A causal interactions indicator between two time series using extreme variations in the first eigenvalue of lagged correlation matrices

1.   Introduction

Let $\mathcal{L}$ be a finite lattice with a least element $0$ and $A(\mathcal{L})$ be the set of all atoms. Before reading the paper, to familiarize with the notation and concepts used here, we strongly recommend the readers to read the first part of this work ^[3]. The annihilating-ideal graph of a lattice $\mathcal{L}$, denoted by $\mathbb{AG}(\mathcal{L})$, and defined by the graph whose vertex set is the set of all non-trivial ideals of $\mathcal{L}$ and two distinct vertices $I$ and $J$ being adjacent if and only if $I \land J = 0$, which was introduced by Afkhami et al. ^[1]. Note that the graph $\mathbb{AG}(\mathcal{L})$ is an $r$-partite graph for some $r\in \mathbb{N}$.

One of the most important topological properties of a graph is its genus. The genus of graphs associated with algebraic structures has been studied by many authors, see ^[2,4,5,11]. The planar and crosscap one annihilating-ideal graph of lattices were characterized by Shahsavar ^[13] and Parsapour et al. ^[10], respectively. Also, whether the line graph associated with the annihilating-ideal graph of a lattice is planar or projective was characterized by Parsapour et al. ^[12]. Moreover, the authors of ^[9] characterized all lattices $\mathcal{L}$ whose line graph of ${\Bbb{AG}(\mathcal{L})}$ is toroidal. Recently, Asir et al. ^[3], provided the classification of lattices with the number of atoms less than or equal to 4 whose annihilating-ideal graph can be embedded in the non-orientable surface of crosscap two.

Note that a graph is planar if and only if it does not contain either of two forbidden graphs $K_5$ and $K_{3, 3}$. An analogous characterization for embeddings of graphs on surfaces is known for the projective plane, which has 103 forbidden subgraphs, see ^[6]. For surfaces in general, it is known that the set of forbidden minors is finite and an explicit upper bound can be given, see ^[14]. In this paper, we have identified a class of minimal $r$-partite graphs that are not crosscap two graphs. So, these graphs may be realized as forbidden subgraphs for crosscap two.

The aim of this paper is to find the lattices with at least 5 atoms whose annihilating-ideal graph has non-orientable genus two embedding. This lead to the addition of $r$-partite graphs, where $r\geq 5$, to the family of crosscap two graphs. First of all, we observe that, by Proposition 3.3 ^[3], $|A(\mathcal{L})| \leq 6$ whenever the crosscap of $\Bbb{AG}(\mathcal{L})$ is two. Therefore, the lattices under consideration has either 5 or 6 atoms, and the corresponding classifications are done in Theorems 2.2 and 3.1. The reader can find an interesting connection between ${\Bbb{AG}(\mathcal{L})}$ and multipartite graphs in Examples 2.1 and 3.1.

Before moving into our main results, we have collected the crosscap lower bound of some graphs which will be used in the subsequent sections. In what follows, the notations $K_4-e$ and $K_{4, 5}-e$ denote graphs with an arbitrary edge removed from $K_4$ and $K_{4, 5}$ respectively. Also, $K_8-3e$ denotes a graph with three arbitrary edges removed from $K_8$. Moreover, $K_{6, 3}\cup (K_4-e)$, denotes a graph, that includes the vertices and edges of $K_{6, 3}$ with partition $(X, Y)$, $|X| = 6$, as well as the edges of $K_4-e$, a subgraph induced by any 4 arbitrary vertices in the partition $X$.

Proposition 1.1. Let $G$ be a graph.

${({\rm{i}})}$ [3, Proof of Theorem 3] If $G$ is isomorphic to $K_{6, 3}\cup (K_4-e)$ or $K_{4, 5}-e$, then $\tilde{\gamma}(G)\geq 3$.

${({\rm{ii}})}$ If $G$ is isomorphic to $K_8-3e$ or $K_{2, 2, 2, 2}$, then $\tilde{\gamma}(G)\geq 3$.

Proof. (ⅱ) The non-embeddability of $K_8-3e$ in the Klein bottle directly follows from Euler's polyhedral equation. The non-embeddability of $K_{2, 2, 2, 2}$ in the Klein bottle is a straightforward consequence of the characterization ^[8] of the graphs that triangulate both the torus and Klein bottle; note that it is well-known that $K_{2, 2, 2, 2}$ triangulates the torus, see ^[7]. □

Let us directly move on to the lattices with 5 atoms.

2.   The case when $|A(\mathcal{L})|=5$

Before going into the characterization of crosscap two ${\Bbb{AG}(\mathcal{L})}$ for number of atoms of size 5, we rectify missing cases in projective characterization given in ^[10]. In particular, the authors have not discussed the sets of the form $U_{ijk}$ for $1\leq i\neq j\neq k\leq 5$ in Theorem 2.7. To be precise, let $|{\mathop{\cup }}\, _{n = 1}^5U_n| = 5$. From the proof of Theorem 2.7(ⅰ) ^[10], the following two possibilities should be considered.

Case 1. $|U_{ij}| = 2$ for unique $1\leq i\neq j\leq 5$. Then, Theorem 2.7(ⅰ) of ^[10] says that ${\Bbb{AG}(\mathcal{L})}$ is projective whenever $|\underset{k\ne i, j}{\mathop{\cup }}\, (U_{ik}\cup U_{jk})|\leq 2$ with $|U_{ik}|, |U_{jk}|\leq1$. But consider any one of $U_{(ij)^c}, U_{(ik)^c}, U_{(jk)^c}$ as non-empty. More generally, let us assume $U_{(pq)^c}\neq \emptyset$ for some $U_{pq}\neq \emptyset$ with $1\leq p, q \leq 5$. Then the sets $X = U_i \cup U_j \cup U_{ij}\cup \{[I_{pq}, I_{(pq)^c}]\}$ and $Y = \underset{k\neq i, j}{\mathop{\cup }}\, U_k$ form $K_{5, 3}$ in ${\Bbb{AG}(\mathcal{L})}$ and so ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 2$. Therefore, $U_{(pq)^c} = \emptyset$ whenever $U_{pq}\neq \emptyset$.

Case 2. For some fixed $i$, $|{\mathop{\cup }}\, _{j = 1}^5 U_{ij}|\leq 3$, $|U_{ij}|\leq 1$ and $|U_{k\ell}|\leq 1$ for at most one pair $k, \ell$ with $1\leq i\neq j\neq k\neq \ell\leq 5$ such that every vertex of $U_{k\ell}$ is adjacent to a maximum of one vertex from ${\mathop{\cup }}\, _{j = 1}^5 U_{ij}$.

If $|U_{(pq)^c}|\geq 2$ for some $U_{pq}\neq \emptyset$ and $1\leq p, q\leq 5$, then the sets $X = U_p \cup U_q \cup U_{pq}$ and $Y = \underset{r\neq p, q}{\mathop{\cup }}\, U_r \cup U_{(pq)^c}$ form $K_{3, 5}$ in ${\Bbb{AG}(\mathcal{L})}$, a contradiction. Also if $|U_{(pq)^c}|, |U_{(p_1q_1)^c}| = 1$ for some $|U_{pq}|, |U_{p_1q_1}|\neq \emptyset$ and $1\leq p, q, p_1, q_1\leq 5$, then the sets $X = U_p \cup U_q \cup U_{pq}\cup \{[I_{p_1q_1}, I_{(p_1q_1)^c}]\}$ and $Y = \underset{r\neq p, q}{\mathop{\cup }}\, U_r \cup U_{(pq)^c}$ form $K_{4, 4}-e$ which has crosscap two, a contradiction. Therefore, $|{\mathop{\cup }}\, U_{(pq)^c}|\leq 1$, where the union is taken over all $U_{pq}\neq \emptyset$.

Suppose $|U_{k\ell}| = 1$ for some $1\leq k, \ell\leq 5$ with a vertex in $U_{k\ell}$ adjacent to exactly one vertex of ${\mathop{\cup }}\, _{j = 1}^5 U_{ij},$ say a vertex in $U_{ij}$. We now claim that ${\mathop{\cup }}\, U_{(pq)^c} = \emptyset$, where the union is taken over all $U_{pq}\neq \emptyset$. In order to prove the claim, when either $U_{(ij)^c}\neq \emptyset$ or $U_{(k\ell)^c}\neq \emptyset,$ let us take $U_{(ij)^c}\neq \emptyset$, then $X = U_i \cup U_j \cup U_{ij}$ and $Y = \underset{m\neq i, j}{\mathop{\cup }}\, U_m \cup U_{k\ell} \cup U_{(ij)^c}$ form $K_{3, 5}$. Further, if $U_{(ij^\prime)^c}\neq \emptyset$ for some $1\leq j^\prime\neq j\leq 5$, then, $X = U_i \cup U_j \cup U_{ij} \cup [I_{ij^\prime}, I_{(ij^\prime)^c}]$ and $Y = \underset{m\neq i, j}{\mathop{\cup }}\, U_m \cup U_{k\ell}$ form $K_{4, 4}-e$. Thus, ${\mathop{\cup }}\, U_{(pq)^c} = \emptyset$.

Based on the addition of above mentioned cases in projective characterization, we can summarize it as follows.

Theorem 2.1. Let $\mathcal{L}$ be a lattice with $|A(\mathcal{L})| = 5$ and let $|{\mathop{\cup }}\, _{n = 1}^5U_n| = 5$. Then, $\tilde{\gamma}(\Bbb{AG}(\mathcal{L})) = 1$ if and only if $|U_{ij}|\leq 2$ for all $1\leq i\neq j\leq 5$ and one of the following conditions hold:

${({\rm{i}})}$ There is a unique $U_{ij}$ such that $|U_{ij}| = 2$ with $U_{i^\prime j^\prime}, U_{(ij)^c} = \emptyset$ for $i^\prime, j^\prime \in \{1, \ldots, 5\}{\setminus} \{i, j\}$, $|\underset{p\in \{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}|\leq 2$ and $\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c} = \emptyset$. Moreover, if $U_{p_{1} q_1}, U_{p_2q_2}\neq \emptyset,$ then $\{p_1, q_1\}\cap \{p_2, q_2\}\neq \emptyset$.

${({\rm{ii}})}$ $|U_{ij}|\leq 1$ for all $1\leq i\neq j\leq 5$. For some fixed $i\in \{1, \ldots, 5\}$, $|{\mathop{\cup }}\, _{j = 1}^5 U_{ij}|\leq 3$, atmost one of the sets $U_{k\ell}$ such that $|U_{k\ell}| = 1$, where $1\leq i\neq k\neq \ell\leq 5$ and $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}|\leq 1$. Moreover, if $U_{k\ell}$ has a vertex, then it is adjacent to atmost one vertex in ${\mathop{\cup }}\, _{j = 1}^5 U_{ij}$ and if such adjacency exists, then $\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c} = \emptyset$.

We are now in a position to state and prove the main result of this section.

Theorem 2.2. Let $\mathcal{L}$ be a lattice with $|A(\mathcal{L})| = 5$ and let $1\leq i\neq j\neq k\neq l\neq m\leq 5$. Then, $\tilde{\gamma}(\Bbb{AG}(\mathcal{L})) = 2$ if and only if one of the following conditions hold:

$({\rm{i}})$ $|{\mathop{\cup }}\, _{n = 1}^5U_n| = 8$, there exists $U_i$ with $|U_i| = 4$ and $U_{pq} = U_{jkl} = U_{jklm} = \emptyset$ for all $1\leq p\neq q\leq 5$.

$({\rm{ii}})$ $|{\mathop{\cup }}\, _{n = 1}^5U_n| = 7$, one of the following cases is satisfied:

[a] There exists $U_i$ such that $|U_i| = 3$ with $|\underset{j\neq i}{\mathop{\cup }}\, U_{ij}|\leq 2$ and $|\underset{k, \ell, m, n\neq i}{\mathop{\cup }}\, U_{k\ell}\cup U_{k\ell m}\cup U_{k\ell mn}|\leq 1$. Moreover, if $|\underset{k, \ell, m, n\neq i}{\mathop{\cup }}\, U_{k\ell}\cup U_{k\ell m}\cup U_{k\ell mn}| = 1$, then $\underset{j\neq i}{\mathop{\cup }}\, U_{ij} = \emptyset$ and if $\underset{k, \ell, m, n\neq i}{\mathop{\cup }}\, U_{k\ell}\cup U_{k\ell m}\cup U_{k\ell mn} = \emptyset$, then $|\underset{j\neq i}{\mathop{\cup }}\, U_{ij}|\in \{1, 2\}$.

[b] There exist $U_i$ and $U_j$ such that $|U_i| = |U_j| = 2$ with $|\underset{k, \ell \notin \{i, j\}}{\mathop{\cup }}\, U_{k\ell}|\leq 1$. Moreover:

[b1] If $|\underset{k, \ell \notin \{i, j\}}{\mathop{\cup }}\, U_{k\ell}| = 1$, then $\underset{m = i, j; pq\neq ij}{\mathop{\cup }}\, (U_{mn}\cup U_{pqr}) \cup U_{(ij)^c} \cup U_{ij}\cup U_{(i)^c}\cup U_{(j)^c} = \emptyset$.

[b2] If $\underset{k, \ell \notin \{i, j\}}{\mathop{\cup }}\, U_{k\ell} = \emptyset$, then $|\underset{m = i, j; mn\neq ij}{\mathop{\cup }}\, U_{mn}\cup U_{(ij)^c}|\leq 1$. Also, if $|\underset{m = i, j; mn\neq ij}{\mathop{\cup }}\, U_{mn}\cup U_{(ij)^c}| = 1$, then $|U_{ij}|\leq 1$ and $\underset{pq\neq ij; pqr \neq (ij)^c}{\mathop{\cup }}\, U_{pqr}\cup U_{(i)^c}\cup U_{(j)^c} = \emptyset$. Moreover, if $\underset{m = i, j; mn\neq ij}{\mathop{\cup }}\, U_{mn}\cup U_{(ij)^c} = \emptyset$, then $|U_{ij}|\leq 2$. In addition, if $|U_{ij}| = 2$, then $\underset{pq\neq ij; pqr \neq (ij)^c}{\mathop{\cup }}\, U_{pqr}\cup U_{(i)^c}\cup U_{(j)^c} = \emptyset$ and if $|U_{ij}|\leq 1$, then $|\underset{p = i, j; pq\neq ij} {\mathop{\cup }}\, U_{pqr}|\leq 2$ together with $|U_{pqr}|\leq 1$, and either $\underset{p = i; q\neq j} {\mathop{\cup }}\, U_{pqr} \cup U_{(j)^c} = \emptyset$ or $\underset{p = j} {\mathop{\cup }}\, U_{pqr} \cup U_{(i)^c} = \emptyset$.

$({\rm{iii}})$ $|{\mathop{\cup }}\, _{n = 1}^5U_n| = 6$. There exists $U_i$ such that $|U_i| = 2$ with $|\underset{j, k\neq i}{\mathop{\cup }}\, U_{jk}|\leq 2$ and $U_{j^\prime k^\prime} = \emptyset$ when $U_{jk}\neq \emptyset$ for all $j^\prime, k^\prime\notin \{j, k\}$. Moreover:

[a] There is $U_{jk}$ such that $|U_{jk}| = 2$ for $j, k\neq i$ with $\underset{\ell\notin \{j, k\}} {\mathop{\cup }}\, U_{i\ell} = \emptyset$, $|\underset{m\in \{j, k\}; p, q, r\ne i} {\mathop{\cup }}\, (U_{im}\cup U_{pqr})|\leq 2$ in which $|U_{im}|, |U_{pqr}|\leq 1$ and $U_{(st)^c} = \emptyset$ if $U_{st}\neq \emptyset$ for all $1\leq s\neq t \leq 5$.

[b] There exist $U_{jk}, U_{j_1 k_1}$ such that $|U_{jk}| = |U_{j_1 k_1}| = 1$ where $j, k, j_1, k_1\neq i$ and $|\{j, k\}\cap\{j_1, k_1\}| = 1$ with $\underset{\ell\notin \{j, k\}\cap\{j_1, k_1\}} {\mathop{\cup }}\, U_{i\ell} = \emptyset$, $|\underset{m = \{j, k\}\cap\{j_1, k_1\}; p, q, r\ne i} {\mathop{\cup }}\, (U_{im}\cup U_{pqr})|\leq 2$ in which $|U_{im}|, |U_{pqr}|\leq 1$ and $U_{(st)^c} = \emptyset$ if $U_{st}\neq \emptyset$ for all $1\leq s\neq t \leq 5$.

