The Ramanujan-Nagell Theorem: Understanding the Proof

The Ramanujan-Nagell Theorem: Understanding the Proof


A crash introduction to BSD conjecture

The pdf version is A crash introduction to BSD conjecture .

We begin with the Weierstrass form of elliptic equation, i.e. look it as an embedding cubic curve in {\mathop{\mathbb P}^2}.

Definition 1 (Weierstrass form) {E \hookrightarrow \mathop{\mathbb P}^2 }, In general the form is given by,

\displaystyle E: y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 \ \ \ \ \ (1)

If {char F \neq 2,3}, then, we have a much more simper form,

\displaystyle y^2=x^3+ax+b, \Delta:=4a^3+27b^2\neq 0. \ \ \ \ \ (2)

Remark 1

\displaystyle \Delta(E)=\prod_{1\leq i,\neq j\leq 3}(z_i-z_j)

Where {z_i^3+az_i+b=0, \forall 1\leq i\leq 3}.

We have two way to classify the elliptic curve {E} living in a fix field {F}. \paragraph{j-invariant} The first one is by the isomorphism in {\bar F}. i.e. we say two elliptic curves {E_1,E_2} is equivalent iff

\displaystyle \exists \rho:\bar F\rightarrow \bar F

is a isomorphism such that {\rho(E_1)=E_2}.

Definition 2 (j-invariant) For a elliptic curve {E}, we have a j-invariant of {E}, given by,

\displaystyle j(E)=1728\frac{4a^3}{4a^3+27b^2} \ \ \ \ \ (3)

Why j-invariant is important, because j-invariant is the invariant depend the equivalent class of {E} under the classify of isomorphism induce by {\bar F}. But in one equivalent class, there also exist a structure, called twist.

Definition 3 (Twist) For a elliptic curve {E:y^2=x^3+ax+b}, all elliptic curve twist with {E} is given by,

\displaystyle E^{(d)}:y^2=x^3+ad^2x+bd^3 \ \ \ \ \ (4)

So the twist of a given elliptic curve {E} is given by:

\displaystyle H^1(Gal(\bar F/ F), Aut(E_{\bar F})) \ \ \ \ \ (5)

Remark 2 Of course a elliptic curve {E:y^2=x^3+ax+b} is the same as {E:y^2=x^3+ad^2x+bd^4}, induce by the map {\mathop{\mathbb P}^1\rightarrow \mathop{\mathbb P}^1, (x,y,1)\rightarrow (x,dy,1)}.

But this moduli space induce by the isomorphism of {F} is not good, morally speaking is because of the abandon of universal property. see \cite{zhang}. \paragraph{Level {n} structure} We need a extension of the elliptic curve {E}, this is given by the integral model.

Definition 4 (Integral model) {s:=Spec(\mathcal{O}_F)}, {E\rightarrow E_s}. {E_s} is regular and minimal, the construction of {E_s} is by the following way, we first construct {\widetilde{E_s} } and then blow up. {\widetilde E_s} is given by the Weierstrass equation with coefficent in {\mathcal{O}_F}.

Remark 3 The existence of integral model need Zorn’s lemma.

Definition 5 (Semistable) the singularity of the minimal model of {E} are ordinary double point.

Remark 4 Semistable is a crucial property, related to Szpiro’s conjecture.

Definition 6 (Level {n} structure)

\displaystyle \phi: ({\mathbb Z}/n{\mathbb Z})_s^2\longrightarrow E[N] \ \ \ \ \ (6)

{P=\phi(1,0), Q=\phi(o,1)} The weil pairing of {P,Q} is given by a unit in cycomotic fields, i.e. {<P,Q>=\zeta_N\in \mu_{N}(s)}

What happen if {k={\mathbb C}}? In this case we have a analytic isomorphism:

\displaystyle E({\mathbb C})\simeq {\mathbb C}/\Lambda \ \ \ \ \ (7)

Given by,

\displaystyle {\mathbb C}/\Lambda \longrightarrow \mathop{\mathbb P}^2 \ \ \ \ \ (8)

\displaystyle z\longrightarrow (\mathfrak{P}(z), \mathfrak{P}'(z), 1 ) \ \ \ \ \ (9)

Where {\mathfrak{P(z)}=\frac{1}{z^2}+\sum_{\lambda\in \Lambda,\lambda\neq 0}(\frac{1}{(z-\lambda)^2}-\frac{1}{\lambda^2})}, and the Weierstrass equation {E} is given by {y^2=4x^3-60G_4(\Lambda)x-140G_6(\Lambda)}. The full n tructure of it is given by {{\mathbb Z}+{\mathbb Z}\lambda} and the value of {P,Q}, i.e.

\displaystyle P=\frac{1}{N}, Q=\frac{\tau}{N} \ \ \ \ \ (10)

Where {\tau} is induce by

\displaystyle \Gamma(N):=ker(SL_2({\mathbb Z})\rightarrow SL_2({\mathbb Z}/n{\mathbb Z})) \ \ \ \ \ (11)

The key point is following:

Theorem 7 {k={\mathbb C}}, the moduli of elliptic curves with full level n-structure is identified with

\displaystyle \mu_N^*\times H/\Gamma(N) \ \ \ \ \ (12)

Now we discuss the Mordell-Weil theorem.

Theorem 8 (Mordell-Weil theorem)

\displaystyle E(F)\simeq {\mathbb Z}^r\oplus E(F)_{tor}

The proof of the theorem divide into two part:

  1. Weak Mordell-Weil theorem, i.e. {\forall m\in {\mathbb N}}, {E(F)/mE(F)} is finite.
  2. There is a quadratic function,

    \displaystyle \|\cdot\|: E(F)\longrightarrow {\mathbb R} \ \ \ \ \ (13)

    {\forall c\in {\mathbb R}}, {E(F)_c=\{P\in E(F), \|P\|<c\}} is finite.

Remark 5 The proof is following the ideal of infinity descent first found by Fermat. The height is called Faltings height, introduce by Falting. On the other hand, I point out, for elliptic curve {E}, there is a naive height come from the coefficient of Weierstrass representation, i.e. {\max\{|4a^3|,|27b^2|\}}.

While the torsion part have a very clear understanding, thanks to the work of Mazur. The rank part of {E({\mathbb Q})} is still very unclear, we have the BSD conjecture, which is far from a fully understanding until now.

But to understanding the meaning of the conjecture, we need first constructing the zeta function of elliptic curve, {L(s,E)}.

\paragraph{Local points} We consider a local field {F_v}, and a locally value map {F\rightarrow F_{\nu}}, then we have the short exact sequences,

\displaystyle 0\longrightarrow E^0(F_{\nu})\longrightarrow E(F_{\nu})=E_s(\mathcal{O}_F)\longrightarrow E_s(K_0)\longrightarrow 0 \ \ \ \ \ (14)

Topologically, we know {E(F_{\nu})} are union of disc indexed by {E_s(k_{\nu})},

\displaystyle |E_s(k_{\nu})| \sim q_{\nu}+1=\# \mathop{\mathbb P}^1(k_{\nu})

. Define {a_{\nu}=\# \mathop{\mathbb P}^1(k_{\nu})-|E_s(k_{\nu})|}, then we have Hasse principle:

Theorem 9 (Hasse principle)

\displaystyle |a_{\nu}|\leq 2\sqrt{q_{\nu}} \ \ \ \ \ (15)

Remark 6 I need to point out, the Hasse principle, in my opinion, is just a uncertain principle type of result, there should be a partial differential equation underlying mystery.

So count the points in {E(F)} reduce to count points in {H^1(F_{\nu},E(m))}, reduce to count the Selmer group {S(E)[m]}. We have a short exact sequences to explain the issue.

\displaystyle 0\longrightarrow E(F)/mE(F) \longrightarrow Sha(E)[m] \longrightarrow E(F)/mE(F)\longrightarrow 0 \ \ \ \ \ (16)

I mention the Goldfold-Szipiro conjecture here. {\forall \epsilon>0}, there {\exists C_{\epsilon}(E)} such that:

\displaystyle \# (E)\leq c_{\epsilon}(E)N_{E/{\mathbb Q}}(N)^{\frac{1}{2}+\epsilon} \ \ \ \ \ (17)

\paragraph{L-series} Now I focus on the construction of {L(s,E)}, there are two different way to construct the L-series, one approach is the Euler product.

\displaystyle L(s,E)=\prod_{\nu: bad}(1-a_{\nu}q_{\nu}^{-s})^{-1}\cdot \prod_{\nu:good}(1-a_{\nu}q_{\nu}^{-s}+q_{\nu}^{1-2s})^{-1} \ \ \ \ \ (18)


Where {a_{\nu}=0,1} or {-1} when {E_s} has bad reduction on {\nu}.

The second approach is the Galois presentation, one of the advantage is avoid the integral model. Given {l} is a fixed prime, we can consider the Tate module:

\displaystyle T_l(E):=\varprojlim_{l^n} E[l^n] \ \ \ \ \ (19)

Then by the transform of different embedding of {F\hookrightarrow \bar F}, we know { T_{l}(E)/Gal(\bar F/F)}, decompose it into a lots of orbits, so we can define {D_{\nu}}, the decomposition group of {w}(extension of {\nu} to {\bar F}). We define {I_{\nu}} is the inertia group of {D_{\nu}}.

