# Is a finite abstract polytope of Euler characteristic 0 Eulerian?

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 $F_1 < F_2$, $\sum_{i = rank(F_1)}^{rank(F_2)} (-1)^i[F_1,F_2]_i = 0$?

(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].)

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.
• Sorry for this question, but I got confused. Although I have checked what you say, and everything is correct, I still don't understand one thing. Geometrically, $Y\times Z$ must be the cylinder of $Y$, so be homotopy equivalent to $Y$, hence have the same Euler characteristic. How come they are different?? – მამუკა ჯიბლაძე Feb 22 '17 at 7:21
• They would have the same Euler characteristic if $Y$ were the face poset of a triangulation of the torus, but we have adjoined an extra element $t$. – Richard Stanley Feb 22 '17 at 14:22