# Simplex

## What is a regular polygon?

Understanding a simplex begins with familiarity with polygons. A polygon is a shape formed by connecting multiple line segments in 2 dimensions to form a single path closed circuit. The line segments are called edges, and the point where 2 edges meet is called a vertex. Each vertex has an angle formed by the two adjacent sides. A regular polygon is one where all its sides are of equal length and all its angles are of equal measure and turning direction. The list of regular polygons have names as follows:

1. equilateral triangle
2. square
3. regular pentagon
4. regular hexagon
5. regular heptagon (or regular septagon)
6. and so forth

Note that I could instead define a regular polygon as one in which all its line segments were equal and all its vertices are equidistant to a common midpoint. In other words, the vertices of a regular polygon must coincide to equally spaced points along the edge of a circle.

## What is a simplex?

Now, what if I create a new sequence of shapes not unlike the regular polygons, but this time I add a new dimension with every new vertex? For instance, if I start with an equilateral triangle in 2 dimensions, and then I add a vertex somewhere above the midpoint of the triangle in a new dimension. I want this midpoint to be just far enough so that the distance between any 2 vertices is the same. Although I cannot visualize 4 and higher dimensions very well, I can consider each vertex as a list of coordinate measures. If I just deal with numbers and distances, the concept of higher dimensions doesn't really seem so scary. The names of the shapes created in this new way are as follows:

1. line segment (1D)
2. equilateral triangle (2D)
3. regular tetrahedron or 3-simplex (3D)
4. 5-cell or 4-simplex (4D)
5. 5-simplex and so forth (5D and more)

Every shape in this list is called a simplex. So a simplex is a different polytope object in every dimension, and each dimension has just one simplex, if I ignore scale, position and orientation. Polytope is the general term for (line segment, polygon, polyhedron, ...) when I do not specify how many dimensions I am measuring for each vertex. A 5-simplex is a 5-polytope because it requires 5 dimensions of measure to establish each vertex.

## Simplex coordinates

As I have hinted, I can find coordinates for each vertex of any n-simplex. These coordinates are not unique, but I can choose or constrain the size, position and orientation for sake of regularity of construction.

For the position of the n-simplex, I set the origin in $$\mathbf{R}^n$$ as the midpoint. For the size, in context of this position choice, I choose the radius of 1 from this origin to any vertex.

For the orientation, I follow a simple pattern for constructing the n+1-simplex coordinates from the n-simplex coordinates. Basically I will shrink the n-simplex vertex coordinates by a fixed multiple $$m_{n+1}$$, and then append a common negative value $$b_{n+1}$$ for the new dimension. Then I can add a new vertex that has the unit measure in the new dimension.

I will create a special tensor $$S$$ with the following notation which holds coordinates for every simplex:

${}_{v}^{n}S_{c}$

$$n$$ is the dimension of the simplex, $$v \in [1, n+1]$$ specifies a vertex of the simplex, and $$c \in [1, n]$$ specifies an element of the vertex.

Let's start with the 1-simplex denoted $${}^{1}S$$, which is a line segment. As just mentioned, I chose my line segment to be centered at the origin, so my vertices must be 1-tuples like this.

${}_{1}^{1}S=(-b_{1})$
${}_{2}^{1}S=(b_{1})$

I want the vertex of each simplex to be a distance or radius of 1 from the origin, so $$b_{1} = 1$$

${}_{1}^{1}S_1 = -1$
${}_{2}^{1}S_1 = 1$

For the 2-simplex denoted $${}^{2}S$$, I need to find a multiplier $$m_2$$ and an offset $$b_2$$ so that my new coordinates have this form:

${}_{1}^{2}S=({}_{1}^{1}S_1 m_2, b_2)$
${}_{2}^{2}S=({}_{2}^{1}S_1 m_2, b_2)$
${}_{3}^{2}S=(0, 1)$

Now, I could start with a prior understanding that this is an equilateral triangle and use trigonometry functions, but a neat bit of logic will get me to my result much quicker. If my simplex is centered at the origin, then the sum of its coodinates in each dimension must be zero. It is easy to see, $$b_2=-\frac{1}{2}$$, so I have the following:

${}_{1}^{2}S=(-m_2, -\frac{1}{2})$
${}_{2}^{2}S=(m_2, -\frac{1}{2})$
${}_{3}^{2}S=(0, 1)$

I want my vertices on the unit circle, so by the pythagorean theorem:

