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6.6 Mobius inversions


Defn:  The Euler function $\phi(n) = |S_n|$ \hfil \break
where $S_n = \{k ~:~ 1\leq k \leq n , GCD(k, n) = 1\}$


Ex: $\phi(1) =  \underline{\hskip 0.5in}$,  $\phi(6) =  \underline{\hskip 0.5in}$, 
 $\phi(15) =  \underline{\hskip 0.5in}$ 

\vskip 10pt
${\cal N}$ is partially ordered by $k \leq n$ iff $k|n$

Let $F: {\cal N} \rightarrow {\cal R}$, $F(n) = \phi(n)$

Then $G(n) = $%%$\Sigma_{\{d ~:~ d|n\}} \phi(d)$.


By thm 6.6.1 $\phi(n) =$%%$ \Sigma_{\{d ~:~ d|n\}} G(d)\mu(d, n)$.



Ex:  $G(6) = \phi(1) + \phi(2) + \phi(3) + \phi(6) = 1 + 1 + 2 + 2 = 6$



Note that $G(d) = d$.  



Hence $\phi(n) = \Sigma_{\{d ~:~ d|n\}} d ~\mu(d, n)$.


Let  $\zeta(x, y) = \cases{1 & if $x \leq y$ \cr 0 & otherwise}$ 
%%$\in \{f: X \times X \rightarrow {\cal R} ~:~$ 
%%\rightline{if $f(x, y) \not= 0$, then $x \leq y \}$}

 $\zeta(x, y) = \cases{1 & if %%  $x | y$ 
\cr 0 & otherwise}$ 

The  Mobius function, $\mu = \zeta^{-1}$
\eject
Suppose $f(x, x) \not= 0 \forall x \in X$

Then $f^{-1}(x, x) = {1 \over f(x, x)}$

$f^{-1}(x, y) = -\Sigma_{\{z~:~ x \leq z < y\}}f^{-1}(x, z){f(z, y) \over f(y, y)}$ for $x < y$,

$f^{-1}(x, y) = 0$ for $x > y$.


Since $\zeta(x, x) = 1 \forall x \in X$,  Mobius fn, $\mu = \zeta^{-1}$
exists.

 $\mu(x, x) = {1 \over \zeta(x, x)} = 1$

Hence $\mu(1, 1) = 1$

$\mu(x, y) = -\Sigma_{\{z~:~ x \leq z < y\}}\mu(x, z){\zeta(z, y) \over \zeta(y, y)}$ 
$ = -\Sigma_{\{z~:~ x \leq z < y\}}\mu(x, y)$ 

  $\mu(1, d) = -\Sigma_{\{z~:~ 1 \leq z < d\}}\mu(1, z)$
= $ -\Sigma_{\{z~:~ z | d, z \not= d \}}\mu(1, z)$

 
$\mu(1, 2) = -\Sigma_{\{z~:~ 1 \leq z < 2\}}\mu(1, z) = -\mu(1, 1) = -1$ 


$\mu(1, 3) = -\Sigma_{\{z~:~ 1 \leq z < 3\}}\mu(1, z) = -\mu(1, 1) = -1$ 

$\mu(1, 4) = -\Sigma_{\{z~:~ 1 \leq z < 4\}}\mu(1, z) =$ 
%%$-\mu(1, 1) - \mu(1, 2) = 0$ 

$\mu(1, 5) = -\Sigma_{\{z~:~ 1 \leq z < 5\}}\mu(1, z) = -\mu(1, 1) =$
 $-1$ 

$\mu(1, 6) = -\Sigma_{\{z~:~ 1 \leq z < 6\}}\mu(1, z) = -\mu(1, 1) - \mu(1, 2)
- \mu(1, 3) =$ 
%%$= 1 -1 -1 = -1$ 
\vfill
$\mu(1, 12) = -\Sigma_{\{z~:~ 1 \leq z < 30\}}\mu(1, z) = $

\line{$-\mu(1, 1) -
\mu(1, 2) - \mu(1, 3) - \mu(1, 4)- \mu(1, 6)= $}
%%1
%%-1 -1 = -1$

\vfill 
$\mu(1, 30) = -\Sigma_{\{z~:~ 1 \leq z < 30\}}\mu(1, z) = $

\line{$-\mu(1, 1) - 
\mu(1, 2) - \mu(1, 3) - \mu(1, 5)- \mu(1, 6)- \mu(1, 10) - \mu(1, 15)= $}
%%1 
%%-1 -1 = -1$ 

\eject

Suppose $d = p_1^{k_1}p_2^{k_2}....p_r^{k_r}$


$\mu(1, d) = \cases{1 & if $d = 1$ \cr
(-1)^r & if $k_i = 1 \forall i$ \cr
0 & otherwise}$

%%$\not= 0$ iff $d = 1$ or $d = p_1p_2... p_k$ where $p_i$ are distinct 
%%primes.  In which case,


%%$\mu(1, d) = -\Sigma_{\{z~:~ 1 \leq z < d\}}\mu(1, z) = (-1)^r$ where 
%%$r$ is the number of distinct primes dividing $d$ 


Suppose $kx = y$, then $\mu(x, y) = \mu(1, y/x)$


Hence $\phi(n) = \Sigma_{\{d ~:~ d|n\}} d ~\mu(d, n)$ =

%%Hence $\phi(n) = \Sigma_{\{d ~:~ d|n\}} d \mu(1, n/d)
%%= \Sigma_{\{d ~:~ d|n\}} n/d \mu(1, d)$.



