AN ABSTRACT OF THE DISSERTATION OF
Tatsuhiko Hatase for the degree of Doctor of Philosophy in Mathematics presented on
August 24, 2011.
Title: Algebraic Pappus Curves
Abstract approved:
Thomas A. Schmidt
We show that Pappus Curves, introduced by R. Schwartz to study his dynamical
system in the real projective plane generated by iterated applications of the classical
Pappus Theorem, are algebraic exactly in the linear case. Our approach is to use properties
of projective curves such as singular points, genus, number of automorphisms and to apply
elementary invariant theory.
As a complement, we study fixed points of projective transformations of order four.
c
Copyright by Tatsuhiko Hatase
August 24, 2011
All Rights Reserved
Algebraic Pappus Curves
by
Tatsuhiko Hatase
A DISSERTATION
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Doctor of Philosophy
Presented August 24, 2011
Commencement June 2012
Doctor of Philosophy dissertation of Tatsuhiko Hatase presented on August 24, 2011
APPROVED:
Major Professor, representing Mathematics
Chair of the Department of Mathematics
Dean of the Graduate School
I understand that my dissertation will become part of the permanent collection of Oregon
State University libraries. My signature below authorizes release of my dissertation to
any reader upon request.
Tatsuhiko Hatase, Author
ACKNOWLEDGEMENTS
Academic
I owe my deepest gratitude to my thesis advisor Thomas A. Schmidt for keeping me
motivated throughout the whole process for years. I would like to thank my thesis committee members. I am grateful for all that I have learned and for years of financial support
that Mathematics Department of Oregon State University has provided me. Many thanks
to David Wing for TeX support and various other advices.
Personal
I would like to thank my friends for all the moral support over the years. Their constant encouragement kept me going. Special thanks to my continuing source of inspiration
without whom I probably would not have come up with most of my best ideas.
Also, I would like to thank my parents. Without them, I almost surely would not
exist.
TABLE OF CONTENTS
Page
1. INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.1.
Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.2.
Statement of the Main Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.3.
Organization of This Thesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
2. BACKGROUND INFORMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
2.1.
2.2.
Projective Geometry. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
2.1.1 Projective Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.2 Projective Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.3 The Modular Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
7
10
Algebraic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2.1
2.2.2
2.2.3
2.2.4
2.3.
Algebraic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Polarities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Riemann Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
15
19
21
Invariant Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3. SCHWARTZ’S PAPPUS CURVES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.1.
Marked Boxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.2.
Box Operations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3.3.
Orbit Ω and the Incidence Graph Γ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.4.
Return of The Modular Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4. ALGEBRAIC PAPPUS CURVES ARE LINEAR . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
4.1.
Linear Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
4.2.
Excluding Higher Degrees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
4.3.
Cubic Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
TABLE OF CONTENTS (Continued)
Page
4.4.
Conic Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
5. GEOMETRICAL SIGNIFICANCE OF FIXED POINTS OF ORDER FOUR
PROJECTIVE TRANSFORMATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
6. CONCLUSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
APPENDICES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
A
APPENDIX An Additional Proposition By Schwartz . . . . . . . . . . . . . . . . . . 75
LIST OF FIGURES
Figure
1.1
Page
Pappus’ Theorem: A pair of collinear triple points gives us new collinear
triple points. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
An Overmarked Box with vertices p, q, r, and s, distinguished points t
and b, and distinguished edges T and B. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
Box Operations: τ1 , τ2 , and ι. The images of τ1 and τ2 are nested inside
the original marked box. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
Incidence Graph Γ: Each edge represents a marked box and its vertices
represent top and bottom of the box. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
3.4
Normalization R3; τ1 ι is realized by an order three rotation. . . . . . . . . . . . . .
42
3.5
Normalization P2; ι is a polarity with respect to the conic x2 + y 2 + z 2 = 0. 42
3.6
Maps θ, Proj, and ν̄ are PSL2 (Z)-equivariant. G, M, and M̄ define group
actions of PSL2 (Z) on their respective spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
A Linear Pappus Curve: The marked box is symmetric with respect to
the Pappus Curve. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
3.1
3.2
3.3
4.1
5.1
Conics Invariant Under T0 : They are concentric circles centered at [0 : 0 : 1]. 68
5.2
At a fixed point of an order four transformation, conics are tangent. . . . . .
70
0.1
Pappus Curve Λ and Its Transverse Linefield L. . . . . . . . . . . . . . . . . . . . . . . . . .
77
ALGEBRAIC PAPPUS CURVES
2
1.
1.1.
INTRODUCTION
Motivation
Pappus’ Theorem is as old as the hills.
R. Schwartz
Given a pair of triples of collinear points in the projective plane P2 k for field k of
characteristic neither two nor three, Pappus’ Theorem determines the location of a new
triple of collinear points.
We denote the line joining two distinct points a and b by ab and the intersection of
two distinct lines A and B by AB. Furthermore, we denote the intersection of lines ab
and cd by (ab)(cd) and the line joining AB and CD by (AB)(CD).
Theorem 1.1.0.1 (Pappus’ Theorem) Suppose the points a, b and c are collinear and
the points a0 , b0 and c0 are collinear in the projective plane P2 k for field k of characteristic
neither two nor three. Then the points a00 = (ab0 )(a0 b), b00 = (ac0 )(a0 c) and c00 = (bc0 )(b0 c)
are collinear.
a’’
a’
c
b
a
b’’
b’
c’’
c’
FIGURE 1.1: Pappus’ Theorem: A pair of collinear triple points gives us new collinear
triple points.
Details for the rest of this section will be given in Chapter 3.
3
Richard Schwartz, in his paper Pappus’s Theorem and the Modular Group [11],
defines an object called a marked box which is a certain finite configuration of points
and lines in the projective plane P2 R. Besides vertices at intersections of lines, each
marked box contains two distinguished points; similarly, it has two distinguished edges, its
“top” and “bottom”. By iterating Pappus’ Theorem on marked boxes, Schwartz generates
collections of distinguished points and edges.
Schwartz defines box operations that map marked boxes to marked boxes. Composition gives a group multiplication for the box operations. The group of operations
is generated by three basic operations; using these Schwartz proved that the box operation group is isomorphic to the classical modular group. We give more detail about the
modular group in Section 2.1.3. See Theorem 3.4.0.6.
Naturally enough, the group of box operations acts on the set of all marked boxes.
Schwartz proved that the set of distinguished points from any orbit of convex marked
boxes is dense in a homeomorphic image of S 1 in the projective plane. He calls this S 1 in
the projective plane a Pappus Curve.
The technical definition of Pappus Curve is Definition 3.4.0.14.
1.2.
Statement of the Main Problem
By Schwartz’s definition, a Pappus Curve is a topological circle. Furthermore, he
shows that a Pappus Curve, in fact, is an analytic curve. So, one naturally may ask
whether a Pappus Curve is an algebraic curve as well. In this thesis, to answer this
question, we prove the following theorem.
Main Theorem A Pappus Curve is algebraic if and only if it is linear.
Algebraic here means that the points of the given Pappus Curve satisfy an irreducible
4
polynomial equation. Such an equation defines an algebraic curve (See Section 2.2.1); note
that, on this algebraic curve, there may be other points on this algebraic curve CΛ not on
the Pappus Curve Λ (and in the complex projective plane, there always are such points).
We define Pappus Curve and algebraic Pappus Curve in Section 3.4.
In this thesis, to prove the main theorem, we break down the problem into smaller
cases and prove a series of propositions.
First, by an example, we show that a linear Pappus Curve exists. Then by applying
projective transformations, we show that any line in P2 R contains a Pappus Curve.
Then, we show that any algebraic Pappus Curve must be smooth. By using this
fact along with Plücker’s Formula and Hurwitz’s Automorphism Theorem, we see that
any algebraic curve of degree greater than or equal to four cannot be a Pappus Curve.
To rule out cubics, we show that a cubic Pappus Curve must have infinitely many
flexes which contradicts the fact that a smooth cubic has at most nine flexes. We define
flexes in Section 2.2.2.
Then, finally, we see that no conic can be a Pappus Curve by using invariant theory
and computation.
1.3.
Organization of This Thesis
In Chapter 2, we review the basic concepts of projective geometry, algebraic curves,
and invariant theory. We define key concepts that are crucial to understand the problem clearly. Here, we sketch a proof that smooth complex algebraic curves are compact
Riemann surfaces then state some well known theorems.
In Chapter 3, we explore Schwartz’s paper in detail to clearly define convex marked
boxes and their distinguished points and edges. In particular, we review the fact that the
5
group of box operations is generated by a projective transformation of order three and a
polarity.
In Chapter 4, we prove our Main Theorem as outlined in Section 1.2.
In Chapter 5, we investigate projective transformations of order four. The inspiration for this section comes from one of the box operations defined by Schwartz. We study
the significance of the fixed points of projective transformations of order four.
In Chapter 6, we give a conclusion and discuss possible extensions of this thesis.
In an appendix, we discuss another aspect of Schwartz’s paper. An orbit of (convex)
marked boxes gives a topological circle in P2 R × (P2 R)∗ that satisfies certain geometric
property. We fill in some details to the proof of Theorem A0.9.
6
2.
2.1.
BACKGROUND INFORMATION
Projective Geometry
In this section, we discuss projective spaces and their properties. Most of the material in this section can be found in Projective Geometry by H. S. M. Coxeter [1] and
Undergraduate Algebraic Geometry by Miles Reid [10].
2.1.1
Projective Space
Definition 2.1.1.1 Let k be a field. Then the projective n-space over k is defined to be
the set of all one dimensional linear subspaces of k n+1 . We denote the projective n-space
over k by Pn k.
In the vector space k n+1 two nonzero vectors (x1 , . . . , xn+1 ) and λ(x1 , . . . , xn+1 )
span the same subspace of dimension one for λ 6= 0 in k. We denote the point p ∈ Pn k
which corresponds to the subspace spanned by (x1 , . . . , xn+1 ) by p = [x1 : . . . : xn+1 ]. We
call this coordinate system for points in Pn k the homogeneous coordinate system where
[x1 : . . . : xn+1 ] = [λx1 : . . . : λxn+1 ] for all λ ∈ k.
The set of points p = [x1 : . . . : xn+1 ] with xn+1 6= 0 in Pn k is called (the standard)
affine space and the set of points with xn+1 = 0 is the hyperplane at infinity.
A projective n-space can be seen as the union of the affine part from above, which is
k n embedded into Pn k and the hyperplane at infinity. Here, in an affine space, we ignore
the vector space structure of k n .
Now, we define more general affine space of a projective space.
7
Definition 2.1.1.2 An affine space in Pn k is a subspace that is isomorphic to k m for
m ≤ n.
When n = 2, a projective 2-space is called a projective plane.
Let p, q ∈ Pn k be points, and vp , vq ∈ k n+1 be vectors corresponding to the respective
points. The line through p and q in Pn k is the collection of points that are represented
by vectors of the form avp + bvq where a, b ∈ k. Each such point corresponds to a one
dimensional subspace in the two dimensional subspace of k n+1 spanned by the vectors vp
and vq .
In k 3 , any two distinct subspaces of dimension two must have a common subspace of
dimension one. Suppose they do not intersect, then we can get four linearly independent
vectors, two from each subspace, and this is not possible in a vector space of dimension
three. Thus we have that, in P2 k, any two distinct lines must intersect.
Because of this, we have that given a pair of distinct lines, there is a unique point
where they intersect. Analogous to the fact that a pair of distinct points determines
a unique line joining them, each pair of distinct lines meets in a unique point. The
uniqueness of the intersection point is guaranteed by the distinctness of the lines; if there
are two points of intersection, then the two subspaces of dimension two corresponding to
our lines share two linearly independent vectors, and that contradicts the subspaces being
distinct.
This idea gives us the duality of a projective plane. Any geometric statement for
P2 k remains true when the roles of points and lines are switched.
2.1.2
Projective Transformations
The group GLn+1 (k), the set of (n + 1) × (n + 1) matrices in k with nonzero
determinants, represents the set of invertible linear transformations of the vector space
8
k n+1 . Invertible linear transformations map one dimensional subspaces to one dimensional
subspaces, and, in general, m dimensional subspaces to m dimensional subspaces. For
each linear transformation in k n+1 , there is a corresponding map in Pn k that takes points
to points, lines to lines, and, in general, dimension d linear subspaces to dimension d
linear subspaces while preserving intersections and joins. We call this map a projective
transformation on Pn k. It is easy to see that the set of projective transformations forms
a group.
If S, T ∈ GLn+1 (k) are such that, for some nonzero λ ∈ k, S = λT , then for any
x ∈ k n+1 ,
Sx = λT x.
Since [x1 : . . . : xn+1 ] = [λx1 : . . . : λxn+1 ] in Pn k, S and T induce the same map on Pn k.
So, we have that PGLn+1 (k), the quotient group GLn+1 (k)/k ∗ is the group of projective
transformations on Pn k.
Given a pair of m dimensional subspaces of k n+1 , there is a linear transformation
in GLn+1 (k) that maps one to the other; given a pair of points or lines in Pn k, there is a
projective transformation that maps one to the other.
Lemma 2.1.2.1 A projective transformation maps an affine plane to an affine plane [5].
Proof. A projective transformation on P2 R maps a line to a line. Hence it maps the
complement of a line to the complement of a line. Therefore, a projective transformation
maps an affine plane to an affine plane.
In particular, the group of the projective transformations on P2 k is (isomorphic to)
PGL3 (k).
9
Definition 2.1.2.1 We say that a quadruple of points in the projective plane are in general position if no triple of them are collinear.
Remark 2.1.2.1 A group of projective transformations act sharply four transitively on
the projective plane; given a pair of quadruples of points in general position, there exists
a unique projective transformation that maps one to the other.
To see this, given a pair of four points in general position, we find the unique projective transformation that maps one to the other by solving for a matrix in PGL3 (k).
Due to its utility, we state an obvious implication of the sharp four transitivity.
Lemma 2.1.2.2 A projective transformation on the projective plane that fixes four points
in general position is the identity map.
We sketch a proof of the following well-known result.
Lemma 2.1.2.3 A projective transformation that fixes three points on a line fixes the
entire line pointwise.
Proof. Suppose that T fixes collinear the points u, v, and w. Then there exists a
projective transformation S (not unique) such that
u 7→ [1 : 0 : 1]
v 7→ [0 : 0 : 1]
w 7→ [−1 : 0 : 1]
10
We first show that any projective transformation that fixes [1 : 0 : 1], [0 : 0 : 1], and
[−1 : 0 : 1] also fixes all the other points on the line y = 0. Then by conjugacy, we have
that T fixes the line containing u, v, and w pointwise as well. Thus, it suffices to show
the claim holds for this specific case.
Suppose that T 0 is a projective transformation that fixes the three points shown
above on the line y = 0. Then computation shows that


