Geodesic X-ray transform ($I_g\beta (\gamma )=\int {\beta }_{\gamma (t)}(\dot{\gamma }(t),\dot{\gamma }(t))dt$)

For a symmetric covariant two-tensor, the geodesic transform integrates its evaluation on the unit velocity along boundary-to-boundary or closed geodesics, as appropriate to the manifold. A potential symmetric 2-tensor gives a total derivative along each geodesic and therefore integrates to zero on closed geodesics; boundary data require the corresponding boundary potential to vanish.

The displayed one-form on the unit round sphere has zero integral on every closed geodesic. Indeed, the antipodal map pulls it back to its negative, and the second half of an oriented great circle is the antipodal image of the first. Nevertheless $d\theta =2x_{3}d{x}_3\wedge d{x}_1$ is nonzero on an open set, so the one-form is not a closed differential form and cannot be an exact differential form. Thus a curvature-free extension of one-form injectivity fails.

On a closed oriented Riemannian surface this integral identity follows from the canonical commutators and integration by parts against Liouville volume of a surface geodesic flow. If $VX\phi =-2H\phi$ and $K<0$, substitution gives $\Vert X\phi {\Vert }^2+5\Vert H\phi {\Vert }^2+\int (-K)(V\phi {)}^2=0$. Each term is nonnegative, forcing all three derivatives to vanish. For $\phi =V^{2}u+u$, the fibre equation then makes $u$ a constant plus a fibrewise linear function, supplying the potential symmetric 2-tensor kernel.

For a vector field $Z$, define $\beta (v,w)=⟨{\nabla }_{v}Z,w⟩+⟨{\nabla }_{w}Z,v⟩$. Its diagonal evaluation is $2⟨{\nabla }_{v}Z,v⟩$. Along a geodesic this equals $2d⟨Z,\dot{\gamma }⟩/dt$, so it is invisible to closed-geodesic integrals. If a fibrewise linear function $u(x,v)={\eta }_x(v)$ solves $Xu=\beta (v,v)$, then $Z={\eta }^{♯}/2$ gives this convention.