The structure of the consecutive pattern poset
aa r X i v : . [ m a t h . C O ] J a n THE STRUCTURE OF THE CONSECUTIVE PATTERN POSET
SERGI ELIZALDE AND PETER R. W. MCNAMARA
Abstract.
The consecutive pattern poset is the infinite partially ordered setof all permutations where σ ≤ τ if τ has a subsequence of adjacent entriesin the same relative order as the entries of σ . We study the structure ofthe intervals in this poset from topological, poset-theoretic, and enumerativeperspectives. In particular, we prove that all intervals are rank-unimodal andstrongly Sperner, and we characterize disconnected and shellable intervals. Wealso show that most intervals are not shellable and have M¨obius function equalto zero. Contents
1. Introduction 22. Preliminaries 42.1. Fundamentals 42.2. The M¨obius function papers 42.3. Initial observations 63. Disconnectivity 63.1. Characterization of disconnected intervals 73.2. Finding disconnected intervals 84. Shellability 114.1. Background on shellability and CL-shellability 114.2. Shellability in the consecutive pattern poset 125. All intervals are rank-unimodal and strongly Sperner 165.1. Rank-unimodality 165.2. Strongly Sperner property 196. The exterior of a permutation 216.1. The length of the exterior 216.2. Asymptotic behavior of the length of the exterior 226.3. Permutations with no carrier, and the M¨obius function of mostintervals 25Acknowledgement 27References 27
Mathematics Subject Classification.
Primary 06A07; Secondary 05A05, 05A16, 05E45,52B22, 60C05.
Key words and phrases.
Consecutive pattern, poset, disconnected, shellable, rank unimodal,Sperner, exterior, M¨obius function.Sergi Elizalde was partially supported by grant Introduction
Consecutive patterns in permutations generalize well-studied notions such asdescents, ascents, peaks, valleys and runs. A permutation σ is said to be contained in another one τ as a consecutive pattern if τ has a subsequence of adjacent entriesin the same relative order as the entries of σ . Otherwise, τ is said to avoid σ .For example, permutations that avoid both 123 and 321 as a consecutive patternare the well-known alternating and reverse-alternating permutations. A systematicstudy of enumerative aspects of consecutive patterns started in [EN03] and, in thelast decade, such patterns have become a vibrant area of research; see [Eli16] fora recent survey. Underlying all these questions is a partial order P on the set ofall permutations, where we define σ ≤ τ if σ is contained in τ as a consecutivepattern. This paper is the first systematic study of intervals in this consecutivepattern poset. See Figure 1.1 for an example of such an interval.Some questions for consecutive patterns are motivated by the analogous prob-lems for so-called classical patterns, one of the most actively studied topics ofcombinatorics in the last three decades. For the definition of containment in thisclassical case, we remove the restriction that the relevant subsequence of τ consistsof adjacent entries. For example, 123 is less than 2314 in the classical pattern posetbut not in the consecutive pattern poset. See [B´on12] for an exposition of some ofthe most important developments in the study of classical patterns, [Kit11] for adetailed treatment, and [Ste13] for a survey of recent developments.Examples of questions for consecutive patterns that are motivated by the classi-cal case include classifying them into equivalence classes determined by the numberof permutations avoiding them [Nak11], finding generating functions for the dis-tribution of occurrences of a fixed pattern in permutations [EN03, MR06, EN12],and determining the asymptotic growth of the number of permutations avoiding apattern [Eli13, Per13]. Consecutive patterns are interesting not only because theanswers to the above questions and the techniques used to solve them are usuallyquite different than for classical patterns, but also because they have importantapplications to dynamical systems [AEK08, Eli09].While most of the aforementioned work is enumerative, our approach also has aposet-theoretic and topological flavor. An early inspiration for research from thisviewpoint was a question of Wilf [Wil02] asking for the M¨obius function of inter-vals in the poset defined by classical pattern containment. This question remainswide open, but has received increasing attention of late [BJJS11, MS15, SV06,Smi14, Smi15, Smi16, ST10]. In contrast, the M¨obius function for intervals in P ,the consecutive pattern poset, has been determined by Bernini, Ferrari and Ste-ingr´ımsson [BFS11] and by Sagan and Willenbring [SW12]. This already gives anindication that the consecutive pattern case is more tractable for certain types ofquestions than the classical case.The precursor in the classical case to the present work is [MS15], where thefocus is on classifying disconnected open intervals and showing that certain specialintervals are shellable. We successfully address these same topological questions ofdisconnectivity and shellability for the consecutive pattern poset P . Furthermore,we consider what poset-theoretic properties might hold for P , e.g., we show that allintervals are rank-unimodal, a statement that is just a conjecture in the classicalcase [MS15]. Recall that a finite graded poset of rank N with a i elements of rank i is called rank-unimodal if the sequence a , a , . . . a N is unimodal, meaning that HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 3 there exists k with 0 ≤ k ≤ N such that a ≤ a ≤ · · · ≤ a k ≥ a k +1 ≥ · · · ≥ a N .Our main results include the following. ◦ A simple classification of those open intervals in P that are disconnected(Theorem 3.2). ◦ Almost all intervals contain a disconnected subinterval of rank at least 3,in a certain precise sense, and are thus not shellable (Theorem 3.6 andCorollary 4.2). ◦ All other intervals are shellable (Theorem 4.3). One motivation for shella-bility is that it completely determines the homotopy type of the order com-plex of the interval as a wedge of spheres. The number of spheres is theabsolute value of the M¨obius function and is thus readily determined using[BFS11, SW12]. ◦ All intervals are rank-unimodal (Corollary 5.2). The highest rank of themaximum cardinality is easily determined and the interval takes on a par-ticularly simple structure above this rank. ◦ All intervals are strongly Sperner (Theorem 5.6), a condition equating thesizes of the union of the k largest ranks and the largest union of k antichains,for all k . ◦ It is clear from the formula for the M¨obius function µ ( σ, τ ) that it dependsheavily on the exterior of τ , which is the longest permutation that is both aproper prefix and proper suffix of τ . We initiate a study of the asymptoticbehavior of the exterior in Section 6. We show that its expected length isbounded between e − e (Theorem 6.5). ◦ Almost all intervals have a zero M¨obius function, in a certain precise sense(Corollary 6.10).Taken together, these topological, poset-theoretic and enumerative results givea rather comprehensive picture of the structure of intervals in P .Preliminaries and some initial observations are the content of the next section.Disconnectivity is the subject of Section 3, shellability is addressed in Section 4,and the properties of rank-unimodality and strongly Sperner are shown in Section5. The exterior of a permutation is considered in Section 6, along with several openproblems about its behavior. 12123 213 1322134 1243 132421354 12435213546 Figure 1.1.
The interval [12 , P . SERGI ELIZALDE AND PETER R. W. MCNAMARA Preliminaries
In this section, we collect together some useful terminology and notation, andmake some initial observations.2.1.
Fundamentals.
Let S n denote the set of permutations of [ n ] := { , , . . . , n } .If τ ∈ S n , we denote the length of τ by | τ | = n , and we write τ in one-line notation as τ = τ τ . . . τ n . Given a sequence a a . . . a k of distinct positive integers, define its reduction ρ ( a a . . . a k ) to be the permutation of { , . . . , k } obtained by replacingthe smallest entry with 1, the second smallest with 2, and so on. For example, ρ (394176) = 263154. For 1 ≤ i ≤ j ≤ n , let τ [ i,j ] = ρ ( τ i . . . τ j ), and note that τ [ i,j ] ∈ S j − i +1 .We can now define our poset of interest. We say that τ contains σ as a consecutivepattern if there exist 1 ≤ i ≤ j ≤ n such that τ [ i,j ] = σ . We write σ ≤ τ , and wesay that τ i . . . τ j is an occurrence of σ . The relation ≤ defines a partial order onthe set S = S n ≥ S n of all permutations, and we denote by P the correspondingpartially ordered set, called the consecutive pattern poset . The poset P is the mainobject of study in this paper. See Figure 1.1 for an example of an interval in P .A related partial order on S is obtained by considering classical pattern contain-ment instead, i.e., the case when τ is said to contain σ if any subsequence of τ (notnecessarily in consecutive positions) has reduction equal to σ . We will refer to theresulting poset as the classical pattern poset .In the rest of this paper we will say that τ contains σ to mean that τ contains σ as a consecutive pattern, that is, σ ≤ τ . If this is not the case, we say that τ avoids σ .We next recall some common operations on permutations. The reversal of thepermutation τ = τ τ . . . τ n is the permutation τ n . . . τ τ . The complement of τ isthe permutation whose i th entry is n + 1 − τ i . Note that both the reversal andcomplement operations preserve the order relation in P . We note, however, thatunlike in the classical pattern poset, the inverse operation does not preserve theorder relation in P ; e.g., the relation 132 ≤ σ ∈ S m and τ ∈ S n , we define their direct sum σ ⊕ τ as the concatenation of σ and the permutation τ + m formed by adding m to every entry of τ . Similarly, the skew sum σ ⊖ τ is the concatenation of σ + n and τ . For example, 1324 ⊕
21 = 132465and 1324 ⊖
21 = 354621.To consider disconnectivity in Section 3 and shellability in Section 4, we needto define a simplicial complex associated with any interval of any poset. Givenan interval [ σ, τ ], its order complex ∆( σ, τ ) is the simplicial complex whose facesare the chains of the open interval ( σ, τ ). For example, the order complex of theinterval [12 , The M¨obius function papers.
It remains a wide open problem to determinethe M¨obius function for the classical pattern poset; see [BJJS11, MS15, SV06,Smi14, Smi15, Smi16, ST10] for special cases. In contrast, the M¨obius function forthe consecutive pattern poset P has been determined independently by Bernini,Ferrari and Steingr´ımsson in [BFS11] and by Sagan and Willenbring in [SW12].We will present this recursive formula as stated in the latter paper here because it HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 5
Figure 2.1.
