So, let’s suppose that we have a splitting of the roots , as before, associated to a semisimple Lie algebra and a Cartan subalgebra . Recall that a vector for a representation of (not necessarily finite-dimensional!) is called a **highest weight vector** if is annihilated by the nilpotent algebra .

Let be a highest weight module, generated by a highest weight vector . We proved before, using a PBW basis for , that is the direct sum of its finite-dimensional weight spaces—in particular, acts semisimply, which is not a priori obvious since is finite-dimensional—and *so is any subrepresentation.* The highest weight space is one-dimensional. Now I am actually going to talk about them in a bit more detail.

Proposition 1is indecomposable and has a unique maximal submodule and unique simple quotient.

Indeed, let be any proper submodules; we will prove . If either contains , then it is all of . So we may assume both don’t contain ; by the above fact that decompose into weight spaces, they have no vectors of weight the same as . So neither does , which means that .

We can actually take the sum of all proper submodules of ; the above argument shows that this sum does not contain (and has no vectors with nonzero -component). The rest of the proposition is now clear.

There is an important category, the **BGG category **, defined as follows: if is a representation of on which acts locally nilpotently (i.e., each is annihilated by some power of in ), acts semisimply, and is finitely generated over the enveloping algebra . I’m hoping to say a few things about category in the future, but for now, what we’ve seen is that highest weight modules belong to it. It is in fact a theorem that any object in has a filtration whose quotients are highest weight modules.

Proposition 2Any simple highest weight modules of the same weight are isomorphic.

Suppose have highest weight vectors with the same weight and are simple. Then has the vector , which is a highest weight vector with the same weight . Let be , i.e. the submodule generated by . Then there are homomorphisms which are both surjective, since the images contain (respectively).

Anyway, are then both simple quotients of . But this means they are isomorphic by the first proposition.

**Verma modules **

Now, I claim that there is such a thing as a universal highest weight module. In other wrods, fix a weight . Now if is a highest weight vector with weight and a homomorphism of -representations, then is a highest weight vector with weight too.

So, define a covariant functor sending to the collection of highest weight vectors in with weight .

Proposition 3is co-representable.

To say that is a highest weight vector with weight is to say that is annihilated by the left ideal in generated by and for . Therefore, we have

where the bijection is functorial. In other words, we see that is the object in question, which proves the proposition.

The object is denoted ; it is called a **Verma module**. Note that it has a highest weight vector: the image of , and the weight is . The Verma module then is the freest possible highest weight module, in a sense; note that it belongs to the category .

Corollary 4For each , there is a unique simple -representation which is a highest weight module with weight .

It is clear that cannot be isomorphic to for , as simple modules can have at most one highest weight vector (up to scalar multiplication).

In general, is infinite-dimensional. Often , in fact. I aim to discuss criteria for this in the sequel.

I should probably say another (clearly equivalent) way the Verma module can be constructed. Let be the Borel subalgebra. We have a -module where the nilpotent subalgebra acts by zero and acts by . Then

One way to see this, for instance, is that the module defined by the tensor product above satisfies the universal product: for any , we have

An object representing a functor is unique by Yoneda’s lemma.

## Leave a Reply