# Bhargava's work on the BSD conjecture

How much would Bhargava's results on BSD improve if finiteness of the Tate-Shafarevich group, or at least its $\ell$-primary torsion for every $\ell$, was known? Would they improve to the point of showing $100$% of elliptic curves over $\mathbf{Q}$ satisfy the BSD conjecture? (which, of course, would still not prove the BSD conjecture)

I've just attended a very nice seminar talk about the topic, and I'm curious to get some expert info.

This was meant to be a comment, but it won't fit, so here we go.

Just a few naive observations.

For completeness:

Weak BSD For any global field $K$ and any abelian variety $A$ defined over $K$, we have: $$\text{ord}_{s=1}L(A/K,s) = \text{rk}_{\mathbf{Z}}A(K).$$

BSD For any global field $K$ and any abelian variety $A$ defined over $K$, we have:

(1) $\text{Sha}(A/K)$ is a finite abelian group.

(2) $\rho :=\text{ord}_{s=1}L(A/K,s) = \text{rk}_{\mathbf{Z}}A(K).$

(3) $\lim_{s\to 1}L(A/K,s)(s-1)^{-\rho} = \frac{R_A\cdot\Omega_A\cdot\#\text{Sha}(A/K)}{|\Delta_K|^{1/2}\cdot \# A(K)_{\rm tor}\cdot\# A^{\vee}(K)_{\rm tor}}.$

Super-weak BSD The proportion of abelian varieties defined over $K$ that satisfy Weak BSD above, is 100%.

If $K$ is the global function field of a smooth projective geometrically irreducible curve over $\mathbf{F}_q$, then it is known by work of Tate, Milne, Bauer, Schneider, Kato, Trihan, that Weak BSD is equivalent to BSD, and BSD is in turn equivalent to finiteness of the $\ell$-primary torsion in the Tate-Shafarevich group for some prime $\ell$ (allowed to be the characteristic of $K$).

If $K$ is a number field, the above stream of equivalences is still expected to be true, but not known yet, likely because there's still no robust cohomological method to put us in the position to mimic the flat cohomology/crystalline-syntomic cohomology methods employed in the positive characteristic case, which Iwasawa theory for abelian varieties was partially meant for. Let us grant for a moment the following:

Expectation Let $K$ be a number field. For any abelian variety $A$ defined over $K$, the following are equivalent:

(1) $\text{Sha}(A/K)$ is finite.

(2) For all primes $\ell$, the $\ell$-primary torsion subgroup in $\text{Sha}(A/K)$ is finite.

(3) For some prime $\ell$, the $\ell$-primary torsion subgroup in $\text{Sha}(A/K)$ is finite.

(4) Weak BSD is true for $A$.

(5) Strong BSD is true for $A$.

This expectation, a Theorem if $K$ were a positive characteristic global function field, reveals the problem is really about showing finiteness of $\text{Sha}(A/K)(\ell)$.

Over number fields, it is not even known that $\text{Sha}(A/K)(\ell)$ vanishes for almost all primes $\ell$.

The takeaway from the Bhargava-Skinner-Zhang paper, and from the answers and comments here, is that knowledge of finiteness $\text{Sha}(A/K)(\ell)$ does not actually help much or at all to improve their progress on Super-weak BSD ($\text{Sha}$-finiteness enters through the parity conjecture, a theorem unconditionally, and definitely a much weaker statement than finiteness of $\text{Sha}(A/K)(\ell)$), which to me just means such methods fail to get to the point of the problem itself, and will not solve it.

In other words, I don't see any trace of the ability to produce interesting algebraic cycles on abelian varieties, into them, and according to the Expectation above, true in char $p$ though open in char $0$ so far, this should be the whole point.

Regardless, clearly Super-weak BSD, which is what the discussion in the question, comments, and around the Bhargava-Skinner-Zhang paper, were about, is not equivalent to weak BSD under any circumstances, no matter that $K = \mathbf{Q}$ and $A$ is $1$-dimensional.

Showing 100% of elliptic curves over $\mathbf{Q}$ satisfy the BSD conjecture does not show weak BSD (as the OP took care to make clear in his/her question). Such result, if ever available, should be regarded as being motivational only.

EDIT: I should also add that several of the averages (both unconditional and conjectural) that are key to the Bhargava-Skinner-Zhang methods, fail in char $p$, while I'd regard a method towards BSD to be "promising", if it were able to make progress or settle its char $p$ counterpart, first.

