= Identity for Dirichlet convolution
{c}
{title2=$\varepsilon$}
The <Identity for Dirichlet convolution> is the <arithmetic function>
$$
\varepsilon(n)=\begin{cases}1&n=1,\\0&n>1.\end{cases}
$$
For every <arithmetic function> $f$, it satisfies $f*\varepsilon=\varepsilon*f=f$. Indeed, in $(f*\varepsilon)(n)=\sum_{d\mid n}f(d)\varepsilon(n/d)$, only $d=n$ survives. For example,
$$
(f*\varepsilon)(12)=f(1)0+f(2)0+f(3)0+f(4)0+f(6)0+f(12)1=f(12).
$$
At $n=1$ the sole term is $f(1)\varepsilon(1)=f(1)$. This gives a useful way to check <Dirichlet convolution> calculations: inserting $\varepsilon$ leaves every value unchanged. The three <arithmetic functions> $\mathbf1$, $\operatorname{id}$, and $\varepsilon$ all take value $1$ at $1$, but at $12$ the <constant-one arithmetic function>, the <identity arithmetic function>, and the <Identity for Dirichlet convolution> take values $1$, $12$, and $0$, respectively. https://darijgrinberg.gitlab.io/floor.pdf[Darij Grinberg, MIT 18.781 lecture notes, Theorem 2.3.4]{ref}
The <Möbius function> is the inverse of the <constant-one arithmetic function> under <Dirichlet convolution>, meaning $\mu*\mathbf1=\mathbf1*\mu=\varepsilon$. The <Möbius divisor-sum identity> gives a short proof: for $n>1$ the nonzero terms expand $\prod_{p\mid n}(1-1)=0$, while at $1$ the sum is $\mu(1)=1$. At $12$, the computation is
$$
(\mu*\mathbf1)(12)=\mu(1)+\mu(2)+\mu(3)+\mu(4)+\mu(6)+\mu(12)=1-1-1+0+1+0=0=\varepsilon(12).
$$
The zeros occur because $4$ and $12$ are divisible by the square of the <prime number> $2$. Consequently, if $F=f*\mathbf1$ is a <divisor sum>, then <associativity> gives $F*\mu=f*(\mathbf1*\mu)=f*\varepsilon=f$: this is <Möbius inversion>.
Back to article page
