# The current status of the Birch & Swinnerton-Dyer Conjecture

[Une traduction française suit la version anglaise.]

The question is only about elliptic curves $E$ over $\mathbb{Q}$ and concerns only the aspect

(order of vanishing of $L(E,s)$ at $s=1$)$\ =\$(rank of $E(\mathbb{Q})$).

Let $r$ be the LHS and $d$ the RHS, so that (a special case of ) the Birch and Swinnerton-Dyer Conjecture is

BSD?. $r=d$.

By the end of the last millenium, we knew

Theorem (1977--2000). If $\ r=0,1$, then $d=r$ (and $\ \operatorname{Sha}(E)$ is finite).

Some years ago, I heard that there was some progress in proving $(r>0)\Longrightarrow (d>0)$ under the assumption of the finiteness of $\operatorname{Sha}(E)$. What is the current status of the

Statement. Suppose that $\operatorname{Sha}(E)$ is finite. If $r>1$, then $d>0$ ?

L'état actuel de la conjecture de Birch et Swinnerton-Dyer

On s'interesse uniquement aux courbes abéliennes $A$ sur $\mathbf{Q}$ et à l'aspect

(ordre d'annulation de $L(A,s)$ en $s=1$)$\ =\$(rang de $A(\mathbf{Q})$).

Désignons par $r$ le membre de gauche et par $d$ le membre de droite, de sorte que la conjecture de Birch et Swinnerton-Dyer prédit (en particulier)

BSD? $r=d$.

Vers la fin du millénaire dernier, on avait démontré le

Théorème (1977--2000). Si $r=0,1$, alors $d=r$ (et $\ \operatorname{Cha}(A)$ est fini).

Il y a quelques années, j'avais entendu dire qu'on a fait des progrès concernant l'implication $(r>0)\Longrightarrow(d>0)$ sous l'hypothèse de la finitude de Cha$(A)$. Quel est l'état actuel de l'

Énoncé. Supposons que $\operatorname{Cha}(A)$ est fini. Si $r>1$, alors $d>0$ ?

-
Fantastically accurate and elegant translation, Chandan. But why did you give it? –  Georges Elencwajg Jan 12 '10 at 12:18
For the sake of diversity. –  Chandan Singh Dalawat Jan 12 '10 at 12:41
Despite all the celebration of diversity, one often sees a University Avenue, but walk one block over, there is never a Diversity Avenue. In Berkeley, one side is Hearst, the other side Addison. –  Will Jagy Jan 30 '12 at 1:41
Something I never expected to hear in commentary for a soccer match, Ray Hudson, announcing the 25th January Copa del Rey match involving Real Madrid, said "they are trying to solve the problem that is Barcelona. They might as well try to solve the Hodge Conjecture." This was on GolTV... en.wikipedia.org/wiki/Ray_Hudson Not the same conjecture, you can't have everything. –  Will Jagy Jan 30 '12 at 3:48
@Will, there is a Diversity Place coming off University Way at the Okanagan campus of the University of British Columbia. –  Gerry Myerson May 11 '12 at 5:58

The parity conjecture is known, i.e. it is known that if the order of vanishing of the $L$-function is even/odd, then the corank of $p$-Selmer is of the same parity (and I think this is known for every $p$ at this point; Nekovar handled the good ordinary or multiplicative case, and B.D. Kim the good supersingular case. T. and V. Dokchitser then gave a new proof that handled the general case). This would imply that if Sha(E) is finite (even after passing to the $p$-part of Sha for some prime $p$) and the $L$-function has odd order vanishing, then $E({\mathbb Q})$ has positive rank.

There has also been recent work on establishing cases of positive even order vanishing, and trying to prove that the Selmer group has rank at least two. This has been investigated by both Bellaiche--Chenevier and Skinner--Urban. I don't know precise statements though, and I'm not sure if either pair of authors can handle elliptic curves. (In both cases, the arguments involve deforming along an eigenvariety, and there are problems at low weights. So they may only have results for modular forms of weight $k > 2$.)

Incidentally, although it wasn't part of your question, I don't think anything new is known about finiteness of Sha, beyond what you recalled in your question.

