New insights about infinite-dimensional polytopes

[mr_e_man]

Anyway, if the orthoplex is x3o3o3o3..., and the simplex is x3o3o3o3..., then it's obvious that they should be the same.

Be careful here. You're making a pretty big assumption that the limit of the sequence of CD diagrams must itself be a valid CD diagram. That may not necessarily be the case.

I'm not taking a limit. It's just directly an infinite CD diagram, whick looks like ℕ; each node k is connected to node k+1 by an edge (labelled '3' if desired).

What could make it invalid?

This is why I keep telling people, generalizing stuff to infinity requires a lot of precision and care about tiny details, because it can make a HUGE difference. The smallest slip-ups get infinitely magnified into big errors. That last step when you leap from the finite to the infinite requires the utmost care, because things happen at infinity that sometimes have no direct correspondence with any of the preceding finite cases.

That is true.

I'm sure if done carefully the Wythoff construction can be made to work. But one has to be careful not to blindly assume without proof that what holds in the finite case necessarily holds in the infinite case. In fact, in this particular case I've discovered that the infinite CD diagram looks more like o4o3o3o...o3x with an infinite gap in the middle, than having the infinite gap at either end.

As I said, if there's an infinite gap in the middle, the two ends aren't really connected. The mirror vectors at one end are orthogonal to all of the mirror vectors at the other end. So any polytope made in this way is trivially a product of a polytope made from marking subsets of ...3o3o3o and a polytope made from marking subsets of o4o3o3o3... .

This is possible thanks to the apacs operator, which allows arbitrary permutations of coordinates. (And BTW, this implies that some of these polytopes have an uncountable number of vertices, including the hypercube itself.

Good point. The cube vertex (+1,+1,+1,+1,...) is connected by edges to only those countably many vertices which have a finite number of '-1' coordinates; it's not connected by edges to the opposite vertex (-1,-1,-1,-1,...).

This is one area where using surreal numbers may help us distinguish between these cases.

Do you think hyperreal numbers would be sufficient?

[quickfur]

Anyway, if the orthoplex is x3o3o3o3..., and the simplex is x3o3o3o3..., then it's obvious that they should be the same.

Be careful here. You're making a pretty big assumption that the limit of the sequence of CD diagrams must itself be a valid CD diagram. That may not necessarily be the case.

I'm not taking a limit. It's just directly an infinite CD diagram, whick looks like ℕ; each node k is connected to node k+1 by an edge (labelled '3' if desired).

What could make it invalid?

The fact that the o4o... end is missing? That's kinda like the fundamental difference between the hypercube family and the simplex family. Pretty important, if you ask me.

And it could very well be that ∞-dimensional polytopes cannot be represented by infinite CD diagrams. That's also a possibility we haven't disproved.

[...]

I'm sure if done carefully the Wythoff construction can be made to work. But one has to be careful not to blindly assume without proof that what holds in the finite case necessarily holds in the infinite case. In fact, in this particular case I've discovered that the infinite CD diagram looks more like o4o3o3o...o3x with an infinite gap in the middle, than having the infinite gap at either end.

As I said, if there's an infinite gap in the middle, the two ends aren't really connected. The mirror vectors at one end are orthogonal to all of the mirror vectors at the other end. So any polytope made in this way is trivially a product of a polytope made from marking subsets of ...3o3o3o and a polytope made from marking subsets of o4o3o3o3... .

So it's possible that CD diagrams don't work in infinite-dimensional space. Or perhaps there are new subtleties that come into play that we didn't realize.

[...]

This is one area where using surreal numbers may help us distinguish between these cases.

Do you think hyperreal numbers would be sufficient?

I think either would suffice. The more pertinent question is, how to map infinite series/sequences to specific members of the hyperreals/surreal numbers. For example, in computing the dot product of the hypercube's vertex, we have the series 1+1+1+ ... . Which infinite number should we assign this product to? Let's say we assign it to some number called X. Then in what case should a series converge to X+1 or X-1? One nice thing about using surreal numbers is that they all have a specific "birthday", and when confronted with a choice when multiple answers seem equally possible, we can always choose the unique surreal with the earliest birthday. On the other hand, though, it's unclear whether such a choice would actually make sense when it comes to geometric questions involving infinite quantities.

Also, once we obtain a surreal number answer, how do we interpret it? Our motivating case was the difacetal angle of the ∞-cross: it converges to 180°, so how do we determine whether it's exactly 180°, or (180 - w)° for some infinitesimal w? And how do we determine the specific value of w? It's not an easy question because the computation of this angle involves the transcendental arccosine function. I've no idea how to compute a transcendental function over the surreals. And even once we determine the value of w, how do we determine what happens when we attach two polytopes to each other at an angle of (180 - w)°? How do we determine whether it closes up into a spherical polytope? (As opposed to, say, something like a paraboloid that's open on one end.) It's not so easy to construct a consistent model of what happens in such cases.

[quickfur]

Here's something interesting I discovered yesterday. I was looking at the ∞-simplex (as one of the ∞-cross's facets):

Code: Select all

   < 1,  0,  0,  0, ...>
   < 0,  1,  0,  0, ...>
   < 0,  0,  1,  0, ...>
   ...


and one of its facets:

Code: Select all

   < 0,  1,  0,  0, ...>
   < 0,  0,  1,  0, ...>
   < 0,  0,  0,  1, ...>
   ...


