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I have been trying to produce an example of two incompatible atlases on $\mathbb R$. But no success. Could someone help me please? All my example seem compatible. For example, $A_1 = \{((-\infty,1), \mathrm{id}_{(-\infty,1)}) , ((-1,\infty), \mathrm{id}_{(-1,\infty)}\}$ and $A_2=\{(\mathbb R, x \mapsto 2x)\}$. This is only my simplest example I also tried with a similar atlas but involving $x \mapsto \sin x$ and a similar one involving $x \mapsto x^2$.

Please: Could someone show me an example of two incompatible topological atlases on $\mathbb R$?

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1 Answers1

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All the topological atlases will be topologically compatible ( that is, the coordinate changes will be homeomorphisms) because the charts are local homeomorphisms with open subsets of $\mathbb{R}$ ( so the topology is already taken care of).

However, if you just consider a chart as a bijection, without regard for the topology of $\mathbb{R}$ then it is easy to find non-equivalent atlases. Take each consisting of one chart. Consider two bijections $f$, $g$ from $\mathbb{R}$ to $\mathbb{R}$ so that the composition $f\circ g^{-1}$ is not continuous. OK, make $f= 1_{\mathbb{R}}$ and $g(x)= x$ if $x \in \mathbb{Q}$ and $-x$ if $x \in \mathbb{R} \backslash \mathbb{Q}$.

Perpahs more interesting is two topological atlases with for the usual topology that are not $C^{1}$ equivalent, like @Najib Idrissi: suggested. Take $f(x) = x$, $g(x) = \sqrt[3]{x}$.

However, one should note that any two $C^{\alpha}$ structures on $\mathbb{R}$ are still diffeomorphic, perhaps not under the map $1_{\mathbb{R}}$.

orangeskid
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  • "However, if you just consider a chart as a bijection, without regard for the topology of ℝ then it is easy to find non-equivalent atlases." How does this make sense? You cannot have a homeomorphism without a topology and a chart for a topological manifold by definition consist of a homeomorphism and an open connected set. – learner Oct 12 '14 at 08:44
  • And what does you last sentence mean? Even after reading 4 times I cannot figure out its meaning. – learner Oct 12 '14 at 08:44
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    Sometime an atlas is defined for a set where we want to introduce a structure. You set could be $\mathbb{R}$. It depends on the definition. You are right, in most definitionsthe charts are assumed homeomorphisms. The second statement is about smooth structures on the topological manifold $\mathbb{R}$. All of them are equivalent. This is not a tautology. Some topological manifold can have two non-equivalent smooth structures. This won't happen for $\mathbb{R}$ and $\mathbb{R}^2$, but it will for $\mathbb{R}^4$. – orangeskid Oct 12 '14 at 08:58
  • Why do you mention smooth structure when the question is about topological manifolds? And I still can't make sense of your last sentence. – learner Oct 12 '14 at 09:14
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    Just for information purposes. The atlases with one chart each $x \mapsto x$ and $x \mapsto \sqrt[3]{x}$ are not $C^{1}$ equivalent so they define distinct $C^{1}$ structures. However, these structures are nevertheless diffeomorphic. Indeed, take the map from ($\mathbb{R}$ with atlas $x\mapsto x$) to ($\mathbb{R}$ with atlas $x \mapsto \sqrt[3]{x}$) given by $t \mapsto t^3$. This map in the two charts has the expression $x \mapsto x$ therefore it is a diffeomorphism. – orangeskid Oct 12 '14 at 09:21
  • @learner & orangeskid : What's your opinion of my answer please re $\mathbb C^0$ atlases ? – BCLC Jun 13 '24 at 18:50