α

T0 = 
0

0
β
γ
δ
0

0
,

α
and, with this, one easily shows that
T 0 : [x : 0 : z] 7→ [αx : 0 : αz].
Thus, T 0 fixes all the points on the line y = 0.
We are most interested in the case where k = R. In some cases, we will study P2 R
as a subset of P2 C.
On P2 R, the collection of projective transformations is represented by the group
PSL3 (R) = SL3 (R)/{±I} where SL3 (R) is the group of 2 × 2 matrices in R with determinant one and T is the identity matrix. This is because PSL3 (R) and PGL3 (R) are
isomorphic as groups; for any M ∈ SL3 (R), there is a matrix M 0 ∈ GL3 (R) and λ ∈ R
such that M = λM 0 .
2.1.3
The Modular Group
The material in this subsection comes from Fuchsian Groups by Svetlana Katok [7].
11
The modular group, PSL2 (Z) is a subset of the collection of all fractional linear
transformations. For each element of PSL2 (R), a fractional linear transformation on C is
defined by
z 7→
az + b
cz + d
where a, b, c, d ∈ R and ad − bc = 1. For the modular group, we have that a, b, c, d ∈ Z.
Fractional linear transformations are isometries for Poincaré’s upper half-plane model of
hyperbolic geometry. The modular group is generated by two transformations S and ST
where
S : z 7→ −
1
z
and
T : z 7→ z + 1.
We have that S 2 = (ST )3 = I.
A fractional linear transformation is said to be hyperbolic if and only if it fixes two
real points at the boundary of H = {z ∈ C : Im(z) > 0}. A hyperbolic map attracts
toward one of the fixed points, and expands from the other with respect to the hyperbolic
metric defined on H. A matrix representing an element of PSL2 (R) is hyperbolic if and
only if its trace is greater than two in absolute value. So, there are infinitely many
hyperbolic elements in PSL2 (Z) since it is very easy to have a + d > 2 and ad − bc = 1 by
letting b = 1 and c = ad − 1 for any choice of a and d.
Lemma 2.1.3.1 The action of PSL2 (Z) is transitive on Q̂ = Q ∪ ∞ [7].
Proof. For any
a
c
∈ Q, with gcd(a, c) = 1, there exists b, d ∈ Z such that ad−bc = 1.
Therefore, a transformation T (z) =
az+b
cz+d
congruent under the action of PSL2 (Z).
sends ∞ to
a
c.
Thus, any two points in Q̂ are
12
Corollary 2.1.3.1 For any z ∈ R̂ = R ∪ {∞}, the PSL2 (Z) orbit of z is dense in R̂.
By the previous lemma, for any q ∈ Q̂, its images under the action of PSL2 (Z) is Q̂.
Since Q̂ is dense in R̂, it follows that the images of q are dense in R̂. For z ∈
/ Q̂, we can
show the same by taking the conjugate of PSL2 (Z) by an element of PSL2 (R) such that
T (z) ∈ Q̂.
We use a single SL2 (Z) orbit to show the following.
Corollary 2.1.3.2 The hyperbolic fixed points of PSL2 (Z) are dense in R̂.
The map T (z) =
2z+1
z+1
is hyperbolic and fixes z =
√
1± 5
2 .
By the previous lemma,
the images of either fixed point under the action of PSL2 (Z) are dense in R̂. Thus, by
conjugation, one easily shows that the hyperbolic fixed points of PSL2 (Z) are dense in R̂.
2.2.
Algebraic Curves
In this section, we discuss some basic ideas of algebraic curves. Most of this section
is from Algebraic Curves by William Fulton [4].
2.2.1
Algebraic Curves
Definition 2.2.1.1 Let R be a ring. Then we let R[x] denote the ring of polynomials in
P
x with coefficients in R. The degree of a polynomial
ai xi is the largest d such that
ad 6= 0. We say that a polynomial of degree d is monic if ad = 1.
We let R[x1 , . . . , xn ] denote the ring of polynomials in n variables over R. For our
ease, when n = 3, we will write R[x, y, z].
13
Definition 2.2.1.2 The monomials in R[x1 , . . . , xn ] are polynomials of the form xi11 · · · xinn
with non-negative integers ij . Here, the degree of the monomial is i1 + · · · + in . A polynomial F is homogeneous or a form of degree d if all the terms in F are monomials of
degree d with coefficients in R.
Now that we have a solid definition of the ring of polynomials, we discuss the
algebraic curves defined by the polynomials.
Definition 2.2.1.3 A point p = [x1 : . . . : xn+1 ] ∈ Pn k is said to be a zero of a homogeneous polynomial f ∈ k[x1 , . . . , xn+1 ] if f (x1 , . . . , xn+1 ) = 0.
Furthermore, the set of all the zeros of a homogeneous polynomial is called an algebraic hypersurface. If the degree of the homogeneous polynomial is d, then we say that
the degree of the algebraic hypersurface is d.
When n = 2, an algebraic hypersurface is called an algebraic curve.
Note that zeros of homogeneous polynomials and algebraic curves are well-defined.
Since f is homogeneous, if for p = [x1 : . . . : xn+1 ], f (x1 , . . . , xn+1 ) = 0, then for p =
[λx1 : . . . : λxn+1 ] we have that
f (λx1 , . . . , λxn+1 ) = λd f (x1 , . . . , xn+1 ) = 0
where d is the degree of f .
Algebraic curves in P2 k are also referred to as projective plane curves. Projective
transformations send algebraic curves to algebraic curves and preserve the degree.
For any set S of polynomials in k[x1 , . . . , xn+1 ], we define
V(S) = {p ∈ Pn k : p is a zero of each f ∈ S}.
If I is the ideal generated by the set S, then V(I) = V(S). We call V(I) an algebraic set.
14
Definition 2.2.1.4 We say that an algebraic set V ⊂ Pn k is irreducible if it is not the
union of two or more distinct algebraic sets. Otherwise, an algebraic set is said to be
reducible. Each algebraic set in a reducible algebraic set is called a component.
We have that V(hf i) is irreducible if and only if f is an irreducible polynomial where
hf i denotes the ideal in k[x1 , . . . , xn+1 ] generated by f .
Let X ⊂ Pn k, then we define
I(X) = {f ∈ k[x1 , . . . , xn+1 ] : every p ∈ X is a zero of f }.
This set I(X) is an ideal in k[x1 , . . . , xn+1 ] since for any f ∈ I(X) and g ∈ k[x1 , . . . , xn+1 ],
for every p ∈ X,
f g(p) = f (p)g(p) = 0 ∗ g(p) = 0,
and so f g ∈ I(X). We call I(X) the ideal of X.
Given an ideal I in a ring R, the radial of I is the set
rad(I) = { r ∈ R : rn ∈ I for some n ∈ Z+ }.
We call an ideal I a radical ideal if I = rad(I) [2].
There is a one-to-one correspondence between the set of radical ideals of k[x1 , . . . , xn+1 ]
and the set of algebraic sets in Pn k.
Definition 2.2.1.5 Given f1 , . . . , fm ∈ k[x1 , . . . , xn ], if there is no nonzero polynomial
F ∈ k[X1 , . . . , Xm ] such that F (f1 , . . . , fm ) ≡ 0, then the set of polynomials {f1 , . . . , fm }
is said to be algebraically independent.
Algebraic curves of degrees one, two, and three are called: lines, conics, and cubics
respectively. Thus our Main Theorem states that only lines can be algebraic Pappus
Curves.
15
Given a pair of algebraic curves, then for each point in the projective plane, there
is a non-negative integer called the intersection number which describes the multiplicity
of their intersection at this point. See Fulton’s Algebraic Curves [4] for details of the
definition.
The following theorem is very important while studying algebraic curves.
Theorem 2.2.1.1 Bézout’s Theorem
Let f and g be projective plane curves of degrees df and dg respectively over an
algebraically closed field. Assume that f and g have no common component. Then f and
g intersect df dg times counted with multiplicities [4].
2.2.2
Singularities
Definition 2.2.2.1 Given an algebraic hypersurface f (x1 , . . . , xn+1 ) = 0, p ∈ Pn k is a
singular point of f if
f (p) =
∂f
∂f
(p) = · · · =
(p) = 0.
∂x1
∂xn+1
A hypersurface with no singular points is said to be smooth or non-singular.
With this definition, we have the following corollaries of Bézout’s Theorem.
Corollary 2.2.2.1 A reducible algebraic curve has singularities.
Proof. Suppose that C is a reducible algebraic curve defined by a reducible homogeneous polynomial equation f g = 0. Let Cf and Cg be the algebraic curves given by
the polynomial equations f = 0 and g = 0 of degrees df and dg respectively. Then, by
Bézout’s Theorem, f and g intersect df dg times counted with multiplicities. One easily
shows by computation that those intersection points are singular points of C.
16
Corollary 2.2.2.2 An irreducible algebraic curve over a field of characteristic 0 has
finitely many singular points.
Proof. By Bézout’s Theorem, unless f ,
∂f ∂f
∂x , ∂y ,
and
∂f
∂z
have a common component,
the system of equations
f (p) =
∂f
∂f
∂f
(p) =
(p) =
(p) = 0
∂x
∂y
∂z
has only finitely many solutions. Now, if f is irreducible, then it has only one component.
∂f
∂f
Since deg( ∂f
∂x ) = deg( ∂y ) = deg( ∂z ) = deg(f ) − 1, they cannot have a common component
with f . Hence, an irreducible curve has only finitely many singular points.
Remark 2.2.2.1 Let T be a projective transformation. Suppose that C is an algebraic
curve defined by a homogeneous polynomial f . Then T (C) is defined by the homogeneous
polynomial g = f ◦ T −1 . Then, for p ∈ C,
g(T (p)) = f ◦ T −1 (T (p)) = f (p) = 0.
Now, suppose that p is a singular point of C. Then by applying the chain rule
∂g
∂f −1
∂f
(T (p)) = ai
(T (T (p))) = ai
(p) = 0
∂xi
∂xi
∂xi
for some ai = ai (T ) ∈ k. Thus, we have that T (p) is a singular point of T (C).
Hence, a projective transformation T maps singular points of a curve C to singular
points of the curve T (C).
Definition 2.2.2.2 In P2 k, the multiplicity of a point p on a curve C = V(f ) is the
smallest integer m such that for some i+j = m,
of the point p.
∂mf
∂xi ∂y j
6= 0. Let mp denote the multiplicity
17
Definition 2.2.2.3 A line L is said to be tangent to a curve C at p if the intersection
number of L and C at p is greater than mp [14].
When the multiplicity of a point is one, we call the point a simple point of the
curve. Note that a simple point is non-singular. Also, C has exactly one tangent line at
any simple point.
A singular point of C at which there are more than one distinct tangent line is called
a multiple point of C.
Definition 2.2.2.4 A simple point p on a curve C is said to be a flex of the curve C if
its tangent line intersects C at p with multiplicity greater than or equal to three.
Flexes can also be defined in terms of the polynomial that defines the curve. For a
given homogeneous polynomial f of degree d,

 fxx

H = det 
 fyx

fzx
define its Hessian to be

fxy fxz 

fyy fyz 
.

