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 $L_X \alpha =0$?
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2$\begingroup$ What did you try already? $\endgroup$– Vladimir DotsenkoCommented Jun 26, 2018 at 8:07
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$\begingroup$ @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 TaghaviCommented Jun 26, 2018 at 8:25
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1 Answer
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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.
answered Jun 26, 2018 at 8:41
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$\begingroup$ 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? $\endgroup$ Commented Jun 26, 2018 at 9:09
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$\begingroup$ or is there a non compact manifold with this property whose total cohomology is not vanished? $\endgroup$ Commented Jun 26, 2018 at 9:19
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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. $\endgroup$ Commented Jun 26, 2018 at 9:53