This week John continued his discussion on triality and generalised hexagons from last week. This is also the first week where I have used Luca Trevisan’s LaTex to WordPress program to write up the post.

Recall that we had an eight-dimensional vector space ${V}$ over ${\mathrm{GF}(q)}$ equipped with a quadratic form ${Q(x)=x_0x_4+x_1x_5+x_2x_6-x_3x_7}$, where ${x=(x_0,x_1,\ldots,x_7)\in V}$. We were able to divide the totally isotropic solids into two classes, greeks and latins, so that two solids are in the same class if their intersection is a line or trivial. The points, lines, greeks and latins give rise to the Oriflamme geometry where incidence is inclusion and a greek is incident with a latin if their intersection is a plane.

Let ${\mathcal{P}^{(0)}}$ be the set of points, ${\mathcal{P}^{(1)}}$ be the set of greeks, ${\mathcal{P}^{(2)}}$ the set of latins and ${\mathcal{P}^{(3)}}$ the set of lines. Now ${|\mathcal{P}^{(0)}|=|\mathcal{P}^{(1)}|=|\mathcal{P}^{(2)}|}$ and so we can represent the greeks by vectors ${y=(y_0,y_1,\ldots,y_7)}$ and the latins by vectors ${z=(z_0,z_1,\ldots,z_7)}$. Define

$\displaystyle \begin{array}{rl} T(x,y,z)=&\left|\begin{array}{ccc} x_0&x_1&x_2 \\ y_0&y_1&y_2\\z_0&z_1&z_2\end{array}\right| +\left|\begin{array}{ccc} x_4&x_5&x_6\\ y_4&y_5&y_6\\z_4&z_5&z_6 \end{array}\right|\\ &-x_3(z_0y_4+z_1y_5+z_2y_6)+x_7(y_0z_4+y_1z_5+y_2z_6)\\ &-y_3(x_0z_4+x_1z_5+x_2z_6)+y_7(z_0x_4+z_1x_5+z_2x_6)\\&-z_3(y_0x_4+y_1x_5+y_2x_6)+z_7(x_0y_4+x_1y_5+x_2y_6)\\ &+x_3y_3z_3-x_7y_7z_7 \end{array}$

Then a point ${x=(x_0,x_1,\ldots,x_7)}$ and greek ${y=(y_0,y_1,\ldots,y_z)}$ are incident if and only if ${T(x,y,z)\equiv 0}$ as a function of ${z}$. Incidence between other combinations of points, latins and greeks is similar.

Define

$\displaystyle \begin{array}{rrll} \theta:& \mathcal{P}^{(i)}&\rightarrow &\mathcal{P}^{(i+1)}\\ & (x_j) &\mapsto &(x_j^{\sigma})\end{array}$

where ${i\in\{0,1,2\}}$, addition of the superscripts is done modulo 3 and ${\sigma\in\mathrm{Aut}(\mathrm{GF}(q))}$ has order 1 or 3. Then ${\theta}$ defines a triality of the Oriflamme geometry. Recall from last week that we can then form a generalised hexagon whose points are those points ${x}$ which are incident with ${x^{\theta}}$ and the lines are those fixed by ${\theta}$. I have just realised that John didn’t define how ${\theta}$ acts on the lines but I will get John to fill in this detail in the comments. When ${\sigma}$ has order 3 the generalised hexagon we get is the one associated with ${{}^3D_4(q)}$ while when ${\sigma=1}$ we obtain the split Cayley hexagon ${\mathcal{H}(q)}$ of order ${(q,q)}$.

The points of ${\mathcal{H}(q)}$ are those for which ${x_3=x_7}$, and so all points in a nondenerate hyperplane of ${V}$, that is, contained in a ${Q(6,q)}$. The lines are some of the lines of ${Q(6,q)}$. The stabiliser of ${\mathcal{H}(q)}$ in ${P\Omega(7,q)}$ is ${G_2(q)}$. We now discuss some configurations in ${\mathcal{H}(q)}$ and their stabilisers in ${G_2(q)}$.