fzy fzz
We have that H is a form of degree 3(d − 2).
Suppose that the curve C and its Hessian intersect at a point p. Then p is either a
multiple point or a flex of C.
Lemma 2.2.2.1 Projective transformations map flexes to flexes.
Proof. Suppose that C = V(f ). By Remark 2.2.2.1, given a projective transformation T , we have that T (fxx )(T (p)) = (f ◦ T −1 )xx (T (p)) = fxx (p). Similar equations hold
for other second partial derivatives. Therefore, it follows that T (H)(T (p)) = H(p) for any
p ∈ P2 R. Thus, we have that if p is a flex of C, then T (p) is a flex of T (C).
18
Corollary 2.2.2.3 (of Bézout’s Theorem)
A smooth cubic over a field of characteristic 0 has at least one but no more than
nine flexes.
Proof. Given a smooth cubic algebraic curve C defined by a cubic polynomial f ,
its Hessian is of degree three. By Bézout’s Theorem, we have that f and H intersect
nine times counting with multiplicities. Since C is smooth, none of these intersections are
multiple points. Hence, f has at least one but no more than nine flexes.
In fact, one can show that there are nine distinct flexes in a smooth cubic.
Another important characteristic an algebraic curve is the genus which we define
later in this section. This value g can be computed using the following proposition.
Proposition 2.2.2.1 Plücker’s Formula
Let C be a smooth projective plane curve over C. Then the genus g of C is
g=
(d − 1)(d − 2)
2
where d is the degree of C [8].
The following is a handy fact about conics.
Proposition 2.2.2.2 If p1 , . . . , p5 ∈ P2 k are distinct points such that no four are collinear,
then there exists exactly one conic through p1 , . . . , p5 [10].
Proof. On P2 k, the equation of a conic is given by
Ax2 + Bxy + Cy 2 + Dxz + Exy + F z 2 = 0.
19
Suppose C is a conic and pi = [xi : yi : zi ] ∈ C for i = 1, . . . , 5 where those points are such
that no quartet of them are collinear. Then we get a system of five linear equations in k 6
Ax2i + Bxi yi + Cyi2 + Dxi zi + Exi yi + F zi2 = 0
where i = 1, . . . , 5. Since we have five equations, this is enough to solve for five of the six
coefficients as linear terms in the six variables. Now, for any λ 6= 0, the equation
λ(Ax2 + Bxy + Cy 2 + Dxz + Exy + F z 2 ) = 0
defines the same conic as the previous equation. Thus, we have that p1 , . . . , p5 define a
unique conic.
As can be seen in the above proof, any five linearly independent equations are
enough to be able to uniquely solve for the coefficients. For example, if we have three
non-collinear points and the equations of tangent lines for two of those points, we can
determine a unique conic that satisfies those conditions.
This gives a proof to the following proposition.
Proposition 2.2.2.3 Five linearly independent conditions define a unique conic.
Corollary 2.2.2.4 For each d, an irreducible algebraic curve of degree d is uniquely determined by finitely many points it passes through.
Proof. For any d, a polynomial of degree d has only finitely many coefficients.
Thus, given enough points in general position, we have sufficeuntly many equations to
solve for the coefficients uniquely.
2.2.3
Polarities
The material in this subsection is from Algebraic Curves by R. Walker [14] and
Plane Algebraic Curves by H. Hilton [6].
20
Definition 2.2.3.1 Suppose that we are given a line l that intersects the conic C at the
points p and q. Then the pole of l is defined to be the point at which the tangent lines of
C at p and q intersect. If p = q, then we have that l is tangent to C and its pole is p.
Now, suppose that we are given a point p. Then we have that the polar of p is the
collection of poles of all the lines through p.
One can show that the polar of p is a line.
Let the polarity with respect to a given base conic to be the operation which takes
a point to its polar and a line to its pole with respect to the conic.
A polarity is an involution on a projective plane. A polarity preserves incidences.
Suppose that P is the polar of p. If p ∈ P , then we say that p and P are self-conjugate.
Example 2.2.3.1 Let the conic f (x, y, z) = x2 + y 2 + z 2 = 0 be given. Let p = [A : B : C]
and q = [A0 : B 0 : C 0 ] be the points at which the conic and the line ax + by + cz = 0
intersect. Since
∂f
∂x
= 2x,
∂f
∂y
= 2y, and
∂f
∂z
= 2z, the equations of the tangent lines at p
and q are Ax + By + Cz = 0 and A0 x + B 0 y + C 0 z = 0 respectively.
Because p and q are on the line ax + by + cz = 0, we have that
Aa + Bb + Cc = A0 a + B 0 b + C 0 c = 0.
This equation also indicates that the point [a : b : c] is on both Ax + By + Cz = 0 and
A0 x + B 0 y + C 0 z = 0. Therefore, [a : b : c] is the point where the two lines intersect.
So we have that the polar of the point [a : b : c] with respect to x2 + y 2 + z 2 = 0 is
the line ax + by + cz = 0. Similarly, the pole of the line ax + by + cz = 0 is the point
[a : b : c].
The set of projective transformations and polarities together generate the group of
projective symmetries. The non-identity elements of projective transformations, polarities,
21
and projective symmetries are also known as collineations, correlations, and projectivities,
respectively [1].
More details about projectivities can be found in Projective Geometry by H. S. M.
Coxeter [1].
2.2.4
Riemann Surfaces
Algebraic curves over C can be viewed as Riemann surfaces. Here, we review some
basic concepts of Riemann surfaces.
The material in this section is mostly from Algebraic Curves and Riemann Surfaces
by R. Miranda [8].
Basically, Riemann surfaces are spaces that locally look like an open set in the
complex plane. We use complex charts, defined below, to show how a space can look like
an open set in the complex plane.
Definition 2.2.4.1 Suppose that f is a complex valued function on C. Then for z ∈ C,
we can express it as z = x + iy where x, y ∈ R. Also, we can write the function f (z) as
f (x + iy) = u(x, y) + iv(x, y)
where u(x, y) and v(x, y) are real valued functions. Then we say that f is differentiable
at z if the Cauchy-Riemann equations
∂v
∂u
=
∂x
∂y
and
∂u
∂v
=−
∂y
∂x
are satisfied.
A function f is holomorphic if it is differentiable at all points in its domain [3].
A holomorphic map whose inverse is also holomorphic is said to be biholomorphic
[8].
22
Definition 2.2.4.2 A complex chart, or simply a chart, on a topological space X, is a
homeomorphism φ : U → V , where U ⊂ X is an open set in X, and V ⊂ C is an open set
in the complex plane. The chart φ is said to be centered at p ∈ U if φ(p) = 0.
Furthermore, let φ1 : U1 → V1 and φ2 : U2 → V2 be two complex charts on X. We
say that φ1 and φ2 are compatible if either U1 ∩ U2 = ∅, or
φ2 ◦ φ−1
1 : φ1 (U1 ∩ U2 ) → φ2 (U1 ∩ U2 )
is biholomorphic.
With charts, we give the space X local complex coordinates.
Definition 2.2.4.3 A complex atlas A on X is a collection
A = {φα : Uα → Vα }
of pairwise compatible complex charts where {Uα } covers X.
Two complex atlases are said to be equivalent if every chart of one is compatible
with every chart of the other.
If two atlases are equivalent, then their union is also an atlas. We call a maximal
atlas, by inclusion, a complex structure of X.
Definition 2.2.4.4 If a space X has a countable basis for its topology, then X is said to
be second countable [9].
Definition 2.2.4.5 A topological space X is called a Hausdorff space is for each pair x1 ,
x2 of distinct points of X, there exist neighborhoods U1 and U2 of x1 and x2 , respectively,
that are disjoint [9].
23
Definition 2.2.4.6 A Riemann surface is a second countable connected Hausdorff topological space X together with a complex structure.
If a Riemann surface X is a compact space, then we say that X is a compact
Riemann surface.
The genus of a Riemann surface X, is the non-negative integer g such that
H1 (X; Z) ∼
= Z2g
where H1 (X; Z) is the first homology group of X with integer coefficients [3].
Informally, the genus of a Riemann surface is the largest number of closed curves
such that removing them keeps the surface connected.
Definition 2.2.4.7 A mapping F : X → Y is holomorphic at p ∈ X if and only if there
exists charts φ1 : U1 → V1 on X with p ∈ U1 and φ2 : U2 → V2 on Y with F (p) ∈ U2 such
that the composition φ2 ◦ F ◦ φ−1
1 is holomorphic at φ1 (p). We say that F is a holomorphic
map if and only if F is holomorphic on all of X.
Definition 2.2.4.8 An isomorphism between Riemann surfaces is a holomorphic map
F : X → Y which is bijective, and whose inverse F −1 : Y → X is holomorphic. A
self-isomorphism F : X → X is called an automorphism of X.
For a compact Riemann surface X with genus g, there is an important result.
Theorem 2.2.4.1 Hurwitz’s Automorphism Theorem
Let G be a group of automorphisms of a compact Riemann surface X of genus g ≥ 2.
Then
|G| ≤ 84(g − 1).
24
Now, in order to apply Hurwitz’s Automorphism’ Theorem, we show that smooth
algebraic curves are compact Riemann surfaces.
Theorem 2.2.4.2 The Implicit Function Theorem
Let f (u, v) ∈ C[u, v] and X = {(u, v) ∈ C2 | f (u, v) = 0}. Let p = (u0 , v0 ) be a point
on X. Suppose that
∂f
∂v (p)
6= 0. Then there exists a function g(u) defined and holomorphic
in a neighborhood of u0 such that, near p, X is equal to the graph v = g(u). Moreover
∂f
g 0 = − ∂f
∂u / ∂v near u0 .
Theorem 2.2.4.3 Any smooth algebraic curve over C is a compact Riemann surface.
Here, we sketch a variant of the proof provided by Miranda in sections 1.2 and 1.3
in [8].
Proof. Suppose that C is a smooth algebraic curve given by a homogeneous polynomial F . Let Az denote the affine plane {[x : y : z] ∈ P2 C : z 6= 0}. Then let Cz be the
affine curve defined by the polynomial f (y, z) = F (x, y, 1). We have that Cz = C ∩ Az .
Now, suppose that T is a projective transformation represented by a matrix


 a b c 



T =
 d e f ,


g h i
then let AT denote the affine plane to which T maps Az . Now, let CT be the affine curve
defined by the polynomial fT (u, v) = F (u, v, 1) where
u=
ax + by + cz
dx + ey + f z
and v =
.
gx + hy + iz
gx + hy + iz
Then, we have that CT = C ∩ AT .
25
We have that a smooth algebraic plane curve is irreducible since otherwise we would
have singularities at the intersection of the components. Thus, it follows that CT , is also
smooth and irreducible.
On a smooth affine plane curve, we can obtain complex charts by the Implicit
Function Theorem. Thus we have that CT is a Riemann surface.
Now, we have that {CT } are open subsets of C by subspace topology since each of
them is an intersection of C and an affine space which is an open subset of P2 C. Also, the
complex charts defined by the Implicit Function Theorem can be shown to be compatible
with each other. Thus, we have that C is also a Riemann surface.
Furthermore, since C is a closed subset of P2 C, it is compact. Thus, any smooth
algebraic plane curve is a compact Riemann surface.
2.3.
Invariant Theory
In this section, we discuss invariant theory. This material can be found in Bernd
Sturmfels’ Algorithms in Invariant Theory [12].
Even though invariant theory is about algebraic curves, it is not used commonly or
as well known as other concepts we see in this thesis. So we dedicate this section for it.
We are interested in the polynomials that remain invariant under the action of a
finite group of projective transformations that is represented by a group G ⊂ PGLn (C).
Let C[x] denote the ring of polynomials in n variables x = (x1 , . . . , xn ). Given a
finite group G, let C[x]G be the ring of invariant polynomials under the action of G in
C[x]. Our aim is to determine a set {I1 , . . . , Im } that generates the invariant subring
C[x]G . When none of the polynomial in the set can be expressed in terms of the others,
we call each polynomial in the set a fundamental invariant.
26
The following operator, called the Reynolds operator, takes a polynomial in C[x]
and gives us a polynomial that is invariant under the action of a finite matrix group G.
The Reynolds operator is defined by
∗ : C[x] → C[x]G ,
f 7→ f ∗ :=
1 X
f ◦π
|G|
π∈G
where f ◦ π is defined to be the polynomial f (πx) with πx being the transpose of the
matrix multiplication of π and the column vector xT .
The Reynolds operator has the following properties.
• The Reynolds operator is a C-linear map.
• The Reynolds operator acts as the identity map on C[x]G .
Since πx has linear terms in each coordinate,
deg(f ) = deg(f ∗ ).
Elementary calculation with the Reynolds operator can be easily performed.
Example 2.3.0.1 Let
G3 = {g1 , g2 , g0 } ⊂ PGL3 (C)
where

1
 −2
 √
3
g1 = 
 2

0
√
−
3
2
− 12
0

0 

0 
,

1
27

1
 −2
 √
3
g2 = 
 − 2

0

√
3
2
− 12
0

0 

0 
,

1

 1 0 0 



g0 = 
 0 1 0 .


