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 1 is 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 2 Any 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.
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 3 is 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 4 For 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.