16
$\begingroup$

It is well-known that there is an isomorphism between $PSp(4,3)$ (the symplectic group of dimension $4$ over $\mathbb F_3$) and $PSU(4,2^2)$ (the unitary group defined by $4\times4$ unitary matrices over $\mathbb F_4$).

Question: Can the exceptional isomorphism be interpreted in (finite) geometry?

Known examples of geometric interpretation:

The isomorphism $PSL(3,2)=PSL(2,7)$ is presented here.

The isomorphism $S_6=PSp(4,2)$ is explained in John Baez's article.

The isomorphism $A_6=PSL(2,9)$ can be explained by the Taylor graph of $PSL(2,9)$ acting on 10 points (i.e. the projective line of $\mathbb F_9$): this is a distance-regular graph on 20 vertices with intersection array $\{9,4,1;1,4,9\}$. Observing that this is also the intersection array of $J(6,3)$ and Johnson graphs are unique, it lets $S_6$ acts on the graph.

The isomorphism $PSU(3,3)=G_2(2)'$ can be explained by the unitary nonisotropics graph of $PΓU(3,q)$: for $q=3$, the intersection array is $\{6,4,4;1,1,3\}$, which is exactly that of a generalized hexagon of order 2. Furthermore, $GH(2,2)$ is unique up to duality, so it's also the Cayley generalized hexagon where $G_2(2)$ acts.

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ What's "finite geometry"?$$ $$ This group appears in three complex reflection groups; one is the Weyl group of $E_6$, whose reductions mod $2$ and $3$ identify it with $O_6^-(2)$ and $O_5(3)$; another is the complex (Eisenstein) $E_8$ lattice, whose reductions mod $2$ and $3$ take us to the same groups in their $U_4$ and $Sp_4$ guises which you asked about. Short vectors in each lattice give rise to several geometrical structures on which the group acts, including the $27$ lines of a cubic surface and the $36$ pairs $\pm r$ of $E_6$ roots. $\endgroup$
    – Noam D. Elkies
    Commented Dec 11, 2019 at 4:17
  • $\begingroup$ You forgot $A_8=GL(4,2)$ (which has a nice and simple geometric interpretation). $\endgroup$
    – abx
    Commented Dec 11, 2019 at 5:35
  • 1
    $\begingroup$ @NoamD.Elkies this question seems your sort of thing, from yesterday: mathoverflow.net/questions/348159/… $\endgroup$
    – Will Jagy
    Commented Dec 13, 2019 at 0:04

1 Answer 1

Reset to default
14
$\begingroup$

I believe that a finite geometric proof is given by Jean Dieudonné here:

Dieudonné, Jean, Les isomorphismes exceptionnels entre les groupes classiques finis, Can. J. Math. 6, 305-315 (1954). ZBL0055.01904.

If you go to Section 7 of the given paper you will see the proof. Dieudonné displays the required isomorphism using a 1-1 correspondence between the 40 lines in $(\mathbb{F}_3)^4$, and the non-isotropic lines in $(\mathbb{F}_4)^4$ (non-isotropic with respect to a non-degenerate hermitian form).

A google-books version is here.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .