Generators and relations for the enveloping algebra of a unipotent radical - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T07:23:57Z http://mathoverflow.net/feeds/question/92030 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92030/generators-and-relations-for-the-enveloping-algebra-of-a-unipotent-radical Generators and relations for the enveloping algebra of a unipotent radical Chuck Hague 2012-03-23T18:49:51Z 2012-03-23T18:49:51Z <p>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 $\mathfrak n_P$.</p> <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 $$E_{\alpha_i}^{(c)}*E_{\alpha_j} = 0 \textrm{ for } c > -\alpha_j(\alpha_i^\vee) .$$</p> <p>Now to my question: Is there a nice presentation of $U(\mathfrak n_P)$ by generators and relations given in a similar fashion?</p>