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Aymen Hafeez

Higher dimensional spheres and the prime numbers

/ 3 min read

View more blogs with the tag mathematics

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

Vn=πn/2Γ(n2+1)=πn/2(n/2)! 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,

ζ(2k)=(−1)k+1(2π)2kB2k2(2k)! \zeta(2k) = \frac{(-1)^{k+1}(2\pi)^{2k}\Beta_{2k}}{2(2k)!}

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

ζ(s)=∑n=1∞1n2=∏p prime11−p−s \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:

ζ(s)=1Γ(s)∫o∞us−1eu−1du \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 nn-ball and the primes.

Recall again that for 44 dimensional nn-ball, the volume is

V4=π2Γ(3)=π22!=π22=3ζ(2) 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=4s = 4:

V4=3∏p11−p−2 V_4 = 3\prod_{p} \frac{1}{1 - p^{-2}}

Let’s also consider the 88 dimensional nn-ball:

V8=π8/2(8/2)!=π424 V_8 = \frac{\pi^{8/2}}{(8/2)!} = \frac{\pi^4}{24}

And since

ζ(4)=π490 \zeta(4) = \frac{\pi^4}{90}

We get

V8=9024ζ(4)=154ζ(4) V_8 = \frac{90}{24}\zeta(4) = \frac{15}{4}\zeta(4)

And using the Euler product again:

V8=154∏p11−p−4 V_8 = \frac{15}{4}\prod_{p}\frac{1}{1 - p^{-4}}

We can get a general expression relating the VnV_n to the primes:

ζ(2)=π26     ⇒     π2=6ζ(2) \zeta(2) = \frac{\pi^2}{6} \ \ \ \ \ \Rightarrow \ \ \ \ \ \pi^2 = 6\zeta(2)

And so, we have

πn/2=(π2)n/4=(6ζ(2))n/4 \pi^{n/2} = (\pi^2)^{n/4} = (6\zeta(2))^{n/4}

Subsituting this back into the volume formula:

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

Since we know that

ζ(s)=∏p11−p−s \zeta(s) = \prod_{p} \frac{1}{1 - p^{-s}}

Which gives us a general formula for an nn-ball in terms of the prime numbers

Vn=1(n/2)!(6∏p11−p−s)n/4 V_n = \frac{1}{(n/2)!}\left( 6 \prod_{p} \frac{1}{1 - p^{-s}} \right)^{n/4}

While this isn’t as clean as the expression for n=4n = 4 it’s a pretty incredible formula. But because when n=4n = 4 we get a single power of the prime product so the expression simplifies significantly.