Heegner-point constructions for curves x^3 + y^3 = k
a talk presented 8 November 2003, by Noam D. Elkies at the
Princeton Workshop on the conjecture of Birch and
Swinnerton-Dyer,
========================================================================
0) General Introduction: How little we still know about the BSD conjecture
1) Specific Introduction: the curves x^3+y^3=k
2) Why the case of prime k (and sign -1) requires a new idea
3) The Heegner-point variant that works 2/3 of the time for k=p and k=p^2
------------------------------------------------------------------------
[Thank organizers. I was asked to speak about my work on x^3+y^3=k --
which serves me right for procrastinating on writing it up during the
past ~10 years. It's a rather small part of the BSD story; but, in
some sense/direction we don't know that much more about BSD now than
we did in 1993. So I'll start by putting the x^3+y^3=k case in this
context.]
0) Let E/Q be an elliptic curve.
["Random" elliptic curves over "random" number fields
are still hopeless.]
We'll heavily use a modular parametrization of E by X_0(N); thanks to
Wiles, Taylor, Breuil, B.Conrad, and Diamond, we now know that such a
parametrization exists. This is essentially the only new result
towards BSD in the past decade -- it shows that the analytic rank
exists! (But for CM families like x^3+y^3=k and Dy^2=x^3-x we knew it
already.)
Recall that the Birch and Swinnerton-Dyer conjecture asserts that
analytic rank = arithmetic rank, and the refined conjecture relates
the leading term of L(E,s) at s=1 with numerical invariants of the
curve.
How to check it for E? First must compute arithmetic and analytic
ranks.
@ Arithmetic: descents and point searches will "eventually"
determine the rank, if (as conjectured) Sha(E/Q) is finite. For
family of curves, generally even this might fail unless we're really
lucky and the upper and lower bounds agree... No method known for
proving that E(Q) contains enough independent points without actually
finding/constructing them.
@ Analytic: generally hard, even for individual E. One can compute
any derivative of L(E,s) at s=1 to arbitrary precision; this can prove
that the derivative isn't zero, but how to prove it *is* zero?
Analytic rank 0: true by Kolyvagin up to a bounded factor; can be
checked effectively (and often practically) for any given case.
Analytic rank 1: L(E,1)=0 by parity. If L'(E,1)>0 then again OK by
Kolyvagin up to an effectively removable finite fudge factor.
[Both of these cases heavily use Heegner points. In particular, in
the rank-1 case, Gross-Zagier formula says L'(E,1) is proportional to
the canonical height of a Heegner point, so if L'(E,1)>0 this point
generates a full-rank subgroup of E(Q). This is effective not just in
theory but also in practice; e.g., M.Watkins recently did a case of
conductor 66157667 and no special structure, finding a rational point
of canonical height 12557+ (thousands of digits) that is almost
certainly a generator of the M-W group, in under 3 days on Ahtlon MP
1600.]
Analytic rank 2: L(E,1)=0 because it's an integer multiple of period/M
for some effective M. If we can find 2 independent points, we've
checked BSD. But this might not be computationally feasible --
without Heegner points, searching for rational points is
exponential-time, and the only recent progress is lowering the base of
the exponent. If we do have enough generators, we can then check
refined BSD numerically to some precision and guess the size of
Sha(E/Q), but we have no idea how to either prove that this analytic
|Sha| is actually the square integer it's approximating, nor prove
that |Sha| actually is that integer, or indeed that Sha(E) is finite.
Analytic rank 3: L'(E,1)=0 because it's proportional to the height of
a Heegner point and there are no non-torsion points of very small
canonical height. (In practice it's sometimes easier to test whether
L'(E,1)=0 by Heegner computation, but that's another story! [ANTS-V])
Difficulty of testing BSD and refined BSD as in the case of analytic
rank 2.
Analytic rank 4 and higher: we can check numerically that L''(E,1) or
L'''(E,1) seems to vanish, but we can't prove it, so we can't prove
BSD, let alone refined BSD (though we can still test numerically).
[Note that even if rank 2 and higher are "rare", they do occur;
e.g. for "congruent number" curves Dy^2=x^3-x:
r D
0 1
1 5
2 34
3 1254
4 29274
5 48272239
6 6611719866
7 797507543735
The r=7 curve was obtained just last week by Nicholas Rogers, and the
r=5 and r=6 examples are also new improvements on his previous
computation. Is r unbounded? I don't think we have enough evidence,
but I've seen no convincing hints that r=O(1). Current record is 24
(Martin-McMillen 2000).]
Families of curves: even worse. For example: if Dy^2=x^3-x has odd
sign (this is known to be equivalent to: squarefree part of |D| is
congruent mod 8 to one of 5,6,7), must it have positive arithmetic
rank? Even if rank is at most 1 by 2-descent, how to prove it? The
difficulty is that we can neither show directly that L'(E,1)>0, nor
prove in general that the Heegner point does not vanish. This has
been done only in a few simple cases, such as D a prime (one of
Heegner's original applications of Heegner points!). Even there one
uses not Heegner points for the modular parametrization of Dy^2=x^3-x,
a curve of conductor 16D^2 or 32D^2 (how to tell that the resulting
point is nontrivial?), but starting from a curve y^2=x^3-4x or
y^2=x^3+4x of conductor 32 or 64, constructing a point P rational over
Q(sqrt(-|D|)), and showing that its image under Galois is *not* -P but
T-P for a suitable 2-torsion point T -- which proves that P is not
itself torsion.
------------------------------------------------------------------------
The next case past Dy^2=x^3-x and other families of quadratic twists
is cubic, quartic, or sextic twists. These share some of the nice
behavior of Dy^2=x^3-x (constant CM j-invariant) but are more
complicated. We'll consider the curves we'll call E(k): x^3 + y^3 =
k. This is another classical family, studied intensively by Sylvester
(1879), Selmer (1951), Birch-Stephens (1966) ... Here too the ranks
can get quite high; the same N.Rogers, working with me on his doctoral
thesis, recently found the first examples known of ranks 8,9,10,11 --
but that too is material for another talk. Here we'll again
concentrate on families of E(k) of odd analytic rank when one can
prove using a descent argument that the rank is at most 1, and ask to
prove it by actually constructing a point. For Dy^2=x^3-x we used the
2-isogenies with Dy^2=x^3+4x; here we'll descend via the 3-isogenies
between E(k) and the curve E'(k): xy(x+y)=k. Again the complexity of
the descent depends on the number of (distinct) prime factors of k.
The simplest case -- past the classical k=1, r=0 -- is k=p or k=p^2,
where p is a prime other than 3 (it's also classical that E(3) and
E(9) have ranks 0 and 1 respectively). In that case, the 3-descent
yields an upper bound on the arithmetic rank of E(k) that depends on
the residue of p mod 9 as follows:
p mod 9 | 1 2 4 5 7 8
--------+------------------
bound | 2 0 1 0 1 1
in particular, when p mod 9 is 4, 7, or 8, the curve has odd sign and
Birch-Stephens (following Selmer, but apparently not Sylvester!)
conjectured that in each of these cases E(p) actually has rank 1. The
same should hold for E(p^2) with p in the same residue classes. To
prove this we must find a recipe for constructing a non-torsion
rational point on E(k) or E'(k). We have such a recipe, of course --
the Heegner construction on E(k), considered as a modular curve in its
own right -- but again we have no idea how to prove in general that
thie resulting point is non-torsion. So we again look to find a point
P on some initial curve like E(1) that's rational over Q(cbrt(p)), or
better yet K(cbrt(p)) where K is the CM field
K = Q(sqrt(-3)) = Q(rho) [rho a generator of mu_3]
and prove that P transforms correctly under Galois. Note that if we
put the origin of E(k) at the obvious point (1:-1:0) then the CM of
E(k) is given by (x:y:z) --> (x:y:rho*z), so P = (x:y:cbrt(k)*z) can
again be expressed in Galois terms.
------------------------------------------------------------------------
But this didn't work. One can write down candidates: start from E(1),
which is the modular curve X_0(27), and use the 27-isogenies between
CM curves of discriminant -9k^2 or -27k^2. But it always seems to
produce torsion points.
Closest approach: Satge' (1987) proved that E(2p) has a rational
point, and thus rank 1, for odd p congruent to 2 mod 9, starting from
E(2) (which happens to be the modular curve X_0(36)=X(6)) and using
36-isogenies between CM curves of discriminant -12p^2.
In retrospect Gross explained: the Gross-Zagier formula actually gives
the height of the Heegner point not just in terms of L'(E,1), but in
terms of the derivative at s=1 of the product of L-functions of two
curves. In the usual setting of quadratic twists, the other
L-function is the L-function of the initial curve, so we're OK if (as
with Dy^2=x^3-x) that initial curve has rank zero. But for cubic
twists -- where there's as yet no G-Z formula but one can formulate a
conjectural analogue -- the companion elliptic curve is a different
twist. If we go from E(k_0) to E(ck_0), the companion is E(c^2 k_0).
So k_0=1, c=p yields the companion curve E(p^2), which also has sign
-1, so the product vanishes to order at least 2 at s=1, its derivative
vanishes, and we expect the Heegner point to be torsion. For Satge's,
k_0=2, c=p, and E(2p^2) has rank zero by 3-descent, so one expects to
succeed -- Satge' proved it.
------------------------------------------------------------------------
For p=9k+4 or 9k+7, I proved in 1993 that E(p) and E(p^2) do have
positive rank, using yet another variant of the Heegner-point method.
Let pi be a factor of p in Z[rho] determined up to conjugation by the
condition that 3|pi-1. So, p = pi pi' where pi' is the conjugate. We
construct a point P on E(pi) or E(pi^2), show that P is rational over
K(cbrt(pi')), and prove that the image of P under the appropriate
generator of Gal(K(cbrt(pi'))/K) is T+rho*P where T is a nontrivial
3-torsion point. This yields a K-rational point on the plane cubic pi
X^3 + pi' Y^3 = Z^3 or pi^2 X^3 + pi'^2 Y^3 = Z^3, which is a
principal homogeneous space for E(p) or E(p^2), and thus a nontrivial
K-rational point on E(p) or E(p^2). Since K is the CM field, it
follows that these curves have a nontorsion point also over Q. (In
fact the point we construct turns out to be Q-rational already,
because X,Y are also conjugate over Q.) In Gross's interpretation,
when c and k_0 are in K but not necessarily in Q, the companion twist
of E(c k_0) is not E(k_0/c) but E(k_0/c') -- so when k_0=pi and c=pi'
the companion is just E(1), which has rank zero as desired.
To construct our point P, we use a modular parametrization of E(pi) or
E(pi^2). Since this is not an elliptic curve over Q, we can't use any
X_0(N). We'll use a curve we'll call X'_0(9p) or X'_0(27p), the X'
meaning that for the p-level structure we use matrices that reduce mod
p to upper triangular _with cubic residues on the diagonal_. [For k=p
or p^2, the 3-part of the level is 9 if k is 4 mod 9, and 27 if k is 7
mod 9.] We then use a point P_0 on this curve parametrizing a cyclic
isogeny of degree 9p or 27p between two CM curves of discriminant -27,
chosen so that the kernel includes ker(pi'). We then use stanadard CM
theory/techniques to prove the point is defined over K(cbrt(pi')), and
that its image P on E(pi) is mapped by Galois to T+rho*P for some
(sqrt(-3))-torsion point T. [NB we don't need to take a Galois trace
of P from some class field to get the desired rational point, which
would be needed for most Heegner-point constructions.] The hard part
is showing that T is nontrivial, which we do by directly computing the
modular symbols that go into T!