Representativity of Cayley maps

European Journal of Combinatorics 39 (2014) 207–222

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European Journal of Combinatorics

journal homepage: www.elsevier.com/locate/ejc

Representativity of Cayley mapsD. Christopher Stephens a, Thomas W. Tucker b, Xiaoya Zha a

a Department of Mathematical Sciences, MTSU, Murfreesboro, TN 37132, USAb Colgate University, Hamilton, NY, USA

a r t i c l e i n f o

Article history:Received 27 January 2012Accepted 26 December 2013Available online 11 February 2014

a b s t r a c t

Themotivating question of this paper is to ask when, given a groupA and generating set X for A, one can find a Cayleymap CM(A, X, ρ)with representativity at least two (also called a strong embedding ofthe graph; note that in such an embedding, each face is bounded bya cycle of the graph).

We first investigate some general consequences of a Cayleymap’s having representativity one, which means that there ex-ist faces that are self-adjacent (and consequently these faces arenot bounded by proper cycles of the graph). We find that thereexist Cayley graphs of Abelian groups all of whose Cayley mapshave representativity one. This indicates that if these graphs havestrong embeddings (orientable or non-orientable), then these em-beddings cannot be obtained by applying the symmetry of Cayleygraphs in the most natural way. This also indicates that the CycleDouble Cover of these graphs, if it exists, cannot be constructed byapplying the symmetry of these graphs in the most natural way.

In addition, we investigate specific consequences in the case|X | = 2 or 3;we provide a few classes of examples of A and X whichmay be forced to have representativity at least two; andwe providea combinatorial formalism for calculating the representativity of aCayley map without appealing to the ‘‘lifted’’ embedding.

© 2014 Elsevier Ltd. All rights reserved.

1. Introduction

The CycleDouble Cover Conjecture, credited independently to Szekeres [8] and Seymour [6], claimsthat every 2-connected graph G has a collection of cycles C1, . . . , Ck such that each edge of G iscontained in exactly two of the cycles.

E-mail addresses: [email protected], [email protected] (D.C. Stephens), [email protected] (T.W. Tucker),[email protected] (X. Zha).

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208 D.C. Stephens et al. / European Journal of Combinatorics 39 (2014) 207–222

An embedding of a 2-connected graph has representativity at least two if and only if all facialwalks are cycles. Thus, the Cycle Double Cover Conjecture would be implied by the Strong EmbeddingConjecture (mentioned, for instance, in Haggard [2]) which claims that every 2-connected graph hasan embedding in which every facial walk is a cycle. Even stronger is the Orientable Strong EmbeddingConjecture, due to Jaeger [3], which claims that every 2-connected graph has an orientable embeddingof representativity at least two.We refer the reader to Jaeger [3] and Zhang [10] for further informationon the Cycle Double Cover Conjecture, which remains open as of this writing.

In this paper we pursue a line of thought related to the Orientable Strong Embedding Conjecture.We restrict the hypothesis so that the graphs under consideration are required to be Cayley graphs.However, we also restrict the conclusion by requiring a specific type of embedding of a Cayley graph—one inwhich all local clockwise rotations are the ‘‘same’’ with respect to the generating set in question.Such embeddings are called Cayley maps. (For an extensive survey on Cayley maps, see [5].)

The results here relate to the Strong Embedding Conjecture in twoways. First,weprovide examplesof groups and generating sets for which no Cayley map has representativity more than one. Thatis, if a strong embedding is to exist for these Cayley graphs, it is not symmetric in the Cayley-mapsense. These groups are not exotic by any means. In fact, in the context of this paper, commutativityis ‘‘worse’’ than noncommutativity, so that it is easy to find Abelian groups and generating sets forwhich no Cayley map is two-representative. One such small example given in Section 4 is Z11 withgenerating set {1, 2, 4}.

In particular, this means that there is a large class of cyclic groups and generating sets such that ifthe underlying Cayley graph is to have a strong embedding, then that embedding cannot be generatedby an embedded one-vertex voltage graph. This is perhaps surprising, since the underlying graphs arehighly symmetric, and representativity two does not seem to be such a strong demand. It would beinteresting to see whether the symmetry of these graphs can be exploited to give strong embeddingsin other, non-Cayley-map ways.

Second, in Section 6 we provide for any graph embeddingM and any integer r a finite, unbranchedregular covering of M with representativity at least r . Thus, every nonplanar graph has a regularcovering satisfying the Orientable Strong Embedding Conjecture. Of course, the Orientable StrongEmbedding Conjecture claims that in the 2-connected case, the number of sheets to the covering canbe bounded above by one. Since this seems to remain out of reach, we ask whether the number ofsheets to the covering can at least be bounded by two.

The paper is organized as follows. In Section 2 we provide definitions. In Section 3 we investigatesome combinatorial and group-theoretic consequences of a Cayleymap’s having representativity one,and we construct embeddings of representativity at least two in certain special cases. In Section 4 weconsider generating sets of sizes two and three. In Section 5 we take a closer look at cyclic groups.Finally, in Section 6 we provide a combinatorial formalism for discussing representativity of Cayleymaps.

2. Preliminaries

Let A be a group, and let X be a generating set for A. The Cayley graph G(A, X) is the directed graphwith vertex set V (G(A, X)) = A and edge set E(G(A, X)) = {(a, ax)| a ∈ A, x ∈ X}. In this paper wewill assume that A is finite unless otherwise noted; that X does not contain the identity of A; and that,for s in A, either s or s−1 may be in X , but not both.

Given a Cayley graph G(A, X), let ρ be a cyclic permutation of X ∪ X−1. Then the Cayley mapCM(A, X, ρ) is the (orientable) 2-cell embedding of G(A, X) with the property that at each vertex v,the clockwise ordering of edges is ρ.

Example 1. LetA = Z3×Z3, and letX = {(0, 1), (1, 0)}. Letρ be the permutation ((0, 1) → (1, 0) →

−(0, 1) → −(1, 0) → (0, 1)). Fig. 1 depicts the Cayley map CM(A, X, ρ) in two ways. On the right,the actual embedding on the torus is shown. On the left, only the rotation ρ is depicted. We will referto figures such as the one on the left-hand side of Fig. 1 as embedded one-vertex voltage graphs, and wewill refer to the corresponding orientable embeddings as their lifts. For background on voltage graphs,see [1].

