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! " F μn n = 3 # $ % ! &' ! ( )* $ " n n (F ) d 2 (F, μn ) + nd−1 Z/n n (F ) Z/n " 2 (F, μn ) , n (F ) ≤ n (F ) + 1 " $ " " n Z/n F p Qp " n (F ) = 3 &- ! )* . " n (F ) = 2 " n = F μ /& ! 0)*1 ! + F = Qp (T ) / = 2, 3 p = 1 (mod )1 n = (F ) + # $ 2 &3 ! 0* F 2 &0 ! 4* ! 2 (F, μ ) Z/ , " 5 2 " (F ) = 2 F + 2 " F
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$ " " Z/n 2 (F, μn ) Z/n " " n = # $ 2 (F, μ ) + /&3 ! 0*1 2 (F, μ ) Z/ 2 00 S + n S 9 " Z/n(r) = Z/n "
r q (S, r) = q (S, Z/n(r)) = , S = A A " " q (A, r) , T S " " κ(T ) " 5 7T , T → S
S|T : q (S, r) → q (T, r) " " βT = S|T (β)
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q(Fv ,r) , α ∈ q (Fv , r) ∂v (α) α v # ξ S F = κ(ξ) α ∈ q (S, r) v F " ∂v (α) = ∂v (αFv ) ∈ q−1 (κ(v), r − 1).
9 α v ∂v (α) = 0 α , α
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z " ∂D (α)(z) ∂D (α)(z) ∂D (α)(z) = ∂D (α)(z) κ(z) "
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αC ∈ q (7C,S , r) α = λ(αC ) /1 α Ci α(Ci )= |κ(C )(αC) /1 α X DS /1 D ∈ DS z = D ∩ C ∂D · λ = κ(z)|κ(D) · ∂z / 1 αC z D z α D α(D) = κ(z)|κ(D) (αC (z)) C,S
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D = Ci , ⎪ ⎨ θi ∂D (γ) = − κ(z)|κ(D) (∂z (θC )) · (π) D ∈ DS z = D ∩ C, vD (π) = 0 ⎪ ⎪ ⎩ 0 Dγ Ci θi D ∈ DS z : ∂z (θC ) = 0 % θ = λ(θC ) D ⊂ X z = D ∩ C !
∂D (γ) = vD (π)θ − ∂D (θ) · (π) + vD (π)∂D (θ) · (−1) FD .
! 1 (κ(D), Z/n) ≤ 1 (FD , Z/n) , D = Ci
C E C / /011 " vD (π) = 1 θC ∈ 1 (7C,S , Z/n) " ∂D (θ) = 0 ! 0)/1/1 ! ∂D (γ) = F |F (θ) = θi ∈ 1 (κ(Ci ), Z/n) , D ∈ DS vD (π) = 0 ∂D (γ) = −∂D (θ) · (π) = − κ(z)|κ(D) (∂z (θC )) · (π) ! 0)/1/1 , " ∂D (γ) = 0 D 5 S D 5 vD (π) = 0 D ∈ DS , D 5
S vD (π) = 0 E S ∂D (θ) = 0 ! 0)/1/1 ∂D (γ) = 0 , D 5 S vD (π) = 0 D E ∂z (θC ) = 0 Ci
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θC (z) = 0 ! ∂D (θ) = 0 θ(D) = 0 ! 0)/1/ 1 ! ∂D (γ) = 0 C D 5 S vD (π) = 0 D ∈ DS ∂D (γ) = −∂D (θ) · (π) = 0 ∂D (θ) = 0 ! 0)/1/1 D ∈ DS vD (π) = 0 ∂D (γ) = − κ(z)|κ(D) (∂z (θC )) · (π) " z = D ∩ C vD (π) = 0 π < C z D C z π κ(D) 5 (π) n 1 (κ(D), μn ) ! ∂D (γ) 5
∂z (θC ) = 0 9 Dγ Ci C " θi 5 " D ∈ DS z " θC D C Dγ z ∈ E ∩C
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"
π ¯ vz (¯ π ) = 1
π κ(D) (π) z ∂z (∂D (γ2 )) = ∂z (θC ) = 0 ∂z (∂D (γ2 )) = 0 z γ2 z ∈ Sγ ∩ Dα ∩ S , z Dα Ci ∂C (α) = 0 " Cj C z ∂C (γ2 )(z) = ∂C (α)(z)− θC (z) ∂C (γ2 )(z) = −θC (z) z γ2 /1/1 , z Dα Ci ∩ Cj θC (z) = 0 z α /1/1 z γ2 α / 1D
" ∂C (γ2 )(z) = ∂C (γ2 )(z) /1/1 z γ2 z ∈ Sγ ∩ Dα ∩ H F z ∈ Ci ∩ D D ⊂ H ! ∂D (γ1 ) = 0 H ∩ H = ∅ , z Dα ∂C (α) = 0 ∂C (γ2 )(z) = −θC (z) ∂D (γ2 ) = ∂D (α) ∂C (γ2 )(z) = ∂D (γ2 )(z) /1/1 z γ2 , z Dα θC (z) = 0 /1/1 z / 1 γ2 , z α ∂C (α)(z) − θC (z) = ∂D (α)(z) /1/1 z γ2 ! 9 γ2 X γ1 γ2 + Δ1 Δ2 F γ1 γ2 [Δ] = [Δ1 ⊗F Δ2 ] (Δ1 ⊗F Δ2 ) = (Δ) = 2 (Δi ) = " Δ1 ⊗F Δ2 Δ Δ1 ⊗F Δ2 2
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