Hypervolume of expanded $n$-simplex

Question

The hypervolume of the expanded $n$-simplex with side $\sqrt{2}$ appears to be $$\displaystyle\frac{\sqrt{\;n+1\;}\;(2n)!}{n!^3}$$ Does anyone know of a published reference to this result? Or can anyone prove it?

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An expanded $n$-simplex can be defined as the vertex figure of an $n$-simplex honeycomb (see the Wikipedia article "simplectic honeycomb") and its coordinates in $(n+1)$-dimensional space can be expressed as all coordinate permutations of $(1,-1,0,0,0,...)$. Low-order examples are the regular hexagon ($n=2$), the cuboctahedron ($n=3$) and the runcinated $5$-cell ($n=4$).

If this is a known result I’d like to cite a reference.

Answer 1

The formula can be derived by decomposing an expanded $n$-symplex in a disjoint union of $*$-rectified symplexes or $*$-rectified hypercubes. Computing the hypervolume of a symplex/hypercube is an easy task, so it is also computing the hypervolume of a $*$-rectified symplex/hypercube.

Rectification is the process of truncating a polytope by marking the midpoints of all its edges, and cutting off its vertices at those points.

Answer 2

I now have a satisfactory proof of this result of which the following is a summary. The bounty has now expired but I'm still interested to know if there's a published proof.

All the polytopes described below have edge length $\sqrt{2}$.

The n-simplex has coordinates which can be expressed in $(n+1)$-space as the permutations of $(1,0,0,...)$. It has$$ N_{mn} =\binom{n+1}{m+1} $$ m-faces $(m=0,1,2,...n-1)$, where an m=face is a vertex $(m=0)$, and edge $(m=1)$, a face $(m=2)$ and so on. Its volume is $$ v_n = \frac{\sqrt{n+1}}{n!} $$

The expanded $n$-simplex has vertices $\textbf{r}$ which can be expressed in $(n+1)$-space as the permutations of $(1,-1,0,...)$. It can be constructed from the expanded $(n-1)$-simplex by adding two $(n-1)$-simplices whose coordinates are the permutations of $(1,0,0,...)$ and $(-1,0,0,...)$, then shifting these simplices respectively by $-1$ and $+1$ units in the $(n+1)$'th dimension.

It can also be formed by expanding the $n$-simplex, a process which creates an $(n-1)$-face for each $m$-face of the $n$-simplex.

The vertices of the facet so created are those which maximise the dot product of $\textbf{r}$ with the position vector of the centroid of the $m$-face. There are $(m+1)(n-m)$ such vertices, whose coordinates are formed from the ways in which the value $-1$ can be assigned to $m+1$ positions in the vector $\textbf{r}$, and the value $1$ to the remaining $n-m$ positions (or vice versa depending on the relative magnitudes of $m+1$ and $n-m$).

The facet is thus the Cartesian product of an $m$-simplex and an $(n-m-1)$-simplex, and its $(n-1)$-volume is therefore $$ A_{mn} = v_{m}\,v_{n-m-1} = \frac{\sqrt{m+1}}{m!} \frac{\sqrt{n-m}}{(n-m-1)!} $$

Its distance from the centre of the expanded n-simplex is $$ D_{mn} = \sqrt{\frac{n+1}{(m+1)(n-m)}} $$

and its contribution to the volume of the expanded n-simplex is the volume of a pyramid having the facet as its base and height $D_{mn}$: $$ V_{mn} = \frac{1}{n} A_{mn} D_{mn} = \frac{1}{n} \frac{\sqrt{m+1}}{m!} \frac{\sqrt{n-m}}{(n-m-1)!} \sqrt{\frac{n+1}{(m+1)(n-m)}} $$ which can be expressed simply as an integer multiple of the $n$-simplex volume: $$ V_{mn} = \binom{n-1}{m} v_n $$

The total volume of the expanded $n$-simplex can now be obtained by summing over the facets: $$ V_n = \sum_{m=0}^{n-1} N_{mn} V_{mn} = \sum_{m=0}^{n-1} \binom{n+1}{m+1} \binom{n-1}{m} v_n \\ = \binom{2n}{n} v_n = \frac{(2n)!}{(n!)^2} v_n = \sqrt{n+1}\frac{(2n)!}{(n!)^3} $$

By growing the halves of the expanded $n$-simplex outwards from the two sides of an expanded $(n-1)$-simplex it can be shown that $V_n$ is also given by $$ V_n = 2\sum_{m=0}^{n-1} \binom{n}{m+1} \binom{n-1}{m} v_n $$