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Problem. To give a concrete example of a geodesically reversible Finsler metric on the two-sphere that is not just the sum of a reversible Finsler metric and an exact $1$-form.

Some background may be useful since the problem is more of a problem in variational calculus than a problem in geometry.

If $M$ is a smooth manifold and $L: TM \rightarrow \mathbb{R}$ is a (Lagrangian) fuction that is smooth outside the zero section and homogeneous of degree one in the velocities (i.e. $L(x,tv) = tL(x,v)$ for every $t > 0$), the variational problem $$ \gamma \mapsto \int_\gamma L $$ is invariant under orientation-preserving reparametrization of the curve $\gamma$. The Lagrangian $L$ is geodesically reversible if changing the orientation of any of its extremals yields another extremal. If the Lagrangian is reversible ($L(x,-v) = L(x,v)$), then it is geodesically reversible, but the converse is not true. For example, an asymmetric norm on $\mathbb{R}^n$ is geodesically reversible, but it is reversible if and only if the norm is symmetric. Another example can be constructed by taking a reversible Lagrangian and adding to it a closed $1$-form considered as a function on the tangent bundle that is linear in the velocities. This last example is somehow "trivial" and I would like to find many examples of geodesically reversible Lagrangians on compact manifolds that are not of this form. On the torus they are easy to construct because we can always compactify the example with the asymmetric norm, but what happens on the sphere?

I can point to a non-solution: If all the geodesics of a geodesically reversible Finsler metric on the $n$-sphere are closed and of the same length, then the metric is the sum of a reversible metric and an exact $1$-form.

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After a few days thinking I understand my own question ... . This is not to say I've solved it, but I think I understood why it may be very hard. In fact, geodesically-reversible Finsler metrics that are not of the form reversible metric + closed 1-form are very rare and there may actually be none in the two-sphere nor in any other compact surface besides the torus and the Klein bottle, both of which admit flat non-revesible Finsler metrics.

Instead of going through some relatively easy partial results (e.g., if the flow of a geodesically reversible Finsler metric is ergodic, then the metric is the sum of a reversible metric and a closed 1-form) I'll explain by an integral-geometric analogy, which is loosely related to the problem, why the problem is probably very hard.

The injectivity of the X-ray transform on the projective plane is not particularly hard, but I don't think it can be called trivial at all. However, the existing proofs use the symmetries of the transform very strongly and, as far as I know, given a Zoll reversible Finsler metric on the projective plane (and there are tons of these)---including one for which all geodesics are projective lines such as solutions of Hilbert's fourth problem---it is not known whether the associated $X$-ray transform is injective. However, in the case of solutions of Hilbert's fourth problem, some of the work of Quinto points in this direction.

Assume for a second you have happily proved this, but someone (maybe yourself) comes along and asks the following

Question. Let $ds$ be the arc-length element of a Riemannian or reversible Finsler metric on the projective plane and suppose a smooth function $f$ on $\mathbb{RP}^2$ is such that the integral of $fds$ over all closed geodesics of $\mathbb{RP}^2$ is equal to zero. Is it true that $f$ must be zero?

Since there are no counter-examples to the affirmative answer of this question, your theorem suddenly looks incredibly small. However, the answer to the new question seems impossibly hard, specially since all results in this direction concern metrics that are negatively curved or Anosov (via Livsic's theorem).

Warning: this note is slightly autobiographical: see here

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