The parabolic highest-weight truncation factors through the cyclotomic quotient

Establish the raw affine action without a downstream dependency. Let $Q_c=Q_{\chi }{\otimes }_W{\mathbb{C}}_c$ and write ${\pi }_c(\eta (u)1_{\chi }\otimes 1_c)=u{1}_c$ for the balanced evaluation of F3. For a tensor basis vector $v_{\mathbf{j}}$ write its actual Whittaker preimage as $z={\sum }_{\mathbf{i}}{z}_{\mathbf{i}}\otimes v_{\mathbf{i}}$, so ${\pi }_c(z_{\mathbf{i}})={\delta }_{\mathbf{i},\mathbf{j}}$. If $E_{ab}\in \mathfrak{p}$, then ${\pi }_c(E_{ab}{z}_{\mathbf{i}})={\delta }_{ab}(c_{\mathrm{col}(a)}+{\sum }_{h\ge \mathrm{col}(a)}{q}_h-n){\delta }_{\mathbf{i},\mathbf{j}}$. If $E_{ab}\in \mathfrak{m}$, the Whittaker equation gives $E_{ab}{z}_{\mathbf{i}}=\chi (E_{ab})z_{\mathbf{i}}-{\sum }_{t:i_t=a}{z}_{\mathbf{i}[a\mapsto b\text{ at }t]}$. Apply these identities to $({\pi }_c\otimes 1){\mathrm{\Omega }}_{01}z$. The nonnegative-column terms give the displayed diagonal scalar before subtraction; the negative character terms give the left neighbour of $j_1$; replacement in the first position subtracts ${\sum }_{h>\mathrm{col}(j_1)}{q}_h$, while replacement in any later position $k$ contributes the negative swap precisely when $\mathrm{col}(j_k)<\mathrm{col}(j_1)$. Thus the resulting operator is exactly F1. Transported places are ordinary place permutations by naturality. F3–F4 therefore give the affine relations for these raw operators, with $x_{j+1}=s_j{x}_j{s}_j+s_j$, independently of the lemma's downstream consumer.

On a tensor basis vector the full local formula is $v_{\mathbf{j}}{x}_1=v_{L_1(\mathbf{j})}+a_{\mathrm{col}(j_1)}{v}_{\mathbf{j}}-{\sum }_{k>1:\mathrm{col}(j_k)<\mathrm{col}(j_1)}{v}_{\mathbf{j}\cdot (1k)}$. Let $C_b$ be the span of tensors whose first box has column at most $b$, with $C_0=0$. The left shift and every negative swap strictly lower the first column; the diagonal term preserves it. Hence every $C_b$ is stable and $C_b(x_1-a_b)\subseteq C_{b-1}$. Starting with $C_l=E$, apply the factors from $b=l$ down to $b=1$. Their product kills $E$, and the factors commute since all are polynomials in $x_1$. Thus $f(x_1)=0$ for $f(t)={\prod }_{b=1}^l(t-a_b)$, including coincident roots with all their repeated factors.

For the ordinary ground module, PBW gives an exact induction functor $U(\mathfrak{g}){\otimes }_{U(\mathfrak{p})}-$, since $U(\mathfrak{g})$ is right-free over $U(\mathfrak{p})$. The tensor identity is the $\mathfrak{g}$-linear map $\theta :U(\mathfrak{g}){\otimes }_{U(\mathfrak{p})}({\mathbb{C}}_{\mu }\otimes V)\to P_c\otimes V$, $u\otimes (1_{\mu }\otimes v)\mapsto \sum u_{(1)}(1\otimes 1_{\mu })\otimes u_{(2)}v$, where $\mathrm{\Delta }(E_{ab})=E_{ab}\otimes 1+1\otimes E_{ab}$. The matrix-unit relations show that $\mathrm{\Delta }$ extends multiplicatively to $U(\mathfrak{g})$, and its formula gives the required balancing over $U(\mathfrak{p})$. On the PBW basis of $U(\mathfrak{m})$, $\theta$ has leading term $u(1\otimes 1_{\mu })\otimes v$; all other terms have smaller PBW degree in the first factor. Finite descending subtraction proves it is bijective. Exact induction and $\theta$ carry the flag ${\mathbb{C}}_{\mu }\otimes V_{\le b}$ to a genuine $\mathfrak{g}$-submodule flag $0=F_0\subset F_1\subset \cdots \subset F_l=P_c\otimes V$. This is the column flag of the tensor product, not a finite flag of $P_c$ itself.

