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arXiv:2410.07423v2 [math.CO] 23 Feb 2025 THE CONSTRUCTION OF A CLASS OF PRESENTATIONS FOR SPECHT MODULES TAMAR FRIEDMANN Abstract. We build on the methods introduced by Friedmann, Hanlon, Stanley, and Wachs, and further developed by Brauner and Friedmann, to construct addi- tional classes of presentations of Specht modules. We obtain these presentations by defining a linear operator which is a symmetrized sum of dual Garnir relations on the space of column tabloids. Our presentations apply to the vast majority of shapes of Specht modules.
- Introduction The Specht modules Sλ, where λ is a partition of n, give a complete set of irre- ducible representations of the symmetric group Sn over a field of characteristic 0, say C. They can be constructed as subspaces of the regular representation CSn or as presentations given in terms of generators and relations, known as Garnir relations. This paper deals primarily with the latter type of construction. Let λ = (λ1 ≥· · · ≥λl) be a partition of n. A Young tableaux of shape λ is a filling of the Young diagram of shape λ with distinct entries from the set [n] := {1, 2, … , n}. Let Tλ be the set of Young tableaux of shape λ. The symmetric group Sn acts on Tλ by replacing each entry of a tableau by its image under the permutation in Sn. To construct the Specht module as a submodule of the regular representation, one can use Young symmetrizers. For t ∈Tλ, the Young symmetrizer is defined by (1.1) et := X α∈Rt α X β∈Ct sgn(β)β, where Ct is the column stabilizer of t and Rt is the row stabilizer of t. The Specht module Sλ is the submodule of the regular representation CSn spanned by {τet : τ ∈Sn}. To construct the Specht module as a presentation, one can use column tabloids and Garnir relations. Let Mλ be the vector space (over C) generated by Tλ subject only to column relations, which are of the form t + s, where s ∈Tλ is obtained from 1
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2 FRIEDMANN t ∈Tλ by switching two entries in the same column. Given t ∈Tλ, let [t] denote the coset of t in Mλ. These cosets, which are called column tabloids, generate Mλ. A Young tableau is column strict if the entries of each of its columns increase from top to bottom. Clearly, {[t] : t is a column strict Young tableau of shape λ} is a basis for Mλ. In [Fu], Fulton introduces dual Garnir relations on the column tabloids and shows that Sλ is isomorphic to the quotient space of Mλ by these relations. There is a dual Garnir relation for each t ∈Tλ, each choice of adjacent columns, and each ℓup to the length of the next column. In particular, each Garnir relation is a weighted sum of a column tabloid [t] and column tabloids obtained from t by exchanging ℓ entries of a column with the top ℓentries of the next column. Fulton then obtains a simplification: it is enough to use only the dual Garnir relations that exchange exactly one entry of a column with the top entry of the next column, i.e. we can just fix ℓ= 1. An analogous simplification is obtained in [FHW3], which improves upon a result in [FHSW]. In the presentation of [FHW3], ℓis also restricted to a single value, but this time it is the maximum possible value: ℓequals the length of the next column, so as many entries are exchanged as the shape of λ allows. This presentation holds for partitions whose conjugate has distinct parts. A different simplification is obtained in [BF], where a symmetrized sum of dual Garnir relations with ℓ= 1 is introduced. The number of relations needed is dra- matically reduced: the construction uses a single relation for every pair of adjacent columns and [t] varies in Mλ, a significantly smaller space than Tλ. The presentation of [BF] holds for all partitions. In the present paper we consider intermediate values of ℓ. For what shapes λ would dual Garnir relations that exchange exactly ℓentries between the columns provide a presentation for Sλ? This is a question posed in [FHW3]. Our results answer that question and generalize both [FHW3], where ℓis maximal, and [BF], where ℓ= 1. Our methods are those envisioned in [FHW3]. The question posed there also inspired [MMS] to address it using a different approach, via representations of the general linear group. See Remark on p. 20 following Theorem 4.2 for a discussion relating the results of [MMS] to the results in the current paper. Our main result is contained in Theorems 4.1 and 4.2. In Theorem 4.1, we pro- vide conditions on the shape of a 2-column partition µ = (n, m)′ for any ℓ, using eigenvalues of an operator, ηℓ, on Mµ. Given a value of ℓ, for any µ for which the
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THE CONSTRUCTION OF A CLASS OF PRESENTATIONS FOR SPECHT MODULES 3 conditions are satisfied, we have obtained a presentation of Sµ. In Theorem 4.2, we use the results for 2-column shapes to state conditions on partitions λ with any number of columns. In Table 4.1, we provide some computer-generated data that tells us which 2-column shapes satisfy these conditions. The data indicates that the conditions are satisfied for the vast majority of shapes µ and values of ℓ. The work in [FHSW] was presented in the language of the generalized Jacobi relations that define the LAnKe or Filippov algebra [BL, DI, DT, Fi, Fr, Gu, Ka, Li, Ta]. An observation in [FHSW], that the restricted class of Garnir relations that fix ℓto be the maximum possible value (for staircase partitions λ) corresponds to the generalized Jacobi relations, motivated the work in the papers [FHSW, FHW3, BF, MMS] and the current paper. This paper is organized as follows. In Section 2, we introduce notation and review the relevant prior results. In Section 3, we introduce the relevant relation and compute its eigenvalues wℓ,i, which appear in Theorem 3.3. A combinatorial identity emerges, given in Corollary 3.4. Section 4 contains our main results, along with a statement about the equivalence of two sets of combinatorial conditions (Corollary 4.4). 2. Notation and prior results The symmetric group Sn acts on Tλ by replacing each entry of a tableau by its image under the permutation in Sn. This induces a representation of Sn on Mλ. In [Fu, Ch. 7.4], Fulton introduces a map α : Mλ →Sλ given by α : [t] 7→et. The map α is Sn-equivariant and surjective. Moreover, ker(α) is generated by a set of relations which Fulton calls the dual Garnir relations. The dual Garnir relation gc,ℓ(t) is (2.1) gc,ℓ(t) = [t] −πc,ℓ(t), where πc,ℓ(t) is the sum of column tabloids obtained from all possible ways of ex- changing the top ℓelements of the (c + 1)st column of t with any subset of size ℓ of the elements of column c, preserving the vertical order of each set of ℓelements. Note that t can be any tableau, not necessarily with increasing columns.
