Genus of the Cartesian Product of Triangles

We investigate the orientable genus of G n , the cartesian product of n triangles, with a particular attention paid to the two smallest unsolved cases n = 4 and 5. Using a lifting method we present a general construction of a low-genus embedding of G n using a low-genus embedding of G n−1. Combining this method with a computer search and a careful analysis of face structure we show that 30 γ(G 4) 37 and 133 γ(G 5) 190. Moreover, our computer search resulted in more than 1300 non-isomorphic minimum-genus embeddings of G 3. We also introduce genus range of a group and (strong) symmetric genus range of a Cayley graph and of a group. The (strong) symmetric genus range of irredundant Cayley graphs of Z n p is calculated for all odd primes p.


Introduction
Finding the minimum genus of a graph is a very difficult problem both from practical and algorithmic perspective.Although in general it is NP-hard to determine the minimum genus [23] even in the class of cubic graphs [24], the minimum genus of various specific families of graphs has been determined in the past.A well-known instance is the determination of the minimum genus of complete graphs [22], which indicates the level of difficulty of determining the minimum genus.The genera of hypercubes have been computed by G. Ringel [21] and by Beineke and Harary [1].Generalizations of these methods were used by A. T. White to calculate the genus of the cartesian products of even cycles [25] and later by T. Pisanski to cartesian products of more general graph classes [15,16].These techniques have been used to determine the genus of most abelian and hamiltonian groups and non-orientable genus of some metacyclic groups [17,18,19].In most cases the developed methods can be adopted to the products where some factors are odd cycles of length at least five.The determination of genus of cartesian products involving triangular factors resisted all attempts in most cases, with the notable exceptions being the Cayley graphs of Z 3 × Z 3 × Z 3 [13,5] and semi-direct product Z 3 ⋊ Z 9 [4].The Cayley graphs of these groups are perhaps the most intriguing being the last groups of order at most 32 whose genus was determined.Moreover, if an abelian group Γ does not contain a Z 3 factor and its rank is at least 4, then the smallest genus of a Cayley graph for Γ follows from a formula in [10].Therefore, it is not very surprising that the determination of the smallest genus of a Cayley graph for an abelian group containing a Z 3 factor is considered to be extremely difficult, see [9,Chapter 6] or [2,Chapter 11].The aim of this paper is to determine good lower and upper bounds on the minimum among genera of Cayley graphs for Z n 3 with a particular attention paid to the smallest unsolved cases n = 4 and 5. Our main result is the following theorem.It can be easily seen that every generating set of Z n 3 contains at least n elements and each Cayley graph of Z n 3 generated by precisely n elements is isomorphic to G n , the cartesian product of n triangles.
Furthermore, motivated by results on genera of Cayley maps we introduce (symmetric) genusrange type invariants for Cayley graphs and groups which in a natural way broaden the concepts of symmetric genus of a Cayley graph and minimum genus of a group.
We assume that the reader is familiar with basics of topological graph theory as covered by, for instance, chapters 2 and 3 of [9] or chapters 5, 6, and 10 of [26].In particular, we assume that the reader is familiar with (regular) voltage graphs.We use standard terminology consistent with [9] and consider only cellular embeddings into orientable surfaces.
The paper is organized as follows.We treat lower bounds in Section 2 and upper bounds in Section 3. Section 3 is additionally divided into three parts, the first discussing genus-range type parameters of groups and Cayley graphs, the second dealing with the computational aspects of the problem, and the third presenting a recursive construction of a low-genus embedding of G n using a low-genus embedding of G n−1 .

Lower bounds
By G n we denote the cartesian product of n triangles, that is, For a prime p, by Z n p we denote the direct product of n copies of the cyclic group Z p ; clearly, G n is a Cayley graph of Z n 3 .For the number of vertices and edges of For an embedding of G n , the number of faces with length i of the embedding is denoted by f i .For instance, f 3 is the number of triangular faces of the embedding.Faces of length three are called triangles, faces of length four are called rectangles, or quadrangular, and faces of length five are called pentagons.For an embedding Π of G n , by F Π we denote the number of faces of Π.An easy counting shows that if G n would have an embedding with all faces being triangles, then the number of faces would be 2n3 n−1 , implying γ(G n ) ≥ 1 + ⌈3 n−1 (n − 3)/2⌉.For n = 3 and 4 this inequality implies γ(G 3 ) ≥ 1 and γ(G 4 ) ≥ 15.However, every edge of G n lies in precisely one triangle.In the union of face boundaries, every edge is traversed twice, once in every direction.Therefore, the maximum number of faces cannot be larger than if the second occurrence of every edge is in a face of length 4.An embedding of G n such that every triangle bounds a face and every other face is quadrangular is called a triangle-quadrangular embedding.The genus of such embedding of G n would be 1 + 3 n−1 (5n − 12)/8 and it is an open problem whether such embeddings exist for n > 1.Clearly, a triangle-quadrangular embedding may exist only if n is congruent to 4 (mod 8).This discussion can be summarized by the following proposition and two open problems.
