Casimir constrained Verma character expansion

Fix a target weight $\tau$. For a subquotient $X$ of $V$, let $p(X)={\sum }_{\eta \ge \tau }\dim X_{\eta }$, a finite nonnegative integer by F1. If nonzero, choose a maximal weight $\mu \ge \tau$ in its finite support window and $0\mathrm{\ne }v\in X_{\mu }$. All positive root operators kill $v$, since a nonzero image would have larger weight in that same window. Thus $v$ is a highest vector, hence primitive in F6's terminology. Its cyclic submodule $H$ is a quotient of $M(\mu )$ by F3 and has simple quotient $L(\mu )$ by F4: the kernel of $M(\mu )\to H$ is proper and is contained in the unique maximal proper submodule. Write $K$ for the kernel of $H\to L(\mu )$. Both $p(K)<p(X)$ and $p(X\mathrm{/}H)<p(X)$, because the nonzero top weight of $H$ survives in $L(\mu )$.

The two exact sequences give $\mathrm{ch}X=\mathrm{ch}K+\mathrm{ch}L(\mu )+\mathrm{ch}(X\mathrm{/}H)$. Induction on $p(X)$ in step 1.1 expresses the character above $\tau$ as a finite sum of characters of simple subquotients, with nonnegative integral multiplicities, plus terms zero on that window. The same applies to every Verma module by F3. These coefficients are unique where needed: order the finite interval $[\tau ,\mathrm{\Lambda }]$ by decreasing height relative to its top; the character of $L(\mu )$ has top coefficient one and all other weights strictly lower. Successively subtracting top coefficients determines each multiplicity. Enlarging the window therefore gives the same answers on the old window. This defines a coefficientwise locally finite simple-character expansion without assuming finite length of $V$.

F5 acts on $V$ as $a=(\mathrm{\Lambda }+2\rho ,\mathrm{\Lambda })$. Its pointwise finite defining operator commutes with passage to a submodule or quotient, so every simple subquotient in step 2.1 also has scalar $a$. On $L(\mu )$, F5 gives scalar $(\mu +2\rho ,\mu )$. Their equality is exactly $\mathrm{\parallel }\mu +\rho {\mathrm{\parallel }}^2=\mathrm{\parallel }\mathrm{\Lambda }+\rho {\mathrm{\parallel }}^2$, where the squares denote the bilinear form, not a positive norm assumption. Applying the same reasoning to $M(\nu )$ shows that its simple-character expansion has coefficient one at $\nu$ and no coefficient across distinct shifted-square values.

On every finite interval, the matrix expressing Verma characters in simple characters is integral unitriangular. Its inverse is integral: if the strictly triangular part is $N$, then $I-N+N^2-\cdots$ stops after the number of interval weights. By step 3.1 it is block diagonal by shifted-square value, and each power and the inverse preserve those same blocks. Inverting the simple expansion of $V$ consequently gives the asserted constrained Verma expansion. The coefficients agree across windows by the same top-down recursion; equivalently they are the coefficients of the uniquely defined series $P\mathrm{ch}V$, where $\mathrm{ch}M(\mu )=e^{\mu }{P}^{-1}$ by F3. The top coefficient is one because a nonzero highest module is a quotient of $M(\mathrm{\Lambda })$ and its top survives. All windows and vector selections above are finite; the unique coefficient recursion defines the global series without AC.