A Formula for $L(A,1)/\Omega _A$

Recall that if $L$ and $M$ are lattices in a real vector space $V$, then the lattice index $[L:M]$ is the absolute value of the determinant of a linear transformation of $V$ taking $L$ onto $M$. The lattice index satisfies the usual properties suggested by the notation, e.g., $[L:M]\cdot [M:N] = [L:N]$.

The canonical real volume of  $A(\mathbf{R})$ is defined as follows. If $\Lambda^*$ is a lattice in the cotangent space

$$T^*=H^0(A_\mathbf{R},\Omega_{A_\mathbf{R}})=S_2(\Gamma_0(N),\mathbf{R})[I_f]$$

of $A_\mathbf{R}$, then $\Lambda^*$ determines a lattice $\Lambda=\Hom(\Lambda^*,\mathbf{Z})$ in the tangent space $T=\Hom(T^*,\mathbf{R})$, and hence a measure on $T$ by declaring that $T/\Lambda$ has measure $1$. The measure of $A(\mathbf{R})^0 = T/H_1(A(\mathbf{R}),\mathbf{Z})$ is then

$$\mu_\Lambda(A(\mathbf{R})^0) = [\Lambda : H_1(A(\mathbf{R}),\mathbf{Z})],$$

we set

$$\mu_\Lambda(A(\mathbf{R})) = \mu(A(\mathbf{R})^0)\cdot c_{\infty},$$

where $c_{\infty} = \char93 (A(\mathbf{R})/A(\mathbf{R})^0)$. The Néron model  $\mathcal{A}$ of $A$ is a canonical model of $A$ over  $\mathbf{Z}$ (see [BLR90]). The global sections $H^0(\mathcal{A},\Omega_{\mathcal{A}})$ define a lattice $\Lambda^*$ in $T^*$, and we define $\Omega_A = \mu_\Lambda(A(\mathbf{R}))$. (We use the symbol $\Omega$ for both the sheaf of differentials and the canonical measure of $A(\mathbf{R})$, but this will not cause any confusion.)

Lemma 4.4   $H_1(A(\mathbf{R}),\mathbf{Z})\cong H_1(A(\mathbf{C}),\mathbf{Z})^+$

Proof. This is well known, but we were unable to locate a suitable reference. Letting $c$ denote conjugation, we we have the commutative diagram

$$\xymatrix @=1.3pc{ 0 \ar[r] & H_1(A(\mathbf{R}),\mathbf{Z}) \ar[r... ...\mathbf{C})^+ \ar[r] & H^1(\langle c \rangle, H_1(A(\mathbf{C}),\mathbf{Z})), }$$

where the upper horizontal sequence is exact, and the lower horizontal sequence is the beginning of the long exact sequence that arises from

$$0\rightarrow H_1(A(\mathbf{C}),\mathbf{Z}) \rightarrow H_1(A(\mathbf{C}),\mathbf{R}) \rightarrow A(\mathbf{C}) \rightarrow 0.$$

The map $i$ is an injection, and the image of $\pi$ in $A(\mathbf{C})^+ = A(\mathbf{R})$ is contained in  $A(\mathbf{R})^0$, since $H_1(A(\mathbf{C}),\mathbf{R})^+$ is connected. Moreover, the image of $\pi$ is $A(\mathbf{R})^0$, since otherwise the cokernel of $\pi$ would not be finite. Hence the map $i$ surjects onto the image of $\pi$, so the first vertical map is an isomorphism. $\qedsymbol$

Theorem 4.5  

$$c_\infty \cdot c_A\cdot \frac{L(A,1)}{\Omega_A} = [ \Phi(H_1(X_0(N),\mathbf{Z}))^+ : \Phi(\mathbf{T}\{0,\infty\})]$$

