In a lecture on differential geometry, we had the following definition of equivalent atlases:
Two atlases $\mathcal A$ and $\mathcal B$ on $M$ are called equivalent if $\mathcal A \cup \mathcal B$ is an atlas on $\mathcal M$.
The definition of atlas we had is the following:
Let $M$ be a second countable Hausdorff topological space. An $n$–dimensional smooth atlas on $M$ is a collection of maps $$\mathcal A = \left\{ \left(\varphi_i, U_i\right) \mid i\in A\right\}, \quad \varphi_i: U_i\rightarrow \varphi_i(U_i)\subset \mathbb R^n,$$ such that all $U_i \subset M$ are open, all $\varphi_i$ are homeomorphisms, and
- $\{U_i, i\in I\}$ is an open covering of $\mathcal M$
- $\varphi_i\circ \varphi_j^{-1}: \varphi_j\left(U_i\cap U_j\right)\rightarrow \varphi_i\left( U_i\cap U_j\right)$ are smooth for all $i, j\in I$.
Now, this thread gives the following "recipe" for constructing non-equivalent atlases:
Here is a very easy way to construct inequivalent atlases on the same differentiable manifold $X$, e.g. $X=\mathbb{R}$ or $X=\mathbb{S}^1$. Pick any homeomorphism $f : X \to X$ which is not a diffeomorphism (one always exists). For each chart in the given atlas $(U,\phi)$, define a chart $(f^{-1}(U),\phi \circ f)$ in the new atlas. The overlap condition holds between charts in this new atlas because the $f$'s cancel out. But an overlap between a chart in the new atlas and one in the old is not smooth, because the $f$ does not cancel out and it would follow that $f$ is smooth which it isn't.
Question: Why exactly cannot $f$ be a diffeomorphism? Or to ask it differently: For example, let's assume that $f$ is "only" a $C^{1}$ diffeomorphism, wouldn't the recipe still hold, because a $C^{1}$ diffeomorphism is in general not smooth, i.e. $C^{∞}$?
EDIT: This is the definition of smoothness that we had for a map $f$ between two smooth manifolds:
Let $M$ and $N$ be two smooth manifolds. A continuous map $f:M\to N$ is called smooth if for all charts $(\varphi, U)$ of $M$, $(\psi, V)$ of $N$, $$\psi\circ f\circ \varphi^{-1}: \varphi(U\cap f^{-1}(V)) \to \psi(V)$$ is smooth.
If I apply this to our case, I get: $$\phi_{\alpha}^{-1}\circ h\circ \phi_{\alpha}: \phi_{\alpha}^{-1}(\phi_{\alpha}^{-1}(U_{\alpha})\cap h^{-1}(\phi_{\alpha}^{-1}(U_{\alpha}))) \to \phi_{\alpha}^{-1}(\phi_{\alpha}^{-1}(U_{\alpha})).$$ We've tried to convince ourselves that $f$ is not differentiable at $\phi_{\alpha}^{-1}(0)$, but is $\phi_{\alpha}^{-1}(0)$ an element of the domain of $\phi_{\alpha}^{-1}\circ h\circ \phi_{\alpha}$? I don't think so for the following reason:
$$h^{-1}(\phi_{\alpha}^{-1}(U_{\alpha})) = h^{-1}(\{x\in U_{\alpha}\mid \phi_{\alpha}(x)\in U_{\alpha}\}) = \{y\in B_{1}(0) \mid h(y)\in\{x\in U_{\alpha}\mid \phi_{\alpha}(x)\in U_{\alpha}\}\}.$$ And here comes my problem: We know that $h: B_{1}(0)\to B_{1}(0)$, so how can $h(y)$ be an element of the set $\{x\in U_{\alpha}\mid \dots\}$, which is a subset of $U_{\alpha}$? After all, $B_{1}(0)$ and $U_{\alpha}$ are in no way related to each other.
Best Answer
From your comment, it appears that your real question is:
(Given such $f$, the pull-back of the smooth atlas on $X$ via $f$ defines a smooth structure on the topological manifold underlying $X$ which is not equivalent to the original smooth atlas.)
Here is a general construction of $f$. Let $X$ be a smooth manifold of dimension $n\ge 1$, let $\phi_\alpha: U_\alpha\to R^n$ be one of the charts (which I assume to be surjective), where $U_\alpha\subset X$ is open. Now, consider the closed unit ball $B=B(0,1)\subset R^n$ with spherical coordinates $(r,\theta), r\in [0,1], \theta\in S^{n-1}$. Define the self-homeomorphism
$$ h: B\to B, h(r,\theta)=(\sqrt{r},\theta). $$ I leave it to you to verify that $h$ is not differentiable at the origin (it does not even have the directional derivative at the origin along any nonzero vector). Transplant $h$ to $X$ via the formula $$ h_\alpha = \phi^{-1}_\alpha \circ h \circ \phi_\alpha. $$ Set $B_\alpha:= \phi_\alpha^{-1}(B)$. Then $h_\alpha$ is a self-homeomorphism $$ B_\alpha\to B_\alpha. $$ The map $h$ restricts to the identity map of the boundary of $B$, hence, $h_\alpha$ restricts to the identity map of the boundary of $B_\alpha$. Thus, extend $h_\alpha$ by the identity to $X\setminus B_\alpha$. I leave it to you to verify that the resulting map $f: X\to X$ is a homeomorphism and that it is not a diffeomorphism (since it is not differentiable at $\phi_\alpha^{-1}(0)$).
Edit 1. I think, I understood your difficulty. When we say that a map between two subsets of $R^n$ is a homeomorphism, it is important to specify both domain and codomain of the map. But when we talk about differentiability of the same map at some point $p$ in the interior of the domain, we by default extend the codomain to be the entire $R^n$. For instance, the definition of the directional derivative $$ D_vf(p)=\lim_{t\to 0} \frac{f(p+tv) - f(p)}{t} $$ requires us to work with vector-valued functions, whose codomains are the entire $R^n$.
Edit 2. Ok, since it is still unclear, let's verify that by map $f$ is not differentiable at the point $p=\phi_\alpha^{-1}(0)$. Consider the open subset $V:=\phi_\alpha^{-1}(int B)\subset X$. The map $f$ that I defined sends $V$ to itself. The map $\psi=\phi:= \phi_\alpha|_V: V\to int B$ is a chart in the smooth atlas of $X$. Consider the composition $$ (\psi\circ f \circ \phi^{-1})|_{int B}= (\phi_\alpha \circ f \circ \phi_\alpha^{-1})|_{int B}.$$ By the very definition of the map $f$, the above composition equals $$ (\phi_\alpha \circ \phi_\alpha^{-1}\circ h \circ \phi_{\alpha}\circ \phi_\alpha^{-1}) |_{int B}= h|_{int B}. $$ If $f$ were differentiable at $p$ then this composition would have been differentiable at $0$ as well. However, as I noted above, $h$ is not differentiable at $0$. Thus, $f$ is not differentiable at $p$.