Algorithm for an Enumaration of Reduced Words of Sn Balanced Labeling Mustafa Türköz∗ Master Student Department of Mathematics Bogazici University Bebek, Istanbul 34432 Turkey June 18, 2015
Abstract Coxeter groups were introduced in 1934 by H.S.M. Coxeter.
Symmetric
groups are one of the crucial example of Coxeter groups. The symmetric group Sn is the group of all permutations of 1, 2, .., n . We know that every permutation can be written as product of adjacent transpositions. A product of this form is called a reduced word. In this paper, we study balanced labeling of the diagram that represents the inversions of a permutation. Moreover, these are natural encodings of reduced words of ω ∈ Σn
Keywords: Reduced word, Rothe diagram, Balanced Labeling ∗
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Introduction Study of reduced word is very crucial in group theory since reduced words lead us to
comprehend Symmetric Group Sn . Symmetric groups are natural encoding of finite group theory because we know that by Cayley theorem, every finite group is isomorphic to subgroup of symmetric group Sn . There are few methods in order to understand reduced words of symmetric groups. These methods divided into two main approach classical algebraic way and algebraic combinatorics. In this paper we will use combinatorics method.
Historically, R.Stanley started the study of reduced words of elements of Sn in 1982. The main result of Stanley was the introduction of a family of a symmetric functions indexed by permutations. Around 1982, Schützenberger and Lascoux founded the theory of Schubert polynomials. In 1990, Alex Kohnert claimed and showed that Schubert polynomials could be constructed by applying a recursive algorithm. In 1992, Bergeron contributed this subject via making another combinatorial algorithm ’RC-graphs’ [5]. In 1993, Biley, Fomin and Kriliov introduced a new set of diagrams that encode the Schubert polynomials. In 1994, Reiner and Shimozono described a method called plactification. In 1997, Fomin, Green, Reiner and Shimozono introduced a new method of enumeration of reduced word decompositions that is called balanced labeling[1]. This theory is developing each passing. A few years ago, Olcay Coskun and Muge Taskin Aydin [6], [7] developed a new method in this direction.
In this paper, we are focusing on Balanced labeling method. Our aim is to show that there is a bijective correspondence between injective balanced labeling of a diagram and reduced word decomposition of the corresponding permutation.
This paper is organised as follows. Section 2 gives the basic definitions and examples. Section 3 provides the main result of balanced labelling method and section 4 gives conclusion. 2
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Reduced Word
Definition 2.1. The symmetric group Sn is the group of all permutations of {1, 2, ..., n}. Definition 2.2. Let A be a set,and σ ∈ SA for fixed a ∈ A the set Oa,σ = {σ n |n ∈ Z} is the orbit of a under σ Definition 2.3. If a permutation has one orbit at least containing more than one element,then it is a called cycle. Definition 2.4. The length of a cycle is the number of elements in its largest orbit. Definition 2.5. A cycle with length two is called a transposition. Definition 2.6. If the transposition is the form (i, i + 1), then it is called adjacent transposition and it is denoted by si . Remark. Every permutation can be expressed as a product of adjacent transpositions. Definition 2.7. Since the set of adjacent transposition {s1 , s2 , ..., sn−1 } generates the symmetric group Sn on [n] = {1, 2, 3, ..., n} ω ∈ Sn can be represented by a product of all si ’s for all i in {s1 , s2 , ..., sn−1 } where si ’s stands for the transposition (i, i + 1). A product of these generators is called a word.
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Definition 2.8. If length of the product is minimal, it is called reduced word. Moreover, this product is called reduced word decomposition. Definition 2.9. Set of all reduced words for any ω is denoted by R(ω). Example 2.1. Let ω = 4132 ∈ S4 then the decomposition is ω = s2 s3 s2 s4 .
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3
Balanced Labeling
Definition 3.1. Diagram means any finite collection of cells D ⊆ Z × Z. Furthermore, meaning of a labeling, in this topic, is filling in the boxes of the diagram with positive integers. Definition 3.2. Let ω ∈ Σn be a permutation. The inversion diagram of ω is defined as
I(ω) = {(i, ωj ) | i < j, ωi > ωj } ⊆ [n] × [n]
where [n] denotes the set {1, 2, ..., n} Definition 3.3. The Rothe diagram is defined as
D(ω) = {(i, ωj ) | i < j, ωi > ωj } ⊆ [n] × [n]
Notice that the inversion diagram and Rothe diagram is equivalent yet Rothe diagram is more useful object. Definition 3.4. To each cell (i, j) of a diagram D, associate the hook Hi,j = Hi,j (D) consisting of those cells (i0 , j 0 ) of D in which either i0 = i and j 0 ≤ j or j 0 = j and i0 ≤ i. The diagram D(ω) of a permutation ω may be obtained from [n] × [n] by removing the hooks corresponding to cells (i, ωi ), i = 1, 2, ..., n.
