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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 $i, j\in I$.

Question: Let $\mathcal A$ and $\mathcal B$ be two atlases on $M$. If $\mathcal A$ consists of charts $(\varphi_i, U_i)$, and $\mathcal B$ constists of charts $(\psi_i, V_i)$, then $\{U_i, V_{i'}, i\in I, i'\in I'\}$ is obviously still an open covering of $\mathcal M$. What might break is that $\varphi_i\circ \psi_j^{-1}: \psi_j(U_i\cap V_j)\rightarrow \varphi_i(U_i \cap V_j)$ is not smooth anymore.

Does anybody happen to have a concrete counterexample of two atlases $\mathcal A$ and $\mathcal B$ that are not equivalent on a manifold $\mathcal M$?

Hermi
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1 Answers1

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The standard example is given already in the comment: let $M = \mathbb R$ with the standard topology and $\mathscr A = \{ (\mathbb R, x)\}$ and $\mathscr B = \{ (\mathbb R, x^3)\}$. Then $\mathscr A, \mathscr B$ are both smooth atlas for $\mathbb R$, but they are not compatible since

$$ x\circ (x^3)^{-1} = x^{1/3}$$

is not smooth.

For a general manifold $M$, let $\mathscr A$ be any atlas on $M$. Let $H: M\to M$ be any homeomorphism. Then $$\mathscr A^H = \{ (H(U), \varphi \circ H^{-1}) : (U, \varphi ) \in \mathscr A\}$$ is another charts on $M$ which is not compatible with $\mathscr A$ in general (here $H(U) = \{ h(p) : p\in U\}$).

For some explicit example of such $H$ so that $\mathscr A$ and $\mathscr A^H$ are not compatible: Pick any fixed $(U_0, \varphi_0 ) \in \mathscr A$ and let $h : \varphi_0(U_0) \to \varphi_0(U_0)$ be any homeomorphism (but not diffeomorphism) which is identity outside a compact set $K$ in $\varphi_0 (U_0)$. Then define $H: M \to M$ by $H(p) =\varphi_0^{-1} \circ h\circ \varphi_0 (p)$ when $p \in U_0$ and $H(p) = p$ otherwise).

Then $\mathscr A$ and $\mathscr A^H$ are not compatible: note that $(U_0, \varphi_0)\in \mathscr A$ and $(U_0, \varphi_0 \circ H^{-1}) \in \mathscr A^H$ and

$$ \varphi_0 \circ (\varphi_0 \circ H^{-1})^{-1} = \varphi_0 \circ H \circ \varphi_0^{-1} = h$$

is not a diffeomorphism.

Arctic Char
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  • Thanks Artctic for your answer. i) Concerning the case with a general manifold: Can we prove that the chart $\mathcal A^H$ you defined is never compatible with $\mathcal A$ for an arbitrary $H: M\rightarrow M$? ii) Also: What do you mean by $H(U)$? So the open covering of $M$ is given by the image of $H$ under $U$? – Hermi Nov 27 '21 at 14:51
  • With some $H$ $\mathscr A$ might be compatible with $\mathscr A^H$: for example when $H = id$. Yes $H(U)$ is just the image. Since ${ U}$ covers $M$ and $H$ is a homemorphism, ${H(U)}$ also covers $M$. @Hermi – Arctic Char Nov 27 '21 at 15:01
  • Thank you. I thought a bit more about the explicit example of $H$ you gave: i) I guess it is not really necessary to define $h$ outside $\varphi(U)$ on a compact set $K$? ii) I am afraid I have not understood yet why we can define the atlas $\mathcal A^H := {(H(U), \varphi\circ H^{-1}) \mid (U, \varphi)\in \mathcal A}$. I though that the homeomorphism $\varphi$ of the chart $(U, \varphi)$ needs to have as domain $U$, but the domain of $\varphi\circ H^{-1}$ is $M$, not $H(U)$. – Hermi Nov 27 '21 at 16:00
  • I hope I'm not too picky, but additionally, couldn't the map $h$ be a diffeomorphism. As a matter of, couldn't it even be a $C^k$ diffeomorphism for arbitrary $k\in \mathbb N$. I suspect that the only thing that is not is allowed is $h\in C^{\infty}$, since then we would have a smooth map (at least this is my interpretation of smoothness...). – Hermi Nov 27 '21 at 16:02
  • My notation is a bit unclear. I meant to pick a fixed chart $(U, \varphi)$ from $\mathscr A$ to construct this homeomorphism $H$. @Hermi – Arctic Char Nov 27 '21 at 16:25
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    And I choose $h$ to be something not a diffeomorphism. Indeed it could still be smooth (but without a smooth inverse) – Arctic Char Nov 27 '21 at 16:27