Research Contribution DOI:10.56994/ARMJ.011.004.006
Received: 25 Dec 2024; Accepted: 9 Aug 2025


Spirals, tic-tac-toe partition, and deep diagonal maps

Zhengyu ZouDepartment of Mathematics,
Brown University
Providence, RI, USA
email: Zhengyu_Zou@brown.edu×
Abstact.
The deep diagonal map Tk acts on planar polygons by connecting the k-th diagonals and intersecting them successively. The map T2 is the pentagram map, and Tk is a generalization. We study the action of Tk on two subsets of the so-called twisted polygons, which we term type-α and type-β k-spirals. For k2, Tk preserves both types of k-spirals. In particular, we show that for k=2 and k=3, both types of k-spirals have precompact forward and backward Tk-orbits modulo projective transformations. We derive a rational formula for T3, which generalizes the y-variables transformation formula of the corresponding quiver mutation by M. Glick and P. Pylyavskyy. We also present four algebraic invariants of T3. These special orbits in the moduli space are partitioned into cells of a 3×3 tic-tac-toe grid. This establishes the action of Tk on k-spirals as a geometric generalization of T2 on convex polygons.

Mathematics Subject Classification:
05E99, 37J70, 51A05
Key words and phrases:
pentagram map, twisted polygons, cluster algebra, polygonal dynamics, projective geometry, integrable systems, Casimir functions, discrete dynamical systems

1 Introduction

1.1 Context and Motivation

Given a polygon P in the real projective plane, let Tk be the map that connects its k-th diagonals and intersects them successively to form another polygon P whose vertices are given by the following formula:

Pi=PiPi+kPi+1Pi+k+1. (1)

Figure 1 demonstrates an example of the action of T2 on a convex heptagon. The map T2 is called the pentagram map, a well-studied discrete dynamical system (see [20, 21, 23, 18]). A well-known result is that T2 preserves convexity.111A projective polygon is convex if some projective transformation maps it to a planar convex polygon in the affine patch The T2-orbit of a convex polygon sits on a flat torus in the moduli space of projective equivalent convex polygons. On the other hand, the geometry of the map Tk is less well-behaved. For k3, the Tk images of convex polygons may not even be embedded. See Figure 1 for an example of T3 taking a convex heptagon to a polygon that is not even embedded.

Refer to caption
Figure 1: Left: The iterative images of a convex heptagon under the action of T2. Right: A convex heptagon whose image under T3 is not even embedded.

Previous results of Tk often had an algebraic and combinatorial flavor, motivated by two branches of studies. The first one was a sequence of works [23, 18, 28, 19] that established that the T2 action on the moduli space of projective convex polygons is a discrete completely integrable system; the second one was M. Glick’s discovery in [5] of the connection between T2 and cluster algebras. In [4], M. Gekhtman, M. Shapiro, S. Tabachnikov, and A. Vainshtein generalized the cluster transformations in [5] to the map Tk acting on so-called “corrugated polygons,” which are polygonal curves in k satisfying certain coplanarity conditions. [4] showed that Tk is a discrete integrable system. There are numerous integrability results for these higher-dimensional analogs. See [13, 15, 16, 14, 11]. These led to many applications and connections of Tk to other fields, such as octahedral recurrence [23, 3], the condensation method of computing determinants [23, 6], cluster algebras [5, 4, 7, 3], Poisson Lie groups [2, 10], T-systems [12, 3], Grassmannians [1], algebraically closed fields [29], Poncelet polygons [22, 25, 9, 26], and integrable partial differential equations [23, 18, 17].

The geometric aspects of Tk and other deep diagonal maps on planar polygons remain underexplored. There are only a few studies on the geometries of Tk that focused on small k or polygons with many symmetries. See [25, 26]. There is no established general framework on the type of geometric properties preserved under Tk for k3 that is analogous to convexity under T2. Even less is known for geometric objects that have precompact orbits under Tk.

The most relevant result to this endeavor is the discovery of k-birds under the map Δk in [27]. A k-bird P is a planar n-gon with n>3k, such that there exists a continuous path of polygons P(t) connecting P to the regular n-gon where the four lines

Pi(t)Pik1(t),Pi(t)Pik(t),Pi(t)Pi+k(t),Pi(t)Pi+k+1(t)

are distinct for all i=1,,n and tI. The map Δk connects the (k+1)-th diagonal of a polygon and intersects the diagonals that are k clicks apart. See Figure 2 for the action of Δ2 on 2-birds. In [27], R. Schwartz showed that the k-birds are invariant under both Δk and Δk1. Experimentally, the k-birds seem to have toroidal orbits under Δk, which highly resembles the orbit of convex n-gons under T2. Schwartz also showed that the k-birds have precompact forward Δk-orbits modulo affine transformations—a property satisfied by convex n-gons under T2.

Refer to caption
Figure 2: Action of Δ2 on two heptagons that are 2-birds.

This paper has two main results. The first one is the discovery of two classes of geometric objects called type-α and type-β k-spirals that are preserved under Tk for all k2. These two classes of objects are subsets of twisted polygons: bi-infinite sequences P:2 such that no three consecutive points are collinear, and Pi+n=ϕ(Pi) for some fixed projective transformation ϕ called the monodromy. The moduli space of projective equivalent twisted n-gons is conventionally denoted by 𝒫n. The type-α and type-β k-spirals are the first discovered classes of geometric constructions of Tk that generalize the pentagram map, which provides crucial evidence for a more general understanding of geometrically preserved classes under Tk.

The second result is the precompactness of both forward and backward Tk-orbits of type-α and type-β k-spirals modulo projective transformations for k=2 and 3, a key property satisfied by convex polygons under the pentagram map discovered by Schwartz in [20]. We first examine the action of T3 on type-α and type-β 3-spirals. We show that one can characterize type-α and type-β 3-spirals via linear constraints on the corner invariants. We also derive a birational formula of T3 for the corner invariants, which is a generalization of the combinatorial formulas developed by [7]. Then, we present four global invariants under T3, which we use to prove the precompactness of T3-orbits modulo projective transformations. For the case k=2, we show that there exists no type-α 2-spirals and that the type-β 2-spirals are distinct from closed convex polygons. We use the Casimir functions of the T2-invariant Poisson structure developed in [23] and [18] to show that type-β 2-spirals have precompact T2-orbits modulo projective transformations.

1.2 The k-Spirals under the Map Tk

Here we describe the geometric picture of a k-spiral. For the formal definition, see §3.1. Geometrically, [P]𝒫n is a k-spiral if for all N, we can find a representative P such that {Pi}iN lies on the affine patch, and the triangles (Pi,Pi+1,Pi+2) and (Pi,Pi+1,Pi+k) have positive orientation for all iN. We call P an N-representative of [P].

Refer to caption
Figure 3: A gallery of 5-spirals. Left: 𝒮5,3α. Middle: 𝒮5,3β. Right: 𝒮5,20β. The red-shaded triangles indicate the defining orientations and containment relations of type-α and type-β k-spirals..

We are mainly interested in two types of k-spirals, which we term type-α and type-β (although there certainly exist many more types of spirals, we only consider these two types here). They are k-spirals with additional constraints on the arrangement of the four points Pi, Pi+1, Pi+k, Pi+k+1. For type-α spirals, we require Pi+k to be contained in the interior of the triangle (Pi,Pi+1,Pi+k+1). For type-β spirals, Pi+k+1 needs to be contained in the interior of (Pi,Pi+1,Pi+k). We say P is a type-α or type-β N-representative. A class of twisted polygons [P] is a type-α k-spiral (resp. β) if and only if it admits a type-α (resp. β) N-representative for all N. Let 𝒮k,nα and 𝒮k,nβ denote the space of type-α and type-β k-spirals modulo projective equivalence. We will see in §3.1 that they are both open in 𝒫n and hence have dimension 2n. Figure 3 illustrates three examples of representatives of 𝒮5,nα for n=3, and 20.

It turns out that 𝒮k,nα and 𝒮k,nβ are invariant under both Tk and Tk1. Figure 4 shows the inward half of a representative P of [P]𝒮5,3β, with the red arc representing P=T5(P). On the right we have five polygonal arcs by joining vertices of P that are 5 clicks apart. We call them the transversals of P. One way to distinguish type-α and type-β spirals is by looking at the orientations of transversals. The transversals of type-α spirals are counterclockwise, whereas those of type-β are clockwise (See Figure 11). In §3, we use the orientations of these transversals to prove the following main theorem.

Theorem 1.1.

For all n2 and k2, we have Tk(𝒮k,nα)=𝒮k,nα. The same is true for type-β.

Refer to caption
Figure 4: Left: T5 acting on a representative P of [P]𝒮5,3β. Right: Transversals of P.

A key property satisfied by convex polygons under the pentagram map is that the forward and backward orbits of any convex polygon under the pentagram map are precompact modulo projective tranformations. See [20, Lemma 3.2]. Experimental results suggest that the k-birds also have precompact Δk-orbits. In [27, Conjecture 8.2] Schwartz conjectured that the k-birds have precompact forward Δk-orbits modulo affine transformations. We observed experimentally that 𝒮k,nα and 𝒮k,nβ behave analogously under Tk.

Conjecture 1.2.

For n2 and k2, the forward and backward Tk-orbit of any [P]𝒮k,nα is precompact in 𝒫n. The same holds for type-β.

In §6 and §7, we prove Conjecture 1.2 for k=2 and k=3.

1.3 Tic-Tac-Toe Partition and Precompact T3 Orbits

Our main focus will be the case k=3, which we prove in §6.2.

Theorem 1.3.

For n2, the forward and backward T3-orbit of any [P]𝒮3,nα is precompact in 𝒫n. The same holds for type-β.

We discovered several interesting properties of the two types of k-spirals and the map T3 along our way to prove Theorem 1.3. One major discovery is that the sets 𝒮3,nα and 𝒮3,nβ fit well with a local parameterization of 𝒫n2n introduced by [20] called corner invariants (See §2.4). The invariant sets of 𝒫n under T3 are partitioned by linear boundaries in the parameter space. The boundary lines give a grid pattern that resembles the board of the game “tic-tac-toe.” Each of the four “side-squares” is invariant under T3.

Refer to caption
Figure 5: The partition of 2 into a 3×3 grid, and the four side-squares of our interest.

To construct the tic-tac-toe board, consider the three intervals I,J,K of given by I=(,0), J=(0,1), K=(1,). The squares are of the form I×I, I×J, I×K, J×I, etc.. We mark the four side-squares Sn(I,J), Sn(J,I), Sn(K,J), Sn(J,K). See Figure 5 for a visualization of the tic-tac-toe grid. Given [P]𝒫n, we say [P]Sn(I,J) if all even corner invariants of [P] are in I, and all odd ones are in J. This means if we plot all n pairs of corner invariants (x2i,x2i+1) onto 2, we would see n points lying in I×J. The other three side squares are defined analogously.

Refer to caption
Figure 6: Left: T3 acting on a representative of [P]S4(K,J). Right: The orbit of P3(m) in 𝔸2 by fixing P2=(0,0), P1=(1,0), P0=(1,1), P1=(0,1).

Figure 6 shows vertices of a representative P of [P]S4(K,J) and the image P=T3(P). On the right, we have the projection of the first 211 iterations of the orbit of P under T3. Each point corresponds to P3(m) after normalizing (P2(m),P1(m),P0(m),P1(m)) to the unit square (here P(m)=T3m(P)). We speculate that the orbit lies on a flat torus, where the map T3 acts as a translation on the flat metric.

Twisted polygons that are assigned to these squares have geometric properties. For example, the closed convex polygons always lie in the center square; two of the side-squares are 𝒮3,nα and 𝒮3,nβ; the other two side-squares are obtained by reverting the indexing of vertices of these two types of k-spirals. These facts will be proved in §4.

The proof of Theorem 1.3 is algebraic. In §5 I show that T3 is a birational map on the corner invariants, which generalizes a direct application of [7, Theorem 1.6]. For the explicit formulas, see Equation (19). In §6, I derive four algebraic invariants of T3, which allow me to show boundedness of the corner invariants of the T3-orbits, thereby proving Theorem 1.3. This approach is reminiscent of Schwartz’s second proof of precompactness of T2-orbits of convex polygons in [21, Section 3B & 3C].

1.4 The Type-β 2-Spirals and Precompact T2 Orbits

We now proceed to the case k=2, where the map T2 is the classical pentagram map. In §3.1 we show that there exist no type-α 2-spirals (so Conjecture 1.2 is vacuously true for type-α 2-spirals). On the other hand, type-β 2-spirals are nontrivial geometric constructions that are distinct from convex polygons. In §7.1, we show that the corner invariants of type-β 2-spirals are also partitioned by linear boundaries, and in particular 𝒮3,nα𝒮2,nβ.

We point out that the type-β 2-spirals are not related to the pentagram spirals in [24]. The latter requires P to be a relabeling of T2m(P) for some positive integer m.

