# Motivation for definition of Mobius function

Why is the Mobius function defined the way it is? \begin{align*} \mu(n) = \begin{cases} (-1)^r & \text{ if $n$ is square-free and is of the form }n=p_1p_2\ldots p_r\\ 0 & \text{ if $n$ is not square-free} \end{cases} \end{align*}

I can see that the function takes $-1$ on all primes. But why is extended in a way it is just multiplicative and not completely multiplicative?

Also, why is this particular function interesting to study? I can understand studying other arithmetic functions like the divisor function, totient function, etc. This function definition seems to be pulled out of thin air.

Thanks

• Have a look at "mobius inversion". Dec 31 '13 at 16:28
• @GerryMyerson Maybe flesh out that comment in your answer? I think the inversion application and the inclusion-exclusion principle connections actually make a better answer than the not-obviously-motivated appearance of the zeta function... Dec 31 '13 at 16:36
• @StevenStadnicki In the spirit of "Mobius inversion", can we think of the Mobius function as what the complex exponential try to do in reconstructing the function from its frequencies? Dec 31 '13 at 16:47
• @Steven, I'm thinking that Mobius Inversion is in so many intro textbooks and on so many websites that it would be superfluous for me to flesh it out here --- although it might be a good exercise for Leslie to follow up on and then post an answer. Inclusion-exclusion is an excellent motivation --- why don't you write it up? Dec 31 '13 at 17:06
• @GerryMyerson That's sensible; I don't know if I'll have a good opportunity to today, but if work lightens up I'll see if I can put together something about inclusion-exclusion. Dec 31 '13 at 17:49

Let $\mathbb Z^+$ be the set of all positive integers and let $\mathbb C$ be the set of all complex numbers. A function $f:\mathbb Z^+ \to \mathbb C$ is called multiplicative if $f(1) = 1$ and $f(ab) = f(a)f(b)$ for all relatively prime $a$ and $b$.

Let $p_i$ represent the $i^\text{th}$ prime number. Then every positive integer $x$ can be written uniquely as an infinite product $\displaystyle x = \prod_{i=1}^\infty p_i^{\alpha_i}$ where we require that all but a finite number of the $\alpha_i$ be equal to $0$. It follows that, if $f$ is a multiplicative function, then $\displaystyle f(x) = \prod_{i=1}^\infty f(p_i^{\alpha_i})$.

The convolution of two multiplicative functions, say $f$ and $g$, is defined as $f*g$ where $$(f*g)(n) = \sum_{ab=n} f(a)g(b)$$

The set $\mathcal F$ of all multiplicative functions is an abelian group with respect to the convolution operator.

Two important multiplicative functions are $\epsilon$ and $\mathbf 1$ defined as $$\epsilon(n) = \left\{ \begin{array}{ll} 1 & \text{If}\; n = 1\\ 0 & \text{If}\; n \ne 1 \end{array} \right.$$

and

$$\mathbf 1(n) = 1$$

It is easy to prove that $\epsilon$ is the multiplicative identity of the group $[\mathcal F, *]$.

For any multiplicative function $f$, note that $$\displaystyle (f*\mathbb 1)(n) = \sum_{a|n} f(a)$$

$\mathbf{Theorem. }$ Let $f$ and $g$ be multiplicative functions. Define $F = f*\mathbf 1$ and $G = g*\mathbf 1$. Define $h = f*g$ and $H = h*1$. Then $H(n) = F(n)G(n)$ for all positive integers $n$.

In a way, $\; f*1$ behaves very much like a Fourier transform of $\; f$.

