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.

Recently I’ve been working with generalised quadrangles which arise from flocks, and in particular, I use the model introduced by Norbert Knarr in his 1992 paper “A geometric construction of generalized quadrangles from polar spaces of rank three”. However, I’m not going to tell you everything because it would be very long. I’m not going to tell you what a generalised quadrangle is, what a flock is or even why we ought to care about such objects. Instead, my purpose is to de-mystify the Knarr model by explaining where it comes from for the classical object, the simplest case. So really, this post is intended for someone who is already familiar with the topic, but maybe later I or someone else will trace backwards to where all this began.

The three-dimensional finite Hermitian variety $H(3,q^2)$ can be constructed in the following way. Consider the following Hermitian form on a four-dimensional vector space over $GF(q^2)$:

$\langle x,y\rangle :=x_1y_1^q+x_2y_2^q+x_3y_3^q+x_4y_4^q.$

Let the points be the one-dimensional subspaces for which this form restricts to the zero form, and let the lines be the two-dimensional subspaces for which this form computes only zero on them. Then this geometry of points and lines gives us a generalised quadrangle, and it is the classical example in the category of flock quadrangles. In other words, this geometry is a classical polar space of rank 2. Now the vectors of the vector space $GF(q^2)^4$ are certainly in one-to-one correspondence with the vectors of $GF(q)^8$, but in fact, more can be said. We can define a bilinear form over $GF(q)$ by $B(u,v):=\gamma(\langle u,v\rangle -\langle u,v\rangle^q)$, where $\gamma$ is some element of $GF(q^2)$ such that $\gamma=-\gamma^q$, and we see that is alternating since

$B(v,u) = \gamma(\langle v,u\rangle -\langle v,u\rangle^q)=\gamma(\langle u,v\rangle^q -\langle u,v\rangle)=-B(u,v)$.

So this form defines a symplectic space $W(7,q)$ on $GF(q)^8$. What is interesting in this correspondence is that the points of $H(3,q^2)$ go to lines of $W(7,q)$, and the lines of $H(3,q^2)$ go to solids of $W(7,q)$. Now take a point P of $H(3,q^2)$. Then P maps to a line P’ of $W(7,q)$. We then take an arbitrary point X on this line P’ and note that P’ is contained in $X^\perp$. Now we project to the quotient polar space $X^\perp/X$ (which is isomorphic to $W(5,q)$) via the map $U\mapsto (X^\perp\cap \langle X, U\rangle)/X$. So we obtain a map from totally isotropic subspaces of $H(3,q^2)$ to totally isotropic subspaces of $W(5,q)$, and something interesting happens with respect to the point P that we started with:

 $P$ a point R of W(5,q) lines on $P$ a set of t.i. planes $\pi_i$ on R such that the projection to $R^\perp/R$ yields a BLT-set of lines of W(3,q) lines not on $P$ t.i. planes which meet some $\pi_i$ in a line not on R points collinear with $P$ lines of W(5,q) contained in some $\pi_i$ points not-collinear with $P$ points of W(5,q) not in the perp of R.

This is the Knarr model of a flock generalised quadrangle, where the input is a BLT-set of lines of $W(3,q)$ (not just the one obtained from a pencil of lines of $H(3,q^2)$). So essentially the Knarr model of a flock generalised quadrangle is the generalisation of the field reduction of $H(3,q^2)$, where we change the BLT-set in the resulting geometry of $W(5,q)$.