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Simple disk, trapped annulus, and conjugate-point cancellation

Example

Simple disk. Let $M = {x \in R^{2} : ∣ x ∣ \leq 1}$ with the Euclidean metric and define

f (x) = {\begin{cases} \exp ⁣ (- \frac{1}{1 - 4 ∣ x ∣^{2}}), & ∣ x ∣ < \frac{1}{2}, \\ 0, & ∣ x ∣ \geq \frac{1}{2} . \end{cases}

This is a nonzero smooth function compactly supported in $M^{\circ}$ . The disk is simple: the outward normal is $ν_{o u t} (x) = x$ on the unit circle, so $g (\nabla_{w} ν_{o u t}, w) = ∣ w ∣^{2} > 0$ for every nonzero tangent $w$ ; Euclidean Jacobi fields satisfy $J^{''} = 0$ , 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 $((- 1, 0), (1, 0))$ ,

I_{0} f ((- 1, 0), (1, 0)) = \int_{- 1}^{1} f (t, 0) d t > 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 $M = [0, 2] \times (R / 2 π Z)$ and equip it with $g = d r^{2} + a (r)^{2} d θ^{2}$ , where $a (r) = 2 + \frac{1}{2} \cos (2 π r) > 0$ . Since $a^{'} (1) = 0$ , the Christoffel symbols $Γ_{θ θ}^{r} = - a a^{'}$ and $Γ_{r θ}^{θ} = a^{'} / a$ vanish at $r = 1$ . Thus $γ (t) = (1, θ_{0} + t / a (1))$ is a unit-speed closed geodesic of period $2 π a (1)$ , 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 $S^{2} \subset R^{3}$ with its induced Riemannian metric, write $x_{3}$ for the third coordinate and define the closed-geodesic transform by $I_{c l} f (c) = \int_{0}^{2 π} f (c (t)) d t$ for every unit-speed great circle $c (t) = a \cos t + b \sin t$ with orthonormal $a, b \in R^{3}$ . 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 $[0, 2 π]$ vanish. Also the antipodes $p = (1, 0, 0)$ and $- p$ are conjugate along $γ (t) = (\cos t, \sin t, 0)$ : for $γ_{s} (t) = \cos t (1, 0, 0) + \sin t (0, \cos s, \sin s)$ , 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 $D_{t}^{2} J + R (J, \dot{γ}) \dot{γ} = 0$ ; the field is nonzero but vanishes at $t = 0$ and $t = π$ . 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 $S^{2}$ .

Example references

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

Facts & Assumptions

Given: The Euclidean unit disk with the displayed bump, the annulus with the displayed warped metric, and the unit sphere with the stated closed-geodesic transform.

[F1]

Simple surface and geodesic X-ray transform: strict convexity means $g (\nabla_{w} ν_{o u t}, w) > 0$ for nonzero boundary-tangent $w$ ; $I_{0} f$ integrates over unit-speed boundary-to-boundary geodesics.

[F2]

Simple surface and geodesic X-ray transform: no conjugate points means every endpoint-zero Jacobi field on a positive-length geodesic segment is zero; nontrapping means every maximal interior unit-speed geodesic exits in finite forward and backward time.

[F3]

Jacobi field: a Jacobi field satisfies $D_{t}^{2} J + R (J, \dot{γ}) \dot{γ} = 0$ ; Curvature of an affine connection: $R (A, B) C = \nabla_{A} \nabla_{B} C - \nabla_{B} \nabla_{A} C - \nabla_{[A, B]} C$ .

[F4]

Covariant derivatives commute up to curvature in a two parameter variation: $D_{s} D_{t} W - D_{t} D_{s} W = R (\partial_{s} γ, \partial_{t} γ) W$ in a smooth two-parameter variation.

[F5]

Riemannian metric and riemannian manifold: a Riemannian metric is a smooth positive-definite symmetric tensor.

[F6]

Levi civita connection: the Levi-Civita connection is metric-compatible and torsion-free; Fundamental theorem of riemannian geometry: it exists uniquely for every supplied smooth metric.

Proof

1.1givenalgebra

The function $s \mapsto e^{- 1 / s}$ for $s > 0$ , extended by $0$ for $s \leq 0$ , is smooth and flat at $0$ : each derivative for $s > 0$ is a polynomial in $1 / s$ times $e^{- 1 / s}$ , which tends to $0$ as $s ↓ 0$ . Composing it with $s = 1 - 4 ∣ x ∣^{2}$ proves the displayed $f$ is smooth and vanishes outside the closed radius- $1 / 2$ disk, a compact subset of $M^{\circ}$ . It is positive inside that disk, so it is nonzero.

