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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.

Hermi
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    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. – user3257842 Dec 21 '21 at 16:23
  • The more structure an atlas has the easier it is to construct charts that are not compatible with it. – user3257842 Dec 21 '21 at 16:31
  • @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)? – Hermi Dec 21 '21 at 16:38
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    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. – user3257842 Dec 21 '21 at 16:47
  • @user3257842 Sorry, but one additional thing came up: In the recipe that I quoted, it says that a homeomorphism $f: X\rightarrow X$, which is not a diffeomorphism, always exists. I first thought that $f = \text{Id}$ is meant, but the identity is a $C^{\infty}$ function. Do you know which $f$ always exists? – Hermi Dec 22 '21 at 15:30
  • There is no canonical difeomorphism that isn't differentiable. You usual construct them by gluing together pieces of other functions. Eg. $f:[-1,1]\rightarrow [-2,3]$ with $f(x) = 2 x $ for $x\leq 0$ and $3x$ for $x>0$. I don't recall a general way to construct them. – user3257842 Dec 22 '21 at 19:19
  • Voting to close. The post to which you have linked already has an answer to this question: the last paragraph of that answer gives an explicit example of a homeomorphism $\mathbb R \mapsto \mathbb R$ that is not a diffeomorphism, and an explicit example of a homeomorphism $\mathbb S^1 \mapsto \mathbb S^1$ that is not a homeomorphism. – Lee Mosher Dec 23 '21 at 00:43
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    @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? – Hermi Dec 30 '21 at 21:57
  • I do not understand what your question really is: Are you asking for a proof that every smooth manifold $X$ of positive dimension admits a self-homeomorphism $X\to X$ which is not a diffeomorphism? – Moishe Kohan Dec 30 '21 at 22:15
  • The self-homeomorphism shouldn't be smooth, but it can a diffeomorphism (also refer to the preceding discussion). But yeah, that is my question. – Hermi Dec 30 '21 at 22:17

1 Answers1

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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 $p$.

