A minimal numerical Serre kernel obeys the quantum Casimir constraint

Let $\tilde{U}$ be the algebra generated by all $E_i,F_i,K_h$ with the toral, toral-action and mixed relations, and no relations within either root half. For every integral label $\lambda$, its universal highest module is freely based by $T{v}_{\lambda }$. To verify this without a presentation freeness assumption, define $F_i$ by left concatenation, $K_h$ on the word of degree $-\gamma$ by $t^{⟨\lambda -\gamma ,h⟩}$, and, for $w=F_{i_1}\cdots F_{i_n}$, define $E_{j}w{v}_{\lambda }$ to be the sum over $r$ with $i_r=j$ of $[⟨\lambda -{\sum }_{s>r}{\alpha }_{i_s},h_j⟩{]}_t$ times the word with its $r$th letter deleted. The deletion formula satisfies $E_i{F}_{j}w=F_j{E}_{i}w+{\delta }_{ij}(K_i-K_i^{-1})w\mathrm{/}(t-t^{-1})$; its labels verify both toral action relations. There are no further root-half relations to check. Thus these are $\tilde{U}$-modules. Conversely the crossing relations move every positive and toral letter to the right in a highest module, so negative words span; the displayed representation makes them independent and proves the universal property. Denote this module by $\tilde{M}(\lambda )$.

The actual algebra $U_t$ is a quotient of $\tilde{U}$: by [F1] its undivided roots and torus generate it and satisfy all the required mixed and toral relations. The canonical module map $\tilde{M}(\lambda )\to M_{U_t}(\lambda )$ is, on negative coordinates, exactly $T\to U_t^{-}$. Its kernel is consequently $J{v}_{\lambda }$ for every $\lambda$, so this subspace is a $\tilde{U}$-submodule. At $\lambda =0$ the augmentation subspace $T_{+}{v}_0$ is also a submodule: the deletion formula cannot produce a scalar except from a single letter, whose coefficient is $[0{]}_t=0$. Each $F_i{v}_0$ is highest of label $-{\alpha }_i$. Therefore the maps $\tilde{M}(-{\alpha }_i)\to T_{+}{v}_0$, $v_{-{\alpha }_i}\mapsto F_i{v}_0$, are module maps, and their direct sum is an isomorphism because $T_{+}={⨁}_{i}T{F}_i$ is the unique last-letter decomposition.

Induce that direct sum along $\tilde{U}\to U_t$. The universal property, or the mutually inverse maps $u\otimes (a{v}_{\lambda })\mapsto ua{v}_{\lambda }$ and $a{v}_{\lambda }\mapsto a\otimes v_{\lambda }$, gives $V:=U_t{\otimes }_{\tilde{U}}(T_{+}{v}_0)\cong {⨁}_i{M}_{U_t}(-{\alpha }_i)$. Define $\mathrm{\ell }:J\to V$ by $\mathrm{\ell }(a)=1\otimes a{v}_0$. Since $J{v}_0$ is a submodule by step 2.1, its image $\mathrm{\ell }(J)$ is a $U_t$-submodule of $V$. For $a={\sum }_i{u}_i{F}_i$, the direct-sum coordinates of $\mathrm{\ell }(a)$ are $(\pi (u_i)v_{-{\alpha }_i}{)}_i$. By [F1], this is zero if and only if each $u_i\in J$. Consequently $\ker \mathrm{\ell }={\sum }_{i}J{F}_i=J{T}_{+}$; the last equality follows by splitting the last letter of a nonempty right factor and using that $J$ is a two-sided ideal. No flatness of tensor induction was assumed: its kernel is computed directly in the known Verma coordinates.

The subspace $S{v}_0$ is a $\tilde{U}$-submodule. Multiplication by a negative generator and toral action preserve it. For a positive generator, the specialized opposite-Serre commutator of [F2] vanishes on every Serre generator $s$. Expanding $[E_i,asb]=[E_i,a]sb+as[E_i,b]$ leaves a Serre factor in each term; moving its toral factors across this homogeneous factor multiplies it only by a scalar. The deletion formula therefore sends $S{v}_0$ into itself. It follows that $\mathrm{\ell }(S)$ is a $U_t$-submodule of $V$, contained in $\mathrm{\ell }(J)$, and $\mathrm{\ell }(J)\mathrm{/}\mathrm{\ell }(S)$ is a submodule of $V\mathrm{/}\mathrm{\ell }(S)$. By [F1], the Casimir acts by $1$ throughout this quotient.

Choose $a\in J[-\beta ]$ with $a\mathrm{\notin }S$. Its image $\mathrm{\ell }(a)+\mathrm{\ell }(S)$ is nonzero. Otherwise $a-s\in \ker \mathrm{\ell }=J{T}_{+}$ for a homogeneous $s\in S[-\beta ]$; writing $a-s={\sum }_i{u}_i{F}_i$ gives $u_i\in J[-\beta +{\alpha }_i]$ of smaller height, hence $u_i\in S$ by minimality. This would give $a\in S$, a contradiction. Moreover $E_{i}a{v}_0$ belongs to $J{v}_0$ in lower degree by step 2.1, and therefore belongs to $S{v}_0$. Thus the nonzero image of $a$ in $\mathrm{\ell }(J)\mathrm{/}\mathrm{\ell }(S)$ is highest of label $-\beta$. Apply [F1] to the containing subquotient $V\mathrm{/}\mathrm{\ell }(S)$: $(\beta ,\beta )=2\mathrm{ht}\beta$. Degree zero and every pure simple-root polynomial component survive $T\to U_t^{-}$ by the integral rank-one half, so in particular $\beta \mathrm{\ne }0$; its positive height makes the right side strictly positive. This proves the claim.