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On the Geometric view of Pentagram Integrals of Polygons
Inscribed in Non-Degenerate Conics.
BY
Esther Gyasi Opoku - (10638386)
BSc(Mathematics), MSc(Mathematical Sciences)
THIS THESIS IS SUBMITTED TO THE UNIVERSITY OF GHANA,
LEGON IN PARTIAL FULFILLMENT OF THE REQUIREMENT FOR
THE AWARD OF MPHIL MATHEMATICS DEGREE
JULY, 2018
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DECLARATION
This thesis was written in the Department of Mathematics, University of Ghana,
Legon from August 2017 to July 2018 in partial fullment of the requirements for
the award of Master of Philosophy degree in Mathematics under the supervision of
Dr. Richard Kena Boadi and Dr. Ralph Twum of the Kwame Nkrumah University
of Science and Technology and the University of Ghana respectively.
I hereby declare that except where due acknowledgement is made, this work has never
been presented wholly or in part for the award of a degree at the University of Ghana
or any other University.
Signature:
Student: Esther Gyasi Opoku
(egopoku001@st.ug.edu.gh)
Signature:
Dr. Richard Kena Boadi
Signature:
Dr. Ralph Twum
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ABSTRACT
The Pentagram map is a well notable integrable system that is dened on the moduli
space of polygons. In 2005, Richard Evan Schwartz introduced certain polynomials
called pentagram integrals (Monodromy invariants) of the pentagram map and
dened certain associated integrals, the analogous rst integrals. Schwartz further
studied in 2011 with S. Tabachnikov on how these integrals behave on inscribed
polygons.
They discovered that the integrals are equal for every given weight of polygons in-
scribed in non-degenerate conics. However, the proof of their outcome was combina-
torial which appeared to be more involving hence there was a need for quite a simple
proof.
Anton Izosimov in 2016 gave quite a simple conceptual geometric proof of these
invariants of polygons inscribed in non-degenerate conics.
In this thesis, we seek to analyse the geometry of these invariants by reviewing Anton's
work. Our core analyses is that for any polygon inscribed in a non-degenerate conic,
the analogous monodromy should satisfy a certain self-duality relation.
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DEDICATION
I humbly dedicate this thesis to God for giving me heavenly knowledge, wisdom,
strength and directions to produce this project. All pages of this work are dedicated
to my parents, Mr. Isaac Opoku Gyasi and Mrs Janet Opoku Gyasi for their
support and all my loved ones.
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ACKNOWLEDGEMENTS
I am most greatful to God for His guidance and for His Grace by which I can still
hope.
I am truely indebted to my supervisors Dr. Richard Kena Boadi and Dr. Ralph
Twum for their devotion, guidance, support, encouragement and helpful constructive
critique throughout this work.
My warmest appreciation to all lecturers in the department of Mathematics for
shapening me to come this far and instilling in me hope for the future.
To my dignied nancial donors, The Ghana National Petroleum Coperation (GNPC),
I sincerely appreciate your great support.
To Dr. Edward Prempeh, Dr. Bernard Oduoku Bainson and Dr. Jeery Ezearn, I
am profoundly grateful for your enormous support.
To my parents and siblings, I appreciate you for the great love and strength to come
this far. And to all my wonderful colleagues in the University of Ghana; Somé and
Cyril and all other sister universities in Ghana, thank you for your immense love and
support.
Glory be to God!
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Contents
Declaration i
Abstract iii
Dedication iv
Acknowledgements v
1 Introduction 2
2 Preliminaries 7
2.1 Projective Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.1.2 Real Projective Line . . . . . . . . . . . . . . . . . . . . . . . 8
2.1.5 Real Projective Plane . . . . . . . . . . . . . . . . . . . . . . . 8
2.2 Projective Transformations . . . . . . . . . . . . . . . . . . . . . . . . 8
2.3 Cross Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.4 Twisted Polygon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.5 Corner Invariant (The Invariant Function) . . . . . . . . . . . . . . . 18
2.6 Projective Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.7 Dual Polygon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.8 Self-dual Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
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2.8.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.9 Non-degenerate Conic . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.10 Inscribed Polygon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.11 Involution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3 Pentagram Map and Monodromy Invariants 26
3.1 Formula for the Pentagram Map . . . . . . . . . . . . . . . . . . . . . 26
3.2 Construction of Monodromy Invariants . . . . . . . . . . . . . . . . . 30
4 Monodromy Invariants of the Spectral Curve and Corner Invariants
in the Monodromy Matrix 33
4.1 Monodromy Invariants and the Spectral Curve . . . . . . . . . . . . . 33
4.2 Monodromy Matrix and the Corner Invariant . . . . . . . . . . . . . 38
5 Inscribed Polygons and their Monodromy Invariants 48
5.1 Proof of Theorem 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
6 Conclusion and Further Discussion 54
6.1 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
6.2 Further Discussion - Degenerate Conics . . . . . . . . . . . . . . . . . 55
A Other Useful Concepts 57
A.1 DEFINITIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
A.2 Integrable System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
A.2.1 Zero locus of set of Polynomial . . . . . . . . . . . . . . . . . 59
References 61
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List of Figures
1.1 A Pentagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Pentagramma Miricum . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Construction of the pentagram . . . . . . . . . . . . . . . . . . . . . . 4
1.4 Extending the sides of a pentagon in the euclidean plane: The red
dash lines represent the extension of the two parallel sides which do
not meet in the Euclidean plane . . . . . . . . . . . . . . . . . . . . . 4
2.1 Cross ratio (denition) . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 (A,B : C,D) is well dened . . . . . . . . . . . . . . . . . . . . . . . 13
2.3 corner invariant; denition. . . . . . . . . . . . . . . . . . . . . . . . . 19
2.4 corner invariant; example. . . . . . . . . . . . . . . . . . . . . . . . . 20
2.5 Construction of corner invariants at dierent vertices on a twisted m-gon. 21
2.6 Projective duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.7 Duality of the pentagon . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.8 Self Dual polygon . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.9 Non-degenerate conics shown in red . . . . . . . . . . . . . . . . . . 24
2.10 A polygon inscribed in a non-degenerate conic (ellipse) . . . . . . . . 25
2.11 Involution of the Pentagram map . . . . . . . . . . . . . . . . . . . . 25
3.1 Projective duals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.2 Duality maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
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3.3 Labelling schemes of the Pentagram Map . . . . . . . . . . . . . . . 30
5.1 Lemma 4.2.2, proof. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
6.1 Degenerate Conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
A.1 Convex hull . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
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NOTATIONS
RP2 ≡ P2(R) Real projective plane
RP1 ≡ P1(R) Real projective line
GL(n,R) General linear group of dimension n over the real numbers
Xk, Yk Monodromy invariants
PGLn Projective general linear group of dimension n
M Pentagram map
T Monodromy
xi, yi Corner invariants
S2 Sphere
(pi) Twisted polygon
(Pi) Lifts of the points (pi)
T
 Monodromy of the lifts (Pi)
Vm Space of twisted m−gons modulo projective transformation
qi Cross ratio of inscribed points projected onto a projective line
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Chapter 1
Introduction
The pentagram is intuitively dened as a ve star-like gure, (see Figure 1.1) mostly
used by the ancient Greeks and now by most religions as a symbol representing an
object they worshiped. For instance Christians used it to represent the ve wounds
of Jesus Christ on the cross. It is said that every man or woman is a pentagram, with
reference to Leonardo da Vinci's relation of Mathematics and Arts. One novelty of
the pentagram is that, it has connections with the number phi (φ ≈ 1.168) which has
several applications in real life. The pentagram, in its real life applications, is also
used in drawing maps (Google maps) with reference to the pentagram that was found
carved into the surface of the earth in Kazakhstan.
In the theory of mathematics, the pentagram was rst introduced by Gauss when he
studied the geometry of a right-angle triangle (A) inside a sphere S2 in 1876. He used
the word Pentagramma Miricum to describe the construction of the pentagram
as shown in Figure 1.2.
Studies showed that the pentagram can be constructed by either extending the edges
of a pentagon or by joining the corresponding diagonals of the vertices of a pentagon
as shown in Figure 1.3.
One can locate a pentagon either inside or outside (by joining the vertices of the
pentagram together) the pentagram. Several pentagons can be constructed from
any given initial pentagon. The map that transforms one pentagon P into another
pentagon P ′ is called Pentagram map as shown in Figure 1.3. M. Glick studied
the pentagram map for only pentagons in 1945 [7]. However, in 1992, Richard Evan
Schwartz introduced the pentagram map and dened it generally for all polygons [11].
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Figure 1.1: A Pentagram
S2
b
b
A
b
b
b
b
Figure 1.2: Pentagramma Miricum
Hence, the pentagram map in general terms is dened as the map that transforms one
polygon into another polygon. Because we obtain an equivalent polygon based on the
above constructions, we say that any two polygons P and P ′ under the pentagram
map are projectively equivalent.
The construction of the Pentagram map is similar to the constructions shown in
Figure 1.3 in the usual Euclidean plane. However, the pentagram map acting on
certain polygons cannot be constructed by the approach in Figure 1.3(a) since they
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b b
b b
P
b b b b
P ′
i
b b
m P j b
b l k b
b b
b (b) The Pentagram obtained from
the vertices of a pentagon by join-
(a) The Pentagram obtained by extending the ing the corresponding diagonals of
edges of a pentagon each vertex
Figure 1.3: Construction of the pentagram
may require the extension of parallel lines as shown in Figure 1.4
b b
b b
Figure 1.4: Extending the sides of a pentagon in the euclidean plane: The red dash lines
represent the extension of the two parallel sides which do not meet in the Euclidean plane
Now, since parallel lines do not meet in the Euclidean plane, we require a much
more convenient space for the ecient construction of the Pentagram map under the
constructions of both gures in Figure 1.3. Hence, the projective plane dened
over any eld is mostly convenient for our constructions since parallel lines meet at
innity in this space.
Schwartz wrote a number of papers such as [11], [15], [12] for general polygons and also
in recent papers [10], [17], [3]. It is discussed that the map commutes with projective
transformations and induces a generally dened map on the space of cyclically n-gons
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modulo projective transformations [18]. A bigger space called the space of twisted
polygons was introduced in [15]. In this space, the invariant polynomials,
O1, · · · , O[n ], n, E1, · · · , E[n ], n : Pn → R,2 2
under the pentagram map were produced for every n−gon (see Chapter 3). These in-
variant polynomials are calledMonodromy Invariants, also known as Pentagram
Integrals in [16].
Several works have been done in recent papers on the Pentagram map and its Inte-
grals. In [15], it was shown that the map is a completely integrable system on Pn,
that is obtained from the construction of an invariant poisson bracket on Pn together
with monodromy invariants. Additionally, [9] mainly considered the integrability of
the map on the space of cyclically ordered polygons (main space of interest in this
paper). The rst result of complete integrability of the map was introduced in [13].
[19] discusses the algebraic-geometric integrability for any monodromy either in twisted
or closed polygons. Here, the notion of spectral data which consists of Riemann sur-
face called the spectral curve was introduced. Due to this, it was shown that the
pentagram map can be written as a zero-curvature equation with a spectral parame-
ter with corresponding spectral curve and the dynamics of the Jacobian.
This approach of integrability was more preferable to that in [10] since it works on
both closed and twisted polygons. Additionally, it is equally t in both continuous
and discrete cases, which can be used to integrate the continuous limit of the map
(the Boussinesq equation).
The Lax representation was the major ingredient used to prove this algebraic inte-
grability since it organises certain invariant functions in the form of a spectral curve.
The Lax representation was used over Arnold integrability since it is useful in both
continuous and discrete cases and also works for both twisted and closed polygons.
A discrete analogue of the monodromy matrix is a monodromy operator. In relation
to [19], we shall represent our monodromy operator as scaled monodromy matrix.
In 2011, Schwartz and S. Tabachnikov had a further research on how the monodromy
invariants behave when polygons are inscribed in non-degenerate conics [18]. Their
studies resulted in the theorem below;
Theorem 1.0.1. For every n-gon P inscribed in a non-degenerate conic, Ok = Ek,
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∀k = 1, · · · , [n ], n.
2
Explicit formulas were used to prove Theorem(1.0.1) combinatorially. However, their
proof seemed to be more involving and complicated.
In 2016, Anton Izosimov gave a quite simple geometric proof of Theorem (1.0.1).
The focus of his study was that, for every given polygon that is inscribed in a non-
degenerate conic, its analogous monodromy should satisfy a particular self-duality
relation.
The core of this work is to discuss in details the geometry of the above theorem by
reviewing Anton's work. The structure of this thesis is shown as follows; Chapter
2 discusses some relevant denitions and their associated examples. In Chapter 3,
the work is discussed in two (2) sections. Whereas the rst section discusses the
formula for the pentagram map in coordinates, the second section discusses how the
monodromy invariants are constructed ( in this work, Ok and Ek are represented by
Xk and Yk respectively). Similarly, Chapter 4 discusses the occurrence of monodromy
invariants as coecients of the spectral curve in the rst section. The second section is
dedicated for the relationship between the monodromy matrix and corner invariants.
Chapter 5 discusses monodromy invariants of inscribed polygons using the self-duality
relation coupled with idea of stereographic projection of a point on a conic onto the
projective line. The results obtained is nally used to prove Theorem(1.0.1). Chapter
6 is devoted to conclusion and Appendix for further discussion.
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Chapter 2
Preliminaries
In this chapter, we dene some relevant concepts with their useful examples.
2.1 Projective Space
Denition 2.1.1. Let X be a vector space. The projective space P(X) is dened
as the set of all 1−dimensional vector subspaces of X. Thus, it is the set of lines
through the origin of the vector space X. The projective space P n(X) can be dened
as
Pn(X) = Pn(R) = Rn ∪ Pn−1(R), for X = R
Pn(R) is called the real projective space. If R is replaced by C, then the projective
space is called the complex projective space.
Suppose X is an (n+1) dimensional vector space, then the projective space P (X) has
dimension n. The 1− dimensional and 2−dimensional projective spaces are termed
as the Projective line and the Projective plane respectively. Elements (points) in the
projective space are called homogeneous coordinates.
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2.1.2 Real Projective Line
Denition 2.1.3. The real projective line is the set of lines in R2 which pass through
the origin and is obtained by the extension of the usual line when a point at innity
is added. Thus
P1(R) = R ∪ point = R ∪ {∞}.
Example 2.1.4. [8] Let n = 1. Since a 0-dimensional projective space is a single
point, we obtain the projective line P1(R) = R ∪ P0, where P0 is a point in the
projective line. Usually, we call this extra point `the point at∞'. When X = C, then
the projective line becomes the extended complex plane (C ∪ {∞}).
2.1.5 Real Projective Plane
Denition 2.1.6. The real projective plane, P2(R) = R3\ ∼ is the set of lines through
the origin of R3. Similar to the projective line where points meet at innity, a plane
in projective geometry contains a single line at innity at which the edges of the
plane meet. Note that points of P2(R) consists of points of R2 and points in P1(R)
whiles lines of P2(R) consist of: lines in R2 with point at innity and points P1(R)
at innity. For any set of lines and points in a projective plane, there exists some
incidence relations that describes the lines and points in the space. Nevertheless,
there exists a unique projective line that passes through any two distinct points in a
projective plane [4].
2.2 Projective Transformations
Points in Pn(R) are just lines which pass through the origin of Rn+1, we therefore
need a group of transformations which map such lines in Rn+1 onto lines which pass
through the origin of Rn+1. The appropriate transformations for such purposes are the
invertible linear transformations which map Rn+1 onto itself. These transformations
map points x ∈ Rn+1 onto pointsNx ∈ Rn+1. Therefore the projective transformation
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with an associated matrix N maps points x of Pn(R) onto points of Pn(R).
Denition 2.2.1. Let N ∈ GL(n+1,R) be a non-singular matrix. Then the function
T : Pn(R)→ Pn(R) dened by the map x 7→ Nx, for x ∈ Rn+1 which maps a line [x]
into another line [Nx] is called a projective transformation of Pn(R).
For every α ∈ K, where K is a eld, αT and T dene the same projective transfor-
mation. Thus,
[αT (x)] = α[T (x)] = [T (x)]
which are of the same equivalence class.
Example 2.2.2. Let us consider the Möbius transformation of the extended complex
plane. These are projective transformations of the complex projective line P1(C) onto
itself. Points in P1(C) are described by homogeneous coordinates. Let us consider
the homogeneous coordinates (z1, z ) ∼ (z′ ′2 1, z2) such that z = λz′ and z = λz′1 1 2 2,
0 6= λ ∈ C. Then the linear transformation with coordinates [z , z ] is
       1 2z1 z a b z7−→  1 =   1 (2.1)
z2 z2 c d z2
 
