Orbits of antichains in certain root posets

This paper gives another proof of Propp and Roby's theorem saying that the average antichain size in any reverse operator orbit of the poset $[m]\times [n]$ is $\frac{mn}{m+n}$. It is conceivable that our method should work for other situations. As a demonstration, we show that the average size of antichains in any reverse operator orbit of $[m]\times K_{n-1}$ equals $\frac{2mn}{m+2n-1}$. Here $K_{n-1}$ is the minuscule poset $[n-1]\oplus ([1] \sqcup [1]) \oplus [n-1]$. Note that $[m]\times [n]$ and $[m]\times K_{n-1}$ can be interpreted as sub-families of certain root posets. We guess these root posets should provide a unified setting to exhibit the homomesy phenomenon defined by Propp and Roby.


Introduction
Recently, the study of orbit structure of the reverse operator, which is also called rowmotion [12] or Fon-Der-Flaass action [10] in the literature, has emerged at the front lines of inquiry. This paper introduces a method that aims to calculate the sizes of antichains in each orbit. Our method turns out to be effective in dealing with certain root posets coming from the classical Lie algebras. Now let us be more precise. A subset I of a finite poset (P, ≤) is called an ideal if x ≤ y in P and y ∈ I implies that x ∈ I. Let J(P ) be the set of ideals of P , partially ordered by inclusion. A subset A of P is an antichain if its elements are mutually incomparable. We collect the antichains of P as An(P ). For any x ∈ P , let I ≤x = {y ∈ P | y ≤ x}. Given an ideal I of P , set Γ(I) := max(I). For an antichain A of P , let I(A) = a∈A I ≤a . The reverse operator Ψ is defined by Ψ(A) = min(P \ I(A)). This operator was introduced on hypergraphs by Duchet in 1974 [1], and studied on [m] × [n]-the product of two chainsby Fon-Der-Flaass in 1993 [3]. Since antichains of P are in bijection with ideals of P , the reverse operator acts on ideals of P as well. For convenience, we may just call an Ψ P -orbit of antichains or ideals an orbit.  [8]).
The first aim of this paper is to give another proof of Theorem A. It is conceivable that the method developed in Sections 3 and 4 can be applied to more general situations. As a demonstration, we report the following.
Theorem B. The average size of antichains in any Ψ-orbit of [m] × K n−1 equals 2mn m+2n−1 . Theorem B follows from Propositions 5.3 and 6.5. In Section 7, we will recall certain root posets ∆ (1). They consists of the infinite families [m]×[n], H n , [m]×K n−1 , and twenty other posets. By Theorems A and B, and by checking the twenty posets directly, one knows that the average size of antichains in any Ψ-orbit of ∆(1) equals #∆ (1) d+1 , where d is the maximum height of the roots in ∆(1).
The paper is outlined as follows: Section 2 prepares necessary preliminaries. Section 3 aims to prove Theorem A via another method, where the proof of a key lemma is postponed to Section 4. Sections 5 and 6 are devoted to proving Theorem B. More precisely, they handle orbits in [m] × K n−1 of type I and II, respectively. Section 7 recalls the root posets ∆(1), and raises two conjectures pertaining to their orbit structure. Our conjecures simply aim to guess that ∆(1) would provide a unified setting to exhibit the homomesy phenomenon invented by Propp and Roby.
As on page 244 of Stanley [8], a finite poset P is said to be graded if every maximal chain in P has the same length. In this case, there is a unique rank function r from P to the positive integers P such that all the minimal elements have rank 1, and r(x) = r(y) + 1 if x covers y. We always assume that the finite poset P is graded. Let d be the maximum value of r on P . For 1 ≤ j ≤ d, let P j be the j-th rank level of P . That is, P j = {p ∈ P | r(p) = j}. Set P 0 to be the empty set. Note that each P j is an antichain. Put L i = i j=1 P j for 1 ≤ i ≤ d, and let L 0 be the empty set. We call those L i rank ideals. Recall that the reverse operator acts on ideals as well. For instance, It is well known that a general ideal of [m] × P can be identified with (I 1 , . . . , I m ), where each I i is an ideal of P and I m ⊆ · · · ⊆ I 1 . We say that the ideal (I 1 , . . . , I m ) is full rank if each I i is a rank ideal in P . Let O(I 1 , . . . , I m ) be the Ψ [m]×P -orbit of (I 1 , . . . , I m ). The following lemma is taken from Section 2 of [2]. One may also verify it directly from the definition of Ψ.
