# Constructing non-equivalent atlases

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.

• 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. Dec 21, 2021 at 16:23
• The more structure an atlas has the easier it is to construct charts that are not compatible with it. Dec 21, 2021 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)? Dec 21, 2021 at 16:38
• 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. Dec 21, 2021 at 16:47
• @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? Dec 30, 2021 at 21:57

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 $$p$$.

• 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? Dec 31, 2021 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. Dec 31, 2021 at 14:41
• As for a homeomorphism, there is only one, standard, definition, the one which you wrote. I am using it. Dec 31, 2021 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? Jan 1, 2022 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 $X\backslash B_{\alpha}$? Jan 1, 2022 at 12:18