D.C. Stephens et al. / European Journal of Combinatorics 39 (2014) 207–222 209

Fig. 1. A Cayley map.

If a graph G is embedded in a surface S, then the representativity of the embedding is the smallestnumber k such that S has a noncontractible closed curve that intersects G in k points. Let rep(A, X, ρ)denote the representativity of the Cayley map CM(A, X, ρ) as an orientable embedding. We defineR(A, X) to be

R(A, X) = maxρ

rep(A, X, ρ).

Before proceeding, we observe that if map M covers map N (possibly with branching) thenrep(M) ≥ rep(N) and equality is possible no matter how many sheets in the covering. Moreover,rep(M) is not necessarily a multiple of rep(N). Indeed, let T (m, n) be the m × n checkerboard on thetorus. Then the representativity of T (m, n) is just min(m, n). On the other hand, T (m, kn) regularlycovers T (m, n) (without branching).

3. Obtaining representativity at least two

We begin this section by discussing a few implications of a Cayley map’s having representativityonly one.

Lemma 2. Let B be the embedded one-vertex voltage graph corresponding to a Cayleymap (A, X, ρ), andlet W be the boundary word for a face F of B . Then the lifts of F are self-adjacent if and only if W ord(W )

contains a proper subword which is reducible to 1 in A.

Proof. The boundarywalk of the lift of F isW ord(W ). If thiswalk is not a cycle, then certainly it containsa proper subwalk which both starts and ends at some vertex v, so that the word corresponding to thissubwalk is equivalent to 1 in the group. The converse is equally straightforward. �

This immediately yields the following.

Corollary 3. Let B be an embedded one-vertex voltage graph. If the boundary word for a face F of B isof length 1, then the lifts of F are not self-adjacent. �

If x1, . . . , xn are elements of the group A, then ⟨x1, . . . , xn⟩will denote the subgroup of A generatedby x1, . . . , xn, i.e., the smallest subgroup of A containing the elements x1, . . . , xn.

Our next lemma characterizes what causes a Cayley map to have representativity one, and is usedfrequently in the remainder of the paper. Before stating it, we will consider an example.

Let A = Z12, and consider the Cayley map given by the embedded one-vertex voltage graph inFig. 2. By standard voltage arguments, the ‘‘middle’’ face bounded by edges labeled 3 and 4 lifts toexactly one face. Call it F . If we trace F beginning at the vertex 0, we get the facial walk depicted inFig. 3.

This face is self-adjacent because, for instance, 0 appears as the first (and last) vertex of F , butappears also as an internal vertex. (Similarly, of course, every other vertex appears twice in F , so theself-adjacency is extensive.)


Fig. 2. A Cayley map having representativity 1.

Fig. 3. A facial walk from Fig. 2.

The reason for the multiple appearances of vertices may be seen by examining the voltages in theCayley map. In the Cayley map, the face whose lift we are examining has net voltage +7. Since 7 and12 are coprime, the lifted face will ‘‘terminate’’ when 12 copies of the +7 voltage have occurred. InFig. 3, the +7 voltage corresponds to the dotted arcs depicted outside the actual facial walk; and eachiteration of the +7 voltage takes us from a solid vertex to a solid vertex.

However, there are also internal voltages of +4 and +3 (depicted as dashed and solid directededges, respectively). We observe that with only 8 copies of +7, we get to +56, which is congruent to8 (modulo 12) (a solid vertex in Fig. 3); and that, moreover, the dashed edge that follows brings usback to 0 before the ‘‘end’’ of the face.

We summarize the preceding example in the following lemma.

Lemma 4. Let B be an embedded one-vertex voltage graph, and let F be a face of B with boundary wordW. Then the lifts of F are self-adjacent if and only if there exists a cyclic shift X of W which may be writtenX = YZ, where neither Y nor Z is the empty word, such that ⟨X⟩ = ⟨Y , Z⟩.

Proof. Self-adjacency is equivalent to X jY = 1 for some initial subword Y of X . So self-adjacencyimplies X jY = 1. Then clearly Y ∈ ⟨X⟩. If Z is the final segment so X = YZ , then Z = Y−1X so Z is alsoin ⟨X⟩. Since X = YZ and thus X ∈ ⟨Y , Z⟩, we have ⟨X⟩ = ⟨Y , Z⟩.

Conversely, suppose that X = YZ , with ⟨X⟩ = ⟨Y , Z⟩. Then Y = X j for some j, giving us XkY = 1for some k, so we have self-adjacency. �


Fig. 4. A Cayley map for minimum-size generating sets.

Fig. 5. A Cayley map for minimum-size generating sets, with involutions.

Corollary 5. If X is a generating set of minimum size for a group A and X contains at most two involutions,then R(A, X) ≥ 2.

Proof. First assume X contains no involutions. Let B be the one-vertex voltage graph with E(B) = X .Embed B in the plane, as in Fig. 4, so that the local rotation is

s1s2 · · · sk−1sksksk−1 · · · s2s1.

The faces of our embedding of B are of two types. The ‘‘innermost’’ and ‘‘outermost’’ faces haveboundary walks of length one, and therefore by Corollary 3, their lifts are not self-adjacent. The otherfaces have boundary walks of the type W = sisj. By Lemma 4, if such a face with boundary word Wlifts to a face that is self-adjacent, then there exists a cyclic shift X ofW such that si and sj are both in⟨X⟩, contradicting minimality of S.

Now assume S contains two involutions and at least one noninvolution. Write S = {s1, . . . ,sk, t1, t2}, where t1 and t2 are involutions. Then we may embed s1, . . . , sk as before, and put t1 andt2 in the innermost and outermost faces, respectively, as in Fig. 5. In this case every face has lengthtwo, and by the argument in the previous paragraph, lifts of such faces may not be self-adjacent bythe fact that S has minimum size.

