# Finding dual forms of a frame field on a sphere

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.

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

## 1 Answer

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

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