AA GENERALIZATION OF THE GREENE-KLEITMANDUALITY THEOREM
FRANK Y. LU
Abstract:
In this paper, we describe and prove a generalization ofboth the classical Greene-Kleitman duality theorem for posets and thelocal version proved recently by Lewis-Lyu-Pylyavskyy-Sen in studyingdiscrete solitons, using an approach more closely linked to the approachof the classical case. 1.
Introduction
The Greene-Kleitman duality theorem for finite posets, first de-scribed in Greene’s paper, [Gre76] (see also [BF99], upon which thefollowing exposition is loosely based) states the following result. Givena poset P, let A k be the maximal possible sum of the lengths of k disjoint increasing sequences of elements (chains), and D k is the max-imal possible sum of the lengths of k disjoint sequences of elementswhere no two elements are pairwise comparable (anti-chains). Then A k , D k are conjugate in the following sense: A + ( A − A ) + · · · and D + ( D − D ) + · · · form conjugate partitions of n. See [BF99][ §
8] fora description of one proof of this result, attributed to A. Frank, usinga graph theoretic construction, which will be relevant to us. Here, wewill refer to this as the classical Greene-Kleitman duality theorem.The duality of these partitions lends itself to applications, which[BF99] discusses in detail. For instance, [BF99][ § σ of n elements by imposing the ordering, on the set ofelements { ( i, σ ( i )) | i = 1 , , . . . , n } , that ( i, σ ( i )) < ( j, σ ( j )) if and onlyif i < j and σ ( i ) < σ ( j ) . Recently, another related duality result was published in [Lew+20],which [Lew] (from which the exposition regarding this theorem is based)calls the localized Greene’s theorem . Here, we start with a permuta-tion σ on a set of elements { , , . . . , n } , and consider the sequence Date : January 28, 2021. a r X i v : . [ m a t h . C O ] J a n FRANK Y. LU σ (1) , σ (2) , . . . , σ ( n ) . From there, the same sort of duality in the origi-nal duality theorem was shown: however, instead of the original valuesfor the classical theorem being conjugate, we have quantities A (cid:48) k and D (cid:48) k being conjugate, which are defined as follows. For A (cid:48) k , we consider,over all sets of k disjoint subsequences, the maximum of the sum ofthe ascents of each subsequence, where the ascent of a subsequence s , s , . . . s m is the number of indices i so that s i < s i +1 , plus one (or 0if the sequence is empty). For D (cid:48) k , this is defined to be the maximum,over all sets of k consecutive subsequences, of the sum of the lengthsof the maximal descending sequences of each subsequence. Note theconsecutive condition here, in contrast with the classical theorem: forinstance, if we have 2 , , , , , . The proof in [Lew+20] of this result differssubstantially from the classical proof, utilizing the study of discretesolitons in the paper.In this paper, we unite these two theorems with a generalization,Theorem 2.1, which we detail in the next section, using an overallstructure of the proof similar to the proof provided by Frank. Here, wetranslate the problem into a problem about direct graphs and flows ondirect graphs; again, see [BF99][ §
8] for one version of Frank’s proof, forinstance, upon which the main ideas for this proof are built. However,the structure of the graph is constructed in a way where flows andpotentials correspond more “naturally” to sequences.Specifically, in Section 2, we introduce the generalized theorem andthe specific information necessary. From there, in section 3 we set upthe required graph theory to allow for the translation of the problem tothis graph theoretic construction, along the lines of the classical proof.From here, we prove some basic properties of the graph constructionwhich will be useful in Section 4, before proceeding on to the corepart of the proof. In Section 5, we link the two desired poset-basedquantities to the graph-theoretic construction and use this to arriveat an inequality, which we sharpen to the desired equality in Section6. Finally, in Section 7, we show that both versions of the Greene-Kleitman duality theorem follow as corollaries of this general theorem,and provide another interesting special case.Thanks to Dr. Pavlo Pylyavskyy for introducing me to this problem,as well as offering suggestions on drafts, including on the exposition insections 1, 2, and 3, and the abstract. Thanks as well to Dr. EmilyGunawan for suggestions on the draft, especially with regards to theexposition of sections 2 and 3, including the example, and thanks toDr. Joel Lewis for comments on the earlier version of the draft, inparticular on the exposition in sections 2 and 3 as well.
GENERALIZATION OF THE GREENE-KLEITMAN DUALITY THEOREM 3 The Generalized Problem
The exposition loosely adapts from [Lew] in generalizing this prob-lem, in that the notation and exposition here generalizes that of thelocalized Greene’s theorem given in [Lew].Given a poset P on elements S P = { e , e , . . . , e n } and a bijection h : S P → { , , . . . , n } , we pick a set C P of pairs of distinct elementsin S P with the following properties:(1) Given x, y ∈ S P , if x < y and h ( x ) < h ( y ) , then ( x, y ) ∈ C P . (2) Given that ( x, y ) ∈ C P , we have that h ( x ) < h ( y ) . (3) Given that ( x, y ) ∈ C P and ( y, z ) ∈ C P , we have that ( x, z ) ∈ C P . In other words, C P is some binary transitive relation on S P that is asubset of the strict total ordering given by h, which also contains theintersection of the relations given by P and the relations given by h. Given the set C P and bijection h, we say that a sequence of distinctelements s , s , . . . , s m is adjacentable if for each j, ≤ j ≤ m − , ( s j , s j +1 ) ∈ C P . In addition, we say that the sequence is h − ordered if it satisfies that h ( s j ) < h ( s j +1 ) for each j. Note that adjacentablesequences are necessarily h − ordered, but the reverse isn’t true if C P isa strictly smaller relation than h. Now, let S be an adjacentable sequence of distinct elements s , s , . . . , s m . Define asc ( S ) to be the number of indices j so that s j < s j +1 , plus one,or to equal 0 if the sequence is empty.In addition, for any h − ordered sequence S of distinct elements, de-fine desc ( S ) to be the length of the longest subsequence of S, say s , s , . . . , s n , so that s i (cid:54) < s j for each i < j. Example 2.1.
