Zhang et al.[1] introduced a concept of $Z$-transformation graph (called by some authors resonance graph) on the set of perfect matchings (or 1-factors) of a hexagonal system, Zhang and Zhang [2] extended the concept to a general plane bipartite graph with a perfect matching, and Zhang [3] surveyed rich theoretical results in several direction. Let $G$ be a plane bipartite graph with a perfect matching. Denote by $\mathcal{M}(G)$ the set of all perfect matchings of $G$. The $Z$-transformation directed graph $\vec{Z}(G)$ is an orientation of $Z$-transformation graph [4].

Lam and Zhang [5] proved that $\mathcal{M}(G)$ equipped with a partial order is a finite distributive lattice and its Hasse diagram is isomorphic to $\vec{Z}(G)$. Recently, Zhang et al.[6] introduced the concept of matchable distributive lattice and got some consequences on matchable distributive lattices, Yao and Zhang [7] obtained some results on non-matchable distributive lattices with a cut element.

In a finite lattice, an element is meet-irreducible if it is covered by exactly one element; from a graphical point of view if and only if it is a vertex of indegree $1$ in Hasse diagram. Consider an arc $f$ with its tail $M$ in $\vec{Z}(G)$. We introduce the concept of meet-irreducible cell with respect to $M$. Furthermore, we have some equivalent characterizations of the concept (i.e. Theorem 3.2). Finally, by Theorem 3.2, we extend a result on non-matchable distributive lattices obtained by Yao and Zhang [7] (i.e. Theorem 2.3), and obtain a class of non-matchable distributive lattices by Kuratowski's Theorem.

A set $P$ equipped with a binary relation $\le$ satisfying reflexivity, antisymmetry and transitivity is said to be a partially ordered set (poset for short). Given any poset $P$, the dual $P^*$ of $P$ by defining $x\le y$ to hold in $P^*$ if and only if $y\le x$ holds in $P$. A poset $P$ is a chain if any two elements of $P$ are comparable, and we write $\bf{n}$ to denote the chain obtained by giving $\{0, 1, \cdots, n-1\}$ the order in which $0<1<\cdots<n-1$. The subset $S$ of the poset $P$ is called convex if $a, b \in S$, $c \in P$, and $a \le c \le b$ imply that $c \in S$. A lattice is nontrivial if it has at least two elements and a finite distributive lattice is irreducible if it cannot be decomposed into a direct product of two nontrivial finite distributive lattices.

The symmetric difference of two finite sets $A$ and $B$ is defined as $A\oplus B := (A\cup B)\setminus(A\cap B)$. If $M$ is a perfect matching of a graph and $C$ is an $M$-alternating cycle of the graph, then the symmetric difference of $M$ and edge-set $E(C)$ is another perfect matching of the graph, which is simply denoted by $M\oplus C$. Let $G$ be a plane bipartite graph with a matching $M$, and the vertices of $G$ be colored properly black and white such that the two ends of every edge receive different colors. An $M$-alternating cycle of $G$ is said to be proper, if every edge of the cycle belonging to $M$ goes from white end-vertex to black end-vertex by the clockwise orientation of the cycle; otherwise improper [8].

For some concepts and notations not explained in the paper, refer to [9, 10] for poset and lattice, [11, 12] for graph theory.

An inner face of a graph is called a cell if its boundary is a cycle, and we will say that the cycle is a cell too. Observe that an $M$-alternating cell intersecting an improper $M$-alternating cell must be proper, vice versa. Obviously, we have the following result.

Lemma 2.1   (see [13]) If $G$ be a plane bipartite graph with a matching $M$, then any two proper resp. improper $M$-alternating cells are disjoint.

Definition 2.1   (see [2]) Let $G$ be a plane bipartite graph. The $Z$-transformation graph $Z(G)$ is defined on $\mathcal{M}(G)$: $M_1, M_2 \in \mathcal{M}(G)$ are joined by an edge if and only if $M_1\oplus M_2$ is a cell of $G$. And $Z$-transformation digraph $\vec{Z}(G)$ is the orientation of $Z(G)$: an edge $M_1M_2$ of $Z(G)$ is oriented from $M_1$ to $M_2$ if $M_1\oplus M_2$ form a proper $M_1$-alternating (thus improper $M_2$-alternating) cell.

