 Solution

Use the given Pestov identity for a geodesic flow for the function $\phi$ constructed above. Since $VX\phi =-2H\phi$, its left side is $-4\Vert H\phi {\Vert }_{L^2(\mu )}^2$. Moving this to the right givesAll terms are nonnegative, and $-K$ is uniformly positive by compactness. The smooth derivatives therefore vanish pointwise:Since $X,H,V$ span the tangent bundle, $\phi =c$ is constant, component by component if necessary.

On each circle fibre, $V^{2}u+u=c$ is the elementary periodic differential equation $u_{\vartheta \vartheta }+u=c$. 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 $\eta$ on $M$ with $u(x,v)=c+{\eta }_x(v)$.

Along a geodesic, the velocity is parallel, so differentiating givesSet $Z={\eta }^{♯}/2$. ThenEquality initially holds on unit vectors, then on every vector by homogeneity. The polarization identity yields the tensor identityThis proves that $\beta$ is a potential symmetric 2-tensor, with the precise factor of two in the requested convention.