Lemma 3

Let M be a $\mathfrak{g}$-module which admits a character ch(M) whose support is contained in a finite union of the sets $\mu-P_+$. Let U be the enveloping algebra of $\mathfrak{g}$, Z the centre of U$,\lambda_0\in \mathfrak{h}^*$, and $\chi_{\lambda_0}$ the corresponding homomorphism from Z to $k($§8, Cor. 1 of Th. 2). Assume that, for all $z\in Z,z_M$ is the homothety with ratio $\chi_{\lambda_0}(z)$. Let $D_M$ be the set of $\lambda \in W(\lambda_0+\rho )-\rho$ such that $\lambda + Q_+$ meets Supp(ch M). Then ch(M) is a $\mathbf{Z}$-linear combination of the ch $Z(\lambda )$ for $\lambda \in D_M$.

If Supp(ch M) is empty, the lemma is clear. Assume that Supp(ch $M$)$\not=\emptyset$. Let $\lambda$ be a maximal element of this support, and put dim $M^{\lambda}=m$. There exists a $\mathfrak{g}$-homomorphism $\varphi$ from $(Z(\lambda ))^m$ to M which maps $(Z(\lambda )^{\lambda})^m$ bijectively onto $M^{\lambda}($§6, no. 3, Prop. 6 (i)). Thus, the central character of $Z(\lambda )$ is $\chi_{\lambda_0}$, so $\lambda \in W(\lambda_0+\rho )-\rho ($§8, no. 5, Cor. 1 of Th. 2). This proves that $D_M\not=\emptyset$, and allows us to argue by induction on Card $D_M$. Let L and N be the kernel and cokernel of $\varphi$. Then we have an exact sequence of $\mathfrak{g}$-homomorphisms:

so

ch(M) $=-$ch(L) $+m$ ch $Z(\lambda ) +$ ch(N)

(§7, no. 7, formula (6)). The sets Supp(ch L) and Supp(ch N) are contained in a finite union of sets $\mu-P_+$. For $z\in Z,z_L$ and $z_N$ are homotheties with ratio $\chi_{\lambda_0}(z)$. Clearly, $D_N\subset D_M$. On the other hand, $(\lambda + Q_+)\cap$Supp(ch $M$) $=\{\lambda \}$, and $\lambda \notin$ Supp(ch N), so $\lambda \notin D_N$ and

Card $D_N<$ Card $D_M$.

On the other hand, L is a submodule of $(Z(\lambda ))^m$; if $\lambda '\in D_L$, then $\lambda '+ Q_+$ meets Supp(ch $L$)$\subset$ Supp ch $Z(\lambda )$, so $\lambda \in \lambda '+ Q_+($§6, no. 1, Prop. 1 (ii)); it follows that $D_L\subset D_M$. Since $L\cap (Z(\lambda )^{\lambda})^m= 0$, we have $\lambda \notin D_L$, so

Card $D_L<$ Card $D_M$.

It now suffices to apply the induction hypothesis.