Decomposition of the ring of functions on the unipotent radical of a Borel - MathOverflow

Decomposition of the ring of functions on the unipotent radical of a Borel

2012-02-15T21:53:27Z

Background

Let $G$ be a semisimple linear algebraic group over an algebraically closed field $k$ of arbitrary characteristic. Let $B \subseteq G$ be a Borel subgroup of $G$ and let $U \subseteq B$ be its unipotent radical. Consider $k[U]$ as a $U$-module under left multiplication in $U$; let's call this module $k[U]_L$. By, say, identifying $U$ with the big cell in the flag variety $G/B^-$, it is not hard to see that $k[U]_L$ is isomorphic as a $U$-module to a direct limit of standard modules $H^0(\lambda)$ for $G$. (In characteristic 0 one can also see this by identifying $k[U]$ with the dual zero Verma module for the enveloping algebra of $G$, cf this paper). In fact there is even a natural ring structure on this direct limit such that this isomorphism is a ring isomorphism.

EDIT: A reference for this fact is Lemma 2.5 and the discussion following it in "The Nil Hecke Ring and Singularity of Schubert Varieties," by Shrawan Kumar. Although the construction is done there over $\mathbb C$, it works over any algebraically closed field, and although Kumar only proves that there is a $T$-equivariant morphism, it is easy to check that the morphism he constructs is $U$-equivariant.

Question

Now consider $k[U]$ as a $U$-module under the conjugation action of $U$ on itself; let's call this module $k[U]_C$. Is there a nice description of $k[U]_C$ as a $U$-module in terms of $G$-modules in a way analogous to the description of $k[U]_L$?

Answer by Alexander Braverman for Decomposition of the ring of functions on the unipotent radical of a Borel

2012-02-15T22:42:00Z

It is probably hard to give a complete description. A lot of partial information about this is contained in this paper of Kostant http://front.math.ucdavis.edu/1201.4494 (in particular there is a description of the subring of invariant elements in $k[U]$; in fact Kostant works with the Lie algebra of $U$ instead of $U$ itself, but clearly they are isomorphic as $U$-varieties with respect to the adjoint action).

Answer by Jim Humphreys for Decomposition of the ring of functions on the unipotent radical of a Borel

2012-02-15T23:12:13Z

At first sight it's unclear how much the question has to do with the ambient semisimple group $G$ : consider the simplest case when the rank is 1 and $U$ is just the additive group. It's true that the conjugation action of $G$ itself on its function algebra has a rich structure. This was shown in characteristic 0 by Kostant and then in a more algebraic setting by Richardson, after which Steve Donkin (Invent. Math. 91, 1988) generalized their results to almost all prime characteristics; his paper is freely available online at

http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002104814

The maximal unipotent subgroup $U$ plays an essential role in the study of a semisimple group, but as a variety it just has the structure of an affine space. So its actions on its own functions won't provide direct information related to the representations of $G$ .

ADDED: The answer to the question as stated is no, based on the rank 1 case where there is no useful connection between the (trivial) conjugation action of $U$ and the representation theory of $G$ . Even with the added references, I haven't been able to figure out what is really being asked in this smallest case. It would help to start with a more explicit formulation when $G = SL_2$ .