观点

Lemma 4. Suppose P=P^*=P^2∈_m([G]) , an odd D>0 clears its denominators, and P=uv factors through r≥1 columns. Let q=2^k>1 satisfy q≡1 D . Choose arbitrary integral lifts U,V of the factors supplied by Lemma 3. Then A= qI_r, V; U, q^ -1 …

Lemma 4. Suppose P=P*=P2∈m([G]) , an odd D>0 clears its denominators, and P=uv factors through r≥1 columns. Let q=2k>1 satisfy q≡1 D . Choose arbitrary integral lifts U,V of the factors supplied by Lemma 3. Then A= qIr, V; U, q-1 (UV-P-q(Im-P)) ∈r+m ([G]) has an inverse inr+m ([G]) .

证明 1证否 0作为观点被引用 3链条图发起推演

发起推演1

证明
Lemma 3. Let G be a group, m,r≥1 , D an odd positive integer, and P∈_m([G])…Proposition 2. Let X,Y∈_d(_2[H_2]) , d≥1 , satisfy XY=I_d≠YX . In the group…观点
The matrix P+q(I_m-P)=qI_m+(1-q)P is integral because D(q-1) . Its reduction modulo q is e=P_q , which is also the reduction of UV . Thus every coefficient of UV-P-q(I_m-P) is an integer divisible by q ; the bottom-right block of Equation is integral. For J and B defined above, the complementary projection identities give J^ -1 =-qP-(I_m-P) . For x∈[0,1] put L_x= I_r, 0; xU/q, I_m , R_x= I_r, xV/q; 0, I_m . Their inverses negate the off-diagonal blocks, and multiplication without interchanging any factors gives M(x):=L_xBR_x= qI_r, xV; xU, x^2UV/q+J . In particular A=M(1) , with the rational group-ring inverse R_1^ -1 B^ -1 L_1^ -1 . For general real x , both M(x) and its inverse lie in the real group ring and depend continuously differentiably on x . No identities PU=U , VP=V , or V=U^* are required of the chosen integer lifts.
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