# The Flapping Birds in the Pentagram Zoo

Richard Evan Schwartz Supported by N.S.F. Grant DMS-21082802

###### Abstract

We study the $(k+1,k)$ diagonal map for $k=2,3,4,...$. We call this map
$\Delta_{k}$. The map $\Delta_{1}$ is the pentagram map and $\Delta_{k}$ is a
generalization. $\Delta_{k}$ does not preserve convexity, but we prove that
$\Delta_{k}$ preserves a subset $B_{k}$ of certain star-shaped polygons which
we call $k$-birds. The action of $\Delta_{k}$ on $B_{k}$ seems similar to the
action of $\Delta_{1}$ on the space of convex polygons. We show that some
classic geometric results about $\Delta_{1}$ generalize to this setting.

##  1 Introduction

###  1.1 Context

When you visit the pentagram zoo you should certainly make the pentagram map
itself your first stop. This old and venerated animal has been around since
the place opened up and it is very friendly towards children. When defined on
convex pentagons, this map has a very long history. See e.g. [15]. In modern
times [19], the pentagram is defined and studied much more generally. The
easiest case to explain is the action on convex $n$-gons. One starts with a
convex $n$-gon $P$, for $n\geq 5$, and then forms a new convex $n$-gon
$P^{\prime}$ by intersecting the consecutive diagonals, as shown Figure 1.1
below.

The magic starts when you iterate the map. One of the first things I proved in
[19] about the pentagram map is the successive iterates shrink to a point.
Many years later, M. Glick [3] proved that this limit point is an algebraic
function of the vertices, and indeed found a formula for it. See also [9] and
[1].

Figure 1.1: The pentagram map iterated on a convex $7$-gon $P$.

Forgetting about convexity, the pentagram map is generically defined on
polygons in the projective plane over any field except for
$\mbox{\boldmath{$Z$}}/2$. In all cases, the pentagram map commutes with
projective transformations and thereby defines a birational map on the space
of $n$-gons modulo projective transformations. The action on this moduli space
has a beautiful structure. As shown in [17] [18], and independently in [23],
the pentagram map is a discrete completely integrable system when the ground
field is the reals. ([23] also treats the complex case.) Recently, M.
Weinreich [24] generalized the integrability result, to a large extent, to
fields of positive characteristic.

The pentagram map has many generalizations. See for example [2], [14], [16],
[10], [11], [6]. The paper [2] has the first general complete integrability
result. The authors prove the complete integrability of the $(k,1)$ diagonal
maps, i.e. the maps obtained by intersecting successive $k$-diagonals. Figure
1.3 below shows the $(3,1)$ diagonal map. (Technically, [2] concentrates on
what happens when these maps act on so-called corregated polygons in higher
dimensional Euclidean spaces.) The paper [6] proves an integrability result
for a very wide class of generalizations, including the ones we study below.
(Technically, for the maps we consider here, the result in [6] does not
establish the algebraic independence of invariants needed for complete
integrability.) The pentagram map and its many generalizations are related to
a number of topics: alternating sign matrices [20], dimers [5], cluster
algebras [4], the KdV hierarchy [12], [13], spin networks [2], Poisson Lie
groups [8], Lax pairs [23], [10], [11], [6], [8], and so forth. The zoo has
many cages and sometimes you have to get up on a tall ladder to see inside
them.

Figure 1.2: The $(3,1)$-diagonal map acting on $8$-gons.

The algebraic side of the pentagram zoo is extremely well developed, but the
geometric side is hardly developed at all. In spite of all the algebraic
results, we don’t really know, geometrically speaking, much about what the
pentagram map and its relatives really do to polygons.

Geometrically speaking, there seems to be a dichotomy between convexity and
non-convexity. The generic pentagram orbit of a projective equivalance class
of a convex polygon lies on a smooth torus, and you can make very nice
animations. What you will see, if you tune the power of the map and pick
suitable representatives of the projective classes, is a convex polygon
sloshing around as if it were moving through water waves. If you try the
pentagram map on a non-convex polygon, you see a crazy erratic picture no
matter how you try to normalize the images. The situation is even worse for
the other maps in the pentagram zoo, because these generally do not preserve
convexity. Figure 1.2 shows how the $(3,1)$-diagonal map does not necessarily
preserve convexity, for instance. See [21], [22] for more details.

If you want to look at pentagram map generalizations, you have to abandon
convexity. However, in this paper, I will show that sometimes there are
geometrically appealing replacements. The context for these replacements is
the $(k+1,k)$-diagonal map, which I call $\Delta_{k}$, acting on what I call
$k$-birds. $\Delta_{k}$ starts with the polygon $P$ and intersects the
$(k+1)$-diagonals which differ by $k$ clicks. (We will give a more formal
definition in the next section.) $\Delta_{k}$ is well (but not perfectly)
understood algebraically [6]. Geometrically it is not well understood at all.

###  1.2 The Maps and the Birds

The Maps: Given a polygon $P$, let $P_{a}$ denote the $(a)$th vertex of $P$.
Let $P_{ab}$ be the line through $P_{a}$ and $P_{b}$. The vertices of
$\Delta_{k}(P)$ are

$P_{j,j+k+1}\cap P_{j+1,j-k}.$ (1)

Here the indices are taken mod $n$. Figure 1.3 shows this for $(k,n)=(2,7)$.

The Birds: We call a polygon $P$ planar if some projective transformation maps
$P$ to a bounded subset of $\mbox{\boldmath{$R$}}^{2}$, when it is considered
as the affine patch of the projective plane. Given an $n$-gon $P$, we let
$P_{a,b}$ denote the line containing the vertices $P_{a}$ and $P_{b}$. We call
$P$ $k$-nice if $n>3k$, and $P$ is planar, and the $4$ lines

$P_{i,i-k-1},\hskip 10.0ptP_{i,i-k},\hskip 10.0ptP_{i,i+k},\hskip
10.0ptP_{i,i+k+1}$ (2)

are distinct for all $i$.

We call $P$ a $k$-bird if $P$ is the endpoint of a path of $k$-nice $n$-gons
that starts with the regular $n$-gon. We let $B_{k,n}$ be the subspace of
$n$-gons which are $k$-birds. Note that $B_{k,n}$ contains the set of convex
$n$-gons, and the containment is strict when $k>1$. As Figure 1.3 illustrates,
a $k$-bird need not be convex for $k\geq 2$. We will show that $k$-birds are
always star-shaped, and in particular embedded. The homotopy part of our
definition looks a bit strange, but it is necessary. For instance, a $2n$-gon
that wraps twice around a convex $n$-gon is $1$-nice but not a $1$-bird.

Figure 1.3 shows $\Delta_{2}$ acting on what we call $2$-birds.

Figure 1.3: $\Delta_{2}$ acting on $2$-birds.

For us, a polygon is both the vertices and the edges. $\Delta_{k}$ only acts
on the vertices but, given the homotopical way we have defined birds, we
automatically have a way to define the edges of our birds. We just extend
continuously from the regular $n$-gon.

###  1.3 The Main Result

Here is the main result.

###### Theorem 1.1

Let $k\geq 2$ and $n>3k$ and $P\in B_{k,n}$. Then

  1. 1.

$P$ is strictly star-shaped.

  2. 2.

$\Delta_{k}(P)$ lies in the interior of the region bounded by $P$.

  3. 3.

$\Delta_{k}(B_{k,n})=B_{k,n}$.

The statement that $n>3k$ is present just for emphasis. $B_{n,k}$ is by
definition empty when $n\leq 3k$. The restriction $n>3k$ is necessary. Figure
1.4 illustrates what would be a counter-example to Theorem 1.1 for the pair
$(k,n)=(3,9)$. The issue is that a certain triple of $4$-diagonals has a
common intersection point. This does not happen for $n>3k$. See Lemma 3.6.

Figure 1.4: $\Delta_{3}$ acting on a certain convex $9$-gon.

###  1.4 The Energy

We will deduce Statements 1 and 2 of Theorem 1.1 in a geometric way. The key
to proving Statement 3 is a natural quantity associated to a $k$-bird. We let
$\sigma_{a,b}$ be the slope of the line $P_{a,b}$ and we define the cross
ratio

$\chi(a,b,c,d)=\frac{(a-b)(c-d)}{(a-c)(b-d)}.$ (3)

We define

$\chi_{k}(P)=\prod_{i=1}^{n}\chi(i,k,P),\hskip
20.0pt\chi(i,k,P)=\chi(\sigma_{i,i-k},\sigma_{i,i-k-1},\sigma_{i,i+k+1},\sigma_{i,i+k})$
(4)

Here we are taking the cross ratio the slopes the lines involved in our
definition of $k$-nice. When $k=1$ this is the familiar invariant
$\chi_{1}=OE$ for the pentagram map $\Delta_{1}$. See [19], [20], [17], [18].
When $n=3k+1$, a suitable star-relabeling of our polygons converts
$\Delta_{k}$ to $\Delta_{1}$ and $\chi_{k}$ to $1/\chi_{1}$. So, in this case
$\chi_{k}\circ\Delta_{k}=\chi_{k}$. Figure 1.4 illustrates this for
$(k,n)=(3,10)$. Note that the polygons suggested by the dots in Figure 1.4 are
not convex. Were we to add in the edges we would get a highly non-convex
pattern.

Figure 1.5: A star-relabeling converts $\Delta_{1}$ to $\Delta_{3}$ and
$1/\chi_{1}$ to $\chi_{3}$.

In general, $\chi_{k}$ is not as clearly related to $\chi_{1}$. Nonetheless,
we will prove

###### Theorem 1.2

$\chi_{k}\circ\Delta_{k}=\chi_{k}$.

Theorem 1.2 is meant to hold for all $n$-gons, as long as all quantities are
defined. There is no need to restrict to birds.

###  1.5 The Collapse Point

When it is understood that $P\in B_{k,n}$ it is convenient to write

$P^{\ell}=\Delta_{k}^{\ell}(P)$ (5)

We also let $\widehat{P}$ denote the closed planar region bounded by $P$.
Figure 1.6 below shows
$\widehat{P}=\widehat{P}^{0},\widehat{P}^{1},\widehat{P}^{2},\widehat{P}^{3},\widehat{P}^{4}$
for some $P\in B_{4,13}$.

Figure 1.6: $\Delta_{4}$ and its iterates acting on a member of $B_{4,13}$.

Define

$\widehat{P}_{\infty}=\bigcap_{\ell\in\mbox{\boldmath{$Z$}}}\widehat{P}^{\ell},\hskip
30.0pt\widehat{P}_{-\infty}=\bigcup_{\ell\in\mbox{\boldmath{$Z$}}}\widehat{P}^{\ell}.$
(6)

###### Theorem 1.3

If $P\in B_{k,n}$ then $\widehat{P}_{\infty}$ is a point and
$\widehat{P}_{-\infty}$ is an affine plane.

Our argument will show that $P\in B_{k,n}$ is strictly star-shaped with
respect to all points in $\widehat{P}^{n}$. In particular, all polygons in the
orbit are strictly star-shaped with respect to the collapse point
$\widehat{P}_{\infty}$. See Corollary 7.3.

As we remarked above, $B_{n,k}$ contains the set of convex $n$-gons. Thus, if
we fix some convex $n$-gon, we get one collapse point for each
$k\in[1,...,n/3)$. (The case $k=1$ gives the pentagram map collapse point.) I
satisfied myself that these collapse points are generally distinct from each
other.

One might wonder if some version of Glick’s formula works for the
$\widehat{P}_{\infty}$ in general. I discovered experimentally that this is
indeed the case for $n=3k+1$ and $n=3k+2$. See §9.2.

###  1.6 The Triangulations

In §7.1 we associate to each $k$-bird $P$ a triangulation
$\tau_{P}\subset\mbox{\boldmath{$P$}}$, the projective plane. Here $\tau_{P}$
is an embedded degree $6$ triangulation of $P_{-\infty}-P_{\infty}$. The edges
are made from the segments in the $\delta$-diagonals of $P$ and its iterates
for $\delta=1,k,k+1$.

Figure 1.7: The triangulation associated to a member of $B_{5,16}$.

Figure 1.7 shows this tiling associated to a member of $B_{5,16}$. In this
figure, the interface between the big black triangles and the big white
triangles is some $\Delta_{5}^{\ell}(P)$ for some smallish value of $\ell$. (I
zoomed into the picture a bit to remove the boundary of the initial $P$.) The
picture is normalized so that the line $P_{-\infty}$ is the line at infinity.
When I make these kinds of pictures (and animations), I normalize so that the
ellipse of inertia of $P$ is the unit disk.

###  1.7 Paper Organization

This paper is organized as follows.

  * •

In §2 we prove Theorem 1.2.

  * •

In §3 we prove Statement 1 of Theorem 1.1.

  * •

In §4 we prove Statement 2 of Theorem 1.1.

  * •

In §5 we prove a technical result called the Degeneration Lemma, which will
help with Statement 3 of Theorem 1.1.

  * •

In §6 we prove Statement 3 of Theorem 1.1.

  * •

In §7 we introduce the triangulations discussed above. They help with the
proof of Theorem 1.3.

  * •

In §8 we prove Theorem 1.3.

  * •

In §9, an appendix, we sketch an alternate proof of Theorem 1.2 which Anton
Izosimov kindly explained. We also discuss Glick’s collapse formula and star
relabelings of polygons.

###  1.8 Visit the Flapping Bird Exhibit

Our results inject some more geometry into the pentagram zoo. Our results even
have geometric implications for the pentagram map itself. See §9.3. There are
different ways to visit the flapping bird exhibit in the zoo. You could read
the proofs here, or you might just want to to look at some images:
http://www.math.brown.edu/$\sim$reschwar/BirdGallery
You can also download and play with the software I wrote:
http://www.math.brown.edu/$\sim$reschwar/Java/Bird.TAR
The software has detailed instructions. You can view this paper as a
justification for why the nice images actually exist.

###  1.9 Acknowledgements

I would like to thank Misha Gekhtman, Max Glick, Anton Izosimov, Boris Khesin,
Valentin Ovsienko, and Serge Tabachnikov for many discussions about the
pentagram zoo. I would like to thank Anton, in particular, for extensive
discussions about the material in §9.

##  2 The Energy

The purpose of this chapter is to prove Theorem 1.2. The proof, which is
similar to what I do in [19], is more of a verification than a conceptual
explanation. My computer program allows the reader to understand the technical
details of the proof better. The reader might want to just skim this chapter
on the first reading. In §9 I will sketch an alternate proof, which I learned
from Anton Izosimv. Izosimov’s proof also uses the first two sections of this
chapter.

###  2.1 Projective Geometry

Let $P$ denote the real projective plane. This is the space of $1$-dimensional
subspaces of $\mbox{\boldmath{$R$}}^{3}$. The projective plane $P$ contains
$\mbox{\boldmath{$R$}}^{2}$ as the affine patch. Here
$\mbox{\boldmath{$R$}}^{2}$ corresponds to vectors of the form $(x,y,1)$,
which in turn define elements of $P$.

Let $\mbox{\boldmath{$P$}}^{*}$ denote the dual projective plane, namely the
space of lines in $P$. The elements in $\mbox{\boldmath{$P$}}^{*}$ are
naturally equivalent to $2$-dimensional subspaces of
$\mbox{\boldmath{$R$}}^{3}$. The line in $P$ such a subspace $\Pi$ defines is
equal to the union of all $1$-dimensional subspaces of $\Pi$.

Any invertible linear transformation of $\mbox{\boldmath{$R$}}^{3}$ induces a
projective transformation of $P$, and also of $\mbox{\boldmath{$P$}}^{*}$.
These form the projective group $PSL_{3}(\mbox{\boldmath{$R$}})$. Such maps
preserve collinear points and coincident lines.