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4 FRIEDMANN If we let Gλ be the subspace of Mλ generated by the Garnir relations in (2.2) {gc,ℓ(t) : c ∈[λ1 −1], ℓ∈[λ′ c+1], t ∈Tλ}, where λ′ is the conjugate partition of λ, then Gλ is invariant under the action of Sn. In [Fu, Ch. 7.4], Fulton shows that Gλ = ker(α), thereby obtaining the following presentation of Sλ: (2.3) Mλ/Gλ ∼=Sn Sλ. As mentioned in the introduction, Gλ contains a dual Garnir relation for each t ∈Tλ, each choice of adjacent columns, and each ℓup to the length of the next column. On page 102 (after Ex. 15) of [Fu], a presentation of Sλ with a smaller set of relations is given. In this presentation, the index ℓin gc,ℓ(t) of (2.2) is restricted to a single value: ℓ= min[λ′ c+1] = 1. More precisely, the presentation is (2.4) Mλ/Gλ,min ∼=Sn Sλ, where Gλ,min is the subspace of Gλ generated by the subset of Garnir relations with ℓ= 1: {gc,1(t) : c ∈[λ1 −1], t ∈Tλ}. This is Fulton’s simplification mentioned in the introduction. In the analogous simplification of [FHW3], the index ℓin gc,ℓ(t) of (2.2) is restricted to the maximum value ℓ= λ′ c+1: Theorem 2.1. [FHW3, Theorem 1.1] Let λ be a partition whose conjugate has distinct parts. Then (2.5) Mλ/Gλ,max ∼=Sn Sλ, where Gλ,max is the subspace of Gλ generated by {gc,λ′ c+1(t) : c ∈[λ1 −1], t ∈Tλ}. Moreover, this set of relations can be further reduced by restricting t to the set of column strict tableaux. The approach in [FHW3], which improves on an earlier result [FHSW] that applied only to staircase partitions, is to define a certain linear operator on the space of column tabloids and study its eigenspaces. This approach has also been used in [BF] to obtain a different presentation of Sλ with a reduced number of relations, which works for all shapes. Rather than using a subset of the Garnir relations, [BF]
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THE CONSTRUCTION OF A CLASS OF PRESENTATIONS FOR SPECHT MODULES 5 introduce a relation consisting of symmetrized sums of the dual Garnir relations that generate Gλ,min: (2.6) ηc,1([t]) = m[t] − X [s] , where the sum ranges over all possible tableaux s obtained from t by swapping one entry in column c + 1 of t with one entry in column c. The relation ηc,1 can be thought of as a sum of gc,1 relations which has the ad- vantage of symmetrizing over all positions of elements in the (c + 1)st column. For t ∈Tλ, let hc,1([t]) be the image of ηc,1 on the c and (c + 1)st columns of [t] that leaves the other columns of [t] fixed. The result obtained in [BF] is: Theorem 2.2. [BF, Theorem 3.5] For any partition λ of n, let Hλ be the space generated by hc,1([t]) for every [t] ∈Mλ and 1 ≤c ≤λ1 −1. Then the kernel of α is Hλ. Thus, Mλ/Hλ ∼=Sn Sλ. We can see that only the single relation ηc,1 is needed for each pair of adjacent columns. In the next section, we address the case of an intermediate value of ℓthat was proposed in [FHW3], generalizing the methods of [FHW3, BF]. As mentioned in the introduction, [MMS] was motivated by [FHW3] to address this same question in a different way, using representations of the general linear group. See Remark on p. 20 following Theorem 4.2 for a statement of their theorem and a discussion relating their results to ours. 3. A linear operator and its eigenvalues We now define a linear operator on the space of column tabloids, and study its eigenspaces. We use a symmetrized sum of gc,ℓ(t) relations, where we symmetrize over all ways of exchanging any ℓelements of the (c+1)st column with any ℓelements of column c, where ℓ∈[λ′ c+1]. This is a generalization of [BF], where ℓ= 1, and of [FHW3], where ℓ= λ′ c+1. In this section we focus on 2-column partitions µ of n + m with shape 2m1n−m, so µ has a column of size n and a column of size m for 1 ≤m ≤n, and µ′ = (n, m). We shall address the implications of these results to partitions with more than two columns in the next section.
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6 FRIEDMANN Definition 3.1. Let µ = 2m1n−m, and let ℓ∈[m]. We define the map ηℓ: Mµ → Mµ to be ηℓ[t] = m ℓ [t] − X [s] where the sum ranges over all possible tableaux s obtained from t by swapping ℓ entries in the second column of t with ℓentries in the first column, preserving the vertical order of each set of ℓentries. Example: Let [t] = 1 5 2 6 3 7 4 . Then η2([t]) = 3 1 5 2 6 3 7 4 − 5 1 6 2 3 7 4 + 5 1 2 3 6 7 4 + 5 1 2 4 3 7 6 + 1 2 5 3 6 7 4 + 1 2 5 4 3 7 6 + 1 3 2 4 5 7 6 − 5 1 7 6 3 2 4 + 5 1 2 6 7 3 4 + 5 1 2 6 3 4 7 + 1 2 5 6 7 3 4 + 1 2 5 6 3 4 7 + 1 3 2 6 5 4 7 − 6 5 7 1 3 2 4 + 6 5 2 1 7 3 4 + 6 5 2 1 3 4 7 + 1 5 6 2 7 3 4 + 1 5 6 2 3 4 7 + 1 5 2 3 6 4 7 . In determining the coefficient of [t] in Definition 3.1, we used the fact that �m ℓ is the number of ways to pick ℓentries from the second column, making ηℓ([t]) a sum of dual Garnir relations. Since ηℓis defined via its action on positions of [t], it is a right action on Mλ. Meanwhile, Sn+m acts on the letters of t, so its action on Tλ and its induced action on Mλ are left actions. Therefore, the actions of ηℓand Sn+m commute and ηℓ is Sn+m-equivariant. Furthermore, it follows from equation (2.3) that Im(ηℓ) ⊆ ker(α), as ηℓ([t]) is a sum of dual Garnir relations. Using techniques employed in [FHSW, FHW3, BF], we will show that for the vast majority of partitions µ, the relations generated by ηℓfor any single value of ℓare all that is needed to generate
