Abstract
In this paper, we associate a simple graph to a lattice \(\mathcal L \), in which the vertex set is being the set of all elements of \(\mathcal L \), and two distinct vertices \(x\) and \(y\) are adjacent if \(x\vee y\in S\), when \(S\) is a multiplicatively closed subset of \(\mathcal L \). We denote this graph by \(\Gamma _S(\mathcal L )\). We study some properties of \(\Gamma _S(\mathcal L )\). Moreover, we investigate the planarity of \(\Gamma _S(\mathcal L )\), whenever \(S\) is a saturated multiplicatively closed subset of \(\mathcal L \).
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1 Introduction
The study of algebraic structures, using the properties of graph theory, tends to an exciting research topic in the last decade. There are many papers on assigning a graph to a ring, see for example [2–4, 6]. One of these graphs is the total graph. Let \(R\) be a commutative ring. The total graph of \(R\) is a simple graph with vertex set \(R\), and two distinct vertices \(a\) and \(b\) are adjacent if \(a+b\) is a zero divisor of \(R\) (see for example [1, 2]). By using this idea, for a multiplicatively closed subset \(S\) of \(R\), in [5], the authors defined a graph on \(R\), denoted by \(\Gamma _S(R)\), with vertices as elements of \(R\), and two distinct vertices \(a\) and \(b\) are adjacent if and only if \(a+b\in S\). In this paper, we generalize the concept of \(\Gamma _S(R)\) for a lattice. Recall that a lattice is an algebra \(\mathcal{L }=( L,\vee ,\wedge )\) satisfying the following conditions: for all \(a,b,c\in L\),
-
1.
\(a\wedge a=a\), \(a\vee a=a\),
-
2.
\(a\wedge b= b\wedge a\), \(a\vee b= b\vee a\),
-
3.
\((a\wedge b)\wedge c= a\wedge (b \wedge c)\), \((a\vee b)\vee c= a\vee (b \vee c)\), and
-
4.
\(a\vee (a\wedge b)= a\wedge (a \vee b)=a\).
There is an equivalent definition for a lattice (see for example [8, Theorem 2.1]). To do this, for a lattice \(\mathcal L \), one can define an order \(\leqslant \) on \(\mathcal L \) as follows: For any \(a,b\in \mathcal{L }\), we set \(a\le b\) if and only if \(a\wedge b=a\). Then \((\mathcal{L }, \le )\) is an ordered set in which every pairs of elements has a greatest lower bound (g.l.b.) and a least upper bound (l.u.b.). Conversely, let \(P\) be an ordered set such that, for every pair \(a,b\in P\), \({\mathrm{g.l.b.}}(a,b)\in P\) and \({\mathrm{l.u.b.}}(a,b)\in P\). For each \(a\) and \(b\) in \(P\), we define \(a\wedge b:={\mathrm{g.l.b.}}(a,b)\) and \(a\vee b:= {\mathrm{l.u.b.}} (a,b)\). Clearly \((P,\wedge , \vee )\) is a lattice.
The lattice \(\mathcal L \) is said to be bounded if there are elements \(0\) and \(1\) such that
for all \(x\in \mathcal L \). Let \(S\) be a nonempty subset of \(\mathcal L \). We say that \(S\) is a multiplicatively closed subset of \(\mathcal L \) if \(x\wedge y\in S\), for all \(x\) and \(y\) of \(S\). Also, we say that a subset \(S\) of \(\mathcal L \) is saturated if \(x\wedge y\in S\) if and only if \(x,y\in S\).
For a multiplicatively closed subset of \(\mathcal L \), we define \(\Gamma _S(\mathcal L )\) as a simple graph, with vertex-set \(\mathcal L \) and two distinct vertices \(x\) and \(y\) being adjacent if and only if \(x\vee y \in S\).
In the second section, we study some basic properties of the graph \(\Gamma _S(\mathcal{L })\) such as connectivity, diameter and completeness whenever \(\mathcal{L }\) is a bounded lattice. In the third section, for bounded lattices, we investigate \(\Gamma _S(\mathcal L )\), whenever \(S\) is a saturated multiplicatively closed subset of \(\mathcal L \) and, in the final section, we study the planarity of \(\Gamma _S(\mathcal L )\) where \(\mathcal L \) is a bounded lattice.
Now, we start to remind a belief necessary background of lattice theory from [8]. Let \(x\) and \(y\) be two distinct elements of \(\mathcal L \). Whenever \(x \leqslant y\) and there is no element \(z\) in \(\mathcal L \) such that \(x \leqslant z \leqslant y\), we say \(y\) covers \(x\). An element \(a\in \mathcal{L }\) is an atom of lattice \(\mathcal L \) if it covers \(0\). Also, an element \(m\in \mathcal L \) is a coatom of lattice \(\mathcal L \) if \(1\) covers it. We denote the set of all coatoms of \(\mathcal L \) by Coatom \((\mathcal L )\) and the set of atoms of \(\mathcal L \) by Atom \((\mathcal L )\). An ideal \(I\) of \(\mathcal L \) is a non-empty subset of \(\mathcal L \) such that
-
(i)
for all \(a\) and \(b\) of \(I\), \(a\vee b\in I\), and
-
(ii)
for any \(a\in I\) and \(b\in \mathcal L \) , \(a\wedge b \in I\).
An ideal \(I\) of \(\mathcal L \) is maximal if \(I\) is proper (\(I\ne \mathcal L \)) and the only ideal having \(I\), as a proper subset, is \(\mathcal L \). We use the notation J\((\mathcal L )\) for the jacobson radical of \(\mathcal L \) which is the intersection of all maximal ideals of \(\mathcal L \). Given a lattice \(\mathcal L \) and \(A\subseteq \mathcal L \). An element \(x \in \mathcal L \) is a lower bound of \(A\) if \(x \le a\) for all \(a\in A\). An upper bound is defined in a dual mannar. The set of all lower bounds of \(A\) is denoted by \(A^\ell \) and the set of all upper bounds by \(A^u\), where
and
Now, we recall some definitions of graph theory from [7] which are needed in this paper. The degree of a vertex \(v\) in the graph \(X\) is the number of edges of \(X\) incident with \(v\) and denoted by \({\mathrm{deg}}(v)\). In a graph \(X\) with vertex-set \(V(X)\), the distance between two distinct vertices \(a\) and \(b\), denoted by \({\mathrm{d}}(a,b)\), is the length of a shortest path connecting \(a\) and \(b\), if such a path exists; otherwise, we set \({\mathrm{d}}(a,b):=\infty \). The diameter of a graph \(X\) is \({\mathrm{diam}}(X)={\mathrm{sup}}\{{\mathrm{d}}(a,b)\) : \(a\) and \(b\) are distinct vertices of \(X\}\). For two distinct vertices \(a\) and \(b\) in \(X\), the notation \(a-b\) means that \(a\) and \(b\) are adjacent. Also the girth of a graph \(X\), denoted by \({\mathrm{gr}}(X)\), is the length of a shortest cycle in \(X\) if \(X\) has a cycle; otherwise, we set \(\mathrm{{gr}}(X):=\infty \). A graph \(X\) is said to be connected if there exists a path between any two distinct vertices, and it is complete if it is connected with diameter one. We use \(K_n\) to denote the complete graph with \(n\) vertices. Also, we say that \(X\) is totally disconnected if no two vertices of \(X\) are adjacent. A clique of a graph is a complete subgraph of \(X\) and the number of vertices in a largest clique of \(G\) is called the clique number of \(G\) and is denoted by \(\omega (G)\).