It follows that there are times when we know what $F = f*\mathbf 1$ is and we need to know what $f$ is. This is where $\mu$, the Mobius inversion function, comes to the rescue. $\mu$ is defined as the inverse of $\mathbf 1$. That is

$$\mathbf 1 * \mu = \epsilon$$

We make a few computations. Let $p$ be a prime number and let $\alpha$ be a non negative integer.

\begin{align} (1*\mu)(1) &= \epsilon(1)\\ \mu(1) &= 1 \end{align}

\begin{align} (1*\mu)(p) &= \epsilon(p)\\ 1 + \mu(p) &= 0\\ \mu(p) &= -1 \end{align}

\begin{align} (1*\mu)(p^2) &= \epsilon(p^2)\\ 1 + \mu(p)+\mu(p^2) &= 0\\ \mu(p^2) &= 0\end{align}

\begin{align} (1*\mu)(p^3) &= \epsilon(p^3)\\ 1 + \mu(p)+\mu(p^2)+\mu(p^3) &= 0\\ \mu(p^3) &= 0\end{align}

We see that

$$\mu(p^\alpha) = \left\{ \begin{array}{rl} 1 & \text{If}\; \alpha = 0\\ -1 & \text{If}\; \alpha = 1\\ 0 & \text{If}\; \alpha \ge 2 \end{array} \right.$$

You can infer the usual definition of $\mu$ from this and the fact that $\mu$ is a multiplicative function.

If you know about the Riemann zeta function, $\zeta(s)$, then $${1\over\zeta(s)}=\sum_{n=1}^{\infty}{\mu(n)\over n^s}$$ for all complex $s$ with real part exceeding 1.

If you don't know about the Riemann zeta function, look it up --- it's the most important function in analytic number theory.

• So the primary motivation is that it allows us to invert stuff? And yes, I know about the zeta function and also a proof of the result. Dec 31 '13 at 16:32
• It is the inverse of the constant sequence 1 under the operation of Dirichlet convolution, and both the zeta formula and the Mobius Inversion formula come from that. Dec 31 '13 at 17:03

I felt the need to add that the definition in the OP isn't the only definition that we have of $\mu(n)$. It is sometimes also defined as the sum of the complex primitive $n^\text{th}$ roots of unity.

It can be shown that this definition is equivalent to the definition in the OP. This is a definition that most people would not consider to have appeared out of thin air.

The Möbius function rightly defined is: $$\underset{n > 1}{\mu(n)} = - \underset{a = n}{\sum_{a \geq 2}} 1 + \underset{ab = n}{\sum_{a \geq 2} \sum_{b \geq 2}} 1 - \underset{abc = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} 1 + \underset{abcd = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} 1 - \cdots$$

The definition you posted is comparable to a reduced fraction: \begin{align*} \mu(n) = \begin{cases} (-1)^r & \text{ if n is square-free and is of the form }n=p_1p_2\ldots p_r\\ 0 & \text{ if n is not square-free} \end{cases} \end{align*}

in the way that this binomial sum also gives the Möbius function: $$\mu(n) = \sum_{k \geq 0} (-1)^{(k-1)}\binom{A001222(n)-1}{k}\binom{A001221(n)-1+k}{k}$$ $$\mu(n) = \sum_{k \geq 0} (-1)^{(k-1)}\binom{\Omega(n)-1}{k}\binom{\nu(n)-1+k}{k}$$ The comparison to reduced fractions is that:

$$k=0$$ $$(-1)^{(k-1)}\binom{\Omega(n)-1}{k}\binom{\nu(n)-1+k}{k} \approx -\underset{a = n}{\sum_{a \geq 2}} 1$$ $$k=1$$ $$(-1)^{(k-1)}\binom{\Omega(n)-1}{k}\binom{\nu(n)-1+k}{k} \approx + \underset{ab = n}{\sum_{a \geq 2} \sum_{b \geq 2}} 1$$ $$k=2$$ $$(-1)^{(k-1)}\binom{\Omega(n)-1}{k}\binom{\nu(n)-1+k}{k} \approx - \underset{abc = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} 1$$

$$k=3$$ $$(-1)^{(k-1)}\binom{\Omega(n)-1}{k}\binom{\nu(n)-1+k}{k} \approx + \underset{abcd = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} 1$$

$$\cdots$$

Somehow this reduces down to the definition you posted. I can't prove it though.

• The values of $k$ might be +-1 off the truth, but it is something like the above I posted. Nov 25 '19 at 15:56