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6.6 Mobius inversions

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