
30.0pt\zeta_{3}=3c+4c^{2}+c^{3}-d+3c^{2}d+3cd^{2}+d^{3}.$ (31)

Here $\zeta_{4}$ is some large polynomial. We note that $\zeta_{3}$ is one of
the factors of $F_{2}(a,b,c,d)$ when $a$ is as in Equation 30 and $b$ is as in
Equation 29. Hence $\zeta_{3}\not=0$. But then $\zeta_{4}=0$ along the
flowline. We compute that ${\rm res\/}(\zeta_{4},N(f_{ac}),c)$ and ${\rm
res\/}(\zeta_{4},N(f_{ac}),d)$ are both nontrivial. This is a contradiction.
Hence $\phi_{2}\not=0$. The symmetry $p\to-p$ interchanges Cases 7 and 8.
Hence $\phi_{3}\not=0$.

Case 9: The only case left is:

$\phi_{4}(a,b,c,d)=\phi_{4}(b,a,d,c)=\phi_{4}(c,d,a,b)=\phi_{4}(d,c,b,a)=0.$

If $bd+1=0$ then $\phi_{4}=(c/d)(1-d^{2})$. This gives $d=\pm 1$, a case we
have ruled out. Hence $bd+1\not=0$. Solving $\phi_{4}(a,b,c,d)=0$ for $c$ and
$\phi_{4}(c,d,a,b)=0$ for $a$ we find that $a$ and $c$ are roots of the same
quadatic polynomial

$Ax^{2}+Bx+C=0,\hskip 20.0ptA=-1-bd,\hskip 10.0ptB=bd(b+d),\hskip
10.0ptC=(1+bd)^{2}.$ (32)

By Case 4, $a\not=c$. Hence $a+b=-B/A$. Expanding this out and clearing
denominators, we get:

$\nu_{1}=\nu(a,b,c,d):=-a-c-abd-cbd+b^{2}d+bd^{2}=0.$ (33)

By symmetry $\nu_{2}=\nu(b,a,d,c)=0$. This gives the defining functions for
$\cal Y$:

$0=\frac{\nu_{1}-\nu_{2}}{-(a-b+c-d)}=ac+bd+1.$
$0=\frac{1}{2}(\nu_{1}+\nu_{2}+(a+b+c+d)(1+ac+bd))=ac^{2}+ca^{2}+bd^{2}+db^{2}.$

##  5 Cylinders in the Positive Part

###  5.1 The Main Result

We carry out Step 2 of the outline in §3.6. Recall that ${\cal X\/}_{+}$ is
the subset of $\cal X$ where all the factors of $F_{1}$, $F_{2}$, and $G$ are
positive. Let $L_{+}$ be an arbitrarily level set of ${\cal X\/}_{+}$. We
prove in this chapter that $L_{+}$ is a (single) cylinder. Here is an outline
of the proof. Recall that a $G$-curve of $L_{+}$ is one that is integral to
$X_{G}$. We consider $G$-curves to be subsets of ${\cal X\/}_{+}$. Let $\cal
U$ be as in Step 2.

  1. 1.

We prove the $L_{+}$ is bounded in $\mbox{\boldmath{$R$}}^{4}$.

  2. 2.

We analyse the ends of $G$-curve in ${\cal X\/}_{+}$ and show that every
$G$-curve intersects $\cal U$. Later in the chapter, we show more strongly
that each $G$-curve of $L_{+}$ intersects $\cal U$ exactly once, and
transversely.

  3. 3.

We use the intersection property to prove that ${\cal X\/}_{+}$ is connected.

  4. 4.

We show that $L_{+}\cap{\cal U\/}$ consists of the nice loop from Step 2 of
the outline in §3.6 and perhaps finitely many extra loops. Using the
continuity, transversality, and the connectivity of ${\cal X\/}_{+}$ we
conclude that the number of extra loops is independent of the choice of level
set.

  5. 5.

We exhibit a choice of parameter values for which there are no extra loops.
Hence $L_{+}\cap\cal U$ is just the nice loop.

  6. 6.

Since $L_{+}$ is foliated by $G$-curves, each of which intersects $\cal U$
once, we see that $L_{+}$ is homeomorphic to the product of an open interval
with $L_{+}\cap\cal U$. This latter set is a loop. Hence $L_{+}$ is a
cylinder.

###  5.2 Bounded Level Sets

###### Lemma 5.1

The level set $L_{+}$ is bounded in $\mbox{\boldmath{$R$}}^{4}$.

Proof: We work with the functions $g_{ab}$ and $g_{cd}$ from Equation 8. (We
repeat the formula for $g_{ab}$ on the next page.) Suppose
$\\{(a_{n},b_{n},c_{n},d_{n})\\}$ is an unbounded sequence in $L_{+}$. By
symmetry it suffices to consider the case when one of $c_{n}$ or $d_{n}$ tends
to $\infty$. Since $|c_{n}-d_{n}|<1$, both of these coordinates tend to
$\infty$. Hence $g_{cd}(a_{n},b_{n},c_{n},d_{n})\to-2$. We will show that
$g_{ab}(c_{n},b_{n},c_{n},d_{n})\to+2$. Since $g_{ab}^{*},g_{cd}^{*}\in(0,2)$
we get $G(a_{n},b_{n},c_{n},d_{n})\to 0$, a contradiction. This function is
nonzero and constant along $L_{+}$.

Case 1: Suppose that $a_{n}+b_{n}\geq 1$ for all $n$. An exercise in Lagrange
multipliers tells us that in this case that $g_{ab}<2$ in this case. Since
$g_{ab}+g_{cd}>0$ and $g_{cd}\to-2$ we must have $g_{ab}\to 2$. This gives us
our contradiction.

Case 2: Suppose $a_{n}+b_{n}<1$ and $a_{n}$ and $b_{n}$ both remain in a
compact subset of the interval $(0,1)$. We have

$a_{n}-d_{n}+a_{n}c_{n}+b_{n}d_{n}>0$ (34)

because this is one of the factors of $F_{2}$. Since $c_{n}/d_{n}\to 1$ and
both terms go to $+\infty$, we must have $a_{n}+b_{n}\to 1$. If $a_{n}$ and
$b_{n}$ remain in a compact subset of the open interval $(0,1)$ then again
$g_{ab}(a_{n},b_{n},c_{n},d_{n})\to 2$. This gives us the same contradiction
as in Case 1.

Case 3: After passing to a subsequence, it only remains to consider what
happens when $a_{n}+b_{n}<1$ and $(a_{n},b_{n})\to(1,0)$ or
$(a_{n},b_{n})\to(0,1)$. We treat the second case; the first then follows from
symmetry.

Let us examine Equation 34 again. Because $c_{n}<d_{n}+1$ we get

$0<a_{n}-d_{n}+a_{n}c_{n}+b_{n}d_{n}<a_{n}+a_{n}(d_{n}+1)-d_{n}+b_{n}d_{n}=(a_{n}+b_{n}-1)(d_{n})+2a_{n}.$

Rearranging the last inequality, we have

$\frac{1-b_{n}}{a_{n}}<1+\epsilon,\hskip 30.0pt\epsilon=\frac{2}{d_{n}}.$

Since $d_{n}\to\infty$ we can take $\epsilon>0$ as small as we like. But then

$\frac{1-a_{n}^{2}-b_{n}^{2}}{a_{n}}<\frac{1-b_{n}^{2}}{a_{n}}=(1+b_{n})\times\frac{1-b_{n}}{a_{n}}<2+2\epsilon.$

Since $b_{n}\to 1$ we have

$g_{ab}(a_{n},b_{n},c_{n},d_{n})=\frac{1-a_{n}^{2}-b_{n}^{2}}{2a_{n}b_{n}}<2+2\epsilon$
(35)

once $n$ is sufficiently large. Since $\epsilon$ is arbitrary, we see that
$g_{ab}\to 2$ and we get the same contradiction. $\spadesuit$

###  5.3 Ends of Flowlines

Recall that the map $I(a,b,c,d)=(c,d,a,b)$ preserves both ${\cal X\/}_{+}$ and
$L_{+}$.

###### Lemma 5.2

Up to the action of $I$, every accumulation point of $L_{+}$ that does not lie
in ${\cal X\/}$ has the following form.

  1. 1.

$(0,\overline{b},\overline{b},0)$ for $\overline{b}\in[0,1)$.

  2. 2.

$(1,0,1,0)$ or $(0,1,1,0)$ or $(0,1,0,1)$.

  3. 3.

$(1,0,\overline{c},\overline{d})$ for $\overline{c},\overline{d}>0$ and
$\overline{c}+\overline{d}>1$.

Proof: The denominators of each invariant $F_{1},F_{2},G$ are of the form
$abcd$ and moreover these invariants are constant along $L_{+}$. Hence every
accumulation point $(\overline{a},\overline{b},\overline{c},\overline{d})$ of
$L_{+}$ outside of $\cal X$ must have at least one zero coordinate. Using the
action of $I$ we reduce to the case where either $\overline{a}=0$ or
$\overline{b}=0$.

Case 1: Suppose $\overline{a}=0$ and $\overline{b}<1$. Let $g_{ab}$ and
$g_{cd}$ be as in Equation 8. Then $g_{ab}\to+\infty$. Since $|c-d|<1$ we have
$(c-d)^{2}=c^{2}+d^{2}-2cd<1$. Rearranging this gives $g_{cd}>-2$. Hence
$g_{ab}+g_{cd}\to+\infty$. Since $G(a,b,c,d)$ is constant on $L_{+}$, we have
$a-b+c-d\to 0$ as we approach our accumulation point. Hence our point is
$(0,\overline{b},\overline{b}+t,t)$ for some $t\geq 0$. The denominator of
$F_{1}$ is positive and the numerator converges to
$-t^{2}(1-\overline{b}^{2})^{2}$. Since $F_{1}>0$ this is only possible if
$t=0$. The same argument works when $\overline{a}<1$ and $\overline{b}=0$.

