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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.

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  • $\begingroup$ Where did your argument use $\omega$ nowhere-vanishing? $\endgroup$
    – Ted Shifrin
    Apr 24, 2020 at 23:58
  • $\begingroup$ @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 $\endgroup$
    – user302934
    Apr 25, 2020 at 11:45

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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 $q \in M$.

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  • $\begingroup$ Can we have $L_X\omega =f\omega$ even when $\omega$ vanishes at some point $p$? $\endgroup$
    – user302934
    Apr 24, 2020 at 20:19
  • $\begingroup$ 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. $\endgroup$
    – Mnifldz
    Apr 24, 2020 at 20:21
  • $\begingroup$ 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? $\endgroup$
    – user302934
    Apr 24, 2020 at 20:31
  • $\begingroup$ @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. $\endgroup$
    – Mnifldz
    Apr 24, 2020 at 20:57
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    $\begingroup$ Alternatively, just pull everything back to the orientation double covering of $M$. $\endgroup$
    – Jason DeVito
    Apr 24, 2020 at 21:21

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