0 0 1
Then, by Reynolds operator, we have that
1
x∗ =
3
1
y∗ =
3
!
√
√
3
1
3
1
y) + (− x +
y) + x = 0,
(− x −
2
2
2
2
!
√
3
3
1
1
x − y) + (−
x − y) + y = 0,
(
2
2
2
2
√
z∗ =
1
(x2 )∗ =
3
1
(x ) =
3
3 ∗
1
(y 3 )∗ =
3
1
(z + z + z) = z,
3
!
√
√
1
1
1
3 2
3 2
(− x −
y) + (− x +
y) + x2 = (x2 + y 2 ),
2
2
2
2
2
!
√
√
1
3 3
1
3 3
1
3
(− x −
y) + (− x +
y) + x
= (x3 − 3xy 2 ),
2
2
2
2
4
√
!
√
3
1 3
3
1 3
1
(
x − y) + (−
x − y) + y 3 = (y 3 − 3x2 y).
2
2
2
2
4
The following proposition shows that every finite subgroup of PGLn (C) has at least
n invariants.
28
Proposition 2.3.0.1 Every finite matrix group G ⊂ PGLn (C) has n algebraically independent invariants, i.e., the ring C[x]G has transcendence degree n over C.
Proof. Let, for each i ∈ {1, . . . , n},
Pi :=
Y
(xi ◦ π − t) ∈ C[x][t]
π∈G
and consider Pi (t) as a monic polynomial in t with coefficients from C[x]. Since Pi is
invariant under the action of G on the x-variables, its coefficients are also invariant, so
Pi ∈ C[x]G [t].
Now, we have that t = xi is a root for Pi (t), because one of the elements in G is
the identity map. This means that the variables x1 , . . . , xn are algebraically dependent on
some invariant polynomials in C[x][t]. Hence the invariant subring C[x]G and C[x] have
the same transcendence degree n over C.
The following theorem of Hilbert guarantees that any invariant ring C[x]G with G
a finite matrix group in PGLn (C) is finitely generated.
Theorem 2.3.0.4 Hilbert’s Finiteness Theorem
The invariant ring C[x]G of a finite matrix group G ⊂ PGLn (C) is finitely generated.
Proof. Let IG = hC[x]G
+ i be the ideal in C[x] that is generated by all homogeneous
invariants of positive degree. Then, it follows that IG is generated by the polynomials
(xe11 · · · xenn )∗ where (e1 , . . . , en ) ranges over all nonzero, nonnegative integers.
By Hilbert’s basis theorem, we have that every ideal in C[x] is finitely generated,
so there are finitely many homogeneous invariants I1 , . . . , Im such that IG = hI1 , . . . , Im i.
Furthermore, we have that this set of invariants generates C[x]G .
29
We continue our discussion of finding finitely many polynomials to generate the
invariant ring.
Let C[x]G
d denote the set of all homogeneous invariants of degree d. We have that
the invariant ring C[x]G is the direct sum of the finite-dimensional C-vector spaces C[x]G
d.
Definition 2.3.0.9 The Hilbert series of C[x]G is the generating function
ΦG (t) =
∞
X
d
dimC ( C[ x ]G
d )t .
d=0
The following theorem by Molien in 1897 provides a formula for the Hilbert series
of C[x]G .
Theorem 2.3.0.5 Molien [12]
The Hilbert series of the invariant ring C[x]G equals
ΦG (t) =
1 X
1
.
|G|
det(id − tπ)
π∈G
The following lemma helps us determine when we have found a set of invariants that
generate the invariant subring C[x]G .
Lemma 2.3.0.1 Let p1 , . . . , pm be algebraically independent elements of C[x] which are
homogeneous of degrees d1 , . . . , dm respectively. Then the Hilbert series of R := C[p1 , . . . , pm ]
is
H(R, t) :=
∞
X
(dimC Rd ) td =
n=0
1
(1 −
td1 ) · · · (1
− tdm )
.
30
Proof. Let Rd be the set of polynomials of degree d in R. Since the pi are algebraically independent, the set
{pi11 · · · pimm | i1 , . . . , im ∈ N and i1 d1 + · · · + im dm = d}
forms a C-basis for the Rd as a vector space. Thus the dimension of Rd is equal to the
cardinality of the set
Ad = {(i1 , . . . , im ) ∈ Nm | i1 d1 + · · · + im dm = d}.
Then
1
1
1
=
···
(1 − td1 ) · · · (1 − tdm )
1 − td1
1 − tdm
!
!
∞
∞
X
X
=
ti1 d1 · · ·
tim dm
i1 =0
=
∞
X
im =0


