See Edward Frenkel's article "Lectures on the Langlands program and conformal field theory" for an exposition of the function fields Langlands correspondence (now a theorem of Drinfel'd, L.Lafforgue & V.Lafforgue). It's available here: https://arxiv.org/abs/hep-th/0512172

My question is about function fields Langlands for $GL_n$ (see page 32 of the article). Drinfeld-Lafforgue's theorem gives a bijection between $n$-dimensional irreducible representations of $Gal(\overline{F}/F)$ and irreducible cuspidal automorphic representations of $GL_n(\mathbb{A}_F)$. Here $X$ is a projective curve over $k = \mathbb{F}_q$ (a finite field with $q$ elements), and $F = k(X)$ is the field of rational functions; it's algebraic closure is $\overline{F}$. See Section 2.2 (on pg 27) for more details. The ring of adeles, $\mathbb{A}_F$ is the restricted product of the fields $F_x$, where $x$ runs over all closed points of $X$; see Section 2.3 (on pg 29) for more details, and a definition of ``irreducible cuspidal automorphic representations".

Let $X$ be a curve with genus either 0 (i.e. $\mathbb{P}^1$), 1 (i.e. an elliptic curve) or 2 (i.e. a hyperelliptic curves). What are some examples of the function fields Langlands correspondence (i.e. an example of a Galois representation, with a computation of the corresponding irreducible cuspidal automorphic representation)?

If there are any papers computing this (for instance, using the theory of Drinfel'd shtukas), please let me know. Thanks.


1 Answer 1


If you really want an example of a representation, there's something funny you will find. Any irreducible cuspidal automorphic representation of $GL_n(\mathbb A_F)$ factors as a restricted tensor product of representations of the group over local fields $GL_n(F_x)$. So giving an example really means giving an example of a local representation at each place. It follows from Lafforgue's theorem (and prior work) that the local representation in this case is determined by the restriction of the Galois representation to the Galois group of $F_x$.

So if you want an example of the global Langlands correspondence, all you really need is a set of examples of the local Langlands correspondence sufficient to contain all the local Galois representations of a global representation. For instance, I could tell you how to construct the representations associated to unramified Galois representations using the Satake isomorphism or unramified principal series, and how to construct the representations associated to unipotent Galois representations using Steinberg representations, and then write down an example of a semistable elliptic curve, whose local representations are all either unramified or unipotent.

However, a problem with this approach is that it doesn't provide any intuition for why the representation actually appears in $GL_n(F) \backslash GL_n (\mathbb A_F)$.

Thus, another approach is to, rather than explicitly describing the representation, describe a function on $GL_n(F) \backslash GL_n (\mathbb A_F)$ which generates the representation. There are some papers doing this in the literature, in the special case of rigid Galois representations, where this is the easiest (because the function can then usually be characterized by certain symmetry properties). For instance, see Heinloth-Ngo-Yun section 2.1 (though that may be a little too general/abstract for your purposes).

Here is what I believe is the simplest example:

Let $F= \mathbb F_q(t)$, with $q$ odd.

Let $E$ be the elliptic curve over $\mathbb F_q(t)$ given by the equation $y^2 = x(x-1)(x-t)$ . It has singular fibers at $0, 1$, and $\infty$.

Let $J$ be the subgroup of $K = \prod_x GL_2 (\mathcal O_{x})$ consisting of tuples of matrices whose elements at the primes $0, 1$, and $\infty$ are congruent modulo the uniformizer to upper-triangular matrices.

Let $\chi$ be the character of $J$ which is defined by restricting to the matrix over the prime at infinity and taking a quadratic character of the ratio of the two diagonal entries mod the uniformizer.

Then it is not too hard to calculate the space of functions on $GL_2(\mathbb A_F)$, left invariant under $GL_2(F)$, right $\chi$-equivariant under $J$, and invariant under $GL_1(\mathbb A_F)$, is two-dimensional.

Indeed, one sees directly that for each double coset of $GL_2(F)$ by $JGL_1(\mathbb A_F)$, generated by an element $g$, the space of such functions on the coset is either zero or one-dimensional, and is one-dimensional if and only if $\chi$ vanishes on $g^{-1} GL_2(F) g \cap J GL_1(\mathbb A_F)$. The cosets are in bijection with rank two vector bundles on $\mathbb P^1$ with a fixed one-dimensional subsace of the fiber at $0,1,$ and $\infty$, up to twisting by line bundles, and that intersection is simply their automorphism group preserving the subspaces. So one is looking to find such vector bundles whose automorphism group does not map surjectively to $\mathbb G_m \times \mathbb G_m$ under its natural action on the one-dimensional space at $\infty$ and its quotient. Classifying, the vector bundles, one sees that there are two such.

This two-dimensional space is generated by two different forms, each of which generates an automorphic representation. One corresponds to the Tate module of $E$, and the other to the Tate module of the quadratic twist of $E$.

It should be possible to check this by calculating the Hecke eigenvalues by hand.

  • $\begingroup$ Thanks for the explanation! Do you understand how this works in genus zero (i.e. for P^1)? Understanding the Galois side is not difficult here; is the automorphic side easy as well? $\endgroup$
    – Vinoth
    May 28, 2018 at 18:16
  • $\begingroup$ @Vinoth What do you mean by that? Galois representations on $\mathbb P^1$ can be quite complicated. I'm guessing you're referring to unramified representations. I know how to do that, but it's not very enlightening. $\endgroup$
    – Will Sawin
    May 28, 2018 at 20:31
  • $\begingroup$ @Vinoth One calculates explicitly the double coset space, then calculates the eigenspace of the Hecke operators at one place. Finding that it is one-dimensional, one determines that the eigenvalues at every other place are determined, and by a formula that matches what you get from the fact that the Frobenius conjugacy class at every place is determined by its value at one place. $\endgroup$
    – Will Sawin
    May 29, 2018 at 5:27
  • $\begingroup$ Thanks Will. Raeez Lorgat & I have been talking about this. I'll e-mail you both soon. $\endgroup$
    – Vinoth
    Jun 1, 2018 at 5:10

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