The Unapologetic Mathematician

Mathematics for the interested outsider

Decomposition of Semisimple Lie Algebras

We say that a Lie algebra L is the direct sum of a collection of ideals L=I_1\oplus\dots\oplus I_n if it’s the direct sum as a vector space. In particular, this implies that [I_i,I_j]\subseteq I_i\cap I_j=0, meaning that the bracket of any two elements from different ideals is zero.

Now, if L is semisimple then there is a collection of ideals, each of which is simple as a Lie algebra in its own right, such that L is the direct sum of these simple ideals. Further, every such simple ideal of L 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 L to that summand, as we expect for any ideal of a Lie algebra.

If I\subseteq L is any ideal then we can define the subspace I^\perp of vectors in L that are “orthogonal” to all the vectors in I with respect to the Killing form \kappa. The associativity of \kappa shows that I^\perp is also an ideal, just as we saw for the radical. Indeed, the radical of \kappa is just L^\perp. Anyhow, Cartan’s criterion again shows that the intersection I\cap I^\perp is solvable, but since L is semisimple this means I\cap I^\perp=0, and we can write L=I\oplus I^\perp.

So now we can use an induction on the dimension of L; if L has no nonzero proper ideal, it’s already simple. Otherwise we can pick some proper ideal I to get L=I\oplus I^\perp, where each summand has a lower dimension than L. Any ideal of I is an ideal of L — the bracket with anything from I^\perp is zero — so I and I^\perp must be semisimple as well, or else there would be a nonzero solvable ideal of L. By induction, each one can be decomposed into simple ideals, so L can as well.

Now, if I is any simple ideal of L, then [I,L] is an ideal of I. It can’t be zero, since if it were I would be contained in Z(L), which is zero. Thus, since I is simple, we must have [I,L]=I. But the direct-sum decomposition tells us that [I,L]=[I,L_1]\oplus\dots\oplus[I,L_n], so all but one of these brackets [I,L_i] must be zero, and that bracket must be I itself. But this means I\subseteq L_i for this simple summand, and — by the simplicity of L_iI=L_i.

From this decomposition we conclude that for all semisimple L we have [L,L]=L. Every ideal and every quotient of L must also be semisimple, since each must consist of some collection of the summands of L.

September 8, 2012 - Posted by | Algebra, Lie Algebras

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