## New insights about infinite-dimensional polytopes

Ideas about how a world with more than three spatial dimensions would work - what laws of physics would be needed, how things would be built, how people would do things and so on.

### New insights about infinite-dimensional polytopes

Hello everyone!

There is something interesting I realized recently about regular polytopes in dimension n= infinity:

If we assume that there are three regular polytopes from dimension 5 and up, we have

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}

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

the dihedral angle or difacetal angle for the simplex reaches 90 degrees at some point..
So we should be able to put four of these together to get a kind of simplex-honeycomb:

5) {3^n-1,4}

...

However, this form also represents a cross polytope!.. so we may, in a sense now have 2 cross polytopes in the same dimension,
which appears to be a contradiction.

This simplex=honeycomb/second cross polytope should also have a dual:

6) {4,3^n-1}

This would give us a second different hypercube in the same dimension!

Notice also that this second hypercube would need to have a vertex figure {3^n-1} which is the same as the simplex in this dimension, another odd contradiction.

So we end up with 2 different cubes, 2 different cross polytopes, a simplex and a simplex honeycomb, all with some seemingly contradictory properties.

I have one proposed alternative to this:

Suppose in the previous dimensions, n= ∞-1 we have the ordinary 3 regular polytopes and 1 cubic honeycomb, as usual

with the property that the simplex, in ∞-1 dimensions is an ∞-cell, {3^n-1}, achieves a dihedral/difacetal angle of 90 degrees in this dimension rather than the next,
reaching this 90 degree limit, no simplex is possible in the next dimension since it has reached its limit.
Keeping in mind that all simplices take the form of an n+1 cell, this would make this simplex the "∞-cell".. it seems natural that this would be the last one,
rather than having one more in the next dimension as an ∞+1 cell... (to me it makes intuitive sense that an "∞-cell" would meet the 90 degree limit)

then, if so,

In n=∞ we may now have:

- NO simplex (!)

1) a hypercube {4,3^n-2}
2) it's dual a hypercross {3^n-2,4} [which, in a sense, is also a simplex-honeycomb]
[and 3) a hypercube honeycomb {4,3^n-2,4} ]

It seems odd to have no simplex,
but in this version,
there is not both a simplex and a simplex honeycomb with a strange dual, there is not 2 different hypercubes in the same dimension, 2 orthoplexes, and there is not a cube with a vertex figure which resides in the same dimension

In either scenario, it seems that having the 3 ordinary polytopes plus one cubic honeycomb may not be logically possible!

My modest conjecture here is that perhaps in n= ∞ there are only 2 regular polytopes and one honeycomb, and the last simplex lives in the second-to-last dimension.
Dionian

Posts: 37
Joined: Wed Aug 18, 2021 2:00 am

### Re: New insights about infinite-dimensional polytopes

The problem with generalizing polytopes to infinite dimensions is that you run into problems of what things like (n-1) or (n-2) mean when n is not a finite quantity. There are ways of rationalizing this, but basically all of them requires sacrificing one or more desirable properties in finite-dimensioned polytopes.

The most straightforward way of generalization would give you an infinite-dimensioned hypercube whose vertex figure is equal to the infinite-dimensional simplex, the same simplex which also serves as facets of the infinite-dimensional cross polytope. The simple-seeming description belies some thorny issues. For example, it is no longer true that the facets of an infinite-dimensional polytope is a polytope of a lower dimension. In fact, the facets of a polytope are of the same (infinite) dimensionality. This has the counterintuitive consequence that the surface area of a polytope has the same units as its bulk (hypervolume). Does that mean that surface polytopes can fill space? I don't know yet. But this is one example of how familiar properties in finite-dimensional space break down in infinite-dimensional space.

Another example is the circumradius of the infinite-dimensional hypercube. It is, unfortunately, not a finite quantity, so things that involve circumscription, for example, become problematic. You cannot (easily) construct a sphere that circumscribes the infinite-dimensional hypercube, for example, because that sphere must have a non-finite radius. Certain angles become immeasurable because the dot product of certain vectors are infinite. So either we have to sacrifice certain closure properties that are taken for granted in finite-dimensional space, or we have to somehow close the arithmetic/geometry by adopting a number system that can manipulate infinite quantities without running into trouble.

The latter quickly leads into a further quagmire. One could, conceivably, adopt an arithmetic system like Conway's surreal numbers in order to allow us to perform calculations involving the hypercube's radius, for example. But that quickly leads to the need to adopt the number system not only in measurements of geometric properties, but also in the coordinate system, because certain geometric operations will bring infinite quantities into the resulting coordinates (e.g., the intersection of the circumsphere of the infinite-dimensional hypercube with the coordinate axes), which, via inversion operations, will require polytopes with infinitesimal (but still distinct!) coordinates. Depending on how you proceed from there, you could end up with entire infinite hierarchies of distinct infinite-dimensional spaces, and you soon land in the quagmire of as-yet unanswered questions such as how to perform integration/differentiation over the surreal numbers. It's a quagmire from which it's virtually impossible to extricate oneself. quickfur
Pentonian

Posts: 2855
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: New insights about infinite-dimensional polytopes

Just to be clear, for what I am proposing in my original post, by "infinity" I mean n = actual infinity, not potential infinity. I don't mean a "never ending" ongoing amount, or something like Hilbert's infinite hotel where the amount changes every step of the way; and not something where you can add to it and subtract and it stays the same amount. Any version of infinity that treats ∞-1 or ∞+1 as the same value as ∞ is not a definite amount, not an exact value- is not an actual infinity, and could not be used as a number of dimensions or number of elements without leading to logical contradictions as far as I can tell. For sure, a number system is needed that can at the very least distinguish between ∞, ∞-1, 2∞, 2^∞, √∞, etc.

Surreal numbers should be able to handle this, for example:

an ω-cube would have 2^ω vertices, 2ω facets which are (ω-1)-cubes, the diagonal would be √ω, the circumradius = (√ω)/2... etc..

But I am not sure whether or not it would all hold together with surreals if it was explored in depth.. if Conway's ω is the same as Cantor's, the transition from very large finite numbers to the first infinite number is not clearly defined and ends up leading to some contradictions.

There are going to be certain calculations that cannot be completed for any large enough number, finite or infinite.. However, there may be ways to do some of the integration and differentiation you mentioned using non-standard analysis, and things like ℓ^2 spaces, but it is not something I am familiar with at all. But even if those are limitations, I don't see why that would stop us from figuring out things like the enumeration of regular polytopes in infinity dimensions, the number of vertices, facets, etc, of regular polytopes, you know?

Infinity and higher dimensions are both subjects that most people consider to be outside of the realm of possible conceptualization by humans. I find inspiration in this forum to try to push further in terms of what we may be able to figure out and understand.
Dionian

Posts: 37
Joined: Wed Aug 18, 2021 2:00 am

### Re: New insights about infinite-dimensional polytopes

The Hilbert hotel is an actual infinity. The only thing is, the arithmetic is weirdly-defined. The thing about arithmetic with infinite quantities is that we cannot carry over all properties of finite arithmetic (if we could, it wouldn't be an infinite quantity anymore!). We can construct systems of infinite arithmetic that preserve some desirable properties, but it will always be at the cost of sacrificing other possibly also desirable properties.

One reason the Hilbert hotel behaves the way it does is because of the difference between cardinals and ordinals. With finite quantities, ordinals and cardinals are identical: the size of a set can also serve as an ordering of the set. You can count elements from 1 to N, and assign a unique index to each element. With infinite quantities, however, cardinals and ordinals diverge. If you retain the same ordering of an infinite set, then N+1 is distinct from N. But if you reorder the set, then you can transform N+1 into N. It doesn't mean that the size of the set changed; it just means that infinite quantities do not behave in the same way as finite quantities. (Again, if they did, they wouldn't be infinite!)

Of course, this leads to a counterintuitive arithmetic that doesn't behave the way we want to. By preserving the quantitative nature of cardinals, we lose the elegance of finite arithmetic. To salvage this, we can turn to a system like surreal arithmetic, that does preserve the distinction between N, N+1, N-1, and even sqrt(N), etc.. However, we should not be misled to think that this somehow changes the meaning of the cardinality of a set. Rather, the nice arithmetic we obtain by adopting the surreal numbers come at the cost of sacrificing the association of numbers with cardinals. That is to say, given a surreal number like sqrt(w), we cannot meaningfully construct a set with sqrt(w) elements, because sqrt(w) is not a cardinality. Some surreals happen to coincide with actual cardinalities, so we can construct a set with w elements, for example. However, it's unclear what it means to construct a set with (w+1) elements, because in terms of cardinality, a set with (w+1) elements is merely a rearrangement of a set with w elements, i.e., they are actually exactly the same size. Similarly, we can define a set with (w-1) elements as a set obtained by removing an element from a set with w elements... but in terms of actual cardinality, they are again exactly the same size. Meaning that if we use surreal numbers to talk about the number of vertices in some infinite-dimensional polytope, we can no longer assume that just because the surreal numbers are different, the actual numbers of vertices are different. They could very well be the same, which can introduce subtle contradictions into our train of thought if we're not careful.

