The Volume of an Even Dimensional Ball

Posted on Categories General, Mathematics

What is the volume V(r,d) of a ball of dimension d and radius r ? This ball is the following subset of \mathbb{R}^d :

B(r,d) = \{ (x_{1},..., x_{d}) \in \mathbb{R}^d \ | \ \sum_{i = 1}^{d} x_{i}^2 \leq r^2 \},

and we measure the volume using the standard Euclidean measure.

If you only need the answer for even dimensions, here’s an easy way to remember: define V(r) to be the sum of the volumes of balls of radius r over all of the even dimensions:

V(r) = \sum_{n = 0}^{\infty} V(r, 2n) .

Remarkably, the formula for this function turns out to be V(r) = e^{\pi r^2} . This is easy to remember since it’s the exponential of the volume (i.e. the area) of a disc. Then, you can read off the volumes V(r,2n) as the terms in the Taylor expansion of V(r) = e^{\pi r^2} . Explicitly:

V(r,2n) = \frac{\pi^n r^{2n}}{n!}.

A nice, and immediate, corollary is that the derivative of V(r) , with respect to the radius r , gives you the sum of the volumes of all of the odd-dimensional spheres:

2 \pi r e^{\pi r^2} = \sum_{n = 0}^{\infty} Vol(S^{2n + 1}) r^{2n + 1}.

By looking at the coefficients of the Taylor expansion, we can conclude that the volume of the unit sphere of dimension 2n + 1 is \frac{2 \pi^{n+1}}{n!} .

There’s a lot more to say about the volumes of balls and spheres, and if you’re interested, the Wikipedia page is a good place to start.

Okay, but why is the formula for V(r) true? And what’s special about even dimensions? That’s what I want to explain in this post.

In everything I’ve told you so far, there are really only two key facts from which everything follows. First, there is the formula for the area of a disc V(r, 2) = \pi r^2. Second, there is the relationship

V(r,2)^n = n! V(r, 2n).

This says that the volume of a product of n copies of a disc B(r,2) is n! times the volume of the corresponding ball. This fact is reminiscent of another interesting volume computation: the volume of an n-simplex \Delta^n_{r} .

Computing the Volume of a Simplex

There are different realizations of the simplex, and they don’t all have the same volume. We will need the following version:

\Delta^n_{r} = \{ (t_{1}, t_{2}, ..., t_{n}) \in \mathbb{R}^n \ | \ 0 \leq t_{1} \leq t_{2} ... \leq t_{n} \leq r \}.

Here’s a picture in two dimensions:

In this picture, it’s quite easy to see that the two-dimensional simplex fills out exactly half of the square, and for this reason, its area is Vol(\Delta^2_{r}) = \frac{r^2}{2}. This fact can be easily generalized to all dimensions, giving the following formula:

Vol(\Delta^n_{r}) = \frac{r^n}{n!}.

Here is the argument: a point in the cube [0,r]^n is a collection of numbers (t_{1}, ..., t_{n}) which all lie between 0 and r . For almost all of these points (namely, those for which none of the coordinates t_{i} are equal to each other), there will be a unique permutation \sigma \in S_{n} that orders the coordinates t_{i} and therefore sends the point (t_{1}, ..., t_{n}) into \Delta^n_{r} . This implies that by applying permutations \sigma \in S_{n} the simplex \Delta^n_{r} is transformed into a collection of n! distinct simplices which completely fill out the cube. In other words, the cube [0,r]^n decomposes into a collection of n! simplices, implying that

Vol([0,r]^n) = n! Vol(\Delta^n_{r}).

Just as for balls, we can package the volumes of the simplices of various dimensions into a single function

S(r) = \sum_{n = 0}^{\infty} Vol(\Delta^n_{r})

and from the above calculation, it follows that S(r) = e^{r} .

Comparing this volume of a simplex to the volume of an even-dimensional ball, we notice the following analogy:

The simplex \Delta^n is to the cube [0,1]^n (which is a product of n intervals), as the ball B(2n) is to the polydisc B(2)^n (which is a product of n discs).

At least, this is true when considering volumes. What I want to explain in this post is that this analogy can be made precise. In fact, the volume of the even-dimensional ball can be derived from the volume of a simplex.

Before we continue, a simplification: since the volume of an n -dimensional shape of ‘radius’ r has the form C r^n , the main interest lies in determining the coefficient C , which is the volume when r = 1. Hence, for the rest of the post I will simplify the formulas by setting r = 1.

