For $q$ an odd prime power, one can construct $H:=O\_4^-(q)<U\_4(q)$ by treating the $H$-invariant bilinear form as Hermitean. On the other hand $U\_4(q)$ is isomorphic to $O\_6^-(q)$; I need to understand the geometric meaning of $H$ in $O\_6^-(q)$. 


For $q \equiv 1\pmod {4}$, it appears that $H$ can be viewed as $O\_3(q^2)$; namely, the relative trace form $x\mapsto \langle x,Bx\rangle+\langle x,Bx\rangle^q$,
with $x\mapsto \langle x,Bx\rangle$ being the $O\_3(q^2)$-invariant form, gives a minus type form in $\mathbb{F}\_q^6$, and thus the desired embedding $H<O\_6^-(q)$.


However, I don't understand how to deal with the case $q \equiv 3\pmod {4}$, for the relative trace form then is of plus type. Any suggestions?



embedding of $O_4^-(q)$ in $U_4(q)$
As is pointed out in the question, our job is to display an (irreducible) copy of $H=O\_3(q^2)$ inside $G=O\_6^-(q)$.


As mentioned in comments, this can be done by taking $H$ to be a field-extension subgroup of $G$. An explicit construction for $q\equiv 3 \pmod 4$ is as follows.


We take $H$ to be the group of isometries of
$$Q: V \to \mathbb{F}\_{q^2}, \,\, (x,y,z) \mapsto xy+z^2,$$
where $V$ is 3-dimensional over $\mathbb{F}\_{q^2}$.


Now, viewing $V$ as $6$-dimensional over $\mathbb{F}\_q$, and letting $L:\mathbb{F}\_{q^2}\to \mathbb{F}\_q$ be any $\mathbb{F}\_q$-linear function, we observe that the elements of $H$ preserve the form $LQ:V\to \mathbb{F}\_q$, and we obtain an embedding of $H$ in the isometry group of $LQ$.


Thus, to answer the question, we must exhibit $L$ so that $LQ$ is a non-degenerate form of minus type. I claim that we can take $L=L\_\mu$ where


* $\zeta$ is a primitive element of $\mathbb{F}\_{q^2}$ and $\mu=\zeta^a$ for some odd number $a$.
* $L\_\mu:\mathbb{F}\_{q^2}\to \mathbb{F}\_q, x \mapsto \mu x + (\mu x)^q$. (In other words $L\_\mu(x)= {\rm Tr}(\mu x)$.)


I won't bother showing that $L\_\mu Q$ is non-degenerate as this is easy. To see that it is minus type, I must show that there exists a 2-dimensional $\mathbb{F}\_q$-subspace of $V$ that is a maximal totally isotropic subspace with respect to $L\_\mu Q$.


Let $W={\rm span}\_{\mathbb{F}\_{q^2}}((1,0,0))$. Since $Q(1,0,0)=0$, it is clear that $W$ is t.i. with respect to $Q$, and hence with respect to $L\_\mu Q$. 


Suppose $\langle W, (x,y,z)\rangle$ is totally isotropic. Let $\beta$ be the polarization of $Q$. Then
$$
\beta((x\_1,y\_1,z\_1), (x\_2,y\_2,z\_2)) = x\_1y\_2+x\_2y\_1+2z\_1z\_2
$$
and we conclude that, for any $\lambda \in \mathbb{F}\_{q^2}$,
$$
L\_\mu\beta((\lambda, 0,0), (x,y,z)) = L\_\mu(\lambda y) = (\mu\lambda y)+(\mu\lambda y)^q = 0.
$$
One concludes, immediately, that $y=0$.


Now, in addition, our supposition implies that $(x,y,z)$ is totally singular. Then
$$
L\_\mu Q(x,y,z)= \mu(xy+z^2)+\left(\mu(xy+z^2)\right)^q=\mu z^2+ \left(\mu z^2\right)^q = 0.
$$
Our choice of $\mu$, along with the fact that $q\equiv 3\pmod 4$, now implies that $z=0$. But this means that $(x,y,z)\in W$ and so $W$ is maximal t.i., as required.


(**Final note**: The same argument shows that ${\rm span}\_{\mathbb{F}\_{q^2}}((0,1,0))$ is maximal t.i. This implies, in particular, that $L\_\mu Q$ is non-degenerate.)


