Living in infinite-dimensional space

[Frisk-256]

one thing you could do in this space is draw anything you want on a peice of ∞D paper, then rotate it through ∞D space into a solid object, you could do it like this
paper
(0,x1,x2,x3,x4,x5,x6,x7,x8,x9...)
rotating it an infinite amount of times in this order
(0,x1,x2,x3,x4,x5,x6,x7,x8,x9...)
x0x1 plane
(x1,0,x2,x3,x4,x5,x6,x7,x8,x9...)
x1x2 plane
(x1,x2,0,x3,x4,x5,x6,x7,x8,x9...)
x2x3 plane
(x1,x2,x3,0,x4,x5,x6,x7,x8,x9...)
x3x4 plane
(x1,x2,x3,x4,0,x5,x6,x7,x8,x9...)
...
you would get
(x1,x2,x3,x4,x5,x6,x7,x8,x9...)
and therfor a solid object
this would mean you could make a slice of bread into a full loaf of bread, meaning infinite bread if you just slice your loaf of bread and then turn each of them into an entire loaf.
∞D is just broken, It seems to be full of paradoxes and almost all measurements seem to be infinity or 0. ∞D space should be banned the same way 0/0 is banned

[quickfur]

There is no contradiction, just a lot of counterintuitive behaviors because we're dealing with infinite quantities, which do not behave like finite quantities.

The rotation you described is exactly what I alluded to in an earlier post about "limit" rotations, which are equivalent to projections, and which in infinite dimensional space makes a lower-dimensional object equal to a higher dimensional one. In fact, the infinite dimensional cube's facets are themselves infinite dimensional cubes, which map to the entire cube exactly by the limit rotation you described. This is a known property of the infinite dimensional cube, there's no surprise here.

Most of the alleged contradictions with infinite dimensional space comes from wrongly expecting infinite dimensional objects to behave like finite dimensional ones, or expecting basic arithmetic rules to apply to infinite quantities, which they do not. When dealing with infinite quantities in infinite dimensional spaces, you gotta be very careful with your assumptions; any unfounded assumptions can quickly unravel the whole thing.

Also, R^∞ is known to be a pathological space in mathematical circles. This is why things like Hilbert spaces and Banach spaces were invented: they are infinite dimensional spaces that have much less pathological properties than R^∞. But this thread is specifically for grappling with these pathological properties head-on. Here be dragons, but we're not afraid of them!

[DonSoreno]

In order to avoid some of the pathologies associated with R^∞,
we could use a system of rotations that conserve both dimension of geometric quantities (vectors,bivectors,planes, hypercubes) and co-dimension.
So that the hypercube (having dimension => Infinity, co-dimension => 0)
cannot be rotated into a "hypercube" of co-dimension 1 --> flat sheet.

I would propose we limit the rotations to rotation matrices R of the following Form:

R = exp(A)



where A is a square-summable matrix & antisymmetric matrix.
whose entry A_ij is the angle, by which the j-th basis vector is turned towards the i-th basis vector.

These are exactly the rotations that have a measurable/finite angle.
Or more importantly, the points v on the surface of a unit-sphere, whose rotation at time t is:

v(t) = exp(t*A) v
|v| = 1
Would have finite velocity (for all points v on the surface), if & only if A is square-summable.


In other words, the Rotation (matrix) R = exp(A) has "finite angle"
(obviously there's more than a single "angle" in 4D and beyond.)

Here is this constraint, expressed in Clifford Algebra language:

The attachment constrained_rotor.png is no longer available

We still have to deal with all the other pathologies though .

[mr_e_man]

Ah! Welcome back!

I was wondering if you were still alive.

[quickfur]

Ah! Welcome back!

I was wondering if you were still alive.

I took a turn along the wrong axis from my last visit to ∞-dimensional space, and ended up a non-finite distance away from the corner of the ∞-cube where the exit was. Took me months to find my way back!

