I was checking problems on differential forms and I found the following one.
Consider the sphere $S^2 \subseteq R^3$ and the map $\omega_p : T_pS^2 \times T_pS^2 \rightarrow \mathbb{R}$ given by $\omega_p(X_p,Y_p) = (X_p \times Y_p) \cdot p$.
Globally, $\omega$ defines a differential $2$-form over $S^2$. And what about if I want to compute $\int_{S^2} w$.
I think that $X_p \times Y_p = \pm p$, so $\omega_p$ would have a simpliest form. Is that true?
How can I write $\omega = f d\theta \wedge d\phi$ in order to compute the integral?
Thanks for any hint.
Just write down the parametric form: $$ p=(x,y,z)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) $$ Now the two derivations (tangent vectors) can be computed by: $$ \begin{split} \partial/\partial\theta&=(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta)\\ \partial/\partial\phi&=(-\sin\theta\sin\phi,\sin\theta\cos\phi,0) \end{split} $$ The only thing left is to compute $f$ in $\omega=fd\theta\wedge d\phi=f(d\theta\otimes d\phi-d\phi\otimes d\theta)$. For this, we plug in $\partial/\partial\theta,\partial/\partial\phi$ to get: $$ \begin{split} f&=(\partial/\partial\theta\times\partial/\partial\phi)\cdot p\\ &=\sin\theta \end{split} $$ And therefore: $$ \int_{S^2}\omega=\int_0^{2\pi} d\phi\int_0^{\pi}\sin\theta d\theta=4\pi $$