Tetrahedra require octahedra; 5-cells require...? It's well known that equilateral triangles tessellate $\Bbb R^2$ but regular tetrahedra do not tessellate $\Bbb R^3$. However, in three dimensions, we can make a a tessellation if we are permitted to also use octahedra to "fill in" the gaps that the tetrahedra leave. ![enter image description here]( **Question:** What happens in four dimensions? Do the 'gaps' again form cross-polytopes? Is a single convex polytope even sufficient? As a stretch question, if we demand the polytopes to be regular, can any finite collection fill such a gap? I'd love an answer to the question in general dimension, but I'd be satisfied for just four.

The figure you drew looks like a big tetrahedron, where you bisected each edge to cut off a smaller tetrahedron at each original edge, leaving the octahedron in the center.

This approach could be generalized to higher dimensions, and would not result in cross polytopes but in 4d leads to something with 10 cells, 5 of them octahedra and 5 tetrahedra, unless I'm mistaken.

That subdivision doesn't directly lead to a tiling of the space, however, unless you allow for filler polytopes of infinitely many different sizes, so you can combine larger simplices and then fill gaps between these again.