a b
having its corresponding invertible linear transformation T : C2 → C2 as  .
c d
Now, if z2 =6 0, then z = z1 which give the coordinates [z, 1]. Following from Equationz2
(2.1), we obtain
z1 = az1 + bz2 and z2 = cz1 + dz2.
Then
z1 az1 + bz2 az + b z1
= z = = , z = .
z2 cz1 + dz2 cz + d z2
is called a Möbius transformation [8].
Example 2.2.3. The projective transformation of the real projective plane, P2(R).
We have that,
P2(R) = R2 ∪ P1(R)
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having its projective line at innity is x0 = 0. A linear transformation
T : R3 → R3
is given by   X
′
1 d b1 b2
   
 X1
X ′2 = c1 a11 a12 X2
X ′3 c2 a21 a22 X3
The components of the vector in R3 are homogeneous coordinates of a point in P 2(R).
The coordinates where X1 = 0 represent the points on the line at innity. All other
points haveX1 =6 0 and a special ch

oice of such coordinates can be made. For example
 1  1X2X1  = x
X3 y
X1
The x and y dened here are ane coordinates. From the above transformation, we
can write that
   X ′ 1  d b1 b2   

1
′ X =2 c1 a11 a12 x
X ′3 c2 a21 a22 y
and so dening x′ = X2 and y′ = X3 in a similar way to x and y, we obtain;
X1 X1
′ c1 + a11x+ a12y c2 + a21x+ a22yx = and y′ =
d+ b1x+ b2y d+ b1x+ b2y
We can write this matrix as
c+ Av
v′ =
d+ b · v
where v is the column 2-vector with components x and y, c is the column 2-vector
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with components c1 and c2 etc. and
 a11 a12A = 
a21 a22
2.3 Cross Ratio
Let A,B,C,D be a set of four collinear points. We consider what happens to these
points when they are projected from an external point O to another set of four
collinear points A′, B′, C ′, D′ as shown in Figure 2.1. This relation is termed as cross
ratio. The cross ratio is described as the most important invariant in projective
geometry which is preserved by all projective transformations and is said to be the
ratio of ratios of the set of four distinct collinear points [8].
A′b
B′b
A C
′
b b
Bb
Cb bD
′
Db
O b
Figure 2.1: Cross ratio (denition)
The cross ratio is dened in terms of homogeneous coordinates in the projective plane.
This denition will be used to prove the two relevant theorems below:
a. The cross ratio is well dened
b. The cross ratio is projectively invariant.
Denition 2.3.1. Let A,B,C,D be four collinear points with homogeneous coordi-
nates (non-zero vectors) [a], [b], [c], [d] respectively in P2(R). Since A,B,C,D are
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collinear, one can express c and d as a linear combination of a and b. Thus ∃
α, β, γ, δ ∈ R,
c = αa+ βb
d = γa+ δb.
We dene the cross ratio of A,B,C,D as the ratio of the ratios β and δ [8]. Thus
α γ
β
[A,B,C,D] = ÷ δ βγ= . (2.2)
α γ αδ
Example 2.3.2. Let us consider the collinear points A = [a] = [1,−1,−1], B = [b] =
[1, 3,−2], C = [c] = [3, 5,−5], D = [d] = [1,−5, 0].
By the denition of cross ratio [A,B,C,D], we rst have to express c and d as linear
combinations of a and b then we nd the real numbers ∀α, β, γ, δ ∈ R, associated to
each of them to compute the ratio. Thus for c = αa+ βb, we have
(3, 5,−5) = α(1,−1,−1) + β(1, 3,−2).
Comparing each corresponding coordinate, we have the following system of equations
3 = α + β (2.3)
5 = −α + 3β (2.4)
− 5 = −α− 2β. (2.5)
Solving equations, (2.3), (2.4) and (2.5), we have α = 1 and β = 2. Next, given that
d = γa+ δb then,
(1,−5, 0) = γ(1,−1,−1) + δ(1, 3,−2).
Comparing each corresponding coordinate, we have the following system of equations
1 = γ + δ (2.6)
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− 5 = −γ + 3δ (2.7)
0 = −γ − 2δ. (2.8)
Solving equations, (2.6), (2.7) and (2.8), we have γ = 2 and δ = −1. Therefore, from
(2.2)
2 2
[A,B,C,D] = × − = −4.1 1
Theorem 2.3.3. The cross ratio, [A,B,C,D] is well dened (that is, it is indepen-
dent of the choices of homogeneous coordinates used to represent the collinear points
A,B,C,D).
Proof. [8] Suppose A = [a], B = [b], C = [c], D = [d] are homogeneous coordinates
of collinear points A,B,C,D. Let
c = αa+ βb
d = γa+ δb.
Now, suppose A = [a′], B = [b′], C = [c′], D = [d′]. Then we have by the equivalence
relation of homogeneous coordinates that a = λa′, b = µb′, c = νc′, d = ξd′, for some
non-zero λ, µ, ν, ξ ∈ R as shown in the gure 2.2.
a = λa′
b = µb′
A
B
a′ ′ c = νc
′
b
C
c′
D d = ξd′
O d′
Figure 2.2: (A,B : C,D) is well dened
Substituting the new values into c and d, we have
νc′ = αλa′
αλ βµ
+ βµb′ =⇒ c′ = a′ + b′.
ν ν
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and
γλ δν
ξd′ = γλa′ + δµb′ =⇒ d′ = a′ + b′
ξ ξ
respectively.
Let, αλ = α′, βµ = β′, γµ = γ′ and δλ = δ′. Then
ν ν ξ ξ
c′ = α′a′ + β′b′ and d′ = γ′a′ + δb′
and we have,
β′ δ′
(A,B : C,D) =
α′
÷
γ′
( ) ( )
βµ ÷ αλ δµ γλ= ÷ ÷
ν ν ξ ξ
βµ
= × γλ
αλ δµ
βγ
= .
αδ
Since the cross ratio of A,B,C,D with coordinates a, b, c, d is the same when coor-
dinates are changed to a′, b′, c′, d′, we say that the cross-ratio is independent of the
choice of homogeneous coordinates and hence it is well-dened.
Example 2.3.4. The cross-ratio for the coordinates A = [1, 2, 3], B = [2, 2, 4], C =
[−3,−5,−8], D = [3,−3, 0] is −1 . Now, If the values of the homogeneous coordinates
3
are changed to A = [1, 2, 3], B = [1, 1, 2], C = [3, 5, 8], D = [1,−1, 0], the cross-ratio
remains the same because the corresponding homogeneous coordinates are equivalent.
We now want to describe how the cross ratio is preserved by projective transformation.
Theorem 2.3.5. Let τ be a projective transformation having the coordinatesA,B,C,D.
If A′ = τ(A), B′ = τ(B) and C ′ = τ(C), then
[A,B,C,D] = [A′, B′, C ′, D′].
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Proof. [8] Recall the denition of projective transformation. If N ∈ GL(3,R) is
an invertible matrix dened by the projective map, then there exist a projective
transformation τ such that [x] 7→ [Nx]. Then we have that A′ = [a′] = [Na],
B′ = [b′] = [Nb], C ′ = [c′] = [Nc] and D′ = [d′] = [Nd]. Since A,B,C,D are
collinear, then we have that
c = αa+ βb and d = γa+ δb
and [A,B,C,D] = β γ . Therefore,
α δ
Nc = αNa+ βNb and Nd = γNa+ δNb
implies
c′ = αa′ + βb′ and d′ = γa′ + δb′.
Hence,
[A′
β γ
,B′, C ′, D′] = .
α δ
Since [A,B,C,D] = [A′, B′, C ′, D′], we conclude that the cross-ratio is projectively
invariant.
Permutations of the Cross Ratio
For every number of points in a projective space, we can nd the number of permu-
tations of its cross ratio. For instance, if there are four points, A,B,C,D, then the
number of permutations will be 4! = 24.
Theorem 2.3.6. Let A,B,C,D be four distinct collinear points in the projective
plane, and let [A,B,C,D] = n. Then
1
[B,A,C,D] = [A,B,D,C] = and [A,C,B,D] = [D,B,C,A] = 1− n.
n
Proof. Suppose [a], [b], [c], [d] are homogeneous coordinates of four collinear points
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A,B,C,D such that A = [a], B = [b], C = [c], D = [d] . Let
c = αa+ βb and d = γa+ δb.
Then by the denition of cross-ratio, we let [A,B,C,D] = βγ = n. Next, we inter-
αδ
change the positions of A and B then we obtain
c = βb+ αa and d = δb+ γa.
Therefore,
α γ α δ 1
[B,A,C,D] = ÷ = × = .
β δ β γ n
Again, we interchange the positions of C and D and obtain
d = γa+ δb and c = αa+ βb
Therefore,
δ β δ α 1
[A,B,D,C] = ÷ = × = .
γ α β γ n
Hence, [B,A,C,D] = [A,B,D,C] = 1 .
n
Now, for [A,C,B,D], we interchange the positions of C and B then we obtain
βb = − ⇒ α 1αa+ c = b = − a+ c
β β
and ( ) ( )
d = γa+ δb =⇒ α 1 γβ − δα δd = γa+ δ − a+ c = a+ .
β β β β
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Therefore,
( ) ( − )1 α δ γβ δα
[A,C,B,D] = ÷− ÷ ÷
β β β β
( )
− 1 ÷ δ= × β
α β γβ − δα
−(γβ − δα)=
αδ
βγ
= 1−
αδ
= 1− n.
Following from [A,C,B,D] = 1− n, we can have that
[A,C,B,D] = 1− n
= 1− [A,B,C,D]
1
= 1−
[A,B,D,C]
= 1− [B,A,D,C]
1
=
[B,D,C,A]
= [D,B,C,A].
Hence, [A,C,B,D] = [D,B,C,A] = 1− n.
