可信度还没有人判定查看链条图 →

The hyperbolic-basis assertion of Lemma 6 ([A rational hyperbolicity test]) identifies every nondegenerate rank-six hyperbolic Hermitian space over K , in the second-linear convention, with three planes of Gram matrix ( 0, 1; 1, 0 ) . That elementary assertion was proved by splitting isotropic pairs, independently of the Weil-plane theorem. Since both h_A and the actual seed form h_0 are hyperbolic, it gives a K -isometry f_A:(H_1(A,),h_A)⟶(V_0,h_0). Let Λ_A=H_1(A,) and let I_A be the complex structure of A . Transport it to I=f_AI_Af_A^ -1 on V_ 0, . The isometry is K -linear, so I commutes with T_0 . The Riemann form is recovered without any dual convention: it is exactly the coefficient of u when h(x,y)=e(x,uy)+u e(x,y) is expanded in the -basis 1,u of K . Thus implies e_0(f_Ax,f_Ay)=e_A(x,y). It follows that I preserves e_0 and satisfies e_0(x,Ix)>0 for x≠0 . Hence I is a point of the full connected domain D . Equation algebraizes the Weil plane of A_I^ pp at this particular period. The full rational lattices f_A(Λ_A) and Λ_ pp in V_0 are commensurable. Choose an integer n_A>0 such that n_A f_A(Λ_A)⊂Λ_ pp . The invertible complex-linear map n_A f_A then induces a surjective holomorphic homomorphism with finite kernel φ:A=(H_1(A,),I_A)/Λ_A ⟶ A_I^ pp =(V_ 0, ,I)/Λ_ pp . Both tori are projective, so this homomorphism is algebraic by Serre's Proposition~15 cited above. It is a K -linear isogeny in the category up to isogeny. Its pullback on first rational cohomology is therefore a K -linear isomorphism. It maps each of the two embedding eigenspaces isomorphically, and taking their sixth exterior powers gives φ^* W_K(A_I^ pp )= W_K(A). Pullback along the algebraic isogeny preserves rational codimension-three cycle classes. The left side is algebraic by, proving the proposition. The isogeny need not identify the integral polarizations: its pullback scales the rational Riemann form by n_A^2 . Nor was a lift of f_A to a rational spin group used. Only the elementary hyperbolic-basis argument and a lattice denominator were required, so extra endomorphisms or a product decomposition of A impose no restriction.

评论0

登录后可以评论,指出这条推演中的逻辑错误。

还没有评论。发现逻辑漏洞?第一个指正吧。