-
It would be nice if someone who has read the Bellaïche--Chenevier and Skinner--Urban papers tells us what their result have to say about the question $r>1?\Rightarrow d>0$. –  Chandan Singh Dalawat Jan 13 '10 at 4:35
Bellaiche and Chenevier have a long paper/book on the topic of eigenvarieties and Selmer groups. There are (at least) a couple of people who post here who are at least somewhat familiar with it; I'm one of them, and if I get a chance, I'll try to figure out what they can prove. As for Skinner and Urban, they have annouced examples where the parity is even, the $L$-function vanishes, and the Selmer group has rank at least $2$, but no details have appeared in print. Also, as I said, I think that for both sets of authors, they may be restricted to the case of modular forms of weight $> 2$. –  Emerton Jan 13 '10 at 4:51

Reading Olivier's comment reminded me that it is always possible to verify rigorously, for a given elliptic curve $E$ over $\mathbb Q$ for which the rank of $E({\mathbb Q})$ is at most $3$, that one has an equality of algebraic and analytic ranks (i.e. in the notation of the question, whether or not $r = d$), assuming that it is indeed true.

The point is the following: one can compute the sign of the functional equation, and hence the parity of the order of vanishing of $L(E,s)$ at $s = 1$ (i.e. the parity of $r$, in the notation of the question).

If this parity is even, one proceeds as follows: the computation of $L(E,1)/\Omega$ (where $\Omega$ is the real period of the curve) is exact (one does it via modular sybmols), and in particular one can determine whether or not $L(E,1)$ vanishes. If the rank of $E({\mathbb Q})$ vanishes, then one expects that in fact $L(E,1)$ is non-zero; if not, one has a counterexample to BSD.

Now if the rank of $E({\mathbb Q})$ equals 2, then one knows that necessarily $L(E,1)$ vanishes, and that in fact it must vanish to order at least 2. On the other hand, one can approximate the 2nd derivative of $L(E,s)$ at $s = 1$ as accurately as one wants, and so in particular, can verify that it doesn't equal zero (again, as must be the case if BSD is true).

If the rank of $E({\mathbb Q})$ is 1 or 3, then one expects that $L(E,1)$ vanishes with odd order, and this can be verified by computing the sign in the functional equation (and if it doesn't hold, again one has a counterexample to BSD). The Gross-Zagier formula lets one compute the derivative of $L(E,s)$ at $s = 1$, by using an explicit modular parameterization of $E$ and seeing if the Heegner point is torsion or not; if it doesn't vanish, then one know that the rank of $E({\mathbb Q})$ must be 1. If it does vanish, then the rank of $E({\mathbb Q})$ will have to be 3 (or else BSD fails), and one knows that $L(E,s)$ vanishes at $s = 1$ to an odd order that is greater than 1. Again, one can verify that the third derivative of $L(E,s)$ at $s = 1$ is non-zero, and so show that the order of vanishing is exactly 3, verifying BSD. (Otherwise, BSD would fail.)

I suppose in practice it could happen that in the case of rank 2 or 3, the non-zero 2nd or 3rd derivative that you have to compute would be so small that it was hard to distinguish from 0. On the other hand, one can use the conjectural formula for the leading term (coming from the full BSD conjecture) to determine an expected lower bound for this term, and again this should be an actual lower bound unless BSD fails. (Here it is useful to note that the regulator and the order of Sha appear in the numerator of the BSD formula, so one doesn't need to know about Sha, or the precise generators of $E({\mathbb Q})$, to determine this lower bound.)

I once (back in 1993) read a masters thesis from Macquarie University (by a student named Chris Daniels, if I remember correctly) which verified a rank 3 example via the above scheme. I don't know how many other examples have been verified.

