© Wuhan University 2024

0 Introduction

In this paper, all graphs are simple and connected. A polyhedron on the surface is called a quadrangulation if each face of the polyhedron is a quadrangle with four distinct vertices. Let$\Phi$ be the current group of a current graph. Two digraphs $G_{\mathrm{1}}$ and $G_{\mathrm{2}}$ are said to be isomorphic if there are bijections $\omega :V(G_{\mathrm{1}})\to V(G_{\mathrm{2}})$ and $\overline{\omega }:A(G_{\mathrm{1}})\to A(G_{\mathrm{2}})$, such that if an arbitrary arc$\delta \in A(G_{\mathrm{1}})$ is directed from a vertex $v$ to a vertex $u$, then the arc $\overline{\omega }(\delta )$ is directed from the vertex $\omega (v)$ to the vertex $\omega (u)$. Two digraphs $G_{\mathrm{1}}$ and $G_{\mathrm{2}}$ are said to be strong isomorphic if there is an isomorphism $(\omega ,\overline{\omega })$ of $G_{\mathrm{1}}$ into $G_{\mathrm{2}}$ and an automorphism $\phi$ of $\Phi$ such that $\overline{\omega }(\delta )=\phi (\delta )$ for every arc $\delta$ of $G_{\mathrm{1}}$^[1].

Embedding theory of graphs has been widely used in very large scale integrated circuits^[2], etc. There are many results on the maximum genus orientable embeddings (orientable surface embeddings with one face), for instance, Xuong^[3], Liu^[2,4,5] and Škoviera^[6] studied this problem. The maximum genus orientable embeddings of current graphs are often used to construct the embedding of complete graphs^[7]. There have been many results on triangular embeddings of complete graphs, for example, the results of Ref. [1] and Refs. [8-11].

Korzhik^[12] generated nonisomorphic quadrangular embeddings of the complete graph $K_{\mathrm{8}s+\mathrm{5}}$. Hartsfield and Ringel^[13] proved that a complete graph $K_n$ has an orientable quadrangular embedding if $n\equiv \mathrm{5}(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{8})$, and these embeddings determine polyhedra with a minimal number of quadrangles. Hartsfield and Ringel^[14] proved that a complete graph $K_n$ has a nonorientable quadrangular embedding if $n\equiv \mathrm{1}(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{4})$, and these embeddings determine polyhedra with a minimal number of quadrangles. In recent years, Liu et al^[15,16] have obtained some new results on quadrangulations. Korzhik^[12] constructed $(\mathrm{2}s)!$ nonisomorphic oritentable quadrangular embeddings of $K_{\mathrm{8}s+\mathrm{5}}$.

In this paper, we first construct the current graph of the complete graph $K_{\mathrm{12}s+\mathrm{9}}$, and prove that $K_{\mathrm{12}s+\mathrm{9}}$ has at least ${\mathrm{6}}^{\mathrm{2}s}\times {\mathrm{3}}^{\frac{s+\mathrm{3}}{\mathrm{2}}}$ nonisomorphic orientable quadrangular embeddings by finding the maximum genus embedding of the current graph of $K_{\mathrm{12}s+\mathrm{9}}$. Then, by finding the proper 4-edge-colors of the current graph of $K_{\mathrm{12}s+\mathrm{9}}$ and constructing a mapping function, we prove that every one of the nonisomorphic orientable quadrangular embeddings has at least twenty-four 4-edge-colors, and each color appears around each face of orientable quadrangular embeddings.

1 Construction of the Current Graph

Let $s$ be an odd number, and let $G^{\mathit{*}}$ be a 4-regular graph with $\mathrm{3}s+\mathrm{2}$ vertices and $\mathrm{6}s+\mathrm{4}$ edges, as shown in Fig. 1.

Fig. 1 ${\mathit{G}}^{\mathit{*}}$: The current graph of ${\mathit{K}}_{\mathrm{12}\mathit{s}+\mathrm{9}}$

Each arc of$G^{\mathit{*}}$ has an orientation and a value called its current. $G^{\mathit{*}}$ can be decomposed into two Hamiltonian cycles, $H_{\mathrm{1}}(G^{\mathit{*}})$and $H_{\mathrm{2}}(G^{\mathit{*}})$ (shown as Fig. 2 and Fig. 3). $H_{\mathrm{1}}(G^{\mathit{*}})$ has current $\mathrm{2,4},\mathrm{6,8},\cdots ,\mathrm{6}s,\mathrm{6}s+\mathrm{2,6}s+\mathrm{4}$. $H_{\mathrm{2}}(G^{\mathit{*}})$ has current $\mathrm{6}s+\mathrm{3,6}s+\mathrm{1,6}s-\mathrm{1,6}s-\mathrm{3},\cdots ,\mathrm{5,3},\mathrm{1}$.

