Honors 1: Undergraduate Math Lab^{1}

Peter Sarnak^{2}, Steven J. Miller^{3}, Alex Barnett^{4}

Courant Institute of Mathematical Sciences New York University New York, NY

November 19, 2002

1Homepage: http://www.math.nyu.edu/∼millerj/
2E-mail: sarnak@math.princeton.edu
3E-mail: millerj@cims.nyu.edu or sjmiller@math.princeton.edu
4E-mail: barnett@nmr.mgh.harvard.edu

Abstract

The purpose of the Undergraduate Mathematics Laboratory is to form a research team of undergraduates, graduate students and faculty to investigate interesting unsolved conjectures theoretically and experimentally. The UML is sponsored by a VIGRE grant of the National Science Foundation.

In addition to the standard lecture-homework classes, we wanted a class where the undergraduates would work on hot conjectures and see what kinds of problems mathematicians study. In the sciences and engineering, undergraduates are often exposed to state of the art projects through experimental labs; we wanted to bring a similar experience to the math majors.

The undergrads often have enough theory to understand the basic framework and proofs of simple cases. Building on this, they then numerically test the conjectures. The undergrads learn a good deal of theory, they learn about coding, simulations and optimization (very marketable skills), and they get to see what is out there. The graduate students and the faculty get a potent calculating force for numerical investigations. A similar course has been run at Princeton (2000 − 2002). Many of the problems investigated by the Princeton students arose from graduate dissertations or current faculty research. It has been very easy finding graduate students and faculty excited about working on this course; at the end of a semester or year, instead of having a folder of solution keys to calculus, a graduate student should be able to co-author an experimental math paper with the undergrads.

Below are the notes from the NYU Fall 2002 class.

Problem List

1. Primality Testing:

In a major theoretical breakthrough, Manindra Agarwal, Nitin Saxena and Neeraj Kayal discovered a deterministic polynomial time algorithm to determine if a number is prime or composite. Previous algorithms are known to be polynomial only under well believed conjectures (GRH), or are probabalistic. Some aspects of current primality testing algorithms will be explored, possibly the distribution of the least primitive root mod p.

2. Ramanujan Graphs:

The construction of graphs that are highly connected but have few edges have many important applications, especially in building networks. To each graph G we can associate a matrix A, where A_{ij} is the number of edges from vertex i to vertex j. Many properties of G are controlled by the size of the second largest eigenvalue of G. One project will be to investigate the distribution of the normalized second largest eigenvalues.

3. Randomness of Arithmetic Maps:

For a prime p, consider the map Inv_{p} which sends x to its inverse mod p, x ↦ \overline{x}. One project will be to compare this map to random maps mod p. For example, let L(p) be the length of the longest increasing subsequence. If p is congruent to 3 mod 4, the inverse map is a fixed-point-free signed involution, and the length of L(p) can be compared to that from random fixedpoint-free signed involutions (studied by Rains and others). Let S(m, n; c) be the Kloosterman sum,

S(m, n; c) = ∑_{xmodp} e^{2πi \frac{mx+n\overline{x}}{c}}  (1)

An additional project will be to investigate ∑_{c} \frac{|S(1,1;c)|^{2}}{c^{2}} , which is related to number variance of the Upper Half Plane mod SL(2, ℤ). Finally, let \sqrt{p} ≤ x ≤ 2\sqrt{p}. Arrange in increasing order the \sqrt{p} numbers \overline{x}, and compare their spacings to Poissonian behavior.

Contents

1 Introduction, Primality Testing, Algebra Review 8 1.1 Primality Testing . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.2 Arithmetic Modulo p . . . . . . . . . . . . . . . . . . . . . . . . 9

1.2.1 Algebra Review . . . . . . . . . . . . . . . . . . . . . . . 10 1.2.2 Using Fermat’s Little Theorem . . . . . . . . . . . . . . . 11 1.2.3 Quadratic Reciprocity . . . . . . . . . . . . . . . . . . . 11

1.3 Lecture Next Week . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.4 WWW Resources . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Notation, Euclid’s Algorithm, Lagrange’s Theorem, Riemann Zeta Function 13 2.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.2 Euclid’s algorithm for GCD . . . . . . . . . . . . . . . . . . . . . 15 2.3 Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 16

2.3.1 Basic group theory . . . . . . . . . . . . . . . . . . . . . 16 2.3.2 Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . . 16