In §7.2, we use the Casimir functions of the T2-invariant Poisson structure on 𝒫n from [23] and [18] to prove Conjecture 1.2 for k=2.

Theorem 1.4.

For n2, the forward and backward T2-orbit of any [P]𝒮2,nβ is precompact in 𝒫n.

1.5 Obstacles for k>3 and Future Directions

Our algebraic method of proving Theorem 1.3 and 1.4 requires a complete characterization of the corner invariants of 𝒮k,nα and 𝒮k,nβ and enough algebraic invariants of Tk that uniformly bound the corner invariants away from the boundaries of 𝒮k,nα and 𝒮k,nβ. However, the corner invariants seem to be not partitioned by linear boundaries for k>3, which makes it difficult to analyze the boundaries of the corner invariants of 𝒮k,nα and 𝒮k,nβ. Moreover, the map Tk for the corner invariants seems not birational from computer algebra. This makes it difficult to algebraically characterize the corner invariants.

One future direction is to look at the cross-ratio of different combinations of points other than the ones involved in the definition of corner invariants. In §8 we present a conjecture on a potential algebraic invariant of Tk, which can be interpreted as a Casimir function of a Poisson structure over the y-parameteris of a quiver QS. The quiver QS is associated to a Y-mesh of type S from [7] and is isomorphic to the quiver in [4], which corresponds geometrically to the map Tk.

Another direction is to analyze the two types of k-spirals geometrically. There are yet many open problems on the geometry of the two types of k-spirals that could hint at the behavior of their Tk-orbits. For open problems, see the end of §3.1. Answering these geometric problems may provide a new approach to tackle Conjecture 1.2.

Finally, for the case k=3, the birational formula for T3 could be applied to other settings such as the action of T3 on Poncelet polygons [26] or discovering T3-compatible Poisson structures on 𝒫n that generalizes the one in [4] for corrugated polygons.

1.6 Accompanying Program

I wrote a web-based program to visualize the orbits of twisted polygons under Tk. Readers can access it from the following link:

https://zzou9.github.io/pentagram-map/spiral.html

When reaching the website, you will see a representative of a twisted polygon displayed in the middle of the screen. You can click on the user manual button for instructions on how to use the program. I discovered most of the results by computer experiments using this program. The paper contains rigorous proofs of the beautiful pictures I observed from it.

1.7 Acknowledgements

This work was supported by a Brown University SPRINT/UTRA summer research program grant. I would like to thank my advisor, Richard Evan Schwartz, for introducing the concept of deep diagonal maps, providing extensive insights throughout the project, and offering guidance during the writing process. I would like to thank Anton Izosimov and Sergei Tabachnikov for their insightful discussions on the tic-tac-toe partition. Finally, I am grateful to the referees for their helpful comments and for highlighting the connections between the birational formula and the work of Glick and Pylyavskyy.

2 Background

2.1 Projective Geometry

The real projective plane 2 is the space of 1-dimensional subspaces of 3. Points of 2 are lines in 3 that go through the origin. We say that [x:y:z] is a homogeneous coordinate of V2 if the vector V~=(x,y,z) is a representative of V. Given two distinct points V1,V22, the line l=V1V2 connecting V1 and V2 is the 2-dimensional hyperplane spanned by the two 1-dimensional subspaces. Let l1,l2 be two lines in 2. The point of intersection l1l2 is the 1-dimensional line given by the intersection of the two 2-dimensional subspaces. In 2, there exists a unique line connecting each pair of distinct points and a unique point of intersection given two distinct lines. We call a collection of points V1,V2,,Vn2 in general position if no three of them are collinear.

The affine patch 𝔸2 consists of points in 2 with homogeneous coordinate [x:y:1]. We call this canonical choice of coordinate (x,y,1) the affine coordinate of a point V𝔸2. There is a diffeomorphism Φ:2𝔸2 given by Φ(x,y)=[x:y:1]. We often identify 𝔸2 as a copy of 2 in 2. The line 2𝔸2 is called the line at infinity.

A map ϕ:22 is a projective transformation if it maps points to points and lines to lines and is bijective. Algebraically, the group of projective transformations is PGL3()=GL3()/I, where we are modding by the subgroup I={λI:λ} and I is the 3×3 identity matrix. We state a classical result regarding projective transformations below with its proof omitted.

Theorem 2.1.

Given two 4-tuples of points (V1,V2,V3,V4) and (W1,W2,W3,W4) in 2, both in general position, there exists a unique ϕPGL3() such that ϕ(Vi)=Wi.

The group of affine transformations Aff2() on 𝔸2 is the subgroup of projective transformations that fixes the line at infinity. It is isomorphic to a semidirect product of GL2() and 2. Elements of Aff2() can be uniquely expressed as a tuple (M,v) where MGL2() and v2. Let Aff2+() denote the subgroup of Aff2+() where (M,v)Aff2+() iff det(M)>0. These are orientation-preserving affine transformations.

2.2 Orientation of Affine Triangles

Given an ordered 3-tuple (V1,V2,V3) of points in 2 or 𝔸2, let int(V1,V2,V3) denote the interior of the affine triangle with vertices V1,V2,V3. There is a canonical way to define the orientation of an ordered 3-tuple. Let V~i be the affine coordinate of Vi. We consider the signed area 𝒪(V1,V2,V3) of the oriented triangle, which can be computed as

𝒪(V1,V2,V3)=det(V~1,V~2,V~3). (2)

The determinant is evaluated on the 3×3 matrix with column vectors V~i. We say an ordered 3-tuple (V1,V2,V3) is positive if 𝒪(V1,V2,V3)>0. Figure 7 shows an example of a positive 3-tuple.

Refer to caption
Figure 7: A positive 3-tuple of affine points (V1,V2,V3).

Here is another way to compute 𝒪 using the 2 coordinates of V1,V2,V3:

𝒪(V1,V2,V3) =det(V1,V2)+det(V2,V3)+det(V3,V1) (3)
=det(ViVi1,Vi+1Vi) for i=1,2,3

where the determinant is evaluated on the 2×2 matrix.

𝒪 interacts with the action of Aff2+() and the symmetric group S3 on planar/affine triangles in the following way: Given MAff2+(), let Vi=M(Vi). One can show that (V1,V2,V3) is positive iff (V1,V2,V3) is positive. On the other hand, for all σS3, 𝒪(Vσ(1),Vσ(2),Vσ(3))=sgn(σ)𝒪(V1,V2,V3), so 𝒪(Vσ(1),Vσ(2),Vσ(3))=𝒪(V1,V2,V3) when σ is a 3-cycle.

Below are useful equivalence conditions for the positivity of (V1,V2,V3). The proof is elementary, so we will omit it.

Proposition 2.2.

Given V1,V2,V32 in general position, and Wint(V1,V2,V3), the following are equivalent:

  1. 1.

    (V1,V2,V3) is positive.

  2. 2.

    (Vi,Vi+1,W) is positive for some i{1,2,3}.

  3. 3.

    (Vi,Vi+1,W) is positive for all i{1,2,3}.

  4. 4.

    det(ViVi1,Vi+1W)>0 for some i{1,2,3}.

  5. 5.

    det(ViVi1,Vi+1W)>0 for all i{1,2,3}.

2.3 The Cross-Ratio

The cross-ratio is used to construct a projective-invariant parametrization of the k-spirals. There are multiple ways to define the cross-ratio of four collinear points on the projective plane, each using its own permutation of the points. We follow the convention used in [20]. Given four collinear points A,B,C,D on a line ω2, we choose a projective transformation ψ that maps ω to the x-axis of 𝔸2. Let a,b,c,d be the x-coordinates of ψ(A), ψ(B), ψ(C), ψ(D). We define the cross-ratio to be the following quantity:

χ(A,B,C,D):=(ab)(cd)(ac)(bd). (4)

If A lies on the line at infinity, we let χ(A,B,C,D)=cdbd. One can check that given any ϕPGL3(),

χ(A,B,C,D)=χ(ϕ(A),ϕ(B),ϕ(C),ϕ(D)).

We also define the cross-ratio for four projective lines. Let l,m,n,k be four lines intersecting at a common point O. Normalize with a projective transformation so that l,m,n,k𝔸2 with slopes sl,sm,sn,sk. We define

χ(l,m,n,k)=(slsm)(snsk)(slsn)(smsk) (5)

with χ(l,m,n,k)=snsksmsk if sl=.

If ω is a line that does not go through O and intersects l,m,n,k at A,B,C,D respectively, we have

χ(l,m,n,k)=χ(A,B,C,D). (6)

See Figure 8 for the configuration. The proof is elementary, so we will omit it.

Refer to caption
Figure 8: The configuration in Equation (6).

2.4 Twisted Polygons, Corner Invariants

Introduced in [23], a twisted n-gon is a bi-infinite sequence P:2, along with a projective transformation MPGL3() called the monodromy, such that every three consecutive points of P are in general position, and Pi+n=M(Pi) for all i. When M is the identity, we get an ordinary closed n-gon. Two twisted n-gons P,Q are equivalent if there exists ϕPGL3() such that ϕ(Pi)=Qi for all i. The two monodromies Mp and Mq satisfy Mq=ϕMpϕ1. Let 𝒫n denote the space of twisted n-gons modulo projective equivalence.

The cross-ratio allows us to parameterize 𝒫n with coordinates in 2n. Given a twisted n-gon P, the corner invariants of P is a coordinate system x0(P),,x2n1(P) given by

{x2i(P)=χ(Pi2,Pi1,Pi2Pi1PiPi+1,Pi2Pi1Pi+1Pi+2);x2i+1(P)=χ(Pi+2,Pi+1,Pi+2Pi+1PiPi1,Pi+2Pi+1Pi1Pi2). (7)
Refer to caption
Figure 9: Left: The corner invariants x2i(P)=χ(Pi2,Pi1,A,O) computed using Equation (7). Right: x2i(P)=χ(l1,2,l1,1,l1,0,l1,2) computed using Equation (8).

See the left side of Figure 9 for a geometric interpretation of the corner invariants. Let la,b=Pi+aPi+b. By Equation (5), the corner invariants can be computed by

{x2i(P)=χ(l1,2,l1,1,l1,0,l1,2);x2i+1(P)=χ(l1,2,l1,1,l1,0,l1,2). (8)

See the right side of Figure 9 for the line configurations.

Since χ is invariant under projective transformations, for all j we have xj(P)=xj+2n(P), so a 2n-tuple of corner invariants is enough to fully determine the projective equivalence class of a twisted n-gon. We use xj(P) to denote the corner invariants of [P]𝒫n without adding square brackets around P. To obtain the corner invariants of [P]𝒫n, one can simply choose an arbitrary representative P and compute its corner invariants. [23, Equation (19) & (20)] showed that one can also revert the process and obtain a representative twisted polygon of the equivalence class given its corner invariants.

3 The Spirals and Tk-Orbit Invariance

In this section, we explore the geometric properties of type-α and type-β k-spirals and prove Theorem 1.1. In §3.1, we give rigorous definitions of the two types of k-spirals and discuss their geometric properties. In §3.2, we introduce a construct associated to the two types of k-spirals called the transversals. In §3.3 and §3.4, we prove Theorem 1.1 using geometric properties of the transversals.

3.1 The Geometry of k-Spirals

Here we give the formal definition of a k-spiral and its two subsets called type-α and type-β. We then explore their geometric properties and present some open problems.

Definition 3.1.

Given integers k2, n2, we say that [P]𝒫n is a k-spiral if for all N, there exists a representative P that satisfies the following: For all iN, Pi lies in 𝔸2, (Pi,Pi+1,Pi+2) is positive, and (Pi,Pi+1,Pi+k) is positive. Such a representative is called an N-representative. Saying that [P] is a k-spiral means that [P] admits an N-representative for all N.

Remark 3.2.

The idea of considering an N-representative for each N is new to the literature and may at first seem superfluous. Readers will see in §4 that this condition is natural when we examine the corner invariants of the two types of k-spirals. See the end of this section for open problems related to the geometry of N-representatives.

In practice, since [P] is a twisted n-gon, it suffices to find a single N0-representative P0 for some N0. One can then obtain other N-representatives for N<N0 by applying the m-th power of the monodromy of [P] to P0, where m>N0Nk+1.

Definition 3.3.

A k-spiral [P]𝒫n is of type-α or type-β if for all N, it has an N-representative P that satisfies the following conditions:

  • [P] is of type-α if Pi+kint(Pi,Pi+1,Pi+k+1) for all iN;

  • [P] is of type-β if Pi+k+1int(Pi,Pi+1,Pi+k) for all iN.

Refer to caption
Figure 10: Left: The inward half of a 0-representative P of a type-α 6-spiral. The red triangle is joined by (Pi,Pi+1,Pi+k+1), which is positive by Proposition 3.4 and contains Pi+k in its interior. Right: The inward half of a 0-representative P of a type-β 6-spiral. The cyan triangle is joined by (Pi,Pi+1,Pi+k), which is positive and contains Pi+k+1 in its interior.