$m_2 = \sqrt{1^2-(b_2)^2} = \sqrt{\frac{3}{4}}$

Finally, I have my 2-simplex coordinates

${}_{1}^{2}S=(-\sqrt{\frac{3}{4}}, -\frac{1}{2})$
${}_{2}^{2}S=(\sqrt{\frac{3}{4}}, -\frac{1}{2})$
${}_{3}^{2}S=(0, 1)$

Let's go on to the 3-simplex (regular tetrahedron), where some patterns begin to emerge. Remember I need to find a specific $$m_3$$ and $$b_3$$ which can extend the 2-simplex coordinates as follows:

${}_{1}^{3}S=({}_{1}^{2}S m_3, b_3)$
${}_{2}^{3}S=({}_{2}^{2}S m_3, b_3)$
${}_{3}^{3}S=({}_{3}^{2}S m_3, b_3)$
${}_{4}^{3}S=(0, 0, 1)$

Applying the same principles again, I find

$b_3 = -\frac{1}{3}$
$m_3 = \sqrt{1^2-(b_3)^2} = \sqrt{\frac{8}{9}}$

Substituting back what I know, I arrive at the following:

${}_{1}^{3}S=(-\sqrt{\frac{3}{4}}\sqrt{\frac{8}{9}}, -\frac{1}{2}\sqrt{\frac{8}{9}}, -\frac{1}{3})$
${}_{2}^{3}S=(\sqrt{\frac{3}{4}}\sqrt{\frac{8}{9}}, -\frac{1}{2}\sqrt{\frac{8}{9}}, -\frac{1}{3})$
${}_{3}^{3}S=(0, \sqrt{\frac{8}{9}}, -\frac{1}{3})$
${}_{4}^{3}S=(0, 0, 1)$

Or equivalently:

${}^{3}S=\begin{bmatrix} -1 \sqrt{\frac{3}{4}}\sqrt{\frac{8}{9}} & -\frac{1}{2}\sqrt{\frac{8}{9}} & -\frac{1}{3} \\ 1 \sqrt{\frac{3}{4}}\sqrt{\frac{8}{9}} & -\frac{1}{2}\sqrt{\frac{8}{9}} & -\frac{1}{3} \\ 0 & 1\sqrt{\frac{8}{9}} & -\frac{1}{3} \\ 0 & 0 & 1 \end{bmatrix}$

Repeating for the 4-simplex, I have

${}^{4}S=\begin{bmatrix} -1 \sqrt{\frac{3}{4}}\sqrt{\frac{8}{9}}\sqrt{\frac{15}{16}} & -\frac{1}{2}\sqrt{\frac{8}{9}}\sqrt{\frac{15}{16}} & -\frac{1}{3}\sqrt{\frac{15}{16}} & -\frac{1}{4} \\ 1 \sqrt{\frac{3}{4}}\sqrt{\frac{8}{9}}\sqrt{\frac{15}{16}} & -\frac{1}{2}\sqrt{\frac{8}{9}}\sqrt{\frac{15}{16}} & -\frac{1}{3}\sqrt{\frac{15}{16}} & -\frac{1}{4} \\ 0 & 1\sqrt{\frac{8}{9}}\sqrt{\frac{15}{16}} & -\frac{1}{3}\sqrt{\frac{15}{16}} & -\frac{1}{4} \\ 0 & 0 & 1\sqrt{\frac{15}{16}} & -\frac{1}{4} \\ 0 & 0 & 0 & 1\end{bmatrix}$

## N-simplex coordinates formula

Finally, a pattern emerges:

${}_{v}^{n}S_{c} = \begin{cases} -\frac{1}{c} M(c, n) & v\lt c+1 \\ M(c, n) & v= c+1 \\ 0 & v\gt c+1 \end{cases}$

where

$M(c, n) = \prod_{i=c+1}^{n} \sqrt{\frac{i^2-1}{i^2}}$

Expanding and factoring reveals a simplification.

