On Embeddings of Circulant Graphs

A circulant of order n is a Cayley graph for the cyclic group Z n , and as such, admits a transitive action of Z n on its vertices. This paper concerns 2-cell embeddings of connected circulants on closed orientable surfaces. Embeddings on the sphere (the planar case) were classified by Heuberger (2003), and by a theorem of Thomassen (1991), there are only finitely many vertex-transitive graphs with minimum genus g, for any given integer g 3. Here we completely determine all connected circulants with minimum genus 1 or 2; this corrects and extends an attempted classification of all toroidal circulants by Costa, Strapasson, Alves and Carlos (2010).


Introduction
A circulant is a Cayley graph for a cyclic group.To be more precise, if n is any positive integer and X is any subset of Z n \ {0} = {1, 2, . . ., n − 1}, then the circulant C n (X) is the undirected simple graph of order n with vertex-set Z n = {0, . . ., n − 1} and edge-set {{i, i + a} : i ∈ Z n , a ∈ X}.Under addition mod n, the group Z n induces a regular group of automorphisms of C n (X); for example, the mapping i → i + 1 (mod n) gives an automorphism that permutes the n vertices of C n (X) in a cycle: (0, 1, 2, . . ., n − 1).Hence in particular, every circulant C n (X) is vertex-transitive, and therefore regular, of Examples include the 2-valent simple cycle C n (with X = {1}), and the (n−1)-valent complete graph K n (with X = Z n \ {0}).Note that the valency k can be even or odd, but is odd if and only if n is even and X contains n 2 .If X = {a 1 , a 2 , . . ., a m }, then we also denote C n (X) by C n (a 1 , a 2 , . . ., a m ).By the above observations, we may assume that 0 < a 1 < a It is well known (and easy to see) that the circulant C n (a 1 , . . ., a m ) is connected if and only if gcd(a 1 , . . ., a m , n) = 1.More generally, the number of connected components of the electronic journal of combinatorics 22(2) (2015), #P2.28 C n (a 1 , . . ., a m ) is d = gcd(a 1 , . . ., a m , n), with each of the vertices 0, 1, . . ., d − 1 lying in different components, and with each component being isomorphic to C n/d (a 1 /d, . . ., a m /d).
Various properties of circulants (and their adjacency matrices, which are circulant matrices) are well known, for example with regard to chromatic number, connectivity, domination properties, factorisations, metric dimension, self-complementarity, and symmetry.
Also a number of things are known about the embeddability of connected circulants on surfaces.All of them are upper-embeddable, which means that they all have an embedding on an orientable surface (of maximum possible genus), with just one or two faces.This is trivial for valency 2 (simple cycles), and for valency greater than 3 it follows from a more general theorem of Škoviera and Nedela about upper-embeddability of connected finite Cayley graphs; see [11,Proposition 7].For valency 3, it follows from Theorem 5 in [11]), since up to isomorphism the only 3-valent connected circulants of girth 3 are C 4 (1, 2) ∼ = K 4 and the triangular prism graph C 6 (2, 3), both of which are upper-embeddable.
The most common question about embeddability is the minimum genus of a graph: the smallest genus of all the orientable surfaces on which the graph has a 2-cell embedding.Circulant graphs that have minimum genus 0 (or equivalently, are planar) were completely classified in 2003 by Heuberger [7]: Proposition 1.Let C n (X) = C n (a 1 , a 2 , . . ., a m ) be a connected circulant with 0 < a 1 < a 2 < • • • < a m n 2 .Then C n (X) is planar if and only if one of the following holds: • m = 1 and a 1 is a unit in Z n , or • m = 2 and n ≡ 2 mod 4 and a 1 is even and a 2 = n 2 , or • m = 2 and n is even and either a 2 = ±2a 1 or a 1 = ±2a 2 in Z n .Equivalently, every planar connected circulant graph is isomorphic to either the simple n-cycle C n (1), or the n-prism C n (2, n 2 ) where n ≡ 2 mod 4, or C n (1,2) where n is even.In a major piece of work [14], Thomassen proved a conjecture of Babai, that for every g 3, there are only finitely many vertex-transitive graphs with minimum genus g.
Recently Strapasson, Costa and Alves [12] announced a classification of circulants embeddable on the torus, but unfortunately they made some errors and omitted many cases.
In Section 2 we give some further background, and in particular, we show that if a connected circulant has minimum genus 1 or 2, then its valency must be 3, 4, 5, 6 or 7.In Section 3 we exhibit embeddings of genus 1 and 2 for the circulants listed in Theorems 2 and 3, and then to establish those theorems, we consider circulants with valencies 7, 5 and 6 in Sections 4, 5 and 6, respectively.Some of the more challenging cases are left to Section 7, where we give explanations of how they can be treated (using a range of approaches), rather than giving full details.

Further background
A 2-cell embedding of a connected graph G on an orientable surface S is a representation (or drawing) of G on S with the property that when the graph is removed from the surface, it breaks it up into simply-connected regions (homeomorphic to disks), called faces.
