# Manifold has uncountable many smooth stuctures if it has one

This is the Problem 1-6 of John Lee's Introduction to smooth manifold:

Let $M$ be a nonempty topological manifold of dimension $n\geq1$. If $M$ has a smooth structure, show that it has uncountably many distinct ones. [Hint: first show that for any $s>0$, $F_s(x)=|x|^{s-1}x$ defines a homeomorphism from $\mathbb{B}^n$ to itself, which is a diffeomorphism if and only if $s=1$.]

What I tried:

It can be proved there is a atlas $\mathcal{A}$ (not maximal) which is compact with the original smooth sturcture of $M$ and has the following property: $\forall(U,\psi)\in\mathcal{A}$, $\psi(U)=\mathbb{B}^n$. I tried to define $\psi'=F_s\circ\psi$ and hope $\{(U, \psi')\}$ to form a new atlas for $M$. But $$\varphi'\circ(\psi')^{-1}=F_s\circ\varphi\circ\psi^{-1}\circ F_s^{-1}$$ may not be diffeomorphism. Any help, thanks.

• I think the idea is to modify only a single chart - find an atlas which contains a ball whose centre is not in any other chart, then compose the corresponding chart with $F_s$. Since $F_s$ is a diffeomorphism when restricted away from $0$ this should work out. Jul 31, 2014 at 6:17
• @AnthonyCarapetis I think you are right. Would you please coppy your comment as an anwser so I can accept it. Thank you! Jul 31, 2014 at 10:10

It took me a while to understand the great idea proposed by Anthony Carapetis since I think that other people may have the same doubt that I had, I decided to write a more detailed answer using his idea.

First of all, remember the Proposition $$1.17$$ (this proposition is straightforward to prove) of John Lee's Introduction to smooth manifold: and note that $$F_s: B(0,1) \rightarrow B(0,1)$$ is an homeomorphism that isn't a diffeomorphism $$\forall s>0$$ and $$s\neq 1$$, and $$F_s: B(0,1)\setminus \{0\} \rightarrow B(0,1)\setminus \{0\}$$ is a smooth diffeomorphism $$\forall s>0$$ ( where $$B(y,r) = \{x\in \mathbb{R}^n;$$ $$|x-y| $$\}$$).

Let $$\mathcal{A} = (\varphi_i,U_i)_{i \in I}$$ be a smooth atlas of $$M$$. We will construct a smooth atlas $$\mathcal{B}$$ such that $$\mathcal{A} \cup \mathcal{B}$$ is not a smooth atlas.

Note that for every $$x$$ $$\in$$ $$M$$, there exists a chart $$(\varphi_x,U_x)$$ $$\in$$ $$\mathcal{A}$$, satisfying $$x$$ $$\in$$ $$U_x$$. Using that $$\varphi(U_x) \subset \mathbb{R}^n$$ is open, $$\exists$$ $$\delta_x >0$$, such that, $$B(\varphi(x), \delta_x) \subset \varphi (U_x)$$.

So, we can define a smooth atlas $$\mathcal{C} = \left\{(\varphi_x, \varphi^{-1}_x \left( B(\varphi(x),\delta_x) \right) )\right\}_{x \in M},$$ which has the same smooth structure of $$\mathcal{A}$$.

Moreover, note that for every $$x\in M$$, there is a function $$\xi_x : B(\varphi(x), \delta_x) \rightarrow B(0,1)$$ such that $$\xi_x$$ is a smooth diffeomorphism between $$B(\varphi(x), \delta_x)$$ and $$B(0,1)$$ (in fact we cand define $$\xi_x(y) = \frac{1}{\delta_x}(y-\varphi(x) )$$ ).

Consequently, we are able to define a new smooth atlas $$\mathcal{D} = \left\{\left(\xi_x \circ \varphi_x, \varphi_x^{-1}\left(B(\varphi(x), \delta_x) \right)\right)\right\}_{x \in M},$$ which has the same smooth structure of $$\mathcal{A}$$ and $$B(0,1) = \xi_x \circ \varphi^{-1}_x(B(\varphi(x),\delta_x ))$$, $$\forall$$ $$x$$ $$\in$$ $$M$$.

Now, fixed $$x_0$$ $$\in$$ $$M$$, and using that $$M$$ is Hausdorff, for every $$y$$ $$\in$$ $$M$$ ($$y$$ $$\neq$$ $$x$$), exists a neighborhood $$V_y$$ of $$y$$, such that $$x$$ $$\notin$$ $$V_y$$.

Therefore, $$\mathcal{E} = \left\{(\xi_{x_0} \circ \varphi_{x_0}, \varphi_x^{-1}(B(\varphi(x_0), \delta_{x_0}) )\right\} \cup \left\{(\xi_y \circ \varphi_y, \varphi_y^{-1}(B(\varphi(y), \delta_y) \cap V_y )\right\}_{y \in M\setminus \{x_0\}}$$ is a smooth atlas which has same smooth structure of $$\mathcal{A}$$.

