aa r X i v : . [ m a t h . C O ] M a y INTERVALS AND FACTORS IN THE BRUHAT ORDER
BRIDGET EILEEN TENNER
Abstract.
In this paper we study those generic intervals in the Bruhat order of the symmet-ric group that are isomorphic to the principal order ideal of a permutation w , and considerwhen the minimum and maximum elements of those intervals are related by a certain propertyof their reduced words. We show that the property does not hold when w is a decomposablepermutation, and that the property always holds when w is the longest permutation. Keywords: permutation, Bruhat order, interval, principal order ideal, reduced word
The interval structure of the Bruhat order on the symmetric group is not well understood.One reason for this is that, sometimes, even those intervals that are isomorphic to principalorder ideals are actually making use of the fact that a reduced word for the minimum elementin the interval can be formed by deleting an arbitrary subword of symbols from a reducedword of the maximum element in the interval. The purpose of this paper is to gain a betterunderstanding of that phenomenon. More precisely, we explore the principal order ideals Λ w with the property that whenever [ x, y ] is isomorphic to Λ w , one may obtain a reduced wordfor x by deleting a consecutive subword from a reduced word for y . Note that deletion ofsome subword will always produce a reduced word for x , but not necessarily the deletion ofa consecutive one. The possibility for a consecutive such word is what we highlight in thiswork, and what we refer to as “forcing” a factor, as defined in Definition 2.7. Structuralanalyses of intervals and principal order ideals are of particular interest because it followsfrom [2] that the Kazhdan-Lusztig polynomial corresponding to the interval [ x, y ] dependsonly on the principal order ideal Λ y .The precise question we answer here is laid out in Section 1, along with the relevantobjects and examples. Section 2 gives additional definitions, and the main results appear inTheorems 3.4 and 4.10. Finally, Section 5 discusses directions for subsequent work.1. Introduction
The symmetric group S n on { , . . . , n } is generated by { s , . . . , s n − } , where s i is thepermutation interchanging i and i + 1, and fixing all other elements. These generators,known as simple reflections , satisfy the Coxeter relations s i = 1 for all i,s i s j = s j s i if | i − j | > , and s i s i +1 s i = s i +1 s i s i +1 for 1 ≤ i ≤ n − . A permutation w can be recorded as a product of simple reflections w = s i s i · · · s i r , Mathematics Subject Classification. of which there are infinitely many representations, or written uniquely in one-line notation w = w (1) w (2) · · · w ( n ) . These two representations of the same class of objects are quite different, and each haveadvantages in certain contexts. The main result of [10] was a way to translate between thetwo options.The order in which we compose maps indicates that s i w interchanges the positions of thevalues i and i + 1 in the one-line notation of w , and ws i interchanges the values in positions i and i + 1 in the one-line notation of w . Example 1.1.
The permutation w ∈ S which maps 1 to 3, 2 to itself, 3 to 4, and 4 to 1,would be written in one-line notation as w = 3241 . A few of the infinite many ways to express w as a product of simple reflections include: w = s s s s = s s s s = s s s s = s s s s s s . As demonstrated in Example 1.1, the number of simple reflections involved in representinga particular permutation is not fixed. There is, however, a minimum value.
Definition 1.2. If w = s i · · · s i ℓ ( w ) where ℓ ( w ) is minimal, then s i · · · s i ℓ ( w ) is a reduceddecomposition of w , and the string of subscripts i · · · i ℓ ( w ) is a reduced word of w . The set ofreduced words of w is denoted R ( w ). The nonnegative integer ℓ ( w ) is the length of w .To avoid confusion with permutations, which are also strings of integers, reduced wordsand their symbols will be written in sans serif.The Coxeter relations among the symbols in a reduced decomposition have obvious ana-logues for the symbols in a reduced word: s i s j = s j s i ←→ ij = ji if | i − j | > , and s i s i +1 s i = s i +1 s i s i +1 ←→ i ( i + ) i = ( i + ) i ( i + ) for 1 ≤ i ≤ n − . A product of simple reflections is reduced if it is a reduced decomposition for some permuta-tion, and a string of integers is reduced if the corresponding product of simple reflections isreduced.
Example 1.3.
