This is a continuation of my last post on this subject. As Gordon remarked in one of his posts, you may need to refresh your browser if some of the embedded gifs do not appear as they should.

### Dualities and isomorphisms of classical groups

Four of the five families of classical generalised quadrangles come in dual pairs: (i) $Q(4,q)$ and $W(3,q)$; (ii) $H(3,q^2)$ and $Q^-(5,q)$. Both can be demonstrated by the Klein correspondence. Recall from the last post that the Klein correspondence maps a line of $\mathsf{PG}(3,q)$ represented as the row space of

$M_{u,v}:=\begin{bmatrix}u_0&u_1&u_2&u_3\\v_0&v_1&v_2&v_3\end{bmatrix}$

to the point $(p_{01}:p_{02}:p_{03}:p_{12}:p_{31}:p_{23})$ where

$p_{ij}=\begin{vmatrix}u_i&u_j\\v_i&v_j\end{vmatrix}$.

Now consider the symplectic generalised quadrangle $W(3,q)$ defined by the bilinear alternating form

$B(x,y):=x_0y_1 - x_1y_0 + x_2y_3 - x_3y_2.$

A totally isotropic line $M_{u,v}$ must then satisfy

$0 = B(u,v) = p_{01} + p_{23}$.

Therefore, the lines of $W(3,q)$ are mapped to points of $Q^+(5,q)$ lying in the hyperplane $\pi:X_0+X_5=0$. Now the quadratic form defining $Q^+(5,q)$ is $Q(x)=x_0x_5+x_1x_4+x_2x_3$ and $\pi$ is the tangent hyperplane at the projective point $(1:0:0:0:0:1)$, which does not lie in the quadric. Hence the hyperplane $\pi$ is non-degenerate and so we see that $W(3,q)$ maps to points of $Q(4,q)$. That this mapping is bijective follows from noting that the number of lines of $W(3,q)$ is equal to the number of points of $Q(4,q)$ (namely, $(q+1)(q^2+1)$).

Hence $\mathsf{P\Gamma Sp}(4,q)\cong \mathsf{P\Gamma O}(5,q)$.

Now we will consider a more difficult situation which reveals that the generalised quadrangles $H(3,q^2)$ and $Q^-(5,q)$ are also formally dual to one another. We could take the approach above, but instead, we will use a cunning way (told to me by Tim Penttila) to produce an Hermitian form on $\mathsf{PG}(3,q^2)$. First we consider $Q^-(5,q)$ as the subspaces of $Q^+(5,q^2)$ that are fixed by a particular involution known as a Baer involution. In order for this to look nice, I will change the form that defines $Q^+(5,q^2)$:

$Q(x) = x_0x_5 + x_1x_4 + f(x_2,x_3)$

where $f$ is an irreducible homogeneous quadratic in two variables over $GF(q)$. So when we restrict this form to the points with entries only in $GF(q)$, we have a form of minus type; a polar space isomorphic to $Q^-(5,q)$. The Baer involution $\tau$ is simply the map which raises each coordinate to its q-th power, and so our $Q^-(5,q)$ is exactly the fixed singular elements of this involution. Moreover, $\tau$ switches the latins and greeks. To see this, note that the latins and greeks form a block-system for the full orthogonal group. Now consider the following singular plane:

$\pi = \{ (a:b: c: d: 0: 0) \mid (a,b,c,d) \ne 0, f(c,d)=0\}$.

Then applying $\tau$ to each point comprising $\pi$ will give us a different plane which intersects $\pi$ in the line

$L = \{(a:b:0:0:0:0)\mid (a,b)\ne 0\}$.