An edge of graph $G$ is allowed if it lies in a perfect matching of $G$. A graph $G$ is said to be elementary if its allowed edges form a connected subgraph of $G$, Let $G$ be a bipartite graph. A subgraph $H$ of $G$ is said to be nice if $G - V(H)$ has a perfect matching [14]; from Theorem 4.1.1 in [14], we have that a bipartite graph is elementary if and only if it is connected and every edge of it is allowed.

Definition 2.2   (see[2]) A bipartite graph $G$ is weakly elementary if the subgraph of $G$ consisting of $C$ together with its interior is elementary for every nice cycle $C$ of $G$.

Let $G$ be a plane bipartite graph with a perfect matching. A binary relation $\le$ on $\mathcal{M}(G)$ is defined as: for $M_1, M_2\in\mathcal{M}(G)$, $M_1\le M_2$ if and only if $\vec{Z}(G)$ has a directed path from $M_2$ to $M_1$ [2]. It is shown that $(\mathcal{M}(G);\le)$ is a poset [5]. For convenience, we write $\mathcal{M}(G)$ for poset $(\mathcal{M}(G), \le)$.

Theorem 2.2   (see[5]) If $G$ is a plane (weakly) elementary bipartite graph, then $\mathcal{M}(G)$ is a finite distributive lattice and its Hasse diagram is isomorphic to $\vec{Z}(G)$.

Definition 2.3   (see[6]) A finite distributive lattice $L$ is matchable if there is a plane weakly elementary bipartite graph $G$ such that $L\cong\mathcal{M}(G)$; otherwise it is non-matchable.

Yao and Zhang [7] had obtained the following result on non-matchable distributive lattices with a cut element.

Theorem 2.3   (see[7]) Let $L$ be a finite distributive lattice, and $L$ have cut element covered by $m$ elements and covering $n$ elements. If $m \ge 3$ and $n \ge 3$, then $L$ is non-matchable.

The proof of Lemma~3.7 in [15] implies the following proposition.

Proposition 3.1   If $G$ is a plane elementary bipartite graph with a perfect matching $M$, then there exists a hypercube in $\vec{Z}(G)$ generated by some pairwise disjoint $M$-alternating cells. In particular, $M$ is the maximum (resp. minimum) element of the corresponding Boolean lattice in $\mathcal{M}(G)$ if these $M$-alternating cells are proper (resp. improper).

It is obvious that the dimension of the hypercube is equal to the number of these pairwise disjoint $M$-alternating cells. In particular, the hypercube is a quadrilateral if and only if it is generated by exactly two disjoint $M$-alternating cells in $G$ [13, 15, 16].

Definition 3.1   Let $G$ be a plane (weakly) elementary bipartite graph with a perfect matching $M$. A proper $M$-alternating cell $f$ is meet-irreducible with respect to $M$ if $M\oplus f$ is meet-irreducible in $\mathcal{M}(G)$.

In fact, by the definition of meet-irreducible element, it follows that $M\oplus f$ is meet-irreducible in $\mathcal{M}(G)$ whenever $f$ is a meet-irreducible cell with respect to $M$. The equivalent characterizations of meet-irreducible elements is given as follows, the thought of which is analogous to that in [17]. A part of Theorem 3.2 is published, however, to make the paper self-contained, we would rather include the proof here than refer to [18].

Theorem 3.2   Let $G$ be a plane (weakly) elementary bipartite graph $G$ with a perfect matching $M$ and let $f$ be a proper $M$-alternating cell.

(1) If $G$ has no improper $M$-alternating cell (namely, $M$ is the maximum element of $\mathcal{M}(G)$), then every (proper) $M$-alternating cell is a meet-irreducible cell with respect to $M$;

(2) If $G$ has some improper $M$-alternating cells, then the following are equivalent:

(a) the cell $f$ is a meet-irreducible cell with respect to $M$;

(b) the cell $f$ intersects every improper $M$-alternating cell;

(c) there is no perfect matching $M'$ in $V(\mathcal{Q})\setminus\{M\}$ such that $f$ is a proper $M'$-alternating cell, where $\mathcal{Q}$ is a corresponding Boolean lattice generated by all improper $M$-alternating cells.

Proof   (1) It is trivial by the definition of $Z$-transformation directed graph.

(2) Firstly suppose that the cell $f$ is a meet-irreducible cell with respect to $M$, but there is at least one improper $M$-alternating cell $f'$ such that $f$ and $f'$ are disjoint. Thus $M\oplus f = ((M\oplus f')\oplus f)\oplus f'$, i.e., $G$ has two improper $M\oplus f$-alternating cells, hence $M\oplus f$ is not meet-irreducible, contradicting the supposition that $f$ is a meet-irreducible cell with respect to $M$.