1. Subhexagons of order ${(1,q)}$: There is a nondegenerate 6-space of ${V}$ where the restriction of the quadratic form is of pluus type, that is a ${Q^+(5,q)}$ which when intersected with ${\mathcal{H}(q)}$ gives ${2(q^2+q+1)}$ points and ${(q+1)(q^2+q+1)}$ lines. The points can be thought of as two copies of ${PG(2,q)}$. This is best realised by taking the incidence graph of ${PG(2,q)}$ and placing another vertex at the centre of each edge. The vertices of the original graph are the points and the new vertices coming from midpoints are the lines. The stabiliser of this configuration in ${G_2(q)}$ is ${SL(3,q):2}$, where the 2 on top comes from interchanging the points and lines in ${PG(2,q)}$.
2. Line regulus: Let ${L}$ be a line of ${\mathcal{H}(q)}$ and let ${M}$ be a line opposite ${L}$, that is, at distance 6 in the incidence graph. Thinking of ${L}$ and ${M}$ as being lines in ${Q(6,q)}$, we have that ${L^{\perp}\cap M^{\perp}}$ is a conic, that is, a ${Q(2,q)}$. Moreover, there are ${q+1}$ mutually disjoint lines in ${\mathcal{H}(q)}$ which are perp to ${L^{\perp}\cap M^{\perp}}$. These lines span a ${Q^+(3,q)}$. We can then find another ${q+1}$ mutually lines in the ${Q^+(3,q)}$ which are not in ${H(q)}$. The picture below demonstrates this for ${q=3}$ with the intersections of lines being points.

The stabiliser of this configuration in ${G_2(q)}$ is ${O^+(4,q)=(SL(2,q)\circ SL(2,q)).2}$.

3. Hermitian spread: A spread of a polar space ${Q^-(5,q)}$ is a set of maximal totally isotropic subspaces which form a partition of the points. A spread of ${\mathcal{H}(q)}$ is a maximal set of lines which are pairwise opposite. Such a set has ${q^3+1}$ lines and is fixed by ${SU(3,q):2}$.

Constructing ${\mathcal{H}(q)}$ from ${H(3,q^2)}$

Let ${V}$ be a 4-dimensional vector space over ${\mathrm{GF}(q^2)}$ equipped with the hermitian form ${\langle x,y\rangle=x_0y_0^q+x_1y_1^q+x_2y_2^q+x_3y_3^q\in\mathrm{GF}(q^2)}$. This defines the hermitian space ${H(3,q^2)}$ which has ${(q^2+1)(q^3+1)}$ points and ${(q+1)(q^3+1)}$ lines and is a generalised quadrangle of order ${(q^2,q)}$. We can then define ${B(x,y)=\mathrm{Tr}_{q^2\rightarrow q}(\langle x,y\rangle)\in\mathrm{GF}(q)}$ which when ${V}$ is interpreted as an 8-dimensional vector space over ${\mathrm{GF}(q)}$ provides a symmetric bilinear form. This provides an embedding of ${H(3,q^2)}$ in ${Q^+(7,q)}$.  The points get mapped to lines and the lines get mapped to solids.

We can take the intersection of the image of this embedding with a ${Q(6,q)}$ section. This is often attributed to Barlotti, Cofman and Segre. If we choose a ${H(2,q^2)}$ as a subspace of ${H(3,q^2)}$, (that is, a nondegenerate hyperplane), then the points in ${H(2,q^2)}$ get mapped to lines in ${Q(6,q)}$ which are contained in a ${Q^-(5,q)}$ while the points of ${H(3,q^2)\backslash H(2,q^2)}$ get mapped onto the points of ${Q(6,q)\backslash Q^-(5,q)}$, (that is, the lines in ${Q^+(7,q)}$ arising from these points meet the ${Q(6,q)}$ in a point.)

John Bamberg together with Nicola Durante has recently been able to show that this embedding can be used to construct ${\mathcal{H}(q)}$ as follows:

Let ${\mathcal{O}}$ be a ${H(2,q^2)}$ in ${H(3,q^2)}$ and let ${\Omega}$ be a set of Baer subgenerators (that is, copies of ${PG(1,q)}$) which span a totally isotropic line of $H(3,q^2)$) with a point in ${\mathcal{O}}$ such that every point of ${H(3,q^2)\backslash \mathcal{O}}$ is on ${q+1}$ elements of ${\Omega}$ and they span a Baer subplane. Then the geometry whose points are

1. (a) the totally isotropic lines of ${H(3,q^2)}$, and
2. (b) the points of ${H(3,q^2)\backslash \mathcal{O}}$.

and whose lines are

1. (a) points of ${\mathcal{O}}$, and
2. (b) the elements of ${\Omega}$.

and incidence is inclusion or the inherited incidence, is a generalised hexagon isomorphic to ${\mathcal{H}(q)}$. In fact, ${\Omega}$ is an orbit under ${SU(3,q^2)}$, that is, the stabiliser of ${\mathcal{O}}$.