Theorem:
If X in the root space little g alpha, Y in the root space little g minus alpha are nonzero then X, Y and H equals X bracket Y will span a subalgebra isomorphic to little s l 2 C.
sl(2,C) subalgebras, 2
Finding sl(2,C) subalgebras
Statement of theorem
This time, we will prove:
Lemma 1
We start by proving a lemma which identifies X bracket Y.
Lemma:
If X in the root space little g alpha and Y in the root space little g minus alpha then X bracket Y equals K of X, Y times alpha sharp.
Remark:
Here alpha is in little h dual and alpha sharp in little is dual to alpha under the Killing form. Recall that:
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little h is the complexification of the Lie algebra of the maximal torus.
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alpha sharp is the unique vector such that K of alpha sharp, v for all v in little h. Uniqueness follows from nondegeneracy of K.
Proof of Lemma 1
For a start, let's show that X bracket Y is in little h. This is because X in the root space little g alpha and Y in the root space little g minus alpha, soX bracket Y is in little g alpha minus alpha, which equals little g_0 and we proved last time that little g_0 equals little h.
To show that X bracket Y equals K of X, Y times alpha sharp, we therefore need to prove that K of X bracket Y, Z equals K of X, Y times alpha of Z for all Z in little h. We have K of X bracket Y, Z equals the trace of little ad of (X bracket Y) times ltitle ad_Z.
Since little ad is a Lie algebra representation, little ad of X bracket Y equals little ad_X times little ad_Y minus little ad_Y times little ad_X. Therefore K of X bracket Y, Z equals the trace of little ad_X times little ad_Y times little ad_Z minus little ad_Y times little ad_X times little ad_Z
Since trace of P Q R equals trace of R P Q equals trace of Q R P for any three matrices P, Q, R, we can cyclically permute little ad_Y times little ad_X times little ad_Z to get little ad_X times little ad_Z times little ad_Y, so overall: K of X bracket Y, Z equals trace of little ad_X times little ad_Y times little ad_Z minus little ad_X times little ad_Z times little ad_Y, which equals trace of little ad_X times little ad of (Y bracket Z).
which is equal to K of X, Y bracket Z. Since Y bracket Z equals minus Z bracket Y, which equals minus little ad_Z Y, which equals alpha of Z times Y because Y is in the root space little g minus alpha and Z is in little h. This proves K of X bracket Y, Z equals alpha of Z times K X, Y as required.
Remark:
Interestingly, the only way that X bracket Y depends on X and Y through the scalar factor K(X, Y). This will become important later.
Proof of theorem
Pick X in the root space little g alpha and Y in the root space little g minus alpha and let H be X bracket Y equals K of X, Y times alpha sharp.
In fact, by rescaling our choice of Y (and hence linearly rescaling K(X, Y)), we can assume that K of X, Y equals 2 over K of alpha sharp, alpha sharp. Note that this makes sense because K of alpha sharp, alpha sharp equals K star of alpha, alpha is nonzero because alpha is in little h R dual and K star is positive definite on little h R dual (because our group is compact and semisimple).
To check that the subalgebra spanned by X, Y and alpha sharp is isomorphic to little s l 2 C, we just need to check that the standard commutation relations hold: H bracket X equals 2 X, H bracket Y equals minus 2 Y and X bracket Y equals H.
X bracket Y equals H holds by definition.
X is in the root space little g alpha so H bracket X equals little ad_H of X, which equals alpha of H times X and alpha of H equals K of X, Y times alpha of alpha sharp, which equals 2 over K of alpha sharp, alpha sharp times alpha of alpha sharp but alpha of alpha sharp equals K of alpha sharp, alpha sharp because alpha sharp was defined by the equation K of alpha sharp, v equals alpha of v for all v in little h.
Therefore H bracket X equals 2 X. The proof of H bracket Y equals minus 2 Y is similar.
Remark:
Here, we used the fact that K is positive definite on little h R, for which we appealed to the fact that our Lie algebra is the Lie algebra of a compact group. We don't need to do this: it is possible to show that K of alpha sharp, alpha sharp is nonzero assuming only semisimplicity, but you need to work harder.