Constrained rotations are all good and everything, but the problem still remains that the vertices of the ∞-cube are infinitely far from the center. This also means that the centroid of the ∞-cross's facets are infinitesimally close to the center of the ∞-cross, meaning that in some weird sense the ∞-cross behaves as if it's a concave object (even though it isn't) with zero volume. Well actually, technically speaking the ∞-cross's facets do not have centroids, but if you leave it at that then they will also have "holes". This means that geometrical quantities like distances and dot products run into the problem of non-finite values, so some angles cannot be computed.

Take, for example, two vertices of the ∞-cube that differ from each other in a non-finite number of coordinates. Draw two vectors from the origin to each of these 2 vertices. What's the angle between these two vectors? Well, to compute that, the most obvious way is to take their dot product... but since they differ in an infinite number of places, this dot product will be a non-finite number. The magnitude of the vectors is also non-finite. How are you supposed to take their ratio, then? And that doesn't even begin to deal with the problem of how to invert that cosine function to solve for the angle, since you end up with the arccos of the ratio of two non-finite quantities.

R^∞ is very weird and pathological. No wonder most mathematicians have shied away from it and stuck with saner constructs like Banach spaces and Hilbert spaces.

[quickfur]

Recently, I was considering the question of bisecting the n-cube diagonally, i.e., cutting the n-cube with a hyperplane orthogonal to the line connecting two antipodal vertices. For example, in 2D, you can cut a square along a diagonal to get two triangles. In 3D, you can cut a cube into two equal halves with a hexagonal cross-section and two truncated triangular trapezohedra. In 4D, you can cut a tesseract in half to get two pieces with an octahedral cross-section. Now, interestingly enough, in odd dimensions, this bisection separates the vertices into two disjoint sets, each of which contains vertices less than n/2 edge hops away from the corresponding antipodal vertex. In even dimensions, such as in 2D and 4D, some of the vertices will lie in on the cutting hyperplane, so you'd get three sets of vertices: those that are "closer" to one antipodal vertex, and those that are equidistant from either antipode.

Why is this interesting? Because the n-cube happens to be the shape of the Hassé diagram of the subset partial order of a set of n elements. One antipode can be identified with the empty set (the bottom node of the Hassé diagram) and the other antipode with the entire set, then the vertices in between are the subsets of the full set. If we cut this n-cube in half, then we get a set of vertices representing "large" subsets (i.e., contains most of the elements of n), another set of vertices representing "small" subsets (contains few of the elements of n), and if n is even, an intermediate set representing subsets of "intermediate size". Now this being the n-cube, the number of vertices in the intermediate set is always even, so in theory we can divide it in half and assign half of its elements to the "large" set and the rest to the "small" set, and we'd have two sets of subsets, the "large" set and the "small" set. This division is arbitrary, since the vertices in the middle are equidistant from the antipodes, so there is no real criteria by which you could decide whether a vertex should belong to the "large" set or the "small" set.

Now, let n go to infinity. (This topic is about the ∞-cube after all! ) Here, we find something very interesting. Take two antipodes, say <0,0,0,...> and <1,1,1,...>, and collect the vertices into three sets: those reachable from <0,0,0,...> in a finite number of steps (let's call this the "small set"), those reachable from <1,1,1,...> in a finite number of steps (let's call this the "large set"), and a third set (the "intermediate set") containing all vertices that are infinitely distant from either antipode. If we interpret the coordinates as indicating membership in the set N of natural numbers, then this describes the Hassé diagram of the powerset of the natural numbers, i.e., <0,0,0,...> is the empty set, and the vertices in the "small set" are the finite sets. <1,1,1,...> is the entire set N, and the vertices in the "large set" are the cofinite sets. The "intermediate set" corresponds with those subsets of N that are both infinite and have infinite complement.