2.4 Introduction to Riemann zeta function . . . . . . . . . . . . . . . 17 2.4.1 Prelude: Euclid’s proof of infinity of primes . . . . . . . . 17 2.4.2 Definition, two forms . . . . . . . . . . . . . . . . . . . . 18 2.4.3 ζ(s)’s Behaviour and the Infinitude of Primes . . . . . . . 19

3 Legendre Symbols, Gary Miller’s Primality Test and GRH 20 3.1 Review of the Legendre Symbol . . . . . . . . . . . . . . . . . . 20 3.2 Gary Miller’s Primality Test, 1976 . . . . . . . . . . . . . . . . . 21

3.2.1 Aside: Finite Abelian Groups . . . . . . . . . . . . . . . 22 3.2.2 Lehmer’s Proposition . . . . . . . . . . . . . . . . . . . . 24

3.3 GRH Implies Just Need To Check Up To O(log^{2} n) in Miller … 25 3.3.1 Fourier Analysis . . . . . . . . . . . . . . . . . . . . . . 25 3.3.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 25 3.3.3 Characters of 𝔽_{p}^{∗}: Dirichlet Characters . . . . . . . . . . . 26 3.3.4 General Riemann Hypothesis (GRH) . . . . . . . . . . . 27 3.3.5 Proof of the Miller Test . . . . . . . . . . . . . . . . . . . 27 3.3.6 Review of Proof . . . . . . . . . . . . . . . . . . . . . . 29

3.4 Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.4.1 Aside: For p Odd, Half the Non-Zero Numbers are Quadratic

Residues . . . . . . . . . . . . . . . . . . . . . . . . . . 30 3.4.2 Chinese Remainder Theorem . . . . . . . . . . . . . . . . 30 3.4.3 Partial Summation . . . . . . . . . . . . . . . . . . . . . 30

4 Cosets, Quotient Groups, and an Introduction to Probability 32 4.1 Quotient groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 4.2 Random walk and discrete probability . . . . . . . . . . . . . . . 34

4.2.1 Probability distribution for a single event, mean, variance . 34 4.2.2 Multiple events . . . . . . . . . . . . . . . . . . . . . . . 35 4.2.3 Simplest random walk . . . . . . . . . . . . . . . . . . . 38 4.2.4 Central Limit Theorem . . . . . . . . . . . . . . . . . . . 38

5 Quadratic Reciprocity, Central Limit Theorem and Graph Problems 40

5.1 Eisenstein’s Proof of Quadratic Reciprocity . . . . . . . . . . . . 40 5.1.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . 40 5.1.2 First Stage . . . . . . . . . . . . . . . . . . . . . . . . . 41 5.1.3 Second Stage . . . . . . . . . . . . . . . . . . . . . . . . 43

5.2 Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . 45 5.3 Possible Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 45

5.3.1 Combinatorics and Probability . . . . . . . . . . . . . . . 45

6 Efficient Algorithms, Probability, Alg+Transcendental, Pidgeon Hole, Chebychev 48 6.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 6.2 Efficient Algorithms . . . . . . . . . . . . . . . . . . . . . . . . 49

6.2.1 Polynomial Evaluation . . . . . . . . . . . . . . . . . . . 49 6.2.2 Exponentiation . . . . . . . . . . . . . . . . . . . . . . . 50 6.2.3 Euclidean Algorithm . . . . . . . . . . . . . . . . . . . . 51

6.3 Probabilities of Discrete Events . . . . . . . . . . . . . . . . . . . 53 6.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 53 6.3.2 Means . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 6.3.3 Variances . . . . . . . . . . . . . . . . . . . . . . . . . . 56 6.3.4 Random Walks . . . . . . . . . . . . . . . . . . . . . . . 59 6.3.5 Bernoulli Process . . . . . . . . . . . . . . . . . . . . . . 59 6.3.6 Poisson Distribution . . . . . . . . . . . . . . . . . . . . 61 6.3.7 Continuous Poisson Distribution . . . . . . . . . . . . . . 62 6.3.8 Central Limit Theorem . . . . . . . . . . . . . . . . . . . 64

6.4 Algebraic and Transcendental Numbers . . . . . . . . . . . . . . 64 6.4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 64 6.4.2 Countable Sets . . . . . . . . . . . . . . . . . . . . . . . 65 6.4.3 Algebraic Numbers . . . . . . . . . . . . . . . . . . . . . 67 6.4.4 Transcendental Numbers . . . . . . . . . . . . . . . . . . 68