$M(c, n)^2 = \frac{(c+1)^2-1}{(c+1)^2} \times \frac{(c+2)^2-1}{(c+2)^2} \times ... \times \frac{(n-2)^2-1}{(n-2)^2} \times \frac{(n-1)^2-1}{(n-1)^2} \times \frac{n^2-1}{n^2}$
$M(c, n)^2 = \frac{c}{c+1} \frac{c+2}{c+1} \times \frac{c+1}{c+2} \frac{c+3}{c+2} \times ... \times \frac{n-3}{n-2} \frac{n-1}{n-2} \times \frac{n-2}{n-1} \frac{n}{n-1} \times \frac{n-1}{n} \frac{n+1}{n}$
$M(c, n)^2 = \frac{c}{c+1} \frac{\bcancel{c+2}}{\cancel{c+1}} \times \frac{\cancel{c+1}}{\bcancel{c+2}} \frac{c+3}{c+2} \times ... \times \frac{n-3}{n-2} \frac{\bcancel{n-1}}{\cancel{n-2}} \times \frac{\cancel{n-2}}{\bcancel{n-1}} \frac{\bcancel{n}}{\cancel{n-1}} \times \frac{\cancel{n-1}}{\bcancel{n}} \frac{n+1}{n}$

And after matching and cancelling from either end, I see all the middle terms cancel too.

$M(c, n)^2 = \frac{c}{c+1} \frac{\bcancel{c+2}}{\cancel{c+1}} \times \frac{\cancel{c+1}}{\bcancel{c+2}} \xcancel{ \frac{c+3}{c+2} \times ... \times \frac{n-3}{n-2}} \frac{\bcancel{n-1}}{\cancel{n-2}} \times \frac{\cancel{n-2}}{\bcancel{n-1}} \frac{\bcancel{n}}{\cancel{n-1}} \times \frac{\cancel{n-1}}{\bcancel{n}} \frac{n+1}{n}$
$M(c, n)^2 = \frac{c (n+1)}{(c+1)n}$
$M(c, n) = \sqrt{\frac{c(n+1)}{n(c+1)}}$

And finally, I have this marvelous result:

${}_{v}^{n}S_{c} = \begin{cases} 0 & c \lt v-1 \\ \sqrt{\frac{c(n+1)}{n(c+1)}} & c = v-1 \\ -\frac{1}{c} \sqrt{\frac{c(n+1)}{n(c+1)}} & c \gt v-1 \end{cases}$

I can factor out part for ultra-simplicity of manual simplex calculations.

