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 2 m-gon. Then the vertices of T m − 2 ( A ) lie alternately on 2 lines. T m − 2

Example (of theorem)

Example (of theorem) 5 7 2 6 4 1 8 3

Example (of theorem) 5 7 2 6 4 1 8 3

Example (of theorem)

Example (of theorem) 6 7 5 2 4 8 3 1

Example (of theorem) 6 7 5 2 4 8 3 1

Example (of theorem)

Example (of theorem)

b b b b bb b b b bb b b b b b b b b b b The space of twisted polygons A twisted polygon is a sequence A = ( A i ) i ∈ Z of points in the projective plane that is periodic modulo some projective transformation φ , i.e., A i + n = φ ( A i ) for all i ∈ Z . A 3 A 7 A 11 A 2 A 6 A 10 A 4 A 8 A 12 A 1 A 5 A 9 A 13 Let P n = { twisted n -gons } / ( projective equivalence ).

b b b b b b b b b b Alternate Indexing Scheme A 4 B 3 . 5 A 0 = A 5 A 3 B 4 . 5 B 0 . 5 B 2 . 5 B 1 . 5 A 1 A 2 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.

b b b b b b b b b b b b b Schwartz’ coordinate system Let A be a twisted polygon. The x -coordinates of A are the cross ratios x 2 k ( A ) = χ ( A k − 2 , A k − 1 , B , D ) x 2 k +1 ( A ) = χ ( A k +2 , A k +1 , C , D ) for B , C , D as below. A k +2 A k +1 C A k D A k − 2 A k − 1 B

Proposition (Schwartz) The map ( x 1 , . . . , x 2 n ) : P n → R 2 n restricts to a bijection between dense open subsets of the domain and range. The same holds with P n replaced by P ∗ n Proposition (Schwartz) Let A be a twisted n-gon indexed by 1 2 + Z . Let x j = x j ( A ) . Then  1 − x j − 3 x j − 2  , x j − 1 j even   1 − x j +1 x j +2 x j ( T ( A )) = 1 − x j +3 x j +2  x j +1 , j odd   1 − x j − 1 x j − 2 Alternately, if A is indexed by Z then  1 − x j +3 x j +2  , x j +1 j even   1 − x j − 1 x j − 2 x j ( T ( A )) = 1 − x j − 3 x j − 2  x j − 1 , j odd   1 − x j +1 x j +2

b b b b b b b b b b b b b b b b b b b The y -parameters The y -parameters of a twisted polygon A are the cross ratios y 2 k ( A ) = − ( χ ( A k − 1 , B , C , A k +1 )) − 1 y 2 k +1 ( A ) = − χ ( D , A k , A k +1 , E ) for B , C , D , E as below. A k +3 A k +2 A k +2 A k +1 E C A k +1 B A k A k A k − 2 A k − 1 D A k − 2 A k − 1

Properties of y -parameters The y -parameters y j = y j ( A ) of a twisted n -gon A are related to its x -coordinates via: y 2 k = − ( x 2 k x 2 k +1 ) − 1 y 2 k +1 = − x 2 k +1 x 2 k +2 It follows that y 1 y 2 · · · y 2 n = 1. Proposition A twisted n-gon A can be reconstructed up to projective equivalence from y 1 , . . . , y 2 n together with additional quantities O n = x 1 x 3 · · · x 2 n − 1 and E n = x 2 x 4 · · · x 2 n .

Properties of y -parameters The y -parameters y j = y j ( A ) of a twisted n -gon A are related to its x -coordinates via: y 2 k = − ( x 2 k x 2 k +1 ) − 1 y 2 k +1 = − x 2 k +1 x 2 k +2 It follows that y 1 y 2 · · · y 2 n = 1. Proposition A twisted n-gon A can be reconstructed up to projective equivalence from y 1 , . . . , y 2 n together with additional quantities O n = x 1 x 3 · · · x 2 n − 1 and E n = x 2 x 4 · · · x 2 n . Lemma (Schwartz) The quantities O n and E n are interchanged by the pentagram map, i.e. O n ( T ( A )) = E n ( A ) and E n ( T ( A )) = O n ( A ) .

