Conics, Unitals and Net Replacement

Daniel Marshall

Thesis submitted for the degree of Doctor of Philosophy in Pure Mathematics at The University of Adelaide (Faculty of Engineering, Computer Science and Mathematics)

School of Mathematical Sciences

May, 2010 Contents

Abstract i

Signed Statement iii

Acknowledgements iv

1 Background 1

1.1 Aﬃne and Projective Planes ...... 2

1.2 Finite Fields ...... 4

1.3 Projective Spaces ...... 8

1.4 Collineations ...... 11

1.5 Substructures of Projective Planes ...... 13

1.5.1 k-arcs and ovals ...... 13

1.5.2 Unitals ...... 14

1.5.3 Baer sublines & Baer subplanes ...... 15

1.6 Derivation ...... 16

1.7 Quadrics ...... 18

1.7.1 Conics of PG(2, q)...... 19

1.7.2 Quadrics of PG(3, q)...... 22

1.7.3 Quadrics of PG(4, q)...... 23

1.7.4 Ovoids ...... 24 2 Bruck-Bose Correspondence 25

2.1 Translation Planes ...... 25

2.2 Spreads ...... 27

2.3 Construction ...... 28

2.4 Baer subplanes and Baer sublines in Bruck-Bose ...... 29

2.5 Derivation in Bruck-Bose ...... 33

2.6 Unitals in Bruck-Bose ...... 34

3 Net Replacement 37

3.1 Replaceable Nets in PG(3, q)...... 37

3.2 t-nests ...... 39

3.2.1 Basic Results ...... 41

3.2.2 Geometric Properties ...... 43

3.2.3 Combinatorial Properties ...... 47

3.2.4 Replaceable t-nests ...... 56

3.3 Net-Derivation ...... 62

3.3.1 Derivation ...... 65

3.3.2 Nest-Derivation ...... 65

3.3.3 Multiple Net-Derivation ...... 65

3.3.4 Net-Derivation on l 6= l∞ ...... 67

3.3.5 Net-Derivation in P0 6= PG(2, q2)...... 68 4 Unitals and Net Replacement 69

4.1 Introduction ...... 70

4.2 Derivation ...... 71

4.3 t-nests ...... 73

4.4 Net Replacement on l∞ ...... 76

4.5 Net-Derivation not on l∞ ...... 84

4.6 Net-Derivation on any line l ...... 89

4.7 Inherited unitals ...... 94

4.7.1 Derivation ...... 95

4.7.2 t-nests ...... 95

4.8 O’Nan conﬁgurations ...... 97

4.8.1 Inherited unitals in PN ...... 98 5 Conics and Net-Derivation 101

5.1 Introduction ...... 102

5.1.1 Nest-derivation ...... 103

5.2 Derivation ...... 104

5.2.1 Basic theorems ...... 104

5.2.2 The derivation set DR ...... 107

5.3 Characterisation ...... 109

5.3.1 Conics of BR ...... 109

5.3.2 Case C ∩ l∞ = {P,Q} with P ∈ DR,Q/∈ DR ...... 111

5.3.3 Derivation ...... 114

5.3.4 Multiple derivation ...... 119

5.4 Previous work ...... 121

5.4.1 Projective equivalence ...... 121

5.4.2 Known results ...... 122

5.5 Derivation with sDR and nDR ...... 124

2 2 2 5.6 The conics Cc,d : x0 − cx1 − dx2 = 0 under multiple derivation . . . . . 131

5.6.1 Preliminaries ...... 132

5.6.2 Derivation with φ(nDR) and φ(sDR)...... 133

5.6.3 Double derivation with DR and σ(DR)...... 137

5.6.4 Double derivation with φ(nDR) and σ(DR)...... 143

5.7 Inherited arcs in Andr´eplanes of odd order ...... 147

Bibliography 154 Abstract

The main concerns of this thesis are inherited unitals and conics in finite translation planes. Translation planes may be constructed from particular incidences in other translation planes. One method for doing this is ”net-derivation” or the corresponding operation ”net replacement”. We consider conics and unitals of the finite projective plane PG(2, q2) and observe the effect of net-derivation on their pointsets. Our aim is to determine when the pointsets of conics and unitals of PG(2, q2) are conics and unitals respectively in the translation planes formed after net-derivation. In particular, we focus on t-nest replacement and the corresponding nest-derivation sets.

Chapter one introduces all the necessary background on finite affine and projective planes. We consider all relevant substructures and concepts. Of major importance are the definitions of unitals, quadrics, Baer subplanes, Baer sublines and derivation.

Chapter two introduces the Bruck-Bose correspondence. We use the Bruck-Bose correspondence extensively in chapters three and four. The Bruck-Bose correspondence is a correspondence between PG(2, q2) and certain incidences in PG(4, q). The key elements are spreads of PG(3, q) as a subspace of PG(4, q). We also detail the known correspondences for Baer sublines, Baer subplanes and unitals as well as the equivalent operation for derivation.

Chapter three is where we begin our main work. Here we define net replacement in spreads and show the equivalence to net-derivation sets in PG(2, q2). We look at t- nests in depth, which are an example of net replacement. We prove several known results as well as a host of new geometric and combinatorial properties about t-nests. We show a detailed example of a known t-nest and also define a particular type of replacement set that is common to most t-nests. We finish with examples of different kinds of net-derivation.

Chapter four looks at unitals of PG(2, q2) and the eﬀect of general net-derivation. Given a unital of PG(2, q2), suppose we perform net-derivation in PG(2, q2) to form a new translation plane. Can we complete the aﬃne points of the unital to a unital in the new translation plane?

i We ﬁrst detail the known results for unitals and derivation. We then prove results for unitals and general net-derivation for all known cases where the net-derivation set lies on l∞. The particular case for t-nests was published separately by the author in [9].

We prove a new result for when the net-derivation set is not on l∞ which is also a new result when just considering derivation.

Next, we generalise several other results about derivation and unitals to include general net-derivation. We show the existence of non-inherited unitals in translation planes formed by t-nest replacement of a type that are not present in translation planes formed by derivation. We ﬁnish by considering O’Nan conﬁgurations contained in unitals in PG(2, q2) and planes formed by net-derivation.

Chapter five considers conics and the effect of multiple derivation. Given a conic of PG(2, q2), suppose we perform net-derivation in PG(2, q2) to form a new translation plane. Can we complete the affine points of the conic to a conic in the new translation plane? In particular, we focus on inherited conics with respect to multiple derivation.

We begin by deﬁning notation and present a new corollary on nest-derivation and conics, followed by several basic theorems on conics and derivation. We then present, in three stages, a novel characterisation of the equations of conics that are not arcs after derivation with the real derivation set.

Next we provide a brief survey of the known results for inherited conics and derivation.

We then restrict our attention to conics contained in a particular family Cc,d. Using this family, we prove several new theorems on the existence of inherited (q2 +1)-arcs in a class of planes formed by double derivation in PG(2, q2), where q is odd. We follow this by computing an example of a complete 24-arc in a particular translation plane of order 25. Finally, we show the existence of a family of inherited arcs in a class of Andr´eplanes which includes the regular Nearﬁeld planes of odd order.

ii Signed Statement

This work contains no material which has been accepted for the award of any other degree or diploma in any university or other tertiary institution to Daniel Marshall and, to the best of my knowledge and belief, contains no material previously published or written by another person, except where due reference has been made in the text.

I give consent to this copy of my thesis, when deposited in the University Library, being made available for loan and photocopying, subject to the provisions of the Copyright Act 1968. I also give permission for the digital version of my thesis to be made available on the web, via the Universitys digital research repository, the Library catalogue, the Australasian Digital Theses Program (ADTP) and also through web search engines, unless permission has been granted by the University to restrict access for a period of time.

SIGNED: ...... DATE: ......

iii Acknowledgements

First I would like to thank my principal supervisor Sue Barwick for tirelessly helping to improve the thesis and being consistently skeptical and correct in the face of my many incorrect paths. Thanks also goes to the different secondary supervisors I have had over the course of the work, Catherine Quinn, Matthew Brown, Alison Wolff and Rey Casse. Their input may have been more important than they realise and the nature of the benefit even not as they would have predicted.

On a personal level, I give thanks to all the other postgraduate students that made life more bearable under the strain of work. Schnitzels, darts and coﬀees characterised the social landscape, there is no need to digress into the details of those involved. I would like to also thank my Mother for supporting me, especially in the beginning, which gave me the freedom of a clear mind.

Finally, I would like to really really thank my now wife Chen Yang for dealing with me through the life of most of the project.

iv Chapter 1

Background

In this thesis, we are primarily interested in projective planes and their substructures. We assume the reader is familiar with undergraduate level algebra, discrete mathematics and number theory. In particular, we assume the reader is familiar with groups, linear algebra and ﬁnite ﬁelds. We also assume familiarity with group actions, including the ideas of transitivity and orbits.

In this chapter we firstly define finite affine and projective planes. We then introduce finite fields and consider the notions of field extensions and square and non-square elements. We move on to look at projective spaces and their homogeneous coordinates. The idea of a collineation is then introduced with emphasis on homographies. We then visit substructures of projective planes such as arcs and unitals which are of major importance later. Further, we look at Baer subplanes and Baer sublines. Next, we consider derivation, a technique for constructing new projective planes from existing ones. Finally, we consider quadrics and in particular conics which are both substructures of a projective space.

For a general introductory treatment on the topic of projective geometry, see Casse [21]. Unless otherwise stated, all the deﬁnitions, properties and results in this chapter can be found in texts such as [21], [33], [34] and [55].

1 1.1 Aﬃne and Projective Planes

We begin by defining affine and projective planes and providing several basic definitions and properties.

Deﬁnition 1.1.1 An aﬃne plane A is a set of objects, called points, together with certain subsets of points, called lines, such that

1. any two distinct points are contained in a unique common line,

2. given a point P and a line l with P/∈ l, there exists a unique line m with P ∈ m and l ∩ m = ∅, and

3. there exist three points which are not contained in a common line.

If two lines l and m have no common point, then l and m are parallel. The set of all lines in an aﬃne plane can be partitioned into parallel classes such that two lines lie in the same class if they are parallel.

Deﬁnition 1.1.2 A projective plane P is a set of objects, called points, together with certain subsets of points, called lines, such that

1. any two distinct points are contained in a unique common line,

2. any two distinct lines meet in a unique point, and

3. there exist four points, no three of which are contained in a common line.

If a set of points lie on a common line, then the points are collinear. If a set of lines all contain a common point, then the lines are concurrent.

Deﬁnition 1.1.3 Let P and P0 be two projective planes. Suppose there exists a bijec- tion θ that maps points of P to points of P0 and lines of P to lines of P0. Suppose also that if P lies on a line l of P then P θ lies on the line lθ of P0. Then we say that P and P0 are isomorphic.

2 We can deﬁne isomorphism between aﬃne planes in a similar way.

Let A be any affine plane. For each parallel class of lines in A, add a point at infinity to each line of that parallel class. Create a line at infinity, denoted l∞, which contains precisely the points at infinity. The affine plane A together with the line at infinity and points at infinity form a projective plane called the projective completion of A.

If P is any projective plane and l is any line of P, then removing the line l and all its points from P will produce an aﬃne plane P\l. In this case, the points P\l are referred to as the aﬃne points of P.

Here, we are concerned with affine and projective planes that contain a finite number of points. Such planes are referred to as finite. The number of points and lines and their incidence in a finite projective plane can be characterised by a single number as in the following theorem.

Theorem 1.1.4 Let P be any ﬁnite projective plane. Then there exists some integer n ≥ 2, called the order of P, such that

1. each line of P contains n + 1 points,

2. each point of P lies on n + 1 lines,

3. the number of points of P is n2 + n + 1, and

4. the number of lines of P is n2 + n + 1.

Next we state an important geometric property for characterising projective planes.

Theorem 1.1.5 (Desargues Theorem) Let ABC, A0B0C0 be two triangles, with distinct vertices, such that the lines AA0,BB0,CC0 are concurrent in the point V . Then the points L = BC ∩ B0C0,M = AB ∩ A0B0,N = AC ∩ A0C0 are collinear.

A projective plane P is called Desarguesian, if Desargues Theorem always holds, otherwise P is called a non-Desarguesian projective plane.

3 1.2 Finite Fields

There is a strong connection between projective geometry and certain algebraic structures. Here, we consider ﬁnite ﬁelds and their relevant properties. The properties we present are stated without proof.

Every finite field has order ph, for p prime and h a positive integer. All finite fields of a given order are isomorphic. We denote the unique field of order ph as GF (q) where h q = p . The intersection of all subfields of GF (q) is isomorphic to Zp. The number p is called the characteristic of GF (q).

The multiplicative group of GF (q) is cyclic, thus GF (q) contains an element ω such that ω, ω2, ω3, ..., ωq−1 are the q − 1 elements of GF (q)\{0}. The element ω is called a primitive element or generator of GF (q). The q − 1 non-zero elements of GF (q) satisfy xq−1 = 1.

Example 1.2.1 Suppose we work with the ﬁeld GF (5) = {0, 1, 2, 3, 4}. Addition and multiplication of elements is calculated modulo 5. Notice that 2 is a generator for GF (5) as 2 ∗ 2 = 4, 4 ∗ 2 = 8 = 3, 3 ∗ 2 = 6 = 1 and 1 ∗ 2 = 2.

An automorphism of a ﬁnite ﬁeld GF (q) is an isomorphism σ : GF (q) → GF (q). The automorphisms of GF (q) are of the form σ : a → apr , where 0 ≤ r ≤ h − 1.

Further, any automorphism of GF (q) acts as the identity on elements of Zp.

Let f be an irreducible polynomial over a field GF (q). Let α be a zero of f, thus necessarily α∈ / GF (q). Using α we can create an extension field GF (q)(α) of GF (q). The field GF (q)(α) is given by elements of the form,

2 n−1 a0 + a1α + a2α + ... + an−1α ,

where a0, . . . , an−1 ∈ GF (q) and n is the degree of f. The extension ﬁeld GF (q)(α) is a ﬁeld of order qn. It is always possible to choose f such that α is a primitive element of GF (q)(α). In this case f is called a primitive polynomial of GF (q).

4 Since all finite fields of a given order are isomorphic, then all finite extension fields of a given order are also isomorphic. We will refer to such extension fields uniquely as GF (qn), where qn is the order of the extension field. The automorphism defined by x → xq where x ∈ GF (qn), fixes GF (q) as a subfield of GF (qn).

Let α ∈ GF (q) such that α = β2, for some β ∈ GF (q)\{0}. All such α are called 1 squares of GF (q) and there are 2 (q − 1) such elements. The remaining elements of GF (q)\{0} are called non-squares. The element {0} is neither a square or a non- square.

Let q be odd and let GF (q)\{0} = {α, α2, α3, ..., αq−1}, where α is a primitive element of GF (q). The squares of GF (q) are the even powers of α and the non-squares are the odd powers of α. The element αq−1 is a square since q − 1 is even.

If q is even, every ﬁeld of order q has characteristic 2. Further, every element of GF (q), where q is even, has a unique square root and is thus a square.

Lemma 1.2.1 Let s1, s2 be squares in GF (q) where q is odd. Let n1, n2 be non-squares in GF (q). Then,

1. s1 × s2 is a square.

2. s1 × n1 and n1 × s1 are non-squares.

3. n1 × n2 is a square.

−1 4. s1 is a square.

−1 5. n1 is a non-square.

2k1 2k2 2k3+1 2k4+1 Proof Let s1 = ω , s2 = ω , n1 = ω , n2 = ω , where k1, k2, k3, k4 ∈ q−1 {0,..., 2 }. Then,

2k1 2k2 2(k1+k2) 1. s1 × s2 = ω × ω = ω which is square.

2k1 2k3+1 2(k1+k3)+1 2. s1 × n1 = n1 × s1 = ω × ω = ω which is a non-square.

2k3+1 2k4+1 2(k3+k4)+2 3. n1 × n2 = ω × ω = ω which is a square.

5 −1 q−1 −1 q−1−2k1 4. s1 × s1 = ω . Therefore s1 = ω which is a square.

−1 q−1 −1 q−1−2k3−1 5. n1 × n1 = ω . Therefore n1 = ω which is a non-square.

Let p be the characteristic of GF (q), then (a + b)p = ap + bp where a, b ∈ GF (q) and (a − b)p = ap − bp.

Suppose we are working in the extension ﬁeld GF (q2) and let α be a generator for GF (q2). The non-zero elements of GF (q) as a subﬁeld of GF (q2) are,

{αq+1, α2(q+1), . . . , α(q−1)(q+1)}.

Here, the squares of GF (q) are the elements α2(q+1), α4(q+1), . . . , α(q−1)(q+1) and the non-squares of GF (q) are the elements α(q+1), α3(q+1), . . . , α(q−2)(q+1). Every element of GF (q) is a square in GF (q2). We also have the following lemma.

Lemma 1.2.2 Let x be a non-zero element of GF (q2), where q is odd. Then, the element xq+1 is an element of GF (q) as a subﬁeld of GF (q2). Further, if n is a non- square in GF (q2), then nq+1 is a non-square in GF (q). Also, if s is a square in GF (q2), then sq+1 is a square in GF (q).

Proof Let α be a generator for GF (q2) and thus x = αk, for some k ∈ 1, . . . , q2. We may write k = k1(q − 1) + k2, where k1 ≥ 0 and 0 ≤ k2 ≤ q − 2. Then,

xq+1 = α(k1(q−1)+k2)(q+1)

= αk1(q−1)(q+1)+k2(q+1)

= αk2(q+1).

The element αk2(q+1) is an element of GF (q), hence xq+1 is an element of GF (q).

2 k1(q−1)+k2 Let n be a non-square in GF (q ), then we may write n = α , where k1 ≥ 0,

q+1 k2(q+1) 0 ≤ k2 ≤ q−2 and k2 odd. Hence n = α , which is a non-square in GF (q) as k2 is odd. Similarly, let s be a non-square in GF (q2), then we may write s = αk1(q−1)+k2 ,

q+1 k2(q+1) where k1 ≥ 0, 0 ≤ k2 ≤ q − 2 and k2 even. Hence s = α , which is a square in

GF (q) as k2 is even.

6 We conclude this section by giving an example of an extension ﬁeld which we will use throughout the document.

Example 1.2.2 Let GF (5) = {0, 1, 2, 3, 4}. The polynomial f = x2 + x + 2 is a primitive polynomial over GF (5) (see Hirschfeld [33]). Let α be a zero of f and thus α2 = 4α + 3. Let GF (52) denote the extension ﬁeld GF (5)(α). There are two ways in which we can represent GF (52).

The ﬁrst representation is using α as a generator, hence

GF (52) = {0, α, α2, . . . , α24}.

In this case, the squares of GF (52) are the even powers of α. The non-squares are the odd powers of α. The GF (5) subﬁeld consists of the elements,

{0, α6, α12, α18, α24}, where α6 = 2, α12 = 4, α18 = 3, α24 = 1.

The second representation is the vector space representation, in this case,

GF (52) = {a + bα| a, b ∈ GF (5)}.

We convert between the two representations of the ﬁeld to perform addition and multiplication as required. Here, we present the calculations needed for later examples.

α21 + α17 = 2α + 1 + α + 1 α21 + α23 = 2α + 1 + 2α + 2 = 3α + 2 = α4 = 4α + 3 = α2 α24 + α8 = 1 + 3α + 1 α24 + α14 = 1 + α + 2 = 3α + 2 = α4 = α + 3 = α15 α21 + α5 = 2α + 1 + 4α + 4 α24 + α20 = 1 + 2α + 4 = α = 2α = α7 α21 + α11 = 2α + 1 + 3α + 3 α24 + α2 = 1 + 4α + 3 = 4 = α12 = 4α + 4 = α5 α + α6 = α + 2 α24 + α13 = 4α + 1 = α14 = α22

7 α + α12 = α + 4 α24 + α19 = 3α + 1 = α10 = α8 α + α18 = α + 3 α24 + α = α + 1 = α15 = α17 α + α24 = α + 1 α24 + α7 = 2α + 1 = α17 = α21

1.3 Projective Spaces

The study of finite projective geometry is not restricted to the plane. Here we define finite projective spaces of dimension greater than or equal to two.

Let V be a (d + 1)-dimensional vector space over GF (q), for any integer d ≥ 2. Let the points be the 1-dimensional subspaces of V excluding the zero vector, the lines be the 2-dimensional subspaces of V , the planes be the 3-dimensional subspaces of V,..., and the hyperplanes be the d-dimensional subspaces of V . Then, using containment of subspaces, the resulting structure is deﬁned as PG(d, q). We say PG(d, q) is a ﬁnite projective space of dimension d and order q.

We can count the total number and incidences of the points, lines, planes and hyper- d d−1 qd+1−1 plane in PG(d, q). In PG(d, q) there are q + q + ... + q + 1 = q−1 points and each line contains q + 1 points.

In PG(3, q) there are,

• q3 + q2 + q + 1 points,

• q3 + q2 + q + 1 planes,

• q + 1 points on each line and q2 + q + 1 lines through each point,

• q2 + q + 1 points on each plane and q2 + q + 1 planes through each point,

• q2 + q + 1 lines in each plane and q + 1 planes through each line,

• q4 + q3 + 2q2 + q + 1 lines.

8 In P G(4, q) there are,

• q4 + q3 + q2 + q + 1 points,

• q4 + q3 + q2 + q + 1 hyperplanes,

• q + 1 points on each line and q3 + q2 + q + 1 lines through each point,

• q2 +q +1 points in each plane and q4 +q3 +2q2 +q +1 planes through each point,

• q3 + q2 + q + 1 points in a hyperplane and q3 + q2 + q + 1 hyperplanes through each point,

• q2 + q + 1 lines in each plane and q2 + q + 1 planes through each line,

• q4 + q3 + 2q2 + q + 1 lines in each hyperplane and q2 + q + 1 hyperplanes through each line,

• q3 + q2 + q + 1 planes in a hyperplane and q + 1 hyperplanes through each plane.

Let h, k and r be positive integers such that h ≤ r and k ≤ r. Let Sr be a projective space of dimension r and let Sh and Sk be two subspaces of Sr with dimension h and k respectively. The span of Sh and Sk, denoted by hSh,Ski, is the intersection of all the subspaces of Sr containing both Sh and Sk. Every projective space satisﬁes the dimension axiom as follows,

dimSh + dimSk = dim(Sh ∩ Sk) + dim(hSh,Ski).

The projective space PG(2, q) is a projective plane of order q and in fact a ﬁnite projective plane is Desarguesian if and only if it is isomorphic to PG(2, q).

Example 1.3.1 Let l be a line of PG(4, q) and let P be a point on l. Any hyperplane containing l also contains P . There are q2 + q + 1 hyperplanes of PG(4, q) containing the line l and there are q3 + q2 + q + 1 hyperplanes through the point P . Hence there are q3 + q2 + q + 1 − (q2 + q + 1) = q3 hyperplanes through P that do not contain l.

9 The underlying field can be used to coordinatise our finite projective spaces. Consider the finite projective space PG(d, q) over the field GF (q), and let V denote the underlying (d + 1)-dimensional vector space over GF (q). Let P be a point of PG(d, q), thus P is a 1-dimensional subspace hvi of V . Any non-zero scalar multiple of v will represent the same point P and we call v the homogeneous coordinates of P .

Points of PG(d, q) have homogeneous coordinates,

{(x0, x1, ..., xd)| x0, ..., xd ∈ F, not all zero},

where (x0, x1, ..., xd) and t(x0, x1, ..., xd) for t ∈ GF (q)\{0} represent the same point.

Any hyperplane H of PG(d, q) is represented by a homogeneous linear equation,

a0x0 + a1x1 + ... + adxd = 0,

where a0, ..., ad ∈ GF (q). The vectors h[a0, . . . , ad]i are the homogenous coordinates of H. Any point (x0, x1, . . . , xd) satisfying the equation of the hyperplane H is a point contained in H.

Example 1.3.2 The points of PG(2, q) have homogeneous coordinates of the form

(x0, x1, x2), where x0, x1, x2 ∈ GF (q). Let l∞ be the line x2 = 0. Points on l∞ are of the form (x0, x1, 0). Since the coordinates are homogeneous, we can write the q + 1 points of l∞ uniquely as,

{(0, 1, 0)} ∪ {(1, x1, 0)| x1 ∈ GF (q)}.

In a similar way, the aﬃne points of PG(2, q) can be written uniquely as (x0, x1, 1), where x0, x1 ∈ GF (q).

If we remove the line l∞ and its points from PG(2, q) we form an affine plane which we denote AG(2, q) whose points are the affine points of PG(2, q). Let P be a non- Desarguesian projective plane of order q. Suppose we remove a line l from P to form an affine plane P\l, then P\l is not isomorphic to AG(2, q). An affine plane that is not isomorphic to AG(2, q) is called non-Desarguesian.

10 1.4 Collineations

Here we consider maps acting on the points of a projective space.

Deﬁnition 1.4.1 A one to one map of an aﬃne or projective space π onto itself, is said to be a collineation if and only if the images of collinear points are collinear.

Let S be a set of points in an aﬃne or projective space π. Let σ be a collineation of π such that σ(s) ∈ S for all s ∈ S. We say σ ﬁxes the pointset S.

Deﬁnition 1.4.2 Let A be a (d + 1) × (d + 1) non-singular matrix over GF (q). Then the map σ : PG(d, q) → PG(d, q) deﬁned by,

0 0 (x0, . . . , xd) → (x0, . . . , xd), such that,     0 x x0  0     .   .  ρ  .  = A  .  , 0 xd xd 0 0 for some ρ 6= 0, where (x0, . . . , xd) and (x0, . . . , xd) are points in PG(d, q), is called a homography of PG(d, q). A homography is a collineation of PG(d, q).

Let X and X0 be two general points of PG(d, q) and let AX = ρX0, where A is the matrix of a homography and ρ 6= 0. Any non-zero multiple λA of A determines the same homography, since (λA)X = λ(AX) = λ(ρX0) ≡ ρX0.

To simplify notation, we will refer to a homography as a map and as a matrix inter- changeably. That is suppose σ is the map deﬁning a homography then we also use σ to refer to the matrix that deﬁnes the homography.

11 The group of homographies of PG(d, q), with respect to composition of functions, is called the projective (linear) group of dimension d+1 over GF (q), and is denoted by P GL(d + 1, q).

Deﬁnition 1.4.3 Let α be an automorphism of GF (q). Then α induces an automorphic collineation σ on PG(d, q), deﬁned by,

α α (x0, . . . , xd) → (x0 , . . . , xd ), where (x0, . . . , xd) is a point in PG(d, q). An automorphic collineation of PG(d, q) is a collineation of PG(d, q).

Theorem 1.4.4 Every collineation of PG(d, q) is the product of a homography and an automorphic collineation.

The set of automorphic collineations of PG(d, q) form a group with respect to the composition of functions. The product of the group of homographies of PG(d, q) and the group of automorphic collineations is called the projective group of dimension d + 1 over GF (q), and is denoted by P ΓL(d + 1, q).

Suppose σ ∈ P GL(d + 1, q) and σ ﬁxes a set of points S in PG(d, q), not necessarily pointwise. The set of all such σ forms a subgroup of the full collineation group and we denote this subgroup by P GL(d + 1, q)S. Similarly, if σ ∈ P ΓL(d + 1, q) and σ ﬁxes a pointset S then we denote the subgroup of such σ as P ΓL(d + 1, q)S.

Example 1.4.1 The group P GL(3, q)l∞ consists of all the homographies that ﬁx l∞ in PG(2, q). This group consists of the homographies with the following matrices,   a b g      e f h  , 0 0 i where a, b, e, f, g, h, i ∈ GF (q) and the determinant of the matrix is non-zero.

12 One useful subgroup of P GL(3, q)l∞ is AGL(3, q). The group AGL(3, q) consists of the homographies with matrices,   a b 0      e f 0  , 0 0 1 where a, b, e, f ∈ GF (q) and the determinant of the matrix is non-zero. This group will be used extensively in a later chapter. It is known that AGL(3, q) is 2-transitive on the points of AG(2, q).

Suppose AG(2, q) = PG(2, q)\l∞, then the full collineation group of AG(2, q) is P ΓL(3, q)l∞ .

The group P ΓL(3, q)l∞ is also denoted by AΓL(3, q).

1.5 Substructures of Projective Planes

Let S be a set of points in a projective space π. A line of π that meets S in k points is called a k-secant.

1.5.1 k-arcs and ovals

Deﬁnition 1.5.1 A k-arc in a projective plane P, is a set of k points where no three are collinear. We call a line of P tangent or secant to a k-arc K depending on whether the line is a 1-secant or a 2-secant to K.

Deﬁnition 1.5.2 Let P be a projective plane of order q.A (q + 1)-arc of P is called an oval. If q is even, then all the tangents to an oval O are concurrent in a point N called the nucleus of O. The oval O together with its nucleus form a (q + 2)-arc called a hyperoval.

13 1.5.2 Unitals

One substructure we use heavily in later chapters is the unital. Here, we deﬁne unitals along with several necessary lemmas and deﬁnitions. For a more complete look at unitals, see Barwick and Ebert [11].

Deﬁnition 1.5.3 Let P be a projective plane of order q2.A unital is a set of q3 + 1 points U in P such that every line of P meets the set in 1 or q + 1 points. We call a line of P tangent or secant to a unital U depending on whether the line is a 1-secant or a (q + 1)-secant to U.

Lemma 1.5.4 [50] Let U be a unital of a projective plane P of order q2. Through a point P of U there passes one tangent and q2 secants. Through a point Q not on U there pass q + 1 tangents and q2 − q secants. It follows that U has q3 + 1 tangents and q2(q2 − q + 1) secants.

Suppose P ∼= PG(2, q2). Let σ be the automorphism of GF (q2) deﬁned by σ(x) = xq, for all x ∈ GF (q2). Let X be the homogeneous coordinates of a general point of PG(2, q2), written as a column vector and let Xt denote the transpose of X. Let A be a 3 × 3 matrix over GF (q).

Deﬁnition 1.5.5 The set of points of P satisfying the equation XtAXσ = 0 is called a Hermitian variety. A Hermitian variety is said to be degenerate if there exists a homography that reduces the equation of the Hermitian variety into an equation involving less than three variables. Otherwise the Hermitian variety is said to be non- degenerate.

A non-degenerate Hermitian variety is called a Hermitian curve. The points of a Hermitian curve form a unital [33], and such a unital we call classical. Later we will mention the existence of non-classical unitals, such unitals are not the point sets of Hermitian curves.

14 1.5.3 Baer sublines & Baer subplanes

Let P be a projective plane. Let P0 be a subset of the points of P together with the subsets of the lines of P containing only the points of P0. The structure P0 is called a subplane of P, if P0 is a projective plane using the incidence of P restricted to the members of P0.

Deﬁnition 1.5.6 Let P be a projective plane of order q2. A Baer subplane B of P is a subplane of order q. A line of P that meets B in 1 point is called a tangent and a line that meets B in q + 1 points is called a secant.

If B is a Baer subplane of a projective plane P, then every line of P is either a tangent or secant to B. If P ∈ P\B, there is a unique line containing P that is secant to B and the other q2 lines containing P are tangent to B.

Deﬁnition 1.5.7 Let P be a projective plane of order q2. Let l be a secant to a Baer subplane B of P. Then the points l ∩ B are a Baer subline of l and of P.

We will mainly be working with Baer sublines and subplanes in PG(2, q2). A Baer subplane B of PG(2, q2) is a subplane of PG(2, q2) that is isomorphic to PG(2, q).

Theorem 1.5.8 Let P , Q and R be three distinct collinear points of PG(2, q2). Then there is a unique Baer subline of PG(2, q2) containing P , Q and R.

Theorem 1.5.9 Let P , Q, R, and S be four distinct points of PG(2, q2), with no three collinear. Then there exists a unique Baer subplane of PG(2, q2) containing the points P , Q, R and S.

Lemma 1.5.10 Let l be a line of PG(2, q2). Then l contains q(q2 + 1) unique Baer sublines.

15 q2+1 Proof There are 3 choices for three distinct points on l and three such points q+1 determine a unique Baer subline. Each Baer subline contains 3 diﬀerent choices for (q2+1)q2(q2−1) 2 three points. Hence the total number of Baer sublines on l is (q+1)q(q−1) = q(q + 1).

Lemma 1.5.11 Let l be a line of PG(2, q2). Let P and Q be two points on l. Then there are q + 1 Baer sublines of l containing both P and Q.

Proof Let R be a point of l such that R 6= P,Q and hence there are q2 − 1 choices for R. By Theorem 1.5.8, there is a unique Baer subplane of l containing P,Q and R. Each Baer subline on l containing P and Q contains q − 1 points of l other than P and q2−1 Q. Hence there will be q−1 = q + 1 distinct Baer sublines on l that contain both the points P and Q.

Let σ be collineation of a projective plane P. If b is a Baer subline of P, then σ(b) is also a Baer subline of P. Likewise, if B is a Baer subplane of P, then σ(B) is a Baer 2 subplane of P. Let l be a line of PG(2, q ), then the collineation group P GL(3, q)l is transitive on all Baer sublines contained in l.

1.6 Derivation

In this section we detail a method for forming new affine and projective planes from existing planes. Suppose A is a finite affine plane of order q2 and let P be the projective completion of A with line at infinity l∞.

Deﬁnition 1.6.1 Let D be a set of q + 1 points on l∞ such that for any distinct points

X,Y of A for which XY meets l∞ in a point of D, there is a unique Baer subplane B of P containing X,Y and D. We call D a derivation set of A and P.

