6
$\begingroup$

This question was posted on MathEducators a few days ago. Users there suggested I post on MSE.

I am seeking an intuitive explanation (that would make sense to U.S. college students) why the number of regular polytopes in dimension $d$ is:

  • $d=2$, number: $\infty$.
  • $d=3$, number: $5$, the five Platonic solids.
  • $d=4$, number: $6$, with the $24$-cell the polytope with no clear $\mathbb{R}^3$ analog.
  • $d \ge 5$, number: $3$, the simplex, hypercube, and its dual the cross-polytope (or orthoplex).

The derivations (Diophantine equations) are convincing without providing clear intuition. Is there some intuitive explanation, perhaps some explanation about how much "room" there is in $\mathbb{R}^d$? I've wondered if there is a connection to the maximum volume of a unit-ball achieved in dimension $5$,


         
          Plot: Dave Richeson's blog, 2010.
but see Bill Thurston's remarks on the unit-ball volume.

$\endgroup$
4
  • 1
    $\begingroup$ Don't forget $d=0$ and $d=1$, each with "number" $1$. $\endgroup$ – Blue Jun 2 '18 at 2:36
  • 1
    $\begingroup$ Which one of the remarks on the MathOverflow question is Bill Thurston's? It's hard at tell, at least on mobile. $\endgroup$ – user856 Jun 2 '18 at 3:16
  • $\begingroup$ @Rahul: Sorry, I had the wrong link. Now fixed. $\endgroup$ – Joseph O'Rourke Jun 2 '18 at 9:43
  • $\begingroup$ I have no suggestions but I would love to find more on this topic because I never gave it a thought before. $\endgroup$ – poetasis Jun 2 '18 at 10:08
2
$\begingroup$

For that reasoning you do not ask for the existance of all those regular polytopes, but for the limitations.

That there are regular polygons {p} with any number p of sides, clearly is obvious from their inscription into a circle, while reducing the side length.

Beyond consider the Schläfli Symbols: regular polyhedra are {p,q} where {p} is the regular face being used, and {q} is the regular vertex figure. In order to have some angular defect, you just can have 3x{3}, 4x{3}, 5x{3}, 3x{4}, and 3x{5} per vertex, i.e. {3,3}, {3,4}, {3,5}, {4,3}, and {5,3} respectively.

The next level then is to use regular polyhedra {p,q} for cells and also regular polyhedra {q,r} for vertex figures. This defines the regular polychoron {p,q,r}. Coxeter rules out first, that whenever a 5 is being used in the numbers of a Schläfli Symbol, then no 4 can occur. Considering for angular defects again, this thus just leaves {3,3,3}, {3,3,4}, {3,3,5}, {3,4,3}, {4,3,3}, and {5,3,3}.

Then, for 5D polytopes, you need some regular polychora {p,q,r} for facets and some regular polychora {q,r,s} for vertex figure. This defines the regular {p,q,r,s}. Again considering the angular defect, you will be left with {3,3,3,3}, {3,3,3,4}, and {4,3,3,3}.

And being restricted to those 3, it becomes obvious that in any higher dimension no more than 3 regular polytopes can exist.

--- rk

$\endgroup$
1
$\begingroup$

I don't know if this is so much of an "explanation" as it is an "observation", or perhaps some other kind of "-ation", but here goes.

There is an interesting phenomenon in mathematics of which this is one example. Namely, sometimes objects that are classified by "dimension" or by some other natural number valued complexity have a stable pattern which kicks in only after a finite period of exceptional behavior in low complexity.

Another good example of this phenomenon is the classification of semisimple Lie algebras: there are four "stable" infinite families $A_n$, $B_n$, $C_n$, $D_n$; and then there are five exceptional examples $E_6$, $E_7$, $E_8$, $F_4$, $G_2$.

Another good example is the list of Lie groups representing the orientation preserving isometry groups of $n$-dimensional hyperbolic space $\mathbb{H}^n$. When $n=2$ this group has two Lie group descriptions: it is isomorphic to either of the Lie groups $SL(2,\mathbb{R})$ or $SO(2,1)$. When $n=3$ this group also has two Lie group descriptions: it is isomorphic to either of the Lie groups $SL(2,\mathbb{C})$ or $SO(3,1)$. When $n \ge 4$ the picture simplifies, and the only Lie group description is $SO(n,1)$.

$\endgroup$

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.