4-colouring of generalized signed planar graphs

Assume G is a graph and S is a set of permutations of positive integers. An S-signature of G is a pair (D,σ), where D is an orientation of G and σ ∶ E(D)→ S is a mapping which assigns to each arc e = (u, v) a permutation σ(e) in S. We say G is S-k-colourable if for any S-signature (D,σ) of G, there is a mapping f ∶ V (G) → [k] such that for each arc e = (u, v) of G, σ(e)(f(u)) ≠ f(v). The concept of S-k-colourable is a common generalization of many colouring concepts. This paper studies the problem as to which subsets S of S4, every planar graph is S-4-colourable. We call such a subset S of S4 a good subset. The Four Colour Theorem is equivalent to saying that S = {id} is good. It was proved by Jin, Wong and Zhu (arXiv:1811.08584) that a subset S containing id is good if and only if S = {id}. In this paper, we prove that, up to conjugation, every good subset of S4 not containing id is a subset of {(12), (34), (12)(34)}. Mathematics Subject Classifications: 05C10, 05C15


Introduction
A signed graph is a pair (G, σ), where G is a graph and σ ∶ E(G) → {1, −1} is a mapping which assigns to each edge e of G a sign σ(e) ∈ {1, −1}. The mapping σ is called a signature of G and an edge e is called positive or negative if σ(e) = 1 or σ(e) = −1, respectively. Colouring of signed graphs was first studied by Zaslavsky [12] in 1980's, and has attracted a lot of recent attention. Let Z k = {0, 1, . . . , k − 1} denote the cyclic group of order k.
Definition 1. Assume (G, σ) is a signed graph and k is a positive integer. A k-colouring of (G, σ) is a mapping f ∶ V (G) → N k such that for any edge e = xy of G, f (x) ≠ σ(e)f (y).
A Z k -colouring of (G, σ) is a mapping f ∶ V (G) → Z k such that for any edge e = xy of G, f (x) ≠ σ(e)f (y).

Generalized signed graph colouring
In colourings of a signed graph, the sign of an edge specify which colour pairs are forbidden to be assigned to its end vertices. It is natural to consider graphs with more variety of edges and hence with more types of restrictions on colour pairs that can be assigned to the end vertices of edges. Indeed, various colourings of graphs corresponding to different restrictions have been studied in the literature. Below are three types of graph colourings that are examples of such constraints. In 1992, Jaeger, Linial, Payan and Tarsi [2] introduced the concept of group colouring. Assume Γ is an Abelian group, D is an orientation of G and σ ∶ E(D) → Γ assigns to each arc e = (u, v) an element σ(e) ∈ Γ. A Γ-colouring of (D, σ) is a mapping f ∶ V (G) → Γ such that for any arc e = (u, v) of D, f (v) − f (u) ≠ σ(e). We say G is Γ-colourable if for any orientation D of G and for any σ ∶ E(D) → Γ, there exists a Γ-colouring of (D, σ). The group chromatic number of G is the minimum integer k such that for any Abelian group Γ of order k, G is Γ-colourable ( see [6,7,10] for some study on this subject).
In 2018, Dvořák and Postle [1] defined DP-colouring of a graph as follows: induces a clique, and for each edge uv of G, edges between {v} × [k] and {u} × [k] is a matching, and there is no other edge. An independent set I of H with I = V (G) is a DP-colouring of G with respect to H. We say G is DP-k-colourable if for any k-cover H of G, there exists a DP-colouring of G with respect to H. It was shown in [1] that every DP-k-colourable graph is k-choosable, and the concept of DP-colouring is used to show that every planar graph without cycles of lengths 4, 5, 6, 7, 8 is 3-choosable, which is a problem that had remained open for 15 years.
In each of the colourings defined above, edges of G are labeled, and certain pairs of colours are forbidden to be assigned to the end vertices of edges with given label. In this sense, the labels play the same role as the sign of the edges in a signed graph. However, instead of + and − signs, we may have many different signs. Adopting this point of view, we introduce the concept of colouring of S-signed graphs as follows.
