A pure geometry problem presented itself as I wondered about the math of compressive sensing.
<p>Consider the nD cube {(x<sub>1</sub>, … x<sub>n</sub>) | 0 &lt; j ≤ n → 0 &lt; x<sub>j</sub> &lt; 1}.
The subset of the cube with coordinates that sum to 1 is an n−1 dimensional regular simplex.
What shape is the subset whose coordinates sum to 2?
In general what of the subset P(n, k) where the coordinate sum is k &lt; n?
<p>It is of course still an n−1 dimensional convex polytope.
P(3, 2) is another regular simplex and in general P(p+q, p) is congruent to P(p+q, q) for p, q > 0.
<p>In general P(n, k) has C(n, k) vertices.
(C(n, k) = the binomial coefficient = n!/(k!(n-k)!).)
What’s P(4, 2)?
The vertices are the 6 points with two 1’s as coordinates and 0’s as the other coordinates.
It is clearly the regular octahedron.
<p>P(5, 2)
Miscellaneous facts<ul>
<li> is a 4D polytope with 10 vertices.
<li>A face of this polytope is 3D.
<li>The 1D edges of this figure are diagonals of the 2D squares that abut the 5D cube.
<li>Considering the 5 1D edges of the cube that emanate from the origin as 5 vectors, sums of two of this set form the vertices of the polytope.  
</ul>
<h3>Meandering in 5D</h3>
As we wander about inside P(5, 2) we watch the 5 coordinates drift between 0 and 1 while their sum remains 2.
When some coordinate bumps up against 1 then we have reached a 3D face for the other 4 coordinates can vary while our pinned coordinate ceases to vary.
At this stage four coordinates vary freely but must sum to 1.
This 3D face that we have reached is thus congruent to P(4, 1) which is a regular tetrahedron.
When some coordinate hits 0 then the other coordinates may continue to vary but now their sum is stuck at 2.
This face is a P(4, 2) which is an octahedron.
<p>We see thus that P(5, 2) is bounded by five regular tetrahedra and five regular octahedra.
I am sure it would be a lovely figure to contemplate, but it is not platonic for its faces are not congruent to each other.


When two of the five coordinates are stuck at 0 the other three sum to 2 and so produce a triangle which bounds two octahedra.
There are ten such triangles between octahedra.
When one coordinate is stuck at 0 and another at 1, the other three sum to 1 and we have a triangle between an octahedron and a tetrahedron.
There are 20 of these triangles.


<p>Another way to see what is happening is to consider P(5, 1.1).
This is nearly the 4D simplex but with its 5 vertices clipped off.
The exposed small face at each clipping is a tetrahedron.
We have a truncated 4 simplex.
As x moves from 0 to 1, the truncation of P(5, 1+x) increases until the exposed faces meet at points such as (1, 1, 0, 0, 0).
At this phase the faces of the 4D simplex, which begin as tetrahedra have also been truncated to become octahedra.
<p>This general process can be extended recursively to yet higher dimensions but with receding possibilities of visualization.
<hr><a href=http://www.jstor.org/pss/2690829>This earlier 1991 article</a> has results such as these.