\end
Let $F: {\cal N} \rightarrow {\cal R}$, $F(n) = \phi(n)$

Then $G(n) = \Sigma{\{d ~:~ d|n\}} \phi(d)$.


$\phi(n) = \Sigma{\{d ~:~ d|n\}} G(d)\mu(d, n)  $.




By thm 6.6.1:








\eject

Let $S_n^d = \{k ~:~ 1\leq k \leq n , GCD(k, n) = d\}$


Note if $d|n$, then $|S_n^d| = \phi(n/d)$

Proof:  Let $f: S_{n \over d} \rightarrow S_n^d$, $f(k) = dk$

Claim:  $f$ is well defined:  \hfil \break
if $GCD(k, n/d) = 1$, then $GCD(dk, n) = d$


Claim $f$ is onto:\hfil \break
Suppose $ GCD(m, n) = d$, then $ GCD(m/d, n/d) = d/d$ and $f(m/d) = m$

Hence $f$ is a bijection and  $|S_n^d| = \phi(n/d)$




${\cal N}$ is partially ordered by $k \leq n$ iff $k|n$


Let  $\zeta(x, y) = \cases{1 & $x \leq y$ \cr 0 & otherwise}$ 
%%$\in \{f: X \times X \rightarrow {\cal R} ~:~$ 
%%\rightline{if $f(x, y) \not= 0$, then $x \leq y \}$}

 $\zeta(x, y) = \cases{1 & $x | y$ \cr 0 & otherwise}$ 

The  Mobius function, $\mu = \zeta^{-1}$
\eject
Suppose $f(x, x) \not= 0 \forall x \in X$

Then $f^{-1}(x, x) = {1 \over f(x, x)}$

$f^{-1}(x, y) = -\Sigma_{\{z~:~ x \leq z < y\}}f^{-1}(x, z){f(z, y) \over f(y, y)}$ for $x < y$,

$f^{-1}(x, y) = 0$ for $x > y$.


Since $\zeta(x, x) = 1 \forall x \in X$,  Mobius fn, $\mu = \zeta^{-1}$
exists.

 $\mu(x, x) = {1 \over \zeta(x, x)} = 1$

Hence $\mu(1, 1) = 1$

$\mu(x, y) = -\Sigma_{\{z~:~ x \leq z < y\}}\mu(x, z){\zeta(z, y) \over \zeta(y, y)}$ 
$ = -\Sigma_{\{z~:~ x \leq z < y\}}\mu(x, y)$ 
 
$\mu(1, 2) = -\Sigma_{\{z~:~ 1 \leq z < 2\}}\mu(1, z) = -\mu(1, 1) = -1$ 


$\mu(1, 3) = -\Sigma_{\{z~:~ 1 \leq z < 3\}}\mu(1, z) = -\mu(1, 1) = -1$ 

$\mu(1, 4) = -\Sigma_{\{z~:~ 1 \leq z < 4\}}\mu(1, z) = -\mu(1, 1) - \mu(1, 2) = 0$ 

$\mu(1, 5) = -\Sigma_{\{z~:~ 1 \leq z < 4\}}\mu(1, z) = -\mu(1, 1) = -1$ 

$\mu(1, n) = -\Sigma_{\{z~:~ 1 \leq z < 4\}}\mu(1, z) = \phi(n)$ 

Let $x < y$, let $d = gcd(x, y)$, then $\mu(x, y) = \mu(x/d, y/d)$


Let $F: {\cal N} \rightarrow {\cal R}$, $F(n) = \phi(n)$

Then $G(n) = \Sigma{\{d ~:~ d|n\}} \phi(d)$.


$\phi(n) = \Sigma{\{d ~:~ d|n\}} G(d)\mu(d, n)$.




By thm 6.6.1:

















\end


Let $X$ be a finite set.

Let ${\cal F} = \{f: X \times X \rightarrow {\cal R} ~:~$ if $f(x, y) \not= 0$, then 
$x \leq y \}$


Define the operation * on ${\cal F}$ by 
$$f*g = \cases{\Sigma_{\{z~:~ x \leq z \leq y \}} f(x, z)g(z, y) & if $x \leq y$ \cr 0 & otherwise}$$

Note $*$ is associative:  $f*(g*h) = (f*g)*h$


For example $\delta = \cases{1 & x = y \cr 0 & otherwise}$



Note $\delta$ acts as the indentity for *:  $f* \delta = \delta * f = f$


Let $\zeta(x, y) = \cases{1 & $x \leq y$ \cr 0 & otherwise}$

If $f(x, x) \not= 0$ for all $x \in X$, then $f$ is invertible:  There exist $f^{-1}$ such that $f*f^{-1} = f^{-1}*f = \delta$.

