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The influential conjectures of Chowla [1] and Hardy–Littlewood [4] assert that for any fixed tuple = h , .., h of distinct integers, H { 1 k} µ(n + h ) µ(n + h ) = o(X), 1 ··· k nX6X Λ(n + h ) Λ(n + h ) = S( )X + o(X), 1 ··· k H nX6X (1 νp/p) for the singular series S( ) = − , where ν = # h , .., h (mod p) . Both conjec- H p (1 1/p)k p { 1 k } tures remain open for any k >2.Q − We establish an average result for Chowla–Hardy–Littlewood correlations. Theorem 1.3. Suppose log H/ log X as X , and define ψ by 2 →∞ →∞ δ log2 X 1/3 δ (1.2) ψδ(X) = min , (log X) − . n log H o Then for any m, k >1, and fixed tuple = a , .., a of disinct integers, we have A { 1 k} m k XHm (1.3) µ(n + hj) Λ(n + ai) m,δ, m . ≪ A ψδ(X) h1,..,hXm 6H nX6X Yj=1 Yi=1 It is worth emphasizing particular aspects of this result. First, (1.3) holds for an arbitrary fixed prime k-tuple. We must average over at least m >1 copies of µ in order to obtain cancellation. Notably, the cancellation becomes quantitatively stronger for larger m, e.g. m/3 δ error savings (log X) − . For the case m =0,k = 2, Matom¨aki-Radziwil l-Tao [10] handled A binary correlations n 6X Λ(n)Λ(n + h) on average with error savings (log X) , though in the much larger regimeP H >X8/33+ε. In particular, the Chowla conjecture holds on average along the subsequence of primes. Corollary 1.4. Suppose log H/ log X as X . Then for any m >1, 2 →∞ →∞ (1.4) µ(p + h ) µ(p + h ) = o π(X)Hm . 1 ··· m m h1,..,hXm 6H pX6X Moreover, using Markov’s inequality we may obtain qualitative cancellation for almost all shifts, with arbitrary log factor savings in the exceptional set. Corollary 1.5. Suppose log H/ log X as X . Then for any A> 0, 2 →∞ →∞ µ(p + h ) µ(p + h ) = o (π(X)), 1 ··· m m pX6X m A m for all except O (H (log X)− ) shifts (h , .., h ) [1,H] . A 1 m ∈ These results build on earlier work of Matom¨aki-Radziwil l-Tao [9], who established an average form of Chowla’s conjecture, (1.5) µ(n + h ) µ(n + h ) = o (XHm), 1 ··· m m h1,..,hXm 6H nX6X for any H = H(X) arbitrarily slowly. Whereas, our results require the faster growth H = (log X)ψ(X) with→ψ ∞(X) arbitrarily slowly. →∞ 2 1.2. Correlations with divisor functions. Consider fixed integers a >1, k >l >2. The well studied correlation of two divisor functions dk,dl is predicted to satisfy k l 2 d (n + h)d (n) = C X + o(X) (log X) − − , k l k,l,h · nX6X for a certain (explicit) constant Ck,l,h > 0. Recently, Matom¨aki-Radziwil l-Tao [11] have shown the following averaged result, in the regime H >(log X)10000k log k, k l 2 k+l 2 dk(n + h)dl(n) Ck,l,h X(log X) − − = ok(HX(log X) − ). − · hX6H nX6X For higher correlations of divisor functions with M¨obius, we obtain Theorem 1.6. For any j >1, k , .., k >2, let k = j k and take any fixed tuple = 1 j i=1 i A a1, .., aj of distinct integers. If log H/ log2 X ,P then { } →∞ j k j µ(n + h) dki (n + ai) = ok, HX(log X) − . A hX6H nX6X Yi=1 Again, we emphasize the need to average over the shift h that inputs to M¨obius µ(n + h), while ai may be fixed arbitrarily. Remark 1.7. For simplicity, the results are stated for the M¨obius function µ, but our results hold equally for its completely multiplicative counterpart, the Liouville function λ. In fact, the proof strategy is to reduce from µ to λ. The main number-theoretic input is the classical Vinogradov-Korobov zero-free region c (1.6) σ + it : 1 σ< − max log q, log( t + 3)2/3 log log( t + 3)1/3 | | | | for L(s, χ), where χ is a Dirichlet character of modulus q 6(log X)A in the Siegel-Walfisz range, see [6, 8]. § 1.3. Beyond M¨obius. We also consider general multiplicative functions f : N C, which do not pretend to be a character f(n) nitχ(n) for some χ (mod q). More precisely,→ we follow Granville and Soundararajan [3] and≈ define the pretentious distance 1 Re(f(p)g(p)) 1/2 D(f,g; X)= − , p pX6X and the related quantity (1.7) M(f; X, Q) = inf D f, n nitχ(n); X 2. t 6X 7→ χ (|q|),q 6Q We may apply recent work of Matom¨aki-Radziwil l-Tao [12] on Fourier uniformity, in order to more directly obtain (qualitative) cancellation for averages of non-pretentious multiplica- tive functions over shifted primes. 