Connection compatible with a volume form Let $M$ be a smooth, orientable $n$-manifold and $\eta$ a volume form on $M$. Does there exist a connection $A$ on $TM$ such that 
$$\tag{$*$}D\eta=0,$$
where $D$ is the appropriate covariant derivative associated to $A$ on $\Omega^n(M)$? Can one assume $A$ to be symmetric?
I have a feeling that if one writes $(*)$ plus the torsion equation and then applies the Frobenius theorem, some local result can be obtained. But that doesn't give global conditions (on $M$ nor $\eta$). A reasonable assumption would be $M$ parallelizable. 

If $\eta$ vanishes somewhere, then the only way it can be parallel (for any connection at all) is if it vanishes identically. Thus I will assume $\eta$ is non-vanishing.
Let $g$ be a Riemannian metric on $M$ and $\omega$ the volume form of $g.$ Since $\omega,\eta$ are smooth non-vanishing sections of a line bundle, there is some positive function $f \in C^\infty(M)$ such that $\eta = f \omega.$ (I'm assuming here that we take the orientation of $\omega$ to match that of $\eta.$)
The conformally related metric $\tilde g = f^{2/n}g$ then has volume form $\tilde \omega = (f^{2/n})^{n/2} \omega = \eta,$ which is made parallel by the Levi-Civita connection of $\tilde g.$ Thus there are no conditions necessary - there is always a symmetric connection that does the job.
