16-cell honeycomb (4D cross-polytope tesselation) A 4-dimensional cross-polytope (also called 16-cell) is a regular polytope whose vertices are all permutations of $(\pm1,0,0,0)$. It is known that this body tiles the space $\mathbb{R}^4$ by translation and that the set of translation vectors describing the tiling can be chosen to form a lattice $L$.
My question is how can we determine the generator matrix of $L$ (or give some equivalent explicit description)?
EDIT: The statement that 16-cell is a lattice-tile is incorrect. I apologize for this. See the answer below.
 A: That's not true, you can't do it by translations alone.
Apparently, two neighboring 16-cells must have a common 3D face, and their 3D faces are tetrahedra. Let's pick one of these, namely the one formed by the points $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, and $(0,0,0,1)$. There must be another cell that shares this face with our initial cell (which is centered at 0). Where would be the center of that cell? Must be beyond the center of the face, at the same distance from it as our cell's center was. Now, the face's center is $\left({1\over4},{1\over4},{1\over4},{1\over4}\right)$, hence the other cell's center must be at $\left({1\over2},{1\over2},{1\over2},{1\over2}\right)$.
What are the vectors from the center of that cell to its vertices? Obviously, $\left(-{1\over2}, -{1\over2},-{1\over2},{1\over2}\right)$, its permutations, and their opposites. (Note that they still have length 1 and are mutually perpendicular, except for those which are opposite, hence this really is an orthoplex, just like the one we started with; we didn't break it in the process.) Does that look like the permutations of $(\pm1,0,0,0)$? No. Hence you can't get either of these two cells from another by translation; there must be some rotation as well.
Looking in some other directions, we might notice an orthoplex in yet another orientation, where its center-to-vertex vectors are $\left({1\over2},{1\over2},{1\over2},{1\over2}\right)$, $\left(-{1\over2},-{1\over2},-{1\over2},-{1\over2}\right)$, and all permutations of $\left(-{1\over2}, -{1\over2},{1\over2},{1\over2}\right)$.
So no, while we surely can tile the 4D space with 16-cells to produce the thing known as 16-cell honeycomb, this is not done by translations alone.
