Hindawi Publishing Corporation International Journal of Combinatorics Volume 2013, Article ID 392437, 19 pages http://dx.doi.org/10.1155/2013/392437 Research Article Initial Ideals of Tangent Cones to the Richardson Varieties in the Orthogonal Grassmannian Shyamashree Upadhyay School of Mathematics, Tata Institute of Fundamental Research, Homi Bhabha Road, Colaba, Mumbai 400005, India Correspondence should be addressed to Shyamashree Upadhyay; shyamashree.upadhyay@gmail.com Received 26 October 2012; Accepted 9 January 2013 Academic Editor: Nantel Bergeron Copyright © 2013 Shyamashree Upadhyay. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. A Richardson variety ππΌπΎ in the Orthogonal Grassmannian is defined to be the intersection of a Schubert variety ππΎ in the Orthogonal Grassmannian and an opposite Schubert variety ππΌ therein. We give an explicit description of the initial ideal (with respect to certain conveniently chosen term order) for the ideal of the tangent cone at any T-fixed point of ππΌπΎ , thus generalizing a result of Raghavan and Upadhyay (2009). Our proof is based on a generalization of the Robinson-Schensted-Knuth (RSK) correspondence, which we call the Orthogonal-bounded-RSK (OBRSK). 1. Introduction The Orthogonal Grassmannian is defined in Section 2. A Richardson variety ππΌπΎ in the Orthogonal Grassmannian (The Richardson varieties in the ordinary Grassmannian are also studied by Stanley in [1], where these varieties are called skew Schubert varieties. Discussion of these varieties in the ordinary Grassmannian also appears in [2].) is defined to be the intersection of a Schubert variety ππΎ in the Orthogonal Grassmannian with an opposite Schubert variety ππΌ therein. In particular, the Schubert and opposite Schubert varieties are special cases of the Richardson varieties. In this paper, we provide an explicit description of the initial ideal (with respect to certain conveniently chosen term order) for the ideal of the tangent cone at any π-fixed point ππ½ of ππΌπΎ . It should be noted that the local properties of the Schubert varieties at π-fixed points determine their local properties at all other points, because of the π΅ action; but this does not extend to the Richardson varieties, since Richardson varieties only have a π-action. In Raghavan and Upadhyay [3], an explicit description of the initial ideal (with respect to certain conveniently chosen term orders) for the ideal of the tangent cone at any π-fixed point of a Schubert variety in the Orthogonal Grassmannian has been obtained. In this paper, we generalize the result of [3] to the case of the Richardson varieties in the Orthogonal Grassmannian. Sturmfels [4] and Herzog and Trung [5] proved results on a class of determinantal varieties which are equivalent to the results of [6–8] for the case of the Schubert varieties (in the ordinary Grassmannian) at the π-fixed point πid . The key to their proofs was to use a version of the Robinson-SchenstedKnuth correspondence (which we will call the “ordinary” RSK) in order to establish a degree-preserving bijection between a set of monomials defined by an initial ideal and a “standard monomial basis.” The difficulty in extending this method of proof to the case of the Schubert varieties (in the ordinary Grassmannian) at an arbitrary π-fixed point ππ½ lies in generalizing this bijection, which is done in the three papers [6–8]; the work done in [8] is slightly more general, since it applies to the Richardson varieties, and not just to the Schubert varieties. These three bijections, when restricted to the Schubert varieties in the ordinary Grassmannian, are in fact the same bijection (This supports the conviction of the authors in [6] that this bijection is natural and that it is in some sense the only natural bijection satisfying the required geometric conditions.), although this is not immediately apparent. In the work of Kreiman in [8], this “generalized 2 bijection” has been viewed for the first time as an extended version of the “ordinary” RSK correspondence, which he calls the Bounded-RSK correspondence; this viewpoint was not present in [6, 7]. Although the formulations of the bijections in [6, 7] are similar to each other, the formulation of the bijection in [8] is in terms of different combinatorial indexing sets. The work done in [3] by Raghavan and Upadhyay does not involve any version of the RSK correspondence, unlike the work done by Herzog and Trung in [5]. The work done in [3] relies on a degree-preserving bijection between a set of monomials defined by an initial ideal and a “standard monomial basis,” and this bijection is proved by Raghavan and Upadhyay in [9]. It is mentioned in [9] that it will be nice if the bijection proved therein can be viewed as a kind of “Bounded-RSK” correspondence, as done by Kreiman in [8] for the case of the Richardson varieties in the ordinary Grassmannian. This paper fulfills the expectation of [9] that one should be able to view the bijection there as a generalized-bounded-RSK correspondence, which we call here the Orthogonal-bounded-RSK correspondence (OBRSK). Kreiman has mentioned in his paper [8] that he believes in the possibility of adapting the methods of his paper ([8]) to the Richardson varieties in the Symplectic and the Orthogonal Grassmannian as well; this paper supports Kreiman’s belief for the Orthogonal Grassmannian case. 2. The Orthogonal Grassmannian and Richardson Varieties in It Fix an algebraically closed field k of characteristic not equal to 2. Fix a natural number π, a vector space π of dimension 2π over k, and a nondegenerate symmetric bilinear form β¨ , β© on π. For π an integer such that 1 ≤ π ≤ 2π, set π∗ := 2π + 1 − π. Fix a basis π1 , . . . , π2π of π such that β¨ππ , ππ β© equals 1 if π = π∗ and is 0 otherwise. Denote by ππ(π) the group of linear automorphisms of π that preserve the bilinear form β¨ , β© and also the volume form. A linear subspace of π is said to be isotropic, if the bilinear form β¨ , β© vanishes identically on it. Denote by Mπ (π)σΈ the closed subvariety of the Grassmannian of π-dimensional subspaces consisting of the points corresponding to maximal isotropic subspaces. The action of ππ(π) on π induces an action on Mπ (π)σΈ . There are two orbits for this action. These orbits are isomorphic: acting by a linear automorphism that preserves the form but not the volume form gives an isomorphism. We denote by Mπ (π) the orbit of the span of π1 , . . . , ππ and call it the (even) Orthogonal Grassmannian. One can define the Orthogonal Grassmannian in the case when the dimension of π is not necessarily even. But it is enough to consider the case when the dimension of π is even, the reason being the following (see also Section 2.1 of [9] for the same ): suppose that the dimension of π is odd; say dimension of π = 2π + 1. Μ be a vector space of dimension πΜ with Let πΜ := 2π + 2 and π a nondegenerate symmetric form. Let Μπ1 , . . . , ΜππΜ be a basis of Μ such that β¨Μππ , Μππ β© equals 1 if π = π∗ and is 0 otherwise. Put π π := Μππ+1 and π := Μππ+2 . Take π to be an element of the field Μ such that π2 = 1/2. We can take π to be the subspace of π International Journal of Combinatorics spanned by the vectors Μπ1 , . . . , Μππ , ππ+ππ, Μππ+3 , . . ., and ΜππΜ , and a basis of π to be these vectors in that order. Μ σΈ to Mπ (π): interThere is a natural map from Mπ+1 (π) Μ of dimension π + 1 secting with π an isotropic subspace of π gives an isotropic subspace of π of dimension π, and we denote this map by ∩. The map ∩ is onto, for every isotropic Μ (and hence of π) is contained in an isotropic subspace of π Μ of dimension π + 1. In fact, more is true: the subspace of π map ∩ is two-to-one. The map ∩ being two-to-one, it is also elementary to see that the two points in any fiber lie one Μ σΈ . We therefore get a natuin each component of Mπ+1 (π) Μ and Mπ (π). Therefore, ral isomorphism between Mπ+1 (π) now onwards we call the (even) Orthogonal Grassmannian Mπ (π) (as defined above for a 2π-dimensional vector space π) the Orthogonal Grassmannian. Let Mπ (π) ⊆ πΊπ (π) σ³¨ → P(∧π π) be the PluΜcker embedding (where πΊπ (π) denotes the Grassmannian of all π-dimensional subspaces of π). Thus Mπ (π) is a closed subvariety of the projective variety πΊπ (π), and hence Mπ (π) inherits the structure of a projective variety. We take π΅ (resp., π΅− ) to be the subgroup of ππ(π) consisting of those elements that are upper triangular (resp., lower triangular) with respect to the basis π1 , . . . , π2π and the subgroup π of ππ(π) consisting of those elements that are diagonal with respect to π1 , . . . , π2π . It can be easily checked that π is a maximal torus of ππ(π); π΅ and π΅− are the Borel subgroups of ππ(π) which contain π. The group ππ(π) acts transitively on Mπ (π), and the π-fixed points of Mπ (π) under this action are easily seen to be of the form β¨ππ1 , . . . , πππ β© for {π1 , . . . , ππ } in πΌ(π), where πΌ(π) is the set of subsets of {1, . . . , 2π} of cardinality π satisfying the following two conditions: (i) for each π, 1 ≤ π ≤ 2π, the subset contains exactly one of π, π∗ , and (ii) the number of elements in the subset that exceed π is even. We write πΌ(π, 2π) for the set of all π-element subsets of {1, . . . , 2π}. There is a natural partial order on πΌ(π, 2π) and so also on πΌ(π): V = (V1 < ⋅ ⋅ ⋅ < Vπ ) ≤ π€ = (π€1 < ⋅ ⋅ ⋅ < π€π ) if and only if V1 ≤ π€1 ,. . ., Vπ ≤ π€π . For π = {π1 , . . . , ππ } ∈ πΌ(π, 2π), π1 < ⋅ ⋅ ⋅ < ππ , define the complement of π as {1, . . . , 2π} \ π, and denote it by π. The π΅-orbits (as well as π΅− -orbits) of Mπ (π) are naturally indexed by its π-fixed points: each π΅-orbit (as well as π΅− orbit) contains one and only one such point. Let πΌ ∈ πΌ(π) be arbitrary, and let ππΌ denote the corresponding π-fixed point of Mπ (π). The Zariski closure of the π΅- (resp., π΅− -) orbit through ππΌ , with canonical reduced scheme structure, is called a Schubert variety (resp., opposite Schubert variety) and denoted by ππΌ (resp., ππΌ ). For πΌ, πΎ ∈ πΌ(π), the schemetheoretic intersection ππΌπΎ = ππΌ ∩ ππΎ is called a Richardson variety. Each π΅-orbit (as well as π΅− -orbit) being irreducible and open in its closure, it follows that π΅-orbit closures (resp., π΅− -orbit closures) are indexed by the π΅-orbits (resp., π΅− orbits). Thus the set πΌ(π) becomes an indexing set for the Schubert varieties in Mπ (π), and the set consisting of all pairs of elements of πΌ(π) becomes an indexing set for the International Journal of Combinatorics Boundary of N(π£) Diagonal Figure 1 Richardson varieties in Mπ (π). It can be shown that ππΌπΎ is nonempty if and only if πΌ ≤ πΎ; that for π½ ∈ πΌ(π), ππ½ ∈ ππΌπΎ if and only if πΌ ≤ π½ ≤ πΎ. 3. Statement of the Problem and the Strategy of the Proof 3.1. Initial Statement of the Problem. The problem that is tackled in this paper is this: given a π-fixed point on a Richardson variety in Mπ (π), compute the initial ideal, with respect to some convenient term order, of the ideal of functions vanishing on the tangent cone to the Richardson variety at the given π-fixed point. The term order is specified in Section 3.4, and the answer is given in Theorem 7. For the rest of this paper, πΌ, π½, and πΎ are arbitrarily fixed elements of πΌ(π) such that πΌ ≤ π½ ≤ πΎ. So, the problem tackled in this paper can be restated as follows. Given the Richardson variety ππΌπΎ in Mπ (π) and the π-fixed point ππ½ in it, find the initial ideal of the ideal of functions vanishing on the tangent cone at ππ½ to ππΌπΎ , with respect to some conveniently chosen term order. The tangent cone being a subvariety of the tangent space at ππ½ to Mπ (π), we first choose a convenient set of coordinates for the tangent space. But for that we need to fix some notation. 3.2. Basic Notation. For this subsection, let us fix an arbitrary element V of πΌ(π, 2π). We will be dealing extensively with ordered pairs (π, π), 1 ≤ π, π ≤ 2π, such that π is not and π is an entry of V. Let R(V) denote the set of all such ordered pairs, and set N(V) := {(π, π) ∈ R(V) | π > π}, OR(V) := {(π, π) ∈ R(V) | π < π∗ }, ON(V) := {(π, π) ∈ R(V) | π > π, π < π∗ } = OR(V) ∩ N(V), dV := {(π, π) ∈ R(V) | π = π∗ }, AR(V) := {(π, π) ∈ R(V) | π > π∗ }, and AN(V) := {(π, π) ∈ R(V) | π > π, π > π∗ }. We will refer to dV as the diagonal. 