So, let’s suppose that we have a splitting of the roots {\Phi = \Phi^+ \cup \Phi^-}, as before, associated to a semisimple Lie algebra {\mathfrak{g}} and a Cartan subalgebra {\mathfrak{h}}. Recall that a vector {v \in V} for a representation {V} of {\mathfrak{g}} (not necessarily finite-dimensional!) is called a highest weight vector if {v} is annihilated by the nilpotent algebra {\mathfrak{n} = \bigoplus_{\alpha \in \Phi^+}}.

Let {V} be a highest weight module, generated by a highest weight vector {v}. We proved before, using a PBW basis for {U\mathfrak{g}}, that {V} is the direct sum of its finite-dimensional weight spaces—in particular, {\mathfrak{h}} acts semisimply, which is not a priori obvious since {V} 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 {V} is indecomposable and has a unique maximal submodule and unique simple quotient.

 

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

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

There is an important category, the BGG category {\mathcal{O}}, defined as follows: {X \in \mathcal{O}} if {X} is a representation of {\mathfrak{g}} on which {\mathfrak{n}} acts locally nilpotently (i.e., each {x \in X} is annihilated by some power of {\mathfrak{n}} in {U\mathfrak{g}}), {\mathfrak{h}} acts semisimply, and {X} is finitely generated over the enveloping algebra {U\mathfrak{g}}. I’m hoping to say a few things about category {\mathcal{O}} 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 {\mathcal{O}} has a filtration whose quotients are highest weight modules.

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

 

Suppose {W_1, W_2} have highest weight vectors {v_1, v_2} with the same weight {\lambda} and are simple. Then {W_1 \oplus W_2} has the vector {(v_1, v_2)}, which is a highest weight vector with the same weight {\lambda}. Let {W' \subset W_1 \oplus W_2} be {U\mathfrak{g} (v_1, v_2)}, i.e. the submodule generated by {(v_1, v_2)}. Then there are homomorphisms {\phi_1: W' \rightarrow W_1, \phi_2: W' \rightarrow W_2} which are both surjective, since the images contain {v_1, v_2} (respectively).

Anyway, {W_1, W_2} are then both simple quotients of {W'}. 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 {\lambda \in \mathfrak{h}^{\vee}}. Now if {v \in V} is a highest weight vector with weight {\lambda} and {f: V \rightarrow V'} a homomorphism of {\mathfrak{g}}-representations, then {f(v)} is a highest weight vector with weight {\lambda} too.

So, define a covariant functor {F_{\lambda} : Rep(\mathfrak{g}) \rightarrow \mathbf{Set}} sending  V to the collection of highest weight vectors in V with weight \lambda

Proposition 3 {F_{\lambda}} is co-representable.

 

To say that {v \in V} is a highest weight vector with weight {\lambda} is to say that {v} is annihilated by the left ideal {I_{\lambda}} in {U\mathfrak{g}} generated by {\mathfrak{n}} and { h - \lambda(h)} for {h \in H}. Therefore, we have

\displaystyle \hom_{Rep(\mathfrak{g})}( U\mathfrak{g}/ I_{\lambda}, V) \simeq F_{\lambda}(V)

where the bijection is functorial. In other words, we see that {U\mathfrak{g}/ I_{\lambda}} is the object in question, which proves the proposition.

The object {U\mathfrak{g}/ I_{\lambda}} is denoted {V(\lambda)}; it is called a Verma module. Note that it has a highest weight vector: the image of {1 \in U\mathfrak{g}}, and the weight is {\lambda}. The Verma module then is the freest possible highest weight module, in a sense; note that it belongs to the category {\mathcal{O}}.

Corollary 4 For each {\lambda \in \mathfrak{h}^{\vee}}, there is a unique simple {\mathfrak{g}}-representation {L(\lambda)} which is a highest weight module with weight {\lambda}.

 

It is clear that {L(\lambda)} cannot be isomorphic to {L(\lambda')} for {\lambda \neq \lambda'}, as simple modules can have at most one highest weight vector (up to scalar multiplication).

In general, {L(\lambda)} is infinite-dimensional. Often {L(\lambda) \simeq V(\lambda)}, 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 {\mathfrak{b} = \mathfrak{h} \oplus \mathfrak{b}} be the Borel subalgebra. We have a {U\mathfrak{b}}-module {1_{\lambda}} where the nilpotent subalgebra {\mathfrak{b}} acts by zero and {\mathfrak{h}} acts by {\lambda}. Then

\displaystyle V_{\lambda} = U\mathfrak{g} \otimes_{U \mathfrak{b}} 1_{\lambda}.

One way to see this, for instance, is that the module {V'_{\lambda}} defined by the tensor product above satisfies the universal product: for any {V \in Rep(\mathfrak{g})}, we have

\displaystyle \hom_{U\mathfrak{g}}( V'_{\lambda}, V) = \hom_{U\mathfrak{b}}( 1_{\lambda}, V) \simeq F_{\lambda}(V).

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

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