6.5 Introduction to Number Theory . . . . . . . . . . . . . . . . . . . 70 6.5.1 Dirichlet’s Box Principle . . . . . . . . . . . . . . . . . . 70 6.5.2 Counting the Number of Primes . . . . . . . . . . . . . . 71

7 More of an Introduction to Graph Theory 76 7.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 7.2 Size of Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . 77

8 Linear Algebra Review, especially Spectral Theorem for Real Symmetric Matrice 78 8.1 Linear Algebra Review . . . . . . . . . . . . . . . . . . . . . . . 78

8.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 78 8.1.2 Spectral Theorem for Real Symmetric Matrices . . . . . . 84

9 Central Limit Theorem, Spectral Theorem for Real Symmetric, Spectral Gaps 88 9.1 Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . 88 9.2 Spectral Theorem for Real Symmetric Matrices . . . . . . . . . . 93 9.3 Applications to Graph Theory . . . . . . . . . . . . . . . . . . . 94 9.4 2\sqrt{k − 1} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

10 Properties of Eigenvalues of Adjacency Matrices of Random Graphs 99 10.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 10.2 ρ_{k}(2n) and λ_{max} . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 10.3 Measure from the Eigenvalues . . . . . . . . . . . . . . . . . . . 102 10.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 11 Spacings of Eigenvalues of Real Symmetric Matrices; Semi-Circle Law105

11.1 Joint density function of eigenvalues of real symmetric matrices (‘GOE’) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 11.1.1 Dirac Notation . . . . . . . . . . . . . . . . . . . . . . . 105 11.1.2 2 × 2 Gaussian Orthogonal Ensemble (GOE) . . . . . . . 106 11.1.3 Transformation to diagonal representation . . . . . . . . . 108 11.1.4 Generalization to n × n case . . . . . . . . . . . . . . . . 110

11.2 Eigenvalue spacing distribution in 2 × 2 real symmetric matrices . 111 11.2.1 Reminder: Integral of the Gaussian . . . . . . . . . . . . 111 11.2.2 Spacing distribution . . . . . . . . . . . . . . . . . . . . 112

11.3 Delta Function(al) . . . . . . . . . . . . . . . . . . . . . . . . . . 113 11.4 Definition of the Semi-Circle Density . . . . . . . . . . . . . . . 114 11.5 Semi-Circle Rule: Preliminaries . . . . . . . . . . . . . . . . . . 114 11.6 Sketch of Proof of the Semi-Circle Law . . . . . . . . . . . . . . 115

11.6.1 Calculation of Moments via Trace Formula . . . . . . . . 115 11.6.2 Calculation of Moments from the Semi-Circle . . . . . . . 118

12 More Graphs, Maps mod p, Fourier Series and n alpha 120 12.1 Kesten’s Measure . . . . . . . . . . . . . . . . . . . . . . . . . . 120 12.2 Generating Functions on k-Regular Trees . . . . . . . . . . . . . 121

12.2.1 R(z) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 12.2.2 Q(z) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 12.2.3 T (z) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

12.3 Recovering the Measure f(x) from R(z) . . . . . . . . . . . . . . 122 12.3.1 Poisson Kernel . . . . . . . . . . . . . . . . . . . . . . . 123 12.3.2 Cauchy Integral Formula . . . . . . . . . . . . . . . . . . 124

12.4 Third Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 12.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 124 12.4.2 Character Sums . . . . . . . . . . . . . . . . . . . . . . . 125 12.4.3 Completing the Square . . . . . . . . . . . . . . . . . . . 126 12.4.4 Weil’s Bound . . . . . . . . . . . . . . . . . . . . . . . . 126 12.4.5 Fourier Expansion of Sums . . . . . . . . . . . . . . . . . 127 12.4.6 Brief Review of Fourier Series . . . . . . . . . . . . . . . 128

12.5 Fourier Analysis and the Equi-Distribution of {nα} . . . . . . . . 128 12.5.1 Inner Product of Functions . . . . . . . . . . . . . . . . . 128 12.5.2 Fourier Series and {nα} . . . . . . . . . . . . . . . . . . 131 12.5.3 Equidistribution . . . . . . . . . . . . . . . . . . . . . . . 134 13 Liouville’s Theorem Constructing Transcendentals 138

13.1 Review of Approximating by Rationals . . . . . . . . . . . . . . 138 13.2 Liouville’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 140 13.3 Constructing Transcendental Numbers . . . . . . . . . . . . . . . 142

