## Infinite-dimensional polytopes?

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

### Infinite-dimensional polytopes?

After doing much work on my n-dimensional vector calculator program, I've started wondering if it is possible to generalize R<sup>n</sup> to infinite dimensions, in a way that vector algebra still holds as in the finite case. In particular, I'd like to know if it is possible to have infinite-dimensional polytopes in such a space, and if it is at all possible to project them into finite dimensions so we can see them! :-)

I've just starting thinking about this a few days ago... and it occurred to me that an easy way to have infinite dimensions is simply to treat vectors in this space as sequences of real numbers. Sequences are well-studied in calculus, so this is still familiar ground. The problem comes with defining vector operations, such as vector norm or the dot (inner) product.

From the little googling that I did, I found out that some people have already considered infinite-dimensional vectors, but I've only ever seen subsets of infinite-dimensional space, such as the set of convergent sequences, etc.. For these sequences, it's easy to generalize the Euclidean norm: you just sum over the square of each element in the sequence and take the square root, and there you have a workable definition of norm (provided the resultant sequence is convergent). Similarly, dot products can be computed as long as the resultant series is convergent.

However, I'm actually thinking of norms and dot products that apply to arbitrary sequences (i.e. arbitrary infinite-dimensional vectors). In such cases, a naive generalization of the Euclidean norm doesn't get anywhere, since it would diverge most of the time and so provide no meaningful use for vector comparison. I'm wondering if it's possible to define the norm or dot product indirectly, such that you can say things like vector X has the same length as vector Y without actually computing the norm, or have a way of deciding when two vectors X and Y are orthogonal (dot product = 0) without actually computing the dot product in the general case. Also interesting would be how to define rotation in an infinite-dimensional space. Would it be possible to actually specify an actual example of the equivalent of an infinite matrix, that performs a specific rotation through infinite-dimensional space?

Alternatively, perhaps it is possible to represent the norm or dot product of infinite-dimensional vectors using an extension of real number algebra that includes infinitesimals and infinite numbers. I actually have a little experience with this, having explored the classification of univariate functions over reals that diverge at x=infinity, by defining an equivalence relation and a partial ordering that classifies them based on divergence rate. The resulting classification has linearly-ordered subsets that behave like real numbers plus infinitesimals and infinities. It may be possible to map infinite-dimensional vectors to this set in a way that behaves like the Euclidean norm.

The motivation behind this, of course, is that I'm interested in finding out the properties of infinite-dimensional equivalents of things like spheres and polytopes. Would an infinite-dimensional sphere have zero (hyper)volume, since the (hyper)volume of n-spheres in (n+1) space approaches 0 as n gets large?

Does anybody have any ideas, or know of any literature that deals with this topic? :-)

(Now, I'm tempted to think about uncountable sequences as a further generalization of vectors... but perhaps I'm going a bit too far there. :-P)
quickfur
Pentonian

Posts: 2200
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

It sounds interesting that you are working on a proggie for vectors. With a little adjustment, it could handle oblique coordinate systems as well. I describe this in my glossary under "matrix dot". If S is a matrix, and u, v are vectors, one multiplies S.u dot v to get the dot product. This works in all dimensions.

See, zb http:\\www.geocities.com\os2fan2\gloss\index.html

Vector theory works in infinite dimensions. i have it working in hyperbolic space as well!

What happens with volumes etc is that these things tend to be exponential, and so drop off fairly fast, even for small dimensions. The maximal packing of spheres in 8d, for example, only covers something like 1/4 of that space.

Dot products do have already a definition that is independent of the Rn representation of space. That the R^n form is more useful to calculate with is why one often thinks of dot products in terms of R^n.

One can for example, use an oblique coordinate system, and do dot products without reduction to an orthogonal system, because the dot product is not bound to a specific coordinate system. The R^n rule here is to use a thing called "matrix dot" product: where the matrix is comprised of the unit vectors v_i . v_j, and one multiplies the matrix by a vector, and dots this with the second vector.

As to spheres, etc, there are useful iterative methods for finding the volumes of these.

sphere(n) = 2.pi.sphere(n-2)/n

whence: volume sphere = (pi)^(n/2) / (n/2)!

This converges on a term dominated by (2pi.e/n)^n/2.

Wendy
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1562
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

wendy wrote:It sounds interesting that you are working on a proggie for vectors. With a little adjustment, it could handle oblique coordinate systems as well. I describe this in my glossary under "matrix dot". If S is a matrix, and u, v are vectors, one multiplies S.u dot v to get the dot product. This works in all dimensions.

See, zb http:\\www.geocities.com\os2fan2\gloss\index.html

Does it work in infinite dimensions too?

Vector theory works in infinite dimensions. i have it working in hyperbolic space as well!

How?

I know that the theory of vector spaces can be applied to infinite vector spaces such as the space of all matrices. What I'm looking for is a bit more constrained, in the sense that I want to define the space S to be the set containing R<sup>n</sup> for all positive integers n, and define its inner product in such a way that if two vectors v and w in S happen to also be in R<sup>n</sup> for some n, then the inner product should be equal to the Euclidean dot product. That case is of course easy, since we already have a definition of dot product that works for all finite n. The tricky part is how to define the dot product for arbitrary elements of S, which generally would have an infinite number of unbounded, non-convergent coordinates, and have it still reduce to the Euclidean dot product when only a finite number of coordinates are non-zero.

What happens with volumes etc is that these things tend to be exponential, and so drop off fairly fast, even for small dimensions. The maximal packing of spheres in 8d, for example, only covers something like 1/4 of that space.

Yep, and judging by the trend, the maximal packing of infinite-dimensional spheres would probably occupy 0% of the volume of the space, even though they would have non-zero volume! :-)

Dot products do have already a definition that is independent of the Rn representation of space. That the R^n form is more useful to calculate with is why one often thinks of dot products in terms of R^n.

Right. Now, I'm interested in actually calculating, for example, the angle between two non-trivial infinite-dimensional vectors. (Non-trivial as in, the corresponding infinite sequences of their coordinates do not converge.) The dot product in any finite dimension, when divided by the norms of the corresponding vectors, is always between -1 and 1, and is equal to the cosine of the angle between them. I wonder if this can be meaningfully generalized to infinite dimensions, so that you could say, for example, that the sequence of coordinates (1,2,3,4,5,....) makes such-and-such an angle with the sequence (1,1,1,1,1,...).

