July 4, 2026: How many point counts determine an abelian variety?

With Shiva Chidambaram, we proved that for an abelian variety $A$ of dimension $g$ over a finite field $\mathbf{F}_q$ , the first $g$ point counts $\#A(\mathbf{F}_{q^i})$ for $1 \le i \le g$ already determine its zeta function, and hence, by Tate’s theorem, its isogeny class, once $q > Q(g)$ for an explicit (but far-from-optimal) constant. Since Kedlaya proved in 2006 that $2g$ counts suffice, this halves the number needed.

Why this matters. The counts $c_n = \#A(\mathbf{F}_{q^n})$ are packaged into the zeta function, which is determined by the $L$ -polynomial $f_A(T) = \det(1 - \operatorname{Frob}_q T \mid {\mathrm{H}}_\mathrm{\acute{e}t}^1(A)) = \sum_{k=0}^{2g} a_k T^k$ of degree $2g$ . Naively one expects $2g$ counts to solve for its $2g$ coefficients. But Poincaré duality pairs up the Frobenius eigenvalues, $\alpha_i \alpha_{i+g} = q$ , giving a functional equation $a_{2g-k} = q^{g-k} a_k$ : the top half of the coefficients mirrors the bottom, so $f_A$ has only $g$ free parameters. Our theorem turns “morally $g$ counts suffice” into a proof.

The idea. We track the power sums $s_n = \sum_i \alpha_i^{-n}$ of the inverse eigenvalues; knowing $s_1,\dots,s_g$ is equivalent to knowing $f_A$ . Since $q^n s_n = \operatorname{Tr}(\operatorname{Frob}_q^n \mid {\mathrm{H}}_\mathrm{\acute{e}t}^1(A))$ is an integer, computing $s_n$ to additive error below $\tfrac12 q^{-n}$ and rounding recovers it exactly, and each count determines a perturbed power sum that the Weil bounds keep close to $s_n$ . For $n \le g/2$ the count $c_{2n}$ is also available and a Möbius inversion isolates $s_n$ ; the hard range $g/2 < n \le g$ needs an induction on the doubled power sums, whose error analysis is the technical heart of the paper.

An open sequence. Let $N(g)$ be the least $N$ such that, for all large $q$ , the first $N$ counts determine the isogeny class of every $g$ -dimensional abelian variety. Our theorem gives $N(g) \le g$ , and a monotonicity argument gives $N(g) \ge 4$ for $g \ge 4$ . The small values surprise: $N(1)=1$ , $N(2)=2$ , but $N(3)=2$ : two counts already determine an abelian threefold, because for $n \le 2$ the counts $c_1 = f_A(1)$ and $c_2 = f_A(1)f_A(-1)$ are linear functionals whose common fibre meets the Weil region in at most one lattice point. This fails for $n \ge 3$ , and indeed $N(4)=4$ , witnessed by an explicit pair of fourfolds sharing $c_1, c_2, c_3$ as polynomials in $q$ but differing at $c_4$ . A Gröbner-basis computation for $g = 5, 6$ leads us to conjecture $N(5) = N(6) = 4$ , but by $g = 8$ candidate collisions reappear and the growth of $N(g)$ is wide open. As proved, the sequence reads $N(g):\qquad 1,\ 2,\ 2,\ 4,\ [4,5],\ [4,6],\ [4,7],\ [4,8],\ \dots,$ with intervals where we cannot yet pin down a single value.

Checked by machine. The recovery is constructive: an accompanying Sage script takes $q$ and $c_1,\dots,c_g$ and returns $f_A$ . Its analytic core, including the inductive error bound, is formalized in the Lean 4 proof assistant on top of Mathlib. The paper and all code are available at https://github.com/TimoKellerMath/AbelianVarietiesFqZetaPointsCount.