A duality from $P$ to $\mbox{\boldmath{$P$}}^{*}$ is an analytic
diffeomorphism $\mbox{\boldmath{$P$}}\to\mbox{\boldmath{$P$}}^{*}$ which maps
collinear points to coincidence lines. The classic example is the map which
sends each linear subspace of $\mbox{\boldmath{$R$}}^{3}$ to its orthogonal
complement.

A PolyPoint is a cyclically ordered list of points of $P$. When there are $n$
such points, we call this an $n$-Point. A PolyLine is a cyclically ordered
list of lines in $P$, which is the same as a cyclically ordered list of points
in $\mbox{\boldmath{$P$}}^{*}$. A projective duality maps PolyLines to
PolyPoints, and vice versa.

Each $n$-Point determines $2^{n}$ polygons in $P$ because, for each pair of
consecutive points, we may choose one of two line segments to join them. As we
mentioned in the introduction, we have a canonical choice for $k$-birds.
Theorem 1.2 only involves PolyPoints, and our proof uses PolyPoints and
PolyLines.

Given a $n$-Point $P$, we let $P_{j}$ be its $j$th point. We make a similar
definition for $n$-Lines. We always take indices mod $n$.

###  2.2 Factoring the Map

Like the pentagram map, the map $\Delta_{k}$ is the product of $2$
involutions. This factorization will be useful here and in later chapters.

Given a PolyPoint $P$, consisting of points $P_{1},...,P_{n}$, we define
$Q=D_{m}(P)$ to be the PolyLine whose successive lines are $P_{0,m}$,
$P_{1,m+1}$, etc. Here $P_{0,m}$ denotes the line through $P_{0}$ and $P_{m}$,
etc. We labed the vertices so that

$Q_{-m-i}=P_{i,i+m}.$ (7)

This is a convenient choice. We define the action of $D_{m}$ on PolyLines in
the same way, switching the roles of points and lines. For PolyLines,
$P_{0,m}$ is the intersection of the line $P_{0}$ with the line $P_{m}$. The
map $D_{m}$ is an involution which swaps PolyPoints with PolyLines. We have
the compositions

$\Delta_{k}=D_{k}\circ D_{k+1},\hskip 30.0pt\Delta_{k}^{-1}=D_{k+1}\circ
D_{k}.$ (8)

The energy $\chi_{k}$ makes sense for $n$-Lines as well as for $n$-Points. The
quantities $\chi_{k}\circ D_{k}(P)$ and $\chi_{k}\circ D_{k+1}(P)$ can be
computed directly from the PolyPoint $P$. Figure 2.1 shows schematically the
$4$-tuples associated to $\chi(0,k,Q)$ for $Q=P$ and $D_{k}(P)$ and
$D_{k+1}(P)$. In each case, $\chi_{k}(Q)$ is a product of $n$ cross ratios
like these. If we want to compute the factor of $\chi_{k}(D_{k}(P))$
associated to index $i$ we subtract (rather than add) $i$ from the indices
shown in the middle figure. A similar rule goes for $D_{k+1}(P)$.

Figure 2.1: Computing the $k$-energy.

Theorem 1.2 follows from the next two results.

###### Theorem 2.1

$\chi_{k}\circ D_{k}=\chi_{k}$.

###### Theorem 2.2

$\chi_{k}\circ D_{k+1}=\chi_{k}$.

These results have almost identical proofs. We consider Theorem 2.1 in detail
and then explain the small changes needed for Theorem 2.2.

###  2.3 Proof of the First Result

We study the ratio

$R(P)=\frac{\chi_{k}\circ D_{k}(P)}{\chi_{k}(P)}.$ (9)

We want to show that $R(P)$ equals $1$ wherever it is defined. We certainly
have $R(P)=1$ when $P$ is the regular $n$-Point.

Given a PolyPoint $P$ we choose a pair of vertices $a,b$ with $|a-b|=k$. We
define $P(t)$ to be the PolyPoint obtained by replacing $P_{a}$ with

$(1-t)P_{a}+tP_{b}.$ (10)

Figure 2.2 shows what we are talking about, in case $k=3$. We have rotated the
picture so that $P_{a}$ and $P_{b}$ both lie on the $X$-axis.

Figure 2.2: Connecting one PolyPoint to another by sliding a point.

The two functions

$f(t)=\chi_{k}(P(t)),\hskip 30.0ptg(t)=\chi_{k}\circ D_{k}(P(t))$ (11)

are each rational functions of $t$. Our notation does not reflect that $f$ and
$g$ depend on $P,a,b$.

A linear fractional transformation is a map of the form

$t\to\frac{\alpha t+\beta}{\gamma t+\delta},\hskip
30.0pt\alpha,\beta,\gamma,\delta\in\mbox{\boldmath{$R$}},\hskip
30.0pt\alpha\delta-\beta\gamma\not=0.$

######  Lemma 2.3 (Factor I)

If $n\geq 4k+2$ and $P$ is a generically chosen $n$-Point, then $f(t)$ and
$g(t)$ are each products of $4$ linear fractional transformations. The zeros
of $f$ and $g$ occur at the same points and the poles of $f$ and $g$ occur at
the same points. Hence $f/g$ is constant.

The only reason we choose $n\geq 4k+2$ in the Factor Lemma is so that the
various diagonals involved in the proof do not have common endpoints. The
Factor Lemma I works the same way for all $k$ and for all choices of (large)
$n$. We write $P\leftrightarrow Q$ if we can choose indices $a,b$ and some
$t\in\mbox{\boldmath{$R$}}$ such that $Q=P(t)$. The Factor Lemma implies that
when $P,Q$ are generic and $P\leftrightarrow Q$ we have $R(P)=R(Q)$. The
result for non-generic choices of $P$ follows from continuity. Any $n$-Point
$Q$ can be included in a finite chain

$P_{0}\leftrightarrow P_{1}\leftrightarrow\cdots\leftrightarrow P_{2n}=Q,$

where $P_{0}$ is the regular $n$-Point. Hence $R(Q)=R(P_{0})=1$. This shows
that Theorem 2.1 holds for $(k,n)$ where $k\geq 2$ and $n\geq 4k+2$. (The case
$k=1$ is a main result of [19], and by now has many proofs.)

###### Lemma 2.4

If Theorem 2.1 is true for all large values of $n$, then it is true for all
values of $n$.

Proof: If we are interested in the result for small values of $n$, we can
replace a given PolyPoint $P$ with its $m$-fold cyclic cover $mP$. We have
$\chi_{k}(mP)=\chi_{k}(P)^{m}$ and
$\chi_{k}(D_{k}(mP))=\chi_{k}(D_{k}(p))^{m}$. Thus, the result for large $n$
implies the result for small $n$. $\spadesuit$

In view of Equation 4 we have

$f(t)=f_{1}(t)...f_{n}(t),\hskip 30.0ptf_{j}(t)=\chi(j,k,P(t)).$ (12)

Thus $f(t)$ is the product of $n$ “local” cross ratios. We call an index $j$
asleep if none of the lines involved in the cross ratio $f_{j}(t)$ depend on
$t$. In other words, the lines do not vary at all with $t$. Otherwise we call
$j$ awake.

As we vary $t$, only the diagonals $P_{0,h}$ change for $h=-k,-k-1,k+1,k$.
From this fact, it is not surprising that there are precisely $4$ awake
indices. These indices are

$j_{0}=0,\hskip 10.0ptj_{1}=k+1,\hskip 10.0ptj_{2}=-k-1,\hskip
10.0ptj_{3}=-k.$ (13)

The index $k$ is not awake because the diagonal $P_{0,k}(t)$ does not move
with $t$.

We define a chord of $P(t)$ to be a line defined by a pair of vertices of
$P(t)$. The point $P_{0}(t)$ moves at linear speed, and the $4$ lines involved
in the calculation of $f_{c_{j}}(t)$ are distinct unless $P_{0}(t)$ lies in
one of the chords of $P(t)$. Thus $f_{c_{j}}(t)$ only has zeros and poles at
the corresponding values of $t$. It turns out that only the following chords
are involved.

$\begin{matrix}-k\cr-k-1\end{matrix}\hskip 15.0pt\begin{matrix}-k\cr
k+1\end{matrix}\hskip 15.0pt\begin{matrix}-k\cr 1\end{matrix}\hskip
15.0pt\begin{matrix}-k\cr-2k-1\end{matrix}\hskip
15.0pt\begin{matrix}-k-1\cr-1\end{matrix}\hskip
15.0pt\begin{matrix}-k-1\cr-2k-1\end{matrix}\hskip 15.0pt\begin{matrix}k+1\cr
1\end{matrix}\hskip 15.0pt\begin{matrix}k+1\cr 2k+1\end{matrix}$ (14)

We call these $c_{0},...,c_{7}$. For instance, $c_{0}$ is the line through
$P_{-k}$ and $P_{-k-1}$. Let $t_{j}$ denote the value of $t$ such that
$P(t_{j})\in c_{j}$.

The PolyPoint $Q(t)=D_{k}(P(t))$ has the same structure as $P(t)$. Up to
projective transformations $Q(t)$ is also obtained from the regular PolyPoint
by moving a single vertex along one of the $k$-diagonals. The pattern of zeros
and poles is not precisely the same because the chords of $Q(t)$ do not
correspond to the chords of $P(t)$ in a completely straightforward way. The
$k$-diagonals of $Q(t)$ correspond to the vertices of $P(t)$ and vice versa.
The $(k+1)$ diagonals of $Q(t)$ correspond to the vertices of
$\Delta_{k}^{-1}(P(t))$. This is what gives us our quadruples of points in the
middle picture in Figure 2.1.

We now list the pattern of zeros and poles. We explain our notation by way of
example. The quadruple $(f,2,4,5)$ indicates that $f_{c_{2}}$ has a simple
zero at $f_{4}$ and a simple pole at $t_{5}$.

$(f,0,0,1),\hskip 20.0pt(f,1,6,7),\hskip 20.0pt(f,2,4,5),\hskip
20.0pt(f,3,2,3).$ (15) $(g,0,6,5),\hskip 20.0pt(g,1,0,3),\hskip
20.0pt(g,2,2,1),\hskip 20.0pt(g,3,4,7).$ (16)

Since these functions have holomorphic extensions to $C$ with no other zeros
and poles, these functions are linear fractional transformations. This pattern
establishes the Factor Lemma I.

Checking that the pattern is correct is just a matter of inspection. We give
two example checks.

  * •

To see why $f_{c_{2}}$ has a simple zero at $t_{4}$ we consider the quintuple

$(-k-1,-2k-1,-2k-2,0,-1).$

At $t_{4}$ the two diagonals $P_{-k-1,0}$ and $P_{-k-1,-1}$ coincide. In terms
of the cross ratios of the slopes we are computing $\chi(a,b,c,d)$ with $a=b$.
The point $P_{0}(t)$ is moving with linear speed and so the zero is simple.

  * •

To see why $g_{c_{2}}$ has a simple pole at $t_{1}$ we consider the $4$ points

$P_{2k+2,k+2}\cap P_{1,k+1},\hskip 10.0ptP_{k+1},\hskip 10.0ptP_{1},\hskip
10.0ptP_{1,k+1}\cap P_{-k,0}.$ (17)

These are all contained in the $k$-diagonal $P_{1,k+1}$, which corresponds to
the vertex $(-k-1)$ of $D_{k}(P)$. At $t=t_{1}$ the three points $P_{0}(t)$
and $P_{-k}$ and $P_{k+1}$ are collinear. This makes the $2$nd and $4$th
listed point coincided. In terms of our cross ratio $\chi(a,b,c,d)$ we have
$b=d$. This gives us a pole. The pole is simple because the points come
together at linear speed.

The other explanations are similar. The reader can see graphical illustrations
of these zeros and poles using our program.

###  2.4 Proof of the Second Result

The proof of Theorem 2.2 is essentially identical to the proof of Theorem 2.1.
Here are the changes. The Factor Lemma II has precisely the same statement as
the Factor Lemma I, except that

  * •

When defining $P(t)$ we use points $P_{a}$ and $P_{b}$ with $|a-b|=k+1$.

  * •

We are comparing $P(t)$ with $D_{k+1}(P(t))$.

This changes the definition of the functions $f$ and $g$. With these changes
made, the Factor Lemma I is replaced by the Factor Lemma II, which has an
identical statement. This time the chords involved are as follows.

$\begin{matrix}-k-1\cr-k\end{matrix}\hskip 15.0pt\begin{matrix}-k-1\cr
k\end{matrix}\hskip 15.0pt\begin{matrix}-k-1\cr-1\end{matrix}\hskip
15.0pt\begin{matrix}-k-1\cr-2k-1\end{matrix}\hskip 15.0pt\begin{matrix}-k\cr
1\end{matrix}\hskip 15.0pt\begin{matrix}-k\cr-2k-1\end{matrix}\hskip
15.0pt\begin{matrix}k\cr-1\end{matrix}\hskip 15.0pt\begin{matrix}k\cr
2k+1\end{matrix}$ (18)

This time the $4$ awake indices are:

$j_{0}=0,\hskip 10.0ptj_{1}=k,\hskip 10.0ptj_{2}=-k-1,\hskip 10.0ptj_{3}=-k.$
(19)

Here is the pattern of zeros and poles.

$(f,0,1,0),\hskip 20.0pt(f,1,7,6),\hskip 20.0pt(f,2,3,2),\hskip
20.0pt(f,3,5,4).$ (20) $(g,0,5,6),\hskip 20.0pt(g,1,3,0),\hskip
20.0pt(g,2,7,4),\hskip 20.0pt(g,3,1,2).$ (21)

The pictures in these cases look almost identical to the previous case. The
reader can see these pictures by operating my computer program. Again, the
zeros of $f$ and $g$ are located at the same places, and likewise the poles of
$f$ and $g$ are located at the same places. Hence $f/g$ is constant. This
completes the proof the Factor Lemma II, which implies Theorem 2.2.

##  3 The Soul of the Bird

###  3.1 Goal of the Chapter

Given a polygon $P\subset\mbox{\boldmath{$R$}}^{2}$, let $\widehat{P}$ be the
closure of the bounded components of $\mbox{\boldmath{$R$}}^{2}-P$ and let
$P^{I}$ be the interior of $\widehat{P}$.

Suppose now that $P(t)$ for $t\in[0,1]$ is a path in $B_{n,k}$ starting at the
regular $n$-gon $P(0)$. We can adjust by a continuous family of projective
transformations so that $P(t)$ is a bounded polygon in
$\mbox{\boldmath{$R$}}^{2}$ for all $t\in[0,1]$. We orient $P(0)$ counter-
clockwise around $P^{I}(0)$. We extend this orientation choice continuously to
$P(t)$. We let $P_{ab}(t)$ denote the diagonal through vertices $P_{a}(t)$ and
$P_{b}(t)$. We orient $P_{a,b}(t)$ so that it points from $P_{a}(t)$ to
$P_{b}(t)$. We take indices mod $n$.

When $P$ is embedded, we say that $P$ is strictly star shaped with respect to
$x\in P^{I}$ if each ray emanating from $x$ intersects $P$ exactly once.

Figure 3.1: The soul of a $3$-bird

Each such $(k+1)$-diagonal defines an oriented line that contains it, and also
the (closed) distinguished half plane which lies to the left of the oriented
line. These $n$ half-planes vary continuously with $t$. The soul of $P(t)$,
which we denote $S(t)$, is the intersection of the distinguished half-planes.
Figure 3.1 shows the an example.

The goal of this chapter is to prove the following result.

###### Theorem 3.1

Let $P$ be a bird and let $S$ be its soul. Then:

  1. 1.

$S$ is has non-empty interior.

