# Are there Maass forms where the expected Galois representation is $\ell$-adic?

Recall that by theorems of Deligne and Deligne--Serre, there is the following dichotomy:

1. Modular forms on the upper half plane of level $N$ and weight $k\geq 2$ correspond to representations $\rho:\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)\to\operatorname{GL}(2,F_\lambda)$ for some number field $F$ and prime $\lambda$ over $\ell$.

2. Modular forms on the upper half plane of level $N$ and weight $k=1$ correspond to representations $\rho:\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)\to\operatorname{GL}(2,\mathbb C)$ satisfying $\det\rho(\sigma)=-1$ ($\sigma$ is complex conjugation).

Now I hear that Maass forms of eigenvalue $\frac 14$ are conjectured to correspond with representations $\rho:\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)\to\operatorname{GL}(2,\mathbb C)$ satisfying $\det\rho(\sigma)=1$ ($\sigma$ is complex conjugation). Is this still true (that is, conjectured) for Maass forms of higher weight? Or do they "turn $\ell$-adic" in higher weight?

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Here's some piece of the bigger picture. Maass forms and holomorphic modular forms are both automorphic representations for $GL(2)$ over the rationals. An automorphic representation is a typically huge representation $\pi$ of an adele group (in this case $GL(2,\mathbf{A})$, with $\mathbf{A}$ the adeles of $\mathbf{Q}$). Because the adeles is the product of the finite adeles and the infinite adeles, this representation $\pi$ is a product of a finite part $\pi_f$ and an infinite part $\pi_\infty$. The infinite part is a representation of $GL(2,\mathbf{R})$ (loosely speaking -- there are technicalities but they would only cloud the water here).

The representation theory of $GL(2,\mathbf{R})$, in this context, is completely understood. The representations basically fall into four categories, which I'll name (up to twist):

1) finite-dimensional representations (these never show up in the representations attached to cusp forms).

2) Discrete series representations $D_k$, $k\geq2$ (these are the modular forms of weight 2 or more).

3) The limit of discrete series representation $D_1$ (these are the weight 1 forms).

4) The principal series representations (these are the Maass forms).

Now what does Langlands conjecture? He makes a conjecture which does not care which case you're in! He conjectures the existence of a "Galois representation" attached to $\pi$, and this is a "Galois representation" in a very loose sense: it is a continuous 2-dimensional complex representation of the conjectural "Langlands group", attached to $\pi$. Note that there should be a map from the Langlands group to the Galois group, and in the case of Maass forms and weight 1 forms Langlands' representation should factor through the Galois group. For modular forms of weight 2 or more Langlands' conjecture has not been proved and in some sense it is almost not meaningful to try to prove it because no-one can define the group. In particular Deligne did not prove Langlands' conjecture, he proved something else.

So Clozel came along in 1990 and tried to put Deligne's work into some context and he came up with the following: he formulated the notion of what it meant for $\pi_\infty$ to be algebraic (in fact there are two notions of algebraic, which differ by a twist in this context, so let me write "$L$-algebraic" to make it clear which one I'm talking about) and conjectured that if $\pi$ were $L$-algebraic then there should be an $\ell$-adic Galois representation $\rho_\pi$ attached to $\pi$. Maass forms with eigenvalue $1/4$, and holomorphic eigenforms, are $L$-algebraic, and the $\ell$-adic Galois representation attached to the Maass forms/weight 1 forms is just the one you obtain by fixing an isomorphism $\mathbf{C}=\overline{\mathbf{Q}}_\ell$. I should say that Clozel worked with $GL(n)$ not $GL(2)$ and also worked over an arbitrary number field base.

Whether or not the image of $\rho_\pi$ is finite is something which is conjecturally determined by $\pi_\infty$: you can read it off from the infinitesimal character of $\pi_\infty$ and also from the local Weil group representation attached to $\pi_\infty$ by the local Langlands conjectures, which are all theorems (of Langlands) for real reductive groups.

Put within this context your question becomes purely local: one has to figure out what Clozel's recipe gives in each case to get a handle on what your question is asking. You're asking about principal series representations. If you work out Clozel's recipe in these cases you find that if $\lambda\not=1/4$ then $\pi_\infty$ is not $L$-algebraic (and so we don't even expect a representation of the Galois group, we just expect a representation of the conjectural Langlands group), and if $\lambda=1/4$ then, up to twist, we expect the image to be always finite, because, well, that's what the calculation gives us.

I learnt this by just doing all these calculations myself. I wrote them up in brief notes at http://www2.imperial.ac.uk/~buzzard/maths/research/notes/automorphic_forms_for_gl2_over_Q.pdf and http://www2.imperial.ac.uk/~buzzard/maths/research/notes/local_langlands_for_gl2R.pdf (both available from http://www2.imperial.ac.uk/~buzzard/maths/research/notes/index.html ).