Continuing Example 1.1, ℓ (3241) = 4 and R (3241) = { , , } .(Note that we have not explained how to compute these values.)The Bruhat order is a partial ordering given to the elements of a Coxeter group. Althoughwe are only concerned with the symmetric group, we give the full definition here. Definition 1.4.
Suppose that x and y are elements of a Coxeter group. Then x ≤ y in theBruhat order if there exists a reduced word of x that can be obtained from a reduced wordof y by deleting symbols and simplifying Coxeter relations as necessary. NTERVALS AND FACTORS IN THE BRUHAT ORDER 3
The Bruhat order gives a graded poset structure to any Coxeter group, where the rankof an element is the element’s length. The Bruhat order of S is depicted in Figure 1. Ofcourse, Definition 1.4 can be stated analogously in terms of reduced decompositions. ∅ s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s Figure 1.
The Bruhat order of the symmetric group S , where each elementis labeled by one of its (possibly many) reduced decompositions.The Bruhat order gives a partial ordering to the permutations of a given set. Another wayto describe it is as a subword order on reduced decompositions/words. Dyer studied intervalsin the Bruhat order for all finite Coxeter groups, showing that there are only finitely manynon-isomorphic intervals of any given length [3]. Jantzen and Hultman have classified allpossible length 4 intervals, as well as the length 5 intervals in the symmetric group [4, 5, 6].The special class of intervals for which the minimal element is the identity, known as principalorder ideals and defined formally in Definition 2.3, were studied by the author in [9]. As isobvious from Table 1, there are intervals in S n that never appear as principal order ideals.Length 0 1 2 3 4 5 S n S n Table 1.
The number of non-isomorphic intervals and principals order idealsof length at most 5 (as defined in Definition 1.2) appearing in the Bruhat orderof symmetric groups.
BRIDGET EILEEN TENNER
Example 1.5.
The interval [2143 , S , depicted in Figure 2,never appears as a principal order ideal in the Bruhat order of any symmetric group.2143423123412431 24133241 4123413231424213 Figure 2.
The interval [2143 , ⊂ S , not isomorphic to any principalorder ideal appearing in the Bruhat order of any S n .The purpose of this paper is to explore how and when generic intervals in S n might beisomorphic to principal order ideals. So that we can refer to it subsequently, we highlight thefollowing discussion in a remark. Remark . Principal order ideals in S n are special cases of intervals, and they lack a certainfreedom that more general intervals possess. In the Bruhat order, x ≤ y if there is an elementof R ( x ) that is a subword of an element of R ( y ). In a principal order ideal, this x is theidentity permutation, and so R ( x ) = {∅} . Of course ∅ is a subword of every word. Thus,for principal order ideals, the entire reduced word for y must be deleted in order to yield thereduced word for x . In particular, note that what is getting deleted is a consecutive subword(in fact, the entire word itself). On the other hand, in a generic interval, what gets deletedfrom the reduced word for y need not be a consecutive subword. This potential yields somegeneric intervals that do not appear as principal order ideals. Example 1.7.
The interval [21543 , R (21543) = { , , , , , , , } and R (52341) = { , , , , , , , , , , , , , , , , , , } . Thus no element of R (21543) can be obtained by deleting a consecutive subword of symbolsfrom an element of R (52341).The purpose of the current work is to examine when a principal order ideal in the symmetricgroup also appears as a more general interval. In particular, we want to understand when such NTERVALS AND FACTORS IN THE BRUHAT ORDER 5 an interval, not necessarily beginning at rank 0 in the poset, must still come from deleting afactor from a reduced word of its maximum element.Before making this property precise, consider the following result, where vexillary permu-tations are exactly those that avoid the pattern 2143.In [10], we showed that if a permutation w contains a vexillary p -pattern, then there is areduced word for w possessing a reduced word f for p as a factor, possibly with a fixed positiveinteger added to each letter in f . If v is the permutation obtained by deleting this subwordfrom w , then this implies that the interval [ v, w ] in the Bruhat order would be isomorphic toΛ p , the principal order ideal for p . Indeed, this would also imply that there is a copy of Λ p sitting as a principal order ideal inside of the principal order ideal Λ w of w , as depicted inFigure 3. ew vew v ∼ = Λ p Figure 3.
