(Hyper)volume of the demi(hyper)cube

The demihypercube is formed by taking the convex hull of the set of points in $$\{0, 1\}^n$$ with an even number of $$1$$s. It can be thought of as taking a hypercube, and cutting off "every other" vertex.

Examples

For example, the $$3$$-demihypercube (which is also a regular tetrahedron) can be described with vertices $$(0,0,0)$$, $$(0,1,1)$$, $$(1,0,1)$$, and $$(1,1,0)$$.

The $$4$$-demicube is the $$16$$-cell.

Question

What is the hypervolume of the demihypercube under this construction?

(I couldn't find this information on Wikipedia or elsewhere, and I'm curious about it in part because it relates to my question "https://math.stackexchange.com/q/3813522/121988".)

• hal-upec-upem.archives-ouvertes.fr/hal-00793781/document Sep 5, 2020 at 1:16
• @JeanMarie, thanks! If you want to put this as a comment with the answer and a link to this, I'll delete my answer and accept yours. Sep 5, 2020 at 5:48
• Let us keep it like this. It has been a pleasure for me to discover this new geometrical "object". Sep 5, 2020 at 15:18

This might be a naive approach.

The demi-cube is the cube but, as you said, with $$2^n/2=2^{n-1}$$ of its vertices "cut off". How much do we cut off each time? We cut off a simplex spanned by the standard basis vectors $$e_1,...,e_n$$, the volume of which is $$1/n!$$.

So the demi-cube (constructed from the cube of edge-length 1) should have volume

$$V_n = 1-\frac{2^{n-1}}{n!}.$$

It is not quite what you got in your answer. But maybe the difference is just about the scale of the initial cube.

• This is because I had misunderstood the paper, and I quoted from the wrong section. Your answer is lovely, and I especially like the illustrations! Sep 7, 2020 at 19:07

As Jean Marie commented, the formula appears at the bottom of page 26 in the the paper "The volume product of convex bodies with many hyperplane symmetries" by Franck Barthe and Matthieu Fradelizi:

The demihypercube is obtained from the cube by cutting off $$2^{n-1}$$ simplicies whose vertices are a vertex $$x \in \{-1; 1\}^n$$ of the cube such that $$\prod x_i = -1$$ and its $$n$$ neighbours. They have a right angle at $$x$$ and the length of the edges through this vertex are $$2$$ thus $$|DH_n| = 2^n - 2^{n-1} \times \frac{2^n}{n!} = 2^n\left(1 - \frac{2^{n-1}}{n!}\right).$$

Note, the authors define the demihypercube on the convex hull of points with coordinates of $$\{-1, 1\}$$, so to get the volume of the construction in the question, divide by a factor of $$2^n$$, yielding $$1 - \frac{2^{n-1}}{n!}.$$

Edit: the original quote that I had taken from the paper gave the volume product, not the volume as I had asked for in the question. This answer has been corrected to reflect this.

• What is the scale of this formula? When I plug in $n=3$, I obtain $8/3>1$, which is apparently not the demi-cube contained in the unit cube of volume 1. Sep 7, 2020 at 9:52
• @M.Winter, I had quoted the volume product formula from the paper instead of the volume formula. The answer has now been updated to reflect this. Sep 7, 2020 at 19:08