Suppose we have a poset P on the set { a, b, c, d, e } , with the cover relations a < b, b < d, c < d, d < e, and the function h that takes on the following values: h ( a ) = 1 h ( b ) = 3 h ( c ) = 5 h ( d ) = 4 h ( e ) = 2 . Let C P be the set { ( x, y ) ∈ { a, b, d, e } × { a, b, d, e }| h ( x ) < h ( y ) } ∪{ ( a, c ) } If we have the sequence S be ( a, e, b, d ) , we have asc ( S ) = 3 , as a < e and b < d. Also desc ( S ) = 2 , by taking the subsequence e, d. Note that this will naturally be 0 if S is empty. FRANK Y. LU
We say that two disjoint h − ordered sequences s , s , . . . , s m and t , t , . . . , t l of P are semi-overlapping if and only if there exist indices i, j, k, l so that ( t j , s i ) and ( s k , t l ) lie in C P . For instance, note that a, e, d and b, c are semi-overlapping (since f ( d ) > f ( b ) , f ( a ) < f ( b )),but a, e and b, c aren’t.From here, define A (cid:48) k to be the maximum value, over all sets of k disjoint adjacentable sequences { S , S , . . . , S k } of asc ( S ) + asc ( S ) + · · · + asc ( S k ) . Similarly, define D (cid:48) k to be the maximum value, over allsets { S , S , . . . , S k } , of k disjoint h − ordered sequences where no twoare semi-overlapping, of desc ( S ) + desc ( S ) + · · · + desc ( S k ) . For Example 2.1, we compute that A (cid:48) = 3 , using the sequence S =( a, e, b, d ) . Similarly, we see that A (cid:48) = 4 , using S = ( a, e, b, d ) and S = ( c ) , and A (cid:48) = 5 , using S = ( a, e ) , S = ( b, d ) and S = ( c ) . Wecompute also that D (cid:48) = 3 , using the sequence ( e, b, c ) , D (cid:48) = 4 using thesequences ( e, b, c ) and ( d ) , and D (cid:48) = 5 using the sequences ( a ) , ( e, b, c ) , and ( d ) . Given these quantities, we have the following theorem.
Theorem 2.1.
Let λ = A (cid:48) , and µ = D (cid:48) , and for k ≥ , let λ k = A (cid:48) k − A (cid:48) k − , µ k = D (cid:48) k − D (cid:48) k − . Then, the sums n = λ + λ + · · · and n = µ + µ + · · · are partitions; moreover, they are conjugate partitions. For instance, as we’ll show in Section 7, if we let P be the naturalordering on the set of elements { , , . . . , n } , if we take C P to be theset { ( x, y ) ∈ S P × S P | h ( x ) < h ( y ) } and h to be the permutation, wewill arrive at the localized Greene’s theorem for permutations from[Lew+20]. Also, if we let P be a poset, h to be a linear extension, and C P = { ( x, y ) | x < y } , we will arrive at the Greene-Kleitman theoremfrom [Gre76]. This latter result, however, will require a little bit morework, as we will do in Section 7.As mentioned before, the general method of proof is similar to [BF99] § §
8, which was used to prove the classical Greene-Kleitman the-orem. 3.
Setup
In this section, we establish a directed graph which reflects the struc-ture of the poset P. The exposition in this section follows [BF99] § The Graph.
Given a poset P on set S P with n elements, andbijection h between elements of P and the set { , , . . . , n } , we nowconstruct a directed graph G P,h,C P = ( V, E ) . Here, the set V consists GENERALIZATION OF THE GREENE-KLEITMAN DUALITY THEOREM 5 of 2 n + 2 elements: a source vertex b , a sink vertex t n +1 , and for eachelement e ∈ P, we have a “top” vertex t h ( e ) and a “bottom” vertex b h ( e ) . The set of edges E is the union of the following four sets, wherewe have the ordered pair ( v, w ) represent a directed edge from vertex v to vertex w :(1) The set { ( b , t i ) | ≤ i ≤ n } of edges from b to each of thevertices t i . (2) The set { ( b i , t n +1 ) | ≤ i ≤ n } from each of the b i to t n +1 . (3) The set { ( t i , b i ) | ≤ i ≤ n } from each t i to its corresponding b i . (4) The set { ( b i , t j ) | ( h − ( i ) , h − ( j )) ∈ C P } . Notice that these four sets of edges are distinct. For Example 2.1,we get a graph like the following graph. a e bdcb t Figure 1. G P,h,C P for Example 2.1Here, green represents the first set, red represents the second set,blue represent the third, and black represent the fourth set. Next toeach pair of vertices t i , b i for i from 1 to 5 is the element in P that itcorresponds to (namely, h − ( i )).3.2. Minimal-Cost Flow.
We now consider imposing a flow onto thegraph, and finding, for a given flow value v (defined as in [BF99] as thesum of the flows assigned to each edges going out of a source node), theminimal cost flow. As mentioned before, the exposition we use here issimilar to that of [BF99] §
7, with some adaptations from [Wil19].
FRANK Y. LU
We use the definition of flow used in [BF99] §
7: a flow on a directedgraph with vertex set V and edge set E, with one source node andone sink node, is a function f : E → R ≥ so that, for each vertex v that isn’t a source or a sink, (cid:80) ( w,v ) ∈ E f (( w, v )) = (cid:80) ( v,w ) ∈ E f (( v, w )) . Thisproperty is also known as flow conservation. The value of a flow is thenjust (cid:80) ( s,w ) ∈ E f (( s, w )) , where s is the source node. Notice that this flowcan be restricted in value; the capacity of a given edge gives us thebounds for what values f can take on the edge. For this discussion, as f is nonnegative, we let the capacity function simply be the maximumvalue that f can take on each edge.Now, for the costs of this graph, define the function c : E → Z , sothat an edge e = ( v, w ) ∈ E has cost − w = t n +1 , or v = b i , w = t j , where h − ( i ) < h − ( j ) and i < j, and all other edges have cost 0 . Defineas well the capacity function u : E → Z that sets the capacity of alledges to be 1 . The exposition from here follows that of [Wil19], as [BF99] doesn’tprovide us with a sufficiently general context, though we will return tothe mechanics of [BF99] afterwards.Working more directly in the context of [Wil19], we have the follow-ing definition:
Definition 1 (Definition 5.2 from [Wil19]) . Given a directed graphwith vertices V and edges E, we add to the edges the reverse of theseedges (so if ( v, w ) ∈ E, we add ( w, v )), and we denote the total setof edges as E (cid:48) . Suppose we are also given a function u : E (cid:48) → Z with u ( e ) ≥ e ∈ E (cid:48) , and cost function c : E (cid:48) → Z , where c (( v, w )) = − c (( w, v )) . Then, a circulation is a function g : E (cid:48) → R ≥ is a function satisfying the following properties: • For all edges e ∈ E (cid:48) , we have that g ( e ) ≤ u ( e ) . • For all vertices i ∈ V, we have that (cid:80) k ∈ V | ( i,k ) ∈ E g (( i, k )) = 0 . • For all vertices v, w so that ( v, w ) ∈ E (cid:48) , g (( v, w )) = − g (( w, v )) . We say that the cost of the circulation is (cid:80) e ∈ E c ( e ) g ( e ) , which we denoteas c ( g ) . We have the following theorem from [Wil19] which corresponds to[BF99][Theorem 7.1], giving us certain criteria for when we have theminimal cost flow. We weaken the theorem to only the needed condi-tions.
