User:Tetracube/Coordinates of uniform polytopes

This article describes the derivation of Cartesian coordinates for the vertices of uniform polytopes having the symmetry of the regular n-simplex, the n-hypercube, or the n-cross polytope.

Cartesian coordinates are useful in analysing these polytopes, since convex hull algorithms can be used to derive the face lattice of the polytope from them.

The n-cube/n-cross family
The coordinates of uniform polytopes with n-cube or n-cross symmetry can be derived directly from their Coxeter-Dynkin diagram as follows.

First, a base point is constructed by reading the Coxeter-Dynkin diagram from left to right, where the edge marked 4 is on the left:
 * 1) If the first node in the diagram is ringed, then the first coordinate of the base point is 1. Otherwise, it is 0.
 * 2) For each subsequent node, if the node is ringed, then the corresponding coordinate is the previous coordinate plus $$\sqrt{2}$$. Otherwise, the corresponding coordinate is a repetition of the previous coordinate.

After the base point is constructed, all permutations of coordinates and sign are taken (corresponding to reflecting them across the mirrors defined by the Coxeter-Dynkin symbol), which yields the vertices of the polytope in question.

For example, the cantellated tesseract has the Coxeter-Dynkin diagram:. Reading from left to right, the first node is ringed, so the corresponding coordinate is 1. The second node is not ringed, so the second coordinate is a repetition of the first, that is, 1. The third node is ringed, so it is $$1+\sqrt{2}$$, and the fourth node is not ringed, so it is a repetition of the third coordinate, $$1+\sqrt{2}$$. Hence, the base point of the cantellated tesseract is $$\left(1,\ 1,\ 1+\sqrt{2},\ 1+\sqrt{2}\right)$$. The coordinates of the vertices of the cantellated tesseract are therefore all permutations of sign and coordinates of this point.

Note that when all permutations of sign are taken, it is to be applied to each coordinate as a whole, and not to individual terms. For example, if a base point coordinate is $$1+\sqrt{2}$$, then there are two permutations of sign: $$\pm\left(1+\sqrt{2}\right)$$, not $$\pm1\pm\sqrt{2}$$.

This derivation produces uniform polytopes of edge length 2, and can be applied to the n-cube and n-cross themselves, yielding the coordinates $$\left(\pm1,\pm1,\pm1,\ldots\right)$$ for the n-cube and all permutations of $$\left(0,\ 0,\ \ldots,\ 0,\ \pm\sqrt{2}\right)$$ for the n-cross.

The n-simplex family
Coordinates for uniform polytopes in the n-simplex family may be obtained by a two-step process:


 * 1) First, they occur as a subset of the coordinates of a uniform (n+1)-polytope in the (n+1)-cube family.
 * 2) These coordinates, which are embedded in (n+1)-space, are mapped back to n-space via a suitable transformation.

Derivation from the (n+1)-cube family
The coordinates of uniform polytopes of the n-simplex family can be derived from the observation that they occur as facets in the uniform polytopes derived from the (n+1)-cube and (n+1)-cross.

More precisely, let C be any uniform (n+1)-polytope having the symmetry of an (n+1)-cube or (n+1)-cross. Remove the single node connected to the edge labelled 4 from its Coxeter-Dynkin diagram. Then the resulting graph is the Coxeter-Dynkin diagram of a uniform n-polytope A having the symmetry of an n-simplex, which occurs as (some of the) facets of C.

Conversely, take any uniform n-polytope A having the symmetry of an n-simplex. If a new node (ringed or otherwise) is joined to a terminal node in its Coxeter-Dynkin diagram with an edge labelled 4, then the new diagram describes a uniform (n+1)-polytope C having the symmetry of an (n+1)-cube (or cross), and having A as (some of) its facets.

As facets of C, instances of A lie in hyperplanes parallel to those bounding an (n+1)-cross. In particular, there is one such facet that lies in the hyperplane orthogonal to the vector $$(1,\ 1,\ 1,\ \ldots)$$.

Now, the coordinates of C are all permutations of sign and coordinates of some base point P. If only permutations of coordinates of P are taken (i. e., take only vertices of C with non-negative coordinates), then the resulting points will all have the same dot product with the vector $$(1,\ 1,\ 1,\ \ldots)$$. In other words, they lie on a hyperplane orthogonal to $$(1,\ 1,\ 1,\ \ldots)$$. Therefore, they are the coordinates of A embedded in (n+1) dimensions.

