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Let $L$ be a lie algebra. Then if $L$ is semisimple, we have $L = L_1 \oplus \cdots\oplus L_n$ for some simple ideals $L_i$. But we can also consider the adjoint representation. In this representation, each $L_i$ will be an irreducible submodule. So we get that the representation is completely reducible.

I don't get why the converse is false. I don't want a counterexample, but rather, an explanation. Since the submodules in the adjoint representation correspond to ideals, the converse should be true, no? If $L$ (when considered as a module with the adjoint representation) is a sum of irreducible modules, then it seems that each of these irreducible submodules will be simple ideals.

nigel
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1 Answers1

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Part of the definition of a simple ideal is that this ideal is nonabelian as a Lie algebra. If $L$ has a one-dimensional ideal $I$, then $I$ is an irreducible submodule of the adjoint representation, but $I$ is abelian and therefore not a simple ideal of $L$.

We say that a Lie algebra $L$ is reductive if its adjoint representation is completely reducible. One can show that $L$ is reductive if and only if it is the direct sum of a semisimple Lie algebra and an abelian Lie algebra.

bradhd
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