Case 2: Suppose that $\overline{a}=0$ and $\overline{b}\geq 1$. Since
$|a-b|<1$ along $L_{+}$ we must have $\overline{b}=1$. The expression
$c-b+ac+bd>0$, being a factor of $F_{2}$, is positive on $L_{+}$. At
$(0,1,\overline{c},\overline{d})$ this expression equals
$\overline{c}+\overline{d}-1$. Hence $\overline{c}+\overline{d}\geq 1$. The
case when $\overline{b}=0$ and $\overline{a}=1$ has a similar treatment, using
the expression $d-a+ca+bd$ instead. Since $|\overline{c}-\overline{d}|\leq 1$
we either have both $\overline{c},\overline{d}>0$ or we get the options on
Item 2 of our list

Case 3: We assume that $\overline{c},\overline{d}>0$. We want to see that
$\overline{c}+\overline{d}>1$. We choose a sequence
$\\{(a_{n},b_{n},c_{n},d_{n})\\}$ converging to
$(1,0,\overline{c},\overline{d})$. Let us write $a_{n}=1-K_{n}b_{n}$. The
denominator of $F_{2}(a_{n},b_{n},c_{n},d_{n})$ is asymptotic to
$(\overline{c}\overline{d})b_{n}$ in the sense that the difference is
vanishingly small in comparison to $b_{n}$. The first factor of the numerator
of $F_{2}(a_{n},b_{n},c_{n},d_{n})$ is $b_{n}(1-K_{n})$. The remaining factors
converge to

$1-\overline{c}+\overline{d},\hskip 15.0pt2\overline{c},\hskip
15.0pt1+\overline{c}-\overline{d}.$

These are all nonzero, Since $F_{1}$ is constant on $L_{+}$, we conclude that
the limit $K=\lim K_{n}$ exists and is finite. From this we compute that

$\lim
F_{1}(a_{n},b_{n},c_{n},d_{n})=\frac{2(1+\overline{c}-\overline{d})(\overline{c}+\overline{d}-1)(1+\overline{d}+K\overline{c})}{\overline{c}\overline{d}}.$

Since this nonzero, we have $\overline{c}+\overline{d}>1$. $\spadesuit$

From the discussion following Lemma 3.3, each $G$-curve $\chi$ exits every
compact subset of ${\cal X\/}_{+}$ at both ends. Since $L_{+}$ is bounded,
each end of $\chi$ accumulates on at least one point of
$\mbox{\boldmath{$R$}}^{4}$.

###### Lemma 5.3

The backwards end of any $G$-curve has a unique accumulation point, and it has
the form $(x,0,0,x)$ or $(0,x,x,0)$ for some $x\in[0,1)$. The forwards end of
any $G$-curve has a unique accumulation point, and it is either $(1,0,1,0)$ or
$(1,0,x,y)$ or $(x,y,1,0)$ with $x,y>0$ and $x+y>1$.

Proof: Suppose the backwards end of $\chi$ accumulates on $(1,0,0,1)$. By
Equation 15, the quantity $a-b$ increases along $\chi$ but remains less than
$1$. This is an immediate contradiction. The same argument, using one of
Equation 15 or 16, rules out $(0,1,1,0)$ and $(1,0,x,y)$ and $(x,y,1,0)$ as
backwards accumulation points. Suppose $(0,1,x,y)$ is a backwards accumulation
point. Since $a-b+c-d>0$ in ${\cal X\/}_{+}$ we must have $x-y\geq 1$. But,
again, this contradicts the fact that $c-d$ increases along $\chi$ and always
satisfies $c-d<1$. The same argument rules out $(x,y,0,1)$ as a limit point.
Lemma 5.2 now says that any accumulation point has the claimed form.

The uniqueness of the backwards accumulation point follows from the
characterization of the backwards accumulation points and from the
monotonicity of $g_{ab}^{*}$ and $g_{cd}^{*}$.

Suppose that the forward end of $\chi$ accumulates on
$(a^{\prime},0,0,a^{\prime})$. This contradicts the monotonicity of $a-b$ and
$c-d$ along $\chi$ and the form of the backwards limit point. The same
argument rules out $(0,b^{\prime},b^{\prime},0)$. If $(0,1,x,y)$ is a forward
accumulation point, then by monotonicity we would have $a-b<-1$ along $\chi$.
This is a contradiction. The same argument rules out $(x,y,0,1)$. Arguments
like we have already given rule out $(0,1,1,0)$ and $(1,0,0,1)$. Lemma 5.2 now
says that any accumulation point has the claimed form.

The uniqueness is more subtle here. If the uniqueness is false, then really we
have a whole continuum of accumulation points. By continuity and symmetry we
can find a single $G$-curve having accumulation points $(1,0,x_{1},y_{1})$ and
$(1,0,x_{2},y_{2})$ with $x_{1},y_{1},x_{2},y_{2}>0$. But then the
monotonicity implied by Equations 16 and 18 says that $g_{cd}$ and
$g_{cd}^{*}$ both are finite and coincide on these points. This forces the
points to coincide. $\spadesuit$

###### Lemma 5.4

Every $G$-curve in $L_{+}$ intersects the set $\cal U$ consisting of points
$(a,b,c,d)$ where $a,b,c,d>0$ and $\max(a+b,c+d)=1$.

Proof: Let $\chi$ be a $G$-curve. The backwards limit point of $\chi$ has the
form $(a,b,c,d)$ where $\max(a+b,c+d)<1$. The forwards limit point has the
form $(a,b,c,d)$ where $\max(a+b,c+d)\geq 1$ and we have equality only if the
forwards limit point is $(1,0,1,0)$. Let $\\{a_{n},b_{n},c_{n},d_{n}\\}$ be a
sequence of points along $\chi$ converging to $(1,0,1,0)$. The function
$g_{ab}^{*}+g^{*}_{cd}=a-b+c-d$ converges to $2$ along this sequence. Hence
the function $g_{ab}+g_{cd}$ also converges to some positive value.

We will assume that $a_{n}+b_{n}<1$ and $c_{n}+d_{n}<1$ for all $n$ and derive
a contradiction. For such points we have $g_{ab},g_{cd}<2$. We conclude that
therefore $g_{ab}$ and $g_{cd}$ are bounded. But they are also monotone along
$\chi$. Hence $g_{ab}$ and $g_{cd}$ converge to a well-defined limit. Writing

$a_{n}=1+\beta_{n}b_{n},\hskip 30.0ptc_{1}=1+\delta_{n}d_{n},$ (36)

we find that $\beta=\lim\beta_{n}$ and $\delta=\delta_{n}$ because

$\lim g_{ab}(a_{n},b_{n},c_{n},d_{n})=-2\beta,\hskip 20.0pt\lim
g_{cd}(a_{n},b_{n},c_{n},d_{n})=-2\delta.$

We compute that

$\lim_{n\to\infty}F_{1}(a_{n},b_{n},c_{n},d_{n})=2\times
2\times(1+\beta)\times(1+\delta).$

The factors are the limits, respectively of

$1+a_{n}-b_{n},\hskip 10.0pt1+c_{n}-d_{n},\hskip
10.0pt\frac{b_{n}-c_{n}+a_{n}c_{n}+b_{n}d_{n}}{b_{n}},\hskip
10.0pt\frac{d_{n}-a_{n}+a_{n}c_{n}+b_{n}d_{n}}{d_{n}}.$

Hence, each of $1+\beta$ and $1+\delta$ is positive. But then eventually
$1+\beta_{n}$ and $1+\delta_{n}$ are positive. This gives us
$a_{n}+b_{n},c_{n}+d_{n}>1$ eventually. $\spadesuit$

###  5.4 Path Connectivity

We use the results in the previous section to prove ${\cal X\/}_{+}$ is path
connected. We let ${\cal U\/}_{ab}$ denote the subset of $\cal U$ where
$a+b=1$. Likewise we define ${\cal U\/}_{cd}$. Finally, let ${\cal
U\/}_{abcd}={\cal U\/}_{ab}\cap{\cal U\/}_{cd}$.

###### Corollary 5.5

The space ${\cal X\/}_{+}$ is path connected.

Proof: Since every $G$-curve of every level set in ${\cal X\/}_{+}$ intersects
$\cal U$. It suffices to prove that ${\cal X\/}_{+}\cap{\cal U\/}$ is
connected. We will show that the $3$ sets ${\cal X\/}_{+}\cap{\cal
U\/}_{abcd}$ and ${\cal X\/}_{+}\cap{\cal U\/}_{ab}$ and ${\cal
X\/}_{+}\cap{\cal U\/}_{cd}$ are path connected. It then follows that the
union is path connected as well. By symmetry we just have to deal with the
first two sets.

Consider the first set. Setting $b=1-a$ and $d=1-c$ we find that the factors
for $F_{1},F_{2},G$ all have the form $a,c,1-a,1-c,a+c-1$. The set of choices
$a,b\in(0,1)^{2}$ making these all positive is clearly connected.