If the cross ratio of [A,B,C,D] is n, then the following identities are true [6]:
a. [A,C,D,B] = (1− n)−1.
b. [A,D,B,C] = 1− n−1.
c. [A,D,C,B] = (1− n−1)−1.
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2.4 Twisted Polygon
Denition 2.4.1. A twisted m-gon is dened as a map
ρ : Z→ P2(R)
and a projective transformation T such that
ρ(i+m) = T (ρ(i)), ∀m,
where Z = {1, 2, · · · , 2m}. Here, T ∈ PGL(3,R) is a projective transformation of
the plane P2(R). T is also called the monodromy of ρ. Suppose f ∈ PGL(3,R) such
that f ◦ ρ1 = ρ2 and let ρ1 and ρ2 be two twisted m-gons then ρ1 and ρ2 are said
to be equivalent. Note that we require none of these three consecutives points ρ(i),
ρ(i+ 1), ρ(i+ 2) to be collinear (they are in general position). Any two monodromies
T1 and T2 will satisfy T
−1
2 = fT1f . Every twisted polygon is uniquely determined
by its m consecutive vertices and monodromy T ∈ PGL(3,R). Furthermore, because
the dimension of a polygon with bi-innite vertices is 2m and that of the monodromy
T ∈ PGL(3,R) is 8. We obtain the dimension of the space of twisted m-gons as
2m+ 8.
2.5 Corner Invariant (The Invariant Function)
Let q1, q2, q3, q4 be four collinear points on a projective line RP 1. We dene the cross
ratio for corner invariants as
(q1 − q3)(q2 − q4)
[q1, q2, q3, q4] = .
(q1 − q2)(q3 − q4)
Let (zj) be a twisted m−gon. Consider the point zj (in red), a vertex of (zj) and let
zj+1,zj+2,zj−1, zj−2, wj−1, wj+1, z̄j, z̃, represent other points as shown in Figure 2.3.
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zj+b 2
zj+1
b P
b
wj+1 z̄j
b
wj b zj b
b
wj−1
b z̃j
b
zj−1
b
zj−2
Figure 2.3: corner invariant; denition.
Denition 2.5.1. We dene,
xj := [wj, wj+1 : zj+1, zj+2] and yj := [wj, wj−1 : zj−1, zj−2]
The coordinates xj, yj are called Corner Invariants of the point zj.
Example 2.5.2. Let zj represent the vertices of a polygon. Consider the vertex j = 6
as shown in Figure 2.4, then
x6 := [w6, w7 : z1, z2], y6 := [w6, w5 : z5, z4].
The corner invariant at j = 6 is ([w6, w7 : z1, z2], [w6, w5 : z5, z4]).
Example 2.5.3. This example describes the corner invariants at dierent vertices
(z2, z3, z4) on a given twisted m−gon as shown in Figure 2.5. The corresponding
invariants are shown in the table below.
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b z2 ≡ z8
z1 ≡ zb 7 P6
b
w7 z̄6
b
w b z6 b z b 36
b w5
b z̃6
b
z5
b
z4
Figure 2.4: corner invariant; example.
zj xj yj
z2 x2 := [z4, z3 : C,D] y2 := [z0, z1 : E,D]
z3 x3 := [z5, z4 : A,B] y3 := [z1, z2 : C,B]
z4 x4 := [z6, z5 : G,F ] y4 := [z2, z3 : A,F ]
2.6 Projective Duality
According to the denition of duality, given any point P in projective plane, there is
a correspondence which carries collinear points to concurrent lines [17]. This corre-
spondence is termed, projective duality. Let A, B be two distinct points in the real
projective plane and C a circle. If two dierent tangent lines are drawn from each
point to meet the circle, then a line is drawn to join the tangential points. Thus, for
A, we have A′A′′ and B′B′′ for point B. Let A′A′′ ∩ B′B′′ = p. It is observed that
A′A′′ → A, B′B′′ → B and p → l as shown in Figure 2.6 . This gives a relation
between points in RP2 and lines in (RP2)∗.
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b
F
bG
b z5
b z6
z4 b
A b
b z3
B b
b
C
b
z b2
b z
b z
0
1
E
b
D
Figure 2.5: Construction of corner invariants at dierent vertices on a twisted m-gon.
Ab
b B
l
B ′b
b bAA ′ ′′b p
b
B ′′
Figure 2.6: Projective duality
2.7 Dual Polygon
We describe the dual polygon P ′ ∈ (RP2)∗ as the cyclically ordered collection of
n−tuple of points (p1, · · · , pn) and lines (l1, · · · , ln) which represent the edges and
the vertices of the polygon respectively.
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Each point lj of the dual is obtained when any two lines in P meet whilst each line
pn is obtained from the points pn of the original polygon P . An example is shown in
Figure 2.7 .
p5 l4 p4
l4
p4
p5
l l5 3 l
l 35
P ′′
p1 p3
p ′1 P
l1 p2 l2 p3
l1 P
l2
p2
Figure 2.7: Duality of the pentagon
2.8 Self-dual Polygons
Let A′ be the dual of an original polygon A. The polygon A is said to be self dual if
A is isomorphic to A′, thus, A ' A′. Intuitively, whenever a polygon is equal to its
dual, then it is called a self dual polygon [20].
2.8.1 Example
Consider a polygon of sides 6 (P6) as shown in the Figure 2.8.
Since P ′6 ' P6, we call P6 a self dual polygon.
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b P ′6
P6
b b
b b
b
b b b
b b b
b b b b
b
b b
Figure 2.8: Self Dual polygon
2.9 Non-degenerate Conic
A conic is obtained when a plane intersects a cone at a given angle. There are two
types of conics: degenerate and non-degenerate conics. When a plane intersects a
cone in any angle apart from its vertex, we obtain a non-degenerate conic. Non-
degenerate conics are the ellipse, parabola and hyperbola as shown in Figure 2.9
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z z z
P y yy
P
x
x x
P
Ellipse Parabola Hyperbola
Figure 2.9: Non-degenerate conics shown in red
2.10 Inscribed Polygon
A polygon is said to be inscribed in a non-degenerate conic if all its vertices lie on
the conic. Consider an example in Figure 2.10
2.11 Involution
Denition 2.11.1. An involution is a transformation or a function that is equal to
its inverse. It is also described as a linear operator T such that T 2 = I.
Example 2.11.2. The identity map is an involution. Let H be a set, then the
identity function f on H is dened as the function with domain and codomain H
which satises f(x) = x,∀x ∈ H. Then f(f(x)) = x, ∀x ∈ H is an involution.
Example 2.11.3. The squared Pentagram map T 2(P ) modulo scaling is an involution
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C
P
Figure 2.10: A polygon inscribed in a non-degenerate conic (ellipse)
since P and T 2(P ) are the same as shown in Figure 2.11
Figure 2.11: Involution of the Pentagram map
Another example of an involution is projective duality. Thus (P ∗)∗ = P for P ∈ RP2
and P ∗ ∈ (RP2)∗ [2]. The composition of two involutions h and g, (g ◦ h) is an
involution if and only if they commute. Thus g ◦ h = h ◦ g.
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Chapter 3
Pentagram Map and Monodromy
Invariants
In this chapter, we rst describe the formula for the pentagram map using the corner
invariants denition and later construct the monodromy invariants from this deni-
tion.
3.1 Formula for the Pentagram Map
The goal of this section is to describe the formula for the pentagram map in coordi-
nates. Nevertheless, we shall rst describe the pentagram map as the composition of
two involutions which are all projective duals.
Recall the denition of twisted polygons. Let Vm be the space of twisted m-gons
modulo projective transformations The dimension of the space Vm is 2m since the
dimension of the space of twisted m−gons is 2m+ 8 [5].
A twisted m-gon depends on 2m variables which represents coordinates of vertices
pi := ρ(i), ∀i = 1, · · · ,m and consists of 8 parameters of the monodromy matrix
T , while the PSL(3,R) equivalence reduces the dimension by 8. The pentagram
map T is generically dened on the space Vm. Namely, for every twisted m-gon the
vertices of its image are the intersections of pairs of consecutive shortest diagonals:
T (pi) := (pi−1, pi+1) ∩ (pi, pi+2). Such intersections are well dened for each generic
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point in Vm .
There are two coordinates on the space Vm called the (ci), (di) coordinates and the
corner invariants xi, yi. Because the corner invariants are already dened in Deni-
tion 2.5.1, we shall now consider the denition of (ci, di) coordinates and state their
relationship with the xi, yi coordinates In what follows, we assume that any three
