Properties of the nonsymmetric Robinson-Schensted-Knuth algorithm

Transcription

1 Properties of the nonsymmetri Roinson-Shensted-Knuth algorithm James Haglund Department of Mathematis University of Pennsylvania, Philadelphia, PA 90. USA Sarah Mason Department of Mathematis Wake Forest University Winston-Salem, NC 709. USA Septemer 7, 0 Jeffrey Remmel Department of Mathematis University of California, San Diego La Jolla, CA USA jremmel@usd.edu Astrat We introdue a generalization of the Roinson-Shensted-Knuth insertion algorithm for semi-standard augmented fillings whose asement is an aritrary permutation σ S n. If σ is the identity, then our insertion algorithm redues to the insertion algorithm introdued y the seond author [Sém. Lothar. Comin. 57 (006/08), Art. B57e, pp.] for semi-standard augmented fillings and if σ is the reverse of the identity, then our insertion algorithm redues to the original Roinson-Shensted-Knuth row insertion algorithm. We use our generalized insertion algorithm to otain new deompositions of the Shur funtions into nonsymmetri elements alled generalized Demazure atoms (whih eome Demazure atoms when σ is the identity). Other appliations inlude Pieri rules for multiplying a generalized Demazure atom y a omplete homogeneous symmetri funtion or an elementary symmetri funtion, a generalization of Knuth s orrespondene etween matries of non-negative integers and pairs of taleaux, and a version of evauation for omposition taleaux whose asement is an aritrary permutation σ. Introdution Let N denote the set of natural numers {0,,,...} and P denote the set of positive integers {,,..., }. We say that γ = (γ, γ,..., γ n ) is a weak omposition of m into n parts if eah γ i N and n i= γ i = m. Letting γ = i γ i, the (olumn) diagram of γ is the figure dg (γ) onsisting of γ ells arranged into olumns so that the i th olumn ontains γ i ells. For example, the diagram of γ = (, 0,, 0, ) is pitured in Figure. The augmented diagram of γ, denoted y dg(γ), onsists of the diagram of γ together with an extra row of n ells attahed elow. These extra ells are referred to as the asement of the augmented diagram. We let λ(γ) e the partition that results y taking the weakly dereasing rearrangement of the parts of γ. Thus if γ = (, 0,, 0, ), then λ(γ) = (,,, 0, 0). Figure : The diagram of γ = (, 0,, 0, ). Madonald [7] defined a famous family of symmetri polynomials P λ (x, x,...,x n ; q, t), whih have important appliations to a variety of areas. In [6], Madonald showed that many of the properties of the P λ, suh as satisfying a multivariate orthogonality ondition, are shared y a family of nonsymmetri polynomials E γ (x,...,x n ; q, t), where γ is a weak omposition with n parts. Haglund, Haiman and Loehr [] otained a ominatorial formula for E γ (x,...,x n ; q, t) in terms of fillings of dg(γ) y positive integers satisfying ertain onstraints. It will e simpler for us to phrase things in terms of a transformed version of the E γ studied y Marshall [8] whih we denote y Êγ(x,...,x n ; q, t). The Êγ an e otained from the

2 E γ y sending q /q, t /t, reversing the x-variales, and reversing the parts of γ. The orresponding ominatorial expression for Êγ(x,...,x n ; 0; 0) from [] involves what the seond author [9], [0] later alled semi-standard augmented fillings. It was previously known that Êγ(x,...,x n ; 0, 0) (hereafter denoted more simply y Êγ(x,..., x n )), equals the standard ases of Lasoux and Shützenerger [5], whih are also referred to as Demazure atoms. The seond author introdued a generalization of the RSK insertion algorithm involving semi-standard augmented fillings, and used this to give ominatorial proofs of several results involving Demazure atoms. For example, this generalized RSK insertion algorithm gives a ijetive proof that for any partition β, s β (x,..., x n ) = γ λ(γ)=β Ê γ (x,..., x n ). () This extended Roinson-Shensted-Knuth insertion algorithm is also instrumental in work of Haglund, Luoto, Mason and van Willigenurg, who developed the theory of a new asis for the ring of quasisymmetri funtions alled quasisymmetri Shur funtions [], []. In partiular these authors use it in proving a generalization of the Littlewood-Rihardson rule, where the produt of a Shur funtion and a Demazure atom (Demazure harater, quasisymmetri Shur funtion) is expanded in terms of Demazure atoms (Demazure haraters, quasisymmetri Shur funtions), respetively, with positive oeffiients. Let ǫ n denote the identity n in S n and ǫ n the reverse of the identity n n. In [9], [0] and in [],[], the asements of the diagrams dg(γ) are always filled y either ǫ n (i.e., i is in the ith olumn of the asement), or y ǫ n. In this artile we show that many of the nie properties of the extended RSK insertion algorithm hold with the asement onsisting of an aritrary permutation σ S n. In partiular we define a weight preserving ijetion whih shows s β (x,...,x n ) = γ Ê σ γ (x,..., x n ) () where the sum is over all weak ompositions γ suh that λ(γ) = β and γ i γ j whenever i < j and σ i > σ j. Here Êσ γ (x,..., x n ) is the version of Êγ(x,..., x n ) with asement σ whih we all a generalized Demazure atom. In the speial ase when σ = ǫ n there is only one term in the sum aove so that s β = E ǫn β, while if σ is ǫ n then () redues to (). Part of our motivation for studying the Êσ γ (x,..., x n ) is an unpulished result of M. Haiman and the first author whih an e desried riefly as follows. Let Êσ γ (x,...,x n ; q, t) denote the polynomial otained y starting with the ominatorial formula from [] for Êγ(x,...,x n ; q, t) involving sums over nonattaking fillings, replaing the asement ǫ n y σ σ σ n, and keeping other aspets of the formula the same. Then if i + ours to the left of i in the asement σ σ σ n, we have T i Ê σ γ (x,...,x n ; q, t) = t A Ê σ γ (x,..., x n ; q, t). () Here A equals one if the height of the olumn of dg(γ) aove i + in the asement is greater than or equal to the height of the olumn aove i in the asement, and equals zero otherwise. Also, σ is the permutation otained y interhanging i and i + in σ. The T i are generators for the affine Heke algera whih at on monomials in the X variales y T i x λ = tx si(λ) + (t ) xλ x si(λ) x αi, with x αi = x i /x i+. See [] for a more detailed desription of the T i and their relevane to nonsymmetri Madonald polynomials. Our Êσ γ (x,...,x n ) an e otained y setting q = t = 0 in Êσ γ (x,...,x n ; q, t), and hene are a natural generalization of the Êγ(x,...,x n ) to investigate. If we set q = t = 0 in the Heke operator T i, it redues to a divided differene operator similar to those appearing in the definition of Shuert polynomials. By (), Êσ γ (x,...,x n ) an e expressed (up to a power of t) as a series of the divided differene operators applied to the Demazure harater Ê ǫn γ (x,..., x n ).

