5
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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)

$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$ Nov 3, 2015 at 15:53
  • $\begingroup$ But that is really what is happening: $W^{\perp}/W$ is indeed the aniso tropic part of the Hermitian form. $\endgroup$ Nov 3, 2015 at 15:55
  • $\begingroup$ 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? $\endgroup$ Nov 3, 2015 at 15:56
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 3, 2015 at 16:36
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 6, 2015 at 11:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.