points pi, pi+1, pi+2 are in general position (not collinear).
Denition 3.1.1. ((ci), (di) Coordinates): Assume m is not divisible by 3 [10]. Then
there should exist a unique lift of points pi = ρ(i) ∈ RP2 to vectors P ∈ R3i such that
the determinant of the lifts, det(Pi, Pi+1, Pi+2) = 1, ∀i ∈ Z. We identify a dierence
equation to the sequence of vectors Pi ∈ R3 by setting
Pi+3 = ciPi+2 + diPi+1 + Pi ∀i ∈ Z. (3.1)
Because Pi+m = T (Pi) ∀i ∈ Z, we say that the sequences (ci), (di) are m−periodic
for any monodromy T ∈ SL(3,R) and ci, di : 0 ≤ i ≤ m− 1. We shall prove this as a
lemma in the next chapter.
The relationship between these two coordinate systems is dened [10] as;
ci−2 − di−1xi = , yi = . (3.2)
di−2di−1 ci−2ci−1
In [15], the pentagram map is described as the composition of two involutions, α(pi)
and β(pi) of which each of them is a kind of projective duality. The idea of projective
duality in RP2 is as a result of the fact that the space (RP2)∗ consists of 1-dimensional
subspaces of RP2.
Lemma 3.1.2. For any given sequence of points pi ∈ RP2. We dene α(pi) ∈ (RP2)∗
as the line (pi, pi+1) and β(pi) ∈ (RP2)∗ as the line (pi−1, pi+1) as shown in Figure 3.1.
Then α2 = τ , β2 = Id and α ◦ β = M , where M is the pentagram map and τ is a
cyclic permutation; τ ∗(pi) = pi+1.
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α(pi) β(pi)
pi+1 pi+1
p
p i+2
pi
i
pi−1 pi−1
Figure 3.1: Projective duals
Proof. From the denitions of α(pi) and β(pi)it implies that
α2(pi) = α(pi, pi+1)
= α(pi) ∩ α(pi+1)
= (pi, pi+1) ∩ (pi+1, pi+2)
and
β2(pi) = β(pi−1, pi+1)
= β(pi−1) ∩ β(pi+1)
= (pi−2, pi) ∩ (pi, pi+2)
and hence,
(α ◦ β)pi = α(β(pi))
= α(pi−1, pi+1)
= α(pi−1) ∩ α(pi+1)
= (pi−1, pi+1) ∩ (pi+1, pi+2)
= β(pi) ∩ β(pi+1)
= M
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Because (α ◦ β)pi = M , we say that the pentagram map is the composition of the
two involutions. From this denition, we state that the pentagram map M preserves
the monodromy T . Geometry of the proof is shown in Figure 3.2
α(p β(p )i+1) α(p ii)
β(pi−1) pi+1
pi+1 pi+1
β(p )
p i+1 pi p
p
− i
pi+2
p i+2
i 2
i β(pi+1)
pi−1 pi−1 pi−1
pi−2 pi−2 pi−2
α2(pi) β2(pi) (α ◦ β(pi)) =M
Figure 3.2: Duality maps
Denition 3.1.3. Given Vm, we say that the map M that transforms one polygon
to the other also descends to a densely dened map
M : Vm → Vm,
called the Pentagram map
.
To dene the Pentagram map in its coordinates, we use the right labelling approach
for our convenience, though, we could seemingly use any of the other two labelling
schemes in Figure 3.3.
In terms of the corner invariants, we dene the map M by
′ 1− xi−1yi−1 1− xi+2yi+2xi = xi − and y
′
1 x i
= yi+1 . (3.3)
i+1yi+1 1− xiyi
The proof of this formula combines results in Lemma 3.1.2 and Equation 3.2 [10].
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1 1 1
2 2 2
7
7 1 7 27 1
2 1 3
2
3
3 3 7
6 37
3 4 6
6 5 46 5 6
4 6 5
4 4 4
Right lab5elling Left La5belling Square5d Pentagram’s natural labelling
Figure 3.3: Labelling schemes of the Pentagram Map
3.2 Construction of Monodromy Invariants
From Section 2.5, we dene
∏m ∏m
Xm = xj and Ym = yj.
j=1 j=1
Following from Equation (3.3), the functions Xm, Ym are invariant under the penta-
gram map.
For any arbitrary equivalence class V ∈ Vm of twisted m−gons, the analogous mon-
odromy T dened up to scalar multiplication and conjugation is a square matrix of
dimension 3.
We dene
trace3(T−1) trace3(T )
ϕ1 = − and ϕ2 = (3.4)det(T 1) det(T )
as well-dened invariants under the Pentagram map since the map preserves T . [15]
By denition,
∑ 3 [m ]1 2  1 ∑ [m 3]2ϕ1 = Xk and ϕ2 = Y k , (3.5)
X2 Y Y 2m m Xmk=0 m k=0
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where,
∑[m ] ∑[m ]2 2
trace(T−1) = Xk, det(T
−1) = X2mYm and trace(T ) = Yk, det(T ) = Y
2
mXm
k=0 k=0
Because the pentagram map M on every m−gon preserves the monodromy T , it
implies that the map M : Vm → Vm, dened on equivalence classes of polygons
preserves ϕ1 and ϕ2 [5]. Hence we also dene
ϕ̃1 := X
2
mYmϕ1 and ϕ̃
2
2 := YmXmϕ2
as invariant polynomials in the corner invariants xi, yi (Proof is shown in [15]). From
Equation 3.4, we will have that
∑ [m ]2ϕ̃ = X 3
∑ [m 3]2 
1 k and ϕ̃2 = Yk .
k=0 k=0
X0 and Y0 are just constant functions X  0
= 0 = Y0 so we obtain
∑ 3  [m ] ∑[m 3]2 2
ϕ̃1 = 1 + X   k and ϕ̃2 = 1 + Yk
k=1 k=1
The remaining monodromy invariants are described as the weighted homogeneous
parts of these polynomials.
The Pentagram map commutes with a rescaling operation
Ru(x1, y1, x2, y2, · · · , xm, ym) 7−→ (ux , u−11 y1, ux2, u−1y2, · · · , ux , u−1m ym). (3.6)
Note that a polynomial in the corner invariants has weight k if R∗u(V ) = u
kV , where
R∗u is the natural action of Ru on the polynomials. For instance, the weight of Xm is
m and that of Ym is −m.
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By the denition of the rescaling operator, we have
∑[m ] ∑[m ]2 2
R∗ (ϕ̃ ) = ( X uk)3 and R∗(ϕ̃ ) = ( Y u−k)3u 1 k u 2 k .
k=0 k=0
Denition 3.2.1. The functions Xk and Yk which represent the weight k polynomials
in each sum where k = {1, · · · , [1 ],m} are called the monodromy invariants
2
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Chapter 4
Monodromy Invariants of the
Spectral Curve and Corner Invariants
in the Monodromy Matrix
This chapter describes two major concepts in two sections.
1. In the rst section, we describe the relationship between monodromy invariants
and the spectral curve.
2. In the second section, we discuss the relationship between the monodromy ma-
trix and the corner invariants. We shall do this by dening the monodromy in
terms of corner invariants.
4.1 Monodromy Invariants and the Spectral Curve
This section basically discuss the occurrence of monodromy invariants as coecients
of the spectral curve.
Suppose V ∈ Vm is an equivalent class of twisted m−gons. Then the map Ru :
Vm → Vm (Rescaling operation) denes every family Vu ∈ Vm, where Vu = Ru(V )
parametrised by u ∈ C∗.
Note that each equivalence class has a corresponding conjugacy class of monodromies.
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Thus if V2 = λV1, V3 = λV2, · · · such that the set V = {V1, V2, V3, · · · },∀V ∈ Vm, then
we could have T = {T1, T2, T3, · · · } as the set of each corresponding equivalence class
of Vm.
We take any representative of {V1, V2, V3, · · · } ∈ V and lift it to GL3. Hence we
obtain a set T (u) of 3× 3 matrices that is parametrized by u ∈ C∗. T (u) is called the
scaled monodromy of the equivalence class V .
We dene the spectral function of scaled monodromy matrix T (u) is given as
det(λ(u)vI − T (u))
R(u, v) :=
λ(u)3
1
where λ(u) := (umdetT (u) 3 ), I is the identity matrix and R(u, v) is the characteristic
polynomial of T (u). The polynomial R(u, v) of T (u) is normalised in a way such that
it resists changes if we multiply T (u) by a scalar function of u. Because T (u) is a
square matrix, we will have the characteristic polynomial in the form
M (λ) = λ3 − tr(T (u))λ2 + Pm(T (u))λ− det(T (u)).
Explicitly, we have
R(u, v) = v3 − Y (u−1)v2 +X(u)u−mv − u−m, (4.1)
where
tr(T (u)) = Y (u−1), Pm(T (u)) = X(u)u−m det(T (u)) = u−m
and 
− 2 − 1 ∑   [m ] [m ]2 2− 2 − 1 ∑
Y (u) := Y 3 3 k 3 3 km Xm 1 + Y ku , X(u) := Xm Ym 1 + X u k .
k=1 k=1
Denition 4.1.1. (Spectral Curve) We dene the spectral curve as the zero locus of
the characteristic equation. Thus, the value of λ such that M (λ) = 0.
For Example
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Let T (u) be a scaled monodromy matrix of the equivalence class of twisted m−gons
dened as
   a(u) b(u) c(u) a b cT (u) = d(u) e(u) f(u) = d e f (u) (4.2)
g(u) h(u) i(u) g h i
Then we will have that    λ(u)v 0  a(u) b(u) c(u)λ(u)vI − T (u) = 0 λ(u)v 0 −d(u) e(u) f(u) (4.3)
0 0 λ(u)v g(u) h(u) i(u)
Let λ(u) = λ, a(u) = a and all other entries in T (u), then we have that
 
