Studying ribbon tableaux is quite a difficult subject, which mainly uses huge numerical experimentations. This is why we are interested in finding efficient algorithms for generating and computing statistics on them. The matricial coding of ribbon tableaux used for numerical experimentations in 8; 9 is deficient because it does not give some elementary properties shape and position of the head of the ribbons for example without additional computations and cannot be generalized to skew shapes.

The aim of this paper is to present a more general algorithm for generating ribbon tableaux and computing spin polynomials, using a different and more transparent coding for ribbon tableaux. A k-ribbon R is a connected skew diagram with k cells which does not contains a square. The first north-west cell is called the head and the last one south-east the tail. Ribbon tableaux are labelled tilings verifying monotonicity conditions similar to these of Young tableaux. The spin of a k-ribbon tableau is the spin of the underlying tiling.

The main basic idea is to apply recursively the algorithm of removing k-ribbon strips from a partition.. By construction, we can immediately read the shape of the corresponding ribbon tableau, and the length of the ribbons is obtained by dividing the number of non negative cells by the number of positive cells. The structure used in was a matrix where the entry i, j is equal to k if there is the tail of a ribbon labelled i in the j-th column of the shape. We show that the main difficulty in carrying out this construction is that we cannot read the shape of the partition without reconstructing the.

Furthermore, this coding become non injective when we try to generalize it on ribbon tableaux of skew shape. Output: The k-ribbon tableau corresponding to the array M. As a k-ribbon strip has a unique tiling by k-ribbons, it is completely defined by the positions of all the ribbons s heads.

Generating algorithm and computation of the spin and cospin polynomials for ribbon tableaux of a given shape and weight We will generalize the generating algorithm of to the case of k-ribbon tableaux of skew shapes. In other words this algorithm permits one to compute the spin and cospin polynomials without additional computation. By studying the progress of this algorithm with huge set of ribbon tableaux we remark that at the bottom of the tree there are a lot of nodes, but in there is only few different shape for the remaining partition. Our previous algorithm search, at step i, all the possibility to retire a k ribbon strip from the remaining partition, but in fact we need to test all the possibility only on the few remaining partitions.

That s why a recursive implementation of the previous algorithm with a remember option seems to be the most efficient way to generate ribbon tableaux. Andrews The theory of Partitions, Vol. Haglund, M. Haiman, N. Loehr, J. Remmel, A. Hivert, N. Krob, A.

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Lascoux, B. Leclerc, J. Thibon, B. Ung, S. Thibon, Ribbon tableaux, Hall-Littlewood functions, quantum affine algebras and unipotent varieties, Journal of Mathematical Physics 8 , [9] B. Leclerc , Adv. Studies in Pure Math. Lothaire Algebraic Combinatorics on Words, Vol. Stanton and D. White, A Schensted algorithm for rim-hook tableaux, J.

Theory A 0, , Van Leeuwen, Spin-preserving Knuth correspondences for ribbon tableaux 00 [] S. Schilling, M. Shimozono, D. Inspired by the spin-inversion statistic of Schilling, Shimozono and White [8] and Haglund et al. We determine the necessary and sufficient combinatorial. Korean J. In [] Schensted constructed the Schensted algorithm,. In previous work Regev used part of the representation theory of Lie superalgebras to compute the values of a character of the symmetric.

An algebraic formula for the Kostka-Foukles polynomials is given. We prove a positive combinatorial formula for the Schur expansion of LLT polynomials indexed by a 3-tuple of skew shapes. Manufactured in The Netherlands. Operators on k-tableaux and the k-littlewood Richardson rule for a special case by Sarah Elizabeth Iveson A dissertation submitted in partial satisfaction of the requirements for the degree of Doctor of.

International Journal of Algebra, Vol. The classical. We give some new formulas about factorizations of K-k-Schur functions, analogous to the k-rectangle factorization formula. Using the action. Stembridge Department. Notes on Creation Operators Mike Zabrocki August 4, This set of notes will cover sort of a how to about creation operators.

Because there were several references in later talks, the appendix will. Combinatorics for algebraic geometers Calculations in enumerative geometry Maria Monks March 17, Motivation Enumerative geometry In the late 18 s, Hermann Schubert investigated problems in what is. Combinatorial Structures Contents 1 Permutations 1 Partitions. Skew diagrams Dominance order Basic Definitions Dyck paths are one of the many combinatorial objects enumerated by the Catalan numbers, sequence A in [2]:. China Email: chenstation yahoo. References 1. Aguiar, M. Discrete Comput. Introduction In statistical mechanics, one studies models based on the interconnections between thermodynamic.

We define a generalization. In this short note we give a new proof of the quantum generalization of Regev s theorems by applying the Murnaghan-Nakayama formula. Please share how this access benefits you. Your story matters. Citation As Published Publisher. A Nekrasov-Okounkov type formula for affine C. James Haglund; Jiang Zeng. Consider the probability that an arbitrary chosen. Hivert a,1, J. Novelli b,1, and J. These notes are intended as a fairly self contained explanation of Fock space and various algebras that act on it, including a Clifford algebras, a Weyl algebra,.

We introduce a quasisymmetric generalization of Berele and Regev s hook Schur functions and prove that these new. Equations 4. Inequality 4. The condition 4. Now we assume 4. Since by Proposition 3. This settles 4. This concludes the inductive proof of 4.

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Now assume that 4. This concludes the proof of 4. Next we prove the reverse direction of the claim. Then 4. Since ci is finite. Because of 4. Using 3. It combines points 1 and 2 of the Definition-Proposition 4. For the proofs the formulation of Definition-Proposition 4.

Suppose x occurs in ZCj in column c0. The vacancy numbers are calculated with respect to this new sequence of rectangles.

