The volume of an Euclidian sphere in $n$-dimensions with radius $r$ equals $$V_n=r^n \frac{\pi^{n/2}}{\Gamma(\frac{n}{2}+1)}.$$ This formula is valid for $n\in\mathbb{N_0}$. Now I am asking myself if I can extend the definition to the rational numbers. Let's define $p=1/2$ and let $r=1$. The formula above, regardless of what it says about spheres, can be evaluated to $$V_{1/2} = 1^{1/2}\frac{\pi^{1/4}}{\Gamma(\frac{1}{4}+1)}=\frac{4\sqrt[4]\pi}{\sqrt{2 \varpi\,\sqrt{2 \pi}}},$$ where $\varpi$ is the lemniscatic constant. (It is equal to $\pi\cdot G$, $G$ is the Gauss constant). So let's assume that the volume of a $1/2$-dimensional unit sphere is equal to the expression above. Keep all of this in mind I will come back to it.

To measure dimensions of "objects" (sorry for the non-rigorosity) we can use the Hausdorff dimension. The Hausdorff dimension of a circle is two, the Hausdorff dimension of a cube is three etc., you get it. The nice thing about the Hausdorff dimension is, that it also can measure non-integer dimensions, for example the Koch snowflake that has the dimension $\frac{\log 4}{\log 3}$. Assume we can define a $1/2$-dimensional sphere with the Hausdorff dimension. With enough axioms we can then find properties of this sphere. So here is my question:

Can we find the volume of a $1/2$-dimensional sphere by using the Hausdorff dimension, and if yes, is it equal to the value $V_{1/2}$ we arrived at earlier?

  • 3
    $\begingroup$ How do you construct a fractional dimensional (hausdorff) sphere? $\endgroup$ Feb 23, 2021 at 20:18
  • 1
    $\begingroup$ That is one of my questions: "Can we find the volume of a $1/2$-dimensional sphere by using the Hausdorff dimension,... " $\endgroup$
    – vitamin d
    Feb 23, 2021 at 21:00
  • 2
    $\begingroup$ As @BenjaminWang suggests, exactly what is a $1/2$-dimensional sphere? I don't see where you've indicated that. Rather, you've taken the formula for the volume of a unit sphere in $\mathbb R^n$, relaxed the condition that $n\in\mathbb N$, and observed that we get a number. If there were a well-known object accepted as the "$1/2$-dimensional sphere", then I suppose it might be interesting if your number returned its $1/2$-dimensional Hausdorff measure. I don't know such an object so I think you need to define it here to yield an interesting question. $\endgroup$ Mar 14, 2021 at 11:14
  • $\begingroup$ But that's exactly my question? What is a $1/2$-dimensional sphere? How could one define it? I also think you understood my question wrong. I'm not saying that the number I got from the formula is the volume of such a sphere. I'm asking if it would be equal to other values we get if we approach this question from, for example, the Hausdorff-dimension. $\endgroup$
    – vitamin d
    Mar 14, 2021 at 17:23

1 Answer 1


Since no one has answered this question yet, let me write what I think about this interesting, but ill posed (in my opinion) question. First of all, In mathematics there are multiple meaning of dimension depending on the object we are working with. Here you are trying to incorporate two different "dimensions" (dimensional of a real vector space and Hausdorff dimension of a metric space) and see if one extend the other while making sense of volume. This may or may not possible, and most likely not. For a simple example, consider the embedding $\mathbb{N}\hookrightarrow\mathbb{Z}.$ Using $\mathbb{N}$ we count thing, like $5$ apples while $-5$ apples makes no sense anymore in the extension $\mathbb{Z}.$ Similarly the meaning of dimension turn out to be different when one looks at more general spaces.

If we merely start with a space with Hausdorff dimension $1/2,$ the volume of a sphere must means its Hausdorff_measure. Consider the generalized Cantor set obtained by removing at the middle $1/2$ segment of $[0,1],$ and in the $n$th iteration remove the central interval of length $1/2L_{n-1}$ from each remaining segment. This is an example for a this kind of space. Now you can test your conjecture $$V_{1/2} =2^{5/4}\varpi^{-1/2}$$ by an explicit computation. I will leave the rest for someone familiar with Geometric measure theory. But keep in mind that there is no canonical space with Hausdorff dimension $1/2$ that we could declare to be $\mathbb{R}^{1/2}.$ So, this is more likely to be a dead-end.

Another less promising way to approach this problem is understand what is a vector spaces of fractional dimension. Clear this will not be a natural generalization of the existing concept. For example, a linear transformation between two vector spaces is a matrix (after choosing a basis for the domain). Then, what would be a factional dimensional matrix?

Under the assumption that there is a some "fractional" analogue vector spaces: A priori, there's no reason to believe that they'd look like typical fractals as this is an algebraic approach oppose to previous geometric approach. In general, the dimension of a vector space is the cardinality of any basis. Rephrasing this definition in other ways (such as trace of the identity operator, etc.) allow us to generalize it in some cases. For example, there is already a well established notion of super vector spaces with negative integer dimensions in this modified sense (and generalizes to dualizable object in a symmetric monoidal category; see). But so far there is no sensible distinguished class of "fractional dimensional vector spaces". Even if one manage to identify them, then we need a nice Lebesgue measure to measure volumes of subsets. So, lot more things are still waiting in this road to be discovered, if possible.

Near the end of writing this answer, I found this question with two well written answers along the same line as mine. Hope it will help to clear the scenario more.

  • $\begingroup$ Thank you. I also asked this question on MO but still no answer.Would you recommend to set a bounty? $\endgroup$
    – vitamin d
    Mar 23, 2021 at 18:09
  • $\begingroup$ @vitamind: Setting a bounty would definitely attract some attention. But I don't think you can get a sensible answer to this question, as others also pointed out. If your primary question is "defining a $1/2$-dimensional sphere", it is interesting but may not be (immediately) useful. As first half of my answer explains, this is may achieve by calculus/analysis on (fractal) metric spaces, but there won't be a such "canonical" space. $\endgroup$
    – Bumblebee
    Mar 23, 2021 at 18:44
  • $\begingroup$ ctd: On the other hand, I don't find the "volume of a $1/2$-dimensional sphere" (whatever that means) is interesting and (more importantly) useful. Ever if there is something like that you can always normalize it to be $1,$ as Spherical measure does. $\endgroup$
    – Bumblebee
    Mar 23, 2021 at 18:44
  • $\begingroup$ Very helpful answer and comments. Thank you. I will give you a checkmark with delight when I worked the problem out (probably have to do it alone). Will take a some time. $\endgroup$
    – vitamin d
    Mar 23, 2021 at 18:48
  • $\begingroup$ @vitamind: Another reason for your question not so well defined is, in fractional settings things don't have the typical meanings and behaviors. For example: first derivative of a function at a point measures slope of tangent lines, second derivative measures convexity, third measure torsion/curvature and so on. All of these are local properties. However fractional derivatives are entirely different species. I remember, I was also wondering about such interpolations as a high schooler. $\endgroup$
    – Bumblebee
    Mar 23, 2021 at 19:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .