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Let $S$ be a vector subspace of the Lie algebra $\mathfrak{g}$.

Is $S$ an ideal of the normalizer $N_\mathfrak{g}(S)$, I would say yes since:

$\forall n\in N_\mathfrak{g}(S), [n,S]\subseteq S\implies [N_\mathfrak{g}(S),S]\subseteq S$ which is the definition of an ideal right?

duffymo
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So many hats
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  • Thanks @duffymo. I didn't know that Which is pretty bad – So many hats Aug 17 '15 at 11:28
  • Do you have any doubt about your argument ? If yes, could you say what exactly is doubtful ? – Dietrich Burde Aug 17 '15 at 11:34
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    @DietrichBurde Since $S$ isn't a subalgebra it shouldn't work, since $i$ is an ideal of $g$ $\implies$ $i$ is a subalgebra of $g$

    But here that isn't the case by assumption, but I can't see why the subalgebra condition is needed.

    $S$ is not necessarily contained in the normaliser, but $[N_\mathfrak{g}(S),S]\subseteq S$ seems to be true?

     – So many hats Aug 17 '15 at 11:41
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    @Lie Algebra - a small thing. Your mathematics are most impressive. No worries. – duffymo Aug 17 '15 at 11:52

1 Answers1

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The normalizer of a subspace $S$ in $\mathfrak{g}$ is the largest subalgebra of $\mathfrak{g}$ containing $S$ as an ideal. Here "containing" means that $S\subseteq N_{\mathfrak{g}}(S)$, i.e., that $[S,S]\subseteq S$. So $S$ is a subalgebra, and of course then also an ideal, as you have shown.

Dietrich Burde
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