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The nontrivial steps of the above algorithm applied to the third tableau of Example 2. A rigged partition is represented by its Ferrers diagram where to the right of each part the corresponding rigging is indicated. The vacancy numbers are given to the left of each part. Example for the bijection algorithm see Example 4. Evacuation Theorem In this section we prove the Evacuation Theorem 5. So it may be assumed that R is a single rectangle with exactly two columns. Lemma 5.

Assume also that the map j is injective. Then the front face must also commute. We wish to show the front face commutes. By Lemma 5. For the remaining case suppose the last rectangle of R is a single column. The back face commutes by induction. The top face obviously commutes. The bottom face commutes by Lemma 5. In the remaining case, the last rectangle of R has more than one column. We have the following diagram: R 98 A. The commutation of the top face is obvious. The bottom face commutes by conjugating the result of Lemma 5.

Proof of the Evacuation Theorem Theorem 5. If R is empty the result holds trivially. Suppose that the last rectangle ev ev of R has more than one column. So by Lemma 5.

This recurrence shall be used to prove some properties of the transpose maps on LR tableaux and rigged configurations. Splitting off the first or last row Let R be obtained from R by splitting off the last row from the last rectangle.

## Ebook A Bijection Between Littlewood Richardson Tableaux And Rigged Configurations 2002

Recall the transpose map trLR of t Definition 2. Note that Vol. Therefore the front face commutes by Lemma 5. Removal of a cell from a single row Suppose RL is a single row.

The top triangle obviously commutes. It suffices to show that the bottom triangle commutes. The result is now clear. If the last rectangle of R has more than one row, then one may apply the commutative diagram 6. In the following we recall its definition and prove the Transpose Theorem 7. By [13, 9. Fortunately equation 7. To see this, observe that it still follows from [13, Section 9] that trRC is an involution on the level of configurations.

But this follows from 7. Hence trRC is an involution for rigged configurations. We shall prove [13, Conjecture 16]. Theorem 7. Lemma 7. Suppose the last rectangle of R consists of a single column. The proof of this Lemma is given in Appendix B. Proof of Theorem 7. The bottom face commutes by Lemma 7. The left face is the commutative diagram 4. Suppose the last rectangle is a single column. The left face commutes by 4.

Embeddings In [21, Section 6. Let R and R0 be two sequences of rectangles. Thus the relation R D R0 is a preorder. It is generated by the following two relations see [24, Section 2. Definition-Proposition 8. If p A. In this case the sets of tableaux are all empty or all singletons by [23, Prop. Let R D R0. Then R0 may be obtained from R by a sequence of transformations of the form E1 and E2 ; fix such a sequence.

## [math/] A bijection between Littlewood-Richardson tableaux and rigged configurations

By [24, Theorem 4] it follows that the embedding iR R is independent of the sequence of transformations E1 and E2 leading from R to R0. This fact also follows immediately from Theorem 8. Clearly it suffices to prove Theorem 8. The following result immediately proves Theorem 8.

Lemma 8. The commutation of the bottom face is obvious. The top face commutes by definition and the left and right faces commute by Theorem 5. The desired commutation of the front face is reduced to the commutation of the back face. But in this case the sets in 8. Proposition 8. Using the well-known fact Sketch of Proof.

Let WR R resp. Again the result follows from the special cases E1 and E2. Consider the case E1. By [23, Prop. The vertical maps are bijections and the horizontal maps are embeddings, so the diagram must commute. Now the evacuation map is applied to the diagram 8. To obtain the desired commutation of the front face, it is enough to show that the back face commutes.

But in this case by [23, Prop. Single rows Let rows R be obtained by slicing all of the rectangles of R into single rows. Theorem 9. By [24, Prop. On the other hand by Theorem 8. By replacing R by rows R it may be assumed that R consists of single rows. The transpose maps are applied to reduce to the case of single columns. Recall the definition of the number n R [13, 2. From these facts, by replacing R with Rt it may be assumed that R consists of single columns. Applying the first reduction again, it may be assumed that R consists of single cells.

It follows immediately from the definition of c Vol. The selected strings have label zero before and after being shortened. By Example 2. Corollary 9. Let T 0 be obtained from T by applying the automorphism of conjugation see Remark 8. By Lemma 8. By [21, Lemma 6. Appendix A. Proof of Lemma 3. After the lemmas e in Lemma 3. Hence e The rest of this section is devoted to the proofs of Lemmas A.

Proof of Lemma A. The proof proceeds by induction on k. By Remark A. This proves in particular A. Hence it may be assumed that the hypotheses of A. This contradicts the assumptions. Again 3. This completes the proof of A. Doubly singular case. Then by A.

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Applying 3. But this is a contradiction. Using A. Let p 0. Then by 3. So or has label zero. This concludes the proof of Case 1. But by A. Suppose it is Case 3. Since is Case 1. Selected strings, generic case. Selected strings, Case 1. Suppose neither holds.

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First A. Also by A. Unselected strings. Again we follow the division into cases of the proof of Lemma A. It follows from the k proof of Lemma A. It follows from A. Proof of Lemma 7. Remark B. Then by [13, j 9. Lemma B. By [13, It suffices to show that the sum over a is zero. Suppose not. Let sm be the cell just east Vol. We shall show that one of three possibilities must occur, which are mutually disjoint by Remarks B.

In particular it follows immediately from 3. Similarly B. By Remark B. By j Lemma B. This finishes the proof of the trichotomy. Follows directly from B. Denote the corresponding partitions by Jn , Jn k n t n t k , Jn and J t n , respectively. Hence, compared to Jk , Jk has one part of less. It follows from Lemma B.