## Some interesting relations involving the zeta function

In this post I will talk about a few identities, and a mysterious question which was solved a few days ago by Joseph Horan and I. For an arithmetic function $f : \mathbb{Z}^+ \to \mathbb{C}$, we write

$\displaystyle D(f,s) = \sum_{n=1}^\infty \frac{f(n)}{n^s}$

for the Dirichlet series associated to $f$. Observe at once that the Riemann zeta function is given by

$\displaystyle \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = D(u,s)$

where $u(n) = 1$ for all $n$. The ubiquity of the zeta function in number-theoretic relationships is due perhaps to this fact — it is the simplest Dirichlet series, just as $u$ is one of the simplest arithmetic functions. Similarly, $\zeta(s-1) = D(N,s)$ where $N(n) = n$ for all $n$. Also, $\frac{1}{\zeta(s)} = D(\mu,s)$ where $\mu$ is the Mobius function, indeed we have for $\mathrm{re}(s) > 1$ that

$\displaystyle \zeta(s) \sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \sum_{k=1}^\infty \frac{1}{(k \ell)^s} \sum_{\ell = 1}^\infty \mu(\ell) = \sum_{n=1}^\infty \frac{1}{n^s} \sum_{d \mid n} \mu(d) = 1.$

This last fact is a consequence of the behaviour of multiplication of Dirichlet series. Namely, if two Dirichlet series $D(f,s)$ and $D(g,s)$ converge absolutely for some value of $s$, then so too does their product, and indeed $D(f,s)D(g,s) = D(f * g, s)$. That is, multiplying Dirichlet series amounts to applying Dirichlet convolution to the corresponding coefficients.

Using the behaviour of multiplication of Dirichlet series, we can derive many interesting identities. For example, because the divisor sum function $\sigma$ satisfies the relation $\sigma = N * u$, and the divisor count function $\tau$ satisfies the relation $\tau = u * u$, we obtain that

$\displaystyle \sum_{n=1}^\infty \frac{\sigma(n)}{n^s} = D(N,s)D(u,s) = \zeta(s-1)\zeta(s)$,

$\displaystyle \sum_{n=1}^\infty \frac{\tau(n)}{n^s} = D(u,s)D(u,s) = \zeta^2(s)$,

and so on. Now, let $\lambda(n)$ be the Liouville function, that is, define $\lambda(n) = (-1)^{\Omega(n)}$, where $\Omega(n)$ is the number of prime divisors of $n$ (counting multiplicity). For example, $\lambda(6) = (-1)^2 = 1$ and $\lambda(12) = (-1)^3 = -1$. Observe that $\lambda = \mu$ when we restrict to the square-free integers.

It turns out that we have the relation $\lambda * |\mu| = I$, where $I(n) = \lfloor 1/n \rfloor$ is the Dirichlet identity. Let us say that $n$ is $k$-free if no prime appears in the factorization of $n$ with a multiplicity of $k$ or more. We usually say “square-free” rather than “$2$-free”. Then $|\mu|$ is nothing more than the indicator function of the square-free integers.

Let $q_k$ denote the indicator function of the $k$-free integers, so that $q_2 = |\mu| = \mu^2$. It is quite clear that $q_k$ is multiplicative. Observe $D(q_k, s)$ can be written as the Euler product

$\displaystyle \prod_{p \text{ prime}} \left( 1 + \frac{1}{p^s} + \ldots + \frac{1}{p^{(k-1)s}} \right)$

and therefore, note that

$\displaystyle \zeta(ks) D(q_k,s) = \prod_{p \text{ prime}} \left( 1 + \frac{1}{p^s} + \ldots + \frac{1}{p^{(k-1)s}} \right) \left( 1 + \frac{1}{p^{ks}} + \ldots \right) = \zeta(s).$

We have therefore demonstrated the relation

$\displaystyle \frac{\zeta(s)}{\zeta(ks)} = D(q_k, s) = \sum_{n=1}^\infty \frac{q_k(n)}{n^s}. \qquad (\dagger)$

In particular,

$\displaystyle \frac{\zeta(s)}{\zeta(2s)} = \sum_{n=1}^\infty \frac{|\mu(n)|}{n^s}.$

As I stated above, the Liouville function $\lambda$, whose definition seems innocently unrelated to the number $2$, is the Dirichlet inverse of $|\mu| = q_2$. What is going on here? Actually, there is one thing we know, particularly that

$\displaystyle \sum_{d \mid n} \lambda(d) = \begin{cases} 1 & \text{if } n \text{ is a perfect square} \\ 0 & \text{otherwise.} \end{cases}$

Let $\alpha_k$ denote the indicator function of the perfect $k$th powers. Then the above reads $\lambda * u = \alpha_2$ (be wary of the important distinction between “perfect $k$th power” and “$k$-free integer”). Notice that $I_k$ has associated Dirichlet series

$\displaystyle D(I_k, s) = \sum_{n=1}^\infty \frac{I_k(n)}{n^s} = \sum_{n=1}^\infty \frac{1}{n^{ks}} = \zeta(ks).$

We want to find, for each $k$, a function $F_k$ such that $F_k * q_k = I$. For example, we will have $F_2 = \lambda$, due to the above discussion. However, notice that because we showed above (see $\dagger$) that $I_k * q_k = u$, we know that

$\displaystyle F_k * q_k = I \iff F_k * I_k * q_k = I_k \iff F_k * u = I_k.$

Hence, suppose that $F_k$ is the Dirichlet inverse of $q_k$. We know that such a thing exists and must be multiplicative, since $q_k$ is multiplicative, and the multiplicative functions form a subgroup (hence are closed under inversion). Then we must have $F_k * u = I_k$. Mobius inversion then allows us to write

$\displaystyle F_k(n) = \sum_{d \mid n} I_k(d) \mu \left(\frac{n}{d} \right).$

Let’s assume $n=p^r$ is a prime power. Note that since $\mu$ vanishes whenever its argument is not squarefree, we can restrict our attentions to the divisors $d$ of $n$ such that $d = p^{r-1}$ or $d = p^r$. A bit of consideration convinces us that the above sum will only ever have one term. Indeed,

$\displaystyle F_k(p^r) = \begin{cases} \phantom{-}1 & \text{if } r = 0 \text{ mod } k \\ -1 & \text{if } r = 1 \text{ mod } k \\ \phantom{-}0 & \text{otherwise}. \end{cases}$

Having determined the behaviour of $F_k$ on prime powers, due to its multiplicativity we now understand it fully. The Dirichlet series associated to $F_k$ are

$\displaystyle D(F_k, s) = \frac{\zeta(ks)}{\zeta(s)}.$