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By Kirillov A.N., Schilling A., Shimozono M.
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Additional resources for A bijection between Littlewood-Richardson tableaux and rigged configurations
2 the front face commutes. 2. 6 (Evacuation Theorem). The following diagram commutes: CLR(λ; R) ev ✲ CLR(λ; Rev ) φR ❄ RC(λt ; Rt ) φRev ❄ ✲ RC(λt ; Revt ). ev θ R Proof. If R is empty the result holds trivially. Suppose that the last rectangle ev ev of R has more than one column. Obviously R∧ = (Rev )∨ and R∨ = (Rev )∧ . Consider the diagram: CLR(λ; R) yyy yyyı∧ yyy yy9 CLR(λ; R∧ ) φR φR ∧ t RC(λt ; R∧ ) oU ooo o o oo ∧ ooo RC(λt ; Rt ) ev ev G CLR(λ; Rev ) nn ı∨nnnn n n n vnnn G CLR(λ; R∧ ev ) φR∧ev φRev G RC(λt ; R∧ evt ) h ∨ G RC(λt ; Revt ) ev ev θR ∧ θR Vol.
For all S ∈ RLR(λ; R) we have (γR (S))ev = P (#γR (S)) = P (#γR (word(S))) = P (γRev (#word(S))) = γRev (P (#word(S))) = γRev (S ev ). 6. Again the result follows from the special cases (E1) and (E2). Consider the case (E1). By the definition of the relabeling map βR R and the fact that in this case both θR and iR R only change the subtableaux corresponding to the first two rectangles, one may reduce to the case that L = 2. By [23, Prop. 33] all sets are either singletons or empty. The vertical maps are bijections and the horizontal maps are embeddings, so the diagram must commute.
8 (2002) Bijection between LR tableaux and rigged configurations 97 Proof. Suppose that the last rectangle of R is also a single column; this case subsumes the base case that R = (R1 ) is a single column. Clearly R = R. D CLR(λ; R) xxx xxx− xxx xxx 8 CLR(λ− ; R) φR φR D t RC(λ−t ; R ) V qqq q q qqq qqq δ RC(λt ; Rt ) G CLR(λ− ; R) oo o − oo o oo ow oo δ δ G G CLR(λ−− ; R) φ φR R t RC(λ−−t ; R ) gyyy yyy yyy yy δ G RC(λ−t ; Rt ) In the above diagram there is a map t φR : CLR(λ−− ; R) → RC(λ−−t ; R ).
A bijection between Littlewood-Richardson tableaux and rigged configurations by Kirillov A.N., Schilling A., Shimozono M.