Cyclic sieving for longest reduced words in the hyperoctahedral group
aa r X i v : . [ m a t h . C O ] M a y CYCLIC SIEVING FOR LONGEST REDUCED WORDSIN THE HYPEROCTAHEDRAL GROUP
T. KYLE PETERSEN AND LUIS SERRANO
Abstract.
We show that the set R ( w ) of reduced expressions for the longestelement in the hyperoctahedral group exhibits the cyclic sieving phenomenon.More specifically, R ( w ) possesses a natural cyclic action given by moving thefirst letter of a word to the end, and we show that the orbit structure of thisaction is encoded by the generating function for the major index on R ( w ). Introduction and main result
Suppose we are given a finite set X , a finite cyclic group C = h ω i acting on X ,and a polynomial X ( q ) ∈ Z [ q ] with integer coefficients. Following Reiner, Stanton,and White [RSW], we say that the triple ( X, C, X ( q )) exhibits the cyclic sievingphenomenon (CSP) if for every integer d ≥
0, we have that | X ω d | = X ( ζ d ) where ζ ∈ C is a root of unity of multiplicitive order | C | and X ω d is the fixed pointset of the action of the power ω d . The sizes of the fixed point sets determine thecycle structure of the canonical image of ω in the group of permutations of X , S X .Therefore, to find the cycle structure of the image of any bijection ω : X → X , it isenough to determine the order of the action of ω on X and find a polynomial X ( q )such that ( X, h ω i , X ( q )) exhibits the CSP.The cyclic sieving phenomenon has been demonstrated in a variety of contexts.The paper of Reiner, Stanton, and White [RSW] itself includes examples involvingnoncrossing partitions, triangulations of polygons, and cosets of parabolic sub-groups of Coxeter groups. An example of the CSP with standard Young tableauxis due to Rhoades [Rh] and will discussed further in Section 4. Now we turn to theCSP of interest to this note.Let w = w ( B n )0 denote the longest element in the type B n Coxeter group. Givengenerating set S = { s , . . . , s n } for B n , ( s being the “special” reflection), we willwrite a reduced expression for w as a word in the subscripts. For example, w ( B )0 can be written as s s s s s s s s s ;we will abbreviate this product by 121323123. It turns out that if we cyclically per-mute these letters, we always get another reduced expression for w . Said anotherway, s i w s i = w for i = 1 , . . . , n . The same is not true for longest elements ofother classical types. In type A, we have s i w ( A n )0 s n +1 − i = w ( A n )0 , and for type D, w ( D n )0 = ( s i w ( D n )0 s i if n even or i > ,s i w ( D n )0 s − i if n odd and i = 1 , . Let R ( w ) denote the set of reduced expressions for w in type B n and let c : R ( w ) → R ( w ) denote the action of placing the first letter of a word at the end. Then the orbit in R ( w ( B )0 ) of the word above is: { → → → → → → → → } . As the length of w is n , we clearly have c n = 1, and the size of any orbitdivides n . For an example of a smaller orbit, notice that the word 213213213 hascyclic order 3.For any word w = w . . . w l , (e.g., a reduced expression for w ), a descent of w is defined to be a position i in which w i > w i +1 .The major index of w , maj( w ), isdefined as the sum of the descent positions. For example, the word w = 121323123has descents in positions 2 ,
4, and 6, so its major index is maj( w ) = 2 + 4 + 6 = 12.Let f n ( q ) denote the generating function for this statistic on words in R ( w ): f n ( q ) = X w ∈ R ( w ) q maj( w ) . The following is our main result.
Theorem 1.
