supercuspidal

English

Etymology

From super- +‎ cuspidal.

Pronunciation

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Adjective

supercuspidal (not comparable)

1. (mathematics) That has a zero Jacquet functor for every proper parabolic subgroup
   • 2015, Manish Mishra, “A Galois side analogue of a theorem of Bernstein”, in arXiv‎:

     A theorem of Bernstein states that for any compact open subgroup ${\displaystyle K}$ of ${\displaystyle G(k)}$, there are, up to unramified twists, only finitely many ${\displaystyle K}$-spherical supercuspidal representations of ${\displaystyle G(k)}$.