I have a triangle on the upper half-plane. One corner is at infinity (0,inf), one corner is at zero (0,0) and one corner is at (0,1). However when displayed using the poincare disk model the triangle is equilateral.
I want to put a gradient on the triangle so that (0,1) has a value of red, while zero and infinity have a value of blue. That is easy, however the hard part is to make a symmetrical gradation.
By gradient I mean a color gradient of course. Going from blue to red, say. Let's set red equal to 1 and blue equal to zero to keep this simple. Gradient is a function of X and Y where X is the real component and Y is the imaginary component.
The best I got was this: gradient = x*(1./(1.+y))
This was close, but as you see below there is more red in the top half then in the bottom half.
Let us denote the Poincaré disk coordinates with $(x_P,y_P)$ and the half-plane coordinates with $(x_H,y_H)$. Some possible solutions:
Inspection of your Poincaré disk picture shows that gradient = $x_P$ clearly gives us the solution. To obtain the same gradient in half-plane coordinates, we need to use the mapping from half-plane to Poincaré disk, i.e., inversion (plus some scaling and shifting). You seem to be using inversion centered at point (0,-1), which gives gradient = $x_P = 2x_H/(x_H^2+(y_H+1)^2)$.
In the solution above the "lines of constant colors" do not seem to have any strong interpretations in hyperbolic geometry (they are equidistants, but not equidistants from the "blue line"). If we want the "lines of constant colors" to be equidistant from the vertical line $(0,0)-\infty$ (or, in other words, the color to be the function of hyperbolic distance from that line), we need to recall that equidistants from such a line are represented in the halfplane model by straight lines passing through $(0,0)$. So, our gradient should be a function of the angle between the vertical line $(0,0)-\infty$ and the line $(0,0)-(x_H,y_H)$, for example, $\frac{2}{\pi}\arctan(x/y)$.
We could also make our "lines of constant colors" to be hyperbolic straight lines -- just like in the first solution, but use the $x$ coordinate in the Klein model for the gradient. It seems this approach yields more complex formulas.