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Example 3.1. If ω ∈ (3, 5, 2, 1, 4) ∈ S5 , then the cells of D(ω) are represented by circles in the illustration below. We represent the permutation ω by placing × in the cells (i, wi ) for all i: ◦ ◦ ◦ ×
◦ ◦ ×
× ◦
×
×
Definition 3.5. If D = D(ω) and (i, j) is cell of D, we define the cell (i, j) is called the corner cell of the hook Hi,j . Definition 3.6. A labelling of the cells of Hi,j (ω) with non- negative integers (possibly repeated) is called balanced if it satisfies the following condition: if one rearranges the labels within the hook so that they weakly increase from right to left and from top to bottom, then the corner label remains unchanged. A hook with a balanced labelling is called a balanced hook. Example 3.2. The hook is balanced: 3 6 1
2 2
4
6
1 3
Definition 3.7. Let D be a diagram with L cells. Then: 1) A labeling of the cells of D with positive integers is said to be balanced if each hook Hi,j (ω), (i, j) ∈ D(ω) is balanced 2) A balanced labeling is said to be injective if each of the labels 1, 2, ..., L appears exactly once. 3) A balanced labeling is said to be column strict if no column contains two equal entries. Example 3.3. Let ω = (3, 5, 2, 1, 4) and labeling D(ω) is injective balanced: 2 5 1 ×
4 6 ×
× 3
×
×
Example 3.4. Let ω = (3, 5, 2, 1, 4) and labeling D(ω) is ithe second is column-strict: 2 4 1 ×
3 4 ×
× 3
×
×
Definition 3.8. Let ω ∈ Σn be a permutation of length L, and let a be a reduced decomposition of ω. Let Ta : D(ω) ←→ [L] be the injective labeling defined by setting Ta (p, q) = i if qi transposes ωp and q where q < ωp . Then Ta is called the canonical labelling of D(ω) induced by a.
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Theorem 3.1. Let ω ∈ Σn , RD(ω) denote the set of reduced decompositions of ω, let RL (D) denote the set of injective balanced labellings of D. Then the map a 7→ Ta defines a bijection from RD(ω) to RL (D(ω)) Example 3.5. Let ω = (3, 5, 2, 1, 4) ∈ Σn and let a = (1, 2, 4, 1, 3, 2). The action of each ai is demonstrated in the table on the left, the canonical labelling Ta is illustrated on the right: a1 a2 a3 a4 a5 a6
=1 =2 =4 =1 =3 =2
1 2 3 2 1 3 2 3 1 2 3 1 3 2 1 3 2 5 3 5 2
4 5 4 5 4 5 5 4 5 4 1 4 1 4
2 4 5 6 1 × ×
× 3
×
×
The proof of theorem 3.1 will be provided as a corollary of more general results. Lemma 3.1.
Let ω ∈ Σn , and let T : D(ω) ←→ [L] be an injective map. Then T is
balanced if and only if, for all 1 ≤ i < j < k, the restriction of T to the subdiagram of D(ω) determined by rows i, j, k and columns ωi , ωj , ωk is balanced. Proof. This is a routine verification. Definition 3.9. If ω is a permutation and α = (i, j) ∈ D(ω), then α is said to be border cell if ωi+1 = j.
