The Unapologetic Mathematician

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 $L_i$ $I=L_i$.

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 $L$.