Uniformity of the Möbius function in Fq[t] · The Möbius randomness principle Recall (n) = ˆ ( 1)k if n = p1p2 pk is a product of k distinct primes, 0 if n is not squarefree. Thus

Uniformity of the Möbius function in Fq[t ]

Thái Hoàng Lê1 & Pierre-Yves Bienvenu2

1University of Mississippi

2University of Bristol

January 13, 2018

Thái Hoàng Lê & Pierre-Yves Bienvenu Uniformity of the Möbius function JMM 2018 1 / 21

The Möbius randomness principle

Recall

µ(n) =

{(−1)k if n = p1p2 · · · pk is a product of k distinct primes,0 if n is not squarefree.

Thus the sequence {µ(n)} is 1,−1,−1,0,−1,1, . . ..

The Möbius randomness principle states that µ is random-like, i.e.for any bounded, “simple” or “structured” function F , we have

N∑n=1

µ(n)F (n) = o(N).



Recall

µ(n) =



The Möbius randomness principle states that µ is random-like,

i.e.for any bounded, “simple” or “structured” function F , we have

N∑n=1

µ(n)F (n) = o(N).



Recall

µ(n) =



The Möbius randomness principle states that µ is random-like, i.e.for any bounded, “simple” or “structured” function F , we have

N∑n=1

µ(n)F (n) = o(N).


Examples:

1 If F (n) = 1, then PNT is equivalent to∑N

n=1 µ(n) = o(N) and RHis equivalent to

N∑n=1

µ(n) = Oε

(N1/2+ε

)for any ε > 0.

2 If F (n) is periodic with period q, then∑N

n=1 µ(n)F (n) = o(N) isequivalent to PNT in arithmetic progressions.


Examples:

1 If F (n) = 1, then PNT is equivalent to∑N

n=1 µ(n) = o(N) and RHis equivalent to

N∑n=1

µ(n) = Oε

(N1/2+ε

)for any ε > 0.

2 If F (n) is periodic with period q, then∑N

n=1 µ(n)F (n) = o(N) isequivalent to PNT in arithmetic progressions.


3 Sarnak’s conjecture: If (X ,T ) is a dynamical system (i.e. X is acompact metric space, T : X → X is continuous) with topologicalentropy zero, then for any x ∈ X and f ∈ C(X ),

N∑n=1

µ(n)f (T nx) = o(N).

4 (Kalai/Green 2011) If F can be computed by bounded depthcircuits, then

N∑n=1

µ(n)F (n) = o(N).

5 (Dartyge-Tenenbaum 2005) If s(n) is the sum of digits of n in baseq, then for any α ∈ R/Z,

N∑n=1

µ(n)e(s(n)α) = o(N).

Here e(x) = e2πix .



N∑n=1



N∑n=1

µ(n)F (n) = o(N).


N∑n=1





N∑n=1



N∑n=1

µ(n)F (n) = o(N).


N∑n=1




Exponential sums

Davenport (1937): for any A > 0,

N∑n=1

µ(n)e(nα)�AN

logA N

uniformly in α ∈ R/Z. The implied constant is ineffective.

Baker-Harman (1991), Montgomery-Vaughan (unpublished):Assuming GRH, we have

N∑n=1

µ(n)e(nα)�ε N3/4+ε

uniformly in α ∈ R/Z, for any ε > 0.


Exponential sums

Davenport (1937): for any A > 0,

N∑n=1

µ(n)e(nα)�AN

logA N

uniformly in α ∈ R/Z. The implied constant is ineffective.

Baker-Harman (1991), Montgomery-Vaughan (unpublished):Assuming GRH, we have

N∑n=1

µ(n)e(nα)�ε N3/4+ε

uniformly in α ∈ R/Z, for any ε > 0.


Since ∫ 1

0

∣∣∣∣∣N∑

n=1

µ(n)e(nα)

∣∣∣∣∣2

=N∑

n=1

|µ(n)|2 � N,

we cannot do better than N1/2.

Green-Tao (2008, 2012): If F (n) is a nilsequence, then for any A > 0,

N∑n=1

µ(n)F (n)�AN

logA N.

Nilsequences include F (n) = e(αn2 + βn

)and F (n) = e (bnαcβn).


Since ∫ 1

0

∣∣∣∣∣N∑

n=1

µ(n)e(nα)

∣∣∣∣∣2

=N∑

n=1

|µ(n)|2 � N,



N∑n=1

µ(n)F (n)�AN

logA N.




Since ∫ 1

0

∣∣∣∣∣N∑

n=1

µ(n)e(nα)

∣∣∣∣∣2

=N∑

n=1

|µ(n)|2 � N,



N∑n=1

µ(n)F (n)�AN

logA N.




Function field analogy

Let Fq be the finite field on q elements. Fq[t ] is similar to Z in manyaspects (both are UFDs).

For f ∈ Fq[t ], define

µ(f ) =

(−1)k if f = cP1P2 · · ·Pk , Pi distinct monic irreducibles,

c ∈ Fq×,

0 if f is not squarefree.

RH is true in Fq[t ]: for n ≥ 2,∑deg f=n,f monic

µ(f ) = 0.

Furthermore, GRH is true in Fq[t ] (Weil 1948).





µ(f ) =


c ∈ Fq×,



µ(f ) = 0.






µ(f ) =


c ∈ Fq×,



µ(f ) = 0.






µ(f ) =


c ∈ Fq×,



µ(f ) = 0.



The field of fractions Fq(t) = {f/g : f ,g ∈ Fq[t ],g 6= 0} is the analog ofQ.

Define |f/g| = qdeg f−deg g . The completion of Fq(t) with respect to | · | is

Fq

((1t

))=

{α =

n∑i=−∞

ai t i : n ∈ Z,ai ∈ Fq for every i

},

the set of formal Laurent series in 1t . This is the analog of R.

Define T = {α ∈ Fq((1

t

)): |α| < 1}. This is the analog of R/Z.




Fq

((1t

))=

{α =

n∑i=−∞


},



t





Fq

((1t

))=

{α =

n∑i=−∞


},



t



Fix eq : Fq → {z ∈ C : |z| = 1} to be an additive character of Fq.

For α =∑n

i=−∞ ai t i ∈ Fq((1t )), define

e(α) = eq(a−1).

Thus e : Fq((1t ))→ {z ∈ C : |z| = 1} is an additive character of

Fq((1t )).



For α =∑n


e(α) = eq(a−1).


Fq((1t )).



For α =∑n


e(α) = eq(a−1).


Fq((1t )).


Problems we want to solve

Problem 1. Let k ≥ 1 and Φ(x) = αkxk + αk−1xk−1 + · · ·+ α1x be apolynomial in Fq((1

t ))[x ]. Show that∑deg f<n

µ(f )e(Φ(f )) = o(qn)

uniformly in Φ of degree k . Better yet, (since we have GRH) obtain abound O(qck n) for some constant ck < 1.

There are differences between this and the integer case. In fact, whenk ≥ p = char(Fq), even the Weyl sum

∑deg f<n e(Φ(f )) is not well

understood, since the Weyl differencing process breaks down (k ! = 0).


Problems we want to solve

Problem 1. Let k ≥ 1 and Φ(x) = αkxk + αk−1xk−1 + · · ·+ α1x be apolynomial in Fq((1

t ))[x ]. Show that∑deg f<n

µ(f )e(Φ(f )) = o(qn)

uniformly in Φ of degree k . Better yet, (since we have GRH) obtain abound O(qck n) for some constant ck < 1.

There are differences between this and the integer case. In fact, whenk ≥ p = char(Fq), even the Weyl sum

∑deg f<n e(Φ(f )) is not well

understood, since the Weyl differencing process breaks down (k ! = 0).


Problem 2. Let k ≥ 1 and Q ∈ Fq[x0, x1, . . . , xn−1] be a polynomial ofdegree k . Show that ∑

deg f<n

µ(f )eq(Q(f )) = o(qn)

uniformly in Q of degree k . Here Q(f ) is Q evaluated at the coefficientsof f . Better yet, (since we have GRH) obtain a bound O(qck n) for someconstant ck < 1.

Problem 2 is more general than Problem 1 since for any polynomialΦ(x), (Φ(f ))−1 is a polynomial in the coefficients of f (recall thate(α) = eq((α)−1)).


