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Example: Consider the partition π := {(1, 2, 7), (3), (4, 6), (5), (8)} ∈ N C(8). For the complement K(π) we get K(π) = {(1), (2, 3, 6), (4, 5), (7, 8)} , as can be seen from the graphical representation: 1 ¯1 2 ¯2 3 ¯3 4 ¯4 5 ¯5 6 ¯6 7 ¯7 8 ¯8 . ], ... 5. 1) Denote by Sn the group of all permutations of (1, 2, . . , n). Let α be a permutation in Sn , and let π = {V1 , . . , Vr } be a partition of (1, 2, . . , n). Then α(V1 ), . . , α(Vr ) form a new partition of (1, 2, . . , n), which will be denoted by α · π.

Rk (r ) (r ) (r ) α1,p1 . . ,rk q p (r ) α1,p1 | = (r ) α ¯ 1,q1 . . α ¯ k,qk (r ) . . αk,pk |2 p ≥ 0. 5. The last part of the proof consisted in showing that the entry-wise product (so-called Schur product) of positive matrices is positive, too. This corresponds to the fact that the tensor product of states is again a state. 6. We call the probability space (A, ϕ) constructed in Theorem ... the free product of the probability spaces (Ai , ϕi ) and denote this by (A, ϕ) = (A1 , ϕ1 ) ∗ . .

Bs ] kτ2 [C1 , . . , Ct ] . We will apply now the induction hypothesis on kτ1 [B1 , . . , Bs ] and on kτ2 [C1 , . . , Ct ]. According to the definition of Aj , both Bk (k = 1, . . , s) and Cl (l = 1, . . , t) are products with factors from (a1 , . . , an ). Put (b1 , . . , bp ) the tuple containing all factors of (B1 , . . , Bs ) and (c1 , . . , cq ) the tuple consisting of all factors of (C1 , . . , Ct ); this means (b1 , . . , bp ) ∪ (c1 , . . , cq ) = (a1 , . . , an ) (and p + q = n).

### Combinatorics of free probability theory [Lecture notes] by Roland Speicher

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