6

The BSD conjecture is usually formulated like this.

If $E/\mathbf{Q}$ is an elliptic curve, then $$ \text{rank }E/\mathbf{Q} = \text{ord}_{s=1} L(E,s), $$ where $L(E,s)$ is the Hasse-Weil $L$-function of $E$.

But in Anthony Knapp's book Elliptic Curves (see Conjecture 1.9 on page 17), there is an another more "elementary" formulation of BSD, given like this:

If $E/\mathbf{Q}$ is an elliptic curve, then $$ \prod_{p<X} \dfrac{\#E(\mathbf{F}_p)}{p} \sim (\text{const}) (\log X)^r, $$ where $r = \text{rank }E/\mathbf{Q}$.

The main difference between the two versions of BSD is that the "analytic" sides are formulated differently. In the first version, the analytic quantity is $\text{ord}_{s=1}L(E,s)$. In the second version, the analytic quantity is the rate of growth of the product $\prod_{p<X} \dfrac{\#E(\mathbf{F}_p)}{p}$.

My question is: why are these two analytic quantities the same? That is, why is the exponent $r$ in the formula $$\prod_{p<X}{} \dfrac{\#E(\mathbf{F}_p)}{p} \sim (\text{const})(\log X)^r$$ actually equal to $\text{ord}_{s=1} L(E,s)$?

The heuristic reason for this is that if we formally evaluate $L(E,s)$ at $s=1$ via the Euler product definition, we get $$L(E,1) "=" \prod_{p<X} \dfrac{p}{\#E(\mathbf{F}_p)},$$ but this does not rigorously make sense because the Euler product definition only converges for $\text{Re }s> 3/2$. So is there a rigorous way to justify this phenomenon?

Adithya Chakravarthy
  • 2,582
  • 1
    I don’t know a reference, but I know the “elementary” version implies (given maybe modularity or at least analytic continuation) the “usual” version and much more (that the zeroes of the $L$-function are concentrated on the critical line). But it’s a theorem, so it’s not trivial. – Aphelli Jun 25 '22 at 08:14

2 Answers2

2

The $L$-function reformulation of BSD was given by Goldfeld here:

Goldfeld, Dorian. Sur les produits partiels eulériens attachés aux courbes elliptiques. (French) [On the partial Euler products associated with elliptic curves] C. R. Acad. Sci. Paris Sér. I Math. 294 (1982), no. 14, 471–474.

The proof is not too long, and it proceeds by showing the exponent $r$ in that product leads to a zero-free region for $L(E,s)$ of $Re(s) > 1+r$. Interestingly, the formal manipulation to relate the BSD product to $L(E,1)$ is off by a factor of $\sqrt 2$! See also

Conrad, Keith. Partial Euler products on the critical line. Canad. J. Math. 57 (2005), no. 2, 267–297.

for more about this issue.

Kimball
  • 3,197
0

The convergence can be established for $s=1$.

From https://en.wikipedia.org/wiki/Elliptic_curve#The_Birch_and_Swinnerton-Dyer_conjecture

we have

$$L(E(\mathbf{Q}), s) = \prod_{p\not\mid N} \left(1 - a_p p^{-s} + p^{1 - 2s}\right)^{-1} \cdot \prod_{p\mid N} \left(1 - a_p p^{-s}\right)^{-1}$$

Hasse Theorem gives $|\# E(K) - (q + 1) | \le 2\sqrt{q}$, hence $s>\frac32$.

However, a further argument (e.g. Sato-Tate conjecture) proves that the $a_p$ are evenly distributed about $0$, so that the function approximates the alternating zeta function instead of the zeta function.

JMP
  • 22,637