  2. 2.

$S\subset P^{I}$.

  3. 3.

$P$ is strictly star-shaped with respect to any point in $S$.

Theorem 3.1 immediately implies Statement 1 of Theorem 1.1.

We are going to give a homotopical proof of Theorem 3.1. We say that a value
$t\in[0,1]$ is a good parameter if Theorem 3.1 holds for $P(t)$. All three
conclusions of Theorem 3.1 are open conditions. Finally, $0$ is a good
parameter. For all these reasons, it suffices to prove that the set of good
parameters is closed. By truncating our path at the first supposed failure, we
reduce to the case when Theorem 3.1 holds for all $t\in[0,1)$.

###  3.2 The Proof

For ease of notation we set $X=X(1)$ for any object $X$ associated to $P(1)$.

###### Lemma 3.2

If $P$ is any $k$-bird, then $P_{0}$ and $P_{2k+1}$ lie to the left of
$P_{k,k+1}$. The same goes if all indices are cyclically shifted by the same
amount.

Proof: Consider the triangle with vertices $P_{0}(t)$ and $P_{k}(t)$ and
$P_{k+1}(t)$. The $k$-niceness condition implies that this triangle is non-
degenerate for all $t\in[0,1]$. Since $P_{0}(t)$ lies to to the left of
$P_{k,k+1}(t)$, the non-degeneracy implies the same result for $t=1$. The same
argument works for the triple $(2k+1,k,k+1)$. $\spadesuit$

###### Lemma 3.3

$S$ is non-empty and contained in $P^{I}$.

Proof: By continuity, $S$ is nonempty and contained in $P\cup P^{I}$. By the
$k$-niceness property and continuity, $P_{1}$ lies strictly to the right of
$P_{0,k+1}$. Hence the entire half-open edge $[P_{0},P_{1})$ lies strictly to
the right of $P_{0,k+1}$. Hence $[P_{0},P_{1})$ is disjoint from $S$. By
cyclic relabeling, the same goes for all the other half-open edges. Hence
$S\cap P=\emptyset$. Hence $S\subset P^{I}$. $\spadesuit$

###### Lemma 3.4

$P$ is strictly star-shaped with respect to any point of $S(1)$.

Proof: Since $P(t)$ is strictly star-shaped with respect to all points of
$S(t)$ for $t<1$, this lemma can only fail if there is an edge of $P(1)$ whose
extending line contains a point $x\in S$. We can cyclically relabel so that
the edge of $\overline{P_{0}P_{1}}$.

Figure 3.2: The diagonal $P_{0,k+1}$ does not separate $1$ from $x$.

Since $x\not\in P$, either $P_{1}$ lies between $P_{0}$ and $x$ or $P_{0}$
lies in between $x$ and $P_{1}$. In the first case, both $P_{1}$ and $x$ lie
on the same side of the diagonal $P_{0,k+1}$. This is a contradiction: $P_{1}$
is supposed to lie on the right and $x$ is supposed to lie on the left. In the
second case we get the same kind of contradiction with respect to the diagonal
$P_{-k,1}$. $\spadesuit$

We say that $P$ has opposing $(k+1)$-diagonals if it has a pair of
$(k+1)$-diagonals which lie in the same line and point in opposite directions.
In this case, the two left half-spaces are on opposite sides of the common
line.

###### Lemma 3.5

$P$ does not have opposing $(k+1)$-diagonals.

Proof: We suppose that $P$ has opposing diagonals and we derive a
contradiction. In this case $S$, which is the intersection of all the
associated left half-planes, must be a subset of the line $L$ containing these
diagonals. But then $P$ intersects $L$ in at least $4$ points, none of which
lie in $S$. But then $P$ cannot be strictly star-shaped with respect to any
point of $S$. This is a contradiction. $\spadesuit$

We call three $(k+1)$-diagonals of $P(t)$ interlaced if the intersection of
their left half-spaces is a triangle. See Figure 3.3.

Figure 3.3: Interlaced diagonals on $P(t)$.

Given interlaced $(k+1)$-diagonals, and a point $x$ in the intersection, the
circle of rays emanating from $x$ encounters the endpoints of the diagonals in
an alternating pattern: $a_{1},b_{3},a_{2},b_{1},a_{3},b_{2}$, where
$a_{1},a_{2},a_{3}$ are the tail points and $b_{1},b_{2},b_{3}$ are the head
points. Here $a_{1}$ names the vertex $P_{a_{1}}(t)$, etc.

###### Lemma 3.6

$P(t)$ cannot have interlaced diagonals for $t<1$.

Proof: Choose $x\in S(t)$. The $n$-gon $P(t)$ is strictly star-shaped with
respect to $x$. Hence, the vertices of $P$ are encountered in order (mod $n$)
by a family of rays that emanate from $x$ and rotates around full-circle.
Given the order these vertices are encountered, we have
$a_{j+1}=a_{j}+\eta_{j}$, where $\eta_{j}\leq k$. Here we are taking the
subscripts mod $3$ and the vertex values mod $n$. This tells us that
$n=\eta_{1}+\eta_{2}+\eta_{3}\leq 3k$. This contradicts the fact that $n>3k$.
$\spadesuit$

It only remains to show that $S$ has non-empty interior. A special case of
Helly’s Theorem says the following: If we have a finite number of convex
subsets of $\mbox{\boldmath{$R$}}^{2}$ then they all intersect provided that
every $3$ of them intersect. Applying Helly’s Theorem to the set of interiors
of our distinguished half-planes, we conclude that we can find $3$ of these
open half-planes whose triple intersection is empty. On the other hand, the
triple intersection of the closed half-planes contains $x$. Since $P$ has no
opposing diagonals, this is only possible if the $3$ associated diagonals are
interlaced for $t$ sufficiently close to $1$. This contradicts Lemma 3.6.
Hence $S$ has non-empty interior.

##  4 The Feathers of the Bird

###  4.1 Goal of the Chapter

Recall that $P^{I}$ is the interior of the region bounded by $P$. We call the
union of shaded triangles in Figure 4.1 the feathers of the bird. the black
region in the center is the soul.

Figure 4.1 The feathers of a $3$-bird.

Each feather $F$ of a $k$-bird $P$ is the convex hull of its base, an edge $e$
of $P$, and its tip, a vertex of $\Delta_{k}(P)$.

The goal of this chapter is to prove the following result, which says that the
simple topological picture shown in Figure 4.1 always holds.

###### Theorem 4.1

The following is true.

  1. 1.

Let $F$ be an feather with base $e$. Then $F-\\{e\\}\subset P^{I}$.

  2. 2.

Distinct feathers can only intersect at a vertex of $P$.

  3. 3.

The line segment connecting two consecutive feather tips lies in $P^{I}$.

When we apply $\Delta_{k}$ to $P$ we are just specifying the points of
$\Delta_{k}(P)$. We define the polygon $\Delta_{k}(P)$ so that the edges are
the bounded segments connecting the consecutive tips of the feathers of $P$.
With this definion, Statement 2 of Theorem 1.1 follows immediately from
Theorem 4.1.

###  4.2 The Proof

There is one crucial idea in the proof of Theorem 4.1: The soul of $P$ lies in
the sector $F^{*}$ opposite any of its feathers $F$. See Figure 4.2.

Figure 4.2 The soul lies in the sectors opposite the feathers.

We will give a homotopical proof of Theorem 4.1. By truncating our path of
birds, we can assume that Theorem 4.1 holds for all $t\in[0,1)$. We then want
to rule out the various ways that Theorem 4.1 can fail for $t=1$. As in the
previous chapter we set $P=P(1)$, etc. Figure 4.3 shows the $2$ ways that
Statement 1 could fail:

  1. 1.

The tip $v$ of the feather $F$ could coincide with some $p\in P$.

  2. 2.

Some $p\in P$ could lie in the interior point of $\partial F-e$.

Figure 4.3: Case 1 (left) and Case 2 (right).

For all $x\in F^{*}$, the ray $\overrightarrow{xp}$ intersects $P$ both at $p$
and at a point $p^{\prime}\in e$. This contradicts the fact that for any $x\in
S\subset F^{*}$, the polygon $P$ is strictly star-shaped with respect to $x$.
This establishes Statement 1 of Theorem 4.1.

Let $F_{1}$ and $F_{2}$ be two feathers of $P$, having bases $e_{1}$ and
$e_{2}$. For Statement 2, it suffices to prove that $F_{1}-e_{1}$ and
$F_{2}-e_{2}$ are disjoint.

The same homotopical argument as for Statement 1 reduces us to the case when
$F_{1}$ and $F_{2}$ have disjoint interiors but $\partial F_{1}-e_{1}$ and
$\partial F_{2}-e_{2}$ share a common point $x$. If $\partial F_{1}$ and
$\partial F_{2}$ share an entire line segment then, thanks to the fact that
all the feathers are oriented the same way, we would have two $(k+1)$
diagonals of $P$ lying in the same line and having opposite orientation. Lemma
3.5 rules this out.

In particular $x$ must be the tip of at least one feather. Figure 4.4 shows
the case when $x=v_{1}$, the tip of $F_{1}$, but $x\not=v_{2}$. The case when
$x=v_{1}=v_{2}$ has a similar treatment.

Figure 4.4: Opposiing sectors are disjoint

In this case, the two sectors $F_{1}^{*}$ and $F_{2}^{*}$ are either disjoint
or intersect in a single point. This contradicts the fact that $S\subset
F_{1}^{*}\subset F_{2}^{*}$ has non-empty interior. This contradiction
establishes Statement 2 of Theorem 4.1.

Recall that $\widehat{P}=P\cup P^{I}$. Let $F_{1}$ and $F_{2}$ be consecutive
feathers with bases $e_{1}$ and $e_{2}$ respeectively. Let $f$ be the edge
connecting the tips of $F_{1}$ and $F_{2}$. Our homotopy idea reduces us to
the case when $f\subset\widehat{P}$ and $f\cap P\not=\emptyset$. Figure 4.5
shows the situation.

Figure 4.5: The problem a common boundary point

Note that $f\cap P$ must be strictly contained in the interior of $f$ because
(by Statement 1 of Theorem 4.1) the endpoints of $f$ lie in $P^{I}$. But then,
$f\cap P$ is disjoint from $F_{1}^{*}\cap F_{2}^{*}$, which is in turn
contained in the shaded region $G$. For any $x\in G$ and each vertex $p$ of
$f$, the ray the ray $\overrightarrow{xp}$ also intersects $P$ at a point
$p^{\prime}\in e_{1}\cup e_{2}$. This gives the same contradiction as above
when we take $x\in S\subset F_{1}^{*}\cap F_{2}^{*}\subset G$. This completes
the proof of Statement 3 of Theorem 4.1.

##  5 The Degeneration of Birds

###  5.1 Statement of the Result

Let $k\geq 2$ and $n>3k$. Let $B_{k,n}$ denote the space of $n$-gons which are
$k$-birds. Let $\chi_{k}$ denote the $k$-energy. In this chapter we will prove
a technical result which will help us prove, in the next chapter, that
$\Delta_{k}(B_{k,n})=B_{k,n}$. The reader should probably just use the
statement as a black box on the first reading. Our argument is a pretty
tedious case-by-case analysis.

We say that a degenerating path is a path $Q(t)$ of $n$-gons such that

  1. 1.

$Q(t)\in B_{k,n}$ for all $t\in[0,1)$ but $Q(1)\not\in B_{k,n}$.

  2. 2.

$\chi_{k}(Q(t))>\epsilon_{0}>0$ for all $t\in[0,1]$.

  3. 3.

$Q_{j}(1)\not=Q_{j+k}(1)$ for all $j=1,...,n$.

  4. 4.

$Q_{j}(1)\not=Q_{j+k+1}(1)$ for all $j=1,...,n$.

######  Lemma 5.1 (Degeneration)

If $Q(\cdot)$ is a degenerating path, then all but at most one vertex of
$Q(1)$ lies in a line segment.

Let us first explain that this kind of degeneration can actually occur.
Consider the case where $Q(t)$ is projectively eqivalent to the regular
$n$-gon for all $t\in[0,1)$. That is $Q(t)=T_{t}(P)$ where $P$ is the regular
$n$-gon and $T_{t}$ is some projective transformation that depends on $t$. We
can choose $T_{t}$ to be quite drastic, so that the points
$Q_{-1}(t),Q_{0}(t),Q_{1}(t)$ are the vertices of a triangle and the remaining
vertices converge to the line segment joining $Q_{\pm 1}(t)$. The limit $Q(1)$
will have the shape of an equilateral triangle, with all the vertices distinct
and all but one vertex contained in the same line.

Remarks: (1) For our application of the Degeneration Lemma, all the vertices
of $Q(1)$ are distinct. However, for a later application to Theorem 1.3 we
prove the result under the weaker hypotheses we have stated.
(2) The Degeneration Lemma also works, with the same proof, in case we have a
sequence $\\{Q^{\ell}\\}\subset B_{k,n}$, rather than a path, which converges
to some $Q^{\infty}\not\in B_{k,n}$ and has the uniform lower bound on
$\chi_{k}$. We will invoke the sequence case of the result when we prove
Theorem 1.3.

###  5.2 Distinguished Diagonals

We orient $Q(t)$ so that it goes counter-clockwise around the region it
bounds. We orient the diagonal $Q_{ab}$ so that it points from $Q_{a}$ to
$Q_{b}$. For $t<1$ the vertices $Q_{1}(t)$ and $Q_{k}(t)$ lie to the right of
the diagonal $Q_{0,k+1}(t)$, in the sense that a person walking along this
diagonal according to its orientation would see that points in the right. This
has the same proof as Lemma 3.2. The same rule holds for all cyclic
relabelings of these points. The rule holds when $t<1$. Taking a limit, we get
a weak version of the rule: Each of $Q_{1}(1)$ and $Q_{k}(1)$ either lies to
the right of the diagonal $Q_{0,k+1}(1)$ or on it. The same goes for cyclic
relabeings. We call this the Right Hand Rule.

Say that a distinguished diagonal of $Q(t)$ is either a $k$-diagonal or a
$(k+1)$-diagonal. There are $2n$ of these, and they come in a natural cyclic
order:

$Q_{0,k}(t)\hskip 10.0ptQ_{0,k+1}(t),\hskip 10.0ptQ_{1,k+1}(t),\hskip
10.0ptQ_{1,k+2}(t),...$ (22)

The pattern alternates between $k$ and $(k+1)$-diagonals. We say that a
diagonal chain is a consecutive list of these.

We say that one oriented segment $L_{2}$ lies ahead of another one $L_{1}$ if
we can rotate $L_{1}$ by $\theta\in(0,\pi)$ radians counter-clockwise to
arrive at a segment parallel to $L_{2}$, In this case we write $L_{1}\prec
L_{2}$. We have

$Q_{0,k+1}(t)\prec Q_{1,k+1}(t)\prec Q_{1,k+2}(t)\prec Q_{2,k+2}(t).$ (23)

Figure 5.1: The turning rule

This certainly holds when $t=0$. By continuity and the Right Hand Rule, this
holds for all $t<1$. Taking a limit, we see that the $k$-diagonals of $Q(1)$
weakly turn counter-clockwise in the sense that either $L_{1}\prec L_{2}$ for
consecutive diagonals or else $L_{1}$ and $L_{2}$ lie in the same line and
point in the same direction. Moreover, the total turning is $2\pi$. If we
start with one distinguished diagonal and move through the cycle then the
turning angle increases by jumps in $[0,\pi]$ until it reaches $2\pi$. We call
these observations the Turning Rule.

