This is a very naive question which really does little more than highlight my ignorance of how converse theorems really work.

Take an algebraic gadget which should be conjecturally associated to an automorphic representation. For example, take a finite image continuous complex representation of the absolute Galois group of a number field. Or take an elliptic curve over a totally real field. There are converse theorems of the form "if the $L$-function of this gadget and the $L$-function of sufficiently many twists of this gadget have analytic continuation, are bounded in vertical strips, and satisfy the expected functional equations, then the gadget is indeed associated to an automorphic representation". For 2-dimensional gadgets (for example 2-dimensional complex Galois representations, or elliptic curves) one only needs to consider abelian twists---this is essentially proved in Jacquet--Langlands.

However it is nowadays pretty standard that these $L$-functions have meromorphic continuation and the expected functional equation. In the finite image case this follows from Brauer's theorem in finite group theory, and in the elliptic curve case this follows from the fact that these curves are now known to be potentially modular, following work of Kisin, Taylor and others.

So one might ask whether meromorphic continuation + functional equation, plus converse theorem techniques, is enough to prove something about the algebraic gadget (other than "it potentially comes from an automorphic representation" -- something which we already know and are using to get the meromorphic continuation). But presumably there is some serious theoretical obstruction to proving anything interesting here, or else it would all have been done for Artin $L$-functions in the 70s.

What is the obstruction?

  • $\begingroup$ See math.uiowa.edu/~mkrishna/Research/converse.pdf "In this paper we show that one may relax the hypotheses of the Jacquet-Langlands theorem over number fields by allowing the twisted L-functions $L(s,\pi\otimes\omega)$ to have essentially arbitrary poles." In 1.1 their condition is: "ii) continue meromorphically to ratios of entire functions of finite order" and "iv) are entire whenever $\omega$ is unramified at every non-archimedean place." $\endgroup$ – Junkie Apr 29 '11 at 10:55
  • $\begingroup$ So it seems to me that instead of "all the twists have analytic continuation" one now has something close to "all the twists have meromorphic continuation, but the $L$-function itself has analytic continuation": this seems to basically be enough. So the question now becomes "why can't one remove this one last assumption of analyticity?". $\endgroup$ – Kevin Buzzard Apr 29 '11 at 11:04
  • $\begingroup$ What is an example that shows the assumption about twists in converse theorems is somehow essential? That is, what is an example of a Dirichlet series which fits the conditions of a converse theorem with the twist aspect not taken into account, and the conclusion of the converse theorem is wrong? $\endgroup$ – KConrad Apr 30 '11 at 3:39
  • $\begingroup$ I'm no expert Keith, but my guess is that such an example would be very hard to come by---see the last para of Kowalski's answer. Note also that Booker proved in the 2-d Artin case over Q that analytic continuation of the function alone is enough. See Booker's Annals paper---MR2031863. The same result is true for modular forms of some small levels I think: e.g. Hecke's work (also cited by Emmanuel) didn't need any twists. Of course you may well know all this already so apologies. $\endgroup$ – Kevin Buzzard Apr 30 '11 at 7:48
  • $\begingroup$ "why can't one remove this one last assumption of analyticity?" I think it's necessary. Kowalski pointed out what I had written/deleted about Hecke's work (see bottom page 5 of Booker/krishnamurthy), essentially that automorphy fails from poles when moving contours. And Emerton's comment has a version as $F(s)=L(E_{37a},s)^2/L(E_{37b},s)$ or something silly (I just chose first $N=37$ with 2 elliptic curves). Here $F(s)=\sum_n b_n/n^s$ with $b_n$ having polyboundedness, converge in right-plane, continue meromorph to entire of finite order, satisfy func equation, but the conclusion is false. $\endgroup$ – Junkie Apr 30 '11 at 9:55

Dear Kevin,

My understanding of the current meromorphic continuation results is that they do something along the lines of expressing a given Galois representation as the induction of a virtual (i.e. positive and negative coefficients) combination of various automorphic Galois representations.

