A family of Condorcet domains that are single-peaked on a circle

Fishburn's alternating scheme domains occupy a special place in the theory of Condorcet domains. Karpov (2023) generalised these domains and made an interesting observation proving that all of them are single-picked on a circle. However, an important point that all generalised Fishburn domains are maximal Condorcet domain remained unproved. We fill this gap and suggest a new combinatorial interpretation of generalised Fishburn's domains which provide a constructive proof of single-peakedness of these domains on a circle. We show that classical single-peaked domains and single-dipped domains as well as Fishburn's alternating scheme domains belong to this family of domains while single-crossing domains do not.


Preliminaries
A Condorcet domain is a set of linear orders on a given set of alternatives such that, if all voters of a certain society are known to have preferences over those alternatives represented by linear orders from that set, the pairwise majority relation of this society is acyclic.Any background information on Condorcet domains can be found in survey papers [Monjardet, 2009, Puppe andSlinko, 2023].
One of the best known Condorcet domains is the domain of single-peaked linear orders on a line spectrum Black [1958].Recently Peters and Lackner [2020] generalised this domain to single-peaked domains on a circle.And, although so generalised domains are not necessarily Condorcet, as will be demonstrated in this paper, they have a role to play in the theory of Condorcet domains too.
Intuitively, a domain is single-peaked on a circle if all the alternatives can be placed on a circle so that, for every order of the domain, we can 'cut' the circle once so that the given order becomes single-peaked on the resulting line spectrum.The location of the cutting point may differ for different orders of the domain.
Let us briefly touch some of the basics of Condorcet domains.More information about them can be found in Monjardet [2009], Puppe and Slinko [2023].
By L(A) we will denote all linear orders on the set of alternatives A which will always be assumed to be finite.For a linear order v ∈ L(A) and two alternatives x, y ∈ A we write x ≻ v y if v ranks x higher than y.The set of alternatives A is often taken as [n] = {1, 2, . . ., n}.Up to an isomorphism, for n = 3 there are only three maximal Condorcet domains: The domain D 1 on the left contains all the linear orders on [3] in which 2 is never ranked first, the domain D 2 in the middle contains all the linear orders on [3] in which 1 is never ranked second, and domain D 3 on the right contains all the linear orders on [3] in which 2 is never ranked last.Following Monjardet [2009], we denote these conditions as 2N {1,2,3} 1, and 1N {1,2,3} 2, and 2N {1,2,3} 3, respectively.Definition 1.Any condition of type xN {a,b,c} i with x ∈ {a, b, c} and i ∈ {1, 2, 3} is called a never condition since it being applied to a domain D requires that in orders of the restriction D| {a,b,c} of D to {a, b, c} alternative x never takes ith position.We say that a subset N of is a complete set of never-conditions if N contains exactly one never condition for every triple a, b, c of elements of A.
The following criterion is a well-known characterisation of Condorcet domains that goes back to Sen [1966].See also Theorem 1(d) in Puppe and Slinko [2019] and references therein.
Criterion 1.A domain of linear orders D ⊆ L(A) is a Condorcet domain if and only if it satisfies a complete set of never conditions.
The following property of Condorcet domains was shown to be very important.
Definition 2 (Slinko [2019]).A Condorcet domain D is copious if for any triple of alternatives a, b, c ∈ A the restriction D| {a,b,c} of this domain to this triple has four distinct orders, that is, |D| {a,b,c} | = 4.We note that, if a Condorcet domain is copious, then it satisfies a unique complete set of never conditions of the form (1). Copiousness is often an important step in proving maximality of the domain.
Proposition 1.Let N be a complete set of never conditions and D(N ) is the set of all linear orders from L(A) that satisfy N .If D(N ) is copious, then D(N ) is a maximal Condorcet domain.
Proof.Suppose D(N ) is copious but not maximal.Then there exists a linear order u such that D ′ = D(N ) ∪ {u} is a Condorcet domain.Since u / ∈ D(N ) for a certain triple of alternatives a, b, c the domain D ′ | {a,b,c} contains an order on a, b, c which is not in D| {a,b,c} .But then D ′ | {a,b,c} contains five orders on a, b, c which is not possible as it would not be a Condorcet domain.
Many Condorcet domains are defined relative to some sort of societal axis, also called spectrum.In politics it is often referred to as left-right spectrum of political opinions.Definition 3. A domain D ⊆ L(A) is said to be (classical) single-peaked if there exists a societal axis (spectrum) such that for every linear order v ∈ L(A) and a ∈ A the upper counter set U (a, v) = {b ∈ A | b ≻ v a} is convex relative to the spectrum.By SP n (◁) we will denote the domain of all single-peaked orders on ◁.
Up to an isomorphism A is often taken as [n] and the societal axis as 1 Never conditions allow us to define useful classes of Condorcet domains as was pioneered by Peter Fishburn [1997] who introduced the so-called alternating scheme of never conditions and constructed Condorcet domains of large order.Karpov [2023] generalised his scheme as follows.
Definition 4 (Karpov [2023]).Let A = [n].A complete set of never conditions is said to be a generalised alternating scheme, if for some subset K ⊆ [2, . . ., n − 1] and for all 1 ≤ i < j < k ≤ n we have jN {i,j,k} 3, if j ∈ K, and jN {i,j,k} 1, if j / ∈ K. (1) The domain which consists of all linear orders satisfying the generalised alternating scheme is called the generalised Fishburn's domain or GF-domain.
The original Fishburn's alternating scheme has K equal to the set of even numbers in [2, . . ., n−1].The GF-domain constructed using a subset K ⊆ [2, . . ., n − 1] will be denoted as F K .Every GF-domain has orders 12 . . .n and n . . .21 as they obviously satisfy conditions (1).Domains with this property are said to have maximal width [Puppe, 2018].

