Platonic Solids

This math worksheet goes with my Platonic Solid SymMath notebook.

Defining our terms:

$n =$ dimension of manifold which our shape resides in.
$\tilde{T}_i \in \mathbb{R}^{n \times n} =$ i'th automorphic transform in the minimal set.
$\tilde{\textbf{T}} = \{ 1 \le i \le p, \tilde{T}_i \} =$ minimum set of automorphic transforms that can be used to recreate all automorphic transforms.
$p = |\tilde{\textbf{T}}| =$ the number of minimal automorphic transforms, which we can assert for n-dimensional shape will be $(n-1)$ dimensions to cover the surface of the shape.
$T_i \in \mathbb{R}^{n \times n} =$ i'th automorphic transform in the set of all unique transforms.
$\textbf{T} = \{ T_i \} = \{ 1 \le k, i_1, ..., i_k \in [1,m], \tilde{T}_{i_1} \cdot ... \cdot \tilde{T}_{i_k} \} =$ set of all unique automorphic transforms.
$q = |\textbf{T}| =$ the number of unique automorphic transforms.

Vertexes / 0-forms:

$v_1 \in \mathbb{R}^n =$ the position of some arbitrary initial vertex.
$\textbf{v} = \{v_i \} = \{ T_i \cdot v_1 \} =$ the set of all vertices.
$m = |\textbf{v}| =$ the number of vertices.
(Notice that $m \le q$, i.e. the number of vertices is $\le$ the number of unique automorphic transforms.)
$V \in \mathbb{R}^{n \times m}=$ matrix with column vectors the set of all vertices, such that $V_{ij} = (v_j)_i$.
$P_i \in \mathbb{R}^{m \times m} =$ permutation transform of vertices corresponding with i'th transformation, such that $T_i V = V P_i$.

And now half starting fresh...
Some more notation:

$S =$ solid. Maybe for convenience write $S^n$ to denote the dimension of the solid. Maybe I'll use a second index to denote which unique polytope of this dimension, like maybe $S^n_m$ is the m-sided polytope of n-dimensions.
$n \in \mathbb{N} =$ dimension of the manifold this solid resides in.
$v_j(S) = $ the j'th vertex of solid S.
$\textbf{v}(S) = \left\{ v_j(S) \right\} =$ set of vertexes of the solid S.
$q_k(S) = $ number of k-forms in solid S. 0-forms are vertexes, so $|\textbf{v}(S)| = q_0(S)$.
$\partial S = $ the solid that is the surface element of solid S. For regular convex polytopes there is only one shape, unlike archimedean solids which have multiple shapes in its surface.
$\star S = $ dual of solid S.
$\textbf{T}(S) = $ the set of automorphic transforms -- rotations only -- associated with this shape, such that the vertexes, edges, and faces are all preserved. Something something about duals means that preserving vertexes means preserving faces right?

In 2D:

Let $S^2_k$ be our regular convex polytope in 2D, i.e. polygon, with k sides, for $k \ge 3, k \in \mathbb{N}$.

A k-sided regular convex polygon has k vertexes (0-forms), i.e. $q_0(S^2_k) = k$.

Vertexes can be defined as $v_j = \left[ \begin{matrix} cos(\frac{2 \pi j}{k}) \\ sin(\frac{2 \pi j}{k}) \end{matrix} \right]$.

It has k edges (1-forms), i.e. $q_1(S^2_k) = k$.

So for 2D regular convex polytopes we see the number of vertexes and edges match, i.e. $q_0(S^2_k) = q_1(S^2_k) = k$.

From this we can use the Euler characteristic equation: $\chi = V - E + F$, where V = the number of vertexes, E = the number of edges, F = number of faces. So $\chi = q_1(S^2_k) + q_0(S^2_k) = - k + k = 0$.

For automorphic transform groups in 2D we also have a group size of $|\textbf{T}(S^2_k)| = k = q_0 = q_1$ rotations that preserve the shape.
For our earlier choice of vertices, the k'th transform takes the form of $T^2_{p,k} = \left[\begin{matrix} cos(\frac{2 \pi k}{p}) & -sin(\frac{2 \pi k}{p}) \\ sin(\frac{2 \pi k}{p}) & cos(\frac{2 \pi k}{p}) \end{matrix}\right]$.

Notice our solids are self-dual, so a k-sided object's dual is another k-sided object, with vertexes replaced with edges and edges replaced with vertexes.
i.e. $\star S^2_k = S^2_k$. But mind you, if you want to split hairs about the locations of the vertexes, then all the sudden we have to distinguish between our solids.

In 3D we only have 5 regular convex polytopes:

I'll keep the same convention from 2D denotation: my denotation of $S^3_m$ will be m-sided regular convex polytopes in 3-dimensions.