Tropicalization

The key to the comparison between even-dimensional balls and simplices, and the answer to the question ‘why even-dimensional balls?’, is tropicalization, by which I mean the comparison between complex numbers and real numbers given by taking the radius squared:

T: \mathbb{C} \to \mathbb{R}_{\geq 0}, \qquad z \mapsto |z|^2.

You should picture this map in the following way:

Applying the tropicalization map to various subsets of \mathbb{C}^n produces subsets of (\mathbb{R}_{\geq 0})^n which are a sort of coarse-graining, in that the phase of the complex numbers have been forgotten.

In fact, this is not quite tropicalization. If you look in a paper, such as this one, you see that tropicalization involves taking the logarithm of the radius, Log_{t}(|z|^2) , and then letting the base t go off to infinity. Using a finite value t , the images of subsets of (\mathbb{C}^*)^n are what some people call amoebas.

In any case, the map T above is what we need. The key fact is that it behaves well with respect to the Euclidean measures on \mathbb{C} and \mathbb{R}_{\geq 0} .

Lemma. The pushforward of the Euclidean volume form dx\wedge dy on \mathbb{C} = \mathbb{R}^2 is the volume form \pi dt on \mathbb{R}_{\geq 0} .

Indeed, writing the volume form dx\wedge dy on \mathbb{C} in polar coordinates (t = |z|^2, \theta) gives

dx\wedge dy = \frac{1}{2} dt \wedge d \theta ,

and integrating out the angle coordinate d \theta over the circle gives \pi dt , where t is the coordinate on \mathbb{R}_{\geq 0} .

Given any subset S \subseteq \mathbb{C}^n which is invariant under rotation of phases, its volume can be computed as

Vol(S) = \pi^n Vol(T(S)),

where Vol(T(S)) is the Euclidean volume of the image T(S) \subseteq (\mathbb{R}_{\geq 0})^n.

Let’s see how this helps. First, the polydisc D(2)^n is given by the following subset of \mathbb{C}^n :

B(2)^n = \{ (z_{1}, ..., z_{n}) \in \mathbb{C}^n \ | \ |z_{i}|^2 \leq 1 \text{ for all } i \}.

Applying the tropicalization map T: \mathbb{C}^n \to (\mathbb{R}_{\geq 0})^n produces the unit cube:

T(B(2)^n) = [0,1]^n.

This implies that the volume of the polydisc is Vol(B(2)^n) = \pi^n .

On the other hand, the 2n -dimensional ball B(2n) is the following subset of \mathbb{C}^n :

B(2n) = \{ (z_{1}, ..., z_{n}) \in \mathbb{C}^n \ | \ \sum_{i = 1}^n |z_{i}|^2 \leq 1 \},

and tropicalization gives

T(B(2n)) = \{ (t_{1}, ..., t_{n}) \in (\mathbb{R}_{\geq 0})^n \ | \ \sum_{i = 1}^n t_{i} \leq 1 \}.

This set is a simplex, but it isn’t quite the simplex \Delta^n we defined earlier. Here’s a picture of what you get in two dimensions:

You can see that this differs from the earlier picture of a simplex. In fact, the two simplices are related by a simple change of coordinates. Define

s_{i} = \sum_{j=1}^{i} t_{j},

for i = 1 ... n . You should convince yourself that a point (t_{1}, ..., t_{n}) lies in T(B(2n)) if and only if the corresponding point (s_{1}, ..., s_{n}) lies in \Delta^n . But now observe that the linear transformation that produces the change of coordinates is a shear transformation. Indeed, it is given by the following matrix

L = \begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 1 & 1 & 0 & \cdots & 0 \\ 1 & 1 & 1 & \cdots & 0 \\ \vdots & & & \ddots & 0 \\ 1 & 1 & 1 & \cdots & 1 \end{pmatrix},

whose determinant is easily seen to be 1 . Therefore, the change of coordinates is volume-preserving and we see that

Vol(T(B(2n))) =Vol(\Delta^n) = \frac{1}{n!}.

Combined with the previous facts, we conclude that

Vol(B(2n)) = \pi^n Vol(T(B(2n))) = \frac{\pi^n }{n!} = \frac{Vol(B(2)^n)}{n!}.

We have derived the volume of even-dimensional balls from the volume of simplices!


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