convex-polytopes
For $d=3$, vertex coordinates of a regular simplex have a simple expression since vertices correspond to four vertices of a cube. Is there a simple expression for higher dimensions? In particular I'm interested in $d=2^n-1$, integer $n$.
Edit: by coordinates I mean points in $\mathbb{R}^d$. Every $d$-simplex has a simple expression for coordinates in $\mathbb{R}^{d+1}$, as Mariano shows below
Here's the answer. The main claims (Claims 1-4) I am fairly sure I got right, but I could easily have missed a case (or counted an extra case) in the later enumeration. If anybody finds a mistake, please comment. Let me remark that I find Claims 1-4 much more interesting than the subsequent enumeration based on them.
The simplex centroid $e=(1,1,1,1,1,1,1,1)$ is included in all our hyperplanes, but I'm generally not counting it as a centroid in the discussion below (so centroid means centroid of a $k$-dimensional face, with $k < 7$).
We'll divide the question into cases.
The first case we deal with is when we don't have any unexpected centroids. We start with three centroids $a$, $b$, and $c$. These will automatically generate $\bar{a}$, $\bar{b}$, $\bar{c}$. We will call any centroid other than these six an unexpected centroid. We represent our centroids as subsets of {$1,2,\ldots,8$}. If we have the centroid {$1,2,3$}$=\langle 1,1,1,0,0,0,0,0\rangle $, then we automatically have the centroid {$4,5,6,7,8$}$=\langle 0,0,0,1,1,1,1,1\rangle$ corresponding to the complement of the set. So, to summarize, the first case consists of hyperplanes which pass through exactly six centroids: $a,b,c,\bar{a},\bar{b},\bar{c}$.
Now, let's represent this case by putting the numbers {$1,2,\ldots,8$} on the vertices of a cube. The cube will have three faces corresponding to $a,b,c$ and the three opposite faces will correspond to $\bar{a},\bar{b},\bar{c}$. For example, if the sets were $a=${$1,2,3$}, $b=${ $1,5,6$} and $c=${$2,5,6,7$}, then the vertex $\bar{a}bc$ would contain {$5,6$}, the vertex $\bar{a}\bar{b}\bar{c}$ would contain {$4,8$}, and the vertex $\bar{a}b\bar{c}$ would be empty. It's not too hard to see that whether there is an unexpected centroid only depends on the positions of the empty vertices. It's also clear that rotations and reflections of this cube give equivalent sections.
Claim 1: If two adjacent vertices are empty, there is an unexpected centroid.
Proof: The two adjacent vertices form an edge. We might as well rotate the cube so that the empty edge is the $ab$ edge. Then we have $a \cap b = \emptyset$. This means that $a \cup b$ is an unexpected centroid (here we have to use the fact $a \neq \bar{b}$). Example: if $a =${$1,2$}, $b = ${$3,4,5$}, then $a \cup b =${$1,2,3,4,5$} is in the linear span of $a$ and $b$.
Claim 2: If two opposite vertices of the cube are empty, then there is an unexpected centroid.
Proof: We can rotate the cube so the empty vertices correspond to $\bar{a}\bar{b}\bar{c}$ and $abc$. Then, for example, if $a=(1,1,1,0,0,0,0,0)$, $b=(0,0,1,1,1,1,0,0)$ and $c=(1,0,0,0,0,1,1,1)$, we can take $a+b+c-e$ where $e$ is the all-ones vector, and get $(1,0,1,0,0,1,0,0)$.
Claim 3: If the odd- or even-parity vertices of the cube are empty, then there is an unexpected centroid.
Proof: Rotate the cube so that $\bar{a}\bar{b}\bar{c}$ is empty. Now, every coordinate is in exactly 1 or 3 of $a,b,c$. Thus, $\frac{1}{2}(a+b+c-e)$ is an unexpected centroid. Example $a=(1,1,1,0,0,0,1,1)$, $b=(0,0,0,1,1,0,1,1)$ and $c=(0,0,0,0,0,1,1,1)$, and $\frac{1}{2}(a+b+c-e) = (0,0,0,0,0,0,1,1)$
Claim 4: If none of the situations in Claims 1,2,3 hold, then there is no unexpected centroid.
Proof: We can rotate the cube so that the empty vertices are a subset of $\bar{a}bc$, $a\bar{b}c$ and $ab\bar{c}$. For there to be an unexpected centroid, you must be able to find $\alpha a + \beta b + \gamma c$ so that the coordinates of this vector take on two values. One of these coordinates is 0 (since $\bar{a}\bar{b}\bar{c}$ is not empty), meaning we can assume wlog that $\alpha, \beta, \gamma$ are either 1 or 0. But for $\alpha + \beta + \gamma$ to also be either 1 or 0, we need two of $\alpha, \beta, \gamma$ to be 0, which means that we don't get an unexpected centroid.
So now, we need to enumerate the number of ways of putting 8 elements onto the vertices of a unit cube so that at least one element is on each of the nonempty vertices. Since sections are equivalent under permutations of the coordinates, we should consider these to be 8 identical elements (so the only thing that matters is how many elements are on a vertex). There are four cases.
Case A: no empty vertices. There is just 1 way of doing this: putting one element on each vertex.
Case B: one empty vertex. There are 3 ways of doing this. Exactly one vertex will have two elements on it, and it can be either Hamming distance one, two, or three from the empty vertex.
Case C: two empty vertices. In this case, these two vertices must have Hamming distance 2. The two extra elements can either be on the same vertex (3 ways) or two different vertices (7 ways).
Case D: three empty vertices. In this case, any pair of these three vertices must have Hamming distance 2, so there's only one way of arranging them. The three extra elements can either be all on the same vertex (3 ways), divided two on one vertex and one on another (7 ways), or on three different vertices (4 ways).
