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Jun 6, 2020 at 2:22 comment added Ben Wieland @SimonHenry Yes, if you want an equivalence of categories, it is going to work everywhere or nowhere, as I said. In particular, it is good to check in easy examples like group algebras. Yes, you need the Hopf structure.
Jun 3, 2020 at 13:09 comment added Simon Henry ... (without considering for e.g. the Hopf algebra structure) and anyway morphisms (bimodule) between k[G] and k[H] are quite different from group morphisms...
Jun 3, 2020 at 13:08 comment added Simon Henry @BenWieland : I'm staying vague because you did not gave enough details on the constructions you have in mind. But If you see algebra as a 2-category using bi-module then given two discrete groupe $G$ and $H$, you can consider the "discrete" stack $BG$ and $BH$ (they exists in all the categories of stacks considered here), morphisms from $BG$ tp $BH$ are group morphisms $G \to H$ up to conjugations. The associated convolution algebras are the groups algebra $k[G]$ and $k[H]$. I don't think you can fully recover a group from its group algebra alone...
Jun 3, 2020 at 3:19 comment added Ben Wieland @SimonHenry what do you mean by recovering a differentiable stack without recovering a stack? Do you mean smooth functions rather than continuous functions? And if you switch from ad hoc constructions to 2-categories, how can the wrong 2-category yield anything without yielding everything?
Jun 2, 2020 at 16:10 comment added Dmitri Pavlov @BenWieland: I know quite a few people who would be extremely interested in seeing a reference for the construction you are alluding to. This would solve a known research problem. For starters, what algebras correspond to Lie groupoids? And how do we characterize bimodules corresponding to bibundles?
Jun 2, 2020 at 2:54 comment added Simon Henry ?? I agree that there are way to organise algebras in a $2$-category and that this allows to reconstruct at least a differentiable stack from the convolution algebra, but I really don't see a 2-category structure that would get you near to what you claim then ! (i.e. recover the stack attached to the groupoid).
Jun 2, 2020 at 1:52 comment added Ben Wieland @SimonHenry You can recover the isotropy in a complementary way. I'm not alluding to ad hoc constructions, but to the fact, mentioned by Dmitri, that algebras form a 2-category and thus it directly yields a prestack on topological spaces.
Jun 1, 2020 at 21:44 comment added Simon Henry I believe @BenWieland refers to the fact that if you remember the Cartan sub-algebra in the convolution algebra then you can at least recover the initial space and the equivalence relation. It is not quite the same as a stack in the sense of stack of groupoids as you do not quite recover the isotropy, but it fits with the idea that you completely understand the "bad quotient".
Jun 1, 2020 at 0:52 comment added Dmitri Pavlov @BenWieland: If it is an exercise, can you tell us what algebras (with a precise list of properties) correspond to Lie groupoids and what bimodules between these algebras (again with a precise list of properties) correspond to bibundles between Lie groupoids?
Jun 1, 2020 at 0:36 comment added Ben Wieland @DmitriPavlov Yes, but it's an exercise.
May 31, 2020 at 18:49 comment added Dmitri Pavlov Groupoid convolution algebras can indeed be used to study sufficiently nice stacks (e.g., those presented by Lie groupoids), but claiming that convolution algebras are stacks is stretching things too far. For instance, is there an actual theorem (with proof) in the literature that establishes an equivalence of bicategories between sufficiently nice stacks and sufficiently nice algebras? Otherwise one simply cannot refer to algebras as a “notion of stack”.
May 31, 2020 at 17:53 comment added Praphulla Koushik I do not understand your last sentence "So you might say that Connes-style noncommutative geometry is (in part) the study of stacks, or you might say it is a reason that stacks are not more popular." Can you explain what it means?
May 31, 2020 at 17:47 history answered Ben Wieland CC BY-SA 4.0