A geometric realization of a 4-polytope with 7 vertices

Question

I am looking for a geometric realization of a 4-polytope with 7 vertices. A list can be found in "An Enumeration of Simplicial 4-Polytopes with 8 vertices" by Grünbaum and V. P. Sreedharan. For an example we have a 4 polytope called $P_1^7$ with a list of facets: 1256, 1245, 1234, 1237, 1345, 1356, 1267, 1367, 2367, 2345, 2356, where a digit represents a vertex of $P_1^7$. My question would be: how should I choose 7 arbitrary, but indexed (what I mean is that I know which one is 1 for example) points on the sphere, so that the 4-polytope given by the convex hull of these points have the same combinatorial type as listed above? Is there any intelligent way to do this?

Answer 1

For a general approach, you can use the technique of Gale diagrams to transform a $d$-dimensional polytope with $n$ vertices into an $(n-d-2)$-dimensional arrangement of (signed) points with $n$ vertices. In your case, your have $n-d-2=1$, so the arrangement is of a very low dimension and can be handled with geoemtric intuition. The process also works the other way around too, and so to construct your $4$-dimensional polytope with seven vertices we can start from a 1-dimensional Gale diagram.

This is a rather technical approach, but I will lay out the general approach (the details can be found in Chapter 6 of Ziegler's "Lectures on Polytopes"). You will need to find a 1-dimensional arrangement of seven signed points:

• indexed by $\{1,...,7\}$,
• where signed means that each point is either colored black or white,
• and so that if $\bar F\subseteq \{1,...,7\}$ is the complement of a facet of your polytope (which in your case is always of size three) then the corresponding poins are two black points surrounding a white point, or vice versa.

The complements of your facets are

$$145,\;147,\;167,\;245,\;247,\;267,\;345,\;347,\;367,\;456,\;567.$$

You can check that the following diagram has the desired properties:

Technically, the diagram has to satisfy some further properties, and this one does, so I will not talk about them.

Now, there is a sequence of steps we have to perform to obtain the vertices of the polytope. I will not explain them but just perform them here:

1. We have to choose some actual coordinates for the shown points. Since the arrangement is 1-dimensional, each position is described by a single number. In the order of the indexing we can choose the coordinates $6,7,1,2,3,4,5$.

2. Now we lift the arrangement by one dimension and get rid of the signs: a black point at position $x$ is replaced by a vector $(x,1)\in\Bbb R^2$, and a white point is replaced by $(-x,-1)\in\Bbb R^2$. If we put all these vectors as columns into a matrix, we get

$$M:=\begin{bmatrix} 6 & 7 & -1 & 2 & -3 & 4 & -5 \\ 1 & 1 & -1 & 1 & -1 & 1 & -1 \end{bmatrix}\in\Bbb R^{2\times 7}.$$

3. Let $U:=\mathrm{span}(M^\top)\subseteq\Bbb R^7$ be the column span of the transpose $M^\top\in\Bbb R^{7\times 2}$. This is a $2$-dimensional subspace of $\smash{\Bbb R^7}$. Thus, if $U^\bot$ denotes its orthogonal complement, then $U^\bot$ is a $5$-dimensional subspace of $\Bbb R^7$ (this is the crucial "duality step"; we transitioned from an arrangement of dimension $2=1+1$ to an arrangement of dimension $5=4+1$).

4. Determine a basis of $U^\bot$. For example \begin{align} u_1 &= [-5,\phantom+6,1,0,0,0,0],\\ u_2 &= [\phantom+4,-5,0,1,0,0,0],\\ u_3 &= [-3,\phantom+4,0,0,1,0,0],\\ u_4 &= [\phantom+2,-3,0,0,0,1,0],\\ u_5 &= [-1,\phantom+2,0,0,0,0,1]. \end{align} We put these as rows into a matrix: $$ \bar M = \begin{bmatrix} -5 & \phantom+6&1&0&0&0&0\\ \phantom+4&-5&0&1&0&0&0\\ -3&\phantom+4&0&0&1&0&0\\ \phantom+2&-3&0&0&0&1&0\\ -1&\phantom+2&0&0&0&0&1 \end{bmatrix}\in\Bbb R^{5\times 7}.$$

5. Read out the columns of that matrix, that is $$v_1=[-5,4,-3,2,-1],\;v_2=[6,-5,4,-3,2]\;\text{and}\;v_i=e_{j-2}\;\text{for $3\le i\le 7$}.$$

6. This set of vectors is "acyclic", that is, there is a vector $c\in\Bbb R^5$ with $\langle c,v_i\rangle >0$ for all $i\in\{1,...,7\}$. This follows from a property of the Gale diagram that I have not mentioned (namely, being totally cyclic). One choice of such a vector is

$$c=(1,7,1,1,17)\in\Bbb R^5.$$

7. Finally, the vertices of your polytopes are the points $p_i$, where $p_i$ is the intersection of the ray $\Bbb R v_i$ with the hyperplane $\langle c,\cdot\rangle=1$ (which is 4-dimensional). Choosing an appropriate basis in this hyperplane gave me the following result:

$$ p_1=\begin{bmatrix} -112\\-14\\0\\-56 \end{bmatrix},\; p_2=\begin{bmatrix} -184\\-16\\-4\\-86 \end{bmatrix},\; p_3=\begin{bmatrix} 7 \\ 0\\ 0 \\ 0 \end{bmatrix},\; p_4=\begin{bmatrix} 0\\7\\0\\0 \end{bmatrix}, $$ $$ p_5=\begin{bmatrix} 0 \\ 0\\7\\0 \end{bmatrix},\; p_6=\begin{bmatrix} 0\\0\\0\\7 \end{bmatrix},\; p_7=\begin{bmatrix} -119\\-7\\-7\\-49 \end{bmatrix}. $$

It is likely that you can find nicer coordinates if you make a better choice in any of the steps above (other coordinates for the diagram or a different basis for $U^\bot$). But at least all of the coordinates here are integers.

---

Here is Mathematica code that automates the Gale transformation:

vec = Transpose@NullSpace[{
   {-3,-2, 1, 0,-1, 2,-3}, 
   { 1, 1,-1, 1,-1, 1,-1}
}]

sol = FindInstance[
   Dot[#, {c1,c2,c3,c4,c5}]>0 &/@ vec, 
   {c1,c2,c3,c4,c5}
]
null = NullSpace[
   {{c1,c2,c3,c4,c5}}] /. sol[[1]
]

pnts = Table[
   Dot[v,n]/Dot[v,{1,1,1,1,1}], 
   {v,vec2}, {n,null}
]