The triple ( R ( w ) , h c i , X ( q )) exhibits the cyclic sieving phenomenon,where X ( q ) = q − n ( n ) f n ( q ) . For example, let us consider the case n = 3. We have X ( q ) = q − X w ∈ R ( w ( B ) q maj( w ) = 1 + q + 2 q + 2 q + 2 q + 4 q + 3 q + 4 q + 4 q + 4 q + 3 q + 4 q + 2 q + 2 q + 2 q + q + q . Let ζ = e πi . Then we compute: X (1) = 42 X ( ζ ) = 6 X ( ζ ) = 6 X ( ζ ) = 0 X ( ζ ) = 0 X ( ζ ) = 0 X ( ζ ) = 0 X ( ζ ) = 0 X ( ζ ) = 0Thus, the 42 reduced expressions for w ( B )0 split into two orbits of size three (theorbits of 123123123 and 132132132) and four orbits of size nine.We prove Theorem 1 by relating it to another instance of the CSP, namelyRhoades’ recent (and deep) result [Rh, Thm 3.9] for the set SY T ( n m ) of rectangularstandard Young tableaux with respect to the action of promotion (defined in Section2). To make the connection, we rely on a pair of remarkable bijections due toHaiman [H1, H2]. The composition of Haiman’s bijections maps to R ( w ) fromthe set of square tableaux, SY T ( n n ). In this note our main goal is to show thatHaiman’s bijections carry the orbit structure of promotion on SY T ( n n ) to the orbitstructure of c on R ( w ).We conclude this section by remarking that this approach was first outlined byRhoades [Rh, Thm 8.1]. One purpose of this article is to fill some nontrivial gapsin his argument. A second is to justify the new observation that the polynomial X ( q ) can be expressed as the generating function for the major index on R ( w ).We thank Brendon Rhoades for encouraging us to write this note. Thanks also toKevin Dilks, John Stembridge, and Alex Yong for fruitful discussions on this andrelated topics, and to Sergey Fomin for comments on the manuscript. YCLIC SIEVING IN THE HYPEROCTAHEDRAL GROUP 3 Promotion on standard Young tableaux
For λ a partition, let SY T ( λ ) denote the set of standard Young tableaux of shape λ . If λ is a strict partition , i.e., with no equal parts, then let SY T ′ ( λ ) denote theset of standard Young tableaux of shifted shape λ . We now describe the action of jeu de taquin promotion , first defined by Sch¨utzenberger [Sch].We will consider promotion as a permutation of tableaux of a fixed shape (resp.shifted shape), p : SY T ( λ ) → SY T ( λ ) (resp. p : SY T ′ ( λ ) → SY T ′ ( λ )). Given atableau T with λ ⊢ n , we form p ( T ) with the following algorithm. (We denote theentry in row a , column b of a tableau T , by T a,b .)(1) Remove the entry 1 in the upper left corner and decrease every other entryby 1. The empty box is initialized in position ( a, b ) = (1 , p ( T ) a,b :=min { T a,b +1 , T a +1 ,b } . Set ( a, b ) := ( a ′ , b ′ ), where ( a ′ , b ′ ) are the coordi-nates of box swapped, and go to 2a).(3) Fill the empty box with n .Here is an example: T = 1 2 4 83 6 75 p ( T ) . As a permutation, promotion naturally splits
SY T ( λ ) into disjoint orbits. Fora general shape λ there seems to be no obvious pattern to the sizes of the orbits.However, for certain shapes, notably Haiman’s “generalized staircases” more can besaid [H2] (see also Edelman and Greene [EG, Cor. 7.23]). In particular, rectanglesfall into this category, with the following result. Theorem 2 ([H2], Theorem 4.4) . If λ ⊢ N = bn is a rectangle, then p N ( T ) = T for all T ∈ SY T ( λ ) . Thus for n × n square shapes λ , p n = 1 and the size of every orbit divides n .With n = 3, here is an orbit of size 3:(1) 1 2 53 6 84 7 9 → → → · · · . There are 42 standard Young tableaux of shape (3 , , R ( w ( B )0 ). Stanley first conjectured that R ( w ) and SY T ( n n )are equinumerous, and Proctor suggested that rather than SY T ( n n ), a more directcorrespondence might be given with SY T ′ (2 n − , n − , . . . , R ( w ) [H2, Theorem 5.12]. Moreover, in [H1, Proposition 8.11], he gives a bijectionbetween standard Young tableaux of square shape and those of doubled staircaseshape that (as we will show) commutes with promotion. T. K. PETERSEN AND L. SERRANO
As an example, his bijection carries the orbit in (1) to this shifted orbit:1 2 4 5 83 6 97 → → → · · · . Both of these orbits of tableaux correspond to the orbit of the reduced word132132132. 3.