2 Basic properties of \(\Gamma _S(\mathcal L )\)
Form now on \(\mathcal L \) is a bounded lattice. In this section \(S\) is a multiplicatively closed subset of \(\mathcal L \). We begin with the following lemma.
Lemma 2.1
If \(0\notin S\), then \(|S\cap \mathrm{{Atom}}(\mathcal{L })|\le 1\).
Proof
Assume in contrary that \(|S\cap \mathrm{{Atom}}(\mathcal{L })|\ge 2\). Let \(a_1\) and \(a_2\) be two distinct elements in \(S\cap \mathrm{{Atom}}(\mathcal{L })\). Since \(S\) is a multiplicatively closed subset of \(\mathcal L \), we have \(a_1\wedge a_2\in S\) which implies that \(0\in S\). This is a required contradiction. Hence \(|S\cap \mathrm{{Atom}}(\mathcal L )|\le 1\). \(\square \)
Lemma 2.2
The following statements hold.
-
(i)
If \(0\in S\), then \(\mathrm{{deg(0)}}=|S|-1\). Otherwise, \(\mathrm{{deg(0)}}=|S|\).
-
(ii)
If \(1\in S\), then \(\mathrm{{deg(1)}}=|\mathcal{L }|-1\). Otherwise, \(\mathrm{{deg(1)}}=0\).
Proof
Since \(0\vee x=x\) and \(1\vee x=1\), for all \(x\in \mathcal{L }\), the statements (i) and (ii) hold.
\(\square \)
In the following theorem, we study the connectivity of \(\Gamma _S(\mathcal L )\).
Theorem 2.3
\(\Gamma _S(\mathcal L )\) is connected if and only if \(1\in S\). Moreover, if \(\Gamma _S(\mathcal L )\) is connected, then \(\mathrm{{diam}}(\Gamma _S(\mathcal L ))\le 2\).
Proof
Suppose that \(1\in S\). By part (ii) of Lemma 2.2, \(\mathrm{{deg(1)}}=|\mathcal L |-1\) which implies that \(\Gamma _S(\mathcal L )\) is connected.
Conversely, if \(\Gamma _S(\mathcal L )\) is connected, then \(\mathrm{{deg(1)}}\ne 0\). Now, by part (ii) of Lemma , we have \(1\in S\).
The last statement is clear. \(\square \)
Proposition 2.4
\(\Gamma _S(\mathcal L )\) is complete if and only if \(S=\mathcal L \) or \(S=\mathcal L \setminus \{0\}\).
Proof
Assume that \(\Gamma _S(\mathcal L )\) is complete. So \(\mathrm{{deg(0)}}=|\mathcal L |-1\). By part (i) of Lemma , we have \(|S|=|\mathcal L |-1\) or \(|S|-1=|\mathcal L |-1\). Hence \(S=\mathcal L \) or \(S=\mathcal L \setminus \{x\}\), for some \(x\in \mathcal L \). We claim that \(x=0\). Assume in contrary that \(x\ne 0\). Since \(0\vee x=x\), the vertex \(0\) is not adjacent to \(x\), which is impossible. So \(x=0\), and thus \(S=\mathcal L \) or \(S=\mathcal L \setminus \{0\}\).
The converse statement follows easily. \(\square \)
Theorem 2.5
The following statements hold.
-
(i)
The graph \(\Gamma _S(\mathcal L )\) is regular if and only if it is either complete or totally disconnected.
-
(ii)
\(\Gamma _S(\mathcal L )\) is a star graph if and only if \(S=\{1\}\) or \(S=\{1,0\}\) and \(|\mathrm{{Coatom}}(\mathcal L )|=1\).
Proof
(i) Suppose that \(\Gamma _S(\mathcal L )\) is regular and that is not totally disconnected. By part (ii) of Lemma , we have \(1\in S\). Then \(\mathrm{{deg}}(1)=|\mathcal L |-1\). Since \(\Gamma _S(\mathcal L )\) is regular, \(\mathrm{{deg}}(x)=|\mathcal L |-1\), for all \(x\in \mathcal L \). Hence \(\Gamma _S(\mathcal L )\) is complete.
The converse statement is clear.
(ii) Suppose that \(\Gamma _S(\mathcal L )\) is a star graph. Since \(\Gamma _S(\mathcal L )\) is connected, we have \(1\in S\), and so \(\mathrm{{deg(1)}}=|\mathcal L |-1\). Hence \(1\) is the center of \(\Gamma _S(\mathcal L )\). On the other hand, since \(0\) is adjacent to all element of \(S\), we have \(S=\{1\}\) or \(S=\{1,0\}\). Assume in contrary that \(|\mathrm{{Coatom}}(\mathcal L )|\ge 2\). Thus there exists distinct vertices \(m\) and \(m^{\prime }\) in \(\mathrm{{Coatom}}(\mathcal L )\) which are adjacent in \(\Gamma _S(\mathcal L )\). This is impossible. Hence \(|\mathrm{{Coatom}}(\mathcal L )|=1\).
The converse statement is clear. \(\square \)
Proposition 2.6
Assume that \(1\in S\). Then \(\mathrm{{gr}}(\Gamma _S(\mathcal{L }))\in \{3,\infty \}\).
Proof
Suppose that \(|\mathrm{{Coatom}}(\mathcal L )|\ge 2\) and consider the cycle \(1-m-m^{\prime }-1\) in \(\Gamma _S(\mathcal L )\), where \(m,m^{\prime }\in \mathrm{{Coatom}}(\mathcal L )\), to deduce that \(\mathrm{{gr}}(\Gamma _S(\mathcal L ))=3\). Otherwise, \(|\mathrm{{Coatom}}(\mathcal L )|=1\). Now we have the following cases:
-
(i)
Suppose that \(|S|=1\). Then \(S=\{1\}\) and so, by Proposition 2.5, the graph \(\Gamma _S(\mathcal L )\) is a star graph. Thus \(\mathrm{{gr}}(\Gamma _S(\mathcal L ))=\infty \).
-
(ii)
Suppose that \(|S|=2\). If \(S=\{0,1\}\), then, by Proposition 2.5, \(\Gamma _S(\mathcal L )\) is a star graph and so \(\mathrm{{gr}}(\Gamma _S(\mathcal L ))=\infty \). Otherwise, \(S=\{1,s\}\) where \(s\ne 0\). Now, the cycle \(1-s-0-1\) is the shortest cycle in the graph \(\Gamma _S(\mathcal L )\) which implies that \(\mathrm{{gr}}(\Gamma _S(\mathcal L ))=3\).