Obviously this facet, let's call it F, must be a simplex of some sort. The most obvious choice is that it's equal to the ∞-simplex itself, since, after all, it has an infinite number of vertices, and there's an obvious isomorphism here mapping <0, x, y, z, ...> to <x, y, z, ...>, preserving the face lattice structure. There's a slight problem, however. In finite dimensions, if two polytopes are identical, there must be an isometry, i.e., a rigid motion, that translates one polytope into the other. However, the mapping <0, x, y, z, ...> → <x, y, z, ...> is not a rigid motion, because if applied to another polytope that has a non-zero first coordinate, it drops that coordinate. I.e., it's a projection rather than a rigid motion. But if the only thing that could map F to the original ∞-simplex is a projection, then they can't possibly be identical?

However, there's something more interesting going on here. Consider the reflection that maps <0, 1, ...> to <1, 0, ...>, i.e., it interchanges the first two coordinates. Obviously, this is a rigid motion. We can compose it with another reflection that interchanges the 2nd and 3rd coordinates. And compose the result with a 3rd reflection that interchanges the 3rd and 4th coordinates. And so on. At each stage, we have a compound reflection through a series of mirrors, that shift the first n coordinates left by 1 position, pushing the first coordinate to the n'th place. Obviously, these are all rigid motions. They are permutations of coordinates (no coordinates are "missing" after the transformation, unlike the projection case above). Interestingly enough, though, if we take the limit as n→∞, we get a transformation that shifts all the coordinates of the point left by 1 position, but the 1st coordinate is pushed to infinity, i.e., the limiting transformation is no longer a rigid motion, but a projection.

IOW, the limit of this series of left-shifting reflections is a projection.

This has very interesting implications, because it implies that there's a connection between rigid motions and projections, namely, the latter is a limit of infinite sequences of the format. Since we're in ∞-dimensional space, it would seem logical that we could perform an infinite sequences of reflections to a polytope. And it appears to be the case that such a limit transformation is required in order to map the ∞-simplex's facets onto itself. Or, to put it another way, the ∞-simplex consists of ∞-simplex facets, but infinitely reflected such that there is no rigid motion between them, but a projection / lifting (lifting = immersing in a space 1 dimension higher than the original one, by prepending a coordinate like in this case to map the ∞-simplex to F -- only, in this case, this higher-dimensional space happens to be identical to the original space).

In finite-dimensional space, the dimension of a polytope P is greater than the dimension of its projection, i.e., dim(proj(P)) ≤ dim(P), and the dimension of a facet of P is less than the dimension of P, i.e., dim(facet(P)) ≤ dim(P). In this case, however, the ∞-simplex is a projection of its facet F; which seems to imply that dim(∞-simplex) ≤ dim(F). But since F is a facet of the ∞-simplex, we also have dim(F) ≤ dim(∞-simplex). So we have dim(∞-simplex) ≤ dim(F) ≤ dim(∞-simplex), meaning that dim(∞-simplex) = dim(F). (Note that this also implies that (∞-1)=∞, because the facets of the ∞-simplex, by dimensional analogy, ought to be (∞-1)-simplices, but since these (∞-1)-simplices project to the ∞-simplex, they cannot be of a smaller dimension than ∞.)

All of this implies that in ∞-dimensional space, there are (at least) two kinds of motions: rigid motions, that behave like their finite-dimensional counterparts, and limit motions, which are projections that, under certain circumstances, can map an ∞-polytope into a "different" copy of itself such that it can become facets of another ∞-polytope or vice versa. This "different" copy is isomorphic (in fact, isometric) to the original polytope, but oriented in an infinitely different way such that there does not exist a rigid motion that maps it back to the original polytope. Furthermore, there's an infinite hierarchy of these "different orientations" (consider, for example, the facets of F and their respective sub-facets), all disconnected from each other in terms of rigid motions, but connected via limit motions.

The corollary is that if you reflect/rotate/etc. an ∞-polytope, you can "change" its dimensionality. Infinite-dimensional space is weird.

(The same argument can be applied to the ∞-cube: its facets are also ∞-cubes, but oriented in a way that's infinitely different from the original ∞-cube such that no rigid motion can transform one to the other. But they are isometric under projection / lifting, i.e., under limit motions.)

[quickfur]

[...]
v₀ = (1, 1/√3, 1/√6, 1/√10, 1/√15, ...)
v₁ = (-1, 1/√3, 1/√6, 1/√10, 1/√15, ...)
v₂ = (0, -2/√3, 1/√6, 1/√10, 1/√15, ...)
v₃ = (0, 0, -3/√6, 1/√10, 1/√15, ...)
v₄ = (0, 0, 0, -4/√10, 1/√15, ...)
v₅ = (0, 0, 0, 0, -5/√15, ...)
:

You could've just asked, this has been solved a long time ago: https://en.wikipedia.org/wiki/User:Tetr ... _n-simplex


Basically, the coordinates of the upper triangle are square roots of the inverse triangular numbers, and the coordinates down the diagonal are the negative square roots of the ratio of square numbers to triangular numbers.

[mr_e_man]

Anyway, if the orthoplex is x3o3o3o3..., and the simplex is x3o3o3o3..., then it's obvious that they should be the same.

Be careful here. You're making a pretty big assumption that the limit of the sequence of CD diagrams must itself be a valid CD diagram. That may not necessarily be the case.

I'm not taking a limit. It's just directly an infinite CD diagram, whick looks like ℕ; each node k is connected to node k+1 by an edge (labelled '3' if desired).

What could make it invalid?