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THE CONSTRUCTION OF A CLASS OF PRESENTATIONS FOR SPECHT MODULES 7 Gµ. To do so, we study the eigenvalues of ηℓ; the rest of this section is devoted to this study. Note that because Mµ ∼= m M i=0 S2i1n+m−2i is multiplicity-free, by Schur’s Lemma ηℓacts as a scalar on each irreducible sub- module of Mµ. Thus, finding the kernel of ηℓis equivalent to finding the irreducible submodules of Mµ on which ηℓacts like the 0 scalar. We proceed by computing the action of ηℓon each irreducible submodule of Mµ. For each T ∈ �[n+m] n , let vT ∈Mµ be the column tabloid with first column T (both columns assumed to be in increasing order). For any v ∈Mµ, let ⟨v, vT⟩be the coefficient of vT in the expansion of v in the basis of all vT. Lemma 3.2. For every S, T ∈ �[n+m] n , ⟨ηℓ(vS), vT⟩= �m ℓ if S = T, 0 if S ̸= T and |S ∩T| ̸= n −ℓ, (−1) Pℓ k=1(ck+dk)+ℓ+1 if |S ∩T| = n −ℓwith S\T = {c1, · · · , cℓ}, T\S = {d1, … , dℓ}. Proof. The first two cases easily follow from the definition of ηℓ. We consider the third case. Let S = {a1, a2, … , an} and [n + m] \ S = {b1, b2, … , bm}. Then the columns of vS, arranged from left to right instead of top to bottom, are a1, a2, … , an and b1, b2, … , bm, where the entries of each column are in increasing order. Let i1, i2, … , iℓ∈[n] and j1, j2, … , jℓ∈[m]. We will exchange the entries ai1, ai2, … , aiℓwith the entries bj1, bj2, … , bjℓto obtain a term that appears in ηℓ(vS). The two columns of such a term look like a1, … , ai1−1, bj1, ai1+1, … , ai2−1, bj2, ai2+1, … , aiℓ−1, bjℓ, aiℓ+1, … , an and b1, … , bj1−1, ai1, bj1+1, … , bj2−1, ai2, bj2+1, … , bjℓ−1, aiℓ, bjℓ+1, … , bm.
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8 FRIEDMANN In order to get an element in the basis of Mµ, we need to reorder the entries of each column so they are increasing. Every exchange within a column results in a sign. What sign will we end up with? We consider each column individually first. Each column will require two stages. Then we will put the resulting signs together. Column 1, Stage 1: We move all the b’s to the end of the column. Moving bjℓto the end gives n −iℓtranspositions. Moving bjℓ−1 to just before bjℓgives n −iℓ−1 −1 transpositions, and so on. The total number of transpositions for this stage is: (3.1) ℓ−1 X k=0 (n −iℓ−k −k) = nℓ−ℓ(ℓ−1) 2 − ℓ X k=1 ik . After these transpositions, the first column is a1, … ai1−1, ai1+1, … , ai2−1, ai2+1, … aiℓ−1, aiℓ+1, … , an, bj1, bj2, … , bjℓ. Column 1, Stage 2: We move the b’s to their correct positions. We begin with bj1. We have that bj1 is larger than j1 −1 b’s and therefore it is larger than (bj1 −1) − (j1 −1) = bj1 −j1 a’s. Some of those a’s, say γ1 of them, are not in the first column (because they were exchanged into the second column). So in the first column, there are bj1 −j1 −γ1 a’s smaller than bj1. We need to move bj1 from after all the n −ℓa’s to after bj1 −j1 −γ1 a’s. That requires (n −ℓ) −(bj1 −j1 −γ1) transpositions. Define γk to be the number of a’s in {ai1, … , aiℓ} that are smaller than bjk. So in the first column, there are bjk −jk −γk a’s smaller than bjk. We have dealt with bj1 already. Now we move bj2 from after all the n −ℓa’s to after bj2 −j2 −γ2 a’s. Since bj1 < bj2, we do not need to transpose through bj1. So we have (n −ℓ) −(bj2 −j2 −γ2) transpositions. Continuing this way gives (3.2) ℓ X k=1 �(n −ℓ) −(bjk −jk −γk) = ℓ(n −ℓ) − ℓ X k=1 �bjk −jk −γk transpositions. This completes the reordering of the first column. Column 2, Stage 1: We move all the a’s to the end of the column. By the same reasoning as in stage 1 of the first column, the number of transpositions required is (3.3) ℓ−1 X k=0 (m −jℓ−k −k) = mℓ−ℓ(ℓ−1) 2 − ℓ X k=1 jk . Now the second column is b1, … , bj1−1, bj1+1, … , bj2−1, bj2+1, … , bjℓ−1, bjℓ+1, … , bm, ai1, … aiℓ.
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THE CONSTRUCTION OF A CLASS OF PRESENTATIONS FOR SPECHT MODULES 9 Column 2, Stage 2: We move the a’s to their correct positions. We have that aik is larger than ik −1 of the a’s and (aik −1) −(ik −1) = (aik −ik) of the b’s. Define δk to be the number of b’s in {bj1, … , bjℓ} that are smaller than aik. Then we need to move ai1 left to just after (ai1 −i1 −δ1) b’s, which requires (m −ℓ) − (aik −ik −δk) transpositions. Continuing with ai2 and so on, we have (3.4) ℓ X k=1 �(m −ℓ) −(aik −ik −δk) = ℓ(m −ℓ) − ℓ X k=1 �aik −ik −δk transpositions. Total from both columns: The total number of transpositions for both columns combined is given by the sum of equations (3.1), (3.2), (3.3), and (3.4). We are only concerned with the parity of the total number of transpositions. Adding the four equations and omitting any obviously even terms gives ℓ X k=1 �γk + δk − ℓ X k=1 �aik + bjk transpositions. We now show that Pℓ k=1 �γk + δk has the same parity as ℓ. For any pair (c, d) ∈[ℓ]×[ℓ], we have either aic < bjd or aic > bjd. Therefore, we can think of the pair (c, d) as contributing 1 to γd in the first case, or as contributing 1 to δc in the second case. So each pair (c, d) contributes exactly 1 to the sum Pℓ k=1 �γk +δk . Hence, the sum equals the number of pairs (c, d), which is ℓ2, which has the same parity as ℓ. Since S \ T = {ai1, … , aiℓ} and T \ S = {bj1, … , bjℓ}, and there is a sign in the definition of ηℓ, the lemma follows. □ We are now ready to compute the scalar action of ηℓon each irreducible submodule of Mµ. Theorem 3.3. On the irreducible submodule of Mµ isomorphic to S2i1(n+m)−2i, the operator ηℓacts like multiplication by the scalar ωℓ,i, where ωℓ,i := m ℓ − ℓ X ℓ1=0 m −i ℓ1 n −i ℓ1 i ℓ−ℓ1 (−1)ℓ1.