X

d=0
(i1 ,...,im )∈Ad
d
t
=
∞
X
|Ad |td
d=0
and this proves the lemma.
Now, with an example, we see how to find a set of invariants that generate C[x]G
by use of its Hilbert series.
Example 2.3.0.2 Let G3 be as in Example 2.3.0.1. Then the Hilbert series of C[x, y, z]G3
is
1
1
1
1
ΦG3 (t) =
+
+
3 det(id − tg1 ) det(id − tg1 ) det(id − tg1 )
=
1
1
1
1
+
+
3 (1 − t)(t2 + t + 1) (1 − t)(t2 + t + 1) (1 − t)3
=
1 + t3
(1 − t)(1 − t2 )(1 − t3 )
31
= 1 + t + 2t2 + 4t3 + 5t4 + 7t5 + · · ·
The coefficients of t, t2 , and t3 of this Hilbert series tells us that the invariant subring
has one polynomial of degree one, one polynomial of degree two, and two polynomials of
degree three in its generating set. In Example 2.3.0.1, using the Reynolds operator, we
found such invariant polynomials
I1 := z
I2 := x2 + y 2
I3 := x3 − 3xy 2
I4 := y 3 − 3x2 y.
We know that there are at most three algebraically independent invariants here and we
have four invariant polynomials, so we know that at least one can be eliminated by an
algebraic relation. We find that the equation
I42 = I23 − I22
is satisfied. One easily shows that there is no algebraic relation between just I1 , I2 , and
I3 . Thus we have that
C[I1 , I2 , I3 , I4 ] = C[I1 , I2 , I3 ] ⊕ I4 C[I1 , I2 , I3 ].
We have that C[I1 , I2 , I3 ] is a subring generated by algebraically independent homogeneous
polynomials, so by the lemma, the Hilbert series of C[I1 , I2 , I3 ] is
1
.
(1−t)(1−t2 )(1−t3 )
Now,
since the elements of degree d in C[I1 , I2 , I3 ] are in one-to-one correspondence with the
elements of degree d + 3 in I4 C[I1 , I2 , I3 ], the Hilbert series of I4 C[I1 , I2 , I3 ] is equal to
t3
.
(1−t)(1−t2 )(1−t3 )
Hence, we have that the Hilbert series of C[I1 , I2 , I3 , I4 ] is
1+t3
(1−t)(1−t2 )(1+t3 )
which is the Hilbert series of C[x, y, z]G3 . Hence we have that C[x, y, z]G3 is generated by
I1 , I2 , I3 , and I4 .
32
Corollary 2.3.0.1 The only conic invariant under the standard rotation of order three
that fixes the point o = [0 : 0 : 1] and passes through t = [1 : 0 : 1] is x2 + y 2 − z 2 = 0.
As we saw in Example 2.3.0.2, any invariant conic under the rotation is of the form
a(x2 + y 2 ) + bz 2 = 0. This equation is satisfied when the equation is a(x2 + y 2 − z 2 ) = 0.
In the previous example, we saw a case where all group elements are in R ⊂ C, and
the invariant polynomials we found have real coefficients. Since invariant theory thus far
has been defined using C, it may not be not obvious that when G ⊆ PGLn (R), the set of
invariants with real coefficients that generates C[x]G also generates R[x]G .
The following proposition guarantees that when G ⊆ PGLn (R), the set of invariants
with real coefficients that generates C[x]G also generates R[x]G .
Proposition 2.3.0.2 When G ⊆ PGLn (R), the set of invariants with real coefficients
that generates C[x]G also generates R[x]G .
Proof. Let G ⊂ PGLn (R) be a finite subgroup. Since G is a finite subgroup
of PGLn (C), by hypothesis there exist I1 , . . . , Im with real coefficients∗ such that they
generate C[x]G and {I1 , . . . , In } is algebraically independent. Thus, we have that
G
C[x] =
m
M
Ik C[I1 , . . . , In ].
k=n+1
Since I1 , . . . , Im ∈ R[x] and they are invariant under G, we have that
R[x]G ⊇
m
M
Ik R[I1 , . . . , In ].
k=n+1
∗
The coefficients are real since we can apply the Reynolds operator to a term with real coefficients and
since our G has matrices with real entries, the output of the operator also has real coefficients.
33
Now, suppose that I ∈ R[x]G . Then, we can express I as
m X
X
(ak,j + ibk,j )Ij
I=
n=n+1
j
where Ij is a monomial in {I1 , . . . , In } and ak,j and bk,j are real. Then, we can express I
as
I = p + iq
where p, q ∈ R[x] with
p=
m X
X
k=n+1
ak,j Ij
j
and
q=
m X
X
k=n+1
bk,j Ij .
j
Since I is a polynomial with real coefficients, we have that q = 0. But then it follows that
m
X
Ik bk,j = 0
k=n+1
for all j since {I1 , . . . , In } was chosen so that the set is algebraically independent. Hence
I=
m X
X
k=n+1
ak,j Ij ∈
j
m
M
Ik R[I1 , . . . , In ]
k=n+1
and
G
R[x] ⊆
m
M
Ik R[I1 , . . . , In ].
k=n+1
Therefore, we have shown that
G
R[x] =
m
M
Ik R[I1 , . . . , In ].
k=n+1
Hence, we may apply the invariant theory with base field R.
34
3.
SCHWARTZ’S PAPPUS CURVES
In this section, we summarize pertinent material from Richard Schwartz’s paper
Pappus’s Theorem and the Modular Group [11]. This section, including the figures, is
adapted from Schwartz’s paper.
3.1.
Marked Boxes
Schwartz uses Pappus’ Theorem (Theorem 1.1.0.1) to create a collection of marked
boxes (defined below); he iterates the theorem by applying it to a new pair of three
collinear points a, b and c and a00 , b00 and c00 (or a0 , b0 and c0 and a00 , b00 and c00 ). Schwartz
defines objects called marked boxes and derives curves of points and lines from Pappus’
Theorem.
Definition 3.1.0.10 An overmarked box is a pair of 6-tuples of points and lines in P2 R
((p, q, r, s; t, b), (P, Q, R, S; T, B))
where p, q, r, and s are the vertices of the box, and P = ts, Q = tr, R = bq, S = bp,
T = pq, and B = rs are lines with t ∈ T and b ∈ B. See Figure 3.1.
Remark 3.1.0.1 The collection of distinct points (p, q, r, s; t, b) uniquely determines an
box.
There is an involution on the set of overmarked boxes defined by
φ((p, q, r, s; t, b), (P, Q, R, S; T, B)) 7→ ((q, p, s, r; t, b), (Q, P, S, R; T, B)).
35
Since the lines pq and qp are the same, we have that this involution maps overmarked
boxes to themselves while exchanging the labeling of the vertices.
Definition 3.1.0.11 A marked box is an equivalence class of overmarked boxes under φ.
Let
Θ = ((p, q, r, s; t, b), (P, Q, R, S; T, B)),
then the pair (t, T ) is the top of Θ and (b, B) is the bottom of Θ where T and B are
distinguished edges and t and b are distinguished points of Θ.
Definition 3.1.0.12 A marked box is said to be convex if p and q separate t and T B on
the line T , and r and s separate b and T B on the line B as shown in Figure 3.1.
T
p
q
t
P
Q
S
s
B
R
b
r
FIGURE 3.1: An Overmarked Box with vertices p, q, r, and s, distinguished points t and
b, and distinguished edges T and B.
This definition of convexity is the reason why we are working specifically with R
and not just any field; we need the field to have an ordering to define the convexity in the
way that works for us.
Informally, a convex marked box is a convex quadrilateral in P2 R with a distinguished top edge, a distinguished bottom edge, a distinguished top point, and a distinguished bottom point.
36
Definition 3.1.0.13 The interior of a convex marked box ((p, q, r, s; t, b), (P, Q, R, S; T, B))
is the convex region bounded by t, B, ps, and qr that contains the intersection pr ∩ qs.
3.2.
Box Operations
There are three natural box operations one can perform on marked boxes defined
using Pappus’ Theorem.
We let ι be the operation where the top vertices, edge, and the distinguished point
and the respective bottom ones in a given marked box are exchanged so that the resulting
box is convex. Let τ1 be the operation where the top vertices, edge, and the distinguished
point are fixed and the bottom ones go to the points and the line defined by Pappus’
Theorem such that the resulting box is convex. Let τ2 be the operation where the bottom
ones are fixed and the top ones go to the points and the line defined by Pappus’ Theorem
so that the resulting box is convex.
Explicitly, those operations are defined by
ι(Θ) = ((s, r, p, q; b, t), (R, S, Q, P ; B, T )),
τ1 (Θ) = ((p, q, QR, P S; t, (qs)(pr)), (P, Q, qs, pr; T, (QR)(P S)))
τ2 (Θ) = ((QR, P S, s, r; (qs)(pr), b), (pr, qs, S, R; (QR)(P S), B)).
We also let I denote the identity map.
Figure 3.2 indicates how the operations work.
With these explicit expressions of the box operations, one easily computes the following relations.
37
ι
p
q
t
m
PS
q
τ1
Q
P
QR
S
t
p
PS
m
QR
τ2
R
s
s
r
r
ι
b
FIGURE 3.2: Box Operations: τ1 , τ2 , and ι. The images of τ1 and τ2 are nested inside
the original marked box.
b
Lemma 3.2.0.2 The following relations hold:
ι2 = I; τ1 ιτ2 = ι; τ2 ιτ1 = ι; τ1 ιτ1 = τ2 ; τ2 ιτ2 = τ1 .
From this lemma, we see that (τ1 ι)3 = I. Now, let G be the group of box operations
generated by ι, τ1 , and τ2 . Then by the lemma, we have that G is generated by ι and τ1 ι.
Proposition 3.2.0.3 The group of box operations G is isomorphic to the modular group.
Proof. By definition, we have that every element in G can be expressed as a word
in ι, τ1 , τ2 and their inverses. But, by Lemma 3.2.0.2, we have that
ι−1 = ι; τ1−1 = ιτ2 ι; τ2−1 = ιτ1 ι.
Thus, we can express any element of G as a word in ι, τ1 , and τ2 .
As Figure 3.2 indicates, given a convex marked box Θ, τ1 (Θ) and τ2 (Θ) are nested
strictly in Θ. Hence, any word of a finite length in the box operations τ1 and τ2 is neither
the identity map nor ι.
Now, given a word in τ1 , τ2 , and ι, we can make substitutions using the relations
ι2 = I; τ1 ιτ2 = ι; τ2 ιτ1 = ι; τ1 ιτ1 = τ2 ; τ2 ιτ2 = τ1
38
from Lemma 3.2.0.2 to rewrite the word as one of (1) a word in τ1 and τ2 , (2) ι, Tι, ιT,
or (3) ι2 = I; where T is a word in τ1 and τ2 . Since, of these, only ι2 can be the identity,
there are no further non-trivial relations.
Furthermore, since G is generated by ι and τ1 ι,
G = hι, τ1 ι : ι2 = (τ1 ι)3 = Ii.
Hence, G is isomorphic to the Modular Group.
3.3.
Orbit Ω and the Incidence Graph Γ
With G and a marked box Θ, we generate an orbit Ω = Ω(Θ) of marked boxes. The
action of G on Ω is a group action that is fixed point free, effective, and transitive.
The structure of Ω can be expressed by an incidence graph Γ which is constructed
in the Poincaré unit disk model of the hyperbolic plane along with the unit circle at the
boundary. The edges of Γ correspond to marked boxes in Ω; the vertices, that sit on
the boundary circle, of each edge correspond to the top and bottom of the marked box
represented by the edge. Each edge is directed from the top to the bottom. Vertices on
distinct edges are identified if the corresponding marked boxes share a distinguished point.
Figure 3.3 shows Γ.
The vertices in Γ are generated by the following construction. Given an edge corresponding to a marked box Θ with a pair of vertices vt and vb which represent the
distinguished points t and b in Θ. Then the boundary of Γ is divided into two arcs. We
may choose the arc that represents the interior of Θ. Then the other arc represents the interior of ι(Θ). Now, we place a vertex which represents the new distinguished point given
by applying Pappus’ Theorem to Θ on the first arc. We construct the rest of the vertices
in a similar manner while making sure that no edges intersect except at the boundary.
39
Given a pair of vertices on each of the two arcs, they define another vertex between
them. Hence, we have that the vertices are dense on the boundary of Γ.
FIGURE 3.3: Incidence Graph Γ: Each edge represents a marked box and its vertices
represent top and bottom of the box.
3.4.
Return of The Modular Group
There is a second group action of the modular group on the unit disk and its
boundary that commutes with the action of G when restricted to Γ. This second action is
given by group M generated by order two rotations with respect to centers of edges and
order three rotations about centers of triangles in Γ.
Here, given an edge of Γ, then the order two rotation with respect to its center is
the same map as ι defined by the box corresponding to the edge. Also, given a triangle
in Γ, the order three rotation with respect to its center is the same map as τ1 ι defined by
the marked boxes which correspond to the three edges of the triangle. So, the actions of
G and M are the same on Γ. Hence M is a Fuchsian triangle group of signature same as
PSL2 (Z). Therefore M is PSL2 (R) conjugate of PSL2 (Z) [7].
A marked box Θ induces a group action M̄ on P2 R corresponding action of M on
the unit disk and its boundary. Details of the action of M̄ is given below in Theorem
40
3.4.0.7.
Given a convex marked box Θ, define ν to be the map which takes the vertices of Γ
to the corresponding distinguished points in the orbit Ω.
Then Schwartz uses convexity to prove the following key theorem.
Theorem 3.4.0.6 The set of distinguished points of the orbit under the box operations,
of a convex marked box, are dense in a homeomorphic image of S 1 in P2 R.
With this theorem, Schwartz defines Pappus Curves.
Definition 3.4.0.14 Given an orbit of marked boxes Ω, the Pappus Curve of Ω is the
topological circle, Λ ⊂ P2 R, in which the distinguished points of marked boxes in Ω are
dense.
Now, we define algebraic Pappus Curves.
Definition 3.4.0.15 An algebraic Pappus Curve is a Pappus Curve Λ which is a subset
of some irreducible complex algebraic curve CΛ .
Lemma 3.4.0.3 Given an algebraic Pappus Curve Λ, the irreducible algebraic curve
which contains Λ is unique.
Proof. Suppose that Λ is an algebraic Pappus Curve. Unless Λ is a line, any given
line intersects it at finitely many points. Hence, one easily finds quadruple of points in
general position. By Corollary 2.2.2.4, we need only finitely many points to determine an
41
algebraic curve uniquely for a given degree. Also if Λ is a line, then it is contained in an
irreducible algebraic curve. Thus, we have that an algebraic Pappus Curve determines
the unique irreducible curve that contains it.
Theorem 3.4.0.7 Given a convex marked box Θ, each box operation is the restriction of
some projective symmetry in M̄ which preserves Ω.
Proof. We show that, given a marked box Θ ∈ Ω, τ1 ι and ι, which generate G, are
restrictions of:
1. An order three projective transformation having the cycle
ι(Θ) 7→ τ1 (Θ) 7→ τ2 (Θ) 7→ ι(Θ).
2. A polarity having the cycle
Θ 7→ ι(Θ) 7→ Θ.
Let Θ = ((b1 , b3 , a3 , a1 ; b2 , a2 ), (. . .)). Furthermore, let c1 , c2 , and c3 be the points
obtained by Pappus’ Theorem with ci = aj bk ∩ ak bj . Then we have that
τ1 ι(τ2 (Θ)) = ι(Θ) = ((a1 , a3 , b1 , b3 ; a2 , b2 ), (. . .)),
τ1 ι(ι(Θ)) = τ1 (Θ) = ((b1 , b3 , c1 , c3 ; b2 , c2 ), (. . .)),
τ1 ι(τ1 (Θ)) = τ2 (Θ) = ((c1 , c3 , a1 , a3 ; c2 , a2 ), (. . .)).
By computation, one shows that τ1 ι is a projective transformation of order three. See
Figure 3.4.
Now, let Ψ be such that t and b are at infinity, and p = [A : 0 : 1], q = [− A1 : 0 : 1],
r = [0 : B : 1], and s = [0 : − B1 : 1] as shown in Figure 3.5. Then the polarity with
42
a2
a1
b3
c2
c1
c3
a3
b1 b
2
FIGURE 3.4: Normalization R3; τ1 ι is realized by an order three rotation.
respect to the equation x2 + y 2 + z 2 = 0, as shown in Example 2.2.3.1, maps p to the line
Ax + z = 0 which is R, b to the line y = 0 which is T , and the line B to t and so on. So,
this polarity is the same as ι. Note that any marked box can be mapped by a projective
transformation to satisfy these conditions.
P
s
q
p
T
Q
r
R
B
S
FIGURE 3.5: Normalization P2; ι is a polarity with respect to the conic x2 + y 2 + z 2 = 0.
Definition 3.4.0.16 We say that any projective transformation that is conjugate to the
order three rotation about [0 : 0 : 1] is an order three rotation.
43
Definition 3.4.0.17 Let a marked box Θ be given. Normalization R3 of Θ is the placement of Θ such that τ1 ι on the three boxes ι(Θ), τ1 (Θ), and τ2 (Θ) is the order three
rotation about [0 : 0 : 1].
This normalization can be apply to any overmarked box Θ unless a2 , b2 , and c2 are
collinear. However, unless the given algebraic Pappus Curve is contained in a line, there
is a marked box Θ ∈ Ω such that the distinguished points of ι(Θ), τ1 (Θ), and τ2 (Θ) are
not collinear. For otherwise, the curve Λ has infinitely many points in common with a
line. But an algebraic Pappus Curve is contained in an irreducible algebraic curve, unless
the curve is the line itself, it is not possible.
Definition 3.4.0.18 Let a marked box Θ be given. Normalization P2 of Θ is when we
have t = [1 : 0 : 0], b = [0 : 1 : 0], p = [A : 0 : 1], q = [− A1 : 0 : 1], r = [0 : B : 1], and
s = [0 : − B1 : 1]. See Figure 3.5.
Lemma 3.4.0.4 If Λ is an algebraic Pappus Curve with a corresponding group of projective symmetries M̄, then the irreducible algebraic curve which contains Λ is also invariant
under M̄.
Proof. By Lemma 3.4.0.3, given an algebraic Pappus Curve Λ, it determines a
unique irreducible algebraic curve CΛ which contains Λ. Hence, since Λ is invariant under
M̄, CΛ is also invariant.
In the proof of Theorem 3.4.0.6, Schwartz shows that the homeomorphism ν̄ conjugates the action of M to M̄ on P2 R × (P2 R)∗ (see appendix). We have the shown in Figure
3.6 commuting diagram.
44
PSL2 (Z)
M̄
G
Proj
B
ν
θ
M
P2 R × (P2 R)∗
ν̄
Γ⊃V
S1
FIGURE 3.6: Maps θ, Proj, and ν̄ are PSL2 (Z)-equivariant. G, M, and M̄ define group
actions of PSL2 (Z) on their respective spaces.
Let
B = {(p, q, r, s, t, b) ∈ (P2 R)6 : ((p, q, r, s; t, b), (. . .)) is a convex marked box},
denote the set of configurations of convex boxes, and (using an obvious simplification of
notation)
Proj : B → P2 R
(p, q, r, s, t, b) 7→ t.
Now, fix a vertex vt in Γ, and suppose that Θ = ((p, q, r, s; t, b), (. . .)) is a convex marked
box. Then define a map
θ = θΘ,vt : B → Γ
such that θ maps an orbit of convex marked boxes Ω to its corresponding abstract graph
Γ such that the top distinguished point t of Θ gets mapped to the vertex vt in Γ.
Let V ⊂ Γ be the set of vertices in Γ. Then define
ν : V → P2 R
vt 7→ t.
45
Now, by Theorem 3.4.0.6, the map ν extends continuously to ν̄ on S 1 in which V is
a dense subset.
On B, we have a group of box operations G. On Γ, we have a group of isometries M.
On P2 R, we have a group of projective symmetries. All these three actions have the group
structure of PSL2 (Z). By Theorem 3.4.0.6, Proj, θ, and ν̄ are PSL2 (Z)−equivariant.
46
4.
ALGEBRAIC PAPPUS CURVES ARE LINEAR
In this chapter, we prove the main theorem.
4.1.
Linear Case
Theorem 4.1.0.8 A Pappus Curve is algebraic if and only if it is linear.
We prove this theorem by a series of properties.
Proposition 4.1.0.4 There exists a linear Pappus Curve.
Proof. Let
p = [−1 : 1 : 1], t = [0 : 1 : 1], q = [1 : 1 : 1],
s = [−1 : −1 : 1], b = [0 : −1 : 1], r = [1 : −1 : 1].
By applying Pappus’ construction, we obtain three new points
1
1
l = [− : 0 : 1], m = [0 : 0 : 1], n = [ : 0 : 1]
2
2
with m being a new distinguished point as can be seen in Figure 4.1.
Now, let T be the projective transformation that fixes p and q and maps l to s and
n to r. Then T maps the line pq onto itself and ln onto rs. Thus, the intersection pq ∩ ln
gets mapped to the intersection pq ∩ rs, but here we have that pq ∩ ln ∩ rs = [1 : 0 : 0].
Hence we have that T fixes [1 : 0 : 0]. Therefore, when viewed as a linear transformation
47
p
q
t
l
s
m
b
n
r
FIGURE 4.1: A Linear Pappus Curve: The marked box is symmetric with respect to the
Pappus Curve.
in R3 , T has the x-axis as an eigenvector. Therefore the yz-plane, which is the subspace of
R3 that is orthogonal to the x-axis, gets mapped onto itself by T . Hence, on the projective
plane, we have that the vertical line x = 0 gets mapped onto itself by T .
Now, since T maps the marked box ((p, q, n, l; t, m), (. . .)) back to the marked box
((p, q, r, s; t, b), (. . .)), the above argument shows that all the distinguished points obtained
by applying τ1 repeatedly lie on the vertical line x = 0. A similar argument shows this is
true for marked points generated by ι and τ2 . Hence we have that all the marked points
lie on a single line. So this line an algebraic Pappus Curve.
Since any line can be mapped to any other line on the projective plane by a projective transformation and the group action G of box operations commutes with projective
transformation, we have the following.
Corollary 4.1.0.2 Any line in P2 R contains a Pappus Curve.
48
4.2.
Excluding Higher Degrees
Now, we show that any algebraic curve of degree greater than or equal to four cannot
contain a Pappus Curve. In order to do this, we first discuss the number of singular points
an algebraic curve may have if it contains a Pappus Curve.
Let H ≤ G be the subgroup, of index two, of projective transformations. Then we
have that |H| = ∞. Also, let CΛ be an algebraic curve that contains a Pappus Curve. We
have that CΛ is invariant under the action of H.
Lemma 4.2.0.5 A conic partitions P2 R into three connected regions. The three regions
are two disjoint connected sets and the conic itself. [13]
Proof. Suppose CΛ = V(f ) is a conic. Then, CΛ = {p ∈ P2 R : f (p) = 0}. We also
have two other disjoint sets of points
RT B (f, +) = {p ∈ P2 R : f (p) > 0},
and
RT B (f, −) = {p ∈ P2 R : f (p) < 0}.
These sets are well-defined since, for all nonzero λ ∈ R, f (λp) = λ2 f (p). Also, we have
that
RT B (cf, +) = RT B (f, +) when c > 0,
and
RT B (cf, +) = RT B (f, −) when c < 0.
49
Definition 4.2.0.19 Given three distinct lines L, M , and N , the interior of a triangle
determined by them is the intersection of three regions
RLM (f, LM ), RM N (g, M N ), RLN (h, LN ),
where each is either 1 or −1 chosen in an obvious manner.
Note that not all the combinations of such intersections give nonempty set of points.
Lemma 4.2.0.6 Let Θ be a marked box and R1 , R2 ∈ M̄ be order three projective transformation such that
R1 : ι(Θ) 7→ τ1 (Θ) 7→ τ2 (Θ) 7→ ι(Θ)
and
R2 : Θ 7→ τ1 ι(Θ) 7→ τ2 ι(Θ) 7→ Θ.
Then the fixed points of R1 and R2 are distinct.
Note that Ri both realize the box operation τ1 ι, but on distinct boxes. Thus, they
are distinct projective transformations.
Proof. Denote the distinguished edges of Θ by T and B. Let m1 and M1 be the
distinguished point and edge given by applying Pappus’ construction on Θ, and m2 and
M2 be those given by ι(Θ). By Lemma 4.2.0.5, the degenerate conic T ∪ B gives two
disjoint open subsets of P2 R. Let the subset that contains m1 be S1 and the second subset
be S2 . Then, we have that m2 is contained in S2 since ι maps points in S1 to S2 .
Now, Figure 3.4 shows that the fixed point of R1 is contained in the interior of a
triangle defined by T , B, and M1 . Furthermore, we have that the interior of this triangle
is contained in S2 . Similarly, we have that the fixed point of R2 is contained in a triangle
50
defined by T , B, and M2 , and its interior is in S1 . Since S1 and S2 disjoint, the fixed
points of R1 and R2 are distinct.
Lemma 4.2.0.7 An order three rotation has only one fixed point.
Proof. Let R be a rotation of order three. Then, this map is conjugate to the
rotation g1 in Example 2.3.0.2. We saw that g1 fixes o = [0 : 0 : 1] and leaves z = 0
invariant. Suppose that g1 fixes a point p 6= o. Then, it follows that g1 leaves op invariant.
However, z = 0 does not contain o, so it is not the line op. Hence, no such p can exist.
Since R is a conjugate of g1 , they have same number of fixed points. Thus, R has only
one fixed point.
Lemma 4.2.0.8 The algebraic curve CΛ which contains a Pappus Curve Λ does not have
exactly one singular point.
Proof. Suppose that there is exactly one singular point on CΛ , say p. Then, since a
projective transformation maps a singularity to a singularity (Corollary 2.2.2.2), it follows
that all the elements in H fix p.
Suppose that there is a projective transformation R of order three in H. Then by
Lemma 4.2.0.7, R has only one fixed point. Suppose that its fixed point o0 is not p. Then
we have a contradiction with every element in H fixing p.
Now, suppose that R fixes p. Then by Lemma 4.2.0.6, there is a projective transformation R0 ∈ H of order three which fixes a point other than p. Thus, we have a
contradiction with every element in H fixing p.
Hence, CΛ cannot have only one singular point.
51
Lemma 4.2.0.9 The algebraic curve CΛ which contains a Pappus Curve Λ does not have
exactly two singular points.
Proof. Suppose that there are exactly two singular points, p and q, on CΛ . Given
a rotation of order three, possible lengths of orbits are either one or three since the length
of an orbit must divide the order of the transformation [2]. If R ∈ H is an order three
rotation, then it maps a singular point to a singular point while leaving C invariant, so
we must have that R fixes both p and q because otherwise there must be at least three
singular points in an orbit. However, as we saw above, an order three rotation fixes only
one point. Therefore, it is not possible to have exactly two singular points.
Definition 4.2.0.20 A projective transformation T ∈ M̄ is said to be hyperbolic if its
corresponding hyperbolic isometry in M is hyperbolic.
Lemma 4.2.0.10 A hyperbolic projective transformation in M̄ is conjugate to the map


