A bunch of infinite posets $P$ with $\hat 0$ have the following property

  1. For every $x\in P$, the principal filter $\{ y\in P : y\ge x\}$ is isomorphic as a poset to $P$ itself.

Examples include ${\bf N}$ under the usual order, ${\bf N}$ ordered by divisibility, and the set of all finite subsets of ${\bf N}$.

Separate from this, there is another property I'm interested in, which is "transitivity". (Perhaps it's called "multiplicativity"?)

  1. For every $x\le y\le z$, we have $\mu(x,y)\mu(y,z) = \mu(x,z)$.

What I'd love to know is whether either/both of 1. and 2. have a name, and whether they are possibly related. I'm pretty sure that 1. does not imply 2., since ${\bf N}$ under the usual order satisfies 1. but we have $$\mu(1,2)\mu(2,3) = (-1)(-1) \ne 0 = \mu(1,3).$$ That said I think that the other two posets I mentioned above have property 2. So perhaps 2. implies 1.? In any case I'm interested in any references (or if there was a name for these properties I could Google on my own!).

Edit: I'd also be interested in any sort of natural conditions that are necessary or sufficient for either 1. or 2. to arise.


1 Answer 1


This looked quite interesting at first – unfortunately, little of it actually pans out.

Your example of $\mathbb N$ ordered by divisibility doesn’t have the “transitivity” property: We have $1\mid2\mid4$ but $\mu(1,2)\mu(2,4)=(-1)(-1)\ne0=\mu(1,4)$. (This is because the classical Möbius function is only a multiplicative function and not a completely multiplicative function, i.e. it’s determined by the product of its values at the prime powers, not the primes).

The transitivity property is a lot more restrictive than it might seem at first sight. Since $\mu(x,y)=-1$ if $y$ covers x, it implies that if $a\le b$ then $\mu(a,b)$ is $\pm1$ according to the parity of the length of paths from $a$ to $b$ in the Hasse diagram (which implies that all paths from $a$ to $b$ have the same parity.) So it might more aptly be said that the Möbius function is “binary” or “determined by parity”.

And it doesn’t imply the uniformity property $1$. For instance, consider the poset of sets that are subsets of one of two copies of $\mathbb N$ (ordered by inclusion). The empty set is the least element. Subsets of different copies are incomparable. If $x\le y\le z$, then $x$, $y$ and $z$ are all subsets of the same copy, and the Möbius function for each copy is the same as for $\mathbb N$ alone, so this poset satisfies the “transitivity” property. But it looks different at the empty set than at any other element. That there is no isomorphism between the entire poset and the principal filters at points other than the empty set follows e.g. from the fact that two singletons from different copies cover the empty set but are not covered by a common set, whereas at all other points the covering sets obtained by adding elements $x$ and $y$ are covered by the set obtained by adding both $x$ and $y$.

By the way, another example of an infinite poset with least element with the uniformity property is the poset of subspaces of the vector space $\mathbb F^\mathbb N$ (where $\mathbb F$ is a finite field), ordered by the subspace relation.

  • 1
    $\begingroup$ Thanks for this super detailed answer! Yes I realised the thing about the classical Möbius function not actually being multiplicative for all triples, but forgot to edit my post. That being said I really appreciate your in-depth analysis of this subject :) $\endgroup$
    – marcelgoh
    Feb 5, 2023 at 17:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .