# Closed 2-form (Symplectic)

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 $$\tilde{\omega}=\tilde{\nu}_1\cdot dy\wedge dz+\tilde{\nu}_2\cdot dz\wedge dx+\tilde{\nu}_3\cdot dx\wedge dy$$

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.