Do (closed) Riemannian manifolds admit anti-self-adjoint vector fields (locally or globally)?

Question

We have the following set up. Let $(M^n,g)$ be a (possibly closed) Riemannian manifold, then a vector field $X\in\mathcal{T}(M)$ is said to be anti-self-adjoint if for any $\phi,\eta\in\mathscr{C}_0^\infty(M)$, we have the identity $$ \int_M\phi(X\eta)\,\mathrm d\mu_g = -\int_M(X\phi)\eta\,\mathrm d\mu_g. $$ Note that $\mathbb T^n$ admits a global tangent frame of anti-self-adjoint vector fields. But can we find any anti-self-adjoint vector fields in general, or at least locally (i.e. imposing that $\phi$ and $\eta$ vanish outside of some neighborhood)?

Answer 1

Perhaps I've missed something, but by the Leibniz rule this is equivalent to $$\int_M X(\phi \eta) d\mu_g = 0$$ for all $\phi,\eta\in C^\infty_0$, which in turn is equivalent to $\int_M X(f) d\mu_g = 0$ for all $f \in C^\infty_0.$ Integrating by parts (i.e. applying $\operatorname{div}(fX) = X(f)+f\operatorname{div}(X)$ and the Riemannian divergence theorem), this is the same as $$\int_M f \operatorname{div}(X) d \mu_g = 0,$$ so by the fundamental lemma of the calculus of variations this is equivalent to $\operatorname{div}(X) = 0.$

Thus you're just considering the divergence-free/volume-preserving/solenoidal vector fields, which are in plentiful supply.