One can for example, use an oblique coordinate system, and do dot products without reduction to an orthogonal system, because the dot product is not bound to a specific coordinate system. The R^n rule here is to use a thing called "matrix dot" product: where the matrix is comprised of the unit vectors v_i . v_j, and one multiplies the matrix by a vector, and dots this with the second vector.

But how does one go about generalizing this to infinite dimensions? As in, the vectors have an actual infinite number of coordinates, not merely a possibility of having an unbounded number of coordinates. In other words, I'm talking about the set of all real-valued functions from N to R (where N is the set of natural numbers and R is the set of real numbers). The idea being, of course, that the function tells you what the i'th coordinate of the infinite vector is.

Is there any meaningful way to assign a "norm" to these functions/vectors? Is there any meaningful definition of "dot product" that would work for these vectors such that when the number of non-zero coordinates are finite, it reduces to the familiar Euclidean dot product?

So far, I've come up with the idea that instead of actually computing the value of the norm or the dot product, which would be infinite in the general case, one could define computable relations on them. For example, given two infinite-dimensional vectors v and w, we can determine whether the infinite series sum<sub>i=0 to infinity</sub>(v(i)<sup>2</sup> - w(i)<sup>2</sup>) = 0. If it is, we conclude that |v| = |w|. If the sum diverges to positive infinity or converges to a positive number, then we say that |v| > |w|; conversely if it diverges to negative infinity or converges to a negative number, then we say that |v| < |w|.

One thing I'm not 100% sure about is whether trichotomy always holds in this case. Ideally, it should, so that we can define an infinite-dimensional sphere as the set S of all vectors (functions) such that for any s in S, |s| = |r| for some reference vector r which defines the "radius" of the sphere. Then all vectors t where |t| < |r| would be in the interior of the sphere, and all vectors u where |u| > |r| would be in the exterior of the sphere.

One can also define orthogonality in a similar way: let v and w be vectors (functions) in our infinite-dimensional space. If the infinite series sum<sub>i=0 to infinity</sub> v(i)w(i) = 0, then v and w are orthogonal.

The next part is where I'm currently stuck at: if the infinite series is not 0, then presumably v and w make some angle, say theta, with each other. But how would one go about computing this angle? I tried to use the same method using quotients of infinite series, but I'm at a loss as to how to prove that it will always converge, let alone be within the range 1 and -1 so that it can be meaningfully interpreted as a cosine.

As to spheres, etc, there are useful iterative methods for finding the volumes of these.

sphere(n) = 2.pi.sphere(n-2)/n

whence: volume sphere = (pi)^(n/2) / (n/2)!

This converges on a term dominated by (2pi.e/n)^n/2.

Oh? So the volume doesn't actually vanish as n approaches infinity? Does that mean that an infinite-dimensional sphere actually has non-zero volume in that space?
quickfur
Pentonian

Posts: 2200
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

quickfur wrote:Oh? So the volume doesn't actually vanish as n approaches infinity? Does that mean that an infinite-dimensional sphere actually has non-zero volume in that space?

n-factorial grows much faster than pi<sup>n</sup>. So, the volume of an n-dimensional sphere approaches zero as n approaches infinity.

The volume is maximized when n is about 7.25695 and is strictly decreasing after that.
pat
Tetronian

Posts: 563
Joined: Tue Dec 02, 2003 5:30 pm
Location: Minneapolis, MN

ANGLE BETWEEN INFINITE VECTORS

From what i recall, it is something like there is a normal distribution based on something like a S.D. of 1/e or something, where the volume of the sphere is that of the cube. Correspondingly, the radius of such a sphere is typically of the order of sqrt(n).

Accordingly, the cosine of the angle between v and w is then a normal distribution based on S.D. of 1/e.sqrt(n). As we let n go to infinity, we see that for a great part, the cosine of the angle between random vectors is 0, and hence, if you can't see all the coordinates of both vectors, then there is a better-than-average shot that v.w = k |v| |w| / sqrt(N) [k small], and the cosine of the angle tends to k / sqrt(N); ie tends to 0.

I did not actually do the calculations here: i am recalling and interpretating some i read by someone who is involved heavily in this, i think it's Neil Sloane.

DOT PRODUCT

It is worth noting that the dot product does not always correspond to the inner product. This only happens when there is mutually orthogonal unit vectors as the frame of reference.

If you are using some different method, then one needs to drop a matrix into the inner product, as S_ij v_i w_j. This is the matrix-dot product.

One can see that for any axis, that 2 v*w must always be less than v^2+w^2 (since (v-w)^2 is positive. Accordingly what we end up with is something that where in general, (v.w)(v.w)/(v.v)(w.w) tends to zero very quickly, as N goes to infinity.

HYPERBOLIC VECTORS

You can use vectors in hyperbolic space, it is just that they have different effects on equidistant lines. Been using it years.

Also, hyperbolic geometry can be made euclidean if one uses crooked rulers (eg ones measured in millimetres). Knowing this, you can bundle all that junk about hyperbolic trig out the door.

SPHERE VOLUME

The iteration-rule is correct, the general n-dimensional volume is akin to

(pi)^n / (n/2)!

which goes to (2.e.pi / n) ^ n/2 as n -> oo

e = 2.71828, pi = 3.14159, 2 = 2.0000,
it gives (17/N)**(N/2). For N=170, we see that this is ~ 1E-85.
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1562
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Infinite-dimensional space is known as "Hilbert space".
Kaku, in Hyperspace, p.105 says "an infinite number of dimensions does not seem to be physically possible". I wonder what his reason for saying that is. It leads one to ask what the highest number could be. String theorists seem to think it is the 9 or 10 or possibly 25 suggested in their theories (in which case, all but 3 are curled up to the Planck length).
I guess it has to do with the fact that infinity is not a real "number", but only a hypothetical concept, so I always wonder how you could have infinite dimensions. You could never really locate anything, because every point requires an infinite number of coordinates. No wonder spheres shrink down to zero, as you all have been saying!
I would think that just like polytopes, where the higher the number, the less difference is between them (a megagon and a megahenagon are basically the same, as is a gigagon, teragon, and a circle!) the same would be with dimensions. 3 and 4 are vastly different, as a triangle and square, but once you get up in number, you get the idea of an limitless number of dimensions, and each higher space is basically the same as the next one. One of the sites even says that the different polytopes lose their significance after about 10 dimensions.
Last edited by Eric B on Thu Oct 23, 2008 12:07 am, edited 2 times in total.
Eric B
Trionian

