# 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 $\Re(s) > 1$,

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

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

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.*