Edge colourings and qualitative representations of chromatic algebras

Conventional Ramsey-theoretic investigations for edge-colourings of complete graphs are framed around avoidance of certain configurations. Motivated by considerations arising in the field of Qualitative Reasoning, we explore edge colourings that in addition to forbidding certain triangle configurations also require others to be present. These conditions have natural combinatorial interest in their own right, but also correspond to qualitative representability of certain nonassociative relation algebras, which we will call chromatic.


Introduction
In an edge n-colouring of a complete graph, each triangle of edges consists of either one colour, two colours or three colours: monochromatic, dichromatic or trichromatic. In this article, we explore edge-colourings determined by disallowed triangle colour combinations, but also requiring others. Thus, disallowing monochromatic triangles restricts to edge-coloured complete graphs within the Ramsey bound R(3, 3, . . . , 3). But what if in addition to disallowing monochromatic triangles, we also impose the dual constraint that all remaining colour combinations (trichromatic and dichromatic) are present: is it possible to find such a network? These are natural combinatorial considerations in their own right, but there is an additional motivation by way of the algebraic foundations of qualitative reasoning, which finds wide application in AI settings around scheduling [1], navigation [20,28] and geospatial positioning amongst others [29]. The constraint language underlying typical qualitative reasoning systems determines a kind of non-associative relation algebra (in the sense of Maddux [23]), which is attracting considerable attention from a theoretical computer science perspective; see [9,14,15,21,32] for example. The inverse problem of deciding if a suitably defined non-associative algebra arises from a concrete constraint network is shown to be NP-complete in [14]. The present work focusses on a natural family of combinatorially intriguing cases, that we find have nontrivial solutions, and provide some novel extensions of classically understood connections between certain associative relation algebras and combinatorial geometries, such as in Lyndon [22].
Let C + = {1 , c 1 , . . . , c n } be a finite set of colours, let U be any set and let λ : U × U → C + be a surjective map such that λ(x, y) = λ(y, x) and λ(x, y) = 1 iff x = y. Then, {λ −1 (1 ), λ −1 (c 1 ), . . . , λ −1 (c n )} is a partition of U × U with every λ −1 (c i ) symmetric and λ −1 (1 ) the identity relation. We call λ an edge n-colouring of the complete graph with the set of vertices U , informally thinking of 1 as an invisible colour. The set C = C + \ {1 } is the set of proper colours. We will consider natural conditions on the colourings, forbidding certain triangles and requiring others. To forbid the occurrence of monochromatic triangles for example, we can impose the following condition where • is the relational composition. Strengthening the condition above to an equivalence ∀a, b, c ∈ C : |{a, b, c}| = 1 ⇔ λ −1 (a) • λ −1 (b) ∩ λ −1 (c) = ∅ we still have all monochromatic triangles forbidden, but moreover all non-monochromatic triangles must occur. As noted earlier, it is not immediately clear whether such colourings exist: the Ramsey Theorem gives an upper bound on the size of the graphs, the occurrence of all non-monochromatic triangles gives a lower bound -conceivably greater than the upper bound.
In general, letting F be any subset of {1, 2, 3} we arrive at 8 natural conditions of the form ∀a, b, c ∈ C : |{a, b, c}| ∈ F ⇔ λ −1 (a) • λ −1 (b) ∩ λ −1 (c) = ∅ and 8 corresponding existence questions; these are tied precisely to the algebraic properties of interest in Lemma 3.2 below. We will also consider the weaker conditions, with forward implication only; these are somewhat less interesting since they do not determine a unique algebraic object (see Lemma 3.3).

