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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.

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