3 As with the extended insertion algorithm, we shall see that our insertion algorithm with general asements also ommutes in a natural way with the RSK insertion algorithm. This useful fat will allow us to extend the results of the seond author to our more general setup. Moreover, we shall give a preise haraterization of how the results of our insertion algorithm vary as the asement σ varies. If σ = ǫ n our algorithm eomes essentially equivalent to the ordinary RSK row insertion algorithm, while if σ = ǫ n, it redues to the extended insertion algorithm. The outline of this paper is as follows. In setion, we formally define the ojets we will e working with, namely permuted asement semi-standard augmented fillings relative to a permutation σ (PBFs). In setions and, we desrie our insertion algorithm for PBFs and derive its general properties. In setion 5, we use it to prove analogues of the Pieri rules for the produt of a homogeneous symmetri funtion h n (x,..., x n ) times an Êσ γ (x,..., x n ) and the produt of an elementary symmetri funtion e n (x,...,x n ) times an Êσ γ (x,..., x n ). In setion 6, we define a generalization of the RSK orrespondene etween N- valued matries and pairs of olumn strit taleaux for permuted asement fillings and prove several of its asi properties. Finally, in setion 7 we study the analogue of evauation for PBF s. Permuted asement semi-standard augmented fillings. The positive integer n is fixed throughout, while γ will always denote a weak omposition into n parts and σ a permutation in S n. We let (i, j) denote the ell in the i-th olumn, reading from left to right, and the j-th row, reading from ottom to top, of dg(γ). The asement ells of dg(γ) are onsidered to e in row 0 so that dg(γ) = dg (γ) {(i, 0) : i n}. The reading order of the ells of dg(γ) is otained y reading the ells in rows from left to right, eginning at the highest row and reading from top to ottom. Thus a ell a = (i, j) is less than a ell = (i, j ) in the reading order if either j > j or j = j and i < i. For example, if γ = (0,, 0,,,, 0, 0, ), then dg(γ) is pitured in Figure where we have plaed the numer i in the i-th ell in reading order. An augmented filling, F, of an augmented diagram dg(γ) is a funtion F : dg(γ) P, whih we piture as an assignment of positive integers to the ells of dg(γ). We let F(i, j) denote the entry in ell (i, j) of F. The reading word of F, read(f), is otained y reording the entries of F in the reading order of dg (γ). The ontent of F is the multiset of entries whih appear in the filling F. Throughout this artile, we will only e interested in fillings F suh that entries in eah olumn are weakly inreasing reading from top to ottom and the asement entries form a permutation in the symmetri group S n Figure : The reading word order of the ells of the augmented oard for γ = (0,, 0,,,, 0, 0, ). Next we define type A and B triples as in [9]. A type A triple in an augmented diagram of shape γ is a set of three ells a,, of the form (i, k), (j, k), (i, k ) for some pair of olumns i < j of the diagram and some row k > 0, where γ i γ j. A type B triple is a set of three ells a,, of the form (j, k + ), (i, k), (j, k) for some pair of olumns i < j of the diagram and some row k 0, where γ i < γ j. Note that asement ells an e elements of triples. As noted aove, in this artile our fillings F have weakly inreasing olumn entries reading from top to ottom, so we always have the entry values satisfying F(a) F(). We say that a triple of either type is an inversion triple if the relative order of the entries is either F() < F(a) F() or F(a) F() < F(). Otherwise we say that the triple is a oinversion triple, i.e. if F(a) F() F(). Figure pitures type A and B triples. A semi-standard augmented filling is a filling of an augmented diagram with positive integer entries so that (i) the olumn entries are weakly inreasing from top to ottom, (ii) the asement entries form a permutation of,,..., n where n is the numer of ells in the asement, and (iii) every Type A or B triple is an inversion triple. We say that ells = (x, y ) and = (x, y ) are attaking if either and lie in the same row, i.e. y = y, or if lies stritly to the left and one row elow, i.e. if x < x and

4 Type A triple Type B triple a a olumn i olumn j olumn i olumn j γ i > γ j γ i < γ j Figure : Type A and B triples. y = y +. We say that filling F is non-attaking if F( ) F( ) whenever and are attaking. It is easy to see from our definition of inversion triples that a semi-standard augmented filling F must e non-attaking. A supersript σ on a filling F, as in F σ, means the asement entries form the permutation σ. We say that a filling F σ is a permuted asement semi-standard augmented filling (PBF) of shape γ with asement permutation σ if (I) F σ is a semi-standard augmented filling of dg(γ), (II) F σ ((i, 0)) = σ i for i =,...,n, and (III) for all ells a = (i, j), = (i, j ) suh that i < i and γ i < γ i, we have F σ () < F σ (a). We shall all ondition (III) the B-inreasing ondition, as pitured in Figure. We note that the fat that a PBF F σ has weakly inreasing olumns, reading from top to ottom, and satisfies the B-inreasing ondition automatially implies that every B-triple in F σ is an inversion triple. That is, suppose that γ i < γ j where i < j and a = (j, k + ), = (i, k) and = (j, k) is B-triple. Then F σ () < F σ (a) F σ () sine the B-inreasing ondition fores F σ () < F σ (a) and the weakly inreasing olumn ondition fores F σ (a) F σ (). Thus {a,, } is an inversion triple. row j row j a F σ () < F σ (a) olumn i olumn i γ i < γ i Figure : The B-inreasing ondition for F σ. Given a PBF F σ of shape γ, we define the weight of F σ, W(F σ ), to e W(F σ ) = x F σ (i,j). () (i,j) dg (γ) We let PBF(γ, σ) denote the set of all PBFs F σ of shape γ with asement σ. We then define Êγ σ (x, x,...,x n ) = W(F σ ). (5) F σ PBF(γ,σ) The following fat aout PBFs will e used frequently in the sequel. Lemma. Let F σ e a PBF of shape γ and assume that i < m. (i) Suppose that F σ (i, j) < F σ (m, j) for some j > 0. Then F σ (i, j ) < F σ (m, j). Moreover, for all 0 k < j, F σ (i, k) < F σ (m, k + ) F σ (m, k).