Fig. 2 Hamiltonian cycle ${\mathit{H}}_{\mathrm{1}}({\mathit{G}}^{\mathbf{*}})$

Fig. 3 Hamiltonian cycle ${\mathit{H}}_{\mathrm{2}}({\mathit{G}}^{\mathbf{*}})$

$G^{\mathbf{*}}$ has the following properties:

C1) Each vertex has valence 4.

C2) Each element $\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{6}s+\mathrm{3,6}s+\mathrm{4}$ of $Z_{\mathrm{12}s+\mathrm{9}}$ appears exactly once as a current on some $\mathrm{a}\mathrm{r}\mathrm{c}.Z_{\mathrm{12}s+\mathrm{9}}$ is an cyclic group, the additive group of integers modulo $\mathrm{12}s+\mathrm{9}$.

C3) At each vertex, the sum of the inward flowing currents equals the sum of the outward flowing currents. This property is known as Kirchhoff's Current Law (KCL).

2 Maximum Genus Embeddings

A spanning tree $T$ in a graph $G$ is called optimal, if the number of odd components, denoted by $\omega (T)$, of $G-T$is the smallest among all the spanning trees of $G$.

Theorem 1^[3,5] The maximum genus of a graph $G$ in orientable surfaces is ${\gamma }_M(G)=\frac{\beta (G)-\omega (T)}{\mathrm{2}}$, where $\beta (G)$ is the Betti-number of $G(\beta (G))=\left|E(G)\right|-\left|V(G)\right|+\mathrm{1})$ and $\omega (T)$ is the number of odd components in an optimal tree$T$ in $G$.

Theorem 2^[2,3] Let $T$ be an optimal tree in a graph $G$ with $\omega (T)$ odd components in $G-T$. Then edges of $E(G)-E(T)$ may be partitioned as follows:

$E(G)-E(T)=\stackrel{n}{\underset{i=\mathrm{1}}{\cup }}\{e_{\mathrm{2}i-\mathrm{1}},e_{\mathrm{2}i}\}\bigcup \{f_{\mathrm{1}},f_{\mathrm{2}},\cdots ,f_k\}$

$e_{\mathrm{2}i-\mathrm{1}}\bigcap e_{\mathrm{2}i}\ne \mathrm{\varnothing }$ and $\{f_{\mathrm{1}},f_{\mathrm{2}},\cdots ,f_k\}$ is a matching of $G$,$n={\gamma }_M(G)$, $k=\omega (T)$.

Theorem 3^[2] Let $\beta (G)$ be the Betti-number of$G$. If $\beta (G)=\mathrm{0}(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{2})$, then $G$ can be embedded in an orientable surface with one face.

Theorem 4   The current graph $G^{\mathit{*}}$ of $K_{\mathrm{12}s+\mathrm{9}}$ (see Fig. 1, $s$ is an odd number) has at least ${\mathrm{6}}^{\mathrm{2}s+\mathrm{1}}\times {\mathrm{3}}^{\frac{s+\mathrm{1}}{\mathrm{2}}}$ orientable surface embeddings with one face, and the orientable maximum genus ${\gamma }_M(G^{\mathit{*}})=\frac{\mathrm{3}s+\mathrm{3}}{\mathrm{2}}$.

Proof   $G^{\mathbf{*}}$ has $\mathrm{3}s+\mathrm{2}$ vertices and $\mathrm{6}s+\mathrm{4}$ edges,$\beta (G^{\mathit{*}})=\mathrm{0}(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{2})$. By Theorem 3, $G^{\mathit{*}}$ can be embedded in an orientable surface with one face. As shown in Fig. 4, $G^{\mathit{*}}$ has an optimal tree $T(G^{\mathit{*}})$. By Theorem 2, $E(G^{\mathit{*}})-E(T(G^{\mathit{*}}))$ may be partitioned as follows: $E(G^{\mathit{*}})-E(T(G^{\mathit{*}}))=\stackrel{\frac{\mathrm{3}s+\mathrm{3}}{\mathrm{2}}}{\underset{i=\mathrm{1}}{\cup }}\{e_{\mathrm{2}i-\mathrm{1}},e_{\mathrm{2}i}\}$, $e_{\mathrm{2}i-\mathrm{1}}\bigcap e_{\mathrm{2}i}=\mathrm{\varnothing }$, and $\frac{\mathrm{3}s+\mathrm{3}}{\mathrm{2}}={\gamma }_M(G^{\mathit{*}})$.

