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So, how does one construct a galois representation from a Maass form?

For a modular cusp eigenform, I am familiar with the work of Eichler-Shimura, Deligne, Deligne-Serre, and realize these are different situations because of the involved geometry. I am also familiar with Maass's construction of Maass forms of weight zero from Hecke characters on real quadratic fields, so I can reverse this to answer a tiny bit of my question. There is also Langlands-Tunnell, which I am not familiar with. Finally, I realize that most Maass forms are not conjectured to be associated to galois representations.

Searching the web did not yield much. But I do want to ask an interesting precise question, so here it is:

Is there an infinite family of Maass eigenforms, such that an irreducible galois representation of infinite image is constructed to each form, and these do not somehow arise from Maass's original construction or Langlands-Tunnell?

If not, is there a conjectural association that has been checked (without proof) computationally?

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  • $\begingroup$ The Maass forms associated to Hecke characters of real quadratic fields have weight zero (not one). $\endgroup$
    – GH from MO
    Commented Feb 29, 2012 at 12:10
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    $\begingroup$ Henniart (Guy), Formes de Maass et représentations galoisiennes. Séminaire Bourbaki, 31 (1988-1989), Exposé No. 711, 26 p. (numdam.org/numdam-bin/fitem?id=SB_1988-1989__31__277_0). $$ $$ Henniart (Guy), Erratum à l'exposé n°711 : «Formes de Maass et représentations galoisiennes». Séminaire Bourbaki, 33 (1990-1991), Art. No. 16, 2 p. (numdam.org/numdam-bin/fitem?id=SB_1990-1991__33__485_0). $\endgroup$
    – Chandan Singh Dalawat
    Commented Feb 29, 2012 at 12:48
  • $\begingroup$ @Dror: I am a bit confused when you say "irreducible Galois representation of infinite image". Usually one assumes that a Galois representation is continuous, hence it factors through a finite extension of the base field, hence it has finite image. $\endgroup$
    – GH from MO
    Commented Mar 1, 2012 at 9:06
  • $\begingroup$ @GH: I didn't know that it is expected that Maass forms are attached to galois representations over $\mathbb{C}$. Over $\mathbb{Q}_\ell$ you can have infinite image. $\endgroup$
    – Dror Speiser
    Commented Mar 1, 2012 at 10:00
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    $\begingroup$ @Dror: Also, just to say what is probably obvious to you: most Maass forms are expected to have little to do with algebra (algebraic number theory, algebraic geometry). I think by a counting argument one can show that most Maass forms are not associated with Galois representations (e.g. the Maass forms constructed by Maass have zero density). $\endgroup$
    – GH from MO
    Commented Mar 1, 2012 at 17:11

1 Answer 1

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Several remarks before answering your questions: (1) Langlands-Tunnell is a result in the other direction: from Galois representation to automorphic forms; it is therefore not relevant. (2) One expects to be able to attach Galois representations only to certain types of Mass forms, those whose component at infinity in algebraic (in the automorphic representation settings) or equivalently, whose eigenvalue for the Laplacian is $1/4$. (3) this Galois representation is expected to take values in Gl${}_2(\mathbb{C})$, hence to have finite image.

So you ask: "how does one construct a galois representation from a Maass form?". The answer is: one still doesn't know how to. It's one of the most striking open problem in the Langlands program. There was 25 years ago an announcement that this problem has been essentially solved (with published articles), but it was soon after retracted: see the two references given by Chandan in comments.

And for your displayed question about the infinite family, stripped of the reference to Langlands-Tunell and of the "infinite image" condition, the answer is no, as far as I know,

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  • $\begingroup$ I wonder, are Maass's forms for real quadratic fields as I mentioned above also not relevant? $\endgroup$
    – Dror Speiser
    Commented Feb 29, 2012 at 15:06
  • $\begingroup$ @Dror: They are relevant, but you need to assume the Hecke character is of finite order so that it can be identified with a 1-dimensional Galois representation. $\endgroup$
    – GH from MO
    Commented Feb 29, 2012 at 15:50
  • $\begingroup$ @GH: algebraic Hecke characters of infinite order can be identified with 1-dimensional ($p$-adic) Galois representations. Of course, for a totally real field, all Hecke characters are cyclotomic twists of the finite order characters, so as Katz says, I've been "speaking prose all along". Perhaps you mean to say that only finite order Hecke characters give weight zero Maass forms? I'm not really familiar with the construction, though I imagine it's just an automorphic induction. $\endgroup$
    – Rob Harron
    Commented Mar 1, 2012 at 1:08
  • $\begingroup$ @Rob: I am not too familiar with $p$-adic Galois representations. At any rate, I had complex representations in mind. Note also that Hecke characters give weight zero Maass forms regardless of their archimedean component, see Theorem 1.9.1 in Bump: Automorphic forms and representations. And yes, this is a special case of automorphic induction. $\endgroup$
    – GH from MO
    Commented Mar 1, 2012 at 9:04
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    $\begingroup$ @GH: Weil used the terminology "Hecke character of type $A_0$" for what we now call "algebraic Hecke characters". One way to define these is to view the Hecke character as a complex (quasi-)character $\chi$ of the ideles $\mathbf{A}^\times_K$ of the number field $K$. Then, on the connected component of the identity of the infinite ideles, we have $\phi((x_v)_v)=\prod x_v^{m_v}$ ($v$ varying over the infinite places). Then, χ is algebraic if $m_v\in\mathbf{Z}$ for all $v$. One can also phrase this in terms of the infinitesimal character of the $(\mathfrak{g},K)$-module of $\chi$, or as ... $\endgroup$
    – Rob Harron
    Commented Mar 2, 2012 at 18:45

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