Problem 2. Let k ≥ 1 and Q ∈ Fq[x0, x1, . . . , xn−1] be a polynomial ofdegree k . Show that ∑

deg f<n

µ(f )eq(Q(f )) = o(qn)

uniformly in Q of degree k . Here Q(f ) is Q evaluated at the coefficientsof f . Better yet, (since we have GRH) obtain a bound O(qck n) for someconstant ck < 1.

Problem 2 is more general than Problem 1 since for any polynomialΦ(x), (Φ(f ))−1 is a polynomial in the coefficients of f (recall thate(α) = eq((α)−1)).


The linear case

When k = 1 then Problem 1 and Problem 2 are the same. This isbecause for any linear form ` ∈ Fq[x0, . . . , xn−1], there is α ∈ T suchthat `(f ) = (αf )−1 for any deg f < n.

TheoremFor any ε > 0, we have∑

deg f<n

µ(f )e(αf )�ε,q q(3/4+ε)n

uniformly in α ∈ T.


The linear case

When k = 1 then Problem 1 and Problem 2 are the same. This isbecause for any linear form ` ∈ Fq[x0, . . . , xn−1], there is α ∈ T suchthat `(f ) = (αf )−1 for any deg f < n.

TheoremFor any ε > 0, we have∑

deg f<n

µ(f )e(αf )�ε,q q(3/4+ε)n

uniformly in α ∈ T.


Recall Baker-Harman and Montgomery-Vaughan’s bound in Z (underGRH)

N∑n=1

µ(n)e(αn)� N3/4+ε.

Porritt independently proved a similar result which gives the sameexponent 3/4 + ε when q is sufficiently large in terms of ε.

Our argument is different from the proof in Z in some respects. This isbecause the topologies of T and R/Z are different and there is nosummation by parts in Fq[t ].



N∑n=1

µ(n)e(αn)� N3/4+ε.





N∑n=1

µ(n)e(αn)� N3/4+ε.




We use Hayes’ L-functions of arithmetically distributed relations, asopposed to Dirichlet L-functions. Let M be the set of monicpolynomials in Fq[t ]. Define an equivalence relation f ≡ g (mod Rl,Q)on M as follows

f ≡ g (mod Q) and the first l + 1 coefficients of f and g are the same.

Then M/Rl,Q is a semigroup, Gl,Q := (M/Rl,Q)× is a group. If λ is anontrivial character of Gl,Q, then

L(s, λ) =∑f∈M

λ(f )

|f |s

satisfies GRH (Rhin 1972). (If l = 0 then λ is a Dirichlet character.)


We use Hayes’ L-functions of arithmetically distributed relations, asopposed to Dirichlet L-functions. Let M be the set of monicpolynomials in Fq[t ]. Define an equivalence relation f ≡ g (mod Rl,Q)on M as follows

f ≡ g (mod Q) and the first l + 1 coefficients of f and g are the same.

Then M/Rl,Q is a semigroup, Gl,Q := (M/Rl,Q)× is a group. If λ is anontrivial character of Gl,Q, then

L(s, λ) =∑f∈M

λ(f )

|f |s

satisfies GRH (Rhin 1972). (If l = 0 then λ is a Dirichlet character.)


TheoremAssuming a quantitative form of the bilinear Bogolyubov theorem andp > 2. Then for any A > 0 and quadratic polynomial Q inFq[x0, . . . , xn−1], ∑

deg f<n

µ(f )e(Q(f ))�q,A qnn−A

uniformly in Q.


The quadratic case

TheoremThere is a constant c < 1 such that∑

deg f<n

µ(f )e(αf 2 + βf )�q qcn

uniformly in α, β ∈ T.


Ideas of proof

The classical circle method deals with sums like∑

deg f<n µ(f )e(αf ) or∑deg f<n µ(f )e(αf 2 + βf ). To estimate such sums we need to

distinguish two cases.

α is close to a rational with small denominator (α is in the majorarcs): use our understanding of the distribution of irreducibles incongruence classes, i.e. L-functions.

α is not in the major arcs (α is in the minor arcs): use Vaughan’sidentity to decompose the sum into Vinogradov’s Type I/Type IIsums.


