Heegner-point constructions for curves x^3 + y^3 = k12a talk presented 8 November 2003, by Noam D. Elkies at the3Princeton Workshop on the conjecture of Birch and4Swinnerton-Dyer,56========================================================================780) General Introduction: How little we still know about the BSD conjecture9101) Specific Introduction: the curves x^3+y^3=k11122) Why the case of prime k (and sign -1) requires a new idea13143) The Heegner-point variant that works 2/3 of the time for k=p and k=p^21516------------------------------------------------------------------------1718[Thank organizers. I was asked to speak about my work on x^3+y^3=k --19which serves me right for procrastinating on writing it up during the20past ~10 years. It's a rather small part of the BSD story; but, in21some sense/direction we don't know that much more about BSD now than22we did in 1993. So I'll start by putting the x^3+y^3=k case in this23context.]24250) Let E/Q be an elliptic curve.2627["Random" elliptic curves over "random" number fields28are still hopeless.]2930We'll heavily use a modular parametrization of E by X_0(N); thanks to31Wiles, Taylor, Breuil, B.Conrad, and Diamond, we now know that such a32parametrization exists. This is essentially the only new result33towards BSD in the past decade -- it shows that the analytic rank34exists! (But for CM families like x^3+y^3=k and Dy^2=x^3-x we knew it35already.)3637Recall that the Birch and Swinnerton-Dyer conjecture asserts that38analytic rank = arithmetic rank, and the refined conjecture relates39the leading term of L(E,s) at s=1 with numerical invariants of the40curve.4142How to check it for E? First must compute arithmetic and analytic43ranks.4445@ Arithmetic: descents and point searches will "eventually"46determine the rank, if (as conjectured) Sha(E/Q) is finite. For47family of curves, generally even this might fail unless we're really48lucky and the upper and lower bounds agree... No method known for49proving that E(Q) contains enough independent points without actually50finding/constructing them.5152@ Analytic: generally hard, even for individual E. One can compute53any derivative of L(E,s) at s=1 to arbitrary precision; this can prove54that the derivative isn't zero, but how to prove it *is* zero?5556Analytic rank 0: true by Kolyvagin up to a bounded factor; can be57checked effectively (and often practically) for any given case.5859Analytic rank 1: L(E,1)=0 by parity. If L'(E,1)>0 then again OK by60Kolyvagin up to an effectively removable finite fudge factor.6162[Both of these cases heavily use Heegner points. In particular, in63the rank-1 case, Gross-Zagier formula says L'(E,1) is proportional to64the canonical height of a Heegner point, so if L'(E,1)>0 this point65generates a full-rank subgroup of E(Q). This is effective not just in66theory but also in practice; e.g., M.Watkins recently did a case of67conductor 66157667 and no special structure, finding a rational point68of canonical height 12557+ (thousands of digits) that is almost69certainly a generator of the M-W group, in under 3 days on Ahtlon MP701600.]7172Analytic rank 2: L(E,1)=0 because it's an integer multiple of period/M73for some effective M. If we can find 2 independent points, we've74checked BSD. But this might not be computationally feasible --75without Heegner points, searching for rational points is76exponential-time, and the only recent progress is lowering the base of77the exponent. If we do have enough generators, we can then check78refined BSD numerically to some precision and guess the size of79Sha(E/Q), but we have no idea how to either prove that this analytic80|Sha| is actually the square integer it's approximating, nor prove81that |Sha| actually is that integer, or indeed that Sha(E) is finite.8283Analytic rank 3: L'(E,1)=0 because it's proportional to the height of84a Heegner point and there are no non-torsion points of very small85canonical height. (In practice it's sometimes easier to test whether86L'(E,1)=0 by Heegner computation, but that's another story! [ANTS-V])87Difficulty of testing BSD and refined BSD as