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

8 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

  7. I do get the proof, but do you have some reference for this?

    Comment by Peter | November 15, 2017 | Reply

    • I don’t have my library immediately at hand, but it’s a pretty standard result in the representation theory of Lie Algebras, and I’d think most textbooks would contain it. Humphreys probably does, but I’d have to look it up to be certain.

      Comment by John Armstrong | November 16, 2017 | Reply


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