I need help with an exam question:

Let $S$ be a regular oriented surface such that for any simple, closed, and positively oriented curve in $S$ the value of the integral of the geodesic curvature along this curve is always the same( that is, independent of the curve). What can be said about the Gaussian curvature of $S$?

I thought about using the Gauss-Bonnet Theorem:

$$\int_C \kappa_g(s) ds + \iint_R K d\sigma + \sum_{l=1}^p\theta_l = 2\pi\chi(R)$$

Since the $\int_C \kappa_g(s) ds$ is always the same, for any curve. Then for a geodesic, the curvature is zero, so $\int_C \kappa_g(s) ds = 0$. So, I can assume that if the integral of the geodesic curvature is always the same, then it must be zero. Is it correct to assume that?

If that is correct, where do I go from there?

  • 2
    $\begingroup$ If this is an exam question, I don't think you should be asking for us to help you. Please clarify that. One mathematical comment here: How you do know you have closed geodesics? $\endgroup$ – Ted Shifrin Jun 29 '14 at 15:22
  • $\begingroup$ I may have expressed myself badly here. I'm actually learning differential geometry by myself, using Do Carmo's book, and got this question test as a form to see if I understand the concepts, which clearly I still don't. lol. Since I don't have a tutor, and I'm not that all familiar with tensor calculus (which is present in all explanations I find on internet) I thought I might ask for some light here. $\endgroup$ – cryptow Jun 29 '14 at 18:58
  • 1
    $\begingroup$ Where did you get the test? This seems like a crazy question. For example, if we take small (geodesic) circles of different radii, we get different answers for $\int_C \kappa_g(s)ds$. $\endgroup$ – Ted Shifrin Jun 29 '14 at 19:09
  • $\begingroup$ Maybe is just not very well formulated, i guess. $\endgroup$ – cryptow Jun 29 '14 at 20:18
  • $\begingroup$ But since the integral of geodesic curvature is, by hypotesis, constant, then I guess I could say that $\iint_R K d\sigma$ is constant too, that is for any region $R$ in the surface. Right? $\endgroup$ – cryptow Jun 29 '14 at 20:21

First of all, when you ask "Is it correct to assume that [the integral is always zero]," you're asking the wrong question. You can assume something if it's a hypothesis, or if it's an axiom. But if you assume something in addition to the theorem's hypotheses, you're proving a different theorem.

I think what you meant to say is "Can we prove that the integral is always zero?" You offered a possible proof -- because the integral around a closed geodesic must be zero, if the integral is always the same it must be zero. But as @Ted pointed out, this argument would only be valid if you could show that $S$ always admits a closed geodesic. Think about the case in which $S$ is a plane -- in that case, there are no closed geodesics on $S$, so your argument would not be valid.

The way to approach this is to think about some special curves that always do exist. (When you know something to be true "for all $x$," it's often most useful to apply it to some specially chosen $x$ that are easy to analyze.)

In this case, one class of curves to which you can apply the hypothesis is small closed curves (with no angles) bounding regions homeomorphic to disks. (You can always find plenty of such curves by taking small circles in the plane, and looking at their images under parametrizations.) For these curves, you can explicitly evaluate every term in the Gauss-Bonnet formula. It turns out that Euler characteristic term is always the same for such curves (what is it?), and the angle-sum term is always zero, so as you suggested in your comment, the fact that the boundary integral is always the same constant implies that the integral of Gaussian curvature is always the same constant: let's call it $C$.

To determine what the constant is, consider a sequence of small circles containing a point $p$ whose radius and area go to zero. Since each integral is equal to $C$, the integrals approach $C$ as a limit. On the other hand, since the curvature is bounded near $p$ and the area is going to zero, the limit must be zero. Ergo we conclude $C=0$.

Now we've shown that $K$ has the property that its integral over every region bounded by a simple closed curve is zero. What function has that property?

Edit: For extra credit, determine what constant $\int_C \kappa_g\, ds$ is always equal to, and exhibit an example of a surface that satisfies the hypotheses.

  • $\begingroup$ That is exacly what i've been thinking after @Ted enlightened me about the existence of closed geodesics. Considering regions homeomorphic to disks, the Euler characteristic of such regions is $\chi(R) = 1$. Then $\int_{\partial{R}}\kappa_g ds + C = 2\pi$. Since $C = \iint_R K d\sigma = 0$ then must be $K \equiv 0$. $\endgroup$ – cryptow Jun 30 '14 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.