16 Example 1.6.1 We coordinatise PG(2, q2) as in Example 1.3.2. The pointset

DR = {(0, 1, 0)} ∪ {(1, x1, 0)| x1 ∈ GF (q)}

2 is a derivation set of PG(2, q ) on l∞.

In PG(2, q2), every Baer subline is a derivation set and every derivation set is a Baer subline.

Lemma 1.6.2 Let D be a derivation set in PG(2, q2). There are q2(q + 1) Baer subplanes of PG(2, q2) that contain D.

Let D be a derivation set in a projective plane P = A ∪ l∞. A derivation set can be used to deﬁne a new incidence structure AD as follows.

• The points of AD are the points of A.

• The lines of AD are

– the lines of A whose projective extensions meet l∞ in a point not in D, and – the Baer subplanes of P that contain D.

• Incidence is inclusion.

Theorem 1.6.3 [1], [45] The incidence structure AD is an aﬃne plane of order q2.

We call AD the derived plane of A. The aﬃne plane AD can be uniquely completed D 0 0 2 to a projective plane P by adding a line l∞. The points of l∞\D are the q − q points D of l∞. The correspondence between the original plane P and the new plane P is shown in Figure 1.1.

Suppose P ∼= PG(2, q2) then PD is the Hall plane of order q2. We denote this plane by Hall(q2).

In PG(2, q2), since every Baer subline is a derivation set, there exists derivation sets on all the lines of PG(2, q2). Thus we can derive with respect to a derivation set on a line l 6= l∞.

17 R R D 0 l∞ l∞ P P Q Q B

P PD

Figure 1.1: Correspondence between P and PD.

If D1,..., Dk are k disjoint derivation sets on l∞, then we can perform multiple derivations in the plane. We denote the resulting plane by PD1...Dk . The order in which we perform the derivation is unimportant, hence if D1 and D2 are disjoint derivation sets in P then PD1D2 ∼= PD2D1 . It is also possible to perform a multiple derivation on a line other than l∞.

1.7 Quadrics

Deﬁnition 1.7.1 The set of points (x0, . . . , xd) of PG(d, q) satisfying the homogeneous polynomial φ in x0, . . . , xd of degree two, X φ : aijxixj = 0, 0≤i≤j≤d where aij ∈ GF (q) is called a quadric of PG(d, q).

The equation of a quadric Q can also be written as a matrix equation XtAX = 0,     a00 . . . a0d x0 h i      . .   .  x0 . . . xd  . ... .   .  = 0.

ad0 . . . add xd We call A the associated matrix of Q.

Suppose the characteristic of GF (q) is not 2, then we can write the equation of the quadric as X 2 φ : aiixi + 2aijxixj = 0. j>i

18 In this case, the quadric has associated matrix,   a00 . . . a0d    . .  A =  . ... .  ,

a0d . . . add where A is symmetric.

Deﬁnition 1.7.2 Let Q be a quadric in PG(d, q).

1. The quadric Q is degenerate, if there exists a homography which transforms φ = 0 into a quadratic with less than d + 1 variables. Otherwise, the quadric Q is non-degenerate.

2. For q odd, let A be the associated matrix of Q. The quadric Q is degenerate if and only if det(A) = 0.

A subspace of maximal dimension contained in a quadric Q is called a generator of Q. At each point P of a non-degenerate quadric Q there exists a tangent hyperplane which meets the quadric in the point P and all the generators of Q that contain P .

Lemma 1.7.3 Let Sr be an r-dimensional subspace of PG(d, q) and let Q be a quadric of PG(d, q). Then Q ∩ Sr is a quadric of Sr, unless Sr lies wholly on Q.

1.7.1 Conics of PG(2, q)

A conic of PG(2, q) is a quadric of PG(2, q). A conic has the general equation,

2 2 2 a00x0 + a11x1 + a22x2 + (a01 + a10)x0x1 + (a12 + a21)x1x2 + (a02 + a20)x0x2 = 0, or in matrix form,     a00 a01 a02 x0 h i         x0 x1 x2  a10 a11 a12   x1  = 0.

a20 a21 a22 x2

19 Notice that for a given conic equation there is more than one corresponding matrix, however for a particular matrix, the matrix form determines a unique conic equation. If the characteristic of GF (q) is not equal to 2, we can write the general equation of a conic as 2 2 2 ax0 + bx1 + cx2 + 2fx1x2 + 2gx2x0 + 2hx0x1 = 0, or in matrix form,     a h g x0 h i         x0 x1 x2  h b f   x1  = 0.

g f c x2

That is XtAX = 0 where A is a symmetric matrix.

The tangent hyperplanes to a conic are lines. A line not contained entirely in the conic C of P G(2, q) meets C in either 0, 1 or 2 points. We call each line of PG(2, q) either external, tangent or secant respectively.

There is one type of non-degenerate conic, the points of this conic form an oval. If q is odd, every oval of PG(2, q) is a non-degenerate conic [51]. If q is even, a non-degenerate conic C has a nucleus N and C ∪ N forms a hyperoval. However not all hyperovals of PG(2, q) are a conic together with its nucleus.

Lemma 1.7.4 There is a unique non-degenerate conic passing through 5 points, no 3 collinear.

Lemma 1.7.5 Let C be the aﬃne points of a non-degenerate conic C of P G(2, q2). If q ≥ 3, then the aﬃne points C are contained in C and no other conic of PG(2, q2).

Proof If q ≥ 3, then C contains at least 10 points, of which a maximum of 2 can be on l∞. Thus the affine pointset C contains at least eight points. Any five points with no three collinear determine a unique conic. Hence any five points of C will uniquely determine C and will be contained in no other conic of PG(2, q2).

20 Suppose Cˆ is a conic of PG(2, q). We can naturally embed PG(2, q) as a Baer subplane of PG(2, q2). The conic Cˆ can be naturally extended to a conic of PG(2, q2) with the same equation which meets the subplane PG(2, q) in the points of Cˆ.

The image of a non-degenerate conic under a collineation is a non-degenerate conic. In fact, we have the following theorem.

Theorem 1.7.6 [33] The group P GL(3, q) is transitive on non-degenerate conics of PG(2, q).

Recall our notation for homographies. Let σ be a homography. We use σ to refer to a homography as a map as well as to the matrix which deﬁnes the homography. We can ﬁnd the equation of the image of a conic as follows.

Lemma 1.7.7 If we apply a homography σ (with associated matrix also denoted by σ) to a conic with associated matrix A, the resulting conic has associated matrix σ−tAσ−1.

Proof Given a conic with associated matrix A, the general point X on the conic satisﬁes XtAX = 0. Suppose we apply the homography σ to the points of the conic, such that say X0 = σX. We wish to determine the equation of the conic after we apply σ, thus.

XtAX = 0 ⇐⇒ (σ−1X0)tA(σ−1X0) = 0 ⇐⇒ X 0t(σ−1)tAσ−1X0 = 0.

Hence the equation of the new conic is given by the matrix σ−tAσ−1.

There are three types of degenerate conics in PG(2, q): a point, a line, and a pair of intersecting lines.

21 1.7.2 Quadrics of PG(3, q)

There are two types of non-degenerate quadrics in PG(3, q).

One type is the elliptic quadric which consists of q2 + 1 points with no three collinear. The tangent hyperplanes are planes and these meet the elliptic quadric in one point. All other planes of PG(3, q) meet the quadric in a non- degenerate conic.

The second type is the hyperbolic quadric which consists of (q + 1)2 points and two sets of q + 1 mutually skew lines, with any line from one set meeting every line of the other set in a unique point of the quadric. The tangent hyperplanes are planes and these meet the hyperbolic quadric in a pair of intersecting lines. Every other plane of PG(3, q) meets the hyperbolic quadric in a non-degenerate conic.

There are four types of degenerate quadrics in PG(3, q): a line, a plane, a pair of intersecting planes (necessarily meet- V ing in a line) and a quadratic cone. A quadratic cone is a 3-dimensional cone whose base is some non-degenerate conic C contained in a plane α and whose vertex V is some point not contained in α. Every line contained in C α a quadratic cone necessarily contains the vertex V . Any plane of PG(3, q) meets a quadratic cone in either a point, a line, a pair of lines, or a conic. If the intersecting plane does not contain V , then the intersection is a conic.

22 1.7.3 Quadrics of PG(4, q)

There is one type of non-degenerate quadric in PG(4, q) which is the parabolic quadric. A parabolic quadric consists of q3 +q2 +q+1 points and contains q3 +q2 +q+1 lines, with q + 1 such lines through each of its points.

The tangent hyperplane at each point meets the quadric in the q + 1 generator lines through that point. These lines form a quadratic cone contained in the quadric. All other hyperplanes of PG(4, q) meet the quadric in a hyperbolic quadric or an elliptic quadric. Every plane of PG(4, q) meets a parabolic quadric in a point, a line, two intersecting lines or a non-degenerate conic. Every line of PG(4, q) meets a parabolic quadric in 0, 1, 2 or q + 1 points.

There are six degenerate quadrics in PG(4, q), one V of which is the elliptic cone. An elliptic cone is a 4-dimensional cone whose base E is some 3- Σ dimensional elliptic quadric contained in a hyper- E plane Σ and whose vertex V is a point not contained in Σ. Every plane of PG(4, q) meets a elliptic cone in either a point, a line, two intersecting lines or a non-degenerate conic.

Example 1.7.1 Let {(x0, x1, x2, x3, x4)|x0, . . . , x4 ∈ GF (q), and not all zero} be the set of homogeneous coordinates for the points of PG(4, q). Let Σ∞ be the subspace of

PG(4, q) deﬁned by x4 = 0. Suppose φ = 0 is the equation of a hyperbolic quadric in

Σ∞, then a non-degenerate equation of the form,

φ + d0x0x4 + d1x1x4 + d2x2x4 + d3x3x4 = 0,

where d0, . . . , d3 ∈ GF (q) is a parabolic quadric of PG(4, q) that meets Σ∞ in the hyperbolic quadric φ. There are many valid choices of d0 . . . , d3, thus there are many parabolic quadrics in PG(4, q) that contain the hyperbolic quadric φ = 0.

23 1.7.4 Ovoids

Deﬁnition 1.7.8 An ovoid in PG(3, q) is a set of q2 + 1 points such that no three are collinear.

If q is odd, then every ovoid is an elliptic quadric. If q is even there is another example of an ovoid, the Suzuki-Tits Ovoid [54]. The existence of other ovoids is an open question.

An ovoidal cone is a 4-dimensional cone in V PG(4, q) whose base is some ovoid O contained in a hyperplane Σ and whose vertex V is some Σ point not contained in Σ. Every plane of PG(4, q) O meets an ovoidal cone in either a point, a line, two intersecting lines or an oval.

24 Chapter 2

Bruck-Bose Correspondence

A major tool we employ for investigating projective planes is the Bruck-Bose correspondence. The Bruck-Bose correspondence is a representation of a translation plane of order q2 as an incidence structure in PG(4, q). Often it is much easier to work with the corresponding incidence structure in PG(4, q) than it is to work in the plane.

First, we review the definition of translation planes. We then define a regulus and a spread in PG(3, q). Next, we present the Bruck-Bose correspondence and information about the construction. We follow on by examining Baer sublines and Baer subplanes in terms of the correspondence. We finish by describing the Bruck-Bose equivalent of derivation before presenting results on what a unital looks like in PG(4, q) using the correspondence.

2.1 Translation Planes

We will be working with a class of projective planes called translation planes. We introduce several deﬁnitions and then give the properties that deﬁne when a projective plane is a translation plane.

25 l1 . . . lq+1

m1 . .

mq+1

0 Figure 2.1: The regulus R = {l1, ..., lq+1} and its opposite regulus R = {m1, ..., mq+1}.

Deﬁnition 2.1.1 Let σ be a collineation of a projective plane P. Then,

1. A line l is ﬁxed pointwise by σ if and only if σ(P ) = P for all points P ∈ l.

2. A point P is ﬁxed linewise by σ if and only if σ(l) = l for all lines l through P .

Deﬁnition 2.1.2 Let σ be a collineation of a projective plane P that ﬁxes a line l pointwise and a point P linewise. Then σ is called a (P, l)-perspectivity.

Deﬁnition 2.1.3 Let P be a point and l a line of a projective plane P.

1. The plane P is (P, l)-transitive, if given any pair of points A, B of P such that

(a) A, B, P are collinear and distinct,

(b) A, B∈ / l,

then there exists a (P, l)-perspectivity σ with σ(A) = B.

2. If P is (X, l)-transitive for all points X on a line m, then the plane P is said to be (m, l)-transitive.

3. If P is (l, l)-transitive, then P is said to be a translation plane with respect to l. The line l is called a translation line.

In the projective plane PG(2, q), every line is a translation line.

26 2.2 Spreads

Next, we deﬁne reguli and spreads. These two structures will feature heavily in our work and appear regularly when we use the Bruck-Bose correspondence.

Deﬁnition 2.2.1 A regulus R in PG(3, q) is a set of q + 1 mutually skew lines such that any line which meets three of the lines of R meets all q + 1 of them.

Theorem 2.2.2 Any three mutually skew lines in PG(3, q) are contained in a unique regulus in PG(3, q).

0 For every regulus R = {l1, ..., lq+1}, there is a unique regulus R = {m1, ..., mq+1} that covers the same points as R, and each line of R meets each line of R0 in a point. The regulus R0 is called the opposite or reverse regulus of R and is shown in Figure 2.1. The points and lines of a regulus and its opposite regulus, together, form a hyperbolic quadric in PG(3, q).

Note that the opposite regulus of R0 is the original regulus R.

Deﬁnition 2.2.3 A spread S in PG(3, q) is a set of lines which partitions the points of PG(3, q). A spread S is regular, if given any three lines of S, the unique regulus containing these three lines is contained in S.

We may coordinatise a regular spread of PG(3, q) as follows (see Bruck [14]),

l∞ : = h(1, 0, 0, 0), (0, 1, 0, 0)i,

la0,a1 : = h(a0, αa1, 1, 0), (a1, a0, 0, 1)i, where a0, a1 ∈ GF (q) and α is a generator of GF (q).

Note that PG(3, q) contains q3 + q2 + q + 1 points, and each line of PG(3, q) contains q + 1 points. Hence a spread of PG(3, q) is a set of q2 + 1 mutually disjoint lines.

27 t T

l∞

Σ∞ P P

P G(4, q)

Figure 2.2: The Bruck-Bose correspondence

2.3 Construction

Now we give the construction for the representation of PG(2, q2) as an incidence structure in PG(4, q). This construction was ﬁrst presented by Bruck and Bose [15], [16] and Andr´e[1].

Let Σ∞ be a hyperplane of PG(4, q) and let S be a spread of Σ∞. We refer to a plane of PG(4, q) that is not contained in Σ∞ as a plane of PG(4, q)\Σ∞. Consider the following incidence structure A(S):

• the points of A(S) are the points of PG(4, q)\Σ∞,

• the lines of A(S) are the planes of PG(4, q)\Σ∞ that contain an element of S,

• the incidence in A(S) is induced by incidence in PG(4, q).

This construction is referred to as the Bruck-Bose construction or representation of A(S) in PG(4, q).

Theorem 2.3.1 [15] The incidence structure A(S) is an aﬃne plane of order q2.

28 The aﬃne plane A(S) can be uniquely completed to a projective plane P(S) by adding a line at inﬁnity, say l∞. Planes of PG(4, q)\Σ∞ through a line t of S correspond to lines of A(S) in the same parallel class. In the completion P(S), the lines in this parallel class meet in the point T of l∞. The point T corresponds to the line t of S, hence the points of l∞ are in one-to-one correspondence with the lines of the spread S. The correspondence for P(S) is pictured in Figure 2.2. It can be shown that the line 2 l∞ is a translation line of P(S) and hence P(S) is a translation plane of order q .

Theorem 2.3.2 [15] The spread S is regular if and only if P(S) is Desarguesian.

2.4 Baer subplanes and Baer sublines in Bruck- Bose

In this section we determine the Bruck-Bose representation of Baer subplanes and Baer sublines of P(S) in PG(4, q). Although the results are well known, we oﬀer proofs to most of the lemmas as the results are fundamental to many of our later arguments.

As before, let Σ∞ be a hyperplane of PG(4, q) and let S be a spread of Σ∞ such that P(S) ∼= PG(2, q2).

Theorem 2.4.1 A plane of PG(4, q) that doesn’t contain an element of S corresponds to a Baer subplane of P(S).

Proof Let α be a plane of PG(4, q) that does not contain a line of S. No plane of

Σ∞ can contain two lines of the spread, since any two lines in a plane intersect in a 2 point. There are q + 1 lines in the spread S and there are q + 1 planes of Σ∞ through 3 2 each line of S. Hence there are q + q + q + 1 planes of Σ∞ which contain a line of the spread and this is the total number of planes in Σ∞. Hence every plane of Σ∞ contains a unique line of S. Hence α∈ / Σ∞, since α does not contain a line of S.

In PG(4, q), a plane is either contained in a given hyperplane or meets the hyperplane in a line. Let π be a plane of PG(4, q) that does not contain an element of S. The

29 plane π meets Σ∞ in a line m. Since m is not a line of the spread, the line m meets q + 1 lines of S. We denote this set of q + 1 lines by R.

Now, using the Bruck-Bose correspondence, the plane α corresponds to a set of points 2 B in P(S). There are q points of α in PG(4, q)\Σ∞ and the plane α meets q + 1 lines of the spread S. Hence in P(S), the set B contains q2 + q + 1 points.

Deﬁne the lines of B to be the intersection of the lines of P(S) with the set B (when the intersection is greater than one point). We will now show that every line of P(S) meets B in 1 or q + 1 points and hence that every line of B contains q + 1 points.

Let l be a line of P(S). In PG(4, q), the line l corresponds to a plane αl of PG(4, q)\Σ∞ through a line of the spread S. We have two cases, either αl contains a line of R or αl contains no line of R.

Suppose αl contains a line of R, hence |αl ∩ α| ≥ 1. Two distinct planes in PG(4, q) meet in a line or a point, thus either |αl ∩ α| = 1 or |αl ∩ α| = q + 1. Hence in P(S) either |l ∩ B| = 1 or q + 1.

Suppose αl contains no line of R. Here, the line αl ∩Σ∞ is disjoint from the line α∩Σ∞ and hence together they span Σ∞. Now, the planes αl and α are not planes of Σ∞, thus dim(αl ⊕ α) = 4. Hence dim(αl ∩ α) = 0. Therefore |αl ∩ α| = 1 and thus in P(S) we have |l ∩ B| = 1.

We now verify the projective plane axioms inside the set B using the lines of B. This will prove that B is a subplane of order q and thus a Baer subplane of P(S).

Let P,Q be two points of B, then P,Q lie on a unique line l of P(S). Since |l ∩ B| ≥ 2 then |l ∩ B| = q + 1. Hence P,Q lie on a unique line of B.

In PG(4, q), the line PQ is a line of α which meets m = α ∩ Σ∞ in a point of a unique line of the spread S, say t. Now, P,Q and t span a plane of PG(4, q)\Σ∞ and this plane corresponds to a line of P(S). Therefore, lines of B correspond to lines of

PG(4, q)\Σ∞.

30 Let a, b be two lines of B. In PG(4, q) the lines a and b correspond to two lines of α and thus meet in a point X of α. If a, b 6= l∞, then X/∈ Σ∞ and X corresponds to a point X of B in P(S). If one of a, b is l∞, then X ∈ Σ∞. Hence X is contained in a unique line of the spread, which corresponds to a unique point on l∞. This point is also contained in both a and b. Thus a and b meet in a unique point of B.

Corollary 2.4.2 A line of PG(4, q) not contained in Σ∞ corresponds to a Baer subline of P(S) with a point on l∞.

Theorem 2.4.3 A regulus in a spread S corresponds to a Baer subline of l∞ in P(S).

Proof Let m be a line of the opposite regulus to a regulus in S. There exists a plane, say α, through m which contains no line of S. By Theorem 2.4.1, the plane α corresponds to a Baer subplane of P(S). Hence the line m corresponds to a Baer subline of l∞.

∼ 2 Theorem 2.4.4 A Baer subline of l∞ in P(S) = PG(2, q ) corresponds to a regulus of the regular spread S.

Proof From Theorem 2.4.3, a regulus corresponds to a Baer subline. Also, any three points of l∞ lie in a unique Baer subline of l∞ and any three lines of S lie in a q2+1 q+1 unique regulus of S. Therefore both sets have size 3 / 3 . Hence the reguli of S correspond exactly to the Baer sublines of l∞.

∼ 2 Theorem 2.4.5 A Baer subplane of P(S) = PG(2, q ) that meets l∞ in q + 1 points corresponds to a plane of PG(4, q) that contains no lines of the regular spread S.

31 Proof Denote the lines of Σ∞\S as the lines of Σ∞ excluding the lines of S. From

Lemma 2.4.1, a plane of PG(4, q) through a line of Σ∞\S corresponds to a Baer subplane of P(S). Hence if the number of such planes equals the number of Baer subplanes of P(S) that meet l∞ in q + 1 points then we have proved the result.

4 3 2 2 There are q + q + 2q + q + 1 lines in Σ∞ and q + 1 lines in the spread S. Hence there 4 3 2 2 are q + q + q + q lines of Σ∞\S. Through each line of Σ∞, there are q + q + 1 planes 2 4 3 2 of PG(4, q) but q + 1 of these are contained in Σ∞. Hence there are q (q + q + q + q) planes of PG(4, q) through a line of Σ∞\S.

2 2 Through a Baer subline of l∞ in PG(2, q ) there are (q + 1)q Baer subplanes. Baer 2 subplanes that contain diﬀerent Baer sublines of l∞ are distinct. There are q(q + 1) 2 2 3 3 2 Baer sublines on l∞. Hence there are q(q + 1)q (q + 1) = q (q + q + q + 1) Baer subplanes that contain a Baer subline of l∞. This is the same number as the number of planes of PG(4, q) through a line of Σ∞\S as required.

Hence for PG(2, q2), we have a direct correspondence between Baer subplanes of 2 PG(2, q ) and planes of PG(4, q) that do not meet Σ∞ in a line of the spread. We also 2 have a direct correspondence between Baer sublines of PG(2, q ) on l∞ and reguli of the spread.

Next we comment on those Baer subplanes and Baer sublines that meet l∞ in exactly one point.

Let t be a line of PG(4, q) and C be a conic contained in some plane of PG(4, q) that is disjoint from t. Consider a homography φ that maps the points of t to the points of

C. Let P1,...,Pq+1 be the points of t and hence φ(P1), . . . , φ(Pq+1) are the points of

C. Let mi, where i = 1, . . . , q + 1, be the line joining φ(Pi) to Pi. This construction is pictured in Figure 2.3.

All the points contained in the lines m1, . . . mq+1 together form a ruled cubic surface

V. The line t is called the line directix of V and the lines m1, . . . , mq+1 are called generators of V.

32 t

mi φ(Pi) Pi

Figure 2.3: Ruled cubic surface.

2 Theorem 2.4.6 [13][46][56] Let B be a Baer subplane in PG(2, q ) that meets l∞ at the unique point T . Then B corresponds to a ruled cubic surface V in PG(4, q).

In this case the ruled cubic surface V has line directix t where t is the line of the spread corresponding to the point T on l∞. We note that not every ruled cubic surface of PG(4, q) that contains a line of the spread corresponds to a Baer subplane of PG(2, q2).

2.5 Derivation in Bruck-Bose

Let Σ∞ be a hyperplane of PG(4, q) and let S be a spread of Σ∞. Let P(S) be the translation plane formed from S using the Bruck-Bose construction as before.

Suppose S contains a regulus R. Then we can construct a new spread S0 = (S\R) ∪ R0 by replacing the regulus R in S with its opposite regulus R0 as shown in Figure 2.4.

Consider the plane P(S0) that corresponds to S0. The regulus R corresponds to a Baer subline b of P(S) which is contained in l∞. Planes of PG(4, q)\Σ∞ that contain a line of R0 correspond to Baer subplanes of P(S) containing b. Hence P(S) and P(S0) have the same aﬃne pointset.

33 R R0

Figure 2.4: Replacing R with R0 in the spread.

The lines of P(S0) are of three types.

• Lines of P(S) that meet l∞ in a point not in b.

• The Baer subplanes containing b.

• The collection of points corresponding to the lines of the spread S0.

In PG(2, q2) all Baer sublines are derivation sets. Hence P(S0) is the plane obtained by deriving P(S) with respect to b. That is P(S0) = P(S)D, where D = b.

Replacing multiple disjoint reguli with their opposite regulus is equivalent to performing multiple derivations with the corresponding derivation sets.

2.6 Unitals in Bruck-Bose

In a later chapter we will be studying unitals using the Bruck-Bose correspondence. Here we give the known correspondences between unitals in translation planes and pointsets in PG(4, q). Let U be a unital in a projective plane and let U be the corresponding point set in PG(4, q).

Theorem 2.6.1 [19] Let U be a classical unital in P(S) ∼= PG(2, q2).

1. If U meets l∞ in the point P∞, then in PG(4, q), the point set U is an elliptic

cone that meets Σ∞ in the line p∞ of S corresponding to P∞. The vertex of U

lies on p∞.

34 P∞

Σ∞

U U

P G(2, q2) P G(4, q)

Figure 2.5: Bruck-Bose correspondence for unitals in P(S).

2. If U meets l∞ in q + 1 points of the Baer subline b, then in PG(4, q), the point set U is a parabolic quadric that meets S in the regulus corresponding to b.

Theorem 2.6.2 [19] Let S be a spread, not necessarily regular, in a hyperplane Σ∞ in PG(4, q).

1. If U is an ovoidal cone of PG(4, q) that meets Σ∞ in a line of S, then U corre-

sponds to a unital in P(S) which is tangent to l∞.

2. If U is a parabolic quadric in PG(4, q) that meets Σ∞ in a regulus of the spread

S, then U corresponds to a unital of P(S) which is secant to l∞.

These two constructions give the only known unitals in PG(2, q2).

A unital U arising from either of these constructions is called a Buekenhout unital. When the unital arises from the ovoidal cone construction, we call U an ovoidal- Buekenhout-Metz unital. If the ovoidal cone is an elliptic cone, we call the unital an orthogonal-Buekenhout-Metz unital. A unital arising from the parabolic quadric construction is called a non-singular-Buekenhout unital.

Theorem 2.6.3 [5] If U is a non-singular-Buekenhout unital in PG(2, q2) then the unital U is classical.

35 In Metz [40] it is shown that there exists orthogonal-Buekenhout-Metz unitals that are not classical. When q is even, there exists ovoids of PG(3, q) that are not elliptic quadrics. If we use such an ovoid to construct an ovoidal cone as in Theorem 2.6.2, the corresponding unital is also non-classical.

Barwick and Quinn [10] showed that, if U is a unital of PG(2, q2) that is ovoidal- Buekenhout-Metz with respect to two tangent lines, then U is classical. This together with Theorem 2.6.3 shows that a non-classical Buekenhout unital of PG(2, q2) is Buekenhout with respect to a unique tangent or secant line.

36 Chapter 3

Net Replacement

∼ Using the Bruck-Bose correspondence, a spread S of Σ∞ = PG(3, q) gives rise to a 2 translation plane P(S) of order q . Different spreads of Σ∞ can give rise to different translation planes. One method for forming new spreads is to replace a net in an existing spread. In this chapter we consider examples of replacing a net in a regular spread and the effect of the replacement on the corresponding projective plane.

Firstly, we define replaceable nets and investigate a desired regulus-based property. Next, we explore t-nests which are examples of a replaceable net. We prove several basic lemmas about t-nests and investigate an assortment of new geometric and combinatorial properties. We define half-regulus replaceable t-nests which are the only known examples of replaceable t-nests. Using the Bruck-Bose correspondence, we identify the corresponding net-derivation set in the plane to a regulus-based replaceable net in the spread. We conclude with examples of different types of net-derivation.

3.1 Replaceable Nets in PG(3, q)

Deﬁnition 3.1.1 Let S be a spread of PG(3, q).A net V in S is any subset of the lines of S. If we can ﬁnd a set V 0 of pairwise skew lines in PG(3, q) that cover the same set of points as the lines of V , then we call V 0 a replacement set for V .

37 An example of a net is a regulus R whose lines are contained in the spread S. The opposite regulus R0 to R is a replacement set for R.

Deﬁnition 3.1.2 Suppose V is a net with replacement set V 0 and no proper subset of V 0 is a replacement set for a proper subset of V , then we call V a replaceable net.

0 Let R1 and R2 be two reguli in the spread S with respective opposite reguli R1 and 0 0 R2. If V = R1, then V is a replaceable net with replacement set R1. If V = R1 ∪ R2 0 0 0 then V is a net and V = R1 ∪ R2 is replacement set for V . However, in this case V is 0 0 not a replaceable net since R1 and R1 are proper subsets of V and V respectively.

Let V be a replaceable net in the spread S and let V 0 be a replacement set for V . Using the Bruck-Bose correspondence, let P(S) be the projective plane given by the spread S. The replacement set V 0 can be used to construct a new spread. Suppose we replace the lines V by the lines V 0 in the spread S, then the lineset S0 = (S\V ) ∪ V 0 is a spread of Σ∞. Using the Bruck-Bose correspondence, we refer to the projective 0 0 0 0 plane corresponding to S as P(S ). We deﬁne l∞ to be the line at inﬁnity of P(S ).

Deﬁnition 3.1.3 Let S be a spread of PG(3, q). A net V in S is said to be regulus- based, if every line of V is contained in a regulus that is also contained in V .

The replaceable net V = R, where R is a regulus, is an example of a regulus-based net. We will see other examples of regulus-based nets in the next section. Next we prove a lemma that shows that in a regular spread, all replaceable nets are regulus-based. In Weida [57] an equivalent result is credited to Oakden’s Ph.D. thesis [41]. The thesis is not readily available and thus there does not appear to be any proof in print at this time.

∼ Lemma 3.1.4 Suppose S is a regular spread of Σ∞ = PG(3, q). Every replaceable net V in S is regulus-based.

38 Proof Suppose V 0 is a replacement set for V . Let l be a line of V 0 and let α be a plane of PG(4, q) not contained in Σ∞. Using the Bruck-Bose correspondence and Theorem 2.4.1, the plane α corresponds to a Baer subplane B of P(S). The line l meets q + 1 lines of V , hence in P(S), the Baer subplane B meets l∞ in q + 1 points. The points

B ∩ l∞ necessarily form a Baer subline, say b.

Since S is a regular spread, by Theorem 2.4.4, the Baer subline b corresponds to a 0 regulus Rb of S. Hence, the line l in V is contained in the opposite regulus to Rb and thus meets every line of Rb. Hence the lines of Rb are contained in V . There was no restriction on the choice of l, thus every line of V 0 is contained in the opposite regulus to some regulus contained in V . Hence every line of V is contained in a regulus contained in V .

There are two known examples of regulus-based replaceable nets. They are derivation as deﬁned in Section 2.5 and t-nest replacement. In the next section we will deﬁne a t-nest and investigate several geometric and combinatorial properties.

3.2 t-nests

In Baker and Ebert [2] a form of net replacement is deﬁned, that is nest replacement. Let S be a regular spread, thus P(S) ∼= PG(2, q2), and let q be odd with q ≥ 5.

Deﬁnition 3.2.1 A nest of reguli in the spread S is deﬁned to be a set N of reguli contained in S such that every line of S is contained in precisely 0 or 2 reguli of N. If N contains t reguli then N is called a t-nest.

A t-nest is a regulus-based net. The lines contained in the reguli of N will be referred to as the lines of N. For a survey of the existence of t-nests and various results, see Ebert [29].

39 Example 3.2.1 Let GF (5) = {0, 1, 2, 3, 4}. From Example 1.2.1 we know 2 is a generator for GF (5). We provide an explicit example of a 5-nest in a regular spread of PG(3, 5). This example of a t-nest comes from the class of q-nests defined in [2]. We have recalculated the elements using our field defined in Example 1.2.2. We refer to this family of q-nests as Nq and we will refer to this example as N5. We use the regular spread for PG(3, 5), as defined in Section 2.3, given by lines of the form,

l∞ : = h(1, 0, 0, 0), (0, 1, 0, 0)i,

la0,a1 : = h(a0, 2a1, 1, 0), (a1, a0, 0, 1)i, where a0, a1 ∈ GF (5).

