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

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

**Example text**

First suppose that (k) Then ≥ (k) (1) = 1. for all 0 ≤ k < c. 16) with k = c − 1. (k) Conversely, if (1) > 1, then < (k) for all 0 ≤ k < c so that c ≤ c− . This (1) proves asc(S) = α1 . (1) Now it is shown that cc(ν, J) − cc(δ(ν, J)) = α1 when R consists of single cells. 1) )). k≥1 Next let us determine the difference of the sums of quantum numbers. 16). The selected strings have label zero before and after being shortened. Then (k) |Jn(k) | − k,n≥1 |Jn | k,n≥1 (mn (ν (k) ) − δn, = (k) )(χ( (k−1) ≤n< (k) ) − χ( (k) ≤n< (k+1) k,n≥1 (α = k≥1 (k) (k−1) −α (k) (k) )− (α k≥1 (k) (k) −α (k) (k+1) χ( )+ k≥1 (k) < (k+1) ).

The case (k−1) − 1 = occur for Case 1 at k − 1. By induction this implies that m For (k) (k−1) (k) −1 = = − 1 and one needs m (k) P −1 (ν) (k) (k−1) (k) = −1. Since (k−1) (k) −1 (ν) (k) −1 (ν) + (k−1) =P So = (k−1) ≥ ≥ . 5), so (k) Suppose P −1 (ν) (k−1) (k−1) = . Hence (k−1) (k−1) = 0. For then −1 (ν (k) (k−1) (k−1) χ( (k−1) = ≤ (k) , and (k−1) ≥ 1. 5), ) ≥ 1. < , and by = − 1. 5) (k−1) (k−1) − 1, so = (k−1) and (k−1) = (k−1) = . By (k−1) . ≤ − 1. This leads to a contradiction ≤ − 1. (k−1) ≥ and (k−1) , by induction the generic case holds for ν (k−1) and (k) (k) ≤ − 1).

7. 6) ❄ ✲ CLR(λ; Rev ). 6. Using the well-known fact Sketch of Proof. βR = γR std ◦Ev = ev ◦ std it is enough to show that the following diagram commutes: RLR(λ; R) ev ✲ γR ❄ CLR(λ; R) RLR(λ; Rev ) γRev ev ❄ ✲ CLR(λ; Rev ). Vol. 8 (2002) Bijection between LR tableaux and rigged configurations 111 By [18] S ev = P (#word(S)) for S ∈ ST(λ) where #w is the reverse of the complement of the word w in the alphabet [1, |λ|]. Let WR(R) (resp. WC(R)) be the set of standard words w such that the standard tableau P (w) is in RLR(λ; R) (resp.

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

by Jason

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