3 Figure 1 shows a drawing of R(V). We think of π and π in (π, π) as row index and column index, respectively. The columns are indexed from left to right by the entries of V in ascending order and the rows from top to bottom by the entries of {1, . . . , 2π} \ V in ascending order. The points of dV are those on the diagonal, the points of OR(V) are those that are (strictly) above the diagonal, and the points of N(V) are those that are to the South-West of the polyline captioned “boundary of N(V)”—we draw the boundary so that points on the boundary belong to N(V). The reader can readily verify that π = 13 and V = (1, 2, 3, 4, 6, 7, 10, 11, 13, 15, 18, 19, and 22) for the particular picture drawn. The points of OR(V) indicated by solid circles form an extended V-chain (see Figure 1); the definition of an extended V-chain is given later in Section 3.5. We will be considering monomials in some of these sets. A monomial, as usual, is a subset with each member being allowed a multiplicity (taking values in the nonnegative integers). The degree of a monomial has also the usual sense: consider the underlying set of the monomial and look at the multiplicity with which each element of this underlying set appears in the monomial; the degree of the monomial is the sum of these multiplicities. The intersection of any two monomials in a set also has the natural meaning; it is the monomial whose underlying set is the intersection of the underlying sets of the given two monomials and the multiplicity of each element being the minimum of the multiplicities with which the element occurs in the two given monomials. Given any two monomials π΄ and π΅ consisting of elements of R(V), let set (π΄) and set (π΅) denote the underlying sets of π΄ and π΅, respectively. We say that π΅ ⊆ π΄ (as monomials) if set (π΅) ⊆ set(π΄), and the multiplicity with which every element occurs in the monomial π΅ is less than or equal to the multiplicity with which the same element occurs in the monomial π΄. Given two monomials π΄ and π΅ consisting of elements of R(V) such that π΅ ⊆ π΄ (as monomials), we can define a monomial called the “monomial minus” of π΅ from π΄ (denoted by π΄\π π΅) as follows. Take any element π₯ of set(π΅). If the multiplicity with which π₯ occurs in π΄ is ππ₯ (π΄) and the multiplicity with which π₯ occurs in π΅ is ππ₯ (π΅), then the multiplicity with which π₯ occurs in the monomial π΄\π π΅ is ππ₯ (π΄) − ππ₯ (π΅). And any element in set(π΄) \ set(π΅) occurs in the monomial π΄\π π΅ with the same multiplicity with which it occurs in π΄. This finishes the description of π΄\π π΅. Remark 1. Note that in this subsection, and V was any element of πΌ(π, 2π), V was not necessarily in πΌ(π). In particular, all the above basic notations will hold true, if we take V ∈ πΌ(π) as well. 3.3. The Ideal of the Tangent Cone to ππΌπΎ at ππ½ . Let π½ be the element of πΌ(π) which was fixed at the beginning of this section. Consider the matrix of size 2π×π whose columns are numbered by the entries of π½ and the rows by {1, . . . , 2π}; the rows corresponding to the entries of π½ form the π × π identity matrix, and the remaining π rows form a skew symmetric 4 International Journal of Combinatorics matrix whose upper half entries are variables of the form π(π,π) , where (π, π) ∈ OR(π½). Let Mπ (π) ⊆ πΊπ (π) σ³¨ → P(∧π π) be the PluΜcker embedding (where πΊπ (π) denotes the Grassmannian of all πdimensional subspaces of π). For π in πΌ(π, 2π), let ππ denote the corresponding PluΜcker coordinate. Consider the affine patch A of P(∧π π) given by ππ½ =ΜΈ 0. The affine patch Aπ½ := Mπ (π) ∩ A of the Orthogonal Grassmannian Mπ (π) is an affine space whose coordinate ring can be taken to be the polynomial ring in variables of the form π(π,π) with (π, π) ∈ OR(π½). For π ∈ πΌ(π), consider the submatrix of the above mentioned matrix given by the rows numbered π \ π½ and columns numbered π½ \ π. Such a submatrix is of even size and is skewsymmetric about antidiagonal, therefore its determinant is a square. The square root, which is determined up to sign, is called the Pfaffian. Let ππ,π½ denote this Pfaffian. Set ππΌπΎ (π½) := ππΌπΎ ∩ Aπ½ . From [10] we can deduce a set of generators for the ideal πΌ of functions on Aπ½ vanishing on ππΌπΎ (π½) (see Section 4.2 of [9] for the special case of the Schubert varieties). The following equation gives the generators: πΌ = (ππ,π½ | π ∈ πΌ (π) , πΌ β° π or π β° πΎ) . Note here that the first 3 conditions above are the same as the conditions put on the total order >1 as mentioned in Section 1.6 of [3]. Recall that in the paper [3], initial ideals of ideals of tangent cones at torus fixed points to the Schubert varieties in Orthogonal Grassmannians were computed, and the paper [3] did not deal with the Richardson varieties. The last 3 conditions above arise in this paper as an addition to the 3 conditions put on the total order >1 (as mentioned in Section 1.6 of [3]), because here we are dealing with the Richardson varieties in Mπ (π). Let β³ be the term order on monomials in OR(π½) given by the homogeneous lexicographic order with respect to >. Remark 2. The total order > on OR(π½) satisfying the 6 properties mentioned above can be realized as a concrete total order on OR(π½) if we put the following extra condition on it. Let π(π), π(π), π(π), and π(π) denote the row index of π, the row index of π, the column index of π, and the column index of π, respectively. If π(π) < π(π), π ∈ OR(π½) \ ON(π½), π ∈ ON(π½), and π(π) < π(π), then π > π when (π(π), π(π)) ∉ N(π½)and π > π when (π(π), π(π)) ∈ N(π½). (1) We are interested in the tangent cone to ππΌπΎ at ππ½ or, what is the same, the tangent cone to ππΌπΎ (π½) at the origin. Observe that ππ,π½ is a homogeneous polynomial. Because of this, ππΌπΎ (π½) itself is a cone and so is equal to its tangent cone at the origin. The ideal of the tangent cone to ππΌπΎ at ππ½ is therefore the ideal πΌ in (1). 3.4. The Term Order. We now specify the term order β³ on monomials in the coordinate functions (of the tangent space to Mπ (π) at the torus fixed point ππ½ ; it is easy to see that coordinate ring of the tangent space to Mπ (π) at ππ½ is the polynomial ring in variables of the form π(π,π) , where (π, π) ∈ OR(π½)) with respect to which the initial ideal of the ideal πΌ of the tangent cone is to be taken. Let > be a total order on OR(π½) satisfying all of the following 6 conditions. (i) π > π, if π ∈ ON(π½), π ∈ OR(π½) \ ON(π½), and the row indices of π and π are equal. (ii) π > π, if π ∈ ON(π½), π ∈ ON(π½), the row indices of π and π are equal, and the column index of π exceeds that of π. (iii) π > π if π ∈ ON(π½), π ∈ OR(π½), and the row index of π is less than that of π. (iv) π > π, if π ∈ OR(π½) \ ON(π½), π ∈ ON(π½), and the column indices of π and π are equal. (v) π > π, if π ∈ OR(π½) \ ON(π½), π ∈ OR(π½) \ ON(π½), the column indices of π and π are equal, and the row index of π exceeds that of π. (vi) π > π, if π ∈ OR(π½) \ ON(π½), π ∈ OR(π½), and the column index of π is less than that of π. 3.5. Extended V-Chains and Associated Elements of πΌ(π). For this subsection, let V be an arbitrarily fixed element of πΌ(π, 2π) (not necessarily an element of πΌ(π), unless otherwise stated). For elements π = (π , πΆ), π = (π, π) of R(V), we write π > π if π > π and πΆ < π (Note that these are strict inequalities.). A sequence π 1 > ⋅ ⋅ ⋅ > π π of elements of OR(V) is called an extended V-chain. Note that an extended V-chain can also be empty. Letting πΆ to be an extended V-chain, we define πΆ+ := πΆ∩ON(V) and πΆ− := πΆ∩(OR(V)\ON(V)). We call πΆ+ (resp., πΆ− ) the positive (resp., negative) parts of the extended V-chain πΆ. We call an extended V-chain πΆ positive (resp., negative) if πΆ = πΆ+ (resp., πΆ = πΆ− ). The extended V-chain πΆ is called nonempty if it has at least one element in it. Note that if we specialize to the case when V ∈ πΌ(π), then whatever is called a V-chain in Section 2.4.1 of [9] is a positive extended V-chain over here. To every extended V-chain πΆ, we will now associate 2 subsets dVπΆ(+) and dVπΆ(−) of dV (each of even cardinality), but for that we first need to fix some notation and recall certain terminology from [3, 9]. Definition 3 (Pr and Pro). Given any subset π· of ON(V), let Pr(π·) denote the monomial whose underlying set is given by {π | π is a vertical or horizontal projection of some element of π·} (both vertical and horizontal, as defined in Section 2.3 of [9]), and the cardinality of any element π of this underlying set inside the monomial Pr(π·) equals the number of elements in π· whose one of the projections (vertical or horizontal) is π. For π = (π, π) in R(V), define π# := (π∗ , π∗ ). The involution π σ³¨→ π# is just the reflection with respect to the diagonal dV . For a subset E of N(V), the symbol E# has the obvious meaning. One calls E π π¦ππππ‘πππ if E = E# . Given any symmetric subset πΈ of N(V), let one denote by πΈ(top) the set πΈ ∩ ON(V), by πΈ(diag) the set πΈ ∩ dV , and by Pro(πΈ) the monomial formed by taking the union of the subset πΈ(diag) with the monomial Pr(πΈ(top)). Let one make the definition of International Journal of Combinatorics 5 Pro(πΈ) more precise. The multiplicity with which any element occurs in the monomial Pro(πΈ) is equal to the sum of the multiplicities with which the element occurs in the subset πΈ(diag) and the monomial Pr(πΈ(top)). So for any symmetric subset πΈ of N(V), Pro(πΈ) is a monomial consisting of elements from the diagonal. Similarly for any subset π· of ON(V), Pr(π·) is also a monomial consisting of elements from the diagonal. Let V ∈ πΌ(π). Let πΆ : πΌ1 = (π1 , π1 ) > πΌ2 = (π2 , π2 ) > ⋅ ⋅ ⋅ > πΌπ = (ππ , ππ ) be a positive extended V-chain in ON(V). Two consecutive elements πΌπ and πΌπ+1 of πΆ are said to be connected if the following conditions are both satisfied: (i) their legs are “intertwined”, that is, ππ∗ ≥ ππ+1 , (ii) the point (ππ+1 , ππ∗ ) belongs to N(V). Consider the coarsest equivalence relation on the elements of πΆ generated by the above relation. The equivalence classes of πΆ with respect to this equivalence relation are called the connected components of the V-chain πΆ. For any πΌ = (π, π) in ON(V), the elements πV (πΌ) := (π∗ , π) and πβ (πΌ) := (π, π∗ ) of the diagonal are called, respectively, the vertical and horizontal projections of πΌ. Given the positive extended V-chain πΆ as above, let SπΆ be the monomial of N(V) associated to πΆ defined as follows. First suppose that πΆ is a connected V-chain in ON(V). Observe that if there is at all an integer π (1 ≤ π ≤ π) such that the horizontal projection πβ (πΌπ ) does not belong to N(V), then π = π. Define {πV (πΌ1 ) , . . . , πV (πΌπ )} if π is even { { { {πV (πΌ1 ) , . . . , πV (πΌπ )} ∪ {πβ (πΌπ )} { { { { if π is odd, πβ (πΌπ ) ∈ N (V) SπΆ := { { { # { { { {πV (πΌ1 ) , . . . , πV (πΌπ−1 )} ∪ {πΌπ , πΌπ } { if π is odd, πβ (πΌπ ) ∉ N (V) . { (2) For a V-chain πΆ that is not necessarily connected, let πΆ = πΆ1 ∪ πΆ2 ∪ ⋅ ⋅ ⋅ be the partition of πΆ into its connected components, and set SπΆ := SπΆ1 ∪ SπΆ2 ∪ ⋅ ⋅ ⋅ . (3) Note that even if we take V to be in πΌ(π, 2π) (and not merely in πΌ(π)) and define SπΆ for any positive extended Vchain πΆ exactly in the same way as we did above, there is no logical inconsistency. Hence we extend the definition of SπΆ to any positive extended V-chain πΆ, where V ∈ πΌ(π, 2π). Clearly SπΆ is a symmetric subset of N(V). Hence the monomial Pro(SπΆ) is well defined for any positive extended V-chain πΆ, where V ∈ πΌ(π, 2π). Definition 4 (the flip map πΉ). For any V ∈ πΌ(π, 2π) and any element π = (π, π) ∈ R(V), let πΉ(π) be the element of R(V∗ ) given by πΉ(π) := (π, π). So πΉ is an invertible map from R(V) to R(V∗ ) (note here that if V ∈ πΌ(π), then V∗ needs not always to belong to πΌ(π)); let one denote the inverse of πΉ by πΉ−1 . The map πΉ naturally induces an invertible map from the set of all monomials in R(V) to the set of all monomials in R(V∗ ). One continues to call the induced map πΉ and its inverse πΉ−1 . Definition 5 (the subsets dVπΆ(+) and dVπΆ(−) of dV ). Given any non-empty extended V-chain πΆ, one will now associate 2 subsets dVπΆ(+) and dVπΆ(−) of dV (each of even cardinality) with it. Let dVπΆ (+) −1 { πΉ ( Pr (πΉ (πΆ)) \π Pro (SπΉ(πΆ) )) := { if πΆ is negative + Pro (S ) if otherwise. πΆ { (4) Similarly, let dVπΆ (−) := { Pr (πΆ) \π Pro (SπΆ) if πΆ is positive πΉ−1 (Pro (SπΉ(πΆ− ) )) if otherwise. (5) It is an easy exercise to check that dVπΆ(+) and dVπΆ(−) thus defined are actually subsets (not monomials) of dV and that each of them has even cardinality. Definition 6 (elements of πΌ(π) associated with dVπΆ(+) and dVπΆ(−)). For this definition, we let V to be an arbitrary element in πΌ(π). Note that given any subset π of dV of even cardinality, one can naturally associate an element of πΌ(π) to it by removing those entries from V which appear as column indices in the elements of the set π and then adding to it the row indices of all the elements of π. It is easy to check that the resulting element actually belongs to πΌ(π). One denotes the resulting element by πΌ(π)(π). If π is empty, then πΌ(π)(π) is taken to be V itself. Let π€πΆ+ (V) := πΌ(π)(dVπΆ(+)) and π€πΆ− (V) := πΌ(π)(dVπΆ(−)). These are the two elements of πΌ(π) that we can naturally associate with the subsets dVπΆ(+) and dVπΆ(−) of dV . 