Timeline for All two-point correlations equal to $0$, three-point correlation not $0$?

6 events

when toggle format	what		by	license	comment
Jul 12, 2021 at 2:03	comment	added	Will Sawin		We can allow $f$ to be valued in $[-1,1]$, instead of $\{-1,1\}$, by interpreting a value $a$ as a probability $\frac{a+1}{2}$ of $+1$ and $\frac{1-a}{2}$ of $-1$. Then choosing $f (\theta/2 \pi)= c_1 \cos (\theta) + c_2 \cos(2\theta)$, the triple correlation is $c_1^2 c_2 /4$, and choosing small enough nonzero $\theta$ gives the desired example.
Jul 11, 2021 at 20:24	comment	added	Will Sawin		Note that $f$ sending $[0,1/2)$ to $-1$ and $[1/2,1)$ to $+1$ doesn't work, since the distribution of triples is symmetric under negation and thus the triple correlation vanishes. But I agree that maybe another $f$ works.
Jul 11, 2021 at 18:12	history	edited	David E Speyer	CC BY-SA 4.0	added 1 character in body
Jul 11, 2021 at 18:02	comment	added	David E Speyer		It's a good question, and I think that Tao and Gowers have thought about things like this, but I don't have the answer. Let's see if someone shows up who does.
Jul 11, 2021 at 17:48	comment	added	H A Helfgott		Nice. But might this in some sense be the only kind of example? How to detect it (see my comment on "magic sauce" above).
Jul 11, 2021 at 17:29	history	answered	David E Speyer	CC BY-SA 4.0