Then {D_{\nu}/I_{\nu}} is generated by some Frobenius elements

\displaystyle Frob{\nu}x\equiv x^{q_{\nu}} (mod w),\forall x\in \mathcal{O}_{\bar Q} \ \ \ \ \ (20)

So we can define

\displaystyle L_{\nu}(s,E)=(1-q_{\nu}^{-s}Frob_{\nu}|T_{l}(E)^{I_{\nu}})^{-1} \ \ \ \ \ (21)

And then {L(s,E)=\prod_{\nu}L_{\nu}(s,E)}.

Faltings have proved {L_{\nu}(s,E)} is the invariant depending the isogenous class in the follwing meaning:

Theorem 10 (Faltings) {L_{\nu}(s,E)} is an isogenous ivariant, i.e. {E_1} isogenous to {E_2} iff {\forall a.e. \nu}, {L_{\nu}(s,E_1)=L_{\nu}(s,E_2)}.

\displaystyle L(s,E)=L(s-\frac{1}{2},\pi ) \ \ \ \ \ (22)

Where {\pi} come from an automorphic representation for {GL_2(A_F)}. Now we give the statement of BSD onjecture. {R} is the regulator of {E}, i.e. the volume of fine part of {E(F)} with respect to the Neron-Tate height pairing. {\Omega} be the volume of {\prod_{v|\infty}F(F_v)} Then we have,

  1. {ord_{s=1}L(s,E)=rank E(F)}.
  2. {|Sha(E)|<\infty}.
  3. {\lim_{s\rightarrow 0}L(s,E)(s-1)^{-rank(E)}=c\cdot \Omega(E)\cdot R(E)\cdot |Sha(E)|\cdot |E(F)_{tor}|^{-2}}

Here {c} is an explictly positive integer depending only on {E_{\nu}} for {\nu} dividing {N}.


SL_2(Z) and its congruence subgroups

The pdf version is SL_2(Z) and its congruence subgroups.

We know we can always do the following thing:

\displaystyle R\ commutative\ ring \longrightarrow \ "general\ linear\ group" \ GL_2(R) \ \ \ \ \ (1)


\displaystyle GL_2(R):=\{\begin{pmatrix} a & b \\ c & d \end{pmatrix}: det \begin{pmatrix} a & b \\ c & d \end{pmatrix}=R^*, a,b,c,d\in R\} \ \ \ \ \ (2)


Remark 1 Why it is {R^*} but not 1? if it is 1, then the action {R/GL_2(R)} distribute is not trasitive on {R}, i.e. every element in unite group present a connected component

Now we consider the subgroup {SL_2(R)\subset GL_2(R)}.

\displaystyle SL_2(R)=\{\begin{pmatrix} a & b\\ c & d \end{pmatrix}: det \begin{pmatrix} a & b\\ c & d \end{pmatrix}=1, a,b,c,d\in R \} \ \ \ \ \ (3)

We are most interested in the case {R={\mathbb Z}, {\mathbb Z}/n{\mathbb Z}}. So how to investigate {SL_2({\mathbb R})}? We can look at the action of it on something, for particular, we look at the action of it on Riemann sphere, i.e. { \hat {\mathbb C}/({\mathbb R})} given by fraction linear map:

\displaystyle g(z):=\frac{az+b}{cz+d}, g(\infty)=\frac{a}{c} \ \ \ \ \ (4)

Remark 2 What is fraction linear map? This action carry much more information than the action on vector, thanks for the exist of multiplication in {{\mathbb C}} and the algebraic primitive theorem. Due to I always looks the fraction linear map as something induce by the permutation of the roots of polynomial of degree 2, this is true at least for fix points, and could natural extension. So how about the higher dimension generate? consider the transform of {k-1} tuples induce by polynomial with degree {k}?

Remark 3

  1. {SL_2({\mathbb R})/\pm I:=PSL_2({\mathbb R})}, then {PSL_2({\mathbb R}) } action faithful on {\hat C}, i.e. except identity, every action is nontrivial. This is easy to be proved, observed,

    \displaystyle \frac{az+b}{cz+d}=z,\forall z\in \hat{\mathbb C}\Longrightarrow \begin{pmatrix} a & b\\ c & d \end{pmatrix}=\begin{pmatrix} 1 & 0\\ 0&1 \end{pmatrix} or \begin{pmatrix} -1&0\\ 0&-1 \end{pmatrix} \ \ \ \ \ (5)

  2. Up half plane {H} is invariant under the action of {PSL_2({\mathbb R})}, i.e. {\forall g\in PSL_2({\mathbb R})}, {gH=H}. The proof is following,
  3. \displaystyle \begin{array}{rcl} Im(\frac{az+b}{cz+d}) & = & Im(\frac{(az+b)(c\bar z+d)}{|cz+d|^2})\\ & = & Im(\frac{ac|z|^2+bc\bar z+adz+bd}{|cz+d|^2})\\ & > & 0,\ due\ to\ ad=bc+1. \end{array}


Now we focus on {SL_2({\mathbb Z})} or the same,{PSL({\mathbb Z})}. All the argument for {SL_2(R)} make sense for

\displaystyle \Gamma:= SL_2({\mathbb Z}), \bar \Gamma:=SL_2({\mathbb Z})/\pm I \ \ \ \ \ (6)

Fix {N\in {\mathbb N}}, define,

\displaystyle \Gamma(N):=\{\begin{pmatrix} a &b\\ c&d \end{pmatrix}, a,d \equiv 1(mod N), b,c\equiv 0(mod N)\} \ \ \ \ \ (7)

Then {\Gamma(N)} is the kernel of map {SL_2({\mathbb Z})\rightarrow SL_2({\mathbb Z}/n{\mathbb Z})}, i.e. we have short exact sequences,

\displaystyle 0\longrightarrow \Gamma(N)\longrightarrow \Gamma\longrightarrow SL_2({\mathbb Z}/n{\mathbb Z})\longrightarrow 0 \ \ \ \ \ (8)

Remark 4 The relationship of {\Gamma(N)\subset \Gamma} is just like {N{\mathbb Z}+1\subset {\mathbb Z}}.

Definition 1 (Congruence group) A subgroup of {\Gamma} is called a congruence group iff {\exists n\in {\mathbb N}}, {\Gamma(N)\subset G}.

Example 1 We give two examples of congruence subgroups here.

  1. \displaystyle \Gamma_1(N)=\{\begin{pmatrix} 1& *\\ 0 &1 \end{pmatrix} mod N\} \ \ \ \ \ (9)

  2. \displaystyle \Gamma_0(N)=\{\begin{pmatrix} * & *\\ 0 & * \end{pmatrix}mod N\} \ \ \ \ \ (10)


Definition 2 (Fundamental domain)

\displaystyle F=\{z\in H:-\frac{1}{2}\leq Re(z)\leq \frac{1}{2}\ and |z|\geq 1\} \ \ \ \ \ (11)

Now here is a theorem charistization the fundamental domain.

Theorem 3 This domain {F} is a fundamental domain of {\hat {\mathbb H}/({\mathbb R})}

Proof: Ths key point is {SL_2({\mathbb Z})} have two generators,

  1. {\tau_a: z\rightarrow z+a, \forall a\in {\mathbb Z}}.
  2. {s: z\rightarrow \frac{1}{z}}.

Thanks to this two generator exactly divide the action of {\Gamma} on {H} into a lots of scales, then {\Omega} is a fundamental domain is a easy corollary. \Box

Remark 5 This is not rigorous, {H} need be replace by {\hat H}, but this is very natural to get a modification to a right one.

Remark 6 {z_1,z_2\in \partial F} are {\Gamma} equivalent iff {Re(z)=\pm \frac{1}{2}} and {z_2=z_1\pm 1} or if {z_1} on the unit circle and {z_2=-\frac{1}{z_1}}

Remark 7 If {z\in F}, then {\Gamma_z=\pm I} expect in the following three case:

  1. {\Gamma=\pm \{\tau,s\}} if {z=i}.
  2. {\Gamma=\pm\{ I,s\tau, (s\tau)^2\}} if {z=w=-\frac{1}{2}+\frac{\sqrt{-3}}{2}}.
  3. {\Gamma=\pm\{I,\tau s, (\tau s)^2\}} if {z=-\bar w=\frac{1}{2}+\frac{\sqrt{-3}}{2}}.

Where {\tau=\tau_1}.

Remark 8 The group {\bar \Gamma=SL_2({\mathbb Z})/\pm I} is generated by the two elements {s}, {\tau}. In other word, any fraction linear transform is a “word” induce by {s,\tau,s^{-1}.\tau^{-1}}. But not free group, we have relationship {s^2=-I,(s\tau)^3=-I}.

The natural function space on {F} is the memorphic function, under the map: {H\rightarrow D-\{0\}}, it has a {q}-expension,

\displaystyle f(q)=\sum_{k\in {\mathbb Z}}a_kq^k \ \ \ \ \ (12)

And there are only finite many negative {k} such that {a_k\neq 0}.

Dirichlet hyperbola method

A pdf version is Dirichlet hyperbola method.

1. Introduction

Theorem 1

\displaystyle \sum_{1\leq n\leq x}d(n)=\sum_{1\leq n\leq x}[\frac{x}{n}]=xlogx+(2\gamma-1) x+O(\sqrt{x}) \ \ \ \ \ (1)


Remark 1 I thought this problem initial 5 years ago, cost me several days to find a answer, I definitely get something without the argument of Dirchlet hyperbola method and which is weaker but morally the same camparable with the result get by Dirichlet hyperbola method.