The order complex of the interval [12 , P from Figure 1.1. Only chains in the open interval ( σ, τ ) are includedin the order complex of [ σ, τ ]; including σ and τ would simplyresult in the join of the order complex with a line segment andhence would be contractible.introduces some key ideas we will need later, and then we will explain the parts ofthe viewpoint of the former paper that will be useful to us.We say that σ is a prefix (respectively suffix ) of τ ∈ S n if σ = τ [1 ,i ] (resp. σ = τ [ n − i +1 ,n ] ) for some i , and is a proper prefix (resp. suffix) if it does not equal τ . For example, 312 is a proper prefix of 51342. We say that σ is a bifix of τ if it isboth a proper prefix and proper suffix of τ . The exterior of τ , denoted x ( τ ), is thelongest bifix of τ . Note that x ( τ ) always exists whenever τ > τ in that case. The interior of τ is τ [2 ,n − and is denoted i ( τ ).For example, if τ = 21435, then x ( τ ) = 213 and i ( τ ) = 132. We say that τ is monotone if it equals 12 . . . n or n . . . Theorem 2.1 ([BFS11, SW12]) . For σ, τ ∈ P with σ ≤ τ , µ ( σ, τ ) = µ ( σ, x ( τ )) if | τ | − | σ | > and σ ≤ x ( τ ) i ( τ ) , if | τ | − | σ | = 2 , τ is not monotone, and σ ∈ { i ( τ ) , x ( τ ) } , ( − | τ |−| σ | if | τ | − | σ | < , otherwise. It is clear from this important theorem that the exterior of a permutation is wor-thy of attention, and yet this attention seems to be absent from the literature. Thismotivates our detailed consideration of the statistic | x ( τ ) | for general permutationsin Subsections 6.1 and 6.2. In particular, we show that the expected value of | x ( τ ) | as | τ | → ∞ is bounded between e − e .In [BFS11], if the exterior x ( τ ) satisfies the condition that σ ≤ x ( τ ) i ( τ ), then x ( τ ) is called the carrier element of [ σ, τ ]. As we see in Theorem 2.1, when [ σ, τ ] hasa carrier element we get the interesting case where the calculation of µ ( σ, τ ) reducesto determining µ ( σ, x ( τ )). It is therefore natural to ask under what conditions [ σ, τ ]has a carrier element. This motivates our consideration in Subsection 6.3 of thepermutations τ such that [1 , τ ] has a carrier element. In this case of σ = 1, x ( τ )can be called the carrier element of τ , and it exists if and only if x ( τ ) i ( τ ). SERGI ELIZALDE AND PETER R. W. MCNAMARA
We determine the asymptotic probability of a permutation not having a carrierelement and, as an application, prove that the M¨obius function is almost always 0,in a particular precise way.2.3.
Initial observations.
One feature of the consecutive pattern poset thatmakes it more tractable than the classical pattern poset is that any permutation τ of length n covers at most two elements, namely τ [1 ,n − and τ [2 ,n ] . The nextlemma will be useful later; it appears in [BFS11, SW12] and is routine to check. Lemma 2.2.
For τ ∈ P of length n , we have τ [1 ,n − = τ [2 ,n ] if and only if τ ismonotone. As a consequence, we get our first structural result about intervals in P . Proposition 2.3.
The interval [ σ, τ ] in P is a chain if and only if either ◦ τ is monotone, or ◦ σ occurs exactly once in τ and does so as a prefix or a suffix.Proof. The “if” direction is clear, while for the “only if” direction we observethat if [ σ, τ ] is a chain, then τ covers just one element, leading to the two givenpossibilities. (cid:3) As previously shown in [BFS11], we can completely determine the structure of[ σ, τ ] when σ has just one occurrence in τ , a result we prove here for the purposesof self-containment. This contrasts with the classical pattern poset, where thecorresponding result is unknown. Proposition 2.4 ([BFS11]) . For an interval [ σ, τ ] in P , if σ occurs exactly oncein τ then [ σ, τ ] is a product of two chains. Moreover, if σ = τ [ i,j ] , then these chainshave lengths i and | τ | − j + 1 .Proof. Any π ∈ [ σ, τ ] is obtained from τ by deleting a entries from the left end of τ and b entries from the right end of τ , where0 ≤ a ≤ i − ≤ b ≤ | τ | − j. (2.1)Thus there is a surjective map φ from such tuples ( a, b ) to the elements π of [ σ, τ ].This map is injective since there is just one occurrence of σ in τ , allowing us touniquely determine a and b from π . We also see that φ is order-reversing in thesense that φ ( a, b ) ≤ φ ( a ′ , b ′ ) in [ σ, τ ] if and only if a ≥ a ′ and b ≥ b ′ as integers.Therefore, [ σ, τ ] is order-isomorphic to the dual of a product of two chains, whichis itself a product of two chains. The lengths of the chains follow from (2.1). (cid:3) Disconnectivity
We say that an interval [ σ, τ ] is disconnected (resp. shellable ) if ∆( σ, τ ) is discon-nected (resp. shellable). Equivalently, [ σ, τ ] is disconnected if the Hasse diagram ofthe open interval ( σ, τ ) is disconnected. For example, in Figure 1.1, the subinterval[213 , P , a natu-ral question is to ask when such intervals [ σ, τ ] are disconnected, preferably givingthe answer in terms of simple conditions on σ and τ . We answer this question inTheorem 3.2 below. Determining when an interval contains a non-trivial discon-nected subinterval is a more difficult task, which we explore following Theorem 3.2.Interestingly, as we show in Theorem 3.6, almost all intervals in P do contain sucha disconnected subinterval. HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 7
Characterization of disconnected intervals.
It will be helpful to deal withposets of rank 2 separately because we can completely classify them, and becausethey sometimes require special treatment. In particular, unlike disconnected inter-vals of rank at least 3, a disconnected interval of rank 2 is shellable; we will studythe topic of shellability in detail in the next section. Since every element of P coversat most 2 elements, a rank-2 interval in P is either a chain or has two elements ofrank 1. Thus the structure of rank-2 intervals is completely determined by Propo-sition 2.3. With rank-2 disconnected intervals now classified, we will follow [MS15]in saying that a disconnected interval is non-trivial if it has rank at least 3. Definition 3.1.
For σ < τ , we say that σ straddles τ if σ is both a prefix and asuffix of τ and has no other occurrences in τ .It is easy to check that σ straddles τ if and only if σ is the carrier element of[ σ, τ ], as defined in the last paragraph of Subsection 2.2. Either way we state it,this is exactly the condition that causes non-trivial disconnectedness of [ σ, τ ], as wenow show. Theorem 3.2.
For σ, τ ∈ P with | τ | − | σ | ≥ , we have that [ σ, τ ] is disconnectedif and only if σ straddles τ .Proof. Let n = | τ | . If σ straddles τ , then the open interval ( σ, τ ) consists of twochains: one containing the elements τ [1 ,j ] for | σ | < j < n , and one containingelements τ [ n +1 − j,n ] for | σ | < j < n . These chains are disjoint since σ appears asa prefix in the elements of the former chain, and as a suffix in the elements of thelatter chain. Thus ( σ, τ ) is disconnected, as required. This argument is very similarto that of [BFS11, Lemma 3.3], where their conclusion is that µ ( σ, τ ) = 1.To prove the converse, suppose now that ( σ, τ ) is disconnected. In P , the per-mutation τ covers at most two elements, namely τ [1 ,n − and τ [2 ,n ] . If these twoelements are equal or if one of them avoids σ , then ( σ, τ ) contains a unique elementof length n −
1, which contradicts the fact that ( σ, τ ) is disconnected. Thus, wehave σ < τ [1 ,n − , σ < τ [2 ,n ] , and τ [1 ,n − = τ [2 ,n ] .In P , the permutation τ [2 ,n − is covered by both τ [1 ,n − and τ [2 ,n ] . Note that τ [2 ,n − = σ , because | τ [2 ,n − | = n − > | σ | . If τ [2 ,n − ∈ ( σ, τ ), then the inter-val ( σ, τ ) would be connected, because from every element there would be a pathin the Hasse diagram to either τ [1 ,n − or τ [2 ,n ] , and thus to τ [2 ,n − . Therefore, τ [2 ,n − / ∈ ( σ, τ ), which implies that τ [2 ,n − avoids σ . Together with the fact that σ is contained in both τ [1 ,n − and τ [2 ,n ] , it follows that σ straddles τ . (cid:3) As we will observe in the next section, an interval is non-shellable if it containsa non-trivial disconnected subinterval. Therefore the following direct consequenceof Theorem 3.2 is important to the classification of shellable intervals.
Corollary 3.3.
An interval [ σ, τ ] contains a non-trivial disconnected subintervalif and only if there exists π ∈ [ σ, τ ] such that there are two adjacent occurrencesof π in τ that are offset from each other by at least 3 positions. Specifically, if τ i . . . τ j is the minimal consecutive subsequence of τ containing these two adjacentoccurrences, then [ π, τ [ i,j ] ] is disconnected. For example, [1 , π = 21in 2143576 start at positions 1, 3 and 6, and we can check that [ π, τ [3 , ] = [21 , SERGI ELIZALDE AND PETER R. W. MCNAMARA
In the rest of the paper, we will use the term disconnected subintervals to meannon-trivial disconnected subintervals.3.2.
Finding disconnected intervals.
Finding a disconnected subinterval usingCorollary 3.3 may require checking all possible π ∈ [ σ, τ ]. Our next result impliesthat, in some cases, it is sufficient to search in the interval [ x ( τ ) , τ ], which maycontain significantly fewer elements than [ σ, τ ] (as in Example 3.5 below). Proposition 3.4.
Suppose that | x ( τ ) | 6 = 2 . If [1 , τ ] contains a disconnected subin-terval, then so does [ x ( τ ) , τ ] . Note that if [ σ, τ ] contains a disconnected subinterval, then so does [1 , τ ], and wecan apply the above proposition. When σ ≤ x ( τ ), this result nicely complementsTheorem 2.1, which says that computing µ ( σ, τ ) often boils down to computing µ ( σ, x ( τ )). So, when determining the M¨obius function, the interesting part of theinterval is often the part below x ( τ ), but when looking for disconnected subintervals,the interesting part is often above x ( τ ). Proof.
