0
$\begingroup$

Let $Z(\mathfrak{g})$ be the centre of $U(\mathfrak{g})$ and let $\chi_\lambda$ be an algebra homomorphism $Z(\mathfrak{g}) \to \mathbb{C}$ such that $ z \cdot v = \chi_\lambda(z)v $ for all $z \in Z(\mathfrak{g})$, $v \in M(\lambda)$, then $M_I(\lambda)$ and its subquotients including $L(\lambda)$ have the same infinitesimal character $\chi_\lambda$.

Why we call the infinitesimal character $\chi_\lambda$ of $L(\lambda)$ regular if $\langle \lambda + \rho,\alpha^\lor\rangle \neq 0$ for all $\alpha\in\Phi$ instead of $\langle \lambda,\alpha^\lor\rangle \neq 0$ for all $\alpha\in\Phi$?

I know that $\chi_\lambda=\chi_\mu$ iff $\lambda=w\cdot\mu$ for some $w\in W$. It makes more sense to define in the former way. But is there any other/fundamental reason for the definition?

$\endgroup$
2
  • 1
    $\begingroup$ Could you please introduce your notation and assumptions ? $\endgroup$ – Paul Broussous Dec 17 '18 at 17:20
  • $\begingroup$ Note that the language "dot-regular" is also used, to clarify the use of the $\rho$-shift. The unmodified "regular" is unfortunately ambiguous. (By the way, I agree with Vit that it's best to use the dot-notation rather than try to modify the highest weight as people earlier did.) " $\endgroup$ – Jim Humphreys Dec 17 '18 at 22:39
1
$\begingroup$

I don't think there is other reason. I think in general group representation theory you consider regular elements as those which have trivial stabilizer. Then you should perhaps speak about "regular with respect to affine action".

People sometimes abuse notation and talk about character $\lambda$ when in fact they mean $\chi_\lambda$. Sometimes you can find definitions of Verma modules that incorporates $\rho$ shift, e.g. $M(\lambda) = \mathfrak{U(g)\otimes_{U(p)}} \mathbb{F}_{\lambda - \rho}$. This can be quite confusing to newcomers. I prefer to work without the shift for weights and for calculations instead of affine action I just add $\rho$ use normal action and call the resulting numerical vector character.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.