If S has only one involution, we may omit t2 and use the same arguments.Finally, if S has two involutions and no noninvolution, then B has only one possible embedding,

with only one face of size two, whose lifts, again, may not be self-adjacent. This completes theproof. �

Note that if X has minimum size, this does not mean that rep(A, X, ρ) > 1 for all ρ. For example,if A is abelian and |X | > 2, then there is always a ρ with rep(A, X, ρ) = 1, even if X has minimumsize for A: to force rep(A, X, ρ) = 1, one need only make the embedded voltage graph have a faceboundary with a commutator as a proper subword, and this can always be done if |X | > 2.


A generating set X for A is called irredundant if no proper subset of X generates A. Minimum sizegenerating sets are irredundant, but irredundant generating sets may fail to be of minimum size. Itis tempting to replace ‘‘minimum size’’ by ‘‘irredundant’’ in the statement of Corollary 5. We observethat this does not work.

Observation. If X is an irredundant generating set for A with no involutions, then R(A, X) may be 1.Consider, for example, A = Z14, X = {7, 4}. �

4. Generating sets of size two or three

In this section we consider R(A, X) when |X | = 2 or |X | = 3. The following theorem belongs morenaturally in Section 5, so we postpone the proof. However, it has a simple application to the case|X | = 2, so we state it here.

Theorem 6. Let S = {s1, . . . , sk} be a set of distinct positive integers such that k is even, n/2 ∈ S, and

ki=1

si < n.

Let xi represent the congruence class of integers (modulo n) containing si, and let X = {x1, . . . , xk}. If Xis a generating set for Zn then R(Zn, X) ≥ 2. �

Theorem 6 allows us to proceed to a conclusion in case |X | = 2:

Corollary 7. If |X | = 2, then R(A, X) = 1 iff either:

1. A = Zn, n ≡ 2(mod 4), and X = {s, t} where t = n/2 and (s, n) = 2; or2. A = Zn, n ≡ 0(mod 4), and X = {s, t} where t = n/2 and (s, n) = 1.

Proof. If X is of minimum size wemay use Corollary 5. If X is not of minimum size, then A is cyclic, soA = Zn for some n. In this case we may write X = {s1, s2}. |X | is even, and s1 + s2 < n (after replacingsi by its inverse, if necessary), so we may apply Theorem 6, provided X contains no involutions.

We are left with the case that A = Zn, where n is even, and X contains an involution. Let X = {s, t},where t = n/2. Only one rotation is possible, and the resulting embedded one-vertex voltage graphhas face boundary words s and s + t . By Corollary 3 the face with boundary word s does not haveself-adjacent lifts, so we need only check the face with boundary word s + t . We consider three cases.

First, suppose gcd(s, n) = 1 but gcd(s + t, n) = 1. In this case s + t does not generate s, thereforeby Lemma 4 the face bounded by s + t does not have self-adjacent lifts.

Next, suppose gcd(s, n) = 1 and also gcd(s + t, n) = 1. Since n is even, both s and s + t must beodd; therefore t is even. But t = n/2, so n ≡ 0(mod 4). Since gcd(s + t, n) = 1, there is a positiveinteger k < n such that k(s + t) = −s; therefore k(s + t) + s = 0, and by Lemma 4 the face boundedby s + t must have lifts that are self-adjacent, and R(A, X) = 1.

Finally, suppose gcd(s, n) = 1 but gcd(s + t, n) = 1. Again, since s + t generates Zn, the lifts ofthat face are self-adjacent and R(A, X) = 1. Now suppose p > 2 divides both s and n. Then p mustalso divide t = n/2, in which case it divides s + t , contradicting gcd(s + t, n) = 1. Therefore theonly common nontrivial divisor of s and n must be 2, and moreover t must be odd. Then we havegcd(s, n) = 2 and n ≡ 2(mod 4), and the proof is complete. �

We will now make a few comments on the case |X | = 3.

Lemma 8. If |X | = 3, no member of X is an involution, and R(A, X) = 1, then there exists a memberof X, say b, such that ⟨bx⟩ contains both b and x for all x in X such that x = b. (Consequently, b is in thecenter of A.)

Proof. Suppose not. Write X = {a, b, c}. Since a may not serve as the special element guaranteed bythe lemma, without loss of generality suppose that ⟨ab⟩ does not contain both a and b. But likewise


a b c d e

Fig. 6. Chord diagrams of the five embeddings on one vertex and three edges.

c may not serve as the special element, so assume ⟨bc⟩ does not contain both b and c. Now embed Bwith rotation abccba.

The faces of B have boundary words a, c, ab, bc . By Corollary 3, the lifts of faces with boundarywords a and c are not self-adjacent. By Lemma 4, if the lift of the face with boundary word ab is self-adjacent, then either ⟨ab⟩ contains a and b, or ⟨ba⟩ contains a and b. In either case a and b commute,so ⟨ab⟩ contains both a and b, contrary to our assumption. The same argument may be made for theboundary word bc.

Thus no face has lifts that are self-adjacent, a contradiction. �

Next we will present two examples of A and X where R(A, X) = 1. Up to isomorphism, there arefive embeddings of a one-vertex graphwith three loops. These five embeddings are shown in Fig. 6. Forthe sake of readability, are depicted in an ‘‘inverted’’ sense, so that the circle represents the perimeterof the vertex v, and the loops are drawn on the inside instead of the outside (see Fig. 8). (Such a figureis sometimes called a ‘‘chord diagram’’.)

Example 9. Let A = Z11 = ⟨a|a11 = 1⟩. Let X = {a, a2, a4}. Consider the possible words of lengththree whichmay be built from pairwise distinct members of X ∪X−1. Up to cyclic shifts and reversals,they are as follows: aa2a4, aa4a2, aa−2a4, aa4a−2, aa2a−4, aa−4a2, aa−2a−4, and aa−4a−2. These wordsare equivalent to a7, a7, a3, a3, a10, a10, a6, and a6, respectively, each of which generates A. Thus byLemma 4, if there is any face F of size three in the one-vertex voltage graph corresponding to a Cayleymap on A, X , then the lifts of F are self-adjacent. This eliminates Fig. 6(a) and (c).