Nonassociative relation algebras and their representations
In this section, we recall basic constructions and ideas from the theory of relation algebras, which we then tie to the combinatorial conditions discussed informally at the start of the article. This culminates in Lemma 3.2 of the next section, which shows that the basic hierarchy of strengths of representability (strong implies qualitative implies feeble) in the algebraic setting correspond precisely to three natural constraints around forbidden colour combinations and required colour combinations. The reader wishing to skip the algebraic connections can survive most of the remainder of the article with only the notation presented in Definition 3.1 and by using Lemma 3.2 as a Rosetta stone to relate algebraic terminology to combinatorial properties on edge-coloured complete graphs.
The equivalences in (4) are known as the triangle laws, or Peircean laws. If the operation ; is associative, then A is a relation algebra (RA), in the sense of Tarski [31]; see a monograph such as Givant [10], Hirsch and Hodkinson [13] or Maddux [24]. For any set U and an equivalence relation E ⊆ U ×U , the powerset P(E) carries a natural relation algebra structure. The Boolean operations are set-theoretical, ; is the relational composition,˘the converse, and 1 the identity relation. Denote this algebra by is the algebra of all binary relations on U . In general, E is always an equivalence relation on U , but not necessarily the full relation. The bottom element of the of the Boolean reduct of P(E) is ∅ and the top element is E, so the Boolean complement is calculated relative to E. A relation algebra A is representable if there is an injective homomorphism φ : A → Re E (U ), for some U (called the base of the representation) and E. In particular, we have where −1 and • are the operations of relational converse and composition, respectively. If E = U × U , the representation is called square. To avoid confusion with 'representation' in a generic sense of 'any kind of representation', we will refer to representations above as strong representations. Two weakenings of this notion of representation, which apply to nonassociative algebras, were defined in [14], and we now recall them.
In any herd H, the smallest element R containing S • T , if it exists, is called the weak composition of S and T and is often denoted by S T . When H is finite (or more generally, complete) as a Boolean algebra, then S T always exists as we may simply intersect the elements of H that fully contain the relation S • T . A nonassociative algebra A is said to be qualitatively representable if there is a bijection φ from A to a herd H such that If (q4) is strengthened to (r4), then the qualitative representation φ is a strong square representation.
On the other hand, if (q4) is weakened to ϕ(a ; b) ⊇ ϕ(a) • ϕ(b), we obtain a still weaker notion of representation, called feeble in [14]. The next lemma combines well known facts on representations of relation algebras with parts of Lemmas 12 and 18 of [14].

Lemma 2.1 Let A be complete and atomic with the set of atoms At(A), and let
Since a, b, c are atoms, the equivalence (a ; b) ∧ c = 0 ⇔ a ; b ≥ c holds in A. Strong representations must reflect this property by (R2), but qualitative and feeble representations do not need to.
y} is the set of consistent triples. Triples (x, y, z) / ∈ T are called forbidden. Conversely, for a relational structure X = (X ,˘, I , T ) where˘is a unary function, I ⊆ X , and T ⊆ X 3 , the complex algebra Cm(X ) of X is the algebra (P(X ), ∩, ∪, ;, ¬,˘, ∅, X , I ), where for R, S ⊆ X , we put S˘= {s˘: s ∈ S}, and S ; R = {u ∈ X : (s, r , u) ∈ T , for some s ∈ S, r ∈ R}.
If A is an atomic nonassociative algebra, then the map x → {a ∈ At(A) : a ≤ x} is an embedding of A into Cm(At(A)). If A is moreover complete, then it is an isomorphism. It is convenient and very common to present complete and atomic nonassociative algebras in terms of their atom structures. The next lemma, proved in [23], characterises structures that are atom structures of nonassociative algebras.

Lemma 2.3
Let X = (X ,˘, I , T ) be a structure such that I is a subset of X , T is a subset of X 3 , and˘is a function satisfying a˘˘= a. The following are equivalent: • X is the atom structure of some nonassociative algebra.
• For all a, b, c ∈ X , we have There are many finite relation algebras that admit strong representations over infinite sets but no finite set, and in general deciding strong representability for finite relation algebras is algorithmically undecidable, by Hirsch and Hodkinson [12]. Qualitative representability is decidable, though is NP-complete [14, Theorem 15]. Much of the simplification is due to the following useful lemma, whose easy proof is a brute-force argument relying on the fact that in a finite algebra A there are at most 3|At(A)| distinct triangles that must be witnessed; so if a qualitative representation exists it can always be pruned to contain only as many points as are needed to witness all the necessary triangles.
Lemma 2.4 [14,Lemma 13] If an atomic nonassociative algebra A admits a qualitative representation, then it admits a qualitative representation as relations on set of size at most 3|At(A)| points.

