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

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

**Example text**

And (k) = (k) . Therefore r = r and r ≤ r. It follows that Case 3 holds for (ν, J)(r) . 116 A. N. Kirillov, A. Schilling and M. Shimozono (r) Sel. , New ser. (r) But then = = ∞ so that r ≤ r < r < r, which is a contradiction. Therefore r = r and r = r as desired. If r > r a similar proof shows r = r and r = r. So it may be assumed (r) (r) (r) = ∞ implies that = = ∞ that r = r = r. Notice that (r) = so that r ≤ r and r ≤ r. Suppose Case 3 does not hold for (ν, J)(r−1) . Then (r−1) (r−1) (r−1) ≤ (r−1) < ∞ and ≤ < ∞ so that r ≥ r and r ≥ r.

It is possible to verify cR (S) = cc(φR (S)) for S ∈ ST(λ) directly. 6]. Let asc(S) denote the number of ascents in S, that is, the number of indices i such that i + 1 is in a later column in S than i. It follows immediately from the definition of c Vol. 8 (2002) Bijection between LR tableaux and rigged configurations 113 that c(S − ) = c(S) − asc(S). By induction it suffices to show that if φR (S) = (ν, J), (1) (1) then asc(S) is equal to α1 and cc(ν, J) − cc(δ(ν, J)) = α1 . (1) (k) It is first shown that asc(S) = α1 .

K−1) ≥ and (k−1) , by induction the generic case holds for ν (k−1) and (k) (k) ≤ − 1). induction, the generic case holds for ν (k−1) (k) (k−1) = = , it follows that m (k) P −1 (ν) (k) ≤ − 1 and = χ( Suppose that can only ) = 0. 16), and P −1 (ν) = 0, P < = 0 as before. The goal is to show that It suffices to show that (k−1) −1 (k) (k−1) ≤ (k−1) (k) P −1 (ν) − 1. Since = (k) (k−1) (k) ≥ ≥ . Therefore = 0 and = = , which contradicts > − 1. To finish Case 1 it suffices to show that m −1 (ν (k+1) ) = 0.