Negative Dimensions

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Voekoevaka
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Negative Dimensions

Postby Voekoevaka » Fri Jun 28, 2013 7:49 pm UTC

I am actually studying Algebraic Number Theory, and when I was messing up with the discrete additive subgroups of ℝn, I had an idea.

I'll recall the "non-formal" definition of Hausdorff-Besicovitch Dimension (that doesn't work in all cases) :

Take a set A, and apply on it an homothetic transformation with a scale k (u(A)). We define n as the quotient between the measure of u(A) and A.
And we define the number ln(n)/ln(k) as the Hausdorff-Besicovitch Dimension of A.

Example 1 : A is a point, and whatever the homotetic transformation we apply to it, it gives a point, that contains only one times itself, so the dimension of a point is ln(1)/ln(whatever)=0.

Example 2 : A is a line. If we multiply the lengths by a factor 2, we get a new lint that can contain two times itself, its dimension is ln(2)/ln(2)=1.

It works also with a square (ln(4)/ln(2)=2), with a cube, with the Kock snowflake, and with the Menger sponge.

So I tried to apply it on a discrete group of ℝ (a group of number of the form xn, where x is a fixed number of ℝ* and n∈ℤ) and we take a definition of the measure of the discrete sets. With a transformation of factor 2, it gives the sets of the 2xn. And the original set can contain two times the new set, so its dimension should be : ln(1/2)/ln(2)=-1 !
So I thought that negative dimensions could be the dimensions of countable infinite sets.

Another example (very strange) : consider a set made of the union of an infinity of parallel lines (that are equidistant), like this :

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Double its size :

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We can cut now the red part, and paste it like this :

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So, doubling its size doens't change its measure, so its dimension is ln(1)=ln(2)=0.

What do you think about this ? Could a solid definition of negative dimensions can be done ?
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Re: Negative Dimensions

Postby Forest Goose » Sat Jun 29, 2013 9:20 am UTC

I don't understand why a countable set of points doesn't end up with dimension 0, what do you refer to when you mention definition of measure of discrete sets? Also, what you're talking about looks a lot like what Wikipedia describes as the Fractal Dimension, is this accurate? It's been a really long time since I've read anything on Fractal Geometry, my apologies if I'm a bit rusty.
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Re: Negative Dimensions

Postby antonfire » Sun Jun 30, 2013 5:10 am UTC

The basic observation here is that if you scale up the integers by a factor of 2, then in some sense you get a set that's twice as small as the integers. So from that point of view Z has dimension -1.

It's a jump to say that every countable set has dimension -1, though. For instance, you should say that Z2, has dimension -2, since when you scale it up by a factor of 2 you get a set that's 4 times as small. If you scale up Q by a factor of 2, you just get Q again, so it should have dimension 0. If you scale up Q n [0,1) by a factor of 2, you get two copies of the original, so it should have dimension 1. And of course the countable "boundary" of the Cantor set has dimension log(2)/log(3).

The fact that Z crossed with a line segment (your "very strange set") has dimension 0 is not surprising. A line segment has dimension 1 and Z has dimension -1, so it makes sense that their product would have dimension 0.

All of this sort of makes sense, but you have some problems. For instance, you get that R and Q are both 0-dimensional. And what about the dyadic rationals? If you scale them up by a factor of 2, you get the original set, but if you scale them up by a factor of 3, you get something "three times as small"! You also have some nasty technicalities since we're no longer considering countable sets to be negligible. For instance, you can't just "cut the red part and paste it" like you said unless all those intervals are half-open.

Basically, in order to make the basic observation at the beginning, you have to drop the notion that measures should be countably additive. This takes away some pretty important measure-theoretic tools, which might make this whole notion pretty limited in the end. But I still think it's a good observation morally speaking even if it doesn't lead to a nice general theory.
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Re: Negative Dimensions

Postby arbiteroftruth » Tue Jul 02, 2013 5:30 pm UTC

For situations like the dyadic rationals, you can use a limiting argument. Suppose you scale the dyadic rationals by some integer n*2m. All the powers of 2 have no effect on the set, so the resulting set is "n times as small" as the original. This gives a dimensionality of ln(1/n)/ln(n*2m). But as n approaches infinity, the calculated dimensionality approaches -1, regardless of the value of m. So if we wish to assign a single number to the dimensionality of the set, the most sensible value is -1. By similar logic, we should be able to say the same about any subset of rationals defined by a restriction on the prime factors in the denominator. The scaling factor becomes n*qm, and the dimensionality becomes ln(1/n)/ln(n*qm), which still has a limit of -1.