$q_k(S^3_m) = $ the number of k-forms in our m-sided 3D regular convex polytope, i.e. platonic solid.
so $q_0(S^3_m) = $ the number of vertexes.
so $q_1(S^3_m) = $ the number of edges.
so $q_2(S^3_m) = $ the number of faces.
$\star S^3_m = $ the solid's dual solid.
$|\textbf{T}(S^3_m)| = $ the number of automorphic transforms associated with our solid.
$\partial S^3_p = S^2_q = $ the 2D q-sided solid that makes up surface of the 3D p-sided solid. TODO later come up with an equation that relates p and q.
$q_k(\partial S^3_m) = $ the number of k-forms of a surface-element of the 3D regular convex polytope. $q_1(\partial S^3_m) = $ the number of sides of the polygons that cover our m-sided polyhedron. For 2D this is also equal to the number of vertexes, and to the size of the automorphic transform group.
$\chi = q_0(S^3_m) - q_1(S^3_m) + q_2(S^3_m) = $ the Euler characteristic.

name	$S^3_m$	$q_0(S^3_m)$	$q_1(S^3_m)$	$q_2(S^3_m)$	$\chi$	$\star S^3_m$	$\|\textbf{T}(S^3_m)\|$	$\partial S^3_m$	$m = q_0(\partial S^3_m) = q_1(\partial S^3_m) = \|\textbf{T}(\partial S^3_m)\|$
tetrahedron	$S^3_4$	4	6	4	2	$S^3_4$	12	$S^2_3$	3
cube	$S^3_6$	8	12	6	2	$S^3_8$	24	$S^2_4$	4
octahedron	$S^3_8$	6	12	8	2	$S^3_6$	24	$S^2_3$	3
dodecahedron	$S^3_{12}$	20	30	12	2	$S^3_{20}$	60	$S^2_5$	5
icosahedron	$S^3_{20}$	12	30	20	2	$S^3_{12}$	60	$S^2_3$	3

If our notation is that $S^3_m$ represents a m-sided regular convex polytope in 3D then of course the number of 2-forms of $S^3_m$ will be m, so $q_2(S^3_m) = m$.
In general I will call a m-polytope in n-dimensions $S^n_m$ and therefore it will have m surface polytopes, i.e. $q_{n-1}(S^n_m) = m$.

One easy rule to see is that the number of faces of the dual ($q_2(\star S^3_m)$) is equal to the number of vertexes of the original shape ($q_0(S^3_m)$), and same vice versa, that $q_0(\star S^3_m) = q_2(S^3_m)$.
In general, $q_k(S^n_m) = q_{n-k}(\star S^n_m)$.

Another rule is that the automorphic transform set size of a shape and its dual match. In fact the groups are identical. Because both groups must preserve both the vertexes and the sides, and between duals these two properties are exchanged.
So $\textbf{T}(S^n_m) = \textbf{T}^n(\star S^n_m)$.

Another rule that will hold true to higher dimensions is that the transform group size $|\textbf{T}(S^3_m)|$ is equal to the number of sides of our shape, multiplied by each side's transform group size. So $|\textbf{T}(S^3_m)| = q_2(S^3_m) \cdot |\textbf{T}(\partial S^3_m)| = m \cdot |\textbf{T}(\partial S^3_m)|$.
In general: $|\textbf{T}(S^n_m)| = q_{n-1}(S^n_m) \cdot |\textbf{T}(\partial S^n_m)| = m \cdot |\textbf{T}(\partial S^n_m)|$.
expanded: $|\textbf{T}(S^n_m)| = \underset{k=0}{\overset{n-2}{\Pi}} q_{n-1-k}(\partial^k S^n_m)$

Another rule relates the number of edges and the number of faces to the number of sides of the surface elements: $2 \cdot q_1(S^3_m) / q_2(S^3_m) = q_1(\partial S^n_m)$.
This is going to hold true in 4D (and further?) as $2 \cdot q_{n-2}(S^n_m) / q_{n-1}(S^n_m) = \frac{2}{m} \cdot q_{n-2}(S^n_m) = q_{n-2}(\partial S^n_m)$.

This rule can be used to determine the number of sides of the $(n-1)$-dimensional solids that cover our n-dimensional solids,
for $\partial S^n_m = S^{n-1}_r$:
$\frac{2}{m} \cdot q_{n-2}(S^n_m) = q_{n-2}(\partial S^n_m) = q_{n-2}(S^{n-1}_r)$,
combined with $q_{n-2}(S^{n-1}_r) = r$,
gives us: $\frac{2}{m} \cdot q_{n-2}(S^n_m) = r$.
so for our n-dimensional m-sided regular convex polytope $S^n_m$, its $(n-1)$-sided surface elements $\partial S^n_m = S^{n-1}_r$ have $r = \frac{2}{m} \cdot q_{n-2}(S^n_m) = 2 \cdot q_{n-2}(S^n_m) / q_{n-1}(S^n_m)$ sides.

Also to calculate the number of edges of a 3D regular convex polytope: if we know a shape (by its number of sides) and its dual (by its number of sides), we know $q_0(S^3_m)$ and $q_2(S^3_m)$, and can calculate from $\chi = 2$ our remaining $q_1(S^3_m) = q_0(S^3_m) + q_2(S^3_m) - \chi = q_2(S^3_m) + q_2(\star S^3_m) - \chi$.

For more on some exact tables of vertexes, transforms, and multiplication tables between transforms and either vertexes or transforms, check out my symmath work on Platonic Solids.