Mapping coordinates back to n-space
The coordinates of the n-simplex family of polytopes derived above are embedded in (n+1) dimensions. In order to map these coordinates back to n dimensions, a series of plane rotations may be used to re-orient the coordinates such that they lie in a hyperplane orthogonal to the vector $$(1,\ 0,\ 0,\ \ldots)$$, and then the first coordinate can be dropped to project them to n-space.

Since the coordinates in question lie in a hyperplane orthogonal to the vector $$(1,\ 1,\ 1,\ \ldots)$$, the required series of rotations has the property that they also map $$(1,\ 1,\ 1,\ \ldots)$$ to $$(\sqrt{n},\ 0,\ 0,\ \ldots)$$. Hence, they can be sequenced as follows:


 * Rotate $$(1,\ 1,\ \ldots,\ 1,\ 1,\ 1)$$ to $$(1,\ 1,\ \ldots,\ 1,\ \sqrt{2},\ 0)$$
 * Then rotate that to $$(1,\ 1,\ \ldots,\ \sqrt{3},\ 0,\ 0)$$
 * ... etc.,
 * Until finally $$(1,\ \sqrt{n-1},\ 0,\ 0,\ \ldots)$$ is rotated into $$(\sqrt{n},\ 0,\ 0,\ \ldots)$$.

In the first step, the rotation involves mapping $$(1,\ 1)$$ to $$(\sqrt{2},\ 0)$$; in the second step, it involves mapping $$(1,\ \sqrt{2})$$ to $$(\sqrt{3},\ 0)$$; and so forth. In general, the k&#x27;th step involves mapping $$(1,\ \sqrt{k})$$ to $$(\sqrt{k+1},\ 0)$$. The requisite rotation for the k&#x27;th step, therefore, can be expressed by the rotation matrix:


 * $$\left[\begin{array}{cc}

\sqrt{\frac{1}{k+1}} & \sqrt{\frac{k}{k+1}} \\ -\sqrt{\frac{k}{k+1}} & \sqrt{\frac{1}{k+1}} \end{array}\right]$$

Writing these rotations as n&times;n matrices and multiplying them according to their sequence yields the following matrix:


 * $$R_n=\left[\begin{array}{ccccccc}

\sqrt{\frac{1}{n}} & \sqrt{\frac{1}{n}} & \sqrt{\frac{1}{n}} & \ldots & \sqrt{\frac{1}{n}} & \sqrt{\frac{1}{n}} & \sqrt{\frac{1}{n}} \\ -\sqrt{\frac{n-1}{n}} & \sqrt{\frac{1}{n(n-1)}} & \sqrt{\frac{1}{n(n-1)}} & \ldots & \sqrt{\frac{1}{n(n-1)}} & \sqrt{\frac{1}{n(n-1)}} & \sqrt{\frac{1}{n(n-1)}} \\ 0 & -\sqrt{\frac{n-2}{n-1}} & \sqrt{\frac{1}{(n-1)(n-2)}} & \ldots & \sqrt{\frac{1}{(n-1)(n-2)}} & \sqrt{\frac{1}{(n-1)(n-2)}} & \sqrt{\frac{1}{(n-1)(n-2)}} \\ 0 & 0 & -\sqrt{\frac{n-3}{n-2}} & \ldots & \sqrt{\frac{1}{(n-2)(n-3)}} & \sqrt{\frac{1}{(n-2)(n-3)}} & \sqrt{\frac{1}{(n-2)(n-3)}} \\ & \vdots & & \ddots & & \vdots & \\ 0 & 0 & 0 & \ldots & -\sqrt{\frac{2}{3}} & \sqrt{\frac{1}{6}} & \sqrt{\frac{1}{6}} \\ 0 & 0 & 0 & \ldots & 0 & -\sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} \end{array}\right]$$

Since the vertices of the uniform polytope in (n+1)-space lie on the hyperplane orthogonal to $$(1,\ 1,\ 1,\ \ldots)$$, they will all have the same first coordinate after transformation by $$R_n$$, corresponding to the first row of the matrix. To project them to n-space, this first coordinate is simply dropped. This is equivalent to discarding the first row of the matrix:


