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I am asked if such a conformal map exists. My strategy has been to find an isothermal parametrization of the helicoid. Moreover I know that we must have $E=G=\phi$, $F=0$ for some $\phi$ and by virtue of Gauss's Egregium Theorem, $\phi$ must be constant. I have tried changing $u$ by some $f(u)$ as Do Carmo sometimes does but have reached nowhere. Is there a simple argument to show such a map does not exists or to construct such parametrization?

Could I use that in such case that the plane has Gaussian curvature $0$ while the Helicoid in such parametrization verifies: $$ K = (EG-F^2)/(eg-f^2)= \phi^2/(eg-f^2)\neq0$$

Thank's in advance.

Sultan
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    You can probably just use the uniformization theorem. – Moishe Kohan Jun 11 '22 at 18:05
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    This should follow immediately from the Weierstrass-Enneper parametrization of a minimal surface, shouldn't it? – Ted Shifrin Jun 11 '22 at 18:23
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    An explicit conformal parametrization of the standard circular helicoid is known, but the goal is presumably to find it rather than simply to exhibit it. One approach is to start with your favorite parametrization (which presumably is not already conformal) and do an explicit change of parameters. It sounds as if this has been your approach so far, though I'd forget the second fundamental form and work only with the metric. If it matters, $\phi$ is not constant. – Andrew D. Hwang Jun 11 '22 at 18:36
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    By the way, the helicoid, being a minimal surface, has $K<0$ everywhere. – Ted Shifrin Jun 11 '22 at 21:16

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There does exist such a (local) conformal map. I'll present a very brief approach which uses the uniformization theorem, as Moishe Kohan suggested. The helicoid is simply connected (because it deformation retracts onto a helix which deformation retracts onto a point), therefore (by the uniformization theorem) it is conformally equivalent to either $\mathbb{H}^2, \mathbb{R}^2$ or $\mathbb{S}^2$. All of the former are (locally) conformally flat.

I can't think of any explicit example of such a map right now, however.

Matheus Andrade
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  • We have been taught only the very basic rudiments of surfaces (up to Gauss's Egregium Theorem) how could I begin proving such a map even exists? Is the proof of the uniformization theorem constructive? – Sultan Jun 11 '22 at 19:10
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    I'm not sure how one could do it (explicitly constructing such a map) using only the basics. I don't know enough about the standard proof of the uniformization theorem to tell you whether it's constructive. I'll keep thinking about it, if I come up with anything I'll update my answer. – Matheus Andrade Jun 13 '22 at 17:29
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    I found it. $X(u, v) =(\sinh u \cos v, \sinh u \sin v, v) $ – Sultan Jun 14 '22 at 08:49