Timeline for Unipotent radical of minimal parabolic subgroup of a unitary group over an arbitrary field
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 7, 2015 at 3:16 | comment | added | Venkataramana | This is really nice; so the unipotent radical of a maximal parabolic is again a two step unipotent group even in the skew field case. | |
| Nov 6, 2015 at 11:10 | vote | accept | Tom De Medts | ||
| Nov 6, 2015 at 11:10 | comment | added | Tom De Medts | 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 3, 2015 at 16:36 | comment | added | Venkataramana | 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 16:28 | comment | added | Tom De Medts | This one-dimensional example sounds helpful. Could you be a little bit more explicit? In particular, could you explicitly describe this Heisenberg group in terms of the hermitian form $h$? (Sorry for insisting for an explicit description, but that is really what I need for my purposes.) | |
| Nov 3, 2015 at 16:02 | comment | added | Venkataramana | that the unipotent radical is not abelian may be seen in several different ways: you may take,instead of $W$, a one dimensional totally isotropic subspace $Ew$ where $w$ is isotropic, and consider the flag $0\subset Ew\subset w^{\perp}\subset E^n$. The group $U'$ which preserves this flag and acts trivially on successive quotients, is the unip radical of a maximal parabolic subgroup, hence $U'\subset U$ where $U$ is a unip radical of minimal parabolic. Now, $U'# is actually a Heisenberg group and hence is not abelian | |
| Nov 3, 2015 at 15:56 | comment | added | Tom De Medts | 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 15:55 | comment | added | Venkataramana | But that is really what is happening: $W^{\perp}/W$ is indeed the aniso tropic part of the Hermitian form. | |
| Nov 3, 2015 at 15:53 | comment | added | Tom De Medts | 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 15:48 | history | answered | Venkataramana | CC BY-SA 3.0 |