λ(u)vI − T (u) = λv − a −b −c−d λv − e − f  (4.4)
−g −h λv − i
det(λ(u)vI − T (u)) = λ3v3 + v2(−λ2i− eλ2 − aλ2) + v(λei+ aλi+ aeλ− λhf − bdλ− cλg)
+(−aei+ ahf + bdi− bgf − cdh+ ceg)
Dividing through by λ(u)3, we obtain
1
det(λ(u)vI − T (u)) = v3 − λ−1v2(i+ e+ a) + λ−2v(ei+ ai+ ae− hf − bd− cg)
λ(u)3
+λ−3(−aei+ ahf + bdi− bgf − cdh+ ceg)
m 1where λ(u) = (v detT (u)) 3 .
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1
det(λ(u)vI − T (u)) = v3 − λ−1v2(i+ e+ a) + λ−2v(ei+ ai+ ae− hf − bd− cg)
λ(u)3
+λ−3(−aei+ ahf + bdi− bgf − cdh+ ceg)
= v3 − [λ−1(i+ e+ a)]v2
+[λ−2(ei+ ai+ ae− hf − bd− cg)]v
+λ−3(−aei+ ahf + bdi− bgf − cdh+ ceg)
= v3 − [λ−1(u)(i+ e+ a)(u)]v2
+[λ−2(u)(ei+ ai+ ae− hf − bd− cg)(u)]v
+λ−3(u)(−aei+ ahf + bdi− bgf − cdh+ ceg)(u)
= v3 − [λ−1(u)trace(T (u))]v2
+[λ−2(u)Pm(T (u))]v + λ−3(u)det(T (u))
3 − trace(T (u)) 2 Pm(T (u)) − det(T (u))= v v + v
λ(u) λ2 λ3(u)
1
However, since λ(u) = (vmdetT (u)) 3 , we will have that
trace(T (u)) Pm(T (u)) det(T (u))
R(u, v) = v3 − 1 v2 + 2 v − (4.5)
(vmdetT (u)) 3 (vmdetT (u)) (vm3 detT (u))
Recall the denition
∑m2
traceT (u) = Y and detT (u) = Y 2m mXm,
k=0
∑m
2 Ym Pm(T (u))
R(u, v) = v3 − k=0 2 −m1 v + 2 v − u
(umY 2mXm) 3 (u
mY 2mXm) 3
m
2
1 − 23 − −m −
1 ∑ 2 4 − 2
= v (u ) Y 3X 3 Y v2 + (u−m)−3 3Y 3X 3m m m m m Pm(T (u))v − u−m
k=0
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Hence,
R(u, v) = v3 − Y (u−1)v2 +X(u)u−m − u−m, (4.6)
where,
m
2
−1 −m 1 −
2 − 1 ∑ 2 4 − 2
Y (u ) = (u ) 3Y 3X 3m m Ym and X(u) = (u
−m)− 3Y 3X 3m m Pm(T (u))
k=0
Therefore, the spectral curve is the value of v such that
v3 − Y (u−1)v2 +X(u)u−mv − u−m = 0.
Corollary 4.1.2. In [5], we have that the spectral curve being invariant under the
involution
(u, v)↔ (u−1, v−1)
and the monodromy invariants satisfying Yk = Xk, ∀k = 1, · · · , [1 ],m are equiva-2
lent.
Proof. From Equation 4.1, we have R(u, v) = 0. Thus,
= v3 − Y (u−1)v2 +X(u)u−mv − u−m = 0. (4.7)
m
Next, we nd −u 3 R(u, v) (involution of R(u, v)) and obtainv
um − umY (u−1)v−1 +X(u)v−2 − v−3 = 0. (4.8)
Hence,
v−3 −X(u)v−2 + umY (u−1)v−1 − um = 0. (4.9)
Additionally,
R(u−1, v−1) = v−3 − Y (u)v−2 + umX(u−1)v−1 − um = 0. (4.10)
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Hence the spectral curve is invariant under the involution
(u, v)↔ (u−1, v−1).
Next, comparing coecients in Equations 4.9 and 4.10, we have that X(u) = Y (u)
and X(u−1) = Y (u−1). Because X(u) = Y (u) we will have that
 ∑   ∑ [m ] [m ]2 2− 2 − 1 − 2 − 1
X 3Y 3 1 + X uk = Y 3 3 km m k m Xm 1 + Y u k ,
k=1 k=1
which implies Xk = Yk. Hence the monodromy invariants satisfy Yk = Xk, ∀k =
1, · · · , [1 ],m .
2
The Newton polygon is used to demonstrate the involution on the spectral curve as
dened in chapter 2.
For instance the Newton polygon of the pentagram spectral curve is obtained as a
parallelogram when it has 7 sides [5].
4.2 Monodromy Matrix and the Corner Invariant
Here, we describe the monodromy in terms of the corner invariants.
We shall use the eld C instead of R though this does not cause any dierence in our
constructions.
We recall the coordinate system of Vm in Chapter 3 as introduced in [10] called the
(ci, di) coordinates where ci, di ∈ C, ∀i ∈ Z such that ci+m = ci, di+m = di together
with the corner invariant coordinates (xi, yi).
In addition, we also recall that these sequence of coordinates are associated by a
dierence equation in Equation (3.1).
The solution to Equation (3.1) are described as the vectors P = Pi, ∀Pi ∈ C
that satises Equation(3.1). Because m is periodic, it implies that there exists V ∈
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SL(3,C) such that
Pi+m = Vi (4.11)
The lemma below and corollary are proved in other to help us view the relationship
between the monodromy invariant and the corner invariant.
Lemma 4.2.1. Given any three non-collinear points pi, pi+1, pi+2 of a twisted polygon
(pi) ∈ SL(3,C), there exists an associated dierence equation of the form:
Pi+3 = ciPi+2 + diPi+1 + Pi. 
(4.12)
0 0 1
Proof.  Suppose Ri = (Pi, Pi+1, Pi+2) andQi = 1 0 di for some sequences ci, di ∈
0 1 ci
C, where det (Pi, Pi+1, Pi+2) = 1 such that
Ri+1 = RiQi. (4.13)
Then we will have that



