Average rank of elliptic curve twists

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Average rank of elliptic curve twists

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Recursos

Suppose someone tells you a family of quadratic twists has rank 0 for half its members and rank 1 for the other half. Can you conclude that the average rank is 1/2? This class gives you the check that answers that question, and walks through how one paper carries it out for every elliptic curve over ℚ.

Count is not mass

Twist E by every signed squarefree d with |d| ≤ Y. A companion theorem shows that ranks 0 and 1 each have density 1/2 [0:10]. That still leaves room for a problem: a set of density zero can carry a lot of rank mass, meaning count times rank. Take Y/log Y twists, each of rank about log Y. That set has density zero and breaks no pointwise Jensen bound, yet its mass is about Y, which would push the mean up by a constant [0:47].

The tail bound and the order of limits

The paper's main step is Theorem 1.2: the rank mass above R, divided by Y, is eventually at most C_E/R [1:36]. With that in hand:

fix R first, so ranks 2 through R contribute at most R·o(Y);
let Y → ∞;
then let R → ∞. The excess goes to zero and the mean is 1/2.

The proof of the tail bound never uses the density theorem. The densities come in only at this last step.

In the think-first moment [2:27], counts of size T(k) ≤ CX/k² add up to a mass of about R·T(R) + Σ_{k≥R} T(k) ≍ X/R. Above K = (log X)^{3/5}, the pointwise bound a ≪ log X takes over and costs only X(log X)^{-1/5}.

Why high ranks are rare

The completed L-function g_d satisfies g_d(−z) = ε_d g_d(z). Moving a contour across the pole at 0 gives an exact weighted Dirichlet series S_d. If a > k and ε_d = (−1)^k, then S_d = 0 [3:20]. Next, split S_d into pieces [4:11]. A mollifier makes the first piece close to a positive main term. Logarithmic second moments (Poisson summation plus prime-square inflation) make the other pieces small for most d. A sum like that cannot be zero, so the exceptions number O(X/k²).

Your test for any family average [4:54]: before trusting the densities, show that the rank mass above R is O(1/R), and only then let R grow.

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