Now consider the possiblewords of length twowhichmay be built frompairwise distinctmembersof X ∪ X−1. Up to cyclic shifts and reversals, they are as follows: aa2, aa−2, aa4, aa−4, a2a4, and a2a−4.These words are equivalent to a3, a10, a5, a8, a6, and a9, respectively. Again each generates A, and byLemma 4 if there is any face F of size two in the one-vertex voltage graph corresponding to a Cayleymap on A, X , then the lifts of F are self-adjacent. This eliminates Fig. 6(b) and (d).

Finally, notice that Fig. 6(e) contains a face whose boundary already has a proper subwordequivalent to 1 (namely, it contains a proper subword which amounts to a commutator), so that nomatter the labeling, the lifts of this face will be self-adjacent. Thus in any case, rep(A, X, ρ) = 1 andtherefore R(A, X) = 1. �

Example 9 suggests the following fact.

Proposition 10. Suppose A = Zn = ⟨a|an = 1⟩ and X = {ax, ay, az}. Suppose further that all of thefollowing conditions hold:

(1) gcd(x + y, n) | x;(2) gcd(x + z, n) | x;(3) gcd(y + z, n) | y;(4) gcd(x + y + z, n) is a divisor of one of x, y, or z;(5) gcd(x + y − z, n) is a divisor of one of x, y, or z;(6) gcd(x − y + z, n) is a divisor of one of x, y, or z; and(7) gcd(x − y − z, n) is a divisor of one of x, y, or z.

Then R(A, X) = 1.

Proof. Assume the embedded one-vertex voltage graph has a face of size three. Then the boundaryword W of this face is of the form axa±ya±z . The subgroup of A generated by W is thus isomorphic to


ab, where b = gcd(x ± y ± z, n); but this group contains one of ax, ay, or az by hypothesis, and so thelifts of this face are self-adjacent by Lemma 4. This eliminates the embeddings in Fig. 6(a) and (c).

Now assume the Cayley map contains a face of size two, and suppose its boundary word W is ofthe form axa±y. IfW = axay, thenW generates ax by condition (1); then by Lemma 4, the lifts ofW areself-adjacent. Thus we may assume W = axa−y. Let d = gcd(x + y, n). Then x + y = dm1; moreoverby condition (1), x = dm2 for some integer m2. Thus we have x + y = dm1 = dm2 + y, whence y isalso a multiple of d, say y = dm3. But in that case, x− y is divisible by d, so gcd(x− y, n) | x. So again,in this case, the lifts ofW are self-adjacent by Lemma 4.

The case thatW is of the form ax±z or ay±z are similar. This eliminates the embeddings in Fig. 6(b)and (d).

Finally, any labeling of the embedding in Fig. 6(e) contains a face of the form sts−1t−1r , which, inan Abelian group, generates r . Thus by Lemma 4, Fig. 6(e) is eliminated. �

While the above conditions are sufficient for R(A, X) = 1 when A is cyclic and |X | = 3, theyare not necessary. The following result provides a slightly different set of conditions guaranteeingR(A, X) = 1, when A is cyclic and |X | = 3. The proof is similar to the proof of Proposition 10, and isomitted.

Proposition 11. Suppose A = Zn = ⟨a|an = 1⟩ and X = {ax, ay, az}. Suppose further that all of thefollowing conditions hold:

(1′) gcd(x + y, n) | z;(2′) gcd(x + z, n) | x;(3′) gcd(y + z, n) | y;(4′) gcd(x + y + z, n) is a divisor of one of x, y, or z;(5′) gcd(x + y − z, n) is a divisor of one of x, y, or z;(6′) gcd(x − y + z, n) is a divisor of one of x, y, or z; and(7′) gcd(x − y − z, n) is a divisor of one of x, y, or z.

Then R(A, X) = 1. �

In fact, these two propositions describe the only ways that R(A, X) may equal 1 when A is cyclicand |X | = 3:

Theorem 12. Suppose that A = Zn = ⟨a|an = 1⟩, and suppose that X = {ax, ay, az}. Then R(A, X) = 1if and only if x, y, z, and n satisfy either (1)–(7) from Proposition 10 or (1′)–(7′) from Proposition 11.

Proof. One direction follows from Propositions 10 and 11. For the other, suppose that R(A, X) = 1. ByLemma 8, if X has no involutions then the word aiaj must generate both ai and aj, for two of the threepossible pairs of distinct elements ai, aj from X . Without loss of generality suppose that {ax, az} and{ay, az} are the pairs satisfying this property. But if aiaj = ai+j generates ai, then certainly gcd(i+ j, n)is a divisor of i, and therefore j. Thus conditions (2)–(3) and (2′)–(3′) are satisfied.

Conditions (4)–(7) and (4′)–(7′) are satisfied by noting that all possible labelings of Fig. 6(a) musthave a face whose lifts are self-adjacent; by Lemma 4 this means that each wordW of the form aiajak,where ai, aj, ak are pairwise distinct elements of X ∪X−1, must always generate one of its own propersubwords.

Finally, if Fig. 6(d) is to have a face whose lifts are self-adjacent, and the third pair of elements{ax, ay} does not satisfy property (1), then labeling Fig. 6(d) so that the face F of size two has boundaryword axay guarantees that the lifts of F are not self-adjacent. Therefore the other face F ′ of Fig. 6(d)must have self-adjacent lifts, and therefore by Lemma 4 the word axazaya−z must generate one of itsown proper subwords: up to isomorphism either ax, or axaz , or axazay. If axazaya−z

= axay generatesaxazay, then it also generates az and thus condition (1′) is satisfied; in this case the theorem is proven.If axay generates ax, then condition (1) is satisfied, a contradiction. Thuswemay assume axay generatesaxaz . But axaz generates az by Conditions (2)–(3) or (2′)–(3′), which are known to be true; thus axayalso generates az , satisfying Condition (1′).

This completes the proof of the theorem. �


In the following example, we consider an Abelian group which is not cyclic.

Example 13. Let A = Z45⊕Z15, and let X = {(0, 1), (15, 0), (1, −1)}. The possible three-letterwordshave sums (16, 0), (16, 2), (31, 0), and (29, 2). Each of these generates (15, 0), so by Lemma 4, if thereis any face F of size three in the one-vertex voltage graph corresponding to a Cayley map on A, X , thenthe lifts of F are self-adjacent. This eliminates Fig. 6(a) and (c).

