Sieve weight smoothings and moments

Transcription

1 Sieve weight smoothings and moments Andrew Granville, with Dimitris Koukoulopoulos & James Maynard Université de Montréal & UCL Workshop in Number Theory and Function Fields Exeter, Friday, January 21, 2016

2 Selberg s sieve Selberg s sieve upper bound: µ(d) d (a,m) d (a,m) for any λ d R with λ 1 = 1. Minimize quadratic form 2 = λ d1 λ d2 #{a A : D a}. a A d (a,m) λ d d 1,d 2 m D=[d 1,d 2 ] Assume λ d supported on squarefree d R, so D R 2. λ d Optimizing (making assumptions on A) yields ( ) log(r/d) κ λ d c µ(d) (d R), log R 2,

3 Selberg s sieve weight & beyond ( ) log(r/d) κ λ d c µ(d) (d R), log R κ sieve dimension Weights λ d decay smoothly to 0. The larger the dimension κ, the smoother as d R. Maynard and Tao: In work on small gaps between primes, used k-dimensional generalization of M f (n; R) := ( ) log d µ(d)f, log R d n f : R R, support on (, 1], suff smooth, f(0) = 1. If n has no prime factors R then M f (n; R) = 1.

4 Why does M f (n; R) := d n Why so lucky? µ(d)f ( ) log d, log R come close enough to recognizing primes for certain f, but not when we have a sharp cut-off? Idea now to study: M f (n; R) k which equals d 1,...,d k 1 D:=[d 1,...,d k ] n x k j=1 µ(d j )f ( ) log dj 1 log R n x D n

5 Why so lucky? M f (n; R) := d n µ(d)f ( ) log d, log R n x M f (n; R) k = d 1,...,d k 1 D:=[d 1,...,d k ] k j=1 µ(d j )f ( ) log dj log R n x D n 1 = x M f,k (R) + O(( f R) k ), where M f,k (R) := d 1,...,d k 1 kj=1 µ(d j )f ( log dj log R ) [d 1,..., d k ]

6 Why super-smooth f work Theorem 1. If f C A (R) with A > 2 + 2k( 1 2kk ), and f, f,..., f (A) uniformly bounded then 1 x n x M f (n; R) 2k c k,f log R as x/rk. where c k,f > 0. Moreover, main contribution from n with all prime factors > R ɛ : 1 M x f (n; R) 2k η 2k k,f log R. n x P (n) R η

7 What about the sharp-cutoff sum? M(n; R) := µ(d). d n d R If n = p 1 p 2 p k with p 1 < p 2 <... < p k then usually log log p j j for each j. If so, this implies log(p 1 p 2 p m ) e+e e m e e 1 em log p m So if p 1 p 2 p m R < p m+1 then the above sum is M(n; R) := µ(d) = 0. d p 1 p 2 p m x #{n x : M(n; R) 0} (log R) δ (log log R) 3/2, 1 + log log 2 where δ := 1 = log 2

8 Proof of proportion of non-zero M(n; R) #{n x : M(n; R) 0} x (log R) δ (log log R) 3/2. Proof: If n is even write n = 2 1 m, and squarefree d n as d = 2 0 or 1 D with D m so that µ(d) + µ(2d) D m d n d R µ(d) = D m D R = D m D R µ(d) 2D R D m D R/2 µ(d) = D m R/2<D R µ(d). Ford (2008) Proportion of integers with a divisor in ( ) [R/p, R] is log p δ log R 1 (log log R) 3/2, and a positive proportion of those have exactly one such divisor.

9 Main number theory problem For M(n; R) = d n, d R 1 x n x M(n; R) 2k x What is Exponent(k)? µ(d), as x, d 1,...,d 2k R c k (log R) Exponent(k). µ(d 1 ) µ(d 2k ) [d 1,..., d 2k ]

10 Main number theory problem For M(n; R) = d n, d R 1 x n x M(n; R) 2k x What is Exponent(k)? µ(d), as x, d 1,...,d 2k R c k (log R) Exponent(k). BRIAN? µ(d 1 ) µ(d 2k ) [d 1,..., d 2k ]

11 Main number theory problem For M(n; R) = d n, d R 1 x n x M(n; R) 2k x What is Exponent(k)? µ(d), as x, d 1,...,d 2k R c k (log R) Exponent(k). BRIAN? Dress, Iwaniec, Tenenbaum (1983): Exponent(1) = 0 Motohashi (2004): Exponent(2) = 2 µ(d 1 ) µ(d 2k ) [d 1,..., d 2k ] de la Bretèche (2001), Balazard, Naimi, Pétermann (2008): Exponent(k) exists for all k.