in the case of analytic88rank 2.8990Analytic rank 4 and higher: we can check numerically that L''(E,1) or91L'''(E,1) seems to vanish, but we can't prove it, so we can't prove92BSD, let alone refined BSD (though we can still test numerically).9394[Note that even if rank 2 and higher are "rare", they do occur;95e.g. for "congruent number" curves Dy^2=x^3-x:9697r D98990 11001 51012 341023 12541034 292741045 482722391056 66117198661067 797507543735107108The r=7 curve was obtained just last week by Nicholas Rogers, and the109r=5 and r=6 examples are also new improvements on his previous110computation. Is r unbounded? I don't think we have enough evidence,111but I've seen no convincing hints that r=O(1). Current record is 24112(Martin-McMillen 2000).]113114Families of curves: even worse. For example: if Dy^2=x^3-x has odd115sign (this is known to be equivalent to: squarefree part of |D| is116congruent mod 8 to one of 5,6,7), must it have positive arithmetic117rank? Even if rank is at most 1 by 2-descent, how to prove it? The118difficulty is that we can neither show directly that L'(E,1)>0, nor119prove in general that the Heegner point does not vanish. This has120been done only in a few simple cases, such as D a prime (one of121Heegner's original applications of Heegner points!). Even there one122uses not Heegner points for the modular parametrization of Dy^2=x^3-x,123a curve of conductor 16D^2 or 32D^2 (how to tell that the resulting124point is nontrivial?), but starting from a curve y^2=x^3-4x or125y^2=x^3+4x of conductor 32 or 64, constructing a point P rational over126Q(sqrt(-|D|)), and showing that its image under Galois is *not* -P but127T-P for a suitable 2-torsion point T -- which proves that P is not128itself torsion.129130------------------------------------------------------------------------131132The next case past Dy^2=x^3-x and other families of quadratic twists133is cubic, quartic, or sextic twists. These share some of the nice134behavior of Dy^2=x^3-x (constant CM j-invariant) but are more135complicated. We'll consider the curves we'll call E(k): x^3 + y^3 =136k. This is another classical family, studied intensively by Sylvester137(1879), Selmer (1951), Birch-Stephens (1966) ... Here too the ranks138can get quite high; the same N.Rogers, working with me on his doctoral139thesis, recently found the first examples known of ranks 8,9,10,11 --140but that too is material for another talk. Here we'll again141concentrate on families of E(k) of odd analytic rank when one can142prove using a descent argument that the rank is at most 1, and ask to143prove it by actually constructing a point. For Dy^2=x^3-x we used the1442-isogenies with Dy^2=x^3+4x; here we'll descend via the 3-isogenies145between E(k) and the curve E'(k): xy(x+y)=k. Again the complexity of146the descent depends on the number of (distinct) prime factors of k.147The simplest case -- past the classical k=1, r=0 -- is k=p or k=p^2,148where p is a prime other than 3 (it's also classical that E(3) and149E(9) have ranks 0 and 1 respectively). In that case, the 3-descent150yields an upper bound on the arithmetic rank of E(k) that depends on151the residue of p mod 9 as follows:152153p mod 9 | 1 2 4 5 7 8154--------+------------------155bound | 2 0 1 0 1 1156157in particular, when p mod 9 is 4, 7, or 8, the curve has odd sign and158Birch-Stephens (following Selmer, but apparently not Sylvester!)159conjectured that in each of these cases E(p) actually has rank 1. The160same should hold for E(p^2) with p in the same residue classes. To161prove this we must find a recipe for constructing a non-torsion162rational point on E(k) or E'(k). We have such a recipe, of course --163the Heegner construction on E(k), considered as a modular curve in its164own right -- but again we have no idea how to prove in general that165thie resulting point is non-torsion. So we again look to find a point166P on some initial curve like E(1) that's rational over Q(cbrt(p)), or167better yet K(cbrt(p)) where K is the CM field168169K = Q(sqrt(-3)) = Q(rho) [rho a generator of mu_3]170171and prove that P transforms correctly