This creates a map, and if we denote the number of faces, edges and vertices of the map by F , E, and V respectively, then by the well known Euler-Poincaré formula we have where χ is the Euler characteristic of the surface S. If g = 0 (and χ = 2) then the map is called planar or spherical, while if g = 1 (and χ = 0) then it is Euclidean or toroidal, and if g > 1 (and χ < 0) then the map is hyperbolic.
A connected graph G can have several different embeddings, and the genus g of each one is determined by the number of faces, since the numbers of vertices and edges are fixed (for given G).The smallest and largest achievable values of g are called the minimum genus (or simply the genus) of G and the maximum genus of G, respectively.The minimum genus γ(G) occurs when the number of faces is maximised.
Before continuing, we will adapt notation a little, and define V (G) and E(G) to be the vertex-set and edge-set of G, and V (M ), E(M ) and F (M ) to be the vertex-set, edge-set and face-set of the map M resulting from an embedding of G. (Note that V (M ) = V (G) and E(M ) = E(G) for all such M , but F (M ) is variable.) As explained in [2] and [6] for example, every embedding of a graph G on an orientable surface is uniquely determined by the cyclic orientations of the edges at the vertices of G.
To make things easier, we will assume the graph G is simple.
Then for each vertex v the embedding defines a cyclic permutation ρ v of the neighbours of v, which indicates (by convention) the anti-clockwise ordering of the edges incident with v when the map is viewed from the 'outside' of the surface.The set {ρ v : v ∈ V (G)} is then called a rotation system.Conversely, any such set of cyclic permutations gives rise to a unique embedding, and so there is a one-to-one correspondence between the orientable 2-cell embeddings of a simple graph G and its rotation systems.
Moreover, the rotation system for an embedding can be used to calculate the number of faces, using the well known (and obvious) face tracing algorithm: from any vertex u, take an edge to one of its neighbours v, then 'turn left' at v (using the rotation ρ v ) and continue to the next vertex, and so on.This process traces out a closed walk beginning with the arc (u, v), and by finiteness must return at some stage to the vertex u along the arc (ρ −1 u (v), u).Each such closed walk traces a face of the map, and all faces can be found by successively choosing the initial arc (u, v) from among those not already used.
(Alternatively, one may 'turn right' at every vertex, instead of turning left, and in that case the same faces will be found, but with each face traced in reverse order.) Note that if the vertex v has valency k, then there are (k − 1)! possibilities for the rotation ρ v , and hence if G has order n and is regular of valency k, then the number of possible rotation systems is ((k−1)!) n .For small n (and k), these can be enumerated by computer, but for large k that becomes infeasible.Accordingly, determining the minimum genus of a given graph G is computationally challenging.
In fact, the question of deciding whether a given connected graph G has minimum genus γ(G) g (for given g) has received considerable attention.This is known as the graph genus problem, and it is computationally challenging.In [5], Filotti, Miller and Reif described an algorithm for finding an embedding of a given connected graph of order n on an orientable surface of genus g, when such an embedding exists.This algorithm runs in n O(g) steps.But then it was shown by Thomassen [13] that the graph genus problem is NP-complete.More generally, finding the minimum genus is NP-hard.
Nevertheless, it is possible to determine the minimum genus of all graphs in particular families; for example, Ringel and Youngs showed that the minimum genus of the complete graph K n is (n−3)(n−4)

12
, as a key to their proof of the Heawood Map Colouring Problem [10].Also Ringel [9] showed earlier that the minimum genus of the complete bipartite graph

4
. But even for some small graphs, determining the minimum genus can be difficult; for example, it took several pages to do this for the Cartesian product C 3 × C 3 × C 3 (on 27 vertices); see [4] (and [8]).
For circulants, the graph genus problem appears to be just as challenging as it does for connected graphs in general.As noted earlier, the connected circulants with minimum genus 0 were completely determined by Heuberger [7], and an attempt to classify those with minimum genus 1 was made by Strapasson, Costa and Alves [12].In the latter paper, it was shown that some of the circulants listed in Theorem 2 have toroidal embeddings and therefore minimum genus 1, but the authors claimed (erroneously) that the cases they gave in their Proposition 5 are the only such cases -and also they did not provide a complete proof in the 3-valent cases.
When we attempted a classification of connected circulants with minimum genus 2, we discovered that the graph C 9 (1, 2, 4) has minimum genus 1, but did not occur in the list given in [12].Then also we found another infinite family of other examples with minimum genus 1, namely the circulants C n (a 1 , 2a 1 , n 2 − a 1 ) where n ≡ 0 mod 4 and n 12; this gives the family in the fourth case of Theorem 2. It turns out there was a flaw in an argument in the proof of Proposition 4 in [12], where it was assumed that the restriction of a 6-valent toroidal embedding to a 4-valent subgraph had quadrangular faces.
Here we note that it is easy to prove that all connected circulants C n (a 1 , . . ., a m ) with m = 1 or 2 (and hence valency 2, 3 or 4) have minimum genus 0 or 1.
In the other direction, we have the following: Lemma 4. All connected circulants with minimum genus 1 or 2 have valency at most 7.