Proposition 2.1.The maximum number of faces in an embedding of , with equality holding if and only if G n has a triangle-quadrangular embedding.
Problem 1.Is there an integer n > 1 such that G n has a triangle-quadrangular embedding?Problem 2. Determine all integers n such that G n has a triangle-quadrangular embedding.
Of a certain interest might be also quadrangular embeddings of G n , that is, embeddings in which every face has length 4. Note that the well-known genus embedding of G 2 in the torus is quadrangular.Problem 3. Is there an integer n > 2 such that G n has a quadrangular embedding?
Our main result concerning lower bounds is Theorem 2.7 below, which asserts that γ(G 4 ) ≥ 30.The proof is based on a method used by Brin and Squier in [5] to prove that γ(G 3 ) ≥ 6.Note that the analysis used in [5] to prove that γ(G 3 ) ≥ 7 is quite involved.Nevertheless, its use may lead to a better lower bound on the genus of G 4 .We start with necessary definitions.A plane is a subgraph of G n obtained from G n by fixing all but two coordinates.Clearly, a plane is isomorphic to G 2 and therefore every plane contains 9 vertices and 18 edges.It is easy to see that any face of length 3 or 4 in any embedding of G n lies in some plane of G n .A cycle of G is called present if it bounds a face, otherwise it is called absent.For a fixed embedding of G n , let a denote the total number of absent triangles, that is a = T (G n ) − f 3 .Let a i denote the number of planes with precisely i absent triangles.The following two results can be proved by an easy counting.Proposition 2.2.Every triangle of G n is lying in n − 1 planes of G n .In particular, (n − 1)a = A calculation analogous to the proof of Proposition 2.5 does not exclude the possible existence of a triangle-quadrangular embedding of G n for n ≥ 5 such that n ≡ 4 (mod 8).
It was observed in [5, Proposition 2] that minimum-genus embeddings of G n with the maximum number of present triangles do not contain faces of length 5.The next proposition is a slight extension of this result.Proposition 2.6.For any integer g such that g ≥ γ(G n ) there is an embedding of G n in an orientable surface of genus at most g without faces of length 5.
Proof.Let g ′ = min{g, γ M (G)} and let Π be any embedding of G n in the orientable surface of genus g ′ ; by Interpolation theorem for orientable surfaces (see [8]) and the choice of g ′ such an embedding exists.If f 5 = 0, then there is nothing to prove.Suppose that f 5 > 0 and that F is a face of length 5.It is easy to see that any pentagon has the form aabab −1 for some generators of G n a and b; denote by v the vertex of F incident with two occurrences of a.It follows that Π does not contain the triangle T of the form aaa incident with v. Let e be the edge of T not contained in F .Moving e into the interior of F splits F into a triangular face bounded by T and a rectangle; denote the resulting embedding by Π ′ .Since the move of e can be replaced by removing e from Π and then adding it back in a different position and a removal of an edge changes the number of faces by at most one, the genus of Π ′ is not larger than the genus of Π.If the genera of Π and Π ′ are equal, then e lies on the boundary of two distinct faces of Π.The removal of e merges these two faces into a pentagon in Π ′ if and only if one of them is triangle and the other is rectangle in Π. Observe that e lies in precisely one triangle, the triangle T , which is absent in Π.Therefore, e does not lie in a face of length 3 in Π and the removal of e cannot merge two faces into a pentagon.It follows that Π ′ has either a smaller genus or a smaller number of pentagons than Π and repeating the process yields the desired embedding of G n .
Theorem 2.7.The genus of G 4 is at least 30.
Proof.Let Π be an embedding of G n .Since the union of face boundaries includes every edge precisely twice and every face that is not triangular has length at least four, we have (1) Let χ denote the Euler characteristic of the underlying surface.Euler formula implies Substituting this equality into (1) and manipulating we get Consequently, By Proposition 2.6 we can assume that f 5 = 0.It follows that and after manipulation we get Lemma 2.4 implies Using Proposition 2.2 for n = 4 on (4) we obtain Note that for any n we have a = T (G n ) − f 3 , substituting this equality into (5) yields Combining the last inequality with Proposition 2.3 for n = 4 we get Two times (7) gives an upper bound on 6f 3 + 2f 4 , while adding three times (2) to (3) bounds 6f 3 + 2f 4 from below.Combining these inequalities gives Relating χ < −58 with the genus of G 4 gives γ(G 4 ) ≥ 30.