Proof. It is easier to compute with $\tilde{\Lambda}^*=S_2(\Gamma_0(N),\mathbf{Z})[I_f]$ than with $\Lambda^*$, so let $\tilde{\Omega}_A = \mu_{\tilde{\Lambda}}(A(\mathbf{R}))$. Note that $\tilde{\Omega}_A \cdot c_A = \Omega_A$, where $c_A$ is the Manin constant. Let

$$\Phi : H_1(X_0(N),\mathbf{Q}) \rightarrow \Hom(S_2(\Gamma_0(N))[I_f],\mathbf{C})$$

be the map induced by integration, scaled so that

$$\Phi(\{0,\infty\}) = L(f,1)$$

(that $\{0,\infty\}\in H_1(X_0(N),\mathbf{Q})$ is the Manin-Drinfeld theorem, and that $\int_0^{\infty} f$ is a multiple of $L(f,1)$ follows from the definition of $L(f,s)$ as a Mellin transform). By Lemma 4.4 and Section 2.2,

$$\tilde{\Omega}_A = c_\infty\cdot [\tilde{\Lambda} : H_1(A(\mathbf{R}),\mathbf{Z})]$$

$$= c_\infty \cdot [\Hom(S_2(\Gamma_0(N),\mathbf{Z})[I_f],\mathbf{Z}) : \Phi(H_1(X_0(N),\mathbf{Z}))^+].$$

For any ring $R$ the pairing

$$\mathbf{T}_R \times S_2(\Gamma_0(N),R) \rightarrow R$$

given by $\langle T_n, f \rangle = a_1(T_n f)$ is perfect, so $(\mathbf{T}/I_f)\otimes R \cong \Hom(S_2(\Gamma_0(N),R)[I_f],R).$ Using this pairing, we may view $\Phi$ as a map

$$\Phi : H_1(X_0(N),\mathbf{Q}) \rightarrow (\mathbf{T}/I_f)\otimes \mathbf{C}.$$

Then

$$\tilde{\Omega}_A = c_\infty \cdot [\mathbf{T}/I_f : \Phi(H_1(X_0(N),\mathbf{Z}))^+],$$

and

$$\det(\Phi(\{0,\infty\})) = \prod L(f^{(i)},1) = L(A,1)$$

since $\Phi(\{0,\infty\})$ sends the projection $\pi_i$ onto the factor of $(\mathbf{T}/I_f)\otimes \mathbf{C}$ corresponding to $f^{(i)}$ to $L(f^{(i)},1)\cdot \pi_i$. Letting $H = H_1(X_0(N),\mathbf{Z})$ and $e=\{0,\infty\}\in H\otimes \mathbf{Q}$, we have

$$[\Phi(H)^+ : \Phi(\mathbf{T}e)] = [\Phi(H)^+ : (\mathbf{T}/I_f) \cdot \Phi(e) ]$$

$$= [\Phi(H)^+ : \mathbf{T}/I_f] \cdot [\mathbf{T}/I_f : \mathbf{T}/I_f\cdot \Phi(e)]$$

$$= \frac{c_\infty}{\tilde{\Omega}_A} \cdot \det(\Phi(e))$$

$$= \frac{c_\infty c_A}{\Omega_A} \cdot L(A,1),$$

which proves the theorem. $\qedsymbol$

Theorem 4.5 was inspired by the case when $A$ is an elliptic curve (see [Cre97, §II.2.8]) or the winding quotient of $J_0(p)$ (see [Aga99]), and it generalizes to forms of weight $>2$ (see [Ste00]).

Theorem 4.5 is true with $\Phi$ replaced by any linear map with the same kernel as $\Phi$. One way to find such a linear map with image in a $\mathbf{Q}$-vector space is to compute a basis $\varphi _1, \ldots \varphi _d$ for $\Hom(H_1(X_0(N),\mathbf{Q}),\mathbf{Q})[I_f]$ and let $\Phi=\varphi _1\times \cdots \times \varphi _d$. Also, since $H_1(X_0(N),\mathbf{Z})^+$ and $\mathbf{T}\{0,\infty\}$ are contained in $H_1(X_0(N),\mathbf{Q})^+$, Theorem 4.5 implies that $L(A,1)/\Omega_A\in\mathbf{Q}$, a fact well known to the experts (see [Gro94, Prop. 2.7] for the statement, but without proof).