Lemma 2.1. Keep the notation as above. Then for any n 0 ∈ N, where 0 ≤ i s < · · · < i 1 < d, L n 0 d denotes n 0 copies of L d and so on.
In view of Lemma 2.1, the ideal (I 1 , . . . , I m ) is full rank if and only if each ideal in its Ψ [m]×P -orbit is full rank. Therefore, it makes sense to say that the orbit O(I 1 , . . . , I m ) is of type I if (I 1 , · · · , I m ) is full rank, and to say it is of type II otherwise. In this paper, we care most about the case that P = K n−1 , whose elements are labelled by 1, 2, . . . , n−1, n, n ′ , n+1, . . . , 2n−2, 2n−1. Fig. 1 illustrates the labeling for K 3 . Note that L i (0 ≤ i ≤ 2n − 1) are all the full rank ideals. For instance, we have L n = {1, 2, . . . , n, n ′ }. Moreover, we put I n = {1, . . . , n − 1, n} and I n ′ = {1, . . . , n − 1, n ′ }. The following lemma is taken from Section 2 of [2]. Again, it can be checked directly.
Keeping the notation of Lemma 2.2, one sees that any non-full-rank ideal of [m] × K n−1 either has the form . . , L ns is , I m 0 n ′ , L m 1 j 1 , . . . , L mt jt ). We say that the above two ideals are dual to each other and use "∼" to denote this relation. It is immediate from Lemma 2.2 that I ∼ I ′ implies that Ψ(I) ∼ Ψ(I ′ ), and thus Ψ i (I 1 ) ∼ Ψ i (I 2 ) for any i ∈ Z. In such a case, we say that the two orbits are dual to each other, and write this relation as O(I) ∼ O(I ′ ).

Another proof of Propp and Roby's theorem
This section aims to give another proof of Theorem A. We use an NE-SW path to separate an ideal I from its complement. Here NE stands for northeast, and SW stands for southwest. Such a path becomes a binary word when we interpret an NE step by 1 and a SW step by 0. This process associates a binary word Θ(I) to each I ∈ J([m] × [n]).
Example 3.1. We present the NE-SW paths in blue lines for three lower ideals of [2] × [3] in Fig. 2. Their associated binary words are 01101, 10110 and 11001, respectively. For any integer a, we put 1 a := 1 . . . 1 a , which is interpreted as the empty word if a ≤ 0.
The notation 0 a is defined similarly. A general binary word w in B(m, n) has the form where s ≥ 2, each a i , b i is positive except for that a 1 and b s may be 0. We define The following lemma is essentially a translation of Lemma 2.1 into the language of binary words. For convenience of reader, we provide a proof.  Proof. Suppose that Θ(I) = w has the form (4). Without loss of generality, we assume that s ≥ 3 and focus on a middle part 1 a i 0 b i , where 1 < i < s. We draw the corresponding portion of the path for I in red line in Fig. 3, where dots stand for elements in I, while circles stand for elements in its complement. Note that the NE (resp. SW) thick red line segment has length a i (resp. b i ). Then two elements of the antichain Γ(Ψ(I)) are shown in blue circles, and the corresponding portion of the path for Ψ(I) is drawn in blue. Now the SW (resp. NE) thick blue line segment has length b i (resp. a i ), as desired. We omit the similar analysis for the first part 1 a 1 0 b 1 and the last part 1 as 0 bs of Θ(I).  Let the binary word of I be given by (4). Assume that a 1 ≥ 1 and b s ≥ 1. The pattern of 0's and 1's are respectively. There is a unique way to combine (7) and (8) in the zig-zag fashion, after which we recover Θ(I).
If b 1 ≥ 2 and a s ≥ 2, then by Lemma 3.2, the pattern of 0's becomes Comparing (9) with (7), one sees that the leftmost 0 has been bumped out, while we add a new 0 from the right. We remember this new 0 as the (m + 1)-th, and show it in bold.
Note that the (m + 1)-th 0 is followed by a −. This is not an accident: if one continues the analysis for one more step, one sees that the (m + 2)-th 0 will be separated from the (m + 1)-th. On the other hand, the pattern of 1's becomes Comparing (10) with (8), one sees that the rightmost 1 has been bumped out, while we add a new 1 from the left. We remember this new 1 as the (n + 1)-th, and show it in bold. Note that the (n + 1)-th 1 is preceded by a −. Again, this is not an accident: if one continues the analysis for one more step, one sees that the (n + 2)-th 1 will be separated from the (n + 1)-th 1. Finally, note that there is a unique way to combine (9) and (10) in the zig-zag fashion, after which we recover Θ(Ψ(I)).
