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I'd like to make sure I'm getting the proof of the following statement right:

Let $S^2\subset \mathbb{R}^3$ be the unit sphere and define a vector field $N(x,y,z)=(x,y,z)$. Define a 2-form $\omega\in \Omega^2(\mathbb{S}^2)$ as $\omega(p)(V,W)=\det(N(p),V,W)$ where we think of $N(p),V,W$ as column vectors. Show that $\mathbb{S}^2$ has an orientation such hat for every oriented basis $\langle V,W \rangle$ of $T_p\mathbb{S}^2$, $\omega(p)(V,W)>0$. Use this to show that $\int_{\mathbb{S}^2} \omega>0$ and that $\omega$ is not exact.

Solution: The orientation that one should take on $\mathbb{S^2}$ is the one induced from $\mathbb{R}^3$. For each $p\in \mathbb{S}^2$, given a basis $\langle V,W\rangle$ of $T_p\mathbb{S}^2$, simply orient it in such a way that $\langle N(p),V,W\rangle$ is an oriented basis of $T_p\mathbb{R}^3=\mathbb{R}^3$. Then one automatically has that $\omega(p)(V,W)>0$.

Now recall that if $\omega$ is an orientation form on a smooth manifold $M$, then $\int_M \omega>0$. Indeed, let $(U_i,\phi_i)$ be a collection of oriented charts and let $\{\rho_i\}$ be a partition of unity subordinate to the cover $\{U_i\}$. Then by definition $$\int_M \omega=\sum_i \int_{U_i}\rho_i\omega= \sum_i \int_{\varphi_i(U_i)} \rho_i (\varphi_i^{-1})^{\ast}\omega$$ where $(\varphi_i^{-1})^{\ast}\omega$ is a positive function times $dx^1\wedge \cdots \wedge dx^n$. The sum on the RHS consists of non-negative terms, and at least one is trictly positive.

Finally to see that $\omega$ is not exact we apply Stokes theorem. If it was, then $\omega=d\eta$ for some 1-form $\eta$. Since $S^2$ bounds a ball $B$ we would have $$\int_{\mathbb{S}^2} \omega = \int_{\mathbb{S}^2=\partial B} d\eta = \int_B d^2\eta=0$$

user54631
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    Perfectly true. Note, however, that the important point is $S^2$ being a surface without boundary. The face that it bounds a ball is stronger and not necessary. Of course, in $\mathbb{R}^3$ the two are equivalent, but in other spaces they are not. – Amitai Yuval Aug 10 '14 at 19:26

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Following Amital Yuval, I think the last paragraph should be rewritten. Indeed, you supposed that $\omega=d\eta$ for some $1$-form $\eta$ on the sphere. It is not immediate that this form $\eta$ can be smoothly extended to the ambient space (it can be, but that's a theorem of its own). Without such extension, integration over $B$ does not make sense.

Better form:
$$\int_{\mathbb{S}^2} \omega=\int_{\mathbb{S}^2} d\eta = \int_{\partial \mathbb{S}^2=\varnothing} \eta = 0$$