On page 263 of this book review appears the following:

Given the centrality of L-functions to the Langlands program, nothing would seem more natural (than a presentation of elementary algebraic number theory from the standpoint of L-functions and their analytic properties), but in fact the properties of L-functions traditionally of interest to analytic number theorists - for example, the location of zeroes in the critical strip (the Generalized Riemann Hypothesis) - have historically had little to do with the preoccupations of the Langlands program. Thanks largely to the efforts of a few charismatic and determined individuals, this is beginning to change and Langlands himself has in recent years turned to methods in analytic number theory in an attempt to get beyond the visible limits of the techniques developed over the last few decades.

I'd like to ask for a big picture exposition of how such questions about the location of zeroes of L-functions appear and interact with the Langlands program. My interest is mainly cultural and the answer should be tailored for the outsider to number theory (I'm viewing Langlands program algebraically as the pursuit of a nonabelian class field theory.)

A more crude question is:

Does the Langlands program say anything about the Grand Riemann Hypothesis or vice versa?

This is almost certainly too crude a question for MO, but Langlands seems to have such an amazing unifying appeal, that I feel a temptation to see how much it subsumes. I fully expect an answer like "It is impossible to coherently discuss this without years of training". Thank you for any attempt to explain things to someone who is not a number theorist, in advance!

  • 1
    $\begingroup$ The Langlands program has always and explicitly had a lot to do with Artin L-functions, and in particular with the Artin conjecture on their meromorphy. So in some sense it is a question of how adjacent the Artin conjecture is to GRH. The obstruction to using Brauer's theorem to prove the Artin conjecture is partly to do with not being able to take rational powers of meromorphic functions at will, absent control of zeroes (and poles). The quote from the book review may have overstated the point, to some extent. $\endgroup$ – Charles Matthews Jan 19 '13 at 14:06
  • 5
    $\begingroup$ ... the Artin conjecture on their holomorphy. $\endgroup$ – Chandan Singh Dalawat Jan 19 '13 at 14:08
  • $\begingroup$ @Charles Matthews: This is precisely the sort of thing I wanted to know. Thanks! $\endgroup$ – Jon Bannon Jan 19 '13 at 14:10
  • 5
    $\begingroup$ @Chandan Singh Dalawat: I am sometimes a trivial character. $\endgroup$ – Charles Matthews Jan 19 '13 at 18:12
  • 2
    $\begingroup$ I am far outside the field, but one observation is that many results of Langlands depend on the Selberg trace formula and its extensions. Moreover, the Selberg zeta functions satisfy a Riemann hypothesis trivially, since the zeroes correspond to the eigenvalues of an elliptic operator (the trace formula relates these eigenvalues to lengths of closed curves on a Riemann surface). I think there is some hope that the zeroes of the zeta function will be associated to some elliptic operator, and maybe there will be a generalized associated Selberg trace formula? $\endgroup$ – Ian Agol Jan 19 '13 at 18:28

One can use Langlands functoriality to eliminate the so-called Siegel zeros of an automorphic $L$-function. For example, Hoffstein-Ramakrishnan (IMRN 1995) proved that the $L$-function of a $GL(n)$ cusp form for $n>1$ has no Siegel zero if all $GL(m)\times GL(n)$ $L$-functions are $GL(mn)$ $L$-functions. There are several unconditional results along this line, e.g. in the same paper it is shown that the $L$-function of a $GL(2)$ cusp form has no Siegel zero.

  • 1
    $\begingroup$ Note: one must assume $n\geq 2$ for this result (so it does not address the zeros of Dirichlet $L$-functions.) $\endgroup$ – Denis Chaperon de Lauzières Jan 19 '13 at 18:07
  • $\begingroup$ @Denis: Thank you, this condition is crucial indeed. $\endgroup$ – GH from MO Jan 19 '13 at 18:17
  • $\begingroup$ This is pretty strong. If I don't get any other answers to this thing, I think I'll accept this one. $\endgroup$ – Jon Bannon Jan 21 '13 at 17:46
  • $\begingroup$ It does answer the question as posed... Perhaps there will be more to say, but this is pretty good. $\endgroup$ – Jon Bannon Jan 21 '13 at 17:47

Number theory can seem to the beginner like a very random collection of results, and it is only fairly recently in its 5000 year history that the larger picture has begin to emerge. A report from the NAS in the early 1990s opened my eyes to the fact that number theory now is centered around three questions, each having to do with $L$-functions.

The first area is the Riemann hypothesis, and its generalizations to more general $L$-functions. Questions about the vertical as well as horizontal distribution of the zeros initiated by Montgomery and influenced by random matrix theory fall in this area.

The second area is the Langlands program. Unlike the Riemann hypothesis, the Langlands program is named for the most advanced part of the theory, not the original question. But the roots of the Langlands program go all the way back to Gauss' Law of Quadratic Reciprocity: Given an odd prime $q$, let $\epsilon=\pm 1$ so that $\epsilon q\equiv 1\bmod 4$. Then for an odd prime $p$, the Legendre symbols $(\frac{p}{q})$ and $(\frac{\epsilon q}{p})$ are equal. More abstractly, the Galois representation arising from the Kronecker symbol $(\frac{\epsilon q}{*})$ has the same $L$ function as the Dirichlet character $(\frac{*}{q})$. Langlands interprets the latter as an automorphic from on $GL(1)$.

The third area is the Bloch-Beilinson conjectures, which include the Birch and Swinnerton-Dyer conjecture as a special case. The simplest manifestation of these is the Dirichlet Class Number Formula.

As GH's answer, different areas relate to each other. The possibility of a Landau-Siegel zero prevents us from getting the lower bound on class numbers we expect. The fact CM elliptic curves were known to be automorphic allowed Birch and Swinnerton-Dyer to be able to compute sufficient examples to make a conjecture.

  • 5
    $\begingroup$ Do you have a reference for the NAS report? $\endgroup$ – Cam McLeman Jan 22 '13 at 3:35
  • $\begingroup$ This is a nice flyover! Thank you for the response. $\endgroup$ – Jon Bannon Jan 22 '13 at 11:53
  • $\begingroup$ @Cam No, I don't. I've since tried to track it down and been unable to. As I recall, the number theory section was just a portion of a report on the state of mathematics generally. $\endgroup$ – Stopple Jan 22 '13 at 16:30
  • 3
    $\begingroup$ The first paragraph of Langlands' 1978 ICM lecture "L-functions and Automorphic Representations" publications.ias.edu/rpl/paper/65 reads "Introduction. There are at least three different problems with which one is confronted in the study of L-functions: the analytic continuation and functional equation; the location of the zeroes; and in some cases, the determination of the values at special points. The first may bethe easiest. It is certainly the only one with which I have been closely involved." $\endgroup$ – Jonah Sinick Jan 23 '13 at 4:04
  • $\begingroup$ The NAS report might be this one: nap.edu/read/1598/chapter/10#103 I've tried to link to section 9 in Appendix B that discusses number theory $\endgroup$ – j.c. Feb 29 '16 at 21:47

This doesn't really address the question, but it seems worth mentioning that in order for the Riemann hypothesis to be well formulated for a motivic L-function, one has to know that the L-function is also an automorphic L-function. For example, a priori all one knows about the L-function of an elliptic curve is that it converges for $Re(s) > 3/2$ (by Hasse's bound). One needs automorphicity to extend the L-function to $Re(s) = 1$, and the Riemann hypothesis for the L-function of an elliptic curve states that all (nontrivial?) zeros of the L-function lie on the line $Re(s) = 1$.

  • $\begingroup$ Are our converse theorems strong enough to say that now? I thought many sorts of motivic L-functions could non-disprovably have analytic continuations without automorphicity. $\endgroup$ – Will Sawin Jan 19 '13 at 19:11
  • $\begingroup$ I don't think that our converse theorems are strong enough for what I wrote to be provably true (in general), but I think it's true. What do you mean when you write "many sorts of motivic L-functions could non-disprovably have analytic continuations without automorphicity"? $\endgroup$ – Jonah Sinick Jan 19 '13 at 19:15
  • 4
    $\begingroup$ Actually one should shift normalize every $L$-function so that $1/2$ is the center, and $Re(s)=1/2$ is the critical line. It is very confusing that elliptic curve $L$-functions are normalized differently than most $L$-functions. In fact most $L$-functions have no algebraic object behind them, so the functional equation (or the size of the coefficients) should be the basis of normalization. And we should also pay respect to Riemann... $\endgroup$ – GH from MO Jan 19 '13 at 20:03
  • $\begingroup$ @ GH – Yes, I'm aware of this, I glossed over this for brevity. Feel free to edit my answer if you'd like. $\endgroup$ – Jonah Sinick Jan 19 '13 at 20:22
  • 4
    $\begingroup$ For what it's worth, I strongly disagree with GH's assertion about how L-functions should be normalized. That normalization is sensible if you only care about analytic properties, but very bad if you are interested in special values, p-adic theory, etc, because in this case one cares about L-functions whose Dirichlet coefficients all lie in some fixed number field; shifting by a half-integer will destroy this. $\endgroup$ – David Loeffler Jan 25 '13 at 9:12

The analogue of the Riemann hypothesis for the Selberg zeta function for $\Gamma(N) \backslash \mathbb{H}$ is known as the Selberg eigenvalue conjecture. It would follow from the Langlands functoriality. However, all but finitely many of these zeros lie on $\Re s =1/2$ beforehand, so this is very different from the Riemann zeta function.

  • $\begingroup$ Any potential counterexample would lie on the real axis, and so would be the analog of a Landau-Siegel zero. $\endgroup$ – Stopple Feb 1 '13 at 20:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.