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Fix an orientation and a positive volume form $\Omega$. Define the divergence by the Lie derivativeA 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 givesThus 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 ${\mathrm{div}}_{\Omega }F=Fu$, the preceding product rule showsConversely, if $e^h\Omega$ is invariant, then ${\mathrm{div}}_{\Omega }F=-Fh=F(-h)$ is a flow coboundary. Vanishing of the Lie derivative is equivalent to flow invariance, since ${\partial }_t{\varphi }_t^{\ast }\Omega ={\varphi }_t^{\ast }({\mathcal{L}}_F\Omega )$. HenceThe 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.

Use the following permitted kernel result, stated in the form needed here. For an Anosov flow generated by a smooth surface thermostat on a closed oriented Riemannian surface, if a smooth function $u$ on $SM$ satisfieswhere $h$ is a smooth base function and $\eta$ is a smooth one-form, then $h=0$ and $\eta$ is an exact differential form. Conversely these two conditions give a smooth solution by taking a base primitive. This is the kernel of the thermostat X-ray transform in degrees zero and one. In particular it applies to Gaussian thermostats without an additional curvature condition; it is stronger than the magnetic or geodesic special case.

Put $\theta =E^{♭}$, so ${\theta }_x(v)=⟨E(x),v⟩$. Suppose the thermostat preserves a smooth volume form. Write it as $\rho \mu$ with smooth $\rho >0$, choosing its sign separately on each component if necessary. By the invariant volume criterion for a smooth flow and the divergence computation in part (b),Thus $u=\log \rho$ is a smooth solution of $Fu={\theta }_x(v)$. Apply the stated kernel theorem with $h=0$ to obtain $\theta =df$ for a smooth base function $f$. The definition of the gradient then gives $E={\mathrm{grad}}_{g}f$.

Conversely, suppose $E={\mathrm{grad}}_{g}f$. Since $d\pi (F)=v$, the vertical part has no effect on a base function andThe divergence of a Gaussian thermostat and the volume-change formula therefore giveConsequentlyOnly the necessary direction uses the Anosov kernel theorem; the displayed invariant density works for any potential Gaussian thermostat. In particular its exponent is positive, as required to cancel the negative divergence found in part (b).