This suggests the funny idea that infinity is an even number. Of course, this isn't actually the case, but let's say that regardless, we want to divide the intermediate set into two "equal" parts. How would we decide how to divide them? As in the finite even-dimensional case, there is no clear criterion we could use, since all the vertices in this set are "equidistant" from either antipode (in a crude sense of the word: they are infinitely distant, and there's no obvious way to differentiate among them). Certainly, cardinality is of no help here, unlike with the small set, where the number of 1's is a finite cardinality and the number of 0's is infinite cardinality (so there are objectively more 0's than 1's), and the large set, where the number of 1's is an infinite cardinality and the number of 0's is a finite cardinality (there are objectively more 1's than 0's).

Interestingly enough, there's something that meets our exact need: a free ultrafilter on N. By definition, an ultrafilter U on N is a set of subsets of N such that (1) no finite subset is a member of U (none of the "small set" elements are in U); (2) all cofinite subsets are members of U (all of the "large set" elements are in U); and (3) for any given subset S of N, either S is in U, or the complement of S (i.e., N \ S) is in U. Furthermore, (4) if a set S is in U, then any superset of S that's a subset of N is also in U; and (5) given any two members x,y of U, there exists another member z of U such that z is a subset of both x and y.

For finite and cofinite subsets, (1) and (2) already tell us that U contains all the "large subsets": all the cofinite subsets are in it, but none of the finite subsets are. For the "intermediate" subsets of N, either the subset is in U (it's assigned to the "large" set) or its complement is in U (it's assigned to the "small" set). Any superset of this member of U is also in U, which corresponds with the idea that if we add more elements to a "large" subset, it remains a "large" subset, thus respecting the ordering of the Hassé diagram.

Even more interestingly (5) tells us that this ordering is total: any two members of U are always comparable, so we can always decide which among them is the "larger" one. This property lets us construct a total ordering on the powerset of N, such that trichotomy continues to hold; we can use this to construct the so-called "hyper-naturals", which are a non-standard model of Peano arithmetic that contain infinitely large "natural numbers". (Which also has many very strange properties, like having a non-finite number that can be infinitely divided by 2, a number that's divisible by all finite natural numbers, and having a dense subset with the order type of the rationals yet not being dense in itself(!). But I digress.)

Anyway, coming back to the ∞-cube: the strange thing about this bisection of the ∞-cube is that the "intermediate set" (before we applied the ultrafilter on it) has uncountable cardinality, whereas the "large" and "small" sets (before the ultrafilter) has finite cardinality. This means that the vast majority of the vertices of the ∞-cube are, in some weird sense, "equidistant" from the two antipodes! I haven't worked out the exact number of such vertices in finite dimensions, but I expect that for (very) large finite even n, the number of vertices that lie on the cutting hyperplane would overwhelm the vertices that are "closer" to either antipode. Applying the ultrafilter, then, corresponds with imposing an arbitrary division of this intermediate set and arbitarily assigning them one to the "large" set and one to the "small" set.

//

Now, all of this ado about ultrafilters suggests one way of perhaps getting past the difficulties with infinite geometric quantities when dealing with the ∞-cube: the hyperreals are constructed as infinite sequences of real numbers, with an ultrafilter used to impose a total ordering on the resulting set. Having a total ordering is very helpful, because it lets us compare two sequences of reals that diverge to infinity and decide unambiguously which is "greater". Which suggests that we can construct such sequences from geometric operations that we might wish to perform in ∞-space: for example, if we want to compute the dot product between <1,1,1,1,...> and <1,2,3,4,...>, which results in the diverging series 1 + 2 + 3 + 4 + ..., we could map it to the ∞-vector <1, 2, 3, 4, ...> and interpret it as a hyperreal, let's say h1. Then take another dot product, say <1,3,5,7,...> and <2,4,6,8,...>, and construct the hyperreal h2 = <1*2, 3*4, 5*6, ...>. The ultrafilter then lets us decide whether h1 < h2 or h1 ≥ h2. So now our infinite dot products become comparable!

