embedding of $O_4^-(q)$ in $U_4(q)$ 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?
 A: 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.)
