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I'm new to differential geometry and reading Lee's book Manifold and Differential Geometry.

In the first chapter, he mentioned the following two maps on $\mathbb{R}^n$:

(1) $id: (x_1,x_2\cdots x_n) \rightarrow (x_1,x_2\cdots x_n)$

(2) $\varphi: (x_1,x_2\cdots x_n) \rightarrow (x_1^3,x_2\cdots x_n)$

Then, $\mathcal{A}_1$= { $(\mathbb{R}^n,id)$ } and $\mathcal{A}_2$= { $(\mathbb{R}^n,\varphi)$ } are two differential structure on $\mathbb{R}^n$, and $\mathcal{M}_1=(\mathbb{R}^n, \mathcal{A}_1)$, $\mathcal{M}_2=(\mathbb{R}^n,\mathcal{A}_2)$ are two manifolds.

It easy to verify that $\mathcal{M}_1$ and $\mathcal{M}_2$ have the same induced topology, the standard topology.

$\mathcal{A}_1$ and $\mathcal{A}_2$ are not compatible, for $id\circ \varphi^{-1}:(x_1,x_2\cdots x_n) \rightarrow (x_1^{\frac{1}{3}},x_2\cdots x_n)$ is not differentiable at origin. Therefore, $\mathcal{M}_1$ and $\mathcal{M}_2$ have different differential structure.

My question is: are they diffeomorphic?

According to Lee, the author, they are diffeomorphic through $\varphi$ (page 27).

But I don't think $\varphi$ is a diffeomorphism between them because $\varphi^{-1}$ is not differentiable at origin.

So are they not diffeomorphic?

But according the result of Donaldson and Freedman, each $\mathbb{R}^n$ except $n=4$ (with standard topology) only have one diffeomorphism class, so for any $\mathbb{R}^n$ except $\mathbb{R}^4$, $\mathcal{M}_1$ and $\mathcal{M}_2$ are diffeomorphic.

But why?

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up vote 11 down vote accepted

In order not to confuse the diffeomorphism with the chart, define $$ u : (\mathbb{R}^n, \mathcal{A}_1) \rightarrow (\mathbb{R}^n, \mathcal{A}_2)$$ by $u(x_1, ..., x_n) = (x_1^3, x_2, ..., x_n)$. It is a homeomorphism (why?). To check that it is a diffeomorphism, you also need to check that $u$ and $u^{-1}$ are smooth. A map is smooth by definition if its local representation in charts are smooth. Here, we have two global charts.

To check that $u$ is smooth, we need to check that $\varphi^{-1} \circ u \circ id$ is smooth as a regular map $\mathbb{R}^n \rightarrow \mathbb{R}^n$. And indeed, $$ (\varphi^{-1} \circ u \circ id) (x_1, ..., x_n) = (\varphi^{-1} \circ u)(x_1, ..., x_n) = \varphi^{-1} (x_1^3, x_2, ..., x_n) = (x_1, ..., x_n) $$ and this is a smooth map.

To check that $u^{-1}$ is smooth, we need to check that $id^{-1} \circ u^{-1} \circ \varphi$ is smooth. Similarly, $$ (id^{-1} \circ u^{-1} \circ \varphi)(x_1, ..., x_n) = (x_1, ..., x_n). $$ Note that it doesn't matter that $u^{-1}(x) = (x_1^{\frac{1}{3}}, x_2, ..., x_n)$ is not smooth as a map $\mathbb{R}^n \rightarrow \mathbb{R}^n$, because you treat $u^{-1}$ as a map between the manifolds $(\mathbb{R}^n, \mathcal{A}_2) \rightarrow (\mathbb{R}^n, \mathcal{A}_1)$, and to check whether it is smooth as a map between the manifolds, you need to compose it with the charts and check. The map $u^{-1}$ is not smooth as a "regular" map or as a map $(\mathbb{R}^n, \mathcal{A}_1) \rightarrow (\mathbb{R}^n, \mathcal{A}_1)$, but is smooth as a map $(\mathbb{R}^n, \mathcal{A}_2) \rightarrow (\mathbb{R}^n, \mathcal{A}_1)$.

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very clear! Thank your very much! – hxhxhx88 Oct 21 '12 at 17:28
So it is an example that the same topological space equipped with different differential structure but inducing the same differential manifold (in the diffeomorphism sense)? – hxhxhx88 Oct 21 '12 at 17:31
Exactly. The topological space $\mathbb{R}$ is endowed with two different differential structures, by the two incompatible charts $id$ and $\varphi$, but the resulting differential manifolds (structures) are diffeomorphic. – levap Oct 21 '12 at 17:34

They are diffeomorphic through $\varphi$, by definition of the structure on $\mathcal{M}_2$. The inverse $\varphi^{-1}$ is not differentiable with respect to the standard structure, i.e., as a map from $\mathcal{M}_1$ to $\mathcal{M}_1$. However, to test whether a map is differentiable as a map from $\mathcal{M}_1$ to $\mathcal{M}_2$, you have to test it in the given charts, in which both $\varphi$ and $\varphi^{-1}$ become the identity. (The same argument would be true if you replace $\varphi$ by any homeomorphism from $\mathbb{R}^n$ to itself.)

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Yeah... I forget the definition....Thx! – hxhxhx88 Oct 21 '12 at 17:34

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