The fact that the o4o... end is missing? That's kinda like the fundamental difference between the hypercube family and the simplex family. Pretty important, if you ask me.

I mean it's a valid CD diagram, regardless of whether it represents anything cubic or orthoplicial.

And it could very well be that ∞-dimensional polytopes cannot be represented by infinite CD diagrams. That's also a possibility we haven't disproved.

We can't prove anything until we specify exactly what an infinite-dimensional polytope is supposed to be.

An infinite CD diagram determines an infinite set of vertices, and taking the convex hull... rather, the closure of the convex hull; the centre of the ∞-simplex is not a finite convex combination of the vertices, so the centre is not in the convex hull! The closed convex hull of an infinite set of vertices (not contained in any finite-dimensional space, like x4o4o or x5o4o) is one possible definition of an ∞-polytope.

Another possibility is to first define abstract ∞-polytopes.

[...]
v₀ = (1, 1/√3, 1/√6, 1/√10, 1/√15, ...)
v₁ = (-1, 1/√3, 1/√6, 1/√10, 1/√15, ...)
v₂ = (0, -2/√3, 1/√6, 1/√10, 1/√15, ...)
v₃ = (0, 0, -3/√6, 1/√10, 1/√15, ...)
v₄ = (0, 0, 0, -4/√10, 1/√15, ...)
v₅ = (0, 0, 0, 0, -5/√15, ...)
:

You could've just asked, this has been solved a long time ago: https://en.wikipedia.org/wiki/User:Tetr ... _n-simplex


Basically, the coordinates of the upper triangle are square roots of the inverse triangular numbers, and the coordinates down the diagonal are the negative square roots of the ratio of square numbers to triangular numbers.

There's no harm in solving it again. It wasn't too difficult; it may have been easier, or faster, than asking and searching. And I wanted to derive the coordinates directly from the simplex CD diagram, not referring to cubic things with an extra dimension.

Also, that page is not specifically about infinite dimensions.

[mr_e_man]

Here's something interesting I discovered yesterday. I was looking at the ∞-simplex (as one of the ∞-cross's facets):

Code: Select all

   < 1,  0,  0,  0, ...>
   < 0,  1,  0,  0, ...>
   < 0,  0,  1,  0, ...>
   ...


Now look at the opposite facet, containing the vertices (-1,0,0,0,...), (0,-1,0,0,...), (0,0,-1,0,...), ... . Both facets have the same orientation; the same set of differences between vertices. (That is the span of f₂, f₃, f₄, ... .) Both facets have the same centre 0. Their affine hulls are different parallel subspaces; one contains (1,0,0,0,...) and the other doesn't, and neither contains 0 exactly, but both come arbitrarily close to 0. Both facets have the same closed affine hull. Does that mean that the two facets are "coplanar", and thus that the orthoplex is degenerate?

[quickfur]

[...]
We can't prove anything until we specify exactly what an infinite-dimensional polytope is supposed to be.

Hmm, that is a very good point. Up to now, I've been assuming the equivalence of a set of vertices with the polytope itself (i.e. the convex hull). But you're right, the convex hull of an infinite set of vertices needs to be properly defined, otherwise we're treading on thin ice, and may be inadvertently talking about things that don't exist.

In order to talk about convex hulls, at least in the traditional sense used in finite-dimensional space, we need to talk about bounding hyperplanes and their intersections. I think it should be relatively safe to define a bounding hyperplane as a tuple containing an ∞-vector N (the normal of the hyperplane), and a constant c (the value of the dot product of N with points that lie on the hyperplane). Then the hyperplane is simply the set of points x for which N·x = c. Turning that into an inequality N·x ≤ c gives us a half-space, which allows us to define a polytope as the set of points contained in a given set of half-spaces.

Is it possible to prove that a given set of vertices will always have a corresponding set of half-spaces that would let us define the interior of the corresponding polytope? I'm not sure how to go about proving this. And in fact, I already have a counterexample: the would-be omnitruncated ∞-cube would have coordinates of the form (1+k√2) for k = 0, 1, 2, ... . Since the coordinates are unbounded, any prospective bounding hyperplane must contain infinite coefficients (otherwise it could not possibly bound the polytope's vertices!). So if we limit ourselves to finite coefficients in our hyperplane equations, then we must exclude the omnitruncated ∞-cube from our definition of ∞-polytopes.

Interestingly enough, just today I've been playing around more with coordinate-based study of the hypercube family of uniform truncates, and have discovered at least 3 distinct classes of such uniform truncates (or should I say, would be uniform truncates -- pending an actual definition of a ∞-polytope). There's a "small" class that have finite outradius; these are contained in Hilbert space and seem to be the most promising candidates for a definition of polytope involving hyperplanes of finite coefficients. After them is a "large" class that do not have a finite outradius, like the ∞-cube itself. This large class can be further split into "large" and "very large", with "very large" meaning the coefficients are unbounded. The "very large" uniforms include things like the would-be omnitruncated ∞-cube and other such beasts, which cannot be bounded by hyperplanes with finite coefficients. They serve as limiting shapes (well, if such shapes exist ) in various sequences of important representatives of the cube/cross uniform polytopes, so they would seem useful to have around, even if as unattainable ideals. They have the peculiar property that their vertices are in R^∞ but their bounding hyperplanes are not.

The large (but not very large) class, which includes the ∞-cube, ostensibly do have bounding hyperplanes with finite coefficients (to be proven!), so they could potentially be proper ∞-polytopes outside of Hilbert space. These may be somewhat problematic, though, because their non-finite outradius may imply that some of them may not have duals in R^∞, since some of them may have hyperplanes with non-finite coefficients.

