Up to duality, there are two known families of finite (thick) generalised hexagons:

1. the Split Cayley hexagons of order $(q,q)$ related to the Dickson exceptional groups $G_2(q)$.
2. the Twisted Triality hexagons of order $(q,q^3)$ related to the Steinberg exceptional groups $\,^3D_4(q)$.

Since I will only consider generalised hexagons up to duality, I will assume that the order $(s,t)$ of a given one has $s\le t$. The parameter $s$ is one less than the number of points on a line, and $t+1$ is the number of lines on a point. To date, we do not know much about the possible values of these positive integers $s$ and $t$. Here is what we know:

• $s\le t^3$ and $t\le s^3$ (Haemers and Roos, 1981)
• $s^2+st+t^2$ divides $s^3(s^2t^2+st+1)$ (from the multiplicities of the eigenvalues of the point graph).
• There are two (up to duality) generalised hexagons with $s=2$ and they have $t\in\{2,8\}$ respectively (Cohen and Tits 1985).

We do not know that $s+1$ divides $t+1$, even though the known examples satisfy this simple divisibility relation.

Here is a cool number theoretic thing I’ve observed recently, but I can’t (quite yet) prove that it is true:

Claim: If $s,t$ are integers greater than 1, satisfying $s+1 \mid t+1$ and $s^2+st+t^2\mid s^3(s^2t^2+st+1)$, then apart from a small finite number of exceptions, we have $t=s$ or $t=s^3$.

I claim that the only exception is $(s,t)=(14,224)$.

We know that the known examples satisfy the extra relation $s+1\mid t+1$, and it is conceivable that such a simple looking relation holds for generalised hexagons and that it arises for some combinatorial reason. It’s simplicity is the only reason to believe that it might hold, but wouldn’t it be cool if every generalised hexagon satisfied it?

The day started badly. I woke at 3am to a sick and miserable one-year-old, I got stuck in a traffic jam on the way to work, and by lunchtime I was feeling down about the usual whackacademic innuendo. But at 3pm, Eric Swartz knocked on my door. Eric has stunningly proved an outstanding conjecture on generalised quadrangles which I’ll report on when I have his blessing.

Rather than let this blog die off due to lack of posts, I thought I’d write about a small but annoying problem that I’ve been thinking about recently, but without success, and a natural conjecture that has arisen from it.

[Note: On some browsers there appears to be a sizing problem with the images WordPress uses for LaTeX formulas, and they appear ten times the right size. If this happens, reloading the webpage seems to fix it]

The problem arose in the context of binary matroids, but because a binary matroid is really just a set of points in a binary vector space, it can be phrased entirely as a linear algebra problem.

So let ${M}$ be a set of non-zero vectors in the vector space ${V = GF(2)^r}$ such that ${M}$ spans ${V}$ and, for reasons to be clarified later, no vector in ${M}$ is independent of the others. In matroid terminology, this just says that ${M}$ is a simple binary matroid of rank ${r}$ with no coloops.

Then define a  basis of ${M}$ to be a linearly independent subset of ${M}$ of rank ${r}$ (in other words, just a basis of ${V}$ and a circuit of ${M}$ to be a minimally dependent set of vectors, i.e. a set of vectors that is linearly dependent but any proper subset of which is linearly independent.

As an example, take ${M = PG(2,2)}$ (a.k.a the Fano plane) — this means to take all the non-zero vectors in ${GF(2)^3}$. This has 28 bases (7 choices for a first vector ${v}$, 6 for a second vector ${w}$ and then 4 for the third vector which cannot be ${v}$, ${w}$ or ${v+w}$ , then all divided by 6 because this counts each basis ${6}$ times). It has 7 circuits of size 3 (each being of the form ${v}$, ${w}$, ${v+w}$) and 7 circuits of size 4, being the complements of the circuits of size 3, for a total of 14 circuits.

Letting ${b(M)}$, ${c(M)}$ denote the numbers of bases and circuits of ${M}$ respectively, the question is about the ratio of these two numbers. More precisely,

Determine a lower bound, in terms of the rank ${r}$, for the ratio ${\frac{b(M)}{c(M)}}$?

A while ago, I used the mixed-integer linear programming software Gurobi to show that there are no Cameron-Liebler line classes with certain parameters, and recently I’ve looked at a similarly difficult computational problem: packings of 3-dimensional projective space. But first, some motivation.