Proof.Let G be any connected circulant, of order n and valency k, such that G has an orientable embedding with genus g 2. Then by the Euler-Poincaré formula where F is the number of faces.Also because each face has 3 or more edges (and each edge lies in at most 2 faces), we have 3F 2E = nk, and by the inequality above Thus k(k − 6) < n(k − 6) 12, and it follows that k < 8.
Hence for minimum genus 1 or 2 we need only consider the cases where the valency k is 7 (and m = 4) or 5 or 6 (and m = 3), as we do in Sections 4 to 6.The sporadic cases dealt with in Section 7 have minimum genus 2 or more, and for some those we give only a short description of how that can be proved.Also in one case we make use of theorems of Battle, Harary, Kodama and Youngs (see [1,Theorem 1 & corollaries]), which show that if a graph G has a subgraph that is the union of two subgraphs H and K (with no vertex in common), then the minimum genus of G is at least equal to the sum of the minimum genera of H and K.

Some (minimal) embeddings
In this Section, we turn to minimum genus, and begin by proving that the circulants listed in Theorems 2 and 3 have embeddings of genus 1 and 2 respectively.Note that some of this was done for the genus 1 case in [12], but certainly not all of it; in particular, the 6-valent family C n (1, 2, n 2 −1) and the 5-valent graph C 9 (1, 2, 4) were missed in Proposition 5 of [12], and also the 5-valent cases of C n (a 1 , a 2 , a 1 +a 2 ) were not properly considered.
Proposition 5.The circulants listed in Theorem 2 all have toroidal embeddings, and hence minimum genus 1.
Proof.We treat each case in turn, showing that it has an orientable embedding of genus 1, and noting that none of the circulants listed in the statement of the theorem is isomorphic to one of those in Heuberger's classification of connected planar circulants.
• C n (a 1 , a 2 ) where a 1 and a 2 do not satisfy the conditions of Proposition 1: As explained in [12], in the 4-valent case where 0 < a 1 < a 2 < n 2 , a genus 1 embedding of C n (a 1 , a 2 ) can be obtained from a quadrangular map on the torus with faces of size 4 bounded by closed walks of the form (i, i+a 1 , i+a 1 +a 2 , i+a 2 ), or equivalently, by taking the rotation (i+a 1 , i+a 2 , i−a 1 , i−a 2 ) at each vertex i.The 3-valent case where 0 < a 1 < a 2 = n 2 (which was not treated properly in [12]) can be dealt with by the same argument: simply double each edge of the form {j, j + n 2 }, then embed the resulting multigraph on the torus taking a rotation of the form (i+a 1 , i+ n 2 , i−a 1 , i+ n 2 ) at each vertex i, and delete one edge from each pair of edges of the form {j, j + n 2 }.If n 2 and a 1 are both odd, then one such toroidal embedding has rotation of the form (i+a 1 , i−a 1 , i+ n 2 ) at vertex i for all even i, and the inverse of this for all odd i, and all faces have length 6.In other cases, the face lengths can vary, with average 6.
• C n (a 1 , a 2 , a 3 ) with a 3 = ±(a 1 +a 2 ): If n 2 ∈ {a 1 , a 2 , a 3 }, then the valency is 6, and as explained in [12], we can take a toroidal embedding of C n (a 1 , a 2 ) with n faces of size 4 bounded by closed walks of the form (i, i+a 1 , i+a 1 +a 2 , i+a 2 ), with rotation ρ i at each vertex i being (i+a 1 , i+a 2 , i−a 1 , i−a 2 ), and then add the edges {i, i+a 1 +a 2 } as diagonals of those quadrangular faces, to give an embedding of C n (a 1 , a 2 , a 1 +a 2 ) in the torus with 2n triangular faces.The rotation ρ i at each vertex i is then (i+a 1 , i+a 1 +a 2 , i+a 2 , i−a 1 , i−a 1 −a 2 , i−a 2 ).On the other hand, if n 2 ∈ {a 1 , a 2 , a 3 } then the valency is 5, and by rearrangement and negation if necessary we can assume that a 1 + a 2 = a 3 = n 2 .This case is more tricky, and was not treated properly in [12].If n 2 is odd, we can add an edge from vertex i to vertex i + n 2 as a diagonal across the quadrangular face of the toroidal embedding of C n (a 1 , a 2 ) bounded by (i, i+a 1 , i+ n 2 , i+a 2 ) whenever i is even; then the rotation at a vertex i is 2 is even, we can add the diagonal whenever i ∈ {0, 1, . . ., n 2 − 1}, and then we have the rotation α i for i ∈ {0, 1, . . ., n 2 −1} and rotation +1, . . ., n−1}.In both of these two sub-cases, exactly half of the n quadrangular faces of the toroidal embedding of C n (a 1 , a 2 ) are split into triangular faces, while the other half are unchanged, so we have n triangular faces and n 2 quadrangular faces, giving a total of 3n 2 , as required.
the electronic journal of combinatorics 22(2) (2015), #P2.28 , where n ≡ 2 mod 4 and n 10: Here a toroidal embedding can be obtained by taking the rotation ρ i at vertex i as (i+1, i−1, i−2, i+ n 2 , i+2) when i is even, and (i+1, i+2, i+ n 2 , i−2, i−1) when i is odd.Then just as in the 5-valent case above, we have n triangular faces and n 2 quadrangular faces.The triangular faces are bounded by 3-cycles of the form (i, i+1, i−1) and (i, i+2, i+1) for all even i, and the quadrangular faces by 4-cycles of the form (i, i−2, i+ n 2 −2, i+ n 2 ) for all even i.This is the same as the embedding described in a different way in [12].