5 (ii) Suppose that F σ (i, j) > F σ (m, j) for some j 0. Then γ i γ m and, for all j k γ m, F σ (i, k) > F σ (m, k). Proof. For (i), we onsider two ases. First if γ i < γ m, then the B-inreasing ondition fores F σ (i, j ) < F σ (m, j). Seond, if γ i γ m, then onsider the A-triple a = (i, j), = (m, j), and = (i, j ). As we are assuming that F σ (a) < F σ (), it must e the ase that F σ (i, j ) = F σ () < F σ () = F σ (m, j) sine otherwise {a,, } would e oinversion triple in F σ. Thus it always the ase that F σ (i, j ) < F σ (m, j). But then we know that F σ (i, j ) < F σ (m, i) F σ (m, j ) so that F σ (i, j ) < F σ (m, j ). Thus we an repeat our argument to show that for all 0 k < j, F σ (i, k) < F σ (m, k + ) F σ (m, k). For (ii), suppose that F σ (i, j) > F σ (m, j). Then we laim that it annot e the ase that γ i < γ m sine otherwise (m, j + ) must e a ell in F σ whih would mean that F σ (i, j) > F σ (m, j) F σ (m, j + ). But then a = (m, j + ) and = (i, j) would violate the B-inreasing ondition. Thus it must e the ase that γ i γ m. We laim that it also must e the ase that F σ (i, k) > F σ (m, k) for all j < k γ m. If this is not the ase, then let k e the smallest l suh that l j and F σ (i, l) F σ (m, l). This implies the triple {(i, k), (m, k), (i, k )} is a type A oinversion triple sine F σ (i, k) F σ (m, k) F σ (m, k ) < F σ (i, k ). Sine we are assuming that F σ has no type A oinversion triples, there an e no suh k. Note that part (ii) of Lemma tells us that the asement permutation σ restrits the possile shapes of a PBF F σ with asement σ. That is, if σ i > σ m, then it must e the ase that height of olumn i in F σ is greater than or equal to the height of olumn m in F σ. We end this setion y onsidering the two speial ases of PBFs where the asement is either the identity or the reverse of the identity. In the speial ase where the asement permutation σ = ǫ n, a PBF is a semi-standard augmented filling as defined in [9]. Next onsider the ase where F ǫn is a PBF of shape γ = (γ,...,γ n ) with asement ǫ n. In that ase, Lemma implies that γ γ γ n and that F σ must e stritly dereasing in rows. Sine the entries of F ǫn must weakly derease in olumns reading from ottom to top, we see that F ǫn is what ould e alled a reverse row strit taleau with asement ǫ n attahed. It follows that for γ a partition, Ê ǫn γ (x, x,..., x n ) is equal to the Shur funtion s γ (x, x,...,x n ). An analogue of Shensted insertion In [9], the seond author defined a proedure k F to insert a positive integer k into a semi-skyline augmented filling, whih is a PBF with asement permutation equal to the identity. In this setion, we shall desrie an extension of this insertion proedure whih inserts a positive integer into a PBF with an aritrary asement permutation. Let F σ e a PBF with asement permutation σ S n. We shall define a proedure k F σ to insert a positive integer k into F σ. Let F σ e the extension of F σ whih first extends the asement permutation σ y adding j in ell (j, 0) for n < j k and then adds a ell whih ontains a 0 on top of eah olumn. Let (x, y ), (x, y ),... e the ells of this extended diagram listed in reading order. Formally, we shall define the insertion proedure of k (x, y ), (x, y ),... of k into the sequene of ells (x, y ), (x, y ),.... Let k 0 = k and look for the first i suh that F σ (x i, y i ) < k 0 F σ (x i, y i ). Then there are two ases. Case. If F σ (x i, y i ) = 0, then plae k 0 in ell (x i, y i ) and terminate the proedure. Case. If F σ (x i, y i ) 0, then plae k 0 in ell (x i, y i ), set k 0 := F σ (x i, y i ) and repeat the proedure y inserting k 0 into the the sequene of ells (x i+, y i+ ), (x i+, y i+ ),.... In suh a situation, we say that F σ (x i, y i ) was umped in the insertion k F σ. The output of k F σ is the filling that keeps only the ells that are filled with positive integers. That is, we remove any ells of F σ that still have a 0 in them. 5

6 The sequene of ells that ontain elements that were umped in the insertion k F σ plus the final ell whih is added when the proedure is terminated will e alled the umping path of the insertion. For example, Figure 5 shows an extended diagram of a PBF with asement permutation equal to 6 5. If we insert 5 into this PBF, then it is easy to see that the first element umped is the in olumn. Thus that will e replaed y 5 and we will insert into the remaining sequene of ells. The first element that an ump is the in olumn. Thus that will replae the in olumn and will e inserted in the remaining ells. But then that will ump the 0 in olumn 5 so that the proedure will terminate. Thus the irled elements in Figure 5 orrespond to the umping path of this insertion. Clearly, the entries of F σ in the umping path must stritly derease as we proeed in reading order = 6 5 Figure 5: The umping path of an insertion into a PBF. We note that if we try to insert 8 in to the PBF pitured in Figure 5, 8 would have no plae to go unless we reated extra olumns with asement entries 7 and 8. Thus in our ase, it is easy to see that inserting 8 into the PBF Figure 5 would give us the PBF pitured in Figure 6. For the rest of this paper, when we onsider an insertion k F σ, we will assume that σ S n where n is greater than or equal to k and all of the entries in F σ = Figure 6: Inserting 8 into the PBF of Figure 5. The following lemmas are needed in order to prove that the insertion proedure terminates and the result is a PBF. Lemma. Let = (i, j ) and = (i, j ) e two ells in a PBF F σ suh that F σ ( ) = F σ ( ) = a, assume appears efore in reading order, and no ell etween and in reading order ontains the entry a. Let = (i, j ) and = (i, j ) e the ells in k F σ ontaining the entries from and respetively. Then j > j. Proof. Consider the ell = (i, j ) immediately elow in the diagram F σ. Note that attaks all ells of F σ to its right that lie in same row as well as all ells to its left that lie one row elow the row of. Sine entries in ells whih are attaked y must e different from F σ ( ), it follows that must appear weakly after in reading order. If = = (i, j ), then the entry in ell annot e umped eause that would require F σ (i, j ) < k 0 F σ (i, j ). Thus either is not umped in whih ase the Lemma automatially holds or is umped in whih ase its entry ends up in a ell whih is later in reading order so that j = j > j j. Thus we may assume that F σ ( ) > F σ ( ) and that follows in reading order. This means that the element = (i, j + ) whih lies immediately aove follows in reading order and the entry in ell must e stritly less than a y our hoie of. If the entry in is not umped, then again we an onlude as aove that the entry in will end up in a ell whih follows in reading order so that again j = j > j j. Finally, suppose that the entry a in ell is umped. Sine F σ ( ) < a = F σ ( ), it follows that F σ ( ) is a andidate to e umped y a. Thus the a that was umped out of ell must end 6

