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.

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    $\begingroup$ 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. $\endgroup$ Commented Jul 31, 2014 at 6:17
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    $\begingroup$ @AnthonyCarapetis I think you are right. Would you please coppy your comment as an anwser so I can accept it. Thank you! $\endgroup$
    – Danielsen
    Commented Jul 31, 2014 at 10:10

3 Answers 3


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: enter image description here

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 $B(y,r) = \{x\in \mathbb{R}^n;$ $|x-y|<r$ $\}$).

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.

  • $\begingroup$ The proof only shows there is a smooth structure different from $\mathcal{D}$. How do you know there are uncountably many distinct smooth structures on $M$? $\endgroup$ Commented Aug 13, 2022 at 10:57
  • $\begingroup$ For every $s$ you get a different one, and there are uncountably many $s$'s to choose from. $\endgroup$ Commented Aug 16, 2022 at 23:50
  • $\begingroup$ You only proved that for each $s$ you get a smooth structure different form $\mathcal{A}$. You need to show the resulted smooth structures for each $s$ are different from each other. $\endgroup$ Commented Aug 17, 2022 at 15:48
  • $\begingroup$ Not really. Notice that $F_{st} = F_s \circ F_t.$ Therefore, if you call $\mathcal B_s$ the smooth structure induced by $F_s$. It follows that showing that $\mathcal B_s \neq \mathcal B_t$ for $t\neq s$, it is the same thing of showing that $B_{ s/t} \neq \mathcal A.$ $\endgroup$ Commented Aug 17, 2022 at 20:39

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.

  • $\begingroup$ 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. $\endgroup$
    – hardmath
    Commented Feb 6, 2017 at 19:27
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    $\begingroup$ F_s is not a diffeomorphism, I don't understand how it yields a smooth structure. $\endgroup$ Commented Mar 3, 2017 at 5:51

I found the other answer here a little difficult to interpret, so I'm posting one in case it helps.

Given a smooth structure $\mathcal{A}$ on $M$ we shall construct a new smooth structure $\mathcal{B}$ on $M$ as follows. Fix a smooth coordinate ball $B$ (in the original atlas $\mathcal{A}$) whose image under its smooth coordinate map $\varphi$ is $\mathbb{B}^{n}$ (without loss of generality). Let $x = \varphi^{-1}(0)$. Since the smooth coordinate balls form a basis for $M$, for each $y\neq x$ we may pick a smooth coordinate ball $B_{y}$ containing $y$ such that $B_{y}\subseteq M\setminus\{x\}$. Let $\varphi_{y}$ denote the corresponding smooth coordinate maps. Define $\mathcal{B}$ to be the smooth structure (the unique maximal smooth atlas) determined by the smooth atlas $$\{(B_{y}, \varphi_{y}): y\neq x\}\cup\{(B, F_{s}\circ\varphi)\}$$ where $s\neq 1$. Now use the idea in Anthony Carapetis's comment.


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