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Define a smooth manifold to be a ringed space (a Hausdorff, second countable, $n$-manifold) that is locally isomorphic (as a ringed space) to the sheaf of smooth real valued functions on some open $U \subseteq \mathbb{R}^n$.

Similarly, a differentiable manifold is a ringed $n$-manifold locally isomorphic to the sheaf of differentiable real valued functions on some open $U \subseteq \mathbb{R}^n$. EDIT: I should also mention the sections of these sheafs are actually real valued functions defined on the open subsets of the space.

I believe I've been told that the two notions are actually equivalent; that is, a differentiable structure can be shown to be smooth. I've heard this is a difficult result.

basket
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    whoops fixed that – basket Jan 07 '16 at 11:02
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    Maybe relevant: http://math.stackexchange.com/questions/225528/differentiable-manifolds-as-locally-ringed-spaces. – Martín-Blas Pérez Pinilla Jan 07 '16 at 11:17
  • Nope that are not the same. We use the term $n$-fold continuously differentiable function, or simply $C^n$-map. And a smooth map is a $C^\infty$ map. And for distinct $n,m\in \mathbb{N}\cup {\infty}$, the category of $C^n$ manifold and the category of $C^m$ manifold are not equivalent. – Z Wu Feb 20 '23 at 15:52

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Here's my understanding of the relationship, not being a geometer I fully expect to be either told off for misuse of language or supplanted through a better explanation of key concepts.

A differentiable manifold equipped with an equivalence class of atlases whose transition functions are all differentiable. A smooth manifold is a differentiable manifold equipped with smooth transition maps. That is to say; the deriviatives of all orders exists (i.e., it is a $C^{k}$ manifold).

For my purposes, the relations rests on this transition map 'equivalence'.

  • Isn't this distinction exactly the same as the one I stressed in my definition, just in atlas parlance? – basket Jan 07 '16 at 09:42
  • Yes it is - I think we have the same understanding. The notion of equivalence between the two is hard to prove if memory serves me correctly; I think is rests on the discovery of every local smooth extension of the transition maps near every point in the domain. –  Jan 07 '16 at 09:58
  • to clarify, "the notion of equivalence" you are referring to is the equivalence between 'smooth' and 'differentiable and not simply the equivalence of the atlas of charts and ringed space definitions, right? – basket Jan 07 '16 at 20:53
  • @basket Yes that is correct, apologies, convoluted use of language! –  Jan 08 '16 at 09:16