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The question is to prove that a regular surface $S \subset \mathbb{R}^3$ can't contain more than two lines passing though a point $p$ if its gaussian curvature at $p$ is $\neq 0$

José
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  • Welcome to MSE! What have you tried yourself so far? – Diglett Feb 09 '19 at 16:10
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    I've tried using the fact that the gaussian curvature is the product of the two principal curvatures to show that there can't be more than two lines. There seems to be something i'm missing, because this seems to lead nowhere. – José Feb 09 '19 at 16:16

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HINT: A line contained in a surface is an asymptotic curve.

Ted Shifrin
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  • I think i get it. Shape operator vanishes on three independent vectors, so it must be null, implying the curvature is $0$. Thanks, i don't have enough reputation to upvote you. – José Feb 09 '19 at 16:32
  • No, be careful. The shape operator doesn't vanish on any nonzero vector if $K(p)\ne 0$. An asymptotic vector $\vec v (\ne \vec 0)$ is one for which $S(p)(\vec v)\cdot \vec v = 0$. (My differential geometry text, linked in my profile, might be of use to you.) – Ted Shifrin Feb 09 '19 at 17:24