To get true automorphy, it seems that one might need to know: a very general result about automorphic induction, and also replace virtual combination by true combination (i.e. only positive coefficients).

Since automorphic induction is supposed to be true, the first issue is only an indiction of the possible difficulty of the issue, not an intrinsic obstruction. But won't it be the case that any virtual Galois representation will give rise to an $L$-function with meromorphic continuation, which in general won't be holomorphic? So isn't this a genuine obstruction to concluding automorphy just from meromorphic continuation? One needs an extra ingredient which measures the true, as opposed to merely virtual, nature of the Galois representation; as far as I understand, at the moment the only such ingredient that people know how to feed into converse theorems is holomorphic continuation, rather than meromorphic continuation, of some $L$-function(s).



  • $\begingroup$ Thanks Matt---that's a super observation. So, for example, if my "algebraic gadget" is a formal difference of an irreducible 3-d Artin rep and a 1-d Artin rep that has nothing to do with it, then I get meromorphic continuation everywhere but this Galois rep isn't even close to being modular. So somehow this is a proof that, whatever the obstruction is, the obstruction is truly there. $\endgroup$ – Kevin Buzzard Apr 29 '11 at 15:19

I think it is also instructive to just look how the entireness condition comes in the proof of the simplest converse theorem: the case of cusp forms for $SL(2,\mathbf{Z})$; there you start with a Dirichlet series $$L(s)=\sum_{n\geq 1}{a_nn^{-s}}$$ and form a suitable completed $L$-function $$\Lambda(s)=(2\pi)^{-s}\Gamma(s)L(s),$$ and a function $$ f(z)=\sum_{n\geq 1}{a_n\exp(2i\pi nz)} $$ which is holomorphic and $1$-periodic on the upper half-plane, provided the coefficients of $L$ grow polynomially, which means $L(s)$ converges on some right half-plane.

Now the statement, which is due to Hecke here is (essentially): $f$ is a cusp form of weight $k$ on $SL(2,\mathbf{Z})$ if and only if $L(s)$ is entire with polynomial growth in vertical strips and satisfies $\Lambda(s)=i^k\Lambda(k-s)$.

The proof does not work if you assume merely that $\Lambda(s)$ is meromorphic: one starts by expressing $f(iy)$ as the inverse Mellin transform of $\Lambda(s)$ when the real part of $s$ is large enough, i.e., an integral on a vertical line with large enough real part that the Dirichlet series converges absolutely; then one proceeds to shift the line of integration to the left, so that it becomes possible to apply the functional equation to replace $s$ by $k-s$ and get the modularity relation (of weight $k$) for $z\mapsto -1/z$ (applied to $f(iy)$). The point is that this shift of contour requires that one does not pass through poles in the middle, and that the function which is integrated (here $\Lambda(s)y^s$ roughly) decays fast enough at infinity (to handle the horizontal segments in shifting the contour).

This being said, it is a fact that it is quite hard to find examples of Euler products which (1) converge absolutely in a half-plane, but not everywhere; (2) have analytic continuation (meromorphic) to the whole complex plane; (3) are not related in any way to automorphic forms/Galois representations...

  • $\begingroup$ Thanks Emmanuel. So here's a vague question: surely if I start with an Eisenstein series for $SL(2,\mathbf{Z})$ then I get meromorphic continuation with genuine poles, but...still the same sort of functional equation? So the "machine" must be able to deal with poles somehow. $\endgroup$ – Kevin Buzzard Apr 29 '11 at 18:56
  • $\begingroup$ Dear Kevin, The $L$-function doesn't see the constant term of the Eisenstein series, and the (very simple) poles put the constant term back in, so to speak. But in the case of a virtual representation, even if we assume that the constituents have well-behaved $L$-functions, the $L$-function of the virtual representation will have all kinds of poles along various critical lines, I guess (coming from the non-trivial zeroes of its denominator). Cheers, Matt $\endgroup$ – Emerton Apr 29 '11 at 22:14

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