 * $$R'_n=\left[\begin{array}{ccccccc}

-\sqrt{\frac{n-1}{n}} & \sqrt{\frac{1}{n(n-1)}} & \sqrt{\frac{1}{n(n-1)}} & \ldots & \sqrt{\frac{1}{n(n-1)}} & \sqrt{\frac{1}{n(n-1)}} & \sqrt{\frac{1}{n(n-1)}} \\ 0 & -\sqrt{\frac{n-2}{n-1}} & \sqrt{\frac{1}{(n-1)(n-2)}} & \ldots & \sqrt{\frac{1}{(n-1)(n-2)}} & \sqrt{\frac{1}{(n-1)(n-2)}} & \sqrt{\frac{1}{(n-1)(n-2)}} \\ 0 & 0 & -\sqrt{\frac{n-3}{n-2}} & \ldots & \sqrt{\frac{1}{(n-2)(n-3)}} & \sqrt{\frac{1}{(n-2)(n-3)}} & \sqrt{\frac{1}{(n-2)(n-3)}} \\ & \vdots & & \ddots & & \vdots & \\ 0 & 0 & 0 & \ldots & -\sqrt{\frac{2}{3}} & \sqrt{\frac{1}{6}} & \sqrt{\frac{1}{6}} \\ 0 & 0 & 0 & \ldots & 0 & -\sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} \end{array}\right]$$

This matrix transforms the coordinates of the uniform polytope from (n+1)-space to n-space.

The matrices $$R'_n$$ have the property that the bottom (n-2) rows of $$R'_n$$ are precisely the rows of $$R'_{n-1}$$ with a column of 0's added to the left.

These matrices may be expressed in a concise form by noting that each i&#x27;th row begins with (i-1) zeroes, followed by $$-\sqrt{\frac{n-i}{n-i+1}}$$, and then repetitions of $$\sqrt{\frac{1}{(n-i)(n-i+1)}}$$ until the end of the row.

The n-simplex
The coordinates of the origin-centered regular n-simplex can be obtained by applying $$R'_{n+1}$$ to all vertices of the n-cross with non-negative coordinates. The coordinates thus derived have some interesting properties.

Firstly, the resulting simplex is highly symmetrical with the coordinate axes. One of its vertices, the apex, lies along the line spanned by $$(1,\ 0,\ 0,\ \ldots)$$. The remaining points, its base, has reflective symmetry across a coordinate plane. The base itself is an (n-1)-simplex also having these same properties.

Secondly, these coordinates are related to the triangular numbers. Let $$T_n = \frac{n(n+1)}{2}$$ be the triangular numbers. Let $$S_n = n^2$$ be the square numbers. Define two sequences:


 * $$A_n=-\sqrt{\frac{S_n}{T_n}}$$

which is the negative square root of the quotient of the square numbers and the triangular numbers; and


 * $$B_n=\sqrt{\frac{1}{T_n}}$$

which is the square root of the reciprocal triangular numbers.

Then these coordinates may be expressed as:


 * $$\begin{array}{cccccc}

(B_{n}, & B_{n-1}, & \ldots & B_{3} & B_{2}, & B_{1}) \\ (B_{n}, & B_{n-1}, & \ldots & B_{3} & B_{2}, & A_{1}) \\ (B_{n}, & B_{n-1}, & \ldots & B_{3} & A_{2}, & 0) \\ (B_{n}, & B_{n-1}, & \ldots & A_{3} &    0, & 0) \\ & \vdots & & & \vdots & \\ (B_{n}, & A_{n-1}, & \ldots &    0 &     0, & 0) \\ (A_{n}, &      0, & \ldots &     0 &     0, & 0) \\ \end{array}$$

The apex of the simplex is $$(A_{n},\ 0,\ 0,\ \ldots)$$, and the reflective symmetry across a coordinate plane is conferred by the first two sets of coordinates, since $$A_{1}=-B_{1}$$.

