On a geodesic mapping of a square

MathOverflow Asked by Logan Fox on August 8, 2020

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)?

One Answer

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, $(theta,phi)=(2spi/3,pi/2)$.

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

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} fleft(0,frac12right) propto, & big(phantom{-sqrt{3}},1phantom{sqrt{3}}, 0 , 1 big) \ fleft(frac12,frac12right) propto, & left(phantom{-sqrt{3}}0phantom{sqrt{3}}, frac{sqrt{3}}2, frac{sqrt{3}}2right)\ fleft(1,frac12right) 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.

Correct answer by Matt F. on August 8, 2020

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