Projecting points from 4d-space into 3d-space in Mathematica

Answer 1

One possibility:

1. Generate four (non-orthoganal) 3-vectors, \vec{v}_i from the center of the tetrahedron toward each vertex.
2. For each four position x = (x_1 .. x_4) form the vector sum \Sum_i x_i*\vec{v}_i.

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Of course this mapping is not unique in general, but you condition that the x_i's sum to 1 constrains things.

Answer 2

You want

   (1,0,0,0) -> (0,0,0)
   (0,1,0,0) -> (1,0,0)
   (0,0,1,0) -> (1/2,sqrt(3)/2,0)
   (0,0,0,1) -> (1/2,sqrt(3)/6,sqrt(6)/3))

And it is a linear transformation so you transform

   (x,y,z,w) - > (y + 1/2 * (z + w), sqrt(3) * (z / 2 + w / 6), sqrt(6) * w / 3)

Edit You want the center at the origin -- just subtract the average of the four points. Sorry

(1/2, sqrt(3)/6, sqrt(6) / 12)