4D regular convex polytopes:

Euler characteristic: $\chi = q_0 - q_1 + q_2 - q_3$

name	$S^4_m$	$q_0(S^4_m)$	$q_1(S^4_m)$	$q_2(S^4_m)$	$q_3(S^4_m)$	$\star S^4_m$	$\|\textbf{T}(S^4_m)\|$	$\partial S^4_m$	$q_0(\partial S^4_m)$	$q_1(\partial S^4_m)$	$q_2(\partial S^4_m)$	$\|\textbf{T}(\partial S^4_m)\|$	$q_0(\partial^2 S^4_m) = q_1(\partial^2 S^4_m) = \|\textbf{T}(\partial^2 S^4_m)\|$
5-cell	$S^4_5$	5	10	10	5	$S^4_5$	60	$S^3_4$	4	6	4	12	3
hypercube, i.e. 8-cell	$S^4_8$	16	32	24	8	$S^4_{16}$	192	$S^3_6$	8	12	6	24	4
4-orthoplex, i.e. 16-cell	$S^4_{16}$	8	24	32	16	$S^4_8$	192	$S^3_4$	4	6	4	12	3
24-cell	$S^4_{24}$	24	96	96	24	$S^4_{24}$	576	$S^3_8$	6	12	8	24	3
120-cell	$S^4_{120}$	600	1200	720	120	$S^4_{600}$	7200	$S^3_{12}$	20	30	12	60	5
600-cell	$S^4_{600}$	120	720	1200	600	$S^4_{120}$	7200	$S^3_4$	4	6	4	12	3

We see our rules still hold true:
$q_k(S^n_m) = q_{n-k}(\star S^n_m)$
$\textbf{T}(S^n_m) = \textbf{T}(\star S^n_m)$
$|\textbf{T}(S^n_m)| = q_{n-1}(S^n_m) \cdot |\textbf{T}(\partial S^n_m)| = m \cdot |\textbf{T}(\partial S^n_m)|$
$2 \cdot q_{n-2}(S^n_m) / q_{n-1}(S^n_m) = \frac{2}{m} \cdot q_{n-2}(S^n_m) = q_{n-2}(\partial S^n_m)$,
i.e. $2 \cdot q_2(S^4_m) / q_3(S^4_m) = \frac{2}{m} \cdot q_2(S^4_m) = q_2(\partial S^4_m)$,
i.e. $\partial S^n_m = S^{n-1}_r$ for $r = \frac{2}{m} \cdot q_{n-2}(S^n_m)$.

We can apply the last rule twice to find the sizes of our 2D surface elements on our 3D surface elements on our 4D regular convex polytopes:
$S^{n-2}_s = \partial S^{n-1}_r = \partial^2 S^n_m$
$s = \frac{2}{r} \cdot q_{n-3}(S^{n-1}_r)$
... idk if i like the look of that, try again...
$2 q_2(S^4_m) / q_3(S^4_m) = q_2(\partial S^4_m)$
$2 q_1(S^3_m) / q_2(S^3_m) = q_1(\partial S^3_m)$
so $2 q_1(\partial S^4_m) / q_2(\partial S^4_m) = q_1(\partial^2 S^4_m) = q_1(S^2_r) = r$

Higher dimensions? They only have 3 different regular convex polytopes, the $(n+1)$-simplex, the $(2n)$-hypercube, and the $(2^n)$-orthoplex.

name	$S^n_m$	$q_0(S^n_m)$	$q_k(S^n_m)$	$q_{n-1}(S^n_m)$	$\star S^n_m$	$\|\textbf{T}(S^n_m)\|$	$\partial S^n_m$
simplex	$S^n_{n+1}$	$n+1$	$\left( \begin{matrix} n+1 \\ k+1 \end{matrix} \right)$	$n+1$	$S^n_{n+1}$	$\frac{1}{2}(n+1)!$	$S^{n-1}_n$
hypercube	$S^n_{2n}$	$2^n$	$2^{n-k} \left( \begin{matrix} n \\ k \end{matrix} \right)$	$2 n$	$S^n_{2^n}$	$2^{n-1} \cdot n!$	$S^{n-1}_{2(n-1)}$
orthoplex	$S^n_{2^n}$	$2 n$	$2^{k+1} \left( \begin{matrix} n \\ k+1 \end{matrix} \right)$	$2^n$	$S^n_{2 n}$	$2^{n-1} \cdot n!$	$S^{n-1}_n$

What is the transformation group size of the (n+1)-simplex?
And $\partial S^n_{n+1} = S^{n-1}_n$. So there's our recursive relation.
$|\textbf{T}(S^n_{n+1})| = (n+1) \cdot |\textbf{T}(\partial S^n_{n+1})| = (n+1) \cdot |\textbf{T}(S^{n-1}_n)|$.
Base-case: $|\textbf{T}(S^2_3)| = 3 = \frac{3!}{2}$.
From there we can calculate $|\textbf{T}(S^n_{n+1})| = \underset{m=2}{\overset{n}{\Pi}} (m+1) = \frac{1}{2} (n+1)!$.

