physical content of the concept of a smooth structure

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Apr 5, 2022 at 14:48	comment	added	Arshak Aivazian		@SebastianGoette Wow, you've moved the question in a very good and specific direction! Apparently, a large list of non-topological invariants of smooth manifolds and their intuition would be a satisfactory answer. Thanks a lot!
Apr 5, 2022 at 14:14	comment	added	Sebastian Goette		@AivazianArshak You ask how two homeomorphic non-diffeomorphic manifolds differ. If you have a homeomorphism at hand, they differ by having different sets of smooth functions. So if you can visualise a non-smooth function, you are done. But if you only know that the two manifolds are abstractly homeomorphic, then you need some kind of differential-topological insight to tell you they cannot be diffeomorphic. In a few cases, one can compute some invariants like the Eells-Kuiper invariant for 7-spheres. But I have no clue how to visualise those. Locally, every manifold is smoothable.
Mar 23, 2022 at 20:29	comment	added	Deane Yang		@LSpice, thanks!
Mar 23, 2022 at 19:10	comment	added	LSpice		MathJax note: As unpleasant as it is, you have to write `$\newcommand\R{\mathbb R}$The purpose …`, not `$\newcommand\R{\mathbb R}$ The purpose …`, or your text will have a spurious space. I have edited accordingly.
Mar 23, 2022 at 19:09	history	edited	LSpice	CC BY-SA 4.0	Delete spurious space
Mar 23, 2022 at 0:23	comment	added	Deane Yang		Alas, I didn't really answer the question you asked. Unfortunately, I don't see any physical intuition for two homeomorphic but non-diffeomorphic manifolds. Notice that this is a global concept. In general, there will exist diffeomorphisms between dense open subsets of the two manifolds that can be extended uniquely as homeomorphisms that are not diffeomorphisms.
Mar 22, 2022 at 23:37	comment	added	Arshak Aivazian		I understand this (and love the language of sheaves), but it does not answer the question of what physical idea the notion of smooth structure captures. Imagine two homeomorphic non-isometric metric spaces. You see how they differ. Imagine two homeomorphic non-diffmomorphic manifolds. Do you see how they differ?
Mar 22, 2022 at 23:26	comment	added	Deane Yang		Oops. I forgot return to functions. You can now say that a function $f: M \rightarrow \R$, where $M$ is a smooth manifold, is smooth if $f\circ\Phi^{-1}: \phi(O) \rightarrow \R$ is smooth for any coordinate map $\phi$. The assumption on the change of coordinate maps is the necessary and sufficient condition that you don't run into any logical inconsistencies in this definition.
Mar 22, 2022 at 23:19	history	answered	Deane Yang	CC BY-SA 4.0