Given a series $p_n(s)=\sum_{k=1}^n a_k $. I'd like to get $\hat{p}_n(s)=\sum_{k=1}^n \mu(k)a_k $. Think of $a_k=k^{-s}$ for example. If you let $n$ go to $\infty$, you'll see the well-known relation between $\zeta$ and $1/\zeta$.
May I apply something like a Möbius operator on $p_n(s)$, such that $\mu\circ p=\hat p$?
The wiki page on the Primon gas mentions the Fermi operator $(-1)^F$, so I thought about the following: Given a $n$-dimensional vector $\vec a=(a_k)^T$. Let $\vec e=(1,\dots,1)^T$, then $p=\vec e \cdot \vec a$ and $\hat p=\vec e \cdot (\mu \vec a)$, with $\mu$ being a diagonal matrix with entries $\mu_{kk}=\mu(k)$. Is this a finite dimensional matrix representation of the Fermi operator?
EDIT: The thing is, I don't see why there is on the one hand something like
Another inversion formula is (where we assume that the series involved are absolutely convergent): $ g(x) = \sum_{m=1}^\infty \frac{f(mx)}{m^s}\quad\mbox{ for all } x\ge 1\quad\Longleftrightarrow\quad f(x) = \sum_{m=1}^\infty \mu(m)\frac{g(mx)}{m^s}\quad\mbox{ for all } x\ge 1. $ (see here)
further on the other
The classic version states that if g(n) and f(n) are arithmetic functions satisfying $ g(n)=\sum_{d\,\mid \,n}f(d)\quad\text{for every integer }n\ge 1 $ then $ f(n)=\sum_{d\,\mid\, n}\mu(d)g(n/d)\quad\text{for every integer }n\ge 1 $ (see there)
and on the last hand
A related inversion formula more useful in combinatorics is as follows: suppose F(x) and G(x) are complex-valued functions defined on the interval [1,∞) such that $ G(x) = \sum_{1 \le n \le x}F(x/n)\quad\mbox{ for all }x\ge 1 $ then $ F(x) = \sum_{1 \le n \le x}\mu(n)G(x/n)\quad\mbox{ for all }x\ge 1. $
None of them fits my needs :-( I want someting like $F(x) = \sum_{1 \le n \le x}\mu(n)G(n)\quad\mbox{ for all }x\ge 1. $
EDIT: Peter Sarnak's Or Gil Kalai's work (e.g. The Depth Of The Möbius Function or The AC0 Prime Number Conjecture) shows the right kind of formulas, but I'm not sure if they are talking about what I need: My function $G(n)$ would sent the natural numbers to $be^{ia}$, with $b$ being an integer and not only to $\{-1,1\}$...