Combinatorial representation of GF-domains
The idea of this representation comes from an example in Danilov et al. [2010].
A set of n vertices on a circle, some white and some black, are numbered by integers 1, 2, . . ., n.We will often identify the vertices with the numbers on them.
Definition 5.An arrangement of black and white vertices on a circle will be called a necklace and the vertices themselves will be called beads.Definition 6.A set of beads X ⊆ [n] is said to be white convex (or simply w-convex) if • X is an arc of the circle; • X does not consist of a single black bead; • There does not exist i < j < k such that i, k ∈ X, j / ∈ X and j is white. where Any flag (2) of w-convex sets defines a linear order v = x 1 x 2 . . .x n on [n], where {x i } = X i \ X i−1 (for convenience we assume that X 0 = ∅).
Definition 8. Given a necklace S the domain D(S) is the set of all linear orders corresponding to flags of w-convex sets.
Our example shows that the construction is promising and generating maximal GFdomains.Let us generalise these examples and offer a new combinatorial representation of GF-domains from which we will deduce their maximality.
Example 3.For n = 3 we have two options: one with K 1 = ∅ and another with K 2 = {2}.
Proof.Since L = ∅ in S K there are no black beads and beads 1 and n are neighbours on the circle.The only w-convex set containing both of them is the longer arc Z with endbeads 1 and n, that is, {1, 2, . . ., n}.Any arc X ⊆ Z is w-convex.So w-convex sets coincide with the upper contour sets of the classical single-peaked domain with the spectrum 1 ◁ 2 ◁ . . .◁ n − 1 ◁ n.
Lemma 1.In domain D(S K ): (i) A black bead a satisfies the never-top condition in any triple {a, b, c} such that b < a < c, that is aN {a,b,c} 1.
(ii) A white bead a satisfies the never-bottom condition in any triple {a, b, c} such that b < a < c, that is aN {a,b,c} 3.
Proof.(i) Suppose a ∈ L ⊆ [2, . . ., n] is black and there is a linear order v in D(S K ) whose restriction to subset {a, b, c} is abc with b < a < c.Let X be the first w-convex set from the flag corresponding to v that contains a. Then a = ℓ i and X = {ℓ i , . . ., ℓ 1 , 1, k 1 , . . ., k j } does not contain b and c (here it is possible that j = 0).We note that b / ∈ L as b < ℓ i and not in X, thus b = k s with s > j.But then 1 < k s < ℓ i and X is not w-convex.
(ii) Suppose a ∈ K ⊆ [2, . . ., n] white and a certain linear order v in D(S K ) has restriction bca to subset {a, b, c} is with b < a < c.Let X be the largest w-convex set in the flag corresponding to v that still does not contain a. Then it contains b and c which contradicts to w-convexity of X.Thus, abc, cba, bac, bca all belong to the restriction of S K onto {a, b, c}.
3. a is white; b, c are black.Then obviously, abc and acb belong to the restriction of S K onto {a, b, c}.But also in the restriction of S K onto {n, a, b, c} we have ncba and ncab, hence cba and cab belong to S K | {a,b,c} , so this restriction has four suborders.
4. b is white; a, c are black.Then bac and bca are in S K | {a,b,c} as well as ncba and 1abc (or 1cba and nabc belong to the restrictions of S K onto {n, a, b, c} and {1, a, b, c}, respectively.Hence cba and abc are in S K | {a,b,c} and this restriction has four suborders as well.
5. a is black; b, c are white.Then bca, cba, bac and 1abc belong to respective restrictions, so four suborders.
6.If a and b are black and c is white.Then cba and cab belong to S K | {a,b,c} together with 1bca and 1bac.These are all possible cases.
Combining Proposition 1 with Lemma 2 we get Theorem 2. For any K ⊆ [2, . . ., n − 1] the domain F K is a maximal Condorcet domain.
The universal domain L(A) has many representations.One of the most useful ones is by the permutohedron of order n [Monjardet, 2009], whose vertices are labeled by the permutations of [n] from the symmetric group S n .Two vertices are connected by an edge if their permutations differ in only two neighbouring places.Domains can be considered as a subgraphs of the permutohedron.Definition 9. A domain D of maximal width is called semi-connected if the two completely reversed orders e and ē from D can be connected by a shortest path (geodesic path) in the permutohedron whose all vertices belong to D. It is directly connected, if any two orders of a domain are connected by a shortest path in the permutohedron that stays within the domain.
Maximality of GF-domains has a number of profound consequences.
Theorem 3. Every GF-domain D is a directly connected domain of maximal width.
Proof.We have already noticed that GF-domains have maximal width containing 12 . . .n and n . . .21.By their definition, they are also the so-called peak-pit domains which means that they satisfy a complete set of never-top and never-bottom conditions.By Theorem 2 of Danilov et al. [2012] maximality of D implies that this is a tiling domain and, in particular, it is semi-connected.It has been observed in Puppe [2016] (Proposition A.1) that maximal semi-connected domains are directly connected.
Let us now give a formal definition of a domain single-peaked on a circle.
Definition 10 (Peters and Lackner [2020]).A linear order v ∈ L(A) is said to be singlepeaked on a circle, if alternatives from A can be placed on a circle in anticlockwise order so that for every alternative a ∈ A the upper counter set U (a, v) = {b ∈ A | b ≻ v a} is a contiguous arc of the circle.
A domain D ⊆ L(A) is said to be single-peaked on a circle if there exists an arrangement of alternatives on that circle such that each order of D is single-peaked on a circle relative to their common arrangement of alternatives.
Our Theorem 1 as a corollary provides a constructive proof of the following theorem.
Corollary 1 (Karpov [2023]).Every GF-domain F K is single-peaked on a circle.Proof.By Theorem 1 F K is isomorphic to D(S K ).The statement now follows from the fact that any upper contour set of this domain is a contiguous arc of the necklace.Original Karpov's proof was based on the characterisation of single-peaked on a circle domains by means of forbidden configurations given in Peters and Lackner [2020].
The question may be asked: Are all peak-pit maximal Condorcet domains of maximal width are single-peaked on a circle?The answer is negative.
Theorem 4. For any n ≥ 4 single-crossing maximal Condorcet domain are not singlepeaked on a circle.
In this proof effectively we spotted in any single-crossing maximal Condorcet domain one of the forbidden configurations described in Peters and Lackner [2020].

