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.