I just discovered an awesomely cool fact about 4D...
The cell-first projection of the tetracube has the same envelope as the cell-first projection of the 16-cell: a 3D cube!
I stumbled upon this curious fact when doodling around with various projections. I observed that the cell-first projection of the octahedron consists of two interpenetrating triangles inside a hexagonal envelope. So I wondered what 4D object would project to two interpenetrating tetrahedra.
To make a long story short, the answer is: the 16-cell, projected cell-first.
This particular projection is very fascinating, since it is closely related to the geometry of the 3D cube. There are precisely two ways to inscribe a tetrahedron inside a cube. The edges of the tetrahedron would lie on the diagonals of the cube's faces: each face would have one diagonal that coincides with one of the tetrahedron's edges. If you take the set of diagonals not coincident with the tetrahedron's edges, you get a dual tetrahedron: precisely the other possible way to inscribe the tetrahedron.
These two possible ways to inscribe a tetrahedron in a cube corresponds with the closest and farthest cell of the 16-cell, in its cell-first projection to 3D. Let's call them F and B (for "front" and "back"). Now here's the interesting bit: there are 4 cells that share a face with F, and these cells project precisely into the volume between the inscribed tetrahedron and the cubical envelope in the projection. This is also true with cell B. This gives us 10 of the 16-cell's cells.
Now notice that the tetrahedra around F don't share a face with the tetrahedra around B. They correspond with two different diagonals on each face of the cubical envelope. But notice that a square with two diagonals is actually a projection of a tetrahedron! This means that the 6 faces of the cubical envelope correspond with 6 tetrahedral cells. So we have precisely 16 cells!
Note furthermore an unusual property of this projection: all of the edges of the 16-cell project onto the surface of the cubical envelope! There are no edges that lie inside the cubical envelope. The edges are projected precisely to the 12 edges of the cubical envelope plus the diagonals on each face.
So, the cell-first projections of the 4D cross polytope and the 4D measure polytope have identical envelopes. This is also true of the 2D cross and measure polytopes (although they happen to be the same thing in 2D), but it's not true in 3D: the cell-first projection of an octahedron is a hexagon, but the cell-first projection of a cube is a square. It seems that there is something special about 4D that makes the cross and the measure project cell-first to the same envelope. Does anyone know if this is true in all even dimensions, or is it really something unique to 4D?