Consider the parametrization below. How can we show that the curve $v=v_{0}= constant\,value$ are geodesics (when parametrized by arc length) ?
The answer talks about the curve being contained in the fixed point set of an isometry of $S$ with reflection in a vertical plane containing the $z-axis$, thus geodesic. Can someone explain this?
A curve is a geodesic if its covariant acceleration is in the same direction as its velocity. Since the curve is fixed by the reflection, its acceleration is too; but the only vectors tangent to $S$ preserved by the reflection are the ones in the direction of constant $v$, i.e. tangent to the curve. Note that this works only because the tangent space is 2-dimensional and thus the subspace fixed by the reflection is one-dimensional.
A more intuitive way of putting this: if a curve on a surface is not a geodesic, it must (at least for some short time) be curving either to the left or to the right. But then the reflection (which is locally through a "vertical" line) would not preserve the curve, since it takes right to left and vice versa.