Ideas of proof

The classical circle method deals with sums like∑

deg f<n µ(f )e(αf ) or∑deg f<n µ(f )e(αf 2 + βf ). To estimate such sums we need to

distinguish two cases.

α is close to a rational with small denominator (α is in the majorarcs): use our understanding of the distribution of irreducibles incongruence classes, i.e. L-functions.

α is not in the major arcs (α is in the minor arcs): use Vaughan’sidentity to decompose the sum into Vinogradov’s Type I/Type IIsums.


In the case of∑

deg f<n µ(f )eq(Φ(f )), where Φ(x) = xT Mx and M is asymmetric matrix, the major arcs and minor arcs correspond to lowrank and high rank matrices M. This is because∣∣∣∣∣∣

∑x∈Fq

n

eq(xT Mx)

∣∣∣∣∣∣ ≤ qn−rank(M)/2.

The low rank can be easily reduced to the linear case.

Suppose for a contradiction that∑

deg f<n µ(f )eq(Φ(f )) ≥ δqn. We wantto show that rank(M) is small.


In the case of∑


∑x∈Fq

n

eq(xT Mx)

∣∣∣∣∣∣ ≤ qn−rank(M)/2.





In the case of∑


∑x∈Fq

n

eq(xT Mx)

∣∣∣∣∣∣ ≤ qn−rank(M)/2.





After using Vaughan’s identity, Vinogradov’s Type I/Type II sums,Cauchy-Schwarz and some combinatorial reasoning, we find that forsome n� k ≤ n, the set of pairs

Ph := {(a,b) : deg a, deg b ∈ Gk+1 ×Gk+1 : rank Ma,b ≤ h}

is large (has size (δ/n)O(1)q2k+2) for some h = O(log(n/δ)).

Here Gm = {f : deg f < m},

Ma,b = LTa MLb + LT

b MLa

and La is the matrix of the map Gn−k → Gn, f 7→ af .


After using Vaughan’s identity, Vinogradov’s Type I/Type II sums,Cauchy-Schwarz and some combinatorial reasoning, we find that forsome n� k ≤ n, the set of pairs


is large (has size (δ/n)O(1)q2k+2) for some h = O(log(n/δ)).

Here Gm = {f : deg f < m},

Ma,b = LTa MLb + LT

b MLa

and La is the matrix of the map Gn−k → Gn, f 7→ af .


Problem: We know


is large, where Ma,b = LTa MLb + LT

b MLa. Show that rank M is small.

If (a,b), (a′,b) ∈ Ph, then (a− a′,b) ∈ P2h sinceMa−a′,b = Ma,b −Ma′,b. Similarly for the second coordinate.

By repeatedly “smoothing” in each coordinate, we find that P64hcontains an interesting structure.


Problem: We know







Problem: We know







Theorem (Bilinear Bogolyubov Theorem)

If |Ph| ≥ ηq2(k+1), then there exist Fp-subspaces W1,W2 of theFp-vector space Gk+1 of codimension r1, r2 and Fp-bilinear formsQ1, . . . ,Qr on W1 ×W2 such that

P64h ⊃ {(x , y) ∈W1 ×W2 | Q1(x , y) = · · · = Qr (x , y) = 0}

and max(r , r1, r2) ≤ c(η).

A similar result was proved independently by Gowers-Milicevic.

We can take c(η) = O(exp(exp(exp(logO(1) η−1)))). This is too large tobe useful. If we can take c(η) = logO(1) η−1 then in our application wecan take δ = n−A and this implies that M has small rank.




P64h ⊃ {(x , y) ∈W1 ×W2 | Q1(x , y) = · · · = Qr (x , y) = 0}







P64h ⊃ {(x , y) ∈W1 ×W2 | Q1(x , y) = · · · = Qr (x , y) = 0}







P64h ⊃ {(x , y) ∈W1 ×W2 | Q1(x , y) = · · · = Qr (x , y) = 0}





Documents

Uniformity of the Möbius function in Fq[t] · The Möbius randomness principle Recall (n) = ˆ ( 1)k if n = p1p2 pk is a product of k distinct primes, 0 if n is not squarefree. Thus