Timeline for What is the strongest, most natural, conjectural form of Langlands?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Sep 7, 2011 at 4:46 | comment | added | James D. Taylor | 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. | |
| Sep 7, 2011 at 4:24 | comment | added | Emerton | ... 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 | |
| Sep 7, 2011 at 4:19 | comment | added | Emerton | 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 ... | |
| Sep 7, 2011 at 3:43 | history | answered | James D. Taylor | CC BY-SA 3.0 |