An infinite CD diagram determines an infinite set of vertices, and taking the convex hull... rather, the closure of the convex hull; the centre of the ∞-simplex is not a finite convex combination of the vertices, so the centre is not in the convex hull! The closed convex hull of an infinite set of vertices (not contained in any finite-dimensional space, like x4o4o or x5o4o) is one possible definition of an ∞-polytope.

Another possibility is to first define abstract ∞-polytopes.

Hmm, yeah, this aspect could prove problematic... I wonder if at some point the use of hyperreals/surreals would become unavoidable, because even with seemingly "simple" polytopes like the ∞-cross we're already seeing some odd behaviours that seem to imply distinctions that can only be made with infinitesimals.

[...]
v₀ = (1, 1/√3, 1/√6, 1/√10, 1/√15, ...)
v₁ = (-1, 1/√3, 1/√6, 1/√10, 1/√15, ...)
v₂ = (0, -2/√3, 1/√6, 1/√10, 1/√15, ...)
v₃ = (0, 0, -3/√6, 1/√10, 1/√15, ...)
v₄ = (0, 0, 0, -4/√10, 1/√15, ...)
v₅ = (0, 0, 0, 0, -5/√15, ...)
:

You could've just asked, this has been solved a long time ago: https://en.wikipedia.org/wiki/User:Tetr ... _n-simplex


Basically, the coordinates of the upper triangle are square roots of the inverse triangular numbers, and the coordinates down the diagonal are the negative square roots of the ratio of square numbers to triangular numbers.

There's no harm in solving it again. It wasn't too difficult; it may have been easier, or faster, than asking and searching. And I wanted to derive the coordinates directly from the simplex CD diagram, not referring to cubic things with an extra dimension.

Well, it's nice to know that those coordinates are confirmed by independent calculations.

Also, that page is not specifically about infinite dimensions.

Yeah, but the pattern is easily extended to infinite dimensions.

[quickfur]

Here's something interesting I discovered yesterday. I was looking at the ∞-simplex (as one of the ∞-cross's facets):

Code: Select all

   < 1,  0,  0,  0, ...>
   < 0,  1,  0,  0, ...>
   < 0,  0,  1,  0, ...>
   ...


Now look at the opposite facet, containing the vertices (-1,0,0,0,...), (0,-1,0,0,...), (0,0,-1,0,...), ... . Both facets have the same orientation; the same set of differences between vertices. (That is the span of f₂, f₃, f₄, ... .) Both facets have the same centre 0. Their affine hulls are different parallel subspaces; one contains (1,0,0,0,...) and the other doesn't, and neither contains 0 exactly, but both come arbitrarily close to 0. Both facets have the same closed affine hull. Does that mean that the two facets are "coplanar", and thus that the orthoplex is degenerate?

In a way, I'm not surprised that the orthoplex is potentially degenerate. Consider the volume of the n-sphere as n→∞, for example, which converges to 0, which seems to imply that in R^∞ a sphere of non-zero radius has zero volume(!). (Or at least, infinitesimal volume.) In the case of the orthoplex, its facets are more "concave" than the surface of the ∞-sphere, so it seems somehow "logical" that the facet centers should be arbitrarily close to the origin, weird as that may be.

The facet centers of the orthoplex, assuming dimensional analogy holds, ought to lie along the line k*<1,1,1,...>. The fact that this normal vector has non-finite length is an indication that something fishy is going on in this direction. My present wild speculation is that ∞-dimensional space is so infinitely large along this direction, that anything smaller than an infinite radius couldn't possibly cover a non-zero distance along it. Remember the Curse of Dimensionality, in which the higher the dimension, the "pointier" the n-cube becomes? And conversely, the more "concave" the n-sphere become? At n=∞ I think something drastic takes place in which this pointiness becomes extreme, such that the "bluntness" of the orthoplex in this direction practically collapses onto the origin somehow. I'm not 100% sure what's going on here, but something is definitely fishy here.

Consider also this: the inradius of the n-cross (i.e., the distance of its facets from the origin) converges to 0 as n increases without bound. This seems to imply that at n=∞, something truly bizarre happens and the facets actually touch the origin? This definitely deserves more study to figure out what exactly is going on here!

[quickfur]

P.S. Further thoughts about the opposite facets of the orthoplex: so we have two parallel hyperplanes that are clearly distinct (one contains <1,0,0,...>, the other doesn't), yet come arbitrarily close to each other. That's totally weird, and somewhat reminiscient of straight lines in hyperbolic space, where parallel lines can have a point of closest approach or in some cases even converge to each other asymptotically.

It really makes me wonder if the fact that R^∞ has an infinite number of coordinate axis somehow endows it with properties that resemble the exponentially-expanding space in hyperbolic geometry. Clearly, something is expanding exponentially here: the number of vertices in the ∞-cube, for example, the infinite length of the vector <1,1,1,...>, and various other circumstantial clues, that all seem to be hinting at something resembling (yet fundamentally different from) hyperbolic space. The space is flat in the sense of curvature, but something about the infinite number of axes is causing some kind of distortion. Maybe it's something to do with trying to move a non-vanishing distance along an infinite number of axes at the same time?

On a tangential note: it would appear that Hilbert space, if I'm understanding this correctly, has zero volume(!) in R^∞... Infinity is weird.