If we let a = (0, 1), b = (15, 0), and c = (1, −1), then a + b = (15, 1) generates a and b;a − b = (30, 1) generates a and −b; b + c = (16, −1) generates b and c; and b − c = (14, 1)generates b and −c. This eliminates Fig. 6(b).

Now a + c = (1, 0) clearly does not generate either a or c . Thus we may attempt to useFig. 6(d), labeled so that it has a 2-face with boundary word a + c and a 4-face with boundary worda+ b+ c − b. The 2-face is guaranteed to have lifts which are not self-adjacent. However, we observethat a + b + c − b = a + c = (1, 0) generates b, so the lifts of this face are self-adjacent. Likewise,a − c = (−1, 2) generates b, so Fig. 6(d) is eliminated.

As in the previous example, since A is Abelian, any labeling of Fig. 6(e) contains a face whose liftsare self-adjacent. Thus R(A, X) = 1. �

Again this example suggests a general theorem for Abelian, noncyclic A and |X | = 3.

Theorem 14. Let A = Zr ⊕ Zs and let X = {(0, 1), (p, 0), (1, −1)} be a generating set for A. ThenR(A, X) = 1 if the following conditions on p, r, s, and n are satisfied:

(1) gcd(p + 1, r) = 1;(2) gcd(s(p − 1), r) | p;(3) gcd(p − 1, r) = 1;(4) gcd(sp, r) | p;(5) gcd(s(p + 1), r) | p; and(6) s | p.

Proof. First name a = (0, 1), b = (p, 0), and c = (1, −1), and assume that conditions (1)–(4) aresatisfied. Condition (1) guarantees that a + b + c = (p + 1, 0) generates b. Condition (2) guaranteesthat s(a + b − c) = s(p − 1, 2) = (s(p − 1), 0) generates b, hence a + b − c does. Condition (3)guarantees that a − b + c = (1 − p, 0) generates b (hence also −b). Finally, condition (5) guaranteesthat s(a − b − c) = s(−p − 1, 2) = −(s(p + 1), 0) generates b (hence also −b), whence a − b − cdoes also. Up to cyclic shifts and reversals, these are the only words of length three that can be formedfrom three pairwise distinct elements of X . Therefore if R(A, X) is to be greater than 1, the rotationused may not contain a face of length three. This eliminates Fig. 6(a) and (c).

Now we consider faces of size two. s(a + b) = s(p, 1) = (sp, 0) generates b by condition (4),whence also a + b does. Likewise s(a − b) = s(−p, 1) = −(sp, 0) generates b by condition (4), soa − b does. s(b + c) = s(p + 1, −1) = (s(p + 1), 0) generates b by condition (5), so b + c generatesb. Finally, s(b − c) = s(p − 1, 1) = (s(p − 1), 0) generates b by condition (2). Thus the only size-twowords that may bound a face in a representativity-2 Cayley map CM(A, X, ρ) are a+ c and a− c . Thiseliminates Fig. 6(b).

The four-face in Fig. 6(d) has one of the boundarywords a+b+c−b = a+c , a+b−c−b = a−c ,a − b + c + b = a + c , or a − b − c + b = a − c. Now a + c = (1, 0) certainly generates b (andtherefore also −b). But s(a− c) = s(−1, 2) = −(s, 0) generates b by condition (6), so a− c generatesb (and hence also −b). Therefore Fig. 6(d) is eliminated. But as observed in the preceding examples,the commutativity of A guarantees that Fig. 6(e) will have a face whose lifts are self-adjacent. ThusR(A, X) = 1. �

5. Representativity two for cyclic Cayley maps

We have seen that it is possible for R(Zn, X) = 1, for certain generating sets X of size three. In thissection, we give some sufficient conditions on X that guarantee R(Zn, X) > 1. We also show that ifwe only specify n and the size s = |X |, we can always find a generating set X such that R(Zn, X) > 1.

We begin by proving Theorem 6, which appeared in Section 4.


Fig. 7. A one-face embedding of a one-vertex graph.

Fig. 8. Chord diagram corresponding to Fig. 7.

Theorem 6. Let S = {s1, . . . , sk} be a set of distinct positive integers such that k is even, n/2 ∈ S, and

ki=1

si < n. (1)

Let xi represent the congruence class of integers (modulo n) containing si, and let X = {x1, . . . , xk}. If Xis a generating set for Zn then R(Zn, X) ≥ 2.

Proof. Order the si so that s1 < s2 < · · · < sk.Let B be the one-vertex voltage graph with E(B) = X . We first claim that B can be drawn so that

the embedding has exactly one face, with facial walk

W = x1x3x5 · · · xk−1xkxk−2 · · · x4x2x1−1x3−1· · · xk−1

−1xk−1xk−2−1

· · · x4−1x2−1.

To see this, consider a one-vertex graph Gwith vertex v and 2m edges e1, . . . , e2m. Draw G so thatthe clockwise rotation is

e1e2e3 · · · e2m−1e2me1e2e3 · · · e2m−1e2m,

as shown in Fig. 7.Now we assign directions to the edges as follows. Orient e1 arbitrarily, and consider the segment

P of the perimeter which runs clockwise from the head of e1 to its tail, as in Fig. 9. If i is even, orient eiso that its tail appears on P; if i is odd, orient ei so that its head appears on P . Now consider the faceF of our embedding of G which contains the positive orientation of ei, in the clockwise sense, wherei < 2m. The successor of ei in F is ei+1. Likewise, if F ′ is the face containing e−1

i (where again i < 2m),then the successor of e−1

i on F ′ is e−1i+1. Moreover, the successor of e2m in face-traversal is e−1

1 , and thesuccessor of e−1

2m is e1. Thus our drawing of G has only one face, with boundary walk

e1e2e3 · · · e2m−1e2me−11 e−1

2 e−13 · · · e−1

2m−1e−12m.


Fig. 9. Orienting the edges ei .

