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# Shifts and characterization of the elements of A

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Shifts and characterization of the elements of A? Matthieu Deneufchatel 21/04/2010 We take the following notations : if A be an algebra and f ? A?, – fx : y 7? f(xy) (left shift) ; – xf : y 7? f(yx) (right shift) ; – xfy : z 7? f(yzx) ; – A? denotes the finite dual (also called Sweedler's dual). Theorem 0.1. Theorem by Abe (extended by Schutzenberger's condition (property (vi) be- low)) - Characterization of the elements of A? : Let A be an algebra and f ? A?. The following properties are equivalent : – (i) tµ(f) ? A? ?A? ; – (ii) The family (fx)x?A is of finite rank ; – (iii) The family (xf)x?A is of finite rank ; – (iv) The family (xfy)x,y?A is of finite rank ; – (v) f(xy) = n∑ i=1 fi(x)gi(y) ; – (vi) ? ? : A ? kn?n and (?, ?) ? k1?n ? kn?1 such that ?x ? A, f(x) = ??(x) ?; – (vii) Ker(f) contains an ideal of finite codimension (i.

• therefore tµ

• argument allows

• now

• argument also applies

• take ?

• finite rank

• among all

• ?fi ?

• therefore

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Shifts and characterization of the elements ofA
MatthieuDeneufchˆatel 21/04/2010
We take the following notations : ifAbe an algebra andf∈ A, fx:y7→f(xy) (left shift); f:y7→f(yx;) (right shift) x fy:z7→f(yzx) ; x Adenotes the ﬁnite dual (also called Sweedler’s dual). Theorem 0.1.u¨hcneztgrebsrebeyAxt(edeenySdbytcmobnrdeietoiTohn(proper(vi)be-low))-Characterization of the elements ofA: LetAbe an algebra andf∈ A. The following properties are equivalent : t∗ ∗ (i)µ(f)∈ A⊗ A; (ii)The family(fx)is of ﬁnite rank; x∈A (iii)The family(f);is of ﬁnite rank x x∈A f;is of ﬁnite rank (iv)The family(x y)y∈A x, n X (v)f(xy) =fi(x)gi(y); i=1 n×n1×n n×1 (vi)α:A →kand(λ, γ)k×ksuch that x∈ A, f(x) =λ α(x)γ; (vii) Ker(f)contains an ideal of ﬁnite codimension (i.e. there exists an idealIsuch that dim(Ker(f)/I)<).(1) Proof :We begin by two lemmas : n X Lemma 0.2.Lettbe an element ofVW. If we decomposetasfigiwithnminimal, then the i=1 fi’s and thegi’s form free families. n1 X Proof :Indeed, if that was not the case, one could write thatyn=αiyiand i=1 n n1 X X t=xnyn+xiyi= (αixn+xi)yi. i=1i=1 But this equation shows thatnis not minimal and it is impossible by hypothesis.Lemma 0.3.LetU, VandWbe vector spaces andφ:U×VWa bilinear map. Then ifzIm(φ) P n and ifz=φ(xi, yi)withnminimal among all decompositions of this form, the families(xi)1in i=1 and(yi)1inare free in their respective spaces. Note that the ﬁrst lemma is a consequence of the second one : takeφ=.
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