[c] There is a unique $U_{jk}$ such that $|U_{jk}| = 1$ for all $j, k\neq i$ with $|\underset{\ell\notin \{j, k\}} {\mathop{\cup }}\, U_{i\ell}\cup U_{(jk)^c}|\leq 1$. Also, if $|\underset{\ell\notin \{j, k\}} {\mathop{\cup }}\, U_{i\ell}\cup U_{(jk)^c}| = 1,$ then $|\underset{m\in \{j, k\}} {\mathop{\cup }}\, U_{im}|\leq 2$ in which each $|U_{im}|\leq 1$ and $\underset{p, q, r, s\ne i} {\mathop{\cup }}\, U_{pqr}\cup U_{pqrs} = \emptyset$. Furthermore, one of the following is satisfied in the case of $\underset{\ell\notin \{j, k\}} {\mathop{\cup }}\, U_{i\ell}\cup U_{(jk)^c} = \emptyset$:

[c1] If $|\underset{m\in \{j, k\}} {\mathop{\cup }}\, U_{im}| = 3$ or $4$, then exactly one of the sets $U_{im}$, for $m = j, k$, has more than one element and $\underset{p, q, r, s\ne i} {\mathop{\cup }}\, U_{pqr}\cup U_{pqrs} = \emptyset$.

[c2] If $|U_{im}| = 2$ and $U_{im^\prime} = \emptyset$ for $m\neq m^\prime\in \{j, k\}$, then $U_{(im)^c} = \emptyset$ and $|\underset{p, q, r\ne i}{\mathop{\cup }}\, U_{pqr}|\leq 1$. Also, $U_{pqrs} = \emptyset$ for $p, q, r, s\ne i$ whenever $|\underset{p, q, r\ne i} {\mathop{\cup }}\, U_{pqr}| = 1$.

[c3] If $|\underset{m\in \{j, k\}} {\mathop{\cup }}\, U_{im}|\leq 2$ in which each $|U_{im}|\leq 1$, then $U_{(im)^c} = \emptyset$ and $1\leq |\underset{m\in \{j, k\}; p, q, r\ne i} {\mathop{\cup }}\, (U_{im} \cup U_{pqr})|\leq 3$. Also, $U_{pqrs} = \emptyset$ for $p, q, r, s\ne i$ whenever $|\underset{m\in \{j, k\}; p, q, r\ne i} {\mathop{\cup }}\, (U_{im} \cup U_{pqr})| = 3$ together with $|\underset{p, q, r\ne i} {\mathop{\cup }}\, U_{pqr}| = 2$.

[d] $\underset{j, k\neq i}{\mathop{\cup }}\, U_{jk} = \emptyset$ and $|\underset{m\neq i} {\mathop{\cup }}\, U_{im}|\leq 4$ in which $|U_{im}|\leq 3$. Also, if two of $U_{im}$'s has 2 elements, then $\underset{p, q, r, s\ne i} {\mathop{\cup }}\, U_{pqr}\cup U_{pqrs} = \emptyset$, and if exactly one set $U_{im}$ has more than 1 element, then $|\underset{p, q, r\ne i} {\mathop{\cup }}\, U_{pqr}|\leq 1$ together with $\underset{U_{im}\neq \emptyset} {\mathop{\cup }}\, U_{(im)^c} = \emptyset$. Furthermore, one of the following is satisfied in case of $|U_{im}|\leq 1$ for all $m\neq i$:

[d1] If $|\underset{m\neq i} {\mathop{\cup }}\, U_{im}| = 4$, then $\underset{U_{im}\neq \emptyset} {\mathop{\cup }}\, U_{(im)^c} = \emptyset$.

[d2] If $|\underset{m\neq i} {\mathop{\cup }}\, U_{im}|\leq 3$, then $|\underset{U_{im}\neq \emptyset} {\mathop{\cup }}\, U_{(im)^c}|\leq 1$. Also, if $|\underset{U_{im}\neq \emptyset} {\mathop{\cup }}\, U_{(im)^c}| = 1,$ then $\underset{p\neq i, pqr\neq (im)^c} {\mathop{\cup }}\, U_{pqr} = \emptyset$. If $\underset{U_{im}\neq \emptyset} {\mathop{\cup }}\, U_{(im)^c} = \emptyset,$ then $|\underset{p, q, r\neq i; pqr\neq (im)^c} {\mathop{\cup }}\, U_{pqr}|\leq 2$ whenever $|\underset{m\neq i} {\mathop{\cup }}\, U_{im}| = 3$, and $2\leq |\underset{p, q, r\neq i; pqr\neq (im)^c} {\mathop{\cup }}\, U_{pqr}|\leq 4$ with atmost one $|U_{pqr}|\in \{2, 3\}$ whenever $|\underset{m\neq i} {\mathop{\cup }}\, U_{im}|\leq 2$. Further, in the last part, if $|\underset{p, q, r\neq i; pqr\neq (im)^c} {\mathop{\cup }}\, U_{pqr}| = 2,$ then exactly one non-empty set exists in the collection $\{U_{pqr}: p, q, r\neq i; pqr\neq (im)^c\}$.

$({\rm{iv}})$ $|{\mathop{\cup }}\, _{n = 1}^5U_n| = 5$ and one of the following cases is satisfied:

[a] There is a $U_{ij}$ such that $|U_{ij}| = 4$, $\underset{\ell, m\notin\{i, j\}}{\mathop{\cup }}\, U_{\ell m} \cup U_{(ij)^c} = \emptyset$, $|\underset{p\in\{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}|\leq 2$ in which $|U_{pq}|\leq 1$ and $U_{p^\prime q^\prime}, U_{(pq)^c} = \emptyset$ when $|U_{pq}| = 1$ where $p^\prime, q^\prime\notin \{p, q\}$.

[b] There is a $U_{ij}$ such that $|U_{ij}| = 3$, $\underset{\ell, m\notin\{i, j\}}{\mathop{\cup }}\, U_{\ell m} \cup U_{(ij)^c} = \emptyset$, $|\underset{p\in \{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}|\leq 3$, where the choice $i$ or $j$ for $p$ is placed at most two times in the union, in which at most one of $U_{pq}$'s has two elements and $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}|\leq 1$. Further, if $|U_{pq}| = 2$ for some $p\in \{i, j\}; q\notin \{i, j\}$, then $\underset{p^\prime, q^\prime\notin \{p, q\}} {\mathop{\cup }}\, U_{p^\prime q^\prime} \cup \underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c} = \emptyset$, and if $|\underset{p\in \{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}| = 3$ with $|U_{pq}|\leq 1$, then the three choices for $q$ is not distinct. Moreover, if $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}| = 1$, then $|\underset{p\in \{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}|\leq 2$ with the choice for two pairs of $p, q$'s are not mutually disjoint.

[c] There is a $U_{ij}$ such that $|U_{ij}| = 2$ with $|\underset{\ell, m\notin\{i, j\}}{\mathop{\cup }}\, U_{\ell m} \cup U_{(ij)^c}|\leq 1$. Further, if $|\underset{\ell, m\notin\{i, j\}}{\mathop{\cup }}\, U_{\ell m} \cup U_{(ij)^c}| = 1$, then $|\underset{p\in\{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}|\leq 2$ with $|U_{pq}|\leq 1$ and $U_{p^\prime q^\prime}, U_{(pq)^c} = \emptyset$ when $|U_{pq}| = 1$, where $p^\prime, q^\prime\notin \{p, q\}$. Moreover, if $\underset{\ell, m\notin\{i, j\}}{\mathop{\cup }}\, U_{\ell m} \cup U_{(ij)^c} = \emptyset$, then $2\leq |\underset{p\in \{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}|\leq 4$ in which at most one of the sets $U_{pq}$ has two elements, where the choice $i$ or $j$ for $p$ is placed at most once in the union, and one of the following is satisfied:

[c1] If $|U_{rs}| = 2$ for some $r\in \{i, j\}, s\notin \{i, j\}$, then $\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c} = \emptyset$ and at most one of the sets $U_{tu}$ is non-empty with the property that $\{r, s\}\cap \{t, u\} = \emptyset$.

[c2] If $|U_{pq}|\leq 1$ for all $p\in \{i, j\}, q\notin \{i, j\}$, then $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}|\leq 1$. Also, if $|U_{(pq)^c}| = 1$ for some $|U_{pq}| = 1$, then every non-empty set $U_{rs}$ should have the property that $\{r, s\}\cap \{p, q\}\neq \emptyset$.

[d] $|U_{ij}|\leq 1$ for all $1\leq i\neq j\leq 5$.

[d1] If $|\underset{1\leq p\neq q\leq 5} {\mathop{\cup }}\, U_{pq}| = 5$, then $\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c} = \emptyset$, and at least one of the sets $U_{p_1q_1}, U_{p_2q_2}, U_{p_3q_3}$ or $U_{p_4q_4}$ must be empty whenever the indices satisfy the condition $\{p_{1}, q_{1}\}\cap \{p_{2}, q_{2}\} = \emptyset$ and $\{p_{3}, q_{3}\}\cap \{p_{4}, q_{4}\} = \emptyset$.

[d2] If $|\underset{1\leq p\neq q\leq 5} {\mathop{\cup }}\, U_{pq}| = 4$, then $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}|\leq 1$. Moreover, if $\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c} = \emptyset$, then the subgraph induced by the set $\underset{1\leq p\neq q\leq 5} {\mathop{\cup }}\, U_{pq}$ has more than one edge and if $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}| = 1$, say $|U_{(rs)^c}| = 1$, then the vertex in $U_{rs}$ is adjacent to at most two vertices of $\underset{U_{pq}\neq \emptyset; pq\neq rs} {\mathop{\cup }}\, U_{pq}$. Further, if there is an adjacency between the vertex of $U_{rs}$ and a vertex of $\underset{U_{pq}\neq \emptyset; pq\neq rs} {\mathop{\cup }}\, U_{pq}$, then the subgraph induced by the set $\underset{U_{pq}\neq \emptyset; pq\neq rs} {\mathop{\cup }}\, U_{pq}$ is an empty graph.

[d3] If $|\underset{1\leq p\neq q\leq 5} {\mathop{\cup }}\, U_{pq}|\in \{2, 3\}$, then $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}|\leq 3$. Moreover:

● If $|\underset{U_{pq\neq \emptyset}}{\mathop{\cup }}\, U_{(pq)^c}| = 3,$ then $|U_{(pq)^c}| = 3$ for some $1\leq p\neq q\leq 5$ and no non-empty set $U_{rs}$ exist with $\{r, s\}\cap \{p, q\} = \emptyset$.

● Let $|\underset{U_{pq\neq \emptyset}}{\mathop{\cup }}\, U_{(pq)^c}| = 2$. If a unique set $U_{(pq)^c}\neq \emptyset$ for $1\leq p\neq q\leq 5$, then at most one non-empty set $U_{rs}$ exists with $\{r, s\}\cap \{p, q\} = \emptyset$. If two sets $U_{(p_1q_1)^c}, U_{(p_2q_2)^c}\neq \emptyset$ for $1\leq p_1, q_1, p_2, q_2\leq 5$, then no non-empty set $U_{rs}$ exists with $\{r, s\}\cap \{p_f, q_f\} = \emptyset$ for $1\leq f\leq 2$.

● If $|\underset{U_{pq\neq \emptyset}}{\mathop{\cup }}\, U_{(pq)^c}| = 1$, then exactly one non-empty set $U_{rs}$ exists with $\{r, s\}\cap \{p, q\} = \emptyset$.

[d4] If $|\underset{1\leq p\neq q\leq 5} {\mathop{\cup }}\, U_{pq}| = 1$, then $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}|\in \{2, 3\}$.

Proof. If $|{\mathop{\cup }}\, _{n = 1}^5U_n|\geq 9$, then ${\Bbb{AG}(\mathcal{L})}$ contains $K_{5, 4}$ as a subgraph so that $|{\mathop{\cup }}\, _{n = 1}^5U_n|\leq 8$.

Case 1. Let $|{\mathop{\cup }}\, _{n = 1}^5 U_n| = 8$. Suppose $|U_1| = 4$. If $\underset{k, p\neq 1} {\mathop{\cup }}\, U_{ij}\cup U_{k\ell m}\cup U_{pqrs} \neq \emptyset$ for some $1\leq i < j \leq 5$, then ${\Bbb{AG}(\mathcal{L})}$ contains $K_{4, 5}-e$ and by Proposition 1.1, we have ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq3$. Therefore, $\underset{k, p\neq 1} {\mathop{\cup }}\, U_{ij}\cup U_{k\ell m}\cup U_{pqrs} = \emptyset$. Now the graph ${\Bbb{AG}(\mathcal{L})}$ (except the vertices of degree one and two) is a subgraph of $H_1$ (as given in [3, Figure 1(a)]) and so by [3, Lemma 3.5], we get ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$. If $|U_1| = 3$, then the subgraph induced by the sets $X = U_1\cup U_5$ and $Y = U_2\cup U_3\cup U_4$ contains $H_4$ in ${\Bbb{AG}(\mathcal{L})}$ and so by [3, Lemma 3.6], ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Also, if $|U_1| = 2$, then ${\Bbb{AG}(\mathcal{L})}$ contains $K_{2, 2, 2, 2}$ as a subgraph and by Proposition 1.1, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq3$.

Case 2. Let $|{\mathop{\cup }}\, _{n = 1}^5U_n| = 7$.

Case 2.1. Suppose $|U_1| = 3$. If the subgraph induced by $\langle V({\Bbb{AG}(\mathcal{L})})-\{{\mathop{\cup }}\, _{n = 1}^5 U_n\}\rangle$ has an edge $(I, J)$, then the vertices $I_1, I_1^\prime, I_1^{\prime\prime}, I_2, I_3, I_4, I_5, [I, J]$ form $K_{8}-3e$ and so by Proposition 1.1, we have ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Therefore each vertex of $U_{1mn}$ is adjacent to exactly two vertices in ${\Bbb{AG}(\mathcal{L})}$ which are also adjacent. Also if $I, J\in \underset{i, p, s\neq 1} {\mathop{\cup }}\, U_{ij} \cup U_{pqr}\cup U_{stuv}$, then the subgraph induced by $I_1, [I_1^\prime, I], [I_1^{\prime\prime}, J], I_2, I_3, I_4$ and $I_5$ form $K_7$ in ${\Bbb{AG}(\mathcal{L})}$, a contradiction. Thus $|\underset{i, p, s\neq 1} {\mathop{\cup }}\, U_{ij} \cup U_{pqr}\cup U_{stuv}|\leq 1$ and among all the remaining sets we have to examine only those sets of the form $U_{1k}$.

Let $I\in \underset{i, p, s\neq 1} {\mathop{\cup }}\, U_{ij} \cup U_{pqr}\cup U_{stuv}$. If $U_{1k}\neq \emptyset$ for some $k\neq 1$, then, the subgraph $G_{21} = {\Bbb{AG}(\mathcal{L})}-\{I, (I_k, I_\ell), (I_k, I_m), (I_k, I_n)\}$ contains $K_{4, 4}-e$ with partite sets $X = U_1\cup U_{1k}$ and $Y = U_k \cup U_\ell \cup U_m\cup U_n$ where $\ell, m, n\in\{2, 3, 4, 5\}{\setminus} \{k\}$ and $e = (I_{1k}, I_k)$. Note that any $N_2$-embedding of $K_{4, 4}-e$ has one hexagonal and six rectangular faces. Since $I$ is adjacent to three vertices $I_1, I_1^\prime$ and $I_1^{\prime\prime}$ of $X$, the vertex $I$ must be inserted into the hexagonal face of the $N_2$-embedding of $K_{4, 4}-e$. If $I_k$ is in the hexagonal face, then $I_k$ is in exactly two distinct rectangular faces so that the three edges incident with $I_k$, namely $(I_k, I_\ell), (I_k, I_m), (I_k, I_n)$, cannot be drawn without edge crossing, a contradiction. If not, $I_{1k}\in X$ must be in the hexagonal face. Therefore, the hexagonal face does not contain all the three vertices of $X$ namely $I_1, I_1^\prime$ and $I_1^{\prime\prime}$. Thus, $I$ cannot be embedded, a contradiction. Hence, $U_{1k} = \emptyset$ for all $2\leq k\leq 5$.