3.6. The Main Theorem and a Strategy of the Proof. Recall that the ideal of the tangent cone to ππΌπΎ at ππ½ is the ideal πΌ given by (1). Let β³ be as in Section 3.4. For any element π ∈ πΌ; let inβ³ π denote the initial term of π with respect to the term order β³. We define inβ³ πΌ to be the ideal β¨inβ³ π | π ∈ πΌβ© inside the polynomial ring π := k[π(π,π) | (π, π) ∈ OR(π½)]. For any monomial π in OR(π½), let us denote by ππ the product of all the elements π(π,π) , where (π, π) runs over the elements in π with multiplicities. Let ChainsπΎπΌ (π½) denote the set {ππΆ |πΆ is a nonempty extended π½-chain in OR(π½) such that either (i) or (ii) below holds}. (i) πΆ− is nonempty, and πΌ β° π€πΆ−− (π½). (ii) πΆ+ is nonempty, and π€πΆ++ (π½) β° πΎ. The main result of this paper is the following. Theorem 7. inβ³ πΌ = β¨ChainsπΎπΌ (π½)β©. Example 8. Let π = 5, πΌ = (1, 2, 3, 6, 7), π½ = (1, 2, 4, 6, 8), and πΎ = (1, 2, 5, 7, 8). Clearly then πΌ ≤ π½ ≤ πΎ. It can be verified that the set ChainsπΎπΌ (π½) consists of elements of the following 3 types. (i) Any extended π½-chain in OR(π½) which contains the element (3, 6). (ii) Any extended π½-chain in OR(π½) which contains the element (3, 4) and a nonempty positive part. 6 International Journal of Combinatorics (iii) All positive extended π½-chain in OR(π½) except the singleton-π½-chain {(5, 4)}. One can therefore determine the corresponding initial ideal inβ³ πΌ using Theorem 7. Remark 9. It follows from the statement of Theorem 7 that the set of all monomials in OR(π½) which contain at least one extended π½-chain πΆ such that ππΆ ∈ ChainsπΎπΌ (π½) form a vector space basis of the initial ideal inβ³ πΌ over the field k; see, for example, Example 8. In the special case when the Richardson variety is a Schubert variety, it is easy to see that the previous statement of this remark says exactly what has been said in the main theorem (Theorem 1.8.1) of [3]. Remark 10. Since inβ³ πΌ = β¨ChainsπΎπΌ (π½)β©, it follows that the dimension as a vector space of the graded piece of π/inβ³ πΌ of degree π equals the cardinality of the set of all monomials in OR(π½) of degree π which do not contain any extended π½-chain πΆ such that ππΆ ∈ ChainsπΎπΌ (π½). Since π/πΌ and π/inβ³ πΌ have the same Hilbert function, the same is true with π/inβ³ πΌ (in the previous statement) replaced by π/πΌ. Hence if π πΌπΎ (π½) denotes the coordinate ring of the tangent cone to ππΌπΎ at ππ½ , then the dimension as a vector space of the homogeneous piece of π πΌπΎ (π½) of degree π equals the cardinality of the set of all monomials in OR(π½) of degree π which are such that for every extended π½-chain πΆ in the monomial, πΌ ≤ π€πΆ−− (π½) and π€πΆ++ (π½) ≤ πΎ. From Remark 10, the following corollary is immediate (the proof is similar to that of Corollary 2.5.2 of [9]) Corollary 11. The multiplicity of ππΌπΎ at ππ½ equals the number of monomials in OR(π½) of maximal cardinality that are squarefree and are such that, given any extended π½-chain πΆ in any of these monomials, one has: πΌ ≤ π€πΆ−− (π½) and π€πΆ++ (π½) ≤ πΎ. We now briefly sketch the proof of Theorem 7 (omitting details, which can be found in Section 8). In order to introduce the main combinatorial objects of interest and outline a strategy of the proof, we will first need to prove that the set ChainsπΎπΌ (π½) ⊆ inβ³ πΌ, and this proof will follow from whatever is said in Remark 12. Remark 12. Let πΆ be a nonempty extended π½-chain in OR(π½) such that ππΆ ∈ ChainsπΎπΌ (π½). If πΆ+ is nonempty and π€πΆ++ (π½) β° πΎ, then it can be proved that ππΆ+ ∈ inβ³ πΌ, the proof being exactly the same as that in Section 4 of [3]. Then since inβ³ πΌ is an ideal and ππΆ = ππΆ− ππΆ+ , it follows that ππΆ ∈ inβ³ πΌ. If πΆ− is nonempty and πΌ β° π€πΆ−− (π½), look at πΉ(πΆ− ), where πΉ is the flip map as defined in Definition 4 from the set of all monomials in R(π½) to the set of all monomials in R(π½∗ ). Then πΉ(πΆ− ) is a positive extended π½∗ -chain in OR(π½∗ ). We need to prove that ππΆ ∈ inβ³ πΌ, for which it is enough to show that ππΆ− ∈ inβ³ πΌ. To prove that ππΆ− ∈ inβ³ πΌ, we will proceed in a way equivalent to the proof done in Section 4 of [3]. But there is a subtle difference between what is proved in Section 4 of [3] and what we are going to prove here; namely, Whatever was proved in Section 4 of [3] can be rephrased in the language of this paper as “Every positive extended π½-chain + (π½) β° πΎ belongs to the initial π· satisfying the property that π€π· ideal of the ideal of the tangent cone,” but here we are going to prove that “Every negative extended π½-chain π· satisfying − (π½) belongs to the initial ideal of the the property that πΌ β° π€π· ideal of the tangent cone.” Because of this subtle difference, we need to construct certain gadgets for negative extended π½-chains, which will play a role similar to the role of the objects like the new forms, Proj and Projπ corresponding to positive extended π½-chains (For positive extended π½-chains, such objects are already defined in [3]). This construction is given in the following paragraph. Consider the positive extended π½∗ -chain πΉ(πΆ− ). We can construct new forms, Proj and Projπ for πΉ(πΆ− ) in the same way as they were constructed in [3], note here the fact that π½∗ may or may not belong to πΌ(π) does not really affect the construction of the new forms, Proj and Projπ for πΉ(πΆ− ). Then we apply the map πΉ−1 to these objects constructed for πΉ(πΆ− ), the resulting objects are the analogues of the new forms, Proj and Projπ for the negative extended π½-chain πΆ− . We apply a similar treatment to any other monomial related to πΉ(πΆ− ) that we happen to encounter if we replace the “V-chain π΄” in Section 4.2 of [3] with “the positive extended π½∗ -chain πΉ(πΆ− ).” In Section 2.4 of [3], an element π¦πΈ of πΌ(π) corresponding to any V-chain πΈ (the notion of a V-chain being as in Section 1.7 of [3]) has been defined. The analogous element of πΌ(π) for the negative extended π½-chain πΆ− (we call it π¦πΆ− here) can be obtained from πΉ−1 (Projπ (πΉ(πΆ− ))) by following the natural process: the column indices of elements of πΉ−1 (Projπ (πΉ(πΆ− ))) occur as members of π½; these are replaced by the row indices to obtain the desired element of πΌ(π) for πΆ− . It is easy to check that π¦πΆ− belongs to πΌ(π) and that π¦πΆ− ≤ π€πΆ−− (π½) ≤ π½. Since we already have that πΌ β° π€πΆ−− (π½), it follows that πΌ β° π¦πΆ− . These facts about π¦πΆ− will be needed to produce an analogue of the main proof of [3] in our present case. To be more precise, these facts about π¦πΆ− give the analogues of Propositions 2.4.1 and 2.4.2 of [3], and these two propositions had been used quite crucially inside the main proof of [3]. With all these analogues constructed for negative extended π½-chains, we can proceed in an “equivalent” manner (Here, by “equivalent” we mean keeping track of the subtle difference as mentioned above and working accordingly.) as in the paper [3] and end up proving the desired fact, namely, ππΆ− ∈ inβ³ πΌ. Since ChainsπΎπΌ (π½) ⊆ inβ³ πΌ, we have β¨ChainsπΎπΌ (π½)β© ⊆ inβ³ πΌ. To prove Theorem 7, we now need to show that β¨ChainsπΎπΌ (π½)β© ⊇ inβ³ πΌ. For this, it suffices to show that, in any degree, the number of monomials of β¨ChainsπΎπΌ (π½)β© is ≥ the number of monomials of inβ³ πΌ. Or equivalently, it suffices to show that, in any degree, the number of monomials of π/β¨ChainsπΎπΌ (π½)β© is ≤ the number of monomials of π/inβ³ πΌ. Consider the affine patch ππΌπΎ (π½) (:= ππΌπΎ ∩ Aπ½ ) of the Richardson variety ππΌπΎ . The following is a well known result. Theorem 13. k[ππΌπΎ (π½)] = π/πΌ, where π = k[π(π,π) | (π, π) ∈ OR(π½)] and πΌ is as in (1). International Journal of Combinatorics 7 Table 1: Two subsets of the ring π = k[π(π,π) |(π, π) ∈ OR(π½)] and their indexing sets. Set of elements in π = k[π(π,π) |(π, π) ∈ OR(π½)] Monomials of π/ β¨ChainsπΎπΌ (π½)β© Standard monomials on ππΌπΎ (π½) Indexing set Pairs of nonempty skew-symmetric lexicographic multisets on π½ × π½ bounded by ππΌ , ππΎ Nonvanishing skew-symmetric notched bitableaux on π½ × π½ bounded by ππΌ , ππΎ Both the monomials of π/inβ³ πΌ and the standard monomials on ππΌπΎ (π½) form a basis for π/πΌ, and thus agree in cardinality in any degree. Therefore, to prove, that in any degree, the number of monomials of π/β¨ChainsπΎπΌ (π½)β© is ≤ the number of monomials of π/inβ³ πΌ, it suffices to give a degree-preserving injection from the set of all monomials in π/β¨ChainsπΎπΌ (π½)β© to the set of all standard monomials on ππΌπΎ (π½). We construct such an injection, the Orthogonal-bounded-RSK (OBRSK), from an indexing set of the former to an indexing set of the later. These indexing sets are given in Table 1. 4. Skew-Symmetric Notched Bitableaux In this section onwards, the terminology and notation of Sections 4 and 5 of [8] will be in force. Recall the definition of a semistandard notched bitableau from Section 5 of [8]. Definition 14 (dual of an element with respect to a semistandard notched bitableau). Let (π, π) be a semistandard notched bitableau. Let ππ,π (resp., ππ,π ) denote the entry in the πth row and πth column of π (resp., π). For any row number π of π (or of π), let ππ denote the total number of entries in the πth row of π (or π). One calls the entry ππ,ππ +1−π of π the dual of the entry ππ,π of π with respect to (π, π). Similarly, we call the entry ππ,ππ +1−π of π the dual of the entry ππ,π of π with respect to (π, π). Note that any entry of π or π can be identified uniquely by specifying 4 coordinates, namely, the entry π₯, the tableau π΄ (π΄ = π or π) in which the entry lies, the row number π of the entry in the tableau π΄, and the column number π of the entry in the tableau π΄. Let set(π, π) denote the set of all 4tuples of the form (π₯, π΄, π, π). Given any 4-tuple (π₯, π΄, π, π) ∈ set(π, π), let us denote by π·(π,π) (π₯, π΄, π, π) the 4-tuple which represents the dual of π₯ with respect to (π, π) (as defined in Definition 14). For (π₯, π΄, π, π), (π₯σΈ , π΄σΈ , πσΈ , πσΈ ) ∈ set(π, π), we say that (π₯, π΄, π, π) βͺ― (π₯σΈ , π΄σΈ , πσΈ , πσΈ ) if π₯ ≤ π₯σΈ . Similarly, we say that (π₯, π΄, π, π) βΊ (π₯σΈ , π΄σΈ , πσΈ , πσΈ ) if π₯ < π₯σΈ and (π₯, π΄, π, π) β (π₯σΈ , π΄σΈ , πσΈ , πσΈ ) if π₯ = π₯σΈ . A semistandard notched bitableau (π, π) is said to be skew-symmetric if the following 2 conditions are satisfied simultaneously. (i) The bitableau (π, π) should be of even size; that is, the number of elements in each row of π and π should be even. (ii) For any (π₯, π΄, π, π), (π₯σΈ , π΄σΈ , πσΈ , πσΈ ) ∈ set(π, π), (π₯, π΄, π, π) βͺ― (π₯σΈ , π΄σΈ , πσΈ , πσΈ ) if and only if π·(π,π) (π₯, π΄, π, π) βͺ° π·(π,π) (π₯σΈ , π΄σΈ , πσΈ , πσΈ ). Moreover, (π₯, π΄, π, π) βΊ (π₯σΈ , π΄σΈ , πσΈ , if and only if π·(π,π) (π₯, π΄, π, π) β» πσΈ ) π·(π,π) (π₯σΈ , π΄σΈ , πσΈ , πσΈ ), and (π₯, π΄, π, π) β (π₯σΈ , π΄σΈ , πσΈ , πσΈ ) if and only if π·(π,π) (π₯, π΄, π, π) β π·(π,π) (π₯σΈ , π΄σΈ , πσΈ , πσΈ ). Property (ii) will be henceforth referred to as the duality property associated to the skew-symmetric notched bitableau (π, π). Note that a skew-symmetric notched bitableau is a semistandard notched bitableau by default. The degree of a skew-symmetric notched bitableau (π, π) is the total number of boxes in π (or π). The notions of negative, positive, and nonvanishing skew-symmetric notched bitableau remain the same as in Section 5 of [8]. The notion of a skew-symmetric notched bitableau (π, π) being bounded by π, π (where π, π are subsets of N2 ), and the notion of negative and positive parts of a skew-symmetric notched bitableau (π, π) remain the same as they were in Section 5 of [8]. If (π, π) is a nonvanishing skew-symmetric notched bitableau, define π(π, π) to be the notched bitableau obtained by reversing the order of the rows of (π, π). One checks that π(π, π) is a nonvanishing skew-symmetric notched bitableau. The map π is an involution, and it maps negative skewsymmetric notched bitableaux to positive ones and vice versa. Thus π gives a bijective pairing between the sets of negative and positive skew-symmetric notched bitableaux. 5. Pairs of Skew-Symmetric Lexicographic Multisets on N2 Recall the notions of a multiset on N2 , finiteness of a multiset on N2 and degree of a multiset on N2 , from Section 4 of [8]. By a lexicographic multiset on N2 , we mean a finite multiset π = {(πΌ1 , π½1 ), . . . , (πΌπ‘ , π½π‘ )} on N2 such that π½π ≥ π½π+1 ∀π, and if π½π = π½π+1 , then πΌπ ≥ πΌπ+1 , π = 1, . . . , π‘ − 1. Given a lexicographic multiset π on N2 , let ππ‘ denote the multiset on N2 (which is not necessarily lexicographic) obtained by switching the two coordinates of π. We call the multiset ππ‘ on N2 the transpose of the multiset π. Consider a pair {π1 , π2 } of multisets on N2 (not necessarily lexicographic), where both π1 and π2 are of the same degree (say, π‘). Let π1 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )}, and π2 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )}. We call the first coordinate of the multiset π1 the π-cell, the second coordinate of π1 the π-cell, the first coordinate of π2 the π-cell, and the second coordinate of π2 the π-cell. Any entry in the pair {π1 , π2 } of multisets on N2 can be identified uniquely by specifying 3 coordinates: the cell ∇ of {π1 , π2 } in which the entry lies (∇ = π, π, π, ππ π), the position π (counting from left to right) of the entry in the cell ∇, and the value β (∇, π) of the entry sitting in the πth position of the cell ∇. 