I’m now aiming to get to the major character formulas for the (finite-dimensional) simple quotients ${L(\lambda)}$ for ${\lambda}$ dominant integral. They will follow from formal manipulations with character symbols and a bit of reasoning with the Weyl group. First, however, it is necessary to express ${\mathrm{ch} L(\lambda)}$ as a sum of characters of Verma modules. We will do this by considering any highest weight module of weight ${\lambda}$ for ${\lambda}$ integral but not necessarily dominant, and considering a filtration on it whose quotients are simple modules ${L(\mu)}$ where there are only finitely many possibilities for ${\mu}$. Applying this to the Verma module, we will then get an expression for ${\mathrm{ch} V(\lambda)}$ in terms of ${\mathrm{ch} L(\lambda)}$, which we can then invert.

First, it is necessary to study the action of the Casimir element (w.r.t. the Killing form). Recall that this is defined as follows: consider a basis ${B}$ for the semisimple Lie algebra ${\mathfrak{g}}$ and its dual basis ${B'}$ under the Killing form isomorphism ${\mathfrak{g} \rightarrow \mathfrak{g}^{\vee}}$. Then the Casimir element is

$\displaystyle \sum_{b \in B} b b^{\vee} \in U \mathfrak{g}$

for ${b^{\vee} \in B'}$ dual to ${b}$. As we saw, this is a central element. I claim now that the Casimir acts by a constant factor on any highest weight module, and that the constant factor is determined by the weight in such a sense as to give information about the preceding filtration (this will become clear shortly).

Central characters

Let ${D \in Z(\mathfrak{g}) := \mathrm{cent} \ U \mathfrak{g}}$ and let ${v_+ \in V(\lambda)}$ be the Verma module. Then ${Dv_+}$ is also a vector with weight ${v_+}$, so it is a constant multiple of ${v_+}$. Since ${v_+}$ generates ${V(\lambda)}$ and ${D}$ is central, it follows that ${D}$ acts on ${V(\lambda)}$ by a scalar ${\mathrm{ch}i_{\lambda}(D)}$. Then ${\mathrm{ch}i_{\lambda}}$ becomes a character ${ Z(\mathfrak{g}) \rightarrow \mathbb{C}}$, i.e. an algebra-homomorphism. There is in fact a theorem of Harish-Chandra that states that ${\mathrm{ch}i_{\lambda}}$ determines the weight ${\lambda}$ up to “linkage” (i.e. up to orbits of the dot action of the Weyl group: ${w \dot \lambda := w(\lambda + \rho) - \rho}$), though I shall not prove this here.

The goal is to compute ${\mathrm{ch}i_{\lambda}(C)}$ for ${C}$ the Casimir element as above. We shall first find a general expression.

By the PBW basis theorem, if we choose a basis ${h_1 \dots h_l }$ of the Cartan subalgebra ${\mathfrak{h}}$ and a ${\mathfrak{sl}_2}$-type basis ${f_1 \dots f_m, e_1 \dots e_m}$ of ${\bigoplus_{\alpha \in \Phi} \mathfrak{g}_{\alpha}}$ (with ${f_i}$ being paired with ${e_i}$ and spanning a subalgebra isomorphic to ${\mathfrak{sl}_2}$ with ${[f_i, e_i]}$, etc.), then we can find a basis for ${U\mathfrak{g}}$ via

$\displaystyle f_1^{i_1} \dots f_m^{i_m} h_1^{j_1} \dots h_l^{j_l} e_1^{k_1} \dots e_m^{k_m} \quad (*).$

There is a vector subspace ${U \mathfrak{h} \subset U\mathfrak{g}}$, and any weight ${\lambda}$ extends to a character ${U \mathfrak{h} \rightarrow \mathbb{C}}$, still denoted ${\lambda}$. With respect to the above basis, there is a projection ${pr: U \mathfrak{g} \rightarrow U \mathfrak{h}}$.

Proposition 1 $\displaystyle \mathrm{ch}i_{\lambda}(z) = \lambda(pr(z)), \quad \forall z \in Z(\mathfrak{g}).$

This computes the character ${\mathrm{ch}i}$.