A set S of permutations of integers is called inverse closed if for each π ∈ S, π −1 ∈ S.
Definition 4. Assume G is a graph and S is an inverse closed set of permutations of positive integers. An S-signature of G is a pair (D, σ), where D is an orientation of G and σ ∶ E(D) → S is a mapping which assigns to each arc e = (u, v) a permutation σ(e) ∈ S. A proper k-colouring of (D, σ) is a mapping f ∶ V (D) → [k] such that for each arc e = (u, v) of D, σ(e)(f (u)) ≠ f (v). We say G is S-k-colourable if every S-signature (D, σ) of G has a proper k-colouring.
To define an S-signature of G, we need an orientation D of G and define a mapping σ ∶ E(D) → S. The orientation D is just for reference. Indeed, if E ′ is a subset of E(G) and D ′ is obtained from D by reversing the orientations of edges in E ′ , and then a proper k-colouring of (D, σ) is the same as a proper k-colouring of (D ′ , σ ′ ). In particular, if all the permutations π ∈ S are involutions, i.e., π −1 = π, then the orientation of the edges are irrelevant, and we simply define a mapping σ ∶ E(G) → S. The k-colouring and the Z k -colouring of signed graphs are all colourings of S-signed graphs for some special sets S of permutations. The sets S consists of involutions only. Thus we do not need orientations of G.
Assume we are considering S-k-colouring of a graph G. For π ∈ S, if i ∉ [k] or π(i) ∉ [k] and e = (x, y) is an arc with σ(e) = π, then colouring x with i (or π(i)) puts no restriction on the colour of y. So for i ∉ [k] or π(i) ∉ [k], the image π(i) is irrelevant. Sometimes it is convenient to allow permutations π ∈ S to have i ∈ [k] and π(i) ∉ [k] or i ∉ [k] and π(i) ∈ [k]. However, in this paper, when we consider S-k-colourings of graphs, we are interested in those permutations π with π(i) ∈ [k] for every i ∈ [k]. In other words, the restriction of π to [k] is a permutation of [k]. As the images π(i) of i ∉ [k] are irrelevant, we simply consider π as a permutation of [k], i.e., S is an inverse closed subset of the symmetric group S k .
The concept of colouring of generalized signed graphs is a common generalization of many colouring concepts: Observation 5. Assume S is a set of permutations of positive integers. Then the following hold: • If S = {id}, then S-k-colourable is equivalent to k-colourable.
• If Γ is an Abelian subgroup of S k , then there is a subset S of S k which is isomorphic to Γ and S-colourable is equivalent to Γ-colourable in the sense defined by Jaeger et al [2]. For example, If S is the subgroup of S k generated by (12 . . . k), then Scolourable is equivalent to Z k -colourable, i.e., for any mapping φ ∶ E(G) → Z k , there is a mapping f ∶ V (G) → Z k such that for each arc e = (x, y), f (y)−f (x) ≠ φ(e). We observe that S-k-colourable not only depends on the group structure of S, but depends on the permutations in S. It may happen that S is a subgroup of S k isomorphic to Γ, but S-k-colourable is different from Γ-colourable.
• Assume Γ is a group with Γ = n and k is a positive integer. Let τ ∶ Γ → [n] be a one-to-one correspondence from Γ to [n]. Let k ′ = kn + 1 and for each π ∈ Γ, let π ′ be the permutation of and j ∈ {0, 1, . . . , k − 1}, and π ′ (kn • A signature σ of a graph G is called k-consistent if for every directed cycle C = (e 1 , e 2 , . . . , e k ) of G, the composition π = σ(e 1 )σ(e 2 ) . . . σ(e k ) of permutations on the edges of the cycle has the following property: for i ∈ [k], either π(i) = i or for some Here it is necessary to consider permutations π whose restriction of [k] is a partial permutation, i.e., it is allowed that Proof. All the statements above follow straightforward from the definition. We give a brief explanation of the last one. Assume (G, σ) is k-colourable for every k-consistent signature σ of G, and L is a klist assignment of G. Without loss of generality, we may assume that colours in L(v) for v ∈ V (G) are positive integers. For each vertex v, let π v be any permutation of positive integers such that It is straightforward to verify that ψ is a k-colouring of (G, σ).
