# Lie derivative of a top-dimensional differential form on a compact manifold

Suppose $$M$$ is a compact, smooth $$n$$-manifold, $$X$$ is a smooth vector field on $$M$$, and $$\omega$$ is a smooth $$n$$-form on $$M$$. Then is it true that the Lie derivative $$L_X \omega$$ is not nowhere vanishing, i.e., $$(L_X \omega)|_p=0$$ for some $$p\in M$$?

I know that an exact $$1$$-form on a compact manifold is not nowhere vanishing, so I tried to prove similarly, but I couldn't.

If $$M$$ is orientable without boundary and $$\omega$$ is nowhere vanishing, then this would be true, because by Cartan's formula we have $$L_X \omega =d \iota_X \omega$$, so by Stokes' theorem it follows that $$\int_M L_X\omega=\int_M d\iota _X\omega=\int_{\partial M} \iota_X \omega=0$$ because $$M$$ is boundaryless.

• Where did your argument use $\omega$ nowhere-vanishing? Apr 24, 2020 at 23:58
• @TedShifrin If $\omega$ is nowhere-vanishing then $L_X\omega =f\omega$ for some $C^\infty$ function $f$, so $0=\int_M L_X\omega=\int_M f\omega$, but since $\int_M \omega >0$ or $<0$ (nowhere vanishing), $L_X\omega$ should have a zero somewhere Apr 25, 2020 at 11:45

You essentially solved this for the case that $$M$$ is orientable. Let's see the non-orientable case. By Cartan's formula we also have that
$$L_X\omega \;\; =\;\; d(i_X\omega) \;\; =\;\; \eta$$
where $$\eta$$ is some other top-dimensional form. However, given that $$M$$ is non-orientable then $$M$$ does not possess a non-vanishing top form. Therefore $$\eta_q = 0$$ for some $$q \in M$$.
• Can we have $L_X\omega =f\omega$ even when $\omega$ vanishes at some point $p$? Apr 24, 2020 at 20:19
• Yes, we can. This result follows from the fact that $\Lambda^n(M)$ is a line bundle. The space of top forms is 1-dimensional, hence any top form is a function-multiple of any other. Apr 24, 2020 at 20:21
• Then for a nonzero (not nowhere vanishing) $n$-form $\omega$ and any $n$-form $\tau$, if $\omega_p=0$ for some $p$ then we must have $\tau _p=0$. Is this generally true? Apr 24, 2020 at 20:31
• @user002233 I undeleted my answer and changed my response. I was wrong in my claim from earlier, but it is taken care of either way by the case if $M$ is non-orientable. Apr 24, 2020 at 20:57
• Alternatively, just pull everything back to the orientation double covering of $M$. Apr 24, 2020 at 21:21