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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.

At unit speed on a Riemannian surface, the right-hand side is $⟨E,iv⟩iv$. This removes the velocity-parallel component of the forcing, so differentiating $|\dot{\gamma }{|}^2$ gives zero. The dynamics need not preserve the ordinary Liouville volume of a surface geodesic flow.

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.

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.

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.

The geodesic flow of a closed negatively curved surface is a topologically transitive Anosov flow. In particular, its Jacobi-field hyperbolicity and the usual negative-curvature transitivity theorem allow the smooth Livsic theorem to be applied. Set $b(x,v)={\beta }_x(v,v)$. The given closed-geodesic integrals are exactly its periodic-orbit obstructions on $SM$. Therefore

We make the differential operators and signs explicit. On the unit tangent bundle, let $V$ generate positive fibre rotation, $X$ be the geodesic vector field and $H=[V,X]$. For the canonical coframe of a surface unit tangent bundle, writeand let $\omega$ be the Levi-Civita connection form with $\omega (V)=1$. This uses $\theta$ for a one-form, keeping $\beta$ for the given tensor. The dual frame is $(X,H,V)$, and the structure equations areThese follow by writing the two horizontal forms as a rotated local orthonormal coframe and $\omega =d\vartheta +{\omega }_0$, with ${\omega }_0$ its connection 1-form. Equivalently, differentiating the sine and cosine components of $X$ twice in the fibre angle givesThe curvature equation also gives $[X,H]=KV$. The Liouville volume of a surface geodesic flow will be taken positive in the base/positive-fibre orientation, namely $\mu =\alpha \wedge \theta \wedge \omega$.

In an oriented orthonormal frame at $x$, put $v=(\cos \vartheta ,\sin \vartheta )$. The symmetric tensor evaluation isSince $V={\partial }_{\vartheta }$, this proves $(V^3+4V)b=0$.

The Lie bracket of vector fields identities give an operator identity that supplies the required equation. First,Using $VH=HV-X$ and $XV=VX-H$, differentiate this expression once more to obtainFor $\phi =V^{2}u+u$ and $Xu=b$, the first term vanishes by the explicit fibre calculation. Hence

Use the canonical coframe of a surface unit tangent bundle $(\alpha ,\theta ,\omega )$ and its dual $(X,H,V)$. In constant curvature $-1$, the decaying normal Jacobi fields have data $(j,j^{\prime })=(a,-a)$, so the stable line is generated by $H-V$. The weak stable bundle is thereforeIt is annihilated by the nonvanishing form $\zeta =\theta +\omega$. The structure equations now giveThus the auxiliary form in the Godbillon-Vey invariant can be taken as $\eta =\alpha$. In the natural orientation given by the oriented base and positive fibre rotation, put $\mu =\alpha \wedge \theta \wedge \omega$. ThenThis sign is important: in these conventions the contact three-form is the negative of the positive Liouville volume of a surface geodesic flow.

Integration around each circle fibre gives ${\int }_{S_{x}M}\omega =2\pi$, while $\alpha \wedge \theta ={\pi }^{\ast }{\Omega }_a$. Hence ${\int }_{SM}\mu =2\pi \mathrm{Area}(M)$. The Gauss-Bonnet theorem and $K=-1$ give $-\mathrm{Area}(M)=2\pi \chi (M)$. ConsequentlyFor genus $g\ge 2$, this is $-8{\pi }^2(g-1)$ in the stated orientation. The opposite orientation of $SM$ reverses the sign.

The forcing is perpendicular to velocity, soStarting at unit speed therefore keeps the motion in $SM$. The horizontal part is the geodesic transport $X$, while the acceleration rotates the unit velocity at angular speed $f$. Thus the magnetic flow generator is $F=X+(f\circ \pi )V$.

For the canonical coframe of a surface unit tangent bundle, set $\mu =\alpha \wedge \theta \wedge \omega$. Its interior product of a differential form with the two relevant fields isThe structure equations giveThe Cartan formula for the Lie derivative consequently gives ${\mathcal{L}}_X\mu ={\mathcal{L}}_V\mu =0$. For the variable vertical coefficient, use the full product rule:Indeed $f$ is pulled back from the base, so $Vf=0$ and $df$ has only $\alpha ,\theta$ components. HenceThe Liouville volume of a surface geodesic flow is therefore preserved by this magnetic modification for every smooth $f$, without requiring the Anosov property.

On a closed oriented Riemannian surface this integral identity follows from the canonical commutators and integration by parts against Liouville volume of a surface geodesic flow. If $VX\phi =-2H\phi$ and $K<0$, substitution gives $\Vert X\phi {\Vert }^2+5\Vert H\phi {\Vert }^2+\int (-K)(V\phi {)}^2=0$. Each term is nonnegative, forcing all three derivatives to vanish. For $\phi =V^{2}u+u$, the fibre equation then makes $u$ a constant plus a fibrewise linear function, supplying the potential symmetric 2-tensor kernel.