[mr_e_man]

In order to talk about convex hulls, at least in the traditional sense used in finite-dimensional space, we need to talk about bounding hyperplanes and their intersections.

No, those are two different definitions of convex polytopes, which happen to be equivalent, at least in finite dimensions. We can talk about convex hulls without half-spaces.

A convex combination of a (finite or infinite) set S of points is any finite linear combination with non-negative coefficients whose sum is 1. That is, a convex combination has the form sum_k=1ⁿ t_kx_k, where n is some natural number, each point x_k is in S, each coefficient t_k≥0, and sum_k=1ⁿ t_k = 1. The convex hull of S is the set of all possible convex combinations of S. (For the affine hull, the coefficients are allowed to be negative.)

The notion of closure (meaning, including boundary points) requires topology, which is provided in Hilbert space, but is not provided (in a nice form) in the larger space of all sequences. https://en.wikipedia.org/wiki/Sequence_space

Here's something interesting I discovered yesterday. I was looking at the ∞-simplex (as one of the ∞-cross's facets):

Code: Select all

   < 1,  0,  0,  0, ...>
   < 0,  1,  0,  0, ...>
   < 0,  0,  1,  0, ...>
   ...


Now look at the opposite facet, containing the vertices (-1,0,0,0,...), (0,-1,0,0,...), (0,0,-1,0,...), ... . Both facets have the same orientation; the same set of differences between vertices. (That is the span of f₂, f₃, f₄, ... .) Both facets have the same centre 0. Their affine hulls are different parallel subspaces; one contains (1,0,0,0,...) and the other doesn't, and neither contains 0 exactly, but both come arbitrarily close to 0. Both facets have the same closed affine hull. Does that mean that the two facets are "coplanar", and thus that the orthoplex is degenerate?

In fact that common closed affine hull is the whole Hilbert space! In other words, the subspace containing one facet comes arbitrarily close to any point, not just 0. This follows from the fact that the closure is a linear subspace (it contains 0), and it contains (1,0,0,...)=f₁, and it also contains f₂,f₃,f₄,..., and these span the Hilbert space. (Here I use "span" to include infinite linear combinations, which are boundary points of the set of finite linear combinations.)

[quickfur]

[...]

Here's something interesting I discovered yesterday. I was looking at the ∞-simplex (as one of the ∞-cross's facets):

Code: Select all

   < 1,  0,  0,  0, ...>
   < 0,  1,  0,  0, ...>
   < 0,  0,  1,  0, ...>
   ...


Now look at the opposite facet, containing the vertices (-1,0,0,0,...), (0,-1,0,0,...), (0,0,-1,0,...), ... . Both facets have the same orientation; the same set of differences between vertices. (That is the span of f₂, f₃, f₄, ... .) Both facets have the same centre 0. Their affine hulls are different parallel subspaces; one contains (1,0,0,0,...) and the other doesn't, and neither contains 0 exactly, but both come arbitrarily close to 0. Both facets have the same closed affine hull. Does that mean that the two facets are "coplanar", and thus that the orthoplex is degenerate?

In fact that common closed affine hull is the whole Hilbert space! In other words, the subspace containing one facet comes arbitrarily close to any point, not just 0. This follows from the fact that the closure is a linear subspace (it contains 0), and it contains (1,0,0,...)=f₁, and it also contains f₂,f₃,f₄,..., and these span the Hilbert space. (Here I use "span" to include infinite linear combinations, which are boundary points of the set of finite linear combinations.)

Hmm, this is both perplexing and interesting at the same time.

So the affine hulls give us distinct parallel subspaces, but the closed affine hull spans the entire Hilbert space? That's... wow, weird and wonderful at the same time. This seems to indicate that closures must be taken with utmost care, as it can fundamentally alter the results. I wonder if using hyperreals/surreals might provide a way to differentiate between these cases. But it seems tricky, though.

[quickfur]

So it turns out that the hyperplane equations for the facets of the ∞-cross are very simple: they are of the form

<1,1,...>·x = 1

with various sign combinations of the normal vector <1,1,...>. This means that the origin does not lie on the hyperplane; the centroid of the facet is must be a point with infinitesimal coordinates, because <0,0,...> does not satisfy this equation, but <1/w, 1/w, ...> where w is the surreal number corresponding with the first infinite ordinal, does.

So it seems inescapable that we have to use hyperreals or surreals if we want to have a nice, closed geometry on ∞-polytopes. It's either that, or we have to live with the counterintuitive conclusion that the facets of the ∞-cross do not have centroids.

[mr_e_man]

Here's something interesting I discovered yesterday. I was looking at the ∞-simplex (as one of the ∞-cross's facets):

Code: Select all

   < 1,  0,  0,  0, ...>
   < 0,  1,  0,  0, ...>
   < 0,  0,  1,  0, ...>
   ...


Now look at the opposite facet, containing the vertices (-1,0,0,0,...), (0,-1,0,0,...), (0,0,-1,0,...), ... . Both facets have the same orientation; the same set of differences between vertices. (That is the span of f₂, f₃, f₄, ... .) Both facets have the same centre 0. Their affine hulls are different parallel subspaces; one contains (1,0,0,0,...) and the other doesn't, and neither contains 0 exactly, but both come arbitrarily close to 0. Both facets have the same closed affine hull. Does that mean that the two facets are "coplanar", and thus that the orthoplex is degenerate?

