Generalized kac moody casimir is central and scalar on highest weight modules

For $z\in {\mathfrak{g}}_{\beta -\alpha }$, the tensors ${\sum }_s{y}_{\alpha ,s}\otimes [z,x_{\alpha ,s}]$ and ${\sum }_t[y_{\beta ,t},z]\otimes x_{\beta ,t}$ agree. Pair with arbitrary $a\otimes b\in {\mathfrak{g}}_{\alpha }\otimes {\mathfrak{g}}_{-\beta }$: their values are respectively $(a,[b,z])$ and $([z,a],b)$, equal by invariance. Perfect finite-dimensional pairings imply the tensor equality. This also covers a missing root space by interpreting the corresponding maps as zero.

Write $S={\sum }_{\alpha >0,s}{y}_{\alpha ,s}{x}_{\alpha ,s}$ and $t_i={\nu }^{-1}({\alpha }_i)=d_i{h}_i$. In $[S,e_i]$, the term $y_{\alpha }[x_{\alpha },e_i]$ cancels the term $[y_{\beta },e_i]x_{\beta }$ with $\beta =\alpha +{\alpha }_i$ by step 1.1. The only unmatched degree is ${\alpha }_i$, where the dual pair is $e_i,d_i{f}_i$, giving $[S,e_i]=-t_i{e}_i$. Terms of mixed root sign vanish, and $2{\alpha }_i$ is absent. For $[S,f_i]$ the same identity with $z=f_i$ pairs $y_{\beta }[x_{\beta },f_i]$ with $[y_{\alpha },f_i]x_{\alpha }$ for $\beta =\alpha +{\alpha }_i$; the unmatched simple term is $d_i{f}_i[e_i,f_i]=f_i{t}_i$. Thus $[S,f_i]=f_i{t}_i$. All cancellations are finite on a fixed vector: root spaces kill that vector, its images under $e_i,f_i$, and all but finitely many shifted degrees.

Let $C={\sum }_a{u}_a{u}^a$. Expanding with $[h,x]=\beta (h)x$ gives $[C,x]=x(2{\nu }^{-1}(\beta )+(\beta ,\beta ))$ for $x\in {\mathfrak{g}}_{\beta }$. Also $[2{\nu }^{-1}(\rho ),x]=2(\rho ,\beta )x$. For $e_i$, these finite terms total $e_i(2t_i+4d_i)$, while $2[S,e_i]=-2t_i{e}_i=-2e_i{t}_i-4d_i{e}_i$. For $f_i$ they total $-2f_i{t}_i$, canceled by $2[S,f_i]=2f_i{t}_i$. Every summand has weight zero, so $[\mathrm{\Omega },h]=0$. Since these elements generate $\mathfrak{g}$, the commutator identity with a product or bracket proves centrality on the entire algebra action.

On a highest vector all positive factors $x_{\alpha ,s}$ vanish. The Cartan terms give ${\sum }_a\lambda (u_a)\lambda (u^a)=(\lambda ,\lambda )$ and $2\lambda ({\nu }^{-1}\rho )=2(\rho ,\lambda )$. This proves the displayed scalar on $v$. By step 3.1, $\mathrm{\Omega }(uv)=u\mathrm{\Omega }v$ for every finite enveloping word $u$, so the scalar holds on the generated module.