Chromatic algebras
Definition 3.1 Let S ⊆ {1, 2, 3}. We define a nonassociative algebra E S n+1 with n + 1 atoms as the complex algebra Cm(C) of the structure C = (C + ,˘, I , T ), where C + = {1 , c 1 , . . . , c n }, c˘= c for any c ∈ C + , I = {1 }, and T is given by We call algebras E S n+1 chromatic. It is easy to see that T is closed under the Peircean transforms. Note that S gives the types of consistent triples rather than forbidden ones, which would be more in line with the remarks in Sect. 1. We stated the definition this way to keep the notation E S n+1 in agreement with [24]. Expressing (2) in terms of the operations in the complex algebra Cm(C), and identifying singletons with their single elements in a set-theoretically incorrect but notationally convenient way, we get Then, putting F = {1, 2, 3} \ S, we arrive at an equivalent version of (2) matching the remarks in Sect. 1 and the statement of Lemma 2.1. With this the next lemma is not difficult to prove: the representation (strong, qualitative or feeble) on the atoms is precisely λ −1 . (1) E S n+1 is strongly representable if and only if there exists an edge n-colouring λ of a complete graph, satisfying (2) E S n+1 is qualitatively representable if and only if there exists an edge n-colouring λ of a complete graph, satisfying Note that the left-hand sides of (a) and (b) above are negations of one another, but the right-hand sides are not. Thus, (a) alone forces ; to be represented as weak composition, but (a) together with (b) force ; to be represented as composition.  For E ∅ n+1 , all triangles are forbidden, so the only possible representation is the colouring of K 2 with a single colour (n = 1). For E {1} n+1 , the only possible representation is K m for m ≥ 3 coloured with a single colour (n = 1). These representations are strong.
The existence of strong (infinite) representations for E

{1,2}
n+1 is Theorem 453 of Maddux [24], and the lack of finite representability for n > 3 is Theorem 455 there. The proof of Theorem 453 can be presented in terms of the representability games of Hirsch and Hodkinson [13]. A finite qualitative representation can be then extracted by Lemma 2.4.
Finite strong representations of E {1,2,3} n+1 are obtained by Jipsen, Maddux and Tuza in [16]. A qualitative representation is much easier to get: take the disjoint union of all possible triangles, and use an arbitrary colour for all the missing edges.