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10 FRIEDMANN Proof. Let T = [n] so that vT = 1 n+1 2 n+2 … … m n+m m+1 … n , and let t be the standard Young tableau of shape 2i1n+m−2i given by t = 1 n+1 2 n+2 … … i n+i i+1 … n n+i+1 … n+m . Recall that the Specht module S2i1(n+m)−2iis spanned by {τet : τ ∈Sn+m}, where et is the symmetrizer of Equation (1.1). In order to study the action of ηℓon this Specht module, we begin by simplifying the action of et on vT by factorizing et as follows. Let rt = P α∈Rt α. Let dt be the signed sum of column permutations stabilizing {1, 2, … , n}, {n + 1, … , n + i}, and {n + i + 1, … n + m}, i.e. the signed
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THE CONSTRUCTION OF A CLASS OF PRESENTATIONS FOR SPECHT MODULES 11 sum of permutations in the subgroup S{1,…,n} × S{n+1,…,n+i} × S{n+i+1,…,n+m} ⊆Ct. Now let ft be the signed sum of left coset representatives of the above subgroup of Ct, that is, permutations σ in Ct that satisfy1 σ(1) < · · · < σ(n), σ(n + i + 1) < · · · < σ(n + m), σ(n + 1) < · · · < σ(n + i). Then etvT = rtftdtvT. The antisymmetry of column tabloids ensures that dtvT is a scalar multiple of vT, because it simply permutes within columns. Therefore we can conclude that rtftvT is a scalar multiple of etvT, and in particular that etvT is nonzero, as the coefficient of vT in rtftvT is 1. Consider ηℓ(rtftvT). In the subspace restricted to S2i1n+m−2i, the fact that ηℓacts on etvT as a scalar implies the same is true of rtftvT. In fact, because the coefficient of vT in rtftvT is 1, we can determine precisely what this scalar is by computing ⟨ηℓ(rtftvT), vT⟩. We have rtftvT = X S∈([n+m] n ) ⟨rtftvT, vS⟩vS. Applying the linear operator ηℓthus gives ηℓ(rtftvT) = X S∈([n+m] n ) ⟨rtftvT, vS⟩ηℓ(vS). Note that when T = S, by Lemma 3.2 we have ⟨ηℓ(vT), vT⟩= �m ℓ . With this, we can compute the coefficient of vT in general by ωℓ,i = ⟨ηℓ(rtftvT), vT⟩= X S∈([n+m] n ) ⟨rtftvT , vS⟩⟨ηℓ(vS), vT⟩ (3.5)
m ℓ + X S∈([n+m] n ){T} ⟨rtftvT, vS⟩⟨ηℓ(vS), vT⟩. The contributions to the sum arise only when S and T differ by ℓelements. In the sum rtftvT, there are four different ways to obtain a vS that fulfills this criterion: 1In [FHSW] and [BF], the representatives used were mistakenly right coset representatives. The results in [FHSW] are unaffected by the error. The main results in [BF] are also unaffected, but see footnote on page 21.
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12 FRIEDMANN (1) vS is obtained by doing row permutations only (possible iffℓ≤i). (2) vS is obtained by doing column permutations only (possible iffℓ≤m −i). (3) vS is obtained by a subset of the column permutations of case (2) above, followed by row permutations. (4) vS is obtained via a combination of column permutations and row permuta- tions: ℓ1 entries are exchanged via column permutations and ℓ2 entries via row permutations, where ℓ= ℓ1 + ℓ2 (possible iffℓ1 ≤m −i and ℓ2 ≤i). In case (4), if we let ℓ1 = 0 we get case (1), and if we let ℓ2 = 0 we get cases (2) and (3). We find it helpful to compute cases (1), (2), and (3) before case (4). Case (1): row exchanges only. Pick u1, u2, … , uℓ∈[i] and let α = (u1, n + u1)(u2, n + u2) · · ·(uℓ, n + uℓ) . Then the columns of αvT, written left to right instead of top to bottom, are 1, 2, … , u1 −1, n + u1, u1 + 1, … , uℓ−1, n + uℓ, uℓ+ 1, … , n and n + 1, … , n + u1 −1, u1, n + u1 + 1, … , n + uℓ+ 1, uℓ, n + uℓ+ 1, … , n + m. For αvT to have ordered entries, we need to move the n + u1, n + u2, … , n + uℓto the end of the first column, and to move the u1, u2, … , uℓin the second column to the beginning of the column. For the first column, we first move n + uℓto the end, then n + uℓ−1, and so on. The number of transpositions for the first column is: ℓ−1 X k=0 (n −uℓ−k −k) = nℓ− ℓ X k=1 uk −ℓ(ℓ−1) 2 . The number of transpositions for the second column is: ℓ X k=1 (uk −k) = ℓ X k=1 uk −ℓ(ℓ+ 1) 2 . Together, the number of transpositions for ordering the columns of αvT is nℓ−ℓ2 .
