Gaussian curvature and mean curvature. [closed]

Define Gaussian curvature for a nonorientable surface. Can you define mean curvature for a nonorientable surface?

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closed as off-topic by Ricardo Andrade, Andrey Rekalo, Daniel Moskovich, Jeremy Rickard, Stefan KohlNov 28 '13 at 10:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, Andrey Rekalo, Daniel Moskovich, Jeremy Rickard, Stefan Kohl
If this question can be reworded to fit the rules in the help center, please edit the question.

Is this a homework question? – Andrew Critch Oct 17 '09 at 1:09
Please phrase this as a question rather than a demand. – David Zureick-Brown Oct 17 '09 at 1:10
This is number 17 in section 3.3 on page 172 of do Carmo's book on the differential geometry of curves and surfaces. Verbatim. – ziggurism Dec 18 '13 at 0:56

Both of these are local notions, while orientability is a global constraint, so there's no need for a different definition when the surface is nonorientable.

A Riemannian metric on a manifold is a nondegenerate section of Sym2(T*M), that is a smooth family of nondegenerate symmetric bilinear forms on TpM, whether the manifold is orientable or not. The Gaussian curvature of a metric on a surface (or more generally, the sectional curvature of a Riemannian manifold) is defined locally, and so the definition you know works for non-orientable surfaces as well. Familiar theorems will hold for non-orientable surfaces as well; for example, the proof of the Gauss-Bonnet theorem goes through verbatim. (Fun exercise: find a trick to conclude the Gauss-Bonnet theorem for non-orientable surfaces directly from Gauss-Bonnet for orientable surfaces.)

The mean curvature is defined for an embedding or immersion of your surface into Euclidean space; again, the definition is local and so does not need to be changed. Underlying both these examples is the fact that the principal curvatures themselves are defined locally.

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Since you ask about mean curvature, we should assume that you mean a surface that is embedded or at least immersed in Euclidean 3-space. Then my favorite definition of Gauss curvature at a given point p is the following:

Position the surface so that p is at the origin and the tangent plane is the xy-plane. By rotating the surface about the z-axis, you can always position it so that the surface is given by

z = -(a x^2 + b y^2)/2 + higher order terms

Then a, b are known as the principal curvatures at p, relative to the unit normal given by (0,0,1) and K = ab is the Guass curvature. The mean curvature is given by H = a + b.

However, there are (at least) two possible ways to position the surface this way. You can also flip the surface upside down. This has the effect of flipping the signs of a and b. This flips the sign of H but not K.

Therefore, the Gauss curvature K is defined independent of the choice of the unit normal. It therefore is well defined on a non-orientable surface.

On the other hand, defining H on the whole surface requires a choice of the unit normal on the entire surface (or, equivalently, a choice of orientation on the surface itself). You can do this, but neither canonically nor continuously. So, I know only how to define mean curvature up to sign.

(EDIT by Deane: Missing minus sign and factor of 1/2 inserted into formula above. I never keep close track of these things. You can always work out details like that by checking a canonical example. In this case, it would be the unit sphere shifted down:

z = sqrt{1 - x^2 - y^2) - 1 = -(x^2 + y^2)/2 + higher order terms)

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If the surface is written as $z = -\frac{1}{2}(ax^2+by^2)+\cdots$ then the usual definition of mean curvature is the mean of the principal curvatures, i.e. $H = \frac{1}{2}(a+b).$ Removing the factor of one half, i.e. setting $H = a + b$ is non-standard and seems to be most prevalent in fluid mechanics. Recall that a surface can be written as $z = \frac{1}{2}(\kappa_1x^2+\kappa_2y^2) + \cdots$ where $\kappa_1$ and $\kappa_2$ are principal curvatures and the $x$-axis and $y$-axis are principal directions. If $\kappa_1 = \kappa_2$ then the point is an umbilic and every direction is principal. – Fly by Night Jun 28 '11 at 14:49

Sure you can. Why not? Nonorientability is irrelevant, since curvature is purely local. You don't need any global conditions like orientability.

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