Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 14 (The X-ray Transform in Geometry and Dynamics)

For a nonsingular smooth flow generated by $F$ on a closed manifold, the Anosov flow condition is a continuous $D{\varphi }_t$-invariant splittingand constants $C,\lambda >0$ such that, for all $t\ge 0$,The norm can come from any auxiliary Riemannian metric; compactness makes the resulting definitions equivalent. The centre direction is the flow direction, with $F$ nowhere zero.

One construction is a suspension flow. The hyperbolic toral automorphism of ${\mathbb{T}}^2$ defined byhas eigenvalues ${\lambda }_{\pm }=(3\pm \sqrt{5})/2$, with $0<{\lambda }_{-}<1<{\lambda }_{+}$. Form the closed three-dimensional mapping torus $({\mathbb{T}}^2\times [0,1])/((x,1)\sim (Ax,0))$ and move in the second coordinate at unit speed. Each crossing of the section applies $A$. The two one-dimensional eigenspaces give invariant transverse bundles, and after $k$ crossings their derivatives multiply by ${\lambda }_{-}^k$ and ${\lambda }_{+}^k$. The difference between $k$ and elapsed time is bounded, so any smooth Riemannian metric gives the required uniform exponential estimates. The vertical direction supplies the centre line.

A different construction is the geodesic flow on the unit tangent bundle of a closed oriented hyperbolic surface. This bundle is a closed three-manifold. In curvature $-1$, a normal Jacobi field along a unit-speed geodesic satisfies $j^{\prime \prime }-j=0$. The initial data $(j,j^{\prime })=(a,-a)$ give $j(t)=a{e}^{-t}$, while $(a,a)$ give $j(t)=a{e}^t$. The linearized geodesic flow transports these initial data as $(j(t),j^{\prime }(t))$, so they give one stable and one unstable direction with precisely those exponential rates. The remaining tangent direction is the geodesic vector field. Thus this also constructs an Anosov flow, using geometric rather than return-map hyperbolicity.

Put $\rho =1/f$ and denote the flow generated by $fF$ by ${\psi }_t$. It follows the same oriented orbits as ${\varphi }_s$, with the clock determined implicitly byBecause $f$ is positive on a compact manifold, $0<f_{\min }\le f\le f_{\max }$ and $f_{\min }t\le s(t,x)\le f_{\max }t$ for $t\ge 0$. Negative times obey the corresponding absolute-value comparison. This makes it possible to transfer exponential estimates once the correct transverse graphs are found.

For $v\in E^s(x)$ defineThe integrand is bounded by a constant times $e^{-\lambda r}\Vert v\Vert$, since $d\rho$ is bounded and $v$ is stable. The integral converges uniformly on bounded bundle vectors, so it is continuous and linear in $v$. Consider the graphWe verify its invariance and contraction directly; the old Anosov stable bundle need not itself remain invariant under a positive time change of a smooth flow.

For a fixed old time $s$, setDifferentiation gives $D_{x}T(v)=I_s(x,v)$ and $D_{x}T(F)=\rho ({\varphi }_{s}x)-\rho (x)$. Differentiating the implicit clock at fixed $t$, then differentiating ${\varphi }_{s(t,x)}(x)$, therefore yields for $w=v+aF(x)$For $a={\lambda }_s(x,v)$ this becomesThe tail coefficient equals ${\lambda }_s({\varphi }_{s}x,D{\varphi }_{s}v)$ by the flow composition law, proving graph invariance. Both the first term and the tail have norm at most $C^{\prime }{e}^{-\lambda s}\Vert v\Vert$. Projection onto $E^s$ in the original splitting is uniformly bounded, hence $\Vert v\Vert \le C^{\prime \prime }\Vert w\Vert$. The clock comparison now givesFor the Anosov unstable bundle, use the convergent past integralThe same differentiation formula gives the tail $-\rho ({\varphi }_{s}x{)}^{-1}{\int }_{-\infty }^{s}d\rho (D{\varphi }_{r}v)dr$ for $s\le 0$, proving invariance and exponential contraction in negative new time.

