# A geometric realization of a 4-polytope with 7 vertices

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?

• It might be easier to consider the dual objects? Objects with larger numbers of vertices can constructed form smaller ones by a process of "slicing off" vertices (or edges, sometimes). – Donald Splutterwit Apr 21 at 14:00
• Are you asking in general, or for this specific example? – M. Winter Apr 22 at 20:49
• Thanks for the answers, I tried it with dual objects, but unfortunately I failed. For the second question, this specific example would be enough, but if there is a general method that would be more than enough. – Csaba Medgyes Apr 23 at 21:14

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

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

1. 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 $$5=4+1$$).

2. 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: $$\bar M = \begin{bmatrix} -5 & \phantom+6&1&0&0&0&0\\ \phantom+4&-5&0&1&0&0&0\\ -3&\phantom+4&0&0&1&0&0\\ \phantom+2&-3&0&0&0&1&0\\ -1&\phantom+2&0&0&0&0&1 \end{bmatrix}\in\Bbb R^{5\times 7}.$$

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

4. 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

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

1. 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}
]