-
(iii)
Suppose that \(|S|\ge 2\). Thus there is \(s\in S\) such that \(s\ne 0,1\). Now, the cycle \(1-s-0-1\) is the shortest cycle in the graph \(\Gamma _S(\mathcal L )\) . So \(\mathrm{{gr}}(\Gamma _S(\mathcal L ))=3\).
\(\square \)
Let \(\mathcal A \) be a chain in \(S\). Since \(x\vee y\in S\) for all \(x,y\in \mathcal{A }\), we have the following proposition.
Proposition 2.7
\(\omega (\Gamma _S(\mathcal{L }))\ge \mathrm{max} \{ |\mathcal{A }|; \mathcal{A }\,\mathrm{is\,a\,chain\,in}\,S\}\).
3 Basic properties of \(\Gamma _S(\mathcal L )\) where \(S\) is a saturated subset of the lattice \(\mathcal L \)
Throughout this section, \(S\) is a saturated subset of the lattice \(\mathcal L \). It is easy to see that if \(x\in S\), then \(\{x\}^u\subseteq S\), and so we always have \(1\in S\). Hence, by Proposition 2.3, one can conclude that \(\Gamma _S(\mathcal L )\) is connect with diameter less than three.
Proposition 3.1
For all \(s\in S\), we have that \(\mathrm{{deg(s)}}=|\mathcal L |-1\).
Proof
Let \(s\) be an arbitrary element in \(S\). Since \(s\wedge (s\vee x)=s\) for all \(x\in \mathcal L \), we have that \(s\wedge (s\vee x)\in S\). Therefore \(\mathrm{{deg(s)}}=|\mathcal L |-1\) for all \(s\in S\). \(\square \)
In the following proposition, we present a lower bound for the clique number of \(\Gamma _S(\mathcal L )\).
Proposition 3.2
In the graph \(\Gamma _S(\mathcal L )\) we have the following inequality.
Proof
Since \(m\vee m^{\prime }=1\) for all \(m\) and \(m^{\prime }\) in \(\mathrm{{Coatom}}(\mathcal L )\), the set \(\mathrm{{Coatom}}(\mathcal L )\cup \{1\}\) forms a clique in \(\Gamma _S(\mathcal L )\). Also, by Proposition 3.1, \(S\cup \{0\}\) is a clique in \(\Gamma _S(\mathcal L )\). This implies that \(\omega (\Gamma _S(\mathcal L ))\ge \mathrm{{max}}\{|S|, |\mathrm{{Coatom}}(\mathcal L )|\}+1\). \(\square \)
Lemma 3.3
Assume that \(0\in S\). Then \(\Gamma _S(\mathcal L )\) is complete.
Proof
Since \(0\in S\), we have that \(\{0\}^u \subseteq S\). Hence \(S=\mathcal L \). Now, by Proposition 2.4, the graph \(\Gamma _S(\mathcal L )\) is complete. \(\square \)
From now on, we assume that \(0\notin S\). So, by Lemma 2.1, we have \(|S\cap \mathrm{{Atom}}(\mathcal L )|\le 1\).
Proposition 3.4
If \(S\cap \mathrm{{Atom}}(\mathcal L )= \{a\}\), then \(S=\{a\}^u\).
Proof
Clearly \(\{a\}^u \subseteq S\). Now, assume in contrary that there exists \(b\in S\setminus \{a\}^u\). Since \(S\) is a multiplicatively closed subset of \(\mathcal L \), we have \(0=a\wedge b\in S\), which is impossible. Thus \(S=\{a\}^u\). \(\square \)
Proposition 3.5
Suppose that \(|\mathcal L |\ge 3\).
-
(i)
If \(|S|\ge 2\), then every vertex of the graph \(\Gamma _S(\mathcal L )\) lies in a cycle of length 3, and so \(\mathrm{{gr}}(\Gamma _S(\mathcal L ))=3.\)
-
(ii)
If \(|S|=1\), then \(\mathrm{{gr}}(\Gamma _S(\mathcal L ))\in \{3, \infty \}.\)
Proof
(i) Since \(|S|\ge 2\), we can choose \(s\ne 1\) in \(S\). Now, let \(x\) be an arbitrary element in \(\mathcal L \). Then we have the cycle \(1-x-s-1\), and so each vertex of the graph \(\Gamma _S(\mathcal L )\) lies in a cycle of length 3 and \(\mathrm{{gr}}(\Gamma _S(\mathcal L ))=3\).
(ii) Since \(|S|=1\), we have that \(S=\{1\}\). The result now follows from Proposition 2.6.\(\square \)
Proposition 3.6
If \(|S|=1\), then \(\mathrm{{deg}}(x)=1\), for all \(x\in J(\mathcal L )\).
Proof
Since \(|S|=1\), we have that \(S=\{1\}\). If \(|\mathrm{{Coatom}}(\mathcal L )|=1\), then, by Proposition 2.5, \(\Gamma _S(\mathcal L )\) is a star graph with center \(1\). Since \(1\notin J(\mathcal L )\), every vertex in \(\mathrm{{J}}(\mathcal L )\) has degree one. Otherwise, \(|\mathrm{{Coatom}}(\mathcal L )|\ge 2\). Let \(x\) be an arbitrary vertex in \(J(\mathcal L )\). Clearly \(1\) is adjacent to every vertex in \(\Gamma _S(\mathcal L )\) which implies that \(\mathrm{{deg}}(x)\ge 1\). Assume in contrary that \(\mathrm{{deg}}(x)\ge 2\). So there is \(y\ne 1\) in \(\mathcal L \) such that \(x\) and \(y\) are adjacent, and so \(x\vee y=1\). If \(y\in \mathrm{{Coatom}}(\mathcal L )\), then \(x\vee y=y\), which is impossible. Hence \(y\notin \mathrm{{Coatom}}(\mathcal L )\), but there is \(m\in \mathrm{{Coatom}}(\mathcal L )\) such that \(y\in \{m\}^{\ell }\). Since \(x\in \cap _{m\in \mathrm{{Coatom}}(\mathcal L )} \{m\}^{\ell }\), we have that \(x\vee y\le m\), which is impossible. So \(\mathrm{{deg}}(x)=1\), for all \(x\in J(\mathcal L )\). \(\square \)
Proposition 3.7
If \(|S|=1\) and \(|\mathrm{{Coatom}}(\mathcal L )|\ge 2\), then every vertex in graph \(\Gamma _S(\mathcal L \setminus \mathrm{J}(\mathcal L ))\) lies in a cycle of length 3.