So, we can finally use Carapetis' idea. Define $$\mathcal{B} = \left\{(F_s \circ \xi_{x_0} \circ \varphi_{x_0}, \varphi_x^{-1}(B(\varphi(x_0), \delta_{x_0}) )\right\} \cup \left\{ (\xi_y \circ \varphi_y, \varphi_y^{-1}(B(\varphi(y), \delta_y) \cap V_y )\right\}_{y \in M\setminus \{x_0\}},$$

Now, we need to prove that $$\mathcal{B}$$ is a smooth atlas, the only nontrivial property that needs to be checked is: if, $$\forall$$ $$y$$ $$\in$$ $$M\setminus \{x_0\}$$

$$F_s \circ \xi_{x_0} \circ \varphi_{x_0} \circ (\xi_{y} \circ \varphi_y)^{-1}: \xi_y \circ \varphi_y (Z_y) \rightarrow F_s \circ \xi_{x_0}\circ \varphi_{x_0} (Z_y),$$

( where $$Z_y = \left( V_y \cap \varphi_y^{-1}(B(\varphi(y), \delta_{y})) \right) \cap \varphi_{x_0}^{-1}(B(\varphi(x_0), \delta_{x_0}))$$ ) is a smooth diffeomorphism.

This follows directly from the fact that $$\xi_{x_0} \circ \varphi_{x_0} \circ (\xi_{y} \circ \varphi_y)^{-1}$$ and $$F_s\vert_{ \xi_{x_0} \circ \varphi_{x_0} \circ (\xi_{y} \circ \varphi_y)^{-1}(Z_y)}$$ are diffeomorphisms, because $$0$$ $$\notin \xi_{x_0} \circ \varphi_{x_0} \circ (\xi_{y} \circ \varphi_y)^{-1}(Z_y)$$, sinse $$x_0$$ $$\notin$$ $$U_y$$.

Then $$\mathcal{B}$$ is a smooth atlas, but $$\mathcal{B} \cup \mathcal{D}$$ isn't a smooth atlas, because

$$F_s = F_s \circ \xi_{x_0} \circ \varphi_{x_0} \circ ( \xi_{x_0} \circ \varphi_{x_0})^{-1} : B(0,1) \rightarrow B(0,1)$$ is not a smooth diffeomorphism.

Using Proposition 1.17, we conclude that the smooth structure determined by $$\mathcal{B}$$ is different of the smooth structure determined by $$\mathcal{A}$$ (because of the smooth structure determined by $$\mathcal{A}$$ $$=$$ smooth structure determined by $$\mathcal{D}$$ $$\neq$$ smooth structure determined by $$\mathcal{B}$$ ). Since this result holds for all $$s>0$$, we can construct uncountable many differents smooth atlas $$\mathcal{B}$$, which all have different smooth structures among them, which completes the proof.

Given a smooth structure, we would like to find a coordinate ball $(U_0,\phi_0)$ such that the center $p$ of $U_0$ is covered by only this chart. It's not hard to see that we need only find a point $p$ covered by one chart. Then we can replace this chart with $(U_0,F_s \circ \phi_0)$ and get a new smooth structure, which is not smoothly compatible with the original one.

By Thm 1.15 and Lemma 1.10, we can find a countable, locally finite open refinement of the smooth structure consisting of precompact coordinate balls. This refinement is also a smooth structure, let's work with it. Then choose an arbitrary point $q$ on the manifold, it has a neighborhood intersects finitely many smooth charts denoted them as $U_1$ ~ $U_k$.

1)If $k=1$, then $q$ is only covered by $U_1$. We can replace $(U_1,\phi_1)$ with $(U_1,F_s\circ\frac{\phi_1 - \phi_1(q)}{r+|\phi_1(q)|})$, where r is the radius of $\phi(U_1)$.

2) If $k>1$, then repeat the following procedure starting from $i=1$: If $U_i$ is covered by the rest charts, then remove it from the refinement and get a new smooth structure otherwise stop the procedure. Eventually, there is going to be a point $q'$ covered by only one precompact coordinate ball. If we stop before $i=k$, then $q' \neq q$ otherwise $q'=q$. Apply 1).

By 1) and 2), we find a smooth structure distinct from the original one. Since there are uncountably many $F_s$, we have proved that given any smooth structure of a topological manifold, there exists uncountably many distinct smooth structures on the manifold.

• You are welcome to post an Answer to this two+ year-old Question, but you've structured your post as a Comment. If you wish, revise your post to benefit future Readers as a self-contained Answer. Review How do I write a good Answer? for guidelines. Feb 6, 2017 at 19:27
• F_s is not a diffeomorphism, I don't understand how it yields a smooth structure. Mar 3, 2017 at 5:51