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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 $k[U]_L$?

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    $\begingroup$ The all-purpose explanation "it is not hard to see" could use a reference here. $\endgroup$
    – Jim Humphreys
    Commented Feb 15, 2012 at 23:14

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It is probably hard to give a complete description. A lot of partial information about this is contained in this paper of Kostant https://arxiv.org/abs/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).

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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 $G = SL_2$.

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  • $\begingroup$ Thanks for the comments; perhaps this is another way of stating my question: The structure of $k[U]_L$ as a $U$-module can be fruitfully analyzed by a construction involving $G$-modules, so can a similar thing be done to understand $k[U]_C$? In fact the paper of Joseph that I link to above shows that $k[U]_L$ even has the structure of a module for the hyperalgebra of $G$ compatible with the $U$-module structure (the dual zero Verma structure), although I don't expect such a thing occurs for $k[U]_C$. (cont'd) $\endgroup$
    – Chuck Hague
    Commented Feb 16, 2012 at 16:11
  • $\begingroup$ (cont'd) In the case of $SL_2$, the construction of Kumar shows that each standard module for $G$ (which in this case is just characterized by its dimension) occurs naturally as a $U$-submodule of $k[U]_L$. I would like to know if something equally nice happens for $k[U]_C$. $\endgroup$
    – Chuck Hague
    Commented Feb 16, 2012 at 16:13
  • $\begingroup$ Look at Kostant's paper that I mentioned above - he does exactly what you want for the $U$-invariants in $k[U]_C$. $\endgroup$
    – Alexander Braverman
    Commented Feb 16, 2012 at 19:41

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