3 Theorem 1.8. Given θ (0, 1) let H = Xθ. Given a multiplicative function f : N C 6 ∈ 1 2 2 ρ → with f 1. There exists ρ (0, 8 ) such that, if M(f; X /H − , Q) as X for each| fixed| Q> 1, then ∈ → ∞ → ∞ f(p + h) = oθ,ρ Hπ(X) . hX6H pX6X In particular, f = µ does not pretend to be a Dirichlet character, a fact equivalent to the prime number theorem in arithmetic progressions. Indeed, 1+Re χ(p)pit 1 M(µ; X, Q) > inf > ε log log X + O(1), t 6X p 3 − 2/3+ε χ (|q|),q 6Q e(log X) X 6p 6X where the latter inequality is well-known to follow from the zero-free region (1.6). 1.4. Overview of the proof of Theorem 1.1. We now indicate the general form of the proof. We pursue a variation on the approach of Matom¨aki-Radziwil l-Tao [9]. Namely, we first restrict (1.1) to ‘typical’ terms µ(n) for n = p + h that have prime factors lying ∈ S in certain prescribed intervals [P1, Q1], [P2, Q2]. The terms with n / are sparse, and thus may be shown to contribute negligibly by standard sieve estimates.∈ (For S higher correlations, one may also use sieve estimates, along with work of Henriot [5] to handle a general class of functions with ‘moderate growth’ that are ‘amenable to sieves.’) Once reduced to numbers with ‘typical factorization,’ we decouple the short interval corre- lation between M¨obius and the indicator for the primes, using a Fourier identity and applying Cauchy-Schwarz (Lemma 2.1). This yields a bound of π(X) X/ log X times a Fourier-type integral for µ, ≪ X sup µ(n)e(nα) dx . α Z 0 x 6nX6x+H n ∈S This decoupling step is a gambit. It has the advantage of only needing to consider µ on its own, but loses a factor of log X from the density of the primes. To make this gambit worthwhile, we must recover over a factor of log X savings in the above Fourier integral for µ. However, Matom¨aki-Radziwil l-Tao [9, Theorem 2.3] bound the above integral with roughly 1 (log X) 500 savings (though their bound holds for any non-pretentious multiplicative function g.) Therefore we must refine Matom¨aki-Radziwil l-Tao’s argument in the special case of g = µ to win back over a full factor of log X. We note this task is impossible unless H is larger than a power of log (this already hints at why we must assume log H/ log log X ). We accomplish this task in the ‘key Fourier estimate’ (Theorem 2.2), which→∞ bounds the above integral with (log X)A savings for any A> 0 (though will implicitly depend on A). As with [9], this bound is proven by reducing to the analogousS estimate with the completely multiplicative λ, and splitting up α [0, 1] into major and minor arcs. The main technical innovation here∈ comes from the major arcs (Proposition 5.1), essen- tially saving a factor (log X)A in the mean values of ‘typical’ Dirichlet polynomials of the form λ(n)χ(n) ns X 6Xn 62X n ∈S 4 for a character χ of modulus q 6(log X)A in the Siegel-Walfisz range. This refines the seminal work of Matom¨aki-Radziwil l [7], who obtained a fractional power of log savings for the corresponding mean values. However, Matom¨aki-Radziwil l’s results apply to the general setting of (non-pretentious) multiplicative functions and appeal to Hal´asz’s theorem, which offers small savings. By contrast, our specialization to the M¨obius function affords us the full strength of Vinogradov-Korobov estimates (Lemma 4.5). The Matom¨aki-Radziwil l method saves roughly a fractional power of P1 in the Dirichlet mean value when Q1 H. So in order to recover from our initial gambit, we are prompted ≈ C to choose P1 = (log X) for some large C > 0. Then by a standard sieve bound the log P1 log log X size of is morally O( ) = OC ( ). This highlights the need for our assumption S log Q1 log H log H/ log log X . →∞ We remark that the Matom¨aki-Radziwil l method requires two intervals [P1, Q1], [P2, Q2] (that define ) in order to handle ‘typical’ Dirichlet polynomials in the regime H = (log X)ψ(X) for ψ(X) S . Note in general [7] the slower H the more intervals we require (though by a neat→∞ short argument [8], only one interval is→∞ needed in the regime H = Xθ for θ> 0). Notation The M¨obius function is defined multiplicatively from primes p by µ(p)= 1 and µ(pk)=0 for k >2. Similarly the Liouville function λ is defined completely multiplicatively− by λ(p)= 1. − We use standard asymptotic notation: X Y and X = O(Y ) both mean X 6CY for some some absolute constant C, and X Y≪means X Y X. If x is| a parameter| tending to infinity, X = o(Y ) means that≍X 6c(x)Y for≪ some≪ quantity