Pairwise Casimirs satisfy infinitesimal braid relations

If $\{i,j\}\cap \{r,s\}=\mathrm{\varnothing }$, every summand of ${\mathrm{\Omega }}_{ij}$ acts on tensor factors disjoint from every summand of ${\mathrm{\Omega }}_{rs}$. Operators on distinct factors commute, so their finite sums commute.

For $y,z,w\in \mathfrak{g}$, contraction of $(\mathrm{ad}y\otimes 1+1\otimes \mathrm{ad}y)C$ against $B(\cdot ,z)\otimes B(\cdot ,w)$ is $B([y,w],z)+B([y,z],w)=0$ by invariance and symmetry of the form $B$ in the Statement. Nondegeneracy of $B$ makes this tensor zero. Its specialization $y=x_b$ is ${\sum }_a[x_a,x_b]\otimes x^a+x_a\otimes [x^a,x_b]=0.$ Multiply the first tensor factor on slot $i$, the second on slot $j$, and sum against $x^b$ acting on slot $k$. Since different slots commute, the first resulting sum is $[{\mathrm{\Omega }}_{ij},{\mathrm{\Omega }}_{ik}]$ and the second is $[{\mathrm{\Omega }}_{ij},{\mathrm{\Omega }}_{jk}]$. Their sum is zero. The inverse Gram tensor is symmetric, so ${\mathrm{\Omega }}_{ij}={\mathrm{\Omega }}_{ji}$; the relation follows for every ordering of distinct indices. All operators act on finite-dimensional tensor products; no completion is involved.