Kac moody relation module embeds in verma modules and obeys the casimir constraint

Put $T=U({\tilde{\mathfrak{n}}}^{-})$, the free associative algebra on the $f_i$ by F4 and its free-Lie construction. In $\tilde{M}(0)=T\tilde{v}$, the augmentation subspace $T_0\tilde{v}$ is a submodule: its quotient is the trivial one-dimensional module. The vectors $f_i\tilde{v}$ are highest of weight $-{\alpha }_i$, because $e_j{f}_i\tilde{v}={\delta }_{ij}{h}_i\tilde{v}=0$. The last-letter decomposition $T_0={⨁}_{i}T{f}_i$ and F5 therefore identify this submodule with ${⨁}_i\tilde{M}(-{\alpha }_i)$, not merely a quotient.

Let $V=U(\mathfrak{g}){\otimes }_{U(\tilde{\mathfrak{g}})}(T_0\tilde{v})$. Associativity of balanced tensor products (the maps $a\otimes (b\otimes 1)\mapsto ab\otimes 1$ and its reverse) and F5 identify $V={⨁}_i{M}_A(-{\alpha }_i)$. Define $\mathrm{\ell }(a)=1\otimes a\tilde{v}$ for $a\in {\mathfrak{r}}^{-}$. For $x\in \tilde{\mathfrak{g}}$, $x\tilde{v}\in T_0\tilde{v}$, since the quotient in step 1.1 is trivial. Therefore $\mathrm{\ell }([x,a])=1\otimes xa\tilde{v}-1\otimes ax\tilde{v}=\pi (x)\otimes a\tilde{v}-\pi (a)\otimes x\tilde{v}=\pi (x)\mathrm{\ell }(a)$. In particular commutators in ${\mathfrak{r}}^{-}$ are killed. The two ideals ${\mathfrak{r}}^{+}$ and ${\mathfrak{r}}^{-}$ commute because their bracket lies in their zero intersection. Hence the source modulo its self-commutator carries the adjoint $\mathfrak{g}$-action, and $\mathrm{\ell }$ factors through a module map on it.

Write $a={\sum }_i{u}_i{f}_i$ using its unique associative last-letter coefficients. The degree-one part of ${\mathfrak{r}}^{-}$ is zero, since the simple $f_i$ survive, so all $u_i\in T_0$. In the PBW identifications, $\mathrm{\ell }(a)=(\pi (u_i)v_{-{\alpha }_i}{)}_i$. By F5, this is zero exactly when each $\pi (u_i)=0$. F1 gives $u_i\in {\mathfrak{r}}^{-}T$, so $a\in {\mathfrak{r}}^{-}{T}_0$, and F1 then gives $a\in {\mathfrak{r}}^{-}\cap {\mathfrak{r}}^{-}{T}_0=[{\mathfrak{r}}^{-},{\mathfrak{r}}^{-}]$. The reverse kernel inclusion was proved in step 2.1. Thus the module map is injective. These are associative coefficients, not adjoint coefficients.

Each summand $M_A(-{\alpha }_i)$ has Casimir scalar $({\alpha }_i,{\alpha }_i)-2(\rho ,{\alpha }_i)=2d_i-2d_i=0$ by F3. Hence $\mathrm{\Omega }=0$ on the embedded relation module and each of its subquotients; the pointwise formula respects submodules. The relation module belongs to $\mathcal{O}$, being a submodule of a finite sum of the Verma modules of F5. A primitive vector of weight $-\alpha$ has a nonzero highest image in a quotient. F3 applied to that image gives $0=(\alpha ,\alpha )-2(\rho ,\alpha )$. Its degree is neither zero nor simple, as ${\mathfrak{r}}^{-}$ has neither component. F2 proves generation of the abelianized relation module by these degrees.

Let $K$ be the ideal of ${\tilde{\mathfrak{n}}}^{-}$ generated by all the indicated full homogeneous spaces of ${\mathfrak{r}}^{-}$. Step 4.1 says ${\mathfrak{r}}^{-}=K+[{\mathfrak{r}}^{-},{\mathfrak{r}}^{-}]$. If the positively regraded Lie algebra $L={\mathfrak{r}}^{-}\mathrm{/}K$ were nonzero, choose its least positive height $m$. Every nonzero bracket in $L$ has height at least $2m$, so $L_m$ cannot lie in $[L,L]$. This contradicts $L=[L,L]$. Thus $K={\mathfrak{r}}^{-}$. The sign-changing involution gives the positive assertion with the identical equation on $\alpha$.