7 up in a ell whih weakly preedes in reading order and hene it ends up in a row whih is higher than the row of. Sine the elements in a umping path stritly derease, the a in ell annot e part of the umping path. Thus the lemma holds. Lemma. Suppose that F σ is a PBF and k is a positive integer. Then every type A triple in k F σ is an inversion triple. Proof. Suppose that F σ is of shape γ = (γ,..., γ n ) where n k. Consider an aritrary type A triple {a = (x, y ), = (x, y ), = (x, y )} in F σ := k F σ. Suppose for a ontradition that {a,, } is a oinversion triple so that F σ (a) F σ () F σ (). Sine the entries in the umping path in the insertion k F σ form a stritly dereasing sequene when read in reading order, only one of {F σ (a), F σ (), F σ ()} an e umped y the insertion proedure k F σ. Let F σ e the extended diagram orresponding to F σ as defined in our definition of the insertion k F σ. We laim that the triple onditions for F σ imply that either F σ () < F σ (a) F σ () or F σ (a) F σ () < F σ (). This follows from the fat that F σ is a PBF if a,, are ells in F σ. Sine the shape of F σ arises from γ y adding a single ell on the outside of γ, we know that is a ell in F σ. However, it is possile that exatly one of a or is not in F σ and is filled with a 0 in F σ. If it is, then we automatially have F σ () < F σ (a) F σ (). If it is a, then the olumn that ontains a is stritly shorter than the olumn that ontains eause in F σ, it must e the ase that the height of olumn x is greater than or equal to the height of olumn x sine {a,, } is a type A triple in F σ. But then the B-inreasing ondition for F σ fores F σ () < F σ () and, hene, F σ (a) F σ () < F σ () must hold. We now onsider two ases. Case. F σ () < F σ (a) F σ (). Note in this ase, 0 < F σ (a) so that a is a ell in F σ. Moreover the entries in a and annot e umped in the insertion k F σ sine their replaement y a larger value would not produe the desired ordering F σ (a) F σ () F σ (). Thus it must e the ase that F σ () was umped in the insertion k F σ. We now onsider two suases. Suase.a. F σ (a) < F σ (). We know that F σ () umps F σ (). We wish to determine where F σ () ame from in the insertion proess k F σ. It annot e that F σ () = k or that it was umped from a ell that omes efore a in the reading order sine it would then meet the onditions to ump the entry F σ (a) in ell a as F σ (a) < F σ () F σ (). Thus it must have een umped from a ell after a ut efore in reading order. That is, F σ () = F σ (d) where d = (x, y ) and x < x < x. Thus we have the situation pitured in Figure 7. row y row y a d olumn x olumn x olumn x Figure 7: Piture for Suase.a. However, this is not possile sine if γ x γ x, then the entries in ells a, d, and would violate the A-triple ondition for F σ and, if γ x < γ x, then the entries in ells and d would violate the B-inreasing ondition on F σ. Suase. F σ (a) = F σ (). Again we must determine where F σ () ame from in the insertion proess k F σ. To this end, let r 7

8 e the least row suh that r > y and F σ (x, r) < F σ (x, r ). Then we will have the situation pitured in Figure 8 where d is the ell in olumn x and row r. Thus all the entries of F σ in the ells in olumn x etween a and d are equal to F σ (a). row r d < = = = = row y row y = a olumn x olumn x Figure 8: Piture for Suase.. Now the region of shaded ells pitured in Figure 8 are ells whih are attaked or attak some ell whih is equal to F σ (a) and hene their entries in F σ must all e different from F σ (a). Hene F σ () annot have ome from any of these ells sine we are assuming that F σ (a) = F σ (). Thus F σ () must have ome from a ell efore d in reading order. But this is also impossile eause F σ () would then meet the onditions to ump F σ (d) whih would violate our assumption that it umps F σ (). Case. F σ (a) F σ () < F σ (). The entry in ell is the only entry whih ould e umped in the insertion k F σ if we are to end up with the relative ordering F σ (a) F σ () F σ (). Sine F σ () is umped, this means that is not in the asement. But if we do not ump either a or in the insertion k F σ and a and are ells in F σ, it must e the ase that a and are ells in F σ and that there is no hange in the heights of olumns x and x. Thus γ x γ x. Let e the ell immediately elow and e the ell immediately elow. Thus we must have F σ () < F σ () F σ (). We now onsider two suases. Suase.a. F σ () = F σ (). Let r e the least row suh that r > y and F σ (x, r) < F σ (x, r ). Then we will have the situation pitured in Figure 9 where d is the ell in olumn x and row r. Thus all the entries of F σ in the ells on olumn x etween and d are equal to F σ (). Now the region of shaded ells pitured in Figure 9 are ells whih are attaked or attak some ell whih is equal to F σ () and hene their entries in F σ must all e different from F σ (). Thus F σ () annot have ome from any of these ells sine we are assuming that F σ () = F σ (). Hene F σ () must have ome from a ell efore d in reading order. But this is also impossile eause F σ () would then meet the onditions to ump F σ (d) whih would violate our assumption that it umps F σ (). Suase.. F σ () < F σ (). First onsider the A-triple,, in F σ. We annot have that F σ () < F σ () F σ () sine that would imply F σ () F σ () < F σ (), whih would violate our assumption that F σ (a) < F σ () < F σ (). Thus it must e the ase that F σ () F σ () < F σ (). But then we would have F σ () < F σ () F σ () < F σ () whih would mean that F σ () satisfies the onditions to ump F σ (). Sine it does not ump F σ (), it must e the ase that F σ () ame from a ell whih is after in the reading order. We now onsider two more 8

