Assume that $\mathbb{N}=\{0,1,2,\ldots\}$ is partitioned into $k\ge 2$ disjoint sets $J(1),\ldots,J(k)$ such that for every $1\le p \le k$ the set $J(p)$ has an asymptotic density $$ d(J(p))=\lim_{n\to\infty}\frac1n\frac{|J(p,n)|}{n}>0,\quad\text{where }J(p,n)=J(p)\cap\{0,1,\ldots,n-1\}. $$ Assume further that $\alpha>0$ is irrational. For $x\in\mathbb{R}$ let $e(x)$ stand for $e^{2\pi i x}$. Is it true that $$ \lim_{n\to\infty}\frac{1}{n^2}\sum_{p\neq q}\left(\left(\sum_{j\in J(p,n)}e(j\alpha)\right)\cdot\left(\overline{\sum_{j\in J(q,n)}e(j\alpha)}\right)\right)=0? $$
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asked Mar 2, 2023 at 13:49
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$\begingroup$ Edit I corrected an error - the summation should be over all pairs $(p,q)$ where $p\neq q$ and $1\le p,q\le k$, not only above those where $p<q$... $\endgroup$– Dominik KwietniakCommented Mar 2, 2023 at 14:03
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1$\begingroup$ The edit history is public, so you don’t have to announce the edits. $\endgroup$– user44143Commented Mar 2, 2023 at 14:10
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1$\begingroup$ What happens if J(p) is the set of all n for which $n \sqrt{2} \in [p/k, (p+1)/k)$ mod $1$? And then take $\alpha = -\sqrt{2}$. $\endgroup$– Ben GreenCommented Mar 2, 2023 at 14:14
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What happens if you take $k=2$, $J(1)=\{p\mid \operatorname{Re}(e(j\alpha)>0\}$ and $J(2)=\{p\mid \operatorname{Re}(e(j\alpha)<0\}$?
It seems to me that $\sum_{j\in J(1,n)}e(j\alpha)\sim n\frac{2}{\pi}$ and $\sum_{j\in J(2,n)}e(j\alpha)\sim -n\frac{2}{\pi}$ (by uniform distribution of the orbit of the irrational rotation). Hence in this case the limit you are looking for would be $-\frac{8}{\pi^2}$.
However, it seems to me that any limit value of the sequence you are looking for should be non positive: again, by uniform distribution, we should have, for any $p$, $$\sum_{q\neq p}\sum_{j\in J(q,n)}e(j\alpha)+\sum_{j\in J(p,n)}e(j\alpha) = o(n^2),$$ so by gathering terms in sums in a clever way you would only have non positive terms up to epsilon.
