Does $(\nabla \times F) \cdot F= (\nabla \times F) \cdot \nabla f$ have a solution? A grad student asked me this question during office hours, and I couldn't for the life of me come up with a proof or counterexample:
For a given $F:\mathbb{R}^3 \to \mathbb{R}^3$, does $(\nabla \times F) \cdot F= (\nabla \times F) \cdot \nabla f$ in $R^3$ always have a solution in $f$?
 A: There exist $F$ for which there is no global solution $f$ to the above equation.  Here is how you can construct an example:
First, regard $F$ as a vector field on $\mathbb{R}^3$ and consider its dual $1$-form $\phi$.  The left hand side of your equation can then be written as the Hodge dual of $\phi\wedge\mathrm{d}\phi$ and the right hand side can be written as the Hodge dual of $\mathrm{d}f\wedge\mathrm{d}\phi$, so your equation becomes
$$
\phi\wedge\mathrm{d}\phi = \mathrm{d}f\wedge\mathrm{d}\phi = \mathrm{d}\bigl(f\,\mathrm{d}\phi\bigr).\tag1
$$
Now suppose that $\phi$ has compact support and that the integral of $\phi\wedge\mathrm{d}\phi$ over $\mathbb{R}^3$ is nonzero.  (See below for a construction of such a $\phi$.)  Then integrating the ends of (1) over $\mathbb{R}^3$ and using Stokes' Theorem will yield a contradiction.
Now, to construct such a $\phi$, let $x,y,z$ be standard coordinates on $\mathbb{R}^3$, and let $h\not\equiv0$ be a smooth function with compact support on $\mathbb{R}^3$.  Set
$$
\phi = h\,(\mathrm{d}z - y\,\mathrm{d}x).\tag2
$$
Then computation shows that $\phi\wedge\mathrm{d}\phi = h^2\,\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z$. Hence, its integral over $\mathbb{R}^3$ is positive, as desired.
