# On a geodesic mapping of a square

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)?

MathOverflow Asked by Logan Fox on August 8, 2020

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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