# Deriving equations for the “Bianchi-Pinkall torus”

I am trying to work out explicit parametric equations for the "Bianchi-Pinkall flat torus" as depicted in this note, but I seem to have gotten stuck in understanding the descriptions given in that note.

Thus far, I figured that the parametric equations in $\mathbb R^4$ should look like this:

\begin{align*} x_1&=\cos(a+b\sin(2p v))\cos(u+v)\\ x_2&=\cos(a+b\sin(2p v))\sin(u+v)\\ x_3&=\sin(a+b\sin(2p v))\cos(u-v)\\ x_4&=\sin(a+b\sin(2p v))\sin(u-v) \end{align*}

where I changed $aa$ in the note to $a$ here, and respectively $bb$ to $b$, and $ee$ to $p$. I have gotten stuck on the part that says that to obtain the $\mathbb R^3$ embedding, one should stereographically project this torus from the point $(\cos c\pi,0,\sin c\pi,0)$, as well as on the part about varying a parameter $ff$ from $0$ to $2\pi$, as this parameter is nowhere to be found in the parametric equations given.

I am familiar with the usual "north pole" stereographic projection,

$$\left(\frac{x_1}{1-x_4},\frac{x_2}{1-x_4},\frac{x_3}{1-x_4}\right)$$

but I do not know how to generalize this formula to the projection point given in the note.

In particular I wanted to replicate the surface depicted here in another surface plotting program, but I have been unsuccessful in doing so. Thanks for any help.

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According to this document, $f\!f$ is just another parameter controlling the surface. Your linked PDF says that it "isometrically rotates $S^3$ so that the Hopf circle $v = 0$ is the rotation axis." This circle is the set of points $(e^{iu}\cos a,e^{iu}\sin a)$ for all $u\in[0,2\pi)$. It lies in the subspace with basis $b_1=(\cos a,0,\sin a,0)$ and $b_2=(0,\cos a,0,\sin a)$, and the orthogonal subspace has basis $b_3=(-\sin a,0,\cos a,0)$ and $b_4=(0,-\sin a,0,\cos a)$. To perform the rotation, express the point to be rotated as a linear combination of $b_i$, and then rotate by an angle of $f\!f$ in the $b_3b_4$ plane.
As for projecting relative to an arbitrary point $P$, you just have to rotate the sphere so that $P$ coincides with $(0,0,0,1)$ and then apply the usual stereographic projection formula. For the particularly simple form $P=(\cos c\pi,0,\sin c\pi,0)$, this is essentially a rotation by $-c\pi$ in the $x_1x_3$ plane, namely $$\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}\mapsto\begin{pmatrix}x_1\cos c\pi+x_3\sin c\pi\\x_2\\x_3\cos c\pi-x_1\sin c\pi\\x_4\end{pmatrix}.$$ Actually, this maps $P$ to $(1,0,0,0)$, so switch $x_1$ and $x_4$ in the projection formula.
By the way, it seems a little too specific to call this sort of surface "the Bianchi-Pinkall torus". A cursory search reveals that what Pinkall proved [1] was that "the inverse image under the Hopf mapping of a simple closed curve on $S^2$ is a flat torus on $S^3$" [2]. So any smooth curve on the familiar 2-sphere gives you a flat torus in the 3-sphere, which you would probably call a Pinkall torus. I haven't been able to find out how Bianchi comes into the picture.
1. Pinkall, "Hopf Tori in $S^3$", Invent. math. 81, 1985.