0 0 1
(Pi+1, Pi+2, Pi+3) = (Pi, Pi+1, Pi+2) 1 0 di
0 1 ci
= (Pi+1, Pi+2, Pi + diPi+1 + ciPi+2) (4.14)
Comparing the right and left hand side of Equation (4.14), we have that Pi+1 = Pi+1,
Pi+2 = Pi+2 and hence Pi+3 = ciPi+2 + diPi+1 + Pi.
In general, lets compute Ri+m. If Pi+3 = ciPi+2 + diPi+1 + Pi, then we can obtain
Ri+n as follows:
Ri = (Pi, Pi+1, Pi+2)
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Ri+1 = (Pi+1, Pi+2, Pi+3)
Ri+2 = (Pi+2, Pi+3, Pi+4)
...
Ri+m = (Pi+m, Pi+m+1, Pi+m+2). (4.15)
Similarly from Equation (4.13) we have that:
Ri+2 = Ri(QiQi+1)
Ri+3 = Ri(QiQi+1Qi+2)
...
Ri+m = Ri(QiQi+1 · · ·Qi+m−1) (4.16)
Lemma 4.2.2. [5] Suppose (pi) ∈ Vm for Vm ∈ P2 having the monodromy T ∈ PGL3
and corner invariants (xi, yi). Then T is conjugate to any of the matrices Ti :=
LiLi+1 · · ·Li+m−1, for any arbitrary i ∈ Z and
  0 0 1
Lj = −xjyj 0 1 . (4.17)
0 −yj 1
Proof. We consider the following steps in our proof.
1. Suppose pi, pi+1, pi+2 are non-collinear points in the real projective plane. Tak-
ing the lifts Pi, Pi+1, Pi+2, we write the monodromy of the lifts T
 i as a conjugate
of the product of the matrices Qi and Di.
2. Use the corner invariants of the lift to describe the relationship between Lj and
Qi.
3. Write Ti in terms of T
 i Di and λi.
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4. We nd λi+m−2 by describing quasi periodicity conditions for λ.
5. We shall use the above result to prove the lemma.
Assume pi, pi+1, pi+2 are in general position. Then we lift the points pi to vectors
P ∈ C3i in the special linear group (thus, det (Pi, Pi+1, Pi+2) = 1 ∀i ∈ Z). From
Lemma (4.2.1) we have that
Pi+3 = ciPi+2 + diPi+1 + Pi,
for some sequences ci, di ∈ C.
Re-writing in matrix form, we obtain Equation (4.13) and in general we have Equation
(4.16).
We consider T
, an arbitrary lift of the monodromy T . Because every twisted polygon
(pi) is of the form pi+1 = T (pi), we can state that
T
Pi = tiPi+n for some sequence ti ∈ C. (4.18)
For Pi ∈ Ri and ti ∈ Di, we have that
T
Ri = Ri+mDi where Di := diag(ti, ti+1, ti+2). (4.19)
Substituting Equation (4.16) into Equation (4.19), we have that
T
Ri = Ri(QiQi+1 · · ·Qi+m−1)Di
which implies
T
 := (QiQi+1 · · ·Qi+m−1)Di
Therefore by the comparison of Equations (4.16) and (4.19), we say that T
i is conju-
gate to any of the matrices
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T
i := (QiQi+1 · · ·Qi+m−1)Di ∀i ∈ Z. (4.20)
Using the denition of corner invariant of the lift Pi of a twisted polygon satisfying
Equation (4.12)
ci−2 di−2
xi = , yi = − , (4.21)
di−2di−1 ci−2ci−1
it is observed that Li and Qi are related by means of the gauge transformation
1
L = λ−1i i−2Qi−2λi−1, λi := diag(1, di, ci). (4.22)ci−1
Verication
    
