Minimal automorphic superpositions of crowns

After an overview of uses and explorations of minimal automorphic ordered sets, we present a criterion when certain superpositions of crowns are minimal automorphic. A key lemma to exclude certain retracts can also be applied to ordered sets recently presented in Schröder (Order, https://doi.org/10.1007/s11083-021-09574-3, 2021).

of [12] states the impression that crowns should play a central role in the description of ordered sets that do not have the fixed point property. This is mirrored in [1], where it is shown that any connected automorphic ordered set is a union of circulants, that is, ordered sets of height 1 such that there is an automorphism that induces a cyclic permutation on the maximal elements and a cyclic permutation on the minimal elements. For any fixed point free automorphism, any pair of orbits that is not discrete necessarily generates a circulant ordered set. However, for a union of circulants, the typically rather complex interaction of, or linking between, the circulants determines whether there is a fixed point free automorphism for the whole ordered set. Problem 2 in [12] did not introduce the terminology "minimal automorphic ordered set," but it can be seen as a, beyond ordered sets of height 1 in [11], first attempt to characterize the fixed point property for a class of ordered sets, namely, ordered sets of height 2, through a listing of forbidden retracts. It should be noted that Section 9.5 in [15] shows that a simple complete characterization of minimal automorphic ordered sets of height 2 is essentially impossible, a fact that was communicated to the author by an anonymous referee in the early 1990s. This observation is consistent with the fact that the decision problem whether an ordered set has a fixed point free order-preserving self map is NP-complete, see [4].
In addition to crowns, there are a few classes of ordered sets which are known to be minimal automorphic and known to be the forbidden retracts for the fixed point property in certain classes of ordered sets. Truncated finite Boolean lattices are minimal automorphic and they are the forbidden retracts in the class of ordered sets discussed in [14]. It is easy to see that 4-crown towers are minimal automorphic. By Theorem 1 in [7], they are the forbidden retracts in the class of ordered sets of width 2, by Theorem 4 in [8], they are the forbidden retracts in the class of N-free ordered sets of dimension 2, and, in Exercise 4-41 in [15], they are the forbidden retracts in the class of collapsible ordered sets, which includes the class of ordered sets of (interval) dimension For part 1, suppose, for a contradiction, that P has an irreducible point a. Without loss of generality, we can assume that a has a unique upper cover b. Because is an automorphism, for all k ∈ N, we have that k (b) is the unique upper cover of k (a). Moreover, { k (a) : k ∈ N} ∩ { k (b) : k ∈ N} = ∅. Because the map that maps every k (a) to k (b) and that leaves all other points fixed is a retraction, we conclude that R := P \ k (a) : k ∈ N is a retract of P. Because maps R to itself, R is an automorphic retract of P, contradicting that P itself is minimal automorphic.
For part 2, let A ⊂ P be a nontrivial maximal order-autonomous subset of P. Suppose for a contradiction, that P is not series decomposable. Because is an automorphism, every k [A] is a maximal order-autonomous subset of P, too.
If we had [A] = A, then, because (see, for example, Proposition 7.11 in [15]) any two maximal orderautonomous subsets of a connected, non-series-decomposable ordered set are either equal or disjoint, for all j, k ∈ N, we would have that j [A] and k [A] are either equal or disjoint. Let m ∈ N be the smallest positive integer such that m [A] = A. Let a ∈ A and define Then r would be a retraction and the function : m−1 (a) := a would be an automorphism that has no fixed points, which cannot be, because P is minimal automorphic. Thus Then r is a retraction and | r [P] is an automorphism that has no fixed points, a contradiction.
we obtain a contradiction as in part 1. This leaves the case b ∈ k (a) : k ∈ N . In this case, let m ∈ N be so that b = m (a).
Then jm (a) : j ∈ N is a nontrivial order-autonomous antichain, which implies via part 2 that P is seriesdecomposable.
We conclude this introduction with an easily proved lemma that is part of the folklore on retractions and which shows that there is no loss of generality if we restrict ourselves to retractions that map minimal elements to minimal elements and maximal elements to maximal elements. For any element x in an ordered set P, we define ↑ x := { p ∈ P : p ≥ x} and ↓ x = {p ∈ P : p ≤ x}. Recall that an element m of an ordered set P is called minimal iff ↓ m = {m}, and that it is called maximal iff ↑ m = {m}. We define min(P) to be the set of all minimal elements of P and we define max(P) to be the set of all maximal elements of P. Lemma 1.2 Let P be a finite connected ordered set and let R ⊆ P be a retract of P with retraction r : P → R. Then there is a retraction s : P → R such that s[min(P)] ⊆ min(R) and s[max(P)] ⊆ max(R).
Proof We define the retraction s : P → R as follows. For every x ∈ P that is not in min(P) ∪ max(P) or that is in min(P) and its image under r is in min(R) or that is in max(P) and its image under r is in max(R), we set s(x) := r (x). For every t ∈ max(P) such that r (t) / ∈ max(R), we can pick a q t ∈ max(R)∩ ↑ r (t) and set s(t) := q t . For every b ∈ min(P) such that r (b) / ∈ min(R), we can pick a q b ∈ min(R)∩ ↓ r (b) and set s(b) := q b . It is easy to check that s is a retraction onto R.

