supercuspidal

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English

Etymology

From super- +‎ cuspidal.

Adjective

supercuspidal (not comparable)

(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 $K$ of $G(k)$ , there are, up to unramified twists, only finitely many $K$ -spherical supercuspidal representations of $G(k)$ .

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