8 International Journal of Combinatorics Set ππ1 ,π2 := {π₯ | π₯ = (∇, π, β (∇, π)), ∇ ∈ {π, π, π, π}, π ∈ {1, . . . , π‘}}. For any π₯ ∈ ππ1 ,π2 , let π·π1 ,π2 (π₯) { { { { { { { { { { { { { { := { { { { { { { { { { { { { { { (π, π‘ + 1 − π, β (π, π‘ + 1 − π)) if π₯ = (π, π, β (π, π)) (π, π‘ + 1 − π, β (π, π‘ + 1 − π)) if π₯ = (π, π, β (π, π)) (π, π‘ + 1 − π, β (π, π‘ + 1 − π)) if π₯ = (π, π, β (π, π)) (π, π‘ + 1 − π, β (π, π‘ + 1 − π)) if π₯ = (π, π, β (π, π)) . ∀π ∈ {1, . . . , π‘} (6) We call π·π1 ,π2 (π₯) the dual of π₯ with respect to the pair {π1 , π2 } of multisets on N2 . Note that, for every π₯ ∈ ππ1 ,π2 , we have π·π1 ,π2 (π₯) ∈ ππ1 ,π2 . For any two elements π₯, π₯σΈ ∈ ππ1 ,π2 , where π₯ = (∇, π, β (∇, π)) and π₯σΈ = (∇σΈ , πσΈ , β (∇σΈ , πσΈ )), we say that π₯ βΎ π₯σΈ if β (∇, π) ≤ β (∇σΈ , πσΈ ). Similarly, we say that π₯β¨π₯σΈ if β (∇, π) < β (∇σΈ , πσΈ ) and π₯ ≈ π₯σΈ if β (∇, π) =β (∇σΈ , πσΈ ). The above pair {π1 , π2 } of multisets on N2 is said to be skew-symmetric lexicographic if the following conditions are satisfied simultaneously. (i) π1 is a lexicographic multiset on N2 . (ii) π2π‘ is a lexicographic multiset on N2 . (iii) ππ < ππ‘+1−π ∀π ∈ {1, . . . , π‘}. (iv) ππ < ππ‘+1−π ∀π ∈ {1, . . . , π‘}. (v) For any π₯, π¦ ∈ ππ1 ,π2 , π₯ βΎ π¦ if and only if π·π1 ,π2 (π₯) βΏ π·π1 ,π2 (π¦). Moreover, π₯ β¨ π¦ if and only if π·π1 ,π2 (π₯) β© π·π1 ,π2 (π¦), and π₯ ≈ 𦑠if and only if π·π1 ,π2 (π₯) ≈ π·π1 ,π2 (π¦) ( → this property is called the duality property associated to the pair {π1 , π2 } of skew-symmetric lexicographic multisets on N2 ). For any pair {π1 , π2 } of skew-symmetric lexicographic multisets on N2 , we define the degree of the pair to be 2 times the degree of π1 (or of π2 , they are the same). A pair {π1 , π2 } of skew-symmetric lexicographic multisets on N2 is said to be negative if ππ < ππ , π = 1, . . . , π‘, positive if ππ > ππ , π = 1, . . . , π‘, and nonempty if ππ =ΜΈ ππ , π = 1, . . . , π‘, where π1 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )} and π2 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )}. Note that condition (v) above will imply that if ππ < ππ ∀π = 1, . . . , π‘, then ππ‘+1−π < ππ‘+1−π ∀π = 1, . . . , π‘. Similarly, if ππ > ππ ∀π = 1, . . . , π‘, then ππ‘+1−π > ππ‘+1−π ∀π = 1, . . . , π‘, and if ππ =ΜΈ ππ ∀π = 1, . . . , π‘, then ππ‘+1−π =ΜΈ ππ‘+1−π ∀π = 1, . . . , π‘. Let {π1 , π2 } be a pair of nonempty skew-symmetric lexicographic multisets on N2 given by π1 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )} and π2 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )}. Let us denote by π1− (resp., π1+ ) the lexicographic multiset on N2 consisting of those elements of π1 such that ππ < ππ (resp., ππ > ππ ). Let us denote by π2− (resp., π2+ ) the lexicographic multiset on N2 consisting of those elements of π2 such that ππ < ππ (resp., ππ > ππ ). We call {π1− , π2− } and {π1+ , π2+ } the negative and positive parts respectively of the pair {π1 , π2 }. Note here that because of condition (v) above, π1− and π2− will have the same degree, and the same holds true for π1+ and π2+ . It is easy to see now that both the pairs {π1− , π2− } and {π1+ , π2+ } of multisets on N2 are skew-symmetric lexicographic in their own right. Given a lexicographic multiset π on N2 , define π(π) to be the lexicographic multiset on N2 obtained by first switching the two coordinates of π and then rearranging the elements so that the new multiset is lexicographic. Let ππ‘ be a map from the set of all lexicographic multisets on N2 to itself given by first switching the two coordinates of a given lexicographic multiset π and then rearranging the elements so that the transpose of the resulting multiset becomes lexicographic. We now define an involution πΏ on the set of all pairs of skew-symmetric lexicographic multisets on N2 by πΏ({π1 , π2 }) := {π(π1 ), ππ‘ (π2 )}. It is easy to check that this map is well defined, and it maps pairs of negative skew-symmetric lexicographic multisets on N2 to positive ones, and vice versa. Thus πΏ gives a bijective pairing between the set of all pairs of negative skew-symmetric lexicographic multisets on N2 and the set of all pairs of positive skew-symmetric lexicographic multisets on N2 . 6. The Orthogonal-BoundedRSK Correspondence We next define the Orthogonal-bounded-RSK correspondence (OBRSK) as a function which maps a pair of negative skew-symmetric lexicographic multisets on N2 to a negative skew-symmetric notched bitableau. Let {π1 , π2 } be a pair of negative skew-symmetric lexicographic multisets on N2 whose entries are labeled as in Section 5. We inductively form a sequence of notched bitableaux (π(0) , π(0) ), (π(1) , π(1) ), . . . , (π(π‘) , π(π‘) ), such that each (π(π) , π(π) ) is of even size, and π(π) is semistandard on ππ for every π = 1, . . . , π‘ as follows. Let (π(0) , π(0) ) = (0, 0), and let π0 = π1 . Assume inductively that we have formed (π(π) , π(π) ), such that the notched bitableau (π(π) , π(π) ) is of even size, π(π) is semistandard on ππ and thus on ππ+1 , since ππ+1 ≤ ππ . (π) Let us first fix some notation and terminology. Let πππ (π) (resp., πππ ) denote the entry in the πth row and πth column of π(π) (resp., π(π) ). Let 2ππ(π) denote the total number of entries (note that it is always even) in the πth row of π(π) (or π(π) ). Given an arbitrary notched tableau π and any row number π of π, we call the entry in the πth box (counting from left to right) the forward πth entry of the πth row of π. Similarly, we call the entry in the πth box (counting from right to left) of π the backward πth entry of the πth row of π. It is now easy to see that the backward πth entry of the πth row of π(π) is actually equal to the forward (2ππ(π) +1−π)th entry of π(π) . We now describe the OBRSK correspondence for the pair {π1 , π2 } of negative skew-symmetric lexicographic multisets on N2 as mentioned above in Section 5. ππ+1 Perform the bounded insertion process π(π) ← ππ+1 as in [8]. In this finite-step process of bounded insertion, suppose that ππ+1 had bumped the “forward π1 th entry” of the 1st row of π(π)<ππ+1 (see Section 3 of [8] for the notation π(π)<ππ+1 , it International Journal of Combinatorics 9 comes in the paragraph right after Example 3.1 of [8]); the “forward π1 th entry” of the 1st row of π(π)<ππ+1 has bumped the “forward π2 th entry” of the 2nd row of π(π)<ππ+1 , . . . and so on until, at some point, a number is placed in a new box at the right end of some row of π(π)<ππ+1 , say this happens at the row number πΎ(π) of π(π)<ππ+1 . Say that the entry of the new box (as mentioned in the previous statement) becomes the (π) ππ+1 forward ππΎ(π) th entry of the πΎ(π) th row of π ← ππ+1 . We then construct a new notched tableau as follows. We let ππ‘−π bump the “backward π1 th entry” of the 1st row of π(π) ; then we let the “backward π1 th entry” of the 1st row of π(π) bump the “backward π2 th entry” of the 2nd row of π(π) , . . . and so on until, at some point, a number is placed in a new box at the backward ππΎ(π) th position of the πΎ(π) th row of π(π) , shifting all entries in the backward 1st . . . up to (and including) the backward (ππΎ(π) − 1)th positions of the πΎ(π) th row of π(π) to the right by one box. Essentially, whatever we did for the bounded insertion process producing ππ+1 π(π) ← ππ+1 , we do a dual version of the same process on π(π) with the integer ππ‘−π . We denote the resulting notched tableau dual by π(π) ← ππ‘−π . ππ+1 dual Note here that the tableaux π(π) ← ππ+1 and π(π) ← ππ‘−π so constructed are of the same shape, but there exists one row in both of them in which the total number of entries is odd. We wanted to construct a notched bitableau (π(π+1) , π(π+1) ) inductively from (π(π) , π(π) ) which should be of even size. We make it possible in the following way. Let πΎ(π) be the row number of π(π) (or of π(π) ) at which the above-mentioned insertion algorithm had stopped. Place ππ‘+1−(π+1) (= ππ‘−π ) in a new box at the rightmost end of the ππ+1 πΎ(π) th row of π(π) ← ππ+1 . We denote the resulting notched tableau by π(π+1) . It is an easy exercise to see that π(π+1) as constructed above will be semistandard on ππ+1 . After this, we place ππ+1 in a new box at the leftmost end of the πΎ(π) th dual row of π(π) ← ππ‘−π , shifting all previously existing entries in dual the πΎ(π) th row of π(π) ← ππ‘−π to the right by one box. We denote the resulting notched tableau by π(π+1) . Clearly π(π+1) and π(π+1) have the same shape. Now we have got hold of a notched bitableau (π(π+1) , π(π+1) ) which is of even size. Then OBRSK ({π1 , π2 }) is defined to be (π(π‘) , π(π‘) ). In the (π+1) (π+1) (π) (π) ππ+1 ,ππ‘+1−(π+1) process above, we write (π ,π ) = (π , π ) ← ππ+1 , ππ‘+1−(π+1) . In terms of this notation, OBRSK ({π1 , π2 }) = π1 ,ππ‘ ππ‘ ,π1 ((0, 0) ← π1 , ππ‘ ) ⋅ ⋅ ⋅ ← ππ‘ , π1 . Lemma 15. With notation as in the definition of the OBRSK correspondence mentioned above, π(π) is row strict for all π ∈ {1, . . . , π‘}. Proof. We will prove the lemma by induction on π. The base case (i.e., when π = 1) of induction is easy to see. Now let π ∈ {1, . . . , π‘ − 1}. Assume inductively that π(π) is row strict. ππ+1 We will now prove that π(π+1) is row strict. That π(π) ← ππ+1 is row strict follows in the same way as in [8]. Note that π(π+1) ππ+1 is obtained from π(π) ← ππ+1 by adding ππ‘−π at the rightmost ππ+1 end of some row of π(π) ← ππ+1 , say the πth row. It now suffices to ensure that ππ‘−π is strictly bigger than all entries in the πth ππ+1 row of π(π) ← ππ+1 . It follows from the defining properties of the pair of negative skew-symmetric lexicographic multisets {π1 , π2 } on N2 that ππ‘−π is bigger than or equal to all entries ππ+1 of π(π) ← ππ+1 . But here we need to prove something sharper; namely, ππ‘−π is strictly bigger than all entries in the πth row of ππ+1 π(π) ← ππ+1 . We will prove this now. ππ+1 Clearly all the entries of π(π) ← ππ+1 are contained in {π1 , . . . , ππ+1 } ∪Μ {ππ‘+1−π , . . . , ππ‘ } (see Section 4 of [8] for the Μ Also it is easy to observe that ππ < ππ‘−π ∀π ∈ notation ∪). {1, . . . , π + 1}. So if the rightmost element of the πth row of ππ+1 π(π) ← ππ+1 equals ππ for some π ∈ {1, . . . , π + 1}, then we are done. Otherwise, the element in the rightmost end of the πth ππ+1 row of π(π) ← ππ+1 is ππ for some π ∈ {π‘ + 1 − π, . . . , π‘} (say π0 ). If ππ0 < ππ‘−π , then we are done. If not, then clearly ππ0 = ππ‘−π . It then follows from duality that ππ‘+1−π0 = ππ+1 , and it is also clear that π‘ + 1 − π0 < π + 1. But it is an easy exercise to check that if π, πσΈ ∈ {1, . . . , π‘} are such that π < πσΈ and ππ = ππσΈ , then the number of the row σΈ in which ππ‘+1−πσΈ lies in π(π ) is strictly bigger than the number σΈ of the row in which ππ‘+1−π lies in π(π ) (Here, the row number is counted from top to bottom.). So ππ0 and ππ‘−π cannot lie in the same row of π(π+1) , a contradiction, hence proved. The proof of the following lemma appears in Section 9. Lemma 16. If {π1 , π2 } is a pair of negative skew-symmetric lexicographic multisets on N2 , then OBRSK({π1 , π2 }) is a negative skew-symmetric notched bitableau. Lemma 17. The map OBRSK is a degree-preserving bijection from the set of all pairs of negative skew-symmetric lexicographic multisets on N2 to the set of all negative skewsymmetric notched bitableaux. Proof. That OBRSK is degree preserving is obvious. To show that OBRSK is a bijection, we define its inverse, which we call the reverse of OBRSK or ROBRSK. Note that the entire procedure used to form (π(π+1) , π(π+1) ) from (π(π) , π(π) ), ππ+1 , ππ+1 , ππ‘−π and ππ‘−π , π = 1, . . . , π‘ − 1, is reversible. In other words, by knowing only (π(π+1) , π(π+1) ), we can retrieve (π(π) , π(π) ), ππ+1 , ππ+1 , ππ‘−π , and ππ‘−π . First, we obtain ππ+1 ; it is the minimum entry of π(π+1) . Look at the lowest row in which ππ+1 appears in π(π+1) , say it is row number π (counting from top to bottom). In the same row (row number π , counting from top to bottom) of π(π+1) , look at the rightmost entry: this entry is precisely ππ‘−π . Remove this entry (which is ππ‘−π ) from the π th row of π(π+1) , that will give us the notched 10 International Journal of Combinatorics ππ+1 tableau π(π) ← ππ+1 . Similarly remove the leftmost entry (which is ππ+1 ) from the π th row of π(π+1) , and all other entries in this row of π(π+1) should be moved one box to the left: this dual will give us the notched tableau π(π) ← ππ‘−π . ππ+1 Then, in the π th row of π(π) ← ππ+1 , select the greatest entry which is less than ππ+1 . This entry was the new box of the bounded insertion. If we begin reverse bounded insertion with this entry, we retrieve π(π) and ππ+1 . Look at the path in ππ+1 π(π) ← ππ+1 starting from the π th row to the topmost row, along which this reverse bounded insertion had happened. dual Trace the “dual path” in π(π) ← ππ‘−π and do a dual of the reverse bounded insertion (which was done originally on ππ+1 dual π(π) ← ππ+1 to retrieve π(π) and ππ+1 ) on π(π) ← ππ‘−π : that (π) dual (π) will give us π and ππ‘−π out of π ← ππ‘−π . We call this process of obtaining (π(π) , π(π) ), ππ+1 , ππ+1 , ππ‘−π , and ππ‘−π from (π(π+1) , π(π+1) ) described in the paragraphs above a reverse step and denote it by (π(π) , π(π) ) = (π+1) (π+1) ππ+1 ,ππ‘−π (π ,π ) σ³¨σ³¨σ³¨σ³¨σ³¨→ ππ+1 , ππ‘−π . We call the process of applying all the reverse steps sequentially to retrieve {π1 , π2 } from (π(π‘) , π(π‘) ) the reverse of OBRSK or ROBRSK. If (π(π‘) , π(π‘) ) is an arbitrary negative skew-symmetric notched bitableau (which we do not assume to be OBRSK ({π1 , π2 }) for some {π1 , π2 }), then we can still apply a sequence of reverse steps to (π(π‘) , π(π‘) ), to sequentially obtain (π(π) , π(π) ), ππ+1 , ππ+1 , ππ‘−π , and ππ‘−π , π = π‘ − 1, . . . , 1. For this process to be well defined, however, it must first be checked that the successive (π(π) , π(π) ) are negative skewsymmetric notched bitableaux. For this, it suffices to prove a statement very similar to that proved in Lemma 16; namely, “If (π, π) is a negative skew-symmetric notched bitableau, π,π then (πσΈ , πσΈ ) := (π, π) σ³¨σ³¨→ π, π is a negative skew-symmetric notched bitableau, π < π, π < π, π < π, π < π are positive integers, π is greater than or equal to all entries of π, and π is less than or equal to all entries of π.” That π < π, π < π, π < π, and π < π are positive integers, π is greater than or equal to all entries of π, and π is less than or equal to all entries of π follows immediately from the definition of a reverse step. That (πσΈ , πσΈ ) is a negative skew-symmetric notched bitableau follows in much the same manner as the proof of Lemma 16; we omit the details. It remains to be shown that the pair of multisets on N2 produced by applying this sequence of reverse steps to the arbitrary skew-symmetric notched bitableau (π(π‘) , π(π‘) ) is skew-symmetric lexicographic. The proof of this uses the duality property of skew-symmetric notched bitableaux, the facts mentioned in the preceding paragraph regarding the integers π, π, π, and π, and the rest of the proof goes similarly as in the proof of Lemma 6.3 of [8]. At each step, OBRSK and the reverse of ROBRSK are inverse to each other. Thus they are inverse maps. The map OBRSK can be extended to all pairs of nonempty skew-symmetric lexicographic multisets on N2 . If {π1 , π2 } is a pair of positive skew-symmetric lexicographic multisets on N2 , then define OBRSK({π1 , π2 }) to be π(OBRSK(πΏ({π1 , π2 }))), where the definition of the map π is given at the end of Section 4. If {π1 , π2 } is a pair of nonempty skew-symmetric lexicographic multisets on N2 , with negative and positive parts {π1− , π2− } and {π1+ , π2+ }, then define OBRSK({π1 , π2 }) to be the skew-symmetric notched bitableau whose negative and positive parts are OBRSK ({π1− , π2− }) and OBRSK ({π1+ , π2+ }). Example 18. Let π1 = {(4, 17), (3, 17), (3, 14), (7, 10), (4, 9)}, and π2 = {(20, 25), (19, 22), (15, 26), (12, 26), (12, 25)}. See Figure 2. Therefore. 3 4 12 19 10 17 25 26 OBRSK(π1 , π2 ) = ( 3 7 12 20 , 9 17 22 26 ). 4 15 14 25 (7) 7. Restricting the OBRSK Correspondence 7.1. Restricting by π½. Let π½ ∈ πΌ(π). We say that a skewsymmetric notched bitableau (π, π) is on π½ × π½, if all entries of π are in π½, all entries of π are in π½, and the sum of any entry in π (or in π) with its dual (with respect to (π, π)) is 2π + 1. Given any monomial π in OR(π½), we can define a monomial π# in AR(π½) as follows: π# := {(π∗ , π∗ )|(π, π) ∈ π}. We say that a pair {π1 , π2 } of skew-symmetric lexicographic multisets on N2 is a pair of skew-symmetric lexicographic multisets on π½ × π½, if π1 is a monomial in OR(π½), π2 is a monomial in AR(π½), a number of elements (counting multiplicities) in π1 and π2 are the same, and π2 = π1# . It is easy to see from the construction of OBRSK that, if {π, π# } is a pair of nonempty skew-symmetric lexicographic multisets on π½ × π½, then OBRSK ({π, π# }) is a nonvanishing skewsymmetric notched bitableau on π½ × π½, and vice versa. Thus we obtain the following. Corollary 19. The map OBRSK restricts to a degree-preserving bijection from the set of all pairs of nonempty (resp., negative, positive) skew-symmetric lexicographic multisets on π½ × π½ to the set of all nonvanishing (resp., negative, positive) skewsymmetric notched bitableaux on π½ × π½. 7.2. Restricting by π and π. Let π and π be negative and positive subsets of N2 , respectively, satisfying the condition that: π(1) , π(2) , π(1) , and π(2) are subsets of N, (8) where the notation π(1) , π(2) , π(1) , and π(2) is explained in Section 4 of [8]. In this subsection, we will define the boundedness of a nonempty pair of skew-symmetric lexicographic multisets on N2 by π, π, and this is given in Definition 26, after some definitions which are necessary to understand it. Definition 20. A dual pair of chains in N2 is a pair of subsets {πΆ1 = {(π1 , π1 ), . . . , (ππ , ππ )}, πΆ2 = {(π1 , β1 ), . . . , (ππ , βπ )}} International Journal of Combinatorics 11 π(0) = ∅ π(0) = ∅ 14 dual π(0) ← 4 = 4 π(1) = (1) 17 π(0) ← 25 = 25 π(1) = 17 25 4 12 17 3 12 ← 3 = 4 12 ← 3 = 4 3 12 π(2) = 4 12 3 12 14 3 12 14 ← 3 = 3 12 π(2) ← 3 = 4 12 4 3 12 π(3) = 3 12 4 15 dual dual π(1) ← 26 = 17 25 ← 26 = 17 26 25 17 26 π(2) = 17 25 17 26 dual 17 26 dual π(2) ← 26 = ← 26 = 17 26 17 25 25 17 26 π(3) = 17 26 14 25 3 12 3 7 12 10 10 π(3) ← 7 = 3 12 ← 7 = 3 12 4 15 4 15 17 26 17 22 26 dual dual π(3) ← 22 = 17 26 ← 22 = 17 26 14 25 14 25 π (4) π 10 17 22 26 π(4) = 17 26 14 25 3 7 12 19 = 3 12 4 15 3 7 12 19 3 4 12 19 9 9 π(4) ← 4 = 3 12 ← 4 = 3 7 12 4 15 4 15 10 17 22 26 10 17 25 26 dual dual π(4) ← 25 = 17 26 ← 25 = 17 22 26 14 25 14 25 3 4 12 19 π(5) = 3 7 12 20 4 15 10 17 25 26 π(5) = 9 17 22 26 14 25 Figure 2 such that πΆ1 is a chain in N2 in the sense of Section 7 of [8], and {π1 , π2 } is a pair of skew-symmetric lexicographic multisets on N2 , where π1 = {(π1 , π1 ), . . . , (ππ , ππ )} and π2 = {(π1 , β1 ), . . . , (ππ , βπ )}. Definition 21. Let {π1 , π2 } be a pair of skew-symmetric lexicographic multisets on N2 . Let {πΆ1 , πΆ2 } be a dual pair of chains in N2 such that πΆπ is contained in the underlying set of ππ for all π = 1, 2. Given any element in πΆ1 (say, the πth element counting from left to right), look at the first (counting from left to right) element in π1 which is entrywise the same as the πth element of πΆ1 . Call this element of π1 the πmin th element. Let π‘ be the total number of elements in π1 . One calls the (π‘ + 1 − πmin )th element of π2 (counting from left to right) the dual element in π2 corresponding to the πth element of πΆ1 . Definition 22. Given any pair {π1 , π2 } of skew-symmetric lexicographic multisets on N2 , where π1 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )} and π2 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )}, one says that the elements (ππ , ππ ) and (ππ‘+1−π , ππ‘+1−π ) of π1 and π2 , respectively, are dual to each other with respect to {π1 , π2 }. Definition 23. Given a pair of {π1 , π2 } of skew-symmetric lexicographic multisets on N2 , one says that a dual pair {πΆ1 , πΆ2 } of chains in N2 is a dual pair of chains in {π1 , π2 }, if the following two conditions are satisfied simultaneously: (i) πΆπ is contained in the underlying set of ππ for all π = 1, 2. (ii) Given any element of πΆ1 (say, the πth element counting from left to right), the dual element in π2 corresponding to it is entrywise the same as the dual element of the πth element of πΆ1 with respect to {πΆ1 , πΆ2 }. Definition 24. Given any row-strict notched bitableau (π, π), we associate with it 2 subsets of N2 as follows. Let π1 and π1 denote the topmost rows of π and π, respectively. Let π11 < ⋅ ⋅ ⋅ < π1π1 and π11 < ⋅ ⋅ ⋅ < π1π1 denote the entries of π1 and π1 , respectively. We denote by (π, π)up the subset of N2 given by {(π11 , π11 ), . . . , (π1π1 , π1π1 )}. We denote by (π, π)down the subset of N2 obtained similarly if we work with the lowermost rows of π and π, instead of the topmost rows. See Example 25 for an illustration. Example 25. Consider the row-strict notched bitableau π· π΄ πΎ π π» π½ π π πΌ π π , πΈ π΅ πΏ π) πΉ π πΆ π (πΊ (9) 12 International Journal of Combinatorics and call it (P, Q). Then (P, Q)up = {(π», π·), (π½, π΄), (π, πΎ), (π, π)}, and (P, Q)down = {(πΉ, πΆ), (π , π)}. Definition 26. A nonempty pair of skew-symmetric lexicographic multisets {π1 , π2 } on N2 is said to be bounded by π, π if for every dual pair {πΆ1 , πΆ2 } of chains in {π1 , π2 } one has up π ≤ (ππΆ1− ,πΆ2− , ππΆ1− ,πΆ2− ) , down (ππΆ1+ ,πΆ2+ , ππΆ1+ ,πΆ2+ ) (10) ≤ π, where one uses the order on multisets on N2 defined in Section 4 of [8]; (ππΆ1− ,πΆ2− , ππΆ1− ,πΆ2− ) is defined to be OBRSK ({πΆ1− , πΆ2− }), and (ππΆ1+ ,πΆ2+ , ππΆ1+ ,πΆ2+ ) is defined as OBRSK ({πΆ1+ , πΆ2+ }). It is worthwhile to note that (ππΆ1− ,πΆ2− , ππΆ1− ,πΆ2− )up , (ππΆ1+ ,πΆ2+ , 2 down , π, and π are subsets of N , and they are not ππΆ1+ ,πΆ2+ ) pairs of subsets! With this definition, the OBRSK correspondence is a bounded function, in the sense that it maps bounded sets to bounded sets. More precisely, we have the following Lemma (to understand the statement of this lemma, we need to recall the notion of a semistandard notched bitableau being bounded by π, π from Section 5 of [8]), whose proof appears in Section 9. Lemma 27. If a pair {π1 , π2 } of nonempty skew-symmetric lexicographic multisets on N2 is bounded by π, π, then OBRSK({π1 , π2 }) is bounded by π, π. Definition 28. A dual pair {πΆ1 , πΆ2 } of chains in N2 is called a dual pair of chains in π½ × π½ if {πΆ1 , πΆ2 } is a pair of skewsymmetric lexicographic multisets on π½ × π½. Remark 29. A general dual pair of chains in π½ × π½ will look like {πΆ, πΆ# } for some extended π½-chain πΆ in OR(π½). Note also that if {π1 , π2 } is a pair of skew-symmetric lexicographic multisets on π½ × π½, that is, if π2 = π1# , then any dual pair of chains {πΆ1 , πΆ2 } in {π1 , π2 } must be a dual pair of chains in π½ × π½; in other words, we must have πΆ2 = πΆ1# , where πΆ1 is an extended π½-chain in OR(π½). Remark 30. Let π and π be negative and positive subsets of N2 , respectively, satisfying (8). A nonempty pair of skewsymmetric lexicographic multisets {π, π# } on π½ × π½ is said to be bounded by π, π if, for every dual pair of chains {πΆ, πΆ# } in π½ × π½ which is contained in the underlying set of {π, π# }, up π ≤ (ππΆ− ,πΆ− # , ππΆ− ,πΆ− # ) , down (ππΆ+ ,πΆ+ # , ππΆ+ ,πΆ+ # ) (11) ≤ π, where we use the order on multisets on N2 defined in Section 4 of [8], and (ππΆ− ,πΆ− # , ππΆ− ,πΆ− # ) (resp., (ππΆ+ ,πΆ+ # , ππΆ+ ,πΆ+ # )) is defined to be OBRSK ({πΆ− , πΆ− }) (resp., OBRSK # ({πΆ+ , πΆ+ })). It is worthwhile to note that (ππΆ− ,πΆ− # , ππΆ− ,πΆ− # )up , (ππΆ+ ,πΆ+ # , ππΆ+ ,πΆ+ # )down , π, and π are subsets of N2 ; they are not pairs of subsets! # Let π and π be negative and positive subsets of π½ × π½, respectively, satisfying (8). Combining Corollary 19 and Lemma 27, we obtain the following. Corollary 31. For any positive integer π, the number of pairs of nonempty skew-symmetric lexicographic multisets on π½ × π½ bounded by π, π of degree 2π is less than or equal to the number of nonvanishing skew-symmetric notched bitableaux on π½ × π½ bounded by π, π of degree 2π. 8. The Initial Ideal In this section, we will prove the main result of the paper, that is, Theorem 7. Proof of Theorem 7. We wish to show that inβ³ πΌ = β¨ChainsπΎπΌ (π½)β©. Since we already know from Remark 12 that π‘ππ₯π‘πΆβππππ πΎπΌ (π½) ⊆ inβ³ πΌ, it follows that β¨ChainsπΎπΌ (π½)β© ⊆ inβ³ πΌ. Hence it suffices to show that inβ³ πΌ ⊆ β¨ChainsπΎπΌ (π½)β© or equivalently that, for any π ≥ 1, cardinality of the set of all degree π monomials in π/β¨ChainsπΎπΌ (π½)β© is less than or equal to cardinality of the set of all degree π monomials in πΎ π/inβ³ πΌ. Now since standard monomials on ππΌ,π½ (standard πΎ monomials on ππΌ,π½ is defined in Definition 32) and the monomials in π/inβ³ πΌ both induce homogeneous bases for π/πΌ, it suffices to show that, for any π ≥ 1, cardinality of the set of all degree π monomials in π/β¨ChainsπΎπΌ (π½)β© is less than or equal to cardinality of the set of all degree π standard πΎ monomials on ππΌ,π½ . We wish to give a different indexing set for the standard πΎ monomials on ππΌ,π½ . Let πΌπ½ (skew-symm) denote the set of all pairs (π , π) such that all of the following conditions are satisfied. (i) π ⊂ π½. (ii) π ⊂ π½. (iii) |π | = |π|, and this cardinality is even. (iv) If π = {π1 < ⋅ ⋅ ⋅ < π2π } and π = {π 1 < ⋅ ⋅ ⋅ < π 2π }, then ππ + π 2π+1−π = 2π + 1 ∀π ∈ {1, . . . , 2π}. Defining π − π := π ∪Μ (π½ \ π) (see Section 4 of [8]), we have the following fact, which is easily verified. The map (π , π) σ³¨→ π − π is a bijection from πΌπ½ (skewsymm) to πΌ(π), (Indeed, the inverse map is given by π σ³¨→ (π \ π½, π½ \ π).). Note that under this bijection, (0, 0) maps to π½. Let (π πΌ , ππΌ ) and (π πΎ , ππΎ ) be the preimages of the elements πΌ and πΎ (of πΌ(π)), respectively. Define ππΌ and ππΎ to be any subsets of π½ × π½ such that (ππΌ )(1) = π πΌ , (ππΌ )(2) = ππΌ , (ππΎ )(1) = π πΎ , and (ππΎ )(2) = ππΎ . Observe that ππΌ and ππΎ satisfy (8). Under this identification of πΌπ½ (skew-symm) with πΌ(π), the inequalities which define nonvanishing skew-symmetric notched bitableaux on π½ × π½ bounded by ππΌ , ππΎ are precisely International Journal of Combinatorics 13 the inequalities which define the standard monomials on πΎ ππΌ,π½ (the inequalities (12), (13), and (14) of this paper). Thus we obtain that the degree 2π nonvanishing skew-symmetric notched bitableaux on π½ × π½ bounded by ππΌ , ππΎ forms an πΎ indexing set for the degree π standard monomials on ππΌ,π½ . Hence it suffices to show that, for any π ≥ 1, cardinality of the set of all degree π monomials in π/β¨ChainsπΎπΌ (π½)β© is less than or equal to cardinality of the set of all nonvanishing skewsymmetric notched bitableaux on π½ × π½ bounded by ππΌ , ππΎ of degree 2π. On the other hand, it is easy to see that the set ChainsπΎπΌ (π½) can be rewritten in the following format: ChainsπΎπΌ (π½) is the set of all ππΆ, where πΆ is any nonempty extended π½-chain in OR(π½) such that either (i)σΈ or (ii)σΈ below holds. σΈ − (i) The chain πΆ is nonempty, and π πΌ − ππΌ up up (ππΆ− ,πΆ− # , ππΆ− ,πΆ− # )(1) − (ππΆ− ,C− # , ππΆ− ,πΆ− # )(2) . β° (ii)σΈ The chain πΆ+ is nonempty, and down (ππΆ+ ,πΆ+ # , ππΆ+ ,πΆ+ # )(1) − (ππΆ+ ,πΆ+ # , ππΆ+ ,πΆ+ # )down β° (2) π πΎ − ππΎ . It is now obvious that the pairs of nonempty skewsymmetric lexicographic multisets on π½ × π½ bounded by ππΌ , ππΎ of degree 2π form an indexing set for the degree π monomials of π/β¨ChainsπΎπΌ (π½)β©. Hence we have that, for any π ≥ 1, cardinality of the set of all degree π monomials in π/β¨ChainsπΎπΌ (π½)β© is equal to cardinality of the set of all pairs of nonempty skew-symmetric lexicographic multisets on π½ × π½ bounded by ππΌ , ππΎ of degree 2π. Again from Corollary 31, it follows that for any π ≥ 1, cardinality of the set of all pairs of nonempty skew-symmetric lexicographic multisets on π½ × π½ bounded by ππΌ , ππΎ of degree 2π is less than or equal to cardinality of the set of all nonvanishing skew-symmetric notched bitableaux on π½ × π½ bounded by ππΌ , ππΎ of degree 2π. Thus β¨ChainsπΎπΌ (π½)β© ⊇ inβ³ πΌ. Definition 32. Let π = k[π(π,π) | (π, π) ∈ OR(π½)]. Recall the concept of a Pfaffian (denoted by ππ,π½ for π ∈ πΌ(π)) from Section 3.3 of this paper. One calls π = ππ1 ,π½ ⋅ ⋅ ⋅ πππ ,π½ ∈ π a standard monomial if π1 , . . . , ππ ∈ πΌ(π), π1 ≤ ⋅ ⋅ ⋅ ≤ ππ (12) and for each π ∈ {1, . . . , π}, either ππ < π½ or ππ > π½. (13) If in addition, for πΌ, πΎ ∈ πΌ(π), πΌ ≤ π1 , ππ ≤ πΎ, (14) πΎ ππΌ,π½ . then we say that π is standard on We define the degree of the standard monomial ππ1 ,π½ ⋅ ⋅ ⋅ πππ ,π½ to be the sum of the π½-degrees of π1 , . . . , ππ , where for any π ∈ πΌ(π); the π½-degree is defined to be one-half the cardinality of π \ π½. Remark 33. In general, a standard monomial is not a monomial in the affine coordinates π(π,π) , (π, π) ∈ OR(π½); rather, it is a polynomial. It is only a monomial in the ππ,π½ ’s. 9. Proofs In this section, we give proofs of Lemmas 16 and 27. To give a proof of Lemma 16, we need Lemma 34 mentioned below whose proof is left to the reader as an exercise. Lemma 34. If π΄ and π΅ are two finite multisets on N of the same degree such that π΄ ≤ π΅ (where the termwise order ≤ on multisets on N and the degree of a multiset on N are as defined in Section 4 of [8]) and π, π are natural numbers such that π ≤ π, then π΄ ∪Μ {π} ≤ π΅ ∪Μ {π}. 9.1. Proof of Lemma 16 Proof. The proof is by induction. The base case of induction is easy to see, and hence we omit the details. Let {π1(π‘−1) , π2(π‘−1) } be a pair of negative skew-symmetric lexicographic multisets on N2 , where π1(π‘−1) = {(π1 , π1 ), . . . , (ππ‘−1 , ππ‘−1 )} and π2(π‘−1) = {(π2 , π2 ), . . . , (ππ‘ , ππ‘ )}. Let (π, π) = OBRSK ({π1(π‘−1) , π2(π‘−1) }). Assume inductively that (π, π) is a negative skew-symmetric notched bitableau. Now let {π1(π‘) , π2(π‘) } be a pair of negative skew-symmetric lexicographic multisets on N2 such that π1(π‘) = π1 and π2(π‘) = π2 , where π1 and π2 are given by π1 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )} and π2 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )}. Let (πσΈ , πσΈ ) = OBRSK ({π1(π‘) , π2(π‘) }). It suffices to show that (πσΈ , πσΈ ) is also a negative skew-symmetric notched bitableau. We will first prove that (πσΈ , πσΈ ) is a negative semistandard notched bitableau. The fact that πσΈ is row strict follows from Lemma 15. It then follows from duality that πσΈ is also row strict. Hence (πσΈ , πσΈ ) is row strict. Let πσΈ be the total number of rows of πσΈ (ππ πσΈ ). Let ππσΈ , ππσΈ , ππ , and ππ denote the multiset consisting of all elements in the πth row of πσΈ , πσΈ , π, and π, respectively. In fact, these multisets (ππσΈ , ππσΈ , ππ , ππ ) happen to be sets because πσΈ , πσΈ , π, and π are row strict. It needs to be σΈ σΈ − ππ+1 for 1 ≤ π ≤ πσΈ − 1, and shown that ππσΈ − ππσΈ ≤ ππ+1 σΈ σΈ ππσΈ − ππσΈ β 0 (see Section 4 of [8] for the notation β). We will prove the former statement first. To prove that ππσΈ − ππσΈ ≤ σΈ σΈ ππ+1 − ππ+1 for 1 ≤ π ≤ πσΈ − 1, there will be three nontrivial possibilities which will be registered below as Cases 1, 2, and 3, respectively. Case 1. ππ‘ and π1 are added to the first row of π, and π1 and ππ‘ are added to the first row of π. All rows other than the 1st row of π and π remain unchanged. In this case, it suffices to show that π1 ∪Μ π2 ∪Μ {ππ‘ , π1 } ≤ π2 ∪Μ π1 ∪Μ {π1 , ππ‘ }. It follows from induction hypothesis that π1 ∪Μ π2 ≤ π2 ∪Μ π1 . Since {π1(π‘) , π2(π‘) } is a pair of negative skew-symmetric lexicographic multisets on N2 , we have ππ‘ < ππ‘ and π1 < π1 . Therefore ππ‘ ≤ ππ‘ and π1 ≤ π1 . These facts together with Lemma 34 (applied twice) imply easily that π1 ∪Μ π2 ∪Μ {ππ‘ , π1 } ≤ π2 ∪Μ π1 ∪Μ {π1 , ππ‘ }. Hence we are done in this case. Case 2. π₯π bumps π¦π from ππ , π¦π bumps π§π from ππ+1 , and the dual bumping happens on ππ and ππ+1 . Let us express the dual bumping by saying that π₯π bumps π¦π from ππ and π¦π bumps π§π from ππ+1 . 14 International Journal of Combinatorics Clearly then, π₯π ≤ π¦π ≤ π§π < ππ‘ , and hence π₯π ≥ π¦π ≥ π§π by the property (ii) in the definition of a skew-symmetric notched bitableau. Again since all entries in π are ≥ ππ‘ , we get that π₯π ≥ π¦π ≥ π§π ≥ ππ‘ . σΈ σΈ − ππ+1 , or, in We need to show that ππσΈ − ππσΈ ≤ ππ+1 σΈ Μ σΈ σΈ σΈ σΈ σΈ = other words, ππ ∪ ππ+1 ≤ ππ+1 ∪Μ ππ . Note that ππ ∪Μ ππ+1 σΈ σΈ Μ π )\ [(ππ ∪Μ ππ+1 )\{π¦π , π§π }] ∪Μ {π₯π , π¦π } and ππ+1 ∪Μ ππ = [(ππ+1 ∪π {π§π , π¦π }] ∪Μ {π₯π , π¦π }. It suffices to show that (ππ ∪Μ ππ+1 ) \ {π¦π , π§π } ≤ (ππ+1 ∪Μ ππ ) \ {π§π , π¦π } because if we can prove this much, then since π₯π ≤ π¦π and π₯π ≥ π¦π we can apply Lemma 34 and get done with the proof as in Case 1. We will now prove that (ππ ∪Μ Qπ+1 ) \ {π¦π , π§π } ≤ (ππ+1 ∪Μ ππ ) \ {π§π , π¦π }. Since π§π < ππ‘ and all entries of ππ are ≥ ππ‘ , therefore π§π is the smallest element in ππ+1 ∪Μ ππ which is ≥ π¦π . It then follows from duality that π§π is the biggest element in ππ ∪Μ ππ+1 , that is, ≤ π¦π . Also since π¦π < ππ‘ ≤ π§π , we have π¦π < π§π , and hence by duality π¦π > π§π . Also by induction hypothesis, ππ ∪Μ ππ+1 ≤ ππ+1 ∪Μ ππ . All these facts put together prove the required thing easily, and we are done in this case. Case 3. π₯π bumps π¦π from ππ , and π¦π along with π1 are added to ππ+1 . The dual phenomenon happens with ππ and ππ+1 ; we express the dual phenomenon by saying that π₯π bumps π¦π from ππ , and π¦π along with ππ‘ are added to ππ+1 . Clearly then, π₯π ≤ π¦π < ππ‘ , and hence π₯π ≥ π¦π by the property (ii) in the definition of a Skew-symmetric notched bitableau. Again since all entries in π are ≥ ππ‘ , we get that π₯π ≥ π¦π ≥ ππ‘ . Also π¦π < π1 since the position of π1 is at the rightmost end of that row of ππ+1 in which π¦π is placed. Hence by duality, we have π¦π > ππ‘ . σΈ σΈ − ππ+1 , or, in We need to show that ππσΈ − ππσΈ ≤ ππ+1 σΈ Μ σΈ σΈ σΈ σΈ σΈ Μ Μ = other words, ππ ∪ ππ+1 ≤ ππ+1 ∪ ππ . Note that ππ ∪ ππ+1 σΈ Μ πσΈ = (ππ+1 ∪Μ ∪π (ππ ∪Μ ππ+1 \ {π¦π }) ∪Μ {ππ‘ } ∪Μ {π₯π , π¦π }, and ππ+1 ππ \ {π¦π }) ∪Μ {π1 } ∪Μ {π₯π , π¦π }. Using Lemma 34, it is enough to prove that (ππ ∪Μ ππ+1 \{π¦π }) ∪Μ {ππ‘ } ≤ (ππ+1 ∪Μ ππ \{π¦π }) ∪Μ {π1 }, which we will do now. Let ππ ∪Μ ππ+1 be given by π1 ≤ ⋅ ⋅ ⋅ ≤ π2π , and let ππ+1 ∪Μ ππ be given by π1 ≤ ⋅ ⋅ ⋅ ≤ π2π (since the bitableau (π, π) is of even size, the cardinality of ππ ∪Μ ππ+1 or of ππ+1 ∪Μ ππ is even). Note that there are no elements of ππ+1 which are ≥ π¦π and < ππ‘ . Also since ππ‘ is ≤ all elements of ππ , therefore we can conclude that there are no elements of ππ+1 ∪Μ ππ which are ≥ π¦π and <ππ‘ . It follows from duality that there are no elements of ππ ∪Μ ππ+1 which are ≤ π¦π as well as > π1 . Let (ππ ∪Μ ππ+1 \ {π¦π }) ∪Μ {ππ‘ } be given by π1 ≤ ⋅ ⋅ ⋅ ≤ ππ−1 ≤ ππ+1 ≤ ⋅ ⋅ ⋅ ≤ ππ ≤ ππ‘ ≤ ππ+1 ≤ ⋅ ⋅ ⋅ ≤ π2π (15) 9.2. Proof of Lemma 27 and let (ππ+1 ∪Μ ππ \ {π¦π }) ∪Μ {π1 } be given by π1 ≤ ⋅ ⋅ ⋅ ≤ π2π−π ≤ π1 ≤ π2π−π+1 ≤ ⋅ ⋅ ⋅ ≤ π2π−π ≤ π2π−π+2 ≤ ⋅ ⋅ ⋅ ≤ π2π . π1 ≤ ⋅ ⋅ ⋅ ≤ π2π ). Also since π¦π = ππ and π¦π < ππ‘ , it follows that π < π. Observe that ππ‘ is the πth element (counting from left to right) in the ordered sequence π1 ≤ ⋅ ⋅ ⋅ ≤ ππ−1 ≤ ππ+1 ≤ ⋅ ⋅ ⋅ ≤ ππ ≤ ππ‘ ≤ ππ+1 ≤ ⋅ ⋅ ⋅ ≤ π2π of integers and π1 (the dual of ππ‘ with respect to the pair {π1 , π2 } of multisets) is the πth element (counting from right to left) in the ordered sequence π1 ≤ ⋅ ⋅ ⋅ ≤ π2π−π ≤ π1 ≤ π2π−π+1 ≤ ⋅ ⋅ ⋅ ≤ π2π−π ≤ π2π−π+2 ≤ ⋅ ⋅ ⋅ ≤ π2π of integers. Consider any element in (ππ ∪Μ ππ+1 \ {π¦π }) ∪Μ {ππ‘ }. It can either be ππ for some π =ΜΈ π or ππ‘ . If it is ππ for some π =ΜΈ π, then we either need to show that ππ ≤ ππ , ππ ≤ π1 , or ππ ≤ ππ−1 (these are the only 3 possibilities; the last possibility ππ ≤ ππ−1 happens if and only if π ∈ {2π − π + 2, . . . , 2π − π + 1}). If it is ππ‘ , then we need to show that ππ‘ ≤ the πth element (counting from left to right) of (ππ+1 ∪Μ ππ \ {π¦π }) ∪Μ {π1 }, which can either be ππ , π1 , or ππ−1 . After all these observation(s), the proof follows easily using the inequalities π < π, π₯π ≤ π¦π < ππ‘ , π₯π ≥ π¦π ≥ ππ‘ , π¦π < π1 , and π¦π > ππ‘ and the facts that there are no elements of ππ+1 ∪Μ ππ which lie in the interval [π¦π , ππ‘ ) and that there are no elements of ππ ∪Μ ππ+1 which lie in the interval (π1 , π¦π ]. We have now proved that (πσΈ , πσΈ ) is a semistandard notched bitableau. To prove that the semistandard notched bitableau (πσΈ , πσΈ ) is negative, it is enough to prove that ππσΈ σΈ − ππσΈ σΈ β 0, where πσΈ denotes the total number of rows of πσΈ (ππ πσΈ ). If ππσΈ σΈ = ππσΈ and ππσΈ σΈ = ππσΈ , then since (π, π) is assumed to be a negative skew-symmetric notched bitableau, it follows immediately that ππσΈ σΈ − ππσΈ σΈ β 0. Otherwise, we do the following: let ππσΈ σΈ and ππσΈ σΈ be given by π 1 < ⋅ ⋅ ⋅ < π π σΈ and πΏ1 < ⋅ ⋅ ⋅ < πΏπ σΈ , respectively. It follows easily from the way the OBRSK algorithm works that there exists at least one entry in ππσΈ σΈ which is < ππ‘ . Let π π denote the largest entry in ππσΈ σΈ which is < ππ‘ . Since ππ‘ ≤ πΏ1 , it now follows easily that π π < πΏπ ∀π ∈ {1, . . . , π}. Since π π is the largest entry in ππσΈ σΈ which is <ππ‘ , therefore π ≥ 1, and it follows from duality that πΏπ σΈ +1−π = πΏπ σΈ −(π−1) > π1 . Hence πΏπ σΈ ≥ πΏπ σΈ −(π−1) > π1 = π π σΈ . It only remains to show that if π + 1 < π σΈ , then π π < πΏπ ∀π ∈ {π + 1, . . . , π σΈ − 1}. Clearly since π ≥ 1 and π + 1 < π σΈ , it follows that, in this case, the last row of π (resp., π) is the πσΈ th row, namely ππσΈ (resp., ππσΈ ), and πΏπ σΈ −(π−1) is the new box of the dual insertion in ππσΈ . Since π π σΈ = π1 , πΏπ σΈ −(π−1) > π1 , ππσΈ − ππσΈ β 0 by induction hypothesis and π π < π π σΈ ∀π ∈ {π + 1, . . . , π σΈ − 1}, it is now easy to see that π π < πΏπ ∀π ∈ {π + 1, . . . , π σΈ − 1}. Hence we have proved that (πσΈ , πσΈ ) is a negative semistandard notched bitableau. The fact that (πσΈ , πσΈ ) is skewsymmetric is easy to see from the very construction of the OBRSK algorithm and from the fact that the pair {π1(π‘) , π2(π‘) } of multisets is skew-symmetric, lexicographic. (16) This means that π¦π = ππ and its dual π¦q = π2π−π+1 (the πth element counting from right to left in the ordered sequence Proof (proof of Lemma 27). Let π be a negative subset of N2 such that