In the following, we concentrate on S that are subsets of S k . Observe that if S is a subset of S k and there is an integer a ∈ [k] such that for any π ∈ S, π(a) ≠ a, then for any S-signed graph (G, σ), the mapping f (v) = a for all v ∈ V (G) is a k-colouring of (G, σ). Definition 6. We call a subset S of S k normal if for each a ∈ [k], there is a permutation π ∈ S such that π(a) = a.
By the observation above, we shall restrict to normal subsets of S k . One particular type of normal subsets S of S k consists of those subsets S containing id.
In this paper, we concentrate on 4-colouring of generalized signed planar graphs. We are interested in the following question.
Question 7. For which subsets S of S 4 , every planar graph is S-4-colourable?
We call a subset S of S 4 good if every planar graph is S-colourable. The Four Colour Theorem says that S = {id} is good.
It is obvious that if S and S ′ are conjugates and S is good, then S ′ is good. The following result was proved in [3].
In this paper, we prove the following result.

Proof of Theorem 10
To prove Theorem 10, we shall show that if S is a normal subset of S 4 not containing id and S is not conjugate to a subset of {(12), (34), (12)(34)}, then there is a planar graph G which is not S-4-colourable.
For convenience, the sets S listed in this section need not be inverse closed. If π ≠ π −1 and π ∈ S, then π −1 is implicitly assumed to be a member of S, but will not be listed explicitly as a member of S. Proof. Assume S is a minimal normal subset of S 4 not containing id. Then each permutation in S fixes one or two colours in [4]. So 2 ≤ S ≤ 4.
If S = 3, then one permutation in S fixes two colours in [4]. Without loss of generality, we may assume that (12) ∈ S. Since S is a minimal normal subset of S 4 , no permutation in S fixes both 1 and 2. So one of the remaining permutation π 1 ∈ S fixes 1, and the other permutation π 2 ∈ S fixes 2. Thus If {(12), (34)} is not a core of S (up to conjugation), then by Lemma 12, the core of S is already conjugate to a member of S .
Thus to prove Theorem 10, it suffices to show that for each S ∈ S , there is a planar graph G and an S-signature (D, σ) of G such that (D, σ) is not 4-colourable.

Lemma 14.
For any S ∈ S , for any a, b ∈ [4] for which π * (a) ≠ b (it is possible that a = b), there is a planar graph H with u, u ′ be two vertices on the boundary of H, such that there is an S-signature (D, σ) of H and there is no 4-colouring φ of (D, σ) with φ(u) = a and φ(u ′ ) = b.
We leave the proof of Lemma 14 to the next section. With this lemma, we can complete the proof of Theorem 10.
Assume S ∈ S . For each pair (a, b) of integers with a, b ∈ [4] and π * (a) ≠ b, let H a,b be a planar graph with u, u ′ be two vertices on the boundary of H a,b , such that there is an S-signature σ of H a,b and there is no 4-colouring φ of (H a,b , σ) with φ(u) = a and φ(u ′ ) = b.
Let G be obtained from the disjoint union of {H a,b ∶ a, b ∈ [4], π * (a) ≠ b} by identifying all the copies of u into a single vertex u * , all the copies of u ′ into a single vertex v * and adding an edge u * v * .
If (12) ∈ S, then let (D, σ) be the S-signature of G whose restriction to each H a,b is as defined above and σ(u * v * ) = π * (note that as (π * ) −1 = π * , the orientation of the edge u * v * is irrelevant). It follows that there is no 4-colouring of (D, σ). Indeed, if φ is a 4-colouring of G with φ(u) = a and φ(v) = b, then since σ(u * v * ) = π * , we have π * (a) ≠ b. Then by Lemma 14, the restriction of φ to H a,b is not a proper 4-colouring of H a,b .
