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Suppose $(M,g)$ is a two dimensional simply connected compact Riemannian manifold with smooth boundary. I want to understand if in general there is a correlation between the following two statements in the sense that is one stronger or one weaker, etc

(1) the manifold is simple. i.e there are no pair of conjugate points

(2) the manifold is a foliation by geodesics.

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  • $\begingroup$ Does "is a foliaton by geodesics" mean "admits a foliation by the geodesics of $g$"? $\endgroup$
    – Ben McKay
    Commented Dec 1, 2017 at 6:37
  • $\begingroup$ What do you mean by "foliation" on a manifold with boundary? Do you just foliate the interior, or does the foliation have to be transverse to the boundary, or tangent to the boundary, or have constant dimension of intersection of the tangent bundle to the foliation and the tangent bundle to the boundary? $\endgroup$
    – Ben McKay
    Commented Dec 1, 2017 at 13:16
  • $\begingroup$ It just means that the leafs intersect the bounday in points.so indeed constant dimension of intersection of fhe tangent bundle to the foliation and tangent bundle to boundary. $\endgroup$
    – Ali
    Commented Dec 1, 2017 at 13:53
  • $\begingroup$ Intersecting in points is not the same as having constant dimension of intersection of the tangent bundle to the foliation and the tangent bunde to the boundary. Consider a closed disk in the Euclidean plane, and take a foliation of the plane by parallel lines. You get two lines tangent to the circle, but all intersections are in one or two points. $\endgroup$
    – Ben McKay
    Commented Dec 1, 2017 at 17:28

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The original question was without boundary: they don't exist, for either (1) or (2). See Chavel, Riemannian Geometry, p. 329 for proof that (1) don't exist: compact manifolds with no conjugate points have integral of scalar curvature at most equal to zero, and your surface must be a topological sphere,so follows from Gauss--Bonnet. For (2), it follows from Poincare-Hopf, since the surface is a sphere. A choice of foliation gives a line field (its tangent lines), and after at most a 2-1 cover, we pick a direction on each line, and a unit vector in that direction i.e. a nowhere zero vector field.

With boundary: delete an open spherical cap around the north pole, and one of the same radius around the south pole, of the unit sphere. The equator still has a conjugate point. There is a geodesic foliation by the geodesics from north to south pole. So (2) does not imply (1). You can quotient by the antipodal map to get an example of (2) but not (1) on the Moebius strip, with only one boundary component.

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  • $\begingroup$ Do you mean (2) implies (1)? $\endgroup$
    – Ali
    Commented Dec 1, 2017 at 7:09
  • $\begingroup$ I mean that (1) cannot happen, and that (2) also cannot happen. $\endgroup$
    – Ben McKay
    Commented Dec 1, 2017 at 7:10
  • $\begingroup$ I don't understand this. Just consider the unit disk with euclidean metric this is clearly a foliation by geodesics $\endgroup$
    – Ali
    Commented Dec 1, 2017 at 7:13
  • $\begingroup$ Also one can construct many examples of riemannian manifolds with conjugate points $\endgroup$
    – Ali
    Commented Dec 1, 2017 at 7:14
  • $\begingroup$ I forgot to mention it is a manifold with boundary! My bad $\endgroup$
    – Ali
    Commented Dec 1, 2017 at 7:18

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