1Function: ellpadicbsd 2Section: elliptic_curves 3C-Name: ellpadicbsd 4Prototype: GGLDG 5Help: ellpadicbsd(E, p, n, {D = 1}): returns [r,Lp] where 6 r is the (conjectural) analytic rank of the p-adic L-function attached 7 to the quadratic twist E_D and Lp is (conjecturally) equal 8 to the product of the p-adic regulator and the cardinal of the 9 Tate-Shafarevich group. 10Doc: Given an elliptic curve $E$ over $\Q$, its quadratic twist $E_D$ 11 and a prime number $p$, this function is a $p$-adic analog of the complex 12 functions \tet{ellanalyticrank} and \tet{ellbsd}. It calls \kbd{ellpadicL} 13 with initial accuracy $p^n$ and may increase it internally; 14 it returns a vector $[r, L_p]$ where 15 16 \item $L_p$ is a $p$-adic number (resp. a pair of $p$-adic numbers if 17 $E$ has good supersingular reduction) defined modulo $p^N$, conjecturally 18 equal to $R_p S$, where $R_p$ is the $p$-adic regulator as given by 19 \tet{ellpadicregulator} (in the basis $(\omega, F \omega)$) and $S$ is the 20 cardinal of the Tate-Shafarevich group for the quadratic twist $E_D$. 21 22 \item $r$ is an upper bound for the analytic rank of the $p$-adic 23 $L$-function attached to $E_D$: we know for sure that the $i$-th 24 derivative of $L_p(E_D,.)$ at $\chi^0$ is $O(p^N)$ for all $i < r$ 25 and that its $r$-th derivative is nonzero; it is expected that the true 26 analytic rank is equal to the rank of the Mordell-Weil group $E_D(\Q)$, 27 plus $1$ if the reduction of $E_D$ at $p$ is split multiplicative; 28 if $r = 0$, then both the analytic rank and the Mordell-Weil rank are 29 unconditionnally $0$. 30 31 Recall that the $p$-adic BSD conjecture (Mazur, Tate, Teitelbaum, Bernardi, 32 Perrin-Riou) predicts an explicit link between $R_p S$ and 33 $$(1-p^{-1} F)^{-2} \cdot L_p^{(r)}(E_D, \chi^0) / r! $$ 34 where $r$ is the analytic rank of the $p$-adic $L$-function attached to 35 $E_D$ and $F$ is the Frobenius on $H^1_{dR}$; see \tet{ellpadicL} 36 for definitions. 37 \bprog 38 ? E = ellinit("11a1"); p = 7; n = 5; \\ good ordinary 39 ? ellpadicbsd(E, 7, 5) \\ rank 0, 40 %2 = [0, 1 + O(7^5)] 41 42 ? E = ellinit("91a1"); p = 7; n = 5; \\ non split multiplicative 43 ? [r,Lp] = ellpadicbsd(E, p, n) 44 %5 = [1, 2*7 + 6*7^2 + 3*7^3 + 7^4 + O(7^5)] 45 ? R = ellpadicregulator(E, p, n, E.gen) 46 %6 = 2*7 + 6*7^2 + 3*7^3 + 7^4 + 5*7^5 + O(7^6) 47 ? sha = Lp/R 48 %7 = 1 + O(7^4) 49 50 ? E = ellinit("91b1"); p = 7; n = 5; \\ split multiplicative 51 ? [r,Lp] = ellpadicbsd(E, p, n) 52 %9 = [2, 2*7 + 7^2 + 5*7^3 + O(7^4)] 53 ? ellpadicregulator(E, p, n, E.gen) 54 %10 = 2*7 + 7^2 + 5*7^3 + 6*7^4 + 2*7^5 + O(7^6) 55 ? [rC, LC] = ellanalyticrank(E); 56 ? [r, rC] 57 %12 = [2, 1] \\ r = rC+1 because of split multiplicative reduction 58 59 ? E = ellinit("53a1"); p = 5; n = 5; \\ supersingular 60 ? [r, Lp] = ellpadicbsd(E, p, n); 61 ? r 62 %15 = 1 63 ? Lp 64 %16 = [3*5 + 2*5^2 + 2*5^5 + O(5^6), \ 65 5 + 3*5^2 + 4*5^3 + 2*5^4 + 5^5 + O(5^6)] 66 ? R = ellpadicregulator(E, p, n, E.gen) 67 %17 = [3*5 + 2*5^2 + 2*5^5 + O(5^6), 5 + 3*5^2 + 4*5^3 + 2*5^4 + O(5^5)] 68 \\ expect Lp = R*#Sha, hence (conjecturally) #Sha = 1 69 70 ? E = ellinit("84a1"); p = 11; n = 6; D = -443; 71 ? [r,Lp] = ellpadicbsd(E, 11, 6, D) \\ Mordell-Weil rank 0, no regulator 72 %19 = [0, 3 + 2*11 + O(11^6)] 73 ? lift(Lp) \\ expected cardinal for Sha is 5^2 74 %20 = 25 75 ? ellpadicbsd(E, 3, 12, D) \\ at 3 76 %21 = [1, 1 + 2*3 + 2*3^2 + O(3^8)] 77 ? ellpadicbsd(E, 7, 8, D) \\ and at 7 78 %22 = [0, 4 + 3*7 + O(7^8)] 79 @eprog 80