For instance, here are the values of the coordinates for $$n = 8$$:


 * $$\begin{array}{cccccccc}

(1/\sqrt{36}, & 1/\sqrt{28}, & 1/\sqrt{21}, & 1/\sqrt{15}, & 1/\sqrt{10}, & 1/\sqrt{6}, & 1/\sqrt{3}, & 1 ) \\ (1/\sqrt{36}, & 1/\sqrt{28}, & 1/\sqrt{21}, & 1/\sqrt{15}, & 1/\sqrt{10}, & 1/\sqrt{6}, & 1/\sqrt{3}, & -1 ) \\ (1/\sqrt{36}, & 1/\sqrt{28}, & 1/\sqrt{21}, & 1/\sqrt{15}, & 1/\sqrt{10}, & 1/\sqrt{6}, & -2/\sqrt{3}, & 0 ) \\ (1/\sqrt{36}, & 1/\sqrt{28}, & 1/\sqrt{21}, & 1/\sqrt{15}, & 1/\sqrt{10}, & -3/\sqrt{6}, & 0, & 0 ) \\ (1/\sqrt{36}, & 1/\sqrt{28}, & 1/\sqrt{21}, & 1/\sqrt{15}, & -4/\sqrt{10}, & 0, & 0, & 0 ) \\ (1/\sqrt{36}, & 1/\sqrt{28}, & 1/\sqrt{21}, & -5/\sqrt{15}, & 0, & 0, & 0, & 0 ) \\ (1/\sqrt{36}, & 1/\sqrt{28}, & -6/\sqrt{21}, & 0, & 0, & 0, & 0, & 0 ) \\ (1/\sqrt{36}, & -7/\sqrt{28}, & 0, & 0, & 0, & 0, & 0, & 0 ) \\ (-8/\sqrt{36}, & 0, & 0, & 0, & 0, & 0, & 0, & 0 ) \\ \end{array}$$

The permutohedron of order n
An omnitruncated uniform polytope is a uniform polytope whose Coxeter-Dynkin diagram has every node ringed. The coordinates of an omnitruncated (n-1)-simplex, therefore, may be derived by adding a node to its Coxeter-Dynkin diagram, joined by an edge labelled 4, and deriving the non-negative coordinates of the resulting uniform n-polytope (which has n-cube symmetry). Assuming the new node is not ringed, the base point of this polytope is $$(0,\ \sqrt{2},\ 2\sqrt{2},\ 3\sqrt{2},\ \ldots\ (n-1)\sqrt{2})$$. The convex hull of all permutations of this point is the omnitruncated (n-1)-simplex.

Since the shape of the hull is unchanged by scaling and translation, we may divide these coordinates by $$\sqrt{2}$$ to obtain $$(0,\ 1,\ 2,\ 3,\ \ldots\ (n-1))$$, and then add $$(1,\ 1,\ 1,\ldots)$$ to obtain $$(1,\ 2,\ 3,\ \ldots\ n)$$. The permutations of this new base point form the permutohedron of order n.

Therefore, the permutohedron of order n is the omnitruncated (n-1)-simplex.

Listing of coordinates
All the following coordinates describe origin-centered polytopes with edge length 2.

In each dimension n, there are precisely $$2^n-1$$ distinct polytopes with hypercubic/cross polytope symmetry except in 2D, where the square is identical to its dual.

Due to the self-duality of the n-simplex, there are only as many uniform polytopes with n-simplex symmetry as half the number of non-palindromic binary strings with n digits plus the number of palindromic binary strings with n digits. In 3D, due to the coincidence of the rectified tetrahedron with the octahedron, some of these uniform polyhedra coincide with those having cubic symmetry. When deriving coordinates for the uniform n-simplicial polytopes, it does not matter whether the new node added to the Coxeter-Dynkin diagram is ringed or not; for simplicity, the following listing will list the base point resulting from adding a non-ringed node.

3D
(&dagger;)Note: these coordinates are not the usual coordinates for the tetrahedral polyhedra because the general rotation scheme does not take into account the coincidence of the rectified tetrahedron with the regular octahedron. As a result, they are have a different orientation more consistent with the general n-simplex. The coordinates for the octahedron (as rectified tetrahedron) is oriented such that four of its edges each bisects a coordinate axis.

(*)Due to the coincidence of the rectified tetrahedron with the octahedron, these starred forms coincide with the polyhedra listed among those with cubic symmetry. Their coordinates are listed here in tetrahedral orientation instead of the more usual cubic/octahedral axis-aligned orientation.

4D
(*) Note: due to the coincidence of the rectified 16-cell with the 24-cell, these starred forms have a higher degree of symmetry than merely hypercubic.