# Prove non-curvature is asymptotic with unit binormal vector

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 $$U$$.