I am interested to know what the status of Local Langlands Conjectures in positive characteristic is? By a positive characteristic local field, I mean a field of the form $\mathbb{F}_q((t))$.

A nice summary of what is known for characteristic zero local fields is given in the first paragraph of


I'm looking for a similar description in the positive characteristic case.

  • $\begingroup$ Why do you not consider ${\mathbf F}_q((t))$ to be a positive characteristic local field, where $q$ is a prime power but not necessarily prime? Do you consider finite extensions of ${\mathbf Q}_p$ to be local fields of characteristic 0? $\endgroup$
    – KConrad
    Feb 4, 2014 at 23:06
  • $\begingroup$ Sorry, that was a typo. It is correct now. Thanks. $\endgroup$
    – Dr. Evil
    Feb 4, 2014 at 23:39
  • $\begingroup$ Lafforgue-Genestier is state-of-the-art, they prove half of the conjecture for arbitrary reductive groups. arxiv.org/abs/1709.00978 $\endgroup$ Sep 6, 2022 at 3:05

1 Answer 1


I am not an expert, but what I think I know on the subject is: For $GL_n$, the local Langlands conjecture has long been known by work of Laumon, Rapoport and Stuhler (Inventiones Math, 1993). Of course, in this case, even the global correspondence is now known, due to the Fields-Medal-winning work of Laurent Lafforgue.

For general reductive groups, the local Langlands correspondence is not known at this date but there is movement right now. Vincent Lafforgue (Laurent's younger brother) has recently released a paper proving the direction "automorphic --> Galois" of the global correspondence. In this paper, he announces a work in preparation of himself and Genestier aiming at establishing the local Langlands correspondence for reductive groups. So when this paper is released, the answer to your question may well be "solved!".

  • $\begingroup$ Laumon pas Laumont! $\endgroup$ Feb 6, 2014 at 8:16
  • 2
    $\begingroup$ Ah oui désolé Lauren :-) Non sérieusement je change... $\endgroup$
    – Joël
    Feb 6, 2014 at 13:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.