Let $G$ be a complex reductive algebraic group (connected, simply connected etc), viewed as a real group. We study the representations of $G$, and we follow the notations in the paper of Barbasch and Vogan:

"Unipotent representations of complex semisimple groups", Annals of Math. Vol.121, 41-110, 1985.

Consider the irreducible representations $\overline{X}(\lambda, \mu)$ of $G$ (following Zhelobenko's notation or the notation in the above paper of Barbasch-Vogan, which means the irreducible constituent of principal series $X(\lambda, \mu)$ containing the extremal weight $\lambda-\mu$) and their wavefront sets, we know their wavefront sets are closures of nilpotent orbits of the Lie algebra.

We have the following results (Definition 1.10 in the above paper):

If $\lambda, \mu$ are integral, and the wavefront set of $\overline{X}(\lambda, \mu)$ is the closure $\overline{O}$ of the nilpotent orbit $O$, then the orbit $O$ is special, i.e. under the Springer correspondence, it correspond to a special representation of the Weyl group $W$ (in the sense of Lusztig).

My question is:

  1. Is there any example of irreducible representation, with the closure of a non-special orbit as its wavefront set? For integral infinitesimal characters and special orbits, I have a bunch of examples. But I have never seen any non-special orbits as wavefront sets.

  2. In Theorem 3.20 of the above paper of Barbasch-Vogan, the authors show an effective way to calculate wavefront sets of any irreducible representations with integral infinitesimal characters. So for non-integral infinitesimal characters, is there any generalization of the theorem 3.20, that we can use to calculate their wavefront sets?



1 Answer 1


1) An example is the metapletic representation in the complex symplectic group. They have half-integral infinitesimal characters, and the wavefront set is the minimal orbit which is non-special.

2) Check out the book by Monty McGovern (here). In Section 5. He mentioned how to find out the wavefront set/associated variety (which are the same by a deep theorem of Schmid and Vilonen) for $U(g)/J(\lambda)$, where $\lambda$ can be any infinitesimal character.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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