# Algebra behind equidistant subdivision of unit simplices

Let $k\in\mathbb Z_+$ and define the $k$-dimensional unit simplex: \begin{align*} \bigtriangleup_k\equiv\left\{(\lambda_1,\ldots\lambda_{k+1})\in\mathbb R^{k+1}\,\Bigg|\,\lambda_1,\ldots,\lambda_{k+1}\geq0\text{ and }\sum_{i=1}^{k+1}\lambda_i=1\right\}. \end{align*} My goal is to describe a rigorous algebraic construction of the equidistant subdivision of this simplex into smaller simplices that meet each other only at vertices, edges, or (hyper)faces.

Formally, let $\ell\in\mathbb N$. (Note: As @LordSharktheUnknown and @EthanBolker have pointed out in their comments, the construction under investigation may fail to work as desired if $\ell$ is too small as compared to $k$. I do not mind imposing additional assumptions on $\ell$, for example, that $\ell$ be “sufficiently” large and/or a multiple of $k+1$. All I care about is that I need the construction to work for infinitely many values of $\ell$ for each fixed $k$.) Consider $$\bigtriangleup_{k,\ell}\equiv\left\{\left(\frac{w_1}{\ell},\ldots,\frac{w_{k+1}}{\ell}\right)\,\Bigg|\,w_1,\ldots,w_{k+1}\in\mathbb Z_+\text{ and }\sum_{i=1}^{k+1}w_i=\ell\right\},$$ those points in $\bigtriangleup_k$ that are gained by dividing each edge of the original simplex into $\ell$ equal segments and then “cutting” $\bigtriangleup_k$ by parallel hyperplanes accordingly. The intuitive idea is pretty simple and is depicted in the following figure for $k=2$ and $\ell=3$.

Define $\mathscr A$ to be the set consisting of collections of the $k+1$ vertices of each of these smaller polyhedra in the equidistant subdivision. That is, $(\boldsymbol{\lambda}^{(1)},\ldots,\boldsymbol\lambda^{(k+1)})$ is in $\mathscr A$ if

• $\boldsymbol\lambda^{(j)}\in\bigtriangleup_{k,\ell}$ for every $j\in\{1,\ldots,k+1\}$;
• the vectors $\boldsymbol\lambda^{(1)},\ldots,\boldsymbol\lambda^{(k+1)}$ are affinely independent—that is, $\alpha_1,\ldots,\alpha_{k+1}\in\mathbb R$, $\sum_{j=1}^{k+1}\alpha_j\boldsymbol\lambda^{(j)}=\mathbf{0}$ and $\sum_{j=1}^{k+1}\alpha_j=0$ together imply $\alpha_1=\cdots=\alpha_{k+1}=0$; and
• $\boldsymbol\lambda^{(1)},\ldots,\boldsymbol\lambda^{(k+1)}$ are “adjacent” in the sense that $$\lambda_i^{(\alpha)}-\lambda_i^{(\beta)}\in\left\{-\frac{1}{\ell},0,\frac{1}{\ell}\right\}$$ for every $i,\alpha,\beta\in\{1,\ldots,k+1\}$.

For each $(\boldsymbol\lambda^{(1)},\ldots,\boldsymbol\lambda^{(k+1)})\in\mathscr A$, let $S_{(\boldsymbol\lambda^{(1)},\ldots,\boldsymbol\lambda^{(k+1)})}$ denote the convex hull of $\{\boldsymbol\lambda^{(1)},\ldots,\boldsymbol\lambda^{(k+1)}\}$.

My goal is to provide an elementary (!) proof showing that the collection $$\mathscr S\equiv\{S_{(\boldsymbol\lambda^{(1)},\ldots,\boldsymbol\lambda^{(k+1)})}\,|\,(\boldsymbol\lambda^{(1)},\ldots,\boldsymbol\lambda^{(k+1)})\in\mathscr A\}$$ constitutes a simplicial subdivision of $\bigtriangleup_k$, that is:

• $\bigtriangleup_k=\bigcup_{S\in\mathscr S}S$; and
• if $S,S'\in\mathscr S$, then $S\cap S'$ is either empty or a common face of both $S$ and $S'$.

