4
$\begingroup$

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)?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.