When a permutation w contains a vexillary pattern p , then Λ w willcontain intervals isomorphic to Λ p as indicated by the shading. (There may, ofcourse, be other intervals in Λ w isomorphic to Λ p as well.) Example 1.8.
The permutation 4213 contains the vexillary pattern 321. The interval[1243 , are both isomorphic to Λ , as shown inFigure 4.We are now able to clarify the main question of this paper: when does a principal orderideal Λ w ⊆ S n have the property that for all intervals [ x, y ] ∼ = Λ w , there is a reduced word i ∈ R ( x ) and a reduced word j ∈ R ( y ) such that i can be formed by deleting a consecutivesubword from j ?The main results of this paper are that no decomposable permutation can force a factor(Theorem 3.4), and that the permutation n ( n − · · ·
321 does force a factor for all n (The-orem 4.10). These are preceded by a discussion of the useful terminology in Section 2, andthe paper concludes with suggestions for future work.2. Definitions
We now define the main objects of the current work. More details and background infor-mation about Coxeter groups can be found in [1] and [7].Two of the most fundamental structural features of a poset are its intervals and its orderideals. In a poset with a unique minimal element, these objects intersect at the concept of aprincipal order ideal.
BRIDGET EILEEN TENNER
Figure 4.
The principal order ideal Λ has a copy of the principal orderideal Λ , both as a principal order ideal (thick dashed lines) and as a genericinterval (thick solid lines).
Definition 2.1.
Consider a poset P and elements x, y ∈ P with x ≤ y . The set of elements { z : x ≤ z ≤ y } ⊆ P is denoted [ x, y ], and is called an interval . Definition 2.2. An order ideal in a poset P is a subset I ⊆ P such that if y ∈ I and x ≤ y ,then x ∈ I . Definition 2.3. A principal order ideal in a poset P is an order ideal with a unique maximalelement. Equivalently, the principal order ideals of P are the subsets Λ y = { z : z ≤ y } foreach y ∈ P . If P has a unique minimal element ˆ0, then the principal order ideals are exactlythe intervals of the form [ˆ0 , y ].To calculate the length of a permutation, we must make a preliminary definition. Definition 2.4.
Given a permutation w , an inversion in w is a pair ( i, j ) such that i < j and w ( i ) > w ( j ).Inversions are easy to see in the one-line notation of a permutation: they consist of a value(the w ( i ) of the definition) appearing somewhere to the left of a smaller value (the w ( j ) ofthe definition). It is well known that ℓ ( w ) is equal to the number of inversions in w , and itis now clear that there is a unique permutation in S n having maximal length. Definition 2.5.
Fix a positive integer n . The unique element of maximal length in S n is w n = n ( n − · · · ℓ ( w n ) = (cid:0) n (cid:1) .As described in Definition 1.4, the relation x ≤ y in the Bruhat order allows any subsetof symbols to be deleted from a reduced word of y in order to form a reduced word of x . Inparticular, these symbols need not be consecutive. For example, as shown in Figure 1,1324 = s < s s s = 2341 . NTERVALS AND FACTORS IN THE BRUHAT ORDER 7
In this paper, we will examine when all of the activity happening within an interval [ x, y ]in the Bruhat order of some symmetric group is actually happening within some consecutivesubstring of the symbols of an element of R ( y ), all of which must be deleted to form anelement of R ( x ). Definition 2.6. A factor in a word is a consecutive substring.This paper is concerned with understanding when intervals [ x, y ] that are isomorphic toΛ w for some w may or may not be formed in “interesting” ways. This is made more precisein the following definition. Definition 2.7.
Fix an element w ∈ S n , and consider its principal order ideal Λ w . If it istrue that for every interval [ x, y ] ∼ = Λ w , there exists an element of R ( x ) formed by deleting afactor from an element of R ( y ), then w forces a factor. Otherwise w does not force a factor .The “interesting” feature noted above was described in Remark 1.6 in the introduction tothis work. The first thing to note about this topic is that determining which permutationsforce a factor is an interesting problem. More precisely, not all permutations do so. Example 2.8.
Consider Λ ⊂ S . Note that the principal order ideal Λ is isomorphicto the interval [1324 , ∅ s s s s s s s s s s s s However, there is no element of R (1324) = { } that can be formed from by deleting a singlefactor from an element of R (2341) = { } . Thus 2314 = s s does not force a factor.3. Permutations that do not force a factor
In this section we describe a large class of permutations that do not force a factor.
