Science Mathematics

P-Adic Monodromy and the Birch and Swinnerton-Dyer by Barry Mazur, Glenn Stevens

By Barry Mazur, Glenn Stevens

Contemporary years have witnessed major breakthroughs within the thought of $p$-adic Galois representations and $p$-adic sessions of algebraic forms. This publication includes papers provided at the Workshop on $p$-Adic Monodromy and the Birch and Swinnerton-Dyer Conjecture, held at Boston college in August 1991. The workshop aimed to deepen knowing of the interdependence among $p$-adic Hodge thought, analogues of the conjecture of Birch and Swinnerton-Dyer, $p$-adic uniformization thought, $p$-adic differential equations, and deformations of Galois representations. a lot of the workshop used to be dedicated to exploring how the targeted values of ($p$-adic and "classical") $L$-functions and their derivatives are proper to mathematics matters, as expected in "Birch-Swinnerton-Dyer-type conjectures", "Main Conjectures", and "Beilinson-type conjectures" à l. a. Greenberg and Coates

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Extra info for P-Adic Monodromy and the Birch and Swinnerton-Dyer Conjecture: A Workshop on P-Adic Monodromy and the Birch and Swinnerton-Dyer Conjecture August 12

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46) we have Δk+ +1 > Δk+ ≤ c4 Δk+ −1 ≤ c24 Δk+ −2 ≤ · · · ≤ c4 Δk . From the fact that 0 < c4 < 1 it now follows readily that limk→ ∞ Δk = 0. From Taylor’s Theorem we obtain aredk = f(x(k) ) − f(x(k) + sk ) = −gTk sk + O(lub2 (∇2 f(yk )) sk = −gTk sk + O( sk 2 2 ) ). Here, yk ∈ R is an appropriate point and we have used the assumption of the theorem that lub2 (f(x)) ≤ M for all x ∈ Rn to obtain the last equality. On the other hand n predk = Φk (0) − Φk (sk ) = −gTk sk − sTk Bk sk = −gTk sk + O(lub2 (Bk ) sk = File: × ÒØ¹Ñ Ø Ó ×ºØ Ü Revision: ½º¿¿ Date: ¾¼¼ »¼ »¿¼ ¼ −gTk sk ½¾ ÅÌ + O( sk 2 ).

8). (iv) limk→ ∞ x(k) = x∗ . Then, the following statements are equivalent: (a) lim x(k+1) − x∗ =0 x(k) − x∗ (b) lim (Bk − F (x∗ ))sk =0 sk k→ ∞ k→ ∞ File: Ò ÛØÓÒºØ Ü Revision: ½º¾¼ Date: ¾¼¼ »¼ »¿¼ ¼ ½¾ ÅÌ 51 52 Newton-Like Methods (c) lim k→ ∞ Bk sk − yk =0 sk Proof: We only prove the equivalence of (a) and (c). 8) reduces to: (k) x(k+1) := x(k) − B−1 ). 11) So, we have Bk sk = −F(x(k) ) and F(x (k+1) ) = yk + F(x(k) ) = yk − Bk sk . 12) Suppose that (a) holds. 13) F (x∗ + t(x(k+1) − x∗ ))dt.

X∗ := limk→ ∞ x(k) exists. Then g(x∗ ) = 0 and ∇2 f(x∗ ) is positive semidefinite. In other words, the limit x ∗ satisfies the necessary first and second order conditions for a local minimum of f. 23. Thus, we only need to show that ∇2 f(x∗ ) is positive definite. Let H(x) := ∇2 f(x) and λ1 (B) denote the smallest Eigenvalue of a matrix B. Then, we must show that λ1 (H(x∗ )) ≥ 0. , λ1 (H(x∗ )) = −2α < 0. 50) λ1 (Bk ) ≤ −α < 0 for all sufficiently large k. 51) −gTk sk + f(x (k) + sk ) − f(x 1 0 (k) ).

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