Remark 2 How to get the formula:

\displaystyle \sum_{1\leq n\leq x}d(x)=\sum_{1\leq n\leq x}[\frac{x}{n}]? \ \ \ \ \ (2)

In fact,

\displaystyle \sum_{1\leq n\leq x}d(x)=\sum_{1\leq ab\leq x}1=\sum_{1\leq n\leq x}[\frac{x}{n}]\ \ \ \ \ (3)

Which is the integer lattices under or lying on the hyperbola {\{(a,b)|ab=x\}}.

Remark 3 By trivial argument, we can bound the quantity as following way,

\displaystyle \begin{array}{rcl} \sum_{1\leq n\leq x}[\frac{x}{n}] & = & \sum_{1\leq ab\leq x}1\\ & = & x\sum_{i=1}^x\frac{1}{i}-\sum_{i=1}^x\{\frac{x}{i}\}\\ & = &xlnx+\gamma x+O(x) \end{array}

The error term is {O(x)}, which is too big. But fortunately we can use the symmetry of hyperbola to improve the error term.


\displaystyle \begin{array}{rcl} \sum_{1\leq n\leq x}d(n) & = & \sum_{ab\leq x}1\\ & = & \sum_{a\geq \sqrt{x}}[\frac{x}{b}]+\sum_{b\geq \sqrt{x}}[\frac{x}{a}]-\sum_{1\leq a,b\leq \sqrt{x}}1\\ & = & xlogx+(2\gamma-1)x+O(\sqrt{x}) \end{array}


Theorem 2

Given a natural number k, use the hyperbola method together
with induction and partial summation to show that

\displaystyle \sum_{n\leq x}d_k(n) = xP_k(log x) + O(x^{1-\frac{1}{k}+\epsilon}), n\leq x \ \ \ \ \ (4)

where {P_k(t)} denotes a polynomial of degree {k-1} with leading term {\frac{t^{k-1}}{(k-1)!}}.

Remark 4 {P_k(x)} is the residue of {\zeta(s)^kx^ss^{-1}} at {s=1}.


We can establish the dimension 3 case directly, which is the following asymptotic formula,

\displaystyle \sum_{1\leq xy\leq n}[\frac{n}{xy}]=xP_2(logx)+O(x^{1-\frac{1}{3}+\epsilon}) \ \ \ \ \ (5)

The approach is following, we first observe that

\displaystyle \sum_{1\leq xy\leq n}[\frac{n}{xy}]=\sum_{xyz\leq n}1 \ \ \ \ \ (6)

The problem transform to get a asymptotic formula for the lattices under 3 dimension hyperbola. The first key point is, morally {([n^{\frac{1}{3}}],[n^{\frac{1}{3}}],[n^{\frac{1}{3}}])} is the central point under the hyperbola.

Then we can divide the range into 3 parts, and try to get a asymptotic formula for each part then add them together. Assume we have:

  1. {A_x=\sum_{1\leq r\leq [n^{\frac{1}{3}}]}\sum_{1\leq yz\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]}.
  2. {A_y=\sum_{1\leq r\leq [n^{\frac{1}{3}}]}\sum_{1\leq xz\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]}.
  3. {A_z=\sum_{1\leq r\leq [n^{\frac{1}{3}}]}\sum_{1\leq xy\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]}.

Then the task transform to get a asymptotic formula,

\displaystyle A_x=A_y=A_z=xQ_2(logx)+O(x^{1-\frac{1}{3}+\epsilon}) \ \ \ \ \ (7)

But we can do the same thing for {\sum_{1\leq yz\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]} and then integral it. This end the proof. For general {k\in {\mathbb N}}, the story is the same, by induction.

Induction on {k} and use the Fubini theorem to calculate {\sum_{x_1...x_r\leq n}\frac{n}{x_1...x_r},\forall 1\leq r\leq k}. \Box

There is a major unsolved problem called Dirichlet divisor problem.

\displaystyle \sum_{n\leq x}d(n) \ \ \ \ \ (8)

What is the error term? The conjecture is the error term is {O(x^{\theta}), \forall \theta>\frac{1}{4}}, it is known that {\theta=\frac{1}{4}} is not right.

Remark 5

To beats this problem, need some tools in algebraic geometry.

2. Several problems

{\forall k\in {\mathbb N}}, is there a asymptotic formula for {\sum_{t=1}^n\{\frac{kn}{t}\}} ?

{\forall k\in {\mathbb N}}, {f(n)} is a polynomial with degree {k}, is there a asymptotic formula for {\sum_{t=1}^n\{\frac{f(n)}{t}\}} ?

{\forall k\in {\mathbb N}}, {g(n)} is a polynomial with degree {k}, is there a asymptotic formula for {\sum_{t=1}^n\{\frac{n}{g(t)}\}} ?

Theorem 3 {k\in {\mathbb N}}, then we have

\displaystyle \lim_{n\rightarrow \infty}\frac{\{\frac{kn}{1}\}+\{\frac{kn}{2}\}+...+\{\frac{kn}{n}\}}{n}=k(\sum_{i=1}^k\frac{1}{i}-lnk-\gamma) \ \ \ \ \ (9)


\displaystyle \begin{array}{rcl} \frac{\{\frac{kn}{1}\}+\{\frac{kn}{2}\}+...+\{\frac{kn}{n}\}}{n} & = & \frac{\sum_{i=1}^k\frac{kn}{i}-\sum_{i=1}^n[\frac{kn}{i}]}{n}\\ & = & k(lnn+\gamma +\epsilon_n)-\frac{\sum_{i=1}^{kn}[\frac{kn}{i}]-\sum_{i=n+1}^{kn}[\frac{kn}{i}]}{n} \end{array}


Now we try to estimate

\displaystyle S_k(n)=\sum_{i=1}^{kn}[\frac{kn}{i}]-\sum_{i=n+1}^{kn}[\frac{kn}{i}] \ \ \ \ \ (10)

In fact, we have,

\displaystyle \begin{array}{rcl} S_k(n) & = & (2\sum_{i=1}^{[\sqrt{kn}]}[\frac{kn}{i}]-[\sqrt{kn}]^2)-(\sum_{i=1}^k[\frac{kn}{i}]-kn)\\ & = & 2\sum_{i=1}^{[\sqrt{kn}]}\frac{kn}{i}-\sum_{i=1}^k\frac{kn}{i}+2\{\sqrt{kn}\}[\sqrt{kn}]+\{\sqrt{kn}\}^2-2\sum_{i=1}^{[\sqrt{kn}]}\{\frac{kn}{i}\}+\sum_{i=1}^k\{\frac{kn}{i}\}\\ & = & 2kn(ln[\sqrt{kn}]+\gamma+\epsilon_{[\sqrt{kn}]})-kn\sum_{i=1}^k\frac{1}{i}+r(n)\\ & = & knln(kn)+kn(2\gamma-\sum_{i=1}^k\frac{1}{i})+r'(n)\\ & = & knln n+kn(2\gamma+lnk-\sum_{i=1}^k\frac{1}{i})+r'(n) \end{array}

Where {-3\sqrt{n}<r(n)<3\sqrt{n}}, {-3\sqrt{n}<r'(n)<3\sqrt{n}}.

So by 1 we know,

\displaystyle \begin{array}{rcl} \frac{\{\frac{kn}{1}\}+...+\{\frac{kn}{n}\}}{n} & = & k(lnn+\gamma+\epsilon_n)-klnn-k(2\gamma+lnk-\sum_{i=1}^k\frac{1}{i})+\frac{r'(n)}{n}\\ & = & k(\sum_{i=1}^k\frac{1}{i}-lnk-\gamma)+\frac{r'(n)}{n}+k\epsilon_n \end{array}

So we have,

\displaystyle \lim_{n\rightarrow \infty}\frac{\{\frac{kn}{1}\}+...+\{\frac{kn}{n}\}}{n} =k(\sum_{i=1}^k\frac{1}{i}-lnk-\gamma)=k\epsilon_k \ \ \ \ \ (11)

Remark 6 In fact we can get {0<k\epsilon_k<\frac{1}{2}, \forall k\in {\mathbb N}}, by combining the theorem 3 and 1.

3. Lattice points in ball

Gauss use the cube packing circle get a rough estimate,

\displaystyle \sum_{n\leq x}r_2(n)=\pi x+O(\sqrt{x}) \ \ \ \ \ (12)


In the same way one can obtain,

\displaystyle \sum_{n\leq x}r_k(n)=\rho_kx^{\frac{k}{2}}+O(x^{\frac{k-1}{2}}) \ \ \ \ \ (13)

Remark 7 Where {\rho_k=\frac{\pi^{\frac{k}{2}}}{\Gamma(\frac{k}{2}+1)}} is the volume of the unit ball in {k} dimension.

