Tate's question, Standard conjecture D, semisimplicity and Dynamical degree comparison conjecture
Fei Hu, Tuyen Trung Truong, Junyi Xie
TL;DR
The paper investigates dynamical questions in positive characteristic under the Standard conjectures, proving a Weaker Dynamical Degree Comparison and establishing an equivalence between the semisimplicity of the Frobenius endomorphism and that of polarized endomorphisms, with a bound on Jordan blocks for the action on $H^k(X)$; these results hinge on Standard Conjecture $D$ on $H^{2n}(X\times X)$ and leverage the interplay between cohomology and numerical cycle dynamics. A central technique combines Frobenius twists with cohomological correspondences, using Liebrman’s lemma to relate $H^*(X)$- and $N^*(X)$-level dynamics, and yields concrete inequalities linking $sp(f^*|_{H^k(X)})$ to $\lambda_j(f)$. The authors extend the theory to blowups and Hilbert schemes, showing preservation of $D$ and semisimplicity, and provide new proofs of Tate’s question/Serre’s conjecture in the polarized-endomorphism setting. They also discuss Conjecture $G_r$, its status in positive characteristic, and heuristic connections to their dynamical framework, highlighting practical routes to generate new examples and understand semisimplicity from geometric constructions.
Abstract
Let $X$ be a smooth projective variety of dimension $n$ over the algebraic closure of a finite field $\mathbb{F}_p$. Assuming the standard conjecture $D$, we prove a weaker form of the Dynamical Degree Comparison conjecture; equivalence of semisimplicity of Frobenius endomorphism and of any polarized endomorphism (a more general result, in terms of the biggest size of Jordan blocks, holds). We illustrate these results through examples, including varieties dominated by rational maps from Abelian varieties and suitable products of $K3$ surfaces. Using the same idea, we provide a new proof of the main result in a recent paper by the third author, including Tate's question/Serre's conjecture that for a polarized endomorphism $f:X\rightarrow X$, all eigenvalues of the action of $f$ on $H^k(X)$ have the same absolute value.
