# How to construct many smooth structures on a differentiable manifold? [duplicate]

In John M. Lee's introduction to smooth manifolds, in Problem 1.6, the question asks to produce infinitely many smooth structures on a smooth manifold.

The hint says that for $s>0$, $F_{s}(x) = |x|^{s-1}x$ is a homoemorphism but not diffeomorphism if $s\neq 1$. I understood the hint, but I don't know how to follow from here.

$\begin{matrix} \text{The point to construct many smooth structures is to find other homeomorphism } \\ \text{which are not }{{\text{C}}^{\infty }}-compatible\text{ with the one you already found}\text{.} \\ \text{If }s=1,{{F}_{1}}(x)=x\\ \text{which means it is an identical map that maps every point }p\text{ to itself at the same location}\text{.} \\ \text{For an (unit) open ball with radius }r=1\text{,we always have }\!\!|\!\!\text{ x}{{\text{ }\!\!|\!\!\text{ }}^{s-1}}\le {{r}^{s-1}}\le 1\\ \text{so }|x{{|}^{s-1}}x\text{ is still in the ball and} \\ {{F}_{s}}(x)=|x{{|}^{s-1}}x\text{ is a homeomorphism because the map }F:x\mapsto |x{{|}^{s-1}}x\\ \text{ is apparently a bijection and }{{C}^{0}}. \\ \text{The atlas consisting of the single chart (}{{\mathbb{B}}^{n}}\text{,}{{F}_{1}}\text{) defines a smooth structure on }{{\mathbb{B}}^{n}}\text{ and so is (}{{\mathbb{B}}^{n}}\text{,}{{F}_{2}}\text{)}\text{.} \\ \text{But they are not smoothly compatible with each other, because the transition map } \\ {{F}_{1}}\circ {{F}_{2}}^{-1}(x)=\frac{x}{|x|}\text{ is not smooth at the origin}\text{.} \\ \text{So (}{{\mathbb{B}}^{n}}\text{,}{{F}_{1}}\text{) and (}{{\mathbb{B}}^{n}}\text{,}{{F}_{2}}\text{) are different smooth structure on }{{\mathbb{B}}^{n}.}\\ \text{I can also pick s=3,4,...}\text{,n,so there are uncountably many distinct smooth structures} \text{.} \end{matrix}$
• Yeah, I got how to get many smooth structures on $\mathbb{B}^{n}$. But how to translate it to $M$, any smooth manifold. I want to put smooth structures on $M$ which are distinct from the given smooth structure. Commented Aug 7, 2018 at 14:47
• The chart $F_{s} \circ f$ is not compatible with the chart $f$, it is clear, but what is not clear to me is that why two charts on different domains $F_{s}\circ f$ and $F_{s}\circ g$ be compatible. To define a smooth structure, I need to cover $M$ by soothly compatible charts. Commented Aug 9, 2018 at 5:49
• Myabe we can make the composite function like $f \circ F_{s}$. Because any two charts $(U,f),(V,g)$ in $\mathscr A_{max}$ are compatible with each other, and their composite function make the charts into $(U,f \circ F_{s}),(V,g \circ F_{s})$. Since $U \cap V \neq \varnothing$ is guaranteed , and the transition map $f \circ F_{s} \circ (g \circ F_{s})^{-1}=f \circ F_{s} \circ F_{s}^{-1} \circ g^{-1}=f\circ g^{-1}$ is a diffeomorphism. But I can't figure out is $F_{s} \circ f \circ ( F_{s} \circ g)^{-1}$ a diffeomorphism,I think it should be. Commented Aug 11, 2018 at 6:02