As part of John’s goal to turn me into a geometer I have been doing some research lately into generalised quadrangles.  To help me in understanding all the various ways of constructing them, I have decided to write a series of posts outlining the known generalised quadrangles and in particular looking at what their automorphism groups are.

Regular readers of this blog and especially followers of the posts from our Buildings, geometries and algebraic groups study group,  will be familar with the basic definitions and the classical examples. They were first introduced in the second post of that series and generalised polygons were introduced in the eighth. Actually the term “generalised quadrangle’ is one of the more prominent tags in the tag cloud.

Recapping, a generalised quadrangle is a point-line geometry satisfying 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

The second axiom implies that there are no triangles in the geometry, that is, there is no triple of points with any pair collinear. A generalised quadrangle is said to have order $(s,t)$ if each line contains $s+1$ points and each point lies on $t+1$ lines. If $s=t$ we usually just say that the generalised quadrangle has order $s$.  If $s,t\geq 2$, the quadrangle is said to be thick.

Given a generalised quadrangle $\mathcal{Q}$ of order $(s,t)$ we can define a new generalised quadrangle whose points are the lines of $\mathcal{Q}$, whose lines are the points of $\mathcal{Q}$ and incidence is inherited from $\mathcal{Q}$. This new GQ has order $(t,s)$ and is called the dual of $\mathcal{Q}$.

The classical generalised quadrangles come from sesquilinear forms and are as follows:

${\mathsf{W}(3,q)}$:

Let ${V}$ be a 4-dimensional vector space over the field ${\mathrm{GF}(q)}$. Let ${B:V\times V\rightarrow \mathrm{GF}(q)}$ be an alternating form, that is, a bilinear form ${B}$ with ${B(v,w)=-B(w,v)}$ and ${B(v,v)=0}$ for all ${v,w\in V}$. (Note that the second condition follows from the first when ${q}$ is odd.) Further we require that ${B}$ is \emph{nondegenerate}, that is the only element ${w\in V}$ such that ${B(w,v)=0}$ for all ${v\in V}$ is the zero vector. We take ${B}$ to be defined by ${B(u,v)=uJv^T}$ where

$\displaystyle J=\begin{pmatrix} 0&0&0&1\\ 0&0&1&0\\ 0&-1&0&0\\-1&0&0&0\end{pmatrix}$

Any nondegenerate alternating form on ${V}$ is equivalent to ${B}$, that is, if $\beta(u,v)=uJ'v^T$  is another such form then there exists ${A\in\mathrm{GL}(4,q)}$ such that ${AJ'A^T=J}$.

For a subspace ${U}$ of ${V}$ we define ${U^{\perp}}$ to be the subspace ${\{w\in V\mid B(u,w)=0 \text{ for all } u\in U\}}$. Note that the codimension of ${U^{\perp}}$ is equal to the dimension of ${U}$. A subspace ${U}$ of ${V}$ is called totally isotropic if ${B(u,w)=0}$ for all ${u,w\in U}$. Note that ${U}$ is totally isotropic if and only if ${U\leqslant U^{\perp}}$. It follows that the totally isotropic subspaces have dimension at most 2.

Let ${\mathcal{Q}}$ be the point-line incidence geometry whose points are the 1-dimensional subspaces of ${V}$ (note that these are all totally isotropic) and whose lines are the totally isotropic 2-spaces. Incidence is the incidence inherited from the vector space.

The lines incident with a point ${\langle v\rangle}$ are all contained in the 3-space ${\langle v\rangle ^{\perp}}$. The form ${B}$ induces a nondegenerate alternating form ${\overline{B}}$ on the quotient space ${\langle v\rangle ^{\perp}/\langle v\rangle}$ via ${\overline{B}(\langle v\rangle +w,\langle v\rangle+u)=B(w,u)}$. Then the lines incident with ${\langle v\rangle}$ correspond to the totally isotropic 1-spaces of ${\langle v\rangle ^{\perp}/\langle v\rangle}$. Since all 1-spaces of this quotient space are totally isotropic, ${\langle v\rangle}$ lies on ${q+1}$ lines. Moreover, each line contains ${q+1}$ points.

Given a line ${\ell}$ and a point ${\langle v\rangle}$ not on ${\ell}$ we have that ${\langle v\rangle^{\perp}}$ is 3-dimensional and so must meet ${\ell}$ in a 1-space ${\langle w\rangle}$. Thus there is a unique line through ${\langle v\rangle}$ which meets ${\ell}$, namely ${\langle v,w\rangle}$. Hence ${\mathcal{Q}}$ is a generalised quadrangle of order ${(q,q)}$, or more simply of order ${q}$. It is commonly denoted by ${\mathsf{W}(3,q)}$, or sometimes just ${\mathsf{W}(q)}$.

