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Multiplicative Functions Functions in in Short Intervals, with With Applications Applications Kaisa Matomäki (University of Turku, Finland) Kaisa Matomäki (University of Turku, Finland)

The understanding of the behaviour of multiplicative func- of prime factors. The result (1) is actually equivalent to the tions in short intervals has significantly improved during the theorem, asserting that past decade. This has also led to several applications, in par- p X : p P lim |{ ≤ ∈ }| = 1, (2) ticular concerning correlations of multiplicative functions. X / →∞ X log X where P denotes the set of prime numbers. In this article, the 1 Introduction letter p will always denote a prime. Let us start by defining the key players. A function f : N C It will be very convenient for us to use o(1) and O(1) no- → = / is said to be multiplicative if f (mn) = f (m) f (n) whenever tations, so that A o(B) means that A B 0 for X = | | → →∞> gcd(m, n) = 1. We define the Liouville function λ: N and A O(B) means that A CB for some constant C 0 → | |≤ 1, 1 by λ(n):= ( 1)k when n has k prime factors (counted depending only on subscripts of O. In this notation (1) and (2) {− } − with multiplicity). For instance, λ(45) = λ(3 3 5) = ( 1)3 = can be written as · · − λ = 1. The function λ(n) is clearly multiplicative. (n) o(X) (3) − n X It is well known that the average value of λ(n) is 0, i.e. ≤ and 1 X X lim λ(n) = 0. (1) 1 = + o . (4) X X log X log X →∞ n X p X ≤ ≤   In other words, about half of the numbers have an odd number Before discussing the Liouville function further, let us define of prime factors and half of the numbers have an even number another important object: write ζ : C C for the Riemann →

EMS Newsletter December 2020 39 Feature zeta function which is defined by ask how slowly H can tend to infinity with X so that we are 1 guaranteed to have ∞ 1 1 − ζ(s):= = 1 for s > 1, (5) ns − ps  λ(n) = o(H), (9) n=1 p P   ∈ X

0, of integers that are not divisible by a square of a prime. (9) holds for H X7/12+ . ≥ The function µ(n):= λ(n)1n square-free is called the Möbius The proofs of these results are based on zero-density es- function and its behaviour is very similar to that of the Liou- timates for the , i.e. estimates that give ville function. Consequently, the zeros of ζ(s) correspond to an upper bound for the number of zeta zeros in the rectangle s the poles of the Dirichlet series n N µ(n)n− that is closely ∈ s C: (s) [σ, 1] and (s) T related to the Liouville function. ∈  ∈ | |≤ One can show through (6) that (3) (and thus also the prime for given σ (1/2, 1] and T 2. ∈ ≥  number theorem (4)) is equivalent to the fact that the Riemann Using the multiplicativity of λ(n) in a crucial way, (9) was zeta function has no zeros with s = 1. The equivalence recently shown to hold for H X0.55+ε for any ε>0 by  ≥ with the prime number theorem stems from the fact that, for Teräväinen and the author [7]. However, this result is still very s > 1, one has far from what is expected to be true – random models suggest  that (9) holds in intervals much shorter than H = Xε. ζ ∞ log p (s) = , One can ask what about (9) in almost all short intervals − ζ pks k=1 p P ∈ rather than all (we say that a statement holds for almost all so the zeros of the zeta-function also correspond to the poles x X if the cardinality of the exceptional set is o(X)). In this ≤ of a Dirichlet series that is closely related to the characteristic case, the techniques based on the proof of Huxley’s prime function of the primes. number theorem get one down to H X1/6+ε for any ε>0; ≥ In general, the Liouville function is expected to behave whereas assuming the , Gao has proved more or less randomly. In particular, we expect that it has so- it (in an unpublished work) for H (log X)A for certain fixed ≥ called square-root cancellation, i.e. one has, for all X 2, A > 0. ≥ 1/2+ε All the results discussed so far, with the exception of the λ(n) = O (X ) for any ε>0. (7) n X very recent work [7], have their counterparts for the primes. ≤ The conjecture (7) is in fact equivalent to the Riemann hy- Given these similarities and the so-called parity phenonenom, 1/2+ε 1 δ it is natural that until recently the problems for the primes and pothesis, and proving (7) even with X replaced by X − with a small fixed δ seems to be a distant dream which would for the Liouville function have been expected to be of equal ffi correspond to the Riemann zeta function having no zeros with di culty. real part 1 δ for some fixed δ>0. The best result currently However, in the recent years this expectation has turned ≥ − is that out to be wrong; in [2], Radziwiłł and the author made a breakthrough on understanding the Liouville function in short C(log X)3/5 λ(n) = O X exp (8) intervals by proving the following. − (log log X)1/5 n X    ≤ Theorem 1. Let X H 2. Assume that H with for some absolute constant C > 0. This follows from the ≥ ≥ →∞ X . Then Vinogradov–Korobov zero-free region for the Riemann zeta →∞ λ(n) = o(H) (10) function that has been essentially unimproved for sixty years. x