Deﬁne the linesets R1,...,R5 as follows,    l = h(0, 0, 1, 0), (0, 0, 0, 1)i  l = h(0, 2, 1, 0), (1, 0, 0, 1)i  0,0  0,1    l4,3 = h(4, 1, 1, 0), (3, 4, 0, 1)i  l1,4 = h(1, 3, 1, 0), (4, 1, 0, 1)i     l0,2 = h(0, 4, 1, 0), (2, 0, 0, 1)i l4,0 = h(4, 0, 1, 0), (0, 4, 0, 1)i R1 : R2 :  l = h(1, 3, 1, 0), (4, 1, 0, 1)i  l = h(4, 3, 1, 0), (4, 4, 0, 1)i  1,4  4,4    l4,4 = h(4, 3, 1, 0), (4, 4, 0, 1)i  l0,3 = h(0, 1, 1, 0), (3, 0, 0, 1)i     l1,3 = h(1, 1, 1, 0), (3, 1, 0, 1)i, l1,0 = h(1, 0, 1, 0), (0, 1, 0, 1)i,

   l = h(1, 2, 1, 0), (1, 1, 0, 1)i  l = h(0, 0, 1, 0), (0, 0, 0, 1)i  1,1  0,0    l0,2 = h(0, 4, 1, 0), (2, 0, 0, 1)i  l1,1 = h(1, 2, 1, 0), (1, 1, 0, 1)i     l4,0 = h(4, 0, 1, 0), (0, 4, 0, 1)i l4,2 = h(4, 4, 1, 0), (2, 4, 0, 1)i R3 : R4 :  l = h(0, 3, 1, 0), (4, 0, 0, 1)i  l = h(1, 4, 1, 0), (2, 1, 0, 1)i  0,4  1,2    l4,1 = h(4, 2, 1, 0), (1, 4, 0, 1)i  l0,3 = h(0, 1, 1, 0), (3, 0, 0, 1)i     l1,0 = h(1, 0, 1, 0), (0, 1, 0, 1)i, l4,1 = h(4, 2, 1, 0), (1, 4, 0, 1)i,

  l = h(0, 2, 1, 0), (1, 0, 0, 1)i  0,1   l4,3 = h(4, 1, 1, 0), (3, 4, 0, 1)i   l4,2 = h(4, 4, 1, 0), (2, 4, 0, 1)i R5 :  l = h(0, 3, 1, 0), (4, 0, 0, 1)i  0,4   l1,2 = h(1, 4, 1, 0), (2, 1, 0, 1)i   l1,3 = h(1, 1, 1, 0), (3, 1, 0, 1)i.

40 Deﬁne the hyperbolic quadrics Q1,..., Q5 of PG(3, 5) as follows,

2 2 Q1 : x0 + 2x1 + 3x0x3 + 2x1x2 = 0, 2 2 2 2 Q2 : x0 + 2x1 + 4x2 + 3x3 + x0x3 + 4x1x2 = 0, 2 2 2 2 Q3 : x0 + 2x1 + 4x2 + 3x3 + 4x0x3 + x1x2 = 0, 2 2 Q4 : x0 + 2x1 + 2x0x3 + 3x1x2 = 0, 2 2 2 2 Q5 : x0 + 2x1 + 2x2 + 4x3 = 0.

For i = 1,..., 5, each set of mutually disjoint lines Ri is contained in the hyperbolic quadric Qi and thus forms a regulus of the regular spread in PG(3, 5). By inspection the lines contained in R1,...R5 are doubly covered by the reguli. Hence N5 = {R1,...,R5} is a 5-nest of the regular spread in PG(3, 5).

3.2.1 Basic Results

Here we prove three well known results about t-nests. The results themselves are not new, but the proofs are original. First, we present a result on how many lines of the spread are covered by the reguli of a t-nest.

q+1 Theorem 3.2.2 A t-nest N in a regular spread S covers t( 2 ) lines of S.

Proof Deﬁne the set,

X = {(R, l)| R is a regulus of N, l is a line of N contained in R}.

We count this set in two ways. Firstly, every regulus of N contains q + 1 lines thus |X| = t(q + 1). Secondly, the lines of N are doubly covered by reguli of N, hence each line is contained in two reguli of N. Thus |X| = 2 × (number of lines in N). Hence,

t(q + 1) = 2 × (number of lines in N).

q+1 Thus the number of lines in N is t( 2 ).

Next we present two results which give the upper and lower bounds on the possible size of a t-nest.

41 Theorem 3.2.3 Let N be a t-nest contained in a regular spread S, then t ≤ 2(q − 1).

2 q+1 Proof The spread S contains q +1 lines. By Theorem 3.2.2, the nest N covers t( 2 ) lines of S. Hence, q + 1 t ≤ q2 + 1 2 ≤ (q + 1)(q − 1) + 2, 4 t ≤ 2(q − 1) + . q + 1

4 Recall a t-nest is deﬁned for q ≥ 5 hence q+1 < 1. Thus t ≤ 2(q − 1) and we have our bound.

In Ebert [25], it is shown that there exists a class of 2(q−1)-nests, which gives sharpness to this bound.

q+3 q+3 Theorem 3.2.4 Let N be a t-nest, then t ≥ 2 . Further, if t = 2 , then every regulus of N meets every other regulus of N in exactly two lines.

Proof Consider the regulus R of N and recall that R contains q + 1 lines. The lines of R are doubly covered by the reguli of N. Since two reguli having three lines in common, necessarily coincide, any other regulus R0 of N can share no more than two lines with R. Thus, in order to doubly cover the lines of R, the t-nest N contains at q+1 q+1 q+3 least 2 reguli not including R. Hence there are at least 2 + 1 = 2 reguli of N.

q+3 Assume t = and let N = {R1,...,R q+3 }. From Theorem 3.2.2, the t-nest N 2 2 q+3 q+1 contains ( 2 )( 2 ) lines. We can count the number of lines in a diﬀerent way. A line l of N is double covered by two reguli Ri and Rj of N, where i 6= j. No other reguli of N can contain l hence each line of N can be regarded as the intersection between two reguli of N. Thus,

q+1 q+3 X2 X2 number of lines of N = |Ri ∩ Rj|. i=1 j>i

42 Equating the two values for the number of lines of N gives, q+1 q+3 q + 3 q + 1 X2 X2 = |R ∩ R |. (3.1) 2 2 i j i=1 j>i

There are q + 1 q + 1 ( q+1 )( q+3 ) + − 1 + ... + 1 = 2 2 (3.2) 2 2 2 such pairs (Ri,Rj), where i < j. No two reguli of N can share more than two lines thus |Ri ∩ Rj| ≤ 2, for i < j. Hence from (3.1) and (3.2), we have |Ri ∩ Rj| = 2, for i < j. Thus every regulus of N shares two lines with every other regulus of N.

q+3 A 2 -nest is also known as a Bruen chain. These were discovered by Bruen [17] and were the ﬁrst known examples of a t-nest. Currently, Bruen chains are known to exist for q = 5, 7, 9, 11, 13, 17, 19, 23, 25, 27, 31 (see [32] for example). Using computer searches, it has been shown that Bruen chains do not exist for q = 29, 37, 41, 43, 47, 49 (see [20] for example). The existence of Bruen chains for higher q is an open question.

3.2.2 Geometric Properties

In this section, we investigate how a regulus in a spread can meet the reguli of a t-nest contained in the same spread. The theorem presented appears in Barwick & Marshall [9]. Further, we obtain a previously known result as a corollary.

Theorem 3.2.5 Let N be a t-nest in a regular spread S and let R be a regulus in S. If R is not a regulus of N, then R contains at most t lines of N.

Proof We wish to compute the maximum number of lines in the intersection of R and the lines of N. Let R1,...,Rt be reguli of S and let N = {R1,R2,...,Rt}.

The process begins as follows, suppose N and R share at least one line l. Then l is covered by two reguli of N, say R1 and R2. The regulus R can share at most two lines with any regulus of N, otherwise R is a regulus of N. Hence R1 can contain at most one further line of R, similarly R2 can contain at most one further line of R. Thus there are four cases to be considered:

43 Case 1: Suppose R1 meets R in two lines l, m and R2 meets R in two lines l, n (m 6= n).

The reguli of N doubly cover m, hence there exists a regulus of N, say R3, that covers m. From here we have two possibilities:

Case 1a: The regulus R3 also covers n.

In this case, there are no more lines of R contained in R1,R2 or R3 and we have doubly covered three lines in R. We have t − 3 further reguli R4,...Rt of N and need to consider whether these meet R. Such a case, when t reguli of N are needed to doubly cover t lines of R, is called a maximal case. We now start the process again, from the beginning of the proof, using the reguli R4,...,Rt. We will do this every time there is a maximal case and there are reguli of N remaining.

Case 1b: There exists another regulus of N, say R4, that covers n. In this case we have four possibilities,

Case 1b(i): The reguli R3 and R4 contain no more lines of the intersection of R and N.

This case is not maximal, as we have con- l ∈ R1, R2 m ∈ R1,R3 sidered four reguli R1,R2,R3,R4 of N but n ∈ R2,R4 only three lines l, m, n of R.

Case 1b(ii): The reguli R3 and R4 both share one further line p ∈ R.

Then there are no more reguli of N sharing lines of R with R1, R2, R3 and R4, l ∈ R1, R2 m ∈ R ,R hence we have t−4 reguli of N remaining 1 3 n ∈ R2,R4 and have considered four lines l, m, n, p of p ∈ R3,R4 R. This is a maximal case. We start the process again, from the beginning of the proof, using the reguli R5,...,Rt.

44 Case 1b(iii): One of R3 and R4, say R3, has another line p in common with R but p∈ / R4.

In this case the line p needs to be doubly covered by reguli of N. Let p also be a line l ∈ R1,R2 m ∈ R1, R3 of the regulus R5, say. If R5 shares no fur- n ∈ R2,R4 ther line with R, then we have considered p ∈ R3 ﬁve reguli to cover four lines of intersec- l5 ∈ R3,R5 tion, which is not a maximal case. Hence ... assume R5 shares a further line with R, li ∈ Ri,Ri+1 say l5. We can continue this case, each ... time covering the new line li shared be- lt−1 ∈ Rt−1,Rt tween some Ri of N and R with a new regulus Ri+1 of N until we consider all the lines of R or all the reguli of N. In the former case, we have q + 1 lines of intersection. In the latter case, when i = t − 1, we have only one more regulus Rt of N to consider and Rt must doubly cover the extra line shared between Rt−1 and R. The regulus Rt can not share any more lines with R, otherwise we would not be able to doubly cover this extra line. Hence we have considered t reguli but only have t − 1 lines of intersection. Again, this is a not a maximal case.

Case 1b(iv): The regulus R3 has a line p in common with R and R4 has a line q in common with R, but p∈ / R4 and q∈ / R3.

In this case, we can repeat the process as q ∈ R4 in Case 1 with two new reguli R5 and R6. l ∈ R1,R2 It is possible to continue with this case m ∈ R1, R3 n ∈ R2,R4 considering Ri and Ri+1 each having dis- p ∈ R tinct lines in common with R. We ﬁnish 3 when either we run out of reguli in N or we reach a point when there is only one remaining regulus of N. In the former case we have considered more reguli of N than lines of R, hence we do not have a maximal case. In the latter case, the ﬁnal regulus

45 Rt of N must complete the double covering of the last two reguli Rt−2 and Rt−1 as in Case 1b(ii).

Case 2: The reguli R1 and R2 both meet R in the same two lines l, m.

Here R1 and R2 can contain no more lines of R. Hence we have used two lines of R and there are a further t − 2 reguli of N left to consider. This is a maximal case. We start the process again, from the beginning of the proof, with the reguli R3,...,Rt.

Case 3: The reguli R1 and R2 only meet R in the same unique line l. Here we have considered one line of R and have t − 2 reguli of N left to consider. This case is not maximal, since we have considered more reguli of N than lines of R.

Case 4: Only one of R1 and R2 meets R in a second line m. This case is not maximal.

We repeat the above process, assuming R and N share additional lines in common until we have considered all the reguli of N or exhausted all the lines of R. At this point we reach the maximum number of shared lines between R and N. From above, the maximal case is that we use one regulus of N for each line of intersection between R and N. Hence R contains at most t lines from the reguli of N.

As a corollary to this lemma, we have an alternative proof to a result that appears in Weida [57]. That is,

Corollary 3.2.6 Let N be a t-nest in the spread S. If t ≤ q, there exists no regulus contained in the lines of N, other than the reguli of N.

Proof Suppose t ≤ q. By Theorem 3.2.5, a regulus R of S that is not a regulus of N can contain at most q lines of N. Since R contains q + 1 lines, the regulus R can not be contained in the lines of N.

46 3.2.3 Combinatorial Properties

In this section we prove several new results on the combinatorial structure of t-nests. First, we deﬁne 1-connections and 2-connections in a t-nest which we will use in our proofs. We ﬁnd bounds on the number of 1-connections in a t-nest and we also consider sets of mutually disjoint reguli contained in t-nests.

Deﬁnition 3.2.7 Let N be a t-nest in a regular spread S. If two reguli of N share only one line, we call the line of intersection a 1-connection. If two reguli of N share two lines, we call the two lines together, a 2-connection. In this case we say two reguli share a 2-connection.

Thus every line of a t-nest N is either a 1-connection or contained in a 2-connection. Now we ﬁnd a relationship between the number of lines in a t-nest and the numbers of 1-connections and 2-connections.

Theorem 3.2.8 Let N be a t-nest in a regular spread S. Let M1 be the number of

1-connections in N and let M2 be the number of 2-connections in N. Then,

number of lines in N = M2 × 2 + M1.

Proof A 2-connection represents two reguli sharing two lines. These two lines are doubly covered by the two reguli, hence no other regulus of N can contain these lines. Thus each 2-connection corresponds to two unique lines in the nest N. Similarly, each 1-connection corresponds to a unique line in N. Since every line of N is contained in either a 1-connection or 2-connection we have our result.

Next we present two results that provide bounds on the number of 1-connections in a t-nest. We then give a small example that veriﬁes these bounds for N5 as deﬁned in Example 3.2.1.

Theorem 3.2.9 Let N be a t-nest that contains an odd number of lines. The number of 1-connections in N is at least three.

47 Proof Suppose the t-nest N contains one 1-connection. Let R be one of the two reguli of N that contain this 1-connection. The regulus R contains 2m + 1 lines, where m is the number of 2-connections in R. Since q is odd, every regulus contains an even number of lines. Now 2m + 1 is not even, hence there is a contradiction. By Theorem 3.2.8, the t-nest N contains an odd number of 1-connections. Hence N contains at least three 1-connections.

Theorem 3.2.10 Let N be a t-nest in a regular spread S.

t(2t−2−(q+1)) 1. If t ≤ q + 2, then N contains at most 2 1-connections.

t(q+1) 2. If t ≥ q + 2, then N contains at most 2 1-connections.

Proof Every regulus contains q + 1 lines, hence the most number of 1-connections in a regulus is q + 1. There are t reguli in N and every 1-connection is contained in two t(q+1) reguli of N, hence there can be at most 2 1-connections in N.

R R Let R be a regulus of N. Let M1 be the number of 1-connections in R and let M2 be the R R number of two connections in R. Since R contains q +1 lines, then M1 +2M2 = q +1. Also, since there are t reguli in N, the regulus R can share at most either a 1-connection R R or a 2-connection with the other t − 1 reguli of N. Hence M1 + M2 ≤ t − 1. Solving R for M1 gives, q + 1 − M R M R + 1 ≤ t − 1, 1 2 R R 2M1 + q + 1 − M1 ≤ 2(t − 1), R M1 ≤ 2(t − 1) − (q + 1).

There are t reguli in the nest N and each of these can contain at most 2(t − 1) − (q + 1) 1-connections. By deﬁnition, every 1-connection is contained in two reguli of N, hence t(2t−2−(q+1)) there are at most 2 1-connections in N.

t(q+1) For a given nest N the most number of 1-connections is the lowest of 2 and t(2t−2−(q+1)) 2 . If t ≤ q + 2 then, t(q + 1) t(2t − 2 − (q + 1)) ≥ . 2 2

48 If t ≥ q + 2, then t(q + 1) t(2t − 2 − (q + 1)) ≤ . 2 2

Example 3.2.2 Consider N5 as deﬁned in Example 3.2.1. There we have t = 5 and q = 5. The nest N5 contains q + 1 6 q = 5 ∗ = 15 2 2 lines. Since N5 contains an odd number of lines, by Theorem 3.2.9 and Theorem 3.2.8, the nest N5 contain an odd number of 1-connections and at least 3 1-connections. By

Theorem 3.2.10, the nest N5 has a maximum of

t(2t − 2 − (q + 1)) 5(10 − 2 − 6) = = 5 2 2

1-connections. Counting shows that N5 has 5 1-connections and 5 2-connections which satisﬁes the results.

Next, we use 1-connections and 2-connections to explore the possible size of sets of mutually disjoint reguli contained in t-nests.

Theorem 3.2.11 Let N be a t-nest in a regular spread S.

q+1 1. If N contains a set of k ≥ 2 mutually disjoint reguli, then t ≥ 2k.

q+1 ∗ ∗ ∗ 2. If N contains a set of k ≥ 2 mutually disjoint reguli N = {R1,...,Rk} and ∗ ∗ t = 2k, then each of the k reguli Ri ∈ N\N meets Rj in two lines, where ∗ j ∈ 1, . . . , k and i 6= j. Further, if Ri and Rj are both in N\N , then Ri meets

Rj in two lines, for i 6= j.

49 ∗ ∗ ∗ ∗ ∗ Proof Let R1,...,Rk be k mutually disjoint reguli of S and let N = {R1,...,Rk}. We wish to build a minimal structure by adding reguli to N ∗ until we form a nest N. We wish to limit the number of reguli we add to N ∗, hence assume each regulus of N ∗ q+1 ∗ contains 2 2-connections. To doubly cover a regulus of N , we need to add reguli to N ∗ to form these 2-connections.

∗ Suppose we can add a regulus R1 to N that shares a 2-connection with each of the ∗ ∗ ∗ reguli R1,...,R q+1 . Next, suppose we can then add a regulus R2 to N that shares a 2 ∗ ∗ q+1 2-connection with each of R2,...,R q+1 . If k = 2 then we wrap around the indices 2 +1 ∗ and the regulus R2 shares a 2-connection with R1 as well. In general, suppose we can ∗ add Ri for i = 1, . . . , k to N such that Ri shares a 2-connection with the following reguli: q − 1 R∗,...,R∗ , if i + ≤ k, or i i+ q−1 2 2 q − 1 R∗,...,R∗,R∗, if i + = k + 1, or i k 1 2 q − 1 R∗,...,R∗,R∗,...,R∗ , if i + > k + 1. i k 1 i+ q−1 −k 2 2 ∗ q+1 In general, the regulus Rj shares a 2-connection with the reguli R q−1 , ...,Rj. 2 j− 2 ∗ ∗ Let N = {R1,...,Rk} ∪ {R1,...,Rk}. By construction, every line of N is contained in a 2-connection. Hence every line of N is doubly covered and thus N is a t-nest.

We have added no new lines to N ∗ to construct N and have utilised the minimum number of connections with each added regulus, hence this structure is minimal. There are 2k reguli in N and we have our result.

As an immediate corollary, we have a bound on the maximum size of a set of mutually disjoint reguli contained in a t-nest.

Corollary 3.2.12 Let N be a t-nest in the regular spread S. The nest N can contain no set of more then k = q − 1 mutually disjoint reguli.

q+1 Proof By Theorem 3.2.11, if k ≥ 2 , then the minimum size for the t-nest N is t = 2k. By Theorem 3.2.3, we have t ≤ 2(q − 1). Hence 2k ≤ 2(q − 1) which implies k ≤ q − 1.

50 For the next theorem we will need some extra definitions and results. We assume the reader is familiar with the idea of a finite undirected graph on n vertices. All graphs we consider will be defined in this way.

Deﬁnition 3.2.13 Let G = (V,E) be a graph with vertex set V and edge set E. The graph G is called k-regular if every vertex is contained in k edges.

Theorem 3.2.14 [22] Let n and k be integers such that 0 ≤ k < n. Then there exists a k-regular graph with n vertices if and only if nk is even.

We also need the following lemma.

Lemma 3.2.15 Let k1, k1 + 1, . . . , k1 + (k1 − 1) be a set of consecutive integers. Any integer greater than or equal to k1 can be written as a sum of these integers.

Proof Let n be an integer such that n ≥ k1. Since n is an integer it can written as k + mk1 for some k < k1. Hence we can write n as k1 + t + (m − 1)k1 where t < k1.

Now we present our result.

Theorem 3.2.16 Let N be a t-nest in a regular spread S.

q+1 q+1 1. If N contains a set of k < 2 mutually disjoint reguli, then t ≥ k + 2 .

q+3 ∗ q+1 2. IF N contains a set of k ≤ 4 mutually disjoint reguli N and t = k + 2 , then q+1 ∗ the 2 reguli N\N pairwise intersect.

∗ ∗ ∗ ∗ ∗ Proof Let R1,...,Rk be k mutually disjoint reguli and let N = {R1,...,Rk}. As in the proof of Theorem 3.2.11, we wish to build a minimal structure by adding reguli to N ∗ until we form a nest N. Initially, to add the minimum number of reguli, we only use 2-connections.

51 ∗ Suppose we add a regulus R1 to N . The regulus R1 can have at most one 2-connection ∗ ∗ q+1 with each of the reguli R1,...,Rk. Since k < 2 , the regulus R1 will add at least (q + 1) − 2k extra lines to N ∗ and we will need ensure these lines are doubly covered in order to obtain a nest.

∗ Hence, suppose we can add the reguli R1,...,R q+1 to N , where each regulus Ri, for 2 q+1 ∗ ∗ i ∈ 1,..., 2 , shares a 2-connection with each of the reguli R1,...,Rk. Each regulus ∗ Ri will contain (q +1)−2k lines not covered by reguli of N . Note that, no two distinct ∗ reguli Ri and Rj, where i 6= j, can share the same 2-connection with a regulus of N , otherwise the lines in the 2-connection will be triply covered. Hence, the minimum ∗ ∗ ∗ q+1 number of reguli we can add to N to doubly cover the lines of {R1,...,Rk} is 2 .

∗ ∗ q+1 We wish to doubly cover the sets of lines Ri\{R1,...,Rk}, for i ∈ 1,..., 2 , by adding ∗ the minimum number of reguli to N ∪ {R1,...,R q+1 }. We will show that it may be 2 possible to do this without adding new reguli. Thus suppose that each of the (q+1)−2k ∗ ∗ q+1 lines in the set Ri\{R1,...,Rk}, for i ∈ 1,..., 2 , is also contained in some regulus

Rj, where i 6= j.

Here we consider two cases. Firstly, we try to ensure that the reguli R1,...,R q+1 2 pairwise intersect. We then consider the more general case with this condition relaxed.

Case 1: The regulus Ri shares at least one line with every regulus Rj where i 6= j.

Suppose that Ri has a 2-connection with m of the reguli in {R1,...,R q+1 }\{Ri}. Also, 2 suppose that Ri has a 1-connection with n of the reguli in {R1,...,R q+1 }\{Ri}. The 2 ∗ ∗ set of lines Ri\{R1,...,Rk} contains q+1−2k lines and each of these lines is contained in a unique Rj where i 6= j. Hence

2m + n = q + 1 − 2k. (3.3)

q+1 There are − 1 reguli in {R1,...,R q+1 }\{Ri}, hence this is possible if, 2 2 q + 1 m + n = − 1, 2 q − 1 = . (3.4) 2

52 Substituting (3.3) into (3.4) gives, q − 1 m + = q + 1 − 2k 2 q − 1 m = q + 1 − 2k − . (3.5) 2

Also from (3.4), q − 1 n = − m 2 q − 1 q − 1 = − q − 1 + 2k + by (3.5) 2 2 = −2 + 2k.

This situation is feasible if n ≥ 0 and m ≥ 0. That is, if k ≥ 1 and q − 1 q + 1 − 2k − ≥ 0, 2 q − 1 −2k ≥ −(q + 1) + , 2 q + 1 q − 1 k ≤ − 2 4 q + 3 ≤ . 4

q+1 Now, we wish to see if it is possible for all the Ri, where i ∈ 1,..., 2 to simultaneously share this property. That is, we require each Ri to share a 2-connection with m distinct

Rj, where i 6= j and share a 1-connection with n distinct Rk, where i 6= k and k 6= j, such that Ri shares at least one line with every regulus in {R1,...,R q+1 }\{Ri}. 2

To do this, we view the reguli and their connections as graphs. Let G2 = (V,E2) be a graph. Define the vertices V to be the reguli R1,...,R q+1 and let an edge e ∈ E2 be 2 defined between two vertices of V if there is a 2-connection between the corresponding reguli. We similarly define the graph G1 = (V,E1), with an edge e ∈ E1 defined between two vertices of V if there is a 1-connection between the corresponding reguli.

Hence, if each Ri simultaneously shares the above property, then every vertex of V in

G2 is connected to m other vertices and thus G2 is a m-regular graph. Similarly, every vertex of V in G1 is connected to n other vertices and thus G1 is a n-regular graph. Hence we wish to determine if such graphs can be constructed.

53 q+1 q+1 The graphs G2 and G1 each contain 2 vertices. If 2 is even, then |V | is even and q+1 q−1 by Theorem 3.2.14, the graphs G1 and G2 exist. Suppose 2 is odd and thus 2 is q−1 even. From above, m = q + 1 − 2k − ( 2 ) and n = −2 + 2k. Now, q + 1 is even and q−1 2k is even and since 2 is even then m is even. Also n is even. Hence, by Theorem

3.2.14, the graphs G1 and G2 can exist.

∗ ∗ q+1 q+3 Hence N = {R1,...,Rk} ∪ {R1,...,R q+1 } is a (k + )-nest deﬁned for k ≤ . The 2 2 4 nest N also possesses the property that the reguli {R1,...,R q+1 } pairwise intersect. 2

Case 2: Suppose the reguli {R1,...,R q+1 } do not necessarily pairwise intersect. 2

q+3 q+1 Assume 4 < k < 2 since the other values of k have been addressed in case one and Theorem 3.2.11.

q+1 Let {R1,...,Rl1 } be a proper subset of {R1,...,R }. Let Ri be a regulus of 2

{R1,...,Rl1 }, where i ∈ 1, . . . , l1. As above, suppose that Ri has a 2-connection with m1 of the reguli in {R1,...,Rl1 }\{Ri}. Also, suppose that Ri has a 1-connection with n1 of the reguli in {R1,...,Rl1 }\{Ri}. Thus Ri shares lines with m1 + n1 reguli of R1,...,Rl1 . Note that this implies m1 ≥ 0 and n1 ≥ 0.

Suppose we enforce that l1 = m1+n1+1. Hence Ri has a 1-connection or a 2-connection with every regulus of {R1,...,Rl1 }\Ri. Also, suppose we enforce that every line in the ∗ ∗ set Ri\{R1,...,Rk} is covered by a regulus of {R1,...,Rl1 }\Ri. Then, as above, we have

2m1 + n1 = q + 1 − 2k.

1 As in case one, we view the reguli and their connections as graphs. Let G2 = (V1,E2) be a graph. Define the vertices V1 to be the reguli R1,...,Rl1 and let an edge e ∈ E2 be defined between two vertices of V1, if there is a 2-connection between the corresponding 1 reguli. We similarly define the graph G1 = (V1,E1), with an edge e ∈ E1 defined between two vertices of V1, if there is a 1-connection between the corresponding reguli.

1 For each Ri ∈ {R1,...,Rl1 } to simultaneously share the above property, we need G2 to 1 be a m1-regular graph and G2 to be an n1-regular graph. Since q + 1 − 2k is even, then 1 1 n1 is even. The graphs G2 and G2 each contain m1 + n1 + 1 vertices. If m1 + n1 + 1 is

54 even, then m1 is odd. If m1 + n1 + 1 is odd, then m1 is even. Hence (m1 + n1 + 1)m1 1 1 is always even. Thus by Theorem 3.2.14, the graphs G1 and G2 can exist.

q+1−2k q+1−2k The number m1 + n1 + 1 can range from 2 + 1, where m1 = 2 and n1 = 0, to q + 1 − 2k + 1, where m1 = 0 and n1 = q + 1 − 2k. Notice that q + 1 − 2k q + 1 − 2k q + 1 − 2k + 1 = + 1 + + 1 − 1. (3.6) 2 2

q+1 q+1 Since {R1,...,Rl1 } is a proper subset of {R1,...,R }, then m1 + n1 + 1 < 2 2

q+1 and thus all the lines of {R1,...,R } are not doubly covered. Let {Rl1+1,...,Rl2 } 2

q+1 be a subset of {R1,...,R }\{R1, ...,Rl1 }. We repeat the same construction with 2

{Rl1+1,...,Rl2 } as with {R1,...,Rl1 }.

2 2 1 1 2 Let the graphs G1 and G2 be deﬁned similarly to G1 and G2, where G1 is a m2-regular 2 2 2 graph and G2 is a n2-regular graph but the vertices of G1 and G2 correspond to the 2 reguli Rl1+1,...,Rl2 . As above, we may choose m2 and n2 such that the graphs G1 and 2 G2 exist.

j We continue this process building the nj-regular graph G1 and the mj-regular graph j G2 until all the lines of {R1,...,R q+1 } are doubly-covered. For this construction to 2 exist, we need to choose the mj and nj in each set of reguli such that we doubly cover the lines of {R1,...,R q+1 } exactly. 2

That is we need

X q + 1 (m + n + 1) = . (3.7) j j 2 j

q+1−2k q+1−2k q+1−2k From (3.6), each mj + nj + 1 ranges between 2 + 1 and 2 + 1 + 2 + 1 − 1 and a valid mj and nj can be chosen for any number within this range. Thus, using

Lemma 3.2.15, we can satisfy (3.7) with valid choices for the mj and nj.

∗ ∗ q+1 q+3 q+1 Hence N = {R1,...,Rk} ∪ {R1,...,R q+1 } is a (k + )-nest, for < k < . 2 2 4 2

This lemma could be useful when constructing new t-nests. Here we see that for k = 1, q+3 the minimum size is t = 2 , verifying the earlier result in Theorem 3.2.4.

55 3.2.4 Replaceable t-nests

In this section, we examine replaceable t-nests with a particular type of replacement set. The internal structure of replaceable t-nests will be considered and we make several observations on extending a known result.

Deﬁnition 3.2.17 Let N be a t-nest in a regular spread S. Let V be the set of lines contained in the reguli of N. If we can ﬁnd a replacement set V 0 for V , then N is called replaceable.

Most known t-nests are replaceable, however in Ebert [26], there is an example of a (q + 2)-nest that is not replaceable.

Suppose S is a regular spread and N is a replaceable t-nest with replacement set V 0. From the spread S, we can form a new spread S0 = (S\V ) ∪ V 0 and hence, using the Bruck-Bose construction, a translation plane P(S0). In Weida [57], it was shown that P(S0) is always non-Desarguesian.

The replacement sets for all known replaceable t-nests have a certain property which we describe in the following deﬁnition.

0 Deﬁnition 3.2.18 Let N = {R1,...,Rt} be a t-nest. Let Ri denote the opposite regulus of Ri where i = 1 . . . , t. Then N is called a half-regulus replaceable t-nest, ˆ q+1 0 if for each regulus Ri, a set Ri of 2 lines from the opposite regulus Ri can be chosen ˆ ˆ such that the lines of Ri and Rj are mutually skew, for i 6= j. See Figure 3.1 for a diagram.

ˆ ˆ ˆ Suppose V is the set of lines contained in the reguli of N then N = R1 ∪ ... ∪ Rt is a ˆ replacement set for V . The linesets Ri will be referred to as half-reguli.

56 Ri

ˆ Ri

ˆ Rj

Figure 3.1: Deﬁnition of half-regulus replaceable.

Example 3.2.3 In [2] a replacement set for the family of nests Nq was found. Here we give the construction for N5 from Example 3.2.1.

Let T1 and T 2 be collineations of PG(3, 5) deﬁned by,

T1 :(x0, x1, x2, x3) → (x0 + x3, x1 + 2x2, x2, x3),

T 2 : (x0, x1, x2, x3) → (2x0 + x1, 2x0 + 2x1, 2x2 + x3, 2x2 + 2x3).

The collineation T1 maps the quadrics Qi to each other. For example, the quadric

T1(Q1) is given by,

2 2 (x0 − x3) + 2(x1 − 2x2) + 3(x0 − x3)x3 + 2(x1 − 2x2)x2 = 0

2 2 2 2 2 2 ⇒ x0 − 2x0x3 + x3 + 2x1 − 8x1x2 + 8x2 + 3x0x3 − 3x3 + 2x1x2 − 4x2 = 0 2 2 2 2 ⇒ x0 + 2x1 + 4x2 − 2x3 + x0x3 − 6x1x2 = 0 2 2 2 2 ⇒ x0 + 2x1 + 4x2 + 3x3 + x0x3 + 4x1x2 = 0, since we are working in GF (5).

This is the equation of the quadric Q2.

Now we consider how T 2 acts on lines of the spread. We have

T 2((a, 2b, 1, 0)) = (2a + 2b, 2a + 4b, 2, 2) = (2a, 4b, 2, 0) + 2(2b, 2a, 0, 2) = 2(a, 2b, 1, 0) + (b, a, 0, 1),

T 2((b, a, 0, 1)) = (2b + a, 2b + 2a, 1, 2) = (a, 2b, 1, 0) + 2(b, a, 0, 1).

57 R1 R1

l l

T2(l)

T1(l)

Figure 3.2: Action of T1 and T2.

Hence T 2(la,b) = la,b and thus T 2 ﬁxes every line of the regular spread S. Deﬁne T2 to 2 be T 2. The group of collineations generated by T2 has order 3 and is given by,

T2 :(x1, x2, x3, x4) → (x1 + 4x2, 3x1 + x2, x3 + 4x4, 3x3 + x4),

2 T2 :(x1, x2, x3, x4) → (3x1 + 3x2, x1 + 3x2, 3x3 + 3x4, x3 + 3x4).

Let l = hP1,P2i = h(0, 0, 1, 0), (2, 0, 0, 1)i. The line l meets l0,0 = h(0, 0, 1, 0), (0, 0, 0, 1)i and l0,2 = h(0, 4, 1, 0), (2, 0, 0, 1)i and also meets l4,3 = h(4, 1, 1, 0), (3, 4, 0, 1)i since (4, 1, 1, 0) + (3, 4, 0, 1) = (2, 0, 1, 1) = (0, 0, 1, 0) + (2, 0, 0, 1). Hence l meets three lines 0 of R1. Since R1 is a regulus, the line l is contained in the opposite regulus R1 to R1.