${}_{v}^{n}S_{c} = \sqrt{\frac{n+1}{n}} \begin{cases} 0 & c \lt v-1 \\ \sqrt{\frac{c}{c+1}} & c = v-1 \\ -\frac{1}{c} \sqrt{\frac{c}{c+1}} & c \gt v-1 \end{cases}$
${}_{v}^{n}S_{c} = \frac{\sqrt{n(n+1)}}{n} \begin{cases} 0 & c \lt v-1 \\ \frac{\sqrt{c(c+1)}}{c+1} & c = v-1 \\ -\frac{\sqrt{c(c+1)}}{c(c+1)} & c \gt v-1 \end{cases}$
Name Coordinates Vertices
$${}^{n}S$$ c = 1 → 2 → … $$\max v$$
$${}^{1}S$$
1-Simplex
Line segment
$$\begin{bmatrix} -1 \\ 1 \end{bmatrix}$$ 2
$${}^{2}S$$
2-Simplex
Equilateral triangle
$$\begin{bmatrix} -\frac{1}{2}\sqrt{3} & -\frac{1}{2} \\ \frac{1}{2}\sqrt{3} & -\frac{1}{2} \\ 0 & 1 \end{bmatrix}$$ 3
$${}^{3}S$$
3-Simplex
Tetrahedron
$$\begin{bmatrix} -\frac{1}{3}\sqrt{6} & -\frac{1}{3}\sqrt{2} & -\frac{1}{3} \\ \frac{1}{3}\sqrt{6} & -\frac{1}{3}\sqrt{2} & -\frac{1}{3} \\ 0 & \frac{2}{3}\sqrt{2} & -\frac{1}{3} \\ 0 & 0 & 1 \end{bmatrix}$$ 4
$${}^{4}S$$
4-Simplex
5-Cell
$$\begin{bmatrix} -\frac{1}{4}\sqrt{10} & -\frac{1}{12}\sqrt{30} & -\frac{1}{12}\sqrt{15} & -\frac{1}{4} \\ \frac{1}{4}\sqrt{10} & -\frac{1}{12}\sqrt{30} & -\frac{1}{12}\sqrt{15} & -\frac{1}{4} \\ 0 & \frac{1}{6}\sqrt{30} & -\frac{1}{12}\sqrt{15} & -\frac{1}{4} \\ 0 & 0 & \frac{1}{4}\sqrt{15} & -\frac{1}{4} \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ 5
$${}^{5}S$$
5-Simplex
$$\begin{bmatrix} -\frac{1}{5}\sqrt{15} & -\frac{1}{5}\sqrt{5} & -\frac{1}{10}\sqrt{10} & -\frac{1}{10}\sqrt{6} & -\frac{1}{5} \\ \frac{1}{5}\sqrt{15} & -\frac{1}{5}\sqrt{5} & -\frac{1}{10}\sqrt{10} & -\frac{1}{10}\sqrt{6} & -\frac{1}{5} \\ 0 & \frac{2}{5}\sqrt{5} & -\frac{1}{10}\sqrt{10} & -\frac{1}{10}\sqrt{6} & -\frac{1}{5} \\ 0 & 0 & \frac{3}{10}\sqrt{10} & -\frac{1}{10}\sqrt{6} & -\frac{1}{5} \\ 0 & 0 & 0 & \frac{2}{5}\sqrt{6} & -\frac{1}{5} \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ 6
$${}^{6}S$$
6-Simplex
$$\begin{bmatrix} -\frac{1}{6}\sqrt{21} & -\frac{1}{6}\sqrt{7} & -\frac{1}{12}\sqrt{14} & -\frac{1}{60}\sqrt{210} & -\frac{1}{30}\sqrt{35} & -\frac{1}{6} \\ \frac{1}{6}\sqrt{21} & -\frac{1}{6}\sqrt{7} & -\frac{1}{12}\sqrt{14} & -\frac{1}{60}\sqrt{210} & -\frac{1}{30}\sqrt{35} & -\frac{1}{6} \\ 0 & \frac{1}{3}\sqrt{7} & -\frac{1}{12}\sqrt{14} & -\frac{1}{60}\sqrt{210} & -\frac{1}{30}\sqrt{35} & -\frac{1}{6} \\ 0 & 0 & \frac{1}{4}\sqrt{14} & -\frac{1}{60}\sqrt{210} & -\frac{1}{30}\sqrt{35} & -\frac{1}{6} \\ 0 & 0 & 0 & \frac{1}{15}\sqrt{210} & -\frac{1}{30}\sqrt{35} & -\frac{1}{6} \\ 0 & 0 & 0 & 0 & \frac{1}{6}\sqrt{35} & -\frac{1}{6} \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ 7
$${}^{\infty}S$$
∞-Simplex
$$\begin{bmatrix} -\frac{1}{2}\sqrt{2} & -\frac{1}{6}\sqrt{6} & -\frac{1}{12}\sqrt{12} & -\frac{1}{20}\sqrt{20} & -\frac{1}{30}\sqrt{30} & -\frac{1}{42}\sqrt{42} & \dots \\ \frac{1}{2}\sqrt{2} & -\frac{1}{6}\sqrt{6} & -\frac{1}{12}\sqrt{12} & -\frac{1}{20}\sqrt{20} & -\frac{1}{30}\sqrt{30} & -\frac{1}{42}\sqrt{42} & \dots \\ 0 & \frac{1}{3}\sqrt{6} & -\frac{1}{12}\sqrt{12} & -\frac{1}{20}\sqrt{20} & -\frac{1}{30}\sqrt{30} & -\frac{1}{42}\sqrt{42} & \dots \\ 0 & 0 & \frac{1}{4}\sqrt{12} & -\frac{1}{20}\sqrt{20} & -\frac{1}{30}\sqrt{30} & -\frac{1}{42}\sqrt{42} & \dots \\ 0 & 0 & 0 & \frac{1}{5}\sqrt{20} & -\frac{1}{30}\sqrt{30} & -\frac{1}{42}\sqrt{42} & \dots \\ 0 & 0 & 0 & 0 & \frac{1}{6}\sqrt{30} & -\frac{1}{42}\sqrt{42} & \dots \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{7}\sqrt{42} & \dots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}$$

## How to check my N-simplex formula

To check that my n-simplex formula is a regular polytope, I need to validate that the distance between any 2 connected vertices is the same. Within each n-simplex, all vertices are connected.

I want to find the edge length $$e_{n}$$ from taking the difference in the first 2 vertices.

