Is there an equation to describe regular polytopes?

Question

I am asking for generalizations of equations for regular polygons. To me the answer by Raskolnikov seems the most elegant, but I can't think how to generalize it to hyperspace since the path taken from vertex to vertex isn't clear cut.

Answer 1

First let $d = 2$ and let $K$ be any convex $d$-polytope (polygon) in $\mathbb{R}^d$ and $q$ is a point in its interior. For any $(d-1)$-face (edge) $F$ of $K$, let

• $\mathcal{A}_F$ be the $(d-1)$-volume (area) of $F$.
• $P_F$ be the hyperplane (line) holding $F$.
• $\alpha_F = (\alpha_1, \ldots, \alpha_d)$ be the outward pointing unit normal vector of $F$.
• $\beta_F$ be the nearest distance between $P_F$ and $o$.

For any $p = (x_1,\ldots,x_d) \in \mathbb{R}^d$, it lies on $P_F$ when and only when

$$\theta_F(p) \stackrel{def}{=} \beta_F - \alpha_F \cdot (p - q) = 0$$

For $p \not\in P_F$, $\theta_F(p)$ will be positive/negative depends on whether $p$ is on the same or opposite site of $K$ with respect to $P_F$. The points on or inside $K$ are precisely those $p$ where $\theta_F(p) \ge 0$ for all faces $F$.

Let $\mathcal{V}$ be the $d$-volume (volume) of $K$. Notice $\frac1d A_F\theta_F(p)$ is the signed hypervolume of the convex hull of $p$ and face $F$. If we sum over the faces, we have an identity independent of $p$.

$$\sum_F \mathcal{A}_F \theta_F(p) = d\mathcal{V}$$ Using this, we can reexpress the condition that all $\theta_F(p) \ge 0$ as $$\sum_F \mathcal{A}_F |\theta_F(p)| - d\mathcal{V} = 0$$ We can and will use this as the equation for solid $K$.

$K's$ boundary $\partial K$ is the intersection of $K$ with $\bigcup_F P_F$. Since $\bigcup_F P_F$ is described by the equation

$$\prod_{F}\theta_F(p) = 0\quad\iff\quad\prod_{F} |\theta_F(p)| = 0$$

We can use following relation as an equation of $\partial K$.

$$\sum_F \mathcal{A}_F |\theta_F(p)| + \gamma \prod_{F} |\theta_F(p)| - \mathcal{V} = 0\tag{*1}$$

where $\gamma$ is any positive constant.

Nothing in above construction really need $d = 2$. We can forget about $d = 2$ and use $(*1)$ to generate a equation for the boundary of any non-degenerate convex $d$-polytope.

As an example, consider the tetrahedron $K$ with vertices $$(-1,-1,-1), (-1,1,1), (1,-1,1),(-1,-1,)$$ and take $q = (0,0,0)$. We have all $\mathcal{A}_F = 2\sqrt{3}$, $\mathcal{V} = \frac{8}{3}$ and the four $\theta_F(p)$ are:

$$ \frac{1-x-y-z}{\sqrt{3}},\; \frac{1-x+y+z}{\sqrt{3}},\; \frac{1+x-y+z}{\sqrt{3}}\;\text{ and }\; \frac{1+x+y-z}{\sqrt{3}} $$ Take the constant $\gamma$ to $18$, the equation for $\partial K$ can be simplifies to:

$$ \begin{array}{rl} \phantom{+}\,|1-x-y-z|\\ +\,|1-x+y+z|\\ +\,|1+x-y+z|\\ +\,|1+x+y-z|\\ \end{array} + \begin{array}r \phantom{+}\,|1-x-y-z|\\ \times\,|1-x+y+z|\\ \times\,|1+x-y+z|\\ \times\,|1+x+y-z|\\ \end{array} = 4$$

Equations for other polytopes (like the regular ones) can be constructed but you need to figure out where the faces are.

Update

It turns out there is a simpler formula which doesn't need us to compute any hypervolume.
For a point $p$ to lie on boundary $\partial K$, what one really need is:

1. For all $F$, $p$ either belongs to $P_F$ or on same side of $K$ with resepct to $P_F$. i.e. $\theta_F(p) \ge 0$.
2. For some face $F$, $p$ is lying on $P_F$, i.e. $\theta_F(p) = 0$.

This two conditions together can be summarized as:

$$\max_F \theta_F(p) = 0$$

For the tetrahedron discussed before, an alternate equation for its boundary is

$$\max\{ 1-x-y-z, 1-x+y+z, 1+x-y+z, 1+x+y-z \} = 0$$