c(x) that tends to | | zero as x . Let logk X = logk 1(log X) denote the kth-iterated logarithm. Unless→∞ otherwise specified, all− sums range over the integers, except for sums over the 2πix variable p (or p1, p2,..) which are understood to be over the set of primes P. Let e(x) := e . We use 1S to denote the indicator of a predicate S, so 1S = 1 if S is true and 1S = 0 if S is false. When is a set, we write 1 (n) = 1n as the indicator function of . Also let 1 f denote the functionS n 1 (n)f(nS). ∈S S S 7→ S 2. Initial reductions We begin with a Fourier-type bound to decouple correlations of arbitrary functions. Lemma 2.1 (Fourier bound). Given f,g : N C, let F (X) := f(n) 2. Then → n 6X | | P 2 X (2.1) f(n) g(n + h) F (X +2H) sup g(n)e(nα) dx . ≪ · α Z hX6H nX6X 0 x 6nX6x+2H | | Proof. First, the lefthand side of (2.1) is 2 2 2 2 2 (2.2) f(n) g(n + h) H− (2H h ) f(n) g(n + h) =: H− Σ. ≪ −| | hX6H nX6X hX62H nX6X | | | | 5 Expanding the square in Σ and letting h = m n = m′ n′, we have − − 2 Σ= (2H h ) f(n)f(n′)g(n + h)g(n′ + h) −| | hX62H n,nX′ 6X | | X X = f(n)f(n′)g(m)g(m′)1m n=m′ n′ 1x 6n,m 6x+2H dx 1x′ 6n′,m′ 6x′+2H dx′ . ′ ′ − − · Z Z n,nX6X m,mX 0 0 1 Then orthogonality 1m n=m′ n′ = e((m n m′ + n′)α) dα gives − − 0 − − 1 X R X Σ= f(n)g(m)e (m n)α dx f(n′)g(m′)e (n′ m′)α dx′ dα Z Z − · Z ′ ′ ′ ′ − 0 0 x 6n,mX6x+2H 0 x 6n ,mX6x +2H 1 X 2 = f(n)g(m)e (m n)α dx dα . Z Z − 0 0 x 6n,mX6x+2H Using Cauchy-Schwarz, we bound Σ as 1 X 2 X 2 Σ 6 g(m)e(mα) dx f(n)e(nα) dy dα Z Z · Z 0 0 x 6mX6x+2H 0 y 6nX6y+2H X 1 X 2 (2.3) H sup g(m)e(mα) dx f(n)e(nα) dy dα . ≪ α Z Z Z 0 x 6mX6x+2H 0 0 y 6nX6y+2H 1 Using 0 e(nα) dα = 1n=0 again, the second integral in (2.3) is 1 RX 2 X 1 f(n)e(nα) dy dα = f(n)f(n′) e (n n′)α dα dy Z Z Z ′ Z − 0 0 y 6nX6y+2H 0 y 6n,nX6y+2H 0 X n = f(n) 2 dy = f(n) 2 dy HF (X +2H). Z0 | | | | Zn 2H ≪ y 6nX6y+2H n 6XX+2H − 2 Hence plugging the bound (2.3) for H− Σ back into (2.2) gives the result. Next we consider numbers with ‘typical factorization.’ For A, δ >0, define ψ via H = (log X)ψ(X) so that (1.2) becomes 1/3 δ ψ (X) = min ψ(X), (log X) − . δ { } Consider the intervals 33A ψ(X) 4A (2.4) [P1, Q1] = [(log X) , (log X) − ], 2/3+δ/2 1 δ/2 [P2, Q2] = [exp (log X) , exp (log X) − ], and define the ‘typical factorization’ set (2.5) = (X, A, δ) := n 6X : prime factors p ,p n with p [P , Q ] . S S { ∃ 1 2 | j ∈ j j } Using the Fourier bound, we shall reduce Theorem 1.1 to the following. 6 Theorem 2.2 (Key Fourier estimate for µ). Given any A > 5, δ > 0, let = (X, A, δ) as in (2.5). Then if log H/ log X , S S 2 →∞ X HX sup µ(n)e(nα) dx A,δ A/5 . α Z ≪ (log X) 0 x 6nX6x+H n ∈S Proof of Theorem 1.1 from Theorem 2.2. By a standard sieve upper bound [2, Theorem 7.1], for each h 6H, j =1, 2 we have log Pj h (2.6) # p 6X : q ∤ p + h q [Pj, Qj] π(X) . { ∀ ∈ } ≪ log Qj ϕ(h) Thus, recalling the choice of [P , Q ] in (2.4), the terms p+h / trivially contribute to (1.1) j j ∈ S A δ 1/3 h µ(p + h) 6 1 π(X) + (log X) − ≪ ψ(X) ϕ(h) hX6H pX6X 1X6j 62 hX6H pX6X hX6H p+h/ q∤p+h q [P ,Q ] ∈S ∀ ∈ j j π(X) h Hπ(X) (2.7) A A . ≪ ψδ(X) ϕ(h) ≪ ψδ(X) hX6H On the other hand for p + h , Lemma 2.1 with f(n)= 1P(n), g(n)= 1 µ(n) gives ∈ S S 2 X HX2 µ(p + h) π(X +2H) sup µ(n)e(nα) dx A,δ A/5+1 , ≪ · α Z ≪ (log X) hX6H pX6X 0 x 6nX6x+2H p+h n ∈S ∈S assuming Theorem 2.2. Thus by Cauchy-Schwarz we obtain 2 1/2 HX (2.8) µ(p + h) H µ(p + h) . ≪ ≪ (log X)A/10+1/2 hX6H pX6X hX6H pX6X p+h p+h ∈S ∈S Hence (2.7) and (2.8) with A = 6 give Theorem 1.1. Let W = (log X)A. Recall Theorem 2.2 asserts that M (X) XH/W 1/5 for H ≪ X MH (X) := sup µ(n)e(nα) dx . α Z 0 x 6nX6x+H n ∈S 1/5 We first note, that, for technical convenience, it suffices to establish MH0 (X) XH0/W with H := min H, exp (log X)2/3 . Indeed, if H>H then by the triangle≪ inequality 0 { } 0 (X + kH )H H XH XH 6 0 0 0 MH (X) MH0 (X + kH0) 1/5 1/5 = 1/5 ≪ W ≪ H0 · W W k 6XH/H0 k 6XH/H0 ⌈ ⌉ ⌈ ⌉ as desired. Hence we may assume H 6 exp (log X)2/3 hereafter. This reduction is not strictly necessary, but will simplify the argument. For example, in this case Q1 Note, using the trivial bound and swapping the order of summation and integration, the contribution of d > W to (2.10) is HX h(d) HX 3/2 HX (2.11) h(d) H | | d− ≪ | | ≪ W 1/4 d3/4 ≪ W 1/4 ≪ W 1/4 WX 6 assuming Proposition 2.3, and noting 1 (md) = 1 d (m) since d W So we may split [0, 1] into major arcs M and minor arcs m, according to the size of denomi- nator q compared to W , M = M(q) and m = [0, 1] M, \ q[6W 8 where M(q)= α : α a/q 61/qQ . Recall the definitions (2.4), (2.9), (a,q)=1{ | − | 1} S 33A ψ(X) 4A [P1, Q1] = [(log X) , (log X) − ], 2/3+δ/2 1 δ/2 [P2, Q2] = [exp (log X) , exp (log X) − ], (X, A, δ)= m 6 X/d : p ,p m with p [P , Q] . Sd { ∃ 1 2 | j ∈ j j } We shall obtain Proposition 2.3 from the following results. Proposition 3.1 (Key minor arc estimate). Given any A > 5, H = (log X)ψ(X) with 2/3 A ψ(X) and ψ(X) 6(log X) , let d 6W = (log X) and d = d(X, A, 0) as in (2.9). Then for→ ∞ any completely multiplicative g : N C with g 61, weS haveS → | | X HX sup g(n) e(nα) dx A 3/4 1/5 . α m Z0 ≪ d W ∈ x 6ndX6x+H n d ∈S Proposition 3.2 (Key major arc estimate for λ). Given any A> 5, δ > 0, H = (log X)ψ(X) 2/3 A with ψ(X) and ψ(X) 6(log X) , let d 6W = (log X) and d = d(X, A, δ) as in (2.9). Then→ we ∞ have S S X HX sup λ(n) e(nα) dx A,δ . α M Z0 ≪ dW ∈ x 6ndX6x+H n d ∈S We remark that the bounds in the minor arc hold for any bounded multiplicative function, whereas in the major arc the specific choice of λ is needed. 3.1. Minor arc. In this subsection, we prove Proposition 3.1. Recall for α m in the minor arc, α a/q < W 4/qH with q [W, H/W 4]. It suffices to show ∈ | − | ∈ log log X 1/2 (3.1) Im := θ(x) 1 (n)g(n) e(nα) dx HX ψ(X), Z Sd ≪ dW R x 6ndX6x+H uniformly for any α m and measurable θ : [0,X] C with θ(x) 61. Letting = p : ∈ → | | P { P1 6p 6Q1 , by definition each n d has a prime factor in , so we use a variant of the Ramar´eidentity} ∈ S P 1 (1) (mp) Sd (3.2) 1 d (n)= , S # q : q m + 1p∤m Xp n=∈Pmp { ∈ P | } where (1) = m 6X/d : p m, p [P , Q ] . As g is completely multiplicative, we obtain Sd { ∃ | ∈ 2 2 } 1 (1) (mp)g(m)g(p)e(mpα) d Im = S θ(x)1x 6mpd 6x+H dx . # q : q m + 1 Z Xp Xm p∤m R ∈P { ∈ P | } Next we split into dyadic intervals [P, 2P ]. It suffices to show for each P [P , Q ], P ∈ 1 1 1 (1) (mp)g(m)g(p)e(mpα) d HX log log X 1/2 (3.3) S θ(x)1x 6mpd 6x+H dx , # q : q m + 1 Z ≪ log P dW Xp Xm { ∈ P | } p∤m R P 6p∈P62P 9 since then (3.1) will follow by (2.4) and the triangle inequality, using 1 1 log Q1 ψ(X) 4A log = log − A ψ(X). log P ≪ j ≪ log P1 33A ≪ P1 XP Q1 log P1 Xj log Q1 ≪P =2≪j ≪ ≪ Fix P . We may replace 1p∤m with 1 in (3.3) at a cost of O(HX/dP ). Indeed, since the 6 integral is R θ(x)1x 6mpd 6x+H dx H, and 1 (1) (mp) = 0 unless m X/dP , the cost of such ≪ d substitutionR is S X HX H P H = . ≪ ≪ dP 2 dP Xp m 6XX/dP P 6p∈P62P p m | Now the left hand side of (3.3) becomes g(m) g(p)e(mpα)1mpd 6X θ(x)1x 6mpd 6x+H dx # q : q m +1 ZR mX(1) { ∈ P | } Xp ∈Sd P 6p∈P62P g(p)e(mpα)1mpd 6X θ(x)1x 6mpd 6x+H dx ≪ Z m 6XX/dP Xp R P 6p∈P62P 2 1/2 1/2 (X/dP ) g(p)e(mpα)1mpd 6X θ(x)1x 6mpd 6x+H dx , ≪ Z m 6XX/dP Xp R P 6p∈P62P by the trivial bound and Cauchy-Schwarz. Hence for (3.3) it suffices to show 2 H2PX log log P (3.4) g(p)e(mpα)1mpd 6X θ(x)1x 6mpd 6x+H dx . Z ≪ W (log P )2 m 6XX/dP Xp R P 6p∈P62P We expand the left hand side of (3.4) and sum the resulting geometric series on m, g(p1)g(p2)θ(x1)θ(x2) e m(p1 p2)α dx1 dx2 ZR2 − p1,pXX2 [P,2P ] m 6X/dpX i 62 ∈P∩ i ∀ xi 6mdpi 6xi+H H 1 HX min , , ≪ dP (p1 p2)α p1,pX2 62P k − k since for given d,p1,p2, there are O(X) choices for x1 and O(H) subsequent choices for x2 since x2 = x1(p2/p1)+ O(H). Note z denotes the distance of z R to the nearest integer. Thus (3.4) reduces to showing k k ∈ H 1 HP log log P (3.5) min , 2 . P (p1 p2)α ≪ W (log P ) p1,pX2 62P k − k The difference of primes is p1 p2 P . Conversely, any integer n P may be written 6 − ≪n 2 log log P ≪ as n = p1 p2 for p1,p2 2P in ϕ(n) P (log P )− P (log P )2 ways by a standard upper − ≪ 10 ≪ bound sieve, see [2, Proposition 6.22]. Hence for (3.5) it suffices to obtain H 1 H min , (α m). n nα ≪ W ∈ 1 6Xn P ≪ k k But this follows by the standard ‘Vinogradov lemma’ [6, p.346]. Lemma 3.3. Given H,P > 1, take α [0, 1] with α a/q 61/q2 for some (a, q)=1. Then ∈ | − | H 1 H H min , + +(P + q) log q. n nα ≪ q P 1 6Xn 6P k k 4 Observe H/q + H/P + (P + q) log q H/W since q [W, H/W ], P [P1, Q1] = [W 24,H/W 4]. This completes the proof in≪ the minor arc. ∈ ∈ 3.2. Major arc. In this subsection, we prove the key major arc estimate assuming the following mean value result for the (twisted) Liouville function. Proposition 