Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 14 / 2

The Livsic theorem for a topologically transitive Anosov flow says that a Hölder function $h:N\to \mathbb{R}$ satisfies the periodic-orbit obstructionif and only if there is a Hölder function $u$ satisfying the flow coboundary identityThe solution is differentiable along orbits and $Fu=h$ in that sense. For a smooth flow and smooth $h$, the smooth regularity version gives $u\in C^{\infty }(N)$, so $Fu=h$ holds smoothly. The solution is unique up to an additive constant: the difference of two solutions is invariant and must be constant on a dense orbit and then everywhere by continuity. Necessity of the periodic obstruction follows immediately by taking $t=T$ in the displayed identity.

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 given Pestov identity for a geodesic flow for the function $\phi$ constructed above. Since $VX\phi =-2H\phi$, its left side is $-4\Vert H\phi {\Vert }_{L^2(\mu )}^2$. Moving this to the right givesAll terms are nonnegative, and $-K$ is uniformly positive by compactness. The smooth derivatives therefore vanish pointwise:Since $X,H,V$ span the tangent bundle, $\phi =c$ is constant, component by component if necessary.

On each circle fibre, $V^{2}u+u=c$ is the elementary periodic differential equation $u_{\vartheta \vartheta }+u=c$. Its solutions areThe first harmonics transform as a covector when the local oriented orthonormal frame changes. Their coefficients are smooth, either directly from the smooth fibre equation or from their sine/cosine integral formulas. Thus they define a smooth one-form $\eta$ on $M$ with $u(x,v)=c+{\eta }_x(v)$.

Along a geodesic, the velocity is parallel, so differentiating givesSet $Z={\eta }^{♯}/2$. ThenEquality initially holds on unit vectors, then on every vector by homogeneity. The polarization identity yields the tensor identityThis proves that $\beta$ is a potential symmetric 2-tensor, with the precise factor of two in the requested convention.