GENERALIZATION OF THE GREENE-KLEITMAN DUALITY THEOREM 7
Theorem 3.1 (part of Theorem 5 . . The following areequivalent for a circulation g, given capacity and cost functions u, c respectively: • g is a minimum-cost circulation. • There exists a potential function p : V → R so that for allvertices v, w where ( v, w ) ∈ E (cid:48) and u (( v, w )) − g (( v, w )) ≥ ,c (( v, w )) + p ( v ) − p ( w ) ≥ . Using this theorem, we prove that a strengthened version of [BF99][Theorem7.1] holds. Suppose that we have a flow f and potential p on G P,h,C P , with the cost function c and capacity function u, so that f always liesbetween 0 and u for each edge in E. We prove the following theorem.
Theorem 3.2 (Modified Theorem 7.1 from [BF99]) . Let G be a directedgraph, with set of vertices V and set of edges E, with a single sourceand a single sink vertex. If we have a flow f and potential p so that p ( w ) − p ( v ) < c (( v, w )) = ⇒ f (( v, w )) = 0 , and p ( w ) − p ( v ) > c (( v, w )) = ⇒ f (( v, w )) = u (( v, w )) for any ( v, w ) ∈ E, then f has minimal cost over all flows of the samevalue; that is, the sum of f ( e ) c ( e ) over all edges e is minimal for thisflow.Proof. Suppose that the flow f satisfies these conditions, with value v. We’ll show that it is minimal by comparison with [Wil19][Theorem5.3]. Denote the source node a and the sink node b. First, as in [Wil19][ § G = ( V, E ) and a desiredflow value v, add to G the vertex s, and two edges, one from s to a, and one from b to s, both with capacity v and cost 0 . Given p as well,extend the potential function so that p ( s ) = p ( a ) as well.In addition, once we’ve added these two edges, perform the modifi-cations in the beginning of definition 1 . Specifically, let E (cid:48) be the newset of edges. Extend the capacity function u to u (cid:48) : E (cid:48) → R that is v on the edges ( s, a ) and ( b, s ) , − v on their reverses, and 0 on all theother edges not in E . Furthermore, extend the cost function to equal0 on the new edges.Now, [Wil19] notes that given this flow, there is a correspondingcirculation with the same cost. We show this more precisely.To do this, given any flow f (cid:48) , construct a function g (cid:48) where thefollowing hold: g (( v, w )) = f (( v, w )) for all ( v, w ) in E.g (( v, w )) = − f (( w, v )) for all ( v, w ) in E (cid:48) − E. FRANK Y. LU g (( s, a )) = g (( b, s )) = v.g (( a, s )) = g (( s, b )) = − v. It is not hard to check that this is a circulation.By construction, notice that the cost of g (cid:48) and the cost of f (cid:48) are thesame. Notice that the two new edges and their respective “reversed”edge have cost 0 and so don’t contribute to the total cost. Also, observethat for every edge ( v, w ) ∈ E in the circulation, the contributionof the cost due to ( v, w ) and ( w, v ) in total is ( c (( v, w )) f (( v, w )) + c (( w, v )) f (( w, v ))) = c (( v, w )) f (( v, w )) , and summing these up yieldsthe same cost.Now, let g be the circulation constructed from the particular flow f mentioned at the beginning of the proof. Notice that for all ( v, w ) ∈ E (cid:48) , we have three cases to consider.(1) First, if ( v, w ) ∈ E, by construction notice that if u (( v, w )) >f (( v, w )) , then by construction we see that p ( w ) − p ( v ) ≤ c (( v, w )) , or that c (( v, w )) − p ( w ) + p ( v ) ≥ . (2) Next, if ( v, w ) is so that ( w, v ) ∈ E, then notice that u (( w, v )) >f (( w, v )) ⇐⇒ f (( v, w )) > , which in turn means that p ( w ) − p ( v ) ≥ c (( v, w )) , or that c (( w, v )) − p ( v ) + p ( w ) ≥ . (3) For the last four edges, notice that their circulation equals theircapacity, so there’s nothing that needs to be checked here.Then, it follows that g is a minimal cost circulation, which means thatthe cost of g is at most the cost of g (cid:48) . But then it follows that the costof f is at most the cost of f (cid:48) , for any flow f (cid:48) with value v, which givesus the desired. (cid:3) In particular, notice that the first condition in [BF99], namely thatthe potential is bounded between its values at the source and sinknodes, is not necessary to maintain for minimality, thus allowing usmore flexibility with the potential function. We are now able to returnback to the notation of [BF99], but now with the possibility of negativepotentials and costs.3.3.
Applying the Algorithm.
