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How can I prove that any regular surface with non zero mean curvature is orientable?

UPDATE: The surface is embedded in $\mathbb{R}^3$.


marked as duplicate by Andrew D. Hwang, Christopher, Davide Giraudo, TravisJ, user147263 May 13 '15 at 0:50

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  • $\begingroup$ Can anyone offer a complete proof? $\endgroup$ – Imanol Pérez Arribas Apr 26 '15 at 6:17

I assume your surface is embedded in $\mathbf{R}^{3}$. At each point of your surface there are precisely two unit normal vectors. Because the mean curvature is non-vanishing, it makes geometric sense to "pick the normal vector for which the mean curvature is positive".

  • $\begingroup$ OK, where is the proof? $\endgroup$ – Han de Bruijn Apr 23 '15 at 17:45
  • $\begingroup$ @HandeBruijn: Given the lack of technical conditions in the OP's question, I assumed their surface is embedded in Euclidean $3$-space (where "orientable" is equivalent to "admits a continuous unit normal field"), and then gave justification that the surface admits a continuous unit normal field (distinguished by the sign of the mean curvature). Since (as I now see) the OP placed a bounty two days after my post, it appears they didn't find my argument convincing, either. I'm disinclined to expand without some indication of what was perceived lacking, but you're of course welcome to answer. :) $\endgroup$ – Andrew D. Hwang Apr 23 '15 at 18:31
  • $\begingroup$ @user86418 Yes, the surface is embedded in $\mathbb{R}^3$. I edited the question. $\endgroup$ – Imanol Pérez Arribas Apr 23 '15 at 19:37

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