I would like to understand how I can visualize the quintic threefold $$ z_1^5 + z_2^5 + z_3^5 + z_4^5 +z_5^5 - 5\psi z_1z_2z_3z_4z_5 = 0$$
For a similar problem, Hanson proposes the following:
These images show equivalent renderings of a 2D cross-section of the 6D manifold embedded in CP4 described in string theory calculations by the homogeneous equation in five complex variables: $$ z_1^5 + z_2^5 + z_3^5 + z_4^5 +z_5^5 = 0$$ The surface is computed by assuming that some pair of complex inhomogenous variables, say $z_3/z_5$ and $z_4/z_5$, are constant (thus defining a 2-manifold slice of the 6-manifold), normalizing the resulting inhomogeneous equations a second time, and plotting the solutions to $z_1^5 + z_2^5 = 1$ The resulting surface is embedded in 4D and projected to 3D using Mathematica (left image)
Let me at least take Hanson's quintic, and try to understand. I first put the equation in inhomogenous form, assuming $z_5 \neq 0$: $$ (z_1/z_5)^5 + (z_2/z_5)^5 + (z_3/z_5)^5 + (z_4/z_5)^5 + 1 = 0$$ then
$z_3/z_5$ and $z_4/z_5$, are constant
and I'll assume equal to $1$. So: $$ (z_1/z_5)^5 + (z_2/z_5)^5 + 3 = 0$$ then
normalizing the resulting inhomogeneous equations a second time,
I guess this just means that we let $z_5=1$ : $$ z_1^5 + z_2^5 + 3 = 0$$
and plotting the solutions to $z_1^5 + z_2^5 = 1$
I guess he just had different values for $z_3/z_5$ and $z_4/z_5$ - right? Or did I misunderstand ?
The resulting surface is embedded in 4D and projected to 3D
what does that projection mean ? I just randomly set one of my 4D components (say, I set the imaginary component of $z_2$ to $\alpha$) ? Is-there something more fancy, or is-there something I misunderstood earlier in my reasoning ?
Addendum Even doing so, this would result in two algebraic equations to solve (one for the real part, one for the imaginary part) : why would this define a surface embedded in 3D rather than a curve ?