Posts: 77
Joined: Wed Jul 20, 2005 11:46 pm
Location: NYC

Well, the number of possible uniform polytopes goes always up - for a given N, the N-dimensional space contains:

2^(n-1)+2^(n/2)-1 simplex-derived figures (for even n)
2^(n-1)+2^((n-1)/2)-1 simplex-derived figures (for odd n)
2^n-1 measure/cross-derived figures
2^(n-2) demicube-derived figures

This not counting prisms. Also, there are various exceptions in lower-dimensional spaces, and there are additional "basic" polytopes up to 8D, but this should work for 9D+.

This leads to interesting question, though: How would an infinite-dimensional Coxeter-Dynkin graph look? For example, infisimplex could be written as
o-o-o-o-o-...
i.e. as a chain with definite beginning, but no end. Basic polytope would then have itself as a vertex figure, and it would have no well-defined facets (or would it?). The vertex figure of any of its rectates would be a prism: finite simplex X infinisimplex.

Would it be a chain
o-o-o-o-...---...-o-o-o-o
with infinite amount of members in between, then any rectate would still be finite simplex X infinisimplex, but the simplex itself would have a dual.

Third option,
o-o-o-o-o-...---...-o-o-o-o-...---...-o-o-o-o
has some nodes with infinite amount of both predeccessors and successors. However, this thing requires the existence of infinisimplex number 2, since rectates taken in one of such nodes are infinisimplex X infinisimplex

The fun is still starting, though. For example, the curvature of an uniform polytope (whether it's a true polytope or Euclidean/hyperbolic tiling) is generally divined by examining whether its vertex figure fits into a unit sphere. But what if the vertex figure equals to the original figure? Then, how can we determine the curvature?

Another thing we might try is to define the infinite-dimensional polytopes directly. The cube will then have vertices as all possible sign choices for infinite set of 1's. This is uncountable amount of vertices. Infinite cross polytope would have permutations of [+-1,0,0,0,0,...] as its vertices. That's countable amount. However, being dual of the cube, it would have uncountable set of faces.
Marek14
Tetronian

Posts: 933
Joined: Sat Jul 16, 2005 6:40 pm

The problem is not so much infinite space is the problem, but infinity itself is. These are really so many things, that calling them all infinity, is calling all non-infinite numbers by the name 'finity', and that because a triangle and a {1728000} have finity sides, they must be the same!

The horizons on H2 and E2 are the same, since they are formed by an intersection of a locally centered horocycle. The radii of H2 and E2 to the horizons must therefore stand as E2 = R, H2 = ln(R). Yet we refer to both as infinity.

So if we say that the horizon is countable ie alef0, what makes the radius of H2-space, which is ln(alef-0).

One ties oneself into all sorts of knots by trying to fit cantor-infinities (which are nonsense in the first instance) onto anything that is infinite.

For all we know, the radius of space might be some number that we can freely write: eg 369.0157.31E9.5007.9016. This is however, the same number of vertices that a cube-polytope has in 64 dimensions.

W
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1562
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### infinite regular polytopes

I'm new to the list, but have spent years studying higher-dimensional spaces. I've even tried to "construct" infinite-dimensional regular polytopes. Here's what I come up with:

1. The infinite-simplex, which include cells (also infinite-simplexes)meeting at 90 degrees. It looks like they could fill infinite space forming an infinite-honeycomb. It's infinite-volume is zero, asuming a finite edge. It's inradius is also zero.

2. The infinite measure polytope, which have a diagonal of infinite length, asuming a finite edge. In fact the circumradius is infinite. These too fill infinite space, creating a honeycomb.

3. The infinite cross polytope. It's cells (infinite-simplexes) meet at 180 degrees, thus the figure IS a honeycomb! (A COMPACT honeycomb?). It's infinite-volume is zero. It's inradius is also zero.

I wonder if it makes any difference if the infinite number of dimentions is even or odd...
bsaucer
Dionian

Posts: 22
Joined: Mon Nov 28, 2005 6:30 pm
Location: Mobile, AL

### Re: infinite regular polytopes

My approach to infinite-dimensional polytopes was more combinatoric:

Infinite simplex can be put in infinite-dimensional space with points that have all coordinates zero except for one that is +1. Assuming countable dimension, it has countable amount of vertices, edges, etc... countable amount of all finite elements. But it contains other infinite simplexes, and their amount is uncountable. However, the infinite simplexes with finite complement (i.e. those that take all but finite amount of dimensions) are countable again.

Infinite cube's vertices can have +1 or -1 as every vertex coordinate and their vertices are all combinations of these numbers - this means that the number of vertices is uncountable. Not sure where the number of sub-polytopes becomes countable - probably with finite-complement parts.

Infinite cross polytope is dual to the cube, and its vertices can be those with all coordinates zero except for one that is +1 or -1. The number of finite parts is once again countable, but it looks as number of infinite parts (including the faces, infinite simplexes), is uncountable.
Marek14
Tetronian

Posts: 933
Joined: Sat Jul 16, 2005 6:40 pm

I've only just noticed this thread, but I've been reading about infinite dimensions for a while. From what I've seen on the net, infinite dimensions are most often used to describe a function with a taylor series. The function is a vector, and each coefficient in the taylor series is a component of the vector.

Is there any meaningful way to assign a "norm" to these functions/vectors? Is there any meaningful definition of "dot product" that would work for these vectors such that when the number of non-zero coordinates are finite, it reduces to the familiar Euclidean dot product?

The dot product is usually a finite sum over the products of the components. But since we have an infinite number of components, you need to integrate over the product. So the dot product of f(x) and g(x) is integral( f(x)*g(x) dx). I think you can also have weighting functions or something, but I don't understand it myself.

