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Use the Levi-Civita connection to represent a tangent vector to the unit tangent bundle at $(x,v)$ by the initial data $(J(0),D_{t}J(0))$ of the Jacobi field generated by a variation of the unit-speed geodesic. The unit-speed constraint gives $⟨D_{t}J(0),v⟩=0$. Along the geodesic, writeHere $a$ is constant, while the perpendicular Jacobi field equation isThe flow direction is represented by $(a,y,\dot{y})=(1,0,0)$. Consequently $(y,z)=(y,\dot{y})$ are coordinates on the transverse quotient. Define its nondegenerate quadratic form by $Q(y,z)=yz$. Its signature of a quadratic form is $(1,1)$, andSince the Gaussian curvature is strictly negative and the surface is a closed manifold, there is $k_0>0$ with $K\le -k_0$. HenceThe quadratic-form criterion for an Anosov flow now proves the geodesic flow is Anosov. This is the geodesic hyperbolicity from negative Gaussian curvature argument. If the surface is not oriented, the local choice of perpendicular unit vector changes both $y$ and $z$ in sign; their product is unchanged, so the quadratic form still exists globally.

When $K=-1$, the perpendicular solutions are $y=A{e}^t+B{e}^{-t}$. The pure contracting data have $(y,z)=(B,-B)$ and the pure expanding data have $(y,z)=(A,A)$. In the canonical coframe of a surface unit tangent bundle, $H$ represents $(iv,0)$ and $V$ represents $(0,iv)$, soEquivalently the Anosov stable bundle consists of pairs $(\eta ,-\eta )$ with $\eta \perp v$, and the Anosov unstable bundle consists of $(\eta ,\eta )$. The derivative multiplies these data by $e^{-t}$ and $e^t$, respectively, using parallel transport along the geodesic to identify the perpendicular directions. Together with $\mathbb{R}X$ they give the required invariant splitting.

Let $H$ be the horizontal lift of $iv$ supplied by the canonical coframe of a surface unit tangent bundle. The frame $(X,H,V)$ has commutatorsSince $F=X+\lambda V$, the Lie bracket product rule givesThese formulas hold for every smooth $\lambda$, with no curvature or Anosov assumption.

Follow a tangent variation $\xi (t)=D{\varphi }_t\xi (0)$ and express it in the moving frame asPulling this identity back by $D{\varphi }_{-t}$ makes its left side constant. The derivative of the pulled-back frame vector $B$ is the pulled-back bracket $[F,B]$, so the coefficients satisfyThis derives the linearized transverse equation for a surface thermostat; its first equation in $a$ disappears on the quotient by $F$.

Near a vertical plane, $z\ne 0$, and the smooth projective coordinate $r=y/z$ describes the plane asThe vertical Maslov cycle of a surface thermostat is exactly $r=0$. Differentiating the quotient with the derived equations yieldsConsequentlySince $r$ is a local defining function for the hypersurface, the nonzero normal derivative proves that the lifted vector field is transverse to the Maslov cycle everywhere. In particular, the sign follows from the forward derivative lift and the positive vertical-rotation convention, rather than from an unspecified projective orientation.

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.