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Let $X$ be a noetherian scheme. I will describe a construction of a simplicial ring which I think is called the Bellinson higher Adeles complex (or something similar).

Consider the augmented simplicial set of chains of points in $X$ denoted by $S(X)$. It is defined as follows:

  • The $-1$-simplices a one point set: $S(X)_{-1}=\{1\}$ and the augmentation $d_{-1}: S(X)_0 \to S(X)_{-1}$ is just the only possible map.

  • An $n\ge0$ -simplex $\xi \in S(X)_{n}$ is a chain of points $\xi = \{x_0,...,x_n\}$ s.t. $x_{i+1} \in \overline{\{x_{i}\}}$.

  • The face maps remove a point in the chain and the degeneracy repeat a point. Explicitly $d_{i}(\xi)=\{x_0,...,x_{i-1},x_{i+1},...,x_n\}$, $s_i(\xi) = \{x_0,...,x_{i-1},x_{i},x_{i},x_{i+1},...,x_n\}$.

We now define localization of a coherent sheaf $M$ w.r.t. to a chain $\xi$. The definition is inductive:

  • If $\xi = {1}$ we define $M_{\xi}=\Gamma(X,M)$.
  • For a chain $\xi \in S(X)_n$ we define $M_{\xi}:= \lim_{j \to \infty} (M_{x_n}/\mathfrak{m}_{x_n}^{j+1} M_{x_n})_{d_n\xi}$

This is functorial (and exact) so when fixing $M=\mathcal{O}_X$ the above localizations fit into a simplicial commutative ring with the face and degeneracies induced from the corresponding maps on chains. Call this simplicial commutative ring $\mathbb{A}_{\bullet}(\mathcal{O}_X)$. The same can be done for any coherent sheaf $M$ to obtain a simplicial module $\mathbb{A}_\bullet(M)$ over the simplicial ring $\mathbb{A}_{\bullet}(\mathcal{O}_X)$.

We get a "localization" functor $\mathbb{A}_\bullet(-): Coh(X) \to \mathbb{A}_\bullet(\mathcal{O}_X)-Mod$.

I think if everything is characteristic $0$ there's no problem in extending this functor via an equivalence so that the target will be $DG$-modules over the $DG$-algebra correpsponding to $\mathbb{A}_\bullet(\mathcal{O}_X)$.

First question: Is this functor $\mathbb{A}_\bullet(-)$ full? faithful? If not can we modify the target category by remembering the topology of the various completions so that it is? Can we modify so that this becomes an equivalence? Basically: Can we reconstruct a noetherian scheme from its complex/simplicial ring of higher adeles?

I found this construction when trying to read about and understand residue complexes, injective hulls, dualizing complexes, differential operators and their various connections. Alas what I found so far was too technical for me to understand or even extract the big picture.

Second question: What is the big picture here regarding the construction of the residue complex from these adeles? How does it interact with differential operators/injective hulls/dualizing complexes? Is their a friendly source that discusses these topics? (By friendly I mean in particular that I prefer it having no proofs at all then being too technical to digest).

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  • $\begingroup$ Could you please tell where to read about this stuff? How is it related to "ordinary" adeles? $\endgroup$ Commented Jan 20, 2017 at 21:11
  • $\begingroup$ @მამუკაჯიბლაძე I got this from papers by yekutieli e.g. math.bgu.ac.il/~amyekut/publications/bcas/journal.pdf. There are others, look up "explicit residue complex" or bellinson completion algebras and you'll find them. As I said they were too dense and technical for me to grasp. I think that in the case of a smooth curve you get a complex which at one degree has the function field and at the other precisely the classical adeles. $\endgroup$ Commented Jan 20, 2017 at 21:17
  • $\begingroup$ @მამუკაჯიბლაძე Correction: To get the adeles (meaning the restricted product and not just the arbitrary product) you should modify this probably. Perhaps some kind of limit over all finite sub-simplicial sets of $S(X)$. I have no idea. $\endgroup$ Commented Jan 20, 2017 at 21:31
  • $\begingroup$ Thanks! In the meanwhile I found a very accurate paper by Annette Huber, she explicitly states that for the $\mathcal O$ sheaf over the spectrum of a Dedekind ring you get exactly the classical adeles. $\endgroup$ Commented Jan 20, 2017 at 21:34
  • $\begingroup$ @მამუკაჯიბლაძე If you could comment about how she does things differently so that it does work it would help a lot. $\endgroup$ Commented Jan 20, 2017 at 21:37

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