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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.

Ali Taghavi
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    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. – Robert Bryant May 27 '18 at 06:30
  • @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. – Ali Taghavi May 27 '18 at 07:29
  • @RobertBryant But there exist another possible obstruction. It is possible that a particular geodesic foliation admits such isocline transversal. – Ali Taghavi May 27 '18 at 07:33
  • @RobertBryant I would appreciate if you consider this particular case. Assume that $P(x,y),Q(x,y)$ are quadratic polynomials with real coefficients. We consider the foliation associated to the vector field $P\partial_x+Q\partial_y$.(We remove the algebraic curve $yP-xQ=0$ ). Now we consider the Riemannian metric associated to the frame $\left{\frac{x^2+y^2}{yP(x,y)-xQ(x,y)}V,\ \frac{1}{x^2+y^2}W\right}.$ (or a possible rescaling of the second vector of this frame). The foliation is adapted to the metric.Is this particular foliation satisfies that local property? – Ali Taghavi May 27 '18 at 08:21
  • The motivation for such a frame is written in this post: https://mathoverflow.net/questions/277495/limit-cycles-as-closed-geodesics2 – Ali Taghavi May 27 '18 at 08:23
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    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. – Robert Bryant May 27 '18 at 09:52
  • @RobertBryant I understand you fox a geodesic $\alpha$ then you inteoduce a folition by geodesics for which $\alpha$ is an isocline transversal. But in my previous comment I am considering the reverse situation: I fix a foliation then I ask for existence of Local isocline transversal. As your example showed the radial foliation does not admit such isocline transversal. So the situatio depends on the foliation we initially fix. So my question is the following:We fix the foliation associated to a quadratic vector field. We also fix the metric we introduced with that frame. So .. – Ali Taghavi May 27 '18 at 17:37
  • ...so we have a foliation by geodesics. Now our question is that : for a given point $x$, is there an isocline geodesic passing x. If the answer would be yes, the above frame(and corresponding metric) is not a godd candidate for the idea of consideration of limit cycles as closed geodesocs in a negatively curved space. May be I am missing some thing to understand your previous comment? – Ali Taghavi May 27 '18 at 17:42
  • @RobertBryant to be honest I do not understand how does "3 parameter" family arise?May I ask you to explain on it?Moreover could I clarify my question in the previous 2 comments?Is the question in my previous comment has a similar mistake as my question in my intiall post? – Ali Taghavi May 28 '18 at 07:06
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    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. – Robert Bryant May 29 '18 at 08:45
  • @RobertBryant But my question is some what different. In your previous comments you fix a geodesic and you introduce a (several parameter)geodesic foliation for which the initial geodesic is an isocline. But after your first comment I am considering a new version of my question: We fix a foliation by geodesics(In our case this the quadratic system for which the frame described above is a compatible metric. Then we ask:"is it true to say for every x in the phase space($\mathbb{R}^2\setminus C$, there exist a local isocline geodesic passing $x$. If t the answer is posit – Ali Taghavi Jun 04 '18 at 10:48
  • positive then we can not hope that the sign curvature is constant. – Ali Taghavi Jun 04 '18 at 10:49
  • @RobertBryant I think that I just understand what you hinted me; For a given geodesic $\alpha$ there is a (3 parameter) geodesic foliation for which $\alpha$ is an isolcline. On the other hand, for a given $x$ there is a one parameter family of geodesics passing $x$. so we have totally 4 parameter family of geodesics which works for our purpose. Am I right? If yes how can we apply this situation to our case? – Ali Taghavi Jun 04 '18 at 13:20
  • @RobertBryant based on your very helpful comments I tried to correct the concept in this new post. then I mentioned the motivation for consideration of "isocline" i would appreciate if you read this new version. thank you for your attention. https://mathoverflow.net/questions/302009/some-possible-obstructions-to-limit-cycles-as-closed-geodesics3 – Ali Taghavi Jun 05 '18 at 00:47

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