Let $q=2^k$. I need to explicitly construct $U\_3(q)$ as a subgroup of $G=GO\_6^-(q)$. It is well-known that
$G\cong U\_4(q)$, and as a subgroup of the latter one has $U\_3(q)$ fixing a non-isotropic point in the $\mathbb{F}\_{q^2}^4$ endowed with a sesquilinear Hermitean form preserved by $G$. 


If I represent $\mathbb{F}\_{q^2}$ by 2x2-matrices over $\mathbb{F}\_{q}$ I should be able to write $U\_3(q)$ as a subgroup of $GL\_6(q)$, but would it be a subgroup of $GO\_6^-(q)$, and if yes, then how do I find out the corresponding invariant quadratic form? (this is inspired by Nick Gill's great answer to my question [embedding of $O\_4^-(q)$ in $U\_4(q)$](https://mathoverflow.net/questions/227540/embedding-of-o-4-q-in-u-4q))



describing embedding $U_3(q)<O_6^-(q)$, $q$ even
This is an example of a much more general embedding. Let $q$ be a prime power and $m$ a positive integer. Let $V$ be an $m$-dimensional vector space over $F=GF(q^2)$ and let $B:V\times V\rightarrow F$ be a nondegenerate hermitian form. Then $V$ is also a $2m$-dimensional vector space over $K=GF(q)$. Also $Q(v)=B(v,v)\in K$ defines a nondegenerate quadratic form on $V$. When $m$ is even this quadratic form is hyperbolic while it is elliptic if $m$ is odd. 


Now, if $U\_m(q)$ is the unitary group of all isometries of $B$ then it also preserves the quadratic form $Q$. Thus for $m$ odd this gives an embedding $U\_m(q) < GO^-\_{2m}(q)$ while for $m$ even this gives an embedding $U\_m(q) < GO^+\_{2m}(q)$.


