Proof of the Main Theorem

The notation in this section is as in previous section.

Lemma 6.2   Let $\pi_*:X_J \rightarrow X_{A^{\vee}}$ and $\alpha: X_J \rightarrow \Hom(\pi^* X_A,\mathbf{Z})$ be as in previous section. Then

$$\ker(\pi_*) = \ker(\alpha ).$$

Proof. Suppose $x\in \ker(\pi_*)$, and let $y=\pi^* z$ with $z\in X_A$. Then

$$\langle x, y \rangle = \langle x, \pi^* z \rangle = \langle \pi_* x, z \rangle = 0,$$

so $x\in\ker(\alpha )$. Next let $x\in\ker(\alpha )$. Then for all $z\in X_A$,

$$0 = \langle x, \pi^* z\rangle = \langle \pi_* x, z\rangle,$$

so $\pi_* x$ is in the kernel of the monodromy map

$$X_{A^{\vee}} \rightarrow \Hom(X_A,\mathbf{Z}).$$

Since $X_{A^{\vee}}$ and $\Hom(X_A,\mathbf{Z})$ are free of the same finite rank and the cokernel is torsion, the monodromy map is injective. Thus $\pi_* x=0$ and $x\in \ker(\pi_*)$. $\qedsymbol$

Let $\pi^*:X_A \rightarrow X_J$ be as in previous section.

Lemma 6.3   The monodromy-pairing map $X_J \rightarrow \Hom(X_J,\mathbf{Z})$ composed with restriction $\Hom(X_J,\mathbf{Z})\rightarrow \Hom(\pi^* X_A,\mathbf{Z})$ gives rise to an exact sequence

$$X_J \rightarrow \Hom(\pi^* X_A,\mathbf{Z}) \rightarrow \Phi_A \rightarrow 0.$$

Proof. Lemma 6.2 gives the following commutative diagram with exact rows

$$\xymatrix{0\ar[r] & X_J/\ker(\alpha )\ar[d]^{\cong } \ar[r] & {\... ...r[r] & X_{A^{\vee}}\ar[r] & {\Hom(X_A,\mathbf{Z})}\ar[r] & {\Phi_A}\ar[r] & 0.}$$

By Lemma 6.2, the first vertical map is an isomorphism. The second is an isomorphism because it is induced by the isomorphism $\pi^*:X_A\rightarrow \pi^* X_A$. It follows that $\coker(\alpha )\cong \Phi_A$, as claimed. $\qedsymbol$

Recall that $\mathcal{L}$ denotes the saturation of $\pi^* X_A$ in $X_J$, and that $L\subset \mathcal{L}$ denotes a subgroup of finite index.

Lemma 6.4   The rational number $$\frac{\char93 \Phi_L}{m_L}$$ is independent of the choice of $L$.

Proof. Suppose $L'$ is another finite index subgroup of  $\mathcal{L}$, and let $n=[L:L']$. Here $n$ is a rational number, the lattice index of $L'$ in $L$. Since $\alpha$ is injective when restricted to $\mathcal{L}$, it follows that

$$m_{L'} = [\alpha (X_J):\alpha (L')] = [\alpha (X_J):\alpha (L)]\cdot[\alpha (L):\alpha (L')] = m_L\cdot n.$$

Similarly, $\char93 \Phi_{L'} = \char93 \Phi_L\cdot n$. $\qedsymbol$

Recall that $m_A = \sqrt{\deg(\theta)}$ and

$$\Phi_A \cong \coker(X_{A^{\vee}}\rightarrow \Hom(X_A,\mathbf{Z})),$$

where $m_A$ is the degree of $A$ and $\Phi_A$ is the component group of $A$.

Proof. [Proof of Theorem 6.1] By Lemma 6.4 we may assume that $L=\pi^*X_A$. With this choice of $L$, Lemma 6.3 asserts that $\Phi_L \cong \Phi_A$. By Lemma 6.2, properties of the index, and Lemma 5.2 we have

$$m_L = [\alpha (X_J):\alpha (L)]$$

$$= [\pi_*(X_J):\pi_*(L)]$$

$$= [X_{A^{\vee}}:\pi_*(\pi^*X_A)]$$

$$= [X_{A^{\vee}}:\theta^* X_A]$$

$$= \char93 \coker(\theta^*)$$

$$= \sqrt{\deg(\theta)} = m_A.$$

$\qedsymbol$

Recall that $\Phi_\mathcal{L}$ denotes the cokernel of the natural map $X_J \rightarrow \Hom(\mathcal{L},\mathbf{Z})$ induced by composing the monodromy map $X_J \rightarrow \Hom(X_J,\mathbf{Z})$ with the natural restriction map $\Hom(X_J,\mathbf{Z})\rightarrow \Hom(\mathcal{L},\mathbf{Z})$.

Proposition 6.5   The group $\Phi_\mathcal{L}$ is canonically isomorphic to the image of the map from $\Phi_J$ to $\Phi_A$ induced by $\pi:J\rightarrow A$. Thus

$$\image(\Phi_J\rightarrow \Phi_A) \cong \Phi_\mathcal{L}.$$

Proof. Since $\pi^*X_A\subset \mathcal{L}\subset X_J$, an application of Lemma 6.3 gives the following commutative diagram with exact rows:

$$\xymatrix{ X_J\ar[r]\ar@{=}[d]& \Hom(X_J,\mathbf{Z})\ar[r]\ar[d]&... ...r]\ar[d] & 0\\ X_J\ar[r]& \Hom(\pi^*X_A,\mathbf{Z})\ar[r]& \Phi_A \ar[r]& 0. }$$

The map $\Hom(\mathcal{L},\mathbf{Z})\rightarrow \Hom(\pi^*X_A,\mathbf{Z})$ is an isomorphism, so the map $\Phi_\mathcal{L}\rightarrow \Phi_A$ is injective. Thus

$$\image(\Phi_J\rightarrow \Phi_A) \cong \image(\Phi_J\rightarrow \Phi_\mathcal{L}).$$

The cokernel of $\Hom(X_J,\mathbf{Z})\rightarrow \Hom(\mathcal{L},\mathbf{Z})$ surjects onto the cokernel of $\Phi_J\rightarrow \Phi_\mathcal{L}$. Using the exact sequence

$$0\rightarrow \mathcal{L}\rightarrow X_J \rightarrow X_J/\mathcal{L}\rightarrow 0,$$

we find that

$$\coker(\Hom(X_J,\mathbf{Z})\rightarrow \Hom(\mathcal{L},\mathbf{Z})) \subset \Ext^1(X_J/\mathcal{L},\mathbf{Z}).$$

Because  $\mathcal{L}$ is saturated, the quotient $X_J/\mathcal{L}$ is torsion free, so the indicated $\Ext^1$ group vanishes. Thus the map $\Phi_J\rightarrow \Phi_\mathcal{L}$ is surjective, from which the proposition follows. $\qedsymbol$

Corollary 6.6   The cokernel of the map from $\Phi_J$ to $\Phi_A$ induced by $\pi:J\rightarrow A$ has order $m_A / m_\mathcal{L}$. Thus

$$\char93 \coker(\Phi_J\rightarrow \Phi_A) = \frac{m_A}{m_\mathcal{L}}.$$

Proof. Combine Theorem 6.1 and Proposition 6.5. $\qedsymbol$