Let $G$ be an algebraic connected subgroup of $GL_n(\mathbb{C})$ and let $\chi : G \to \mathbb{C}^*$ a character. Consider $d_e\chi : \mathfrak{g} \to \mathbb{C}$ the differential of $\chi$ at the identity of $G$. Is it true that $d_e\chi(\nu)=0$ if $\nu$ is a nilpotent element?If not, is it true under some assumptions? Can we say something if $\nu$ is semisimple?


1 Answer 1


The question should be formulated more precisely. The assumption seems to be that $G$ is an algebraic group and that $\chi$ is a morphism of algebraic groups (?) If so, assuming as we may that $\chi$ is nontrivial, $\chi$ induces an isomorphism of a 1-dimensional quotient of $G$ (a torus) onto the image . The corresponding 1-dimensional quotient of the Lie algebra is then isomorphic to the Lie algebra of the 1-torus and thus consists of semisimple elements. In particular, all nilpotent elements of $\mathfrak{g}$ must lie in the kernel of $d_e\chi$. (This uses the algebraic theory, with Chevalley's version of Jordan decomposition in both the group and its Lie algebra along with good behavior of quotients. There are some similar ideas in prime characteristic, but with added complications.)

P.S. If the setting is supposed to be algebraic, the tag algebraic-groups is appropriate here.

SOURCES: In characteristic 0, Chevalley first investigated how the traditional Lie group correspondence between the groups and their Lie algebras would work for algebraic groups. This got combined with Jordan decomposition ideas by Borel (partly in collaboration with Springer), working over more general fields. In the expanded second edition of his older lecture notes, published as Springer GTM 126, see Section 7 for a discussion of characteristic 0, and combine this with Remark 4.9 in the section on Jordan decomposition. The remark points forward to two general results 11.8 and 14.26, which characterize the semisimple and nilpotent elements in the Lie algebra of a connected algebraic group.

  • $\begingroup$ $\chi$ is a character attached to a semi-invariant of the action of $G$ on $\mathbb{C}^n$. thank you. $\endgroup$ Feb 18, 2011 at 14:38
  • $\begingroup$ where can I found material about that? $\endgroup$ Feb 18, 2011 at 14:40

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