I have read in a paper that $\mu(0,1)\neq 0$, for the Möbius function $\mu$ on an atomistic (finite, modular) lattice with $0$ and $1$. Could someone provide me with a reference for this claim? Many thanks in advance.
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2 Answers
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More generally, the Möbius function of a geometric lattice is never 0. Although not stated explicitly in the text of Enumerative Combinatorics, vol. 1, 2nd ed., it is a simple consequence of equation (3.33) on page 277, as mentioned in the solution to Exercise 3.100(b). It also follows from Exercise 3.98 (where the displayed $\leq$ should be $\geq$) and from Exercise 3.99.
answered Nov 29, 2017 at 17:01
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$\begingroup$ Great! I took a quick look through your book but didn't find it explicitly, so these pointers are very helpful. $\endgroup$– Todd Trimble ♦Nov 29, 2017 at 18:13
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(Disclaimer: I am not an expert in this area.) I expect there is more than one way to see this, but the literature on calculating $\mu(0, 1)$ for finite geometric lattices $L$ (which includes your case: a geometric lattice is a finite atomistic semi-modular lattice) is pretty extensive. In this paper by Anders Björner, it is shown that $|\mu(0, 1)|$ counts the number of neat base-families for $L$ (see section 3 for the definition), so in view of that it's just a matter of showing that neat base-families for $L$ exist, and I think that follows pretty readily from the atomisticity.
answered Nov 29, 2017 at 15:04