One way to think of harmonic maps is to think of one manifold as rubber, being stretched over another manifold that is marble. So here my rubber manifold is the unit square and my marble manifold is $T^2 \to \Bbb R^3.$ The map is called harmonic if it is a critical point of the Dirichlet energy functional:
$$E(\phi)=\int_M || d\phi ||^2 dVol. $$
Q: Does there exist a coordinate grid system comprising of only algebraic equations in the unit square, which when mapped, via harmonic map, onto $T^2 \to \Bbb R^3$ maps directly onto the geodesics of $T^2 \to \Bbb R^3?$
Here's an example of what I mean by a grid system comprising of algebraic equations:
$x,y\in \Re(0,1)$ and $\Re(s) \ge1:$
$\zeta:= \{ (x, y) \in \Bbb R^2 | x^s + y^s = 1 \}.$
$\tau:= \{ (x, y) \in \Bbb R^2 | (1-x)^s + y^s = 1 \}$
$\psi:= \{ (x, y) \in \Bbb R^2 | x^s + (1-y)^s = 1 \}$
$\phi:= \{ (x, y) \in \Bbb R^2 | (1-x)^s + (1-y)^s = 1 \}.$
If I change an assumption and allow the coordinate chart to be comprised of an infinitude of geodesic grid lines in the unit square (horizontal and vertical grid lines), then I can see that I can just stretch this grid system over $T^2 \to \Bbb R^3$ and see that all chart lines map to geodesics on $T^2$ (not a proof of course).
I don't see exactly how to construct or disprove a grid system in the unit square comprising of an infinite number of these algebraic curves that when mapped to $T^2$ are all mapped to geodesics on $T^2.$
The goal is to be able to map the coordinate grid system built from algebraic equations in the unit square, from the unit square to $T^2$ so that chart lines match up with the green geodesics in the picture and back again.
