Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 17 (The X-ray Transform in Geometry and Dynamics)

Let $F$ be the generator of the smooth flow. The Anosov flow property requires $F$ to be nowhere zero and a continuous splitting of the tangent bundlewhich is invariant under $D{\varphi }_t$. There must be constants $C\ge 1$ and $\kappa >0$ such that, for every $x$, every $t\ge 0$ and every vector in the indicated fibre,The first inequality defines the Anosov stable bundle; the second defines the Anosov unstable bundle and is equivalent to uniform forward expansion there. The remaining one-dimensional direction is the flow direction. The norm can come from any Riemannian metric, since norms are uniformly equivalent on a closed manifold. The essential requirements are a continuous invariant splitting and estimates uniform in both the base point and time, rather than separate exponential rates along individual orbits.

Here is the quadratic-form criterion for an Anosov flow. Assume that the generator $F$ is nowhere zero and form the transverse vector bundleSince $D{\varphi }_{t}F(x)=F({\varphi }_{t}x)$, the derivative induces a linear cocycle $P_t:{\mathcal{E}}_x\to {\mathcal{E}}_{{\varphi }_{t}x}$. Suppose there is a continuous nondegenerate indefinite quadratic form $Q_x$ on each fibre, with constant signature of a quadratic form, such that $t\mapsto Q_{{\varphi }_{t}x}(P_t\xi )$ is differentiable andis continuous and positive definite on every transverse fibre. Then ${\varphi }_t$ is an Anosov flow. An equivalent way to specify $Q$ is a quadratic form on $TN$ whose radical is exactly $\mathbb{R}F$ and whose induced form and derivative satisfy these conditions on the quotient.

On the closed manifold, compactness makes the positivity uniform: for some $c>0$, $\mathcal{D}{Q}_x(\xi )\ge c\Vert \xi {\Vert }^2$. Thus the positive and negative cones are strictly separated by the linearized flow, yielding uniform unstable and stable transverse subbundles and their invariant lifts. The positive index is the unstable dimension and the negative index is the stable dimension. Strict positivity must be imposed transversely; imposing it on the flow direction would be impossible because that direction is carried into itself.

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.

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.

A simple Riemannian manifold has strictly convex boundary, no conjugate points, and a unique connecting geodesic depending smoothly on its endpoints. In particular its unit-speed geodesics are nontrapping. Write $\nu$ for the inward unit normal, and identify inward and outward parts of its boundary unit tangent bundle byThe signs are a convention; here $+$ means inward. If $(x,v)$ is strictly inward, let $\tau (x,v)>0$ be its first exit time. The geodesic scattering relation isThus it records both the exit point and the outward unit velocity, not just the exit point. On strictly outward vectors extend $\alpha$ by following the same geodesic flow backwards to its previous entry point, retaining the velocity orientation. On tangential vectors set $\alpha (x,v)=(x,v)$. The extended scattering relation is an involution on the full boundary unit bundle, since its forward and backward constructions undo one another. On a simple Riemannian manifold, the travel time equals the boundary distance function between the two endpoints; travel time is often recorded separately as lens data.

First recover the induced boundary metric from the boundary distance function. If $c(t)$ is a smooth boundary curve with $c(0)=x$ and $\dot{c}(0)=w\in T_x\partial M$, the local metric expansion of distance givesOne can obtain this expansion in local coordinates from the fact that the metric tends uniformly to its value at $x$; distances to nearby points have that constant metric's first-order norm. Equality of the two boundary distance functions makes the two quadratic values equal for every boundary-tangent $w$. Polarization therefore recovers the full bilinear induced metric:This does not yet identify the metrics on normal vectors or mixed tangent-normal pairs.

Let ${\nu }_i$ be the inward unit normal for $g_i$. The boundary normal coordinatesare collar neighborhoods for a common sufficiently small $\epsilon$. The Gauss lemma gives $c_i^{\ast }{g}_i=d{r}^2+h_i(r)$ with $h_i(0)=h$. Hence the local boundary-fixing map $c_2\circ c_1^{-1}$ already satisfiesbecause both pulled-back metrics there are $d{r}^2+h$.

To justify a global diffeomorphism, rather than assume a local collar map extends, interpolate the positive metrics by $g_s=(1-s)g_1+s{g}_2$, $0\le s\le 1$. Their inward unit normals and exponential maps give smoothly varying collarsCompactness of $[0,1]\times \partial M$ gives a uniform collar size after shrinking $\epsilon$. On the image of each collar define the time-dependent vector fieldMultiply it by a smooth cutoff which is one for small $r$ and zero before the collar's outer edge, and extend it by zero to the rest of $M$. This gives a smooth global time-dependent vector field. It vanishes on the boundary, because $C_s(x,0)=x$ for all $s$. Its flow ${\Psi }_s$ exists for the full interval and fixes the boundary. For sufficiently small fixed $r$, the curve $s\mapsto C_s(x,r)$ solves this flow equation, so uniqueness givesHere $C_0=c_1$ and $C_1=c_2$. Thus the final map $\psi ={\Psi }_1$ agrees with the local map from the $g_1$ collar to the $g_2$ collar near the boundary. In particular $D\psi$ fixes tangential vectors and sends ${\nu }_1$ to ${\nu }_2$ there. Since the tangent-normal splittings are orthogonal and the normals have unit length,The construction only needs equality of the induced boundary metrics; the equal-distance hypothesis supplies that equality.

