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My question is similar to this one but about finite field case.

So, the set up is the following:

Let $G$ be $GU_n(q)$ acting on unitary space $(V, {\bf f})$, where $V=\mathbb{F}_{q^2}^n$ and ${\bf f}:V \times V \to \mathbb{F}_{q^2}$ is non-degenerate unitary form. Let $H \le G$ be irreducible and imprimitive with a system of imprimitivity: $$V=V_1 \oplus \ldots \oplus V_k.$$

Now, since $H$ is irreducible, all $V_i$-s are either non-degenerate or totally isotropic. Let us, for simplicity, stick with the first option, so assume that all $V_i$-s are non-degenerate subspaces of $V.$

My question is: is it true that ${\bf f}(V_i,V_j)=0$ for $i \ne j$?

An additional question is: If the statement in the first question is false, can I find another imprimitivity system of $H$ for which it is true?

The proof in here works only for algebraically closed fields. In particular, if $H$ is absolutely irreducible, then the answer is yes.

Improve this question
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    $\begingroup$ I don't know the answer. The answer to the corresponding question for the symplectic groups with $V_1$ a totally isotropic subspace is no. An example is ${\rm Sp}(6,13)$, which has an imprimitive subgroup $H \cong {\rm SL}(2,5)$ which acts imprimitively on six 1-dimensional subspaces, but does not preserve a symplectic system of imprimitivity. $\endgroup$
    – Derek Holt
    Commented Feb 14, 2020 at 17:34

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