Surreal numbers may, of course, be exactly what we need for certain geometric operations like assigning a quantity to an angle between infinite vectors. But we have to be careful not to confuse surreal arithmetic with cardinal arithmetic, or cardinal arithmetic with ordinal arithmetic. Otherwise we will end up in contradiction, and our whole system will collapse.
quickfur
Pentonian

Posts: 2855
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: New insights about infinite-dimensional polytopes

Quickfur- I think we just need to distinguish between the type of infinite-dimensional polytopes I am proposing here, and what you are proposing. I am trying to approach this in a way that does not lead to a contradiction.

If a notion of infinity is used which is an exact, definite, distinct number, one that doesn't change when you rearrange the elements, an actual, ordinal, infinity, which is one more than the dimension before it, which can have 2∞ facets, which is a different amount from ∞, and 2^∞ vertices, and a diagonal which is the square root of infinity wide, where the square root of infinity is a distinct amount much smaller than ∞, etc, I am suggesting that we might find that dimension ∞-1 might have three regular polytopes, while dimension ∞ might only have 2.

If we allow for paradoxical, self-contradictory polytopes similar to Hilbert's hotel and the Banach Tarski paradox.. and use a definition of cardinal infinity which can be ∞, ∞-1, ∞+1, 2∞, ∞/2, etc all at the same time and can change depending on how we rearrange elements and that sort of thing, you could have anything from objects with impossible, paradoxical effects like having facets that are made up of the object itself, like you have suggested, to a wide variety of "different" but the "same" polytopes with as many or as few elements as you'd like, with numbers of facets that can be rearranged to equal any number, in dimension ∞ and ∞-1 and ∞-2, etc, which can be treat as all the same dimension, or higher, or lower, all the same time.

Personally, I think the notion based on cardinal infinity has some major problems, and that the foundations of accepting such a notion of cardinal infinity, redefining equivalency according to a flawed notion of bijection, improper diagonalization arguments, etc, are foundational errors of mathematics that need to be fixed and moved beyond.

Any source that I have ever come across suggests that from dimension 5 onward, all dimensions should have 3 regular polytopes, and that all dimensions have a simplex following the same pattern. My novel insight here is that maybe this does not hold up for all dimensions since it seems to lead to a contradiction in n=∞. Dionian

Posts: 37
Joined: Wed Aug 18, 2021 2:00 am

### Re: New insights about infinite-dimensional polytopes

If a notion of infinity is used which is an exact, definite, distinct number, one that doesn't change when you rearrange the elements

Actually, that sentence already has a contradiction, albeit one that isn't apparent at first glance.

Suppose I take a point <1,2,3,4,...> from an ∞-dimensional space. I can, ostensibly, construct a lower-dimensional point by dropping one coordinate, obtaining the point <2,3,4,...>. This point ought to belong to (∞-1)-dimensional space, right? Well, not really. Suppose I translate this point by subtracting <1,1,1,1...> from it (<1,1,1,1...> is certainly in ∞-dimensional space, it's one of the vertices of the ∞-dimensional hypercube). What do I get? I get <1,2,3,4...> again. Oops. Meaning that dropping one coordinate from my ∞-dimensional point did not create a lower-dimensional point; it merely moved it somewhere else in the same space!

For a more general case: take any point P from ∞-dimensionsal space, let's represent its coordinates as <x1,x2,x3,x4,...>. Drop any one of them, say x2, to make the vector <x1,x3,x4,...>, ostensibly of a lower dimension. Now construct the vector <0, (x2-x3), (x3-x4), (x4-x5), ...>, which is a perfectly valid vector in ∞-dimensional space. Add this vector to <x1,x3,x4,...>, and we get <x1,x2,x3,x4,...>, which is our original point again. So again, dropping a coordinate from P didn't make a point from a lower-dimensional space; it merely moved it somewhere else in the same ∞-dimensional space!

What about adding more coordinates to make a higher-dimensional point? Let's say we add a coordinate 0 to <1,2,3,4,...> to make the point <0,1,2,3,4,...>, ostensibly of (∞+1)-dimensional space. Now add the vector <1,1,1,...>, our hypercube vertex, to this point, and we get: <1,2,3,4,...>. OK, so adding a new coordinate didn't move the point into a higher dimension; it merely translated it to another place in the same ∞-dimensional space.

Note that this does not mean that the value of ∞ changes; it merely means that adding or subtracting 1 from ∞ does not change its value. Just like multiplying any number by 1 doesn't change its value. No finite number has such a property that adding 1 to it does not change its value; but the above examples prove that ∞ does have this property.

Furthermore, you can extend the above exercise to any (finite) number of coordinates, and there's always a way to get the original point back just by some elementary geometric operations: translations, rotations, reflections. It even works if you drop an infinite number of certain subsets of coordinates, such as every other coordinate of <1,2,3,4,...> to obtain <1,3,5,7,...>. A rotation across a suitably-oriented plane maps that back to <1,2,3,4,...>, meaning that we have merely moved the point, we didn't drop to a lower dimension at all. Only when we drop, say, every coordinate after the 10th coordinate, say, do we end up in an actual, lower dimension (a finite one).

Working with infinity is very tricky business. As I already said, infinite quantities do not behave like finite quantities; expecting them to do so leads to instant contradiction.
quickfur
Pentonian

Posts: 2855
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: New insights about infinite-dimensional polytopes

You are equivocating between different types of infinities.

1+1+1+1...

is not the same as

1+1+1...+1

The first is a potential infinity, it does not have an exact value. It is not a number.

The second represents a specific, exact number with a last "+1" that reaches some kind of limit.

This limit is an exact, specific number. Adding or subtracting 1 results in a different number.

In geometry, we can think of this as being comparable to a sequence of regular polygons with 1 additional side added each time. The angles between two edges get closer and closer to 180°, and an ∞-gon is the last in the sequence, as it reaches 180°. A polygon with fewer than ∞ sides, even just one less, will not have edges that meet at 180°. An ∞-gon (or "apeirogon") has different properties than other regular polygons, since the edges meet at 180°, and every other polygon in the sequence will have different symmetry since odd numbered polygons will lack reflective symmetry. An apeirogon has an even number of sides, an (∞-1)-gon has an odd number of sides, and for example, every 3rd and 5th edge of an (∞-1)-gon could be colored symmetrically while you could not do the same for an ∞-gon. From the apeirogon, with side ∞ next to side 1, meeting at 180°, we learn that there is a modular property to infinite lines, moving to the right will eventually lead to the left side and back to where you started.

The ∞th dimension has certain similar modular and limiting properties, and as I have conjectured, dimension ∞-1 may have a limiting case for the simplex.

Geometry gives us a clear view of the properties of infinity, set theory leads us astray. In set theory, the axiom of infinity states that every finite number has a finite successor, meaning 1 added to any finite number will remain in the closed group of "N", but that there is at least one infinite number that is defined as being "not finite", which is not a useful definition. The amount of finite numbers in set theory is considered an infinite amount, which is a contradiction. There is nothing new or different in set theory that happens, to indicate that adding one should eventually reach some special number that is different from the others, and the nature of any kind of transition between the finite and infinite, if there even is one in set theory, is not clearly defined and different sources give contradictory information on the matter. A limit cardinal is added in axiomatically without evidence, with no explanation of its properties other than it being "not finite". An endless array of higher orders of infinity are represented by symbols, but with no defined properties other than which is larger than another, and an unfounded 2^n operation which cannot be used for much of anything since an exact value of infinity is never defined to begin with.

In geometry, there is a natural limit to a first and last infinity, and from the geometric case, we can deduce, for example, whether infinity is even or odd, its prime factorization and even a representation as a sequence of digits.

(∞ = ...432948736)

I personally believe that working with actual infinity is not all that tricky, besides limitations that apply to all extremely large numbers, finite and infinite. Most people believe that it is impossible to understand infinity and that it is inherently paradoxical, so they don't try to understand it.

Working with flawed concepts like potential infinity and cardinal infinity is tricky because potential, cardinal infinity is not a specific number, and is not even a number at all. It would be just as tricky to work with a collection of finite numbers, defined in a contradictory manner, but to treat them as if they are all one exact number called "finity" that has special properties that we should not count as being impossible in this special case even though these properties are considered impossible for all actual numbers.

The argument you present is similar to the foundations of set theory- you claim to be talking about something that is actually infinite, but you only show the very beginning of it. You don't show the end. Then you do something that modifies the whole thing, and again, you don't show the end, and claim by showing the beginning that they are the same thing. This argument would not work for even a very low finite numbered dimension.

