Suppose to have a Lie algebra L with a reductive lie subalgebra G. Let l an element of L such that [l,g] is in G for every g in G, is it true that l is an element of G?if not, there are some restriction on G that makes it true?
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$\begingroup$ The answer to your first question is rather obviously no, since if $L = G \times H$ with H abelian then every element $l\in H$ has $[l,g]=0\in G$. $\endgroup$– Robin ChapmanCommented Apr 20, 2010 at 14:26
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$\begingroup$ It is true.But If L doesn't decompose like that? $\endgroup$– Michele TorielliCommented Apr 20, 2010 at 14:33
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$\begingroup$ The fact is that I have in mind a specific example: L is the set of weight zero vector fields in C^n and G is the set of vector fields tangent to an hypersurface. $\endgroup$– Michele TorielliCommented Apr 20, 2010 at 14:39
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2 Answers
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Sorry, this started as a comment, but got too long.
If $G$ is semisimple, then every derivation of $G$ is inner, so that the normalizer $N_L(G)=C_L(G)+G$ where $C_L(G)$ is the centralizer. In this situtaion Michele's condition holds if and only if the centralizer is trivial.
However the case where $G$ is reductive isn't so easy. A reductive $G$ is the direct sum of an Abelian and a semisimple Lie algebra. The Abelian part can have (if at least two-dimensional) non-trivial derivations. These will lift to non-inner derivations of $G$. If we let $L$ be the semidirect product of $G$ with a one-dimensional Lie algebra using this derivation, then $L$ normalizes $G$, $L\ne G$ but $C_L(G)$ is trivial.
answered Apr 20, 2010 at 16:09
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EDIT: Realized I had misread the question.
The set of $l$ such that $[l,g]\in G$ for all $G$ is called the normalizer of $G$. It's not particularly common for the normalizer to be the same as G, though it does happen sometimes. There are a few theorems I know about specific Lie algebras being self-normalizing, such as Cartan and Borel subalgebras in reductive Lie algebras, but it's not very common and I don't know any general condition which guarantees it.
answered Apr 20, 2010 at 14:40
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$\begingroup$ L/G is not necessarily a Lie algebra, right? Do you mean that the centralizer of G in L should be trivial? $\endgroup$– damianoCommented Apr 20, 2010 at 14:59
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$\begingroup$ thanks!are there restriction on L and G to make the centralizer to be zero? $\endgroup$ Commented Apr 20, 2010 at 15:02