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Let $M$ be a compact Riemannian manifold with metric $g$ and associated Riemannian volume $\nu$ and geodesic flow $G_t : UTM \rightarrow UTM$, where the unit tangent bundle is indicated. Let $X_j \subset UTM$ for $1 \le j \le n$ be open disjoint codimension one submanifolds transversal to $G_t$, i.e., local cross sections (a global cross section does not exist).

Is it possible to choose a metric $g'$ on $M$ with geodesic flow $G'_t = G_t$ and $\nu'_1(X_j) \equiv 1$?

NB. Here $\nu'_1$ denotes the induced codimension one [relative] measure on $UTM$.

This question was prompted by a helpful comment to this one.

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  • $\begingroup$ What is the definition of $\nu_1'$? Is it the restriction of the metric to $X_j$? in that case, I don't know why should $g$ be changed, do you know an example where $g$ does not work? $\endgroup$
    – rpotrie
    Commented Jun 3, 2010 at 19:41
  • $\begingroup$ $\nu_1'$ is the restriction of the Riemannian measure induced by $g'$. The idea is to tweak stuff transversal to the flow to achieve local cross sections with uniform induced measure. $\endgroup$
    – Steve Huntsman
    Commented Jun 3, 2010 at 19:47
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    $\begingroup$ Ok, so you want the local cross section to be connected? Otherwise you can take "enough" different small cross sections in order to get as much measure as you like. For 3-dimensional flows which are Anosov geodesic flows, you can look at Fried's work (he constructs "almost"-global cross sections) D. FRIED: Transitive Anosov flows and pseudo-Anosov maps, Topology 22,3 (1983) (sorry I couldn't make an hyperlink, but google takes you there) $\endgroup$
    – rpotrie
    Commented Jun 3, 2010 at 19:56
  • $\begingroup$ I want the local cross sections to be generic and given in advance, and to then manipulate the metric in order to uniformize their measures while preserving the geodesic flow. $\endgroup$
    – Steve Huntsman
    Commented Jun 3, 2010 at 20:08
  • $\begingroup$ Thanks. Finally understood, nice question. Sorry (I don't have an answer). $\endgroup$
    – rpotrie
    Commented Jun 3, 2010 at 20:11

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Unless I misunderstand this reference* after a cursory look, it appears that the answer is generally no: "geodesic conjugacy" often implies isometry.

* and anways, why is it that references seem so much easier to find once I've posted a question on MO?

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