Fig. 4 $\mathit{T}({\mathit{G}}^{\mathbf{*}})$: Optimal tree of ${\mathit{G}}^{\mathbf{*}}$

$T(G^{\mathbf{*}})$ has $s$ vertices of valence 4 and $\mathrm{2}s+\mathrm{2}$ vertices of valence 1. A vertex of valence 4 can have one of the six different rotations. Every rotation scheme determines a planar embedding of $T(G^{\mathit{*}})$ with a single region (face) whose boundary is $W_{\mathrm{0}}$. Therefore, $T(G^{\mathit{*}})$ has ${\mathrm{6}}^s$ orientable surface embeddings with one face.

As shown in Fig. 5, $e_{\mathrm{1}}=(v_{\mathrm{1}},p)$, $e_{\mathrm{2}}=(v_{\mathrm{1}},u_s)$. Consider the vertex $v_{\mathrm{1}}\in e_{\mathrm{1}}\bigcap e_{\mathrm{2}}$ and fix it at a copy of $v_{\mathrm{1}}$ on $W_{\mathrm{0}}$. We choose a copy of $p$ and $u_s$ on $W_{\mathrm{0}}$, respectively. Note that there are $d_{G_{\mathrm{0}}}(v_{\mathrm{1}}),d_{G_{\mathrm{0}}}(p)$ and $d_{G_{\mathrm{0}}}(u_s)$ ways of doing so, where $G_{\mathrm{0}}=T(G^{\mathit{*}})$. Then, we find a new facial walk $W_{\mathrm{1}}$, which contains $W_{\mathrm{0}}$ and the edges $e_{\mathrm{1}}$, $e_{\mathrm{2}}$.

Fig. 5 $\mathit{T}({\mathit{G}}^{\mathbf{*}})+\{{\mathit{e}}_{\mathrm{1}},{\mathit{e}}_{\mathrm{2}}\}$

We choose the labeling of $p$ and $u_s$ so that, in $W_{\mathrm{0}}$, the selected copies of vertices $v_{\mathrm{1}},p,$ and $u_s$ occur in the order of $v_{\mathrm{1}},p,u_s$. Between the two consecutive edges at the current rotation at$v_{\mathrm{1}},W_{\mathrm{0}}$ goes $e,v_{\mathrm{1}},e\text{'}$. We extend the rotation $W_{\mathrm{0}}$ at $v_{\mathrm{1}}$ from $\cdots ,e,e\text{'},\cdots$to$\cdots ,e,e_{\mathrm{1}},\cdots ,e_{\mathrm{2}},e\text{'},\cdots$. Likewise, we insert the edge $(v_{\mathrm{1}},p)$ between the consecutive (at that copy of $p$) edges and the same for $(v_{\mathrm{1}},u_s)$ at $u_s$. Now we obtain $G_{\mathrm{1}}=G_{\mathrm{0}}+\{e_{\mathrm{1}},e_{\mathrm{2}}\}$ to have its orientable surface embeddings with one face on the torus, with a single region (face) bounded by edges of $W_{\mathrm{1}}$. It is clear that there are at least $d_{G_{\mathrm{0}}}(v_{\mathrm{1}})d_{G_{\mathrm{0}}}(p)d_{G_{\mathrm{0}}}(u_s)=\mathrm{1}\times \mathrm{1}\times \mathrm{1}=\mathrm{1}$ ways to construct$W_{\mathrm{1}}$.

As shown in Fig. 6,$e_{\mathrm{3}}=(v_{\mathrm{2}},v_{\mathrm{1}})$,$e_{\mathrm{4}}=(v_{\mathrm{2}},u_{s-\mathrm{1}})$. Consider the vertex $v_{\mathrm{2}}\in e_{\mathrm{3}}\bigcap e_{\mathrm{4}}$ and fix it at a copy of $v_{\mathrm{2}}$ on $W_{\mathrm{1}}$. Then choose a copy of $v_{\mathrm{1}}$ and $u_{s-\mathrm{1}}$ on $W_{\mathrm{1}}$, respectively. Note that there are $d_{G_{\mathrm{1}}}(v_{\mathrm{2}}),d_{G_{\mathrm{1}}}(v_{\mathrm{1}})$ and $d_{G_{\mathrm{1}}}(u_{s-\mathrm{1}})$ ways of doing so, where $G_{\mathrm{1}}=T(G^{\mathit{*}})+\{e_{\mathrm{1}},e_{\mathrm{2}}\}$. Then we find a new facial walk $W_{\mathrm{2}}$, which contains $W_{\mathrm{1}}$ and the edges $e_{\mathrm{3}},e_{\mathrm{4}}$.