13.3.1 ∑_{m} 10^{−m!} . . . . . . . . . . . . . . . . . . . . . . . . . 142 13.3.2 [10^{1!}, 10^{2!}, …] . . . . . . . . . . . . . . . . . . . . . . . 143 13.3.3 Buffon’s Needle and π . . . . . . . . . . . . . . . . . . . 144

14 Poissonian Behavior and {n^{k}α} 146 14.1 Equidistribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 14.2 Point Masses and Induced Probability Measures . . . . . . . . . . 146 14.3 Neighbor Spacings . . . . . . . . . . . . . . . . . . . . . . . . . 149 14.4 Poissonian Behavior . . . . . . . . . . . . . . . . . . . . . . . . 150

14.4.1 Nearest Neighbor Spacings . . . . . . . . . . . . . . . . . 151 14.4.2 k^{th} Neighbor Spacings . . . . . . . . . . . . . . . . . . . 153

14.5 Induced Probability Measures . . . . . . . . . . . . . . . . . . . 155 14.6 Non-Poissonian Behavior . . . . . . . . . . . . . . . . . . . . . . 156

14.6.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . 156 14.6.2 Proof of Theorem 14.6.2 . . . . . . . . . . . . . . . . . . 157 14.6.3 Measure of α ∉ ℚ with Non-Poissonian Behavior along

a sequence N_{n} . . . . . . . . . . . . . . . . . . . . . . . 158

15 More Graphs, Kloosterman, Randomness of x → \overline{x} mod p 160 15.1 Kloosterman Sums . . . . . . . . . . . . . . . . . . . . . . . . . 160 15.2 Projective Geometry . . . . . . . . . . . . . . . . . . . . . . . . 161 15.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 15.4 Stereographic Projections and Fractional Linear Transformations . 162 15.5 More Kesten . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 15.6 SL_{2}(ℤ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 15.7 Is x → \overline{x} mod p Random? . . . . . . . . . . . . . . . . . . . . . 165

15.7.1 First Test . . . . . . . . . . . . . . . . . . . . . . . . . . 165 15.7.2 Second Test . . . . . . . . . . . . . . . . . . . . . . . . . 165 15.7.3 Third Test: Hooley’s R^{∗} . . . . . . . . . . . . . . . . . . 166

15.8 Note on Non-trivial Bound of Fourth Powers of Kloosterman Sums 166

16 Introduction to the Hardy-Littlewood Circle Method 168

16.1 Problems where the Circle Method is Useful . . . . . . . . . . . . 168 16.1.1 Waring’s Problem . . . . . . . . . . . . . . . . . . . . . . 168 16.1.2 Goldbach’s Problem . . . . . . . . . . . . . . . . . . . . 169 16.1.3 Sum of Three Primes . . . . . . . . . . . . . . . . . . . . 169

16.2 Idea of the Circle Method . . . . . . . . . . . . . . . . . . . . . . 169 16.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 169 16.2.2 Useful Number Theory Results . . . . . . . . . . . . . . 171 16.2.3 Average Sizes of (f_{N}(x))^{s} . . . . . . . . . . . . . . . . 171 16.2.4 Definition of the Major and Minor Arcs . . . . . . . . . . 173

16.3 Contributions from the Major and Minor Arcs . . . . . . . . . . . 174 16.3.1 Contribution from the Minor Arcs . . . . . . . . . . . . . 174 16.3.2 Contribution from the Major Arcs . . . . . . . . . . . . . 175

16.4 Why Goldbach is Hard . . . . . . . . . . . . . . . . . . . . . . . 177 16.4.1 s = 3 Sketch . . . . . . . . . . . . . . . . . . . . . . . . 177 16.4.2 s = 2 Sketch . . . . . . . . . . . . . . . . . . . . . . . . 179

16.5 Cauchy-Schwartz Inequality . . . . . . . . . . . . . . . . . . . . 180 16.6 Partial Summation . . . . . . . . . . . . . . . . . . . . . . . . . 181

17 Multiplicative Functions, Kloosterman, p-adic Numbers, and Review of the Three Problems: Germain Primes, λ_{1}(G) for Random Graphs, Randomness of x → \overline{x} mod p 185 17.1 Multiplicative Functions, Kloosterman and p-adic Numbers …. 185

