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If $[A,B]$ and $A$ are in Lie algebra $\mathcal{L}$ then does this imply that $B \in \mathcal{L}$?

Matta
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  • Do you mean $\mathcal{L}$ to be a subalgebra of some larger Lie algebra? In that case, no, and otherwise, the question does not really make sense. – Tobias Kildetoft Feb 02 '15 at 14:26
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    The keyword for a counterexample is "ideal". – Jason DeVito - on hiatus Feb 02 '15 at 14:31
  • So in my head I have my Lie algebra, say for the orthogonal group. I find a random matrix and I know that it satisfies the condition on the left hand side. I would like to know if the right hand side holds. I felt that the answer was no but I couldn't think of a counter-example. – Matta Feb 02 '15 at 14:32
  • @JasonDeVito - Thank you very much, I knew I was missing something obvious...admittedly very obvious. – Matta Feb 02 '15 at 14:38
  • @Uther: It always seems obvious in retrospect ;-). If you'd like to write up an answer yourself, please feel free. – Jason DeVito - on hiatus Feb 02 '15 at 14:46
  • My keyword for counterexample is "$0$". –  Mar 17 '16 at 15:59

1 Answers1

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An ideal $I$ of a Lie algebra $\mathfrak{g}$ is a subspace of $\mathfrak{g}$ such that \begin{equation} [\mathfrak{g},I] \subseteq I. \end{equation} The ideal $I$ forms a subalgebra of $\mathfrak{g}$.

Therefore if the Lie algebra $\mathcal{L}$ in the question is some ideal and one chooses any $A \in \mathcal{L}$ and some $B$ in the larger algebra then we would get \begin{equation} [A,B] \in \mathcal{L}. \end{equation} even for $B \notin \mathcal{L}$.

This provides a counter example and so the answer to the question is a (somewhat resounding) NO!

Matta
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