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Example 3.6. Let ω = (3, 4, 6, 1, 5, 2) and D(ω), the border cells are illustrated by solid dots: ◦ ◦ • ×
◦ × ◦ ◦
× ◦
•
×
×
Lemma 3.2. If α = (i, j) is border cell of D(ω), then
D/α = D(ω 0 )
where ω 0 = ω(i, i + 1). Proof. Proof is immediate. Example 3.7. If α = (3, 1) with ω = (3, 4, 6, 1, 5, 2) then D/α is ◦ ◦ ◦ ◦ × ◦ • ×
× × ◦ × ×
×
Then easily, the diagram D(ω 0 ) where ω 0 = (3, 4, 1, 6, 5, 2) = ω(3, 4)
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Lemma 3.3. If T is a balanced column-strict labeling of D(ω) with largest label M , then every row containing an M must contain an M in a border cell. In particular, if i is the index of such a row, then i must be a descent of ω. Proof. Suppose that row i of T contains an occurrence of the largest label M . We must first show that i is a descent of ω. Assume not, ωi < ωi+1 there exist some cell(s) in D(ω) below (i, j) in the same column. This implies that the hook Hij is unbalanced, because there are no other occurrences of M to the right of column j in this row, and there are no other occurrences of M in column i by column-strictness. Therefore i must be a descent of ω. It follows that row i contains a unique border cell α = (i, j). We must show that T (i, j) = M . Since (i, j) is a border cell we have ωi+1 = j, which implies that the hook Hij is entirely horizontal and lies in row i. Balancedness of T then implies that this hookâĂŹs leftmost cell (i, j) contains the maximal entry of the hook. The rightmost occurrence of M in row i cannot lie to the left of (i, j), since then there would exist some cell(s) in D(ω) below it in the same column, and the corresponding hook is unbalanced. Hence M must occur in Hij , and it follows that T (i, j) = M . Theorem 3.2. Let T be any labeling of D(ω), and assume that some border cell α contains the largest label M in T . Then T is balanced if and only if T /α is balanced. Proof. Let α = (i, j) be the border cell, and ω 0 = ω(i, i + 1) by lemma 3.1 it is sufficient to show that for all 1 ≤< a < b < c the restriction Tabc of T to the subdiagram of D(ω) determined by rows a, b, c and columns ωa , ωb , ωc is balanced if and only if (T /α)abc is balanced. There are several cases to check depending on how the sets {a, b, c} and {i, i + 1} intersect, but many of these cases are trivial. Since, for any r, s the (r, ωs )-entry of T coincides with 10
the (r, ωs0 ) entry of T /α if it is (r, ωs ) = (i, j), the verification is trivial if it is not i ∈ {a, b, c} and j ∈ {ωa , ωb , ωc }. Hence we may assume that we are in this case, so that (T /αabc ) will have one fewer entry than Tabc . Moreover, if Tabc has at most two entries then the verification is trivial from the fact that M occurs in a border cell. Thus, we may assume that Tabc ωa > ωb > ωc and their (i, i+1, , j) = (a, b, ωb ) or (i, i+1, , j) = (b, c, ωc ). In the former case, one can check that Tabc being balanced and (T /α)abc being balanced are both equivalent to the condition that T (a, ωc ) ≥ T (b, ωc ). In the latter case, one can check that Tabc being balanced and (T /α)abc being balanced are both equivalent to the condition that T (a, ωc ) ≥ T (a, ωb ).
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Algortihms
Algorithm 1. Decoding Algortihm We give a ’decoding rule ’ which associates to each balanced diagram RD(D(ω)) a pair of sequences
a1 a2 ......aL−1 aL b1 b2 .......bL−1 bL where a = (a1 , a2 , ..., aL ) ∈ RD(D(ω)) and b = (b1 , b2 , ..., bL ) satisfies all sequence 1 ≤ b1 ≤ b2 ≤ .... ≤ bn such that bi < bi+1 whenever ai < ai+1 . The rule is as follows: (i) set M equal to the maximum label in T , with α = (I, J) the border cell containing the leading entry in T ; (ii) set aL = I and bL = M (iii) set T = T /α, L = L − 1, and repeat until T is empty. Finally, note that if bi = bi+1 , this means we have deleted the same label twice in succession. It follows that ai > ai+1 , since in this case the leading entry of T /α must lie in a strictly lower row than the leading entry of T. Hence ai < ai+1 implies bi < bi+1 Next, construct a labeling T : D(ω) −→ P by assigning bi to the cell corresponding to the pair inverted by ai , for i = 1, 2, ...., L We claim that T is a balanced column strict labeling, and that the correspondence is inverse to the one defined above. To see this, observe that if
bj = bj+1 = .... = bL = M
, then aj > aj+1 .... > aL
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. This means that the leading entry M in T lies in the border cell α determined by aL . Inductively, we may assume that a1 , a2 , ...aL−1 and b1 , b2 , ...bL−1 determine a balanced columnstrict labeling of D(ω 0 ) = D(ω)/α, where ω 0 = sa1 , sa2 , ..., saL−1 . By theorem 3.2, T is balanced. In order to see that T is column-strict, we need only check that if α = (i, j), then there are no occurrences of M in column j of T /α below row i, since there are no cells of D/α here, and there are no occurrences of M in a row above i in T /α because
aj > aj+1 > ... < aL
Lastly, note that this construction clearly inverts the one defined earlier, so we have the desired bijection. Example 4.1. If ω = (3, 5, 2, 1, 4) and T is corresponding balanced diagram. such that: 2 5 1 ×
×
4 6 ×
3
×
× the decoding algorithm described above yields
a=1 2 4 1 3 2 b = 1 2 3 3 4 4.