12 Function field version Monic irreducible polynomials in F q [t]: Unique factorization, and µ. Poly q (n, m; k) := 1 q n = M 1,...,M 2k deg M j =m and when n 2km, = N F q [t] deg N=n µ(m 1 ) µ(m 2k ) 1 q n M 1,...,M 2k deg M j =m M N deg M=m µ(m 1 ) µ(m 2k ) q deg[m 1,...,M 2k ] µ(m) F : deg F =n [M 1,...,M 2k ] N. 2k 1,

13 Möbius for permutations Every σ S N is unique product of cycles, C 1... C r. Divisors are analogous to sub-products of cycles of σ : These cycles act on subsets T [N], so T is fixed by σ. If σ T = C 1... C l, then µ(σ T ) = ( 1) l. Signature ɛ(σ) = ( 1) #{transpositions to create σ}. For cycle C, ɛ(c) = ( 1) C 1, so ɛ(σ) = ( 1) N r. If T = m then ɛ(σ T ) = ( 1) m l, and so µ(σ T ) = ( 1) m ɛ(σ T ). T [N], T =m, σ(t )=T T [N], T =m, σ(t )=T Eberhart, Ford, Green (2015) implies this sum is non-zero for a proportion 1/(log m) δ (log log m) 3/2 of the permutations in S N.

14 We now wish to study Perm(N, m; k) := 1 N! Sieve weights The permutation version σ S N Try k = 1, expand sum to obtain 1 N! T,U [N], T = U =m, σ S N σ(t )=T, σ(u)=u T [N], T =m, σ(t )=T ɛ(σ T ) ɛ(σ T )ɛ(σ U ) Let R = T U, then T = R P, U = R Q, and partition [N] = P Q R V. Hence σ(v ) = V, σ(q) = Q, σ(p ) = P, σ(r) = R Think of σ as an arbitrary permutation on each piece: 2k.

15 Permutation version; second moment R = T U, T = R P, U = R Q, [N] = P Q R V so that ɛ(σ T ) = ɛ(ρ)ɛ(π) and ɛ(σ U ) = ɛ(ρ)ɛ(κ). Perm(N, m; 1) := ρ S R ɛ(ρ) 2 m r=0 1 N! π S P ɛ(π) [N]=P Q R V R =r, P = Q =m r, V =N +r 2m κ S Q ɛ(κ) ν S V 1 As S N = (1, 2)S N, we have ξ S N ɛ(ξ) = 0 for N 2, so P, Q 1. Hence sum is m 1 N! N! r!(m r)! 2 = 2. (N + r 2m)! r!(n+r 2m)! r=m 1

16 Permutation version; 2kth moment If subsets are T 1,... T 2k, partition into sets R I, defined recursively (largest first): T i = R J ; and R = [N] T i, i i I J [2k] I J So {R I : I [2k]} partitions [N], and Perm(N, m; k) equals 1 ɛ(ρ N! I ) I. ρ I S RI r I 0 I I: i I r I=m [N]= I R I R I =r I I I [2k] Outer sums = r I! unless I odd, r I > 1, to get 0. This equals c(m, k), the number of sets of 2 2k 1 non-negative integers {r I : I [2k]} for which r I = m for each i, and r I = 0 or 1 if I is odd. I: i I

17 Permutation version; 2kth moment We have if N 2mk then Perm(N, m; k) = c(m, k), the #{(r I : I [2k])} N 22k 1 for which r I = m for each i, and r I = 0 or 1 if I is odd. I: i I One has c(m, k) m 22k 1 2k Also coeff of (x 1 x 2... x 2k ) m in I odd (1 i I x i) I even, >0 (1 i I x i) = I : I [2k] 1 i I ( - not + in numerator terms ok for parity reasons) x i ( 1) I 1

18 If n 2km Poly q (n, m; k) = Sieve weights Back to Function field version M 1,...,M 2k deg M j =m µ(m 1 ) µ(m 2k ) q deg[m 1,...,M 2k ] Lcm a pain. Break up M i s into gcds: Given M i s define M I for non-empty I [2k], by induction on 2k I : M [2k] = gcd(m 1,..., M 2k ). Otherwise M I = gcd(m i : i I) / I J [2k]. M J. Then M I are pairwise coprime (as M i are squarefree) and each M i = I: i I M I

19 Function field version, 2 We have M I are pairwise coprime (as M i are squarefree) and each M i = I: i I M I, so Poly q (n, m; k) = µ(m 1 ) µ(m 2k ) q deg[m 1,...,M 2k ] = M I M 1,...,M 2k deg M j =m I [2k] j I deg M I=m j µ(m I ) I q deg M I ; which is the coefficient of (x 1 x 2... x 2k ) m in ( µ(m I ) I i I x i q M I I [2k] ) deg MI

20 FTA then gives M I P irreducible I [2k] Function field version, q deg P µ(m I ) I ( i I x i q I : I [2k] ) deg MI = ( 1) I i I x i deg P. Main term is I : I [2k] P irreducible ( 1 ( i I x i q ) deg P ) ( 1) I 1.

21 Function field version, 4 Now using the function field zeta function identity ( ) 1 x deg P 1 = (1 qx) 1, P irreducible we are looking for the coefficient of (x 1 x 2... x 2k ) m in ( ( i I 1 x ) ) deg P ( 1) I 1 i q I : I [2k] = P irreducible I : I [2k] 1 i I which is how c(m, k) was defined! x i ( 1) I 1, We have therefore proved that if n 2mk then Poly q (n, m; k) = c(m, k)(1 + O k (1/q)).