Dirchlet’s hyperbola method works nicely for the lattic points in a ball of dimension {k\geq 4}. Langrange proved that every natural number can be represented as the sum of four squares, i.e. {r_4(n)>0}, and Jacobi established the exact formula for the number of representations

\displaystyle r_4(n)=8(2+(-1)^n)\sum_{d|n,d\ odd}d. \ \ \ \ \ (14)

Hence we derive,

\displaystyle \begin{array}{rcl} \sum_{n\leq x}r_4(n) & = & 8\sum_{m\leq x}(2+(-1)^m)\sum_{dm\leq x, d\ odd}d\\ & = & 8\sum_{m\leq x}(2+(-1)^m)(\frac{x^2}{4m^2}+O(\frac{x}{m}))\\ & = & 2x^2\sum_1^{\infty}(2+(-1)^m)m^{-2}+O(xlogx)\\ & = & 3\zeta(2)x^2+O(xlogx) = \frac{1}{2}(\pi x)^2+O(xlogx) \end{array}

This result extend easily for any {k\geq 4}, write {r_k} as the additive convolution of {r_4} and {r_{k-4}}, i.e.

\displaystyle r_k(n)=\sum_{0\leq t\leq n}r_4(t)r_{k-4}(n-t) \ \ \ \ \ (15)

Apply the above result for {r_4} and execute the summation over the remaining {k-4} squares by integration.

\displaystyle \sum_{n\leq x}r_k(n)=\frac{(\pi x)^{\frac{k}{2}}}{\Gamma(\frac{k}{2}+1)}+O(x^{\frac{k}{2}-1}logx) \ \ \ \ \ (16)


Remark 8 Notice that this improve the formula 12 which was obtained by the method of packing with a unit square. The exponent {\frac{k}{2}-1} in 16 is the best possible because the individual terms of summation can be as large as the error term (apart from {logx}), indeed for {k=4} we have {r_4(n)\geq 16n} if {n} is odd by the Jacobi formula. The only case of the lattice point problem for a ball which is not yet solved (i.e. the best possible error terms are not yet established) are for the circle({k=2}) and the sphere ({k=3}).

Theorem 4

\displaystyle \sum_{n\leq x}\tau(n^2+1)=\frac{3}{\pi}xlogx+O(x) \ \ \ \ \ (17)

4. Application in finite fields

Suppose {f(x)\in {\mathbb Z}[x]} is a irreducible polynomial. And for each prime {p}, let

\displaystyle \rho_f(p)=\# \ of \ solutions\ of f(x)\equiv 0(mod\ p) \ \ \ \ \ (18)

By Langrange theorem we know {\rho_f(p)\leq deg(f)}. Is there a asymptotic formula for

\displaystyle \sum_{p\leq x}\rho_f(p)? \ \ \ \ \ (19)

A general version, we can naturally generated it to algebraic variety.

\displaystyle \rho_{f_1,...,f_k}(p)=\#\ of \ solutions\ of f_i(x)\equiv 0(mod\ p) ,\ \forall 1\leq i\leq k \ \ \ \ \ (20)

Is there a asymptotic formula for

\displaystyle \sum_{p\leq x}\rho_{f_1,...,f_k}(p)? \ \ \ \ \ (21)

Example 1 We give an example to observe what is involved. {f(x)=x^2+1}. We know {x^2+1\equiv 0 (mod \ p)} is solvable iff {p\equiv 1 (mod\ 4)} or {p=2}. One side is easy, just by Fermat little theorem, the other hand need Fermat descent procedure, which of course could be done by Willson theorem. In this case,

\displaystyle \sum_{p\leq n}\rho_f(p)=\# \ of \{primes \ of \ type\ 4k+1 \ in \ 1,2,...,n\} \ \ \ \ \ (22)

Which is a special case of Dirichlet prime theorem.

Let {K} be an algebraic number field, i.e. the finite field extension of rational numbers, let

\displaystyle \mathcal{O}_K=\{\alpha\in K, \alpha \ satisfied \ a\ monic \ polynomial\ in\ {\mathbb Z}[x]\} \ \ \ \ \ (23)


Dedekind proved that,

Theorem 5

  1. {\mathcal{O}_K} is a ring, we call it the ring of integer of {K}.
  2. He showed further every non-zero ideal of {\mathcal{O}_K} could write as the product of prime ideal in {\mathcal{O}_k} uniquely.
  3. the index of every non-zero ideal {I} in {\mathcal{O}_K} is finite, i.e. {[\mathcal{O}_K:I]<\infty}, and we can define the norm induce by index.

    \displaystyle N(I):=[\mathcal{O}_K:I] \ \ \ \ \ (24)

    Then the norm is a multiplication function in the space of ideal, i.e. {N(IJ)=N(I)N(J), \forall I,J \in \ ideal\ class\ group\ of\ \mathcal{O}_K}.

  4. Now he construct the Dedekind Riemann zeta function,

    \displaystyle \zeta_K(s)=\sum_{N(I)\neq 0}\frac{1}{N(I)^s}=\prod_{J\ prime \ ideal\ }\frac{1}{1-\frac{1}{N(J)^s}},\ \forall Re(s)>1 \ \ \ \ \ (25)


Now we consider the analog of the prime number theorem. Let {\pi_K(x)=\{I,N(I)<x\}}, does the exist a asymptotic formula,

\displaystyle \pi_K(x)\sim \frac{x}{ln x}\ as\ x\rightarrow \infty? \ \ \ \ \ (26)

Given a prime {p}, we may consider the prime ideal

\displaystyle p\mathcal{O}_K=\mathfrak{P}_1^{e_1}\mathfrak{P}_2^{e_2}...\mathfrak{P}_k^{e_k} \ \ \ \ \ (27)

Where {\mathfrak{P}_i } is different prime ideal in {\mathcal{O}_K}. But the question is how to find these {\mathfrak{P}_i}? For the question, there is a satisfied answer.

Lemma 6 (existence of primitive element) There always exist a primetive elements in {K}, such that,

\displaystyle K={\mathbb Q}(\theta) \ \ \ \ \ (28)

Where {\theta} is some algebraic number, which’s minor polynomial {f(x)\in {\mathbb Z}[x]}.

Theorem 7 (Dedekind recipe) Take the polynomial {f(x)}, factorize it in the polynomial ring {{\mathbb Z}_p[x]},

\displaystyle f(x)\equiv f_1(x)^{e_1}...f_{r}(x)^{e_r}(mod \ p) \ \ \ \ \ (29)

Consider {\mathfrak{P}_i=(p, f_i(\theta)) \subset \mathcal{O}_K}. Then apart from finite many primes, we have,

\displaystyle p\mathcal{O}_K=\mathfrak{P}_1^{e_1}\mathfrak{P}_2^{e_2}...\mathfrak{P}_k^{e_k} \ \ \ \ \ (30)

Where {N(\mathfrak{P}_i)=p^{deg{f_i}}}.

Remark 9 The apart primes are those divide the discriminant.

Now we can argue that 4 is morally the same as counting the ideals whose norm is divide by {p} in a certain algebraic number theory.

And we have following, which is just the version in algebraic number fields of 2.

Theorem 8 (Weber) {\#} of ideals of {\mathcal{O}_K} with norm {\leq x} equal to,

\displaystyle \rho_k(X)+O(x^{1-\frac{1}{d}}), where \ d=[K:Q] \ \ \ \ \ (31)


Diophantine approximation

I explain some general ideal in the theory of diophantine approximation, some of them is original by myself, begin with a toy model, then consider the application on folklore Swirsing-Schmidt conjecture.


1. Dirichlet theorem, the toy model

The very basic theorem in the theory of Diophantine approximation is the well known Dirichlet approximation theorem, the statement is following.

Theorem 1 (Dirichlet theorem) for all {\alpha} is a irrational number, we have infinity rational number {\frac{q}{p}} such that:

\displaystyle |\alpha-\frac{q}{p}|<\frac{1}{p^2} \ \ \ \ \ (1)


Remark 1 It is easy to see the condition of irrational is crucial. There is a best constant version of it, said, instead of {1}, the best constant in the suitable sense for the theorem 1 should be {\frac{1}{\sqrt{5}}} and arrive by {\frac{\sqrt{5}+1}{2}} at least. The strategy of the proof of the best constant version involve the Frey sequences.

Now we begin to explain the strategies to attack the problem.

\paragraph{Argument 1, boxes principle} We begin with a easiest one, i.e. by the argument of box principle, the box principle is following,

Theorem 2 (Boxes principle) Given {n\in {\mathbb N}} and two finite sets {A={a_1,a_2,...,a_n,a_{n+1}}}, set {B={b_1,...,b_{n}}}, if we have a map:

\displaystyle f:A\longrightarrow B \ \ \ \ \ (2)

Then there exists a element {b_k\in B} such that there exist at least two element {a_i,a_j\in A}, {f(a_i)=f(a_j)=b_k}.

Proof: The proof is trivial. \Box

Now consider, {\forall N\in {\mathbb N}}, the sequences {x,2x,...,Nx}, then {\{ix\}\in [0,1], \forall i\in \{1,2,...,n\}}. Divide {[0,1]} in an average way to {N} part: {[\frac{k-1}{N},\frac{k}{N}]}. Then the linear structure involve (which, in fact play a crucial role in the approach). And the key point is to look at {\{nx\}} and integers.

\paragraph{Argument 2, continue fractional} We know, for irrational number {x}, {x} have a infinite long continue fractional:

\displaystyle x=q_0+\frac{1}{q_1+\frac{1}{q_2+\frac{1}{q_3+....+\frac{1}{q_k+...}}}} \ \ \ \ \ (3)



\displaystyle |x-q_0+\frac{1}{q_1+\frac{1}{q_2+\frac{1}{q_3+....+\frac{1}{q_k}}}}|\sim \frac{1}{(q_1q_2...q_{k-1})^2q_k} \ \ \ \ \ (4)

And we have,

\displaystyle \frac{1}{q_1+\frac{1}{q_2+\frac{1}{q_3+....+\frac{1}{q_k}}}}=\frac{a_n}{b_n}, (a_n,b_n)=1 \ \ \ \ \ (5)

Then {b_n=O(q_1...q_k)}.