Proof.Again we treat the cases in turn, but this time with not so much detail, leaving calculation of the faces to the reader.

These rotations give an embedding with 3n
2 −2 faces, of which n−2 are triangular, n 2 −2 are quadrangular, and two have length 7. The Euler characteristic is n the electronic journal of combinatorics 22(2) (2015), #P2.28 , where n ≡ 2 mod 4 and n 10: ) for i = 0 and n 2 .These rotations give an embedding with 2n−2 faces, of which 2n−4 are triangular and two have length 6.The Euler characteristic is 2 ), where n ≡ 2 mod 4 and n 10: In this case we let n = 4k + 2 and consider two sub-cases: k even, and k odd.If k is even, take If k is odd, take In both sub-cases, again the rotations give an embedding with 3n 2 −2 faces, of which n−2 are triangular, n 2 −2 are quadrangular, and two have length 7, and the Euler characteristic These rotations give an embedding with 22 faces, of which 16 are triangular and 6 are quadrangular, giving Euler characteristic 12 − 36 + 22 = −2.
the electronic journal of combinatorics 22(2) (2015), #P2.28 The rest of this paper is devoted to showing that the circulants treated above are the only connected circulants of genus 1 or 2. Recall that every such circulant has valency at most 7 (by Lemma 4), and that all connected circulants with valency k 4 have minimum genus 0 or 1. Accordingly, we need only consider valencies 7, 5 and 6, which we do in the next three sections.
For each of these, we have 3F 2E = 84, so a genus 1 embedding is impossible, and a genus 2 embedding requires F = 28 faces, all of which would have length 3, giving what we may call a triangular embedding.Both  (2,3,4,6).Each of these can be shown to have no triangular embedding, and so has minimum genus at least 3.

Valency 5
In this section, we consider circulants of valency 5, which have the form C n (a 1 , a 2 , n 2 ), where n is even and n 6.If such a connected circulant has an embedding of genus 1 or 2, then the Euler-Poincaré formula gives −2  (1,3,4), and each of these has a genus 1 embedding (by Proposition 5), so we may suppose that at least one face is triangular, and that n > 8.
For there to be a triangular face, some relation of the form a i ± a j ± a k ≡ 0 mod n must be satisfied.If i, j, k are distinct, then it is easy to see that a 3 = n 2 = a 1 + a 2 in Z n , in which case we have a genus 1 embedding, again by Proposition 5, so we may assume that at least two of i, j, k coincide.Then since a 3 = n 2 , one (or more) of the following must occur: either 2a 1 ≡ ±a 2 mod n, or a 1 ≡ ±2a 2 mod n, or n 2 = a 3 = 2a 1 or 2a 2 , or n = 3a 1 or 3a 2 .Here we will temporarily drop the assumption that a 1 < a 2 , and then by interchanging a 1 and a 2 if necessary, we may suppose that 2a 1 ≡ ±a 2 mod n, or n 2 = a 3 = 2a 2 , or n = 3a 1 .We will consider these three cases in turn.Case 5a: 2a 1 ≡ ±a 2 mod n Note that connectedness implies that gcd(a 1 , 2a 1 , n 2 , n) = 1.If n ≡ 0 mod 4 then n 2 is even so a 1 is odd, and gcd(a 1 , n) = 1.Thus a 1 is invertible mod n, and we can multiply ), which has genus 1 for n = 8, and a genus 2 embedding for n 12. Similarly, if n ≡ 2 mod 4, then gcd(a 1 , n 2 ) = 1, but in this case there are two possibilities: either a 1 is odd, and again 2 ), which has minimum genus 1, or a 1 is even, in which case gcd(a 2 ), which has a genus 2 embedding.We now prove that C n (1, 2, n 2 ) has no genus 1 embedding when n ≡ 0 mod 4 and n 12, and that C n (2, 4, n 2 ) has no genus 1 embedding when n ≡ 2 mod 4 and n 10.
