观点

Lemma 27 ([A translation preserving two directions]). Let E be a field of characteristic zero, let n≥1 be an integer, let m=11 , and index the blocks of M_ mn (E) by ℤ/mℤ . Suppose Y(s_1,s_2)∈M_ mn (E) ⟨\!⟨s_1,s_2⟩\!⟩^u has constant tuple …

Lemma 27 ([A translation preserving two directions]). Let E be a field of characteristic zero, let n≥1 be an integer, let m=11 , and index the blocks of Mmn (E) by ℤ/mℤ . Suppose Y(s1,s2)∈Mmn (E) ⟨\!⟨s1,s2⟩\!⟩u has constant tuple C=diag(Ci) generating ∏iMn(E) . Suppose its two linear coefficients have the following supports: (Y1)ij =0 unless i-j∈1,2, (Y2)ij =0 unless i-j∈3,4. Assume every displayed edge coefficient is nonzero modulo the cross-inner image CiT-TCj:T∈Mn(E) . For a new central Laurent parameter γ , define Y'(s'1,s'2)=Y(γ+s'1,s'2). Then Y'0=Y(γ,0) generates Mmn (E((γ))) , and Y'1,Y'2 are independent modulo J(Y'0) .

证明 1证否 0作为观点被引用 0链条图发起推演

发起推演1

证明
The substitution is well-defined. In the coefficient of a fixed new word, a contribution from an old word of length r contains r minus the new word length factors of γ . Thus only finitely many terms contribute at each γ degree. Coefficients converge in E[[γ]] , and the substitution respects products. Let B=∏_iM_n(E) and let I_C be the kernel of evaluation of the free algebra at C . First-letter differentiation on I_C has image a B -bimodule in M_ mn (E) . Its projection onto every first-color edge is nonzero. Otherwise the derivative into that cross bimodule would descend to a derivation δ:B→M_n(E) , where the left and right actions are through the distinct endpoint factors i,j . Such a derivation is inner. In fact, for the central idempotent e_i of factor i , differentiating e_i a=a e_i gives δ(a)=a_iT-Ta_j, T=δ(e_i). Applied to the generators, this contradicts the transverse edge hypothesis. Since each full rectangular block is a simple bimodule for its two matrix factors, the derivative image contains the full block on every first-color edge. Divide the corresponding evaluations of elements of I_C at Y(γ,0) by γ . Their reductions supply all these edge blocks. Ordinary evaluations already supply the full diagonal product. The offset-one edges form a directed cycle through all vertices, so products supply every matrix block. A finite set of such elements with independent reductions therefore proves that Y(γ,0) generates the full matrix algebra. It remains to check the two directions. Put N=mn and J_0=J(C) . Since C generates B , its commutant consists of the m block scalars. Hence J_0=N^2-m . Choose N^2-m inner columns with independent reductions spanning J_0 . For a block scalar c=diag(c_i I_n) , the additional column γ^ -1 [c,Y(γ,0)] is integral and reduces to [c,Y_1] . These columns have rank m-1 modulo J_0 . Indeed, [c,Y_1]∈J_0 forces c_i=c_j on each offset-one edge, by its transverse hypothesis, and the cycle forces all c_i to agree. The two shifted linear coefficients reduce to Y_1,Y_2 . Suppose aY_1+bY_2∈J_0+ [c,Y_1]:c is block scalar . At a second-color edge, only bY_2 contributes outside J_0 , so b=0 . On the directed offset-one cycle, the relation then forces a=c_i-c_j on each edge. Summing gives ma=0 , whence a=0 in characteristic zero. Thus these two reductions are independent modulo the enlarged inner-column space. We have found N^2-m+(m-1)+2=N^2+1 independent reductions among inner columns and the two tangents. They lift to independent columns over E((γ)) . Full generation of Y'_0 implies J(Y'_0)=N^2-1 , so the two tangent classes are independent.