π(1) , π(2) are subsets of N. Suppose that {π1 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )}, π2 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )}} is a pair of negative skew-symmetric lexicographic multisets on N2 . For each π = 1, . . . , π‘, let π1(π) := {(π1 , π1 ), . . . , (ππ , ππ )} and π2(π) := {(ππ‘+1−π , ππ‘+1−π ), . . . , (ππ‘ , ππ‘ )}. Let (π(π) , π(π) ) := International Journal of Combinatorics 15 OBRSK({π1(π) , π2(π) }). For proving Lemma 27, it suffices to show that if {π1 , π2 } is bounded by π, 0, then (π(π) , π(π) ) is bounded by π, 0 for all π = 1, . . . , π‘ because then the proof of Lemma 27 will follow similarly as the proofs of parts (iii), (iv), and (v) of Lemma 11.1 in [8]. Assume that {π1 , π2 } is bounded by π, 0. For every π ∈ {1, . . . , π‘}, let π1(π) (resp., π1(π) ) denote the topmost row of π(π) (resp., π(π) ). We need to show that π(1) − π(2) ≤ π1(π) − π1(π) . Equivalently, we need to show that for all positive integers π§, |(π(1) − π(2) )≤π§ | ≥ |(π1(π) − π1(π) )≤π§ |, where we use the definition π΄ − π΅ := π΄ ∪Μ (N \ π΅) for any two subsets π΄ and π΅ of N (see Section 4 of [8]). We will prove this by induction on π. Let us first see the base case of induction. When π = 1, π1(π) = {(π1 , π1 )} and π2(π) = {(ππ‘ , ππ‘ )}. Let πΆ1 = {(π1 , π1 )} and πΆ2 = {(ππ‘ , ππ‘ )}. Since {π1 , π2 } is bounded by π, 0, it follows that for the dual pair {πΆ1 , πΆ2 } of chains in {π1 , π2 }, we have π ≤ (ππΆ1− ,πΆ2− , ππΆ1− ,πΆ2− )up ≤ 0. But it can be easily seen that (ππΆ1− ,πΆ2− , ππΆ1− ,πΆ2− )up = {(π1 , π1 ), (ππ‘ , ππ‘ )}. This proves the base case of induction. Fix an integer π ∈ {1, . . . , π‘−1}, and assume that (π(π) , π(π) ) is bounded by π, 0 for all π ≤ π. We will now show that (π(π+1) , π(π+1) ) is bounded by π, 0. Before we begin the proof, let us fix some notation for (π) } to be the the rest of this subsection. Define {π1(π) , . . . , π2π π (π) (π) {π1(π) , . . . , π2π } π the topmost row topmost row of π and (π) of π , both listed in increasing order. Let (π, π) = (π(π) , π(π) ), π = ππ+1 , π = ππ+1 , π = ππ‘−π , π = ππ‘−π , (π) (πσΈ , πσΈ ) = (π(π+1) , π(π+1) ), {π1 , . . . , π2Μπ} = {π1(π) , . . . , π2π }, and π (π) }. Note that {π1 , . . . , π2Μπ} ⊂ {π1 , {π1 , . . . , π2Μπ} = {π1(π) , . . . , π2π π . . . , ππ } ∪Μ {ππ‘+1−π , . . . , ππ‘ } and {π1 , . . . , π2Μπ} ⊂ {π1 , . . . , ππ } ∪Μ {ππ‘+1−π , . . . , ππ‘ }. Let π1 (resp., π1 ) denote the topmost row of π (resp., π). Similarly let π1σΈ (resp., π1σΈ ) denote the topmost row of πσΈ (resp., πσΈ ). We consider two cases corresponding to the two ways in which (π1σΈ , π1σΈ ) can be obtained from (π1 , π1 ). Case 1. π1σΈ is obtained by π bumping ππ in π1 , for some 1 ≤ π ≤ 2Μπ, that is, π1σΈ = (π1 \ {ππ }) ∪Μ {π} and π1σΈ = (π1 \ {π2Μπ+1−π }) ∪Μ {π}. Since π is less than or equal to all elements of {π1 , . . . , ππ } and ππ < ππ‘+1−π ∀π ∈ {1, . . . , π‘}, it follows that π ≤ all elements of {π1 , . . . , ππ } ∪Μ {ππ‘+1−π , . . . , ππ‘ }. Therefore π < π ≤ π1 , and hence by duality π > π ≥ π2Μπ. The fact that π bumps ππ implies that π ≤ ππ and ππ < π. Hence π ≤ ππ < π ≤ π1 , and therefore by duality, we have π ≥ π2Μπ+1−π > π ≥ π2Μπ. For π§ < π, ππ ≤ π§ < π2Μc+1−π , or π§ ≥ π, σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨(π(1) − π(2) )≤π§ σ΅¨σ΅¨σ΅¨ ≥ σ΅¨σ΅¨σ΅¨(π1 − π1 )≤π§ σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ ≤π§ σ΅¨σ΅¨σ΅¨ σΈ = σ΅¨σ΅¨(π1 − π1σΈ ) σ΅¨σ΅¨σ΅¨ , σ΅¨ σ΅¨ (17) where the first inequality follows from induction hypothesis, and second equality follows from the facts that ππ−1 < π ≤ ππ < π ≤ π1 ≤ π2Μπ+1−π ≤ π and π2Μπ ≤ π < π. If π = ππ , then we are done. Thus we assume that π < ππ (and hence by duality that π > π2Μπ+1−π ). We now need to consider only two possible positions of π§, namely, π ≤ π§ < ππ and π2Μπ+1−π ≤ π§ < π. We claim that, for π§ such that π ≤ π§ < ππ or π2Μπ+1−π ≤ π§ < π, there exists a dual pair (1) (2) of chains {πΆπ+1 , πΆπ+1 } in {π1(π+1) , π2(π+1) } such that up up |((ππΆ(1)− ,πΆ(2)− , ππΆ(1)− ,πΆ(2)− )(1) − (ππΆ(1)− ,πΆ(2)− , ππΆ(1)− ,πΆ(2)− )(2) )≤π§ | ≥ π+1 π+1 π+1 π+1 π+1 π+1 π+1 π+1 |(π1σΈ − π1σΈ )≤π§ |. Assuming the claim and using the fact that π ≤ (ππΆ(1)− ,πΆ(2)− , ππΆ(1)− ,πΆ(2)− )up (which is because {π1 , π2 } π+1 π+1 π+1 π+1 is bounded by π, 0), we are done. The claim follows from Lemmas 41 and 42. Case 2. π1σΈ is obtained by adding π to π1 in position π from the left and adding π to π1 at the rightmost end (after π2Μπ); π1σΈ is obtained from π1 by adding π to the leftmost end of π1 and adding π at the backward πth position of π1 . That is, π1σΈ = π1 ∪Μ {π, π} = {π1 < ⋅ ⋅ ⋅ < ππ−1 < π < ππ < ⋅ ⋅ ⋅ < π2Μπ < π}, and π1σΈ = π1 ∪Μ {π, π} = {π < π1 < ⋅ ⋅ ⋅ < π2Μπ+1−π < π < π2Μπ+1−(π−1) < ⋅ ⋅ ⋅ < π2Μπ}. Note that ππ−1 < π < π < π1 , π < π ≤ ππ (since π > ππ would require that π bumps ππ in the bounded insertion process), and π < π < π < π. For π§ < π, ≤π§ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨(π(1) − π(2) )≤π§ σ΅¨σ΅¨σ΅¨ ≥ σ΅¨σ΅¨σ΅¨(π1 − π1 )≤π§ σ΅¨σ΅¨σ΅¨ = σ΅¨σ΅¨σ΅¨σ΅¨(π1σΈ − π1σΈ ) σ΅¨σ΅¨σ΅¨σ΅¨ , σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ (18) where the first inequality follows from induction hypothesis and the second equality follows from the facts that ππ−1 < π < ππ and π < π < π1 . For π§ such that π ≤ π§ < π, note that σ΅¨σ΅¨ σΈ σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨(π − πσΈ )≤π§ σ΅¨σ΅¨σ΅¨ = σ΅¨σ΅¨σ΅¨(πσΈ ∪Μ (N \ πσΈ ))≤π§ σ΅¨σ΅¨σ΅¨ 1 1 σ΅¨σ΅¨ 1 σ΅¨σ΅¨ σ΅¨σ΅¨ 1 σ΅¨σ΅¨ σ΅¨σ΅¨ ≤π§ σ΅¨σ΅¨ σ΅¨σ΅¨ ≤π§ σ΅¨σ΅¨ = σ΅¨σ΅¨σ΅¨(π1σΈ ) σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨(N \ π1σΈ ) σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨ ≤π§ ≤π§ σ΅¨ = (σ΅¨σ΅¨σ΅¨(π1 ) σ΅¨σ΅¨σ΅¨σ΅¨ + 1) + (σ΅¨σ΅¨σ΅¨σ΅¨(N \ π1 ) σ΅¨σ΅¨σ΅¨σ΅¨ − 1) σ΅¨ ≤π§ σ΅¨ σ΅¨ ≤π§ σ΅¨ = σ΅¨σ΅¨σ΅¨σ΅¨(π1 ) σ΅¨σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨σ΅¨(N \ π1 ) σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ ≤π§ σ΅¨ = σ΅¨σ΅¨σ΅¨σ΅¨(π1 − π1 ) σ΅¨σ΅¨σ΅¨σ΅¨ . (19) Hence for π ≤ π§ < π, we have ≤π§ σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨(π(1) − π(2) )≤π§ σ΅¨σ΅¨σ΅¨ ≥ σ΅¨σ΅¨σ΅¨(π1 − π1 )≤π§ σ΅¨σ΅¨σ΅¨ = σ΅¨σ΅¨σ΅¨σ΅¨(π1σΈ − π1σΈ ) σ΅¨σ΅¨σ΅¨σ΅¨ . σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ (20) For π§ ≥ π, σ΅¨σ΅¨ σΈ σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨(π − πσΈ )≤π§ σ΅¨σ΅¨σ΅¨ = σ΅¨σ΅¨σ΅¨(πσΈ ∪Μ (N \ πσΈ ))≤π§ σ΅¨σ΅¨σ΅¨ 1 1 σ΅¨σ΅¨ 1 σ΅¨σ΅¨ σ΅¨σ΅¨ 1 σ΅¨σ΅¨ σ΅¨σ΅¨ σΈ ≤π§ σ΅¨σ΅¨ σ΅¨σ΅¨ ≤π§ σ΅¨σ΅¨ = σ΅¨σ΅¨σ΅¨(π1 ) σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨(N \ π1σΈ ) σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ ≤π§ σ΅¨σ΅¨ ≤π§ σ΅¨ = (σ΅¨σ΅¨σ΅¨(π1 ) σ΅¨σ΅¨σ΅¨ + 2) + (σ΅¨σ΅¨σ΅¨(N \ π1 ) σ΅¨σ΅¨σ΅¨σ΅¨ − 2) σ΅¨ ≤π§ σ΅¨ σ΅¨ ≤π§ σ΅¨ = σ΅¨σ΅¨σ΅¨σ΅¨(π1 ) σ΅¨σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨σ΅¨(N \ π1 ) σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ ≤π§ σ΅¨ = σ΅¨σ΅¨σ΅¨σ΅¨(π1 − π1 ) σ΅¨σ΅¨σ΅¨σ΅¨ . (21) Hence for π§ ≥ π, we have ≤π§ σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨(π(1) − π(2) )≤π§ σ΅¨σ΅¨σ΅¨ ≥ σ΅¨σ΅¨σ΅¨(π1 − π1 )≤π§ σ΅¨σ΅¨σ΅¨ = σ΅¨σ΅¨σ΅¨σ΅¨(π1σΈ − π1σΈ ) σ΅¨σ΅¨σ΅¨σ΅¨ . σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ (22) 16 International Journal of Combinatorics It now remains to show that for π§ such that π ≤ π§ < π or π ≤ π§ < π, ≤π§ σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨(π(1) − π(2) )≤π§ σ΅¨σ΅¨σ΅¨ ≥ σ΅¨σ΅¨σ΅¨σ΅¨(π1σΈ − π1σΈ ) σ΅¨σ΅¨σ΅¨σ΅¨ . (23) σ΅¨ σ΅¨ σ΅¨ σ΅¨ We claim that for π§ such that π ≤ π§ < π or π ≤ π§ < π, there (1) (2) exists a dual pair of chains {ππ+1 , ππ+1 } in {π1(π+1) , π2(π+1) } such that σ΅¨σ΅¨ σ΅¨σ΅¨({topmost row of π (1) (2) } σ΅¨σ΅¨ ππ+1 ,ππ+1 ≤π§ σ΅¨σ΅¨ (24) ∪Μ (N \ {topmost row of ππ(1) ,π(2) })) σ΅¨σ΅¨σ΅¨σ΅¨ π+1 π+1 σ΅¨ σ΅¨σ΅¨ ≤π§ σ΅¨σ΅¨ ≥ σ΅¨σ΅¨σ΅¨(π1σΈ − π1σΈ ) σ΅¨σ΅¨σ΅¨ . σ΅¨ σ΅¨ Assuming the claim, we are done as in Case 1. But the proof of the claim is similar to the proof of the claim made in Case 1, so we omit it. Hence we are done with the required proof modulo, the proof of the claim made in Case 1. As mentioned in Case 1, the claim follows from Lemmas 41 and 42. It now remains to prove Lemmas 41 and 42. But before we go to prove these two lemmas, let us have a close look at their statements and try to understand the reason behind generating such complicated lemmas for the purpose of proving the claim. It is clear that once we understand the reason behind the statement of Lemma 41 being so complicated, Lemma 42 is a natural thing to be proved for finishing the proof of the claim. Observe that Lemma 41 looks like a Generalization of Lemma 11.1 (i) of [8] in the present case of the Orthogonal grassmannian. In Lemma 11.1 (i) of [8], Kreiman says that, for 1 ≤ π ≤ π(π), there exists a chain πΆπ,π in π(π) which has π elements, the last of which has first component ππ(π) . We need to compare this statement of Kreiman with the statement of Lemma 41 here. The fact that we need a dual pair of chains here (in Lemma 41) instead of “a chain” (as mentioned in Lemma 11.1 (i) of [8]) is quite natural to believe. But it is not apparently clear as to why we need the dual pair of chains in Lemma 41 to satisfy a “strange” condition which is condition (##) (this condition is explained in Definition 40). We will now illustrate by an example (see Example 35 and the illustration associated to it below) that asking for the dual pair of chains in Lemma 41 to satisfy condition (##) is not at all strange, in fact it is the only natural thing to be asked for in the present case of the Orthogonal grassmannian. Another explanation for the same is provided in Remark 36. Example 35. Let π = 5 and π½ = (5, 7, 8, 9, 10). Consider the multiset π = {(1, 9), (3, 7)} in N2 contained in the underlying set of π½ × π½. Applying the map BRSK (which is defined in Section 6 of [8]) to the multiset π, we get BRSK (π) = ( 1 3 , 7 9 ) . Set πΆ = πΆ 1,1 2,1 = {(1, 9)} and πΆ2,2 = π. It is easy to see that the chains πΆ1,1 , πΆ2,1 , and πΆ2,2 satisfy the required conditions of lemma 11.1 (i) of [8]. On the other hand, let us take the pair of negative skewsymmetric lexicographic multisets on N2 given by π1 , π2 , where π1 = π and π2 = π# = {(4, 8), (2, 10)}. Observe that OBRSK ({π, π# }) = ( 1 2 3 4 , 7 8 9 π ) , where π = 10 (we have denoted 10 by the alphabet π here because the young tableau package used here does not work with 2(1) (2) (1) = {(1, 9)}, πΆ1,1 = {(2, 10)}, πΆ1,2 = digit integers). Set πΆ1,1 (2) (1) (2) {(1, 9)}, πΆ1,2 = {(2, 10)}, πΆ2,1 = {(1, 9)}, πΆ2,1 = {(2, 10)}, (1) (2) (1) (2) = {(1, 9)}, πΆ2,2 = {(2, 10)}, πΆ2,3 = π1 , πΆ2,3 = π2 , and πΆ2,2 (1) (2) πΆ2,4 = π1 , πΆ2,4 = π2 . It is easy to see that according to the notation of Lemma 41, we have π(π) = 2 for π = 1 and π(π) = 4 for π = 2. It can also be easily verified that, for each π ∈ {1, 2} and each π ∈ {1, . . . , π(π)}, the dual (1) (2) , πΆπ,π } of chains (as mentioned above) satisfies the pair {πΆπ,π required condition (which is condition (##) with respect to the bitableau (π(π) , π(π) ) and the integer π) of Lemma 41. In fact, more is true; these dual pairs of chains satisfy condition (##)-strong (this condition is explained in Definition 39) with respect to the bitableau (π(π) , π(π) ) and the integer π for every π ∈ {1, 2} and π ∈ {1, . . . , π(π)}. Illustration Associated to Example 35. A possible analog of Lemma 11.1 (i) of [8] in the present case of the Orthogonal grassmannian would be to demand for the dual pair of chains (1) (2) (1) , πΆπ,π } in Lemma 41 to have exactly π elements in πΆπ,π {πΆπ,π (2) (1) (as well as in πΆπ,π ) and the last element of πΆπ,π to have first coordinate ππ(π) . But it is easy to see from Example 35 that such a demand is meaningless. Instead, it would be natural (1) (2) and meaningful to demand for the dual pair {πΆπ,π , πΆπ,π } of chains in Lemma 41 to satisfy condition (##)-strong (this is introduced in Definition 39) with respect to the bitableau (π(π) , π(π) ) and the integer π. But in Lemma 41, we demand the (1) (2) dual pair of chains {πΆπ,π , πΆπ,π } to satisfy condition (##) (this is introduced in Definition 40) which is little weaker a condition to be satisfied than condition (##)-strong. The reason behind (1) (2) , πΆπ,π } of this is that if we demand for the dual pair {πΆπ,π chains to satisfy condition (##)-strong instead of condition (##), then the proof of Lemma 41 becomes more difficult. This difficulty (in the proof of Lemma 41) arises at the point(s) (1) (2) , πΆπ+1,π } where we need to show that the dual pair {πΆπ+1,π (π+1) (π+1) of chains in {π1 , π2 } satisfies condition (##)-strong. (1) (2) , πΆπ+1,π } of It is easier to show that the dual pair {πΆπ+1,π chains satisfies condition (##). This completes the explanation behind the statement of Lemma 41 being so complicated. Remark 36. Note that the properties of the dual pair (1) (2) {πΆπ,π , πΆπ,π } of chains in Lemma 41 are a Generalization of the properties of the chain πΆπ,π of Lemma 11.1 of [8]. In fact, this Generalization precisely reveals