9 d < row r a = = = olumn x olumn x Figure 9: Piture for Suase.a. suases. Suase..i. F σ () is in the same row as F σ (). Assume that F σ () = F σ (d) where d = (x, y ) and x < x. It annot e that γ x < γ x sine then the B-inreasing ondition would fore that F σ () < F σ (d) = F σ (). But that would mean that F σ () < F σ () F σ () whih violates the fat that F σ () F σ () < F σ (). Thus it must e the ase that γ x γ x and, hene,,, d is a type A triple. As we annot have F σ () < F σ (d) = F σ (), it must e the ase that F σ () = F σ (d) < F σ () F σ (). But this is also impossile eause we are assuming that F σ () < F σ (). Suase..ii. F σ () is in the same row as F σ (). In this ase, let e,..., e s, e s+ = e the ells in the umping path of the insertion of k F σ in row y, reading from left to right. Thus we are assuming that F σ () = F σ (e s ). For eah e i, we let e i e the ell diretly elow e i and e i e the ell diretly aove e i. Thus we have the piture in Figure 0 where we are assuming that s = and we have irled the elements in the umping path. e e e e e e a e e e Figure 0: Piture for Suase..ii. Sine the elements in the umping path stritly derease, we have that F σ (e ) > > F σ (e s ) > F σ (e s+ ) = F σ () and that for eah i, F σ (e i+ ) < F σ (e i ) F σ (e i+ ). Let e j = (z j, y ). Thus z s+ = x. By Lemma, we must have γ z γ zs γ x. This means that the e i s are ells in F σ so that Lemma also implies that F σ (e ) > > F σ (e s ). Note that in this ase, we have γ x γ x so that we know that γ z γ zs γ x γ x. Now onsider the A triples {e i, e i, }. We are assuming that F σ () = F σ (e s+ ) F σ (e s+ ) = F σ () < F σ (). But sine F σ (e s+ ) < F σ (e s ) F σ (e s+ ), the {e s, e s, } A-triple ondition must e that F σ (e s ) F σ (e s ) < F σ (). Now if e s exists, then we know that F σ (e s ) < F σ (e s ) F σ (e s ) and, hene, the {e s, e s, } A-triple ondition must also e that F σ (e s ) F σ (e s ) < F σ (). If e s exists, then we know that F σ (e s ) < F σ (e s ) F σ (e s ) and, hene, the {e s, e s, } A-triple ondition must also e that F σ (e s ) F σ (e s ) < F σ (). Continuing on in this way, we an onlude that for all j, F σ (e j ) F σ (e j ) < F σ (). Next onsider the e i, e i, 9

10 A-triple onditions. We are assuming that F σ () < F σ () = F σ (e s ). Thus it must e the ase that F σ () < F σ (e s ) F σ (e s ). Sine F σ (e s ) < F σ (e s ) < < F σ (e ), it must e the ase that for all j, F σ () < F σ (e j ) F σ (e j ). Thus in this ase, we must have F σ () < F σ (e ) F σ (e ) F σ (e ) < F σ (). Now the question is where an the element z whih umps F σ (e ) ome from? We laim that z annot equal k or ome from a ell efore in reading order sine it satisfies the ondition to ump and is not umped. Thus it must have ome from a ell d = (x, y ) whih lies in the same row as ut omes after in reading order. In that ase, we must have F σ (e ) < F σ (d) F σ (e ) < F σ (). Thus it annot e that γ x < γ x sine the B-inreasing ondition would fore F σ () < F σ (d). Thus γ x γ x. But in that ase, we would have F σ () < F σ (d) < F σ () whih would e a oinversion A triple in F σ. Thus we have shown that in Suase, ould not have een umped and, hene, there an e no oinversion A triples in k F σ. It is ovious that our insertion algorithm ensures that the olumns of k F σ are weakly inreasing when read from top to ottom. Thus if we an show that k F σ satisfies the B-inreasing ondition, we know that all B triples in k F σ will e inversion triples. Lemma. If F σ is a PBF, then F σ = k F σ satisfies the B-inreasing ondition. Proof. Suppose that F σ is of shape γ = (γ,..., γ n ) where n k. Suppose that F σ does not satisfy the B-inreasing ondition. Thus there must e a type B triple { = (x, y ), a = (x, y + ), = (x, y )} in F σ := k F σ as depited in Figure suh that F σ () F σ (a). Assume that we have piked a and so that is as far left as possile. Let denote the ell immediately aove and denote the ell immediately elow. Then there are two possiilities, namely, it ould e that γ x < γ x so that {a,, } forms a type B triple in F σ or it ould e that γ x = γ x and we added an element on the top of olumn x during the insertion k F σ so that in F σ, the height of olumn x is stritly less than the height of olumn x. a row y olumn x olumn x Figure : A type B triple. Case. γ x < γ x. In this ase, the B-inreasing ondition for F σ implies that F σ () < F σ (a) and F σ () < F σ (). As the elements in the umping path stritly derease, it must e the ase that F σ () is umped and F σ (a) is not umped. Thus we must have that F σ () F σ () > F σ (). First we laim that we annot e the ase that F σ () = F σ (a). Otherwise, let r e the least row suh that r > y + and F σ (x, r) < F σ (x, r ). Then we will have the situation pitured in Figure where d is the ell in olumn x and row r. Thus all the entries of F σ in the ells in olumn x etween a and d are equal to F σ (a). Now the region of shaded ells pitured in Figure are ells whih are attaked or attak some ell whih is equal to F σ (a) and hene their entries in F σ must all e different from F σ (a). Hene F σ () annot have ome from any of these ells sine we are assuming that F σ (a) = F σ (). Thus F σ () must e either equal to k or have ome from a ell in F σ whih preedes d in reading order. But this is also impossile eause F σ () would then meet the onditions to ump F σ (d) whih would violate our assumption that it umps F σ (). Thus we an assume that F σ (a) < F σ (). Now the question is where did F σ () ome from? 0

11 row r d < = = = = row y row y = a olumn x olumn x Figure : The ells whih are attaked y ells equal to F σ (a). First it annot e that F σ () was either equal to k or was equal to F σ (d) where d omes efore a in reading order sine then we have that F σ (a) < F σ () < F σ () F σ () < F σ (). But this would mean that F σ () meets the ondition to ump F σ (a) whih would violate our assumption that F σ () umps F σ (). Similarly, it annot e the ase that F σ () = F σ (d) where d is a ell to the right of a and in the same row as a. That is, if d = (x, y + ) where x < x, then either (i) γ x < γ x in whih ase the fat that F σ () > F σ (d) = F σ () would mean that ells d and violate the B-inreasing ondition for F σ or (ii) γ x γ x in whih ase the triple {a,, d} would e a type A oinversion triple in F σ. Thus it must e the ase that F σ () ame from a ell to the left of and in the same row as in F σ. So let e,...,e s, e s+ = e the ells in the umping path of the insertion of k F σ in row y, reading from left to right. Thus we are assuming that F σ () = F σ (e s ). For eah e i, we let e i e the ell diretly elow e i. Thus we have the situation pitured in Figure where we are assuming that s = and we have irled the elements in the umping path. a e e e e e e Figure : Piture for F σ () is in the same row as. Sine the elements in the umping path stritly derease, we have that F σ (e ) > > F σ (e s ) > F σ (e s+ ) = F σ () > F σ (a). Moreover, for eah i s, we have F σ (e i+ ) < F σ (e i ) F σ (e i+ ). Let e j = (z j, y ) for j =,...,s +. Thus z s+ = x. By Lemma, we must have γ z γ zs γ x. Note that the fat that we hose to e as far left as possile means that it must e the ase that γ zj γ x for j s. That is, if for some j s, γ zj < γ x, then the entries in ells a and e j would violate the B-inreasing ondition in F σ whih would violate our hoie of. Thus {e j, e j, } is a type A triple for j s. Sine F σ () > F σ () = F σ (e s+ ) F σ (e s ), it must e the ase that the {, e s, e s } A triple ondition is F σ (e s ) F σ (e s ) < F σ (). Now assume y indution that we have shown that F σ (e j ) F σ (e j ) < F σ (). Then sine F σ (e j ) F σ (e j ), the {a, e j, e j } A triple ondition must e that F σ (e j ) F σ (e j ) < F σ (). It thus follows that F σ (e ) F σ (e ) < F σ (). Now the question is where did F σ (e ) ome from? Note that we have shown that F σ (a) < F σ (e ) < F σ (e ) F σ (e ) < F σ ().

