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This is inspired by my previous question: 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.

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    $\begingroup$ Although this may be a negative answer, my experience is that either you stick to $GL(n)$ in which case you seem to know what is going on, or you work with an arbitrary connected reductive group, in which case, perhaps surprisingly, the stronger you get, the less natural things become ($L$-packets, $A$-packets, multiplicities greater than one, even failure of weak multiplicity one, no precise statement of local Langlands yet, endoscopy, and all sorts of other funny phenomena that don't exist for $GL(n)$). $\endgroup$
    – Kevin Buzzard
    Commented Aug 3, 2011 at 12:41

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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.

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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:

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.

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    $\begingroup$ Dear James, But to a large extent the Langlands program is about justifying the existence of the Langlands group $\mathcal L_F$ (for a number field $F$) and its relationship to automorphic forms --- this is part of the problem of functoriality, and you completely bypass it when you simply posit the existence of $\mathcal L_F$. The problem of relating $\mathcal L_F$ to the motivic Galois group is then the problem of reciprocity. The fact that this reciprocity map should be compatible with the local reciprocity maps arising from local Langlands is then the equality of $L$-functions ... $\endgroup$
    – Emerton
    Commented Sep 7, 2011 at 4:19
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    $\begingroup$ ... that you objected to in the statement of your question (more-or-less); but it this requirement of compatibility that gives reciprocity its content. I guess what I am saying is that the formulation stated in your other question leaves much of the actual content of the problems in the Langlands program (extremely) implicit, which is one reason why it didn't appear as an answer to your question here. (See e.g. the third para. of Paul Garrett's answer.) Regards, Matthew $\endgroup$
    – Emerton
    Commented Sep 7, 2011 at 4:24
  • $\begingroup$ Thanks a lot for your commentary. I guess I was looking for a natural end goal. I'm sure that in due time I will understand more of the mathematics involved in having a Langlands group, and I will probably have better insight. $\endgroup$
    – James D. Taylor
    Commented Sep 7, 2011 at 4:46

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