1.2F1givenalgebra

On $\partial M$ , $ν_{o u t} (x) = x$ , and its Euclidean derivative on tangent $w$ is $w$ , so $g (\nabla_{w} ν_{o u t}, w) = ∣ w ∣^{2} > 0$ . This verifies strict convexity.

1.3F2F3F5F6givenalgebra

Cartesian metric coefficients are constant, so their Levi-Civita Christoffel symbols vanish; the curvature definition [F3] gives $R = 0$ . Along any Euclidean geodesic the Jacobi equation from [F3] is then $J^{''} = 0$ , and an affine field vanishing at two distinct endpoints is identically zero, verifying absence of conjugate points.

1.4F6givenalgebra

For $g_{r r} = 1$ , $g_{θ θ} = a^{2}$ , $g_{r θ} = 0$ , the Levi-Civita coordinate formula gives $Γ_{θ θ}^{r} = - \frac{1}{2} \partial_{r} (a^{2}) = - a a^{'}$ and $Γ_{r θ}^{θ} = Γ_{θ r}^{θ} = \frac{1}{2} a^{- 2} \partial_{r} (a^{2}) = a^{'} / a$ . At $r = 1$ , $a^{'} = 0$ , so the geodesic equations are satisfied by $r (t) = 1$ and $θ (t) = θ_{0} + t / a (1)$ . Its speed is $\sqrt{a (1)^{2} (1 / a (1))^{2}} = 1$ , and its periodic angular coordinate keeps it in the interior for every real $t$ . Thus it is trapped.

1.5F5givenalgebra

For any orthonormal $a, b \in R^{3}$ , $c^{'} (t) = - a \sin t + b \cos t$ has norm one in the induced sphere metric. Direct integration gives $I_{c l} f (c) = a_{3} \int_{0}^{2 π} \cos t d t + b_{3} \int_{0}^{2 π} \sin t d t = 0$ . Since $f (0, 0, 1) = 1$ , this is a nonzero invisible function for the explicitly defined closed-geodesic transform.

2.1F2step 1.2step 1.3algebra

A Euclidean line intersects the compact convex disk in a closed segment of length at most $2$ , so every interior unit geodesic exits in finite positive and negative time. This verifies nontrapping and, with steps 1.2–1.3, proves the disk is simple.

3.1F1givenstep 1.1step 2.1

The diameter transform is the integral of a continuous nonnegative function which is strictly positive for $∣ t ∣ < 1 / 2$ ; therefore its value is strictly positive, proving the disk detects this $f$ . The geodesic is entering at $(- 1, 0)$ and exits at $(1, 0)$ , as required by the stated incoming convention.

4.1F2F3F4F5F6givenstep 1.5∎

The vectors $p = (1, 0, 0)$ and $v_{s} = (0, \cos s, \sin s)$ are orthonormal for every $s$ , so $γ_{s} (t) = p \cos t + v_{s} \sin t$ is a unit-speed great circle: its ambient acceleration is $- γ_{s} (t)$ , which is normal to $S^{2}$ , and its tangential covariant acceleration is zero. Its variation field at $s = 0$ is $J (t) = \partial_{s} γ_{s} (t) ∣_{s = 0} = \sin t (0, 0, 1)$ . For each fixed $s$ , the $t$ -curve is geodesic, so $D_{t} \partial_{t} γ_{s} = 0$ ; torsion-freeness gives $D_{s} \partial_{t} γ_{s} = D_{t} \partial_{s} γ_{s}$ , and the curvature identity [F4] then yields $D_{t}^{2} J + R (J, \dot{γ}) \dot{γ} = 0$ . The field is nonzero at $t = π / 2$ and vanishes at $0$ and $π$ , proving those antipodes conjugate by [F2]. This conclusion concerns the closed sphere and its closed-geodesic transform only.

Source locator

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.

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Sources

Gabriel P. Paternain, Mikko Salo, and Gunther Uhlmann, Geometric inverse problems, with emphasis on two dimensions (standard reference, not scraped)
Joonas Ilmavirta, Geometry of geodesics, full lecture notes (standard reference, not scraped)