# Is there a conformal diffeomorphism between R3 minus a line and R2 x S1?

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.

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.

• 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? Jan 20, 2018 at 20:55
• On my opinion, this question should be closed, not answered here. This is not an MO question. Jan 20, 2018 at 23:31
• @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... Jan 21, 2018 at 8:37
• 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. Jan 21, 2018 at 13:06
• @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. Jan 21, 2018 at 13:08

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 $\phi:\mathbb{R}^{3}\rightarrow \mathbb{R}^{2}\times \mathbb{S}^{1}.$
$\textbf{IGNORE}$

• 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. Jan 20, 2018 at 20:07
• 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. Jan 20, 2018 at 20:12
• Well, I guess I can't delete this answer since it was accepted, but I'll do an edit. Jan 20, 2018 at 20:13