6
$\begingroup$

Let $F$ be a nonarchimedian local field. Since the Weil group $W_F$ is a dense subgroup of $G_F=Gal(\bar{F}/F)$, it's clear that restriction gives an injection $Irr(G_F)\rightarrow Irr(W_F)$ of irreducible (complex) representations.

In some notes of Prasad and Raghuram (found here: http://www.math.tifr.res.in/~dprasad/ictp2.pdf), they assert that the representations are in fact the same "perhaps after a twist." This fact is crucial for their argument that reduces establishing LLC for $GL_n$ to the case of supercuspidal representations. But I'm not sure I understand what it means.

So my question is: how does one take a representation of $W_F$ and twist it in such a way that it can be extended to a representation of $G_F$?

$\endgroup$

1 Answer 1

6
$\begingroup$

Pick $V \in Irr(W_F)$. Due to Schur's lemma and the fact that inertia in $W_F$ acts through a finite quotient, some power of (a fixed choice of) Frobenius acts as a scalar on $V$. Take an appropriate root of that scalar to make up an unramified character $\chi$ of $W_F$ such that the same power of Frobenius acts trivially on $U := V \otimes \chi^{-1}$. I claim that $U \in Irr(G_F)$, i.e., that $W_F$ acts through a finite quotient on $U$. Indeed, that quotient is generated by a finite quotient of the inertia and the finite order image of the chosen Frobenius.

$\endgroup$
1
  • $\begingroup$ To elaborate: Let the $I_F$-action on $V$ factor through a finite quotient $G$. Then any lift $\varpi\in W_F$ of the Frobenius acts on $G$ by automorphisms, so some power $\varpi^N$ acts trivially on $G$. Now the original representation factors through $G\rtimes\langle\varpi\rangle$, and $\varpi^N$ is central, so it acts as a scalar on $V$. $\endgroup$ Jul 12, 2023 at 2:51

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.