# Primons, supersymmetry, and the Möbius function

February 22, 2016. Primons are fictitious particles connected to prime numbers. We see how their statistical-mechanical properties are related to results in number theory via supersymmetry.

## Introduction

Prerequisites: basic theory of the Riemann zeta, the Möbius function, quantum statistical mechanics.

In recent years, some interesting connections between number theory and physics have come to light. The Riemann zeta function provides a nice example. Recall that, for $\Re(s) > 1$,

$\zeta(s) = \sum_{n\geq 1}\frac{1}{n^s} = \prod_{p}\frac{1}{1-p^{-s}}.$

The last form is called the Euler product, and relates the Riemann zeta to prime numbers.

 Bosons and fermions sitting in a nice structure. (Wikimedia Commons)

We’ll see a way to view the Riemann zeta in terms of statistical mechanics, and turning things around, use supersymmetry to think about a related number-theoretic object called the Möbius function.

## Bosons

To begin with, imagine we have a gas of non-interacting bosonic “primons”, with energy levels

$\epsilon_n = \epsilon \ln(p_n), \quad n \in \mathbb{N}.$

Here, $p_n$ denotes the $n$th prime. In Fock space, we can encode the occupation numbers for each energy level in the prime factors of a single integer $m$,

$m = p_1^{d_1}\cdots p_k^{d_k}$

with $d_k$ the occupation number of energy level $k$. (This neat trick is due to Gödel, who used it in his proof of the first incompleteness theorem.) For the state $|m\rangle$, the total energy (since primons are non-interacting) is

$E_m = \epsilon \sum_{i=1}^k d_i\ln(p_i) = \epsilon \ln m.$

Thus, for $\epsilon > kT = \beta^{-1}$, in the canonical ensemble we have

$Z_B[\beta] = \sum_{m \geq 1} e^{-E_m \beta} = \sum_{m \geq 1} \frac{1}{m^{\epsilon \beta}} = \zeta(\epsilon \beta).$

Let’s calculate the energy. First, note that from the Euler product representation,

$\ln\zeta(s) = -\sum_{p} \ln(1 - p^{-s}).$

Recalling that $\langle E \rangle = -\partial_\beta \ln Z_B[\beta]$, we have

$\langle E \rangle = -\frac{\partial}{\partial \beta}\ln \zeta(\epsilon\beta) = \frac{\partial}{\partial \beta}\sum_{p} \ln(1 - p^{-\epsilon \beta}) = \sum_{n\geq 1} \frac{\epsilon_n}{p_n^{\epsilon \beta} - 1}.$

So, the expected occupancy at level $n$ is $(e^{\epsilon_n \beta} - 1)^{-1}$, a Bose-Einstein distribution. As $s = \epsilon \beta \to 1^+$, there is the familiar $s = 1$ singularity of the zeta function, and the energy blows up. The temperature $T = \epsilon/k$ is called the Hagedorn temperature, and represents a phase transition. (You would expect to use the analytic continuation of the zeta function on the other side of the singularity. Apparently you run into difficulties, but I haven’t investigated.)

## Fermions

We can do the same thing for fermionic primons, taking the Pauli exclusion principle into account. This means the $n$th occupation number $d_n \in {0,1}$, and hence, each Fock space state is represented by a square-free natural number in Gödel numbering. As before, for a square-free integer $m$, $E_m = \epsilon \ln m$, but now the partition function is harder to calculate. Let’s use the Möbius function $\mu(n)$, defined by

$\mu(m) = \begin{cases} (-1)^k & \text{m = p_1\cdots p_k is square free} \\ 0 & \text{else}.\end{cases}$

By reciprocating the Euler product, it’s easy to show that

$\frac{1}{\zeta(s)}=\prod_{p}\left(1-p^{-s}\right)=\sum_{m\geq 1} \frac{\mu(m)}{m^s}.$

Moreover,

$\frac{\zeta(s)}{\zeta(2s)} = \prod_{p}\frac{1-p^{-2s}}{1-p^{-s}} = \prod_{p}\left(1+p^{-s}\right)=\sum_{m\geq 1} \frac{|\mu(m)|}{m^s}.$

Finally, we can evaluate the fermionic partition function:

$Z_F[\beta] = \sum_{m \geq 1} |\mu(m)|e^{-E_m \beta} = \sum_{m \geq 1} \frac{|\mu(m)|}{m^{\beta\epsilon}} = \frac{\zeta(\beta\epsilon)}{\zeta(2\beta\epsilon)}.$