Consider the second set. Here we have $b=1-a$ and $c+d\leq 1$. If $c+d=1$ we
just perturb slightly so that $c+d<1$ and we remain in our intersection, by
continuity. So, we take $c+d<1$. In addition to the factors
$a,1-a,1-c+d,1+c-d$, which are all satisfied when $(a,c,d)\in(0,1)$, we get
the new factors

$\phi_{1}=-a+ac+2d-ad,\hskip 10.0pt\phi_{2}=-1+a+c+ac+d-ad,\hskip
10.0pt\phi_{3}=-1+2a+c-d.$

We compute

$\frac{\partial\phi_{1}}{\partial a}=-1+c+d<0,\hskip
20.0pt\frac{\partial\phi_{2}}{\partial a}=2c>0,\hskip
20.0pt\frac{\partial\phi_{2}}{\partial a}=1+c-d>0.$

We also note that when $a=1$ we have $\phi_{1}=c+d-1<0$. So, there is some
$a_{0}\in(0,1)$ such that $\phi_{1}(a_{0},1-a_{0},c,d)=0$ and the other
factors are positive. Solving for $a$ we find that

$a_{0}=\frac{2d}{1-c+d}.$

For this value we compute that

$\phi_{2}(a_{0},1-a_{0},c,d)=\phi_{3}(a_{0},1-a_{0},c,d)=\frac{(1-c)^{2}+d(2+2c-d)}{1-c+d}.$
(37)

This is a quadratic function which is positive on a convex subset of
$(0,1)^{2}$. Hence, by increasing $a$ to $a_{0}$, and stopping just before, we
can join two points of ${\cal X\/}_{+}\cap{\cal U\/}$ to two points which then
can be connected by a straight line segment that remains in our intersection.
$\spadesuit$

###  5.5 A Closer Look at the Intersection

We take a closer look at the intersection $L_{+}\cap\cal U$.

###### Lemma 5.6

$L_{+}$ is transverse to ${\cal U\/}_{ab}$ and to ${\cal U\/}_{cd}$. Each
$G$-curve in $L_{+}$ intersects $\cal U$ exactly once.

Proof: By symmetry it suffices to consider ${\cal U\/}_{ab}$. Since $L_{+}$ is
smooth it suffices to prove that $X_{G}$ is not tangent to ${\cal U\/}_{ab}$.
The vector $(1,1,0,0)$ is perpendicular to this space, and

$X_{G}\cdot(1,1,0,0)=\frac{4a(1-a)(1+c-d)(1-c+d)}{cd}\not=0.$ (38)

This does it. Here we have used $b=1-a$ to simplify the expression.

We have already proved that each $G$-curve intersects $L_{+}$ at least once.
The forward direction of a $G$-curve has a positive dot product with the
normal to $\cal U$ whenever it hits. But then it cannot come back and hit a
second time because it would be returning in such a way as to make negative
dot product with the same vector. For intersections which lie in ${\cal
U\/}_{ab}\cap{\cal U\/}_{cd}$ we can use either $(1,1,0,0)$ or $(0,0,1,1)$ for
this argument. $\spadesuit$

By transversality, the intersection $L_{+}\cap\cal U$ is a smooth $1$-manifold
away from ${\cal U\/}_{abcd}$. and overall a topological $1$-manifold. A
smooth arc which starts in ${\cal U\/}_{ab}$ and hits ${\cal U\/}_{abcd}$
simply continues across the itersection into ${\cal U\/}_{cd}$ as another
smooth arc.

We say that a nice loop in $\cal U$ is one which intersects ${\cal
U\/}_{abcd}$ exactly twice and is preserved by the symmetry $I$. See the right
side of Figure 4 below.

###### Lemma 5.7

The intersection $L_{+}\cap{\cal U\/}$ is compact. It consists of a nice loop
and perhaps finitely many additional smooth loops.

Proof: It is convenient to work with the invariants $G$ and $H=F_{1}/F_{2}$.
We concentrate on $L_{+}\cap{\cal U\/}_{ab}$. Thus, we are interested in
points $(a,1-a,c,d)$ with $c,d>0$ and $c+d<1$. This is a triangle $\tau$ in
the $(c,d)$-plane. It is yellow in Figure 4 below.

Solving the equation $H(a,1-a,c,d)=h$, we find that

$a=\frac{2d+h-ch-dh}{1-c+d+h+ch-dh}.$ (39)

Let us check that this equation is valid on $\tau$. Setting $a=0,1,\infty$
respectively, we find that

$h=\frac{2h}{c+d-1}<0,\hskip 30.0pth=\frac{c+d-1}{2c}<0,\hskip
30.0pth=\frac{-1+c-d}{1+c-d}<0.$

But $h>0$. So, whenever $(c,d)$ are in $\tau$, the formula gives $a\in(0,1)$
and it does not blow up. We write

$\Gamma=G(a(c,d),1-a(c,d),c,d)-g=\frac{N}{D},$ (40) $D=cd(1-c+d+h+ch-dh)$
$N=2-4c+4c^{3}-2c^{4}-4d-4c^{2}d+8c^{3}d-4cd^{2}-12c^{2}d^{2}+4d^{3}+8cd^{3}-$
$2d^{4}+cdg-c^{2}dg+cd^{2}g-2h+4ch-4c^{3}h+2c^{4}h+4dh+4c^{2}dh-$
$8c^{3}dh+4cd^{2}h+12c^{2}d^{2}h-4d^{3}h-8cd^{3}h+2d^{4}h+cdgh+c^{2}dgh-
cd^{2}gh$

Setting $D=0$ and solving for $h$ we find that $h=\frac{(-1+c-d)}{(1+c+d)}<0.$
Hence $G=g$ on $\tau$ iff $N=0$. The intersections of $L_{+}\cap{\cal
U\/}_{ab}$ correspond to points $(c,d)$ where $N(c,d)=0$. We let $V_{N}$ be
this set.

Figure 4: A contour plot (left) and a schematic picture (right).

To give the reader a feel for what is going on, and as a sanity check, we plot
an example. Figure 4 shows a contour plot of $N$ for $g=1$ and $h=3/4$. This
case corresponds to $(F_{1},F_{2})=(3,4)$. In Figure 4 (left), the open yellow
region $\tau$ is the region of interest to us. The set $\Gamma=0$ has $4$
planar components. The portion of the component in $\tau$ tis exactly
$L_{+}\cap{\cal U\/}_{ab}$ The boundary points of this arc are swapped by the
map $(a,b,c,d)\to(c,d,a,b)$ and together the two arcs $L_{+}(3,4)\cap{\cal
U\/}_{ab}$ and arcs $L_{+}(3,4)\cap{\cal U\/}_{cd}$ join together at these
points to make a nice embedded loop which is precisely $L_{+}(3,4)\cap\cal U$.
Figure 4 (right) shows this schematically. Note that the right half of this
schematic picture is not the continuation of the level set from Figure 4
(left).

Here are some observations about $N$ which hold for all $g>0$ and all
$h\in(0,1)$ which correspond to level sets in ${\cal X\/}_{+}$. It is
important to remark that we only look at pairs $(g,h)$ corresponding to such
level sets.

  1. 1.

$\Gamma$ is quartic in $c$ and $d$. Hence any straight line intersects $V_{N}$
at most $4$ times by Bezout’s Theorem.

  2. 2.

We compute that $\Gamma(0,d)=2(1-h)(d-1)^{3}(d+1).$ From this we see that the
line $c=0$ intersects $V_{N}$ a total of $4$ times counting multiplicity. The
point $(0,1)$ is a point of multiplicity $3$ and the point $(0,-1)$ is a point
of multiplicity $1$. This is why we have added the number $3$ in several
places in Figure 4.

  3. 3.

For at least one parameter choice, say the one in the plot, the level set
through $(0,1)$ goes from region $N$ to region $W$, bypassing $\tau$ entirely.
This cannot change because then at some parameter we would have an inflection
point corresponding to multiplicity $4$.

  4. 4.

The level set through $(-1,0)$ goes from $S$ to $W$, completely bypassing
$\tau$. One can see this for a single parameter in a plot and then the same
multiplicity argument shows that it is always true.

  5. 5.

Similar statements holds for the line $d=0$.

Using continuity, transversality, and the observations above, we see that
$\partial\tau$ is a barrier for any component of $V_{N}$ that starts in
$\tau$. The component is trapped in $\tau$. Hence $L_{+}\cap{\cal U\/}$ is
compact. To finish the proof of this lemma, we just need to see that $V_{N}$
always interesects the line $c+d=1$ at $2$ distinct points on the open segment
$\sigma$ bounded by $(1,0)$ and $(0,1)$. Once we know this, the two strants of
$V_{N}$ emanating from these points have no choice to connect to each other
and make an arc.

By Bezout’s Theorem, there can be at most $2$ intersections, because the line
extending $\sigma$ contains $(1,0)$ and $(0,1)$. Plugging in a single choice
of $(g,h)$, say the ones for the plot, we see that there are $2$ intersection
points. What could go wrong as we vary the level set continuously? If the two
intersection points coalesce into $1$ point $(a,1-c,c,1-c)$ then the map

$(a,b,c,d)\to(c,d,a,b)$

fixed this point. This forces $c=a$. But such points belong to level sets in
${\cal P\/}_{8,4}$. This is a contradiction.

The only thing we need to rule out is the possibility that our two
intersection points move into the boundary. We compute

$\Gamma(c,1-c)=-2c(1-c)\Gamma_{0}(c),$
$\Gamma_{0}(c)=16c-16c^{2}-g+cg-16ch+16c^{2}h-cgh.$

The factor $2c(1-c)$ accounts for the roots at $(1,0)$ and $(0,1)$. We compute
$\Gamma_{0}(0)=-g$ and $\Gamma_{0}(1)=-gh$. These quantities do not vanish, so
the intersection points cannot move into the boundary. $\spadesuit$

###### Lemma 5.8

$L_{+}\cap\cal U$ is just a single nice loop.

