By Kirillov A.N., Schilling A., Shimozono M.

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Since m −1 (ν ) = 0. But then − 1 = (k−1) (k) ) = 1 and ≥ (k−1) (k−1) > (k−1) = > − 2, so = − 1. However there are no strings of length − 1 in ν (k−1) , which is a contradiction. (k) Suppose m −1 (ν (k−1) ) = 0. 15) it is enough to show that (k−1) (k−1) ≤ = (k−1) ≤ − 2 and − 2. Suppose (k) (k−1) = −1. But m contradicting (k−1) (k−1) −1 (ν = − 1. > ≤ − 2. In this subcase − 2. Now (k−1) ) = 0 and −1 = (k−1) (k) ≥ (k−1) ≤ (k−1) ≤ −2, so m > −1 (ν − 2 and − 2 so (k−1) ) = 0, Vol. 8 (2002) Bijection between LR tableaux and rigged configurations 125 Selected strings, Case 2.

16) P (k) −1 (ν) =P (k) −1 (ν) (k−1) − χ( ≤ − 1) − χ( (k−1) ≤ − 1). 9) m (ν (k−1) ) ≤ 2. We divide into cases for the choices of m (ν (k−1) ) ∈ {0, 1, 2}. (k) Suppose m (ν (k−1) ) = 2. 9) P −1 (ν) = 0. It follows immediately from (k) −1 (ν) = 0. 18) that P Suppose m (ν (k−1) ) = 1. 18) it is enough to − 1. Suppose neither holds. Then = . 1) (k−1) = . Now m (ν (k−1) ) = 1 and = = (k−1) (k) ≥ (k−1) > −1 so m (ν (k−1) ) = 0, = . contradicting (k) Suppose m (ν (k−1) ) = 0. 1), it follows that (k−1) ≤ − 1 and ≤ − 1.

Write ( , 0) ∈ (ν, J)(k) is also singular. (k) = (k) = . The string Selected strings, Case 1. Here only the string ( , 0) ∈ (ν, J)(k) and its images (k) (k) (k) (k) = = −1. under δ and δ are selected. Moreover P −1 (ν) = P −1 (ν) = 0 and The string ( , 0) is sent to a string of length − 2 and singular label under δ ◦ δ and (k) zero label under δ ◦ δ. Hence it must be shown that P −2 (ν) = 0. 16), P (k) −2 (ν) =P (k) −2 (ν) − χ( (k−1) ≤ − 2) − χ( (k−1) ≤ − 2). 2. Suppose first that (k) P −1 (ν) = 0.

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A bijection between Littlewood-Richardson tableaux and rigged configurations by Kirillov A.N., Schilling A., Shimozono M.


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