# sl(2,C) subalgebras, 2

## Finding sl(2,C) subalgebras

### Statement of theorem

This time, we will prove:

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.

### 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:

• little h is the complexification of the Lie algebra of the maximal torus.

• 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.

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.