I don’t know that we must characterize the intersection of the boundaries of our hedrons, but here goes anyway.
In 2D the boundaries are 1D and their intersection is 0D—a set of oriented points. (At some points polygon X enters Y and Y leaves X.) In n dimensions the intersection is a (n−2)-D complex.
In 3D the intersection of boundaries is a 1D complex whose vertices are at the piercings of a 2D face (boundary element) by a 1D edge of the other hedron. These do not alternate.
I stretch my intuition to 4D. The boundary intersection is 2D and its vertices (0D elements) are:
Alas this is false.
In 4D consider the two tetrahedra which are the convex hulls of
{[1, 1, 1, 1], [1, −1, −1, 1], [−1, 1, −1, −1], [−1, −1, 1, −1]} and
{[1, −1, 1, −1], [1, 1, −1, −1], [−1, 1, 1, 1], [−1, −1, −1, 1]}.
If the coordinates are called [w, x, y, z] note that w=z in the first simplex and w=−z in the second.
This must thus be true as well of the respective hulls and thus w=z=0 for their intersection.
The points [0, 1, 0, 0], [0, 0, 1, 0], [0, −1, 0, 0], [0, 0, −1, 0]
are mid points of edges of each of the two hulls.
The square that is the convex hull of these four points is thus the intersection of the two original tetrahedra.
Bahh.