I wonder what the plot of $x^2 + y^2 = 1$, with $x,y \in \CC^2$ looks like?

It's a 2-dimensional surface inside a four dimensional $\RR^4$, so to visualize it, we need to map it to $\RR^3$ somehow.

How? There are of course many ways, so let's just do one arbitrary way first, e.g., setting the imaginary
part of the second complex number to 0, i.e., we are *projecting* to that space...

In any case, the surface in $\RR^4$ is $(x_0+i x_1)^2 + (y_0 + iy_1)^2 = 1$

$\displaystyle \left(2 x_{0} x_{1} + 2 y_{0} y_{1}\right) i + x_{0}^{2} - x_{1}^{2} + y_{0}^{2} - y_{1}^{2}$

[x0^2 - x1^2 + y0^2 - y1^2, 2*x0*x1 + 2*y0*y1]

So the equations of the surface in $\RR^4$ is:

$x_0^2 - x_1^2 + y_0^2 - y_1^2 = 1$ $x_0x_1 + y_0y_1 = 0$

Hmmm, it's tempting to eliminate one variable, and end up with a surface in $\RR^3$, so might as well do that. We have $y_1 = -x_0 x_1 / y_0$ so $x_0^2 - x_1^2 + y_0^2 - (-x_0 x_1 / y_0)^2 = 1$

Multiply through by $y_0^2$:

$x_0^2 y_0^2 - x_1^2 y_0^2 + y_0^4 -x_0^2x_1^2 = y_0^2.$

(x, y, z)

$\displaystyle x^{2} y^{2} + y^{4} - x^{2} z^{2} - y^{2} z^{2} = y^{2}$

3D rendering not yet implemented