# Completely Multiplicative Functions

## Multiplicative and Completely Multiplicative Functions

An arithmetic function is a function defined on the positive integers. Such a function

$$
f:\mathbb{N}\to\mathbb{C}
$$

is called multiplicative if

$$
f(ab)=f(a)f(b)
$$

whenever

$$
\gcd(a,b)=1.
$$

It is called completely multiplicative if the same identity holds for all positive integers $a$ and $b$, without assuming coprimality:

$$
f(ab)=f(a)f(b).
$$

Thus complete multiplicativity is stronger than multiplicativity.

## Basic Examples

The constant function

$$
f(n)=1
$$

is completely multiplicative.

The identity function

$$
f(n)=n
$$

is also completely multiplicative, since

$$
f(ab)=ab=f(a)f(b).
$$

For a fixed real or complex number $s$, the function

$$
f(n)=n^s
$$

is completely multiplicative.

The Liouville function is another important example:

$$
\lambda(n)=(-1)^{\Omega(n)}.
$$

Since

$$
\Omega(ab)=\Omega(a)+\Omega(b),
$$

we have

$$
\lambda(ab)=\lambda(a)\lambda(b)
$$

for all positive integers $a,b$.

## Determination by Prime Values

A completely multiplicative function is determined entirely by its values on primes.

If

$$
n=p_1^{\alpha_1}\cdots p_r^{\alpha_r},
$$

then complete multiplicativity gives

$$
f(n)=f(p_1)^{\alpha_1}\cdots f(p_r)^{\alpha_r}.
$$

Thus once the values

$$
f(p)
$$

are known for all primes $p$, the function is known on every positive integer.

For example, if

$$
f(p)=-1
$$

for every prime $p$, then

$$
f(n)=(-1)^{\Omega(n)}=\lambda(n).
$$

## Difference from Ordinary Multiplicativity

For an ordinary multiplicative function, one only has

$$
f(ab)=f(a)f(b)
$$

when $a$ and $b$ are coprime.

Thus the values on prime powers must be specified separately:

$$
f(p), f(p^2), f(p^3),\ldots
$$

For a completely multiplicative function, however,

$$
f(p^\alpha)=f(p)^\alpha.
$$

This is a strong restriction.

For example, the Möbius function is multiplicative but not completely multiplicative. We have

$$
\mu(2)=-1,
$$

so complete multiplicativity would imply

$$
\mu(4)=\mu(2)^2=1.
$$

But actually

$$
\mu(4)=0.
$$

Therefore $\mu$ is not completely multiplicative.

## Dirichlet Series

Completely multiplicative functions have especially simple Dirichlet series.

Suppose $f$ is completely multiplicative and the series converges absolutely. Then

$$
\sum_{n=1}^{\infty}\frac{f(n)}{n^s} =
\prod_p
\left(1+\frac{f(p)}{p^s}+\frac{f(p)^2}{p^{2s}}+\cdots\right).
$$

Since the local series is geometric,

$$
1+\frac{f(p)}{p^s}+\frac{f(p)^2}{p^{2s}}+\cdots =
\frac{1}{1-f(p)p^{-s}}.
$$

Therefore

$$
\sum_{n=1}^{\infty}\frac{f(n)}{n^s} =
\prod_p
\frac{1}{1-f(p)p^{-s}}.
$$

This Euler product is simpler than the general multiplicative case because each local factor is determined by a single value $f(p)$.

## Characters as Examples

Dirichlet characters are important examples of completely multiplicative functions, after being extended periodically and with zeros on non-coprime integers.

A Dirichlet character modulo $n$ is a function $\chi$ satisfying

$$
\chi(ab)=\chi(a)\chi(b)
$$

for all integers $a,b$, together with periodicity modulo $n$.

Such functions are central in the study of primes in arithmetic progressions and Dirichlet $L$-functions.

They show that complete multiplicativity is not merely a formal property. It encodes arithmetic symmetries.

## Cancellation

Many completely multiplicative functions take values on the unit circle or among signs.

For such functions, sums like

$$
\sum_{n\le x} f(n)
$$

measure cancellation. If $f(n)$ behaves randomly, positive and negative or complex values should partly cancel.

For the Liouville function, this sum is

$$
\sum_{n\le x}\lambda(n).
$$

Understanding its cancellation is connected to deep questions about prime factorization and the zeros of the zeta function.

## Role in Number Theory

Completely multiplicative functions are basic objects in multiplicative number theory. They translate multiplication of integers directly into multiplication of function values.

Their structure is rigid: values on primes determine everything. This makes them algebraically simple and analytically useful.

They appear in Euler products, Dirichlet characters, $L$-functions, sign patterns of prime factors, and cancellation problems. Complete multiplicativity is therefore one of the cleanest ways to encode prime factorization into arithmetic functions.