Definition 3.1.
A permutation w is decomposable if Λ w ∼ = Λ u × Λ v , where neither Λ u nor Λ v is itself isomorphic to Λ w . If w is not decomposable, then it is indecomposable . Example 3.2.
The permutation 4213 ∈ S is decomposable because Λ ∼ = Λ × Λ .This is depicted in Figure 4. Proposition 3.3 ([8]) . A permutation w ∈ S n is decomposable if and only if there exists m ∈ [ n − and a reduced word a a ∈ R ( w ) , where a and a are nonempty, such that a i consists only of letters less than or equal to m , and a − i consists only of letters strictly greaterthan m . Example 3.2 (continued).
The reduced words of 4213 ∈ S are { , , } . Thenin the language of Proposition 3.3, we can let m = 2, and a = and a = . Thus 4213 isdecomposable. BRIDGET EILEEN TENNER
We now show, constructively, that no decomposable permutation forces a factor. Thismeans that for any decomposable permutation w , we must produce an interval [ x, y ] ∼ = Λ w such that no reduced word for x can be obtained by deleting a factor from any reduced wordfor y . Theorem 3.4. If w is decomposable, then w does not force a factor.Proof. Suppose that w is decomposable. Consider the value m and the reduced word a a ∈ R ( w ) guaranteed by Proposition 3.3. We can assume, without loss of generality, that i = 1in the proposition. Let k ≤ m be the largest value appearing in a , and k ≥ m + 1 be thesmallest value appearing in a .Let a ′ be the string obtained from a by increasing each symbol by 1. Also, define b = ( k + )( k + ) · · · ( m + ) · · · ( k − ) k , a string of consecutive increasing letters. Nowdefine w − = s k +1 s k +2 · · · s m +1 · · · s k − s k ; that is, b ∈ R ( w − )It is not hard to see that a b a ′ is reduced, by construction. Define w + so that a b a ′ ∈ R ( w + ) . It is not hard to see that Λ w ∼ = [ w − , w + ] . Neither a nor a ′ contain any of the letters { k + 1 , k + 2 , . . . , k } . Moreover, k ∈ a and k + 1 ∈ a ′ . Thus the Coxeter relations prohibit k ∈ a from commuting into or across b from the left, and similarly k + 1 ∈ a ′ cannot do so from the right. Thus it will be impossibleto get k and k + 1 into the same factor whose deletion would yield b .Therefore w does not force a factor. (cid:3) Example 2.8 depicts the procedure outlined in Theorem 3.4. In that case, w = s s = 2314,and so m = 1, a = , and a = . Then k = 1 and k = 2, and so a ′ = and the resultingstring a b a ′ = . Therefore w − = s = 1324 and w + = s s s = 2341. This yields exactlythe demonstrative interval [1324 , Example 3.5.
Consider the permutation w = 3412 = s s s s ∈ S . Because R ( w ) = { , } , we see that this w is indecomposable. It is not hard to check that[12543 , ∼ = Λ . To show that w does not force a factor, note that R (12543) = { , } , and recall R (52341)from Example 1.7. There is no element of R (52341) from which a single factor could be deletedto yield either or . Thus w does not force a factor. NTERVALS AND FACTORS IN THE BRUHAT ORDER 9 Permutations that do force a factor
In this section we prove that the longest permutation n ( n − · · ·
321 always forces a factor.With the understanding that the symmetric group is given a poset under the Bruhat order,we will abuse notation slightly and write Λ w n ∼ = S n , henceforth. Thus we now show that forany interval [ x, y ] appearing in the Bruhat order of a symmetric group satisfying [ x, y ] ∼ = S n ,there exists some i ∈ R ( x ) that can be obtained from some j ∈ R ( y ) by deleting a factor.The main result will be proved inductively, and its proof will benefit from some preliminaryresults. The first of these is about generic intervals in the Bruhat order of the symmetricgroup, not of any fixed isomorphism class. The proposition concerns the coatoms in an interval[ x, y ], that is, the elements w of the interval that are covered by y (denoted x ≤ w ⋖ y ). Proposition 4.1.