Now the $(2n)$-hypercube:
$\partial S^n_{2n} = S^{n-1}_{2(n-1)}$.
$|\textbf{T}(S^n_{2n})| = 2n \cdot |\textbf{T}(\partial S^n_{2n})| = 2n \cdot |\textbf{T}(S^{n-1}_{2(n-1)})|$.
Base-case: $|\textbf{T}(S^2_4)| = 4$.
From there we get $|\textbf{T}(S^n_{2n})| = \underset{m=2}{\overset{n}{\Pi}} 2 m = 2^{n-1} \cdot n!.$

Last is the $(2^n)$-orthoplex:
$\partial S^n_{2^n} = S^{n-1}_n$
$|\textbf{T}(S^n_{2^n})| = 2^n \cdot |\textbf{T}(\partial S^n_{2^n})| = 2^n \cdot |\textbf{T}(S^{n-1}_n)|$.
No need to do a recursive definition, the transform group size is just based on the previous dimension's simplex transform group:
$|\textbf{T}(S^n_{2^n})| = 2^n \cdot \frac{1}{2} \cdot n! = 2^{n-1} \cdot n!$
Duh. It's dual to the hypercube, of course it has the same sized automorphic transform group.

Ok the number of k-forms I got from https://en.wikipedia.org/wiki/List_of_regular_polytopes_and_compounds#Convex_4. Maybe I'll try to prove them later.

It's interesting that the hypercube and orthoplex are duals (they have matching automorphic transform groups), but the simplex and the orthoplex have matching surface elements (of simplexes).

Mind if I also included reflections then my automorphism rotation group size becomes $|\textbf{T}(S^n_{n+1})| = (n+1)!$ and $|\textbf{T}(S^n_{2^n})| = |\textbf{T}(S^n_{2n})| = 2^n n!$. This looks a lot cleaner. But I'm trying to preserve my handedness.

Let's look at a graph of surface elements across dimension n.
Duals will be marked with $\underleftrightarrow{ \ \ \ \star \ \ \ }$. If no dual is present you can assume it is self-dual.

				$S^n_{n+1}$		$S^n_{2^n}$	$\underleftrightarrow{ \ \ \ \star \ \ \ }$	$S^n_{2n}$
				$\partial \downarrow$	$\swarrow \partial$			$\partial \downarrow$
				$S^{n-1}_n$		$S^{n-1}_{2^{n-1}}$	$\underleftrightarrow{ \ \ \ \star \ \ \ }$	$S^{n-1}_{2(n-1)}$
				$\partial \downarrow$	$\swarrow \partial$			$\partial \downarrow$
...
				$\partial \downarrow$	$\swarrow \partial$			$\partial \downarrow$
$S^4_{120}$	$\underleftrightarrow{ \ \ \ \star \ \ \ }$	$S^4_{600}$		$S^4_5$		$S^4_{16}$	$\underleftrightarrow{ \ \ \ \star \ \ \ }$	$S^4_8$	$S^4_{24}$
$\partial \downarrow$			$\partial \searrow$	$\partial \downarrow$	$\swarrow \partial$			$\partial \downarrow$
$S^3_{12}$	$\underleftrightarrow{ \ \ \ \star \ \ \ }$	$S^3_{20}$		$S^3_4$		$S^3_8$	$\underleftrightarrow{ \ \ \ \star \ \ \ }$	$S^3_6$
	$\partial \searrow$		$\partial \searrow$	$\partial \downarrow$	$\swarrow \partial$		$\swarrow \partial$
		$S^2_5$		$S^2_3$		$S^2_4$		$S^2_{k \ge 6}$ ...
			$\partial \searrow$	$\partial \downarrow$	$\swarrow \partial$		$\swarrow \partial$
$S^1$

So the big question: Is there a number theory reason that there are only 5 Platonic solids, and 6 regular convex polytopes of 4D?

If a regular convex polyhedron must have a dual then it and its dual must have transform groups that are equal to their surface polygon number of sides times the number of surface polygons, then what arrangements are possible simply by prime factorization?

Vertex Central Angles:

Dimension $n=1$:
$S^1 =$ line segment. Vertices $v = \{[1], [-1]\}$

Dimension $n=2$:
$S^2_k, k \ge 3$ = a k-sided regular convex polytope.
Number of vertices and edges: $q(S^2_k) = \{ k, k \}$
The cosine of angle between adjacent vertexes is $cos(2 \pi / k)$.
A set of all cosines of angles between any two normalized vertices is $\mathbf{c}(S^2_k) = \{ cos (2 \pi j / k) \}$ for $0 \le j \le m$.
The vertex central angle of the polytope will be the 2nd largest of the set. The largest will always be 1, corresponding with the inner product of any normalized vertex with itself.
Surface element $\partial S^2_k = S^1$ = line segment.
Our 1D surface polytopes are scaled by $a(S^2_k) = sin(\pi / k)$, translated by $b(S^2_k) = -cos(\pi / k)$, and rotated by $2 \pi j / k$ to fit in our 2D polytope.