Using the method from the proof of Theorem 2.7 for bounding the genus of G 5 gives γ(G 5 ) ≥ 133, which is the same as the bound from Proposition 2.1.

Upper bounds
The determination of the genus of G n in general seems to be a very difficult problem.On the contrary, when we concentrate only on symmetric embeddings of G n , it is possible to determine not only the symmetric genus, but also the complete set of genera of surfaces upon which G n admits a symmetric embedding.This fact is our motivation for discussing several natural variants of genus range for groups and Cayley graphs which were not investigated before.

Genus range and symmetric genus range of groups and Cayley graphs
We start by presenting definitions of symmetric and strongly symmetric embeddings of a Cayley graph.We follow [2,Chapter 11] to call an embedding of a Cayley graph G of a group Γ symmetric if the natural action of Γ by left-multiplication on the vertices of G can be extended to an action on the underlying surface.An embedding of a Cayley graph is called strongly symmetric if it is symmetric and the extended action preserves orientation of the surface.We introduce the symmetric genus range of a Cayley graph G as the set of genera of surfaces upon which G admits a symmetric embedding and strong symmetric genus range of G as the set of all genera of surfaces upon which G has a strong symmetric embedding.Note that the symmetric genus range and strong symmetric genus range parameters are analogous to the genus range parameter, thus extending the correspondence between symmetric and all embeddings beyond the well-known (strong) symmetric genus of a Cayley graph.
Our main result in this direction is Theorem 3.2 which completely determines the strong symmetric genus range and symmetric genus range of G n .Additionally, the theorem shows that, unlike the genus range, the (strong) symmetric genus range can contain arbitrarily large gaps.
For a set X of elements of a group, denote by X the union of elements of X and their inverses.Recall that a Cayley map of a Cayley graph G with a generating set X is an embedding of G in which every local rotation induces the same cyclic order of X.The proof of Theorem 3.2 is based on the following correspondence between Cayley maps and symmetric embeddings, see Chapter 10, Theorem 4.1, and Chapter 11 of [2].An embedding of a Cayley graph G is strongly symmetric if and only if it is a Cayley map of G.An embedding of a Cayley graph of a group Γ with generating set X is symmetric, but not strongly symmetric, if and only if there is an index-two subgroup Γ ′ of Γ such that the local rotations of all vertices corresponding to Γ ′ induce the same cyclic order of X and the local rotations of all vertices corresponding to Γ − Γ ′ induce the reverse cyclic order.Therefore, the problem of determining the strong symmetric genus range of G n is equivalent with the problem of determining the genera of all Cayley maps of G n .Moreover, if Γ does not have an index-two subgroup, then every symmetric embedding of G is strongly symmetric.Consequently, the fact that Z n 3 does not have an index-two subgroup for any nonnegative integer n implies that the symmetric genus range and the strong symmetric genus range of G n coincide.
Recall that a generating set X of a group Γ is irredundant if no proper subset of X generates Γ.We call a Cayley graph irredundant if it is generated by an irredundant generating set of the group.Let B n denote the bouquet of n circles; that is, a single vertex incident with n loops.It is well known that every Cayley graph of a group Γ with generating set not containing involutions is the derived graph of B n for some integer n and a (regular) voltage assignment in Γ, see [9].The derived embedding Π ′ is a Cayley map of G n , whose genera was determined by [3].In the case of irredundant Cayley graphs of Z n p it is given by formula where t is the number of faces of Π and m i , the period of the i-th face of Π, is the group order of the sum of group elements (voltages) assigned to the edges on the boundary of the face.All non-zero elements of the voltage group Z n p have order p.The following lemma characterises the possible periods of an embedding of B n with voltages in Z n p such that the derived graph is an irrendudant Cayley graph of Z n p .
Lemma 3.1.Let Π be an embedding of B n with a voltage assignment from Z n p for some odd prime p and positive integer n such that the derived graph is an irredundant Cayley graph of Z n p .If Π has one face, then the period of that face is 1.If Π has at least two faces, then the period of every face of Π is p.
