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.

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September 11, 2012 - Posted by | Algebra, Lie Algebras

6 Comments »

  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


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