See Figure 10 for 0-representatives of type-α and type-β 6-spirals. For the type-α k-spirals, we show that positivity of (Pi,Pi+1,Pi+k) is equivalent to positivity of (Pi,Pi+1,Pi+k+1). The latter condition turns out to be more convenient for showing Tk invariance.

Proposition 3.4.

[P]𝒫n is a type-α k-spiral if and only if for all N, there exists a representative P that satisfies the following: for all iN, Pi lies in 𝔸2, (Pi,Pi+1,Pi+2) is positive, (Pi,Pi+1,Pi+k+1) is positive, and Pi+kint(Pi,Pi+1,Pi+k+1).

Proof.

Since Pi+kint(Pi,Pi+1,Pi+k+1), we see that int(Pi,Pi+1,Pi+k+1) is nonempty, so the three points Pi, Pi+1, Pi+k+1 are in general position. It then follows from Proposition 2.2 that (Pi,Pi+1,Pi+k) is positive iff (Pi,Pi+1,Pi+k+1) is positive. ∎

Corollary 3.5.

There exists no type-α 2-spirals.

Proof.

It suffices to show that there exists no configuration of four points A,B,C,D𝔸2 such that (A,B,D), (B,C,D) are both positive and Cint(A,B,D). If (A,B,D) is positive and Cint(A,B,D), then Proposition 2.2 implies (B,D,C) is positive, but that contradicts (B,C,D) positive because 𝒪(B,C,D)=𝒪(B,D,C). ∎

On the other hand, type-β 2-spirals do exist. Geometrically, their N-representatives look like triangular spirals. See §7 for a more thorough discussion on type-β 2-spirals.

Remark 3.6.

One may attempt to define the two types of k-spirals on bi-infinite sequences of points in 2 with no periodicity constraints. The results in this section hold true for this more general definition. We restrict our attention to twisted polygons because it’s a finite-dimensional space, which allows us to more easily keep track of the Tk-orbits.

We now proceed to discuss some geometric properties of type-α and type-β k-spirals. A twisted polygon P is called k-nice if the four points Pi,Pi+1,Pi+k,Pi+k+1 are in general position for all i. The k-nice condition is projective invariant. Let 𝒫k,n denote the space of k-nice twisted n-gons modulo projective equivalence.

Proposition 3.7.

For all k2, 𝒫k,n is open in 𝒫n, so it has dimension 2n.

Proof.

The condition that four points Pi,Pi+1,Pi+k,Pi+k+1 are in general position remains true if we perturb one of the points in a small enough neighborhood of 2. The dimension of 𝒫k,n comes from the fact that 𝒫n has dimension 2n, which is shown in [18, Lemma 2.2]. ∎

Proposition 3.8.

Both type-α and type-β k-spirals are k-nice.

Proof.

We give a proof to the type-α case. The type-β case is analogous, so we will omit it. Given a type-α k-spiral [P] and an integer i, let P be an i-representative of [P]. Since (Pi,Pi+1,Pi+k+1) is positive, these three points cannot be collinear. Also, since Pi+kint(Pi,Pi+1,Pi+k+1), Pi+k does not lie in any of the lines joined by two of the three vertices Pi,Pi+1,Pi+k+1. This shows that Pi,Pi+1,Pi+k,Pi+k+1 are in general position. ∎

As stated in §1.2, we let 𝒮k,nα and 𝒮k,nβ denote the space of type-α and type-β k-spirals (By Corollary 3.5, 𝒮2,nα= for all n2 ).

Proposition 3.9.

Both 𝒮k,nα and 𝒮k,nβ are open in 𝒫k,n, so they both have dimension 2n.

Proof.

The positivity conditions of (Pi,Pi+1,Pi+2) and (Pi,Pi+1,Pi+k) are open conditions from continuity of the determinant function. The condition Pi+kint(Pi,Pi+1,Pi+k+1) for type-α (or Pi+k+1int(Pi,Pi+1,Pi+k) for type-β) is equivalent to the positivity of certain determinants by Proposition 2.2, so this is also an open condition. Finally, 𝒮k,nα𝒫k,n and 𝒮k,nβ𝒫k,n follows from Proposition 3.8. ∎

A twisted polygon P is closed if there exists some positive integer n such that Pi+n=Pi, or [P]𝒫n with identity monodromy. We show that neither type-α nor type-β k-spirals are closed.

Proposition 3.10.

For all k2 and n2, if [P]𝒮k,nα, then [P] is not closed. The same holds for 𝒮k,nβ.

Proof.

Given any closed n-gon P on 𝔸2, let C be the convex hull of the vertices of P. Since P has finitely many vertices, there exists a vertex Pi such that Piint(C). Then, since int(Pik,Pik+1,Pi+1)int(C), we must have Piint(Pik,Pik+1,Pi+1). It follows that P is not an N-representative of type-α k-spiral for any N or k. The proof for type-β is similar, so we omit it. ∎

The two types of k-spirals seem to possess rich geometric properties. We will present some open problems. In the discussion below, [P] denotes a type-α or type-β k-spiral.

Problem 3.11.

For all N, is it always possible to find N-representatives P such that for all j>i+1, (Pi,Pi+1,Pj) is positive (in other words, Pj always lies on the same side of the line PiPi+1)?

Problem 3.12.

Let P be an arbitrary representative of [P]. Is there a minimal N such that P is an N-representative on some affine patch of 2? Does there exist P that is an N-representative for all N?

Problem 3.13.

Given an N-representative P, does Pi converge to a point in 𝔸2 as i?

3.2 Transversals of the Spirals

In this section, we prove our remark in §1.2 that transversals for type-α spirals are oriented counterclockwise, whereas transversals for type-β are oriented clockwise. Recall that the transversals of an N-representative P of a k-spiral are k polygonal arcs joined by vertices Pi,Pi+k,Pi+2k, for i=N,,N+k1. See Figure 11 for one of the k transversals of the two representatives from Figure 10.

Refer to caption
Figure 11: Transversals of two representatives from Figure 10.
Lemma 3.14.

Given O,A,B,C,D𝔸2 (See Figure 12) such that (A,O,B), (A,O,D), (B,O,C), (C,O,D) are all positive. Then, (A,O,C) is positive iff (B,O,D) is positive.

Proof.

For the forward direction, normalize with Aff2+() so that O=(0,0) and A=(1,0). Let B=(xb,yb), C=(xc,yc), and D=(xd,yd). Since (A,O,B) is positive, Equation (3) gives us

𝒪(A,O,B)=det(OA,BO)=det(A,B)=yb>0.

Similarly, positivity of (A,O,C) and (A,O,D) give us yc,yd>0. Next, observe that

𝒪(B,O,C)=det(B,C)=xbyc+xcyb;
𝒪(B,O,D)=det(B,D)=xbyd+xdyb;
𝒪(C,O,D)=det(C,D)=xcyd+xdyc.

Since yb,yc,yd>0, we have ybyc,ydyc>0, which implies

𝒪(B,O,D)=xbyd+xdyb=ybyc𝒪(C,O,D)+ydyc𝒪(B,O,C)>0.

This shows positivity of (B,O,D).

The proof for the backward direction is analogous. Normalize so that O=(0,0) and D=(1,0). Let A=(xa,ya), B=(xb,yb), C=(xc,yc). Positivity of (A,O,D), (B,O,D), and (C,O,D) implies ya,yb,yc>0. One can then check that

𝒪(A,O,C)=xayc+xcya=ycyb𝒪(A,O,B)+yayb𝒪(B,O,C)>0.

This shows positivity of (A,O,C). ∎

Refer to caption
Figure 12: Examples of O,A,B,C,D in Lemma 3.14.

The next proposition formalizes our claim on the orientation of transversals.

Proposition 3.15.

Let P be an N-representative of a k-spiral [P]. For all i>N, if [P] is type-α, then (Pi,Pi+k,Pi+2k) is positive; if [P] is type-β, then (Pi+2k,Pi+k,Pi) is positive.

Refer to caption
Figure 13: Left: 𝒮k,nα configuration. Right: 𝒮k,nβ configuration.
Proof.

The proof applies Lemma 3.14 with suitable choices of O,A,B,C,D. See Figure 13 for the configuration of points involved.

We start with P of type-α. Consider the following choices of vertices:

O=Pi+k;A=Pi;B=Pi+k1;C=Pi+2k;D=Pi+k+1.

It follows immediately from the definition of a type-α N-representative that (B,O,C) and (B,O,D) are positive. The other conditions follow from applications of Proposition 2.2. Apply Proposition 2.2 with (Pi1,Pi,Pi+k) positive and Pi+k1int(Pi1,Pi,Pi+k) to get positivity of (A,O,B). Apply Proposition 2.2 with (Pi,Pi+1,Pi+k+1) positive and Pi+kint(Pi,Pi+1,Pi+k+1) to get positivity of (A,O,D). Apply Proposition 2.2 with (Pi+k,Pi+k+1,Pi+2k+1) positive and Pi+2kint(Pi+k,Pi+k+1,Pi+2k+1) to get positivity of (C,O,D). Then, the backward direction of Lemma 3.14 implies (Pi,Pi+k,Pi+2k) is positive.

The proof for type-β is analogous. Consider the following choices of vertices:

O=Pi+k;A=Pi+k1;B=Pi+2k;C=Pi+k+1;D=Pi.

Positivity of (A,O,C) and (B,O,C) follows from the definition of a type-β N-representative. A similar application of Proposition 2.2 as in the case of type-α gives positivity of (A,O,B), (A,O,D), and (C,O,D), which we will omit. Finally, the forward direction of Lemma 3.14 implies (Pi+2k,Pi+k,Pi) is positive. ∎

3.3 Invariance of Forward Orbit

In this section, we prove that 𝒮k,nα and 𝒮k,nβ are Tk-invariant. We will use Equation (1) for our labeling convention. See Figure 14.

Refer to caption
Figure 14: The labeling convention of the map Tk from Equation (1).

If P is k-nice, then P is always well-defined. In particular, Proposition 3.8 implies Tk is well-defined on 𝒮k,nα and 𝒮k,nβ.

Remark 3.16.

Tk doesn’t necessarily send k-nice twisted polygons to k-nice twisted polygons. Here is an example provided by the anonymous referee: Fix r(0,1). Consider the function P:2 mapping zrzexp(zπi/k). One can check that P is a k-nice twisted n-gon for any n2 with monodromy that is a scale-rotation, but Tk(P) is the zero function and hence not k-nice. What we will show is that in the case of type-α and type-β k-spirals, Tk does preserve k-niceness. This is a direct consequence of Theorem 1.1 and Proposition 3.8.

We proceed to prove the Tk-invariance of 𝒮k,nα and 𝒮k,nβ separately. We start with the following lemma.

Lemma 3.17.

Given four points A,B,C,D in 2 in general position with Dint(A,B,C). Let O=ABCD. There exist s(0,1) and t(1,) such that

O=(1s)A+sB=(1t)C+tD.
Proof.

Since Dint(A,B,C), there exists λ1,λ2,λ3(0,1) such that

λ1+λ2+λ3=1;D=λ1A+λ2B+λ3C.

Taking s=λ21λ3 and t=11λ3 gives us the desired result. ∎

Proposition 3.18.

For all k2 and n2, Tk(𝒮k,nα)𝒮k,nα.

Proof.

Given an N-representative P of some [P]𝒮k,nα, we will show that P=Tk(P) is a type-α N-representative of [Tk(P)] by proving that for all iN, (Pi,Pi+1,Pi+2) is positive, (Pi,Pi+1,Pi+k+1) is positive, and Pi+kint(Pi,Pi+1,Pi+k+1). See the left side of Figure 15 for configurations of relevant vertices of P and P.

Let iN be fixed. Since P is a type-α N-representative, Pj+kint(Pj,Pj+1,Pj+k+1) for all jN. Applying Lemma 3.17 with Equation (1) on Pj for j{i,i+1,i+2,i+k,i+k+1} gives us

Pi=(1s1)Pi+1+s1Pi+k+1; Pi+1=(1t1)Pi+1+t1Pi+k+1; (9)
Pi+1=(1s2)Pi+2+s2Pi+k+2; Pi+2=(1t2)Pi+2+t2Pi+k+2;
Pi+k=(1s3)Pi+k+1+s3Pi+2k+1; Pi+k+1=(1t3)Pi+k+1+t3Pi+2k+1,

where s1,s2,s3(0,1) and t1,t2,t3(1,). In particular, this shows Pi+k+1PiPi+1, so the three points Pi,Pi+1,Pi+k+1 are in general position.

To see that (Pi,Pi+1,Pi+2) is positive, Equation (3) and (9) give us

𝒪(Pi,Pi+1,Pi+2) =det(Pi+1Pi,Pi+2Pi+1) (10)
=det((s1t1)Pi+1+(t1s1)Pi+k+1,(s2t2)Pi+2+(t2s2)Pi+k+2)
=(t1s1)(t2s2)det(Pi+k+2Pi+2,Pi+1Pi+k+1).