Proof.Since the generating set is irredundant, the order of every voltage is strictly greater than 1 and the voltages are pairwise independent.As the order of every element of Z n p is p, it follows that the order of every voltage is exactly p.Clearly, if Π has precisely one face, then every edge is traversed twice by the face, once in each direction, implying that the period is the group identity.Suppose that Π has at least two faces.If F is a face of Π, then the boundary of F contains an edge e that is traversed precisely once by F .The voltage assigned to e has order p and is independent from all other voltages assigned to the edges of F , thus the period of F is at least p.The fact that Z n p does not contain elements of order strictly greater than p implies that the period of F is exactly p. Theorem 3.2.Let G be an irredundant Cayley graph of Z n p for some prime p and a positive integer n.Then the symmetric genus range and the strong symmetric genus range of G coincide and are given by Proof.It is not difficult to see that any Cayley map of G is the derived embedding of an embedding of B n with voltages from Z n p .Clearly, B n is a planar graph, the maximum genus of B n is ⌊n/2⌋, and Interpolation theorem for orientable surfaces implies that B n has a cellular embedding in the orientable surface of genus g if and only if 0 ≤ g ≤ ⌊n/2⌋.Suppose that an embedding Π of B n has at least two faces.By Lemma 3.1 the period of each face is p and therefore the genus of the derived embedding is determined by the number of faces of the base embedding alone.To calculate the genus of the derived embedding Π ′ , we can substitute f /p for the sum in (8), where f is the number of faces of Π.Additionally, expressing f from the Euler formula for Π we get f = n + 1 − 2g and again substituting gives γ(Π A straightforward calculation shows that if the embedding of B n has one face, then the genus of the derived embedding coincides with the largest value in the previous case.
As a consequence of Theorem 3.2, we get that the strong symmetric genus range of a Cayley graph can contain arbitrarily large gaps, as opposed to the genus range, which is always a contiguous interval by Interpolation theorem for orientable surfaces [8].
In particular, the lowest of genera of Cayley maps of G n gives an upper bound on γ(G n ).
Concerning the genera of (strong) symmetric embeddings of Cayley graphs, a large part of the existing results deal with a determination of the genus -the minimum integer in the symmetric genus range.The focus is usually on a specific group or a family of groups, or a specific surface, see for example [7,12].Another important direction in the study of symmetric embeddings of Cayley graphs is aimed at regular maps arising from Cayley maps, see for example [20].The following problem offers a slightly different perspective on symmetric embeddings of Cayley graphs.Moreover, Theorem 3.2 indicates that this problem may be more approachable in the case of irredundant Cayley graphs of groups with relatively simple structure such as Z n p .
Problem 4. For a given Cayley graph G, determine the symmetric genus range and the strong symmetric genus range of G.
The genus range of a graph G is the set of integers g such that G admits a cellular embedding in the orientable surface of genus g.By the Interpolation theorem for orientable surfaces ( [8]), an integer g lies in the genus range of G if and only if γ(G) ≤ g ≤ γ M (G), where γ M (G) is the maximum genus of the graph G.However, for groups and their Cayley graphs, only the minimum genus parameter was introduced: the minimum genus of a group Γ is the minimum among genera of all Cayley graphs of Γ.Therefore, we formally propose the concept of genus range of a group.For a group Γ, the genus range of Γ is the set of all integers g such that there is a Cayley graph for Γ having a cellular embedding in the orientable surface of genus g.The maximum value in the genus range of a group Γ is called the maximum genus of Γ and is denoted by γ M (Γ).Theorem 3.2 suggests that the problem of determining the genus range of a group may have different characteristics when restricted to (strong) symmetric embeddings of irredundant Cayley graphs of the group.Therefore, we introduce also the symmetric and irrendundant variants of the genus range of a group.For a group Γ, the sharp genus range of Γ is the set of all integers g such that there is an irredundant Cayley graph for Γ having a cellular embedding in the orientable surface of genus g.The maximum value in the sharp genus range of a group Γ is called the sharp maximum genus of Γ and is denoted by γ ′ M (Γ).The (strong) symmetric genus range of Γ is the set of integers g such that there is a Cayley graph for Γ having a (strong) symmetric cellular embedding in the orientable surface of genus g.The sharp (strong) symmetric genus range of Γ is the set of integers g such that there is an irredundant Cayley graph for Γ having a (strong) symmetric cellular embedding in the orientable surface of genus g.Nonorientable analogues of the genus range parameters of a group may be introduced straightforwardly.