If (12) ∉ S, i.e., S = {(234), (134), (124), (123)}, then let G ′ be obtained from the disjoint union of three copies of G by identifying the three copies of u * into a single vertex, which is still named u * . Let (D, σ) be the S-signature of G whose restriction to each H a,b is as defined above and in the first copy of G, the edge u * v * is oriented as (u * , v * ) and σ(u * , v * ) = (123). In the second copy of G * , the edge u * v * is oriented as (v * , u * ) and σ(v * , u * ) = (123). In the third copy of G * , the edge u * v * is oriented as (v * , u * ) and σ(v * , u * ) = (124). Now we show that there is no 4-colouring of (D, σ). Assume to the contrary that φ is a proper 4-colouring of (D, σ). If φ(u * ) = 1, then φ(v * ) ≠ 2 for v * in the first copy of G. The same argument as above leads to a contradiction. If φ(u * ) = 2, then φ(v * ) ≠ 1 in the second copy of G. Similarly, we arrive at a contradiction. If φ(u * ) = 3, then φ(v * ) ≠ 3 in the third copy of G. Again a contradiction. If φ(u * ) = 4, then φ(v * ) ≠ 4 in the first copy of G, which also leads to a contradiction. This completes the proof of Theorem 10.

Proof of Lemma 14
For different sets S ∈ S , the constructions of H and the S-signature (D, σ) are different, but some of them are very similar to each other. By grouping similar constructions together, we divide the proof of Lemma 14 into four claims.
The construction will use five graphs H 1 , H 2 , H 3 , H 4 , H 5 as gadgets. These five graphs are depicted in Figures 1 and 2, where in Figure 2, each 3-face of H 5 containing a * contains a copy of T whose boundary triangle is identified with the 3-face. Proof. We consider three cases.  • If a = 1, b = 3, then let (D, σ) be the S-signature of H 1 defined as in Figure 3(b).
• If a = 1, b = 4, then let (D, σ) be the S-signature of H 1 defined as in Figure 3(c).
• If a = 3, b = 4, then let (D, σ) be the S-signature of H 1 defined as in Figure 3(d).
Since e = vw has σ(e) = (12), we conclude that φ(v) ≠ φ(w). Assume first that φ(v) = 3 and φ(w) = 4. Then none of x, y, z can be coloured by 3 or 4. So x, y, z are all coloured by 1 and 2. However, the three edges e connecting x, y, z have σ(e) = (34). Hence no two vertices of x, y, z can be coloured by the same colour, a contradiction.
For these cases, the corresponding S-signatures are given in Figures 8(a), 8(b) and 8(c), respecitvely.
• If a = b = 1, then let (D, σ) be the S-signature of H 1 defined as in Figure 9(a).
• If a = 1, b = 3, then let (D, σ) be the S-signature of H 1 defined as in Figure 9(b).
The permutations on the edges uv, uw, u ′ v, u ′ w in both of Figure 9(a) and 9(b) lead to φ(v) = 3, φ(w) = 4 or φ(v) = 4, φ(w) = 3. This implies that there are only two idenditical colours, 1 and 2, left for either the triangle xyz or x ′ y ′ z ′ , which is a contradiction.
The permutations on the edges uv, uw, u ′ v, u ′ w in both of Figure 9(c) and 9(d) lead to φ(v) = φ(w) = 1 or φ(v) = φ(w) = 2. In the former case (resp. the later case, in which the permutations on the edges u ′ v ′ , vv ′ , wv ′ force φ(v ′ ) = 3), there are only two identical colours left for the triangle xyz (resp. x ′ y ′ z ′ ), which is impossible. Proof. By symmetry, it suffices to consider the cases that a = b = 1 or a = 1, b = 3 or a = 1, b = 4.
• If a = b = 1, then let (D, σ) be the S-signature of H 3 defined as in Figure 10(a).
• If a = 1, b = 3, then let (D, σ) be the S-signature of H 3 defined as in Figure 10(b).
• If a = 1, b = 4, then let (D, σ) be the S-signature of H 4 defined as in Figure 10(c).