This is a question on a previous comprehensive exam that I am currently using to study. I can answer the first part, but am not quite sure what to do with the second.

Let $a > 0$ and let $M$ be the catenoid given by the parametrization $$X(u,v) = \left(\sqrt{u^2 + a^2} \cos v, \sqrt{u^2 + a^2} \sin v, a \ln \left(u + \sqrt{u^2 + a^2}\right) \right) $$ and let $N$ be the helicoid given by the parametrization $$Y(u,v) = (u\cos v, u \sin v, av).$$ Show that the map $X(u,v) \mapsto Y(u,v)$ is a local isometry that takes principal curves in $M$ to principal curves in $N$.

I have shown that the first fundamental forms are the same, so the map $X(u,v) \mapsto Y(u,v)$ is in fact a local isometry. However, I am not sure how to show that it takes principal curves in $M$ to asymptotic curves in $N$.

I assume that I should start with a principal curve on $M$ and push it forward using the $\Phi$ given by $\Phi(X(u,v)) = Y(u,v)$, but am not quite sure how to do that.


In this particular example, the coordinate curves of the catenoid clearly map to the coordinate curves of the helocoid. Morover:

  • The coordinate curves on the catenoid are principal because the catenoid is a surface of rotation (whose principal curves are latitudes and meridians);

  • The coordinate curves are asymptotic for the helicoid because the $u$-coordinate curves are rulings (acceleration tangent to the surface), the $v$-coordinate curves are orthogonal (image of mutually-orthogonal curves under a local isometry), and the helicoid is a non-planar minimal surface (so it has two orthogonal asymptotic directions at each point).

For all these properties, see for example Elementary Differential Geometry, second revised edition, by Barrett O'Neill, pp. 242–244. (By freakish coincidence, I opened right to page 244.)

  • $\begingroup$ Thank you! I will mark as answered as soon as I get a chance to flip through the text and understand it. $\endgroup$ – swygerts Dec 29 '16 at 18:07
  • $\begingroup$ You're welcome! Incidentally, a slightly less magical argument for the helicoid would be to check that the helices ($v$-coordinate curves) have acceleration tangent to the surface. $\endgroup$ – Andrew D. Hwang Dec 29 '16 at 18:16
  • $\begingroup$ In general, is there a good way to do this? For instance, if these surfaces were not ruled or of revolution, would there be a good method using a given local isometry to determine what the images of special curves are under the local isometry? $\endgroup$ – swygerts Dec 30 '16 at 3:59
  • $\begingroup$ The fact that the respective parametrizations have the same first fundamental form means the map $i(X(u, v)) = Y(u, v)$ is a local isometry, as you note. If $\gamma(t) = (u(t), v(t))$ parametrizes a curve in $X$ (e.g., a line of curvature), the image $i(\gamma(t)) = Y(u(t), v(t))$ parametrizes the image. Here you can check explicitly that $t \mapsto Y(t, v_{0})$ and $t \mapsto Y(u_{0}, t)$ are asymptotic curves. $\endgroup$ – Andrew D. Hwang Dec 30 '16 at 12:50

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.