A graph G is called upper-embeddable if its maximum genus reaches the natural upper bound ⌊β(G)/2⌋, where β(G) is the cycle rank of G. Equivalently, G is upper-embeddable if it has an embedding with one face (if its cycle rank is even) or with two faces (if its cycle rank is odd).While the calculation of the minimum genus of a group does not depend on the precise definition of the arising Cayley graphs, in the case of genus range and maximum genus the involutions (elements of order 2) in the generating sets need to be treated specifically.It is customary to define the Cayley graph to have cycles of length 2 corresponding to involutions and to define the alternative, or reduced, Cayley graph in which every cycle of length two is replaced by a single edge, see for instance [9].Nedela and Škoviera [14] proved that every Cayley graph is upper-embeddable and that a reduced Cayley graph G is upper-embeddable unless the generating set consists from two elements r and s such that r 2 = s 3 = 1 and |V (G)| ≥ 18, in which case an exact formula for the maximum genus is given.
The restriction to irredudant generating sets in the definition of genus of a group is justified by following observation: if X and X ′ are generating sets for a group Γ such that X X ′ , then the Cayley graph G of Γ generated by X is a subgraph of the Cayley graph G ′ generated by X ′ and consequently γ(G) ≤ γ(G ′ ).Without the restriction to irredundant generating sets also in the definition of the genus range, the whole group (without the identity element) may be chosen as the generating set, giving rise to a graph on |Γ| vertices with |Γ|(|Γ| − 1)/2 edges.Thus the calculation of the maximum genus of Γ reduces to finding a generating set of Γ with least involutions in the case of reduced Cayley graphs and is trivial otherwise.The problem of determining the sharp maximum genus of a group reduces to finding an irredundant generating set X such that |X| − i X is maximized, where i X is the number of involutions in X.In this context it would be interesting to know whether there is a group with non-upper-embeddable reduced Cayley graph for any irredundant generating set.
Since Z n p have essentially only one irredundant generating set, we get the next results.
Theorem 3.4.For any odd prime p, the sharp maximum genus of Z n p is given by Theorem 3.5.For any odd prime p, the sharp symmetric genus range and the sharp strong symmetric genus range of Z n p coincide and are given by Theorem 3.6.The sharp genus range of Z 3 , Z 2 3 , and Z 3 3 is equal to {0}, {1, 2, 3, 4}, and {g; 7 ≤ g ≤ 26}, respectively.
The following problem suggests itself for both redundant and irredundant Cayley graphs of the group.Problem 5.For a given group Γ, determine the genus range, the (strong) symmetric genus range, sharp genus range, and sharp (strong) symmetric genus range of Γ.
In spite of the Interpolation theorem for orientable surfaces and Theorem 3.2 it is natural to ask whether the genus range and the (strong) symmetric genus range of a group and their sharp analogues can contain gaps.
Finally, note that while the case of G n is probably the most difficult in determining the (nonsymmetric) genus among the Cayley graphs of abelian groups, most likely it is one of the easiest for the sharp (strong) symmetric genus.

Computer search
The first author wrote a series of computer programs for experimenting with the embeddings of G n .The second author wrote an independent program for checking the validity of results.The data are available at www.dcs.fmph.uniba.sk/~kotrbcik/suppl/and this section briefly summarizes the main results.
We start by introducing the invariant used to distinguish nonisomorphic embeddings of a graph.A face distribution of an embedding Π is the sequence {f i }, where f i is the number of faces of Π with length i.The concept of face distribution appears as 'region distribution' in [26], where all possible face distributions of K 5 and K 6 are presented, in the case of K 6 reporting an earlier work [11].An extended face distribution of an embedding Π is the face distribution of Π together with the sequence {r i }, where r i is the number of faces of length i that contain some vertex more than once.Clearly, if two embeddings of the same graph have different extended face distribution, then they are nonisomorphic.Theorem 3.7.For the genus of G 4 we have γ(G 4 ) ≤ 37.Moreover, there are more than 10.000 nonisomorphic embeddings of G 4 into the orientable surface of genus 37 with pairwise distinct extended face distributions.
An embedding of G 4 in the orientable surface of genus 37 was obtained by computer search; the extended face distribution of the embedding is presented in Table 3, where a face is called repetitive if it contains some vertex more than once.The rotation schemes for more than 10.000 nonisomorphic embeddings of G 4 in the orientable surface of genus 37 and their extended face distributions can be found on the web pages containing the supplementary material.For the sake of completeness we present the corresponding rotations scheme of one such embedding in Table 4. length of the face 3 4 5 6 7 8 9 number of faces 88 59 8 10 2 2 2 number of repetitive faces 0 0 0 0 2 0 0 Table 3: Extended face distribution of an embedding of G 4 with genus 37 presented in Table 4.