12 Thus it annot e that F σ (e ) is equal to k or is equal to F σ (d) for some ell d whih preedes a in reading order sine then F σ (e ) would ump F σ (a). By our hoie of e, the only other possiility is that F σ (e ) = F σ (d) for some ell d to the right of a and in the same row as a. Say d = (x, y +) where x < x. Then it annot e that γ x < γ x sine then the ells d and would violate the B-inreasing ondition in F σ and it annot e that γ x γ x sine then the triple {a,, d} would e a type A oinversion triple in F σ. Thus we have shown that γ x < γ x is impossile. Case. γ x = γ x = y Thus we must have added an element on the top of olumn x during the insertion k F σ so that in F σ, the height of olumn x is stritly less than the height of olumn x. In this ase, neither nor were involved in the umping path of k F σ so that F σ () = F σ () and F σ () = F σ (). We laim that it must e the ase that F σ (x, y) F σ (x, y + ). That is, if y = y, then F σ (x, y) = F σ () F σ (a) = F σ (x, y + ) sine we are assuming that F σ () F σ (a). If y > y, then the triple {, a, } is a type A triple in F σ and F σ (a) = F σ (a). We now have two possiilities, namely, either (i) F σ (a) < F σ () F σ () or (ii) F σ () F σ () < F σ (a). Note that (ii) is inonsistent with our assumption that F σ () F σ (a) so that it must e the ase that F σ () > F σ (a). But then we know y part (ii) of Lemma that F σ (x, y) > F σ (x, y). Our insertion algorithm ensures that F σ (x, y) F σ (x, y + ) so that F σ (x, y) > F σ (x, y + ) in this ase. Now onsider the question of where F σ (x, y + ) ame from in the umping proess. It annot e the ase that F σ (x, y + ) = k or was umped from a ell efore (x, y + ) in the reading order eause then F σ (x, y + ) ould e plaed on top of F σ (x, y) and F σ (x, y + ) = 0 in this ase. Thus F σ (x, y + ) must have een umped from some ell d etween (x, y + ) and (x, y + ) in reading order. But this is impossile sine F σ (x, y) F σ (d) = F σ (x, y + ) would mean that (x, y) and d do not satisfy the B-inreasing ondition in F σ. Thus we have shown that the assumption that F σ () F σ (a) leads to a ontradition in all ases and, hene, F σ must satisfy the B-inreasing ondition. Proposition 5. The insertion proedure k F σ is well-defined and produes a PBF. Proof. Let F σ e an aritrary PBF of shape γ and asement σ S n and let k e an aritrary positive integer less than or equal to n. We must show that the proedure k F σ terminates and that the resulting filling is indeed a PBF. Lemma implies that at most one ourrene of any given value will e umped to the first row. Therefore eah entry i in the first row will e inserted into a olumn at or efore the olumn σ (i). This means that the insertion proedure terminates and hene is well-defined. Lemmas and imply that k F σ is a semi-standard augmented filling whih satisfies the B-inreasing ondition. Thus k F σ is a PBF. Before proeeding, we make two remarks. Our first remark is onerned with the proess of inverting our insertion proedure. That is, the last ell or terminal ell in the umping path of k F σ must e a ell that originally ontained 0 in F σ. Suh a ell was not in F σ so that the shape of F σ is the result of adding one new ell on the top of some olumn of the shape of F σ. However, there are restritions as to where this new ell may e plaed. That is, we have the following proposition whih says that if is the top ell of a olumn in a sequene of olumns whih have the same height in k F σ, then must e in the rightmost of those olumns. Proposition 6. Suppose that σ S n and F σ is a PBF with asement σ and k n. Suppose that F σ has shape γ = (γ,..., γ n ), k F σ has shape δ = (δ,...,δ n ), and (x, y) is the ell in δ/γ. Then it must e ase that if x < n, then + γ x γ x+j for j n x. In partiular, if x < n, then δ x δ x+j for j n x. Proof. Arguing for a ontradition, suppose that x < n and + γ x = γ x+j = y for some j suh that j n x. Let G σ = k F σ. and let F σ and Ḡσ e the fillings whih result y plaing 0 s on top of the olumns of F σ and G σ respetively. Thus we would have the situation pitured in Figure for the end of the umping path in the insertion k F σ.

13 In F σ In G σ a... row y ol. x ol. x+j ol. x ol. x+j Figure : The end of the umping path in k F σ. Hene is at the top of olumn x + j in oth F σ and G σ and neither F σ () nor F σ () are umped during the insertion of k F σ. Note that B-inreasing ondition in F σ fores that F σ () < F σ (). Thus the {a,, } A-triple ondition in G σ must e that G σ (a) G σ () < G σ (). Now onsider the question of where G σ (a) ame from in the umping path of the insertion k F σ. It annot e that G σ (a) = k or G σ (a) was umped from a ell efore (x + j, y + ) eause of the fat that G σ (a) < G σ () = F σ () would allow G σ (a) to e inserted on top of ell. Thus either (i) G σ (a) = F σ (z, y+) for some z > x + j or (ii) G σ (a) = F σ (z, y) for some z < x. Case (i) is impossile sine then we would have γ x+j < γ z and the B-inreasing ondition in F σ would fore G σ () = F σ () < F σ (z, y + ) = G σ (a). If ase (ii) holds, let e,..., e s, e s+ = (x, y) e the ells in row y of the umping path of the insertion of k F σ, reading from left to right. Thus we are assuming that G σ (a) = F σ (e s ). For eah e i, we let e i e the ell diretly elow e i. Thus we have the piture in Figure 5 where we are assuming that s = and we have irled the elements in the umping path. e e e 0 (x,y) 0 (x+j,y) e e e (x,y ) Figure 5: Piture for ase (ii). Sine the elements in the umping path stritly derease, we have that F σ (e ) > > F σ (e s ) = G σ (a) and that for eah i, F σ (e i+ ) < F σ (e i ) F σ (e i+ ). Let e j = (z j, y). Thus z s+ = x. It follows from Lemma that γ z γ zs > γ x. Now onsider the A-triples {e i, e i, (x + j, y)} for i =,...,s in F σ. We have estalished that F σ (e s ) = G σ (a) G σ () < G σ () = F σ (x + j, y). Thus it follows from the {e s, e s, (x + j, y)} A-triple ondition that F σ (e s ) F σ (e s ) < F σ (). But then F σ (e s ) < F σ (e s ) F σ (e s ) so that the {e s, e s, (x+j, y)} A-triple ondition also implies that F σ (e s ) F σ (e s ) < F σ (). Continuing on in this way, we an onlude from the {e i, e i, (x + j, y)} A-triple ondition that F σ (e i ) F σ (e i ) < F σ () for i =,...,s. Now onsider the element z that umps F σ (e ) in the insertion k F σ. We must have F σ (e ) < z F σ (e ) < F σ (). Thus it annot e that z = k or z = F σ (d) for some ell d whih preedes (x + j, y + ) in reading order eause that would mean that z meets the onditions to e plaed on top of. Thus it must e that z = F σ (d) for some ell d whih follows (x + j, y + ) in reading order. Suppose that d = (t, y + ) where t > x + j. But we are assuming that (x + j, y) is the top ell in olumn x + j. Thus it must e the ase that γ x+j < γ t. But then the B-inreasing ondition in F σ would fore F σ () < F σ (d) = z whih is a ontradition. Thus ase (ii) annot hold either whih implies + γ x γ x+j. Exept for the restritions determined y Proposition 6, we an invert the insertion proedure. That is, to invert the proedure k F σ, egin with the entry r j ontained in the new ell appended to F σ and read