0 0 11 0 0 
Qj−2λj−1 = 1 0 di0 di−1 0  
0 0 ci−1
=  1 0 dici−1 .
0 1 ci 0 0 ci−1 0 di−1 cici−1
 −1    1 0 0  di−2ci−2 0 0  1 0 0−1 1 λi−2 = 0 d 1i−2 0  =  0 ci−2 0  = 0 0 d di−2ci−2    i−2 
0 0 ci−2 0 0 di−2 0 0
1
ci−2
    1 0 0 0 0 ci−1   0 0 ci−1 λ−1 i−2Qi−2λi− 11 = 0 0 1 0 d c 1 dici−1d − i i−1 =  0 i 2 di−2 di−2 
0 0 1 0 d c di−1 cici−1
c − i−1 ici−1 0i 2 ci−2 ci−2
   
0 0 ci−1 0 0 1
1 −1 1  d c   λi−2Qi−2λi− =  1 0 i i−1 =  1 0 di1  (4.23)ci−1 c di−1 i−2 di−2   ci−1di−1 di−2
0 di−1 cici−1 0 di−1 di
ci−2 ci−2 ci−1ci−2 di−2
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We observe that Equation (4.23) is equal to Li since
− 1 − di−1xiyi = and yi = .
ci−1di−1 ci−1ci−2
This transformation is called the gauge transformation.
From Equation (4.20), we have that
(Q 
 −1iQi+1 · · ·Qi+m−1) = TiDi . (4.24)
Substituting Equation 4.24 into Equation 4.22, we obtain the product of our matrices
Ti := LiLi+1 · · ·Li+m−1 as
T = a λ−1 T
 −1i i i−2 i−2Di−2λi+m−2 (4.25)
In what is left, we nd λi+m−2 from Equation (4.25).
From the left, we multiplyT
 by Equation (4.13) to obtain
T
Ri+1 = T
RiQi. (4.26)
Substituting Equation (4.19) into Equation (4.26) we have that
Ri+m+1Di+1 = Ri+mDiQi (4.27)
We now compare Equation (4.27) with Equation (4.13).
From Equation (4.13), if Ri+1 = RiQi, then Ri+m+1 = Ri+mQi+m (same as Ri+nR =
Ri+mQi+m) and hence
R = Qi+m (4.28)
Substituting this into Equation (4.27)(which can be re-written as Ri+mRDi+1 =
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Ri+mDiQi), we have
Ri+mQi+mDi+1 = Ri+mDiQi
Qi+mDi+1 = DiQi,
and hence
Q −1i+m = DiQiDi+1. (4.29)
Because Di = diag(ti, ti+1, ti+2), then Di+1 = diag(ti+1, ti+2, ti+3) and therefore
 −1ti+1 0 0−1  Di+1 = 0 ti+2 0
0 0 ti+3 ti+2ti+3 0 01  = 0 tt t i+1ti+3 0i+1 i+2ti+3
 0 0 ti+1ti+2
 1 0 0ti+1=  1 0 0 ti+2 
0 0 1
ti+3
where det(Di+i) = ti+1ti+2ti+3.
Next we have that
 
 
 
 
 1 
 0 0 1ti+3
0 0 1 0 0  ti+1−1   1   QiDi+1 = 1 0 b 0 0  =  1 0 dii  ,ti+2 ti+1 ti+3
0 1 ai 0 0
1  
ti+3  
0 1 ci
ti+2 ti+3
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and
   0 0
1
ti+3
  
−1 

ti+1 0 0 
   1 d DiQiD ii+1 = 0 ti+2 0  0 ti+1 ti+30 0 ti+3  
 0 1 citi+2 ti+3

  
0 0 ti 0 0 ti
ti+3  
  
ti+3
  

=  t i+1 0 diti+1 diti+1 ti+1 ti+3 
 = 1 0 ti+3  
0 ti+2 citi+2 0 1 citi+2
ti+2 ti+3 ti+3
From Equation (4.29), we have that
   0 0 titi+3 
0 0 1  
 
1 0 d diti+1i+m = 1 0  . (4.30)ti+3 
0 1 ci+m  
0 1 citi+2
ti+3
By comparison, we have that
ti
= 1 implies ti = ti+3,
ti+3
diti+1 ti+1 citi+2 ti+2
di+m = = di and ci+m = = ci .
ti+3 ti ti+3 ti
Using the above results, we obtain the quasi periodicity condition for λi := diag(1, di, ci)
in the equation below.
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1
λi+n = Diλi (4.31)
ti
From Equation(4.31), we have that
    0 0 1  ti 0 0  1  
1 0 0
λi+m = 1 0 di+m = 0 ti+1 0 ti 0 di 0
0 1 ci+m 0 0 ti+2 0 0 ci


ti 0 0
1 
=
t 0 diti+1 0i 
0 0 citi+2
 1 0 0  
= 

0 d t

i+1
 i 0ti 


0 0 c ti+2i ti
From Equation (4.31), we have that
1
λi+m−2 = Di−2λi−2 (4.32)
ti−2
Now, substituting Equation (4.32) into 4.25, we have that
T = a λ−1 
 −1
1
i i i−2Ti−2Di−2 Di−2λi−2ti−2
ai
= λ−1 T
i−2λi−2 (4.33)
t i−2i−2
Thus, T
i−2 is conjugate to Ti and this completes the proof.
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From the above Lemma, we obtain that for any given equivalence class V ∈ Vm
with corner invariants (xi, yi), the scaled monodromy is given by any of the matrices
Ti(u) := Li(u)Li+1(u) · · ·Li+n−1(u) with
 0 0 1
Lj = −xjyj 0 1 . (4.34)
0 −u−1yj 1
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Chapter 5
Inscribed Polygons and their
Monodromy Invariants
The main goal of this chapter is demonstrate that the scaled monodromy satises a
certain self-duality relation for inscribed polygons. We shall prove two Lemmas and
nally use these to prove Theorem 1.0.1.
Suppose (· · · , p−1, p0, p1, · · · ) is an inscribed polygon. Considering the points (pi) as
points on the projective line, we describe their cross ratios and set
qi := 1− [pi−2, pi−1, pi, pi+1] (5.1)
By a way of Stereographic projection right from a point of the conic, we can identify
a non-degenerate conic with RP 1.
An arbitrary dierent choice of the center of projection gives a projective transforma-
tion of RP 1. This implies that the identication of this projection can be described
as unique up to projective transformation of the projective line.
Lemma 5.0.1. [18] Suppose
xi = [pi−2, pi−1, pi, pi+2] and yi = [pi−2, pi, pi+1, pi+2] (5.2)
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then the map T mapping (qi)→ (xi, yi) is given by (pi)→ (xi, yi) such that
1− qi 1− qi+1
xi = and yi = (5.3)
qi+1 qi
Proof. Considering projection from the point pi+1 in Figure 5.1, we obtain
xi = [pi−2, pi−1, E, F ]
= [(pi+1pi−2), (pi+1pi−1), (pi+1E), (pi+1F )]
= [pi−2, pi−1, pi, pi+2]
Now, expanding xi in terms of qi's, we obtain
1− [1− [pi−2, pi−1, pi, pi+1]]
[pi−2, pi−1, pi, pi+2] =
1− [pi−1, pi, pi+1, pi+2]
[pi−2, pi−1, pi, pi+1]
=
1− [pi−1, pi, pi+1, pi+2]
Similarly for y, we have that
1− [1− [pi−1, pi, pi+1, pi+2]]
[pi−2, pi−1, pi, pi+2] =
1− [pi−2, pi−1, pi, pi+1]
[pi−1, pi, pi+1, pi+2]
=
1− [pi−2, pi−1, pi, pi+1]
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F b
b Epi−1
b
pi−2
b
b pi
C b
pi+1
b
pi+2
Figure 5.1: Lemma 4.2.2, proof.
Lemma 5.0.2. Suppose V ∈ Vm is an equivalence class of polygons inscribed in a
non-degenerate conic. Then the analogous scaled monodromy matrix can be chosen
to satisfy the self-duality relation
T (u) = (T (u−1)−1)t.
Proof. Considering any polygon (pi) in the equivalence class of Vm inscribed in a non-
degenerate conic C, from Lemma 5.0.1 and results of scaled monodromy associated
to Lemma 4.2.2 (thus, Equation (4.34)). We can express the scaled monodromy in
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terms of qi's where Ti(u) := Li(u)Li+1(u) · · ·Li+m−1(u) with 
L (u) =  0 0 1 (1− q−1 j j )(qj+1−1 − 1) 0 1 . (5.4)
0 u−1q−1j (qj+1 − 1) 1
Verication
From Equation (5.3),
1− qj 1− qi+1
xjyj = ·
qi+j qi
1− qi+1 − qi + qiqi+1
=
qi+1qi
= q−1 −1 −1 −1i+1qi − qi − qi+1 + 1
= q−1(q−1 − 1)− 1(q−1i i+1 i+1 − 1)
= (q−1 − 1)(q−1i i+1 − 1),
and hence
−x y = (1− q−1)(q−1j j i i+1 − 1), yj = q−1 −1j (1− qj+1), −yj = qj (qj+1 − 1).
It is observed that Lj(u) and ˆj(u) are related by gauge transformation:
q
ˆ jj(u) := − H L
−1
j jHj+1, Hj = diag(q
−1
j − 1, 1− qj, 1), (5.5)1 qj
where, Hj+1 = diag(q
−1
j+1 − 1, 1− qj+1, 1),
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


1
−1 0 0q
 j+1   
H−1 =   
0 01 
0 1j+1  0 , LjH
−1 −1
j+1 = (1− qj ) 0 1 (5.6)q j+1  0 −u−1q−1j 1
0 0 1
and 


 0 0 (q−1j − 1)  
H L H−1 = (1− q )(1− q−1j j j+1  j j ) 0 (1− qj)  . 
0 −u−1q−1j 1
Hence Equation (5.5) becomes
  0 0 1 q 