The statement is trivially true if x ( τ ) = 1, so let us assume that | x ( τ ) | ≥ n = | τ | and k = | x ( τ ) | .Suppose that [ x ( τ ) , τ ] contains no disconnected subintervals. Then, by Corol-lary 3.3, any two adjacent occurrences of x ( τ ) in τ must be offset by one or twopositions. If x ( τ ) were monotone, then this would force τ to be monotone as well,but we know that in this case [1 , τ ] is a chain, which contains no disconnectedsubintervals.Thus, we will assume that x ( τ ) is not monotone. By Lemma 2.2, no two oc-currences of x ( τ ) in τ can be offset by one, so all adjacent occurrences of x ( τ ) in τ must be offset by two positions. Equivalently, τ [ j,j + k − = x ( τ ) for every odd j with 1 ≤ j ≤ n − k + 1, and note that n and k must have the same parity. Thisimplies that for every i , the order relationship between τ i and τ i +1 is the same asbetween τ i +2 and τ i +3 . By considering the complement of τ if necessary, we willassume without loss of generality that τ < τ , and thus τ i < τ i +1 for every odd i .Since τ is not monotone, we must then have τ i > τ i +1 for every even i , and so τ isan up-down permutation, that is, τ < τ > τ < τ > . . . .By Corollary 3.3, the only way for [1 , τ ] but not [ x ( τ ) , τ ] to contain a disconnectedsubinterval would be if there exists a permutation π with x ( τ ) (cid:2) π ≤ τ which hastwo adjacent occurrences in τ offset by 3 or more. Note that we must also have π (cid:2) x ( τ ), since otherwise the offsets between adjacent occurrences of π would betwo. Thus, since π is incomparable with τ [ j,j + k − for every odd j , we must havethat | π | = k and any occurrence of π in τ is of the form τ [ i,i + k − for some even i . Let a < b be even positions so that τ [ a,a + k − and τ [ b,b + k − are two adjacentoccurrences of π offset by 3 or more, that is, π straddles τ [ a,b + k − .Suppose first that τ < τ . Since k ≥
3, the fact that the patterns τ [ j,j + k − areequal for every odd j implies that τ < τ < τ < τ < . . . , that is, the valleysof τ are increasing. Let σ = 1 ⊕ π . Since τ a − and τ b − are valleys, we have that τ [ a − ,a + k − = τ [ b − ,b + k − = σ . It follows that σ straddles τ [ a − ,b + k − , since thislatter permutation cannot have a third occurrence of σ without τ [ a,b + k − havinga third occurrence of π . Additionally, x ( τ ) = τ [ a − ,a + k − ≤ σ , since a − σ, τ [ a − ,b + k − ] is a disconnected subinterval contained in [ x ( τ ) , τ ],contradicting our original assumption. HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 9
Now consider the case τ > τ . We now have that the valleys of τ are decreasing.If k is odd, then so is n , and we can instead consider the reversal of τ , whose firstthree entries form the pattern 132, and apply the above argument to it. If k is even,then k ≥
4, and so the order relationship between τ and τ determines whether thepeaks of τ are increasing or decreasing. If they are increasing, then τ is the uniqueup-down permutation of length n with decreasing valleys and increasing peaks, and[1 , τ ] does not have disconnected subintervals since any adjacent occurrences of apattern π will be offset by at most two positions. (That case will be discussed indetail below when studying the permutations with | x ( τ ) | = | τ | − τ are decreasing, then the pattern τ [ i,i + k − is the same for every even i : it startswith the largest entry k , followed by the pattern determined by the first k − x ( τ ). It follows that adjacent occurrences of π are offset by 2 and not by3, which is a contradiction. (cid:3) Example 3.5.
Consider the interval [ σ, τ ] = [1 , x ( τ ) =2413, so by Proposition 3.4, we only need to look at [2413 , σ, τ ]. Using Corollary 3.3 and its notation, the onlynon-trivial candidates for π are 2413, 35241 and 52413. In each case, the adjacentoccurrences of π in τ are offset by just two positions, so [ σ, τ ] has no disconnectedsubintervals. To use Corollary 3.3 without applying Proposition 3.4, we would havehad eight non-trivial candidates for π to check.The condition | x ( τ ) | 6 = 2 is used in the proof of Proposition 3.4 to argue that alladjacent occurrences of x ( τ ) are offset by two positions. To see an example of howthe statement may fail when | x ( τ ) | = 2, take τ = 1325746. Its exterior is x ( τ ) = 12,but the only disconnected subinterval in [1 , τ ] is [21 , x ( τ ) , τ ].An immediate consequence of Proposition 3.4 is that [1 , τ ] has no (non-trivial)disconnected subintervals if | τ | − | x ( τ ) | ≤
2. In fact, we can precisely describe thestructure of any interval [ σ, τ ] where τ has this property, as follows.When | τ | − | x ( τ ) | = 1, Lemma 2.2 implies that τ is monotone, and in this case[ σ, τ ] is always a chain.When | τ | − | x ( τ ) | = 2, we have that x ( τ ) = τ [1 ,n − = τ [3 ,n ] , where n = | τ | . Itfollows that τ [ i,j ] = τ [ i +2 ,j +2] for every 1 ≤ i ≤ j ≤ n − . (3.1)Thus, for every 2 ≤ k ≤ n −
1, there are exactly two patterns of length k contained in τ , namely τ [1 ,k ] and τ [2 ,k +1] . These patterns are different from each other, becauseotherwise τ would be monotone, implying that | τ | − | x ( τ ) | = 1. For any σ ≤ τ , theinterval [ σ, τ ] has two elements of each length k with | σ | < k < | τ | . Additionally,any two elements in [ σ, τ ] of different lengths are comparable. The Hasse diagramof such an interval is drawn in Figure 3.1. Note that the two elements of length n − x ( τ ) and i ( τ ).We conclude this section with a result that states that, in a certain precise sense,almost all intervals in P contain disconnected subintervals. The approach belowfollows that in [MS15], where the analogous result is shown for the classical patternposet. Theorem 3.6.
Given a permutation σ , let P n ([ σ, τ ] contains a disconnected subinterval ) σx ( τ ) i ( τ ) τ Figure 3.1.
The interval [ σ, τ ] in the case | τ | − | x ( τ ) | = 2. denote the probability that [ σ, τ ] contains a non-trivial disconnected subinterval,where τ is chosen uniformly at random from S n . Then lim n →∞ P n ([ σ, τ ] contains a disconnected subinterval ) = 1 . Before proving Theorem 3.6, let us give a quick preliminary result which willalso be useful in Subsection 6.3.
Lemma 3.7.
Given a permutation σ , let P n ( σ ≤ τ ) denote the probability that σ ≤ τ , where τ is chosen uniformly at random from S n . Then lim n →∞ P n ( σ ≤ τ ) = 1 . Proof.
Let k = | σ | . Let us break the first k ⌊ n/k ⌋ entries of τ into ⌊ n/k ⌋ disjointblocks of length k , starting at the beginning. The probability that none of theseblocks is an occurrence of σ is (cid:18) − k ! (cid:19) ⌊ n/k ⌋ . Since k is fixed, this probability clearly goes to 0 as n → ∞ , implying the desiredresult. (cid:3) Proof of Theorem 3.6 . Letting k = | σ | , we can assume that k ≥
3, since estab-lishing the result for such permutations will also clearly prove it for all σ . Assumefirst that σ > σ k . In this case, we see that σ straddles σ ⊕ σ and so, by Theo-rem 3.2, [ σ, σ ⊕ σ ] is disconnected. Applying Lemma 3.7 tells us that τ contains σ ⊕ σ almost surely as n → ∞ . We conclude that as n → ∞ , the interval [ σ, τ ] containsthe disconnected interval [ σ, σ ⊕ σ ] with probability approaching 1, as required.If σ < σ k , then σ straddles σ ⊖ σ and we proceed similarly. (cid:3) In contrast with Theorem 3.6, we note that there are infinite classes of intervalsthat do not contain disconnected subintervals. For a simple example, any time σ satisfies σ > σ | σ | , we can take monotone increasing sequences α and β and let τ = α ⊕ σ ⊕ β . Then any element of [ σ, τ ] will occur only once in τ , so there areno disconnected subintervals. For examples of intervals [ σ, τ ] that are not just a HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 11 product of two chains but contain no disconnected subintervals, let σ = 21 and τ = α ⊕ ⊕ ⊕ · · · ⊕ ⊕ β , where the number of copies of 21 is at least 2 andwhere α and β are monotone increasing as before.4. Shellability
Our goal for this section is to prove that all intervals in P are shellable, exceptfor those which are not shellable for a straightforward reason. We begin by givingthe necessary background on shellability. We refer the interested reader to [Wac07]for further details and a wealth of other information about poset topology.4.1. Background on shellability and CL-shellability.
When considering acombinatorially defined simplicial complex, such as the order complex ∆( σ, τ ) of aninterval [ σ, τ ] of P , it is common to ask if the simplicial complex is shellable. Sinceour intervals [ σ, τ ] are graded, we can restrict our discussion to order complexesthat are pure, i.e., all the facets have the same dimension. A shelling of a pure d -dimensional simplicial complex ∆ is a linear ordering F , F , . . . , F s of its facetssuch that the intersection k − [ j =1 F j ∩ F k (4.1)is pure and ( d − k = 2 , , . . . , s . A simplicial complex is shellable if it has a shelling. Figure 2.1 is an example of a non-shellable complexsince in any ordering of the facets, the point 213 will be a ( d − σ, τ ] is shellable if ∆( σ, τ ) is.Much of the the interest in shellability arises from the fact that if an interval[ σ, τ ] is shellable, then this tells us that the homotopy type of ∆( σ, τ ) is a wedgeof spheres of the top dimension. The number of spheres is | µ ( σ, τ ) | which, in thecase of intervals in P , can be determined from Theorem 2.1. In the shellable case,we deduce that either the number of spheres is one or ∆( σ, τ ) is contractible.These topological considerations are also why we take the permutation 1 as thebottom element of P , rather than the empty permutation ∅ : since ∅ is only coveredby the permutation 1, any complex of the form ∆( ∅ , π ) will be contractible.We will prove shellability using Bj¨orner and Wachs’s theory of CL-shellability,which is a generalization of Bj¨orner’s theory of EL-shellability [Bj¨o80]. The originaldefinition of CL-labeling was given in [BW82] under the name “L-labeling”; the nowcustomary definition below appears in [BW83]. We will partially follow [Wac07] inthe exposition here.A poset Q is said to be bounded if it has a unique minimal and a unique maximalelement, which we denote ˆ0 and ˆ1 respectively. When x is covered by y , we willwrite the corresponding edge as x ← y ; the head of the arrow reflects the < symbol,and this notation will seem natural later when we primarily follow chains from topto bottom. Rather than just labeling edges x ← y as is the case for EL-labelings, weconsider the set ME ( Q ) of pairs ( C, x ← y ) consisting of a maximal chain C and anedge along that chain. A chain-edge labeling of Q is a map λ : ME ( Q ) → Λ, whereΛ is some poset, satisfying the following condition: if two maximal chains coincide Actually, it remains open whether there exist posets that are CL-shellable but not EL-shellable. The point of CL-shellability is that it is more flexible than EL-shellability, makingCL-shellings sometimes easier to find. ab cd ef
12 2 23 1321
Figure 4.1.