Assume $\underset{i, p, s\neq 1} {\mathop{\cup }}\, U_{ij} \cup U_{pqr}\cup U_{stuv} = \emptyset$. Suppose $|{\mathop{\cup }}\, _{k = 2}^{5} U_{1k}|\geq 3$ and let $I_{1\ell}, I_{1m}, I_{1n}\in {\mathop{\cup }}\, _{k = 2}^{5} U_{1k}$. Then, the subgraph ${\Bbb{AG}(\mathcal{L})}-\{I_{1m}, I_{1n}\}$ contains $K_{4, 4}-e$ with partitions $X = U_1\cup U_{1\ell}$ and $Y = U_2\cup U_3 \cup U_4\cup U_5$. Clearly, any $N_2$-embedding of $K_{4, 4}-e$ has one hexagonal and six rectangular faces. The vertices $I_{1m}$ and $I_{1n}$ are adjacent to three vertices of $Y$ in ${\Bbb{AG}(\mathcal{L})}$. So it requires at least two hexagonal faces in a $N_2$-embedding of $K_{4, 4}-e$, a contradiction. Thus, $|{\mathop{\cup }}\, _{k = 2}^{5} U_{1k}|\leq 2$.

Further, ${\Bbb{AG}(\mathcal{L})}$ is projective whenever $\underset{1\leq i < j\leq 5} {\mathop{\cup }}\, U_{ij} = \emptyset$ with $\underset{p, s\neq 1} {\mathop{\cup }}\, U_{pqr}\cup U_{stuv} = \emptyset$. Thus, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ if $|\underset{i, p, s\neq 1} {\mathop{\cup }}\, U_{ij} \cup U_{pqr}\cup U_{stuv}| = 1$ with ${\mathop{\cup }}\, _{k = 2}^{5} U_{1k} = \emptyset$ or $\underset{i, p, s\neq 1} {\mathop{\cup }}\, U_{ij} \cup U_{pqr}\cup U_{stuv} = \emptyset$ with $|{\mathop{\cup }}\, _{k = 2}^{5} U_{1k}|\in \{1, 2\}$.

Case 2.2. Suppose $|U_1| = 2$. Then, $|U_2|$ must be $2$.

If $|U_{ij}|\geq 2$ for some $i\neq 1, 2$, then the contraction of ${\Bbb{AG}(\mathcal{L})}$ induced by the set $\{I_1, [I_1^\prime, I_{ij}], I_2, [I_2^\prime, I_{ij}^\prime], I_3, I_4, I_5\}$ form $K_7$. So, $|U_{ij}|\leq 1$ for all $i, j\notin \{1, 2\}$.

Suppose $|U_{ij}| = 1$ for some $i\neq 1, 2$. If $I\in \underset{k = 1, 2;pq\neq 12}{\mathop{\cup }}\, (U_{k\ell}\cup U_{pqr}) \cup U_{(ij)^c} \cup U_{1345}\cup U_{2345},$ then $I$ is adjacent to one of $I_1$, $I_2$ or $I_{ij}$. The latter case, that is $(I, I_{ij})\in E({\Bbb{AG}(\mathcal{L})})$, is not possible because $\langle {\mathop{\cup }}\, _{n = 1}^5 U_n \cup \{[I, I_{ij}]\} \rangle\cong K_{8}-2e$. Also if either $(I, I_{1})\in E({\Bbb{AG}(\mathcal{L})})$ or $(I, I_{2})\in E({\Bbb{AG}(\mathcal{L})})$, then we can merge such an edge so that $K_8-3e$ is a minor subgraph of ${\Bbb{AG}(\mathcal{L})}$. Thus, in this case, $V({\Bbb{AG}(\mathcal{L})})\setminus \{{\mathop{\cup }}\, _{n = 1}^5 U_n \cup U_{12i} \cup U_{12j}\} = \emptyset$.

Suppose $U_{ij} = \emptyset$ for all $i\neq 1, 2$. Let $I, J\in \underset{k = 1, 2;k\ell\neq 12}{\mathop{\cup }}\, U_{k\ell}\cup U_{345}$. If $I, J\in U_{345}$, then the partition sets $\{I, J, I_3, I_4, I_5\}$ and $\{U_1\cup U_2\}$ form $K_{5, 4}$ in ${\Bbb{AG}(\mathcal{L})}$. If not, we have $|U_{k\ell}|\geq 1$ for some $k\in \{1, 2\}$ and $k\ell\neq 12$ so that the partition sets $U_k\cup U_{\ell} \cup \{I, J\}$ and $\underset{m\neq k, \ell} {\mathop{\cup }}\, U_m$ form $K_{5, 4}-e$ in ${\Bbb{AG}(\mathcal{L})}$. Thus, $|\underset{k = 1, 2;k\ell\neq 12}{\mathop{\cup }}\, U_{k\ell}\cup U_{345}|\leq 1$.

Let $I\in \underset{k = 1, 2;k\ell\neq 12}{\mathop{\cup }}\, U_{k\ell}\cup U_{345}$.

● In the case of $|U_{12}|\geq 2$, note that $I$ is adjacent to either the vertices of $U_1$ or $U_2$, say $U_1$. Here, the contraction of ${\Bbb{AG}(\mathcal{L})}$ contains $K_{6, 3}\cup (K_4-e)$ with partite sets $\{I_1, [I_1^\prime, I], I_2, I_2^\prime, I_{12}, I_{12}^\prime\}$ and $\underset{m\neq 1, 2}{\mathop{\cup }}\, U_{m}$.

● In the case of $\underset{pq\neq 12;pqr \neq 345}{\mathop{\cup }}\, U_{pqr}\cup U_{1345}\cup U_{2345}\neq \emptyset$, by contracting a single edge in ${\Bbb{AG}(\mathcal{L})}$, we get the contraction of ${\Bbb{AG}(\mathcal{L})}$ contains $H_4$.

Thus, $|\underset{k = 1, 2;k\ell\neq 12}{\mathop{\cup }}\, U_{k\ell}\cup U_{345}| = 1$, $|U_{12}|\leq 1$ and $\underset{pq\neq 12;pqr \neq 345}{\mathop{\cup }}\, U_{pqr}\cup U_{1345}\cup U_{2345} = \emptyset$. For this case, with the help of Figure 1, we get ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$.

Let $\underset{k = 1, 2;k\ell\neq 12}{\mathop{\cup }}\, U_{k\ell}\cup U_{345} = \emptyset$.

● In the case of $|U_{12}|\geq 3$, ${\Bbb{AG}(\mathcal{L})}$ contains $K_{3, 7}$ with partite sets $U_1\cup U_2\cup U_{12}$ and $U_3\cup U_4\cup U_5$.

● In the case of $|U_{12}| = 2$, we have $\underset{pq\neq 12;pqr \neq 345}{\mathop{\cup }}\, U_{pqr}\cup U_{1345}\cup U_{2345} = \emptyset$. If not, there exists some $J\in \underset{pq\neq 12;pqr \neq 345}{\mathop{\cup }}\, U_{pqr}\cup U_{1345}\cup U_{2345}$, then $J$ is adjacent to either $I_1$ or $I_2$, say $(J, I_1)\in E({\Bbb{AG}(\mathcal{L})})$ so that the contraction of ${\Bbb{AG}(\mathcal{L})}$ contains $K_{6, 3}\cup (K_4-e)$ with partite sets $\{[J, I_1], I_1^\prime, I_2, I_2^\prime, I_{12}, I_{12}^\prime\}$ and $U_3\cup U_4\cup U_5$.

● In the case of $|U_{12}|\leq 1$:

(a) If $|U_{pqr}|\geq 2$ for $pq\neq 12$ and $pqr\neq 345$, then ${\Bbb{AG}(\mathcal{L})}$ contains $K_{3, 6}\cup(K_4-e)$ with partite sets $U_p\cup U_q\cup U_r \cup U_{pqr}$ and $\underset{m\neq p, q, r}{\mathop{\cup }}\, U_{m}$. Therefore, $|U_{pqr}|\leq 1$ for $pq\neq 12$ and $pqr\neq 345$.

(b) If $J\in \underset{p = 1;q\neq 2} {\mathop{\cup }}\, U_{pqr} \cup U_{1345}$ and $K\in \underset{p = 2} {\mathop{\cup }}\, U_{pqr} \cup U_{2345}$, then the contraction of ${\Bbb{AG}(\mathcal{L})}$ induced by the set $\{I_1, [I_1^\prime, K], I_2, [I_2^\prime, J], I_3, I_4, I_5\}$ form $K_7$. Therefore, either $\underset{p = 1;q\neq 2} {\mathop{\cup }}\, U_{pqr} \cup U_{1345} = \emptyset$ or $\underset{p = 2} {\mathop{\cup }}\, U_{pqr} \cup U_{2345} = \emptyset$.

(c) If $|\underset{p = 1;q\neq 2} {\mathop{\cup }}\, U_{pqr}|\geq 3$, that is $|U_{134}| = |U_{135}| = |U_{145}| = 1$, then consider the subgraph ${\Bbb{AG}(\mathcal{L})}-\{I_{135}, I_{145}, (I_1, I_4), (I_1^\prime, I_4), (I_3, I_4), (I_1, I_3), (I_1^\prime, I_3)\}$ which contains $K_{5, 3}$ with partite sets $X = U_1\cup U_3\cup U_{4}\cup U_{134}$ and $Y = U_2\cup U_5$. Notice that any $N_2$-embedding of $K_{5, 3}$ has one hexagonal face and six rectangular faces. Also in ${\Bbb{AG}(\mathcal{L})}$, the vertex $I_4$ is adjacent to the vertices of $\{I_1, I_1^\prime, I_{3}\}\subseteq X$ and $I_{135}$ is adjacent to $I_4$ as well as $\{I_2, I_2^\prime\}\subseteq Y$. Since $deg_{K_{5, 3}}(I_4) = 3$, three rectangular faces cannot adopt all these edges incident with $I_3$ together with the edges incident with $I_{135}$. Therefore, $I_4$ must be in the hexagonal face. A similar technique also proves that $I_3$ is a part of the hexagonal face. To an extent, the hexagonal face can adopt the vertices $I_{135}$ and $I_{145}$ with its edges together with an edge incident to either $I_4$ or $I_3$. We let the edge $(I_1, I_4)$ be embedded in the hexagonal face. Here the two other edges incident with $I_4$, namely $(I_1^\prime, I_4)$ and $(I_3, I_4)$ can be embedded in two rectangle faces that contains $I_4$. Now we have to embed two more edges incident with $I_3$, namely $(I_1, I_3)$ and $(I_1^\prime, I_3)$ but we are left-out with only one rectangular face that contains $I_3$, a contradiction. Therefore, $|\underset{p = 1;q\neq 2} {\mathop{\cup }}\, U_{pqr}|\leq 2$; likewise $|\underset{p = 2} {\mathop{\cup }}\, U_{pqr}|\leq 2$.

Thus, in the case of $\underset{k = 1, 2;k\ell\neq 12}{\mathop{\cup }}\, U_{k\ell}\cup U_{345} = \emptyset$, either $|U_{12}| = 2$ with $\underset{pq\neq 12;pqr \neq 345}{\mathop{\cup }}\, U_{pqr}\cup U_{1345}\cup U_{2345} = \emptyset$ or $|U_{12}|\leq 1$ with $|\underset{p = 1, 2;pq\neq 12} {\mathop{\cup }}\, U_{pqr}|\leq 2$ and $|U_{pqr}|\leq 1$ together with either $\underset{p = 1;q\neq 2} {\mathop{\cup }}\, U_{pqr} \cup U_{1345} = \emptyset$ or $\underset{p = 2} {\mathop{\cup }}\, U_{pqr} \cup U_{2345} = \emptyset$. For all these cases, by using Figure 2, we get ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$.

Thus, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ if $|\underset{i\neq 1, 2}{\mathop{\cup }}\, U_{ij}| = 1$ with $\underset{k = 1, 2;pq\neq 12}{\mathop{\cup }}\, (U_{k\ell }\cup U_{pqr}) \cup U_{(ij)^c} \cup U_{12}\cup U_{1345}\cup U_{2345} = \emptyset$ or $\underset{i\neq 1, 2}{\mathop{\cup }}\, U_{ij} = \emptyset$ with $|\underset{k = 1, 2;k\ell\neq 12;pq\neq 12}{\mathop{\cup }}\, U_{k\ell}\cup U_{pqr}|\leq 1$. Also, if $|\underset{k = 1, 2;k\ell\neq 12;pq\neq 12}{\mathop{\cup }}\, U_{k\ell}\cup U_{pqr}| = 1$, then $|U_{12}|\leq 1$ and $U_{1345} = U_{2345} = \emptyset$. Moreover, if $\underset{k = 1, 2;k\ell\neq 12;pq\neq 12}{\mathop{\cup }}\, U_{k\ell}\cup U_{pqr} = \emptyset$, then $|U_{12}|\leq 2$. In addition, $U_{1345} = U_{2345} = \emptyset$ whenever $|U_{12}| = 2$ and either $U_{1345} = \emptyset$ or $U_{2345} = \emptyset$ whenever $|U_{12}|\leq 1$.

Case 3. Let $|{\mathop{\cup }}\, _{n = 1}^5 U_n| = 6$. Then $|U_1| = 2$. If $|\underset{i\neq 1}{\mathop{\cup }}\, U_{ij}|\geq 3$, then the graph $G_{21}$ is contained in ${\Bbb{AG}(\mathcal{L})}$ and so ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Therefore, $|\underset{i\neq 1}{\mathop{\cup }}\, U_{ij}|\leq 2$. Further, if $|U_{ij}| = |U_{k\ell}| = 1$ for some $\{i, j\}\cap \{k, \ell\} = \emptyset$, then the graph $(H_4 \cup (u_1, u_2)) - (v_2, v_4)$ is contained in ${\Bbb{AG}(\mathcal{L})}$ and by [3, Lemma 3.6], we have ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. That is, $E(\langle {\mathop{\cup }}\, _{i\neq 1} U_{ij}\rangle) = \emptyset$.

Case 3.1. Assume $|\underset{i\neq 1}{\mathop{\cup }}\, U_{ij}| = 2$. Let $|U_{ij}| = 2$ for some $i\neq 1$. If $I\in (\underset{k\ne i, j}{\mathop{\cup }}\, U_{1k}) \cup U_{(ij)^c}$, then the sets $X = U_i\cup U_j\cup U_{ij}$ and $Y = U_1\cup U_k \cup U_{k^\prime}\cup I$, where $k^\prime\in \{2, 3, 4, 5\}{\setminus} \{i, j, k\}$ form $K_{4, 5}$ in ${\Bbb{AG}(\mathcal{L})}$, a contradiction. Therefore, $\underset{k\ne i, j}{\mathop{\cup }}\, U_{1k} = U_{(ij)^c} = \emptyset$.

If $U_{1i}$ or $U_{1j}$ has two elements, say $|U_{1i}|\geq 2$, then the subgraph $G_{22} = {\Bbb{AG}(\mathcal{L})}-\{I_{1i}, I_{1i}^\prime, (I_k, I_i), (I_k, I_j), (I_k, I_{ij}), (I_k, I_{ij}^\prime)\}$ contains $K_{5, 3}$ with partite sets $X = U_i\cup U_j\cup U_{k}\cup U_{ij}$ and $Y = U_1\cup U_{k^\prime}$, where $k\in\{2, 3, 4, 5\}{\setminus}\{i, j\}$ and $k^\prime\in \{2, 3, 4, 5\}{\setminus} \{i, j, k\}$. Note that any $N_2$-embedding of $K_{5, 3}$ has one hexagonal, six rectangular faces and out of which three faces contains the vertex $I_k$ because $deg_{K_{5, 3}}(I_{k}) = 3$. Since $I_{1i}$ and $I_{1i}^\prime$ are adjacent to $I_j, I_k, I_{k^\prime}$ in ${\Bbb{AG}(\mathcal{L})}$, it requires two distinct faces that contains $I_k$ to embed the vertices $I_{1i}$ and $I_{1i}^\prime$. So, after embedding $I_{1i}$ and $I_{1i}^\prime$ in any $N_2$-embedding of $K_{5, 3}$, it may adopt at most three distinct edges with one end in $I_k$ and another end in one of the vertices of $X$. But, $I_k$ is adjacent to $\{I_i, I_j, I_{ij}, I_{ij}^\prime\}\subset X$, a contradiction. Thus, $|U_{1i}|, |U_{1j}| \leq 1$.

Suppose $|U_{1i}|, |U_{1j}| = 1$ and $\underset{p\neq 1}{\mathop{\cup }}\, U_{pqr}\neq \emptyset$. Then, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Suppose $|U_{1i}| = 1$ and $U_{1j} = \emptyset$. If $U_{(1i)^c}\neq \emptyset$, then the sets $X = U_i \cup U_j \cup U_{ij}\cup\{[I_{1i}, I_{(1i)^c}]\}$ and $Y = U_1 \cup U_m \cup U_n$, where $m, n\in \{1, \ldots, 5\}{\setminus} \{1, i, j\}$ form $K_{5, 4}$ in ${\Bbb{AG}(\mathcal{L})}$, a contradiction. Also, if $I, J\in \underset{p\neq 1}{\mathop{\cup }}\, U_{pqr}$, then it is not difficult to verify that ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$.

