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?

  • $\begingroup$ 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. $\endgroup$ Nov 28, 2013 at 10:47
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    $\begingroup$ Such polytope is called a hypersimplex (it depends on $n$ and $k$ of course). Google it to find out more. $\endgroup$ Nov 28, 2013 at 11:56
  • $\begingroup$ Thank you! I looked this up and it helped a bunch. See further clarification in the main question. $\endgroup$ Nov 30, 2013 at 4:07

2 Answers 2


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.)

To expand on achille hui's comment using this mental model: It's a simplex near the corners because when you cut close enough to the corner, one of the homothets is small enough to fit entirely inside the other, so the cross-section is just the smaller homothet.

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 $n=4$.

  • $\begingroup$ 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. $\endgroup$ Nov 30, 2013 at 4:14
  • $\begingroup$ Ah, no, sorry, in my little corner of convex geometry we say "symmetric" to mean "centrally symmetric"; I often forget that this is parochial. As for equal area or equal circumference: not in general, I think; e.g., the tetrahedron with tips cut off has facets of two different types, some tiny triangles and some big irregular hexagons. $\endgroup$
    – user21467
    Nov 30, 2013 at 13:04
  • $\begingroup$ At least you have convex geometry in your corner! ...re: the article above I mentioned which says all vertices have the same number of adjacent vertices, does this apply for general k? If so, how does that reconcile with combos of tiny triangles and hexagons? Any chance you can use your skills to help me visualize this? $\endgroup$ Dec 2, 2013 at 2:07
  • $\begingroup$ 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. $\endgroup$ Dec 11, 2013 at 23:16

If you intersect a hypercube perpendicular to its body diagonal then the various sections at vertex layers describe nothing but the quasiregular members of the simplex group $A_n$. In fact $$o3o...o3o4x = \sqrt2\cdot hull(\ o3o...o3o\ ||\ o3o...o3x\ ||\ o3o...x3o\ ||\ …\\ …\ ||\ o3x...o3o\ ||\ x3o...o3o\ ||\ o3o...o3o\ )$$ E.g. the cube can be seen along its body diagonal (up to the common scaling) as point (vertex, $o3o$) atop triangle ($x3o$) atop dual triangle ($o3x$) atop antipodal point ($o3o$).

If you'd intersect the hypercube inbetween 2 such consecutive non-extremal layers, then the edges, corresponding to the marked node of the one vertex layer side, continuously decrease in size, while the edges, corresponding to the marked node of the next vertex layer side, continuously increase in size. In fact, the formerly diagonal section of a square runs down to its tip, respectively from a tip down to the diagonal. Thus a general section could be described (up to general rescaling) by the Dynkin diagram $o3o...o3x3y3o...o3o$. Esp., when being midways inbetween 2 consecutive vertex layers, you would get $x=y$, i.e. these special midway sections are being described by the Wythoffian (multi)truncates $o3o...o3x3x3o...o3o$.

E.g. the equatorial section of the cube is just a regular hexagon ($x3x$).

An intersection within the extremal segments OTOH provides nothing but a mere scaling of the base figure, while running up to the tip of the corresponding vertex pyramid.

--- rk


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