Algebras
The algebra E {3} 4 has a unique strong representation on 4 points. For n > 3, the algebras E {3} n+1 are not associative, so strong representability is impossible, making qualitative representability of particular interest. Our construction for representability will be based around quasigroups, and we direct the reader to a text such as Smith [30] for further background.
Proof It is clear that λ 1 and λ 2 are surjective, so to establish that both determine feeble representations we must ensure that there are no monochromatic or dichromatic triangles. For λ 1 , this is just the cancellativity property of quasigroups: showing that no triangle contains two edges of the same colour. Thus, λ −1 1 is a feeble representation. Now, observe that under λ 1 , each point i is incident to edges of all colours c 0 , . . . , c n−1 except colour c i (this is where idempotency of (Q, ·) is used). The colouring λ 2 extends λ 1 to the extra vertex −1 by adding the missing colour for each vertex: an edge of colour c i between −1 and i, for each i ∈ {0, . . . , n − 1}. It follows immediately that no non-trichromatic triangles are added under this extension: edges leaving −1 are coloured according to the name of the vertex at the other end, so cannot be the same colour; and such an edge does not coincide in colour with any other edge incident to that vertex. Thus, λ −1 2 also is a feeble representations of E {3} n+1 . For the three conditions: (1) implies (2) because λ 2 extends λ 1 , while according to the definition of λ 1 , the 3-cycle condition in (3) is precisely the condition that λ 1 provides instances of each trichromatic triangle, so (1) is equivalent to (3).
It is well known that commutative idempotent quasigroups exist for all and only odd orders, though we omit details because it emerges naturally from the proof of our main classification of qualitative representability E {3} n+1 ; Theorem 5.3 below. We next recall a standard example of a commutative idempotent quasigroup, for every odd order, and verify that it satisfies the 3-cycle condition.
On universe Q n = {0, . . . , n − 1} for an odd n, define a multiplication · by where x 2 stands for the unique integer u ∈ Z n such that 2u = x (mod n), which exists because n is odd.
Proof Let |{i, j, k}| = 3 (equivalently |{c i , c j , c k }| = 3). We claim that it is possible to choose a, y a , x a ∈ Q n , so that we have i = a · x a , j = a · y a , k = x a · y a , verifying the 3-cycle condition. By the unique solution property of (Q n , ·), for each a ∈ Q n , there exists x a , y a ∈ Q n , where a · x a = i a · y a = j.
Then, a simple calculation shows that holds. It follows that choosing a to be k −i − j (mod n) provides the desired solution.
n+1 is qualitatively representable for some n. Then, there exists a finite representation, over a complete graph K m . Equivalently, there is a colouring map λ : K m × K m → C, with |C| = n. Consider an arbitrarily chosen vertex v ∈ K m . Since dichromatic and monochromatic triangles are forbidden, all adjacent edges must be of different colours, so m − 1 ≤ n. As there are n 3 different trichromatic triangles to realise, we must have m ≥ n. So, n ≤ m ≤ n + 1.
If m = n + 1, then every vertex has adjacent edges of all colours. Since edges of the same colour cannot be adjacent, the edges of any given colour form a disjoint union of copies of K 2 and there are no isolated vertices. So the number of vertices is even, hence n is odd.
Next, assume m = n. Since there are n 3 trichromatic triangles to realise, each necessary triangle appears exactly once in the representation. By symmetry of the colouring condition, each edge of a given colour appears the same number of times, and as there are n(n−1) 2 edges and n colours, each edge appears n−1 2 times, so n is odd. Thus, qualitative representations can exist only for odd orders, and Lemma 5.2 and Lemma 5.1 show that all odd orders of size at least 3 are possible.
For the second part, first consider a representation on n vertices. Each vertex has adjacent edges of all but one colour, call it the missing colour. Suppose there are vertices v and w such that they miss the same colour, say m, so that no edge coloured m is adjacent to either v or w. As each edge appears n−1 2 times, K n \{v, w} contains n−1 2 edges coloured m. These edges are disjoint, so K n \{v, w} must contain n − 1 vertices, in contradiction to |K n \ {v, w}| = n − 2. It follows that each vertex has a different missing colour. It follows further that adding one vertex, say v −1 to K n and letting λ(v −1 , v i ) be the missing colour of v i , we obtain a representation on n + 1 vertices. So, every representation on n vertices is uniquely extendable to a representation on n + 1 vertices. Equivalently, every representation on n vertices can be obtained from a representation on n + 1 vertices by removing one vertex. Now, consider a representation on n + 1 vertices. Numbering the vertices It is then routine to verify that {0, . . . , n − 1} is a commutative idempotent quasigroup, and that the colouring coincides with λ 2 (restricting to coincide with λ 1 amongst the vertices v 0 , . . . , v n−1 ).
The two qualitative representations of E where v j is the complex conjugate of v j . Rotate and repeat with a different colour, as in Fig. 2. To get a representation on n points remove the origin and its outgoing edges.
For feeble representations, we only need to consider E 6 Algebras E

{2} n+1
The algebras E Then, any triangle that occurs is dichromatic, and all colours are used, so we have a feeble representation.
Qualitative representations do not exist for n > 2, but proving this is considerably more difficult than for strong representations. We will work our way towards a contradiction. Assume a qualitative representation over a base B exists, and let v ∈ B. The chromatic degree deg χ (v) of v is the number of colours of its adjacent edges. If a vertex v has deg χ (v) = n, we call v chromatically saturated. To distinguish clearly between triangles given as sets of vertices, and triangles given as sets of edges, we will use parentheses (v i , v j , v k ) for vertices and square brackets [ , p, q] for edges/colours.
n+1 has a qualitative representation over a base B, and v 0 ∈ B, then it also has a qualitative representation over a base B , extending B, in which v 0 is chromatically saturated.
n+1 has a qualitative representation over B. Pick an arbitrary v 0 ∈ B. We can assume B = {v 0 , . . . , v m }. If deg χ (v) = n, there is nothing to prove, so assume deg χ (v) = k < n. Let d be a colour not adjacent to v 0 . Add a vertex v and extend the labelling by We will show that this labelling is consistent. As the only possible inconsistency Repeating the extension procedure sufficiently many times produces the desired qualitative representation.
By Lemma 6.2, we can assume that E  Proof Antireflexivity is immediate from the definition. To show transitivity, assume v p v q and v q v r . By definition of , only p can be 0, and in this case the claim follows immediately, as λ(v 0 , v r ) = r by definition. Assume Linearity is also immediate. Take arbitrary v p and v q . If p = 0 or q = 0, the claim holds trivially. Otherwise, we have n+1 , then G = B. Consider a vertex u / ∈ G. As |G| = n + 1, by the pigeonhole principle we must have λ(u, v p ) = λ(u, v q ) for some distinct vertices v p , v q ∈ G. Assume that denotes the value λ(u, v p ) = λ(u, v q ).
Renumbering the colours, we can assume that coincides with the usual strict order on natural numbers. From now on, we will work under this assumption, using < for the induced order on colours. We will refer to G as a naturally ordered subgraph of B. While there may be many subgraphs of this form, it suffices to work with a single chosen one and only vary the choice of u / ∈ G, as the values p, q, depend on the choice of u.
We can assume p < q, so λ(v p , v q ) = q, and thus, = q. We can also assume that q is the largest vertex (amongst v 0 , . . . , v n ) with a repeat, that is, such that λ(u, v s ) = λ(u, v q ) holds for some s = q. Thus, for every distinct s, t ≥ q we have λ(u, v s ) = λ(u, v t ). We will keep these assumptions fixed throughout the rest of the section and invite the reader to draw pictures while reading. Lemma 6.4 Let G be a naturally ordered subgraph of B, and let u / ∈ G. The following hold: (1) If p = , then, for every r ≤ p we have λ(u, v r ) = .
For (6), we have λ(u, v r ) ∈ { , r }, but λ(u, v r ) = contradicts the assumption that v q is the largest vertex with a repeat. Hence, λ(u, v r ) = r . Lemma 6.5 Let G be a naturally ordered subgraph of B, and let u / ∈ G. Then, v = v q−1 or v = v q+1 . Therefore, = q − 1 or = q + 1.
Proof We will apply Lemma 6.4 repeatedly, and referring to numbers (1)-(6) under the tacit understanding that these will be the items in Lemma 6.4. Firstly, assume that p = and p < r < q. From (5), we get λ(u, v r ) = . By assumption, This contradicts (2), so such a configuration cannot occur. As p > 0, such an i always exists. So, when p = , we require that no such r exists, i.e. that = p = q − 1. We can now assume that p = .
Assume 0 < < p. Then, by (1) which is a contradiction, as monochromatic triangles are forbidden.
Finally, the case q = is impossible due to [λ(u, v q ), λ(u, v p ), λ(v p , v q )], so the results cover all cases. Lemma 6.6 Let G be a naturally ordered subgraph of B, and let u / ∈ G. Let p, q and be as in Lemmas 6.4 and 6.5. Then, exactly one of the following three cases occurs: Proof By Lemma 6.5, we have either = q + 1 or = q − 1. Thus, = q + 1 implies λ(u, v q ) = q + 1, and then cases (1), (5), and (6) of Lemma 6.4 apply, showing that the first and the third items of (1) above hold. For the second item, note that λ(u, v q+1 ) ≥ q + 1 contradicts the assumption of v q being the largest vertex with a repeat. If = q − 1, then we have λ(u, v q ) = q − 1 = λ(u, v p ). Recalling that p < q, we will discover that case (3) above will result from p = q − 1 and case (2) from p < q − 1.

Corollary 6.7 Let G be a naturally ordered subgraph of B, and let u /
∈ G. Then, the following hold: Proof Follows by inspecting the cases of Lemma 6.6.
It will be useful to state separately one special case of Lemma 6.6, covering the case where the largest repeat occurs somewhere at the last three vertices.

Lemma 6.8 Let G be a naturally ordered subgraph of B, and let u /
∈ G. Assume λ(u, v n ) = n. The exactly one of the following three cases occurs:  λ(a, b) = n and λ(a, c) = 1 = λ(b, c). We have several cases to consider.

Lemma 6.9 Let a, b, c / ∈ G. Then, G ∪ {a, b, c} is an inconsistent configuration.
Proof Consider λ(a, v n ) and λ(b, v n ). They cannot be both equal to n, so assume λ(a, v n ) = n. Taking u = a, we see that the following possibilities can occur. We begin with case (2) of Lemma 6.8.
Next, cases (1) and (3) of Lemma 6.8 will be dealt with together. For case (1), we have λ(a, v i ) = n for all i < n. As λ(a, b) = n, it follows that λ(b, v i ) = n for all i < n. For case (3), we have λ(a, v n−1 ) = λ(a, v n ) = n − 1, and λ(a, v i ) = n for all i < n − 1. Then, λ(b, v i ) = n for all i < n − 1. Therefore, in both cases we have

Lemma 6.10 Let a, b / ∈ G and c ∈ G. Then, G∪{a, b} is an inconsistent configuration.
Proof By the assumption, c = v j for some j, so λ(a, v j ) = 1 = λ(b, v j ). As in the previous lemma, we can have at most one of λ(a, v n ), λ(b, v n ) equal to n, so without loss assume λ(a, v n ) = n. We analyse possible cases.