A chain-edge labeling that is a CL-labeling. Noticethat the edge d ← f receives two possible labels: the label 2 as anelement of the chain a ← b ← d ← f , and the label 1 as an elementof the chain a ← c ← d ← f . The bottom-rooted interval [ b, f ] a ← b has a unique increasing chain, namely b ← d ← f which has labelsequence (2 , k edges, then their labels also coincide along these edges. Thekey point is that the label on an edge depends on the maximal chain along whichwe arrive at that edge. See Figure 4.1 for an example.It follows that to restrict a chain-edge labeling of Q to an interval [ u, v ] in Q , weneed to record how we arrived at the bottom element u of the interval. Therefore,we introduce the idea of a bottom-rooted interval [ u, v ] r , which is the interval [ u, v ]together with a maximal chain r of [ˆ0 , u ]. This rooting allows the restriction of λ to an interval [ u, v ] to be consistent with λ defined over all of Q . More precisely,for an edge x ← y in [ u, v ] r the label received by x ← y when considered as beingalong the maximal chain C of [ u, v ] r is equal to the label received by x ← y whenconsidered as begin along the maximal chain r ∪ C ∪ s of Q , where s is any maximalchain of [ v, ˆ1].When we say that a maximal chain M of [ u, v ] r is weakly increasing or lexico-graphically precedes another, we are referring to the sequence of labels along M aswe read from u up to v . Definition 4.1.
Let Q be a bounded poset. A chain-edge labeling λ is a CL-labeling (chain-lexicographic labeling) if in each bottom-rooted interval [ u, v ] r of Q ,there is a unique weakly increasing maximal chain, and this chain lexicographicallyprecedes all other maximal chains in [ u, v ] r . A poset which has a CL-labeling issaid to be CL-shellable .An example of a CL-labeling is shown in Figure 4.1. As the terminology suggests,a bounded poset that is CL-shellable is shellable [BW82, Theorem 3.3] and, alongwith EL-shellability, exhibiting CL-shellability is one of the most common ways toprove shellability of order complexes.4.2.
Shellability in the consecutive pattern poset.
It is easy to see that dis-connected intervals of rank at least 3 are not shellable. Moreover, a result of Bj¨orner[Bj¨o80] states that if a poset is shellable, then all of its subintervals are shellable. Inparticular, if an interval [ σ, τ ] contains a disconnected subinterval of rank at least
HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 13
3, then it is certainly not shellable. As an example, the disconnected subinterval[213 , P arenot shellable, in a particular precise sense. Corollary 4.2.
Given a permutation σ , let τ be chosen uniformly at random from S n . Then the probability that [ σ, τ ] is shellable tends to 0 as n tends to infinity. Our main result of this section is that the converse of Bj¨orner’s result is truein P : if an interval [ σ, τ ] does not contain a disconnected subinterval, then it isshellable. So, roughly speaking, all intervals in the consecutive pattern poset thathave any hope of being shellable are in fact shellable. Recall that we have alreadyclassified those intervals that contain disconnected subintervals in Corollary 3.3 interms of the entries of σ and τ , so the theorem we are about to state classifies those σ and τ that yield a shellable interval [ σ, τ ]. Theorem 4.3.
The interval [ σ, τ ] in P is shellable if and only if it contains nonon-trivial disconnected subintervals. It remains to prove the “if” direction, which we will do using CL-shellability.Because we will read the labels on our maximal chains from top to bottom, wewill actually show that the dual of [ σ, τ ] is CL-shellable and hence shellable; thisimplies the shellability of [ σ, τ ] since the order complex of [ σ, τ ] is clearly isomorphicto that of its dual. The analogue of Theorem 4.3 in the classical pattern poset isactually false ([123 , π ′ has length ℓ and covers π along some maximal chain C of [ σ, τ ] and π = π ′ [2 ,ℓ ] , then we label the edge π ′ → π by 0. Otherwise, we assignthe label 1. Referring to Lemma 2.2, note that we assign the label 0 whenever π ′ ismonotone. However, the resulting labeling, which is actually an edge-labeling (i.e.,each edge label is independent of the chain C on which the edge appears), is nota dual CL-labeling. For example, the disconnected rank-2 intervals [21 , , ε > ε ≤ / ( | τ | − | σ | ) will suffice. For every maximal chain C , work along C from top to bottom, pausing at any triple π ′′ → π ′ → π suchthat π straddles π ′′ , as in the examples of the previous paragraph. We see that thissituation of straddling for rank-2 intervals is exactly the condition that results inchains with labels (0,0) and (1,1). Do nothing to the two labels that are already 0.For the other chain, letting λ C ( π ′ , π ) denote the label of the edge π ′ → π along C ,decrease the label on π ′ → π so that λ C ( π ′ , π ) = λ C ( π ′′ , π ′ ) − ε . After completionof these modifications, every label will either be 0 or 1 − kε for some k ≥
0. SeeFigure 4.2 for an example. The edge 213 →
21 there shows that this labeling is achain-edge labeling rather than just an edge labeling.
21 213132 213413242143 1324521435 214356 ∗∗∗ − ε − ε − ε Figure 4.2.
The interval [21 , , → and [21 , → → have had their labels modified inthe particular way we describe. The edge labeled 1 ∗ takes the label1 − ε along the chain that arrives via 2143 and otherwise takesthe label 1. All other labels are independent of the chain on whichthey appear. Proof of Theorem 4.3 . As already mentioned in the first paragraph of this sub-section, the interval [ σ, τ ] is not shellable if it contains a non-trivial disconnectedsubinterval. For the converse, we will prove dual CL-shellability. So suppose [ σ, τ ]contains no non-trivial disconnected subintervals, and let [ α, β ] r be a top-rootedinterval in [ σ, τ ], the obvious analogue of bottom-rooted interval. Using the la-beling λ C defined above and reading labels along chains from top to bottom, wewish to show that there is a unique increasing maximal chain in [ α, β ] r and thatthis increasing chain has the lexicographically first labels of all maximal chains in[ α, β ] r .Every maximal chain D from β to α can be said to “end” at a particular occur-rence of α in β as follows. Each edge of D corresponds to deleting an entry fromthe left or from the right of β , with the convention that the outgoing edge of amonotone permutation corresponds to deleting from the left (this is consistent withthe label 0 assigned in our definition of λ C ); the entries not deleted from β aftertraversing D give an occurrence of α .Define a particular maximal chain D of [ α, β ] r in the following way: starting at β , delete the leftmost entry if the result will still be in [ α, β ] r , and otherwise deletethe rightmost entry. In other words, D will delete from the left of β until it reachesthe rightmost occurrence of α in β , at which point it will switch to deleting fromthe right. None of the resulting labels will be of the form 1 − kε for k > D will never delete from the right when there is also the option of deleting fromthe left; see also the technical note in the next paragraph. Thus the sequence oflabels will take the form 0 , , . . . , , , , . . . , D , these labels are the lexicographically least of anymaximal chain. HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 15
It is necessary to make one technical note at this point. One might wonder aboutwhether the deletions along r could cause any of the labels along D to be of theform 1 − kε with k >
0. Let us show that this never happens. We see that theonly possibility is that the first label within [ α, β ] r along D is of this form. Bydefinition of D , this implies that D always deletes from the right, and so α onlyoccurs in β as a prefix. Thus there is only one maximal chain D in [ α, β ] r and ithas label sequence (1 − kε, , , . . . ,
1) from top to bottom, which is increasing asrequired. While this shows that labels of the form 1 − kε with k > D do notcause any difficulty, it does not fulfill our promise to show that such labels neveroccur. So let β + be the element covering β on r , and let β − be the element β covers on D . If β → β − gets the label 1 − kε with k >
0, this means by definitionof our label modifications that β − = β +[1 ,ℓ − = β +[3 ,ℓ ] where ℓ = | β + | . Since α onlyoccurs in β and hence β − as a prefix, the element covered by β − along D is then β +[1 ,ℓ − = β +[3 ,ℓ − . But β +[3 ,ℓ − = β [3 ,ℓ − , contradicting the fact that α only occursin β as a prefix. We conclude that all of the labels along D are indeed 0 or 1.Now suppose [ α, β ] r contains another increasing maximal chain D ′ . Our ap-proach will be to consider the possible locations of the occurrence of α at which D ′ ends, and to show some contradiction in each case. If D ′ ends at the rightmostoccurrence β i . . . β j of α in β , then D ′ must equal D to be increasing, since thepath along D ′ must make all its left deletions before making its right deletions.Therefore, assume D ′ ends at an occurrence of α in β that is not rightmost, namely β i ′ . . . β j ′ . Since D ′ is increasing, its last deletion is on the right and thus receivesa label of the form 1 − kε for k ≥
0. This already yields a contradiction in thecase when σ = 1 since any permutation covering the permutation 1 is monotone.Therefore, assume that | σ | > D , choose D ′ so that i ′ is as large as possible. Thereare three cases to consider.First suppose i ′ = i − β [ i ′ ,j ′ +1] is monotone.Since the last deletion along D ′ is on the right, the element covering α on D ′ is thismonotone permutation β [ i ′ ,j ′ +1] . But our convention for monotone permutationsthen implies that the last edge in D ′ is labeled 0, which is a contradiction.Next suppose that i ′ = i −
2. Thus β [ i − ,j − = β [ i,j ] = α . If it is also the casethat β [ i − ,j − = α , then applying Lemma 2.2 three times tells us that β [ i − ,j ] ismonotone. Since the last deletion of D ′ is on the right, the element covering α on D ′ will be β [ i − ,j − , obtaining a contradiction as in the previous paragraph. Thus β [ i − ,j − = α , and so α straddles β [ i − ,j ] . Since D ′ ends at β [ i − ,j − , the last twoedges on D ′ must receive the label 1 − kε followed by the label 1 − ( k + 1) ε for some k ≥
0. This contradicts D ′ having increasing labels.For the third and final case, let i ′ < i −
2. Suppose first that, except for β i . . . β j ,the rightmost occurrence of α in β is β i ′ . . . β j ′ . Then consider the subinterval[ α, β [ i ′ ,j ] ] of [ α, β ] r . Since α straddles β [ i ′ ,j ] , Theorem 3.2 implies that [ α, β ] r hasa non-trivial disconnected subinterval, contradicting our hypothesis. Thus theremust be some other occurrence β i ′′ . . . β j ′′ of α with i ′ < i ′′ < i and, among all suchoccurrences, let us choose i ′′ to be as close to i ′ as possible. If i ′′ = i ′ + 1, then β [ i ′ ,j ′ +1] is monotone, yielding a contradiction as before. If i ′′ = i ′ + 2, then we canapply the argument of the previous paragraph with i ′′ and j ′′ in place of i and j to yield a contradiction. Finally, suppose i ′′ > i ′ + 2, and consider the subinterval[ α, β [ i ′ ,j ′′ ] ] of [ α, β ] r . By our choice of i ′′ , we see that α straddles β [ i ′ ,j ′′ ] , yielding by Theorem 3.2 a non-trivial disconnected subinterval of [ α, β ] r , contradicting ourhypothesis. (cid:3) We conclude this section with a comparison of Theorem 4.3 to related results inthe literature.