2 ) with n ≡ 0 mod 4 and n 12, we assume the contrary.Then there must be an embedding with 3n  2 faces.If F 3 is the number of triangular faces, and F is the number of faces of length greater than 3, then counting edge-face incident pairs gives Also by counting pairs (v, ∆) where ∆ is a triangular face at the vertex v, we find that the average number of triangular faces at each vertex is at least the electronic journal of combinatorics 22(2) (2015), #P2.28 2 ) with n 12, the only possible triangular faces at the vertex 0 are bounded by vertex-triples {0, 1, 2}, {0, 1, −1} and {0, −2, −1}, and so the only way to have three or more triangular faces at vertex 0 is to take ρ 0 = (1, −1, −2, n 2 , 2) or its inverse, both of which give exactly three such faces.It follows that F 3 = n, and so 2 , and all non-triangular faces must have length 4, with two at each vertex.In particular, there must be two faces of length 4 at vertex 0, and by considering common neighbours of n 2 and ±2, it is easy to see that these faces must be bounded by the 4-cycles (0, 2, n 2 + 2, n 2 ) and (0, −2, n 2 − 2, n 2 ).By reflecting the surface if necessary, we can assume ρ 0 = (1, −1, −2, n 2 , 2).Then ρ 1 must have the form (.., −1, 0, 2, ..), and so ρ 1 = (2, 3, n 2 +1, −1, 0), which is the analogue of ρ −1 0 obtained by adding 1 mod n to each point of the rotation.By induction, we find that But also the choice of ρ 0 and the boundaries of the two quadrangular faces at 0 imply that the rotation ρ n 2 has the form (.., n 2 +2, 0, n 2 −2), and it follows that n 2 must be odd.
If gcd(a 1 , n) = 1, then we can multiply X = {a 1 , a 2 , a 3 } = {a 1 , n 4 , n 2 } by the inverse of a 1 mod n and find that C n (a 1 , a 2 , a 3 ) ).Now let F 3 be the number of triangular faces, and let F be the number of faces of length greater than 3. Then counting edge-face incident pairs and using the inequality F 3n−4 2 (shown at the beginning of this section) gives 5n = 2E 3F 3 + 4F = 4F − F 3 2(3n−4) − F 3 , so that n F 3 + 8.But the only possible triangular faces at a vertex i are bounded by vertex-triples of the form {i, , and so there can be at most two at each vertex, and therefore n − 8 F 3 2n 3 , which gives n 24.Thus n = 8, 12, 16, 20 or 24.We know that the graph C 8 (1, 2, 4) has genus 1, and also that C 12 (1,3,6) has a genus 2 embedding.On the other hand, C n (1, n 4 , n 2 ) has no genus 1 embedding for n 12, since that would require 3n 2 faces, with at least n being triangular, so at least three at each vertex.Also we will show in Section 7 that the minimum genus of each of C 16 (1,4,8), C 20 (1, 5, 10) and C 24 (1,6,12) is greater than 2.
If gcd(a 1 , n) = 2, then a 1 /2 is coprime to n and we can multiply X = {a 1 , n 4 , n 2 } by its inverse mod n and find that C n (a 1 , a 2 , a 3 ) ∼ = C n (2, n 4 , n 2 ).But now n 4 must be odd, so the same argument about triangular faces as used in the previous paragraph gives n = 12 or 20, leaving only C 12 (2,3,6) and C 20 (2,5,10) to consider.The first of these has a genus 2 the electronic journal of combinatorics 22(2) (2015), #P2.28 embedding, but no genus 1 embedding since it can have at most two triangular faces at each vertex, and the second has genus greater than 2, as we explain in Section 7.
Similarly, if gcd(a 1 , n) = 4, we can multiply X by the inverse of a 1 /4 mod n and find that C n (a 1 , a 2 , a 3 ) ∼ = C n (4, n 4 , n 2 ), with n 4 odd, and so (4,5,10).The graph C 12 (3, 4, 6) has a genus 2 embedding but no genus 1 embedding, while the genus of C 20 (4, 5, 10) is greater than 2, as we explain in Section 7. Case 5c: n = 3a 1 In this case, n ≡ 0 mod 6 (since the valency is odd), and X = {a 1 , a 2 , a 3 } = { n 3 , a 2 , n 2 }.Also we may assume that none of the earlier equations (in 5a or 5b) holds.Under that assumption, the only possible triangular face at each vertex i is bounded by the vertextriple {i, i+a 1 , i−a 1 }, and so the argument used in case 5b about triangular faces gives n − 8 F 3 n 3 , which implies that n 12. Thus n = 6 or 12.

Valency 6
In this section we consider connected circulants C n (a 1 , a 2 , a 3 ) with 0 < a 1 < a 2 < a 3 < n 2 .By the Euler-Poincaré formula, the number of faces in an embedding of genus 1 or 2 must be 2n or 2n−2, respectively.Also if we let F 3 be the number of triangular faces and F the number of faces of length greater than 3, then counting edge-face incident pairs gives 6n = 2E 3F 3 + 4F = 4F − F 3 = 8n − F 3 or 8n − 8 − F 3 , so that F 3 2n or 2n − 8 respectively.Moreover, if the average number of triangular faces at a vertex is at most 3, then 2n − 8 F 3 n, so n 8, and the only possibilities would be C 7 (1, 2, 3) and C 8 (1, 2, 3), both of which have genus 1 (by Proposition 5).Hence we may suppose there are at least four triangular faces at some vertex, and by vertex-transitivity, we might as well assume that this happens at vertex 0.
As earlier, we observe that having a triangular face implies some relation of the form a i ± a j ± a k ≡ 0 mod n.Moreover, if i, j, k are distinct, then by the proof of Proposition 5, there exists a genus 1 embedding, covered by the second item of Theorem 2.
the electronic journal of combinatorics 22(2) (2015), #P2.28 Each such relation gives rise to at most three triangular faces at a vertex, so to get an embedding of genus 1 or 2 (not already covered by the second item of Theorem 2), we need to assume that at least two of these relations hold.
