Poncelet's Theorem in Finite Projective Planes and Beyond

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Doctoral Thesis

Poncelet's Theorem in finite projective planes and beyond

Author(s): Kusejko, Katharina

Publication Date: 2016

Permanent Link: https://doi.org/10.3929/ethz-a-010682687

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ETH Library Diss. ETH No. 23366

Poncelet’s Theorem in ﬁnite projective planes and beyond

A thesis submitted to attain the degree of DOCTOR OF SCIENCES of ETH ZURICH (Dr. sc. ETH Zurich)

presented by Katharina Kusejko MSc Math. ETH Zurich born on 01.10.1989 citizen of Austria

accepted on the recommendation of Prof. Dr. Norbert Hungerb¨uhler,advisor Prof. Dr. Alexander Pott, co-advisor

2016

”Perhaps I could best describe my experience of doing mathematics in terms of en- tering a dark mansion. One goes into the first room and it is dark, entirely dark. One stumbles around, bumping into the furniture. Gradually you learn, where each piece of furniture is. And finally, after six months or so, you find the light switch, you turn it on and suddenly, everything is illuminated and you can see exactly where you are.”

Andrew Wiles

Zusammenfassung

Die vorliegende Dissertation handelt von Kegelschnitten in endlichen projektiven Ebenen sowie von Kreisketten in endlichen Möbiusebenen. Die Hauptresultate wur- den bereits in [22], [23], [26] und [24] veröffentlicht. Ein besonderer Fokus liegt auf dem Schließungssatz des französischen Mathematikers Jean-Victor Poncelet, der 1813 fur¨ Kegelschnitte in der reellen projektiven Ebene eine interessante Beobachtung zu Polygonzugen,¨ dessen Kanten tangential zu einem Kegelschnitt sind und dessen Ecken auf einem anderen Kegelschnitt liegen, forma- lisiert und bewiesen hat. Dieses klassische Resultat, welches zu Beginn der These im Detail erklärt wird, wurde in den letzten 200 Jahren wiederholt in verschiedenen Teilgebieten der Mathematik - unter anderem in der Analysis, der Geometrie und der Theorie elliptischer Kurven - aufgegriffen. Endliche projektive Ebenen, welche uber¨ einem endlichen Körper definiert werden, erfullen¨ eine endliche Version des Poncelet’schen Satzes. Diese Tatsache wird in meh- reren Kapiteln mit Hilfe unterschiedlicher Techniken, darunter kombinatorische, syn- thetische und algebraische Methoden, bewiesen. Im Falle jener Ebenen, welche nicht uber¨ endlichen Körpern konstruiert werden, ist die Poncelet’sche Schließungseigen- schaft nicht zwangsläufig gegeben. Dies wird anhand endlicher projektiver Ebenen der Ordnung 9 im Detail erläutert. Weiterfuhrende¨ Fragen zum Schließungssatz von Poncelet, wie etwa ein Poncelet’sches Schließungskriterium, werden fur¨ Spezialfälle von Kegelschnittpaaren betrachtet. Dies liefert unter anderem einen neuen Zugang zu trigonometrischen Identitäten in endlichen projektiven Ebenen, fur¨ welche ein geometrischer Zugang zur Trigonometrie a priori nicht gegeben ist. Im Vordergrund des behandelten Spezialfalles stehen Kegelschnitte in Diagonalform, was in der Folge Fragen zur Diagonalisierbarkeit von Kegelschnittpaaren im Endlichen aufwirft. Mit Hilfe kombinatorischer Uberlegungen¨ werden Bedingungen fur¨ die Diagonalisierbar- keit präsentiert und Unterschiede zum reellen Fall erklärt. Eine dem Poncelet’schen Satz verwandte Aussage ist jene des Schweizer Mathema- tikers Jakob Steiner, welche von Ketten sich beruhrender¨ Kreise handelt. Hier wird eine analoge Aussage fur¨ Steiner-Ketten in endlichen Möbiusebenen, welche uber¨ endlichen Körpern konstruiert werden, im Detail studiert. Ahnlich¨ wie im Satz von Poncelet fuhren¨ auch hier die Berechnungen eines Schließungskriteriums fur¨ Kreis- ketten zu einer Interpretation trigonometrischer Identitäten in endlichen Ebenen. Durchgehend werden die jeweiligen Eigenschaften reeller und endlicher Ebenen ver- glichen, und vor allem erstaunliche Parallelen und kontraintuitive Unterschiede her- vorgehoben.

Abstract

This dissertation is mainly about conics in finite projective planes and chains of circles in finite Möbiusplanes. The main results are already published in [22], [23], [26] and [24]. A particular focus lies on Poncelet’s Theorem, a famous result by the French mathematician Jean-Victor Poncelet. In 1813, he made an interesting observation about two conics in the real projective plane and a polygonal chain with vertices on one conic and the edges tangent to the other conic. This classical result, which is discussed in detail at the beginning of the thesis, was studied repeatedly in the last 200 years in various areas of mathematics - such as analysis, geometry and in the theory of elliptic curves. Finite projective planes defined over a finite field satisfy a finite version of Pon- celet’s Theorem. This is proven in several chapters of this thesis by using different mathematical tools, such as combinatorial, synthetic and algebraic methods. In the case of finite projective planes not defined over finite fields, the finite version of Poncelet’s Theorem is proven to be wrong. This is shown by examining the finite projective planes of order 9. Further questions related to Poncelet’s Theorem, such as a criterion for a polygonal chain to close up, are discussed for special cases of conics. This leads to a new approach of trigonometric identities in finite projective planes, for which a geometric interpretation of trigonometry is a priori not given. A particular focus lies on conics in diagonal form and therefore questions about the diagonalisation of conics in finite projective planes are discussed. Conditions on the simultaneous diagonalisation of conic pairs are presented by using combinatorial methods. These conditions are then compared to the real case. A result related to Poncelet’s Theorem deals with chains of touching circles and is due to the Swiss mathematician Jacob Steiner. In this dissertation, we discuss a similar statement for Steiner chains in finite Möbiusplanes constructed over finite fields. Again, by investigating a criterion for the chain of circles to close, we get an interpretation of trigonometric identities in finite Möbiusplanes. Throughout, the properties of real and finite planes are compared, with an emphasis on surprising parallels and counter-intuitive differences.

Contents

1 Introduction 1

2 Preliminaries 7 2.1 Finite projective planes ...... 7 2.2 Finite Fields ...... 9 2.3 The Desarguesian Plane ...... 10 2.4 Conics ...... 12 2.5 Pencil of conics ...... 13

3 Poncelet’s Theorem 15 3.1 The classical version ...... 15 3.2 Interpretation in ﬁnite planes ...... 17 3.3 Degenerate Cases ...... 19 3.4 Examples ...... 21

4 Poncelet’s Theorem in finite projective planes of order 9 25 4.1 Poncelet’s Theorem in PG(2, 9)...... 25 4.2 Poncelet’s Theorem in the finite projective planes over S ...... 33 4.2.1 The miniquaternion near-field S ...... 33 4.2.2 The plane Ω ...... 35 4.2.3 The plane ΩD ...... 38 4.2.4 The plane Ψ ...... 40

5 Some proofs for Poncelet’s Theorem 45 5.1 A synthetic proof ...... 45 5.2 A combinatorial proof ...... 53 5.2.1 Preliminaries and preparation ...... 53 5.2.2 Poncelet’s Theorem for triangles ...... 55 5.2.3 Proof of Poncelet’s Theorem ...... 57 Contents

6 A Poncelet Criterion 63 6.1 Construction and properties ...... 63

6.2 Poncelet’s Theorem for conics Ok ...... 64 6.3 Relations for pairs of conics ...... 66 6.4 A Poncelet Criterion ...... 72 6.4.1 Poncelet coeﬃcients ...... 72 6.4.2 Poncelet polynomials ...... 80 6.5 Comparison to other methods ...... 82 6.5.1 Comparison to the Euclidean Plane ...... 82 6.5.2 Comparison to Cayley’s Criterion ...... 85

7 Simultaneous Diagonalization 87 7.1 The disjoint case ...... 87 7.2 The nondisjoint case ...... 94 7.3 Summary ...... 98

8 Steiner’s Theorem in the finite Möbiusplane M(q) 99 8.1 Miquelian Möbius planes ...... 100 8.2 Steiner’s Theorem in M(q)...... 100 8.3 The plane M(5) ...... 103 8.4 Existence and length of Steiner chains ...... 105 8.5 Generalization ...... 110 8.5.1 A Möbiusinvariant for pairs of circles ...... 110 8.5.2 Transformation of non-intersecting circles into concentric circles111 8.5.3 General criterion for Steiner chains ...... 113 8.6 Comparison to the Euclidean plane ...... 113

Bibliography 117

Dank 121

Curriculum Vitae 123 Chapter 1

Introduction

This thesis is about a branch of geometry called finite projective geometry. In a nutshell, projective geometry is like Euclidean geometry with additional infinite elements. Imagine standing on a train track and looking at the horizon. Although the tracks are parallel, they seem to meet at infinity. 1

This is exactly what projective geometry deals with. In particular, there are no parallel lines in a projective plane. This explains the idea of the term projective. Now for the finiteness. In our continuous world there are infinitely many points on every line. But one can describe a concept of geometry also by taking only finitely many points. In particular, a line will be a collection of finitely many points. This branch of geometry is relatively new and has for example applications in coding theory and cryptography. In this thesis, we are mainly interested in one particular property of a finite projective plane, namely the property of being a Poncelet plane. Jean-Victor Poncelet (1788 - 1867) [36] was the first mathematician who formalized projective geometry, and it was him, in 1813, to state one of the most beautiful results in this field. One version reads as follows. 1Picture taken from http://www.thoughtsupport.net/how-to-get-back-on-track/

1 Introduction

Theorem (Poncelet’s Theorem). Let C and C0 be two conics generally situated in the real projective plane. If it is possible to find an m-sided polygon, m 3, such that the vertices lie on C0 and the sides are tangent to C, then there are≥ infinitely many other such m-sided polygons. Moreover, one cannot find such an m0-sided polygon for m = m0 for the same pair of conics C and C0. 6

Figure 1.1: Example of Poncelet’s Theorem

In Chapter 2, we will start with some preliminaries about finite projective planes. Finite projective Desarguesian planes, denoted by PG(2, q) for q a prime power, are discussed in more detail. Such planes PG(2, q) can be seen as a finite equivalent to classical projective planes. We will see several times in this thesis that a finite version for Poncelet’s Theorem can be proved for all finite projective Desarguesian planes. This brings us to Chapter 3, where we will have a closer look at the statement of Poncelet’s Theorem itself. We consider Poncelet’s Theorem in finite projective planes and a first challenge is to suitably reformulate the statement of Poncelet’s Theorem. For this, we introduce the terms Poncelet pairs and Poncelet planes, i.e. planes in which a finite version of Poncelet’s Theorem is proved for every pair of conics. Some examples will give a first impression of Poncelet’s Theorem in finite projective planes. In Chapter 4, we will analyze all projective planes of order 9 and decide, which of them are Poncelet planes. The material covered in Chapter 4 is published in [22]. The order 9 is the smallest order for which nonisomorphic finite projective planes exist, in particular, there are exactly four different planes of order 9. Besides PG(2, 9) there are three non-Desarguesian planes, all of them constructed over a near-field of order 9. These planes behave rather different than classical planes. Our main observation for finite projective planes of order 9 will be the following. Theorem. The only Poncelet plane of order 9 is the finite projective Desarguesian plane PG(2, 9).

There are numerous proofs for Poncelet’s Theorem in classical projective geometry arising from diﬀerent areas of mathematics. For example, there is a synthetic proof by Berger [4] and a combinatorial proof by Halbeisen and Hungerb¨uhler[18]. In

2 Introduction

Chapter 5, we will have a closer look at those two well-known proofs and adapt them for finite projective Desarguesian planes. Aside from proving Poncelet’s Theorem itself, much work has been done to find criteria for the existence of Poncelet polygons, i.e. a closed polygonal chain with vertices on one conic C0 and the sides tangent to another conic C. The most advanced result is due to Arthur Cayley in 1853 [8]. In Chapter 6 we will look at Poncelet’s Theorem for a specific family of conics in finite projective planes PG(2, q). The results can also be found in [23]. In particular, we will look at pairs of conics Oα and Oβ which lie in a nested position, i.e. they lie inside each other, where we of course explain what this means for a finite projective plane. For such pairs, we will show the following finite version of Poncelet’s Theorem in PG(2, q).

Theorem (cf. Theorem 6.2.1). Let (Oα,Oβ) be a pair of conics in PG(2, q) given by the zeros of the polynomials

x2 + ky2 + ckz2, k α, β , ∈ { } where α, β GF (q) 0 and c a nonsquare in GF (q) are ﬁxed. If an n-sided ∈ \{ } − Poncelet polygon with vertices on Oβ and the sides tangent to Oα can be constructed starting with a point P Oβ, then such an n-sided Poncelet Polygon can be con- ∈ structed starting with any other point Q Oβ as well. ∈ Moreover, we will describe a criterion for the existence of Poncelet polygons in such planes, which turns out to be a number theoretic condition. Theorem (cf. Theorem 6.3.2). A Poncelet polygon can be constructed for a pair of ovals (Oα,Oβ) with vertices on Oβ and the sides tangent to Oα if and only if β(β α) is a nonsquare in GF (q). − In case that a Poncelet polygon exists, we are interested in the number n of vertices of such a polygon. For example, a ﬁrst result will be that if Oα and Oβ carry a Poncelet triangle, we get the condition 4β = α.

We are able to derive an algorithm to deduce for all such pairs (Oα,Oβ) whether an n-sided Poncelet polygon can be constructed for a given n. For that, so-called Poncelet polynomials and Poncelet coeﬃcients need to be introduced ﬁrst. Theorem (cf. Corollary 6.4.16). The following four steps give a complete description of n-sided Poncelet polygons for conic pairs (Oα,Oβ) in PG(2, q).

φ(n) 1. Deduce all n 3 with n (q + 1). For every such n, calculate 2 , which gives the number≥ of indices| k, such that an n-sided Poncelet Polygon can be constructed for (Ok,O1).

2. For all values n obtained in Step 1, look up the Poncelet polynomial Pn.

3. For every Pn deduced in Step 2, solve Pn(k) = 0 in GF (q), the finite field with q elements. This gives the corresponding Poncelet coefficients k, such that an n-sided Poncelet polygon can be constructed for (Ok,O1). 4. By using coordinate transformations, the information obtained in Step 3 can be transferred to all pairs (Oα,Oβ).

3 Introduction

In addition, we will also take a brief look at the Euclidean plane and investigate some parallels to the formulas derived for finite planes. For example the half-angle formula which can henceforth be interpreted in finite projective planes. The conics considered in Chapter 6 are all in diagonal form, which means that the matrix corresponding to the conic equation is a diagonal matrix. For that, recall the following well-known problem in linear algebra: Is it possible, for two given Hermitian (symmetric) 3 3 matrices A and B over the complex (real) numbers, to find a matrix S such that×S∗AS and S∗BS are both in diagonal form, where S∗ is the conjugate transpose of S? If such a matrix S exists, A and B are said to be simultaneously diagonalizable. See for example [35] or [21] for the complex (real) case. In Chapter 7, we will show that the situation over finite fields is for some cases rather different to the situation over the real or complex numbers. The results are published in [26]. For example, if A and B are two real symmetric matrices, a sufficient but not necessary condition for A and B being simultaneously diagonalizable is that vT Av and vT Bv have no nontrivial common zeros, where v is a column vector. When viewing vT Av and vT Bv as polynomials, which define conics in the real projective plane, the condition above would be equivalent with the conics being disjoint. We will see that for finite projective Desarguesian planes, not all pairs of disjoint conics are simultaneously diagonalizable. A criterion can be obtained by analyzing the pencil of two conics C and C˜, which is the set of conics obtained by considering GF (q)-linear combinations of C and C˜. In particular, we obtain the following condition for two proper disjoint conics in PG(2, q) to be simultaneously diagonalizable.

Theorem. Two proper disjoint conics C and C˜ in PG(2, q) are simultaneously diagonalizable if and only if their pencil (C, C˜) leads to a partition of the form P P, g, C1,...,Cq−1 or P, P˜ , gg,˜ C1,...,Cq−2 , where P , P˜ are points, g is a line, { } { } gg˜ a pair of lines and Ci proper conics.

We are interested in geometric conditions such that the pencil of two proper disjoint conics is of the form P, g, C1,...,Cq−1 or P, P˜ , gg,˜ C1,...,Cq−2 . The main result concerning this question{ is the following} theorem{ about conics C} and C˜ in nested position, i.e. all points of C are external points of C˜ or all points of C are inner points of C˜ and vice versa.

Theorem. Consider two proper disjoint conics C and C˜ in PG(2, q). The pencil (C, C˜) is of the form P, g, C1,...,Cq−1 if and only if all pairs of proper conics Plie in nested position. { }

For two proper conics which intersect in exactly one or exactly three points, we will prove that simultaneous diagonalization is never possible. Two proper conics which intersect in exactly four points, however, can always be diagonalized simultaneously. For two proper conics which intersect in exactly two points, the property of being simultaneously diagonalizable depends again on their pencil. In Chapter 8, we will leave the topic of Poncelet’s Theorem and conics in ﬁnite projective planes, but move to another equally beautiful result in geometry. In

4 Introduction the 19th century, the Swiss mathematician Jakob Steiner (1796 - 1863) discovered a result about chains of tangent circles in the Euclidean plane, also known as Steiner’s Porism. One version reads as follows.

Theorem (Steiner’s Porism). Let 1 and 2 be disjoint circles in the Euclidean B B plane. Consider a sequence of circles 1,..., k, which are tangent to both 1 and T T B 2. Moreover, let i and i+1 be tangent for all i = 1, . . . , k 1. If 1 and k are Btangent as well, thenT thereT are inﬁnitely many such chains,− called SteinerT chains.T In particular, every chain of consecutive tangent circles closes after k steps.

Figure 1.2: Two examples of Steiner chains in the Euclidean plane.

Steiner thoroughly investigated such chains and discovered many interesting properties. For example, he could prove that the consecutive tangent points of the circles 1,..., k lie on a conic, and of course on a circle in the case of starting with two T T concentric circles 1 and 2. Steiner also studied conditions for such a chain to close B B after k steps in terms of the radii and the distance between the centers of 1 and 2. We refer the interested reader to [9], [34] or [10] for more information onB Steiner’sB original result. In recent years, some refinements and generalizations of Steiner’s Porism were studied. For example in [44], Steiner chains with rational radii are discussed and in [7], a 3-dimensional analogue of Steiner’s Porism is presented. In Chapter 8, we will discuss Steiner’s Porism in a class of finite Möbiusplanes, denoted by M(q), for q a prime power. The material of this chapter is published in [24]. Such planes consist of points P and circles B, which satisfy three axioms.

We will start with two disjoint concentric circles a and b in M(q) and look for conditions and properties of their potential commonB tangentB circles. We are interested in finding a condition for the existence of Steiner chains, i.e. a chain of successively tangent circles which are also tangent to both a and b. We will see that the finiteness directly implies that all chains close up, i.e.B the firstB and the last circle in the chain are tangent as well. In particular, our finite version of Steiner’s Porism reads as follows.

Theorem. (cf. Theorem 8.2.3) Consider the circles 1 and b in M(q) with common center 0 and with radii 1 and b, respectively. AssumeB we canB ﬁnd two circles g and

5 Introduction

h tangent to b which are tangent to 1 in 1 and P . Let g and h be tangent as well. Then a SteinerB chain of length k canB be constructed, where k 1, . . . , q + 1 is minimal with P k = 1. Moreover, such a Steiner chain of length k∈can { be constructed} starting with any point Q on 1. B Moreover, we will ask for conditions on the length of a Steiner chain and obtain the following result.

Theorem. Let b = µ2 for µ in GF (q). A proper Steiner chain of length k can be constructed if and only if the following conditions are satisﬁed.

1. k divides pm + 1,

2. µ is a nonsquare in GF (q), − k 3. µ solves P1 = 1 for P1 given by

µ2 + 6µ 1 + 4(µ 1)√ µ P1 = − − − − , (1.1) (1 + µ)2

but P l = 1 for all 1 l k 1. 1 6 ≤ ≤ − Finally, the results will be compared to the conditions on Steiner chains in the Euclidean plane. The main observation is that in the Euclidean plane, for every k N, k 3, a Steiner chain of length k can be constructed. In M(q), however, only certain∈ divisors≥ of q can serve as the length of a Steiner chain.

6 Chapter 2

Preliminaries

In this chapter we recall the most important definitions and theorems used later on. First, we will define what finite projective planes are and mention some basic properties. After that, we will give some information about finite fields, which we need to describe finite projective Desarguesian planes PG(2, q). Finally, we will have a closer look at conics and pencils of conics in finite projective Desarguesian planes. All these basic definitions and results can for example be found in [20] and [30].

2.1 Finite projective planes

Deﬁnition 2.1.1. The triple (P, B, I) with I P B is called projective plane, if the following axioms are satisﬁed. ⊂ ×

(A1) For any two elements P,Q P, P = Q, there exists a unique element g B with (P, g) I and (Q, g) ∈I. 6 ∈ ∈ ∈ (A2) For any two elements g, h B, g = h, there exists a unique element P P with (P, g) I and (P, h) ∈I. 6 ∈ ∈ ∈ (A3) There are four elements P1,P2,P3,P4 P such that for all g B we have ∈ ∈ (Pi, g) I and (Pj, g) I with i = j implies (Pk, g) / I for k = i, j. ∈ ∈ 6 ∈ 6 Elements of P are called points and elements of B are called lines. By (P, g) I we indicate that the point P is incident with the line g. A more convenient way∈ of describing this incidence relation is P g. ∈ Deﬁnition 2.1.2. Three points are said to be collinear if they are incident with the same line and three lines are said to be concurrent if they are incident with the same point.

So far, the definitions apply for all projective planes, also the classical ones. Definition 2.1.3. A projective plane is called finite, if the sets P and B are finite.

In the case of a finite projective plane, it turns out that each line is incident with n + 1 points and each point is incident with n + 1 lines, for some n 1. By using ≥ 7 2.1. Finite projective planes the axioms in Definition 2.1.1 it can be shown that there are n2 + n + 1 lines and n2 + n + 1 points in a finite projective plane, for some n 2. According to that, we define the following. ≥ Definition 2.1.4. A finite projective plane (P, B, I) is said to be of order n, if 2 P = B = n + n + 1. Such a plane is denoted by n. | | | | P It is well-known how to construct a finite projective plane of order q, for q a prime power. But it is an unsolved question whether finite projective planes of order m, m not a prime power, exist. Given one projective plane of order n, we can easily construct another one.

Definition 2.1.5. For a finite projective plane n = (P, B, I) of order n, we define P the dual projective plane of n by P D := (B, P, I∗), Pn with (P, g) I (g, P ) I∗. ∈ ⇔ ∈ D Then n is called self-dual, if n = , i.e. if there exists a bijective map P P ∼ Pn φ :(P, B) (B, P) → such that (P, g) I φ(P, g) I∗. ∈ ⇐⇒ ∈ Incidence statements where the sets P and B are interchanged are said to be dual to each other.

Later in Chapter 4 we will work with self-dual planes, as for example PG(2, 9), as well as with planes which are not self-dual. Next we introduce ovals, the main objects when reformulating Poncelet’s Theorem for ﬁnite projective planes.

Deﬁnition 2.1.6. An oval in n is a set of n + 1 points, no three of which are collinear. P

Ovals can be seen as a combinatorial equivalence of conics. Hence, the following deﬁnitions come naturally.

Deﬁnition 2.1.7. Let O be an oval in n. A line which intersects O in two points is called a secant, a line which intersectsP O in one point is called a tangent and a line which does not intersect O is called an external line of O.

Lemma 2.1.1. Let O be an oval in n. Every point of O is incident with exactly P one tangent and n secants of O. In particular, every oval O in n has n+1 tangents, n+1 = 1 n(n + 1) secants and 1 n(n 1) external lines. P 2 2 2 − Deﬁnition 2.1.8. Let O be an oval. The points of O are called oval points or on- points. A point, which is not an on-point and incident with a tangent of O is called external point, all other points are called inner points of O.

The lemma shows that ﬁnite projective planes of even order behave rather diﬀerent from planes of odd order.

Lemma 2.1.2. Let n be a ﬁnite projective plane of order n. If n is even, then all tangents of an oval meetP in one point, the so-called nucleus. If n is odd, then there are exactly two tangents through every external point of an oval.

8 2.2. Finite Fields

2.2 Finite Fields

We recall the most important properties and definitions of finite fields GF (pm), for p a prime and m 1. ≥ Definition 2.2.1. An element α GF (pm) is called a primitive element of GF (pm), if the set ∈ αk, 0 k pm 1 { ≤ ≤ − } equals GF (pm).

It is well-known that such a primitive element, i.e. an element which generates GF (pm), exists for every finite field. Definition 2.2.2. An element a GF (pm) is called a square in GF (pm) if there exists some b GF (pm) with a = ∈b2. Otherwise, a GF (pm) is called a nonsquare in GF (pm). ∈ ∈

Exactly half of the elements in GF (pm) 0 are squares. Note that the squares of GF (pm) form a subgroup of GF (pm), but\{ the} nonsquares do not form a subgroup of GF (pm). Multiplying two nonsquares in GF (pm) gives a square and multiplying a square and a nonsquare in GF (pm) gives a nonsquare. For any nonsquare x in GF (pm), we can construct an extension ﬁeld of GF (pm) by adjoining some element α with α2 = x to GF (pm). This extension ﬁeld is denoted by GF (pm)(α) and is of the form GF (pm)(α) = a + αb a, b GF (pm) . { | ∈ } Lemma 2.2.1. All elements of GF (pm) are squares in GF (pm)(α).

Proof. To see this, take some element a GF (pm). If a is a square in GF (pm), it is clearly a square in GF (pm)(α) as well. If∈ a is a nonsquare in GF (pm), then ax is a square in GF (pm) and hence ax = b2 which leads to a = α−2b2 in GF (pm)(α), i.e. a is a square in GF (pm)(α).

Since GF (pm)(α) is isomorphic to any field with p2m elements, we denote it by GF (p2m). Definition 2.2.3. For z GF (p2m) the conjugate element of z over GF (pm) is defined by ∈ m z := zp .

Note that z = z if and only if z GF (pm). ∈ Deﬁnition 2.2.4. For z GF (p2m) the trace of z over GF (pm) is deﬁned by ∈ T rGF (p2m)/GF (pm)(z) := z + z and the norm of z over GF (pm) by

NGF (p2m)/GF (pm)(z) := zz, where we omit the subscript GF (p2m)/GF (pm) for notational convenience.

9 2.3. The Desarguesian Plane

Recall that T r(z) and N(z) are always in GF (pm). Moreover, we have

z1 + z2 = z1 + z2 as well as z1z2 = z1 z2.

2.3 The Desarguesian Plane

A particular class of ﬁnite projective planes are so-called Desarguesian planes. The set of points P is deﬁned as

P = [x, y, z]T GF (q)3, [x, y, z]T = [0, 0, 0]T / ∈ 6 ∼ where is an equivalence relation given by ∼ [λx, λy, λz]T [x, y, z]T ∼ for all λ GF (q)∗ := GF (q) 0 . ∈ \{ } Using the same equivalence relation, the set of lines B is deﬁned as

B = [a, b, c] GF (q)3, [a, b, c] = [0, 0, 0] / . ∈ 6 ∼ The incidence relation I is given by the scalar product, i.e. P = [x, y, z]T P is incident with g = [a, b, c] B if and only if ∈ ∈ P T g = xa + yb + zc = 0 in GF (q).

The ﬁnite projective Desarguesian plane of order q = pn, p a prime and n N, constructed in this way is unique up to isomorphisms and is denoted by PG(2∈, q). Since we have coordinates, there is an easy way to perform calculations in ﬁnite Desarguesian planes. In particular, recall the following two results.

Lemma 2.3.1. Let g = [g1, g2, g3] and h = [h1, h2, h3] be two diﬀerent lines in PG(2, q). The unique intersection point P of g and h is given by the vector product of g and h, i.e.

T P = [g2h3 g3h2, g3h1 g1h3, g1h2 g2h1] . − − − T T Similarly, for two points P = [P1,P2,P3] and Q = [Q1,Q2,Q3] in PG(2, q), the unique line g through P and Q is given by

g = [P2Q3 P3Q2,P3Q1 P1Q3,P1Q2 P2Q1]. − − − T T T Lemma 2.3.2. Let P = [P1,P2,P3] , Q = [Q1,Q2,Q3] and R = [R1,R2,R3] be three diﬀerent points in PG(2, q). Then P, Q, R are collinear if and only if   P1 Q1 R1 det P2 Q2 R2 = 0. P3 Q3 R3

10 2.3. The Desarguesian Plane

All points, lines, pairs of lines and conics in PG(2, q) can be described as varieties

V = V(ax2 + by2 + cz2 + dxy + exz + fyz) (2.1) with a, b, c, d, e, f GF (q), i.e. the set of zeros of ∈ ax2 + by2 + cz2 + dxy + exz + fyz. (2.2)

Another way to look at the quadratic form (2.2) is to consider the matrix representation

vT Av = 0 (2.3) for v = [x, y, z]T and  d e  a 2 2 d f A =  2 b 2  . e f 2 2 c The variety (2.1) then corresponds to a proper conic if and only if the corresponding matrix A is regular. If A is a singular matrix and vT Av is irreducible, then V(vT Av) is one point only. Otherwise, if the quadratic form vT Av splits into two linear factors, the variety V(vT Av) corresponds to one or two lines. In this case, we talk about degenerate conics (or nonproper conics). It is a well-known result that for a ﬁnite projective Desarguesian plane over a ﬁeld of odd characteristic, ovals coincide with conics (see for example [40]). Lemma 2.3.3. Let C be any conic in PG(2, q) and P be any point in PG(2, q). Then

P is an on-point of C if and only if CP is a tangent of C, • P is an exterior point of C if and only if CP is a secant of C, • P is an inner point of C if and only if CP is an external line of C. • An important tool are collinear maps of PG(2, q). Lemma 2.3.4. If S is a regular 3 3 matrix with coeﬃcients in GF (q), then × φS : PG(2, q) PG(2, q),P SP → 7→ is bijective and collinear, i.e. a set of collinear points is mapped to a set of collinear points. T T Lemma 2.3.5. For any three noncollinear points P = [p1, p2, p3] , Q = [q1, q2, q3] T and R = [r1, r2, r3] in PG(2, q), there exists a collineation φS which maps those T T T points to [1, 0, 0] , [0, 1, 0] and [0, 0, 1] , namely the map φS with T T T φS(P ) = [1, 0, 0] , φS(Q) = [0, 1, 0] , φS(R) = [0, 0, 1] where   q2r3 q3r2 q3r1 q1r3 q1r2 q2r1 − − − S = p3r2 p2r3 p1r3 p3r1 p2r1 p1r2 . − − − p2q3 p3q2 p3q1 p1q3 p1q2 p2q1 − − − 11 2.4. Conics

An even stronger statement is true for ﬁnite projective Desarguesian planes as well.

Lemma 2.3.6. Any quadrilateral can be chosen arbitrarily in PG(2, q).

2.4 Conics

In this section, we make some combinatorial considerations about proper conics. In particular, we count the number of proper conics, which intersect in a particular number of points and use this to count pairs of proper conics which have a certain number of points in common. The numbers obtained in this section are used in Chapter 7 to count the number of pencils of conics in a particular shape. This is the main ingredient to solve the question whether or not two conics can be diagonalized simultaneously. First, we deduce how many proper conics there are through a given number of points.