A formula for T Proposition Let A be a twisted n-gon indexed by 1 2 + Z . Let y j = y j ( A ) . Then  (1 + y j − 1 )(1 + y j +1 ) (1 + y j − 3 )(1 + y j +3 ) ,  y j − 3 y j y j +3 j even def y ′ = y j ( T ( A )) = j  y − 1 , j odd j If A is indexed by Z then  y − 1 , j even  j def y ′′ = y j ( T ( A )) = (1 + y j − 1 )(1 + y j +1 ) j y j − 3 y j y j +3 (1 + y j − 3 )(1 + y j +3 ) , j odd 

A recursive formula for T k Let α 1 : ( y 1 , . . . , y 2 n ) �→ ( y ′ 1 , . . . , y ′ 2 n ) and α 2 : ( y 1 , . . . , y 2 n ) �→ ( y ′′ 1 , . . . , y ′′ 2 n ) be the rational maps defined on the previous slide. If A ∈ P n then it follows that the y -parameters of T k ( A ) can be expressed in terms of y 1 , . . . , y 2 n by the rational map . . . ◦ α 2 ◦ α 1 ◦ α 2 . � �� � k

Y -patterns [Fomin, Zelevinsky] A Y -seed is a pair ( y , Q ) where y = ( y 1 , . . . , y n ) 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 µ k in direction k results in a new Y -seed µ k ( y , Q ) = ( y ′ , Q ′ ), where

Y -patterns [Fomin, Zelevinsky] A Y -seed is a pair ( y , Q ) where y = ( y 1 , . . . , y n ) 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 µ k in direction k results in a new Y -seed µ k ( y , Q ) = ( y ′ , Q ′ ), where ◮ The vector y ′ is obtained from y via the following steps: 1. For each j → k in Q , multiply y j by 1 + y k . y k 2. For each k → j in Q multiply y j by 1+ y k . 3. Invert y k

Y -patterns [Fomin, Zelevinsky] A Y -seed is a pair ( y , Q ) where y = ( y 1 , . . . , y n ) 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 µ k in direction k results in a new Y -seed µ k ( y , Q ) = ( y ′ , Q ′ ), where ◮ The vector y ′ is obtained from y via the following steps: 1. For each j → k in Q , multiply y j by 1 + y k . y k 2. For each k → j in Q multiply y j by 1+ y k . 3. Invert y k ◮ 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 1+ y 2 , 1 y 2 ( y 1 , y 2 , y 3 , y 4 ) ( y 1 y 2 , y 3 (1 + y 2 ) , y 4 ) 4 3 4 3 µ 2 1 2 1 2

A Y -pattern related to the pentagram map Let Q 0 be the quiver of rank 2 n with arrows 2 i → (2 i ± 3) and (2 i ± 1) → 2 i for all i = 1 , . . . , n (indices are taken modulo 2 n ). 4 6 2 5 3 7 1 8 16 9 15 11 13 10 14 12

Theorem Let µ even and µ odd be the compound mutations µ even = µ 2 n ◦ . . . ◦ µ 4 ◦ µ 2 µ odd = µ 2 n − 1 ◦ . . . ◦ µ 3 ◦ µ 1 Let − Q 0 denote the quiver Q 0 with all its arrows reversed. Then µ even ( y , Q 0 ) = ( α 2 ( y ) , − Q 0 ) µ odd ( y , − Q 0 ) = ( α 1 ( y ) , Q 0 )

Theorem Let µ even and µ odd be the compound mutations µ even = µ 2 n ◦ . . . ◦ µ 4 ◦ µ 2 µ odd = µ 2 n − 1 ◦ . . . ◦ µ 3 ◦ µ 1 Let − Q 0 denote the quiver Q 0 with all its arrows reversed. Then µ even ( y , Q 0 ) = ( α 2 ( y ) , − Q 0 ) µ odd ( y , − Q 0 ) = ( α 1 ( y ) , Q 0 ) Corollary Let A ∈ P n and let y 0 = ( y 1 , . . . , y 2 n ) be its y-parameters. Compute the y k = ( y 1 , k , . . . , y 2 n , k ) for k ≥ 1 by the sequence of mutations µ even µ odd µ even µ odd ( y 0 , Q 0 ) − − − → ( y 1 , − Q 0 ) − − → ( y 2 , Q 0 ) − − − → ( y 3 , − Q 0 ) − − → · · · Then, y j , k = y j ( T k ( A )) .