Geodesic X-ray transforms on simple surfaces

In a boundary chart the metric coefficients are smooth up to the boundary, so its Levi-Civita Christoffel coefficients extend locally to smooth coefficients on an open chart. The geodesic equation is the smooth first-order system $({\dot{x}}^i,{\dot{v}}^k)=(v^i,-{\mathrm{\Gamma }}_{ij}^k(x)v^i{v}^j)$ on the chart tangent coordinates.

On an oriented coordinate patch choose a positive orthonormal frame $e_1,e_2$ and write $\mathrm{\nabla }{e}_1=\omega e_2$, $\mathrm{\nabla }{e}_2=-\omega e_1$. With $v=\cos \theta e_1+\sin \theta e_2$ and $v^{\perp }=-\sin \theta e_1+\cos \theta e_2$, the fibre density is $d\theta$ and $d\mathrm{\Sigma }=d{V}_{g}d\theta$. A change of positive orthonormal frame changes $\theta$ by a smooth rotation angle depending on $x$; its coordinate Jacobian is triangular with fibre derivative $1$, so this density is independent of the frame.

The local ODE result in [F4] gives unique geodesic arcs with smooth dependence on initial state. The coordinate equations agree on overlaps because they are the covariant equation $D_t\dot{\gamma }=0$; metric compatibility gives $\frac{d}{dt}g(\dot{\gamma },\dot{\gamma })=2g(D_t\dot{\gamma },\dot{\gamma })=0$. Thus the local arcs patch uniquely on their common intervals and unit-speed initial states remain in $SM$ until exit.

The horizontal lifts are $H_i=e_i-\omega (e_i)V$ and $X=\cos \theta H_1+\sin \theta H_2$. At fixed $\theta$, the base divergence of $v$ is ${\mathrm{div}}_{g}v=\omega (v^{\perp })$, while the vertical coefficient of $X$ is $-\omega (v)$ and its $\theta$-derivative is $-\omega (v^{\perp })$. Hence ${\mathrm{div}}_{d\mathrm{\Sigma }}X=0$. Also ${\mathrm{div}}_{d\mathrm{\Sigma }}V=0$ because $V={\mathrm{\partial }}_{\theta }$, and ${\mathrm{div}}_{d\mathrm{\Sigma }}{X}^{\perp }=0$ because $X^{\perp }=[X,V]$ and $\mathrm{div}[X,V]=X(\mathrm{div}V)-V(\mathrm{div}X)$. The local flow equation for the density is $\frac{d}{dt}{\phi }_t^{\ast }d\mathrm{\Sigma }={\phi }_t^{\ast }({\mathrm{div}}_{d\mathrm{\Sigma }}Xd\mathrm{\Sigma })=0$, so the geodesic flow preserves $d\mathrm{\Sigma }$ until boundary exit.

Write ${\omega }_i=\omega (e_i)$. The curvature definition gives $R(e_1,e_2)e_1=d\omega (e_1,e_2)e_2$ and $R(e_1,e_2)e_2=-d\omega (e_1,e_2)e_1$, so $d\omega =-Kd{V}_g$ for $K=g(R(e_1,e_2)e_2,e_1)$. First-slot skewness follows from [F6]; last-slot skewness follows by applying metric compatibility in [F3] to $g(R(A,B)C,D)+g(C,R(A,B)D)$. Under a positive rotation of $(e_1,e_2)$, each alternating pair in $g(R(e_1,e_2)e_2,e_1)$ is multiplied by the determinant $1$, so $K$ is frame-independent. Torsion-freeness gives $[e_1,e_2]=-{\omega }_1{e}_1-{\omega }_2{e}_2$ and direct expansion of $H_i=e_i-{\omega }_{i}V$ gives $[H_1,H_2]=-{\omega }_1{H}_1-{\omega }_2{H}_2+KV$. Put $c=\cos \theta$, $s=\sin \theta$, $X=c{H}_1+s{H}_2$, and $X^{\perp }=s{H}_1-c{H}_2$. Since $V(c)=-s$ and $V(s)=c$, the definitions give $[X,V]=X^{\perp }$ and $[V,X^{\perp }]=X$. In $[X,X^{\perp }]$, the horizontal-frame bracket contributes $-[H_1,H_2]={\omega }_1{H}_1+{\omega }_2{H}_2-KV$; differentiating the coefficients $c,s$ along $H_i$ contributes $-{\omega }_1{H}_1-{\omega }_2{H}_2$, leaving $[X,X^{\perp }]=-KV$.

