Abstract

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.

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