###  5.3 Subdivision into Cases

We set $X=X(1)$ for each object $X$ associated to $Q(1)$. The situation is
that $Q(t)$ is $k$-nice for all $t<1$ but $Q$ is not. Figure 5.2 shows the
distinguished diagonals incident to $Q_{0}$. We always take indices mod $n$.
Thus $-k-1=n-k-1$ mod $n$.

Figure 5.2: The $4$ distinguished diagonals incident to $Q_{0}(t)$.

We say that $Q$ has collapsed diagonals at $Q_{k}$ if the $4$ distinguished
diagonals incident to $Q_{k}$ do not all lie on distinct lines.

###### Lemma 5.2

If $Q$ has collapsed diagonals at $Q_{0}$ then $Q_{-k-1,0}$ and $Q_{0,k+1}$
point in opposite directions or $Q_{-k,0}$ and $Q_{0,k}$ point in the same
direction.

Proof: Associated to each diagonal incident to $Q_{0}$ is the ray which starts
at $Q_{0}$ and goes in the direction of the other endpoint of the diagonal.
(Warning: The ray may have the opposite orientation than the diagonal it
corresponds to.) If the angle between any of the rays tends to $\pi$ as $t\to
1$ then the angle between the outer two rays tends to $\pi$. In this case
$Q_{-k,0}$ and $Q_{0,k}$ point in the same directions. If the angle between
non-adjacent rays tends to $0$ then $Q_{-k-1,0}$ and $Q_{0,k+1}$ are squeezed
together and point in opposite directions.

Suppose that the angle between adjacent rays tends to $0$. If the two adjacent
rays are the middle ones, we have the case just considered. Otherwise, either
the angle between the two left rays tends to $0$ or the angle between the two
right rays tends to $0$. In either case, the uniform lower bound on the cross
ratio forces a third diagonal to point either in the same or the opposite
direction as these adjacent diagonals when $t=1$. Any situation like this
leads to a case we have already considered. $\spadesuit$

###  5.4 The Case of Aligned Diagonals

We say that $Q$ has aligned diagonals if there are $2$ parallel distinguished
diagonals of $Q$ such that both diagonal chains which start and stop with
these have length at least $2k$. The is one of the cases of Lemma 5.2.

###### Lemma 5.3

If $Q$ has aligned diagonals, then $2k+1$ consecutive points of $Q$ are
collinear.

Proof: The total turning of the diagonals is $2\pi$, so one of the two chains
defined by our diagonals turns $2\pi$ and the other turns $0$. Hence, we can
find $2k$ consecutive parallel distinguished diagonals. We will suppose that
our chain starts with $Q_{-k,0}$ and ends with $Q_{0,k}$. The proof is
essentially the same if the chain starts with a $(k+1)$-diagonal rather than a
$k$-diagonal.

Given that $Q_{-k,0}$ and $Q_{-k,1}$ are parallel, $Q_{-k},Q_{0},Q_{1}$ are
collinear. In other words, $(-k,0,1)$ is a triple of indices for collinear
points. Likewise $(-k,-k+1,0)$ is such an index. Continuing this way, we get
collinear triples

$(-k,0,1),\hskip 10.0pt(-k,-k+1,1),\hskip 10.0pt(-k+1,1,2),\hskip
10.0pt...,\hskip 10.0pt(-1,0,k).$

This implies that the points $Q_{-k},...,Q_{0},...,Q_{k}$ are all collinear.
$\spadesuit$

Now we forget about the aligned diagonals and we just use the property that
$Q$ has a consecutive run of $2k+1$ collinear points. Let $L$ be the line
containing these points. If all points of $Q$ lie in $L$, we are done.
Otherwise there is some smallest index $i>k$ such that $Q_{i}\not\in L$ but
the preceding $2k+1$ points are in $L$. Cyclically relabelling, we can assume
that $i=k+1$. Once we make this relabeling, we lose control over where our
aligned diagonals are. Now we regain some control.

###### Lemma 5.4

The length $2k$-diagonal chain $Q_{-k,0}\to...\to Q_{0,k}$ consists entirely
of parallel diagonals. There is no turning at all.

Proof: The diagonals $Q_{-k,0}$ and $Q_{0,k}$. are either parallel or anti-
parallel. If they are anti-parallel, then the angle between the corresponding
rays incident $Q_{0}(t)$ tends to $0$ as $t\to 1$. But these are the outer two
rays. This forces the angle between all $4$ rays incident to $Q_{0}(t)$ to
tend to $0$. The whole picture just folds up like a fan. But one or these rays
corresponds to $Q_{0,k+1}(t)$. This picture forces $Q_{k+1}\in L$. But this is
not the case.

Now we know that $Q_{-k,0}$ and $Q_{0,k}$ are parallel. All the diagonals in
our chain are either parallel or anti-parallel to the first and last ones in
the chain. If we ever get an anti-parallel pair, then the diagonals in the
chain must turn $2\pi$ around. But then none of the other distinguished
diagonals outside our chain turns at all. That is,
$Q_{0,k},Q_{1,k+1},...,Q_{n-k,n}$ are all parallel. In this situation the
argument proving Lemma 5.3 shows that $Q\subset L$, a contradiction.
$\spadesuit$

We rotate the picture so that $L$ coincides with the $X$-axis and so that
$Q_{0,k}$ points in the positive direction. Since we are already using the
words left and right for another purpose, we say that $p\in L$ is forward of
of $q\in L$ if $p$ has larger $X$-coordinate. Likewise we say that $q$ is
backwards of $p$ in this situation. We say that $Q_{0,k}$ points forwards. We
have established that $Q_{-k,0},...,Q_{0,k}$ all point forwards.

###### Lemma 5.5

$Q_{k+2}\in L$ and both $Q_{1,k+2}$ and $Q_{2,k+2}$ point backwards.

Proof: Let us first justify the fact that $Q_{k+1}$ lies above $L$. This
follows from Right Hand Rule applied to $Q_{0,k+1}$ and $Q_{k}$ and the fact
that $Q_{0,k}$ points forwards. Since $Q_{-k-1},Q_{-k},Q_{1}$ are collinear,
$Q$ has collapsed diagonals at $Q_{1}$. But $Q$ cannot have aligned diagonals
because $Q_{1,k+1}$ is not parallel to $Q_{-k,1}$. Hence $Q$ has folded
diagonals at $1$. Since $Q_{-k,1}$ points forwards $Q_{1,k+2}$ points
backwards.

We have $Q_{2}\in L$ because $2\leq k$. Suppose $Q_{2,k+2}$ points forwards.
We consider the $3$ distinguished diagonals

$Q_{0,k},\hskip 10.0ptQ_{1,k+2},\hskip 10.0ptQ_{2,k+2}.$

These diagonals respectively point forwards, backwards, forwards. But then, in
going from $Q_{0,k}$ to $Q_{2,k+2}$, the diagonals have already turned $2\pi$.
Since the total turn is $2\pi$, the diagonals $Q_{2,k+2},\
Q_{3,k+3},...,Q_{n,n+k}$ are all parallel. But then $Q_{2},...,Q_{n}\in L$.
This contradicts the fact that $Q_{k+1}\not\in L$. $\spadesuit$

###### Lemma 5.6

For at least of the two indices $j\in\\{2k+2,2k+3\\}$ we have $Q_{j}\in L$ and
$Q_{k+2,j}$ points forwards.

Proof: Since $Q_{1},Q_{2},Q_{k+2}$ are collinear, $Q$ has collapsed diagonals
at $Q_{k+2}$. So, by Lemma 5.2, we either have folded diagonals at $Q_{k+2}$
or aligned diagonals at $Q_{k+2}$. The aligned case gives $Q_{2k+2}\in L$ and
the folded case gives $Q_{2k+3}\in L$. We consider the cases in turn.

Consider the aligned case. Suppose $Q_{k+2,2k+2}$ points backwards, as shown
in Figure 5.3.

Figure 5.3: Violation of the Right Hand Rule

This violates the Right Hand Rule for $Q_{k+2}$ and $Q_{k+1,2k+2}$ because
$Q_{k+1}$ lies above $L$.

Consider the folded case. Since $Q_{k+2,2k+3}$ and $Q_{1,k+2}$ point in
opposite directions, and $Q_{1,k+2}$ points backwards (by the previous lemma),
$Q_{k+2,2k+3}$ points forwards. $\spadesuit$

Let $j\in\\{2k+2,2k+3\\}$ be the index from Lemma 5.6. Consider the $3$
diagonals

$Q_{0,k},\hskip 10.0ptQ_{1,k+1},\hskip 10.0ptQ_{k+2,j}.$

These diagonals are all parallel to $L$ and respectively point in the
forwards, backwards, forwards direction. This means that the diagonals in the
chain $Q_{0,k}\to...\to Q_{k+2,j}$ have already turned $2\pi$ radians. But
this means that the diagonals

$Q_{k+2,2k+3},\hskip 10.0ptQ_{k+3,2k+3},\hskip 10.0ptQ_{k+3,2k+4},\hskip
10.0pt...\hskip 10.0ptQ_{0,k}=Q_{n,n+k}$

are all parallel and point forwards along $L$. Hence
$Q_{k+2},Q_{k+3},...,Q_{n}\in L$. Hence all points but $Q_{k+1}$ lie in $L$.

###  5.5 Separating the Soul from the Polygon

Recall that $Q=Q(1)$ and $Q_{0}=Q_{0}(1)$, etc. It remains to analyze the case
of folded diagonals, but before doing that we discuss some properties of the
limiting soul. We define $S$ to be the set of all accumulation points of
sequences $\\{p(t_{n})\\}$ where $p(t_{n})\in S(t_{n})$ and $t_{n}\to 1$.
Since $S(t)$ is non-empty and closed for all $t<1$, we see by compactness that
$S$ is also a non-empty closed subset of the closed region bounded by $Q$.
Also, $S$ lies to the left of all the half-planes defined by the oriented
$(k+1)$ diagonals.

###### Lemma 5.7

The Degeneration Lemma is true for $Q$ if $S\cap Q\not=\emptyset$.

We prove this result through several smaller results.

###### Lemma 5.8

The Degeneration Lemma is true for $Q$ if $S\cap Q$ contains a point in the
interior of an edge of $Q$.

Proof: Suppose this happens. We relabel so that $S$ contains an interior point
of the edge $\overline{Q_{0}Q_{1}}$. We rotate and scale so that $Q_{0}=(0,0)$
to $Q_{1}=(1,0)$. So $(x,0)\in S$ for some $x\in(0,1)$. By the Right Hand
Rule, the vertex $Q_{k+1}$ lies either on or above the $X$-axis. The only way
for $(x,0)$ to lie on or to the left of $Q_{0,k+1}$ is if $Q_{k+1}=(x_{0},0)$
for some $x_{0}>0$. Similar considerations show that $Q_{-k}=(x_{1},0)$ for
some $x_{1}<1$. Figure 5.4 shows the relevant points of $Q(t)$ for $t$ very
near $1$.

Figure 5.4: The relevant points and lines.

The diagonals $Q_{-k,1}$ and $Q_{0,k+1}$ are parallel. Given the indices
involved, they are aligned. The proof in §5.4 finishes the proof. $\spadesuit$

###### Lemma 5.9

The Degeneration Lemma is true for $Q$ if $S\cap Q$ contains a vertex of $Q$.

Proof: We again relabel so that $Q_{0}\in S$. We first give the proof when
$Q_{-1},Q_{0},Q_{1}$ are all distinct. Figure 5.5 below shows the situation.
This picture is meant to depict $Q(t)$ for $t$ near $1$. Figure 5.5 shows two
cases, depending on whether the interior angle of $Q$ at $Q_{0}$ is acute or
obtuse. The interior angle of $Q$ could also be $0$ or $2\pi$. We explain the
various degenerate cases at the end. The shaded region in Figure 5.5 contains
the soul, though this fact is not relevant to our argument.

The same analysis as in the previous lemma shows that $Q_{1},Q_{0},Q_{-k}$ are
collinear and $Q_{1}$ is an extreme point. In other words, $Q_{1}$ is not
between $Q_{0}$ and $Q_{-k}$. Likewise $Q_{-1},Q_{0},Q_{k}$ are collinear and
$Q_{-1}$ is an extreme point. We have drawn the cases when $Q_{0}$ is between
$Q_{\pm 1}$ and $Q_{\mp k}$. This is not essential for the argument.

Figure 5.5: two pairs of folded diagonals

Since $Q_{-1},Q_{0},Q_{k}$ are collinear, $Q$ has collapsed diagonals at
$Q_{k}$. If $Q$ has aligned diagonals at $Q_{k}$ then the proof in §5.4
finishes the job. So, we can assume $Q$ has folded diagonals at $Q_{k}$.
Likewise $Q$ has folded diagonals at $Q_{-k}$. Now we look at the diagonal
chain. We have

$Q_{-1,k}\longrightarrow Q_{k,2k+1}\longrightarrow Q_{-2k-1,-k}\longrightarrow
Q_{-k,1}.$ (24)

If $n=3k+1$ then the inner two diagonals in Equation 24 coincide, and we have
an impossible situation. So, we must have $n>3k+1$ in this case. But then
these diagonals come in order in the diagonal chain. In going fron successive
diagonals on our list we turn respectively by $\pi,\theta,\pi$ degrees, where
$\theta>0$. This gives us $2\pi+\theta$ turning before we have completed the
chain, violating the Turning Rule.

We now deal with the degenerate cases. When $Q_{-1},Q_{0},Q_{1}$ are distinct
and lie in order on a single line, we have $\theta=0$ above, and now we
observe that the diagonals in the chain $Q_{-2k-1},...,Q_{k,2k+1}$ are all
parallel. This gives us more than $2k+1$ consecutive parallel points on $L$,
and the proof in §5.4 finishes the job.

if $Q_{-1}=Q_{0}$ then we directly see that the diagonals at $Q_{k}$ are
collapsed. This gives us our folded diagonals at $Q_{k}$. The same goes for
$Q_{-k}$ if $Q_{0}=Q_{1}$. So, in all cases, we get the chain in Equation 24.
Either we have $\theta>0$ and an outright contradiction, or we have $\theta=0$
and the proof finishes as above. $\spadesuit$

###  5.6 Confining the Soul

We need one more lemma about the soul $S$ of $Q$. This time we use the
assumption about the folded diagonals. The diagonals $Q_{-k-1,0}$ and
$Q_{0,k+1}$ point in opposite directions. We normalize, as above, so that both
are contained in the $X$-axis and $Q_{0,k+1}$ points forwards. We treat the
case when $Q_{k+1}$ does not lie forwards of $Q_{-k-1}$. (The points could
coincide.) We can always get to this case by dihedrally relabeling and then
reflecting.

###### Lemma 5.10

Suppose that $Q$ does not have aligned diagonals. Then $S$ is a subset of the
closed line segment joining $Q_{0}$ to $Q_{k+1}$.

Proof: Since $S$ lies to the left of (or on) each $(k+1)$ diagonal, $S$ is a
subset of the line $L$ common to the folded diagonals.

Now we consider the picture for $t<1$. We rotate so that $Q_{0,k+1}(t)$ is
horizontal and points forwards. We make a counter-clockwise turn of less than
$\pi$ to get from the ray $\overrightarrow{Q_{0}(t)Q_{k+1}(t)}$ to the ray
$\overrightarrow{Q_{0}(t)Q_{-k-1}(t)}$. Hence $Q_{-k-1}(t)$ lies above $L$, as
drawn in Figure 5.6.

Let $e(t)$ be the edge joining $Q_{k}(t)$ to $Q_{k+1}(t)$. The vertex
$Q_{k}(t)$ lies below $Q_{0,k+1}(t)$ by the Right Hand Rule. Let $F(t)$ be the
feather based at $e(t)$. The tip $v(t)$ of $F(t)$ lies on $Q_{0,k+1}(t)$, and
$S(t)$ lies in the sector opposite $F(t)$ across $v(t)$.

Figure 5.6: $S(t)$ is trapped in a triangle.

Hence $S(t)$ lies in the triangle $\Delta(t)$ bounded the lines

$\overline{Q_{k}(t)v(t)},\hskip 30.0pt\overline{Q_{0}(t)Q_{k+1}(t)},\hskip
30.0pt\overline{Q_{0}(t)Q_{-k-1}}(t).$

This structure gives us the claim of the lemma. The only way $\Delta(t)$ can
accumulate on points not on $\overline{0,k+1}$ as $t\to 1$ is when
$Q_{k,2k+1}$ is parallel to $Q_{0,k+1}$ in the limit. But then $Q$ has aligned
diagonals. $\spadesuit$

###  5.7 Folded Edges

We say $Q(1)$ has a fold if two consecutive edges of $Q(1)$ are on the same
line.

######  Lemma 5.11 (Folding)

$Q(1)$ cannot have a fold.

Proof: Consider the lines extending the edges of $Q(t)$ incident to
$Q_{a}(t)$. These lines bound $4$ acute sectors. We let $C(t)$ be the sector
which locally intersects $Q(t)$, as shown in Figure 5.7

Figure 5.7: The cone

It follows from the $k$-niceness property and continuity that the $(k+1)$
diagonals incident to $Q_{a}(t)$ lie in $C(t)$ for all $t<1$. This is true
even when the interior angle of $Q(t)$ at $Q_{a}(t)$ exceeds $\pi$. If this
interior angle tends to either $0$ or $2\pi$ then the angle of $C(t)$ tends to
$0$. This forces our two $(k+1)$ diagonals to lie in the same line, violating
the $k$-niceness of $Q(1)$. $\spadesuit$

###  5.8 Good Folded Diagonals

Henceforth we assume that $Q$ does not have aligned diagonals and also that we
have $S\cap Q=\emptyset$.

We say that the folded diagonals $Q_{-k-1,0}$ and $Q_{0,k+1}$ are good if all
the points $Q_{k+1},Q_{k+2},...,Q_{n-k-1}$ are collinear. We make a similar
definition for other folded diagonals. Note that there are $n-2k-1$ of these
points. If $n\geq 4k+2$ then $Q$ has at least $2k+1$ consecutive collinear
points, and the proof in Lemma 5.4 finishes this case. However, we might have
$3k<n<4k+2$. In that case, we need the following lemma.

###### Lemma 5.12

If $n>3k+2$ and some pair of folded diagonals of $Q$ is good, then the
Degeneration Lemma is true for $Q$.

Proof: We normalize as in the previous section, so that $Q_{-k-1,0}$ and
$Q_{0,k+1}$ are folded, and good.

Suppose $Q_{1}\in L$. Since $Q_{0},Q_{1},Q_{k+1}$ are collinear, $Q$ has
collapsed diagonals at $Q_{k+1}$. To avoid the proof in §5.4, we can assume
that $Q$ has folded diagonals at $Q_{k+1}$. Since $Q_{0,k+1}$ points forwards,
$Q_{k+1,2k+2}$ points backwards. Hence $Q_{-k-1,0}$ and $Q_{k+1,2k+2}$ are
parallel. Since $n>3k+2$ these two diagonals are aligned. This is a
contradiction.

So, to finish the proof of this lemma, we show that $Q_{1}\in L$. We first
give the proof when all edges between $Q_{k+1}$ and $Q_{n-k-1}$ are
nontrivial. We claim that $Q_{k+2}$ is fowards of $Q_{k+1}$. Suppose not. Then
there is some index $a\in\\{k+2,...,-k-2\\}$ such that $Q_{a}$ is backwards of
$Q_{a\pm 1}$. This gives us folded edges at $Q_{a}$.

By Lemma 5.11, we have folded diagonals at $Q_{a}$. But then $Q_{a,a+k+1}$
points forwards. Hence $Q_{0,k+1}$ and $Q_{a,a+k+1}$ are parallel. Given the
indices, these diagonals are aligned. This is a contradiction.

Now we know that $Q_{k+2}$ is forwards of $Q_{k+1}$. Suppose $Q_{1}\not\in L$.
by the Right Hand Rule applied to the diagonal $Q_{0,k+1}$, the point $Q_{1}$
lies beneath $L$, as shown in Figure 5.8.

Figure 5.8: A contradiction involving $Q_{1}$.

But then $Q_{k+1}$ lies to the left of the diagonal $Q_{1,k+2}$. This violates
the Right Hand Rule. Now we know that $Q_{1}\in L$.

Now we give the proof that $Q_{1}\in L$ when some edges are collapsed. That
is, there is an index $a\in\\{k+2,...,n-k-2\\}$ such that $Q_{a}=Q_{a+1}$.
Then the diagonals at $Q_{a-k-1}$ are collapsed. In order to avoid the proof
in §5.4 we must have folded diagonals at $Q_{a-k-1}$. Let $L^{\prime}$ be the
line containing these folded diagonals. The proof in Lemma 5.10 shows that
$S\subset L^{\prime}$. But $S\subset L$. If $L^{\prime}\not=L$ then $S=L\cap
L^{\prime}=\\{Q_{a-k-1}\\}$. But then $S$ contains a vertex of $S$. This is a
contradiction. Hence $L^{\prime}=L$.

Since $S\cap Q=\emptyset$, the point $Q_{a}$ lies forwards of $Q_{0}$. If both
$Q_{a+k+1}$ and $Q_{a-k-1}$ lie forwards of $Q_{a}$ then Lemma 5.10 says that
does not $S$ lie backwards of $Q_{a}$. This is not the case. Hence $Q_{a-k-1}$
lies backwards of $Q_{a}$. Hence $Q_{0,k+1}$ and $Q_{a-k-1,a}$ are parallel.
Consider the diagonal chain

$Q_{0,k+1}\to...\to Q_{a-k-1,a}.$

This chain either consists of parallel diagonals or else it twists $2\pi$
radians. The latter does not happen because the complementary diagonal chain
contains a diagonal that is not parallel to $Q_{0,a}$, namely $Q_{-k-1,0}$.
Hence $Q_{0,k+1}$ and $Q_{1,k+1}$ are parallel. Hence $Q_{1}\in L$ in this
case as well. $\spadesuit$

###### Lemma 5.13

If $n=3k+2$ and all folded diagonals of $Q$ are good, then the Degeneration
Lemma is true for $Q$.

Proof: Suppose $n=3k+2$. We have all the same arguments as in the previous
lemma. In particular, $Q_{1}\in L$ and hence $Q_{0,k+1}$ and $Q_{k+1,2k+2}$
are folded. We need a different endgame because now the parallel diagonals
$Q_{-k-1,0}$ are $Q_{k+1,2k+2}$ are not sufficiently well separated to be
called aligned.

Again, $Q_{0,k+1}$ and $Q_{k+1,2k+2}$ are folded diagonals. Since these folded
diagonals are good, the points $Q_{2k+2},Q_{2k+3},...,Q_{0}$ are collinear. We
already know that $Q_{2k+2},Q_{0}\in L$. The collinearity gives
$Q_{2k+2},...,Q_{0}\in L$. Since $n=3k+2$ we have $(2k+2)=(-k)$ mod $n$ we get
$Q_{k+1},...,Q_{0}\in L$, which is a run of more than $2k+1$ collinear points.
The proof in §5.4 now shows that the Degeneration Lemma is true for $Q$.
$\spadesuit$

###### Lemma 5.14

If $n=3k+1$ and all folded diagonals of $Q$ are good, then the Degeneration
Lemma is true for $Q$.

Proof: We have the same set-up as in the previous result but this time all we
can say is that the points

$Q_{k+1},...,Q_{2k},Q_{2k+2},...,Q_{0}\in L.$

Note that $Q$ has collapsed diagonals at all these points except perhaps for
$Q_{0}$ and $Q_{k+1}$. In order to avoid the proof in §5.4, we see that $Q$
must have folded diagonals at all these points. Since all these folded
diagonals are good, this suffices to show $Q_{1},...,Q_{k}\in L$. $\spadesuit$

###  5.9 Ungood Folded Diagonals

The only case left to consider is when $Q$ has a pair of folded diagonals
which are not good. Also, we can assume that $Q$ has no aligned diagonals and
$S\cap Q=\emptyset$. We normalize as above, so that $Q_{0},Q_{k+1},Q_{-k-1}$
lie in forward order on $L$, which is the $X$-axis. Here we list some
information we have.

  * •

Not all of $Q_{k+1},...,Q_{-k-1}$ lie in $L$.

  * •

Not all of $Q_{-k-1},...,Q_{0},...,Q_{k+1}$ lie in $L$. Otherwise we’d have
$2k+1$ consecutive collinear points, and the proof in §5.4 would finish the
job.

  * •

$Q_{-k-1}$ and $Q_{k+1}$ divide $Q$ into $2$ arcs, both of which start and end
on $L$ to the right of $x\in S$. This point $x$ does not lie in $Q$.

We call an edge of $Q$ escaping if $e\cap L$ is a single point. We call two
different edges of $Q$, in the labeled sense, twinned if they are both
escaping and if they intersect in an open interval. Even if two distinctly
labeled edges of $Q$ coincide, we consider them different as labeled edges.

###### Lemma 5.15

$Q$ cannot have twinned escaping edges.

Proof: Consider $Q(t)$ for $t$ near $1$. This polygon is strictly star shaped
with respect to a point $x(t)$ near $x$.

Figure 5.9: Rays intersecting the twinned segments

There is a disk $D$ about $x$ such that every $p\in D$ contains a ray which
intersects the twinned edges in the middle third portion of their
intersection. Figure 5.11 shows what we mean. Once $t$ is sufficiently near
$1$, the soul $S(t)$ will intersect $D$, and for all points $p\in D$ there
will be a ray which intersects $Q(t)$ twice. This contradicts the fact that
$Q(t)$ is strictly star-shaped with respect to all points of $S(t)$.
$\spadesuit$

We say that an escape edge rises above $L$ if it intersects the upper half
plane in a segment.

###### Lemma 5.16

$Q$ cannot have two escape edges which rise above $L$ and intersect $Q$ on the
same side of the point $x$.

Proof: This situation is similar to the previous proof. In this case, there is
a small disk $D$ about $x$ such that every point $p\in D$ has a ray which
intersects both rising escape edges transversely, and in the middle third of
each of the two subsegments of these escape edges that lie above $L$. Figure
5.10 shows this situation.

Figure 5.10: Rays intersecting the rising segments.

In this case, some part of $Q(t)$ closely shadows our two escape edges for $t$
near $1$. But then, once $t$ is sufficiently near $1$, each ray we have been
talking about intersects $Q(t)$ at least twice, once by each escaping edge.
This gives the same contradiction as in the previous lemma. $\spadesuit$

We define falling escape segments the same way. The same statement as in Lemma
5.16 works for falling escape segments. Since $x\not\in Q$ we conclude that
$Q$ can have at most $4$ escaping segments total.

But $Q=Q_{+}\cup Q_{-}$, where $Q_{\pm}$ is an arc of $Q$ that starts at
$Q_{k+1}$ and ends at $Q_{-k-1}$. Since both these arcs start and end on $L$,
and since both do not remain entirely on $L$, we see that each arc has at
least $2$ escape edges. and none of these are twinned. This means that both
$Q_{+}$ and $Q_{-}$ have exactly two escape edges.

Now for the moment of truth: Consider $Q_{+}$. Since $Q_{+}$ just has $2$
escape edges, they both have to be either rising or falling. Also, since
$Q_{+}$ starts and ends on the same side of $x$, and cannot cross $x$ (By
Lemma 3.1) we see that $Q_{+}$ has $2$ of the same kind of escape arc on the
same side of $x$. This is a contradiction. The same argument would work for
$Q_{-}$ but we don’t need to do it twice.

##  6 The Persistence of Birds

In this chapter we prove Statement 3 of Theorem 1.1, namely the fact that
$\Delta_{k}(B_{n,k})=B_{n,k}$. First we use the Degeneration Lemma to prove
that $\Delta_{k}(B_{n,k})\subset B_{n,k}$. Then we deduce the opposite
containment from projective duality and the factoring we discussed in §2.2.

###  6.1 Containment

Suppose for the sake of contradiction that there is some $P\in B_{k,n}$ such
that $\Delta(P)\not\in B_{k,n}$. Recall that there is a continuous path $P(t)$
for $t\in[0,1]$ such that $P(0)$ is the regular $n$-gon.

Define $Q(t)=\Delta_{k}(P(t))$. There is some $t_{0}\in[0,1]$ such that
$Q(t_{0})\not\in B_{k,n}$. We can truncate our path so that $t_{0}=1$. In
other words, $Q(t)\in B_{n,k}$ for $t\in[0,1)$ but $Q(1)\not\in B_{k,n}$.

###### Lemma 6.1

$Q(\cdot)$ is a degenerating path.

Proof: Note that $Q(\cdot)$ satisfies Property 1 for degenerating paths. The
energy $\chi_{k}$ is well-defined and continuous on $B_{k,n}$. Hence, by
compactness, $\chi_{k}(P(t))>\epsilon_{0}$ for some $\epsilon_{0}>0$ and all
$t\in[0,1]$. Now for the crucial step: We have already proved that

$\chi_{k}\circ\Delta_{k}=\chi_{k}.$ (25)

Hence $\chi_{k}(Q(t))>\epsilon_{0}$ for all $t\in[0,1]$. That is, $Q(\cdot)$
satisfies Property 2 for degenerating paths. Finally, if any two vertices of
$Q(1)$ coincide, we violate Theorem 4.1 for the $k$-bird $P(1)$. Hence
$Q(\cdot)$ satisfies Properties 3 and 4 for degenerating paths. In short,
$Q(\cdot)$ is a degenerating path. $\spadesuit$

We conclude that all but at most $1$ vertex of $Q(1)$ lies in a line $L$.
Stating this in terms of $P(1)$, we can say that all but at most one of the
feathers of $P(1)$ have their tips in a single line $L$. Call an edge of
$P(1)$ ordinary if the feather associated to it has its tip in $L$. We call
the remaining edge, if there is one, special. Thus, all but at most one edge
of $P$ is ordinary.

Let $S(t)$ be the soul of $P(t)$. We know that $S(1)$ has non-empty interior
by Theorem 3.1. The contradiction in our argument is to be that the
arrangement of tips of $P(1)$ is going to force $S(1)$ to either be empty or
to have empty interior.

###### Lemma 6.2

$P(1)$ cannot have ordinary edges $e_{1}$ and $e_{2}$ that lie on opposite
sides of $L$ and are disjoint from $L$.

Proof: Suppose this happens. Let $F_{1}$ and $F_{2}$ be the two associated
feathers. Then the opposite sector $F_{1}^{*}$ lies above $L$, and the
opposite sector $F_{2}^{*}$ lies below $L$ and the two tips are distinct. But
then $S(1)$ is empty. $\spadesuit$

###### Lemma 6.3

$P(1)$ cannot have more than $2$ ordinary edges crossing $L$.

Proof: As we trace along $L$ in one direction or the other we encounter the
first crossing edge and then the last one. Let $v$ and $v^{\prime}$ be the
tips of the corresponding feathers. Then the line segment from $v$ to
$v^{\prime}$ crosses all the other edges crossing $L$. But this contradicts
Statement 3 of Theorem 4.1. $\spadesuit$

We know that $P(1)$ cannot have ordinary edges on both sides of $L$ and
disjoint from $L$. We know also that at most $2$ ordinary edges can cross $L$
by Lemma 6.3. Finally, an ordinary edge cannot lie in $L$ because then the tip
would not. Hence, all but at most $2$ of the ordinary edges of $P(1)$ lie on
one side of $L$. Call this the abundant side of $L$. Call the other side the
barren side. From this structure, we see that $P(1)$ has at most $2$ vertices
on the barren side of $L$ that are not contained in $L$. At the same time,
each ordinary edge on the abundant side contributes two vertices to the barren
side: We just follow the diagonals comprising the corresponding feather. These
diagonals cross $L$ from the abundant side into the barren side. Two different
ordinary edges contribute at least $3$ distinct vertices to the barren side.
This is a contradiction.

