# 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

1. I have a basic understanding of the nature of (finite) groups. I have played around with S3 for a few hours and days, and have respect for the depth of its properties. I experience both fear and awe when trying to think about larger and larger S groups.

Algebra seems like an infinite maze. The fact there are only countably many possible algebraic expressions is some comfort, but not that much, because my brain feels decidedly finite.

I am trying to get a grip on implications and applications. Does this theorem lead one (eventually) to a better understanding of polynomial equations? Is there something geometrical one can infer? Can I use it to write interesting computer programs?

Comment by Ralph Dratman | September 11, 2012 | Reply

2. It helps simplify the project of classifying Lie algebras and their representations, which turns out to be of use on quite a lot of theoretical physics, for one thing.

Comment by John Armstrong | September 11, 2012 | Reply

3. To me this seems like breaking large rocks for small change, but I guess you have to enjoy it.

Still — to contradict myself — I actually do find this tempting. I wish someone would give me a few hints about those uses in physics. QCD or something like that?

Comment by Ralph Dratman | September 12, 2012 | Reply

4. I’ve mentioned before — though quite a while ago, now — that Lie algebras arise as the “infinitesimal” versions of Lie groups. That is, if you look at a continuously-varying collection of symmetries, if you want to do calculus on it you’re going to end up using Lie algebras. Since quite a lot of modern physics is about symmetries, this comes up a lot.

As for large rocks and small change, I understand the frustration given how hard I’ve twisted your arm to force you to read this stuff.

Comment by John Armstrong | September 12, 2012 | Reply

5. As for me, I find the algebraic approach much easier to wrap my brain around than the other presentations of Lie theory I’ve struggled with. (Still not _easy_, just considerably easier :-) I’ve seen all the topics that have been covered so far in this series many times before, but until now have never had any clue what the heck they meant. I *really* appreciate the way John has presented this material, it’s finally starting to make a bit of sense to me.

Comment by Joe English | September 14, 2012 | Reply

6. I apologize. I certainly did not mean to denigrate your work. On the contrary, I admire it. Otherwise, of course, I would not be reading and asking questions.

Comment by Ralph Dratman | September 14, 2012 | Reply