Upper bounds for ranks of modular jacobians - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T22:44:19Z http://mathoverflow.net/feeds/question/54255 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54255/upper-bounds-for-ranks-of-modular-jacobians Upper bounds for ranks of modular jacobians Barinder Banwait 2011-02-03T22:53:04Z 2011-03-24T04:08:23Z <p>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:</p> <p>Consider the genus 32 curve $X_0(389)$, and denote its Jacobian variety as $J_0(389)$.</p> <p>I am interested in finding an upper bound for the Mordell-Weil rank of $J_0(389)(\mathbb{Q}(i))$. </p> <p>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</p> <p>$\mbox{rank } J_0(q)(\mathbb{Q}) \leq C \mbox{ dim} J_0(q)$. </p> <p>[Ideally this equation would be in the center]</p> <p>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 $C = \frac{99}{100}$". </p> <p>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.</p> <blockquote> <p>Does anyone know how to get a better upper bound? Or am I wrong in hoping for a smaller upper bound?</p> </blockquote> <p>Furthermore, since I'm interested in $\mathbb{Q}(i)$-rank, there is the following:</p> <blockquote> <p>What is the biggest the rank can jump by when going from $J_0(389)(\mathbb{Q})$ to $J_0(389)(\mathbb{Q}(i))$?</p> </blockquote> <p>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?</p> <p>It's my bedtime now, so I'll pick this thread up in 9 or so hours. </p> <p>[1]: "Explicit Upper Bound for the (Analytic) rank of $J_0(q)$". E. Kowalski, P. Michel. Israel J. Math, 2000. <a href="http://www.math.u-bordeaux1.fr/~kowalski/explicit-rank.pdf" rel="nofollow">Preprint Available here</a></p> http://mathoverflow.net/questions/54255/upper-bounds-for-ranks-of-modular-jacobians/54262#54262 Answer by Joe Silverman for Upper bounds for ranks of modular jacobians Joe Silverman 2011-02-03T23:36:33Z 2011-02-03T23:36:33Z <p>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 <code>$\hbox{rank} (A^\chi(\mathbf{Q}))-\hbox{rank} (A(\mathbf{Q}))$</code> 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.</p> http://mathoverflow.net/questions/54255/upper-bounds-for-ranks-of-modular-jacobians/54273#54273 Answer by William Stein for Upper bounds for ranks of modular jacobians William Stein 2011-02-04T00:20:07Z 2011-02-04T00:20:07Z <p>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...</p> http://mathoverflow.net/questions/54255/upper-bounds-for-ranks-of-modular-jacobians/59369#59369 Answer by David Hansen for Upper bounds for ranks of modular jacobians David Hansen 2011-03-23T23:27:57Z 2011-03-24T04:08:23Z <p>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</p> <p><strong>a.</strong> $L(s,f)$ 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;</p> <p><strong>b.</strong> $L(s,f\otimes \chi_{-4})$ 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.</p> <p>The existence of such a newform follows from the asymptotic evaluation </p> <p>$\sum_{f\in S_2^{new}(\Gamma_0(p^3)),\ \varepsilon(f)=+1}L(1/2,f)L'(1/2,f\otimes \chi_{-4}) \sim cp^3$ </p> <p>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 <strong>a.</strong> guarantees that $\mathrm{rank}A_f(\mathbf{Q})=0$ by the work of Kolyvagin-Logachev/Zhang/Longo/Tian-Zhang, while <strong>b.</strong> 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. <a href="http://mathoverflow.net/questions/24923/galois-orbits-of-newforms-with-prime-power-level" rel="nofollow">this question</a>), so we're done. </p> <p>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.</p>