In this post I wish to discuss flock generalised quadrangles. As mentioned in the first of this series, John has already discussed these a bit in a previous post so my main aim will be to flesh that out and provide more background. I have relied heavily on Maska Law’s PhD thesis which is available from Ghent’s PhD theses in finite geometry page.

Recall that a conic ${\mathcal{C}}$ is the set of zeros of a nondegenerate quadratic form on ${\mathrm{PG}(2,q)}$. Embed ${\mathrm{PG}(2,q)}$ as a hyperplane ${\pi}$ in ${\mathrm{PG}(3,q)}$ and take a point ${P}$ not on ${\pi}$. For each of the ${q+1}$ points of ${\mathcal{C}}$ there is a unique line through such a point and ${P}$. Let ${\mathcal{K}}$ be the set of all ${q^2+q+1}$ points on these ${q+1}$ lines. The set ${\mathcal{K}}$ is called a quadratic cone with vertex ${P}$. Now ${\mathsf{P}\Gamma\mathsf{L}(4,q)}$ acts transitively on the set of pairs ${(P,\pi)}$ of points ${P}$ and hyperplanes ${\pi}$ of ${\mathrm{PG}(3,q)}$ where ${\pi}$ does not contain ${P}$, and the stabiliser of such a pair induces ${\mathrm{GL}(3,q)}$ on ${\pi}$ and so acts transitively on the set of conics contained in ${\pi}$ . Thus all quadratic cones of ${\mathrm{PG}(3,q)}$ are equivalent.

An easy way to construct a quadratic cone is to take the zeros of the degenerate quadratic form ${Q(x)=x_2^2-x_1x_3}$, where ${x=(x_1,x_2,x_3,x_4)}$. Here we take ${P}$ to be ${\langle (0,0,0,1)\rangle}$ and ${\pi=\langle (1,0,0,0),(0,1,0,0),(0,0,1,0)\rangle}$. The zeros of ${Q}$ on ${\pi}$ form the conic ${\{\langle (1,t,t^2,0)\rangle\mid t\in\mathsf{GF}(q)\}\cup \{\langle (0,0,1,0) \rangle\}}$. Note that for any plane ${\pi'}$ of ${\mathsf{PG}(3,q)}$ the set of zeros of ${Q}$ on ${\pi'}$ forms a conic. This is all reminiscent of the classical case of a cone in ${\mathbb{R}^3}$, where the intersections of a plane with the cone are the conic sections and are either a point, a circle, an ellipse, a parabola or a hyperbola.

A flock of a quadratic cone ${\mathcal{K}}$ with vertex ${P}$ is a partition of ${\mathcal{K}\backslash\{P\}}$ into ${q}$ disjoint conics. Each conic is the intersection of ${\mathcal{K}}$ with a plane. Let ${\ell}$ be a line of ${\mathrm{PG}(3,q)}$ which intersects ${\mathcal{K}}$ trivially. Then ${\ell}$ is contained in ${q+1}$ planes, one of which contains ${P}$. Each of the remaining ${q}$ planes containing ${\ell}$ meets each of the lines which make up ${\mathcal{K}}$ and hence meets ${\mathcal{K}}$ in ${q+1}$ points. Such a set of ${q+1}$ points is a conic. Morever, since the intersection of two planes through ${\ell}$ is ${\ell}$, the ${q}$ conics we obtain are all disjoint and so we get a flock. Such a flock is called a linear flock.

Here is the next instalment in our Phan systems study group which was held yesterday.  I forgot to bring my laptop  so I took notes by hand. Blogging the study group has already had one benefit as Gordon is now attending. Having been initially put off by the title he has realised that we are actually doing lots of stuff that he is interested in learning.

John began the discussion by working through the geometry of the symplectic polar space of rank 2.  Let $J=\left(\begin{array}{cc} 0_2 &I_2\\ -I_2& 0_2\end{array}\right)$ and let $\beta$ be the bilinear form on $V=K^4$, for some field K, given by $\beta(v,w)=vJw^T$. Note we are using row vectors, and if $v=(v_1,v_2,v_3,v_4)$, $w=(w_1,w_2,w_3,w_4)$ we have $\beta(v,w)=v_1w_3-v_3w_1+v_2w_4-v_4w_2$. Then $\beta$ is an alternating form, that is $\beta(v,w)=-\beta(w,v)$ and $\beta(v,v)=0$ for all $v,w\in V$. The totally isotropic subspaces of V, are those subspaces W for which the restriction $\beta_{\mid W}$ is zero, that is, $\beta(v,w)=0$ for all $v,w\in W$. We can then form a point-line geometry whose points are the totally isotropic 1-spaces of V and the lines are the totally isotropic 2-spaces of V.  Note that all 1-spaces of V are totally isotropic, but V does not contain any totally isotropic 3-spaces. A hyperbolic pair is a pair of vectors $(e,f)$ such that $\beta(e,f)=1$.

Let $e_1=(1,0,0,0), e_2=(0,1,0,0), f_1=(0,0,1,0)$ and $f_2=(0,0,0,1)$. Then $(e_1,f_1)$ and $(e_2,f_2)$ are hyperbolic pairs, while $\langle e_1,e_2\rangle,\langle f_1,f_2\rangle,\langle e_1,f_2\rangle$ and $\langle e_2,f_1\rangle$ are totally isotropic. We have $V=\langle e_1,f_1\rangle\perp\langle e_2,f_2\rangle$, that is, V can be written as an orthogonal direct sum of two hyperbolic lines. One of the chambers in this geometry is $\langle e_1\rangle\subset \langle e_1,e_2\rangle$.

The point-line geometry we have just defined is an example of a generalised quadrangle. That is, it satisfies the following two axioms:

• Any two points lie on at most one line.
• Given  a line $\ell$ and a point p not incident with $\ell$, there is a unique point q on $\ell$ which is incident with p.

Such a geometry is often referred to as a geometry of type $B_2$  and is denoted by the diagram