If $\alpha$ is a planar geodesic on surface $M$, show that $\alpha$ is a line of curvature.
My try:
$\alpha$ planar imply torsion=0, and binomial vector is constant. Since $0=\kappa_g=\kappa_\alpha B\cdot U$, ($U$ is normal vector of $M$), so B perpendicular to $U$. Here's where I stuck. Please help
WLOG, suppose $\textbf{$\alpha$}$ is arc-length parametrized. Since $\textbf{$\alpha$}$ is a geodesic, we know its normal vector $\textbf{N}$ is collinear with the surface normal $\textbf{n}$. That is, $\textbf{N} = \pm \textbf{n}$.
From this, we have $S_p(\textbf{T}) = -D_{T}\textbf{n} = \pm D_T\textbf{N}= \pm \frac{d\textbf{N}}{ds} = \pm (-\kappa \textbf{T} + \tau \textbf{B})$
Can you finish up and conclude that $\textbf{T}$ is an eigenvector of the shape operator at all points along $\textbf{$\alpha$}$?
Unfortunately, differential geometry notation differs widely from text to text, so I will take a moment to clarify the above:
$\textbf{$\alpha$}$ is our given curve.
$\textbf{T}$, $\textbf{N}$, and $\textbf{B}$ are the three vectors in the Frenet frame of $\textbf{$\alpha$}$. They represent the unit tangent, normal, and binormal vectors to $\textbf{$\alpha$}$.
$\textbf{n}$ is the unit normal to $T_pM$ at a given point. I.e. it is the 'surface normal'.
$S_p$ denotes the shape operator of $M$.
$D_\mathbf{v}X$ denotes the directional derivative of $X$ in the direction $\textbf{v}$.
Suppose $\alpha(s)$ is parametrized by its arc-length $s$; set, in the usual Frenet-Serret notation, $T(s) = \alpha'(s)$, so that $T(s)$ is the unit tangent vector to $\alpha(s)$. It is indeed true that, since $\alpha(s)$ is a planar curve, we must have the torsion of $\alpha(s)$, $\tau(s) = 0$ as long as the curvature $\kappa(s)$ of $\alpha(s)$ is non-zero, $\kappa(s) \ne 0$. This follows from the fact that $\alpha(s)$ lies in some plane $P$; thus $T(s)$ must also be tangent to $P$ for all $s$; hence $N(s) = (\kappa(s))^{-1} T'(s)$ must be tangent to $P$ as well. These assertions may be readily verified by expressing $\alpha(s)$ in the form $\alpha(s) = \beta + \alpha_1(s) \vec e_1 + \alpha_2(s) \vec e_2$, where the $\vec e_i$, $i = 1, 2$ are constant orthonormal vectors which span $P$ and $\beta$ itself is a constant vector; thus $T(s) = \alpha'(s) = \alpha_1'(s) \vec e_1 + \alpha_2'(s) \vec e_2$ and $N(s) = (\kappa(s))^{-1}T'(s) = (\kappa(s))^{-1}(\alpha_1''(s) \vec e_1 + \alpha_2''(s) \vec e_2)$ are both tangent to $P$. We have in turn that $B(s) = T(s) \times N(s)$ is a continuous unit vector field along $\alpha(s)$ which is normal to $P$; thus it must be constant whence $0 = B'(s) = -\tau(s) N(s)$ forces $\tau = 0$. Since $\tau(s) = 0$, the Frenet-Serret equations give $N'(s) = - \kappa(s) T(s)$. Now the fact that $\alpha(s)$ is geodesic in $M$ implies that the component of its normal vector tangent to $M$ vanishes, which in turn allows us to conclude the $N(s)$ is in fact a surface normal vector field to $M$ along $\alpha(s)$. But, denoting the shape operator of $M$ by $S$, we have
$S(T(s)) = \nabla_{T(s)}N(s) = \dfrac{N(s)}{ds} = \kappa(s) T(s), \tag{1}$
that is, $T(s)$ is an eigenvector of $S$ with eigenvalue $\kappa(s)$. We thus see that $\alpha(s)$ is a line of curvature of the surface $M$. QED.
PLEASE NOTE once again that I have added the assumption $\kappa(s) \ne 0$; this is tantamount to the hypothesis that $N(s) = (\kappa(s))^{-1} T'(s)$ is defined along $\alpha(s)$; I don't at present know how to be rid of this assumption, and would be interested in seeing a generalization of this result which lifts the restriction $\kappa(s) \ne 0$. Furthemore, I take $S(T(s)) = \nabla_{T(s)}N(s)$ whereas Kaj takes $S$ to be of opposite sign. But the result is the same in any event.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
