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This may be as much a question about computers as about math.

Let $C=\{f(r,s,t)=0\}$ be a curve in $\mathbb{CP}^2.$ By forgetting about the points at infinity we can view $C$ as a surface in $\mathbb{R}^4,$ given by the equations $\mathrm{Re}(f(r,s,1))=0$ and $\mathrm{Im}(f(r,s,1))=0.$ Letting $r=x+iy$ and $s=z+iw$, we now have two equations $u(x,y,z,w)$ and $v(x,y,z,w)$ cutting out $C$. Of course we can project our surface form $\mathbb{R}^4$ to $\mathbb{R}^3$ simply by forgetting the coordinate $w$.

I am interested in seeing how things like the genus of $C$ affect the resulting picture. The problem is, I'm not sure of the best way to obtain this picture. I need to tell the computer (say, Mathematica) to color all of the points $(x_0,y_0,z_0)$ so that the equations $u(x_0,y_0,z_0,w)=0$ and $v(x_0,y_0,z_0,w)=0$ have a simultaneous solution. If I were to pick some $w_0\in\mathbb{R},$ I could plot $u(x,y,z,w_0)=0$ and $v(x,y,z,w_0)=0$ to get an intersection curve $A_{w_0}.$ Then the shape I want is the union of all $A_{w_0}$s. But I'm not sure how to tell Mathematica to do this, or if it is computationally feasible, or if it is the best way to do it. (Presumably not.)

(I've heard of using ContourPlot3D in Mathematica for simpler versions of this task, but didn't really understand.)

If people think that this belongs in the Mathematica stack exchange, feel free to move it. I thought there might be a better mathematical formulation than the one I have above, which is why I posted here.


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Duplicate of this – John Habert Jan 30 '14 at 20:03
Thanks - I was having a computer issue and thought my first version had not successfully posted. – Rob Silversmith Jan 31 '14 at 19:51

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