Generally, a polytope is just a shape formed by lots of intersecting hyperplanes. A triangle is three lines, a tetrahedron is four planes, etc. In n dimensions, you need at least (n+1) intersecting hyperplanes to describe a polytope.

Now, a (n-1)D hyperplane in nD space can be defined by its normal vector, so an inf-D hyperplane is just a single vector, which is represented by a function. So all you need for an infinite dimensional polytope is an infinite sequence of functions.

Take the cube, as the simplest example. In 3D, we have six planes, with the following normal vectors:
(1,0,0) (-1,0,0) (0,1,0) (0,-1,0) (0,0,-1) (0,0,1)
If we intepret each vector as a set of coeffecients for a polynomial, we get this:
1, -1, x, -x, x^2, -x^2

If we extend this to infinite dimensions, we get:
1, -1, x, -x, x^2, -x^2, x^3, -x^3, ...

Can anyone work out what the infinite simplex would be?

PWrong
Pentonian

Posts: 1573
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

I think you're losing something in going from planes to polynomials...

You're only using the plane normals... not the plane offsets. As such, your plane with normal (1,0,0) and your plane with (-1,0,0) are just constant multiples of each other....

You could probably correct this by just adding the offset into the coordinate representation.... if the normal is (a,b,c) and the offset is d, then convert (d,a,b,c) to your polynomial.

So, your cube would then be: 1 + x, 1 - x, 1 + x<sup>2</sup>, 1 - x<sup>2</sup>, 1 + x<sup>3</sup>, 1 - x<sup>3</sup>.

Of course, your polynomials are n-dimensional whilst mine are (n+1)-dimensional.

Simpler might be using convex hull expressions. In that case, the infinite dimensional simplex is just the linear combinations of:
(1,0,0,0,...)
(0,1,0,0,...)
(0,0,1,0,...)
(0,0,0,1,...)
...
such that all coefficients are non-negative and they sum to one.
pat
Tetronian

Posts: 563
Joined: Tue Dec 02, 2003 5:30 pm
Location: Minneapolis, MN

The trouble with looking at infinite polytopes is that there's so many out there to meet. The largest practical dimension i played around with is 124 dimensions, believed to be the home of the extended polytope derived from the group 4f. It's a very efficient non-lattice packing of spheres.

I have been able to show in 120d that the leech-lattice is more dense than the semicubic product allows, and thus you can not treat the leech-lattice as a superdense cluster of 8192 semicubics.

There are supposed to be interesting tilings in some high dimensions, like 52 and 129 dimensions. Still, if i can read 'quaterions and octonions' clearly, i prolly would understand this :S

W
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1562
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Of course, I was restricting my list to regular polytopes, of which there are only three in any dimension higher than four.
bsaucer
Dionian

Posts: 22
Joined: Mon Nov 28, 2005 6:30 pm
Location: Mobile, AL

My new linear algebra textbook (Linear Algebra with Applications 3E, Otto Bretscher) has something to say about infinite dimensional (linear) space as well:

"Example 19 implies that the space P of all polynomials does not have a finite basis f_1, ... f_n.

Here we are faced with an issue that we did not encounter in Chapter 3, when studying R^n and its subspaces (they all have finite bases). This state of affairs calls for some new terminology.

Definition 4.1.8 Finite dimensional linear spaces

A linear space V is called finite dimensional if it has a (finite) basis f_1, ... f_n, so that we can define its dimension dim(V) = n... Otherwise, the space is called infinite dimensional.

As we have just seen, the space P of all polynomials is infinite dimensional (as was known to Peano in 1888)....

The basic theory of infinite dimensional spaces of functions was established by David Hilbert (1862-1943) and his student Erhard Schmidt (1876-1959), in the first decade of the twentieth century, based on their work on integral equations. A more general and axiomatic approach was presented by Stefan Banach (1892-1945) in his 1920 doctoral thesis. These topics (Hilbert spaces, Banach spaces) would be discussed in a course on Functional Analysis rather than linear algebra..."

And here is one of the possible homework questions:

"Here is an infinite-dimensional version of Euclidean space: In the space of all infinite sequences, consider the subspace l_2 of square-summable sequences [i.e., those sequences (x_1, x_2, ...) for which the infinite series (x_1)^2 + (x_2)^2 + ... converges]. For x and y in l_2, we define

|x| = sqrt((x_1)^2 + (x_2)^2 + ...)
x . y = (x_1)(y_1) + (x_2)(y_2) + ...

(Why does the series (x_1)(y_1) + (x_2)(y_2) + ... converge?)

a. Check that x = (1, 1/2, 1/4, 1/8, 1/16, ...) is in l_2, and find |x|. Recall the formula for the geometric series:
1 + a + a^2 + a^3 = 1/(1 - a), if -1 < a < 1.

b. Find the angle between (1, 0, 0, ...) and (1, 1/2, 1/4, 1/8, ...).

c. Give an example of a sequence (x_1, x_2, ...) that converges to 0 (i.e. lim_(n ---> infinity) x_n = 0) but does not belong to l_2.

d. Let L be the subspace of l_2 spanned by (1, 1/2, 1/4, 1/8, ...). Find the orthogonal projection of (1, 0, 0, ...) onto L.

The Hilbert space l_2 was initially used mostly in physics: Werner Heisenberg's formulation of quantum mechanics is in terms of l_2. Today, this space is used in many other applications, including economics. (See, for example, the work of the economist Andreu Mas-Colell of the University of Barcelona)."
jinydu
Tetronian

Posts: 721
Joined: Thu Jun 10, 2004 5:31 am

jinydu wrote:[...]"Here is an infinite-dimensional version of Euclidean space: In the space of all infinite sequences, consider the subspace l_2 of square-summable sequences [i.e., those sequences (x_1, x_2, ...) for which the infinite series (x_1)^2 + (x_2)^2 + ... converges]. [...]

This is easy enough; as long as the sum of squares converges, we can take the square root and call it the "norm".

The limitation with this approach is that you are restricted to a very small subset of infinite vectors (there are many more infinite vectors whose norms do not converge).