We have ruled out all possible behavior for $P(1)$ assuming that $Q(1)$ is
degenerate. Hence, $Q(1)$ is not degenerate. This means that $Q(1)$ is a bird.
This completes the proof that

$\Delta_{k}(P_{k,n})\subset P_{k,n}.$ (26)

###  6.2 Equality

Now we show that $\Delta_{k}(B_{k,n})=B_{k,n}$. We follow the setup and
notation from §2.2. A polygon is a PolyPoint together with additional data
specifying an edge in $P$ joining each consecutive pair of points. Dually, we
get a polygon from a PolyLine by specifying, for each pair of consecutive
lines $L_{j},L_{j+1}$, an arc of the pencil of lines through the intersection
point which connects $L_{j}$ to $L_{j+1}$.

Equation 8 implies that

$\Delta_{k}^{-1}=D_{k+1}\circ\Delta_{k}\circ D_{k+1}.$ (27)

Since the dual projective plane $\mbox{\boldmath{$P$}}^{*}$ is an isomorphic
copy of $P$, it makes sense to define $B_{k.n}^{*}$. This space is just the
image of $B_{k,n}$ under any projective duality. Below we prove

###### Theorem 6.4

$D_{k+1}(B_{k,n})\subset B^{*}_{k,n}$.

It follows from Theorem 6.4 and projective duality that
$D_{k+1}(B_{k,n}^{*})\subset B_{k,n}$. This combines with Equation 27 and
Equation 26 to show $\Delta_{k}^{-1}(B_{k,n})\subset B_{k,n}$.

Implicit in the statement that $D_{k+1}(B_{k,n})\subset B_{k,n}^{*}$ is that
statement that we have a way to enhance $D_{k+1}$ so that it maps polygons to
polygons. We will explain this below. Theorem 6.4 combines with what we know
already to prove that $\Delta(B_{k,n})=B_{k,n}$.

Remark: In view of Equation 8, an alternate way of proving Statement 3 of
Theorem 1.1 would be to show that $D_{k}(B_{k,n})\subset B_{k,n}^{*}$. If we
knew this, we could bypass the painful Degeneration Lemma. I couldn’t see how
to prove directly that $D_{k}(B_{k,n})\subset B_{k,n}$.

###### Lemma 6.5

$D_{k+1}$ maps a member of $B_{k,n}$ to an $n$-gon which is $k$-nice.

Proof: Let $Q=D_{k+1}(P)$. A $(k+1)$-diagonal of $Q$ is just a vertex of $P$.
A $k$ diagonal of $Q$ is a vertex of $\Delta_{k}(p)$. Thus, to check the
$k$-nice property for $Q$ we need to take $n$-collections of $4$-tuples of
points and check that they are distinct. In each case, the points are
collinear because the lines of $Q$ are coincident.

Figure 6.1 One of the $n$ different $4$-tuples we need to check.

Once we make this specification, there is really combinatorially only
possibility for which collections we need to check. Figure 6.1 shows one such
$4$-tuple, $a,b,c,d$. The shaded triangles are the two feathers of $P$ whose
tips are $b,c$. But $a,b,c,d$ are distinct vertices of $P\cup\Delta_{k}(P)$
and so they are distinct. That is all there is to it. $\spadesuit$

###### Lemma 6.6

If $P\in B_{k,n}$, then we can enhance $D_{k+1}(P)$ in such a way that
$D_{k+1}(P)$ is a planar polygon in $\mbox{\boldmath{$P$}}^{*}$. The
enhancement varies continuously.

Proof: Specifying an enhancement of $D_{k+1}(P)$ is the same as specifing, for
each consecutive pair $L_{1},L_{2}$ of $(k+1)$ diagonals of $P$, an arc of the
pencil through their intersection that connects $L_{1},L_{2}$. There are two
possible arcs. One of them avoids the interior of the soul of $P$ and the
other one sweeps through the soul of $P$. We choose the arc that avoids the
soul interior. Figure 6.2 shows that we mean for a concrete example.

Figure 6.2: Enhancing a PolyLine to a polygon: Avoid the soul.

Since the soul of $P$ has non-empty interior, there exists a point $x\in P$
which is disjoint from all these pencil-arcs. Applying duality, this exactly
says that there is some line in $\mbox{\boldmath{$P$}}^{*}$ which is disjoint
from all the edges of our enhanced $D_{k+1}(P)$. Hence, this enhancement makes
$D_{k+1}(P)$ planar. Our choice also varies continuously on $B_{n,k}$.
$\spadesuit$

To show that $Q=D_{k+1}(P)$ is a $k$-bird, we consider a continuous path
$P(t)$ from the regular $n$-gon $P(0)$ to $P=P(1)$. We set $Q(t)=P(t)$. By
construction, $Q(0)$ is a copy of the regular $n$-gon in
$\mbox{\boldmath{$P$}}^{*}$, and $Q(t)$ is $k$-nice for all $t\in[0,1]$, and
$Q(t)$ is a planar polygon for all $t\in[0,1]$. By definition $Q=Q(1)$ is a
$k$-bird.

This completes the proof that $D_{k+1}(B_{k,n})\subset B_{k,n}^{*}$, which in
turn completes the proof that $\Delta_{k}(B_{k,n})=B_{k,n}$. Our proof of
Theorem 1.1 is done.

##  7 The Triangulation

###  7.1 Basic Definition

In this section we gather together the results we have proved so far and
explain how we construct the triangulation $\tau_{P}$ associated to a bird
$P\in B_{k,n}$.

Since $\Delta_{k}(B_{k,n})\subset B_{k,n}$, we know that $\Delta_{k}(P)$ is
also a $k$-bird. Combining this with Theorem 3.1 and Theorem 4.1 we can say
that $\Delta_{k}(P)$ is one embedded $n$-gon contained in $P^{I}$, the
interior of the region bounded by the embedded $P$. The region between $P$ and
$\Delta_{k}(P)$ is a topological annulus. Moreover, $\Delta_{k}(P)$ is
obtained from $P$ by connecting the tips of the feathers of $P$. The left side
Figure 7.1 shows how this region is triangulated. The black triangles are the
feathers of $P$ and each of the white triangles is made from an edge of
$\Delta_{k}(P)$ and two edges of adjacent feathers.

Figure 7.1: The triangulation of the annulus

###### Lemma 7.1

For every member $P\in B_{k,n}$, the associated $2n$ triangles have pairwise
disjoint interiors, and thus triangulate the annular region between $P$ and
$\Delta_{k}(P)$.

Proof: As usual, we make a homotopical argument. If this result is false for
some $P$, then we can look at path which starts at the regular $n$-gon (for
which it is true) and stop at the first place where it fails. Theorem 4.1
tells us that nothing goes wrong with the feathers of $P$. The only thing that
can go wrong is $\Delta_{k}(P)$ fails to be an embedded polygon. Since this
does not happen, we see that in fact there is no counter-example at all.
$\spadesuit$

We can now iterate, and produce $2n$ triangles between $\Delta_{k}(P)$ and
$\Delta_{k}^{2}(P)$, etc. The right side of Figure 7.1 shows the result of
doing this many times. The fact that $\Delta_{k}(B_{k,n})=B_{k,n}$ allows us
to extend outward as well. When we iterate forever in both directions, we get
an infinite triangulation of a (topological) cylinder that has degree $6$
everywhere. This is what Figure 1.6 is showing. We call this bi-infinite
triangulation $\tau_{P}$.

###  7.2 Some Structural Results

###### Theorem 7.2

Let $P\in B_{n,k}$. Let $S$ be the soul of $B$. Then for $\ell\geq n$ we have
$\Delta^{\ell}_{k}(P)\subset S$.

Proof: We first note the existence of certain infinite polygonal arcs in
$\tau_{P}$. We start at a vertex of $P$ and then move inward to a vertex of
$\Delta_{k}(P)$ along one of the edges. We then continue through this vertex
so that $3$ triangles are on our left and $3$ on our right. Figure 7.2 below
shows the two paths like this that emanate from the same vertex of $P$.

Figure 7.2: The spiral paths.

The usual homotopical argument establishes the fact that the spiral paths are
locally convex. One can understand their combinatrics, and how they relate to
the polygons in the orbit, just by looking at the case of the regular $n$-gon.
We call the two spiral paths in Figure 7.2 partners. In the regular $n$-gon
the partners intersect infinitely often. So this is true in general. Each
spiral path has an initial segment joining the initial endpoint on $P$ to the
first intersection point with the partner. We define a petal to be the region
bounded by the initial paths of the two partners.

It is convenient to write $P^{\ell}=\Delta_{k}^{\ell}(P)$. In the regular
case, $P^{\ell}$ is contained in the petal for $\ell>n-1$.. Hence, the same
goes in the general case. Because the initial segments are locally convex, the
petal lies to the left of the lines extending the edges $e_{1}$ and $e_{2}$
when these edges are oriented according to the $(k+1)$-diagonals of $P$. But
this argument works for every pair of partner spiral paths which start at a
vertex of $P$. We conclude that for $\ell\geq n$, the polygon $P^{\ell}$ lies
to the left of all the $(k+1)$-diagonals of $P$. But the soul of $P$ is
exactly the intersection of all these left half planes. $\spadesuit$

Theorem 7.2 in turn gives us information about the nesting properties of birds
within an orbit. Let $S_{\ell}$ denote the soul of $P^{\ell}$. Let

$S_{\infty}=\bigcap_{\ell\in\mbox{\boldmath{$Z$}}}S_{\ell},\hskip
30.0ptS_{-\infty}=\bigcup_{\ell\in\mbox{\boldmath{$Z$}}}S_{\ell}.$ (28)

It follows from Theorem 7.2 that $\widehat{P}_{\infty}=S_{\infty}$ and
$\widehat{P}_{-\infty}=S_{-\infty}$, because

$S_{\ell+n}\subset P^{\ell+n}\subset S_{\ell}\subset P^{\ell}.$ (29)

Hence these sets are all convex subsets of an affine plane.

###### Corollary 7.3

Any $P\in B_{k,n}$ is strictly star-shaped with respect to all points in the
convex hull of $\Delta_{k}^{n}(P)$.

Proof: Since $P^{\ell+n}\subset S_{\ell}$, and $P^{\ell}$ is strictly star
shaped with respect to all points of $S^{\ell}$, we see that $P^{\ell}$ is
strictly star shaped with respect to all points of $P^{\ell+n\/}$. Since
$S_{\ell}$ is convex, we can say more strongly that $P^{\ell}$ is strictly
star-shaped with respect to all points of the convex hull of $P^{\ell+n}$. Now
we just set $\ell=0$ and recall the meaning of our notation, we get the exact
statement of the result. $\spadesuit$

An immediate corollary is that $P$ is strictly star-shaped with respect to
$\widehat{P}_{\infty}$. (Theorem 1.3 says that this is a single point.)

##  8 Nesting Properties of Birds

In this chapter we prove Theorem 1.3. We just keep peeling away cases until
the proof is done.

###  8.1 Discussion

This discussion is not part of our proof, but some people who are used to the
pentagram map and its connection to projective geometry might appreciate it.
Other readers can safely ignore this discussion.

Usually when one studies the pentagram map, say on convex polygons, one takes
the quotient by the group of projective transformations. One can also take the
quotient by the group of affine transformations, but this seems like a
somewhat different thing. I want to relate the two concepts.

There is an asymptotic sense in which the affine quotient and projective
quotient are quite closely related. Let $\Delta_{1}$ be the pentagram map. We
start with a convex polygon $P$ and look at $P^{\ell}=\Delta_{1}^{\ell}(P)$
for very large $\ell>0$. Suppose we choose an affine transformation $T_{\ell}$
so that $Q^{\ell}=T_{\ell}(P^{\ell})$ remains bounded and in some sense
uniformly fat. For instance, we could normalize so that a certain three
vertices make an equilateral triangle. For large $\ell$, the map $T^{\ell}$
has projective significance and not just affine significance.

Recall that as $\ell\to\infty$ the union $\bigcup\widehat{P}^{-\ell}$
converges to an affine plane $\widehat{P}_{-\infty}$. Here
$\widehat{P}^{-\ell}$ is the closure of the region bounded by $P^{-\ell}$. The
projective significance is that $T^{\ell}(\widehat{P}_{\infty})$ is very
nearly the affine patch of $P$. Making such a normalization is usually
something you would do with a projective transformation, but here we are doing
it in an automatic way with an affine transformation. In the asymptotic
forward direction, the best affine normalization really coincides with the
best projective normalization.

To put this another way, the compactness of the full orbit $\\{P^{\ell},\
\ell\in\mbox{\boldmath{$Z$}}\\}$ modulo projective transformations is very
closely related to the compactness of the forward orbit modulo affine
transformations. In the case of convex polygons, we can freely convert between
one kind of compactness and the other. For the birds, I do not know how to do
this and I find it easier to work with affine transformations.

For our proof, we will concentrate on Statement 1 of Theorem 1.3, and then
bring back the projective geometry, in the form of projective duality, to
deduce Statement 2 from Statement 1.

###  8.2 The Compact Case

Suppose that $S_{1}\subset S_{2}$ are two compact convex sets. The ratio of
diameters, ${\rm diam\/}(S_{1})/{\rm diam\/}(S_{2})$, is not affine invariant.
We come up with a replacement notion. For each direction $v$ in the plane, we
let $\|S\|_{v}$ denote the maximum length of $L\cap S$ where $L$ is a straight
line parallel to $v$. We then define

$\delta(S_{1},S_{2})=\sup_{v}\frac{\|S_{1}\|_{v}}{\|S_{2}\|_{v}}\in[0,1].$
(30)

The quantity $\delta(S_{1},S_{2})$ is affine invariant, and (choosing a
direction $\mu$ which realizes the diamater of $S_{1}$) we have

$\frac{{\rm diam\/}(S_{1})}{{\rm
diam\/}(S_{2})}\leq\frac{\|S_{1}\|_{\mu}}{\|S_{2}\|_{\mu}}\leq\delta(S_{1},S_{2}).$
(31)

Let $P\in B_{n,k}$ and suppose that $P$ is normalized so as to be a subset of
$\mbox{\boldmath{$R$}}^{2}$. Let $P^{\ell}=\Delta_{k}^{\ell}(P)$. We define

$\delta(P)=\delta(S(P^{n}),S(P)).$ (32)

Here $S(P)$ is the soul of $P$. By Theorem 5.8, we have $S(P^{n})\subset
P^{n}\subset S(P)$, so our definition makes sense. Also, $\delta(P)<1$ because
$S(P^{n})$ is contained in the interior of $P^{n}$.

We equip the space of compact convex subsets of $\mbox{\boldmath{$R$}}^{2}$
with the Hausdorff metric: The distance between two such subsets $A_{0},A_{1}$
is the infimal $\epsilon$ such that $A_{j}$ is contained in the
$\epsilon$-tubular neighborhood of $A_{1-j}$ for $j=0,1$. This metric lets us
talk about the convergence in an easy way. The function $\delta(\cdot,\cdot)$
is continuous with respect to this metric.

