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It is known that the natural functor of smooth functions from the category of smooth manifolds into the category of locally ringed spaces is a full embedding (see, for example, here).

  1. Is a similar functor from the category of diffeological spaces to the category of locally ringed spaces a full embedding?
  2. If so, what can be said about the abstract characterization of its image? An answer to a similar question for smooth manifolds can be found here
Arshak Aivazian
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    Almost certainly not. Diffelofical strucutre can capture very singular spaces, like the non-commutative torus, that have no smooth function defined on them. I really don't see a way to attach to them a locally ringed space in a way that capture any interesting information – Simon Henry Mar 26 '22 at 20:11
  • @SimonHenry Thanks! In fact, I'm looking for a small extension of the category of smooth manifolds to include the important naturally occurring infinite-dimensional manifolds (such as spaces of smooth functions), but still being a full subcategory of locally ringed spaces. Apparently, diffiological spaces are not suitable for my purposes. Now I asked this question separately. – Arshak Aivazian Mar 26 '22 at 23:12
  • @SimonHenry: what do you mean by "have no smooth function defined on them"? Every diffeological space has smooth functions, for example all constant ones. – Konrad Waldorf Mar 28 '22 at 11:24
  • I mean't no non-constant smooth functions to $\mathbb{R}$. – Simon Henry Mar 28 '22 at 11:32
  • And did you mean the "irrational" torus instead of "non-commutative"? Because the D-topology of the irrational torus is discrete, so it seems to fit very well that it doesn't have non-constant smooth functions. – Konrad Waldorf Mar 28 '22 at 11:39
  • @SimonHenry: Ok, now I see your point. The irrational torus has only constant smooth functions and has the discrete D-topology. Hence, it can not be distinguished from the discrete torus as locally ringed spaces. However, it is not diffeomorphic to the discrete torus. This shows that the functor is not full. Correct? – Konrad Waldorf Mar 28 '22 at 11:59
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    Essentially yes. I wouldn't say it "shows" because the question didn't really specify how you build a locally ringed space out of a diffeological spaces - but at least it shows that Naive constructions using smooth function don't distinguish between the different non-commutative torus (or Irrational if you prefer). It is different from the discrete space though : the discrete space has plenty of non-constant smooth functions on it. – Simon Henry Mar 28 '22 at 14:02
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    I am not aware of a "nice" embedding functor of diffeological spaces into locally ringed spaces. However, maybe you find the framework of $C^\infty$-rings and locally $C^\infty$-ringed spaces helpful; see, for instance, the book "Smooth infinitesimal analysis" by Moerdijk and Reyes, as well as Dominic Joyce's https://arxiv.org/abs/1001.0023v7. – Severin Bunk Mar 29 '22 at 07:45
  • @SeverinBunk Thank you very much for the link to this book! I have long been interested in synthetic differential geometry - I think now that this is the best approach. But unfortunately it doesn't help me now -- see this comment of mine (correct me if I'm wrong) – Arshak Aivazian Mar 29 '22 at 13:49

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