We now apply [BF99][Algorithm 7.2]to the graph G P,h,C P , with the 2 n + 2 vertices b , t , b , . . . , t n +1 , butwith a few modifications (specifically to the initial conditions), whichare produced below. Let V be the set of vertices and E the set of edgesin this graph. Algorithm 1 (Modified Algorithm 7.2 from [BF99]) . The algorithmis as follows:
GENERALIZATION OF THE GREENE-KLEITMAN DUALITY THEOREM 9 (1)
To initialize the flow and potential, set f to be so that f ( e ) = 0 for every edge e ∈ E. We also declare that p ( b i ) = − i = p ( t i ) . (2) Let G (cid:48) be the modified graph with the same vertices and edges ¯ E = { ( v, w ) : ( v, w ) ∈ E, p ( w ) − p ( v ) = c (( v, w )) , f (( v, w )) } . From here, let X be the set of vertices v where a path existsfrom the source s to v using the edges in ¯ E. If t ∈ X, then goto step 3. Otherwise, go to step 4. (3) There exists a path through vertices s, v , v , . . . , v k , t, where allthese vertices lie in X and the edges are in ¯ E. Increase the flowof each edge along here by , then go to step 5. (4) Otherwise, for every vertex not in X, increase the potential ofthat vertex by . Go to step 5 next. (5)
If we have maximal flow, stop. Otherwise, go to step 2 again. [BF99][Theorem 7.3] says that the above algorithm maintains a min-imum cost flow for each flow value at each step, comparing with theconditions in [BF99][Theorem 7.1]. We will explicitly prove that thistheorem holds even if we strip the potential bounding condition, forthe sake of completeness.
Theorem 3.3 (Modified Theorem 7.3, [BF99]) . The above algorithmproduces, for each flow value, a minimal-cost flow, as the two conditionsdescribed in Theorem 3.2 are preserved after each step.Furthermore, the algorithm terminates when we reach a maximalflow value.Proof.
We prove that the initial conditions have the desired propertiesin Theorem 3.2, and then that, after running through the algorithm,the desired properties hold, assuming that they held initially. This willprove the desired claim by induction, and hence Theorem 3.2. For easeof notation, let the index of v be the value i so that either v = t i or v = b i ; initially, we see that the index of v is just − p ( v ) by construction.First, for the initial conditions, notice that the flow everywhere is0 , by construction, so the first condition is vacuously true. As for thesecond, notice that p ( w ) − p ( v ) > c (( v, w )) = ⇒ f (( v, w )) = u (( v, w )) , means that the index of w is smaller than that of v. But then we haveno edges from w to v, which means that this vacuously holds for alledges ( v, w ) ∈ E. Now, for the algorithm. The only issues we need to check are for steps3 and 4 . Suppose G P,h,C P initially satisfied the conditions in Theorem3.2. If we reach step 3, then by the algorithm we have a sequence of vertices s, v , . . . , v k , t, where each consecutive pair of vertices in thesequence has an edge in ¯ E, and we’ve increased the flow along theseedges by 1 . But notice that, by construction, the potentials between every pairof consecutive vertices equals the cost. This means that the conditionsstill hold, since the only pairs of vertices ( v, w ) where the flow changesare those where p ( w ) − p ( v ) = c (( v, w )) , so the conditions remain sat-isfied.Now, suppose we reached step 4. Consider any edge ( v, w ) ∈ E. If p ( w ) − p ( v ) < c (( v, w )) , notice then that, since p, c are always integers, p ( w ) − p ( v ) remains at most c (( v, w )) , and similarly for the > symbol.The only thing we need to check is when p ( w ) − p ( v ) = c (( v, w )) initially,and where exactly one of the potentials changes.Suppose that p ( w ) increases by 1 . Then, it follows that w is not in X, but v is in X. But this means that, since we have a path from s to v along edges in ¯ E, there is no edge between v and w in ¯ E. This meansthat, as ( v, w ) ∈ E, we have that f (( v, w )) = u (( v, w )) , since the flowmust remain at most the capacity. But then notice that this satisfiesthe condition.Similarly, if p ( v ) increases by 1 , this means that w ∈ X, v (cid:54)∈ X. Butagain, this means that we have no edge from w to v. But this meansthat f (( v, w )) = 0 , as ( v, w ) ∈ E. This means that the condition issatisfied for that edge too.For maximality, we will prove this at the end of the next section. (cid:3)
This allows us to notice that, at every stage of the algorithm, evenwith a different potential function, we still output a minimal cost flowfor a given flow value v. Basic Properties
First, we prove some properties of the flow on G P,h,C P in general,throughout the algorithm. We say that a vertex is “reachable by b ,” orjust “reachable,” if it lies in the set X (as per the notation of [BF99][ § t i or b i , where1 ≤ i ≤ n, can have at most one edge with nonzero flow going in, andat most one edge with nonzero flow going out. To see this, notice that t i has only one edge that flows out, and b i has only one edge goinginto it, and all edges in this case have capacity 1 . Since all flows areintegral, by the algorithm, it follows that there can only be one edgefor the other side that has nonzero flow.We now have two lemmas that we’d like to prove.
GENERALIZATION OF THE GREENE-KLEITMAN DUALITY THEOREM 11
Lemma 4.1.
For any edge from b i to t j , if there is a flow along thatedge, then p ( t j ) − p ( b i ) equals the cost of the edge.Proof. Suppose for the sake of contradiction that this fails at somepoint during Algorithm 1. Consider the first step at which this fails,after making the change in flow or potential.Note that this can’t be the first time that there is a flow betweenthe two edges, since by construction we only add the flow if the costequals the potential change. So this must mean that this occurs whilepotential drops; in other words, one of b i , t j is reachable by s along thisnew graph (in the sense that it lies in X ) and the other isn’t.Suppose that t j is reachable by b . Then, by construction, since be-fore the change in potential the cost of flow along the edge equals thedifference in potentials, we must have that b i is also reachable.Similarly, if b i is reachable by b , then there had to exist some pointbefore it that allowed us to reach it. But this means that either we hadto reach it via an unused edge (going forwards), or a used edge goingbackwards. The former, however, is impossible, since by the fact thatthere is flow out of b i there is flow into b i , and there is only one edgeflowing into b i . This means that we had to have reached t j to get to b i . Hence, thesupposed situation is impossible, which proves that the condition inthe lemma always holds, as desired. (cid:3)
In addition, we have the following property:
Lemma 4.2.
For any i ∈ { , , . . . , n } , p ( t i ) ≥ p ( b i ) − . Proof.