One of the most confounding and fundamental open problems in finite geometry is the question of classifying the hyperovals of the Desarguesian projective planes. Beyond this problem, there are many other related interesting avenues of research: what about hyperovals of non-Desarguesian planes, maximal arcs or linear codes with minimum distance 3?

An arc of a projective plane is a set of points with no three lying on a common line. For $q$ odd, the largest an arc can be in the Desarguesian projective plane $PG(2,q)$ is $q+1$, an oval, and it is exhibited by a conic: for example, the solutions to the quadratic equation $XY=Z^2$ (where our points are represented in homoegenous coordinates $(X,Y,Z)$). To see that there can be no larger arc, choose a point $P$ not in the arc; if there were exactly $q+2$ points in the arc, then there would be exactly $(q+2)/2$ lines on $P$ meeting the arc, and this number is not an integer for $q$ odd. A bigger arc would have a sub-arc of size $q+2$, so this argument rules out all larger sizes. For $q$ even, we can have $q+2$ points in an arc, a so-called hyperoval. But then, a similar counting argument shows that there is no larger arc. Segre showed that an oval of $PG(2,q)$ for $q$ odd, must be projectively equivalent to a conic. So the ovals here are completely classified. Does a similar result hold for hyperovals when $q$ is even?

Here is another problem in finite geometry that I personally find intriguing, but I’m sure many others in the community find is also one of the big-ish problems in our area. It concerns spreads of Hermitian varieties, and I guess this question goes back to Segre’s 200-page manuscript “Forme e geometrie hermitiane, con particolare riguardo al caso finito” (Forms and hermitian geometries, with particular regard to the finite case).

A Hermitian variety is one of the central objects in finite and algebraic geometry. We begin with a field $F$ having an involutory automorphism $\sigma$, and an Hermitian matrix $A$; so $A^\sigma=A^T$ (the transpose of $A$). Now consider the variety defined by the following equation

$X^\sigma A X^T=0,\quad X=(X_1,X_2,\ldots,X_n).$

So for example, we could have $F=\mathbb{C}$ and $\sigma$ being complex conjugation. We could take $A$ to be the identity matrix and we would obtain the simplest Hermitian variety. We will now consider non-degenerate Hermitian varieties, where $A$ is invertible, and let us suppose now that $F$ is a finite field. So now $F$ has square order $q^2$ and $\sigma$ is nothing other than the map $x\mapsto x^q$. It turns out that for a given dimension and order of the finite field, any pair of non-degenerate Hermitian varieties are isometric; that is, if they are defined by the matrices $A_1$ and $A_2$, then there is an invertible matrix $P$ such that $P^\sigma A_1 P^T=A_2$. So from now on, we will denote the non-degenerate Hermitian variety of $PG(n-1,q^2)$ by $H(n-1,q^2)$.

A unital in the projective plane $PG(2, q^2)$ is a set $U$ of $q^3+1$ points of $PG(2,q^2)$such that every line meets $U$ in 1 or q+1 points. These lines are known as tangent and secant lines respectively. If we take the points of the unital together with the secant lines, we obtain a $3-(q^3+1, q+1, 1)$ design.

### An example of a unital; the Hermitian curve.

Consider the set of points $(x_0,x_1,x_2)$ of $PG(2,q^2)$ which satisfy

$x_0^{q+1}+x_1^{q+1}+x_2^{q+1}=0.$

Then these points form a unital known as a Hermitian curve or classical unital. A classical unital is defined to be a set of points which is the image of this example under the group $P\Gamma L(3,q^2)$. Continue reading “Open problems in Finite Geometry: Unitals of Desarguesian projective planes”

An ovoid of a finite 3-dimensional projective space $PG(3, q)$ is a set of $q^2 + 1$ points such that no three are on a common line. Besides the case $q=2$, the bound $q^2+1$ is the maximum size of a set of points with no three collinear. Ovoids are central to finite geometry, and two examples of their significance are the constructions of inversive planes and generalised quadrangles from ovoids. Peter Dembowski (1963) proved that every inversive plane of even order arises from an ovoid of $PG(3,q)$, $q$ even, and Jacques Tits (see Dembowksi’s book) gave a construction of a generalised quadrangle of order $(q,q^2)$ from an ovoid. Here is a simple example of an ovoid: Continue reading “Open problems in “Finite Geometry”: Ovoids of projective spaces”