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I think of the 2-torus as $\mathbb T^2=\mathbb R^2/\mathbb Z^2$. Suppose that I have constructed some new charts on it (with $C^{1+\alpha}$ transition maps) that are not smooth differentiable with respect to the standard coordinate charts. I will call the manifold with the new charts $M$. The identity map from $\mathbb T^2$ to $M$ is a homeomorphism. Is there a nice way to construct a diffeomorphism from $\mathbb T^2$ to $M$?

EDIT: I think I can ask the question I really want to ask more succinctly as

Given a $C^1$ manifold that is homeomorphic to the standard 2-torus, must it be $C^1$-diffeomorphic to the standard 2-torus?
anthonyquas
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    What do you mean by "diffeomorphism"? $M$ is not a smooth manifold. – Kajelad Nov 10 '21 at 03:05
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    @Kajelad :Yes it is a smooth manifold. It has two distinct atlases on it. I am looking for a diffeomorphism from the torus with the standard atlas to the torus with the non-standard atlas. – anthonyquas Nov 10 '21 at 06:12
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    You are saying "Suppose that I have constructed some new charts on it (with $^{1+\alpha}$- transition maps": An atlas whose transition maps are merely $^{1+\alpha}$ is usually not regarded as smooth: Smoothness in differential topology is usually understood as $C^\infty$. Please, clarify. I also do not understand what do you mean by "a nice way". It is known that two homeomorphic surfaces which are $C^k$-smooth, are also $C^k$-diffeomorphic. This is a hard theorem without easy proofs. – Moishe Kohan Nov 11 '21 at 01:09
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    @MoisheKohan: Thanks for this. I can probably re-phrase my question as: suppose that I have a $C^{1+\alpha}$ manifold that is homeomorphic to the standard 2-dimensional torus. Is it $C^{1+\alpha}$ diffeomorphic to the torus also? Two follow up questions suggested by your answer: 1) is there a reference for the hard theorem you cite? 2) Since we're dealing with a very familiar topological space here, could there be a more bare hands proof? [ I will edit the post to correct the reference to smoothness.] – anthonyquas Nov 11 '21 at 01:15
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    The answer to your question in the edit is indeed positive. One proves it in several steps. The first one is that every maximal $C^1$-atlas contains a $C^\infty$ subatlas (true in all dimensions). Then you use any proof you know of the classification of smooth surfaces to conclude that you have a $C^\infty$-diffeomorphism. Having $T^2$ does not help with either one of these two steps. – Moishe Kohan Nov 11 '21 at 14:53
  • Take a look here if you want to see a reference to smoothing of $C^1$-manifolds. – Moishe Kohan Nov 11 '21 at 15:19
  • Thanks very much for the info. – anthonyquas Nov 11 '21 at 21:06

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