Thus, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ if $\underset{k\notin \{i, j\}} {\mathop{\cup }}\, U_{1k} = \emptyset$, $|\underset{k\in \{i, j\}, p\ne 1} {\mathop{\cup }}\, (U_{1k}\cup U_{pqr})|\leq 2$ with $|U_{1k}|, |U_{pqr}|\leq 1$ and $U_{(mn)^c} = \emptyset$ if $U_{mn}\neq \emptyset$ for all $1\leq m, n \leq 5$.

Moreover, it is not difficult to verify that the same argument is also valid for $|U_{ij}| = |U_{mn}| = 1$ for some $i, m\neq 1$. Since $E(\langle {\mathop{\cup }}\, _{i\neq 1} U_{ij}\rangle) = \emptyset,$ let $\{i, j\}\cap \{m, n\} = \ell$. So ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ whenever $\underset{k\neq \ell} {\mathop{\cup }}\, U_{1k} = \emptyset$, $|U_{1\ell}|\leq 1$, $|U_{1\ell} \cup \underset{p\ne 1} {\mathop{\cup }}\, U_{pqr}|\leq 2$ and $U_{(mn)^c} = \emptyset$ if $U_{mn}\neq \emptyset$ for all $1\leq m, n \leq 5$.

Case 3.2. Suppose $|U_{ij}| = 1$ for some unique $i\neq 1$. If $I, J \in \underset{k\ne i, j}{\mathop{\cup }}\, U_{1k}\cup U_{(ij)^c}$, then the sets $X = U_i\cup U_j\cup U_{ij}$ and $Y = U_1\cup I \cup J \cup \underset{k\neq i, j}{\mathop{\cup }}\, U_k$ form $K_{3, 6}\cup(K_4-e)$ in ${\Bbb{AG}(\mathcal{L})}$ and so by Proposition 1.1, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Therefore, $|\underset{k\neq i, j}{\mathop{\cup }}\, U_{1k} \cup U_{(ij)^c}|\leq 1$.

Case 3.2.1. Suppose $I \in \underset{k\neq i, j}{\mathop{\cup }}\, U_{1k} \cup U_{(ij)^c}$. If $|U_{1i}|\geq 2$, then the graph induced by sets $X = \{I_1, I_1^\prime, I_i, I_{1i}, I_{1i}^\prime, [I_{ij}, I]\}$ and $Y = \underset{k\neq i, j}{\mathop{\cup }}\, U_k$ contains a minor $K_{6, 3}\cup (K_4-e)$ in ${\Bbb{AG}(\mathcal{L})}$ and so by Proposition 1.1, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Therefore, $|U_{1i}|, |U_{1j}|\leq 1$. Further, if $J\in U_{pqr}\cup U_{pqrs}$ for some $p\neq 1$, then the set $\{I_1, [I_1^\prime, J], I_2, I_3, I_4, I_5, [I_{ij}, I]\}$ form $K_7$ in ${\Bbb{AG}(\mathcal{L})}$, a contradiction. For the remaining cases, by using Figure 3(a), we have ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$.

Case 3.2.2. Suppose $\underset{k\neq i, j}{\mathop{\cup }}\, U_{1k} \cup U_{(ij)^c} = \emptyset$. Let $\max\{|U_{1i}|, |U_{1j}|\} = |U_{1i}|$ and ${\ell}, m\in \{2, 3, 4, 5\}{\setminus} \{i, j\}$. Clearly $|U_{1i}|\leq 3$, otherwise, the sets $X = U_1\cup U_i\cup U_{1i}$ and $Y = U_j\cup U_\ell\cup U_{m}$ form $K_{7, 3}$ in ${\Bbb{AG}(\mathcal{L})}$. If $|U_{1i}|\geq 2$ and $|U_{1j}|\geq 2$, then ${\Bbb{AG}(\mathcal{L})}-\{I_{1j}, I_{1j}^\prime, I_{ij}, (I_j, I_\ell), (I_j, I_\ell^\prime)\}$ contains $K_{5, 3}$ with $X = U_1\cup U_i \cup U_{1i}$ and $Y = U_j\cup U_\ell\cup U_{m}$ which is similar to the graph $G_{15}$ (refer Case 4.2.2 of [3, Theorem 5.2]) so that ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$.

Let $|U_{1i}| = 3$. If $I\in \underset{p\neq 1} {\mathop{\cup }}\, U_{pqr} \cup U_{pqrs}$, then the subgraph ${\Bbb{AG}(\mathcal{L})}-\{I_{ij}, I, (I_1, I_i), (I_1^\prime, I_i)\}$ contains $K_{6, 3}$ with $X = U_1\cup U_i \cup U_{1i}$ and $Y = U_j\cup U_\ell\cup U_{m}$. Note that any $N_2$-embedding of $K_{6, 3}$ has only rectangular faces. Further, in ${\Bbb{AG}(\mathcal{L})}$, $I_{ij}$ is adjacent to $I_1, I_1^\prime, I_\ell, I_m$ and $I$ is adjacent to $I_1, I_1^\prime$. So, to embed the vertices $I_{ij}$ and $I$, it requires two distinct rectangular faces that contain both $I_1$ and $I_1^\prime$. Next, to embed the edges $(I_1, I_i)$ and $(I_1^\prime, I_i),$ it requires two more distinct rectangular faces with diagonals $I_1, I_i$ and $I_1^\prime, I_i$. In such a case, one cannot construct the remaining five distinct rectangular faces by using the existing vertices and edges, a contradiction. Therefore, $\underset{p\neq 1}{\mathop{\cup }}\, U_{pqr} \cup U_{pqrs} = \emptyset$. In this case, that is $|U_{1i}| = 3$, $|U_{1j}|\leq 1$ with $\underset{p\neq 1} {\mathop{\cup }}\, U_{pqr} \cup U_{pqrs} = \emptyset$, and by the help of Figure 3(b), we get ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$.

Let $|U_{1i}| = 2$. Then, $U_{(1i)^c} = \emptyset$; otherwise, the minor subgraph induced by the set $\{I_1, [I_1^\prime, I_{ij}], I_2, I_3, I_4, I_5, [I_{1i}, I_{(1i)^c}]\}$ form $K_7$ in ${\Bbb{AG}(\mathcal{L})}$. Suppose $|U_{1j}| = 1$. If $I\in \underset{p\neq 1} {\mathop{\cup }}\, U_{pqr} \cup U_{pqrs}$, then ${\Bbb{AG}(\mathcal{L})}-\{I_{1j}, I_{ij}, I, (I_1, I_i), (I_1^\prime, I_i)\}$ contains $K_{5, 3}$ with $X = U_1\cup U_i \cup U_{1i}$ and $Y = U_j\cup U_\ell\cup U_{m}$. By using the structure of the graph $G_{15}$ (refer Case 4.2.2 of [3, Theorem 5.2]), we have ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Suppose $U_{1j} = \emptyset$. If $|\underset{p\neq 1; pqr \neq (1i)^c} {\mathop{\cup }}\, U_{pqr}|\geq 2$ or $\underset{p\neq 1; pqr \neq (1i)^c} {\mathop{\cup }}\, U_{pqr}, U_{2345}\neq \emptyset$, then the reader can verify that ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Therefore, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ whenever $|U_{1j}| = 1$ with $\underset{p\neq 1} {\mathop{\cup }}\, U_{pqr} \cup U_{pqrs} = \emptyset$ or $U_{1j} = \emptyset$ with $|\underset{p\neq 1;pqr \neq (1i)^c} {\mathop{\cup }}\, U_{pqr}|\leq 1$ and $U_{2345} = \emptyset$ while $|\underset{p\neq 1;pqr \neq (1i)^c} {\mathop{\cup }}\, U_{pqr}| = 1$.

Let $|U_{1i}|, |U_{1j}|\leq 1$. Then, $U_{(1i)^c}, U_{(1j)^c} = \emptyset$ whenever $U_{1i}, U_{1j}\neq \emptyset$. Note that by Theorem 2.8 ^[10], ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 1$ if $\underset{p\neq 1} {\mathop{\cup }}\, U_{pqr} = \emptyset$.

Thus, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ if $|U_{1i}\cup U_{1j}| = 2$ with $|\underset{p\neq 1;pqr\neq {(1i)^c}, {(1j)^c}} {\mathop{\cup }}\, U_{pqr}| = 1$ (or) $|U_{1i}\cup U_{1j}| = 1$ with $|\underset{p\neq 1;pqr\neq {(1i)^c}, {(1j)^c}} {\mathop{\cup }}\, U_{pqr}| = 2$, $U_{2345} = \emptyset$ or $|\underset{p\neq 1;pqr\neq {(1i)^c}, {(1j)^c}} {\mathop{\cup }}\, U_{pqr}| = 1$ (or) $|U_{1i}\cup U_{1j}| = \emptyset$ with $1\leq |\underset{p\neq 1} {\mathop{\cup }}\, U_{pqr}|\leq 2$.

Case 3.3. Suppose $U_{ij} = \emptyset$ for all $i\neq 1$. Then, the subgraph induced by the neighborhood set of each vertex in $U_{1mn}$ for all $2\leq m, n\leq 5$ is an edge and so it does not play any role in determining the value of the crosscap. If $|\underset{2\leq k\leq 5} {\mathop{\cup }}\, U_{1k}|\geq 5$ or $|U_{1k}|\geq 4$ or $|U_{pqr}|\geq 4$ for $p\neq 1$, then $K_{3, x}$ where $x\geq 7$ is a subgraph of ${\Bbb{AG}(\mathcal{L})}$ and so ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$.

Case 3.3.1. Let $|U_{1k}|\in \{2, 3\}$. Clearly, $U_{(1m)^c} = \emptyset$ whenever $U_{1m}\neq \emptyset$ for all $2\leq m\leq 5$; otherwise, $K_{3, 6}\cup (K_4-e)$ is a subgraph of ${\Bbb{AG}(\mathcal{L})}$. If $|U_{pqr}| = 2$ for $p\neq 1$, then $pqr\neq (1k)^c$ and so $\{p, q, r\}\cap \{k\}\neq \emptyset$. Therefore, we assume that $p = k$. Now the subgraph $G_{23} = {\Bbb{AG}(\mathcal{L})}-\{(I_p, I_q), (I_p, I_r), (I_p, I_\ell)\}$ contains $K_{6, 4}-4e$ with partite set $X = U_p\cup U_q \cup U_r \cup U_{\ell}\cup U_{pqr}$ and $Y = U_{1} \cup U_{1k}$, where $\ell\in \{2, 3, 4, 5\}{\setminus} \{p, q, r\}$. Clearly, every face in any $N_2$-embedding of $K_{6, 4}-4e$ is rectangular. So to embed the edges $(I_p, I_q), (I_p, I_r)$ and $(I_p, I_\ell)$, it requires three rectangular faces which contains $I_p$, a contradiction to $deg_{K_{6, 4}-4e}(I_p) = 2$.

Suppose $|U_{1k}| = 3$ for some $2\leq k\leq 5$. Then ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ provided $|\underset{\ell\neq k} {\mathop{\cup }}\, U_{1\ell}|\leq 1$ with $U_{(1k)^c}, U_{(1\ell)^c} = \emptyset$ and $|\underset{p\neq 1, pqr\neq (1k)^c, (1\ell)^c} {\mathop{\cup }}\, U_{pqr}|\leq 1$.

Suppose $|U_{1k}| = 2$ for some $2\leq k\leq 5$. If $|U_{1 \ell}| = 2$ for some $\ell \in \{2, 3, 4, 5\} {\setminus} k$, then $|\underset{p\neq 1, pqr\neq (1k)^c, (1\ell)^c} {\mathop{\cup }}\, U_{pqr}\cup U_{2345}| = \emptyset$, otherwise, ${\Bbb{AG}(\mathcal{L})}$ contains $G_{23}$. Therefore, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ provided $|\underset{\ell\neq k} {\mathop{\cup }}\, U_{1\ell}|\leq 2$ with $U_{(1k)^c}, U_{(1\ell)^c} = \emptyset$. Moreover, if $|U_{1\ell}| = 2$, then $|\underset{p\neq 1, pqr\neq (1k)^c, (1\ell)^c} {\mathop{\cup }}\, U_{pqr}\cup U_{2345}| = \emptyset$ and if $|U_{1\ell}|\leq 1$, then $|U_{pqr}|\leq 1$ for all $p\neq 1$ and $pqr\neq (1k)^c, (1\ell)^c$.

Case 3.3.2. Suppose $|U_{1k}|\leq 1$ for all $k\in \{2, 3, 4, 5\}$.

Suppose $|\underset{U_{1k}\neq \emptyset} {\mathop{\cup }}\, U_{(1k)^c}|\geq 2,$ say $I, J\in \underset{U_{1k}\neq \emptyset} {\mathop{\cup }}\, U_{(1k)^c}$. Then, the contraction of ${\Bbb{AG}(\mathcal{L})}$ induced by $\{I_1\}\cup \{[I_1^\prime, I_{(1k^c)}^\prime]\}\cup \{[I_{1k}, I_{(1k)^c}]\}\cup \underset{\ell\neq 1}{\mathop{\cup }}\, U_\ell$ in ${\Bbb{AG}(\mathcal{L})}$ form $K_7$, a contradiction. Thus $|\underset{U_{1k}\neq \emptyset} {\mathop{\cup }}\, U_{(1k)^c}|\leq 1$.

Claim A: If $|\underset{2\leq k\leq 5} {\mathop{\cup }}\, U_{1k}| = 4,$ then $\underset{p\neq 1} {\mathop{\cup }}\, U_{pqr} = \emptyset$; equivalently, $\underset{2\leq k\leq 5} {\mathop{\cup }}\, U_{(1k)^c} = \emptyset$.

Let $|\underset{2\leq k\leq 5} {\mathop{\cup }}\, U_{1k}| = 4$. Assume on the contrary that $U_{pqr}\neq \emptyset$ for some $p\neq 1$. Then $pqr = (1m)^c$ where $m\in\{2, 3, 4, 5\}{\setminus}\{p, q, r\}$. Now the subgraph ${\Bbb{AG}(\mathcal{L})}-\{I_{1q}, I_{1r}, I_{1m}, (I_p, I_1), (I_p, I_1^\prime)\}$ contains $K_{4, 4}$ with partite sets $X = U_1\cup U_p \cup U_{1p}$ and $Y = U_q \cup U_{r}\cup U_{m} \cup U_{(1m)^c}$. Note that each face of any $N_2$-embedding of $K_{4, 4}$ is rectangular, the vertices $I_{1q}, I_{1r}, I_{1m}$ are adjacent to $I_p\in X$ and two vertices of $Y$, and $I_p$ is adjacent to $I_1, I_1^\prime\in X$. So to embed the remaining three vertices and two edges, it requires five rectangular faces which contains $I_p$ but $deg_{K_{4, 4}}(I_p) = 4$, a contradiction. Therefore, the claim holds.

Claim B: If $|\underset{2\leq k\leq 5} {\mathop{\cup }}\, U_{1k}|\leq 3$ and $|\underset{U_{1k}\neq \emptyset} {\mathop{\cup }}\, U_{(1k)^c}| = 1$, then $U_{pqr} = \emptyset$ for all $p\neq 1, pqr\neq (1k)^c$.

If $U_{(1k)^c}\neq \emptyset$ for $2\leq k\leq 5$ and $U_{pqr}\neq \emptyset$ for $p\neq 1$ and $pqr \neq (1k)^c$, then the contraction of ${\Bbb{AG}(\mathcal{L})}$ induced by $\langle \{I_1\}\cup \{[I_1^\prime, I_{pqr}]\}\cup \{[I_{1k}, I_{(1k)^c}]\}\cup {\mathop{\cup }}\, _{\ell\neq 1} U_\ell\rangle$ contains $K_7$, a contradiction.

Claim C: In case of $|\underset{2\leq k\leq 5} {\mathop{\cup }}\, U_{1k}|\leq 3$ and $\underset{U_{1k}\neq \emptyset} {\mathop{\cup }}\, U_{(1k)^c} = \emptyset$.

Claim C1: If $|\underset{2\leq k\leq 5} {\mathop{\cup }}\, U_{1k}| = 3$, then $|U_{pqr}|\leq 2$ for $p\neq 1, pqr\neq (1k)^c$.

