Decomposition of Semisimple Lie Algebras
We say that a Lie algebra is the direct sum of a collection of ideals if it’s the direct sum as a vector space. In particular, this implies that , meaning that the bracket of any two elements from different ideals is zero.
Now, if is semisimple then there is a collection of ideals, each of which is simple as a Lie algebra in its own right, such that is the direct sum of these simple ideals. Further, every such simple ideal of is one in the collection — there’s no way to spread another simple ideal across two or more summands in this decomposition. And the Killing form of a summand is the restriction of the Killing form of to that summand, as we expect for any ideal of a Lie algebra.
If is any ideal then we can define the subspace of vectors in that are “orthogonal” to all the vectors in with respect to the Killing form . The associativity of shows that is also an ideal, just as we saw for the radical. Indeed, the radical of is just . Anyhow, Cartan’s criterion again shows that the intersection is solvable, but since is semisimple this means , and we can write .
So now we can use an induction on the dimension of ; if has no nonzero proper ideal, it’s already simple. Otherwise we can pick some proper ideal to get , where each summand has a lower dimension than . Any ideal of is an ideal of — the bracket with anything from is zero — so and must be semisimple as well, or else there would be a nonzero solvable ideal of . By induction, each one can be decomposed into simple ideals, so can as well.
Now, if is any simple ideal of , then is an ideal of . It can’t be zero, since if it were would be contained in , which is zero. Thus, since is simple, we must have . But the direct-sum decomposition tells us that , so all but one of these brackets must be zero, and that bracket must be itself. But this means for this simple summand, and — by the simplicity of — .
From this decomposition we conclude that for all semisimple we have . Every ideal and every quotient of must also be semisimple, since each must consist of some collection of the summands of .
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