22 Theorems We have if N 2mk then Perm(N, m; k) = c(m, k), where c(m, k) m 22k 1 2k If n 2mk then Poly q (n, m; k) = c(m, k)(1 + O k (1/q)). What is the exponent for the question in the integers?

23 Approaching the number theory problem For M(n; R) = d n, d R µ(d), as x/r2k, 1 x n x M(n; R) 2k d 1,...,d k R µ(d 1 ) µ(d 2k ). [d 1,..., d 2k ] Lcm a pain. Break up d i s into gcds: Given d i s define d I for non-empty I [2k], by induction on 2k I : d [2k] = gcd(d 1,..., d 2k ). Otherwise d I = gcd(d i : i I) / I J [2k] d J. Then d I are pairwise coprime (as d i are squarefree) and each d i = I: i I d I

24 1 x n x Sieve weights Approaching our problem, 2 M(n; R) 2k = d 1,...,d k R d I 1 I [2k] µ(d 1 ) µ(d 2k ) [d 1,..., d 2k ] µ(d I ) I where d I are pairwise coprime and d i = I: i I d I R. Use Perron to get (changing d I to m I ) 1 µ(m (2iπ) 2k I ) I R(s j )=λ j / log R 1 j 2k I [2k] m I 1 d I m 1+s I I where s I := i I s i. We see ζ(1 + s I ) ±1 factors: + if I is even, - if I is odd. j R s j s j ds j

25 Being precise about the zeta-function µ(m I ) I I [2k] m I 1 m 1 pairwise coprime = p prime 1 + I [2k] m 1+s I I ( 1) I. p 1+s I The factor at each prime p therefore looks like = ( 1 1 ) ( 1) I p 1+s I I [2k] plus terms 1/p Hence our zeta-function equals I even ζ(1 + s I) I odd ζ(1 + s I) E(s 1,..., s 2k ) where E(s 1,..., s 2k ) is abs cvgent if each Re(s i ) > ɛ.

26 Approaching our problem, 3 Ignore gcd of M I s (2ndary terms); integral becomes 1 I even (2iπ) 2k ζ(1 + s I) I odd ζ(1 + s I) Rs [2k] j R(s j )=λ j / log R 1 j 2k ds j s j where the products are over non-empty subsets I of [2k]. The plan is to move contours to the left and pick up the (2k-dimensional) poles. If some s j = 0 at the pole then: ζ(1 + s I )/ζ(1 + s I {j} ) 1 whenever j I, and s j ζ(1 + s j ) 1, as s j 0, so no contribution. The integrand is now i j Rs i/s i, and so the pole must be at (0, 0,..., 0), a multi-pole of order 2k 1. (Power of log R) = (Order of Pole) 2k = 1. So this contributes a c/ log R term. Where else?

27 Approaching our problem, 4 Need to find highest order pole of I even ζ(1 + s I) I odd ζ(1 + s I) R s [2k]. s 1 s 2k with all s i 0. Order of pole equals #{I [2k], I even : s I = 0} #{I [2k] I odd : s I = 0} 1. Maximum is achieved if and only if s i = s( 0) for k values of i, and s j = s for the other k values of j. Then the power of log R is ( ) ( ) 2k 2k 0 1 (2k 1) = 2k k k (the poles come from shitfing all but s 2k ).

28 Number theoretic results One can use this to prove 1 M(n; R) 2k c x k (log R) (2k k ) 2k n x where c k is some non-zero constant. That is, Exponent(k) = ( 2k k ) 2k. More generally: Theorem 2. Let f A (t) := { (1 t) A for t 1; 0 otherwise, which is A-times differentiable on R. Then 1 M x fa (n; R) 2k c k,a (log R) (2k k ) 2k(A+1) + c k,a log R, n x where c k,a is some non-zero constant.

29 Comparing results for Möbius moments For the permutation moments, and the polynomials in finite field moments, the exponent is 2 2k 1 2k 1. For the integer moments, the exponent is ( ) 2k 2k. k???

30 Comparing results for Möbius moments For the permutation moments, and the polynomials in finite field moments, the exponent is 2 2k 1 2k 1. For the integer moments, the exponent is ( ) 2k 2k. k??? Much seems to depend on whether ζ-poles in the denominator of the zeta-function cancel those in the numerator. For quadratic χ (mod q), try 1 2k x χ(d) n x d n, d R

31 Quadratic character χ (mod q), moments 1 x n x d n, d R χ(d) Integrand product look for poles in L(1 + s I, χ 0 ) L(1 + s I, χ). I even,>0 I odd Pole at (0,..., 0) order 2 2k k

32 Quadratic character χ (mod q), moments 1 x n x d n, d R χ(d) Integrand product look for poles in L(1 + s I, χ 0 ) L(1 + s I, χ). I even,>0 I odd Pole at (0,..., 0) order 2 2k 1 1. If χ has a Siegel zero then L(1 + s I, χ) ζ(1 + s I ) 1 in a limited range. Asymptotic: c 1 L(1, χ) 22k 1 ( ) φ(q) 2 2k 1 1 log R +c q 2 (q)(log R) (2k k ) 2k. 2k