3
$\begingroup$

The classification of the regular polytopes in any finite amount of dimensions is well known. In 2D, 3D and 4D, there are quite a few exotic shapes, but from $5$ dimensions up, every regular polytope lies in one of three infinite families:

  • Simplexes, the analogs of tetrahedra.
  • Hypercubes, the analogs of cubes.
  • Orthoplexes, the analogs of octahedra.

What strikes me is that when we consider infinite-dimensional space $\mathbb{R}^\mathbb{N}$, all of these shapes still have clear (?) analogs. Here's the constructions I propose:

  • For an $\infty$-simplex, we consider as vertexes the points $(1,0,0,\ldots),$ $(0,1,0,\ldots),$ $(0,0,1,\ldots),$ $\ldots,$ and we create a $k$-face out of the $k$-simplex formed by every $k$ vertices.
  • For an $\infty$-hypercube, we consider as vertexes the points of the form $(\pm1,\pm1,\pm1,\ldots),$ and we create a $k$-face out of the $k$-hypercube formed by every set of $2^k$ vertexes in which all but $k$ coordinates remain constant.
  • For an $\infty$-orthoplex, we consider as vertexes the points $(\pm1,0,0,\ldots),$ $(0,\pm1,0,\ldots),$ $(0,0,\pm1,\ldots),$ $\ldots,$ and we create a $k$-face out of the $k$-simplex formed by every set of $k$ vertexes in which the non-zero coordinate has a different index.

I have a few very related questions:

 a) Is there a formal definition of an "infinite-dimensional regular polytope"?

 b) Do my constructions satisfy it?

 c) Are there any other infinite-dimensional regular polytopes?

Any reference or any reasonable definition would suffice for a), and based on it, it'd be nice to have proofs for b) and c).

I've looked through the internet, found nothing related. I think that we could simply take the usual regular polytope definition and just not require a maximal element, but I have no idea of how we could talk about symmetries afterwards, when there's barely a distance notion. Also, I'd guess that no other regular infinite-dimensional polytopes are possible, since their possible facets (excluding low-dimensional ones) would be restricted to the three aforementioned families, which is kind of a strict condition. But again, I have no idea how to formalize my intuition.

Edit: For question a), we could slightly redefine isometries, so that they need to preserve distance between points at finite distance. That way, the normal definition of a regular polytope would work. I'm almost sure that, under this definition, we could answer question b) in the positive, by combining simpler transformations, but I'm a bit sketchy on those details. I'm still completely stuck on c).

$\endgroup$
  • 1
    $\begingroup$ I know of now way to generalize polytopes to infinite dimensions in some metric way. But what seems very possible is taking the definition of an abstract (regular) polytope and just dropping the axiom for the maximal element (as you said). I would imagine that you would not run into any problem at that point. Regularity then just means that the symmetry group of the abstract polytope acts regularly on the flags (maximal chains of mutually incident faces) ... $\endgroup$ – M. Winter Jan 7 at 19:17
  • $\begingroup$ ... However, abstract regular polytopes are much wilder than the usual convex ones. You already have more than five 3-dimensional ones, and a lot of other infinite examples (not infinite in dimension, but in e.g. vertex count). So, I would expect that you find more than the standard-three regular infinite dimensional polytopes. $\endgroup$ – M. Winter Jan 7 at 19:17
  • 1
    $\begingroup$ You can always try to generalize the standard ways of defining polytopes: convex hull of finitely many points (but this will yield a finite dimensional polytope since the convex hull of $n$ points is at most $(n-1)$-dimensional), or the intersection of finitely many half-spaces. Latter one is defined in infinite dimensional spaces too. Also, just googling "infinite dimensional polytope" brought up several links discussing this topic, e.g. this MO question. $\endgroup$ – M. Winter Jan 7 at 20:05
3
+100
$\begingroup$

Using unit edge sizes you can calculate various other measure properties of any (finite dimensional) simplex, orthoplex, and hypercube simply as a function of the dimension. E.g.

Circumradius of $D$-dimensional (unit-edged) simplex = $$\sqrt\frac D{2(D+1)}$$ Circumradius of $D$-dimensional (unit-edged) orthoplex = $$\frac1{\sqrt2}$$ Circumradius of $D$-dimensional (unit-edged) hypercube = $$\frac{\sqrt D}2$$

or

Inradius of $D$-dimensional (unit-edged) simplex = $$\frac1{\sqrt{2D(D+1)}}$$ Inradius of $D$-dimensional (unit-edged) orthoplex = $$\frac1{\sqrt{2D}}$$ Inradius of $D$-dimensional (unit-edged) hypercube = $$\frac12$$

or

Dihedral angle of $D$-dimensional simplex = $$\arccos\left(\frac1D\right)$$ Dihedral angle of $D$-dimensional orthoplex = $$\arccos\left(\frac2D-1\right)$$ Dihedral angle of $D$-dimensional hypercube = $$\arccos(0)=\frac{\pi}2$$

or

Volume of $D$-dimensional (unit-edged) simplex = $$\frac1{D!}\sqrt{\frac{D+1}{2^D}}$$ Volume of $D$-dimensional (unit-edged) orthoplex = $$\frac1{D!}\sqrt{2^D}$$ Volume of $D$-dimensional (unit-edged) hypercube = $$1$$

etc. And for all these terms you could try to evaluate the limit at $D\to\infty$. Thus e.g.