\paragraph{Argument 3, Bohr set argument} We begin with some kind of Bohr set:

\displaystyle B_p=I-\cup_{q\in \{0,1,...,p-1\}}(\frac{q}{p}-\frac{1}{p^2},\frac{q}{p}+\frac{1}{p^2}) \ \ \ \ \ (6)

The key point is the shift of Bohr set, on the vertical line i.e. {|B_p\cap B_{p+1}|} is very slow, and can be explained by

\displaystyle \frac{k}{p+1}+\frac{1}{(p+1)^2}>\frac{k}{p}-\frac{1}{p^2} \ \ \ \ \ (7)


\displaystyle \frac{1}{p^2}+\frac{1}{(p+1)^2}>\frac{k}{p(p+1)} \ \ \ \ \ (8)

in {|B_p \cap B_{p+1}|\sim \frac{1}{p(p+1)}} But in fact they are not really independent, as the number of Bohr sets increase, then you can calculate the correlation, thanks to the harmonic sires increasing very slowly, wwe can get something non trivial by this argument, but it seems not enough to cover the whole theorem 1.

\paragraph{Argument 4, mountain bootstrap argument} This argument is more clever than 3, although both two arguments try to gain the property we want in 1 from investigate the whole space {[0,1]} but not {x}, this argument is more clever.

Now I explain the main argument, it is nothing but sphere packing, with the set of balls

\displaystyle \Omega=\{B_{p,q}:=(\frac{q}{p}-\frac{1}{q^2},\frac{q}{p}+\frac{1}{p^2})| \forall p\in {\mathbb N}, 1\leq q\leq p-1 \} \ \ \ \ \ (9)

and define its subset

\displaystyle \Omega_l=\{B_{p,q}:=(\frac{q}{p}-\frac{1}{q^2},\frac{q}{p}+\frac{1}{p^2})| \forall 1\leq p\leq l, 1\leq q\leq p-1 \} \ \ \ \ \ (10)

Then {\Omega_l\subset \Omega}, and {\Omega =\cup_{l\in {\mathbb N}}\Omega_l}. If we can proof,

Lemma 3 For all {l\in {\mathbb N}}, there is a subset {A_l} of {\Omega-\Omega_l} such that {\cup_{i\in A_l}B_i=[0,1]}.

Remark 2 If we can proof 3, it is easy to see the theorem 1 follows.

Proof: The proof follows very standard in analysis, may be complex analysis? Key point is we start with a ball {B_{p,q}}, whatever it is, this is not important, the important thing is we can take some ball {B_{p',q'}} with the center of {B_{p',q'}} in {B_{p,q}}, then try to consider {B_{p',q'}\cup B_{p,q}} to extension {B_{p,q}} and then we find the boudary is also larger then we can extension again, step by step just like mountain bootstrap argument. So we involve in two possible ending,

  1. The extension process could extension {B{p,q}} to whole space.
  2. we can not use the extension argument to extension to the whole space.

If we are in the first situation, then we are safe, there is nothing need proof. If we are in second case, anyway we take a ball {B_{p,q}=(\frac{q}{p}-\frac{1}{p^2},\frac{q}{p}+\frac{1}{p^2})}. Then try to find good ball {B_{p',q'}} to approximate {B_{p,q}}, but this is difficult… \Box

Remark 3 Argument 1 is too clever to be true in generalization, argument 2 is standard, by the power of renormalization. argument 3 and argument 4 have gap… I remember I have got a proof similar to argument 4 here many years ago, but I forgot how to get it…

2. Schimidt conjecture

The Schimidt conjecture could be look as the generalization of Dirchlet approximation theorem 1 to algebraic number version, to do this, we need define the height of a algebraic number.

Definition 4 We say a number {\alpha\in {\mathbb C}} is a {k-}order algebraic number if and only is the minimal polynomial of {\alpha}, {f(x)=a_nx^n+...+a_1x+a_0, a_n\neq 0} have degree {deg(f)=n, f\in {\mathbb Z}[x]}.

Definition 5 (Height) Now we define the height of a {k-}th order algebraic number as {H(\alpha):=\max\{\|a_n\|_h,\|a_{n-1}\|_h,...,\|a_0\|_h\}}, Where

\displaystyle h(a_i)=\|a_i\|_{\infty} \ \ \ \ \ (11)

Now we state the conjecture:

Theorem 6 (Swiring-Schimidt conjecture) For all transendental number {x\in {\mathbb C}}, there is infinitely {\alpha} are {k-}th algebraic number such that:

\displaystyle |x-\alpha|<\frac{c_k}{H(\alpha)^{k+1}} \ \ \ \ \ (12)

Where {c_k} is a constant only related to {k} but not {x}.

I point out the conjecture is very related to the map:

\displaystyle F:(x_1,...,x_n) \longrightarrow (\sigma_1(x_1,...,x_n),\sigma_2(x_1,...,x_n),...,\sigma_n(x_1,...,x_n)) \ \ \ \ \ (13)

Where {\sigma_k(x_1,...,x_n)=\sum_{1\leq i_1<...<i_k\leq n}\Pi_{j=1}^kx_{i_1}x_{i_2}...x_{i_k}} is the {k-}th symmetric sum.

Remark 4 {F} is a map {{\mathbb C}^n\rightarrow {\mathbb C}^n}, what we consider is its inverse, {G=F^{-1}}, but {G} is not smooth, it occur singularity when {x_i=x_j} for some {i\neq j}. And the map, as we know, the singularity depend on the quantity {\Pi_{1\leq i< j\leq n}(x_i-x_j)}.

Remark 5 I then say something about the geometric behaviour of the map {G}, as we know, what we have in mind is consider the map {G} as a distortion {{\mathbb C}^n\rightarrow {\mathbb C}^n}, Then {H(\alpha)} is just the pullback of the canonical metric on {{\mathbb C}}(morally) to {{\mathbb C}}.


Discrete harmonic function in Z^n

There is some gap, in fact I can improve half of the argument of Discrete harmonic function , the pdf version is Discrete harmonic function in Z^n, but I still have some gap to deal with the residue half…


1. The statement of result

First of all, we give the definition of discrete harmonic function.

Definition 1 (Discrete harmonic function) We say a function {f: {\mathbb Z}^n \rightarrow {\mathbb R}} is a discrete harmonic function on {{\mathbb Z}^n} if and only if for any {(x_1,...,x_n)\in {\mathbb Z}^n}, we have:

\displaystyle f(x_1,...,x_n)=\frac{1}{2^n}\sum_{(\delta_1,...,\delta_n )\in \{-1,1\}^n}f(x_1+\delta_1,...,x_n+\delta_n ) \ \ \ \ \ (1)


In dimension 2, the definition reduce to:

Definition 2 (Discrete harmonic function in {{\mathbb R}^2}) We say a function {f: {\mathbb Z}^2 \rightarrow {\mathbb R}} is a discrete harmonic function on {{\mathbb Z}^2} if and only if for any {(x_1,x_2)\in {\mathbb Z}^2}, we have:

\displaystyle f(x_1,x_2)=\frac{1}{4}\sum_{(\delta_1,\delta_2)\in \{-1,1\}^2}f(x_1+\delta_1,x_2+\delta_2 ) \ \ \ \ \ (2)


The result establish in \cite{paper} is following:

Theorem 3 (Liouville theorem for discrete harmonic functions in {{\mathbb R}^2}) Given {c>0}. There exists a constant {\epsilon>0} related to {c} such that, given a discrete harmonic function {f} in {{\mathbb Z}^2} satisfied for any ball {B_R(x_0)} with radius {R>R_0}, there is {1-\epsilon} portion of points {x\in B_R(x_0)} satisfied {|f(x)|<c}. then {f} is a constant function in {{\mathbb Z}^2}.

Remark 1 This type of result contradict to the intuition, at least there is no such result in {{\mathbb C}}. For example. the existence of poisson kernel and the example given in \cite{paper} explain the issue.

Remark 2 There are reasons to explain why there could not have a result in {{\mathbb C}} but in {{\mathbb Z}^2},

  1. The first reason is due to every radius {R} there is only {O(R^2)} lattices in {B_R(x)} in {{\mathbb Z}^2} so the mass could not concentrate very much in this setting.
  2. The second one is due to there do not have infinite scale in {{\mathbb Z}^2} but in {{\mathbb C}}.
  3. The third one is the function in {{\mathbb Z}^2} is automatically locally integrable.


The generation is following:

Theorem 4 (Liouville theorem for discrete harmonic functions in {{\mathbb R}^n}) Given {c>0,n\in {\mathbb N}}. There exists a constant {\epsilon>0} related to {n,c} such that, given a discrete harmonic function {f} in {{\mathbb Z}^n} satisfied for any ball {B_R(x_0)} with radius {R>R_0}, there is {1-\epsilon} portion of points {x\in B_R(x_0)} satisfied {|f(x)|<c}. then {f} is a constant function in {{\mathbb Z}^n}.

In this note, I give a proof of 4, and explicit calculate a constant {\epsilon_n>0} satisfied the condition in 3, this way could also calculate a constant {\epsilon_n} satisfied 4. and point the constant calculate in this way is not optimal both in high dimension and 2 dimension.