Lemma 6.11 Let b, c /
∈ G and a ∈ G. Then, G∪{b, c} is an inconsistent configuration.
Proof We have a = v j for some j. First, we show that we can assume j = n. If j = n, then λ(c, v n ) = 1, and since n > 2 only the case (1) of Lemma 6.8 applies. It follows that λ(c, v i ) = n for all i < n.  λ(b, c). Now, we can assume a = v j for some j = n. (2) and (3) of Lemma 6.8, as n > 2. The remaining case (1) gives λ(b, v i ) = n for all i < n, and then, it follows that

. This excludes cases
producing an inconsistent, monochromatic triangle.
The so-called Ramsey algebras E

{2,3}
n+1 (sometimes also known as Monk algebras and as Comer relation algebras) have received particularly ardent attention in the literature, with the obvious close connection to classical Ramsey-theoretic considerations of avoidance of monochromatic triangles, as discussed in the introduction; [11] and [17] for example. The problem of determining when E {2,3} n+1 has a strong representation was posed by Maddux in [23] and by Comer in [8], and remains open. Strong representability presents very tightly constrained condition (see Lemma 3.2) and proves to be particularly challenging. Ramsey algebras appear in both of the texts [13] and [24], with strong representability for 1 ≤ n ≤ 5 established by Comer [8] and later n = 6, 7 by Maddux [25]. The limits of strong representability were subsequently pushed upward via a series of efforts of Kowalski [18], Alm and Manske [4] and finally Alm [2], which is the current state of knowledge: strong representability is achievable for all n ≤ 2000 except possibly n = 8, 13. The cases of n = 8, 13 and n > 2000 remain tantalisingly unresolved, though the finite field constructions in the above articles have been computationally verified as not providing solutions for n = 8, 13 (see A263308 in the Online Encyclopaedia of Integer Sequences [3]). In [19], Kramer and Maddux develop a version of Ramsey theory where monochromatic and trichromatic triangles must both be avoided. Their results give the existence and an upper bound for the sizes of feeble representations. Qualitative representability presents an interesting intermediate condition, and in this section, we show that qualitative representability is possible for all n.
We employ a variant of Walecki's construction (see, e.g. [5]) originally used to partition K 2k into k Hamiltonian paths. Let the vertices of K 2n be distributed evenly on a circle. Colour a zigzagging path in one colour, as in Fig. 3. Then, rotate the picture one step, change the colour and repeat. Rotating n times and using n different colours gives the desired colouring of K 2n .
It is clear that no monochromatic triangles occur, so it remains to be shown that all non-monochromatic triangles are realised. To avoid cluttering subscripts, we will use 1, . . . , n as the colours, instead of c 1 , . . . , c n . For each n ∈ N, define a map s n : Z → {1, . . . , n} by x → R n (x − 1) + 1, where R n (k) is k (mod n). In particular, we have s n (s 2n (a) + s 2n (b)) = s n (a + b) for all a, b ∈ Z and n ∈ N. We will use this fact repeatedly below to omit s 2n within parentheses.
It can be seen from the construction that the colour of (u 1 , u 1+s ) is s+1 2 , for each s ∈ {1, 2, . . . , 2n −1}. Rotating i steps clockwise, we get that for each i ∈ {1, . . . , 2n} and s ∈ Z\{2mn : m ∈ Z}, the colour of (u i , u s 2n (i+s) ) is To see this, note that the colour of (u 1 , u s 2n (1+s) ) is s n ( s+1 2 ) and then, shift the colour i − 1 steps forward using the rotational symmetry and the fact that the colours cycle with period n.

