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 $q \equiv 1\ mod\ {4}$, 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 $q \equiv 3\ mod\ {4}$, for the relative trace form then is of plus type. Any suggestions?

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?