Let S be an oriented regular surface. A regular curve $\gamma:I\rightarrow S$ is called asymptotic if $II_{\gamma(t)}(\gamma'(t))=0$ for all $t \in I$.

If $\gamma$ has non-zero curvature, prove that it is asymptotic if and only if its unit binormal vector $b(t)$ is parallel to $N(\gamma(t))$ for all $t\in I$.

So I think $II_{\gamma(t)}(\gamma'(t))$ here represents the normal curvature in some way but the notation is confusing me.

Not sure if this is on the right track or not, but I have:

Assuming non-zero curvature, $\gamma'(t)$ is orthogonal to $N(\gamma(t))$ for all $t$. For every $\gamma$, we want $\gamma''\times N(\gamma(t))=II_{\gamma(t)}(\gamma'(t))=0$ to be asymptotic.

$b(t)=$unit tangent$\times$unit normal$=\frac{\gamma'}{\left \| \gamma' \right \|}\times N(\gamma(t))$

What does having $b(t)$ parallel to the unit normal do towards solving this proof?


First some remarks: $II_{\gamma(t)}(\gamma'(t))$ here means the normal curvature of at $\gamma(t)$ in the direction $\gamma'(t)$. Also note that $\gamma'' = T'=\kappa N$, so $\gamma''\times N(\gamma(t))$ is trivially zero. And finally, $\gamma''\times N(\gamma(t))=II_{\gamma(t)}(\gamma'(t))$ doesn't mean anything, since the LHS is a vector and the RHS a scalar.

Hint. What is the definition of sectional curvature? The expression should involve the surface normal, let's call it $U$. From $II_{\gamma(t)}(\gamma'(t))=0$, you can find a relation between the normal $N$ of the curve and the surface normal $U$.

If you want to have a peek at the answer, see this question.

  • $\begingroup$ Does asymptotic make sense for non-oriented surfaces? $\endgroup$
    – Two Fears
    Nov 19 '19 at 19:18
  • $\begingroup$ Yes, it does. A curve is asymptotic if its tangent vector is asymptotic, i.e. the normal curvature of the surface in that direction is zero. A surface has asymptotic directions at a point iff the Gauss curvature at that point is non-positive. But it has nothing to do with being oriented or not. $\endgroup$
    – Ernie060
    Nov 19 '19 at 20:27

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