Specific cases that are used as 3D regular convex polytope surface elements:
$S^2_3 =$ triangle.
$\mathbf{c}(S^2_3 ) = \{1, -\frac{1}{2}\}$
$a(S^2_3) = \frac{\sqrt{3}}{2}$
$b(S^2_3) = -\frac{1}{2}$

$S^2_4 =$ square
$\mathbf{c}(S^2_4) = \{1, 0, -1\}$
$a = \frac{1}{\sqrt{2}}$
$b = -\frac{1}{\sqrt{2}}$

$S^2_5 =$ pentagon
$\mathbf{c}(S^2_5) = \{1, \frac{\sqrt{5} - 1}{4}, -\frac{1+\sqrt{5}}{4} \}$
$a = \frac{\sqrt{10 - 2 \sqrt{5}}}{8}$
$b = -\frac{1 + \sqrt{5}}{4}$

Dimension $n=3$:
$S^3_4 =$ tetrahedron
$q = \{4, 6, 4\}$ = number of vertexes, edges, and faces.
$\mathbf{c}(S^3_4) = \{1, -\frac{1}{3}\}$ = cos angles between vertexes.
$\partial S^3_4 = S^2_3$ = surface polytope
$a = \frac{\sqrt{8}}{3}$ = surface element scale
$b = -\frac{1}{3}$ = surface element translation

$S^3_6 =$ cube
$q = \{8, 12, 6\}$
$\mathbf{c}(S^3_6) = \{1, \frac{1}{3}, -\frac{1}{3}, -1\}$ = cos angles vertexes
$\partial S^3_6 = S^2_4 =$ surface polytope
$a = \sqrt{\frac{2}{3}} =$ surface element scale
$b = -\frac{1}{\sqrt{3}} =$ surface element translate

$S^3_8 =$ octahedron $q = \{6, 12, 8\}$
$\mathbf{c}(S^3_8) = \{1,0,-1\}$ = cos angles
$\partial S^3_8 = S^2_3$ which has cos angles $\mathbf{c}(S^2_3) = \{1, -\frac{1}{2}\}$
$a=$
$b=$

Nontrivial 3D:
$S^3_{12} =$ dodecahedron
$q = \{ 20, 30, 12 \}$
$\mathbf{c}(S^3_{12}) = \{1, \frac{\sqrt{5}}{3}, \frac{1}{3}, -\frac{1}{3}, -\frac{\sqrt{5}}{3}, -1 \}$ = cos angles
$\partial S^3_{12} = S^2_5 =$ surface which has cos angles $\mathbf{c}(S^2_5) = \{1, \frac{\sqrt{5} - 1}{4}, -\frac{1+\sqrt{5}}{4} \}$
$a=$
$b=$

$S^3_{20} =$ icosahedron
$q = \{12, 30, 20\}$
$\mathbf{c}(S^3_{20}) = \{1, \frac{1}{\sqrt{5}}, -\frac{1}{\sqrt{5}}, -1\} =$ cos angles
$\partial S^3_{20} = S^2_3 =$ surface, which cos angles $\mathbf{c}(S^2_3) = \{1, -\frac{1}{2} \}$
$a=$
$b=$

Dim $n=4$

$S^4_5$ $q = \{5, 10, 10, 5\}$
$c = \{1, -\frac{1}{4} \}$
$a = \frac{\sqrt{8}}{3}$
$b = -\frac{1}{3}$

$S^4_8$
$q = \{16, 32, 24, 8\}$
$c = \{1, \frac{1}{2}, 0, -\frac{1}{2}, -1\}$
$a = \frac{\sqrt{3}}{2}$
$b = -\frac{1}{2}$

$S^4_{16}$
$q = \{8, 24, 32, 16\}$
$c = \{1, 0, -1\}$

(nontrivial:)
$S^4_{24}$
$q(S^4_{24}) = \{24, 96, 96, 24\}$
$\mathbf{c}(S^4_{24}) = \{ 1, \frac{1}{2}, 0, -\frac{1}{2}, -1 \}$

$S^4_{120}$
$q(S^4_{120}) = \{600, 1200, 720, 120\}$
$\mathbf{c}(S^4_{120}) = \{ 1, \frac{1+3\sqrt{5}}{8}, \frac{5+\sqrt{5}}{8}, \frac{1+\sqrt{5}}{4}, \frac{3}{4}, \frac{-1+3\sqrt{5}}{8}, \frac{3+\sqrt{5}}{8}, \frac{\sqrt{5}}{4}, \frac{1}{2}, \frac{1+\sqrt{5}}{8}, \frac{5-\sqrt{5}}{8}, \frac{-1+\sqrt{5}}{4}, \frac{1}{4}, \frac{-1+\sqrt{5}}{8}, \frac{3-\sqrt{5}}{8}, 0, \frac{-3+\sqrt{5}}{8}, \frac{1-\sqrt{5}}{8}, -\frac{1}{4}, \frac{1-\sqrt{5}}{4}, \frac{-5+\sqrt{5}}{8}, -\frac{1+\sqrt{5}}{8}, -\frac{1}{2}, -\frac{\sqrt{5}}{4}, -\frac{3+\sqrt{5}}{8}, \frac{1-3\sqrt{5}}{8}, -\frac{3}{4}, -\frac{1+\sqrt{5}}{4}, -\frac{5+\sqrt{5}}{8}, -\frac{1+3\sqrt{5}}{8}, -1 \}$