Let $\rho$ be a boundary defining function, positive in $M^{\circ }$. At a boundary point, ${\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}=-\mathrm{\nabla }\rho \mathrm{/}\mathrm{\mid }d\rho {\mathrm{\mid }}_g$. For tangent $w$, differentiation gives $\mathrm{Hess}\rho (w,w)=-\mathrm{\mid }d\rho {\mathrm{\mid }}_{g}g({\mathrm{\nabla }}_w{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}},w)<0$. Along a geodesic tangent to the boundary, $\frac{d^2}{d{t}^2}\rho (\gamma (t))=\mathrm{Hess}\rho (\dot{\gamma },\dot{\gamma })<0$, so the geodesic lies outside $M$ on both sides near tangency. Therefore a geodesic arriving from the interior exits transversely. For an interior initial state $z_0$, nontrapping gives a finite first exit time $t_0={\tau }_{+}(z_0)$. In a local chart of $SM$ near $z_0$, choose a boundary defining function $\rho$ near the exit and extend the geodesic ODE locally across the boundary in a boundary chart. A finite composition of the local flows from [F4] gives the smooth map $F(z,t)=\rho ({\phi }_t(z))$ near $(z_0,t_0)$. Transversality gives $F(z_0,t_0)=0$ and ${\mathrm{\partial }}_{t}F(z_0,t_0)=d{\rho }_{{\gamma }_{z_0}(t_0)}({\dot{\gamma }}_{z_0}(t_0))\mathrm{\ne }0$. For each finite $k\ge 1$, [F8] gives a local $C^k$ root $t=T(z)$; applying this for every $k$ makes $T$ smooth. For each $ϵ>0$ with $ϵ<t_0$, the compact arc ${\gamma }_{z_0}([0,t_0-ϵ])$ and compact boundary are disjoint, so their positive distance in the metric topology [F7] keeps nearby arcs interior on that interval. Transversality gives a unique crossing near $t_0$; shrinking the neighborhood therefore makes the IFT root the first exit for nearby interior states. Thus the exit time is smooth near every interior initial state.

For $Y\in \{X,V,X^{\perp }\}$, the local density formula for divergence gives ${\mathrm{div}}_{d\mathrm{\Sigma }}(abY)=(Ya)b+a(Yb)$ because ${\mathrm{div}}_{d\mathrm{\Sigma }}Y=0$. Integrating in coordinate charts and applying one-variable integration by parts gives the boundary flux term ${\int }_{\mathrm{\partial }SM}ab{\iota }_{Y}d\mathrm{\Sigma }$. It vanishes for $X$ and $X^{\perp }$ when $ab=0$ on $\mathrm{\partial }SM$; for $V$ it vanishes for arbitrary $a,b$ because $V$ rotates only the fibre and is tangent to $\mathrm{\partial }SM$. A finite chart cover and a partition of unity sum these local identities to yield the asserted global formulas.

Fix a unit-speed segment $\gamma :[0,L]\to M$. Since $M$ is oriented, let $Q$ be positive rotation by $90$ degrees on each tangent plane. In any local positive orthonormal frame $e_1,e_2=Q{e}_1$, metric compatibility gives ${\mathrm{\nabla }}_A{e}_1=\omega (A)e_2$ and ${\mathrm{\nabla }}_A{e}_2=-\omega (A)e_1$; for $e_1$, $({\mathrm{\nabla }}_{A}Q)e_1={\mathrm{\nabla }}_A{e}_2-Q(\omega (A)e_2)=-\omega (A)e_1+\omega (A)e_1=0$, and for $e_2$, $({\mathrm{\nabla }}_{A}Q)e_2={\mathrm{\nabla }}_A(-e_1)-Q(-\omega (A)e_1)=-\omega (A)e_2+\omega (A)e_2=0$. Thus $\mathrm{\nabla }Q=0$ and $n(t)=Q\dot{\gamma }(t)$ is a parallel unit normal along the whole segment. By [F5], the scalar coefficient $j$ in the Jacobi field $jn$ solves $j^{\mathrm{\prime }\mathrm{\prime }}+K(\gamma (t))j=0$. Let $j(0)=0$, $j^{\mathrm{\prime }}(0)=1$. This solution is positive for $0<t\le L$: it is positive near $0$, and a later zero would give a nonzero Jacobi field vanishing at both endpoints of a positive-length subsegment, contradicting [F1].

