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. Continue reading “The Klein Correspondence II”

One of the most useful things to know about the finite general linear group $GL(d, q)$ are the Singer cycles, what they do and what they are about. A subgroup $G$ of $GL(d, q)$ is irreducible if it fixes no proper nontrivial subspace of the vector space $GF(q)^d$, in the natural action of $G$ on this vector space. Consider $GF(q^d)$ as a vector space over its subfield $GF(q)$. Then the multiplicative group $S:=GF(q^d)\backslash\{0\}$ acts on the nonzero vectors of $GF(q^d)$ by right multiplication, and this action is clearly regular. But what about similar elements in the classical groups?
An important map in the following will be the relative trace map $Tr_{q^d\to q^i}:GF(q^d)\to GF(q^i)$, where $i$ divides $d$, and it is defined by
$Tr_{q^d\to q^i}(x) = \sum_{j=0}^{d/i-1}x^{(q^i)^j}.$ Continue reading “Irreducible cyclic groups of classical groups”