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Thanks to the gamma function the formula for the surface of a unit http://mathworld.wolfram.com/Hypersphere.html $$ S(n) = \frac{2 \pi^{n/2}}{\Gamma(n/2)} $$ allows to calculate the surface of a hypersphere of non-integer dimensions. I wanted to know, what is the number of dimensions I need, so that the surface of a n-sphere (with radius 1) equals the area of a square (with "radius" 1), which means solving the equation $$ 4 = \frac{2 \pi^{n/2}}{\Gamma(n/2)} $$ for $n$. Since $S(n)$ has a maximum at $n=7.256...$ one get't two positive solutions: $$ n=1.534...\\ n=15.86... $$ (see https://www.wolframalpha.com/input/?i=2Pi%5E(n%2F2)%2FGamma(n%2F2)+%3D%3D+4).

Now my questions is: Since the number of dimensions of the n-sphere from this equation is a non-integer, does that mean such a sphere would be a fractal? If so, is it possible to construct a n-sphere with 1.534 dimensions somehow and draw it?

asmaier
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  • See: http://arxiv.org/pdf/physics/0001035.pdf – Jacob May 13 '16 at 13:47
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    @Jacob Can you explain a bit why you think this paper is relevant to the question? – asmaier May 14 '16 at 08:46
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    This paper talks about the extension of time and space to fractional dimensions which may be of use to you. However, fractal dimensions are completely different from spatial dimensions so I believe the answer to your question is no. I've tried to find information on fractional spatial dimensions for quite a while but to no avail. It appears that there is no established concept of fractional spatial dimensions as of now. – Jacob May 16 '16 at 17:19
  • @Jacob Thanks for your comment. Are fractional spatial dimensions maybe related to https://en.wikipedia.org/wiki/Continuous_geometry ? – asmaier May 16 '16 at 20:53
  • I asked a tangentially related question here that has some commentary about continuous geometry and its relation to the problem at hand. – Zach466920 May 17 '16 at 02:35
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    This question is not really well-posed. On one hand, for each $\alpha$ and each $n\ge \alpha$ there are fractal topological $n$-dimensional spheres in $R^N$ whose Hausdorff dimension is $\alpha$. By rescaling such a topological shpere, one can make its $\alpha$-dimensional area to be any positive real number. However, the formula you started with is for the areas of spheres of the unit radius. What does it mean for a fractal sphere to be of the unit radius is totally unclear. Lastly, you have to decide on the definition of a fractal. One viewpoint is that a metric space is fractal if... – Moishe Kohan May 17 '16 at 19:57
  • ... its Hausdorff dimension is strictly larger than its topological dimension. If this is all what you are asking about then indeed a sphere equipped with a metric whose Hausdorff dimension is not an integer, then such a sphere is indeed fractal. But this question is essentially void of the mathematical content, it is just the matter of semantics. – Moishe Kohan May 17 '16 at 20:00
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    @studiosus Your objection seems to be void of content as well. You seem to have an annoyance with the fact the OP hasn't properly defined what metric to use or how to scale the objects involved. However, the question is about how to construct a n-sphere with fractional $n$. If the OP knew the answer to those other questions, there'd be no problem with answering the main question. Essentially, all you have done is broken the one main question into its component parts and then complained that if the component parts aren't answered the main question is mathematically void... – Zach466920 May 18 '16 at 16:21

2 Answers2

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Since no one has posted anything yet, I'll at least give you something to think about.

The d-dimensional generalization of the Euclidean length formula is,

$$(1) \quad L=\sqrt{\sum_{n=1}^d x^2_n}$$

For fractional dimensions, we need to be able to evaluate the quantities inside the square root in a consistent manner.

In the pleasant case that $x_i=x$, we have,

$$(2) \quad \sum_{n=1}^d x^2_n=d \cdot x^2$$

If we assume $(2)$ holds for non-integer $d$ then we can substitute $(2)$ into $(1)$ and obtain,

$$(3) \quad L=|x| \cdot \sqrt{d}$$

Now, for a d-dimensional sphere, the points that make up the object are found by finding the set of all spatial points such that $(1)$ holds. Here, we are solving a subset of that problem. Namely, we are looking for spatial points, such that the coordinates can be permuted and still satisfy $(1)$.

Solving for $x$, we obtain,

$$(4) \quad |x|=\cfrac{L}{\sqrt{d}}$$ $$\Rightarrow x=\pm \cfrac{L}{\sqrt{d}}$$

For a unit sphere, $L=1$, and $d=1.534$, we have,

$$x=\pm 0.80739...$$

We also have the scenario where $x_i=L$ and $x_j=0$ for $j \not =i$. In this case also we have the conditions of $(1)$ satisfied still without having fully defined what coordinates are in fractional dimensions.

What does any of that mean? IDK...but with appropriate axioms there's nothing that prevents interpretation.

Extra Bit: I should mention that if a proper formalism was developed, we'd be seeing these "fractals" in their natural habitat. There's no reason to assume that they'd look like typical fractals, if viewed from this vantage point.

Zach466920
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    I don't really understand the meaning of your $x$. In e.g. two dimensions $x=1/\sqrt{2}$. But what does that mean? – asmaier May 21 '16 at 20:37
  • @asmaier Idk...I even say that in the post. For 2-d the interpretation is straightforward. I've worked off and on a theory for a couple years now where dimensionality is built up from "micro" dimensions. Basically, each typical degree of freedom inherent to integer dimensions is treated as the union of a infinite number of orthogonal vectors. Intuitive? Idk...I haven't worked it all out. I'm lazy, it's not sure the theory would work out, and the problem is difficult. – Zach466920 May 21 '16 at 20:55
  • I think your $x$ is basically just the ratio of the side vs. the diagonal of a d-dimensional unit cube. This is a characteristic number for a d-dimensional space (e.g. for 2d it is $\frac{1}{\sqrt{2}}$, for 3d it is $\frac{1}{\sqrt{3}}$). As you showed, one can generalize this number to non-integer dimension and you calculated this number for 1.534-dimensional space. It doesn't really answer the question, but it gives a better intuition about what non-integer dimensions could mean. – asmaier May 23 '16 at 18:20
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As I remember works of Benua Mandelbrot, the fractal object must have not only the non-integer dimension but also the property of the self-similarity. For example, any part of a straight line segment also is a line segment, so in this case we have the self-similarity but do not have the non-integer dimension. The hypersphere in a space with the non-integer dimension can be a fractal if it is self-similar, but I doubt that is a fact.

Peter Nazarenko
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