The collineation T1 applied i times to l acts as follows,

i T1(l) = h(0, 2i, 1, 0), (2 + i, 0, 0, 1)i, for i ∈ 1,... 5. The group generated by T1 has order 5. Since T2 ﬁxes the lines of the 0 spread, T2(l) is a line of R1.

ˆ We claim that the orbit of T = T1T2 on l is a replacement set N5 for N5. We need to i j i j show that T1(l) and T1 (l) are disjoint for i 6= j. Suppose T1(l) meets T1 (l), then

(2 + i, 0, 0, 1) + k1(1, 2i, 1, 0) = (2 + j, 0, 0, 1) + k2(1, 2j, 1, 0), for some k1, k2, ∈ GF (5).

This happens when k1 = k2 and hence 2i = 2j. Thus i = j.

58 i j We also need to show that T2(T (l)) and T (l) are disjoint for all i, j ∈ 1,..., 5. Suppose i j T2(T (l)) meets T (l), then

(8 + i, 3 + 2i, 1, 3) + k1(2 + i, 1 + 3i, 4, 1) = (2 + j, 0, 0, 1) + k2(1, 2j, 1, 0)), for some k1, k2, λ ∈ GF (5) and λ 6= 0.

This happens when,

3 + k1 = 1, (3.8)

1 + 4k1 = k2, (3.9)

8 + i + 2k1 + k1i = 2 + j + k2, (3.10)

3 + 2i + k1 + 3k1i = 2jk2. (3.11)

From (3.8) we have k1 = −2 = 3. From (3.9) we have k2 = 1 + 4(3) = 3. Also (3.10) gives us 1 + 8i + 2(3) + 3i = 2 + j + 3 and hence 2 + i = j. From (3.11), we get 3 + 2i + 3 + 3(3)i = 2j(3) thus 6 + 11i = 6j which is a contradiction.

The group T has order 15 hence we have constructed a set of 15 mutually disjoint lines that cover the same points as the lines of N5. It follows that the orbit of T on l is a replacement set for N5.

Next, we quote without a proof, a combined theorem due to Weida [57]. The theorem establishes conditions on when a replaceable t-nest is half-regulus replaceable and provides further information when t ≤ q.

Theorem 3.2.19 [57] Let N be a replaceable t-nest in a regular spread S with t ≤ q. Let V denote the lines of N and let V 0 be the replacement set for V . Let S0 = (S\V )∪V 0 denote the resulting spread of Σ∞. Then

0 q+1 1. V consists of 2 lines from the opposite regulus to each regulus of N,

2. V 0 contains no reguli,

3. V contains no reguli other than those of N, and

4. any regulus of S0 is contained in S\V if t < q − 1.

59 The theorem states that, if t ≤ q then any replaceable t-nest N is half-regulus replaceable and further that a set of opposite half-reguli is the only possible replacement set for N. It has been conjectured that all t-nests, where t ≤ q, are replaceable and thus, by Theorem 3.2.19, half-regulus replaceable.

There are no known examples of replaceable t-nests that are not half-regulus replaceable. However, for t ≥ q + 1, a replaceable t-nest N is not necessarily half-regulus replaceable. We give two examples of how such a case may occur.

Firstly, from Theorem 3.2.5, if t ≥ q + 1 the lines of N may contain a regulus R that is not a regulus of N. A line of the replacement set V 0 could potentially come from the opposite regulus to R.

Secondly, suppose that every line of the replacement set V 0 is contained in a opposite regulus to a regulus of N. The following theorem shows that, if t ≥ q + 1, then V 0 can contain less than half the lines of an opposite regulus to a regulus of N.

Theorem 3.2.20 Let N be a replaceable t-nest in the regular spread S. Suppose every line of the replacement set V 0 for N is contained in the opposite regulus to some regulus q+1 of N. Suppose there exists a regulus R in N with less then 2 lines from its opposite regulus in V 0. Then t ≥ q + 1.

q+1 Proof Suppose there are reguli R1, ...,R q+1 of N which share lines with R. Let 2 2 0 0 ˆ ˆ R1,...,R q+1 denote the corresponding opposite reguli and let R1,..., R q+1 denote the 2 2 corresponding sets of lines in the replacement set V 0.

q+1 0 q+1 The regulus R contains 2 − k lines of V , where 1 ≤ k < 2 . Hence, to cover all the 0 q+1 points contained in the lines of R, the replacement set V must contain 2 + k lines 0 0 from each opposite regulus R1,...,R q+1 . Suppose two of these reguli Ri and Rj share 2 ˆ ˆ q+1 a line l. This implies the lines of Ri and the lines of Rj all meet l. There are 2 + k ˆ ˆ 0 0 lines in Ri and Rj hence there are q + 1 + 2k lines of V that meet l. The lines of V are pairwise disjoint, however l contains q + 1 points, hence we have a contradiction. q+1 Thus Ri and Rj are disjoint, for 1 ≤ i < j ≤ 2 .

60 Hence, to cover the lines of the regulus R1 say, at a minimum we will need to add 0 to N = {R,R1,...,R q+1 } at least one regulus for each pair of lines of R1 other than 2 q−1 0 R1 ∩ R. Hence, at a minimum, we must add at least 2 more reguli to N . Thus the q+1 q−1 t-nest N contains at least 1 + 2 + 2 = q + 1 reguli.

This proof veriﬁes that if t ≤ q, then any replaceable t-nest N is half-regulus replaceable. We have also shown that for t ≥ q + 1, it is possible for a replaceable t-nest to not be half-regulus replaceable.

Now, we consider several replaceable t-nests and the reguli formed in the new spread after replacement. Note the fourth property in Theorem 3.2.19, that for t < q − 1, the resulting spread S0 contains no regulus that was not a regulus of S\V . If t ≥ q − 1, this is not always the case.

In Baker and Ebert [2], a replaceable q-nest Nq is deﬁned. In Example 3.2.1 we have deﬁned N5 from this family. It is shown in [2] that the lines of the replacement set 0 q+1 V for Nq can be partitioned into 2 sets of size q such that each set together with 0 q+1 l∞ forms a regulus. Note that for Nq, the line l∞ ∈/ V . Hence there are 2 reguli in S0 = (S\V ) ∪ V 0 that share lines between V 0 and S\V . Also, contained in the lines q−1 S\V there are 2 reguli disjoint from V .

In Ebert [27] the replaceable (q + 1)-nest Nq+1 is deﬁned. It was shown that the lines 0 of the replacement set V for Nq+1 can be partitioned in two diﬀerent ways as a union q+1 0 of 2 mutually disjoint reguli. Hence there are at least q + 1 reguli contained in V . q−3 It is also shown that S\V can be partitioned into two lines and 2 mutually disjoint reguli. It is conjectured that these reguli described are the only reguli of S0.

The two families of nests Nq and Nq+1 will become important later, when we look at inherited unitals in projective planes formed by t-nest replacement.

61 3.3 Net-Derivation

In this section, we wish to determine how replacing a net in a regular spread S aﬀects the corresponding plane P(S) ∼= PG(2, q2). Recall the Bruck-Bose correspondence for Baer sublines and subplanes from Section 2.4.

∼ 0 Let V be a replaceable net in the regular spread S of Σ∞ = PG(3, q). Let V be a replacement set for V . By Lemma 3.1.4, every replaceable net is regulus-based and every line of V 0 is contained in the opposite regulus to some regulus contained in V . Using the Bruck-Bose correspondence, every regulus of the regular spread S corresponds to a Baer subline of l∞. Let bV be the set of Baer sublines of P(S) on l∞ corresponding to the reguli of V that have a line of their opposite regulus in V 0.

Let l be a line of V 0. There are q2 planes of PG(4, q) containing l, not contained in 2 0 Σ∞. These planes correspond to q Baer subplanes in P(S). Suppose V contains M 2 lines, then there are q M Baer subplanes of P(S) containing a Baer subline of bV . We shall denote this set of Baer subplanes as BV = {B1,..., Bq2M }.

Suppose we replace V by the replacement set V 0 to form a new spread S0. The lines of P(S0) are of three types.

0 • If a line l of P(S) meets l∞ outside the set bV , then l is a line of P(S ).

• The Baer subplanes of BV .

0 • The line at inﬁnity l∞ which is the collection of points corresponding to the lines of the spread S0.

∼ In a projective plane P = P(S) we deﬁne a net-derivation set as N = (bV ,BV ) where bV and BV have the above correspondence. We say we perform a net-derivation with N when we replace the corresponding net V in the spread. We refer to the points contained in the Baer sublines of bV as points of bV . When we perform the net- derivation with N we refer to the new plane as PN = P(S0).

0 Let V be a replaceable net with replacement set V in a regular spread S of Σ∞. Let

N = (bV ,BV ) be the corresponding net-derivation set. We have shown the following:

62 X x l∞ b Σ∞

αl l

Figure 3.3: Bruck-Bose correspondence for Theorem 3.3.1.

• A point of bV corresponds to a line of V .

• A Baer subline of bV corresponds to a regulus contained in the lines of V .

• A Baer subplane of BV corresponds to a plane of PG(4, q) not contained in Σ∞ that contains a line of V 0.

The next theorem is fundamental to generalising results about derivation to results about net-derivation. The second theorem shows several properties of net-derivation sets that will be useful later.

Theorem 3.3.1 Let N = (bV ,BV ) be a net-derivation set on l∞ in the projective plane ∼ 2 P(S) = PG(2, q ). Let X be a point of a Baer subline of bV . Let l be any line of P(S) through X, where l 6= l∞. Then every Baer subline of l containing X lies in a unique

Baer subplane of BV .

∼ Proof Recall that V is a replaceable net in the regular spread S of Σ∞ = PG(3, q). Using the Bruck-Bose correspondence in PG(4, q), let the line l correspond to the plane

αl and the point X of l∞ correspond to the line x in the spread S. A Baer subline b of l containing X corresponds to a line b in the plane αl as shown in Figure 3.3.

The line b meets the line x in a point, say P . There exists a unique line m of V 0 that meets x in the point P . There is a unique plane containing m and b, hence there is a unique Baer subplane of BV containing b.

63 Theorem 3.3.2 Let N = (bV ,BV ) be a net-derivation set on l∞ in the projective plane ∼ 2 P(S) = PG(2, q ). Let l be a line of P(S) that meets l∞ at the point X, where l 6= l∞. Then, the following holds.

2 1. If b is a Baer subline of bV that contains X then there are at least q − q Baer

subplanes of BV containing b that meet l in precisely the point X.

3 2. In total, there are q − q Baer subplanes of BV that meet the line l in precisely the point X.

∼ Proof Recall that V is a replaceable net in the regular spread S of Σ∞ = PG(3, q).

Using the Bruck-Bose correspondence, the spread S corresponds to the line l∞ of P(S). Let V be the replaceable net and V 0 be the replacement set for V . Let x be the line of S corresponding to the point X. The line l of P(S) corresponds to a plane αl of

PG(4, q)\Σ∞ through the line x.

From Lemma 3.1.4, a replaceable net in a regular spread S is regulus-based. If X is 0 contained in a Baer subline of bV , then in PG(4, q), there exists q + 1 lines of V that meet x in a point. Let one such line be denoted by m and let m ∩ x be the point P .

Every plane through m, not contained in Σ∞, corresponds to a Baer subplane of BV .

2 There are q lines of αl that contain the point P , not including x. There are q planes containing m, not contained in Σ∞. Let r be a line of αl that contains P . There exists a unique plane which is the span of the line m and the line r and these planes are 2 distinct for each choice of r. Hence there are q − q planes containing m that meet αl in just the point P . These planes correspond to Baer subplanes of BV that meet l in only the point X.

Let b be a Baer subline of bV that contains X. Since S is regular, the Baer subline b corresponds to a regulus Rb of S. Any Baer subplane of BV containing b corresponds 0 to a plane of PG(4, q)\Σ∞ containing a line of the opposite regulus to Rb. Hence V contains at least one line from the opposite regulus to Rb. If we choose m to be one 2 such line, then we have shown that there are at least q − q Baer subplanes of BV that contain b and meet l in only the point X. This proves part one.

64 There are q + 1 choices for the line m of V 0. Hence there are (q + 1)(q2 − q) = q3 − q 3 planes of PG(4, q)\Σ∞ that meet l in only the point P . Hence there are q − q Baer subplanes of BV that meet l in only the point X.

In the following subsections, we provide examples of known net-derivation sets and information about other types of net-derivation.

3.3.1 Derivation

Derivation as deﬁned in Section 1.6 is a form of net-derivation. Let N = (bV ,BV ) be a net-derivation set for derivation set D. The set bV is just the single Baer subline D.

The Baer subplanes BV are all the Baer subplanes that contain D.

3.3.2 Nest-Derivation

Replaceable t-nests in a regular spread S from Section 3.2.4 have a corresponding net- derivation set in P(S). The net-derivation set N = (bV ,BV ) for a replaceable t-nest N with replacement set V 0 is constructed as follows.

Using the Bruck-Bose correspondence, the t reguli of the nest correspond to t Baer sublines on l∞. These Baer sublines are bV and every point contained in a Baer subline 2 of bV is doubly covered by Baer sublines of bV . There are q t planes of PG(4, q)\Σ∞ 0 that contain a line of V . These planes correspond to the Baer subplanes of BV .

We refer to a net-derivation set that corresponds to a t-nest replacement as a nest- derivation set.

3.3.3 Multiple Net-Derivation

It is possible to perform multiple net-derivations in the same plane. This can be done using distinct net-derivation sets that are disjoint on l∞. One example is multiple derivation as deﬁned in Section 1.6. We may also ”mix” diﬀerent types of net- derivations, which is possible due to the following results.

65 2 Lemma 3.3.3 Let P be a projective plane of order q . Let B1 and B2 be two Baer subplanes of P that meet l∞ in q + 1 points and further suppose that B1 and B2 are disjoint on l∞. Then B1 and B2 can meet in no more than one point.

Proof Suppose B1 and B2 intersect in two points P and Q of P\l∞. There exists a unique line l in P that contains P and Q. The line l meets l∞ in a unique point R.

This is a contradiction since B1 and B2 are disjoint on l∞. Hence B1 and B2 intersect in at most one point.

2 Theorem 3.3.4 Let P be a projective plane of order q and suppose N1 = (bV1 ,BV1 ) and N2 = (bV2 ,BV2 ) are two net-derivation sets on l∞. Suppose bV1 and bV2 are disjoint on l∞. Then every Baer subplane of BV1 meets every Baer subplane of BV2 in exactly one point.

Proof Let B1 be a Baer subplane of BV1 . Let P be a point contained in a Baer subline of bV2 . Let Q be a point of B1\l∞. Let R 6= P,Q be a point on the line PQ. There is a unique Baer subline of PQ containing the points R, P and Q. By Theorem 3.3.1, 2 there is a unique Baer subplane of BV2 containing such a Baer subline. There are q −1 choices for R and q − 1 of these will lie in the same Baer subline containing P and Q. Hence there are q + 1 unique Baer sublines on PQ containing P and Q.

Let M be the number of points contained in the Baer sublines bV2 . Hence there are M choices for the point P in bV2 . There are M(q + 1) unique Baer sublines containing Q and a point of bV2 . Let B2 be a Baer subplane of BV2 containing Q. There is a unique

Baer subline of B2 containing Q and each of the q + 1 points B2 ∩ l∞. Hence each Baer subplane of BV2 containing Q also contains q + 1 Baer sublines that contain Q and a point of bV2 . Hence there are M unique Baer subplanes of BV2 containing Q.

2 Now, there are q points contained in B1\l∞. By Lemma 3.3.3, a Baer subplane of BV2 can meet a Baer subplane of BV1 in at most one point. Hence, each point of B1\l∞ will 2 be contained in a distinct set of M Baer subplanes in BV2 . There are q M total Baer subplanes in BV2 hence every Baer subplane of BV2 meets B1 in exactly one point. This will be true for every Baer subplane in BV1 .

66 As a consequence, suppose we perform a net-derivation with two net-derivation sets

N1 = (bV1 ,BV1 ) and N2 = (bV2 ,BV2 ) that are disjoint on l∞. Suppose we treat the Baer

N1N2 subplanes of both BV1 and BV2 as lines in the new plane P . By Theorem 3.3.4, these lines intersect uniquely as required.

Example 3.3.1 Let N5 be deﬁned as in Example 3.2.1. Let the lineset G1 be deﬁned as,   l = h(1, 0, 0, 0), (0, 1, 0, 0)i  1,0   l3,2 = h(3, 4, 1, 0), (2, 3, 0, 1)i   l3,1 = h(3, 2, 1, 0), (1, 3, 0, 1)i G1 :  l = h(3, 3, 1, 0), (4, 3, 0, 1)i  3,4   l3,0 = h(3, 0, 1, 0), (0, 3, 0, 1)i   l3,3 = h(3, 1, 1, 0), (3, 3, 0, 1)i.

Let the hyperbolic quadric G1 be deﬁned as,

G1 : x0x1 + 2x0x3 + 2x1x2 = 0.

The lineset G1 is contained in the hyperbolic quadric G1 and thus forms a regulus. The lines of G1 are disjoint from the lines of N5 hence the net-derivation set corresponding to G1 and the net-derivation set corresponding to N5 together form a multiple net- derivation.

3.3.4 Net-Derivation on l 6= l∞

Suppose P ∼= PG(2, q), then in P every line is a translation line. Suppose we treat a line l of P, where l 6= l∞, as the line at inﬁnity. Using the Bruck-Bose correspondence, the line l corresponds to a regular spread. Hence we can construct net-derivation sets on any line l of P.

Theorem 3.3.1 and Theorem 3.3.2 both apply to net-derivation sets not on l∞. We restate them here in this context.

67 Theorem 3.3.5 Let N = (bV ,BV ) be a net-derivation set on a line l in the projective ∼ 2 plane P(S) = PG(2, q ). Let X be a point of a Baer subline of bV . Let m be any line of P(S) through X, where m 6= l. Then every Baer subline of m containing X lies in a unique Baer subplane of BV .

Lemma 3.3.6 Let N = (bV ,BV ) be a net-derivation set on a line l in the projective plane P(S) ∼= PG(2, q2). Let m be a line of P(S) that meets l at the point X, where m 6= l. Then, the following holds.

2 1. If b is a Baer subline of bV that contains X then there are at least q − q Baer

subplanes of BV containing b that meet m in precisely the point X.

3 2. In total, there are q − q Baer subplanes of BV that meet the line m in precisely the point X.

3.3.5 Net-Derivation in P0 6= PG(2, q2)

Let P0 be a projective plane of order q2 that is not isomorphic to PG(2, q2), with line 0 0 at inﬁnity l∞. It may be possible to ﬁnd a net-derivation set on the line l∞.

∼ 2 For example, let P = PG(2, q ) and let l∞ be the line at inﬁnity of P. Let N1 and N2 be two nest-derivation sets on l∞ and suppose N1 and N2 are disjoint on l∞. Suppose

N1 we perform a net-derivation with N1 to form the plane P . We know from Section 3.2.4 that PN1 is non-Desarguesian.

0 ∼ N1 0 Suppose P = P , then let l∞ be the line l∞ after net-derivation with N1. Since N1 0 0 and N2 are disjoint, the set N2 is a net-derivation set on l∞ in P .

In general we will not focus on this, however some of our later results on unitals could possibly be generalised to further include net-derivation and unitals in non- Desarguesian planes.

68 Chapter 4

Unitals and Net Replacement

In this chapter, we consider unitals in PG(2, q2) and whether or not their aﬃne point sets can be completed to unitals in projective planes formed by net replacement in PG(2, q2).

∼ Let S be a regular spread of Σ∞ = PG(3, q). We use the Bruck-Bose correspondence, thus the plane P(S) is PG(2, q2). Let V be a replaceable net in the spread S, with replacement set V 0. Suppose we perform a net replacement in the spread S using the replacement set V 0. We can form a new spread S0 = (S\V ) ∪ V 0. Suppose U is a unital of P(S). The question is, can the aﬃne points of U be completed to a unital in the new plane P(S0)?

In Section 4.1, we deﬁne the distinct cases for unitals and how they meet a net- derivation set. We also introduce notation that will be used throughout the chapter. In Section 4.2, we survey the known results to our question with regards to derivation.

Our original material for the chapter begins in Section 4.3. We prove a result in the t-nest setting and in Section 4.4, we prove a theorem for all the distinct cases when we replace a net and the unital is classical. These results generalise the known results for derivation.

From here, we use net-derivation sets instead of replaceable nets. In Section 4.5, we prove a new result for non-classical unitals that are Buekenhout with respect to l∞ and

69 where the net-derivation set is on a line l 6= l∞. This also gives a new result in the case of derivation.

In Section 4.6, we generalise two more theorems from the derivation case which characterise inherited unitals based on the number of Baer subplanes and Baer sublines that meet the unital.

In Section 4.7, we look at the inherited unitals and also those unitals which have not been inherited. We comment that for two particular t-nest replacements there are non- singular-Buekenhout unitals in the plane formed after the replacement, that were not inherited from PG(2, q2).

Finally, in Section 4.8, we consider O’Nan configurations in unitals. We generalise a result for non-classical Buekenhout-Metz unitals. We show that after net-derivation, if the unital contained O’Nan configurations in the original plane, then if the pointset is still a unital after net-derivation, the derived unital also contains O’Nan configurations.

4.1 Introduction

Recall the two unital constructions from Section 2.6 and that they are the only known unitals in PG(2, q2). Using the Bruck-Bose correspondence, we deﬁne them here again as follows.

If U is an ovoidal cone of PG(4, q) that meets Σ∞ in a line of S, then U corresponds to a unital U of P(S) which is tangent to l∞. These unitals are called ovoidal-Buekenhout- Metz unitals.

If U is a parabolic quadric of PG(4, q) that meets Σ∞ in a regulus of S, then U corresponds to a unital U of P(S) which is secant to l∞. These unitals are called non-singular-Buekenhout unitals.

Let V be a replaceable net in the spread S. There is a corresponding net-derivation set N = (bV ,BV ) on l∞ in P(S) (using the notation deﬁned in Section 3.3). Let PV denote the pointset of bV .

70 If we wish to observe the eﬀect of net-derivation with N on a unital U of P(S) then there are the following possibilities which we need to consider.

• The unital U is a non-singular-Buekenhout unital and thus U meets l∞ in q + 1 points. Then either,

– U ∩ l∞ is contained in PV ,

– U ∩ l∞ is partially contained in PV , that is 0 < |U ∩ PV | < q + 1,

– U ∩ l∞ is disjoint from PV .

• The unital U is a ovoidal-Buekenhout-Metz unital and thus U meets l∞ in one point. Then either,

– U ∩ l∞ is contained in PV ,

– U ∩ l∞ is disjoint from PV .

Suppose N is a net-derivation set on a line l 6= l∞. There are a similar set of possibilities for U with respect to l.

The following notation will be used throughout this chapter. Let U be a unital and let U be the aﬃne points of U. Let S be a spread of Σ∞. Using the Bruck-Bose correspondence, let U be the set of points in PG(4, q) corresponding to U. Let U be the pointset of PG(4, q) corresponding to the aﬃne points U.

Suppose we replace a replaceable net V by V 0 in S to form the spread S0 = S\V ∪ V 0. In P(S0), we refer to the aﬃne points U as U 0 and we refer to the points U of PG(4, q), that correspond to U, as U 0. We use U 0 to refer to the projective completion of U 0 in P(S0). We use U 0 to denote the pointset of PG(4, q) that corresponds to U 0.

4.2 Derivation

Suppose our replaceable net is a regulus R of S with replacement set the opposite regulus R0. Suppose we replace the regulus R to form the new spread S0 = (S\R) ∪ R0. In the plane P(S), this corresponds to performing a derivation with a derivation set D which forms the Hall plane P(S)D ∼= P(S0).

71 Thus suppose we perform a derivation in the plane P ∼= PG(2, q2) with a derivation set D on l∞. What eﬀect does this have on the unitals of P?

The ﬁrst author to look at this problem was Gr¨uning[31] who showed the following.

Theorem 4.2.1 [31] Let U be a non-singular-Buekenhout unital of PG(2, q2). Suppose 0 D U meets l∞ in precisely the points of D. Then U can be completed to a unital of P .

In later work, Barwick ([7],[8]) and Rinaldi ([48], [47]) independently addressed the other cases for derivation on l∞.

Theorem 4.2.2 [7][8][48][47]

1. If the unital U is a non-singular-Buekenhout unital in PG(2, q2), then

(a) if 1 ≤ |U ∩ D| < q + 1, then U 0 can not be completed to a unital of PD.

(b) if U ∩ D = ∅, then U 0 can be completed to a unital of PD.

2. If the unital U is a Buekenhout-Metz unital in PG(2, q2), then

(a) if |U ∩ D| = 1, then U 0 can not be completed to a unital of PD.

(b) if U ∩ D = ∅ then U 0 can be completed to a unital of PD.

Using two diﬀerent methods, Dover [24] and Barlotti and Lunardon [4] considered the case where we derive on a line l 6= l∞. They both showed the following theorem.

Theorem 4.2.3 [24][4] Suppose U is tangent to l∞ at the point T . Let l 6= l∞ be a line of P through T , that is a secant of U. Suppose D = {l ∩ U} and we derive with respect to D. Then U 0 can be completed to a unital of the plane PD.

The case where U is disjoint from D was also investigated by Blokhuis and O’Keefe

[12]. We state their results here which are valid for D on both l∞ and a line l 6= l∞.

72 Theorem 4.2.4 [12] Suppose U is a unital of P such that U ∩ D = ∅. Suppose each Baer subplane containing D meets U in at least 1 point and

1. if l is a secant of U, then at least (q + 1)2 Baer subplanes containing D meet U in exactly one point.

2. if l is tangent to U, then at least q(q + 1) Baer subplanes containing D meet U in exactly one point.

0 Then, if we perform a derivation using D, the pointset U is a unital in the plane PD.

Theorem 4.2.5 [12] If there is a unital U in P such that P ∩ D = ∅ and each Baer subplane containing D meets U in 1 or at least q + 1 points. Then, if we perform a 0 derivation with D, the pointset U is a unital in the plane PD.

In the following sections we will extend some of these results to the more general concept of net-derivation. We will also extend the case for net-derivation on a line l 6= l∞ from which we will obtain new insight into the derivation case.

4.3 t-nests

Recall the deﬁnition of t-nests in Section 3.2 and the deﬁnition of half-regulus replaceable t-nests from Section 3.2.4.

Suppose N = {R1,...,Rt} is a half-regulus replaceable t-nest in the regular spread S. ˆ ˆ ˆ Let N = {R1,..., Rt} be a replacement set for N. Recall that N is a set of t reguli ˆ that doubly cover the lines contained in N and each Ri is an opposite half-regulus to the regulus Ri for i = 1, . . . , t.

Suppose we form the new spread S0 = (S\N) ∪ Nˆ. We will answer the question of whether the aﬃne points of unitals in P(S) can be completed to unitals of P(S0). These results have been published in [9] by the author. Since writing this publication, we have generalised the results to the more general setting of net-derivation which are presented in Section 4.4.

73 Here, we present one lemma in the t-nest setting to give a ﬂavour for the general net-derivation setting.

2 Let U be a non-singular-Buekenhout unital in PG(2, q ) secant to l∞. Using the Bruck- Bose correspondence, the unital U corresponds to a parabolic quadric U. The quadric U meets the spread S in a regulus.

Lemma 4.3.1 If U meets S in a regulus of the t-nest N, then the pointset U 0 in P(S0) can not be completed to a unital in P(S0).

ˆ ˆ ˆ Proof Consider N = {R1,..., Rt}, the replacement set of opposite half-reguli to

N = {R1,...,Rt}. Suppose the quadric U meets S in the regulus Rt of N, then U q+1 ˆ contains the 2 lines of Rt. ˆ Let m be a line of Rt and let P be a point of U. The plane hP, mi meets U in the q + 1 points of m and the point P . A plane in PG(4, q) intersects a quadric in 1, q + 1 or 2q + 1 points. Hence hP, mi meets U in 2q + 1 points. Thus hP, mi meets the aﬃne points U in q collinear points.

Suppose we replace the nest N with Nˆ to form the new spread S0 = (S\N) ∩ Nˆ. The 0 0 plane hP, mi corresponds to a line mP in P(S ) and mP is a q-secant of U . Every line of P(S0) intersects a unital in 1 or q + 1 points. Hence, to complete U 0 to a unital of 0 0 0 P(S ), we need to add the point of inﬁnity mP ∩ l∞ to U . Hence, for each line m of ˆ 0 0 q+1 Rt, there is a corresponding point of l∞ that we need to add to U . There are 2 lines ˆ q+1 0 0 in Rt, hence we need to add 2 points of l∞ to U .

Suppose Rt shares lines with k other reguli of N, say {R1,...,Rk}. Let P be a point ˆ of U. Let l be a line of Ri, for some i ∈ 1, . . . , k, thus l is a line in an opposite half- regulus to a regulus of N that shares lines with Rt. Denote the particular half-regulus ˆ ˆ ˆ to which l belongs by Rl. Every line of a half-regulus Ri, where Ri shares a line with

Rt, intersects U in either one or two points. Hence, the line l intersects U in one or two points. Thus the plane hP, li meets U in at least two points, P and l ∩ U.

In P G(4, q), a plane meets a parabolic quadric in 1, q + 1 or 2q + 1 points, hence hP, li meets U in either q + 1 or 2q + 1 points. There are three cases we need to consider.

74 Case 1: Suppose |hP, li ∩ U| = 2q + 1.

Then the plane hP, li meets U in 2q or 2q − 1 points depending on whether l intersects

Rt in one or two points. If we replace the nest N, then hP, li corresponds to a line 0 0 0 lP of P(S ) which is either a (2q)- or (2q − 1)-secant of U . This means U cannot be completed to a unital in P(S0) as every line of P(S0) intersects a unital of P(S0) in 1 or q + 1 points.

Case 2: Suppose |hP, li ∩ U| = q + 1 and l intersects Rt in two points.

In this case, the plane hP, li meets U in q − 1 points. If we replace the nest N, the 0 0 plane hP, li corresponds to a line lP of P(S ) that is a (q − 1)-secant of U . Adding the 0 0 0 point lp ∩ l∞ to U does not make lP a (q + 1)-secant. Hence U can not be completed to a unital in P(S0).

Case 3: Suppose then that |hP, li ∩ U| = q + 1 and l intersects Rt in one point.

We also assume that this is the case for every line l contained in the reguli Ri, where i = 1, . . . , k, since we have considered the other possible cases above. That is, every regulus of N that contains a line of Rt, contains exactly one line of Rt. In this case there are q + 1 reguli of N, say R1,...,Rq+1 that share one line with Rt. The line l ˆ ˆ belongs to one of the corresponding half-reguli R1,..., Rq+1.

Now, the plane hP, li meets U in q points. If we replace the nest N then hP, li corre- 0 0 0 sponds to a line lP in P(S ) which is a q-secant of U . For U to be completed to a 0 unital in P(S ), the line lP needs to be completed to a (q + 1)-secant. To do this, we 0 0 must add the point of inﬁnity lp ∩l∞ to U and this point corresponds to the line l in the ˆ ˆ spread. This will be the case for every such line l contained in the reguli R1,..., Rq+1. ˆ ˆ Now, there are q + 1 half-reguli R1,..., Rq+1 whose lines intersect lines of Rt and each q+1 q+1 0 0 of these half-reguli contains 2 lines. Hence we add (q + 1) 2 points of l∞ to U to attempt to complete U 0 to a unital.

q+1 0 From above, we have already added at least 2 points of l∞, hence we add a total q+1 q+1 0 q+1 q+1 of 2 + (q + 1) 2 points. This means l∞ is a ( 2 + (q + 1) 2 )-secant. Since q+1 q+1 0 2 + (q + 1) 2 is greater than q + 1, the line l∞ is not a tangent or a (q + 1)-secant to U 0. Hence, the pointset U 0 can not be completed to a unital of P(S).

75 4.4 Net Replacement on l∞

Let V be a replaceable net in the regular spread S. Let V 0 be a replacement set for V . Using the Bruck-Bose correspondence, we can form the plane P(S) associated with the spread S. Since S is regular, P(S) ∼= PG(2, q2)

Suppose we replace V by V 0 to form a new spread S0 and thus the translation plane 0 P(S ). We proceed by looking at the ﬁve diﬀerent cases for U ∩ l∞ as stated in Section 4.1 and see if the inherited pointsets can be completed to unitals in P(S0).

Case I: non-singular-Buekenhout unitals

2 Let U be a non-singular-Buekenhout unital in PG(2, q ). That is, U is secant to l∞ and corresponds to a parabolic quadric U in the Bruck-Bose representation in PG(4, q) such that U meets Σ∞ in a regulus of the spread S.

The first lemma is equivalent to Theorem 4.2.1. We offer an alternative proof to that given in Grüning[31].

Lemma 4.4.1 [31] If U meets S in a regulus R = V , then U 0 can be completed to a unital of P(S0).

Proof Let R0 be the opposite regulus to R and thus the replacement set for V . The 0 quadric U meets Σ∞ in the hyperbolic quadric containing the lines of both R and R . If we replace R by R0, the quadric U is unchanged. Hence, after the replacement, the aﬃne points of U are contained in the same quadric U. Using the Bruck-Bose correspondence, the quadric U will correspond to a unital in P(S0).