Now we apply this drawing to the graph B, with 2m = k; ei = x2i−1 when i ≤ k/2; andei = x2(k−i−1) when i > k/2. This yields the desired embedding having one face with boundary word

W = x1x3x5 · · · xk−1xkxk−2 · · · x4x2x−11 x−1

3 · · · x−1k−1x

−1k x−1

k−2 · · · x−14 x−1

2 .

Now W is equivalent to the identity in A, because A is abelian and W contains each element of Xand its inverse exactly once each. Therefore it only remains to show that no proper subword of anycyclic shift of W is equivalent to the identity in A.

Suppose to the contrary that some cyclic shift W ′ of W has initial segment Y which is equivalentto the identity in A. Write W ′

= YZ , and observe that if Y is equivalent to the identity in A then so isZ . Therefore (by swapping Y and Z if necessary) we may assume the length of Y is at most k. Now ifY contains only elements of the form xi, then by (1) it follows that Y is not equivalent to the identity.Similarly, if Y contains only elements of the form x−1

i then Y is not equivalent to the identity. Thus wemay assume that Y is either of the form

xi · · · x4x2x−11 x−1

3 · · · x−1j

or

x−1i · · · x−1

4 x−12 x1x3 · · · xj.

By symmetry we need only consider one of the two cases, so we may assume

Y = xi · · · x4x2x−11 x−1

3 · · · x−1j .

We shall now think of A as an additive group, and for any finitewordD = xϵ1i1xϵ2i2

· · · xϵmim on X , where

ϵl ∈ {−1, 1} for 1 ≤ l ≤ m, we set

σD = ϵ1si1 + ϵ2si2 + · · · + ϵmsim .

Observe that D is equivalent to the identity in A if and only if σD ≡ 0(mod n).Let Yp,q be the subword ofW given by

Yp,q = x2p · · · x4x2x−11 x−1

3 · · · x−12q−1,

where 2p ≤ k and 2q + 1 ≤ k − 1. Then

σYp,q = s2p + · · · + s4 + s2 − s1 − s3 − · · · − s2q−1.

By (1) and the ordering of the si, we have immediately that 0 < σp,q < n whenever p ≥ q, and−n < σp,q < 0 whenever p < q.

Consequently, we have that if Y is of the form Y = Yp,q, then Y is not equivalent to the identityin A. But the length of Y is at most k, so that Y is forced to have either an initial segment of the formYp,p+1 or a terminal segment of the form Yp,p. In the former case 0 < σY < n, and in the latter case−n < σY < 0. Thus there is no proper segment Y of any cyclic shift of W which is equivalent to theidentity in A, and consequently R(Zn, X) ≥ 2. �


Theorem 15. Let S = {s1, . . . , sk} be a set of distinct positive integers with

ki=1

si = n. (2)

Let xi represent the congruence class of integers (modulo n) containing si, and let X = {x1, . . . , xk}. If Xis a generating set for Zn then R(Zn, X) ≥ 2.

Proof. Letρ be the rotation x1, −x1, x2, −x2, . . . , xk, −xk. The only facewith boundaryword of lengthgreater than one has word s1 + · · · + sk, which is 0 mod n, and which clearly has no proper subwordwith that property. �

Theorem 16. For any n ≥ 4 and for any 3 ≤ s ≤ n/2, there is a generating set X for Zn with |X | = s,such that R(Zn, X) ≥ 2.

Proof. Let k be the largest integer such that 1 + 2 + · · · + k = k(k + 1)/2 ≤ n.Now we may assume s > k. Indeed, if s ≤ k, then we may define t = n − [1 + 2 + · · · + (s − 1)],

and take S = {1, 2, . . . , s − 1, t}; by Theorem 15 we are done.Let r = s − k > 0. First suppose r is even. Begin by choosing m with k ≤ m < k + (k + 1) such

that 1 + 2 + · · · + (k − 1) + m = n. It is straightforward to check that m < n/2 when n ≥ 13 andwhen n = 11; m = n/2 when n = 6 and n = 12; and m > n/2 when n = 4, 5, 7, 8, 9, 12. Thereforewe define t = n − m when n = 4, 5, 7, 8, 9, 12, and t = m otherwise.

Let A = {k, k + 1, . . . , t − 1}, and let B = {t + 1, t + 2, . . . , ⌊n/2⌋}. If |A| and |B| areboth even, then we may obtain r/2 pairs of consecutive integers from A ∪ B. If |A| and |B| haveopposite parities, then r < |A| + |B|, since r is even. Therefore again we may obtain r/2 pairs ofconsecutive integers from A∪ B. In either case, we let X0 = {1, 2, . . . , k − 1, t}, and form the rotation1, −1, 2, −2, . . . , k − 1, −k − 1,m, −m, as in Theorem 15. Next we write r/2 = i1 + i2 + · · · + il,where ij = ij′ if j = j′ and each ij ∈ {1, 2, . . . , k−1, t}. Thenweembed i1 of our r/2pairs of consecutiveintegers sequentially inside the i1 loop in such a way that the common face boundary sums to zero;i2 pairs inside the i2 loop; and so on. Now we have formed a rotation with s loops, labeled with sdistinct residues of Zn, such that every face boundary sums to 0 or n, and no face boundary has aproper subword summing to 0 or n, as desired.

Now suppose both |A| and |B| are odd. If t < n/2, define t ′ = t + 1. If t = n/2, define t ′ = t − 1.Let X0 = {2, . . . , k − 1, t ′}, and form the rotation 2, −2, 3, −3, . . . , k − 1, −k − 1, t ′, −t ′. Now welet A′

= {k, k+ 1, . . . , t ′ − 1}, and let B′= {t ′ + 1, t ′ + 2, . . . , ⌊n/2⌋}. Since |A| and |B| were both odd

and we have shifted both by one, we have that |A′| and |B′

| are both even. Thus we may find r/2 pairsof consecutive integers from A ∪ B. We embed the corresponding residues, along with the element 1,inside the loops labeled with elements of X0 as in the preceding paragraph.

The proof is similar in the case that r = s − k is odd. Begin by choosing m with k − 1 ≤ m <k − 1 + k + (k + 1) such that 1 + 2 + · · · + (k − 2) + m = n. It is straightforward to check thatm < n/2 when n ≥ 45. We define t = n − mwhen m > n/2, and t = m otherwise.