Recall that in the above analysis, we have assumed that a 1 ≥ 1, b s ≥ 1, b 1 ≥ 2 and a s ≥ 2. The important thing here is that the assumption "a 1 ≥ 1 and b s ≥ 1" guarantees that Θ(I) starts with 1 and ends with 0; while the assumption "b 1 ≥ 2 and a s ≥ 2" guarantees that Θ(Ψ(I)) starts with 0 and ends with 1. Of course, there are situations where these assumptions do not hold. However, in any case, guided by Lemma 3.2, the Join-Separate rule always applies: • if Θ(I) ends with 1, separate the (m + 1)-th 0 from the m-th 0 by inserting a "−" between them; otherwise, join the (m + 1)-th 0 from right with the m-th 0 of Θ(I).
This rule will be the basic component in the proof of Lemma 3.5.
Given an ideal I of [m] × [n], recall that Γ(I) := max(I) is the antichain corresponding to I. The following lemma is immediate.
Suppose that where a k = m, b k = n, a 1 ≥ 1 and b 1 ≥ 1. Now let us define two functions for I on the interval [1, m + n]. They will be crucial in calculating the sizes of antichains in the Ψ-orbit The following lemma will play a key role in forthcoming discussion. We postpone its proof to the next section so that one can quickly grasp the sketch.
Lemma 3.5. Suppose that I is the lower ideal of [m] × [n] given by (11). Then for i ∈ [1, m + n], we have Let us give an eample to illustrate the formula (14).
with binary word 0011101111. Then one finds that Proposition 3.7. Suppose that I is the ideal whose binary word Θ(I) is given by (11). Then Proof. For convenience, we temporarily put N = m + n. We have that where the first step uses (14), the third step cites (12) and (13) When Θ(I) starts with 0 and ends with 1, say of the form (11), this has been done in Proposition 3.7. Our argument works for the other three cases as well.

Proof of Lemma 3.5
This section is devoted to proving Lemma 3.5. We adopt the notations in Section 3. We always suppose that (11) holds. Namely, let I be the lower ideal of [m] × [n] corresponding to the binary word Note that 0 − 0 1 = 0. The notation 1 − 1 a is defined similarly. We associate the long 0− sequence to I as follows: Note that this sequence contains 2m + n 0's in total, and ends with −. Fix any i ∈ [1, m + n]. Cut out the consecutive segment of (17) starting with the (i + 1)-th 0 and ending with the (i + m)-th 0, and include the "−" left (resp. right) to the (i + 1)-th 0 (resp. (i + m)-th 0) if there is such a "−". We call this segment the i-th 0− sequence for I. In a similar fashion, we associate the long 1− sequence to I as follows: Note that this sequence contains m + 2n 1's in total, and ends with −. Here we always read the long 1− sequence and its consecutive segments from right to left. For instance, cut out the consecutive segment of (18) starting with the first 1 and the (n + a 1 )-th 1, we get Fix any i ∈ [1, m + n]. Cut out the consecutive segment of (18) starting with the (i + 1)-th 1 and ending with the (i + n)-th 1, and include the "−" left (resp. right) to the (i + n)-th 1 (resp. (i + 1)-th 1) if there is such a "−". We call this segment the i-th 1− sequence for I.
The following lemma reads Θ(Ψ i (I)) from (17) and (18)   There is a unique way to combine the i-th 0− sequence and the i-th 1− sequence for I in the zig-zag fashion, and one gets Θ(Ψ i (I)).
Proof. It amounts to check that (17) and (18)  (A). Firstly, let us analyze the pattern of the next n 0's. Since the first n 1's (counted from right to left) are of the form the (m + 1)-th 0 to the (m + n)-th 0 are of the form by the Join-Separate rule. This agrees with (17) up to the (m + b k )-th 0.
(B). Secondly, let us switch to analyze the pattern of the next m 1's. Since the first m 0's are of the form 0 a 1 − 0 a 2 −a 1 − · · · − 0 a k −a k−1 −, the (n + 1)-th 1 to the (n + m)-th 1 (counted from right to left) are of the form by the Join-Separate rule. This agrees with (18) up to the (n + a k )-th 1.
