4
$\begingroup$

Let $\pi, \pi'$ be a unitary, irreducible, supercuspidal representations of $GL_2(F)$. Does an equality of roots numbers $\epsilon(\pi, \psi, s) = \epsilon(\pi', \psi, s)$ for all $s \in \mathbb{C}$ imply an isomorphism $\pi \cong \pi'$?

$\endgroup$

1 Answer 1

8
$\begingroup$

No -- there are far too many supercuspidals $\pi$ for it to be possible to distinguish them by a single $\varepsilon$-factor -- but this is true if you also consider root numbers of twists; this is called the "local converse theorem" and is in the Jacquet-Langlands book (SLN 114). See also Andrew Snowden's thesis. There is a generalization to $GL(n)$ if you allow twists by $GL(n-2)$ -- this is a result of Henniart.

$\endgroup$
1
  • $\begingroup$ Now, I feel stupid. Of course, converse theorems is the key word. Thank you! $\endgroup$
    – Marc Palm
    Aug 28, 2013 at 17:30

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.