Let $\gamma$ be a simple closed curve in the unit sphere $\mathbb{S}^{2} \subset \mathbb{R}^{3}$. I wish to understand how the property of being simple and closed affects the geodesic curvature of $\gamma$.
Clearly, the geodesic curvature can be constant (e.g., geodesics are simple and closed). In general, I suppose that being simple and closed must result in a minimum number of critical points of the geodesic curvature. In fact, my question is the following. If the geodesic curvature $\kappa_{g}$ of $\gamma$ is nonconstant, does it then follow that its derivative $\kappa_{g}'$ changes sign at least once?
Hints on how to prove/disprove this statement would also be helpful.
Let me assume that $\gamma : S^1 \to S^2$ is smooth, and that a transverse orientation of $\gamma$ has been chosen. The geodesic curvature is therefore defined as a smooth function $\kappa_\gamma : S^1 \mapsto \mathbb R$.
If the geodesic curvature is not constant, there are points $c,C \in \gamma$ at which $\kappa_\gamma(c) < \kappa_\gamma(C)$.
Applying the Mean Value Theorem, in the arc of $\gamma$ from $c$ to $C$ there is a point $d_+ \in \gamma$ where $\kappa'_\gamma(d_+) > 0$, and in the opposite arc from $C$ to $c$ there is a point $d_- \in \gamma$ where $\kappa'_\gamma(d_-) < 0$. The function $\kappa'_\gamma$ must therefore change from positive to negative in the arc of $\gamma$ from $d_+$ to $d_-$, and back again from negative to positive in the opposite arc from $d_-$ to $d_+$.