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For $F=X+\lambda V$, the canonical fields $X,V$ preserve Liouville volume of a surface geodesic flow, so the divergence is $V\lambda$. For a Gaussian thermostat, $\lambda =⟨E,iv⟩$ and $V(iv)=-v$, giving the stated sign. If $E=\nabla f$, then the density $e^{f\circ \pi }$ makes this divergence zero.

A smooth real function $\lambda$ on the unit tangent bundle determines this unit-speed smooth flow. Its base velocity is $v$, so its generator never vanishes. In the canonical frame its divergence in Liouville volume of a surface geodesic flow is $V\lambda$. The special case where $\lambda$ depends only on the base is a magnetic flow; when $\lambda (x,v)=⟨E(x),iv⟩$ it is a Gaussian thermostat.

This is a Gaussian thermostat, with generator and angular accelerationThe canonical fields $X$ and $V$ preserve Liouville volume of a surface geodesic flow. To check this directly, use the canonical coframe of a surface unit tangent bundle $(\alpha ,\beta ,\omega )$, with dual frame $(X,H,V)$ andFor the positive volume $\mu =\alpha \wedge \beta \wedge \omega$, contraction gives ${\iota }_X\mu =\beta \wedge \omega$ and ${\iota }_V\mu =\alpha \wedge \beta$. Both two-forms are closed by the displayed structure equations, so the Cartan formula for the Lie derivative gives ${\mathcal{L}}_X\mu ={\mathcal{L}}_V\mu =0$.

The product rule for divergence now yieldsThe vertical vector field of a surface unit tangent bundle fixes the base point and satisfies $V(v)=iv$, $V(iv)=-v$. Since $E$ depends only on the base point,This is the divergence of a Gaussian thermostat. Reversing the overall sign of the chosen volume form does not change its divergence, so the answer is also valid if Liouville volume is written as the oppositely oriented contact volume.

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).