Suppose $\underset{U_{1k}\neq \emptyset} {\mathop{\cup }}\, U_{(1k)^c} = \emptyset$ and $|U_{pqr}|\geq 3$ for some $p\neq 1, pqr\neq (1k)^c$. This implies that $|U_{1p}| = |U_{1q}| = |U_{1r}| = 1$. Now consider the graph ${\Bbb{AG}(\mathcal{L})}-\{I_{r}, I_{1p}, I_{1q}, I_{1r}\}$ which contains $K_{5, 3}$ with partite sets $X = U_p\cup U_{q}\cup U_{pqr}$ and $Y = U_1\cup U_\ell$ where $\ell \notin \{1, p, q, r\}$. Notice that any $N_2$-embedding of $K_{5, 3}$ has one hexagonal face and six rectangular faces. Label the hexagonal face as $F_1$. Now, try to embed the left-out vertices of ${\Bbb{AG}(\mathcal{L})}$ into a $N_2$-embedding of $K_{5, 3}$. In ${\Bbb{AG}(\mathcal{L})}$, $I_{r}$ is adjacent to $I_1, I_1^\prime, I_p, I_q$ and $I_\ell$ so that $I_r$ should be embedded into the face $F_1$. Since $I_{1p}$ and $I_{1q}$ are adjacent to both $I_{r}$ and $I_\ell$, we have both $I_{1p}$ and $I_{1q}$ embedded together with $I_r$ in $F_1$ and further the face $F_1$ should have the path $I_p - I_\ell -I_q$. The point to remember here is the other neighbors of $I_p$ and $I_q$ in $F_1$ are $I_1$ and $I_1^\prime$. Also, $I_{1r}$ is adjacent to $I_\ell, I_p$ and $I_{q}$, so to embed $I_{1r}$ it requires a rectangular face, say $F_2$, that contains the path $I_p - I_\ell -I_q$. The point here is the fourth vertex of $F_2$ must be either $I_1$ or $I_1^\prime$. At last, since $deg_{K_{3, 5}}(I_p) = 3$, there must be another rectangular face, say $F_3$, in any $N_2$-embedding of $K_{5, 3}$ that should have $I_p$. But, the edge $(I_p, I_\ell)$ is already used twice for forming the faces $F_1$ and $F_2$ so that the two neighbors of $I_p$ in $F_3$ must be $I_1$ and $I_1^\prime$. This contradicts the fact that at least one of the edges $(I_p, I_1)$ or $(I_p, I_1^\prime)$ was used twice in $F_1$ and $F_2$. Thus, the claim holds true.

Claim C2: If $|\underset{2\leq k\leq 5} {\mathop{\cup }}\, U_{1k}|\leq 2$, then $2\leq |\underset{p\neq 1, pqr\neq (1k)^c} {\mathop{\cup }}\, U_{pqr}|\leq 4$ with at most one $|U_{pqr}|\in \{2, 3\}$. Further, if $|\underset{p\neq 1, pqr\neq (1k)^c} {\mathop{\cup }}\, U_{pqr}| = 2,$ then there exists $U_{pqr}$ such that $|U_{pqr}| = 2$.

First recall that if $|\underset{2\leq k\leq 5} {\mathop{\cup }}\, U_{1k}|\in \{1, 2\}$, $\underset{U_{1k}\neq \emptyset} {\mathop{\cup }}\, U_{(1k)^c} = \emptyset$ and $|\underset{p\neq 1, pqr\neq (1k)^c} {\mathop{\cup }}\, U_{pqr}|\leq 2$ with $|U_{pqr}|\leq 1$, then by Theorem 2.8 of ^[10], ${\Bbb{AG}(\mathcal{L})}$ is projective.

● If $|U_{pqr}|\geq 4$ for some $p\neq 1$, then ${\Bbb{AG}(\mathcal{L})}$ contains $K_{7, 3}$ with partite sets $U_{p}\cup U_{q}\cup U_r \cup U_{pqr}$ and $U_1 \cup U_m$ where $m \in \{2, \ldots, 5\}{\setminus} \{p, q, r\}$.

● Suppose there exist two sets $U_{p_1q_1r_1}$ and $U_{p_2q_2r_2}$ from the collection $\{U_{pqr}: p\neq 1, \; pqr\neq (1k)^c \}$ each having more than two elements. Clearly $|\{p_1, q_1, r_1\}\cap \{p_2, q_2, r_2\}| = 2$. So let us take $p_1 = p_2 = p$ and $q_1 = q_2 = q$. Now, consider the graph ${\Bbb{AG}(\mathcal{L})}-\{I_{pqr_2}, I_{pqr_2}^\prime, (I_p, I_q), (I_p, I_{r_1}), (I_{q}, I_{r_1}), (I_1, I_{r_2}), (I_1^\prime, I_{r_2})\}$ which is isomorphic to $K_{5, 3}$ with partite sets $X = U_p\cup U_q \cup U_{r_1}\cup U_{pqr_1}$ and $Y = U_1\cup U_{r_2}$. Any $N_2$-embedding of $K_{5, 3}$ has one hexagonal face and six rectangular faces. Note that $I_{pqr_2}$ and $I_{pqr_2}^\prime$ are adjacent to $I_1, I_1^\prime$ and $I_{r_1}$. To embed the vertices $I_{pqr_2}$ and $I_{pqr_2}^\prime$ into a $N_2$-embedding of $K_{5, 3}$, we have two possibilities; (ⅰ) both $I_{pqr_2}$ and $I_{pqr_2}^\prime$ together with its edges are embedded in two rectangular faces, or (ⅱ) $I_{pqr_2}$ and $I_{pqr_2}^\prime$ together with its edges are embedded in hexagonal and rectangular faces respectively.

(ⅰ) In this case, both rectangular faces must have $I_1, I_1^\prime$ and $I_{r_1}$. Now, embedding of the edges $(I_{r_1}, I_p)$ and $(I_{r_1}, I_q)$ together with the fact that $deg_{K_{5, 3}}(I_{r_1}) = 3$ implies that the hexagonal face must contain $I_{r_1}$. So, the edges either $(I_{r_1}, I_1)$ or $(I_{r_1}, I_1)$ belong to the hexagonal face, a contradiction because it is used twice in two rectangular faces.

(ⅱ) In this case, to embed the edges $(I_p, I_q), (I_p, I_{r_1})$ and $(I_{q}, I_{r_1})$, at least two rectangular faces are required. Finally, to embed the edges $(I_1, I_{r_2})$ and $(I_1^\prime, I_{r_2})$, it requires two more rectangular faces in which the diagonals are $I_1, I_{r_2}$ and $I_1^\prime, I_{r_2}$ respectively. But, such a case does not exist and so ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$.

● Suppose $|\underset{p\neq 1, pqr\neq (1k)^c} {\mathop{\cup }}\, U_{pqr}|\geq 5$. Then, the possibilities from the collection $\{U_{pqr}: p\neq 1, \; pqr\neq (1k)^c \}$ are (ⅰ) one set with three elements and two singleton sets, and (ⅱ) one set with two elements and three singleton sets. For case (ⅰ), the graph ${\Bbb{AG}(\mathcal{L})}-\{I_{4}, I_{245}, I_{345}, (I_2, I_3)\}$ has $K_{5, 3}$ with partite sets $X = U_2\cup U_{3}\cup U_{234}$ and $Y = U_1\cup U_5$ which behave as a similar structure of the graph given in Claim C1. So ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. We leave it to the reader to prove ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$ for case (ⅱ).

Thus, the claim holds true.

Case 4. Let $|{\mathop{\cup }}\, _{n = 1}^5U_n| = 5$ and let $\underset{1\leq p < q\leq 5} \max |U_{pq}| = |U_{ij}|$. Clearly ${\Bbb{AG}(\mathcal{L})}$ is projective when $U_{ij} = \emptyset$. If $|U_{ij}|\geq 5$, then the sets $X = U_i\cup U_j \cup U_{ij}$ and $Y = \underset{k\neq i, j}{\mathop{\cup }}\, U_k$ form $K_{3, 7}$ in ${\Bbb{AG}(\mathcal{L})}$, a contradiction.

Case 4.1. Assume that $|U_{ij}|\in \{3, 4\}$. If $U_{\ell m}\neq \emptyset$ or $U_{\ell mn}\neq \emptyset$ for some $\ell, m, n\notin \{i, j\}$, then the sets $X = U_i\cup U_j \cup U_{ij}$ and $Y = U_\ell \cup U_m \cup U_n\cup U_{\ell m}\cup U_{\ell mn}$ form $K_{5, 4}$ in ${\Bbb{AG}(\mathcal{L})}$, a contradiction. So, any non-empty two index sets $U_{pq}$ and three index sets $U_{pqr}$ must have either $i$ or $j$ as one of their indices. Therefore, every vertex in $U_{ijk}$, for any $k$, is adjacent to exactly two vertices in ${\Bbb{AG}(\mathcal{L})}$, hence, these vertices do not play any role in finding the crosscap value. Thus, we avoid the sets $U_{ijk}$ for all $k$ from $V({\Bbb{AG}(\mathcal{L})})$. Also, if $U_{ik}, U_{i\ell}, U_{im}\neq \emptyset$ for $k, \ell, m\notin \{i, j\}$, then $G_{24} = {\Bbb{AG}(\mathcal{L})}-\{I_{ik}, I_{i\ell}, I_{im}, (I_i, I_j)\}$ contains $K_{5, 3}$ with partite sets $X = U_i\cup U_j\cup U_{ij}$ and $Y = U_k\cup U_\ell \cup U_{m}$. Notice that any $N_2$-embedding of $K_{5, 3}$ has one hexagonal and six rectangular faces. Also, in ${\Bbb{AG}(\mathcal{L})}$, $I_{ik}$ is adjacent to $I_j, I_\ell, I_{m}$; $I_{i\ell}$ is adjacent to $I_j, I_k, I_{m}$ and $I_{im}$ is adjacent to $I_j, I_k, I_{\ell}$. So to embed the vertices $I_{ik}, I_{i\ell}, I_{im}$ in a $N_2$-embedding of $K_{5, 3}$, it requires either one hexagon with a rectangular face or three rectangular faces which contains $I_j$. If $I_{ik}, I_{i\ell}, I_{im}$ are embedded in three rectangular faces, then since $deg_{K_{5, 3}}(I_j) = 3$, no other face contains $I_j$ so the edge $(I_i, I_{j})$ cannot be drawn without crossing. If not, two vertices must be inserted in the hexagonal face and the other vertex should be inserted in a rectangle face. In such cases, the third face which contains $I_j$ does not exist because each edge occurs in exactly two faces. So ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Therefore one of the sets $U_{ik}$ or $U_{i\ell}$ or $U_{im}$ must be empty for $k, \ell, m\notin \{i, j\}$. A slight modification of the proof would show that one of the sets $U_{jk}$ or $U_{j\ell}$ or $U_{jm}$ must be empty for $k, \ell, m\notin \{i, j\}$.

Case 4.1.1. Suppose $|U_{ij}| = 4$. If $|U_{pq}|\geq 2$ for $p\in \{i, j\}$ and $q\notin \{i, j\},$ then the subgraph ${\Bbb{AG}(\mathcal{L})}-\{I_{pq}, I_{pq}^\prime, (I_i, I_j)\}$ has a similar structure to the graph $G_{16}$ (refer Case 5.1 [3, Theorem 5.2]) so that ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} \geq 3$. Therefore, at most two sets from $\{U_{ik}, U_{i\ell}, U_{im}\}$ and two sets from $\{U_{jk}, U_{j\ell}, U_{jm}\}$, where $k, \ell, m\notin \{i, j\}$, may have an element.

Further, if $U_{pq}\neq \emptyset$ for $p\in \{i, j\}$ and $q\notin \{i, j\}$, then we claim that the set $U_{p^\prime q^\prime} \cup U_{(pq)^c} = \emptyset$, where $p^\prime\in \{i, j\}{\setminus} \{p\}$ and $q\neq q^\prime\notin \{i, j\}$. Suppose not, $I\in U_{p^\prime q^\prime} \cup U_{(pq)^c}$, then the graph ${\Bbb{AG}(\mathcal{L})}-\{[I_{pq}, I]\}$ contains $K_{6, 3}$ with partite sets $X = U_i\cup U_j\cup U_{ij}$ and $Y = \underset{k\ne i, j}{\mathop{\cup }}\, U_k$. Since the merged vertex $[I_{pq}, I]$ is adjacent to all the five vertices of ${\mathop{\cup }}\, _{n = 1}^{5} U_n$, it requires a face of length at least five in an $N_2$-embedding of $K_{6, 3}$. A contradiction to the fact that every face in any $N_2$-embedding of $K_{6, 3}$ is a rectangle.

Thus, in the case of $|U_{ij}| = 4$, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ provided $\underset{\ell, m\notin\{i, j\}}{\mathop{\cup }}\, U_{\ell m} \cup U_{(ij)^c} = \emptyset$, $|\underset{p\in\{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}|\leq 2$, in which $|U_{pq}|\leq 1$ and $U_{p^\prime q^\prime}, U_{(pq)^c} = \emptyset$ when $|U_{pq}| = 1$, where $p^\prime, q^\prime\notin \{p, q\}$.

Case 4.1.2. Suppose $|U_{ij}| = 3$. In every part of the case, let $k, \ell, m\in \{1, \ldots, 5\}{\setminus} \{i, j\}$. If $|U_{pq}| = 3$ for $p\in \{i, j\}$ and $q\in \{k, \ell, m\}$, then the graph ${\Bbb{AG}(\mathcal{L})}- \{I_{pq}, I_{pq}^\prime, I_{pq}^{\prime\prime}, (I_{p^{\prime}}, I_{q^{\prime}}), (I_{p^{\prime}}, I_{q^{\prime \prime}})\}$, where $p^\prime\in \{i, j\}{\setminus} \{p\}$ and distinct $q^\prime, q^{\prime \prime} \in \{k, \ell, m\}{\setminus} \{q\}$, contains a similar structure of the graph $G_{15}^\prime$ (refer Case 5.2 [3, Theorem 5.2]) so that ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} \geq 3$. Also, if $|U_{pq}| = |U_{p^\prime q^\prime}| = 2$ for distinct $p, p^\prime\in \{i, j\}$ and distinct $q, q^\prime\in \{k, l, m\}$, then ${\Bbb{AG}(\mathcal{L})}-\{I_{pq}, I_{pq}^\prime, I_{p^\prime q^\prime}, I_{p^\prime q^\prime}^\prime, (I_p, I_{p^\prime})\}$ contains a similar structure of the graph $G_{15}$ (refer to Case 4.2.2 [3, Theorem 5.2]) so that ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} \geq 3$. Therefore, at most one set from the collection $\{U_{ik}, U_{i\ell}, U_{im}, U_{jk}, U_{j\ell}, U_{jm}\}$ has two elements.

(ⅰ) Without loss of generality, let us take $|U_{ik}| = 2$. Then, by Case 4.1, we have $|U_{i\ell}\cup U_{im}|\leq 1$ and at most two sets from the collection $\{U_{jk}, U_{j\ell}, U_{jm}\}$ have an element. But, our next claims are:

Claim A: $|U_{i\ell}\cup U_{im}\cup U_{jk}|\leq 1$ and $U_{j\ell}\cup U_{jm} \cup U_{(ik)^c} = \emptyset$.

Suppose $|U_{i\ell}\cup U_{im}\cup U_{jk}| = 2$. Since $|U_{i\ell}\cup U_{im}|\leq 1$ and $|U_{jk}|\leq 1,$ we choose $I\in U_{i\ell}\cup U_{im}$ and $J\in U_{jk}$. Now the graph ${\Bbb{AG}(\mathcal{L})}-\{I_{ik}, I_{ik}^\prime, (I_i, I_j), (I_j, [I, J])\}$ contains $K_{6, 3}$ with partite sets $X = U_i\cup U_j\cup U_{ij}\cup \{[I, J]\}$ and $Y = U_k\cup U_\ell \cup U_{m}$. Note that all the faces in any $N_2$-embedding of $K_{6, 3}$ are rectangular. Since $I_{ik}$ and $I_{ik}^\prime$ are adjacent to $I_j, I_{\ell}$ and $I_m$, to embed the vertices $I_{ik}$ and $I_{ik}^\prime$, it requires two distinct rectangular faces that should have the vertex $I_j$ in its boundary. Since $deg_{K_{6, 3}}(I_j) = 3$, there is exactly one more rectangular face containing $I_j$, in which the two edges $(I_j, I_i)$ and $(I_j, [I, J])$ cannot be embedded so that ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$.

