Simple Riemannian background

Let $n\ge 2$ and let $(M,g)$ be a compact, connected, nonempty smooth $n$-dimensional Riemannian manifold with smooth boundary. Let ${\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}}$ be the outward unit normal and let $\mathrm{\nabla }$ be the Levi–Civita connection of $g$. The boundary is strictly convex when

$$g({\mathrm{\nabla }}_w{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}},w)>0$$

for every boundary point and every nonzero boundary-tangent vector $w$. Use the curvature convention

$$R(X,Y)Z={\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z-{\mathrm{\nabla }}_{[X,Y]}Z\mathrm{.}$$

A Jacobi field along a unit-speed geodesic $\gamma$ is a smooth vector field $J$ satisfying $D_t^{2}J+R(J,\dot{\gamma })\dot{\gamma }=0$, where $D_{t}J={\mathrm{\nabla }}_{\dot{\gamma }}J$. The manifold has no conjugate points if every positive-length unit-speed geodesic segment $\gamma :[a,b]\to M$ has no nonzero Jacobi field with $J(a)=J(b)=0$. It is nontrapping if every maximal unit-speed geodesic starting in $M^{\circ }$ reaches $\mathrm{\partial }M$ in finite forward and backward time. The pair $(M,g)$ is a simple Riemannian background if its boundary is strictly convex, it has no conjugate points, and it is nontrapping.

When $n=2$, this is the simple-surface convention used for the scalar geodesic X-ray transform on the required inverse-problems page.

Riemannian metric and riemannian manifold

Levi civita connection

Fundamental theorem of riemannian geometry

Jacobi field

Simple surface and geodesic X-ray transform