So why is there this asymmetry? Well actually this asymmetry is not surprising because it is predicted on the Galois side as well. If you look at an irreducible mod $p$ ordinary Galois representation which is odd then its universal ordinary deformation is often known to be isomorphic to a Hecke algebra of the type defined by Hida (so in particular we get lots of interesting $\ell$-adic Galois representations with infinite image). In particular its Krull dimension should be 2 (and this was already known to Mazur in the 80s). But the calculations for these Krull dimensions involve local terms, and the local term at infinity depends on whether the representation is odd or even. If you consider deformations of an even Galois representation then the calculations come out differently and the Krull dimension comes out one smaller. In particular one only expects to see finite image lifts, plus twists of such lifts by powers of the cyclotomic character.

So in summary you see differences on both sides -- the automorphic side and the Galois side -- and they match up perfectly! You don't expect $\ell$-adic representations to show up in the Maass form story and yet things are completely consistent anyway.

Toby Gee and I recently tried to figure out the complete conjectural picture about how automorphic representations and Galois representations were related. Our conclusions are at http://www2.imperial.ac.uk/~buzzard/maths/research/papers/bgagsvn.pdf . But for $GL(n)$ this was all due to Clozel over 20 years ago (who would have known all those calculations that I linked to earler; these are all standard).

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At work today I realised that in some sense I could also offer a more low-level answer: here it is. The basic yoga for functional equations is well-understood. So here's the experiment you might want to do. Think of the functional equation for a Maass form. If you understand what's going on then I guess it's clear that the factor at infinity is not going to be the factor at infinity for the functional equation for an $\ell$-adic Galois representation, unless the eigenvalue is $1/4$, and in this case it's going to be the functional equation for a Galois representations whose H-T weights are... –  Kevin Buzzard Jun 6 '12 at 17:57
...the same! In particular there's no room for some "rich" theory like in the holomorphic case. Oh -- by "functional equation for a Maass form" I mean "functional equation for the $L$-function of a Maass form" and similarly for Galois reps. –  Kevin Buzzard Jun 6 '12 at 17:58
@Kevin: Is your "Algebraic automorphic representation" a generalization of Weil's "characters of type $(A_0)$" ? If so, could you explain the above picture ($\mathbb C$-adic vs $l$-adic) in this $GL_1$ case ? Thank you –  user4245 Jun 16 '12 at 8:09
In the abelian case the two notions of algebraic (C and L) that Toby Gee and I introduce coincide. It's only in the non-abelian case that one sees two "normalisations" occurring in the literature, which Gee and I explicitly disentangle. –  Kevin Buzzard Jun 18 '12 at 22:06
To point 3): I am not sure whether, one should distinguish limit of dicrete series from principal series here, since they are not a subquotient in the case of GL(2), only for SL(2). –  Marc Palm Feb 19 '13 at 9:39

@Kevin: The explanation of the asymmetry from the Galois side does not seem to work since by recent work of Calegari, "EVEN GALOIS REPRESENTATIONS AND THE FONTAINE MAZUR CONJECTURE", he has shown that there exist universal deformation of even Galois representations with universal deformation ring of large dimension such that none of the corresponding Galois representations are geometric.

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You must be mistaken. What result of Calegari are you exactly referring to, and how does it contradict Kevin's argument? –  Joël Apr 4 '13 at 17:33
Dear Marcelo, I think you have misread Calegari's cor. 5.2. Firstly, he works in the context of a totally real field, not just $\mathbb Q$. Secondly, he looks at deformations of a $\rhobar$ which is odd at some places and even at other places (but even at least one place). The dimension of the deformation ring is bounded below by (and presumably equals) $1 + 2r$ (plus the Leopoldt defect, which conjecturally equals $0$), where $r$ is the number of odd places. In particular, if $r = 0$ (which is necessarily the case when we are working with $G_{\mathbb Q}$, since $\mathbb Q$ has ... –  Emerton Apr 6 '13 at 2:27
... only one real place), then the dimension we get is one, which is just given by twisting. So indeed the def. ring of an even $\overline{\rho}: G_{\mathbb Q} \to \GL_2(\overline{\mathbb F}_p)$ is expected to equal $1$, and typically contains no geometric points at all (e.g. because maybe $\overline{\rho}$ has image too big to be lift to a finite subgroup of $\GL_2(\mathbb C)$). The only way to get big dimension is to take higher degree totally real fields, and then to take $r$ larger. But again, the dimension Calegari computes is completely consistent with Kevin's discussion above; ... –  Emerton Apr 6 '13 at 2:29
P.S. In the above, $r$ is the number of real places where $\overline{\rho}$ is odd. –  Emerton Apr 6 '13 at 2:31