The Modular Degree

Since $A_f$ is an optimal quotient, the dual map $A_f^{\vee}\rightarrow J_0(N)$ is injective and the composite $\theta_f:A_f^{\vee} \rightarrow A_f$ has finite degree. It can be shown that $\theta_f$ arises from a polarization, so $\deg(\theta_f)$ is a perfect square. The modular degree of $A_f$ is the square root of the degree of $\theta_f$:

$$\moddeg(A_f) = \sqrt{\deg(\theta_f)}.$$

When $\dim A_f=1$, $\moddeg(A_f)$ is the usual modular degree, i.e., the degree of $X_0(N)\rightarrow A_f$.

If $M$ is an abelian group, let $M^* = \Hom_\mathbf{Z}(M,\mathbf{Z})$. The Hecke algebra acts in a natural way on $H_1(X_0(N),\mathbf{Z})$ and $H_1(X_0(N),\mathbf{Z})^*$, and we have a natural restriction map

$$r_f: H_1(X_0(N),\mathbf{Z})^*[I_f] \rightarrow (H_1(X_0(N),\mathbf{Z})[I_f])^*.$$

The following proposition leads to an algorithm for computing the modular degree.

Proposition 3.1   $\coker(r_f)\cong \ker(\theta_f)$, so $\moddeg(A_f) = \sqrt{\char93 \coker(r_f)}$.

The proposition is proved in [KS00]. The proof makes use of the Abel-Jacobi theorem, which realizes the Jacobian $J_0(N)_\mathbf{C}$ as a complex torus:

$$0\rightarrow H_1(X_0(N),\mathbf{Z}) \rightarrow \Hom(S_2(\Gamma_0(N)),\mathbf{C}) \rightarrow J_0(N)(\mathbf{C}) \rightarrow 0,$$

where $H_1(X_0(N),\mathbf{Z})$ is embedded as a lattice of full rank in the complex vector space $\Hom(S_2(\Gamma_0(N)),\mathbf{C})$ using the integration pairing, and this description of $J_0(N)(\mathbf{C})$ is compatible with the action of Hecke operators.