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This is taking from Tu's Introduction to Manifolds book. We have defined $\mathbb{R}$ as the real line with the differentiable structure given by the maximal atlas of the chart $(\mathbb{R},\phi=\operatorname{Id}_\mathbb{R}:\mathbb{R}\to\mathbb{R})$ and $\mathbb{R}'$ as the real line with the differentiable structure given by the maximal atlas of the chart $(\mathbb{R},\psi:\mathbb{R}\to\mathbb{R})$ where $\psi(x)=x^{1/3}$.

We are then instructed to show that there is a diffeomorphism between $\mathbb{R}$ and $\mathbb{R}'$, followed by a hint that it's not the identity map because it is not smooth.

My question(s):

1) Why is the identity map not a diffeomorphism/not smooth?

2) Does anyone have a hint/suggestion for an approach to finding the diffeomorphism that they are looking for?

Many thanks in advance!

Ayman Hourieh
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ThatOneMathGuy
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2 Answers2

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1) When composed with the given charts, the identity map gives $(\psi \circ \operatorname{id}_\Bbb R \circ\ \phi^{-1})(x) = x^{1/3}$. This map is not smooth at $0$. Hence the identity map is not a diffeomorphism.

2) Think of a bijection that "cancels out" with the cubic root in $\psi$.

Ayman Hourieh
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  • So, I instantly thought of the cubic function, but that can't be used as a diffeomorphism because it doesn't have a smooth inverse (the problematic function $x^{1/3}$), or am I completely interpreting diffeomorphisms incorrectly? – ThatOneMathGuy Feb 18 '14 at 00:34
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    @ThatOneMathGuy Smoothness depends on the given charts. Let $f(x) = x^3$. When composed with the given charts, we have $(\psi \circ f \circ \phi^{-1})(x) = x$, and $(\phi \circ f^{-1} \circ \psi^{-1})(x) = x$. Both are smooth. – Ayman Hourieh Feb 18 '14 at 00:37
  • Alright. Thanks! That completely clears it up for me. – ThatOneMathGuy Feb 18 '14 at 01:17
  • @ThatOneMathGuy you're welcome. – Ayman Hourieh Feb 18 '14 at 13:23
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The key is that a diffeomorphism is a map $f\colon\ \mathbb R_a\longrightarrow \mathbb R_b$ such that $f$ and $f^{-1}$ is smooth, but the definition of what it means for a map between manifolds to be smooth is in general dependent on the differentiable structure in such a way that if $(U_a,\psi_a)$ is the differentiable structure for $\mathbb R_a$ and $(U_b,\psi_b)$ is the differentiable structure for $\mathbb R_b$, then $f$ is called smooth if the function $\psi_b \circ f \circ \psi_a^{-1}$ is smooth.

1) In your case, $(\psi \circ \mathrm{id} \circ \phi^{-1})(x) = x^{1/3}$ is not smooth at 0.

2) Try choosing $f(x) = x^3$ as your diffeomorphism and work out the details.

Wisław
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