Finally, these two graphs and $\mathbb{R}F=\mathbb{R}(fF)$ form a direct splitting. Projecting onto $TN/\mathbb{R}F$ identifies the graphs with the old complementary stable and unstable subspaces, so they intersect neither each other nor the centre. We have established the continuous invariant splitting and both uniform estimates, and therefore the flow of $fF$ is Anosov.

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.

A simple Riemannian manifold has strictly convex boundary and a unique geodesic between each pair of points, smoothly dependent on its endpoints; there are no conjugate points and no trapped geodesics. Let $\nu$ denote the outward unit normal. For a boundary unit vector with $⟨v,\nu ⟩<0$, follow its unit-speed geodesic ${\gamma }_{x,v}$ until the next boundary intersection at time $\tau (x,v)>0$. Its geodesic scattering relation isThus it maps incoming boundary unit vectors to outgoing ones, recording both exit point and direction.

If the scattering relation is defined on the full boundary unit bundle, extend it on outgoing vectors by following the same oriented geodesic backwards to its previous boundary intersection, and fix tangential vectors. This extension is an involution, because the second application follows the same segment back to its starting point with the same velocity. The exit time is not a component of the scattering relation itself; together with it one obtains lens data. For a simple Riemannian metric, $\tau$ is the boundary distance between the endpoints of the segment.

First the common boundary distance function determines the induced tangential Riemannian metric. For a boundary curve $c(t)$ with $c(0)=x$ and $c^{\prime }(0)=w\in T_x\partial M$, the local first-order distance expansion givesFor clarity, the upper bound follows from the length of the boundary curve. In a small coordinate neighbourhood the Riemannian metric is arbitrarily close to its value at $x$, so the length of any sufficiently short joining path is bounded below by that constant-metric displacement norm with an arbitrarily small relative error. A minimizing short path cannot leave that neighbourhood. These two bounds prove the limit. Equality of the boundary distance functions then gives $|w{|}_{g_1}=|w{|}_{g_2}$. The polarization identity yieldsThis alone fixes no choice of normal coordinates, so a boundary-fixing gauge is needed to compare the full tangent-space Riemannian metrics.

Let $n_i$ be the inward unit normals and letbe sufficiently small boundary normal coordinates. Their pullbacks have the form $C_i^{\ast }{g}_i=d{r}^2+h_i(r)$, with $h_i(0)=h$. On a collar define $\psi =C_2\circ C_1^{-1}$. It fixes the boundary, has $D\psi (w)=w$ for tangential $w$, and sends $n_1$ to $n_2$. For tangent-space vectors $w+a{n}_1$ and $z+b{n}_1$ at the boundary, therefore,This proves the desired full Riemannian metric equality locally at the boundary.

We also justify extending this local identification to a global diffeomorphism. Interpolate the positive Riemannian metrics by $g_s=(1-s)g_1+s{g}_2$, for $0\le s\le 1$. Let $C_s$ be their inward unit-normal exponential collars. Compactness of the boundary and parameter interval gives a common small collar width for all $s$. The intermediate Riemannian metrics need not be simple; only these local collars are used.

On the image of $C_s$, define the time-dependent vector fieldwhere $\chi$ is one near zero and vanishes before the outer edge of the collar; extend $Y_s$ by zero elsewhere. This is a smooth vector field on $M$, vanishing at the boundary because $C_s(x,0)=x$. Its time-dependent flow ${\Psi }_s$ is a global diffeomorphism fixing the boundary. For every sufficiently small fixed $r$, the curve $s\mapsto C_s(x,r)$ solves its equation, since $\chi (r)=1$. Uniqueness gives ${\Psi }_s(C_0(x,r))=C_s(x,r)$. Consequently ${\Psi }_1$ agrees near the boundary with $C_2{C}_1^{-1}$. Taking $\psi ={\Psi }_1$ proves

Full equality of the Riemannian metrics at the boundary identifies their boundary unit bundles, outward normals and induced boundary Riemannian metrics. We show directly that the common boundary distance function determines both endpoint directions.