Superpositions of crowns
Superpositions of crowns (see Definition 2.1 below) were defined in [3], but they already occurred in Figure  4 in [12]. In [10], three straight superpositions of 3 crowns with 10 elements each are first modified and then merged in a certain way to construct an ordered set with the fixed point property, but not the strong fixed point property. 1 The ordered sets in Example 9 in [16] are obtained by merging two straight superpositions . . , h}} and, for all (x, y) ∈ S with y < h, we have that (x, y) < (x, y + 1), (x + 1 mod n, y + 1), and the only additional comparabilities are those dictated by transitivity.
We will sometimes omit the specification that the superposition of the crowns is straight. Section 5 will indicate that any other way to vertically merge crowns is not likely to produce minimal automorphic ordered sets. For the remainder of this paper, we will not explicitly indicate that these computations are modulo n. Remark 2.3 (Also see Fig. 1.) Clearly, for any superposition S of h crowns with 2n elements each and for any (a, b), (e, f ) ∈ S, we have that (a, b) < (e, f ) iff there are q, t ≥ 0 such that 0 < q + t ≤ h − b, e = a + q and f = b + q + t.
Equally clearly, the function (x, y) → (x + 1, y) is a fixed point free automorphism.
In this section, Proposition 2.7 will show that, under certain circumstances, a superposition of crowns can indeed retract onto a smaller superposition of crowns. Theorem 4.2 will show that, for superpositions of crowns with 3h < n, the converse of Proposition 2.7 holds. The idea for the retractions in Lemma 2.5 below is displayed in Fig. 1: Essentially, if there is a way to partition S into the right set of triangles and diamonds, then there is a retraction onto a smaller superposition of crowns. The idea for the retractions in Lemma 2.6 below is similar, but a little more complicated because wraparounds modulo n need to be considered.

Definition 2.4
Let S be a superposition of h crowns with 2n elements each.
1. We define a set C ⊆ S to be a regular surrounding cycle 2 in S iff there is an x such that max(C) = {(x + ih, h) ∈ S : i ∈ N} and min(C) = {(x + ih, 0) ∈ S : i ∈ N}. 2. Let C 1 , . . . , C m be regular surrounding cycles. We define U to be a filled union of regular surrounding cycles induced by C 1 , . . . , C m iff U is the set of all (x, y) ∈ S for which there are, not necessarily distinct, The boxed elements in Fig. 1 show a filled union of one surrounding cycle C (a union of a single surrounding cycle is allowed) with x = 0.

Lemma 2.5 Let S be a superposition of h crowns with 2n elements each and let U be a filled union of regular surrounding cycles. Then U is a retract of S.
Proof Let C 1 , . . . , C m be the regular surrounding cycles that induce U . Without loss of generality, we can assume that ( We define x z+1 := n, and, for every ∈ {1, . . . , z}, we define y 0, To prove that r is order-preserving, Let (a, b) < (e, f ). Without loss of generality, we can assume that and ) and r is order-preserving.

Lemma 2.6
Every superposition of h = kn crowns with 2n elements each retracts to a 4-crown tower.
Proof First, we consider the case that S is a superposition of h crowns with 2n elements each such that h = n. We let We claim that r is a retraction. Because the sets in the definition of r are pairwise disjoint, r is well-defined. Because the union of these sets is S, r is totally defined. Because r [S] = R and r is the identity on R, we have that r is idempotent.
To prove that r is order-preserving, let x < y. Note that this means, in particular, that x is not maximal and that y is not minimal.
We have thus proved that r is a retraction. Now consider a superposition S of h = kn crowns with 2n elements each.
For h n, let d := gcd(h, n) > 1. Then C := (ih, 0), (ih, h) : i = 0, . . . , n d is a regular surrounding cycle in S that does not contain the element (1, 0). The claim follows via application of Lemma 2.5 to the filled union of regular surrounding cycles induced by the single regular surrounding cycle C.

