Why is this hyperbolic background invariant?

I have a mathematical question for you, because it is a mystery for me.

Look at this animation:

I get it as follows. Each frame corresponds to a value of $t \in [0,3[$ (I take $160$ values of $t$ subdivising $[0,3[$). Here is how I get the frame corresponding to one value of $t$:

• for each point in the unit square $S = {[0,1]}^2$, I take its complex affix $z$, and I send $z$ to the open upper half-plane $\mathbb{H} = \bigl\{z | \Im(z) > 0\bigr\}$ with a conformal map $\psi$ from $S$ to $\mathbb{H}$;

• I attribute a color to $R^t\Bigl(\lambda\bigl(\psi(z)\bigr)\Bigr)$ where $R$ is the Möbius transformation of order $3$ defined by $R(z) = -\dfrac{1}{z+1}$ and $\lambda$ is the modular lambda function.

The modular lambda function and the conformal map $\psi$ are implemented in my R package jacobi.

My question is: why does the hyperbolic tessellation that we can see as the “background” of the animation not move? Why is it invariant?

I observed the same phenomenon for other modular functions, for example the Klein j-invariant function.

The color mapping $\mathcal{C}$ is defined with the help of the HSI color space. The color $\mathcal{C}(z)$ depends on the phase of $z$ only. Precisely, $\mathcal{C}(z)$ is the HSI color with:

• hue $\in [0, 360[$ given by the phase $\varphi(z) \in [0, 2\pi[$ of $z$ converted to degrees;

• saturation given by $\sqrt{\bigl(1 + \sin(w)\bigr) / 2}$ where $w = 2\pi\log\bigl(1+\varphi(z)\bigr)$;

• intensity given by $\bigl(1 + \cos(w)\bigr) / 2$ with $w$ as above.

Why do we get such a result? I really don’t know.

Before leaving you, let me show you a 3D version of this animation that I made with the isocuboids R package:

I hope you like it.