Today’s is going to be a long post, but an important one. It tells us precisely what weights are allowed to occur as highest weights in finite-dimensional representations of a semisimple Lie algebra.

**Dominant integral weights **

Let be a finite-dimensional simple representation of a semisimple Lie algebra , with Cartan subalgebra , and root space decomposition . Suppose given a base and a corresponding division .

For each , choose such that and generate a subalgebra isomorphic to .

Consider the weight space decomposition

where denotes the set of weights of . Then if is the weight associated to a highest weight vector, is necessarily a nonnegative integer by the representation theory of . In other words,

Any weight satisfying that identity for all is called **dominant integral**. We have shown that the highest weight of a finite-dimensional simple -representation is necessarily dominant integral. In fact, given a dominant integral weight, we can actually construct such a finite-dimensional simple module.

The set of merely integral weights—those with for —form a lattice, spanned by vectors such that , where the last is the Kronecker delta.

Theorem 1The unique simple quotient of the Verma module is finite-dimensional if and only if is dominant integral.

We will show that if for all , then finite-dimensionality follows.

Let the simple quotient be denoted . We shall prove the theorem in a series of lemmas.

Lemma 2Let be the maximal vector in . Then is annihilated by a power of .

To see this, we recall the commutation relations in :

and

The first is just a fact about and was proved earlier. For the second, it reduces (because bracketing is a derivation) to checking that .

We can now prove the lemma. It is already known that acts locally nilpotently (because it does on the Verma module). The hard part is to show that does so. Let the highest weight vector be . Then

by the first identity above. So it follows that is actually annihilated by . It is also annihilated by by the second commutation relation.

In other words, is a highest weight vector—of a weight lower than ! This cannot happen unless .

Lemma 3is the sum of finite-dimensional -modules.

Indeed, let be the sum of all finite-dimensional -modules contained in . Then , so . Next, I claim that , which will imply by irreducibility. Indeed, is the sum of finite-dimensional objects, and

By maximality of , we have .

Corollary 4act locally nilpotently on .

This is because local nilpotence is true for finite-dimensional representations (this is the theory again).

The next step is to prove that the weights of are invariant under the Weyl group. This is nontrivial.

**Another realization of reflections **

Fix a positive root . We will now define an automorphism of that preserves and induces the reflection on , via the isomorphism from the Killing form.

This is

The exponential of a derivation is an automorphism, so is an automorphism too. Let us compute how it acts on . If is orthogonal to (or equivalently, ), then

It can also be checked directly, using the commutation relations for , that

It now follows that indeed has the appropriate properties.

Now, consider a -module (with the Lie algebra representation ) on which act locally nilpotently, so that

is well-defined. We will use heavily below.

**The weights are invariant under the Weyl group **

Lemma 5Suppose act locally nilpotently on . Then for any weight.

Let .

First, we have some linear algebra. Let be nilpotent, arbitrary. Then:

which is

This is because as a nilpotent operator on is given by for left nad right translation. Then commute so

As a consequence, we may prove the lemma:

(Whoops. Looks like the end of this isn’t appearing. It’s supposed to end . I.e., the ad part acts on the , which itself is a transformation of .)

In this huge mess, is a homomorphism of Lie algebras, so this becomes

which, finally, proves the lemma; we have the appropriate weight.

**Proof of finite-dimensionality **

So, we now know that ‘s weights are invariant under the Weyl group, because the simple reflections generate it, and we have the previous lemma. I now claim there are only finitely many weights; this, when proved, will imply that is finite-dimensional, since each weight space is finite.

Now recall that any weight of must satisfy that is a sum (with nonnegative integral coefficients) of positive roots.

There is a closed Weyl chamber , of **dominant** weights. Any weight of can be conjugated into this Weyl chamber, so it suffices to prove that the number of weights of in this Weyl chamber is finite.

Suppose is a dominant weight such that for each nonegative. Then I claim . Indeed,

because is dominant, and is a sum with each .

There are thus only finitely many weights which are dominant, is integral, and such that is eligible to be weights of ; this is because has to lie in a closed ball and in a discrete set.

Phew! Next time I’m going to aim for character formulas.

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