For distinct boundary points $x,y$, let $\gamma$ be the unique joining unit-speed geodesic, with initial velocity $v$ and final velocity $w$. The first variation of its length gives, for $a\in T_x\partial M$ and $b\in T_y\partial M$,Indeed, integration by parts in the length variation leaves just the endpoint terms, because the interior term contains ${\nabla }_{\dot{\gamma }}\dot{\gamma }=0$. The Riemannian metric is simple, so the off-diagonal endpoint dependence is smooth and these derivatives are valid.

Thus the tangential components are $v_T=-{\mathrm{grad}}_x^{\partial M}{d}_g(x,y)$ and $w_T={\mathrm{grad}}_y^{\partial M}{d}_g(x,y)$. Unit speed and the incoming/outgoing signs supply the missing normal components:All quantities on the right use only boundary distance and the common boundary Riemannian metric. Every strictly incoming vector for $g_1$ has some exit point $y$; the geodesic joining the same pair for $g_2$ has exactly the same reconstructed initial and final vectors. Hence both geodesic scattering relations give the same exit for that input. Tangential vectors are fixed, and the outgoing extensions agree as inverses. ThereforeIn fact their exit times also agree, since each is $d_g(x,y)$.

A coorientation gives a globally nonvanishing $C^2$ one-form $\zeta$ with $E=\ker \zeta$. For tangent fields $A,B$ in the integrable distribution,since integrability makes their bracket tangent to $E$. Thus $\zeta \wedge d\zeta =0$. Choose a transverse field $Y$ with $\zeta (Y)=1$. The $C^1$ one-form $\eta =-{\iota }_{Y}d\zeta$ then satisfies $d\zeta =\eta \wedge \zeta$, as is checked on $E$ and $Y$.

The Godbillon-Vey invariant in the unnormalized convention isThe three-form is continuous at the specified regularity, so its integral is defined. We prove independence of both choices by producing exact differences with $C^1$ primitives, allowing ordinary Stokes theorem.

First, for a fixed $\zeta$, another admissible auxiliary form has the form ${\eta }^{\prime }=\eta +a\zeta$. From $d^2\zeta =0$ and $d\zeta =\eta \wedge \zeta$, one gets $d\eta \wedge \zeta =0$. Expanding the two three-forms, using this relation and $\zeta \wedge d\zeta =0$, givesIts integral on the closed manifold vanishes.

Next any other defining form is ${\zeta }^{\prime }=h\zeta$ with $h$ nowhere zero. Let $g=\log |h|$. Thenso one may use ${\eta }_0^{\prime }=\eta +dg$. Its three-form differs byAgain the difference integrates to zero; any subsequent auxiliary choice is handled by the first calculation. This includes reversing the chosen coorientation, since $h$ can be negative. Thus the invariant, and the corresponding de Rham cohomology class, depends only on the oriented manifold and the foliation defined by $E$. Reversing the orientation of $N$, unlike reversing the coorientation of $E$, changes the sign of its numerical value.

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.

We use the following precise degree-zero/one kernel of the magnetic X-ray transform in degrees zero and one theorem, as permitted here: for a smooth Anosov magnetic flow on a closed oriented surface, ifwith $q$ a smooth base function and $\xi$ a smooth one-form, then $q=0$ and $\xi$ is exact. Exact one-forms give the converse, because $F(w\circ \pi )=dw(v)$. No assumption of pointwise negative Gaussian curvature is needed for this theorem. We now derive the contact obstruction from it rather than assume the desired rigidity conclusion.

Use the usual connected-surface convention. Since ${\int }_{M}f{\Omega }_a=0$, the top-degree de Rham cohomology class of $f{\Omega }_a$ vanishes: integration identifies $H_{\mathrm{dR}}^2(M)$ with $\mathbb{R}$. Choose a smooth one-form $\eta$ on $M$ satisfying $d\eta =f{\Omega }_a$, and setThe structure equations and the previous volume calculation giveThis primitive exists because of the zero-flux hypothesis, and need not itself be contact.

