# Volumes of n-balls: what is so special about n=5?

I am reposting this question from math.stackexchange where it has not yet generated an answer I had been looking for.

• The volume of an $n$-dimensional ball of radius $R$ is given by the classical formula $$V_n(R)=\frac{\pi^{n/2}R^n}{\Gamma(n/2+1)}.$$ It is not difficult to see that the "dimensionless" ratio $V_n(R)/R^n$ attains its maximal value when $n=5$.

• The "dimensionless" ratio $S_n(R)/R^n$ where $S_n(R)$ is the $n$-dimensional volume of an $n$-sphere attains its maximum when $n=7$.

Question. Is there a purely geometric explanation of why the maximal values in each case are attained at these particular values of the dimension?

[EDIT. Thanks to all for the answers and comments.]

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I find this question interesting, and hope the following comment is not considered as inaproriate / too off-topic. There is an article, which I fail to find at the moment, making an argument that $2 \pi$ should be the 'special constant' rather than $\pi$, considering the diameter as central not the radius. If this were so it seems this question would dissolve, as the (modified) ratio for the volume would be decreasing right from the start. –  quid Jan 24 '11 at 21:18
Yes - this is close to the point I make below in my answer. –  Marty Jan 24 '11 at 21:27
I can think of two choices of n-cubes whose volumes might be related to the volume of an n-sphere in a geometrically interesting way: the inscribed one and the circumscribed one. R^n is not the volume of either n-cube. –  Qiaochu Yuan Jan 24 '11 at 21:28
@unknown: Bob Palais, "$\pi$ Is Wrong", Opinion column in Math Intelligencer, Vol. 23, No. 3, 2001. –  Hans Lundmark Jan 24 '11 at 22:04
@unknown: It doesn't matter whether we call it $\pi$, $2\pi$, $\pi/2$, or whatever, the number in the formula is still our old friend 3.14... . –  John Bentin Jan 24 '11 at 22:23

## 4 Answers

There are many "dimensionless" ratios: choosing $R$ as the linear measurement is arbitrary. For instance, the ratio of the volume of the sphere to the volume of the circumscribed cube has a maximum at $n=1$. The ratio to the volume of the inscribed cube never attains a maximum. There are intermediate geometrically-related "midscribed" cubes, where all faces of some dimension are tangent to the unit sphere. Here is the graph for the ratio of volumes when the codimension 2 faces of a cube are tangent. It attains the maximum for $n = 12$ (just barely more than for $n=11$). There are many other reasonable dimensionless comparisons, for instance comparing to a simplex, etc. etc. Since the Gamma function grows super-exponentially, these simple geometric variations tend to shift the maximum --- there's nothing special about 5 or 7.

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Brian Hayes has very nice article about the volume of the n-ball in the current issue of American Scientist (Nov 2011). In particular, he discusses the surprising fact that the maximum volume occurs at $n=5$.

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A very celebrated result on a related subject is the following, which was a very major advance in the Busemann-Petty problem (don't worry, the math review has all you need to know). EDIT by popular demand, the review can be seen here: http://dl.dropbox.com/u/5188175/BallReview.pdf

@incollection {MR950983,

AUTHOR = {Ball, Keith},

TITLE = {Some remarks on the geometry of convex sets},


BOOKTITLE = {Geometric aspects of functional analysis (1986/87)},

SERIES = {Lecture Notes in Math.},

VOLUME = {1317},

PAGES = {224--231},


PUBLISHER = {Springer},

ADDRESS = {Berlin},

  YEAR = {1988},


MRCLASS = {52A40},

MRNUMBER = {950983 (89h:52009)},

MRREVIEWER = {G. D. Chakerian},

   DOI = {10.1007/BFb0081743},

URL = {http://dx.doi.org/10.1007/BFb0081743},


}

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It never hurts to provide at least a description of the theorem. Not everyone has access to MathSciNet! –  Mariano Suárez-Alvarez Jan 24 '11 at 21:32
@Mariano: Now you do :) –  Igor Rivin Jan 24 '11 at 22:46

In my opinion, nothing is special about $n = 5$.

The "dimensionless ratio" $V_n(R) / R^n$ is the ratio of the volume of the $n$-ball of radius $1$ to the volume of the $n$-cube of side-length $1$. So this is maximized at $n =5$, but bluntly, so what?

More interesting geometrically might be the equally dimensionless ratio $V_n(R) / (2R)^n$, which is the ratio of the volume of the $n$-ball to the volume of the smallest $n$-cube containing it. This is monotonic decreasing (for $n \geq 1$), showing that balls decrease in volume relative to their smallest cubical container, as the dimension increases. This has more geometric content, since there is a simple geometric relationship between the sphere and cube here.

One could consider many similar problems, involving inscribing a cube inside a sphere (instead of the other way around), or using an orthoplex or polycylinder or other figure instead of a cube. All of these have some geometric content, and are expressed as a sequence of dimensionless ratios.

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