Let $v:\Bbb R^n\to\Bbb R^n$ be a smooth vector field with an isolated zero at $z\in\Bbb R^n$. Suppose that $(\operatorname{div}v)(z)>0$. Can we say anything about the index of $v$ at $z$?
I know that if a vector field is a "source" in the sense that the field always points away from the zero, then its index will be $1$. But as far as I know this understanding of positive divergence is a heuristic, and in general one can not find a sphere $S_\epsilon$ centered at $z$ such that $v$ points out of this sphere. However, if positive divergence indeed guarantees the existence of such an $S_\epsilon$, then the vector field on $S_\epsilon$ is smoothly homotopic to the Gauss mapping, and so has index $1$.