Suppose we have a sequence $\\{Q^{\ell}\\}$ of $k$-birds which converges to
some other bird $Q^{\ell}$ in the sense that the vertices and edges converge.
Then $\Delta_{k}^{n}(Q^{\ell})$ converges to $\Delta_{k}^{n}(Q^{\infty})$ and
the corresponding souls converge in the Hausdorff metric. This means that
$\delta(Q^{\ell})\to\delta(Q^{\infty})=\delta_{0}<1$ for some $\delta_{0}$. We
conclude that there is some $\delta_{1}<1$ so that
$\delta(Q^{\ell})<\delta_{1}<1$ for all $\ell>0$.

If our forward orbit $\\{P^{\ell}\\}$ is compact modulo affine
transformations, then we can find a sequence of affine tranformations
$\\{T_{\ell}\\}$ such that $Q^{\ell}=T_{\ell}(P^{\ell})$ converges to another
bird $Q^{\infty}$. The affine invariance of $\delta$ then tells us that
$\delta(P^{\ell})<\delta_{1}<1$. But then

${\rm diam\/}(S(P^{\ell+n}))<\delta_{1}\ {\rm diam\/}(S(P^{\ell})).$

This shows that $\widehat{P}_{\infty}=\bigcap S(P^{\ell})$ is a single point.

###  8.3 Normalizing by Affine Transformations

Henceforth we assume that the forward orbit $\\{P^{\ell}\\}$ of $P$ under
$\Delta_{k}$ is not compact modulo affine transformations. Our first step is
to normalize as much as we can.

###### Lemma 8.1

There is a sequence $\\{T_{\ell}\\}$ of affine transformations such that

  1. 1.

$T_{\ell}(P^{\ell})$ has (the same) $3$ vertices which make a fixed
equilateral triangle.

  2. 2.

$T_{\ell}$ expands distances on $P^{\ell}$ for all $\ell$.

  3. 3.

$T_{\ell}(P^{\ell})$ is contained in a uniformly bounded subset of
$\mbox{\boldmath{$R$}}^{2}$.

Proof: To $P^{\ell}$ we associate the triangle $\tau_{\ell}$ made from $3$
vertices of $P^{\ell}$ and having maximal area. The diameter of $\tau_{\ell}$
is uniformly small, so we can find a single equilateral triangle $T$ and an
expanding affine map $T_{\ell}:\tau_{\ell}\to T$. Let $d$ be the side length
of $T$. Every vertex of $T_{\ell}(P^{\ell})$ is within $d$ of all the sides of
$T$, because otherwise we’d have a triangle of larger area. The sequence
$\\{T_{\ell}\\}$ has the advertised properties. $\spadesuit$

Let $Q^{\ell}=T_{\ell}(P^{\ell})$. By compactness we can pass to a subsequence
so that the limit polygon $Q$ exists, in the sense that the vertices and the
edges converge. Note that some vertices might collapse in the limit. This does
not bother us. Let $Q_{0},Q_{1}$, etc. be the vertices of $Q$.

Each distinguished diagonal of $Q^{\ell}$ defines the unit vector which is
parallel to it. Thus $Q^{\ell}$ defines a certain list of $2n$ unit vectors.
We can pass to a subsequence so that all these unit vectors converge. Thus, to
each distinguished diagonal of $Q$ we still have a well-defined direction,
even if the diagonal is trivial. We are keeping track of the $1$-jet. Given a
point and a unit direction, we have a well-defined oriented line which
contains the point and is oriented along the direction, and a corresponding
left half-plane. Thus we associate to each $(k+1)$-diagonal, trivial or not, a
left-half plane. We define $\widehat{S}$ to be the intersection of all these
half-planes.

We also define the limiting souls. We set $S^{\ell}=S(Q^{\ell})$. We define
the soul $S$ of $Q$ just as we defined it in §5.5: It is the set of
accumulation points of sequences $\\{p^{\ell}\\}$ with $p^{\ell}\in S^{\ell}$.
We will see that $S\subset\widehat{S}$.

###  8.4 The Limiting Soul

###### Lemma 8.2

$S\subset\widehat{S}$.

Proof: Fix $\epsilon>0$. If this is not the case, then by compactness we can
find a convergent sequence $\\{p^{\ell}\\}$, with $p^{\ell\/}\in S^{\ell}$,
which does not converge to a point of $\widehat{S}$. But $p^{\ell}$ lies in
every left half plane associated to $Q^{\ell}$. But then, by continuity, the
accumulation point $p$ lies in every left half plane associated to $Q$. Hence
$p\in\widehat{S}$. This is a contradiction. $\spadesuit$

###### Corollary 8.3

Suppose that $\widehat{P}_{\infty}$ is not a single point. Then
$\delta(S,H_{Q})=1$. Here $H_{Q}$ is the convex hull of $Q$.

Proof: Suppose not. Note that $H_{Q^{\ell}}\subset S^{\ell-n}$ by Theorem 7.2
and convexity. Then for $\ell$ large we have

$\delta(Q^{\ell-n})=\delta(S^{\ell},S^{\ell-n})\leq\delta(S^{\ell},H_{Q^{\ell}})<\delta(S,H_{Q})+\epsilon,$

and we can make $\epsilon$ as small as we like. This gives us a uniform
$\delta<1$ such that $\delta(Q^{\ell})<\delta$ once $\ell$ is large enough.
The argument in the compact case now shows that $\widehat{P}_{\infty}$ is a
single point. $\spadesuit$

Corollary 8.3 gives us a powerful structural result. It says in particular
that $S$ and $Q$ have the same diameter. Hence there is a chord $S^{*}\subset
S$ which has the same diameter as $Q$. Since $Q$ is a polygon, this means that
$Q$ must have vertices at either endpoint of $S^{*}$. We normalize so that
$S^{*}$ is the unit segment joining $(0,0)$ to $(1,0)$.

###### Lemma 8.4

Let $Q^{\prime}\subset Q$ be an arc of $Q$ that joins $(0,0)$ to $(1,0)$.

  1. 1.

The vertices of $Q^{\prime}$ must have non-decreasing $x$-coordinates.

  2. 2.

If consecutive vertices of $Q^{\prime}$ have the same $x$-coordinate, they
coincide.

  3. 3.

Either $Q^{\prime}\subset S^{*}$ or $Q^{\prime}$ intersects $S^{*}$ only at
$(0,0)$ and $(1,0)$.

Proof: Suppose the Statement 1 is false. Then we can find a vertical line
$\Lambda$ which intersects $S^{*}$ at a relative interior point and which
intersects $Q^{\prime}$ transversely at $3$ points. But then once $\ell$ is
sufficiently large, $Q^{\ell}$ will intersect all vertical lines sufficiently
close to $\Lambda$ in at least $3$ points and moreover some of these lines
will contain points of $S^{\ell}$. This contradicts the fact that $Q^{\ell}$
is strictly star-shaped with respect to all points of $Q^{\ell}$.

For Statement 2, we observe that $Q^{\prime}$ does not contain any point of
the form $(0,y)$ or $(1,y)$ for $y\not=0$. Otherwise $Q$ has larger diameter
than $1$. This is to say that once $Q^{\prime}$ leaves $(0,0)$ it immediately
moves forward in the $X$-direction. Likewise, once $Q^{\prime}$ (traced out
the other way) leaves $(1,0)$ it immediately moves backward in the
$X$-direction. If Statement 2 is false, ten we can find a non-horizontal line
$\Lambda^{\prime}$ which intersects $S^{*}$ in a relative interior point and
which intersects $Q^{\prime}$ transversely at $3$ points. The slope is
$\Lambda^{\prime}$ depends on which of the two vertices of $Q^{\prime}$ lies
above the other. Once we have $\Lambda^{\prime}$ we play the same game as for
the first statement, and get the same kind of contradiction.

Suppose Statement 3 is false. We use the kind of argument we had in §5.9. By
Statements 1 and 2 together, $Q^{\prime}$ must have an escape edge which
touches $S^{*}$ in a relative interior point. Moreover, this one escape edge
is paired with another escape edge. Thus we can find a point $x\in S^{*}$
which strictly lies on the same side of both of these same-type escape edges.
The argument in §5.9 now shows that $Q^{\ell}$ is not strictly star-shaped
with respect to points of $S^{\ell}$ very near $x$. $\spadesuit$

###### Corollary 8.5

Up to adding repeated indices, $Q$ is embedded.

Proof: Lemma 8.4 implies that up to adding repeated vertices, $Q$ is the union
of two embedded arcs which connect $(0,0)$ to $(1,0)$, both of which are
graphs of functions on $[0,1]$. Call these functions $f$ and $g$. Since
$Q\not\subset S^{*}$, at least one of these functions is strictly nonzero. If
it ever happens that $f(p)=g(p)$ for $p\in(0,1)$, then the two arcs intersect
at some point not on $S^{*}$. But then some vertical ray through a relative
interior point of $S^{*}$ intersects $Q$ transversely at two points, giving
the same contradiction. Hence our two arcs are disjoint. $\spadesuit$

###### Corollary 8.6

Suppose $0\leq a<b<n$ and $Q_{a}=Q_{b}$. Then either we have
$Q_{a}=Q_{a+1}=...=Q_{b}$ or else we have $Q_{b}=Q_{b+1}=...=Q_{a+n}$.

###  8.5 The Triangular Limit Case

Here we prove a more general result that covers the triangular limit case.
Suppose that there is a line $L$ such that $Q_{0}\not\in L$ and $Q_{j}\in L$
unless $j\in\\{-k+1,...,0,...,k-1\\}$. This is a run of $2k-1$ consecutive
indices.

Figure 8.1: The triangular limit $Q$.

The cross ratio of the lines

$Q_{0,k},\ Q_{0,k+1},\ Q_{n-k-1,0},\ Q_{n-k,0}$

is at least $\epsilon_{0}$. Also, these lines are cyclically ordeded about $0$
as indicated in Figure 8.1, thanks to the $k$-niceness property and
continuity. Also, the two lines containing $Q_{0,k}$ and $Q_{-k,0}$ are not
parallel because $Q_{0}\not\in L$. Hence $S$ is contained in the shaded region
in Figure 8.1, namely the triangle with vertices $Q_{0}$ and $Q_{\pm(k+1)}$.
But this shaded region has diameter strictly smaller than the triangle $\tau$
with vertices $Q_{0}$ and $Q_{\pm k}$. Hence ${\rm diam\/}(S)<{\rm
diam\/}(\tau)\leq{\rm diam\/}(Q)$. This is a contradiction.

###  8.6 The Case of Folded Diagonals

In this section we suppose $Q$ has a pair of folded diagonals. We relabel so
that the folded diagonals are $Q_{-k-1,0}$ and $Q_{0,k+1}$.

The left half planes defined by these folded diagonals intersect in a line.
Hence $\widehat{S}$ is contained in a line $L$. Since $S\subset\widehat{S}$,
we see that $S\subset L$. Indeed, we must have $S^{*}=S$, where $S^{*}$ is as
in §8.4. We will repeatedly use the fact that $S$ realizes the diameter of
$Q$, so that points of $Q$ not in $S$ do not lie in the line $L$ containing
$S$. There are $3$ cases.

Case 1: Suppose that $Q_{k+1}$ is not an endpoint of $S$. Then by Lemma 3.6
the arc $Q_{0}\to...\to Q_{k+1}$ lies in $S$. Likewise the arc
$Q_{k+1}\to...\to Q_{n-k-1}$ also lies in $S$. Hence $Q_{j}\in L$ unless
$j\in\\{-k,...,-1\\}$. We can cyclically relabel so that this case is covered
by the result in §8.5. Exactly the same argument works when $Q_{-k-1}$ is not
an endpoint of $S$.

Case 2: Suppose $Q_{k+1}$ and $Q_{-k-1}$ are the same endpoint of $S$ and that
$Q_{0}\not=Q_{k+1}$. In this case, Corollary 8.6 says that
$Q_{k+1}=...=Q_{-k-1}\in S$. If $Q_{\pm k}\in S$ then the case in §8.5 would
cover us. So, one of these points does not lie in $S$. We consider the case
when $Q_{k}\not\in S$. The other case has the same treatment.

Suppose $n>3k+1$. Then $2k,2k+1\in\\{k+1,...,.n-k-1\\}$. Hence $Q_{k,2k}$ and
$Q_{k,2k+1}$ are nontrivial and contained in some line $L^{\prime}\not=L$. The
notions of collapsed diagonals, folded diagonals, and aligned diagonals from
§5 make sense for $Q$ because the concepts just involve the directions of the
diagonals. Likewise, Lemmas 5.2 and 5.3 hold for $Q$. By construction $Q$ has
collapsed diagonals at $Q_{k}$. If $Q$ has folded diagonals at $Q_{k}$ then
$S\subset L^{\prime}$, a contradiction. Hence $Q$ has aligned diagonals at
$Q_{k}$. Lemma 5.3 applies either to the short diagonal chain and gives us
$Q_{0},...,Q_{2k}\in L^{\prime}$ or else it applies to the long diagonal chain
and gives us all points of $Q$ in $L^{\prime}$. In either case $Q_{0}\subset
L^{\prime}$. But then $L^{\prime}$ contains both $Q_{0}$ and
$Q_{2k+1}=Q_{k+1}\not=Q_{0}$. So does $L$. Hence $L=L^{\prime}$. This is a
contradiction.

Suppose $n=3k+1$. In the next section we will show that when $Q$ has a folded
diagonal, $Q$ always has a run of $k+1$ repeating points (which is stronger
than the $k$ repeating points we could get from the argument above in this
case.) We cyclically relabel so that $Q_{k+1}=...=Q_{2k+1}$. If $Q_{k}\in S$
then §8.5 covers the case. Otherwise $Q_{k}\not\in S$ and we can run the
argument as above.

Case 3: Suppose that $Q_{k+1}$ and $Q_{-k-1}$ are each endpoints of $S$.
Either $Q_{k+1}=Q_{-k-1}$ or $Q_{0}=Q_{k+1}$ or $Q_{0}=Q_{-k-1}$. We cannot
have $Q_{0}$ strictly between these points because then $Q_{-k-1,0}$ and
$Q_{0,k+1}$ are not folded. If $Q_{k+1}=Q_{-k-1}$ then to avoid Case 2 we have
$Q_{0}=Q_{k+1}=Q_{-k-1}$. In all cases, Lemma 8.6 then gives us at least $k+1$
consecutive points of $Q$ which coincide. So, we have some index $a$ such that
$\nu:=Q_{a}=...=Q_{a+k}\in S$. Using Lemma 3.6 we can assume that either
$Q_{a-1}\not\in L$ or $Q_{a+k+1}\not\in L$. Again $L$ is the line containing
$S$. We consider the second case. The first case has the same treatment.

Let $L^{\prime}\not=L$ denote the line containing $\nu$ and $Q_{a+k+1}$. Two
of the distinguished diagonals connect $Q_{a+k+1}$ to $\nu$ and hence point
along $L^{\prime}$. Hence $Q$ has collapsed diagonals at $Q_{a+k+1}$. If $Q$
has folded diagonals at $Q_{a+k+1}$ then $S\subset L^{\prime}$. This is
contradiction. Hence $Q_{a+1,a+k+1}$ and $Q_{a+k+1,a+2k+1}$ are aligned and
point along $L^{\prime}$. The same argument as in Case 2 now tells gives us
$2k$ parallel diagonals $Q_{a+1,a+k+1},...,Q_{a+k+1,a+2k+1}$ which point along
$L^{\prime}$.

Let us now cyclically relabel so that $a=0$. This means that our parallel
diagonal chain is $Q_{0,k},...,Q_{k,2k}$. This is all we are going to use in
our proof for the rest of this case. Lemma 5.3 tells us that
$Q_{0},...,Q_{2k}\in L^{\prime}$.