Proof
Since \(|S|=1\), we have that \(S=\{1\}\). Let \(y\) be an arbitrary element in \(\mathcal L \setminus J(\mathcal L )\). We need to show that \(y\) lies in a cycle of length 3 in \(\Gamma _S(\mathcal L \setminus J(\mathcal L ))\). If \(y\in \mathrm{Coatom}(\mathcal L )\), since \(|\mathrm{Coatom}(\mathcal L )|\ge 2\), there exists \(m\) with \(m\ne y\) in \(\mathrm{Coatom}(\mathcal L )\). Thus, we can consider the cycle \(y-m-1-y\) in \(\Gamma _S(\mathcal L \setminus J(\mathcal L ))\). Otherwise \(y\notin \mathrm{Coatom}(\mathcal L )\). Since \(y\notin J(\mathcal L )\), there exists \(m\in \mathrm{Coatom}(\mathcal L )\) such that \(y\notin \{m\}^{\ell }\). Therefore \(y\vee m=1\), which implies that \(y\) and \(m\) are adjacent in \(\Gamma _S(\mathcal L \setminus J(\mathcal L ))\). So we can consider the cycle \(y-m-1-y\) and the result follows. \(\square \)
4 Planarity of \(\Gamma _S(\mathcal L )\) when \(S\) is a saturated subset of \(\mathcal L \)
In this section, we characterize all planar graphs \(\Gamma _S(\mathcal L )\), where \(S\) is a saturated subset of \(\mathcal L \). Recall that a planar graph is a graph that can be embedded on the plane, that is, it can be drawn on the plane in such a way that it’s edges intersect only at their endpoints. Kuratowski provided a nice characterization of planar graphs, which now is known as Kuratowski’s Theorem:
A finite graph is planar if and only if it does not contain a subdivision of \(K_5\) or \(K_{3,3}\).
Proposition 4.1
If \(0\in S\), then \(\Gamma _S(\mathcal L )\) is planar if and only if \(|\mathcal L |\le 4\).
Proof
If \(0\in S\), then \(S=\mathcal L \). Hence \(\Gamma _S(\mathcal L )\) is a complete graph, and so the result follows from Kuratowski’s Theorem. \(\square \)
Proposition 3.2 in conjunction with Kuratowski’s Theorem implies the following lemma.
Lemma 4.2
If \(|\mathrm{Coatom}(\mathcal L )|\ge 4\) or \(|S|\ge 4\), then \(\Gamma _S(\mathcal L )\) is not planar.
In view of Lemma 4.2, we investigate to study the planarity of \(\Gamma _S(\mathcal L )\), in the case that \(|S|\leqslant 3\) and \(|\mathrm{Coatom}(\mathcal L )|\leqslant 3\). We start with the following proposition.
Proposition 4.3
Suppose that \(|S|=3\). Then \(\Gamma _S(\mathcal L )\) is planar if and only if \(|\mathcal L |\le 5\).
Proof
Suppose that \(\Gamma _S(\mathcal L )\) is planar. Assume in contrary that \(|\mathcal L |\ge 6\). Now, put \(V_1:= S\) and \(V_2:=\{x_1,x_2,x_3\}\subseteq \mathcal L \setminus S\). Clearly one can find a copy of \(K_{3,3}\) in \(\Gamma _S(\mathcal L )\). Therefore, by Kuratowski’s Theorem, \(\Gamma _S(\mathcal L )\) is not planar, which is impossible. Hence \(|\mathcal L |\le 5\).
Conversely, assume that \(|\mathcal L |\le 5\). It is clear that if \(|\mathcal L |\le 4\), then \(\Gamma _S(\mathcal L )\) is planar. Also, if \(|\mathcal L |=5\), then, in view of Proposition 2.4, \(\Gamma _S(\mathcal L )\) is not \(K_5\). So \(\Gamma _S(\mathcal L )\) is planar. \(\square \)
Note that, if \(|S|=2\), then \(S=\{1,m\}\) for some \(m\in \mathrm{Coatom}(\mathcal L )\). Also, if \(|S|=1\), then \(S=\{1\}\).
For the rest of the paper, we need the following definition.
Definition 4.4
Let \(x\) be an arbitrary element in \(\mathcal L \). We define the lower neighbors of \(x\) as the set \(B_{\ell }(x):=\{y\in \mathcal L ; \ x\ covers \ y\}\). Also, for every subsets \(A\) and \(B\) of \(\mathcal L \), we put \(L_{A}^{B}:=\{B\}^{\ell }\setminus \{A\}^{\ell }\). Moreover, for \(x\in \mathcal L \), we denote the sets \(L_{A}^{\{x\}}\) and \(L_{\{x\}}^{B}\), by \(L_{A}^{x}\) and \(L_{x}^{B}\), respectively.
Lemma 4.5
Assume that \(S=\{1,m\}\), where \(m\in \mathrm{Coatom}(\mathcal L )\). If \(|B_{\ell }(m)|\ge 3\), then \(\Gamma _S(\mathcal L )\) is not planar.
Proof
Since \(|B_{\ell }(m)|\ge 3\), we can choose the subset \(A=\{x_1,x_2,x_3\}\) of the set \(B_{\ell }(m)\). It is clear that \(x_i\vee x_j=m\) for all \(1\le i\ne j\le 3\). So, the induced subgraph of \(\Gamma _S(\mathcal L )\) on \(A\cup S\) is isomorphic to \(K_5\). Thus, by Kuratowski’s Theorem, \(\Gamma _S(\mathcal L )\) is not planar. \(\square \)
Proposition 4.6
Let \(S=\{1,m\}\), where \(m\in \mathrm{Coatom}(\mathcal L )\). If \(\Gamma _S(\mathcal L )\) is planar, then one of the following statements holds:
-
(i)
\(|B_{\ell }(m)|=1\),
-
(ii)
\(|B_{\ell }(m)|=2\), and \(|L_{x}^{y}|\le 2\), for all \(x,y\in B_{\ell }(m)\), and if \(|L_{x}^{y}|= 2\), then \(|L_{y}^{x}|=1\), and if \(|L_{y}^{x}|= 2\), then \(|L_{x}^{y}| =1\).
Proof
Since \(\Gamma _S(\mathcal L )\) is planar, by Lemma 4.5, \(|B_{\ell }(m)|\le 2\). Suppose that \(|B_{\ell }(m)|\ne 1\). Thus, we have that \(|B_{\ell }(m)|=2\). Set \(B_{\ell }(m):=\{x,y\}\). If \(|L_{y}^{x}|\ge 3\) or \(|L_{x}^{y}|\ge 3\), then it is easy to see that \(\Gamma _S(\mathcal L )\) has a subgraph which is isomorphic to \(K_{3,3}\), and so \(\Gamma _S(\mathcal L )\) is not planar, which is impossible. Hence \(|L_{y}^{x}|\le 2\) and \(|L_{x}^{y}|\le 2\).
