The Pentagram Map and Y -patterns Max Glick University of Michigan - - PowerPoint PPT Presentation

The Pentagram Map and Y -patterns

Max Glick University of Michigan April 20, 2011

The pentagram map

Given a polygon A,

The pentagram map

Given a polygon A, draw its shortest diagonals

The pentagram map

Given a polygon A, draw its shortest diagonals and use them as the sides of a new polygon T(A).

The pentagram map

Given a polygon A, draw its shortest diagonals and use them as the sides of a new polygon T(A). T is known as the pentagram map.

Related work

◮ R. Schwartz, The pentagram map, Experiment. Math. 1

(1992), 71–81.

◮ R. Schwartz, Discrete monodromy, pentagrams, and the

method of condensation, J. Fixed Point Theory Appl. 3 (2008), 379–409.

◮ V. Ovsienko, R. Schwartz, and S. Tabachnikov, The

pentagram map: a discrete integrable system, Comm. Math.

• Phys. 299 (2010), 409-446.

◮ S. Morier-Genoud, V. Ovsienko, and S. Tabachnikov, 2-frieze

patterns and the cluster structure of the space of polygons, arXiv:1008.3359

Goals

• 1. Define coordinates on the space of n-gons and express the

pentagram map T in these coordinates.

• 2. Give a non-recursive formula for T k = T ◦ T ◦ . . . ◦ T
• k

.

• 3. Use this formula to better understand the long run behavior of

the pentagram map.

Theorem (R. Schwartz (m even))

Let A be a closed, axis-aligned 2m-gon. Then the vertices of T m−2(A) lie alternately on 2 lines. T m−2

Example (of theorem)

Example (of theorem)

1 2 3 4 5 6 7 8

Example (of theorem)

1 2 3 4 5 6 7 8

Example (of theorem)

Example (of theorem)

1 2 3 4 5 6 7 8

Example (of theorem)

1 2 3 4 5 6 7 8

Example (of theorem)

Example (of theorem)

The space of twisted polygons

A twisted polygon is a sequence A = (Ai)i∈Z of points in the projective plane that is periodic modulo some projective transformation φ, i.e., Ai+n = φ(Ai) for all i ∈ Z.

b b b

b b b b bb b b b bb b b b b

b b b

A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 A11 A12 A13 Let Pn = {twisted n-gons}/(projective equivalence).

Alternate Indexing Scheme

b b b b b

A1 A2 A3 A4 A0 = A5

b b b b b

B1.5 B2.5 B3.5 B4.5 B0.5 It is convenient to also allow twisted polygons to be indexed by

1 2 + Z. Let P∗ n be the space of twisted polygons indexed in this

manner, modulo projective equivalence.

Cross Ratios

The cross ratio of 4 real numbers a, b, c, d is defined to be χ(a, b, c, d) = (a − b)(c − d) (a − c)(b − d)

◮ The cross ratio of 4 collinear points in the plane is defined

similarly using signed distances along the common line.

◮ Cross ratios are invariant under projective transformations.

Schwartz’ coordinate system

Let A be a twisted polygon. The x-coordinates of A are the cross ratios x2k(A) = χ(Ak−2, Ak−1, B, D) x2k+1(A) = χ(Ak+2, Ak+1, C, D) for B, C, D as below.

b b b b b

Ak−2 Ak−1 Ak Ak+1 Ak+2

b b b b b b b b

B C D

Proposition (Schwartz)

The map (x1, . . . , x2n) : Pn → R2n restricts to a bijection between dense open subsets of the domain and range. The same holds with Pn replaced by P∗

n

Proposition (Schwartz)

Let A be a twisted n-gon indexed by 1

2 + Z. Let xj = xj(A). Then

xj(T(A)) =        xj−1 1 − xj−3xj−2 1 − xj+1xj+2 , j even xj+1 1 − xj+3xj+2 1 − xj−1xj−2 , j odd Alternately, if A is indexed by Z then xj(T(A)) =        xj+1 1 − xj+3xj+2 1 − xj−1xj−2 , j even xj−1 1 − xj−3xj−2 1 − xj+1xj+2 , j odd

