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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 $\rho (x,y)$.

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 $d_g(x,y)$.