generators of the ideal of an unipotent-generated algebraic group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:06:33Z http://mathoverflow.net/feeds/question/16988 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16988/generators-of-the-ideal-of-an-unipotent-generated-algebraic-group generators of the ideal of an unipotent-generated algebraic group yell 2010-03-03T18:31:31Z 2010-03-04T18:07:33Z <p>Given any affine algebraic group $G$ over an algebraically closed field $\mathbb{F}$ of characteristic $0$ with a faithfull representation in $GL_n(\mathbb{F})$ . If one knows the generators of the corresponding ideal, what can be said about the generators of $G^u$. Here $G^u$ shall denote the group generated by all unipotent elements of $G$. (Unlike the case where $G$ is irreducible and solvable, this group is not necessarily unipotent).</p> <p>I am particular interested in bounds on the degrees of the generators; also any reference, which deals with unipotent generated groups is welcome.</p> http://mathoverflow.net/questions/16988/generators-of-the-ideal-of-an-unipotent-generated-algebraic-group/17049#17049 Answer by Pavel Etingof for generators of the ideal of an unipotent-generated algebraic group Pavel Etingof 2010-03-04T03:20:48Z 2010-03-04T03:31:33Z <p>Suppose we are over an algebraically closed field, and $G$ is connected. Then, we have an exact sequence <code>$$1\to U\to G\to G_r\to 1,$$</code> where $U$ is the unipotent radical of $G$, and $G_r$ is a reductive group. Since a semisimple or unipotent group is generated by unipotent elements, this implies that $G^u$ is the intersection of the kernels of all the characters of $G$. Characters of $G$ are grouplike elements of the Hopf algebra ${\mathcal O}(G)$. So the additional relations are that some grouplike elements <code>$g_j\in {\mathcal O}(G)$</code> generating the group of characters of $G$ are equal to $1$. </p>