The Mobius function $\mu$ of a poset $P$ is the function $\mu:P^2\to \mathbb{Z}$ defined by the recursion $$ \mu(x,y) = \begin{cases} 1, &x=y \\ -\sum_{x\leq z\leq y} \mu(x,z), & x< y \\ 0, &\text{else}. \end{cases} $$
My understanding is that it is possible to use the recursion to prove that $|\mu(x,y)| \leq (|P|-1)!$, where $|P|$ is the number of elements in the poset $P$.
I am interested in other bounds for the Mobius function in the case that $P$ has some more structure. For example, if $P$ is the poset formed by taking subsets of a finite set using inclusion as the partial ordering, one has $|\mu(x,y)|\leq 1$ for every $x,y\in P$.
This makes me suspect for example that if $P$ is a poset formed by taking some subset of the power set of a finite set ($P\subset \mathcal{P}([n])$), there might be a more reasonable bound for $\max_{x,y\in P} |\mu(x,y)|$.
My question is whether there are any special classes of partially ordered sets for which a substantially better bound for the maximum (absolute) value of the mobius function is known (perhaps there are bounds for the mobius function of a lattice?)
I found a bound that is useful enough for me. The starting point is the identity due to Phillip Hall: $$ \mu(x,y) = \sum (-1)^{\ell(C)}, $$ where the sum is over chains $C$ in the poset with minimal element $x$ and maximal element $y$. As a trivial consequence, $|\mu(x,y)|$ is bounded by the number of chains in the poset.
There are at most $2^n n!$ chains in the boolean algebra on $n$ elements, so if for example $P$ can be embedded as a subset of this poset one has the bound $|\mu_P(x,y)| \leq 2^nn!$.