Lemma 2.4.1. In PG(2, q), there are (q 2) proper conics through four given points, (q 1)2 proper conics through three given− points, q2(q 1) proper conics through two− given points and q2(q2 1) proper conics through one− given point. Furthermore, there are q5 q2 proper conics− in total. − Proof. To count the number of proper conics through four given points, we use Lemma 2.3.6 and suppose, without loss of generality, that [1, 0, 0]T , [0, 1, 0]T , [0, 0, 1]T and [1, 1, 1]T lie on the proper conics. It follows that the matrices representing the conics are, up to scaling, of the form

0 1 e  A = 1 0 1 e . e 1 e − 0− − − We need A to be regular, which leads to the conditions e = 0 and e = 1. This gives exactly q 2 choices for e. For the second statement,6 let [1, 0, 0]6 T ,− [0, 1, 0]T and [0, 0, 1]T be− on the proper conics. Again,

0 1 e A = 1 0 f e f 0 needs to be regular, i.e. ef = 0, which gives (q 1)2 choices. Similarly can be proceeded for the number of6 conics through two or− one ﬁxed points, as well as for the number of proper conics in total.

Now we can use Lemma 2.4.1 to compute the following quantity.

Deﬁnition 2.4.1. The number of proper conics in PG(2, q) which intersect any given proper conic C˜ in exactly k points is denoted by n o Nk(q) := # C C a proper conic, C C˜ = k for C˜ ﬁxed . | | ∩ |

12 2.5. Pencil of conics

Lemma 2.4.2. In PG(2, q), for q odd, we have

N5(q) = 1, q + 1 N4(q) = (q 3), 4 − q + 1 2 N3(q) = ((q 1) 1) 4N4(q), 3 − − − q + 1 2 N2(q) = (q (q 1) 1) 6N4(q) 3N3(q), 2 − − − − 2 2 N1(q) = (q + 1)(q (q 1) 1) 4N4(q) 3N3(q) 2N2(q), 5 2 − − − − − N0(q) = (q q ) N5(q) N4(q) N3(q) N2(q) N1(q). − − − − − − Proof. Since any proper conic is uniquely determined by five of its points, we get N5(q) = 1. Let us consider the proper conic C˜ and fix four points on C˜. There are q 3 proper conics, which intersect C˜ in exactly these four points, since the total number− of proper conics through four points is given by q 2, as seen in Lemma 2.4.1. − q+1 This is true for any set of four points on C˜, which gives us N4(q) = (q 3). 4 − For the third statement, with the same argument as before, we get q+1((q 1)2 1) 3 − − proper conics through any three points of the fixed proper conic C˜, not counting C˜ itself. But now, we also count those proper conics intersecting C˜ in four points, we 4 even count these conics 3 = 4 times. Hence, we have to subtract 4N4(q) to get the desired result. Similarly for the fourth statement, recall that there are q2(q 1) 1 proper conics through any two points on C˜, not counting C˜ itself. Again,− we− also count those proper conics intersecting C˜ in exactly four points and now, we even count them 4 ˜ 2 = 6 times. In addition, we count those proper conics intersecting C in exactly 3 three points, namely 2 = 3 times. By subtracting these expressions, we obtain the claimed number.

To obtain N1(q), a similar discussion can be made. Finally, to ﬁnd the number of proper conics which are disjoint to C˜, we just have to subtract the number of all proper conics which intersect C˜ in exactly one, two, three or four points, as well as C˜ itself, of the number of all proper conics, which is q5 q2. −

2.5 Pencil of conics

We consider the homogeneous polynomials Ei of degree two with coeﬃcients ai, bi, ci, di, ei, fi in GF (q), not all zero, i.e.

2 2 2 Ei = aix + biy + ciz + dixy + eixz + fiyz. (2.4)

We study the structure of conics Vi := V(Ei) obtained by considering the pencil of such objects over GF (q). For this, let us deﬁne a pencil of conics.

Deﬁnition 2.5.1. Let Vi = V(Ei) and Vj = V(Ej) be two conics in PG(2, q). Let α

13 2.5. Pencil of conics

be any primitive element of GF (q). Then we deﬁne the pencil of Vi and Vj as

k (Vi,Vj) := Vi Vj 0≤k≤q−2 V(Ej + α Ei) P { } ∪ { } ∪ { } i.e. the pencil consists of the zero sets of all polynomials obtained by GF (q)-linear combinations of Ei and Ej.

Our goal is to show that the pencil of two disjoint conics Vi and Vj is a partition of the plane PG(2, q). Note that we do not assume that the starting conics are proper, i.e. Vi and Vj can correspond to points, lines and pairs of lines as well as to proper conics. As a direct consequence of the deﬁnition, we mention that if a point P lies in two conics Vi and Vj, it lies in every element of their pencil (Vi,Vj). P Since a pencil of two conics is constructed by GF (q)-linear combinations of two conic equations, the following statement is immediate.

Lemma 2.5.1. The pencil (Vi,Vj) is independent of the representatives Vi and Vj. P The next result deals with the question whether all points of the plane PG(2, q) are in some element of (Vi,Vj). P

Lemma 2.5.2. Consider the pencil (Vi,Vj). Let P be any point in PG(2, q). If P P lies in some W (Vi,Vj), then it is either only in W or in every element of ∈ P (Vi,Vj). Moreover, every point P of the plane PG(2, q) is contained in at least P one element of (Vi,Vj). P Proof. We already showed the ﬁrst statement above. For the second statement, let P be an arbitrary point in PG(2, q) and assume that P does not lie in every element of the pencil (Vi,Vj). By the ﬁrst statement, P can lie in at most one element of P (Vi,Vj). Suppose P neither lies in Vi nor in Vj, so Ei(P ) = 0 and Ej(P ) = 0. But P 6 Ej (P ) 6 then, P lies in the element of (Vi,Vj) given by the equation Ej Ei. P − Ei(P )

Corollary 2.5.3. Let V1 and V2 be disjoint conics. Then the pencil (Vi,Vj) gives a partition of all points in PG(2, q). P

Proof. Since V1 and V2 are disjoint, no point of PG(2, q) can be in more than one element in (Vi,Vj) since otherwise, by Lemma 2.5.2, such a point would lie in P every conic of (Vi,Vj) and hence would be a common point of V1 and V2 as well. P Moreover, every point of PG(2, q) is contained in at least one element of (Vi,Vj) by Lemma 2.5.2 again. P

14 Chapter 3

Poncelet’s Theorem

3.1 The classical version

Figure 3.1: Jean-Victor Poncelet 1

Jean-Victor Poncelet (1788 - 1867) was a french mathematician and the founder of projective geometry. In 1813, he was a war prisoner in Saratov, Russia, after Napoleon’s war against Russia. During his captivity, he was able to prove one of the most important and beautiful results in projective geometry. In 1822, Poncelet published his results in Traitédes propriétésprojectives des figures [36]. One version of Poncelet’s Theorem reads as follows. Theorem 3.1.1 (Poncelet’s Porism). Let C and C0 be two conics. If it is possible to find an m-sided polygon, m 3, such that the vertices lie on C0 and the sides are tangent to C, then there are infinitely≥ many other such m-sided polygons. Moreover,

1http://www-history.mcs.st-andrews.ac.uk/PictDisplay/Poncelet.html

15 3.1. The classical version

Figure 3.2: Example for Poncelet’s Theorem with two ellipses and 5-gons one cannot ﬁnd such an m0-sided polygon for m = m0 for the same pair of conics C and C0. 6

In Figure 3.2 we see the most common and natural way to visualize Poncelet’s Theorem. But the Theorem does not restrict to disjoint ellipses. Figure 3.3 gives some examples of Poncelet’s Theorem involving ellipses, hyperbolas and parabolas.

Figure 3.3: More examples for Poncelet’s Theorem

Until today, Poncelet’s Theorem plays an important role in projective geometry. It inspired many mathematicians to look for generalizations and further questions arising with Poncelet’s Theorem. In particular, there are many diﬀerent proofs for Poncelet’s Theorem using tools from various areas in mathematics. Poncelet himself could give an analytic proof as well as a synthetic proof of his theorem.

16 3.2. Interpretation in ﬁnite planes

Later, in 1828, Carl Gustav Jacob Jacobi (1804 - 1851) presented a proof using elliptic functions [25]. Another interesting question concerning Poncelet’s Theorem is, whether it is possible at all to construct a Poncelet n-gon and if so, what is the expected length n of the polygon. Based on this analytic view of Poncelet’s Theorem, Arthur Cayley (1821 - 1895) was able to give a criterion on the existence of a Poncelet polygon of length n, published in [8]. There is a section in the book Geometry 2 by Marcel Berger [4], which presents a synthetic proof of a similar statement of Poncelet’s Theorem. We have a closer look at that approach in Chapter 5.1. Recently, due to the bicentannial of Pon- celet’s theorem in 2013, a combinatorial proof was published by Halbeisen and Hungerbühler[18]. This is discussed in detail in Chapter 5.2. Moreover, there are purely geometric proofs of some special cases of Poncelet’s Theorem (for example if the conics do not intersect), which use only properties of the Euclidean plane such as angle and length. In 1978, a deep connection between Poncelet’s Theorem and the theory of elliptic curves has been established by Griffith and Harris [16], which mainly shows that Poncelet’s Theorem is in a sense equivalent to the group law of an elliptic curve. See the recent book by V. Dragovićand M. Radnović[15] for an overview of Pon- celet’s Theorem, applications and similar statements.

3.2 Interpretation in ﬁnite planes

Our main aim is to investigate Poncelet’s Theorem in finite projective planes. Since ovals are the combinatorial equivalent of conics in the finite projective plane, these are the objects we need to study. The first challenge is to think about an interpretation of Poncelet’s Theorem for the finite case. Of course, we cannot find infinitely many polygons which are inscribed in one oval and circumscribed around another oval, as formulated in Theorem 3.1.1. For the reformulation of Poncelet’s Theorem for finite projective planes, let us consider two ovals Ot and Os.

Deﬁnition 3.2.1. An m-sided Poncelet polygon is a polygon with m sides, 3 m ≤ ≤ n + 1, such that the vertices are on Os and the sides are tangent to Ot. According to that, we call Ot the tangent oval and Os the secant oval of the Poncelet polygon.

Deﬁnition 3.2.2. For 3 m n + 1 ﬁxed, (Ot,Os) is said to form a Poncelet ≤ ≤ m-pair, if there exists at least one m-sided Poncelet polygon for Ot and Os, but no m0-sided Poncelet polygon, m0 = m, 3 m0 n + 1, for the same pair can be constructed. 6 ≤ ≤

We say that (Ot,Os) forms a Poncelet 0-pair, if there is no secant of Os, which is a tangent of Ot, i.e. no point of Os is an external point of Ot.

We say that (Ot,Os) forms a Poncelet -pair, if there exists at least one secant of ∞ Os, which is a tangent of Ot, but no m-sided Poncelet polygon for 3 m n + 1 can be constructed. ≤ ≤

17 3.2. Interpretation in ﬁnite planes

Before we define what a Poncelet plane is, let us take some time to point out diffi- culties which may occur in finite projective planes. Such finite planes exhibit rather unintuitive phenomena compared to the real projective plane. For example, we can have the following situations.

1. A pair of ovals can be located such that no point of one oval is incident with a tangent of the other oval and vice versa. To interpret this in the real projective plane, that would mean that one oval O lies inside another oval 0, but 0 lies inside O as well. One example for such a pair of ovals is O O O = V(x2 + y2 + z2) and 0 = V(x2 + 6y2 + 6z2) O in PG(2, 7). In Chapter 6, we will discuss such properties in more detail and also the above example appears there. 2. A pair of ovals can be located such that every point of one oval is incident with a tangent of the other oval and vice versa. Similar to before, this would mean that one oval O lies outside 0, but 0 lies outside O as well. One example for such a pair of ovals is O O O = V(x2 + y2 + z2) and 0 = V(x2 + 3y2 + 3z2) O in PG(2, 7). 3. A pair of disjoint ovals O and 0 can be positioned such that some points of 0 are incident with a tangentO of O, but some are not. Note that two conics withO this property can never be disjoint in the classical projective plane. 4. In ﬁnite projective planes of even order it may happen that two ovals have all their tangents in common.

With that in mind, we are ready to deﬁne a Poncelet plane.

Definition 3.2.3. We call a finite projective plane n a Poncelet plane, if every P pair of ovals (Ot,Os) is a Poncelet m-pair, for 3 m n + 1, m = 0 or m = . ≤ ≤ ∞ Note that in planes of even order all tangents of an oval meet in one point, the so-called nucleus of the oval. Because of that, only Poncelet 0-pairs and Poncelet -pairs can be constructed in such planes. This leads immediately to the following statement.∞ Theorem 3.2.1. All finite projective planes of even order are Poncelet planes.

There are some other trivial cases as well. Look for example at the plane 3. There are only four points on every oval, so necessarily, every pair of ovals is aP Poncelet 3-pair, a Poncelet 4-pair, a Poncelet 0-pair or a Poncelet -pair. Similarly for ∞ 5, where we have six points on every oval. So immediately, we get the following statement.P

Theorem 3.2.2. All ﬁnite projective planes 3 and 5 are Poncelet planes. P P

The first interesting plane is therefore the plane 7, since a pair of ovals which carry a Poncelet 3-pair as well as a Poncelet 4-pairP might exist. However, in the next chapters we prove that a plane 7 is actually a Poncelet plane. The smallest finite projective planes which are notP Poncelet planes are three nonisomorphic finite projective planes of order 9. This is discussed in Chapter 4.

18 3.3. Degenerate Cases

3.3 Degenerate Cases

According to the classical formulation of Poncelet’s Theorem, if there exists one Poncelet n-gon with vertices on the conic C0 and the sides tangent to the conic C, then there are inﬁnitely many such Poncelet n-gons. Let us again have a look at the example in Figure 3.2. In that case, the two conics C0 and C are two ellipses, one lying inside the other one. We see that starting with any point on C0 leads to a polygonal chain in the Poncelet fashion which closes up after 5 steps. In general, however, it is not the case that we can pick any starting point we want. Of course, if some points on C0 are inner points of C, then no Poncelet polygon can be constructed using such inner points. That is clear, since inner points of C0 are not incident with any tangent of C. But these inner points are not the only points on C0 which need to be excluded as starting points for the construction of a Poncelet polygon. Let us have a look at the following example. We have an ellipse and a hyperbola which carry Poncelet triangles (see Figure 3.4).

Figure 3.4: The two conics carry Poncelet triangles

We can vary the starting point (i.e. one of the three vertices of one Poncelet triangle) to create smaller and smaller Poncelet triangles. At some point, the triangle gets inﬁnitely small, which means that it degenerates into a line segment. In Figure 3.5, we see that starting with the point P or Q on the hyperbola leads to a degenerate Poncelet polygon. This happens since in the process of constructing a Poncelet polygon we hit an intersection point of the two conics. Note that P is incident with two tangents of the ellipse C. One tangent is incident with an intersection point of the two conics and the other tangent is a common tangent of the two conics. Now let us move to the next example. We consider two ellipses C and C0 which carry a Poncelet 4-gon. Figure 3.6 visualizes the possible degenerate versions of the Poncelet 4-gon. First, consider the point P on C0 which is incident with two tangents of C, but both tangents are incident with intersection points of C and C0. This can be seen as a degenerate 4-gon with P a double point. Moreover, consider the points S and T on C. The line connecting S and T is tangent to C, but the tangents of S and T are both tangent to C as well. Hence, this gives another degenerate 4-gon

19 3.3. Degenerate Cases

Figure 3.5: A degenerate Poncelet triangle with S and T counting as double points.

Figure 3.6: A degenerate Poncelet 4-gon

The following two ellipses carry Poncelet 5-gons. Of course, starting the Poncelet construction with one of the intersection points of the two ellipses leads to a degenerate Poncelet polygon again. We can find four more points in the Poncelet construction until we end up with a common tangent of the ellipses (see Figure 3.7). The procedure for constructing degenerate examples for n 6 is similar. Imagine a pair of intersecting conics which carries a proper Poncelet≥n-gon. Then start with a common intersection point and construct a polygon in the Poncelet fashion until the process terminates by hitting an intersection point of the conic or a common tangent of both conics. These degenerate cases correspond to the Poncelet -pairs defined before. This means that we can start constructing a polygonal chain∞ such that the vertices are on one conic C0 and the sides are tangent to C, but the polygonal chain does not close up. In the classical case, the property of the polygonal chain not closing up does not necessarily indicate a degenerate case. It could just be the case, that the polygonal chain finally lies dense, as Figure 3.8 indicates (see also [15, Section 2.6]). In the finite case, however, that is of course not possible due to the limited number of points. The only reason for a polygonal chain not to close up is therefore that

20 3.4. Examples

Figure 3.7: A degenerate Poncelet 5-gon.

Figure 3.8: This Poncelet conﬁguration will ﬁnally lie dense. during the construction, we either hit an intersection point of the two ovals or a common tangent. This corresponds to the degenerate cases described above.

3.4 Examples

In this section, we present some examples of Poncelet m-pairs in the plane PG(2, 7). For this, recall that the points in this plane are given by P = [x, y, z]T GF (7)3, [x, y, z]T = [0, 0, 0]T / . { ∈ 6 } ∼ It is enough to consider one element in every equivalence class, so we can rewrite P as P = [1, y, z]T , y, z GF (7) [0, 1, z]T , z GF (7) [0, 0, 1]T { ∈ } ∪ { ∈ } ∪ { } and see directly that we obtain 72 + 7 + 1 = 57 points. Similar for the lines of PG(2, 7).

For all our examples we consider secants of the oval Os given by 2 2 2 Os := V(x + y + z ),

21 3.4. Examples which is

T T T T T T T T Os := [1, 2, 3] , [1, 2, 4] , [1, 3, 2] , [1, 3, 5] , [1, 4, 2] , [1, 4, 5] , [1, 5, 3] , [1, 5, 4] . { }

Poncelet -pair: Consider the oval Ot given by ∞ 2 2 2 Ot := V(x + 5y + 6z ).

The tangents of Ot are given by t ([1, 0, 6]T ]) = [0, 1, 0]T , [1, 0, 6]T , [1, 1, 6]T , [1, 2, 6]T , [1, 3, 6]T , [1, 4, 6]T , [1, 5, 6]T , [1, 6, 6]T Ot { } t ([1, 0, 1]T ]) = [0, 1, 0]T , [1, 0, 1]T , [1, 1, 1]T , [1, 2, 1]T , [1, 3, 1]T , [1, 4, 1]T , [1, 5, 1]T , [1, 6, 1]T Ot { } t ([1, 3, 5]T ]) = [0, 1, 3]T , [1, 0, 3]T , [1, 1, 6]T , [1, 2, 2]T , [1, 3, 5]T , [1, 4, 1]T , [1, 5, 4]T , [1, 6, 0]T Ot { } t ([1, 3, 2]T ]) = [0, 1, 4]T , [1, 0, 4]T , [1, 1, 1]T , [1, 2, 5]T , [1, 3, 2]T , [1, 4, 6]T , [1, 5, 3]T , [1, 6, 0]T Ot { } t ([1, 2, 0]T ]) = [0, 0, 1]T , [1, 2, 0]T , [1, 2, 1]T , [1, 2, 2]T , [1, 2, 3]T , [1, 2, 4]T , [1, 2, 5]T , [1, 2, 6]T Ot { } t ([1, 5, 0]T ]) = [0, 0, 1]T , [1, 5, 0]T , [1, 5, 1]T , [1, 5, 2]T , [1, 5, 3]T , [1, 5, 4]T , [1, 5, 5]T , [1, 5, 6]T Ot { } t ([1, 4, 5]T ]) = [0, 1, 4]T , [1, 0, 3]T , [1, 1, 0]T , [1, 2, 4]T , [1, 3, 1]T , [1, 4, 5]T , [1, 5, 2]T , [1, 6, 6]T Ot { } t ([1, 4, 2]T ]) = [0, 1, 3]T , [1, 0, 4]T , [1, 1, 0]T , [1, 2, 3]T , [1, 3, 6]T , [1, 4, 2]T , [1, 5, 5]T , [1, 6, 1]T Ot { } Note that the two ovals intersect in four points, namely in [1, 3, 2]T , [1, 3, 5]T , [1, 4, 2]T and [1, 4, 5]T . Now we start to construct a polygonal chain in the Poncelet fashion. The line T T connecting the two points [1, 4, 2] and [1, 2, 3] on Os is the tangent of Ot in the point [1, 4, 2]T . Next, the line connecting the two points [1, 2, 3]T and [1, 2, 4]T on T Os is the tangent of Ot in the point [1, 2, 0] . The line connecting the two points T T T [1, 2, 4] and [1, 4, 5] on Os is again a tangent of Ot, namely in the point [1, 4, 5] . T But in [1, 4, 5] , which is an intersection point of Ot and Os, we cannot continue the polygonal chain. The same happens when starting with the point [1, 3, 2]T . This means, that we only get the two polygonal chains [1, 4, 2]T [1, 2, 3]T [1, 2, 4]T [1, 4, 5]T → → → → [1, 3, 2]T [1, 5, 3]T [1, 5, 4]T [1, 3, 5]T → → → → which do not close up. This makes the pair (Ot,Os) a Poncelet -pair. The corresponding situation in the real case is visualized in Figure 3.9. ∞

Poncelet 8-pair: Let us consider the oval Ot given by

2 2 2 Ot := V(x + 5y + 5z ).

The tangents of Ot are given by t ([1, 0, 2]T ]) = [0, 1, 0]T , [1, 0, 2]T , [1, 1, 2]T , [1, 2, 2]T , [1, 3, 2]T , [1, 4, 2]T , [1, 5, 2]T , [1, 6, 2]T Ot { } t ([1, 0, 5]T ]) = [0, 1, 0]T , [1, 0, 5]T , [1, 1, 5]T , [1, 2, 5]T , [1, 3, 5]T , [1, 4, 5]T , [1, 5, 5]T , [1, 6, 5]T Ot { } t ([1, 3, 3]T ]) = [0, 1, 6]T , [1, 0, 6]T , [1, 1, 5]T , [1, 2, 4]T , [1, 3, 3]T , [1, 4, 2]T , [1, 5, 1]T , [1, 6, 0]T Ot { } t ([1, 3, 4]T ]) = [0, 1, 1]T , [1, 0, 1]T , [1, 1, 2]T , [1, 2, 3]T , [1, 3, 4]T , [1, 4, 5]T , [1, 5, 6]T , [1, 6, 0]T Ot { } t ([1, 2, 0]T ]) = [0, 0, 1]T , [1, 2, 0]T , [1, 2, 1]T , [1, 2, 2]T , [1, 2, 3]T , [1, 2, 4]T , [1, 2, 5]T , [1, 2, 6]T Ot { } t ([1, 5, 0]T ]) = [0, 0, 1]T , [1, 5, 0]T , [1, 5, 1]T , [1, 5, 2]T , [1, 5, 3]T , [1, 5, 4]T , [1, 5, 5]T , [1, 5, 6]T Ot { } t ([1, 4, 3]T ]) = [0, 1, 1]T , [1, 0, 6]T , [1, 1, 0]T , [1, 2, 1]T , [1, 3, 2]T , [1, 4, 3]T , [1, 5, 4]T , [1, 6, 5]T Ot { } t ([1, 4, 4]T ]) = [0, 1, 6]T , [1, 0, 1]T , [1, 1, 0]T , [1, 2, 6]T , [1, 3, 5]T , [1, 4, 4]T , [1, 5, 3]T , [1, 6, 2]T Ot { }

22 3.4. Examples

Figure 3.9: The corresponding degenerate conﬁguration in the real case.

The Poncelet 8-gon is [1, 2, 3]T [1, 4, 5]T [1, 3, 5]T [1, 5, 3]T [1, 5, 4]T → → → → [1, 3, 2]T [1, 4, 2]T [1, 2, 4]T [1, 2, 3]T . → → → → Since all eight points of Os are used in the Poncelet 8-gon, the pair (Ot,Os) is clearly a Poncelet 8-pair.

Poncelet 4-pair: Let us consider the oval Ot given by 2 2 2 Ot := V(x + 4y + z ).

The tangents of Ot are given by t ([1, 2, 2]T ]) = [0, 1, 3]T , [1, 0, 3]T , [1, 1, 6]T , [1, 2, 2]T , [1, 3, 5]T , [1, 4, 1]T , [1, 5, 4]T , [1, 6, 0]T Ot { } t ([1, 2, 5]T ]) = [0, 1, 4]T , [1, 0, 4]T , [1, 1, 1]T , [1, 2, 5]T , [1, 3, 2]T , [1, 4, 6]T , [1, 5, 3]T , [1, 6, 0]T Ot { } t ([1, 6, 3]T ]) = [0, 1, 6]T , [1, 0, 2]T , [1, 1, 1]T , [1, 2, 0]T , [1, 3, 6]T , [1, 4, 5]T , [1, 5, 4]T , [1, 6, 3]T Ot { } t ([1, 6, 4]T ]) = [0, 1, 1]T , [1, 0, 5]T , [1, 1, 6]T , [1, 2, 0]T , [1, 3, 1]T , [1, 4, 2]T , [1, 5, 3]T , [1, 6, 4]T Ot { } t ([1, 1, 3]T ]) = [0, 1, 1]T , [1, 0, 2]T , [1, 1, 3]T , [1, 2, 4]T , [1, 3, 5]T , [1, 4, 6]T , [1, 5, 0]T , [1, 6, 1]T Ot { } t ([1, 1, 4]T ]) = [0, 1, 6]T , [1, 0, 5]T , [1, 1, 4]T , [1, 2, 3]T , [1, 3, 2]T , [1, 4, 1]T , [1, 5, 0]T , [1, 6, 6]T Ot { } t ([1, 5, 2]T ]) = [0, 1, 4]T , [1, 0, 3]T , [1, 1, 0]T , [1, 2, 4]T , [1, 3, 1]T , [1, 4, 5]T , [1, 5, 2]T , [1, 6, 6]T Ot { } t ([1, 5, 6]T ]) = [0, 1, 3]T , [1, 0, 4]T , [1, 1, 0]T , [1, 2, 3]T , [1, 3, 6]T , [1, 4, 2]T , [1, 5, 5]T , [1, 6, 1]T Ot { } We can construct two Poncelet 4-gons, namely [1, 2, 3]T [1, 3, 2]T [1, 5, 3]T [1, 4, 2]T [1, 2, 3]T , → → → → [1, 2, 4]T [1, 3, 5]T [1, 5, 4]T [1, 4, 5]T [1, 2, 4]T . → → → →

23 3.4. Examples

Since all eight points of Os are used in one of the two Poncelet 4-gons, the pair of ovals (Ot,Os) is clearly a Poncelet 4-pair.

Poncelet 0-pair: Consider the oval Ot given by

2 2 2 Ot := V(x + y + 2z ) which consists of the points

[1, 2, 1]T , [1, 2, 6]T , [1, 3, 3]T , [1, 3, 4]T , [1, 4, 3]T , [1, 4, 4]T , [1, 5, 1]T , [1, 5, 6]T . { } T The tangent of Ot in the point [1, 2, 1] is given by

1 0 0 1 1 0 1 0 2 = 2 , 0 0 2 1 2 so all points [x, y, z]T in PG(2, 7) which satisfy x + 2y + 2z = 0 mod 7. Since this is T not fullﬁlled by any of the points on Os, [1, 2, 1] is an inner point of Os. It can be checked that all points of Os are inner points of Ot, which makes (Ot,Os) a Poncelet 0-pair.

24 Chapter 4

Poncelet’s Theorem in ﬁnite projective planes of order 9

The results presented in this section are published in [22].

4.1 Poncelet’s Theorem in PG(2, 9)

The main goal in this section is to show the following statement. Theorem 4.1.1. The ﬁnite projective Desarguesian plane of order 9 is a Poncelet plane.

In planes of order 9, ovals consist of 10 points. Recall that in the case of a Desargue- sian plane, every oval is in fact a conic. We want to emphasize that we are working with the same objects in all four planes, i.e. with ovals, hence we stick to the term oval rather than conic even in this section about PG(2, 9). To prove that PG(2, 9) is a Poncelet plane, it is enough to show that if a 3-sided or a 4-sided Poncelet polygon can be constructed for a pair of ovals (Ot,Os), then this pair is a Poncelet 3-pair or 4-pair, respectively. The ﬁrst step is to show the following result for Poncelet triangles. n Theorem 4.1.2. Let (Ot,Os) be a pair of ovals in PG(2, p ) such that a 3-sided Poncelet polygon can be constructed. Then no m-sided Poncelet polygon for 4 m n + 1 for the same pair of ovals can be constructed. ≤ ≤ To see this, we need some preliminary results. Lemma 4.1.3. Let A, B, C, D, E and F be the six vertices of a nondegenerate hexagon. Assume that the points P = AB DE,Q = BC EF,R = CD AF ∩ ∩ ∩ are collinear. Then the points A, B, C, D, E and F lie on an oval.

Note that the theorem above, also known as the Braikenridge-Maclaurin theorem, is the converse statement of Pascal’s Theorem. All these classical, well-known results can for example be found in [11].

25 4.1. Poncelet’s Theorem in PG(2, 9)

Theorem 4.1.4 (Pascal’s Theorem). Let A, B, C, D, E and F be six points lying on an oval in PG(2, pn), such that they form a nondegenerate hexagon, i.e. no three of these points are collinear. Then the three intersection points of the opposite edges, i.e. the points

P = AB DE,Q = BC EF,R = CD AF ∩ ∩ ∩ are collinear. The line connecting the three points P , Q and R is called Pascal’s Line (see Figure 4.1).

Figure 4.1: Pascal’s Theorem

Proof of Lemma 4.1.3. We choose coordinates such that

A = [1, 0, 0]T ,C = [0, 1, 0]T ,E = [0, 0, 1]T .

Since the points A, B, C, D, E and F lie on a (nondegenerate) hexagon, no three of the points are collinear. Because of this, all coordinates of the remaining three points are nonzero, so we have

T T T B = [1,B2,B3] ,D = [1,D2,D3] ,F = [1,F2,F3] with B2,B3,D2,D3,F2,F3 = 0. Moreover, we have B2 = D2, D2 = F2, B2 = F2, 6 6 6 6 B3 = D3, D3 = F3 and B3 = F3. To see this, assume B2 = D2. In this case, the points6 6 6 T T T B = [1,B2,B3] ,D = [1,B2,D3] ,E = [0, 0, 1] are collinear, since they are all incident with the line g = [ B2, 1, 0]. This leads to a contradiction because of the assumption that the six points− lie on a (nondegenerate) hexagon. The same can be shown analogously for the other coordinates. Using Lemma 2.3.1, we get the connecting lines

AF = [0,F3, F2], AB = [0,B3, B2], BC = [B3, 0, 1] − − − CD = [D3, 0, 1], DE = [D2, 1, 0], EF = [F2, 1, 0]. − − − Using Lemma 2.3.1 once more, we obtain

T T T P = [B2,B2D2,B3D2] ,Q = [1,F2,B3] ,R = [F3,D3F2,D3F3] .

26 4.1. Poncelet’s Theorem in PG(2, 9)

By assumption, these points are collinear, hence we have

det(P, Q, R) = 0. (4.1)

Since no three of the six points A, B, C, D, E and F are collinear, we can deﬁne an oval using ﬁve of them. Let us look for the oval through A, B, C, D and E. Since A, C and E lie on this oval, we get O

= V(xy + exz + fyz). O Using B and D, we ﬁnd for e and f the expressions

B2(B3D2 D2D3) B3D2 + B2D3 e = − and f = − . − B3D3(B2 D2) B3D3( B2 + D2) − − These two parameters e and f are well deﬁned, since B3,D3 = 0 and B2 = D2. It remains to show that F lies on this oval as well, hence we have6 to show 6

F2 + eF3 + fF2F3 = 0.

Using (4.1), we deduce

B2B3D2F3 B2D2D3F3 F2 = − , B2B3D3 B3D2D3 + B3D2F3 B2D3F3 − − which is again well-deﬁned, since

B2B3D3 B3D2D3 + B3D2F3 B2D3F3 = 0 − − ⇔ B2D2F3(B3 D3) = 0. −

As B2,D2,F3 = 0 and B3 = D3, the above expression is not equal to zero. We 6 6 indeed get F2 + eF3 + fF2F3 = 0.

