Consider a unit sphere $S^2 \subset R^3$ and a map $\omega_p : T_pS^2 \times T_pS^2 \to \mathbb{R}$ defined by
$$\omega_p(u,v) = (u \times v) \cdot p$$
How do I know is this 2-form (on $S^2$) closed and exact?
What I have done so far is to parametrize $S^2$ by
$$p = (\sin\theta\cdot\cos\phi,\sin\theta\cdot\sin\phi,\cos\theta)$$
And, $u = \partial /\partial\theta$ and $v = \partial /\partial\phi$
Therefore $\omega_p(u,v) = \sin\theta \cdot d\theta \wedge d\phi$
Than how do I compute $d\omega$? And for exact, should I find (or show no such) $f\in \Omega^1(S^2)$ such that $\omega = df$?
Since $S^2$ is 2-dimensional so it is closed
And $dxdydz$ is volume for ${\bf R}^3$, then $$ a=i_N dxdydz $$ is volume for $S^2$ where $N=p$ is unit normal. If $e_i$ is orthonormal frame for $T_p S^2$, then $$ a(e_1,e_2)=i_N dxdydz (e_1,e_2)=(p\times e_1)\cdot e_2 = e_1\times e_2\cdot p=\omega (e_1,e_2) $$
Hence $\omega$ is volume for $S^2$ so that if $\omega$ is exact i.e., $db=\omega$, $$ 4\pi=\int_{S^2} \omega =\int_{\partial S^2} b= 0 $$
Hence it is not exact