Since $\ln Z_F[\beta] = \ln \zeta(\beta\epsilon) - \ln\zeta(2\beta\epsilon)$, our calculations for the bosonic case show that, for fermions,

$\langle E \rangle = \sum_{n\geq 1} \left[\frac{\epsilon_n}{p_n^{\epsilon \beta} - 1}-\frac{2\epsilon_n}{p_n^{2\epsilon \beta} - 1}\right] = \sum_{n\geq 1} \frac{\epsilon_n}{e^{\epsilon_n \beta} + 1},$

again, the right occupancy for a Fermi gas. In fact, we could turn things around: from the Fermi distribution, it follows that

$\sum_{m\geq 1} |\mu(m)|m^{-s} = \zeta(s)/\zeta(2s)$

without having to use the Euler product! Also, we should note that since $\zeta$ has a simple pole singularity as $s = 1$, naively taking the limit in the result above gives

$\sum_{m\geq 1} \frac{\mu(m)}{m} = 0.$

In fact, it’s clear that if the series converges, it must vanish, and showing it converges is straightforward. Remarkably, this turns out to be equivalent to the celebrated Prime Number Theorem!

$\pi(x) \sim \frac{x}{\log x}.$

See Spector (1990) for references.

## Supersymmetry

Let’s explore the role of the Möbius function a bit more. We set up a supersymmetric quantum field theory, which can create boson primons with $b_n^\dagger$ and fermion primons with $f_n^\dagger$. For states with nonzero energy, these are interchanged by a supercharge $Q$. We can label Fock states as $|m_B, m_F\rangle \equiv |m_B\rangle \otimes |m_B\rangle$, tensoring the bosons and fermion spaces together.

It’s useful to define the operator $(-1)^F$, where $F$ is the number of fermions in a Fock state. The Witten index $\Delta$ for the theory is the number of bosonic zero energy states minus the number of fermionic zero energy states. Since bosonic and fermionic contributions cancel out for nonzero energy, $\Delta$ is computed by

$\Delta = \mbox{tr} [(-1)^F e^{-\beta H}].$

Denote $m \equiv m_Bm_F$. Let’s calculate $\Delta$ for our supersymmetric primon gas:

\begin{align*} \Delta & = \mbox{tr} [(-1)^F e^{-\beta H}] \\ & = \sum_{m_B, m_F}\mbox{tr} [(-1)^F e^{-\beta H}|m_B, m_F\rangle\langle m_B, m_F|]\\ & = \sum_{m_B, m_F}e^{-\beta E_m}\langle m_B, m_F|(-1)^F|m_B, m_F\rangle\\ & = \sum_{m\geq 1}\frac{1}{m^{\beta\epsilon}}\sum_{m_F|m}\mu(m_F). \end{align*}

On the fourth line, something very important happened: we replaced the expectation of $(-1)^F$ by the Möbius function acting on $m_F$, since the contribution vanishes unless the state’s Gödel number is square-free, and if it is square free, $(-1)^F = \mu(m_F)$.

To proceed further we can use either physics or number theory. Using properties of the Möbius function, we see that the last sum over $m_F$ vanishes for $m > 1$ and equals $1$ for $m = 1$. Explicitly,

$\sum_{d|m}\mu(d) = \begin{cases}0 & m>1 \\ 1 & m = 1.\end{cases}$

Hence,

$\Delta = e^{-\beta E_1} = 1.$

If you know about the Witten index, you can go the other way and read these fundamental identities for $\mu$ off the physics! For instance, accepting that $\Delta$ is an invariant and requiring $\beta$ independence forces $\sum_{d|m}\mu(d) = 0$ for $m > 1$. The remaining identity for $\mu(m)$ follows provided $\Delta = 1$. We can see this directly, since the unique vacuum state has Gödel number $m =1$ and is bosonic (it contains no fermions).

The basic moral is that physical intuition is a great labour-saving device. Of course, it comes at the cost of rigour, but what distinguishes a theoretical physicist from a mathematical physicist is that this seems like a good trade!

For more details on this argument and other results of a similar nature, see Spector (1990). For more on primon gases and the physics of the Riemann zeta function, see Schumayer and Hutchinson (2011).

### References

• “Physics of the Riemann hypothesis” (2011), Daniel Schumayer and David Hutchinson. Reviews of Modern Physics.
• “Supersymmetry and the Möbius inversion function” (1990), Donald Spector. Communications in Mathematical Physics.
Written on February 22, 2016