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Re: Negative Dimensions

Postby Qaanol » Tue Jul 02, 2013 11:03 pm UTC

I think sets of the form n for a fixed n>0 are dim=0 here, because for any r>0, we have

r·nN = nN+logn(r)

which is tantamount to a horizontal shift by logn(r), and exactly equal to the original set when r is a power of n. So a multiple doesn't change the size.
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Re: Negative Dimensions

Postby skeptical scientist » Sat Jul 06, 2013 7:01 pm UTC

Another interesting example like those in the original post would be a tesselated fractal, like this Sierpinski triangle tesselation. If you scale that set by a factor of 1/3, that gives the same result as filling in all of the largest-sized triangular holes in the original set with a Sierpinski's triangle. Since each basic triangle-shaped region in the original tesselation starts by containing 3 Sierpinski triangles that size, and ends up with 4, you could say that scaling by a factor of 1/3 increases the content by a factor of 4/3. So by the logic of the original post, the "dimension" would be log(4/3)/log(1/3). Voila, negative fractional dimension. Although I don't think it makes sense to call this dimension (more on that below).

* * *

I think this is just another of the many examples when notions of size start behaving weirdly when size is infinite.

In this particular instance, what is going on is that all of the sets in your examples (and mine) have infinite d-dimensional Hausdorff measure, where d is the Hausdorff dimension of the set. If you think about it, that's exactly when you might expect the informal definition of dimension you are using to break down. The reason it works when Hausdorff measure is finite (and nonzero) is that each an d-dimensional object scaled down by a factor 1/k has measure scaled down by (1/k)d. If the original measure was finite and nonzero, that means it should take exactly n=kd If these objects to equal the content of the original object. So if you know n and k, you can use this to compute d. But if the original measure is infinite or zero, this doesn't work.
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Re: Negative Dimensions

Postby Voekoevaka » Mon Jul 08, 2013 12:00 pm UTC

Thanks for your replies, they all made me think about.

@Forest Goose :

Hausdorff-Besicovitch Dimension is the same tool as Fractal Dimension. About how to measure the sets with a countable infinity of points, I had an idea, and I will present it at the end of my post.

@antonfire :

antonfire wrote:The fact that Z crossed with a line segment (your "very strange set") has dimension 0 is not surprising. A line segment has dimension 1 and Z has dimension -1, so it makes sense that their product would have dimension 0.


It is just a bit surprising when you are not introduced to the concept of negative dimensions. But yes, reasoning with roduct of sets is clever here.

antonfire wrote:And what about the dyadic rationals? If you scale them up by a factor of 2, you get the original set, but if you scale them up by a factor of 3, you get something "three times as small"!

Maybe we can make a definition of the dimension of such a set by integrating the relative size of the set by the different factors. You'll get a divergent integral. But I think this is not a problem : I've made a thread about regularization of divergent series, it can be applied for divergent integrals

@arbiteroftruth :
I think what you said is helpful, but I think the dimension should be zero. But also we should discuss about why taking the 2n scales, and not the others ?

@skeptical scientist :

I have read one day an article aboute "pseudo fractals" that have their dimension changing according to how much you have zoomen in. In the case of your set, the dimension should be log(4)/log(2)=2 (because if you take the image twice larger, you get a figure with four times more triangles), but, by zooming, you found the dimension of the Sierpinski triangle, log(3)/log(2). So there are two asymptotes for the dimension of your figure.

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Here is my method for measuring countable infinite sets : if you take the set ℤ, and you multiply its size by two, you get a set that is "virtually" twice smaller, but it can be put in bijection with the original set, so it have the same number of points.
My idea is not to count the number of points, but the density of points, by placing the set into ℝn. To make it, you have to consider the set sequences of convex open sets of ℝn Ui, where Uj⊂Ui if j>i, and where lim(Ui, i→∞)=ℝn. We call A the set we want to measure, and we consider the sequences of μ(A∩Ui) and μ(u(A)∩Ui), where μ is the measure of the bounded set and u an homothetic transformation with scale k.
So the dimension of A should be log(lim(μ(u(A)∩Ui)/μ(A∩Ui), i→∞))/log(k).
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Re: Negative Dimensions

Postby skeptical scientist » Mon Jul 08, 2013 3:43 pm UTC

Voekoevaka wrote:I have read one day an article aboute "pseudo fractals" that have their dimension changing according to how much you have zoomen in.

