# Möbius Inversion

## Recovering a Function from Divisor Sums

Many arithmetic functions are defined through sums over divisors. For example,

$$
\sigma(n)=\sum_{d\mid n} d
$$

and

$$
\tau(n)=\sum_{d\mid n}1.
$$

A natural question arises: if a function $F(n)$ is defined by summing another function $f(n)$ over the divisors of $n$, can one recover $f$ from $F$?

Möbius inversion gives the answer.

## Statement of Möbius Inversion

Suppose $f$ and $F$ are arithmetic functions satisfying

$$
F(n)=\sum_{d\mid n} f(d).
$$

Then

$$
f(n)=\sum_{d\mid n}\mu(d)F\left(\frac nd\right).
$$

Equivalently,

$$
f(n)=\sum_{d\mid n}\mu\left(\frac nd\right)F(d).
$$

This formula inverts divisor summation.

## Convolution Form

Using Dirichlet convolution, the hypothesis

$$
F(n)=\sum_{d\mid n}f(d)
$$

can be written compactly as

$$
F=f*\mathbf{1},
$$

where $\mathbf{1}(n)=1$.

Since

$$
\mu*\mathbf{1}=\varepsilon,
$$

the Möbius function is the convolution inverse of $\mathbf{1}$.

Convolving both sides with $\mu$,

$$
\mu*F =
\mu*(f*\mathbf{1}).
$$

By associativity,

$$
\mu*F =
(\mu*\mathbf{1})*f =
\varepsilon*f =
f.
$$

Thus

$$
f=\mu*F.
$$

Written explicitly,

$$
f(n)=\sum_{d\mid n}\mu(d)F\left(\frac nd\right).
$$

This is Möbius inversion.

## First Example

Let

$$
F(n)=\tau(n).
$$

Since

$$
\tau(n)=\sum_{d\mid n}1,
$$

we have

$$
F=f*\mathbf{1}
$$

with

$$
f(n)=1.
$$

Applying Möbius inversion,

$$
1=\sum_{d\mid n}\mu(d)\tau\left(\frac nd\right).
$$

Thus the constant function $1$ can be recovered from the divisor-counting function through Möbius inversion.

## Recovering Euler's Totient Function

Earlier we proved

$$
\sum_{d\mid n}\varphi(d)=n.
$$

Thus

$$
\operatorname{id}=\varphi*\mathbf{1},
$$

where

$$
\operatorname{id}(n)=n.
$$

Applying Möbius inversion gives

$$
\varphi=\operatorname{id}*\mu.
$$

Therefore

$$
\varphi(n) =
\sum_{d\mid n}\mu(d)\frac nd.
$$

For example, if

$$
n=12,
$$

then the divisors are

$$
1,2,3,4,6,12.
$$

Only squarefree divisors contribute:

$$
\varphi(12) =
12 -
6 -
4
+
2 =
4.
$$

This matches the direct calculation.

## Inclusion-Exclusion Interpretation

Möbius inversion is an arithmetic form of inclusion-exclusion.

The Möbius function alternates signs according to the number of prime factors:

$$
\mu(n)=(-1)^r
$$

for squarefree integers with $r$ distinct prime factors.

Thus Möbius inversion systematically adds and subtracts contributions from divisibility conditions.

For example, suppose we wish to count integers up to $x$ that are coprime to $n$. We remove multiples of primes dividing $n$, add back multiples of products of two such primes, subtract multiples of products of three such primes, and so on.

This alternating process is encoded exactly by the Möbius function.

## Counting Primitive Objects

Möbius inversion often separates primitive objects from nonprimitive ones.

Suppose $A(n)$ counts all objects of size $n$, including repetitions arising from smaller periods, while $P(n)$ counts primitive objects. Frequently one has a relation

$$
A(n)=\sum_{d\mid n}P(d).
$$

Then Möbius inversion gives

$$
P(n)=\sum_{d\mid n}\mu(d)A\left(\frac nd\right).
$$

This principle appears in combinatorics, dynamical systems, and finite field theory.

## Multiplicative Form

If $f$ and $F$ are multiplicative and

$$
F=f*\mathbf{1},
$$

then Möbius inversion preserves multiplicativity.

Since $\mu$ is multiplicative and convolution of multiplicative functions is multiplicative,

$$
f=\mu*F
$$

is multiplicative whenever $F$ is.

Thus Möbius inversion is compatible with prime factorization.

## Generalized Form

More generally, suppose $g$ is an arithmetic function with Dirichlet inverse $g^{-1}$. If

$$
F=f*g,
$$

then

$$
f=F*g^{-1}.
$$

Möbius inversion is the special case

$$
g=\mathbf{1},
\qquad
g^{-1}=\mu.
$$

Thus Möbius inversion is part of a larger algebraic theory of arithmetic functions under Dirichlet convolution.

## Dirichlet Series Interpretation

Dirichlet convolution corresponds to multiplication of Dirichlet series. Since

$$
\mu*\mathbf{1}=\varepsilon,
$$

their Dirichlet series satisfy

$$
\left(\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}\right)
\left(\sum_{n=1}^{\infty}\frac1{n^s}\right)
=1.
$$

But

$$
\sum_{n=1}^{\infty}\frac1{n^s} =
\zeta(s).
$$

Therefore

$$
\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s} =
\frac1{\zeta(s)}.
$$

This analytic identity reflects the algebraic fact that $\mu$ is the inverse of $\mathbf{1}$ under convolution.

## Role in Number Theory

Möbius inversion is one of the central tools of multiplicative number theory. It reverses divisor summation and transforms global divisor information into local arithmetic data.

It appears in counting problems, coprimality arguments, inclusion-exclusion formulas, analytic number theory, finite fields, and combinatorics.

The key principle is that divisor sums are invertible because the Möbius function provides the arithmetic analogue of alternating cancellation.