Pulling bipartites
Visually, the main idea behind minimal automorphic superpositions of crowns and behind minimal automorphic ordered sets as in Example 9 in [16] is that the diagrams of these sets, when drawn in 3 dimensions, surround a "central void," and that fixed point free maps can only be achieved through rotations around (or, in the case of superpositions of crowns, reflections through) this "central void." This section shows that retracts that surround the "central void" must contain certain sets of maximal or minimal elements. Lemma 3.5 and the definitions leading up to it may look technical, but it provides the appropriate tool for superpositions of crowns in Sect. 4, for ordered sets as in Example 9 in [16] in Sect. 6, and possibly for other classes of ordered sets. Convention 3.1 Throughout this section, all arithmetic is modulo n. We therefore will forgo any explicit indication of this fact. n, k(0), . . . , k(n − 1) ∈ N. An ordered set P of height 1 is called a pulling (n, n)-bipartite with signature (k(0), . . . , k(n − 1)) (also see Fig. 2) iff it has the following properties. Informally, pulling bipartites are ordered sets of height 1 such that, for the j th maximal element, there is an offset k j such that all minimal elements below it are at most j steps "behind" it. Example 3. 3 We find pulling (n, n)-bipartites as the union of the set of minimal elements with the set of maximal elements in the following ordered sets.
Lemma 3.5 Let P be a pulling (n, n)-bipartite with signature (k(0), . . . , k(n − 1)) such that 2K < n. Let R ⊆ P be a retract of P such that there are g, d ≥ 0 such that g + d + K < n and such that, for every retraction r : P → R that maps minimal elements to minimal elements and that maps maximal elements to maximal elements, the following hold.
Then, for every maximal element t j ∈ R with a circular train T (t j ) behind it, the train T (t j ) behind t j is contained in R, too.
Proof We start with a sequence of claims and then prove the result. Claim 2: Let t j be a maximal element such that there is an ∈ {0, . . . , d} such that r (t j ) = t j+ . Then there is an ∈ { , . . . , d} such that r (b j−k( j) ) = b j−k( j)+ .
By Claim 2, there is an ∈ { , . . . , d} such that r (b j−k( j) ) = b j−k( j)+ . Now, by Claim 1, there is an ∈ { , . . . , d} such that r (t j−k( j) ) = t j−k( j)+ . Now we can finish the proof of this lemma. Let t j ∈ R have a circular train T (t j ). First suppose, for a contradiction, that there is a maximal element t u ∈ T (t j ) \ R. Without loss of generality, we can assume that u is chosen such that there is a t v ∈ T (t j ) ∩ R such that u = v − k(v). By Claim 3 applied to t v with = 0, and because t u / ∈ R, there is an ∈ {1, . . . , d} such that r (t u ) = t u+ . Because T (t j ) is circular, repeated application of Claim 3 yields that r (t j ) = t j+ for some ∈ {1, . . . , d}, a contradiction to r (t j ) = t j . Thus all maximal elements of T (t j ) are in R. Now, for every maximal element t u of T (t j ), because 2K < n, every minimal element b u−k(u) is the unique common lower bound of t u−k(u) and t u . Because r (b u−k(u) ) is a common lower bound of r (t u−k(u) ) = t u−k (u) and

Superpositions of crowns with 3h < n
We can now provide a characterization of minimal automorphic superpositions of h crowns with 2n elements each in the case that 3h < n. Recall that an ordered set P is dismantlable iff the elements of P can be enumerated as p 1 , . . . , p z such that, for j = 1, . . . , z − 1, we have that p j is irreducible. Also recall (see Corollary 2 in [11]) that dismantlable ordered sets have the fixed point property and can therefore not be automorphic.
By symmetry, the claim is proved if it can be proved in the case that a = 0 and (n − 1, 0) ∈ R. Suppose, for a contradiction, that the claim does not hold in this case.
Let s be the smallest index such that there is a maximal element (s, h) ∈ R that is not above (h − 1, 0). Then s > t + h, because otherwise R would contain a minimal common lower bound (z, 0) of (s, h) and (t, h) and, because 3h < n, the index z would be in {0, . . . , t}, which cannot be. Now R∩ ↑ (s − h, 1) = ∅ and that leads, via symmetry and Claim 1, to R being dismantlable, a contradiction.
The claim now follows from application of Lemma 3.5 to P = min(S) ∪ max(S). Now consider the case that A is an arbitrary automorphic retract of S. Let a : S → A be a retraction. We define a map e : A → S as follows.
Then r := e • a is a retraction from S to R := e[A], which is an automorphic retract of S that is isomorphic to A (the isomorphism is e and its inverse is a| e [A] ) such that max(R) ⊆ max(S) and min(R) ⊆ min(S).
Proof The direction "⇒" follows from the contrapositive of Proposition 2.7.
For the direction "⇐," let A ⊆ S be an automorphic retract of S. By Lemma 4.1, every maximal element M of A is maximal in S and the regular surrounding cycle that contains M is contained in A. Because gcd(h, n) = 1, every regular surrounding cycle consists of all minimal elements of S and all maximal elements of S, which implies that all elements of S are in A.
The following conjecture is now natural, tempting and beyond the reach of the author's capabilities.