Fig. 6 $\mathit{T}({\mathit{G}}^{\mathbf{*}})+\{{\mathit{e}}_{\mathrm{1}},{\mathit{e}}_{\mathrm{2}}\}+\{{\mathit{e}}_{\mathrm{3}},{\mathit{e}}_{\mathrm{4}}\}$

We choose the labeling of $v_{\mathrm{1}}$ and $u_{s-\mathrm{1}}$ so that, in $W_{\mathrm{1}}$, the selected copies of vertices $v_{\mathrm{2}},v_{\mathrm{1}}$, and $u_{s-\mathrm{1}}$ occur in the order of$v_{\mathrm{2}},v_{\mathrm{1}},u_{s-\mathrm{1}}$. Between the two consecutive edges at the current rotation at $v_{\mathrm{2}}$, $W_{\mathrm{1}}$ goes $e,v_{\mathrm{2}},e\text{'}$. We extend the rotation $W_{\mathrm{1}}$ at $v_{\mathrm{2}}$ from $\cdots ,e,e\text{'},\cdots$ to $\cdots ,e,e_{\mathrm{3}},\cdots ,e_{\mathrm{4}},e\text{'},\cdots$. Likewise, we insert the edge $(v_{\mathrm{2}},v_{\mathrm{1}})$ between the consecutive (at that copy of$v_{\mathrm{1}}$) edges and the same for $(v_{\mathrm{2}},u_{s-\mathrm{1}})$ at $u_{s-\mathrm{1}}$, as shown in Fig.7. Now we obtain $G_{\mathrm{2}}=G_{\mathrm{1}}+\{e_{\mathrm{3}},e_{\mathrm{4}}\}$ to have its orientable surface embeddings with one face on $S_{\mathrm{2}}$, with a single region (face) bounded by edges of $W_{\mathrm{2}}$. It is clear that there are at least $d_{G_{\mathrm{1}}}(v_{\mathrm{2}})d_{G_{\mathrm{1}}}(v_{\mathrm{1}})d_{G_{\mathrm{1}}}(u_{s-\mathrm{1}})=\mathrm{1}\times \mathrm{3}\times \mathrm{1}=\mathrm{3}$ ways to construct $W_{\mathrm{2}}$.

Fig. 7 $\mathit{T}({\mathit{G}}^{\mathit{*}})+\{{\mathit{e}}_{\mathrm{1}},{\mathit{e}}_{\mathrm{2}}\}+\cdots +\{{\mathit{e}}_{\mathrm{2}\mathit{s}+\mathrm{1}},{\mathit{e}}_{\mathrm{2}\mathit{s}+\mathrm{2}}\}$

Repeat the above procedure for $i=\mathrm{3,4},\cdots ,\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}$, we add all $\{e_{\mathrm{2}i-\mathrm{1}},e_{\mathrm{2}i}\}$. We can obtain $G_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}=G_{\mathrm{2}}+\{e_{\mathrm{5}},e_{\mathrm{6}}\}+\cdots +\{e_{\mathrm{3}s},e_{\mathrm{3}s+\mathrm{1}}\}$ to have its orientable surface embeddings with one face on $S_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}$, with a single region (face) bounded by edges of $W_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}$, as shown in Fig.8.

Fig. 8 $\mathit{T}({\mathit{G}}^{\mathbf{*}})+\{{\mathit{e}}_{\mathrm{1}},{\mathit{e}}_{\mathrm{2}}\}+\cdots +\{{\mathit{e}}_{\mathrm{3}\mathit{s}},{\mathit{e}}_{\mathrm{3}\mathit{s}+\mathrm{1}}\}$

As shown in Fig. 9, $e_{\mathrm{3}s+\mathrm{2}}=(p,u_s)$, $e_{\mathrm{3}s+\mathrm{3}}=(p,q)$. Consider the vertex $p\in e_{\mathrm{3}s+\mathrm{2}}\bigcap e_{\mathrm{3}s+\mathrm{3}}$ and fix it at a copy of $p$ on $W_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}$. Then choose a copy of $u_s$ and $q$ on $W_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}$, respectively. Note that there are $d_{G_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}}(p),d_{G_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}}(u_s)$ and $d_{G_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}}(q)$ ways of doing so, where $G_{\frac{\mathrm{3}s+\mathrm{3}}{\mathrm{2}}}=G_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}+\{e_{\mathrm{3}s+\mathrm{2}},e_{\mathrm{3}s+\mathrm{3}}\}$. Then we find a new facial walk $W_{\frac{\mathrm{3}s+\mathrm{3}}{\mathrm{2}}}$, which contains $W_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}$ and the edges $e_{\mathrm{3}s+\mathrm{2}},e_{\mathrm{3}s+\mathrm{3}}$.