Next suppose that the cell $f$ intersects every improper $M$-alternating cell, but there is a perfect matching $M'$ in $V(\mathcal{Q})\setminus\{M\}$ such that $f$ is a proper $M'$-alternating cell. In fact, by Proposition 3.1, there is at least one improper $M$-alternating cell $f'$ that is a proper $M'$-alternating cell. Hence $f$ and $f'$ are disjoint by Lemma 2.1, a contradiction.

Finally, suppose that there is no perfect matching $M'$ in $V(\mathcal{Q})\setminus\{M\}$ such that $f$ is a proper $M'$-alternating cell, but $f$ is not a meet-irreducible cell with respect to $M$. Thus $G$ has at least one improper $M\oplus f$-alternating cell $f'$ except $f$, by Lemma 2.1, hence $f$ and $f'$ are disjoint. Therefore $f'$ is an improper $M$-alternating cell, which implies that $f$ is a proper $M\oplus f'$-alternating cell, i.e., there is a perfect matching $M'=M\oplus f'$ in $V(\mathcal{Q})\setminus\{M\}$ such that $f$ is a proper $M'$-alternating cell, a contradiction.

Assume that every proper $M$-alternating cell is a meet-irreducible cell with respect to $M$. If $G$ has an improper $M$-alternating cell, then every proper $M$-alternating cell intersects every improper $M$-alternating cell, hence $M$ is a cut element [13]. And $M$ is the maximum element of $\mathcal{M}(G)$ otherwise. Moreover we obtain the following consequence of Theorem 3.2.

Corollary 3.3   (see[4, 13]) If $G$ is a plane elementary bipartite graph with a perfect matching $M$, and has both proper and improper $M$-alternating cells, then $M$ is a cut vertex of $Z(G)$ if and only if every proper $M$-alternating cell intersects every improper $M$-alternating cell; i.e., every proper $M$-alternating cell is a meet-irreducible cell with respect to $M$.

Note that duality of lattice, meet-irreducible cell, Theorem 3.2 and Corollary 3.3 could be treated in dual.

Subdividing an edge $e$ is to delete $e$, add a new vertex $v$, and join $v$ to the ends of $e$. Any graph derived from a graph $G$ by a sequence of edge subdivisions is called a subdivision of $G$.

Given a plane graph $G$, its (geometric) dual $G^*$ is constructed as follows: place a vertex in each face of $G$ (including the exterior face) and, if two faces have an edge $e$ in common, join the corresponding vertices by an edge $e^*$ crossing only $e$ [12]. It is easy to see that the dual $G^*$ of a plane graph $G$ is itself a plane graph [11].

Theorem 4.1   (Kuratowski's Theorem) A graph is planar if and only if it contains no subdivision of either $K_5$ or $K_{3, 3}$.

Similar to the proof of Lemma 4.2 in [7] and Theorem 3.2, we have Theorem 4.2.

Theorem 4.2   Let $L$ be a finite distributive lattice and $x\in L$. If $x$ is covered by at least three elements and covers at least three meet-irreducible elements, then $L$ is non-matchable.

Proof   Suppose to the contrary that $L$ is matchable. Then there is a plane (weakly) elementary bipartite graph $G$ such that $\mathcal{M}(G) \cong L$ [6], which implies that a perfect matching $M_x$ of $G$ corresponds to $x \in L$. According to the premise, $G$ has at least three improper $M_x$-alternating cells, and at least three proper $M_x$-alternating cells $f_1$, $f_2$ and $f_3$ that are meet-irreducible. By Theorem 3.2, such three proper $M_x$-alternating cells intersect all improper $M_x$-alternating cells. This shows that the dual $G^*$ of $G$ contains a $K_{3, 3}$ as subgraph. But it is impossible by Kuratowski's theorem.

If $x$ is a cut element, Theorem 4.2 implies Theorem 2.3 (i.e., Theorem 4.3 in [7]); on the other hand, Theorem 4.2 can determine some non-matchable distributive lattice that can not be determined only by Theorem 2.3. For instance, it is easy to see that each distributive lattice in Figure 1 is non-matchable by Theorem 4.2, but not determined only by Theorem 2.3.

Obviously, a dual version of Theorem 4.2 could be obtained easily.

