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I'm trying to understand the definition of a differentiable structure. Is it correct that $x\rightarrow x$ and $x \rightarrow x^3$ doesn't form a diffeomorphism, since $x\rightarrow x^{1/3}$ isn't differentiable in $0$?

marco31
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1 Answers1

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You're talking about a common example when introducing the notion of differentiable structure in $\mathbb{R}$. We will see that even though the underlying topological space is the same, what is usually seen as a differentiable mapping with one structure need not be differentiable when considering two different structures.

To be explicit, I am assuming you are considering two different differentiable structures in $\mathbb{R}$. One of them would be given by the atlas $$A_1 = \{(\mathbb{R}, f)\}$$ where $f(x) = x$ and the other by another atlas $$A_2 = \{(\mathbb{R}, g)\}$$ where $g(x) = x^3$. Now, the identity $i: \mathbb{R}\rightarrow \mathbb{R}$ $i(x)=x$ certainly is a homeomorphism of topological spaces. However, as a map between differentiable manifolds, it depends on the structures whether or not it is a diffeomorphism. In particular, $$ i:(\mathbb{R},A_1) \rightarrow (\mathbb{R},A_2) $$ as seen in coordinates $$ g^{-1} \circ i \circ f (s) = s^{1/3}$$ is not a differentiable mapping (not differentiable at the origin). Note that however the inverse $$ i:(\mathbb{R},A_2) \rightarrow (\mathbb{R},A_1) $$ is differentiable, as it can be seen in coordinates $f^{-1}\circ i \circ g (\xi) = \xi^3$.

topolosaurus
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