And BTW, according to the standard (har har) construction of the hyperreals, sequences that diverge to infinity correspond with non-finite hyperreals, whereas sequences that do not diverge correspond with finite hyperreals. Which makes sense in terms of dealing with geometry in ∞-space, giving us a way to say that a dot product that looks like 1 + 1 + 1 + ... is, in some sense, "less than" a dot product that looks like 1 + 2 + 3 + ... .

Funnily enough, sequences whose terms diminish to zero correspond with infinitesimal hyperreals. Which means that dot products with finite values are in some sense infinitesimally small compared to the infinite quantities you get in ∞-space. (And the nice thing about hyperreals is that it actually lets you compare these infinitesimals under a total order. So you'll never run into the problem of dot products being incomparable, for example.)

Of course, all of this applies not only to dot products, but to anything involving an infinite number of terms, like computing the length of an ∞-vector, etc.. So that means geometry in ∞-space is now more tractible, without having to worry about most quantities being infinite!

[DonSoreno]

The number of vertices, that are incident, to this diagonal hyperplane can be computed using the binomial coefficient:
BinomialCoefficient( n ; n/2 )


n | vertices on "space" diagonal of a hypercube
----------------------------------------------------------
2 | 2
4 | 6
6 | 20
8 | 70



So a large amount of vertices will lie on the diagonal, but the fraction #vertices on diagonal / #total vertices, will go to 0, as n --> infinity.



The construction of an ultrafilter on an uncountable set (e.g. the corners of an countably-infinite dimensional hilbert cube) requires the Axiom of choice, right?

A

Also, the same argument, regarding the distance between vertices, also applies to the countably-infinite dimensional Clifford Algebra, whose basis elements:

e1,e2,e3,... , e1e2,e1e3,e1e4, ..., e1e2e3, e1e2e4, ... ,e1e2e3e4, ... ... , e1e2e3e4e5e6e7...
can be treated as subsets of {e1,e2,e3,e4,...}.

If we allow for infinite grade elements, we also have the problem that said Clifford Algebra is not graded anymore

In this Clifford algebra, the "intermediate size" elements (hypervolumes of infinite dimension, and infinite codimension),
also cause problems, regarding the grading and should probably be avoided entirely.

I think working purely on finite, and cofinite elements is the best way.

For all finite elements F, cofinite elements C, and intermediate elements I (not to be confused with the identity element), we have:

F * F = F
F * C ---> some mixture of finite and cofinite grade elements.
C * C ---> some mixture of finite and cofinite graded elements.
F * I --> any mix of the three.
C * I --> mixture of finite, cofinite, and intermediate elements.
* being the geometric product

If we choose to use intermediate elements, the "finite" neigbhorhood N(X) = {X xor Y | Y is finite subset of {e1,e2,e3,...}}
forms an equivalence class,
and we could probably fix a grading (meaning a plus-minus convention), for each of these neighborhoods.
The finite elements are then simply the finite neighborhood around the empty subset (corresponding to scalar elements of the Clifford Algebra.)
The cofinite elements are the finite neighborhood around the pseudoscalar --> hypervolume, of codimension 0.

There are uncountable many such neighborhoods, so




I was also looking hyperreals as a way to model infinitesimal forces, steming from the 1/r^∞

[quickfur]

The number of vertices, that are incident, to this diagonal hyperplane can be computed using the binomial coefficient:
BinomialCoefficient( n ; n/2 )


n | vertices on "space" diagonal of a hypercube
----------------------------------------------------------
2 | 2
4 | 6
6 | 20
8 | 70



So a large amount of vertices will lie on the diagonal, but the fraction #vertices on diagonal / #total vertices, will go to 0, as n --> infinity.

Hmm. That's weird, because the vertices finitely-reachable from either antipode in the ∞-cube are countable, but the vertices in the middle (where the coordinates have an infinite number of 0's and 1's, so they are neither finite nor cofinite) are uncountable. Which seems to contradict this.