Then, since 𝒪(Pi+1,Pi+2,Pi+k+2)>0 and Pi+k+1int(Pi+1,Pi+2,Pi+k+2), Proposition 2.2 implies det(Pi+k+2Pi+2,Pi+1Pi+k+1)>0, so 𝒪(Pi,Pi+1,Pi+2)>0.

Next, we show that Pi+kint(Pi,Pi+1,Pi+k+1). Let r1=1s1t1s1 and r2=s3t3. (9) implies r1,r2(0,1) and

Pi+k =(1s3)Pi+k+1+s3Pi+2k+1
=(1s3)(t11)t1s1Pi+(1s3)(1s1)t1s1Pi+1+s3(t31)t3s3Pi+k+s3(1s3)t3s3Pi+k+1.

It follows that

Pi+k =t3s3t3(s31)((1s3)(t11)t1s1Pi+(1s3)(1s1)t1s1Pi+1+s3(1s3)t3s3Pi+k+1)
=(t3s3)(1t1)t3(t1s1)Pi+(t3s3)(s11)t3(t1s1)Pi+1+s3t3Pi+k+1
=(1r2)(1r1)Pi+(1r2)r1Pi+1+r2Pi+k+1.

Observe that the coefficients (1r2)(1r1), (1r2)r1, r2 are all in (0,1) and sum up to 1, so Pi+kint(Pi,Pi+1,Pi+k+1).

Finally, using Equation (3) and (9), we have

det(Pi+1Pi,Pi+k+1Pi+k) =det((t1s1)(Pi+k+1Pi+1),(t3s3)(Pi+2k+1Pi+k+1)) (11)
=(t1s1)(t3s3)det(Pi+k+1Pi+1,Pi+2k+1Pi+k+1)
=(t1s1)(t3s3)𝒪(Pi+1,Pi+k+1,Pi+2k+1).

Proposition 3.15 implies 𝒪(Pi+1,Pi+k+1,Pi+2k+1)>0, so det(Pi+1Pi,Pi+k+1Pi+k)>0 Since Pi,Pi+1,Pi+k+1 are in general position and Pi+kint(Pi,Pi+1,Pi+k+1), Proposition 2.2 and Equation (11) imply 𝒪(Pi,Pi+1,Pi+k+1)>0. We conclude that P is a type-α N-representative. ∎

Refer to caption
Figure 15: Left: Proposition 3.18 configuration. Right: Proposition 3.19 configuration.
Proposition 3.19.

For all k2 and n2, Tk(𝒮k,nβ)𝒮k,nβ.

Proof.

The proof is analogous to the one for Proposition 3.18. Replacing α with β, we may work with the setup in the proof of Proposition 3.18. See the right side of Figure 15.

The key difference between type-α and type-β is that conditions for type-β k-spirals give us the following linear relations when we apply Lemma 3.17 with (1) on Pj for j{i,i+1,i+2,i+k,i+k+1}:

Pi=(1t1)Pi+1+t1Pi+k+1; Pi+1=(1s1)Pi+1+s1Pi+k+1; (12)
Pi+1=(1t2)Pi+2+t2Pi+k+2; Pi+2=(1s2)Pi+2+s2Pi+k+2;
Pi+k=(1t3)Pi+k+1+t3Pi+2k+1; Pi+k+1=(1s3)Pi+k+1+s3Pi+2k+1,

where s1,s2,s3(0,1) and t1,t2,t3(1,). We can see that Pi+kPiPi+1, so the three points Pi,Pi+1,Pi+k are in general position.

A very similar computation as Equation (10) shows positivity of (Pi,Pi+1,Pi+2), so we will omit it. Next, let r1=t11t1s1 and r2=s3t3. Notice that (1r2)(1r1), (1r2)r1, and r2 are all in (0,1) and sum up to 1. Also, Equation (12) implies

Pi+k+1=(1r2)(1r1)Pi+(1r2)r1Pi+1+r2Pi+k.

This shows Pi+k+1int(Pi,Pi+1,Pi+k). Finally, positivity of (Pi.Pi+1,Pi+k) follows from a similar computation as Equation (11), Pi+k+1int(Pi,Pi+1,Pi+k), the points Pi,Pi+1,Pi+k are in general position, and Proposition 2.2. ∎

3.4 Invariance of Backward Orbit

In this section, we complete the proof of Theorem 1.1 by showing that 𝒮k,nα and 𝒮k,nβ are Tk1-invariant. One can derive a formula for Tk1 from Equation (1). Given any k-nice twisted n-gon P, P=Tk1(P) is given by

Pi=Pik1PikPi1Pi. (13)

Proposition 3.8 implies Tk1 is well-defined on 𝒮k,nα and 𝒮k,nβ. In general, Tk1 needs not preserve k-niceness of twisted polygons.

Proposition 3.20.

For all k2 and n2, Tk1(Sk,nα)Sk,nα.

Proof.

Given P a type-α N-representative, we will show that P=Tk1(P) is a type-α (N+k+1)-representative by proving that for all iN+k+1, (Pi,Pi+1,Pi+2) is positive, (Pi,Pi+1,Pi+k+1) is positive, the four points Pi,Pi+1,Pi+k,Pi+k+1 are in general position, and Pi+kint(Pi,Pi+1,Pi+k+1). See the left side of Figure 16 for configurations of relevant vertices of P and P.

Let iN+k+1 be fixed. Since P is a type-α N-representative, we must have Pj+kint(Pj,Pj+1,Pj+k+1) for all jN. Applying Lemma 3.17 with Equation (13) on Pj for j{i,i+1,i+2,i+k,i+k+1} gives us

Pi=(1s1)Pik+s1Pik1; Pi=(1t1)Pi+t1Pi1; (14)
Pi+1=(1s2)Pik+1+s2Pik; Pi+1=(1t2)Pi+1+t3Pi;
Pi+2=(1s3)Pik+2+s3Pik+1; Pi+k=(1s4)Pi+s4Pi1;
Pi+k+1=(1s5)Pi+1+s5Pi,

where s1,s2,s3,s4,s5(0,1) and t1,t2(1,).

We first show that (Pi,Pi+1,Pi+k+1) is positive. From Equation (14) we have

𝒪(Pi,Pi+1,Pi+k+1)=(t1t2(1s5)t1(1t2)s5)𝒪(Pi1,Pi,Pi+1).

It follows that 𝒪(Pi,Pi+1,Pi+k+1)>0, so (Pi,Pi+1,Pi+k+1) is positive.

Next, we show that Pi+kint(Pi,Pi+1,Pi+k+1). Let r1=t21t2s5 and r2=s4t1. Equation (14) implies r1,r2(0,1) and

Pi+k=(1r2)(1r1)Pi+1+(1r2)r1Pi+k+1+r2Pi.

Observe that the coefficients (1r2)(1r1), (1r2)r1, and r2 are all in (0,1) and sum up to 1, so Pi+kint(Pi,Pi+1,Pi+k+1).

Finally, we check (Pi,Pi+1,Pi+2) is positive. We aim to invoke Lemma 3.14 with the following choices of vertices:

O=Pi+1;A=Pi;B=Pi+k+1;C=Pi+2;D=Pik+1. (15)

Positivity of (A,O,B) is a direct consequence of the above argument. Positivity of (B,O,C) follows from positivity of (Pi+1,Pi+2,Pi+k+2), Pi+k+1int(Pi+1,Pi+2,Pi+k+2), and Proposition 2.2. Next, observe that

𝒪(A,O,D)=s1s2𝒪(Pik1,Pik,Pik+1); (16)
𝒪(C,O,D)=(1s3)s2𝒪(Pik,Pik+1,Pik+2);
𝒪(B,O,D)=s2(1s5)𝒪(Pik,Pik+1,Pi+1)+s2s5𝒪(Pik,Pik+1,Pi);

Then, positivity of (A,O,D) and (C,O,D) follows from positivity of (Pik1,Pik,Pik+1) and (Pik,Pik+1,Pik+2). To see that (B,O,D) is positive, apply Proposition 2.2 on (Pik,Pik+1,Pi+1) positive and Piint(Pik,Pik+1,Pi+1) to get (Pik,Pik+1,Pi) positive. The backward direction of Lemma 3.14 then implies (Pi,Pi+1,Pi+2) is positive. We conclude that P is a type-α (N+k+1)-representative. ∎

Refer to caption
Figure 16: Left: Proposition 3.20 configuration. Right: Proposition 3.21 configuration.
Proposition 3.21.

For all k2 and n2, Tk1(Sk,nβ)Sk,nβ.

Proof.

The proof is similar to that of Lemma 3.20 (See right side of Figure 16). We will point out some key differences. Replacing α with β, we may work with the setup in the proof of Proposition 3.20. Applying Lemma 3.17 with (13) on Pj for j{i,i+1,i+2,i+k,i+k+1} gives us

Pi=(1s1)Pik+s1Pik1; Pi=(1t1)Pi1+t1Pi; (17)
Pi+1=(1s2)Pik+1+s2Pik; Pi+1=(1t2)Pi+t3Pi+1;
Pi+2=(1s3)Pik+2+s3Pik+1; Pi+k=(1s4)Pi1+s4Pi;
Pi+k+1=(1s5)Pi+s5Pi+1,

where s1,s2,s3,s4,s5(0,1) and t1,t2(1,). Positivity of (Pi,Pi+1,Pi+k) follows from a similar computation as in (3.4). Next, let r1=1s4t1s4 and r2=s5t2. Equation (17) implies

Pi+k+1=(1r2)(1r1)Pi+k+(1r2)r1Pi+r2Pi+1.

Observe that the coefficients (1r2)(1r1), (1r2)r1, and r2 are all in (0,1) and sum up to 1, so Pi+k+1int(Pi,Pi+1,Pi+k).

Finally, assign O,A,B,C,D to be the same vertices as in (15). Positivity of (A,O,B), (B,O,C), (C,O,D), (A,O,D), and (B,O,D) follows from a very similar proof as that of Proposition 3.20, with (16) replaced by

𝒪(A,O,D)=s1s2𝒪(Pik1,Pik,Pik+1);
𝒪(C,O,D)=(1s3)s2𝒪(Pik,Pik+1,Pik+2);
𝒪(B,O,D)=s2(1s5)𝒪(Pik,Pik+1,Pi)+s2s5𝒪(Pik,Pik+1,Pi+1).

The backward direction of Proposition 3.14 then implies (Pi,Pi+1,Pi+2) is positive. ∎

We conclude this section by stating that Proposition 3.18, 3.19, 3.20, 3.21 together prove Theorem 1.1.

4 Coordinate Representation of 3-Spirals

4.1 The Tic-Tac-Toe Grids

Recall the intervals I=(,0), J=(0,1), K=(1,) from §1.3. One can partition 2 into a 3×3 grid. See Figure 5. We make the following definition:

Definition 4.1.

For n2, let Sn(I,J) be the subset of 𝒫n that satisfies the following: given [P]Sn(I,J), for all i{0,,n1}, (x2i,x2i+1)I×J. We similarly define Sn(K,J), Sn(J,I), and Sn(J,K).

The following symmetries of the four grids follow directly from Definition 4.1.

Proposition 4.2.

For i, define the map σi: by σi(x)=x+i. Define the map ι: by ι(x)=x. Given [P]𝒫n, the following are true:

  • If [P]Sn(I,J), then [Pσi]Sn(I,J) for all i. This also holds for Sn(K,J), Sn(J,I), and Sn(J,K).

  • [P]Sn(I,J) if and only if [Pι]Sn(J,I).

  • [P]Sn(K,J) if and only if [Pι]Sn(J,K).

To understand the geometry implied by the corner invariants, we need to examine what happens when the corner invariants take value from 0,1,.

Proposition 4.3.

For all [P]𝒫n with corner invariants xj=xj(P) and i, we have the following correspondence between the position of Pi+2 and the values of x2i and x2i+1:

Configuration Coordinates Configuration Coordinates
Pi+2Pi+1Pi x2i=0 Pi+2Pi1Pi+1 x2i+1=0
Pi+2Pi+1Pi2 x2i=1 Pi+2Pi1Pi2 x2i+1=1
Pi+2Pi+1Pi1 x2i= Pi+2Pi1Pi x2i+1=
Refer to caption
Figure 17: Configurations of points and lines in the proof of Proposition 4.3.
Proof.

Consider the following lines:

l1 =Pi+1Pi2; l2 =Pi+1Pi1; l3 =Pi+1Pi; l4 =Pi+1Pi+2;
m1 =Pi1Pi+2; m2 =Pi1Pi+1; m3 =Pi1Pi; m4 =Pi1Pi2.

