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Evaluate the smooth one-form on unit velocities to obtain $f(x,v)={\theta }_x(v)$. Its integral along a periodic orbit of the geodesic flow is exactly the one-form integral along the corresponding closed geodesic. Negative curvature makes this geodesic flow an Anosov flow; the geodesic flow of a connected closed negatively curved surface is also topologically transitive. Apply the smooth Livsic theorem to obtain $u\in C^{\infty }(SM)$ withIf there is more than one connected component, apply this argument to each component separately.

The vertical vector field of a surface unit tangent bundle differentiates rotation of $v$, soFor each $x$, writing ${\theta }_x(v)=A\cos \vartheta +B\sin \vartheta$ shows explicitly thatThe fibre disintegration of Liouville volume of a surface geodesic flow is $d\mu =d{A}_{g}d\vartheta$. Integrating the displayed equality over the base therefore gives $\Vert VXu{\Vert }_{L^2}^2=\Vert Xu{\Vert }_{L^2}^2$. Substitution into the supplied Pestov identity for a geodesic flow yieldsBoth terms are nonnegative and $-K\ge k_0>0$. Thus $Vu=0$ everywhere. Since every unit-circle fibre is connected, $u$ is independent of velocity and descends to a smooth function $h$ on $M$: $u=h\circ \pi$. Its geodesic derivative isComparison with $Xu={\theta }_x(v)$, first on unit vectors and then on all vectors by linearity, provesThus the one-form is an exact differential form. This proves the degree-one geodesic X-ray transform conclusion directly from the given integral identity and the stated smooth coboundary theorem.

Let $\pi :SM\to M$ be the unit tangent bundle projection. Write $X$ for the geodesic flow generator and $V$ for the vertical vector field of a surface unit tangent bundle, positively rotating $v$ toward $iv$. The acceleration prescribed in the question changes only this velocity angle, so the surface thermostat generator isIt never vanishes: $d\pi (F)=v\ne 0$. Also $F$ and $V$ are linearly independent, since $d\pi (V)=0$.

The vertical Maslov cycle of a surface thermostat isThere is exactly one such plane in each fibre, so ${\Lambda }_V$ is the image of a smooth section. A two-plane containing $F$ corresponds to a line in the rank-two quotient $T(SM)/\mathbb{R}F$; thus $\Lambda (SM)$ is a projective bundle with fibre ${\mathbb{RP}}^1$. Its total dimension is four and the cycle has dimension three, making it a smooth hypersurface. In this two-dimensional transverse space the Maslov cycle means the projective line meeting the distinguished vertical line, which here is just that distinguished line itself in each fibre.

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.

For a surface thermostat, the planes containing $F$ form the projective bundle of $T(SM)/\mathbb{R}F$. The vertical Maslov cycle is the section selecting the image of the vertical vector field of a surface unit tangent bundle. It is a smooth hypersurface. In the local projective chart $\mathbb{R}F+\mathbb{R}(V+rH)$, it is $r=0$; the lifted flow satisfies $\dot{r}=1$ on the cycle, hence crosses it transversely.