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Lets we have a Riemannian metric on an open subset of the plane which satisfies the following local property.

Local Property: For every point $x$ and every foliation by geodesics around $x$, there exist a geodesic $\alpha$ passing $x$ which is locally transverse to the foliation and is an isocline, that is its angle with leaves it intersect is a constant(independent of leaves intersected by $\alpha$).

Does this local property imply that the metric is necessarily a flat metric?

The motivation: For the moment assume that all $2$ dimensional Riemannian metric satisfy this local property. Then the answer to the question "Limit cycles as closed geodesics" would be negative, that is, we would not be able to count a limit cycle as a closed geodesic of a negatively curved space.

The argument: Assume that we have a Riemannian metric with negative curvature compatible to a vector field $X$, that is all orbits of $X$ are unparametrized geodesics for the Riemanian metric. Assume that $\gamma$ is a limit cycle of $X$. We choose a point $x\in \gamma$ and an isocline geodesic $\alpha$ passing $x$ as a local section(local transversal section). We choose a point $y\in \alpha$ different from $x$. We denote by $p$ the Poincare return map associated to the limit cycle which is defined on $\alpha$. Consider the simple closed curve starting $y$ moving along the orbit of the vector field passing $p(y)$ then joining $p(y)$ to $y$ via isocline $\alpha$. Then applying the Gauss Bonnet theorem we obviously get a contradiction. In fact the isocline property enable us to consider $\gamma '$ as a smooth closed geodesic whose interior contain another clised geodesic $\gamma$. So it is obvious that a negatively curved space in the plane can not posses two nested closed geodesic.

Since I can not drow a picture I use the following picture: I mean that, in the following picture,if the $x$ axis is an isocline geodesic and the curvature is negative then the geodesic in the picture which is spirali g can not surround a closed geodesic. It can not be contained in a closed geodesic, too:

https://commons.m.wikimedia.org/wiki/File:Limit_cycle_Poincare_map.svg

In fact the matter is that the Gauss Bonnet formula for the sum of angles of a $n$-polygon is valid for a $2$ polygon. We used this fact in the above argument. Our $2$-polygon has $2$ edges. The first edge the flow-orbit starting $x$ and ending $p(x)$. The second edge is a sub segment of isocline $\alpha$ starting $p(x)$ and ending at $x$.

Note: Of course the local property of this question is satisfied by the Euclidean metric and does not satisfied by the Hyperbolic metric on the upper half plane with vertical foliation.

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    $\begingroup$ I don't understand your question. Even the Euclidean metric in the plane doesn't have this property. Look at the foliation $\mathcal{F}$ consisting of lines through the origin $O$ and let $x$ be any point other than the origin. Then $\mathcal{F}$ is a foliation of a neighborhood of $x$ for which no such 'isocline' geodesic $\alpha$ exists. In fact, no metric could have this property: The set of geodesics $\alpha$ through $x$ is a 1-dimensional family, and, for any such geodesic $\alpha$, there is only a 1-dimensional family of geodesic folliations that meet $\alpha$ at a constant angle. $\endgroup$
    – Robert Bryant
    Commented May 27, 2018 at 6:30
  • $\begingroup$ @RobertBryant I am sorry that my question was incorrect. Thanks for your comment which help me to realize my mistake. on the other hand your valuable comment help me to be calm that this "local property", which does not exist at all, is not an obstruction for consideration of limit cycles as closed geodesic of a negatively or positively curved space. $\endgroup$
    – Ali Taghavi
    Commented May 27, 2018 at 7:29
  • $\begingroup$ @RobertBryant But there exist another possible obstruction. It is possible that a particular geodesic foliation admits such isocline transversal. $\endgroup$
    – Ali Taghavi
    Commented May 27, 2018 at 7:33
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    $\begingroup$ As I explained in my first comment, every embedded geodesic $\alpha$ on a Riemannian surface is an 'isocline transversal' of a $1$-parameter family of geodesic foliations; you just need to specify the constant angle that it makes. There's nothing special about these, except that, for the flat metric, this only generates a 1-parameter family of geodesic foliations that admit 'isoclines'. For most metrics, this generates a 3-parameter family of (local) geodesic foliations, but I don't see how they would be useful for your problem. $\endgroup$
    – Robert Bryant
    Commented May 27, 2018 at 9:52
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    $\begingroup$ Here's an example: Consider the hyperbolic plane $H$ (i.e., $K\equiv-1$). Given an oriented geodesic $\alpha$ and angle $\theta\in(0,\pi)$, there is a unique foliation $\mathcal{F}_(\alpha,\theta)$ of $H$ whose leaves are the oriented geodesics that meet $\alpha$ in an angle of $\theta$. If $\mathcal{F}_(\alpha,\theta)=\mathcal{F}_(\beta,\psi)$, then $\alpha=\beta$ and $\theta=\psi$ (up to orientation and some 'signs'). Thus, this is a $3$-parameter family of geodesic foliations of $H$ (since geodesics depend on $2$ parameters). This holds for the generic suitably convex Riemannian surface. $\endgroup$
    – Robert Bryant
    Commented May 29, 2018 at 8:45

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