4
$\begingroup$

I'm attempting to calculate the Gaussian curvature of the sphere of radius $r$, but I'm not sure how to find the dual forms of the frame field.

I start with the parametrization $X(\phi, \theta) = (r \sin \phi \cos \theta, r \sin \phi \sin \theta, r \cos \phi)$. Then this gives me the tangent frame field

$E_1 = r \cos \phi \cos \theta \frac{\partial}{\partial x} + r \cos \phi \sin \theta \frac{\partial}{\partial y} - r \sin \phi \frac{\partial}{\partial z}$

$E_2 = -r \sin \phi \sin \theta \frac{\partial}{\partial x} + r \sin \phi \cos \theta \frac{\partial}{\partial y}$

I also compute

$dx = r \cos \phi \cos \theta d\phi - r \sin \phi \sin \theta d\theta$

$dy = r \cos \phi \sin \theta d\phi + r \sin \phi \cos \theta d\theta$

$dz = -r \sin \phi\ d\phi$

Now we must find the dual forms of the frame field ${E_1, E_2}$, but how do we do that? I thought we would just substitute $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}$ with $dx, dy, dz$ in their expressions, but this doesn't get me the right answer. I have searched for an explanation of this but couldn't find anything clear. Any help is appreciated.

$\endgroup$
  • $\begingroup$ Which definition of Gauss curvature are you using? You don't have to find the dual basis. $\endgroup$ – user99914 Nov 2 '15 at 1:50
  • $\begingroup$ I'm using $d \omega_{12} = -K d \eta_1 \wedge d \eta_2$ where $\{\eta_1, \eta_2\}$ is the dual basis and $\omega_{12}$ the connection form. $\endgroup$ – Pedro Nov 2 '15 at 1:56
  • $\begingroup$ Do you need to assume that $E_1, E_2$ are orthonormal? $\endgroup$ – user99914 Nov 2 '15 at 1:57
  • $\begingroup$ Yes, I think we do $\endgroup$ – Pedro Nov 2 '15 at 2:02
  • 2
    $\begingroup$ The dual frame is defined by $\theta^i (E_j) = \delta^i_j$, so you need to find a left-inverse for the matrix $[E_1 E_2]$. $\endgroup$ – Anthony Carapetis Nov 2 '15 at 2:12
1
$\begingroup$

Note that all the calculations does not depend a lot on the ambient space. Once you find the first fundamental form, you can forget $(x, y, z)$ and work solely on $(\phi, \theta)$.

You have

$$\begin{split} \frac{\partial}{\partial \phi} &= (r \cos \phi \cos \theta, r \cos \phi \sin \theta, -r \sin \phi),\\ \frac{\partial}{\partial \theta} &= (-r \sin \phi \sin \theta, r \sin \phi \cos \theta, 0). \end{split}$$

So the metric is given by $$\begin{bmatrix} g_{\phi\phi} & g_{\phi\theta} \\ g_{\theta\phi} & g_{\theta\theta} \end{bmatrix} = \begin{bmatrix} r^2 & 0 \\ 0 & r^2\sin^2\phi \end{bmatrix}$$

Then

$$E_1 = \frac 1r \frac{\partial}{\partial \phi},\ \ \ E_2 = \frac{1}{r\sin\phi} \frac{\partial}{\partial \theta}$$

and local orthonormal frame. Then the dual of these two vectors are

$$\eta_1 = r d\phi,\ \ \ \eta_2 = r\sin\phi d\theta.$$

Note that we are using

$$d\phi \left( \frac{\partial}{\partial \phi}\right) = 1,\ \ d\phi \left( \frac{\partial}{\partial \theta}\right) =0, \ \ d\theta \left( \frac{\partial}{\partial \phi}\right) = 0,\ \ d\theta \left( \frac{\partial}{\partial \theta}\right) = 1.$$

$\endgroup$

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.