2. some element properties with discrete harmonic function

We warm up with some naive property with discrete harmonic function. The behaviour of bad points could be controlled, just by isoperimetric inequality and maximum principle we have following result.

Definition 5 (Bad points) We divide points of {{\mathbb Z}^n} into good part and bad part, good part {I} is combine by all point {x} such that {|f(x)|<c}, and {J} is the residue one. So {A\amalg B={\mathbb Z}^n}.

For all {B_R(0)}, we define {J_R:=J\cap B_R(0), I_R=I\cap B_R(0)} for convenient.

Theorem 6 (The distribution of bad points) For all bad points {J_R} in {B_R(0)}, they will divide into several connected part, i.e.

\displaystyle J_R=\amalg_{i\in S_R}A_i \ \ \ \ \ (3)

and every part {A_i} satisfied {A_i\cap \partial B_R(0)\neq \emptyset}.

Remark 3 We say {A} is connected in {{\mathbb Z}^n} iff there is a path in {A} connected {x\rightarrow y, \forall x,y\in A}.

Remark 4 the meaning that every point So the behaviour of bad points are just like a tree structure given in the gragh.

Proof: A very naive observation is that for all {\Omega\subset {\mathbb Z}^n} is a connected compact domain, then there is a function

\displaystyle \lambda_{\Omega}: \partial \longrightarrow {\mathbb R} \ \ \ \ \ (4)

such that {\lambda_{\Omega}(x,y)\geq 0, \forall (x,y)\in \Omega \times \mathring{\Omega}}. And we have:

\displaystyle f(x)=\sum_{y\in\partial \Omega}\lambda_{\Omega}(x,y)(y) \ \ \ \ \ (5)


This could be proved by induction on the diameter if {\Omega}. Then, if there is a connected component of {\Omega} such that contradict to theorem 6 for simplify assume the connected component is just {\Omega}, then use the formula 5we know

\displaystyle \begin{array}{rcl} \sup_{x\in \Omega}|f(x)| & = & \sup_{x\in \Omega}\sum_{y\in\partial \Omega}\lambda_{\Omega}(x,y)(y) \\ & \leq & \sup_{\partial \Omega}|f(x)| \\ & \leq & c \end{array}

The last line is due to consider around {\partial \Omega}. But this lead to: {\forall x\in \Omega, |f(x)|<c} which is contradict to the definition of {\Omega}. So we get the proof. \Box

Now we begin another observation, that is the freedom of extension of discrete harmonic function in {{\mathbb Z}^n} is limited.

Theorem 7 we can say something about the structure of harmonic function space of {Z^n}, the cube, you will see, if add one value, then you get every value, i.e. we know the generation space of {Z^n}

Proof: For two dimension case, the proof is directly induce by the graph. The case of {n} dimensional is similar. \Box

Remark 5 The generation space is well controlled. In fact is just like n orthogonal direction line in n dimensional case.

3. sktech of the proof for \ref


The proof is following, by looking at the following two different lemmas establish by two different ways, and get a contradiction.

\paragraph{First lemma}

Lemma 8 (Discrete poisson kernel) the poisson kernel in {{\mathbb Z}^n}. We point out there is a discrete poisson kernel in {{\mathbb Z}^n}, this is given by:

\displaystyle f(x)=\sum_{y\in \partial B_R(z)}\lambda_{B_R(z)}(x,y)(y) \ \ \ \ \ (6)

And the following properties is true:

  1. {\lambda_{B_R(z+h)}(x+h,y+h)=\lambda_{B_R(z)}(x,y)} , {\forall x\in \Omega, h\in {\mathbb Z}^n}.
  2. \displaystyle \lambda_{B_R(z)}(x,y)\rightarrow \rho_R(x,y) \ \ \ \ \ (7)

Remark 6 The proof could establish by central limit theorem, brown motion, see the material in the book of Stein \cite{stein}. The key point why this lemma 8 will be useful for the proof is due to this identity always true {\forall x\in B_R(0)}, So we will gain a lots of identity, These identity carry information which is contract by another argument.

\paragraph{Second lemma} The exponent decrease of mass.

Lemma 9 The mass decrease at least for exponent rate.

Remark 7 the proof reduce to a random walk result and a careful look at level set, reduce to the worst case by brunn-minkowski inequality or isoperimetry inequality.

\paragraph{Final argument} By looking at lemma 1 and lemma 2, we will get a contradiction by following way, first the value of {f} on {\partial B_R(0)} increasing too fast, exponent increasing by lemma2, but on the other hand, it lie in the integral expresion involve with poisson kernel, but the pertubation of poisson kernel is slow, polynomial rate in fact…



\bibitem{stein} Functional analysis


Log average sarnak conjecture


This is a note concentrate on the log average Sarnak conjecture, after the work of Matomaki and Raziwill on the estimate of multiplication function of short interval. Given a overview of the presented tools and method dealing with this conjectue.


1. Introduction

Sarnak conjecture \cite{Sarnak} assert that for any obersevable {\{f(T^n(x_0))\}_{n=1}^{\infty}} come from a determination systems {(T,X),T:X\rightarrow X}, where {h(T)=0}, {x_0\in X, f\in C(X)}. The correlation of it and the Liuvillou function is 0, i.e. they are orthongonal to each other, more preseicesly it is just to say,

\displaystyle \sum_{n<x}\mu(x)f(T^n(x_0))=o(x) \ \ \ \ \ (1)


This is a very natural raised conjecture, Liuville function is the presentation of primes, due to we always believe the distribution of primes in {\mathbb N} should be randomness.

It has been known as observed by Landau \cite{Laudau} that the simplest case,

\displaystyle \sum_{n<x}\mu(n)=o(x)

already equivalent to the prime number theorem. It is not difficult to deduce the spetial case of Sarnak conjecture when with the obersevation in $latex {(1)}&fg=000000$ come from finite dynamic system is equivalent to the prime number theorem in athremetic progress by the similar argument. Besides this two classical result, may be the first new result was established by Davenport,

Theorem 1 Let {T:S_1\rightarrow S_1, T(x)=x+\alpha}, {\alpha} is a inrational, then the obersevation come from {(T,S_1)} is orthogonal to Mobius function. due to {\{e^{2\pi ikx}\}_{k\in \mathbb Z}} is a basis of {C(S_1)}, suffice to proof,

\displaystyle \sum_{n<x}e^{2\pi ikn\alpha}\mu(n)=o(x), \forall k\in \mathbb N

There is a lots of spetial situations of Sarnak’s conjecture have been established, The parts I mainly cared is the following:

  1. Interval exchange map.
  2. Skew product flow.
  3. Obersevable come from One dimensional zero entropy flow.
  4. Nilsequences.

But in this note, I do not want to explain the tecnical and tools to establish this result, but considering an equivalent conjecture of Sarnak conjecture, named Chowla conjecture, and explain the underlying insight of the suitable weak statement, i.e. the log average Chowla conjecture and the underlying insight of it.

The note is organized as following way, in the next section $latex {(2)}&fg=000000$, we give a self-contained introduction on the tools called Bourgain-Sarnak-Ziegler critation, explain the relationship of this critation and the sum-product phenomenon, also given some more general critation along the philosephy use in establish the Bourgain-Sarnak-Ziegler critation, which maybe useful in following development combine with some other tools. The key point is transform the sum from linear sum to bilinear sum and decomposition the bilinear sum into diagonal part and off-diagonal part, use the assume in the critation to argue the off-diagonal part is small and on the orther hand the diagonal part is also small by the trivial estimate and the volume of diogonal is small, this is very similar to a suitable Caderon-Zugmund decomposition.

In section $latex {(4)}&fg=000000$, I try to give a proof sketch of the result of Matomaki and Raziwill, which is also a key tools to understanding the Sarnak conjecture, or equivalent the Chowla conjecture. The key points of the proof contains following:

  1. Find a suitable fourier indentity
  2. Construct a multiplication-addition dense subset {S}, and proof that the theorem MR hold we need only to proof it hold for {S\cap [1,2,...,n]} instead of {[1,2,...,n]}
  3. Involve the power of euler product formula. divide the whole interval into a lot of small interval with smaller and smaller scale and a residue part. We look the part come from every small scale as a major term and look the residue part as minor term.
  4. Deal with the major term at every scale, by a combitorios identity and second moments method.
  5. find a enough decay estimate from a scale to the next smaller scale.
  6. Deal with the minor term by the H… lemma.

Due to the theorem of MR do not exausted the method they developed, we trying to make some more result with their method, Tao and Matomaki attain the average version of Chowla conjecture is true by this way, and combine this argument and the entropy decresment argument they established the 2 partten of the log average Chowla conjecture is true. Very recently Tao and his coperator proved the odd partten case of log average chowla conjecture is true, combine an argument of frustenberg crresponding principle and entopy decresment argument. But it seems the even and large than 2 case is much difficult and seems need something new to combine with the method of MR and entropy decresment and frunstenberg corresponfing principle to make some progress.

So, in section $latex {(5)}&fg=000000$, we give a self-contain introduction to the entropy decresment argument of Tao, and combine with the frustenberg corresponding principle.

In the last section $latex {(6)}&fg=000000$, I state some result and method and phylosphy of them I get on nilsequences and wish to combine them with the previous method to make some progress on log average Chowla conjecture on the even partten case.


