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What is the name of the polytope $\Sigma\cap (-\Sigma)$ for $\Sigma$ a $d-$simplex with barycenter at the origin?

In dimension $2$, one gets a hexagon, in dimension $3$ an octahedron (given by the $6$ midpoints of edges), in dimension $4$ a polytope with 30 vertices (given for example by all permutations of $(0,1,1,-1,-1)$).

(More generally, this polytop has ${2n\choose n}$ vertices in dimension $d=2n-1$ and $(2n+1){2n\choose n}$ vertices in dimension $d=2n$.)

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    $\begingroup$ Based on the answer of Vince, I suspect the 4-dimensional intersection to be the bitruncated 5-cell. $\endgroup$
    – M. Winter
    Jun 16, 2019 at 14:06

5 Answers 5

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Bacher writes in the comments: "In odd dimension $(2n-1)$, I get simply the polytope with all vertices of $[-1,1]^n$ having coordinate-sum $0$."

This means that in the odd dimensions, this polytope is a hypersimplex (actually a special case of it). This goes back to Laplace. Its volume is the Eulerian number $A_{2n-1,n}$ times $2^n/n!$. I don't immediately see what happens in the even case, but if you figure out explicitly what are the vertices, there is probably a good chance these polytopes are also related to hypersimplices. For refs and related results, see e.g. here.

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Just to supplement Roland Bacher's description, here is the octahedron in $\mathbb{R}^3$:


              polytope


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I don't think there's a standard or widely used name for this polytope, although I don't have the expertise needed to say that with great confidence. The dual polytope is the union of two antipodal simplices, and it's sometimes called the diplo-simplex, at least by Conway and Sloane and people influenced by them (see J. Conway and N. J. A. Sloane, The cell structures of certain lattices, in Miscellanea mathematica, Springer, Berlin, 1991, pp. 71–107). In this naming scheme, your polytope would be the dual diplo-simplex.

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  • $\begingroup$ In odd dimension $2n-1$, I get simply the polytope with all vertices of $[-1,1]^n$ having coordinate-sum $0$. Is this consistent with this terminology? $\endgroup$ Mar 31, 2011 at 15:44
  • $\begingroup$ Yes, and not just in odd dimensions: the dual of the diplo-simplex is always the cross section of the hypercube by the hyperplane on which the coordinate sum is zero. (The difference between even and odd dimensions is that when the cross section is even dimensional, its vertices are not actually vertices of the hypercube.) This is dual to the fact that the $n$-dimensional diplo-simplex is the orthogonal projection of the $(n+1)$-dimensional cross polytope onto the coordinate-sum-zero hyperplane. There ought to be a good name for this cross section of the hypercube, but I don't know of one. $\endgroup$
    – Henry Cohn
    Mar 31, 2011 at 18:58
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    $\begingroup$ (But you are right that describing it as a cross section of the hypercube is probably clearer than specifying its dual.) $\endgroup$
    – Henry Cohn
    Mar 31, 2011 at 19:03
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I do not know of any specific name for this family of polytopes, but their Coxeter diagrams are of a recognizable form.

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In fact, every second one is a hyper-simplex. In general, you can look them up on Wikipedia's list of uniform polytopes.

The intersection $\Sigma \cap(-\Sigma)$ of $d$-dimensional simplices can be constructed as

$$\Big\{\sum_{i=1}^{d+1} \lambda_i e_i \in\Bbb R^{d+1} \;\Big|\; \sum_{i=1}^{d+1}\lambda_i=1,\lambda_i\in\big[0,\frac{2}{d+1}\big]\Big\}.$$

This is almost the convex hull of the $e_i$, but the coefficients of the convex combinations are more restricted. With this, it is possible to prove what Vince's suspected about the coordinates of the vertices.

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Some insight might be gained by considering barycentric coordinates for the dual diplo-simplex. In two dimensions, we have the coordinates of the hexagon as the six permuations of (0, 1/3, 2/3). In three dimensions, the vertices of the octahedron are the six permutations of (0, 0, 1/2, 1/2).

Let d - 1 represent the number of dimensions. This simplifies some formatting, since barycentric coordinates will then have d terms.

In general, for even d, we have for vertices all permutations of (0, ..., 0, 2/d, ...,2/d), where there are d/2 copies each of 0 and 2/d. When d is odd, the vertices are all permutations of (0, ..., 0, 1/d, 2/d, ..., 2/d), where 0 and 2/d each occur (d-1)/2 times. (This was suggested by some work with Mathematica; the number of vertices works out, though I don't have a rigorous proof at the moment.)

So in four dimensions, the vertices are permutations of (0, 0, 1/5, 2/5, 2/5). This polytope has 10 truncated tetrahedra (the Archimedean truncation) as cells; the dual simplex truncates the original simplex in a nice way.

In five dimensions, we have permutations of (0, 0, 0, 1/3, 1/3, 1/3), so that the faces are 12 completely truncated 4-simplices (in the sense that the octahedron is the complete truncation of the tetrahedron).

The advantage of using barycentric coordinates is that the vertices are easy to describe, the symmetry is evident, and so the combinatorics are relatively easy to discern.

Was there a specific context in which this polytope arose? Are there any specific combinatorial properties of this polytope which are needed?

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  • $\begingroup$ If I am not mistaken, the 4-dimensional intersection is knows as bitruncated 5-cell. $\endgroup$
    – M. Winter
    Jun 16, 2019 at 14:02

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