Haiman’s bijections
We first describe the bijection between reduced expressions and shifted standardtableaux of doubled staircase shape. This bijection is described in Section 5 of [H2].Let T in SY T ′ (2 n − , n − , . . . , T , (i.e., n ),occupies one of the outer corners. Let r ( T ) denote the row containing this largestentry, numbering the rows from the bottom up. The promotion sequence of T isdefined to be Φ( T ) = r · · · r n , where r i = r ( p i ( T )). Using the example above of T = 1 2 4 5 83 6 97 , we see r ( T ) = 2 , r ( p ( T )) = 1, r ( p ( T )) = 3, and since p ( T ) = T , we haveΦ( T ) = 132132132 . Haiman’s result is the following.
Theorem 3 ([H2], Theorem 5.12) . The map T Φ( T ) is a bijection SY T ′ (2 n − , n − , . . . , → R ( w ) . By construction, then, we haveΦ( p ( T )) = c (Φ( T )) , i.e., Φ is an orbit-preserving bijection( SY T ′ (2 n − , n − , . . . , , p ) ←→ ( R ( w ) , c ) . Next, we will describe the bijection H : SY T ( n n ) → SY T ′ (2 n − , n − , . . . , H commutes with promotion.We assume the reader is familiar with the Robinson-Schensted-Knuth insertion algorithm (RSK). (See [Sta, Section 7.11], for example.) This is a map betweenwords w and pairs of tableaux ( P, Q ) = ( P ( w ) , Q ( w )). We say P is the insertiontableau and Q is the recording tableau .There is a similar correspondence between words w and pairs of shifted tableaux( P ′ , Q ′ ) = ( P ′ ( w ) , Q ′ ( w )) called shifted mixed insertion due to Haiman [H1]. (Seealso Sagan [Sa] and Worley [W].) Serrano defined a semistandard generalization ofshifted mixed insertion in [Ser]. Throughout this paper we refer to semistandardshifted mixed insertion simply as mixed insertion . Details can be found in [Ser,Section 1.1]. YCLIC SIEVING IN THE HYPEROCTAHEDRAL GROUP 5
Theorem 4 ([Ser] Theorem 2.26) . Let w be a word. If we view Q ( w ) as a skewshifted standard Young tableau and apply jeu de taquin to obtain a standard shiftedYoung tableau, the result is Q ′ ( w ) (independent of any choices in applying jeu detaquin). For example, if w = 332132121, then( P, Q ) = , , ( P ′ , Q ′ ) = ′ ′ ′ , . Performing jeu de taquin we see:1 2 53 6 84 7 9 → → → . Haiman’s bijection is precisely H ( Q ) = Q ′ . That is, given a standard squaretableau Q , we embed it in a shifted shape and apply jeu de taquin to create astandard shifted tableau. That this is indeed a bijection follows from Theorem 4,but is originally found in [H1, Proposition 8.11]. Remark 5.