Moreover, those relations must involve all three of a 1 , a 2 and a 3 , for if they involve only a i and a j , say, then only possible triangular faces at vertex 0 are bounded by triples made up of 0 and two of ±a i and ±a j , and no more than three of these can be taken together in a candidate for the rotation ρ 0 .Also some of the combinations of these relations (involving all three of a 1 , a 2 and a 3 ) are inconsistent.
For n > 12, there is no way to arrange these to give six triangular faces at vertex 0, and so there cannot be six triangular faces at any vertex.Moreover, the only way to get five triangular faces at vertex 0 is to take ρ 0 as either (4, 2, 1, −1, −2, −4) or its inverse, but in that case ρ 2 must be of the form (.., 1, 0, 4, ..) or (.., 4, 0, 1, ..), so ρ 2 cannot be (6, 4, 3, 1, 0, −2) or its inverse.It follows that if there are five five triangular faces at a vertex v, then there are at most four triangular faces at vertex v +2.Hence if r is the number of vertices that lie in five triangular faces, and s is the number that do not, we have r s, and so r n 2 .Now using the inequality F 3 2n−8 (shown at the beginning of this section) and counting pairs (v, ∆) where ∆ is a triangular face at the vertex v, we have 3(2n−8) 3F 3 5r + 4s = 4(r+s) + r = 4n + r 4n + n 2 , and therefore n 16.Hence we need only consider C n (1, 2, 4) for 8 n 16.
Finally, for n ∈ {11, 13, 14, 15, 16} a genus 1 embedding would require 2n faces, all triangular, but that is impossible since there can be only one triangular face containing the edge {0, 4}, namely one that has 2 as its third vertex.In Section 7 we explain why none of these five graphs can have an embedding of genus 2. Case 6b: {T1, T7}, with a 2 = 2a 1 and 2a 2 + a 3 = n Here C n (a 1 , a 2 , a 3 ) = C n (a 1 , 2a 1 , n − 4a 1 ) ∼ = C n (a 1 , 2a 1 , 4a 1 ), which was met in case 6a.Case 6c: {T1, T8}, with a 2 = 2a 1 and a 1 + 2a 3 = n In this case, we have a 3 = n−a 1 2 , and also 1 = gcd(a a 1 is a unit mod n, and since n − a 1 = 2a 3 is even, both a 1 and n are odd, and therefore we can multiply by twice the inverse b of a 1 mod n and find that On the other hand, if d = 2, then both a 1 and n are even, but a 3 = n−a 1 2 is odd, and so n+a 1 2 = n−a 1 2 + a 1 is odd, and therefore gcd(n, n+a 1  2 ) = gcd( n 2 , n+a 1 2 ) = gcd( n 2 , a 1 2 ) = 1.Letting x be n+a 1  2 , and y be its inverse mod n, we have 2x = n + a 1 ≡ a 1 mod n and a 3 = x − a 1 , so multiplying by y we have 2 ≡ ya 1 mod n and ya 3 ≡ 1 − ya 1 mod n, and therefore ), again as in 6a.Case 6d: {T1, T9}, with a 2 = 2a 1 and a 2 + 2a 3 = n Here n is even and a 1 + a 3 = n 2 , with a 2 = 2a 1 ≡ −2a 3 mod n, and it follows that gcd(a 1 , n) = 1 or gcd(a 3 , n) = 1.If gcd(a 1 , n) = 1, then a 1 is odd, and we can multiply by its inverse mod n and find that C n (a 1 , a 2 , a 3 ) ∼ = C n (1, 2, n 2 − 1).On the other hand, if gcd(a 3 , n) = 1, then a 3 is odd and we can multiply by its inverse mod n to show that . If n ≡ 0 mod 4, this circulant has genus 1, by Proposition 5, and if n ≡ 2 mod 4, then it has a genus 2 embedding, by Proposition 6.Also the graph C 10 (1, 2, 4) was shown to have genus 2 in case 6a.

Remaining cases
In this section we deal with the remaining unresolved cases, some of which are rather more difficult than others.

• C 12 (3, 4, 6) has no genus 1 embedding
In this case, a genus 1 embedding needs 18 faces, at least 12 of which are triangular, so the average number of triangular faces at each vertex is at least 3. On the other hand, there are only four vertex-triples that can bound a triangular face at vertex 0, namely {0, 3, 6}, {0, 3, 9}, {0, 6, 9} and {0, 4, 8}, and at most three of these can be taken together.Hence there must be 12 triangular faces, with three at each vertex.Moreover, two of the three triangular faces at a given vertex share a common edge, and in each case, the vertices of those two triangles must all differ by multiples of 3, while the other triangular face is bounded by one of the triples {0, 4, 8}, {1, 5, 9}, {2, 6, 10} and {3, 5, 11}.
