# Unipotent radical of minimal parabolic subgroup of a unitary group over an arbitrary field

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

When one starts with a hermitian form, the same considerations apply, except that one gets a root system of type $\mathbf{BC}_q$.

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?

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.
• Thanks for your answer, but it doesn't quite answer my question. I know indeed that the minimal parabolic is the stabilizer of a maximal flag, but what I'm really after is an explicit parametrization of the unipotent radical, in the same style as Borel's description for $SO(q)$ where the unipotent radical is parametrized by the anisotropic part of the vector space on which $q$ lives. For $SU(h)$, we should somehow find a parametrization by $V_{\mathrm{an}} \times S$, where $S$ is the subset of the defining skew field consisting of the skew-symmetric elements. Nov 3, 2015 at 15:53
• But that is really what is happening: $W^{\perp}/W$ is indeed the aniso tropic part of the Hermitian form. Nov 3, 2015 at 15:55
• I agree, but then how do we see that the relative root system is $\mathbf{BC}_q$? In other words, how do we see that the unipotent radical is non-abelian in this case? Nov 3, 2015 at 15:56
• Full disclosure: I have a paper where I had to deal with this! Apart from that , in the one dimensional case, the uni radical is an extension of $E^{n-2}$ by $F$ (the smaller field), and the commutator $[x,y]$ with $x,y \in E^{n-1}$ is just the imaginary part of $h(x,y)$. The part $F$ is central in the unipotent radical, so this completely describes the commutator: hence the unip radical is a two step unipotent group. Nov 3, 2015 at 16:36
• Thanks! With your examples at hand, I have now been able to find the general description I was looking for. For what it's worth: in the general case, i.e. over arbitrary skew fields, we get that the commutator $[x,y]$ is equal to $h(x,y)-h(y,x) = h(x,y)-h(x,y)^\sigma$, which is indeed a skew-symmetric element (as I expected in my first comment, and which corresponds to the imaginary part in your description). This also reveals the difference between $O(f)$ and $U(h)$, since for $O(f)$, the similar expression $f(x,y)-f(y,x)$ vanishes. Nov 6, 2015 at 11:10