0
$\begingroup$

Shpere

First of all, I take the surface patch for sphere

$$\sigma(u,v)=(\sin u\cos v, \sin u\sin v, \cos u)$$

And then I calculated its second fundametal form. And I got the following result

$$ -du^2-\sin^2 u \ dv^2$$

Second of all, I take different surface patch for the sphere.

$$\sigma(u,v)=\cos u\sin v, \sin u\sin v, \cos v)$$

And again I calculated the second fundamental form for it.

I obtained that $$ \sin^2 v du^2+ dv^2$$

The second fundamental forms are different since the surface patches are different. However, my instructor says that although the results are different, in fact, they are not completely different.

I think its reason. But I cannot find. Please explains why they are not completely different? Thanks a lot.

$\endgroup$
1
  • $\begingroup$ Mathematically the question is "how does the 2nd fundamental form transform under reparametrization of the surface?" This boils down to a chain rule computation. Please check Zhang's answer below. $\endgroup$
    – Avitus
    Dec 2, 2013 at 14:05

1 Answer 1

Reset to default
3
$\begingroup$

The second fundamental form has different forms in different coordinate representations, but they are essentially the same. It's just like a linear operator has different matrix representations in different set of basis. For example, if we transform $(u,v)$ representation to $(x,y)$ presentation, we obtain $$\begin{align}II(u,v)&=Ldu^2+2Mdudv+Ndv^2=L'dx^2+2M'dxdy+N'dy^2\end{align}$$ where $L', M', N'$ can be obtained by chain rule $$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy\\ dv=\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial y}dy$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .