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1         Heegner-point constructions for curves  x^3 + y^3 = k
2
3          a talk presented 8 November 2003, by Noam D. Elkies at the
4          Princeton Workshop on the conjecture of Birch and
5          Swinnerton-Dyer,
6
7========================================================================
8
90) General Introduction: How little we still know about the BSD conjecture
10
111) Specific Introduction: the curves x^3+y^3=k
12
132) Why the case of prime k (and sign -1) requires a new idea
14
153) The Heegner-point variant that works 2/3 of the time for k=p and k=p^2
16
17------------------------------------------------------------------------
18
19[Thank organizers.  I was asked to speak about my work on x^3+y^3=k --
20which serves me right for procrastinating on writing it up during the
21past ~10 years.  It's a rather small part of the BSD story; but, in
22some sense/direction we don't know that much more about BSD now than
23we did in 1993.  So I'll start by putting the x^3+y^3=k case in this
24context.]
25
260) Let E/Q be an elliptic curve.
27
28  ["Random" elliptic curves over "random" number fields
29  are still hopeless.]
30
31We'll heavily use a modular parametrization of E by X_0(N); thanks to
32Wiles, Taylor, Breuil, B.Conrad, and Diamond, we now know that such a
33parametrization exists.  This is essentially the only new result
34towards BSD in the past decade -- it shows that the analytic rank
35exists!  (But for CM families like x^3+y^3=k and Dy^2=x^3-x we knew it
37
38Recall that the Birch and Swinnerton-Dyer conjecture asserts that
39analytic rank = arithmetic rank, and the refined conjecture relates
40the leading term of L(E,s) at s=1 with numerical invariants of the
41curve.
42
43How to check it for E?  First must compute arithmetic and analytic
44ranks.
45
46@ Arithmetic: descents and point searches will "eventually"
47determine the rank, if (as conjectured) Sha(E/Q) is finite.  For
48family of curves, generally even this might fail unless we're really
49lucky and the upper and lower bounds agree...  No method known for
50proving that E(Q) contains enough independent points without actually
51finding/constructing them.
52
53@ Analytic: generally hard, even for individual E.  One can compute
54any derivative of L(E,s) at s=1 to arbitrary precision; this can prove
55that the derivative isn't zero, but how to prove it *is* zero?
56
57Analytic rank 0: true by Kolyvagin up to a bounded factor; can be
58checked effectively (and often practically) for any given case.
59
60Analytic rank 1: L(E,1)=0 by parity.  If L'(E,1)>0 then again OK by
61Kolyvagin up to an effectively removable finite fudge factor.
62
63[Both of these cases heavily use Heegner points.  In particular, in
64the rank-1 case, Gross-Zagier formula says L'(E,1) is proportional to
65the canonical height of a Heegner point, so if L'(E,1)>0 this point
66generates a full-rank subgroup of E(Q).  This is effective not just in
67theory but also in practice; e.g., M.Watkins recently did a case of
68conductor 66157667 and no special structure, finding a rational point
69of canonical height 12557+ (thousands of digits) that is almost
70certainly a generator of the M-W group, in under 3 days on Ahtlon MP
711600.]
72
73Analytic rank 2: L(E,1)=0 because it's an integer multiple of period/M
74for some effective M.  If we can find 2 independent points, we've
75checked BSD.  But this might not be computationally feasible --
76without Heegner points, searching for rational points is
77exponential-time, and the only recent progress is lowering the base of
78the exponent. If we do have enough generators, we can then check
79refined BSD numerically to some precision and guess the size of
80Sha(E/Q), but we have no idea how to either prove that this analytic
81|Sha| is actually the square integer it's approximating, nor prove
82that |Sha| actually is that integer, or indeed that Sha(E) is finite.
83
84Analytic rank 3: L'(E,1)=0 because it's proportional to the height of
85a Heegner point and there are no non-torsion points of very small
86canonical height.  (In practice it's sometimes easier to test whether
87L'(E,1)=0 by Heegner computation, but that's another story! [ANTS-V])
88Difficulty of testing BSD and refined BSD as in the case of analytic
89rank 2.
90
91Analytic rank 4 and higher: we can check numerically that L''(E,1) or
92L'''(E,1) seems to vanish, but we can't prove it, so we can't prove
93BSD, let alone refined BSD (though we can still test numerically).
94
95[Note that even if rank 2 and higher are "rare", they do occur;
96e.g. for "congruent number" curves Dy^2=x^3-x:
97
98r   D
99
1000   1
1011   5
1022   34
1033   1254
1044   29274
1055   48272239
1066   6611719866
1077   797507543735
108
109The r=7 curve was obtained just last week by Nicholas Rogers, and the
110r=5 and r=6 examples are also new improvements on his previous
111computation.  Is r unbounded?  I don't think we have enough evidence,
112but I've seen no convincing hints that r=O(1). Current record is 24
113(Martin-McMillen 2000).]
114
115Families of curves: even worse.  For example: if Dy^2=x^3-x has odd
116sign (this is known to be equivalent to: squarefree part of |D| is
117congruent mod 8 to one of 5,6,7), must it have positive arithmetic
118rank?  Even if rank is at most 1 by 2-descent, how to prove it?  The
119difficulty is that we can neither show directly that L'(E,1)>0, nor
120prove in general that the Heegner point does not vanish.  This has
121been done only in a few simple cases, such as D a prime (one of
122Heegner's original applications of Heegner points!).  Even there one
123uses not Heegner points for the modular parametrization of Dy^2=x^3-x,
124a curve of conductor 16D^2 or 32D^2 (how to tell that the resulting
125point is nontrivial?), but starting from a curve y^2=x^3-4x or
126y^2=x^3+4x of conductor 32 or 64, constructing a point P rational over
127Q(sqrt(-|D|)), and showing that its image under Galois is *not* -P but
128T-P for a suitable 2-torsion point T -- which proves that P is not
129itself torsion.
130
131------------------------------------------------------------------------
132
133The next case past Dy^2=x^3-x and other families of quadratic twists
134is cubic, quartic, or sextic twists.  These share some of the nice
135behavior of Dy^2=x^3-x (constant CM j-invariant) but are more
136complicated. We'll consider the curves we'll call E(k): x^3 + y^3 =
137k. This is another classical family, studied intensively by Sylvester
138(1879), Selmer (1951), Birch-Stephens (1966) ... Here too the ranks
139can get quite high; the same N.Rogers, working with me on his doctoral
140thesis, recently found the first examples known of ranks 8,9,10,11 --
141but that too is material for another talk. Here we'll again
142concentrate on families of E(k) of odd analytic rank when one can
143prove using a descent argument that the rank is at most 1, and ask to
144prove it by actually constructing a point.  For Dy^2=x^3-x we used the
1452-isogenies with Dy^2=x^3+4x; here we'll descend via the 3-isogenies
146between E(k) and the curve E'(k): xy(x+y)=k.  Again the complexity of
147the descent depends on the number of (distinct) prime factors of k.
148The simplest case -- past the classical k=1, r=0 -- is k=p or k=p^2,
149where p is a prime other than 3 (it's also classical that E(3) and
150E(9) have ranks 0 and 1 respectively).  In that case, the 3-descent
151yields an upper bound on the arithmetic rank of E(k) that depends on
152the residue of p mod 9 as follows:
153
154  p mod 9 | 1  2  4  5  7  8
155  --------+------------------
156  bound   | 2  0  1  0  1  1
157
158in particular, when p mod 9 is 4, 7, or 8, the curve has odd sign and
159Birch-Stephens (following Selmer, but apparently not Sylvester!)
160conjectured that in each of these cases E(p) actually has rank 1. The
161same should hold for E(p^2) with p in the same residue classes. To
162prove this we must find a recipe for constructing a non-torsion
163rational point on E(k) or E'(k).  We have such a recipe, of course --
164the Heegner construction on E(k), considered as a modular curve in its
165own right -- but again we have no idea how to prove in general that
166thie resulting point is non-torsion.  So we again look to find a point
167P on some initial curve like E(1) that's rational over Q(cbrt(p)), or
168better yet K(cbrt(p)) where K is the CM field
169
170  K = Q(sqrt(-3)) = Q(rho)  [rho a generator of mu_3]
171
172and prove that P transforms correctly under Galois.  Note that if we
173put the origin of E(k) at the obvious point (1:-1:0) then the CM of
174E(k) is given by (x:y:z) --> (x:y:rho*z), so P = (x:y:cbrt(k)*z) can
175again be expressed in Galois terms.
