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Let $\mathfrak{g}$ be a semisimple finite dimensional Lie algebra and $V_\lambda$, resp. $V_\mu$ its finite dimensional highest weight modules with highest weights $\lambda$, resp. $\mu$. Let $\chi_\lambda, \chi_\mu : C(\mathcal{U}(\mathfrak{g})) \rightarrow \mathbb{C}$ be the corresponding central characters. Harish-Chandra theorem asserts that $\chi_\lambda = \chi_\mu$ if and only if $w(\lambda+\delta)-\delta = \lambda$ for some $w$ in the Weyl group, where $\delta$ is the Weyl vector, i.e. the sum of all fundamental weights.

Is it also true in this setting, that $V_\lambda \simeq V_\mu$ if and only if $\chi_\mu = \chi_\nu$ ?

How is this version of the theorem related to the fact that Harish-Chandra homomorphism is an isomorphism ?

Thank you very much for your answers. (I am studying a program in mathematical physics and trying to figure out how general is the procedure of labeling irreducible representations by values of Casimir operators, as physicists do so often)

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    For $V_\lambda$ to be finite dimensional, we need $\lambda$ to be dominant. Then $\mu=w(\lambda+\delta)-\delta$ is not dominant unless $w=id$, $\mu=\lambda$. Therefore I don't quite understand the question. My recollection of enveloping algebras is a bit rusty, but isn't there stuff other than the Casimir operator in there as well? For example, isn't Casimir operator stable under the graph automorphism of the Dynkin diagram, so e.g. for type $A_n$ the Casimir operator would act as the same constant as on the dual of an irreducible rep?2012-06-14
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    Ah, thank you. I should have revised the theory of Lie algebras more before I asked. So it is clear from this argument with dominant weights that Harish-Chandra theorem implies directly what I asked, i.e. that we can distinguish irreps by their central characters.2012-06-14
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    We can distinguish **finite-dimensional** irreps by their central characters. IIRC when you extend that to category $\mathcal{O}$ you get irreducible reps with non-dominant highest weights, but they are infinite dimensional, but there are indecomposable reps with composition factors (belonging to same central character), one f.d. the others not. That's what Kazhdan-Lusztig business was about.2012-06-15

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