′ j −1 − 

Lj(u) := − HjLjHj+1 = (qj 1) 0 qj 1 qj  
0 u−1(qj − 1)−1 qj(1− q )−1j
Because ˆi(u) is conjugate to Li(u), it implies that the matrix Ti(u) is also conjugate
to T
(u) := ˆi(u)ˆi+1(u) · · ·ˆi+m−1(u).
From the results of the proof of the above Lemma in [5], it is observed that ˆj(u) and
T
(u) satisfy the relations Equations (20) and (21) in [5]. By the equivalency in L(u)
and T (u), since L(u) satises the self duality relation, one can conclude that T (u)
satises the self-duality relation. This completes the proof.
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5.1 Proof of Theorem 1.1
Proof. Following from Lemma 5.0.2, suppose (pi) is a polygon inscribed in a non-
degenerate conic. Then the analogous equivalence class V ∈ Vm exhibits a scaled
monodromy associated with the self-duality relation.
With reference to the self-duality relation also is the normalised form of the charac-
teristic polynomial satisfying the relation R(u−1, v−1) = −umv−3R(u, v).
This relation implies that the spectral curve is invariant under the involution dened
earlier. Furthermore by the equivalency in the spectral curve and Lemma 5.0.2, we
conclude that ∀k Xk = Yk for every m-gon inscribed in a non-degenerate conic.
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Chapter 6
Conclusion and Further Discussion
This chapter gives the summary of this thesis and also discusses some future works.
6.1 Conclusion
The core of this thesis which discusses the geometry of pentagram integrals of polygons
inscribed in non-degenerate conics began with a historical view of the pentagram, its
integrals and some useful concepts mainly in projective geometry.
Some concepts such as cross ratio, corner invariant, projective plane etc aid in our
construction of both the monodromy invariants and the formula for the pentagram
map. The pentagram map was dened in terms of corner invariants. This gave a
result that the analogous rst integrals for the pentagram map are invariant. It was
also observed that the invariants are certain polynomials in the corner invariant.
Further discussions on the relationship between the spectral curve and monodromy
invariants were also considered. The spectral curve dened as the zero locus of the
characteristic polynomial was introduced for our algebraic-geometric integrability. We
were able to dene the monodromy invariants as coecients of the spectral curve. It
was also proved that the spectral curve being invariant under an involution and that
the corresponding monodromy invariants being equal.
It was further discussed that the corner invariants can be used to dene the mon-
odromy matrix and the scaled monodromy matrix. We proved a lemma that for every
three points in general position, the monodromy should be conjugate to any of the
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some matrices dened in terms of the corner invariants
We nally disused the behavior of monodromy invariants on polygons inscribed in
non-degenerate conics. By the method of stereographic projection, we established
a formula between the corner invariants and the cross ratio of points on inscribed
polygons. Additionally, we showed that the scaled monodromy matrix were chosen
to satisfy a self-duality relation.
Using the results in the non-degenerate conic and the normalisation of the character-
istic polynomial of the spectral curve we proved that the invariants are equal when
polygons are inscribed in non-degenerate conics.
6.2 Further Discussion - Degenerate Conics
Degenerate conics are obtained when a plane intersect the vertex of a cone. The
various types of degenerate conics are discussed below (also see Figure 6.1).
1. If an ellipse degenerates, we obtain a single point
2. If a parabola degenerates, a straight line is obtained.
3. If a hyperbola degenerates, two intersecting lines are obtained
b b b
Ellipse
Parabola Hyperbola
Figure 6.1: Degenerate Conics
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1. How does Theorem 1.1 generalise to polygons that are inscribed in degenerate
conics?
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Appendix A
Other Useful Concepts
This Chapter talks about other concepts that were quite useful to the work.
A.1 DEFINITIONS
Denition A.1.1 (Group Action). Suppose G is a group and X is a set. We denote
(g, x) −7 → gx,
a mapping fromG×X intoX is called action of G on X if the following two conditions
are satised:
a. ex = x∀x ∈ X
b. (gh)x = g(hx), ∀x ∈ X and g, h ∈ G.
Denition A.1.2 (Orbits). Assume G is a group acting on a set X. The orbit of an
element x ∈ X is the set
Gx = {gx|g ∈ G}.
Intuitively, the orbit of an element means taking a single element and acting on it
with everything in the group. The set of all orbits is denoted x|G.
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Denition A.1.3. Let G be a group and gxg−1 = gx, ∀g, x ∈ G. The resulting map
G×G→ G, (g, x) 7−→ gx
is called Conjugation. We describe the conjugation as a group action G onto itself.
Denition A.1.4 (Conjugacy Classes). Let G be a group and x ∈ G. The orbit of
the element x under conjugation is called the conjugacy class of x and is denoted
Gx.
Denition A.1.5. (Convex Hull) Let x1, x2, · · · , xn be the set of points in the plane,
RP2. The smallest polygon that contains all points in the plane (including all its
interior points) as shown in Figure A.1
b
b
b b
b
b b b
b b
b
b
b
b b b
b b b b
b b b
b b
b
b b b b
b b b
b
b
b
(b) points in RP2 that generating a smallest
(a) set of n points in RP2 polygon
Figure A.1: Convex hull
is called Convex Hull.
Denition A.1.6. (Newton Polygon) Newton polygon of an algebraic curve
∑
a ziwjij = 0,
is the convex hull of
{(i, j) ∈ Z2|aij 6= 0}.
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A.2 Integrable System
An integrable system is a system with the property of concrete model used to solve
dierential equations in a closed form or expressions in terms of quadratures. In
dierential systems, we have Frobenius integrability and in classical theory of Hamil-
tonian dynamical system and also the notion of Liouville integrability. Solutions to
equations are expressed using integrals.
Example
The Lax equation is an example of an integrable system. Let I = I(t) and L = L(t)
be two time dependent matrices. Lax pair is a pair of time dependent matrices that
satises a corresponding dierential equation
dL
= [I, L]
dt
called the Lax equation where [I, L] = IL− LI is a commutator.
A.2.1 Zero locus of set of Polynomial
Suppose X is a eld. Let B = X[y1, y2, · · · , yn] be the polynomial ring in n variables
where n ≥ 1 ∈ N such that A ∈ B is a set. We dene the zero locus of A as the set
V (A) = {a ∈ Xn : g(A) = 0,∀g ∈ A}
Denition A.2.2. (Spectral Curve) Given a classical integrable system, the spectral
curve is dened as the locus of zeros of the characteristic polynomial of the Lax matrix
of the integrable system.
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References
[1] Bea, G. M. (2014). On integrable generalizations of the pentagram map. Inter-
national Mathematics Research Notices, 2015(12):36693693.
[2] Fuchs, D. and Tabachnikov, S. (2009). Self-dual polygons and self-dual curves.
Functional analysis and other mathematics, 2(2-4):203220.
[3] Glick, M. (2011). The pentagram map and Y-patterns. Advances in Mathematics,
227(2):10191045.
[4] Hitchin, N. (2003). Projective geometry. b3 Course.
[5] Izosimov, A. (2016). Pentagrams, inscribed polygons, and Prym varieties. arXiv
preprint arXiv:1607.03558.
[6] Mohr, R. and Triggs, B. (1996). Projective geometry for image analysis. In XVIIIth
International Symposium on Photogrammetry & Remote Sensing (ISPRS'96).
[7] Motzkin, T. (1945). The pentagon in the projective plane, with a comment on
Napier's rule. Bulletin of the American Mathematical Society, 51(12):985989.
[8] Opoku, E. G. (2017). Projective Geometry. MSc thesis, AIMS-Ghana.
[9] Ovsienko, V., Schwartz, R., and Tabachnikov, S. (2009). Quasiperiodic motion
for the pentagram map. arXiv preprint arXiv:0901.1585.
[10] Ovsienko, V., Schwartz, R., and Tabachnikov, S. (2010). The pentagram map: a
discrete integrable system. Communications in Mathematical Physics, 299(2):409
446.
[11] Schwartz, R. (1992). The pentagram map. Experimental Mathematics, 1(1):71
81.
[12] Schwartz, R. and Tabachnikov, S. (2000). Recurrence of the pentagram map.
60
University of Ghana  http://ugspace.ug.edu.gh
[13] Schwartz, R. E. (2001). The pentagram map is recurrent. Experimental Mathe-
matics, 10(4):519528.
[14] Schwartz, R. E. (2007). The Poncelet grid. Advances in Geometry, 7(2):157175.
[15] Schwartz, R. E. (2008). Discrete monodromy, pentagrams, and the method of
condensation. Journal of Fixed Point Theory and Applications, 3(2):379409.
[16] Schwartz, R. E. (2015). The pentagram integrals for Poncelet families. Journal
of Geometry and Physics, 87:432449.
[17] Schwartz, R. E. and Tabachnikov, S. (2010). Elementary surprises in projective
geometry. The Mathematical Intelligencer, 32(3):3134.
[18] Schwartz, R. E. and Tabachnikov, S. (2011). The pentagram integrals on inscribed
polygons. the electronic journal of combinatorics, 18(1):171.
[19] Soloviev, F. et al. (2013). Integrability of the pentagram map. Duke Mathematical
Journal, 162(15):28152853.
[20] Zakharevich, V. (2010). (m)-self-dual polygons. Polythecnic institute of New
York.
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