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The question:

Prove that the function $\varphi : \mathbb{R} \rightarrow \mathbb{R}$, $\varphi(x) = x^3$, defines a $C^\infty$ differentiable structure on $\mathbb{R}$ different from the usual one $(\mathbb{R},Id_{\mathbb{R}})$.

It is easy to see that $(\mathbb{R},\varphi)$ forms an atlas. That is, $\mathbb{R} \subset \mathbb{R}$ is connected and $\varphi$ is a homeomorphism that is also $C^\infty$ compatible with itself. To see that $(\mathbb{R},\varphi)$ is not equivalent to $(\mathbb{R},Id_{\mathbb{R}})$ its enough to verify that $Id_{\mathbb{R}} \circ \varphi^{-1} = \frac{1}{3(x)^{2/3}}$ is not defined on $x = 0$. And so both atlases are different.

Question: However, in order for $(\mathbb{R},\varphi)$ to be a differentiable structure/maximal atlas it must satisfy the following:

Let $\mathcal{F} = \{(U_\alpha,\varphi_\alpha)\}$ be an atlas. If $(U,\varphi)$ is a coordinate system such that $\varphi \circ \varphi_\alpha^{-1}$ and $\varphi_\alpha \circ \varphi^{-1}$ are $C^\infty$ on $\varphi_{\alpha}(U \cap U_\alpha)$ and $\varphi(U \cap U_\alpha)$, respectively, then $(U,\varphi) \in \mathcal{F}$.

Is there any way to check this property? Perhaps there's a theorem that I don't know about and can't find it.

Thank you very much!

user57
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  • I would probably try proving maximality by contradiction. I don't know of a condition for maximality that one could use to avoid simply using the definition. – LSK21 Feb 15 '23 at 12:00
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    Your chart defines a non-maximal atlas. See this question for a similar confusion: Any non-maximal atlas gives a maximal atlas. – student91 Feb 15 '23 at 12:11
  • @student91 It does not define a maximal atlas because we can find another chart that is compatible with it and so their union would be "bigger". However, the book says it defines a differentiable structure? What am I supposed to understand from that? – user57 Feb 15 '23 at 12:52
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    Let $\mathcal{A}$ be the collection of all charts that are compatible with your chart. Then $\mathcal{A}$ is a maximal atlas, and it is uniquely determined by your first chart. Therefore, a non-maximal atlas gives (but is not equal to) a maximal atlas. In this sense, an atlas defines a differentiable structure, because it uniquely determines a maximal atlas. – student91 Feb 15 '23 at 12:57
  • @student91 Thank you very much. – user57 Feb 15 '23 at 17:59

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