2. Bourgain-Sarnak-Zieglar creation

We begin with the easiest one, this is the main result established in \cite{BSZ}, I try to give the main ideal under the proof, but with a no quantitative version is the following,

Theorem 2 (Bourgain-Sarnak-Zieglar creation, not quantitative version) if for all primes {p,q>>1} we have:

\displaystyle \sum_{n=1}^Nf(T^{pn}(x))\overline{f(T^{qn}(x))}=o(N) \ \ \ \ \ (2)


Then for multiplication function {g(n)} we have

\displaystyle \sum_{n=1}^Ng(n)\overline{ f(T^n(x))}=o(N) \ \ \ \ \ (3)


Remark 1 For simplify we identify {f(T^n(x)):=F(n)}.

Remark 2

The idea is following, break the sum into a bilinear one, so, of course, we multiplication it with itself. i.e. we consider to control,

\displaystyle |\sum_{i=1}^Ng(n)\overline{ F(n)}|^2=\sum_{n=1}^N\sum_{m=1}^Ng(n)g(m)\overline{F(n)F(m)} \ \ \ \ \ (4)


To control 4, we need exhausted the mutiplication property of {g(n)}, we have {g(mn)=g(n)g(m),\forall\ m,n\in {\mathbb N}}. We can not get good estimate for all term,

\displaystyle g(n)g(m)\overline{F(n)F(m)} \ \ \ \ \ (5)

The condition in our hand if following,

\displaystyle \sum_{n=1}^NF(pn)\overline{F(qn)}=o(N), \forall \ p,q\in \mathop{\mathbb P} \ \ \ \ \ (6)

So, just like the situation of Cotlar-Stein lemma \cite{Cotlar-Stein lemma}, we wish to estimate like following:

\displaystyle \begin{array}{rcl} |\sum_{p\in W}\sum_{n\in V}F(pn)g(pn)| & \leq & \sum_{n\in V}|g(n)|\cdot |\sum_{p\in W}F(pn)g(p)| \\ & \leq &\sum_{n\in V}|\sum_{p \in W}F(pn)g(p)|\\ & \overset{Cauchy-Schwarz}\leq & |V|^{\frac{1}{2}}[\sum_{n\in V}|\sum_{p\in W}F(pn)g(p)|^2]^{\frac{1}{2}}\\ & = & |V|^{\frac{1}{2}}[\sum_{p_1,p_2\in W}\sum_{n\in V}F(p_1n)\overline{F(p_2n)}g(p_1)\overline{g(p_2)}]^{\frac{1}{2}}\\ \end{array}

Then we consider divide the sum into diagonal part and non-diagonal part, as following,

\displaystyle |V|^{\frac{1}{2}}[\sum_{p_1\neq p_2\in W}\sum_{n\in V}F(p_1n)\overline{F(p_2n)}g(p_1)\overline{g(p_2)}]^{\frac{1}{2}}+|V|^{\frac{1}{2}}[\sum_{p\in W}\sum_{n\in V}|F(pn)|^2]^{\frac{1}{2}} \ \ \ \ \ (7)

But the first part is small, i.e.

\displaystyle |V|^{\frac{1}{2}}[\sum_{p_1\neq p_2\in W}\sum_{n\in V}F(p_1n)\overline{F(p_2n)}g(p_1)\overline{g(p_2)}]^{\frac{1}{2}} =o(|W||V|) \ \ \ \ \ (8)

Because of

\displaystyle \sum_{n\in V}F(p_1n)\overline{F(p_2n)}=o(V), \forall p_1\neq p_2\in W \ \ \ \ \ (9)

and the second part is small, i.e.

\displaystyle |V|^{\frac{1}{2}}[\sum_{p\in W}\sum_{n\in V}|F(pn)|^2]^{\frac{1}{2}}=o(|W||V|) \ \ \ \ \ (10)

Because diagonal part is small in {W\times W} and trivial inequality

\displaystyle \sqrt{\sum_{n\in V}|F(pn)|^2}\leq |V|^{\frac{1}{2}} \ \ \ \ \ (11)

But the method in remark 2 is not always make sense in any situation, we need to construct two suitable sets {W,V} and then break up {\{1,2,...,n{\mathbb N}\}} into {W\times V}, this mean,

\displaystyle \{1,2,...,N\}\sim W\times V+o(N) \ \ \ \ \ (12)

But this {W,V} could be construct in this situation, thanks to the prime number theorem,

Theorem 3 (Prime number theorem)

\displaystyle \pi(n)\sim \frac{n}{ln(n)} \ \ \ \ \ (13)

Morally speaking, this is the statement that the primes, which is the generator of multiplication function, is not very sparse.

3. Van der curpurt trick

There is the statement of Van der carport theorem:

Theorem 4 (Van der curpurt trick) Given a sequences { \{x_n\}_{n=1}^{\infty}} in { S_1}, if { \forall k\in N^*}, { \{x_{n+k}-x_n\}} is uniformly distributed, then { \{x_n\}_{n=1}^{\infty}} is uniformly distributed.

I do not know how to establish this theorem with no extra condition, but this result is true at least for polynomial flow. \newpage Proof:

\displaystyle \begin{array}{rcl} |\sum_{n=1}^Ne^{2\pi imQ(n)}|& = &\sqrt{(\sum_{n=1}^Ne^{2\pi imQ(n)})(\overline{\sum_{n=1}^Ne^{2\pi imQ(n)}})}\\ & = &\sqrt{\sum_{h_1=1}^N\sum_{n=1}^{N-h_1}e^{2\pi imQ(n+h_1)-Q(n)}}\\ & = &\sqrt{\sum_{h_1=1}^N\sum_{n=1}^{N-h_1}e^{2\pi im \partial^1_{h_1}Q(n)}}\\ & \leq & \sqrt{\sum_{h_1=1}^N|\sum_{n=1}^{N-h_1}e^{2\pi \partial^1_{h_1}Q(n)}|}\\ & = &\sqrt{\sum_{h_1=1}^N\sqrt{ (\sum_{n=1}^{N-h_1}e^{2\pi \partial^1_{h_1}Q(n)} )(\overline{\sum_{n=1}^{N-h}e^{2\pi \partial^1_{h_1}Q(n)})}}}\leq\sqrt{\sum_{h_1=1}^N\sqrt{ \sum_{h_2=1}^N|\sum_{n=1}^{N-h_1}e^{2\pi\partial^1_{h_2} \partial^1_hQ(n)} |}}\\ & \leq ....\leq & \\ & = & \sqrt{\sum_{h_1=1}^N\sqrt{ \sum_{h_2=1}^N \sqrt{....\sqrt{\sum_{h_{k-1}=1}^{N-h_{k-2}}|\sum_{n=1}^{N-h_{k-1}}e^{2\pi\partial_{h_1h_2...h_{k-1}Q(n)}}|}}}} =o(1) \end{array}


This type of trick could also establish the following result, which could be understand as a discretization of the Vinegradov lemma.

Remark 3

Uniformly distribution result of { F_p}: Given {Q(n)=a_kn^k+...+a_1n+a_0}, {\{Q(0),Q(1),...,Q(p-1)\}} coverages to a uniformly distribution in {\{0,1,...,p-1\}} as {p \rightarrow \infty}.

Remark 4 But I definitely do not know how to establish the similar result when {Q(n)=n^{-1}}.

Remark 5

This trick could also help to establish estimate of correlation of low complexity sequences and multiplicative function, such as result:

\displaystyle S(x)=\sum_{n\le x}\left(\frac{n}{p}\right)\mu(n)=o(n)

Maybe with the help of B-Z-S theorem.


4. Matomaki and Raziwill’s work

In this section we explain the main idea underlying the paper \cite{KAISA MATOMA 虉KI AND MAKSYM RADZIWILL}. But play with a toy model, i.e. the corresponding corollary of the original result on Liouville鈥檚 function.

Definition 5 (Lioville’s function)

\displaystyle \lambda(n)=(-1)^{\alpha_1+\alpha_2+...+\alpha_k}, \forall \ n=p_1^{\alpha_1}...p_k^{\alpha_k}. \ \ \ \ \ (14)

Remark 6

\displaystyle |\int_{X}^{2X}\lambda(n)dx|=o(x) \ \ \ \ \ (15)

is equivalent to the prime number theorem 3.

The most important beakgrouth of analytic number theory is the new understanding of multiplication function on share interval, this result is established by Kaisa Matom盲ki and Maksym Radziwill. Two very young and intelligent superstars.

The main theorem in them article is :

Theorem 6 (Matomaki,Radziwill) As soon as {H\rightarrow \infty} when {x\rightarrow \infty}, one has:

\displaystyle \sum_{x\leq n\leq x+H}\lambda(n)= o(H) \ \ \ \ \ (16)

for almost all {1\leq x\leq X} .

In my understanding of the result, the main strategy is:

  1. Parseval indetity, transform to Dirchelet polynomial.
  2. Involved by multiplication property, spectral decomposition.
  3. From linear to multilinear , Cauchy schwarz inequality.
  4. major term estimate.
  5. Estimate the contribution of area which is not filled.