The y-parameters

The y-parameters of a twisted polygon A are the cross ratios y2k(A) = −(χ(Ak−1, B, C, Ak+1))−1 y2k+1(A) = −χ(D, Ak, Ak+1, E) for B, C, D, E as below.

b b b b b

Ak−2 Ak−1 Ak Ak+1 Ak+2

b b b b

B C

b b b b b b

Ak−2 Ak−1 Ak Ak+1 Ak+2 Ak+3

b b b b

D E

Properties of y-parameters

The y-parameters yj = yj(A) of a twisted n-gon A are related to its x-coordinates via: y2k = −(x2kx2k+1)−1 y2k+1 = −x2k+1x2k+2 It follows that y1y2 · · · y2n = 1.

Proposition

A twisted n-gon A can be reconstructed up to projective equivalence from y1, . . . , y2n together with additional quantities On = x1x3 · · · x2n−1 and En = x2x4 · · · x2n.

Properties of y-parameters

The y-parameters yj = yj(A) of a twisted n-gon A are related to its x-coordinates via: y2k = −(x2kx2k+1)−1 y2k+1 = −x2k+1x2k+2 It follows that y1y2 · · · y2n = 1.

Proposition

A twisted n-gon A can be reconstructed up to projective equivalence from y1, . . . , y2n together with additional quantities On = x1x3 · · · x2n−1 and En = x2x4 · · · x2n.

Lemma (Schwartz)

The quantities On and En are interchanged by the pentagram map, i.e. On(T(A)) = En(A) and En(T(A)) = On(A).

A formula for T

Proposition

Let A be a twisted n-gon indexed by 1

2 + Z. Let yj = yj(A). Then

y ′

j def

= yj(T(A)) =    yj−3yjyj+3 (1 + yj−1)(1 + yj+1) (1 + yj−3)(1 + yj+3), j even y −1

j

, j odd If A is indexed by Z then y ′′

j def

= yj(T(A)) =    y −1

j

, j even yj−3yjyj+3 (1 + yj−1)(1 + yj+1) (1 + yj−3)(1 + yj+3), j odd

A recursive formula for T k

Let α1 : (y1, . . . , y2n) → (y ′

1, . . . , y ′ 2n)

and α2 : (y1, . . . , y2n) → (y ′′

1 , . . . , y ′′ 2n)

be the rational maps defined on the previous slide. If A ∈ Pn then it follows that the y-parameters of T k(A) can be expressed in terms of y1, . . . , y2n by the rational map . . . ◦ α2 ◦ α1 ◦ α2

• k

.

Y -patterns [Fomin, Zelevinsky]

A Y -seed is a pair (y, Q) where y = (y1, . . . , yn) is a collection of rational functions and Q is a quiver, i.e. a directed graph on vertex set {1, 2, . . . , n} without oriented 2-cycles. Given a Y -seed (y, Q) and some k ∈ {1, . . . , n}, the mutation µk in direction k results in a new Y -seed µk(y, Q) = (y′, Q′), where

Y -patterns [Fomin, Zelevinsky]

A Y -seed is a pair (y, Q) where y = (y1, . . . , yn) is a collection of rational functions and Q is a quiver, i.e. a directed graph on vertex set {1, 2, . . . , n} without oriented 2-cycles. Given a Y -seed (y, Q) and some k ∈ {1, . . . , n}, the mutation µk in direction k results in a new Y -seed µk(y, Q) = (y′, Q′), where

◮ The vector y′ is obtained from y via the following steps:

• 1. For each j → k in Q, multiply yj by 1 + yk.
• 2. For each k → j in Q multiply yj by

yk 1+yk .

