Divergence-free vector field on a 2-sphere.

Question

What is the solution of the differential equation

$$\text{div}X=0$$

on the 2-sphere?

Answer 1

$\def\div{\operatorname{div}}$Let $\nu$ be the volume form on the sphere. The map $X\mapsto\nu(X,\mathord-)$ gives a bijection from vector fields to $1$-forms and, in particular, if $h:S^2\to\mathbb R$ is a smooth function, there is a unique vector field $X_h$ such that $\nu(X_h,Y)=dh(Y)$ for all vector fields $Y$. Using Cartan's magic formula it is easy to see that $\def\L{\mathcal L}\L_{X_h}\nu$, the Lie derivative of $\nu$ in the direction of $X_h$, vanishes. Since $\L_{X_h}\nu=\div(X_h)\nu$ and $\nu$ vanishes nowhere, we see that $\div(X_h)=0$.

In this way, we find a divergence-free vector on the sphere for each function $h$. In fact, they all arise in this way. This is a more complicated result —it depends on the simply-connectedness of the sphere and is a symplectic thing. Let's see: let $X$ be a vector field on the sphere with zero divergence, so that $\L_X\nu=0$. Using Cartan's formula, we see that the $1$-form $i_X\nu$ which we get by contracting $\nu$ with $X$ is closed. It therefore has a class in the degree $1$ de Rham cohomology vector space $[i_X\nu]\in H^1(S^2)$. As this vector space is zero, because the sphere is simply connected, $i_X\nu$ has to be exact, that is, there exists a function $h:S^2\to\mathbb R$ such that $i_X\nu=dh$. If you consider what this means, you see that $X$ is in fact equal to the vector field $X_h$.