Remark 4.4.
It follows from Theorem 4.3 that the homotopy type of an interval[ σ, τ ] without non-trivial disconnected subintervals is a wedge of spheres. As wementioned in Subsection 4.1, the number of spheres is | µ ( σ, τ ) | so, by Theorem 2.1,we conclude that ∆( σ, τ ) is homotopic to a single sphere or is contractible. Thissame conclusion is given as [SW12, Theorem 2.8] but they do not show shellabil-ity and instead use discrete Morse theory. In fact, their conclusion applies to allintervals [ σ, τ ] in P , even those containing disconnected subintervals. An exampleof this more general setting is the 2-dimensional simplicial complex in Figure 2.1,which contracts to a sphere of dimension 1.It also follows from Theorem 4.3 that the single spheres that arise must be of thesame dimension as ∆( σ, τ ), but this is no longer true in the non-shellable cases, asshown by the aforementioned example in Figure 2.1. In summary, Theorem 4.3 hasa stronger hypothesis but also a stronger conclusion than [SW12, Theorem 2.8]. Remark 4.5.
Billera and Myers [BM99] have shown a general result of a similarflavor to Theorem 4.3: a bounded graded poset is shellable if it is (2 + 2)-free,meaning that it contains no induced subposet that is isomorphic to a disjoint sumof two chains of length 2. In [Wac99], Wachs shows that such (2+2)-free posets areCL-shellable. We note, however, that these results are not enough to imply The-orem 4.3, since there exist intervals in P that contain no non-trivial disconnectedsubintervals but fail to be (2+2)-free. One example is in Figure 4.2, where the twochains 2143 < < All intervals are rank-unimodal and strongly Sperner
Since intervals in P are graded, it is natural to ask about the structure of theseintervals with regard to the number of elements at each rank. It is clear fromFigure 1.1 that the intervals are not rank-symmetric in general. It is also the casethat the sequence of rank sizes is not log-concave in general. For example, theinterval [1 , P have: they are rank-unimodal and strongly Sperner. Our goal in this section isto prove this assertion. We note that neither result is known for the classical patternposet [MS15]. Even though the definitions of rank-unimodal and strongly Spernerare quite different, the two results appear together because they both rely heavilyon the same injection between rank levels (Lemma 5.1), and because Theorem 5.5below of Griggs connects the two properties.5.1. Rank-unimodality.
For a finite graded poset Q , the set of elements of rank i will be called a rank level of Q , and the cardinality a i of this rank level will bereferred to as the size of rank i . Recall that a finite graded poset of rank N is called rank-unimodal if the sequence a , a , . . . , a N of rank sizes is unimodal.Before proving rank-unimodality for intervals in P , it will be helpful and infor-mative to examine an explicit 3-step method for constructing such intervals, upto isomorphism. Fix σ ≤ τ , and let n = | τ | . Consider the poset P of intervals HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 17 [ i, j ] = { i, i + 1 , . . . , j } with 1 ≤ i ≤ j ≤ n , ordered by set inclusion. Each interval[ i, j ] is associated with a subsequence τ i . . . τ j of τ . The poset P is a graded join-semilattice of rank n −
1, having n − r elements or rank r for all r , namely [ i, i + r ]for 1 ≤ i ≤ n − r . See Figure 5.1 for an example.[1,1] [2,2] [3,3] [4,4] [5,5] [6,6][1,2] [2,3] [3,4] [4,5] [5,6][1,3] [2,4] [3,5] [4,6][1,4] [2,5] [3,6][1,5] [2,6][1,6] Figure 5.1.
For [ σ, τ ] = [12 , P , with P given by the elements outlined with bold lines. Fig-ure 5.2 shows P . In this example, the breaking rank is b = 1because [4 , ∼ [1 , P be the poset obtained from P by deleting all theelements [ i, j ] such that τ [ i,j ] avoids σ . For the final step, let P be the quotientposet P / ∼ , where ∼ is the equivalence relation defined by [ i, j ] ∼ [ i ′ , j ′ ] whenever τ [ i,j ] = τ [ i ′ ,j ′ ] . For equivalence classes in the quotient poset, denoted [ i, j ], recallthat one defines [ i, j ] ≤ [ i ′ , j ′ ] if there exist [ i , j ] ∈ [ i, j ] and [ i , j ] ∈ [ i ′ , j ′ ] suchthat [ i , j ] ⊆ [ i , j ]. It is clear that the map [ i, j ] τ [ i,j ] is a poset isomorphismbetween P and the interval [ σ, τ ] in P ; see Figure 5.2 for an example. In whatfollows, we will work with P instead of [ σ, τ ] when helpful.From this construction of [ σ, τ ], we can make our first observation of this sectionabout the structure of [ σ, τ ]: the upper ranks of [ σ, τ ] are “grid-like.” More precisely,let N = | τ | − | σ | denote the rank of [ σ, τ ] (or of P ). By comparing the structure of P and P , we know there are at most N + 1 − r elements of rank r in [ σ, τ ]. Let b denote the largest value of r such that the number of elements of rank r is strictlyless than N + 1 − r . We will call b the breaking rank of [ σ, τ ]. In the context of P , b is the rank of [ i, i + j ] where j is the largest value such that either an element of theform [ i, i + j ] was deleted when going from P to P , or there is an equivalence classof the form [ i, i + j ] in P containing more than one element of P . For example,for [12 , b = 1 since [4 , ∼ [1 , P and P that happen at the breaking rank There is more than one notion of quotient poset in the literature. Our construction followsHallam and Sagan [HS15]: given an equivalence relation ∼ on a poset P , the quotient P/ ∼ is theset of equivalence classes, with X ≤ Y in P/ ∼ if and only if x ≤ y in P for some x ∈ X and some y ∈ Y . Hallam and Sagan show that P/ ∼ is a poset if whenever X ≤ Y in P/ ∼ , we have thatfor all y ∈ Y there exists x ∈ X such that x ≤ y in P . We see that P / ∼ satisfies this conditionand so P is indeed a poset. ↔ [2 , ↔ [2 ,
4] 213 ↔ [1 ,
3] 132 ↔ [3 , ↔ [1 ,
4] 1243 ↔ [2 ,
5] 1324 ↔ [3 , ↔ [1 ,
5] 12435 ↔ [2 , ↔ [1 , Figure 5.2.
The interval [12 , P has already appearedin Figure 1.1, but here it is adorned with the corresponding ele-ments of P . The bold edges show two examples of the injectionof Lemma 5.1. For the injection from m = 2 elements of E into E , we chose k = 2 in the notation of the proof of Lemma 5.1. Forthe injection from m = 2 elements of E into E , our only validoption is k = 3.trickle down P , meaning that if r is a rank less than the breaking rank then therewill again be strictly less than the maximum N + 1 − r elements of rank r .By the “grid-like” structure of the upper ranks of [ σ, τ ] we mean that, by defi-nition of the breaking rank b , the rank levels of rank r with b < r ≤ N of [ σ, τ ] areessentially the same as the top N − b rank levels of P . For example, in Figure 5.1,the grid-like form of the top three rank levels gives rise to the grid-like form of thetop three rank levels in Figure 5.2. This observation about the upper ranks directlyrelates to the rank-unimodality of [ σ, τ ] since it implies a b +1 > a b +2 > · · · > a N , (5.1)which will be the basis of the “downward portion” of the unimodality.The following lemma will be used in both the proof of rank-unimodality and ofthe strongly Sperner property, and allows us to pair up elements of adjacent ranks. Lemma 5.1.