3.4. Given A > 5, δ > 0, let q 6W = (log X)A, d < W 33, χ (mod q), h [H/W 5,H], and = (X, A, δ) as in (2.9). Then for all Y [X/W 4,X], we have ∈ Sd Sd ∈ 2Y 1 2 Y (3.6) Jd,h,q(Y ; χ) := λ(m)χ(m) dx A,δ . Z h ≪ W 10 Y x 6mX6x+h m d ∈S Proof of Proposition 3.2 from Proposition 3.4. To obtain the key major arc estimate we shall prove the stronger bound, X HX (3.7) IM := sup 1 d (n)λ(n) e(nα) dx . α M Z0 S ≪ dW ∈ x 6ndX6x+H 4 In the major arc recall α = a + θ with q 6W and θ 6 W . By partial summation with q | | qH an = 1>x/d(n)1 (n)λ(n) e(na/q), and A(t)= 6 an, we have Sd n t P (x+H)/d 1 (n)λ(n) e(nα)= e( x+H θ)A( x+H ) e( x θ)A( x ) 2πiθ e(tθ)A(t) dt Sd d d − d d − Z x 6ndX6x+H x/d H/d 1 (n)λ(n) e(an/q) + θ 1 (n)λ(n) e(na/q) dh . ≪ Sd | | Z Sd x 6ndX6x+H 0 x/d 6Xn 6x/d+h Thus taking the maximizing h and integrating over x [0,X], we obtain ∈ H W 4 (3.8) IM IH/d + θ sup Ih IH/d + sup Ih, ≪ | | d h 6H/d ≪ qd h 6H/d where X (3.9) Ih := 1 (n)λ(n) e(an/q) dx . Z Sd 0 x/d 6Xn 6x/d+h 11 Then splitting into residues b (mod q) gives X X Ih 6 e(ab/q) 1 (n)λ(n) dx = 1 (n)λ(n) dx . | | Z Sd Z Sd Xb (q) 0 x/d 6Xn 6x/d+h Xb (q) 0 x/d 6Xn 6x/d+h n b (q) n b (q) ≡ ≡ Now suppose we have the bound qhX (3.10) I for h [qH/W 5,H/d]. h ≪ W 5 ∈ 5 Then, combining with the trivial bound Ih 6hX when h 6qH/W , (3.8) becomes W 4 IM IH/d + sup Ih + sup hX ≪ qd qH/W 5 6h 6H/d h 6qH/W 5 qHX W 4 qHX qHX HX 5 + 5 + 5 , ≪ dW qd dW W ≪ dW for q 6W in the major arc. Hence it suffices to show (3.10). Now to bound I , we extract the gcd. Let c := (b, q) so that c n, and we let b′ = b/c, h | q′ = q/c, h′ = h/c, m = n/c. Thus since λ is completely multiplicative, we have X ∗ Ih 6 λ(c) 1 (cm)λ(m) dx | | Z Sd Xc q bX′ (q′) 0 x/cd 6mX6x/cd+h/c | m b′ (q′) ≡ X/cd 6 ∗ cd 1 cd (m)λ(m) dy , Z ′ S Xc q bX′ (q′) 0 y 6mX6y+h | m b′ (q′) ≡ using the substitution y = x/cd, and noting 1 (cm) = 1 (m) since c 6q So plugging back into (3.12), we obtain Y h′ hd qhX ′ ′ Ih cdϕ(q ) 5 = 5 ϕ(q ) Y 5 . ≪ W W ′ ≪ W Xc q YX=2j Xq q YX=2j | X 6Y 6 2X | X 4. Preparatory lemmas We collect some standard lemmas on Dirichlet polynomials. The first is the integral mean value theorem [6, Theorem 9.1]. s Lemma 4.1 (mean value). For D(s)= n 6N ann− , we have T P 2 2 D(it) dt =(T + O(N)) an . Z T | | | | − nX6N One may discretize the mean value theorem by replacing the intergal over [ T, T ] with a sum over a well-spaced set [ T, T ]. − W ⊂ − Definition 4.2. A set R is well-spaced if w w′ >1 for all w,w′ . W ⊂ | − | ∈ W Next is the Hal´asz-Montgomery inequality [6, Theorem 9.6], which offers an improvement to the (discretized) mean value theorem when the well-spaced set is ‘sparse.’ s Lemma 4.3 (Hal´asz-Montgomery). Given D(s)= a n− and a well-spaced set n 6N n W ⊂ [ T, T ]. Then P − D(it) 2 (N + √T ) log 2T a 2. | | ≪ |W| | n| tX nX6N ∈W We also need a bound on the size of well-spaced sets in terms of the values of prime Dirichlet polynomials on 1 + i [7, Lemma 8]. W W Lemma 4.4. Let a C be indexed by primes, with a 61, and define the prime polynomial p ∈ | p| a P (s)= p . ps L 6Xp 62L Suppose a well-spaced set [ T, T ] satisfies P (1 + it) >1/U for all t . Then W ⊂ − | | ∈ W U 2 T 2(log U+log log T )/ log L. |W| ≪ 2 6 A Lemma 4.5. Given A,K > 0, θ> 3 , and a Dirichlet character χ mod q (log X) . Assume θ s exp (log X) 6P 6Q 6X, and let P (s, χ)= χ(p)p− . Then for any t 6X, P 6p 6Q | | P log X K P (1+ it, χ) + (log X)− . | | ≪A,K,θ 1+ t 13 | | Proof. This follows as with [8, Lemma 2], except that the Vinogradov–Korobov zero-free region for ζ(s) is replaced by that of L(s, χ). We also use a Parseval-type bound. This shows that the average of a multiplicative function in almost all short intervals can be approximated by its average on a long interval, provided the mean square of the corresponding Dirichlet polynomial is small. 1/15 1/4 Lemma 4.6 (Parseval bound). Given T0 [(log X) ,X ], and take a sequence (am)m∞=1 with a 61. Assume 1 6h 6h 6X/T 3.