By Stuart Martin, A Makelov
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Extra info for Representation theory
We construct this involution such that the number of i’s and (i + 1)’s get exchanged when passing from T to T (and all the other multiplicities stay the same). Given T , each column contains either: an i, i + 1 pair; exactly one of i or i + 1; or neither. Call the pairs fixed, and all other occurrences of i, i + 1 or neither free. e. if the row consists of k free i’s followed by l free i + 1’s, replace them by l free i’s followed by k free i + 1’s. Example Take T to be 1 1 1 1 2 2 2 2 2 3 2 2 3 3 3 3 3 and let i = 2.
Km π (i) = 1 where (k1 , l1 ), . . , (km , lm ) are the coordinates of the NE corners of the l1 . . l m (i) shadow diagram of π (i−1) . The shadow lines for π (i) are denoted by Lj . 15 (Viennot). Let π −−−→ (P, Q). Then π (i) is a ‘permutation’ such that RS π (i) −−−→ (P (i) , Q(i) ), where P (i) (respectively Q(i) ) comprises rows i and below of P (respectively Q). 16 (Schutzenberger). Given π ∈ Sn , P (π −1 ) = Q(π), Q(π −1 ) = P (π). Proof. Taking the inverse of a permutation corresponds to reflecting the shadow diagram in the line y = x.
So when we switch them, we get 1 1 1 1 2 2 2 3 3 3 2 2 2 3 3 3 3 It’s easy to see T is semistandard by the definition of ‘free’. Since fixed i’s and i + 1’s come in pairs, the map we get has the desired exchange property, and moreover it’s easy to see it’s an involution. Knuth’s generalization of R-S. Instead of starting with a permutation π ∈ Sn , begin with some r × s matrix A = (aij ) of non-negative integers only finitely many = 0 entries. Stanley calls these N-matrices of finite support. 7.