14 akward through the reading order eginning with this ell until an entry is found whih is greater than r j and immediately elow an entry less than or equal to r j. Let this entry e r j, and repeat. When the first ell of k F σ is passed, the resulting entry is r = k and the proedure has een inverted. Our seond remark onerns the speial ase where σ = ǫ n and k n. In that ase, we laim that our insertion proedure is just a twisted version of the usual RSK row insertion algorithm. That is, we know that F σ must e of shape γ = (γ,...,γ n ) where γ γ γ n and that F σ is weakly dereasing in olumns, reading from ottom to top, and is stritly dereasing in rows, reading from left to right. Now if k F σ (, γ ), then we just add k to the top of olumn to form k F σ. Otherwise suppose that F σ (, y ) k > F σ (, y + ). Then all the elements in F σ that lie weakly aove row y + and stritly to the right of olumn must e less than or equal to F σ (, y +). Thus the first plae that we an insert k is in ell (, y + ). Thus it will e that ase that k umps F σ (, y + ). Sine elements in the umping path are dereasing and all the elements in olumn elow row y + are stritly larger than F σ (, y + ), it follows that none of them an an e involved in the umping path of the insertion k F σ. It is then easy to hek that sine F σ (, y + ) n, the result of the insertion k F σ is the same as the result of the insertion of F σ (, y + ) into the PBF formed from F σ y removing the first olumn and then adding ak olumn of F σ with F σ (, y + ) replaed y k. Thus our insertion proess satisfies the usual reursive definition of the RSK row insertion algorithm. Hene, in the speial ase where the asement permutation is ǫ n and k n, our insertion algorithm is just the usual RSK row insertion algorithm sujet to the ondition that we have weakly dereasing olumns and stritly dereasing rows. General Properties of the insertion algorithm In this setion, we shall prove several fundamental properties of the insertion algorithm k F σ. In partiular, our results in this setion will allow us to prove that our insertion algorithm an e fatored through the twisted version of RSK row insertion desried in the previous setion. For any permutation σ, let E σ e the empty filling whih just onsists of the asement whose entries are σ,...,σ n reading from left to right. Let s i denote the transposition (i, i + ) so that if σ = σ... σ n, then s i σ = σ... σ i σ i+ σ i σ i+... σ n. Our next lemma will desrie the differene etween inserting a word w into E σ versus inserting w into E siσ. If w = w... w t, then let w E σ = w t (... (w (w E σ ))...). Theorem 7. Let w e an aritrary word whose letters are less than or equal to n and suppose that σ = σ...σ n is a permutation in S n suh that σ i < σ i+. Let F σ = w E σ, γ = (γ,...,γ n ) e the shape of F σ, F siσ = w E siσ, and δ = (δ,...,δ n ) e the shape of F siσ. Then. {γ i, γ i+ } = {δ i, δ i+ } and δ i δ i+,. F siσ (i, j) > F siσ (i +, j), for j δ i where we let F siσ (i +, j) = 0 if (i +, j) is not a ell in F σ.. F siσ (j, k) = F σ (j, k) for j i, i + so that γ j = δ j for all j i, i +,. for all j, {F siσ (i, j), F siσ (i +, j)} = {F σ (i, j), F σ (i +, j)}. Proof. Note that sine (s i σ) i = σ i+ > σ i = (s i σ) i+, Lemma implies () and (). Thus we need only prove () and (). We proeed y indution on the length of w. The theorem learly holds when w is the empty word. Now suppose that the theorem holds for all words of length less than t. Then let G = w... w t E σ and H = w... w t E siσ and suppose G has shape α = (α,..., α n ) and H has shape β = (β,..., β n ). Let Ḡ and H e the fillings with 0 s added to the tops of the olumns of G and H respetively. Let G = w t G and H = w t H and suppose that G has shape γ = (γ,...,γ n ) and H has shape δ = (δ,..., δ n ). We ompare the umping path of w t H to the umping path in w t G. That is, in the insertion proess w t H, suppose we ome to a point were we are inserting some element whih is either w t or some

