Page 29 - Textos de Matemática Vol. 39

(4.4)
666 ˜5 5 15
Θ : 3 −→ 3 −→ 4 . 1 42 1 42 32
4Θ4 ˜34 34 3 3 3 −→ 3 3 Θ: 1 3 −→ 1 3 −→2 4 . 1212 2213
JEU DE TAQUIN, INVARIANT FACTORS 19
(4.3)
555555
Θ˜:3 −→ 3−→1 4 Θ˜:3 4−→3 4−→3 4. 1 24 1 24 32 1 2 12 21
(4.5)
e r a t i o n θ˜ a s a t r a n s l a t i o n o f t h e o p e r a t i o n Θ˜ . R e c a l l t h e o p e r a t i o n θ i , d e s c r i b e d
Using RSK∗-correspondence, as explained in Section 3, we define the op- at the end of Section 2. Then θ˜ denotes a variant of θ based on a non standard
ii
pairing of parentheses procedure. Considering ω restricted to the two-letter al- phabet {i, i + 1}, written ω|{i,i+1}, we remove consecutively subwords i + 1 i and as a result we obtain a subword of the form:
(1) ir−l(i+1i)l, with 0 ≤ l ≤ r, [(i+1i)li+1s−l, 0 ≤ l ≤ s] which is replaced with i + 1r+l [is+l]. In this case, P(θ˜ ω ) ≤ P(θ ω ) with
i |{i,i+1} i |{i,i+1} respect to the cyclage order, and, in particular, P (θ˜ ω ) = P (θ ω ),
if l = 0. i |{i,i+1} i |{i,i+1}
(2) ir−l(i + 1 i)li + 1s, 0 ≤ l ≤ r − s, [ir(i + 1 i)li + 1s−l, 0 ≤ l ≤ s − r,]
which is replaced by isi + 1r+l [is+li + 1r]. In this case, we may have either
P ( θ˜ ω ) ≤ P ( θ ω ) o r P ( θ˜ ω ) ≥ P ( θ ω ) b u t i f l = 0 , i |{i,i+1} ˜ i |{i,i+1} i |{i,i+1} i |{i,i+1}
we have always P (θiω|{i,i+1}) ≥ P (θiω|{i,i+1}) with respect to the cyclage order. For instance, from (4.1), θ1(21112) = 21122 is based on the standard pairing procedure; from (4.2), θ˜1(21112) = 21212, based on the nonstandard p a i r i n g ( 2 1 1 1 ) 2 w h i c h l e a v e s 1 2 2 f r e e a n d i s r e p l a c e d w i t h 1 2 2 , a n d θ˜ 1 ( 2 1 1 1 2 ) = 22112 is based on the nonstandard pairing (211)12 which leaves 122 free and is replaced with 212; from (4.3) θ˜1(21212) = 21112 is based on the pairing (2121)2, or (21)212, which leaves 10(21)121 free and is replaced with 11+120+1, and θ˜1(21212) = 11212 is based on the pairing 21(21)2 which leaves 10(21)121 free and is replaced with 11+120+1; from (4.4) we have θ˜1(212122) = 211112 is
based on the pairing (21)2122 which leaves 10(21)122 free and is replaced with 12+120+1.