Is there a vector field such that one differential form is the Lie derivative of the other? 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.
 A: 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) $\DeclareMathOperator{\dive}{div}$
$$
\dive_\mu X=\dive_g X=\ast d\ast  \omega_X\Longleftrightarrow \nu=\ast \dive X=d\ast\omega_X.
$$
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.