Note that all ﬁnite projective Desarguesian planes are self-dual (see [20]) and hence, we may consider the dual form of Lemma 4.1.3, which is known as Brianchon’s Theorem. Corollary 4.1.5 (Brianchon’s Theorem). Let A, B, C, D, E and F be the six vertices of a nondegenerate hexagon. Moreover, let the lines AD, BE and CF meet in one point, the Brianchon point. Then there exists an oval such that the lines AB, BC, CD, DE, EF and FA are tangents of the oval.

Lemma 4.1.6. Let Os be an oval with two inscribed triangles, denoted by ACE and BDF , such that no three of the vertices are collinear. Then there exists4 an 4 oval Ot, such that the sides of the two triangles are tangents of Ot.

This result was used in [41] to prove Poncelet’s Theorem in the real projective plane for triangles. Since the arguments used there cannot be applied to the ﬁnite projective plane, we have to give an alternative proof here. Proof of Lemma 4.1.6. Let

A = [1, 0, 0]T ,C = [0, 1, 0]T ,E = [0, 0, 1]T

27 4.1. Poncelet’s Theorem in PG(2, 9)

be on Os, i.e Os = V(xy + exz + fyz), with e = 0 and f = 0. For every other point on this oval, all three coordinates are nonzero.6 In particular,6 we have (by scaling if necessary)

B = [1,B2,B3]T ,D = [1,D2,D3]T ,F = [1,F 2,F 3]T with B2,B3,D2,D3,F2 and F3 nonzero. The sides of the triangles ACE and BDF are denoted by 4 4

ACE : g1 = AC, g3 = CE, g5 = EA, 4 BDF : g2 = BD, g4 = DF , g6 = FB. 4 Explicitly, we have

g1 = [0, 0, 1], g2 = [ B3D2 + B2D3,B3 D3, B2 + D2], − − − g3 = [1, 0, 0], g4 = [ D3F2 + D2F3,D3 F3, D2 + F2], − − − g5 = [0, 1, 0], g6 = [B3F2 B2F3, B3 + F3,B2 F2]. − − − The intersection points of these lines are given by

A1 = g6 g1,A2 = g1 g2,A3 = g2 g3, ∩ ∩ ∩ A4 = g3 g4,A5 = g4 g5,A6 = g5 g6. ∩ ∩ ∩ This leads to

T T A1 = [B3 F3,B3F2 B2F3, 0] ,A2 = [ B3 + D3, B3D2 + B2D3, 0] , − − T − − T A3 = [0, B2 + D2, B3 + D3] ,A4 = [0,D2 F2,D3 F3] , − − T − − T A5 = [D2 F2, 0, D3F2 + D2F3] ,A6 = [B2 F2, 0, B3F2 + B2F3] . − − − −

We would like to ﬁnd an oval Ot, such that the lines g1, . . . , g6 are tangents of it. By Brianchon’s Theorem (Corollary 4.1.5), we know that this is equivalent to showing that A1A4, A2A5 and A3A6 meet in one point. Using Lemma 2.3.1, we obtain

2 A1A4 = [B3D3F2 B2D3F3 B3F2F3 + B2F3 , − − 2 B3D3 + B3F3 + D3F3 F ,B3D2 B3F2 D2F3 + F2F3] − − 3 − − 2 2 A2A5 = [B3D2D3F2 B2D3F2 B3D2F3 + B2D2D3F3, − 2 − B3D3F2 + D3F2 + B3D2F3 D2D3F3, − 2 − B3D B2D2D3 B3D2F2 + B2D3F2] 2 − − 2 A3A6 = [B2B3F2 B3D2F2 B2 F3 + B2D2F3, − − 2 B2B3 + B2D3 + B3F2 D3F2,B B2D2 B2F2 + D2F2]. − − 2 − − Observe, that the points B,D and F lie on the original oval, which means that they satisfy eB3 eD3 eF3 B2 = − ,D2 = − ,F2 = − . 1 + fB3 1 + fD3 1 + fF3

28 4.1. Poncelet’s Theorem in PG(2, 9)

Note that we have 1 + fB3 = 0, 1 + fD3 = 0 and 1 + fF3 = 0. To see this, assume 6 1 6 6 1 + fB3 = 0. It follows B3 = f and using the equation xy + exz + fyz = 0 which − e determines the oval, we obtain f = 0, contradicting the fact that e = 0 and f = 0. Finally it follows that the three− lines are concurrent, since 6 6

det(A1A4, A2A5, A3A6) = 0. 2 Proof of Theorem 4.1.2. Let (Ot,Os) be a pair of ovals, such that there exists a triangle which consists of tangents of Ot and the vertices are on Os. Let the lines of this triangle be denoted by t1, t2 and t3.

Assume that there exists another closed polygon using tangents of Ot with vertices on Os. Since we need at least three vertices on Os, such that the lines connecting these are tangent to Ot for the new polygon, we can assume the existence of at least three such lines. Hence, we start with s1, which is a tangent of Ot and joins two points of Os, denoted by S1 and S2.

By assumption, there exists another line s2, which joins S2 with another point of Os and which is a tangent of Ot. Let s2 intersect Os in S2 and S3. The claim is now, that the line connecting S1 and S3, denoted by s3, is a tangent of Ot as well. Assume the opposite (Figure 4.2), i.e. assume that s3 is not a tangent of Ot.

Figure 4.2: We assume the opposite assumption, i.e. that the line connecting S1 and S3 is not tangent to Ot.

By Lemma 4.1.6, there exists an oval O˜ such that t1, t2, t3, s1, s2 and s3 are tangents of O˜. But t1, t2, t3, s1 and s2 are tangents of Ot as well. Since an oval is uniquely determined by ﬁve of its tangents, we have Ot = O˜. Hence, every other polygon which can be closed is a 3-sided Poncelet polygon as well. This shows that (Ot,Os) is a Poncelet 3-pair.

Let us quickly have a look at what happens in the case of a degenerate Poncelet triangle. Given two ovals Ot and Os such that a Poncelet triangle can be constructed with vertices A, B and C on Os and the sides tangent to Ot. Now assume that Ot and Os intersect in some point P and the tangent of Ot in P intersects Os in some

29 4.1. Poncelet’s Theorem in PG(2, 9)

other point Q. The goal is to show that the tangent of Ot through Q is a tangent of Os as well. In that case, the point Q should be counted double and the points P and Q form a degenerate triangle (see Figure 4.3).

Figure 4.3: The tangent of Os in Q is tangent to Ot.

So, assume the opposite, i.e. let the tangent of Ot through Q intersect the oval Os in some other point R different from Q. Now the points A, B, C, P , Q and R are six different points on Os. The points A, B and C form a Poncelet triangle and PQ and QR are tangents of Ot. Since we are given five tangents of Os, with the same argument as above, we can show that PR is a tangent of Ot. But there is only one tangent of Ot through the point P and therefore, P , Q and R are collinear, which leads to a contradiction. Note that we did not restrict ourselves to PG(2, 9) in the proof of Theorem 4.1.2, and hence it applies to all finite projective Desarguesian planes PG(2, pn). Now we turn our attention to 4-sided Poncelet polygons. Since there are only 10 points on an oval in PG(2, 9), it suffices to prove the following.

Theorem 4.1.7. Let (Ot,Os) be a pair of ovals in PG(2, 9) such that a 4-sided Poncelet polygon can be constructed. Then no m-sided Poncelet polygon for m = 5 or m = 6 for the same pair of ovals can be constructed.

Proof. First, recall the following fundamental theorem for PG(2, pn) [20].

Theorem 4.1.8. Let P1,P2,P3,P4 and Q1,Q2,Q3,Q4 be sets of four points, such that no three points{ of the same} set are{ collinear. Then} there exists a unique projective map T , such that T (Pi) = Qi, for 1 i 4. ≤ ≤

We may assume that the pair of ovals (Ot,Os) carries the Poncelet quadrilateral

A = [1, 1, 0]T ,B = [1, 0, 1]T ,C = [1, 1, 0]T ,D = [1, 0, 1]T . − −

Then Os is given by

2 2 2 Os = V(ax + by + cz + dxy + exz + fyz), (4.2)

30 4.1. Poncelet’s Theorem in PG(2, 9) for a, b, c, d, e, f GF (9) and the associated matrix is nonsingular. We want the ∈ points A, B, C and D to lie on the oval Os, which gives the four conditions a + b d = 0, a + c e = 0, a + b + d = 0, a + c + e = 0. − − Since a = 0 would lead to a singular matrix, we may scale a = 1 and get

2 2 2 Os(f) = V(x y z + 2fyz) (4.3) − − for f = 1, which ensures that the associated matrix is nonsingular. It is enough 6 ± to consider ovals of the above form for Os. Now, the four lines AB = [1, 1, 1], BC = [1, 1, 1], CD = [1, 1, 1], DA = [1, 1, 1] − − − − need to be tangents of Ot. To find the corresponding ovals, we first find the ovals which contain the four points [1, 1, 1]T , [1, 1, 1]T , [1, 1, 1]T and [1, 1, 1]T . We have to solve the system of equations − − − −

a + b + c + d + e + 2f = 0 a + b + c d + e 2f = 0 − − a + b + c d e + 2f = 0 − − a + b + c + d e 2f = 0. − − We immediately obtain d = e = f = 0 and after scaling a = 1, we end up with

V(x2 + by2 (1 + b)z2). − Since we need an oval with tangents [1, 1, 1], [1, 1, 1], [1, 1, 1] and [1, 1, 1] rather than points, we have to take the oval which corresponds− − to− the inverse matrix− of the matrix associated to V(x2 + by2 (1 + b)z2) which is − 2 1 2 1 2 Ot(b) = V x + y z , (4.4) b − 1 + b for b = 0, 1. To exclude the simultaneous existence of a 4-sided Poncelet polygon and a6 5-sided− or 6-sided Poncelet polygon, respectively, it is enough to consider pairs of the ovals described above. Before we start analyzing these oval pairs, we quickly collect some facts about the ﬁeld we are working in. We can write GF (9) as

GF (9) = 0, 1, a, a2, a3, a4, a5, a6, a7 , where a is a root of the polynomial f(x) = x2 + x 1, which is irreducible over GF (3). Addition and multiplication obey the rules −

a2 + a = 1, aiaj = ai+j, a8 = 1.

Therefore, we consider pairs of ovals of the form (Ot(b),Os(f)) for b 1, a, a2, a3, a5, a6, a7 and f 0, a, a2, a3, a5, a6, a7 . ∈ ∈

By inspecting the pair (Ot(b),Os(f)) we obtain exactly the same results as for (Ot(b),Os( f)), because changing the sign of the y-coordinate has the eﬀect − T T [x, y, z] Os(f) [x, y, z] Os( f) ∈ ⇐⇒ − ∈ − 31 4.1. Poncelet’s Theorem in PG(2, 9) and T T [x, y, z] Ot(b) [x, y, z] Ot(b). ∈ ⇐⇒ − ∈ Hence, it is enough to consider f 0, a, a2, a3 . ∈ { } 1 1 Moreover, when calculating the coeﬃcients b and 1+b for all values of b above, we obtain

2 2 2 Ot(1) = V(x + y + z ) 2 7 2 5 2 Ot(a) = V(x + a y + a z ) 2 2 6 2 2 Ot(a ) = V(x + a y + az ) 3 2 5 2 7 2 Ot(a ) = V(x + a y + a z ) 5 2 3 2 2 2 Ot(a ) = V(x + a y + a z ) 6 2 2 2 3 2 Ot(a ) = V(x + a y + a z ) 7 2 2 6 2 Ot(a ) = V(x + ay + a z ).

By symmetry, we have

T T [x, y, z] Ot(b) [x, z, y] Ot(b) ∈ ⇔ ∈ and T T [x, y, z] Os(f) [x, z, y] Os(f). ∈ ⇔ ∈ Therefore, it is enough to consider b 1, a, a2, a5 . ∈ { } All we have to do is to exclude the existence of a 5-sided and a 6-sided Poncelet polygon for the following 16 oval pairs

2 5 2 3 (Ot(b),Os(f)), b 1, a, a , a , f 0, a, a , a . ∈ ∈

By direct inspection, we can count the number of all points on Os(f) that are incident with a tangent of Ot(b), which are called exterior points of Ot(b) on Os(f). Table 4.1 contains these numbers.

2 5 Ot(1) Ot(a) Ot(a ) Ot(a ) Os(0) 8 6 4 4 Os(a) 4 6 4 8 2 Os(a ) 8 4 5 5 3 Os(a ) 4 6 8 4

Table 4.1: Number of points on Os(f) which are incident with a tangent of Ot(b)

Since by construction all of these pairs already form one 4-sided Poncelet polygon, the condition of 9 or 10 exterior points of Ot(b) on Os(f) is necessary to ﬁnd a 5-sided or 6-sided Poncelet polygon, respectively. Since there are at most 8 exterior points of Ot(b) on Os(f), we can exclude their existence. This completes the proof of PG(2, 9) being a Poncelet plane.

As in the case of Poncelet triangles, let us quickly turn our attention to degenerate cases. We see for example, that Os(0) contains six external points of Ot(a). By construction, we expect four of these points to form a Poncelet 4-gon.

32 4.2. Poncelet’s Theorem in the ﬁnite projective planes over S

But what happens to the other two external points? Let us analyze this situation. The ovals intersect in the four points

[1, a2, a2]T , [1, a2, a6]T , [1, a6, a2]T , [1, a6, a6]T { } and all remaining six points on Os(0) are external points of Ot(a). Clearly, the points [1, 0, 1]T , [1, 0, 1]T , [1, 1, 0]T , [1, 1, 0]T { − − } form a Poncelet 4-gon. 2 T Now, take the point [0, 1, a ] on Os(0). There are two tangents of Ot(a) through [0, 1, a2]T , namely the tangents in the points [1, a2, a6]T and [1, a6, a2]T , respectively, 6 T which are intersection points of Os(0) and Ot(a). Similarly for the point [0, 1, a ] . Hence, we end up with one proper Poncelet 4-gon and two degenerate Poncelet 4-gons. The corresponding situation in the real case is visualized in Figure 3.6, Section 3.

4.2 Poncelet’s Theorem in the ﬁnite projective planes over S

4.2.1 The miniquaternion near-ﬁeld S

We describe the near-field we use to construct the three non-Desarguesian finite projective planes, denoted by Ω, ΩD and Ψ. All notations and well-known properties are based on [42]. A finite near-field is a system (S, +, ), such that

(i) S is ﬁnite, (ii)( S, +) is a commutative group with identity 0, (ii) the multiplication is a group operation on S 0 with identity 1 and \{ } (iv) the multiplication is right distributive over the addition, i.e.

(m + n)l = ml + nl, m, n, l S. ∀ ∈ Note that we do not necessarily need the multiplication to be commutative, hence the left distribution law does not have to be valid for all elements in the near-field. This is exactly the property used in the construction of the non-isomorphic planes of order 9. We need to describe addition and multiplication for a near-field with nine elements. For this, consider S = 0, 1, 1, i, i, j, j, k, k { − − − − } where we define j := 1 + i, k := 1 i. −

33 4.2. Poncelet’s Theorem in the ﬁnite projective planes over S

We can view the nine elements as elements over the basis 1, i and call { } D := 0, 1, 1 { − } the real elements and S∗ := i, i, j, j, k, k { − − − } the complex elements. By the definition of j and k above and taking the coefficients of 1 and i modulo 3, we are able to add any two elements in the near-field (Table 4.2). For the multiplication in S, we want to end up with a non-commutative operation. We use the relations

i2 = j2 = k2 = 1, ij = k = ji, jk = i = kj, ki = j = ik − − − − which again enables us to multiply any two elements in S (Table 4.2).

Table 4.2: Addition and multiplication of elements in S

+ 0 1 1 i i j j k k − − − − 0 0 1 1 i i j j k k 1 1 1− 0 j− k k −i j− i 1 1− 0 1 k j− i− k −i j − i − i j k − i −0 k 1− 1 j i i k −j −0 i 1 −k j −1 −j −j k− i k 1 j −0 1 − i j j − i k 1 k −0 j− i −1 −k −k −j i −1 −j 1 i k 0 k k− i− j j 1 − i 1− 0 k − − − − −

0 1 1 i i j j k k − − − − 0 0 0 0 0 0 0 0 0 0 1 0 1 1 i i j j k k 1 0 1− 1 i− i j− j k− k − i 0 − i i −1 1 −k k −j j i 0 i− i −1 1 k− k− j j −j 0 −j j k− k −1 1 i −i j 0 j− j− k k −1 1 i− i −k 0 −k k j −j i− i −1 1 k 0 k− k j− j− i i −1 1 − − − − −

Note that this is the multiplication law of the quaternion group, which explains the name ‘miniquaternion near-ﬁeld’. Note also, that in this near-ﬁeld the left distribution law does not hold in general, as for example i(j + k) = i( 1) = i but ij + ik = k j = i. − − −

34 4.2. Poncelet’s Theorem in the ﬁnite projective planes over S

4.2.2 The plane Ω

In order to construct the finite projective plane Ω of order 9 using the near-field S we start with an affine plane and extend it to a projective plane. We distinguish between so-called proper points on Ω, which are in affine form (x, y), and ideal points, which connect the parallel lines. More precisely, the points of Ω are given by

- 81 proper points of the form (x, y), for x, y S, ∈ - 9 ideal points of the form (1, y, 0) for y S and ∈ - one ideal point of the form (0, 1, 0).

The lines of Ω are given by

- 81 proper lines of the form y = xµ + ν for µ, ν S, denoted by (µ, ν), ∈ - 9 proper lines of the form x = λ for λ S, denoted by (λ), and ∈ - one ideal line, denoted by I.

For the incidence relation, we have:

- On the line y = xµ + ν there are nine proper points (x, y) and the ideal point (1, µ, 0).

- On the line x = λ there are nine proper points (λ, y) and the ideal point (0, 1, 0).

- All 10 ideal points are on the ideal line I.

It is crucial to consider y = xµ + ν instead of y = µx + ν, since multiplication is not commutative. It can be shown that the above defined points and lines with the incidence relation give indeed a finite projective plane of order 9 (see [42]). Now we want to find ovals in the plane Ω, i.e. we want to find sets of 10 points, no three of which are collinear. Compared to finite projective coordinate planes, it is much harder to find ovals in this plane, since ovals cannot be described by quadratic forms. Hence, for a set of 10 points, we have to search all lines connecting them to be sure that no three of them are collinear. For any point on the oval, we end up with nine lines connecting this point to the other points on the oval, and all these secants have to be different. Similarly it is much harder to find the tangent in a point of the oval.

Nevertheless, the set O1 given by

 0 1  1 0 1 i i j j k  , , , , , , , , 1 , 0 −i 1 0 −j j −1 k −i      − − − 0 0  is an example of an oval in Ω. To see this, we have to calculate all secants and check, whether they are diﬀerent. Table 4.3 shows all secants and tangents of O1.

35 4.2. Poncelet’s Theorem in the ﬁnite projective planes over S

Table 4.3: The diagonal entries are the tangents of O1 and the other entries are the secants of the points of O1 listed in the ﬁrst row and column.

0 1 1 0 1 i i j j k O1 − − − − 1 0 i 1 0 j j 1 k i     − − − 0 0 1 j k i 1 j k i 1 0 − − − − − ( 1) i k 1 i k 1 j 0 j − i − − − − 0 i 1 j 1 j k i 0 − − − − (0) 1 1 1 1 1 1 1 1 1 1 j 1 i k j k 0 − − − (1) 0 j 1 i k j k 0 − − − i i k i k j 0 − − − ( i) j k 0 i j i − j − − − i j 1 i k 0 − (i) j i k k j j − − − j i 1 j 0 − − ( j) 1 j i 0 − 1 − − j j 1 0 − − (j) k i 1 k − − k i 0 − − ( k) i k − i − − − 0 1 (k) I 0 1 0 0   k 0 −

Recall that in PG(2, 9), Pascal’s Theorem plays a central rˆolein the proof of Pon- celet’s Closure Theorem. We will see that Pascal’s Theorem is not true in general in the plane Ω. For this, take for example the six points

A = ( 1, i),B = (0, 1),C = (1, 0),D = (i, j),E = ( k, i),F = (0, 1, 0). − − −

These points all lie on O1, hence they lie indeed on a non-degenerate hexagon. We have

AB : y = xk + 1, BC : y = x + 1, CD : y = xi + 1 − − DE : y = xk j, EF : x = k, FA : x = 1. − − − The intersection points we need in Pascal’s Theorem are given by

P = (1, k, 0),Q = ( k, j),R = ( 1, i). − − − − These are not collinear, as the line through P and Q is given by y = xk + k and the line through P and R is y = xk + j.

Theorem 4.2.1. The ﬁnite projective plane Ω of order 9 is not a Poncelet plane.

36 4.2. Poncelet’s Theorem in the ﬁnite projective planes over S

Proof. We have to ﬁnd a pair of ovals (Ot,Os) which carries at the same time an n-sided and an m-sided Poncelet polygon with m = n and m, n 3. For Ot we take 6 ≥ O1, and for Os we choose the oval

  1  1 0 i i j j j k k  , , , − , , , − , − ,  j , j j i k j 0 j i k −  − − − 0 0  Again, we have to ensure that all secants of this set are diﬀerent. For this, see Table 4.4.

Table 4.4: The diagonal entries indicate the tangents of Os, the other entries indicate the secants.  1  1 0 i i j j j k k Os − − −  j j j i k j 0 j i k − − − − 0 0 0 i i 0 1 1 k k j j (0) − − − − j j j j j j j j j j i 1 k i k 0 1 j j − (i) − − − i i k k 1 i 0 1 j − − − − i 1 k k i 1 0 j j − − − k 1 1 i i j k 0 k − − − − − j 1 1 i i j j − ( j) − − − j − j 0 1 i k i − − − − j 0 i k j j (j) − − 0 0 k i i 1 − j 0 k i k j − − j j k 1 k i − − − − k 1 j j − ( k) − i 1 − i 0 − k 1 j j − − − k k j 1 − − −  1  j  j − I − i 0 1 j j   k 0 −

For this oval pair (Ot,Os), we can now ﬁnd simultaneously a 5-sided Poncelet polygon and a 4-sided Poncelet polygon. To see this, start with the point (0, j) on Os. The line joining (0, j) and (1, j, 0) is the line y = xj + j, as given in Table 4.4. − − This line is a tangent of oval Ot, namely the tangent in the point (1, 0). Continuing with (1, j, 0), we see that the line joining (1, j, 0) and ( k, i) is y = xj i, − − − − − which is the tangent of Ot in the point (j, k). Moreover, the line joining ( k, i) and − (j, 0) is y = xi+k, which is the tangent of Ot in ( i, j). The line joining (j, 0) and − − (i, i) is y = xi k, which is again a tangent of Ot, namely in the point ( k, i). For the last− step,− we see that the line joining (i, i) and (0, j) is y = xi + j−, which− is the tangent of Ot in the point ( j, 1). This gives a 5-sided Poncelet polygon − − 37 4.2. Poncelet’s Theorem in the ﬁnite projective planes over S

for this pair. Similarly, by starting with (i, k) on Os, a 4-sided Poncelet polygon occurs. To summarize the result, we have the− 5-sided Poncelet polygon   1 0 (1,0) (j,k) k (−i,−j) j (−k,−i) i (−j,−1) 0 j j   −i 0 i j −−→ −0 −−→ −−−−→ −−−−→ −−−−→ and the 4-sided Poncelet polygon   1 i (−1,i) (i,j) j (0,1) k (1,0,0) i j − . k −−−→ −−→ j −−→ k −−−→ k − 0 − − −

The remaining point ( j, j) on Os is an inner point of Ot, which means that it is not − incident with any tangent of Ot. This pair (Ot,Os) is therefore no Poncelet m-pair for any possible value of m, which shows that Ω is not a Poncelet plane.

This pair of ovals gives even one more proof of Ω not being a Poncelet plane, namely by changing the rôlesof Ot and Os. If we consider points on Ot and tangents of Os, we find simultaneously a 4-sided and a 3-sided Poncelet polygon, namely   0 0 (−k,i) i (−k,−k) j (−j,j) (0,j) 0 − 1 1 −−−→ j −−−−→ 1 −−−→ −−→ 1 − 0 and   1 1 (i,−k) i (j,−j) (j,0) 1 − 0 . 0 −−−→ j −−−→ −−→ 0 − 0 This shows that in Ω, Poncelet’s Theorem is not even partially true for only 3-sided polygons or only 4-sided polygons, as we could find counter examples in both cases.

4.2.3 The plane ΩD

Now, we want to look at the dual plane of Ω, that is, we want to change the rˆole of the points and lines constructed for Ω to obtain the plane ΩD. Note that ΩD is indeed not isomorphic to Ω (see [42]). Recall the incidence relation for Ω, given by (x, y) (µ, ν) y = xµ + ν. ∈ ⇔ By changing the rˆolesof points and lines, (x, y) denotes a line in ΩD and (µ, ν) denotes a point in ΩD. Hence, the incidence relation becomes (µ, ν) (x, y) ν = xµ + y. ∈ ⇔ − For the line x = λ, the incidence relation stays the same. Moreover, the ideal line is not changed either. By adjusting the notation above by taking x instead of x, we obtain the incidence relations for ΩD. Namely, on the line y = µx + ν, there− are nine proper points (x, y) and the ideal point (1, µ, 0). On the line x = λ, there are nine proper points (λ, y) and the ideal point (0, 1, 0). All 10 ideal points are on the ideal line I.

38 4.2. Poncelet’s Theorem in the ﬁnite projective planes over S

Theorem 4.2.2. The ﬁnite projective plane ΩD of order 9 is not a Poncelet plane.

Proof. In order to prove that ΩD is not a Poncelet plane either we can dualize the ovals from the previous section. For this, recall the oval Ot given by  0 1  1 0 1 i i j j k  , , , , , , , , 1 , 0 −i 1 0 −j j −1 k −i      − − − 0 0  in Ω. The set of tangents of this oval is given by the lines j i j i j i j i 0 , , , , , , , , (k), . k −1 −j k i j −i −k k − − − − Note that for the proper lines (µ, ν), we have to take the minus sign for the x- D coordinate. This gives the oval Ot j i j i j i j i 0 , , , , , , , , (k), −k 1 j −k −i −j i k k − − − − in ΩD. D D Similarly, the dualization of the oval Os leads to another oval in Ω , namely Os given by 1  1    1 1 0 0 1 1 j j 0 , , − ,  j , , , − , , , − . i 1 − 0 j 1 k i k  0 − − 0 − − 

Now we can dualize the n-sided Poncelet polygons as well. Recall that for Ot and Os in Ω, we had the 4-sided Poncelet polygon   1 i (−1,i) (i,j) j (0,1) k (1,0,0) i j − k −−−→ −−→ j −−→ k −−−→ k − 0 − − − for vertices on Os.

The tangent of Ot in ( 1, i) is (j, k), hence we can start with the corresponding D − point ( j, k) on Ot . The connection of two points of Os in Ω is now the same − D D as the intersection of two tangents of Os . Hence, in Ω , we have to take vertices D D on Ot and tangents of Os . The next tangent we consider, namely the tangent D in (i, j), corresponds to ( j, i), a point on Ot . The line connecting these two − − D points is the tangent of Os in (1, j, 0) in Ω, which gives ( j, k) interpreted in Ω . When performing this operation with all points obtained− before,− we get the 4-sided Poncelet polygon j (−j,−k) j (0,−j) i (1,k) 0 (−1,−1) j − − − k −−−−→ i −−−→ 1 −−→ k −−−−→ k − − and 5-sided Poncelet polygon j (j,i) j (−1,1) i (0,0) i (1,−i) i (1,0,0) j − − . j −−→ i −−−→ k −−→ k −−−→ j −−−→ j − − Hence, ΩD is not a Poncelet plane either.

39 4.2. Poncelet’s Theorem in the ﬁnite projective planes over S

4.2.4 The plane Ψ

Similarly to the construction of the plane Ω, we can define the plane Ψ using again the fact that S is not left distributive. We make use of the homogeneous approach, unlike the affine approach before. A point is defined as the set of vectors P κ, κ S, κ = 0 , P S3 (0, 0, 0) . A point is called real, if there exists a non-zero{ ∈ κ in S6 , such} that∈ all\{ coordinates} of P κ are in D. Otherwise, the point is called complex. Note that there are 13 real points and 78 complex points. The line through P and Q is defined by

P P κ + Q, κ S . { } ∪ { ∈ } A line is called real if at least two real points are on the line, otherwise complex. We can choose the line at inﬁnity z = 0. All points not on this line can be parameterized by P = (x, y, 1) and all points on the line z = 0 can be seen as Q = (1, κ, 0). We get 13 real lines, namely

- 9 lines of the form y = mx + c, m, c D, denoted by (m, c, 1), ∈ - 3 lines of the form x = c, c D, denoted by (c, 1, 0), and ∈ - one line z = 0, denoted by (0, 0, 0).

The 78 complex lines are given by

- 54 lines of the form y s = κ(x r), r, s D, κ S∗, denoted by (s, r, κ), − − ∈ ∈ - 18 lines of the form y = mx + κ, m D, κ S∗, denoted by (m, κ, 1), and ∈ ∈ - 6 lines x = κ, κ S∗, denoted by (κ, 1, 0). ∈ Note that we have parameterized the lines and points in a diﬀerent way, since for example the vectors (1, 1, i) and ( 1, 1, i) do not represent the same line, but they do represent the same point.− It− can− be shown that these points and lines together with the incidence relations form indeed a ﬁnite projective plane of order 9 which is not isomorphic to Ω or ΩD, and Ψ is self-dual (see [42]).

Theorem 4.2.3. The ﬁnite projective plane Ψ of order 9 is not a Poncelet plane.

Proof. Look at the two ovals Ot

 1  1  0  0 1 1  j  j  k  k  − −   1 ,  1  ,  i , i , 0 , 0 ,  k , k ,  j , j − − − −  1 1 1 1 1 1 1 1 1 1  and Os

 1  1  0  0  1  1 1 1 i k  − −   i  ,  k  ,  1 , 0 ,  1 , 1 , i , k , k , i . − −  1 1 1 1 1 1 1 1 1 1 

Similarly to the approach before, in Table 4.5 and 4.6 we just list all secants and tangents to check that these sets are indeed ovals.

