The question in the title naturally breaks up in two parts, namely the torsion part and the rank part. I already read about some results on both the torsion and the rank part. And I want to know whether this is currently still the state of the art, or whether there has been some recent progress in these questions.

First the torsion part, as far as I know the only results on the torsion parts are for the case $N$ is a prime, and for prime $N$ the results are quite good.

The structure of $J_0(N)(\mathbb Q)_{tors}$ is determined by Bary Mazur in his 'Modular Curves and the Eisenstein Ideal' paper. For $J_1(N)(\mathbb Q)_{tors}$ there exists a conjecture by Conrad, Edixhoven and Stein which states that it can be generated by the cusps. This conjecture was recently proved (up to two torsion) by Ohta in his 'Eisenstein ideals and the rational torsion subgroups of modular Jacobian varieties' paper.

Are there any results on the structure of $J_0(N)(\mathbb Q)_{tors}$ or $J_1(N)(\mathbb Q)_{tors}$ for composite N?

Now for the rank part I am only interested in for which $N$ it is $0$. Barry Mazur already proved in the same paper as mentioned earlier that for $N = 37, 43, 54, 61, 67$ and all primes $N \geq 73$ one has that $J_0(N)$ has positive rank. Now this makes me suspect that there will be only finitely many $N$ such that $J_0(N)$ has rank zero.

Are there only finitely many $N$ such that the rank of $J_0(N)(\mathbb Q)$ is $0$, and if so is this list known?

I have the feeling that I can answer the rank question myself by doing some computations, but I would much rather give a reference to the literature if it exists.


1 Answer 1


For torsion subgroups, one can consider the "rational cuspidal subgroup", which is the subgroup of degree zero divisors generated by $\mathbb{Q}$-rational divisors coming from cusps. (The cusps themselves need not be $\mathbb{Q}$-rational, but linear combinations of them can be.) This gives a way of constructing lots of torsion points on $J_{0}(\mathbb{Q})$ and $J_{1}(\mathbb{Q})$, but it seems little is known (when the level isn't prime) about whether this gives the entire torsion subgroup.

For the case of $J_{0}(N)$, one can compute the structure of the rational cuspidal subgroup from the structure of level-$N$ eta-quotients. It appears that an implementation of this algorithm should be functional in Sage sometime soon. Also, there are some theoretical results too. (There's a paper by Toshikazu Takagi that handles squarefree $N$, a paper by San Ling that handles the case of $N = p^{r}$ for $p$ prime, $p > 2$, and I wrote a paper with my postdoc John Webb that deals with the case that $N$ is a power of $2$.)

About the rank: The $L$-function of $J_{0}(N)$ factors as a product of $L(f_{i},s)$ where the $f_{i}$ are newforms of weight $2$ and level dividing $d$ for some divisor $d$ of $N$ (and occurring more than once if $d < N$). A newform $f_{i}$ is fixed by the Fricke involution $W_{N}$ if and only if the sign of the functional equation for $f_{i}$ is $-1$. The work of Gross and Zagier shows that the abelian variety associated to $f_{i}$ has rank one if $L(f_{i},1) = 0$ and $L'(f_{i},1) \ne 0$. If $J_{0}(N)(\mathbb{Q})$ has rank zero, then the same is true of $J_{0}(d)(\mathbb{Q})$ for all $d | N$, and this implies that all primes factors of $N$ are $\leq 71$. A finite (but somewhat lengthy computation) should produce all $N$ so that $J_{0}(N)(\mathbb{Q})$ has rank zero.

  • $\begingroup$ Thanks for the answer, yes I am aware of the Manin Drinfeld theorem that cusps are torsion. The three papers on the structure of the cuspidal group are new to me and seem interesting. Something along the lines of the lengthy computation that you describe was indeed what I had in mind when I said that I could probably tackle that question on my own. But I didn't want to waste my time on this if not necessary. I already have some sage code that computes the structure of the cuspidal class group for modular quotients of X_1(N) using modular symbols at github.com/koffie/mdsage. $\endgroup$ Commented May 19, 2014 at 19:47
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    $\begingroup$ Combining Jeremy's last paragraph with mathoverflow.net/a/66016/2, the complete list is [ 11, 14, 15, 17, 19, 20, 21, 23, 24, 26, 27, 29, 31, 32, 35, 36, 39, 41, 47, 49, 50, 59, 71 ] $\endgroup$ Commented Apr 8, 2016 at 20:53

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