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I came across the following result (mentioned on Pg. 3 of this talk) that states that

If $D$ is an open connected subset of a complete Riemannian manifold with smooth metric then $\partial D$ convex iff $D$ is geodesically convex.

Here $D$ is geodesically convex if each two points of D can be joined by a non-necessarily unique geodesic which minimizes the distance in D and $\partial D$ is convex if the second fundamental form of the boundary with respect to the interior normal is positive semidefinite.

If we apply this theorem to a thin strip around the equator on the sphere, then that should imply that it is geodesically convex since the boundary of the thin strip is convex (it has positive curvature). However, clearly this is not true if we take any two points that are on the opposite sides of the equator (say at a distance $\pi$) from each other. Then the geodesic joining them is not contained in the thin strip. Where am I going wrong here?

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    $\begingroup$ In your example, the geodesic curvature of the boundary of the thin strip is $negative$ definite, since the outward normal points in the opposite direction of the spherical cap (whose second fundamental form is positive). $\endgroup$
    – Gabe K
    Commented Apr 19 at 1:11

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