# The Unapologetic Mathematician

## 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