Conjecture 4.3 Theorem 4.2 also holds for 3h > n.
The effort required in Sect. 5 of [6] to prove that superpositions of h crowns with 6 elements each are minimal automorphic iff 3 h, which is Conjecture 4.3 for n = 3, seems to indicate that Conjecture 4.3 is quite challenging.

Some comments on ordered sets of height 2
Recall that a fence (of length n) is an ordered set with pairwise distinct elements f 0 , . . . , f n such that, for all i, j ∈ {0, . . . , n} we have that f i is comparable to f j iff |i − j| ≤ 1. Further recall that an ordered set is called connected iff, for any two elements x, y ∈ P, there is a fence that starts at x and ends at y, and that connected components are maximal connected subsets of disconnected sets. Necessary conditions for the fixed point property are few and far in between. Lemma 5.2 below shows that an ordered set P with the fixed point property must satisfy a stronger condition than the known necessary condition of connectivity: If b is minimal and P \ {b} is connected, then {x ∈ P : x > b} must be connected, too. Proposition 5.4 then indicates that the only way to superimpose two crowns on each other that yields a minimal automorphic set is via a straight superposition that satisfies Theorem 4.2.
Definition 5.1 Let P be a connected ordered set. For two elements x, y ∈ P, we define the distance d(x, y) from x to y as the length of a shortest fence from x to y. For two subsets X, Y ⊆ P, we define the distance d(X, Y ) := min{d(x, y) : x ∈ X, y ∈ Y }. A fence F is said to realize the distance from X to Y iff its length is d(X, Y ), its first element is in X and its last element is in Y .

Lemma 5.2
Let P be a finite ordered set, let b ∈ P be a minimal element of P such that {x ∈ P : x > b} is disconnected and such that P \ {b} is connected. Then P retracts onto a crown.
Proof When a fence realizes the distance between two connected components of {x ∈ P : x > b}, then the only elements of the fence that are above b are the endpoints. Thus, for any two connected components of {x ∈ P : x > b}, there is a fence that realizes the distance between them such that both endpoints are maximal. In particular, this fence has even length.
Let C 1 , . . . , C m be the connected components of {x ∈ P : x > b}, labeled such that, for all distinct We claim that r is a retraction to the crown F ∪ {b}. By definition of the R j , for i = 0, . . . , 2k, we have that f i ∈ R i , and hence the function r is indeed well-defined. Now it is clear that r is idempotent. To prove that r is order-preserving, let p < q.
. Thus, we can assume that p = b. Because P \ {b} is connected, every element of P \ {b} is in a set R j . If both p and q are in the same set R j , then r ( p) = r (q). Thus we can assume that p and q are in two different sets R j . Let p ∈ R i . By definition of the R j , if i were even, we would have q ∈ R i , which was already excluded. Thus i must be odd and, again by definition of the R j , q ∈ R i−1 ∪ R i+1 . Now r ( p) = f min{i,2k} , which is less than or equal to both f min{i−1,2k} and f min{i+1,2k} }, and one of which is r (q). Remark 5.3 Lemma 5.2 explains why the ordered sets in Example 9 in [16] cannot be chosen to be significantly simpler without either losing the fixed point property or losing the property that addition or removal of a single comparability leads to an ordered set that does not have the fixed point property. Let P be an ordered set of height 2 such that all points have at most 2 upper or lower covers. Adding a point or an adjacency to P either leads to a minimal element whose set of strict upper bounds is disconnected (which, by Lemma 5.2, means the resulting ordered set does not have the fixed point property), or, dually, to a maximal element whose set of strict lower bounds is disconnected, or, it leads to the presence of retractable minimal or maximal elements, which means that a comparability can be added without losing the fixed point property. Thus to generate examples of height 2 as in [16] there must be points with more than 2 upper or lower covers.
In fact, the author believes that all points would need to have at least 3 upper or lower covers in a simplest example, which would make the examples in [16] optimal in a sense. Proving such a result seems highly, and likely needlessly, specialized, as well as very technical. However, good intuition thus developed could help determine the largest n such that the covering graph of the set of orders on {1, . . . , n} that have the fixed point property is connected.
We now also have the tools to show that, when crowns are merged in an ordered set of height 2, then the straight superposition is the only arrangement that can yield a minimal automorphic ordered set.

