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?
Best Answer
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,