$S^4_{600}$
$q(S^4_{600}) = \{120, 720, 1200, 600\}$
$\mathbf{c}(S^4_{600}) = \{ 1, \frac{1 + \sqrt{5}}{4}, \frac{1}{2}, \frac{-1 + \sqrt{5}}{4}, 0, \frac{1 - \sqrt{5}}{4}, -\frac{1}{2}, \frac{-1 - \sqrt{5}}{4}, -1 \}$

Dim $n \gt 4$

$S^n_{n+1}$
$\partial S=S^{n-1}_n$
$\mathbf{c}(S^n_{n+1}) = \{1, -\frac{1}{n} \}$
$a = \frac{\sqrt{n^2-1}}{n}$
$b = -\frac{1}{n}$

$S^n_{2n}$
$\partial S=S^{n-1}_{2(n-1)}$
$\mathbf{c}(S^n_{2n}) = { 2k/n-1 }, 0 \le k \le n$
$a = \sqrt{\frac{n-1}{n}}$
$b = -\frac{1}{\sqrt{n}}$

$S^n_{2^n}$
$\partial S=S^{n-1}_n$
$\mathbf{c}(S^n_{2^n}) = \{1, 0, -1\}$
$a =$
$b =$

I'll go class by class through the trivial cases of regular convex polytopes: the simplex, the cube, and the orthoplex.
Since these three polytopes consistently exist in all dimensions past 4, let's formulaically describe their vertexes.
Let $V^n_{i,j} = $ the j'th component of the i'th vertex of a regular convex polytope of its type in dimension n.

Vertexes of a n-simplex:
n-simplexes have n+1 vertexes, so $V^n_{i,j}$ exists for $1 \le i \le n+1, 1 \le j \le n$.
We will assume the norm of each vertex is 1, i.e. $\langle V^n_i, V^n_k \rangle = 1$ for $i=k$, $c_n$ otherwise.

Assume that, relative to the vertexes of the (n-1)-simplex, the vertexes of the n-simplex are scaled by some factor $a_n$ and shifted by some factor $b_n$

Base case:
$n=1, V^1_1 = [1], V^1_2 = [-1]$
Next base case (in case the last one is too simple):
$ n=2, V^2_1 = \left[\begin{matrix} \frac{\sqrt{3}}{2} \\ -\frac{1}{2} \end{matrix}\right], V^2_2 = \left[\begin{matrix} -\frac{\sqrt{3}}{2} \\ -\frac{1}{2} \end{matrix}\right], V^2_3 = \left[\begin{matrix} 0 \\ 1 \end{matrix}\right] $
So we can assert that $a_2 = \frac{\sqrt{3}}{2}, b_n = -\frac{1}{2}, c_n = -\frac{1}{2}$

I'm going to assert that $b_n = -\frac{1}{n}$ and then prove this by induction. Also, thanks to normalization, we can deduce $(a_n)^2 + (b_n)^2 = 1$, so $a_n = \frac{\sqrt{n^2 - 1}}{n}$, which I'll show later.
Same with the fact that $b_n = c_n$, i.e. the amount that we have to translate the previous vertices into the new dimension is equal to the inner product of all vertices of that dimension's n-simplex.

Inductive case:
Let's find our rescaling factor for our n-simplex recursive case:

Assume our (n-1)-simplex has unit-length vertexes, i.e.:
$(V^{n-1}_{i,1})^2 + ... + (V^{n-1}_{i,n-1})^2 = 1$ for $1 \le i \le n$
Assume they all form equal angles, therefore their inner products are all equal, and equal to the induction assertion:
$(V^{n-1}_{i,1}) (V^{n-1}_{k,1}) + ... + (V^{n-1}_{i,n-1}) (V^{n-1}_{k,n-1}) = c_{n-1} = -\frac{1}{n-1}$ for $i \ne k$.

Now to find the vertexes of the n-simplex, first we will take the vertexes of the (n-1)-simplex, rescale them by factor $a_n$, and translating them into the new dimension by $b_n$:
Let $V^n_{i,j} = a_n V^{n-1}_{i,j}$ for $1 \le 1 \le n, 1 \le j \lt n$.
Let $V^n_{i,n} = b_n$ for $1 \le i \le n$
Our unit norm vertexes will be:
$(V^n_{i,1})^2 + ... + (V^n_{i,n})^2 = 1$ for $1 \le i \le n$
...substitute our definition to find...
$(a_n V^{n-1}_1)^2 + ... + (a_n V^{n-1}_{n-1})^2 + (b_n)^2 = 1$
$(a_n)^2 ((V^{n-1}_1)^2 + ... + (V^{n-1}_{n-1})^2) + (b_n)^2 = 1$
...substitute our (n-1)-simplex unit length definition to find...
$(a_n)^2 + (b_n)^2 = 1$
$b_n = \pm \sqrt{1 - (a_n)^2}$
So for vertexes $1 \le i \le n$, we will take from $V^{n-1}$, scale by $a_n$, and offset into our new dimension by $b_n = \pm \sqrt{1 - (a_n)^2}$.

