Higher dimensional spheres and the prime numbers

In a previous post we showed how to derive an expression for the volume of an $n$-ball:

$$V_n = \frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2} + 1\right)} = \frac{\pi^{n/2}}{(n/2)!}$$

In another post we’ve also looked at the Riemann zeta function at even values,

$$\zeta(2k) = \frac{(-1)^{k+1}(2\pi)^{2k}\Beta_{2k}}{2(2k)!}$$

as well how the deriving Euler’s prime product formula,

$$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \prod_{p \text{ prime}} \frac{1}{1 - p^{-s}}$$

We’ve also looked at showing how the zeta function is related to the Gamma function:

$$\zeta(s) = \frac{1}{\Gamma(s)}\int_{o}^{\infty} \frac{u^{s-1}}{e^u - 1} \text{d}u$$

In this post we’ll see how all these tie together to give a surprising relation between the volume of an $n$-ball and the primes.

Recall again that for $4$ dimensional $n$-ball, the volume is

$$V_4 = \frac{\pi^2}{\Gamma(3)} = \frac{\pi^2}{2!} = \frac{\pi^2}{2} = 3\zeta(2)$$

And then substituting in the prime product formula for $s = 4$:

$$V_4 = 3\prod_{p} \frac{1}{1 - p^{-2}}$$

Let’s also consider the $8$ dimensional $n$-ball:

$$V_8 = \frac{\pi^{8/2}}{(8/2)!} = \frac{\pi^4}{24}$$

The Riemann zeta function also gives us an expression for $\pi$ raised to fourth power:

$$\zeta(4) = \frac{\pi^4}{90}$$

Putting these together and then equation to the prime product formula at $s = 4$:

$$V_8 = \frac{90}{24}\zeta(4) = \frac{15}{4}\zeta(4)$$ $$V_8 = \frac{15}{4}\prod_{p}\frac{1}{1 - p^{-4}}$$

We can get a general expression relating $V_n$ to the primes:

$$\zeta(2) = \frac{\pi^2}{6} \ \ \ \ \ \Rightarrow \ \ \ \ \ \pi^2 = 6\zeta(2)$$ $$\pi^{n/2} = (\pi^2)^{n/4} = (6\zeta(2))^{n/4}$$

Substituting this back into the volume formula:

$$V_n = \frac{(6\zeta(2))^{n/4}}{(n/2)!}$$

And then using the prime product formula at $s = 2$:

$$V_n = \frac{1}{(n/2)!}\left( 6 \prod_{p} \frac{1}{1 - p^{-2}} \right)^{n/4}$$

This is a general formula for a unit $n$-ball in terms of the prime numbers. While this isn’t as clean as the expression for $n = 4$ it’s still a pretty incredible formula. But because when $n = 4$ we get a single power of the prime product so the expression simplifies significantly.