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Let $f:\mathbb{Z}^+\to \mathbb{C}$ be bounded. Say we are interested in studying how $f$ behaves in short three-term arithmetic progressions. It is very well-known that we can bound $$\sum_{h\leq H} \sum_{n\leq N} f(n) f(n+h) f(n+2 h)$$ by $o(H N)$ (to set ourselves a low bar...) provided that we can bound the Gowers $U^2$ norm $|f|_{U^2}$ by $o(1)$, where $|f|_{U^2}$ is defined by $$|f|_{U^2}^4=\frac{1}{H^2 N} \sum_{n\leq N} \sum_{h_1,h_2\leq H} f(n) \overline{f(n+h_1) f(n+h_2)} f(n+h_1+h_2).$$ The reduction is simple: change variables letting $n'=n+h$, put the sum on $n'$ on the outside and apply Cauchy-Schwarz.

What happens if what we want to show is rather that $$\sum_{h\leq H} \left|\sum_{n\leq N} f(n) f(n+h) f(n+2 h)\right|^2 = o(H N^2)?$$Is the $U^2$ norm enough? Are higher $U^k$ norms enough? Is, say, the $U^3$ norm needed and enough?

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If you expand square as a double sum, you get a sum over copies of some configuration of 6 points. By the usual Cauchy--Schwarz argument, this sum is going to be controlled by some $U^d$ norm. The smallest $d$ such that this is controlled by $U^{d+1}$ is called ``true complexity'' of the system. For your case of 6 linear forms in 3 variables, the most relevant paper is probably https://arxiv.org/abs/1705.06801 (see also references therein).

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  • $\begingroup$ Thanks! Looking at arxiv.org/pdf/0711.0185.pdf , I see that the smallest $d$ such that the {\em $U^{d+1}$} (not $U^d$) norms control the sum is called the true complexity. $\endgroup$
    – Nell
    Mar 22, 2019 at 19:24
  • $\begingroup$ An application of Cauchy-Schwarz reduces the sum of squares I asked about to a sum of the form $\sum_{h,h'\lesssim H, n\lesssim N} f(n) f(n+h) f(n+2h) \overline{f(n+h') f(n+h'+h) f(n+h'+2h)}$. Now, that sum can be controlled by a $U^3$ norm after some more Cauchy-Schwarzing, since the "Cauchy-Schwarz complexity" of the system is $2$ (see arxiv.org/pdf/0711.0185.pdf). By the same source, we could just use the $U^2$ norm instead of the "true complexity" were $1$, but, by Section 3 of the same, the "true complexity" of the system is $2$, so, yes, using $U^3$ is both necessary and enough. $\endgroup$
    – Nell
    Mar 22, 2019 at 19:30
  • $\begingroup$ @Nell Thanks for catching the mistake. I corrected the answer. $\endgroup$
    – Boris Bukh
    Mar 22, 2019 at 19:32

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