Therefore, $\pi^\tau$ is in a different class to $\pi$. Furthermore, if we take a line of $Q^-(5,q)$, it is fixed by $\tau$ and the two singular planes incident with this line are interchanged by $\tau$. If we relate what $\tau$ is doing back to $\mathsf{PG}(3,q^2)$, we see that it is inducing a polarity of $\mathsf{PG}(3,q^2)$ having $(q^2+1)(q^3+1)$ absolute lines. By the classification of polarities, this is a unitary polarity. So we see that $Q^-(5,q)$ corresponds to the hermitian surface $H(3,q^2)$. Hence $P\Gamma O^-(5,q)\cong P\Gamma U(4,q)$.

An ovoid of $Q^+(5,q)$ is a set of $q^2+1$ points such that no two are collinear in a line of $Q^+(5,q)$. So under the Klein correspondence, we have a set of $q^2+1$ lines of $\mathsf{PG}(3,q)$ such that no two are concurrent in a point of $\mathsf{PG}(3,q)$ (and equivalently, they do not span a plane of $\mathsf{PG}(3,q)$). So an ovoid of $Q^+(5,q)$ is equivalent to a spread of $\mathsf{PG}(3,q)$.

We also mention here a beautiful observation of Shult and Thas:

any 1-system of $Q^+(5,q)$, q odd, is equivalent to the classical 1-system (arising from the Klein image of an elliptic quadric).

A 1-system of $Q^+(5,q)$ is a set of $q^2+1$ lines of $Q^+(5,q)$ that are pairwise opposite: there is no singular plane on one of the lines meeting the other in a point. Under the Klein correspondence, we end up with a set of point/plane incident pairs, such that no two will intersect in a line that passes through their given points. Such a configuration defines an ovoid of $\mathsf{PG}(3,q)$: a set of $q^2+1$ points with no 3 collinear. By a result attributed to the independent work of Barlotti and Panella (1955), this ovoid is projectively equivalent to en elliptic quadric. So the 1-system is the Klein image of an elliptic quadric.

### Isomorphisms of simple gorups

The Klein Correspondence yields a simple way to explain why $A_8 \cong \mathsf{PSL}(4,2)$ and $\mathsf{PSL}(3,2)\cong \mathsf{PSL}(2,7)$. First, consider an eight dimensional vector space over $\mathbb{F}_2$ equipped with the quadratic form

$Q(x):=x_0x_5+x_1x_4+x_2x_3+x_6x_7$.

Think of the vectors as eight-tuples, so that we have a natural action of the symmetric group $S_8$ on the vectors: simply by permutation of coordinates. Then $S_8$ fixes $(1,1,1,1,1,1,1,1)$ and the subspace $W$ of vectors whose coordinates sum to $0$. Then $S_8$ induces an action on the quotient six-dimensional vector space

$W/\langle(1,1,1,1,1,1,1,1)\rangle$.

We should also observe that $S_8$ preserves the quadratic form $Q$, and the inherited quadratic form on $W/\langle(1,1,1,1,1,1,1,1)\rangle$. So $S_8$ has a representation into $\mathsf{O}^+(6,2)$, and then by orders, we have an isomorphism. The Klein correspondence then yields $A_8 \cong \mathsf{PSL}(4,2)$.

Now for $\mathsf{PSL}(3,2)\cong \mathsf{PSL}(2,7)$, we observe that we can use the same setting since $\mathsf{PSL}(3,2)$ can be realised inside the stabiliser of a point or a plane of $\mathsf{PG}(3,2)$. Take the following singular plane of $Q^+(5,2)$:

$\pi:=\{(a+b+c,a+b,a+c,a,b+c,b,c,0)\mid 0\ne(a,b,c)\in\mathbb{F}_2^3\}$

Now $S_8$ does not fix this plane by acting on coordinates, but we do have a subgroup of $S_8$, namely $\mathsf{PSL}(2,7)$ that acts 2-transitively on these coordinates if we think of its natural action on the projective line labelled in the order $(0,1), (1,0),(1,1), (1,3), (1,2), (1,6), (1,4), (1,5)$. Then by comparing orders, we see that $\mathsf{PSL}(2,7)$ is isomorphic to $\mathsf{PSL}(3,2)$.