Haiman’s bijection H applies more generally between rectangles and“shifted trapezoids”, i.e., for m ≤ n , we have H : SY T ( n m ) → SY T ′ ( n + m − , n + m − , . . . , n − m + 1) . All the results presented here extend to this generality, withsimilar proofs. We restict to squares and doubled staircases for clarity of exposition. We will now fix the tableaux P and P ′ to ensure that the insertion word w hasparticularly nice properties. We will use the following lemma. Lemma 6 ([Ser], Proposition 1.8) . Fix a word w . Let P = P ( w ) be the RSKinsertion tableau and let P ′ = P ′ ( w ) be the mixed insertion tableau. Then the setof words that mixed insert into P ′ is contained in the set of words that RSK insertinto P . Now we apply Lemma 6 to the word w = n · · · n | {z } n · · · · · · | {z } n · · · | {z } n . If we use RSK insertion, we find P is an n × n square tableau with all 1s in rowfirst row, all 2s in the second row, and so on. With such a choice of P it is notdifficult to show that any other word u inserting to P has the property that in anyinitial subword u · · · u i , there are at least as many letters ( j + 1) as letters j . Suchwords are sometimes called (reverse) lattice words or (reverse) Yamanouchi words .Notice also that any such u has n copies of each letter i , i = 1 , . . . , n . We call thewords inserting to this choice of P square words .On the other hand, if we use mixed insertion on w , we find P ′ as follows (with n = 4): 1 1 1 1 2 ′ ′ ′ ′ ′ ′ . T. K. PETERSEN AND L. SERRANO
In general, on the “shifted half” of the tableau we see all 1s in the first row, all 2sin the second row, and so on. In the “straight half” we see only prime numbers,with 2 ′ on the first diagonal, 3 ′ on the second diagonal, and so on. Lemma 6 tellsus that every u that mixed inserts to P ′ is a square word. But since the sets ofrecording tableaux for P and for P ′ are equinumerous, we see that the set of wordsmixed inserting to P ′ is precisely the set of all square words. Remark 7.
Yamanouchi words give a bijection with square standard Young tableauxthat circumvents insertion completely. In reading the word from left to right, if w i = j , we put letter i in the leftmost unoccupied position of row n + 1 − j . (See [Sta, Proposition 7.10.3(d)] .) We will soon characterize promotion in terms of operators on insertion words.First, some lemmas.For a tableau T (shifted or not) let ∆ T denote the result of all but step (3) ofpromotion. That is, we delete the smallest entry and perform jeu de taquin, butwe do not fill in the empty box. The following lemma says that, in both the shiftedand unshifted cases, this can be expressed very simply in terms of our insertionword. The first part of the lemma is a direct application of the theory of jeu detaquin (see, e.g., [Sta, A1.2]); the second part is [Ser, Lemma 3.9]. Lemma 8.
For a word w = w w · · · w l , let b w = w · · · w l . Then we have Q ( b w ) = ∆ Q ( w ) , and Q ′ ( b w ) = ∆ Q ′ ( w ) . The operator e j acting on words w = w · · · w l is defined in the following way.Consider the subword of w formed only by the letters j and j + 1. Consider every j + 1 as an opening bracket and every j as a closing bracket, and pair them upaccordingly. The remaining word is of the form j r ( j + 1) s . The operator e j leavesall of w invariant, except for this subword, which it changes to j r − ( j +1) s +1 . (Thisoperator is widely used in the theory of crystal graphs .)As an example, we calculate e ( w ) for the word w = 3121221332. The subwordformed from the letters 3 and 2 is 3 · · · , which corresponds to the bracket sequence ()))((). Removing paired brackets, oneobtains ))(, corresponding to the subword · · · · · · · . We change the last 2 to a 3 and keep the rest of the word unchanged, obtaining e ( w ) = 31212 Lemma 9.
Recording tableaux are invariant under the operators e i . That is, Q ( e i ( w )) = Q ( w ) , YCLIC SIEVING IN THE HYPEROCTAHEDRAL GROUP 7 and Q ′ ( e i ( w )) = Q ′ ( w ) . Let e = e · · · e n − denote the composite operator given by applying first e n − ,then e n − and so on. It is clear that if w = w · · · w n is a square word, then e ( b w )1is again a square word. Theorem 10.
Let w = w · · · w n be a square word. Then, p ( Q ( w )) = Q ( e ( b w )1) , and p ( Q ′ ( w )) = Q ′ ( e ( b w )1) . In other words, Haiman’s bijection commutes with promotion: p ( H ( Q )) = H ( p ( Q )) . Proof.