The problem of determining the complete genus distribution of a graph G asks for the number of embeddings of G in every surface, where two embeddings are considered to be different if their rotation schemes differ.Therefore, the following theorem does not take into account any symmetries of G 2 or the embedding.
Theorem 3.8.The embedding range of G 2 is [1, 5], that is, G 2 admits a cellular embedding into the surfaces of genus 1, 2, 3, 4, and 5, and the complete genus distribution is given in Table 5.In particular, there are 330 genus embeddings into torus with only 7 distinct extended face distributions, presented in Table 6, and 46.908 embeddings in the double torus with 146 distinct extended face distribution." On the web pages containing the supplementary material we list the rotations schemes and extended face distributions of all embeddings of G 2 with genus at most two.Furthermore, we provide all distinct face distributions and the corresponding rotation schemes for embeddings of genus at most two.Perhaps surprisingly, G 2 embedded in the torus admits only 7 distinct extended face distributions, they are listed in Table 6 along with the number of such embeddings.Three of these distributions contain exactly one repetitive face and four of them do not contain a repetitive face.Finally, note that the two embeddings with 9 quadrangles are mirror images of each other.Theorem 3.9.There are at least 1319 genus embeddings of G 3 with pairwise distinct extended face distributions.
The rotation schemes for the nonisomorphic genus embeddings of G 3 can be found on the web pages containing the supplementary material.Table 7 contains several particularly interesting face distributions of genus embeddings of G 3 ; the corresponding embeddings can be also found as a separate part of the supplementary material.Although all these embeddings have all faces nonrepetitive, it is interesting that for all of them except the first two, there is an embedding with the same face distribution and one of the longest faces repetitive.Note also that the last embedding in Table 7 has the same face distribution as the embedding constructed in [13] to show that γ(G 3 ) ≤ 7. length of the face 3 4 5 6 length of the remaining face number of faces 22 12 0 8 number of faces 24 7 4 7 number of faces 24 9 0 9 number of faces 26 9 0 6 12 number of faces 27 6 0 8 9 number of faces 27 8 0 6 13 number of faces 27 9 0 5 15 Table 7: Face distributions of some of the 1319 genus embeddings of G 3 from Theorem 3.9.

Recursive construction
Let G ′ n denote G n with a loop attached to every vertex.Clearly, G n+1 is the derived graph of G ′ n with respect to Z 3 with a non-zero element of Z 3 assigned to an edge e if and only if e is a loop.It is easily seen that if a loop bounds a face (of length 1) and the voltage assigned to the loop has order 3, then the face lifts to a triangle.Therefore, if every loop bounds a face and a non-zero element of Z 3 is assigned to every loop, then the derived embedding has at least |V (G ′ n )| triangles.The main idea of the proof of Theorem 3.11 is that if we can embed loops inside faces of an embedding Π in such a way that every face contains either zero or at least two loops, then there is a voltage assignment to loops such that every face of Π with length at least 2 lifts to three faces.The fact that the loops can be distributed appropriately is captured by the following definition.Definition 3.10.Let Π be an embedding of a graph G.A face-covered partition of Π is a partition of the vertex set V (G) into sets P i , i = 1, . . ., k such that each P i contains at least two vertices and for any i there is a face of Π whose boundary contains all vertices from P i .
Our method does not rely on the fact that the base graph is G n and we state the result in a more general form.Theorem 3.11.Let Π be an embedding of a graph G in an orientable surface S. Let G ′ denote G with a loop attached to every vertex.If Π admits a face-covered partition of G, then there is an embedding of G ′ in S and a voltage assignment from Z 3 to G ′ such that the derived graph is G K 3 embedded with 3F Π + |V (G)| faces.
Proof.Let P = {P 1 , . . ., P k } be a face-covered partition of Π for G and let F i be a face of Π that covers P i for i ∈ {1, . . ., n}.We arbitrarily choose a prefered direction for each loop e and denote it by e; the opposite direction is denoted by e −1 .For any vertex v ∈ P, embed the loop e based at v into F i in such a way that the rotation at v is (ee −1 a . ..),where a is an arc traversed by F i .Denote the resulting embedding by Π ′ .Denote by L the set of loops added to G and by F L the set of faces of length 1 bounded by a loop from L. We say that a face F of Π ′ contains a loop l if the boundary of F traverses l.Our goal is to prescribe a voltage assignment ζ from the arcs of G ′ to the elements of Z 3 such that the derived embedding has 3F Π + |V (G)| faces.To achieve this property, we choose ζ such that the period of each face in F L is 3 and the period of each face of Π ′ not in F L is 1.