This gives 28 essentially different sections with no unexpected centroids.
We now must count the cases with unexpected centroids corresponding to Claims 1-3. We'll deal with the situation in the Claims 1,2,3 separately.
Case of Claim 3 Let's start with the situation in Claim 3. First, we can assume that there are no empty vertices of the cube other than the two opposite ones (if this happens, we are in the Claim 1 situation, and we take care of it there).
We now can choose another centroid $d$ in the linear span of $a,b,c,e$ so that $a \cup b \cup c \cup d$ covers every coordinate exactly twice. By the criterion that there are six non-empty vertices, none of the six intersections $a \cap b$, $a \cap c$, etc. can be empty. We need to put the eight elements into these six intersections. This corresponds to putting 8 elements on the edges of a tetrahedron so that every edge corresponds to at least one element. There are 3 ways to do this (two extra on one edge, one extra on each of two opposite edges, and one extra on each of two adjacent edges). Claim 3 thus gives 3 more non-equivalent sections. Notice that if we had analysed Claim 3 by just looking at the symmetries of the cube (as we did for the cases without unexpected centroids), we would have obtained four non-equivalent sections.
Case of Claim 2 What I'd like to claim here is that this is really the situation in Claim 1 disguised. Maybe the best way to do this is by example. If we have $a=${$1,2,7,8$}, $b=${$3,4,7,8$}, $c=${$5,6,7,8$}, then the centroid {$7,8$} is in our hyperplane, and the hyperplane is thus generated by $a'=${$1,2$}, $b'=${$3,4$}, $c'=${$5,6$}, which is covered by Claim 1.
Case of Claim 1 Here, there are three possibilities. In the first one, there are four centroids $a$, $b$, $c$, $d$, with pairwise empty intersections so that $a \cup b \cup c \cup d =${$1,2,\ldots,8$}. The number of ways of doing this is the number of partitions of 8 into four non-empty parts, which is 5: {$(5,1,1,1), (4,2,1,1), (3,3,1,1),(3,2,2,1),(2,2,2,2)$}.
In the second possibility, we have three pairwise disjoint centroids $a$, $b$, $c$, with $a \cup b \cup c = e$, and also another centroid $d$ so that both $d\cap x$ and $\bar{d} \cap x$ are non-empty for $x=a,b,c$. The cardinalities of $a,b,c$ could be {$4,2,2$} or {$3,3,2$}. In either case, we get two non-equivalent sections, giving 4 total non-equivalent sections.
For the third possibility, we have three pairwise disjoint centroids $a$, $b$, $c$, with $a \cup b \cup c = e$, and we have two more pairwise disjoint centroids $f$ and $g$ so that $f \cup g = a \cup b$. In this case, the cardinality of $c$ can range from 1 to 4. I'll just list representative vectors for these possibilities. The coordinates considered are those not in $c$.
$c=4$
$(1,1,0,0),(0,1,1,0),(0,0,1,1),(1,0,0,1)$
$c=3$
$(1,1,1,0,0),(0,0,1,1,0),(0,0,0,1,1),(1,1,0,0,1)$
$c=2$
$(1,1,1,0,0,0),(0,0,1,1,1,0),(0,0,0,1,1,1),(1,1,0,0,0,1)$
$(1,1,1,1,0,0),(0,0,1,1,1,0),(0,0,0,0,1,1),(1,1,0,0,0,1)$
$(1,1,1,1,0,0),(0,1,1,1,1,0),(0,0,0,0,1,1),(1,0,0,0,0,1)$
$c=1$
$(1,1,1,1,1,0,0),(0,1,1,1,1,1,0),(0,0,0,0,0,1,1),(1,0,0,0,0,0,1)$
$(1,1,1,1,1,0,0),(0,0,1,1,1,1,0),(0,0,0,0,0,1,1),(1,1,0,0,0,0,1)$
$(1,1,1,1,0,0,0),(0,0,1,1,1,1,0),(0,0,0,0,1,1,1),(1,1,0,0,0,0,1)$
$(1,1,1,1,0,0,0),(0,1,1,1,1,0,0),(0,0,0,0,1,1,1),(1,0,0,0,0,1,1)$
This gives 9 more non-equivalent sections, making 49 altogether.
It does follow from Euler's theorem that every graph of a 3-polytope has either a triangle face or a vertex of degree 3. To see this note that if every face has 4 or more edges then $4F \le 2E$ (double count pairs (e,f) where e is an edge f is a face and f contains e.) and now look at Euler's theorem: $2V-2E+2F=4$ so $2V-E \ge 4$ and $E \le 2V-4$ and this imples that there is a vertex of degree 3.
You may want to look at the question and answer on Eberhard's theorem Characterizing faces of 3-dimensional polyhedra. (Related to Victor Eberhard's Theorem [1890]:)
High dimensional analogues of the result I mentioned in the spirit of your question are very interesting.
Best Answer
It is known that there is a regular simplex of side length $\sqrt{(d+1)/2}$ whose vertices are vertices of the cube $[-1,1]^d$ in $\Bbb{R}^d$ if and only if there exists a Hadamard matrix of order $d+1$; this is a square matrix of $\pm 1$-entries with pairwise orthogonal columns.
In particular, there exist Hadamard matrices of order $2^n$, one of which can be constructed using the recursive Sylvester's construction as explained on the above linked wikipedia page:
Let $H_0=[1]$ and $H_{n+1}=\left[\array{H_n & H_n \\\\ H_n & -H_n}\right].$
Note that the first column of $H_n$ consists only of ones. Delete it to obtain $2^n$ row vectors in $\Bbb{R}^{2^n-1}$. These are the coordinates of a regular simplex.