Now, suppose that \(|L_{x}^{y}|= 2\) and \(|L_{y}^{x}|= 2\). Set \(V_1:=L_{x}^{y}\cup \{m\}\) and \(V_2:=L_{y}^{x} \cup \{1\}\). It is easy to see that \(\Gamma _S(\mathcal L )\) has a subgraph isomorphic to \(K_{3,3}\) with parts \(V_1\) and \(V_2\), which is impossible. So if \(|L_{x}^{y}|= 2\), then \(|L_{y}^{x}|=1\). Also, if \(|L_{y}^{x}|= 2\), then \(|L_{x}^{y}| =1\).
\(\square \)
Assume that \(S=\{1,m\}\), where \(m\in \mathrm{Coatom}(\mathcal L )\). By Lemma 4.2, we have \(|\mathrm{Coatom}(\mathcal L )|\le 3\). In the following proposition, we present a necessary and sufficient condition for planarity of \(\Gamma _S(\mathcal L )\) whenever \(|\mathrm{Coatom}(\mathcal L )|=1\).
Theroem 4.7
Let \(S=\{1,m\}\) and \(\mathrm{Coatom}(\mathcal L )=\{m\}\). Then the graph \(\Gamma _S(\mathcal L )\) is planar if and only if one of the conditions (i) or (ii) in Proposition 4.6 holds.
Proof
Let \(\Gamma _S(\mathcal L )\) is planar. Then, by Proposition 4.6, we are done.
Conversly, we show that if one of the conditions (i) or (ii) in Proposition 4.6 occurs, then \(\Gamma _S(\mathcal L )\) is planar. One can easily check that, if \(|B_{\ell }(m)|=1\), then \(\Gamma _S(\mathcal L )\) is isomorphic to \(K_{1,1, |\mathcal L |-2}\), and so \(\Gamma _S(\mathcal L )\) is planar. Also, if condition (ii) in Proposition 4.6 occurs, then it is not hard to see that \(\Gamma _S(\mathcal L )\) is planar. \(\square \)
In the next theorem, we provide a necessary and sufficient condition for planarity of \(\Gamma _S(\mathcal L )\) whenever \(|\mathrm{Coatom}(\mathcal L )|=2\).
Theroem 4.8
Suppose that \(S=\{1,m\}\) and \(\mathrm{Coatom}(\mathcal L )=\{m , m^{\prime }\}\). Then the graph \(\Gamma _S(\mathcal L )\) is planar if and only if \(\mathcal L \) is one of the following lattices:
-
(i)
\(\mathcal L =S \cup L_{m^{\prime }}^{x} \cup \mathrm{J}(\mathcal L )\), where \(B_{\ell }(m)=\{x\}\) such that \(x\in \{m^{\prime }\}^{\ell }\).
-
(ii)
\(\mathcal L =S \cup L_{x}^{m^{\prime }} \cup L_{m^{\prime }}^{x} \cup \mathrm{J}(\mathcal L )\), where \(B_{\ell }(m)=\{x\}\) such that
-
(a)
\(x\notin \{m^{\prime }\}^{\ell }\).
-
(b)
The numbers \(| L_{x}^{m^{\prime }} |\) and \(| L_{m^{\prime }}^{x}|\) are less than 3, and if \(| L_{x}^{m^{\prime }} |=2\), then \(| L_{m^{\prime }}^{x}|=1\), and if \(| L_{m^{\prime }}^{x} |=2\), then \(| L_{x}^{m^{\prime }}|=1\).
-
(a)
-
(iii)
\(\mathcal L =S \cup \{m^{\prime }\} \cup B_{\ell }(m) \cup \mathrm{J}(\mathcal L )\), where \(|B_{\ell }(m)|=2\) and \(|B_{\ell }(m)\setminus \{m^{\prime }\}^{\ell }|= 1\).
Proof
Assume that \(\Gamma _S(\mathcal L )\) is a planar graph. Then, by Lemma 4.6, \(|B_{\ell }(m)\setminus \{m^{\prime }\}^{\ell }|\le 2\). Now, suppose that \(|B_{\ell }(m)\setminus \{m^{\prime }\}^{\ell }|=2\), and set \(B_{\ell }(m)\setminus \{m'\}^{\ell }:=\{x,y\}\). Since the induced subgraph of \(\Gamma _S(\mathcal L )\) on \(\{1,m,m^{\prime },x,y\}\) is isomorphic to \(K_5\), by Kuratowski’s Theorem, \(\Gamma _S(\mathcal L )\) is not planar, which is impossible. Thus \(|B_{\ell }(m)\setminus \{m^{\prime }\}^{\ell }|\le 1\). Also, by Proposition 4.6, one of the conditions (i) or (ii) in Proposition 4.6 must holds. So we have the following cases:
Case 1. Suppose that the condition (i) in Proposition 4.6 holds and put \(B_{\ell }(m):=\{x\}\). At first note that \(\mathcal L =S \cup L_{x}^{m^{\prime }} \cup L_{m^{\prime }}^{x} \cup \mathrm{J}(\mathcal L )\). Now, assume that \(|B_{\ell }(m)\setminus \{m^{\prime }\}^{\ell }|= 1\). Thus \(B_{\ell }(m)\setminus \{m^{\prime }\}^{\ell }=\{x\}\). If \(|L_{x}^{m^{\prime }}|\ge 3\), then it is easy to see that the graph \(\Gamma _S(\mathcal L )\) has a copy of \(K_{3,3}\), which is impossible. Therefore \(|L_{m}^{m^{\prime }}|\le 2\). Similarly \(|L_{m^{\prime }}^{x}|\le 2\). Now, if \(|L_{x}^{m^{\prime }}|=|L_{m^{\prime }}^{x}|= 2\), then consider the sets \(V_1:= L_{x}^{m^{\prime }} \cup \{1\}\) and \(V_2:= L_{m^{\prime }}^{x} \cup \{m\}\), to deduce that \(\Gamma _S(\mathcal L )\) has a copy of \(K_{3,3}\), which is impossible. Therefore, if \(|L_{x}^{m^{\prime }}|=2\), then \(|L_{m^{\prime }}^{x}|=1\). Similarly, if \(|L_{m^{\prime }}^{x}|=2\), then \(|L_{x}^{m^{\prime }}|=1\). So \(\mathcal L =S \cup L_{x}^{m^{\prime }} \cup L_{m^{\prime }}^{x} \cup \mathrm{J}(\mathcal L )\), where \(B_{\ell }(m)=\{x\}\) and \(| L_{x}^{m^{\prime }} |\), and \(| L_{m^{\prime }}^{x}|\) are less than 3, and if \(| L_{x}^{m^{\prime }} |=2\), then \(| L_{m^{\prime }}^{x}|=1\), and also, if \(| L_{x}^{m^{\prime }} |=2\), then \(| L_{m^{\prime }}^{x}|=1\).
Now, let \(|B_{\ell }(m)\setminus \{m^{\prime }\}^{\ell }|= 0\). Since \(|B_{\ell }(m)\setminus \{m^{\prime }\}^{\ell }|= 0\), we have that \(L_{m^{\prime }}^{x}=\emptyset \). So \(\mathcal L =S \cup L_{x}^{m^{\prime }} \cup \mathrm{J}(\mathcal L )\).