Let A = {k − 1, k, k + 1, . . . , t − 1}, and let B = {t + 1, t + 2, . . . , ⌊n/2⌋}, and proceed preciselyas above.

The only problem in this case is that k− 2 might be 1, in which casem cannot be used. This occursin two cases: when n = 8 and s = 4, and when n = 9 and s = 4. In the first case, we may useX = {2, 3, 4, 7}, with a straightforward rotation. In the second case, wemay use X = {2, 3, 5, 8}. �

In case the order of the group is prime and congruent to 1 (mod 4), we may find a very niceembedding using all possible generators. Namely, the base embedding has only one face as well asonly one vertex.

Theorem 17. If A = Zp (p prime) and X = {1, 2, . . . , (p − 1)/2} (so X is a generating set for A ofmaximum possible size), and if p = 1(mod 4), then R(A, X) ≥ 2; moreover, we may embed the one-vertex voltage graph whose edges correspond to X so that the embedding has only one face.


Proof. By the previous theorem, we already know that R(A, X) ≥ 2.Since p = 1(mod 4), (p− 1)/2 is even. Therefore there exists an embedding of B (namely, the one

discussed in the proof of Theorem 6) such that

(i) B has exactly one face F , and(ii) Any generator x always appears exactly ‘‘halfway’’ around F from its inverse x−1.

Since p is prime, it has a primitive root, say x. Let k = (p − 1)/2. Since x is a primitive root, we havexk = p − 1(mod p). Thus xi+k

= xi(p − 1) = −xi(mod p). Now label the edges of B so that theboundary walkW of F is

W = x, x2, x3, . . . , xk, xk+1, . . . , x2k.

This makes sense with respect to property (ii) of our embedding, because as we noted above, xi+k=

−xi in Zp. Now, since x is a primitive root of p, the net voltage of F is

x + x2 + · · · + xp−1= (the sum of all distinct moduli of p)= 0(mod p),

so the lifts of F have size p − 1.On the other hand, if the lift is self-adjacent, then there are some i, j, where i > 0, j < p − 2 such

that

xi + xi+1+ · · · + xi+j

= 0(mod p). (3)

Factoring out xi, we have,

xi(1 + x + · · · + xj) = 0(mod p)

⇒ x + · · · + xj = −1(mod p). (4)

On the other hand, factoring out xi−1 from (3), we have

xi−1(x + · · · + xj+1) = 0(mod p)

=> x + · · · xj+1= 0(mod p). (5)

Subtracting (4) from (5), we get

xj+1= 1(mod p), ⇒ xj+2

= 0(mod p)⇒ j + 2 = p

contradicting our assumption that j < p − 2. �

6. Representativity without lifting

In this section we will provide a method of calculating the representativity of a Cayley mapCM(A, Xρ) purely combinatorially; i.e., without appealing to the surface of the embedding itself.

Let A = ⟨X |RA⟩ be a presentation for A, and let CM(A, X, ρ) be a Cayley map. Define

RF = {W (ordAW )|W is the boundary word for a facial walk of CM(A, X, ρ)}.

Now consider the group Γ (A, X, ρ) = ⟨X |RF ⟩.

Proposition 18. If a word W on X is equivalent to 1 in Γ (A, X, ρ), then it is equivalent to 1 in A.

Proof. The boundary word for any face in CM(A, X, ρ) must be equal to 1 in A. Consequently, allrelators in RF are equivalent to 1 in A and the result follows immediately. �

Thus A is a homomorphic image of Γ (A, X, ρ) under the obvious homomorphism which sendsRA − RF to the identity.


It is straightforward to show that the underlying surface of the (possibly infinite) Cayley mapCM(Γ (A, X, ρ), X, ρ) is simply connected. Indeed, this fact is nearly tautological basedon the Poincarémethod for computing a presentation for the fundamental group of a complex (cf. [7,4,9]). Thus thehomomorphism from Γ (A, X, ρ) to Awhich sends RA \RF to the identity induces a universal coveringmap from CM(Γ (A, X, ρ), X, ρ) to CM(A, X, ρ).

The following fact is immediate:

Proposition 19. If a cycle C in CM(A, X, ρ) has boundary wordW, then C is contractible if and only if Wis equivalent to 1 in Γ (A, X, ρ).

A word W in X is a cycle in CM(A, X, ρ) if and only if it is equivalent to 1 in A; by Proposition 19,it is noncontractible in CM(A, X, ρ) if and only if it is not equivalent to 1 in Γ (A, X, ρ). Therefore wehave the following:

Fact 20. A word W on X represents a noncontractible cycle of edges in CM(A, X, ρ) if and only if Wis equivalent to 1 in A, but not equivalent to 1 in Γ (A, X, ρ). �

We are now prepared to introduce our method of calculating representativity.Let S(RF ) be the set containing all cyclic shifts of elements of RF . Given a word W on X which is

equivalent to 1 in A, say that W = Y1Y2 · · · Yk is a k-face decomposition of W if each Yi is a subwordof some element of S(RF ). Define the jump number jump(W ) of a word W to be the minimum k suchthatW has a k-face decomposition. Then let

jump(A, X, ρ) = minW

jump(W ),

where theminimum is to be taken over all wordsW that are equivalent to 1 in A but not inΓ (A, X, ρ).

Theorem 21. jump(A, X, ρ) = rep(A, X, ρ).

Proof. Each wordW which is equivalent to 1 in A but not in Γ (A, X, ρ) represents a noncontractiblecycle. A k-face decomposition of W is simply a method for traversing the edges indicated by W inthe interior of the faces of the Cayley map CM(A, X, ρ). That is, let W = Y1Y2 · · · Yk be a k-facedecomposition of W . Since each Yi is a subword of a cyclic shift of the boundary word of a face Fi,we traverse the edges indicated by Yi in the interior of Fi, then we cross a vertex and traverse Yi+1 inthe interior of Fi+1, and so on. The jump number ofW , then, gives away to traverse the edges indicatedby W with the minimal number of such vertex-crossings.

