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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 ?

Thanks!

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    $\begingroup$ Have you, by any chance, already seen this? $\endgroup$ Apr 11 '13 at 15:36
  • $\begingroup$ yep, I saw it, but they compute a parameterization of the surface. I would like to render these manifolds using raytracing (i.e., this only requires to numerically compute the intersection between a line and the manifold), and in this context, I'd like to understand why I would need 2 algebraic equations (+the equation of my line) to compute an (or several) intersection point(s). I didn't really understand how the "projection in 3D" worked in his paper as well :s Thanks! $\endgroup$
    – WhitAngl
    Apr 11 '13 at 15:58
  • $\begingroup$ (also, according to the images they show, their visualization resulting from their paper is quite different from the usual one that they displayed with Mathematica on the left) $\endgroup$
    – WhitAngl
    Apr 11 '13 at 18:31
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The surface Hanson visualized is a three-dimensional object in a two-dimensional complex space consisting of four real dimensions. Don't think of it as separate real and imaginary parts, just four variables with one constraint to define the surface.

The renormalization $ z_5=(-1/3)^{1/5} $ will produce Hanson's equation from what you have. Then he chooses three variables for the projection, in his case the two real parts and a linear combination of the imaginary parts. The fourth variable, a linearly independent combination of imaginary parts, is ignored. This is an orthogonal projection, like how the sun casts shadows when directly overhead.

As to why the two images on Hanson's website look different, it's because one is rotated relative to the other. You can see that in a fully interactive visualization like the one here.

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