= Solution
Fix an orientation and a positive <volume form> $\Omega$. Define the <divergence> by the <Lie derivative>
$$
\mathcal L_F\Omega=(\operatorname{div}_\Omega F)\Omega.
$$
A smooth function is a <flow coboundary> when it equals $Fu$ for some smooth $u$. Every other positive <volume form> is $e^h\Omega$ for a smooth $h$. The product rule for the <Lie derivative> gives
$$
\mathcal L_F(e^h\Omega)=e^h\bigl(Fh+\operatorname{div}_\Omega F\bigr)\Omega,
\qquad
\operatorname{div}_{e^h\Omega}F=\operatorname{div}_\Omega F+Fh.
$$
Thus changing the reference <volume form> changes its <divergence> by a <flow coboundary>. If volume forms of either orientation are allowed, take the logarithm of the absolute ratio on each connected component; multiplying a reference form by a constant sign leaves its divergence unchanged.
If $\operatorname{div}_\Omega F=Fu$, the preceding product rule shows
$$
\mathcal L_F(e^{-u}\Omega)=0.
$$
Conversely, if $e^h\Omega$ is invariant, then $\operatorname{div}_\Omega F=-Fh=F(-h)$ is a <flow coboundary>. Vanishing of the <Lie derivative> is equivalent to flow invariance, since $\partial_t\phi_t^*\Omega=\phi_t^*(\mathcal L_F\Omega)$. Hence
$$
\boxed{\phi_t\text{ preserves a smooth volume form}
\iff\operatorname{div}_\Omega F\text{ is a smooth flow coboundary}.}
$$
The equivalence holds for every choice of reference <volume form>, proving the <invariant volume criterion for a smooth flow>. The smoothness requirement matters: a merely measurable transfer function does not give a smooth invariant volume form.
Back to article page