Suppose I take a point <1,2,3,4,...> from an ∞-dimensional space. I can, ostensibly, construct a lower-dimensional point by dropping one coordinate, obtaining the point <2,3,4,...>. This point ought to belong to (∞-1)-dimensional space, right? Well, not really. Suppose I translate this point by subtracting <1,1,1,1...> from it (<1,1,1,1...> is certainly in ∞-dimensional space, it's one of the vertices of the ∞-dimensional hypercube). What do I get? I get <1,2,3,4...> again

take the point <1,2,3,4... ∞-3,∞-2,∞-1,∞>, with a total of ∞ terms.

drop one coordinate to get: <2,3,4,...-3,∞-2,∞-1,∞>, with a total of ∞-1 terms.

A) Translate this by subtracting <1,1,1,1...1,1,1,1> with a total of ∞ terms, what do you get?

B) Or, translate it instead by subtracting <1,1,1,1...1,1,1> with a total of ∞-1 terms, what do you get?

You are getting paradoxical results because you are using potential infinity without an end, instead of actual infinity with a defined end and a distinction between n and n-1.
Dionian

Posts: 37
Joined: Wed Aug 18, 2021 2:00 am

### Re: New insights about infinite-dimensional polytopes

adam ∞ wrote:You are equivocating between different types of infinities.

1+1+1+1...

is not the same as

1+1+1...+1

Actually, it is exactly the same, you just rearranged the terms a little. Consider this: if you take a point <1,2> in the 2D plane, you can construct a mirror on the line x=y which will reflect <1,2> to <2,1>. If you take a point <1,2,3> in 3D, you can construct a mirror on the plane x=y which will reflect <1,2,3> to <2,1,3>. You can also construct a mirror on the plane y=z to get the point <1,3,2>. Or a mirror on the plane x=z to get the point <3,2,1>. By applying several reflections in a row, you can get any rearrangements of coordinate of <1,2,3>. In higher dimensions it is the same: given any point <1,2,3,...n>, you can obtain any permutation of coordinate by a suitable arrangement of diagonal mirrors.

So one would expect that in finite-dimensional space, you can also construct a series of mirrors (possibly an infinite number of them) to obtain any rearrangement of coordinates for any given point. So 1+1+1+... under a suitable rearrangement is identical to 1+1+1+...+1.

[...]
Geometry gives us a clear view of the properties of infinity, set theory leads us astray. In set theory, the axiom of infinity states that every finite number has a finite successor, meaning 1 added to any finite number will remain in the closed group of "N", but that there is at least one infinite number that is defined as being "not finite", which is not a useful definition.

You appear to be confused because of reading unreliable sources on the subject. There is no mathemetics that defines an infinite number as being "not finite", which is indeed a useless definition if that's how it's defined!

Rather, the set of numbers is built up in a way that's entirely self-consistent. We start with the empty set {}, which we call 0 (because, well, it contains nothing). Then we define 1 to be the set that contains 0. Or, in set notation, 1 = { 0 }. This set contains exactly 1 element, so it sorta makes sense to name it 1. So now we have at least 2 different sets: {} and {0}. We can thus construct a 3rd set by putting these two things in it: {{}, {0}}. Or, if we use our numerical names for these elements, {0, 1}. This set has exactly 2 elements, so we can call it 2. Now we have 3 different sets: {}, {0}, and {0,1}, or, using our numerical names, 0, 1, 2. We can thus put all of these into yet another set {0, 1, 2}. This set has 3 elements, which we can name 3. Which gives us 4 different sets that we can construct new sets with. We can keep going with such a construction, and each time we obtain a number 1 larger than the one before. Notice also that each set we construct contains everything that came before it, and is strictly larger than everything that came before it.

Of course, we will never reach infinity by repeating this process, because such a process will never terminate. Instead, we construct a new set in a somewhat different way: we take all finite numbers as stuff them all into a new set N. That is, we define this set to include every set we can possibly obtain by repeating the above process. Such a set is guaranteed to contain every finite number there is, and is therefore larger than any finite number (since it contains all of them). Since this is the smallest set with such a property (if you removed anything from it, there would be at least 1 finite number it does not contain, so we can no longer say this set is larger than that number), we can consider this set to be the first infinite number.

It doesn't stop there, of course. We can keep going by constructing the set N ∪ {N}. I.e., this set contains all the finite numbers plus N itself. This new set can therefore be called N+1, since it obviously contains N and everything before it, so it is larger than N. And so on.

Continuing in the same way gives us the ordinal numbers. It is actually extremely useful, because it lets us perform computations with infinite sets in a way that we couldn't have done in a consistent way before. By defining the "less than" relation in terms of set containment, we leave no room for doubt that we're performing any sleight of hand in the definition of "<". If A is a member of B, then A < B. Otherwise, it's not. It's that simple.

The amount of finite numbers in set theory is considered an infinite amount, which is a contradiction.

How is that a contradiction? Are you saying that the set of finite numbers is finite? Then there would be a contradiction, because then there would be at least one finite number, that is the size of the set itself, that isn't in the set, so the set does not contain all finite numbers after all.

There is nothing new or different in set theory that happens, to indicate that adding one should eventually reach some special number that is different from the others, and the nature of any kind of transition between the finite and infinite, if there even is one in set theory, is not clearly defined and different sources give contradictory information on the matter.

You cannot add one to some finite number to get an infinite number, no matter how many times you do it. That's obvious, because given any finite number X, adding 1 merely gives you another finite number. To say otherwise is the real contradiction, because if X is finite and X+1 is infinite, then it means that you can reduce infinity to 0 in a finite number of steps (namely, X+1). So it's not infinite after all.

A limit cardinal is added in axiomatically without evidence, with no explanation of its properties other than it being "not finite". An endless array of higher orders of infinity are represented by symbols, but with no defined properties other than which is larger than another, and an unfounded 2^n operation which cannot be used for much of anything since an exact value of infinity is never defined to begin with.

From what you're saying, it appears that you have been reading informal sources on the subject, and getting frustrated because it doesn't make sense. And I'd agree with you, because informal descriptions do not do justice to what is actually defined in set theory. You need to work with the actual, formal definitions in order for it to make any sense. Don't rely on hand-waving descriptions that often are inaccurate or have subtle errors in them.

[...]

I'll reply to the rest of your post later, since I'm out of time and have to go.
quickfur
Pentonian

Posts: 2855
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: New insights about infinite-dimensional polytopes

Actually, it is exactly the same, you just rearranged the terms a little.

Well, there is a distinction between "1+1+1+1..." and "1+1+1...+1", they mean something different. It does not indicate a rearrangement in any way. It is used as a way to indicate two different types of series, one potentially infinite and one actually infinite.

"1+1+1+1..." = potential infinity
"we will never reach infinity by repeating this process, because such a process will never terminate"
This is the definition of potential infinity. It is synonymous with "such a process will never terminate". That is what potential infinity means. When a distinction between the two is made, the "..." without a 1 at the end is used to make the distinction.

"1+1+1...+1" = actual infinity. Such a process that does terminate. The totality of this process with a last successor. The "..." does not continue, it comes to an end. The previous 1's lead to the last 1 with no discontinuities.

You appear to be confused because of reading unreliable sources on the subject. There is no mathematics that defines an infinite number as being "not finite"

How did Zermello and Fraenkel define "I" when introducing the axiom of infinity?

Have you ever asked a set theorist to define infinity in a way that is not just a negation of being finite? This is what they tell you. This is what the symbols mean. They define "counting" using a bijection function, which is their new definition of "equals" and they say if a set does not equal a finite set, it is an infinite set, and that that is what an infinite set is defined as. As simply being a set that is not finite.

-

Starting with an "empty set", which is meaningless, defined as "nothing", which is meaningless, and having "sets" of "nothing", and "sets" of "sets" of "nothing" is meaningless.

Of course, we will never reach infinity by repeating this process, because such a process will never terminate. Instead, we construct a new set in a somewhat different way: we take all finite numbers as stuff them all into a new set N. That is, we define this set to include every set we can possibly obtain by repeating the above process.

Why do you presume repeating the process will never reach infinity? Why do you assume the process will never terminate? Geometry informs us otherwise. And you have a major contradiction here: if the process never terminates, meaning it is only potentially infinite, there is no such thing as "all" of them to "stuff" into a new set "N". If the process does not terminate, there is no set of all of them. Logically, it is impossible to have a set that includes "every" set obtained in the process above. There is no such thing as "every" one of something that is not an exact, definite completed amount.

Such a set is guaranteed to contain every finite number there is, and is therefore larger than any finite number (since it contains all of them).

No, the definition is illogical and self-contradictory. There is no such thing as "every finite number" if you have defined finite numbers to be a potentially infinite, non terminating process without a last term! I don't mean to sound like a broken record, but I don't think you comprehend what is meant by "potential infinity" and "actual infinity".

