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.