Assume that $M$ is a manifold and $X$ is a vector field on $M$.

Is it true to say that every closed form is De Rham-cohomologue to a closed form $\alpha$ with $L_X \alpha =0$?

  • 2
    $\begingroup$ What did you try already? $\endgroup$ – Vladimir Dotsenko Jun 26 '18 at 8:07
  • $\begingroup$ @VladimirDotsenko For open orientable manifolds the question has obviously positive answer in their last cohomology. So I am curious if the question has trivial answer in all cases, compact or non compact, in all cohomology dimension. $\endgroup$ – Ali Taghavi Jun 26 '18 at 8:25

You could try the vector field $X=x\partial_x$ on the real projective line, so with affine coordinate $x$. Then in another affine chart $y=1/x$, $X=-y\partial_y$, so $X$ is smooth everywhere. Every 1-form on the real projective line is closed. An invariant 1-form has to be $\alpha=C dx/x$. To be defined near $x=0$, it has to be $0$. But on the real projective line, there is cohomology in dimension 1, as the real projective line is the circle.

  • $\begingroup$ Thank you very much for your answer and your attention to my question. Is there a compact manifold which satisfy the property mentioned in the question? $\endgroup$ – Ali Taghavi Jun 26 '18 at 9:09
  • $\begingroup$ or is there a non compact manifold with this property whose total cohomology is not vanished? $\endgroup$ – Ali Taghavi Jun 26 '18 at 9:19
  • 3
    $\begingroup$ If a compact Lie group $G$ acts smoothly on a manifold $M$, then every smooth closed form on $M$ is cohomologous to a $G$ invariant closed form, by averaging over the biinvariant volume form on $G$. So there are interesting examples, like the circle with a rotation vector field. $\endgroup$ – Ben McKay Jun 26 '18 at 9:53

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.