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Let $M=\{(x,y,|x|) \in \Bbb R^3 \mid (x,y) \in \Bbb R^2 \}$. Construct a $C^\infty$ atlas $\mathcal{A}$ so that $(M, \overline{\mathcal{A}})$ (where $\overline{\mathcal{A}}$ is the maximal atlas) becomes a differentiable $2$-manifold. Is it diffeomorphic to $\Bbb R^2$? If so, what is the diffeomorphism $f: M \to \Bbb R^2$?

Define $\varphi : M \to \Bbb R^2$ as $\varphi(x,y, |x|) = (x,y)$. Then $\varphi$ is a homeomorphism. And if we define $\mathcal{A}=\{(M, \varphi)\}$ this is an atlas, but is it the maximal atlas?

Also I don't think this is diffeomorphic to $\Bbb R^2$ since the inverse $\varphi^{-1}(x,y)=(x,y,|x|)$ has the last coordinate function as the absolute value function which isn't differentiable at the origin? We need $\varphi^{-1}$ to be differentiable also?

Jasper
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1 Answers1

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In order to test whether a particular function $f : M \to \mathbb R^2$ is a diffeomorphism, using your atlas $\mathcal A$ which only has a single coordinate function, the requirement is that the composition $$\mathbb R^2 \xrightarrow{\phi^{-1}} M \xrightarrow{f} \mathbb R^2 $$ must be a diffeomorphism.

So, now do this with $f=\phi$. Is the composition $$\mathbb R^2 \xrightarrow{\phi^{-1}} M \xrightarrow{\phi} \mathbb R^2 $$ a diffeomorphism?

Lee Mosher
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  • The composition is the identity which is a diffeomorphism. Why consider $f = \phi$? – Jasper Feb 03 '22 at 19:51
  • You want to know that that a certain function $M \xrightarrow{f} \mathbb R^2$ is smooth, right? You test smoothness by using each coordinate chart in your atlas. What coordinate charts do you have in your atlas? Just $\phi$. – Lee Mosher Feb 03 '22 at 20:32
  • And so what function $f$ shall you test for being a diffeomorphism? You said so in your own post, you decided to test the function $f(x,y,|x|)=(x,y)$. – Lee Mosher Feb 03 '22 at 20:34
  • That makes sense. This definition/test you are mentioning for a map to be diffeomorphic, I cannot find it anywhere online. The definition I have is that the map $f$ has to be differentiable bijection with differentiable inverse. Will your result somehow imply this? – Jasper Feb 03 '22 at 20:35
  • I typed "definition of a smooth maps" into google and one of the first hits was these lecture notes, which have the definition I'm using near the top of the first page, "Definition 2". Basically, in any book which expresses the abstract definition of manifolds in terms of atlases and coordinate functions, the definition of smooth functions will also be expressed in the same terms. – Lee Mosher Feb 03 '22 at 21:04
  • Here is an MSE link to the same definition. – Lee Mosher Feb 03 '22 at 21:05
  • I might not have been clear, but it seems that you're using the definition for smoothnes to test if a function is diffeomorphic? Yes I also have this definition at my disposal that a map $f : M \to \Bbb R$ is smooth if for all $p \in M$ there exists ${U, \varphi}$ such that $p \in U$ and $f \circ \varphi^{-1}$ is smooth. – Jasper Feb 03 '22 at 21:26
  • So you're okay now? Because a diffeomorphism is nothing other than a bijection which is smooth with smooth inverse... – Lee Mosher Feb 03 '22 at 21:30