• 3. Invert yk

Y -patterns [Fomin, Zelevinsky]

A Y -seed is a pair (y, Q) where y = (y1, . . . , yn) is a collection of rational functions and Q is a quiver, i.e. a directed graph on vertex set {1, 2, . . . , n} without oriented 2-cycles. Given a Y -seed (y, Q) and some k ∈ {1, . . . , n}, the mutation µk in direction k results in a new Y -seed µk(y, Q) = (y′, Q′), where

◮ The vector y′ is obtained from y via the following steps:

• 1. For each j → k in Q, multiply yj by 1 + yk.
• 2. For each k → j in Q multiply yj by

yk 1+yk .

• 3. Invert yk

◮ The quiver Q′ is obtained from Q via the following steps:

• 1. For every length 2 path i → k → j, add an arc from i to j.
• 2. Reverse the orientation of all arcs incident to k.
• 3. Remove all oriented 2-cycles.

An example of a Y -seed mutation

(y1, y2, y3, y4) 1 2 4 3 (y1

y2 1+y2 , 1 y2, y3(1 + y2), y4)

1 2 4 3 µ2

A Y -pattern related to the pentagram map

Let Q0 be the quiver of rank 2n with arrows 2i → (2i ± 3) and (2i ± 1) → 2i for all i = 1, . . . , n (indices are taken modulo 2n).

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Theorem

Let µeven and µodd be the compound mutations µeven = µ2n ◦ . . . ◦ µ4 ◦ µ2 µodd = µ2n−1 ◦ . . . ◦ µ3 ◦ µ1 Let −Q0 denote the quiver Q0 with all its arrows reversed. Then µeven(y, Q0) = (α2(y), −Q0) µodd(y, −Q0) = (α1(y), Q0)

Theorem

Let µeven and µodd be the compound mutations µeven = µ2n ◦ . . . ◦ µ4 ◦ µ2 µodd = µ2n−1 ◦ . . . ◦ µ3 ◦ µ1 Let −Q0 denote the quiver Q0 with all its arrows reversed. Then µeven(y, Q0) = (α2(y), −Q0) µodd(y, −Q0) = (α1(y), Q0)

Corollary

Let A ∈ Pn and let y0 = (y1, . . . , y2n) be its y-parameters. Compute the yk = (y1,k, . . . , y2n,k) for k ≥ 1 by the sequence of mutations (y0, Q0)

µeven

− − − → (y1, −Q0)

µodd

− − → (y2, Q0)

µeven

− − − → (y3, −Q0)

µodd

− − → · · · Then, yj,k = yj(T k(A)).

A formula for T k

General cluster algebra theory (see [CA-IV]) applied in this setting proves that yj(T k(A)) =        Mj,k Fj−1,kFj+1,k Fj−3,kFj+3,k , j + k even M−1

j,k−1

Fj−3,k−1Fj+3,k−1 Fj−1,k−1Fj+1,k−1 , j + k odd where the Mj,k are monomials in y1, . . . , y2n and the Fj,k are polynomials in y1, . . . , y2n.

Components of the formula

◮ The monomials in the formula for T k are given by

Mj,k =

k

• i=−k

yj+3i

◮ The Fj,k are the so called F-polynomials of the corresponding

cluster algebra, and are defined recursively by Fj,−1 = 1 Fj,0 = 1 Fj,k+1 = Fj−3,kFj+3,k + Mj,kFj−1,kFj+1,k Fj,k−1 , for k ≥ 0 We will identify the F-polynomials as generating functions of the order ideals of a sequence of posets denoted Pk.

