By Jürgen Müller

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**Extra resources for Algebraic combinatorics**

**Example text**

N ], where 1 ≤ r < s ≤ n such that µs > µs+1 and µr−1 > µr , if r > 1, and such that either s = r + 1, or s > r + 1 and µr = µs : If µ <· λ, then let r := min{i ∈ {1, . . , n}; µi = λi } and s := min{k ∈ {r + k k 1, . . , n}; i=1 µi = i=1 λi }, thus 1 ≤ r < s ≤ n. Hence we have µr < λr , and λr ≤ λr−1 = µr−1 if r > 1, as well as µs > λs ≥ λs+1 ≥ µs+1 . This yields µ ν := [µ1 , . . , µr−1 , µr + 1, µr+1 , . . , µs−1 , µs − 1, µs+1 , . . , µn ] λ, hence ν = λ. It remains to show µr = µs whenever s > r + 1: Assume to the contrary that µr > µs , and let r < t := min{i ∈ {r + 1, .

I}. Thus for M ∈ Nλ , where λ = [nan , . . , 1a1 ] n, we have n n n n n rk(M k ) = i=k+1 (i − k)ai = i=k+1 j=i aj = i=k+1 j=i (λi − λi+1 ) = n i=k+1 λi , for all k ∈ {0, . . , n}, implying n − rk(M k ) = ki=1 λi . Hence Nλ is uniquely determined by the rank sequence [n − ki=1 λi ∈ N0 ; k ∈ {0, . . , n}]; note that we have rk(M 0 ) = n and rk(M n ) = 0 anyway. Moreover, for µ n and N ∈ Nµ we have µ λ if and only if λ µ , which holds if and only if rk(M k ) ≥ rk(N k ), for all k ∈ {0, . . , n}.

S−1 , µs − 1, µs+1 , . . , µn ], where 1 ≤ r < s ≤ n such that µs > µs+1 and µr−1 > µr , if r > 1, and such that either s = r + 1, or s > r + 1 and µr = µs : If µ <· λ, then let r := min{i ∈ {1, . . , n}; µi = λi } and s := min{k ∈ {r + k k 1, . . , n}; i=1 µi = i=1 λi }, thus 1 ≤ r < s ≤ n. Hence we have µr < λr , and λr ≤ λr−1 = µr−1 if r > 1, as well as µs > λs ≥ λs+1 ≥ µs+1 . This yields µ ν := [µ1 , . . , µr−1 , µr + 1, µr+1 , . . , µs−1 , µs − 1, µs+1 , . . , µn ] λ, hence ν = λ.