Suppose that x, y ∈ S n with x < y . Fix i satisfying x ( i ) = y ( i ) . Thenthere exists a permutation w with x ≤ w ⋖ y and w ( i ) = y ( i ) .Proof. We prove the result by induction on ℓ ( y ) − ℓ ( x ).If ℓ ( y ) − ℓ ( x ) = 1, then set w = x . Now consider ℓ ( y ) − ℓ ( x ) >
1, and suppose inductivelythat the result is true for all intervals of length less than ℓ ( y ) − ℓ ( x ).Fix some v satisfying x ≤ v ⋖ y . If v ( i ) = y ( i ), then set w = v and we are done. If, instead, v ( i ) = y ( i ), then we can apply the inductive assumption to the interval [ x, v ]. This yields apermutation u with x ≤ u ⋖ v and u ( i ) = v ( i ). Consider the interval [ u, y ]. As described inTable 1, this has only one possible form, and includes a fourth element which will denote w . uyv w Because ℓ ( y ) − ℓ ( u ) = 2, the permutations u and y differ, as strings, in either three or fourpositions, one of which is necessarily position i .If u and y differ in four positions, then u and v differ in two positions ( i and j , for some j ), and v and y differ in two other positions. The two transpositions commute, and so w isobtained from y by swapping the values in positions i and j . In other words, w ( i ) = y ( i ).Suppose, on the other hand, that u and y differ in just three positions: i , j , and j . Then,because v ( i ) = y ( i ), we have that v and y differ in positions j and j . Because w = v , thetwo positions in which w and y differ, which must be a subset { i, j , j } , cannot be both j and j . Thus, one of them must be i , meaning that w ( i ) = y ( i ). (cid:3) We now focus on a particular kind of interval in the symmetric group, and look at whatsuch an interval implies for the reduced words of its minimum and maximum elements.
Definition 4.2.
Let s be a string of integers, and t ∈ Z . The shift of s by t is the string s t obtained by adding t to each of the values in s . Example 4.3. ( − ) = and ( − ) − = − − . Proposition 4.4.
Fix a positive integer k . Suppose that x, y ∈ S n have reduced words ac ∈ R ( x ) and abc ∈ R ( y ) , and that [ x, y ] ∼ = S k . Then there exists an integer t ≥ such that b − t ∈ R ( w k ) . Proof.
The length of b is equal to ℓ ( y ) − ℓ ( x ) = ℓ ( w k ) = (cid:0) k (cid:1) . Because b is a reduced word, itcannot contain fewer than k − k − x, y ] would not be isomorphic to Λ w k . Thus b contains exactly k − ℓ ( w k ) out of k − t of a reduced word for w k , where t + 1 be the smallest symbol in b . (cid:3) Example 4.5.
In the language of Proposition 4.4, let x = 21534 and y = 24531, with reducedwords ∈ R ( x ) and ∈ R ( y ). The interval [ x, y ] ⊂ S is isomorphic to S , as drawnin Figure 5. In this example, t = 1 because − ∈ R (321).2351423541 21543245132153424531 Figure 5.
The interval [21534 , ⊂ S , which is isomorphic to S .The next two propositions require an additional definition. Definition 4.6.
Fix a string s . Consider a monotonic substring s ′ of s with smallest value a and largest value b (the endpoints of the monotonic substring). This s ′ is thin if no value c s ′ , with a < c < b , appears between a and b in s . Example 4.7.
Let s = . The monotonic substring is thin, while the monotonicsubstring is not thin because of the appearing between and in s . Proposition 4.8.
Fix a positive integer k . Suppose that x, y ∈ S n have reduced words ac ∈ R ( x ) and abc ∈ R ( y ) , with b − t ∈ R ( w k ) for some integer t ≥ . Then x and y , asstrings, are identical outside of a thin monotonic substring of length k , which appears inincreasing order in x and decreasing order in y . To ease the discussion, we will call this monotonic substring that distinguishes x from y the swap-string . Note that if k is odd, then x and y will also be identical in the central positionof the swap-string. Proof.
