What is the strongest, most natural, conjectural form of Langlands? - MathOverflow

What is the strongest, most natural, conjectural form of Langlands?

2011-08-02T23:33:25Z

This is inspired by my previous question: http://mathoverflow.net/questions/71743/what-is-the-precise-relationship-between-langlands-and-tannakian-formalism

As well as the excellent link that Tom Leinster put in a comment to that thread: http://golem.ph.utexas.edu/category/2010/08/what_is_the_langlands_programm.html

It seems that people are reluctant to say a form of Langlands that is too strong, but as consequence the statement is less natural, and more convoluted. So here I prefer that the statement be natural and bold rather than unnatural (for example, I consider the statement that each $L$ function coming from Galois representations is the $L$ function of some automorphic form to be unnatural).

Question

What is the strongest, most natural statement of Langlands? It would be nice if you can give a short definition of the words you use, but I am mostly interested in the narrative (each this has a blah, to each blah is a this, this is associated to this category by blah, and this is conjectured to be an equivalent category to blah, and so forth)

Words like: stack, motive, Tannakian, motivic Galois group, L-packets are encouraged. (of this list $L$-packets are by far the thing I know the least about)

This is subjective, so community wiki it is.

Answer by paul garrett for What is the strongest, most natural, conjectural form of Langlands?

2011-08-03T00:18:31Z

I am confused by your negative reaction to a putative assertion that every Galois repn's L-function is automorphic (perhaps meaning that it is the standard L-function attached to a cuspidal automorphic form on some $GL_n$ over $\mathbb Q$, and, thus, has the expected analytic continuation and functional equation). To me, various forms of the assertion that every motivic L-function is automorphic (with what is implied...) is one of the best capsulizations of "L's conjectures".

Ok, yes, this does mostly disregard the obvious intuitive senses of "functoriality", refering to putative maps/correspondences of afms on one group to another. And, yes, we know (potential modularity etc: Harris-Taylor's Sato-Tate, et alia) that just a lil' bit of "modularity" goes a long way...

I think that combining the "raw" conjectural ideas with the other dose of conjecture, namely, about existence of a group whose tensor/Tannakian/whatever category is ... automorphic forms...?... may add enough ambiguity to leave it all tooo ambiguous. Or, perhaps, those things will provoke someone's imagination?

But, seriously, two operational components come to mind: motivic L-functions are automorphic, and, "functoriality" is valid for automorphic L-functions.

Answer by James D. Taylor for What is the strongest, most natural, conjectural form of Langlands?

2011-09-07T03:43:13Z

Now that I've been exposed for a little longer to literature about Langlands (this answer is a month after I asked this question), let me submit my own non-expert answer to this question. I did enjoy the generality and naturality of the formulation of Langlands described in a newer question of mine:

http://mathoverflow.net/questions/74698/how-does-the-conjectural-langlands-group-fit-into-the-tannakian-point-of-view

As you can see from that question, I'm still learning about this. But I humbly suggest that this formulation might be what I was looking for, and so for the sake of completeness I add this as an answer.