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This question is in a sense the continuation of my previous one, Non-commutative duality I: Which C*-algebras are (isomorphic to a) convolution algebra?. Reading the great answers and the equally good comments, I have recognized several key points.

Here I would like to focus on the first of them, namely the Cartan subalgebra.

What is it? At some basic level, it is almost a trivial idea:

a groupoid is made of two ingredients, the underlying space, and the "path space".

If we start from a locally compact groupoid and we assume a minimum of regularity, the space is also a compact one, and carries its own commutative algebra, namely the functions on points. Now, the obvious observation is that this algebra is (or better can be modeled as) a distinguished commutative subalgebra of the convolution algebra .

If one thinks of the pair of C* algebras, the enlightening counterexample provided by Simon Henry breaks down: $\mathbb{BZ}$ and $\mathbb{U}$ are not anymore the same (in the first case the "space" subalgebra is simply $\mathbb{C}$ (one point) , whereas in the other case it is the entire convolution algebra).

The above consideration seems to suggest that to establish the golden non-commutative duality, extending Pontryagin duality, we should consider the category of * -algebras pairs (I skip the formal definition, but it should be obvious -maps between such creatures would respect the inclusion of the subalgebra-.

Now, I am pretty sure there are (possibly several) obstructions to carry out this simple idea. For instance, there may be C* algebras which admit no non trivial Cartan sub-algebra in the first place. Or, there may be more then one candidate for the "algebra of the underlying space", making it difficult or impossible to reconstruct the groupoid from the pair.

What kind of obstructions would one find to the schema above? And where?

I was thinking that there could be non commutative algebras whose space has no points, but I wondered if in that case one could move to locally compact locales to fix the situation

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    $\begingroup$ It works well in good situation, but not in general. an example where it fails is simply for group $C^*$-algebras: there the sub-algebra representing the object space are simply $\mathbb{C}$, so you are left with just the $C^*$-algebras and the information it is a group $C^*$-algebra, and in general you can't quite recover a group from its $C^*$-algebra alone. But there are many situation where the idea works and it has been studied a lot. Some example have been given at your previous question. I'll leave to people that knows the literature on the topic better than I do to answer. $\endgroup$
    – Simon Henry
    Oct 10, 2020 at 20:45
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    $\begingroup$ The problem I mentioned is generally avoided by restricting to topologically free groupoids. $\endgroup$
    – Simon Henry
    Oct 10, 2020 at 20:46
  • $\begingroup$ @SimonHenry thanks again for chiming in. So, looks like I must really read the refs you and the others gave :). As to your point: ok, these data are not enough to capture all C* algebra. But I still wonder if there is some interesting adjunction in place, which in general is not a duality, unless some stricter requirement is imposed on the relation between the two algebras. I suspect there is an additional ingredient here, and it has to do with your "fourier" transform., But that is for the next question. $\endgroup$
    – Mirco A. Mannucci
    Oct 10, 2020 at 21:05

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