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Let $M$ a manifold. Two atlas $A_1=\{(U_i,\varphi_i)\}$ and $A_2=\{(V_i,\psi_i)\}$. They define the same smooth structure if whenever $f$ is smooth wrt $A_1$ then it will be smooth wrt $A_2$. For example $$A_1=\{(\mathbb R^n,id_{\mathbb R^n})\}$$ and $$A_2=\{(B_1(x),Id_{B_1(x)})\mid x\in \mathbb R^n\}$$ are two different atlas but they define the same smooth structure.

Do you have an example (on $\mathbb R^n$ or any easy smooth manifold) that describe two different smooth structure ? I don't really see it.

Peter
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  • If you google “different smooth structures” you get a few links to expository papers. This slideshow might be useful: http://users.jyu.fi/~eejohaka/talk/exotic_structures.pdf – Matthew Leingang Aug 27 '18 at 13:50
  • @MatthewLeingang: In your document, it's written (page 6) that $M=\mathbb R$ with the chart $\varphi:M\to \mathbb R$ defined by $\varphi(x)=Sgn(x)\sqrt{|x|}$ is smooth. How can it be true since even $\varphi$ is not smooth ? (not derivable at $0$). – Peter Aug 27 '18 at 19:05
  • $\phi$ only needs to be a homeomorphism to be a chart. Any pair of charts $\phi$ and $\psi$ must satisfy the condition that $\phi\circ \psi^{-1}$ is smooth. Since there is only one chart, that condition is automatically satisfied. When you say $\phi$ isn't smooth, you're comparing it to the usual atlas on $\mathbb{R}$. – Matthew Leingang Aug 27 '18 at 19:14
  • So in fact every homeomorphism $\varphi: M\to \mathbb R^n$ is a smooth structure, right ? @MatthewLeingang – Peter Aug 27 '18 at 19:16

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Here are two different smooth structures on $\mathbb{R}$ : $\{(\mathbb{R},\operatorname{id})\}$ and $\{(\mathbb{R},x\mapsto x^3)\}$.

Using the same idea, you can prove that there exists an uncountable set of smooth structures on $\mathbb{R}$.

If $M$ is a topological manifold, its homeomorphisms group acts on its set of smooth structures by: $$f\cdot\{(U_i,\varphi_i)\}_{i\in I}=\{f^{-1}(U_i),\varphi_i\circ f\}_{i\in I},$$ and generally an orbit of a given smooth atlas is uncountable.

C. Falcon
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