Convexity of curves in Riemannian surfaces It is known that a curve $f:[0,2\pi]\to \mathbf{R}^2$ is convex if $\partial_t (\arg f'(t))\ge 0$. My question is: does this statement have an analogue in the setting of Riemannian surfaces instead of $\mathbf{R}^2$?
 A: First, I think you have to define convex curves in Riemannian surfaces. We may say a convex curve is a connected piece of the boundary of a convex set. However, convex sets themselves in Riemannian surfaces have many definitions. A good reference for such definitions is:
Alexander, S. Local and global convexity in complete Riemannian manifolds. Pacific J. Math. 76 (1978), no. 2, 283--289. PDF
Convexity of a set depends on the minimizing uniqueness properties of geodesics connecting its points. For example, the great circle in the $2-$dimensional sphere $S^2$, which is a geodesic, is not convex since antipodal points have two geodesics joining them.
On the other side, if we consider complete simply connected Riemannian surfaces without conjugate points $W^2$, then points are connected by unique and hence minimal geodesics. Every geodesic divides the plane to two convex sets and every convex set is supported by a geodesic at its boundary points. One may say a closed convex set is the intersection of all closed half spaces containing it. Hyperbolic halfspace model, Euclidean plane and surfaces with non-positive Gaussian curvature are surfaces without conjugate points.
This discussion leads us to say that a convex curve in $W^2$ is globally supported at every point by a geodesic and hence its geodesic curvature should be greater than or equal to the geodesic curvature of the supporting geodesic i.e.
$$K_g \ge 0$$

Finally, if my guess is true, the required formula is
$$K_g=\frac{\gamma ''. ( n\times \gamma ')}{|\gamma '|^3}\ge 0$$
where $\gamma $ is a curve on a surface with normal $n$.
