Find a projectivity to create a graph.

Question

I have the tetrahedron {xyzt=0} in projective space with homogeneous coordinate (x,y,z,t). I need to create a graph but the tetrahedron in affine coordinate is {xyz=0} and I can't visualize the figure. So I need a projectivity to pass from {xyzt=0} to another tetrahedron, preferably equilateral. What can I do? Thanks

Answer 1

$xyzt=0$ indicates the union of four planes. The corresponding corners could be represented by the four canonical basis vectors of $\mathbb R^4$. These are the corners of the standard simplex. They satisfy the equation $x+y+z+t=1$. So the plane at infinity in your original setup would be the plane $x+y+z+t=0$, i.e. the plane with coordinate vector $(1,1,1,1)$. If you want to eventually drop the $t$ coordinate, you have to apply a projective transformation which maps that plane to the plane $(0,0,0,1)$, i.e. to $t=0$. Then you can dehomogenize your points by dividing by $t$, then drop the last coordinate.

If you want the result to be equilateral, then I'd suggest you write your transformation as a matrix whose columns are the Cartesian coordinates of a regular tetrahedron, homogenized by adding a $1$.

E.g.:

$$\begin{pmatrix}x'\\y'\\z'\\t'\end{pmatrix}= \begin{pmatrix} 1 & 1 & -1 & -1 \\ 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \begin{pmatrix}x\\y\\z\\t\end{pmatrix} \qquad\longrightarrow\qquad \begin{pmatrix}x''\\y''\\z''\end{pmatrix}= \frac1{t'}\begin{pmatrix}x'\\y'\\z'\end{pmatrix} $$

Or written excplicitely as a single step:

$$ \begin{pmatrix}x\\y\\z\\t\end{pmatrix}\mapsto \frac{1}{x+y+z+t} \begin{pmatrix} x+y-z-t\\ x-y+z-t\\ x-y-z+t \end{pmatrix} $$