Def. The  Mobius function, $\mu = \zeta^{-1}$


Example:  if $X = X_n = \{1, 2, ..., n\}$  and ${\cal P}(X_n) $ is partially ordered by the relation $\subset$, then $\mu(A, B) = (-1)^{|B|-|A|}$


Thm 6.6.1.  Let $(X, \leq)$ be a partially ordered set with a smallest element 0.  If $F: X \rightarrow {\cal R}$, define $G: X  \rightarrow {\cal R}$ by $G(x) = \Sigma_{\{z ~:~ z \leq x\}}F(z)$.  Then

$F(x) = \Sigma_{\{y ~:~ y \leq x\}}G(y)\mu(y, x) $

Proof:  Evaluate $\Sigma_{\{y ~:~ y \leq x\}}G(y)\mu(y, x) $

\end

















Let $X_n = \{1, 2, ..., n\}$ 

${\cal P}(X_n) = \{A \subset X_n \}$

${\cal P}(X_n) $ is partially ordered by the relation $\subset$.

Suppose $F: {\cal P}(X_n)  \rightarrow {\cal R}$

Define $G: {\cal P}(X_n)  \rightarrow {\cal R}$ by $G(K) = \Sigma_{L \subset K} F(L)$


Claim:  we can invert this equation to recover $F$ from $G$:  $$F(K) = \Sigma_{L \subset K} (-1)^{|K| - |L| }G(L)$$


Example:  Let $S$ be any finite set.  Let $A_i \subset S$ and \hfil \break
 let $A_i$ be indexed by elements of $X_n$  \hfil \break
(ie, we have $A_1$, $A_2$, ..., $A_n$).


Let $F: {\cal P}(X_n)  \rightarrow {\cal R}$ be defined by \hfil \break
$F(K) = |\{ s ~:~ s \not\in A_i \forall i \in K, s \in A_j \forall j \not\in K \}|$


$\{ s ~:~ s \not\in A_i ~\forall i \in K, s \in A_j ~\forall j \not\in K \}$
\vskip -5pt
=
$\{ s ~:~ s \in \overline{A_i} ~\forall i \in K, s \in A_j ~\forall j \in \overline{K} \}$
\vskip -5pt
$= (\cap_{i \in K}\overline{A_i}) \cap  (\cap_{i \in \overline{K}}A_i)$ 
\vfill
\eject

Suppose $S = \{a_1, a_2, a_3, a_4\}$.  Suppose $X_n = \{1, 2\}$

Let $A_1 =  \{a_1, a_2, a_3\}$. Let $A_2 = \{a_1, a_2\}$. 

$F(\emptyset) =  |(\cap_{i \in \emptyset}\overline{A_i}) \cap  (\cap_{i \in {X_n}}A_i)| = |A_1 \cap A_2| =  |\{a_1, a_2\}| = 2$ 

$F( \{1\} ) =  |(\cap_{i \in  \{1\} }\overline{A_i}) \cap  (\cap_{i \in  \{2\}} A_i)| = |\overline{A_1} \cap A_2| =  |\emptyset| = 0$ 

$F( \{2\} ) =  |(\cap_{i \in  \{2\} }\overline{A_i}) \cap  (\cap_{i \in  \{1\}} A_i)| = |\overline{A_2} \cap A_1| =  |\{a_3\}| = 1$ 

$F( \{1, 2\} ) =  |(\cap_{i \in  \{1, 2\} }\overline{A_i}) \cap  (\cap_{i \in  \emptyset} A_i)| = |\overline{A_1} \cap \overline{A_2}| =  |\{a_4\}| = 1$ 

 $G(K) = \Sigma_{L \subset K} F(L) = |\cap_{i \in \overline{K}}A_i)|$

$G(\emptyset)  = F(\emptyset) = 2 $

$G( \{1\} ) = F(\emptyset)  +  F( \{1\} ) = 2 $

$G( \{2\} ) = F(\emptyset)  +  F( \{2\} ) = 3 $

$G( \{1, 2\} ) = F(\emptyset)  +  F( \{1\} ) + F( \{2\} ) + F( \{1, 2\} ) = 4$

By claim, $F(K) = \Sigma_{L \subset K} (-1)^{|K| - |L| }G(L)$

Hence $F(X_n) = \Sigma_{L \subset X_n} (-1)^{ |X_n| - |L| }G(L)$

%%\hskip 0.8in $ = \Sigma_{L \subset X_n} (-1)^{n - |L| }G(L) $

%%\hskip 0.8in $=  \Sigma_{L \subset X_n} (-1)^{ |X_n|  - |L| } |\cap_{i \in \overline{L}}A_i)| $

\hskip 0.8in $=  \Sigma_{L \subset X_n} (-1)^{ |\overline{L}| } |\cap_{i \in \overline{L}}A_i)| $

\hskip 0.8in $=  \Sigma_{\overline{L} \subset X_n} (-1)^{ |\overline{L}| } |\cap_{i \in \overline{L}}A_i)| $

\hskip 0.8in $=  \Sigma_{K \subset X_n} (-1)^{ |K| } |\cap_{i \in K}A_i)| $

\eject

\end