The challenge that I am trying to tackle is how to deal with infinite vectors whose norms do not converge. Obviously, for such vectors, values like the norm would not (generally) be a real number; but it is interesting to ponder whether it is possible to map the norm to a larger set, say, of the hyperreals, or some such set. If this could be done in a consistent way, we can actually compare arbitrary infinite vectors, and/or take infinite dot products without worrying whether or not the sum of products converge. Since the hyperreals are linearly-ordered, the resulting vector algebra would still be consistent.

Of course, the lingering question remains, suppose the dot product of two infinite vectors A and B is C, and suppose D is not a real number. By analogy with finite-dimensional geometry, we'd expect D/(|A||B|) to be equal to the cosine of the angle between A and B. Can we expect D/(|A||B|) to always be within the range [-1,1], such that we would get a meaningful angle?

Having said all that, here's a surmised possible way to work with infinite norms/dot products:

Let A and B be infinite-dimensional vectors. (We're restricting "infinite" to mean "countable", to keep things manageable.)

(1) We define |A| to mean the hypothetical square root of the sum of squares of each of A's components: |A|<sup>2</sup> = A<sub>1</sub><sup>2</sup> + A<sub>2</sub><sup>2</sup> + ... . (We don't care if this series diverges.)

(2) We can compare vectors with infinite norms by this circumlocution using quotients:
(a) If |A|/|B| diverges, or is greater than 1, then we say that A has a larger norm than B, and write |A| < |B|.
(b) If |A|/|B| converges to 1, then we say that A and B are of equal norms (have the same length), and write |A|=|B|.
(c) If |A|/|B| converges and is less than 1, then we say that A has a smaller norm than B, and write |A| > |B|.
(d) Now, since we haven't actually defined what |A| and |B| are (because our definition involves |A|<sup>2</sup>, and we don't know how to take the square root of a diverging series), we observe that in the finite case, (|A|/|B|)<sup>2</sup> = |A|<sup>2</sup>/|B|<sup>2</sup>, and hence, we revise the above definitions to read |A|<sup>2</sup>/|B|<sup>2</sup> instead. One would note that since 1<sup>2</sup> = 1, we need not change any other part of the definitions.

(3) Dot products. Since we can't directly deal with an infinite dot product, we will use the same circumlocution to define comparisons:
(a) We define the dot product of A and B to be the infinite series: A.B = A<sub>1</sub>B<sub>1</sub> + A<sub>2</sub>B<sub>2</sub> + ...
(b) Suppose we have two dot products, A.B and C.D. Again, if A.B/C.D converges to 1, we say that A.B = C.D.
(c) If A.B/C.D diverges, or converges to a number greater than 1, then we say that A.B > C.D.
(d) Similarly, if A.B/C.D converges to a number less than 1, then we say that A.B < C.D.

(4) Angles. We'd like to define the angle X between A and B as that number which satisfies cos X = A.B / |A||B|; but how can we prove that the right-hand side always converges, and if it does, how do we go about computing it? As a first step, we'd probably have to square the right-hand side, so that we deal in terms of |A|<sup>2</sup> and |B|<sup>2</sup>, which are well-defined. Unfortunately, a naive algebraic expansion of this expression leads to a very unwieldy double-sum which I don't know how to handle.

The upshot of all this, is that we are dealing with convergences and divergences of quotients of (more-or-less arbitrary) infinite series. Surely there is literature out in the field about this?
quickfur
Pentonian

Posts: 2200
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: Infinite-dimensional polytopes?

I would be more interested in Pn as n tends to infinity. Partly because I find projective spaces neater and partly because it is needed to perform certain calculations in string theory.

Axiomatically, projective spaces are the simplest to deal with. Traditionally we begin with a small number of dimensions and work upwards. I am wondering what would happen if you start with infinitely many and work down.

Characterising the topologies of infinite-dimensional polytopes would be a challenge - for starters each would have an infinite series of Betti numbers and torsion coefficients (things I struggle to understand in 4 dimensions).
steelpillow
Mononian

Posts: 6
Joined: Sat Jan 15, 2011 7:06 pm
Location: England

### Re: Infinite-dimensional polytopes?

Heh, I wrote this 3 years ago... since that time I've come to realize that infinite dimensional polytopes with geometric properties analogous to their finite-dimensional counterparts are impossible unless we also extend the scalar field to include infinite/infinitesimal quantities, or at least have some way of dealing with unbounded limits. Here are some of the difficulties one encounters:

- Restricting the set of vectors to those with a finite norm, say, eliminates full-dimensioned polytopes such as the unit hypercube.
- The infinite-dimensional unit hypercube has volume 1, but the radius of its vertices are infinite, being all permutations of sign of (1,1,1,1,...), which implies a norm of sqrt(1+1+1+...).
- Rotation is not closed in the space. For example, geometrically speaking it should be possible to rotate one corner of the unit hypercube (1,1,1,1,...) such that it is parallel to (1,0,0,0,...). This may be accomplished by a suitable infinite-dimensional rotation matrix, for example. However, the only non-zero coordinate of the result has an infinite value. So, in a sense, the line stretching from (-infinity,0,0,0,...) to (+infinity,0,0,0,...) is shorter than the line stretching from (-1,-1,-1,...) to (1,1,1,...), yet the latter is in the space, but the former isn't.
- Similarly, the inverse rotation matrix that rotates (1,0,0,0,...) to be parallel to (1,1,1,1,...), which geometrically speaking should exist, will transform it into the zero vector instead, because any non-zero vector parallel to (1,1,1,1,...) will have infinite norm. In other words, the only way for a unit hypercube to have a finite radius is for its vertices to have infinitesimal coordinates.

There are many other such difficulties. This indicates that either (1) infinite-dimensional polytopes do not behave like finite-dimensional ones, and therefore infinite-dimensional geometry has rather bizarre properties markedly different from finite-dimensional geometry, or (2) a full infinite-dimensional geometry is possible only if we extend the scalar field to include infinite and infinitesimal quantities.

Personally, I prefer (2), because (1) is in some sense incomplete, having such strange properties as having two mutually-exclusive classes of vectors (infinite-normed and finite-normed, that cannot be interconverted with a rotation), polytopes with infinite radius but finite volume, hyperspheres that have "holes" through which certain lines can pass through the center without intersecting the boundary (e.g., the set of points commensurate with (1,1,1,1,...)), and other such pathological properties.
quickfur
Pentonian

Posts: 2200
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: Infinite-dimensional polytopes?