We now prove each of the remaining cases as lemmas and summarise them in a subse- quent theorem.

Lemma 4.4.2 If U meets S in a regulus R and R is properly contained in the lines of the replaceable net V , then the aﬃne pointset U 0 can not be completed to a unital in P(S0).

76 Proof Let R0 be the opposite regulus to R. Let x be the number of lines of V 0 that 0 0 belong to R . We denote this set of lines by VR0 .

Suppose the replacement set V 0 contains all the lines of the opposite regulus R0. This implies that V \R is a replaceable net with V 0\R0 as a replacement set. Since V is a replaceable net, by deﬁnition, no proper subset of V can be a replaceable net and no proper subset of V 0 can be a replacement set for V . Hence x < q + 1.

0 Let m be a line of VR0 and let P be a point of U. Now, the plane hP, mi meets U in at least the q + 1 points of m and the point P . A plane in PG(4, q) intersects a quadric in 1, q + 1 or 2q + 1 points, hence hP, mi meets U in 2q + 1 points. We remove the points of m when looking at the aﬃne part of U, hence hP, mi meets U in q points,

Suppose we replace the nest V with V 0 to form the new spread S0. The plane hP, mi 0 0 corresponds to a line mP in P(S ) and the line mP is a q-secant of U . Now, every line of P(S0) intersects a unital in 1 or q + 1 points. Hence, to complete U 0 to a unital of 0 0 0 P(S ), we need to add the point of inﬁnity mP ∩ l∞ to U . This point corresponds to a 0 0 0 line of the spread S . Hence, for each line of VR0 , there is a corresponding point of l∞ 0 0 that we need to add to U . There are x lines in VR0 , hence we need to add x points of 0 0 l∞ to U .

0 0 Since there are x lines of R in VR0 , there are y = q + 1 − x points on each line of R 0 0 that are not covered by lines of VR0 . Any other line of V can intersect the lines of R 0 0 in at most two points, otherwise it will be a line of R . We denote, by VR, the lines of V 0 that meet R but are not contained in R0.

0 Let l be a line of VR, thus we have |l ∩ U| = 1 or 2. The plane hP, li meets U in at least two points, namely P and l ∩ U.

In P G(4, q), a plane meets a quadric in 1, q + 1 or 2q + 1 points, hence the plane hP, li meets U in either q + 1 or 2q + 1 points. From here, we have three cases to consider.

Case 1: Suppose |hP, li ∩ U| = 2q + 1.

Then hP, li meets U in 2q or 2q − 1 points depending on whether l intersects R in one or two points. If we replace the net V , then the plane hP, li corresponds to a line lP

77 of P(S0) which is either a (2q)- or (2q − 1)-secant of U 0. Since every line intersects a unital in 1 or q + 1 points, the pointset U 0 cannot be completed to a unital in P(S0) in this case.

Case 2: Suppose |hP, li ∩ U| = q + 1 and l intersects the lines of R in two points.

In this case, the plane hP, li meets U in q − 1 points. If we replace the net V , the 0 0 plane hP, li corresponds to a line lP of P(S ) that is a (q − 1)-secant of U . Adding the 0 0 0 point lP ∩ l∞ to U does not give us the required (q + 1)-secant. Hence U can not be completed to a unital of P(S0) in this case.

Case 3: Suppose |hP, li ∩ U| = q + 1 and l intersects the lines of R in one point.

0 We also assume that this is the case for every line l ∈ VR, since we have considered the other possible cases above.

Now, the plane hP, li meets U in q points. If we replace the net V then this plane 0 0 0 corresponds to a line lP in P(S ) which is a q-secant of U . For U to be completed to 0 a unital in P(S ), the line lP needs to be completed to a (q + 1)-secant. To do this, 0 0 we must add the point of inﬁnity lP ∩ l∞ to U . We will need to add such a point at 0 inﬁnity for every l ∈ VR. There are (q + 1)y such lines l, hence we add (q + 1)y points 0 0 0 of l∞ to U to attempt to complete U to a unital.

0 0 From above, we have already added at least x points of l∞ to U , hence we add at least 0 0 a total of x + (q + 1)y points. Thus l∞ is at least a (x + y + qy)-secant to U . Since 0 x + y = q + 1 and y > 0 then x + y + qy is greater than q + 1. Thus l∞ is not a tangent or a (q + 1)-secant to U 0. Hence, the pointset U 0 can not be completed to a unital of P(S0).

Lemma 4.4.3 If U meets S in a regulus R that shares lines with V but is not contained in V , then U 0 can not be completed to a unital of P(S0).

78 Proof Let S0 = (S\V ) ∪ V 0 be the spread formed by replacing V with V 0. Suppose U contains exactly r lines of V , where 1 ≤ r < q + 1. Through every point on a line of V , there is a line of V 0. There are q + 1 points on each line of V , thus U meets at least (q + 1) lines of V 0. Also, the quadric U meets exactly q + 1 − r lines of S0\V 0 since these lines are unchanged from those of S. Now, the quadric U meets at least (q + 1) + q + 1 − r = 2q + 2 − r lines of S0, which is greater than q + 1 since r < q + 1. By a similar argument as in the proof of Lemma 4.4.2, we must add too many points 0 0 0 0 of l∞ to U and hence U can not be completed to a unital of P(S ).

Lemma 4.4.4 If U meets S in a regulus R and R contains no lines of V , then U 0 can be completed to a unital in P(S0).

Proof Let PV denote the set of points in P(S) that correspond to the lines of V . The lines of R correspond to the points U ∩ l∞, in P(S). Since R contains no lines of V , in

P(S) the points U ∩ l∞ are disjoint from the points PV . If we replace the net V , then 0 0 0 the q +1 points of U ∩l∞ remain unchanged as q +1 points of l∞ in P(S ). In P(S ) we denote U 0 together with these q + 1 points as the set U 0. We denote the corresponding set in PG(4, q), under the Bruck-Bose representation, as U 0.

We want to show that U 0 is a unital of P(S0). The pointset U 0 contains q3 + 1 points hence we need to show that every line of P(S0) meets U 0 in either 1 or q + 1 points.

0 0 The line l∞ meets U in q + 1 points since l∞ meets U in q + 1 points.

0 0 Let l be a line of P(S ) that meets l∞ in the point P . The point P corresponds to a line p of the spread S. If p is not a line of V 0, then l is a line of P(S) and thus l contains 1 or q + 1 points of U. Hence l contains 1 or q + 1 points of U 0.

0 0 Suppose P does correspond to some line p of V , then P/∈ U since U ∩ PV is empty. Now, using the Bruck-Bose correspondence, let α be the plane of PG(4, q) correspond-

0 ing to the line l, thus α ∩ Σ∞ = p. Now, the plane α meets the quadric U in either a point, a line, a conic or two lines.

0 Suppose α ∩ U contains a line lα. Since α is a plane through p, the line lα meets p in a point. This implies p ∩ U 0 is not empty. Thus, in P(S0), the point P is a point of

79 U 0, which is a contradiction. Hence α meets U 0 in either a point or a conic. Thus, in P(S0), the line l meets U 0 in 1 or q + 1 points and the pointset U 0 is a unital of P(S0).

Case II: ovoidal-Buekenhout-Metz unitals

Let U be an ovoidal-Buekenhout-Metz unital of PG(2, q2). That is, U is tangent to l∞ and corresponds to an ovoidal cone U in the Bruck-Bose representation in PG(4, q) such that U meets Σ∞ in exactly one line p∞ of the spread S.

There are two possibilities for p∞, either p∞ is a line of V or it is not. The next two lemmas consider these two cases separately.

0 Lemma 4.4.5 If U meets S in one line p∞ and p∞ is a line of V , then U can not be completed to a unital of P(S0).

0 Proof Let l be a line of V such that l meets p∞ in one point. Let P be a point of U. In PG(4, q) a plane meets an ovoidal cone in 1, q + 1 or 2q + 1 points. Here, the plane hP, li meets U in at least two points, hence hP, li meets U in q + 1 or 2q + 1 points. We consider the two cases separately.

Case 1: Suppose hP, li meets U in 2q + 1 points.

In this case, the plane hP, li meets U in 2q points. If we replace the net V , then the 0 0 plane hP, li corresponds to a line lP of P(S ) which is a (2q)-secant to U . Hence in this case U 0 can not be completed to a unital in P(S0) as every line of P(S0) intersects a unital in 1 or q + 1 points.

Case 2: Suppose hP, li meets U in q + 1 points.

In this case, the plane hP, li meets U in q points. If we replace the net V , then hP, li 0 0 corresponds to a line lP that is a q-secant of U . Each line of P(S ) intersects a unital in either 1 or q + 1 points. Hence, if we wish to complete U 0 to a unital of P(S), we 0 0 need to add the point lP ∩ l∞ to U . This point corresponds to the line l of the spread S0. Thus suppose the plane hP, li meets U in q + 1 points for all choices of P .

80 The lines of V 0 are pairwise skew and cover every point of the lines of V . Thus there 0 are q + 1 lines of V that meet p∞ and hence there are q + 1 diﬀerent choices for l. 0 0 0 3 Thus we need to add q +1 points of l∞ to U . The aﬃne pointset U contains q points, 0 3 hence the extra q + 1 points from l∞ gives us q + q + 1 points. A unital contains only q3 + 1 points, hence we can not complete U 0 to a unital in P(S0).

0 Lemma 4.4.6 If U meets S in one line p∞ and p∞ is not a line of V , then U can be completed to a unital in P(S0).

Proof Using the Bruck-Bose representation, let PV denote the set of points in P(S) that correspond to the lines of V . The line p∞ corresponds to a point P∞ on l∞ in

P(S). Since p∞ ∈/ V , then P∞ ∈/ PV .

0 0 If we replace the net V , then the point P∞ remains unchanged as a point of l∞ in P(S ). 0 0 0 In P(S ), we denote U together with the point P∞ as U . We denote the corresponding set in PG(4, q) as U 0.

We wish to show that U 0 is a unital in P(S0). The pointset U 0 contains q2 + 1 points, hence we need to show that every line of P(S0) meets U 0 in either 1 or q + 1 points.

0 0 0 The line l∞ meets U in one point since l∞ meets U in one point.

0 0 Let l be a line of P(S ) that meets l∞ in the point P . The point P corresponds to a line p of the spread S0. If p is not a line of V 0, then l is a line of P(S) and thus l contains 1 or q + 1 points of U 0.

0 0 Now, suppose p is a line of V . Since P∞ ∈/ PV , then P/∈ U . Using the Bruck- Bose correspondence, let α be the plane of PG(4, q) corresponding to the line l, thus

0 α ∩ Σ∞ = p. The plane α meets the ovoidal cone U in either a point, a line, an oval 0 or two lines. Suppose α ∩ U contains a line lα. Since the plane α contains p, the line 0 0 lα meets p in a point. This implies that p ∩ U is not empty. Thus, in P(S ), the point P is a point of U 0, which is a contradiction. Hence, the plane α meets U 0 in either a point or an oval.

Thus, in P(S0), every line meets U 0 in either 1 or q + 1 points and hence the pointset U 0 is a unital in P(S0).

81 In summary, we have shown the following:

Theorem 4.4.7 Let U be a unital of P(S) ∼= PG(2, q2). Let S be a regular spread corresponding to P(S) via the Bruck-Bose representation. Let V be a replaceable net in S. Let PV be the point set on l∞ in P(S) corresponding to the lines of V . Suppose we replace the net V by V 0 to form the new spread S0 = (S\V ) ∪ V 0. Let U 0 be the points of P(S0) corresponding to the aﬃne points of U. Then we have,

1. if U is a non-singular-Buekenhout unital of P(S), then

(a) if |U ∩ PV | = q + 1, then

(i) if V is a regulus, the pointset U 0 can be completed to a unital in P(S0). (ii) otherwise, the pointset U 0 can not be completed to a unital in P(S0).

0 (b) if 1 ≤ |U ∩ PV | < q + 1 then the pointset U can not be completed to a unital in P(S0).

0 0 (c) if |U ∩ PV | = 0 then the pointset U can be completed to a unital in P(S ).

2. if U is an ovoidal-Buekenhout-Metz unital of P(S), then

0 (a) if U ∩ l∞ ∈ PV , then the pointset U can not be completed to a unital in P(S0).

0 0 (b) if U ∩ l∞ ∈/ PV then the pointset U can be completed to a unital in P(S ).

Since every classical unital of PG(2, q2) is a Buekenhout unital, we have described the case for all classical unitals with the net-derivation set on l∞. Suppose the net- 2 derivation set is on a line l 6= l∞. Since every line in PG(2, q ) is a translation line, a classical unital of PG(2, q2) will also be Buekenhout with respect to l. Hence we have also described the case for a classical unital and net-derivation on a line l 6= l∞.

We have described the case for all known non-classical unitals with the net-derivation set on l∞. However, for a non-classical unital and a net-derivation set on a line l 6= l∞ we do not have any new information. We address this case in Section 4.5.

82 Since Theorem 4.4.7 is valid for all replaceable nets in a regular spread S, the theorem includes the known results on derivation in PG(2, q2). Hence Theorem 4.4.7 generalises Theorem 4.2.2 and Theorem 4.2.1 from Section 4.3. Theorem 4.4.7 also generalises the case for t-nests as presented by the author in [9] and partially shown in Lemma 4.3.1.

Note that for replacing multiple disjoint replaceable nets, the theorem can be used by combining the cases for each replaceable net.

Example 4.4.1 Here we let our replaceable net be N5 as deﬁned in Example 3.2.1.

Let G1 be deﬁned as in Example 3.3.1. Let Q1 be deﬁned as in Example 3.2.1.

Recall G1 is a hyperbolic quadric in Σ∞. A parabolic quadric of PG(4, q) that contains

G1 is,

2 G1 : x0x1 + 2x0x3 + 2x1x2 + x4 = 0.

Since the lines of G1 are disjoint from the lines of N5, by Theorem 4.4.7, the parabolic 2 quadric G1 corresponds to a unital of PG(2, q ) whose aﬃne points can be completed to a unital of P(S0).

Recall Q1 is a hyperbolic quadric in Σ∞. A parabolic quadric of PG(4, q) which contains

Q1 is,

2 2 2 Q1 : x0 + 2x1 + 3x0x3 + 2x1x2 + x4 = 0.

Since Q1 contains a regulus of N5, by Theorem 4.4.7, the parabolic quadric Q1 corresponds to a unital of PG(2, q2) whose aﬃne points can not be completed to a unital of P(S0).

83 4.5 Net-Derivation not on l∞

A non-classical unital that is Buekenhout with respect to l∞ is not necessarily Bueken- hout with respect to any other secant or tangent line. Hence, if we perform a net- derivation on a line l 6= l∞, Theorem 4.4.7 does not necessarily apply.

As mentioned in Section 4.1, with regards to unitals and a net-derivation on a line l 6= l∞ we have similar cases as to those for l∞.

In Theorem 4.2.3 it was shown that if U is a non-classical ovoidal-Buekenhout-Metz unital and U ∩ l = D, then U 0 can be completed to a unital in PD. Here we oﬀer a new result for when we perform a net-derivation for a particular case as stated in the following theorem. We ﬁrst introduce some extra notation.

0 When we are performing a net-derivation on a line l 6= l∞, we refer to l as the new line in PN . We refer to U as the pointset U\l. Using the Bruck-Bose correspondence, the line l corresponds to a plane αl of PG(4, q)\Σ∞. Hence the pointset U corresponding 0 0 to U is U\αl. The pointsets U and U are deﬁned as before but in terms of U = U\l and U = U\αl.

Theorem 4.5.1 Suppose U is a non-classical ovoidal-Buekenhout-Metz unital in the ∼ 2 projective plane P = PG(2, q ). Suppose l∞ is tangent to U at the point T . Let l be a line of P through T but l 6= l∞. Let N = (bV ,BV ) be a net-derivation set on l such that T is a point contained in bV , the Baer subline l ∩ U is not a Baer subline of bV and either or both,

1. the number of points |l ∩ U| ≤ q − 1,

2. two Baer sublines of bV that both contain T , also share a second point of bV .

Then, if we perform a net-derivation in P with N , then U 0 can not be completed to a unital in the resulting plane PN .

84 ∼ Proof Let S be a regular spread of Σ∞ = PG(3, q). Using the Bruck-Bose correspondence, let t be the line of S that corresponds to the point T . The line l of P corresponds to a plane αl of PG(4, q)\Σ∞. The unital U corresponds to an ovoidal cone U whose vertex X lies on the line t. Let the lines of U through X be t, l1, . . . , lq2 .

2 2 The lines t and li span a plane ht, lii of PG(4, q)\Σ∞, for each i ∈ 1, . . . , q . There are q 2 planes of PG(4, q)\Σ∞ that contain t and there are q lines of U that are distinct from t. Hence, every plane of PG(4, q)\Σ∞ through t meets U in t and one further line li for 2 some i ∈ 1, . . . , q . The plane αl corresponds to one of these planes, say αl = ht, lki, 2 2 for some k ∈ 1, . . . , q . Hence, we have U = U\l in P G(2, q ) and U = U\ht, lki in PG(4, q).

Since U ∩ l is not a Baer subline of bV , then every Baer subline of bV containing T meets the pointset U ∩ l in either one or two points. Let b be one such Baer subline.

The Baer subline b corresponds to a line d of PG(4, q)\Σ∞ that contains some point of t. The line d is contained in the plane ht, lki as b is a Baer subline of l.

Let B be a Baer subplane containing b which contains only the point T of l∞. From Theorem 3.3.6, there are at least q2 − q such Baer subplanes. The Baer subplane B corresponds to a ruled cubic surface V with line directix t and line generators d, m1, m2, . . . , mq.

The lines m1, . . . , mq are lines of PG(4, q)\Σ∞, hence each of the lines m1, . . . , mq lies 2 in a plane ht, lii, for some i ∈ 1, . . . , q . Since the lines m1, . . . mq are mutually disjoint, each mj, for j ∈ 1, . . . , q, lies in a separate plane ht, lii.

There are four cases for how the line d meets the lines t and lk in PG(4, q). We consider each case and determine whether the pointset U 0 of PN can be a completed to a unital.

Case 1: The line d is the line lk.

In this case, the line d corresponds to the Baer subline b = U ∩l which is a contradiction as bV is deﬁned such that U ∩ l is not a Baer subline of bV .

Case 2: The line d is the line t.

This case is not possible as t is a line of Σ∞ and d is a line of PG(4, q)\Σ∞.

85 t t

d d l1 l1 . m. 1 . m1 . X . X . . mj li = mj . . . . lq2 . mq mq

lq2

Figure 4.1: Case 3a and Case 3b for Theorem 4.5.1.

Case 3: The line d does not contain X.

In this case, the line d meets lk in a point not on t, hence d meets U in two points.

Consider the ruled cubic surface V, then one of the lines m1, . . . , mq, say mj, contains

X. There are two cases for mj as depicted in Figure 4.1.

Case 3a: The line mj is not one of the li.

2 In this case the line mj lies in the plane ht, lii, for some i ∈ 1, . . . , q , but only meets the line li in the point X which is not a point of U. Each ms, where s ∈ 1, . . . , q and s 6= j will meet U in one point. The line d does not meet U as d is contained in the plane αl = ht, lki.

Hence V ∩ U contains a total of q − 1 points. If we perform a net-derivation with N then V corresponds to a (q −1)-secant to U 0. We can only add a maximum of one point to each of these secants from the line l0, hence these secants can not be completed to (q + 1)-secants. Thus U 0 can not be completed to unital in PN .

Case 3b: The line mj is one of the li 6= lk.

In this case, the pointset V ∩ U contains the q points of mj not including X. Each line 2 containing a point of t is contained in a plane ht, lsi, for some s ∈ 1, . . . , q , thus every line containing a point of t meets a line of U. Hence the q − 1 remaining mi, where i 6= j, also meet U in one point. The line d does not meet U as it is contained in the plane αl = ht, lki.

86 Hence V meets U in a total of 2q − 1 points. If we perform a net-derivation with N , then V corresponds to a (2q − 1)-secant to U 0 and hence U 0 can not be completed to a unital in PN .

Case 4: The line d contains X.

Suppose at least two Baer sublines b1 and b2 of bV that contain T also meet in another point P of bV . Using the Bruck-Bose correspondence, the Baer sublines b1 and b2 correspond to two lines d1 and d2 of PG(4, q), contained in the plane αl, that meet in a point P not on the line t. Suppose, without loss of generality, that d1 = d. Since d1 contains X, then d2 meets t in a point other than X as d1 and d2 meet in P . Hence we may apply case 3 above to the line d2.

Thus, suppose all the Baer sublines of bV that contain T are pairwise disjoint excluding the point T . The line d contains X but is not a line of U. Hence d meets U in only the point X. Each line mj meets a unique li and hence V contains q points of U. If we perform a net-derivation with N , then V corresponds to a q-secant to U 0 in PN .

3 By Theorem 3.3.6, there are q − q Baer subplanes of BV that meet l in just the point T . These Baer subplanes correspond to q-secants of U 0, hence there are at least q3 − q q-secants to U 0.

0 N The points U ∩ bV correspond to new points on l∞ in P . Let |U ∩ bV | = k, where 1 ≤ k ≤ q + 1. There can be at most q2 lines of PN , other than l0 that contain a new point of PN . Hence, we can complete up to kq2 of the q-secants to (q + 1)-secants.

Now, if 1 ≤ k ≤ q − 1, then

kq2 ≤ (q − 1)q2 = q3 − q2 < q3 − q.

Hence, if 1 ≤ k ≤ q − 1, then not all q-secants can be completed to (q + 1)-secants and thus U 0 can not be completed to a unital of PN .

Note that this lemma gives new information about the derivation case, as stated in the following corollary.

87 Corollary 4.5.2 Suppose U is a non-classical ovoidal-Buekenhout-Metz unital in the ∼ 2 projective plane P = PG(2, q ). Suppose U is tangent to l∞ at the point T . Let l be a line of P through T but l 6= l∞. Let D be a derivation set on l such that T is a point contained in D but D 6= l ∩ U. If we perform a derivation in P with D, then U 0 can not be completed to a unital in the resulting plane PD.

We also get new information with respect to bruen chains.

Corollary 4.5.3 Suppose U is a non-classical ovoidal-Buekenhout-Metz unital in the ∼ 2 projective plane P = PG(2, q ). Suppose U is tangent to l∞ at the point T . Let l be a line of P through T but l 6= l∞. Let N be a nest-derivation set corresponding to a q+3 2 -nest on l such that T is a point contained in N , but the Baer subline l ∩ U is not 0 in bV . Then, if we perform a net-derivation in P with N , the pointset U can not be completed to a unital in the resulting plane PN .

q+3 Proof From Theorem 3.2.4, for a 2 -nest N, every line of N is contained in two reguli of N and every regulus of N shares two lines with every other regulus of N.

From Section 3.3.2, two reguli of N correspond to two Baer sublines of bV in the corresponding net-derivation set N = (bV ,BV ). Hence, for N corresponding to a bruen chain, every Baer subline of bV meets every other Baer subline of bV in two points and each point of bV is contained in two Baer sublines of bV .

Thus, there exist two Baer sublines of bV that both contain the point T and the two

Baer sublines also share a second point of bV not equal to T . Hence Theorem 4.5.1 applies.

The remaining general cases for l 6= l∞ are an open question.

88 4.6 Net-Derivation on any line l

In this section we will construct unitals in projective planes formed by performing a net-derivation in translation planes of order q2. We determine when an inherited pointset is a unital based on the intersection of Baer subplanes with the unital in the original plane.

The ﬁrst result is a generalisation of Theorem 4.2.4. We use a similar technique as to that used by Blokhuis and O’Keefe [12] in their proof for the case of derivation.

2 Theorem 4.6.1 Let P be a translation plane of order q . Let N = (bV ,BV ) be a net- derivation set on a translation line l of P. Let M be the number of points contained in the Baer sublines of bV . Suppose U is a unital of P such that U ∩ bV = ∅. Suppose each Baer subplane of BV meets U in at least 1 point and

1. if l is a secant of U then at least (q + 1)M Baer subplanes of BV meet U in exactly one point.

2. if l is tangent to U then at least qM Baer subplanes of BV meet U in exactly one point.

Then, if we perform a net-derivation using N , the pointset U 0 can be completed to a unital in the plane PN .

2 2 Proof Let BV = {B1,..., Bs}, where s = Mq . Let ti = |Bi ∩ U|, where i = 1, . . . , q .

Since every Baer subplane of BV meets U in at least one point, then ti ≥ 1, for all i.

In both cases, our method will be to compute the variance of the diﬀerence between ti and q + 1 for all i. We use this to prove the possible numbers of points in the intersection between a Baer subplane of BV and U.

To perform the calculation of the variance we need two sums. First, we calculate Ps i=1 ti. This is equivalent to counting the number of ordered pairs (P, B) where Ps P ∈ U, B ∈ BV and P ∈ B. Second, we calculate the sum i=1 ti(ti − 1). This is

89 equivalent to counting the number of triples (P, Q, B) where P,Q ∈ U, B ∈ BV and P,Q ∈ B.

Case 1: Assume l is secant to U and at least (q + 1)M Baer subplanes of BV meet U in exactly one point.

First we count the number of ordered pairs (P, B) where P ∈ U, B ∈ BV and P ∈ B.

Since bV is disjoint from U, none of the points U ∩ l∞ are contained in a Baer subplane of BV . Hence, if P ∈ B ∩ U, then P ∈ U, where U is the aﬃne pointset of U.

There are q3 + 1 points in U. Since U is secant to l then |U ∩ l| = q + 1. Hence U contains q3 − q points.

Let P be a point of U. Using the Bruck-Bose correspondence for l, let V be the replaceable net with replacement set V 0 that corresponds to the net-derivation set N . 0 A line of V corresponds to a Baer subline b of bV . A point P of P\l corresponds to a point P of PG(4, q)\Σ∞.

0 Let m be a line of V . There is a unique plane hP, mi of PG(4, q)\Σ∞ that contains m and P and the plane hP, mi corresponds to a Baer subplane of BV . There are M 0 3 lines of V , hence there are M Baer subplanes of BV that contain P . There are q − q choices for P , thus there are M(q3 − q) ordered pairs (P, B) where P ∈ U, P ∈ B and

B ∈ BV . Hence, Xs 3 ti = (q − q)M. i=1

Let P and Q be two points of P. Next we count the number of ordered triples (P, Q, B) where P,Q ∈ U, B ∈ BV and P,Q ∈ B. Since bV is disjoint from U, if the points P,Q ∈ B ∩ U, then P,Q ∈ U.

Note that, if P,Q ∈ U, then the line PQ is a secant to U and thus meets U in q + 1 points.

2 Let X be a point of bV . By Lemma 1.5.4, there are q − q secants to U that contain X. Since l is secant to U, there are q2 − q − 1 secants to the pointset U that contain X. On each secant there are (q + 1)q choices for an ordered pair of points that are both contained in U. Let P,Q be one such pair of points.

90 The points P,Q together with the point X deﬁne a unique Baer subline containing P,Q and X. By Theorem 3.3.1, this Baer subline is contained in a unique Baer subplane of

BV . Hence P and Q are contained in the same Baer subplane of BV .

Thus for any choice of points P,Q ∈ U that lie on a secant to U containing X, the points P and Q are also contained in a Baer subplane of BV . Hence for a given point 2 X in bV there are (q − q − 1)(q + 1)q distinct triples (P, Q, B) where P,Q ∈ U, B ∈ BV and P,Q ∈ B. There are M choices for X, thus Xs 2 ti(ti − 1) = M(q − q − 1)(q + 1)q. i=1

Now, we calculate the variance of the diﬀerence between the ti and q + 1, Xs Xs 2 2 2 (ti − (q + 1)) = (ti − 2(q + 1)ti + (q + 1) ) i=1 i=1 Xs Xs Xs 2 2 = (ti − ti) − (2q + 1) ti + (q + 1) i=1 i=1 i=1 = M(q2 − q − 1)(q + 1)q − (2q + 1)(q3 − q)M + q2M(q + 1)2 = M((q2 − q − 1)(q2 + q) − (2q4 − 2q2 + q3 − q) + q2(q2 + 2q + 1)) = M(q4 + q3 − q3 − q2 − q2 − q − 2q4 + 2q2 − q3 + q + q4 + 2q3 + q2) = M(q3 + q2) = M(q + 1)q2.

There are at least (q + 1)M Baer subplanes in BV that meet U in exactly one point and thus ti = 1 for the corresponding i. Each of these Baer subplanes contributes (1 − (q + 1))2 = q2 to the variance and, in total, for all such Baer subplanes we add at least q2(q + 1)M to the variance. The total variance is M(q + 1)q2, hence there are exactly (q + 1)M Baer subplanes in BV that meet U in exactly one point. Thus the remaining Baer subplanes contribute zero to the variance of the diﬀerence between the ti and q + 1. Hence these remaining Baer subplanes intersect U in exactly q + 1 points.

Hence every Baer subplane of BV meets U in 1 or q + 1 points. When we perform a net-derivation with N , every new line of PN is either a tangent or a secant to U 0. 0 N 0 Since U is disjoint from bV , we may add the points U ∩ l to U in P to complete U to a unital of PN .

91 Case 2: Assume l is tangent to U and at least qM Baer subplanes of BV meet U in exactly one point.

We proceed as in Case 1. First we count the number of ordered pairs (P, B) where

P ∈ U, B ∈ BV and P ∈ B.

There are q3 +1 points in U. Since U is tangent to l then |U ∩l| = 1. Hence U contains q3 points. By a similar argument to that in part (1), we have

Xs = q3M. i=1

Then we count the number of ordered triples (P, Q, B) where P,Q ∈ U, B ∈ BV and P,Q ∈ B.

Since l is tangent to U, there are q2 − q secants to U through a point of l not contained in U. Hence, by a similar argument to that in Case 1, we have Xs 2 ti(1 − ti) = M(q − q)(q + 1)q. i=1

Again, we wish to calculate the variance of the diﬀerence between the ti and q + 1, Xs Xs 2 2 2 (ti − (q + 1)) = (ti − 2(q + 1)ti + (q + 1) ) i=1 i=1 Xs Xs Xs 2 2 = (ti − ti) − (2q + 1) ti + (q + 1) i=1 i=1 i=1 = M(q2 − q)(q + 1)q − (2q + 1)q3M + q2M(q + 1)2 = M((q2 − q)(q2 + q) − (2q4 + q3) + q2(q2 + 2q + 1)) = M(q4 + q3 − q3 − q2 − 2q4 − q3 + q4 + 2q3 + q2) = Mq3.

There are at least qM Baer subplanes in BV with ti = 1. Each of these Baer subplanes contributes (1 − (q + 1))2 = q2 to the variance and hence, in total, they contribute at least q3M to the variance. The total variance is q3M, hence there are exactly qM

Baer subplanes in BV that meets U in exactly one point. Thus the remaining Baer

92 subplanes contribute zero to the variance of the diﬀerence between the ti and q + 1. Hence these remaining Baer subplanes intersect U in exactly q + 1 points.

The proof ﬁnishes using a similar argument as to that in Case 1.

The second result is a generalisation of Theorem 4.2.5 to the case of net-derivation. The proof follows similarly to the proof for the derivation case as presented in [12]. First, we need several deﬁnitions and a theorem.

Deﬁnition 4.6.2 A blocking set in a projective plane P is a set of points which meet every line of P. A blocking set B is said to be minimal, if B\{P } is not a blocking set for any point P ∈ B.A tangent of a blocking set B is a line which meets B in exactly one point; thus a blocking set is minimal if and only if each of its points lie on a tangent.

Theorem 4.6.3 [18] Let B be a minimal blocking set in a projective plane P of order √ q. Then |B| ≤ q q + 1 with equality if and only if B is a unital.

Now we present the result.

2 Theorem 4.6.4 Let P be a translation plane of order q . Let N = (bV ,BV ) be a net- derivation set on a line l in P. Suppose there is a unital U in P such that U ∩ bV = ∅ and each Baer subplane of BV meets U in 1 or at least q + 1 points. Then, if we perform a net-derivation using N , the pointset U 0 can be completed to a unital in the plane PN .

Proof Let U = U\l. Suppose we perform a net-derivation with N in P to form the N 0 N plane P with line at inﬁnity l . The points l ∩ U are unchanged in P as U ∩ bV = ∅. 0 0 0 Let Ul be the set of points on l that correspond to the points l ∩ U and thus |Ul | = 1 0 0 0 0 3 N or q + 1. Let U = U ∪ Ul , then U is a set of q + 1 points in P .

N Now, the lines of P are either lines of P or Baer subplanes of BV . The lines of P 0 0 meet U in 1 or q + 1 points and the Baer subplanes of BV meet U in 1 or at least

93 q + 1 points. Hence, every line of PN meets U 0 in 1 or at least q + 1 points and thus U 0 is a blocking set.

Suppose there exists a point P ∈ U 0 such that P lies on no tangent of U 0. Hence, every line of PN that contains P meets U 0 in at least q + 1 points. There are q2 + 1 lines containing P , and no two lines containing P can share a further point of U 0, thus |U 0| ≥ (q2 + 1)(q + 1), which is a contradiction. Hence, all points of U 0 lie on a tangent of U 0 and thus U 0 is minimal.

Hence U 0 is a minimal blocking set of size q3 +1, thus from Theorem 4.6.3, the pointset U 0 is a unital of PN .

4.7 Inherited unitals

∼ Let S be a spread of Σ∞ = PG(3, q), where Σ∞ is a subspace of PG(4, q). Using the Bruck-Bose correspondence, let P(S) be the projective plane corresponding to S.

