MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

I am particular interested in bounds on the degrees of the generators; also any reference, which deals with unipotent generated groups is welcome.

share|cite|improve this question
up vote 2 down vote accepted

Suppose we are over an algebraically closed field, and $G$ is connected. Then, we have an exact sequence $$ 1\to U\to G\to G_r\to 1, $$ 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 $g_j\in {\mathcal O}(G)$ generating the group of characters of $G$ are equal to $1$.

share|cite|improve this answer
I didn't get yet, how one can use the levi decomposition to see (the non-trivial part of) $G^u=$intersection of kernels$=:H$. Aside from that, that was the answer i was looking for, thanks. Let me fill in some more details, which were helpful to me: There are finitely many of those $g_j$ , since $G/H$ is diagonalizable. (Assuming the representation from above there are n those characters.) Using the isomorphism $X(G)=X(G/H)$ , one should see that the relations necessary are of degree $1$. – yell Mar 4 '10 at 18:06
It suffices to show that when there are no nontrivial characters of $G$, then $G^u=G$. In this case, $G_r$ is semisimple, so generated by unipotent elements. Thus, $G$ is generated by elements whose projection to $G_r$ is unipotent. But such elements are clearly unipotent themselves, since an extension of unipotent groups is unipotent. – Pavel Etingof Mar 4 '10 at 21:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.