Geodesic X-ray transforms on simple surfaces — Examples and Counterexamples

Simple disk. Let $M=\{x\in {\mathbb{R}}^2:\mathrm{\mid }x\mathrm{\mid }\le 1\}$ with the Euclidean metric and define

This is a nonzero smooth function compactly supported in $M^{\circ }$. The disk is simple: the outward normal is ${\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}(x)=x$ on the unit circle, so $g({\mathrm{\nabla }}_w{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}},w)=\mathrm{\mid }w{\mathrm{\mid }}^2>0$ for every nonzero tangent $w$; Euclidean Jacobi fields satisfy $J^{\mathrm{\prime }\mathrm{\prime }}=0$, so endpoint-zero fields vanish; and every straight geodesic in the compact disk exits in finite forward and backward time. Along the diameter entering at $((-1,0),(1,0))$,

so this nonzero compactly supported function is detected by the transform. The diagram locates the support and one measured chord; the integral calculation proves the claim.

diagram failed to render:
\begin{tikzpicture}[scale=2]
  \draw[thick] (0,0) circle (1);
  \fill[blue!18] (0,0) circle (0.5);
  \draw[red,very thick,->] (-1,0) -- (0.65,0);
  \draw[red,very thick] (0.65,0) -- (1,0);
  \node[below] at (-1,0) {$(-1,0)$};
  \node[below] at (1,0) {exit};
  \node[blue!60!black] at (0,0.72) {$\operatorname{supp}f$};
\end{tikzpicture}

A diameter chord crosses the compact support of the bump.

Trapped annulus. Let $M=[0,2]\times (\mathbb{R}\mathrm{/}2\pi \mathbb{Z})$ and equip it with $g=d{r}^2+a(r{)}^{2}d{\theta }^2$, where $a(r)=2+\frac{1}{2}\cos (2\pi r)>0$. Since $a^{\mathrm{\prime }}(1)=0$, the Christoffel symbols ${\mathrm{\Gamma }}_{\theta \theta }^r=-a{a}^{\mathrm{\prime }}$ and ${\mathrm{\Gamma }}_{r\theta }^{\theta }=a^{\mathrm{\prime }}\mathrm{/}a$ vanish at $r=1$. Thus $\gamma (t)=(1,{\theta }_0+t\mathrm{/}a(1))$ is a unit-speed closed geodesic of period $2\pi a(1)$, entirely in $M^{\circ }$. It never reaches the boundary, so this compact geometry fails the nontrapping condition in the theorem. This example makes no assertion about the boundary transform's nullspace.

A parallel circle at a critical radius is a closed trapped geodesic.

Conjugate points and closed-geodesic visibility failure. On the unit sphere $S^2\subset {\mathbb{R}}^3$ with its induced Riemannian metric, write $x_3$ for the third coordinate and define the closed-geodesic transform by $I_{\mathrm{c}\mathrm{l}}f(c)={\int }_0^{2\pi }f(c(t))dt$ for every unit-speed great circle $c(t)=a\cos t+b\sin t$ with orthonormal $a,b\in {\mathbb{R}}^3$. The nonzero function $f(x)=x_3$ is invisible to this transform, since $f(c(t))=a_3\cos t+b_3\sin t$ and both trigonometric integrals over $[0,2\pi ]$ vanish. Also the antipodes $p=(1,0,0)$ and $-p$ are conjugate along $\gamma (t)=(\cos t,\sin t,0)$: for ${\gamma }_s(t)=\cos t(1,0,0)+\sin t(0,\cos s,\sin s)$, each $t$-curve is a unit-speed great circle and its variation field at $s=0$ is $J(t)=\sin t(0,0,1)$. Differentiating the geodesic equation in $s$ and using the curvature commutator gives $D_t^{2}J+R(J,\dot{\gamma })\dot{\gamma }=0$; the field is nonzero but vanishes at $t=0$ and $t=\pi$. The closed sphere has no boundary, so this closed-geodesic example is not a counterexample to the boundary-to-boundary theorem; it exhibits an exact invisible mode in a conjugate-point setting.

A great semicircle joining a conjugate pair on $S^2$.

Definition of the simple-surface boundary transform

Injectivity theorem on simple surfaces

Definition of a Jacobi field and conjugate points

Paternain, Salo, and Uhlmann, Geometric inverse problems, with emphasis on two dimensions, Chapter 2, §2.5

Ilmavirta, Geometry of geodesics, Chapters 8 and 13

The source comparisons and item-specific dispositions are recorded in research/inverse-problems-knot-theory-all-109-2026-10-19-batch-38.coverage.json and the batch notes.