A product geometry with an injective simple-surface transversal

The boundary of $M$ is the level set $x_1^2+4\mathrm{\mid }{x}^{\mathrm{\prime }}{\mathrm{\mid }}^2=1$. Its defining function has gradient $(2x_1,8x_2,8x_3)$, which cannot vanish on that level set, so $M$ is a compact smooth three-dimensional domain. Every point of $M$ has $\mathrm{\mid }{x}^{\mathrm{\prime }}\mathrm{\mid }\le 1\mathrm{/}2<1$, hence $M⋐\mathbb{R}\times D^{\circ }$; the ambient metric is exactly $d{x}_1^2\oplus g_D$. By [F1], this is a transversally anisotropic product domain.

The disk has strictly convex boundary: its outward unit normal is $x^{\mathrm{\prime }}$ on $\mathrm{\mid }{x}^{\mathrm{\prime }}\mathrm{\mid }=1$ and $g({\mathrm{\nabla }}_w{\nu }_{\mathrm{o}\mathrm{u}\mathrm{t}},w)=\mathrm{\mid }w{\mathrm{\mid }}^2>0$ for every nonzero tangent $w$. Its curvature is zero, so a normal Jacobi field has scalar coefficient $a+bt$; if it vanishes at two distinct times then $a=b=0$, proving no conjugate points. Every unit-speed straight line exits the bounded disk in finite forward and backward time, and the segment between two points in the disk is unique. Thus $D$ is simple by [F2].

Apply [F3] to this simple surface. Its unweighted scalar geodesic ray transform is injective on the stated compactly supported smooth real functions. This verifies the injective-transversal condition for $M$.