Normal sections

Question

Let $S$ be a surface and let $p \in S$ with normal $n_{p}$. Let $q \in S$ nearby $p$, say within the injectivity radius of the exponential map at $p$. Consider the plane $\Pi = \operatorname{span}\{n_p, \vec{pq}\}$. The intersection of $\Pi$ with $S$ gives a curve $\gamma$ from $p$ to $q$. Is $\gamma$ a geodesic? Equivalently is $\Pi \cap S$ a normal section?

Answer 1

This is far too good to be true in general - you're trying to use only $p$-local ($n_p$) and averaged ($\vec{pq}$) data to determine the global behaviour of a geodesic, which is determined uniquely by the tangent to $\Pi \cap S$ at $p$ and depends highly on the geometry of $S$ between there and $q$. I expect this is only true when you have some kind of symmetry of $S$ about $\vec{pq}$, and only true for all admissible $q$ when $S$ is a sphere or plane. Here's an explicit counterexample:

Let $S$ be the cylinder $x^2 + y^2 = 1$, $p=(1,0,0)$ and $q=(0,1,1)$. Then $n_p =e_x$ and $\vec{pq}=(-1,1,1)$, so $\Pi = \left\{(1+s-t, t, t) \mid s,t \in \Bbb R\right\}$. The intersection of this with $S$ is the subset where $$(1+s-t)^2 + t^2 = 1,$$ which has local solution (between $p$ and $q$)

$$ S \cap \Pi \supset \left\{ (\sqrt{1-t^2}, t, t) \mid t \in \Bbb [0,1] \right\}.$$

Plotting this curve on the cylinder shows it's clearly curved in the surface, particularly near $q$:

Indeed, changing to cylindrical coordinates $(\theta,z)$ via $x= \cos \theta, y = \sin \theta$ we see that $$S \cap \Pi = \left\{ (\theta, \sin \theta)\mid \theta \in \Bbb R\right\}.$$

Since the metric in these coordinates is $d\theta^2 + dz^2$, the geodesics are "straight lines" $a \theta + b z = c$, and $S \cap \Pi$ is clearly not one of these. You can replace $q$ with closer points $(\cos \theta_0, \sin \theta_0, \sin \theta_0)$ to get counterexamples with $d(p,q)$ arbitrarily small.