The representativity of an embedding is the minimum number of intersections of a cycle withthe graph, taken over all noncontractible surface cycles; the jump number is the minimum numberof intersections of a cycle with the graph, taken over a special class of surface cycles (namely, thosedescribed by k-face decompositions of qualifying words W ). Therefore jump(A, X, ρ) takes the sameminimum as rep(A, X, ρ), over a smaller set of cycles, and we have rep(A, X, ρ) ≤ jump(A, X, ρ).

On the other hand, it is easy to see that a noncontractible surface cycle C with minimum numberof intersections with an embedded graph G may be homotopically shifted, without changing thecardinality of C ∩G, so that intersections of C with G occur at vertices of G. Furthermore, if a (non-self-intersecting) surface path P has its endpoints at vertices ofG, and otherwise lies entirely in the interiorof some face F , then P is homotopic to a path P ′ having the same endpoints as P , but traversing theinterior of F along the boundary of F . Therefore in calculating the representativity of an embedding ofG, it suffices to consider noncontractible walks along the sides of edges of G, which only intersect G atvertices. In other words, rep(A, X, ρ) ≥ jump(A, X, ρ). �

The group Γ (A, X, ρ) is a familiar object for Riemann surfaces. It is a presentation for acrystallographic group, either spherical, Euclidean, or non-Euclidean [1]. Such groups are residuallyfinite [5] in the non-spherical case: given any finite collection S of elements of Γ (A, X, ρ), there is afinite quotient group B of Γ (A, X, ρ) such that the image of each element of S is not the identity.

This has interesting consequences for representativity. If one takes for S all words of length atmost rf , where f is the maximum face size for CM(Γ (A, X, ρ), X, ρ), then the resulting Cayley map


CM(B, X, ρ) has representativity at least r . Thus low representativity of a Cayley map CM(A, X, ρ) isas much the fault of the relations RA − RF as it is the choice of rotation ρ.

To be more precise, given a Cayley map CM(A, X, ρ), define the quotient data to be the abstractpermutation ρ together with the orders of the net voltages on the faces of one-vertex graph with |X |

loops, embedded with rotation ρ. The quotient data gives the relations RF defining Γ (A, X, ρ). Callthe data non-spherical if Γ (A, X, ρ) is infinite. For example, if the embedding of a one-vertex graph ρis in a surface other than the sphere or the projective plane, then the quotient data is non-spherical.

Theorem 22. Given CM(A, X, ρ)with non-spherical quotient data and given any positive integer r, thereis a finite Cayley map having the same quotient data as CM(A, X, ρ) and representativity at least r. �

Corollary 23. Given any map M on a surface other than the sphere or projective plane and given anypositive integer r, there is a finite, unbranched, regular covering of M having representativity at least r.

Proof. Choose a spanning tree T for G and contract it to a single vertex v in the map M to obtain anembedding of a bouquet of circles. Use this embedding to give non-spherical quotient data by requir-ing all net voltages on faces to be the identity (order 1). Then by Theorem 22 there is a finite regularcovering with representativity at least r . For this regular covering, one can expand each vertex lyingabove v to a copy of T and thus obtain a regular covering of the map M . The covering is unbranchedsince all net voltages on faces have order one. �

Example 24. To raise representativity from one to three or more may take several sheets. To see this,let n be odd and consider the Cayleymap for Z2n with generating set {x, y}where x ≡ 2 (mod 2n) andy = n. (Note that there is only one possible rotation since y is an involution, and hence correspondsto a semi-edge.) Then there are two faces of size n (bounded by xn) and one, F , of size 4n (bounded byxy). Now F has n self-adjacencies along the n edges labeled y. If we want representativity greater thantwo, we need any pair of faces in the covering lying over F to be not doubly adjacent. Since there are nself-adjacencies on the face F , we will need at least n copies of F in the covering (given a lift F ′, as onegoes around the face, one should never encounter the same lift of F twice as a neighbor of F ′ along ay edge). So here representativity three can be achieved only by a cover of at least n sheets, whereas acover of only two sheets will do to achieve representativity two. �

Corollary 25. Every nonplanar graph G has a regular covering satisfying the Orientable Strong EmbeddingConjecture.

Proof. If G is not planar, it has some embedding in a surface other than the projective plane, and wemay use Corollary 23. �

7. Conclusion

Ideally, one would like to bound the number of sheets in the covering of the above two corollaries.For example, we might ask:

Question 26. Does every nonplanar graph have a 2-sheeted covering graph satisfying the StrongOrientable Embedding Conjecture? �

It is perhaps surprising that such a large class of highly symmetric Cayley graphs of Abelian (evencyclic) groups have no Cayley map with representativity more than one. It would be interesting toknow if the symmetry of such graphs can be exploited at all in the context of strong embeddings:

Problem 27. Find strong embeddings for the Cayley graphs (as in, for instance, Propositions 10, 11,and Theorem 14) which do not have 2-representative Cayley maps.


Acknowledgment

The first author’s research was supported in part by NSA grant H98230-07-1-0006. The thirdauthor’s research was supported in part by NSA grant H98230-09-1-0041.

References

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Math. 27 (1985), 1–12.[4] H. Poincaré, Analysis situs, J. de l’ École Polyt. 1 (2) (1895) 1–123.[5] R.B. Richter, J. Širáň, R. Jaycay, T.W. Tucker, M.E. Watkins, Cayley maps, J. Combin. Theory Ser. B 95 (2005) 189–245.[6] P.D. Seymour, Sums of circuits, in: J.A. Bondy, U.S.R. Murty (Eds.), Graph Theory and Related Topics, Academic Press, New

York, 1979, pp. 341–355.[7] J. Stillwell, Classical Topology and Combinatorial Group Theory, second ed., Springer, 1993.[8] G. Szekeres, Polyhedral decompositions of cubic graphs, Bull. Aust. Math. Soc. 8 (1973) 367–387.[9] H. Tietze, Über die topologischen Invarianten mehrdimensionaler Mannigfaltigkeiten, Monatsh. Math. Phys. 19 (1908).

[10] C.-Q. Zhang, Integer Flows and Cycle Covers of Graphs, Marcel Dekker, New York, 1996.

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