However, the reason may be because our classification of the middle vertices as those having an infinite number of 0's and 1's is too coarse, and does not actually capture the limiting shape. After all, the distance between any two vertices in the finitely-reachable set from either antipode is finite, but the distance between these vertices and any vertex in the "middle" set is infinite. If the finite Hassé diagram picture is any reliable indicator of behaviour at n=∞, this would mean that the middle vertices, using this coarse criterion, must span an infinite number of Hassé diagram levels, so the "diagonal" vertices must be a strict subset of them.

How to characterize this subset, though, is not so simple, since cardinality alone does not give us a sufficiently fine criterion for isolating them.

The construction of an ultrafilter on an uncountable set (e.g. the corners of an countably-infinite dimensional hilbert cube) requires the Axiom of choice, right?

Yes, the Axiom of choice is needed because you need to make an infinite number of arbitrary choices between a set and its complement. Which, unfortunately, means that the (non-principal, or free) ultrafilter cannot be fully described in finite space.

However, it seems that most people just stop at this: we needed AC to give us the ultrafilter, it has the characteristics we need, good enough, move on, don't worry about the details of the ultrafilter.

I wasn't satisfied with this, though. Perhaps we cannot fully describe a free ultrafilter, but what stops us from attempting to arrive at a partial description? Can it not be possible to at least describe enough of it such that we can form some idea of what the full thing might look like? So far, I haven't made too much progress yet, but I did at least make a first step: let U be a free ultrafilter over N. By definition, we know that no finite subset of N is in U, and all cofinite subsets are in U. We also know that some of the "intermediate" subsets (i.e., infinite subsets of N that are not cofinite) must be in U whilst their complements are not. This is rather non-specific, but the definition of ultrafilter does impose criteria on what subsets can or cannot be in U.

For starters, N can be arbitrarily rearranged (or relabelled) via a bijection. Which means we can postulate that the set of odd numbers A = {1, 3, 5, 7, ...} is in U. (If this is not the case, simply apply a relabelling of N such that this is true -- we know there must be at least one infinite non-cofinite subset of N in U; with appropriate relabelling, this set can be made to be A.) This can be done without loss of generality, thanks to relabelling.

Now, we can apply the definition of an ultrafilter to find other sets that must be in U: for example, U being a filter requires that if B is any subset of N such that A is a subset of B, then B must also be in U. What does this look like? Simple: B is any set that consists of the odd numbers with 1 or more even numbers sprinkled in. If the resulting set is missing only a finite number of even numbers, then it's a cofinite set that's already in U. But if the resulting set has an infinite number of missing even numbers, then it's another distinct member of U. So that gives us a good picture of what the elements higher up in the Hassé diagram from A look like.

Now, A obviously cannot be the least element of U, because if so, it would contradict U being an ultrafilter (it would be a filter, but not ultra, because you'd be able to add a subset of A to U, fill in the missing upper parts of the partial order, and get a larger filter than U, which contradicts ultra). So there must be elements of U that are less than A, i.e., subsets of A. These subsets cannot be finite (otherwise it contradicts non-principal); they must all be infinite. What do they look like? Well, following the intuition of U deciding between "large" and "small" subsets of N, we can postulate that if we remove a finite number of odd numbers from A, it should still be in U (because A is infinite, removing a finite number of elements should not diminish its size significantly; only if you removed an infinite number of elements you could argue that it's in some sense "smaller"). If we proceed with this idea, we see that subsets of A that are cofinite in A (but not necessarily in N) should also be in U. I.e., these are the subsets you obtain by deleting a finite number of elements from A. We never delete an infinite number of elements from A, so any such subset would still have a non-empty, non-finite intersection with A, thus preserving the properties of ultrafilter.

