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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    $\begingroup$ Second question: Schur's lemma. $\endgroup$ – Qiaochu Yuan Jul 6 '11 at 17:06
  • $\begingroup$ My copy is in my office, but IIRC the answer to your first question has something to do with normalization. Was that base normalized with respect to the Killing form or something? Using that basis made it somehow a bit more universal, but the other would work as well (scalar multiplier difference). Qiaochu nailed your second question. $\endgroup$ – Jyrki Lahtonen Jul 6 '11 at 18:12
  • $\begingroup$ also see math.stackexchange.com/questions/44152/… $\endgroup$ – wildildildlife Jul 6 '11 at 22:08
  • $\begingroup$ @Qiaochu, thank you. Here $\phi: L \to gl(V)$ where $V$ is irriducible. Since $\phi(c_L): V\to V$ is irriducible, $\phi(c_L)$ acts as by scalar or acts as $0$. Why $\phi(c_L)$ do not act as $0$. Where do we use the condition that $\phi(c_L)$ commutes with $\phi(g)$ for each $g \in L$? $\endgroup$ – LJR Jul 7 '11 at 17:23
  • $\begingroup$ @user9791: $0$ is a scalar. The condition is one of the hypotheses of Schur's lemma. $\endgroup$ – Qiaochu Yuan Jul 7 '11 at 17:42

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.

  • $\begingroup$ thank you very much. Does $\bigotimes {}^{\cdot}g$ mean the tensor algebra of $g$? $\endgroup$ – LJR Jul 7 '11 at 17:26
  • $\begingroup$ Yes, that tensor-with-superscript-bullet it meant to be "tensor algebra" of g. Perhaps better put, it is the "universal associative algebra" attached to the vector space g. The fact that it can be constructed in terms of tensors isn't its defining characteristic, categorically. $\endgroup$ – paul garrett Jul 7 '11 at 20:33
  • $\begingroup$ Is this still true for semisimple Lie algebras, which are not necessarily simple? I guess it is, because the Killing form remains non-degenerate. And what do you mean by G, if g is the Lie algebra? $\endgroup$ – red_trumpet Oct 16 '17 at 12:03
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    $\begingroup$ 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 $G$. $\endgroup$ – paul garrett Oct 16 '17 at 12:28

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