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I would appreciate if you consider the following two questions on $1$ dimensional foliations whose leaves are geodesic.

1)Assume that $M$ is a Riemannian manifold which is either an open manifold or is a compact manifold with zero Euler characteristic. Does $M$ admit a foliation by geodesics?

2)Assume that $M$ is a Riemannian surface which admit at least one foliation by geodesics. Does there necessarily exist a foliation of $M$ by geodesics which satisfy the "Isocline Locale property"?

The Isocline local property is defined as follows:

For every $x\in M$ there is locally a geodesic $\alpha $ which is transverse to the foliation and it intersect all leaves with the same angle.

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No to the first question. Let $M$ be a Riemannian $2$-manifold whose universal cover is the hyperbolic plane $H$, and whose fundamental group is not cyclic. For any foliation of $H$ by geodesics (lines), it seems to me that the endpoints of these lines on the boundary circle of $H$ will fill up the whole circle except for two points. These two points will have to be preserved by deck transformations, so any discrete group of isometries of $H$ that preserves the foliation must be cyclic.

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    $\begingroup$ But for these surfaces the Euler characteristic is not zero (which is indeed a necessary condition for a codimension 1 foliation, be it geodesic or not). $\endgroup$
    – ThiKu
    Commented Jun 16, 2018 at 16:02
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    $\begingroup$ @Thiku: The $M$ in this answer is an open manifold. $\endgroup$
    – Lee Mosher
    Commented Jun 16, 2018 at 17:59
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    $\begingroup$ Yes, I meant to say choose it to be open, for example the sphere minus three points (with constant curvature metric). $\endgroup$
    – Tom Goodwillie
    Commented Jun 16, 2018 at 18:58
  • $\begingroup$ To be honest I can not understand this answer. For example consider $M=H\setminus \{p,q\}$. Then is not $H$ the universal covering space of $M$? But M admit a foliation by geodesic. The vertical foliation. Am I mistaken? $\endgroup$
    – Ali Taghavi
    Commented Jun 17, 2018 at 10:54
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    $\begingroup$ This $M$ is not complete, as it would have to be if its universal covering space were to be metrically isomorphic to $H$. $\endgroup$
    – Tom Goodwillie
    Commented Jun 17, 2018 at 11:44

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