4.1. Parseval indetity, transform to Dirchelet polynomial

We wish to establish the equality,

\displaystyle \frac{1}{X}\int_{X}^{2X}|\sum_{x\leq n\leq x+H}\lambda(n)|dx=o(H) \ \ \ \ \ (17)

This is the {L^1} norm, by Chebyschev inequality, this could be control by {L^2} norm, so we only need to establish the following,

\displaystyle \frac{1}{X}\int_X^{2 X}|\sum_{x\leq n\leq x+H}\lambda(n)|^2dx=o(H^2) \ \ \ \ \ (18)


We wish to transform from the discretization sum to a continue sum, that is,

\displaystyle \int_{{\mathbb R}}|\sum_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(n)1_{X\leq n\leq 2X}|^2\frac{dx}{x} \ \ \ \ \ (19)


Remark 7 There are two points to understand why 19 and 18 are the same.

  1. {[xe^{-\frac{1}{T}},xe^{\frac{1}{T}}]\sim [x-H,x+H]}.
  2. {1_{x\leq n\leq 2x}} and {\frac{1}{x}} is to make that {x=O(X)}.

So the Magnitude of 18 and 19 are the same. i.e.

\displaystyle \int_{{\mathbb R}}|\sum_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(n)1_{X\leq n\leq 2X}|^2\frac{dx}{x}\sim \frac{1}{X}\int_X^{2 X}|\sum_{x\leq n\leq x+H}\lambda(n)|^2dx \ \ \ \ \ (20)

Now we try to transform 19 by Parseval indetity, this is something about the {L^2} norms of the quality we wish to charge. It is just trying to understanding 19 as a quantity in physical space by a more chargeable quality in frequency space. Image,

\displaystyle \int_{{\mathbb R}}|\sum_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(n)1_{X\leq n\leq 2X}|^2\frac{dx}{x}:=\int_{{\mathbb R}}|f_X(x)|^2dx \ \ \ \ \ (21)

Then {f_X(x)=\int_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(x)1_{X\leq n\leq 2X}}. Note that,

\displaystyle \begin{array}{rcl} \widehat{f_X(\xi)} & = & \int_{{\mathbb R}}f_X(x)e^{2\pi ix\xi}dx\\ & = & \sum_{x\leq n\leq 2x}\lambda(x)\int_{logn-\frac{1}{T}}^{logn+\frac{1}{T}}e^{2\pi ix\xi}dx, \ T=\frac{X}{H}\\ & = & \sum_{X\leq n\leq 2X}\lambda(x)e^{2\pi ilog(n)\cdot \xi}\cdot\frac{e^{2\pi i\frac{\xi}{T}}-e^{2\pi i-\frac{\xi}{T}}}{2\pi i\xi}\\ \end{array}

So by Parseval identity, we have,

\displaystyle \begin{array}{rcl} \int_{{\mathbb R}}|f_X(x)|^2dx & = & \int_{{\mathbb R}}|\widehat{f_X(\xi)}|^2d\xi \\ & = & \int_{{\mathbb R}}|\sum_{X\leq n\leq 2X}\lambda(n)\cdot n^{2\pi i\xi}|^2(\frac{e^{2\pi i\frac{\xi}{T}}-e^{2\pi i\frac{-\xi}{T}}}{2\pi i\xi})^2d\xi\\ & \sim & \int_{{\mathbb R}}|\sum_{X\leq n\leq 2X}\lambda(n)\cdot n^{2\pi i\xi}|^2\frac{1}{T^2}1_{|\xi|^2\leq T}\\ \end{array}

Remark 8 We know the Fejer kernel satisfied,

\displaystyle (\frac{e^{2\pi i\frac{\xi}{T}}-e^{2\pi i\frac{-\xi}{T}}}{2\pi i\xi})^2\sim \frac{1}{T^2}1_{|\xi|\leq T} \ \ \ \ \ (22)

So morally speaking, we get the following identity.

\displaystyle \frac{1}{X}\int_{X}^{2X}|\sum_{x\leq n\leq x+H}\lambda(n)|^2dx\sim \frac{1}{(x/H)^2}\int_{0}^{\frac{X}{H}}|\sum_{x\leq n\leq 2x}\lambda(x)x^{2\pi i\xi}|^2d\xi \ \ \ \ \ (23)

In fact we do a cutoff, the quality we really consider is just:

\displaystyle \frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt \ \ \ \ \ (24)

established the monotonically inequality:

Theorem 7 (Paserval type identity)

\displaystyle \frac{1}{X}\int_{X}^{2X}|\frac{1}{H}\sum_{x\leq n\leq x+H}\lambda(n)|^2dx \sim聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt \ \ \ \ \ (25)


Remark 9

In my understanding, This is a perspective of the quality, due to the quality is a multiplicative function integral on a domain { \mathbb N^*} with additive structure, it could be looked as a lots of wave with the periodic given by primes, so we could do a orthogonal decomposition in the fractional space, try to prove the cutoff is a error term and we get such a monotonically inequality.

But at once we get the monotonically inequality, we could look it as a聽compactification process and this process still carry most of the information so lead to the inequality.

It seems something similar occur in the attack of the moments estimate of zeta function by the second author. And it is also could be looked as something similar to the 聽spectral decomposition with some basis come from multiplication generators, i.e. primes.

4.2. Involved by multiplication property, spectral decomposition

I called it is “spectral decomposition”, but this is not very exact. Anyway, the thing I want to say is that for multiplication function {\lambda(n)}, we have Euler-product formula:

\displaystyle \Pi_{p,prime}(\frac{1}{1-\frac{\lambda(p)}{p^s}})=\sum_{n=1}^{\infty} \frac{\lambda(n)}{n^s} \ \ \ \ \ (26)


But anyway, we do not use the whole power of multiplication just use it on primes, i.e. {\lambda(pn)=\lambda(p)\lambda(n)} leads to following result:

\displaystyle \lambda(n)=\sum_{n=pm,p\in I}\frac{\lambda(p)\lambda(m)}{\# \{p|m, p\in I\}+1}+\lambda(n)1_{p|n;p\notin I} \ \ \ \ \ (27)

This is a identity about the function {\lambda(n)}, the point is it is not just use the multiplication at a point,i.e. {\lambda(mn)=\lambda(m)\lambda(n)}, but take average at a area which is natural generated and compatible with multiplication, this identity carry a lot of information of the multiplicative property. Which is crucial to get a good estimate for the quality we consider about.

4.3. From linear to multilinear , Cauchy schwarz

Now, we do not use one sets {I}, but use several sets {I_1,...,I_n } which is carefully chosen. And we do not consider [X,2X] with linear structure anymore , instead reconsider the decomposition:

{[X,2X]=\amalg_{i=1}^n (I_i\times J_i) \amalg U}

On every {I_i\times J_i} it equipped with a bilinear structure. And {U} is a very small set, {|U|=o(X)} which is in fact have much better estimate.

{\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt =\sum_{i=1}^n\int_{I_i\times J_i}聽聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt +\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt}

Now we just use a Cauchy-Schwarz:

{\sum_{i=1}^n\int_{I_i\times J_i}聽聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt +\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt}

4.4. major term estimate

{=\sum_{i=1}^n\int_{I_i\times J_i}聽聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt}

{\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt}

4.5. estimate the contribution of area which is not filled


5. Entropy dcrement argument


6. Correlation with nilsequences

I wish to establish the following estimate: {\lambda(n)} is the liouville function we wish the following estimate is true.

\displaystyle \int_{0\leq x\leq X}|\sup_{f\in \Omega^m}\sum_{x\leq n\leq x+H}\lambda(n)e^{2\pi if(x)}|dx =o(XH). \ \ \ \ \ (28)

Where we have { H\rightarrow \infty} as { x\rightarrow \infty},

\displaystyle \Omega^m=\{a_mx^m+a_{m-1}x^{m-1}+...+a_1x+a_0 | a_m,...,a_1,a_0\in [0,1]\}

is a compact space.

I do not know how to prove this but this is result is valuable to consider, because by a Fourier identity we could transform the difficulty of (log average) Chowla conjecture to this type of result.

There is some clue to show this type of result could be true, the first one is the result established by Matomaki and Raziwill in 2015:

Theorem 8 (multiplication function in short interval)

{f(n): \mathbb N\rightarrow \mathbb C} is a multiplicative function, i.e. { f(mn)=f(n)f(m), \forall m,n\in \mathbb N}. {H\rightarrow \infty} as {x\rightarrow \infty}, then we have the following result,

\displaystyle \int_{1\leq x\leq X}|\sum_{x\leq n\leq x+H}f(n)|=o(XH). \ \ \ \ \ (29)

And there also exists the result which could be established by Vinagrodov estimate and B-S-Z critation :

Theorem 9 (correlation of multiplication function and nil-sequences in long interval)

{f(n): \mathbb N\rightarrow \mathbb C} is a multiplicative function, i.e. { f(mn)=f(n)f(m), \forall m,n\in \mathbb N}. {g(n)=a_n^m+...+a_1n+a_0} is a polynomial function then we have the following result,

\displaystyle \int_{1\leq n \leq X}|f(n)e^{2\pi i g(n)}|=o(X) \ \ \ \ \ (30)

\newpage {9} \bibitem{Sarnak} Peter Sarnak, Mobius Randomness and Dynamics.

\texttt{ }. \bibitem{Laudau} JA 虂NOS PINTZ (BUDAPEST). LANDAU鈥橲 PROBLEMS ON PRIMES.

\texttt{} \bibitem{BSZ} Knuth: Computers and Typesetting,


\bibitem{Cotlar-Stein lemma} Almost orthogonality