First note that any loop e from L lies on the boundary of two distinct faces in Π ′ , one of them is a face of length 1 bounded by e, and the second is a face of length at least 2. For any arc a of G not contained in G ′ let ζ(a) = 0. Let F be a face of Π ′ not in F L and containing a loop.The choice of P and Π ′ implies that F contains at least two loops.If In both cases the period of the face in Z 3 is 3.If F is a face of Π ′ not in F L and not containing a loop, then all arcs on the boundary of F are assigned 0 by ζ and the period of F is 1.If F is a face of Π ′ not in F L containing two or three loops, then clearly the period of the face is 1.The fact that the period of F is 0 if F contains at least four loops can be estabilished by an easy inductive argument which is ommited.
Clearly, the derived graph of Our next aim is to show that every minimum-genus embedding of G n admits a face-covered partition.The main idea of our method of constructing face-covered partitions is that if we take a matching in G n , then every pair of matched vertices is covered by some face.As G n − v contains a perfect matching for any vertex v, the problem reduces to covering the exceptional vertex v.
To cover this vertex we then use the fact that the embedding has a face of length at most 5.In the proofs of the following two auxiliary results we use the fact that the vertices of G n can be bijectively identified with words of length n over {0, 1, 2} with two vertices being adjacent if and only if their representations differ at exactly one position.Lemma 3.12.Let S be a set of three pairwise adjacent vertices of G n for some n ≥ 1.Then G n − S has a perfect matching.Consequently, G n − v has a perfect matching for any vertex v.
Proof.First observe that if G n − S has a perfect matching for any triangle S, then G n − v has a perfect matching for any vertex v, as every vertex lies in some triangle.We proceed by induction on n.For n = 1 the claim is obvious.For n ≥ 2 we show how to construct the desired matching.From the fact that S forms a triangle it follows that there is a unique position such that the representations of the vertices of S pairwise differ only in this position.Restricting G n − S to all but this one position yields three disjoint copies of G n−1 , each of them with one vertex removed.By the induction hypothesis the copies of G n−1 have a perfect matching; union of these perfect matchings is a perfect matching of G n − S. Proposition 3.13.Let Π be an embedding of G n that contains a face of length at most 5. Then Π admits a face-covered partition of G.
Proof.We distinguish three cases according to the length of the shortest face of Π. i) The length of a shortest face of Π is 3. Let F be a face of length 3 and let S be the set of vertices incident with S. By Lemma 3.12, there is a perfect matching M of G n − S. Denote the edges of M by m 1 , . . ., m k and let P i = {u i , v i }, where u i and v i are endpoints of m i for i ∈ {1, . . ., k}.Finally, let P k+1 = S. Since every edge m i is traversed by some face of Π, for any i, 1 ≤ i ≤ k, there is a face of Π that covers both vertices of P i .Moreover, the vertices of P k+1 are covered by F .It follows that the system of sets P i for i ∈ {1, . . ., k + 1} is a face-covered partition of Π.
ii) The length of a shortest face of Π is 4. Let F be a face of length 4 in Π.Without loss of generality suppose that the vertices of F are represented by 00x, 10x, 11x, and 01x, where x is arbitrary, but fixed word of length n − 2 over {0, 1, 2}.Consider the graph G ′ = G n − {abx; a, b ∈ {0, 1, 2}}.Restricting G ′ to positions 2, . . ., n yields three disjoint copies of G n−1 , each of them with the triangle at position 2 removed.By Lemma 3.12, each copy of G n−1 with a triangle removed admits a perfect matching; denote by M the union of these perfect matchings.We construct a facecovered partition of G n using M and a partition covering the three removed triangles.Suppose that the edges of M are u 1 v 1 , u 2 v 2 , . . ., u k v k , where k = 3 n−1 − 3. Let P i = {u i , v i } for i ∈ {1, . . ., k}.Clearly, these sets cover all vertices of M. To cover the vertices of the removed triangles, let P k+1 = {00x, 10x, 11x}, P k+2 = {20x, 21x}, P k+3 = {01x, 02x}, and P k+4 = {12x, 22x}.Since the set P k+1 is covered by the face F and any set P i for i = k + 1 contains exactly two vertices joined by an edge, the system of sets P i , i ∈ {1, . . ., k + 4} forms the desired face-covered partition.
iii) The length of a shortest face of Π is 5. Every pentagon in G n has form aabab −1 for some generators of G n a and b.In particular, every pentagon contains vertices of some triangle S. To get the desired face-covered partition it suffices to take the face-covered partition for Π covering the vertices of S which is constructed in the proof of case i).