a
0
0





1
T0 = 
 0 a 0 


0 0 1
with real a 6= 1.
Proof. Suppose that T is a hyperbolic projective transformation and p and q are
its fixed points. Then, since the vertices of Γ represents both distinguished points and
edges of marked boxes, T leaves two lines invariant. Let the intersection of them be r.
Therefore, T fixes three point.
Now, map p, q, and r to [1 : 0 : 0], [0 : 1 : 0] and [0 : 0 : 1], respectively. Then, by
computation, one shows that any projective transformation T 0 fixing those three points is
52
represented by a matrix

 a 0

1
T0 = 
 0 a

0 0

0 

0 


1
for |a| =
6 1. Hence, we have that any hyperbolic projective transformation is conjugate to
T 0.
When |a| > 1, T 0 attracts to [1 : 0 : 0] and expands from [0 : 1 : 0]. Thus, we have
the following corollary.
Corollary 4.2.0.3 A hyperbolic projective transformation has three distinct fixed points
in P2 (C). Furthermore, any non-fixed points are attracted toward one of the fixed points.
Sketch of Proof. By Lemma 4.2.0.10, we have that any hyperbolic projective transformation is conjugate to T 0 defined above. Since T 0 has three fixed points, so does any
other hyperbolic projective transformation. Two of the fixed points are the ones on Λ
from the definition. Let p and q denote them. The third fixed point is the intersection of
the invariant lines through p and q.
Suppose that w = [x : y : z] is not a fixed point of T 0 . Then
T 0m (w) = [am x : a−m y : z].
Suppose that |a| > 1, then we have that am x → ∞ and a−m y → 0 as m → ∞. Thus, w
attracts to [1 : 0 : 0]. Remark 4.2.0.2 Since the group M is PSL2 (R) conjugate of the modular group, hyperbolic fixed points of M are dense in S 1 by Corollary 2.1.3.2. Hence, hyperbolic fixed points
of M̄ are dense in Λ.
53
Lemma 4.2.0.11 The algebraic curve CΛ does not have exactly three singular points.
Proof. Suppose the algebraic curve CΛ of degree d which contains a Pappus Curve
Λ has exactly three singular points s1 , s2 , and s3 . Then let T ∈ M̄ be a hyperbolic map
such that its fixed points p and q are not on any of the lines defined by a pair of singular
points. This is possible because there are only finitely many of these lines and, by Bézout’s
Theorem, each line intersects CΛ d times, and the hyperbolic fixed points are dense in the
Pappus Curve by Remark 4.2.0.2.
Since elements of M̄ map CΛ onto itself and map singular points to singular points,
the set {s1 , s2 , s3 } is invariant under T . Thus, for some n > 0, we have that T n (si ) = si
for i = 1, 2, 3. Thus, by the exact four transitivity, Remark 2.1.2.1, T n is the identity map.
However, the hyperbolic map T is not of a finite order. Hence, we have a contradiction.
Thus, CΛ does not have exactly three singular points.
Recall that a projective transformation of order three leaves exactly one line invariant as we saw in Example 2.3.0.2.
Corollary 4.2.0.4 Let R1 and R2 be defined as in 4.2.0.6. Then the lines invariant under
them are distinct.
Suppose that the fixed point of and the line invariant under Ri are ri and Li for
i = 1, 2. We have that R2 = Pι R1 Pι where Pι denotes the polarity map which realizes ι
on Θ. If L1 = L2 , then it follows that ι(L1 ) = r1 and ι(r1 ) = L1 . However, this implies
that Pι R1 Pι fixes r1 . But, by Lemma 4.2.0.6, r1 6= r2 . Hence, the lines invariant under
R1 and R2 are distinct.
Proposition 4.2.0.5 If there are four or more singular points on CΛ , then there are at
least four singular points in general position.
54
Proof. Suppose an algebraic curve CΛ contains a Pappus Curve and has four or
more singular points.
Take a singular point s1 , then by Corollary 4.2.0.4, there exists an order three
rotation R1 ∈ M̄ such that R1 (s1 ) 6= s1 . Since there are four or more singular points in
P2 C, there is a singular point s2 such that s2 ∈
/ {s1 , R1 (s1 ), R12 (s1 )}.
Now, let L1 be the unique line invariant under R1 whose existence is proven in
Example 2.3.0.2. Suppose that a singular point s1 is not on L1 and s2 is a singular point
that is not in the orbit of s1 under R1 .
We have that the two lines s1 s2 and R1 (s1 )R1 (s2 ) are distinct, since s1 is not on
L1 . Suppose that s1 , s2 , R1 (s1 ), and R1 (s2 ) are not in general position. By symmetry of
argument to follow, s1 ∈ R1 (s1 )R1 (s2 ). Then on the union of the two lines, there are five
singular points s1 , R1 (s1 ), R12 (s1 ), s2 , and R1 (s2 ). It follows that R1 (s1 ), R12 (s1 ), s2 , and
R1 (s2 ) are in general position.
Now, suppose that all the singular points are on L1 . Then by Corollary 4.2.0.4,
there exists an order three projective transformation R2 with invariant line L2 6= L1 .
Thus, we have at least one singular point, say s1 , which is not on L2 . Thus, by applying
a similar argument from above, one easily shows that there are four singular points in
general position.
Proposition 4.2.0.6 Given an algebraic Pappus Curve Λ, the algebraic curve CΛ has at
most three singular points.
Proof. Suppose that CΛ is an algebraic curve containing a Pappus Curve Λ with
at least four singular points. Then, by Proposition 4.2.0.5, there are four singular points
p, q, r, and s in general position.
55
Now, let H(p) be the orbit of p under H and Hp be the stabilizer of p in H. Then,
by the orbit-stabilizer formula [2], we have that
|H| = #H(p) |Hp |.
We know that |H| = ∞ and we want to show that #H(p) = ∞. Thus, if we can show
that |Hp | is finite, we are done. Suppose that |Hp | = ∞. Then we have that
|Hp | = #Hp (q) |Hpq |
where Hp (q) is the orbit of q under the action of Hp , and Hpq is the stabilizer of q in Hp .
Now, if we have that #Hp (q) = ∞, then we will have shown that there are infinitely
many singularities, and for that, it suffices to show that |Hpq | is finite. Suppose not, then
we have
|Hpq | = #Hpq (r) |Hpqr |.
Once again, if we have #Hpq (r) = ∞ then we are done, and for that, we want |Hpqr |
to be finite. Again, suppose that it is not finite. Then it follows that
|Hpqr | = #Hpqr (s) |Hpqrs |.
But, due to the exact four-transitivity of projective transformations, we have that Hpqrs =
{id}, and so |Hpqrs | = 1. So, #Hpqr (s) = ∞. Therefore, we have shown that at least one
of #H(p), #Hp (q), #Hpq (r), or #Hpqr (s) is infinite. Thus, we have shown that if CΛ has
four or more distinct non-collinear singular points, then there are infinitely many of them.
Lemma 4.2.0.12 Projective transformations in M̄ are automorphisms of the corresponding CΛ .
56
Proof. Let T ∈ M̄ be a projective transformation and Λ be the Pappus Curve which
corresponds to the action of M̄. Then since Λ is invariant under the projective transformations in M̄, T maps Λ onto itself. Therefore, T maps CΛ onto itself by Lemma 3.4.0.3
Furthermore, since the linear transformation in C3 that corresponds to T is invertible, T
is injective and has an inverse. Thus, T is a bijection on CΛ .
Furthermore, as we saw in the proof of Theorem 2.2.4.3, any projective transformation T ∈ M̄ maps from an affine plane to an affine plane, and complex charts defined on the
affine planes by the Implicit Function Theorem are compatible. Thus T is holomorphic.
Hence, T is an automorphism on CΛ .
Proposition 4.2.0.7 Given an algebraic Pappus Curve Λ, the corresponding algebraic
curve CΛ is smooth.
Proof. By Lemma 4.2.0.8, Lemma 4.2.0.9, Lemma 4.2.0.11, and Proposition 4.2.0.6,
any algebraic curve containing a Pappus Curve does not have any singular points.
Proposition 4.2.0.8 No algebraic curve of degree four or greater contains a Pappus
Curve.
Proof. Let C be a smooth algebraic curve over C with genus g > 1. Suppose that
CΛ contains a Pappus Curve Λ. By Hurwitz’s Automorphism Theorem (Theorem 2.2.4.1),
|Aut(CΛ )| ≤ 84(g − 1)
where Aut(C) denotes the set of automorphisms on C. But, we know that for each Pappus
Curve, there are infinitely many projective transformations that map CΛ onto itself in H.
Thus CΛ cannot be of genus two or greater.
57
Furthermore, we have that, for a smooth algebraic curve of degree d and genus g,
Plücker Formula (Proposition 2.2.2.1)
g=
(d − 1)(d − 2)
2
is satisfied. If CΛ is algebraic, then its genus is either zero or one, and by the above
equation, the only possible degrees are d = 1, 2 and 3.
We have already seen that d = 1 is possible in Proposition 4.1.0.4.
4.3.
Cubic Case
Proposition 4.3.0.9 There is no cubic which contains a Pappus Curve.
Proof. Suppose a cubic CΛ contains a Pappus Curve Λ. Then, by Proposition
4.2.0.7, CΛ is smooth and thus irreducible. By Corollary 2.2.2.3, CΛ has k flexes f1 , . . . , fk
for 1 ≤ k ≤ 9. Then, since an order three rotation R ∈ M̄ maps flexes to flexes and, by
Lemma 4.2.0.6, there is a flex of CΛ that is not the fixed point of R, there are at least
three flxes.
Let T ∈ M̄ be a hyperbolic map such that its fixed points p, q ∈ Λ are not on any of
the lines defined by a pair of flexes. This is possible because there are only finitely many
of these lines and the hyperbolic fixed points are dense in the Pappus Curve by Remark
4.2.0.2.
Since elements of M̄ map CΛ onto itself and maps flexes to flexes, the set {f1 , . . . , fk }
is invariant under T . Thus, for some n > 0, we have that T n (fi ) = fi for i = 1, . . . , k.
Thus, by the sharp four transitivity Remark 2.1.2.1, T n is the identity map. However, the
hyperbolic map T is not of a finite order. Hence, we have a contradiction.
Thus, a cubic CΛ cannot contain a Pappus Curve.
58
4.4.
Conic Case
Proposition 4.4.0.10 There is no conic which contains a Pappus Curve.
Proof. We use invariant theory to show that any conic containing a Pappus Curve
cannot be invariant under an order three projective transformation in M̄. This is enough
to show that there is no conic that contains a Pappus Curve since such conic must be
invariant under every projective transformation in M̄.
First, we look at the group generated by a rotation of order three to find a set of
fundamental invariants.
For any given orbit of marked boxes, we may normalize it to the Normalization
R3 (Figure 3.4), so that τ1 ι is the standard rotation of order three that fixes the point
o = [0 : 0 : 1]. We have that this projective transformation can be expressed as a matrix
by

1
 −2
√
 3
 2
R=

0
√
−
3
2
− 21
0

0

0
.

