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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.

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    @user9791: $\phi(c_L)$ is an endomorphism of $V$ as a vector space, but it is not guaranteed to be an endomorphism of $V$ as a representation unless it commutes with $\phi(g)$ for each $g \in L$, and that is the hypothesis of Schur's lemma.2011-07-07

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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.

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    Yes, semi-simplicity is equivalent to non-degeneracy of the Killing form. And by $G$ I meant the Lie group... but/and the same invariance applies to the small-a adjoint action of $\mathfrak g$ if one doesn't have $G$.2017-10-16