# Root spaces are 1-dimensional

## Recap

If little g is the complexified Lie algebra of a compact semisimple group K then little g splits little h direct sum a direct sum over roots alpha of subspaces little g alpha where:

• little h is the complexified Lie algebra of a maximal torus T\subset K,

• little g alpha equals the set of X in little g such that little ad_H X equals alpha of H times X for all H in little h is the root space corresponding to the root alpha in little h Z star inside little h R star.

The figure below shows the root diagram of little s l 3 C. We've been gradually discovering that various nice features of this Lie algebra and its root diagram carry over to root diagrams in this more general context. These nice features include:

• If alpha is a root then so is minus alpha.

• If X is in little g alpha and Y is in little g minus alpha then X, Y and H_alpha equals X bracket Y span a subalgebra S alpha isomorphic to little s l 2 C.

• Partway through the proof of the previous point, we saw that X bracket Y equals K of X, Y times alpha sharp in little h, where alpha sharp is Killing-dual to alpha. This depends on X and Y only through their Killing pairing K(X, Y). We also saw that if we pick X and Y suitably (so that K of X, Y equals 2 over K of alpha sharp, alpha sharp) then H alpha equals X bracket Y equals 2 alpha sharp over K alpha sharp, alpha sharp satisfies the commutation relations H alpha bracket X equals 2 X, H alpha bracket Y equals minus 2 Y, and X bracket Y equals H alpha.

• In this video, we will show that:

Proposition:

All the root spaces little g alpha are 1-dimensional.

## Proof

Fix a root alpha and consider the line through alpha. We will sum some of the root spaces along this line:

• over 0, we take the span of H_alpha.

• we take the root spaces little g alpha, little g 2 alpha, etc and little g minus alpha, little g minus 2 alpha etc, that is root spaces corresponding to integer multiples of alpha.

The result is a subspace V equals the span of H_alpha direct sum the direct sum of little g k alphas where k runs over the nonzero integers.

Lemma:

V is preserved by the action of the subalgebra S alpha. Moreover, it is irreducible as a representation of S alpha.

Remark:

Irreducibility is the key thing, because this implies that the root spaces little g k alpha are all 1-dimensional.

Proof:

We need to show that little ad_X, little ad_Y and little ad H alpha preserve V. Let's just do little ad_X (the others are similar/easier).

• We have little ad_X H alpha equals X bracket H alpha, which equals minus H alpha bracket X, which equals minus 2 X, which is in little g alpha, which is contained in V, so little ad_X sends the span of H alpha to something in V.

• We have little ad_X from little g k alpha to little g (k plus 1) times alpha because X is in little g alpha.

• If k + 1 is nonzero we have little g (k + 1) times alpha is contained in V and we're done.

• If k + 1 = 0 then little g (k + 1) alpha equals little g_0, which is not contained in V. In fact, we have V intersect little g_0 equals the span of H_alpha, so we need to show that little ad_X of Y is a multiple of H alpha for any Y is in little g minus alpha. But we saw that if X is in little g alpha and Y is in little g minus alpha then X bracket Y equals K of X, Y times alpha sharp, and alpha sharp is just a multiple of H alpha, so little ad_X applied to little g minus alpha is contained in the span of H_alpha, which is contained in V and we're done.

How do we see that this representation is irreducible? Let's understand the weight space decomposition of V under the action of S alpha. I claim that little g k alpha is a weight space with weight 2 k for the action of H alpha.

This is because if Z is in little g k alpha, we have little ad H alpha of Z equals k alpha of H alpha times Z and, since H alpha equals 2 alpha sharp over K alpha sharp, alpha sharp, we have alpha of H alpha equals 2 alpha of alpha sharp over K alpha sharp, alpha sharp, but alpha of alpha sharp equals K of alpha sharp, alpha sharp, so little ad H alpha of Z equals 2 k Z and Z has weight 2 k under the action of H alpha as required.

So our weight diagram looks like this: dot dot dot, little g minus 2 alpha, little g minus alpha, the span of H alpha, little g alpha, little g 2 alpha, etc in weights dot dot dot, minus 4, minus 2, 0, 2, 4 etc respectively. In particular, the weight space with weight zero is the span of H alpha, which is 1-dimensional. If we decompose V into irreducibles then these irreducibles all have even highest weight (there are only even weights) and in particular they will all have 1-dimensional weight space in weight zero. But there is only room for one such irreducible piece, so V must itself be irreducible.