# Galois representations attached to Maass form

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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The Maass forms associated to Hecke characters of real quadratic fields have weight zero (not one). – GH from MO Feb 29 '12 at 12:10
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). – Chandan Singh Dalawat Feb 29 '12 at 12:48
@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. – GH from MO Mar 1 '12 at 9:06
@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. – Dror Speiser Mar 1 '12 at 10:00
@Dror: OK, I see now. At any rate, even 2-dimensional complex Galois representations are expected to correspond to weight zero Hecke-Maass forms of Laplacian eigenvalue 1/4. Note that you can regard any complex Galois representation as an $\ell$-adic one. – GH from MO Mar 1 '12 at 15:02

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.
@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. – Rob Harron Mar 1 '12 at 1:08
@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. – GH from MO Mar 1 '12 at 9:04
@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 ... – Rob Harron Mar 2 '12 at 18:45