Suppose $I\in U_{j\ell}\cup U_{jm} \cup U_{(ik)^c}$. Then, $K_{5, 3}$ is a subgraph of the graph ${\Bbb{AG}(\mathcal{L})}-\{I_{ik}, I_{ik}^\prime, I\}$ with partite sets $X = U_i\cup U_j\cup U_{ij}$ and $Y = U_k\cup U_\ell \cup U_{m}$. Note that $\tilde{\gamma}(K_{5, 3}) = 2$. Since $I_{ik}-I-I_{ik}^\prime$ is a path in ${\Bbb{AG}(\mathcal{L})}$, these three vertices should be embedded into a single face of a $N_2$-embedding of $K_{5, 3}$. First, embed the path $I_{ik}-I-I_{ik}^\prime$ together with the edges $(I_{ik}, I_j), (I_{ik}, I_\ell), (I_{ik}, I_m), (I_{ik}^\prime, I_j), (I_{ik}^\prime, I_\ell)$ and $(I_{ik}^\prime, I_m)$ into a face. Then, clearly the middle vertex $I$ of the path cannot adopt any edge incident with $I$, so the edges $(I, I_i)$ and $(I, I_k)$ cannot be embedded. Therefore, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Thus, the claim holds true.

Claim B: If $|U_{i\ell}\cup U_{im}\cup U_{jk}| = 1,$ say $|U_{i\ell}| = 1$, then $U_{(i\ell)^c} = \emptyset$.

If $|U_{i\ell}| = 1$ with $U_{(i\ell)^c}\neq \emptyset,$ then just replace $I$ by $I_{i\ell}$ and $J$ by $I_{(i\ell)^c}$ in the proof of the case $|U_{i\ell}\cup U_{im}\cup U_{jk}| = 2$, and we get ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$.

(ⅱ) Let $|U_{ik}| = 1$. Here, our claim is:

Claim C: $|U_{i\ell}\cup U_{im} \cup U_{jk} \cup U_{j\ell}\cup U_{jm} \cup U_{(ik)^c}|\leq 2,$ $|U_{j\ell}\cup U_{jm} \cup U_{(ik)^c}|\leq 1$ and $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}|\leq 1$. Further, if $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}| = 1$, then $U_{j\ell} \cup U_{jm} = \emptyset$ and $|U_{i\ell}\cup U_{im} \cup U_{jk}|\leq 1$.

A slight modification of the proof of Claim B would show that $|U_{j\ell}\cup U_{jm} \cup U_{(ik)^c}|\leq 1$ and $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}|\leq 1$.

Suppose $|U_{i\ell}\cup U_{im} \cup U_{jk} \cup U_{j\ell}\cup U_{jm} \cup U_{(ik)^c}| = 3$. Since $|U_{i\ell}\cup U_{im}|\leq 1$ and $|U_{j\ell}\cup U_{jm} \cup U_{(ik)^c}|\leq 1$, let $I\in U_{i\ell}\cup U_{im}$, $J\in U_{jk}$ and $K\in U_{j\ell}\cup U_{jm} \cup U_{(ik)^c}$. Then the sets $X = U_i\cup U_j\cup U_{ij}\cup \{[I, J]\}\cup \{K, I_{ik}\}$ and $Y = U_k\cup U_\ell\cup U_{m}$ form $K_{7, 3}$.

Let $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}| = 1$. If $U_{(ik)^c}\neq \emptyset$, then clearly $U_{j\ell} \cup U_{jm} = \emptyset$ and $|U_{i\ell}\cup U_{im} \cup U_{jk}|\leq 1$. If $U_{pq}, U_{(pq)^c}\neq \emptyset$ for $pq\neq ik$, then we have to show that $pq\neq j\ell, jm$. If not, then $X = U_i\cup U_j\cup U_{ij}$ and $Y = U_k\cup U_\ell\cup U_{m}$ form $K_{5, 3}$ in the subgraph ${\Bbb{AG}(\mathcal{L})}-\{I_{ik}, [I_{pq}, I_{(pq)^c}]\}$. Note that $\tilde{\gamma}(K_{5, 3}) = 2$. Since the vertex $I_{ik}$ is adjacent to the vertex $[I_{pq}, I_{(pq)^c}]$, the vertices $I_{ik}$ and $[I_{pq}, I_{(pq)^c}]$ have to be embedded in a single face of any $N_2$-embedding of $K_{5, 3}$. Here, the vertex $[I_{pq}, I_{(pq)^c}]$ is adjacent to all of the five vertices of ${\mathop{\cup }}\, _{n = 1}^{5} U_n$ and the vertex $I_{ik}$ is adjacent to exactly three vertices of ${\mathop{\cup }}\, _{n = 1}^{5} U_n$. Clearly, a single face cannot embed such eight edges onto it, a contradiction. Therefore, the claim holds true.

Thus, in the case of $|U_{ij}| = 3$, we have ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ provided $\underset{\ell, m\notin\{i, j\}}{\mathop{\cup }}\, U_{\ell m} \cup U_{(ij)^c} = \emptyset$, $|\underset{p\in \{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}|\leq 3$, where the choice $i$ or $j$ for $p$ is placed at most two times in the union, in which at most one of the sets $U_{pq}$ has two elements and $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}|\leq 1$. Further, if $|U_{pq}| = 2$ for some $p\in \{i, j\}; q\notin \{i, j\}$, then $\underset{p^\prime, q^\prime\notin \{p, q\}} {\mathop{\cup }}\, U_{p^\prime q^\prime} \cup \underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c} = \emptyset$, and if $|\underset{p\in \{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}| = 3$ with $|U_{pq}|\leq 1$, then the three choices for $q$ are not distinct. Moreover, if $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}| = 1$, then $|\underset{p\in \{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}|\leq 2$ with the choice for two pairs of $p, q$'s are not mutually disjoint.

Case 4.2. Assume $|U_{ij}| = 2$. If $|\underset{\ell, m\notin\{i, j\}}{\mathop{\cup }}\, U_{\ell m} \cup U_{(ij)^c}|\geq 2$, then the sets $X = U_i\cup U_j\cup U_{ij}$ and $Y = V({\Bbb{AG}(\mathcal{L})}){\setminus} X$ form $K_{4, 5}$ in ${\Bbb{AG}(\mathcal{L})}$, a contradiction.

Further, if four non-empty sets exist other than $U_{ij}$ in such a way that two sets of the form $U_{ik}$ and two other sets of the form $U_{jk}$ for $k\neq i, j$ are non-empty, say $U_{i\ell}, U_{im}\neq \emptyset$ and $U_{j \ell}, U_{j n}\neq \emptyset$ for distinct $\ell, m, n\notin \{i, j\}$. Then, the sets $X = \{I_i, I_j, I_{ij}, [I_{i\ell}, I_{jn}]\}$ and $Y = \{I_\ell, I_m, I_{n}, [I_{im}, I_{j\ell}]\}$ form $(H_{3}\cup (u_2, u_3))-(u_2, v_1)$. Therefore, by [3, Lemma 3.6], we have ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$.

Case 4.2.1. Suppose $|\underset{\ell, m\notin\{i, j\}}{\mathop{\cup }}\, U_{\ell m} \cup U_{(ij)^c}| = 1$ and let $I\in \underset{\ell, m\notin\{i, j\}}{\mathop{\cup }}\, U_{\ell m} \cup U_{(ij)^c}$. If $|U_{pq}|\geq 2$ for some $p\in \{i, j\}$ and $q\notin \{i, j\}$, then clearly ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Let $|U_{pq}| = 1$ for some $p\in \{i, j\}$ and $q\notin \{i, j\}$. If $J \in U_{p^\prime q^\prime} \cup U_{p^\prime q^{\prime\prime}}$ for $p^\prime\in \{i, j\}{\setminus} \{p\}$ and $q^\prime, q^{\prime\prime} \notin \{i, j, q\}$, then the sets $X = \{I_i, I_j, I_{ij}, I_{ij}^\prime, [I_{pq}, J]\}$ and $Y = \{I_q, I_{q^\prime}, I_{q^{\prime\prime}}, I\}$ form $K_{5, 4}-e$ in ${\Bbb{AG}(\mathcal{L})}$ so that ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Further, similar verification gives us ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$ whenever $U_{(pq)^c}\neq\emptyset$ or there exist $p\in \{i, j\}$ such that $U_{pq}\neq \emptyset$ for all $q\notin \{i, j\}$.

Therefore, in the case of $|U_{ij}| = 2$ with $|\underset{\ell, m\notin\{i, j\}}{\mathop{\cup }}\, U_{\ell m} \cup U_{(ij)^c}| = 1$, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ provided $|\underset{p\in\{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}|\leq 2$ with $|U_{pq}|\leq 1$ and $U_{p^\prime q^\prime}, U_{(pq)^c} = \emptyset$ when $|U_{pq}| = 1$ where $p^\prime, q^\prime\notin \{p, q\}$.

Case 4.2.2. Suppose $\underset{\ell, m\notin\{i, j\}}{\mathop{\cup }}\, U_{\ell m} \cup U_{(ij)^c} = \emptyset$. If there exist $p, p_1\in \{i, j\}$ and $q, q_1\notin \{i, j\}$ with $pq\neq p_1q_1$ such that $|U_{pq}|, |U_{p_1 q_1}| = 2$, then ${\Bbb{AG}(\mathcal{L})}$ contains a structure of $K_{3, 8}-4e$ and it can be verified that ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} \geq 3$. So assume $|U_{pq}| = 2$ for unique $p\in \{i, j\}$ and $q\notin \{i, j\}$. If $|U_{p^\prime q^\prime}\cup U_{p^\prime q^{\prime\prime}}| > 1$ for $p^\prime\in \{i, j\}{\setminus} \{p\}$ and $q^{\prime}, q^{\prime\prime}\in \{1, \ldots, 5\}{\setminus} \{i, j, q\}$, then the sets $X = U_p \cup U_q \cup U_{pq}$ and $Y = U_{p^\prime} \cup U_{q^{\prime}}\cup U_{q^{\prime\prime}} \cup U_{p^\prime q^\prime}\cup U_{p^\prime q^{\prime\prime}}$ form $K_{4, 5}$ in ${\Bbb{AG}(\mathcal{L})}$, a contradiction. If $|U_{p^\prime q^\prime}\cup U_{p^\prime q^{\prime\prime}}| = 1$, then $|U_{pq^\prime}\cup U_{p q^{\prime\prime}}\cup U_{p^\prime q}|\leq 1$ with $U_{pq^\prime} = \emptyset$ if $|U_{p^\prime q^{\prime\prime}}| = 1$ and $U_{pq^{\prime\prime}} = \emptyset$ if $|U_{p^\prime q^\prime}| = 1$ because of the facts that no two sets of the form $U_{ik}$ and no two sets of the form $U_{jk}$ are non-empty, and the edges $(I_{p^\prime q^\prime}, I_{p q^{\prime\prime}})$ and $(I_{p^\prime q^{\prime\prime}}, I_{p q^{\prime}})$ are in ${\Bbb{AG}(\mathcal{L})}$. Similarly, if $U_{p^\prime q^\prime}, U_{p^\prime q^{\prime\prime}} = \emptyset$, then $|U_{pq^\prime}\cup U_{p q^{\prime\prime}}|\leq 1$ and $|U_{p^\prime q}|\leq 1$. Moreover, $U_{(pq)^c} = \emptyset$ whenever $U_{pq}\neq \emptyset$.

Finally, assume $|U_{pq}|\leq 1$ for all $p\in \{i, j\}$ and $q\notin \{i, j\}$. Since no two sets from $U_{ik}$ and no two sets from $U_{jk}$ are non-empty, $|\underset{p\in \{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}|\leq 4$, where the choice $i$ or $j$ for $p$ is placed at most once in the union. If the subgraph induced by $\langle \underset{p\in \{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq} \rangle$ in ${\Bbb{AG}(\mathcal{L})}$ has an edge, then $U_{(pq)^c} = \emptyset$ for all $U_{pq}\neq \emptyset$. If not, $|\underset{p\in \{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{(pq)^c}|\leq 1$, where the union is taken over all non-empty $U_{pq}$. Also, by Theorem 2.1, ${\Bbb{AG}(\mathcal{L})}$ is projective whenever $|\underset{p\in \{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}|\leq 2$ with $|U_{pq}|\leq1$ and $\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c} = \emptyset$. Further, if two non-empty sets $U_{p_1q_1}$ and $U_{p_2q_2}$ exists in the collection $\{U_{pq}:p\in \{i, j\}; q\notin \{i, j\}\}$, then $\{p_1, q_1\}\cap \{p_2, q_2\}\neq \emptyset$.

Therefore, in the case of $|U_{ij}| = 2$ with $\underset{\ell, m\notin\{i, j\}}{\mathop{\cup }}\, U_{\ell m} \cup U_{(ij)^c} = \emptyset$, we have ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ provided $2\leq |\underset{p\in \{i, j\}; q\notin \{i, j\}}{\mathop{\cup }}\, U_{pq}|\leq 4$ in which at most one of the sets $U_{pq}$ has two elements, where the choice $i$ or $j$ for $p$ is placed at most once in the union. Moreover, one of the following is satisfied:

(a) If $|U_{p_fq_f}| = 2$ for some $p_f\in \{i, j\}, q_f\notin \{i, j\}$, then $\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c} = \emptyset$. Further, at most one of the sets $U_{p_gq_g}$ is non-empty with the property that $\{p_f, q_f\}\cap \{p_g, q_g\} = \emptyset$.

(b) If $|U_{pq}|\leq 1$ for all $p\in \{i, j\}, q\notin \{i, j\}$, then $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}|\leq 1$. Further, if $|U_{(pq)^c}| = 1$ for some $|U_{pq}| = 1$, then every non-empty set $U_{p_f, q_f}$ should have the property that $\{p_f, q_f\}\cap \{p, q\}\neq \emptyset$.

Case 4.3. Suppose $|U_{ij}| = 1$. If $|\underset{1\leq p\neq q\leq 5}{\mathop{\cup }}\, U_{pq}|\geq 6$, then $K_8-3e$ is a minor of ${\Bbb{AG}(\mathcal{L})}$ and so by Proposition 1.1, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Therefore, $|\underset{1\leq p\neq q\leq 5}{\mathop{\cup }}\, U_{pq}|\leq 5$.

(ⅰ) Let $|\underset{1\leq p\neq q\leq 5}{\mathop{\cup }}\, U_{pq}|\in \{4, 5\}$. Suppose non-empty sets $U_{p_1q_1}, U_{p_2q_2}, U_{p_3q_3}, U_{p_4q_4}$ exists which satisfies the conditions $\{p_{1}, q_{1}\}\cap \{p_{2}, q_{2}\} = \emptyset$ and $\{p_{3}, q_{3}\}\cap \{p_{4}, q_{4}\} = \emptyset$. If $\{p_1, q_1\}\cap \{p_3, q_3\} = \emptyset$ or $\{p_2, q_2\}\cap \{p_4, q_4\} = \emptyset$, then the vertex subset $\{{\mathop{\cup }}\, _{n = 1}^{5} U_n \cup [I_{p_1q_1}, I_{p_2q_2}] \cup [I_{p_3q_3}, I_{p_4q_4}]\}$ form $K_7$. So, let us take $p_1\in \{p_1, q_1\}\cap \{p_3, q_3\}$ and $p_2\in \{p_2, q_2\}\cap \{p_4, q_4\}$. Then, the sets $X = U_{p_1}\cup U_{q_1}\cup U_t\cup U_{p_1q_1}\cup U_{p_3q_3}$ and $Y = U_{p_2}\cup U_{q_2}\cup U_{p_2q_2}\cup U_{p_4q_4}$, where $t\in\{1, \ldots, 5\}{\setminus} \{p_1, p_2, q_1, q_2\}$, form $K_{5, 4}-4e$ in ${\Bbb{AG}(\mathcal{L})}$. Note that $\tilde{\gamma}(K_{4, 5}-4e) = 2$. Now it is not hard to demonstrate that the edges $(I_{p_1}, I_{q_1}), (I_{q_1}, I_{p_3q_3}), (I_{p_1}, I_t), (I_{q_1}, I_t), (I_{p_1q_1}, I_t), (I_{p_3q_3}, I_t), (I_{p_2}, I_{q_2})$ and $(I_{q_2}, I_{p_4q_4})$ of ${\Bbb{AG}(\mathcal{L})}$ cannot be embedded into any $N_2$-embedding of $K_{4, 5}-4e$.

Also, notice that the set $\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c} = \emptyset$ when $|\underset{1\leq p\neq q\leq 5} {\mathop{\cup }}\, U_{pq}| = 5$. Otherwise, either $K_7$ or a graph similar to the structure of the graph $G_{24}$, given in Case 4.1 of Theorem 2.2, will be a subgraph of ${\Bbb{AG}(\mathcal{L})}$.