Fig. 9 Underlying graph of the current graph of${\mathit{K}}_{\mathrm{12}\mathit{s}+\mathrm{9}}$

We choose the labeling of$u_s$ and $q$ so that, in$W_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}$, the selected copies of vertices $p,u_s$ and$q$ occur in the order of$p,u_s,q$. Between the two consecutive edges at the current rotation at $p,W_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}$ goes $e,p,e\text{'}$. We extend the rotation $W_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}$ at $p$ from $\cdots ,e,e\text{'},\cdots$to$\cdots ,e,e_{\mathrm{3}s+\mathrm{2}},\cdots ,e_{\mathrm{3}s+\mathrm{3}},e\text{'},\cdots$. Likewise, we insert the edge $(p,u_s)$ between the consecutive (at that copy of $u_s$) edges and the same for $(p,q)$ at $q$. Now we obtain $G_{\frac{\mathrm{3}s+\mathrm{3}}{\mathrm{2}}}=G_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}+\{e_{\mathrm{3}s+\mathrm{2}},e_{\mathrm{3}s+\mathrm{3}}\}$ to have its orientable surface embeddings with one face on $S_{\frac{\mathrm{3}s+\mathrm{3}}{\mathrm{2}}}$, with a single region (face) bounded by edges of $W_{\frac{\mathrm{3}s+\mathrm{3}}{\mathrm{2}}}$. It is clear that there are at least $d_{G_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}}(p)d_{G_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}}(u_s)d_{G_{\frac{\mathrm{3}s+\mathrm{1}}{\mathrm{2}}}}(q)=\mathrm{2}\times \mathrm{3}\times \mathrm{3}=\mathrm{18}$ ways to construct $W_{\frac{\mathrm{3}s+\mathrm{3}}{\mathrm{2}}}$.

From all the steps above, the current graph $G^{\mathit{*}}$ of $K_{\mathrm{12}s+\mathrm{9}}$ has at least ${\mathrm{6}}^s\times {\mathrm{3}}^s\times {\mathrm{2}}^{\frac{s-\mathrm{1}}{\mathrm{2}}}\times {\mathrm{2}}^{\frac{s+\mathrm{1}}{\mathrm{2}}}\times {\mathrm{3}}^{\frac{s+\mathrm{1}}{\mathrm{2}}}\times \mathrm{2}\times \mathrm{3}={\mathrm{6}}^{\mathrm{2}s+\mathrm{1}}\times {\mathrm{3}}^{\frac{s+\mathrm{1}}{\mathrm{2}}}$ orientable surface embeddings with one face. The orientable maximum genus ${\gamma }_M(G^{\mathit{*}})=\frac{\mathrm{3}s+\mathrm{3}}{\mathrm{2}}$.

3 Quadrangular Embeddings

Rule $Q^{\mathit{*}}$: If in row $i$ has: $\cdots j,k\cdots$, and in row$k$ has: $\cdots i,l\cdots$; then in row$l$ has: $\cdots k,j\cdots$, and in row$j$ has:$\cdots l,i\cdots$.

Theorem 5   If the rotation scheme of the complete graph $K_n$ satisfies the Rule $Q^{\mathbf{*}}$, then it is an orientable quadrangular embedding, and the orientable genus is $\frac{n(n-\mathrm{5})}{\mathrm{8}}+\mathrm{1}$.

