# Intersection of hypercube and hyperplane - features of resulting polytope?

Consider a set in $R^n$ defined as the intersection of the unit hypercube $[0,1]^n$ with a hyperplane defined by $\sum x_i = k$. Assume $k \in (0,n)$ so the intersection has positive volume.

Can we characterize the resulting (n-1)-polytope? In particular, does there exist a condition where the polytope is a simplex? I am looking at an application of Stokes Theorem, and trying to figure out if its 2-faces are guaranteed symmetrical.

Edit: (New material below) Thanks for the comments. I looked up hypersimplex in much more detail.

Further reading and thinking has definitely clarified that the 2-faces are not guaranteed triangular, but are they perhaps symmetrical? This arXiv article seems to indicate that each vertex will have the same number of adjacent vertices... but I can't tell if integer k's are implicitly assumed. If correct for arbitrary k, then does this give symmetry of the 2-faces? Say, all equal area/circumference shapes, depending on k?

Intuitively, it seems there would be symmetry because all vertices are combinations of the same number of zeros, ones, and one k-1 element. This arXiv article seems to indicate that the polytope can be broken down into lower dimensional simplices, which also suggests some symmetry? I am having a hard time understanding the definitions in this one, though... maybe they have already solved my problem for me?

• It is a simplex when $k \in (0,1] \cup [n-1,n)$. You can gain a lot of insight by looking at the $3-dim$ case. When $n = 3$ and $k \in (1,2)$, the intersection is a hexagon. The hexagon is a regular one when $k = 3/2$. For other $k$, the edge lengths are alternating between two numbers. – achille hui Nov 28 '13 at 10:47
• Such polytope is called a hypersimplex (it depends on $n$ and $k$ of course). Google it to find out more. – Moishe Kohan Nov 28 '13 at 11:56
• Thank you! I looked this up and it helped a bunch. See further clarification in the main question. – user1166202 Nov 30 '13 at 4:07

The nicest description I know of these polytopes is as intersections of a positive and a negative homothet of the regular simplex (having the same centre). The easiest way to see this is to think of the cube as the intersection of two orthants, namely the usual positive orthant and its reflection in the point $(\frac12,\dotsc,\frac12)$; the intersection of the cube with a hyperplane is the intersection of the respective intersections of these orthants with that hyperplane. The orthants are cones whose cross-sections are regular simplices of the next lower dimension. (For example, the convex hull of the standard basis vectors is a regular simplex of the next lower dimension.)
The 2-faces are not in general centrally symmetric: a suitable cross-section will give a simplex intersected with a large negative copy which is almost but not quite large enough to contain the first simplex, so it cuts off the vertices slightly. All the 2-faces touch some vertex, so they'll be irregular hexagons — again, as noted in achille hui's comment for the case $n=3$; you can also visualize the tetrahedron with its tips cut off to get the case $n=4$.
• Steven, thank you for this response. Maybe I am being imprecise? See my edits to the original question as well. I am looking for a way to integrate over the surface of this hypersimplex, and in my specific application, the function I am integrating is separable and identical over each dimension. That is, $F(\boldsymbol{x}) = F(G(x_1),G(x_2),...,G(x_n))$. So, what I am looking for is symmetry in terms of equal area/circumference of the 2-faces, not their central symmetry about a midpoint. – user1166202 Nov 30 '13 at 4:14
• To add to this, I've been able to show (I think) that all the 2-faces are triangles or hexagons for arbitrary dimensions. We get at least some hexagons as long as $k>1$ or $k<n-1$, otherwise the shape is just a simplex and it's all triangles. Any hexagons are the same size as each other, but there is variation in the triangles. – user1166202 Dec 11 '13 at 23:16