Again we proceed by contradiction. Suppose that at some pointthat p ( t i ) − p ( b i ) was less than − , for some i. Then, since p ( t i ) − p ( b i )can only increase or decrease by 1 at each point, at some point, then, p ( t i ) − p ( b i ) = − . Furthermore, at this point, only t i was reachableby b , and t n +1 wasn’t reachable, to cause the potential difference tochange.By the conditions given in Theorem 3.2, there has to be a flow from t i to b i . Then, note that, since there is flow into t i , there has to beanother vertex, b k , with k < i, where there is nonzero flow along theedge from b k to t i . If k = 0 , then we can’t reach i directly from s viaan unused edge; this means that there is some other vertex b h with h < i and where p ( t i ) − p ( b h ) = c (( b h , t i )) . We take that vertex instead.Otherwise, if k (cid:54) = 0 , we just take b k . In either case, notice that wehave that p ( t i ) − p ( b k ) = c (( b k , t i )) , with the case k (cid:54) = 0 following fromLemma 4.1. In addition, consider the next vertex along the flow line, say t j , j > i, after b i . Notice that there can’t be any flow from b k to t j , as the edgefrom b i to t j has nonzero flow, and k < i. We now do casework:(1) p ( b k ) = p ( t i ) . Since the cost of an edge is either − , wehave that, by Lemma 4.1, p ( t j ) − p ( b i ) = c (( b i , t j )) ≥ − , orthat p ( t j ) ≥ p ( t i ) = p ( b k ) . But this means that the cost of theedge between b k and t j is at most p ( t j ) − p ( b k ). Since therecan’t be any flow between them, the cost must be at least thepotential difference, so their potential difference is the sameas the cost of the edge between them. However, since b k isreachable, this means that t j is too, which means that b i isreachable, contradiction.(2) p ( b k ) = p ( t i ) + 1 . By a similar logic as above, we have that p ( t j ) ≥ p ( t i ) = p ( b k ) − . But also, since there can’t be nonzeroflow in the edge between b k and t j , notice that p ( t j ) − p ( b k ) ≤ c (( b k , t j )) ≤ . Hence, either p ( b k ) = p ( t j ) , or p ( b k ) = p ( t j ) + 1 . The former gives us the same logic as the first case. For thelatter, note that for this to occur, p ( t j ) = p ( t i ) = p ( b i ) − , or that p ( t j ) − p ( b i ) = − . But by Lemma 4.1, as we haveflow on the edge from b i to t j , this potential difference equals c (( b i , t j )) . But this means that either h − ( k ) < h − ( i ) < h − ( j ) , or t j = t n +1 . In either case, note that this means that the cost ofthe edge between b h and t j is − t j , and hence b i , is reachable.In either case, we run into a contradiction, which proves the lemma. (cid:3) From here, we can now prove that we eventually get maximality fromTheorem 3.3.
Proof of Theorem 3.3, continued.
Suppose for the sake of contradic-tion that this doesn’t ever reach maximal flow. Then, Algorithm 1doesn’t terminate, and so eventually reaches a point where step 4 isconstantly repeated, as step 3 increases flow and this maximal flow iswell-defined; see the Ford-Fulkerson theorem, which is, for instance,[Wil19][Theorem 2.6].In fact, here we can be more precise: notice that the maximal flowvalue is n. To see this, notice that the value of the flow is the sum ofthe flows of the edges coming out of b ; with n edges with capacity 1 , this is at most n. But to see maximality, notice that taking the edgesbetween b and t i , t i and b i , and b i to t n +1 , for each i ∈ { , , . . . , n } , GENERALIZATION OF THE GREENE-KLEITMAN DUALITY THEOREM 13 gives a flow with value n. Hence, maximal flow is n. Therefore, for thesake of contradiction, we see that the flow value we reach is v < n.
Now, notice that, in general, step 4 cannot make | X | fall; indeed,notice that step 4 alters potentials of vertices outside of X, and doesn’talter flows, so every vertex in X remains in X. This means that, for us to never have t ∈ X, eventually X reachessome maximal set X (cid:48) , since the number of elements is at most 2 n + 2 . Furthermore, beyond this point, all of the flows of edges in G P,h,C P remain constant. Consider the elements that must lie in this set X (cid:48) . Given that the only edges from b are to vertices of the form t i , andthat furthermore by construction in Algorithm 1 flows for each edgeare integers (either 0 or 1), it follows that there is some t i , i an integerbetween 1 and n, inclusive, so that the edge from b to t i has flow0 (since we are assuming non-maximal flow). By the second part ofTheorem 3.3, it follows that p ( t i ) − p ( b ) = p ( t i ) ≤ . If t i wasn’t in X (cid:48) it would follow that the potential of t i would repeatedly increase by 1 , contradicting this inequality.From here, we have two cases. If the edge between t i and b i doesn’thave a flow, then it follows that p ( b i ) − p ( t i ) ≤ , which using the abovemeans that p ( b i ) − p ( b ) = p ( b i ) − p ( t i ) + p ( t i ) − p ( b ) ≤ . But again, bythe same argument above, b i must lie in X (cid:48) , as otherwise its potentialwill be unbounded as we continually repeat step 4 in Algorithm 1 (with X never changing from X (cid:48) ).Now, notice that, since the only edge that points to b i is from t i , byconstruction, and since we assumed that the flow on the edge was 0 , the edge between b i and t n +1 has flow zero too. But the exact sameargument shows that t n +1 ∈ X (cid:48) , which contradicts the fact that we didstep 4.Otherwise, there is a flow on the edge between t i and b i . But thismeans that, by flow conservation, there exists an edge pointing into t i with flow, say from b j . But notice that all of the flow values areintegers, and since the capacities are 1 , this edge has flow 1 . But noticethen that the only edge pointing into b j is from t j , and it has capacity1 . This means that the edge from b j to t n +1 , by flow conservation, hasflow 0 , meaning that p ( t n +1 ) − p ( b j ) ≤ . However, notice that, byLemma 4.1, we have that p ( b j ) = p ( t i ) − c (( b j , t i )) ≤ p ( t i ) + 1 the latterby construction of the costs.This means, however, that p ( t n +1 ) − p ( b j ) + p ( b j ) ≤ p ( t i ) ≤ , which again means that p ( t n +1 ) is bounded, so t n +1 has to lie in X (cid:48) , contradiction. This means that step 4 isn’t used here, provingmaximality, as desired. (cid:3) Relating Graph and Poset Quantities
We now take G P,h,C P and relate it back to A (cid:48) k and D (cid:48) k . We begin bytranslating the poset quantities to the quantities on the graph, specif-ically flows and potentials, which [BF99][ §
8] also does. However, theway these quantities are related to the poset quantities is somewhatdifferent here compared to the corresponding version in [BF99][ § Proposition 5.1. In G P,h,C P given a fixed flow volume v, the minimalcost of the flow is equal to − A (cid:48) v . Proof.
