New PDF release: A bijection between Littlewood-Richardson tableaux and

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

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Since − 1 in ν (k) and (k) = −1. By definition, = −1, which contradicts the assumption > . (k) (k) (k) Therefore P −1 (ν) ≥ 1. 11) it follows that P −1 (ν) = 1, P +1 (ν) = 0 and m (ν (k+1) ) = 0. (k) (k) Finally consider m (ν (k−1) ) = 2. 11), P −1 (ν) = 0, P +1 (ν) = 0, and (k+1) m (ν ) = 0. 10) are proved. 11, that if (k) p > and mn (ν (k) ) = 0 for all < n < p, then Pn (ν) = 0 for ≤ n ≤ p. 10) furthermore implies that mn (ν (k+1) ) = 0 for < n < p. Suppose ν (k) has a string longer than . Let p be minimal such that p > and mp (ν (k) ) ≥ 1.

Appendix A. 3. 2. Vol. 8 (2002) Bijection between LR tableaux and rigged configurations 115 In this section the following notation is used. Let R = (R1 , . . , RL ) be a sequence of rectangles with Ri = (ηiµi ), such that R1 and RL are single columns and |R| ≥ 2 and let λ be a partition and (ν, J) ∈ RC(λt ; Rt ). Write δ(ν, J) = (ν, J) δ(ν, J) = (ν, J) δ ◦ δ(ν, J) = (ν, J) δ ◦ δ(ν, J) = (ν, J). (k) (k) (k) , (k) , and denote the lengths of the strings that are Furthermore, let shortened in the transformations (ν, J) → (ν, J), (ν, J) → (ν, J), (ν, J) → (ν, J) and (ν, J) → (ν, J), respectively.

So for Then in ν Vol. 8 (2002) Bijection between LR tableaux and rigged configurations ν (k−1) we are in Case 1 or Case 3. Suppose it is Case 3. Then (k−1) = (k−1) = is Case 1. Then (k) = ,m (k−1) −1 (ν and (k−1) (k) (k) ≥ (k−1) = (k−1) > −1 = 123 (k−1) = (k−1) > which is a contradiction. Suppose it − 1. 14), P (k) −1 (ν) ) ≥ 1. So there is a singular string of length (k) = 0. Since − 1 in ν (k) and (k) = −1. By definition, = −1, which contradicts the assumption > . (k) (k) (k) Therefore P −1 (ν) ≥ 1.

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