Proposition 5.4
Let P be a connected finite ordered set of height 2 without covering relations between minimal and maximal elements and such that every connected component of P \ min(P) and of P \ max(P) is a crown. Then P does not have the fixed point property. Moreover, P is minimal automorphic iff P is a superposition of 2 crowns with n minimal elements each and n being 2 or an odd number.
Proof If there is a minimal element b such that {x ∈ P : x > b} is disconnected, then, by Lemma 5.2, P retracts onto a crown, which means P does not have the fixed point property and is not minimal automorphic.
This leaves the case that, for every minimal element b, we have that {x ∈ P : x > b} must be connected, that is, it must be a fence with 5 elements or a 4-crown. Let } be a crown that consists of minimal elements b i and non-maximal elements c j . Then, for any two consecutively labeled (modulo z) elements c j and c j+1 , we have that the union ↑ c j ∪ ↑ c j+1 is a fence with 5 elements or a 4-crown. Consequently z j=1 ↑ c j is a crown, and, because P is connected, P is a straight superposition of 2 crowns with 2z elements each.
Thus, in any case, P does not have the fixed point property, and, if P is minimal automorphic, it must be a superposition of 2 crowns. The remaining claims now follow from Theorem 4.2 and, for n ≤ 6, from Lemma 2.5 and by direct inspection.
The following conjecture seems to be sensible, but the author has little intuition how it could be approached beyond height 2.

Conjecture 5.5
Let P be a minimal automorphic ordered set such that every connected component of any two consecutive ranks is a crown. Then P is a straight superposition of crowns.

The ordered sets from Example 9 in [16]
As another application of Lemma 3.5, and possibly as an indication that there may be a more general principle lurking behind it, we consider the ordered sets from Example 9 in [16]. Proposition 6.4 below is a step towards proving that these ordered sets are minimal automorphic for all large enough m that are not divisible by 8: 3 It shows that the only minimal automorphic retract that contains a crown that "surrounds the central void," see Definition 6.3 below for a more precise formulation, is X m itself.

Conclusion
The work in [10,16] shows that minimal automorphic ordered sets can be used to directly create interesting examples that refute various conjectures. Note that, in each case, superpositions of crowns are modified and merged to produce the examples. The work in [9] shows that minimal automorphic ordered sets, and in particular superpositions of crowns, can serve in the proof of the fixed point property for ordered sets that are related to them through more sophisticated constructions involving simplicial complexes.
In this paper, Theorem 4.2 gives a significant partial answer to the question when a superposition of crowns is minimal automorphic, and Proposition 5.4 indicates that the "straight" superposition may well be the only way to obtain minimal automorphic sets through superpositions of crowns. Interesting further work remains, for example, in the form of Conjectures 4.3 and 5.5.
A key tool in proving our results is Lemma 3.5.
Although numerous examples show that things change significantly when we transition from combinatorics to topology, be it algebraic or otherwise, the author wonders whether there is some principle in algebraic topology lurking behind the combinatorics and "action at a distance" in Lemma 3.5.
The application of Lemma 3.5 to the ordered sets in Example 9 in [16] seems to be a first step in proving Conjecture 12 of [16]. Recent further developments (see [2]) that involve characterizations when a given ordered set retracts to a crown, indicate that, with the right new tools, these rather complicated ordered sets might be manageable after all.
We conclude by revisiting a conjecture by Corominas. In [3], it is conjectured that every minimal automorphic ordered set P is projective, that is, if an order-preserving function f : P × P → P satisfies f (x, x) = x for all x, then f (x, y) = x for all (x, y) or f (x, y) = y for all (x, y). This conjecture was a possible way to establish whether the fixed point property for finite ordered sets is preserved by products. Although it was established in [13] that, for finite ordered sets, the fixed point property is indeed preserved by products, Corominas' conjecture still looks interesting and might lead to further interesting insights on minimal automorphic ordered sets.