supercuspidal

English

Etymology

super- +‎ cuspidal

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^[1]:

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