Proof   We can demonstrate this using Fig. 10. Given the part $i.\cdots j,k\cdots$ of the rotation scheme, this means, in the rotation of the vertex $i$, the arc joining$i$ and$k$ follows the arc joining $i$ and $j$ (the first part of Fig. 10), and the part $k.\cdots i,l\cdots$ of the rotation scheme, this means, in the rotation of the vertex $k$, the arc joining $k$ and$l$ follows the arc joining $k$ and $i$ (the second part of Fig. 10). Using Rule $Q^{\mathbf{*}}$, one gets the part $l.\cdots k,j\cdots$ of the rotation scheme, this means in the rotation of the vertex$l$, the arc joining$l$ and $j$ follows the arc joining$l$ and $k$ (the third part of Fig. 10), and the part $j.\cdots l,i\cdots$ of the rotation scheme, this means, in the rotation of the vertex$j$ ,the arc joining$j$ and$i$ follows the arc joining $j$ and $l$ (the fourth part of Fig. 10). When Rule $Q^{\mathbf{*}}$ is used, the circuit is closed (the fifth part of Fig. 10).

Fig. 10 The construction of quadrangle

Each circuit (face) determined by this rotation scheme is quadrilateral, and the order of each vertex's rotation of the circuit (face) is the same. By Mohar and Thomassen, see P92-93 of Ref.[17], this rotation scheme is an orientable quadrangular embedding.

If $G$ is embedded into an orientable surface$S$, by Euler's formula: $\left|V(G)\right|-\left|E(G)\right|+\left|F(G)\right|=\mathrm{2}-\mathrm{2}p$,$p$ is the orientable genus of surface $S$. Therefore, $n-\frac{n(n-\mathrm{1})}{\mathrm{2}}+\frac{n(n-\mathrm{1})}{\mathrm{4}}=\mathrm{2}-\mathrm{2}p$,$p=\frac{n(n-\mathrm{5})}{\mathrm{8}}+\mathrm{1}$.

The orientable genus $\frac{n(n-\mathrm{5})}{\mathrm{8}}+\mathrm{1}$ is a natural number, then $n\equiv \mathrm{0}$ or $\mathrm{5}(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{8})$. Therefore, we have

Corollary 1^[15] If there exists a quadrangular rotation of the complete graph $K_n$, then $n\equiv \mathrm{0}$ or $\mathrm{5}(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{8})$.

Theorem 6   Let $G^{\mathit{*}}$ (see Fig. 1, $s$ is an odd number) be the current graph of the complete graph $K_{\mathrm{12}s+\mathrm{9}}$, for every one of the orientable surface embeddings with one face of $G^{\mathit{*}}$, there exists a quadrangular embedding of $K_{\mathrm{12}s+\mathrm{9}}$ into an orientable surface, and the orientable genus is $\frac{(\mathrm{12}s+\mathrm{9})(\mathrm{12}s+\mathrm{4})}{\mathrm{8}}+\mathrm{1}$.

Proof   From Chapter 2 and 4 of Ref. [7], we know that for each $\sigma \in \Pi$ ($\Pi$ is a set of rotation systems), the pair $(f^{\mathit{*}},\sigma )$ determines an embedding of $K_{\mathrm{12}s+\mathrm{9}}$ ($f^{\mathit{*}}$ is the current assignment of $G^{\mathit{*}}$). Here, we first find the set $\Pi$ of rotation systems of $K_{\mathrm{12}s+\mathrm{9}}$.

We use the current graph $G^{\mathit{*}}$ to construct quadrangular rotations of $K_{\mathrm{12}s+\mathrm{9}}$ in the following way. Consider the circuit induced by the rotation of the orientable surface embedding with one face of $G^{\mathbf{*}}$. Record the currents sequentially as they occur in the circuit, but if the direction of the circuit is opposite to the direction of the considered arc, record it with a minus sign and take this as row 0 of the rotation scheme of $K_{\mathrm{12}s+\mathrm{9}}$. Applying Additive Rule^[7], from the row 0, we can obtain all rows of the rotation scheme for $K_{\mathrm{12}s+\mathrm{9}}$, in which the row$i$ is the rotation of the vertex $i(i=\mathrm{0,1},\mathrm{2},\cdots ,\mathrm{12}s+\mathrm{8})$ of $K_{\mathrm{12}s+\mathrm{9}}$.

A vertex of valence 4 can have one of the six different rotations. Assume that in the rotation scheme, one finds $i.\cdots j,k\cdots$. The Additive Rule^[7] says that$\mathrm{0}.\cdots j-i,k-i\cdots$ appears in row 0. $j-i$ and $k-i$ are the currents on the arc of $G^{\mathit{*}}$, the two arcs incident to the same vertex of $G^{\mathit{*}}$ (Fig. 11(a)). The valence of each vertex of $G^{\mathit{*}}$ is 4, therefore, there are six cases. The local picture of the current graph is shown in Fig. 11(a).