Claim A: $|\underset{U_{pq\neq \emptyset}}{\mathop{\cup }}\, U_{(pq)^c}|\leq 1$ when $|\underset{1\leq p\neq q\leq 5} {\mathop{\cup }}\, U_{pq}| = 4$. Moreover, if $\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c} = \emptyset$, then the subgraph induced by the set $\underset{1\leq p\neq q\leq 5} {\mathop{\cup }}\, U_{pq}$ has more than one edge and if $|\underset{U_{pq}\neq \emptyset} {\mathop{\cup }}\, U_{(pq)^c}| = 1$, say $|U_{(rs)^c}| = 1$, then the vertex in $U_{rs}$ is adjacent to at most two vertices of $\underset{U_{pq}\neq \emptyset; pq\neq rs} {\mathop{\cup }}\, U_{pq}$. Further, if there is an adjacency between the vertex of $U_{rs}$ and a vertex of $\underset{U_{pq}\neq \emptyset; pq\neq rs} {\mathop{\cup }}\, U_{pq}$, then the subgraph induced by the set $\underset{U_{pq}\neq \emptyset; pq\neq rs} {\mathop{\cup }}\, U_{pq}$ is an empty graph.

Assume that $|\underset{1\leq p\neq q\leq 5} {\mathop{\cup }}\, U_{pq}| = 4$. Since $|U_{pq}|\leq 1$ for all $1\leq p\neq q\leq 5,$ let us take $U_{p_1q_1}, U_{p_2q_2}, U_{p_3q_3}, U_{p_4q_4}\neq \emptyset$.

Let $|\underset{U_{pq\neq \emptyset}}{\mathop{\cup }}\, U_{(pq)^c}|\geq 2$. If $|U_{(p_1q_1)^c}|\geq 2$, then the graph ${\Bbb{AG}(\mathcal{L})}-\{\underset{pq\neq p_1q_1}{\mathop{\cup }}\, U_{pq}, (p_1, q_1)\}$ is similar to the graph $G_{24}$ (Case 4.1 of Theorem 2.2) with partite sets $X = U_{p_1} \cup U_{q_1}\cup U_{p_1q_1}$ and $Y = \underset{r\neq p_1, q_1}{\mathop{\cup }}\, U_r\cup U_{(p_1q_1)^c}$ so that ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Suppose $|U_{(p_1q_1)^c}|, |U_{(p_2q_2)^c}|\geq 1$. If $\{p_3, q_3\}\cap \{p_4, q_4\} = \emptyset$, then the graph induced by the set $\{{\mathop{\cup }}\, _{n = 1}^{5} U_n \cup [I_{p_1q_1}, I_{(p_1q_1)^c}]\cup [I_{p_2q_2}, I_{(p_2q_2)^c}]\cup [I_{p_3q_3}, I_{p_4q_4}]\}$ contains $K_8-3e$ so that ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Also, if $\{p_3, q_3\}\cap \{p_4, q_4\}\neq \emptyset$, say $p_3\in \{p_3, q_3\}\cap \{p_4, q_4\}$, then the sets $X = U_{p_3}\cup U_{p_3q_3}\cup U_{p_4q_4}\cup \{[I_{p_1q_1}, I_{(p_1q_1)^c}]\}\cup \{[I_{p_2q_2}, I_{(p_2q_2)^c}]\}$ and $Y = \underset{r\neq p_3}{\mathop{\cup }}\, U_r$ contains $K_{5, 4}-2e$ in ${\Bbb{AG}(\mathcal{L})}$. Note that $\tilde\gamma{(K_{5, 4}-2e)} = 2$ and every face in any $N_2$-embedding of $K_{5, 4}-2e$ is rectangular. Since $\langle Y\rangle = K_4$ and $K_4$ cannot be embedded in $N_2$ along with rectangular embedding, we have ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$.

Let $|U_{(p_1q_1)^c}| = 1$. If $I_{p_1q_1}$ is adjacent to all vertices of $U_{p_2q_2}\cup U_{p_3q_3}\cup U_{p_4q_4}$, then the sets $X = U_{p_1}\cup U_{q_1}\cup U_{p_1q_1}$ and $Y = \underset{r\neq p_1, q_1}{\mathop{\cup }}\, U_{r}\cup U_{p_2q_2}\cup U_{p_3q_3}\cup U_{p_4q_4} \cup U_{(p_1q_1)^c}$ form $K_{3, 7}$ in ${\Bbb{AG}(\mathcal{L})}$. Suppose $I_{p_1q_1}$ is adjacent to $I_{p_2q_2}$. As mentioned earlier, $I_{p_3q_3}$ is not adjacent to $I_{p_4q_4}$. If $I_{p_2q_2}$ is adjacent to either $I_{p_3q_3}$ or $I_{p_4q_4}$, say $I_{p_3q_3}$, then the set $\{{\mathop{\cup }}\, _{n = 1}^5 U_n \cup [I_{p_1q_1}, I_{(p_1q_1)^c}]\cup [I_{p_2q_2}, I_{p_3q_3}]\}$ form $K_7$. Therefore, $E\left(\langle U_{p_2q_2}\cup U_{p_3q_3}\cup U_{p_4q_4} \rangle\right) = \emptyset$. Thus, the claim holds true.

Claim B: $|\underset{U_{pq\neq \emptyset}}{\mathop{\cup }}\, U_{(pq)^c}|\leq 3$ when $|\underset{1\leq p\neq q\leq 5} {\mathop{\cup }}\, U_{pq}|\leq 3$.

Assume that $|\underset{1\leq p\neq q\leq 5} {\mathop{\cup }}\, U_{pq}|\leq 3$. If $|U_{(pq)^c}|\geq 4$ for $1\leq p\neq q\leq 5$, then ${\Bbb{AG}(\mathcal{L})}$ contains $K_{3, 7}$ with partite sets $X = U_p \cup U_q \cup U_{pq}$ and $Y = \underset{k\neq p, q}{\mathop{\cup }}\, U_k \cup U_{(pq)^c}$. If three sets $U_{(p_1q_1)^c}, U_{(p_2q_2)^c}, U_{(p_3q_3)^c}$ are non-empty, then the three merged vertices $[I_{p_fq_f}, I_{(p_fq_f)^c}]$, for all $1\leq f\leq 3$, together with the vertices in ${\mathop{\cup }}\, _{n = 1}^{5} U_n$ form $K_8-3e$. If $|U_{(p_1q_1)^c}|, |U_{(p_2q_2)^c}|\geq 2$, then the sets $X = U_{p_1}\cup U_{q_1}\cup U_{p_1q_1}$ and $Y = \underset{r\neq p_1, q_1}{\mathop{\cup }}\, U_{r}\cup U_{(p_1q_1)^c}$ form $K_{3, 5}$ which has crosscap 2. Clearly, the path $I_{(p_2q_2)^c}-I_{p_2q_2}-I^\prime_{(p_2q_2)^c}$ together with its edges, could not be embedded into any $N_2$-embedding of $K_{3, 5}$.

Claim C: [C1] If $|\underset{U_{pq\neq \emptyset}}{\mathop{\cup }}\, U_{(pq)^c}| = 3,$ then $|U_{(pq)^c}| = 3$ for $1\leq p\neq q\leq 5$ and no non-empty set $U_{rs}$ exists with $\{r, s\}\cap \{p, q\} = \emptyset$.

[C2] Let $|\underset{U_{pq\neq \emptyset}}{\mathop{\cup }}\, U_{(pq)^c}|\in \{2, 1\}$. If there exists a unique set $U_{(pq)^c}\neq \emptyset$ for $1\leq p\neq q\leq 5$, then at most one non-empty set $U_{rs}$ exist with $\{r, s\}\cap \{p, q\} = \emptyset$. If two sets $U_{(p_1q_1)^c}, U_{(p_2q_2)^c}\neq \emptyset$ for $1\leq p_1, q_1, p_2, q_2\leq 5$ and $\{p_1, q_1\}\cap \{p_2, q_2\}\neq \emptyset$, then no non-empty set $U_{rs}$ exist with $\{r, s\}\cap \{p_f, q_f\} = \emptyset$ for $1\leq f\leq 2$.

The proofs of claims [C1] and [C2] are merely verifications that can be done by the reader.

Now, to determine conditions for ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ (given in the statement), we have to eliminate projective conditions of ${\Bbb{AG}(\mathcal{L})}$ which was given in Theorem 2.1. □

Example 2.1. As an illustration, we consider the case (ⅲ)[c] in Theorem 2.2. Let $|U_1| = 2, |U_i| = 1$ for $i = 2, 3, 4, 5$, and $|U_{14}| = |U_{23}| = 1$. If $|U_{12}| = |U_{13}| = 1$, then the corresponding $6$-partite graph, given in Figure 4(a), is a crosscap two graph. Also, if $|U_{2345}| = 1$, then the crosscap of corresponding $6$-partite graph, given in Figure 4(b), is not equal to two. Moreover, the $6$-partite graph $G$ in Figure 4(b) is minimal with respect to $\tilde{\gamma}(G)\neq 2$.

3.   The case when $|A(\mathcal{L})|=6$

Finally we look into the lattice with 6 atoms. We close the paper by presenting its statement and proof.

Theorem 3.1. Let $\mathcal{L}$ be a lattice with $|A(\mathcal{L})| = 6$. Then, $\tilde{\gamma}(\Bbb{AG}(\mathcal{L})) = 2$ if and only if one of the following conditions hold:

$({\rm{i}})$ $|{\mathop{\cup }}\, _{n = 1}^6U_n| = 7$, there is $U_i$ such that $|U_i| = 2$ with $\underset{1\leq p\neq q\leq 6} {\mathop{\cup }}\, U_{pq} = \emptyset$, $\underset{j, k, \ell, m, n\neq i}{\mathop{\cup }}\, (U_{jk\ell} \cup U_{jk\ell m} \cup U_{jk\ell mn}) = \emptyset$ and at most three sets $U_{ipq}$'s has exactly one element.

$({\rm{ii}})$ $|{\mathop{\cup }}\, _{n = 1}^6U_n| = 6$, $|\underset{1\leq i\neq j\leq 6}{\mathop{\cup }}\, U_{ij}|\leq 2$, $\underset{U_{ij}, U_{ijk}\neq \emptyset}{\mathop{\cup }}\, \left(U_{(ij)^c} \cup U_{(ijk)^c} \right) = \emptyset$ and one of the following cases is satisfied:

[a] In case of $\underset{1\leq i\neq j\leq 6} {\mathop{\cup }}\, U_{ij} = \emptyset$:

[a1] If there is a unique $|U_{i jk}| = 3$ for $1\leq i\neq j\neq k \leq 6$ or $|U_{ijk}| = |U_{\ell mn}| = 2$ for some $ijk\ne \ell mn$ and $1\leq i, j, k, \ell, m, n \leq 6$, then there exist at most eight distinct non-empty $U_{pqr}$'s (including $U_{ijk}, U_{\ell mn}$) in which at most two distinct $U_{pqr}$'s are non-empty such that the intersection of all the sets at their indices has exactly two elements, where $1\leq p\neq q\neq r\leq 6$.

[a2] If there is a unique $|U_{i jk}| = 2$ for $1\leq i\neq j\neq k\leq 6$, then there exist at most nine distinct non-empty $U_{pqr}$'s (including $U_{ijk}$) in which at most three distinct $U_{pqr}$'s are non-empty such that the intersection of all the sets at their indices has exactly two elements, where $1\leq p\neq q\neq r\leq 6$.

[a3] If $|U_{ijk}|\leq 1$ for all $1\leq i\neq j\neq k\leq 6$, then there exist at most ten distinct non-empty $U_{ijk}$'s in which exactly three distinct $U_{ijk}$'s are non-empty such that the intersection of all the sets at their indices has exactly two elements, where $1\leq i\neq j\neq k\leq 6$.

[b] If $|U_{ij}| = 1$ for some unique $1\leq i\neq j\leq 6$, then $\underset{\{\ell, m, n\}\cap \{i, j\} = \emptyset}{\mathop{\cup }}\, U_{\ell mn} = \emptyset$, and $\left| \underset{\{\ell, m, n\}\cap \{i, j\}\neq \emptyset}{\mathop{\cup }}\, U_{\ell mn} \right| \leq 6$ with at most two distinct non-empty $U_{\ell mn}$'s in which at most one set $U_{\ell mn}$ has two elements such that the intersection of all the sets at their indices has exactly two elements.

[c] If $|U_{ij}| = 2$ for some unique $1\leq i\neq j\leq 6$, then $\underset{\{\ell, m, n\}\cap \{i, j\} = \emptyset}{\mathop{\cup }}\, U_{\ell mn} = \underset{1\leq k\leq 6}{\mathop{\cup }}\, U_{ijk} = \emptyset$, and $\left| \underset{|\{\ell, m, n\}\cap \{i, j\}| = 1}{\mathop{\cup }}\, U_{\ell mn} \right| \leq 4$ with $|U_{\ell mn}|\leq 1$ in which at most two distinct $U_{\ell mn}$'s are non-empty such that the intersection of all the sets at their indices has exactly two elements.

[d] If $|U_{ij}| = |U_{ik}| = 1$ for some $1\leq i\neq j\neq k\leq 6$, then $\underset{\{i\} {\text or } \{j, k\}\not\subseteq \{\ell, m, n\}\cap \{i, j, k\}} {\mathop{\cup }}\, U_{\ell mn} = \emptyset$, and $\left| \underset{\{i\} {\text or } \{j, k\}\subseteq \{\ell, m, n\}\cap \{i, j, k\}}{\mathop{\cup }}\, U_{\ell mn} \right|\leq 5$ with $|U_{\ell mn}|\leq 1$ in which at most two distinct $U_{\ell mn}$'s are non-empty such that the intersection of all the sets at their indices has exactly two elements.

Proof. Suppose that $|{\mathop{\cup }}\, _{n = 1}^6U_n|\geq 8$. Then ${\Bbb{AG}(\mathcal{L})}$ contains $K_{2, 2, 2, 2}$ or $K_{8}-3e$ as a subgraph so that by Proposition 1.1, we have ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Thus, $|{\mathop{\cup }}\, _{n = 1}^6U_n|\leq 7$.

Case 1. Let $|{\mathop{\cup }}\, _{n = 1}^6U_n| = 7$. Then, $|U_1| = 2$ and $|U_2| = \ldots = |U_6| = 1$. Clearly, $K_7-e$ is a subgraph of ${\Bbb{AG}(\mathcal{L})}$. If $I\in \underset{i\neq 1}{\mathop{\cup }}\, (U_{ij} \cup U_{ijk} \cup U_{ijk\ell} \cup U_{ijk\ell m})$, then merge the vertices $I$ and $I_1$ so that $K_7$ is a subgraph of ${\Bbb{AG}(\mathcal{L})}$, a contradiction. If $U_{1j}\neq \emptyset$ for some $j = 2, \ldots, 6$, then $X = U_1\cup U_j\cup U_{1j}$ and $Y = \underset{j^\prime\neq 1, j}{\mathop{\cup }}\, U_{j^\prime}$ form a subgraph $H_3$ (refer [3, Lemma 3.6]) in ${\Bbb{AG}(\mathcal{L})}$ and so ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} \geq 3$. Therefore, $\underset{i, j}{\mathop{\cup }}\, U_{ij} = \emptyset$.

If $|U_{1jk}|\geq 2$ for some $2\leq j, k\leq 6$, then ${\Bbb{AG}(\mathcal{L})}$ contains $K_{3, 6}\cup (K_4-e)$ as a minor, so by Proposition 1.1 we have ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. So, $|U_{1jk}|\leq 1$ for all $2\leq j, k\leq 6$.

Suppose $|{\mathop{\cup }}\, U_{1jk}|\geq 4$. Then, it is not hard to verify that $K_{8, 3}-4e$ is a subgraph of ${\Bbb{AG}(\mathcal{L})}$ with the partition set $X\supset \{{\mathop{\cup }}\, U_{1jk} \cup U_1\}$. Further, assume that a vertex in $X$ is adjacent to at least six vertices of $X$ or, two vertices in $X$ is adjacent to at least five vertices of $X$. Note that any $N_2$-embedding of $K_{8, 3}-4e$ has two hexagonal and seven rectangular faces. Since the maximum degree of vertices of $X$ in $K_{8, 3}-4e$ is 3, at most one vertex of $X$ may adopt 5 distinct edges from $\langle X\rangle$. But, $\langle X\rangle$ contains either a vertex of degree 6 or two vertices of degree 5, a contradiction.

Thus, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ whenever ${\mathop{\cup }}\, _{k\neq 1} (U_{ij}\cup U_{k\ell m} \cup U_{k\ell mn} \cup U_{k\ell mnp}) = \emptyset$ with at most three $U_{1\ell m}$'s having one element.

