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

202We now have to prove that this map induces a map ζ :e ∈ E. We have to prove that(∑ ∑ [ζl i,j ⊗ h i,j] ) (∑ ∑Q b · e ⊗ g i = ζijijL⊗H(L⊗H)L +⊗ E H → L. Let[l i,j ⊗ h i,j] b Q⊗ e · g i ).Since e ∈ E, there exist x k , y k ∈ H such that e = [∑ k xk ⊗ y k] . Hence we have toEprove that(∑ ∑ [ζl i,j ⊗ h i,j] ] )bi jQ ·[∑k xk ⊗ y k ⊗ g iE(∑ ∑ [= ζl i,j ⊗ h i,j] ] )Q b ⊗[∑k xk ⊗ y k · g iand by using definition of E µb Qand µ E Qi.e.∑i∑j∑i(∑ ∑ ∑ζi j(∑ ∑ ∑= ζijk li,j S ( h i,j) x k S ( y k) g i = ∑ ijkkwe have to prove that[l i,j S ( h i,j) x k ⊗ y k] )Q b ⊗ g i[l i,j ⊗ h i,j] b Q⊗ x k S ( y k) g i )∑j∑which is true, so that we can conclude that the map ̂ζ :Ek li,j S ( h i,j) x k S ( y k) g iL⊗H(L⊗H)L +⊗ E H → L iswell-defined. Now, we want to prove that ̂ζ is bijective. The inverse of ̂ζ is given byΞ : L →L ⊗ H(L ⊗ H) L ⊗ + E Hl ↦→ [1 H ⊗ 1 H ]b Q⊗ E l.Now we compute(Ξ ◦ ̂ζ) (∑ ∑ [l i,j ⊗ h i,j] ) (∑bi jQ⊗ E g i = Ξi∑j li,j S ( h i,j) )g i∑ ∑= [1 H ⊗ 1 H ]b Q⊗ Ei j li,j S ( h i,j) g i= ∑ ∑[1 [H ⊗ 1 H ]bi j Q⊗ E l i,j ⊗ h i,j] · E gi = ∑ ∑i= ∑ ∑ [l i,j ⊗ h i,j] bi jQ⊗ E g iand ) ()(̂ζ ◦ Ξ (l) = ζ [1 H ⊗ 1 H ]b Q⊗ E l = 1 H S (1 H ) l = l.j [1 H ⊗ 1 H ]b Q[l i,j ⊗ h i,j] E ⊗ E g iLet us show that ̂ζ is an isomorphism of left L-modules. Let a ∈ L and let usconsider̂ζ(a · ∑ ∑ [l i,j ⊗ h i,j] )bi jQ⊗ E g i = ̂ζ(∑ ∑ [( ) a · li,j⊗ h i,j] )bi jQ⊗ E g i= ∑ ∑i j ali,j S ( h i,j) (∑g i = a ·i∑j li,j S ( h i,j) )g i= a · ̂ζ(∑ ∑ [l i,j ⊗ h i,j] )Q b ⊗ E g i .ij