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As the title says.

In particular, I am interested in the story for a general reductive group $G$, say defined over $\mathbb{Q}$. I can imagine that mod-$\ell$ (algebraic) automorphic representation should correspond to conjugacy classes of homomorphisms from the motivic Galois group into $\widehat{G}(\overline{\mathbb{F}_\ell})$, but this does not seem like a very workable definition. Is there a more pragmatic approach?

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This is in my mind a central open problem.

Here is an explicit example which I believe is still wide open. Serre's conjecture (the Khare-Wintenberger theorem) says that if I have a continuous odd irreducible 2-dimensional mod p representation of the absolute Galois group of the rationals then it should come from a mod p modular form. We have a perfectly good definition of mod p modular forms (sections of the usual line bundles on mod p modular curves).

But what is the "even" analogue of Serre's conjecture?

One problem is that in characteristic zero (i.e., for 2-dimensional continuous complex representations, which must have finite image) these guys are expected to come from algebraic Maass forms, which have, as far as anyone knows, no algebraic definition: they are genuinely analytic objects and it is a complete miracle that their Hecke eigenvalues are algebraic. So there's a big open problem: give an algebraic definition of an algebraic Maass form which doesn't use analytic functions on the upper half plane which are eigenvectors for the Laplacian with eigenvalue 1/4. I have no idea how to do this and I don't think that anyone else has either. The problem is that the forms are not holomorphic (so you can't use GAGA and alg geom) and they're not cohomological (so you can't use group cohomology either), and those are in some sense the only tricks we have (other than Langlands transfer, but I don't know any functorial transfer of these guys which (a) loses no information and (b) gives rise to a form which is cohomological or holomorphic). It might be an interesting PhD problem to check this out in fact. Blasius and Ramakrishnan once tried to go up to an im quad field and then to Sp_4 over Q but the resulting form isn't holomorphic. One might define an algebraic automorphic rep to be "accessible" if it's cohomological or perhaps limit of discrete series (perhaps these are the ones for which there's a chance of proving they're arithmetic), and then try and find examples of algebraic auto reps which should never transfer to an "accessible" rep on any other group.

But even after that, there's another problem, which is that as far as I know one does not expect a continuous even irreducible mod p representation of this Galois group to lift to a de Rham representation in characteristic zero (and indeed perhaps Frank Calegari might be able to give explicit counterexamples to this, after his recent observation that oddness sometimes follows from other assumptions). So even if one could give an algebraic definition of a Maass form, one wouldn't have enough Maass forms---they would all (at least conjecturally) give rise to Galois representations with images all of whose Jordan-Hoelder factors are cyclic or A_5 (by the classification of finite subgroups of PGL_2(C)). So there's another non-trivial sticking point.

In summary---nice question but I don't think that mathematics has a good answer yet (unless you're willing to just cheat and say that an automorphic representation "is" a Galois rep with some properties).

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I'm a novice and don't understand most of what you write but I feel like there is a ton of beautiful math in your answer. –  Anonymous Mar 9 '10 at 17:09
    
I completely agree that this is a central open problem, as the case of Maass forms illustrates. I might add, however, that the paper of Gross on "Algebraic Modular Forms" provides an excellent answer when $G$ is a reductive group which is anisotropic mod center -- analogous to the "definite quaternion algebra" case. –  Marty Mar 9 '10 at 17:14
    
Yes absolutely. You're right in that I should have mentioned this case. Here we get a huge source of mod p automorphic forms and a big class of open questions of the form "can we attach a Galois representation to this function?". Didn't Gross and Pollack try to work out an example for G_2 or E_8 or some such thing? They could write down a form and then looked for a Galois representation and found one for which the first few Frobenii matched up perfectly. So there is a positive side to the story, which I should have mentioned. –  Kevin Buzzard Mar 9 '10 at 17:17
    
Yes, I was motivated to ask this question after reading Gross's paper and looking at some of Emerton's papers on p-adic Langlands. Thanks for the beautiful answer, Kevin! Aren't there already strange phenomena with odd mod-2 representations (certain weights/levels which for which the rep'n doesn't come from a char. 0 modular form with that data)? –  David Hansen Mar 9 '10 at 17:28
    
@KB: I can imagine other lifting tricks as well, for example: take an even 2-dimensional mod-p rep $\rho$, and look at $Ad(\rho) \otimes \tau$ where $\tau$ is an odd 2-dimensional mod- rep. This will be an odd 6-dimensional representation. Perhaps this guy comes from a char. zero algebraic aut rep on GL6? If $\tau$ is an induced-from-character representation it may be possible to deduce from this that $Ad(\rho)$ is automorphic. Of course, this would transfer the difficulty to the problem of proving that odd six-dimensional mod-p representations are automorphic! Ugh. –  David Hansen Mar 9 '10 at 17:43
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If $G$ is a connected, linear, reductive group over $\mathbb Q$, let $q_0$ be the dimension so denoted in Borel--Wallach, i.e. $q_0 = (d - l_0)/2,$ where $d = \text{dim } G_{\infty} - \text{ dim } A_{\infty}K_{\infty}$ and $l_0 = \text{rank } G_{\infty} - \text{ rank } A_{\infty}K_{\infty},$ where $G_{\infty} = G(\mathbb R)$, $K_{\infty}$ is a maximal compact subgroup of $G_{\infty}$, and $A_{\infty} = A(\mathbb R)$ with $A$ being the maximal $\mathbb Q$-split torus in the centre of $G$. (So $q_0$ is the first interesting dimension of homology in the locally symmetric arithmetic quotients attached to $G$.)

Then the inductive limit of the cohomology groups $H^{q_0}(G(\mathbb Q)\backslash G(\mathbb A)/A_{\infty}K_{\infty}K_f,\mathbb F_{\ell})$, as $K_f$ runs over all compact open subgroups of $G(\mathbb A_f)$, is an admissible smooth representation of $G(\mathbb G_f)$, which is (I believe) a reasonable substitute for a space of mod $\ell$ automorphic forms.

In the case when $G$ satifies the conditions of Gross's "Algebraic modular forms" paper, $q_0 = 0$, the locally symmetric quotients are just finite sets, and this is precisely the space of mod $\ell$ algebraic modular forms.

This doesn't deal with Kevin's problems; this space will (conjecturally) only contain automorphic forms whose systems of Hecke eigenvalues correspond to odd Galois representations. Nevertheless, it is a space, and one can (and does) try to work with it.

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Right! So this definition massively generalises Gross, and is the best we have, but in some sense the "problem" with it is that it's only interpolating cohomological forms so it can only see representations which are in some sense limits of cohomological forms. On the other hand, as Emerton has been showing over the last few years, these spaces are so rich that we shoulid currently be counting our blessings (e.g. we now have proofs of cases of Fontaine-Mazur etc) rather than moaning that we're not seeing any even Galois representations for GL_2. –  Kevin Buzzard Mar 9 '10 at 18:48
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