40 4.2. Poncelet’s Theorem in the ﬁnite projective planes over S

Table 4.5: Oval Ot in Ψ  1  1  0  0 1 1  j  j  k  k − − Ot  1  1   i i 0 0  k k  j j −1 1 −1 1 1 1 −1 1 −1 1  1 1  1  1  1  0   1  1  1  1  1 − − − − − − − − −  1 0  1   1  1  1  1   1  1  1  1 −1 1 0 −k −j −1 1 −j −i −k −i − − −  1  1  1   1  0  1   1   1   1   1  − −  1   0   1  1 1  1  1  1  1  1 1 1 −j −k 1 −1 −i −j −i −k − − −  0   0  0  0  0  1  1 1  1  − −  i  1 1  i 1  i  1  1  i −1 −i 0 −1 i −1 k j −1 − − 0  0  0  0  1 1  1  1 − − i  1 i  1  1 i  i   1  1 −i 1 i k 1 1 j − − 1 0 0  0  0  0  0 0 0 0  k k  j j 0 0 1 −1 1 −1 1 1 1 0  0  0  0  0 1 1  1  1  1  1 0 j j k k − −  j   1 j 1  0  −  k  0  1 1  0  −1 k 0 1 i − − j 1 0  1 − k 0 0  1 1 k i −1  k   1 k −  j  0  1 −1 j 0 − k 1 j 0 1 j

41 4.2. Poncelet’s Theorem in the ﬁnite projective planes over S

Table 4.6: Oval Os in Ψ  1  1  0  0  1  1 1 1 i k − − Os  i   k   1 0  1 1 i k k i 1 1 −1 1 −1 1 1 0 1 1  1  1  1  1  0   1  1  0 1 1 0 − − − − −  i   k  1   0   0   1   1  1 0 j i 1 −1 0 j i j k i k 1 1 − − −  1  1  1  0   1  1  1 0 0  1  − − − −  k   i  0   0   1   1  0 1 k  j 1 −1 j k j i i k 1 −1 − − −  0   1  0  0   1  1  1  1  1 − − − − −  1  1 1  1  1  0   0   0   0  −1 −1 0 −1 −1 i k k 1 − − 0 0  1 1 0 0  0  0 − 0 0  0  0 0 0  0  0 1 1 1 1 i k j j −  1  1 1  1  1  1  1 − − − −  1 1 1  1   1   1   1  −1 1 0 i k i k − − 1 0 1 1 1  1  1 1 1 1 1  1  1 1 i k j j − 1  1 0  0   1  − i  1 0  1  1 0 −i 0 −i −i 1  1  1   0  − k  1  1  1 0 −k −k −k i i  1 − k 1  1  1 0 1 k k i 1 1 0

42 4.2. Poncelet’s Theorem in the ﬁnite projective planes over S

Using these tables, we will see that in Ψ we can ﬁnd pairs of ovals which carry an n-sided and an m-sided Poncelet polygon for m = n. Indeed, we are able to ﬁnd the 5-sided Poncelet polygon 6

 1  0  i 1 1  1 − (k,−j,1) (j,−k,1) (0,i,1) (1,0,0) (j,k,1) −  i   1 k i k  i  −−−−−→ − −−−−−→ −−−−→ −−−−→ −−−−→ 1 1 1 0 0 1 and the 3-sided Poncelet polygon

0 1  1  0 (−1,−1,1) (1,0,1) (−1,1,1) 0 1  1 0 . −−−−−−→ −−−−→ − −−−−−→ 1 1 1 1

This shows that Ψ is not a Poncelet plane.

Chapter 5

Some proofs for Poncelet’s Theorem

5.1 A synthetic proof

We present a synthetic proof of Poncelet’s Theorem following the idea of Marcel Berger in [4, Section 16.6]. We work in coordinates to emphasize the subtleties one has to face when working in finite fields. In all what follows, our objects are in PG(2, q) for q an odd prime power. Let us start with some assumptions. We consider C to be a proper conic and we fix a pencil of conics with C . All points are considered to be in the complement of the base of ,F i.e. any point∈ F lies in exactly one element of the pencil. Moreover, all lines consideredF are good with respect to , which means that the line does not contain any points of the base of and is notF contained in an element of . F F Here is the first crucial Lemma to prove Poncelet’s Theorem.

Lemma 5.1.1. Let a, b, c C be distinct points and Γ, Γ0 two conics in , such that Γ is tangent to ab in α∈and Γ0 is tangent to ac in β. Set ∆ = αβ. F

1. There exists an element d C, such that Γ is tangent to cd in cd ∆ and Γ0 is tangent to bd in bd ∆.∈ Moreover, there exists an element Γ00 in∩ tangent to ad in ad ∆ and to∩ bc in bc ∆. (see Figure 5.1) F ∩ ∩ 2. Let a, b, c, d C be distinct points and let Γ be tangent to ab in α and to cd in γ. There∈ exists an element Γ0 tangent∈ F to ac in αγ ac (and so, by the results before, also tangent to bd∈at Fαγ bd). ∩ ∩ Proof. We start with the ﬁrst statement. Without loss of generality, let a, b, c be given by a = [1, 0, 0]T , b = [0, 1, 0]T , c = [0, 0, 1]T , which leads to C = V(d1xy + e1xz + f1yz).

45 5.1. A synthetic proof

Figure 5.1: Existence of d

Since C is assumed to be a proper conic, we have d1 = 0, e1 = 0 and f1 = 0. Let F be any other (possibly degenerated) conic, i.e. F is given6 by 6 6

2 2 2 F = V(a2x + b2y + c2z + d2xy + e2xz + f2yz).

The points a, b, c are assumed not to be in the base of , i.e. they do not lie in any F element of other than C. This implies a2 = 0, b2 = 0 and c2 = 0 and we scale F 6 6 6 a2 = 1. So, we have = (C,F ) and all elements of the pencil are given by F P

C = V(d1xy + e1xz + f1yz), and

2 2 2 Fi = V(x + b2y + c2z + (kd1 + d2)xy + (ke1 + e2)xz + (kf1 + f2)yz) (5.1) for k ranging through GF (q). We are now looking for the element Γ , such that the line ab, given by V(z), is a tangent of some Γ (C,F ). Intersecting∈ F z = 0 with some element in the pencil gives the condition ∈ P 2 2 x + b2y + (kd1 + d2)xy = 0 for k GF (q). As y = 0 would lead to a contradiction, we can assume y = 0 and scale∈ to y = 1. We have to solve a quadratic equation in x and since z =6 0 is a 2 tangent of Γ, the discriminant (kd1 + d2) 4b2 needs to vanish. This gives us the parameters − d2 2√b2 k = − ± . d1

46 5.1. A synthetic proof

Note that this condition shows that for the existence of such an element Γ in the pencil, we need b2 to be a square in GF (q). For Γ we obtain two solutions, depending on the sign, namely

2 2 2 2 p Γ± = V(x + b2y + c z 2 b2xy + eΓ xz + fΓ yz) ± ± ± √ √ −d2±2 b2 −d2±2 b2 for eΓ = e1 + e2 and fΓ = f1 + f2. So, ab is tangent to Γ± in ± d1 ± d1

p T α± := Γ± ab = [ b2, 1, 0] . ∩ ∓ Similarly for ac tangent to Γ0, we obtain

0 2 2 2 2 Γ = V(x + b2y + c z + dΓ0 xy 2√c2xz + fΓ0 yz) ± ± ± ± √ √ −e2±2 c2 −e2+2 c2 0 for d 0 = d1 + d2 and f 0 = f1 + f2. The line ac is tangent to Γ Γ± e1 Γ± e1 ± in 0 T β± := Γ ac = [√c2, 0, 1] . ± ∩ ∓ We obtain p ∆± = αβ = [1, b2, √c2]. ± ±

We claim that the tangent in the intersection point of ∆± with Γ± other than α± goes through c. For this, we look at the polar of c with respect to Γ±, i.e.       2 2√b2 eΓ 0 eΓ ± 2 √b2 2b2 fΓ  0 =  fΓ  . ± eΓ fΓ 2c2 1 2c2

We intersect this polar line with ∆±, which gives       eΓ 1 fΓ√c2 2c2√b2 ± ∓ γ± :=  fΓ   √b2 =  2c2 eΓ√c2  . × ± ∓ 2c2 √c2 eΓ√b2 fΓ ± ± −

Indeed, we have γ± Γ±, which shows that the tangent of Γ± in Γ± ∆± goes through c. ∈ ∩

It remains to show that cγ± is not a tangent of C.

The line   2c2 eΓ√c2 ∓ cγ± =  fΓ√c2 2c2√b2 ∓ 0± is a tangent of C if and only if there exists a P C, such that CP = cd±. ∈ T For any P = [1, y, z] C we have CP = cd± if and only if ∈       0 d1 e1 1 2c2 eΓ√c2 ∓ d1 0 f1 y =  fΓ√c2 2c2√b2 ∓ ± e1 f1 0 z 0

47 5.1. A synthetic proof

Considering the third coordinate leads to e y = 1 . −f1

We have to check whether the point P = [1, e1 , z]T can lie on C. For this, we f1 would need − d e 1 1 = 0, f1 which is a contradiction, since d1 = 0 and e1 = 0. Note that we assumed above that the x-coordinate of P is nonzero.6 In the case6 of a zero x-coordinate, we have P = [0, 1, 0]T or P = [0, 0, 1]T , which leads to a similar contradiction.

Hence, we showed that there exists some element d± C, such that cd± is tangent ∈ to Γ± in γ±. Now, consider the elements a, c, d C. We know that ac is tangent to Γ0 in β and cd is tangent to Γ in γ. By the same∈ argument as before, we obtain an element ˜b on C, such that ˜bd is tangent to Γ0 in Γ0 βγ and ˜ba is tangent to Γ in Γ ˜ba, which must be α. But by our assumptions, ab∩ is tangent to Γ in α. We therefore∩ conclude b = ˜b. Next, we need an element Γ00 , such that bc is tangent to Γ00 in bc ∆ and ad is tangent to Γ00 in ad ∆. Similar∈ F to the discussion above, we know that∩bc is tangent ∩ to Fk in bc ∆ for ∈ F ∩ f2 2√b2c2 k = − ± . f1

Since b2 and c2 are both squares, such an element Fk exists indeed. It remains to show that ad is tangent to Fk in ad ∆. ∩ Recall that we already know that ab is tangent to Γ in α and bc is tangent to Γ00 in δ := bc ∆, where ∆ = αβ = αδ. ∩ By the ﬁrst statement of this lemma, we know that there exists an element d˜ such that ad˜ is tangent to Γ00 in ad˜ ∆ and cd˜ is tangent to Γ in cd˜ ∆. But we already know that for γ = ∆ Γ, we get∩ cγ = cd tangent to Γ in γ. We∩ conclude d = d˜. ∩ For the second statement let us start with a, b, c, d C and Γ . Let us choose a = [1, 0, 0]T , b = [0, 1, 0]T , c = [0, 0, 1]T and d = [1, 1∈, 1]T . Hence,∈ F we obtain

C = V(d1xy + e1xz + f1yz) for d1, e1, f1 = 0 and d1 + e1 + f1 = 0. Let Γ be tangent to ab in α and to cd in γ. 6 0 The claim is that there exists a Γ tangent to ac in αγ ac. We√ have already ∈ F i ∩ −d2±2 b2 seen before that ab tangent to Γ implies that Γ = Fi for a = , i.e. b2 is a d1 square. Moreover, we have the assumption that cd = [1, 1, 0] is tangent to Γ as well, so − 2 d2 2√b2 p 2 − ± (e1 + f1) + (e2 + f2) = 4c2(1 b2) , d1 ± 0 which already implies that c2 is a square. This ensures the existence of Γ such that ac is tangent to Γ0 at ac αγ. ∩

48 5.1. A synthetic proof

Corollary 5.1.2. Let a, b, c C be distinct points and Γ0, Γ00 such that ab is 0 ∈ 00 ∈ F tangent to Γ in γ and ac is tangent to Γ in β. There exist exactly two conics Γ1 and Γ2 , tangent to bc and the tangent point α1 of Γ1 is collinear with β and γ ∈ F and the tangent point α2 of Γ2 is such that aα2, bβ, cγ are concurrent.

Proof. Consider again a = [1, 0, 0]T , b = [0, 1, 0]T and c = [0, 0, 1]T . By assumption, there exists an element Γ0 and Γ00 in the pencil such that ab is tangent to Γ0 and ac is tangent to Γ00. We have seen in the proof ofF Lemma 5.1.1 that this necessarily leads to b2 and c2 being quadratic residues in GF (q). We want to ﬁnd the two elements Γ1 and Γ2 in such that bc, which is given by x = 0, is a tangent to both of them. Hence, we intersectF (5.1) with x = 0, which leads to

2 2 b2y + c2z + (kf1 + f2)yz = 0.

As y = 0, we can scale y = 1 and obtain a quadratic equation in z. The line V(x) is tangent6 to the conic if the discriminant of this quadratic equation vanishes, i.e.

2 (kf1 + f2) 4b2c2 = 0. −

Note that this gives indeed two diﬀerent solutions for k, since b2 and c2 are two nonzero squares in GF (q). This ensures the existence of Γ1 and Γ2 given by 2 2 2 f2 + 2√b2c2 Γ1 = V(x + b2y + c2z + − d1 + d2 xy f1 f2 + 2√b2c2 p + − e1 + e2 xz + 2 b2c2yz) f1 and 2 2 2 f2 2√b2c2 Γ2 = V(x + b2y + c2z + − − d1 + d2 xy f1 f2 2√b2c2 p + − − e1 + e2 xz 2 b2c2yz). f1 −

By intersecting Γ1 with bc, we easily deduce that bc is tangent to Γ1 in α1 = h q iT h q iT b2 b2 0, 1, and similarly, that bc is tangent to Γ2 at α2 = 0, 1, . Re- − c2 c2 0 T 00 member that ab is tangent to Γ in γ = [√b2, 1, 0] and ac is tangent to Γ in T − β = [√c2, 0, 1] as deduced in Lemma 5.1.1. Indeed, α1, β and γ are collinear, since −   0 √c2 √b2  1 0 1  det  q −  = 0. b2 1 0 − c2 −

Now, we look at aα2, bβ, cγ, given by

 0      q 1 1 b2 aα2 =   , bβ =  0  , cγ = √b2 . − c2  1 √c2 0

49 5.1. A synthetic proof

Indeed, these lines are concurrent, since

 0 1 1  q b2 det  0 √b2 = 0, − c2  1 √c2 0 which completes the proof. Deﬁnition 5.1.1. A 6-tuple (a, b, c, Γ, Γ0, Γ00), where a, b, c C are distinct and Γ, Γ0, Γ00 in are tangent to bc, ac, ab at α, β, γ respectively,∈ is called positive if aα, bβ, cγ areF concurrent (see Figure 5.2).

Figure 5.2: positive 6-tuple

Now we consider a 2n-tuple (a1, . . . , an, Γ1,..., Γn) where the points ai C are ∈ distinct and aiai+1 is tangent to Γi for all i = 1, . . . , n and an+1 := a1. We define a new set of conics Γ0 , i = 1, . . . , n 3 as follows. i ∈ F − We know that a1a2 is tangent to Γ1 in some point α and a2a3 is tangent to Γ2 in 0 some point β. By Corollary 5.1.2, there exists an element Γ1 such that a1a3 is 0 ∈ F tangent to Γ1 in γ and a1β, a2γ, a3α are concurrent. In other words, there exists a 0 0 Γ such that (a1, a2, a3, Γ2, Γ , Γ1) is positive. 1 ∈ F 1 0 For Γ2, we proceed similarly. We know that a3a4 is tangent to Γ3 and by definition 0 0 of Γ1, we know that a1a3 is tangent to Γ1. By Corollary 5.1.2 there exists an element 0 0 0 0 Γ such that a1a4 tangent to Γ and the 6-tuple (a1, a3, a4, Γ3, Γ , Γ ) is positive. 2 ∈ F 2 2 1 By induction, define Γ0 for i = 2, . . . , n 3 to be the conic in such that i − F 0 0 (a1, ai+1, ai+2, Γi+1, Γi, Γi−1)

50 5.1. A synthetic proof

0 0 is positive. For the same reasons as explained for Γ1 and Γ2 above, existence is guaranteed by Corollary 5.1.2. Note that for i = n 3, we have the positive 6-tuple − 0 0 (a1, an−2, an−1, Γn−2, Γn−3, Γn−4).

For the next step, we consider the three points a1, an−1, an and here, we already know that an−1an is tangent to Γn−1, but also, that a1an is tangent to Γn as well 0 as a1an−1 tangent to Γn−3 by the last induction step before. Hence the following deﬁnition comes naturally. Deﬁnition 5.1.2. Let n 4. We call the 2n-tuple ≥ (a1, . . . , an, Γ1,..., Γn) positive if the 6-tuple 0 (a1, an−1, an, Γn−1, Γn, Γn−3) is positive.

Now we are able to state the Great Poncelet Theorem. 0 00 Theorem 5.1.3. Let (a1, . . . , an, Γ1,..., Γn) be a positive 2n-tuple and b , b C 0 00 ∈ two distinct points such that b b is tangent to Γ1. Then there exists a positive 0 00 2n-tuple (b1, . . . , bn, Γ1,..., Γn) such that b1 = b and b2 = b .

Proof. We first show the statement for n = 3. This can be done in four steps by applying the results of the previous section. 0 00 By assumption, we know that a1a2 as well as b b are tangent to Γ1 at some points α1 and β1 respectively. By the second statement of Lemma 5.1.1, we know that 0 0 00 0 there exists an element Γ such that a1b and a2b are tangent to Γ in the ∈ F 0 intersection points of α1β1 with Γ . 00 0 We know that a2b is tangent to Γ and a2a3 is tangent to Γ2 by assumption. By the first statement of Lemma 5.1.1, we know that there exists an element b000 C, 00 000 000 0 ∈ such that b b is tangent to Γ2 and a3b is tangent to Γ . Hence, we are again in the situation to apply the last statement of Lemma 5.1.1, 000 0 0 namely a3b and a1b are both tangent to Γ . Hence, there exists an element Γ˜ such 0 000 that a1a3 and b b are tangent to Γ.˜ By assumption for the ai’s, we know that 0 00 000 Γ˜ = Γ3. So, (b , b , b , Γ1, Γ2, Γ3) is positive. For the induction step, we only need to use the case n = 3. 0 00 Let (a1, . . . , an, Γ1,..., Γn) be a positive 2n-tuple and b , b C two distinct points 0 00 ∈ 0 such that b b is tangent to Γ1. By definition (a1, a2, a3, Γ2, Γ1, Γ1) is positive and 0 0 00 hence (b1, b2, b3, Γ2, Γ1, Γ1) is, where b1 = b and b2 = b . 0 0 0 0 Moreover, since (a1, a3, a4, Γ3, Γ2, Γ1) is positive, we have (b1, b3, b4, Γ3, Γ2, Γ1) positive. 0 We can continue this until (a1, an−1, an, Γn−1, Γn, Γn−3) is positive, which implies 0 again that (b1, bn−1, bn, Γn−1, Γn, Γn−3) is positive, since we have shown the case n = 3.

Hence, by deﬁnition, (b1, . . . , bn, Γ1,..., Γn) is positive.

51 5.1. A synthetic proof

Now we can choose Γ1 = ... = Γn = Γ to prove the more famous version of Poncelet’s Theorem, which reads as follows.

Theorem 5.1.4. Let (a1, . . . , an) be an n-sided polygon inscribed in C and circumscribed around the conic Γ. For any two points b0, b00 C, such that b0 = b00 0 00 0 00 ∈ 6 and b b is tangent to Γ, there exists a polygon (b , b , b3, . . . , bn) inscribed in C and circumscribed around Γ.

Proof. The only thing to show is that

(a1, . . . , an, Γ,..., Γ) is always positive. By Corollary 5.1.2, we know that a positive 2n-tuple

0 (a1, . . . , an, Γ,..., Γ, Γ ) always exists. We just choose a suitable Γ0 such that the last condition in the deﬁnition of positive is satisﬁed. 0 We know that a2a3 is tangent to Γ and hence, by applying Theorem 5.1.3 to b = a2, 00 b = a3, we obtain the positive 2n-tuple

0 (a2, a3, . . . , an, a1, Γ,..., Γ, Γ ).

Note that the points have to be the ai’s since a2 can only be obtained in two tangents 0 of Γ. Hence, a1an is tangent to Γ . Repeating the argument by shifting the n-gon, 0 we get that a2a3, a3a4,..., an−2an−1 are tangent to Γ as well. For n 5, it follows Γ = Γ0, since a conic is uniquely determined by ﬁve of its tangents.≥

For n = 3, we have to show that (a1, a2, a3, Γ, Γ, Γ) is always positive. Let C be a conic and a = [1, 0, 0]T , b = [0, 1, 0]T , c = [0, 0, 1]T on C, i.e. once more, C is given by C = V(d1xy + e1xz + f1yz) for d1, e1, f1 = 0. Then have to intersect the tangents of C in a, b and c to obtain 6 T T T a1 = [f1, e1, 1] , a2 = [f1, e1, 1] , a3 = [f1, e1, 1] . − − − −

The claim is, that a1c, a2a and a3b are concurrent. Indeed,   0 1 e1 − det  1 0 f1  = 0. e1 f1 0 − − For n = 4, a similar direct computation can be made.

52 5.2. A combinatorial proof

5.2 A combinatorial proof

In this section, we prove that Poncelet’s Theorem is a purely combinatorial consequence of Pascal’s Theorem. The original proof for the real case is due to Halbeisen and Hungerbühler[18]. We will follow the proof presented in [18] closely but use our own notation. As just mentioned, Pascal’s Theorem is assumed at all steps in the proof. In 1966, Buekenhout [6] showed that if Pascal’s Theorem is valid for one oval, it is actually a Pappus plane, i.e. the Theorem of Pappus is thoroughly satisfied. In every Pappus plane, the Theorem of Desargues proves to be true [43]. Because of this, the assumption of Pascal’s Theorem already reduces our case to all finite projective planes with coordinates in a finite field. In other words, this section gives another proof of Poncelet’s Theorem in PG(2, q). In particular, ovals and conics are the same objects. In comparison to the synthetic proof presented in the Section 5.2, we do not use the coordinates of the field here.

5.2.1 Preliminaries and preparation

All points are denoted by indexed capital letters and for two points Pi and Qi, PiQi denotes the line connecting those two points. All lines are denoted by lower-case letters and for two lines gi, hi, gi hi denotes the intersection point of those two lines. A triangle will be denoted by∩ ∆P QR, where P , Q and R are the vertices of the triangle. We will prove the following version of Poncelet’s Theorem.

Theorem 5.2.1 (Poncelet’s Theorem). Let P1,...,Pn,Q1,...,Qn be 2n distinct points on an oval s in PG(2, q). Let the 2n 1 lines O − P1P2,..., Pn−1Pn, PnP1, Q1Q2,..., Qn−1Qn be tangent to an oval t. Then QnQ1 is tangent to t as well. (see Figure 5.3) O O

Figure 5.3: Poncelet’s Theorem for n = 5

The main tools we use for proving Poncelet’s Theorem is Pascal’s Theorem, Brian- chon’s Theorem and Carnot’s Theorem. All these classical result can for example be found in [14].

53 5.2. A combinatorial proof

Theorem 5.2.2 (Pascal’s Theorem). Let P1, P2, P3, P4, P5 and P6 be distinct points in PG(2, q). Then these six points lie on an oval if and only if the three points

P1P5 P2P6, P2P4 P3P5, P1P4 P3P6 ∩ ∩ ∩ are collinear. (see Figure 5.4)

Figure 5.4: Pascal’s Theorem

Note that the ‘only if statement’ above is also known as the Braikenridge-Maclaurin Theorem. An equivalent version of Pascal’s Theorem, known as Carnot’s Theorem, reads as follows.

Theorem 5.2.3 (Carnot’s Theorem). Let P1, P2, P3, P4, P5 and P6 be pairwise distinct points in PG(2, q). Then these six points lie on an oval if and only if the three lines (P1P5 P2P6)(P2P4 P3P5), P1P4, P3P6 ∩ ∩ are concurrent.

The dual version of Pascal’s Theorem is known as Brianchon’s Theorem and reads as follows.

Theorem 5.2.4 (Brianchon’s Theorem). Let g1, g2, g3, g4, g5 and g6 be pairwise distinct lines in PG(2, q). Then these six lines are tangents of an oval if and only if the three lines

(g1 g5)(g2 g6), (g2 g4)(g3 g5), (g1 g4)(g3 g6) ∩ ∩ ∩ ∩ ∩ ∩ are concurrent. (see Figure 5.5)

Theorem 5.2.5 (Dual version of Carnot’s Theorem). Let g1, g2, g3, g4, g5 and g6 be pairwise distinct lines. Then these six lines are tangents of an oval if and only if the three points

(g1 g5)(g2 g6) (g2 g4)(g3 g5), g1 g4, g3 g6 ∩ ∩ ∩ ∩ ∩ ∩ ∩ are collinear.

54 5.2. A combinatorial proof

Figure 5.5: Brianchon’s Theorem

5.2.2 Poncelet’s Theorem for triangles

We now present a proof of Poncelet’s Theorem for triangles, i.e. the case n = 3 of Theorem 5.2.1. In particular, we prove the following statement.

Theorem 5.2.6 (Poncelet’s Theorem for triangles). Let P1, P2 and P3 be distinct points on an oval s in PG(2, q) such that P1P2, P2P3 and P1P3 are tangent to O some other oval t. Let Q1 be another point on s such that there exist two more O O distinct points Q2 and Q3 on s with Q1Q2 and Q1Q3 tangent to t. Then Q2Q3 O O is tangent to t as well. (see Figure 5.6) O

Figure 5.6: Poncelet’s Theorem for triangles

We present here a combinatorial proof of Poncelet’s theorem without using the coordinates for PG(2, q). The strategy, however, is similar to the procedure in Chapter 4 where we already proved Poncelet’s Theorem for triangles in PG(2, q). The following result will easily imply Poncelet’s Theorem for triangles.

Theorem 5.2.7. Let P1, P2, P3, P4, P5 and P6 be distinct points on an oval s. O Consider the triangles ∆P1P3P5 and ∆P2P4P6. Then the six sides of the two triangles P1P3, P2P4, P3P5, P4P6, P1P5 and P2P6 are tangent to another oval t. O

55 5.2. A combinatorial proof

Proof. We apply Pascal’s Theorem. The points P1, P2, P3, P4, P5 and P6 are by assumption distinct and lie on an oval s. Hence, by Theorem 5.2.2, the points O S = P1P5 P2P6,R = P2P4 P3P5,Q = P1P4 P3P6 ∩ ∩ ∩ are collinear. Let us label the sides of the two triangles ∆P1P3P5 and ∆P2P4P6 by

g1 = P1P3, g3 = P3P5, g5 = P1P5 and g2 = P2P4, g4 = P4P6, g6 = P2P6.

By the dual version of Carnot’s Theorem, these six distinct lines are tangent to some oval t if and only if the three points O 0 Q = (g1 g5)(g2 g4) (g1 g3)(g4 g6) 0 ∩ ∩ ∩ ∩ ∩ R = g2 g3 0 ∩ S = g5 g6 ∩ are collinear. But we actually have Q = Q0, R = R0 and S = S0 (see Figure 5.7) and by Pascal’s Theorem before, the points P , Q, R are collinear.

Figure 5.7: Applying the dual version of Carnot’s Theorem

Hence, the lines g1, g2, g3, g4, g5 and g6 are indeed tangent to some oval t. O Now we are able to prove Poncelet’s Theorem for triangles, i.e. Theorem 5.2.6.

Proof of Poncelet’s Theorem for triangles. Let P1, P2 and P3 be distinct points on an oval s such that P1P2, P2P3 and P1P3 are tangent to some other oval t. Let O O Q1 be another point on s such that there exist two more distinct points Q2 and O Q3 on s with Q1Q2 and Q1Q3 tangent to t. By the result before, we know that O 0 O there exists some oval such that the sides of ∆P1P2P3 and ∆Q1Q2Q3 are tangent 0 O to . Since five of the six sides of ∆P1P2P3 and ∆Q1Q2Q3 are already tangent to O 0 t, we necessarily have t = . O O O Note that we are actually working in PG(2, q), hence every oval is a conic. In particular, every oval in PG(2, q) is indeed determined by five of its points or five of its tangents.

56 5.2. A combinatorial proof

5.2.3 Proof of Poncelet’s Theorem

In the previous section, we have already seen the case n = 3 of Theorem 5.2.1. In this section, we show Poncelet’s Theorem for n 4. ≥ Before we start with a preliminary result, let us introduce the following notation.

Ri,j := PiPj QiQj ∩ Si,j := PiQj PjQi ∩ Theorem 5.2.8. Let P1,...,Pn,Q1,...,Qn be 2n distinct points on an oval s in O PG(2, q). Let the 2n 1 lines P1P2,..., Pn−1Pn, PnP1, Q1Q2,..., Qn−1Qn be tangent − to an oval t. Then the three points O R1,2,S2,n−1,Rn−1,n are pairwise distinct and collinear. (see Figure 5.8).

Figure 5.8: R1,2,S2,n−1,Rn−1,n are collinear

Before we can prove this, we need the following two results. Lemma 5.2.9. Let 1 i < j n 1. If ≤ ≤ − Ri,i+1,Si,j,Rj−1,j are distinct and collinear, then

Ri−1,i,Si,j,Rj,j+1 are distinct and collinear as well.

Proof. a) The six lines

PiPi+1, Pj−1Pj, PjPj+1, Qi−1Qi, QiQi+1, Oj−1Qj are all tangent to t. By Brianchon’s Theorem, the three lines given by O g1 := (Pj−1Pj PjPj+1)(Qi−1Qi QiQi+1) = PjQi ∩ ∩ g2 := (PiPi+1 QiQi+1)(Pj−1Pj Qj−1Qj) = Ri,i+1Rj−1,j ∩ ∩ g3 := (PiPi+1 PjPj+1)(Qi−1Qi Qj−1Qj) ∩ ∩

57 5.2. A combinatorial proof

Figure 5.9: Brianchon’s Theorem for the lines PiPi+1, Pj−1Pj, PjPj+1, Qi−1Qi, QiQi+1, Oj−1Qj. are distinct and concurrent. (see Figure 5.9). b) By assumption, the points Ri,i+1, Si,j, Rj−1,j are collinear and hence, the lines g1, g2 and g4 := PiQj meet in Si,j. For that, note that by a), g1 and g2 are distinct and hence, by symmetry (or by contraction, as otherwise Pi, Pi+1, Qj and Qj−1 are collinear), g2 and g4 are distinct as well (see Figure 5.10). Since no three points of Pi, Pj, Qi, and Qj are on one line (they all lie on the same oval), g1 and g4 are distinct as well.

Figure 5.10: g2 and g4 are distinct. c) Let us now have a look at the lines g3, g4 and

g5 := Ri−1,iRj,j+1.

58 5.2. A combinatorial proof

In particular, we look at the lines

g3 := (PiPi+1 PjPj+1)(Qi−1Qi Qj−1Qj) ∩ ∩ g4 := (Pi−1Pi PiPi+1)(Qj−1Qj QjQj+1) = PiQj ∩ ∩ g5 := (Pi−1Pi Qi−1Qi)(PjPj+1 QjQj+1) = Ri−1,iRj,j+1. ∩ ∩

So there are six tangents involved, namely Pi−1Pi, PiPi+1, PjPj+1, Qi−1Qi, Qj−1Qj and QjQj+1. By Brianchon’s Theorem, the lines g3, g4 and g5 are pairwise distinct and concurrent (see Figure 5.11).

Figure 5.11: By Brianchon’s Theorem

To summarize, note that by step a) the lines g1, g2 and g3 are distinct and concurrent, by assumption and step b), the lines g1, g3 and g4 are distinct and concurrent in Si,j and ﬁnally, by step c), the lines g3, g4 and g5 are distinct and concurrent. Hence, the lines g1, g4 and g5 are distinct and meet in Si,j.

Lemma 5.2.10. Let 1 i < j n 1. If Ri−1,i,Si,j,Rj,j+1 are pairwise distinct and collinear, then ≤ ≤ − Ri−1,i,Si−1,j+1,Rj,j+1 are pairwise distinct and collinear as well.

Proof. Let Ri−1,i,Si,j,Rj,j+1 are pairwise distinct and collinear. First, by Pascal’s Theorem, the points Ri−1,i, Si,j and Ri−1,j are pairwise distinct and collinear (see Figure 5.12).

Once more by Pascal’s Theorem, the points Si−1,j+1, Ri−1,j and Rj,j+1 are pairwise distinct and collinear (see Figure 5.13).

59 5.2. A combinatorial proof

Figure 5.12: Ri−1,i, Si,j and Ri−1,j are pairwise distinct and collinear

Figure 5.13: Si−1,j+1, Ri−1,j and Rj,j+1 are pairwise distinct and collinear

Summarizing, we have that

Ri−1,i, Si,j, Rj,j+1 are pairwise distinct and collinear, •

Ri−1,i, Si,j, Ri−1,j are pairwise distinct and collinear, •

Si−1,j+1, Ri−1,j and Rj,j+1 are pairwise distinct and collinear. •

Hence, Ri−1,i, Si−1,j+1 and Rj,j+1 are pairwise distinct and collinear as well.

Now we are able to prove Theorem 5.2.8.

Proof of Theorem 5.2.8. Let us quickly recollect the two statements we need to use.