If $\mathrm{supp}f=\mathrm{\varnothing }$, then $f=0$ and $u_f=0$, so the transport claim is immediate; assume henceforth that the support is nonempty. The nontrapping hypothesis and existence of an interior unit geodesic from step 2.1 imply $\mathrm{\partial }M\mathrm{\ne }\mathrm{\varnothing }$. The unit circle bundle $SM$ is compact: cover compact $M$ by finitely many smaller coordinate neighborhoods with closures inside charts carrying smooth orthonormal frames; each part of $SM$ over such a closure is homeomorphic to the product of that compact closure with $S^1$, so finitely many compact sets cover $SM$. Define $D=\min \{d_g(x,y):x\in \mathrm{\partial }M,y\in \mathrm{supp}f\}$. These disjoint sets are compact in the metric topology by [F7], so the minimum is attained and is positive because $d_g$ is a metric. For a subset $A$, write $d_g(x,A)={\inf }_{a\in A}{d}_g(x,a)$.

Every interior state has a unique backward geodesic to its first boundary hit, and transversality in step 3.1 puts that state in ${\mathrm{\partial }}_{+}SM$. Thus $(z,t)\mapsto {\phi }_t(z)$ maps $z\in {\mathrm{\partial }}_{+}SM$, $0<t<{\tau }_{+}(z)$ bijectively onto $S{M}^{\circ }$. In a boundary coordinate and the fibre angle, adjoining the flow-time column to the two boundary-state tangent columns gives the volume determinant $\mathrm{\mid }g(v,{\nu }_{\mathrm{i}\mathrm{n}})\mathrm{\mid }=\mu (z)$ at $t=0$: the fibre column supplies $d{S}_x(v)$, the boundary column supplies $d{S}_{\mathrm{\partial }M}(x)$, and the component of the flow column normal to $\mathrm{\partial }M$ is $g(v,{\nu }_{\mathrm{i}\mathrm{n}})$. Since the flow preserves $d\mathrm{\Sigma }$ by step 2.2, that determinant is constant along each flow line. The change-of-variables formula therefore gives Santaló's identity; tangential boundary states form a measure-zero set and contribute nothing.

Write an endpoint-zero normal field as $Y=yn$ and set $q=y\mathrm{/}j$ on $(0,L]$. Since $y(0)=j(0)=0$ and $j^{\mathrm{\prime }}(0)=1$, both functions factor smoothly as $t$ times a smooth function and $q$ extends smoothly to $0$; also $q(L)=0$. Expanding $y=jq$ and using $j^{\mathrm{\prime }\mathrm{\prime }}+Kj=0$ gives $(y^{\mathrm{\prime }}{)}^2-K{y}^2=j^2(q^{\mathrm{\prime }}{)}^2+(j{j}^{\mathrm{\prime }}{q}^2{)}^{\mathrm{\prime }}$. The boundary term $j{j}^{\mathrm{\prime }}{q}^2$ vanishes at $0$ because $j(0)=0$ and at $L$ because $q(L)=0$. Consequently ${\mathcal{I}}_{\gamma }(Y,Y)={\int }_0^L{j}^2(q^{\mathrm{\prime }}{)}^{2}dt\ge 0$, with equality only if $q$ is constant and hence zero from $q(L)=0$. Thus the index form is positive definite.

If $d_g(x,\mathrm{\partial }M)<D\mathrm{/}4$, the triangle inequality gives $d_g(x,\mathrm{supp}f)>3D\mathrm{/}4$; hence a unit-speed curve from $x$ cannot reach the support before time $D\mathrm{/}2$, since its length equals elapsed time and endpoint distance is at most length. Fix any $z=(x,v)\in \mathrm{\partial }SM$ and a boundary chart with defining coordinate $\rho$ positive inside. If $v$ is transverse, choose the one of $v,-v$ pointing outward; after shrinking a neighborhood, ${\rho }^{\mathrm{\prime }}(t)<-c<0$ up to some $t_0<D\mathrm{/}2$, and nearby interior states have $\rho (0)<c{t}_0$, so this ray exits by $t_0$. If $v$ is tangent, step 3.1 gives ${\rho }^{\mathrm{\prime }\mathrm{\prime }}(0)<0$. Choose $t_0<D\mathrm{/}2$ small enough to remain in the chart; after shrinking a neighborhood, ${\rho }^{\mathrm{\prime }\mathrm{\prime }}(t)<-c<0$ for $0\le t\le t_0$, while $\rho (0)<c{t}_0^2\mathrm{/}8$ and $\mathrm{\mid }{\rho }^{\mathrm{\prime }}(0)\mathrm{\mid }<c{t}_0\mathrm{/}4$. Taylor's formula yields $\rho (t_0)\le \rho (0)+t_0{\rho }^{\mathrm{\prime }}(0)-c{t}_0^2\mathrm{/}2<0$, so the geodesic has exited by $t_0$. In either case there is a neighborhood of $z$ and one fixed orientation whose forward ray exits before reaching $\mathrm{supp}f$. Shrink each such neighborhood so every state in it also has base point at distance less than $D\mathrm{/}4$ from $\mathrm{\partial }M$. Compactness of $\mathrm{\partial }SM$ from step 3.4 gives a finite subcover, whose union contains an open neighborhood of $\mathrm{\partial }SM$ with the same property.