(C). Thirdly, let us come back to analyze the pattern of the further next m 0's. Since the (n + 1)-th 1 to the (n + m)-th 1 are given by (20), the (m + n + 1)-th 0 to the (m + n + a k )-th 0 are of the form 0 a 1 − 0 a 2 −a 1 − · · · − 0 a k −a k−1 by the Join-Separate rule. This agrees with (17) up to the (m + b k + a k )-th 0.
(D). Fourthly, let us switch to analyze the pattern of the further next n 1's. Since the (m+1)-th 0 to the (m+n)-th 0 are of the form (19), the (n+m+1)-th 1 to the (n+m+n)-th 1 (counted from right to left) are of the form by the Join-Separate rule. This agrees with (18) up to the (n + a k + b k )-th 1.
(E). Finally, the (m + n)-th 1− sequence starts with 1. Thus we should add "−0" to the (2m + n)-th 0 according to the Join-Separate rule. This agrees with the last − of (17). Similarly, the (m + n)-th 0− sequence starts with 0. Thus we should add "1−" to the left side of the (m + 2n)-th 1 in view of the Join-Separate rule. This agrees with the last − of (18), and the proof finishes.   Proof. In view of Lemmas 3.4 and 4.1, we should analyze the difference between the number of "10"s in Θ(Ψ i (I)) and that in Θ(Ψ i−1 (I)). Going from Θ(Ψ i−1 (I)) to Θ(Ψ i (I)), we shall delete the i-th 1 in (18) and add the (n + i)-th 1 in (18) to form the i-th 1− sequence for I. Deleting the i-th 1 decreases the number of "10"s by one if and only if the i-th 1 has the form 1− in (18). This is measured precisely by the value Q I (i) defined in (13). Similarly, adding the (n + i)-th 1 increases the number of "10"s by one if and only if the (n + i)-th 1 has the form 1− in (18). This is measured precisely by the value P I (i) defined in (12). Now (21) follows.
Remark 4.5. The proof above tells us that P I (i) = −Q I (i + n).

Now (14) follows directly from (21) and Lemma 3.5 is established.
At the end of this section, let us briefly compare our approach to Theorem A with that of Propp and Roby [6]. On one hand, Propp and Roby cleverly associate a Stanley-Thomas word to each antichain of [m] × [n] and then Ψ becomes C R -the rightward cyclic shift. The equivariance of the bijection is proved in Proposition 26 of [6]. Then Theorem A follows quickly.
On the other hand, we naively associate a binary word to each antichain of [m] × [n] accroding to the direction (which is either NE or SW) of the path cutting out the corresponding ideal. Then Ψ becomes ψ and the equvariance quickly follows (see Lemma 3.2). However, compared with C R , the map ψ lacks an easy interpretation. Instead, we separate the zeros and ones in the binary word, and use the Join-Separate rule (which comes from the map ψ) to move them forward. In this way, we obtain the long 0−sequence and the long 1−sequence. This process is analyzed in Lemma 4.1. Then we do some counting using the functions P I (i) and Q I (i), and arrive at Theorem A.

Type I orbits in [m] × K n−1
This section aims to consider type I orbits in [m] × K n−1 . Note that any full rank ideal of [m] × K n−1 must have the form Lemma 2.1). By using NE-SW paths, one sees that these ideals are in bijection with B(m, 2n − 1)-all the binary words with m 0's and 2n − 1 1's. We still denote this bijection by Θ. Recall from (5) that there is a map ψ : B(m, 2n − 1) → B(m, 2n − 1). Similar to Lemma 3.2, we have the following. Proof. Note that ǫ n (Θ(I)) = 1 if and only if L n occurs at some place of I, and that the antichain of K n−1 corresponding to L n = {1, · · · , n, n ′ } contains two elements. Proof. In view of Lemmas 5.1 and 5.2, we can work on [m]× [2n − 1], with the exception that whenever the n-th 1 is followed immediately by 0, we should add one to the antichain size. Therefore, let us focus on the long-1-sequence (18) with a k = m, while b k = 2n − 1 instead. Recall that we always count the long-1-sequence from right to left. In view of Proposition 3.7, it remains to prove that there are exactly m 1's which are preceded by "−" in a period of (18). It is easy to observe that the indices of these 1's are as follows:  To focus on a period, let us count those living on the interval [n, m + 3n − 2]. The total number is as desired.