Counting edge-face incident pairs gives the average face-size of the larger faces as 4, and so the six non-triangular faces must be quadrangular.The rotation at vertex 0 can be taken as (4, 8, x, y, z), where (x, y, z) is some permutation of (3, 6, 9), and then the two quadrangular faces at 0 are bounded by 4-cycles of the form (x, 0, 8, u) and (4, 0, z, v).Also by considering common neighbours of 4 and z, we see that (z, v) = (3, 7), (6,10) or (9, 1), and note that in each case the difference z −v is ±4.
Now we consider what happens at the vertex y.The edge {0, y} lies in two triangular faces bounded by the triples {0, x, y} and {0, y, z}, so the other face containing the edge {y, z} must be quadrangular, bounded say by the 4-cycle (z, y, p, q).Then at vertex z, the triangular face bounded by the triple {0, y, z} sits in between the two quadrangular faces bounded by the 4-cycles (4, 0, z, v) and (z, y, p, q).It follows that the other two faces at vertex z must be triangular, and adjacent, but that is impossible, since the difference v − z is not divisible by 3. Hence no genus 1 embedding exists, and so C 12 (3, 4, 6) has genus 2.
Next, recall that we showed in case 5b of Section 5 that C n (1, n 4 , n 2 ) has no genus 1 embedding for n 12, but has one of genus 2 for n = 12, and we claimed that it has no genus 2 embedding for n = 16, 20 or 24.We verify that claim now.
) has no embedding of genus 2, for n = 16, 20 or 24 In each case, a genus 2 embedding requires 3n−4 2 faces, but at any vertex there are at most two triangular faces, so if F 3 is the number of triangular faces, then n−8 F 3 2n 3 .Moreover, if there are two triangular faces at the vertex 0, then the rotation ρ 0 must be of the form (x, y, z, 1, −1) or (x, y, z, −1, 1), where {x, y, z} = { n 4 , n 2 , − n 4 }, and then since 0 is the only common neighbour of 1 and −1, the face at 0 containing the edges {0, 1} and {0, −1} must have length at least 5.
Now if n = 24, then the conditions n−8 F 3 2n 3 force F 3 to be 16, and also there must be exactly two triangular faces at each vertex, and therefore at least one face of length 5 or more.But on the other hand, the remaining 72−4 2 −16 = 18 faces use at most 2E − 3F 3 = 120 − 48 = 72 edges, so they must all be quadrangular, contradiction.
Similarly, if n = 20, then there are 60−4 2 = 28 faces, with 12 or 13 being triangular, and so there must two triangular faces at 16 or more vertices.If F 3 = 12, then the 16 non-triangular faces use at most 2E − 3F 3 = 100 − 36 = 64 edges, so they must all be quadrangular, which is impossible.Hence F 3 = 13, and the 15 non-triangular faces use at most 100 − 39 = 61 edges, so there must be 14 quadrangular faces, plus one of length 5. This, however, implies that there can be at most five vertices that are incident with two triangular faces, again impossible.
Finally, suppose n = 16.Then there are 22 faces, with 8, 9 or 10 of these being triangular, and the number of vertices incident with two triangular faces is at least 8, 11 or 14 respectively.If F 3 = 8 then the other 14 faces use at most 2E − 3F 3 = 80 − 24 = 56 edges, so they must all be quadrangular, contradiction.Similarly, if F 3 = 9 then the other 13 faces use at most 80 − 27 = 53 edges, so there must be 12 quadrangular faces, plus one of length 5, and hence there can be at most five vertices that are incident with two triangular faces, again impossible.Thus F 3 = 10.But now the other 12 faces use at most 80 − 30 = 50 edges, so there must be either 10 quadrangular faces and two of length 5, or 11 quadrangular faces and one of length 6, and then the number of of vertices incident with two triangular faces is at most 2 • 5 = 10 or 1 • 6 = 6 (respectively), so we reach a contradiction in both cases.
The same kind of argument works for the next two 5-valent circulants as well.
A genus 1 embedding would require 30 faces, with at least 20 being triangular, which is impossible.On the other hand, a genus 2 embedding requires 28 faces, and just as in the case of C 20 (1, 5, 10) above, needs 12 or 13 of these to be triangular, with two triangular faces occurring at 16 or more vertices; but again there are either 12 triangular and 16 quadrangular faces, or 13 triangular and 14 quadrangular faces and one of length 5, and in both cases we obtain a contradiction.
Next, we deal with the only remaining 7-valent case.
• C 10 (1, 2, 4, 5) has no embedding of genus 1 or 2 A genus 1 embedding would require 25 faces, which is impossible since these faces would use at least 25 • 3 2 = 75 2 edges, which is too many.(Also C 10 (1, 2, 4, 5) has C 10 (1, 2, 4) as a subgraph, and we know the genus of C 10 (1, 2, 4) is 2.) Now let us suppose C 10 (1, 2, 4, 5) has an embedding of genus 2. This must have 23 faces, of which 22 are triangular and one has length 4. In contrast to earlier situations, there are lots of possibilities (in fact a total of 66) for the rotation at a given vertex that can make all faces at that vertex triangular, so instead, we can focus on the single face of length 4. By vertex-transitivity, we may suppose this face contains vertex 0.Even then, there are 160 possibilities for the 4-cycle bounding the face, but up to reflection and automorphisms of C 10 (1, 2, 4, 5) this reduces to 41 possibilities.We were not able to find a short argument to eliminate all of them, and resorted to using a computer (and specifically, the Magma system [3]) to help prove there is no genus 2 embedding.
