Expansion, divisibility and parity.
2021
Let $\mathbf{P} \subset [H_0,H]$ be a set of primes, where $\log H_0 \geq (\log H)^{2/3 + \epsilon}$. Let $\mathscr{L} = \sum_{p \in \mathbf{P}} 1/p$. Let $N$ be such that $\log H \leq (\log N)^{1/2-\epsilon}$. We show there exists a subset $\mathscr{X} \subset (N, 2N]$ of density close to $1$ such that all the eigenvalues of the linear operator $$(A_{|\mathscr{X}} f)(n) = \sum_{\substack{p \in \mathbf{P} : p | n \\ n, n \pm p \in \mathscr{X}}} f(n \pm p) \; - \sum_{\substack{p \in\mathbf{P} \\ n, n \pm p \in \mathscr{X}}} \frac{f(n \pm p)}{p}$$ are $O(\sqrt{\mathscr{L}})$. This bound is optimal up to a constant factor. In other words, we prove that a graph describing divisibility by primes is a strong local expander almost everywhere, and indeed within a constant factor of being "locally Ramanujan" (a.e.). Specializing to $f(n) = \lambda(n)$ with $\lambda(n)$ the Liouville function, and using an estimate by Matom\"aki, Radziwi{\l}{\l} and Tao on the average of $\lambda(n)$ in short intervals, we derive that \[\frac{1}{\log x} \sum_{n\leq x} \frac{\lambda(n) \lambda(n+1)}{n} = O\Big(\frac{1}{\sqrt{\log \log x}}\Big),\] improving on a result of Tao's. We also prove that $\sum_{N
Keywords:
- Correction
- Source
- Cite
- Save
- Machine Reading By IdeaReader
50
References
0
Citations
NaN
KQI