# The rank of elliptic curves and related quadratic twists

Let $$E/\mathbb{Q}$$ be an elliptic curve, and let $$k_1, k_2$$ be square-free integers. Can anything be said about the related elliptic curves

$$\displaystyle E/\mathbb{Q}, E^{(k_1)}/\mathbb{Q}, E^{(k_2)}/\mathbb{Q}, E^{(k_1 k_2)}/\mathbb{Q}?$$

Here $$E^{(d)}$$ is the quadratic twist of $$E$$ by $$d$$. If $$E$$ has the short Weierstrass equation

$$\displaystyle E: y^2 = x^3 + Ax + B, A, B \in \mathbb{Z}$$

then $$E^{(d)}$$ has the short Weierstrass equation

$$\displaystyle E^{(d)}: dy^2 = x^3 + Ax + B.$$

In particular, let $$r(E)$$ denote the algebraic rank of $$E$$ and $$r_{\text{an}}(E)$$ denote the analytic rank. Are there any relations between the integers

$$\displaystyle r(E), r(E^{(k_1)}), r(E^{(k_2)}), r(E^{(k_1 k_2)})$$

or

$$\displaystyle r_{\text{an}}(E), r_{\text{an}}(E^{(k_1)}), r_{\text{an}}(E^{(k_2)}), r_{\text{an}}(E^{(k_1 k_2)})?$$

Edit:

Observe that if we write

$$\displaystyle L(E,s) = \prod_p \left(1 - \frac{\alpha_p}{p^s} \right) \left(1 - \frac{\beta_p}{p^s} \right)$$

for the Hasse-Weil $$L$$-function of $$E$$, then we have

$$\displaystyle L(E,s) L(E^{(k_1 k_2)},s) = \prod_p \left(1 - \frac{\alpha_p}{p^s} \right) \left(1 - \frac{\chi_{k_1 k_2}(p) \alpha_p}{p^s} \right) \left(1 - \frac{\beta_p}{p^s} \right) \left(1 - \frac{\chi_{k_1 k_2}(p) \beta_p}{p^s} \right)$$

and

$$\displaystyle L(E^{(k_1)}, s) L(E^{(k_2)}, s) = \prod_p \left(1 - \frac{\chi_{k_1}(p) \alpha_p}{p^s} \right)\left(1 - \frac{\chi_{k_2}(p) \alpha_p}{p^s} \right)\left(1 - \frac{\chi_{k_1}(p) \beta_p}{p^s} \right)\left(1 - \frac{\chi_{k_2}(p) \beta_p}{p^s} \right).$$

In particular, their Euler factor at $$p$$ coincides whenever $$\chi_{k_1 k_2}(p) = -1$$, and whenever $$\chi_{k_1 k_2}(p) = \chi_{k_1}(p) = \chi_{k_2}(p) = 1$$. Therefore, the Euler factors match for $$3/4$$'s of primes $$p$$.

• $k_1$ and $k_2$ could be prime numbers, say $p$ and $q$, so then you are asking for relations between $E^{(p)}$, $E^{(q)}$ and just $E$, right? Commented Aug 12 at 13:19
• @AriyanJavanpeykar and also the twist $E^{(pq)}$. In particular, one might hope to say something about a relation between the pairs $\{r(E), r(E^{(pq)})\}, \{r(E^{(p)}), r(E^{(q)})\}$. Commented Aug 12 at 13:20
• Apologies, I misread the (k_1k_2), I thought it was the gcd $(k_1,k_2)$. Commented Aug 12 at 13:21

You are looking at the biquadratic extension $$K:=\mathbb Q(\sqrt{k_1},\sqrt{k_2})$$ of $$\mathbb Q$$ and want to relate properties of $$E(F)$$ where $$F$$ is $$\mathbb Q$$ or one of the quadratic subfields $$F_1,F_2,F_3$$ of $$K$$. If you're also willing to throw $$K$$ into the mix, then you can use the (unique) idempotent relation on $$\text{Gal}(K/\mathbb Q)$$ to get a relation on the ranks of $$E(K)$$, $$E(F_1)$$, $$E(F_2)$$, $$E(F_3)$$, and $$E(\mathbb Q)$$. This is part of a much larger theory. See for example Kani, Ernst; Rosen, Michael, Idempotent relations among arithmetic invariants attached to number fields and algebraic varieties, J. Number Theory 46, No. 2, 230-254 (1994). ZBL0853.14011.

Let $$E, E^{(k)}$$ be defined over a number field $$K$$ and let $$k \in K^\times \setminus K^2$$. Then it is well-known that $$\operatorname{rk}(E/K(\sqrt d)) = \operatorname{rk}(E/K) + \operatorname{rk}(E^{(k)}/K).$$ Your question is related to the parity conjecture, which asserts that $$w(E/K) = (-1)^{\operatorname{rk}(E/K)},$$ where $$w(E/K)$$ is the root number of $$E$$ over $$K$$. The root number of an elliptic curve can be computed relatively easily. The following two results are taken from this survey paper and might be of interest to you.

Let $$E$$ be an elliptic curve over $$\mathbb Q$$. Then the function $$k \mapsto w(E^{(k)}/\mathbb Q)$$ defined on squarefree integers is periodic, with period depending on the conductor of $$E$$. $$50\%$$ of quadratic twists of $$E$$ have root number $$1$$ and $$50\%$$ of quadratic twists of $$E$$ have root number $$-1$$.

Here is some experimental data.

For positive integer $$k$$ let $$E_k: y^2=x^3+k x$$ and $$k_1=2,k_2=3$$.

According to computations with sage, for $$0 < k < 2000$$:

1. At least one of $$\displaystyle r_{\text{an}}(E_k), r_{\text{an}}(E_k^{(2)}), r_{\text{an}}(E_k^{(3)}), r_{\text{an}}(E_k^{(6)})$$ is positive.

2. At least one of $$\displaystyle r_{\text{an}}(E_k), r_{\text{an}}(E_k^{(2)}), r_{\text{an}}(E_k^{(3)}), r_{\text{an}}(E_k^{(6)})$$ is odd.

3. For $$0 < k < 10^5$$ at least one of the root numbers of $$\displaystyle (E_k), (E_k^{(2)}), (E_k^{(3)}), (E_k^{(6)})$$ is $$-1$$.

Here is sage code

#Author Georgi Guninski, Mon Aug 12 04:42:25 PM UTC 2024
lim=100 #upper bound for k
for k in range(2,lim):
Ei=[EllipticCurve([k*m^2,0]) for m in [1,2,3,6]]
ra=[E.analytic_rank() for E in Ei]
rap=[i%2 for i in ra]
print(k,ra,rap)
assert any(ra) and any(rap)
print("end")

• Confirmed for $0 < k < 10^4$, which automatically gives the same result for all nonzero $k$ with $|k| < 10^4$ because for these curves the $d$ and $-2d$ twists are isogenous. Moreover, in each case at least one of the four ranks is even. But it's not just a parity obstruction: each of 1, 2, or 3 odd-rank twists arises about 1/3 of the time. Commented Aug 12 at 21:35
• @NoamD.Elkies Many thanks for the verification. If someone else is verifying, computing only the root number is significantly faster than computing the analytic (or algebraic) rank.
– joro
Commented Aug 13 at 6:30
• @NoamD.Elkies I asked stronger result conjecture: mathoverflow.net/questions/476863/…
– joro
Commented Aug 13 at 14:59