We prove the result by induction on the length of a .If a = ∅ , then y = (cid:0) · · · t ( t + k ) · · · ( t + 2)( t + 1) (cid:1) x. NTERVALS AND FACTORS IN THE BRUHAT ORDER 11
Thus x and y only differ, as strings, in the subsequence involving { t + 1 , . . . , t + k } , whichnecessarily appears in increasing order in x (because bc is reduced, so multiplying c bythe permutation corresponding to b cannot undo any inversions) and decreasing order in y .Because there are no values between t + 1 and t + k that are not already in the swap-string,the result holds.Now suppose that a = ua ′ where u is a single letter. Let x ′ = s u x and y ′ = s u y , and assumeinductively that the result holds for a ′ c ∈ R ( x ′ ) and a ′ bc ∈ R ( y ′ ). This means that the strings x ′ and y ′ are identical except for a substring of length k whose values appear in increasingorder in x ′ and decreasing order in y ′ , and these swap-strings are thin.Because ac and abc are both reduced words, the value u must appear to the left of u + 1in both x ′ and y ′ . Thus at most one of { u, u + 1 } appears in the swap-string for x ′ and y ′ , soswapping the positions of the two values cannot change the monotonicity of the differentiatingsubstrings. In other words, lengthening the prefix might change the specific values that differin the two strings, but cannot alter the swap-string phenomenon.Similarly, the thinness of the swap-string is maintained by this operation. (cid:3) The converse to Proposition 4.8 is also true. Its proof is similar to the main result of [10],and we offer the broad strokes of it here.
Proposition 4.9.
Suppose that x, y ∈ S n are identical, as strings, outside of a thin mono-tonic substring of length k , which appears in increasing order in x and decreasing order in y .Then there exist reduced words ac ∈ R ( x ) and abc ∈ R ( y ) , with b − t ∈ R ( w k ) for some integer t ≥ .Proof. We will transform both x and y into the identity permutation, by minimally manysimple reflections, thus obtaining reduced words for each.Look for any values sitting in between the endpoints of the swap-string that are not actuallyin the swap-string themselves. Because the swap-strings are thin, each of these values is eithersmaller than the minimum value of the swap-string or larger than the maximum value of theswap-string. Identically multiply x and y on the right by a succession of simple reflections(thus swapping the values in adjacent positions in the strings) to move all of the too-large(respectively, too-small) values to the right (respectively, left) of the swap-string. The orderin which these values are moved can be chosen so that each multiplication removes exactlyone inversion. Let C be the reduced word corresponding to the product of these multipliedsimple reflections.We now have two permutations x ′ ≤ x and y ′ ≤ y that are identical, as strings, outsideof a swap-string of length k , which appears in increasing order in x ′ and decreasing order in y ′ . Moreover, the swap-string is a factor in each of x ′ and y ′ . We can now multiply y ′ on theright by a succession of simple reflections, each of which removes exactly one inversion fromthe permutation, to put this decreasing factor into increasing order and thus yield x ′ . Thiswill correspond to a reduced word b . Moreover, b necessarily satisfies b − t ∈ R ( w k ), wherethe leftmost symbol in the swap-string appears in the ( t − a ∈ R ( x ′ ).Let c be the string obtained by writing C in reverse order. Then ac ∈ R ( x ) and abc ∈ R ( y ),with b − t ∈ R ( w k ) for some integer t ≥
0, as desired. (cid:3)
We are now able to prove the main result of this section, describing a family of permutationsthat force factors.
Theorem 4.10.