By Lemma 8, we see that Q ( b w ) is only one box away from p ( Q ( w )). Further,repeated application of Lemma 9 shows that Q ( b w ) = Q ( e n − ( b w )) = Q ( e n − ( e n − ( b w ))) = · · · = Q ( e ( b w )) . The same lemmas apply show Q ′ ( e ( b w )) is one box away from p ( Q ′ ( w )).All that remains is to check that the box added by inserting 1 into P ( e ( b w ))(resp. P ′ ( e ( b w ))) is in the correct position. But this follows from the observationthat e ( b w )1 is a square word, and square words insert (resp. mixed insert) to squares(resp. doubled staircases). (cid:3) Rhoades’ result
Rhoades [Rh] proved an instance of the CSP related to the action of promotion onrectangular tableaux. His result is quite deep, employing Kahzdan-Lusztig cellularrepresentation theory in its proof.Recall that for any partition λ ⊢ n , we have that the standard tableaux of shape λ are enumerated by the Frame-Robinson-Thrall hook length formula : f λ = | SY T ( λ ) | = n ! Q ( i,j ) ∈ λ h ij , where the product is over the boxes ( i, j ) in λ and h ij is the hook length at the box( i, j ), i.e., the number of boxes directly east or south of the box ( i, j ) in λ , countingitself exactly once. To obtain the polynomial used for cyclic sieving, we replace thehook length formula with a natural q -analogue. First, recall that for any n ∈ N ,[ n ] q := 1 + q + · · · + q n − and [ n ] q ! := [ n ] q [ n − q · · · [1] q . Theorem 11 ([Rh], Theorem 3.9) . Let λ ⊢ N be a rectangular shape and let X = SY T ( λ ) . Let C := Z /N Z act on X via promotion. Then, the triple ( X, C, X ( q )) exhibits the cyclic sieving phenomenon, where X ( q ) = [ N ] q !Π ( i,j ) ∈ λ [ h ij ] q is the q -analogue of the hook length formula. Now thanks to Theorem 10 we know that H preserves orbits of promotion, andas a consequence we see the CSP for doubled staircases. T. K. PETERSEN AND L. SERRANO
Corollary 12.
Let X = SY T ′ (2 n − , n − , . . . , , and let C := Z /n Z act on X via promotion. Then the triple ( X, C, X ( q )) exhibits the cyclic sieving phenomenon,where X ( q ) = [ n ] q ![ n ] nq Q n − i =1 ([ i ] q · [2 n − i ] q ) i is the q -analogue of the hook length formula for an n × n square Young diagram. Because of Theorem 3 the set R ( w ) also exhibits the CSP. Corollary 13 ([Rh], Theorem 8.1) . Let X = R ( w ) and let X ( q ) as in Corol-lary 12. Let C := Z /n Z act on X by cyclic rotation of words. Then the triple ( X, C, X ( q )) exhibits the cyclic sieving phenomenon. Corollary 13 is the CSP for R ( w ) as stated by Rhoades. This is nearly our mainresult (Theorem 1), but for the definition of X ( q ).In spirit, if ( X, C, X ( q )) exhibits the CSP, the polynomial X ( q ) should be some q -enumerator for the set X . That is, it should be expressible as X ( q ) = X x ∈ X q s ( x ) , where s is an intrinsically defined statistic for the elements of X . Indeed, nearly allknown instances of the cyclic sieving phenomenon have this property. For example,it is known ([Sta, Cor 7.21.5]) that the q -analogue of the hook-length formula canbe expressed as follows:(2) f λ ( q ) = q − κ ( λ ) X T ∈ SY T ( λ ) q maj( T ) , where κ ( λ , . . . , λ l ) = P ≤ i ≤ l ( i − λ i and for a tableau T , maj( T ) is the sum ofall i such that i appears in a row above i + 1. Thus X ( q ) in Theorem 11 can bedescribed in terms a statistic on Young tableaux.With this point of view, Corollaries 12 and 13 are aesthetically unsatisfying.Section 5 is given to showing that X ( q ) can be defined as the generating function forthe major index on words in R ( w ). It would be interesting to find a combinatorialdescription for X ( q ) in terms of a statistic on SY T ′ (2 n − , n − , . . . ,
1) as well,though we have no such description at present.5.
