How to prove the space of divergence-free vector fields on a manifold is infinite dimensional?

Question

Let $M$ be a closed manifold, equipped with a volume form $\omega$. I understand that the vector space

$$ \{ X \in \Gamma(TM) \, | \, L_X\omega=0 \}$$

is always infinite-dimensional.

Is there an elementary argument showing this?

We can assume of course $\omega$ comes from a Riemannian metric $g$, and then $L_X\omega=0$ if and only if $\text{trace}(\nabla X)=0$.

Answer 1

Let us show first that the space of compactly supported, divergence-free vector fields on $\mathbb{R}^n$ is infinite dimensional. Let $\rho:\mathbb{R}^+\to\mathbb{R}^+$ be a smooth, compactly supported function, such that $\rho\equiv1$ on a neighborhood of $0$. Let $A=(a_i^j)\in M_{n\times n}(\mathbb{R})$ be anti-symmetric. In particular, all the diagonal entries of $A$ vanish. Define a vector field $X_{\rho,A}$ on $\mathbb{R}^n$ by $$X_{\rho,A}(x)=\rho\left(|x|\right)A(x).$$(Here we think of a vector field as a function $\mathbb{R}^n\to\mathbb{R}^n$). This is compactly supported, as $\rho$ is. We compute the divergence: $$\begin{align}\mathrm{div}X_{\rho,A}&=\frac{\partial}{\partial x^1}\left(\rho(|x|)a_1^ix_i\right)+\ldots+\frac{\partial}{\partial x^n}\left(\rho(|x|)a_n^ix_i\right)\\&=\dot{\rho}(|x|)\frac{x_1}{|x|}a_1^ix_i+\ldots+\dot{\rho}(|x|)\frac{x_n}{|x|}a_n^ix_i\\&=0.\end{align}$$ This clearly provides an infinite-dimensional family of vector fields, due to the freedom in the choice of $\rho$.

We return to the closed manifold $M$. Recall that a volume form is always integrable. This means that there is a coordinate neighborhood $U$, in which we have $\omega=dx^1\wedge\ldots\wedge dx^n.$ Now, we may choose some $X_{\rho,A}$ in $U$ and extend it by $0$ to a global vector field.

Answer 2

Note that this result is FALSE in dimension 1. A vector field on $S^1$ is divergence free iff it is of the form $a{\partial \over \partial \theta}$.

A different proof is the following. As $n\geq 2$, the vector space of $n-2$ form is huge. Let $\alpha$ such a form, $d\alpha$ is a closed $n-1$ form, and there exists a field $X$ such that $i_X \omega= d\alpha$, and $X$ is divergent free.

So the result is just to say that derivative of $n-2$ forms is an infinite dimensional space.

To prove this, you can consider a chart with coordinates $x_1,...x_n, \vert x\vert \leq a$ and consider $\alpha = f dx_{1}\wedge dx_{2}...\wedge d x_{n-2}$,where $f=f(x_{n-1})$ is a compactly supported function in the segment $\vert x_{n-1}\vert <a$, for $a$ small enough. Then $d\alpha = f'(x_2) dx_2...dx_n$. So the space of closed form contains a space isomorphic to the space of derivative of function with support in teh segment $\vert x\vert <a$