I'm looking for a reference or answer for the following question:


Let $M$ be an (compact and orientable, if it helps) smooth manifold and $\nu$ and $\mu$ two differential forms. I'm looking for conditions that allow me to find a vector field $X$ on $M$ such that $\nu=\mathcal{L}\_X\mu$, where $\mathcal{L}$ denotes the Lie derivative on differential forms.


A special case in which I'm particularly interested in is if $\mu$ is a positive volume form with $\int\_M\mu=1$ and $\nu$ is a volume form with $\int\_M\nu=0$. Then the above question can be rephrased as: given a smooth function $\varphi$ with $\int\_M\varphi\mu=0$, can we find a vector field $X$ such that $\varphi=\text{div}\_{\mu}X$?


I've read somewhere that under certain assumptions one can always find such a vector field, but I cannot find that statement anymore and so I would appreciate if someone knew a short answer or a reference for this.



Is there a vector field such that one differential form is the Lie derivative of the other?
If $\mu$ is a volume form and $\nu$ is a top degree form, then there exists a vector field $X$ such that $L\_X\mu=\nu$ if and only if $\nu$ is exact. 


You can always fix a metric $g$ on $M$ such that $\mu$ is the volume form 
 determined by the metric and the orientation, $\mu=dVol\_g$.


Denote by $\DeclareMathOperator{\Vect}{Vect}$ $\Vect(M)$ the space of smooth vector fields on $M$ and by $\ast$ the Hodge star operator determined by $g$ and the orientation. Then $\mu=\ast 1$. 


For any vector field $X$ on $M$ denote by $\omega\_X\in \Omega^1(M)$ the $1$-form dual to $X$, i.e.,
$$
\omega\_X(Y)=g(X,Y),\;\;\forall Y\in \Vect(M). 
$$
Then (see Proposition 4.1.48 of [these notes](https://www3.nd.edu/~lnicolae/Lectures.pdf)) $\DeclareMathOperator{\dive}{div}$
$$
\dive\_\mu X=\dive\_g X=\ast d\ast \omega\_X\Longleftrightarrow \nu=\ast \dive X=d\ast\omega\_X.
$$
Thus if there exists a vector field $X$ such that $L\_X\mu=\nu$ then $\nu$ is exact $\nu=d\ast\omega\_X$, i.e., $\nu$ is exact. Conversely, if $\nu$ is exact, $\nu=d\beta$, then the vector field $X$ uniquely characterized by the equality $\ast\omega\_X=\beta$ satisfies $L\_X\mu=\nu$.


**Remark.** If $M$ is compact and connected then $\nu$ is exact if ad only if $\int\_M\nu=0$. If $M$ is noncompact and connected then $\nu$ is automatically exact since $H\_{DR}^{\dim M}(M)=0$. If $M$ has several connected components solve this problem on each component separately.