176
177------------------------------------------------------------------------
178
179But this didn't work.  One can write down candidates: start from E(1),
180which is the modular curve X_0(27), and use the 27-isogenies between
181CM curves of discriminant -9k^2 or -27k^2.  But it always seems to
182produce torsion points.
183
184Closest approach: Satge' (1987) proved that E(2p) has a rational
185point, and thus rank 1, for odd p congruent to 2 mod 9, starting from
186E(2) (which happens to be the modular curve X_0(36)=X(6)) and using
18736-isogenies between CM curves of discriminant -12p^2.
188
189In retrospect Gross explained: the Gross-Zagier formula actually gives
190the height of the Heegner point not just in terms of L'(E,1), but in
191terms of the derivative at s=1 of the product of L-functions of two
192curves.  In the usual setting of quadratic twists, the other
193L-function is the L-function of the initial curve, so we're OK if (as
194with Dy^2=x^3-x) that initial curve has rank zero.  But for cubic
195twists -- where there's as yet no G-Z formula but one can formulate a
196conjectural analogue -- the companion elliptic curve is a different
197twist.  If we go from E(k_0) to E(ck_0), the companion is E(c^2 k_0).
198So k_0=1, c=p yields the companion curve E(p^2), which also has sign
199-1, so the product vanishes to order at least 2 at s=1, its derivative
200vanishes, and we expect the Heegner point to be torsion.  For Satge's,
201k_0=2, c=p, and E(2p^2) has rank zero by 3-descent, so one expects to
202succeed -- Satge' proved it.
203
204------------------------------------------------------------------------
205
206For p=9k+4 or 9k+7, I proved in 1993 that E(p) and E(p^2) do have
207positive rank, using yet another variant of the Heegner-point method.
208
209Let pi be a factor of p in Z[rho] determined up to conjugation by the
210condition that 3|pi-1.  So, p = pi pi' where pi' is the conjugate. We
211construct a point P on E(pi) or E(pi^2), show that P is rational over
212K(cbrt(pi')), and prove that the image of P under the appropriate
213generator of Gal(K(cbrt(pi'))/K) is T+rho*P where T is a nontrivial
2143-torsion point.  This yields a K-rational point on the plane cubic pi
215X^3 + pi' Y^3 = Z^3 or pi^2 X^3 + pi'^2 Y^3 = Z^3, which is a
216principal homogeneous space for E(p) or E(p^2), and thus a nontrivial
217K-rational point on E(p) or E(p^2).  Since K is the CM field, it
218follows that these curves have a nontorsion point also over Q.  (In
219fact the point we construct turns out to be Q-rational already,
220because X,Y are also conjugate over Q.)  In Gross's interpretation,
221when c and k_0 are in K but not necessarily in Q, the companion twist
222of E(c k_0) is not E(k_0/c) but E(k_0/c') -- so when k_0=pi and c=pi'
223the companion is just E(1), which has rank zero as desired.
224
225To construct our point P, we use a modular parametrization of E(pi) or
226E(pi^2).  Since this is not an elliptic curve over Q, we can't use any
227X_0(N).  We'll use a curve we'll call X'_0(9p) or X'_0(27p), the X'
228meaning that for the p-level structure we use matrices that reduce mod
229p to upper triangular _with cubic residues on the diagonal_.  [For k=p
230or p^2, the 3-part of the level is 9 if k is 4 mod 9, and 27 if k is 7
231mod 9.]  We then use a point P_0 on this curve parametrizing a cyclic
232isogeny of degree 9p or 27p between two CM curves of discriminant -27,
233chosen so that the kernel includes ker(pi').  We then use stanadard CM
234theory/techniques to prove the point is defined over K(cbrt(pi')), and
235that its image P on E(pi) is mapped by Galois to T+rho*P for some
236(sqrt(-3))-torsion point T. [NB we don't need to take a Galois trace
237of P from some class field to get the desired rational point, which
238would be needed for most Heegner-point constructions.]  The hard part
239is showing that T is nontrivial, which we do by directly computing the
240modular symbols that go into T!
241