Proof: By transversality, continuity, and the connectivity of ${\cal
X\/}_{+}$, the number of components to $L_{+}\cap\cal U$ is constant. Thus we
just have to exhibit a pair of parameters for which there are no components
besides the nice loop. We set $g=t$ and $h=(4-t)/4$. We have

$N(c,d,t)=(t/4)\times(\lambda(c,d)+t\mu(c,d)),\hskip
30.0pt\mu(c,d)=cd(1+c-d),$ (41)
$\lambda(c,d)=2-4c+4c^{3}-2c^{4}-4d+8cd-4c^{2}d+8c^{3}d-4cd^{2}-12c^{2}d^{2}+4d^{3}+8cd^{3}-2d^{4},$

We compute $\Gamma(1/2,1/2,t)=-t/4<0$. In the square $Q=[2/5,3/5]^{2}$ we have
$\nabla\lambda=0$ only at $(1/2,1/2)$. The point there is a local minimum and
the Hessian is positive definite. Moreover, an easy calculation shows that
$\lambda_{0}>0$ on the region outside $Q$ and inside $[0,1]^{2}$. By
compactness and continuity, the same goes for $N(c,d,t)$ for $t>0$
sufficiently small. Now, $N(c,d,t)$ has a unique local minimum in $Q$ and the
value is negative. The location of the minimum converges to $(1/2,1/2)$ as
$t\to 0$. From all this structure we see that the solution to $V_{N}$ is only
the one arc which is one half of a nice loop. $\spadesuit$

##  6 Intrinsic Boundedness and Concavity

We carry out Step 3 of the outline in §3.6. We show that each level set
$L_{+}$ of ${\cal X\/}_{+}$ is intrinsically bounded and has boundaries which
are locally concave everywhere except $2$ points.

###  6.1 The Reversal Lemmas

######  Lemma 6.1 (Reversal I)

$\Delta(L_{+})=L_{+}$ and $\Delta$ swaps the ends of $L_{+}$.

Proof: We already know that ${\cal X\/}_{+}$ is path connected. A single
evalution, say for the point $(1/4)(2,1,2,1)$, suffices to show that
$\Delta({\cal X\/}_{+})\cap{\cal X\/}_{+}\not=\emptyset$. Since the component
functions of $\Delta$ are positive and since $F_{j}\circ\Delta=F_{j}$ and
$G\circ\Delta=G$ we see that $\Delta$ cannot map any point of ${\cal X\/}_{+}$
into the boundary of ${\cal X\/}_{+}$. This would cause some invariant to
vanish. Since ${\cal X\/}_{+}$ is connected, this implies that $\Delta({\cal
X\/}_{+})\subset{\cal X\/}_{+}$. Since $\Delta$ is an involution we
$\Delta({\cal X\/}_{+})={\cal X\/}_{+}$. Since $\Delta$ preserves both
invariants we must have $\Delta(L_{+})=L_{+}$. Since $\Delta$ preserves both
invariants and negates the symplectic form, we see that $\Delta$ is an
isometry of $L_{+}$ which reverses the direction of the $G$-curves.
$\spadesuit$

Let $L_{+}^{\prime}$ denote the subset of $L_{+}$ consisting of arcs of
$G$-curves which join points on the nice loop $L_{+}\cap\cal U$ to the
backwards end of $L_{+}$. This definition makes sense because each $G$-curve
intersects $L_{+}^{\prime}$ once (and transversely).

######  Lemma 6.2 (Reversal II)

$\iota_{5}(L_{+}^{\prime})=L_{+}^{\prime}$ and $\iota_{5}$ swaps the ends of
$L_{+}^{\prime}$.

Proof: We compute

$\iota_{5}(a,1-a,c,d)=(0,\beta,\beta,0),\hskip
30.0pt\beta=\frac{1-(c+d)}{(c+d)-(c-d)^{2}}.$ (42)

Note that $\beta>0$ when our points lie in ${\cal U\/}_{ab}$.

Conversely, we compute that if $\iota_{5}(a,b,c,d)=(0,\beta,\gamma,0)$ then
one of the following happens:

$cd=0,\hskip 30.0pta+b=0,\hskip 30.0pta-b+c-d=0.$

The last equation comes from the explicit solution

$a=\frac{d(1+c-d)}{1-(c+d)},\hskip 30.0ptb=\frac{c(1-c+d)}{1-(c+d)}.$

In the same way we compute

$\iota_{5}(a,b,c,1-c)=(\alpha,0,0,\alpha),\hskip
30.0pt\alpha=\frac{1-(a+b)}{(a+b)-(a-b)^{2}}.$ (43)

Again, $\alpha>0$ when our points lie in ${\cal U\/}_{cd}$. Conversely, we
compute that if $\iota_{5}(a,b,c,d)=(\alpha,0,0,d)$ then one of the following
happens:

$ab=0,\hskip 30.0ptc+d=0,\hskip 30.0pta-b+c-d=0.$

Now let $\chi$ be a $G$-curve of $L_{+}^{\prime}$. We claim that $\iota_{5}$
cannot map an interior point of $\chi$ to either to $L_{+}\cap\cal U$ to the
back end of $L_{+}$. To see this, let ${\cal H\/}_{ab}$ denote the hyperplane
given by $a+b=1$. The same argument as in Lemma 5.6 shows that each $G$-curve
in $L_{+}$ intersects ${\cal H\/}_{ab}$ at most once. Hence the interior of
$\chi$ contains no points of the form $(a,b,c,d)$ with $a+b=1$. Likewise, the
interior of $\chi$ contains no points of the form $(a,b,c,d)$ with $c+d=1$.
Furthermore, the interior of $\chi$ contains no points where a coordinate is
$0$ and no points where $a-b+c-d=0$. The calculations above now show that
$\iota_{5}$ cannot map an interior point of $\chi$ to the back end of $L_{+}$.
If $\iota_{5}$ maps some interior point of $\chi$ to some point of
$L_{+}\cap\cal U$ then (being an involution) $\iota_{5}$ maps a point of
$L_{+}\cap\cal U$ to an interior point of $\chi$. This contradicts Equations
42 or 43.

Call a $G$-curve $\chi$ of $L_{+}^{\prime}$ good if $\iota_{5}(\chi)$
intersects $L_{+}^{\prime}$. Our claim above combines with the fact that
$\iota_{5}$ is an involution to show that $\iota_{5}(\chi)$ is precisely a
$G$-curve in $L_{+}^{\prime}$. If some $G$-curve in $L_{+}^{\prime}$ is good
then by continuity all nearby $G$-curves in $L_{+}^{\prime}$ are good.
Similarly, if we have a sequence of good $G$-curves in $L_{+}^{\prime}$ then
then their limit is also good.

This point needs a bit more explanation. If we have a sequence
$\\{\chi_{n}\\}$ of good $G$-curves, then the image sequence is just some
other sequence of $G$-curves in $L_{+}^{\prime}$, from what we have said in
the previous paragraph. But then, by transversality, such a sequence has
points which lie just a bit inside $L_{+}^{\prime}$ and near $L_{+}\cap\cal
U$. The limit of such points belongs to $L_{+}^{\prime}$.

Now we know that if $L_{+}^{\prime}$ contains a good $G$-curve then every
$G$-curve of $L_{+}^{\prime}$ is good and hence
$\iota_{5}(L_{+}^{\prime})=L_{+}^{\prime}$. Since $\iota_{5}$ negates the
invariant symplectic form, $\iota_{5}$ must reverse the directions of the
$G$-curves in $L_{+}^{\prime}$. Hence $\iota_{5}$ swaps the ends of
$L_{+}^{\prime}$.

It now follows from the same continuity argument that every $L_{+}^{\prime}$
is good provided that one of them is. Finally, we just make an explicit
calculation to show that $L_{+}^{\prime}(3,4)$ is good. Hence all the level
sets in ${\cal X\/}_{+}$ are good. $\spadesuit$

###  6.2 The Main Argument

We choose a flat metric on $L_{+}$ coming from the integrable structure. Our
results below hold for any choice. Since $L_{+}\cap\cal U$ is actually a
subset of the cylinder $L_{+}$, every point of $p\in L_{+}$ is less than
$D_{p}$ to every point of $L_{+}\cap{\cal U\/}$ for some constant $D_{p}$
which depends on $p$. But now we apply $\iota_{5}$ and conclude that the same
boundedness result holds for the backwards end of $L_{+}$. Next, we apply
$\Delta$ and conclude that the same boundedness result holds for the forwards
end of $L_{+}$. Putting these results together we see that $L_{+}$ is
intrinsically bounded.

Now we show that $L_{+}$ is locally concave near $\partial L_{+}$ except for
$2$ points on each boundary component. Given the properties of $\iota_{5}$ and
$\Delta$, and symmetry, it suffices to prove that $L_{+}^{\prime}$ is locally
concave along the nice loop at all points of ${\cal U\/}_{ab}-{\cal
U\/}_{cd}$. Let

$Y=\alpha_{1}X_{1}+\alpha_{2}X_{2},\hskip 30.0pt\alpha_{1}=1+c-d,\hskip
30.0pt\alpha_{2}=1-c+d,$ (44)

Let

$\phi(a,b,c,d)=a+b,\hskip 30.0pt\psi=Y\cdot\nabla\phi.$ (45)

We compute that $\psi=0$ on ${\cal U\/}_{ab}$. In other words, $Y$ is tangent
to the nice loop along ${\cal U\/}_{ab}$. The local convexity in the intrinsic
metric is equivalent to the statement that

$q=Y\cdot\nabla\psi\bigg{|}_{a,1-a,c,d}\not=0.$ (46)

Geometrically, what this says is that we travel along the straight line
parallel to $Y$ the second derivative of the defining function is nonzero.

