# Mobius Function for 2^X

• Apr 11th 2010, 07:31 PM
redsoxfan325
Mobius Function for 2^X
We have to find the Mobius function $\displaystyle \mu$ for the set $\displaystyle 2^X$ such that $\displaystyle F(Y) = \sum_{Z\subset Y}f(Z)$ for any $\displaystyle Y\subset X\Longleftrightarrow f(Y)=\sum_{Z\subset Y}\mu(Z)F(Y\backslash Z)$.

I started doing this for a few basic sets. If $\displaystyle X=\{a,b\}$, then

$\displaystyle \mu(\emptyset) = 1$
$\displaystyle \mu(\{a\}) = \mu(\{b\}) = -1$
$\displaystyle \mu(\{a,b\}) = 1$

I know the $\displaystyle \mu$-function for integers is

$\displaystyle \mu(n)=\bigg\{\begin{array}{lr}0:&\mbox{n~is~not~s quarefree}\\(-1)^{p(n)}:&\mbox{n~is~squarefree}\end{array}$

where $\displaystyle p(n)$ is the number of prime factors of $\displaystyle n$.

EDIT: I've had a recent breakthrough (I think) and I believe that $\displaystyle \mu(Y)=(-1)^{|Y|}$. So unless that's incorrect, I don't need any help.
• Apr 11th 2010, 09:17 PM
aliceinwonderland
Quote:

Originally Posted by redsoxfan325
We have to find the Mobius function $\displaystyle \mu$ for the set $\displaystyle 2^X$ such that $\displaystyle F(Y) = \sum_{Z\subset Y}f(Z)$ for any $\displaystyle Y\subset X\Longleftrightarrow f(Y)=\sum_{Z\subset Y}\mu(Z)F(Y\backslash Z)$.

I started doing this for a few basic sets. If $\displaystyle X=\{a,b\}$, then

$\displaystyle \mu(\emptyset) = 1$
$\displaystyle \mu(\{a\}) = \mu(\{b\}) = -1$
$\displaystyle \mu(\{a,b\}) = 1$

I know the $\displaystyle \mu$-function for integers is

$\displaystyle \mu(n)=\bigg\{\begin{array}{lr}0:&\mbox{n~is~not~s quarefree}\\(-1)^{p(n)}:&\mbox{n~is~squarefree}\end{array}$

where $\displaystyle p(n)$ is the number of prime factors of $\displaystyle n$.

EDIT: I've had a recent breakthrough (I think) and I believe that $\displaystyle \mu(Y)=(-1)^{|Y|}$. So unless that's incorrect, I don't need any help.

Should that be $\displaystyle F(Y) = \sum_{Z\subset Y}f(Z)$ for any $\displaystyle Y\subset X\Longleftrightarrow f(Y)=\sum_{Z\subset Y}\mu(Z, Y)F(Z)$, where $\displaystyle \mu(Z, Y)= (-1)^{l(Z,Y)}$ and $\displaystyle l(Z, Y)$ is the length of the interval of [Z,Y] in the order-theoretic term?

In your example, I think if F(Y) is the cardinality of the set Y = {a, b}, then we can apply
$\displaystyle \mu (\emptyset, \{a,b\})= (-1)^2=1$, $\displaystyle \mu(\emptyset, \{a\}) = -1$, $\displaystyle \mu(Y, Y) = 1$, etc.