1
Let R be the cyclic subgroup generated by R. Then, of course, |R| = 3. Now, we have
that the Hilbert series of the invariant ring C[x]R is
ΦR (t) =
1 X
1
.
|R|
det(id − tπ)
π∈R
In this case as shown in Example 2.3.0.2, we have that
ΦR (t) =
1 + t3
.
(1 − t)(1 − t2 )(1 − t3 )
This tells us that the ring of invariant curves under R is generated by one line, one conic,
and two cubics.
59
By Corollary 2.3.0.1, the conic that passes through the point t = [1 : 0 : 1] is
x2 + y 2 − z 2 = 0. Hence, if the Pappus Curve of the box in Normalization R3 is contained
in a conic, then it has to be this conic.
Now, we identify the equation of a conic containing an algebraic Pappus Curve.
In Normalization P 2 (Figure 3.5), any Pappus Curve Λ must pass through distinguished points t = [1 : 0 : 0] and b = [0 : 1 : 0], and lines T and B intersect the curve
exactly once at t and b. If CΛ were a conic, then CΛ would have the form
xy − αz 2 = 0.
Note that here the coefficients of x2 and y 2 are 0 and that of xy is not.
In order to apply invariant theory to show that no conic of the form above can be
invariant under our τ1 ι, we need to conjugate R by a projective transformation T , which
takes p, q, r, and s in Normalization P2 to p, q, r, and s in Normalization R3. so that we
have an order three projective transformation in Normalization P 2. In order to find such
a T , in Normalization P 2, we let p = [A : 0 : 1] and r = [0 : B : 1], and in Normalization
R3, we let p = [µ : ν : 1]. This is enough to determine the corner points of the box pqrs
in both normalizations and they give us the transformation T that maps Normalization
P 2 to Normalization R3. This transformation can be expressed as a matrix

1

T = 
0

1
√
√ √
3( 3A+3Aν− 3Aµ)
B(1+2µ)
√
√
1 3A+3Aν− 3Aµ
−2
B(1+2µ)
√ √
√
3(
3A+3Aν−
3Aµ)
− 13
B(1+2µ)
1
6
A(−A2 +B 2 +A2 µ+2B 2 µ+3µ)
A2 −B 2 +2A2 µ−2B 2 µ
3Aν(A2 +1)
A2 −B 2 +2A2 µ−2B 2 µ
2
2
2
+2A µ−2B
− A(−BA2−3−2A
−B 2 +2A2 µ−2B 2 µ
2 µ)



.


Let R0 = T −1 RT ; this is the order three rotation that acts as τ1 ι in Normalization P 2.
Note that, since R0 maps p to s, this gives equations that relate the values of A and
B.
60
Now, by applying the Reynolds operator, we find the conic that is invariant under
R0 , and solve for the values of A, B, µ, and ν when the coefficients of x2 and y 2 are 0 and
the relation between A and B, as mentioned above, are satisfied. Then, we get that
q
√
−U ( 3ν + µ − 1)2
√
√
A=
−2 3ν + 3ν 2 + 2 3µν − 2µ + µ2 + 1
where
√
√
√
U = −3µ + 2B 2 µ2 − B 2 µ − 3 3ν − B 2 − 3B 2 ν + 3µ2 − 2 3B 2 µν + 3ν 2
q
√
√
√
√
ν(4ν 2 + (16 + 8 3)ν + 21 + 12 3)((1 + 3)ν − 2 − 3)
√
√
B=
4ν 2 + (16 + 8 3)ν + 21 + 12 3
2
µ = (2 −
√
3)ν + 1.
We now have everything expressed in terms of ν; with the relation between µ and ν, we
find a distinguished point that cannot be any of the ai , bj , or ck of Figure 3.4. In that
figure, the distinguished point we get by applying Pappus’ Theorem to ιΨ is not given, so
we find this point. For this distinguished point to be on x2 + y 2 − z 2 = 0, we have that ν
must satisfy the equation
√
2(−7 + 4 3)V
−
=0
(2ν)2 + 1
where
√
√
√
√
V = 8ν 4 + (16 + 4 3)ν 3 + (28 + 14 3)ν 2 + (11 + 6 3)ν + 26 + 15 3.
Thus, we need V = 0. One easily shows that this quartic V has no real solution by finding
the minimum of the function V of ν using basic differential calculus. Hence, we have that
61
there is no possible point p = [µ : ν : 1] that gives us a conic that contains a Pappus
Curve.
We have now proven Theorem 4.1.0.8. The only algebraic Pappus Curves are linear.
62
5.
GEOMETRICAL SIGNIFICANCE OF FIXED POINTS OF
ORDER FOUR PROJECTIVE TRANSFORMATIONS
In Chapters 3 and 4, the box operation ι on a marked box is order two. But if we
look at what it does on an overmarked box, we have that
ι2 ((p, q, r, s; t, b), (. . .)) = ((q, p, s, r; t, b), (. . .)).
This operation is order two because a marked box was defined to be the equivalence class
where the two overmarked box in the above equation are the same. If we look further into
this ι, we have that
ι4 ((p, q, r, s; t, b), (. . .)) = ((p, q, r, s; t, b), (. . .)).
This fact was the inspiration to investigate the order four projective transformations.
Let k be an algebraically closed field of characteristic not two.
We first make a few remarks that will be helpful along the way.
Proposition 5.0.0.11 Any two projective transformations of order four are conjugates
in the group of projective transformations on P2 k.
Proof. Suppose that T1 and T2 are projective transformations of order four. Then,
there exist p, p0 ∈ P2 k such that
T1 : p 7→ q 7→ r 7→ s 7→ p
where p, q, r, and s are in general position, and
T2 : p0 7→ q 0 7→ r0 7→ s0 7→ p0
63
with p0 , q 0 , r0 and s0 in general position. Then there is a projective transformation S such
that
p 7→ p0
q 7→ q 0
r 7→ r0
s 7→ s0
Then, we have that T1 = S −1 T2 S.
Corollary 5.0.0.5 Any two projective transformations of order four have the same number of fixed points.
Suppose that T1 and T2 are projective transformations of order four. By the previous
proposition, there exists a projective transformation S such that T1 = S −1 T2 S. Suppose
that T2 fixes p, then T1 fixes S −1 (p). Hence, we have that T1 and T2 have the same number
of fixed points.
Proposition 5.0.0.12 A projective transformation of order four has exactly three fixed
points.
Proof. By the previous corollary, it suffices to show that the claim holds for one
projective transformation of order four.
64
Let
p0 = [1 : 0 : 1]
q0 = [0 : 1 : 1]
(5.0..1)
r0 = [−1 : 0 : 1]
s0 = [0 : −1 : 1]
and T 0 be the projective transformation such that
T0 : p0 7→ q0 7→ r0 7→ s0 7→ p0 .
Then, we have that