∈ For x [X, 2X], define | m| 1 2 0 ∈ a S (x)= a , and A(s)= m . j m ms x 6mX6x+hj X 6Xm 64X Then 1 2X 2 1 X/h1 1 S (x) 1 S (x) dx + A(1 + it) 2 dt h1 1 h2 2 X ZX − ≪ T0 ZT0 | | X/h 2T + max 1 A(1 + it) 2 dt . T >X/h1 T ZT | | 1/15 Proof. This follows as in [7, Lemma 14] with (log X) replaced by general T0. We have a general mean value of products, via the Ramar´eidentity [7, Lemma 12]. Lemma 4.7. For V,P,Q >1, denote = [P, Q] P. Let am, bm,cp be bounded sequences for which a = b c when q ∤ m and q P . Let ∩ mq m q ∈ P c Q (s)= q , v,V qs Xq ∈P ev/V 6q 6e(v+1)/V b 1 R (s)= m , v,V ms · # p m : p +1 Xe−v/V 6Xm 62Xe−v/V { | ∈ P} and take a measurable set [ T, T ]. Then for = [ V log P ,V log Q] Z, we have T ⊂ − I ⌊ ⌋ ∩ 2 an Q 2 dt V log Qv,V (1 + it) Rv,V (1 + it) dt Z n1+it ≪ P Z | | X 6Xn 62X Xv T ∈I T T 1 1 a 2 + +1 + + | n| X V P n X 6Xn 62X p∤n p ∀ ∈P In the next result we employ the Fundamental Lemma of the sieve, along with the Siegel- Walfisz theorem. Lemma 4.8. Given A,K > 0, q 6(log x)A, Dirichlet character χ (mod q), and let = 1/ log log x D p p for any set of primes (q, x ). Then ∈P P ⊂ Q x λ(m)χ(m) . ≪A,K (log x)K mX6x (m, )=1 D 14 Proof. First partition the sum on m by the values of λ(m), χ(m), (4.1) S0 := 1(m, )=1λ(m)χ(m)= νχ(b) 1(m, )=1 D D mX6x b (q)X,ν 1 m X(b,ν) ∈{± } ∈A for the set (b,ν) = m 6x : m b (q),λ(m)= ν . Now it sufficesA to{ prove ≡ } x 1 K (4.2) 1(m, )=1 = 1 + OA,K x(log x)− D 2q − p m X(b,ν) Yp ∈A |D uniformly in b, ν, from which it will follow x 1 K x S = 1 νχ(b) + O x(log x)− , 0 2qd − p ≪ (log x)K Yp b (q)X,ν 1 |D ∈{± } by pairing up terms ν = 1. This will give the lemma. ± (b,ν) Now to show (4.2), write = . For d the set of multiples d = m : d m has size A A |D A { ∈A | } νλ(nd)+1 νλ(d) x = 1 = = λ(n) + + O(q) |Ad| λ(m)=ν 2 2 2qd mX6x nX6x/d nX6x/d d m, m b (q) nd b (q) n bd−1 (q) | ≡ ≡ ≡ n 6y 2K noting (q,d)=1=(q, ). Moreover maxc (q) n c (q) λ(n) A,K y(log y)− by Siegel- D ≡ ≪ 6 P A Walfisz, which is valid by the assumption q (log X) . Thus x x 2K (4.3) = + R , where R log(x/d)− + q. |Ad| 2qd d | d|≪ d d (m, ) Now for any D > 1 the indicator 1(m, )=1 is bounded in between d x λd± λ± = + λ±R . d |Ad| 2q d d d Xd Xd Xd d 1/ log2 x s Let z = x so that (q, z). Then choosing D = z for s = 2K log2 x/ log3 x, the P ⊂ K above error is R x(log x)− by (4.3). And by the Fundamental Lemma [2, ≪ d λd± s 1 = (1+ O(s− )) 1 . d − p Xd Yp d Sl(x) := λ(m)χ(m). x 6Xm 6x+l m ∈Sd K First we claim S (0) x(log x)− for all K > 0. To this, recall from (2.9) that each x ≪ m d has prime factors p1,p2 m with pj [Pj, Qj]. So by inclusion-exclusion, the ∈ S 3 j | ∈ indicator is 1 d (m)= j=0( 1) 1(m, j )=1 where j = p p for the sets of primes 0 = , S − D D ∈Pj P ∅ = [P , Q ], = [PP, Q ], = . Hence applyingQ Lemma 4.8 to each gives P1 1 1 P3 2 2 P2 P1 ∪ P3 Dj 3 j K (5.3) Sx(0) = ( 1) 1(m, )=1λ(m)χ(m) A,K x(log x)− . − Dj ≪ Xj=0 mX6x In particular, letting B = 11A we have x h S (x)= S (0) S (0) 2 , h2 x+h2 − x ≪A (log x)7B ≪A (log x)B 6B where h x(log x)− , and so 2 ≍ 2Y 2 2Y 2 1 1 B 1 1 1 (5.4) J = S (x) dx (log X)− + S (x) S (x) dx . h1 h1 h1 h1 h2 h2 Y ZY ≪ Y ZY − 2B 3 6B Then Lemma 4.6 (Parseval) with T0 = (log X) and h2 = Y/T 0 = Y/(log X) gives Y/h1 2T B 2 Y/h1 2 (5.5) J (log X)− + G(1 + it) dt + max G(1 + it) dt . T >Y/h ≪ ZT0 | | 1 T ZT | | 16 Now for the latter integral over [T, 2T ], we apply the Lemma 4.1 (mean value) if T >X/2, and apply Proposition 5.1 if T [Y/h ,X/2]. Doing so, (5.5) becomes ∈ 1 Y/h1 B Y/h1 J +1 (log X)− + max (T/X + 1) ≪ Y/Q1 T >X/2 T Y/h1 Q1T B + max Y +1 (log X)− Y/h1 6T 6X/2 T Q1 B Y B A B (5.6) ( + 1)(log X)− + W (log X)− = (log X) − . ≪ h1 h1X ≪ 5 4 4 Here we used h1 >H/W , Q1 = H/W , (and Y [X/W ,X]). Hence recalling B = 11A gives Proposition 3.4 as claimed. ∈ 5.1. Mean value of Dirichlet polynomials. In this subsection, we prove Proposition 5.1. Recall the definitions (2.4), (2.9), 33A ψ(X) 4A [P1, Q1] = [(log X) , (log X) − ], 2/3+δ/2 1 δ/2 [P2, Q2] = [exp (log X) , exp (log X) − ], (X, A, δ)= m 6 X/d : p ,p m with p [P , Q] . Sd { ∃ 1 2 | j ∈ j j } 1/3 B Let B = 11A, and α =1/5. Let V = P1 = (log X) and define the prime polynomial λ(p)χ(p) (5.7) Q (s) := . v,j ps Pj 6Xp 6Qj ev/V 6p 6e(v+1)/V v/V Note Qv,j(s) = 0 only if v j := v Z : Pj 6e 6Qj = [ V log Pj ,V log Qj]. We decompose6 [T , T ]=∈ I {as∈ a disjoint union, where} ⌊ = [0, 1]⌋ and 0 T1 ∪ T2 T2 \ T1 αv/V (5.8) = t : Q (1 + it) 6e− v . T1 { | v,1 | ∀ ∈ I1} (j) For j =1, 2 denote by d the integers containing a prime factor in the interval [Pi, Qi] with i = j and possibly, butS not necessarily, with i = j. That is, 6 (j) = m 6X/d : p m with p [P , Q ] for i = j . Sd { ∃ | ∈ i i 6 } Also define the polynomial λ(m)χ(m) 1 (5.9) Rv,j(s)= s . m · # p m : Pj 6p 6Qj +1 Ye−v/V 6Xm 62Ye−v/V { | } m (j) ∈Sd Now Lemma 4.7 (Ramar´e) applies with V = V,P = Pj, Q = Qj and am = λ χ 1 (m), S cp = λ χ(p), bm = λ χ 1 (m), giving Sj 2 2 G(1 + it) dt V log Qj Qv,j(1 + it) Rv,j(1 + it) dt Z j | | ≪ Z j | | vXj T ∈I T 1 1 1 (n) + + + Sd . V Pj n Y 6Xn 62Y p∤n p [P ,Q ] ∀ ∈ j j 17 We crucially note the latter sum vanishes since each n d has a prime factor p [Pj, Qj]. Summing over j =1, 2, the second and third terms above∈ S contribute ∈ 1 1 1 B + = (log X)− . ≪ V P ≪ V 1X6j 62 j Hence the desired integral is T 2 2 B (5.10) G(1 + it) dt = G(1 + it) dt E1 + E2 + (log X)− , ZT0 | | Z 1 2 | | ≪ T ∪T where 2 (5.11) Ej = V log Qj Qv,j(1 + it) Rv,j(1 + it) dt . Z j | | vXj ∈I T B Hence it suffices to bound E1, E2 (Q1T/Y + 1)(log X)− . ≪ αv/V Bound for E : By definition of t , we have Q (1 + it) 6e− for all v , so 1 ∈ T1 | v,1 | ∈ I1 2αv/V 2 2αv/V T E1 V log Q1 e− Rv,1(1 + it) dt V log Q1 e− v/V +1 ≪ Z 1 | | ≪ Y/e vX1 vX1 ∈I T ∈I by Lemma 4.1 (mean value). Summing the resulting geometric series gives 2α P − Q T E V log Q 1 1 +1 1 ≪ 1 1 e 2α/V Y − − 1 6α Q1T B Q1T (5.12) (log X) − +1 (log X)− +1 , ≪ Y ≪ Y 2α/V noting V/(1 e− )= O(1) and 1 6B/5 < B. Bound for− E : We choose the maximizing− v− for E . Thus since 2 2 2 E2 = V log Q2 Qv,2 Rv,2(1 + it) dt (V log Q2) Qv,2 Rv,2(1 + it) dt Z 2 | · | ≪ Z 2 | · | vX2 ∈I T T 2 2 6(V log Q2) sup Qv,2 Rv,2(1 + itn) tn [n,n+1] 2 | · | Xn ∈ ∩T 62(V log Q )2 Q R (1 + it) 2, 2 | v,2 · v,2 | tX ∈W for a well-spaced set 2. For instance, one may take as the even or odd integers in (choose the parityW that ⊂ T gives a larger contribution). WeW shall see is easier to analyze T2 W than 2 itself. T 2/3+δ Now is the critical step for the choice g = λχ and log P2 = (log X) : by Lemma 4.5 (Vinogradov-Korobov), we have for all t [T , T ] ∈ 0 log X B B (5.13) Qv,2(1 + it) δ,A + (log X)− (log X)− , | | ≪ 1+ T0 ≪ 18 2B for T0 = (log X) and B = 11A. So by Lemma 4.3 (Hal´asz-Montgomery), we have 2 2 4B 2 E (V log Q ) (log X) − R (1 + it) 2 ≪ 2 | v,2 | tX ∈W v/V 2 3 4B v/V e (V log Q ) (log X) − (Y e− + √T ) ≪ 2 |W| Y B √TQ2 (log X)− (1 + ), ≪ |W| Y B v/V recalling log T log X, V = (log X) , and e 6Q2. Thus it suffices≍ to bound . We shall obtain |W| B T B (5.14) E2 (log X)− 1+ A,δ (log X)− ≪ Y ≪ 1/2 provided we show T /Q2. To prove this, by definition of 2 , we first partition = (u) where|W| ≪ T ⊃ W u 1 W ∈I W S uα/V (u) Q (1 + it) > e− for all t . | u,1 | ∈ W Hence for each u , we may apply Lemma 4.4 to the prime polynomial Q with U = euα/V ∈ I1 u,1 and L = eu/V , so that (u) 2 2 log U+log log T 2 2α+ 2 log log T 2/5+2/5A+o(1) (5.15) = U T log L U T log L T , |W| |W |≪ ≪ |I1| ≪ uX1 uX1 ∈I ∈I o(1) 2 2α o(1) since 1 < V log Q1 T , U 6Q1 T , log L > log P1 >5A log log T , by recalling |I | 33A ≪ ψ(X) 4A ≪ 1/4 2 [P1, Q1] = [(log X) , (log X) − ] and T [X ,X ]. Hence A> 5 gives T 1/2/Q , and completes∈ the proof of Proposition 5.1. |W| ≪ 2 6. Average Chowla-type correlations In this section, we establish the results for higher correlations stated in the introduction. We first exhibit quantitative cancellation among a broad class of correlations with a ‘typ- ical’ factor 1 µ. We use a standard ‘van der Corput’ argument and then apply the key Fourier estimate.S Lemma 6.1. Given any A > 5, δ > 0, let = (X, A, δ) as in (2.5). Assume H Proof. Let g = 1 µ. By Cauchy-Schwarz it suffices to show S HX2 2 g(n + h)G(n) = G(n)G(n′) g(n + h)g(n′ + h). (log X)A/20 ≫ hX6H nX6X n,nX′ 6X hX6H Using Cauchy-Schwarz again, the right hand side above is bounded by 1 2 2 2 G(n) g(n + h)g(n′ + h) . | | · ′ nX6X Xn,n hX6H 19 recalling g is supported on [1,X]. By assumption G(n) 2 X(log X)A/20, so it n 6X | | ≪ suffices to prove P H2X2 2 2 A/5 g(n + h)g(n′ + h) = ( H h ) g(n)g(n + h) . (log X) ≫ ′ ⌊ ⌋−| | Xn,n hX6H hX We now prove the main technical result of the article, which handles averaged Chowla-type correlations with m >1 copies of the M¨obius function µ and with any function G : N C of ‘moderate growth’ which is ‘amenable to sieves.’ → Theorem 6.2. Given any A > 5, δ > 0, let = (X, A, δ) as in (2.5). Assume H m mXHm A,δ G(n) 1 (n + hj) + . ≪ | | S (log X)A/40 h1,..,hXm 6H nX6X Yj=1 Proof. We observe from Lemma 6.1 that any correlation with a factor 1 µ exhibits strong cancellation. So we split up µ = 1 µ + 1 µ until each term has a factor 1S µ, except for one term with m factors of 1 µ. ThusS the productS in (6.1) becomes S S m m m µ(n + hj)= 1 µ(n + hj) + 1 µ(n + hi) 1 µ(n + hj) µ(n + hj). S S S Yj=1 Yj=1 Xi=1 1Y6j Hence we bound the left hand side of (6.1) by Σ1 + Σ2, where m (6.2) Σ1 = G(n) 1 (n + hj), | | S h1,..,hXm 6H nX6X Yj=1 m (6.3) Σ2 = 1 µ(n + hi)Gi(n) , S Xi=1 h1,..,hXm 6H nX6X where Gi(n)= G(n) 1 6j j Proof of Theorem 1.6. Let G(n) = i=1 dki (n + ai) for the tuple = a1, .., aj and recall j A { } k = i=1 ki. Using work of HenriotQ [5, Theorem 3], we may obtain P j j HX 1 (n) dki (n) 1 S (n + h) dki (n + ai) j+1 . S ≪A (log X) n n hX6H nX6X Yi=1 nX6√X Yi=1 nX6√X By the divisor bound d (n)/n X(log X)ki , and by Mertens’ product theorem n 6√X ki ≪ 1 (Pn) 1 log X S log X + 1 δ . n ≪ − p ≪ ψδ(X) nX6√X p [YP1,Q1] p [YP2,Q2] ∈ ∈ Thus since G(n) 2 X(log X)k, Theorem 6.2 with A = 20k gives n 6X | | ≪ P j HX k j µ(n + h) dki (n + ai) δ, (log X) − . ≪ A ψδ(X) hX6H nX6X Yi=1 6.2. Almost all shifts. Corollary 1.5 follows from the following result by the triangle in- equality for gj = µ. Theorem 6.3. Suppose log H/ log2 X as X . Let g1 = µ and take any gj : N C with g 61 for 1 < j 6k. Then for any→∞ fixed shifts→∞h , ..., h 6H, K > 0 we have → | j| 2 k k (6.5) gj(p + hj) = o(π(X)), pX6X Yj=1 K for all except OK(H(log X)− ) shifts h1 6H. K Proof. Given ε> 0 and fixed shifts h2, .., hk 6H, we aim to show ε H(log X)− for the exceptional set |E| ≪ k (6.6) = h 6H : µ(p + h) gj(p + hj) > 2επ(X) . E n o pX6X Yj=2 21 To this, by Markov’s inequality we have k (επ(X)) µ(p + h) g (p + h ) |E| ≪ j j Xh pX6X Yj=2 ∈E k 6 1 (p + h) + 1 µ(p + h) gj(p + hj) S S Xh pX6X hX6H pX6X Yj=2 ∈E π(X) Hπ(X) (1 + 1 ) + , ≪A ψ(X) p (log X)A/40 Xh Yp h ∈E | p>P1 using Proposition 6.1 when p + h , and a standard sieve upper bound [2, Theorem 7.1] when p + h / . Here = (X, A,∈ δ S) as in (2.5) with A = 80K and δ =1/10, say. ∈ S S S Observe for any h 6H = (log X)ψ(X) the above product is at most (1+ 1 ) log z P1 π(X) (1 + 1 ) = o π(X) . ψ(X) p |E| Xh Yp h ∈E | p>P1 1 K Hence we conclude H(log X)− as desired. |E| ≪ ε 7. Non-pretentious multiplicative functions In this section we prove Theorem 1.8, which we restate below. Theorem 1.8 Let H = Xθ for θ (0, 1), and take a multiplicative function f : N C with 6 1 ∈ 2 2 ρ → f 1. There exists ρ (0, 8 ) such that, if M(f; X /H − , Q) as X for each fixed| | Q, then ∈ → ∞ → ∞ f(p + h) = oθ,ρ Hπ(X) . hX6H pX6X Proof. Consider the exponential sum Fx(α) = x 6m 6x+H f(m)e(mα). The hypotheses of our theorem are made in order to satisfy [12, TheoremP 1.4], which in this case gives X (7.1) sup Fx(α) dx = oθ,ρ(HX). Z0 α | | We critically note the supremum is inside the integral. Now on to the proof, it suffices to show Sf = o(HX) where Sf := Λ(n)f(n + h) (2H h) Λ(n)f(n + h) . ≪ − hX6H nX6X hX62H nX6X 22 We introduce coefficients c(h) to denote the phase of n 6X Λ(n)f(n + h), so that P 1 S (H h)c(h) Λ(n)f(n + h) f ≪ H − hX6H nX6X 1 X = c(h) Λ(n) f(m)1m=n+h 1x 6n,m 6x+H dx H · Z hX6H nX6X m 6XX+H 0 1 X 1 = c(h)e(hα) Λ(n)f(m)e (n m)α dα dx , H Z Z − 0 0 hX6H x 6n,mX6x+H 1 by orthogonality 0 e(nα) dα = 1n=0. That is, we have the following triple convolution R 1 X 1 (7.2) Sf C0(α)Lx( α)Fx(α) dα dx , ≪ H Z0 Z0 − denoting the sums C0(α)= h 6H c(h)e(hα) and Lx(α)= x 6n 6x+H Λ(n)e(nα). We shall split the inner integralP on α according to the sizeP of Lx. Specifically, for each x let = α [0, 1] : L (α) >δH . Then by Markov’s inequality, has measure Tx { ∈ | x | } Tx 1 1 6 4 (7.3) dα 4 Lx(α) dα 4 , Z x (δH) Z x | | ≪ δ H T T since the Fourier identity implies 1 4 Lx(α) dα = Λ(n1)Λ(n2)Λ(n3)Λ(n4)1n1+n2=n3+n4 Z0 | | x 6n1,n2X,n3,n4 6x+H (log X)4 1 H3, ≪ ≪θ x 6p1,p2X,p3,p4 6x+H p1+p2=p3+p4 by a standard sieve upper bound [2, Theorem 7.1]. Thus plugging (7.3) into (7.2), we obtain 1 X 1 X Sf C0(α)Lx( α)Fx(α) dα dx + 4 2 sup C0(α)Lx( α)Fx(α) dx . ≪ H Z0 Z[0,1] x − δ H Z0 α x | − | \T ∈T 4 X Denote the two integrals above by I and I . Observe I δ− sup F (α) dx, 1 2 2 ≪θ 0 α | x | using C0(α) 6H trivially and Lx(α) θ H by the Brun–TitchmarshR theorem. Then by definition| of | , Cauchy-Schwarz| implies| ≪ Tx X X 1 1 1/2 2 2 I1 6δ C0(α)Fx(α) dα dx 6δ C0(α) dα Fx(α) dα dx 6 δHX, Z0 Z[0,1] x | | Z0 Z0 | | · Z0 | | \T by Parseval’s identity applied to C0 and Fx. 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