15 element umped in the insertion w t H into the ells (i, j) and (i+, j). Assume y a seond inner reverse indution on the size of i, that the insertion of w t G will also insert into the ells (i, j) and (i+, j). This will ertainly e true the first time the umping paths interat with elements in olumns i and i + sine our indution assumption ensures that Ḡ restrited to olumns,...,i equals H restrited to olumns,...,i. Let x = H(i, j), y = H(i +, j), x = H(i, j ), and y = H(i +, j ). (See Figure 6.) Our indutive assumption implies that if x > 0, then x > y and if x > 0, then x > y. Our goal is to analyze how the insertion of interats with elements in ells (i, j) and (i +, j) during the insertions w t H and w t G. We will show that either (A) the umping path does not interat with ells (i, j) and (i +, j) during either the insertions w t H or w t G, (B) the insertion of into ells (i, j) and (i +, j) results in inserting some into the next ell in reading order after (i +, j) in oth w t H and w t G, or (C) oth insertions end up terminating in one of (i, j) or (i +, j). This will ensure that w t H and w t G are idential outside of olumns i and i+ thus proving ondition () of the theorem and that {H(i, j), H(i +, j)} = {G(i, j), G(i +, j)} whih will prove ondition () of the theorem. Now suppose that the elements H are in ells (i, j), (i +, j), (i, j ), and (i +, j ) are x, y, x and y, respetively as pitured on the left in Figure 6. If Ḡ and H agree on those ells, then there is nothing to prove. Thus we have to onsider three ases (I), (II), or (III) for the entries in Ḡ in those ells whih are pitured on the right in Figure 6. We an assume that x 0. Now if y = y = 0, then it is easy to see that one of (A), (B), or (C) will hold sine the insertion proedure sees the same elements possily in different olumns. Thus we an assume that y 0 and hene, x > y. I II III row j row j x x y y y x x y y x x y y x y x ol. ol. i i+ ol. ol. i i+ ol. ol. i i+ ol. ol. i i+ in H in G in G in G Figure 6: Possiilities in Ḡ. We now onsider several ases. Case A. x = y = 0. This means that x and y sit on top of olumns i and i + respetively in H. First suppose that x. Then in w t H, the insertion will terminate y putting on top of x. In ase (I), the insertion w t G will terminate y plaing on top of x and in ases (II) and (III), the insertion w t G will terminate y plaing on top of y if y, or y plaing on top of x if y < x. In either situation, (C) holds. Next suppose that x <, Then in w t H, will not e plaed in either ell (i, j) or (i+, j) so that the result is that will end up eing inserted in the next ell in reading order after (i+, j). But then in ases (I), (II), and (III), will not e plaed in either ell (i, j) or (i +, j) in the insertion w t G so that the result is that will end up eing inserted in the next ell in reading order after (i+, j). Thus (B) holds in all ases. Case B. x > 0, and y = 0. Note that ase (I) is impossile sine then x and x would violate the B-inreasing ondition in G. First onsider the ase where does not ump x in the insertion w t H, and the insertion terminates with eing plaed on top of y. Then it must e the ase y. Moreover, it is the ase that x > y Lemma. Hene in ase (II), will not ump x and instead will e plaed on top of x sine y < x and in ase (III), will e put on top of y. However, this is not possile eause then the insertion would violate 5

16 Proposition 6. Thus, we know that ondition (C) holds. Next onsider the ase where in the insertion w t H, umps x and x terminates the insertion y eing plaed on top of y. Thus we know that x < x and x y. This rules out ase (III) sine then x and y would violate the B-inreasing ondition and ase (I) sine then x and x would violate the B-inreasing ondition. Now in ase (II), will ump x and x will e plaed on top of x if y. If > y, then will not ump x and will e plaed on top of x. In either situation, ondition (C) holds. Next onsider the ase where in the insertion w t H, umps x and x annot e plaed on top of y so that x is inserted in the next ell in reading order after (i +, j). Then we must have y < x < x. This rules out ases (I) and (II) sine x annot sit on top of y. In ase (III), annot sit on top of y so will ump x. Thus ondition (B) holds in this ase. Finally onsider the ase where does not ump x and annot e plaed on top of y in the insertion w t H so that is inserted in the next ell in reading order after (i +, j). The fat that does not ump x means that either > x or x. The fat that annot e plaed on top of y means that > y. If > x > y, then in ases (II) and (III), does not meet the onditions for the entries in ells (i, j) and (i+, j) to hange so that the result is that will e inserted in the next ell in reading order after (i+, j). If x, then we know that y < x. This rules out ases (I) and (II) sine x annot sit on y. In ase (III), annot e plaed on y and annot ump x so that Case A holds in either situation. Case C. x, x, y, y > 0. Then we know that x > y and x > y. First suppose that in the insertion w t H, umps x, ut x does not ump y so that the result is that x will e inserted into the ell following (i +, j) in reading order. Sine y < x, the reason that x does not ump y must e that x > y. Thus it must e the ase that x x > y y. This means that ases (I) and (II) are impossile sine x annot sit on top of y in G. But then > x > y so that in the insertion w t G, annot ump y in ase (III). Thus in ase (III), will ump x so that the result is that x will e inserted into the ell following (i +, j) in reading order as desired. Hene, ondition (B) holds in this ase. Next onsider the ase where does not ump x ut umps y. Sine does not ump x then we either have (i) > x or (ii) x. If (i) holds, then > x > y whih means that annot ump y. Thus (ii) must hold. Sine umps y, y < y. Thus we have two possiilities, namely, y < y < x or y < x y. First suppose that y < y < x. Then ases (I) and (II) are impossile sine x annot sit on top of y. In ase (III), will ump y ut y annot ump x sine y < x so that y is inserted in the next ell after (i+, j). Next suppose that y < x y. Then in ase (I), will ump ut y annot ump x sine y < x so that y is inserted in the next ell after (i +, j). In ase (II), does not ump x sine x so that will ump y and y will e inserted in the next ell after (i +, j). In ase (III), will ump y ut y annot ump x sine y < x so that y is inserted in the next ell after (i +, j). Thus in every, y will e inserted in the next ell after (i +, j). Hene, ondition (B) holds in this ase. Next onsider the ase where in the insertion w t H, umps x and then x umps y so that the result is that y will e inserted into the ell following (i +, j) in reading order. In this ase we must have y < x y < x and x < x. In ase (I), it is easy to see that in the insertion w t G, will ump y sine y < x < x, ut y will not ump x so that the result is that y will e inserted into the ell following (i +, j) in reading order. In ase (II), will ump x and then x will ump y if y. However if > y, then will not ump x ut it will ump y. Thus in either situation, the result is that y will e inserted into the ell following (i +, j) in reading order. Finally onsider ase (III). If y, then will ump y ut y will not ump x so that again the result is that y will e inserted into the ell following (i +, j) in reading order. Now if > y, then we must have that y < x y < x. We laim that this is impossile. Reall that α i and α i+ are the heights of olumn i and i + in G, respetively. Now if α i α i+, then {G(i, j) = y, G(i, j ) = y, G(i +, j) = x} would e a type A oinversion triple in G and if α i < α i+, then {G(i, j ) = y, G(i +, j) = x, G(i +, j ) = x} would e a type B oinversion triple in G. Finally onsider the ase where does not ump either x or y in the insertion w t H so that is inserted into the ells following (i +, j) in reading order. Then either y < x so that annot ump either x or y in ases (I)-(III) or > x > y so again annot ump either x or y in ases (I)-(III). Thus in all ases, the result is that will e inserted into the ells following (i +, j) in reading order. Thus we have shown that onditions (A), (B), and (C) always holds whih, in turn, implies that onditions 6