under Galois. Note that if we172put the origin of E(k) at the obvious point (1:-1:0) then the CM of173E(k) is given by (x:y:z) --> (x:y:rho*z), so P = (x:y:cbrt(k)*z) can174again be expressed in Galois terms.175176------------------------------------------------------------------------177178But this didn't work. One can write down candidates: start from E(1),179which is the modular curve X_0(27), and use the 27-isogenies between180CM curves of discriminant -9k^2 or -27k^2. But it always seems to181produce torsion points.182183Closest approach: Satge' (1987) proved that E(2p) has a rational184point, and thus rank 1, for odd p congruent to 2 mod 9, starting from185E(2) (which happens to be the modular curve X_0(36)=X(6)) and using18636-isogenies between CM curves of discriminant -12p^2.187188In retrospect Gross explained: the Gross-Zagier formula actually gives189the height of the Heegner point not just in terms of L'(E,1), but in190terms of the derivative at s=1 of the product of L-functions of two191curves. In the usual setting of quadratic twists, the other192L-function is the L-function of the initial curve, so we're OK if (as193with Dy^2=x^3-x) that initial curve has rank zero. But for cubic194twists -- where there's as yet no G-Z formula but one can formulate a195conjectural analogue -- the companion elliptic curve is a different196twist. If we go from E(k_0) to E(ck_0), the companion is E(c^2 k_0).197So k_0=1, c=p yields the companion curve E(p^2), which also has sign198-1, so the product vanishes to order at least 2 at s=1, its derivative199vanishes, and we expect the Heegner point to be torsion. For Satge's,200k_0=2, c=p, and E(2p^2) has rank zero by 3-descent, so one expects to201succeed -- Satge' proved it.202203------------------------------------------------------------------------204205For p=9k+4 or 9k+7, I proved in 1993 that E(p) and E(p^2) do have206positive rank, using yet another variant of the Heegner-point method.207208Let pi be a factor of p in Z[rho] determined up to conjugation by the209condition that 3|pi-1. So, p = pi pi' where pi' is the conjugate. We210construct a point P on E(pi) or E(pi^2), show that P is rational over211K(cbrt(pi')), and prove that the image of P under the appropriate212generator of Gal(K(cbrt(pi'))/K) is T+rho*P where T is a nontrivial2133-torsion point. This yields a K-rational point on the plane cubic pi214X^3 + pi' Y^3 = Z^3 or pi^2 X^3 + pi'^2 Y^3 = Z^3, which is a215principal homogeneous space for E(p) or E(p^2), and thus a nontrivial216K-rational point on E(p) or E(p^2). Since K is the CM field, it217follows that these curves have a nontorsion point also over Q. (In218fact the point we construct turns out to be Q-rational already,219because X,Y are also conjugate over Q.) In Gross's interpretation,220when c and k_0 are in K but not necessarily in Q, the companion twist221of E(c k_0) is not E(k_0/c) but E(k_0/c') -- so when k_0=pi and c=pi'222the companion is just E(1), which has rank zero as desired.223224To construct our point P, we use a modular parametrization of E(pi) or225E(pi^2). Since this is not an elliptic curve over Q, we can't use any226X_0(N). We'll use a curve we'll call X'_0(9p) or X'_0(27p), the X'227meaning that for the p-level structure we use matrices that reduce mod228p to upper triangular _with cubic residues on the diagonal_. [For k=p229or p^2, the 3-part of the level is 9 if k is 4 mod 9, and 27 if k is 7230mod 9.] We then use a point P_0 on this curve parametrizing a cyclic231isogeny of degree 9p or 27p between two CM curves of discriminant -27,232chosen so that the kernel includes ker(pi'). We then use stanadard CM233theory/techniques to prove the point is defined over K(cbrt(pi')), and234that its image P on E(pi) is mapped by Galois to T+rho*P for some235(sqrt(-3))-torsion point T. [NB we don't need to take a Galois trace236of P from some class field to get the desired rational point, which237would be needed for most Heegner-point constructions.] The hard part238is showing that T is nontrivial, which we do by directly computing the239modular symbols that go into T!240241