Case 2. Let $|{\mathop{\cup }}\, _{n = 1}^6U_n| = 6$. Then $|U_1| = \ldots = |U_6| = 1$. If the subgraph induced by $V({\Bbb{AG}(\mathcal{L})}){\setminus} \{{\mathop{\cup }}\, _{n = 1}^6 U_n\}$ has an edge $(I, J)$, then merge the vertices $I$ and $J$ so that the resulting vertex is adjacent to all the vertices of ${\mathop{\cup }}\, _{n = 1}^6 U_n$. Therefore, $K_7$ is a minor of ${\Bbb{AG}(\mathcal{L})}$, a contradiction. Thus, $\langle V({\Bbb{AG}(\mathcal{L})}){\setminus} \{{\mathop{\cup }}\, _{n = 1}^6 U_n\}\rangle$ is an empty graph.

If $|\underset{i, j}{\mathop{\cup }}\, U_{ij}|\geq 3$, then the structure given for $G_{21}$ is a subgraph of ${\Bbb{AG}(\mathcal{L})}$, so that ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Therefore, $|\underset{i, j}{\mathop{\cup }}\, U_{ij}|\leq 2$.

Case 2.1. Assume $U_{ij} = \emptyset$ for all $1\leq i, j\leq 6$. As mentioned earlier, we have $U_{(ijk)^c} = \emptyset$ when $U_{ijk}\neq \emptyset$.

If $|U_{ijk}\cup U_{ij\ell}\cup U_{ijm}\cup U_{ijn}|\geq 4$, then the subgraph ${\Bbb{AG}(\mathcal{L})} -\{ (I_i, I_k), (I_j, I_k), (I_k, I_{ij\ell}), (I_k, I_{ijm}), (I_k, I_{ijn})\}$ contains $K_{3, 7}-3e$ with partite sets $X = U_i \cup U_j \cup U_k \cup U_{ijk}\cup U_{ij\ell}\cup U_{ijm}\cup U_{ijn}$ and $Y = U_\ell \cup U_m \cup U_n$ (take $\ell, m, n\in \{1, \ldots, 6\}{\setminus} \{i, j, k\}$ in case of $\ell, m, n$ does not exist in the union of the assumption). Note that any $N_2$-embedding of $K_{3, 7}-3e$ have two hexagonal and six rectangular faces. The edges $(I_i, I_k), (I_j, I_k), (I_k, I_{ij\ell}), (I_k, I_{ijm})$ can be inserted without crossing in two hexagonal faces. In such a case one cannot find a rectangular face with diagonals $I_k$ and $I_{ijn}$ and so ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$.

Also, if $|U_{ijk}|\geq 4$ for some $1\leq i\neq j\neq k\leq 6$, then the sets $X = U_i \cup U_j \cup U_k \cup U_{ijk}$ and $Y = \underset{\ell\neq i, j, k}{\mathop{\cup }}\, U_\ell$ form $K_{7, 3}$ in ${\Bbb{AG}(\mathcal{L})}$.

(ⅰ) Assume $|U_{ijk}| = 3$ for some $1\leq i\neq j\neq k\leq 6$. If $|U_{\ell mn}|\geq 2$ for some $\ell mn\neq ijk$, then the the subgraph ${\Bbb{AG}(\mathcal{L})} -\{ I_{\ell mn}, I_{\ell mn}^\prime, (I_i, I_j), (I_i, I_k), (I_j, I_k)\}$ is similar to the structure of $G_{16}$ (refer Case 5.1 of [3, Theorem 5.2]) so that ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. If $|U_{ij\ell}| = |U_{ijm}| = 1$, then the graph ${\Bbb{AG}(\mathcal{L})} -\{ I_{ij\ell}, I_{ijm}, (I_k, I_i), (I_k, I_j)\}$ contains $K_{6, 3}$ with partite sets $X = U_i \cup U_j \cup U_k \cup U_{ijk}$ and $Y = U_\ell \cup U_m \cup U_n$. Note that every face in any $N_2$-embedding of $K_{6, 3}$ is rectangular, $I_{ij\ell}$ is adjacent to $I_k, I_m, I_n$ and $I_{ijm}$ is adjacent to $I_k, I_\ell, I_n$. So, to embed the vertices $I_{ij\ell}$ and $I_{ijm}$, it requires two rectangular faces that contains $I_k$. Now the edges $(I_k, I_i)$ and $(I_k, I_j)$ cannot be embedded into a single rectangular face which has $I_k$, a contradiction. Similarly, we get ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$ whenever $|U_{pqr}| = |U_{pqs}| = | U_{pqt}| = 1$ for $1\leq p\neq q\neq r\neq s\neq t\leq 6$ with $pq\neq ij$.

(ⅱ) Assume $|U_{ijk}| = 2$ for some $1\leq i\neq j\neq k\leq 6$. If $|U_{\ell mn}| = |U_{pqr}| = 2$ for $\ell mn, pqr\neq ijk$, then clearly ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. If $|U_{ij\ell}| = 2$ with $U_{ijm}\neq \emptyset$ for $1\leq \ell \neq m \leq 6$, then the graph ${\Bbb{AG}(\mathcal{L})} -\{(I_k, I_i), (I_k, I_j), (I_k, I_{ij\ell}), (I_k, I_{ij\ell}^\prime), (I_k, I_{ijm})\}$ contains $K_{8, 3}-3e$ with partite sets $X = U_i \cup U_j \cup U_k \cup U_{ijk}\cup U_{ij\ell}\cup U_{ijm}$ and $Y = \underset{n\neq i, j, k}{\mathop{\cup }}\, U_n$. Note that $\tilde{\gamma}(K_{8, 3}-3e) = 2$, and any $N_2$-embedding of $K_{8, 3}-3e$ have one hexagonal and nine rectangular faces. Now, to recover a $N_2$-embedding of ${\Bbb{AG}(\mathcal{L})}$ from a $N_2$-embedding of $K_{8, 3}-3e$, we have to embed five edges in which one end is at $I_k$ and the another end at a vertex of $X$. So it is required that $I_k$ has to be part of at least four faces of a $N_2$-embedding of $K_{8, 3}-3e$, a contradiction to $deg_{K_{8, 3}-3e}(I_k) = 3$.

Suppose $|U_{pqr}| = 2$ for $pq\neq ij$. If $|U_{ij\ell}| = |U_{ijm}| = 1$ for $1\leq \ell \neq m \leq 6$, then the subgraph ${\Bbb{AG}(\mathcal{L})} -\{ I_{pqr}, I_{pqr}^\prime, (I_k, I_i), (I_k, I_j), (I_k, I_{ij\ell}), (I_k, I_{ijm}), (I_i, I_j)\}$ contains $K_{3, 7}-2e$ with partite sets $X = U_i \cup U_j \cup U_k \cup U_{ijk}\cup U_{ij\ell}\cup U_{ijm}$ and $Y = U_\ell \cup U_m \cup U_n$. Note that any $N_2$-embedding of $K_{3, 7}-2e$ have one hexagonal and $8$ rectangular faces. Clearly, each rectangular face can adopt at most one new edge and so to embed the edges $(I_k, I_i), (I_k, I_j), (I_k, I_{ij\ell}), (I_k, I_{ijm})$ it requires one hexagonal and two rectangular faces which contains $I_k$. In addition, we have to embed the vertices $I_{pqr}$ and $I_{pqr}^\prime$. Here, the vertices $I_{pqr}$ and $I_{pqr}^\prime$ are adjacent to at least one of $I_i$ or $I_j$ or $I_k$. If it is $I_k$, then clearly the edge $(I_k, I_{pqr})$ cannot be embedded. So, let $I_{pqr}, I_{pqr}^\prime$ be adjacent to $I_i$. But, $I_i$ is used in embedding the first five edges. So, the remaining two rectangular faces are required to embed the edges $(I_i, I_{pqr})$ and $(I_i, I_{pqr}^\prime)$. In such a case, the edge $(I_i, I_j)$ could not be embedded in $N_2$, a contradiction. For all the remaining cases, one can retrieve a $N_2$-embedding of ${\Bbb{AG}(\mathcal{L})}$ from Figure 5.

If $|U_{ijk}|\leq 1$ for all $1\leq i\neq j\neq k\leq 6$, then by eliminating the projective cases, we will get the result as in the statement.

Case 2.2. Assume $1\leq |\underset{i, j}{\mathop{\cup }}\, U_{ij}|\leq 2$. Let $U_{ij}\neq \emptyset$ for some $1\leq i\neq j \leq 6$. Clearly, $U_{(ij)^c} = \emptyset$ and $U_{pqr} = \emptyset$ for all $\{p, q, r\}\cap \{i, j\} = \emptyset$.

Choose $p$ and $q$ such that $\{p, q\}\cap \{i, j\}\neq \emptyset$. If $U_{pqr_1}, U_{pqr_2}, U_{pqr_3}\neq \emptyset$ with $r_1, r_2, r_3\notin \{i, j\}$, then, the subgraph ${\Bbb{AG}(\mathcal{L})}- \{ I_{pqr_2}, I_{pqr_3}, (I_{r_1}, I_p), (I_{r_1}, I_q), (I_{r_1}, I_{ij})\}$ contains $K_{5, 3}$ with partite sets $X = U_p \cup U_q \cup U_{r_1} \cup U_{ij}\cup U_{pqr_1}$ and $Y = U_{r}\cup U_{r_2}\cup U_{r_3}$, where $r \notin \{p, q, r_1, r_2, r_3\}$. Any $N_2$-embedding of $K_{5, 3}$ have one hexagonal and six rectangular faces. Note that $I_{pqr_2}, I_{pqr_3}$ are adjacent to $I_{r_1}$ and two vertices of $Y$. So, to embed the vertices $I_{pqr_2}, I_{pqr_3}$ together with edges $(I_{r_1}, I_p), (I_{r_1}, I_q)$ and $(I_{r_1}, I_{ij})$, it requires a hexagon with three rectangular faces or five rectangular faces which contains $I_{r_1}$, a contradiction to $deg_{K_{5, 3}}(I_{r_1}) = 3$.

(ⅰ) Assume $|U_{ij}| = 1$ for some unique $1\leq i\neq j \leq 6$. Choose $\ell, m$ and $n$ in such a way that $\{\ell, m, n\}\cap \{i, j\}\neq \emptyset$. If $|U_{\ell mn}|\geq 3$, then ${\Bbb{AG}(\mathcal{L})}$ contains $K_{3, 7}$ as a subgraph, a contradiction. Also, if $|U_{\ell mn}| = |U_{\ell_1 m_1 n_1}| = 2$, then the graph ${\Bbb{AG}(\mathcal{L})} -\{I_{\ell_1 m_1 n_1}, I_{\ell_1 m_1 n_1}^\prime, (I_i, I_j) \}$ contains a subgraph similar to the structure of $G_{16}$ (refer Case 5.1 of [3, Theorem 5.2]) and so ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$. Thus, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ whenever $\underset{\{\ell, m, n\}\cap \{i, j\} = \emptyset}{\mathop{\cup }}\, U_{\ell mn} \cup U_{(ij)^c} = \emptyset$ and, $\left| \underset{\{\ell, m, n\}\cap \{i, j\}\neq \emptyset}{\mathop{\cup }}\, U_{\ell mn} \right| \leq 6$ in which at most one set $U_{\ell mn}$ has two elements with at most two distinct sets $U_{\ell mn}$'s non-empty such that the intersection of all the sets at their indices has exactly two elements.

(ⅱ) Assume $|\underset{i, j}{\mathop{\cup }}\, U_{ij}| = 2$.

Suppose $|U_{ij}| = 2$ for $1\leq i\neq j \leq 6$. If $U_{ijk}\neq \emptyset$ for some $1\leq k \leq 6,$ then $K_{6, 3}$ with partite sets $X = U_i \cup U_j\cup U_k \cup U_{ij}\cup U_{ijk}$ and $Y = U_p\cup U_q\cup U_r$, where $p, q, r\in \{1, \ldots, 6\}{\setminus}\{i, j, k\}$. Every face in any $N_2$-embedding of $K_{6, 3}$ is rectangular. So, to embed the edges $(I_k, I_i), (I_k, I_j), (I_k, I_{ij}), (I_k, I_{ij}^\prime)$, it requires four rectangular faces all of which contains $I_k$ but $deg_{K_{3, 6}}(I_k) = 3$, a contradiction. Thus, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ when $\underset{\{\ell, m, n\}\cap \{i, j\} = \emptyset}{\mathop{\cup }}\, U_{\ell mn} \cup \underset{k}{\mathop{\cup }}\, U_{ijk} \cup U_{(ij)^c} = \emptyset$, and $\left| \underset{|\{\ell, m, n\}\cap \{i, j\}| = 1}{\mathop{\cup }}\, U_{\ell mn} \right| \leq 4$ in which $|U_{\ell mn}|\leq 1$ with at most two distinct sets $U_{\ell mn}$'s non-empty such that the intersection of all the sets at their indices has exactly two elements.

Suppose distinct $U_{ij}, U_{pq}\neq \emptyset$ for $1\leq i, j, p, q \leq 6$. Then, $\{i, j\}\cap \{p, q\}\neq \emptyset$. So let $|U_{ij}| = |U_{ik}| = 1$ for some $1\leq i\neq j\neq k\leq 6$. Let us take $\ell, m$ and $n$ such that $\{\ell, m, n\}\cap \{i, j, k\}$ contains either $\{i\}$ or $\{j, k\}$. If $|U_{\ell mn}|\geq 2$, then the subgraph ${\Bbb{AG}(\mathcal{L})}-\{I_{\ell mn}, I_{\ell mn}^\prime, (I_i, I_j), (I_i, I_k), (I_j, I_{ik}), (I_k, I_{ij}) \}$ contains $K_{5, 3}$ with partite sets $X = U_i \cup U_j\cup U_k \cup U_{ij}\cup U_{ik}$ and $Y = U_p\cup U_q\cup U_r$ where $p, q, r\in \{1, \ldots, 6\}{\setminus}\{i, j, k\}$. Note that $I_{\ell mn}, I_{\ell mn}^\prime$ are adjacent to three vertices of $K_{5, 3}$ and so to embed the vertices $I_{\ell mn}, I_{\ell mn}^\prime$ together with edges $(I_i, I_j), (I_i, I_k)$, it requires one hexagonal and one rectangular face. Thereafter one cannot find two rectangular faces with diagonal vertices $I_j, I_{ik}$ and $I_k, I_{ij}$ respectively. Therefore, either $(I_j, I_{ik})$ or $(I_k, I_{ij})$ cannot be embedded without crossing. Therefore, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))}\geq 3$.

Thus, ${\tilde{\gamma}(\Bbb{AG}(\mathcal{L}))} = 2$ when $\underset{\{i\} {\text or } \{j, k\}\not\subseteq \{\ell, m, n\}\cap \{i, j, k\}} {\mathop{\cup }}\, U_{\ell mn}\cup U_{(ij)^c} \cup U_{(ik)^c} = \emptyset$, and $\left| \underset{\{i\} {\text or } \{j, k\}\subseteq \{\ell, m, n\}\cap \{i, j, k\}}{\mathop{\cup }}\, U_{\ell mn} \right|\leq 5$ in which $|U_{\ell mn}|\leq 1$ with at most two distinct sets $U_{\ell mn}$'s non-empty such that the intersection of all the sets at their indices has exactly two elements. □

Example 3.1. As an illustration, we consider the case (ii)[a1] in Theorem 3.1. Let $|U_i| = 1$ for $i = 1, \ldots, 6$, $|U_{123}| = |U_{134}| = 2$ and $|U_{126}| = |U_{145}| = |U_{245}| = 1$. If $|U_{356}| = 1$, then the corresponding crosscap two $8$-partite graph is given in Figure 6. Also, if $|U_{125}| = 1$, then the crosscap of corresponding $8$-partite graph, as in Figure 7, is not equal to two. Moreover, the $8$-partite graph $G$ in Figure 7 is minimal with respect to $\tilde{\gamma}(G)\neq 2$.

4.   Conclusions

Note that no complete set of forbidden subgraphs for the two-crosscap surface (Klein bottle) is known yet. In this regard, an open problem will be to determine a family of graphs that has crosscap number two. This series of papers provides a class of $r$-partite graphs, where $3\leq r\leq 8$, that (1) can be embedded or (2) cannot be embedded in the two-crosscap surface. This was done by using the complete classification of all lattices whose annihilating-ideal graph has crosscap number two.

For the future work, one can determine all forbidden $r$-partite, $r\geq 4$, subgraphs for the crosscap two surface. Also, it would be interesting to classify all lattices whose annihilating-ideal graph can be embedded in the non-orientable surfaces of crosscap three.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This research project was funded by the Deanship of Scientific Research (DSR) at King Abdulaziz University, under grant no. KEP-44-130-42. The first, second and fourth authors, therefore, acknowledge with thanks DSR for technical and financial support.

Conflict of interest

The authors declare no conflicts of interest regarding the publishing of this paper.