At every non-tangent boundary state, the relevant first boundary hit is transverse: it is the time-zero hit for an outgoing state and the finite future exit for an incoming state. In local coordinates for the extension across the boundary, choose a boundary defining function near the hit and extend the geodesic ODE. The finite-time smooth dependence established in step 3.1 and [F8] give a smooth local root of $F(z,t)=\rho ({\phi }_t(z))=0$. For an outgoing state the nonzero outward derivative makes this root the first exit for nearby interior states. For an incoming state the initial inward derivative gives a short interior segment; each compact middle segment before the final hit has positive distance from $\mathrm{\partial }M$ by [F7], and the final transverse crossing is the IFT root, so it remains the first exit for nearby states. Therefore the integration endpoint and integral defining $u_f$ depend smoothly on the initial state near every non-tangent boundary state. In the neighborhood from step 4.3, one of $u_f(x,v)$ and $u_f(x,-v)$ is zero by the support separation. Their sum is the full boundary-to-boundary integral of $f$ along the same unoriented geodesic, which is $I_{0}f=0$ by [F1]; therefore both vanish throughout that neighborhood. Thus $u_f$ is zero near $\mathrm{\partial }SM$, removing the glancing singularity and making it smooth with compact support in $S{M}^{\circ }$, since $SM$ is compact by step 3.4. Along the flow, $u_f({\phi }_s(x,v))={\int }_s^{{\tau }_{+}(x,v)}f({\phi }_t(x,v))dt$, so differentiation gives $X{u}_f=-f$. The incoming boundary value is $I_{0}f=0$ and the outgoing value is zero because its future exit time is zero.

By [F2], $\pi$ is a two-sheeted covering; applying [F3] to compact $M$ shows that $\tilde{M}$ is compact. In each boundary chart the two sheets are copies of the same half-space chart, so ${\pi }^{-1}(\mathrm{\partial }M)=\mathrm{\partial }\tilde{M}$ and ${\pi }^{-1}(M^{\circ })={\tilde{M}}^{\circ }$.

Every connected component $C$ of the smooth manifold $\tilde{M}$ is open, since each point has a connected chart neighborhood contained in its component, and it is closed, since connected components are closed. Thus $C$ is compact. Smooth boundary charts have path-connected smaller half-ball neighborhoods, so $M$ is locally path-connected and [F8] makes connected $M$ path-connected. Fix $\tilde{x}\in C$ and any $y\in M$; lift a path from $\pi (\tilde{x})$ to $y$ starting at $\tilde{x}$ by [F6]. Its connected image stays in $C$, so its endpoint lies over $y$ and $\pi (C)=M$. The canonical orientation from [F2] restricts to $C$.

The pullback $\tilde{g}={\pi }^{\ast }g$ is smooth and positive definite because $d\pi$ is a smooth isomorphism on each sheet and $g$ is smooth positive definite by [F4]. In the sheet coordinates of [F2], $\pi$ is the identity and the metric coefficient matrices of $\tilde{g}$ and $g$ coincide. The unique Levi-Civita connections therefore have identical coordinate coefficients; the geodesic equations and curvature coefficients also coincide by [F4, F5]. Boundary normals and their second fundamental forms correspond under $d\pi$, so strict convexity of $\mathrm{\partial }M$ gives strict convexity of $\mathrm{\partial }C$.

If $\tilde{J}$ is a Jacobi field along a geodesic segment in $C$, then [F5] and the identical local connection and curvature coefficients in step 2.1 show that $d\pi (\tilde{J})$ is a Jacobi field along the projected geodesic. If $\tilde{J}$ vanishes at both endpoints, its projection vanishes there; [F1] makes the projected field zero, and invertibility of $d\pi$ gives $\tilde{J}=0$. Thus $C$ has no conjugate points.