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THE CONSTRUCTION OF A CLASS OF PRESENTATIONS FOR SPECHT MODULES 13 To compute ⟨ηℓ(vS), vT⟩, note that T \S = {u1, … , uℓ} and S\T = {n+u1, … , n+ uℓ} in Lemma 3.2, giving ⟨ηℓ(vS), vT⟩= (−1) Pℓ k=1(uk+n+uk)+ℓ+1 = (−1)nℓ+ℓ+1 . Hence, ⟨rtftvT, vS⟩⟨ηℓ(vS), vT⟩= (−1)(nℓ+ℓ+1)+(nℓ−ℓ2) = (−1), which is independent of the choice of uk’s. There are �i ℓ ways to pick the uk’s. So case (1) gives an overall contribution of (3.6) − i ℓ . This expression appropriately gives 0 when ℓ> i. Case (2): column exchanges only. The permutations σ in ft satisfy σ(1) < · · · < σ(n), σ(n + i + 1) < · · · < σ(n + m), σ(n + 1) < · · · < σ(n + i). Since σ ∈Ct, it follows that σ fixes {n+1, … , n+i}. Since we need |S ∩T| = n−ℓ, and since T = [n] and S = {σ(1), … , σ(n)}, we need to exchange ℓof the elements in [n] with ℓof the elements n + i + 1, … , n + m. Let Qℓ:= {q1 < … < qℓ} ∈ �[n] ℓ , and Pℓ:= {p1 < … < pℓ} ∈ �{n+i+1,…,n+m} ℓ be the sets of elements exchanged. In one-line notation, σ restricted to the first column of t is the concatenation of the sequences 1, … , q1 −1, q1 + 1, … , q2 −1, q2 + 1, … , qℓ−1, qℓ+ 1, … , n p1, p2, … , pℓ q1, q2, … , qℓ n + i + 1, … , p1 −1, p1 + 1, … , p2 −1, p2 + 1, … pℓ−1, pℓ+ 1, … , n + m. The number of inversions in σ restricted to the first column of t is then ℓ−1 X k=0 (n −qℓ−k −k) + ℓ X k=1 (pk −k −n + i) + ℓ2 = iℓ+ ℓ X k=1 (pk −qk). Now we consider σvT. Its first column is the same as the first n entries of the first column of σt, and does not need further reordering. Its second column is the concatenation of the sequences n + 1, … , n + i
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14 FRIEDMANN q1, q2, … , qℓ n + i + 1, … , p1 −1, p1 + 1, … , p2 −1, p2 + 1, … pℓ−1, pℓ+ 1, … , n + m. To reorder this column, we need to move the qk’s all the way to the left. That takes i transpositions for each qk, giving the sign (−1)ℓi. So ⟨rtftvT, vS⟩= (−1) Pℓ k=1(pk−qk). It remains to compute ⟨ηℓ(vS), vT⟩. Since S \ T = {p1, … , pℓ} and T \ S = {q1, … , qℓ}, in Lemma 3.2 we have ⟨ηℓ(vS), vT⟩= (−1) Pℓ k=1(pk+qk)+ℓ+1. The total sign of this contribution is then (−1) Pℓ k=1(pk−qk)(−1) Pℓ k=1(pk+qk)+ℓ+1 = (−1)ℓ+1. There are �n ℓ choices for the qk’s and �m−i ℓ choices for the pk’s, giving the total contribution from this case of (3.7) n ℓ m −i ℓ (−1)ℓ+1. Case (3): column permutations from case (2) followed by row permuta- tions. If in case (2), qj ≤i and qj+1 > i for some 1 ≤j ≤ℓ, then after σ from case (2), we can apply row swaps (qγ, n + qγ) for any γ ≤j and still have |S ∩T| = n −ℓ, with S \ T = {p1, … , pℓ} and T \ S = {q1, … , qℓ}. The row group is not signed, so all that remains is to reorder the columns of ασvT to be increasing and compute the corresponding sign. Suppose we do the row swaps for qγ1, … , qγr for some r ≤j and 1 ≤γ1 < γ2 < · · · < γr ≤j, so that α = (qγ1, n + qγ1) · · · (qγr, n + qγr). The first column of ασvT is as in Case (2) and needs no further reordering. The sign for the reordering of second column of ασvT can be obtained from the sign of reordering of the second column of σvT as follows. Suppose ρ achieves the reordering of the second column of σvT . Then ρα−1 achieves the reordering of the second
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THE CONSTRUCTION OF A CLASS OF PRESENTATIONS FOR SPECHT MODULES 15 column of ασvT. The contribution from case (2) is therefore modified only by sgn α−1 = (−1)r, so we have in this case: (−1)ℓ+1+r. How many ways are there to have r row swaps? In picking the qk’s, let us pick j of them to be in [i] and ℓ−j of them to be in [n] \ [i]. Then we can pick r of the first j qk’s for the row swaps. There are i j n −i ℓ−j j r ways to do this. There are still �m−i ℓ ways to pick the pk’s. So we have the contri- bution (3.8) m −i ℓ (−1)ℓ+1 ℓ X r=1 ℓ X j=0 i j n −i ℓ−j j r (−1)r. Note that if we set r = 0 in the above equation, we get the contribution of case (2). Case (4): column permutations to exchange ℓ1 elements and row permu- tations to exchange ℓ2 = ℓ−ℓ1 elements. We begin similarly to case (2), replacing ℓwith ℓ1, where 1 ≤ℓ1 < ℓ. We use a column permutation σ to exchange the elements of Qℓ1 = {q1 < · · · < qℓ1} ∈ �[n] ℓ1 with the elements of Pℓ1 = {p1 < · · · < pℓ1} ∈ �{n+i+1,…,n+m} ℓ1 . Then we pick ℓ2 = ℓ−ℓ1 elements of [i], say 1 ≤u1 < · · · < uℓ2 ≤i such that {u1, … , uℓ2} ∩ {q1, … , qℓ1} = ∅and carry out a row exchange on them, α = (u1, n + u1)(u2, n + u2) · · · (uℓ2, n + uℓ2). The permutation ασ exchanges a total of ℓentries between the first and second columns of vT, as desired. The sign of the column permutation is the same as in case (2), except that ℓis replaced by ℓ1: sgn(σ) = (−1)−ℓ1i+Pℓ1 k=1(pk−qk). We now consider the sign of the transposition that orders the two columns of ασvT in increasing order. We begin by applying only σ, so the first row of σvT is 1, … , q1 −1, q1 + 1, … , qℓ1 −1, qℓ1 + 1, … n, p1, p2, … , pℓ, and the second row of σvT is n+1, … , n+i, q1, q2, … , qℓ1, n+i+1, … , p1 −1, p1 +1, … pℓ1 −1, pℓ1 +1, … n+m.