Now, there's another property of ultrafilters that say that for any pair of elements x, y of U, there exists an element z such that z≤x and z≤y (since the partial order here is subsets, ≤ in this context means subset). If we apply this to the new subsets we postulated (the subsets of A that are cofinite in A), we can discover more elements of U. In particular, the intersection of any two such subsets is always another subset of A with an infinite number of odd numbers remaining -- because we only ever remove a finite number of odd numbers to arrive at the subset, the intersection of any two of them will contain at least those odd numbers that we haven't removed, and there are still an infinite number of them left. So this fulfills the requirement that if x,y are in U then there exists a z≤x, z≤y that's also in U. (IOW their intersection is never finite.) Then of course, we can add any subset of even numbers to these sets, and they would also be in U (by the part of the definition that says that any subset B of N that's a superset of our set is also in U).

Of course, we still haven't covered all the bases yet -- we haven't considered subsets of A where an infinite number of odd numbers are removed, but still leaving an infinite number of odd numbers -- e.g., if we removed every other odd number. But the picture we have so far already gives us a good idea of what is in U and what isn't in U. It's already good enough that we could actually name explicit sets that are in U, and we can actually use this knowledge to state explicit results of comparisons under the total ordering imposed by U for a good number of cases. There are still cases that are undecided, but by repeating the above analyses we could make good progress of explicating the contents of the ultrafilter U. We may never exhaust all the cases (since otherwise we wouldn't need to invoke the Axiom of Choice!), but at least we could get a fairly good glimpse of what our ultrafilter looks like!

[...]
If we allow for infinite grade elements, we also have the problem that said Clifford Algebra is not graded anymore

In this Clifford algebra, the "intermediate size" elements (hypervolumes of infinite dimension, and infinite codimension),
also cause problems, regarding the grading and should probably be avoided entirely.

I think working purely on finite, and cofinite elements is the best way.

Yeah, these intermediate elements are the source of much of the pathology of the ∞-cube. Unfortunately, they are uncountable, and so form the majority of the structure of the ∞-cube! Which means that if we exclude them, we exclude most of the ∞-cube. In fact, we are reduced to just the finitely-reachable vertices from either antipode. These reachable vertices form what in earlier posts I called a "piece" of the ∞-cube. Out of the uncountable number of pieces, we are left with 2, and each piece fits into a Hilbert space.

If we're going to work with such a constricted subset of the ∞-cube, we might as well just go back to Hilbert space and work with objects of finite extent. Hilbert space is a lot more well-behaved and we don't need ultrafilters or wrangling with uncountable numbers of vertices. But that wouldn't tell us very much about the full-sized ∞-cube.

For all finite elements F, cofinite elements C, and intermediate elements I (not to be confused with the identity element), we have:

F * F = F
F * C ---> some mixture of finite and cofinite grade elements.
C * C ---> some mixture of finite and cofinite graded elements.
F * I --> any mix of the three.
C * I --> mixture of finite, cofinite, and intermediate elements.
* being the geometric product

If we choose to use intermediate elements, the "finite" neigbhorhood N(X) = {X xor Y | Y is finite subset of {e1,e2,e3,...}}
forms an equivalence class,
and we could probably fix a grading (meaning a plus-minus convention), for each of these neighborhoods.
The finite elements are then simply the finite neighborhood around the empty subset (corresponding to scalar elements of the Clifford Algebra.)
The cofinite elements are the finite neighborhood around the pseudoscalar --> hypervolume, of codimension 0.

There are uncountable many such neighborhoods, so

See my earlier posts on the "pieces" of the ∞-cube: each piece is a finitely-reachable subset of the ∞-cube from a single vertex, and fits in a suitable-translated Hilbert space. Which is what you describe here.

I was also looking hyperreals as a way to model infinitesimal forces, steming from the 1/r^∞

That's an interesting idea!

[DonSoreno]

I guess without the Axiom of (uncountable: 2^aleph_0) Choice, we wont get far regarding the Hilbert cube.

Fixing a grading for each equivalence class, in the clifford algebra: N(X)= {X xor Y | Y finite}
also requires at least such an uncountable choice function.

[Hugh]

I asked ChatGPTo3 if it had any roads to further insights about this and it came up with 7 that might be at least a little helpful.