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Apr 12, 2022 at 22:46 comment added David Ben-Zvi I agree there are functors involved, but the question is how rich the categories are (eg what do you take for a category of automorphic representations), of which I'm ignorant.. if you're talking about a category say of irreducible (or more generally semisimple) reps of some group, there's not a lot of structure there beyond matching sets.
Apr 12, 2022 at 17:40 comment added curious math guy @DavidBen-Zvi I mean yes... but...the correspondence $$\pi\mapsto \text{Hom}_{ \text{Gl}_2(\mathbb{A}_f) }(\pi, \text{colim}_N H^1_c(Sh(N),\overline{\mathbb{Q}_p})$$ is so... tempting to think of as a functor? Well, it is one for sure, no? That's a large reason for why I want there to be a functor in general..
Apr 12, 2022 at 17:16 comment added David Ben-Zvi You can also up the categorical level by varying the group, so you have a category of groups and homomorphisms - or more generally a “Morita” category of groups and joint actions (eg bi- Hamiltonian actions). This setup captures Langlands functoriality and the theory of periods (the relative Langlands program)
Apr 12, 2022 at 17:08 comment added David Ben-Zvi I don’t know a direct answer to your question but it feels somewhat unnatural - the global Langlands correspondence is most naturally a statement on the level of vector spaces (automorphic forms vs “functions on” (eg deformation rings of) Langlands parameters..), not of categories. OTOH Local Langlands and global geometric Langlands are (conjectural) statements about equivalences of categories (while local geometric Langlands concerns an equivalence of 2-categories)..
Apr 11, 2022 at 18:29 comment added Kimball In the classical case, there is the conjectural Langlands group, e.g., see mathoverflow.net/q/74698/6518
Apr 11, 2022 at 16:20 history edited curious math guy CC BY-SA 4.0
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Apr 11, 2022 at 14:11 history asked curious math guy CC BY-SA 4.0