By Lemma 5.2.9, if Ri,i+1, Si,j, Rj−1,j are pairwise distinct and collinear, then • Ri−1,i, Si,j, Rj,j+1 as well,

by Lemma 5.2.10, if Ri−1,i, Si,j, Rj,j+1 are pairwise distinct and collinear, then • Ri−1,i, Si−1,j+1, Rj,j+1 as well.

60 5.2. A combinatorial proof

Our goal is to show that R1,2, S2,n−1 and Rn−1,n are pairwise distinct and collinear. For this, let us distinguish the cases n even and n odd.

First, let n = 2k 1 odd. By Pascal’s Theorem, the three points Rk−1,k, Sk−1,k+1 − and Rk,k+1 are pairwise distinct and collinear. Applying Lemma 5.2.9, we know that Rk−2,k−1, Sk−1,k+1 and Rk+1,k+2 are pairwise distinct and collinear and applying Lemma 5.2.10 gives that Rk−2,k−1, Sk−2,k+2 and Rk+1,k+2 are pairwise distinct and collinear. Now we repeat this procedure until we have that R2,3, S2,n−1 and Rn−2,n−1 are pairwise distinct and collinear. We apply Lemma 5.2.9 one last time and indeed obtain that R1,2, S2,n−1 and Rn−1,n are pairwise distinct and collinear. Now, let us consider the even case n = 2k. By Carnot’s Theorem, the three points Rk−1,k, Sk,k+1 and Rk+1,k+2 are collinear. As in the odd case, we just apply Lemma 5.2.9 and Lemma 5.2.10 until we end up with R1,2, S2,n−1 and Rn−1,n being collinear.

Finally, we are able to give a proof of Poncelet’s Theorem.

Proof of Poncelet’s Theorem. Let P1,...,Pn,Q1,...,Qn be 2n distinct points on an oval s. Let the 2n 1 lines P1P2,..., Pn−1Pn, PnP1, Q1Q2,..., Qn−1Qn be tangent O − to an oval t. We have to show that QnQ1 is tangent to t as well. O O By Theorem 5.2.8 we know that

R1,2,S2,n−1,Rn−1,n are pairwise distinct and collinear. By Pascal’s Theorem, we know that R1,2, S2,n−1 and Rn−1,1 are pairwise distinct and collinear (see Figure 5.14).

Figure 5.14: R1,2, S2,n−1 and Rn−1,1 are pairwise distinct and collinear

Similarly, by Pascal’s Theorem, we know that R1,n−1, Sn−1,n and Rn−1,n are pairwise distinct and collinear. Combining these results, we know that

R1,2,S1,n,Rn,n−1 are pairwise distinct and collinear.

61 5.2. A combinatorial proof

Now, let us apply Carnot’s Theorem to the six lines

P1P2, Pn−1Pn, P1Pn, Q1Q2, Qn−1Qn, Q1Qn.

These six lines are all tangent to t if and only if the points O

S1,n = (P1Pn P1P2)(Qn−1Qn QnQ1) (Q1Qn Q1Q2)(Pn−1Pn PnP1) ∩ ∩ ∩ ∩ ∩ R1,2 = P1P2 Q1Q2 ∩ Rn−1,n = Qn−1Qn Pn−1Pn ∩ are collinear.

But R1,2, S1,n and Rn,n−1 are indeed pairwise distinct and collinear and hence, all the lines involved above are tangent to t. In particular, Q1Qn is tangent to t as well. O O

62 Chapter 6

A Poncelet Criterion

The results presented in this section are published in [23].

6.1 Construction and properties

In all of the following, we only consider conics of the form

2 2 2 Ok = V(x + ky + ckz ), (6.1) for k GF (q) 0 and c a nonsquare in GF (q). ∈ \{ } − The following results will give an explanation for our choice of conics. Also compare to Remark 6.3.1 later on.

To understand the properties of a pair of such conics Ok above, we ﬁrst have a closer look at a speciﬁc partition of the plane PG(2, q). The idea is to start with the point P = [1, 0, 0]T and the line g through [0, 1, 0]T and [0, 0, 1]T . We look at the pencil generated by P and g. We have already seen in the preliminaries that such a pencil consists of q+1 objects, namely P and g as well as q 1 conics. Moreover, these q+1 objects partition the plane PG(2, q). The following− results discuss some properties of this partition. Lemma 6.1.1. The point P = [1, 0, 0]T in the plane PG(2, q) can be described

P = V(y2 + cz2), (6.2) for c a nonsquare in GF (q). − Proof. P = [1, 0, 0]T clearly solves y2 +cz2 = 0. As the associated matrix is singular, it describes a point, a line or a pair of lines. It is a point, if the polynomial y2 + cz2 is irreducible over GF (q), which is the case if and only if c is a nonsquare in GF (q). −

In the construction which follows, we start with any point P and any line g, P/ g. Since there exists a collineation of PG(2, q) which maps any three noncollinear∈ points to any other three noncollinear points, we can restrict the proofs without loss of generality to P = [1, 0, 0]T and g the line through [0, 1, 0]T and [0, 0, 1]T .

63 6.2. Poncelet’s Theorem for conics Ok

Lemma 6.1.2. Let P be a point and g a line in PG(2, q), such that P/ g. Then the pencil generated by P and g forms a partition of the plane PG(2, q).∈ Moreover, P is the unique point in PG(2, q) which is an inner point of all q 1 conics in the pencil generated by P and g. −

Proof. Without loss of generality, take P = [1, 0, 0]T and g the line through [0, 1, 0]T and [0, 0, 1]T . By Lemma 6.1.1, we have

g = V(x2) and P = V(y2 + cz2), for c a nonsquare. Considering the pencil of P and g leads to q 1 conics Ok, determined− by V(x2 + ky2 + kcz2) for k GF (q) 0 . As P/ g, all−q + 1 elements ∈ \{ } ∈ of the pencil are disjoint and since there are q + 1 points on every conic Ok and on g, the q + 1 elements of the pencil form a partition of PG(2, q). The second statement is a straight forward calculation using Lemma 2.3.3.

The next result is also well-known (compare to [20, Theorem 8.3.3]) and can easily be shown by using Lemma 2.3.3.

Lemma 6.1.3. Let P be a point and g a line in PG(2, q) with P/ g. Let Ok, k GF (q) 0 , be the q 1 pairwise disjoint conics in the pencil generated∈ by P ∈ \{ } − and g. Each line through P is a secant of all Oi, if i is a square in GF (q) and an external line of all Oj, if j is a nonsquare in GF (q), or vice versa.

As an easy consequence for the q 1 conics in the pencil generated by P and g we mention the following. −

Corollary 6.1.4. None of the conics Ok, k GF (q) 0 , have tangents in common. ∈ \{ } Remark 6.1.5. Note that the parameter c can indeed be chosen arbitrarily among − all nonsquares, without any changes of incidence relations. In particular, let c1 and − c2 be nonsquares in GF (q). Then c1c2 is a square in GF (q) and the collineation − φS given by 1 0 0  S = 0 1 0  −1 0 0 √c2c1 2 2 2 2 2 2 maps all conics Ok given by V(x +ky +kc1z ) to conics given by V(x +ky +kc2z ).

6.2 Poncelet’s Theorem for conics Ok

Recall that we are only interested in pairs of conics of the form (6.1) described in the previous section. Since these conics Ok are all disjoint and have no common tangents as mentioned in Corollary 6.1.4, we are in the particular situation that if we can ﬁnd one line which is a tangent to Oα and a secant of Oβ, this leads necessarily to a Poncelet polygon. The ﬁnite version of Poncelet’s Theorem we are going to prove here reads as follows.

64 6.2. Poncelet’s Theorem for conics Ok

Theorem 6.2.1. Let (Oα,Oβ) be a pair of conics in PG(2, q) given by

2 2 2 Ok = V(x + ky + ckz ), k α, β , ∈ { } k GF (q) 0 and c a nonsquare. If an n-sided Poncelet polygon can be con- ∈ \{ } − structed starting with a point P Oβ, then an n-sided Poncelet polygon can be ∈ constructed starting with any other point Q Oβ as well. ∈ Proof. We start with the result in planes PG(2, q), q 3(4). By Remark 6.1.5, it suﬃces to consider conics of the form ≡

2 2 2 Ok = V(x + ky + kz ). (6.3)

Since the conics in this case can be interpreted as concentric circles, we apply collineations which rotate the n-sided Poncelet polygons suitably, i.e. we look at members of the dihedral group of q + 1 elements or a subgroup of it.

Let P1,P2,...,Pn be points on Oβ which form an n-sided Poncelet polygon with the sides tangent to some other conic Oα. The goal is to ﬁnd a collineation τ = τ(P1,Q) which maps P1 to Q, but Oα and Oβ are invariant under this collineation. This can be obtained by a suitable rotation

1 0 0  τ = 0 a b  0 b a − 2 2 with a + b 1(q). Hence, applying τ(P,Q) to all points Pi for i = 1, . . . , n yields a new Poncelet≡ polygon using the point Q. Note that a and b can be computed explicitly by using τP = Q. It remains to prove the result in planes PG(2, q), q 1(4). Let c be an arbitrary but ﬁxed nonsquare in GF (q) and consider the ﬁeld≡ extension GF (q)(√c) which is 2 −1 2 2 2 2 isomorphic to GF (q ). We embed the conics Ok = V(k x + y + cz ) in PG(2, q ) with the natural embedding ε : PG(2, q) PG(2, q2),P P . Then we choose the 2 → 7→ collinear transformation φs in PG(2, q ) given by the matrix

1 0 0  S = 0 1 0  0 0 √c

2 in GF (q ). All conics Ok in PG(2, q) are mapped by φS ε to conics S(Ok) in 2 −1 2 2 ◦ 2 PG(2, q ). Then S(Ok) is given by S(Ok) = V(k x + y + z ). It is easy to check that a tangent of Ok is mapped to a tangent of S(Ok) by φS ε. Now, on the conics 2 ◦ S(Ok) in PG(2, q ), we can operate similarly as before. Namely, let P1,...,Pn be points on Oβ which form an n-sided Poncelet polygon with the sides tangent to some 2 other conic Oα. Let Q be any other point on Oβ. We look at P1 and Q in PG(2, q ) and the goal is again to ﬁnd a collineation τ = τ(SP,SQ) which maps SP to SQ and leaves the conics S(Oα) and S(Oβ) invariant. This can be obtained by

1 0 0  τ = 0 a √cb 0 √cb a − 65 6.3. Relations for pairs of conics for a2 + cb2 = 1. We apply τ to all points on the embedded Poncelet n-gon given by SP1,...,SPn, which gives a new Poncelet n-gon for S(Oβ) and S(Oα). Mapping these transformed points back to PG(2, q) gives points on the original conic Oβ. Hence, we end up with a Poncelet n-gon for Oβ and Oα starting with the point Q.

6.3 Relations for pairs of conics

In this section, we study the position of pairs of conics with regard to the existence of a Poncelet polygon.

Deﬁnition 6.3.1. Let O and O0 be two conics in PG(2, q). We say that O lies inside O0 if O0 consists of external points of O only. Notation: O < O0. Moreover, we say that O lies outside O0 if O consists of external points of O0 only. Notation: O > O0.

Note that O < O0 does not imply O0 > O. In particular, in a ﬁnite projective plane we can have the unintuitive situation that O < O0 and O0 < O at the same time. We continue to consider conics described in the previous section, i.e. conics given 2 2 2 2 2 2 by Oα = V(x + αy + cαz ) and Oβ = V(x + βy + cβz ).

Remark 6.3.1. At a ﬁrst glance, this choice of conic pairs Oα and Oβ seems to be rather restrictive. In Chapter 7 the structure of all possible pencils of conics is discussed. For two disjoint conics, there are only three diﬀerent pencils up to collineations. It turns out that only the class of pencils studied in the present paper has the property that O < O0 or O > O0 for all pairs of conics. A similar result was also obtained by Abatangelo et al. [1, Theorem 6.1] for q 17. ≥

Theorem 6.3.2. If one point P Oβ is an external point of Oα, then Oα < Oβ. ∈ Moreover, we have Oα < Oβ if and only if β(β α) is a nonsquare in GF (q). − Proof. Recall that all points of PG(2, q) with a zero x-coordinate lie on the line 2 g = V(x ) and hence due to the partition not on any conic Ok, k GF (q) 0 . T ∈ \{ } A point P of Oβ can therefore be considered as P = [1, p2, p3] , which leads to 2 −1 2 p = β cp . By Lemma 2.3.3, the conic Oα lies inside Oβ if for all such 2 − − 3 points P , the line OαP is a secant of Oα. This is the case if there exist two points T T Q1 = [x1, y1, z1] and Q2 = [x2, y2, z2] on Oα satisfying

−1 α xi + p2yi + cp3zi = 0 (6.4)

p −1 2 T for i = 1, 2. We can rewrite those points as Qi = [1, α czi , zi] and plug it into (6.4). The quadratic equation in z given by ± − −

2 −1 −1 −1 −1 2 −2 −1 z 2α βp3z + α c + α βp α βc = 0 − 3 − needs to have two solutions in z. This is the case, if and only if its discriminant is a nonzero square in GF (q), i.e. if and only if

α−2β2p2 α−1c−1 α−1βp2 + α−2βc−1 3 − − 3 66 6.3. Relations for pairs of conics is a nonzero square in GF (q). Multiplying by α2 and factorizing leads to the condition of (βp2 + c−1)(β α) 3 − being a nonzero square in GF (q). Using P Oβ gives the equivalent expression ∈ ( βp2c−1)(β α). − 2 − Since c is a nonsquare by assumption, we need − β(β α) − to be a nonsquare in GF (q). Since this expression is independent of the point P , we are done.

As a direct consequence of Theorem 6.3.2, we can easily construct chains of nested conics.

Corollary 6.3.3. Consider two conics Oα and Oβ. Then

Oα < Oβ Oβ < O 2 −1 . ⇔ β α

When calculating the relation < for every pair (Oα,Oβ) in a given plane, it is useful to apply the following result.

Lemma 6.3.4. Let (Ok,O1) be a pair of conics in PG(2, q). Then there exists a collinear transformation mapping (Ok,O1) to (Oβk,Oβ), for all β GF (q) 0 . In ∈ \{ } particular, Ok < O1 implies Oβk < Oβ.

Proof. We have to ﬁnd a collinear transformation φS with

φS(Ok) = Oβk.

In the case of β being a square in GF (q) the matrix S given by

1 0 0  −1 S = 0 √β 0  −1 0 0 √β works, which can be checked immediately. For β a nonsquare, we have to distinguish between PG(2, q) with q 3(4) and q 1(4). By Remark 6.1.5, we can choose the parameter c = 1 in the case≡ q 3(4). Now,≡ choose any nonzero square s in GF (q) such that β s is a square as well.≡ Note that we can always ﬁnd such an s, because β s runs for−s a square through exactly q−1 q−1 − 2 values, as there are 2 squares in GF (q) 0 . Then β s = β, since we take q−1 \{ } − 6 s not to be zero. Hence, not all of those 2 values can be nonsquares. Moreover, β s = 0 as s is a square and β a nonsquare. Hence, there is a nonzero square β s. Using− 6 the matrix − 1 0 0  −1 S = 0 √s √β s 0 √β s − √s− − 67 6.3. Relations for pairs of conics we obtain the coordinate transformation we are looking for. In planes PG(2, q), q 1(4), we choose a c such that β c is a nonzero square. To ensure the existence it≡ can be proceeded similarly as before.− For such a c, look at the matrix 1 0 0  −1 S = 0 √β c c  . 0 1− √β c − − We have det(S−1) = β = 0 and it can be checked that this is the collinear transformation we are looking− 6 for. Example 6.3.5. We want to investigate the relation < in PG(2, 7). By looking at β = 1 and shifting the result by using Lemma 6.3.4, we obtain the table of relations for the whole plane PG(2, 7).

O1 O2 O3 O4 O5 O6 O1 < < < O2 < < < O3 < < < O4 < < < O5 < < < O6 < < <

Using Corollary 6.3.3, we detect the following closed chains of conics Oα Oβ → → O 2 −1 ..., namely: β α →

O1 O3 O2 O6 O4 O5 O1 → → → → → → O1 O4 O2 O1 → → → O3 O5 O6 O3 → → → Note that starting with two squares α and β results in a chain of conics with just squares as indices. Similarly, starting with two nonsquares as indices results in a chain of conics with only nonsquares as indices.

Since exactly half of all nonzero elements in GF (q) are squares, the following is immediate.

q−1 Corollary 6.3.6. For every conic Oβ in PG(2, q), there are 2 conics Oα such that Oα < Oβ.

Next, we have a closer look at the relations of the points on Oα and Oβ.

T Lemma 6.3.7. Let P = [1, p2, p3] be any point on Oβ and Oα < Oβ. Then for the T T contact points A1 = [1, y1, z1] and A2 = [1, y2, z2] on Oα of the tangents through P we have −1 p −2 −1 z1,2 = α βp3 p2 α ( c β)(β α) ± − − and ( −1 −1 p2 ( α cp3z1,2), if p2 = 0 q− − 6 y1,2 = −1 2 α cz , if p2 = 0. ± − − 1,2 68 6.3. Relations for pairs of conics

Proof. To see this, we just have to solve the quadratic equation derived in Theorem 6.3.2. Since Oα < Oβ, we indeed get two solutions.

Lemma 6.3.8. Let P and Q be two points on Oβ such that the line connecting P and Q is tangent to Oα in the point A. Then P + Q = A, where addition is coordinate-wise.

T T Proof. Let P = [1, p2, p3] and Q = [1, q2, q3] be two points on Oβ, so we have

2 2 1 + βp2 + cβp3 = 0 and 2 2 1 + βq2 + cβq3 = 0. There are q + 1 points on the line through P and Q, obtained by nontrivial GF (q)- linear combinations of P and Q. Note that P Q has a zero x-coordinate and hence 2 − gives the intersection with the line g : x = 0, which is clearly not a point on Oα. So we know that there exists exactly one k GF (q) 0, 1 such that ∈ \{ − } P + kQ = (1 + k, p2 + kq2, p3 + kq3) = A Oα. (6.5) ∈ So, we need (k + 1, p2 + kq2, p3 + kq3) Oα which is the case if and only is ∈ 2 + 2α(p q + cp q ) k2 + 2 2 3 3 k + 1 = 0. 1 αβ−1 − Note that 1 αβ−1 = 0, since otherwise α = β. Solving for k yields − 6 s 1 + α(p q + cp q ) 1 + α(p q + cp q )2 k = 2 2 3 3 2 2 3 3 1. − 1 αβ−1 ± 1 αβ−1 − − − Note that by (6.5), there can only be one solution to our equation, since we are looking for a tangent of Oα. Because of this, the radicand has to be zero, which is if and only if 1 + α(p q + cp q )2 2 2 3 3 = 1. 1 αβ−1 − Hence k = 1. Since we already excluded k = 1 we obtain k = 1, which indeed gives P + Q±= A. − T Corollary 6.3.9. Let P,Q Oβ such that [1, 0, 0] / PQ. Then there exists an ∈ ∈ α GF (q) 0 , α = β, such that PQ is a tangent of Oα. The contact point is P +∈ Q. \{ } 6

T T T Proof. For P = [1, p2, p3] and Q = [1, q2, q3] we have P + Q = [2, p2 + q2, p3 + q3] . As the characteristic of GF (q) is odd in our case, P + Q is not on the unique line g through [0, 1, 0]T and [0, 0, 1]T . Since we have a partition of the plane PG(2, q), P + Q must be a point on a conic Oα. We have to exclude the possibility of PQ being a secant of Oα. For this, note that there are q + 1 points on PQ, where P,Q Oβ and P Q g. All the other q 2 points must lie on conics and there are at∈ most two points− ∈ on the same conic. Since− q 2 is odd and by Lemma 6.1.4, there is exactly one conic with PQ as a tangent. By− Lemma 6.3.8, we are done.

69 6.3. Relations for pairs of conics

In the following results, an n-sided Poncelet polygon for Oα < Oβ with vertices Bi on Oβ and contact points Ai on Oα is denoted by

A1 A2 A3 An−1 An B1 B2 B3 ... Bn B1, −→ −→ −→ −→ −→ Ai where Bi Bi+1 means that the line connecting Bi and Bi+1 is the tangent of Oα −→ in the point Ai. By Lemma 6.3.8, the following relations are immediate:

B1 + B2 = A1,B2 + B3 = A2,...,Bn−1 + Bn = An−1,Bn + B1 = An (6.6)

Note that by combining Lemma 6.3.7 and Lemma 6.3.8, we are now able to calculate all points of a Poncelet polygon by starting at one point on Oβ. Before we analyze Poncelet polygons for diﬀerent numbers of sides, we need some more properties of the points on Ok and their relations.

Lemma 6.3.10. The conics Oα in PG(2, q), q 3(4), consist of the q + 1 points ≡          1   1   1   1   1 1 1 1  y1 ,  y1 ,  y1  ,  y1 ,..., y q+1  ,  y q+1  ,  y q+1  ,  y q+1  − −  4  − 4   4  − 4   z1 z1 z1 z1 z q+1 z q+1 z q+1 z q+1  − − 4 4 − 4 − 4 if α is a square, and otherwise

         1   1           1 1 1 1 1 1 1 1  y1 ,  y1 ,  y1  ,  y1 ,...,  y q−3  ,  y q−3  , y ,  y , 0 ,  0  . − −  4  − 4  −  z1 z1 z1 z1 z q−3 z q−3 0 0 z z  − − − 4 − 4 −

T Proof. Of course, for y = 0 and z = 0, we have that [1, y, z] Oα implies that T 6 T 6 T ∈ [1, y, z] Oα, [1, y, z] Oα and [1, y, z] Oα. So we just have to check − ∈ T− ∈ T − − ∈ whether or not [1, 0, z] and [1, y, 0] are on Oα. We have

T 2 −1 −1 [1, 0, z] Oα z = α c ∈ ⇔ − and T 2 −1 [1, y, 0] Oα y = α . ∈ ⇔ − As q 3(4), c is a square in GF (q) and 1 is not. Hence these points lie on Oα if and only≡ if α is not a square. −

Lemma 6.3.11. The conics Oα in PG(2, q), q 1(4), consist of the q + 1 points ≡          1   1   1   1       1 1 1 1 1 1  y q 1 y q 1 y q 1 y q 1 y1 ,  y1 ,  y1  ,  y1 ,...,  −  ,  −  ,  −  ,  −  , y ,  y − − 4 − 4 4 − 4 −  z1 z1 z1 z1 z q 1 z q 1 z q 1 z q 1 0 0  − − −4 −4 − −4 − −4 if α is a square, and otherwise

         1   1   1   1       1 1 1 1 1 1  y q 1 y q 1 y q 1 y q 1 y1 ,  y1 ,  y1  ,  y1 ,...,  −  ,  −  ,  −  ,  −  , 0 ,  0  − − 4 − 4 4 − 4  z1 z1 z1 z1 z q 1 z q 1 z q 1 z q 1 z z  − − −4 −4 − −4 − −4 − Proof. The proof is similar to the proof of Lemma 6.3.10.

70 6.3. Relations for pairs of conics

T Corollary 6.3.12. The sum of all points on the conic Oα is [1, 0, 0] .

Proof. This can be seen by checking all possible cases above.

A1 A2 A3 An−1 An Lemma 6.3.13. Let B1 B2 B3 ... Bn B1 be an n-sided Poncelet polygon. Then −→ −→ −→ −→ −→

T B1 + B2 + ... + Bn = [1, 0, 0] = A1 + A2 + ... + An.

Moreover, for n even, we have

T A1 + A3 + ... + An−1 = [1, 0, 0] = A2 + A4 + ... + An.

Proof. By adding the equations (6.6) in diﬀerent ways, we obtain three conditions for n even, namely

B1 + B2 + ... + Bn = A1 + A3 + ... + An−1, (6.7)

B1 + B2 + ... + Bn = A2 + A4 + ... + An, (6.8)

2(B1 + B2 + ... + Bn) = A1 + A2 + ... + An−1 + An. (6.9)

Combining (6.7) and (6.9) gives

T A2 + A4 + ... + An = [1, 0, 0] and combining (6.8) and (6.9) gives

T A1 + A3 + ... + An−1 = [1, 0, 0] .

Hence, T B1 + B2 + ... + Bn = A1 + A2 + A3 + ... + An = [1, 0, 0] . The case n odd is similar.

Lemma 6.3.14. All lines joining opposite vertices Bi and Bn+i of a 2n-sided Pon- T T celet polygon meet in [1, 0, 0] . In particular, Bi + Bn+i = [1, 0, 0] , for i = 1, . . . , n.

Proof. Again, we can use the relation Bi + Bi+1 = Ai for the 2n-sided Poncelet polygon given by points Bi Oβ and Ai Oα. We have ∈ ∈

B1 + Bn+1 = A2n B2n + An Bn − − and B1 + Bn+1 = A1 B2 + An+1 Bn+2. − − Combining these equations gives

2(B1 + Bn+1) = B1 + Bn+1 = A1 B2 + An+1 Bn+2 + A2n B2n + An Bn. − − − −

Taking all Bi to the left and all Ai to the right side gives

B1 + B2 + Bn + Bn+1 + Bn+2 + B2n = A1 + An + An+1 + A2n.

71 6.4. A Poncelet Criterion

T To apply Lemma 6.3.13, we add the remaining Bi to obtain [1, 0, 0] on the left side, i.e.

T [1, 0, 0] = (A1 + An + An+1 + A2n) + (B3 + ... + Bn−1 + Bn+3 + ... + B2n−1).

Rewriting the Ai in terms of Bi gives

T T [1, 0, 0] = [1, 0, 0] + B1 + Bn+1.

Hence we get T [1, 0, 0] = B1 + Bn+1.

Similarly can be proceeded for all remaining Bi + Bn+i, 1 < i n. ≤ Note that Lemma 6.3.14 can bee seen as a generalization of Brianchon’s Theorem [18].

6.4 A Poncelet Criterion

6.4.1 Poncelet coeﬃcients

Here is a ﬁrst result concerning the existence of n-sided Poncelet polygons, namely Poncelet triangles. It has already been observed by Luisi in [31] that one cannot expect to ﬁnd Poncelet triangles in PG(2, q) for all q.

Lemma 6.4.1. Let Oα < Oβ be two conics in PG(2, q) which carry a Poncelet triangle. Then 4β = α in GF (q).

Proof. Let A1 A2 A3 B1 B2 B3 B1 −→ −→ −→ be a Poncelet triangle. By Lemma 6.3.8, we therefore have B1 +B2 = 2A1, B2 +B3 = 2A2 and B3 + B1 = 2A3. Moreover, by Lemma 6.3.13, we have B1 + B2 + B3 = [1, 0, 0]T , which gives the relations

T T T B1 + 2A2 = [1, 0, 0] ,B2 + 2A3 = [1, 0, 0] ,B3 + 2A1 = [1, 0, 0] . (6.10)

As all lines are given by linear combinations of two points, the conditions in (6.10) T T T translate to [1, 0, 0] B1A2, [1, 0, 0] B2A3 and [1, 0, 0] B3A1. Since there are no tangents through∈ [1, 0, 0]T , as seen∈ in Lemma 6.1.2, these∈ lines are secants of Oα and Oβ. With Theorem 6.1.3 we know that α and β are either both squares or both nonsquares. To ﬁnd the remaining intersection points of B1A2, B2A3 and T B3A1 with Oα and Oβ, consider the points A˜i and B˜i, where P˜ := [x, y, z] for a T T T − − point P = [x, y, z] . Since [1, 0, 0] BiB˜i and [1, 0, 0] AiA˜i, for i = 1, 2, 3, these are exactly the intersection points∈ we are looking for. Note∈ that this construction yields another Poncelet triangle. In particular, the second triangle is

A˜1 A˜2 A˜3 B˜1 B˜2 B˜3 B˜1 −→ −→ −→ as visualized in Figure 6.1.

72 6.4. A Poncelet Criterion

B 2 B˜1

A ˜ 2 A3 Oα B3 A˜1 P A3 A1 A˜2 B˜2

Oβ B˜3 B1

Figure 6.1: The triangle B1,B2,B3 induces another triangle B˜1, B˜2, B˜2 via P = [1, 0, 0]T .

T Now, look at B1 = [1, y1, z1] . The secant of Oβ through B1 and B˜1 is given by

s1 = V(z1y y1z). − y1 In the case z1 = 0 we get the relation y = z. Intersecting this line with the conic 6 z1 Oα gives 2 2 z1 z = 2− 2 αy1 + cαz1 and using B1 Oβ gives ∈ 2 −1 2 z = α βz1.

With this, we can calculate the intersection points for Oα, namely

p −1 p −1 T A2 = [1, y1 α β, z1 α β] and p −1 p −1 T A˜2 = [1, y1 α β, z1 α β] . − − Using (6.10), we obtain the condition

p −1 (1 + 2 α β)z1 = 0.

p −1 Since we are in the case z1 = 0 it follows 1 + 2 α β = 0, which implies α = 4β. 6 In the case z1 = 0, we directly deduce z = 0 for the secant through B1 and B˜1. Intersecting with Oα gives the two points

−1 T A2 = [1, √ α , 0] − and −1 T A˜2 = [1, √ α , 0] . − − −1 Applying (6.10) we get the condition y1 2√ α = 0 and using B1 Oβ yields again 4β = α. ± − ∈

Remark 6.4.2. Recall that for Oα < Oβ we have to check whether or not β(β α) is a nonsquare. Hence, in the case 4β = α we have to check whether or not −3β2 is a nonsquare, which is the same as checking whether or not 3 is a nonsquare.− − 73 6.4. A Poncelet Criterion

For p an odd prime, we can compare this to well-known results from number theory (see [19]). For p 1(4), we have that 3 is a nonsquare if and only if 3 (p + 1) and for p 3(4), we≡ have that 3 is a square if and only if 3 (p + 1). In both| cases we therefore≡ have | 3 nonsquare 3 (p + 1). − ⇔ | This gives already a necessary condition for the existence of Poncelet triangles for pairs (Oα,Oβ) in PG(2, p), p an odd prime. By Poncelet’s Theorem for such pairs, as seen in Theorem 6.2.1, the existence of a Poncelet triangle implies 3 (p + 1), as | there are p+1 points on the conic Oβ. This is exactly the condition given by number theoretic results as well.

Using arguments as above, one easily checks the following result.

Lemma 6.4.3. Let Oα < Oβ be two conics in PG(2, q), such that a 4-sided Poncelet polygon can be constructed. Then 2β = α in GF (q).

The main goal is to ﬁnd such a relation for all possible n-sided Poncelet polygons. For this, we ﬁrst investigate which Poncelet n-gons occur in a given plane PG(2, q). Note that this can be done just by applying Poncelet’s Theorem and the Euler divisor sum formula, since we are dealing with a very special family of conics.

Lemma 6.4.4. For a given conic Oβ in PG(2, q) and every n (q + 1) there are φ(n) | exactly 2 conics Oα, such that Oα < Oβ carries a Poncelet n-gon.

q−1 Proof. By Lemma 6.3.6, there are exactly 2 conics Oα, such that Oα < Oβ. More- over, we know that once Oα < Oβ, starting with any point of Oβ leads to a Poncelet polygon. Because of Theorem 6.2.1, the length of this Poncelet polygon has to divide q + 1, i.e. the number of points on Oβ. Recall now Euler’s divisor sum formula for the totient function (see [19]), which states X φ(n) = m n|m for any integer m. Applied to the points of the conic, we have X φ(n) = q 1 − n|(q+1), n≥3 which is the same as X φ(n) q 1 = − . 2 2 n|(q+1), n≥3

φ(n) Hence, there are exactly 2 conics Oα such that Oα < Oβ carries a Poncelet n-gon for every divisor n of q + 1.

The next result reduces the problem of ﬁnding necessary relations for all n-sided Poncelet polygons, such as 4β = α for n = 3, to those with n odd.