0

T0 = 
1

0
−1
0
0

0

0


1
is the matrix representation of this projective transformation.
For any point w = [x : y : z] ∈ P2 k, we have
T0 w = [−y : x : z],
and the only fixed affine point is [x : y : 1] such that x = −y and y = x. Thus, we have
that x = y = 0. Let
O = [0 : 0 : 1].
The fixed points on the line at infinity are of the form [x : 1 : 0] since [1 : 0 : 0] is not a
fixed point. Thus, our x has to satisfy the equation
x = −x−1 .
65
(In order to have a solution to this equation, we need the characteristic of the underlining
field to not be two.) Let i be a solution to such equation, then we have that −i is a
solution as well. Thus, we have that
a0 = (i, 1, 0)
and
b0 = (−i, 1, 0)
are fixed points of T0 . Thus, as claimed, any order four projective transformation T has
three fixed points.
Note that this i is not necessarily the same i as in C. It is used here as the solution
to the equation x2 = −1 in any algebraically closed field of characteristic not equal to
two.
As can be seen in the above proof, we have that O is the intersection of lines p0 r0
and q0 s0 . Thus, we have that given a projective transformation T of order four and if a
point p is such that p, q = T (p), r = T 2 (p) and T 3 (p) are in general position, then the
intersection of line pr and qs is a fixed point of T . Let o denote this fixed point of T , and
a and b denote the other two.
In what follows, we show the geometrical significance of the fixed points a and b.
First, we identify exactly those points for which we do not get orbits of length four
being in general position under T . Let DT denote this set.
Proposition 5.0.0.13 The set DT is ao ∪ bo ∪ ab. Furthermore, T 2 fixes ab pointwise.
Proof. With notation as above, suppose that p is such that p, q, r, and s are in
general position. Then since T 2 (p) = r and T 2 (r) = p, the line pr is T 2 -invariant. Also,
66
we have that ab is T 2 -invariant. Hence, their intersection c = pr ∩ ab is fixed by T 2 .
Therefore, we have that T 2 fixes three points a, b, and c on the line ab. Hence by Lemma
2.1.2.3 T 2 fixes ab pointwise. From this, it follows that the only fixed points of T 2 are
the line ab and the point o, since having any more would give us four points in general
position. That forces T 2 to be the identity map, and that is a contradiction with the order
of T being four.
Now, suppose that p ∈
/ ao ∪ bo ∪ ab. If also p, q, r, and s are collinear, then we have
that pq = qr. Thus, it follows that pq ∩ ab = qr ∩ ab, and so pq ∩ ab is a fixed point of T
since
T (pq ∩ ab) = T (pq) ∩ T (ab) = qr ∩ ab.
Thus, pq ∩ ab is either a or b. Say, without loss of generality, it is a. Then we also have
that pq ∩ bo = qr ∩ bo and so this intersection is either b or o. Either way, we would have
that pq = ab or ao and p ∈ ab ∪ ao, and we have a contradiction. Thus, we have that
DT ⊆ ao ∪ bo ∪ ab.
Now, one easily shows that any point p ∈ ao ∪ bo ∪ ab gives us a quartet p, q, r, and s
that is collinear. Hence we have that
DT = ao ∪ bo ∪ ab.
Now, we begin discussing a partition of a projective plane by a family of conics
(both non-degenerate and degenerate) defined by T .
Proposition 5.0.0.14 For p ∈
/ DT , there is a unique T -invariant conic passing through
p, q, r, and s.
67
Proof. Suppose that N is a projective transformation such that
p 7→ p0
q 7→ q0
r 7→ r0
s 7→ s0
where p0 , q0 , r0 and s0 are as defined above in (5.0.1), and that T0 is the projective
transformation that acts so that
p0 7→ q0 7→ r0 7→ s0 7→ p0 .
Then, we have that T0 = N T N −1 , and if C is T -invariant, then N (C) is T0 -invariant.
Hence, if we can show that there is only one conic through p0 , q0 , r0 and s0 that is
T0 -invariant, then we will have shown the same for p, q, r, and s with T .
Now, we claim that the equation x2 + y 2 − z 2 = 0 gives the unique T0 -invariant conic
through p0 , q0 , r0 and s0 .
The general conic (nondegenerate or not) is of the form
Ax2 + Bxz + Cy 2 + Dyz + Exy + F z 2 = 0.
Since our conic passes through the points p0 , q0 , r0 and s0 , we have that our conic is given
by
x2 + y 2 + Exy − z 2 = 0
for some E ∈ k. Now, T0 maps x to −y and y to x, so for this conic to be T0 -invariant,
the solution set also has to satisfy
x2 + y 2 − Exy − z 2 = 0.
68
Thus E = 0. The only conic that satisfies all of the given conditions is
x2 + y 2 − z 2 = 0.
We have shown that there is a unique T -invariant conic associated to p ∈
/ DT .
q
0
r0
O
p
0
s0
FIGURE 5.1: Conics Invariant Under T0 : They are concentric circles centered at [0 : 0 : 1].
Let Cp denote the T -invariant conic through the point p for p ∈
/ DT .
Remark 5.0.0.3 In the above proof, the fixed points a0 and b0 of T0 are contained in the
conic Cp0 but O is not. Thus, the fixed points a and b of T are contained in Cp for any
p∈
/ DT and o is not. Hence, any T -invariant conic contains fixed points a and b of T .
Remark 5.0.0.4 Suppose that a conic C is T -invariant, then we have that C ∩ DTc is
non-empty. Therefore, there is p ∈ C with p ∈
/ DT . For such p, C = Cp .
Proposition 5.0.0.15 All the T -invariant, non-degenerate conics are tangent to each
other at a and b. In other words, they all intersect each other at a and b with multiplicity
two.
69
Proof. Suppose that C is T -invariant. Then a ∈ C. Let us consider the line ao.
Suppose that ao ∩ C = {a, c}. Now, both ao and C are T -invariant, so the set {a, c} must
be as well. Since T (a) = a, it follows that T (c) = c. Therefore, c is either a or o since
these are the only fixed points on the line ao. But, o ∈
/ C. So, c = a. Therefore, ao is a
tangent line of C at a. Similarly, bo is a tangent line of C at b.
Since any two T -invariant conics share tangent lines at a and b, they are tangent
to each other at those points. Since, by Bézout’s Theorem (Theorem 2.2.1.1), two conics
intersect four times, we have that they intersect at a and b each with multiplicity two.
Now, we have that for any p ∈
/ DT , there is a unique T -invairant conic Cp . With the
degenerate conic ao ∪ bo and the double line conic ab in the family, we have that for any
p 6= a, b, there is a unique T -invariant conic associated to it, and all the conics intersect
at a and b with multiplicity two.
Thus, we have exhibited a geometrical significance of the fixed points a and b of T
that was not as obvious as the one for o.
Furthermore, let us consider a line L through o that is neither ao nor bo. Then, we
have that L intersects C twice, say at c and d. Also, L intersects ab at some point z 6= a, b.
So, L = zo, and we have that L is T 2 -invariant since T 2 fixes both z and o. Because z is
neither a nor b as mentioned above, we have that z ∈
/ C. Thus, it follows that c, d ∈
/ ab,
and of course c 6= o 6= d. Therefore, c and d are not fixed points of T 2 . However, T 2
preserves the intersections, so we have T 2 (c) = d and T 2 (d) = c.
Thus, we have that each conic C is uniquely represented by a pair of points on L,
that are related by T 2 . Also, any point on L, that is not z or o, along with its T 2 -twin,
represents a conic C that passes through those two points and gets fixed by our T .
Here is an example in order to see all of what has been discussed explicitly.
70
Example 5.0.0.3 Suppose that
a = [0 : 0 : 1]
b = [0 : 1 : 0]
o = [1 : 0 : 0]
Then each Cm is of the form zy = mx2 with some m ∈ k where m 6= 0. When we have
that m = 0, then the conic is zy = 0 which is the set ao ∪ bo, and the set x = 0 is the line
ab.
In this setting, our line Lα through o is of the form y = αz with α ∈ k and α 6= 0
since Lα may not pass through a. Of course, Lα intersects ab at (0, α−1 , 1). We may
obtain our pseudo-two-to-one function from P1 onto P1 by mapping the x coordinates of
the intersections of Lα and Cm to m.
b
Lα
ao
a
o
ab
FIGURE 5.2: At a fixed point of an order four transformation, conics are tangent.
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6.
CONCLUSION
The main problem for Chapters 2 through 4 in this thesis was the identification of
all algebraic Pappus Curves. We showed that the only algebraic Pappus Curves are linear.
Furthermore, Corollary 4.1.0.2 showed that any line in P2 R can be an algebraic Pappus
Curve. We also ruled out Pappus Curves in any algebraic curves of degrees greater than
or equal to two by showing that they contradicted properties of algebraic curves.
In Proposition 4.1.0.4, we saw a particular marked box which gives a linear Pappus Curve. However, we may not have identified all the marked boxes which generate
linear Pappus Curves. This is an obvious next step that can be taken to complete the
identification of all algebraic Pappus Curves.
We may study algebraic Pappus Curves in subfields of R. Also, distinguished points
in the orbit of a marked box which is not convex.
In Chapter 5, we studied the fixed points of projective transformations of order
four. While exploring this topic, we saw conics which are invariant under a given projective
transformation of order four. Much like how Schwartz used Pappus’ Theorem to construct
marked boxes, it is possible to apply Pascal’s Theorem and invariant conics of order four
transformations to create a curves of points and of lines that mimic Pappus Curves. This
is a possible extension of this thesis. Pappus’ Theorem can be viewed as Pascal’s Theorem
in the reducible case.
One may ask whether we can investigate the fixed points of other projective transformations with respect to algebraic curves other than conics. However, we should note
that studying projective transformations of order four with conics was a reasonable thing
to do because of the four transitivity of projective transformations and the fact that five
points in general position determine a unique conic. It seems less likely that we get a sim-
72
ilar relationship with other projective transformations and other algebraic curves. But it
might be a good extension of the study of the fixed points of projective transformations.
73
BIBLIOGRAPHY
1. Coxeter, H.S.M., Projective Geometry, 2nd ed., Springer-Verlag New York, Inc., 1987
2. Dummit, David S., Foote, Richard M., Abstract Algebra, 3rd ed., John Wiley and
Sons, 2004
3. Farkas, Hershel M., Kra, Irwin, Riemann Surfaces, 2nd ed., Springer-Verlag, 1992
4. Fulton, William, Algebraic Curves, W.A. Benjamin, Inc., 1969
5. Griffiths, Phillip A., Introduction to Algebraic Curves, American Mathematical Society, 1989
6. Hilton, Harold, Plane Algebraic Curves, 2nd ed., Oxford University Press, 1932
7. Katok, Svetlana, Fuchsian Groups, The University of Chicago Press, 1992
8. Miranda, Rick, Algebraic Curves and Riemann Surfaces, American Mathematical
Society, 1995
9. Munkres, James R., Topology, 2nd ed., Prentice Hall, Inc., 2000
10. Reid, Miles, Undergraduate Algebraic Geometry, Cambridge University Press, 1988
11. Schwartz, Richard E., Pappus’s Theorem and the Modular Group, Publications
mathématiques de l’I.H.É.S., 78 1993, 187-206.
12. Sturmfels, Bernd, Algorithms in Invariant Theory, 2nd ed., Springer-Verlag/Wien,
2008
13. Vulakh, L. Ya, The Markov Spectra For Fuchsian Groups, Trans. Amer. Math. Sco.
352 2000, 4067-4094
14. Walker, Robert J., Algebraic Curves, Princeton University Press, 1950
74
APPENDIX
75
A
APPENDIX An Additional Proposition By Schwartz
In this appendix, our aim is to fill in details of Schwartz’s proof of Theorem A0.9;
we discuss Schwartz’s exploration of a relationship between Pappus Curve Λ and its corresponding curve of lines L.
In the body of the dissertation, we work only on P2 R, but here we work also on its
dual space (P2 R)∗ and their product.
In Section 3.4, by Theorem 3.4.0.6, we saw the definition of a Pappus Curve in
Definition 3.4.0.14.
Schwartz also defines a corresponding topological circle in the projective dual plane
with distinguished edges in Ω.
Definition A0.21 A curve of lines is a topological circle in (P2 R)∗ .
Remark A0.5 We have that P2 R is the quotient space obtained from S 2 by identifying
each point x of S 2 with its antipodal point −x. Hence, there is a quotient topology on
P2 R [9]. We define the topology on (P2 R)∗ analogously. With this topology, a polarity is
a homeomorphism from a projective plane to its dual plane.
Lemma A0.13 Distinguished edges of Ω are dense in a homeomorphic image of S 1 in
(P2 R)∗ .
Proof. As in the proof of Theorem 3.4.0.7, distinguished points in Ω are mapped
to distinguished edges in Ω by polarities in M̄. Since polarities are homeomorphisms, the
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image of distinguished points are dense in the image of Λ. Thus, distinguished edges of Ω
are dense in a homeomorphic image of S 1 .
With this lemma, we define the Pappus Curve of lines of an orbit Ω of convex marked
boxes.
Definition A0.22 Given an orbit of marked boxes Ω, the Pappus Curve of lines is the
topological circle LΩ of lines in the projective dual space (P2 R)∗ in which the distinguished
edges of marked boxes in Ω are dense.
The denseness of the distinguished edges in LΩ is guaranteed by the following lemma.
Since both Λ and L are dense in respective homeomorphic image of S 1 , we have the
following corollary.
Corollary A0.6 The set of pairs (t, T ) in an orbit Ω lies densely on a homeomorphic
copy of S 1 in P2 R × (P2 R)∗ .
Definition A0.23 A transverse linefield to a curve X is a continuous curve Y of lines
such that each line of Y intersects X in exactly one point, and that each point of X is
contained in some line of Y .
Definition A0.24 A section of a curve of lines Y is a curve X such that each point of
X is contained in exactly one line of Y , and each line of Y contains exactly one point of
X.
Recall that a set of lines is said to be coincident if all the lines intersect at one point.
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Remark A0.6 By duality and Remark 2.1.2.1, a projective transformation that leaves
four distinct, non-coincident lines invariant is the identity.
Theorem A0.9 If ΛΩ is not a line, then LΩ is the unique transverse linefield to ΛΩ .
Dually, if LΩ is not a curve of coincident lines, then ΛΩ is the unique section of LΩ [11].
Before proving this theorem, we state a lemma.
Lemma A0.14 Let Ω be an orbit of convex marked boxes be given. Suppose T ∈ M̄ fixes
two distinct points p, q ∈ ΛΩ where ΛΩ is a Pappus Curve. Furthermore, let p be the
attracting fixed point of T . Let lp and lq be lines in LΩ , that are not pq and pass through
p and q, respectively. Suppose m is a line through p distinct from lp and pq. Then T n (m)
converges to lp in (P2 R)∗ . See Figure 0.1 [11].
Proof. Say the line m intersects lq at a point x ∈ lq . By hypothesis on m, this
intersection point is neither q nor lp ∩ lq . Then T n (x) converges to p by Corollary 4.2.0.3.
Hence, T n (m) converges to lp .
q
m
lq
p
lp
FIGURE 0.1: Pappus Curve Λ and Its Transverse Linefield L.
Proof of Theorem A0.9. Let L = LΩ and Λ = ΛΩ .
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We begin by proving that L is a transverse linefield to Λ.
Suppose d ∈ Λ is a distinguished point and ld ∈ L is a distinguished edge containing
d. Then, there exists a convex marked box Θ = ((p, q, r, s; t, b), (. . .)) such that ld is the
line given by applying Pappus construction to Θ. Two of the edges of ι(Θ) are on the lines
ps and qr. Hence, ι(Θ) is contained in a region defined by the degenerate conic ps ∪ qr.
Let R1 denote the region which contains ι(Θ) and R2 denote the other.
By the nesting property of box operations, it follows that a subset of Λ is contained
in the interior of ι(Θ) and the complement is contained in the interior of Θ. Let Λ1 be
the part contained in the interior of ι(Θ) and Λ2 be the part contained in the interior of
Θ. As can be seen in Figure 3.1, ld is contained in the region R2 . Thus, we have that ld
does not intersect Λ1 .
Now, we divide Λ2 into two parts. Let the part of Λ2 that is also contained in the
interior of τ1 (Θ) be Λ2,T and the part contained in the interior of τ2 (Θ) be Λ2,B . Thus,
we divide Λ2 at d.
The projective transformation τ22n defined by τ1 (Θ) maps ld onto itself and maps
a part of Λ1 onto Λ2,T . By letting n go to infinity, one shows that ld does not intersect
Λ2,T since ld does not intersect Λ2,T ∩ τ22n (ι(Θ)) for n > 0. Similarly, by applying τ12n
defined by τ2 (Θ), we have that ld does not intersect Λ2,B . Hence, a distinguished edge ld
intersects Λ only at d.
Given a line in l0 ∈ L and a marked box in Θ0 ∈ Ω, there is a line l in the M̄-orbit
of l0 such that l intersects the interior of ι(Θ0 ) only at its corners if they intersect at all.
Now, by applying τ1−2m and τ2−2n to Θ0 , we can find a marked box Θ ∈ Ω such that l
intersects the interior of Θ and not ι(Θ) since τ1−2m and τ2−2n move the corners of ι(Θ).
Suppose that l ∈ L and Θ ∈ Ω are such that l intersects the interior of Θ and not
the interior of ι(Θ).
Suppose that l intersects Λ at more than one point, and let α and β be two distinct
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points of intersection. Suppose Θ = ((p, q, r, s; t, b), (. . .)) with T = pq. Since l does not
intersect the interior of ι(Θ), α and β are contained in the interior of Θ. By the nesting
properties of box operations, we have that P n (l) → T as n → ∞ where P is the projective
transformation in M̄ that represents τ12 on Θ. Hence P n (α), P n (β) → t since P maps Λ
onto itself.
Now, suppose that l0 ∈ L is such that it intersects Λ at α0 and β 0 given by applying
the box operation ι on Θ. Thus l0 intersects with ι(Θ) and not Θ. If P 0 is the projective
transformation in M̄ that represents τ22 on ι(Θ), then P 0n (α0 ), P 0n (β 0 ) → t as n → ∞.
By the denseness implied by the nesting properties of box operations, the images of
α, β, α0 and β 0 are dense in Λ near t. This implies that Λ intersects itself at t. But, this
contradicts Λ being a homeomorphic image of S 1 . Hence, l cannot intersect Λ at more
than one point.
Therefore, L is a transverse linefield of ΛΩ .
We now prove the uniqueness of L. We do this by showing that any transverse
linefield of Λ contains a dense subset of lines which are also in L, so by continuity, the two
curves of lines must be the same.
Suppose that T ∈ M̄ is hyperbolic which fixes p and q and attracts to p. Then lines
lp , lq ∈ LΩ and pq are invariant under T .
Let Rp,+ and Rp,− denote the two disjoint regions of P2 R defined by the conic pq ∪lp .
Now, suppose that Λ has points in both regions. Take a point r ∈ Λ which is in Rp,+ .
Thus, the line pr intersects Λ at two points. Then by Lemma A0.14 and the continuity of
Λ, any line contained in Rp,+ and passing through p intersects Λ at some point distinct
from p. Similarly, we have that any line in Rp,− that goes through p also intersects Λ at
a point other than p. Thus, lp is the only line through p that intersects Λ at only p.
Let Rq,+ and Rq,− denote the two disjoint regions of P2 R defined by the conic pq ∪lq .
Now, suppose that Λ has points only in Rp,+ . Then Λ has point in Rq,+ and Rq,− since Λ
80
is a topological circle and in order for Λ to be contained in Rp,+ , it has to cross pq. Thus,
by a similar argument as above, lq is the only line through q that intersects Λ only at q.
By Remark 4.2.0.2, hyperbolic fixed points are dense in Λ. Hence any transverse
linefield of Λ has a dense subset of lines that is also dense in L. Therefore, by the continuity,
this transverse linefield is L. Thus L is unique.