Let $\rho (x,y)$ be the common boundary distance function, with $x\ne y$. For each metric, the simple Riemannian manifold property gives a unique unit-speed geodesic from $x$ to $y$, smoothly depending on the endpoint pair. Denote its initial and terminal velocities by $v_i,w_i$.

The first variation of geodesic energy, or equivalently length variation at unit speed, gives the endpoint formulaFor completeness, if $J$ is an endpoint variation along a geodesic, differentiating its length and integrating the covariant derivative by parts givesThe integral vanishes by the geodesic equation, leaving precisely these two boundary terms and their opposite signs.

Because the full metrics agree on the boundary tangent spaces $T_{x}M$, the induced boundary metric $h$, unit spheres and inward unit normals agree there. The displayed covectors therefore determine the same tangential velocity components for the two metrics:Their normal components are fixed by unit length and the entry or exit sign. With inward unit normals ${\nu }_x,{\nu }_y$,Strict convexity makes these signs strict for distinct endpoints. Thus $v_1=v_2$ and $w_1=w_2$ for the same endpoint pair.

Now start with any strictly inward $(x,v)$ and let its $g_1$ exit point be $y$. The $g_2$ geodesic joining $x$ to this same $y$ has the same initial velocity $v$, so uniqueness of the geodesic equation identifies it with the $g_2$ trajectory launched from $(x,v)$. Its exit velocity is also the same. HenceThis is the boundary distance determines geodesic scattering argument. The extension to outward vectors follows by the inverse scattering relation, and tangential vectors are fixed for both metrics. Equality of the travel times follows as well, since both equal $\rho (x,y)$.

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.

Use the derivative to transport the planes:The derivative is an invertible linear map, so it sends a two-dimensional subspace to a two-dimensional subspace. The flow identity ${\varphi }_t\circ {\varphi }_s={\varphi }_{t+s}$ givesHence a plane containing the generator is sent to another plane containing the generator; the displayed formula is a well-defined smooth lift to $\Lambda (SM)$. The chain rule gives ${\varphi }_{t+s}^{\ast }={\varphi }_t^{\ast }\circ {\varphi }_s^{\ast }$, and ${\varphi }_0^{\ast }$ is the identity. Thus it is a genuine lifted smooth flow, with inverse ${\varphi }_{-t}^{\ast }$, rather than merely a family of maps of individual fibres. Its generator is $F^{\ast }={{\partial }_t{\varphi }_t^{\ast }|}_{t=0}$.

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.

Fix an orientation and a positive volume form $\Omega$. Define the divergence by the Lie derivativeA smooth function is a flow coboundary when it equals $Fu$ for some smooth $u$. Every other positive volume form is $e^h\Omega$ for a smooth $h$. The product rule for the Lie derivative givesThus changing the reference volume form changes its divergence by a flow coboundary. If volume forms of either orientation are allowed, take the logarithm of the absolute ratio on each connected component; multiplying a reference form by a constant sign leaves its divergence unchanged.

If ${\mathrm{div}}_{\Omega }F=Fu$, the preceding product rule showsConversely, if $e^h\Omega$ is invariant, then ${\mathrm{div}}_{\Omega }F=-Fh=F(-h)$ is a flow coboundary. Vanishing of the Lie derivative is equivalent to flow invariance, since ${\partial }_t{\varphi }_t^{\ast }\Omega ={\varphi }_t^{\ast }({\mathcal{L}}_F\Omega )$. HenceThe equivalence holds for every choice of reference volume form, proving the invariant volume criterion for a smooth flow. The smoothness requirement matters: a merely measurable transfer function does not give a smooth invariant volume form.

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.

Use the following permitted kernel result, stated in the form needed here. For an Anosov flow generated by a smooth surface thermostat on a closed oriented Riemannian surface, if a smooth function $u$ on $SM$ satisfieswhere $h$ is a smooth base function and $\eta$ is a smooth one-form, then $h=0$ and $\eta$ is an exact differential form. Conversely these two conditions give a smooth solution by taking a base primitive. This is the kernel of the thermostat X-ray transform in degrees zero and one. In particular it applies to Gaussian thermostats without an additional curvature condition; it is stronger than the magnetic or geodesic special case.

Put $\theta =E^{♭}$, so ${\theta }_x(v)=⟨E(x),v⟩$. Suppose the thermostat preserves a smooth volume form. Write it as $\rho \mu$ with smooth $\rho >0$, choosing its sign separately on each component if necessary. By the invariant volume criterion for a smooth flow and the divergence computation in part (b),Thus $u=\log \rho$ is a smooth solution of $Fu={\theta }_x(v)$. Apply the stated kernel theorem with $h=0$ to obtain $\theta =df$ for a smooth base function $f$. The definition of the gradient then gives $E={\mathrm{grad}}_{g}f$.

Conversely, suppose $E={\mathrm{grad}}_{g}f$. Since $d\pi (F)=v$, the vertical part has no effect on a base function andThe divergence of a Gaussian thermostat and the volume-change formula therefore giveConsequentlyOnly the necessary direction uses the Anosov kernel theorem; the displayed invariant density works for any potential Gaussian thermostat. In particular its exponent is positive, as required to cancel the negative divergence found in part (b).