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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?