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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.

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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 $1$.

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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

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