Let $\Sigma\subset \mathbb{R}^3$ be a smooth 2-dimensional submanifold of $\mathbb{R}^3$ and $\nu:\Sigma\to \mathbb{R}^3$ a smooth unit normal vector field. We define $\omega\in \Omega^2(\Sigma)$ to be
$$\omega_p(v,w)=\left\langle\nu(p),v\times w\right\rangle$$
where $\left\langle\cdot,\cdot\right\rangle$ is the inner product and $\times$ is the cross product.
How can I prove that this is a closed 2-form?
I know that the cross product in $\mathbb{R}^3$ is $(dy\wedge dz, dz\wedge dx, dx\wedge dy)$ but I cannot say that $\omega$ is the restriction of a 2-form on $\mathbb{R}^3$ of the form
$$\tilde{\omega}=\tilde{\nu}_1\cdot dy\wedge dz+\tilde{\nu}_2\cdot dz\wedge dx+\tilde{\nu}_3\cdot dx\wedge dy$$
Best Answer
Every $2$-form on a surface is closed. :)
If you extend it in a natural way to be a $2$-form on $\Bbb R^3$, then it will not be closed in general. Consider the example of $\Sigma$ the unit sphere. Then the $2$-form $$\omega = \frac{x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy}{(x^2+y^2+z^2)^{1/2}},$$ as you can check, is definitely not closed.