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I stumpled upon the following question in one of my exercise-sheets:

Justify that there are no minimal surfaces in $\mathbb{R}^3$ that are diffeomorphic to the 2-sphere $S^2$

I have no idea though. Can someone elaborate?

What's the approach i am supposed to take on this?

I highly appreciate any hints! Thank you very much.

Zest
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  • Since $S^2$ is compact, such a surface must have a point that is furthest from the origin. What can you say about the curvatures at this extreme point? – Anthony Carapetis Apr 14 '18 at 23:45
  • You know that $H^2-K\geq 0$. If $H=0$ we have $K\leq 0$. Then it suffices to check that if $S\subseteq \Bbb R^3$ is a compact surface, there is $p\in S$ with $K(p)>0$. – Ivo Terek Apr 15 '18 at 00:02
  • @IvoTerek thank you very much for the hint! If i manage to find a point $p \in S$ with $K(p) >0$, does that already mean that there is no minimal surface that is diffeomorphic to $S$?

    I havent fully understood how exactly the properties of curvature are transfered between two diffeomorphic surfaces.

    If i have a minimal surface, does that mean that for any surface $\tilde{S}$ that is diffeomorphic to the minimal surface we must demand that $H \equiv 0$ at every point $p \in \tilde{S}$?

     – Zest Apr 15 '18 at 00:54
  • @AnthonyCarapetis thank you, too of course. The curvature at such point is $0$, isn't it? Did you mean $S^2$ with "such a surface must have a point that is furthest from the origin" or did you mean the diffeomorphic minimal surface? Like mentioned in my other response, i haven't fully understand how diffeomorphisms transfer properties of curvature of each surface. – Zest Apr 15 '18 at 00:58
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    In general, geometric properties are not preserved by diffeomorphisms, only by isometries. In this regard, talking about $\Bbb S^2$ is a red herring. The (stronger) statement which is key here is "there are no compact minimal surfaces in $\Bbb R^3$" (the hint from my previous comment is for proving this). In particular, a minimal surface can't be diffeomorphic to $\Bbb S^2$, because it cannot be compact. – Ivo Terek Apr 15 '18 at 00:58
  • Ah, okay now i start understanding. That would basically mean that only compact surfaces can be diffeomorphic to other compact surfaces since diffeomorphisms are bijective mappings. Is that the general idea? I really appreciate your help! Thank you very much. – Zest Apr 15 '18 at 01:05
  • Because diffeomorphisms are in particular bijective continuous maps (i.e., homeomorphisms), yes. – Ivo Terek Apr 15 '18 at 01:33
  • @IvoTerek awesome! thank you very much for all the help. – Zest Apr 15 '18 at 01:36

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