In the interval [ σ, τ ] in P , let E r denote the set of elements of rank r for ≤ r ≤ N , where N = | τ | − | σ | . For ≤ r ≤ N − , pick any m r ≤ min {| E r | , N − r } . Then for any selection of m r elements of E r , there exists aninjection f r from these elements into E r +1 with the property that π < f r ( π ) foreach such element π .Proof. We will work in the setting of P as defined above, and the result will followby the isomorphism with [ σ, τ ]. Fix m r elements of rank r . These elements ofrank r in P take the form [ i, i + s ], where s = | σ | − r . For each such element,choose the representative [ i, i + s ] with smallest i . Let I s ⊂ { , , . . . , n − s } be theset of left endpoints of these m r representatives, where n = | τ | as before. Since m r ≤ N − r = n − s −
1, there is some k ∈ { , , . . . , n − s } \ I s . Fix such a k , and HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 19 define, for each i ∈ I s , f ( i ) = ( i if i < k,i − i > k. We claim that the map [ i, i + s ] [ f ( i ) , f ( i ) + s + 1], defined for i ∈ I s , is aninjection from elements of rank r to elements of rank r + 1 in P . See Figure 5.2for examples.To see that this map is well defined, note first that 1 ≤ f ( i ) ≤ n − s − i ∈ I s , and so τ [ f ( i ) ,f ( i )+ s +1] is defined. Additionally, both τ [ i,i + s +1] and τ [ i − ,i + s ] (when defined) contain τ [ i,i + s ] , which translates to the desired property that π Every interval [ σ, τ ] in P is rank-unimodal.Proof. Since we have already shown (5.1), it remains to show that a ≤ a ≤· · · ≤ a b ≤ a b +1 . To see this, fix r with 0 ≤ r ≤ b and consider the elements ofrank r in [ σ, τ ]. By definition of the breaking rank b , there are strictly less than N + 1 − r elements of rank r . The result now follows directly from the injection ofLemma 5.1. (cid:3) Remark 5.3. In the case of the classical pattern poset, it is an open questionwhether every interval is rank-unimodal [MS15]. It is known to be true for intervals[ σ, τ ] with | τ | ≤ Strongly Sperner property. This subsection is devoted to a proof that allintervals in P are strongly Sperner. Let us first give the necessary background.Recall that a graded poset Q is said to be Sperner if the largest rank size equalsthe size of the largest antichain. Since every rank level of Q consists of an antichain,the Sperner property is equivalent to the condition that some rank level of Q is anantichain of maximum size. For any positive integer k , a k -family of Q is a unionof k antichains. We are now ready for the definition of one of the well-knowngeneralizations of the Sperner property. Definition 5.4. Let Q be a poset of rank N . For an integer k with 1 ≤ k ≤ N + 1,we say that Q is k -Sperner if the sum of the sizes of the k largest ranks equals thesize of the largest k -family. In addition, Q is said to be strongly Sperner if it is k -Sperner for all k .Again, being k -Sperner is equivalent to having a set of k ranks that is a k -familyof maximum size. Our technique for showing that every interval of P is stronglySperner will make heavy use of the injections f r from Lemma 5.1 and of Theorem5.5 below, due to Griggs [Gri80], with shorter proofs appearing in [GSS80, Lec97]. Letting Q be a poset of rank N , for all 1 ≤ i ≤ N + 1, let us denote the i thlargest rank level by L i and its size by ℓ i , where we break ties in size arbitrarily.For example, for the interval [12 , L , both of which give ℓ = 3, while ℓ = ℓ = 1. We will say that Q is i -rankintersecting if there exist ℓ i disjoint chains which intersect each of the rank levels L , L , . . . , L i . In particular, note that these disjoint chains must cover L i . Inaddition, we say that Q is strongly rank intersecting if it is i -rank intersecting forall i with 1 ≤ i ≤ N + 1. Theorem 5.5 ([Gri80]) . If a poset Q is rank-unimodal, then Q is strongly Spernerif and only if it is strongly rank intersecting. We now have the tools necessary to prove the promised result. Theorem 5.6. Every interval [ σ, τ ] in P is strongly Sperner.Proof. By Theorem 5.5, it suffices to show that Q = [ σ, τ ] is strongly rank inter-secting, and throughout this proof we will follow the notation of the paragraphimmediately preceding Theorem 5.5. Thus, fixing i with 1 ≤ i ≤ N + 1, considerthe i largest rank levels L , L , . . . , L i . Since [ σ, τ ] is rank-unimodal and we canbreak ties among rank sizes arbitrarily, we can assume that L , L , . . . , L i form aset of consecutive rank levels in some order, with L i occurring at one of the twoextremes of the consecutive sequence. Let us suppose that these rank levels occupyranks r through r in [ σ, τ ].As we will see, the injections f r of Lemma 5.1 will be exactly what we need tocomplete the proof. Since we wish to construct ℓ i disjoint chains, we will be picking ℓ i elements of rank r for all r ≤ r < r . In order to apply the lemma, we need toshow that ℓ i ≤ N − r for all such r , meaning that we need ℓ i ≤ N − r + 1. Bydefinition of L i and hence ℓ i , we have ℓ i = min { a r , a r } , where a j again denotesthe number of elements of rank j . We already showed when considering the grid-likestructure of the upper levels of [ σ, τ ] that a r ≤ N +1 − r . Thus ℓ i ≤ a r ≤ N − r +1,as required.Since ℓ i = min { a r , a r } , we can start by picking ℓ i elements of rank r . Applying f r to these elements, we construct ℓ i disjoint chains of length 1 from rank r to rank r + 1. Continuing in this fashion, applying f r +1 , f r +2 , . . . , f r − , we construct ℓ i disjoint chains that intersect each of the rank levels r through r , as required. Seethe bold edges in Figure 5.2 for an example with i = 3. (cid:3) Remark 5.7. Generalizing the Sperner property in a different way, a poset is saidto be strictly Sperner if every antichain of maximum size is a rank level. To see thatthis property does not hold for intervals in P , consider the antichain { , } inthe interval [12 , σ occurs just once in τ ,we know from Proposition 2.4 that [ σ, τ ] is a lattice. Problem 5.8. Characterize those intervals [ σ, τ ] in P that are lattices. As usual,we would prefer our characterization to be in terms of simple conditions on σ and τ . For example, for | τ | = 4, [1 , τ ] is a lattice if and only if either τ [1 , or τ [2 , (orboth) is monotone, since otherwise both will be upper bounds for 12 and 21. HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 21 The exterior of a permutation The length of the exterior. As we have seen in the previous sections, par-ticularly in Theorem 2.1 and Proposition 3.4, the exterior of a permutation playsan important role in the study of P . From an enumerative perspective, a naturalproblem is to study the distribution of the statistic length of the exterior on per-mutations. Table 6.1 shows the number of permutations according to this statistic.The first column (after the one indexing n ) counts permutations whose exteriorhas length 1. Permutations with this property are usually called non-overlappingpermutations in the literature. It was shown by B´ona [B´on11] that the number ofnon-overlapping permutations of length n is approximately 0 . n !, but no exactformula is known. Next we show a simple divisibility property of these numbers. n \ k Table 6.1. The number of permutations τ ∈ S n with | x ( τ ) | = k . Lemma 6.1. For every n ≥ , the number of non-overlapping permutations in S n is divisible by 4.Proof. If τ ∈ S n is non-overlapping, then so are its reversal, its complement, and itsreverse-complement, and these four permutations are all different. Indeed, if twoof these were the same, then τ would be equal to its reverse-complement, whichwould imply that τ starts with a descent if and only if it ends with a descent, andso τ would be overlapping. Thus, we have partitioned the set of non-overlappingpermutations into disjoint sets of size 4. (cid:3) While the above property of the first column of Table 6.1 is straightforward toprove, there seem to be other less obvious congruence relations. For example, thedata in the second column suggests that the number of permutations in S n withexterior of length 2 might always be congruent to 2 modulo 4.The entries in the rightmost nonzero diagonal of Table 6.1 count permutations τ with | x ( τ ) | = | τ | − 1. It is clear that the number of such permutations in S n is2 for every n ≥ 2, since by Lemma 2.2, only the monotone permutations 12 . . . n and n . . . 