This must be a standard exercise but I have a shape $[0,1]^3$ and I must express it has the union of tetrahedra joined at the faces. The vertices and edges are clear:
$V = \{ 0, 1 \} \times \{ 0, 1 \} \times \{ 0, 1 \}$
$E = \\ \{ \big((x,y,0)-(x,y,1) \big) : x,y \in \{ 0,1\} \} \cup \\ \{ \big((x,0,y)-(x,1,y) \big) : x,y \in \{ 0,1\} \} \cup \\ \{ \big((0,x,y)-(1,x,y) \big) : x,y \in \{ 0,1\} \} $
The first two tetrahedra are easy:
- $T_0 = \{ (0,0,0), (0,0,1),(0,1,0), (1,0,0) \} $
- $T_1 = \{ (1,1,1), (1,1,0), (1,0,1), (0,0,1) \}$
The remaining shape is a triangular prism. And I'm sure that can be split into tetrahedra. For this exercise, I need a careful enumeration of the remaning 2 or 3 pyramids, because this will get placed into another calculation (e.g. the construction of a 3-manifold).
I've narrowed it down to two possible shapes. It is a cone (or pyramid which is a "cone" with a polygonal base ) of an equilateral triangle + another vertex.
- $T_2 = \{ ?\; , (0,0,1),\,(0,1,0), \,(1,0,0) \} $
- $T_3 = \{ ? \;, (1,1,0), \,(1,0,1), \,(0,0,1) \}$
This discussion is not terribly exciting without pictures and I apologize, I can add a few images when I have time.
My partial progress
Another question is how many ways are there to decompose a cube into tetrahedrons. This is certainly not the only starting point. These can be connected by Pachner moves which generalize the "flip" of splitting a square into two triangles.
Here's another decomposition:
Let the coordinate axes be $x, y, z$. The cube can be split into 6 tetrahedra by taking all points such that $x \le y \le z$, all points such that $x \le z \le y$, and so on for each of the 6 possible orderings. These tetrahedra are not the same shape as your $T_0$ and $T_1$, however.