The posets Pk

These posets were defined by Elkies, Kuperberg, Larsen, and Propp in their study of domino tilings of the Aztec diamond:

◮ Pk = {(a, b, c, d) ∈ Z≥0 : a + b + c + d ∈ {k − 2, k − 1}}. ◮ (a, b, c, d) ≤ (a′, b′, c′, d′) if and only if a ≥ a′,b ≥ b′, c ≤ c′,

and d ≤ d′. (0,0,1,0) (0,0,0,1) (0,0,0,0) (1,0,0,0) (0,1,0,0) P2 (0,0,2,0) (0,0,1,1) (0,0,0,2) (0,0,1,0) (0,0,0,1) (1,0,1,0) (1,0,0,1) (0,1,1,0) (0,1,0,1) (1,0,0,0) (0,1,0,0) (2,0,0,0) (1,1,0,0) (0,2,0,0) P3

Formula for the F-polynomials

Theorem

Fj,k =

• I∈J(Pk)
• (a,b,c,d)∈I

y3(b−a)+(d−c)+j where J(Pk) denotes the set of order ideals of Pk.

Example

y−1 y1 y0 y−3 y3 F0,2 = 1 + y−3 + y3 + y−3y3 + y−3y0y3(1 + y−1 + y1 + y−1y1)

Summary

◮ If j + k is even then

yj(T k(A)) = Mj,k Fj−1,kFj+1,k Fj−3,kFj+3,k where Mj,k =

k

• i=−k

yj+3i and Fj,k =

• I∈J(Pk)
• (a,b,c,d)∈I

y3(b−a)+(d−c)+j

◮ If j + k is odd then

yj(T k(A)) = M−1

j,k−1

Fj−3,k−1Fj+3,k−1 Fj−1,k−1Fj+1,k−1

More axis-aligned polygons

Our formula lets us prove a twisted analogue of Schwartz’s theorem about axis-aligned polygons:

Theorem

Let A be a twisted, axis-aligned 2m-gon with Ai+2m = φ(Ai). Suppose that φ fixes every point at infinity. Then the vertices of T m−1(A) lie alternately on 2 lines.

Lemma

Let A and B be twisted polygons, each with no 3 consecutive vertices collinear. Then

• 1. ←

− − − − → Ai−2Ai−1, ← − − − → AiAi+1, ← − − − − → Ai+2Ai+3 are concurrent if and only if y2i+1(A) = −1.

• 2. Bi−2, Bi, Bi+2 are collinear if and only if y2i(B) = −1.

Proof outline of theorem.

A ∈ P2m axis-aligned, φ fixes every point at infinity = ⇒ y1 = y3 = · · · = y4m−1 = −1 y2y6 · · · y4m−2 = y4y8 · · · y4m = (−1)m = ⇒ 0 = Fj,m for all j ≡ m − 1 (mod 2) = ⇒ 0 = Fj−3,m−1Fj+3,m−1 + Mj,m−1Fj−1,m−1Fj+1,m−1 for all j ≡ m − 1 (mod 2) = ⇒ −1 = Mj,m−1Fj−1,m−1Fj+1,m−1 Fj−3,m−1Fj+3,m−1 for all j ≡ m − 1 (mod 2) = ⇒ −1 = yj(T m−1(A)) for all j ≡ m − 1 (mod 2) = ⇒ The vertices of T m−1(A) lie alternately on 2 lines.

Example (m=2)

The claim is that y1 = y3 = y5 = y7 = −1 y2y6 = y4y8 = 1

• =

⇒F2j+1,2 = 0 For instance F5,2 = 1 + y2 + y8 + y2y8 + y2y5y8(1 + y4 + y6 + y4y6) = 1 + y2 + y8 − y2y4y8 − y2y6y8 − y2y4y6y8 = (1 − (y2y6)(y4y8)) + y2(1 − y4y8) + y8(1 − y2y6) = 0 since y5 = −1 and y2y6 = y4y8 = 1.

Open questions

◮ How can singularities of the pentagram map, which are

described algebraically by the vanishing of some F-polynomial, be understood geometrically?

◮ Are other geometric constructions like the pentagram map

related to cluster algebras in a similar way?

◮ Is there some good reason why results like the one concerning

axis-aligned polygons should hold?