“…Since V is irreducible, we have V = V (α) for some α ∈ C (if we have used the isomorphism σ in (2.8) in the above proof, then V is the module V (α)). Following the proof of Lemma 3.2 (now G m is not necessarily diagonal), we have U = U (1) ⊕U (2) , and both U (1) and U (2) are the natural gl N -modules. Since D(U ) ⊂ U , [D, gl N ] = 0 and V is not irreducible, the subspace U = {u ∈ U | Du ∈ Cu} of eigenvectors of D is a proper (and thus simple) gl N -submodule of U (isomorphic to C N as a gl N -module) and D| U is a scalar map λ for some λ ∈ C. Thus U = U ⊕U , where U is another copy of U such that Du = λu +u for u ∈ U , where u ∈ U is the corresponding copy of u .…”