In a problem in a differential geometry (where I prove that a surface such that all of its geodesics are planar, is either contained in a plane or in a sphere) I need to consider the shape operator $S_p$ at some point $p$, evaluated in a direction where the curvature is zero. I.e. let $\gamma$ be a geodesic curve (parametrised by arclength) on the surface with $\gamma(0)=p$, $Z=\gamma'(0)$ and $\gamma''(0)=0$. I then need to evaluate $S_p(Z)$. I would like to claim this is zero, but I am unsure if this is generally true. Since $\gamma$ is a geodesic, I know that the curvature $\kappa$ is equal to (up to sign) the normal curvature $\kappa_n$, which implies that $0=\kappa_n(Z)=\langle S_p(Z),Z \rangle$. Certainly this is consistent with $S_p(Z)=0$, but it is not implied. We could also have $S_p(Z) \perp Z$. Any input on this?
EDIT
To clarfiy, my question was just about the case when $\gamma''(0)=0$. The other cases I can handle. Indeed, I argue as follows. Let $\gamma$ be a geodesic parametrised by arclength. As the geodesic curvature is zero, we have $\gamma''(s) = \langle \gamma''(s), N(\gamma(s)) \rangle N(\gamma(s)) = \kappa_n(\gamma'(s))$, where $N$ is the Gauss map and $\kappa_n(s)$ is the normal curvature in the direction $\gamma'(s)$ at the point $\gamma(s)$.
On the other hand, from the Frenet equations, we have $\gamma''(s) = t'(s) = \kappa(s)n(s)$, where $t(s)$ is the tangent and $n(s)$ is the principal normal. In fact, as the geodesic curvature is zero, we have $\kappa_n(s) = \pm \kappa(s)$. For convenience, let's say $\kappa_n(s)=\kappa(s)$ (this is just a matter of choosing the sign for the Gauss map).
If $\kappa(0) \neq 0$ (where $s=0$ is just an arbitrary choice), then by continuity of $\kappa$, $\kappa \neq 0$ on some neighbourhood of $0$. On this neighbourhood, we then have $N(\gamma(s))=n(s)$. If $p=\gamma(0)$ we then get $$S_p(\gamma'(0)) = -dN_p(\gamma'(0)) = -\frac{d}{ds}\left(N(\gamma(s)) \right)|_{s=0} = -\frac{d}{ds}\left(n(s) \right)|_{s=0} = \kappa(0)\gamma'(0)$$ where in the last line we used the Frenet equations with torsion $\tau=0$ (as the geodesic is assumed to be planar).
I would now like to be able to claim that the equation $S_p(\gamma'(0)) = \kappa(0)\gamma'(0)$ holds even when $\kappa(0)=0$. However, in that case, the above argument doesn't work, since the principal normal isn't even well-defined.
If the equation $S_p(\gamma'(0)) = \kappa(0)\gamma'(0)$ holds in general, we can argue as follows. For any direction $Z$, we can construct a geodesic through $p$ with $\gamma'(0)$. We then have $S_p(Z)=|\gamma''(0)|Z$, so $Z$ is an eigenvector of the shape operator, and thereby a principal direction. As $Z$ was arbitrary, all directions are principal directions. By continuity of $S_p$, the principal curvatures must be the same (otherwise the eigenvalue would change abruptly between nearby directions). Hence $p$ is umbilic. As $p$ was arbitrary, all points are umbilic.