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16 FRIEDMANN Once we apply α, the entries n + uk are in the first column mixed in between 1 and i where the uk used to be, and the entries uk are in the second column mixed in between n + 1 and n + i where the n + uk used to be. To count the number of transpositions required for reordering, we define γk to be the number of elements in Qℓ1 larger than uk. In the first column, we need to move the n + uk’s to just after n. This requires ℓ2 X k=1 (n −γk −uk −(k −1)) transpositions. In the second column, we need to move the uk’s and the qk’s to the left and keep them in order. This requires iℓ1 + ℓ1 X k=1 (uk −k + γk) transpositions. So turning ασvT into vS gives the sign (−1)nℓ2+iℓ1+ℓ2. Finally, since S\T = {p1, … , pℓ1, n+u1, … , n+uℓ2} and T\S = {p1, … , pℓ1, u1, … , uℓ2}, Lemma 3.2 gives ⟨ηℓ(vS), vT⟩= (−1) Pℓ1 k=1(pk+qk)+Pℓ2 k=1(n+uk+uk)+ℓ+1. The overall sign simplifies to (−1)ℓ1+1. How many of these cases are there? If we choose j of the ℓ1 q’s to come from [i], there will be i −j elements in [i] to choose the uk’s from. Summing these and remembering that there are still �m−i ℓ1 ways to choose the pk’s gives m −i ℓ1 (−1)ℓ1+1 ℓ1 X j=0 i j n −i ℓ1 −j i −j ℓ2 . As in case (3), we can also have row swaps using the subset of the qk’s that are in [i], and if there are r such row swaps, this merely introduces a factor of (−1)r, giving (3.9) m −i ℓ1 (−1)ℓ1+1 j X r=0 ℓ1 X j=0 i j n −i ℓ1 −j i −j ℓ2 j r (−1)r. We start the sum from r = 0 to include the case we just computed where no additional row swaps are done.
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THE CONSTRUCTION OF A CLASS OF PRESENTATIONS FOR SPECHT MODULES
17
Note that setting ℓ1 = 0 in the above formula gives us the contribution from case
(1) and setting ℓ1 = ℓin the above formula gives us the contributions from cases
(2) and (3) combined. Therefore, the total eigenvalue from all four cases is given by
summing the above formula from ℓ1 = 0 to ℓ1 = ℓ.
Now, reordering the sum to carry out the sum over r first reveals that we have a
term
(3.10)
j
X
r=0
j
r
(−1)r =
1
j = 0
0
j > 0
.
So the only contribution to the sum above is the j = r = 0 term, and the theorem
is proved.
Note that the cancellation that appears in equation (3.10) means in essence that
we can limit the Qℓ1 to only
�[n][i]
ℓ1
and not consider the contributions from the type
of row swaps discussed in case (3) (where the r originates), since those cancel the
contributions from the Qℓ1 in
�[n]
ℓ1
�[n][i]
ℓ1
.
□
Example. Let n = 8, m = 7, ℓ= 3. We will demonstrate case (4) with ℓ1 = 2,
ℓ2 = 1, i = 4.
Pick q1 = 2, q2 = 5, p1 = 14, p2 = 15, u1 = 3. In this case, j = 1 (because q1 ≤4
and q2 > 4). We have:
vT =
1
9
2
10
3
11
4
12
5
13
6
14
7
15
8
, t =
1
9
2
10
3
11
4
12
5
6
7
8
13
14
15
, σt =
1
9
3
10
4
11
6
12
7
8
14
15
2
5
13
, ασvT =
1
9
11
10
4
3
6
12
7
2
8
5
14
13
15
, vS =
1
2
4
3
6
5
7
9
8
10
11
12
14
13
15
.
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18 FRIEDMANN Following the counting in the proof for sgn(σ), we have sgn(σ) = (−1)−2·4+(14−2)+(15−5) = +1. Alternatively, we see that in cycle notation, σ = (2, 3, 4, 6, 8, 15, 13)(5, 7, 14), which is even. Now following the counting in the proof for ordering the columns of ασvT, we have (−1)8·1+4·2+1 = (−1). Alternatively, to order the first column we can use the permutation (11, 4, 6, 7, 8), which is even, and to order the second column we can use the permutation (9, 2, 10, 3, 5, 12), which is odd, giving the overall sign (−1). Finally, since T \ S = {2, 3, 5} and S \ T = {11, 14, 15}, the sign we get from Lemma 3.2 for ⟨η2(vS), vT⟩is (−1)(2+3+5)+(11+14+15)+3+1 = +1. So the contribution of this case to the eigenvalue is (+1)(−1)(+1)) = −1 = (−1)ℓ1+1, where ℓ1 = 2. The trace: The operator η2 for the shape (8, 7)′ is a �15 8 × �15 8 matrix. Its diago- nal entries all equal �7 3 , so its trace is �7 3 �15 8 . A computation of the sum of the eigenvalues of η2 with multiplicity, where the multiplicity of ω2,i is the dimension of S2i115−2i for i = 0, 1, … , 7, confirms this trace. The computation of the trace in these two different ways for any n ≥m ≥ℓgives the following result. Corollary 3.4 (A combinatorial identity). For n ≥m ≥ℓ, the following identity holds: m X i=0 ℓ X ℓ1=0 m −i ℓ1 n −i ℓ1 i ℓ−ℓ1 n + m i n + m −2i + 1 n + m −i + 1 (−1)ℓ1 = 0. Proof. Given n, m, and ℓ, the diagonal entries of ηℓare equal �m ℓ , and its size is dim M(n,m)′ × dim M(n,m)′. Since dim M(n,m)′ = �n+m n , we have Tr (ηℓ) = m ℓ n + m n .
