Timeline for Does the convolution $C^*$-algebra of locally compact Hausdorff groupoids recover back the respective groupoid?
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| Mar 25, 2022 at 21:07 | comment | added | user40276 | @TheoJohnson-Freyd Thanks for the clarifications. For some reason, I believed that the internal construction of the Gelfand spectrum in the Bohr topoi had some deeper meaning and $C^*(X)$ was actually a good model for algebras over stacks. Instead, one must remember how $C^* (X_1)$ coacts on $C^* (X_0)$, so I suppose some sort of homotopy limit instead would be enough (bth, are you aware of any work on derived constructions on $C^*$-algebras?). You may post your comments as an answer if you want. | |
| Mar 24, 2022 at 23:26 | comment | added | Theo Johnson-Freyd | On the other hand, if you are a tried and true noncommutative geometer, then you will think that remembering a Hopfish structure is too much. (It is arguably a type of commutativity.) Indeed, it is an interesting feature that some invariants of a groupoid, e.g. its K-theory, factor through the C* algebra. | |
| Mar 24, 2022 at 23:22 | comment | added | Theo Johnson-Freyd | But you can absolutely enhance the C* algebra with extra data that knows the (equivalence class of the) groupoid. For example, the C* algebra of any groupoid is naturally cocommutative-Hopfish, corresponding to the ability to tensor representations of the groupoid. (Do I mean Hopfish? Or weak Hopf? Or...? I can never remember which generalization of Hopf is which.) A version of Tannakian duality says you can recover the groupoid from its Hopfish-C*-algebra. Essentially the groupoid is the grouplike-elements in its algebra. | |
| Mar 24, 2022 at 23:16 | comment | added | Theo Johnson-Freyd | The point is, the (isomorphism class of the) convolution C* algebra of a (1-)groupoid knows the number of, and dimension of, the irreducible representations of that groupoid, and not much more. So for example if $A$ is a finite abelian group, then the C* algebra of $BA$ knows the order of $A$, but not the group structure, nor does it know that $BA$ was connected. | |
| Mar 24, 2022 at 21:17 | comment | added | David Roberts♦ | I don't know! Sounds interesting, though... | |
| Mar 24, 2022 at 14:16 | comment | added | user40276 | @MaoWao Thanks for the comment. So my question is trivial then. Even if I restrict to not $1$-truncated groupoids $BG$ and $BG'$ say for $G$ and $G'$ finite groups will have the same algebra. So I guess there's nothing else here to add. | |
| Mar 24, 2022 at 14:04 | comment | added | user40276 | @DavidRoberts Thanks for the suggestion. By the way, does the Bohr topoi determines the algebra completely? Saying in another way: I don't really know whether knowing the lattice of commutative subalgebras (plus their algebra structure) recover the possibly non-commutative algebra. I mean I can likely recover $X_0$ and maybe take the pullback of this sub algebra with itself (like in the atlas of a stack). I don't know if that works, though. | |
| Mar 24, 2022 at 14:03 | comment | added | MaoWao | @TheoJohnson-Freyd That's correct. Every finite-dimensional $C^\ast$-algebra is $\ast$-isomorphic to a direct sum of matrix algebras, and there is only one way to write $2$ as a sum of squares. | |
| Mar 24, 2022 at 13:54 | comment | added | user40276 | @TheoJohnson-Freyd No idea. They are surely isomorphic as vector spaces, but I'm not sure about the product structure. Sorry if that's stupid, but could say what's the isomorphism explicitly in your example ($G = \mathbb{Z}/2$ and $BG$)? Maybe that's trivial, but I'm not seeing it... | |
| Mar 24, 2022 at 13:38 | comment | added | Theo Johnson-Freyd | @user40276 How many 2D C* algebras are there? I think they are all isomorphic to $\mathbb{C} \times \mathbb{C}$. | |
| Mar 24, 2022 at 13:36 | comment | added | David Roberts♦ | Excellent :-). Now I can add one data point, though I am not an expert: a groupoid $C^*$-algebra comes with canonical extra structure, namely the commutative $C^*$-algebra associated to the space of objects (the phrase "Cartan subalgebra" gets used, I see), which interacts with the original algebra in certain ways. I believe one can do quite strong reconstruction results given all this extra info. | |
| Mar 24, 2022 at 13:31 | comment | added | user40276 | @DavidRoberts Done. See if that's reasonable now. | |
| Mar 24, 2022 at 13:29 | history | edited | user40276 | CC BY-SA 4.0 |
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| Mar 24, 2022 at 6:48 | comment | added | David Roberts♦ | Yeah, I get that :-) But I think leading with the concrete, known version (ordinary loc. compact groupoids), and then branching to more speculative is better than posing what looks like (from the outside) a very speculative question, and then inside saying "it's ok if you can only answer for n=1". The people who know the n=1 case will not like the n-groupoid title, and the people who are comfortable with n-groupoids will likely not know the n=1 answer. Getting an n=1 answer here and then seeing where it goes it more useful imho. | |
| Mar 24, 2022 at 1:42 | comment | added | user40276 | @DavidRoberts I'm being intentionally vague. As I recall, for Lie $2$-groupoids, there was a notion of convolution algebra (I don't remember the paper now, though; I will search later) and nLab seems to suggest that one can continue further. In any case, I was aware on how vague the question is. This is why I said that I was ok with the case $n = 1$ and only conjectural suggestions for higher $n$. Do you think the $n$ there will ward off people from even reading the question? I'm willing to change that... | |
| Mar 24, 2022 at 0:11 | comment | added | David Roberts♦ | What is the notion of Haar measure on a topological $n$-groupoid? And what is the $C^*$-algebra associated to it? Have you any references? The page you link to was something that someone was working on at some point but never finished, never extended past $n=2$, and also claimed that "Hence the groupoid convolution algebra constructiuon is a 2-functor" (sic), except that I don't think it is functorial! That's not to say the $n=1$ case is not already an excellent question. I would focus on that, for now. | |
| Mar 23, 2022 at 23:33 | comment | added | user40276 | @TheoJohnson-Freyd I deleted my previous comment, which was totally wrong. Anyway, I guess I don't understand your comment. Say $G =\mathbb{Z}/2$, the algebra will be given by ordered pairs of complex numbers. Now $BG$ is again given by pairs, but the product won't be pointwise multiplication. Say $(f \star g) (0) = f (0)g(0) + f(1) g(1)$. Am I missing something? | |
| Mar 23, 2022 at 21:50 | comment | added | Theo Johnson-Freyd | The convolution algebras $B(\mathbb Z/2)$ and (2 points) are isomorphic, but the groupoids are equivalent. Or did I misunderstand? | |
| Mar 23, 2022 at 21:00 | history | asked | user40276 | CC BY-SA 4.0 |