2
$\begingroup$

Let $X$ be a proper geodesic space which is uniquely geodesic. Let $\phi:[0,1]\times[0,1] \to X$ be a function which satisfies the following:

The maps $\phi(0,\cdot)$, $\phi(\cdot,0)$, $\phi(1,\cdot)$, and $\phi(\cdot,1)$ are all (linearly parametrized) geodesics. Furthermore, for each fixed $s$, the map $\phi(s,\cdot)$ is a (linearly parametrized) geodesic connecting $\phi(s,0)$ to $\phi(s,1)$.

Given the above conditions, is it true that that for any fixed $t$, the map $\phi(\cdot,t)$ is a geodesic connecting $\phi(0,t)$ to $\phi(1,t)$? If not, is there a condition we can apply for which this is true (e.g. the space must be Hadamard)?

$\endgroup$
3
$\begingroup$

This is not true. Let $X$ be the unit sphere, or some hemisphere thereof, which we describe first in spherical coordinates.

Let $f(s,0)$ go east along the equator, $(\theta,\phi)=(2s\pi/3,\pi/2)$.

Let $f(s,1)$ go south from the North Pole, $(\theta,\phi)=(\pi,s\pi/3)$

Let $f(s,t)$ be $t$ of the way from $f(s,0)$ to $f(s,1)$.

Then $f(s,1/2)$ is not a geodesic.

Each $f(s,1/2)$ is the midpoint of $f(s,0)$ and $f(s,1)$, so it is proportional to $f(s,0)+f(s,1)$ in $\mathbb{R}^3$. Thus in Cartesian coordinates:

\begin{align} f\left(0,\frac12\right) \propto\, & \big(\phantom{-\sqrt{3}}\,1\phantom{\sqrt{3}}, \ \ \ \ 0 \ \ ,\ \ 1\ \ \ \big) \\ f\left(\frac12,\frac12\right) \propto\, & \left(\phantom{-\sqrt{3}}0\phantom{\sqrt{3}},\ \frac{\sqrt{3}}2, \frac{\sqrt{3}}2\right)\\ f\left(1,\frac12\right) \propto\, & \left(\frac{-1-\sqrt{3}}2, \frac{\sqrt{3}}2,\ \frac12\ \ \right) \end{align} These three vectors have non-zero determinant, so they are not in the same plane through the origin, and $f(1/2,1/2)$ is not on the geodesic between the other two.

$\endgroup$
3
  • $\begingroup$ Thank you, this is a very helpful example. Do you happen to know if it's possible to construct a similar counterexample on a hyperbolic surface? (If not, that's perfectly fine, just thought I'd ask) $\endgroup$ – Logan Fox Jul 24 '20 at 0:07
  • $\begingroup$ Yes, a hyperbolic surface and almost any surface with non-zero curvature would provide counterexamples. But their midpoint formulas are harder to write down, or not given by closed forms, so showing that a particular quadrilateral is a counterexample would require more work. $\endgroup$ – Matt F. Jul 24 '20 at 1:08
  • 1
    $\begingroup$ Update: in the hyperbolic plane, you can get an example by using the Beltrami-Klein model, picking generic points for the four corners and computing midpoints as above. The advantage of that model is that its equations are straightforward for both "M is on the line through A and B" and "M is equidistant from A and B", so you can solve them to find the midpoints explicitly. $\endgroup$ – Matt F. Aug 27 '20 at 10:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.