Since this is the smallest set with such a property (if you removed anything from it, there would be at least 1 finite number it does not contain, so we can no longer say this set is larger than that number), we can consider this set to be the first infinite number.

There can be no smallest set collecting "all" of the numbers in a non-terminating process. There is no such thing as "all" of them. "All" only applies to actual, definite amounts. There can be no smallest or largest set with this property- it is not an amount.

A set of finite numbers cannot be infinite. That is self-contradictory. Any number in the set of finite number must be a finite number, and any highest finite number will be the total amount of finite numbers. If you claim some infinite number is in the set of finite numbers, it is very clearly a contradiction. How can you say that if you removed the first infinite number from the set, there would be one finite number the set didn't contain?

How is that a contradiction? Are you saying that the set of finite numbers is finite?

Take any finite number.
The amount of finite numbers up to that point is exactly the same as that number.
The amount of numbers up to some number is the number itself.
The nth finite number is n.
If it is claimed that the nth finite number is infinite, this cannot be true-
it cannot be a finite number because infinite numbers are not finite.
If the amount of numbers up to some number is the number itself,
there cannot be an infinite amount of finite numbers, because infinity cannot equal the number itself if it is not finite.

If you believe there can be infinite numbers, there has to be some kind of transition between the finite and the infinite.

If you don't believe that adding 1 over and over again can ever reach infinity, how can you claim that there are an infinite amount of finite numbers in the "set" "N", relying on the axiom of infinity which uses the successor function of continually adding 1 to finite numbers to produce the set "N"?

You cannot add one to some finite number to get an infinite number, no matter how many times you do it. That's obvious, because given any finite number X, adding 1 merely gives you another finite number. To say otherwise is the real contradiction, because if X is finite and X+1 is infinite, then it means that you can reduce infinity to 0 in a finite number of steps (namely, X+1). So it's not infinite after all.

This is not what I am proposing, but a last finite number that, when you add 1 to it gives the first infinite number is much closer to what I think is actually happening than having no definition of the transition between the finite and the infinite, and claiming that it is logically possible to have an infinite amount of finite numbers.

I don't understand why you think that by definition there can be no finite number that can be added to to equal infinity. Where does this idea come from? Who defined infinity in this way? I see it repeated often and taken for granted by a lot of people, but I don't understand where they get this idea. If X is finite, and X+1 is infinite, you just now said that X+1 is infinite, so no, you cannot reduce it to 0 in a finite amount of steps. You have defined the amount of steps, X+1, as being infinite.

I am not just looking at informal sources. Set theory has a million problems, and virtually everything about it is self contradictory. These paradoxes and contradictions were so bad that they lead to Cantor killing himself! I think a lot of Cantor's ideas about higher orders of infinity are really cool and interesting, but I really do not like the reasoning used in set theory. No offense toward you!   Dionian

Posts: 37
Joined: Wed Aug 18, 2021 2:00 am

### Re: New insights about infinite-dimensional polytopes

This is why the infinite dimensional Hilbert space once was introduced, consisting only of those vectors with finitely many non-zero components.
Then no contradictions would occur and vector geometry works out well.
Eg. adding vectors, scalar multiplication, etc. And you don't even need the choice axiom.

Within this space an infinite orthoplex well can be defined, as its vertex coordinates (up to a global scalling) could be taken to be just those which are everywhere zero and at a single position +/-1.
Likewise thie infinite dimensional simplex can be defined, as a simplex always is a facet of the orthoplex, ie. its coordinates could be taken to be everywhere zero and at a single position +1.
But the infinite dimensional hypercube cannot be defined within that Hilbert space! As it would require coordinates which are non-zero throughout.

--- rk
Klitzing
Pentonian

Posts: 1621
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: New insights about infinite-dimensional polytopes

The thing about infinite-dimensional polytopes, is that even something like {4,6} will expand faster than it.

Something like Hilbert's polytope, or the cube, would never be constructable, because of the diameter. The tegum-products can be constructed, but the bulk of these polytopes would lie between vertices. Note also that nearly all of the volume of a sphere would be on the diameter plane.

Even 124D (where there is supposed to be a magical lattice in the line of the Leech lattice), is a tad hard to play with.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger wendy
Pentonian

Posts: 1996
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: New insights about infinite-dimensional polytopes

Klitzing wrote:This is why the infinite dimensional Hilbert space once was introduced, consisting only of those vectors with finitely many non-zero components.

--- rk

No, it's vectors with a well-defined magnitude. That is, vectors of finite length.
PatrickPowers
Tetronian

Posts: 379
Joined: Wed Dec 02, 2015 1:36 am

### Re: New insights about infinite-dimensional polytopes

This got me thinking. In geometric algebra the basis of a Hilbert space would be uncountable.
PatrickPowers
Tetronian

Posts: 379
Joined: Wed Dec 02, 2015 1:36 am

### Re: New insights about infinite-dimensional polytopes

Actually, it is exactly the same, you just rearranged the terms a little.

Well, there is a distinction between "1+1+1+1..." and "1+1+1...+1", they mean something different. It does not indicate a rearrangement in any way. It is used as a way to indicate two different types of series, one potentially infinite and one actually infinite.

"1+1+1+1..." = potential infinity
"we will never reach infinity by repeating this process, because such a process will never terminate"
This is the definition of potential infinity. It is synonymous with "such a process will never terminate". That is what potential infinity means. When a distinction between the two is made, the "..." without a 1 at the end is used to make the distinction.

"1+1+1...+1" = actual infinity. Such a process that does terminate. The totality of this process with a last successor. The "..." does not continue, it comes to an end. The previous 1's lead to the last 1 with no discontinuities.

Actually, there is a discontinuity, right there in the "...". To see this, consider: where in the sequence 1+1+1+...+1 does the sum, which is finite in the beginning, become infinite? It can't be one term before the end, because if it was, that means ∞ = 1 + some large finite number, which is a contradiction. It can't be two terms before the end either, for the same reason. Or 3 terms before, or 4, or 5, or 5, or ....

So where does the leap from finite to infinite take place? In the middle of the sequence? What's the value of the sum at the exact middle of the sequence? There are two possibilities: either it's some fixed finite number X, or it's some infinite quantity like (a hypothetical) ∞/2. The first choice immediately leads to a contradiction, because it implies that X+X = ∞, which means that ∞ is a finite number (namely, the sum of two finite numbers X and X).

So only the second choice is viable. But consider: what exactly is the value of ∞/2? Since we concluded it cannot be finite, it must therefore be infinite. But that merely leaves us with the question: where in the sequence from 0 to ∞/2 did we get from finite to infinite? Again there are two possibilities: it's either some finite number Y that lies between 0 and ∞/2, or it must be another infinite number, which we may call ∞/4 (assuming it's halfway between 0 and ∞/2). Again, by the above reasoning, there cannot be any such Y, because if there were, we'd have the contradiction that ∞/2 is a finite number, because Y is finite, and therefore Y+Y must also be finite.

So again that leaves us with the only possibility that ∞/4 must also be infinite. If we repeat this line of reasoning, we end up with a sequence ∞, ∞/2, ∞/4, ∞/8, .... How long does this sequence go for? Does it stop after some number Z of iterations? Or does it continue forever? Is Z finite or infinite? If Z is finite, we have the contradiction that 2Z = ∞, meaning that ∞ must be finite. So Z can only be infinite. But what's the value of Z? Is Z = ∞? If so, we have the contradiction that 2 = ∞. So obviously, Z < ∞ (otherwise we get an even worse contradiction that raising 2 to the power of Z produces a number that's less than Z).

But now we have another problem: since 0 < Z < ∞, that means Z must occur somewhere in your sequence from 0 to ∞. Which means we're nowhere close to the answer: where in the sequence did we leap from finite to infinite? We can repeat the above argument to conclude that there must be another infinite number, smaller than Z but larger than all finite numbers, which lies between 0 and Z. If we continue in this way, we get the relation 0 < ... < Zn < Zn-1 < ... < Z2 < Z1 where Zi is each infinite number that we construct in the above manner. Again, one may ask, how long is this chain of <'s? Does it ever end? If it does, then there must be a final Zfinal that's the smallest of all infinite numbers, and larger than all finite numbers. Then one may ask, what's the number that comes just before Zfinal in the original sequence? It cannot be finite -- because we get a contradiction that Zfinal must also be finite. But neither can it be infinite: we get a contradiction that Zfinal isn't the smallest infinite number after all.

So the only conclusion is, that sequence never ends. Which means that there's a discontinuity between finite numbers and infinite numbers: no matter how far back you go from the end of the sequence, you can never reach the finite numbers; and no matter how far forward you from the beginning, you can never reach the infinite numbers. The sequence 1+1+1+...+1 has an unremovable discontinuity hidden behind the "...". Contrary to what you claimed, it does not tell us how the leap from finite to infinite took place. No matter where you try to look for the transition, it cannot be found.