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Definition 4.1. Let T be an injective balanced labeling of D(ω) where ω has length L. For each k = 1, 2, ..., L, let
I(k) = row index of the cell containing k, R+ (k) = number of entries k 0 > k in the same row, U + (k) = number of entries k 0 > k above kin the same column.
Algorithm 2. Encoding Algorithm Theorem 4.1. If T is an injective balanced labeling of D(ω) and a is a reduced decomposition such that T = Ta , then for k = 1, 2, ..., L, we have ak = I(k) + R+ (k) − U + (k) (†)
Proof. Rewrite the above equation as
I(k) = ak + R+ (k) − U + (k)
We will show that this formula is valid for all k by induction on I(ω). Let a = a1 a2 ...aL , ω b = ωsaL is shorter than ω, and and the above formula holds for a1 a2 ...aL−1 ∈ RD(b ω ) i.e b = ak + U b + (k) − R b+ (k), I(k)
where the hatted expressions correspond to the word a1 a2 ...aL−1 If k = L, then U + (L) = R+ (L) = 0 and obviously I(L) = aL If k < L and k does not occur in row aL or aL + 1 of D(b ω ), then none of the quantities change. 14
b b+ (k)+1. If k < L and k occurs in row aL +1 of D(b ω ), then I(k) = I(k)−1 and R+ (k) = R b + because the only change within the column s containing k is the Also U + (k) = U exchange of k and the entry k 0 in the cell directly above it in D(b ω ), and we know that k 0 < k by applying lemma 3.1 to the rows aL , aL + 1 and ω b −1 (s). b + 1 and R+ (k) = R b+ (k), If k < L and k occurs in row aL of D(b ω ), then I(k) = I(k) b + + 1 because the only change within the column s containing k is also U + (k) = U the exchange of k and the entry k 0 in the cell directly below it in D(b ω ) we know that k 0 > k by Lemma 3.1 applied to the rows aL , aL + 1 and ω b −1 (s).
Example 4.2. If T is the balanced diagram 2 × 5 7 ×
4 × ×
3 1
×
6 × then we have k I(k) R+ (k) U + (k) ak
1 3 1 1 3
2 1 0 0 1
3 2 2 0 4
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4 2 1 0 3
5 2 0 0 2
6 5 0 0 5
7 3 0 0 3
Remark. A similar decoding procedure can be defined by interchanging the roles of rows and columns, and reversing the entries (this amounts essentially to using ω −1 instead of ω). More precisely, if we define
J(k) = column index of the cell containing k, C − (k) = number of entries k 0 < k in the same column, L− (k) = number of entries k 0 < k to the lef t of k in the same row,
then ak = J(k) + C − (k) − L− (k). (•) For the example illustrated above, we have k J(k) C − (k) L− (k) ak
1 3 0 0 3
2 1 0 0 1
3 3 1 0 4
4 4 0 1 3
5 1 1 0 2
6 3 2 0 5
7 1 2 0 3
Remark. Surprisingly, fomulae † and • lead the same result such that
ak = I(k) + R+ (k) − U + (k) = J(k) + C − (k) − L− (k)
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Conclusion In this paper, we examined to an algorithm of enumeration of reduced words.This algo-
rithm is called balanced labeling. Balanced labeling is one of the most important method to count reduced words. Our study based on introducing decoding and encoding algorithms produced by Fomin and his coauthors [1], at the same time, we also provides proof of the existence of these algorithms. What is more, it is shown that there is a bijection between injective balanced labeling of rothe diagram and reduced expression of the corresponding permutation. Moreover, combinatorial aspect of enumeration reduced word is crucial for representation of symmetric groups.
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References
[1] S. Fomin, C. Greene, V. Reiner and M.Shimozono, Balanced Labellings and Schubert PolynomialsEurop. J. Combinatorics 18 (1997), 373-389 [2] P. Edelman and C. Greene, Balanced tableaux, Adv. Math.63 (1987), 42-99. [3] R. P. Stanley, On the number of reduced decompositions of elements of Coexter groups, Europ. J. Combinatorics 5 (1984), 359-372. [4] G. James and A. Kerber, The Representation Theory of the Symmetric Group, AddisonWesley, Reading, MA, (1981). [5] N. Bergeron, S. Billey, RC-graphs and Schubert polynomials, Experiment Math. 2, (1993), 257-269. [6] O. Coskun, M. Taskin Aydin, Tower Tableaux, Journal of Combinatorial Theory Series A, 120, (2013), 843-871. [7] O. Coskun, M. Taskin Aydin, Sorting and generating reduced words, Archiv der Mathematik, 101, (2013), 1427-436. 17