Conclusion and future work
Now we know that the single-peaked domain, Fishburn's domain and single-dipped domain are members of the same family with single-peaked and single-dipped being the two extremes.It would be interesting to investigate how the size of domain F K depends on K.It is well-known that when K = [2, . . ., n − 1] or K = ∅ (the case of classical single-peaked and single-dipped domains) we have |F K | = 2 n−1 and when K is the set of even numbers in [2, . . ., n − 1] (the case of classical Fishburn's domain) Galambos and Reiner [2008] gave the exact formula for the cardinality of F K : for even n; It is reasonable to conjecture that these are the most extreme cases and the cardinality of |F K | for various K must be somewhere in between.However, since Fishburn's domain is not the largest peak-pit domain of maximal width for at least n ≥ 34 [Karpov and Slinko, 2023], we do not even know if Fishburn's domains are the largest among all GF-domains.

Acknowledgements
The author thanks A. Karpov for bringing the paper Danilov et al. [2010] to his attention.

Theorem 1 .
D(S K ) = F K .Proof.By Lemma 1 we know that D(S K ) ⊆ F K .Let us prove the converse.Let v = a 1 . . .a n ∈ F K .We define the k-th ideal of v as Id k (v) = {a 1 , . . ., a k }.It is enough to show that for any k ∈ [n] the set Id k (v) is w-convex.Suppose not.Then there exist a, b, c ∈ [n] with a, c ∈ Id k (v) and b / ∈ Id k (v)satisfying a < b < c and b ∈ K is white.Then the restriction of v onto {a, b, c} is acb or cab in violation of bN {a,b,c} 3.This contradiction proves the theorem.Lemma 2. For any K ⊆ [2, . . ., n − 1] the domain F K is copious.Proof.We will use Theorem 1 and consider D(S K ) instead.Let a, b, c ∈ [n] with a < b < c.We need to consider several cases.1. a, b, c are all white.Then a = k p , b = k s , c = k r with p < s < r.The following sets are w-convex: {k p }, {k s }, {k r }, {k p , . . ., k s }, {k s , . . ., k r }, {k p , . . ., k s , . . ., k r }.