Equidistribution of sequences on the p-adic unit ball · Equidistribution of sequences on the p-adic unit ball Naveen Somasunderam Department of Mathematics Oregon State Univeristy,

Equidistribution of sequences on the p-adic unit ball

Naveen SomasunderamDepartment of Mathematics

Oregon State Univeristy, Corvallis

September 5, 2018

Naveen Somasunderam Department of Mathematics Oregon State Univeristy, CorvallisEquidistribution of sequences on the p-adic unit ball September 5, 2018 1 / 39

Outline

1 Introduction: Equidistribution on R/Z

2 Weyl’s Criterion

3 Quantitative behavior

4 p-adic fields: A quick intro

5 Equidistribution on the p-adic ball

6 Fourier Analysis on the p-adic ball

7 Discrepancy and quantitative behavior on Zp

8 Conclusions and Future work

9 References


Introduction

Consider a sequence {xn} in R/Z.

How well does this sequence distribute itself ? i.e. Does it hit everysubinterval ‘propotionally’ or ‘evenly’ as n increases, or is it moreconcentrated in certain parts and less so on others ? Are there gaps ?

Example: :

i. {xn = 1/n} is concentrated towards zero.ii. {xn = nα}, α-irrational is dense in R/Z (Ergodic theory). But how

does the sequence distribute itself as n increases ? We want a deepernotion than density.


Introduction

Definition

The sequence {xn} is equidistributed (or uniformly distributed) in R/Z ifgiven any interval A = [a, b] we have

limN→∞

#|{x1, x2, ..., xN} ∩ [a, b]|N

= b − a.

i.e. The proportion of the first N elements lying in A is equal to thelength of A in the limit of large N.

Question: Should this definition be extended to more general Borelmeasurable sets ? No hope: Think rationals and the set of irrationals.


Introduction

Question: Why would this be important ?

1. It has connections to fourier sums.

2. Theory of Riemann Integration.

3. Independent interest in number theoryExample: Consider the sequence xn = nα. This is dense iffα ∈ [0, 1)−Q.


Introduction: Equidistribution and Density

1. Any sequence has a rearrangement that is not equidistributed.

2. If {xn} is dense, then there is a rearrangement that is equidistributed(In particular, there is a rearrangement of the rationals that isequidistributed).

Proof. Pigeon Hole Principle.


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Weyl’s Criterion

We have the following connection to Fourier analysis

Theorem

(Weyl’s Criterion)The sequence {xn} is equidistributed in R/Z if and only if

limN→∞

1

N

N∑n=1

e2πikxn = 0,

for every k ∈ Z− {0}.

Note that ∫ 1

0e2πikx dx = 0,

for all k 6= 0.


Weyl’s Criterion

As a consequence we have the following corollary

Corollary

The sequence {nα} is equidistributed if and only if α is irrational.

Proof. ( =⇒ ) If α is rational, then there exists a k such that e2πiknα = 1 forevery n, and hence the sum

1

N

N∑n=1

e2πikxn = 1.

(⇐= ) Using the geometric sum formula we have

1

N

∣∣∣∣∣N∑

n=1

e2πiknα

∣∣∣∣∣ =1

N

∣∣∣∣1− e2πikNα

1− e2πikα

∣∣∣∣≤ 1

N

∣∣∣∣sin(2πkNα/2)

sin(2πkα/2)

∣∣∣∣≤ 1

N

1

| sin(2πkα/2)|,

and the latter term goes to zero as N tends to infinity. Hence, by Weyl’s criterionthe sequence {xn} is equidistributed.


Weyl’s Criterion

The definition of equidistribution implies the following:

Observation: Suppose E is an interval in [0, 1) and XE be the characteristicfunction of E . If {xn} is equidistributed we have

limN→∞

1

N

N∑n=1

XE (xn) =

∫ 1

0XE dx .

This leads us to a general theorem (Weyl’s criterion),

Theorem

Let {xn} be a sequence in R/Z. Then the following are equivalent

i. For all Riemann integrable f : [0, 1]→ C we have

limN→∞

1

N

N∑n=1

f (xn) =

∫ 1

0f dx .

ii. {xn} is equidistributed.


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Quantitative behavior

One also wants to measure how well a sequence distributes itself. Tothis end we define the notion of Discrepancy DN .

Definition

The discrepancy of a finite sequence {x1, x2, .., xN} is defined to be

DN = sup0≤a≤b≤1

∣∣∣∣#|{x1, x2, ..., xN} ∩ [a, b]|N

− (b − a)

∣∣∣∣ .Fact. A sequence {xn} is equidistributed iff limN→∞DN = 0.

Discrepancy can be used to quantify the idea that some sequences arebetter equidistributed than others.


Quantitative behavior: LeVeque Inequality

The sequence {x1, .., xN} satisfies

D3N ≤

6

π2

∞∑k=1

1

k2

∣∣∣∣∣ 1

N

N∑n=1

e2πi kxn

∣∣∣∣∣2

.

