Let $f: \mathbb{R^2}\to \mathbb{R^2}$ be a Schwartz function. If the eigenvalues of $Df$ all vanish, must $f$ be constant?
It is clear that the condition on $Df$ forces $\nabla \cdot f =\text{Tr } Df=0$, so we may be write $f = \nabla \times g$, if that helps. We also have $\text{Det } Df = 0$. Can any more be said?
Lastly, does an analogous result hold when we replace $2$ by $n$?
