The Degree of a Symmetric Isogeny

We next relate the degree of the isogeny $A^{\vee}\rightarrow A$ defined at the end of Section 3 to the order of the cokernel of the induced map on the character groups of tori defined in Section 4.1. Let $K$ be as in Section 2, and let $A$ be an abelian variety over $K$.

Definition 5.1 (Symmetric isogeny)   A symmetric isogeny $\varphi :A^{\vee}\rightarrow A$ is an isogeny such that the map

$$\varphi ^{\vee}:A^{\vee}\rightarrow (A^{\vee})^{\vee}=A$$

is equal to $\varphi$.

If $J$ and $A$ are as in Section 3 then the principal polarization $\theta_J$ of $J$ is symmetric, so the natural map $A^{\vee}\rightarrow A$ is a symmetric isogeny.

Lemma 5.2   Suppose that $A$ is a purely toric abelian variety over $K$ and that $\varphi :A^{\vee}\rightarrow A$ is a symmetric isogeny. Let $\varphi _a:X_A\rightarrow X_{A^{\vee}}$ denote the induced map on character groups. Then

$$\deg(\varphi ) = \char93 \coker(\varphi _a)^2.$$

Proof. By Corollary 8.7 applied to our isogeny $\varphi$ (so what we are presently calling $A^{\vee}$ and $A$ are respectively called $A$ and $B$ in the discussion surrounding Theorem 8.6), we deduce that

$$\deg(\varphi) = \char93 \ker(\varphi) = \char93 \ker(\varphi_t) \cdot \char93 \ker(\varphi^{\vee}_t)$$

where $\varphi _t$ and $\varphi^{\vee}_t$ are the maps induced by $\varphi$ and $\varphi^{\vee}$ on closed fiber tori.

Since the character group $X_A$ is, by definition, $\Hom_{\overline{k}}(\mathcal{T}_{\overline{k}},{\mathbf{G}_m}_{\overline{k}})$, where $\mathcal{T}$ is the toric part of the closed fiber of $A$, it follows that $\char93 \ker(\varphi _t) = \char93 \coker(\varphi _a)$. Since $\varphi =\varphi ^{\vee}$, this proves the lemma. $\qedsymbol$