See Figure 17 for a visualization of the configurations of points and lines. Equation (8) implies x2i=χ(l1,l2,l3,l4) and x2i+1=χ(m1,m2,m3,m4). This yields

Configuration Lines Coordinates Configuration Lines Coordinates
Pi+2Pi+1Pi l4=l3 x2i=0 Pi+2Pi1Pi+1 m1=m2 x2i+1=0
Pi+2Pi+1Pi2 l4=l1 x2i=1 Pi+2Pi1Pi2 m1=m4 x2i+1=1
Pi+2Pi+1Pi1 l4=l2 x2i= Pi+2Pi1Pi m1=m3 x2i+1=

which is precisely the relationship described in the proposition. ∎

Remark 4.4.

Proposition 4.3 also gives us a way to determine the position of Pi+2 when neither x2i nor x2i+1 takes value in 0,1,. Suppose the four points Pi2,Pi1,Pi,Pi+1 are in general position. For i,j,k{1,2,3} distinct, we define Ui,j to be the connected component of 2(lilj) that does not intersect lk. For i,j,k{2,3,4} distinct, we define Vi,j to be the connected component of 2(mimj) that does not intersect mk. See Figure 18 for a visualization of the Ui,j’s and Vi,j’s using the point configurations given in Figure 17. By Proposition 4.3 and continuity of χ, we have the following:

Configuration Coordinates Configuration Coordinates
Pi+2U2,3 x2i=I Pi+2V2,3 x2i+1=I
Pi+2U1,3 x2i=J Pi+2V2,4 x2i+1=J
Pi+2U1,2 x2i=K Pi+2V3,4 x2i+1=K
Corollary 4.5.

Given [P]𝒫n with corner invariants xj=xj(P), if xj{0,1,} for all j, then P is 3-nice. Moreover, every four consecutive points of P are in general position.

Proof.

Using Proposition 4.3 we may check that

Collinearity Coordinates Collinearity Coordinates
Pi2, Pi1, Pi+1 x2i1= Pi1, Pi, Pi+2 x2i+1=
Pi2, Pi1, Pi+2 x2i+1=1 Pi1, Pi+1, Pi+2 x2i+1=0
Pi2, Pi+1, Pi+2 x2i=1 Pi, Pi+1, Pi+2 x2i=0
Pi1, Pi, Pi+1 x2i2=0

All seven cases contradict the assumption in the corollary. Therefore, the four points Pi2, Pi1, Pi+1, Pi+2 are in general position, and the four consecutive points Pi1, Pi, Pi+1, Pi+2 are in general position for all i. This shows P is 3-nice, and every four consecutive points of P are in general position. ∎

Refer to caption
Figure 18: The connected components Ui,j’s and Vi,j’s in Remark 4.4. The corner invariants value in I if Pi+2 lies in the black-shaded region, J if Pi+2 lies in the red-shaded region, and K if Pi+2 lies in the cyan-shaded region.

Our goal of this section is to prove the following correspondence theorem:

Theorem 4.6.

For all n2, 𝒮3,nα=Sn(J,I), 𝒮3,nβ=Sn(K,J).

This theorem immediately produces the following important corollary.

Corollary 4.7.

For all n2, the four cells Sn(I,J), Sn(K,J), Sn(J,I), Sn(J,K) are both forward and backward invariant under T3.

Proof.

The case Sn(J,I) and Sn(K,J) follows immediately from Theorem 1.1 and 4.6. We will prove the case Sn(I,J). The case Sn(J,K) is completely analogous, so we will omit.

Fix [P]Sn(I,J). Recall the maps σi and ι from Proposition 4.2. Equation (1) implies T3(Pι)=T3(P)ισ4. Then, Proposition 4.2 implies [Pι]Sn(J,I), so [T3(Pι)]Sn(J,I). Finally, observe that

T3(P)=(T3(P)ισ4)(σ4ι)=T3(Pι)(σ4ι).

It follows that [T3(P)]Sn(I,J). We omit the proof of [T31(P)]Sn(I,J). ∎

4.2 The Correspondence of 𝒮3,nα and Sn(J,I)

Here we show that 𝒮3,nα is equivalent to Sn(J,I). We will first show that the corner invariants of a 0-representative P of some [P]𝒮3,nα satisfies Sn(J,I). Then, we will show that we can find type-α N-representatives for all N given any [P]Sn(J,I).

Lemma 4.8.

If P is an N-representative of [P]𝒮3,nα, then Pi+2int(Pi1,Pi,Pi+1) for all i>N+1.

Proof.

Since (Pi1,Pi,Pi+1) is positive, we may normalize with Aff2+() so that Pi1=(1,0), Pi=(0,0), and Pi+1=(0,1). Let Pi+2=(x,y). It suffices to show that x<0, y>0, and yx<1. We get x<0 from positivity of (Pi,Pi+1,Pi+2), and we get y>0 from positivity of (Pi1,Pi,Pi+2). Finally, since (Pi2,Pi1,Pi+2) is positive and Pi+1int(Pi2,Pi1,Pi+2), Proposition 2.2 implies (Pi+1,Pi1,Pi+2) is positive, which gives us yx<1 as desired. ∎

Proposition 4.9.

For all n2, 𝒮3,nαSn(J,I).

Proof.

Fix i. Let P be an (i3)-representative of [P]𝒮3,nα with corner invariants xj=xj(P). Normalize with Aff2+() so that Pi1=(1,0), Pi=(0,0), and Pi+1=(0,1). Let sa,b denote the slope of the line Pi+aPi+b. See Figure 19 for the configuration of points.

We want to show that (x2i,x2i+1)I×J. By Lemma 4.8, Pi+1int(Pi2,Pi1,Pi). This implies s1,2>s1,2>1. On the other hand, since Pi+1int(Pi2,Pi1,Pi+2), we have s1,2>s1,2>1, and s1,2(0,1). This gives us

x2i =(s1,2s1,1)(s1,0s1,2)(s1,2s1,0)(s1,1s1,2)=s1,21s1,21Jand
x2i+1 =(s1,2s1,1)(s1,0s1,2)(s1,2s1,0)(s1,1s1,2)=s1,2(s1,21)s1,2(s1,21)I.

This concludes the proof. ∎

Refer to caption
Figure 19: Configuration of Proposition 4.9 and 4.10.
Proposition 4.10.

For all n2, 𝒮3,nα=Sn(I,J).

Proof.

Proposition 4.9 implies we only need to show 𝒮3,nαSn(I,J). Given [P]Sn(I,J), let P be a representative that satisfies P1=(1,4), P0=(1,0), P1=(0,0), P2=(0,1). Say that P satisfies condition ()i if the three triangles (Pi1,Pi,Pi+1), (Pi,Pi+1,Pi+2), (Pi1,Pi,Pi+2) are all positive, Pi+2int(Pi1,Pi,Pi+1), and Pi+2int(Pi2,Pi1,Pi+1).

We show that for all i>0, if P satisfies ()i1, then P satisfies ()i. Since (Pi1,Pi,Pi+1) is positive, we can normalize with Aff2+() so that Pi1=(1,0), Pi=(0,0), and Pi+1=(0,1). Let sa,b denote the slope of Pi+aPi+b. Since Pi+1int(Pi2,Pi1,Pi), we know that s1,2>s1,2>1. Then, x2iJ implies 0<s1,21s1,21<1. This gives us s1,2>s1,2>1. On the other hand, x2i+1I implies s1,2(s1,21)s1,2(s1,21)<0. Since s1,2>1, this is equivalent to 11s1,2<0, which implies s1,2(0,1). Thus, the two lines Pi1Pi+2 and Pi+1Pi+2 must meet in the shaded triangle in Figure 19, which implies (Pi,Pi+1,Pi+2), (Pi1,Pi,Pi+2) are positive, Pi+2int(Pi1,Pi,Pi+1), and Pi+2int(Pi2,Pi1,Pi+1), so P satisfies ()i. Finally, since P clearly satisfies ()0, by induction P satisfies ()i for all i0, so P is a type-α 0-representative of a 3-spiral. We conclude that [P]𝒮3,nα. ∎

4.3 The Correspondence of 𝒮3,nβ and Sn(K,J)

Here we show that 𝒮3,nβ is equivalent to Sn(K,J). The ideas behind the proofs are essentially the same as the ones in §4.2. We will focus on explaining how to modify the details of the proofs in §4.2 for type-β 3-spirals and Sn(K,J).

Lemma 4.11.

If P is an N-representative of [P]𝒮3,nβ, then the quadrilateral joined by vertices (Pi,Pi+1,Pi+2,Pi+3) is convex for all i>N.

Proof.

Normalize with Aff2+() so that Pi=(1,0), Pi+1=(0,0), Pi+2=(0,1), and Pi+3=(x,y). Positivity of (Pi+1,Pi+2,Pi+3) and (Pi,Pi+1,Pi+3) implies x<0 and y>0. Positivity of (Pi1,Pi,Pi+2), Pi+3int(Pi1,Pi,Pi+2), and Proposition 2.2 shows yx>1. ∎

Proposition 4.12.

For all n2, 𝒮3,nβSn(K,J).

Proof.

Let P be a (3)-representative of [P]𝒮3,nβ with corner invariants xj=xj(P). Lemma 4.11 implies the quadrilateral (Pi2,Pi1,Pi,Pi+1) is convex. Next, since P is a type-β (3)-representative, Piint(Pi2,Pi1,Pi+1) for all i0 (See Figure 20). Referring back to Remark 4.4, convexity of (Pi2,Pi1,Pi,Pi+1) implies PiPi+1 doesn’t go through Pi2,Pi1,Pi+1, so (x2i,x2i+1)K×J whenever Pi+2int(Pi2,Pi1,Pi+1). ∎

Refer to caption
Figure 20: Configuration of Proposition 4.12 and Lemma 4.13.
Lemma 4.13.

Given a 3-nice sequence P:2 and an integer i, let x2i=x2i(P) and x2i+1=x2i+1(P) be the corner invariants of P. If the following conditions are true:

  • (Pi2,Pi1,Pi) and (Pi1,Pi,Pi+1) are both positive;

  • The quadrilateral (Pi2,Pi1,Pi,Pi+1) is convex;

  • (x2i,x2i+1)K×J.

Then, the following hold:

  • Pi+2int(Pi2,Pi1,Pi+1);

  • The quadrilateral (Pi1,Pi,Pi+1,Pi+2) is convex;

  • (Pi,Pi+1,Pi+2) and (Pi1,Pi,Pi+2) are both positive.

Proof.

Recall that from the proof of Proposition 4.12, we claimed that if the quadrilateral (Pi2,Pi1,Pi,Pi+1) is convex, then the line PiPi+1 doesn’t go through (Pi2,Pi1,Pi+1). Since (x2i,x2i+1)K×J, Remark 4.4 implies Pi+2int(Pi2,Pi1,Pi+1), in which case all conclusions of this lemma will hold. See Figure 20 for a visualization of the five points. ∎

Proposition 4.14.

𝒮3,nβ=Sn(K,J).

Proof.

Proposition 4.12 gives us 𝒮3,nβSn(K,J), so we show the other containment. Given [P]Sn(K,J), we can find a representative P that satisfies PN=(0,0), PN+1=(1,0), PN+2=(1,1), PN+3=(0,1). Corollary 4.5 shows that P is 3-nice. To see that (Pi,Pi+1,Pi+2), (Pi,Pi+1,Pi+3) are positive, and Pi+4int(Pi,Pi+1,Pi+3), we may inductively apply Lemma 4.13. This implies [P]𝒮3,nβ. ∎

5 A Birational Formula for T3

Given two spaces X and Y, a rational map f:XY is an equivalence class of maps fU:UY where U is a dense open in X, and the equivalence relation is given by fUfV if they restrict to the same map on UV. A map f:XY is birational if there exists a rational map g:YX such that gf restricts to the identity on a dense open of X and fg restricts to an identity on a dense open of Y.

In this section, we show that T3:𝒫n𝒫n is a birational map by finding an explicit formula using the corner invariants.

5.1 The Formula

Let P be a twisted n-gon, and P=T3(P). In this section, we use a different labeling convention:

Pi=Pi2Pi+1Pi1Pi+2. (18)

We let xj=xj(P) and xj=xj(P) denote the corner invariants of P and P respectively. Our goal is to show that T3 is a birational map over the corner invariants. I discovered it using computer algebra and the reconstruction formula in [23, Equation (19)].

Proposition 5.1.

Given [P]𝒫3,n, the following formula holds (indices taken modulo 2n):

{x2i=x2i2(x2i4+x2i11)x2i2x2i1(1x2i+1)(1x2i4);x2i+1=x2i+3(x2i+2+x2i+51)x2i+2x2i+3(1x2i+5)(1x2i). (19)

One can verify Equation (19) with the following procedure: Given the corner invariants of [P], use the reconstruction formula from [23, Equation (19)] to obtain a representative P. Apply T3 on P as in Equation (18) to get P=T3(P). Then, compute the corner invariants of P. We present a geometric proof of Equation (19) using cross-ratio identities. We start with the following lemma, which is a classical observation in projective geometry called “quadrangular sets.”

