Casimir separates comparable dominant primitive weights

The formula F4 on $v\in V_{\mu }$ gives $(\mathrm{\Omega }-c(\mu ))v=2{\sum }_{\alpha >0,s}{y}_{\alpha ,s}{x}_{\alpha ,s}v$. It preserves the weight and the sum is finite by F5. Every nonzero $x_{\alpha ,s}v$ lies at a strictly higher support weight. Induct on the finite number of support weights at or above $\mu$. At a maximal weight the displayed right side is zero, so $t-c(\mu )$ annihilates $v$ under $t=\mathrm{\Omega }$. Inductively choose for each of the finitely many nonzero vectors on the right an annihilating polynomial that is a product of linear factors. Their product $q$ annihilates the right side because F1 commutes $\mathrm{\Omega }$ through each $y_{\alpha ,s}$. Thus $q(\mathrm{\Omega })(\mathrm{\Omega }-c(\mu ))v=0$. A general vector has finitely many weight components; the product of their annihilators works for it. Zero vectors allow the constant polynomial $1$. Therefore every vector has an annihilator already split into linear factors, without using the fundamental theorem of algebra.

Let $P(t)={\prod }_{a\in E}(t-a{)}^{r_a}$ annihilate $v$, with distinct finite $E$ and $r_a\ge 1$. Put $p_a=(t-a{)}^{r_a}$, $Q_a=P\mathrm{/}{p}_a$, and $R_a=Q_a\mathrm{/}{Q}_a(a)-1$. Since $Q_a(a)\mathrm{\ne }0$ and $R_a$ is divisible by $t-a$, the polynomial $E_a=(Q_a\mathrm{/}{Q}_a(a)){\sum }_{j=0}^{r_a-1}(-R_a{)}^j$ is $1$ modulo $p_a$ and zero modulo each $p_b$ for $b\mathrm{\ne }a$. These congruences imply ${\sum }_a{E}_a-1$ is divisible by $P$: powers of distinct linear factors are coprime, as the same finite inverse construction shows, so divisibility by each factor can be combined successively. Set $v_a=E_a(\mathrm{\Omega })v$. Then $p_a{E}_a$ is divisible by $P$, hence $v_a\in V^{(a)}$, and $v={\sum }_a{v}_a$. If a finite sum of generalized eigenvectors at distinct eigenvalues is zero, take a product with exponents large enough to annihilate every summand and apply its $E_a(\mathrm{\Omega })$ to that relation. It extracts precisely the $a$ summand, proving uniqueness and that the sum of all $V^{(a)}$ is direct. Uniqueness makes the components independent of the chosen annihilator.

Each $V^{(a)}$ is a linear subspace: a maximum of two exponents annihilates a sum. Centrality F1 makes it stable under every generator. By F5 it is in $\mathcal{O}$ with its induced weights, and simple local nilpotence restricts from $V$. The same Casimir formula preserves every submodule and induces the identical operator on a quotient, because its terms are pointwise finite algebra actions. If a primitive vector of weight $\lambda$ in $V^{(a)}$ has nonzero highest class $\bar{v}$ in a quotient, F1 gives $\mathrm{\Omega }\bar{v}=c(\lambda )\bar{v}$, while some power of $\mathrm{\Omega }-a$ kills it. Therefore $(c(\lambda )-a{)}^N\bar{v}=0$ forces $c(\lambda )=a$. F3 gives dominant integrality of this weight.

If two such dominant primitive weights satisfy $\lambda >\mu$, write $\lambda -\mu ={\sum }_i{n}_i{\alpha }_i$ with nonnegative integers $n_i$, at least one positive. Symmetry of the form and F2 give $c(\lambda )-c(\mu )=(\lambda +\mu +2\rho ,\lambda -\mu )={\sum }_i{n}_i{d}_i(\lambda (h_i)+\mu (h_i)+2)>0.$ Every summand is a nonnegative real number and at least one is strictly positive, although the separate Casimir eigenvalues may be complex. This contradicts their common value $a$ from 3.1. Hence the primitive weights form an antichain. For equal weights the difference is zero; zero simple labels still leave the positive $2d_i$ term. The zero module has the empty direct sum, a single eigenvalue requires only one projector, and all choices of annihilators or bounds in 1.1–2.1 are finite. No AC or positivity of the full bilinear form was used.