The symplectic group ${\mathrm{Sp}(4,q)}$ is the group of all linear transformations of ${V}$ which preserve ${B}$, that is all ${A\in\mathrm{GL}(4,q)}$ such that ${B(uA,vA)=B(u,v)}$ for all ${u,v\in V}$. This is equivalent to all matrices ${A}$ such that ${AJA^T=J}$. (Note that different choices for the form ${B}$ give conjugate subgroups of ${\mathrm{GL}(4,q)}$ so we don’t use ${B}$ in the notation for the group.) The group ${\mathrm{Sp}(4,q)}$ acts as a group of automorphisms of the GQ. The kernel of this action is the group of all scalar matrices in ${\mathrm{Sp}(4,q)}$. Letting ${Z=\{\lambda I\mid \lambda\in\mathrm{GF}(q)\backslash\{0\}\}}$, we define ${\mathrm{PSp}(4,q)=\mathrm{Sp}(4,q)/(Z\cap \mathrm{Sp}(4,q))}$, which is isomorphic to the permutation group induced on the set of points of the GQ by ${\mathrm{Sp}(4,q)}$.

We can also define ${\mathrm{GSp}(4,q)}$ to be the group of all ${A\in\mathrm{GL}(4,q)}$ such that there exists ${\lambda\in\mathrm{GF}(q)}$ for which ${B(uA,vA)=\lambda B(u,v)}$ for all ${u,v\in V}$, that is all linear transformations which preserve ${V}$ up to a scalar. Note that ${Z\leqslant \mathrm{GSp}(4,q)}$ and we define ${\mathrm{PGSp}(4,q)=\mathrm{GSp}(4,q)/Z}$.

The group ${\Gamma\mathrm{L}(4,q)}$ is the group of all transformations ${g}$ of ${V}$ such that ${(v+u)^g=v^g+w^g}$ for all ${u,v\in V}$ and there exists a field automorphism ${\sigma}$ of ${\mathrm{GF}(q)}$ such that ${(\lambda v)^g=\lambda^{\sigma}v^g}$ for all ${\lambda\in\mathrm{GF}(q)}$ and ${v\in V}$. Note that ${\mathrm{GL}(4,q)}$ is the subgroup of ${\Gamma\mathrm{L}(4,q)}$ consisting of those transformations whose associated field automorphism is the trivial automorphism. Given a field automorphism ${\sigma}$, the transformation mapping ${(v_1,v_2,v_3,v_4)}$ to ${(v_1^\sigma,v_2^\sigma,v_3^\sigma,v_4^\sigma)}$ is an element of ${\Gamma\mathrm{L}(4,q)}$.

Finally, we let ${\Gamma\mathrm{Sp}(4,q)}$ to be the set of all elements ${g\in\Gamma\mathrm{L}(4,q)}$ such that there exists ${\lambda\in\mathrm{GF}(q)}$ and ${\sigma\in\mathrm{Aut}(\mathrm{GF}(q)(}$ such that ${B(u^g,v^g)=\lambda B(u,v)^{\sigma}}$ for all ${u,v\in V}$. All such elements induce automorphisms of the GQ and in fact the full automorphism group of ${\mathsf{W}(3,q)}$ is ${\mathrm{P}\Gamma\mathrm{Sp}(4,q)=\Gamma\mathrm{Sp}(4,q)/Z}$. This group acts transitively on the points and the lines of the GQ as well as on the set of all point-line incident pairs.

The dual of the generalised quadrangle ${\mathsf{W}(3,q)}$ is another generalised quadrangle of order ${(q,q)}$ and for $q$ even is in fact isomorphic to ${\mathsf{W}(3,q)}$.

${\mathsf{H}(3,q^2)}$:

For our next generalised quadrangle, let ${V}$ be a 4-dimensional vector space over ${\mathrm{GF}(q^2)}$ and let ${B:V\times V\rightarrow \mathrm{GF}(q^2)}$ be a nondegenerate hermitian form on ${V}$, that is, a form which is linear in the first coordinate and ${B(u,v)=(v,u)^q}$. (As before, nondegenerate means that the zero vector is the unique vector which is perpendicular to the whole space.) Note that ${B}$ preserves addition in the second coordinate but ${B(u,\lambda v)=\lambda^qB(u,v)}$. We take ${B}$ to be defined by ${B(u,v)=uv^{qT}}$, where by ${v^q}$ we mean ${(v_1^q,v_2^q,v_3^q,v_4^q)}$. All nondegenerate hermitian forms on ${V}$ are equivalent to ${B}$.

We can define totally isotropic subspaces as before. We let ${\mathcal{Q}}$ be the point-line incidence geometry whose points are the totally isotropic 1-spaces and whose lines are the totally isotropic 2-spaces.