Lemma 5.2.

Let Q1,Q2,Q3,Q4 be four points in general position, and let ω be a line that contains none of the four points. For all ij, let lij=QiQj and Sij=ωlij Then,

χ(S12,S13,S14,S24)=χ(S23,S13,S34,S24).
Refer to caption
Figure 21: Point configurations of Lemma 5.2
Proof.

Let O=l13l24. See Figure 21 for an example of the point configurations. Applying Equation (6) on (l12,l13,l14,Q1D) with respect to ω and Q2Q4 gives us

χ(S12,S13,S14,S24)=𝜔χ(l12,l13,l14,Q1D)=l24χ(Q2,O,Q4,S24).

Next, applying Equation (6) twice on (l23,l13,l34,Q3D) with respect to l24 and ω gives us

χ(Q2,O,Q4,S24)=l24χ(l23,l13,l34,Q3D)=𝜔χ(S23,S13,S34,S24).

Combining the above two equations completes the proof. ∎

Refer to caption
Figure 22: Visualization of Points Assigned in Equation (21). The thick black line segments are edges connecting vertices of P, and the thick red line segments are edges connecting vertices of P.
Proof of Proposition 5.1.

From the symmetry of Equation (19), it suffices to prove the formula for x0. That is,

x0=x2(x4+x11)x2x1(1x4)(1x1). (20)

Let li,j=PiPj and O=l3,2l1,0. We label points as follows:

A=P2;B=P1;C=l3,0l2,1;D=P0;E=P3;F=l3,0l1,1;G=l3,0l2,1;H=l3,0OP1. (21)

Since [P]𝒫3,n, every five consecutive points of [P] are in general position. This ensures that point O and the points in Equation (21) are all distinct. See Figure 22 for a visualization of the assignment of labels to these points.

It follows from Equation (7) that x0=χ(A,B,C,D). Using Equation (8), we have

x4 =χ(l1,4,l1,3,l1,2,l1,0)=l3,0χ(A,E,G,D); (22)
x2 =χ(l0,3,l0,2,l0,1,l0,1)=l3,0χ(E,B,H,D);
x1 =χ(l2,1,l2,0,l2,1,l2,3)=l3,0χ(B,D,G,E);
x1 =χ(l1,2,l1,1,l1,0,l1,2)=l3,0χ(C,F,D,G).

We may further invoke Lemma 5.2 with Q1=P2, Q2=O, Q3=P1, Q4=P1, and ω=l3,0. This gives us

x2=χ(E,B,H,D)=χ(G,B,F,D). (23)

The rest of the proof is just algebraic verification. Normalize with a projective transformation so that l3,0 is the x-axis of 𝔸2. Let a, b, c, d, e, f, g, h be coordinates of A, B, C, D, E, F, G, H respectively. Plugging (22) and (23) into the numerator of (20) gives us

x2(x4+x11) =χ(G,B,F,D)(χ(A,E,G,D)+χ(B,D,G,E)1)
=(gb)(fd)(gf)(bd)((ae)(gd)(ag)(ed)+(be)(gd)(bg)(ed))
=(ab)(gd)(eg)(df)(ag)(bd)(ed)(gf).

The denominator can be computed similarly. We skip the computation and list the results:

x2x1(1x4)(1x1)=(ac)(gd)(df)(eg)(ag)(cd)(de)(fg).

Combining the above two equations gives us

x2(x4+x11)x2x1(1x4)(1x1)=(ab)(cd)(ac)(bd)=χ(A,B,C,D)=x0,

which is precisely Equation (20). ∎

Next, we provide a formula for the inverse of T3.

Proposition 5.3.

The map T3:𝒫n𝒫n is birational. Its inverse is given by

{x2i=x2i+2(x2i+4+x2i+11)x2i+1x2i+2(1x2i1)(1x2i+4);x2i+1=x2i1(x2i3+x2i1)x2ix2i1(1x2i+2)(1x2i3). (24)

We will give an algebraic proof. Consider two families of rational maps {μ(s,t):2n2n}(s,t)2 and {ν(s,t):2n2n}(s,t)2. Write (a0,,a2n1)=μ(s,t)(x0,,x2n1) and (b0,,b2n1)=ν(s,t)(x0,,x2n1). Then, we set

{a2i=1x2i+sx2i+s+ta2i+1=1x2i+1sx2i+1st;{b2i=1x2i+s1x2i+sx2i+s+tb2i+1=1x2i+1s1x2i+1sx2i+1st. (25)
Lemma 5.4.

Let φ:22 be the map given by

φ(s,t)=((1)s+1s,(1)s(2s+t)). (26)

Then, φ is an involution. Moreover, when t is odd, μ(s,t)1=νφ(s,t) and ν(s,t)1=μφ(s,t).

Proof.

To see φ is an involution, a direct computation shows that

φ2(s,t) =φ((1)s+1s,(1)s(2s+t))
=((1)(1)s+1s+s+2s,(1)(1)s+1s(2(1)s+1s+(1)s(2s+t)))=(s,t).

Next, we show that when t is odd, μ(s,t)1=νφ(s,t). We will show by direct computation that μ(s,t)νφ(s,t) is the identity on the 2i-th coordinate when s is even. First, when s is even, φ(s,t)=(s,2s+t). The 2i-th coordinate of μ(s,t)νφ(s,t) is given by

(11x2i+s+(s)1x2i+s+(s)x2i+s+(s)+(2s+t))(1x2i+s+t(s)1x2i+s+t(s)x2i+s+t(s)(2s+t))1
=(11x2i1x2ix2i+2s+t)(1x2i+2s+tx2i1x2i+2s+t)=x2i.

This is precisely what we want. One can similarly carry out the computation of νφ(s,t)μs,t for the (2i+1)-th coordinate, and s odd. We will omit these heavy computations and conclude that μ(s,t)1=νφ(s,t). Finally, to see ν(s,t)1=μφ(s,t), observe that (1)s(2s+t) is odd iff t is odd. Therefore, νs,tμφ(s,t)=νφ2(s,t)μφ(s,t) is the identity map by the previous argument. The same argument shows that μφ(s,t)ν(s,t) is the identity. ∎

The following corollary is immediate. We omit the proof.

Corollary 5.5.

For all (s,t)2 such that t is odd, μ(s,t) and ν(s,t) are birational maps.

Proof of Proposition 5.3.

We first claim that T3=ν(1,1)μ(3,3). We will provide the computation for even coordinates. Let (a0,,a2n1) denote the image of (x0,,x2n1) under μ(3,3), and let (b0,,b2n1) denote the image of (a0,,a2n1) under ν(1,1). Then, we have

b2i =1a2i11a2i1a2i2=(11x2i4x2i1)(1(1x2i4)(1x2i+1)x2i1x2i2)1
=x2i2(x2i1+x2i4+1)x2i1x2i2(1x2i4)(1x2i+1).

Observe that this is precisely the first line of (19). The computation for b2i+1 is analogous, thus omitted. Then, by Corollary 5.5, T31=ν(3,3)μ(1,3). Finally, Equation (24) follows from a direct computation of ν(3,3)μ(1,3) using Equation (25), which we will omit. ∎

5.2 Conjugated Corner Invariants and Its T3 Formula

To relate Equation (19) to parameters (yr)r2 in [7], it is convenient to consider another coordinate system of 𝒫n, which we define below.

Definition 5.6.

Given [P]𝒫n, define the conjugated corner invariants to be coordinate functions x~0(P),,x~2n1(P) given by x~j(P)=xj(P)xj(P)1.

The conjugated corner invariants can be viewed as the image of the corner invariants under a birational map λ:2n2n sending each coordinate xjxjxj1. Observe that λ2 restricted to the dense open set ({0,1})2n is the identity map, so x~j(P) is also a coordinate system for 𝒫n. Geometrically, the map λ corresponds to a different choice of permutation in the cross-ratio.

Throughout this section, we will use x~j=x~j(P) and x~j=x~j(P) to denote the conjugate corner invariants of P and P. We start by observing some symmetries of conjugating our factorization maps μ(s,t) and ν(s,t) from Equation (25).

Lemma 5.7.

For all (s,t)2, we have λμ(s,t)λ=ν(s+t,t).

Proof.

We can check this by direct computation. We show that the equation holds on even coordinates. The 2i-th coordinate of μ(s,t)λ is given by

1x2i+s(x2i+s1)1x2i+s+t(x2i+s+t1)1=1x2i+s+tx2i+s+t(x2i+s1)

The 2i-th coordinate of λμ(s,t)λ is given by

(1x2i+s+tx2i+s+t(x2i+s1))(1x2i+s+tx2i+s+t(x2i+s1)1)1=1x2i+s+t1x2i+s+tx2i+s,

which is precisely the 2i-th coordinate of ν(s+t,t). The computation for the odd coordinates is similar. ∎

Since λ is an involution, it immediately follows that λν(s,t)λ=μ(s+t,t). This allows us to obtain a formula for T3 with respect to the conjugated corner invariants.

Proposition 5.8.

Given any 3-nice twisted n-gon P, the following formula holds (indices taken modulo 2n):

{x~2i=x~2i2(1x~2i1x~2i4)(1x~2i+1)(1x~2i+1x~2i2)(1x~2i1);x~2i+1=x~2i+3(1x~2i+2x~2i+5)(1x~2i)(1x~2ix~2i+3)(1x~2i+2). (27)
Proof.

From the proof of Proposition 5.3, we saw that the formula for T3 on the corner invariants is given by ν(1,1)μ(3,3). It follows that the formula for conjugated corner invariants is λ(ν(1,1)μ(3,3))λ. By Lemma 5.7,

λ(ν(1,1)μ(3,3))λ=(λν(1,1)λ)(λμ(3,3)λ)=μ(2,1)ν(0,3).

It remains to check that μ(2,1)ν(0,3) agrees with Equation (27). The 2i-th coordinate of μ(2,1)ν(0,3) is given by

(11x~2i21x~2i2x~2i+1)(1x~2i11x~2i1x~2i4)1=x~2i2(1x~2i1x~2i4)(1x~2i+1)(1x~2i2x~2i+1)(1x~2i1).

This is precisely x~2i from Equation (27). The computation for odd coordinates is omitted. ∎

Using Lemma 5.4, we can easily compute the formula of T31 with respect to the conjugated corner invariants. The proof is again a direct computation, so we omit it.

Corollary 5.9.

The formula for T31 with conjugated corner invariants is given by μ(0,3)ν(2,3). More specifically,

{x~2i=x~2i+2(1x~2i+1x~2i+4)(1x~2i1)(1x~2i1x~2i+2)(1x~2i+1);x~2i+1=x~2i1(1x~2ix~2i3)(1x~2i+2)(1x~2i+2x~2i1)(1x~2i). (28)

5.3 Relation to Y-Variables

In this section, we discuss how Equation (18) generalizes the results from [7]. The propositions in this section hold for all four cells Sn(I,J), Sn(J,I), Sn(K,J), Sn(J,K). For notational convenience, our statements will only mention Sn(J,I). The readers may assume that the propositions hold for the other three cells with the same proof.

The map T3 along with the labeling convention of Equation (18) corresponds to the following construction in [7]. Let a,b,c,d2 be distinct and assume a2b2c2d2. Say that S={a,b,c,d} is a Y-pin if b2<c2 and the vectors ba, ca, da generate all of 2.

Definition 5.10 ([7, Definition 1.4]).

Let S={a,b,c,d} be a Y-pin and suppose D2. A Y-mesh of type S and dimension D is a grid of points P^i,j in D with i,j which together span all of D and such that

  • P^r+a, P^r+b, P^r+c, P^r+d are distinct for all r2.

  • Let Lr=P^r+aP^r+b. Then, P^r+a, P^r+b, P^r+c, P^r+d all lie on Lr for all r2.

  • The four lines Lra, Lrb, Lrc, Lrd (all of which contain P^r) are distinct for all r2.

Let S={(1,0),(2,0),(0,1),(1,1)}, which is a Y-pin. Given a representative P of some [P]Sn(J,I), we can consider a grid (P^i,j)(i,j)2 where P^i,j is the i-th vertex of T3j(P).

Proposition 5.11.

(P^i,j) is a Y-mesh of type S and dimension 2.

Proof.

The first two conditions of Definition 5.10 are straightforward to verify using the identification Sn(J,I)=𝒮3,nα from Proposition 4.10. For the third condition, let P(j)=T3j(P). Then, we have

Lra=Pi1(j)Pi+2(j),Lrb=Pi1(j)Pi4(j),Lrc=Pi1(j)Pi(j),Lrd=Pi1(j)Pi2(j).