Case 2. Suppose that the condition (ii) in Proposition 4.6 holds. Set \(B_{\ell }(m):=\{x,y\}\). Note that in this case, we have that
Since \(\mathcal L \) is a lattice, we have that \(|B_{\ell }(m)\setminus \{m^{\prime }\}^{\ell }|= 1\). Without loss of generality, we may assume that \(B_{\ell }(m)\setminus \{m^{\prime }\}^{\ell }=\{y\}\). If \(|L_{\{x,y\}}^{m^{\prime }}|\ge 2\), then, by setting \(V_1:= L_{\{x,y\}}^{m^{\prime }} \cup \{y\}\) and \(V_2:= \{1, m, x\}\), one can find a copy of \(K_{3,3}\) in the graph \(\Gamma _S(\mathcal L )\), which is impossible. Hence \(L_{\{x,y\}}^{m^{\prime }}=\{m^{\prime }\}\). Since \(x\in \{m^{\prime }\}^{\ell }\), we have that \(L_{\{y,m^{\prime }\}}^{x}=\emptyset \). Also, if \(|L_{\{x,m^{\prime }\}}^{y}|\ge 2\), then one can easily to find a copy of \(K_{3,3}\), which is impossible. So \(L_{\{x,m^{\prime }\}}^{y}=\{y\}\). On the other hand, since \(x\in \{m^{\prime }\}^{\ell }\), \(L_{m^{\prime }}^{\{x,y\}}=\emptyset \). Thus \(\mathcal L =S \cup \{m^{\prime }\} \cup B_{\ell }(m) \cup \mathrm{J}(\mathcal L )\).
The converse statment is clear. \(\square \)
In the following, we investigate the planarity of the graph \(\Gamma _S(\mathcal L )\) whenever \(|S|=2\) and \(\mathrm{Coatom}(\mathcal L )=\{m, m^{\prime }, m^{\prime \prime }\}\).
Proposition 4.9
Suppose that \(S=\{1,m\}\), \(\mathrm{Coatom}(\mathcal L )=\{m, m^{\prime }, m^{\prime \prime }\}\) and that the graph \(\Gamma _S(\mathcal L )\) is planar. Then the following statements hold.
-
(i)
\(L_{\{m^{\prime },m^{\prime \prime }\}}^{m} =\{m\}\),
-
(ii)
\(|L_{\{m,m^{\prime \prime }\}}^{m^{\prime }}|\le 2\) and \(|L_{\{m,m^{\prime }\}}^{m^{\prime \prime }}|\le 2\), and if \(|L_{\{m,m^{\prime \prime }\}}^{m^{\prime }}|= 2\) then \(|L_{\{m,m^{\prime }\}}^{m^{\prime \prime }}|=1\), and if \(|L_{\{m,m^{\prime }\}}^{m^{\prime \prime }}|= 2\), then \(|L_{\{m,m^{\prime \prime }\}}^{m^{\prime }}|=1\)
Proof
(i) Suppose that \(\Gamma _S(\mathcal L )\) is planar. Assume in contrary that \(|L_{\{m^{\prime },m^{\prime \prime }\}}^{m}|\ge 2\). Note that, for every element \(x\in L_{\{m^{\prime },m^{\prime \prime }\}}^{m}\), we have \(x\vee m^{\prime }=1\) and \(x\vee m^{\prime \prime }=1\). So the graph \(\Gamma _S(\mathcal L )\) has a copy of \(K_5\), which is impossible. Thus \(L_{\{m^{\prime },m^{\prime \prime }\}}^{m} =\{m\}\).
(ii) Assume in contrary that \(|L_{\{m,m^{\prime \prime }\}}^{m^{\prime }} |\ge 3\) or \(|L_{\{m,m^{\prime }\}}^{m^{\prime \prime }}|\ge 3\). One can easily see that the graph \(\Gamma _S(\mathcal L )\) has a subgraph isomorphic to \(K_{3,3}\), and so \(\Gamma _S(\mathcal L )\) is not planar, which is impossible. Therefore \(|L_{\{m,m^{\prime \prime }\}}^{m^{\prime }}|\le 2\) and \(|L_{\{m,m^{\prime }\}}^{m^{\prime \prime }}|\le 2\). Now, if \(| L_{\{m,m^{\prime \prime }\}}^{m^{\prime }}|=2\) and \(|L_{\{m,m^{\prime }\}}^{m^{\prime \prime }}|=2\), then, by setting \(V_1:= L_{\{m,m^{\prime }\}}^{m^{\prime \prime }}\cup \{1\}\) and \(V_2:=L_{\{m,m^{\prime \prime }\}}^{m^{\prime }} \cup \{m\}\), we can find a copy of \(K_{3,3}\) in the graph \(\Gamma _S(\mathcal L )\), which is impossible. Hence if \(|L_{\{m,m^{\prime \prime }\}}^{m^{\prime }}|= 2\), then \(|L_{\{m,m^{\prime }\}}^{m^{\prime \prime }}|=1\). Similarly if \(|L_{\{m,m^{\prime }\}}^{m^{\prime \prime }}|= 2\), then \(|L_{\{m,m^{\prime \prime }\}}^{m^{\prime }}|=1\). \(\square \)
Theroem 4.10
Let \(S=\{1,m\}\) and \(\mathrm{Coatom}(\mathcal L )=\{m , m^{\prime }, m^{\prime \prime }\}\). Then the graph \(\Gamma _S(\mathcal L )\) is planar if and only if \(\mathcal L \) is one of the following lattices:
-
(i)
\( \mathcal L =\{1,m,m^{\prime },m^{\prime \prime },x\}\cup L_{x}^{\{m^{\prime },m^{\prime \prime }\}}\cup \mathrm{J}(\mathcal L )\), where \(B_{\ell }(m)=\{x\}\) and \(x\in \{m^{\prime }\}^{\ell }{\setminus } \{m^{\prime \prime }\}^{\ell }\) or \(x\in \{m^{\prime \prime }\}^{\ell }{\setminus } \{m^{\prime }\}^{\ell }\).