After much more thought about this whole thing, I decided that, perhaps, we don't need an infinite number of dimensions. In a certain sense, it's "good enough" to choose N dimensions for a sufficiently large N. When N is small, the difference between N dimensions and N+1 dimensions is very significant; but when N is large (and by large I mean EXTREMELY large... in the sense I'll describe below), N-space and (N-1)-space behaves more-or-less the same.

For us poor low-dimensional human beings, it's too easy to count 1, 2, 3, and then just take that quantum leap into infinity, as though we had any idea what infinite actually means. In fact, I have this suspicion that for most people, their perception of infinity is actually far smaller than certain finite numbers.

What do I mean? If you are a child, you would "count to infinity" something along these lines: 1, 2, 3, ... infinity! If you had primary school education, your "count to infinity" might be more along the lines of: 1, 2, 3, 100, 200, 300, 1000, infinity! If you had high school education, your count might be: 1, 2, 3, 100, 1000, 1000000, infinity! If you had college education, your count might be: 1, 2, 3, 1000, 1000000, 1 billion, 1 trillion, 1 quadrillion, infinity! And if you had a graduate degree, your count might be, 1, 2, 3, 1000, 1 million, 1 billion, 1 googol (10^100), 1 googolplex (10^10^100), infinity!

But actually, these numbers (excepting the "infinity!"'s) are FAR FAR FAR FAR SMALLER than a LOT of finite numbers. As a first step, we might transcend the commoner's count to infinity as follows:

Start with a number like 1 million or 1 billion. While their value is huge, we can still easily write them down as only 7 or 10 digits. So the next "logical" step is a number that has so many digits that it's hard to write down: 1 googol (10^100) has 100 digits. While in theory it *can* be written down, it's too long to be feasible, and we adopt scientific notation to indicate how many digits it has. Well, the next step, 1 googolplex (10^10^100) is so huge, that the number of digits it has cannot be easily written down; the number of digits in its number of digits is 100. But this is still very small... for example, we can easily construct the number 10^10^10^10^10^10^10^10^10^10 (a power tower of 10 levels). This number is so huge that the number of digits in the number of digits in the number of digits in the number of digits in ... (10 levels) ... of its number of digits is 10. Think about that for a moment. That means, starting from a googolplex, we can construct a number of a googolplex digits. Then we construct another number with that many digits. Then we construct another number of that many digits. Then ... we have to repeat this 10 times in order to reach the value of that power tower. This has probably already exceeded most people's perception of infinity.

But we barely got started. The giveaway clue is that even though this number is already past most people's concept of infinity, we can easily write it down as a power tower of 10 levels. But why stop there? What if we now construct a number with a power tower, whose number of levels is that number? That is to say, our number is 10^10^10^.....^10 where the number of levels is 10^10^10^...^10 (10 levels high). This number surely has burst just about everyone's perception of infinity. In order to deal with such a number, we have to invent a new notation: let's denote a power tower of height H by N[4]H, where N is the "base". So 10[4]10 is 10^10^10^...^10 (10 levels high). Using this notation, the number we just made is 10[4](10[4]10). For convenience let's agree on right-to-left associativity so that we can simply write 10[4]10[4]10. Now, the reason I write [4] is because this notation is intended to be extensible: going backwards, we define 10[1]10 = 10+10, 10[2]10 = 10*10, and 10[3]10 = 10^10. Since exponentiation = repeated multiplication, and multiplication = repeated addition, we can build a hierarchy of operators this way, indexed by their level. Exponentiation is 3rd order, so power towers are 4th order. That's why we write [4].

But now, armed with such a notation, we are ready to actually take the first step towards large numbers. (Yes, this is only a first step. Get ready for the ride of your life!) That last number we just made was merely 10[4]10[4]10, which is a power tower of (power tower of 10 levels) levels deep. It's easy, using our new notation, to now take this resulting number, and use THAT as the number of levels in a power tower of the next number: 10[4]10[4]10[4]10. While it looks innocent in our new powerful notation, this beast is so huge, that the number of levels in its exponential tower cannot be described in digits, scientific notation, or even a power tower of a height that can be written down. The height of its power tower is a number whose power tower is so high that it also cannot be described in scientific notation (i.e., the number of digits in the height of its power tower is too large to feasibly describe in terms of digits). You have to go 4 levels deep before you can actually describe how many levels this exponential tower has. And remember, every level of the exponential tower means a number is so large that its number of digits can only be described by an exponential tower.

Boggled yet? We've barely begun! Even though we're already far outside the realm of perception, our new notation lets us step on the gas real hard and shoot wayyy farther now. Since [4] is repeated exponentiation, [5] is repeated [4]. That is, [5] is repeated (repeated exponentiation). So our last number is actually merely (har har) 10[5]4. We can easily construct 10[5]10, which is 10[4]10[4]...[4]10 (10 levels deep). But why stop there? We can easily go to 10[6]10, which is 10[5]10[5]10[5]...[5]10 (10 levels deep). But then we might as well go to 10[7]10, or, for that matter, 10[10]10.

By this point, even though 10[10]10 is finite, it might as well be infinity for all we can perceive. I mean, even 10[4]10 has already FAR exceed the number of particles in the universe, and 10[4]11 the number of states in the universe, but that is less than a speck compared to 10[10]10. I mean, 10[10]9 is so miniscule compared to 10[10]10, that for all intents and purposes, 10[10]9 might as well be equal to 0. Arithmetic involving these things is *entirely* dominated by the second operand to the [10] operator. For example, even though 2*10[10]10 is in theory twice as large as 10[10]10, for all intents and purposes they are both numbers so large that we can hardly tell the difference between them. Even though 10[10]9 is just as unimaginably huge, we find that 10[10]9 + 10[10]10 is practically equal to 10[10]10, because the former is so small compared to the latter that it's just roundoff error. For that matter, 10[10]9 * 10[10]10 might as well be equal to 10[10]10, simply because 10[10]11 is so unimaginably larger that anything less than or equal to 10[10]10 multiplied by itself is still practically zero by comparison. In fact, 10[10]10 ^ 10[10]10 is still pretty much "approximately equal to" 10[10]10, in comparison with 10[10]11. That is to say, exponentiating 10[10]10 to itself has barely even budged in the direction of 10[10]11. So we can see that at this level, even exponentiation itself barely makes a difference, needless to say any lower operations like * or +. They might as well be no-op for all we care.

