General – Francis Bischoff

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.

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Consider the exponential map for matrices ${\exp : \text{End}(\mathbb{C}^{m}) \rightarrow \text{Gl}(m, \mathbb{C})}$ , which can be defined by the formula

$\displaystyle e^{x} = \sum_{k = 0}^{\infty} \frac{(2 \pi i)^k}{k!} x^{k},$

where ${x}$ is a square matrix. A matrix logarithm is defined to be a right-inverse to the exponential; that is to say, it is a function ${\log: \text{Gl}(m, \mathbb{C}) \rightarrow \text{End}(\mathbb{C}^{m})}$ such that ${\exp \circ \log = \text{id}_{ \text{Gl}(m, \mathbb{C})}}$ . It is a standard result of Lie theory that the exponential is a diffeomorphism in a neighbourhood of ${0}$ in ${\text{End}(\mathbb{C}^{m})}$ , and hence there is a well-defined choice of logarithm in a neighbourhood of the identity in ${\text{Gl}(m, \mathbb{C})}$ . However, it is not at all clear that a logarithm can be globally defined, or that there is a unique way of doing so. This problem also shows up in the more simple case of the exponential for complex numbers. Indeed, there are several choices for the complex logarithm, and none of them are analytic (or even continuous) on all of ${\mathbb{C}^{\times}}$ . However, it turns out that choosing a complex logarithm is enough to determine a unique choice of a matrix logarithm in any dimension. More precisely, we have the following:

Theorem. Consider the exponential sequence for complex numbers

$\displaystyle 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{C} \mathop{\rightarrow}^{\exp} \mathbb{C}^{\times} \rightarrow 0,$

and let ${\tau : \mathbb{C}^{\times} \rightarrow \mathbb{C}}$ be a set-theoretical splitting of this sequence, in other words, a (not-necessarily continuous) branch of the complex logarithm. Then for any positive integer ${m}$ , there is a uniquely defined matrix logarithm