Computing the Hessian, we have

$q=\sum_{i,j}Q_{ij}\alpha_{i}\alpha_{j},\hskip
30.0ptQ_{ij}=X_{i}\cdot\nabla(X_{j}\cdot h).$ (47)

In other words, we evaluate the quadratic form $Q$ on the vector
$(\alpha_{1},\alpha_{2})$.

We compute in Mathematica that

$q=\frac{8a(1-a)(1+c-d)^{2}(1-c+d)^{2}(c+d-1)(c+d-(c-d)^{2})}{c^{2}d^{2}}<0.$

Since $c+d<1$ we have $c+d-1<0$. Since $c,d\in(0,1)$ we have $|c-d|<c+d$. So,
the last factor in the numerator is positive. A single plot is enough to show
that the positive sign determines concavity rather than convexity with respect
to the side of the arc lying in $L_{+}^{\prime}$.

##  7 Analytic Continuation

###  7.1 First Stage

We carry out Step 4 of the outline in §3.6. Let $L_{+}$ be a level set of
${\cal X\/}_{+}$. Let $L$ denote the level set in $\mbox{\boldmath{$R$}}^{4}$
containing $L_{+}$. We will assemble a torus $\widehat{L}$ out of some of the
components of $L$ and show that the group $\langle A,\Delta\rangle$ naturally
acts on $\widehat{L}$. With some more effort we could probably prove that
$\widehat{L}$ uses all the components of $L$. However, this is not necessary
for our proof of Theorem 1.2. Recall that $\iota_{5}=A\Delta A\Delta A$.
Recall that ${\cal U\/}$ is the set where $\max(a+b,c+d)=1$ and ${\cal
U\/}_{ab}$ is the subset where $a+b=1$.

Define

$(V_{1},V_{2},V_{3},V_{4})=d\iota_{5}(X_{G}).$ (48)

###### Lemma 7.1

$V_{1}=V_{4}<0$ for points of $L_{+}\cap{\cal U\/}_{ab}$.

Proof: The fact that $V_{1}=V_{4}$ is a direct computation. We will show that
$V_{1}<0$. We compute

$V_{1}=\frac{2c(c-2)(1-2c+c^{2}-2d-2cd+d^{2})}{(1+c-d)((c-d)^{2}-(c+d))}$ (49)

Since $c,d\in(0,1)$ the only term that can potentially vanish is

$f=1-2c+c^{2}-2d-2cd+d^{2}.$ (50)

The solutions to $f=0$ are given by $c=1+d\pm 2\sqrt{d}.$ Since $c\in(0,1)$
only the $-$ choice is valid. That is

$c=1+d-2\sqrt{d}.$ (51)

We claim that this variable choice forces $F_{1}(a,1-a,c,d)<0$. To see this we
note that this expression is negative when $a=3/4$ and $c=1/2$. The value is
about $-2.343$. Solving the equation $F_{1}=0$ gives the further information
that $d=a^{2}$. This gives $c=(1-a)^{2}$. But for these choices we see that
$G(a,1-a,c,d)=0$. Hence, these points do not lie in ${\cal X\/}_{+}$. Since
${\cal X\/}_{+}$ is connected, and since the nice loops $L_{+}\cap{\cal
U\/}_{ab}$ vary continuously for level sets in ${\cal X\/}_{+}$, it follows
from continuity that we never have $F_{1}>0$ when $c=1+d-2\sqrt{d}$. This
proves $V_{1}<0$, as desired. $\spadesuit$

Recall that the $G$-curves are not defined at the coordinate hyperplanes.
However, it may happen that there are $2$ different $G$-curves $\gamma_{1}$
and $\gamma_{2}$ and a point $p$ such that $p$ is the forwards endpoint of
$\gamma_{1}$ and the backwards endpoint of $\gamma_{2}$. We say that
$\gamma_{1}$ and $\gamma_{2}$ are analytic partners if there is a regular
analytic curve $\gamma=\gamma_{1}\cup\gamma_{2}\cup p$. Here regular means
that it has an analytic parametrization with nonzero derivative everywhere. We
call $p$ the join point for $(\gamma_{1},\gamma_{2})$.

We say that two disjoint components $L_{1},L_{2}$ of $L$ are analytic partners
if all but finitely many common accumulation points of $L_{1}$ and $L_{2}$ are
join points for a pairs of analytic partners, one in $L_{1}$ and one in
$L_{2}$.

###### Lemma 7.2

$L_{+}$ has $2$ disjoint analytic partners across its back end.

Proof: We will show that $L_{+}$ is joined to an analytic partner across the
subset of accumulation points of the form $(0,b,b,0)$. Here $b\in[0,1]$. We
omit the case $b=0$ from discussion. (This is why we throw away finitely many
accumulation points.) By symmetry, $L_{+}$ is joined to another analytic
partner across the subset of accumulation points of the form $(a,0,0,a)$ with
$a\in[0,1)$. These two analytic partners are distinct because the coordinates
of points in them have different signs.

Consider a small open neighborhood $U$ of $L_{+}\cap{\cal U\/}_{ab}$. We have
already shown that $\iota_{5}$ carries $U\cap L_{+}^{\prime}$ to a
neighborhood of the back end of $L_{+}$ that accumulates on points of the form
$(0,b,b,0)$. Now we look at the image of $U-L_{+}^{\prime}$, namely the other
side of $L_{+}\cap\cal U$, under $\iota_{5}$. We argue that $\iota_{5}$
carries this set into the same component of $L$, and that this component is an
analytic partner. This comes down to a calculation.

The map $\iota_{5}$ is well-defined and analytic in $U$. Hence the image of a
$G$-curve in $L$ is analytic. The subset of the image not contained in the
coordinate hyperplanes is a union of $G$-curves, because
$\iota_{5}(X_{G})=-X_{G}$ whenever the domain and range are not in the
coordinate hyperplanes.

We define $(V_{1},V_{2},V_{3},V_{4})=d\iota_{5}(X_{G})$, as above. We have
already proved that $V_{1}=V_{4}<0$ when we are doing the computation based at
a point of $L_{+}\cap{\cal U\/}_{ab}$. Thus, we see that when $U$ is
sufficiently small, points of $L_{+}$ on the other side of $L_{+}\cap{\cal
U\/}_{ab}$ from $L_{+}^{\prime}$ map into the subset of
$\mbox{\boldmath{$R$}}^{4}$ consisting of points
$(a^{\prime},b^{\prime},c^{\prime},d^{\prime})$ where
$a^{\prime},d^{\prime}<0$ and $b^{\prime},c^{\prime}>0$. This remains true as
we move around in $L_{+}\cap{\cal U\/}_{ab}$ as long as we stay away from the
point that $\iota_{5}$ maps to $(0,0,0,0)$. Being connected,
$\iota_{5}(U-L_{+}^{\prime})$ lies in one component of $L$. $\spadesuit$

###  7.2 Second Stage

Recall that $A(a,b,c,d)=(-b,-a,-d,-c)$. This is an everywhere defined linear
map of $\mbox{\boldmath{$R$}}^{4}$. Define

$L_{-}=A(L_{+}).$ (52)

The coordinates of all points on $L_{-}$ are negative. Hence $L_{+}$ and
$L_{-}$ are disjoint. On the other hand, both $L_{+}$ and $L_{-}$ have
$(0,0,0,0)$ as an accumulation point. Since $L_{+}$ has $2$ disjoint analytic
partners across its back end, $L_{-}$ has $2$ disjoint analytic partners
across its front end.

###### Lemma 7.3

The $2$ back-end analytic partners of $L_{+}$ coincide with the $2$ front-end
analytic partners of $L_{-}$.

Proof: Since every component of $L$ is a smooth surface, and since the
analytic partners of $L_{+}$ and $L_{-}$ discussed above vary continuously
with the parameters, and since the space of level sets of ${\cal X\/}_{+}$ is
connected, it suffices to show for some choice of parameters that the lemma is
true.

Let ${\cal H\/}_{ab}$ be the hyperplane given by $a+b=1$. We have ${\cal
U\/}_{ab}\subset{\cal H\/}_{ab}$. Define ${\cal V\/}_{ab}={\cal
H\/}_{ab}-{\cal U\/}_{ab}$. Likewise define ${\cal H\/}_{cd}$ and ${\cal
V\/}_{cd}$. Say that the level set $L$ is clean if $\gamma_{ab}=L_{+}\cap{\cal
H\/}_{ab}$ and $\gamma_{cd}=L_{+}\cap{\cal H\/}_{cd}$ are closed loops.

The same proof as in the previous chapter shows that $L(4-t,4)$ is clean for
$t>0$ sufficiently small. In this case we have a small cylinder passing
transversely through $2$ transverse hyperplanes. From this picture we see that
$\gamma_{ab}$ and $\gamma_{cd}$ intersect exactly twice. Figure 5 shows
schematically what this looks like in the flat structure. The left and right
boundaries are supposed to be identified by translation.

Figure 5: The cylinder $L_{+}$ near $\gamma_{ab}$ and $\gamma_{cd}$.

The various boundary curves are always locally convex but their exact shape
depends on the parameters. The reader should compare Figure 5 with Figure 3 in
§3.6.