Type II orbits in [m] × K n−1
This section aims to consider type II orbits in [m] × K n−1 . Recall that any non-full-rank ideal of [m] × K n−1 either has the form (2) or the form (3), and that we have introduced the duality "∼" between these ideals and type II orbits at the end of Section 2. It is obvious that two dual non-full-rank ideals have the same size, and their anticains have the same size as well. Thus it does no harm to study non-full-rank ideals and their orbits up to duality.
We let L * to be I n or I n ′ , and view the subscript * as a number between n − 1 and n. Interpreted in this way, one sees that L * behaves like a rank ideal by comparing Lemmas 2.1 and 2.2. More precisely, the two cases in Lemma 2.2 can be written uniformly as · · · L ns max{ * , j 1 +1} L m 0 min{ * , j 1 +1} · · · . Here we only present the middle part. Thinking in this way allows us to use a NE-SW path to represent a non-full-rank ideal of [m] × K n−1 up to duality: Instead of [m] × [2n − 1], now we use [m] × [2n], with the rank * inserted between rank n − 1 and rank n. See Fig. 4 for an example.
Let B(m, 2n) be the set of binary words with m 0's and 2n 1's, and that the n-th 1 is followed immediately by 0. Since the integer m 0 in (2) or (3) is always positive, the process gives a bijection For instance, the equivalence class of (2) is mapped to the following binary word: A general binary word w in B(m, 2n) has the form where s ≥ 2, each a i , b i is positive except for that a 1 and b s may be 0. By the definition of B(m, 2n), there must exist an index i such that a 1 + · · · + a i = n. We define The following result is a translation of Lemma 2.2 into the language of binary words. This time we omit the details. Note that although there are five cases in the definition of (25), the pattern of 0's is always the same as that of (5). Therefore, guided by Lemma 6.1, the Join-Separate rule always applies: • if Θ(I) ends with 1, separate the (m + 1)-th 0 from the m-th 0 by inserting a "−" between them; otherwise, join the (m + 1)-th 0 from right with the m-th 0 of Θ(I).
As in Section 4, the long-0-sequence can be formed by the above Join-Separate rule, and it records the zero patterns in Θ(Ψ i (I)) for i ≥ 0. Given any binary word w in B(m, 2n), the n-th 1 always indicates the occurrence of L * in the ideal I of [m] × K n corresponding to w. Replacing the n-th 1 in w by * establishes a bijection between B(m, 2n) and B * (m, 2n − 1). Here B * (m, 2n − 1) consists of binary words with m 0's, 2n − 1 1's and a "*"; moreover, we require that there are n − 1 1's before *, and * is followed immediately by 0. For instance, the element in B * (m, 2n − 1) corresponding to the binary word (24) in B(m, 2n) is where 1 a i −1 is viewed as the empty word if a i = 1. Separating 1's and the * from the above equation, we get its 1 − * pattern: (27) 1 a 1 − · · · − 1 a i −1 * − · · · − 1 as .
Here we omit the last possible −. Moving from Θ(I) to Θ(Ψ(I)), it is direct to check via Lemma 6.1 that the 1 − * pattern becomes (28) p(1 a 1 +1 − · · · − 1 a i −1 * − · · · − 1 as−1 ) if a 1 ≥ 1, and it becomes (29) p(1 − 1 a 1 − · · · − 1 a i −1 * − · · · − 1 as−1 ) if a 1 = 0. Here in both cases, −1 as−1 is interpreted as the empty word if a s = 1. Moreover, the operator p firstly exchanges * and the n-th 1, then it moves the 1 immediately following * (if there exists) to the next segment. For instance, when n = 4, we have Therefore, after adding a new 1, the operator p pushes the n-th 1 through the *. Note that this process does not change the relative patterns of the 1's preceding *. For instance, after adding n − 1 1's to (27) one by one, in the form 1 or 1−, and applying p once after adding a 1, then the n − 1 1's before * will be pushed after *. However, their relative pattern is still Therefore, we should pay attention to how the new 1's are added to (27). By the analysis around (28) and (29), this process obeys the Join-Separate rule: • if Θ(I) starts with 0, add the new 1 from left in the form "1−"; otherwise, add it from left in the form "1".
