By Jürgen Müller

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**Additional info for Algebraic combinatorics**

**Sample text**

We look for a holomorphic map b, on a suitable open disc of positive radius centered at x = 0, fulﬁlling the functional equation b(x)2 − b(x) + x = 0. √ Solving the quadratic equation yields b(x) = 12 (1 − 1 − 4x), where the sign is chosen so that b(0) = 0. Hence b indeed is holomorphic for all x ∈ C such that |x| < 14 , and has the Taylor series expansion b(x) = n≥0 bn xn around x = 0. 1 1 n 2 n≥0 n (−4x) for all x ∈ C such that |x| < 4 . For n n n 1 (−1) · i=1 (2i−1) ·(2n)! −2 2n −1 n = = (−1) n 2n ·n!

Moreover, for the map δ1 : X → Q : x → δ(x, 1) we have (f ∧ δ1 )(y) = x,x ∈X,x∧x =y f (x)δ(x , 1) = f (y), for all y ∈ X, hence δ1 ∈ A(X) is a neutral element with respect to M¨obius multiplication. Since we have distributivity, and c(f ∧ g) = (cf ) ∧ g ∈ F (X), for all c ∈ Q, we conclude that F (X) is a commutative Q-algebra, called the M¨ obius algebra associated with the ﬁnite lattice X. More generally, for y ∈ X let δy : X → Q : x → δ(x, y) and µy : X → Q : x → µ(x, y). Then for z ∈ X we have y≤z µ(x, y) = x≤y≤z µ(x, y) = δ(x, z), for all x ∈ X, saying that y≤z µy = δz ; in particular we have y∈X µy = δ1 .

Thus M¨obius inversion yields f + ∗ µ = (f ∗ ζ) ∗ µ = f ∗ (ζµ) = f ∗ δ = f ∈ F (X), hence f (x) = z≤x f + (z)µ(z, x), for all x ∈ X. Similarly, if X has a one element, using the left A(X)-module structure on F (X), we let f+ := ζ ∗ f ∈ F (X), hence f+ (x) = z≥x f (z), for all x ∈ X; note that (f+ )• = f+ ∈ A(X). Thus M¨obius inversion yields µ∗f+ = µ∗(ζ ∗f ) = (µζ) ∗ f = δ ∗ f = f ∈ F (X), hence f (x) = z≥x µ(x, z)f+ (z), for all x ∈ X. 3) M¨ obius functions of products. Let X and X be tially ordered sets.