  1. for the (unit-edged) simplex wrt. dimensional limit you get

    • Circumradius $\to\frac1{\sqrt2}$
    • Inradius $\to 0$
    • Dihedral angle $\to\arccos(0)=\frac{\pi}2$
    • Volume $\to 0$
  2. for the (unit-edged) orthoplex wrt. dimensional limit you get

    • Circumradius $=\frac1{\sqrt2}$
    • Inradius $\to 0$
    • Dihedral angle $\to\arccos(-1)=\pi$
    • Volume $\to 0$
  3. for the (unit-edged) hypercube wrt. dimensional limit you get

    • Circumradius (diverges)
    • Inradius $=\frac12$
    • Dihedral angle $=\arccos(0)=\frac{\pi}2$
    • Volume $=1$

Thus e.g. the orthoplex would become flat like an honeycomb! But still having finite size! - And the simplex, even so ultimately becoming right angled, still will become as flat as possible: with vanishing inradius!

$$\ $$

Probably you know of several of these things already. And therefore asking about some foundation.

You might look for Hilbert spaces in this context. Those are defined by their inner product, i.e. their scalar product. So, when considering some vector $\vec{v}=(v_1, v_2, v_3, …)$ wrt. its base, you would get its squared length by $<\vec{v}, \vec{v}>=\sum_{i=1}^{\infty}v_i^2$, i.e. it converges only if nearly all addends will (approximately) vanish. Therefore esp. any vertex coordinates of a (unit-edged) hypercube will NOT conform to a Hilbert space description. The simplex and the orthoplex however can be considered there strictly.

--- rk

$\endgroup$
  • $\begingroup$ That’s quite some nice insight! I’m not sure it really answers my questions, but it does make my mysterious objects make some more sense. $\endgroup$ – URL Dec 2 '19 at 22:33
  • 2
    $\begingroup$ You even get a further counter-intuitive property of these $\infty$-dimensional regular solids: the dual of the simplex (in fact a gyrated copy of itself) does surely exist within the Hilbert space description, but the dual of the (there existing) orthoplex does NOT! $\endgroup$ – Dr. Richard Klitzing Dec 4 '19 at 12:13
  • $\begingroup$ Perhaps this was a much harder question than I anticipated. I'll award you the bounty, since there doesn't seem to be any sign that someone else could come up with a better answer. $\endgroup$ – URL Dec 9 '19 at 21:30
  • 1
    $\begingroup$ very interesting indeed ! may I ask you to add also the formulas for the solid angle at the vertices vs. D ? $\endgroup$ – G Cab Jan 7 at 22:56
  • $\begingroup$ @GCab: By means of the hypersine function of Wolfram Alpha (mathworld.wolfram.com/Hypersine.html) you could calculate the according measures of the simplicial corners of the hypercube and the regular simplex, cf. mathoverflow.net/questions/350222/…. But the orthoplex eludes this possibility, as its vertex corner isn't simplicial, and addition theorems are not known for this generalized hypersine function (as far as I know). $\endgroup$ – Dr. Richard Klitzing Jan 11 at 11:34
1
$\begingroup$

I decided to write up my comments as an answer. I am not familiar with any generalization of polytopes to infinite dimensions, so I just write down what I can think of.


First of all, there are two modern standard ways to define a (convex) polytope:

  • as the convex hull of finitely many point, or
  • as the intersection of finitely many half spaces.

Both are unsuited to generalize to infinite dimensions. In the first case, we obtain essentially a finite dimensional polytope, because the convex hull of finitely many points is always finite dimensional. In the second case, we obtain the "infinite dimensional" cylinder over a finite dimensional polytope. Still not too interesting.

Crossing out the "finite" in "finitely many points/hyperplanes" will make matters worse, as now we talk about all closed convex sets (as explained here).

But what might works is the following: allow for infinitely many isolated points (or go further, and exclude limit points). The examples you gave are polytopes under this definition.

What still persists is the problem that in a space like $\Bbb R^{\Bbb N}$ there is no direct way to talk about distance, inner products, angles, hence no direct meaning of isometries (distance preserving transformations) and symmetries. You might want to make your space smaller, e.g. work in the $\ell^2$-space in which an inner product exists. I have no idea what interesting symmetric sets of points (and polytopes) might live there.


There seems to be another solution: abstract (regular) polytopes. Abstract polytopes are mostly studied in the case of highly symmetric polytopes (which is exactly what you want). They do not live in any Euclidean space per se, but they still have a well defined notion of dimension.

An abstract polytope is just a partially ordered set that satisfies some further axioms. Among these, the existence of a smallest and largest element. Drop the "largest element" axiom and we can probably talk about infinite dimensional polytopes. Note however, that in the abstract case, there are already more than five $3$-dimensional regular polytopes, and among these, some with an infinite number of vertices. So you will find much more than the three well known families.

I am not aware of any result listing the "infinite dimensional abstract regular polytopes".

$\endgroup$
  • $\begingroup$ I think we could first define an infinite abstract polytope as an abstract polytope, but instead requiring no maximal face. The notion of regularity would still be well-defined. We could then talk about realizations in $\mathbb R^\mathbb N$, not defining them as intersections of half-planes, but instead as sets of finite-dimensional polytopes satisfying the requirements (in a similar manner as to how we define non-convex polytopes). All and all, this is a very satisfying answer to part a). $\endgroup$ – URL Jan 7 at 21:53

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.