To see this, we will first show that this is attainable. To do this,suppose that we have sequences S , S , . . . , S v that give the value A (cid:48) v . Ifone of the sequences consists of elements s , s , . . . , s l we add the flowline going from b to t h ( s ) , then t h ( s ) to b h ( s ) , then b h ( s ) to t h ( s ) , andso forth, until b h ( s l ) to t n +1 . By construction, notice that we may dothis, since the edges from b h ( s i ) to t h ( s i +1 ) exist by construction, as wedemanded ( s i , s i +1 ) to lie in C P for the sequences.Doing this for each sequence gives us the flow. Note that this satisfiesthe flow requirements, since at each vertex, the in and out flows arethe same for all the vertices besides b , t n +1 . In addition, we only useeach edge once, since the vertices are all distinct in the sequences (fromconstruction).As for the cost of this flow, note that along each flow line, if itcorresponds to sequence S of elements s , s , . . . , s l we see that all theedges have cost 0 except those edges from b h ( s i ) to t h ( s i +1 ) , where s i
Now, we introduce another quantity. Let p = | p ( t n +1 ) | . We saythat P p is the number of i ∈ { , , . . . , n } such that p ( t i ) = p ( b i ) ∈{− p + 1 , − p + 2 , . . . , } . Proposition 5.2.
At any point along Algorithm 1, P p + A (cid:48) v ≥ n + vp. Proof.
This method is very similar to that of [BF99][ § , the only thing that changesis the cost of the flow, which decreases (by construction of this newflow line from the algorithm) by p, since along this new flow-line, foreach edge, the cost is equal to the difference in potential.If there are no flow lines, then note that if t i is reachable, so is b i , and if t i has potential 0 , it is reachable, so again we have no problemshere (potential drop of t doesn’t change P (cid:48) p )Now, suppose that the potential of t n +1 increases by 1 . Consider agiven flow-line, say reaching top and bottom pairs with indices i , i , . . . , i k . Then, note that if t i j is reachable by b , so is b i j − . We can consider consecutive blocks of vertices reachable by b alongthis flow line. Suppose we have a block running from b i c to t i d . Notethat, among these, their potentials stay the same. Furthermore, notethat this sequence cannot cause P (cid:48) p to drop; the only place where oneis no longer counted was if initially t i d and b i d had the same potentials.But note that, from un-reachability, p ( t i c ) increases by 1 , which meansthat it matches up with p ( b i c ) now.Hence, the only way for there to be a drop would be either if a pairof vertices had potential 0 and went up to 1 , or if there is a blockthat went directly to t i . But these are mutually distinct events, for apotential of 0 going up to 1 can only mean that t i and b i had potential0 and weren’t reachable (if any other pair of vertices had potential 0 , the top would be reachable).This means that, when p rises by 1 , P (cid:48) p drops by at most v. But thisthe establishes the inequality. (cid:3)
Note that D (cid:48) p ≥ P (cid:48) p . To see this, we let the sequences be so that the i th sequence has the indices of those whose potentials of the top andbottom vertices are all − i + 1 . This is a valid sequence for two reasons. First, for the actual non-increasing part, note that between the bottom vertex of one and the topvertex of the next, the cost can’t be less than the potential difference,which is 0 (either there is no flow, or there is flow, which means that this follows from Lemma 4.1). Hence, we see that this forms a non-increasing sequence.Now, we claim that if the potential of t a , b a are i, and that for t c , b c is i − , then ( h − ( c ) , h − ( a )) (cid:54)∈ C P . To see this, if not we would have anedge from b c to t a . But then notice that Lemma 4.1 and the condition1 from Theorem 3.2 requires that p ( t a ) − p ( b c ) ≤ c (( b c , t a )) ≤ , con-tradiction. This means that the sequences can’t be semi-overlapping,so this is a valid choice of sequences, giving a value of the sum of the desc over these sequences as P (cid:48) p . The main result, that A (cid:48) v and D (cid:48) p are conjugate in the sense wedescribed, will follow in the next section.6. Establishing Equality
This section follows [BF99][ §
5] in concept, though the actual methodof calculation is slightly different, due to different conditions on theascending and non-ascending sequences.We use the same idea of considering intersections, however. Sup-pose that we are given sequences d , d , . . . , d p as the non-increasingsequences that meet the condition for D (cid:48) p , and a , a , . . . , a v for A (cid:48) v . Notice that if the d i are contained in sequences that are not semi-overlapping, then the d i are not semi-overlapping either.Fixing some a i , note that a i ∩ d , a i ∩ d , . . . , a i ∩ d p (the subsequencesof a i that are also part of d , d , . . . , d p , respectively) are also not pair-wise semi-overlapping, from construction.In fact, notice that if element x ∈ a i ∩ d m and y ∈ a i ∩ d j are so that h ( x ) < h ( y ) , then notice that, by construction, ( x, y ) ∈ C P . But thismeans that all elements in a i ∩ d m occur before those in a i ∩ d j . Now, notice that for pair of consecutive elements within a i ∩ d j , say x and x (cid:48) , there exists a non-ascent in-between x and x (cid:48) in a i , as otherwise x < x (cid:48) , contradiction. Furthermore, in-between these elements, by theargument above, no other a i ∩ d k can have elements, meaning thateach element of a i ∩ d j , letting j vary, other than the last for each,corresponds uniquely to a non-ascent.This means that we have p (cid:80) j =1 | a i ∩ d j | ≤ p + ( des ( a i )) , where des ( a i )is the number of “non-ascents,” which by definition we can see satisfies des ( a i ) = | a i | − asc ( a i ) . Note that this isn’t desc ( a i ) . This in turn yields that p (cid:80) j =1 | a i ∩ d j | ≤ p +( des ( a i )) ≤ p + | a i |− asc ( a i ) . But then we have that v (cid:80) i =1 p (cid:80) j =1 | a i ∩ d j | ≤ vp + v (cid:80) i =1 | a i | − A (cid:48) v . GENERALIZATION OF THE GREENE-KLEITMAN DUALITY THEOREM 17
