Möbius function

The Möbius function is multiplicative (i.e. μ(ab) = μ(a) μ(b) whenever a and b are coprime).

The sum of the Möbius function over all positive divisors of n (including n itself and 1) is zero except when n = 1:

${\displaystyle \sum _{d\mid n}\mu (d)={\begin{cases}1&{\text{if }}n=1,\\0&{\text{if }}n>1.\end{cases}}}$

The equality above leads to the important Möbius inversion formula and is the main reason why μ is of relevance in the theory of multiplicative and arithmetic functions.

Other applications of μ(n) in combinatorics are connected with the use of the Pólya enumeration theorem in combinatorial groups and combinatorial enumerations.

There is a formula[4] for calculating the Möbius function without directly knowing the factorization of its argument:

${\displaystyle \mu (n)=\sum _{\stackrel {1\leq k\leq n}{\gcd(k,\,n)=1}}e^{2\pi i{\frac {k}{n}}},}$

i.e. μ(n) is the sum of the primitive n-th roots of unity. (However, the computational complexity of this definition is at least the same as that of the Euler product definition.)

Using

${\displaystyle \mu (n)=\sum _{\stackrel {1\leq k\leq n}{\gcd(k,\,n)=1}}e^{2\pi i{\frac {k}{n}}},}$

the formula

${\displaystyle \sum _{d\mid n}\mu (d)={\begin{cases}1&{\text{if }}n=1,\\0&{\text{if }}n>1\end{cases}}}$

can be seen as a consequence of the fact that the nth roots of unity sum to 0, since each nth root of unity is a primitive dth root of unity for exactly one divisor d of n.

However it is also possible to prove this identity from first principles. First note that it is trivially true when n = 1. Suppose then that n > 1. Then there is a bijection between the factors d of n for which μ(d) ≠ 0 and the subsets of the set of all prime factors of n. The asserted result follows from the fact that every non-empty finite set has an equal number of odd- and even-cardinality subsets.

This last fact can be shown easily by induction on the cardinality |S| of a non-empty finite set S. First, if |S| = 1, there is exactly one odd-cardinality subset of S, namely S itself, and exactly one even-cardinality subset, namely ∅. Next, if |S| > 1, then divide the subsets of S into two subclasses depending on whether they contain or not some fixed element x in S. There is an obvious bijection between these two subclasses, pairing those subsets that have the same complement relative to the subset {x}. Also, one of these two subclasses consists of all the subsets of the set S \{x}, and therefore, by the induction hypothesis, has an equal number of odd- and even-cardinality subsets. These subsets in turn correspond bijectively to the even- and odd-cardinality {x}-containing subsets of S. The inductive step follows directly from these two bijections.

A related result is that the binomial coefficients exhibit alternating entries of odd and even power which sum symmetrically.

In number theory another arithmetic function closely related to the Möbius function is the Mertens function, defined by

${\displaystyle M(n)=\sum _{k=1}^{n}\mu (k)}$

for every natural number n. This function is closely linked with the positions of zeroes of the Riemann zeta function. See the article on the Mertens conjecture for more information about the connection between M(n) and the Riemann hypothesis.

From the formula

${\displaystyle \mu (n)=\sum _{\stackrel {1\leq k\leq n}{\gcd(k,\,n)=1}}e^{2\pi i{\frac {k}{n}}},}$

it follows that the Mertens function is given by:

${\displaystyle M(n)=-1+\sum _{a\in {\mathcal {F}}_{n}}e^{2\pi ia},}$

where F_n is the Farey sequence of order n.

This formula is used in the proof of the Franel–Landau theorem.[5]

The Dirichlet series that generates the Möbius function is the (multiplicative) inverse of the Riemann zeta function; if s is a complex number with real part larger than 1 we have

${\displaystyle \sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}={\frac {1}{\zeta (s)}}.}$

This may be seen from its Euler product

${\displaystyle {\frac {1}{\zeta (s)}}=\prod _{p\in \mathbb {P} }{\left(1-{\frac {1}{p^{s}}}\right)}=\left(1-{\frac {1}{2^{s}}}\right)\left(1-{\frac {1}{3^{s}}}\right)\left(1-{\frac {1}{5^{s}}}\right)\cdots }$

The Lambert series for the Möbius function is:

${\displaystyle \sum _{n=1}^{\infty }{\frac {\mu (n)q^{n}}{1-q^{n}}}=q,}$

which converges for |q| < 1. For prime ${\displaystyle \alpha \geq 2}$, we also have

${\displaystyle \sum _{n=1}^{\infty }{\frac {\mu (\alpha n)q^{n}}{q^{n}-1}}=\sum _{n\geq 0}q^{\alpha ^{n}},|q|<1.}$

Gauss[2] proved that for a prime number p the sum of its primitive roots is congruent to μ(p − 1) (mod p).

If F_q denotes the finite field of order q (where q is necessarily a prime power), then the number N of monic irreducible polynomials of degree n over F_q is given by:[6]

${\displaystyle N(q,n)={\frac {1}{n}}\sum _{d\mid n}\mu (d)q^{n/d}.}$

In combinatorics, every locally finite partially ordered set (poset) is assigned an incidence algebra. One distinguished member of this algebra is that poset's "Möbius function". The classical Möbius function treated in this article is essentially equal to the Möbius function of the set of all positive integers partially ordered by divisibility. See the article on incidence algebras for the precise definition and several examples of these general Möbius functions.

Constantin Popovici[8] defined a generalised Möbius function μ_k = μ ∗ ... ∗ μ to be the k-fold Dirichlet convolution of the Möbius function with itself. It is thus again a multiplicative function with

${\displaystyle \mu _{k}(p^{a})=(-1)^{a}{\binom {k}{a}}\ }$

where the binomial coefficient is taken to be zero if a > k. The definition may be extended to complex k by reading the binomial as a polynomial in k.[9]