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Let $M$ be a surface in $\Bbb R^3$ with non-zero mean curvature for every point. How could I show that this implies that $M$ is orientable? By our definition, orientable means that an unitary, normal vector can be defined continuously for every point in the surface.

This is a homework question, so just hints would be appreciated :)

DashDotDashDot
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2 Answers2

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Think through all the bits and pieces required to define mean curvature.

Ted Shifrin
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  • I have been working at it since you replied but I've still got nowhere - I mean, my definition of H is the trace of the second fundamental matrix, which then will not be zero, so...? I guess I should eventually find a way to "decide" a sign globally for the normal vector, but so far I can't see how to relate that to the trace of the 2FF matrix... Thank you, anyway! – DashDotDashDot May 12 '13 at 18:33
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    Well, is the sign of $H$ dependent on which normal vector you pick? What would happen to $H$ if you followed a curve (like in a Möbius strip) along which orientation reverses? – Ted Shifrin May 12 '13 at 19:36
  • Do you mean that $H$ is well-defined on the $S^0$ bundle of the normal bundle of $M$, and if $M$ is not orientable, then this $S^0$ bundle is a connected degree 2 covering space of $M$? – Yuchen Liu May 13 '13 at 08:56
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    @jerrysciencemath, yes. I'm trying to give an answer that a beginning diff geo student could understand, rather than one for a second-year grad student. For you, I'll say that the second fundamental form is actually a section of $\text{Sym}^2(T^*M)\otimes N(M)$. This viewpoint is important for higher codimension and for projective varieties. – Ted Shifrin May 14 '13 at 03:58
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Eventually, I found a way to prove it. I leave a hint here in case anybody finds it useful:

If $N$ is the normal vector, $N$ is not canonical (it can be "+" or "-" $N$, identifying its sign with the orientation of the base $\{D_1f, D_2f, N\}$). In the same way, $H$ is not canonical either (it can be positive or negative for each parametrization). Which operation on both would be canonical, i.e., would be uniquely determined with independence of the map?

DashDotDashDot
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    Yes, this was the intent of my leading questions. If you have a non-orientable surface, no matter how you try to define $N$, there's a closed path along which $N$ reverses, and hence $H$ must pass through $0$. – Ted Shifrin May 27 '13 at 15:39
  • Thank you very much, Ted - and sorry I didn't get it only with your hint! I thought it should be along those lines, but orientability along a path was not a concept I had at the time. I have upvoted your answer to thank you for taking some time to help me! – DashDotDashDot May 27 '13 at 15:44
  • @TedShifrin (I will refer to you since Carlos appears to have left the site), could you develop a little, please? I don't understad, how can we a priori talk about constant mean curvature without assuming that the surface is orientable? – Danilo Gregorin Afonso Apr 07 '20 at 18:24
  • Ok, I think I got the intuition, but I am uncertain on how to write a rigorous proof. Does this suffices? Assume that $S$ is not orientable. Choose $N$ at a point $p$. Then there is a closed curve through $p$ such that the sign of $N(p)$ is changed as we go along the curve. Hence the signal of $H$ changes, what is a contradiction. – Danilo Gregorin Afonso Apr 07 '20 at 18:49
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    @DaniloGregorin As I suggested, on a non-orientable surface there must be points with mean curvature $0$, so constant mean curvature implies orientability. The case of $H=0$ requires a separate argument (e.g., having a complex structure). – Ted Shifrin Apr 07 '20 at 19:20
  • Dear @TedShifrin I used the continuity of the curvature and the fact that it changes sign to imply that there where points where $H=0$. Does this completes the proof? I forgot to say so, but I also used that $H$ is not identically zero. The context is the proof of Jellett-Liebmann Theorem. – Danilo Gregorin Afonso Apr 07 '20 at 19:25
  • Also, the correct is "does this complete the proof?" or "does this completeS the proof?" – Danilo Gregorin Afonso Apr 07 '20 at 19:27
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    LOL, @DaniloGregorin, because of the “does” it is the infinitive — no s. Yes, I had in mind the intermediate value theorem. – Ted Shifrin Apr 07 '20 at 19:35