But by PIE, since the a i are disjoint and the d j are disjoint, we havethat D (cid:48) p = | (cid:83) pj =1 d j | = | (cid:83) pj =1 d j ∪ (cid:83) vi =1 a i | − | (cid:83) vi =1 a i | + v (cid:80) i =1 p (cid:80) j =1 | a i ∩ d j | ≤ n − | (cid:83) vi =1 a i | + vp + v (cid:80) i =1 | a i | − A (cid:48) v = n + vp − A (cid:48) v . For equality, now note, for each pair ( p, v ) that are reachable for | p ( t n +1 ) | and flow value, respectively, we have that n + vp − A (cid:48) v ≥ D (cid:48) p ≥ P (cid:48) p ≥ n + vp − A (cid:48) v . To get the desired conjugacy, the exact argument at the end of[BF99][ §
8] allows us to finish. Specifically, we know now that D (cid:48) p + A (cid:48) v = n + vp, where p, v are values that are attained for p ( t n +1 ) and flow value,respectively, during Algorithm 1. We just need to check that we canapply the argument in that section here to all of the indices.Now, notice that, by Theorem 3.3, the algorithm terminates whenflow is maximal for the graph, which is when v = n (taking, for each i ∈ { , , . . . , n } a flow from b to t i to b i to t n +1 ). Furthermore, notethat v starts at 0 . Therefore, notice that, at this ending point, we have flow value n and some potential p . When this occurs, notice that A (cid:48) n + D (cid:48) p = n + np = ⇒ D (cid:48) p = np . But notice that, by construction, we see that D (cid:48) p ≤ D (cid:48) n = n meaning that p = 0 or p = 1 , so by a similar argumentwe see that the value of | p ( t n +1 ) | attains all values between 1 and n. Thus, for each i ∈ { , , . . . , n } when flow value increases from i − i in the algorithm, λ i = A (cid:48) i − A (cid:48) i − = p. Notice that we can show that λ i and µ i are partitions, from the samelogic as in [BF99, § p is weakly decreasing, giving us that the λ i are weaklydecreasing. As for the µ i , notice that, by a similar logic, when thepotential goes from p to p −
1, we have that µ p = D (cid:48) p − D (cid:48) p − = ( n + vp − A (cid:48) v ) − ( n + v ( p − − A (cid:48) v ) = v. But then, observe that, throughout theprocess, p falls and v rises, so again the µ i are also weakly decreasingif we start from i = 1 . This gives us that these are partitions.This yields us the desired conjugacy of λ i = A (cid:48) i − A (cid:48) i − and µ i = D (cid:48) i − D (cid:48) i − , as desired, which proves Theorem 2.1.7. Corollaries
Theorem 2.1 gives us both the localized Greene’s theorem for per-muation posets and the original Greene-Kleitman duality theorem. Weprove each of these results using Theorem 2.1 in this section.
Corollary 7.1 (Localized Greene’s Theorem, Lemma 2.1 [Lew+20]) . Let σ be a permutation on n elements, { , , . . . , n } . Then, with A ∗ k as the maximal sum of the ascents of k disjoint sequences, and D ∗ k as themaximal sum of the longest descending subsequences in k consecutivesequences (as we noted in the introduction, Section 1, which are definedas per [Lew]), if λ k = A ∗ k − A ∗ k − and µ k = D ∗ k − D ∗ k − , then λ + λ + · · · and µ + µ + · · · form conjugate partitions of n. Proof.
Take the poset of 1 , , . . . , n with the natural ordering, and sup-pose that h is the inverse of the permutation σ, which is a bijection.Let C P just be the set { ( x, y ) | ≤ x, y, ≤ n, h ( x ) < h ( y ) } ; in thiscase, h − ordering and adjacentable are the same. Apply Theorem 2.1,obtaining A (cid:48) k and D (cid:48) k . Then, notice that A (cid:48) k is the same as A ∗ k since asc is defined thesame way. To see this, notice that any sequence S, with elements s , s , . . . , s l , where σ ( s j ) < σ ( s j +1 ) for each index j, can be thought ofas a subsequence of elements from σ (1) , σ (2) , . . . , σ ( n ) , as the abovetells us that σ − ( s ) , σ − ( s ) , . . . , σ − ( s l ) is a strictly increasing se-quence. This means we may re-write the sequence as σ ( x ) , σ ( x ) , . . . , σ ( x l )for an increasing sequence x , . . . , x l . But then asc ( S ) is just the num-ber of indices j where σ ( x j ) < σ ( x j +1 ) plus one (or 0 if S is empty),which matches. This means that A (cid:48) k , as the maximum of the sum of asc of k disjoint sequences, is the same as A ∗ k . As for D (cid:48) k , first notice that desc is defined the same way as well,since the condition that s i (cid:54) < s j for each i < j, with the totally orderedset, just means that the sequence must be strictly decreasing. Now,suppose that we have sequences S , S , . . . , S k that give the maximalvalue, such that no two are semi-overlapping.Now, since C P is just h − ordering, notice that for each pair of ele-ments x, y ∈ { , , . . . , n } , either ( x, y ) ∈ C P or ( y, x ) ∈ C P . We maythus re-index the sequences so that ∀ i < j, ∀ a ∈ S i , b ∈ S j , h ( a ) < h ( b )(the semi-overlapping condition allows us to do this re-indexing).From here, suppose that some element x ∈ { , , . . . , n } not in anyof the S i . Let j be the largest index so that ∃ a ∈ S j where h ( a ) < h ( x ) , and suppose that a is chosen so that h ( a ) is the maximum value of { h ( b ) | b ∈ S j , h ( b ) < h ( x ) } . We may then add x to S j right after a ; by construction, this preserves all of the conditions of non semi-overlapping. Furthermore, notice that the (cid:80) ki =1 desc ( S i ) cannot de-crease; indeed, we may take the same descending sequence within S j . By maximality, this value also can’t increase.We may thus assume that maximal S , S , . . . , S k covers all of theelements in { , , . . . , n } . But notice then that, as required in [Lew], S | S | . . . | S k is the sequence h − (1) , h − (2) , . . . , h − ( n ) , or σ (1) , σ (2) , . . . , σ ( n ) . GENERALIZATION OF THE GREENE-KLEITMAN DUALITY THEOREM 19
This means that the value of D (cid:48) k , as defined here, is the same as D ∗ k .This proves the desired. (cid:3) Corollary 7.2 (Classical Greene-Kleitman Duality Theorem, Theorem1.6 [Gre76]) . Given a poset P, let A k be the maximal number of elementswithin k disjoint chains, and D p the maximal number of elements within p disjoint anti-chains. Then, if λ i = A i − A i − and µ i = D i − D i − for i ≥ , with A = D = 0 , then λ + λ + . . . and µ + µ + · · · areconjugate partitions of n. Proof.
Let P be the poset, and h any linear extension of P. From here,let C P be just the set { ( x, y ) | x < y } ; notice that this satisfies theproperties given.Then, notice that any adjacentable sequence, by construction, mustconsist solely of elements where any two adjacent are increasing; inother words, they must be chains. Therefore, it follows that A (cid:48) k in The-orem 2.1 just corresponds to the maximal length of k disjoint chains,which is just A k . As for D (cid:48) p , we need to do a little more work. Notice that D (cid:48) p ≤ D p . To see this, suppose that sequences S , . . . , S p had subsequences d , . . . , d p , whose sum of lengths was D (cid:48) p . By construction, for eachsequence d j , if the elements in order were s ,j , . . . , s l j ,j , then notice thatthe condition that s a,j (cid:54) < s b,j for each a, b, combined with the ordering h, thus demands that, in fact, s a,j and s b,j are not comparable. Thismeans that each of the d i are anti-chains.To show the other direction: suppose that we have p anti-chains by d , d , . . . , d p so that their sum has maximal size. Consider the orderedtuple obtained by taking the elements for d in order, followed by theelements for d in order, and so forth, and order these lexicographicallyusing the linear extension. For instance, if we have the poset on fiveelements a, b, c, d, e, with relations a < b, b < d, c < d, and d < e, with h ( a ) = 1 , h ( b ) = 2 , h ( c ) = 3 , h ( d ) = 4 , and h ( e ) = 5 , taking d to bethe sequence a, c and d to be b yields the tuple ( a, c, b ) . Now, consider the following operation: given d i and d j , where i < j, let A = { x ∈ d i |∃ y ∈ d j so that y < x } . Similarly, let B = { y ∈ d j |∃ x ∈ d i so that y < x } . Then, take the elements from A, and movethem to d j , and take the elements from A, and move them to d i . Callthese new anti-chains d (cid:48) i , d (cid:48) j . First, note that the new d i and d j are both anti-chains. Suppose forthe sake of contradiction this wasn’t the case; then, since d i , d j wereanti-chains, the relations that occur afterwards must have one elementin one of the sets A, B and the other not (since, by anti-chain, all the elements in A are pairwise incomparable, and similarly for B ). Thisyields four cases:(1) If there exists an a ∈ d (cid:48) i , b ∈ B so that b < a, then a ∈ d (cid:48) i meansthat a (cid:54)∈ A. But a (cid:54)∈ B, so a ∈ d i , and a ∈ A, contradiction.(2) If there exists an a ∈ d (cid:48) i , b ∈ B so a < b, then there exists anelement x in d i so that b < x, so then a < x. But a (cid:54)∈ B, so a ∈ d i , contradicting anti-chain.(3) If there exists an a ∈ A, b ∈ d (cid:48) j so that b < a, then notice that b ∈ d (cid:48) j means that b (cid:54)∈ B. But a ∈ A ⊆ d i , meaning that b ∈ B, contradiction.(4) If there exists an a ∈ A, b ∈ d j so a < b, then there exists a y ∈ d j so that y < a < b, or y < b. But b (cid:54)∈ A, so thus b ∈ d j , contradicting anti-chain.Therefore, we end up still with anti-chains, the sum of whose lengthsis the same.Furthermore, notice that the result we get is an element that islexicographically earlier; let x be so that h ( x ) is minimal, among allelements of A, B.
Then, notice that, by construction, x ∈ B, otherwisewe see that there is a y ∈ d j so that h ( y ) < h ( x ) , meaning that y ∈ B as x ∈ A ⊆ d i , contradicting minimality. Then, notice that this movesfrom the list of j s to the list of i s, and by construction no other elementsare moved other than those in A or B. But i < j means that this meansit is lexicographically earlier.Since we only have a finite number of these tuples, we can only applythis process a finite number of times before we end up with a resultwhere, for any i, j, the resulting
A, B are empty. But if
A, B are empty,notice then that these anti-chains are all not semi-overlapping, sincethe semi-overlapping condition for d i , d j here requires that, for i < j, that there exists x ∈ d i , y ∈ d j so y < x, or that the resulting A, B aren’t empty.Therefore, we see that we can re-arrange the anti-chains in a way sothat they are not semi-overlapping, so D (cid:48) p ≥ D p ≥ D (cid:48) p , and these areequal.But this means that the conjugate partitions in this theorem areprecisely those given in Theorem 2.1, as desired. (cid:3) Note that Example 2.1 yields a case that doesn’t fall under either ofthese corollaries. In particular, we can view Corollary 7.1, the localizedGreene’s theorem, as being the case when C P is as large as possible,and poset P is just { , , . . . , n } . On the other hand, Corollary 7.2occurs when C P is as small as possible, and h is a linear extension. EFERENCES 21
References [Gre76] Curtis Greene. “Some partitions associated with a partiallyordered set”. In:
Journal of Combinatorial Theory, SeriesA issn : 0097-3165. doi : https :/ / doi . org / 10 . 1016 / 0097 - 3165(76 ) 90078 - 9 . url : .[BF99] Thomas Britz and Sergey Fomin. Finite Posets and FerrersShapes . 1999. arXiv: math/9912126 [math.CO] .[Wil19] David P. Williamson.
Network Flow Algorithms . CambridgeUniversity Press, 2019. doi : .[Lew+20] Joel Lewis et al. Scaling limit of soliton lengths in a multi-color box-ball system . 2020. arXiv: .[Lew] Joel Lewis.
A localized version of Greene’s theorem . url : https://realopacblog.wordpress.com/2019/11/24/a-localized-version-of-greenes-theorem/ . Department of Mathematics, Princeton University, Princeton, NJ08544, USA
Email address ::