But we haven't even gotten really started yet!!

The hierarchy of [M] operators is called the Grzegorczyk hierarchy. They belong to a class of functions called the "primitive recursive" functions, which are a relatively slow-growing class of functions that can be computed by a programming language with a limited subset of recursion. It's a bit irrelevant to explain this statement here, but the point is that they are actually considered slow-growing. Yes, slow!! And you thought we've already gone beyond the Land of Oz. Well, let's show you something that will blow your mind:

Why restrict ourselves to small numbers with these [M] operators? We can, for example, write 3[64]3, which if I'm not mistaken is in the vicinity of Graham's Number, an insanely huge number that actually got used in a real math proof. But this is just a puny non-attempt at going higher. Why use poor numbers with only a few digits here? Let's recurse the [M] operator! For example, take this number: 10[10[10]10]10. That is, it's 10[M]10 where M = 10[10]10. Now we're talking!

This last number that we just wrote has just shot past the famed Ackermann function, which one inevitably encounters in the quest for large numbers. But why stop at one nesting level? What about 10[10[10[10]10]10]10? That's taking the number we last made, and sticking that inside our [M] operator. Now, not only the number of digits or the number of levels in the power towers are undescribably huge, even the order of the operation we're using is so huge it cannot be feasibly described in any common notation.

But have no fear! That was only 3 levels of nestings inside the [M] operator. What if the number of nestings is so huge it can only be described by another such number? For example, what is the number of levels of nestings is 10[10[10[... (10 levels deep) ... ]10]10]10?

We can keep going, of course. At which point we'd inevitably encounter Jonathan Bowers' Array Notation. Now we're talking real large numbers! Look up Jonathan Bowers' "infinity scrapers" sometime. His names for these things are quite whimsical, but there is real mathematical weight to these things. They are HUGE!!!! They are so huge they explode your perception of infinity several (insert unimaginably large number here) times, and then some!

OK, the whole point of this is to articulate my point, that we don't need to jump to an infinite-dimensional space so quickly. Simply pick the number of dimensions N to be a suitably-huge number, such as one of Jonathan Bowers' pentational arrays, for example. The number of coordinates a vector in such a space would have is, for all practical purposes, infinite. (Even though it is really still finite. I hope you now have a better idea of how huge a finite number can be!) You can enumerate its elements for several lifetimes of the universe, and you wouldn't even be close to getting started. An N-hypercube in such a space would have (N-1)-hypercubes as its facets, of course, but for all practical purposes, they might as well be N-hypercubes, since the difference between N and N-1 is not even worth an afterthought when N is so huge. So you see that, already, N-space is reflecting the behaviour of infinite-dimensional space pretty closely, even though N is still finite.

We can then do some actual geometry, and have the confidence that the results would reflect infinite-dimensional space very closely; in fact, so closely that we almost never see the difference. Such unmanageable things in an actual infinite-dimensional space as the radius of the N-hypercube's vertices are no longer a problem, since the radius would simply be sqrt(N), a finite number, which, given the magnitude of N, can be regarded pretty much as approximately equal to N. Taking dot products, measuring angles, etc., likewise would not run into the problem of actual infinities; we simply work out using finite algebra the numbers concerned, expressed in terms of N, and in the final result, simplify anything containing N to N itself (since pretty much NO geometrical operation will even begin to approach the magnitude of operations that you need to significantly affect the value of N), and anything containing N in the denominator to be essentially equal to zero (since practically NO ordinary geometrical operation would be able to result in a numerator large enough to have a meaningful quotient after being divided by the insanely huge N, excepting expressions involving N itself).

In other words, even though N is a finite number, it essentially behaves like an infinite number, so in a sense we don't even need to specify its value, simply that it's a suitably large finite number. We can just treat it as a special "almost infinite" number that for all practical intents and purposes behave like an infinite number.

With this technique, then, we can actually do some geometry that, for all practical intents and purposes, is "infinite"-dimensional, except without the actual infinities that cause us grief when we draw analogies from the lower dimensions.
quickfur
Pentonian

Posts: 2200
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: Infinite-dimensional polytopes?

This is pretty much the same argument I used in explaining why a circle was effectively a ∞-gon. When you have a polygon, fix the radius, and increase the number of sides (in which case the length of the sides descreses), and as soon as the sides decrease to the point they are too small to be seen by the eye, you basically have a perfect circle. If you fix the length of sides, the radius increases, and the hull flattens out.
Now, you increase to those numbers you just discussed, it is still going to be a perfect circle, and the projection with the fixed length of sides now flattens out to a perfect straight line tiled with the line segments--Basically, the so-called "apeirogon"; though n is still finite.

So it seems like the property of nature is that as you increase any value towards infinity, the distinction between separate numbers fades away. It becomes totally abstract, as no one can do anything concrete with such numbers anyway.
Like is 10[10[10[10]10]10]10 divisible by 3? Well, as a power of 10, n-1 is always what is divisible by 3, but what really is n-1 to a number such as that?

(This also reminds me of some sermons I hear from time to time, trying to remind us how "long" eternity (infinite time) really is).

Actually, as for dimensions, I had read somewhere that after 10 dimensions, the distinction between n dimensions becomes much less relevant.
Eric B
Trionian

Posts: 77
Joined: Wed Jul 20, 2005 11:46 pm
Location: NYC

### Re: Infinite-dimensional polytopes?

Well, a distinction must still be drawn between a very large finite number (as an approximation to infinity) and an actual infinite quantity. While in general conclusions drawn using very large numbers match the limiting behaviour at infinity, this is not always true. A classic example is the alternating function: f(x) = { 0 if odd, 1 if even }. Choose x to be a very large number. Evaluate f(x). You have a 50% chance of getting a 1, and 50% chance of getting 0. What is the value of f(x) when x is infinite? It is unprovable. In fact, we customarily treat such cases as "undefined".

Of course, this is a contrived example, but when using large number approximations, you may run into cases where something completely non-obvious may give rise to non-limiting behaviour, and you risk ending up with completely wrong results. So we should be very careful when using very large numbers as approximations of infinity, because they are just that: mere approximations. They are useful for getting around nasty problems when dealing with actual infinities, but one always needs to take care that one isn't obtaining invalid results due to carelessness.
quickfur
Pentonian

Posts: 2200
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: Infinite-dimensional polytopes?

OK, I don't understand what f is. (I'm sure had alternating functions in school, but that was a long time ago, and I would have forgotten). Is f something you're dividing x by? Or is it determining the divisibility of x (with 1 or 0 being a "remainder").
Eric B
Trionian

Posts: 77
Joined: Wed Jul 20, 2005 11:46 pm
Location: NYC

### Re: Infinite-dimensional polytopes?

f is just a function whose value is 1 when x is even, and 0 when x is odd.

So f(1)=0, f(3)=0, f(5)=0, but f(2)=1, f(4)=1, etc..
quickfur
Pentonian

Posts: 2200
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: infinite regular polytopes

bsaucer wrote:I'm new to the list, but have spent years studying higher-dimensional spaces. I've even tried to "construct" infinite-dimensional regular polytopes. Here's what I come up with:

1. The infinite-simplex, which include cells (also infinite-simplexes)meeting at 90 degrees. It looks like they could fill infinite space forming an infinite-honeycomb. It's infinite-volume is zero, asuming a finite edge. It's inradius is also zero.

2. The infinite measure polytope, which have a diagonal of infinite length, asuming a finite edge. In fact the circumradius is infinite. These too fill infinite space, creating a honeycomb.

3. The infinite cross polytope. It's cells (infinite-simplexes) meet at 180 degrees, thus the figure IS a honeycomb! (A COMPACT honeycomb?). It's infinite-volume is zero. It's inradius is also zero.

I wonder if it makes any difference if the infinite number of dimentions is even or odd...

Considering the statement above that user "bsaucer" wrote in November 2005, I remember I always thought about how would be the infinite-dimensional polytopes. The first thing I realize is that there were no more 3 regular polytopes, but 2.
Once I read an "informal" thoery on how to create a regular polytope. It was something like this:
- Join together 3 or more poly-(n-1)-topes. If their difacetal angles sum less than 2*Pi radians, then they can be joined in a higher dimension. E.g. The sum of 3 triangle digonal angles is Pi, so one can join these 3 triangles in the third dimension. 3 hexagon digonal angles, on the other hand, sum 2*Pi, so one can only create a honeycomb. The sum of 4 pentagonal digonal angles is larger than 2*Pi (12/5*Pi), then only a hyperbolic honeycomb can be created this way.

In addition to this "informal" theory, I would like to remind three things:
1. In 5 or higher dimensions, regular polytopes are only composed of either (n-1)-simplices or (n-1)-measure polytopes;
2. n-simplex difacetal angle is arccos(1/n);
3. n-measure polytope difacetal angle is always Pi/2.

When n gets to infinity, something weird happens.
n-simplex difacetal angle gets to arccos(0) = Pi/2. Thus, the sum of 3 "infinity"-simplex difacetal angles is yet lower than 2*Pi (it is 3*Pi/2), but, on the other hand, the sum of 4 "infinity"-simplex difacetal angles is exactly 2*Pi.
One could be able to build an infinite-dimensional simplex, but the infinite-dimensional cross-polytope would be honeycomb and not a polytope.

Additionaly, 3 "infinite"-measure polytopes can be joined together as the sum of their difacetal angles is lower than 2*Pi (it is 3*Pi/2).

Altogether, it is interesting to note that the dual of the "infinite"-measure polytope (which is a polytope) is the "infinite"-cross-polytope (which is a honeycomb).
How can a honeycomb be the dual of a polytope?
hcesarcastro
Mononian

Posts: 6
Joined: Fri Apr 06, 2012 8:01 pm

### Re: Infinite-dimensional polytopes?

One should always approach infinity with considerable caution. One element is that infinity does not just 'happen', but relies on the diffusion of teelons.

A power of 10 is never a multiple of 3, because the construction does not allow it. Here, 'power of 10' is a definite route, but one finds it ever harder to follow this route in a non-decimal base, like 120. Likewise, 'power of 120' is a different route which is difficult to follow in decimal. That is, very large powers of 120 require some work to get a decimal expression to. None the same, we can say every power of 120 leaves a remainder of 1 when divided by 7 or 17.

A number that has a specific construction is not a general large number, and one must understand that pretending that the properties of a general number apply to it. You simply can not construct a general number. You could construct a number in the form of 120 ** n * pi, which has no specific integral construction, but even these are 'specific constructions', which might lead to incorrect assumptions down the track.

The simplex and the cross-polytope have angles that assymtope on 90 and 180 degrees. However, these asymtopes kick in long before one reach very large numbers. For example, the inradius of a simplex is frightfully small, and despite the large size of these polytopes, the path to rotate over 180 degrees is abidingly finite. That is, on a cross polytope, the distance to travel to be at something like 90 degrees, is not all that far.

The volume of a simplex, of edge sqrt(2), is sqrt(n+1) on the tegmic scale. The volume of a cross-polytope of diagonal sqrt(2), is 1 on this same scale. Since we have the distance between the centre and face-centre of a simplex of edge sqrt(n) is sqrt(n+1)-sqrt(n), which is pretty small. One might travel over n such faces to get from one side to the other, so we get n (sqrt(n+1) - sqrt(n-1)).

I never had any great love for 'large infinity' since it leads to wild and often stupid conclusions. One is better off taking smaller infinities, like 71, to get closer to the truth.
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1562
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Infinite-dimensional polytopes?

I understand that working with infinity is a little bit risky.

I realized that when n gets to infinity, the simplex difacetal angle gets to Pi/2- (the "minus sign" meaning it gets from the left). Therefore, the sum of 4 simplex difacetal angles gets to 2*Pi from the left. The "infinity"-cross-polytope lies on an edgy definition. It is somewhere between a polytope and a honeycomb, and this edgy shape is what fascinates me in infinite-dimension polytopes.
hcesarcastro
Mononian

Posts: 6
Joined: Fri Apr 06, 2012 8:01 pm