-
On the last point, Buhler, Gross, and Zagier already did the $N=5077$ case for rank 3 in a mid80s Math Comp article ams.org/mcom/1985-44-170/S0025-5718-1985-0777279-X/… , though showing that $L'$ was truly zero was not detailed with voluminous error analysis, just invoking Gross-Zagier. I think it is essentially mechanical to do it for any rank 3 curve for which the generators are known and the conductor is small enough to compute $L^{'''}$. –  Junkie Apr 9 '11 at 7:55

There is an article of J.Parson and B.Gross (On the local divisibility of Heegner points) from which I think some very very special instance of r=2, d=2 can be deduced. The argument is a combination of Kolyvagin's work, level-raising and parity results as follows: start with an elliptic curve $E$ over $\mathbb Q$ with analytic rank 1 and level $N$, study its Heegner points, level-raise the attached modular form to a modular form $g$ of level $Np$ such that the sign of the functional equation is now $-1$, compare Heegner points for $E$ and for the abelian varieties $A(g)$. WHen $A(g)$ is again an elliptic curve (and when the Heegner points involved are actually rational), some cases of the question asked can be solved (this very brief description obscures the very important role played by an auxiliary imaginary quadratic field).

An example (from the paper and due to N.Elkies and W.Stein) is the curve $y^2+xy+y=x^3-121830x+341716424$ which has level 78, comes from level-raising from an elliptic curve of level 26 and is known (by the procedure outlined above) to have analytic and algebraic rank equal to 2.

-
Wait, there's no curve of conductor $78$ and rank $2$, and this curve $y^2+xy+y=x^3−121830x+341716424$ has a huge conductor not divisible by either $3$ or $13$. Googling the last coefficient, I gather you made a sign error and meant $y^2+xy+y=x^3−121830x-341716424$, with conductor $79^2\cdot 78$ and rank $2$, which is the quadratic twist by $(-79)^{1/2}$ of a curve of conductor $78$. But I don't think William Stein and I wrote about this curve... –  Noam D. Elkies Jan 30 '12 at 1:37

Alice Silverberg has a "cheat sheet" on all things currently known related to the rank of an elliptic curve. The link is below.

http://math.uci.edu/~asilverb/connectionstalk.pdf

-

This is just an expansion of the remark in Emerton's first answer above:

Skinner-Urban (the preprint is here) offer the following theorem (Theorem 2(b)):

Theorem: Let $E/\mathbb{Q}$ be a semi-stable elliptic curve and $p\geq 11$ a prime of good ordinary reduction. If $L(E,1)=0$, then the corank of the $p^\infty$-Selmer group is at least one.

Combining this theorem with the standard exact sequence:

$$0\to (\mathbb{Q}_p/\mathbb{Z}_p)^d=E(\mathbb{Q}) \otimes (\mathbb{Q}_p/\mathbb{Z_p})\to Sel_{p^\infty}(E/\mathbb{Q})\to Sha_p(E/\mathbb{Q})\to 0,$$ and the assumption that $Sha(E)$ is finite, we see that $r > 0$ implies $d>0$.

I have not read the details of the proof of this Theorem, but my understanding is that it does not make any assumptions about the finiteness of $Sha(E)$.

-

I have technically no clue about this subject but what I glean from "coffee-table discussions" is that this paper is a recent progress about BSD. Or may be that can be said collectively about all the 3 papers that have been written by this friend/college-mate of mine and Manjul Bhargava.

-
Bhargava, as a joke, wrote to the Clay institute, asking that since he had shown (in conjunction with Kolyvagin's work, eg) a positive proportion of ell curves over Q, have BSD true, whether he could have a positive proportion of $1 million. The hard work being though, in the higher rank case. – Junkie Jan 29 '12 at 23:58 @Junkie I have heard that in some seminar he actually specified "10%" for what you say as "positive proportion". – Anirbit Jan 30 '12 at 1:14 François Brunault asks the same question (images.math.cnrs.fr/Le-rang-des-courbes-elliptiques.html) : La conjecture de Birch et Swinnerton-Dyer est l’un des problèmes du Clay Mathematical Institute, dont la solution est récompensée par un prix d’un million de dollars. Sachant que Bhargava et Shankar ont démontré cette conjecture pour une proportion strictement positive de courbes elliptiques, pourront-ils prétendre à une proportion strictement positive de la récompense ? – Chandan Singh Dalawat Jan 30 '12 at 6:33 More seriously, this is certainly a very interesting development, but I was interested in what happens when the order of vanishing of the$L$-function is$>1\$. –  Chandan Singh Dalawat Jan 30 '12 at 8:22