The Unapologetic Mathematician

Mathematics for the interested outsider

All Derivations of Semisimple Lie Algebras are Inner

It turns out that all the derivations on a semisimple Lie algebra L are inner derivations. That is, they’re all of the form \mathrm{ad}(x) for some x\in L. We know that the homomorphism \mathrm{ad}:L\to\mathrm{Der}(L) is injective when L is semisimple. Indeed, its kernel is exactly the center Z(L), which we know is trivial. We are asserting that it is also surjective, and thus an isomorphism of Lie algebras.

If we set D=\mathrm{Der}(L) and I=\mathrm{Im}(\mathrm{ad}), we can see that [D,M=I]\subseteq I. Indeed, if \delta is any derivation and x\in L, then we can check that

\displaystyle\begin{aligned}\left[\delta,\mathrm{ad}(x)\right](y)&=\delta([\mathrm{ad}(x)](y))-[\mathrm{ad}(x)](\delta(y))\\&=\delta([x,y])-[x,\delta(y)]\\&=[\delta(x),y]+[x,\delta(y)]-[x,\delta(y)]\\&=[\mathrm{ad}(\delta(x))](y)\end{aligned}

This makes I\subseteq D an ideal, so the Killing form \kappa of I is the restriction of I\times I of the Killing form of D. Then we can define I^\perp\subseteq D to be the subspace orthogonal (with respect to \kappa) to I, and the fact that the Killing form is nondegenerate tells us that I\cap I^\perp=0, and thus [I,I^\perp]=0.

Now, if \delta is an outer derivation — one not in I — we can assume that it is orthogonal to I, since otherwise we just have to use \kappa to project \delta onto I and subtract off that much to get another outer derivation that is orthogonal. But then we find that

\displaystyle\mathrm{ad}(\delta(x))=[\delta,\mathrm{ad}(x)]=0

since this bracket is contained in [I^\perp,I]=0. But the fact that \mathrm{ad} is injective means that \delta(x)=0 for all x\in L, and thus \delta=0. We conclude that I^\perp=0 and that I=D, and thus that \mathrm{ad} is onto, as asserted.

September 11, 2012 Posted by | Algebra, Lie Algebras | 6 Comments