Key idea: Use geometric arguments along with Parseval’s theorem.


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p-adic fields

The p-adic norm on Q is defined as follows (for p-prime)

Definition

Given x ∈ Q, we first represent it as

x = par

s

where p 6 |r , p 6 |s. We define the p-adic absolute value of x to be

|x |p = p−a,

for x 6= 0 and |x |p = 0 for x = 0.

Example:

1. |1|3 = 1, |2|3 = 1, |3|3 = 1/3, |9|3 = 1/32.

2. |1/3|3 = 3, |1/9|3 = 32.

3. |10|5 = 1/5 (since 10 = 5× 2).


p-adic unit ball Zp

Let ai take values in {0, 1, 2, ..., p − 1}.Define the set of Canonical expansions

Zp ={x | x = a0 + a1p + a2p

2 + ....},

with the p-adic metric.

Since | . |p takes on only discrete values, we have r = 1/pn, n ∈ Z.

The metric topology is totally disconnected due to the Strongtriangle inequality:

|x + y |p ≤ max(|x |p, |y |p).


p-adic unit ball Zp

An open ball or disk D(a, r) centered at a and of radius 1/pr in Zp is theset

D(a, r) = {x | |x − a|p < p−r}.

Properties:

1. If b ∈ D(a, r), then D(b, r) = D(a, r). That is, every point of a ballis also its center .

2. The open ball D(a, r) is both open and closed in Zp.

3. Two balls is Zp have a non-empty intersection if and only if one iscontained in the other.


p-adic unit ball Zp

Note. Keep this picture in mind when we define Equidistribution in Zp.


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Equidistribution on Zp

We make the following definitions (See Kuipers [3]),

Definition

1 Given a ball of raidus 1/pk centered at a and a sequence {xn} we denote byA(a, k ,N)

A(a, k ,N) =#(D(a, k) ∩ {x1, ..., xN})

N.

2 A sequence {xn} is said to be equidistributed in Zp if

limN→∞

∣∣∣∣A(a, k ,N)− 1

pk

∣∣∣∣ = 0,

for every a in Zp and every positive integer k .

Note. The Haar measure of a ball is its radius, hence the 1/pk termabove.


Equidistribution on Zp: Weyl’s criterion

What about a Weyl’s criterion for sequences in Zp ? We have thefollowing theorem

Theorem

(Weyl’s Criterion)A sequence {xn} in Zp is equidistributed if and only if for every Riemannintegrable function f on Zp we have

limN→∞

1

N

N∑n=1

f (xn) =

∫Zp

f (x) dµ.

Note. By a Riemann integral we mean taking partitions of Zp (usingdisjoint open balls) and not the y-axis.


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Fourier analysis on Zp

Fact.

Zp is an LCA group, and hence has a Haar measure. Let µ be thenormalized Haar measure. Then the measure of a ball is its radius

µ(D(a, r)) =1

pr.

We can do Fourier analysis.

Definition

(Dual Group)The set of all continuous group homomorphisms (Characters)

Zp = {γ : Zp → T | γ(x + y) = γ(x)γ(y)},

where T = {z | |z | = 1} ⊂ C, is a group under multiplication called theDual Group.



Fact.

1. Let f ∈ L1(Zp). The fourier transform f : Zp −→ C of f is given by

f (γ) =

∫Zp

f (x)γ(x) dµ,

where µ is the normalized Haar measure on Zp.

2. The fourier inversion formula gives

f (x) =∑γ∈Zp

f (γ)γ(x).



The dual group Zp of Zp is the set of all characters (continuousgroup homomorphisms from Zp to the circle T). A characterγm,n : Zp → T is of the form

γm,n : Zp −→ Tx 7−→ e

2πimbxcnpn .

for m, n ∈ N, and 0 < m < pn, with p - m.

bxcn indicates the truncation of x to the first pn terms of itscanonical expansion, x = a0 + a1p + a2p

2 + .... .

The number n is called the depth of γ and we shall denote it as δ(γ)(It indicates the largest ball centered at zero of radius 1/pn on whichγ is identically one).



Example: The Fourier coefficients of a disk.Denote a disk of radius 1/pr centered at a by D(a; r). The fouriercoefficients of the characteristic function of D are

XD(γ) =

γ(a)p−r δ(γ) ≤ r ,

0 else.

Hence, the fourier inversion formula gives us

XD(x) = p−r∑δ(γ)≤r

γ(a)γ(x).



Using the fourier inversion formula, one can derive the following expression

A(a, k ,N) =#(D(a, k) ∩ {x1, ..., xN})

N,

=1

N

N∑n=1

XD(xn)

=1

N

N∑n=1

∑γ∈Zp

XD(γ)γ(xn).