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THE CONSTRUCTION OF A CLASS OF PRESENTATIONS FOR SPECHT MODULES 19 The trace also equals the sum of the eigenvalues of ηℓwith multiplicity, i.e. Tr (ηℓ) = m X i=0 ωℓ,i dim(S2i1n+m−2i). Hooke’s Law formula gives dim S2i1n+m−2i = n + m i n + m −2i + 1 n + m −i + 1 . So we have m ℓ n + m n
m X i=0 m ℓ − ℓ X ℓ1=0 m −i ℓ1 n −i ℓ1 i ℓ−ℓ1 (−1)ℓ1 n + m i n + m −2i + 1 n + m −i + 1 . Since M(n,m)′ ∼= m M i=0 S2i1n+m−2i, we have the identity n + m n
m X i=0 dim(S2i1n+m−2i) = m X i=0 n + m i n + m −2i + 1 n + m −i + 1 , which results in the corollary. □ 4. A class of presentations of Specht modules It is straightforward to check that for all 2-column partitions µ = 2m1n−m, we have wℓ,m = 0. It follows that ker ηℓ⊇Sµ. The Specht modules for which wℓ,i ̸= 0 for 0 ≤i < m are those for which we have obtained a new presentation. That is, in those cases, ker(ηℓ) ∼= Sµ and Im(ηℓ) = ker(α) for α : Mµ →Sµ. This is the content of our central theorem. Theorem 4.1. Let µ = 2m1n−m and let Hµ,ℓbe the subspace of Mµ generated by ηℓ([t]) for [t] ∈Mµ. Then Mµ/Hµ,ℓ∼= Sµ as Sn+m-modules iff m ℓ − ℓ X ℓ1=0 m −i ℓ1 n −i ℓ1 i ℓ−ℓ1 (−1)ℓ1 ̸= 0 for i = 0, 1, … , m −1.
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20 FRIEDMANN Proof. The theorem is a direct consequence of Theorem 3.3. □ We now state the conditions for presentations of partitions with two or more columns. Let λ be a partition, let ℓc ∈[λ′ c+1], and let hc,ℓc([t]) be the image of ηℓc on the c and (c + 1)st columns of [t] that leaves the other columns of [t] fixed. Theorem 4.2. Let λ be a partition of n and let λ′ be its conjugate partition. Let ˆℓ= (ℓ1, ℓ2, … , ℓλ1−1) and let Hλ,ˆℓbe the space generated by hc,ℓc([t]) for every [t] ∈ Mλ, 1 ≤c ≤λ1 −1, and one choice of ℓc ∈[λ′ c+1] for each c. Then Mλ/Hλ,ˆℓ∼=Sn Sλ iffλ satisfies the conditions λ′ c+1 ℓc − ℓc X k=0 λ′ c+1 −i k λ′ c −i k i ℓc −k (−1)k ̸= 0 for all 1 ≤c ≤λc −1 and i = 0, 1, … , λ′ c+1 −1. Proof. This theorem follows from Theorem 4.1 and the definition of hc,ℓc([t]). □ Remark Motivated by the same question proposed in [FHW3] that led to the present paper, [MMS] addressed the case of intermediate values of ℓ, approach- ing it via representations of the general linear group. In that different context, but using the same generalization of the symmetrized sum of [BF] (Equation (2.6)) as we do in Definition 3.1, they state the following sufficient (but not necessary, except in the 2-column case) condition for the corresponding quotient space to be a Specht module. We modify their notation to match ours. Theorem 4.3. [MMS, Theorem 6.2]2 Let λ be a partition of n and ℓ1, … , ℓλ1−1 positive integers satisfying ℓc ≤λc+1 , c = 1, … , λ1 −1. Let ˆℓ= (ℓ1, … , ℓλ1−1). Then as Sn-modules, we have Mλ/Hλ,ˆℓ∼= Sλ if j X t=1 (−1)t−1 λ′ c+1 −t λ′ c+1 −ℓc j t λ′ c −λ′ c+1 + j + t t ̸= 0 for all c = 1, … , λ1 −1 and j = 1, … , λ′ c+1. 2Due to a typo, the statement of this theorem in the published version of [MMS] erroneously limits the values of j to j = 1, … , ℓc; the values should be j = 1, … , λ′ c+1, as stated here.
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THE CONSTRUCTION OF A CLASS OF PRESENTATIONS FOR SPECHT MODULES 21 For the two-column case of Theorem 4.3, i.e. for λ1 = 2, [MMS] state that this condition is also necessary ([MMS, Corollary 6.1]). Putting their result together with Theorem 4.1 gives Corollary 4.4 (Equivalence of two sets of combinatorial conditions). Let n ≥m ≥ ℓ. Then m ℓ − ℓ X ℓ1=0 m −i ℓ1 n −i ℓ1 i ℓ−ℓ1 (−1)ℓ1 ̸= 0 for i = 0, 1, … , m −1 iff j X t=1 (−1)t−1 m −t m −ℓ j t n −m + j + t t ̸= 0 for j = 1, … , m. As in [BF], Theorem 4.2 dramatically reduces the number of generators needed to obtain Gλ. The original construction leading to Equation (2.3) required enumerating over every 1 ≤k ≤λ′ c+1 for every pair of columns c and c + 1 of every t ∈Tλ. Even Fulton’s simplification using only gc,1 relations requires enumerating over t ∈Tλ for every pair of columns c and c+1. By contrast, our construction uses a single relation for every pair of adjacent columns, and [t] varies in Mλ, a significantly smaller space than Tλ. We now discuss the shapes µ and values ℓfor which the conditions of Theorem 4.1 hold. When ℓ= 1, the condition w1,i ̸= 0 in Theorem 4.1 simplifies to (m−i)(n−i+1) ̸= 0, which is achieved in all cases, i.e. whenever 0 ≤i < m ≤n. This is the case of [BF],3 cited earlier as Theorem 2.2. When ℓ= m, the condition wm,i ̸= 0 in Theorem 4.1 simplifies to 1 − n −i m −i (−1)m−i ̸= 0, which holds for 0 ≤i < m whenever n ̸= m or n = 1. This is the case of [FHW3], cited earlier as Theorem 2.1. In fact, when ℓ= m, no symmetrization occurs and ηm is equal to the Garnir relation g1,m used in [FHW3]. 3 In [BF], the error described in the footnote on page 11 led to the incorrect formula (m−i)(n+
- ̸= 0 which, like the correct formula obtained above, is also achieved in all cases; that error has been corrected here, and Theorem 2.2, the main result of [BF], holds.
