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Math-2784 · mathematics · Number theory

Mean analytic rank of quadratic twists over Q

Status OpenLLM status Open unverified

Statement

Definitions and mathematical statement

Statement recorded in the supplied source.
For every elliptic curve E over Q, does the mean of the analytic ranks of $E^{d}$, over nonzero signed squarefree d with $|d|<=X$, tend to one half as X tends to infinity?

Short English statement

For every elliptic curve E over Q, does the mean of the analytic ranks of $E^{d}$, over nonzero signed squarefree d with $|d|<=X$, tend to one half as X tends to infinity?

Sources

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-2784-A1 · Literature report · LLM-generated · unknown ·

OpenAI claim summary for Math-2784

The result supplies the conclusion in this problem with its displayed hypotheses and quantifiers. The broader statements and applications in the manuscript are kept separate from the particular question above. The supporting result is manuscript 1 (family 006; 2026-09-23).

Mean analytic rank of quadratic twists over Q

OpenAI's claim

OpenAI presents the following claims relevant to this question, with manuscript numbers following paper order in Sources. In manuscript 1 (family 006; 2026-09-23), the asserted result is that the mean analytic rank of signed squarefree quadratic twists tends to $1/2$ for each rational elliptic curve.

Mathematical statement of the claimed result

Manuscript 1.
Let $E/\mathbb Q$ be any elliptic curve, let $E^{(d)}$ denote its quadratic twist, and put $a(E^{(d)})=\operatorname{ord}_{s=1}L(E^{(d)},s)$. For the signed squarefree twisting parameters $\mathcal D(Y)=\{d\in\mathbb Z\setminus\{0\}:d\text{ squarefree},\ |d|\le Y\}$, the claimed limit is \[ \lim_{Y\to\infty}\frac1{|\mathcal D(Y)|} \sum_{d\in\mathcal D(Y)}a(E^{(d)})=\frac12. \] This is an ordinary average over both signs for each fixed curve, without a restriction on its conductor or rational torsion.

Outline of the argument

For manuscript 1 (family 006; 2026-09-23), block mollifiers compare twist families with a finite independent model. Poisson summation and an auxiliary square multiplier send dual twists to smaller height, allowing induction on logarithmic second moments and control of the contribution from large ranks.

Formal verification

No Lean formalization is listed at this snapshot for manuscript 1 (family 006; 2026-09-23).

This site has not executed the Lean code. Described statement coverage is not an independent proof check.

Status caveat

The manuscripts were produced by an internal OpenAI model. OpenAI cautions that results without formalization can contain mistakes. This site has not independently checked the proofs; the label records the scope of OpenAI's claim.

Sources

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