Given a totally isotropic 1-space ${\langle v\rangle}$, as before, the totally isotropic 2-spaces containing it correspond to the totally isotropic 1-spaces of the quotient space ${\langle v\rangle ^{\perp}/\langle v\rangle}$, upon which ${B}$ induces a nondegenerate hermitian form. This space contains ${q+1}$ totally isotropic 1-spaces. Moreover, a totally isotropic 2-space contains ${(q^4-1)/(q^2-1)=q^2+1}$ totally isotropic 1-spaces. The same arguement as in the symplectic case shows that given a point ${P}$ not on a line ${\ell}$ there is a unique line through ${P}$ which meets ${\ell}$. Hence ${\mathcal{Q}}$ is a GQ of order ${(q^2,q)}$. It is usually denoted by ${\mathsf{H}(3,q^2)}$.

The full automorphism group of ${\mathsf{H}(3,q^2)}$ is ${\mathrm{P}\Gamma \mathrm{U}(4,q)}$, that is all ${gZ\in\mathrm{P}\Gamma\mathrm{L}(4,q)}$ for which there exists ${\lambda\in\mathrm{GF}(q)}$ and ${\sigma\in\mathrm{Aut}\mathrm{GF}(q^2)}$ such that ${B(u^g,v^g)=\lambda B(u,v)^\sigma}$. Again, this group acts transitively on the points, lines and point-line incident pairs of ${\mathcal{Q}}$. Due to my group theoretic upbringing I denote this group by ${\mathrm{P}\Gamma \mathrm{U}(4,q)}$ and not ${\mathrm{P}\Gamma \mathrm{U}(4,q^2)}$.

${\mathsf{Q}^-(5,q)}$:

Let ${V}$ be a 6-dimensional vector space over ${\mathrm{GF}(q)}$ and ${Q:V\rightarrow \mathrm{GF}(q)}$ be a nondegenerate quadratic form on ${V}$, that is, ${Q(\lambda v)=\lambda^2Q(v)}$ for all ${v\in V}$ and ${\lambda\in\mathrm{GF}(q)}$, and ${B(v,w):=Q(v+w)-Q(v)-Q(w)}$ is a nondegenerate bilinear form. A subspace ${U}$ of ${V}$ is called totally singular if ${Q(u)=0}$ for all ${u\in U}$.

Up to equivalence there are two quadratic forms on ${V}$: elliptic and hyperbolic. They are distinguished by the dimension of the maximal totally singular subspaces. In the elliptic case they have dimension ${2}$ and in the hyperbolic case they have dimension ${3}$.

To define our GQ, we choose ${Q}$ to be an elliptic quadratic form. The points are then the totally singular 1-spaces and the lines are the totally singular 2-spaces. The GQ obtained is isomorphic to the dual of  ${\mathsf{H}(3,q^2)}$, and has order ${(q,q^2)}$.

${\mathsf{H}(4,q^2)}$:

This is similar to ${\mathsf{H}(3,q^2)}$ except here we take ${V}$ to be a 5-dimensional vector space over ${\mathrm{GF}(q^2)}$. We take ${B}$ to be a nondegenerate hermitian form on ${V}$ (again ${B(u,v)=uv^{qT}}$ can be taken as the form). Then the maximal totally isotropic subspaces have dimension 2. We can then define a point-line geometry ${\mathcal{Q}}$ with points the totally isotropic 1-spaces and lines the totally isotropic 2-spaces. Each point lies on ${q^3+1}$ lines (there are ${q^3+1}$ totally isotropic 1-spaces in ${\langle v\rangle^\perp/\langle v\rangle}$) and each line contains ${q^2+1}$ points. We obtain a GQ of order ${(q^2,q^3)}$ which is usually denoted by ${\mathsf{H}(4,q^2)}$.

The full automorphism group of this GQ is ${\mathrm{P}\Gamma \mathrm{U}(5,q)}$. This acts transitively on points, lines and point-line incident pairs.

${\mathsf{Q}(4,q)}$, $q$ odd:

This is similar of ${\mathsf{Q}^-(5,q)}$. In this case we take ${V}$ to be a 5-dimensional vector space over ${\mathrm{GF}(q)}$ and ${Q}$ to be a nondegenerate quadratic form on ${V}$. Since the dimension of the vector space is odd there is only one type of quadratic form on ${V}$.

In this case our points are the totally singular 1-spaces of ${V}$ and the lines are the totally singular 2-spaces of ${V}$. We obtain a generalised quadrangle of order ${(q,q)}$ which is isomorphic to the dual of ${\mathsf{W}(3,q)}$.

The five generalised quadrangles above  are referred to as the classical generalised quadrangles. There is also the dual of $\mathsf{H}(4,q^2)$.  There are several other constructions of generalised quadrangles. John has already blogged about the Knarr model for the flock generalised quadrangles. This construction seems typical of many constructions in geometry where there are several types of points and several types of lines. My next post will look at some other constructions of generalised quadrangles.