In Loring Tu's Introduction to Manifolds, problem 6.1 (pg. 70):

Let $\mathbb R$ be the real line with the differentiable structure given by the maximal atlas of the chart $(\mathbb R,\phi=\textrm{id}:\mathbb R \rightarrow \mathbb R)$, and let $\mathbb R'$ be the real line with the differentiable structure given by the maximal atlas of the chart $(\mathbb R, \psi:\mathbb R\rightarrow \mathbb R)$, where $\psi(x)=x^{1/3}$. Show that these two differentiable structures are distinct.

In my understanding, the differentiable structure is just the maximal atlas. What does it mean for two maximal atlas to be distinct?

In one of the solutions I found online, it was said that for $\mathbb R$ and $\mathbb R'$ to have the same differentiable structure, there must be a map $F:\mathbb R\rightarrow \mathbb R'$ where $F$ is the identity map and $\psi \circ F\circ \textrm{id}^{-1}$ must be a diffeomorphism. Is this the definition for same differentiable structure?

  • $\begingroup$ Is $\psi$ differentiable at $x=0$? $\endgroup$
    – John Douma
    Commented Jan 20, 2022 at 5:34

1 Answer 1


I know it's been a little while since this question was asked, but let me add a full answer to this question. We have to note that maximal atlases $M_1, M_2$ are just sets (consisting of charts). They are distinct if $M_1$ has a chart that $M_2$ does not or vice versa (i.e. if $M_1 \neq M_2$ as sets). Let $M_1$ be the differentiable structure containing $(\mathbb{R}, \phi)$, and let $M_2$ be the one containing $(\mathbb{R}, \psi)$ (where $\phi$ and $\psi$ are the maps as stated in the question). Note that by definition, all charts in an atlas must be mutually $C^{\infty}$-compatible. If we assume, for a contradiction, that $M_1 = M_2$, it must be the case that the charts $(\mathbb{R}, \phi)$ and $(\mathbb{R}, \psi)$ are $C^{\infty}$- compatible (since they are charts in the same atlas, by assumption). In other words, each of the transition functions $\phi \circ \psi^{-1}$ and $\psi \circ \phi^{-1}$ must be $C^\infty$ on their domains. However, observe that although the domain of $\psi \circ \phi^{-1}(x) = x^{1/3}$ is all of $\mathbb{R}$, it is not differentiable at $x=0$, so it can't be $C^{\infty}$. This is a contradiction.

  • $\begingroup$ Doesn't $M_1 \neq M_2$ follow simply from that fact that $\psi \neq \phi$? Why do we have to bother with compatibility of the two charts? $\endgroup$
    – Dhiraj Rao
    Commented Jun 10 at 13:02
  • $\begingroup$ It can definitely be the case that $\phi \neq \psi$ but yet $\phi, \psi$ may have the same maximal atlas $\endgroup$
    – Michael
    Commented Jun 11 at 23:31

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