The serre quotient has weyl symmetry and no residual kac moody kernel

F2 puts the Serre ideal $S$ inside $\mathfrak{r}$, yielding the surjection and kernel $\bar{\mathfrak{r}}=\mathfrak{r}\mathrm{/}S$. Thus the Cartan embeds in $\bar{\mathfrak{g}}$; its other weights have one sign because it is a homogeneous quotient of $\tilde{\mathfrak{g}}$. Pure multiples of a simple root are absent beyond $\pm {\alpha }_i$, since each free half has that property. The simple components map injectively to $\mathfrak{g}$, so the kernel has no simple or zero weights.

On $\bar{\mathfrak{g}}$, $D=\mathrm{ad}{e}_i$ is nilpotent on every $e_j$ by the defining Serre relations, on $f_j$ by $D{f}_j={\delta }_{ij}{h}_i$, $D^2{f}_i=-2e_i$, $D^3{f}_i=0$, and on $h$ by $D^{2}h=0$. The binomial identity $D^m[x,y]={\sum }_k\left(\genfrac{}{}{0px}{}{m}{k}\right)[D^{k}x,D^{m-k}y]$ proves local nilpotence on all bracket words. The same calculation holds for $\mathrm{ad}{f}_i$ and in $\mathfrak{g}$. The finite exponentials and their negative-exponent inverses preserve brackets. Their product $T_i=\exp (\mathrm{ad}{f}_i)\exp (-\mathrm{ad}{e}_i)\exp (\mathrm{ad}{f}_i)$ sends $(e_i,h_i,f_i)$ to $(-f_i,-h_i,-e_i)$ and fixes $\ker {\alpha }_i$ in the Cartan, by the three-term simple-triple expansions. Thus $T_{i}h=h-{\alpha }_i(h)h_i$, and $[h,T_{i}x]=(s_i\beta )(h)T_{i}x$ for $x$ of weight $\beta$. The quotient map commutes with these finite polynomials, so $T_i$ and its inverse preserve the kernel.

If the positive kernel is nonzero, let $\alpha =\sum k_i{\alpha }_i$ have the smallest height among its nonzero weights. By F1 every element of ${\mathfrak{r}}_{\alpha }^{+}$ is a finite sum of iterated positive adjoints of constrained homogeneous generators. A generator of smaller height maps to zero in the positive kernel by minimality, as do all of its adjoints. Any surviving term at degree $\alpha$ must therefore be a generator in degree $\alpha$ itself. Consequently $(\alpha ,\alpha )=2(\rho ,\alpha )=2{\sum }_i{k}_i{d}_i>0$.

For every $i$, step 2.1 produces a nonzero kernel vector at $s_i\alpha$. Since $\alpha$ is not simple, it has a positive coefficient at some index other than $i$; otherwise it would be a forbidden pure multiple. That coefficient is unchanged by $s_i$, so the one-sign property forces $s_i\alpha >0$. Minimality gives $\mathrm{ht}(s_i\alpha )=\mathrm{ht}(\alpha )-\alpha (h_i)\ge \mathrm{ht}(\alpha )$, hence $\alpha (h_i)\le 0$. F4 now gives $(\alpha ,\alpha )={\sum }_i{k}_i{d}_i\alpha (h_i)\le 0$, contradicting step 2.2. The positive kernel vanishes. The sign-changing involution preserves S and r and interchanges signs, so the negative kernel also vanishes. There is no zero kernel component by step 1.1.