We define the Hopf map as a function from $S^3$ into $S^2$, $f(z_1, z_2) = \frac{z_2}{z_1}$, where $S^3=\{(z_1, z_2) \in \mathbb C^2 | |z_1|^2 + |z_2|^2 = 1\}$ and $S^2$ is the Reimann sphere $z=\frac{z_2}{z_1}$.
Now, we want to find the inverse map $f^{-1}(z)$, (in order to find the latitudinal tori later.)
$f: (z_1, z_2) \to z$
$f^{-1}: z \to (z_1, z_2)$
If we use $z_1 = \frac{z_2}{z}$ and $z_2 = zz_1$, then we should be able to replace $z_1$ and $z_2$ to find the inverse map, and so we get $f^{-1}(z) = \frac{zz_1}{\frac{z_2}{z}} \Rightarrow f^{-1}(z) = z$, which helps little.
It is understandable that a function with two variables as input and one variable as output is so unlikely to be invertible. However, the input variables $z_1$ and $z_2$ are related since $|z_1|^2 + |z_2|^2 = 1$. Moreover, we will need an inverse Hopf map for showing that the tori $\sigma(f^{-1}(|z|=r))$ are tori of revolution. So there should be an algebraic way for getting an $S^1 \times S^1$ surface using the inverse Hopf map.
We know that in the Reimann sphere, $|z|=r$ defines a circle with the center point at the origin and radius $r$. On the other hand, by sending $z$ back to $\mathbb R^4$ using the inverse stereographic projection and applying $S^1$ action we can get a circle worth of unit quaternions in $S^3$. So, we have our $S^1 \times S^1$ in this way here, but how do they relate in an algebraic way to get a torus of revolution such that $(x_1^2 + x_2^2 + x_3^2 + R^2 - r^2)^2 = 4R^2(x_1^2 + x_2^2)$, where $S^3=\{(x_1, x_2, x_3, x_4) \in \mathbb R^4 | x_1^2 + x_2^2 + x_3^2 + x_4^2 = 1\}$, $r=x_3^2 + x_4^2$ which is the smaller radius and $R=(x_1^2 + x_2^2)^2 \pm \sqrt {r^2 - x_3^2}$ which is the bigger radius.
Found $r$ starting from $z = \frac{z_2}{z_1}$ and using $|z| = r$, next, found $R$ starting from $(|z_1| + x_3^2 + R^2 - r^2)^2 = 4R^2|z_1|$ and using $r = |z_2|$.
Reference:
- Treisman, Z., a young person's guide to the Hopf fibration, exercise 4.7., (2009), arXiv:0908.1205 [math.HO]
- The Hopf fibration visual model using Geogebra 3D calculator
Well, yes, but they're not that related. You can't solve for $z_2$ in terms of $z_1$, only its magnitude not its phase. Consider the real variables involved; the domain $S^3$ is three-dimensional and the range $S^2$ is two-dimensional and the Hopf map is continuous so you can expect (and by "invariance of domain" in topology) the function cannot be invertible.
That's not correct. The notation $f^{-1}$ here is being used for inverse image (rather than "image under inverse function"). The fiber of a point $z_1/z_2$ is the set $\{(e^{i\theta}z_1,e^{i\theta}z_2)\}$ which is indeed the orbit of $(z_1,z_2)$ under the action of $S^1$; the fibers are all circles. However, if you take a circle in $S^2$ and take the inverse image, you get a circle of circles, which is a torus. It turns out the fibers are "diagonal" Hopf circles in the torus, not your usual poloidal or toroidal circles.
Such a circle in the Riemann sphere $S^2$ we can say is $\{e^{i\phi}z\}$ so the inverse image will be the torus orbit $\{(uz_1,vz_2)\mid u,v\in S^1\}$ (this is an orbit of $S^1\times S^1$ acting on $S^3$ within $\mathbb{C}^2$). I'll let you verify under stereographic projection this is a torus of revolution.