Let $E/F$ be a Galois extension of number fields, and $G$ a reductive group over $F$. If Langlands Base Change is known for $G/F$ and $G/E$, and moreover the eigenvarieties for $G/F$ and $G/E$ have been constructed, is there a rigid map between the eigenvarieties which interpolates the base change?

Assuming the answer to the previous question is yes, will the map be a closed immersion? What is known about the image with respect to the subvariety of Gal($E/F$)-invariants?

Finally, are there any instances where we have such a map between eigenvarieties but base change is not known?

Sorry for the barrage of questions; I would be happy with even a partial answer to the first.

  • $\begingroup$ The paper springerlink.com/content/56827481262h2186 by Flicker indicates that people have thought about this question, but I haven't been able to find any concrete results. Perhaps Theorem 1 of arxiv.org/abs/math/0301032 could be of some use? $\endgroup$ – Kevin Ventullo Jun 20 '11 at 20:57
  • $\begingroup$ Hm, it seems Theorem 1 only appears in the published version. $\endgroup$ – Kevin Ventullo Aug 1 '11 at 1:07
  • 1
    $\begingroup$ Dear Kevin, if $G=\mathrm{GL}_n$, $F$ is totally real, and $E/F$ is cyclic, then the answer to your question is "yes", at least under the mild assumption that the tame levels are chosen "coprime to the relative different of $E/F$." One does not get a closed immersion of the whole $F$-eigenvariety, but only of its "spine" (which is roughly the union of the irreducible components which contain a dense set of classical points). I can email you if you'd be interested in hearing more details. Cheers, Dave $\endgroup$ – David Hansen Jan 10 '13 at 22:48
  • $\begingroup$ Dear David, that's great! Sure, my email is (mylastname) at math.ucla.edu. Can you remove the condition that $E/F$ is cyclic when $G=GL_2$, using the recent results of Dieulefait? $\endgroup$ – Kevin Ventullo Jan 17 '13 at 0:50

This is all a bit complicated -- the theory is still in its infancy and some arguments aren't quite as smooth as they should be.

If all you know is that "the eigenvarieties have been constructed" then you're in a hopeless situation -- in fact in some sense I don't even know what this statement means. You need to know some facts about the eigenvarieties before you can prove anything. Let's say for example that you're in the happy situation where classical points are dense (for example, $G$ compact mod centre, or perhaps you've defined an eigenvariety to have that property by taking the closure of the classical points in an a priori bigger object). Then you might be able to get somewhere (see next para). However there are examples of Hida families over non-totally real bases where there appear to be natural $p$-adic objects where classical points do not seem to be dense, so then things might be really tough.

If classical points are dense, then there is still another big problem: as far as I know Langlands Base Change (and functoriality in general) only transfers packets to packets. However, eigenvarieties don't parametrise packets, they parametrise systems of Hecke eigenvalues. If you are dealing with inner forms of $GL(2)$ then packets have size 1 so Chenevier doesn't see this problem and he can prove the best possible theorem (this is one of the papers you cite). If however there are packets then one runs into combinatorial issues -- Flicker finds himself with such problems in the other paper you cite. So, even if classical points are dense, things might not be so easy.

I have a student, Judith Ludwig, who is making some headway with issues of this nature. The questions can be quite delicate.

  • $\begingroup$ Dear Kevin: I did mean to assume that classical points are dense, at least in the source. In fact, my question was motivated by the case of $GL_2$ over $\mathbb{Q}$/tot. real $F$, but I naively phrased it in more general terms. If you have any comments on this case, it would be great to hear (read?) them. Anyway, thank you for the illuminating answer! $\endgroup$ – Kevin Ventullo Jun 20 '11 at 22:31
  • $\begingroup$ @Kevin Buzzard: Are there known instances where something in a global automorphic packet which isn't classically automorphic turns out to be p-adically automorphic in an appropriate sense? Best wishes, Dave $\endgroup$ – David Hansen Jun 20 '11 at 23:20
  • $\begingroup$ @Kevin V: what I have to say to you about the $GL(2)/F$ situation would fill about three more comment boxes at least. In short I'm not so sure Emerton's spectral sequence degenerates in this setting so probably you want to use Andreatta-Iovita-Stevens or Pilloni's work on overconvergent sheaves in the Hilbert case, and build the EV a la Coleman. But now you have to prove classical points are dense. Pilloni and Stroh have just announced this in the case $p$ unramified but the paper is technical and I've not finished reading it yet. One might be cautiously optimistic though. $\endgroup$ – Kevin Buzzard Jun 21 '11 at 6:47
  • 1
    $\begingroup$ PS @Kevin V: For $GL(1)$ over a field with both real and complex places, there is a very natural definition of an eigenvariety (p-adically continuous group homomorphisms from $A_k^*/k^*$) and classical points are not dense. I would imagine that one can prove base change in this situation relatively easily. My opinion is that density of classical points should not be taken as an axiom in these settings and should instead be regarded as a mysterious but happy coincidence! $\endgroup$ – Kevin Buzzard Jun 21 '11 at 7:15
  • 1
    $\begingroup$ I can only agree with Kevin B. For eigenvarieties with density of classical points Chenevier's theorem is the natural tool to use, and work without any problem to define a rigid map from the eigencurve to the eigenvariety of Gl_2/F (F a totally real field) constructed by Pilloni and al, if we admit at least all the results in their recent preprints (a student of mine has written this down as a lemma recently). Similarly for a map from the eigencurve attached to a definite quaternion algebra D/Q to the eigenvariety attached to D_F/F, where both eiegnvarieties have been constructed long ago. $\endgroup$ – Joël Jun 21 '11 at 14:39

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.