I am looking for an explicit description of the unipotent radical of a minimal parabolic subgroup of a unitary group, i.e. the group of isometries of a hermitian form, over an arbitrary field.


In his notes "Linear algebraic groups" (Boulder, 1966), Section 6.6 "Examples", Borel gives such a description for an orthogonal group $SO(F)$ for a non-degenerate quadratic form $F$, and subsequently writes



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> When one starts with a hermitian form, the same considerations apply, except that one gets a root system of type $\mathbf{BC}\_q$.
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Working out the details seems rather tricky to me, in particular because a hermitian form is defined over a skew field with involution and not just over a commutative field.


Has this been worked out somewhere in the literature?



Unipotent radical of minimal parabolic subgroup of a unitary group over an arbitrary field
I deal with the simpler case: $G=U(h)$ where $E/F$ is a separable quadratic extension of fields and $h: E^n\times E^n\rightarrow E$ is Hermitian with respect to $E/F$. Suppose $W$ is a maximal isotropic subspace of $E^n$ with respect to the Hermitian form $h$. WE have the partial flag (where $W^{\perp}$ is the orthogonal complement of $W$ w.r.t. the Hermitian form $h$)


$$0\subset W \subset W^{\perp} \subset E^n.$$ The unipotent radical of the minimal parabolic is precisely the subgroup of $G=U(h)$ which preserves this flag and acts trivially on successive quotients. 


When we have a Hermitain form over a skew field, a similar description obtains, but sometimes, you may get that the unitary group becomes a symplectic or orthogonal group (these details are in Tits' article on classification of algebraic groups in the same Boulder conference volume)


