The Casimir operator is basis-independent and intertwining

The form isomorphism $b:\mathfrak{g}\to {\mathfrak{g}}^{\ast }$ sends $x\mapsto B(x,-)$. Under $\mathfrak{g}\otimes {\mathfrak{g}}^{\ast }\cong \mathrm{End}(\mathfrak{g})$, the identity corresponds to ${\sum }_i{x}_i\otimes b(x^i)$. Hence ${\sum }_i{x}_i\otimes x^i$ is the inverse tensor of $B$ and is independent of the basis. Multiplication into $U(\mathfrak{g})$ proves the same for ${\mathrm{\Omega }}_B$.

For $z\in \mathfrak{g}$, the diagonal adjoint action on the inverse tensor is ${\sum }_i([z,x_i]\otimes x^i+x_i\otimes [z,x^i])$. Pairing its second factor with an arbitrary vector and using [L2] shows that this tensor is zero. Multiplication sends it to $[z,{\mathrm{\Omega }}_B]$, hence that commutator is zero. Since $\mathfrak{g}$ generates $U(\mathfrak{g})$, ${\mathrm{\Omega }}_B$ is central.

By [L3], any module action extends to $U(\mathfrak{g})$. Centrality gives $\rho (x)\rho ({\mathrm{\Omega }}_B)=\rho ({\mathrm{\Omega }}_B)\rho (x)$ for every $x$, precisely the intertwining condition. For $\mathfrak{g}=0$, the inverse tensor and Casimir are empty sums and all assertions reduce to $0=0$.