Lemma 7.1 The colouring defined above realises all non-monochromatic triangles.
Proof Take i, j, k ∈ {1, . . . , n} with i < j. We will show that there is a triangle coloured (i, j, k). We consider two cases: k = i and k = j. Assume first that k = i. Put Note that s, t ∈ Z \ {2mn : m ∈ Z}. Hence, the colour of (u , u s 2n ( +s) ) is The colour of (u , u s 2n ( +t) ) is As i < j, we have s < t. Now, which is not a multiple of 2n. Hence, the colour of (u s 2n ( +s) , u s 2n ( +t) ) is So, (u , u s 2n ( +s) , u s 2n ( +t) ) is a triangle with edges coloured by i, j, and k. Next, assume k = j and put := s 2n (i + j − k), s := 2k − 2 j, and t := 2k − 2i + 1.
Again, s, t ∈ Z \ {2mn : m ∈ Z}. By calculations similar to the first case, we get that the colour of (u , u s 2n ( +s) ) is i, and the colour of (u , u s 2n ( +t) ) is j. Moreover, so the colour of (u s 2n ( +s) , u s 2n ( +t) ) is Choosing between the two cases, as needed for (i) k = i, k = j, (ii) k = i, k = j, or (iii) k = i, j we can construct all non-monochromatic triangles.
Proof Immediate by Lemma 7.1.

{1,3} n+1
These algebras are better known as Lyndon algebras. Lyndon [22] showed that for n ≥ 4, strong representations of the algebra E {1,3} n+1 precisely correspond to affine planes of order n−1, so that by the Bruck-Ryser Theorem [7] and standard field constructions, there are infinitely many n for which E {1,3} n+1 is strongly representable, and infinitely many for which it is not strongly representable. This was used by Monk to show that the strongly representable relation algebras admit no finite axiomatisation [26]. We now explore how qualitative and even feeble representations of E {1,3} n+1 relate similarly to geometric objects. The natural starting point will be feeble representations, which we find correspond to parallelisms of linear spaces, in the sense of finite geometry. We direct the reader to a text on linear spaces such as [6] for further background on linear spaces, though the connection with feeble representations of Lyndon algebras is presented here for the first time.
A linear space is a system (G, L), where G is a set of points, and L a family of subsets of G called lines satisfying the following axioms: (LS1) Every pair of distinct points is contained in a line; (LS2) Every pair of distinct lines have intersection that is either empty or a singleton; (LS3) Every line contains at least two points.
An obvious consequence of LS1 and LS2 is that each pair of points p, q are contained within a unique line, which we denote by pq. As usual, points lying on the same line are said to be collinear, two lines are incident if they intersect to a common point.
A parallelism of a linear space (G, L) is an equivalence relation on L with the property that lines within a common block do not intersect. The blocks of are referred to as parallel classes and lines within a block are parallel. The identity relation on L is always an example of a parallelism, and we refer to this as the trivial parallelism. But in general there may be many other parallelisms, with various numbers of parallel classes. The next theorem shows that n-block parallelisms of linear spaces precisely capture feeble representations of the algebra E {1,3} n+1 , modulo the choice of bijection between diversity atoms and parallel classes. n+1 to parallel classes as given. For convenience, we allow the parallel classes to share the same name as the colour of the atom to which they are matched. So the (attempted) feeble representation a θ of a diversity atom a labels the edge from p to q if p is collinear to q by way of a line in a (the identity atom is represented, as required, by the identity relation). By Axioms LS1 and LS2, all pairs receive a unique colour. By LS3, all colours appear, so that θ is a feeble representation of E n+1 on a set G. To define the lines L, for each diversity atom a we will include maximal cliques with respect to edges a θ as lines in L. To define the parallelism , we define lines to be parallel if they are cliques with respect to a common atom. Because each pair of points in G are related by a θ for some atom a, we have that Axiom LS1 holds. Because no edge is coloured by more than one atom, we have that Axiom LS2 holds. It is trivial that Axiom LS3 holds. Because θ is a feeble representation, each atom labels at least one edge, so that the number of blocks in is exactly n. Because there are no dichromatic triangles in the feeble representation θ , we have that the lines within common blocks of do not intersect, so that is a parallelism of (G, L). Finally, it is clear that the representation of E {1,3} n+1 over (G, L), agrees with the one defined in the first half of the proof, with the obvious matching of atoms to parallel classes.
An example of a linear space with parallelism is given in Fig. 4. It is not a Lyndon geometry, but its resemblance to the projective plane of order 3 is not accidental: it illustrates a construction to be used shortly for projective planes of order at least 4.
We mention in passing that an alternative presentation here is to begin with a linear space (G, L) in which there is a distinguished line L ∞ (the line at infinity) that is incident with all other lines and contains precisely n points. Provided that every line other than L ∞ contains at least 3 points, we may obtain a linear space (G\L ∞ , L\{L ∞ }) and define an n-block parallelism by letting two lines be parallel if they were incident in (G, L) to a common point on L ∞ . Conversely, any parallelism of a linear space gives rise to a new linear space by letting the blocks of be treated as additional new points, and the set of blocks of be considered as an additional new line ("at infinity"). This of course is a general case of the familiar interplay between projective planes and affine planes.
Proof This is immediate, as it simply states the condition on representations in the geometric formulation shown equivalent by Theorem 8.1.
We will refer to a linear space with parallelism (G, L, ) as an affine Lyndon geometry if it satisfies the extra axioms LS4 and LS5. The linear space obtained from an affine Lyndon geometry by adjoining the line at infinity and associated directions will be called a Lyndon geometry; note that a Lyndon geometry is a linear space with a distinguished line. As with affine planes and projective planes, these two concepts in some sense differ only in that parallel classes and the parallelism is given the name "points at infinity" and "line at infinity". In order to match the case for affine planes, we let the order of affine Lyndon geometry (G, L, ) be the number n such that has n + 1 blocks.
We now show that affine Lyndon geometries of every order exist.
Let p be the unique point in L 1 ∩ L 2 . As there are n lines in L ∩ o 3 , and Since L 1 , L 2 and L 3 are lines from L, not all through the same point, they intersect pairwise at 3 distinct points. Hence, p ∈ L 1 ∩ L 2 , q ∈ L 1 ∩ L 3 and r ∈ L 2 ∩ L 3   n+1 is qualitatively representable for all n ≥ 3. Proof By Theorem 8.3, it suffices to cover all n ≥ 3 using Theorem 8.4. There exist affine planes of every prime power order, so orders 3 and 4 are covered, and Theorem 8.4 shows that there are affine Lyndon geometries of all orders p, p + 1, . . . , 2( p − 1), for all p ≥ 5. All numbers n ≥ 2 lie in the interval [ p, 2( p − 1)] for some prime p: this is trivial to verify for small n, and follows, for example, from Nagura's variant of Bertrand's postulate [27], showing that for m ≥ 25 there is a prime between m and 5m/4.

