Inductive and Divisional Posets

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Problem: The question asks whether there exists a finite ranked poset $P$ with unique minimum $\hat0$, locally geometric in the sense that every interval $P_{\le x}$ is a geometric lattice, which is not a lattice, and such that $P$ is divisional but not inductive under Definitions 3.6–3.7 of Pagaria–Pismataro–Tran–Vecchi.

Result: Yes. Let $G$ be the graph with vertices $\{0,1,2,3,4,5\}$ and edges

$$01,05,12,13,14,23,24,34 .$$

Let $P$ be the face poset of this graph:

$$P=\{\varnothing\}\cup \{\{i\}:0\le i\le 5\}\cup \{\{i,j\}:ij\in E(G)\},$$

ordered by inclusion.

Each principal ideal below a vertex is a chain, and each principal ideal below an edge is the Boolean lattice $B_2$. Hence $P$ is locally geometric. It is not a lattice, since, for example, the atoms $\{0\}$ and $\{2\}$ have no common upper bound.

For a graph face poset of rank $2$ with $n$ vertices and $m$ edges,

$$\chi_P(t)=t^2-nt+m.$$

Here $n=6,m=8$, so

$$\chi_P(t)=t^2-6t+8=(t-2)(t-4).$$

Take the atom $a=\{0\}$. The restriction $P''=P_{\ge a}$ is a rank-one poset with two atoms, so

$$\chi_{P''}(t)=t-2,$$

which divides $\chi_P(t)$. Since every rank-one poset is divisional, $P$ is divisional.

We show $P$ is not inductive. The vertex degrees are

$$d(0)=2,\quad d(1)=4,\quad d(2)=d(3)=d(4)=3,\quad d(5)=1.$$

For an atom $v$, $\chi_{P_{\ge v}}(t)=t-d(v)$, while deletion gives the induced graph $G-v$. Directly:

• for $v=2,3,4$, $\chi_{P'}(3)=3^2-5\cdot3+5=-1$, so $t-3\nmid \chi_{P'}$;
• for $v=5$, $\chi_{P'}(1)=1-5+7=3$, so $t-1\nmid \chi_{P'}$.

Thus only $v=0$ or $v=1$ could possibly be an inductive deletion atom.

If $v=0$, then $G-v$ is $K_4$ plus one isolated vertex. This poset is not inductive: deleting the isolated vertex leaves $K_4$, which is not inductive, and deleting any $K_4$-vertex leaves $K_3$ plus an isolated vertex, also not inductive.

If $v=1$, then $G-v$ is a triangle $K_3$ disjoint union an edge. This is not inductive: deleting an endpoint of the isolated edge leaves $K_3$ plus an isolated vertex, not inductive; deleting a triangle vertex gives divisibility failure.

Finally, $K_3$ is not inductive because for each vertex the restriction has characteristic $t-2$, while the deletion has characteristic $(t-1)^2$, not divisible by $t-2$. Similarly, $K_4$ is not inductive since each deletion is $K_3$ with characteristic $t^2-3t+3$, not divisible by $t-3$.

Therefore $P\in \mathrm{DP}\setminus \mathrm{IP}$, and $P$ is a non-lattice locally geometric poset.

Citation: Definitions and the open question are from Pagaria, Pismataro, Tran, and Vecchi, “Inductive and divisional posets,” J. London Math. Soc. 109 (2024), DOI: 10.1112/jlms.12829. The counterexample above is not cited there.