I have never heard of "pseudo fractals" and googling them doesn't turn up anything. I'm not quite sure what you mean about dimension changing according to how much you zoom in. The Hausdorff dimension won't change, because the definition does not depend on a "zoom level" at all. But perhaps you mean a set whose Hausdorff dimension is smaller than its packing dimension? The usual examples of such sets do have behavior which can be thought of in some sense as dimension changing as you zoom in (although what's changing isn't really dimension, but an approximation to dimension which is valid in the limit). See, for example, the construction here.

In the case of your set, the dimension should be log(4)/log(2)=2 (because if you take the image twice larger, you get a figure with four times more triangles), but, by zooming, you found the dimension of the Sierpinski triangle, log(3)/log(2). So there are two asymptotes for the dimension of your figure.

There was a mistake in my calculation before. Scaling by a factor of 1/2 increases the content by a factor of 4/3, not 1/3 as I said. Oops. The dimension should be log(4/3)/log(1/2). This is equal to log(3)/log(2) - 2, which is negative.

However, I'm not seeing how you are getting either of the values log(3)/log(2) or log(4)/log(2)=2. Perhaps you misunderstood the set I was describing? Take an infinite tessalation of the plane by triangles, as shown:
Image
Now, replace each triangle by a copy of the Sierpinski triangle of the same size. Call this set A. A has infinite extent, but lots of holes. If you make the set A "twice larger", zooming by a factor of 2 (i.e. take the set A' = {(2x,2y) : (x,y) ∈ A}), then the resulting set is actually a strict subset of A, with more holes. In fact, in every region where A had 4 triangles joining to form a single larger triangle, A' will have 3. So in a sense, the resulting set is 3/4 as big. Calculating dimension this way we get the exact same result as before: dim(A) = log(3/4)/log(2) = (log(3) - 2 log(2))/log(2) = log(3)/log(2) - 2.

Of course, this is not the true Hausdorff dimension of A, which is simply log(3)/log(2), same as the usual Sierpinski triangle.
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Re: Negative Dimensions

Postby Voekoevaka » Tue Jul 09, 2013 3:36 pm UTC

OK, I have just misunderstood the aim of your post. Thanks for putting me in the right direction.

skeptical scientist wrote:There was a mistake in my calculation before. Scaling by a factor of 1/2 increases the content by a factor of 4/3, not 1/3 as I said. Oops. The dimension should be log(4/3)/log(1/2). This is equal to log(3)/log(2) - 2, which is negative.

I see no problems here, as your set is une product of a set of dimension -2 and a set of dimention log(3)/log(2). The dimension of a product of sets is the sum of the dimensions of the factor sets.
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Re: Negative Dimensions

Postby arbiteroftruth » Thu Aug 01, 2013 5:18 pm UTC

I wonder if this scheme for defining dimensions of countable sets might be useful for expanding measure theory. Consider this: replace the property of countable additivity with the property of additivity as long as the result is the same dimensionality as the operands. This way, you still get things like intervals of 1, 1/2, 1/4, and so on summing to an interval of 2, because all the objects involved are 1-dimensional. But adding up finite sets of points for each rational number until you fill the rational unit interval doesn't work because the finite sets of points are 0-dimensional objects while the resulting rational interval is 1-dimensional.

Applying the same reasoning allows you to start defining, for example, infinite uniform probabilities on integers. The measure of the even integers relative to the integers would be 1/2, the measure of integers congruent to 0mod4 would be 1/4, and so on, and you could similarly add up infinitely many of these sets of linearly-spaced points and say that the result has measure equal to the integers, because all the objects involved are -1 dimensions. Meanwhile the measure of a single point would be 0, and additivity fails not because countably-many summands are forbidden, but because a single point is a 0-dimensional object and the result is -1-dimensional.


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