The Stembridge Equality for Skew Stable Grothendieck Polynomials and Skew Dual Stable Grothendieck Polynomials
aa r X i v : . [ m a t h . C O ] F e b The Stembridge Equality for Skew Stable GrothendieckPolynomials and Skew Dual Stable Grothendieck Polynomials
Fiona Abney-McPeek, Serena An, and Jakin NgJanuary 15, 2021
Abstract
The Schur polynomials s λ are essential in understanding the representation theory of thegeneral linear group. They also describe the cohomology ring of the Grassmannians. For ρ = ( n, n − , . . . ,
1) a staircase shape and µ ⊆ ρ a subpartition, the Stembridge equality statesthat s ρ/µ = s ρ/µ T . This equality provides information about the symmetry of the cohomologyring. The stable Grothendieck polynomials G λ , and the dual stable Grothendieck polynomials g λ , developed by Buch, Lam, and Pylyavskyy, are variants of the Schur polynomials and describethe K -theory of the Grassmannians. Using the Hopf algebra structure of the ring of symmetricfunctions and a generalized Littlewood-Richardson rule, we prove that G ρ/µ = G ρ/µ T and g ρ/µ = g ρ/µ T , the analogues of the Stembridge equality for the skew stable and skew dual stableGrothendieck polynomials. In this paper, we prove a Stembridge-type equality for skew stable Grothendieck polynomials andskew dual stable Grothendieck polynomials, namely G ρ/µ = G ρ/µ T , g ρ/µ = g ρ/µ T , where ρ = ( n, n − , . . . ,
1) is the staircase partition.The stable Grothendieck polynomials G λ are K -theoretic analogues of the Schur polynomials s λ ,i.e., they provide information about the K -theory of the Grassmanian. These formal power serieswere introduced by Fomin and Kirillov [3].In [2], Buch gave a combinatorial definition of the skew stable Grothendieck polynomials G λ/µ using set-valued tableaux of shape λ/µ , which are certain fillings of the skew Young diagram ofthe shape λ/µ with sets of positive integers. The dual stable Grothendieck polynomials g λ , firstintroduced by Lam and Pylyavskyy in [6], are dual to the G λ ’s under the Hall inner product. Theformal power series g λ/µ is defined using reverse plane partitions, which are certain fillings of theskew Young diagram of the shape λ/µ with positive integers.The skew stable and skew dual stable Grothendieck polynomials can be viewed as deformationsof the Schur polynomials in that their lowest and highest degree parts, respectively, are the Schurpolynomials. Thus, it is natural to ask whether certain identities for the Schur polynomials canbe extended to this context. For instance, it was conjectured in [1, Conjecture 6.2] that there areanalogues for g ρ/µ and G ρ/µ of the Stembridge equality [7, Corollary 7.32], which states that s ρ/µ = s ρ/µ T , for ρ = ( n, n − , . . . , K -theory of the Grassmanians.1 .1 Outline of the paper In Section 2, we begin by going over the basics of symmetric functions, Schur polynomials, and skewstable and skew dual stable Grothendieck polynomials. Then, we state the problem and introducethe Hopf algebraic structure of the ring of symmetric functions Λ.In Section 3.1, we give a combinatorial proof of Theorem 1.1, the Stembridge equality for skewdual stable Grothendieck polynomials g ρ/µ in the special case where µ = ( k ), using a generalizedLittlewood-Richardson rule for the stable Grothendieck polynomials proven by Buch [2]. Theorem 1.1.
Let ρ = ( n, n − , . . . , be the staircase partition, and µ = ( k ) where k ≤ n . Then, g ρ/µ = g ρ/µ T . In Section 3.2, we prove the Stembridge equality for skew dual stable Grothendieck polynomialsfor general µ , as stated in the following theorem. Theorem 1.2.
Let ρ = ( n, n − , . . . , be the staircase partition, and µ ⊆ ρ any subpartition.Then g ρ/µ = g ρ/µ T . We extend Theorem 1.1, the case for µ = ( k ), to Theorem 1.2, the general case, by utilizing theskewing operator ⊥ coming from the Hopf algebraic structure of Λ, along with an involution τ ofthe completion ˆΛ constructed by Yelliusizov in [9, Theorem 1.1] sending G µ to G µ T .In Section 4, we use a similar strategy to prove the Stembridge equality for skew stableGrothendieck polynomials G ρ/µ . Theorem 1.3.
Let ρ = ( n, n − , . . . , be the staircase partition, and µ ⊆ ρ any subpartition.Then G ρ/µ = G ρ/µ T . First, we prove the identity combinatorially for the case µ = ( k ) in Section 4.1, and thengeneralize it to arbitrary µ in Section 4.2 using the skewing operator and an involution τ of Λ, ananalogue of τ sending g µ to g µ T , introduced by Yelliusizov [9]. Remark 1.4.
In Section 2.8, we prove that the converses of Theorem 1.2 and Theorem 1.3 aretrue. That is, if G ρ/µ = G ρ/µ T (respectively, g ρ/µ = g ρ/µ T ) for all µ ⊆ ρ , then ρ = ( n, n − , . . . , n . This follows from Theorem 2.28, the converse statement in the case ofSchur polynomials, which we show in since s λ/µ is the bottom degree component of G λ/µ and thetop degree component of g λ/µ . A partition λ of a nonnegative integer n is a weakly decreasing sequence of positive integers( λ , λ , . . . , λ ℓ ) whose sum is n We write λ ⊢ n to indicate that λ is a partition of n. The in-teger λ i is the i th part of λ . The number of parts of λ is the length of λ , denoted ℓ ( λ ). Denote theset of all partitions of n by Par( n ), and let Par := S n ≥ Par( n ). Definition 2.1.
