## Upper bounds for ranks of modular jacobians

The following question came to me earlier as a "side question"; something I'd like to know, but which is not totally necessary for what I'm thinking about or doing:

Consider the genus 32 curve $X_0(389)$, and denote its Jacobian variety as $J_0(389)$.

I am interested in finding an upper bound for the Mordell-Weil rank of $J_0(389)(\mathbb{Q}(i))$.

After thinking about this for some time, I turned to Google, which threw up [1]. Apparently, assuming the Birch-Swinnerton-Dyer conjecture, there is an absolute constant $C > 0$ such that for all primes $q$ sufficiently large, we have

$\mbox{rank } J_0(q)(\mathbb{Q}) \leq C \mbox{ dim} J_0(q)$.

[Ideally this equation would be in the center]

The point of that paper is to show that $C = 6.5$ will do (existence of $C$ having been proved in an earlier paper by the same authors), but, "assuming [also] the Riemann Hypothesis for automorphic $L$-functions, Iwaniec, Luo and Sarnak have recently proved that one could take $C = \frac{99}{100}$".

Now I reckon 389 is "sufficiently large", which means (if you believe those conjectures) an upper bound for the rank over $\mathbb{Q}$ is 31. But this feels like it is way too big, maybe because there are generally so few rational points on modular curves.

Does anyone know how to get a better upper bound? Or am I wrong in hoping for a smaller upper bound?

Furthermore, since I'm interested in $\mathbb{Q}(i)$-rank, there is the following:

What is the biggest the rank can jump by when going from $J_0(389)(\mathbb{Q})$ to $J_0(389)(\mathbb{Q}(i))$?

I guess this last question can be asked in greater generality, replacing the Js with any abelian variety $A/\mathbb{Q}$. Can the rank jump be arbitrarily large when passing from $\mathbb{Q}$ to $\mathbb{Q}(i)$? Or is there a bound in terms of the dimension of $A$, say?

It's my bedtime now, so I'll pick this thread up in 9 or so hours.

[1]: "Explicit Upper Bound for the (Analytic) rank of $J_0(q)$". E. Kowalski, P. Michel. Israel J. Math, 2000. Preprint Available here

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The rank of J0(389) over Q is 13. The Jacobian splits into factors of dimensions 1,2,3,6,20. Combining explicit computation with Kolyvagin-Gross-zagier implies that the factors of dim 20 has rank 0. The ones of dim 2,3,6 have ranks 2,3,6. The one of dim 1 has rank 2. (this is all from memory, so....) Similar techniques could be applied to get the quadratic twist by -1 of everything...

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That's an impressive memory! +1 – David Hansen Feb 4 2011 at 5:05
Thanks William. Is there a reference for this combination of explicit computation with Kolyvagin-Gross-Zagier that I can read? – Barinder Banwait Feb 4 2011 at 8:04
Barinder: the "explicit computation" is probably this: you compute the Jacobian in a concrete way via modular symbols, and then you explicitly compute the biggest quotient where the $L$-function doesn't vanish at 1 (the point being that you can compute $L$-function by integrating from 0 to infinity and this is easy with modular symbols), and now by Kolyvagin-Gross-Zagier analytic rank 0 implies algebraic rank zero. – Kevin Buzzard Feb 4 2011 at 20:04
Kevin's exactly right. Here's a Sage worksheet that shows how to do these computations in Sage: nt.sagenb.org/home/pub/31 The quadratic twist by chi of conductor 4 switches the sign in the functional equation, so the ranks of the twists are very likely 1,0,0,0,21. So it's very likely the rank of J0(389)(Q(i)) is 21 + 13 = 34. – William Stein Feb 4 2011 at 20:19

Regarding the last question, if $A/\mathbf{Q}$ is an abelian variety and $K/\mathbf{Q}$ is a quadratic extension, then the rank of $A(K)$ is the sum of the ranks of $A(\mathbf{Q})$ and $A^\chi(\mathbf{Q})$, where $A^\chi/\mathbf{Q}$ is the quadratic twist of $A$ associated to the quadratic extension $K/\mathbf{Q}$. For example, if $A$ is the elliptic curve $y^2=x^3+Ax+B$ and $K=\mathbf{Q}(\sqrt{D})$, then $A^\chi$ is the curve $y^2=x^3+D^2Ax+D^3B$. I see no reason why the difference $\hbox{rank} (A^\chi(\mathbf{Q}))-\hbox{rank} (A(\mathbf{Q}))$ couldn't become arbitrarily large if you fix $K$ and vary $A$ over (say) all elliptic curves defined over $\mathbf{Q}$. However, since we don't know how to create elliptic curves of arbitrarily large rank over $\mathbf{Q}$, there's no way to say anything definitive.

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 Here is an interesting experiment based on Silverman's answer to try: take Elkies rank 28 curve, compute its quadratic twist by Q[i] and compute its rank. It would be nice if it has rank 0! I tried this with GP (to compute the L-series to see if it vanishes) but the computed has been running for a day so far with no luck. – A. Pacetti Feb 11 2011 at 19:02

To partly answer your last two questions: It's not too hard to write down a sequence of abelian varieties $A_i/\mathbf{Q}$ such that $\mathrm{rank}A_i(\mathbf{Q})=0$ but $\mathrm{rank}A_i(\mathbf{Q}(\sqrt{-1}))\to \infty$ as $i\to\infty$. More precisely, if $p\equiv 1\ \mathrm{mod}\ 4$ is large enough, there is some newform $f$ of weight $2$ on $\Gamma_0(p^3)$ such that

a. $L(s,f)$ has root number $+1$, and $L(1/2,f)\neq0$, which in fact implies $L(1/2,f^{\sigma})\neq0$ for all Galois conjugates of $f$ by some work of Shimura;

b. $L(s,f\otimes \chi_{-4})$ has root number $-1$ (which is automatic from a. and my choice of $p$) and $L'(1/2,f \otimes \chi_{-4})\neq 0$, which again implies $L'(1/2,f^{\sigma} \otimes \chi_{-4})\neq 0$ for all Galois conjugates of $f$ by the Gross-Zagier formula and its generalizations.

The existence of such a newform follows from the asymptotic evaluation

$\sum_{f\in S_2^{new}(\Gamma_0(p^3)),\ \varepsilon(f)=+1}L(1/2,f)L'(1/2,f\otimes \chi_{-4}) \sim cp^3$

for some real $c>0$, which nowadays is a standard application of the Petersson formula (see e.g. the final chapter of Iwaniec and Kowalski's book). Given such an $f$, let $A_f$ be the corresponding optimal quotient of $J_0(p^3)$. Then a. guarantees that $\mathrm{rank}A_f(\mathbf{Q})=0$ by the work of Kolyvagin-Logachev/Zhang/Longo/Tian-Zhang, while b. guarantees that $\mathrm{rank}A_f(\mathbf{Q}(\sqrt{-1}))=\mathrm{dim}A_f$ by the same group of people, and $\mathrm{dim}A_f \geq \frac{p-1}{2}$ (see e.g. this question), so we're done.

This whole argument works with $\mathbf{Q}(\sqrt{-1})$ replaced by any fixed imaginary quadratic field. If you demand the same dichotomy for a sequence of $A$'s of bounded dimension, I imagine it's not known.

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