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While our method fundamentally fails for the primes, it Theorem 6. Let α R and let X H 2. Assume that ∈ ≥ ≥ does work for more general multiplicative functions (see [2, H with X . Then →∞ →∞ Theorem 1]): λ(n)e(αn) = o(H) Theorem 2. Let X H 2. Let f : N [ 1, 1] be multi- x = ε Corollary 3. For any 0, there exists a constant C C( ) the long-standing Erdos˝ discrepancy problem from combina- , + √ such that, for all large x, the interval (x x C x] contains torics: xε-smooth numbers (i.e. numbers whose all prime factors are xε). Theorem 8. For any f : N 1, 1 , one has ≤ → {− } Corollary 4. There exists a constant δ>0 such that the Li- sup f (kn) = . ouville function has δX sign changes up to X. ∞ ≥ k,N N n N ∈ ≤ Starting from [4] our work [2] has led to several appli- Later Tao and Teräväinen [16, 17] managed to solve all cations concerning correlations of multiplicative functions. the odd order cases of the logarithmically averaged Chowla Chowla’s conjecture from the 1960s concerning correlations conjecture (without needing Theorem 6). of the Liouville function asserts that, whenever h1,...,hk are Theorem 9. Let k be odd and h ,...,hk Z. Then distinct, one has 1 ∈ λ(n + h ) λ(n + h ) = o(X). λ(n + h1) λ(n + hk) 1 ··· k ··· = o(log X). n X n ≤ n X This is in line with the general philosophy that additive and ≤ multiplicative structures are independent of each other. (When k is odd, one can trivially dispose of the condition Chowla’s conjecture can be equivalently stated as saying about h j being distinct.) k The even cases k 4 of the logarithmic Chowla con- that, for any k 1, each sign pattern (ε1,...,εk) 1, 1 ≥ ≥ ∈ {− } jecture remain open. Tao [15] has shown that the complete appears in the sequence (λ(n+1),...,λ(n+k))n N with density ∈ 1/2k. resolution is equivalent to two other conjectures, the logarith- Given the analogues between the primes and the Liou- mically averaged Sarnak conjecture and the logarithmically ville function, Chowla’s conjecture can be seen as a “Liou- averaged local higher order uniformity conjecture for the Li- ville variant” of the notoriously difficult prime k-tuple conjec- ouville function. ture asserting an asymptotic formula for the number of prime Sarnak’s conjecture roughly asserts that, for a bounded k-tuples (n + h ,...,n + h ) Pk. sequence a(n), one has 1 k ∈ Since already the twin prime conjecture that n and n + 2 a(n)λ(n) = o(X) are both primes infinitely often is completely open, a natural n X starting point is to try to show that, for any h 0, one has ≤ whenever a(n) is of “low complexity”, whereas the higher λ(n)λ(n + h) = o(X). order uniformity conjecture is a vast generalisation of Theo- n X ≤ rem 6 that allows α to depend on x and also replaces the linear In [4] Radziwiłł, Tao and the author managed to show this for phase e(αn) by much more general nilsequences that are the almost all shifts h from a very short range, i.e. characters of the higher order Fourier analysis. The definition Theorem 5. Let X H 2. Assume that H with of nilsequence is so involved that we do not give it here, but ≥ ≥ →∞ X . Then instead we mention two special cases. →∞ First, the higher order uniformity conjecture includes the λ(n)λ(n + h) = o(HX). claim that the Liouville function is locally orthogonal to poly- h H n X ||≤ ≤ nomial phases. More precisely, for any k N it asserts that, ∈ To prove this, we extended Theorem 1 to twists by linear for almost all x X, 2πix ≤ phases e(αn) where e(x):= e . More precisely, we showed sup λ(n)e P(n) = o(H), (11) that P(y) Poly k(R R) ∈ ≤ → x

EMS Newsletter December 2020 41 Feature

where Poly k denotes the set of polynomials of degree at most Hence 1p = 0 for essentially half of the primes, which im- ∈N k. Secondly,≤ the conjecture includes the claim that, for almost plies that the density of is asymptotically C/(log X)1/2, i.e. N all x X, ≤ X X = + sup λ(n)e αn βn = o(H). (12) 1 C o α,β   n X log X  log X  x0. In other words, the average gap the problem much more difficult, and indeed in the recent of two elements of is of size log X/C. Consequently, one ε N progress [5, 6] on this conjecture (in the range H X ) the cannot expect to be regularly distributed in intervals shorter ≥ N  main ingredient is to show that if, e.g., (11) failed for many x, than this. then the corresponding polynomials yielding the supremum In [2] we obtained a quantitative version of Theorem 2 but must be related to each other in certain way. even it is completely trivial for sparsely supported functions In [6] Radziwiłł, Tao, Teräväinen, Ziegler and the author such as f (n) = 1n and so does not tell us anything about ∈N were able to establish the higher order uniformity conjecture the behaviour of 1n in short intervals. But fortunately, the ∈N for H Xε; consequently for instance (11) and (12) hold for method can be adapted to this situation and we proved ≥ almost all intervals of length H Xε. ≥ Theorem 12. As soon as h with X , one has Unfortunately, in order to deduce the logarithmic Chowla →∞ →∞ conjecture, one would need to establish the higher order uni- 1n Ch = o(h) (13) ∈N − formity conjecture in much shorter intervals of length H x

EMS Newsletter December 2020 43 Feature

11/100 Hence by Lemma 13 it suffices to show that, for almost all Now we are left with t, for which P (1 + it) > P− . | 2 | 2 x X, one has Continuing the recursion, we are eventually left with t for ≤ 1/8 which PJ 1(1 + it) P− , say. But now PJ 1 is so large λ(mp) = o(H log log X). (21) | − |≥ J 1 − − P s P

P1− and P2(1+it) 11/100 | | 1/10 2k | |≤ P− we note that 1 ( P (1+it) P ) with k = log P 2 ≤ | 1 | 1  1 and in this case it suffices to show that Kaisa Matomäki [ksmato@utu.fi] is an Academy X/H Research Fellow at University of Turku, Finland. k/5 11/50 k 2 1 − + + = P1 P2 P1(1 it) M2(1 it) dt o 2 She received an EMS Prize in 2020 and is best X/H | | (log P2) −   known for her work with Maksym Radziwiłł on which follows from the mean value theorem which is efficient multiplicative functions. k as M2(s)P1(s) has length about X.

44 EMS Newsletter December 2020