Notice also that Lra=Pi(j+1)Pi+1(j+1) and Lrb=Pi2(j+1)Pi1(j+1), so 3-niceness of P(j+1) implies they are distinct. The other pairings are distinct because of 3-niceness of P(j). ∎

[7] then introduces the parameters yr(P^) associated to a Y-mesh. Fix a Y-pin S={a,b,c,d} and a Y-mesh P^ of type S and dimension D. For all r2, consider

yr(P^)=χ(P^r+a,P^r+c,P^r+d,P^r+b). (29)

See the left side of Figure 23 for the setup using the Y-mesh from Proposition 5.11. [7, Theorem 1.6] give us the following relation on yr:

yi+1,jyi+1,j+2=(1+yi1,j+1)(1+yi+3,j+1)(1+yi,j+11)(1+yi+2,j+11). (30)
Refer to caption
Figure 23: Left: Definition of yr(P^) for the Y-mesh from Proposition 5.11. Right: Relationship between yr(P^) and conjugated corner invariants.
Lemma 5.12.

Given a representative P of [P]Sn(I,J) with conjugated corner invariants x~j=x~j(P). Let (P^i,j) be its corresponding Y-mesh with yr=yr(P^) for all r2. Then, for all i,

yi,0=x~2ix~2i+3. (31)
Proof.

Let la,b=Pi+aPi+b. See right side of Figure 23 for the setup. Equation (8) gives us

x~2i=x2ix2i1=χ(l1,2,l1,1,l1,2,l1,0);x~2i+3=x2i+3x2i+31=χ(l0,3,l0,2,l0,1,l0,1).

Notice that (l1,1l0,1)(l1,2l0,2)=Pi1Pi+2=l1,2. Then, from elementary cross ratio identities we have

x~2ix~2i+3 =χ(l1,1l1,2,l1,2l1,2,l0,3l1,2,l0,2l1,2)
=χ(P^i1,0,P^i,1,P^i+1,1,P^i+2,0)=yi,0,

which is precisely Equation (31). ∎

Remark 5.13.

Equation (31) is very similar to the correspondence of yr and corner invariants in the T2 case. Let P be an arbitrary twisted n-gon with P=T2(P). If we use the labeling convention Pi=Pi1Pi+1PiPi+2, then the T2-orbit (P^i,j)(i,j)2, where P^i,j is the i-th vertex of T2j(P), is a Y-mesh of type S={(1,0),(1,0),(1,1),(0,1)}. Denote by xj=xj(P) the corner invariants of P. Then, for all i,

yi,0=x2i+1x2i+2. (32)

For the proof of Equation (32), see [5, Equation (2.2)].

Theorem 5.14.

For the Y-pin S={(1,0),(2,0),(0,1),(1,1)}, the transformation formula of yr from [7, Theorem 1.6] is a direct consequence of the birational formula for the conjugated corner invariants under T3.

Proof.

It suffices to show that we can use Equation (31) to derive (30) for j=1. We first compute yi+1,1 and yi+1,1 using Equation (27) and (28):

yi+1,1 =x~2i+4(1x~2i+3x~2i+6)(1x~2i+1)(1x~2i+1x~2i+4)(1x~2i+3)x~2i+3(1x~2i+1x~2i+4)(1x~2i+6)(1x~2i+3x~2i+6)(1x~2i+4) (33)
=x~2i+3x~2i+4(1x~2i+1)(1x~2i+6)(1x~2i+3)(1x~2i+4);
yi+1,1 =x~2i(1x~2i2x~2i+1)(1x~2i+3)(1x~2ix~2i+3)(1x~2i+1)x~2i+7(1x~2i+6x~2i+9)(1x~2i+4)(1x~2i+4x~2i+7)(1x~2i+6)
=x~2ix~2i+7(1+yi1,0)(1+yi+3,0)(1x~2i+3)(1x~2i+4)(1+yi,0)(1+yi+2,0)(1x~2i+1)(1x~2i+6).

It follows that

yi+1,1yi+1,1 =x~2ix~2i+3x~2i+4x~2i+7(1+yi1,0)(1+yi+3,0)(1+yi,0)(1+yi+2,0)
=yi,0yi+2,0(1+yi1,0)(1+yi+3,0)(1+yi,0)(1+yi+2,0)=(1+yi1,0)(1+yi+3,0)(1+yi,01)(1+yi+2,01).

This concludes the proof. ∎

6 The Precompactness of T3 Orbits

In this section, we establish four algebraic invariants of T3. We then use them to prove Theorem 1.3. Having Theorem 4.6 in hand, we may fully work with Sn(J,I) and Sn(K,J). Our strategy is to use the algebraic invariants to show that the corner invariants are uniformly bounded.

6.1 The Four Invariants

Proposition 6.1.

Given [P]𝒫n with corner invariants xj=xj(P), consider the following four quantities i=i(P):

1=i=0n1x2ix2i1;2=i=0n1x2i+1x2i+11;3=i=0n1x2ix2i+1;4=i=0n11x2i1x2i+1. (34)

Then, i is invariant under T3 for i=1,2,3,4.

Proof.

We first show that 3 is invariant under T3. Let 3 denote the invariants obtained by plugging in xi from Equation (24). Observe that

3 =3i=0n1x2i4+x2i11x2i+2+x2i+51i=0n1x2i+2x2i+3(1x2i+5)(1x2i)x2i2x2i1(1x2i+1)(1x2i4)
=3i=3n4(x2i+2+x2i+51)i=0n1(x2i+2+x2i+51)i=2n+1(x2i2x2i1(1x2i+1)(1x2i4))i=0n1(x2i2x2i1(1x2i+1)(1x2i4))=3.

This shows 3=3. Next, we show that 1 and 2 are invariant. Using conjugated corner invariants, we see that 1=i=0n1x~2i and 2=i=0n1x~2i+1. We let 1=i=0n1x~2i be the first invariant of T3(P). Equation (27) gives us

1=1i=0n1(1x~2i1x~2i4)(1x~2i+1)(1x~2i+1x~2i2)(1x~2i1)=1,

where the last equality follows from cyclically permuting the numerator. This shows 1=1. The proof for 2 goes through the same computation, so we omit it.

Finally, observe that 4=231, so by invariance of 1,2,3, we know that 4 must also be invariant. This concludes the proof. ∎

Remark 6.2.

As shown in the proof of Proposition 6.1, 1 and 2 correspond to the product of conjugated corner invariants. 3 is the ratio of the two Casimirs OnEn of the T2 invariant Poisson structure on 𝒫n. For discussions on 3 and the Casimirs, see [26, §2.3]. Also, since 14=23, the four T3 invariants are not algebraically independent.

Below is a direct consequence of the invariance of the i’s. Since the i’s are preserved by the forward action, it must also be preserved by the backward action.

Corollary 6.3.

The four invariants 1,2,3,4 are also invariant under T31.

6.2 Proof of Theorem 1.3

Recall that a subset A of a topological space X is precompact if the closure of A is compact. To show that the T3-orbit is precompact, it suffices to show that the corner invariants of the orbit are uniformly bounded away from the singularities 0,1,.

In this section, we let [n]:={1,,n}. Given [P]𝒫n, for all j,m, let xj,m=xj(T3m(P)) whenever T3m(P) exists. Let i,m=i(T3m(P)) for i=1,2,3,4. By Proposition 6.1, i,m is independent of m. All sequences are indexed by 0 unless specified otherwise. Finally, when we say “{am} converges/diverges on a subsequence, and {bm} converges/diverges on the same subsequence,” we mean that a subsequence of {bm} with the same choice of indices as the subsequence of {am} converges/diverges.

Lemma 6.4.

Given [P]Sn(J,I), there exist a,bJ such that x2i,m[a,b] for all i[n] and m0.

Proof.

We first claim that for each i, the sequence {x2i,m} is bounded above uniformly by some biJ. If not, then x2i,m1 on a subsequence, which implies 1x2i,m0 on the same subsequence. Since [T3m(P)]Sn(J,I) for all m0, we must have 1x2j,m(0,1) and (1x2j+1,m)1(0,1) for all j[n]. This implies 4,m0 on the same subsequence, but that contradicts invariance of 4,m. Therefore, {x2i,m} is bounded above by bi=supm{x2i,m}J. Taking b=maxi[n]bi satisfies the condition in the lemma.

Next, we show {x2i,m} is bounded below uniformly by some ai>0. If not, then x2i,m0 on a subsequence, so x2i,m(x2i,m1)10 on the same subsequence. From the argument above, x2j,mb for all m0 and j[n], which gives us |x2j,m(x2j,m1)1||bb1|, so the sequences are uniformly bounded for all ji. This together with |x2i,m(x2i,m1)1|0 on a subsequence implies |1,m|0 on the same subsequence, but that contradicts invariance of 1,m. Therefore, {x2i,m} is bounded below by ai=infm{x2i,m}J. Taking a=mini[n]ai completes the proof. ∎

Lemma 6.5.

With the same notation as in Lemma 6.6, there exist c,dI such that x2i+1,m[c,d] for all i[n] and m0.

Proof.

We first claim that for each i,the sequence {x2i+1,m} is bounded above uniformly by some diI. If not, then, x2i+1,m(x2i+1,m1)10 on a subsequence. Since x2j+1,m(x2j+1,m1)1(0,1) for all j[n], we must have 2,m0 on the same subsequence, but that contradicts invariance of 2,m.

Next, we show that {x2i+1,m} is bounded below uniformly by some ciI. If not, then a subsequence of {x2i+1,m} must diverge, so the same subsequence of {1x2i+1,m} also diverges. Lemma 6.4 and x2i+1,mdi<0 together implies 4,m diverges on the same subsequence, but that contradicts invariance of 4,m. Finally, taking c=mini[n]ci and d=maxi[n]di completes the proof. ∎

The proofs of the following two lemmas are analogous to Lemma 6.4 and 6.5. We will omit the details and point out which invariants to use in each step.

Lemma 6.6.

Given [P]Sn(K,J), there exist a,bJ such that x2i+1,m[a,b] for all i[n] and m0.

Proof.

For each i, the sequence {x2i+1,m} is bounded below by some aiJ, for otherwise 3,m diverges on a subsequence. Next, since {|3,m|} is bounded below by j=0n1aj>0, {x2i+1,m} is bounded above by some biJ. Taking a=mini[n]ai and b=maxi[n]bi completes the proof. ∎

Lemma 6.7.

With the same notation as in Lemma 6.6, there exist c,dK such that x2i,m[c,d] for all i[n] and m0.

Proof.

For each i, the sequence {x2i,m} must be bounded below by some ciK, for otherwise 1,m on a subsequence. It’s also bounded above by some di. To see this, Lemma 6.6 implies all corner invariants are bounded away from 0, so if {x2i,m} is not bounded above, then 3,m diverges on a subsequence. Taking c=minici and d=maxidi completes the proof. ∎

Proof of Theorem 1.3.

We will show that the forward T3 orbit of [P]𝒮3,nα=Sn(J,I) has uniformly bounded corner invariants. By Proposition 4.10, [P]Sn(J,I). Let [a,b]J, [c,d]I be compact intervals derived from Lemma 6.4 and 6.5. Then, the sequence {(x0,m,,x2n1,m)} is contained in i=0n1[a,b]×[c,d], so it is uniformly bounded. To show precompactness of the backward T3 orbit of 𝒮3,nα, one can adapt the proofs of Lemma 6.4 and 6.5 with very few changes. We omit the details. The case 𝒮3,nβ follows from Lemma 6.6 and 6.7 by essentially the same argument, which we again omit. ∎

7 Type-β 2-Spirals and Precompact T2 Orbits

7.1 The Corner Invariants of Type-β 2-Spirals

We finish this paper by discussing the type-β 2-spirals. Proposition 3.10 implies 𝒮2,nβ is disjoint from the moduli space of closed convex polygons, so 𝒮2,nβ is a new invariant geometric construction under the pentagram map by Theorem 1.1. In this section, we analyze the corner invariants of 𝒮2,nβ and show that just like the type-α and type-β 3-spirals, it is cut out by linear boundaries.

Proposition 7.1.

For all n2, given any [P]𝒮2,nβ with corner invariants xj=xj(P), we have x2i>0 and x2i+1<0 for all i[n].

Proof.

Let P be an (i2)-representative of [P]. Normalize by Aff2+() so that Pi1=(1,0), Pi=(0,0), Pi+1=(0,1) on the affine patch, which is possible because (Pi1,Pi,Pi+1) is positive. Let sa,b{} denote the slope of Pi+aPi+b. Positivity of (Pi2,Pi1,Pi) and Pi+1int(Pi2,Pi1,Pi) implies s1,2>1 and s1,2>1. Similarly, since Pi+2int(Pi1,Pi,Pi+1), we have s1,2(0,1) and s1,2>1. It follows that

x2i =(s1,2s1,1)(s1,0s1,2)(s1,2s1,0)(s1,1s1,2)=s1,211s1,2>0; (35)
x2i+1 =(s1,2s1,1)(s1,0s1,2)(s1,2s1,0)(s1,1s1,2)=s1,2(s1,21)s1,2(1s1,2)<0.

