Conjugate points and totally geodesic foliations 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.
 A: 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.