Give $E$ the total tensor grading of F5. The left shift raises this degree by one. The diagonal terms have degree zero, and $E_{ij}^{(1)}{E}_{ji}^{(k)}$ has degree $(\mathrm{col}(j)-\mathrm{col}(i))+(\mathrm{col}(i)-\mathrm{col}(j))=0$, regardless of the column gap. Therefore ${\mathrm{gr}}_1{x}_1=e^{(1)}$. For a nonzero polynomial $g$ of degree $r<l$ with leading coefficient $\gamma$, the degree-$r$ symbol of $g(x_1)$ is $\gamma (e^{(1)}{)}^r\mathrm{\ne }0$: a top vector in a longest row has a nonzero $r$th left shift, and the remaining tensor positions can be any nonzero basis vectors. This includes $r=0$, whose symbol is a nonzero scalar identity. Consequently no polynomial of degree less than $l$ annihilates $E$. Together with step 1.2, this proves that $f$ is the exact minimal polynomial, even when roots coincide.

Put $m_0=1\otimes 1_{\mu }\in P_c$ and choose a box $j$ in column $b$. By the matrix-unit action, $\mathrm{\Omega }(m_0\otimes v_j)={\sum }_k{E}_{jk}{m}_0\otimes v_k$. For $\mathrm{col}(k)\ge b$, $E_{jk}\in \mathfrak{p}$, so only $k=j$ survives and contributes ${\mu }_b{m}_0\otimes v_j$. For each earlier-column box $k$, the primitive coproduct formula in step 1.3 gives $E_{jk}{m}_0\otimes v_k=\theta (E_{jk}\otimes (1_{\mu }\otimes v_k))-m_0\otimes E_{jk}{v}_k$. The first term belongs to $F_{b-1}$ and the second is $-m_0\otimes v_j$. There are exactly $Q_{b-1}$ such boxes. Hence $\mathrm{\Omega }(m_0\otimes v_j)\equiv ({\mu }_b-Q_{b-1})m_0\otimes v_j=a_b{m}_0\otimes v_j(\mathrm{m}\mathrm{o}\mathrm{d}{F}_{b-1})$. Since $\mathrm{\Omega }$ is $\mathfrak{g}$-linear and $F_b$ is generated by these ground tensors in columns at most $b$, this also proves that it preserves the full flag and acts by $a_b$ on $F_b\mathrm{/}{F}_{b-1}$.

The raw affine action in step 1.1 kills $f(x_1)$ by step 1.2, so its right representation kills the two-sided ideal generated by this polynomial and descends to $H_d(\mathbf{a})$. The generator-fixing anti-involution fixes $f(x_1)$, so the quotient statement respects the right-action convention. The preceding proof uses only the raw first-column flag and total grading, not a purported Skryabin image of an ordinary category-$\mathcal{O}$ object.

Apply $\mathrm{\Omega }-a_l,\dots ,\mathrm{\Omega }-a_1$ successively to the flag of step 2.2. Each factor moves the relevant flag piece down one column, so their product annihilates $P_c\otimes V$. The factors commute and their order is immaterial. On $P_c\otimes V^{\otimes d}$ the first ordinary Hecke/Casimir operator is ${\mathrm{\Omega }}_{01}$ tensored with the identity on the remaining $d-1$ natural factors. Tensoring the annihilating operator identity therefore gives $f({\mathrm{\Omega }}_{01})=0$ on this entire module. No irreducibility, integral-origin identification or ordinary-to-Whittaker transfer was used.

The raw minimal-polynomial statement and quotient factorization follow from steps 1.1, 1.2, 2.1 and 3.1, and the ordinary induced-module annihilation follows separately from steps 1.3, 2.2 and 3.2, for arbitrary complex $c$ and every nonempty pyramid. For $d=0$ there are no $x_j$, so there is no $x_1$ or minimal polynomial to assert; the Hecke algebra is $\mathbb{C}$. The integral-root weight and arbitrary-complex coset conventions remain those of the cyclotomic definition and play no role in either local calculation.