74 6.4. A Poncelet Criterion

Lemma 6.4.5. Let (Oβk,Oβ) be a pair of conics in PG(2, q) which carries an n- sided Poncelet polygon for k a square in GF (q). Then (Oβk˜,Oβ) carries a 2n-sided Poncelet polygon for 2 k˜ = 1 √1 − k where only those roots are taken such that k˜ = k. 6

Proof. Let Oα < Oβ be some pair of conics which carries a 2n-sided Poncelet polygon. To calculate the relation between α and β we use that Bi + Bi+1 = Ai for two consecutive vertices of the polygon, as seen in Lemma 6.3.8. Hence,

B1 + B2 Oα, ∈ T i.e. [2, y1 + y2, z1 + z2] Oα which gives immediately ∈ 4 α = 2 − 2 . (y1 + y2) + c(z1 + z2)

2 2 −1 Since B1 Oβ and B2 Oβ, we know that yi + czi = β for i = 1, 2 and we obtain ∈ ∈ − 2β α = . 1 β(y1y2 + cz1z2) − The claim is α = βk˜, hence we have to show the equality 2β 2β = 1 1 β(y1y2 + cz1z2) 1 √ − − k which is equivalent to 1 + β( y1)y2 + cβ( z1)z2 = 0. √k − − The expression above can be interpreted as the incidence relation

1 T [1, βy2, cβz2] , y1, z1 = 0 · √k − − which is the same as 1 T , y1, z1 Tβ(B2), √k − − ∈ where Tβ(B2) denotes the tangent of Oβ in B2. This can be done for all pairs of points Bi,Bi+1 Oβ. We get the conditions ∈ 1 T 1 T , y2`−1, z2`−1 , , y2`+1, z2`+1 Tβ(B2`) √k − − √k − − ∈ for ` = 1, . . . , n where indices are taken cyclically. Exactly n tangents of the conic Oβ are involved. The conditions above are equivalent to showing that the n intersection points are on some conic Oγ and form an n-sided Poncelet polygon with Oβ. Observe

75 6.4. A Poncelet Criterion

T T h 1 i h 1 i that, by Lemma 6.3.14, Bi+n = B˜i, and hence √ , yi+n, zi+n = √ , yi, zi . k − − k Therefore, we have to verify that

1 T , yi, zi Oγ √k ± ± ∈ for i = 1, . . . , n and β = γk. For γ, we directly obtain

x2 O = V( + γy2 + cγz2). γ k

T Since all the points [1, yi, zi] lie on Oβ, we indeed get β = γk. By Lemma 6.3.4 , since (Oβk,Oβ) carries an n-sided Poncelet polygon, so does (Oγk,Oγ) which is what we wanted to show.

Corollary 6.4.6. Let Oα and Oβ be conics in PG(2, q) such that a 2n-sided Poncelet polygon exists. Then there exists another conic Oγ such that the pair (Oγ,Oβ) carries an n-sided Poncelet polygon.

Proof. Let Oα < Oβ such that a 2n-sided Poncelet polygon can be constructed and α = hβ. By Lemma 6.3.2 we know that β(β α) is a nonsquare, so 1 h is a nonsquare in our case. To show the statement− above, we only have to show− that for γ = kβ, 1 h is a nonsquare if and only if 1 k is a nonsquare. This follows immediately by− our formula for 2n-sided Poncelet− polygons seen in Lemma 6.4.5, namely 2 (√k + 1)2 1 h = 1 = . − − 1 √1 1 k − k − This gives us (1 h)(1 k) = (√k + 1)2 − − and hence (1 h) is a nonsquare if and only if (1 k) is a nonsquare. − − Example 6.4.7. We have already seen in Lemma 6.4.3 that if (Ok,O1) forms a 4- sided Poncelet polygon, we immediately have k = 2. Hence by Lemma 6.4.5, we are able to compute the index h such that (Oh,O1) carries an 8-sided Poncelet polygon, namely 2 2 h = = = 4 2√2. 1 √1 1 √1 ± − k ± 2 This is only well deﬁned if 2 is a square. For GF (p), p an odd prime, we know from number theory (see [19]) that

2 is a square in GF(p) p 1(8). (6.11) ⇔ ≡ ± By Poncelet’s Theorem, the existence of an 8-gon already implies 8 (p + 1). Hence, the condition p 1(8) is again equivalent to a purely number theoretic| result. ≡ − The next goal is to deduce such relations for all n-sided Poncelet polygons, n odd. The main idea how to proceed lies already in the next result.

76 6.4. A Poncelet Criterion

Lemma 6.4.8. Let Ok < O1 carry an n-sided Poncelet polygon for the points

B1,...,Bn on O1, n odd. Then O k2 < O1 carries an n-sided Poncelet polygon as (k−2)2 well, for the same points B1,...,Bn on O1.

Proof. Let Ok < O1 such that an n-sided Poncelet polygon can be constructed for n T odd. By Lemma 6.3.8, we have Bi+Bi+1 Ok for all i = 1, . . . , n and Bi = [1, yi, zi] T ∈ and hence [2, yi + yi+1, zi + zi+1] Ok which gives ∈ 2 2 4 + k(yi + yi+1) + ck(zi + zi+1) = 0.

2 2 Using 1 + y + cz = 0 for all Bi O1 gives i i ∈ 2 k = , 1 (yiyi+1 + czizi+1) − which is equivalent to k k2 k2 + yi( yi+1) + c zi( zi+1) = 0. k 2 (k 2)2 − (k 2)2 − − − − This relation is the incidence relation 2 2 k k k T , yi+1, c zi+1 [1, yi, zi] = 0. k 2 −(k 2)2 − (k 2)2 · − − − Hence we need T k [1, yi, zi] T k2 , yi+1, zi+1 ∈ (k−2)2 k 2 − − − as well as T k [1, yi+1, zi+1] T k2 , yi, zi . ∈ (k−2)2 k 2 − − − Summarizing gives the condition T T k [1, yi+1, zi+1] , [1, yi−1, zi−1] T k2 , yi, zi . ∈ (k−2)2 k 2 − − −

This can be done for all i = 1, . . . , n and since n is odd, for O k2 < O1, an n-sided (k−2)2 Poncelet polygon is given via the same points B1,...,Bn.

Note that by Lemma 6.3.4 the conics O1 < Oβ2 carry an n-sided Poncelet polygon if and only if O 1 < O1 carries an n-sided Poncelet polygon. β2

Remark 6.4.9. We have seen that for triangles there is only one conic Ok such that Ok < O1 form a 3-sided Poncelet polygon, namely O4. In this case, we should therefore have k2 k = , (k 2)2 − which is equivalent to k2 5k + 4 = 0. − The only solutions are k = 1, which can be excluded, and k = 4, which we already computed in Lemma 6.4.1 by using other methods.

77 6.4. A Poncelet Criterion

The procedure shown in the proof above can be iterated. To avoid long expressions, we use 2 ti ti+1 := 2 , (6.12) (ti 2) − for t0 := k. Recall that for a given Poncelet n-gon using the points B1,...,Bn on φ(n) O1 and tangents of some Oα, there are 2 1 other conics Oγ such that (Oγ,O1) carries an n-sided Poncelet polygon. −

Example 6.4.10. We know that for Oα < O1 a 5-sided Poncelet polygon for the same φ(5) ﬁve points B1,...,B5 Oα can be constructed in two diﬀerent ways, since = 2. ∈ 2 Fix an ordering of the points B1,...,B5 and start with the polygon

A1 A2 A3 A4 A5 B1 B2 B3 B4 B5 B1. −→ −→ −→ −→ −→

The other 5-gon is then given by connecting Bi and Bi+2, namely

C1 C2 C3 C4 C5 B1 B3 B5 B2 B4 B1. −→ −→ −→ −→ −→

Note that connecting Bi and Bi+3 gives in fact the same polygon again, since we can read the above polygon by reversing the direction.

Figure 6.2: Two diﬀerent 5-sided Poncelet polygons can be constructed using the same ﬁve points on the outer conic.

For 5-sided Poncelet polygons, we therefore get the conditions t0 = t1 and t0 = t2. We have to solve 6 k4 k = , (k2 2(k 2)2)2 − − which is equivalent to

(k 1)(k 4)(16 12k + k2) = 0. − − − We obtain the four solutions n o k 1, 4, 6 + 2√5, 6 2√5 . ∈ −

Since k = 1 and k = 4 solve t0 = t1, we ﬁnd that k = 6 2√5 implies that if ± Ok < O1, then (Ok,O1) carries a 5-sided Poncelet polygon. For GF (p), p and odd prime, a result by Gauss about quadratic residues (see [19]) can be used, namely

5 is a square in GF(p) p 1(5). (6.13) ⇔ ≡ ± 78 6.4. A Poncelet Criterion

Hence in all planes PG(2, p), in which 5 actually divides p + 1, the square root of 5 is well-deﬁned and the indices of the Poncelet 5-gons given by 6 2√5 can be computed. ±

Finally, we can prove the theorem how to find the indices k, such that (Ok,O1) carries an n-sided Poncelet polygon for n odd. Theorem 6.4.11. Let n be any odd number, n 3. Then the indices k such that ≥ (Ok,O1) carries an n-sided Poncelet polygon in a plane PG(2, q) are given by the solutions of t0 = t φ(n) , (6.14) 2 φ(n) where we need t0 = ti in GF (q) for all i < 2 . For a fixed plane PG(2, q), these solutions are called6 Poncelet coefficients for n-sided Poncelet polygons and denoted i φ(n) by kn, i = 1,..., 2 .

Proof. Let Ok < O1 carry an n-sided Poncelet polygon for the points B1,...,Bn, n odd. Let the points be ordered such that for Ot0 < O1 we have

B1 B2 B3 ... Bn B1. → → → → → We have seen in the proof of Lemma 6.4.8 that the n-sided Poncelet polygon of

Ot1 < O1 is given by the order

B1 B3 B5 ... Bn B2 ... Bn−1 B1. → → → → → → → →

Iterating this, we see that the n-sided Poncelet polygon given by Oti < O1 has the order B1 B i B i B i ... B1, → 1+2 → 1+2∗2 → 1+3∗2 → φ(n) where the indices are taken cyclically. We already know that there are exactly 2 diﬀerent Poncelet n-gons. Since we are only working with n odd, we can apply Fermat’s little Theorem (see [19]) and use

φ(n) 2 2 1(n). ≡ ±

This shows directly that for Ot φ(n) we start the polygon by B1 B2 or B1 Bn and 2 → → hence the polygon is equivalent with the very first one. To deduce the coefficients k such that (Ok,O1) carries an n-sided Poncelet polygon, we therefore indeed have to solve (6.14). Remark 6.4.12. Note that for some values of n, the iteration needs fewer steps than φ(n) φ(n) 2 , as the order of 2 modulo n can be smaller than 2 . In these cases, not all indices can be constructed by starting with one Poncelet n-gon only. Nevertheless, the condition (6.14) stays the same but the same coefficients could be derived by computing less, i.e.

t0 = tsn and t0 = ti for all i < sn, where 6 s sn := min s 2 1(n) . { | ≡ ± } φ(n) φ(17) 4 The smallest example for = sn is n = 17, where we have = 8 but 2 2 6 2 ≡ 1(17), i.e. s17 = 4. − 79 6.4. A Poncelet Criterion

Example 6.4.13. We want to deduce the indices k such that Ok < O1 carries a 9-sided φ(9) Poncelet polygon in PG(2, 53). Since 2 = 3, we have to solve

t0 = t3, t1 = t3, t2 = t3 6 6 in GF (53). So, we need solutions of k8 t0 t3 = k = 0 − − (128 + k( 256 + k(160 + ( 32 + k)k)))2 − − Rewriting this equation, we have to solve

k8 k(128 + k( 256 + k(160 + ( 32 + k)k)))2 = 0 − − − We obtain the solutions k 1, 4, 13, 36, 40 . ∈ { } Since we can exclude the solutions 1 and 4, as they also solve t2 = t3, we deduce that O13 < O1,O36 < O1,O40 < O1 are the pairs of conics in PG(2, 53) such that a 9-sided Poncelet polygon can be constructed.

6.4.2 Poncelet polynomials

We are now able to give an algorithm to deduce for all pairs (Oα,Oβ) in PG(2, q), whether an n-sided Poncelet polygon can be constructed for a given n. We use the iteration method described before to ﬁnd polynomials Pn(k) such that the zeros belong to the coeﬃcients k of conics Ok, such that if Ok < O1, then (Ok,O1) carries an n-sided Poncelet polygon. By Lemma 6.3.4, this gives information about all pairs (Oα,Oβ).

Definition 6.4.1. A polynomial Pn with integer coefficients is called Poncelet polynomial for n-sided Poncelet polygons, if the zeros in GF (q) correspond to the coefficients k, such that Ok < O1 carries an n-sided Poncelet polygon in PG(2, q).

Example 6.4.14. We have already seen in Lemma 6.4.1 that P3(k) = k 4 and in 2 − Example 6.4.10 that P5(k) = 16 12k + k . − φ(n) By Lemma 6.4.4, we know that all these polynomials Pn are of degree 2 , as the existence of one conic Ok, such that (Ok,O1) carries an n-sided Poncelet polygon φ(n) in PG(2, q) leads to 2 such conics Ok. Until now, we only know how to produce Poncelet polynomials Pn for n odd, but similar to the Poncelet coefficients k, a doubling process can be applied for finding Pn with n even. Note that to find the coefficients for an odd n-sided Poncelet polygon, we look for indices k, such that

Pn(k) = 0 in GF (q). Applying Lemma 6.4.5 gives

φ(n) k2 (k 2) Pn 2 − (k−2) P2n(k) = Pn(k)

80 6.4. A Poncelet Criterion for n odd and iterating once more, we get

2 φ(n) k P2n(k) = (k 2) Pn − (k 2)2 − for n even.

Example 6.4.15. We have P3(k) = 4 + k and φ(3) = 2. Hence we have to calculate − 2 2 2 k 2 k (k 2) P3 = (k 2) 4 + = ( 4 + k)( 4 + 3k). − (k 2)2 − − (k 2)2 − − − − − Dividing by P3(k) gives P6(k) = 4 3k. − For the general case, note that for numbers n and m which have the same value φ(n) φ(n) = φ(m), it has to be checked by hand, which polynomials of degree 2 given by the iteration belong to the n-gons and which to the m-gons. For example, the φ(n) iteration for 2 = 3 gives the polynomial ( 4 + k)( 1 + k)k( 64 + 96k 36k2 + k3)( 64 + 80k 24k2 + k3). − − − − − − − Excluding the factors (k 4) and (k 1) which already occur at the ﬁrst iteration, we ﬁnd checking by hand− −

2 3 2 3 P7(k) = 64 + 80k 24k + k and P9(k) = 64 + 96k 36k + k . − − − − With some computational effort, we are now able to create a list of all Poncelet polynomials Pn up to a chosen value of n. In Section 6.5 we will compare our findings to results in the real projection plane. This will finally lead to an explicit formula for the Poncelet polynomials (see Theorem 6.5.2).

Using the list of all Poncelet polynomials Pn, n 3, we can write down an algorithm ≥ to ﬁnd the numbers k, such that (Ok,O1) form an n-sided Poncelet polygon, for all Ok < O1 in a plane PG(2, q). Hence, by recollecting the results described in this discussion, we obtain the algorithm we were looking for. Corollary 6.4.16. The following four steps give a complete description of n-sided Poncelet polygons for conic pairs (Oα,Oβ) in PG(2, q).

φ(n) 1. Deduce all n 3 with n (q + 1). For every such n, calculate 2 , which gives the number≥ of indices| k, such that an n-sided Poncelet polygon can be constructed for (Ok,O1).

2. For all values n obtained in Step 1, look up the Poncelet polynomial Pn.

3. For every Pn deduced in Step 2, solve Pn(k) = 0 in GF (q). This gives the corresponding Poncelet coeﬃcients k, such that an n-sided Poncelet polygon can be constructed for (Ok,O1). 4. By using the coordinate transformation described in Lemma 6.3.4, transform the information obtained in Step 3 to all pairs Oα < Oβ.

Example 6.4.17. We want to deduce all relations of conic pairs Oα < Oβ in the plane PG(2, 11) by using the algorithm above.

81 6.5. Comparison to other methods

Step 1: The values n, such that an n-sided Poncelet polygon can be con- • structed, are given by n = 3, 4, 6, 12. Moreover:

n 3 4 6 12 φ(n) 2 1 1 1 2 Step 2: We have the following Poncelet polynomials: • P3(k) = 4 + k − P4(k) = 2 + k − P6(k) = 4 + 3k − 2 P12(k) = 16 + 16k k − − Step 3: The zeros of the Poncelet polynomials in GF (11) are given by: • P3(k) = 0 k = 4 ⇔ P4(k) = 0 k = 2 ⇔ P6(k) = 0 k = 5 ⇔ P12(k) = 0 k = 6, 10 ⇔ Step 4: By suitable collinear transformations, we obtain the complete relation • table:

O1 O2 O3 O4 O5 O6 O7 O8 O9 O10

O1 12 3 4 6 12 O2 4 12 3 6 12 O3 6 12 4 12 3 O4 3 4 6 12 12 O5 6 3 12 4 12 O6 12 4 12 3 6 O7 12 12 6 4 3 O8 3 12 4 12 6 O9 12 6 3 12 4 O10 12 6 4 3 12

6.5 Comparison to other methods

6.5.1 Comparison to the Euclidean Plane

2 2 2 Recall that any point on Ok = V(x + ky + ckz ) has a nonzero x-coordinate. Because of this, we can project these conics on the aﬃne plane by setting x = 1. Moreover, we can look at real solutions of the equations. In the proof of Poncelet’s Theorem for this family of conics, we have seen that there is an aﬃne transformation which maps the whole family into a family of concentric circles. Let us therefore consider pairs of such circles in the Euclidean plane, i.e. 2 2 E1 : x + y = 1 2 2 2 Er : x + y = r , r > 1

82 6.5. Comparison to other methods

We try to find a suitable radius r for Er, such that a regular n-sided polygon which is inscribed in Er and circumscribed around E1 can be constructed. It is elementary that one solution to this problem, namely the circumcircle radius r of a simple, regular n-sided polygon is given by 1 r = π . cos( n ) In terms of Poncelet coefficients as defined for the finite case, this gives 1 kn = 2 π . cos ( n ) Example 6.5.1. The radius r for a simple, regular 5-gon is therefore given by r = 1 √ √ 2 √ cos( π ) = 1 + 5. Note that ( 1 + 5) = 6 2 5, which is exactly one of the 5 − − − zeros of the Poncelet polygon for 5-gons we obtained over finite fields (see Example 6.4.10). The second radiusr ˜, which corresponds to the complex 5-gon circumscribed 1 about E1, can be calculated as well, namely byr ˜ = 2π , which leads tor ˜ = 1+√5. cos( 5 ) Hence we obtainr ˜2 = 6 + 2√5, which belongs to the second coefficient for 5-gons obtained in the finite case.

Now we turn our attention to the formula deduced for the coeﬃcients k˜ for 2n-sided Poncelet polygons in Lemma 6.4.5. For this, note that

φ 1 + cos(φ) cos2 = . 2 2

Hence we get

˜ 1 2 2 2 k = 2 π = π = 1 = 1 , (6.15) cos ( ) 1 + cos( ) 1 r 1 √ 2n n 1 k − 2 π − cos ( n ) which is exactly the formula derived for the ﬁnite case. π Since there does not exist a radical expression for cos( n ) for all integers n, it is con- 1 venient to look again at polynomials with roots 2 kπ . These are closely connected cos ( n ) to the n-th cyclotomic polynomials Φn(x). Recall, that those polynomials can be written as Y 2πik Φn(x) = (x e n ). − 1≤k≤n,(k,n)=1

It is immediate that the degree of Φn is φ(n), the Euler totient function. The zeros 2πik 1 n of Φn(x) are given by e for (k, n) = 1. For a zero x of Φn, also x = x is a zero. Deﬁne 1 − φ(n) q x + := Φ (x)x 2 . n x n

2πik 2πk The zeros of qn are then given by 2 (e n ) = 2 cos( ). Next, deﬁne < n

rn(x) := qn(2x)

83 6.5. Comparison to other methods

2πk which has zeros cos( n ). In the next step, we consider

sn(x) := rn(2x 1) − 1+cos( 2πk ) which has zeros n = cos2( πk ) for k = 1, . . . , n 1. Finally, consider 2 n − φ(n) 1 P˜ (x) = x 2 s n n x

1 with zeros 2 πk , which is exactly the polynomial we wanted. Summarizing, we cos ( n ) have the following explicit formula for the Poncelet polynomials.

Theorem 6.5.2. The Poncelet polynomial P˜n for n 3 is given by ≥ φ(n) − φ(n) P˜n(x) = x 2 Φn(z)z 2 (6.16) √ for 2−2 1−x . Moreover, the zeros of ˜ are 1 for . z = x 1 Pn cos2( πk ) (k, n) = 1 − n 4 3 Example 6.5.3. For n = 5, the cyclotomic polynomial is given by Φ5(x) = x + x + x2 + x + 1, which leads to

2 P˜5(x) = 16 12x + x − which indeed has the zeros 6 + 2√5 and 6 2√5. Note that this is the Poncelet polynomial for 5-gons derived in the ﬁnite case.−

It remains to show that the polynomials P˜n derived in (6.16) are the same as the Poncelet polynomials derived in the chapter before. Remember that we started by iterating the result of Lemma 6.4.8, which states that if Ok < O1 carries a Poncelet n-gon, then O k2 < O1 as well. Recall (k−2)2

2 ti t0 := k, ti+1 := 2 . (ti 2) − Note that 2 ti 2 ti+1 = 2 ti = 1 (ti 2) ⇔ 1 √ − − ti+1 and hence, by the result in (6.15), we actually double our angle at each step. We φ(n) ˜ 1 already know that the 2 zeros of Pn are given by 2 kπ for (k, n) = 1. Hence, cos ( n ) 1 we only have to show that 2 kπ solves t0 = t φ(n) . This is equivalent in showing cos ( n ) 2 1 1 = (6.17) 2 kπ φ(n) cos ( n ) 2 2 2 kπ cos n for all odd n and k < n,(k, n) = 1. For this, note ﬁrst the following two immediate equations for any integer k, namely π π cos = cos 2kπ n ± n 84 6.5. Comparison to other methods and π π cos = cos (2k + 1)π . n − ± n Hence, for all integers k, we know π (kn 1)π cos = cos ± n n By Fermat’s little Theorem, we know that for any odd integer n, we have

φ(n) 2 2 1(n). ≡ ± φ(n) Hence, there exists a k, such that 2 2 = kn 1, which implies (6.17). For n odd, ˜ ± φ(n) φ(n) 1 Pn and Pn are both monic polynomials of degree 2 with the 2 zeros 2 kπ for cos ( n ) (k, n) = 1. Hence, they are indeed the same.

6.5.2 Comparison to Cayley’s Criterion

The criterion deduced by Cayley in 1853 (see [15]) reads as follows. Theorem 6.5.4. Let C and D be the matrices corresponding to two conics generally situated in the projective plane. Consider the expansion

p 2 3 det(tC + D) = A0 + A1t + A2t + A3t + ...

Then an n-sided Poncelet polygon with vertices on C exists if and only if for n = 2m + 1, we have   A2 ...Am+1  . .  det  . ... .  = 0 Am+1 ...A2m and for n = 2m, we have   A3 ...Am+1  . .  det  . ... .  = 0 Am+1 ...A2m

In the discussion above, we were mainly interested in pairs of conics (Ok,O1) with equations 2 2 2 Ok = V(x + ky + ckz ) 2 2 2 O1 = V(x + y + cz ) To apply Cayley’s criterion, we therefore have to look at the expansion of the square root of t + 1 0 0  det  0 t + k 0  0 0 c(t + k) which is given by

(k + 2)√ck2 (k 4)√ck2 (k 2)√ck2 (5k 8)√ck2 √ck2 + t − t2 + − t3 − t4 + O(t)5 2k − 8k 16k − 128k 85 6.5. Comparison to other methods

Example 6.5.5. The condition for a 3-sided√ Poncelet polygon is given by vanishing 2 (k−4) ck2 of the coefficient of t which is A2 = 8k . This expression is zero if and only if k 4 = 0, which is exactly the condition derived in Lemma 6.4.1 for the finite case. − 2 Example 6.5.6. The condition for 5-sided Poncelet polygons is given by A2A4 A3 = c((k−12)k+16) 2 − 0, which is the same as 1024 = 0. This is equivalent to k 12k + 16 = 0, so again, we obtain the same condition as for the finite case (compare− to Example 6.4.10).

86 Chapter 7

Simultaneous Diagonalization

Recall that we are interested in conics given by the zero set of polynomials E of degree two, E = ax2 + by2 + cz2 + dxy + exz + fyz (7.1) where a, b, c, d,e, f GF (q). ∈ For these polynomials, let us first clarify what we mean by diagonal: Definition 7.0.1. We call a polynomial (7.1) diagonal, if the coefficients of the mixed terms xy, xz and yz are zero. We call a polynomial diagonalizable if there is a collineation which maps the zeros of (7.1) to the zeros of a diagonal polynomial.

Let C be a proper conic with matrix representation A and S any regular matrix. For any point P on C, SP lies on C˜, which is given by the matrix representation (S−1)T AS−1, since

T P T AP = 0 (SP )T (S−1) AS−1SP = 0. ⇔ Therefore, a symmetric matrix A is called diagonalizable, if there exists a regular matrix S, such that ST AS has diagonal form, which means that all entries off the diagonal are zero. In this chapter we are interested in the existence of a regular matrix S such that for two given matrices A and B, the matrices ST AS and ST BS are both in diagonal form. Remark 7.0.1. Note that this definition of diagonalizable differs from the definition one might expect from linear algebra, since we are interested in finding a nonsingular matrix S with ST AS in diagonal form, but S is not necessarily orthogonal, i.e. in general, we have ST = S−1. What we are doing is to think about the existence of a collineation which takes6 two conics simultaneously into standard form.

7.1 The disjoint case

We have seen in Corollary 2.5.3 that a pencil given by two disjoint conics leads to a partition of the plane. We are now interested in the possible shapes of these partitions. We know that any conic given by the zero set of a polynomial (7.1) corresponds either to a proper conic, a point, a line or a pair of lines. Recall that

87 7.1. The disjoint case any two distinct lines always intersect in a unique point, hence when starting with two disjoint conics, at most one line or one pair of lines can occur in the whole partition. Since every point has to be contained in exactly one of the q + 1 conics in the pencil, this observation leads us to exactly three possible shapes for the pencil. In particular, we end up with either q, q 1 or q 2 proper conics. According to this, we deﬁne: − −

Deﬁnition 7.1.1. There are three possible shapes of a partition of the plane PG(2, q) given by a pencil of disjoint conics. They are denoted by q-form for the partition P,C1,...,Cq , by (q 1)-form for the partition P, g, C1,...,Cq−1 and by (q 2)- { } − { } − form for the partition P, Q, gg,˜ C1,...,Cq−2 , where P , Q are points, g is a line, { } gg˜ a pair of lines and Ci proper conics, for 1 i q. ≤ ≤ By analyzing these three shapes, we ﬁnd the following condition for simultaneous diagonalization:

Theorem 7.1.1. Two disjoint conics V1 and V2 in PG(2, q) are simultaneously diagonalizable if and only if their pencil (V1,V2) leads to a partition of (q 1)- form or (q 2)-form. P − − Proof. In the ﬁrst part of the proof, we will see that starting with a pencil of two diagonal polynomials yields into a (q 1)-form or a (q 2)-form. For this, let E1 − − and E2 be the polynomials deﬁning V1 and V2, respectively. Consider

2 2 2 2 2 2 E1 = x + b1y + c1z and E2 = x + b2y + c2z where b1, b2, c1, c2 = 0 and b1 = b2 or c1 = c2, hence we start with two distinct proper conics. Note that6 this is no restriction6 for6 q 5, since in every possible form, there are at least two proper conics and by Lemma≥ 2.5.1, the pencil is independent of the choice of representatives. For q = 3, all pencils in 1-form can be analyzed by hand.

Considering the pencil (V1,V2), we see that V(E1 E2) leads to a degenerate conic. b1 P c1 − Moreover, V(E1 E2) and V(E1 E2) lead to one or two more degenerate conics, − b2 − c2 since b1 = b2 or c1 = c2. Hence, we indeed end up with a pencil in (q 1)-form or (q 2)-form.6 6 − − For the other direction, we have to show that all pairs of disjoint conics with a pencil in (q 1)-form or (q 2)-form can be diagonalized simultaneously. For this, let us ﬁrst look− at a pencil in− (q 1)-form. − We start with two disjoint conics such that their pencil is of (q 1)-form, which means that exactly two degenerate conics occur, namely one point− P and one line g. We can now apply a collineation to the whole pencil, such that, without loss of generality, the point P is given by [1, 0, 0]T and the line g is incident with [0, 1, 0]T and [0, 0, 1]T . So we can consider the pencil given by the two elements g = V(x2) and P = V(y2 + cz2 + fyz), where c and f are chosen such that y2 + cz2 + fyz is irreducible. The elements of the pencil (P, g) are then given by: P

88 7.1. The disjoint case

P = V(y2 + cz2 + fyz) g = V(x2) 2 2 2 C0 = V(x + y + cz + fyz) 2 2 2 C1 = V(x + αy + αcz + αfyz) 2 2 2 2 2 2 C2 = V(x + α y + α cz + α fyz) . . 2 q−2 2 q−2 2 q−2 Cq−2 = V(x + α y + α cz + α fyz)

Now we can apply another collineation such that all these conics are in diagonal form, namely φS−1 , for the following matrix

1 0 0  f S = 0 1 2  . 0 0− 1

Indeed, we have 0 0 0  T S AS = 0 1 0  , 2 0 0 c f − 4 where A is the matrix representation of P . Note that applying φS−1 to g gives g again. Since φS−1 (Ei + Ej) = φS−1 (Ei) + φS−1 (Ej), we indeed end up with a pencil of diagonal conics, namely:

P˜ = V(y2 +cz ˜ 2) g˜ = V(x2) 2 2 2 C˜0 = V(x + y +cz ˜ ) 2 2 2 C˜1 = V(x + αy + αcz˜ ) 2 2 2 2 2 C˜2 = V(x + α y + α cz˜ ) . . 2 q−2 2 q−2 2 C˜q−2 = V(x + α y + α cz˜ )

2 forc ˜ = c f . − 4 Now we look at a pencil in (q 2)-form. Let us start with two proper disjoint conics C and C˜. The pencil (C, C˜−) contains a pair of lines and two points. By applying a suitable collineation,P we can assume one of the points to be P = [1, 0, 0]T , i.e. P = V(y2 + cz2 + fyz) for c and f chosen such that y2 + cz2 + fyz is irreducible. Moreover, by Lemma 2.3.6, we can assume the pair of lines to be given by gg˜ with g through [1, 1, 0]T and [0, 0, 1]T andg ˜ through [1, 1, 0]T and [0, 0, 1]T , so we have gg˜ = V(x2 y2). With the same collineations as before,− we can start with diagonal conics for P− and gg˜ and obtain the following pencil:

89 7.1. The disjoint case

P = V(y2 + cz2) gg˜ = V(x2 y2) 2 − 2 C0 = V(x + cz ) 2 2 2 C1 = V(x + (α 1)y + αcz ) 2 2− 2 2 2 C2 = V(x + (α 1)y + α cz ) − . . 2 q−2 2 q−2 2 Cq−2 = V(x + (α 1)y + α cz ) −

Note that C0 corresponds to the second point in the pencil. Again, we end up with diagonal forms only. Remark 7.1.2. It is well known that all proper conics in PG(2, q2) are projectively equivalent to x2 + y2 + z2 = 0, see [20, Section 5]. In particular, any conic can be diagonalized. This can also be seen as a consequence of the above result. To see this, just choose any line g disjoint from C and look at the pencil (C, g), which is by construction of (q 1)-form. P − Example 7.1.3. Let V1 = C and V2 = C˜ be two proper disjoint conics in PG(2, 5) given by C = V(xy + 3xz + yz) and C˜ = V(x2 + y2 + z2 + xz). We obtain:

2 2 2 T V3 = V(x + y + z + xy + 4xz + yz) = [1, 4, 1] 2 2 2 V4 = V(x + y + z + 2xy + 2xz + 2yz) = [0, 1, 4]T , [1, 0, 4]T , [1, 1, 3]T , [1, 2, 2]T , [1, 3, 1]T , [1, 4, 0]T 2 2 2 V5 = V(x + y + z + 4xy + 3xz + 4yz) 2 2 2 V6 = V(x + y + z + 3xy + 3yz)

Since there are four proper conics in this pencil, we can diagonalize C and C˜ simul- T taneously. The ﬁrst step is to perform a collineation φS which maps V3 to [1, 0, 0] 2 and V4 to the line V(x ), i.e. we apply 1 0 1 −1 S = 4 1 0 1 4 4 to V3 and V4, which leads to P = V(y2 + 3z2 + 4yz) and g = V(x2). As the polynomial deﬁning P is not in diagonal form, we have to perform one more collineation φT , given by the matrix 1 0 0 −1 T = 0 1 1 . 0 0 1

90 7.1. The disjoint case

Combining these two steps gives the following pencil:

P˜ = V(y2 + 2z2) g˜ = V(x2) 2 2 2 C˜1 = V(x + y + 2z ) 2 2 2 C˜2 = V(x + 2y + 4z ) 2 2 2 C˜3 = V(x + 4y + 3z ) 2 2 2 C˜4 = V(x + 3y + z )

Hence, we ﬁnd a collineation φM with

1 0 1 1 0 0 1 0 1 −1 −1 −1 M = S T = 4 1 0 0 1 1 = 4 1 1 1 4 4 0 0 1 1 4 3 and 1 0 0 1 0 0 ˜ φM (C) = 0 4 0 and φM (C) = 0 3 0 . 0 0 3 0 0 1

Our next goal is to show that not all pairs of proper disjoint conics in PG(2, q) can be diagonalized simultaneously, i.e. we have to show that pencils in q-form actually exist.

