In Terry Tao's notes on the Chowla conjecture, the said conjecture states that for any fixed integer $m>0$ and nonzero $(a_1,\ldots,a_m)\in\{0,1\}^m$, $$ \lim_{x\rightarrow\infty}\frac{1}{x}\sum_{n\leq x}\mu(n+1)^{a_1}\cdots\mu(n+m)^{a_m}=0, $$ where $\mu$ denotes the Möbius function. I would like information about the rate of this conjectured convergence. Are there any widely believed conjectures along these lines?
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asked Jun 16, 2013 at 22:35
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$\begingroup$ The original statement of Chowla's conjecture (by Chowla; p. 96 of his book) is that $\mu(P(n))$ averages to 0 for any non-constant polynomial $P$. That more general form seems natural to me (though it is even further out of reach than the case of $P$ a product of linear factors). There are applications (if you like the stone age, see my thesis). $\endgroup$– H A HelfgottCommented Aug 27 at 19:01
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I imagine the "correct" conjecture is the Möbius $s$-tuples conjecture in the form that if $s \in \mathbb{N}$, $\alpha_1, \ldots, \alpha_s \in \mathbb{N}$ with at least one $\alpha_i$ odd, and $d_1, \ldots, d_s \in \mathbb{Z}$ distinct, then for all $\varepsilon > 0$ we have that $$\sum_{n \leq x}{\mu(n + d_1)^{\alpha_1} \mu(n + d_2)^{\alpha_2} \cdots \mu(n + d_s)^{\alpha_s}} \ll_{\varepsilon} x^{1/2 + \varepsilon}$$ uniformly for all $|d_i| \leq x$. However, I don't imagine there is any good evidence for this other than the fact that the Riemann hypothesis implies the case $s = 1$. I don't think this problem has really been studied enough for there to be a widely-believed generalisation of Chowla's conjecture.
A form of this conjecture is assumed by Ng in this paper on the distribution of the Möbius function in short intervals; he assumes that this conjecture holds with "for all $\varepsilon > 0$" replaced by "for some $0 < \beta_0 < 1/2$".
answered Jun 16, 2013 at 22:53
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1$\begingroup$ Note Carmon and Rudnick's recent paper on a function field analogue: The autocorrelation of the Mobius function and Chowla's conjecture for the rational function field. The bound they get is quite far from the analogue of $x^{1/2+\epsilon}$, and by analysing their proof, you can see it is actually wrong. But it is only an specific analogue ($\lim q\rightarrow\infty$, "$x$" a fixed degree). $\endgroup$ Commented Jun 17, 2013 at 1:59