An **abstract polytope** is a poset $X$ (here finite), whose elements are called faces, satisfying these 4 conditions:


1. There is a least face and a greatest face.
2. All flags (i.e. maximal chains) have the same number of faces.
3. $X$ is strongly connected, i.e. for every interval $[F\_1,F\_2]$ in $X$ and $F, F' \in (F\_1,F\_2)$ there is a way to get from $F$ to $F'$ via adjacent faces in $(F\_1,F\_2)$.
4. If $F\_1 < F\_2$ differ in rank by 2, then there are exactly two intermediate faces $F, F'$ in the interval $(F\_1,F\_2)$.


Denote the set of faces of rank $i$ in $X$ by $X\_i$ and the rank of $X$ by $n$.
My question is:



> 
> If $X$ is an abstract polytope with $\chi(X) = 0$, i.e. $\sum\_{i=-1}^n (-1)^i X\_i = 0$, then is $X$ necessarily Eulerian, i.e. is it the case that for every $F\_1 < F\_2$, $\sum\_{i = rank(F\_1)}^{rank(F\_2)} (-1)^i[F\_1,F\_2]\_i = 0$?
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(**Edit:** the $X$'s I have in mind are lattices, i.e. any two distinct faces have a unique join [least upper bound] and meet [greatest lower bound].)



Is a finite abstract polytope of Euler characteristic 0 Eulerian?
The answer is no. Let $Y$ be the face poset of a triangulation of a
torus, with a top element $t$ adjoined. Let $\emptyset$ (the empty
face) be the bottom element of $Y$. Let $Z$ be the chain $0<1$. Then
the product $Y\times Z$ satisfies your conditions. However, the
interval from $(\emptyset,0)$ to $(t,0)$ does not satisfy the Eulerian
condition.


