  Let $\phi:X\dashrightarrow X$ be a dominant rational map of a smooth variety
and let $x\in X$, all defined over $\bar{\mathbb Q}$. The dynamical degree
$\delta(\phi)$ measures the geometric complexity of the iterates of $\phi$, and
the arithmetic degree $\alpha(\phi,x)$ measures the arithmetic complexity of
the forward $\phi$-orbit of $x$. It is known that
$\alpha(\phi,x)\le\delta(\phi)$, and it is conjectured that if the $\phi$-orbit
of $x$ is Zariski dense in $X$, then $\alpha(\phi,x)=\delta(\phi)$, i.e.,
arithmetic complexity equals geometric complexity. In this note we prove this
conjecture in the case that $X$ is an abelian variety, extending earlier work
in which the conjecture was proven for isogenies.
