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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.

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    $\begingroup$ That really depends on how much structure you want to impose on the atlas. For example, if your have a $C^{1}$ atlas, (your charts $U_i$ are of class $C^{1}$), then passing it trough a $C^{1}$ difeomorphism preserves all the relevant structure and essentially gives you the same atlas. So the rule of thumb is, to make a new atlas, you need a difeomorphism that's at least $1$ degree less differentiable than the atlas. For $C^{\infty}$ atlases any difeomorphism that's not $C^{\infty}$ will give you a new atlas, but for $C^{1}$ atlases you need a difeomorphism that not $C^{1}$ to get a new one. $\endgroup$ Dec 21, 2021 at 16:23
  • $\begingroup$ The more structure an atlas has the easier it is to construct charts that are not compatible with it. $\endgroup$ Dec 21, 2021 at 16:31
  • $\begingroup$ @user3257842 Okay, so if I understand you correctly, if we have $C^{\infty}$ atlases, then the recipe can still hold if $f$ is a $C^{1}$ diffeomorphism, e.g. (always assuming that one exists, of course)? $\endgroup$
    – Hermi
    Dec 21, 2021 at 16:38
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    $\begingroup$ Yes. This is correct. As long as the diffeomorphism is not $C^{\infty}$ . But a $C^1$ difeomorphism won't work for $C^{1}$ atlases. $\endgroup$ Dec 21, 2021 at 16:47
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    $\begingroup$ @LeeMosher In this post you are writing: "Pick any homeomorphism $f: X\rightarrow X$ which is not a diffeomorphism (one always exists)." Now, you might have given two examples for different $X$ that are homeomorphisms, but not diffeomorphisms. However, you wrote quite generally that "one always exists", and my question is: Which homeomorphism always exists? $\endgroup$
    – Hermi
    Dec 30, 2021 at 21:57

1 Answer 1

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From your comment, it appears that your real question is:

Suppose that $X$ is a smooth manifold of positive dimension. Is there a self-homeomorphism $f: X\to X$ which is not a diffeomorphism?

(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 $p$.

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  • $\begingroup$ Thanks! A few questions: a) The definition of (total) derivative that I know assumes that the domain of $h$ is an open set, but you're considering $\overline{B(0, 1)}$ as the domain, which is closed... b) What exactly are your definitions of homeomorphisms and diffeomorphisms? Here are mine: A homeomorphism is an invertible between two topological spaces which is continuous and whose inverse is also continuous. For a diffeomorphism, we need an invertible mapping that is continuously differentiable and whose inverse is also continuously differentiable. Do you agree? $\endgroup$
    – Hermi
    Dec 31, 2021 at 13:04
  • $\begingroup$ For diffeomorphisms one usually requires infinite differentiability of the map and its inverse (there are many reasons to do so). In any case, either regard $h$ as a map of a smooth manifold with boundary, $B\to B$, or restrict $h$ to the interior of $B$: It is not a diffeomorphism (since it fails to be differentiable at the origin). The main thing is the $f$ is not differentiable, the boundary of $B$ is irrelevant here. $\endgroup$ Dec 31, 2021 at 14:41
  • $\begingroup$ As for a homeomorphism, there is only one, standard, definition, the one which you wrote. I am using it. $\endgroup$ Dec 31, 2021 at 14:57
  • $\begingroup$ Thanks for clarifying your definition of diffeomorphism! I was always referring to a $C^{1}$ diffeomorphism, I think that's where my confusion came from. I'm afraid that two more things came up, sorry! a) Where do we know from that the resulting map $f$ is not differentiable at $\phi_{\alpha}^{-1}(0)$? After all, we only know - by construction - that $h$ is not differentiable at $0$, don't we? $\endgroup$
    – Hermi
    Jan 1, 2022 at 12:18
  • $\begingroup$ b) The definition of smooth manifold that I know assumes that we have a second countable Hausdorff topological space. And for an open set - as you use it in your proof - we need the set to be a subset of a metric space. But then, if $X$ is a second countable Hausdorff space and $B$ is a subset of a metric space, how can we in general consider the space $X\backslash B_{\alpha}$? $\endgroup$
    – Hermi
    Jan 1, 2022 at 12:18

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