Let be a semisimple Lie algebra over and a Cartan subalgebra.

Given , we can define a subspace of

The nonzero that occur with are called **roots**, and they form a set . Because acts on by commuting diagonalizable operators (by semisimplicity of the elements of ), it follows by simultaneous diagonalization, that

Recall that , because a Cartan subalgebra is maximal abelian.

This is called the root space decomposition. A simple but important property is that ; this is checked because the are derivations.

The root space decomposition is highly useful in studying simple representations of .

I shall collect here a few facts about it.

Proposition 1are orthogonal under the Killing form unless .

This follows by a familiar argument, in view of .

As a result, we find that if , then too, because there has to be something with root to dualize under the Killing form.

We now consider the commutators .

Proposition 2We have , where is the dual to under the Killing form.

The definition of is that

This turns out to be relatively easy to prove. Let ; then I claim

To see this, we will show that the two quantities above (which are in , on which is nondegenerate) have the same “inner products” (via ) with any other vector in . But by invariance and symmetry of the Killing form:

This proves the claim.

Proposition 3We have .

Choose nonzero with . Suppose . Now the subspace is a subalgebra by the commutation relations. Moreover by hypothesis. Therefore, this is actually a solvable Lie algebra, and its adjoint action on can be made to happen by upper-traingular matrices. In particular must have nilpotent on . However, is also semisimple, which means , contradiction.

Therefore . So, choose an appropriate multiple of such that

Choose scaled appropriately so that . Then satisfy:

These are precisely the relations for . Indeed, this explains a remark I made way back when about being a highly important example. Let the subalgebra of so defined be denoted ; is isomorphic to .

Proposition 4When , is one-dimensional.

We will prove this next time by a more careful look at the representation theory of .

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