The two diagonals $Q_{0,k+1}$ and $Q_{1,k+1}$ are on our list and hence
parallel. Hence $Q$ has collapsed diagonals at $Q_{k+1}$. To avoid $S\subset
L^{\prime}$, as in Case 2, we must have $Q_{1,k+1}$ and $Q_{k+1,2k+1}$
parallel to each other and also parallel to the $2k$ diagonals we already
have. this extends our parallel diagonal chain from length $2k$ to length
$2k+2$. We now repeat this argument indefinitely, showing that all
distinguished diagonals point along $L^{\prime}$. Hence $Q\subset L^{\prime}$,
a contradiction.

###  8.7 No Folded Diagonals

If $Q$ has a trivial distinguished diagonal, then after relabeling, we can say
that $Q_{0}=Q_{k+i}$ for one of $i=0,1$. In all cases, Lemma 8.6 gives us a
run of $k+1$ repeated points. Cyclically relabeling again, if necessary, we
can arrange that $Q_{0}=...=Q_{k}$ but $Q_{k+1}\not=Q_{0}$. The only case we
have not considered is when $Q$ has no folded diagonals. Let $L^{\prime}$ be
the line through $Q_{0}$ and $Q_{k+1}$. Then $Q$ has collapsed diagonals at
$Q_{k+1}$. Since $Q$ has no folded diagonals, $Q$ has aligned diagonals at
$Q_{k+1}$. These diagonals point along $L^{\prime}$. As in Case 3 above, we
get a parallel diagonal chain of length $2k$. Since we have no folded
diagonals, we can repeat the argument at the end of Case 3 indefinitely to
show $Q\subset L^{\prime}$, a contradiction.

###  8.8 Applying Duality

We have finished proving that $\widehat{P}_{\infty}$ is a point. Now we prove
that $\widehat{P}_{-\infty}$ is an affine plane. We take $\ell\geq 0$ and
consider $P^{-\ell}=\Delta_{k}^{-\ell}(P)$.

###### Lemma 8.7

There is a line $L\subset\mbox{\boldmath{$P$}}$ which is disjoint from
$\widehat{P}^{-\ell}$ for all $\ell$.

Proof: Let $\Omega_{\ell}$ be the set of lines in $P$ which are disjoint from
the interior of $P^{-\ell}$. Since $P^{-\ell}$ is planar, this set is
nonempty. The sets $\\{\Omega_{\ell}\\}$ are nested and hence have a non-empty
intersection. Let $L$ be a line in the intersection. Since the interior of the
region bounded by $P^{-\ell-1}$ contains $P^{-\ell}$ we see that $L$ is
actually disjoint from $P^{-\ell}$ as well. $\spadesuit$

We normalize so that $L$ is the line at infinity and $P_{\infty}$ is the
origin. Let

$\Pi^{\ell}=\Delta_{k}^{\ell}(D_{k+1}(P)).$ (33)

Then $P^{\ell}$ is planar polygon that is strictly star-shaped with respect to
the origin. The map $D_{k+1}$ conjugates $\Delta_{k}$ to $\Delta_{k}^{-1}$ and
maps $k$-birds to $k$-birds. $\\{\Pi^{\ell}\\}$ shrinks to a point in the dual
projective plane $\mbox{\boldmath{$P$}}^{*}$. Because the vertices of
$\Pi^{\ell}$ shrink to a single point, all the $(k+1)$-diagonals of
$P^{-\ell}$ converge to a single line $L^{\prime}$. This is enough to prove
that $\widehat{P}_{-\infty}$ is either all of $\mbox{\boldmath{$R$}}^{2}$ or
an infinite strip in $\mbox{\boldmath{$R$}}^{2}$.

To rule out the strip case we note that the edges of $\Pi^{\ell}$ shrink to a
point as well. We rotate so that the supposed strip is vertical. Then
$L^{\prime}$ is a vertical line. Consider the soul $S^{-\ell}$ of $P^{-\ell}$.
Let $v_{\ell}$ be the vertex of $S^{-\ell}$ with the largest $y$-coordinate.
Let $\Upsilon_{\ell}$ be the set of lines through $v_{\ell}$ which avoid the
interior of $S^{-\ell}$. Compare Figure 6.6. This set corresponds to an edge
of $\Pi^{\ell}$. Hence $\Upsilon_{\ell}$ converges to $L^{\prime}$ as well.
But $\Upsilon_{\ell}$ always contains a horizontal line. This horizontal line
either converges on a subsequence to the line $L$ at infinity or else to some
horizontal line in $\mbox{\boldmath{$R$}}^{2}$. In either case we get a
limiting line that does not equal $L^{\prime}$. This is a contradiction.

This completes the proof of Statement 2 of Theorem 1.3. Our proof of Theorem
1.3 is done.

##  9 Appendix

###  9.1 The Energy Invariance Revisited

In this section we sketch Anton Izosimov’s proof that
$\chi_{k}\circ\Delta_{k}=\chi_{k}$. This proof is more conceptual than the one
in §2 but it is not self-contained. It requires the machinery from [6]. (The
perspective comes from [8], but the needed result for $\Delta_{k}$ is in the
follow-up paper [6].)

Let $P$ be an $n$-gon. We let $V_{1},...,V_{n}$ be points in
$\mbox{\boldmath{$R$}}^{3}$ representing the consecutive vertices of $P$. Thus
the vertex $P_{j}$ is the equivalence class of $V_{j}$. We can choose periodic
sequences $\\{a_{i}\\}$, $\\{b_{i}\\}$, $\\{c_{i}\\}$, $\\{d_{i}\\}$ such that

$a_{i}V_{i}+b_{i}V_{i+k}+c_{i}V_{i+k+1}+d_{i}V_{i+2k+1}=0,\hskip 30.0pt\forall
i.$ (34)

Recall from §2.2 that $\Delta_{k}=D_{k}\circ D_{k+1}$.

###### Lemma 9.1

One of the cross ratio factors of $\chi_{k}\circ D_{k+1}$ is
$(a_{0}d_{-k})/(c_{0}b_{-k})$.

Proof: One of the factors is the cross ratio of $P_{0},y,x,P_{k+1}$, where

$x=P_{0,k+1}\cap P_{k,2k+1},\hskip 30.0pty=P_{-k,1}\cap P_{0,k+1}.$

(Compare the right side of Figure 2.1, shifting all the indices there by
$k+1$.)

The points $x$ and $y$ respectively are represented by vectors

$X=a_{0}V_{0}+c_{0}V_{k+1}=-b_{0}V_{k}-d_{0}V_{2k+1},$
$Y=-a_{-k}V_{-k}-c_{-k}V_{1}=b_{-k}V_{0}+d_{-k}V_{k+1}.$

The point here is that the vector $X$ lies in the span of
$\\{V_{0},V_{k+1}\\}$ and in the span of $\\{V_{k},V_{2k+1}\\}$ and
projectively this is exactly what is required. A similar remark applies to
$Y$.

Setting $\Omega=V_{0}\times V_{k+1}$, we compute the relevant cross ratio as

$\frac{V_{0}\times Y}{V_{0}\times X}\cdot\frac{X\times V_{k+1}}{Y\times
V_{k+1}}=\frac{d_{-k}\Omega}{c_{0}\Omega}\times\frac{a_{0}\Omega}{b_{-k}\Omega}=\frac{d_{-k}a_{0}}{b_{-k}c_{0}},$
(35)

which is just a rearrangement of the claimed term. $\spadesuit$

The other cross ratio factors are obtained by shifting the indices in an
obvious way. As an immediate corollary, we see that

$\chi_{k}(D_{k+1}(P))=\prod_{i=1}^{n}\frac{a_{i}d_{i}}{b_{i}c_{i}}.$ (36)

Let us call this quantity $\mu_{k}(P)$.

###### Lemma 9.2

If $\mu_{k}\circ\Delta_{k}=\mu_{k}$ then $\chi_{k}\circ\Delta_{k}=\chi_{k}$.

Proof: If $\mu_{k}\circ\Delta_{k}=\mu_{k}$ then
$\mu_{k}\circ\Delta_{k}^{-1}=\mu_{k}$. Equation 36 says that

$\chi_{k}\circ D_{k+1}=\mu_{k},\hskip 30.0pt\mu_{k}\circ D_{k+1}=\chi_{k}.$
(37)

The first equation implies the second because $D_{k+1}$ is an involution.
Since $D_{k+1}$ conjugates $\Delta_{k}$ to $\Delta_{k}^{-1}$ we have

$\chi_{k}\circ\Delta_{k}=\chi_{k}\circ D_{k+1}\circ\Delta_{k}^{-1}\circ
D_{k+1}=\mu_{k}\circ\Delta_{k}^{-1}\circ D_{k+1}=\mu_{k}\circ
D_{k+1}=\chi_{k}.$

This completes the proof. $\spadesuit$

Let $\widetilde{P}=\Delta_{k}(P)$. Let $\\{\widetilde{a}_{i}\\}$, etc., be the
sequences associated to $\widetilde{P}$. We want to show that

$\prod_{i=1}^{n}\frac{a_{i}d_{i}}{b_{i}c_{i}}=\prod_{i=1}^{n}\frac{\widetilde{a}_{i}\widetilde{d}_{i}}{\widetilde{b}_{i}\widetilde{c}_{i}}.$
(38)

This is just a restatement of the equation $\mu_{k}\circ\Delta_{k}=\mu_{k}$.

Now we use the formalism from [6] to establish Equation 38. We associate to
our polygon $P$ operator $D$ on the space $\cal V$ of bi-infinite sequences
$\\{V_{i}\\}$ of vectors in $\mbox{\boldmath{$R$}}^{3}$. The definition of $D$
is given coordinate-wise as

$D(V_{i})=a_{i}V_{i}+b_{i}T^{k}(V_{i})+c_{i}T^{k+1}(V_{i})+d_{i}T^{2k+1}(V_{i}).$
(39)

Here $T$ is the shift operator, whose action is $T(V_{i})=V_{i+1}$. If we take
$\\{V_{i}\\}$ to be a periodic bi-infinite sequence of vectors corresponding
to our polygon $P$, then $D$ maps $\\{V_{i}\\}$ to the $0$-sequence.

Next, we write $D=D_{+}+D_{-}$ where coordinate-wise

$D_{+}(V_{i})=a_{i}V_{i}+c_{i}T^{k+1}(V_{i}),\hskip
30.0ptD_{-}(V_{i})=b_{i}T^{k}(V_{i})+d_{i}T^{2k+1}(V_{i}).$ (40)

The pair $(D_{+},D_{-})$ is associated to the polygon $P$.

Let $\widetilde{D}$ and $(\widetilde{D}_{+},\widetilde{D}_{-})$ be the
corresponding operators associated to $\widetilde{P}$. One of the main results
of [6] is that the various choices can be made so that

$\widetilde{D}_{+}D_{-}=\widetilde{D}_{-}D_{+}.$ (41)

This is called refactorization. Equating the lowest (respectively highest)
terms of the relation in Equation 41 gives us the identity
$\widetilde{a}_{i}b_{i}=\widetilde{b}_{i}a_{i+k}$ (respectively
$\widetilde{c}_{i}d_{i+k+1}=\widetilde{d}_{i}c_{i+2k+1}$.) These relations
hold for all $i$ and together imply Equation 38.

###  9.2 Extensions of Glick’s Formula

Let me first review Glick’s formula for $\Delta_{1}$, the pentagram map. Let
$P$ be a convex $n$-gon. Let $(x^{*},y^{*})$ denote the accumulation point of
the forward iterates of $P$ under $\Delta_{1}$. Let
$\widehat{P}_{\infty}=(x^{*},y^{*},1)$ be the collapse point. In somewhat
different notation, Glick introduces the operator

$T_{P}=nI_{3}-G_{P},\hskip
30.0ptG_{P}(v)=\sum_{i=1}^{n}\frac{|P_{i-1},v,P_{i+1}|}{|P_{i-1},P_{i},P_{i+1}|}P_{i}.$
(42)

Here $|a,b,c|$ denotes the determinant of the matrix with rows $a,b,c$ and
$I_{3}$ is the $3\times 3$ identity matrix. It turns out $T_{P}$ is a
$\Delta_{1}$-invariant operator, in the sense that $T_{\Delta_{0}(P)}=T_{P}$.
Moreover $P_{\infty}$ is an eigenvector of $T_{P}$. This is Glick’s formula
for $\widehat{P}_{\infty}$. Actually, one can say more simply that $G_{P}$ is
a $\Delta_{0}$-invariant operator and that $\widehat{P}_{\infty}$ is an
eigenvector of $G_{P}$. The more complicated expression $nI_{3}-G_{P}$ is
easier to work with geometrically.

Define $G_{P,a,b}$ by the formula

$G_{P,a,b}(v)=\sum_{i=1}^{n}\frac{|P_{i-a},v,P_{i+b}|}{|P_{i-a},P_{i},P_{i+b}|}P_{i}.$
(43)

Let $\widehat{P}_{\infty,k}$ be the limit point of the forward iterates of $P$
under $\Delta_{k}$. It seems that when $k\geq 1$ and $n=3k+1$ the operator
$G_{P,k,k}$ is $\Delta_{k}$ invariant and has $\widehat{P}_{\infty,k}$ for an
eigenvector. It seems that when $k\geq 1$ and $n=3k+2$ the operator
$G_{P,k+1,k+1}$ is $\Delta_{k}$ invariant and has $\widehat{P}_{\infty,k}$ for
an eigenvector. I was not able to find any similar formulas when $n>3k+2$.

Anton Izosimov kindly explained the following lemma.

###### Lemma 9.3

These operators are $\Delta_{k}$-invariant.

Proof: These operators are Glick’s operator in disguise. When $n=3k+1$ we can
relabel our $n$-gons in a way that converts $\Delta_{k}$ to the pentagram map.
The corresponding space of birds $B_{n,k}$ corresponds to some strange set of
“relabeled $k$-birds”. This relabeling converts $G_{P,k,k}$ respectively to
Glick’s original operator. This proves the invariance of $G_{P,k,k}$ under
$\Delta_{k}$ when $n=3k+1$. A similar thing works for $n=3k+2$, but this time
the relabeling converts $\Delta_{k}$ to the inverse of the pentagram map.
$\spadesuit$

This result does not explain why the collapse point $\widehat{P}_{\infty}$ is
an eigenvector, but there is plenty of food for thought here. Glick’s formula
is an analytic expression for the collapse point, and perhaps what is going on
here is some kind of analytic continuation. I wonder if this means that the
collapse point exists for all starting points of the pentagram map. Even if
the iterations go completely crazy under the map, perhaps they still collapse
to the point predicted by Glick’s operator. The idea of a completely general
collapse point has always seemed absurd to me, but maybe it is not. Nobody
knows. Even though the algebra of the pentagram is quite well understood, the
geometry is not.

###  9.3 Star Relabelings

Let us further take up the theme in the proof of Lemma 9.3. Given an $n$-gon
$P$ and and some integer $r$ relatively prime to $n$, we define a new $n$-gon
$P^{*r}$ by the formula

$P^{*r}_{j}=P_{rj}.$ (44)

Figure 1.5 shows the $P^{*(-3)}$ when $P$ is the regular $10$-gon.

As we have already mentioned, the action of $\Delta_{1}$ on the $P^{*(-k)}$ is
the same as the action of $\Delta_{k}$ on $P$ when $n=3k+1$. So, when
$n=3k+1$, the pentagram map has another nice invariant set (apart from the set
of convex $n$-gons), namely

$B_{k,n}^{*(-k)}=\\{P^{*(-k)}|\ P\in B_{k,n}\\}.$

The action of the pentagram map on this set is geometrically nice. If we
suitably star-relabel, we get star-shaped (and hence embedded) polygons. A
similar thing works when $n=3k+2$.

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