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Let $S$ be a Riemann surface smoothly embedded in $\mathbb{R}^3$.

Is there necessarily a smooth embedding of $S$ in $\mathbb{R}^3$ such that the Gauss normal map $n:S \to S^2$ would be a holomorphic map?Under what conditions on $S$ is the answer affirmative?

The question asks:can we change smoothly the position of a Riemann surface to have a holomorphic Gauss map?

Note: The sphere is considered as a Riemann surface with its standard structure

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    $\begingroup$ Critical points of the Gauss map are where the Gaussian curvature of the surface vanishes. I think typically the vanishing locus will be a one-(real)-dimensional subset of the surface. An analytic map has isolated critical points so I think this is unlikely $\endgroup$
    – Vik78
    Commented Sep 2, 2023 at 17:40
  • $\begingroup$ @Vik78 thank you. so the first condition we can pose is thst the curvature never vanish example sphere. $\endgroup$
    – Ali Taghavi
    Commented Sep 2, 2023 at 17:45
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    $\begingroup$ A necessary condition is existence of a metric whose curvature has zero dimensional vanishing locus. Whether such a metric exists is an interesting question in itself $\endgroup$
    – Vik78
    Commented Sep 2, 2023 at 18:11
  • $\begingroup$ @Vik78 yes that is very interesting question. BTW in case of identically zero curvature(cylinder)the Gauss map may fail to be holomorphic $\endgroup$
    – Ali Taghavi
    Commented Sep 2, 2023 at 18:22

2 Answers 2

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Gauss map is holomorphic (as a map to the Riemann sphere) if the surface is minimal. This is Lemma 8.3 in the book of Osserman, A Survey of Minimal Surfaces. In fact, if you replace "embedded" by "immersed" then the converse is also true (see Wikipedia article, https://en.wikipedia.org/wiki/Minimal_surface scroll down to "Gauss map definition".

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  • $\begingroup$ Thank you very much for this iinteresting interpretation in terms of minimal surfaces $\endgroup$
    – Ali Taghavi
    Commented Sep 2, 2023 at 20:06
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    $\begingroup$ I think the book is “A Survey of Minimal Surfaces”. What Wikipedia article do you mean? $\endgroup$
    – Vik78
    Commented Sep 2, 2023 at 21:02
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    $\begingroup$ Also, this paper (projecteuclid.org/journals/acta-mathematica/volume-223/issue-1/…) indicates that there are topological obstructions to a surface having a minimal embedding into $\mathbb{R}^3$. Every orientable surface has a complex structure and an embedding into $\mathbb{R}^3$ (mathoverflow.net/questions/112538/…), so this should give a negative answer to the noncompact case of the original question $\endgroup$
    – Vik78
    Commented Sep 2, 2023 at 21:10
  • $\begingroup$ @Vik78 Thank you very much for very valuable answers of both of you which now completly solve my question. $\endgroup$
    – Ali Taghavi
    Commented Sep 2, 2023 at 21:22
  • $\begingroup$ @Vik78: I had a link to Wikipedia in my answer but somehow it does not show. Now I inserted it again. $\endgroup$
    – Alexandre Eremenko
    Commented Sep 2, 2023 at 22:10
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Let $S$ be the embedded surface (assumed compact and not homeomorphic to a sphere) and let $S_1$ and $S_2$ be the subsets of $S$ where the Gaussian curvature is positive and negative, respectively. By continuity both sets are open, and both are nonempty (https://www.math.colostate.edu/~clayton/courses/501/501_9.pdf). Let $S’$ be their union. If the vanishing locus of the curvature of $S$ consists of isolated points, then $S’$ is connected, but $S_1$ and $S_2$ give a partition of $S’$ into nonempty open subsets so this is impossible. The vanishing locus of curvature of $S$ is precisely the critical locus of the Gauss map, and since the critical locus of the Gauss map does not consist of isolated points the Gauss map cannot be analytic.

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  • $\begingroup$ Thank you very much for this perfect answer in compact case. Is it true that evey open Riemann surface in space admit an embedding with holomorphic Gauss map? $\endgroup$
    – Ali Taghavi
    Commented Sep 2, 2023 at 19:03
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    $\begingroup$ Right I assumed compactness. For open surfaces it sounds much more complicated. By the same argument you will need a metric of either positive or negative curvature $\endgroup$
    – Vik78
    Commented Sep 2, 2023 at 19:12
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    $\begingroup$ but your argument incompact case is very interesting. Thanks again for that. I admit that the non compact case is very general and wide. Howevrr I think that it is unlike that a counter example of non compact csse exist $\endgroup$
    – Ali Taghavi
    Commented Sep 2, 2023 at 19:18
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    $\begingroup$ for example what can be said about pseudo sphere in space(the space image of Hyperbolic plane)? $\endgroup$
    – Ali Taghavi
    Commented Sep 2, 2023 at 19:21

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