Moishe Kohan
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  • 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? – Hermi Dec 31 '21 at 13:04
  • 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. – Moishe Kohan Dec 31 '21 at 14:41
  • As for a homeomorphism, there is only one, standard, definition, the one which you wrote. I am using it. – Moishe Kohan Dec 31 '21 at 14:57
  • 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? – Hermi Jan 01 '22 at 12:18
  • 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 $X\backslash B_{\alpha}$? – Hermi Jan 01 '22 at 12:18
  • As for the first question, I suggest you think for 10 more minutes about the definition of differentiable maps on manifolds and then see why $h$ is not differentiable at $\phi_\alpha^{-1}(0)$. As for the second question, I do not understand what you are asking at all. The topological requirements on a manifold are correct of course. However, what metric space are you talking about, what "the subset" are you talking about? What does it mean "how can we in general consider"? Do you need help proving continuity of $h$? – Moishe Kohan Jan 01 '22 at 15:05
  • Wait, why are you writing that $h$ is not differentiable at $\phi_{\alpha}^{-1}(0)$? I reread your answer, and I think what you meant is that $f$ is not differentiable at $\phi_{\alpha}^{-1}(0)$, didn't you? We have: $f(\phi_{\alpha}^{-1}(0)) = h_{\alpha}(\phi_{\alpha}^{-1}(0)) = \dots = \phi_{\alpha}^{-1}(0)$, but I still don't see why $f$ (or $h_{\alpha}$) is not differentiable at $\phi_{\alpha}^{-1}(0)$. I'm not that familiar with differentiable maps on manifolds yet. – Hermi Jan 02 '22 at 11:52
  • As for the second question: I think I totally confused myself. My problem was that I thought the expression $X\backslash B_{\alpha}$ is ill-defined. However, I think I see now why it is well-defined: $B_{1}(0)$ is an open ball in $\mathbb R^n$ and $\phi_{\alpha}: U_{\alpha}\rightarrow \mathbb R^n$. Thus, $\phi_{\alpha}^{-1}(B_{1}(0))$ is an open set, which is a subset of $U$. Since $U$ itself is a subset of $X$, it follows that $\phi_{\alpha}^{-1}(B_{1}(0))\subset X$, and thus the expression is well-defined. – Hermi Jan 02 '22 at 12:10
  • @Hermi: That was a typo, I meant to write $f$ instead of $h$. The notation $X\setminus B_\alpha$ is a set-theoretic difference, where $B_\alpha$ is a subset of $X$. Now, think: Suppose that $X, Y$ are differentiable manifolds, $f: X\to Y$ is a map. $f$ is called differentiable at a point $x\in X$ if for every pair of charts $\phi: U\to R^n$, $\psi: V\to R^m$ (where $x\in U$, $f(x)\in V$) the composition $g=\psi\circ f\circ \phi^{-1}$ is differentiable at $\phi(x)$. Now, apply this definition to the map $f: X\to X$ in my example. What can you take as $\phi$ and $\psi$? What is $g$? – Moishe Kohan Jan 02 '22 at 16:03
  • Thank you, I didn't know this definition. I looked it up in our lecture notes, we'll have it a bit later. So, applying the definition you gave to our case: $\phi = \psi = \phi_{\alpha}^{-1}$. The $g$ that you wrote is in our case $h_{\alpha}$ (or $f$ is we only restrict ourselves to the set $B_{\alpha}$). If we require $g$ to be differentiable at $\phi(x)$, we have: $g(\phi(x)) =(\psi\circ f)(x)$. Applying this to our case, it means $f(\phi_{\alpha}^{-1}(x)) = h_{\alpha}(\phi_{\alpha}^{-1}(x)) = (\phi_{\alpha}^{-1}\circ h)(x)$. – Hermi Jan 02 '22 at 16:50
  • For $x = 0$, we have: $f(\phi_{\alpha}^{-1}(0)) = (\phi_{\alpha}^{-1}\circ h)(0)$. However, we already know that $h$ is not differentiable at $0$, so I guess that the entire map $(\phi_{\alpha}^{-1}\circ h)$ is not differentiable at $0$, then? There's one more thing (which has to do with the domain and codomain of the function $g$ that you introduced) that I have to think about, I'll get back to you about it once I had the chance to think a bit more. – Hermi Jan 02 '22 at 16:55
  • OK, I made an edit to my original question, where I added the definition of a smooth map between manifolds and explained my confusion concerning the codomain of $\phi_{\alpha}^{-1}\circ h\circ \phi_{\alpha}$. – Hermi Jan 02 '22 at 21:56
  • @Hermi: I really do not understand what your difficulty is. I think you are simply making mistakes computing compositions of maps and their domains. Writing $f|B_{\alpha}: B_\alpha\to B_\alpha$ using the chart $\phi_\alpha$ you get $\phi_\alpha \circ f \circ \phi_{\alpha}^{-1}|B= h|B$ and then observe that $h(B)=B$ (the latter is optional). Why are you worried about the codomain? Maybe you are misreading the definition of differentiability for maps defined on subsets of $R^n$ and taking values in $R^m$. – Moishe Kohan Jan 03 '22 at 16:53
  • In my last comment, it should have read that I'm worried about the domain, not the codomain, sorry. However, I cannot really follow your last comment, I'm afraid.

    Let's take your general definition of differentiability on manifolds you had provided. And please let's consider the charts $\phi$ of a $n$-dimensional smooth atlas to have as codomain $\phi(U)\subset \mathbb R^n$ instead of $\mathbb R^n$ (because then we can require the maps $\phi$ to be homeomorphisms).

     – Hermi Jan 03 '22 at 21:43
  • Now: If $g := \psi\circ f\circ \phi^{-1}$ (as you had defined the map in a previous comment), do we agree that $g$ is a mapping from $ \phi(U\cap f^{-1}(V))$ to $\psi(V)$, where $\phi: U\rightarrow \phi(U)$ and $\psi: V\rightarrow \psi(V)$? $U$ is a subset of the smooth manifold $M$, $V$ is a subset of the smooth manifold $N$, and $f:M\to N$. – Hermi Jan 03 '22 at 21:43
  • @Hermi: Sure, but I suggest you take the codomain to be the entire $R^n$ (it is still a codomain): After all, when you took a vector calculus class, you learned how to differentiate functions with values in $R^n$ rather than functions with values in a subset of $R^n$. The difference is small, but it will help your understanding of the matter since the definition of derivative requires one to use quantities with values in the entire $R^n$. – Moishe Kohan Jan 03 '22 at 21:48
  • @Hermi: I hope, I addressed the issue that bothered you in the edit. Incidentally, I suggest, you try to verify rigorously that the map $f: X\to X$ I defined is continuous. You will need to use the Hausdorff property of $X$ in the proof. – Moishe Kohan Jan 03 '22 at 22:16
  • To be honest, I'm afraid my issue is not 100% resolved yet. So: Let's assume that the codomains of the charts $\phi$ and $\psi$ are $\mathbb R^n$ and $\mathbb R^m$, respectively. Then we have: $g: \phi(U\cap f^{-1}(V))\to \mathbb R^m$. Now, comparing this to our case, we had already identified that $\phi=\psi = \phi_{\alpha}^{-1}$. I'd argue that in our case, the sets $U = \mathbb R^n$ and $V = \mathbb R^m$. So putting it together: The domain of our $f$ is $\phi_{\alpha}^{-1}(\mathbb R^n \cap h^{-1}(\mathbb R^m))$. But is $h^{-1}(\mathbb R^m)$ well-defined? You had defined $h: B\to B$. – Hermi Jan 03 '22 at 22:40
  • Also, I had convinced myself that $f: X\to X$ is a homeomorphism: On the set $X\backslash B_{\alpha}$, we defined it as the identity, so no problem there. Inside the domain of $B_{\alpha}$, we have $f = h_{\alpha}$, and since $h_{\alpha}$ is a composition of homeomorphisms, there is no problem either. So the only problem could occur at the boundary of $B_{\alpha}$, but I remember I had briefly calculated that there, the map $h_{\alpha}$, if extended, gives us the identity as well. But I'm unsure about the inverse right now, maybe we need the Hausdorff property for the inverse at the boundary? – Hermi Jan 03 '22 at 22:46
  • First of all, $\phi=\psi=\phi_\alpha$, not $\phi_\alpha^{-1}$. And $U\ne R^n$, $U$ is a subset of $X$, equal to $int(B_\alpha)$. But this is not really relevant, $h^{-1}(R^m)$ is, of course, well-defined, this is simply the preimage of a subset under a map (the map does not have to be invertible for that). As for continuity of $f$, you have to refer to a lemma from general topology, guaranteeing continuity of a piecewise-defined function. In that lemma, you have a topological space $X$ represented as a union of two closed subsets of $X$. – Moishe Kohan Jan 03 '22 at 22:52
  • Could you please explain why $\phi=\psi=\phi_{\alpha}$? You had generally defined $g = \psi\circ f\circ \phi^{-1}$, and in our case, we have $\phi_{\alpha}^{-1}\circ h\circ\phi_{\alpha}$. Further, $h^{-1}(\mathbb R^m) = {x\in B \mid h(x)\in \mathbb R^m}$. Now, we already know that $h(x)\in B$ (by your construction), but $B$ is a subset of $\mathbb R^n$ and not of $\mathbb R^m$!? Could you please send me a reference to which lemma you're referring. I thought that for the map $f$ to be continuous at $p\in X$, the two piecewise functions just need to coincide at $p$. – Hermi Jan 04 '22 at 21:09
  • @Hermi: I really do not understand what's unclear, I think you are miscalculating compositions of some of the maps. In the 2nd edit I wrote a detailed proof of non-differentiability. I am done with this question. – Moishe Kohan Jan 05 '22 at 00:22
  • @Hermi: And I already answered this, see the 4th comment from the bottom. All in all, I suggest you review set-theoretic background required for the class you are taking. – Moishe Kohan Jan 05 '22 at 17:36