Let l be a line of the spread S and choose a point P on l. Let Σ be a 3-dimensional subspace of PG(4, q) containing P but not containing l. We know Σ exists due to

Example 1.3.1. The subspace Σ meets Σ∞ in a plane α.

Let E be an elliptic quadric through P contained in Σ with tangent plane α at P. If we choose a point Q 6= P on l, then we can form an elliptic cone of PG(4, q) with base Σ and vertex Q. There are many choices for Σ and E hence there are many such elliptic cones through a line of the spread. Such an elliptic cone corresponds to a 2 orthogonal-Buekenhout-Metz unital of PG(2, q ) that is tangent to l∞.

From Example 1.7.1, a regulus of the spread S (together with its opposite regulus) is contained in many parabolic quadrics of PG(4, q). These quadrics corresponds to non-singular-Buekenhout unitals that meet l∞ in q + 1 points.

Let V be a replaceable net in S with replacement set V 0. Suppose we form the translation plane P(S0) using the spread S0 = (S\V ) ∪ V 0. Let U 0 be a unital of P(S0). If the aﬃne points of U 0 are the aﬃne points of a unital in P(S) then U 0 is said to be inherited from P(S). Otherwise we refer to U 0 as non-inherited.

94 4.7.1 Derivation

Suppose we perform a derivation by replacing the regulus R in S with its opposite regulus R0 to form a new spread S0 = (S\R) ∪ R0.

Every line of S0\R0 is unchanged, hence by Theorem 4.4.7, every elliptic cone that 0 0 0 meets Σ∞ in a line of S \R corresponds to a unital of P(S ) that is inherited from 0 P(S). Any line l of R is also contained in many elliptic cones that meet Σ∞ in l and these elliptic cones correspond to unitals of P(S0). From Theorem 4.4.7, these elliptic cones do not correspond to unitals of P(S) as they contain a line of R0. Hence these unitals in P(S0) are non-inherited.

Every regulus of S0\R0 is unchanged and these reguli are contained in many parabolic quadrics. These quadrics correspond to unitals of P(S0) that are inherited from unitals of P(S).

In Barwick [8], it was shown that there are no reguli of S0 other than R0 and the reguli of S0\R0. Hence there do not exist non-singular-Buekenhout unitals of P(S0) that are not inherited from non-singular-Buekenhout unitals in P(S).

4.7.2 t-nests

Suppose we replace a nest N with a replacement set Nˆ in the spread S to form the new spread S0 = (S\N) ∪ Nˆ.

Every line of S0\Nˆ is unchanged, hence by Theorem 4.4.7 there are many inherited orthogonal-Buekenhout-Metz unitals in P(S0). Also, there are many non-inherited orthogonal-Buekenhout-Metz unitals through the points of P(S0) corresponding to the lines of Nˆ.

Every regulus of S0\Nˆ is unchanged and these reguli are contained in many parabolic quadrics. These quadrics correspond to unitals of P(S0) that are inherited from unitals of P(S).

95 For t < q, by Theorem 3.2.19, the spread S0 contains no reguli other than those in S0\Nˆ, hence there are no non-inherited non-singular-Buekenhout unitals in P(S0) in this case.

Suppose the nest is either Nq or Nq+1 as deﬁned in Section 3.2.4. In either case, there exist reguli of S0 that were not reguli of the original spread. Hence in P(S0) there will exist many non-inherited non-singular-Buekenhout unitals.

This construction of non-inherited non-singular-Buekenhout unitals in planes formed by t-nest replacement is a new discovery. We ﬁnish this section by giving an example of the construction.

Example 4.7.1 Take our example of N5 from Example 3.2.1 and Example 3.2.3. ∗ 0 Let T be the group generated by T1. It is shown in [2] that for a line l in R1, ∗ then the line-set H1 consisting of the orbits of the group T acting on l and the line h(1, 0, 0, 0), (0, 1, 0, 0)i is a regulus contained in the new spread S0. Thus in our example we have,    h(0, 0, 1, 0), (2, 0, 0, 1)i    h(0, 2, 1, 0), (3, 0, 0, 1)i   h(0, 4, 1, 0), (4, 0, 0, 1)i H : 1   h(0, 1, 1, 0), (0, 0, 0, 1)i    h(0, 3, 1, 0), (1, 0, 0, 1)i   h(1, 0, 0, 0), (0, 1, 0, 0)i.

The lines of H1 are contained in the hyperbolic quadric H1 of PG(3, q) where,

H1 : x0x2 + 2x1x3 + 3x2x3 = 0.

Hence the quadric H1 of P G(4, q) given by,

2 x0x2 + 2x1x3 + 3x2x3 + x4 = 0,

0 0 meets Σ∞ in a regulus of the spread S and thus corresponds to a unital of P(S ) that is not inherited from a unital of P(S).

96 Y

B X A

C D

Figure 4.2: Example of an O’Nan conﬁguration.

4.8 O’Nan conﬁgurations

Here we examine the existence of O’Nan configurations contained in unitals. We consider unitals in P ∼= PG(2, q2) as well as those in the plane PN formed by net-derivation in P with a net-derivation set N . We first define an O’Nan configuration.

Deﬁnition 4.8.1 Let {A, B, C, D} be a quadrangle. An O’Nan conﬁguration is the four lines and six points {A, B, C, D, X, Y }, as given in Figure 4.2, where the point X = AB ∩ CD and the point Y = AD ∩ BC.

It is well known that this conﬁguration cannot exist in the classical unital.

Theorem 4.8.2 [44] A classical unital contains no O’Nan conﬁgurations.

Not much is known about the O’Nan conﬁgurations contained in other unitals of PG(2, q2), however the following is known.

Lemma 4.8.3 [6] If U is a Buekenhout-Metz unital with respect to a point P , then U contains no O’Nan conﬁgurations that contain P .

97 4.8.1 Inherited unitals in PN

∼ 2 Let U be a unital of P = PG(2, q ) and let N = (bV ,BV ) be a net-derivation set on l∞ of P. Suppose we perform a net-derivation with N , such that the pointset U 0 can be completed to a unital U 0 of PN . We may ask, does U 0 contain O’Nan conﬁgurations?

For the derivation case, Barwick [7] proved the following.

Theorem 4.8.4 [7] Let U be a classical unital of P ∼= PG(2, q2) and let D be a derivation set on the line at inﬁnity l∞ of P. Suppose U is disjoint from D. Then, if q > 5, the unital U 0 of PD contains O’Nan conﬁgurations. In particular, if H is a

0 0 0 point of U \l∞ and l a secant of U through H that meets D, then there is an O’Nan conﬁguration of U 0 that contains H and l.

It is an open question as to whether Theorem 4.8.4 can be generalised to the case of net-derivation.

Now, let us suppose U is a non-classical unital of PG(2, q2). Barwick [7] partially addressed this case for derivation, and here we generalise that result for net-derivation.

Lemma 4.8.5 Let U be a non-classical orthogonal-Buekenhout-Metz unital in P ∼= 2 PG(2, q ). Let N = (bV ,BV ) be a net-derivation set on l∞ that is disjoint from U. √ Suppose the Baer sublines of bV contain M points of l∞ and that q > 2 M. Suppose we perform a net-derivation with the net-derivation set N , then if U contains an O’Nan conﬁguration, then so does U 0 in PN .

Proof Let U be a non-classical Buekenhout-Metz unital with respect to the point T of P and suppose that U contains on O’Nan conﬁguration. By Lemma 4.8.3, the unital U cannot contain an O’Nan conﬁguration with T as a vertex, thus let the lines of the

O’Nan conﬁguration meet l∞\T in the points A, B, C, D. In [3], [28] it is shown that the group of collineations ﬁxing an orthogonal Buekenhout-Metz unital is transitive on the points of l∞\T .

98 Hence, if there exists one O’Nan configuration in U, then through each point of l∞\T there is a line contained in an O’Nan configuration of U. Hence U contains at least q2 4 O’Nan configurations. We want to prove that one of these O’Nan configurations is disjoint from bV . This will be true if there are more O’Nan configurations than there q2 are points of bV . That is, if 4 > M. √ Thus, if q > 2 M, the unital U contains an O’Nan configuration disjoint from bV . The lines of such a configuration are unchanged by net-derivation with respect to N and hence they are also secant lines of U 0 in PN . Thus we have constructed an O’Nan configuration in U 0.

Lemma 4.8.5 remains true for derivation as for N a derivation set. Since in that case √ M = q + 1 and q > 2 q + 1, for q ≥ 5.

99 100 Chapter 5

Conics and Net-Derivation

In this chapter, we observe the effect of net-derivation on non-degenerate conics of PG(2, q2). We seek to determine whether the affine point sets of such conics are arcs and further can be completed to ovals in planes formed by performing a net-derivation. That is, suppose C is a non-degenerate conic of P ∼= PG(2, q2) and suppose N is a N net-derivation set on l∞ in P. Are the affine points of C an arc in P and further can this affine pointset be completed to an oval of PN ?

In Section 5.1 we deﬁne notation and present a new corollary on nest-derivation and conics. In Section 5.2 we present several basic theorems on derivation and conics and introduce the real derivation set DR. We then give in Section 5.3, in three stages, a novel characterisation of the equations of conics that are not arcs after derivation with the derivation set DR.

2 Next, in Section 5.4, we divide the derivation sets on l∞ and the conics of PG(2, q ) into projectively equivalent cases. We then provide a brief survey of the previously known results for inherited arcs in the Hall planes of odd order from conics of PG(2, q2), where q is odd.

In Section 5.5, we restrict our attention to conics contained in the family Cc,d. We give an alternative proof to calculating a known homography group and prove a theorem on conics and derivation with sets other than DR.

101 Following this, in Section 5.6, we prove several theorems on the existence of inherited (q2 + 1)-arcs in a class of planes formed by double derivation in PG(2, q2), where q is odd. We also calculate examples to verify existence. We ﬁnish the section by computing an example of a complete 24-arc in a particular translation plane of order 25.

Finally, in Section 5.7, we show the existence of a family of inherited arcs in a class of Andréplanes which includes the Regular Nearfield planes of odd order. These arcs can be completed to (q2 + 1)-arcs in all cases. We conjecture that in the Regular Nearfield planes of odd order, this family of inherited arcs are examples of the Rosati Ovals, a previously discovered family of ovals.

5.1 Introduction

In this chapter, all conics are assumed to be non-degenerate.

We use the following notation, let C be a non-degenerate conic of P ∼= PG(2, q2) and let C denote the aﬃne points of C. Let N = (bV ,BV ) be a net-derivation set on l∞ in P. We use C0 to denote the pointset C when viewed as a pointset of PN .

The following lemma is used extensively throughout this chapter and will be assumed.

∼ 2 Lemma 5.1.1 Let C be a conic in P = PG(2, q ). Let N = (bV ,BV ) be a net- derivation set on l∞ in P and let B be a Baer subplane of BV . Suppose C shares at 0 least three points with the aﬃne pointset B\l∞ then C does not form an arc in the plane PN .

Proof If we perform a net-derivation with N , then B is a line in PN . At least three N points of C are contained in B\l∞ and hence lie on a line of P . Therefore the pointset C0 does not form an arc in PN .

In the next subsection we will present a small result on conics and nest-derivation. However, for the rest of this chapter, we will focus exclusively on derivation and multiple derivation as deﬁned in Section 1.6.

102 5.1.1 Nest-derivation

Let q be odd. Suppose l∞ is tangent to a conic C, then we call the aﬃne points of C a parabola.

Let Σ = Σ(q), be the family of all aﬃne planes of order q. Let A be any plane belonging to Σ. We can identify the points of A with those of AG(2, q), and regard the lines of A as pointsets in AG(2, q).

Korchmaros [37] determined the behaviour of parabolas with respect to derivation in planes of odd order with the following theorem.

Theorem 5.1.2 [37] Let q be odd. Let C be a parabola in AG(2, q2). If C is an arc in an aﬃne translation plane A, then A coincides with AG(2, q2).

We apply the theorem as follows.

∼ 2 Corollary 5.1.3 Let N = (bV ,BV ) be a derivation set on l∞ in P = PG(2, q ), where q is odd. Let C be a conic of P such that l∞ is a tangent of C. If we perform a derivation with N , then C0 will not be an arc in PN .

Proof The plane PN is the Hall plane and every Hall plane is non-Desarguesian. N N 0 2 Hence A = P \l∞ is non-Desarguesian and thus does not coincide with AG(2, q ). 0 Since l∞ is tangent to C in P, by Theorem 5.1.2, the aﬃne pointset C is not an arc in the plane PN .

For t-nests we can say something similar. Recall a nest-derivation set as deﬁned in Section 3.3.

∼ 2 Corollary 5.1.4 Let N = (bV ,BV ) be a nest-derivation set on l∞ in P = PG(2, q ), where q is odd. Let C be a conic of P such that l∞ is a tangent of C. If we perform a nest-derivation with N , then C0 will not be an arc in PN .

103 Proof From Section 3.2.4, for a nest-derivation set N , the plane PN is always non- N N 0 Desarguesian. Hence A = P \l∞ is non-Desarguesian and thus does not coincide 2 0 with AG(2, q ). Since l∞ is tangent to C in P, by Theorem 5.1.2, the aﬃne pointset C is not an arc in the plane PN .

5.2 Derivation

For the rest of this chapter, we will adopt the notation for derivation from Section 1.6 as opposed to using the more general form of net-derivation.

∼ 2 Suppose D is a derivation set on l∞ in P = PG(2, q ). Let C be a non-degenerate conic in P. In this section, we present several known theorems about derivation and conics.

We then explore the real derivation set DR and present several properties we will make use of in later sections.

5.2.1 Basic theorems

The ﬁrst result in this section shows that, for any conic C in P we can always ﬁnd a 0 D derivation set on l∞ such that the pointset C is not an arc in the plane P .

Theorem 5.2.1 Suppose P ∼= PG(2, q2), where q > 3. For any conic C in P, there exists a derivation set D on l∞ such that when a derivation is performed using D, the pointset C0 is not an arc in PD.

Proof Since q > 3, the aﬃne pointset C of the conic C contains at least three points.

Label three of these points as X, Y and Z. The line XY intersects l∞ in a point, say 0 0 0 X , the line YZ intersects l∞ in Y and the line XZ intersects l∞ in Z . By Theorem 0 0 1.5.8, there exists a unique Baer subline D of l∞ containing the three points X , Y and Z0. In PG(2, q2), every Baer subline is a derivation set, hence D is a derivation 0 0 0 set on l∞ containing X , Y and Z .

104 From Theorem 1.5.9, there exists a unique Baer subplane B of P containing the derivation set D and the points X, Y and Z. Hence, if we derive with respect to D, the points X, Y and Z will lie on the line of PD corresponding to B and hence C0 in not an arc in PD.

As a result of Theorem 5.2.1, we will always work with a particular derivation set, say D, and investigate conics with respect to derivation with D.

Suppose the aﬃne pointset C of a conic C is an arc in the plane PD. One question we may ask is, can C0 be completed to a (q2 + 1)-arc in PD by adding points on the line at inﬁnity? In some cases we can be certain, as in the following theorem.

∼ 2 Theorem 5.2.2 Let P = PG(2, q ). Let D be a derivation set on the line l∞ in P.

Let C be a conic of P such that the set of points {C ∩ l∞} is disjoint from D. Suppose the aﬃne pointset C0 forms an arc in the aﬃne plane AD, then C0 can be completed to a (q2 + 1)-arc in PD.

0 2 Proof Suppose {C ∩ l∞} is empty, then C contains q + 1 points and hence is a 2 D (q + 1)-arc in P . Now suppose |{C ∩ l∞}| > 0. Let P be a point of {C ∩ l∞} in P.

Each line through a point of {C ∩ l∞} meets the aﬃne pointset C in at most one point. D If we derive with respect to D, the points {C ∩ l∞} are points of P and the lines of D 0 P, containing a point of {C ∩ l∞}, are lines of P . Hence the points C ∪ {C ∩ l∞} form a (q2 + 1)-arc in PD.

Projective equivalence is a concept we exploit in this chapter. Let φ be a collineation of PG(2, q2). The next theorem shows that the eﬀect of derivation with respect to D on a conic C is equivalent to deriving with respect to the derivation set φ(D) and observing the eﬀect on the conic φ(C).

105 Theorem 5.2.3 Suppose C is a conic in P ∼= PG(2, q2). Let D be a derivation set on a line l of P. Let φ be a collineation of P.

1. If C0 is not an arc in PD, then φ(C0) is not an arc in Pφ(D).

2. If C0 is an arc in PD, then φ(C0) is an arc in Pφ(D).

Proof Let B be a Baer subplane of P that contains D. The Baer subplane φ(B) contains the derivation set φ(D). If B meets C in x points, then φ(B) meets the aﬃne points φ(C) in x points.

Since C is an arc in P, if C0 is not an arc in PD, then some Baer subplane B of P meets C in three or more points. Hence, the Baer subplane φ(B) will meet φ(C) in three or more points. The second case follows similarly.

Disjoint derivation sets act in an independent way with regards to inherited arcs. Let C00 denote the aﬃne points of C after double derivation.

∼ 2 Theorem 5.2.4 Suppose P = PG(2, q ). Let C be a conic of P and let D1 and D2 be two disjoint derivation sets on l∞ in P.

1. Suppose the pointset C0 is an arc in the plane PD1 and the plane PD2 . Then C00 is an arc in the plane PD1D2 .

2. Suppose the pointset C0 is not an arc in one or both of the planes PD1 and PD2 . Then C00 is not an arc in the plane PD1D2 .

Proof The lines of PD1D2 are of three types: the Baer subplanes of PG(2, q2) that 2 2 contain D1; the Baer subplanes of PG(2, q ) that contain D2; and the lines of PG(2, q ) that meet l∞\{D1 ∪ D2}.

Suppose the aﬃne pointset C0 is an arc in the plane PD1 and an arc in the plane PD2 .

0 D1 2 Since C is an arc in P , each Baer subplane of PG(2, q ) that contain D1 meets C 2 in at most two points. Lines of PG(2, q ) that meet l∞\D1 also meet C in at most two points. Likewise, since C0 is an arc in PD2 , each Baer subplane of PG(2, q2) that

106 2 contain D2 meets C in at most two points. Lines of PG(2, q ) that meet l∞\D2 also meet C in at most two points.

Hence, if C0 is an arc in both of the planes PD1 and PD2 , then the lines of PD1D2 each meet C in at most two points and thus C00 is an arc in PD1D2 .

Suppose C0 is not an arc in one or both of the planes PD1 and PD2 . Suppose, without loss of generality, the pointset C0 is not an arc in PD1 . Then at least one Baer subplane 2 0 of PG(2, q ) containing D1 meets C in three or more points. Hence, if C is not an arc in one or both of the planes PD1 and PD2 then at least one line of PD1D2 meets C in three or more points. Hence C00 is not an arc in PD1D2 .

5.2.2 The derivation set DR

2 We coordinatise PG(2, q ) as in Example 1.3.2. That is, let l∞ be the line of equation 2 2 x2 = 0 and the aﬃne points of PG(2, q ) are given by (x0, x1, 1), where x0, x1 ∈ GF (q ).

Recall the real derivation set,

DR = {(0, 1, 0)} ∪ {(1, x1, 0)| x1 ∈ GF (q)}, from Section 1.6.

The aﬃne points of each Baer subplane containing DR can be represented as,

R(a, b, c) = {(ua + b, va + c, 1)| u, v ∈ GF (q)}, where a, b, c ∈ GF (q2) and a 6= 0 (see for example [35], [39]).

The aﬃne Baer subplane R(1, 0, 0) contains aﬃne points of the form (u, v, 1), where u, v ∈ GF (q). We refer to R(1, 0, 0) together with the points of DR, as PG(2, q), the real Baer subplane of PG(2, q2).

Let BR denote the set of aﬃne Baer subplanes R(a, b, c). We refer to an aﬃne Baer subplane of BR together with the points of DR as a Baer subplane of BR.

In the following lemma we verify that a particular group of homographies of PG(2, q2) is transitive on the Baer subplanes of BR.

107 Lemma 5.2.5 The group of homographies, denoted GR, given by the matrices of the form,   k 0 g      0 k h  , 0 0 1

2 where g, h, k ∈ GF (q ), is transitive on the Baer subplanes of BR.

Proof Let B1 be an aﬃne Baer subplane of BR with points given by,

{(ua1 + b1, va1 + c1, 1)| u, v ∈ GF (q)},

2 where a1, b1, c1 ∈ GF (q ). Also, let B2 be an aﬃne Baer subplane of BR with points given by,

{(ua2 + b2, va2 + c2, 1)| u, v ∈ GF (q)},

2 where a2, b2, c2 ∈ GF (q ). We wish to ﬁnd γ ∈ GR such that γ(B1) = B2.

Suppose γ has matrix,   a2 a2b1 0 b2 −  a1 a1   a2 a2c1  0 c2 − ,  a1 a1  0 0 1 and thus γ ∈ GR. Then γ(B1) is,       a2 a2b1 a2 a2b1 0 b2 − ua1 + b1 (ua1 + b1) + b2 −  a1 a1     a1 a1   a2 a2c1     a2 a2c1  0 c2 − va1 + c1 = (va1 + c1) + c2 −  a1 a1     a1 a1  0 0 1 1 1   ua2 + b2     =  va2 + c2  , 1 as required. It follows that the group GR is transitive on the Baer subplanes of BR.

108 5.3 Characterisation

In this section, we characterise the equations of all non-degenerate conics of PG(2, q2) that are not arcs after we perform a derivation with DR. Our method is to ﬁnd the equations of all non-degenerate conics that share at least three points with an aﬃne

Baer subplane of BR.

In Section 5.3.1, we proceed by observing the general associated matrix for a conic that shares q + 1 points with a Baer subplane of BR. In Section 5.3.2, we build a family of conics that share three points with some Baer subplane of BR. Such conics will not be arcs after derivation. In Section 5.3.3, we give a general characterisation for the equations of conics that are not arcs after derivation. We generalise the characterisation for derivation to multiple derivation in Section 5.3.4.

Let C be a conic of PG(2, q2) with associated matrix A and suppose γ is a homography of PG(2, q2). Recall from Lemma 1.7.7, that the conic we obtain when γ is applied to C has associated matrix γ−tAγ−1 and that we use γ to represent both the homography and the associated matrix of the homography as detailed in Section 1.7.

5.3.1 Conics of BR

ˆ Any conic C contained in a Baer subplane of BR can be extended to a conic CR of P G(2, q2). If q ≥ 5, by Lemma 5.1.1, the aﬃne points of such a conic of PG(2, q2) will not be an arc in the plane PDR . We can characterise these conics as follows, ﬁrstly for the case of q odd, and then followed by the case for q even.

Theorem 5.3.1 Let q be odd with q ≥ 5. Suppose C is a conic of P ∼= PG(2, q2) with −t −t −1 −1 associated matrix γ (ω Aω )γ , where ω ∈ P GL(3, q), γ ∈ GR and   0 1 0     A =  1 0 0  . 0 0 −2

0 If we perform a derivation with respect to the derivation set DR, then C is not an arc of PDR .

109 ˆ 2 Proof Let C be a conic of PG(2, q) with equation x0x1 − x2 = 0. An associated matrix for Cˆ is,   0 1 0     A =  1 0 0  . (5.1) 0 0 −2

From Theorem 1.7.6, the group P GL(3, q) is transitive on non-degenerate conics, thus ˆ for a given conic C1 of PG(2, q), we can ﬁnd a homography ω ∈ P GL(3, q) such that ˆ ˆ ω(C) = C1. Hence all conics of PG(2, q) have associated matrix of the form,

ω−tAω−1, where ω ∈ P GL(3, q) and A is deﬁned as in (5.1).

From the discussion in Section 1.7.1, the conics of PG(2, q) can be extended to conics of PG(2, q2) and such conics meet the real Baer subplane PG(2, q) in q + 1 points. Since q ≥ 5, these conics will share at least four points with the aﬃne points of PG(2, q).

ˆ 2 Let CR be the conic C extended to a conic of PG(2, q ). The group GR (as deﬁned in Lemma 5.2.5) is transitive on Baer subplanes of BR, hence for γ ∈ GR, the image 2 C = γ(CR) is a conic of PG(2, q ) that meets a Baer subplane of BR in q + 1 points.

Thus a conic C of P G(2, q2) with associated matrix,

γ−t(ω−tAω−1)γ−1,

where ω ∈ P GL(3, q), γ ∈ GR and A deﬁned as in (5.1), will meet a Baer subplane B of BR in q + 1 points.

The affine pointset C of C meets the affine points of B in at least q − 1 points. Since q ≥ 5, the pointset C meets the affine points of B in at least four points. Hence, if we 0 perform a derivation with respect to the derivation set DR, then the affine pointset C is not an arc in the plane PDR .

110 Theorem 5.3.2 Let q be even with q ≥ 4. Suppose C is a conic of P ∼= PG(2, q2) with −t −t −1 −1 associated matrix γ (ω Aω )γ , where ω ∈ P GL(3, q), γ ∈ GR and   0 1 0     A =  0 0 0  . 0 0 1

0 If we perform a derivation with respect to the derivation set DR, then C is not an arc of PDR .

0 2 Proof Let C be a conic of PG(2, q) with equation x0x1 − x2 = 0. Choose the associated matrix A for C0 as,   0 1 0     A =  0 0 0  . 0 0 1

The rest of the proof follows as in the proof of Theorem 5.3.1.

For q = 2 or 3, a conic of PG(2, q) may only contain two aﬃne points, hence the extension of such a conic to PG(2, q2) will meet two aﬃne points of PG(2, q). Thus, for q = 2 or 3 we can not use the statements of Theorem 5.3.1 and Theorem 5.3.2.

5.3.2 Case C ∩ l∞ = {P,Q} with P ∈ DR,Q/∈ DR

∼ 2 Assume q is odd. Let C be a conic of P = PG(2, q ) such that C meets l∞ in two points P and Q. Here we will ﬁnd a family of conics that satisfy this case with the property that, if we perform a derivation with DR, the aﬃne points of these conics will not be

DR arcs in the plane P . Note that this family is such that P ∈ DR and Q/∈ DR.

This will give a ﬂavour for the full characterisation and further insight into a case that has not been fully explored as we shall see in Section 5.4.

111 Theorem 5.3.3 Let q be odd. Suppose C is a conic of P ∼= PG(2, q2) with associated −t −1 matrix γ Aγ , where γ ∈ GR and   4β −2 −β + 2     A =  −2 0 β + 2  , (5.2) −β + 2 β + 2 −2β − 4 where β ∈ GF (q2)\GF (q). If we perform a derivation with respect to the derivation

0 DR set DR, then C is not an arc of P .

2 Proof Let l∞ be the line of infinity for PG(2, q ). We begin by choosing a quadrangle in PG(2, q) such that three points are contained in AG(2, q) = PG(2, q)\l∞ and one point is contained in DR. We then select another point on l∞, not contained in DR, which is not collinear with any two points of the quadrangle. These five points uniquely determine a non-degenerate conic that shares three points with AG(2, q) and thus the affine points of this conic will not be an arc after derivation with respect to DR.

First, choose the quadrangle, say P = (0, 1, 0),A = (1, 0, 1),B = (0, 1, 1),C = (1, 2, 2) in PG(2, q). Note that A, B and C are points of AG(2, q) and P is on l∞. The lines of the quadrangle P ABC are,

PA = h(0, 1, 0), (1, 0, 1)i : x0 − x2 = 0,

PB = h(0, 1, 0), (0, 1, 1)i : x0 = 0,

PC = h(0, 1, 0), (1, 2, 2)i : 2x0 − x2 = 0,

AB = h(1, 0, 1), (0, 1, 1)i : −x0 − x1 + x2 = 0,

AC = h(1, 0, 1), (1, 2, 2)i : −2x0 − x1 + 2x2 = 0,

BC = h(0, 1, 1), (1, 2, 2)i : x1 − x2 = 0.

The lines of the quadrangle intersect l∞ in points of DR. Hence, if we choose another point Q on l∞, not contained in DR, then P, A, B, C, Q are a set of ﬁve points, no three collinear. Thus, let Q = (1, β, 0), where β ∈ GF (q2)\GF (q).

112 To ﬁnd the unique conic containing the points P, A, B, C, Q, we substitute their coordinates into the general equation,

2 2 2 CQ : ax0 + bx1 + cx2 + 2fx1x2 + 2gx2x0 + 2hx0x1 = 0.

From CQ we get the conditions,

P = (0, 1, 0) ∈ CQ ⇒ b = 0,

A = (1, 0, 1) ∈ CQ ⇒ a + c + 2g = 0, (5.3)

B = (0, 1, 1) ∈ CQ ⇒ b + c + 2f = 0, (5.4)

C = (1, 2, 2) ∈ CQ ⇒ a + 4b + 4c + 8f + 4g + 4h = 0, (5.5)

2 Q = (1, β, 0) ∈ CQ ⇒ a + β b + β2h = 0. (5.6)

Suppose a = 0, then from (5.3) and (5.4) we have c = −2g = −2f. Substituting into (5.5) gives −8f + 8f + 4f + 4h = 0. From (5.6) we also have β2h = 0. Since β ∈ GF (q2)\GF (q), then β 6= 0 and hence h = 0. This implies that 4f = 0 and thus f = 0 and from this we get c = g = 0.

Thus if a = 0, all the coeﬃcients of CQ are zero and hence no such conic exists. Thus, suppose a 6= 0.

Since the equation is homogeneous, let a = 1. Now, from (5.4) we have c = −2f and substituting into (5.3) gives 1 − 2f = −2g. If we then substitute into (5.5) we get, 1 − 8f + 8f − 2 + 4f + 4h = −1 + 4f + 4h = 0. From (5.6), we also have the condition that 1 + β2h = 0. Thus,

−1 h = , 2β 1 − 4h 1 + 2 β + 2 f = = β = , 4 4 4β β+2 β+2 1 − 2f 1 − 2 4β −2β + 4β 4β −β + 2 g = = = = , −2 −2 4β 4β β + 2 −2β − 4 c = −2 = , 4β 4β

113 and since the equation is homogeneous, we can multiply through by 4β to give,

a = 4β, c = −2β − 4, f = β + 2, g = −β + 2, h = −2.

Hence, the conic CQ has associated matrix,   4β −2 −β + 2     A =  −2 0 β + 2  , (5.7) −β + 2 β + 2 −2β − 4 where β ∈ GF (q2)\GF (q).

Suppose γ ∈ GR. The group of homographies GR ﬁxes the derivation set DR, hence

γ(PG(2, q)) is a Baer subplane of BR and also the point γ(P ) ∈ DR and the point

γ(Q) ∈/ DR. The conic C = γ(CQ) shares three points with the aﬃne points of γ(PG(2, q)). Hence, if we perform a derivation with respect to the derivation set

DR, the aﬃne pointset of a conic C with associated matrix,

γ−tAγ−1,

DR where γ ∈ GR and A deﬁned as in (5.7), will not be an arc in P .

5.3.3 Derivation

Here we give a full characterisation of the equations of non-degenerate conics that share at least three points with any of the aﬃne Baer subplanes of BR.

We first find all the orbits in AG(2, q) of triangles under AGL(3, q) and find the equation of the general conic through any representative of these orbits. The orbit of AGL(3, q) on these general conics is all the conics that share three points with AG(2, q). We then apply the group GR to obtain the full characterisation.

114 Theorem 5.3.4 Let q be odd. Suppose C is a conic of P ∼= PG(2, q2) with associated −t −t −1 −1 matrix γ (ω Aω )γ , where ω ∈ AGL(3, q), γ ∈ GR and   1 h − 1  2    A =  h b f  , 1 − 2 f 0 where f, b, h ∈ GF (q2), u2 + bv2 + 2fv − u + 2huv = 0 for some u, v ∈ GF (q), where v 6= 0 and the determinant of A is non-zero. If we perform a derivation with the

0 DR 2 derivation set DR, then C is not an arc of P . Further, any such conic of PG(2, q ) that is not an arc of PDR is of this form.

Proof First, we require a way to represent triangles in AG(2, q). The group AGL(3, q) is 2-transitive on points of AG(2, q), hence we can ﬁx a line in AG(2, q). Let this line be x1 = 0 and take two points on the line, (0, 0, 1) and (1, 0, 1). Any point not on the line x1 = 0 can be written as (u, v, 1), for some u, v ∈ GF (q) and v 6= 0. Thus, without loss of generality, a triangle of AG(2, q) can be written as T = {(0, 0, 1), (1, 0, 1), (u, v, 1)}, for some u, v ∈ GF (q) and v 6= 0.

All conics of PG(2, q2) containing a triangle T , where u, v ∈ GF (q) and v 6= 0, can be computed as follows. Start with a conic given by the general equation,

2 2 2 CT : ax0 + bx1 + cx2 + 2fx1x2 + 2gx2x0 + 2hx0x1 = 0,

and substitute the points of the triangle T into CT ,

(0, 0, 1) ∈ CT ⇒ c = 0, (5.8)

(1, 0, 1) ∈ CT ⇒ a + c + 2g = 0, (5.9)

2 2 (u, v, 1) ∈ CT ⇒ au + bv + c + 2fv + 2gu + 2huv = 0. (5.10)

If we substitute (5.8) into (5.9) we get a = −2g. If we then substitute this into (5.10) we get the condition au2 + bv2 + 2fv − au + 2huv = 0, for some u, v ∈ GF (q), where v 6= 0.

