Universal Casimir element On page 118 of J.E. Humphreys' book Introduction to Lie algebras and representation theory, paragraph 3 of section 22.1, what is the motivation of the definition of $c_{ad}$ in this way? Why we use the basis $x_{\alpha}, z_{\alpha}, t_{\alpha}$ but not  $x_{\alpha}, y_{\alpha}, h_{\alpha}$? 
The last line of paragraph 3 of section 22.1, how to show that $\phi(c_L)$ acts as a scalar using the fact that $\phi(c_L)$ commutes with $\phi(L)$ and $\phi$ is irriducible? Thank you very much.
 A: It doesn't matter what basis is used to define Casimir, as long as the Killing form is used properly: In any simple Lie algebra, by Cartan's criterion the Killing form $\langle,\rangle$ is non-degenerate. Given any basis $x_i$ for the Lie algebra, let $x_i'$ be the dual basis with respect to $\langle,\rangle$. Then the Casimir element is $\sum_i x_i x_i'$ in the universal enveloping algebra.
Many sources give an exercise to prove that this expression is independent of basis and that it is central in the enveloping algebra, but there is a direct argument (probably found in Serre, for example): the natural maps with simple Lie algebra $g$,
$$End(g) \approx g \otimes g^* \approx g\otimes g \subset \bigotimes{}^\bullet g \rightarrow Ug$$
are G-equivariant, where the identification of $g^*$ with $g$ is exactly via the Killing form. The identity map among endomorphisms of $g$ commutes with $G$, so its image at the other end does, as well. Consideration of the second and third items yields expressions as in the first paragraph. Poincare-Birkhoff-Witt proves that the resulting expressions are non-zero.
