The Pentagram Map and Y -patterns

Max GlickUniversity 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 asthe sides of a new polygon T (A).

The pentagram map

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

map.

Related work

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

(1992), 71–81.

I R. Schwartz, Discrete monodromy, pentagrams, and themethod of condensation, J. Fixed Point Theory Appl. 3

(2008), 379–409.

I V. Ovsienko, R. Schwartz, and S. Tabachnikov, Thepentagram map: a discrete integrable system, Comm. Math.

Phys. 299 (2010), 409-446.

I S. Morier-Genoud, V. Ovsienko, and S. Tabachnikov, 2-friezepatterns and the cluster structure of the space of polygons,arXiv:1008.3359

Goals

1. Define coordinates on the space of n-gons and express thepentagram 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 ofthe pentagram map.

Theorem (R. Schwartz (m even))

Let A be a closed, axis-aligned 2m-gon. Then the vertices of

Tm−2(A) lie alternately on 2 lines.

Tm−2

Example (of theorem)

Example (of theorem)

1

2

3

45 67

8

Example (of theorem)

1

2

3

45 67

8

Example (of theorem)

Example (of theorem)

1

2

3

4

56 7

8

Example (of theorem)

1

2

3

4

56 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 theprojective plane that is periodic modulo some projectivetransformation φ, i.e., Ai+n = φ(Ai ) for all i ∈ Z.

b b b

b

b

b

b

b b

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

bb

B1.5

B2.5

B3.5

B4.5B0.5

It is convenient to also allow twisted polygons to be indexed by12 + Z. Let P∗

n be the space of twisted polygons indexed in thismanner, 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)

I The cross ratio of 4 collinear points in the plane is definedsimilarly using signed distances along the common line.

I Cross ratios are invariant under projective transformations.

Schwartz’ coordinate system

Let A be a twisted polygon. The x-coordinates of A are the crossratios

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

bb

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

xj(T (A)) =

xj−11− xj−3xj−2

1− xj+1xj+2, j even

xj+11− xj+3xj+2

1− xj−1xj−2, j odd

Alternately, if A is indexed by Z then

xj(T (A)) =

xj+11− xj+3xj+2

1− xj−1xj−2, j even

xj−11− 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

bb

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 toits 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 toits 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 12 + Z. Let yj = yj(A). Then

y ′jdef= yj(T (A)) =

yj−3yjyj+3(1 + yj−1)(1 + yj+1)

(1 + yj−3)(1 + yj+3), j even

y−1j , j odd

If A is indexed by Z then

y ′′jdef= yj(T (A)) =

y−1j , 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

Letα1 : (y1, . . . , y2n) 7→ (y ′1, . . . , y

′2n)

andα2 : (y1, . . . , y2n) 7→ (y ′′1 , . . . , y

′′2n)

be the rational maps defined on the previous slide. If A ∈ Pn thenit follows that the y -parameters of T k(A) can be expressed interms 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 ofrational functions and Q is a quiver, i.e. a directed graph on vertexset {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 ofrational functions and Q is a quiver, i.e. a directed graph on vertexset {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

I 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

yk1+yk

.3. Invert yk

Y -patterns [Fomin, Zelevinsky]

A Y -seed is a pair (y,Q) where y = (y1, . . . , yn) is a collection ofrational functions and Q is a quiver, i.e. a directed graph on vertexset {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

I 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

yk1+yk

.3. Invert yk

I 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

(y1y2

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

1 2

4 3µ2

A Y -pattern related to the pentagram mapLet 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

TheoremLet 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)

TheoremLet 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(Tk(A)).

A formula for T k

General cluster algebra theory (see [CA-IV]) applied in this settingproves that

yj(Tk(A)) =

Mj ,k

Fj−1,kFj+1,k

Fj−3,kFj+3,k, j + k even

M−1j ,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 arepolynomials in y1, . . . , y2n.

Components of the formula

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

Mj ,k =k∏

i=−k

yj+3i

I The Fj ,k are the so called F -polynomials of the correspondingcluster 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 ofthe order ideals of a sequence of posets denoted Pk .

The posets Pk

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

I Pk = {(a, b, c , d) ∈ Z≥0 : a+ b + c + d ∈ {k − 2, k − 1}}.

I (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

I If j + k is even then

yj(Tk(A)) = Mj ,k

Fj−1,kFj+1,k

Fj−3,kFj+3,k

where

Mj ,k =

k∏

i=−k

yj+3i

andFj ,k =

∑

I∈J(Pk )

∏

(a,b,c,d)∈I

y3(b−a)+(d−c)+j

I If j + k is odd then

yj(Tk(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’stheorem about axis-aligned polygons:

TheoremLet 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

Tm−1(A) lie alternately on 2 lines.

LemmaLet 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 = −1y2y6 · · · 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−1for all j ≡ m − 1 (mod 2)

=⇒ −1 = yj(Tm−1(A)) for all j ≡ m − 1 (mod 2)

=⇒ The vertices of Tm−1(A) lie alternately on 2 lines.

Example (m=2)

The claim is that

y1 = y3 = y5 = y7 = −1y2y6 = 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

I How can singularities of the pentagram map, which aredescribed algebraically by the vanishing of some F -polynomial,be understood geometrically?

I Are other geometric constructions like the pentagram maprelated to cluster algebras in a similar way?

I Is there some good reason why results like the one concerningaxis-aligned polygons should hold?