# Combinatorial mathematics IX : proceedings of the Ninth by Elizabeth J Billington; Sheila Oates-Williams; Anne Penfold

By Elizabeth J Billington; Sheila Oates-Williams; Anne Penfold Street

Read or Download Combinatorial mathematics IX : proceedings of the Ninth Australian Conference on Combinatorial Mathematics, held at the University of Queensland, Brisbane, Australia, August 24-28, 1981 PDF

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Additional info for Combinatorial mathematics IX : proceedings of the Ninth Australian Conference on Combinatorial Mathematics, held at the University of Queensland, Brisbane, Australia, August 24-28, 1981

Example text

Wp ) est la d´ecomposition descendante de w ; comme γwj est un r´earrangement de wj , on voit que ∆(w) est un r´earrangement. Par ailleurs, ´etant donn´es x et y dans X, nous noterons ξx,y (w) le nombre d’entiers i tels que 1 ≤ i ≤ m − 1, xi = x et xi+1 = y. Enfin, si c est une application de X × X dans un mono¨ıde commutatif Ω, nous poserons ξc (w) = 0 si m est 30 ´ CHAPITRE IV : REARRANGEMENTS DE SUITES ´egal a ` 0 ou 1 et (13) ξc (w) = c(x1 , x2 ) + c(x2 , x3 ) + · · · + c(xm−1 , xm ) si m ≥ 2.

Kn ) = u(xi , xi+1 ). i=1 On a alors (10) kn (xσ(1) , . . , xσ(n) ) = Per(u(xi , xj ))1≤i,j≤n . σ∈Sn (10 ) La d´ efinition du permanent conserve un sens pour des matrices a ` coefficients dans un anneau commutatif quelconque. 4. ´ 3. FONCTIONS CARACTERISTIQUES 49 En raisonnant comme dans le lemme pr´ec´edent, on voit qu’il suffit de prouver l’identit´e (10) dans le cas d’une suite croissante (x1 , . . , xn ). Posons alors aij = u(xi , xj ) pour 1 ≤ i, j ≤ n d’o` u aij = 1 lorsque i ≥ j. On est donc ramen´e a ` prouver la relation n−1 (11) n−1 aσ(i),σ(i+1) = σ∈Sn i=1 ai,σ(i) σ∈Sn i=1 (noter que l’on a n ≥ σ(n) d’o` u an,σ(n) = 1).

D´eterminent de mani`ere unique la suite (an )n≥1 par r´ecurrence. Autrement dit, la relation de Spitzer ´equivaut au syst`eme des relations (3). Notre but est maintenant de calculer explicitement le coefficient an en fonction de b1 , . . , bn (pour n ≥ 1 donn´e). On a B = B1 + B2 avec B1 = b 1 t + ∞ b2 2 bn t + · · · + tn , 2 n B2 = k=n+1 bk k t . k Comme la s´erie B2 est divisible par tn+1 , il en est de mˆeme de la s´erie (exp B2 ) − 1 = ∞ (B2 )m /m! et a fortiori de m=1 (exp B1 )((exp B2 ) − 1) = (exp B1 )(exp B2 ) − exp B1 = exp B − exp B1 .