Eulerian poset

A graded poset $P$ is Eulerian if for any $x\le y$ in $P$, theMöbius function (http://planetmath.org/MoebiusInversionFormula)of the interval $[x,y]$ is given by

$${\mu }_P(x,y)={(-1)}^{\rho (y)-\rho (x)},$$

where $\rho$ is the rank function of $P$. If $P$ is finite andgraded with strictlypositive rank, then we can equivalently and more simply say that $P$ isEulerian if and only if each nontrivial interval of $P$ has the same number ofelements of even rank as odd rank.

For example, the lattice (http://planetmath.org/Lattice) $M_3$ is not Eulerian, because ithas 3 elements of rank 1 but only 2 elements of ranks 0 or 2.

On the other hand, the Boolean lattice $B_n$ is always Eulerian.

Each interval of $B_n$ is isomorphic to a smaller Boolean lattice, sowe can give a proof by induction that $B_n$ is Eulerian. Now $B_0$ isa single point and so there is only one interval, which has Möbiusfunction 1. So to show that $B_n$ is Eulerian, we just need to showthat the Möbius function of $B_n$ is ${(-1)}^n$. Recall that by definition,

$${\mu }_{B_n}(B_n)=-\sum _{S\subset \{1,\mathrm{\dots },n\}}\mu (\mathrm{\varnothing },S)$$(*)

Thus ${\sum }_{S\subseteq \{1,\mathrm{\dots },n\}}{\mu }_{B_n}(\mathrm{\varnothing },S)=0$. We can applythe binomial theorem to $0$ to see that$0={(-1+1)}^n={\sum }_{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^k$and thus

${(-1)}^n$$=-{\displaystyle \sum _{k=0}^{n-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^k$

$=-{\displaystyle \sum _{S\subset \{1,\mathrm{\dots },n\}}}\mu (\mathrm{\varnothing },S)$(**)

It then follows from Equation $(*)$ and Equation $(**)$ that $\mu (B_n)={(-1)}^n$.

Eulerian posets are generalizations of certain special posets, such asface lattices of polytopes.

Eulerian posets of small rank

In this section we classify bounded Eulerian posets of rank less than 4.

For , there is exactly one bounded poset of rank $n$, the chain,and it is Eulerian.

A bounded poset of rank $2$ is determined precisely by its number ofcoatoms. The ab-index of a rank $2$ bounded Eulerian poset with $r$coatoms is $a+(r-1)b$, which is a cd-index if and only if $r=2$.Since every Eulerian poset has a cd-index, this implies that everyrank $2$ bounded Eulerian poset has exactly two coatoms. Thus up toisomorphism $B_2$ is the only rank $2$ bounded Eulerian poset.

Let $P$ be a rank $3$ bounded Eulerian poset. Since $P$ has as manyelements of odd rank as even rank, it has exactly the same number ofatoms as coatoms. Since every rank $2$ interval in $P$ is a copy of$B_2$, it follows that $P$ is not a chain and so has $n\ge 2$ atoms.Moreover, every atom is covered by exactly two coatoms, and everycoatom covers exactly two atoms. That is, the Hasse diagram of theinterior of $P$ is a 2-regular bipartite graph. This graph admits adecomposition into even cycles. In each cycle, half of the nodes areatoms and half are coatoms. The non-induced subposet of $P$consisting of the points on a cycle of length $2k$ plus the minimumand maximum elements of $P$ is the face poset of a $k$-gon. Hence $P$itself is the face poset of the disjoint union of one or more $k$-gonsfor various $k\ge 2$.