The
Young diagram of a partition λ , denoted Y ( λ ), is a left-aligned array with λ i cells in the i th row from the top. 2or example,is the Young diagram of λ = (5 , , , λ and µ are two partitions such that µ i ≤ λ i for all i , then we write µ ⊆ λ and say that µ is a subpartition of λ . We may additionally consider the skew partition λ/µ whose skew Young diagram consists of the cells belonging to Y ( λ ) but not to Y ( µ ). For example,is the Young diagram of λ/µ = (5 , , , / (2 , , conjugate of a partition λ , denoted λ T , is the partition whose i th part is the number ofentries of λ that are at least i . Equivalently, Y ( λ T ) is obtained from Y ( λ ) by a reflection overthe main diagonal. For example, (4 , ,
1) and (3 , , ,
1) are conjugates, as seen from their Youngdiagrams below. ⇐⇒ For a sequence of nonegative integers α = ( α , α , . . . ) with P α i = n, define x α := x α x α · · · . A homogeneous symmetric function of degree n is a formal power series f ( x ) = X α c α x α such that the c α are elements of some commutative ring R and for each permutation ω of thepositive integers, f ( x , x , . . . ) = f ( x ω (1) , x ω (2) , . . . ). For our purposes, we will take R = Q andlet Λ n denote the set of all homogeneous symmetric functions of degree n over Q . Additionally,Λ = Λ ⊕ Λ ⊕ · · · , the set of all symmetric functions, is a graded algebra over Q . Definition 2.2.
The elementary symmetric function e n is given by e n := X i < ···
The complete homogeneous symmetric function h n is given by h n := X i ≤···≤ i n x i · · · x i n . In particular, h n is the sum of all monomials with degree n . For a partition λ = ( λ , λ , . . . ), let h λ = h λ h λ · · · . The set { h λ } for all partitions λ forms a basis for Λ. Definition 2.4. A semistandard Young tableau (SSYT) of shape λ/µ is a filling of the cells of Y ( λ/µ ) with positive integers such that the entries weakly increase within each row and strictlyincrease within each column. A semistandard Young tableau T has type α = ( α , α , . . . ) where α i is the number of entries of T equal to i .For example, 2 41 1 41 2 23 46is a SSYT of shape (5 , , , , / (3 ,
1) and type (3 , , , , , T of type α = ( α , α , . . . ), let x T denote x α x α · · · . Definition 2.5.
For a skew shape λ/µ , the skew Schur function s λ/µ in the variables x =( x , x , . . . ) is given by s λ/µ = X T x T , where the sum is over all SSYT T of shape λ/µ . When µ = ∅ , then s λ is the Schur function of λ . Example 2.6.
Every SSYT T of shape λ/µ = (2 , , / (1) is of one of the following forms for somepositive integers i < j < k . iij jij ijk jik kij Thus, s (2 , , / (1) = X i For all skew partitions λ/µ , the skew Schur function s λ/µ is a symmetric function. Theorem 2.8. The set { s λ : λ ∈ Par( n ) } forms a basis for Λ n , and the set { s λ : λ ∈ Par } formsa basis for Λ . Definition 2.9. The Hall inner product h· , ·i on Λ is defined so that the Schur functions areorthonormal; that is, h s λ , s µ i = δ λµ , the Kronecker delta.Additionally, under the Hall inner product, e λ and h λ are dual bases; that is, h e λ , h µ i = δ λµ . Definition 2.10. A reverse plane partition of shape λ/µ is a filling of the cells of Y ( λ/µ ) withpositive integers such that the entries weakly increase within each row and column. A reverse planepartition P has weight w = ( w , w , . . . ), where w i is the number of columns of P containing i .For example, 1 2 2 41 2 51 2 2is a reverse plane partition of shape (5 , , / (1 , 1) and weight (2 , , , , P of weight w = ( w , w , . . . ), let x P denote x w x w · · · . Definition 2.11. For a skew shape λ/µ , define the skew dual stable Grothendieck polynomial g λ/µ to be g λ/µ = X P x P , where the sum is over all reverse plane partitions P of shape λ/µ . Example 2.12. Every reverse plane partition P of shape λ/µ = (2 , / (1) takes on one of thefollowing forms, for some positive integers i < j < k . ii i ii j ji j ij j ij k ji k Thus, g (2 , / (1) = X i x i + X i The terms of highest degree in g λ/µ are achieved by reverse plane partitions inwhich there are no numbers repeated in any column; that is, the columns are strictly increasing.In other words, the reverse plane partition must also be a semi-standard Young tableau. Thus, theterms of highest degree in g λ/µ form s λ/µ . 