# Manifold with different differential structure but diffeomorphic

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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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