The generalized formula for the volume and surface area of n-sphere allows to evaluate volumes and areas of negative-dimentional n-spheres.

$$\begin{array}{ll} S_{n-1}(R) &= \displaystyle{\frac{n\pi^{n/2}}{\Gamma(\frac{n}{2}+1)}R^{n-1}} \\[1 em] V_n(R) &= \displaystyle{\frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)}}R^n \end{array}$$

I wonder what the negative-dimentional n-spheres are? Are they like pseudosphere or hyperboloid?

In general, what a negative-dimentional vector space is?

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    Why stop at integers? You can take any real number. – Yuval Filmus Mar 5 '15 at 2:45
  • @Yuval Filmus what non-integer positive dimentional space and n-sphere is clear: a fractal. But what about negative? – Anixx Mar 5 '15 at 3:17
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    A fractal sphere sounds like a contradiction in terms. So I can't picture that any more than I can picture a sphere in a negative-dimensional setting. – MvG Mar 5 '15 at 9:39
  • @MvG if distance is defined on a fractal (for instance, on Serpinsky triangle en.wikipedia.org/wiki/Hausdorff_dimension#mediaviewer/… ), then all points equally distant from a given one form a sphere. – Anixx Mar 5 '15 at 9:46
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    @Anixx: If $S$ is a sphere in a space $X$ of fractal dimension $d$, do you have any reason to believe: 1. The dimension of $S$ is independent of (a) the fractal $X$ (i.e., is a function of $d$ alone)? (b) the radius $R$? 2. The fractal dimension of $S$ is $d - 1$? 3. The volume of $S$ is given by the formula in your post? Unless all these preliminary questions have affirmative answers, it seems to me the question is on shaky ground. (Then we can haggle about what "negative-dimensional" might mean.) – Andrew D. Hwang Jul 7 '15 at 13:05

There is a way of making sense of all integer dimensions, although I'm not sure if it will be useful to you. Let us start with the collection of Euclidean spaces, $$ E=\{\mathbb R^n;n\in\mathbb Z,n\geq0\}, $$ for example. There two operations for spaces in $E$, the direct sum and the tensor product. If we identify spaces of the same dimension, we have $\mathbb R^n\oplus\mathbb R^m=\mathbb R^{n+m}$ and $\mathbb R^n\otimes\mathbb R^m=\mathbb R^{nm}$. These operations are associative, commutative and distributive.

These operations don't make $(E,\oplus,\otimes)$ into a ring, but that can be remedied. One way is to formally let $n$ range over all of $\mathbb Z$ and define things the same way. If we call this new collection of spaces $F$, then $(F,\oplus,\otimes)$ is (in an obvious way) ring-isomoprhic to $(\mathbb Z,+,\times)$. Then objects like $\mathbb R^{-7}\in F$ don't have any other meaning than being in some kind of an extension of the set of Euclidean spaces. I don't see any obvious way of doing geometry in an object of $F$, especially in a way that would include those volume formulas.

In this case the construction leads to a relatively trivial object (the ring of integers), but it at least gives some way of attaching a meaning to negative dimensional spaces. That is, $\mathbb R^{-4}$ is the thing (not a vector space!) that satisfies $\mathbb R^{-4}\oplus\mathbb R^4=0$ (something like the direct sum of vector spaces), where $0=\mathbb R^0$ is the zero space. Negative dimension cancels positive dimension, whatever that means.

(This answer was inspired by the concept of representation ring, a way of forming a ring out of vector spaces in way that is useful in representation theory.)

Sidenote: $1/\Gamma(1+n/2)=0$ for $n\in\{-2,-6,-10,-14,\dots\}$. This would be really weird behaviour for something that resembles the geometrical idea of volume.

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