So at the end of the day, you have not eliminated the discontinuity in Cantor's transfinite number system at all; you've merely rearranged it to a different place (in the middle of the sequence rather than the end), but the discontinuity between finite and infinite remains.

[...]
Have you ever asked a set theorist to define infinity in a way that is not just a negation of being finite? This is what they tell you. This is what the symbols mean. They define "counting" using a bijection function, which is their new definition of "equals" and they say if a set does not equal a finite set, it is an infinite set, and that that is what an infinite set is defined as. As simply being a set that is not finite.

So what's your definition of infinity?

[...]
Why do you presume repeating the process will never reach infinity? Why do you assume the process will never terminate? Geometry informs us otherwise. And you have a major contradiction here: if the process never terminates, meaning it is only potentially infinite, there is no such thing as "all" of them to "stuff" into a new set "N". If the process does not terminate, there is no set of all of them. Logically, it is impossible to have a set that includes "every" set obtained in the process above. There is no such thing as "every" one of something that is not an exact, definite completed amount.

And why do you assume that your sequence 1+1+...+1 terminates? As I proved above, hidden inside the "..." is an unremovable discontinuity. No matter how we tried to find the transition between finite numbers and the smallest of infinite numbers, we can't find it without running into a contradiction. Meaning that if you start from the finite end of your sequence, you never get past the "..." without making a leap, and if you start from the infinite end of your sequence, you can never get back to the finite side without making a leap.

If, as you claim, geometry proves that the process must end somewhere, then I'd say, show me where the transition from finite to infinite takes place. What's the exact number X in your 1+1+...+1 sequence where the sum stops being finite and becomes infinite? Is X finite or infinite? If you say it's finite, then it's an instant contradiction: X+1 cannot be infinite (since subtracting a finite number from (X+1), namely X, makes it 1, a finite number). If you say it's infinite, it's also a contradiction: (X-1) is a finite number (otherwise X is not the number where the finite/infinite transition takes place), so X must also be a finite number. The only possible answer is, there is no such number. X doesn't exist.

Therefore, your sequence actually doesn't terminate. It appears to terminate because of the last terms, but actually it doesn't because there's a discontinuity in the middle, hidden behind the "...".

Such a set is guaranteed to contain every finite number there is, and is therefore larger than any finite number (since it contains all of them).

No, the definition is illogical and self-contradictory. There is no such thing as "every finite number" if you have defined finite numbers to be a potentially infinite, non terminating process without a last term! I don't mean to sound like a broken record, but I don't think you comprehend what is meant by "potential infinity" and "actual infinity".

OK, so please tell me, what's the last term in the set of finite numbers? What's its value?

It's certainly not ∞, since that's not a finite number. So is it ∞-1? Or perhaps ∞-2? Or maybe ∞/2? What are the digits of the last finite number? How many digits are there? There must be a finite number of digits, since otherwise it's not a finite number. But if it has a fixed number of digits, then neither can it be the last finite number, because you can, for example, add 1 to it to make a larger finite number. Or, for that matter, multiply it by 2. Or multiply it by itself. Etc.. Which is a contradiction.

[...]
A set of finite numbers cannot be infinite. That is self-contradictory. Any number in the set of finite number must be a finite number, and any highest finite number will be the total amount of finite numbers.

So you think that the set of finite numbers must be finite? That instantly leads to a contradiction: what's the (finite) size of the set of finite numbers? What's the value of this size? What are its digits? There must be a finite number of digits, after all, otherwise it would not be a finite number. And it must have a fixed, finite value. So what is this value?

No matter what the value is, let's denote it by X, I can construct a larger finite number from it, say X+1. Or 2*X, or whatever. Then that's a contradiction: (X+1) isn't in your set of finite numbers, so your set does not contain all finite numbers.

If you claim some infinite number is in the set of finite numbers, it is very clearly a contradiction. How can you say that if you removed the first infinite number from the set, there would be one finite number the set didn't contain?

Hold it right there. I never said an infinite number is in the set of finite numbers. That would indeed be a contradiction. And I never said that if you removed the first infinite number from the set, there would be one finite number the set didn't contain. The set of finite numbers, by definition, does not contain any infinite elements. You're chasing a red herring here.

That doesn't mean the size of that set is finite, though. The set of finite numbers cannot be finite, because if it were, there would be a largest finite number X. Which is a contradiction, since X+1 would also be finite, which means X isn't the largest finite number.

How is that a contradiction? Are you saying that the set of finite numbers is finite?

Take any finite number.
The amount of finite numbers up to that point is exactly the same as that number.
The amount of numbers up to some number is the number itself.
The nth finite number is n.
If it is claimed that the nth finite number is infinite, this cannot be true-
it cannot be a finite number because infinite numbers are not finite.
If the amount of numbers up to some number is the number itself,
there cannot be an infinite amount of finite numbers, because infinity cannot equal the number itself if it is not finite.

Very good. So tell me, what's the value of the largest finite number?

And how do you get around the contradiction that, whatever value X this largest finite number must have, X+1 is a larger finite number, which means that X is not the largest finite number?

If you believe there can be infinite numbers, there has to be some kind of transition between the finite and the infinite.

Very good. So tell me, where's the transition between the finite and the infinite in your sequence 1+1+...+1? What's the sum of the sequence at that point? Is it finite or infinite? What's its value? What are the digits?

If you don't believe that adding 1 over and over again can ever reach infinity, how can you claim that there are an infinite amount of finite numbers in the "set" "N", relying on the axiom of infinity which uses the successor function of continually adding 1 to finite numbers to produce the set "N"?

Hold it right there.

The axiom of infinity does NOT use the successor function to produce the entire set. That would indeed be a contradiction. The axiom of infinity defines the set N to contain every number you can obtain by applying the successor function. Please pay attention to the very fine but extremely crucial distinction here. It does NOT, in itself, apply the successor function. It simply posits that there exists a set N such that no matter you many times you apply the successor function, the number you obtain would be included in that set. It does NOT create the set N by applying the successor function some undefined number of times, because N cannot be created in this way. This is why the axiom of infinity is an axiom: because the only way you can obtain a completed set N is to define its existence via an axiom. N is not constructible by applying the successor function, no matter how many times. In other words, N cannot be obtained by applying the other axioms. That's why we need an axiom of infinity. If we could obtain N any other way, there would have been no need for such an axiom.

The only way to obtain N is to accept the axiom of infinity as fact, i.e., the existence of N cannot be proven, it must be assumed. If you reject the axiom of infinity, then there is nothing else to discuss: your mathematics would not contain any infinity (except potential ones), and this whole discussion is moot.

You cannot add one to some finite number to get an infinite number, no matter how many times you do it. That's obvious, because given any finite number X, adding 1 merely gives you another finite number. To say otherwise is the real contradiction, because if X is finite and X+1 is infinite, then it means that you can reduce infinity to 0 in a finite number of steps (namely, X+1). So it's not infinite after all.

This is not what I am proposing, but a last finite number that, when you add 1 to it gives the first infinite number is much closer to what I think is actually happening than having no definition of the transition between the finite and the infinite, and claiming that it is logically possible to have an infinite amount of finite numbers.

The problem is that there is no such last finite number. Because if there is, you get an instant contradiction that adding 1 to a finite number produces an infinite number. The most you can get with this approach is an extremely large finite number that behaves somewhat like an infinite number, but isn't actually infinite. See, for example, Graham's number, as a candidate of such a number. A lot of interesting things can be obtained by using such extremely large finite numbers, but please don't mistake that for being the same thing as an infinite number.
quickfur
Pentonian

Posts: 2855
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: New insights about infinite-dimensional polytopes

PatrickPowers wrote:
Klitzing wrote:This is why the infinite dimensional Hilbert space once was introduced, consisting only of those vectors with finitely many non-zero components.

--- rk

No, it's vectors with a well-defined magnitude. That is, vectors of finite length.

That comes out to be the same: a vector can have finitely many non-zero components only if it is required to have a finite length.
--- rk
Klitzing
Pentonian

Posts: 1621
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: New insights about infinite-dimensional polytopes

Hi Quickfur,

OK, so please tell me, what's the last term in the set of finite numbers? What's its value?

It takes a while to explain what I mean by all of this, but the number system I use has a distinction between finite numbers, infinite numbers, and something in between I call "quasi-infinite" numbers. By "infinite", I mean a proper, actual infinity. These three types of numbers have different properties, comparable to topological properties like flatness, positive curvature, openness and closedness. I can explain in more detail at some point, but briefly, one can think of finite numbers as being flat and open, quasi-infinite numbers as being curved and open, and the first infinite number as being the first that is closed. Any infinite number beyond that is a higher dimensional number, maxing out at the last infinite number in dimension infinity. That is a slight simplification, it is a little more subtle than that when you get into the details of it.

( Some of the terminology is just a matter of convention since people started defining the meaning of "finite" and "infinite" before it was known how finite and infinite numbers and forms actually work. "Infinity" has been a misnomer since at leas the time of Aristotle when he made a distinction between potential and actual infinity. ) In this system, then, the last finite number is the square root of infinity. On a zoomed out circular number line, you cannot see a distinction between the square root of infinity and any of the previous numbers up to that point. The square root of infinity plus one is the first "quasi-infinite" number. The last quasi-infinite number is infinity minus one. I have some of these numbers calculated more than others. The last digits of infinity minus one are ...4342948735. The last digits of the square root of infinity, from memory, are ...856. The last digits of the square root of infinity plus one are ...857.

I believe Conway has the square root of infinity as a transition between the finite and infinite in his system, but I can't find the citation I thought I had right here for it.

I started a new topic in the "General" section to discuss this further, since we have been getting further from the subject of higher dimensional polytopes.

viewtopic.php?f=11&t=2576#p28141

See you there! Dionian

Posts: 37
Joined: Wed Aug 18, 2021 2:00 am

### Re: New insights about infinite-dimensional polytopes

Klitzing wrote:This is why the infinite dimensional Hilbert space once was introduced, consisting only of those vectors with finitely many non-zero components.
Then no contradictions would occur and vector geometry works out well.
Eg. adding vectors, scalar multiplication, etc. And you don't even need the choice axiom.

Within this space an infinite orthoplex well can be defined, as its vertex coordinates (up to a global scalling) could be taken to be just those which are everywhere zero and at a single position +/-1.
Likewise thie infinite dimensional simplex can be defined, as a simplex always is a facet of the orthoplex, ie. its coordinates could be taken to be everywhere zero and at a single position +1.
But the infinite dimensional hypercube cannot be defined within that Hilbert space! As it would require coordinates which are non-zero throughout.
[...]

Indeed! The Hilbert space has many "gaps": points that do not belong to it because they have a non-finite number of non-zero components. In fact, it can be shown that the Hilbert space actually cannot contain any full-dimensioned polytopes, because it does not contain an infinite-dimensional sphere of non-zero radius. Consider w.l.o.g. a sphere of radius 1. Its surface contains points where the root-square-sum of its coordinates converge to 1. There are many infinite series of non-zero terms whose rss converge to 1, but these points cannot belong to the Hilbert space since they have an infinite number of non-zero terms. (For instance, take the series 1/2 + 1/4 + 1/8 + ..., which converges to 1. Take the square root of each term as a coordinate, and we obtain the point <1/√2, 1/2, 1/√8, 1/4, 1/√32, ...> which lies on this sphere, but is not in the Hilbert space.) Therefore, the Hilbert space does not contain an infinite-dimensional sphere of non-zero radius. Consequently, it also cannot contain any polytopes with a non-zero in-radius.

I didn't prove it rigorously but I suspect the orthoplex actually isn't entirely contained in the Hilbert space. Its vertices happen to be, but I'm almost certain some interior points would not be. The fact that its dual, the hypercube, isn't in Hilbert space is very telling.

For this reason, I consider the Hilbert space to be defective as far as polytope geometry is concerned. You can only deal with what's essentially finite-dimensional subspaces of it at a time, but you cannot manipulate full-dimensioned objects in the ambient infinite-dimensional space.

So far, the only way around this that I can see is to use coefficient field that contains non-finite elements (because the computation of the out-radius of the hypercube, for example, results in an infinite magnitude). But once we admit infinite out-radii, we must also admit infinite-valued coordinates -- because, for example, one might conceivably rotate the hypercube such that two opposite vertices lie along a coordinate axis: it follows that their coordinates must contain non-finite values. So if we want closure under typical geometric operations, we must admit vectors with non-finite coordinates. In order for vector arithmetic to proceed without problems in this situation, the only conceivable system I can think of is Conway's surreal numbers (or a suitable closed subset thereof).

Non-trivial computations under such a system would be very cumbersome, though: not only you have to work with potentially-diverging series, you also have to deal with convergence issues of series of possibly-non-finite surreal numbers. Not a task for the faint-hearted!  quickfur
Pentonian

Posts: 2855
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: New insights about infinite-dimensional polytopes

Klitzing wrote:
PatrickPowers wrote:
Klitzing wrote:This is why the infinite dimensional Hilbert space once was introduced, consisting only of those vectors with finitely many non-zero components.

--- rk

No, it's vectors with a well-defined magnitude. That is, vectors of finite length.

That comes out to be the same: a vector can have finitely many non-zero components only if it is required to have a finite length.
--- rk

Consider the infinite series 1, 1/2, 1/4, 1/8.... The sum converges to 2.
PatrickPowers
Tetronian

Posts: 379
Joined: Wed Dec 02, 2015 1:36 am

### Re: New insights about infinite-dimensional polytopes

PatrickPowers wrote:
Consider the infinite series 1, 1/2, 1/4, 1/8.... The sum converges to 2.

If a vector has an infinite number of dimensions with magnitudes 1, 1/2, 1/4, 1/8.... then its length is the square root of 4/3.
PatrickPowers
Tetronian

Posts: 379
Joined: Wed Dec 02, 2015 1:36 am

### Re: New insights about infinite-dimensional polytopes

Klitzing wrote:This is why the infinite dimensional Hilbert space once was introduced, consisting only of those vectors with finitely many non-zero components.
Then no contradictions would occur and vector geometry works out well.
Eg. adding vectors, scalar multiplication, etc. And you don't even need the choice axiom.

Within this space an infinite orthoplex well can be defined, as its vertex coordinates (up to a global scalling) could be taken to be just those which are everywhere zero and at a single position +/-1.
Likewise thie infinite dimensional simplex can be defined, as a simplex always is a facet of the orthoplex, ie. its coordinates could be taken to be everywhere zero and at a single position +1.
But the infinite dimensional hypercube cannot be defined within that Hilbert space! As it would require coordinates which are non-zero throughout.

--- rk

PatrickPowers wrote:No, it's vectors with a well-defined magnitude. That is, vectors of finite length.

Right. The space of vectors with finitely many non-zero coordinates is only a subspace of the Hilbert space. But the cube, with vertices at (±1,±1,±1,...), is in neither of these spaces; those vectors don't have a finite magnitude, as 1²+1²+1²+... diverges.

There is an infinite-dimensional box shape in Hilbert space, with vertices (±1, ±1/2, ±1/3, ±1/4, ±1/5, ...), and a different box with vertices (±1, ±1/2, ±1/4, ±1/8, ...), and of course many other such boxes. Those vectors do have a finite magnitude: in the first case it's π/√6, and in the second case it's 2/√3 as Patrick noted. https://en.wikipedia.org/wiki/Hilbert_cube
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
Tetronian

Posts: 404
Joined: Tue Sep 18, 2018 4:10 am

### Re: New insights about infinite-dimensional polytopes

Haha, I made a fool of myself by assuming that vectors in the Hilbert space must have finite non-zero elements. It actually does include vectors with infinite non-zero coordinates, as long as the norm of the vector converges. So it would include infinite-dimensional spheres of non-zero radius. However, only the "more pointy" members of the hypercube family would be included, i.e., the cross polytope and some of its truncations. However, the "fatter" members of the hypercube family have to be excluded because they contain vectors whose norms diverge. The cube itself diverges, for example, as well as its larger truncates.

Interesting side-question: which members of the uniform hypercube family belong to Hilbert space, and which members don't? Clearly, any members with CD symbol that starts with x4... would diverge, because the marked node with the degree 4 edge would generate a vector whose norm is at least that of the hypercube's out-radius, which diverges. But what about the other CD nodes? Can anything meaningful be said about which combinations of marked nodes would have convergent coordinates, and which would not?
quickfur
Pentonian

Posts: 2855
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: New insights about infinite-dimensional polytopes

For the symmetry group o4o3o3o3o3o..., the mirrors have normal vectors

f₁ = (1, 0, 0, 0, 0, 0, ...),
f₂ = (-1/√2, 1/√2, 0, 0, 0, 0, ...),
f₃ = (0, -1/√2, 1/√2, 0, 0, 0, ...),
f₄ = (0, 0, -1/√2, 1/√2, 0, 0, ...),
f₅ = (0, 0, 0, -1/√2, 1/√2, 0, ...),
:
:

The notation a4b3c3d3... means that the group is applied to some point v = (v₁, v₂, v₃, v₄, ...) to generate the vertices of a polytope. The coordinates in the skewed system are related to the standard orthogonal coordinates by

a = f₁•v = v₁,
b = f₂•v = (v₂ - v₁)/√2,
c = f₃•v = (v₃ - v₂)/√2,
d = f₄•v = (v₄ - v₃)/√2,
:
:

Solving these equations for the latter, we get

v₁ = a,
v₂ = a + b√2,
v₃ = a + b√2 + c√2,
v₄ = a + b√2 + c√2 + d√2,
:
:

For convex polytopes we have the skewed coordinates all non-negative: a,b,c,d,...≥0 . This implies that the orthogonal coordinates are increasing: 0 ≤ v₁ ≤ v₂ ≤ v₃ ≤ v₄ ≤ ... , so if any coordinate is non-zero, then the sum v² = v₁² + v₂² + v₃² + v₄² + ... necessarily diverges!