Fig. 11 The local picture of the current graph of Case 1

Here, all the operations are modulo $\mathrm{12}s+\mathrm{9}$. $-x\equiv \mathrm{12}s+\mathrm{9}-x(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{12}s+\mathrm{9})$, $-y\equiv \mathrm{12}s+\mathrm{9}-y(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{12}s+\mathrm{9})$. $i,j,k\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$, and $i\ne j,j\ne k,i\ne k$.

Case 1 As shown in Fig. 11(a), $j-i$ and$k-i$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$.

Case 1.1 See Fig. 11(a) and Fig. 11(b), $j-i$ and$k-i$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$, i.e.,$\cdots j-i$,$k-i\cdots$ appear in the row 0 of the rotation scheme; $i-k$ and $x$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$, i.e.,$\cdots i-k$, $x\cdots$ appear in the row 0 of the rotation scheme;$-x$ and $-y$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots -x$, $-y\cdots$ appear in the row 0 of the rotation scheme; $y$ and $i-j$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots y$,$i-j\cdots$ appear in the row 0 of the rotation scheme.

Let $l=k+x$ (if$x+k>\mathrm{12}s+\mathrm{9}$, let $l=k+x-(\mathrm{12}s+\mathrm{9})$). Because $x\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$, then $l\ne k$. Because$x=l-k$, $i-k$ are the currents of $G^{\mathbf{*}}$, and$x\ne i-k$, then $l\ne i$.

According to the Kirchhoff's Current Law (property C3),$j-i+i-k=l-k+(-y)$ and consequently $y=l-j$. $y$ is the current of $G^{\mathit{*}}$ and $y\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$. Hence $l\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$ and$l\ne j$.

$\cdots j-i,k-i\cdots$ appear in the row 0 of the rotation scheme. By Additive Rule^[7], we get row $i$ as follows:$i.\cdots j,k\cdots$.

$\cdots i-k,l-k\cdots$ appear in the row 0 of the rotation scheme. By Additive Rule^[7], we get row $k$ as follows: $k.\cdots i,l\cdots$.

$\cdots k-l,j-l\cdots$ appear in the row 0 of the rotation scheme. By Additive Rule^[7], we get row$l$ as follows:$l.\cdots k,j\cdots$.

$\cdots l-j,i-j\cdots$. appear in the row 0 of the rotation scheme. By Additive Rule^[7], we get row $j$ as follows: $j.\cdots l,i\cdots$.

Therefore, Rule $Q^{\mathbf{*}}$ holds. By Theorem 5, we really have constructed an orientable quadrangular rotation of $K_{\mathrm{12}s+\mathrm{9}}$.

Case 1.2 See Fig. 11(a) and Fig. 11(c), $j-i$ and$k-i$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots j-i,k-i\cdots$ appear in the row 0 of the rotation scheme; $i-k$ and $-y$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots i-k,-y\cdots$ appear in the row 0 of the rotation scheme;$y$ and $x$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots y,x\cdots$ appear in the row 0 of the rotation scheme; $-x$ and $i-j$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e., $\cdots -x,i-j\cdots$ appear in the row 0 of the rotation scheme.

Let $l=k+(-y)$ (if $k+(-y)>\mathrm{12}s+\mathrm{9}$, let $l=k+(-y)-(\mathrm{12}s+\mathrm{9})$). Because $-y\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$, we have $l\ne k$. Because $-y=l-k$ and $i-k$ are the currents of $G^{\mathit{*}}$, and $-y\ne i-k$, we have $l\ne i$.

According to the Kirchhoff's Current Law (property C3), we have $j-i+i-k=l-k+x$ and consequently $x=j-l$. $x$ is the current of$G$, and $x\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$. Hence $l\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$ and $l\ne j$.

$\cdots j-i,k-i\cdots$ appear in the row 0 of the rotation scheme. By Additive Rule^[7], we get row $i$ as follows:$i.\cdots j,k\cdots$.

$\cdots i-k,l-k\cdots$ appear in the row 0 of the rotation scheme. By Additive Rule^[7], we get row k as follows: $k.\cdots i,l\cdots$.

$\cdots k-l,j-l\cdots$ appear in the row 0 of the rotation scheme. By Additive Rule^[7], we get row l as follows: $l.\cdots k,j\cdots$.

$\cdots l-j,i-j\cdots$ appear in the row 0 of the rotation scheme. By Additive Rule^[7], we get row $j$ as follows: $j.\cdots l,i\cdots$.