Lemma 7.1.4. Let C be a proper conic in PG(2, q). Then there are 1 (2q 3q2 +q3) 2 − proper conics C˜, such that (C, C˜) is in (q 1)-form, there are 1 ( 6q + 5q2 + P − 4 − 5q3 5q4 + q5) proper conics C˜ such that (C, C˜) is in (q 2)-form and there are 1 (6q− 3q2 7q3 + 3q4 + q5) proper conicsPC˜ such that (C,−C˜) is in q-form. 8 − − P Proof. Let C be any fixed proper conic. To find the number of proper conics C˜ such that (C, C˜) is in (q 1)-form, we can also look at the number of pencils (C, g), P − q(q−1) P for g an external line of C. There are exactly 2 external lines of C. Every such pencil (C, g) leads to (q 2) proper conics C˜. Note that by Lemma 2.5.1, C˜ (C, g)P and C˜ (C, h) for− two lines g and h implies g = h. Hence, there are ∈ P ∈ P q(q−1)(q−2) = 1 (2q 3q2 + q3) proper conics C˜, such that (C, C˜) is in (q 1)-form. 2 2 − P − Similarly, to find the number of proper conics C˜ with (C, C˜) in (q 2)-form, we can also look at the number of pencils (C, gg˜), for gg˜ a pairP of external− lines of C. There P are q(q−1) ( q(q−1) 1) choices for gg˜ and in every such pencil there are q 3 proper 2 2 − − conics C˜. Hence, there are q(q−1) ( q(q−1) 1)(q 3) = 1 ( 6q + 5q2 + 5q3 5q4 + q5) 2 2 − − 4 − − proper conics C˜, such that (C, C˜) is in (q 2)-form. P − By Lemma 2.4.2, there are 1 (2q 5q2 + 7q3 7q4 + 3q5) proper conics C˜, which 8 − − are disjoint from a given conic C. Subtracting the number of pencils (C, C˜) in (q 1)-form and in (q 2)-form, we obtain exactly 1 (6q 3q2 7q3 +P 3q4 + q5) − − 8 − − proper conics C˜ such that (C, C˜) is in q-form. P Corollary 7.1.5. For every conic C, there exists a conic C˜, such that C and C˜ cannot be diagonalized simultaneously in PG(2, q).

91 7.1. The disjoint case

Proof. By Lemma 7.1.4, for every ﬁxed proper conic C, there are 1 (6q 3q2 7q3 + 8 − − 3q4 + q5) proper conics C˜ such that (C, C˜) is in q-form, a number which is strictly greater than zero, for all q 3. ByP Theorem 7.1.1, these pairs of conics C and C˜ cannot be diagonalized simultaneously.≥ Remark 7.1.6. In the real projective plane, all pairs of disjoint conics can be diagonalized simultaneously (see [35]). Note that in this case, two disjoint conics C and C˜ have the property that either all points of C are external points of C˜ or all points of C are internal points of C˜, as they do not intersect. In PG(2, q) however, the two conics can be disjoint without having this property, i.e. C can contain external points of C˜ as well as internal points. We will show in the remaining part of this subsection, that if all conics in a pencil are disjoint and for each pair of proper conics in this pencil, one conic consists of only external points or only internal points of the other one, simultaneous diagonalization is still possible.

We are therefore interested in the following property of conic pairs:

Deﬁnition 7.1.2. Let C and C˜ be two proper conics in PG(2, q). We say that C and C˜ are in nested position if every point of C is an external point of C˜ or every point of C is internal point of C˜ and vice versa, i.e. every point of C˜ is an external point of C or every point of C˜ is internal point of C.

Remark 7.1.7. Note that if C consists of internal points of C˜, that does not imply that C˜ consists of external points of C. In particular, in the ﬁnite projective plane PG(2, q) it is possible that two conics C and C˜ are such that C consists of internal points of C˜ and C˜ consists of internal points of C as well. Because of that, we need to write down both conditions in the deﬁnition above.

The interested reader can refer to [1] for information about the maximal number of external points on C with respect to another given conic C˜ as well as the number of conics C and C˜ which lie in nested position. Moreover, about the construction of conics consisting only of external points of a given conic, one can have a look at [32]. We will use the following well-known result about nested conics:

Lemma 7.1.8. Let C and C˜ be in nested position. Then C and C˜ are disjoint and have no tangents in common.

Proof. The points of C are either all external or all internal to C¯. Thus, C does not contain a point of C˜ and hence, they are disjoint. For the second property, we have to distinguish two cases. We start with C consisting of internal points of C˜ only. In this case, no point of C is incident with a tangent of C˜, i.e. the two conics have no tangents in common. For C consisting of external points of C˜ only, we know that through every point of C there are exactly two tangents of C˜ . There are q + 1 points on C and there are q + 1 tangents of C˜. As every point of C is incident with two tangents of C˜, also every tangent of C˜ need to be incident with two points of C, hence every tangent of C˜ is a secant of C. Remark 7.1.9. As a direct consequence of the deﬁnitions, we know that the property of two proper conics lying in nested position is invariant under collineations.

92 7.1. The disjoint case

Theorem 7.1.10. The only pencil of two proper disjoint conics C and C˜ in PG(2, q) for q odd, where all pairs of proper conics lie in nested position, is a pencil (C, C˜) in (q 1)-form. Moreover, a pencil (C, C˜) in (q 1)-form has the propertyP that all pairs− of proper conics lie in nestedP position. −

Proof. The ﬁrst statement can be proven by combinatorial considerations only, namely by excluding that such a pencil can be in q-form or (q 2)-form. There q(q+1) − are 2 external points for any ﬁxed proper conic C. Assume that all pairs of proper conics in a pencil in q-form are in nested position. For this, we have to consider two cases, namely the unique point P in the pencil being an external point or an internal point of C. Let P be an external point of C. By our assumption, this gives k(q + 1) + 1 external points of C, for an integer k, depending on the number of proper conics in the pencil consisting of external points of C only. This means

q(q + 1) k(q + 1) + 1 = 2

q 1 and therefore k = 2 q+1 , which is not an integer, because q is odd. Hence, this case is not possible. − Now, let P be an internal point of C. Again, by assumption, this would imply q(q+1) k(q + 1) = 2 for an integer k, which is not possible for q odd. Because of this, not all pairs of proper conics in a pencil of q-form can be in nested position. To exclude that such a pencil can be in (q 2)-form, assume again that all pairs of conics lie in nested position. Here, we have− to consider diﬀerent cases of how many points of gg,˜ P, P˜ are external points of C. The lines g andg ˜ intersect in exactly one point, say R. If R is an external point of C, there are two tangents of C through R. Since every two lines intersect, every other of the remaining q 1 tangents of C must intersect both g andg ˜ in diﬀerent points, which gives 1 + (q− 1) = q external points on gg˜. For the case that R is an internal point of C, by the− same argument, we obtain q + 1 external points on gg˜. Moreover, the two points P and P˜ can be either external points or internal points of C, which gives us in total q, q + 1, q + 2 or q + 3 external points not on any proper conic of the pencil. Hence, we obtain the condition q(q + 1) k(q + 1) + i = 2 for some integer k and i 1, 0, 1, 2 , which is not possible for q odd and q 5. Hence, not all pairs of proper∈ {− conics} in a pencil of (q 2)-form can be in nested≥ position. Note that the statement is trivially true for q−= 3 as well, since a pencil in q-form for q = 3 consists of only four elements, which are two points, one pair of lines and only one proper conic. Hence, we do not even have two proper conics in this case. To show the second statement, remember that by Lemma 7.1.9, the property of lying in nested position is invariant under collineations. By assumption, the pencil is in (q 1)-form and hence can be transformed into a pencil of diagonal conics, as shown− in Theorem 7.1.1. Therefore, it is enough to show that in the following pencil, where c is chosen to be a nonsquare, all pairs of proper conics lie in nested position:

93 7.2. The nondisjoint case

P = V(y2 + cz2) g = V(x2) 2 2 2 C0 = V(x + y + cz ) 2 2 2 C1 = V(x + αy + αcz ) 2 2 2 2 2 C2 = V(x + α y + α cz ) . . 2 q−2 2 q−2 2 Cq−2 = V(x + α y + α cz )

So, let Ci and Cj be given by

2 i 2 i 2 2 j 2 j 2 Ci = V(x + α y + α cz ) and Cj = V(x + α y + α cz ).

Since all points of PG(2, q) with a zero x-coordinate lie on the line g = V(x2), every T point P of Ci can be written in the form P = [1, p2, p3] and in particular, we have 2 −i 2 p2 = α cp3. The conics Ci and Cj lie in nested position if either for all such − − T points P of Ci,(AjP ) is a secant of Cj or this happens for no such point P , where Aj is the matrix representation of Cj, i.e.

1 0 0  j Aj = 0 α 0  . 0 0 cαj

i i −j We know that [1, α p2, cα p3] is a secant of Cj if and only if α x + p2y + cp3z = 0 T has two solutions for points [x, y, z] in Cj. Since the x-coordinate of such points is not zero, we are looking for solutions of the form [1, √ α−j cz2, z]T . Hence, we have to ﬁnd solutions of ± − −

2 i−j −j −1 i−j 2 i−2j −1 z 2α p3z + α c + α p α c = 0. − 3 − This quadratic equation is solvable for z if and only if its discriminant is a square 2i−2j 2 −j −1 2 i−j i−2j −1 in GF (q), i.e. if and only if α p3 α c p3α + α c is a square in i 2 −−1 i −j GF (q), which is the same as (α p3 + c )(α α ) being a square in GF (q). Since T i 2 −1 − −1 i 2 [1, p2, p3] lies on Ci, we know α p3 + c = c α p2. The condition is therefore that ( αic−1)(αi αj) needs to be a square,− which is independent of the point P , and hence− C and −C˜ are nested.

7.2 The nondisjoint case

Note that two nondisjoint proper conics can intersect in exactly one, two, three or four points. If they intersect in more than four points, they are actually the same, since any proper conic is uniquely defined by five of its points. To study the question whether or not two proper nondisjoint conics can be diagonalized simultaneously, we look at the following results, each concerned with a different number of intersection points. For all cases, we assume simultaneous diagonalization for the two proper conics we start with. Hence, in all proofs, we have to consider the following pencil:

94 7.2. The nondisjoint case

2 2 2 C1 = V(x + by + cz ) 2 2 2 C2 = V(x + ˜by +cz ˜ ) 2 2 2 V0 = V((1 + 1)x + (b + ˜b)y + (c +c ˜)z ) 2 2 2 V1 = V((1 + α)x + (b + α˜b)y + (c + αc˜)z ) 2 2 2 2 2 2 V2 = V((1 + α )x + (b + α ˜b)y + (c + α c˜)z ) . . 2 2 V q−1 = V((b ˜b)y + (c c˜)z ) 2 − − . . q−2 2 q−2 2 q−2 2 Vq−2 = V((1 + α )x + (b + α ˜b)y + (c + α c˜)z )

for b, ˜b, c, c˜ = 0 and b = ˜b or c =c ˜. Note that V q−1 is a degenerate conic, since 2 q−1 6 6 6 α 2 = 1. Since b + k˜b and c + kc˜ runs through all elements of GF (q) for k = 0, . . . , q −1, there are exactly two or three degenerate conics in this pencil. − Remark 7.2.1. Note that in a pencil of conics which contains more than three degenerate conics, all conics are degenerate. This can for example be seen by setting b = 0 andc ˜ = 0 in the above pencil, i.e. we start with two degenerate conics. Of course, V q−1 is still a degenerate conic. To obtain more than three degenerate conics, we 2 necessarily need c = 0 or ˜b = 0 as well, hence all conics in the pencil are degenerate.

Theorem 7.2.2. No two proper conics C1 and C2 in PG(2, q), which intersect in exactly one point, can be diagonalized simultaneously.

Proof. Let C1 and C2 intersect in exactly one point, say P , and assume that C1 and C2 can be diagonalized simultaneously. We know that all elements in the pencil (C1,C2) must contain P as well and we have to distinguish the cases of P itself occurringP as an element in the pencil or not. Let us start by assuming that P itself is a conic of the pencil. Since we only have q + 1 elements in the pencil and every point of the plane PG(2, q) needs to occur in one of these, there has to be a degenerate conic corresponding to a pair of lines as well. Indeed, this gives 1 + 2q + (q 1)q = q2 + q + 1 diﬀerent points in the q + 1 conics of the pencil, as no point except− P occurs in more than one element of the pencil. As there are equally many points on a proper conic and on a line, the remaining q 1 conics can correspond to lines or proper conics. There are two − or three degenerate conics in the pencil of C1 and C2, hence we have to distinguish further.

First, assume that the pencil is of the form P, gg,˜ C1,...,Cq−1 . In this case, we have exactly two degenerate conics and two of{ the elements of the} pencil above must correspond to a point and a pair of lines, respectively. This means b ˜b = 0 and − 6 c c˜ = 0 as otherwise, V q−1 would correspond to a line. As the degenerate conics − 6 2 are exactly two, there exists a k such that b + αk˜b = 0 and c + αkc˜ = 0. But then, the conic Vk is a line, which is a contradiction.

95 7.2. The nondisjoint case

0 Now, let the pencil be of the form P, gg,˜ g ,C1,...,Cq−2 . Here, we have three { } 2 2 degenerate conics. Assume that V q−1 is a line. Since each zero of x +ty corresponds 2 to a point or a pair of lines, depending on t = 0, we need b = ˜b or c =c ˜ to obtain the line V(z2) or V(y2). In both cases, there can6 only be one more degenerate conic, which is a contradiction. So V q−1 must correspond to the point or the pair of lines. 2 In both cases, we have b = ˜b and c =c ˜. Since we need three degenerate conics, there 0 exists a k such that b + 6αk˜b = 0 and6 a k0 with c + αk c˜ = 0. For k = k0, the conics 0 6 Vk and Vk0 both correspond to a point or a pair of lines. As k = k leads to only one more degenerate conic corresponding to a line, we have a contradiction. Now assume that P does not occur as an element of the pencil. In this case, assume we have the pencil g1, g2,C1,...,Cq−1 or g1, g2, g3,C1,...,Cq−2 . This means { } { } that V q−1 must correspond to a line, i.e. b = ˜b or c =c ˜. This gives only one more 2 degenerate conic, which corresponds to a point or a pair of lines, hence we cannot produce the pencils we assumed. Remark 7.2.3. Note that the same result is true in the real projective plane as well. A shorter proof for that is to look at two proper conics in diagonal form, namely C = V(x2 + by2 + cz2) and C˜ = V(x2 + ˜by2 +cz ˜ 2). Note that if P = [x, y, z]T lies on C and C˜, then so do [x, y, z]T ,[x, y, z]T and [x, y, z]T . Since [1, 0, 0]T , [0, 1, 0]T and [0, 0, 1]T do not− lie on a proper− conic of this− pencil,− we obtain exactly two or four intersection points. This argument holds for the real projective plane as well as for PG(2, q). For this reason, we expect simultaneous diagonalization to fail for conics with exactly one or exactly three common points. The proof above shows that these results can be obtained by considering the pencil of conics as well, although the proof is much longer than the simple argument just mentioned.

Lemma 7.2.4. Two proper conics C1 and C2 which intersect in exactly two points can be diagonalized simultaneously if and only if their pencil is of the form

g1g˜1, g2,C1,C2,...,Cq−1 . { }

Proof. Let C1 and C2 intersect in exactly two points P and Q. None of the elements of the pencil can correspond to a point, since both P and Q lie in all elements of the pencil. Moreover, since every point of the plane except P and Q needs to be in exactly one element of the pencil, we need one pair of lines, since we need (2q + 1) + (q 1)q = q2 + q + 1 points in total. As there is exactly one line through P and−Q, we only have two possible forms for the pencil, namely g1g˜1, g2,C1,C2,...,Cq−1 or g1g˜1,C1,C2,...,Cq . The second pencil has only {one degenerate conic, so} simultaneous{ diagonalization} is not possible in this case, which shows one direction of our claim.

For the other direction, let the pencil be of the form g1g˜1, g2,C1,C2,...,Cq−1 . Note that for two diagonal conics, if they intersect in a point{ [1, y, z]T they intersect} as well in the points [1, y, z]T , [1, y, z]T and [1, y, z]T , which gives potentially four intersection points.− Hence, if the− two conics we− started− with are diagonalizable simultaneously and intersect in exactly two points, they have to intersect in a point with a zero coordinate, e.g. [1, αk, 0]T , which gives only one more intersection point. As we can transform any three noncollinear points to any other three noncollinear k T points, we search for a transformation φS such that φS(P ) = [1, α , 0] , φS(Q) =

96 7.2. The nondisjoint case

k T T [1, α , 0] and φS(g1 g˜1) = [0, 0, 1] . Note that the intersection of the pair of lines, − ∩ i.e. g1 g˜1 cannot be P or Q, since then g1 org ˜1 would be the line through P and ∩ Q, namely g2, and the elements g1g˜1 and g2 would have q + 1 common points. So, 2 2k 2 2 we indeed obtain, without loss of generality, g2 = V(z ) and g1g˜1 = V(α x y ). All elements of the pencil intersect in exactly two points, namely [1, αk, 0]−T and [1, αk, 0]T and hence the two conics we started with are mapped into two diagonal conics,− which intersect in exactly those two points as well.

Lemma 7.2.5. Two proper conics C1 and C2 which intersect in exactly three points, cannot be diagonalized simultaneously.

Proof. Let C1 and C2 intersect in the points P , Q and R. Note that none of the conics in the pencil corresponds to a point. Moreover, none of the conics corresponds to a line, as P , Q and R are not collinear by the deﬁnition of a proper conic. Hence, the only degenerate conics which occur correspond to pairs of lines. The question therefore is how many such pairs there are and if there is only one possible pencil. For this, note that on a pair of lines, there are always 2q + 1 points. As all conics in question intersect in the same three points, we have to be careful not to count any point too often. So, we have to solve the equation

3 + i(2q 2) + (q + 1 i)(q 2) = q2 + q + 1, 0 i 3, − − − ≤ ≤ which gives immediately i = 2, so there are always two degenerate conics which correspond to a pair of lines. Hence, the pencil of two proper conics which intersect in three points is always of the form g1g˜1, g2g˜2,C1,...,Cq−1 . { } Therefore, the conic V q−1 corresponds to a pair of lines, which gives b = ˜b and c =c ˜. 2 6 6 There is only one more degenerate conic, hence there exists a k such that b+αk˜b = 0 k and c + α c˜ = 0. But then Vk corresponds to a line, which is a contradiction. Remark 7.2.6. We already mentioned in Remark 7.2.3 that this happens in the real projective plane as well.

Lemma 7.2.7. Two proper conics C1 and C2 which intersect in exactly four points can always be diagonalized simultaneously.

Proof. In the pencil of two proper conics C1 and C2 which intersect in exactly four points, the only possible degenerate conics correspond to pairs of lines by the same argument as before. Again, we have to think about the number i of pairs of lines needed to ensure that every point of the plane is in an element of this pencil. For this, we have to solve

4 + i(2q 3) + (q + 1 i)(q 3) = q2 + q + 1 − − − which gives us i = 3. Hence, the pencil of two proper conics which intersect in exactly four points is always given by g1g˜1, g2g˜2, g3g˜3,C1,...,Cq−2 . These four intersection points are such that no three{ are on one line and hence} by Lemma 2.3.6, we can choose those points as

P = [1, 1, 1]T ,Q = [1, 1, 1]T ,R = [1, 1, 1]T ,S = [1, 1, 1]T . − − − −

97 7.3. Summary

There are exactly three diﬀerent pairs of lines through these four points P , Q, R and S, given by

2 2 2 2 2 2 g1g˜1 = V(y z ), g2g˜2 = V(x y ), g3g˜3 = V(x z ). − − − This leads to the following pencil:

2 2 g1g˜1 = V(y z ) 2 − 2 g2g˜2 = V(x y ) 2 − 2 g3g˜3 = V(x z ) 2 − 2 2 C1 = V(x + (α 1)y αz ) 2 2− 2− 2 2 C2 = V(x + (α 1)y α z ) − − . . 2 q−2 2 q−2 2 Cq−2 = V(x + (α 1)y α z ) − −

Therefore, the two conics we started with are in diagonal form as well. Remark 7.2.8. In the real projective plane, two conics which intersect in exactly four points can always be diagonalized simultaneously as well. This can for example be shown by proving the existence of a triangle which is self-polar with respect to both conics (see for example [37]).

7.3 Summary

To ﬁnish the discussion about simultaneous diagonalization of two symmetric 3 3 matrices over GF (q) using pencils of conics, we summarize our results in the× following table.

Case Pencil diagonalizable? C1 C2 = 0 P, g, C1,...,Cq−1 yes | ∩ | { } P, P˜ , gg,˜ C1,...,Cq−2 yes { } P,C1,...,Cq no { } C1 C2 = 1 P, gg,˜ C1,C2,C3/g3,...,Cq−1/gq−1 no | ∩ | { } C1,C2,C3/g3,...,Cq+1/gq+1 no { 0 } C1 C2 = 2 gg,˜ g ,C1,...,Cq−1 yes | ∩ | { } gg,˜ C1,...,Cq no { } C1 C2 = 3 g1g˜1, g2g˜2,C1,...,Cq−1 no | ∩ | { } C1 C2 = 4 g1g˜1, g2g˜2, g3g˜3,C1,...,Cq−2 yes | ∩ | { }

Note that the results obtained in this chapter are published in [26].

98 Chapter 8

Steiner’s Theorem in the ﬁnite M¨obiusplane M(q)

M¨obiusplanes consist of points P and circles B, which satisfy three axioms. First, there needs to be a unique circle through three given points. Second, there exists a unique tangent circle through a point on a given circle and a point not on this circle. Finally, a richness axiom ensures that the plane is not trivial. More precisely, the three axioms read as follows.

(M1) For any three elements P, Q, R P, P = Q, P = R and Q = R, there exists a unique element g B with P ∈g, Q 6 g and R6 g. 6 ∈ ∈ ∈ ∈ (M2) For any g B, P,Q P with P g and Q/ g, there exists a unique element h B such∈ that P ∈h and Q ∈h, but for∈ all R P with R g, P = R, we have∈ R/ h. ∈ ∈ ∈ ∈ 6 ∈

(M3) There are four elements P1,P2,P3,P4 P such that for all g B, we have ∈ ∈ Pi / g for at least one i 1, 2, 3, 4 . ∈ ∈ { }

In this chapter, we look at properties of chains of consecutive tangent circles, such as for example Steiner’s Porism in Miquelian M¨obiusplanes, which reads as follows.

Theorem (Steiner’s Porism in finite Miquelian Möbiusplanes). Let 1 and 2 be B B disjoint circles in M(q). Consider a sequence of circles 1,..., k, k 3, which T T ≥ are tangent to both 1 and 2. Moreover, let i and i+1 be tangent for all i = B B T T 1, . . . , k 1. In the case of 1 and k being tangent as well, we call the sequence − T T 1,..., k a Steiner chain of length k for 1 and 2. If there exists one Steiner chainT ofT length k, then every chain of consecutiveB tangentB circles, which are tangent to both 1 and 2, is a Steiner chain of length k for 1 and 2. B B B B Miquelian Möbiusplanes are the classical finite models for the Möbiusaxioms and are constructed over the finite field GF (q) of order q := pm, for p an odd prime and m 1. The resulting plane is denoted by M(q), the details are explained below. ≥

99 8.1. Miquelian M¨obiusplanes

8.1 Miquelian M¨obiusplanes

We are now going to describe finite Miquelian Möbius planes constructed over the pair of finite fields GF (q) and GF (q2). Such planes will be denoted by M(q) and q is called the order of M(q). The q2 + 1 points of M(q) are given by all elements of GF (q2) together with a point at infinity, denoted by . We distinguish between two different types of circles. For circles of the first type,∞ we consider solutions of the equation N(z s) = c, i.e. − 1 :(z s)(z s) = c (8.1) B(s,c) − − for s GF (q2) and c GF (q) 0 . It can easily be seen that there are q + 1 points in GF∈(q2) on every circle∈ (8.1).\{ } Moreover, there are q2(q 1) circles of the first type. − For circles of the second type, we consider the equation T r(sz) = c, i.e.

2 : sz + sz = c (8.2) B(s,c) for s GF (q2) 0 and c GF (q). For every such choice of s and c, equation (8.2) has∈ q solutions\{ } in GF (q∈2). To obtain circles of the second type, we take those q solutions together with . There are (q2 1)q choices for s and c, but scaling with any element of GF (q∞) 0 leads to the− same circle. Hence, there are q(q + 1) circles of the second type. There\{ } are q3 + q circles in total and on each circle there are q + 1 points. This can also be seen by (M1), as three points uniquely define a circle. Now, let a, b, c, d GF (q2) such that ad bc = 0. The map µ defined by ∈ − 6  az+b if z = and cz + d = 0  cz+d 6 ∞ 6  if z = and cz + d = 0 µ : M(q) M(q), µ(z) = ∞a 6 ∞ →  c if z = and c = 0  ∞ 6  if z = and c = 0 ∞ ∞ is called a Möbiustransformation of M(q). Every Möbiustransformation is an au- 1 tomorphism of M(q). A Möbiustransformation of the form µ(z) = z is an inversion in the unit circle zz = 1, which means that the unit circle is fixed under µ. Möbius planes in which for every circle there exists an inversion are called inversive Möbius planes. In [13] it is shown that inversive Möbiusplanes are exactly the Miquelian Möbiusplanes. Note that a Möbiustransformation operates three times sharply transitive, i.e. there is a unique Möbiustransformation mapping any three points into any other three given points. For more background information on finite Möbiusplanes, one can refer to [14].

8.2 Steiner’s Theorem in M(q)

1 1 For a circle of the ﬁrst type (s,c), we refer to s as the center of (s,c), and c is the square of the radius. Note thatB the radius, which is either squareB root of c, is not

100 8.2. Steiner’s Theorem in M(q) necessarily an element of GF (q). Two circles of the ﬁrst type are called concentric, 1 1 2 0 if they are of the form (s,c) and (s,c0) for s GF (q ) and c, c GF (q) 0 . Without loss of generality,B we canB always assume∈ that two concentric∈ circles\{ have} center 0, since the M¨obius transformation

µ: M(q) M(q), µ(z) = z s → − maps any two concentric circles with center s to two concentric circles with center 0. In this section we henceforth consider two concentric circles with center 0, i.e. 1 1 circles (0,a) and (0,b) for a, b GF (q) 0 . For notational convenience, let us deﬁne B B ∈ \{ } 1 a := B B(0,a) for all a GF (q) 0 . ∈ \{ } To obtain more insight into the geometrical properties of the circles, we use Cartesian coordinates in this section, where z = x + αy represents the point (x, y) andz ¯ the point (x, y). The circle 2 is therefore given by − Ba 2 2 2 2 : x δy = a Ba − for δ a nonsquare in GF (q). We see directly, that any such circle intersects the x-axis, i.e. the circle of the second type given by y = 0, in the two points ( a, 0). 2 2 ± Now consider the unit circle 1 given by x δy = 1 and the point (1, 0). It is an easy exercise to show that thereB are exactly− two circles which are tangent to both a+1 a−1 1 and a2 in the point (1, 0). The ﬁrst circle has center 2 , 0 and radius 2 , B B −a+1 a+1 the other one has center 2 , 0 and radius 2 . Figure 8.1 shows those two common tangent circles.

Figure 8.1: Common tangent circles of 1 and 2 B Ba Now consider any (q +1)th root of unity P , i.e. an element in GF (q2) which satisﬁes P q+1 = 1. Since P P = P q+1, the rotation given by

z P z → 101 8.2. Steiner’s Theorem in M(q)

ﬁxes both circles 1 and 2 . Moreover, this rotation is a M¨obiustransformation B Ba and hence takes common tangent circles of 1 and 2 again into common tangent B Ba circles of 1 and 2 . In particular, there are two circles tangent to 1 at the point B Ba B P that are also tangent to 2 , one in the point aP and the other in the point aP . Ba − Note that we use the parameter a2 as subscript, as we need it to be a square in GF (q) for the circles 1 and 2 to have common tangent circles. B Ba For a GF (q) 0 , let ∈ \{ } τ(a) := g B : a g = 1 { ∈ |B ∩ | } denote the set of all tangent circles of a and B τ(a, b) := τ(a) τ(b) ∩ the set of all common tangent circles of a and b. B B The following lemma summarized what we just discussed. Lemma 8.2.1. If b = 1 is a square in GF (q), then a 6 τ(a, b) = 2(q + 1) | | otherwise τ(a, b) = 0. | |

The 2(q + 1) common tangent circles of 1 and a2 partition into two sets. Let P again be a fixed (q + 1)th root of unity.B There isB one set of q + 1 common tangent a−1 j circles with radius 2 and tangent to 1 in the points P and tangent to a2 in the j B B points aP , for j = 0, . . . , q. The other q + 1 common tangent circles of 1 and a2 a+1 j B B have radius 2 and are tangent to 1 in the points P and tangent to a2 in the points aP j, for j = 0, . . . , q. B B − Now we want to construct Steiner chains using the common tangent circles of two concentric circles 1 and 2 . B Ba Note again that the following discussion already covers all cases for two concentric circles a and b, a, b GF (q). To see this, look at the Möbiustransformation B B ∈ z µ(z) = η for η in GF (q2) such that ηη = a. Note that such an η always exists by the properties of the norm map. Then µ maps the circles a and b to 1 and b , respectively. B B B B a A Steiner chain of length k for 1 and a2 , k 3, is a chain of k different circles 2 B B ≥ T1,...,Tk in τ(1, a ) such that Ti Ti+1 = 1 for i = 1, . . . , k 1 and Tk T1 = 1. | ∩ | − | ∩ | Moreover, all circles Ti, i = 1, . . . , k need to be of the form 1 Ti = . B(si,c)

Note that all these circles Ti have the same radius, only their center is shifted. One could deﬁne degenerate Steiner chains as well by allowing circles with diﬀer- ent radius. In this case, we always obtain a degenerate chain for 1 and a2 by considering B B 1 1 1 1 1 ( a+1 , a−1 ) ( a−1 , a+1 ) ( −a−1 , a−1 ) ( −a+1 , a+1 ) ( a+1 , a−1 ). B 2 2 → B 2 2 → B 2 2 → B 2 2 → B 2 2 102 8.3. The plane M(5)

Theorem 8.2.2. If there are two common tangent circles of 1 and a2 with the same radius which are tangent to each other, a Steiner chainB of lengthB k 3 for some k (q + 1) can be constructed. ≥ |

Proof. We start with two such common tangent circles T1 and T2 of 1 and 2 with B Ba the same radius and T1 T2 = 1. For some root of unity P , the rotation z P z | ∩ | → takes the pair (T1,T2) to the pair (T2,T3), which is again a pair of common tangent circles of 1 and 2 which are tangent to each other. We can repeat this rotation k B Ba times, for k a divisor of q + 1 and see that we ﬁnally end up with the pair (Tk,T1), which is a pair of common tangent circles of 1 and a2 with T1 Tk = 1. In other words, we just constructed a Steiner chain ofB lengthBk. | ∩ |

Now we are able to state and prove our ﬁnite version of Steiner’s Theorem.