the difference between the OBRSK algorithm (when applied to a dual pair of chains) in the present case of the Orthogonal Grassmannian and the BRSK algorithm of [8] (when applied to a chain) in the case of the ordinary Grassmannian. Equivalently, this Generalization reveals the difference between the concept of “O-domination of a chain by an element of πΌ(π)” in the case International Journal of Combinatorics of the Orthogonal Grassmannian (as in Section 2.4.4 of [9]) and the concept of “domination of a chain by an element of πΌ(π, 2π)” in the case of the ordinary Grassmannian as in [6]. Let us now prove Lemmas 41 and 42, but for the proofs we need some definitions which are the following. Definition 37. Let (π, π) and π1 be as in Definition 24. Let π ∈ {1, . . . , π1 }. With notation as in Definition 24, one calls (π1π , π1π ) the jth element of (π, π)up , and we denote by up (π, π)≤π the subset {(π11 , π11 ), . . . , (π1π , π1π )} of (π, π)up . See Example 38 for an illustration. Example 38. Consider the row-strict notched bitableau (P, Q) of Example 25. If we take π = 3, then the πth element of up (P, Q)up is (π, πΎ) and (P, Q)≤π equals {(π», π·), (π½, π΄), (π, πΎ)}. We can work out similarly for other possible values of π. Definition 39. Let (π, π) be a negative skew-symmetric notched bitableau. Let {π1 , . . . , π2π } and {π1 , . . . , π2π } be the topmost row of π and π, respectively, both listed in increasing order. Let π ∈ {1, . . . , 2π}. We say that a dual pair {πΆ1 , πΆ2 } of chains in N2 satisfies condition (##)-strong with respect to the bitableau (π, π) and the integer π if all the following conditions are satisfied simultaneously. (i) |πΆ1 |, |πΆ2 | ≤ π. (ii) The first coordinate of the πth element of (ππΆ1− ,πΆ2− , ππΆ1− ,πΆ2− )up is ππ . (iii) All the entries which occur as first coordinates of elements of πΆ1 form a subset of the set of all entries which occur as first coordinates of elements up of (ππΆ1− ,πΆ2− , ππΆ1− ,πΆ2− )≤π . Definition 40. Let (π, π) be a negative skew-symmetric notched bitableau. Let {π1 , . . . , π2π } and {π1 , . . . , π2π } be the topmost row of π and π, respectively, both listed in increasing order. Let π ∈ {1, . . . , 2π}. One says that a dual pair {πΆ1 , πΆ2 } of chains in N2 satisfies condition (##) with respect to the bitableau (π, π) and the integer π if all the following conditions are satisfied simultaneously. (i) |πΆ1 |, |πΆ2 | ≤ π. (ii) There exists an integer ππ which is ≥ π such that the first coordinate of the ππ th element of (ππΆ1− ,πΆ2− , ππΆ1− ,πΆ2− )up is ππ . (iii) All the entries which occur as first coordinates of elements of πΆ1 form a subset of the set of all entries which occur as first coordinates of elements up of (πC−1 ,πΆ2− , ππΆ1− ,πΆ2− )≤ππ . Lemma 41. Let {π1 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )}, π2 = {(π1 , π1 ), . . . , (ππ‘ , ππ‘ )}} be a pair of negative skew-symmetric lexicographic multisets on N2 . For π = 1, . . . , π‘, let π1(π) := {(π1 , π1 ), . . . , (ππ , ππ )} and π2(π) := {(ππ‘+1−π , ππ‘+1−π ), . . . , (ππ‘ , ππ‘ )}. (π) } Let (π(π) , π(π) ) := OBRSK({π1(π) , π2(π) }). Define {π1(π) , . . . , π2π π (π) } the topmost to be the topmost row of π(π) and {π1(π) , . . . , π2π π 17 row of π(π) , both listed in increasing order. Let π(π) := (π) (π) max{π ∈ {1, . . . , 2ππ } | ππ < π1(π) } = |(π1(π) )<π1 |. Then for (1) (2) , πΆπ,π } 1 ≤ π ≤ π(π), there exists a dual pair of chains {πΆπ,π in {π1(π) , π2(π) } which satisfies condition (##) with respect to the bitableau (π(π) , π(π) ) and the integer π. Let one denote by ππ,π the integer corresponding to condition (##) (item number (ii), which is ≥ π). Proof. We prove this by induction on π, with π = 1 the starting point of induction. When π = 1, π1(1) = {(π1 , π1 )} and π2(1) = {(ππ‘ , ππ‘ )}. There are two possible cases, namely, when π(π) = 1 and π(π) = 2. In both the cases, for every (1) (2) = {(π1 , π1 )} and πΆ1,π = {(ππ‘ , ππ‘ )}. π ∈ {1, . . . , π(π)}, take πΆ1,π This proves the base case of induction. Let (π, π) = (π(π) , π(π) ), π = ππ+1 , π = ππ+1 , π = ππ‘−π , π = ππ‘−π , (πσΈ , πσΈ ) = (π(π+1) , π(π+1) ), {π1 , . . . , π2Μπ} = (π) (π) {π1(π) , . . . , π2π }, and {π1 , . . . , π2Μπ} = {π1(π) , . . . , π2π }. Note π π that {π1 , . . . , π2Μπ} ⊂ {π1 , . . . , ππ } ∪Μ {ππ‘+1−π , . . . , ππ‘ } and {π1 , . . . , π2Μπ} ⊂ {π1 , . . . , ππ } ∪Μ {ππ‘+1−π , . . . , ππ‘ }. Let π1 (resp., π1 ) denote the topmost row of π (resp., π). Similarly let π1σΈ (resp., π1σΈ ) denote the topmost row of πσΈ (resp., πσΈ ). Since π is less than or equal to all elements of {π1 , . . . , ππ } and ππ < ππ‘+1−π ∀π ∈ {1, . . . , π‘}, it follows that π ≤ all elements of {π1 , . . . , ππ } ∪Μ {ππ‘+1−π , . . . , ππ‘ }. Therefore π < π ≤ π1 , and hence by duality π > π ≥ π2Μπ. We consider two cases corresponding to the two ways in which (π1σΈ , π1σΈ ) can be obtained from (π1 , π1 ). Case 1. π1σΈ is obtained by π bumping ππ in π1 , for some 1 ≤ π ≤ 2Μπ; that is, π1σΈ = (π1 \ {ππ }) ∪Μ {π} and π1σΈ = (π1 \ {π2Μπ+1−π }) ∪Μ {π}. The fact that π bumps ππ implies that π ≤ ππ and ππ < π. Hence π ≤ ππ < π ≤ π1 , and therefore, by duality, we have π ≥ π2Μπ+1−π > π ≥ π2Μπ. This implies that π(π + 1) = π(π). (1) (1) (2) (2) For π ∈ {1, . . . , π(π)} \ {π}, set πΆπ+1,π = πΆπ,π and πΆπ+1,π = πΆπ,π (Note that in these cases ππ(π) = ππ(π+1) .). We now consider (1) = {(π, π)} the case when π = π. If π = 1, then set πΆπ+1,π (2) and πΆπ+1,π = {(π, π)}. Otherwise consider the dual pair of (1) (2) (1) (1) , πΆπ,π−1 } in {π1(π) , π2(π) }. Let πΆπ+1,π := πΆπ,π−1 ∪ chains {πΆπ,π−1 (2) (2) {(π, π)} and πΆπ+1,π := πΆπ,π−1 ∪ {(π, π)}. Note that the dual (1) (2) , πΆπ+1,π } of chains in {π1(π+1) , π2(π+1) } satisfies the pair {πΆπ+1,π required conditions. Case 2. π1σΈ is obtained by adding π to π1 in position π from the left and adding π to π1 at the rightmost end (after π2Μπ); π1σΈ is obtained from π1 by adding π to the leftmost end of π1 and adding π at the backward πth position of π1 . That is, π1σΈ = π1 ∪Μ {π, π} = {π1 < ⋅ ⋅ ⋅ < ππ−1 < π < ππ < ⋅ ⋅ ⋅ < π2Μπ < π}, and π1σΈ ∪Μ {π, π} = {π < π1 < ⋅ ⋅ ⋅ < π2Μπ+1−π < π < π2Μπ+1−(π−1) < ⋅ ⋅ ⋅ < π2Μπ}. Since ππ−1 < π < π < π1 , it follows that π(π) ≥ (π − 1). Note that π < π ≤ ππ (since π > ππ would require that π bumps ππ in the bounded insertion process). Thus π(π + 1) = (1) (1) (2) (2) π. For π ∈ {1, . . . , π − 1}, set πΆπ+1,π = πΆπ,π and πΆπ+1,π = πΆπ,π . 18 International Journal of Combinatorics (1) (2) Consider the dual pair {πΆπ,π−1 , πΆπ,π−1 } of chains in {π1(π) , π2(π) }. (1) (1) (2) (2) ∪{(π, π)}. Note Let πΆπ+1,π := πΆπ,π−1 ∪{(π, π)} and πΆπ+1,π := πΆπ,π−1 (1) (2) that the dual pair {πΆπ+1,π , πΆπ+1,π } of chains in {π1(π+1) , π2(π+1) } satisfies the required conditions. Lemma 42. Let (π, π) be a negative skew-symmetric notched π,π bitableau. Let π, π, π, and π be integers such that (π, π) ← π,π π, π is well defined. Let (πσΈ , πσΈ ) := (π, π) ← π, π. Let π1 (resp., π1 ) denote the topmost row of π (resp., π). Similarly let π1σΈ (resp., π1σΈ ) denote the topmost row of πσΈ (resp., πσΈ ). Let {π1 , . . . , π2Μπ} (resp., {π1 , . . . , π2Μπ}) denote π1 (resp., π1 ), both listed in increasing order. Suppose π1σΈ is obtained by π bumping ππ in π1 , for some 1 ≤ π ≤ 2Μπ; that is, suppose π1σΈ = (π1 \ {ππ }) ∪Μ {π} and π1σΈ = (π1 \ {π2Μπ+1−π }) ∪Μ {π}. Let {πΆ1 , πΆ2 } be a dual pair of negative chains in N2 which satisfies condition (##) with respect to the bitableau (πσΈ , πσΈ ) and the integer π. Let πΆ1 = {(π1 , π1 ), . . . , (ππ , ππ ), (π, π)} and πΆ2 = {(π, π), (ππ , βπ ), . . . , (π1 , β1 )}, where π1 < ⋅ ⋅ ⋅ < ππ < π < ππ < π < ππ < ⋅ ⋅ ⋅ < π1 and β1 > ⋅ ⋅ ⋅ > βπ > π > π2Μπ+1−π > π > ππ > ⋅ ⋅ ⋅ > π1 . Then for π§ such that π ≤ π§ < ππ or π2Μπ+1−π ≤ π§ < π, one has σ΅¨σ΅¨ up up ≤π§ σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σΈ σΈ ≤π§ σ΅¨σ΅¨ σ΅¨σ΅¨((π σ΅¨σ΅¨ πΆ1 ,πΆ2 , ππΆ1 ,πΆ2 )(1) − (ππΆ1 ,πΆ2 , ππΆ1 ,πΆ2 )(2) ) σ΅¨σ΅¨σ΅¨ ≥ σ΅¨σ΅¨σ΅¨σ΅¨(π1 − π1 ) σ΅¨σ΅¨σ΅¨σ΅¨ . σ΅¨ σ΅¨ (25) Proof. Note that for π ≤ π§ < ππ , we have |((ππΆ1 ,πΆ2 , up up ππΆ1 ,πΆ2 )(1) – (ππΆ1 ,πΆ2 , ππΆ1 ,πΆ2 )(2) )≤π§ | = |({topmost row of ππΆ1 ,πΆ2 } − {topmost row of ππΆ1 ,πΆ2 })≤π§ | = |({topmost row of ππΆ1 ,πΆ2 } ∪Μ (N \ {topmost row of ππΆ1 ,πΆ2 }))≤π§ | = |{topmost row of ππΆ1 ,πΆ2 }≤π§ ∪Μ {N}≤π§ | ≥ ππ + π§ ≥ π + π§, where the last equality (not inequality!) is because π ≤ all entries in ππΆ1 ,πΆ2 and π ≤ π§ < ππ < π (All the other inequalities and equalities being obvious.). Also, π1 < ⋅ ⋅ ⋅ < ππ−1 < π < ππ < π ≤ π1 < ⋅ ⋅ ⋅ < π2Μπ, and π < π. Thus for π ≤ π§ < ππ , |(π1σΈ −π1σΈ )≤π§ | = |(π1σΈ ∪Μ (N\π1σΈ ))≤π§ | = |(π1σΈ ∪Μ N)≤π§ | = π + π§. Hence we are done in the case π ≤ π§ < ππ . Now for π§ such that π2Μπ+1−π ≤ π§ < π, we need to show that σ΅¨σ΅¨ σ΅¨σ΅¨ ({topmost row of ππΆ ,πΆ } σ΅¨σ΅¨ 1 2 ≤π§ σ΅¨σ΅¨ ∪Μ (N \ {topmost row of ππΆ1 ,πΆ2 })) σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ σΈ σ΅¨ ≤π§ σ΅¨ ≥ σ΅¨σ΅¨σ΅¨(π1 ∪Μ (N \ π1σΈ )) σ΅¨σ΅¨σ΅¨ . σ΅¨ σ΅¨ (26) Recall that π1σΈ = (π1 \ {ππ }) ∪Μ {π} and π1σΈ = (π1 \ π,π {π2Μπ+1−π }) ∪Μ {π}. Since (πσΈ , πσΈ ) = (π, π) ← π, π, therefore π ≥ all entries of πσΈ . Hence π ≥ all entries of π1σΈ . On the other hand, since π < ππ < π, it follows from duality that π > π2Μπ+1−π > π. So we have π1 < ⋅ ⋅ ⋅ < ππ−1 < π < ππ+1 < ⋅ ⋅ ⋅ < π2Μπ ≤ π < π2Μπ+1−π < π < π2Μπ+1−(π−1) < ⋅ ⋅ ⋅ < π2Μπ. It is now easy to observe that for π§ such that π2Μπ+1−π ≤ π§ < π, the number of elements in π1σΈ which are ≤ π§ is 2Μπ − π. Hence the number of elements in N \ π1σΈ which are ≤ π§ will be π§ − (2Μπ − π). It is also clear that all the elements of π1σΈ are ≤ π§, and there are 2Μπ many elements in π1σΈ . Therefore, σ΅¨σ΅¨ σΈ σ΅¨ σ΅¨σ΅¨(π ∪Μ (N \ πσΈ ))≤π§ σ΅¨σ΅¨σ΅¨ = 2Μπ + (π§ − (2Μπ − π)) 1 σ΅¨σ΅¨ 1 σ΅¨σ΅¨ (27) = 2Μπ + π§ − 2Μπ + π = π§ + π. Let πΌ1 < ⋅ ⋅ ⋅ < πΌ2Μπ and π½1 < ⋅ ⋅ ⋅ < π½2Μπ denote the topmost rows of ππΆ1 ,πΆ2 and ππΆ1 ,πΆ2 , respectively. It follows from the algorithm of OBRSK applied on {πΆ1 , πΆ2 } that π ≥ πΌ2Μπ > ⋅ ⋅ ⋅ > πΌ1 . On the other hand, since π < ππ < π, it follows from duality that π > π2Μπ+1−π > π. Hence combining all these, we have π > π2Μc+1−π > π ≥ πΌ2Μπ > πΌ1 . So for π§ such that π2Μπ+1−π ≤ π§ < π, the number of elements in the topmost row of ππΆ1 ,πΆ2 which are ≤ π§ is 2Μπ. Since {πΆ1 , πΆ2 } satisfies condition (##) with respect to the bitableau (πσΈ , πσΈ ) and the integer π, it follows that there exists an integer ππ (≥ π) such that the ππ th entry (counting from left to right) of the topmost row of ππΆ1 ,πΆ2 is π. Hence it follows from duality that the backward ππ th entry (i.e., the ππ th entry counting from right to left) of the topmost row of ππΆ1 ,πΆ2 is π. Therefore for π§ such that π2Μπ+1−π ≤ π§ < π, the number of elements in the topmost row of ππΆ1 ,πΆ2 which are ≤π§ is equal to π0 , where π0 is some nonnegative integer such that π0 ≤ 2Μπ−ππ . But ππ ≥ π, hence −ππ ≤ −π, and therefore π0 ≤ 2Μπ−ππ ≤ 2Μπ − π. Therefore, the number of elements in (N \ {topmost row of ππΆ1 ,πΆ2 }) which are ≤ π§ is π§ − π0 . Hence |({topmost row of ππΆ1 ,πΆ2 } ∪Μ (N \ {topmost row of ππΆ1 ,πΆ2 }))≤π§ | = 2Μπ + π§ − π0 ≥ 2Μπ + π§ − (2Μπ − π) = π§ + π = |(π1σΈ ∪Μ (N \ π1σΈ ))≤π§ |. References [1] R. P. Stanley, “Some combinatorial aspects of the Schubert calculus,” in Combinatoire et repreΜsentation du groupe symeΜtrique, vol. 579 of Lecture Notes in Mathematics, Springer, Berlin, Germany, 1977. [2] W. V. D. Hodge and D. Pedoe, Methods of Algebraic Geometry, vol. 1 of Cambridge Mathematical Library, Cambridge University Press, Cambridge, UK, 1994. [3] K. N. Raghavan and S. Upadhyay, “Initial ideals of tangent cones to Schubert varieties in orthogonal Grassmannians,” Journal of Combinatorial Theory Series A, vol. 116, no. 3, pp. 663–683, 2009. [4] B. Sturmfels, “GroΜbner bases and Stanley decompositions of determinantal rings,” Mathematische Zeitschrift, vol. 205, no. 1, pp. 137–144, 1990. [5] J. Herzog and N. V. Trung, “GroΜbner bases and multiplicity of determinantal and Pfaffian ideals,” Advances in Mathematics, vol. 96, no. 1, pp. 1–37, 1992. [6] V. Kodiyalam and K. N. Raghavan, “Hilbert functions of points on Schubert varieties in Grassmannians,” Journal of Algebra, vol. 270, no. 1, pp. 28–54, 2003. [7] V. Kreiman and V. Lakshmibai, Monomial bases and applications for Richardson and Schubert varieties in ordinary and affine Grassmannians [Ph.D. thesis], Northeastern University, 2003. [8] V. Kreiman, “Local properties of Richardson varieties in the Grassmannian via a bounded Robinson-Schensted-Knuth correspondence,” Journal of Algebraic Combinatorics, vol. 27, no. 3, pp. 351–382, 2008. International Journal of Combinatorics [9] K. N. Raghavan and S. Upadhyay, “Hilbert functions of points on Schubert varieties in orthogonal Grassmannians,” Journal of Algebraic Combinatorics, vol. 31, no. 3, pp. 355–409, 2010. [10] V. Lakshmibai and C. S. Seshadri, “Geometry of πΊ/π—II. 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