115 Suppose a = 0, then CT has associated matrix,   0 f 0     A =  f b h  , (5.11) 0 h 0 where f, b, h ∈ GF (q2), bv2 + 2fv + 2huv = 0, for some u, v ∈ GF (q) and v 6= 0 and the determinant of A is non-zero.

The determinant of A is −f(f.0 + 0.h) = 0 which is a contradiction, hence to ensure the determinant of A is non-zero, we need to enforce a 6= 0.

Thus, since the equation is homogeneous, let a = 1. In this case CT has associated matrix,   1 h − 1  2    A =  h b f  , (5.12) 1 − 2 f 0 where f, b, h ∈ GF (q2), u2 + bv2 + 2fv − u + 2huv = 0, for some u, v ∈ GF (q), where v 6= 0 and the determinant of A is non-zero.

The orbit of the group AGL(3, q) acting on the set of triangles,

{(0, 0, 1), (1, 0, 1), (u, v, 1)| u, v ∈ GF (q) and v 6= 0}, will be the full set of triangles in AG(2, q). Hence, the orbit of AGL(3, q) acting on the 2 set of conics that satisfy the equation of CT is the full set of conics of PG(2, q ) that meet AG(2, q) in a triangle.

Let B be an affine Baer subplane of BR. Since GR is transitive on affine Baer subplanes of BR, any affine triangle of AG(2, q) can be mapped to some triangle of B, using a 2 homography from GR as shown in Figure 5.1. Hence any conic of PG(2, q ) meeting PG(2, q) in at least an affine triangle can be mapped to some conic that meets B in at least an affine triangle.

Conversely, any conic of PG(2, q2) meeting B in at least an aﬃne triangle can be mapped to a conic that meets BR in at least an aﬃne triangle, using GR. Such a conic can then be mapped to the conic CT using AGL(3, q).

116 DR l∞

B P G(2, q)

CT γ

Figure 5.1: The action of γ on Baer subplanes of BR.

2 Thus, using GR and AGL(3, q), we can map a conic CT of P G(2, q ) that meets AG(2, q) in an affine triangle to any conic of PG(2, q2) that meets any other affine Baer subplane 2 B of BR in a triangle. Thus the set of conics of PG(2, q ) that meet an affine Baer subplane of BR in a triangle is given by the set of associated matrices,

γ−t(ω−tAω−1)γ−1, where γ ∈ GR, ω ∈ AGL(3, q) and A is deﬁned as in (5.12).

If we perform a derivation with respect to the derivation set DR, then the aﬃne pointset 2 of any conic of PG(2, q ) that meets an aﬃne Baer subplane of BR in three or more points will not be an arc in PDR . Hence these matrices are the associated matrices of all the conics of PG(2, q2) that are not arcs in PDR .

Theorem 5.3.5 Let q be even. Suppose C is a conic of P ∼= PG(2, q2) with associated −t −t −1 −1 matrix γ (ω Aω )γ , where ω ∈ AGL(3, q), γ ∈ GR and   a00 a01 a00     A =  a10 a11 a12  ,

0 a21 0

2 2 2 where a01, a10, a11, a21, a22 ∈ GF (q ), a00u +a11v +(a01+a10)uv+(a12+a21)v+a00u = 0, for some u, v ∈ GF (q), where v 6= 0 and such that A does not give rise to a degenerate 0 equation. If we perform a derivation with respect to the derivation set DR, then C is not an arc of PDR . Further, any such conic of PG(2, q2) that is not an arc of PDR is of this form.

117 Proof Let CT be a conic with general equation,

2 2 2 a00x0 + a11x1 + a22x2 + (a01 + a10)x0x1 + (a12 + a21)x1x2 + (a02 + a20)x0x2 = 0, where aij ∈ GF (q) for i, j ∈ 0,..., 2. As in the proof of Theorem 5.3.4, we can represent an aﬃne triangle of AG(2, q) by three aﬃne points then substitute these points into

CT ,

(0, 0, 1) ∈ CT ⇒ a22 = 0, (5.13)

(1, 0, 1) ∈ CT ⇒ a00 + a22 + a02 + a20 = 0, (5.14)

2 2 (u, v, 1) ∈ CT ⇒ a00u + a11v + a22 + (a01 + a10)uv +

(a12 + a21)v + (a02 + a20)u = 0. (5.15)

Substituting (5.13) into (5.14) gives a00 = a02 + a20. If we then substitute this into 2 2 (5.15), we have the condition a00u + a11v + (a01 + a10)uv + (a12 + a21)v + a00u = 0, for some u, v ∈ GF (q), where v 6= 0.

We then choose an associated matrix for CT ,   a00 a01 a00     A =  a10 a11 a12  ,

0 a21 0

2 2 2 where a01, a10, a11, a21, a22 ∈ GF (q ), a00u +a11v +(a01+a10)uv+(a12+a21)v+a00u = 0, for some u, v ∈ GF (q), where v 6= 0 and such that A does not give rise to a degenerate equation.

Using a similar process to the proof in Theorem 5.3.4 we can characterise the equations of all conics that are not arcs after derivation.

In Theorem 5.3.4 and Theorem 5.3.5 we have characterised the full set of conics of PG(2, q2) whose aﬃne pointsets do not form arcs in the plane PDR . Hence we have also characterised those conics whose aﬃne pointsets do form an arc in the plane PDR . These conics will be all the conics of PG(2, q2) that do not satisfy the equations given in the statements of the theorems.

118 5.3.4 Multiple derivation

Here we seek to characterize the equations of conics that are not arcs after multiple derivation when q is odd. We do this by using orbits on triangles as in Theorem 5.3.4 and then extending to multiple derivation.

2 2 Let D be a derivation set on the line l∞ in PG(2, q ). The group P GL(3, q ) is transitive 2 on derivation sets on a line, hence we can write D as σ(DR) for some σ ∈ P GL(3, q )l∞ .

∼ 2 Theorem 5.3.6 Let q be odd and suppose P = PG(2, q ). Let σ1(DR), . . . , σn(DR) be 2 n disjoint derivation sets on l∞ in P, for some σ1, . . . , σn ∈ P GL(3, q )l∞ . Suppose C is a conic of P with associated matrix either,

−t −t −t −1 −1 −1 σ1 (γ (ω Aω )γ )σ1 or

−t −t −t −1 −1 −1 σ2 (γ (ω Aω )γ )σ2 or . .

−t −t −t −1 −1 −1 σn (γ (ω Aω )γ )σn ,

where ω ∈ AGL(3, q), γ ∈ GR and   1 h − 1  2    A =  h b f  , 1 − 2 f 0 where f, b, h ∈ GF (q2), u2+bv2+2fv−u+2huv = 0 for some u, v ∈ GF (q), where v 6= 0 and the determinant of A is non-zero. If we perform a multiple derivation with respect

0 σ1(DR)...σn(DR) to the derivation sets σ1(DR), . . . , σn(DR), then C is not an arc of P . Further, any such conic of PG(2, q2) that is not an arc of Pσ1(DR)...σn(DR) is of this form.

Proof The Baer subplanes containing each σi(DR) can be identiﬁed by applying σi to the real Baer subplane PG(2, q) and then applying homographies from GR. Hence, a Baer subplane B that contains σi(DR) can be written as γ(σi(PG(2, q))), where

γ ∈ GR. This is shown in Figure 5.2 for σ1 and letting γ be the identity.

119 DR σ1(DR) l∞

P G(2, q) σ1 B = σ1(PG(2, q))

Figure 5.2: The action of σ1 on the real Baer subplane.

In Theorem 5.3.4, we found a general form for the associated matrices of conics through any triangle contained in an aﬃne Baer subplane of BR. That is,

γ−t(ω−tAω−1)γ−1,

where γ ∈ GR, ω ∈ AGL(3, q) and A is deﬁned as in (5.12) in Theorem 5.3.4.

Suppose we derive with respect to the derivation set σ1(DR). The Baer subplanes containing σ1(DR) are given by γ(σ1(PG(2, q))), where γ ∈ GR. If we perform a multiple derivation with respect to the derivation set σ1(DR), the associated matrices of the conics of PG(2, q2) whose aﬃne pointsets are not arcs in Pσ1(DR) are given by,

−t −t −t −1 −1 −1 σ1 (γ (ω Aω )γ )σ1 , where γ ∈ GR, ω ∈ AGL(3, q) and A is deﬁned as in (5.12) in Theorem 5.3.4.

Hence, if we derive with respect to the disjoint derivation sets σ1(DR), . . . , σn(DR), those conics of PG(2, q2) that are not arcs after the multiple derivation are those with matrix of one of the following forms,

−t −t −t −1 −1 −1 σ1 (γ (ω Aω )γ )σ1 ,

−t −t −t −1 −1 −1 σ2 (γ (ω Aω )γ )σ2 , . .

−t −t −t −1 −1 −1 σn (γ (ω Aω )γ )σn ,

where γ ∈ GR, ω ∈ AGL(3, q) and A is deﬁned as in (5.12) in Theorem 5.3.4.

120 5.4 Previous work

We next approach the problem of deriving planes containing conics from a diﬀerent angle. We will build on results by other authors, so in this section we will outline the current state of the work done on the problem of conics and derivation. Most of these details are from the survey in O’Keefe and Pascasio [42] and it is the work in that paper which we are seeking to extend.

First, we divide the problem into projectively inequivalent cases. We then present two combined theorems on the known results for each of these cases, where q is odd.

5.4.1 Projective equivalence

Let C be a conic of P ∼= PG(2, q2) and let D be a derivation set on l in P. Let C be the aﬃne points of C and let C0 be the pointset C viewed as points of PD. Let H(q2) denote the collineation group of the Hall plane PD = Hall(q2).

We wish to define the orbits of H(q2), in PD, on the affine pointsets of all conics of P. 2 We do this by showing these orbits correspond to orbits of AΓL(3, q )D acting on the affine pointsets of all conics of P.

If γ ∈ AΓL(3, q2) ﬁxes a derivation set D on a line l, then γ induces a collineation of the D ∼ 2 2 2 Hall plane P = Hall(q ). Thus AΓL(3, q )D is a subgroup of H(q ), the collineation 2 2 2 group of Hall(q ). If q > 3, then H(q ) = AΓL(3, q )D (for details, see [23], [35]).

Suppose AΓL(3, q2) is the collineation group of the plane PG(2, q2)\l, then the group 2 2 2 2 AΓL(3, q ) = P ΓL(3, q )l. Since P ΓL(3, q )D is a subgroup of P ΓL(3, q )l, then the 2 2 2 group AΓL(3, q )D = P ΓL(3, q )D. Thus, if two sets of points in PG(2, q ) are in the 2 2 same orbit of P ΓL(3, q )D, then they are in the same orbit of H(q ).

2 0 For instance, suppose γ ∈ P ΓL(3, q )D. The collineation γ induces a collineation γ in H(q2). Let C be the aﬃne points of a conic C of P ∼= P G(2, q2). The pointset γ(C) is the same as γ0(C0) when viewed as a pointset of PD.

121 Thus if we wish to investigate a representative from each orbit of H(q2) on the aﬃne points of conics of PG(2, q2) as points of Hall(q2), it is suﬃcient to investigate a 2 representative from each orbit of P ΓL(3, q )D on non-degenerate conics.

2 2 To ﬁnd the orbits of P ΓL(3, q )D on non-degenerate conics in PG(2, q ) we ﬁnd the 2 orbits of P GL(3, q )D on C and l∞, where C is a non-degenerate conic. The orbits of 2 2 P ΓL(3, q )D will be a union of orbits of P GL(3, q )D.

From Theorem 1.7.6, the group P GL(3, q2) is transitive on the set of non-degenerate 2 2 conics in PG(2, q ), hence we may ﬁx a conic C. Further, the group P GL(3, q )C which ﬁxes C is transitive on the sets of secant lines of C, tangent lines of C and external lines of C (see for example [33], [34]). Hence, we consider three cases as to whether l is a secant, a tangent or an external line of a conic C.

5.4.2 Known results

∼ 2 ∼ 2 ∼ 2 Let P = PG(2, q ) and let A = AG(2, q ) = PG(2, q )\l∞. Suppose C is a conic of P and let C be the aﬃne points of C in A. The plane ADR is the aﬃne plane after

DR ∼ 2 derivation in the plane P with respect to DR and P = Hall(q ) is the projective completion of ADR .

As discussed in Section 5.4.1, we consider three cases, whether the line l∞ is either an external line, a tangent or a secant to a conic C of P G(2, q2). We examine what is previously known about each of the cases for q odd. We divide the cases into those for q = 3 and those for q ≥ 5.

Theorem 5.4.1 For q = 3.

1. The line l∞ is secant to C. Let P and Q be the two points of intersection between

l∞ and C.

0 DR (a) The points P,Q ∈ DR. In [42], it is shown that C is an 8-arc in A . In this case there are the following two possibilities.

122 2 (i) The conic C is equivalent to the conic with equation x0x1 − x2 = 0 in AG(2, 9). In this case, C0 is an incomplete 8-arc in Hall(9) and can be completed to an oval.

2 (ii) The conic C is equivalent to the conic with equation x0x1 − dx2 = 0, where d is a ﬁxed non-square in GF (9), in AG(2, 9). In this case, C0 is a complete 8-arc in Hall(9).

(b) Exactly one of P or Q is in DR then using a computer search, it is shown in [42], that if C0 is an arc in ADR then C0 completes to a 9-arc in PDR .

2. The line l∞ is tangent to C. As mentioned in Theorem 5.1.2, in [37], it is proved that C0 is not an arc in ADR .

3. The line l∞ is external to C. Using a computer, in [42] it is shown that the pointset C0 is not an arc in ADR .

Theorem 5.4.2 For q odd, where q ≥ 5,

1. The line l∞ is secant to C. Let P and Q be the two points of intersection between

l∞ and C.

(a) The points P,Q ∈ DR. This case was addressed in [42] and we have the following two cases.

2 0 DR (i) The conic C is equivalent to x0x1 − x2 = 0 and C is not an arc in A . 2 (ii) The conic C is equivalent to x0x1 − dx2 = 0, where d is a non-square in GF (q2) and C0 is a complete (q2 − 1)-arc in PDR .

(b) The points P,Q/∈ D. This case has been considered in both [36] and [42], however no full classiﬁcation is known. There exists one known family of 2 2 2 2 conics Cc,d : x0 − cx1 − dx2 = 0, where d is a non-square in GF (q ) and c is 0 a non-square in GF (q). For C ∈ Cc,d, the pointset C is an arc and can be completed to an oval in PDR .

2. The line l∞ is tangent to C. As mentioned in Theorem 5.1.2, in [37], it is proved that C0 is not an arc in the aﬃne plane ADR .

123 3. The line l∞ is external to C. In this case nothing is known.

Note that the result of part 1(a)(i) of Theorem 5.4.2 has been partially addressed in Theorem 5.3.1. The case for part 1(c) of Theorem 5.4.2 has been partially addressed in Theorem 5.3.3. In Theorem 5.3.3, it shown that the aﬃne points of conics with equation of the form,

2 2 4βx0 + (−2β − 4)x2 − 4x0x1 + (−2β + 4)x1x2 + (2β + 4)x1x2 = 0,

2 where β ∈ GF (q )\GF (q), are not arcs after derivation with respect to DR.

Certain cases for q even have been addressed in [42], [43] and [30], however we are mainly concerned with q odd in this thesis so we do not list the known results for q even.

5.5 Derivation with sDR and nDR

In Section 5.4 the results consider derivation with respect to DR. In this section we observe the eﬀect of derivation on a ﬁxed conic using other derivation sets on l∞.

We coordinatise PG(2, q2) as in Example 1.3.2. Suppose C is the conic with equation 2 x0x1 − x2 = 0 and l∞ is the line x2 = 0, then l∞ is secant to C at the points (1, 0, 0) and (0, 1, 0).

We begin by calculating the derivation sets on l∞ that contain both the points (1, 0, 0) and (0, 1, 0). We then oﬀer an alternative method for computing the orbits of P ΓL(3, q2) on triples (C, l∞, DR) to that given in [42, Theorem 3.2] for q odd.

2 Next, we calculate the orbits of P ΓL(3, q )C,l∞ acting on derivation sets that contain both (1, 0, 0) and (0, 1, 0). We ﬁnish with Theorem 5.5.4 which determines when C is an arc after derivation with respect to the derivation sets that contain both (1, 0, 0) and (0, 1, 0).

Our ﬁrst step is to determine the general form for all derivation sets of PG(2, q2) that contain both (0, 1, 0) and (1, 0, 0).

124 Theorem 5.5.1 The derivation sets kDR = {(0, 1, 0)}∪{(1, kx, 0)| x ∈ GF (q)}, where k ∈ GF (q2)\{0} are the only derivation sets of PG(2, q2) that contain both the points (0, 1, 0) and (1, 0, 0).

Proof Let GF (q2) = {0, α, . . . , αi, . . . , αq2−1} and thus α is a generator for GF (q2). Recall from Section 1.2, that GF (q) as a subﬁeld of GF (q2), is given by the elements q+1 2(q+1) (q−1)(q+1) 0, α , α , . . . , α . Hence we can write the points of kDR as,

i(q+1) kDR = {(1, 0, 0)} ∪ {(0, 1, 0)} ∪ {(1, kα , 0)| i = 1, . . . , q − 1}, where k ∈ GF (q2)\{0}.

Let k = αm, for some m ∈ 0, . . . , q. Hence kαi(q+1) = αmαi(q+1) = αiq+i+m. For each m ∈ 0, . . . , q, the set {αiq+i+m|i = 1, . . . , q − 1} is a unique set of q − 1 elements of GF (q2).

2 q Thus, we obtain q + 1 distinct derivation sets DR, α(DR), α (DR), . . . , α (DR), each containing the points (0, 1, 0) and (1, 0, 0).

By Lemma 1.5.11, any two points of l∞ are contained in q + 1 Baer sublines of l∞ and hence q + 1 distinct derivation sets. Thus we have found all the derivation sets of l∞ that contain both (0, 1, 0) and (1, 0, 0).

2 Next, we compute the group of homographies of PG(2, q ) that ﬁxes C and l∞. This 2 group is denoted by P GL(3, q )C,l∞ .

2 2 Theorem 5.5.2 Suppose q is odd. Let C be the conic x0x1 − x2 = 0 of PG(2, q ) and 2 suppose l∞ is the line x2 = 0. The group P GL(3, q )C,l∞ consists of the homographies with matrix of the form,     1 0 0 0 1 0          0 δ 0  or  δ 0 0  , √ √ 0 0 ± δ 0 0 ± δ where δ is a non-zero square of GF (q2).

125 2 Proof To ﬁnd an element γ of P GL(3, q )C,l∞ , we ﬁrst let γ be a general element of 2 P GL(3, q )l∞ . Hence γ has matrix of the form,   a b g      e f h  , 0 0 k where a, b, e, f, g, h, k ∈ GF (q2) and such that the matrix has non-zero determinant.

If γ is to also ﬁx C, then γ−tAγ−1 = δA, where δ ∈ GF (q2)\{0}. Since γ is a general element of a group, without loss of generality, we can choose the same form for γ−1.

Since q odd, an associated matrix for C is,   0 1 0     A =  1 0 0  . 0 0 −2

Now solve for the coeﬃcients of γ−1,         a e 0 0 1 0 a b g 0 δ 0                  b f 0   1 0 0   e f h  =  δ 0 0  , g h k 0 0 −2 0 0 k 0 0 −2δ where δ ∈ GF (q2), δ 6= 0. Thus,       e a 0 a b g 0 δ 0              f b 0   e f h  =  δ 0 0  , h g −2k 0 0 k 0 0 −2δ     2ea be + af ge + ah 0 δ 0          be + af 2bf gf + bh  =  δ 0 0  . ha + ge gf + bh 2gh − 2k2 0 0 −2δ

Equating the coeﬃcients gives 2ae = 0 and 2bf = 0. This implies that a = 0 and/or e = 0 and also that b = 0 and/or f = 0.

126 Suppose e = 0 and a 6= 0, then since be + af = δ we have af = δ. Since at least one of b and f is zero, this implies that b = 0 and f 6= 0. Also ge + ah = 0 and since e = 0 then ah = 0 which implies h = 0. Also gf + bh = 0 then gf = 0 and hence g = 0. The √ last condition is 2gh − 2k2 = −2δ. Since g = 0 and h = 0, this simpliﬁes to k = ± δ, where δ is a non-zero square of GF (q2). Thus we have the following matrices satisfying the conditions on γ−1,   1 0 0      0 δ 0  , (5.16) √ 0 0 ± δ where δ is a non-zero square in GF (q2).

Suppose a = 0 and e 6= 0, then since be + af = δ, we have be = δ. Since at least one of b and f is zero, this implies that f = 0 and b 6= 0. Also ge + ah = 0 and since a = 0 then ge = 0 which implies g = 0. Also gf + bh = 0 then bh = 0 and hence h = 0. √ Finally, we have k = ± δ, where δ is a non-zero square of GF (q2) as above. Thus we also have the following matrices satisfying the conditions on γ−1,   0 1 0      δ 0 0  , (5.17) √ 0 0 ± δ where δ ∈ GF (q2) and δ is a non-zero square.

Suppose both a = 0 and e = 0. Since be + af = δ, this implies δ = 0 which is a contradiction, hence there is no valid solution in this case.

Since we are computing elements in the entire group, the general form for γ−1 gives 2 2 us the general form for an element γ ∈ P GL(3, q )C,l∞ . That is, if γ ∈ P GL(3, q )C,l∞ then, without loss of generality, we can represent γ by a matrix as given in (5.16) or (5.17).

2 Next, we calculate the orbits of P ΓL(3, q )C,l∞ on the derivation sets that contain both (0, 1, 0) and (1, 0, 0).

127 2 2 Theorem 5.5.3 Let P ΓL(3, q )C,l∞ be the group of homographies of PG(2, q ), where 2 q is odd, from Theorem 5.5.2. There are two orbits of P ΓL(3, q )C,l∞ on the derivation 2 sets kDR, where k ∈ GF (q )\{0}, namely the derivation sets of the form,

sDR = {(0, 1, 0)} ∪ {(1, sx, 0)| x ∈ GF (q)}, where s is a non-zero square in GF (q2) and the derivation sets of the form,

nDR = {(0, 1, 0)} ∪ {(1, nx, 0)| x ∈ GF (q)}, where n is a non-square in GF (q2).

2 Proof The points of l∞ are given by {(0, 1, 0)} ∪ {(1, x1, 0)| x1 ∈ GF (q )}. Since 2 P ΓL(3, q )C,l∞ fixes C and the conic C meets l∞ in the points (0, 1, 0) and (1, 0, 0), 2 then P ΓL(3, q )C,l∞ fixes the pointset {(0, 1, 0), (1, 0, 0)}. Hence, to find the orbits of 2 2 P ΓL(3, q )C,l∞ on the derivation sets kDR, we look at the orbits of P ΓL(3, q )C,l∞ on the points l∞\{(1, 0, 0), (0, 1, 0)}.

2 We begin with the group P GL(3, q )C,l∞ which was calculated in Theorem 5.5.2. We 2 start by calculating the orbits of P GL(3, q )C,l∞ on the points l∞\{(1, 0, 0), (0, 1, 0)},       1 0 0 1 1              0 δ 0   x1  =  δx1  , √ 0 0 ± δ 0 0         0 1 0 1 x1 1             =   =  δ  ,  δ 0 0   x1   δ   x  √ 1 0 0 ± δ 0 0 0

2 where x1 ∈ GF (q)\{0} and δ is a non-zero square in GF (q ).

Recall the relationships between squares and non-squares from Theorem 1.2.1. Since δ 2 2 2 is a square in GF (q ), if x1 is a square in GF (q ), then δx1 is a square in GF (q ). If x1 2 2 is a non-square in GF (q ), then δx1 is a non-square in GF (q ). Also, if x1 is a non-zero 2 1 2 square in GF (q ), then is a square in GF (q ) and similarly, if x1 is a non-square in x1 GF (q2), then 1 is a non-square in GF (q2). x1

128 2 Hence, the orbits of the group P GL(3, q )C,l∞ on l∞\{(1, 0, 0), (0, 1, 0)} are the pointsets {(1, s, 0)| s is a non-zero square in GF (q2)} and {(1, n, 0)| n is a non-square in GF (q2)}.

2 k Let ζ be an automorphic collineation of PG(2, q ), thus ζ((1, x1, 0)) = (1, x1, 0), for some k ∈ p, p2, . . . , pn = q2. Since q is odd, then k is odd. From Lemma 1.2.1, a square raised to an odd power is a square and a non-square raised to an odd power 2 is a non-square. Hence the orbits of P ΓL(3, q )C,l∞ on l∞\{(1, 0, 0), (0, 1, 0)} are the pointsets {(1, s, 0)| s is a non-zero square in GF (q2)} and {(1, n, 0)| n is a non-square in GF (q2)}.

Recall the real derivation set DR = {(0, 1, 0)} ∪ {(1, x1, 0)| x1 ∈ GF (q)} and consider the derivation set,

D = {(0, 1, 0)} ∪ {(1, 0, 0)} ∪ {(1, d1, 0),..., (1, dq−1, 0)},

2 where d1, . . . , dq−1 ∈ GF (q ).

2 Suppose d1 is a non-zero square in GF (q ). Since every element of GF (q) is a square 2 2 in GF (q ), we can write sd1 = x1, for some square s in GF (q ) and x1 in GF (q). The derivation set DR and sD share three points and hence coincide. It follows, that di is a square in GF (q2), for each i ∈ 1, . . . , q −1. Hence the elements of D which are diﬀerent from (1, 0, 0) and (0, 1, 0) are either all of the form (1, s, 0), where s is a non-zero square in GF (q2) or all of the form (1, n, 0), where n is a non-square in GF (q2).

2 Let C be a conic of PG(2, q ). Suppose l∞ is a secant to C and the points C ∩ l∞ are contained in DR. If q is odd and q ≥ 5, then from Theorem 5.4.2, there are two projectively distinct cases for C when deriving with respect to DR.

2 2 These two cases are represented by the two conics x0x1 − x2 = 0 and x0x1 − dx2 = 0, where d is a non-square in GF (q2). In the next theorem, we consider the problem from 2 another angle. We ﬁx the conic x0x1 − x2 = 0 and vary the derivation set, using the 2 group P ΓL(3, q )C,l∞ .

129 2 That is, we ﬁx the conic x0x1 − x2 = 0, which meets l∞ in the points (1, 0, 0) and (0, 1, 0), and consider the eﬀect of derivation using a derivation set containing (1, 0, 0) and (0, 1, 0). By Theorem 5.5.1, the only such derivation sets are of the form kDR, where k ∈ GF (q2)\{0}. By Theorem 5.5.3, there will be two cases two consider, depending on whether k is a non-zero square in GF (q2) or a non-square in GF (q2).

Theorem 5.5.4 Suppose P ∼= PG(2, q2), where q is odd and q ≥ 5. Let C be the conic 2 of P with equation x0x1 − x2 = 0.

1. If we derive with respect to a derivation set of the form sDR, where s is a non-zero square in GF (q2), then C0 is not an arc in PsDR .

2. If we derive with respect to a derivation set of the form nDR, where n is a non- square in GF (q2), then C0 is a complete (q2 − 1)-arc in PnDR .

Proof The derivation set DR is of the form sDR, where s = 1. From Theorem 5.5.3, 2 the derivation sets of the form sDR, where s is a non-zero square in GF (q ), are an 2 orbit under P ΓL(3, q )C,l∞ acting on derivation sets of l∞. Hence, from Theorem 5.2.3 and Theorem 5.4.2 part 1(a)(i), if we derive with respect to a derivation set of the form

0 sDR sDR, then the aﬃne pointset C will not be an arc in P .

2 2 Let C1 be the conic x0x1 − dx2 = 0, where d a non-square in GF (q ). Both C1 and C meet l∞ at the points (1, 0, 0) and (0, 1, 0). Let A be the associated matrix for C1. Let 2 φ be a homography of P GL(3, q )l∞ such that φ(C1) = C. Hence the associated matrix for C is φ−tAφ−1.

The matrix for a suitable φ−1 is given by,   1 0 0      0 d 0  , 0 0 1 since,           1 0 0 0 1 0 1 0 0 0 d 0 0 1 0                      0 d 0   1 0 0   0 d 0  =  d 0 0  ≡  1 0 0  . 0 0 1 0 0 −2d 0 0 1 0 0 −2d 0 0 −2

130 The matrix for φ is,   1 0 0    −1   0 d 0  . 0 0 1

−1 Hence φ(DR) = d DR. The inverse of a non-square is a non-square, hence the deriva- −1 −1 tion set d DR is of the form nDR, where n = d . From Theorem 5.5.3, the deriva- 2 tion sets of the form nDR, where n is a non-square in GF (q ), are an orbit under 2 P ΓL(3, q )C,l∞ acting on derivation sets of l∞.

Hence, from Theorem 5.2.3 and Theorem 5.4.2 part 1(a)(i), if we derive with respect

0 nDR to a derivation set of the form nDR, then the aﬃne pointset C will be an arc in P .

2 2 2 5.6 The conics Cc,d : x0 − cx1 − dx2 = 0 under multiple derivation

2 2 2 In this section, we will consider the family of conics Cc,d : x0 − cx1 − dx2 = 0 where c is a non-square in GF (q) and d is a non-square in GF (q2).

∼ 2 Let P = PG(2, q ) where q is odd and q ≥ 5, also recall DR, the real derivation set on l∞ in P. Recall from Theorem 5.4.2, that the aﬃne points of a conic from the family

DR Cc,d form an arc in the plane P .

Here we look at how conics of Cc,d are affected by multiple derivation on l∞ and whether their affine points are arcs after we perform different examples of multiple derivation. We will assume q is odd and q ≥ 5 for the entire section.

Let C be a conic from the family Cc,d. Firstly, in Section 5.6.1, we introduce some preliminaries. In Section 5.6.2 we consider derivation, where the derivation set contains the two points of the conic C on l∞.

131 In Section 5.6.3, we then consider a double derivation where the two derivation sets are disjoint from the points C ∩ l∞. From this we obtain examples of inherited arcs in non-Desarguesian planes and give an example to verify existence. Building on the work in the previous sections, in Section 5.6.4, we show the existence of complete 24-arcs in certain projective planes of order 25.

5.6.1 Preliminaries

We coordinatise PG(2, q2) as in Example 1.3.2, thus the points of PG(2, q2) on the line 2 at inﬁnity l∞ are represented by {(0, 1, 0)} ∪ {(1, x1, 0)| x1 ∈ GF (q )}.

2 2 2 Let C be the conic with equation x0 − cx1 − dx2 = 0, where c is a non-square in GF (q) and d is a non-square in GF (q2). The conic C has associated matrix,   1 0 0     A =  0 −c 0  . 0 0 −d

The points of C on l∞ are given by,

2 2 1 − cx1 = 0 1 x2 = 1 c 1 x = ±√ . 1 c

√1 That is, the conic C meets l∞ in the two points (1, ± c , 0).

Example 5.6.1 Let P ∼= PG(2, 52) over the ﬁeld GF (52) as deﬁned in Example 1.2.2, 2 2 thus α is a generator of GF (5 ), with α = 4α + 3. Let C be a conic of Cc,d in P with 2 6 2 2 2 equation x0 − α x1 − dx2 = 0, where d is a non-square in GF (5 ). The conic C meets l in the points (1, ± √1 , 0) = (1, ±α21, 0). ∞ α6

Later in this section, we will also use the following lemma.

Lemma 5.6.1 Let c be an element of GF (q2). If c is a non-square in the subﬁeld √ √ GF (q), then ( c)q = − c.

132 √ √ √ √ Proof We have (( c)q)2 = (( c)2)q = c since c ∈ GF (q) and hence ( c)q = ± c. √ √ √ Now, if ( c)q = c then c ∈ GF (q). This implies that c is a square in GF (q), which √ √ is a contradiction. Thus ( c)q = − c.

5.6.2 Derivation with φ(nDR) and φ(sDR).

2 2 2 Let C be a conic from the family Cc,d, thus C has equation x0 − cx1 − dx2 = 0, where c is a non-square in GF (q) and d is a non-square in GF (q2). Let D be a derivation set

√1 that contains the two points (1, ± c , 0) of C on l∞.

We consider derivation with respect to D and under what conditions the affine points of C remain an arc in the Hall plane PD ∼= Hall(q2). Our method will be to find a 2 set of homographies that map C to the conic x0x1 − x2 = 0 and that also fix the line at infinity l∞. We then use our knowledge of derivation sets containing the points of 2 x0x1 − x2 = 0 on l∞ from Theorem 5.5.4.

2 Lemma 5.6.2 A set of homographies of PG(2, q ), where q is odd, that ﬁxes l∞ and 2 2 2 2 maps the conic C1 : x0x1 − x2 = 0 to the conic C : x − cx1 − dx2 = 0, where c is a non-square in GF (q) and d is a non-square in GF (q2), is given by matrices of the form,     1 b 0 1 b 0          √1 − √b 0  or  − √1 √b 0  , (5.18)  c c q   c c q  4b 4b 0 0 ± d 0 0 ± d where b is a non-square in GF (q2).

Proof Suppose C has associated matrix A and C1 has associated matrix A1. Let φ be 2 −1 a homography of PG(2, q ) such that φ(C1) = C, then φ (C) = C1. By Theorem 1.7.7, −1 t the conic φ (C) has associated matrix φ Aφ = A1. Hence to solve for the matrix of φ we solve using the image of φ−1 as shown in Figure 5.3.

