2
$\begingroup$

Is there a simple method for determining the number of vertices, edges, faces, and cells of the 6 regular convex 4D polytopes?

For the 3D platonic solids, it can be shown using combinatorial logic that for a polyhedron with Schläfli symbol {p,q}

pF = 2E = qF

This gives the proper ratios of the numbers of each element. Then Euler's formula can be used to solve the system of equations.

I was able to use similar logic to find a method for calculating the relative number of vertices, edges, faces, and cells of a 4D polytope. However, the Euler characteristic adds no new information because it is 0 so the actual numbers cannot be found.

Is there a simple way to generalize the method for the platonic solids or is more complex math required?

$\endgroup$
0
$\begingroup$

I do not have a complete answer, but I will post what I have so far. The first thing we will need is formulas for the V, E, and F for a 3d polyhedron. These can be derived very simply from the combinatorial identities you stated yourself, that $$ p F = 2 E = q V$$ Since we know that in 3D we have Euler's Characteristic Formula, $$\begin{align} V - E + F &= 2 \\ \frac{2 E}{q} - E + \frac{2 E}{p} &= 2 \\ \frac1{p} + \frac1{q} &= \frac12 + \frac1E \end{align}$$ doing some simple algebra gives us the following formulas for 3d polyhedron $\{p,q\}$ $$ E = \frac{2 p q}{4 - (p-2) (q-2)} $$ $$ V = \frac{4 p}{4 - (p-2) (q-2)} $$ and $$ F = \frac{4 q}{4 - (p-2) (q-2)} $$

Now we move on to 4d polyhedra, defined by $\{p,q,r\}$. First of all we know that $\{p,q\}$ and $\{q,r\}$ must both exist by the recursive definition of Schlafli notation, which limits us to 11 possible polyhedra (at this point, I don't know a non-geometric way to show only 6 of these actually exist). We know that each edge has $r$ $\{p,q\}$ polyhedra "around" it, so motivated by this we consider the vertices, $\overline{V}$, edges, $\overline{E}$, and faces, $\overline{F}$ in each cell. Conveniently, we have already developed formulas for these above. We will also need the number of cells each vertex is a part of. While this seems very difficult to find, if we think about the dual polyhedron, $\{r,q,p\}$, then it is just the number of vertices of the dual polyhedron per cell, or $\tilde{V} = \overline{V}(r,q)$. So we now have formulas for $\overline{V},\overline{E},\overline{F}$, and $\tilde{V}$. We are going to end up writing $V$, $E$, and $F$ in terms of $C$, the total number of cells of our 4d polyhedron.

First, we use a similar argument to the 3d case to count vertices. $\tilde{V}$ counts the number of cells each vertex is a part of, meaning that if we multiply $\tilde{V}$, and $V$, we will be counting cells. This product however overcounts $C$, since each cell is counted by its $\overline{V}$ vertices. Therefore we arrive at $$V \tilde{V} = C \overline{V}$$

The next two arguments are much simpler. Each face is part of 2 cells, so $$ 2 F = C \overline{F}$$

Finally we know that each edge is part of $r$ cells, so $$r E = C \overline{E}$$

We now know $V$, $E$, and $F$, in terms of $C$, similar to how we wrote $V$ and $F$ in terms of E for the 3d case. In 4-dimensions however, the Euler Characteristic Formula takes the form $$V - E + F - C = 0$$ (as can be seen through thinking about the Euler characteristic as being defined as the alternating sum of the Betti Numbers).

As we come to our final step, it might seem that we are nearly there and should be able to solve for C, just as we solved for E in the 3d case, but we run into a problem! Since the right side is 0 instead of 2, we can't divide by C and still have a C present in the equation to solve for, an moreover the left side is equal to 0 for all $p$, $q$, and $r$, meaning that we don't gain any information. If we were able to find any one of $V$, $E$, $F$, or $C$, and given $\{p,q,r\}$, we could determine all of them, but it seems that we can't proceed further.

TLDR: We are able to find the ratio of any pair of $\{V,E,F,C\}$, but not the numbers themselves. If we were able to somehow determine one of the numbers by some other method, we would be able to get all of them. I'm pessemistic about the changes that some combinatorial argument exists, and think any general method would either have to rely on specific geometry of the cells or on some group theoretic argument about the size of the symmetry group of the solid.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.