${}_{2}^{n}S_{c} - {}_{1}^{n}S_{c} = \begin{cases} \sqrt{\frac{2(n+1)}{n}} & c=1 \\ 0 & c \gt 1 \end{cases}$

With all coordinates being equal except the first one, I can see the edge length between the first 2 vertices is

$e_{n} = \sqrt{\frac{2(n+1)}{n}}$

I can consider the difference of all the pairs of vertices $$1 \leq v \lt w \leq n+1$$

${}_{w}^{n}S_{c} - {}_{v}^{n}S_{c} = \begin{cases} 0 & c+1 \lt v \\ -\sqrt{\frac{c(n+1)}{n(c+1)}} & c+1 = v \\ \sqrt{\frac{n+1}{cn(c+1)}} & v \lt c+1 \lt w \\ \sqrt{\frac{(c+1)(n+1)}{nc}} & c+1 = w \\ 0 & c+1 \gt w \end{cases}$

Next I perform a sum of squares over $$c$$ to get

${e_{n}}^{2}= \frac{(v-1)(n+1)}{nv} + \sum_{i=v+1}^{w-1}\frac{n+1}{in(i-1)} + \frac{w(n+1)}{n(w-1)}$

And then I can pull out a common factor

${e_{n}}^{2}=\frac{n+1}{n}\left( \frac{v-1}{v} + \sum_{i=v+1}^{w-1}\frac{1}{i(i-1)} + \frac{w}{w-1} \right)$

I can pull 2 from the fractions to make them smaller

${e_{n}}^{2}=\frac{n+1}{n}\left( 2 - \frac{1}{v} + \sum_{i=v+1}^{w-1}\frac{1}{i(i-1)} + \frac{1}{w-1} \right)$

And then I can pull out the last term of the sum and combine it with the last term

${e_{n}}^{2}=\frac{n+1}{n}\left( 2 - \frac{1}{v} + \sum_{i=v+1}^{w-2}\frac{1}{i(i-1)} + \frac{1}{(w-1)(w-2)} + \frac{w-2}{(w-1)(w-2)} \right)$
${e_{n}}^{2}=\frac{n+1}{n}\left( 2 - \frac{1}{v} + \sum_{i=v+1}^{w-2}\frac{1}{i(i-1)} + \frac{1}{w-2} \right)$

I can continue until the sum has only 1 term left, and then add all three fractions:

${e_{n}}^{2}=\frac{n+1}{n}\left( 2 - \frac{1}{v} + \sum_{i=v+1}^{v+1}\frac{1}{i(i-1)} + \frac{1}{v+1} \right)$
${e_{n}}^{2}=\frac{n+1}{n}\left( 2 - \frac{v+1}{v(v+1)} + \frac{1}{v(v+1)} + \frac{v}{v(v+1)} \right)$

And I see all three fractions cancel so that for any $$v \lt w$$, the edge length is

$e_{n} = \sqrt{\frac{2(n+1)}{n}}$

This proves my edge length is consistent for each simplex.

## N-simplex angles

Now that I have established coordinates for each n-simplex, I can answer some questions such as, what is the angle made between two simplex vertices from the origin? For discussing an angle directly, I now resort to trigonometry by this vector dot product formula.

$\vec{v} \cdot \vec{w} = \lVert \vec{v} \rVert \lVert \vec{w} \rVert cos \theta$

Finding the dot product with the last vertex $$(0, 0, ... , 0, 1)$$ and any other vector is tough, but after some work, I see that

$\theta_n = cos^{-1} \left(-\frac{1}{n}\right)$

What is the angle $$\alpha_n$$ at vertex $$v$$ and two other vertices $$v \lt w \lt x$$?

$\frac{2(n+1)}{n} cos (\alpha_n) = ({}_{w}^{n}S_{c} - {}_{v}^{n}S_{c}) \cdot ({}_{x}^{n}S_{c} - {}_{v}^{n}S_{c}) = \frac{(v-1)(n+1)}{nv} + \sum_{i=v+1}^{w-1} \frac{n+1}{(i-1)ni} + \sqrt{\frac{n+1}{(w-1)nw}} \sqrt{\frac{w(n+1)}{n(w-1)}}$

Divide and reduce

$2 cos (\alpha_n) = 1 - \frac{1}{v} + \sum_{i=v-1}^{w+1} \frac{1}{(i-1)i} + \frac{1}{w-1}$

And as seen before, this complicated expression falls out, so I can see that a calm 60 degrees prevails consistently throughout every triple of vertices for any simplex.

$\alpha_n = cos^{-1}\frac{1}{2}$