This concludes the proof. ∎

Refer to caption
Figure 24: Configuration of Proposition 7.1 and 7.2.
Proposition 7.2.

For all n2, if [P]𝒫n has corner invariants xj=xj(P) such that x2i>0 and x2i+1<0 for all i[n], then [P]𝒮2,nβ.

Proof.

Fix N. Let P be a representative of [P] such that PN2=(13,32), PN1=(1,0), PN=(0,0), and PN+1=(0,1). We say P satisfies condition ()i if (Pi1,Pi,Pi+1) is positive and Pi+2int(Pi1,Pi,Pi+1). Then, P is a type-β N-representative of 2-spirals iff P satisfies ()i for all i>N. Notice that P satisfies ()N, so by an induction argument, it suffices to show that for iN, if P satisfies ()i, then P satisfies ()i+1.

If P satisfies ()i, then (Pi1,Pi,Pi+1) is positive. Normalize by Aff2+() so that Pi1=(1,0), Pi=(0,0), and Pi=(0,1). We will use the same notation la,b and sa,b as Proposition 7.1. Since Pi+1int(Pi2,Pi1,Pi), we have s1,2>1 and s1,2>1. Then, since x2i>0 and x2i+1<0, Equation (35) gives us

s1,21s1,21>0 and s1,2(s1,21)s1,2(s1,21)<0.

It follows that s1,2>1 and 11s1,2<0. The latter inequality implies 1s1,2>1, so in particular s1,2>0 and hence s1,2(0,1). The two conditions s1,2>1 and s1,2(0,1) implies Pi+2int(Pi1,Pi,Pi+1) and (Pi,Pi+1,Pi+2) positive, so P satisfies ()i+1 as desired. We conclude that [P]𝒮2,nβ. ∎

7.2 The Precompactness of T2 Orbits

We adapt the argument for Theorem 1.3 to give a quick proof of Theorem 1.4 using the Casimir functions of the T2-invariant Poisson structure on 𝒫n that were developed in [23, Theorem 1.2]. One can find the proof of the following lemma in [23, §2.2].

Lemma 7.3.

For the map T2 acting on a twisted n-gon P with corner invariants xj=xj(P), one has the following four invariant quantities.

O1(P) =i=0n1(x2i+1+x2i1x2ix2i+1); On(P) =i=0n1x2i+1;
E1(P) =i=0n1(x2i+x2i2x2i1x2i); En(P) =i=1nx2i.

We continue to use the notation from §6.2. In addition, we write O1,m=O1(T2m(P)). We define On,m, E1,m, and En,m analogously. By Lemma 7.3, the values of these four quantities are independent of the choice of m.

Lemma 7.4.

For all n2, given [P]𝒮2,nβ, there exists a,b>0 such that x2i,m[a,b] for all i[n] and m0.

Proof.

Fix i[n]. We first show that x2i,m is uniformly bounded above by some b>0. Since T2m(P)𝒮2,nβ for all m0, we must have E1,m<x2i,m<0. Then, if x2i,m on a subsequence, E1,m also diverges on the same subsequence, but that contradicts invariance of E1,m. This implies x2i,m<bi for some bi>0. Taking b=maxi[n]bi satisfies the condition in the lemma.

Next, we show that x2i,m is uniformly bounded below by some a>0. We first notice that En,m<bin. This implies if x2i,m0 on a subsequence, then En,m0 on the same subsequence, but that contradicts invariance of En,m. Therefore, x2i,m>ai for some ai>0. Taking a=mini[n]ai completes the proof. ∎

Lemma 7.5.

For all n2, given [P]𝒮2,nβ, there exists c,d<0 such that x2i+1,m[c,d] for all i[n] and m0.

Proof.

The argument is analogous to the proof of Lemma 7.4. Fix i[n]. To find ci that bounds {x2i+1,m} uniformly from below, we use the fact that O1,m>x2i+1>0. We then set c=mini[n]ci. To find di that bounds {x2i+1,m} uniformly from above, we use the fact that |On(P)|<|cn|. We then set d=maxi[n]di to complete the proof. ∎

Lemma 7.4 and 7.5 together implies that the forward T2-orbit of any [P]𝒮2,nβ is precompact in 𝒫n. One can use the same argument to show that the backward T2-orbit is also precompact. We have thus completed the proof of Theorem 1.4.

8 Appendix

8.1 Conjectures for Invariants

Given [P]𝒫k,n, we may consider the following quantity:

yi(k)(P)=χ(Pi,PiPi+kPi1Pi+k1,PiPi+kPi+1Pi+k+1,Pi+k). (36)

When Tkj(P) is well-defined, we write yi,j(k)=yi(k)(Tkj(P)), or simply yi,j if the value of k is clear from the context. Let Yj(k) (or simply Yj) denote the product i=0n1yi,j(k).

Proposition 8.1.

For all k,n2, given [P]𝒮k,nα, there exists C, C0, such that

Yj+1(k)(Yj(k))1=C (37)

for all j. The same holds for [P]𝒮k,nβ.

Proof.

The grid (P^i,j)(i,j)2 where P^i,j is the i-th vertex of Tkj(P) is a Y-mesh of type S={(0,0),(k,0),(1,1),(0,1)} with yr=yr(P^) for r2. The proof is essentially the same as the one for Proposition 5.11, so we will omit it. Then, from [7, Theorem 1.6], we have

yi+k,jyi1,j+2=(1+yi1,j+1)(1+yi+k,j+1)(1+yi,j+11)(1+yi+k1,j+11). (38)

It follows that

YjYj+2 =i=0n1(yi+k,jyi1,j+2)=i=0n1(1+yi1,j+1)(1+yi+k,j+1)(1+yi,j+11)(1+yi+k1,j+11) (39)
=(i=0n1(yi,j+1yi+k1,j+1))(i=0n1(1+yi1,j+1)(1+yi+k,j+1)(1+yi,j+1)(1+yi+k1,j+1))=Yj+12.

This implies Yj+2/Yj+1=Yj+1/Yj for all j. Taking C=Y1/Y0 completes the proof. ∎

Remark 8.2.

Combining the results of [7] and [4], we see that Proposition 8.1 is equivalent to [4, Theorem 2.1]. Specifically, the quantity in Equation (37) is shown to be a Casimir function with respect to a Poisson structure that is invariant under the y-variable transformation of a quiver Qk, which we will define below.

Consider the infinite directed graph Qk whose vertices are indexed by ×{0,1}, with directed edges (i,0)(i1,1), (i,0)(ik,1), (i,1)(i,0), and (ik1,1)(i,0) for all i. See Figure 25 for a visual representation of this quiver. We refer the readers to [7, §9] for the construction of this quiver and the proof that the y-variable transformations satisfy (38).

Refer to caption
Figure 25: The quiver Qk. Only edges into and out of the vertex (i,0) are shown.

For all n2, the y-variables corresponding to [P]𝒫k,n are periodic modulo n. We may then identify vertices of Qk via (i,j)(i+n,j), and similarly identify the corresponding edges. The resulting directed graph Qk,n is isomorphic to the quiver 𝒬k,n from [4] by applying a translation to the first entry of the vertices (i,1). Moreover, the y-variables (yi,0)i[n] of the quiver in [7] transforms in the same way as the p-variables (pi)i[n] of 𝒬k,n under the map Tk¯ (see [4, §2]), and the q-variables (qi)i[n] of 𝒬k,n correspond to the multiplicative inverse of yi,1. As a result, Y0(k)/Y1(k)=i=1npiqi, which by [4, Theorem 2.1] is invariant under Tk¯ and forms a Casimir function with respect to a Poisson structure that is invariant under Tk¯.

Both [7] and [4] demonstrate that the quiver Qk,n is a bipartite graph that can be embedded into a torus. For further details, see [7, §9] and [4, §3]. This connection links Qk,n to the Goncharov-Kenyon Dimer Integrable Systems in [8], where a more general definition of Casimir functions is provided.

Conjecture 8.3.

The constant C in Proposition 8.1 equals 1 for all k2.

We prove Conjecture 8.3 for k=2 and k=3. Let xj=xj(P) be the corner invariants of [P]. The case k=2 follows from yi(2)(P)=x2i+1x2i+2 (see Equation (32)), so

i=1nyi(2)(P)=(1)ni=1nx2i+1x2i+2=(1)nOn(P)En(P),

which is T2-invariant by Lemma 7.3.

For the case k=3, Equation (31) implies

i=1nyi(3)(P)=(1)ni=1nx2ix2i+3(x2i1)(x2i+31)=(1)n1(P)2(P),

which is T3-invariant by Proposition 6.1.

References

  • [1] R. Felipe and G. Marí Beffa. The pentagram map on Grassmannians. Annales de l’Institut Fourier, 69(1):421–456, 2019.
  • [2] V. V. Fock and A. Marshakov. Loop groups, clusters, dimers and integrable systems, pages 1–65. Springer International Publishing, Cham, 2016.
  • [3] P. Di Francesco and R. Kedem. T-systems with boundaries from network solutions, 2012.
  • [4] M. Gekhtman, M. Shapiro, S. Tabachnikov, and A. Vainshtein. Higher pentagram maps, weighted directed networks, and cluster dynamics. Electronic Research Announcements, 19(0):1–17, 2012.
  • [5] M. Glick. The pentagram map and y-patterns. Advances in Mathematics, 227(2):1019–1045, 2011.
  • [6] M. Glick. The limit point of the pentagram map. International Mathematics Research Notices, 2020(9):2818–2831, 2018.
  • [7] M. Glick and P. Pylyavskyy. Y-meshes and generalized pentagram maps. Proceedings of the London Mathematical Society, 112(4):753–797, 2016.
  • [8] A. B. Goncharov and R. Kenyon. Dimers and cluster integrable systems. Annales scientifiques de l’École Normale Supérieure, Ser. 4, 46(5):747–813, 2013.
  • [9] A. Izosimov. The pentagram map, poncelet polygons, and commuting difference operators. Compositio Mathematica, 158(5):1084–1124, 2022.
  • [10] A. Izosimov. Pentagram maps and refactorization in poisson-lie groups. Advances in Mathematics, 404:108476, 2022.
  • [11] A. Izosimov and B. Khesin. Long-diagonal pentagram maps. Bulletin of the London Mathematical Society, 55(3):1314–1329, 2023.
  • [12] R. Kedem and P. Vichitkunakorn. T-systems and the pentagram map. Journal of Geometry and Physics, 87:233–247, 2015. Finite dimensional integrable systems: on the crossroad of algebra, geometry and physics.
  • [13] B. Khesin and F. Soloviev. Integrability of higher pentagram maps. Mathematische Annalen, 357(3):1005–1047, 2013.
  • [14] B. Khesin and F. Soloviev. The geometry of dented pentagram maps. Journal of the European Mathematical Society, 018(1):147–179, 2016.
  • [15] G. Marí Beffa. On generalizations of the pentagram map: discretizations of agd flows. Journal of Nonlinear Science, 23(2):303–334, 2013.
  • [16] G. Marí Beffa. On integrable generalizations of the pentagram map. International Mathematics Research Notices, 2015(12):3669–3693, 2014.
  • [17] D. Nackan and R. Speciel. Continuous limits of generalized pentagram maps. Journal of Geometry and Physics, 167:104292, 2021.
  • [18] V. Ovsienko, R. E. Schwartz, and S. Tabachnikov. The pentagram map: a discrete integrable system. Communications in Mathematical Physics, 299(2):409–446, 2010.
  • [19] V. Ovsienko, R. E. Schwartz, and S. Tabachnikov. Liouville-arnold integrability of the pentagram map on closed polygons. Duke Mathematical Journal, 162(12):2149 – 2196, 2013.
  • [20] R. E. Schwartz. The pentagram map. Experimental Mathematics, 1(1):71 – 81, 1992.
  • [21] R. E. Schwartz. The pentagram map is recurrent. Experimental Mathematics, 10:519–528, 2001.
  • [22] R. E. Schwartz. The poncelet grid. Advances in Geometry, 7(2):157–175, 2007.
  • [23] R. E. Schwartz. Discrete monodromy, pentagrams, and the method of condensation. Journal of Fixed Point Theory and Applications, 3(2):379–409, 2008.
  • [24] R. E. Schwartz. Pentagram spirals. Experimental Mathematics, 22(4):384–405, 2013.
  • [25] R. E. Schwartz. A textbook case of pentagram rigidity, 2021.
  • [26] R. E. Schwartz. Pentagram rigidity for centrally symmetric octagons. International Mathematics Research Notices, 2024(12):9535–9561, 2024.
  • [27] R. E. Schwartz. The flapping birds in the pentagram zoo. Arnold Mathematical Journal, page to appear, 2025.
  • [28] F. Soloviev. Integrability of the pentagram map. Duke Mathematical Journal, 162(15):2815–2853, 2013.
  • [29] M. H. Weinreich. The algebraic dynamics of the pentagram map. Ergodic Theory and Dynamical Systems, 43(10):3460–3505, 2023.