Similar to Section 4, we can form the long-1-sequence by consecutively applying the above rule. However, we emphasize that unlike the [m] × [n] situation, now the long-1-sequence no longer aims to record the ones pattern in Θ(Ψ i (I)) for i ≥ 0. Instead, it simply records how the new ones are added in.
The following lemma is immediate.
(A). The 1 − * pattern of (30) is The last n 1's will be bumped out one by one in the form of Therefore, by the Join-Separate rule, the next n zeros will be added to (32) in the form (B). Now let us handle the next n − 1 step. As noted earlier, the n − 1 1's before * in (33) will be pushed through *, and they will be bumped out in the relative form Therefore, by the Join-Separate rule, the next n − 1 zeros will be added in the form (C). Now let us analyze the last m 0's. Since the first m 0's are in the form (32), the first m 1's will be added to (33) in the form Again they will be pushed through *, and then bumped out in the above relative form. Therefore, by the Join-Separate rule, the last m 0's will be added in the form 0 a 1 − 0 a 2 − · · · − 0 as . This finishes the proof. Remark 6.4. In the setting of Lemma 6.3, the binary word Θ(Ψ m+2n−1 (I)) has the same zeros pattern as that of Θ(I), namely, (32). On the other hand, the long-1-sequence goes like Since pushing a sequence of ones through * does not change their relative pattern, one sees that Θ(Ψ m+2n−1 (I)) has the same 1 − * pattern as that of Θ(I), namely, (33). Thus we conclude that Ψ m+2n−1 (I) = I. Then it is immediate that Ψ [m]×K n−1 has order m + 2n − 1. This result should already be known by Rush and Shi. Indeed, in Theorem 10.1 of [10], the cyclic sieving phenomenon defined in [9] has been established for [m] × K n−1 . Proof. Without loss of generality, let I be the non-full-rank ideal of [m]×K n−1 corresponding to the binary word (30). According to Remark 6.4, Ψ [m]×K n−1 has order m + 2n − 1. One can consecutively cut off m + 2n − 1 segments of zeros from (31), each having length m. It remains to show that the total number of −0's in these m + 2n − 1 segments is 2mn. On the other hand, one can consecutively cut off m + 2n segments of zeros, each having length m, from the following sequence: The number we want is the same as the number of −0's in the m + 2n segments. By Proposition 3.7, the latter number is 2mn. This finishes the proof.

Certain root posets
In this section, we shall introduce a bit Lie theory, and illustrate that the several infinite families of posets in Section 1 occur as sub-families of certain root posets.
Let g be a finite-dimensional simple Lie algebra over C. Fix a Cartan subalgebra h of g. The associated root system is ∆ = ∆(g, h) ⊆ h * . Fix a set of positive roots ∆, and let Π = {α 1 , . . . , α l } be the corresponding simple roots. Let [α i ] be the set of all positive roots α such that the coefficient of α i in α is one. Note that [α i ] inherits a poset structure from the usual one of ∆ + : let α and β be two roots of [α i ], then α ≤ β if and only if β − α is a nonnegative integer combination of simple roots. We call the posets [α i ] by ∆(1). The latter notation reflects the fact that they come from Z-gradings of g, see [2] and references therein for more background.
Previously, by using Ringel's paper [7], we have analyzed the structure of ∆(1) in Section 4 of [2] The above posets all come from exceptional Lie algebras. Here we label the simple roots as Knapp [4].
Finally, let us raise two conjectures pertaining to the orbit structure of ∆(1). Let w i 0 be the longest element of the subgroup s α j | j = i of the Weyl group. Here s α j is the reflection according to the simple root α j . Then  Remark 7.3. We mention that the previous two conjectures can be formulated in the language of homomesy invented by Propp and Roby [6]. Indeed, let χ p be the function on J(∆(1)) such that χ p (I) = 1 if p ∈ I and χ p (I) = 0 otherwise. Then Conjecture 7.1 amounts to the claim that the function χ p +χ p * is 1-mesic for any p ∈ ∆(1) under the reverse operator. Similarly, in the antichain setting, let χ ′ p be the function on An(∆(1)) such that χ ′ p (A) = 1 if p ∈ A and χ ′ p (A) = 0 otherwise. Then Conjecture 7.2 is exactly the claim that the function χ ′ p − χ ′ p * is 0-mesic for any p ∈ ∆(1) under the reverse operator. Therefore, the two conjectures amount to guess that ∆(1) could be a good place to exhibit homomesy.