The same analysis as in the previous section shows that $d\iota_{5}$ is
nonsingular on points of $L_{+}\cap{\cal H\/}_{ab}$. We start with a $G$-curve
$\chi$ in $L_{+}$ that is near one of the points where $L_{+}$ intersects
${\cal H\/}_{ab}\cap{\cal H\/}_{cd}$. Figure 5 shows what this curve looks
like intrinsically. Given the properties of $\iota_{5}$, the image
$\iota_{5}(\chi)$ is a regular analytic curve with $2$ break points, one an
accumulation point of the back end of $L_{+}$ and the second one an
accumulation point on the front end of $L_{-}$.

Figure 5 indicates the signs of the coordinates of points along
$\iota_{5}(\chi)$. Once all the coordinates of $\iota_{5}(\chi)$ are negative,
they must lie in $L_{-}$ because the level set $L$ has only one connected
component on $A({\cal X\/}_{-})$. This follows from the connectivity of level
sets in ${\cal X\/}_{+}$ and symmetry. Thus, the existence of
$\iota_{5}(\chi)$ shows that one and the same component is a back-end analytic
partner of $L_{+}$ and a front-end analytic partner of $L_{-}$. A similar
argument works for the other pair of analytic partners. $\spadesuit$

Let $D_{1}$ and $D_{2}$ denote the two back-end analytic partners of $L_{+}$.
As we have just seen $D_{1}$ and $D_{2}$ are also front-end analytic partners
of $L_{-}$.

The analysis above works even for the $G$-curves which go through the points
of $L_{+}\cap{\cal H\/}_{ab}\cap{\cal H\/}_{cd}$. In this case $\iota_{5}$
maps these $G$-curves to regular analytic curves which pass directly from
$L_{+}$ to $L_{-}$. There are $2$ such $G$-curves, corresponding to the $2$
intersection points on the nice loop discussed in the previous chapter. From
all this structure, we see that the development of $L_{+}\cup L_{-}\cup
D_{1}\cup D_{2}$ looks like Figures 3 and 6. The two cylinders $L_{+}$ and
$L_{-}$ abut at two cusp points and $D_{1},D_{2}$ fit in between. The
boundaries of $D_{1}$ and $D_{2}$ are convex because the boundaries of $L_{+}$
and $L_{-}$ are concave away from the two cusps.

###### Lemma 7.4

$A(D_{1})=D_{1}$ and $A(D_{2})=D_{2}$.

Proof: Since $A$ swaps $L_{+}$ and $L_{-}$, we have $A(D_{1}\cup
D_{2})=D_{1}\cup D_{2}$. One of the disks $D_{1}$ has accumulation points of
the form $(a,0,0,a)$ and $(0,-b,-b,0)$ for $a,b>0$. The other disk $D_{2}$ has
accumulation points of the form $(0,b,b,0)$ and $(-a,0,0,-a)$ for $a,b>0$.
Given that $A(a,b,c,d)=(-b,-a,-d,-c)$ we see that $A(D_{1})$ has accumulation
points of the same kind that $D_{1}$ does. Hence $A(D_{1})=D_{1}$. Likewise
$A(D_{2})=D_{2}$. $\spadesuit$

###  7.3 Third Stage

In this section we prove that $L_{+}$ has $2$ disjoint analytic partners
across its front end.

###### Lemma 7.5

$\Delta$ is well-defined and nonzero on $D_{1}\cup D_{2}$.

Proof: For convenience we repeat the equation for $\Delta$.

$\Delta=\bigg{(}\frac{b(c+d+1)}{c(E+a+c+1)},\frac{d(E+b+c)}{c(E+a+c+1)},\frac{d(a+b+1)}{a(E+a+c+1)},\frac{b(E+a+d)}{a(E+a+c+1)}\bigg{)}$
(53)

Since $D_{1}$ is an analytic partner of the back end of $L_{+}$ there are
points $(a,b,c,d)\in D_{1}$ satisfying $|a-b|<1$. Given that $a$ and $b$ have
opposite signs, this forces $|a|+|b|<1$ as well. Likewise $|c|+|d|<1$. The
denominators of the coordinates of $\Delta(a,b,c,d)$ vanish only if

$d=-\frac{(1+a)(1+c)}{b}.$

Since $|a|+|b|<1$ we have $|1+a|/|b|>1$. Thus $|d|>1$. This is a contradition.

Item $k$ in the following list describes the variable choices which lead to
the $k$th coordinate of $\Delta(a,b,c,d)$ vanishing.

  1. 1.

$b=0$ or $c+d=-1$.

  2. 2.

$d=0$ or $bd+ac=-b-c$.

  3. 3.

$a+b=-1$ or $d=0$.

  4. 4.

$b=0$ or $ac+bd=-a-d$.

Our constraints above rule out all these possibilities. To explain one of the
two non-obvious ones, suppose $bd+ac=-b-c$. Then

$d=\frac{-b-c-ac}{b}=-1+\frac{c(-1-a)}{b}.$ (54)

There are two cases and in either case $|a|<1$ and $b/c>0$. This means that
the second term on the right side of Equation 54 is negative and $d<-1$. This
is a contradiction. The other non-obvious impossibility has the about the same
treatment. $\spadesuit$

###### Lemma 7.6

$\Delta(D_{1})$ and $\Delta(D_{2})$ are front-end analytic partners of
$L_{+}$.

Proof: By Lemma 7.5, the sets $\Delta(D_{1})$ and $\Delta(D_{2})$ are
contained in components of $L$. The map $\Delta$ is not defined on $\partial
D_{1}$ or $\partial D_{2}$ but the last two coordinates of $\Delta(a,0,0,a)$
are defined and the first two coordinates of $\Delta(0,b,b,0)$ are well
defined. We have

$\Delta(a,0,0,a)=(*,*,1,0),\hskip 30.0pt\Delta(0,b,b,0)=(1,0,*,*).$ (55)

The starred components would require more information to determine. However,
the above equations imply that $\Delta$ maps the boundaries of $D_{1}$ and
$D_{2}$ into the coordinate hyperplanes. Hence $\Delta(D_{1})$ and
$\Delta(D_{2})$ are precisely components of $L$.

Now we want to see that $\Delta(D_{1})$ and $\Delta(D_{2})$ are analytic
front-end partners of $L_{+}$. We cannot directly compute the value of
$\Delta$ on the boundary of $D_{j}$ but we recall that $\iota_{5}$ maps
$L_{+}\cap\cal U$ to the back end of $L_{+}$, the very spot we are interested
in. When we compute $\Delta\circ\iota_{5}(a,b,c,d)$ we get an expression of
the form

$\bigg{(}*,*,\frac{N_{3}}{D_{3}},\frac{N_{4}}{D_{4}}\bigg{)}$ (56)

We write things this way to make sure we are only dealing with polynomials.
Since $b=1-a$ we find that (as expected) the first two values are $1$ and $0$.
We also have

$N_{3}=-2a^{2}d(1-a)^{3}(1+c-d)^{3}(1-c+d)(c^{2}-2cd-2c+d^{2}-2d+1).$
$D_{3}=-2a^{2}c(1-a)^{3}(1+c-d)^{2}(1-c+d)^{2}(c^{2}-2cd-2c+d^{2}-2d+1).$

We have $D_{3}/D_{4}=1$ and $N_{3}/N_{4}=d(1-c+d)/(1-(c+d))$. Hence we have
the well defined values

$\frac{N_{3}}{D_{3}}=\frac{d(1+c-d)}{c(1-c+d)},\hskip
30.0pt\frac{N_{4}}{D_{4}}=\frac{1-(c+d)}{c(1-c+d)}.$ (57)

All the factors in $N_{3},N_{4},D_{3},D_{4}$ are nonzero except perhaps the
last factor. But these last factors appear identically in the numerator and
the denominator. Hence $\Delta\circ\iota_{5}$ has an analytic extension to
$L_{+}\cap\cal U$. Moreover, $\Delta\circ\iota_{5}$ maps one side of
$L_{+}\cap\cal U$ to a neighborhood of the front end of $L_{+}$ and the other
side to a neighborhood of $\Delta(D_{1})\cup\Delta(D_{2})$. $\spadesuit$

###  7.4 Fourth Stage

Define

$\iota_{3}=A\Delta A,\hskip 30.0ptM=\iota_{3}(L_{+}^{\prime}).$ (58)

Let $D_{3},D_{4}$ be the front-end analytic partners of $L_{+}$ and let
$D_{5},D_{6}$ be the back-end analytic partners of $L_{-}$. In this section we
will identify $M$ as the common analytic partner to all of
$D_{3},D_{4},D_{5},D_{6}$.

Figure 6: A schematic picture in the clean case.

Figure 6 shows a schematic summary of the action of the various involutions we
have shown so far, as well as the some of the structure we determine in this
section. We represent our disks as shaded rectangles and the cylinders as
white rectangles. Note that $\iota_{5}(D_{1}\cup D_{2})\subset L_{+}$ only in
the clean case.

###### Lemma 7.7

$M$ is a component of the level set $L$.

Proof: Letting $e=ac+bd$ we compute

$\iota_{3}(a,b,c,d)=\left(\frac{c(a+d-e)}{d(e-b-d+1)},\frac{a(c+d-1)}{d(e-b-d+1)},\frac{a(b+c-e)}{b(e-b-d+1)},\frac{c(a+b-1)}{b(e-b-d+1)}\right).$
$\iota_{3}(a,1-a,c,d)=\bigg{(}\frac{c(1-c+d)}{d(1+c-d)},\frac{(c+d)-1}{d(1+c-d)},1,0\bigg{)}.$
(59)

The action of $\iota_{3}(A)$ on the other end of $L_{+}^{\prime}$ is the same
as the action of $\Delta$ on one end of $D_{1}\cup D_{2}$, which we have
already discussed. Hence, $\iota_{3}$ maps the ends of $L_{+}^{\prime}$ into
the coordinate hyperplanes.

