I found that both supercuspidal representation and cuspidal representation are defined as representations which are not subrepresentations of induced representations. Is supercuspidal representation the same as cuspidal representation? Thank you very much.


1 Answer 1


I suppose extremely late is still better than never?

First, I'm sure you realise this, but to make sure, the notion of supercuspidal representations only makes sense for algebraic groups over nonarchimedean local fields. Let $F$ be such a field of residual characteristic $p$, let $G=\mathbf{G}(F)$ be the $F$-points of some (connected, reductive) algebraic group over $F$, and let $k$ be some algebraically closed field over which we'll consider representations.

If $k=\Bbb{C}$ (or $k=\bar{\Bbb{Q}}_\ell$, $\ell\neq p$, it's basically the same thing here), then, for any smooth representation $\pi:G\rightarrow\mathrm{Aut}_k(V)$, the following are equivalent:

  • All Jacquet modules are trivial;
  • $\pi$ is not isomorphic to an irreducible quotient of any parabolically induced representation of $G$;
  • $\pi$ is not isomorphic to an irreducible subquotient of any parabolically induced representations of $G$;
  • All matrix coefficients of $\pi$ are compactly supported modulo the centre of $G$.

In this case, any representation satisfying the above is said to be (super)cuspidal -- they're the same thing here. The difference between the two comes when $k$ is of positive characteristic, say $k=\bar{\Bbb{F}}_\ell$. In this case, subquotients aren't necessarily quotients, and so the second and third conditions are no longer equivalent. Cuspidal representations are then the ones satisfying the quotient property, while supercuspidals satisfy the subquotient property.

(Also, be careful in your definition: it's important that you say they don't appear in parabolically induced representations!)

  • 2
    $\begingroup$ Could you explain why are subquotients same as quotients when characteristic is 0? Thanks! $\endgroup$
    – Hebe
    Nov 3, 2015 at 18:07
  • $\begingroup$ You want $\pi$ to be irreducible, or perhaps admissible in that equivalence? $\endgroup$
    – mnr
    Jul 2, 2019 at 15:19

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