5 .5 Stable Grothendieck Polynomials Definition 2.14. For two sets A and B of positive integers, we say that A ≤ B if max A ≤ min B and A < B if max A < min B . A set-valued tableau of shape λ/µ is then a filling of the boxes of Y ( λ/µ ) with nonempty sets of positive integers such that the sets weakly increase along rows andstrictly increase along columns. Definition 2.15. Let | T | denote the sum of the sizes of the sets appearing in T . Example 2.16. The following is a set-valued tableau of shape (5 , , / (2 , 1) and size 15 . , , , , , , Definition 2.17. Let m i be the number of times that i appears in a the set-valued tableau T , andlet x T = x m x m · · · . Then the skew stable Grothendieck polynomial G λ/µ is given by G λ/µ := X T ( − | T |−| λ | x T , where the sum is over all set-valued tableaux T of shape λ/µ. Remark 2.18. A set-valued tableau of shape λ/µ filled with sets of size one is a semi-standardYoung tableau, corresponding to the monomials in G λ/µ of lowest degree. Thus, the terms of lowestdegree in G λ/µ form s λ/µ . The stable Grothendieck polynomial has terms of arbitrarily large degree. Remark 2.19. Let ˆΛ be the completion of Λ, given by allowing infinite linear combinations of agiven basis (e.g. the Schur polynomials). The Hall inner product h· , ·i : Λ × Λ → Z can be extendedto a pairing h· , ·i : ˆΛ × Λ → Z by linearly extending h s λ , s µ i = δ λ,µ as in [9]. The G λ are symmetric functions and they forma basis for ˆΛ . The G λ are also dual to the g λ under the (extended) Hall inner product; that is, h G λ , g µ i = δ λµ . The ring of symmetric functions Λ has a Hopf algebraic structure, as described in [5]. The comul-tiplication ∆ sends f ( x , x , · · · ) f ( x , x , · · · , y , y , · · · ) . Lemma 2.20. The comultiplication acts on g λ as follows: ∆( g λ ) = X µ ⊆ λ g µ ⊗ g λ/µ . roof. From the combinatorial definition, g λ = X P x P , summed over all reverse plane partitions P of shape λ with entries in the alphabet x < x < · · · < y < y < · · · . Since the rows and columns of P are weakly increasing, the restriction of thereverse plane partition to the alphabet x gives a reverse plane partition P x of shape µ ⊆ λ , and therestriction to the alphabet y gives a reverse plane partition P y of shape λ/µ .Then, g λ ( x, y ) = X P x P x · y P y = X µ ⊆ λ X P x x P x ! X P y y P y = X µ ⊆ λ g µ ( x ) g λ/µ ( y ) . Thus, we indeed have that ∆( g λ ) = X µ ⊆ λ g µ ⊗ g λ/µ . We next define the skewing operator ⊥ (see for instance [5], Section 2.8) which we will usethroughout the paper. Definition 2.21. Given ∆( a ) = P ( a ) a (1) ⊗ a (2) , written in Sweedler notation, the skewing operator ⊥ is defined as f ⊥ ( a ) := X ( a ) h f, a (1) i a (2) . Theorem 2.22. For any partition µ ⊆ λ , G ⊥ µ g λ = g λ/µ . Proof. Recall from Lemma 2.20 that ∆( g λ ) = P µ ⊆ λ g µ ⊗ g λ/µ . Then by the definition of the skewingoperator, G ⊥ µ g λ = X ν h G µ , g ν i g λ/ν = X ν δ µ,ν g λ/ν = g λ/µ . The skewing operator also has the following useful property; see, for instance, Section 2.8 of [5]. Lemma 2.23. Let A • be a graded Hopf algebra. For any f, g ∈ A • , h g, f ⊥ ( a ) i = h f g, a i . .7 The Stembridge Equality The Stembridge equality describes an important symmetry for the Schur polynomials and can beproved in a number of different ways (e.g. Corollary 7.32 in [7] and Exercise 2.9.25 in [5]). Theorem 2.24 (Stembridge Equality) . Let ρ = ( n, n − , . . . , be the staircase partition, and µ ⊆ ρ . Then s ρ/µ = s ρ/µ T . In this paper, we extend the Stembridge equality to the skew stable and skew dual stableGrothendieck polynomials.In addition, the converse is true. That is, if s λ/µ = s λ/µ T for all µ ⊆ λ , then λ = ( n, n − , . . . , n . To prove this, we use Pieri’s rule, a well-known fact described in [8,Theorem 7.15.7], for example. Definition 2.25. A skew shape λ/ν is a horizontal strip if it has no two squares in the samecolumn, or a vertical strip if no two squares are in the same row. Theorem 2.26 (Pieri’s Rule) . We have s λ/ ( k ) = X ν s ν , where ν ranges over all partitions ν ⊆ λ for which λ/ν is a horizontal strip of size k . Similarly, s λ/ (1 k ) = X ν s ν , where ν ranges over all partitions ν ⊆ λ for which λ/ν is a vertical strip of size k . Theorem 2.27. If s λ/ ( k ) = s λ/ (1 k ) for all positive integers k , then λ = ρ = ( n, n − , . . . , forsome positive integer n .Proof. Note that s ρ/ ( k ) is zero if and only if k is greater than the number of columns in the Youngdiagram of ρ , and s ρ/ (1 k ) is zero if and only if k is greater than the number of rows in the Youngdiagram of ρ . So we require that Y ( ρ ) has the same number of rows as columns; let this numberbe n .Since the s ν form a basis of Λ, using Pieri’s rule, we require that ρ/ν is a horizontal strip ofsize k if and only if it is a vertical strip of size k . For the sake of contradiction, suppose that ρ contains two consecutive parts of the same size. Then there exists some ν such that ρ/ν consistsof the rightmost box of these two rows. However, then ρ/ν forms a vertical strip of length 2 butnot a horizontal strip, which is a contradiction.Combining the fact that ρ has n rows and n columns and that no two rows have the same size,we have that ρ must be ( n, n − , . . . , 1) for some positive integer n , as desired. Corollary 2.28. If s ρ/µ = s ρ/µ T for all partitions µ , then ρ = ( n, n − , . . . , for some positiveinteger n . In Section 2.8, we will extend this converse to the skew stable and skew dual stable Grothendieckpolynomials. 