Bounding the Torsion From Below

A cusp $\alpha \in{} \Gamma_0(N)\backslash\P ^1(\mathbf{Q}) \subset X_0(N)$ defines a point $(\alpha)-(\infty) \in J_0(N)(\overline{\mathbf{Q}})_{\tor}$. The rational cuspidal subgroup $C$ of $J_0(N)(\mathbf{Q})_{\tor}$ generated by $\mathbf{Q}$-rational cusps is of interest because the order of the image of $C$ in $A_f(\mathbf{Q})_{\tor}$ provides a lower bound on $\char93 A_f(\mathbf{Q})_{\tor}$. Stevens [Ste82, §1.3] computed the action of $\Gal(\overline{\mathbf{Q}}/\mathbf{Q})$ on the subgroup of $J_0(N)(\overline{\mathbf{Q}})$ generated by all cusps (and for other congruence subgroups besides $\Gamma_0(N)$). He found that $\Gal(\overline{\mathbf{Q}}/\mathbf{Q})$ acts on the cusps through $\Gal(\mathbf{Q}(\zeta_N)/\mathbf{Q})\cong (\mathbf{Z}/N\mathbf{Z})^*$, and that $d\in(\mathbf{Z}/N\mathbf{Z})^*$ acts by $x/y\mapsto x/(d'y)$, where $dd'\equiv 1\pmod{N}$. Thus, e.g., $(0)-(\infty)\in J_0(N)(\mathbf{Q})_{\tor}$, and if $N$ is square-free then all cusps are rational.

To compute the image of $C$ in $A_f(\mathbf{Q})_{\tor}$, first make a list of inequivalent cusps using, e.g., the method described in [Cre97, §2.2, pg. 17]. Keep only the $\mathbf{Q}$-rational cusps, which can be determined using the result of Stevens above and [Cre97, Prop. 2.2.3] (when $N$ is squarefree all cusps are rational). Next compute the subgroup  $\mathcal{C}$ of $\mathfrak{M}_2(\Gamma_0(N))$ generated by modular symbols $\{\alpha,\infty\}$, where $\alpha$ is a $\mathbf{Q}$-rational cusp. The image of $C$ in $A_f(\mathbf{Q})_{\tor}$ is isomorphic to the image of  $\mathcal{C}$ in

$$P = \Phi_f(\mathfrak{M}_2(\Gamma_0(N)))/\Phi_f(\mathfrak{S}_2(\Gamma_0(N))),$$

where $\Phi_f : \mathfrak{M}_2(\Gamma_0(N))\rightarrow \Hom(S_2(\Gamma_0(N))[I_f],\mathbf{C})$ is defined by the integration pairing. To keep everything rational, note that $P$ can be computed using any map with the same kernel as $\Phi_f$; a better map can be constructed by finding a basis for $\Hom(\mathfrak{M}_2(\Gamma_0(N)),\mathbf{Q})[I_f]$ (see the end of Section 4.2).

Example 3.6   Let the notation be as in Example 3.4. The cusps on $X_0(39)$ are represented by 0, $\infty$, $-1/9$, and $-4/13$, and since $N=39$ is squarefree, these cusps are all rational. Using MAGMA we find that the image of $C$ in $A_f(\mathbf{Q})_{\tor}$ is isomorphic to $\mathbf{Z}/14\mathbf{Z}$. Thus $A_f(\mathbf{Q})_{\tor}$ is isomorphic to one of $\mathbf{Z}/14\mathbf{Z}$, $\mathbf{Z}/28\mathbf{Z}$, or $\mathbf{Z}/14\mathbf{Z}\times \mathbf{Z}/2\mathbf{Z}$, but we do not know which.

Example 3.7   Let

$$f= q + \frac{1+\sqrt{5}}{2}q^2 + \frac{1-\sqrt{5}}{2}q^3 + \frac{5+\sqrt{5}}{2}q^4 + \cdots \in S_2(\Gamma_0(175))$$

be the form 175D. The cusps of $X_0(175)$ are represented by

$$0, \infty, \frac{1}{25}, \frac{1}{28}, \frac{1}{30}, \... ...frac{1}{60}, \frac{1}{65}, \frac{1}{70}, \frac{1}{105}, \frac{1}{140}.$$

The $\mathbf{Q}$-rational cusps in this list are $0, \infty, \frac{1}{25}, \frac{1}{28}$, and these generate a subgroup of $A_f(\mathbf{Q})_{\tor}$ of order $2$. (Incidentally, the group generated by all cusps, both rational and not, is isomorphic to $\mathbf{Z}/32\mathbf{Z}$.) Using $a_p$ for $p\leq 17$ and the method of the previous section, we see that $\char93 A_f(\mathbf{Q})_{\tor} \mid 4$. The authors do not know if the cardinality is $2$ or $4$.

Example 3.8   The form 209C is

$$f = q + \alpha{} q^2 + (1/2\alpha^4 - \alpha^3 - 5/2\alpha^2 + 4\alpha + 1)q^3 + (\alpha^2 - 2)q^4 + \cdots,$$

where $\alpha^5 - 2\alpha^4 - 6\alpha^3 + 10\alpha^2 + 5\alpha - 4=0$. As above, we find that $\char93 A_f(\mathbf{Q})_{\tor}$ divides $5$. The image of the (rational) cuspidal subgroup in $A_f(\overline{\mathbf{Q}})_{\tor}$ is isomorphic to $\mathbf{Z}/5\mathbf{Z}$, so $A_f(\mathbf{Q})_{\tor}\approx \mathbf{Z}/5\mathbf{Z}$.