# Surface with non-zero mean curvature means orientable

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 :)

• Non-vanishing top forms... – rondo9 May 12 '13 at 16:36

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

• 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! – Carlos May 12 '13 at 18:33
• 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 8:56
• @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 3:58

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?

• 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! – Carlos May 27 '13 at 15:44