Theorem 8.2.3. Consider the circles 1 and 2 . Assume we can construct a B Ba Steiner chain starting with any point P on 1, then a Steiner chain of the same B length can be constructed starting with any other point on 1. B Proof. This is immediate by using again a rotation by a root of unity.

8.3 The plane M(5)

We have a closer look at the M¨obiusplane M(5) constructed over GF (5)(α) with 2 1 α = 3, as described in the preliminaries. Consider the two circles B1 = , B(0,1) 1 = 1, 3 + α, 2 + α, 4, 2 + 4α, 3 + 4α B { } 1 and B4 = , B(0,4) 1 4 := = 2, 1 + 2α, 4 + 2α, 3, 4 + 3α, 1 + 3α . B B(0,4) { } 4 2 Since 1 = 2 is a square, the two circles have exactly 12 common tangent circles. which are given by 1 1 := = 1, 2, 2α, 3 + 2α, 3α, 3 + 3α T B(4,4) { } 1 2 := = α, 4 + α, 1 + 3α, 3 + 3α, 1 + 4α, 3 + 4α T B(2+α,4) { } 1 3 := = α, 1 + α, 2 + 3α, 4 + 3α, 2 + 4α, 4 + 4α T B(3+α,4) { } 1 4 := = 3, 4, 2α, 2 + 2α, 3α, 2 + 3α T B(1,4) { } 1 5 := = 2 + α, 4 + α, 2 + 2α, 4 + 2α, 4α, 1 + 4α T B(3+4α,4) { } 1 6 := = 1 + α, 3 + α, 1 + 2α, 3 + 2α, 4α, 4 + 4α T B(2+4α,4) { } and 1 7 := = 3 + 2α, 4 + 2α, 3α, 2 + 3α, 3 + 4α, 4 + 4α T B(1+3α,1) { } 1 8 := = 1 + 2α, 2 + 2α, 3α, 3 + 3α, 1 + 4α, 2 + 4α T B(4+3α,1) { } 1 9 := = 2, 4, α, 1 + α, 4α, 1 + 4α T B(3,1) { } 1 10 := = 1 + α, 2 + α, 2α, 3 + 2α, 1 + 3α, 2 + 3α T B(4+2α,1) { } 1 11 := = 3 + α, 4 + α, 2α, 2 + 2α, 3 + 3α, 4 + 3α T B(1+2α,1) { } 1 12 := = 1, 3, α, 4 + α, 4α, 4 + 4α . T B(2,1) { }

103 8.3. The plane M(5)

Note that 1 is tangent to 1 in 1 and tangent to 4 in 2. Next, consider 2, which T B B T is tangent to 1 in 3 + 4α and tangent to 4 in 1 + 3α. Note that 1 and 2 only B B T T intersect in 3+3α, i.e. they are tangent. Having a closer look at 2, we see that only T two of the 12 circles above are tangent to 2 in points not on 1 or 4, namely 1, T B B T which we already considered, and 3, which is tangent to 2 in α. Apparently, from now on, there is a unique way ofT constructing a chain ofT common tangent circles of 1 and 4. Proceeding, we ﬁnd that 4 is tangent to 3 in 2 + 3α. Then 5 is B B T T T tangent to 4 in 2 + 2α. Finally, we ﬁnd that 6 is tangent to 5 in 4α and also 1 T T T T is tangent to 6 in 3 + 2α, which closes the chain of circles. T Note that those six tangent points lie on a circle itself, namely on

1 2 := = α, 2 + 2α, 3 + 2α, 2 + 3α, 3 + 3α, 4α . B B(0,2) { } Summarized, we denote this chain by

3 + 3α α 2 + 3α 2 + 2α 4α 3 + 2α 1 2 3 4 5 6 1. T −−−−−→T −−−−−→T −−−−−→T −−−−−→T −−−−−→T −−−−−→T Note that for the above chain, we only use six out of the twelve common tangent circles of 1 and 4, so let us start with a tangent circle not used so far, e.g. 7. We ﬁnd the chainB B T

3α 1 + 4α 1 + α 2α 4 + α 4 + 4α 7 8 9 10 11 12 7. T −−−−−→T −−−−−→T −−−−−→T −−−−−→T −−−−−→T −−−−−→T Again, the six tangent points form a circle, namely

1 3 := = 1 + α, 4 + α, 2α, 3α, 1 + 4α, 4 + 4α . B B(0,3) { } Now, let us proceed from here and look at the circles through two consecutive (where the order is deﬁned by the chain before) points of 1 and the corresponding tangency B point on 2. We obtain six new circles, which are all tangent to 2, given by: B B 1 1 := = 1, 1 + 2α, 3 + 3α, 4 + 3α, 3 + 4α, 4 + 4α S B(1+α,2) { } 1 2 := = 2, 3, α, 3α, 2 + 4α, 3 + 4α S B(2α,2) { } 1 3 := = 4, 4 + 2α, 1 + 3α, 2 + 3α, 1 + 4α, 2 + 4α S B(4+α,2) { } 1 4 := = 4, 1 + α, 2 + α, 1 + 2α, 2 + 2α, 4 + 3α S B(4+4α,2) { } 1 5 := = 2, 3, 2 + α, 3 + α, 2α, 4α S B(3α,2) { } 1 6 := = 1, 3 + α, 4 + α, 3 + 2α, 4 + 2α, 1 + 3α S B(1+4α,2) { } These six circles form a chain of tangent circles as well. Moreover, there is a unique circle, except 2, which is tangent to all of those six new circles, namely 3. B B We can perform the same procedure once more, i.e. we consider circles through two consecutive points on 3 and the corresponding tangency point of the chain in B consideration. We obtain six common tangent circles of 1 and 4, i.e. we have two B B chains including all twelve common tangent circles of of 1 and 4. B B Figure 8.2 summarizes the above discussion, with B4 appearing in two places because the conﬁguration showing the two Steiner chains of six circles each cannot exist in the Euclidean plane.

104 8.4. Existence and length of Steiner chains

4 + 3α 1 + 3α

B4 α

2 + 3α 2 + 4α 3 + 4α 3 + 3α

1 B3 B B4

4 3 2 3 1 2

2 + α 2 + 2α 3 + α 3 + 2α

4α

4 + 2α 1 + 2α

Figure 8.2: An example in M(5).

8.4 Existence and length of Steiner chains

In what follows, we are mainly interested in the existence of Steiner chains as well as their possible lengths.

Lemma 8.4.1. Let a be a nonsquare in GF (q). Then a Steiner chain for 1 and − 1 1 B 2 can be constructed starting with and for Ba B(s,c) B(sP,c) a + 1 a 12 s = , c = − 2 2 and ( a2 + 6a 1) + 4(a 1)√ a P = − − − − . (8.3) (1 + a)2

For a a square in GF (q), no Steiner chain can be constructed for the pair 1 and − B 2 . Ba

105 8.4. Existence and length of Steiner chains

Proof. By the finite version of Steiner’s Theorem (i.e. Theorem 8.2.3) we can start w.l.o.g. with the point 1 on 1 and the point a on a2 . The parameters s and c 1 B B for the circle (s,c) touching 1 and a2 in these two points are straight forward. It B B B 2 1 is well-known that if there is another circle h τ(1, a ) with h (s,c) = 1, then there is a common tangent line l of those two circles∈ through the| ∩ origin. B | Moreover, 1 h is obtained by a reflection of (s,c) at l. By a straight forward calculation, we find 1 B 1 that the tangent line of (s,c) through the origin touches (s,c) in the point Q, given by B B 2a a 1 Q = + √ a − . a + 1 − a + 1 Reflecting the point 1 at the line through 0 and Q gives indeed ( a2 + 6a 1) + 4(a 1)√ a P = − − − − . (1 + a)2

For a a square in GF (q), P lies in GF (q). Since 1 and 1 are the only elements − − in GF (q) on 1, we need a to be a nonsquare in GF (q). B − Theorem 8.4.2. Let 1 and 2 be circles of M(q). If q 1(4), exactly one B Ba ≡ − Steiner chain can be constructed with 1 and 2 . If q 1(4), either two or zero B Ba ≡ Steiner chains can be constructed with 1 and a2 , depending on whether or not a is a square in GF (q). B B

Proof. In the proof of Lemma 8.4.1 we have seen that for a a nonsquare, a Steiner − 2 chain can be constructed starting with the circle 1 for s = a+1 and c = a−1 . B(s,c) 2 2 Of course, the whole proof can be done by replacing a by a. For q 1(4), a is a square if and only if a is a nonsquare and hence, exactly− one Steiner≡ − chain can be constructed. For q −1(4), a is a square if and only if a is a square. Therefore, we can either construct≡ two Steiner chains or none. −

Before we state our general result about the length of a Steiner chain, we discuss some speciﬁc cases for M(q) in detail. First, let us examine a criterion for Steiner chains of length 3.

Corollary 8.4.3. A Steiner chain of length 3 can be constructed for 1 and a2 in M(q) if and only if B B a = 7 + 4√3 GF (q) ∈ and a a nonsquare in GF (q). In particular, this is only possible if q = pm for some prime− p with p 1(12). ≡ ± Proof. By Lemma 8.4.1, we know that a necessarily needs to be a nonsquare in GF (q). Moreover, if a is a nonsquare in−GF (q), we can ﬁnd two circles in τ(1, a2) which are tangent to− 1 , namely 1 and 1 , for P given by (8.3). For a B(s,c) B(sP,c) B(sP ,c) Steiner chain of length 3, also 1 and 1 need to be tangent. So, we need B(sP,c) B(sP ,c) P 3 = 1, or similar P 2 = P . (see Figure 8.3). Solving this equation for a leads to

a = 7 + 4√3.

106 8.4. Existence and length of Steiner chains

2 Ba

P B1 1 P 2

Figure 8.3: A Steiner chain of length 3.

This is only possible for 3 a square in GF (q). It is well-known (see for example [19]) that 3 is a square in GF (q) p 1 (12), ⇒ ≡ ± which gives a necessary condition for q.

Note that this already excludes the existence of a Steiner chain of length 3 in M(5). Indeed, we have seen in Section 8.3 that only Steiner chains of length 6 occur for two concentric circles in M(5). In M(11), however, we can ﬁnd Steiner chains of length 3. For this, calculate ( 7 + 9 = 5 (11) a = 7 + 4√3 = 7 + 2 = 9 (11).

Moreover, 5 = 6 as well as 9 = 2 are nonsquares in GF (11). So, Steiner chains − − of length 3 can be constructed for 1 and 3 as well as for 1 and 4. B B B B Now, let us have a look at Steiner chains of length 4.

Corollary 8.4.4. A Steiner chain of length 4 can be constructed for 1 and a2 in M(q) if and only if B B a = 3 + 2√2 GF (q) ∈ and a a nonsquare in GF (q). In particular, this is only possible if q = pm for some prime− p with p 1(8). ≡ ± Proof. Again, by Lemma 8.4.1, a needs to be a nonsquare in GF (q). Moreover, if a is a nonsquare in GF (q), we− have two circles 1 and 1 in τ(1, a2), which − B(sP,c) B(sP ,c) are tangent to 1 with P given by (8.3). For a Steiner chain of length 4, we need B(s,c) P 4 = 1 or similar, P 2 = 1. (see Figure 8.4) − Solving this equation for a gives

a = 3 + 2√2.

107 8.4. Existence and length of Steiner chains

2 Ba

1 1 B 1 − P 3

Figure 8.4: A Steiner chain of length 4.

This is only possible for 2 a square in GF (q). Again from [19], we have that 2 is a square in GF (q) q = pm prime power, p 1 (8). ⇒ ≡ ±

Let us have a look at M(7) and calculate ( 3 + 6 = 2 (7) a = 3 + 2√2 = 3 + 1 = 4 (7).

Moreover, 2 = 5 as well as 4 = 3 are nonsquares in GF (7). So, Steiner chains − − of length 4 can be constructed for 1 and 2 as well as 1 and 4. Note that 1, 2 and 4 are the only squares in GF (7),B hence,B only SteinerB chainsB of length 4 can be constructed using 1 and b. Moreover, 2 is not a square in GF (11), so in M(11), there are no SteinerB chainsB of length 4. Similarly, we can calculate a criterion for Steiner chains of length 6.

Corollary 8.4.5. A Steiner chain of length 6 can be constructed for 1 and a2 in M(q) if and only if B B 1 a = 3 or 3 and 3 a nonsquare in GF (q). − This criterion is different from the criterion for Steiner chains of length 3 and 4, since no square root appears in the expression for a above. By the finite version of Steiner’s Theorem it follows directly that the existence of a Steiner chain of length 6 in GF (q) implies 6 (q + 1). For p prime, the condition 6 (p + 1) is actually equivalent with 3 being a| nonsquare in GF (p), which can be seen| by number theoretic calculations− only. 2 1 2 Note that in M(5), this gives 3 = 4 = ( 3 ) , i.e. only for 1 and 4 a Steiner chain of length 6 can be constructed. Compare these results alsoB to SectionB 8.3. Now, let us have a look at the conditions for Steiner chains of length 5 and 8. The expression for a becomes more and more difficult, since equations of higher order need to be solved. In particular, for Steiner chains of length 5, we need to solve P 5 = 1 and for Steiner chains of length 8, we need to solve P 4 = 1. − 108 8.4. Existence and length of Steiner chains

Corollary 8.4.6. A Steiner chain of length 5 can be constructed for 1 and a2 in M(q) if and only if B B q a = 11 4√5 + 2 50 22√5 GF (q) − − ∈ and a a nonsquare in GF (q). − We know that

5 is a square in GF (q) q = pm prime power, p 1 (5) ⇒ ≡ ± which gives a necessary, but not suﬃcient condition for the existence of Steiner chains of length 5.

Corollary 8.4.7. A Steiner chain of length 8 can be constructed for 1 and a2 in M(q) if and only if B B q a = 7 4√2 + 2 2(10 7√2) GF (q) − − ∈ and a a nonsquare in GF (q). − Note that 2 needs necessarily to be a square, similar to Steiner chains of length 4. This is not surprising, since 8 is a multiple of 4. Now we are ready to give a condition for Steiner chains of length k, for k 3. ≥ Theorem 8.4.8. Let a GF (q). A Steiner chain of length k 3 can be constructed ∈ ≥ for 1 and 2 in GF (q) if and only if the following conditions are satisﬁed. B Ba 1. a is a nonsquare in GF (q), − 2. a solves P k = 1 for P given by

a2 + 6a 1 + 4(a 1)√ a P = − − − − (8.4) (1 + a)2

but P l = 1 for all 1 l k 1. 6 ≤ ≤ − Proof. Let us assume the existence of a Steiner chain of length k. By Lemma 8.4.1, 1 1 a needs to be a nonsquare in GF (q) to obtain two circles (sP,c) and (sP ,c), which − 1 B B are both tangent to (s,c). Again by Theorem 8.2.3, we know that starting with 1 B 1 2 two such circles (sP,c) and (s,c) in τ(1, a ), which are tangent, we end up with a proper Steiner chain.B Moreover,B the length of the Steiner chain is then given by the smallest integer k, such that P k = 1, i.e. we are back at the starting point. Now let us assume that the above three conditions are satisﬁed. Since a is a nonsquare, we can apply Lemma 8.4.1 to obtain a Steiner chain. Since −k is by assumption such that P k = 1 but P l = 1 for all 1 l k 1, the length of the Steiner chain is indeed k. 6 ≤ ≤ −

109 8.5. Generalization

8.5 Generalization

Two disjoint circles in M(q) define a non-intersecting pencil. Such a pencil consist of q 1 circles and two limiting points. A Möbiustransformation that sends the limiting− points to 0 and , and a point on one of the given circles to 1, will take the given circles to a pair of∞ circles centered at 0, one of which is describes by zz¯ = 1 (see [13], [14] for details, and Section 8.5.2). This is the same procedure, as in the usual proof of Steiner’s porism in the classical Möbiusplane. It allows, also in M(q), to transform a general pair of disjoint circles into the standard pair treated in the previous sections, and to apply the corresponding results. However, it is convenient, also from a computational point of view, to skip the transformation step, and to apply the results in the previous theorems directly to a given pair of disjoint circles. This is done by introducing a Möbiusinvariant for arbitrary pairs of circles.

8.5.1 A M¨obiusinvariant for pairs of circles

In the Euclidean plane two non-intersecting circles form a capacitor. The capacitance is a conformal invariant and therefore in particular invariant under M¨obius transformations. Here we present a discrete analogue of this fact which will be used later to decide whether any two non-intersecting circles carry a Steiner chain of length k. The capacitance associates a number in GF (q) to any pair of circles in M(q) and is deﬁned as follows:

1 1 1 2 cap( (s1,c1), (s2,c2)) := c1 + c2 (s1 s2)(¯s1 s¯2) B B c1c2 − − − 1 2 2 1 1 2 cap( (s1,c1), (s2,c2)) = cap( (s2,c2), (s1,c1)) := (s1s¯2 +s ¯1s2 c2) B B B B c1s2s¯2 − 2 2 1 2 cap( (s1,c1), (s2,c2)) := (s1s¯2 +s ¯1s2) B B s1s¯1s2s¯2 Then it turns out that this quantity is indeed invariant under M¨obiustransformations:

Theorem 8.5.1. Let , ˜ be two circles and µ a M¨obiustransformation. Then B B cap( , ˜) = cap(µ( ), µ( ˜)) B B B B Proof. It is easy to check, that cap is invariant under translations z ζ = z + a, a GF (q2), and similarity transformations z ζ = bz, b GF (q2) 7→0 . ∈ 7→ ∈ \{ } The only tedious part of the proof is to check, that cap is invariant under the reciprocation µ : z ζ = 1/z, since in this case, circles may change from ﬁrst type to second type and7→ vice versa: It is elementary to check that

 1  s¯ c if ss¯ = c 1 B ss¯−c , 2 6 µ( (s,c)) = (ss¯−c) B  2 if ss¯ = c B(¯s,1)

110 8.5. Generalization and ( 1 s¯ ss¯ if c = 0 2 ( c , 2 ) µ( (s,c)) = B c 6 B 2 if c = 0 B(¯s,0) 1 We only carry out the invariance proof for one prototypical case of two circles (s1,c1) 1 B with s1s¯1 = c1 and with s2s¯2 = c2. Then, in this case, B(s2,c2) 6 cap( 1 , 1 ) = B(s1,c1) B(s2,c2) 1 2 = c1 + c2 (s1 s2)(¯s1 s¯2) c1c2 − − − 1 2 = c2 + s1s¯2 + s2(¯s1 s¯2) (8.5) c1c2 − where we have used s1s¯1 = c1. On the other hand, capµ( 1 ), µ( 1 ) = B(s1,c1) B(s2,c2) 2 1 s¯ c = cap (¯s1,1), 2 2 ( , 2 ) B B s2s¯2−c2 (s2s¯2−c2) 2 1 s¯2 s2 = c2 s1 + s¯1 1 2 s1s¯1 s2s¯2 c2 s2s¯2 c2 − (s2s¯2−c2) − − 2 2 (s2s¯2 c2) s¯2 s2 = − s1 + s¯1 1 (8.6) c1c2 s2s¯2 c2 s2s¯2 c2 − − − again since s1s¯1 = c1. Obviously, the expressions in (8.5) and (8.6) agree. The other cases are similar.

In a next step we show that, as it is the case in the classical M¨obiusplane, it is possible to transform any two non-intersecting M¨obiuscircles into concentric circles.

8.5.2 Transformation of non-intersecting circles into concentric circles

Theorem 8.5.2. Any two disjoint circles in M(q) can be mapped to concentric circles using a suitable M¨obiustransformation.

We give a combinatorial proof which uses the following results.

1 3 2 Lemma 8.5.3. For any given circle there are 2 (q 3q + 2q) circles disjoint to . B − B Proof. By axiom (M2), for a point P on and any other point Q not on , there is a unique circle tangent to through P andB Q. There are q2 + 1 points inB total and q + 1 points on . So, forB any of the q2 q points not on , there is such a unique circle through aB given point P on . Since− there are q +B 1 points on every circle, exactly q of these tangent circles areB the same. This can be done for every point on 2 , which leads to (q+1)(q −q) = q2 1 circles which are tangent to . B q − B For the circles intersecting , note that by axiom (M1), there is a unique circle through two points on andB any other point not on . So for two ﬁxed points on B B 111 8.5. Generalization

q2−q , there are q−1 = q circles intersecting in those two points. This can be done B B 2 for any pair of points on , which leads to q (q+1) circles which intersect . B 2 B Since there are q(q2 + 1) circles in total, the number of circles disjoint to is given by B q2(q + 1) 1 q(q2 + 1) (q2 1) 1 = (q3 3q2 + 2q) − − − 2 − 2 −

Lemma 8.5.4. There are exactly q3 q M¨obiustransformations which map the unit circle zz = 1 to itself. In particular,− they are given by bz ba µ1(z) = − az + 1 − for bb = 1 and aa = 1 and by 6 b µ (z) = 2 z for bb = 1.

Proof. Recall that M¨obiustransformations act sharply three times transitive. So, q+1 for P1,P2,P3 on the unit circle zz = 1, there are 3 choices for mapping it to any{ three points} on zz = 1 again. Moreover, for any choice of three points, there are 3! = 6 such transformations. Hence, there are q + 1 3! = q3 q 3 − such M¨obius transformations.

Let us have a closer look at µ1. Of course, we have

bz ba bz ba − − = 1 az + 1 az + 1 − − whenever bb = 1 and zz = 1. The condition aa = 1 ensures that we do not divide by 0. There are q + 1 choices for b and q2 (q +6 1) choices for a. Hence, there are 3 − q 2q 1 transformations µ1. − − Similar, µ2 maps the unit circle to itself whenever bb = 1, so there are q + 1 such transformations. Since (q3 2q 1) + (q + 1) = q3 q, these are indeed all such transformations. − − − Now we are ready to prove Theorem 8.5.2:

Proof of Theorem 8.5.2. There are q 2 circles concentric to zz = 1, namely all those circles zz = c for c = 2, . . . , q 1. We− now apply all the Möbiustransformations, which map the unit circle to itself,− to those concentric circles. Note that every image occurs exactly 2(q + 1) times. To see this, let us first think about the circle zz = c for c 2, . . . , q 1 fixed. ∈ { − } Clearly, µ1 for choosing a = 0 maps zz = c to zz = c, for all bb = 1. Moreover,

112 8.6. Comparison to the Euclidean plane

1 applying µ2 to zz = c gives zz = c as well, for all bb = 1. Since for any other choice of a in µ1, the center of zz = c is translated, the circle zz = c occurs 2(q + 1) 3 times when applying all q q M¨obiustransformations µ1 and µ2 described above to zz = c. Similarly can be− proceeded for other circles (z s)(z s) = c. − − So, we apply the q3 q M¨obiustransformations to the q 2 circles concentric to zz = 1. Every image− occurs 2(q + 1) times, i.e. we get −

(q 2)(q3 q) 1 − − = (q3 3q2 + 2q) 2(q + 1) 2 − which is, by Lemma 8.5.3, exactly the number of circles disjoint to zz = 1.

8.5.3 General criterion for Steiner chains

Let and ˜ be non-intersecting circles in M(q). As we have seen in Section 8.5.2, B B 1 ˜ 1 it is possible to transform them into µ( ) = (0,1), µ( ) = (0,b) by using a suitable M¨obiustransformation. By Theorem 8.5.1,B weB have B B 1 κ := cap( , ˜) = cap(µ( ), µ( ˜)) = cap( 1 , 1 ) = (1 + b)2 B B B B B(0,1) B(0,b) b

1 p Solving for b gives b = 2 (κ 2 κ(κ 4)). Changing between the two possible signs corresponds to applying− an± additional− inversion z 1/z to both circles, and we may take the plus sign by convention. Then, the general7→ criterion follows from Theorem 8.4.8:

Theorem 8.5.5. Let and ˜ be non-intersecting circles, κ := cap( , ˜) and b := 1 p B B 2 ˜ B B 2 (κ 2 + κ(κ 4)). Then, if b = a , for a in GF (q), and carry a Steiner chain− of length k− 3 if and only if the following conditionsB are satisﬁed.B ≥ 1. a is a nonsquare in GF (q), − 2. a solves P k = 1 for P given by

a2 + 6a 1 + 4(a 1)√ a P = − − − − (8.7) (1 + a)2

but P l = 1 for all 1 l k 1. 6 ≤ ≤ −

8.6 Comparison to the Euclidean plane

Steiner’s Porism is well understood in the Euclidean plane. For two concentric circles of radius 1 and R, a Steiner chain of length k which wraps around the inner circle once can be constructed if and only if 1 + sin(φ) R = 1 sin(φ) − π where φ = k .

113 8.6. Comparison to the Euclidean plane

180 π k k sin k R √ 3 √ 3 60 2 7 + 4 3 √ 2 √ 4 45 2 3 + 2 2 √ p √10−2 5 √ √ 5 36 4 11 4 5 + 2 50 22 5 1 − − 6 30 2 3 √ √2− 2 p 8 22.5 7 4√2 + 2 20 14√2 2 − −

Table 8.1: Some values for Steiner chains of length k.

π For some values of k, we can express sin( k ) explicitly in terms of radicals. In the following table, we calculate R for some values of k. Note that our values for R coincide with the values for a calculated in the Lemmas 8.4.3, 8.4.4, 8.4.6, 8.4.5 and 8.4.7. Let us have a closer look at why this is the case. Recall that for two common tangent circles of 1 and 2 , we calculated in Lemma 8.4.1 the point in which the second B Ba circle of the chain is tangent to 1, namely B ( a2 + 6a 1) + 4(a 1)√ a P = − − − − . (8.8) (1 + a)2

For the construction of a Steiner chain, we needed a to be a nonsquare in GF (q). So, let us rewrite P as −

a2 + 6a 1 4(a 1) P = − − + √ a − . (8.9) (1 + a)2 − (1 + a)2

2 Note that −a +6a−1 and 4(a−1) are both in GF (q). By assumption, √ a is not in (1+a)2 (1+a)2 − GF (q), so let us consider P to be an element of GF (q)(√ a), which is isomorphic to GF (q2). So, we can write all elements of GF (q)(√ a) in− the form z = x+y√ a. Let us refer to x as the real part of z, denoted by −(z), and to y as the imaginary− part of z, denoted by (z). < = Now, recall that for a Steiner chain of length k the tangent points on 1 are given by 1, P , P 2, ..., P k−1. The real part of P 2 satisﬁes B

(P 2) = 2 (P )2 1 < < − and the imaginary part of P satisﬁes

(P 2) = 2 (P ) (P ). = < = Note that those equations are the same as for the sine and cosine in the Euclidean plane, namely cos(2φ) = 2 cos(φ)2 1 − and sin(2φ) = 2 cos(φ) sin(φ).

114 8.6. Comparison to the Euclidean plane

Hence, calculating P 2 is in a sense the same as doubling the angle between 1 and P . The results of this chapter are also published in [24].

115

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Dank

Mein besonderer Dank gilt meinem Doktorvater Prof. Dr. Norbert Hungerbühler, der mit seiner Idee ”Endlicher Poncelet” den Anstoß zu dieser Arbeit gegeben hat, mir aber dennoch genügendFreiheit gelassen hat, viele meiner eigenen Ideen und Fragestellungen in die Arbeit einfließen zu lassen. Vielen Dank fürDein Engagement und Dein stets offenes Ohr! Ebenfalls möchte ich mich gerne bei Prof. Dr. Alexander Pott fürdie Annahme des Koreferates und einigen sehr guten Anregungen zur vorliegenden Arbeit bedanken. Ein großer Dank gilt auch meiner Familie, meinen Freunden und meinen Kollegen, die mich auf dem Weg zur Fertigstellung dieser Arbeit unterstützthaben.

121

Curriculum Vitae

PERSONAL DATA:

Date of birth: 01/10/1989 • Place of birth: Oberndorf bei Salzburg, Austria • Citizenship: Austria • EDUCATION:

2011 to 2013: M.Sc. in Mathematics (with distinction), ETH Z¨urich • 2008 to 2011: B.Sc. in Mathematics (with distinction), Paris Lodron Univer- • sit¨atSalzburg (Austria)

2004 - 2008: Matura, Musik- und Sport RG Akademiestrasse Salzburg • 2000 - 2004: Christian Doppler Gymnasium Salzburg • ACADEMIC AND WORK EXPERIENCE:

2013 - 2016: Ph.D student and teaching assistant, ETH Z¨urich, D-Math (by • the supervision of Prof. Dr. Norbert Hungerb¨uhler)

09/2015 - 11/2015: Short internship (part time) at IRsweep • 09/2012 - 12/2012: Teaching assistant, ETH Z¨urich, Department Mathematics • 2007 - 2010: Summer Internships at Sigmatek GmbH and CO KG • 2006: Social Internships at Private clinic Wehrle Salzburg and Retirement • home Oberndorf bei Salzburg

AWARDS AND SCHOLARSHIPS:

2014: ETH medal for outstanding Master Thesis • 2011: Austrian government scholarship for excellent Bachelor degree • PUBLICATIONS:

Kusejko, K.; Simultaneous Diagonalization of Conics in PG(2,q), Des. Codes • Cryptogr. 79(2016) no.3, 565-581

Kusejko, K.; Parapatits, L: A Valuation-Theoretic Approach to Translative • Equidecomposability, Adv. Math. 297 (2016), 174-195

Hungerb¨uhler,N.; Kusejko, K.; Poncelet’s Theorem in the four nonisomorphic • ﬁnite projective planes of order 9, Ars Combinatoria (to appear), http://arxiv.org/abs/1406.7857

Hungerbühler,N.; Kusejko, K.; Steiner’s Porism in finite Miquelian Möbius • planes, Adv. Geom (to appear) http://arxiv.org/abs/1507.01377

PREPRINTS:

Hungerb¨uhler,N.; Kusejko, K.; A Poncelet Criterion for special pairs of conics • in PG(2,p), http://arxiv.org/abs/1409.3035

TALKS:

June 18, 2015: A Valuation-Theoretic Approach to Translative Equidecompos- • ability, ETH Zurich

April 21, 2015: What is... a partition of integers?, Zurich Graduate Collo- • quium, ETH Zurich

March 19, 2015: Simultaneous diagonalization of conics in PG(2, q), Kloster • Banz, Germany

December 16, 2014: A remarkable pencil of conics in PG(2, q), Magdeburg, • Germany

September 18, 2014: Pairs of conics in diagonal form and their application to • Poncelet’s Theorem, Irsee, Germany

May 20, 2014: Poncelet’s Theorem in ﬁnite projective planes, Centre de Re- • cerca Matematica, Barcelona, Spain