Let $G$ be a semisimple algebraic group over an algebraically closed field $k$ of characteristic 0 and let $B \subseteq G$ be a Borel subgroup with unipotent radical $U$. Let $P \supseteq B$ be a parabolic subgroup with unipotent radical $U_P \subseteq U$. Set $\mathfrak n := \textrm{Lie}(U)$ and $\mathfrak n_P := \textrm{Lie}(U_P)$. Then we have the enveloping algebras $U(\mathfrak n)$ and $U(\mathfrak n_P)$ associated to $\mathfrak n$ and $\mathfrak n_P$.

Let $\alpha_1, \ldots, \alpha_\ell$ denote the simple roots of $G$. It is well-known that $U(\mathfrak n)$ is generated as a $k$-algebra by generators $E_{\alpha_i}^{(n)}$ for $1 \leq i \leq \ell$ and $n \geq 0$. (Here we are using the divided-power notation $E_{\alpha_i}^{(n)} = \frac 1 {n!} E_{\alpha_i}^n$). Furthermore, the relations between these generators are given by the Serre relations $$ \sum_{a+b = -\alpha_j(\alpha_i^\vee)+1} (-1)^a E_{\alpha_i}^{(a)} E_{\alpha_j} E_{\alpha_i}^{(b)} = 0 . $$ Succinctly, using the adjoint action $*$ of $U(\mathfrak n)$ on itself, we can write the Serre relations as $$E_{\alpha_i}^{(c)}*E_{\alpha_j} = 0 \textrm{ for } c > -\alpha_j(\alpha_i^\vee) .$$

Now to my question: Is there a nice presentation of $U(\mathfrak n_P)$ by generators and relations given in a similar fashion?

  • $\begingroup$ @Chuck: Is there an ulterior motive for bringing in the divided powers in characteristic 0? Aside from that, you can't expect generators and relations to be all that "similar" for a smaller nilradical, since you're no longer dealing with the set of positive roots in a root system (where the simple roots play a special role). The question is reasonable but needs more in the way of special case evidence to build on. $\endgroup$ – Jim Humphreys Mar 24 '12 at 13:21
  • $\begingroup$ The reason for divided powers is that I want the result in positive characteristic for hyperalgebras, but I figured it would make the question simpler if I stated it for enveloping algebras in characteristic 0. You're probably right that a general statement would be messy, so in some sense this is a fishing expedition. $\endgroup$ – Chuck Hague Mar 24 '12 at 15:25

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