8 .8 Statement of the Problem Now, we are ready to introduce our first main result, Theorem 1.2, an analogue of the Stembridgeequality for the dual stable Grothendieck polynomials, which states that for ρ = ( n, n − , . . . , ,g ρ/µ = g ρ/µ T . We first prove a special case of this theorem, Proposition 1.1, for when µ is the partition ( k ) or( k ) T = (1 k ) , in Section 3.1 using a bijection between set-valued tableaux. Then we extend this togeneral µ using the stable Grothendieck polynomials and Hopf algebraic structure of the symmetricfunctions in Section 3.2.Our second main result, Theorem 1.3, is an analogue of the Stembridge equality for the stableGrothendieck polynomials, stating that G ρ/µ = G ρ/µ T . Similar to the dual stable Grothendieck polynomial case, we will first prove this theorem for µ = ( k )or (1 k ) in Section 4.1 by finding a bijection between set-valued tableaux. Then, we use the Hopfalgebraic structure to extend to general µ in Section 4.2. Example 2.29. When µ = µ T , then ρ/µ = ρ/µ T and g ρ/µ = g ρ/µ T . Example 2.30. Consider ρ = (3 , , 1) and µ = (2). The diagrams for ρ/µ and ρ/µ T are below. ⇐⇒ . For ρ/µ , the top right section does not share any columns with the rest of the diagram, so thenumber occupying the top right square is unconstrained by the remainder of the diagram. Then g ρ/µ = g · g , the product of the two symmetric functions. The same argument holds for ρ/µ T , since the bottomleft section is independent of the top right section, and so g ρ/µ T = g · g , and the two polynomials are equal.We can also prove the converses of our main results (Theorems 1.3 and 1.2) by extendingCorollary 2.28. Theorem 2.31. If g ρ/µ = g ρ/µ T or G ρ/µ = G ρ/µ T , for all µ, then ρ = ( n, n − , . . . , for somepositive integer n .Proof. As stated in Remarks 2.13 and 2.18, the equalities g ρ/µ = g ρ/µ T and G ρ/µ = G ρ/µ T bothrequire the Stembridge equality s ρ/µ = s ρ/µ T , which in turn requires ρ = ( n, n − , . . . , 1) for somepositive integer n . 9 Proof for Dual Stable Grothendieck Polynomials In this section we prove Theorem 1.2, by first proving with the special case when µ = ( k ) or (1 k )combinatorially, and then generalizing to arbitrary µ using the Hopf algebraic structure of thesymmetric functions. µ = ( k ) or (1 k ) Our proof for Theorem 1.1 makes use of the skewing operator ⊥ , described in Section 2.6, andTheorem 3.4 by Buch [2]. Definition 3.1. Let w ( T ) denote the reverse reading word of a set-valued tableau T , read top tobottom along a column, starting with the rightmost column and moving left, and with the elementswithin a box read largest to smallest.For example, the following set-valued tableau has a reverse reading word of 743252153636542.1 , , , , , , Definition 3.2. A reverse reading word is a lattice word if the i th instance of a + 1 comes afterthe i th instance of a for all positive integers i and a .For example, 1121322 is a lattice word, but 121221 is not. Definition 3.3. Let ν ∗ µ denote the skew shape formed by joining the partitions ν and µ suchthat the top right corner of µ touches the bottom left corner of ν .For example, we have (2 , ∗ (4) = (6 , , / (4 , . ∗ =Next, we have the following theorem, a Littlewood-Richardson rule for stable Grothendieckpolynomials, as shown by Buch ([2], Theorem 5.4). Theorem 3.4 (Buch) . The Littlewood-Richardson rule for stable Grothendieck polynomials statesthat G ν G µ = X ρ ( − | ρ |−| ν |−| µ | c ρνµ G ρ , where c ρνµ is the number of set valued tableaux T of shape ν ∗ µ such that w ( T ) is a lattice wordwith content ρ . Let a valid tableau T be a tableau of shape ν ∗ µ such that w ( T ) is a lattice word with content ρ . 10 emma 3.5. In a valid filling of ν ∗ µ , all boxes in the i th row of ν contain the set { i } .Proof. The rightmost box in the first row of ν must contain the set { } , because a lattice wordmust begin with 1, so all boxes in the first row contain the set { } , as rows are increasing left toright. The rightmost box in the second row may only contain numbers greater than 1, and in orderfor ν ’s reading word to be a lattice word, this box must contain the set { } . Thus, all boxes of thesecond row contain the set { } . Analogously, we may inductively show that all boxes in the i throw of ν contain the set { i } .Observe that there are n − j + 1 j ’s in the partition ρ = ( n, n − , . . . , j in each column of ν ∗ µ . Inparticular, in a valid filling of ν ∗ (1 k ), (1 k ) contains at most one of each of the numbers 1 , , . . . , n .In fact, the same holds for a valid filling of ν ∗ ( k ). Lemma 3.6. In a valid filling of ν ∗ ( k ) , ( k ) contains at most one of each of the numbers , , . . . , n .Proof. For