While this seems intuitively obvious (see the figure), I have trouble:

$\phantom{\text{i}}$(i) finding the adjacent grid points in whose convex hull a given $\boldsymbol\lambda\in\bigtriangleup_k$ lies; and

(ii) showing that if $(\boldsymbol\lambda^{(1)},\ldots,\boldsymbol\lambda^{(k+1)}),(\boldsymbol\mu^{(1)},\ldots,\boldsymbol\mu^{(k+1)})\in\mathscr A$, then $$\operatorname{hull}\left(\{\boldsymbol\lambda^{(1)},\ldots,\boldsymbol\lambda^{(k+1)}\}\right)\cap\operatorname{hull}\left(\{\boldsymbol\mu^{(1)},\ldots,\boldsymbol\mu^{(k+1)}\}\right)=\operatorname{hull}\left(\{\boldsymbol\lambda^{(1)},\ldots,\boldsymbol\lambda^{(k+1)}\}\cap\{\boldsymbol\mu^{(1)},\ldots,\boldsymbol\mu^{(k+1)}\}\right).$$

Any suggestion on what proof strategy could make the perhaps “obvious” yet tedious algebra work efficiently would be appreciated.

• In three dimensions, I'm not sure all the cells you get are simplices. – Lord Shark the Unknown Sep 15 '18 at 18:36
• As @LordSharktheUnknown notes. Slicing a tetrahedron with four planes parallel to the faces and bisecting the altitudes leaves an octahedron in the middle. – Ethan Bolker Sep 15 '18 at 18:39
• @EthanBolker I see. However, doesn’t requiring that every collection in $\mathscr A$ be affinely independent ultimately generate simplices along the grid points? – triple_sec Sep 15 '18 at 18:47
• @EthanBolker OK, what if $\ell$ is also assumed to be a multiple of $k+1$? In that case, it seems to me that it can be arranged that the barycenter of the original simplex $\bigtriangleup_k$ be a member of some collection in $\mathscr A$, right? I will make this extra assumption more explicit in the hope it helps. – triple_sec Sep 15 '18 at 18:56
• Did you draw the picture (in principle) for cutting the tetrahedron with four slices parallel to each face (or three - not sure how you are counting)? If that works then you're probably OK, but I suspect it won't. I've no time to try it now. – Ethan Bolker Sep 16 '18 at 0:05

## 1 Answer

Disclaimer: I don't really know much about polytopes; this answer is mostly just compiling facts and references, since even though this is something I "know", I found it hard to pin down anywhere that proves enough for me to believe it.

The desired goal is not possible, regardless of any tweaks you make to it, if indeed it follows from your conditions that this would provide a decomposition of the regular simplex into smaller regular simplices. Intuitively this follows immediately from the fact that vertices, which would follow from a "corrected" version of the adjacency condition (as it stands you have lots of simplices intersecting each other "badly", e.g. for $$k=4$$ and $$\ell=5$$ notice that conv(2111,1200,1020,1002) is probably not one of the simplices you wanted...).

See the third section (that is, after the second line of stars) in this post. This spins off into a number of references, many of which are broken links, but the long and short of it is that there should be an elementary proof using ditopal angles (the generalizations of angles, dihedral angles, dichoral angles,...) Based on this page, it seems that this proof is contained, at least implicitly, in Coxeter's book Regular Polytopes (see Chapters 7 and 13, perhaps?). The Wolfram MathWorld page links to a Gardner book which cites the fact in three dimensions.

A fact which may help if you try to pin down an exact reference for yourself: breaking a polytope $$X$$ into smaller copies of $$X$$, each of the same size, is only possible if $$X$$ [and its isometric images] can fill space. The latter is already a rare property for regular polytopes to possess, and there are four-dimensional regular polytopes which do fill space but still cannot be broken down into smaller copies.

• Thank you for the detailed explanation! I abandoned this attempt for a proof after others pointed out the elusiveness of the geometric intuition gleaned from certain simple low-dimensional cases, which does not generalize at all to higher-dimensional cases. As I mentioned in a comment above, all I needed was to show that any simplex can be divided into arbitrarily small subsimplices. The barycentric subdivision, which at first seemed less intuitive to me than the attempted equidistant division, works well for this purpose. – triple_sec Oct 24 '18 at 17:39