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If $\mathfrak{g}$ is a semisimple Lie algebra, is it true that $$ [\mathfrak{g}, \mathfrak{g}] = \mathfrak{g}.$$

Obviously the derived ideals must stabilize somewhere, but I don't see why $D^{1}(\mathfrak{g}) = \mathfrak{g}$. It seemed my professor was stating this result as a consequence of Weyl's Theorem, but I am not sure why this should follow (or even if this is true).

seeker_after_truth
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  • The answer assumes characteristic $0$ so that semisimple = sum of simples. For a counterexample in positive characteristic, see https://math.jhu.edu/~sakellar/automorphic-files/liestructure.pdf, remark 2.2. – Torsten Schoeneberg Oct 17 '22 at 17:51

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Yes, it is true. A semisimple Lie algebre is a direct sum $\bigoplus_{j=1}^n\mathfrak g_j$ of simple Lie algebras. And $[\mathfrak g_j,\mathfrak g_j]$ is an ideal of $\mathfrak g_j$. Since $\mathfrak g_j$ is simple, this ideal can only be $\mathfrak g_j$ itself. So$$[\mathfrak g,\mathfrak g]=\bigoplus_{j=1}^n[\mathfrak g_j,\mathfrak g_j]=\bigoplus_{j=1}^n\mathfrak g_j=\mathfrak g.$$

José Carlos Santos
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