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.