Geodesic X-ray transform 2026-10-14
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.
Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 14 2 c Solution Created 2026-10-03 Updated 2026-10-14
Use the given Pestov identity for a geodesic flow for the function constructed above. Since , its left side is . Moving this to the right givesAll terms are nonnegative, and is uniformly positive by compactness. The smooth derivatives therefore vanish pointwise:Since span the tangent bundle, is constant, component by component if necessary.
On each circle fibre, is the elementary periodic differential equation . Its solutions areThe first harmonics transform as a covector when the local oriented orthonormal frame changes. Their coefficients are smooth, either directly from the smooth fibre equation or from their sine/cosine integral formulas. Thus they define a smooth one-form on with .
Along a geodesic, the velocity is parallel, so differentiating givesSet . ThenEquality initially holds on unit vectors, then on every vector by homogeneity. The polarization identity yields the tensor identityThis proves that is a potential symmetric 2-tensor, with the precise factor of two in the requested convention.
Pestov identity for a geodesic flow 2026-10-14
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 and , substitution gives . Each term is nonnegative, forcing all three derivatives to vanish. For , the fibre equation then makes a constant plus a fibrewise linear function, supplying the potential symmetric 2-tensor kernel.