The Flapping Birds in the Pentagram Zoo

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].

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.

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.

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. As we mentioned above, we use the homotopic definition of a $k$-bird, to define the edges of $\Delta_{k}(P)$ when $P$ is a $k$-bird.

Example: The homotopy part of our definition looks a bit strange, but it is necessary. To illustrate this, we consider the picture further for the case $k=1$. In this case, a $1$-bird must be convex, though the $1$-niceness condition just means planar and locally convex. Figure 1.4 shows how we might attempt a homoropy from the regular octagon to a locally convex octagon which essentially wraps twice around a quadrilateral. The little grey arrows give hints about how the points are moved. At some times, the homotopy must break the $1$-niceness condition. The two grey polygons indicate failures and the highlighted vertices indicate the sites of the failures. There might be other failures as well; we are taking some jumps in our depiction.

One could make similar pictures when $k\geq 1$, but the pictures might be harder to understand.

Given an embedded planar polygon $P$, let $P^{I}$ denote the interior of region bounded by $P$. 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. More simply, we say that $P$ is strictly star shaped if it is strictly star shaped with respect to some point $x\in P^{I}$. Here is the main result.

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

We define

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 1.5 illustrates this for $(k,n)=(3,10)$. Note that the polygons suggested by the dots in Figure 1.5 are not convex. Were we to add in the edges we would get a highly non-convex pattern.

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

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.

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

We also let $\widehat{P}$ denote the closed planar region bounded by $P$. Figure 1.7 below shows $\widehat{P}=\widehat{P}^{0},\widehat{P}^{1},\widehat{P}^{2},\widehat{P}^{3},\widehat{P}^{4}$ for some $P\in B_{4,13}$.

Define

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.

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 for a discussion of this and related matters.

Here is a corollary of our results that is just about convex polygons.

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 1.8 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.

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. Our Theorem 7.2 will 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.

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.

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.

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 $n$.

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

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

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 $D_{k+1}(P)$.

Theorem 1.2 follows from the next two results.

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

We study the ratio

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

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

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

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

where $P_{0}$ is the regular $n$-Point. Hence $R(Q)=R(P_{0})=1$. This shows that Lemma 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.)

In view of Equation 4 we have

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

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.

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 $t_{5}$.

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.

The proof of Lemma 2.2 is essentially identical to the proof of Lemma 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.

This time the $4$ awake indices are:

Here is the pattern of zeros and poles.

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 Lemma 2.2.

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 $\widehat{P}$. (Eventually we will see that birds are embedded, so $\widehat{P}$ will be a closed topological disk and $P^{I}$ will be an open topological disk.)

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 $n$.

We now recall a definition from the introduction: 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.

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 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 $t\in[0,1)$.

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

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.

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.

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.

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.

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

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.

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.

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.

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 set $P=P(1)$, etc.

Statement 1: 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 $\partial F-e$.

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.

Statement 2: 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.

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.

Statement 3: 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.

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.

Let $B_{k,n}$ denote the space of $n$-gons which are $k$-birds. Let $\chi_{k}$ denote the $k$-energy. With the value of $k$ fixed in the background, we say that a degenerating path is a path $Q(t)$ of $n$-gons such that

1. 1.

   $Q(t)$ is planar for all $t\in[0,1]$.

2. 2.

   All vertices of $Q(t)$ are distinct for all $t\in[0,1]$.

3. 3.

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

4. 4.

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

In this chapter we will prove the following result, which will help us prove that $\Delta_{k}(B_{k,n})\subset B_{k,n}$ in the next chapter. The reader should probably just use the statement as a black box on the first reading.

Here $s$ ranges from $1$ to $0$ as $t$ ranges from $0$ to $1$. We have labeled some of the slopes to facility the calculation (which we leave to the reader) that $\chi_{1}(Q(t))$ remains uniformly bounded away from $0$.

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:

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

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.