A useful elementary fact is that every smooth invariant function of an Anosov flow is constant on a connected manifold. If $Fh=0$, then $d{h}_x(v_s)=d{h}_{{\varphi }_{t}x}(D{\varphi }_t{v}_s)$, and boundedness of $dh$ with exponential contraction makes this zero as $t\to \infty$. Negative time does the same for an unstable vector. Also $dh(F)=0$. The Anosov splitting spans the tangent space, so $dh=0$. This avoids any extra ergodicity assumption.

Suppose a smooth contact form $\tau$ is preserved. Invariance and contraction imply $\tau (v_s)=\tau (D{\varphi }_t{v}_s)=0$ in the limit; negative time gives the same on unstable vectors. Thus $\tau$ annihilates both transverse bundles. Its value $\tau (F)$ cannot vanish anywhere, since otherwise $\tau$ would be zero at that point, contradicting the contact condition. It is a smooth invariant function and therefore a nonzero constant. Divide $\tau$ by that constant to arrange $\tau (F)=1$.

The Cartan formula for the Lie derivative now gives ${\iota }_{F}d\tau =0$. In dimension three, all two-forms annihilating $F$ are scalar multiples of ${\iota }_F\mu$, soThe scalar $k$ is smooth and nowhere zero by the contact condition. Both sides without $k$ are flow-invariant; hence $Fk=0$ and the preceding Anosov argument makes $k$ constant. It follows thatis a closed differential one-form.

We require the Liouville mean of such a closed differential form, and derive it explicitly. Let $R_s$ be fibre rotation and average $\zeta$ over $0\le s\le 2\pi$. Since $d\zeta =0$, the Cartan formula for the Lie derivative givesConsequently its rotational average differs from $\zeta$ by an exact form. For a closed rotationally invariant one-form $\overline{\zeta }$, the identity $d(\overline{\zeta }(V))={\mathcal{L}}_V\overline{\zeta }-{\iota }_{V}d\overline{\zeta }=0$ makes its vertical value a constant $a$. Subtracting $a\omega$ leaves a rotationally invariant horizontal form, which descends to a base form $\xi$. We have therefore proved the averaging closed one-forms on a surface unit tangent bundle decompositionThe last relation follows from closedness and $d\omega =-K{\pi }^{\ast }{\Omega }_a$.

Evaluation at $F$ gives $\zeta (F)={\xi }_x(v)+af+Fh$. The average of the linear expression ${\xi }_x(v)$ around each unit circle is zero. The second term has total integral $2\pi a{\int }_{M}f{\Omega }_a=0$. Finally ${\int }_{SM}Fhd\mu =0$ by volume preservation and Stokes theorem. Thusfor every closed differential one-form $\zeta$ in this zero-flux system. Apply this to $\tau +k\sigma$: the directional average of ${\eta }_x(v)$ also vanishes, so its integral is $(1+k){\mathrm{Vol}}_{\mu }(SM)$. Positivity of the volume forces $k=-1$. Hence $d\tau =d\sigma$ and $\zeta =\tau -\sigma$ is closed.

Using its proved decomposition and $\tau (F)=1$, we obtainThe stated magnetic transform kernel theorem gives $af=0$ and exactness of $\eta -\xi$, hence $d\eta =d\xi$. If $a=0$, then $d\xi =0$ and $f{\Omega }_a=d\eta =0$, so $f=0$. If $a\ne 0$, the equation $af=0$ already gives $f=0$. Necessity is proved.

Conversely, if $f=0$, then $F=X$. The canonical form $\alpha$ is contact because $\alpha \wedge d\alpha =-\mu$ never vanishes, andThus the flow preserves $\alpha$, provingIf disconnected surfaces are admitted, the flux condition must hold on each component for the exactness step; a single cancelling total flux does not imply componentwise exactness. The result above uses the standard connected interpretation of a surface.