Write ${z}$ as a sum of terms as in (*). Those where all the ${i,k}$ are zero contribute precisely ${\lambda(pr(z))}$; it must be seen that the others contribute nothing. Any term with some ${k_r>0}$ must annihilate the vector ${v_+}$ because these correspond to positive weights. I claim now that no term of the form

$\displaystyle f_1^{i_1} \dots f_m^{i_m} h_1^{j_1} \dots h_l^{j_l}$

can even appear in the expansion for ${z}$ in this basis. Then bracketing with any ${h \in \mathfrak{h}}$ is an operator on ${U \mathfrak{g}}$ diagonalizable with respect to this basis. Moreover:

$\displaystyle [h, f_1^{i_1} \dots f_m^{i_m} h_1^{j_1} \dots h_l^{j_l}] = \left( \sum_q i_q \alpha_q(h) \right) f_1^{i_1} \dots f_m^{i_m} h_1^{j_1} \dots h_l^{j_l}$

if ${\alpha_q}$ ranges over the weights corresponding to ${f_q}$. If we choose ${h}$ so that the thing in brackets is nonzero, we find that this is impossible, because anything in the kernel of bracketing with ${h}$ (e.g. in the center of ${U \mathfrak{g}}$) cannot have any term in this basis with nonzero eigenvalue.

This proves the proposition.

Action of the Casimir

It is now necessary to compute ${\lambda(pr(C))}$ for ${C}$ the Casimir.

First, choose a basis ${h_i}$ for ${\mathfrak{h}}$ and let ${k_j}$ be the dual basis. Next, choose for each root ${\alpha \in \Phi}$ some ${u_{\alpha} \in \mathfrak{g}_{\alpha}}$ and a dual ${v_{\alpha} \in \mathfrak{g}_{-\alpha}}$; then ${B(u_{\alpha}, v_{\beta}) = \delta_{\alpha \beta}}$, of course. We can arrange things such that ${u_{\alpha} = v_{-\alpha}, v_{\alpha} = u_{-\alpha}}$.

We have

$\displaystyle C = \sum h_i k_i + \sum u_{\alpha} v_{\alpha} = \sum h_i k_i + \sum_{\alpha \in \Phi^+} u_{\alpha} v_{\alpha} + v_{\alpha} u_{\alpha}$

by the way we arranged things. However, the term ${u_{\alpha} v_{\alpha}}$ is not in the right form for the basis. So we write this as

$\displaystyle \sum h_i k_i + 2 \sum_{\alpha \in \Phi^+} v_{\alpha} u_{\alpha} + \sum_{\alpha \in \Phi^+} [ u_{\alpha}, v_{\alpha}].$

Now, in the projection ${pr}$, the middle term disappears, so we find that

$\displaystyle \mathrm{ch}i_{\lambda}(C) = \sum \lambda(h_i) \lambda(k_i) + \lambda\left( \sum_{\alpha \in \Phi^+} t_{\alpha} \right)$

where ${t_{\alpha}}$ is the dual to ${\alpha}$ under the identification ${\mathfrak{h} \simeq \mathfrak{h}^{\vee}}$. (This is a general fact about brackets between ${\mathfrak{g}^{\alpha}, \mathfrak{g}^{-\alpha}}$; I proved it here.)

Consider the bilinear form ${(., .)}$ on ${\mathfrak{h}^{\vee}}$ obtained by duality, which makes the dual bases of ${h_i, k_i}$ in ${\mathfrak{h}^{\vee}}$ biorthogonal; then

$\displaystyle (\lambda, \lambda) = \sum \lambda(h_i) \lambda(k_i)$

because this is checked for those dual bases in ${\mathfrak{h}^{\vee}}$. Moreover, by assumption ${(\lambda, \alpha) = (t_{\lambda}, t_{\alpha}) = \lambda(t_{\alpha})}$, so in total we find:

Proposition 2 $\displaystyle \mathrm{ch}i_{\lambda}(C) = (\lambda, \lambda) + 2(\lambda, \rho) = (\lambda + \rho, \lambda+\rho) - (\rho, \rho)$

for ${\rho}$ one-half the sum of the positive roots.

Next: to show that the character of simple quotients is a sum of Verma module characters, for integral weights anyway.