Casimir element

Remarks

The element $\Omega_B$ does not depend on the choice of the basis, and it lies in the centre of $U(\mathfrak{g})$. For a semisimple $\mathfrak{g}$ over a field of characteristic $0$, the usual choice of $B$ is the Killing form. For a faithful representation $\rho$ of a semisimple $\mathfrak{g}$ in characteristic $0$ one also uses the trace form $B_\rho(x, y) = \tr(\rho(x)\rho(y))$, and calls the image $\rho(\Omega_{B_\rho})$ the Casimir operator of $\rho$. Example: let $F$ have characteristic other than $2$, and let $\mathfrak{g} = \mathfrak{sl}_2(F)$ with basis $e, f, h$ and $B(x, y) = \tr(xy)$. Then $B(e, f) = 1$, $B(h, h) = 2$ and the other pairings vanish, so the dual basis of $e, f, h$ is $f, e, h/2$ and $\Omega_B = ef + fe + \tfrac{1}{2} h^2$. On $F^2$ one has $ef = E_{11}$, $fe = E_{22}$ and $h^2 = I$, so $\Omega_B$ acts as $\tfrac{3}{2} I$, a scalar as expected. Non-example: the element $e^2 + f^2 + h^2$ uses the basis itself instead of the dual basis, and it is not central, since $[e, e^2 + f^2 + h^2] = hf + fh - 2eh - 2he \ne 0$ in $U(\mathfrak{g})$.