The Monodromy Pairing on the Character Group

Definition 4.2 (Character group of torus)   The character group

$$X_A = \Hom_{\overline{k}}(\mathcal{T}_{\overline{k}},{\mathbf{G}_m}_{\overline{k}})$$

is a free abelian group of rank $t$ contravariantly associated to $A$.

As discussed in [9], if $A$ is semistable there is a monodromy pairing $X_A\times X_{A^{\vee}}\rightarrow \mathbf{Z}$ and an exact sequence

$$0 \rightarrow X_{A^{\vee}} \rightarrow \Hom(X_{A},\mathbf{Z}) \rightarrow \Phi_A \rightarrow 0.$$

Also, the canonical isomorphism $(A^{\vee})^{\vee}\cong A$ induces an isomorphism

$$X_{A^{\vee}} \times X_{(A^{\vee})^{\vee}} \cong X_{A}\times X_{A^{\vee}},$$

which identifies the monodromy pairing associated to $A^{\vee}$ with that associated to $A$.

Example 4.3 (Tate curve)   Suppose $E=\mathbf{G}_m/q^{\mathbf{Z}}$ is a Tate curve over $\mathbf{Q}_p^{\ur}$. The monodromy pairing on $X_E=q^{\mathbf{Z}}$ is

$$\langle q, q\rangle = \ord_p(q)=-\ord_p(j).$$

Thus $\Phi_E$ is cyclic of order $-\ord_p(j)$.

Suppose $J$ is an abelian variety equipped with a symmetric principal polarization. Since $J$ is self dual via the given symmetric principal polarization, we can view the monodromy pairing on $J$ as a pairing $X_J\times X_J \rightarrow \mathbf{Z}$. Because the principal polarization on $J$ is symmetric the resulting pairing $X_J\times X_J \rightarrow \mathbf{Z}$ is symmetric, so there is no ambiguity about left versus right definitions of $X_J \rightarrow \Hom(X_J,\mathbf{Z})$. The above exact sequence then becomes

$$0 \rightarrow X_{J} \rightarrow \Hom(X_J,\mathbf{Z}) \rightarrow \Phi_J \rightarrow 0.$$