Our computation took less than an hour using Magma on a laptop, and showed that all other possibilities either lead to a similar contradiction, or force choices for rotations at other vertices that in turn are impossible.(It also found that many possibilities provide an embedding of genus 3, so the minimum genus of C 10 (1, 2, 4, 5) is 3.) This brings us to the most challenging family, namely the circulants C n (1, 2, 4) for n ∈ {11, 13, 14, 15, 16}.We know these have no embedding of genus 1, since we cannot have all faces triangular.In particular, there are at most five triangular faces at any vertex.Also a genus 2 embedding requires 2n−2 faces, with at least 2n−8 being triangular.Now let V 5 be the number of vertices incident with 5 triangular faces.Then there are at most 4 triangular faces at each of the other n − V 5 vertices, so counting pairs (v, ∆) pairs as previously gives V 5 + 4n = 5V 5 + 4(n − V 5 ) 3F 3 3(2n − 8) = 6n − 24, and therefore V 5 2n−24.In particular, for n ∈ {13, 14, 15, 16}, at least one vertex must be incident with 5 triangular faces.
By vertex-transitivity, it follows that if a vertex v is incident with 5 triangular faces, then v ± 2 and v ± 3 are not.We can now proceed, by considering possibilities for the sequence Y = (y 0 , y 1 , y 2 , . . ., y n−1 ), where 1 if vertex i is incident with 5 triangular faces, or 0 if not.
At least 2n−24 terms of this sequence must be 1, but also these two conditions must hold: (a) if y i = 1, then y j = 0 whenever i and j differ by ±2 or ±3 mod n, and (b) if y i = y i+1 = 1 then we cannot have y j = y j+1 = 1 when i and j differ by ±5 mod n.
• C 16 (1,2,4) has no embedding of genus 2 Here the sequence Y = (y 0 , y 1 . . . ., y 15 ) must contain at least eight 1s, but condition (a) cannot be satisfied when there are more than six 1s.
In each of these four cases, the weighted sum T is exactly 54, and so the number of triangular faces is exactly 18, and then from 54+24 = 3F 3 +4F 2E = 78 we deduce that the six non-triangular faces must all have length 4.
We can now show that if y i = 1 then y j = 0 whenever j −i ≡ ±4 or ±6 mod n.
• C 11 (1,2,4) has no embedding of genus 2 This is by far the most challenging case, and so left until last.It is another one for which we used a computer, and we give only a partial description of how we handled it.
We assume that it has a genus 2 embedding, which must have 20 faces, with at least the electronic journal of combinatorics 22(2) (2015), #P2.28 14 being triangular.Again we consider the sequence Y = (y 0 , y 1 , y 2 , . . ., y n−1 ) as defined above.This time we cannot show that at least one y i is 1, but condition (a) still holds, namely that if y i = 1 then y j = 0 whenever i and j differ by ±2 or ±3 mod n.Up to cyclic rearrangement and reversal we find there are eight possibilities for Y .One of them is (1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0), and this can be eliminated using condition (b), but also by showing that if y i = 1 then y j = 0 whenever i and j differ by ±4 or ±5 mod n (when n = 11).Doing that is not as easy to do as it is in the case n = 13, as there are many more possibilities to check and eliminate.
Hence we have 2 possibilities for the rotations at each of the vertices 1, 4, 7 and 9, and 6 possibilities for the rotations at each of the vertices 2, 3, 6 and 10, and 120 possibilities for the rotation at vertex 8.These combine to give a total of 2 4 • 6 4 • 120 = 2488320.Clearly further observations can be made to reduce this number, but even so, it is a small enough number that all possibilities can easily be checked by computer.We carried out a 35-minute computation using Magma on a laptop, and found that the smallest genus achievable from such choices of rotations is 3.In particular, genus 2 is impossible when y 0 = y 5 = 1.
The same kind of computational approach as described immediately above can also be adapted to show that we cannot have y 0 = y 1 = 1.This takes rather longer (73 hours by computer), but again shows that genus 3 is the smallest possible, and so it eliminates the first of the remaining three possibilities for Y .
In the first case, which involves the rotation we took at vertex 0 when we assumed 0 lay in five triangular faces, one of the two non-triangular faces is partially bounded by the 2-arc (7, 0, 4), and the other is bounded by a 4-cycle of the form (u, 0, v, w) where u and v are adjacent in the graph.In this situation, we can change the rotations at vertices u and v so that the edge {u, v} becomes a diagonal, splitting the latter face into two faces of length 3, while the two faces that previously contained the edge {u, v} are merged into a single face.This achieves a different embedding with the same genus 2, but with five triangular faces at vertex 0, which we know is impossible.Hence ρ 0 is one of π 2 to π 8 .
The same arguments show that the only possibilities for the rotation at each of the vertices 9, 10, 1 and 2 are the analogues of π 2 to π 8 , obtained by adding respectively 9, 10, 1 or 2 (mod 11) to the points of each π i , and their inverses.