# Geometric interpretation of the exceptional isomorphism $PSp(4,3)=PSU(4,2^2)$

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.

• 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. Dec 11 '19 at 4:17
• You forgot $A_8=GL(4,2)$ (which has a nice and simple geometric interpretation).
– abx
Dec 11 '19 at 5:35
• @NoamD.Elkies this question seems your sort of thing, from yesterday: mathoverflow.net/questions/348159/… Dec 13 '19 at 0:04

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).