Consider the generalised cylinder in $\mathbb{R}^3$ given by the parametrisation

$g(u,v) = (f(u)cos(v),f(u)sin(v),u)$

for $u \in (a,b) \text{ and } v \in (0,\pi).$ Find the matrix representation of the second fundamental form $II_x$ for any $x$ on the surface.

I know that the second fundamental form is given by $I(W_x X, Y)$, where $W$ means the Weingarten map. To compute this I need to get the Gauss map first, since the Weingarten map is it's derivative. To compute the Gauss map I suppose I should start like this:

$p\longmapsto \dfrac{\partial_1\times\partial_2}{||\partial_1\times\partial_2||}$, where $\partial_1=\frac{\partial g}{\partial u}$ and $\partial_2=\frac{\partial g }{\partial v}$.

Unfortunately I do not get any further, as I do not know how to concretely go forward. Could you help me?

  • 1
    $\begingroup$ I think you may want $g(u,v) = (f(u)cos(v),f(u)sin(v),u)$ instead of $g(u,v) = (f(u,v)cos(v),f(u)sin(v),u)$; what do you think? Cool question, +1!!!. Cheers! $\endgroup$ – Robert Lewis Nov 15 '18 at 0:01
  • 1
    $\begingroup$ You are right I will make an edit. $\endgroup$ – 3nondatur Nov 15 '18 at 0:03
  • $\begingroup$ In any differential geometry text, you'll find a recipe for the second fundamental form in terms of second-order partial derivatives of $g$. You most definitely do not want to differentiate the Gauss map. $\endgroup$ – Ted Shifrin Nov 15 '18 at 0:14
  • 1
    $\begingroup$ By the way I think you might want $\partial_1 = \partial g / \partial u$ and so forth, not $\partial_1 = \partial f / \partial u$ etc. $\endgroup$ – Robert Lewis Nov 15 '18 at 1:25
  • 1
    $\begingroup$ Again you are right I made an edit. $\endgroup$ – 3nondatur Nov 15 '18 at 7:11

Your Answer

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

Browse other questions tagged or ask your own question.