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.
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have
Disclaimer: All third-party content on this website/platform is and will remain the property of their respective owners and is provided on "as is" basis without any warranties, express or implied. Use of third-party content does not indicate any affiliation, sponsorship with or endorsement by them. Any references to third-party content is to identify the corresponding services and shall be considered fair use under The CopyrightLaw.