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22 FRIEDMANN Generally speaking, the conditions in Theorem 4.1 seem to hold for the vast majority of cases. Considering n ≥m ≥ℓfor 1 ≤n ≤50, only 391 of the possible 22,100 combinations of n, m, and ℓhave values of i < m for which S2i1n+m−2i is in the kernel of ηℓ. In those cases, there is only one such value of i in all but 12 of the cases, in which the number of such values is 2. Data for 1 ≤n ≤28 is recorded in Table 4.1. Acknowledgements The author is grateful to Phil Hanlon and Michelle Wachs for related collaboration and helpful discussions; to Sarah Brauner for helpful discussions and comments on an earlier draft; to Thomas McElmurry for helpful discussions; and to Tony Yan for writing the computer program for calculating wℓ,i. References [BL] J. Bagger and N. Lambert, Three-Algebras and N=6 Chern-Simons Gauge Theories, Phys. Rev. D 79 (2009), 025002, 8 pp.. [BF] S. Brauner and T. Friedmann, A simplified presentation of Specht modules, J. Pure Appl. Algebra, 226 (2022), Paper No. 106979. [DI] J. A. de Azcarraga and J. M. Izquierdo, n-ary algebras: a review with applications, J. Phys. A 43 2010, p. 293001. [DT] Yu.L. Daletskii and L.A. Takhtajan, Leibniz and Lie algebra structures for Nambu algebra, Lett. Math. Phys. 39 (1997), 127–141. [Fi] V.T. Filippov, n-Lie algebras, Sibirsk. Math. Zh. 26 (1985), 126–140. [Fr] T. Friedmann, Orbifold Singularities, Lie Algebras of the Third Kind (LATKes), and Pure Yang-Mills with Matter, J. Math. Phys. 52 (2011), 022–304. [FHW3] Tamar Friedmann, Phil Hanlon, and Michelle L Wachs. A new presentation for Specht modules with distinct parts, The Electronic Journal of Combinatorics 31(4) (2024) P4.41. [FHSW] Tamar Friedmann, Phil Hanlon, Richard P Stanley, and Michelle L Wachs. On a gener- alization of Lie(k): a CataLAnKe theorem, Advances in Mathematics, 380:107570, 2021. [Fu] William Fulton. Young tableaux: with applications to representation theory and geometry, volume 35. Cambridge University Press, 1997. [Gu] A. Gustavsson, Algebraic structures on parallel M2-branes, Nucl. Phys. B 811 (2009), 66–76. [Ka] S. M. Kasymov, On a theory of n-Lie algebras, Algebra i Logika 26 (1987), 277–297. [Kr] Witold Kraskiewicz. Reduced decompositions in Weyl groups, Eur. J. Comb., 16:293–313, 1995. [Li] W. X. Ling, On the structure of n-Lie algebras, PhD Thesis, Siegen, 1993. [MMS] M. Maliakas, M. Metzaki, and D.-D. Stergiopoulou, Presentations of Schur and Specht modules in characteristic zero, J. Pure Appl. Algebra, 229 (2025), Paper No. 107774.
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THE CONSTRUCTION OF A CLASS OF PRESENTATIONS FOR SPECHT MODULES 23 [Sa] B. E. Sagan, The Symmetric Group, 2nd Ed., Graduate Texts in Mathematics 203, Springer- Verlag, New York, 2001. [Ta] L.A. Takhtajan, On foundation of the generalized Nambu mechanics, Comm. Math. Phys. 160 (1994), 295–315. Department of Mathematics, Colby College Email address: tfriedma@colby.edu
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24 FRIEDMANN Table 4.1. Values of (n, m, ℓ, i) for which wℓ,i = 0 for n ≤28, 0 ≤ i < m, and 1 ≤ℓ< m. n m ℓ i 5 4 2 1 6 4 3 2 6 6 4 3 6 6 5 3 7 3 2 1 7 5 3 1 7 6 5 3 7 6 5 4 7 7 5 4 8 5 2 2 8 5 4 2 8 8 2 3 9 5 3 3 9 7 2 3 9 7 3 2 9 7 5 5 9 9 4 4 9 9 4 5 9 9 5 5 10 9 2 4 10 9 5 5 10 9 5 6 10 9 7 6 10 10 5 6 11 6 2 3 11 7 4 5 11 7 5 4 11 8 5 6 11 11 2 5 11 11 9 5 12 4 2 2 12 6 3 2 12 6 3 4 12 10 2 5 12 10 3 6 12 10 7 8 12 10 9 6 12 11 9 9 12 12 4 6 12 12 8 6 n m ℓ i 13 6 4 3 13 9 5 7 13 10 4 7 13 10 7 7 13 12 7 9 14 7 2 4 14 11 2 6 14 12 3 5 14 12 6 8 15 7 3 5 15 10 5 8 15 11 6 9 15 13 9 11 15 14 10 11 15 14 11 8 15 15 13 10 15 15 14 10 16 9 3 4 16 10 2 6 16 12 2 7 16 14 3 9 16 15 7 11 16 15 14 10 16 15 14 11 16 16 4 5 16 16 14 11 17 5 2 3 17 8 2 5 17 11 5 9 17 13 7 11 17 15 3 7 17 15 13 7 17 16 2 9 17 16 10 13 17 16 11 13 17 16 13 14 18 8 3 6 18 9 5 6 18 10 4 6 18 10 4 8 n m ℓ i 18 12 7 9 18 13 2 8 18 14 5 11 18 15 9 13 18 16 11 13 18 16 11 14 18 16 14 12 18 17 11 10 18 17 11 14 19 9 6 3 19 12 5 3 19 12 5 10 19 15 8 13 20 9 2 6 20 14 2 9 20 16 2 10 20 17 7 14 20 17 11 14 21 9 3 7 21 13 5 11 21 15 3 11 21 17 9 15 21 19 13 17 22 6 2 4 22 15 2 10 22 15 4 12 22 15 8 12 22 16 7 14 22 19 2 12 22 21 17 19 23 10 2 7 23 12 9 7 23 13 2 9 23 14 5 12 23 16 11 12 23 19 10 17 23 21 9 17 23 23 2 14 24 10 3 8 24 15 4 10 n m ℓ i 24 16 2 11 24 16 6 14 24 19 9 17 24 22 2 14 24 22 3 12 24 22 11 19 24 22 15 20 24 22 20 12 24 24 4 15 25 13 4 11 25 15 5 11 25 15 5 13 25 21 11 19 25 22 13 20 25 23 9 19 26 11 2 8 26 17 2 12 26 17 3 10 26 22 8 19 26 22 13 19 26 24 3 17 26 25 2 16 27 7 2 5 27 11 3 9 27 16 5 14 27 19 7 17 27 21 9 19 27 22 2 15 27 23 12 21 27 25 17 23 27 26 21 24 28 18 2 13 28 26 3 15 28 28 2 18 28 28 26 18 28 28 26 21 28 28 27 21
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