Corollary 4.3   If $L$ is a matchable distributive lattice, then for every element of $L$, it either is covered by at most two elements or covers at most two meet-irreducible elements in both $L$ and $L^*$.

Let $P$ be a poset and $F \subseteq P$. The subposet $F$ is a filter if, whenever $x \in F$, $y \in P$ and $x \le y$, we have $y \in F$ [9]. The set of all filters of a poset $P$ is denoted by $\mathcal{F}(P)$, and carries the usual anti-inclusion order; and the filter lattice $\mathcal{F}(P)$ is a distributive lattice [9, 10]. The Figure 2(a) shows the Hasse diagram of a poset $\Delta$; and the Figure 2(b) is the highest five layers of Hasse diagram of $\mathcal{F}(\Delta)$, where every filter is labeled by its minimal element(s). Combining Theorem 3.2 and Kuratowski's Theorem, we have another class of non-matchable distributive lattices.

Theorem 4.4   The distributive lattice $\mathcal{F}(\Delta)$ is non-matchable, where $\Delta$ is the poset as shown in Figure 2(a).

Proof   Recall that $\mathcal{F}(\Delta)$ is a finite distributive lattice. Suppose that $\mathcal{F}(\Delta)$ is matchable. Since $\mathcal{F}(\Delta)$ has a cut element labeled by $0$ (see Figure 2), and is irreducible, there exists a plane elementary bipartite graph $G$ such that $\vec{Z}(G)\cong\mathcal{F}(\Delta)$ [6].

Consider a part of $\mathcal{F}(\Delta)$ as drawn in Figure 2(b). The vertices $\emptyset$, $0$, $1$, $\cdots$, $a$ correspond to the perfect matchings $M_{\emptyset}$, $M_0$, $M_1$, $\cdots$, $M_a$ of $G$, respectively. Let $f_0=M_{\emptyset}\oplus M_0$, $f_1=M_0\oplus M_1$, $f_5=M_{12}\oplus M_5$, $f_6=M_{13}\oplus M_6$, $\cdots$, and $f_a=M_{34}\oplus M_a$. By definition of $Z$-transformation graph, then $f_0$ is a nice cell, so are $f_1$, $\cdots$, $f_a$. Since the cells $f_0$, $f_1$, $\cdots$, $f_a$ are meet-irreducible cells, by Theorem 3.2 (2), the cell $f_0$ intersects $f_1$, $f_2$, $f_3$ and $f_4$; the cell $f_5$ intersects $f_1$ and $f_2$, but it does not intersect $f_3$ or $f_4$, because $f_5$, $f_3$ and $f_4$ are proper $M_{12}$-alternating cells. Thus $f_0$ and $f_5$ are distinct; analogously, $f_0$ and $f_i\ (i\in\{6, 7, 8, 9, a\})$ are distinct too.

Next, consider the dual $G^*$ of $G$, as drawn in Figure 3 (a), vertex $f_0^*$ is adjacent with $f_1^*, f_2^*, f_3^*\ \text{and}\ f_4^*$, and $f_5^*$ is adjacent with $f_1^*$ and $f_2^*$, etc. Therefore, let $V'=\{f_0^*, \ \cdots, \ f_a^*\}$, thus $G^*$ contains a subgraph $S^* := G^*[V']$. Clearly $S^*$ (see Figure 3 (b)) is a subdivision of $K_5$. By Kuratowski's Theorem, hence $S^*$ is non-planar, contradicting the planarity of $G$.

If a poset $P$ contains $\Delta$ as a convex subposet, then there are 11 elements in $P$ cover relations of which are identical in $\Delta$. Similar to proof of Theorem 4.4, we prove the following result.

Corollary 4.5   Let $P$ be a poset containing $\Delta$ as a convex subposet. Then distributive lattice $\mathcal{F}(P)$ is non-matchable. In addition, for any finite distributive lattice $L$, the Cartesian product, linear sum and vertical sum [9] of $\mathcal{F}(P)$ and $L$ are non-matchable.

Note that $\Delta$ is a filter of $\bf{2}^4$, the following corollary is immediate.

Corollary 4.6   The distributive lattice $\mathcal{F}(\bf{2}^4)$ is non-matchable. In addition, the distributive lattice $\mathcal{F}\left(\prod_{j=1}^k\bf{n}_j\right)$ is non-matchable, where $k\ge 4$, $\bf{n}_j$ is a chain of length $n_j$ and $n_j\ge 2$ for every $j=1, 2, \cdots, k$.