To finish the proof we just have to see that $\iota_{3}$ is well-defined and
nonzero on $L_{+}^{\prime}$. On $L_{+}^{\prime}$ we have the constraints
$a,b,c,d>0$ and $\max(a+b,c+d)<1$. Here are all the ways that a numerator or a
denominator of a coordinate of $\iota_{3}(a,b,c,d)$ can vanish when
$a,b,c,d\not=0$.

$-ac=(1-b)(1-d),\hskip 10.0pta+b=1,\hskip 10.0ptc+d=1,\hskip
10.0pta+d=ac+bd,\hskip 10.0ptb+c=ac+bd.$

In all situations, the constraints mentioned prevent these equalities.
$\spadesuit$

###### Lemma 7.8

$M$ is an analytic partner of $D_{3}$ and of $D_{4}$.

Proof: We can view $\iota_{5}(D_{1}\cup D_{2})$ as a union of analytic
partners for $L_{+}^{\prime}$, even though there is no actual barrier here. We
then compute

$\iota_{3}\circ\iota_{5}(D_{1}\cup D_{2})=\Delta\circ A(D_{1}\cup
D_{2})=\Delta(D_{1}\cup D_{2})=D_{3}\cup D_{4}.$

This $M$ lies on the other side of $D_{3}\cup D_{4}$ from $L_{+}$. Moreover,
one end of $M$, namely $\iota_{3}(L_{+}\cap{\cal U\/})$ has common
accumulation points with the far end of $D_{4}\cup D_{4}$. To see that $M$ is
really an analytic partner to these disks, we let

$(W_{1},W_{2},W_{3},W_{4})=d\iota_{3}(X_{G}).$

We compute

$W_{4}=\frac{4c(2-c)}{1+c-d}>0.$

From this we conclude that the $G$-curves exiting $D_{3}\cup D_{4}$ on the far
side, the side opposite the one abutting $L_{+}$, are joined to $G$-curves in
$M$ by regular analytic arcs. This shows that $M$ is an analytic partner of
each of $D_{3}$ and $D_{4}$. $\spadesuit$

###### Lemma 7.9

$M$ is an analytic partner of $D_{5}$ and of $D_{6}$.

Proof: Since $L_{-}=A(L_{+})$, the map $\iota_{3}=A\Delta A$ acts on $L_{-}$
just as $\Delta$ acts on $L_{+}$. Thus $\iota_{3}(D_{1}\cup D_{2})=D_{5}\cup
D_{6}$. The rest of the proof is very much like that in Lemma 7.6. The main
point is that want to understand the action of $\Delta$ on the back end of
$L_{+}^{\prime}$, but this is the same as the action of $\iota_{3}\iota_{5}$
on $L_{+}\cap\cal U$. Working similarly to Lemma 7.6 we find that

$\iota_{3}\circ\iota_{5}(a,1-a,c,d)=\delta\circ
A(a,b,c,d)=\bigg{(}\frac{1-(c+d)}{d(1+c-d)},\frac{-1+c-d}{d(1+d-c)},0,-1\bigg{)}.$
(60)

The rest of the argument is the same as in Lemma 7.6.

###  7.5 Behold the Torus

The components of $L$ we have described above fit together as in Figures 3 and
6 to make a flat torus. Call this object $\widehat{L}$.

###### Lemma 7.10

$A(\widehat{L})=\widehat{L}$.

Proof: We have already seen that the action of $A$ permutes all the pieces we
have defined except perhaps for $M$. But $A$ swaps the union $D_{3}\cup D_{4}$
with the union $D_{5}\cup D_{6}$ and preserves analytic partnership. Also
$A(L_{\pm})=L_{\mp}$. Hence $A(M)=M$. Since $A$ permutes all the pieces of $L$
we have $A(\widehat{L})=\widehat{L}$. $\spadesuit$

###### Lemma 7.11

There is an isometric action $\widehat{\Delta}$ on $\widehat{L}$ which agrees
with the action of $\Delta$ on $L$ at all points where $\Delta$ is well
defined.

Proof: Since everything in sight is an intrinsic isometry we just have to
prove that $\Delta$ maps all the components of $L$ involved in $\widehat{L}$
back into $\widehat{L}$, Let us examine all the pieces. Since $\Delta$
preserves $L_{+}\cup D_{1}\cup D_{2}\cup D_{3}\cup D_{4}$ we have

$\Delta(L_{+}\cup D_{1}\cup D_{2}\cup D_{3}\cup D_{4})\subset\widehat{L}.$

Since $L_{-}=A(L_{+})$ and $A(\widehat{L})=\widehat{L}$, we can show that
$\Delta(L_{-})\subset\widehat{L}$ by showing that
$\iota_{3}(L_{+})\subset\widehat{L}$. We have

$\iota_{3}(L_{+})=\iota_{3}(L_{+}^{\prime})\cup\iota_{3}(L_{+}^{\prime\prime})\cup\iota_{3}\circ\iota_{5}(D_{1}\cup
D_{2})=M\cup L_{+}^{\prime\prime}\cup D_{3}\cup D_{4}\subset\widehat{L}.$

Hence $\Delta(L_{-})\subset\widehat{L}$.

Now we want to show that $\Delta(D_{5}\cup D_{6})\cup\widehat{L}$. Since
$A(\widehat{L})=\widehat{L}$ and $A$ swaps $D_{3}\cup D_{4}$ with $D_{5}\cup
D_{6}$, it suffices to show that $\iota_{3}(D_{5}\cup
D_{6})\subset\widehat{L}$. But

$\iota_{3}(D_{5}\cup D_{6})=D_{1}\cup D_{2}\subset\widehat{L}.$

This does the job.

Finally, since $\Delta$ is an involution, we have
$\Delta(M)=L_{+}^{\prime}\subset\widehat{L}.$ $\spadesuit$

##  8 The Final Step

###  8.1 The General Case

We carry out Step 5 of our outline from §3.6. Recall that $\cal C$ is the set
of convex points. Let $\cal P$ be the positive orthant in
$\mbox{\boldmath{$R$}}^{4}$. Say that $p=(a,b,c,d)\in{\cal P\/}$ is a strange
point if every word in the semigroup $\langle A,\Delta\rangle$ is defined on
$p$ and if either the forwards or backwards $T_{3}$ orbit of $p$ lies in $\cal
P$.

###### Lemma 8.1

If $p$ is a counterexample to Theorem 1.2 then $p$ is strange.

Proof: The hypothesis imply that $T_{3}^{k}$ is defined on $p$ for all $k$,
and $T^{k}(p)\in\cal C$ for all $k\in\mbox{\boldmath{$Z$}}$. So, we just have
to see that every word of length less than $4$ in $\langle A,\Delta\rangle$ is
defined on $p$. Recall that $A(a,b,c,d)=(-b,-a,-d,-c)$. Certainly $A$ is
defined on $p$. Also, $A\Delta A$ is defined on $p$ if and only if $\Delta A$
is defined on $p$. Since only $(+)$ appears in the expression for $\Delta$, we
see that $\Delta$ is defined on $p$. Hence so is $A\Delta$. This leaves us to
check the two words $A\Delta A=\Delta T_{3}^{-1}$ and $\Delta A\Delta=AT^{3}$.
Since both $T_{3}(p)$ and $T_{3}^{-1}(p)$ are convex points, we see that $A$
and $\Delta$ are defined on these points. Hence $A\Delta A$ and $\Delta
A\Delta$ are defined on $p$. $\spadesuit$

###### Lemma 8.2

There are no strange points in ${\cal X\/}_{+}$.

Proof: We suppose that $p\in{\cal X\/}_{+}$ is a strange point and then derive
a contradicton. Let $L_{+}$ be the level set in ${\cal X\/}_{+}$ containing
$p$. The work in the previous chapter tells us that the group $\langle
A,\widehat{\Delta}\rangle$ gives an isometric action on the torus
$\widehat{L}$ extending the action of $\langle A,\Delta\rangle$ on $L_{+}$.
Given the action of $T_{3}$ on $\widehat{L}$ we established in Step 4 of the
proof, the $T_{3}$-orbit of $p$ lies in $\widehat{L}$ but both the forwards
and backwards $T_{3}$-orbit exit $L_{+}$. Since $\widehat{L}\cap{\cal
P\/}=L_{+}$, we have a contradiction. $\spadesuit$

###### Lemma 8.3

If $p$ is a strange point then so if $\gamma(p)$ for any word
$\gamma\in\langle J,\Delta\rangle$.

Proof: It suffices to prove this for $\gamma=J$ and $\gamma=\Delta$. Since $p$
and $J(p)$ represent isometric $8$-gons, related by reflection in the line
$y=x$. we see that $J(p)$ is also strange. Since $\Delta$ preserves the
positive orthant in $\mbox{\boldmath{$R$}}^{4}$ and conjugates $T_{3}$ to
$T_{3}^{-1}$ we see that $\Delta(p)$ is also strange. $\spadesuit$

###### Lemma 8.4

Suppose that $p=(a,b,c,d)$ is a convex point. Then $\gamma(p)\in{\cal
X\/}_{+}$ for some word $\gamma\in\langle J,\Delta\rangle$.

