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Let $S\subseteq\mathbb{R}^3$ be a regular surface and let $p$ be a point of $S$. If $p$ lies in a segment contained in $S$ show that $p$ is either parabolic or planar.

Well, I think that an idea is to show that the differential of the Gauss map is zero if calculated on a tangent vector parallel to the segment, but I don't know how to formalize it...

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Since the issue is formal (on the intuitive level the statement should be clear), the question becomes: what exactly is your definition of a parabolic / planar point? – user53153 Jan 4 '13 at 5:21
From my definition I have to show that $K=0$ (where $K$ is the gaussian curvature of the surface) in $p$. For me a parabolic point is a point in which one and only one of the principal curvature is zero and a planar point is a point in which both the two principal curvatures vanishes. – Frankenstein Jan 5 '13 at 11:07
What is a segment here? For me a segment is a length minimizing geodesic. But every point on every regular surface lies on a length minimizing geodesic, so it is of course impossible to derive your desired conclusion from that. – treble Jan 6 '13 at 10:45
With segment I mean part of a straight line in $\mathbb{R}^3$. There exist two points $a,b \in S$ such that $(1-t)a+tb \in S$ $\forall t \in (0,1)$ and there is $t^* \in (0,1)$ such that $(1-t^*)a+t^*b=p$ – Frankenstein Jan 6 '13 at 12:05
up vote 0 down vote accepted

The statement is false. The hyperbolic paraboloid $z=x^2-y^2$ has negative curvature at $p=(0,0,0)$ even though $p$ lies on two lines contained in the surface: $y=\pm x, z=0$. More generally, there are lots of ruled surfaces.

It is correct that $K\le 0$ at such a point.

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Oh, you are right... Maybe I misunderstood the text of the problem. Thank you! – Frankenstein Jan 7 '13 at 19:38
By the way, to prove that $K\le0$ at $p$ it suffices to show that the normal curvature of $S$ at $p$ in the direction of the segment is $0$ and so at least one of the principal curvatures is $\le0$ at $p$, right? – Frankenstein Jan 7 '13 at 22:07
@Frankenstein Basically yes, but "at least one of the principal curvatures is $\le 0$" would not yield the right conclusion: if both are negative, then $K>0$. The careful reasoning is: $K=k_1k_2$ where $k_1$ and $k_2$ are max and min of curvatures of sections by normal planes. Since there is a section with curvature $0$, it follows that $k_1\ge 0$ and $k_2\le 0$. Hence, $k_1k_2\le 0$. – user53153 Jan 7 '13 at 22:10

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