# Simplexity of the n-cube

 Importance: High ✭✭✭
 Author(s):
 Subject: Geometry
 Keywords: cube decomposition simplex
 Recomm. for undergrads: yes
 Posted by: mdevos on: August 6th, 2008

\begin{question} What is the minimum cardinality of a decomposition of the $n$-cube into $n$-simplices? \end{question}

A \emph{decomposition} of a polytope $P$ into $n$-simplices is a set of $n$-simplices which have pairwise disjoint interiors and have union equal to $P$. This is also known as a (generalized) triangulation.

Let $T(n)$ be the minimum cardinality of a decomposition of the $n$-cube into $n$-simplices (the answer to our question). It is trivial that $T(1) = 1$ and easy to see that $T(2) = 2$. A 3-dimensional cube may be decomposed into five simplices by cutting off every other corner as shown in the figure (from [JW]). This division is optimal, so $T(3) = 5$.

\Image{TetrahedronCube_1000.gif}

Chopping off every other corner of a 4-cube leaves a 16-cell (the 4-dimensional cross-polytope) which can then be decomposed into eight simplices (fix a vertex $x$ and then take each of the eight 4-simplices formed as the convex hull of $x$ and a facet which is not incident with $x$). This is also optimal, so $T(4) = 16$. Computer assisted searches have yielded other good decompositions in low dimensions (see [S]).

The decompositions of the 3 and 4-dimensional cubes described here do not generalize to higher dimensions. However, there is a naive decomposition of an $n$-cube into $n!$ simplices. Take the cube to be $[0,1]^n$ and let $S$ be the set of all points $(x_1,\ldots,x_n)$ for which $0 \le x_1 \le x_2 \ldots \le x_n \le 1$. Then $S$ is a simplex contained in our cube which contains the main diagonal from the origin to $(1,1,\ldots,1)$. Further, by permuting the terms $x_1,\ldots,x_n$ in the chain of inequlities, we get a total of $n!$ simplices which form a decomposition of the cube.

This naive decomposition is not optimal in dimensions 3 and 4 since our constructions show $T(3) \le 5 < 3!$ and $T(4) \le 16 < 4!$. Haiman [H] found a clever way to lift efficient lower dimensional decompositions to high dimensions thus achieving a significant improvement on our $n!$ upper bound. To state his result precisely, we require another parameter. Let $T^*(n)$ be the minimum cardinality of a decomposition of an $n$-cube into $n$-simplices with the following additional constraints: \begin{itemize} \item Every vertex of a simplex is a vertex of the cube. \item The intersection of any two simplices is a face of both of them. \end{itemize} It is immediate that $T(n) \le T^*(n)$, but to the best of our knowledge these parameters may always be identical. Indeed, this is a separate interesting question. Anyway, back to Haiman's bound. He proved that $T^*(kn) \le (T^*(n)/n!)^k (kn)!$. Using this inequality with either the 3 or 4-dimensional example from above would give an improvement on the $n!$ upper bound. However, best known is to plug in $T^*(7) \le 1493$, which gives a general upper bound of $T(7n) \le T^*(7n) < .840463^{7n}(7n)!$.

A natural lower bound on $T(n)$ can be obtained by a volume argument. Clearly, $T(n)$ must be at least the volume of an $n$-dimensional cube divided by the volume of the largest simplex it contains. Smith [S] improved upon this by moving the argument to hyperbolic space (where the volume of a cube is comparatively much larger than that of a simplex). His volume estimate here yields $T(n) \ge \frac{1}{2} \cdot 6^{n/2}(n+1)^{- \frac{n+1}{2} } n!$.

## Bibliography

[H] M. Haiman, A simple and relatively efficient triangulation of the n-cube, Discr. Comp. Geom. 6, 4 (1991) 287-289.

[JW] Jackson, Frank and Weisstein, Eric W. "Tetrahedron." From MathWorld--A Wolfram Web Resource.

[S] W. Smith, A lower bound for the simplexity of the n-cube via hyperbolic volume.

* indicates original appearance(s) of problem.