133 (φ−1)−1 = φ

l∞

2 2 2 2 C1 : x0x1 − x2 = 0 C : x0 − cx1 − dx2 = 0 φ−1

Figure 5.3: The action of φ.

First, we need to ensure that l∞ is ﬁxed. We begin with φ as the matrix of a general homography that ﬁxes l∞ (from Example 1.4.1),   a b g      e f h  , 0 0 i where a, b, e, f, g, h, i ∈ GF (q2) and i 6= 0, such that the matrix is non-singular.

2 To ﬁnd the homographies that map C to C1 : x0x1 − x2 = 0, we solve for the matrices −1 −t −1 −1 (φ ) A(φ ) = A1,         a e 0 1 0 0 a b g 0 δ 0                  b f 0   0 −c 0   e f h  =  δ 0 0  , g h i 0 0 −d 0 0 i 0 0 −2δ where δ ∈ GF (q2)\{0}. We proceed by evaluating coeﬃcients,       a −ce 0 a b g 0 δ 0              b −cf 0   e f h  =  δ 0 0  , g −ch −id 0 0 i 0 0 −2δ     a2 − ce2 ba − cef ga − ceh 0 δ 0      2 2     ba − cfe b − cf bg − cfh  =  δ 0 0  . ga − ceh bg − cfh g2 − ch2 − di2 0 0 −2δ

134 Suppose a = 0, then solving for the coeﬃcients gives ce2 = 0 which implies e = 0. We also have −cef = δ, but δ 6= 0, hence there is a contradiction.

Hence, since the equations are homogeneous, suppose a = 1, then solving for coeﬃcients gives, 1 1 − ce2 = 0 ⇒ 1 = ce2 ⇒ e = ±√ c b b2 − cf 2 = 0 ⇒ b2 = cf 2 ⇒ f = ±√ c 1 b b − cef = δ ⇒ b − c(±√ )(±√ ) = δ. (5.19) c c

b There are two possibilities for (5.19), either b − c( c ) = δ and hence δ = 0 which is a b contradiction or b + c c = δ which implies 2b = δ.

Thus, 1 b 1 b e = √ , f = −√ or e = −√ , f = √ . (5.20) c c c c Solving for g and h gives,

g − ceh = 0 ⇒ g = ceh, (5.21) bg − cfh = 0 ⇒ bg = cfh. (5.22)

cfh f Hence in (5.21) and (5.22), if h 6= 0 then ceh = b which implies e = b . This is a contradiction from (5.20) and thus h = 0 and hence g = 0. Also, r 2δ g2 − ch2 − di2 = −2δ ⇒ di2 = 2δ ⇒ i = ± . d

Thus, the matrix of φ = (φ−1)−1 is either,     1 b 0 1 b 0          √1 − √b 0  or  − √1 √b 0  .  c c q   c c q  4b 4b 0 0 ± d 0 0 ± d .

4b 2 2 From this, we see that d is a square of GF (q ). Since d is a non-square in GF (q ), then 4b is a non-square in GF (q2), which implies that b is a non-square of GF (q2). For these matrices to be a non-singular, we need b 6= 0 and b 6= −b which are both true.

135 Using the homographies from Lemma 5.6.2, we can now use our knowledge of conics and derivation with respect to DR.

Theorem 5.6.3 Suppose P ∼= PG(2, q2), where q is odd and q ≥ 5. Suppose C is the 2 2 2 conic given by the equation x0 − cx1 − dx2 = 0, where c is a non-square in GF (q) and d is a non-square in GF (q2). Let φ be a homography given by a matrix as in (5.18) of Lemma 5.6.2.

1. If we perform a derivation with the derivation set φ(sDR), where s is a non-zero square in GF (q2), then the pointset C0 is not an arc in the plane Pφ(sDR).

0 2. If we perform a derivation with the derivation set φ(nDR) then the pointset C forms a complete (q2 − 1)-arc in the plane Pφ(nDR)

−1 2 Proof The conic φ (C) has equation x0x1 −x2 = 0 as in Lemma 5.6.2 and Figure 5.3.

The conic C and the derivation sets φ(sDR) and φ(nDR) are projectively equivalent −1 to the conic φ (C) and the derivation sets sDR and nDR respectively. Hence from

Theorem 5.2.3 and Theorem 5.5.4, if we perform a derivation with φ(sDR), then the pointset C0 is not an arc in the plane Pφ(sDR). Likewise, if we perform a derivation with 0 2 the derivation set φ(nDR), then the pointset C forms a complete (q − 1)-arc after derivation.

We will use Theorem 5.6.3 later in Section 5.6.4 where we compute an example of double derivation. We ﬁnish this section with an example of Theorem 5.6.3 use.

√1 21 Example 5.6.2 Let C be deﬁned as in Example 5.6.1 and hence c = α . Let φ be a 1 homography with matrix as in (5.18) of Lemma 5.6.2, with b = α and d = 4α . Thus √b 21 22 10 − c = −αα . By Example 1.2.2, we have −α = α . Hence φ has matrix,   1 α 0    21 10   α α 0  . 0 0 1

Hence, if we derive with respect to the derivation set φ(DR), then the aﬃne points of

φ(DR) C will not be an arc in P . However, if we derive with respect to φ(αDR), then the aﬃne points of C will be a complete (q2 − 1)-arc in Pφ(αDR).

136 σ

DR σ(DR) l∞

2 2 2 C : x0 − cx1 − dx2 = 0

Figure 5.4: Action of σ.

5.6.3 Double derivation with DR and σ(DR).

One method for finding multiple derivation sets that will leave the affine points of a conic C ∈ Cc,d as an arc is to compute a group of collineations that fixes C and apply these collineations to DR to form new derivation sets. As long as the resulting derivation sets are disjoint, we will have the required multiple derivation.

We proceed by computing a group of homographies that ﬁxes C and l∞. The action of such a homography is shown in Figure 5.4.

∼ 2 Lemma 5.6.4 Let P = PG(2, q ), where q is odd. Let C be a general conic in Cc,d, 2 2 2 thus C has equation x0 − cx1 − dx2 = 0, where c is a non-square in GF (q) and d is a 2 non-square in GF (q ), A group of homographies of P that ﬁxes C and l∞ is given by matrices of the form,       1 ce 0 1 −ce 0 0 ±c 0              e 1 0  or  e −1 0  or  1 0 0  , 0 0 i 0 0 i 0 0 i for e ∈ GF (q2), i 6= 0 and ce2 6= 1.

Proof We begin with the general matrix,   a b 0      e f 0  , 0 0 i

137 2 where a, b, f, i ∈ GF (q ), from a subgroup of the homographies that fix l∞ and find when such a homography fixes the conic C. This occurs when,         a e 0 1 0 0 a b 0 δ 0 0                  b f 0   0 −c 0   e f 0  =  0 −cδ 0  , 0 0 i 0 0 −d 0 0 i 0 0 −dδ       a −ce 0 a b 0 δ 0 0              b −cf 0   e f 0  =  0 −cδ 0  , 0 0 −di 0 0 i 0 0 −dδ     a2 − ce2 ba − cef 0 δ 0 0      2 2     ba − cfe b − cf 0  =  0 −cδ 0  , 0 0 −di2 0 0 −dδ where δ ∈ GF (q2)\{0} and a, b, e, f, i ∈ GF (q2).

Suppose a = 0, then equating coeﬃcients gives,

−ce2 = δ, (5.23) −cef = 0, (5.24) b2 − cf 2 = −cδ. (5.25)

From (5.23), we have c, e 6= 0. Hence from (5.24), we have f = 0. Also, from (5.25), we see that b2 = −cδ. Thus, since the equations are homogeneous, let e = 1. This gives −c = δ in (5.23) and substituting back into (5.25) gives b2 = δ2 which implies b = ±c. If i = 0, then the matrix would have zero determinant, thus i 6= 0.

Hence when a = 0, the matrix of our general homography is,   0 ±c 0      1 0 0  , 0 0 i where i ∈ GF (q2)\{0}.

138 Now suppose a 6= 0, then since the equations are homogeneous, let a = 1. This gives the conditions,

1 − ce2 = δ, (5.26) b − cef = 0, (5.27) b2 − cf 2 = −cδ, (5.28) −di2 = −dδ. (5.29)

From (5.26), we have 1 − ce2 = δ. From (5.27), we have b − cef = 0 hence b = cef. Now from (5.28),

(cef)2 − cf 2 = −cδ ⇒ f 2 − ce2f 2 = f 2(1 − ce2) = δ.

Thus, from (5.26), we have f 2(δ) = δ and hence f = ±1. Further, from (5.29) we have i2 = δ which implies i 6= 0 as δ 6= 0.

Thus, for a = 1, the matrix of our general homography is,     1 ce 0 1 −ce 0          e 1 0  or  e −1 0  , 0 0 i 0 0 i

2 1 for e, δ ∈ GF (q ). For these matrices to be non-singular, we need δ 6= 0 and e 6= ce. Hence ce2 6= 1.

Now using these matrices, we construct derivation sets that are disjoint from DR. This will permit us to perform a double derivation involving DR.

Lemma 5.6.5 Let σ be a homography of PG(2, q2), where q is odd, with associated matrix of the form,     1 ce 0 1 −ce 0          e 1 0  or  e −1 0  , (5.30) 0 0 i 0 0 i

2 2 where e ∈ GF (q )\GF (q), i 6= 0, ce 6= 1 and c is a non-square in GF (q). Then σ(DR) is disjoint from DR.

139 Proof Suppose σ has associated matrix,   1 ce 0      e 1 0  , (5.31) 0 0 i then the derivation set σ(DR) is given by the points,         1 ce 0 0 ce 1             =   =  1  ,  e 1 0   1   1   ce  0 0 i 0 0 0         1 ce 0 1 1 + cex 1             =   =  e+x  ,  e 1 0   x   e + x   1+cex  0 0 i 0 0 0 where x ∈ GF (q).

1 For the derivation set σ(DR) to be disjoint from DR, we need ce ∈/ GF (q) and also e+x 1+cex ∈/ GF (q), for all x ∈ GF (q).

1 e+x Since e∈ / GF (q) and c ∈ GF (q). then ce ∈/ GF (q). Also 1+cex ∈/ GF (q), for all x ∈ GF (q), is equivalent to, e + x q e + x 6= , (5.32) 1 + cex 1 + cex for all x ∈ GF (q). This condition holds when,

(e + x)q(1 + cex) 6= (e + x)(1 + cex)q, (eq + x)(1 + cex) 6= (e + x)(1 + ceqx), since c, x ∈ GF (q), eq + ceq+1x + x + cex2 6= e + ceq+1x + x + ceqx2, eq + cex2 6= e + ceqx2, eq − e 6= ceqx2 − cex2, eq − e 6= x2, ceq − ce 1 6= x2. c

140 1 Thus, the condition (5.32) is equivalent to c being a non-square in GF (q). Now c is 1 a non-square in GF (q) and hence c is a non-square in GF (q). Thus the condition is always satisﬁed.

Suppose σ has associated matrix,   1 −ce 0      e −1 0  , 0 0 i then the derivation set σ(DR) is given by the points,         1 −ce 0 0 −ce 1             =   =  1  ,  e −1 0   1   −1   ce  0 0 i 0 0 0         1 −ce 0 1 1 − cex 1             =   =  e−x  ,  e −1 0   x   e − x   1−cex  0 0 i 0 0 0 where x ∈ GF (q).

1 For the derivation set σ(DR) to be disjoint from DR, we need ce ∈/ GF (q) and also e−x 1−cex ∈/ GF (q), for all x ∈ GF (q).

1 e−x From above, we know that ce ∈/ GF (q). The condition 1−cex ∈/ GF (q), for all x ∈ GF (q), is equivalent to, e − x q e − x 6= , (5.33) 1 − cex 1 − cex for all x ∈ GF (q). This condition holds when,

(e − x)q(1 − cex) 6= (e − x)(1 − cex)q, (eq − x)(1 − cex) 6= (e − x)(1 − ceqx), since c, x ∈ GF (q), eq − ceq+1x − x + cex2 6= e − ceq+1x − x + ceqx2, eq + cex2 6= e + ceqx2, eq − e 6= ceqx2 − cex2, eq − e 6= x2, ceq − ce 1 6= x2. c

141 Thus, the condition (5.33) is equivalent to (5.32) above, and thus is always satisﬁed.

Suppose we apply homographies, with associated matrices from Lemma 5.6.5, to the derivation set DR. Our ﬁrst theorem in this section shows how derivation with the image of DR aﬀect conics from the family Cc,d.

Theorem 5.6.6 Let P ∼= PG(2, q2), where q is odd and q ≥ 5. Let C be a conic of

Cc,d. Let σ be deﬁned as in (5.30) of Lemma 5.6.5. Then, if we perform a derivation

0 2 σ(DR) with respect to the derivation set σ(DR), the pointset C is a (q − 1)-arc in P . Furthermore, the pointset C0 can be completed to a (q2 + 1)-arc in Pσ(DR).

Proof The pointset C0 is a (q2 − 1)-arc in PDR by Theorem 5.4.2. Since σ ﬁxes C, from Theorem 5.2.3, the pointset C0 is a (q2 − 1)-arc in Pσ(DR).

Since C is disjoint from σ(DR), by Theorem 5.2.2, the points C ∩ l∞ can be added to C0 to form a (q2 + 1)-arc in Pσ(DR).

Our main theorem in this section now follows. We bring together the other lemmas and theorem to show the existence of (q2 + 1)-arcs in projective planes of odd order formed by a particular double derivation.

∼ 2 Theorem 5.6.7 Let P = PG(2, q ), where q is odd and q ≥ 5. Let C be a conic of Cc,d. Let σ be deﬁned as in (5.30) of Lemma 5.6.5. Then, if we perform a double derivation 00 2 with respect to the derivation sets DR and σ(DR), the pointset C is a (q − 1)-arc in PDRσ(DR). Furthermore, the pointset C00 can be completed to a (q2 +1)-arc in PDRσ(DR).

Proof The pointset C0 is a (q2 − 1)-arc in PDR by Theorem 5.4.2. The pointset C0 is also a (q2 − 1)-arc in Pσ(DR), from Theorem 5.6.6. By Lemma 5.6.5, we may perform a double derivation with the derivation sets DR and σ(DR). Thus, from Theorem 5.2.4, the pointset C00 is a (q2 − 1)-arc in PDRσ(DR).

Since C is disjoint from DR and σ(DR), by Theorem 5.2.2, the points C ∩ l∞ can be added to C00 to form a (q2 + 1)-arc in PDRσ(DR).

142 Example 5.6.3 Let P ∼= PG(2, 52) over the field GF (52) as defined in Example 1.2.2. Let C be defined as in Example 5.6.1. Let σ be defined as in (5.31) of Lemma 5.6.5 with e = α and i = 1. Thus ce = α6α = α7. Also, we have ce2 = α8 6= 1, hence the matrix for σ is non-degenerate.

Thus σ has matrix of the form,   1 α7 0      α 1 0  . 0 0 1

Hence, by Theorem 5.6.7, if we perform a double derivation with respect to the deriva- 00 2 tion sets DR and σ(DR), then the pointset C can be completed to a (q + 1)-arc in PDRσ(DR).

5.6.4 Double derivation with φ(nDR) and σ(DR).

In Theorem 5.6.7, we use homographies ﬁxing a conic C ∈ Cc,d to ﬁnd derivation sets together with DR that give rise to inherited arcs after double derivation. One may wish to perform a similar operation with the derivation sets and conics described in Theorem 5.6.3. The following lemma shows that it is not possible since the resulting derivation sets can never be disjoint.

Lemma 5.6.8 Let P ∼= PG(2, q2), where q is odd and q ≥ 5. Let φ be a homography with associated matrix as in (5.18) of Lemma 5.6.2. Let n be a non-square in GF (q2).

The derivation sets DR and φ(nDR) are never disjoint.

Proof Recall that the derivation set nDR = {(0, 1, 0)}∪{(1, nx, 0)| x ∈ GF (q), where 2 n is a non-square in GF (q ). We proceed by determining when DR and φ(nDR) are disjoint. Let φ be given by,   1 b 0      √1 − √b 0  ,  c c q  4b 0 0 ± d

143 where b and d are non-squares in GF (q2) and c is a non-square in GF (q). The other case for φ in Lemma 5.6.2 will follow similarly.

√1 The homography φ maps the points (1, 0, 0) and (0, 1, 0) to the points (1, ± c , 0). Since c is a non-square in GF (q), these sets of points are disjoint. Hence we now only consider the other points of φ(nDR).

If we apply φ to a derivation set nDR we have,       1 b 0 1   1        √1 − √b nx   √1 − √b 0   nx  =  c c  ,  c c q     1+bnx  4b 0 0 ± d 0 0 where x ∈ GF (q)\{0}.

If the derivation sets φ(nDR) and DR are disjoint, then for all x ∈ GF (q)\{0}, the 1−√bnx c equivalent condition is 1+bnx ∈/ GF (q). In other words, 1 − bnx q 1 − bnx √ 6= √ , (5.34) c(1 + bnx) c(1 + bnx) for all x ∈ GF (q)\{0}.

From (5.34) we have, √ √ (1 − bnx)q c(1 + bnx) 6= (1 − bnx)( c(1 + bnx))q, √ √ (1 − bqnqx) c(1 + bnx) 6= (1 − bnx)( c)q(1 + bqnqx), since x ∈ GF (q), √ √ (1 + bnx − bqnqx − bq+1nq+1x2) c 6= (1 + bqnqx − bnx − bq+1nq+1x2)( c)q. √ √ From Lemma 5.6.1, we have ( c)q = − c, thus √ √ (1 + bnx − bqnqx − bq+1nq+1x2) c 6= (1 + bqnqx − bnx − bq+1nq+1x2)(− c), √ √ (1 + bnx − bqnqx − bq+1nq+1x2) c 6= (−1 − bqnqx + bnx + bq+1nq+1x2) c, 1 − bq+1nq+1x2 6= −1 + bq+1nq+1x2, 2 6= 2bq+1nq+1x2, 1 6= bq+1nq+1x2, 1 6= x2. bq+1nq+1

144 1 2 Now x ∈ GF (q), thus the condition (5.34) is equivalent to bq+1nq+1 ∈ GF (q ) but not a square in GF (q).

Using Lemma 1.2.2, both bq+1 and nq+1 are non-square elements of GF (q). A non- square multiplied by a non-square is a square, hence bq+1nq+1 is a square in GF (q). 1 The inverse of a square is a square, thus bq+1nq+1 is a square in GF (q). Hence the condition for DR and φ(nDR) to be disjoint is never satisﬁed.

However, we can ﬁnd an example of a double derivation involving φ(nDR). The following lemma shows an example of when such a derivation set is disjoint from the derivation set σ(DR) in Lemma 5.6.5.

Lemma 5.6.9 Suppose P ∼= PG(2, 52) over the field GF (52) defined as in Example 1.2.2. Let φ be defined as in Example 5.6.2. Let σ be defined as in Example 5.6.3.

Then the derivation set φ(αDR) is disjoint from σ(DR).

Proof We proceed by calculating all the points contained in the two derivation sets.

The points of φ(αDR) are given by,         1 α 0 0 α 1          21 10     10   9   α α 0   1  =  α  =  α  , 0 0 1 0 0 0

        1 α 0 1 1 + α2x 1          21 10    =  21 11  =  α21+α11x  ,  α α 0   αx   α + α x   1+α2x  0 0 1 0 0 0 where x ∈ GF (q).

145 Evaluating, for each x ∈ GF (q) gives, α21 + α11α6 α21 + α17 α21 + α11α12 α21 + α23 = = α24 + α2α6 α24 + α8 α24 + α2α12 α24 + α14 α4 α2 = = α24, = = α11, α4 α15

α21 + α11α18 α21 + α5 α21 + α11α24 α21 + α11 = = α24 + α2α18 α24 + α20 α24 + α2α24 α24 + α2 α α12 = = α18, = = α7, α7 α5

α21 + α11.0 α21 = α24 + α2.0 α24 = α21.

The points of σ(DR) are given by,         1 α7 0 0 α7 1              24   17   α 1 0   1  =  α  =  α  , 0 0 1 0 0 0         1 α7 0 1 1 + α7x 1             =   =  α+x  ,  α 1 0   x   α + x   1+α7x  0 0 1 0 0 0 where x ∈ GF (q).

Evaluating, for each x ∈ GF (q) gives, α + α6 α + α6 α + α12 α + α12 = = α24 + α7α6 α24 + α13 α24 + α7α12 α24 + α19 α14 α10 = = α16, = = α2, α22 α8

α + α18 α + α18 α + α24 α + α24 = = α24 + α7α18 α24 + α α24 + α7α24 α24 + α7 α15 α17 = = α22, = = α20, α17 α21

α + 0 α = α24 + α7.0 α24 = α.

146 Thus we have,

17 16 2 22 20 σ(DR) = {(1, α , 0), (1, α, 0), (1, α , 0), (1, α , 0), (1, α , 0), (1, α , 0)},

9 24 11 18 7 21 φ(αDR) = {(1, α , 0), (1, α , 0), (1, α , 0), (1, α , 0), (1, α , 0), (1, α , 0)}.

The two derivation sets have no point in common, hence they are disjoint.

Using this lemma and our previous results on the derivation sets αDR and σ(DR), we can now ﬁnd a family of complete 24-arcs in the plane Pφ(αDR)σ(DR).

Theorem 5.6.10 Suppose P ∼= PG(2, 52) over the field GF (52) defined as in Example 1.2.2. Let φ be defined as in Example 5.6.2. Let σ be defined as in Example 5.6.3. Let C be a conic of Cc,d. Then, if we perform a double derivation with respect to the derivation

00 φ(αDR)σ(DR) sets φ(αDR) and σ(DR), the pointset C is a complete 24-arc in P

Proof From Theorem 5.6.3, the pointset C0 is a complete 24-arc in Pφ(αDR). Also, by Theorem 5.6.6, the pointset C0 is a 24-arc in Pσ(DR). From Lemma 5.6.9, we may perform a double derivation with the derivation sets φ(αDR) and σ(DR).

Hence, if we perform a double derivation with respect to the derivation sets φ(αDR)

00 φ(αDR)σ(DR) and σ(DR), by Theorem 5.2.4, the pointset C is a complete 24-arc in P .

5.7 Inherited arcs in Andr´eplanes of odd order

Here we prove the existence of a family of inherited arcs in a class of Andr´eplanes of odd square order. We do this by bringing together many of the ideas in the previous sections.

The Andréplanes are a class of non-Desarguesian projective planes. We do not give a general definition here, however we will define a class of planes that are known to be Andréplanes. The following definitions can be found in Hughes & Piper [35].

147 ∼ 2 Let P = P G(2, q ). Deﬁne the following derivation sets on l∞ in P,

2 q+1 Dt = {(1, x1, 0)| x1 ∈ GF (q ) and x1 = t}, where t ∈ GF (q)\{0}.

The derivation sets Dt are pairwise disjoint for distinct choices of t. They partition the points of l∞\{(0, 1, 0), (1, 0, 0)} into q − 1 sets of q + 1 points. The points (0, 1, 0) and

(1, 0, 0) are not contained in Dt, for any t.

Dt ,...,Dt It is known that any plane of the form P 1 k , where t1, . . . , tk ∈ GF (q)\{0} and k ∈ {1, . . . , q − 1}, is an Andr´eplane. In particular, if {t1, . . . , tk} is precisely the set D ,...,D of squares of GF (q), then P t1 tk is known as a regular Nearﬁeld plane.

We begin by deﬁning a homography which we will use to exploit projective equivalence.

∼ 2 Lemma 5.7.1 Let P = PG(2, q ), where q is odd and q ≥ 5. Let C1 be a conic of Cc,d in P, where c is a non-square in GF (q) and d is a non-square in GF (q2). Let γ be the homography given by the matrix,  √  1 c 0    √   1 − c 0  . (5.35) 0 0 1

2 Then C = γ(C1) is the conic with equation x0x1 − dx2 = 0.

Proof Let A be the associated matrix of C. The equation of γ(C1) is given by γ−tAγ−1 = 0. The inverse of the homography γ is,   1 1 0  2 2   √1 √1  ,  2 c − 2 c 0  0 0 1 since,

     √ √    1 1 √ 1 1 c c 2 2 0 1 c 0 2 + 2 − 2 + 2 0 1 0 0    √       √1 √1    =  √1 √1 1 1  =   .  2 c − 2 c 0   1 − c 0   2 c − 2 c 2 + 2 0   0 1 0  0 0 1 0 0 1 0 0 1 0 0 1

148 Thus γ−tAγ−1 is,       1 √1 0 1 0 0 1 1 0  2 2 c     2 2   1 √1     √1 √1   2 − 2 c 0   0 −c 0   2 c − 2 c 0  0 0 1 0 0 −d 0 0 1       1 − √c 0 1 1 0 1 − c 1 + c 0  2 2 c   2 2   4 4c 4 4c  =  1 √c   √1 √1  =  1 c 1 c   2 2 c 0   2 c − 2 c 0   4 + 4c 4 − 4c 0  0 0 −d 0 0 1 0 0 −d   0 1 0  2  =  1  .  2 0 0  0 0 −d

2 This is the associated matrix for the conic with equation x0x1 − dx2 = 0.

2 Next, we verify a group of homographies that ﬁx the conic x0x1 − dx2 = 0. We will use these homographies to construct derivation sets in a later lemma.

Lemma 5.7.2 Let P ∼= PG(2, q2), where q is odd and q ≥ 5. Let C be a conic of 2 2 P with equation x0x1 − dx2 = 0, where d is a non-square in GF (q ). A group of homographies that ﬁx the conic C and the line l∞ is given by matrices of the form,   1 0 0      0 δ 0  , (5.36) √ 0 0 ± δ where δ is a non-zero square in GF (q2).

Proof Suppose C has associated matrix A. Let ζ be a homography with matrix as in (5.36), then ζ(C) has associated matrix ζ−tAζ−1. For a given ζ, since the homographies form a group, the inverse ζ−1 is a general element of the same form, hence without loss of generality, let ζ−1 be given by,   1 0 0      0 δ 0  , √ 0 0 ± δ

149 where δ is a non-zero square in GF (q2).

Thus ζ−tAζ−1 is,       1 0 0 0 1 0 1 0 0    2       1     0 δ 0   2 0 0   0 δ 0  √ √ 0 0 ± δ 0 0 −d 0 0 ± δ   0 δ 0  2  =  δ  .  2 0 0  0 0 −dδ

Hence ζ(C) = C.

Lemma 5.7.3 Let γ be a homography deﬁned as in (5.35) in Lemma 5.7.1. The derivation set γ(DR) is the derivation set,

2 q+1 D1 = {(1, x1, 0)| x1 ∈ GF (q ) and x1 = 1}.

Proof First calculate the derivation set γ(DR),  √     √    1 c 0 1 1 + x c 1        √  √ √ 1−x c     =   =  √  ,  1 − c 0   x   1 − x c   1+x c  0 0 1 0 0 0  √     √    1 c 0 0 c 1          √     √     1 − c 0   1  =  − c  =  −1  , 0 0 1 0 0 0 where x ∈ GF (q). Hence, √ 1 − x c γ(D ) = {(1, √ , 0)| x ∈ GF (q)} ∪ {(1, −1, 0)}. R 1 + x c

Now, √ √ √ (1 − x c)q+1 = (1 − x c)q(1 − x c) √ √ = (1q − xq cq)(1 − x c) √ √ = (1 − x cq)(1 − x c), since x ∈ GF (q) √ √ = (1 + x c)(1 − x c), by Lemma 5.6.1.

150 √ √ √ Similarly, we have (1 + x c)q+1 = (1 − x c)(1 + x c). Hence, √ 1 − cx q+1 √ = 1. 1 + cx

√ 1−x c √ q Thus (1, 1+x c , 0) ∈ D1, for all x ∈ GF (q). Now −1 = −1 as −1 ∈ GF (q), hence q+1 −1 = 1 and thus (1, −1, 0) ∈ D1. Hence γ(DR) = D1.

We can now show the existence of an inherited arc in an Andr´eplane formed by multiple 2 2 derivation. The conic of PG(2, q ) with equation x0x1 − dx2 = 0 is the conic that is inherited.

∼ 2 Theorem 5.7.4 Suppose P = PG(2, q ), where q is odd and q ≥ 5. Let C1 be the 2 2 2 conic given by the equation x0 − cx1 − dx2 = 0, where c is a non-square in GF (q) and d is a non-square in GF (q2). Let γ be a homography deﬁned as in (5.35) in Lemma 2 5.7.1, hence C = γ(C1) is the conic with equation x0x1 − dx2 = 0. Then, if we perform 0 2 a derivation with respect to the derivation set D1, the aﬃne pointset C is a (q −1)-arc in the plane PD1 and can be completed to a (q2 + 1)-arc in PD1 .

Proof Using Theorem 5.4.2, if we derive with respect to DR, then the aﬃne pointset

0 2 DR C1 is a (q − 1)-arc in the plane P . Hence, from Theorem 5.2.3, if we derive with

0 2 γ(DR) respect to γ(DR) then the aﬃne pointset C is a (q − 1)-arc in the plane P .

γ(DR) D1 From Lemma 5.7.3, the derivation set γ(DR) = D1. Hence P = P .

2 Here we use the homographies that ﬁx the conic with equation x0x1 − dx2 = 0. We calculate the orbit of these homographies on the derivation set D1.

Lemma 5.7.5 Let ζ be a homography deﬁned as in (5.36) in Lemma 5.7.2, where δ is 2 non-zero square in GF (q ). The derivation set ζ(D1) is given by the points,

0 0 2 0 q+1 Ds = {(1, x1, 0)| x1 ∈ GF (q ) and (x1) = s}, where s = δq+1 is a non-zero square in GF (q).

151 Proof The derivation set D1 is given by the points

2 q+1 {(1, x1, 0)| x1 ∈ GF (q ) and x1 = 1}.

Hence, 2 q+1 ζ(D1) = {(1, δx1, 0)| x1 ∈ GF (q ) and x1 = 1}, where δ is a non-zero square in GF (q2).

q+1 q+1 q+1 q+1 Now (δx1) = δ , since x1 = 1. From Lemma 1.2.2, we know δ is a square in GF (q), since δ is a square in GF (q2). Hence,

0 0 2 0 q+1 ζ(D1) = {(1, x1, 0)| x1 ∈ GF (q ) and (x1) = s},

q+1 where s = δ is a non-zero square in GF (q). Hence ζ(D1) = Ds.

If we put all this together, we can now prove the existence of inherited arcs in a class of Andr´eplanes of odd square order.

Theorem 5.7.6 Suppose P ∼= PG(2, q2), where q is odd and q ≥ 5. Let C be the conic 2 with equation x0x1 − dx2 = 0. Suppose we derive with respect to the derivation sets

Ds1 ,..., Dsk , where s1, . . . , sk are non-zero squares in GF (q). If we perform a multiple 0 derivation with respect to the derivation sets Ds1 ,..., Dsk , then the aﬃne pointset C 2 D ,...,D 0 is a (q − 1)-arc in the plane P s1 sk . Furthermore, the aﬃne pointset C can be 2 D ,...,D completed to a (q + 1)-arc in P s1 sk .

Proof Let α be a generator of GF (q2). Recall from Section 1.2, that GF (q) as a subﬁeld of GF (q2) is {αq+1, α2(q+1), . . . , α(q−1)(q+1)}. Hence, if s ∈ GF (q), there exists δ ∈ GF (q2) such that s = δq+1. Note that since s is a square in GF (q), then 2i(q+1) q−1 2 δ = α , for some i = 1,..., 2 . Thus δ is a square in GF (q ).

q+1 Let ζ be a homography deﬁned as in (5.36) in Lemma 5.7.2, where δ = s1 as discussed. From Lemma 5.7.5, the derivation set ζ(D1) is the derivation set Ds1 .

From Lemma 5.7.2, the homography ζ ﬁxes C. Thus, using Theorem 5.2.3 and Theorem 0 2 D 5.7.4, the aﬃne pointset C is a (q − 1)-arc in the plane P s1 . A similar argument holds, for each of the derivation sets Ds2 ,..., Dsk .

152 The derivation sets Ds1 ,..., Dsk are pairwise disjoint. Thus, if we perform a multiple derivation with respect to the derivation sets Ds1 ,..., Dsk , using Theorem 5.2.4, the 0 2 D ,...,D aﬃne pointset C is a (q − 1)-arc in the plane P s1 sk .

Further, the derivation sets Ds1 ,..., Dsk are disjoint from the conic C, hence by The- orem 5.2.2, the aﬃne pointset C0 can be completed to a (q2 + 1)-arc in the plane D ,...,D P s1 sk .

There have been previous discoveries of inherited ovals in Andréplanes of odd order. Rosati [49] discovered a family of ovals in all nearfield planes of odd order known as the Rosati Ovals. Later, these ovals were explored in more detail in [52] and [38]. We omit the details of the construction, however it was shown that the Rosati Ovals share two points with l∞. A special case of Theorem 5.7.6 gives ovals in the regular Nearfield planes of odd order. It is an open question as to whether the ovals in this case are Rosati Ovals or not.

Szõnyi [53] also discovered inherited ovals in a class of Andréplanes of odd order. The inherited ovals have the same equation in PG(2, q2) as those in Theorem 5.7.6. The class of Andréplanes featured in [53] may contain the class of Andréplanes in Theorem 5.7.6, although the technique used here is different. The author was not aware of the ovals constructed in [53] until after this work had been completed.

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