In fact that common closed affine hull is the whole Hilbert space! In other words, the subspace containing one facet comes arbitrarily close to any point, not just 0. This follows from the fact that the closure is a linear subspace (it contains 0), and it contains (1,0,0,...)=f₁, and it also contains f₂,f₃,f₄,..., and these span the Hilbert space. (Here I use "span" to include infinite linear combinations, which are boundary points of the set of finite linear combinations.)

But this implies, shifting the vertex (1,0,0,...) to the origin, that the span of f₂,f₃,f₄,... comes arbitrarily close to (-1,0,0,...). That is, -f₁ is in the closure of the span of f₂,f₃,f₄,..., though it's not in the span itself. Indeed, we can see that more directly:

√2 (f₂ + (f₂+f₃) + (f₂+f₃+f₄) + ... + (f₂+f₃+...+f_n))
= (-1,1,0,0,0,...) + (-1,0,1,0,0,...) + (-1,0,0,1,0,...) + ... + (-1,0,...,0,1,0,0,...)
= (-(n-1), 1, 1, ... , 1, 0, 0, ...)

√2/(n-1) (f₂ + (f₂+f₃) + (f₂+f₃+f₄) + ... + (f₂+f₃+...+f_n))
= (-1, 1/(n-1), 1/(n-1), ... , 1/(n-1), 0, 0, ...)

(There are n non-zero coordinates.) The difference from -f₁ is

(0, 1/(n-1), 1/(n-1), ... , 1/(n-1), 0, 0, ...)

which has magnitude 1/√(n-1), and therefore approaches 0 as n increases.

But does this mean that -f₁ is an infinite sum of the others? The coefficients depend on n:

√2/(n-1) (f₂ + (f₂+f₃) + (f₂+f₃+f₄) + ... + (f₂+f₃+...+f_n))
= √2/(n-1) ((n-1)f₂ + (n-2)f₃ + (n-3)f₄ + ... + 1f_n)
= √2 f₂ + √2(n-2)/(n-1) f₃ + √2(n-3)/(n-1) f₄ + ... + √2/(n-1) f_n

We can't just take the limit of the individual coefficients, because that sum doesn't converge:

√2 f₂ + √2 f₃ + √2 f₄ + ... + √2 f_n
= (-1, 0, 0, ... , 0, 1, 0, 0, ...)

The difference from -f₁ always has magnitude 1; it doesn't approach 0.

So I can't tell whether the cubic mirror vectors are a basis; whether they're independent. Certainly they're a basis for the subspace where each point has finitely many non-zero coordinates, if we consider only finite sums. They may or may not be a Schauder basis for Hilbert space.

Hilbert space is supposed to be the best-behaved infinite-dimensional space. But it's already too confusing.

[adam ∞]

More fun questions: since the difacetal angle of the orthoplex converges to 180°, does that mean it tiles space? On the other hand the simplex difacetal angle converges to 90°, which appears to imply that folding 4 simplices to a peak ought to tile space. But instead, it only forms an orthoplex. How can that be??

The ∞-orthoplex tiles ∞-space and has the ∞-cube as a dual as far as I can tell.

The thing about the simplex attaining a 90 degree angle, meeting in 4's to form the orthoplex is one of the reasons I suggested that maybe the ∞-cell is the last simplex, occurring in dimension ∞-1, not ∞. Because if in dimension ∞ you have a hypercube, an orthoplex, a simplex, and simplices that can combine to form a second, different orthoplex, composed of simplices of the same dimensions, which would have a dual which is a second, different hypercube, paradoxically composed of hypercubes of the same dimension, it leads to so many weird contradictions, and it just makes more sense that the ∞-cell is the last simplex, rather than an ∞+1-cell in dimension ∞.

My conjecture is that dimension ∞ has no simplex, which solves all of those problems.

[quickfur]

[...]Hilbert space is supposed to be the best-behaved infinite-dimensional space. But it's already too confusing.

I just realized something even more surprising about Hilbert space: it does not contain anything along the lines of the form k*<±1,±1,...> except the origin. So in the ∞-cross contained in Hilbert space or not? If it is, we're forced to conclude that its facets have no centroids (because its facet centroids must lie along the line k*<±1,±1,...>, and the origin does not satisfy the corresponding hyperplane equations). But if the ∞-cross is not in Hilbert space, that means we have to sacrifice all the "nice" properties of Hilbert space even for polytopes that finite outradius!

[...]My conjecture is that dimension ∞ has no simplex, [...]

Of course it does. Here's a ∞-simplex:

Code: Select all

<±1,  1/√3,  1/√6,  1/√10,  1/√15, ...>
< 0, -1/√3,  1/√6,  1/√10,  1/√15, ...>
< 0,     0, -2/√6,  1/√10,  1/√15, ...>
< 0,     0,     0, -3/√10,  1/√15, ...>
< 0,     0,     0,      0, -4/√15, ...>
...


where the square roots are of the triangular numbers.

The center of this simplex is the origin, and all edge lengths are 2.

[adam ∞]

[...]My conjecture is that dimension ∞ has no simplex, [...]

Of course it does. Here's a ∞-simplex:

Code: Select all

<±1,  1/√3,  1/√6,  1/√10,  1/√15, ...>
< 0, -1/√3,  1/√6,  1/√10,  1/√15, ...>
< 0,     0, -2/√6,  1/√10,  1/√15, ...>
< 0,     0,     0, -3/√10,  1/√15, ...>
< 0,     0,     0,      0, -4/√15, ...>
...


where the square roots are of the triangular numbers.

The center of this simplex is the origin, and all edge lengths are 2.

