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

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Extra info for A bijection between Littlewood-Richardson tableaux and rigged configurations

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If δ and δ select the same string in ν (k) (that is, both select a string with the same length and same label), say that the doubly singular case holds for ν (k) . Otherwise say that the generic case holds for ν (k) . The next lemma shows that ν = ν. 2. If the generic case holds for ν (k) , then = (k) Otherwise suppose the doubly singular case holds for ν . Let the common string length. Then 1. If 2. If 3. If (k) (k) (k) < (or = (or > (k) (k) (or Moreover, if (k) (k) < ), then = ), then (k) (k) > ), then = ∞, then λtk = = (k) = = (k) (k) (k) = − 1, and m (k) and (k) = .

Which implies that (k−1) ≤ −1 ≤ − 1. 2. 2, we distinguish the three cases given by m −1 (ν (k−1) ) ∈ {0, 1, 2}. (k) (k) Suppose m −1 (ν (k−1) ) = 2. 15) P −2 (ν) = 0 and P −2 (ν) = 0. Suppose m −1 (ν (k−1) that either (k−1) − 2 and (k−1) ) = 1. 7) P (k−1) ≤ − 2 or > − 2. Thus − 1, it follows that m (k−1) −1 (ν ≤ 1. 15) it suffices to show ≤ − 2. Suppose neither holds. Then (k−1) (k−1) (k) −2 (ν) = − 1. Since m −1 (ν ) = 0. But then − 1 = (k−1) (k) ) = 1 and ≥ (k−1) (k−1) > (k−1) = > − 2, so = − 1.

Recall that in Case 3 one has Otherwise (r−1) (r−1) = (r−1) λtr−1 = λtr . Suppose that (r−2) = (r−1) (r−1) . If < ∞, then r = r = r as desired. = ∞ so that r ≤ r − 1, r ≤ r − 1, and furthermore (r−2) = ∞. 2 that either (r−2) (r−1) = ∞ or = ∞, contradicting the definition of r. Similarly =∞ leads to a contradiction. Hence r = r = r − 1 as desired. 3. 2. The proof proceeds by induction on k. There is nothing (k) and (k) is finite. If one is finite and the to prove unless at least one of (k) other infinite, then obviously δ and δ choose different strings and (k) (k) = (k) and (k) = .

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