= Solution
The <geodesic flow> of a closed negatively curved surface is a topologically transitive <Anosov flow>. In particular, its Jacobi-field hyperbolicity and the usual negative-curvature transitivity theorem allow the smooth <Livsic theorem> to be applied. Set $b(x,v)=\beta_x(v,v)$. The given closed-geodesic integrals are exactly its periodic-orbit obstructions on $SM$. Therefore
$$
\boxed{Xu=b,\qquad u\in C^\infty(SM)}.
$$
We make the differential operators and signs explicit. On the <unit tangent bundle>, let $V$ generate positive fibre rotation, $X$ be the geodesic <vector field> and $H=[V,X]$. For the <canonical coframe of a surface unit tangent bundle>, write
$$
\alpha(\xi)=\langle v,d\pi\xi\rangle,\qquad
\theta(\xi)=\langle iv,d\pi\xi\rangle,
$$
and let $\omega$ be the <Levi-Civita connection> form with $\omega(V)=1$. This uses $\theta$ for a <one-form>, keeping $\beta$ for the given tensor. The dual frame is $(X,H,V)$, and the structure equations are
$$
d\alpha=\omega\wedge\theta,\qquad d\theta=\alpha\wedge\omega,
\qquad d\omega=-K\alpha\wedge\theta.
$$
These follow by writing the two horizontal forms as a rotated local orthonormal <coframe> and $\omega=d\vartheta+\omega_0$, with $\omega_0$ its <connection 1-form>. Equivalently, differentiating the sine and cosine components of $X$ twice in the fibre angle gives
$$
[V,X]=H,\qquad [V,H]=-X.
$$
The curvature equation also gives $[X,H]=KV$. The <Liouville volume of a surface geodesic flow> will be taken positive in the base/positive-fibre orientation, namely $\mu=\alpha\wedge\theta\wedge\omega$.
In an oriented orthonormal frame at $x$, put $v=(\cos\vartheta,\sin\vartheta)$. The <symmetric tensor> evaluation is
$$
b(x,v)=\frac{\beta_{11}+\beta_{22}}2
+\frac{\beta_{11}-\beta_{22}}2\cos(2\vartheta)
+\beta_{12}\sin(2\vartheta).
$$
Since $V=\partial_\vartheta$, this proves $(V^3+4V)b=0$.
The <Lie bracket of vector fields> identities give an operator identity that supplies the required equation. First,
$$
X V^2=V^2X-2HV+X.
$$
Using $VH=HV-X$ and $XV=VX-H$, differentiate this expression once more to obtain
$$
\begin{aligned}
VX(V^2u+u)
&=V^3Xu-2HV^2u+2XVu+2VXu\\
&=(V^3+4V)Xu-2H(V^2u+u).
\end{aligned}
$$
For $\varphi=V^2u+u$ and $Xu=b$, the first term vanishes by the explicit fibre calculation. Hence
$$
\boxed{VX\varphi=-2H\varphi}.
$$
Back to article page