-
(ii)
\(\mathcal L =\{1,m,x\}\cup L_{\{x,m^{\prime \prime }\}}^{m^{\prime }}\cup L_{\{x,m^{\prime \prime }\}}^{m^{\prime }} \cup L_{x}^{\{m^{\prime },m^{\prime \prime }\}}\cup \mathrm{J}(\mathcal L )\), where \(B_{\ell }(m)=\{x\}\) and
-
(a)
\(x\in \{m^{\prime }\}^{\ell } \cap \{m^{\prime \prime }\}^{\ell }\),
-
(b)
\(|L_{\{x,m^{\prime \prime }\}}^{m^{\prime }}|\le 2\) and \(|L_{\{x,m^{\prime \prime }\}}^{m^{\prime }}|\le 2\), and if \(|L_{\{x,m^{\prime \prime }\}}^{m^{\prime }}|= 2\), then \(|L_{\{x,m^{\prime }\}}^{m^{\prime \prime }}|=1\), and if \(|L_{\{x,m^{\prime }\}}^{m^{\prime \prime }}|= 2\), then \(|L_{\{x,m^{\prime \prime }\}}^{m^{\prime }}|=1\)
-
(a)
Proof
Suppose that \(\Gamma _S(\mathcal L )\) is planar. Firstly, by Proposition 4.9 , we have \(L_{\{m^{\prime },m^{\prime \prime }\}}^{m} =\{m\}\) which implies that \(|B_{\ell }(m)\setminus \{m^{\prime },m^{\prime \prime }\}^{\ell }|=0\). Now, by Proposition 4.6, we have the following cases:
Case 1. Suppose that the condition (i) of Proposition 4.6 holds. Then set \(B_{\ell }(m):=\{x\}\). Since \(|B_{\ell }(m){\setminus } \{m^{\prime },m^{\prime \prime }\}^{\ell }|=0\) , we have the following subcases:
-
(a)
\(B_{\ell }(m){\setminus } \{m^{\prime }\}^{\ell }= \{x\}\) and \(B_{\ell }(m){\setminus } \{m^{\prime \prime }\}^{\ell }= \emptyset \). In this case, if \(|L_{\{x ,m^{\prime \prime }\}}^{m^{\prime }} |\ge 2\), then one can easily find a copy of \(K_{3,3}\), which is impossible. So \(|L_ {\{x ,m^{\prime \prime }\}}^{m^{\prime }} |=1\). Similarly \(| L_ {\{x ,m^{\prime }\}}^{m^{\prime \prime }}|=1\). Also \(L_ {\{m^{\prime } ,m^{\prime \prime }\}}^{x}\), \(L_{m^{\prime \prime }}^{\{x,m^{\prime }\}}\) and \(L_{m^{\prime }}^{\{x,m^{\prime \prime }\}}\) are empty sets, because \(x\in \{m^{\prime }\}^{\ell }\). So
$$\begin{aligned} \mathcal{L }=\{1,m,m^{\prime },m^{\prime \prime },x\}\cup L_{x}^{\{m^{\prime },m^{\prime \prime }\}}\cup \mathrm{J}(\mathcal L ). \end{aligned}$$ -
(b)
\(B_{\ell }(m){\setminus } \{m^{\prime \prime }\}^{\ell }= \{x\}\) and \(B_{\ell }(m){\setminus } \{m^{\prime }\}^{\ell }= \emptyset \). It is similar to (a).
-
(c)
\(B_{\ell }(m){\setminus } \{m^{\prime \prime }\}^{\ell }= \emptyset \) and \(B_{\ell }(m){\setminus } \{m^{\prime }\}^{\ell }= \emptyset \). In this case, by Proposition 4.9, we have that \(|L_{\{x,m^{\prime \prime }\}}^{m^{\prime }}|\le 2\) and \(|L_{\{x,m^{\prime \prime }\}}^{m^{\prime }}|\le 2\), and if \(|L_{\{x,m^{\prime \prime }\}}^{m^{\prime }}|= 2\), then \(|L_{\{x,m^{\prime }\}}^{m^{\prime \prime }}|=1\) and vis versa. Also, since \(x\in \{m^{\prime }\}^{\ell }\) and \(x\in \{m^{\prime \prime }\}^{\ell }\), the sets \(L_ {\{m^{\prime } ,m^{\prime \prime }\}}^{x}\), \(L_{m^{\prime \prime }}^{\{x,m^{\prime }\}}\) and \(L_{m^{\prime }}^{\{x,m^{\prime }\}}\) are empty. Thus \(\mathcal L =\{1,m,x\}\cup L_{\{x,m^{\prime \prime }\}}^{m^{\prime }}\cup L_{\{x,m^{\prime \prime }\}}^{m^{\prime }} \cup L_{x}^{\{m^{\prime },m^{\prime \prime }\}}\cup \mathrm{J}(\mathcal L )\), where \(|L_{\{x,m^{\prime \prime }\}}^{m^{\prime }}|\le 2\) and \(|L_{\{x,m^{\prime \prime }\}}^{m^{\prime }}|\le 2\), and if \(|L_{\{x,m^{\prime \prime }\}}^{m^{\prime }}|= 2\) and \(|L_{\{x,m^{\prime }\}}^{m^{\prime \prime }}|=1\) and if \(|L_{\{x,m^{\prime }\}}^{m^{\prime \prime }}|= 2\) and \(|L_{\{x,m^{\prime \prime }\}}^{m^{\prime }}|=1\)
Case 2. Suppose that the condition (ii) of Proposition 4.6 holds. Set \(B_{\ell }(m):=\{x,y\}\). Since \(\mathcal L \) is a lattice and \(|B_{\ell }(m)\setminus \{m',m''\}^{\ell }|=0\), one can conclude that \(|B_{\ell }(m)\setminus \{m'\}^{\ell }|=1\) and \(|B_{\ell }(m)\setminus \{m''\}^{\ell }|=1\). So, without loss the generality, we may assume that \(x\in L_{m''}^{m'}\) and \(y\in L_{m'}^{m''}\). Now if we set \(V_1:=\{1,m',x\}\) and \(V_1:=\{m,m'',y\}\), then one can find a subgraph isomorphic to \(K_{3,3}\) in \(\Gamma _S(\mathcal L )\) which is impossile. So this case never happens.
The converse statment is clear. \(\square \)
Now, the only remaining case is that \(|S|=1\). The following lemmas are useful.
Lemma 4.11
Assume that \(|S|=1\). Then the graph \(\Gamma _S(\mathcal L )\) is planar if and only if \(\Gamma _S(\mathcal L \setminus J(\mathcal L ))\) is planar.
Proof
By Proposition 3.6 we have that \(\mathrm{{deg}}(x)=1\), for all \(x\in J(\mathcal L )\). So, the result holds. \(\square \)
Lemma 4.12
Let \(|S|=1\) and \(|\mathrm{{Coatom}}(\mathcal L )|\ge 2\). If \(| L_{\mathrm{{Coatom}}(\mathcal L )\setminus \{m\}}^{m}|\ge 3\) and \(|L_{m}^{m'}|\ge 2\), for some distinct \(m, m' \in \mathrm{{Coatom}}(\mathcal L )\), then \(\Gamma _S(\mathcal L )\) is not planar.
Proof
Since \(| L_{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}^{m}|\ge 3\), we set \(V_1:=\{m,x_1,x_2\} \subseteq L_{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}^{m}\). Also, since \(|L_{m}^{m'}|\ge 2\), we set \(V_2:=\{1,m', y_1\}\) where \(y_1\in L_{m}^{m'}\). It is easy to see that the subgraph of \(\Gamma _S(\mathcal L )\) on \(V_1\cup V_2\) is isomorphic to \(K_{3,3}\). Thus, by Kuratowski’s Theorem, \(\Gamma _S(\mathcal L )\) is not planar. \(\square \)
In the next theorem, we provide a necessary and sufficient condition for planarity of \(\Gamma _S(\mathcal L )\) when \(|S|=1\).
