What does the surface |z₁|² = |z₂|² look like?

Question

In a quaternionic plane there are 2 axes and each point corresponds to $q \in \mathbb C^2$. Now, $|z_1|^2 = |z_2|^2$ should define a surface that divides a 3-sphere $|z_1|^2 + |z_2|^2 = 1$ into two pieces, where $z_1, z_2 \in \mathbb C$. What does this surface look like? Describe the two pieces.

We know that a point say $q \in \mathbb C^2$ defines a circle, and in the special case where $q \in S^3$ then it should be a fiber in Hopf fibration.

We also know that the intersection of $|z_1|^2 = |z_2|^2$ and $|z_1|^2 + |z_2|^2 = 1$ consists of 4 unit quaternions, which under stereographic projection also defines the 16 vertices of a 4-cube in $\mathbb R^3$.

It is too difficult for me to see how the surface $|z_1|^2 = |z_2|^2$ cuts an $S^3$ into 2 parts, if anything it should cut it into 4 parts, (referring to the quaternionic plane where $S^3$ is represented as a circle.)

Reference: Zachary Treisman (2009) exercise 2.23. arXiv:0908.1205 [math.HO]

Answer 1

A number of misconceptions:

• The intersection of $|z_1|^2=|z_2|^2$ and $|z_1|^2+|z_2|^2=1$ is not just four quaternions, it's an entire $2$-torus worth of quaternions, defined by $|z_1|=1/\sqrt{2}$ and $|z_2|=1/\sqrt{2}$.
• Stereographic projection turns four points into four points, not sixteen.
• The $4$-cube resides in $\mathbb{R}^4$, not $\mathbb{R}^3$. (Unless you're talking about a projection of it, which is typical of depictions of the tesseract.)
• Four points would not cut $S^3$ into four pieces, any more than three would cut $S^2$ into three; the sphere will remain connected no matter how (finitely) many points you delete.

The two pieces $S^1\times S^1$ separates $S^3$ into are defined by $|z_1|<|z_2|$ and $|z_1|>|z_2|$. Can you figure out what these two $3$-dimensional shapes these are?