Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 17 / 2

The Livsic theorem says that for a topologically transitive Anosov flow on a closed manifold, a Hölder continuous function $f$ is a flow coboundary if and only if it has zero integral over every periodic orbit. In the integral formulation there is a Hölder continuous function $u$ satisfyingEquivalently, $u$ is differentiable along the flow and $Fu=f$. The periodic obstruction isIt is necessary because the left-hand side of the coboundary identity is zero at a period. The substantive theorem is its sufficiency. For a smooth flow and smooth $f$, the smooth regularity version gives a smooth $u$. Two such solutions differ by a constant: their difference is flow invariant and hence constant on a dense orbit and, by continuity, everywhere. This smooth version is the one needed in part (b).

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.

No: the round sphere supplies a counterexample. On the unit round $S^2\subset {\mathbb{R}}^3$, restrict the smooth one-formThe antipodal map $A(x)=-x$ satisfies $A^{\ast }\theta =-\theta$, since $x_3^2$ is unchanged and $d{x}_1$ changes sign. Every unit-speed closed geodesic is a great circle, with $\gamma (t+\pi )=A(\gamma (t))$. Therefore its two half-circle integrals cancel:The same holds for all repeated traversals and either orientation.

However, its exterior derivative iswhich is not zero on the sphere. For example, at $p=(0,1/\sqrt{2},1/\sqrt{2})$ the tangent vectors $u=(1,0,0)$ and $w=(0,1/\sqrt{2},-1/\sqrt{2})$ give $d{\theta }_p(u,w)=1$. Hence $\theta$ is not a closed differential form, and cannot be an exact differential form, because $d(dh)=0$. This is the counterexample to geodesic one-form injectivity on the round sphere; periodic geodesic integrals alone do not imply exactness without an additional geometric hypothesis.