Pic plays the role of $\operatorname{End}(A)$

$$\operatorname{Pic}(A)/\operatorname{Pic}^0(A) = \operatorname{NS}(A)$$

$$\bigl(\operatorname{Pic}(A)/\operatorname{Pic}^0(A)\bigr)_{\Bbb{Q}} \simeq \{\phi \in \operatorname{End}(A)_{\Bbb{Q}} : \phi^{\dagger} = \phi\}, \qquad \dagger\text{ the Rosati involution}$$