There are different ways we can express what to expect if it does exist, but that doesn't prove that it exists.

Do you agree with my assessment that this seems to lead to not only a simplex, but also an additional orthoplex made of same-dimensional facets, with a new hypercube with same-dimensional facets? I consider these to be impossible, but I think you are more open minded about what can happen with infinity and paradoxical expectations. Even so, in my estimation, we would be dealing with at least 5 regular polytopes in n=∞ if there is a simplex there. Having a hard time seeing how the conventional interpretation of 3 regular polytopes could be possible here. Also, it is very fitting that once simplices reached ∞ hypercells it could max out, which is what occurs in dimension ∞-1

[quickfur]

Since we have different definitions of what ∞ means, I don't think we'll ever come to an agreement.

As I mentioned in one of my earlier posts analysing the coordinates of a facet of the ∞-cross and one of its facets, a facet of the ∞-simplex could in some sense be considered of a distinct dimensionality from the ∞-simplex itself, because no rigid motion can map it back to the parent ∞-simplex (even though the two are, ostensibly, isomorphic and isometric). However, it does project back to the ∞-simplex, with projection as a limit of a particular series of rigid motions. Since projection in finite dimensions decreases the dimension of the projected object, we have the inequalities dim(facet of ∞-simplex) ≤ dim(∞-simplex) = dim(projection of facet of ∞-simplex) ≤ dim(facet of ∞-simplex). So inevitably, dim(facet of ∞-simplex) = dim(∞-simplex). I don't see any paradox here, but I do see a very interesting feature of ∞-dimensional space that's not present in spaces of finite dimension, and that is, that there exist isometric and isomorphic objects that cannot be transformed to each other by rigid motions alone. To me, this is far more interesting to study than some purported infinite-dimensional space that's nothing more than a facsimile of a finite-dimensional space with some labels renamed.

[adam ∞]

Quickfur, that's super interesting. I will have to wrap my mind around it.. I don't fully understand what you are saying here.

I think regardless of our own views on which interpretation of infinity is most valid, we should be able to come to some kind consensus as to whether, for n=infinity, where infinity means a definite, actual infinity, where infinity cannot behave in a way that is impossible for finite numbers, etc, without polytopes that are made up of facets of the same dimension, etc, what we should expect,
versus n= infinity for a indefinite, potential infinity, where infinity can have properties that are impossible for finite numbers, such as n+1 = n and n-1 = n, where facets of polytopes can be made up of the same dimensional polytope or higher dimensional polytopes, etc.. what we should expect.

In the later case, don't you end up with more than 3 regular polytopes? Or are you counting some of them as the same object?

[quickfur]

How can there be any consensus when we can't even agree on the same axioms? That's a logical impossibility.

[adam ∞]

How can there be any consensus when we can't even agree on the same axioms? That's a logical impossibility.

We don't have to agree to one preferred set of axioms to come to a consensus about the different outcomes of different sets of axioms.

For the thing you are talking about, according to your axioms, do you interpret the number of regular polytopes in n=∞ to be 3? Because from what I am getting from it, it seems that there are at least two different types of orthoplexes, and unless you believe that you can have a lone polytope with no corresponding dual, it seems to imply that there is also a second hypercube. Is there more than one orthoplex in what you are proposing, or do they somehow end up being the same shape in a way that I am not seeing? I was trying to understand the interpretation you are presenting, but that part was not clear to me.

[adam ∞]

Quickfur, put another way, in terms of what you are investigating, according to your conception of this, you agree that there are these three polytopes in the infinitieth dimension, right?

1) the simplex {3^n-1}, which is self-dual
2) the hypercube, {4,3^n-2} and its dual
3) the cross polytope {3^n-2,4}

and there is also the cubic honeycomb:

4) the self-dual cubic honeycomb {4,3^n-2,4} ]

Are you coming up with also a simplex honeycomb

5) {3^n-1,4}

Which (from what I am interpretting) is a second orthoplex?

And then from that, if you have a dual of that, should be like a second hypercube

6) {4,3^n-1}


Do you have something like this, or are some of these that I am counting as more than one shape really the same thing in your system?

In other words, you may have {4,3^n-2} = {4,3^n-1}, and {3^n-2,4}= {3^n-1,4} since you are using n = ∞, and ∞ -1 = ∞

so you have 1) the simplex {3^∞}, which is self-dual
2) the hypercube, {4,3∞} and its dual
3) the cross polytope {3^∞,4}

which is also true of the previous dimension, with these same identical shapes, since you have ∞ = ∞-1

is that correct?

Also, do you also have the diagonal of the ∞-cube with edge length = 1 as ∞ ?

When I try to wrap my mind around these, it seems as if the hypercube is in some sense infinitely wide diagonally, and it's dual being a honeycomb might somehow be connected to this. Does the orthoplex fill all of space, or just part of it?


Have you given any thought to what actual renderings of these might look like?

[quickfur]

So far I have not seen any evidence for there to be more than 3 regular polytopes in infinite dimensional space. The difacetal angle of the orthoplex converges to 180°, but whether or not that implies a tiling is not to be regarded as a given, because we're dealing with limits here, and we've already seen signs that things are not quite what they seem. Similarly, whether the simplex tiles space is yet to be proven; the angle converging to 90° is insufficient proof (even in finite dimensions, it takes more than a 90° difacetal angle to tile space; you need to also consider the global properties of such a tiling and whether it can actually exist).

