Prove
Lemma 15 ([All-sequence evaluation]). Fix K and a sequence of the finite…Proposition 12 ([Finite-precision ring-class characters]). Fix finitely many…Proposition
Use tame inertia and unramified Frobenius, or the Hochschild--Serre complex for these two procyclic directions. At r , inertia minus one is multiplication by t times a unit. It is invertible on all the indicated fields except the characteristic-zero fiber t=0 . There, r→1 , and the two remaining Frobenius differentials have nonzero determinant by, including either sign of the quadratic twist. The bound also ensures that the integral local torsion at this central fiber has bounded length. At an active odd conductor prime, inertia of the quadratic character acts by -1 . Its differential is thus -2 and is invertible in characteristic zero. In characteristic two the quadratic character disappears. At a split prime the Frobenius action is zB , where B_2( F_2) and z=(1+t)^a has odd a∈ Z_2 . Its coefficient of t is one, so z is nonconstant. No nonzero polynomial over the finite constant field vanishes at z . Thus both Frobenius complexes, with operators zB-1 and q^ -1 zB-1 , are invertible over Ω ; here q=1 in the residue field. This argument applies also to a reducible residual representation. The orthogonal full and strict conditions agree when full local cochains are acyclic. Equation then gives self-duality. The global degree-zero assertion will follow from the evaluation maps in Lemma 15 ([All-sequence evaluation]).