A unit-speed geodesic in $C^{\circ }$ projects by step 2.1 to a unit-speed geodesic in $M^{\circ }$. By [F1] that projected geodesic reaches $\mathrm{\partial }M$ in finite forward and backward time. Lift each finite projected segment from the specified initial point by [F6]. In every sheet the lift solves the same geodesic ODE as the original curve, so uniqueness from [F7] makes it an extension of that curve to the exit time; the endpoint lies in $\mathrm{\partial }C$ by step 1.1. Hence $C$ is nontrapping in both directions. Together with steps 1.2–2.1, $C$ is simple.

Since $f$ is supported in the interior, $\tilde{f}=f\circ \pi$ is smooth and vanishes near $\mathrm{\partial }\tilde{M}$. Its support is a closed subset of compact $\tilde{M}$, hence compact, and it lies in $C^{\circ }$ after restriction to $C$. For an incoming $\tilde{z}\in {\mathrm{\partial }}_{+}SC$, local isometry maps its inward normal component to that of $z=\pi (\tilde{z})\in {\mathrm{\partial }}_{+}SM$. The projected unit geodesic and the lifted unit geodesic have the same parameter and first boundary-exit time by steps 1.1 and 4.1; hence $I_{0,C}\tilde{f}(\tilde{z})={\int }_0^{{\tau }_{+}(z)}f({\gamma }_z(t))dt=I_{0,M}f(z)$. This proves the transform identity and the zero-data implication.

Since $V$ rotates the vector within its fixed fibre, it is tangent to $\mathrm{\partial }SM$. The restriction of $u$ to $\mathrm{\partial }SM$ is zero, so its derivative along every tangent vector there is zero; hence $Vu=0$ on $\mathrm{\partial }SM$. In particular, the boundary products $(Vu)(X^{\perp }u)$ and $(Vu)(Xu)$ vanish, as needed for the $X$ and $X^{\perp }$ integration-by-parts formulas in [F3].

Put $P=XVu$ and $Q=X^{\perp }u$. From $[X,V]=XV-VX=X^{\perp }$ we have $VXu=P-Q$, and therefore $\mathrm{\parallel }VXu{\mathrm{\parallel }}^2=\mathrm{\parallel }P{\mathrm{\parallel }}^2-2(P,Q)+\mathrm{\parallel }Q{\mathrm{\parallel }}^2$.

Integrating $X$ by parts with boundary factor $(Vu)(X^{\perp }u)=0$, then using $X{X}^{\perp }=X^{\perp }X-KV$, gives $(P,Q)=-(Vu,X{X}^{\perp }u)=-(Vu,X^{\perp }Xu)+(KVu,Vu)$. Integrating $X^{\perp }$ by parts with boundary factor $(Vu)(Xu)=0$ turns the first term into $(X^{\perp }Vu,Xu)$.

The commutator $[V,X^{\perp }]=X$ implies $X^{\perp }Vu=V{X}^{\perp }u-Xu$. Thus $(X^{\perp }Vu,Xu)=(V{X}^{\perp }u,Xu)-\mathrm{\parallel }Xu{\mathrm{\parallel }}^2=-(X^{\perp }u,VXu)-\mathrm{\parallel }Xu{\mathrm{\parallel }}^2$, where the last equality is integration by parts for $V$, whose vector field is tangent to $\mathrm{\partial }SM$. As $VXu=P-Q$, this gives $(P,Q)=-(Q,P-Q)-\mathrm{\parallel }Xu{\mathrm{\parallel }}^2+(KVu,Vu)=-(P,Q)+\mathrm{\parallel }Q{\mathrm{\parallel }}^2-\mathrm{\parallel }Xu{\mathrm{\parallel }}^2+(KVu,Vu)$. Solving gives $2(P,Q)=\mathrm{\parallel }Q{\mathrm{\parallel }}^2-\mathrm{\parallel }Xu{\mathrm{\parallel }}^2+(KVu,Vu)$.

Substitute this value in step 2.1 and cancel $\mathrm{\parallel }Q{\mathrm{\parallel }}^2$. The result is $\mathrm{\parallel }VXu{\mathrm{\parallel }}^2=\mathrm{\parallel }P{\mathrm{\parallel }}^2+\mathrm{\parallel }Xu{\mathrm{\parallel }}^2-(KVu,Vu)=\mathrm{\parallel }XVu{\mathrm{\parallel }}^2-(KVu,Vu)+\mathrm{\parallel }Xu{\mathrm{\parallel }}^2$, which is the asserted identity.