TITLE: Value of shape operator in direction where the curvature is zero
QUESTION [1 upvotes]: 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.

REPLY [2 votes]: You cannot show that $S_p(Z)$ is zero, you have to show that $S_p(Z)$ is a multiple of $Z$, i.e. $S_p(Z) = \lambda(p,Z) Z$, where $\lambda$ is a function that depends on the direction $Z$ and the point $p$. Once you have this fact, you have to show that $\lambda$ does not depend on the direction $Z$ and that $\lambda$ is constant on the surface. Then you have proven that the surface is umbilical.
Hint: Consider a point $p$ on the surface. Show that, if the curvature of $\gamma$ is not $0$ at $p$, then the normal vector of $\gamma$ is equal to the surface normal. 
This first hint should suffice as a starting point. If you need more input, leave a comment.
Addendum 1: further explanation. Your specific question (the case $\kappa(0)=0$) has already been asked before and a hint can be found in this question. 
The hint actually shows that we can "work around" the directions in which the normal curvatures are zero.
A point is parabolic by definition if there is only one direction in which $\kappa_n$ is zero. A point is hyperbolic, by definition, if $K< 0$, or equivalently, if there are two directions in which the normal curvature is zero. So in case of a parabolic or hyperbolic point, we already know that in every direction, except in one or two directions, $S_p(Z) = \lambda(p,Z) Z$ holds.
Now take two linearly independent vectors $v$ and $w$, such that the normal sections in the directions $v$ and $w$ and $v+w$ are not zero. By the argument hereabove, we know that  $S_p(v)=\lambda(p,v) v$, $S_p(w) = \lambda(p,w) w$ and $S_p(v+w) = \lambda(p,v+w) v+w$. Then by linearity
$$
    \lambda(p,v+w)(v+w) = S_p(v+w) = S_p(v) + S_p(w) 
          = \lambda(p,v) v + \lambda(p,w) w.
$$
We may conclude that $\lambda(p,v)=\lambda(p)$ does not depend on the directions, except maybe in the one or two directions where the normal curvature is zero.
Consider now a direction $Z$ in which $\kappa_n = 0$. Write $Z=av+bw$ with $a,b\in \mathbb{R}$. Then by linearity
$$
 S_p(Z) = a S_p(v) + b S_p(w) = a \lambda(p) v + b \lambda(p) w = \lambda(p) Z.
$$
So we conclude that the normal curvature section $\kappa_n$ must be equal in every direction, so the point is umbilical. Hence, in conclusion, we see that $p$ cannot be a parabolic or hyperbolic point.
Addendum 2: asymptotic directions. Consider a point at a surface. If $e_1$ and $e_2$ are the two principal directions with principal curvatures $\kappa_1$ and $\kappa_2$ respectively, then the normal curvature in the direction of $\cos \theta e_1 + \sin\theta e_2$ is 
$$
 \kappa_n  =  \kappa_1 \cos^2 \theta + \kappa_2 \sin^2 \theta.
$$
This is sometimes called Euler's formula. If $K = \kappa_1 \kappa_2$ is negative at the point, then $\kappa_1 > 0$ and $\kappa_2 < 0$. Then there are exactly two solutions $\theta$ such that $\kappa_n = 0$, which correspond to two linearly independent directions with zero normal curvature. These directions are called asymptotic directions. Also note that  the point is planar (every normal curvature is zero) if and only if $\kappa_1=\kappa_2=0$.
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