Proposition 3.14.Every embedding of G n with genus at most 1 + ⌊3 n−1 (2n − 3)/2⌋ contains a face of length at most 5.In particular, every minimum-genus embedding of G n contains a face of length at most 5.
Proof.If an embedding Π of G n has the length of a shortest face at least 6, then Π contains at most n3 n−1 faces.Using Euler formula we get that the genus of Π is at least 1 + ⌈3 n−1 (2n − 3)/2⌉, which justifies the first claim.By Theorem 3.3 γ(G) ≤ 1 + n3 n−1 − 2 • 3 n−1 .Therefore, the inequality 1 + n3 n−1 − 2 • 3 n−1 < 1 + ⌈3 n−1 (2n − 3)/2⌉ proves the the second claim.Theorem 3.11 and Lemma 3.13 can be applied also to genus embeddings of G 2 and G 3 to construct low-genus embeddings of G 3 and G 4 , respectively.In these cases we get that the derived embeddings have genera 10 and 46, respectively, yielding γ(G 3 ) ≤ 10 and γ(G 4 ) ≤ 46.Note that both these bounds have been superseded by ad-hoc and computer-search methods of this paper.

Table 6 :
length of the face 3 4 5 6 length of the repetitive face frequency number of faces 6 All distinct extended face distributions of G 2 embedded in the torus.

F
contains two loops e and f , let ζ(e) = 1 and ζ(f ) = 2.If F contains three loops e, f , and g, let ζ(e) = ζ(f ) = ζ(g) = 1.Finally, if F contains at least four loops, choose any two of them, say e and f , and let ζ(e) = 1 and ζ(f ) = 2.By repeating the above process the value of ζ can be defined for all loops contained in F and consequently for all arcs of G ′ .Now we show that the periods of all faces of Π ′ under ζ have the required values.The boundary of any face of F L contains precisely one arc a corresponding to a loop in L and ζ(a) = 1 or ζ(a) = 2.
them with period 3 and |V (G)| of them with period 1.Therefore, the derived embedding of Π ′ under ζ in Z 3 has F Π + |V (G)| faces, which completes the proof.

Theorem 3 . 15 .Corollary 3 . 16 .
For any n ≥ 1 we have γ(G n+1 ) ≤ 3γ(G n ) + 3 n − 2. Consequently, for n ≥ 5 we have γ(G n ) ≤ 3 n−4 [γ(G 4 ) + 27n − 109] + 1 ≤ 3 n−2 [3n − 8] + 1. Proof.Let Π be a genus embedding of G n for some n ≥ 2. By Proposition 3.13 and Proposition 3.14, the embedding Π has a face-covered partition P. By Theorem 3.11 applied to Π and P, there is an embedding Π ′ of G n+1 with 3F Π + 3 n faces.Using Euler formula on the number of faces of Π and Π ′ and the number of vertices and edges of the corresponding graphs gives the first claim.The first closed form from the second claim can be obtained by solving the recurrence relation g 4 = γ(G 4 ) and g n+1 = 3g n + 3 n − 2 for n ≥ 4. The second closed form follows from Theorem 3.7 as γ(G 4 ) ≤ 37.For the genus of G 5 we have γ(G 5 ) ≤ 190.Proof.By Theorem 3.7, there is an embedding Π of G 4 with genus 37.The embedding Π contains a triangular face and therefore, by Proposition 3.13, it admits a face-covered partition P. The result follows by applying Theorem 3.11 to Π and P. Proof of Theorem 1.1.The lower bounds follow from Theorem 2.7 and Proposition 2.1.The upper bounds are proved in Theorem 3.7 and Theorem 3.15.

Table 2 :
If a plane of an embedding of G n has i absent triangles, then it has at most m i present rectangles, where the values of m i are in the Table2.The values of m i .The graph G 4 does not have a triangle-quadrangular embedding.Proof.First note that every rectangle lies in precisely one plane of G n .Assume that Π is a triangle-quadrangular embedding of G 4 .As Π has all triangles present, by Lemma 2.4 every plane has at most one present rectangle.By Proposition 2.3 the embedding contains at most 54 rectangles, which is a contradiction.

Table 4 :
Rotation scheme for an embedding of G 4 with genus 37.

Table 5 :
Complete genus distribution of G 2 .