For all integers n > , the permutation w n = n ( n − · · · forces a factor.Proof. First note that Λ ∼ = S is the following poset.If [ x, y ] ∼ = S , then ℓ ( y ) = ℓ ( x ) + 1, and so a reduced word for x must be obtained from areduced word for y by deleting a single letter. A single letter is necessarily a factor, and sothe result holds for n = 2.Now consider some integer n >
2, and suppose, inductively, that the result holds for w n − .Let [ x, y ] ∈ S m be an interval that is isomorphic to S n . In S n , there are two intervals(1) [23 · · · n , w n ] and [ n · · · ( n − , w n ] , each of which is isomorphic to S n − . Thus, there must be two such intervals [ x , y ] and [ x , y ]in [ x, y ]. Recall the results of Propositions 4.4 and 4.8. Let the swap-string for [ x i , y ] ∼ = S n − have values h ( i )1 < · · · < h ( i ) n − .Because [ x , y ] and [ x , y ] overlap extensively in [ x, y ], as do the intervals of (1) in S n , theirrespective swap-strings must share many values. In particular, at the second highest rank in[ x, y ], the intervals [ x , y ] and [ x , y ] overlap in n − n − x i , y ], the n − { h ( i )1 , . . . , h ( i ) n − } or { h ( i )2 , . . . , h ( i ) n − } .Note that [ x , y ] ∪ [ x , y ] includes all of the coatoms of [ x, y ]. By Proposition 4.1, x and y cannot differ in any positions outside the union of the two swap-strings. This union encom-passes exactly n positions. Because the swap-strings are thin, and because ℓ ( y ) − ℓ ( x ) = ℓ ( w n ),we must have the n positions form an increasing substring in x and a decreasing substring in y , and these two monotonic substrings must be thin in their respective permutations. Propo-sition 4.9 now implies that some i ∈ R ( x ) can be obtained from some j ∈ R ( y ) by deleting afactor, completing the proof. (cid:3) To illustrate the proof of Theorem 4.10, we present the following example.
Example 4.11.
Let x = 321456 and y = 361542 in S , for which [ x, y ] ∼ = S . In thelanguage of the proof of Theorem 4.10, then, n = 4. Let x = 341562 and x = 361245. For i ∈ { , } , the interval [ x i , y ] ⊂ [ x, y ] is isomorphic to S , as depicted in Figure 6. Note thatthese intervals share 4 − ∈ S ), and the swap-string { , , } for [ x , y ] shares 4 − { , , } for [ x , y ].In fact, Proposition 4.9 also tells us the form of the factor forced by w n . Corollary 4.12.
For all integers n > , if [ x, y ] ∼ = S n , then some i ∈ R ( x ) can be obtainedfrom some j ∈ R ( y ) by deleting a factor b , where b t ∈ R ( w n ) for some t ∈ Z . Open questions
We have now documented a family of permutations that do not force factors and a secondfamily of permutations that do force factors. Completely characterizing those permutations
NTERVALS AND FACTORS IN THE BRUHAT ORDER 13 x = 341562351462 341652351642 361452 y = 361542 x = 361245361425 361254361452 361524 y = 361542 Figure 6.
The intervals [ x , y ] and [ x , y ] described in Example 4.11, illus-trating the proof of Theorem 4.10.that do (or do not) force factors is still an open question, and one which could shed significantlight on the interval structure of the Bruhat order for the symmetric group.In a different direction, the present work studies only the finite Coxeter group of type A ,although the analogous question can be asked for Coxeter groups of other types as well. Acknowledgements
I am grateful for the thoughtful suggestions of an anonymous referee.
References [1] A. Bj¨orner and F. Brenti,
Combinatorics of Coxeter Groups , Graduate Texts in Mathematics 231,Springer, New York, 2005.[2] F. Brenti, A combinatorial formula for Kazhdan-Lusztig polynomials,
Invent. Math. (1994), 371–394.[3] M. J. Dyer, On the “Bruhat graph” of a Coxeter system,
Compos. Math. (1991), 185–191.[4] A. Hultman, Bruhat intervals of length 4 in Weyl groups, J. Combin. Theory, Ser. A (2003), 163–178.[5] A. Hultman,
Combinatorial complexes, Bruhat intervals and reflection distances , Ph.D. thesis, KTH,2003.[6] J. C. Jantzen,
Moduln mit Einem H¨ochsten Gewicht , Lecture Notes in Mathematics 750, Springer-Verlag,Berlin, 1979.[7] I. G. Macdonald,
Notes on Schubert Polynomials , Laboratoire de combinatoire et d’informatiquemath´ematique (LACIM), Universit´e du Qu´ebec `a Montr´eal, Montreal, 1991.[8] B. E. Tenner,
The combinatorics of reduced decompositions , Ph.D. thesis, MIT, 2006.[9] B. E. Tenner, Pattern avoidance and the Bruhat order,
J. Combin. Theory Ser. A (2007), 888–905.[10] B. E. Tenner, Reduced decompositions and permutation patterns,
J. Algebraic Combin. (2006), 263–284. Department of Mathematical Sciences, DePaul University, Chicago, IL 60614
E-mail address ::