Our last additional new vertex that is orthogonal in the new dimension must be unit length, so:
$V^n_{n+1,j} = \delta_{j,n}$, where $\delta_{j,k}$ is the Kronecher delta, i.e. the vertex component will be 1 for $j=n$ and 0 otherwise.

How to find the rescaling factor? Let's ensure that all our angles are equal, and therefore all our inner products are equal:
$(V^n_{i,1}) (V^n_{k,1}) + ... + (V^n_{i,n}) (V^n_{k,n}) = c_n$ for $i \ne k$, for some constant $c_n$ to be determined.

Let's look at angles between vertexes $1 \le i \le n$ and $n+1$:
$(V^n_{i,1}) (V^n_{n,1}) + ... + (V^n_{i,n-1}) (V^n_{n,n-1}) + (V^n_{i,n}) (V^n_{n,n}) = c_n$ for $1 \le i \le n$.
...substitute $V^n_i, 1 \le i \le n$ and final vertex $V^n_{n+1}$ definitions:
$(a_n V^{n-1}_{i,1}) \cdot 0 + ... + (a_n V^n_{i,n-1}) \cdot 0 + b_n \cdot 1 = c_n$ for $1 \le i \le n$.
$b_n = c_n$ for $1 \le i \le n$.
So now we know that our inner product of our normalized vertexes is equal to the amount that we have shifted our previous dimension's vertexes into the new dimension.

Now consider the case for both $i,k \le n$:
$(a_n V^{n-1}_{i,1}) (a_n V^{n-1}_{k,1}) + ... + (a_n V^{n-1}_{i,n-1}) (a_n V^{n-1}_{k,n-1}) + (b_n)^2 = c_n$
$(a_n)^2 \left( (V^{n-1}_{i,1}) (V^{n-1}_{k,1}) + ... + (V^{n-1}_{i,n-1}) (V^{n-1}_{k,n-1}) \right) + (b_n)^2 = c_n$
...substitute induction case, and substitute $(b_n)^2$:
$(a_n)^2 c_{n-1} + (b_n)^2 = c_n$
...substitute our definition for $c_n$:
$(a_n)^2 c_{n-1} + 1 - (a_n)^2 = \pm \sqrt{1 - (a_n)^2}$

Now let's prove that, for the base case of $\langle V^{n-1}_i, V^{n-1}_k \rangle = -\frac{1}{n-1}$, that the induction case is $\langle V^n_i, V^n_k \rangle = -\frac{1}{n}$:
Assume $\langle V^{n-1}_i, V^{n-1}_k \rangle = -\frac{1}{n-1}$:
$\langle V^n_i, V^n_k \rangle = (V^n_{i,1}) (V^n_{k,1}) + ... + (V^n_{i,n-1}) (V^n_{k,n-1}) + (V^n_{i,n}) (V^n_{k,n}) = -\frac{1}{n}$
$= (a_n V^{n-1}_{i,1}) (a_n V^{n-1}_{k,1}) + ... + (a_n V^{n-1}_{i,n-1}) (a_n V^{n-1}_{k,n-1}) + (b_n) (b_n) = -\frac{1}{n}$
$= (a_n)^2 \left( (V^{n-1}_{i,1}) (V^{n-1}_{k,1}) + ... + (V^{n-1}_{i,n-1}) (V^{n-1}_{k,n-1}) \right) + (b_n) (b_n) = -\frac{1}{n}$
$= (a_n)^2 ( -\frac{1}{n-1} ) + (b_n) (b_n) = -\frac{1}{n}$
$= \frac{n^2 - 1}{n^2} ( -\frac{1}{n-1} ) + \frac{1}{n^2} = -\frac{1}{n}$
$= -\frac{n + 1}{n^2} + \frac{1}{n^2} = -\frac{1}{n}$
$= -\frac{1}{n} = -\frac{1}{n}$
true.

So, in summary:
$V^n_{i,j} = \frac{\sqrt{n^2-1}}{n} \cdot V^{n-1}_{i,j}$ for $1 \le i \le n, 1 \le j \le n-1$; $V^n_{i,n} = -\frac{1}{n}$ for $1 \le i \le n$.
$V^n_{n,j} = \delta_{n,j}$

From here we can calculate all vertex elements of any n-simplex:
$V^1_{1,1} = 1$
$V^1_{2,1} = -1$
$V^n_{n,n+1} = 1$
$V^n_{1,1} = \Pi_{j=1}^n \frac{\sqrt{j^2 - 1}}{j}$
$V^n_{2,1} = \Pi_{j=1}^n \frac{\sqrt{j^2 - 1}}{j}$

