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There exists a conformal diffeomorphism between $\mathbb{R}^3$ and $S_3$ (less a point): $$ g = dr^2+r^2\left(d\theta^2 + \sin^2\theta\ d\phi^2\right) $$ $$ r = R \tan \frac{\alpha}{2} $$

$$ g = \frac{R^2}{4\cos^4\frac{\alpha}{2}}\left[d\alpha^2+\sin^2\alpha\left(d\theta^2 + \sin^2\theta\ d\phi^2\right)\right] $$

There also exists a conformal diffeomorphism between $\mathbb{R}^3$ (less a point) and the cylinder $\mathbb{R}\times S_2$:

$$ r = R\ e^{x/R} $$

$$ g=e^{2x/R}\left[dx^2+R^2\left(d\theta^2 + \sin^2\theta\ d\phi^2\right)\right] $$

There does not appear to be a conformal diffeomorphism between $\mathbb{R}^3$ (less a line) and the cylinder $\mathbb{R^2}\times S_1$. At least, my attempts to find one starting from $\mathbb{R^3}$ in cylindrical coordinates: $$ g = dz^2+ d\rho^2 +\rho^2 d\phi^2 $$ and remapping $(z, \rho)$ have failed. So my questions are:

1) Is it true that there is no such diffeomorphism?

2) If so, how do you show that? I think that local invariants like the Cotton tensor have nothing to say, because both $\mathbb{R}^3$ and $\mathbb{R^2}\times S_1$ are flat, and instead there is some kind of global obstruction.

Edited:

My original question was not sufficiently precise, it asked about existence of a conformal map, but I really meant existence of a conformal diffeomorphism. As Ben pointed out, there is the conformal map

$$ \mathbb{R}^3 \to \mathbb{R}^2 \times S_1 $$ $$ (x, y, z) \mapsto (x, y, \phi= \mathrm{mod}(z, 2\pi)) $$ but that is not injective and so not a diffeomorphism.

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The map $\phi(x,y,\theta)=(x,y,e^{i \theta})$ is a locally isometric covering map, as it is given by taking the usual covering map $\Phi(\theta)=e^{i \theta}$, $\Phi \colon \mathbb{R} \to S^1$, $2 \pi$ periodic, and throwing in $x,y$. In particular, this map is a conformal map, but not a conformal diffeomorphism. Its lift is $\tilde\phi(x,y,\theta)=(x,y,\theta+2\pi)$, a conformal diffeomorphism. The metric is $dx^2+dy^2+dz^2$ where $z=2\pi \theta$.

Update: the question is now whether $\mathbb{R}^3$ minus a line is conformally diffeomorphic to $\mathbb{R}^2 \times S^1$. It is not: $\mathbb{R}^3$ minus a line has developing map taking it to $S^3$ minus a circle. (The map is your usual conformal map to $S^3$ as given in the question above.) On the other hand, $\mathbb{R}^2 \times S^1$ admits a conformal covering map (as given in my previous remarks) by $\mathbb{R}^3$, so its developing map has image $S^3$ minus a point. The same argument proves that $\mathbb{R}^3$ minus any closed set is not conformal to $\mathbb{R}^2 \times S^1$. See Richard Sharpe's book Differential Geometry to read about developing maps of conformal geometries.

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  • $\begingroup$ I am, in fact, looking for a diffeomorphism R3 -> R2 x S1, (excluding from the domain at most a line, say the z axis), similarly to the diffeomorphism R3 -> R x S2 that excludes from the domain the origin of R3. I don't think it exists, are you claiming the contrary? $\endgroup$ – Andrea Allais Jan 20 '18 at 20:55
  • $\begingroup$ On my opinion, this question should be closed, not answered here. This is not an MO question. $\endgroup$ – Alexandre Eremenko Jan 20 '18 at 23:31
  • $\begingroup$ @AlexandreEremenko - Can you say a few words about why you feel that way (here or on meta)? My instinct is that this question is fine... $\endgroup$ – Sam Nead Jan 21 '18 at 8:37
  • $\begingroup$ Because this answer is completely trivial, it is this answer which I had in mind at the moment when I read the question and voted to close it. $\endgroup$ – Alexandre Eremenko Jan 21 '18 at 13:06
  • $\begingroup$ @AlexandreEremenko: I agree that this question should be migrated. I answered it only to correct the other answer, which I couldn't do in a comment. $\endgroup$ – Ben McKay Jan 21 '18 at 13:08
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As Ben noted above and in the comments, my answer doesn't make any sense. Please ignore what follows and I apologize for wasting your time.

$\textbf{IGNORE:}$

Given a conformal map $\phi: \mathbb{R}^{3}\rightarrow \mathbb{R}^{2}\times \mathbb{S}^{1},$ we may lift to a conformal map $\widetilde{\phi}: \mathbb{R}^{3}\rightarrow \mathbb{R}^{3}.$ By Liouville's theorem (https://en.wikipedia.org/wiki/Liouville%27s_theorem_(conformal_mappings)), the map $\phi$ is a Mobius transformation, in particular it is the restriction of a diffeomorphism $\mathbb{S}^{3}\rightarrow \mathbb{S}^{3}$ which maps $\infty$ to $\infty.$ But, the map $\widetilde{\phi}$ satisfies, in particular, $\widetilde{\phi}(0,0,1)=\widetilde{\phi}(0,0,n),$ as a result of the fact that it is a lift. Here, we have normalized the deck group of the cover $\mathbb{R}^{3}\rightarrow \mathbb{R}^{2}\times \mathbb{S}^{1}$ to be generated by the action $(x,y,z)\mapsto (x,y,z+1).$

This proves that $\widetilde{\phi}$ can not be the restriction of a diffeomorphism $\mathbb{S}^{3}\rightarrow \mathbb{S}^{3}$ (or even a bijection), and therefore, there is no conformal map $\phi:\mathbb{R}^{3}\rightarrow \mathbb{R}^{2}\times \mathbb{S}^{1}.$
$\textbf{IGNORE}$

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    $\begingroup$ The lift doesn't have to satisfy $\tilde\phi(0,0,1)=\tilde\phi(0,0,n)$. It only has to satisfy that $\tilde\phi(0,0,n)$ differs from $\tilde\phi(0,0,1)$ by a conformal map with no fixed points. For example, the identity. $\endgroup$ – Ben McKay Jan 20 '18 at 20:07
  • $\begingroup$ Dear Ben, I've realized this right after writing, and you are absolutely correct. I'm going to delete this answer, and I'm sorry for muddying the waters. $\endgroup$ – Andy Sanders Jan 20 '18 at 20:12
  • $\begingroup$ Well, I guess I can't delete this answer since it was accepted, but I'll do an edit. $\endgroup$ – Andy Sanders Jan 20 '18 at 20:13

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