Infinite cardinalities
The definition of chromatic algebras E S n easily extends to allow for the case n = κ for an infinite cardinal κ. Strong representations for E S κ are known for S = {1, 3} (by way of affine planes of order κ), while strong representations for E S κ in the case of S = {1, 2}, {2, 3}, {1, 2, 3} are easily achievable using the game-theoretic methods of Hirsch and Hodkinson [13], but they are somewhat convoluted, being subdirect products of representations of countable subalgebras of E S κ (even in the case κ = ω).
We will not investigate them here. The remaining cases do not admit strong representations: for ∅ and {1} this is trivial, while for S = {3} and S = {2}, the situation is discussed in the relevant Sects. 5 and 6. For qualitative representations, there emerges an interesting contrast with the finite colour results, namely, E Proof For S = ∅ or S = {1}, it is trivial that no representations exist. Now, assume that 3 ∈ S, and let C κ be the set of all colours. There are κ triangles (T α : α < κ) to be witnessed in order to achieve a qualitative representation of E S κ . We will construct a chain of complete networks (N α : α < κ) and a chain of sets of colours (C α : α < κ) such that C α is the set of all colours used in N α . We use ⊗ and ⊕ for cardinal multiplication and addition, and + for ordinal addition. Let N 0 be the empty network. Assume inductively that a chain of complete networks (N β : β < α) has been constructed, such that (i) each N β witnesses the triangles T γ for all γ < β, (ii) |N β | ≤ 3 ⊗ |β|, and (iii) |C β | ≤ 9 ⊗ |β| ⊗ |β|. Construct N α as follows.
The union α<κ N α is a well-defined qualitative representation of E S κ . If 3 / ∈ S but 2 ∈ S, we can apply same methodology as for 3 ∈ S, except that now to complete N α to N α for a successor α we select a single colour from κ unused ones for all the missing edges.

Summary
The