Example: The sequence xn = nα+ β, with α, β ∈ Zp is equidistributed if andonly if α is a unit in Zp.Proof. We can give a fourier analytic proof. It is the reverse direction that ishard to prove

∣∣∣A(a, k,N)− 1/pk∣∣∣ =

1

N

∣∣∣∣∣∣N∑

n=1

∑γ∈Zp

XDk(γ)γ(xn)− N/pk

∣∣∣∣∣∣ ,≤ 1

N

∑′

δ(γ)≤k

p−k

∣∣∣∣∣N∑

n=1

γ(xn)

∣∣∣∣∣ ,=

1

N

∑′

δ(γ)≤k

p−k

∣∣∣∣∣N∑

n=1

γ(α)n

∣∣∣∣∣ ,=

1

N

∑′

δ(γ)≤k

p−k∣∣∣∣1− γ(α)N

1− γ(α)

∣∣∣∣ ,≤ 1

Npk

pk−1∑l=1

∣∣∣∣ 1

sin(2πlαpk)

∣∣∣∣ ,where l = δ(γ), and lα 6≡ 0 (mod ()pk) if α is a unit. The term on the righthand side goes to zero as N goes to infinity.


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Sequences in p-adic fields: Discrepancy

We define discrepancy as follows

Definition

The discrepancy of a finite sequence {x1, x2, .., xN} is

DN = supa,k

∣∣∣A(a, k ,N)− 1/pk∣∣∣

Question: Can one obtain an upper bound on the discrepancy DN

for a given sequence {xn} in terms of N ?

Question: : What about a LeVeque type inequality ?


Sequences in p-adic fields: A Levque Type Inequality

Theorem

(with C. Petsche)The discrepancy of a sequence {xn} is bounded by

D4N ≤ C (p)

∑γ∈Zpγ 6=γ0

1

p3δ(γ)|CN(γ)|2,

where C (p) is a constant and

CN(γ) =1

N

N∑n=1

γ(xn).

Corollary

The sequence na + b where a is a unit in Zp has discrepancy

DN ≤ C (p)1√N.


Classical LeVeque Inequality in R/Z

Recall: The sequence {x1, .., xN} on R/Z satisfies

D3N ≤

6

π2

∞∑k=1

1

k2

∣∣∣∣∣ 1

N

N∑n=1

e2πi kxn

∣∣∣∣∣2

.


LeVeque type Inequality: Idea of Proof

Let {xn} be a sequence in Zp. Define the functionf (x , y) : Zp × Zp −→ R as follows

f (x , y) =#∣∣∣∪Nj=1xj ∩ D(x , |y |p)

∣∣∣N

− |y |p,

= A(x , |y |p,N)− |y |p,

where D(x , |y |p) is a disc of radius |y |p centered at x .

The discrepancy is then

DN = supx ,y|f (x , y)| ,

for a fixed positive integer N.



Pick a ∆ = f (x0, y0) > 0.

Such an (x0, y0) exits since the average of f is zero∫f (x , y) dµx dµy = 0.

Now consider the straight line through the points (|y0|,∆) and(∆ + |y0|, 0) given by

l(|y |) = ∆− (|y | − |y0|).

In the region |y0| ≤ |y | ≤ ∆ + |y0|, and |x − x0| ≤ |y |, we have

f (x , y) ≥ l(|y |),

so that∫|y0|≤|y |≤∆+|y0|

f 2(x , y)dµy ≥∫|y0|≤|y |≤∆+|y0|

(∆− (|y | − |y0|))2 dµy .



We also have

‖f ‖22 ≥

∫|y0|≤|y |≤∆+|y0|

f 2(x , y)dµy ,

and ∫|y0|≤|y |≤∆+|y0|

(∆− (|y | − |y0|))2 dµy ≥ C (p)∆4.

A similar argument holds for ∆ = f (x0, y0) < 0.

We can now apply Parseval’s formula to the LHS ‖f ‖22.

The actual details are of course more intricate, but this is thegeometric idea.


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Conclusions and future work

Distribution of other interesting sequences.

Other interesting avenues of research: Looking at fourier analysis onlocal fields (some work done by Mitchell H. Taibleson in 60’s,questions that are viable for developing a good REU program! ).

Thank you!


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References

Pete. Clark.Foundations of the theory of uniform distribution,http://alpha.math.uga.edu/ pete/expositions2012.html

Hugh L. Montgomery.Ten lectures on the interface between Analytic number theory and HarmonicAnalysis,American Mathematical Society, 1994.

L. Kuipers, H. Niederreiter.Uniform distribution of sequences,Wiley, 1974.

Aditi. Karr.Weyl’s Equidistribution Theorem,Resonance, May, 2003.

Giancarlo Travaglini.Number theory, Fourier analysis, and Geometric discrepancy,Cambridge Univeristy Press , 2014.

Peter Walters.An introduction to Ergodic Theory, Springer Graduate Texts in Mathematics,2000.,Oxford University Press, Second Edition, 2009.


Equidistribution of sequences on the p-adic unit ball · Equidistribution of sequences on the p-adic unit ball Naveen Somasunderam Department of Mathematics Oregon State Univeristy,

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