21 satisfy this condition. The entries in the diagonal immediately belowthese 2s count permutations τ with | x ( τ ) | = | τ | − 2. In Subsection 3.2 we describedthe structure of the intervals under such permutations. Next we enumerate thesepermutations. Lemma 6.2. For every n ≥ , |{ τ ∈ S n : | x ( τ ) | = n − }| = 2 n + 2 . Proof. Since the length of the exterior of any permutation is the same for its com-plement, we can count permutations with τ < τ and multiply the final number by2. We have shown that every τ ∈ S n with | x ( τ ) | = n − τ i < τ i +1 for every odd i and, since τ is not monotone (otherwise | x ( τ ) | = n − τ i > τ i +1 for every even i , so τ is alternating.By Equation (3.1), the relative order of τ and τ determines whether the se-quence of valleys of τ is increasing or decreasing, and similarly the relative orderof τ and τ determines it for the sequence of peaks. When one of these sequencesis increasing and the other decreasing, which happens when τ [1 , equals 1423 or2314, then τ is completely determined.If both sequences are increasing, that is, τ [1 , = 1324, then τ is determinedonce we choose the value of τ relative to the valleys of τ , that is, for which index2 ≤ i ≤ ⌊ n +12 ⌋ we have τ i − < τ < τ i +1 (where we define τ n +1 = τ n +2 = ∞ forconvenience). This choice forces the relative order of all peaks with respect to thevalleys. There are ⌊ n − ⌋ possibilities for τ in this case.Similarly, if both sequences are decreasing, that is, τ [1 , equals 2413 or 3412,then τ is determined once we choose the value of τ relative to the peaks of τ ,that is, for which index 1 ≤ i ≤ ⌊ n ⌋ we have τ i > τ > τ i +2 (where we define τ n +1 = τ n +2 = −∞ for convenience). There are ⌊ n ⌋ possibilities for τ in this case,with just the case i = 1 corresponding to τ [1 , = 3412.In total, the number of τ ∈ S n with | x ( τ ) | = n − (cid:18) (cid:22) n − (cid:23) + j n k(cid:19) = 2 n + 2 . (cid:3) We do not have formulas for the other entries of Table 6.1. Problem 6.3. Find a formula for the entries of Table 6.1. Asymptotic behavior of the length of the exterior. Even without havingexact formulas, we can study the behavior of | x ( τ ) | as the length of τ goes to infinity.For given n , we will write P n and E n to denote the probability and the expectationof events involving τ ∈ S n chosen uniformly at random. In other words, τ isa random variable with uniform distribution over S n . We also use the notation g ( n ) = o ( f ( n )) or g ( n ) ≪ f ( n ) to mean that lim n →∞ g ( n ) /f ( n ) = 0.The main result in this subsection is a constant asymptotic upper bound on theexpected length of the exterior of a permutation. We show that, as the length of thepermutation tends to infinity, the limit of this expected value exists and is between e − e . We start with a lemma that bounds the probability that the length ofthe exterior is large. A similar result appears in [Per13, Lemma 18]. Lemma 6.4. For ≤ m ≤ n − , P n ( | x ( τ ) | ≥ m ) ≤ ⌊ n/ ⌋ X i = m i ! + o ( n − ) , where the summation is defined to be zero when m > ⌊ n/ ⌋ . HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 23 Proof. For 1 ≤ i ≤ n − 1, let B i be the event “ τ has a bifix of length i ,” that is, τ [1 ,i ] = τ [ n − i +1 ,n ] . By definition, | x ( τ ) | is the largest i such that B i holds, and P n ( | x ( τ ) | ≥ m ) = P n n − [ i = m B i ! ≤ n − X i = m P n ( B i ) . When i ≤ ⌊ n/ ⌋ , P n ( B i ) = 1 i ! , since among all the possible permutations of the last i entries, B i holds for exactlyone of them.When i > ⌊ n/ ⌋ , we have the upper bound P n ( B i ) ≤ n − i )! ⌊ nn − i ⌋− . (6.1)This is because if we break up τ into ⌊ nn − i ⌋ blocks of n − i entries starting at thebeginning: τ [1 ,n − i ] , τ [ n − i +1 , n − i )] , . . . , then, for B i to hold, the entries in all of theseblocks have to be in the same relative order. For ⌊ n/ ⌋ < i ≤ n − ⌊√ n ⌋ , we havethat n − i ≥ ⌊√ n ⌋ and ⌊ nn − i ⌋ ≥ 2, and so expression (6.1) gives P n ( B i ) ≤ ⌊√ n ⌋ ! . For n − ⌊√ n ⌋ < i ≤ n − 2, we have that n − i ≥ ⌊ nn − i ⌋ ≥ ⌊√ n ⌋ , and so P n ( B i ) ≤ ⌊√ n ⌋− . When i = n − 1, it is easy to see directly that P n ( B n − ) = 2 n ! , since by Lemma 2.2, B n − only holds for the two monotone permutations.Combining the above bounds, we get P n ( | x ( τ ) | ≥ m ) ≤ ⌊ n/ ⌋ X i = m i ! + n −⌊√ n ⌋ X i = ⌊ n/ ⌋ +1 ⌊√ n ⌋ ! + n − X i = n −⌊√ n ⌋ +1 ⌊√ n ⌋− + 2 n ! . Clearly, the terms other than the first summation approach 0 faster than n − k forany positive constant k , as n tends to infinity. (cid:3) Theorem 6.5. The limit of E n ( | x ( τ ) | ) as n goes to infinity exists and e − ≤ lim n →∞ E n ( | x ( τ ) | ) ≤ e. Proof. By definition of expectation, E n ( | x ( τ ) | ) = ⌊ n/ ⌋ X i =1 i P n ( | x ( τ ) | = i ) + n − X ⌊ n/ ⌋ +1 i P n ( | x ( τ ) | = i )= ⌊ n/ ⌋ X m =1 P n ( m ≤ | x ( τ ) | ≤ ⌊ n/ ⌋ ) + o (1) , (6.2)where we have bounded the second summation from above by n P n ( | x ( τ ) | ≥ ⌊ n/ ⌋ + 1) = o (1) , using Lemma 6.4.To prove that the limit exists, we will show that E n ( | x ( τ ) | ) behaves asymptoti-cally similarly to an increasing sequence, and that it is bounded from above. For n with 1 ≤ n < n , consider the bijection S n −→ S n × S n − n × (cid:18) [ n ] n (cid:19) τ ( σ , π , A )defined as follows. Let σ = τ [1 , ⌊ n / ⌋ ] ∪ [ n −⌈ n / ⌉ +1 ,n ] be the permutation obtained by concatenating the first ⌊ n / ⌋ and the last ⌈ n / ⌉ entries of τ , let A be the set of values of these entries before applying the reductionmap ρ , and let π = τ [ ⌊ n / ⌋ +1 ,n −⌈ n / ⌉ ] be the reduction of the remaining entries. If | x ( σ ) | ≤ ⌊ n / ⌋ , then clearly | x ( τ ) | ≥| x ( σ ) | , since the bifix of σ of length | x ( σ ) | is also a bifix of τ . This means that for anygiven m , the bijection associates to every triple ( σ, π, A ) with m ≤ | x ( σ ) | ≤ ⌊ n / ⌋ a permutation τ with | x ( τ ) | ≥ m . It follows that |{ σ ∈ S n : m ≤ | x ( σ ) | ≤ ⌊ n / ⌋}| ( n − n )! (cid:18) nn (cid:19) ≤ |{ τ ∈ S n : | x ( τ ) | ≥ m }| . Dividing both sides by n !, we get P n ( m ≤ | x ( τ ) | ≤ ⌊ n / ⌋ ) ≤ P n ( | x ( τ ) | ≥ m ) = P n ( m ≤ | x ( τ ) | ≤ ⌊ n/ ⌋ ) + o ( n − )by Lemma 6.4. Summing over m , we have ⌊ n / ⌋ X m =1 P n ( m ≤ | x ( τ ) | ≤ ⌊ n / ⌋ ) ≤ ⌊ n/ ⌋ X m =1 P n ( m ≤ | x ( τ ) | ≤ ⌊ n/ ⌋ ) + o (1) . Using now Equation (6.2), it follows that E n ( | x ( τ ) | ) ≤ E n ( | x ( τ ) | ) + o ′ (1) , (6.3)where here o ′ (1) is an expression that tends to 0 when both n and n go to infinity.To show that the sequence E n ( | x ( τ ) | ) is bounded from above by a constant, weagain use Lemma 6.4 to conclude that E n ( | x ( τ ) | ) = n − X m =1 P n ( | x ( τ ) | ≥ m ) ≤ ⌊ n/ ⌋ X m =1 ⌊ n/ ⌋ X i = m i ! + o (1)= ⌊ n/ ⌋ X i =1 ii ! + o (1) ≤ ∞ X i =1 i − o (1) = e + o (1) . The above inequality implies that L := lim sup n { E n ( | x ( τ ) | ) } exists, and that L ≤ e . Let us show that L is in fact the limit of the sequence. For any ǫ > n so that it satisfies the following two conditions. First, take n largeenough so that for n ≥ n , we have E n ( | x ( τ ) | ) ≤ E n ( | x ( τ ) | ) + ǫ/ n exists by Equation (6.3)). Secondly, pick n so that E n ( | x ( τ ) | ) > L − ǫ/ n exists by definition of L ). Then, for all n ≥ n , E n ( | x ( τ ) | ) ≥ E n ( | x ( τ ) | ) − ǫ > L − ǫ. HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 25 This proves that lim n →∞ E n ( | x ( τ ) | ) = L. To prove the lower bound, recall that B i is the event “ τ has a bifix of length i .”Clearly P n ( | x ( τ ) | ≥ m ) ≥ P n ( B m ), which equals m ! when 1 ≤ m ≤ ⌊ n/ ⌋ . Thus, E n ( | x ( τ ) | ) = n − X m =1 P n ( | x ( τ ) | ≥ m ) ≥ ⌊ n/ ⌋ X m =1 P n ( B m ) = ⌊ n/ ⌋ X m =1 m ! . The limit of the right-hand side as n → ∞ is e − (cid:3) Problem 6.6. Find the exact value of lim n →∞ E n ( | x ( τ ) | ) . When n = 10, the value of E n ( | x ( τ ) | ) is approximately 1.909. Some relativelycrude computations (for each n with 11 ≤ n ≤ Problem 6.7. Find the limiting distribution of | x ( τ ) | , that is, find lim n →∞ P n ( | x ( τ ) | = k ) for all k . It was shown by B´ona [B´on11] that when k = 1, the sequence P n ( | x ( τ ) | = 1) isdecreasing and so its limit exists, and that 0 . ≤ lim n →∞ P n ( | x ( τ ) | = 1) ≤ . k the sequences P n ( | x ( τ ) | = k ) are not monotonein general, it is possible to obtain recurrences similar to those in [B´on11] to showthat the corresponding limits exist for all k and are positive.6.3. Permutations with no carrier, and the M¨obius function of most in-tervals. Recall that τ is said to have a carrier element if x ( τ ) (cid:2) i ( τ ). Carrierelements, defined in [BFS11], play a crucial role in determining the M¨obius func-tion of intervals in P , as we pointed out in Subsection 2.2. A question that arisesis how many permutations have a carrier element. In this subsection we study thisquestion, and we use our findings to deduce that most intervals (in a precise sensethat we will describe) have a zero M¨obius function. Since this subsection involvesinequalities in both R and P , we will use the symbol ≤ P in the latter case as adistinguisher.For n from 2 to 10, the number of permutations τ ∈ S n with no carrier element,that is, satisfying x ( τ ) ≤ P i ( τ ), is given by the sequence0 , , , , , , , , , . . . . (6.4) Problem 6.8. Find a formula for the number of permutations with no carrierelement. We point out that the displayed terms of the sequence (6.4) are all divisible byfour, although we have not proved if this is the case for all terms. Note that permu-tations with n ≥ n ! suggests that, as n grows, mostpermutations in S n have no carrier element. Next we show that this is indeed thecase. Theorem 6.9. lim n →∞ P n ( x ( τ ) P i ( τ )) = 0 . Proof. Let n = | τ | and k = | x ( τ ) | . If k < n/ 2, let x ( τ ) = τ [ k +1 ,n − k ] be thepermutation