Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

)Let(Ãà ∈ M. We have to prove that Φ = a = (C U C F Aε C) ( )◦C Uγ C à C F ∈ D.We have(Cm A ) ◦ (ΦA) ◦ (AΦ) =(Cm A ) ◦ (C U C F Aε C A ) ()◦C Uγ C à C F A ◦ ( A C U C F Aε C) ()◦ A C Uγ C à C FAlift= (C ) (U C F m A ◦ CU C F Aε C A ) () (C ) ()◦C Uγ C à C F A ◦ UÃC F Aε C ◦C UÃγC à C F[ (C ) (= C U F m A ◦ CF Aε C A ) ( ) (ÃC ))]◦ γ C à C F A ◦ F Aε C ◦(Ãγ C à C F[γ= C (C ) ( C U F m A ◦ CF Aε C A ) (C ) () ( )]◦ F C UÃC F Aε C ◦C F C UÃγC à C F ◦ γ C ÃÃC F[Alift (C ) (= C U F m A ◦ CF Aε C A ) ◦ (C F A C U C F Aε C) () ( )]◦ C F A C Uγ C à C F ◦ γ C ÃÃC F[ε= C (C ) ( C U F m A ◦ CF AAε C) ◦ (C F Aε C AC ) () ( )]◦ C F A C Uγ C à C F ◦ γ C ÃÃC F[Alift (C ) (= C U F m A ◦ CF AAε C) () () ( )]◦ C F Aε C C UÃC F ◦C F A C Uγ C à C F ◦ γ C ÃÃC Fso that we get(ε C ,γ C )adj,m A[ (C=C U F Aε C) ◦ (C F m A C ) ( )]◦ γ C ÃÃC F[Alift (C= C U F Aε C) ◦ (C F C Um A e C F ) ( )]◦ γ C ÃÃC F[γ= C (C C U F Aε C) ( )◦ γ C à C F ◦ ( m A e C F )]= (C U C F Aε C) ( )◦C Uγ C à C F ◦ (C Um A e C F ) Alift= Φ ◦ (m A C)(Cm A ) ◦ (ΦA) ◦ (AΦ) = Φ ◦ (m A C) .Moreover we haveΦ ◦ (u A C) = (C U C F Aε C) ( )◦C Uγ C à C F ◦ (u A C)Alift= (C U C F Aε C) ( )◦C Uγ C à C F ◦ (C Uu A e C F )[ (C= C U F Aε C) ( )◦ γ C à C F ◦ ( u A e C F )]γ C = C U [(C F Aε C) ◦ (C F C Uu e A C F ) ◦ ( γ C C F )]Alift= C U [(C F Aε C) ◦ (C F u A C ) ◦ ( γ C C F )]u A= C U [(C F u A)◦( CF ε C) ◦ ( γ C C F )]87so that we get(ε C ,γ C )adj=C U C F u A = Cu AΦ ◦ (u A C) = Cu A .Therefore Φ is a mixed distributive law.Conversely let Φ ∈ D. Then we know that à = b (Φ) (with notations of Proposition