Theroem 4.13
Suppose that \(|S|=1\). Then the graph \(\Gamma _S(\mathcal L )\) is planar if and only if one of the following statements holds:
-
(i)
\(|\mathrm{Coatom}(\mathcal L )|=1\),
-
(ii)
\(|\mathrm{Coatom}(\mathcal L )|=2\) and if there is an element \(m\in \mathrm{Coatom}(\mathcal L )\) such that \(| L_{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}^{m}| \ge 3\), then \(|L_{m}^{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}|=1\)
-
(iii)
\(|\mathrm{Coatom}(\mathcal L )|=3\), and
-
(a)
For all \(m\in \mathrm{Coatom}(\mathcal L )\), \(|L_{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}^{m}|\le 2\) and there is at most one element \(m\) in \(\mathrm{Coatom}(\mathcal L )\) such that \(|L_{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}^{m}|=2\).
-
(b)
If there is \(m\in \mathrm{Coatom}(\mathcal L )\) such that \(|L_{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}^{m}|=2\), then \(|L_{m}^{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}|=0\).
-
(a)
Proof
Assume that the graph \(\Gamma _S(\mathcal L )\) is planar. Then, by Proposition 3.2, \(|\mathrm{Coatom}(\mathcal L )| \le 3\). Suppose that \(|\mathrm{Coatom}(\mathcal L )|\ne 1\). So, we have the following cases:
Case 1. Suppose that \(|\mathrm{Coatom}(\mathcal L )|= 2\) and put \(\mathrm{Coatom}(\mathcal L )=\{m,m'\}\). If there is one element \(c\in \mathrm{Coatom}(\mathcal L )\) such that \(|L_{\mathrm{Coatom}(\mathcal L )\setminus \{c\}}^{c}| \ge 3\), then we show that \(|L_{c}^{\mathrm{Coatom}(\mathcal L )\setminus \{c\}}|=1\). Without loss the generality, we may assume that \(c=m\). Assume in contrary that \(|L_{m}^{m'}|\ge 2\). Then, by setting \(V_1:=\{1,m',x\}\) and \(V_2:=\{m,y_1,y_2\}\), where \(x\in L_{m}^{m'}\) and \(y_1, y_2\in L_{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}^{m}\), one can find a copy of \(K_{3,3} \) in \(\Gamma _S(\mathcal L )\), which is impossible. Therefore if there is one element \(m\in \mathrm{Coatom}(\mathcal L )\) such that \(| L_{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}^{m}| \ge 3\), then \(L_{m}^{m'}=\{m'\}\).
Case 2. Suppose that \(|\mathrm{Coatom}(\mathcal L )|=3\) and put \(\mathrm{Coatom}(\mathcal L )=\{m,m',m''\}\). It is easy to see that if \(| L_{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}^{m} | \ge 3\), then the graph \(\Gamma _S(\mathcal L )\) is not planar, which is impossible. So \(| L_{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}^{m}| \le 2\) for all \(m\in \mathrm{Coatom}(\mathcal L )\). Now, Assume in contrary that \(| L_{\mathrm{Coatom}(\mathcal L )\setminus \{c\}}^{c}| = 2\) and \(|L_{\mathrm{Coatom}(\mathcal L )\setminus \{c'\}}^{c'} |= 2\) for some \(c\) and \(c'\) in \(\mathrm{Coatom}(\mathcal L )\). Without loss of generality, we may assume that \(c=m\) and \(c'=m'\). Put \(V_1:=\{1,m,x\}\) and \(V_2:=\{m',m'',y\}\), where \(x\in L_{\{m',m''\}}^{m}\) and \(y\in L_{\{m,m''\}}^{m'} \). Then one can find a subgraph of \(\Gamma _S(\mathcal L )\) which is isomorphic to \(K_{3,3} \), which is impossible. So, there is at most one element \(c\) in \(\mathrm{Coatom}(\mathcal L )\) such that \(|L_{\mathrm{Coatom}(\mathcal L )\setminus \{c\}}^{c}|=2\). Now, assume that \(|L_{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}^{m}|=2\), and assume in contrary that \(|L_{m}^{\{m',m''\}}|\ge 1\). Then, by setting \(V_1:=\{1\}\cup L_{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}^{m}\) and \(V_2:=\{m',m'',y\}\) where \(y\in L_{m}^{\{m',m''\}}\), we can find a subgraph in the graph \(\Gamma _S(\mathcal L )\) which is isomorphic to \(K_{3,3}\), which is impossible. So, \(|L_{m}^{\{m',m''\}}|=0\).
Conversely, if \(|\mathrm{Coatom}(\mathcal L )|=1\), then, by part (ii) of Proposition 2.5, \(\Gamma _S(\mathcal L )\) is a star graph, and so it is planar.
Now, assume that \(|\mathrm{Coatom}(\mathcal L )|=2\) and that there is one element \(m\in \mathrm{Coatom}(\mathcal L )\) such that \(|L_{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}^{m}| \ge 3\). Then \(|L_{m}^{\mathrm{Coatom}(\mathcal L )\setminus \{m\}}|=1\). It is not hard to see that \(\Gamma _S(\mathcal L\setminus J(\mathcal L\mathcal ))\) is a complete 3-partite graph such that at least two parts of it has exactly one element. Therefore \(\Gamma _S(\mathcal L\setminus J(\mathcal L\mathcal ))\) is planar. Now, by Proposition 4.11, \(\Gamma _S(\mathcal L )\) is planar.
Now, suppose that \(|\mathrm{Coatom}(\mathcal L )|=3\) and put \(\mathrm{Coatom}(\mathcal L )=\{m,m',m''\}\). Also, assume that the conditions (a) and (b) hold. It is easy to see that the subgraph \(\Gamma _S(\mathcal L\setminus J(\mathcal L\mathcal ))\) on \(\{1\}\cup L_{\{m',m''\}}^{m}\cup L_{\{m,m''\}}^{m'}\cup L_{\{m,m'\}}^{m''}\) is a complete 4-partite graph such that at least three parts of it has exactly one element. Now, by condition (b), it is easy to see that \(\Gamma _S(\mathcal L\setminus J(\mathcal L\mathcal ))\) is planar. Hence, by Proposition 4.11, \(\Gamma _S(\mathcal L )\) is planar.
\(\square \)
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The authors are deeply grateful to the referees for careful reading of the manuscript and helpful suggestions.
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Communicated by Salvatore Rionero.
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Afkhami, M., Barati, Z. & Khashyarmanesh, K. A graph associated to a lattice. Ricerche mat. 63, 67–78 (2014). https://doi.org/10.1007/s11587-013-0164-6
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DOI: https://doi.org/10.1007/s11587-013-0164-6