Vertices as column vectors, Varying dimension:
$\left[\begin{matrix} 1 & -1 \end{matrix}\right], \left[\begin{matrix} \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} & 0 \\ -\frac{1}{2} & -\frac{1}{2} & 1 \end{matrix}\right], \left[\begin{matrix} \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{8}}{3} & -\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{8}}{3} & 0 & 0 \\ -\frac{1}{2} \cdot \frac{\sqrt{8}}{3} & -\frac{1}{2} \cdot \frac{\sqrt{8}}{3} & \frac{\sqrt{8}}{3} & 0 \\ -\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} & 1 \end{matrix}\right], ..., \left[\begin{matrix} \Pi_{j=2}^n \frac{\sqrt{j^2-1}}{j} & -\Pi_{j=2}^n \frac{\sqrt{j^2-1}}{j} & 0 & 0 & ... & 0 & 0 & 0 \\ -\frac{1}{2} \cdot \Pi_{j=3}^n \frac{\sqrt{j^2-1}}{j} & -\frac{1}{2} \cdot \Pi_{j=3}^n \frac{\sqrt{j^2-1}}{j} & \Pi_{j=3}^n \frac{\sqrt{j^2-1}}{j} & 0 & ... & 0 & 0 & 0 \\ -\frac{1}{3} \cdot \Pi_{j=4}^n \frac{\sqrt{j^2-1}}{j} & -\frac{1}{3} \cdot \Pi_{j=4}^n \frac{\sqrt{j^2-1}}{j} & -\frac{1}{3} \cdot \Pi_{j=4}^n \frac{\sqrt{j^2-1}}{j} & \Pi_{j=4}^n \frac{\sqrt{j^2-1}}{j} & ... & 0 & 0 & 0 \\ ... & ... & ... & ... & ... & ... & ... & ... \\ -\frac{1}{n-2} \cdot \frac{\sqrt{(n-1)^2-1}}{n-1} \cdot \frac{\sqrt{n^2-1}}{n} & -\frac{1}{n-2} \cdot \frac{\sqrt{(n-1)^2-1}}{n-1} \cdot \frac{\sqrt{n^2-1}}{n} & -\frac{1}{n-2} \cdot \frac{\sqrt{(n-1)^2-1}}{n-1} \cdot \frac{\sqrt{n^2-1}}{n} & -\frac{1}{n-2} \cdot \frac{\sqrt{(n-1)^2-1}}{n-1} \cdot \frac{\sqrt{n^2-1}}{n} & ... & \frac{\sqrt{(n-1)^2-1}}{n-1} \cdot \frac{\sqrt{n^2-1}}{n} & 0 & 0 \\ -\frac{1}{n-1} \cdot \frac{\sqrt{n^2-1}}{n} & -\frac{1}{n-1} \cdot \frac{\sqrt{n^2-1}}{n} & -\frac{1}{n-1} \cdot \frac{\sqrt{n^2-1}}{n} & -\frac{1}{n-1} \cdot \frac{\sqrt{n^2-1}}{n} & ... & -\frac{1}{n-1} \cdot \frac{\sqrt{n^2-1}}{n} & \frac{\sqrt{n^2-1}}{n} & 0 \\ -\frac{1}{n} & -\frac{1}{n} & -\frac{1}{n} & -\frac{1}{n} & ... & -\frac{1}{n} & -\frac{1}{n} & 1 \end{matrix}\right], $

inner product of normalized vertices is $-\frac{1}{n}$

Vertexes of a n-hypercube:
normalized vertices:
$V^n_{i,j} = \frac{1}{\sqrt{n}} (-1)^{(j'th-digit-of-(i-minus-1))}$ for $1 \le i \le 2^n$
norm of vertices is $\Sigma_{j=1}^n (\frac{1}{\sqrt{n}})^2 = \Sigma_{j=1}^n \frac{1}{n} = 1$
inner product between neighboring vertices: $\Sigma_{j=1}^{n-1} (\frac{1}{\sqrt{n}})^2 + (-\frac{1}{\sqrt{n}}) (\frac{1}{\sqrt{n}}) = \frac{n-1}{n} - \frac{1}{n} = \frac{n-2}{n}$

Vertexes of a n-orthoplex:
$V^n_{i,j} = (-1)^{\lfloor (i-1) / n \rfloor} \cdot \delta_{((i-1) \backslash n)+1,j}$ for $1 \le i \le 2n$
norm of vertices is 1.
inner product between neighboring vertices is 0.

Maybe later I'll prove this covers all regular convex polytopes...
Which is just a matter of showing surface nets and deficit angles and enumerating how many matching surface elements can fit in the deficit angles, then throwing duals in the mix ...
In 3D:
Start with triangle surface elements, which have 60-degree internal angle and form a plane with 6 of them.
Remove one to get 5, deficit angle of 60, surface dual of pentagon, continue tiling our triangles to get an icosahedron (20-sided), tile the surface dual to get the icoashedron dual of the dodecahedron (12-sided).
Remove one more to get 4, deficit angle of 120, surface dual of square, continue tiling our triangles to get an octahedron (8-sided), tile the surface dual to get the octahedron dual of the cube (6-sided).
Remove one more to get 3, deficit angle of 180, surface is triangle and therefore self-dual, continue tiling our triangles to get a tetrahedron (4-sided).

In 4D:
Start with tetrahedron surface elements. Honestly this is mind-bending enough for me.
Our surface net is a tetrahedron lattice of $\mathbb{R}^3$. Now start removing them and examining deficit angles.