Math-1770 · mathematics · Combinatorics

Math-1770

Cancellation thresholds and a prime-repunit obstruction for multiplicative subgroups

Status Partially solved · unsolvedmath.comLLM status Open unverified

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Statement

Definitions and mathematical statement

For a prime $p$, define the standard additive character $e_p(x)=\exp(2\pi i a/p)$, where $a$ is any integer representative of $x\in\mathbb F_p$; this is independent of the representative. For a multiplicative subgroup $H\leq\mathbb F_p^*$, set \[ \eta_p(H)=\max_{a\in\mathbb F_p^*} \frac1{|H|}\left|\sum_{x\in H}e_p(ax)\right|. \] The maximum excludes the trivial character $a=0$, whose sum always has absolute value $|H|$. Thus $0\leq\eta_p(H)\leq1$ measures the largest normalized nontrivial additive Fourier coefficient of $H$.

The problem asks how small a subgroup can be while still guaranteeing uniform additive cancellation. More precisely, characterize size thresholds $M(p)$, with $1\leq M(p)\leq p-1$, for which \[ \sup_{\substack{H\leq\mathbb F_p^*\\|H|\geq M(p)}} \eta_p(H)\longrightarrow0 \qquad(p\to\infty\text{ through primes}). \] Seek the weakest possible growth condition on $M(p)$ and quantitative decay estimates in terms of $p$ and $|H|$. Equivalently, for every $\varepsilon>0$, all sufficiently large eligible primes and all their subgroups above the threshold should have every nontrivial character sum at most $\varepsilon|H|$. Uniformity over the character and the subgroup is part of the requirement; a bound for one selected subgroup sequence or for most characters would not suffice. The threshold concerns actual subgroup orders, which must divide $p-1$.

Short English statement

How large must a multiplicative subgroup of a prime field be to force cancellation in every nontrivial additive character sum?

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