As for the diagonal width of the infinite dimensional cube, it's obvious that it cannot be finite; the length of the vector <1,1,1...> diverges, and that's only half of its diagonal width.

I'm wary of drawing conclusions about infinite dimensional space based on unproven generalizations of finite devices like Wythoff constructions or CD diagrams, because things can happen at the limit that may not have any correspondence with any finite case. Such generalizations need to be proven sound first, before one can reliably draw conclusions from it. And we've already seen signs that things like CD diagrams may not be an adequate representation of objects in infinite dimensional space.

[mr_e_man]

I just realized something even more surprising about Hilbert space: it does not contain anything along the lines of the form k*<±1,±1,...> except the origin. So in the ∞-cross contained in Hilbert space or not? If it is, we're forced to conclude that its facets have no centroids (because its facet centroids must lie along the line k*<±1,±1,...>, and the origin does not satisfy the corresponding hyperplane equations). But if the ∞-cross is not in Hilbert space, that means we have to sacrifice all the "nice" properties of Hilbert space even for polytopes that finite outradius!

The ∞-simplex doesn't contain its centre in its convex hull (or its affine hull). Hilbert space is irrelevant. Each vertex

Code: Select all

<±1,  1/√3,  1/√6,  1/√10,  1/√15, ...>
< 0, -1/√3,  1/√6,  1/√10,  1/√15, ...>
< 0,     0, -2/√6,  1/√10,  1/√15, ...>
< 0,     0,     0, -3/√10,  1/√15, ...>
< 0,     0,     0,      0, -4/√15, ...>
...


has an infinite tail of non-zero coordinates, so any combination of them also has an infinite tail, unlike 0.

It's similar to a hyperbolic ∞-gon, inscribed in a horocycle, whose centre is a point at infinity, which is not really part of the hyperbolic plane.

[quickfur]

I just realized something even more surprising about Hilbert space: it does not contain anything along the lines of the form k*<±1,±1,...> except the origin. So in the ∞-cross contained in Hilbert space or not? If it is, we're forced to conclude that its facets have no centroids (because its facet centroids must lie along the line k*<±1,±1,...>, and the origin does not satisfy the corresponding hyperplane equations). But if the ∞-cross is not in Hilbert space, that means we have to sacrifice all the "nice" properties of Hilbert space even for polytopes that finite outradius!

The ∞-simplex doesn't contain its centre in its convex hull (or its affine hull). Hilbert space is irrelevant. Each vertex

Code: Select all

<±1,  1/√3,  1/√6,  1/√10,  1/√15, ...>
< 0, -1/√3,  1/√6,  1/√10,  1/√15, ...>
< 0,     0, -2/√6,  1/√10,  1/√15, ...>
< 0,     0,     0, -3/√10,  1/√15, ...>
< 0,     0,     0,      0, -4/√15, ...>
...


has an infinite tail of non-zero coordinates, so any combination of them also has an infinite tail, unlike 0.

It's similar to a hyperbolic ∞-gon, inscribed in a horocycle, whose centre is a point at infinity, which is not really part of the hyperbolic plane.

Oops, I made a mistake in the coordinates, the negative coordinates are off by 1 in the numerator. It should be:

Code: Select all

<±1,  1/√3,  1/√6,  1/√10,  1/√15, ...>
< 0, -2/√3,  1/√6,  1/√10,  1/√15, ...>
< 0,     0, -3/√6,  1/√10,  1/√15, ...>
< 0,     0,     0, -4/√10,  1/√15, ...>
< 0,     0,     0,      0, -5/√15, ...>
...


These points are supposed to sum up to 0.

For example, the sum of the first 2 points has 0 in the first coordinate and a second coordinate of 2/√3; add the 3rd point and the 2nd coordinate becomes zero. And so on. So the infinite sum should converge to 0.

[mr_e_man]

Yes, it converges to 0 (depending on the topology), but that's not a (finite) affine combination of the vertices.

The smallest subspace containing all of the vertices does not contain the centre.

[quickfur]

So clearly, we need infinite sums in our definition of ∞-polytopes.

I prefer using hyperplane equations, 'cos then we don't run into this problem.

[quickfur]

[...]
I prefer using hyperplane equations, 'cos then we don't run into this problem.

Hmm, actually, I was wrong. Even if we use only hyperplane equations, we have this strange problem where the normal vector of the hyperplane does not intersect the hyperplane: the normal would be <1,1,1...> and the hyperplane equation would be <1,1,1...>*x = k for some k>0. But none of the points along the line i*<1,1,1...> satisfy this equation, because for i=0 the dot product is zero, but for i≠0 the dot product is infinite, so it cannot be equal to k. So there's a "hole" in the hyperplane where you can pass through the hyperplane from the origin to the other side of the plane without intersecting the plane at all. IOW, we have here an ostensibly convex polytope where where there exists a line that intersects an interior point without intersecting its surface. Very weird!

It sounds contradictory but actually it isn't; it's in some sense analogous to the case where an n-dimensional polytope can have a line that intersects an interior point without intersecting its boundary -- if the line exists in (n+1) dimensions. Here, the normal vector <1,1,1...> has infinite length, so even for very small values of i, we're already infinitely-far away from the polytope. In fact, for all arbitrarily-small value of i>0, we're still infinitely far away from the polytope. Only when i is exactly equal to 0 do we get close -- in fact coincident to its center. So it's almost as if all the other points of the line lie along some new dimension that's orthogonal to all the others, just like in the finite case of the n-dimensional polytope with the (n+1)-dimensional line.