Define Gaussian curvature for a nonorientable surface. Can you define mean curvature for a nonorientable surface?
3
-
3$\begingroup$ Is this a homework question? $\endgroup$– Andrew CritchCommented Oct 17, 2009 at 1:09
-
3$\begingroup$ Please phrase this as a question rather than a demand. $\endgroup$– David Zureick-BrownCommented Oct 17, 2009 at 1:10
-
2$\begingroup$ 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. $\endgroup$– ziggurismCommented Dec 18, 2013 at 0:56
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
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.
answered Oct 17, 2009 at 1:07
$\begingroup$
$\endgroup$
1
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)
answered Oct 18, 2009 at 1:12
-
$\begingroup$ 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. $\endgroup$ Commented Jun 28, 2011 at 14:49
$\begingroup$
$\endgroup$
Sure you can. Why not? Nonorientability is irrelevant, since curvature is purely local. You don't need any global conditions like orientability.
answered Oct 17, 2009 at 1:00