Combinatorial description of X ( q )As stated in the introduction, we will show that X ( q ) = q − n ( n ) X w ∈ R ( w ) q maj( w ) . If we specialize (2) to square shapes, we see that κ ( n n ) = n (cid:0) n (cid:1) and X ( q ) = q − n ( n ) X T ∈ SY T ( n n ) q maj( T ) . Thus it suffices to exhibit a bijection between square tableaux and words in R ( w )that preserves major index. In fact, the composition Ψ := Φ H has a strongerfeature.Define the cyclic descent set of a word w = w · · · w l to be the set D ( w ) = { i : w i > w i +1 } (mod l ) YCLIC SIEVING IN THE HYPEROCTAHEDRAL GROUP 9
That is, we have descents in the usual way, but also a descent in position 0 if w l > w . Then maj( w ) = P i ∈ D ( w ) i . For example with w = 132132132, D ( w ) = { , , , , , } and maj( w ) = 0 + 2 + 3 + 5 + 6 + 8 = 24.Similarly, we follow [Rh] in defining the cyclic descent set of a square (in general,rectangular) Young tableau. For T in SY T ( n n ), define D ( T ) to be the set of all i such that i appears in a row above i + 1, along with 0 if n − n in p ( T ).Major index is maj( T ) = P i ∈ D ( T ) i . We will see that Ψ preserves cyclic descentsets, and hence, major index. Using our earlier example of w = 132132132, one cancheck that T = Ψ − ( w ) = 1 2 53 6 84 7 9has D ( T ) = D ( w ), and so maj( T ) = maj( w ). Lemma 14.
Let T ∈ SY T ( n n ) , and let w = Ψ( T ) in R ( w ) . Then D ( T ) = D ( w ) .Proof. First, we observe that both types of descent sets shift cyclically under theirrespective actions: D ( p ( T )) = { i − n ) : i ∈ D ( T ) } , and D ( c ( w )) = { i − n ) : i ∈ D ( w ) } . For words under cyclic rotation, this is obvious. For tableaux under promotion,this is a lemma of Rhoades [Rh, Lemma 3.3].Because of this cyclic shifting, we see that i ∈ D ( T ) if and only if 0 ∈ D ( p i ( T )).Thus, it suffices to show that 0 ∈ D ( T ) if and only if 0 ∈ D ( w ). (Actually, it iseasier to determine if n − S = Φ − ( w ) be the shifted doubled staircase tableau corresponding to w .We have n − ∈ D ( w ) if and only if n is in a higher row in p − ( S ) than in S . Butsince n occupies the same place in p − ( S ) as n − S , this is to say n − n in S . On the other hand, n − ∈ D ( T ) if and only if n − n in T . It is straightforward to check that since S is obtained from T by jeude taquin into the upper corner, the relative heights of n and n − n is below or not) are the same in S as in T . This completes the proof. (cid:3) This lemma yields the desired result for X ( q ). Theorem 15.
The q -analogue of the hook length formula for an n × n squareYoung diagram is, up to a shift, the major index generating function for reducedexpressions of the longest element in the hyperoctahedral group: X w ∈ R ( w ) q maj( w ) = q n ( n ) · [ n ] q ![ n ] nq Q n − i =1 ([ i ] q · [2 n − i ] q ) i . Theorem 15, along with Corollary 13, completes the proof of our main result,Theorem 1. Because this result can be stated purely in terms of the set R ( w )and a natural statistic on this set, it would be interesting to obtain a self-containedproof, i.e., one that does not appeal to Haiman’s or Rhoades’ work. Why must aresult about cyclic rotation of words rely on promotion of Young tableaux? References [EG]
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