Proof: By Lemma 3.2, the set ${\cal C\/}-{\cal I\/}-{\cal I\/}^{*}$ has $4$
connected components, depending on the signs of $g_{ab}+g_{cd}$ and
$g_{ab}^{*}+g_{cd}^{*}$. Let $\gamma$ be some word in $J$ and $\Delta$ and let
$p$ be some point in one of these components. Let $q=\gamma(p)$. The
quantities $F_{1}(q)$, $F_{2}(q)$, $G(q)$ are the same as
$F_{1}(p),F_{2}(p),G(p)$ up to permuting the first two quantities and negating
the sign of the third. So, by the Positivity Lemma, $F_{1}(q)>0$ and
$F_{2}(q)>0$ and $G(q)\not=0$.

Since $q$ has all positive coordinates, we see that no factor of
$F_{1},F_{2},G$ vanishes at $q$. In particular, the signs of these various
factors cannot change as we vary $q$ continuously within its connected
component. This gives us the following principle: If $\gamma(p)\in{\cal
X\/}_{+}$ for some $p$, then $\gamma(p^{\prime})\in{\cal X\/}_{+}$ for all
points $p^{\prime}$ in the same component. To finish the proof we just have to
exhibit a word which works for a single point in each component. Three
evaluations check that the words $J$, $J\Delta$ and $J\Delta J$ respectively
work for the $(-,-)$, $(-,+)$, and $(+,-)$ components. $\spadesuit$

Now we finish Step 5 of our outline in §3.6. Suppose $p\in{\cal C\/}-{\cal
P\/}_{8,4}-{\cal I\/}-{\cal I\/}^{*}$. If $p$ is a counterexample to Theorem
1.2 then $p$ is strange. But then by Lemma 8.4 there is some $\gamma\in\langle
J,\Delta\rangle$ so that $\gamma(p)\in{\cal X\/}_{+}$. By Lemma 8.3, the point
$\gamma(p)$ is strange. This contradicts Lemma 8.2. The work in [17] deals
with the case when $p\in{\cal P\/}_{8,4}-{\cal I\/}-{\cal I\/}^{*}$. For the
reader who does not want to consult [17], we have indicated briefly in §3.6
how our proof here can be adapted to this case.

###  8.2 The Inscribed and Circumscribed Cases

We carry out Step 6 of our outline in §3.6 and thereby finish the proof of
Theorem 1.2. We also prove Theorem 1.3. Let ${\cal CI\/}$ and ${\cal
CI\/}^{*}$ respectively be the set of inscribed and circumscribed convex
points.

###### Lemma 8.5

${\cal CI\/}^{*}$ is forward $T_{3}$-invariant and ${\cal CI\/}$ is backwards
$T_{3}$-invariant.

Proof: We prove the first statement. The second statement follows from the
first statement and duality. The defining equation for ${\cal I\/}^{*}$ is
$a-b+c-d=0$. A calculation shows that this is $T_{3}$ invariant. Hence ${\cal
I\/}^{*}$ is $T_{3}$-invariant. (Again, this is a special case of a result in
[18].) Let $P_{1},...,P_{8}$ be the points of $\xi$. Let
$\sigma_{j}=\overline{P_{j}P_{j+3}}$. Suppose that
$\sigma_{1},\sigma_{2},\sigma_{3}$ are coincident. That is, the lines
extending them intersect at a single point. Then the points $P_{1},...,P_{6}$
are the vertices of an inscribed hexagon, by the converse of Brianchon’s
Theorem. This is impossible for a circumscribed $8$=gon: $P_{1}P_{6}$ is not
tangent to the circumscribing ellipse. Now we know that
$\sigma_{1},\sigma_{2},\sigma_{3}$ are never coincident. By symmetry the same
goes for all $\sigma_{i}$. Hence, as we vary $P\in{\cal CI\/}^{*}$ the image
$T_{3}(P)$ is never degenerate. When $P$ is regular, $T_{3}(P)$ is convex. So,
by continuity this holds for all $P\in{\cal CI\/}^{*}$. Also, $T_{3}(\xi)$ is
again circumscribed because ${\cal I\/}^{*}$ is $T_{3}$-invariant.
$\spadesuit$

Let $\Pi_{k}$ denote the plane consisting of points $(x+k,x-k,y-k,y+k)$. The
space ${\cal I\/}^{*}$ is partitioned into these planes. Let

$h=-g_{ab}\circ(A\Delta)|_{\Pi_{k}}=\frac{4k^{3}-x+y-4kxy}{(k-x)(k+y)}.$ (61)

Let $L(k,\ell)\subset\Pi_{k}$ be the level set $h^{-1}(\ell)$.

###### Lemma 8.6

$T_{3}^{4}$ preserves $L(k,\ell)$ and the action is conjugate to a linear
fractional transformation acting on the Riemann sphere.

Proof: Calculations show that $T_{3}^{2}(\Pi_{k})=\Pi_{-k}$ and, when
$p\in\Pi_{k}$,

$\psi\circ T_{3}(p)=-i\psi(p),\hskip
20.0pt\psi(a,b,c,d)=h(a,b,c,d)+ih(b,a,d,c).$ (62)

Hence $T_{3}^{4}(L(k,\ell))=L(k,\ell)$. Solving $h(x,y)=\ell$ and clearing
denominators, we get the quadratic equation

$(\ell-4k)xy+(1+k\ell)x-(1+k\ell)y-(4k^{3}-k^{2}\ell).$ (63)

This defines a projective curve in
$\mbox{\boldmath{$C$}}\mbox{\boldmath{$P$}}^{2}$ biholomorphic to the Riemann
sphere. Under this identification, the automorphism $T_{3}^{4}$ is a linear
fractional transformation. $\spadesuit$

Let $\phi_{k,\ell}$ be the linear fractional transformation conjugate to the
action extending $T_{3}^{4}$ on $L(k,\ell)$. Note that $L(k,\ell)$ contains
convex points iff $|k|<1/2$ and $|\ell|<2$. Let $\cal K$ denote this set of
$(k,\ell)$.

###### Lemma 8.7

If $(k,l)\in\cal K$ then $\phi_{k,\ell}$ is a hyperbolic linear
transformation. The attracting fixed point is convex Poncelet and the
repelling fixed point is (up to rotation) the star-reordering of the
attracting fixed point.

Proof: In $L(k,\ell)$ we can directly solve the equation $\phi_{k,l}(z)=z$ for
$x$ and $y$. There are (formally) two solutions $(x_{0},-y_{0})$ and
$(y_{0},-x_{0})$, where

$x_{0}+y_{0}=\frac{2(4k-\ell)(k\ell-1)}{(2+\ell)(2-\ell)},\hskip
30.0ptx_{0}y_{0}=\frac{-2-4k+4k\ell-k^{2}\ell^{2}}{(2+\ell)(2-\ell)}.$ (64)

The polynomial $t^{2}-(x_{0}+y_{0})t+x_{0}y_{0}$ has discriminant

$D=4(1+4k^{2}-2k\ell)((8-\ell^{2})-8k\ell+4(k\ell)^{2}).$ (65)

We have $D>0$ when $(k,\ell)\in\cal K$. Thus we get $2$ distinct real
solutions. A calculation shows that $g_{ab}+g_{cd}=0$ for the solutions. Since
$g_{ab}^{*}+g_{cd}^{*}=0$ as well, any non-degenerate fixed point is Poncelet.

The map $\phi_{k,\ell}$ has $2$ real fixed points and preserves the circle in
$\mbox{\boldmath{$C$}}\cup\infty$ corresponding to the real points. Hence
$\phi_{k,\ell}$ is either hyperbolic or elliptic of order $2$. Since $\cal K$
is connected the trivial elliptic option cannot occur. Hence $\phi_{k,\ell}$
is hyperbolic.

The fixed points have the form $(a,b,c,d)$ and $-(c,d,a,b)$. Up to rotation,
the corresponding polygons are star-reorderings of each other. Given the
forward invariance of ${\cal CI\/}^{*}$, one of the fixed points $(a,b,c,d)$
is the limit of convex $8$-gons in $L(k,\ell)$ This gives us

$|a-b|=|k|<1,\hskip 20.0pt|c-d|=k<1,\hskip 20.0pta+b\geq 1,\hskip
20.0ptc+d\geq 1.$

If $a+b=1$ then $g_{ab}(a,b,c,d)=2$. But then $g_{cd}(a,b,c,d)=-2$. This is
not possible for a finite pair $(c,d)$. Hence $a+b>1$. Likewise $c+d>1$. Hence
$(a,b,c,d)$ gives a nontrivial convex $8$-gon. From the calculation above,
this $8$-gon is Poncelet. The other fixed point is thus star-convex Poncelet.
Given the forward invariance of ${\cal CI\/}^{*}$, the convex fixed point must
be attracting. $\spadesuit$

Thus, the backwards $T_{3}^{4n}$ iterates of any point of ${\cal
CI\/}^{*}-{\cal CI\/}$ either accumulate on a star-convex $8$-gon or become
undefined. In either case, the backwards $T_{3}$-orbit of and point of ${\cal
CI\/}^{*}-{\cal CI\/}$ eventually exits ${\cal CI\/}^{*}$. Dually, the
forwards $T_{3}$-orbit of any point of ${\cal CI\/}-{\cal CI\/}^{*}$
eventually exits $\cal CI$. This completes Step 6 of the outline in §3.6. Our
proof of Theorem 1.2 is done.

If we forget about the labelings, we can say more. Since $T_{3}$ fixes
Poncelet polygons up to relabelings, we see that actually $T_{3}^{n}(\xi)$
converges to a convex Poncelet $8$-gon as $n\to+\infty$. That is, we don’t
have to restrict our iterates to multiplies of $4$ to get the convergence.
This (and duality) finishes the proof of Theorem 1.3 .

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