• Someone in my dept saw this post and said "Yess! He/she read my mind". – user97068 Mar 14 '18 at 17:16

For a prime $p$ and an elliptic curve $E/\mathbb{Q}$, we have the exact sequence

$$\displaystyle 0 \rightarrow E(\mathbb{Q})/p E(\mathbb{Q}) \rightarrow S_p(E) \rightarrow \text{Sha}_E[p]\rightarrow 0,$$

where $S_p(E)$ is the $p$-Selmer group of $E$ and $\text{Sha}_E$ is the Tate-Shafarevich group of $E$ and $\text{Sha}_E[p]$ is the $p$-part of it. This gives rise to the equation

$$\displaystyle r_p(S_p(E)) = r(E) + r_p(E(\mathbb{Q}[p]) + r_p (\text{Sha}_E[p]).$$

Here $E(\mathbb{Q})[p]$ is the $p$-torsion subgroup of $E$ (it is always trivial for $p > 11$ (by Mazur's theorem) and usually trivial for $p = 2,3,5,7,11$ (by a density argument)), $r_p$ denotes the $p$-rank of a finite group, and $r(E)$ denotes the Mordell-Weil rank of $E$. On average (by my earlier remark) $r_p(E(\mathbb{Q}[p])$ is zero, and therefore the average rank of $r(E)$ is equal to the average of $r_p(S_p(E)) - r_p(\text{Sha}_E[p])$.

Therefore, if you know the average of $\text{Sha}_p[E]$ for $p = 2,3,5$, then you would know the exact average of the Mordell-Weil rank, since the corresponding average for $p$-Selmer is already known due to work of Bhargava and Shankar. Of course, estimating the average size of Sha is not easier (as far as I know) than estimating the Mordell-Weil rank itself.

However, even if the average comes out to be $1/2$, as expected from conjectures of Goldfeld and Katz-Sarnak, it still doesn't follow that the 'obvious' distribution of half of all curves having rank 0 and the other half having rank 1 would hold. One would need to further control the density of curves having rank $\geq 2$. I am not sure if knowing the size of Sha helps with this question.

I should clarify the connection for those who are unfamiliar with the best known-results on BSD. Essentially, due to important work of many people, it is known that BSD holds for all elliptic curves of rank 0 and 1. Further, it is expected that when ordered by any 'reasonable' height, 50% of elliptic curves have rank 0 and the other half have rank 1. Thus, the latter conjecture implies that 100% of elliptic curves satisfy BSD.

Another remark is that if one assumes the parity conjecture (that is, half of all elliptic curves have odd rank and the other half have even rank, again with respect to a reasonable height), then the Goldfeld/Katz-Sarnak conjecture asserting that the average rank is $1/2$ will be enough to prove that 100% of elliptic curves satisfy BSD. To see this, note that the smallest possible contribution from odd rank curves to the average is $1/2$; which is equivalent to the assertion that 100% of odd rank curves have rank 1 and all larger ranks together occupy 0% of all odd rank curves. This forces 100% of all even rank curves to be rank 0, and then we can conclude that 100% of elliptic curves satisfy BSD.

• Shouldn't the first display have ${\rm Sha}_E[p]$, not ${\rm Sha}_E[2]$? – Noam D. Elkies Mar 14 '18 at 0:38
• @NoamD.Elkies you are right, thanks for spotting the error; I have fixed it. – Stanley Yao Xiao Mar 14 '18 at 0:39
• Knowing finiteness of the $\ell$-primary torsion of Sha tells you the Mordell-Weil rank equals the corank of the Selmer group. Does this help at all? In his BSD paper Bhargava averages on sets of elliptic curves that are assumed to have finite Sha. – user92332 Mar 14 '18 at 1:54
• $E(\mathbb{Q})[p]$ is never empty – Daniel Loughran Mar 14 '18 at 8:33
• Why do we know the average of $r_p(S_p(E))$? I thought that Bhargava-Shankar computed the average size of the $p$-Selmer group, not the average rank. – Will Sawin Mar 14 '18 at 16:00

Bhargava and Shankar have conjectured that the average size of the $n$-Selmer group $S_n(E)$ is the sum of the divisors of $n$. They proved this for $n \leq 5$. If you assume

• Equidistribution of root number,
• The parity conjecture, and
• The Bhargava-Shankar conjecture for infinitely many $n$,

then you can deduce that 50% of $E$ have rank 0 and 50% have rank 1. See Corollary 6 here.

Finiteness of Sha is therefore (potentially) helpful, since it implies the parity conjecture. This implication follows from the p-parity conjecture, which was proved by the Dokchitser brothers (and Nekovar as well, though I forget if his proof works for every $E$).

But assuming these three bullets still does not quite imply (weak) BSD for 100% of $E$, because it is not known that BSD holds for all elliptic curve of algebraic rank 0 or 1. BSD is known if analytic rank is 0 or 1, and there are "converse theorems" of Skinner and W. Zhang which give many cases where algebraic rank 0 or 1 implies analytic rank 0 or 1, but the current state of the art does not give this implication for every $E$ (as far as I'm aware). Indeed, the key hypothesis in these theorems is stronger than "algebraic rank is 0 or 1": you must assume $E$ has $p^\infty$-Selmer corank 0 or 1 for some large enough $p$.

If you assume the Bhargava-Shankar conjecture for a sequence of $n$ consisting of primes, then maybe you could use the current converse theorems to deduce BSD for 100% of $E$. But given the recent work of Alex Smith, it is perhaps more tantalizing to imagine proving Bhargava-Shankar for all $n$ which are powers of 2. For this sequence of $n$, it seems that the state-of-the-art converse theorems would not apply, since Skinner-Urban's proof of the Main Conjecture (which is the Iwasawa-theoretic input for the proof of converse theorems) does not allow $p = 2$.