Modular Symbol and Measures

Let

$$f_E(z) = \sum_{n=1}^{\infty} a_n e^{2\pi i n z} \in S_2(\Gamma_0(N))$$

be the modular form associated to $E$, which is a holomorphic function on the extended upper half plane $\mathfrak{h}\cup \mathbb{Q}\cup \{\infty\}$. Let

$$\Omega_E = \int_{E(\mathbb{R})} \frac{dx}{2y + \underline{a}_1 x + \underline{a}_3} \in \mathbb{R}$$

be the real period associated to a minimal Weierstrass equation

$$y^2 + \underline{a}_1 xy + \underline{a}_3 y = x^3 +\underline{a}_2 x^2 + \underline{a}_4 x + \underline{a}_6$$

for $E$.

The plus modular symbol map associated to the elliptic curve $E$ is the map $\mathbb{Q}\to \mathbb{Q}$ given by sending $r\in \mathbb{Q}$ to

$$[r]= [r]_E= \frac{2 \pi i }{\Omega_E} \left( \int_r^{i\infty} f_E(z) dz + \int_{-r}^{i\infty} f_E(z) dz \right).$$

Question 1.13   Let $E$ vary over all elliptic curve over $\mathbb{Q}$ and $r$ over all rational numbers. Is the set of denominators of the rational numbers $[r]_E$ bounded? Thoughts: For a given curve $E$, the denominators are bounded by the order of the image in $E(\overline{\mathbb{Q}})$ of the cuspidal subgroup of $J_0(N)(\overline{\mathbb{Q}})$. It is likely one can show that if a prime $\ell$ divides the order of the image of this subgroup, then $E$ admits a rational $\ell$-isogeny. Mazur's theorem would then prove that the set of such $\ell$ is bounded, which would imply a ``yes'' answer to this question. Also, for any particular curve $E$, one can compute the cuspidal subgroup precisely, and hence bound the denominators of $[r]_E$.

Let $a_p$ be the $p$th Fourier coefficient of $E$ and note that the polynomial

$$x^2 - a_p x + p \equiv x(x-a_p) \pmod{p}$$

has distinct roots because $p$ is an ordinary prime. Let $\alpha$ be the root of $x^2-a_px+p$ with $\vert\alpha\vert _p = 1$, i.e., the lift of the root $a_p$ modulo $p$, which exists by Lemma 1.12.

Define a measure on $\mathbb{Z}_p^*$ by

$$\mu_E(a+p^n \mathbb{Z}_p) = \frac{1}{\alpha^n} \left[\frac{... ...right] -\frac{1}{\alpha^{n+1}} \left[\frac{a}{p^{n-1}}\right].$$

That $\mu_E$ is a measure follows from the formula for the action of Hecke operators on modular symbols and that $f_E$ is a Hecke eigenform. We will not prove this here.