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 $L\_X \alpha =0$?



Can every De Rham cohomology class be represented by a closed form $\alpha$ with $L_X \alpha=0$
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.


