TITLE: Is a hypersphere of non-integer dimension a fractal?
QUESTION [8 upvotes]: 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?

REPLY [2 votes]: 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.