The orthoplex is allowed to exist because it has a different symmetry group, o3o3o3o3o3o..., the same as the simplex. (There could conceivably be an "other end", like ...o3o4o; but since the two ends aren't really connected, that's just a direct product of the simplicial group and the cubic group.) For a polytope with simplicial symmetry, a3b3c3d3..., we can take the same mirror normal vectors as for cubic symmetry, except f₁. So the skewed coordinates are related to the orthogonal coordinates by

a = (v₂ - v₁)/√2,
b = (v₃ - v₂)/√2,
c = (v₄ - v₃)/√2,
d = (v₅ - v₄)/√2,
:

or equivalently

v₂ = v₁ + a√2,
v₃ = v₁ + a√2 + b√2,
v₄ = v₁ + a√2 + b√2 + c√2,
v₅ = v₁ + a√2 + b√2 + c√2 + d√2,
:

The only difference from the cubic case is that v₁ is not determined, and is allowed to be negative. Still we have v₁ ≤ v₂ ≤ v₃ ≤ v₄ ≤ ... , so if any coordinate is positive, then the sum v² = v₁² + v₂² + v₃² + v₄² + ... diverges. Thus, either all coordinates are negative and approaching 0, or finitely many coordinates are negative and the rest are exactly 0.

Let's assume that the skewed coordinates a,b,c,d,... have only finitely many being non-zero. This implies that the orthogonal coordinates are eventually constant, and thus eventually 0 (for convergence). For example, if a=b=1 and c=d=...=0, then

v₂ = v₁ + √2,
v₃ = v₁ + 2√2,
v₄ = v₁ + 2√2,
v₅ = v₁ + 2√2,
:

which requires v₁ + 2√2 = 0, so v = (-2√2, -√2, 0, 0, 0, ...). In general, if the skewed coordinates are all 0 or 1 (o or x), then v₁ should be a negative multiple of √2, to ensure convergence.

In conclusion, we see that anything cubic doesn't exist in Hilbert space, and anything simplicial does exist in Hilbert space, and in the subspace with finitely many non-zero coordinates.
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
Tetronian

Posts: 404
Joined: Tue Sep 18, 2018 4:10 am

### Re: New insights about infinite-dimensional polytopes

Relatedly, given the cubic mirror vectors as a skewed basis for Hilbert space,

f₁ = (1, 0, 0, 0, 0, 0, ...),
f₂ = (-1/√2, 1/√2, 0, 0, 0, 0, ...),
f₃ = (0, -1/√2, 1/√2, 0, 0, 0, ...),
f₄ = (0, 0, -1/√2, 1/√2, 0, 0, ...),
:

there is no dual basis. That would require a vector g such that gf₁ = 1 and gf₂ = gf₃ = gf₄ = ... = 0, which could only be g = (1,1,1,1,...) which is not in the Hilbert space.

For the simplicial mirror vectors, there is a dual basis:

g₂ = (-√2, 0, 0, 0, 0, ...),
g₃ = (-√2, -√2, 0, 0, 0, ...),
g₄ = (-√2, -√2, -√2, 0, 0, ...),
g₅ = (-√2, -√2, -√2, -√2, 0, ...),
:
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
Tetronian

Posts: 404
Joined: Tue Sep 18, 2018 4:10 am

### Re: New insights about infinite-dimensional polytopes

Whoa, hold on a second here. So what you're saying is that the orthoplex is isomorphic to the simplex?? How could that be, if the number of surtopes per ridge is 4 in the orthoplex but 3 in the simplex?
quickfur
Pentonian

Posts: 2855
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: New insights about infinite-dimensional polytopes

But what is a ridge, in infinite dimensions? It would have to be an (∞ - 2)-dimensional face.

I think the orthoplex has only finite-dimensional faces. And, dually, though the cube doesn't have finite vertices, it could exist with (∞ - k)-dimensional faces for finite values of k, if that makes sense. (Duality works in the usual way: Replace each vertex v with the subspace {x | vx=1}, and similarly replace each k-face F with the subspace {x | ∀v∈F, vx=1}.)

It does appear that the orthoplex is the same as the simplex, but I'm not sure. It probably depends on the definitions, and on which kinds of infinity we're using.
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
Tetronian

Posts: 404
Joined: Tue Sep 18, 2018 4:10 am

### Re: New insights about infinite-dimensional polytopes

mr_e_man wrote:But what is a ridge, in infinite dimensions? It would have to be an (∞ - 2)-dimensional face.

I made a mistake, ridge is wrong, it should be peak, i.e., (n-3)-face, not (n-2)-face (which should always have an incidence of 2 in an orthoplex).

Consider the orthoplex:
Code: Select all
`        <±1,  0,  0,  0, ...>        < 0, ±1,  0,  0, ...>        < 0,  0, ±1,  0, ...>        ...`

If we restrict coordinates to be non-negative, we obtain one of its facets, ostensibly an infinitely-dimensional simplex:

Code: Select all
`        < 1,  0,  0,  0, ...>        < 0,  1,  0,  0, ...>        < 0,  0,  1,  0, ...>        ...`

If we negate one of the coordinates, say the first one, we obtain a different, adjacent facet of the orthoplex:

Code: Select all
`        <-1,  0,  0,  0, ...>        < 0,  1,  0,  0, ...>        < 0,  0,  1,  0, ...>        ...`

The ridge between these two facets can be obtained by dropping the points not in common between them:

Code: Select all
`        < 0,  1,  0,  0, ...>        < 0,  0,  1,  0, ...>        < 0,  0,  0,  1, ...>        ...`

Negating a different coordinate, say the 2nd, gives us a third facet, adjacent to the first:

Code: Select all
`        < 1,  0,  0,  0, ...>        < 0, -1,  0,  0, ...>        < 0,  0,  1,  0, ...>        ...`

Negating both coordinates, the 1st and 2nd, gives us a 4th facet, adjacent to the second and third:

Code: Select all
`        <-1,  0,  0,  0, ...>        < 0, -1,  0,  0, ...>        < 0,  0,  1,  0, ...>        ...`

The peak common to all these 4 facets is:

Code: Select all
`        < 0,  0,  1,  0,  0, ...>        < 0,  0,  0,  1,  0, ...>        < 0,  0,  0,  0,  1, ...>        ...`

which is obtained by dropping the 4 points that are not shared by all 4 facets.

It should be obvious from this description that there are exactly 4 facets incident on this peak, because there are exactly 1 combinations of signs for the first two coordinates: ++, +-, -+, --, which corresponds with the 4 points we dropped from the orthoplex to obtain this peak. There cannot be any other incident facets, because other facets of the orthoplex would contain points that do not match the coordinates of this peak besides the 1st two coordinates. I.e., this peak is not a subset of any other facet of the orthoplex besides these four.

//

Now consider the simplex. For simplicity, just take the first facet of the orthoplex as our canonical example of a simplex:
Code: Select all
`        < 1,  0,  0,  0, ...>        < 0,  1,  0,  0, ...>        < 0,  0,  1,  0, ...>        ...`

What are some of its facets? We already know one of them: the ridge between the first two facets of the orthoplex, which is a facet of the simplex:

Code: Select all
`        < 0,  1,  0,  0, ...>        < 0,  0,  1,  0, ...>        < 0,  0,  0,  1, ...>        ...`

Notice that we obtain this facet by dropping a point from the simplex (a comparison with finite-dimensional simplices should show the dimensional analogy clearly here).

Similarly, if we drop another point, we obtain one of the ridges of the simplex, which happens to be the peak of the orthoplex that we just saw above:
Code: Select all
`        < 0,  0,  1,  0,  0, ...>        < 0,  0,  0,  1,  0, ...>        < 0,  0,  0,  0,  1, ...>        ...`

What about peaks? By dimensional analogy it should be clear that dropping a 3rd point from the simplex would yield a peak. For example, here's a peak that the above ridge is incident on:

Code: Select all
`        < 0,  0,  0,  1,  0,  0, ...>        < 0,  0,  0,  0,  1,  0, ...>        < 0,  0,  0,  0,  0,  1, ...>        ...`

How many simplex facets and ridges are incident on this peak?

Clearly, there are 3 incident ridges: each of the 3 points we dropped from the simplex to obtain this peak, if added back, would yield an incident ridge. So there are at least 3. But there cannot be more than 3, because any other ridge of the simplex (obtained by dropping 2 points) would not contain at least one of the points of this peak, so it cannot be incident (this peak would not be a subset of it).

Similarly, there are 3 incident facets, because there are 3 possible pairs of the 3 points we dropped to obtain this peak. Adding each pair back to the set of points would yield a facet of the simplex. There cannot be any other incident facets, because they would not contain at least one of the points in this peak, so this peak is not a subset of their points; i.e., they are not incident on this peak.

Conclusion: the infinite-dimensional orthoplex is not isomorphic to the infinite-dimensional simplex.
quickfur
Pentonian

Posts: 2855
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: New insights about infinite-dimensional polytopes

P. S. the difacetal angle of the cross polytope converges to 180° as the dimension increases without bound; but the difacetal angle of the simplex converges to 90° as dimension increases without bound. So the two can't possibly be the same.
quickfur
Pentonian

Posts: 2855
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: New insights about infinite-dimensional polytopes

quickfur wrote:If we restrict coordinates to be non-negative, we obtain one of its facets, ostensibly an infinitely-dimensional simplex:

Code: Select all
`        < 1,  0,  0,  0, ...>        < 0,  1,  0,  0, ...>        < 0,  0,  1,  0, ...>        ...`

That may be a problem. At least, it's counter-intuitive, that the centre of that simplex is 0, as the limit of (1/n, 1/n, ..., 1/n, 0, 0, ...) which has magnitude 1/√n.

Anyway, if the orthoplex is x3o3o3o3..., and the simplex is x3o3o3o3..., then it's obvious that they should be the same. So there must be some essential difference between the Wythoff construction and your coordinate construction. Perhaps the Wythoff construction is incomplete, giving only vertices (and finite hulls of vertices), not facets. Perhaps the orthoplex could be constructed dually, by applying the group (cubic or simplicial?) to one facet to generate the other facets.
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
Tetronian

Posts: 404
Joined: Tue Sep 18, 2018 4:10 am

### Re: New insights about infinite-dimensional polytopes

mr_e_man wrote:Relatedly, given the cubic mirror vectors as a skewed basis for Hilbert space,

f₁ = (1, 0, 0, 0, 0, 0, ...),
f₂ = (-1/√2, 1/√2, 0, 0, 0, 0, ...),
f₃ = (0, -1/√2, 1/√2, 0, 0, 0, ...),
f₄ = (0, 0, -1/√2, 1/√2, 0, 0, ...),
:

there is no dual basis. That would require a vector g such that gf₁ = 1 and gf₂ = gf₃ = gf₄ = ... = 0, which could only be g = (1,1,1,1,...) which is not in the Hilbert space.

For the simplicial mirror vectors, there is a dual basis:

g₂ = (-√2, 0, 0, 0, 0, ...),
g₃ = (-√2, -√2, 0, 0, 0, ...),
g₄ = (-√2, -√2, -√2, 0, 0, ...),
g₅ = (-√2, -√2, -√2, -√2, 0, ...),
:

Actually the vectors f₂, f₃, f₄, ... don't form a basis, because f₁ can't be gotten from the others; so there's not much sense in saying that g₂, g₃, g₄, ... form the dual basis.

So let's rebuild the simplicial mirror vectors, starting with h₁ = (1,0,0,0,...), so that they span the whole space. We should have hkhk = 1, hkhk+1 = -1/2, and otherwise hkhj = 0 .

h₁ = (1, 0, 0, 0, 0, ...)
h₂ = (-1/2, √3/2, 0, 0, 0, ...)
h₃ = (0, -√3/3, √6/3, 0, 0, ...)
h₄ = (0, 0, -√6/4, √10/4, 0, ...)
h₅ = (0, 0, 0, -√10/5, √15/5, ...)
:
hk = (0, ... , 0, -√(k(k-1)/2) / k, √(k(k+1)/2) / k, 0, 0, 0, ...)
hk+1 = (0, ... , 0, 0, -√(k(k+1)/2) / (k+1), √((k+1)(k+2)/2) / (k+1), 0, 0, ...)
:

The notation x3o3o3o3... means there is a point v with h₁•v=1 and h₂•v=h₃•v=h₄•v=...=0; the solution is

v = (1, 1/√3, 1/√6, 1/√10, 1/√15, ...);
vv = 2;

and this point is recursively reflected along each hk, according to the formula v - 2(vhk) hk, to generate the vertices of the simplex. Most of the reflections have no effect on the first few points, or don't produce any new points. In fact, at each step, there's only one reflection that gives a new point:

v₀ = v
v₁ = v₀ - 2(v₀•h₁) h₁ = v - 2h
v₂ = v₁ - 2(v₁•h₂) h₂ = v - 2h₁ - 2h
v₃ = v₂ - 2(v₂•h₃) h₃ = v - 2h₁ - 2h₂ - 2h
v₄ = v₃ - 2(v₃•h₄) h₄ = v - 2h₁ - 2h₂ - 2h₃ - 2h
:

The reflection along hk swaps vk-1 with vk, and leaves the other points (vj, for j<k-1 or j>k) unchanged. Explicit orthogonal coordinates for these points, the vertices of an origin-centred ∞-simplex with edge length 2, whose affine hull is the whole space:

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, ...)
:
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
Tetronian

Posts: 404
Joined: Tue Sep 18, 2018 4:10 am

### Re: New insights about infinite-dimensional polytopes

mr_e_man wrote:
quickfur wrote:If we restrict coordinates to be non-negative, we obtain one of its facets, ostensibly an infinitely-dimensional simplex:

Code: Select all
`        < 1,  0,  0,  0, ...>        < 0,  1,  0,  0, ...>        < 0,  0,  1,  0, ...>        ...`

That may be a problem. At least, it's counter-intuitive, that the centre of that simplex is 0, as the limit of (1/n, 1/n, ..., 1/n, 0, 0, ...) which has magnitude 1/√n.

The center of that simplex ought to be, if dimensional analogy doesn't betray us and break down at d=infinity, the corresponding hypercube vertex radially scaled to a distance equal to the in-radius of the orthoplex. Unfortunately, given that the length of the hypercube's vertex <1,1,1,...> is infinite, the only radius that could make this length finite must be infinitesimal. Whether it's exactly zero or not is up for debate, because we're dealing with a degenerate case here (infinitely long vector scaled to a finite length). But it's certainly not a normal positive real number.

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.

As I showed with the coordinates of the orthoplex, the distinguishing feature of the orthoplex in having 4 facets incident per peak is retained even at d=infinity, so a sequence of CD diagrams that more likely represents the real situation would be o4x, o4o3x, o4o3o3x, ... . The limit, unfortunately, does not appear to be directly representable as a CD diagram because it has a ringed node at the end, but the notation o4o3o3o... does not convey this.

So there must be some essential difference between the Wythoff construction and your coordinate construction. Perhaps the Wythoff construction is incomplete, giving only vertices (and finite hulls of vertices), not facets. Perhaps the orthoplex could be constructed dually, by applying the group (cubic or simplicial?) to one facet to generate the other facets.

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.

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. The reason for this is that certain members of the uniform polytope family of the hypercube/orthoplex are constructible with finite outradius, whereas certain others have infinite outradius.

The rectified orthoplex, for example, can be constructed as sign and coordinate permutations of the point <1,1,0,0,...>. This corresponds with ringing the second last node of the infinitely long CD diagram. The truncated orthoplex can also be constructed as permutations of <2,1,0,0,...>, corresponding with ringing the last 2 nodes of the CD diagram.

On the other hand, you have things like apacs<0,0,0,1,1,1,...> which is one of the deeply truncated orthoplexes: infinitely deeply, in fact, that its outradius is infinite. Or <0,0,1,1,2,2,2,2,...> which has two ringed nodes on the o4o... end of the CD diagram

There are also truncates that have rings on both ends of the CD diagram, like <3,0,0,1,1,2,2,2,2,2,...>, which has 2 ringed nodes on the o4... end and one ringed node on the other end. 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.

One open question I have is whether a mesotruncate exists, since I cannot think of any direct representation in terms of CD diagrams. Coordinate-wise, though, apacs<0,1,0,1,0,1,...> appears to be a candidate. Either that, or apacs<2,1,0,1,0,1,0,...> could also be a candidate mesotruncate.

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?? (Current wild speculation: the difacetal angles are infinitesimally close to 180° / resp. 90° but not exactly equal to 180° / resp 90°, so when folded up in infinite dimensional space, it folds up into a bounded polytope instead of a tessellation. This is one area where using surreal numbers may help us distinguish between these cases.)
quickfur
Pentonian

Posts: 2855
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Next 