$\DeclareMathOperator\GL{GL}$Apologies if this question has already been dealt with on MO. I am wondering about the status of the global Langlands conjectures for $\GL_2$ over the rational numbers. How close is humanity to the proof of these conjectures?

I guess Langlands picture is related to (or, should I say, includes?) the Taniyama-Shimura, Fontaine-Mazur, and Serre Conjectures. There has been great progress on these latter conjectures recently. Even assuming these conjectures are settled, how much closer are we to understanding the Langlands program for $\GL_2$?

It would be helpful if somebody points to a paper where a modern and precise version of the global Langlands conjecture for $\GL_2$ is stated. For the local Langlands conjecture, we have the nice paper of Vogan. Is there an analogous paper for the global theory?

  • $\begingroup$ Hi Masoud! I added some dollar signs to your post so the math renders properly. $\endgroup$
    – MTS
    Apr 11, 2013 at 0:02
  • $\begingroup$ +1. And I had been thinking of asking the exact same question! $\endgroup$ Apr 11, 2013 at 0:49
  • $\begingroup$ By "the nice paper of Vogan", do you mean www-math.mit.edu/~dav/md.pdf ? Perhaps you should also mention the Bushnell-Henniart book on The local Langlands conjecture for $GL(2)$. $\endgroup$ Apr 11, 2013 at 4:51

2 Answers 2


This question has already been discussed here, though perhaps not exactly on these terms (but see What makes Langlands for n=2 easier than Langlands for n>2?).

So first there is an ambiguity of what id global Langlands in general. Langlands was interested in all cuspidal automorphic representations for $Gl_2$: he conjectured they should be in natural bijection with complex two-dimensional representations of a large group called the Langlands group of $\mathbb Q$. Problem: the Langlands group of $\mathbb Q$ is a conjectural object, whose very existence depends on a hypothetical structure of Tanakian category on the category of all cuspidal automorphic representations for $Gl_n$ when $n$ varies. For this reason, it does not really makes sense of Langlands' conjecture for $Gl_2$: it is really a conjecture concerning all $Gl_n$ together.

Therefore, when people refers to global Langlands for $Gl_2$ they mean a more restrictive conjecture which deal with only a subclass of the set of cuspidal automorphic representation, namely the one which are algebraic at infinity. The conjecture is then as follows:

Conjecture: Fix an auxiliary prime $\ell$. There is a natural bijection between

(1) the set of algebraic at infinity cuspidal automorphic representations for $Gl_2$

(2) the set of 2-dimensional continuous irreducible Galois representation of $G_{\mathbb Q}$ over the algebraic closure of $ \mathbb Q_\ell$ which are unramified at almost all primes and de Rham at $\ell$.

Moreover, this bijection should be compatible to the local Langlands correspondence at almost every prime $p \neq \ell$ (or better, at every place). (This condition fixes the bijection uniquely).

So what is proved in this conjecture. Well there are three type of representation $\pi_\infty$ which are algebraic, which leads to a tripartition of the set (1): (1a) : the set of cuspidal automorphic forms for $Gl_2$ for which $\pi_\infty$ is a discrete series which is essentially the set of new cuspidal modular forms of weight $k\geq 2$. (1b) : the set of cuspidal automorphic forms for which $\pi_\infty$ is a limit of discrete series, which is essentially the set of new cuspidal modular forms of weight $k=1$. (1c) : the set of cuspidal automorphic forms for which $\pi_\infty$ is a specific principal series, which corresponds to the set of Mass form with Eigenvalue for the Laplace-beltrami operator (a.k.a. the Casimir) 1/4.

To prove the conjecture, one needs (a) to construct one map from (1) to (2) which is compatible with Local Langlands, and (b) prove that it is surjective (the injectivity will be easy by strong multiplicity 1 for $Gl_2$). Now this map is known only in the case (1a) and (1b), not at all in the case (1c). Worse: the map was already constructed in 1970 in the cases (1a) and (1b) by Deligne and Deligne-Serre, and we have made no progress since. So it is very hard to say how far we are from constructing the map in general. There is no publicly voiced ideas to prove it, but maybe some one will come tomorrow and prove it. What about surjectivity? Well, the image of (1a) and (1b) together should be (2ab), the set of Galois rep. as above which in addition are odd. The image of (1a) should be the set (2a) of odd representations with distinct Hodge-Tate weights, and actually we know that there is a surjection (1a) to (2a): this is the part the conjecture of Fontaine-Mazur proved by Kisin and Emerton. We don't know yet the surjectivity of (1b) to (2b) (defined as (2ab)-(2a)), which is essentially Artin's conjecture but special cases have been done by Taylor, Buzzard, and Calegari.

  • $\begingroup$ I hope I don't sound too ignorant, but doesn't this answer ignore the construction from (2) to (1c) for even solvable galois groups? $\endgroup$ Apr 11, 2013 at 14:05
  • $\begingroup$ Technically the limit of dicrete series reps is not a quotient as in the case $SL(2)$, so is it still common to call it that way? It is a principal series representation of $\GL_2(\mathbb{R})$. $\endgroup$
    – Marc Palm
    Apr 11, 2013 at 14:36
  • $\begingroup$ @Dror: it does ignore it, and so do I. You are saying that from an even solvable galois representation one can construct an maass form? $\endgroup$
    – Joël
    Apr 11, 2013 at 15:49
  • $\begingroup$ @Joel: Yeah, I think so, isn't this a part of the Langlands-Tunnel theorem? $\endgroup$ Apr 12, 2013 at 0:37
  • $\begingroup$ @Joel: Thank you for the extensive reply. From what I can gather from Marc's and your reply, the issue of being algebraic at infinity is a condition about the restriction of the automorphic representation to the Archimedean place. So, in other words, if we include all automorphic representations we need Langlands group $L(\mathbb{Q})$, but if we want only algebraic at infinite automorphic representations, then the absolute Galois group $G_{\mathbb{Q}}$ suffices. Is there anyway one can see how much bigger this Langlands group is? $\endgroup$
    – Masoud
    Apr 15, 2013 at 3:48

Additionally to what Joel was saying, Langlands conjectured the existence of a universal group $\widehat{G}$ (depending on the number field only) whose 2-dim'l representation correspond to automorphic representation of GL(2) in a suitable way. These would include automorphic forms with the archimedean factors not being algebraic, e.g., even Maass forms with Laplace eigenvalue $\neq 1/4$. From what I understand, Arthur suggests a definition of $\widehat{G}$ here: http://www.claymath.org/cw/arthur/pdf/automorphic-langlands-group.pdf. The situation is similar to case (1c) described above.


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