Therefore, Rule $Q^{\mathbf{*}}$ holds. By Theorem 5, we really have constructed an orientable quadrangular rotation of $K_{\mathrm{12}s+\mathrm{9}}$.

Case 2 As shown in Fig. 12(a), $j-i$ and $k-i$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of$G^{\mathit{*}}$.

Fig.12 The local picture of the current graph of Case 2

Case 2.1 As shown in Fig. 12(a) and Fig. 12(b),$j-i$ and $k-i$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots j-i$,$k-i\cdots$ appear in the row 0 of the rotation scheme; $i-k$ and$x$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots i-k,x$$\cdots$ appear in the row 0 of the rotation scheme; $-x$ and $-y$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots -x,$$-y\cdots$ appear in the row 0 of the rotation scheme; $y$ and $i-j$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots y,i-j\cdots$ appear in the row 0 of the rotation scheme.

Let $l=k+x$ (if $x+k>\mathrm{12}s+\mathrm{9}$, let $l=k+x-(\mathrm{12}s+\mathrm{9})$). Because $x\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$, we have $l\ne k$. Because $x=l-k$ and $i-k$ are the currents of $G^{\mathit{*}}$, and $x\ne i-k$, we have $l\ne i$.

According to the Kirchhoff's Current Law (property C3), $j-i+i-k=l-k+(-y)$ and consequently $y=l-j$. $y$ is the current of $G^{\mathit{*}}$, and $y\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$. Hence $l\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$ and $l\ne j$.

Similar to the Case 1.1, Rule $Q^{\mathbf{*}}$ holds. By Theorem 5, we really have constructed an orientable quadrangular rotation of $K_{\mathrm{12}s+\mathrm{9}}$.

Case 2.2 As shown in Fig. 12(a) and Fig. 12(c), $j-i$ and$k-i$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots j-i,k-i\cdots$ appear in the row 0 of the rotation scheme; $i-k$ and $-y$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots i-k,-y\cdots$ appear in the row 0 of the rotation scheme; $y$ and $x$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots y,x\cdots$ appear in the row 0 of the rotation scheme; $-x$ and $i-j$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$, i.e.,$\cdots -x,i-j\cdots$ appear in the row 0 of the rotation scheme.

Let $l=k+(-y)$ (if $k+(-y)>\mathrm{12}s+\mathrm{9}$, let $l=k+(-y)-(\mathrm{12}s+\mathrm{9})$). Because $-y\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$, then $l\ne k$. Because $-y=l-k$ and $i-k$ are the currents of $G^{\mathbf{*}}$, and $-y\ne i-k$, then $l\ne i$.

According to the Kirchhoff's Current Law (property C3),$j-i+i-k=l-k+x$ and consequently$x=j-l$. $x$ is the current of $G^{\mathit{*}}$ and$x\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$. Hence $l\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$ and $l\ne j$.

Similar to the Case 1.2, Rule $Q^{\mathbf{*}}$ holds. By Theorem 5, we really have constructed an orientable quadrangular rotation of $K_{\mathrm{12}s+\mathrm{9}}$.

Case 3 As shown in Fig. 13(a),$j-i$ and $k-i$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$.

Fig. 13 The local picture of the current graph of Case 3

Case 3.1 As shown in Fig. 13(a) and Fig. 13(b), $j-i$ and $k-i$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots j-i,k-i\cdots$ appear in the row 0 of the rotation scheme; $i-k$ and$x$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots i-k,x\cdots$ appear in the row 0 of the rotation scheme; $-x$and$-y$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$, i.e.,$\cdots -x,-y\cdots$ appear in the row 0 of the rotation scheme;$y$ and$i-j$ are consecutive currents of the circuit which is induced by the rotation of the orientable surface embedding with one face of $G^{\mathit{*}}$,i.e.,$\cdots y,i-j\cdots$ appear in the row 0 of the rotation scheme.

Let $l=k+x$ (if $x+k>\mathrm{12}s+\mathrm{9}$, let $l=k+x-(\mathrm{12}s+\mathrm{9})$). Because $x\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$, then$l\ne k$. Because $x=l-k$ and$i-k$ are the currents of $G^{\mathbf{*}}$, and $x=i-k$, then $l\ne i$.

According to the Kirchhoff's Current Law (property C3),$j-i+i-k=l-k+(-y)$ and consequently$y=l-j$.$y$ is the current of$G$, and $y\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$. Hence $l\in \{\mathrm{1,2},\mathrm{3},\cdots ,\mathrm{12}s+\mathrm{8}\}$, and $l\ne j$.