the sake of contradiction, suppose that the shape ( k ) contains at least two i ’s. Notethat i ≤ n − ρ contains only one n . Then in the reverse reading word of ν ∗ k , all i + 1’sare listed before the second-to-last i . In other words, the ( n − i )th i + 1 is listed before the ( n − i )th i , contradicting the assumption that the reverse reading word of ν ∗ ( k ) is a lattice word. Theorem 3.7. We have c ρ ( k ) ν = c ρ (1 k ) ν for all positive integers k ≤ n .Proof. Considering Lemma 3.5 and Lemma 3.6, there is at most one i in ( k ) (resp. (1 k )) in a validfilling of ν ∗ ( k ) (resp. ν ∗ (1 k )), so the i th row of ν contains exactly n − i + 1 or n − i i ’s. If i + 1is in ( k ) or (1 k ), it must be the ( n − i )th j + 1 in ν ∗ ( k ) or ν ∗ (1 k ), respectively. Since there areat least n − i i ’s found in ν , any arrangement of the numbers in ( k ) concatenated after the reversereading word of ν will form a lattice word.So given a valid filling of ν ∗ ( k ), there exists a corresponding valid filling of ν ∗ (1 k ) upon rotating( k ) by 90 degrees clockwise. Similarly, given a valid filling of ν ∗ (1 k ), there exists a correspondingvalid filling of ν ∗ ( k ) upon rotating (1 k ) by 90 degrees counterclockwise. Therefore, there is abijection between valid fillings of ν ∗ ( k ) and ν ∗ (1 k ), and c ρ ( k ) ν = c ρ (1 k ) ν .For example, the following are corresponding set-valued tableau under this bijection, for ρ =(5 , , , , ν = (4 , , , k ) = (3).1 1 1 12 2 2 23 341 , ⇐⇒ , Lemma 3.8. Fix ρ = ( n, n − , . . . , and some partition µ ⊆ ρ . If c ρνµ = c ρνµ T for all partitions ν , then g ρ/µ = g ρ/µ T . roof. Assume that c ρνµ = c ρνµ T for all partitions ν . Recall from Theorem 3.4 that G ν G µ = X ρ ( − | ρ |−| ν |−| µ | c ρνµ G ρ , so h G ν G µ , g ρ i = ( − | ρ |−| ν |−| µ | c ρνµ = ( − | ρ |−| ν |−| µ t | c ρνµ T = h G ν G µ T , g ρ i . Additionally, recall that G ⊥ µ g ρ = g ρ/µ . Using the definition of the skewing operator ⊥ andTheorem 2 . 23, we may also write g ρ/µ = G ⊥ µ g ρ = X ν h G µ , g ρ/ν i g ν = X ν h G µ , G ⊥ ν g ρ i g ν = X ν h G ν G µ , g ρ i g ν . Analogously, we have that g ρ/µ T = P ν h G ν G µ T , g ρ i g ν , so g ρ/µ = g ρ/µ T , as desired.Now, we give the proof of Theorem 1 . Proof. By Theorem 3.7, we have that c ρν ( k ) = c ρν (1 k ) . Then Lemma 3.8 gives g ρ/k = g ρ/ k . Recall from Section 2.5 that ˆΛ is the completion of Λ, the ring of symmetric functions.We take the linear map τ : ˆΛ → ˆΛ, given by linearly extending G λ G λ T , as defined byYeliussizov in [9]. He also shows that it is a ring homomorphism and an involution of ˆΛ.We will begin by proving the following theorem. Theorem 3.9. For ρ = ( n, n − , . . . , and all positive integers k , we have e ⊥ k g ρ = τ ( e k ) ⊥ g ρ . In order to do so, we shall first prove the following lemmas and propositions. Lemma 3.10. The stable Grothendieck polynomial of shape (1 k ) can be written as G (1 k ) = X n ≥ k ( − n − k (cid:18) n − k − (cid:19) e n . Proof. The stable Grothendieck polynomial G (1 k ) is a sum over set-valued tableaux T of shape(1 k ). All set-valued tableaux are strictly increasing along the columns, so all entries are distinct ina set-valued tableau of shape (1 k ). Therefore, the monomial x T for each set-valued tableau T is ofthe form x i x i · · · x i n for positive integers i < · · · < i n , where n = | T | . Given n distinct entries,the number of ways to fill in T is equal to the number of compositions of n into k nonempty parts,which is (cid:0) n − k − (cid:1) by stars and bars. 12herefore, G (1 k ) = X T ( − | T |−| k | x T = X n X | T | = n ( − n − k x T = X n ≥ k ( − n − k (cid:18) n − k − (cid:19) X i < ···
We can write the elementary symmetric function e k as an infinite sum of stableGrothendieck polynomials: e k = X n ≥ k (cid:18) n − k − (cid:19) G (1 n ) . Proof. By Theorem 3.10, X n ≥ k (cid:18) n − k − (cid:19) G (1 n ) = X j ≥ n X n ≥ k ( − j − n (cid:18) n − k − (cid:19)(cid:18) j − n − (cid:19) e j . (1)As a result, for a given j ≥ k, the coefficient of e j is X k ≤ n ≤ j ( − j − n (cid:18) n − k − (cid:19)(cid:18) j − n − (cid:19) = X k ≤ n ≤ j ( − j − n (cid:18) j − k − (cid:19)(cid:18) j − kn − k (cid:19) = (cid:18) j − k − (cid:19) δ kj = δ kj , where the first simplification comes from trinomial revision and the second comes from the factthat the alternating sum of a row of binomial coefficients (besides the first row) is 0. This meansthat the coefficient of e j in (1) is 0 for all j = k and 1 for j = k , so e k = X n ≥ k (cid:18) n − k − (cid:19) G (1 n ) . Lemma 3.12. The skewing operator ⊥ distributes over infinite sums; that is, for any a ∈ Λ and f , f , · · · ∈ ˆΛ , (cid:16) X i ≥ f i (cid:17) ⊥ ( a ) = X i ≥ f ⊥ i ( a ) . Proof. Let ∆( a ) = P ( a ) a (1) ⊗ a (2) . Then (cid:16) X i ≥ f i (cid:17) ⊥ ( a ) = X ( a ) D X i ≥ f i , a (1) E a (2) . × Λ Z and is linear, we expand to get (cid:16) X i ≥ f i (cid:17) ⊥ ( a ) = X ( a ) X i ≥ h f i , a (1) i a (2) = X i ≥ X ( a ) h f i , a (1) i a (2) = X i ≥ f ⊥ i ( a ) . Finally, we can give the proof of Theorem 3.9. Proof. Observe that τ is a ring homomorphism of ˆΛ that distributes over infinite sums, as doesthe skewing operator by Lemma 3.12. Applying Proposition 3.11 to decompose e k as a linearcombination of G (1 n ) , and applying Theorem 1.1 to write G ⊥ (1 n ) ( g ρ ) = G ⊥ ( n ) ( g ρ ) gives: τ ( e k ) ⊥ ( g ρ ) = τ (cid:16) X n ≥ k (cid:18) n − k − (cid:19) G (1 n ) (cid:17) ⊥ ( g ρ )= X n ≥ k (cid:18) n − k − (cid:19) τ ( G (1 n ) ) ⊥ ( g ρ )= X n ≥ k (cid:18) n − k − (cid:19) G ⊥ ( n ) ( g ρ )= X n ≥ k (cid:18) n − k − (cid:19) G ⊥ (1 n ) ( g ρ )= (cid:16) X n ≥ k (cid:18) n − k − (cid:19) G (1 n ) (cid:17) ⊥ ( g ρ )= e ⊥ k g ρ . Now, in order to prove Theorem 1.2, we need the following lemmas, which are inspired byExercises 2.9.24 and 2.9.25 in [5]. Lemma 3.13. Let ψ be an arbitrary ring homomorphism of ˆΛ . Then for any given a ∈ Λ , theset A = { f ∈ ˆΛ : f ⊥ ( a ) = ψ ( f ) ⊥ ( a ) } is closed under finite multiplication and (possibly) infiniteaddition of its elements.Proof. First, we show that if f , f , · · · ∈ A , then P i ≥ f i ∈ A . By considering Lemma 3.12 andusing the fact that ψ is a ring homomorphism of ˆΛ, we have (cid:16) ψ (cid:16) X i ≥ f i (cid:17)(cid:17) ⊥ ( a ) = (cid:16) X i ≥ ψ ( f i ) (cid:17) ⊥ ( a )= X i ≥ ψ ( f i ) ⊥ ( a )= X i ≥ f ⊥ i ( a )= (cid:16) X i ≥ f i (cid:17) ⊥ ( a ) , 14o indeed, P i ≥ f i ∈ A .Next, we show that if f , f ∈ A, then f f ∈ A. First, notice that if h k, f i = h k, f i for all k ∈ Λ,then f = f . This is because each f and f can be written uniquely as a (possibly infinite) linearcombination of Schur functions s λ , and taking k to be s λ for all λ in turn, gives us that f and f havethe same coefficient for all s λ . Thus, it suffices to show that h k, ( f f ) ⊥ ( a ) i = h k, ( ψ ( f f )) ⊥ ( a ) i for all k ∈ Λ.Repeatedly applying the property from Lemma 2.23, and using the fact that f , f ∈ A , we cando the following manipulation: h k, ( f f ) ⊥ ( a ) i = h f f k, a i = h f k, f ⊥ ( a ) i = h f k, ψ ( f ) ⊥ ( a ) i = h ψ ( f ) f k, a i = h f ψ ( f ) k, a i = h ψ ( f ) k, f ⊥ ( a ) i = h ψ ( f ) k, ψ ( f ) ⊥ ( a ) i = h ψ ( f ) ψ ( f ) k, a i = h ψ ( f f ) k, a i = h k, ψ ( f f ) ⊥ ( a ) i . This means that ( f f ) ⊥ ( a ) = ψ ( f f ) ⊥ ( a ) , so indeed f f ∈ A . Thus A is closed under finitemultiplication and (possibly) infinite addition. Corollary 3.14. The set A = { f ∈ ˆΛ : f ⊥ ( g ρ ) = τ ( f ) ⊥ ( g ρ ) } , where τ is the ring homomorphismdefined earlier by linearly extending G λ G λ T , is closed under finite multiplication and infiniteaddition. Now we are ready to prove Theorem 1.2. Proof. By Theorem 3.9, we have e ⊥ k g ρ = τ ( e k ) ⊥ g ρ , so this means e k ∈ A = { f ∈ ˆΛ : f ⊥ ( g ρ ) = τ ( f ) ⊥ ( g ρ ) } . By Corollary 3.14, A is closed under multiplication and possibly infinite sums, so forall λ we have e λ = e λ e λ · · · e λ n ∈ A . Since the e λ form a basis for ˆΛ (if we allow infinite linearcombinations), any symmetric function f = P λ a λ e λ is in A . In particular, G µ ∈ A , so g ρ/µ = G ⊥ µ g ρ = τ ( G µ ) ⊥ g ρ = G ⊥ µ T g ρ = g ρ/µ T . In this section we prove Theorem 1.3, by first proving with the special case when µ = ( k ) or (1 k )combinatorially, and then generalizing to arbitrary µ using the Hopf algebraic structure of thesymmetric functions. 15 .1 Proof for µ = ( k ) or (1 k ) Throughout this section, we denote the partition ( n, n − , . . . , 1) by ρ n . The following theorem byBuch ([2], Theorem 6.9), allows us to prove the special case when µ = ( k ) or (1 k ) combinatorially. Theorem 4.1 (Buch) . For a skew partition λ/µ , G λ/µ = X ν ( − | ν |−| λ/µ | α λ/µ,ν G ν , where the coefficient α λ/µ,ν is the number of set-valued tableaux T of shape λ/µ such that w ( T ) isa lattice word with content ν . Now, we have the following recurrence for the α λ/µ,ν coefficients when λ = ρ n and µ = ( k ) forsome positive integer k . Lemma 4.2. Fix a partition ν = ( ν , ν , . . . , ν m ) , and let ν − = ( ν , . . . , ν m ) . For a given positiveinteger n , we have α ρ n / ( k ) ,ν = α ρ n − / ( k ) ,ν − + 2 α ρ n − / ( k − ,ν − . Proof. Consider a set-valued tableau T of shape ρ n / ( k ) such that w ( T ) is a lattice word withcontent ν . The rightmost box in the first row of T must contain the set { } , so all boxes in thefirst row contain the set { } . The rightmost box in the second row must contain the set { } , andsimilarly the rightmost n − k − { } .1 1 1 B k th box B in thesecond row: { } , { } , or { , } .Let the function f map the tableau T to the tableau T − by deleting all 1’s from the boxes of T , deleting all now-empty boxes, and subtracting 1 from all remaining numbers. Assuming that w ( T ) is a lattice word, w ( T − ) is also a lattice word: the i th a + 1 coming before the i th a + 2 in w ( T ) corresponds to the i th a , which comes before the i th a + 1 in w ( T − ). Case 1: B contains { } Then, the leftmost k boxes in the second row must all contain the set { } , since the rows areweakly increasing. 