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