obtained from τ by deleting the initial and final occurrences of x ( τ ).If k ≥ n/ 2, define x ( τ ) to be the empty permutation (and let us add the emptypermutation to P as its new bottom element for convenience). Since x ( τ ) ≤ P i ( τ ),we have P n ( x ( τ ) P i ( τ )) ≤ P n ( x ( τ ) P x ( τ )) . For every m , conditioning on whether | x ( τ ) | ≤ m or not, we can write P n ( x ( τ ) P x ( τ )) as a sum of two terms, the first being P n ( | x ( τ ) | ≤ m ) P n (cid:0) x ( τ ) P x ( τ ) (cid:12)(cid:12) | x ( τ ) | ≤ m (cid:1) ≤ P n ( | x ( τ ) | ≤ m ) P n (cid:0) τ [1 ,m ] P τ [ m +1 ,n − m ] (cid:12)(cid:12) | x ( τ ) | ≤ m (cid:1) ≤ P n (cid:0) τ [1 ,m ] P τ [ m +1 ,n − m ] (cid:1) , (6.5)and the second being P n ( | x ( τ ) | > m ) P n (cid:0) x ( τ ) P x ( τ ) (cid:12)(cid:12) | x ( τ ) | > m (cid:1) ≤ P n ( | x ( τ ) | > m ) ≤ ∞ X i = m +1 i ! + o ( n − ) , using Lemma 6.4.To bound (6.5), notice that τ [1 ,m ] and τ [ m +1 ,n − m ] are independent random per-mutations, and split τ [ m +1 ,n − m ] into ⌊ n − mm ⌋ disjoint blocks of size m (with pos-sibly some leftover entries that do not belong to any of the blocks). The proba-bility that each individual block is not an occurrence of τ [1 ,m ] is 1 − /m !. For τ [1 ,m ] P τ [ m +1 ,n − m ] to hold, none of the blocks can be an occurrence of τ [1 ,m ] .Since these are independent events, we have that P n (cid:0) τ [1 ,m ] P τ [ m +1 ,n − m ] (cid:1) ≤ (cid:18) − m ! (cid:19) ⌊ nm ⌋− . Combining all these bounds, we get P n ( x ( τ ) P i ( τ )) ≤ (cid:18) − m ! (cid:19) ⌊ nm ⌋− + ∞ X i = m +1 i ! + o ( n − ) (6.6)for every m .Finally, it is enough to choose m to be a slowly-growing function of n with m ( n ) → ∞ as n → ∞ , such as m = m ( n ) = ⌊ log log n ⌋ . For such choice of m ,writing (cid:18) − m ! (cid:19) nm = "(cid:18) − m ! (cid:19) m ! nmm ! , HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 27 the term inside the square brackets approaches e − as n → ∞ . Using Stirling’sformula, we see that(log log n )(log log n )! ≪ (log log n ) (log log n ) log log n e log log n (6.7) ≪ e (log log log n )(log log n ) ≪ e log n = n, and so the exponent nmm ! goes to infinity as n grows. It follows that (cid:18) − m ! (cid:19) ⌊ nm ⌋− ≤ (cid:18) − m ! (cid:19) nm − −→ n → ∞ . Clearly, the summation of 1 /i ! in (6.6) tends to 0 as m → ∞ . Thus,the whole right-hand side of (6.6) tends to 0 as n → ∞ , proving the statement. (cid:3) A consequence of Theorem 6.9 is that for any fixed σ , most intervals of the form[ σ, τ ] will have zero M¨obius function. Corollary 6.10. For fixed σ , let P σn ( µ ( σ, τ ) = 0) denote the probability that µ ( σ, τ )= 0 when τ is chosen uniformly at random among permutations in S n that con-tain σ . Then, for every σ , lim n →∞ P σn ( µ ( σ, τ ) = 0) = 1 . Proof. When | τ |−| σ | > 2, Theorem 2.1 states that µ ( σ, τ ) = 0 unless σ ≤ P x ( τ ) P i ( τ ). By Theorem 6.9, the probability that x ( τ ) P i ( τ ) when τ is chosen uniformlyat random from S n tends to 0 as n → ∞ . It follows that, for any given σ ,lim n →∞ P n ( µ ( σ, τ ) = 0) = 1 . To restrict to permutations τ containing σ , note that P σn ( µ ( σ, τ ) = 0) = P n (cid:0) µ ( σ, τ ) = 0 (cid:12)(cid:12) σ ≤ P τ (cid:1) ≤ P n ( µ ( σ, τ ) = 0) P n ( σ ≤ P τ ) . By Lemma 3.7, lim n →∞ P n ( σ ≤ P τ ) = 1, so lim n →∞ P σn ( µ ( σ, τ ) = 0) = 0 asclaimed. (cid:3) We point out that Bernini et al. [BFS11] give several conditions on σ and τ thatimply µ ( σ, τ ) = 0. For example, they show that this happens when the first twoentries of τ are not involved in any occurrence of σ . While the results in [BFS11]show that the probability that µ ( σ, τ ) = 0 is large, their conditions depend on σ , and thus they cannot be used to prove that for fixed σ this probability getsarbitrarily close to one as the length of τ increases, as we do in Corollary 6.10. Acknowledgement We thank anonymous referees for suggestions that improved the exposition. References [AEK08] Jos´e Mar´ıa Amig´o, Sergi Elizalde, and Matthew B. Kennel. Forbidden patterns and shiftsystems. J. Combin. Theory Ser. A , 115(3):485–504, 2008.[BFS11] Antonio Bernini, Luca Ferrari, and Einar Steingr´ımsson. The M¨obius function of theconsecutive pattern poset. Electron. J. Combin. , 18(1):Paper 146, 12 pp., 2011.[BJJS11] Alexander Burstein, V´ıt Jel´ınek, Eva Jel´ınkov´a, and Einar Steingr´ımsson. The M¨obiusfunction of separable and decomposable permutations. J. Combin. Theory Ser. A ,118(8):2346–2364, 2011. [Bj¨o80] Anders Bj¨orner. Shellable and Cohen-Macaulay partially ordered sets. Trans. Amer.Math. Soc. , 260(1):159–183, 1980.[BM99] Louis J. Billera and Amy N. Myers. Shellability of interval orders. Order , 15(2):113–117,1998/99.[B´on11] Mikl´os B´ona. Non-overlapping permutation patterns. Pure Math. Appl. (PU.M.A.) ,22(2):99–105, 2011.[B´on12] Mikl´os B´ona. Combinatorics of permutations . Discrete Mathematics and its Applica-tions (Boca Raton). CRC Press, Boca Raton, FL, second edition, 2012. With a forewordby Richard Stanley.[BW82] Anders Bj¨orner and Michelle Wachs. Bruhat order of Coxeter groups and shellability. Adv. in Math. , 43(1):87–100, 1982.[BW83] Anders Bj¨orner and Michelle Wachs. On lexicographically shellable posets. Trans. Amer.Math. Soc. , 277(1):323–341, 1983.[Eli09] Sergi Elizalde. The number of permutations realized by a shift. SIAM J. Discrete Math. ,23(2):765–786, 2009.[Eli13] Sergi Elizalde. The most and the least avoided consecutive patterns. Proc. Lond. Math.Soc. (3) , 106(5):957–979, 2013.[Eli16] Sergi Elizalde. A survey of consecutive patterns in permutations. In Recent trends incombinatorics , volume 159 of IMA Vol. Math. Appl. , pages 601–618. Springer, [Cham],2016.[EN03] Sergi Elizalde and Marc Noy. Consecutive patterns in permutations. Adv. in Appl.Math. , 30(1-2):110–125, 2003. Formal power series and algebraic combinatorics (Scotts-dale, AZ, 2001).[EN12] Sergi Elizalde and Marc Noy. Clusters, generating functions and asymptotics for con-secutive patterns in permutations. Adv. in Appl. Math. , 49(3-5):351–374, 2012.[Gri80] Jerrold R. Griggs. On chains and Sperner k -families in ranked posets. J. Combin. TheorySer. A , 28(2):156–168, 1980.[GSS80] Jerrold R. Griggs, Dean Sturtevant, and Michael Saks. On chains and Sperner k -familiesin ranked posets. II. J. Combin. Theory Ser. A , 29(3):391–394, 1980.[HS15] Joshua Hallam and Bruce Sagan. Factoring the characteristic polynomial of a lattice. J.Combin. Theory Ser. A , 136:39–63, 2015.[Kit11] Sergey Kitaev. Patterns in permutations and words . Monographs in Theoretical Com-puter Science. An EATCS Series. Springer, Heidelberg, 2011. With a foreword by JeffreyB. Remmel.[Lec97] Bruno Leclerc. Families of chains of a poset and Sperner properties. Discrete Math. ,165/166:461–468, 1997. Graphs and combinatorics (Marseille, 1995).[MR06] Anthony Mendes and Jeffrey Remmel. Permutations and words counted by consecutivepatterns. Adv. in Appl. Math. , 37(4):443–480, 2006.[MS15] Peter R. W. McNamara and Einar Steingr´ımsson. On the topology of the permutationpattern poset. J. Combin. Theory Ser. A , 134:1–35, 2015.[Nak11] Brian Nakamura. Computational approaches to consecutive pattern avoidance in per-mutations. Pure Math. Appl. (PU.M.A.) , 22(2):253–268, 2011.[Per13] Guillem Perarnau. A probabilistic approach to consecutive pattern avoiding in permu-tations. J. Combin. Theory Ser. A , 120(5):998–1011, 2013.[Smi14] Jason P. Smith. On the M¨obius function of permutations with one descent. Electron. J.Combin. , 21(2):Paper 2.11, 19, 2014.[Smi15] Jason P. Smith. A formula for the M¨obius function of the permutation poset based ona topological decomposition. arXiv:1506.04406 [math.CO], 2015.[Smi16] Jason P. Smith. Intervals of permutations with a fixed number of descents are shellable. Discrete Math. , 339(1):118–126, 2016.[ST10] Einar Steingr´ımsson and Bridget Eileen Tenner. The M¨obius function of the permutationpattern poset. J. Comb. , 1(1):39–52, 2010.[Ste13] Einar Steingr´ımsson. Some open problems on permutation patterns. In Surveys in com-binatorics 2013 , volume 409 of London Math. Soc. Lecture Note Ser. , pages 239–263.Cambridge Univ. Press, Cambridge, 2013.[SV06] Bruce E. Sagan and Vincent Vatter. The M¨obius function of a composition poset. J.Algebraic Combin. , 24(2):117–136, 2006. HE STRUCTURE OF THE CONSECUTIVE PATTERN POSET 29 [SW12] Bruce E. Sagan and Robert Willenbring. Discrete Morse theory and the consecutivepattern poset. J. Algebraic Combin. , 36(4):501–514, 2012.[Wac99] M. L. Wachs. Obstructions to shellability. Discrete Comput. Geom. , 22(1):95–103, 1999.[Wac07] Michelle L. Wachs. Poset topology: tools and applications. In Geometric combinatorics ,volume 13 of IAS/Park City Math. Ser. , pages 497–615. Amer. Math. Soc., Providence,RI, 2007.[Wil02] Herbert S. Wilf. The patterns of permutations. Discrete Math. , 257(2-3):575–583, 2002.Kleitman and combinatorics: a celebration (Cambridge, MA, 1999). Department of Mathematics, Dartmouth College, Hanover, NH 03755, USA E-mail address : [email protected] Department of Mathematics, Bucknell University, Lewisburg, PA 17837, USA E-mail address ::