1 1 11 1 1 2 216pplying f gives a tableau T − of shape ρ n − / ( k ) and content ν − , as illustrated below.1 1 11 1 1 2 22 2 2 , ⇐⇒ , T − of shape ρ n − / ( k − 1) has lattice reverse reading word, thenit necessarily corresponds to exactly one tableaux T of shape ρ n / ( k ) with lattice reverse readingword where B contains { } . Then these two sets are in bijection, so there are α ρ n − / ( k ) ,ν − tableaux T of shape ρ n / ( k ) with lattice reverse reading word where B contains { } . Case 2: B contains { } Then the ( k − { } , as otherwise the ( n − k +1)th2 would come before the ( n − k + 1)th 1 in w ( T ) and it would not be a lattice word.1 1 11 1 2 2 2Applying f gives a tableau T − of shape ρ n − / ( k − 1) with content ν − , as shown below.1 1 11 1 2 2 22 3 3 33 4 44 56 ⇐⇒ f is a bijection here, so there are α ρ n − / ( k − ,ν − tableaux T such that B contains { } . Case 3: B contains { , } Then the leftmost k − { } .1 1 11 1 1 , f gives a tableau of shape ρ n − / ( k − 1) with content ν − which has reverse columnword a lattice word. Note that in this case, the inverse map will additionally insert an extra 1 inthe k th box of the second row, as illustrated below.1 1 11 1 1 , ⇐⇒ α ρ n − / ( k − ,ν − tableaux T such that B contains { , } . Combining all three cases, we have α ρ n / ( k ) ,ν = α ρ n − / ( k ) ,ν − + 2 α ρ n − / ( k − ,ν − . There exists a similar recurrence for the α λ/µ,ν coefficients when λ = ρ n and µ = (1 k ) for somepositive integer k . Lemma 4.3. Fix a partition ν = ( ν , ν , . . . , ν m ) , and let ν − = ( ν , . . . , ν m ) . For a given positiveinteger n , we have α ρ n / (1 k ) ,ν = α ρ n − / (1 k ) ,ν − + 2 α ρ n − / (1 k − ) ,ν − . Proof. Consider a set-valued tableau T of shape ρ n / (1 k ) such that w ( T ) is a lattice word withcontent ν . The rightmost box in the first row of T must contain the set { } , so all boxes in thefirst row contain the set { } . The rightmost box in the second row must contain the set { } , soall boxes in the second row must contain the set { } . Analogously, all boxes in the j th row for1 ≤ j ≤ k must contain the set { j } .There are three possibilities for the leftmost box B in the k + 1th row: B contains { } , a setwithout a 1, or a set containing 1 of size at least 2, as illustrated below. Define f as in the previouslemma. 1 1 1 1 12 2 2 23 3 3 B Case 1: B contains { } There is a single 1 in each column, so applying f gives a tableau T − of shape ρ n − / ( k ), asillustrated below.Notice that if a tableau of shape T − of shape ρ n − / (1 k ) has lattice reverse reading word, thenit necessarily corresponds to exactly one tableau T of shape ρ n / (1 k ) with lattice reverse readingword where B contains { } (given by adding 1 to each number and adding a box containing 1 tothe top of each column). Then these two sets are in bijection, so there are α ρ n − / (1 k ) ,ν − tableaux T of shape ρ n / (1 k ) with lattice reverse reading word where B contains { } .18hen this case contributes α ρ n − / (1 k ) ,ν − tableaux.1 1 1 1 12 2 2 23 3 31 4 43 , ⇐⇒ , Case 2: B does not contain a 1 Then the only 1’s in the entire set-valued tableau are in the first row, so f takes T to a tableau ofshape ρ n − / ( k − α ρ n − / (1 k − ) ,ν − tableaux T where B does not contain a 1.1 1 1 1 12 2 2 23 3 32 , ⇐⇒ , Case 3: B contains a set of size ≥ with a 1 The 1’s in the tableau lie either in the top row or in B . Since the box containing B is notdeleted by f , T is mapped to a tableau of shape ρ n − / ( k − α ρ n − / (1 k − ) ,ν − tableaux for this case, since f provides a bijection.1 1 1 1 12 2 2 23 3 31 , ⇐⇒ α ρ n / (1 k ) ,ν = α ρ n − / (1 k ) ,ν − + 2 α ρ n − / (1 k − ) ,ν − . Using induction and combining the two previous lemmas, we have the following equality between α coefficients. Lemma 4.4. We have α ρ n / ( k ) ,ν = α ρ n / (1 k ) ,ν for all positive integers n and nonnegative integers k . roof. We use induction on n and only consider k ≤ n , because otherwise α ρ n / ( k ) ,ν = α ρ n / (1 k ) ,ν = 0.The base case of n = 1 is true because ρ / ( k ) = ρ / (1 k ) for k = 0 , 1. Now suppose that for a given n , α ρ n / ( k ) ,ν = α ρ n / k ,ν for all k ≤ n . Note that ρ n +1 / ( k ) = ρ n +1 = ρ n +1 / (1 k ) for k = 0. For1 ≤ k ≤ n , α ρ n +1 / ( k ) ,ν = α ρ n / ( k ) ,ν − + 2 α ρ n / ( k − ,ν − = α ρ n / (1 k ) ,ν − + 2 α ρ n / (1 k − ) ,ν − = α ρ n +1 / (1 k ) ,ν . In addition, for k = n + 1, we have ρ n +1 / ( n + 1) = ρ n = ρ n +1 / (1 n +1 ) by a translation. Alltogether, α ρ n +1 / ( k ) ,ν = α ρ n +1 / (1 k ) ,ν for all k ≤ n + 1. Thus, by induction, α ρ n / ( k ) ,ν = α ρ n / (1 k ) ,ν forall positive integers n and nonnegative integers k .As a result of this relation between the α coefficients, we can now prove a Stembridge-typeequality for G ρ/µ in the special case where µ = ( k ) for some positive integer k . Theorem 4.5. There is a Stembridge-type equality for the skew stable Grothendieck polynomial inthe case µ = ( k ) , i.e. G ρ/ ( k ) = G ρ/ (1 k ) . Proof. Combining Lemma 4.4 with Theorem 4.1, G ρ/ ( k ) = X ν ( − | ν |−| ρ/ ( k ) | α ρ/ ( k ) ,ν G ν = X ν ( − | ν |−| ρ/ (1 k ) | α ρ/ (1 k ) ,ν G ν = G ρ/ (1 k ) . Now, we will use the Hopf algebraic structure to extend this result to all µ, proving Theorem 1.3,an analogue of the Stembridge equality for G ρ/µ . First, we introduce two definitions and a usefultheorem from Buch [2]. Definition 4.6. A rook strip is a skew partition µ/σ that contains at most one box in each rowand column.The following definition ([2], Equation 6.4), which will allow us to utilize the Hopf algebraicstructure of Λ. Definition 4.7 (Buch) . Define G λ//µ as G λ//µ = X σ ( − | µ/σ | G λ/σ , where the sum is over all σ such that µ/σ is a rook strip.The polynomials G λ/µ are related to the polynomials G λ//µ by the following theorem, as char-acterized by Buch ([2], Equation 7.4). 20 heorem 4.8 (Buch) . We have G λ/µ = X σ ⊆ µ G λ//σ . From the above definitions and theorem, we may prove the following lemma. Lemma 4.9. If ρ is a staircase shape ρ = ( n, n − , . . . , then G ρ// ( k ) = G ρ// (1 k ) .Proof. In order for ( k ) /σ to be a rook strip, we need σ = ( k ) or ( k − G ρ// ( k ) = G ρ/ ( k ) − G ρ/ ( k − . Similarly, G ρ// (1 k ) = G ρ/ (1 k ) − G ρ/ (1 k − ) . Combining with Theorem 4.5, we have G ρ// ( k ) = G ρ// k .Buch ([2], Example 6.8) states that ∆( G ρ ) = P ν ⊆ ρ G ν ⊗ G ρ//ν . Then, using the skewing operatorfrom Definition 2.21 and the identity h g λ , G µ i = δ λµ , we have g ⊥ µ ( G ρ ) = X ν h g µ , G ν i G ρ//ν = h g µ , G µ i G ρ//µ = G ρ//µ . In [9], Yeliussizov constructs τ by linearly extending g λ g λ T , and he shows that τ is a ringhomomorphism and an involution. We use τ in order to extend Lemma 4.9 to all µ. Lemma 4.10. Let ψ be an arbitrary ring homomorphism of Λ . Then the set A = { f ∈ Λ : f ⊥ ( a ) = ψ ( f ) ⊥ ( a ) } is a subalgebra of Λ . Proof. From a similar argument as Lemma 3.13, we can show that for f , f ∈ A, we have f + f ∈ A and f f ∈ A. Corollary 4.11. The set A = { f ∈ Λ : f ⊥ ( G ρ ) = τ ( f ) ⊥ ( G ρ ) } is a subalgebra of Λ . Lemma 4.12. We have G ρ//µ = G ρ//µ T . Proof. Let A = { f ∈ Λ : f ⊥ ( G ρ ) = τ ( f ) ⊥ ( G ρ ) } . The polynomials g ( k ) are elements of A, since g ⊥ ( k ) ( G ρ ) = G ρ// ( k ) = G ρ// (1 k ) = g ⊥ (1 k ) ( G ρ ) = τ ( g ( k ) ) ⊥ ( G ρ ) . By definition, g ( k ) = X P x P , summed over reverse plane partitions P of shape ( k ). For a given reverse plane partition P, thehorizontal strip of length k is filled with numbers i ≤ · · · ≤ i k .i i · · · i k g ( k ) = X P x P = X i ≤···≤ i k x i · · · x i k = h k . Now, since g ( k ) = h k , we have that h k ∈ A. The set A is closed under addition and multiplicationby Lemma 4.11, so this means h λ = h λ h λ · · · h λ i ∈ A. Since the h λ form a basis for Λ , anysymmetric function f = P a λ h λ is in A as well. In particular, g µ ∈ A for any partition µ .Therefore, G ρ//µ = g ⊥ µ ( G ρ )= τ ( g µ ) ⊥ ( G ρ )= g ⊥ µ T ( G ρ )= G ρ//µ T . Lastly, we can use these results from the Hopf algebraic structure of Λ to prove Theorem 1.3. Proof. Combining all the above results, we have G ρ/µ = X σ ⊆ µ G ρ//σ = X σ ⊆ µ G ρ//σ T = X σ T ⊆ µ T G ρ//σ T = G ρ/µ T . We would like to thank Adela (YiYu) Zhang for guiding us on our research. We would also like tothank Professor Darij Grinberg for proposing the project and providing helpful suggestions. Finally,we would like to thank the MIT PRIMES-USA program, under which this research was conducted. References [1] Alwaise, E., Chen, S., Clifton, A., Patrias, R., Prasad, R., Shinners, M., Zheng, A. (2017).Coincidences among skew stable and dual stable Grothendieck polynomials. Involve, a Journalof Mathematics, 11 (1), 143-167.[2] Buch, A. S. (2002). A Littlewood-Richardson rule for the K-theory of Grassmannians. Actamathematica, 189 (1), 37-78.[3] Fomin, S., Kirillov, A. N. (1994). Grothendieck polynomials and the Yang-Baxter equation. In Proc. Formal Power Series and Alg. Comb (pp. 183-190).[4] Galashin, P. (2017). A Littlewood–Richardson rule for dual stable Grothendieck polynomials. Journal of Combinatorial Theory, Series A , 151, 23-35.225] Grinberg, D., Reiner V. (2020) Hopf Algebras in Combinatorics. https://arxiv.org/abs/1409.8356 [6] Lam, T., Pylyavskyy, P. 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