0
$\begingroup$

Let $U=\{ (\theta,\phi,r):\theta \in \mathbb{R}, \phi \in ]0,\pi[,r\gt 0\}$, how I can find the moving frame. I thougt: Consider the parametization for $U$ $$(\theta,\phi,r)\mapsto (r\cos(\theta)\sin(\phi),r\sin(\theta)\sin(\phi),r\cos(\phi)) $$

So we have a basis for $U$:

$$\{\frac{\partial}{\partial \theta},\frac{\partial}{\partial \phi},\frac{\partial}{\partial r}\}$$

Where: $$\frac{\partial}{\partial \theta}= (-r\sin(\theta)\sin(\phi),r\cos(\theta)\sin(\phi),0))$$

$$\frac{\partial}{\partial \phi}= (r\cos(\theta)\cos(\phi),r\sin(\theta)\cos(\phi),-r\sin(\phi))$$

$$\frac{\partial}{\partial \theta}= (-\cos(\theta)\sin(\phi),\sin(\theta)\sin(\phi),\cos(\phi))$$

We can see that basis is orhtogonal, therefore, we only need to divide each vector for its norm.Therefore we have:

$$\{\frac{1}{r\sin(\phi)}\frac{\partial}{\partial \theta},\frac{1}{r}\frac{\partial}{\partial \phi},\frac{\partial}{\partial r}\}$$

It's a orthonormal frame for $U$. But I don't how to find the dual frame, neither the conecctions forms. And I need to use that in the unity sphere, how to restrict that referencial in the sphere? Could someone help me? Thank you!

$\endgroup$
1
$\begingroup$

Hint: Start by showing $\theta^1 = dr$, $\theta^2=r\,d\phi$, $\theta^3=r\sin\phi\,d\theta$ is the dual basis (if you switch your first and third tangent vectors).

Next, compute $d\theta^i$ and find $\omega^i_j=-\omega^j_i$ satisfying $d\theta^i=\sum\theta^j\wedge\omega^i_j$. (I do not know which way you're being taught this material.)

$\endgroup$

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.