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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.

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    $\begingroup$ Two comments: (a) You are evaluating $S_p$ in a direction where the (normal?) curvature is zero. Such a direction might not exist, for instance, when the surface is a sphere. (b) You assume that the geodesic $\gamma$ also satisfies $\gamma''(0)=0$. You cannot assume this without justification. The only thing you know is that $\gamma''(0)$ is normal to the surface, since $\gamma$ is a geodesic. $\endgroup$ – Ernie060 Nov 1 '19 at 19:21
  • $\begingroup$ @Ernie060 Thanks! Please see the edit to my question. I have (I believe) most of the solution correct. My trouble lies in dealing with the case specifically when the curvature is zero. $\endgroup$ – Étienne Bézout Nov 1 '19 at 21:24
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    $\begingroup$ Everything seems largely ok. I will put an addendum in my answer. One comment: $S_p(Z)=\gamma''(0)Z$ is a typo. $\endgroup$ – Ernie060 Nov 1 '19 at 22:32
  • $\begingroup$ @Ernie060 Thanks! I have fixed the typo. $\endgroup$ – Étienne Bézout Nov 2 '19 at 13:09
  • $\begingroup$ You might find my differential geometry text helpful :) $\endgroup$ – Ted Shifrin Nov 2 '19 at 16:41
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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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  • $\begingroup$ Thanks! Please see the edit to my question. I have (I believe) most of the solution correct. My trouble lies in dealing with the case specifically when the curvature is zero. $\endgroup$ – Étienne Bézout Nov 1 '19 at 21:24
  • $\begingroup$ Thanks for the pointer! However, I'm afriad I do not completely understand the hint. Ok, so if the point is planar, there is no issue since all directions have normal curvature zero, so the point is umbilic. But if the point is parabolic or hyperbolic, I do not see why at most one ur two directions have zero curvature, nor do I see how I would complete the proof. Basically, my question is whether, and if so how, $S_p(Z) = 0\cdot Z$ is true whenever $Z=\gamma'(0)$ is a direction corresponding to zero curvature, i.e. $\gamma''(0)=0$. $\endgroup$ – Étienne Bézout Nov 2 '19 at 13:07
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    $\begingroup$ I have added some details. I guess there is a quicker argument, but the argument shows that, once you have enough vectors $v$ such that $S_p(v)=\lambda(p,v)v$, then $S_p$ is a multiple of the identity operator. $\endgroup$ – Ernie060 Nov 2 '19 at 14:44
  • $\begingroup$ Thank you very much! This may be a bit off-topic, but I would still like to ask why, in general, we can claim that e.g. for a hyperbolic point, the curvature is zero in exactly two directions. I can see, by the intermediate value theorem, that it would have to be zero in at least two directions (since the principal curvatures have different signs), but why could it not be zero on a whole neighbourhood of directions? $\endgroup$ – Étienne Bézout Nov 2 '19 at 15:02
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    $\begingroup$ No problem, I have added the explanation in the answer. $\endgroup$ – Ernie060 Nov 2 '19 at 15:40

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