17.1.1 Multiplicative Functions . . . . . . . . . . . . . . . . . . 185 17.1.2 Kloosterman Sums . . . . . . . . . . . . . . . . . . . . . 186 17.1.3 p-adic numbers . . . . . . . . . . . . . . . . . . . . . . . 187

17.2 Germain Primes . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 17.3 Randomness of x → \overline{x} . . . . . . . . . . . . . . . . . . . . . . . 188 17.4 Random Graphs / Ramanujan Graphs . . . . . . . . . . . . . . . 189

18 Random Graphs, Autocorrelations, Random Matrix Theory and the Mehta-Gaudin Theorem 190 18.1 Random Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 18.2 Baire Category . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 18.3 Autocorrelation . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 18.4 Gaudin’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . 192

18.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 192 18.4.2 Vandermonde Determinants . . . . . . . . . . . . . . . . 193 18.4.3 Orthonormal Polynomials . . . . . . . . . . . . . . . . . 193 18.4.4 Kernel K_{N}(x, y) . . . . . . . . . . . . . . . . . . . . . . 194 18.4.5 Gaudin-Mehta Theorem . . . . . . . . . . . . . . . . . . 195 18.4.6 Example . . . . . . . . . . . . . . . . . . . . . . . . . . 197

19 Increasing Length Subsequences and Tracy-Widom 198 19.1 Increasing Length Subsequences . . . . . . . . . . . . . . . . . . 198 19.2 Tracy-Widom . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

20 Circle Method and Germain Primes 200

20.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 20.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 200 20.1.2 Partial Summation . . . . . . . . . . . . . . . . . . . . . 201 20.1.3 Siegel-Walfisz . . . . . . . . . . . . . . . . . . . . . . . 201 20.1.4 Germain Integral . . . . . . . . . . . . . . . . . . . . . . 202 20.1.5 Major and Minor Arcs . . . . . . . . . . . . . . . . . . . 202 20.1.6 Reformulation of Germain Integral . . . . . . . . . . . . 204

20.2 f_{N}(x) and u(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 20.2.1 f(\frac{a}{q}) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 20.2.2 u(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

20.3 f_{N}(α) − \frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)}u(α − \frac{a}{q}), α ∈ M_{a,q} . . . . . . . . . . . . . . 206 20.3.1 Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 20.3.2 S_{1 ∑;a,q} . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 20.3.3 S_{1 ∫;a,q} . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

20.4 Integrals of u(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 20.4.1 Formulations . . . . . . . . . . . . . . . . . . . . . . . . 210

20.4.2 ∫− \frac{1}{2}^{\frac{1}{2}} u(x)e(−x)dx . . . . . . . . . . . . . . . . . . . . . . 211

20.4.3 ∫_{− \frac{1}{2}}^{− \frac{Q}{N}} + ∫ \frac{1}{2} ; \frac{Q}{N} u(x)e(−x)dx . . . . . . . . . . . . . . . . . . 211

20.4.4 Integral over I_{2}, I_{3} . . . . . . . . . . . . . . . . . . . . . 212 20.4.5 Integral over I_{1}, I_{4} . . . . . . . . . . . . . . . . . . . . . 212 20.4.6 Collecting the Pieces . . . . . . . . . . . . . . . . . . . . 215

20.5 Determination of the Main Term . . . . . . . . . . . . . . . . . . 215 20.5.1 Properties of C_{q}(a) and ρ_{q} . . . . . . . . . . . . . . . . . 217 20.5.2 Determination of 𝔖_{N} and 𝔖 . . . . . . . . . . . . . . . . 223 20.5.3 Number of Germain Primes and Weighted Sums . . . . . 224

Chapter 1

Introduction, Primality Testing, Algebra Review

We introduce basic number theory concepts and primality algorithms. Lecture by Peter Sarnak; notes by Steven J. Miller.

1.1 Primality Testing

Given n, is n prime or composite? How difficult is this? How long does it take? Brute force: try factors up to \sqrt{n}, so can do in \sqrt{n} steps.

P = NP problem. Deep central problem in theoretical Computer Science. P is problems solvable in polynomial number of steps (in terms of input); if equals NP, a lot of problems are solvable quickly.

Telling when n is prime: isn’t supposed to be hard, but until two weeks ago, wasn’t known to be a P problem.

Notation: A(x) = O(B(x)) if there exists a C > 0 (which can be computed; if it cannot be computed, we say so) such that |A(x)| ≤ CB(x).

Example: One could show every sufficiently large odd number is the sum of three primes. However, we didn’t know how large sufficiently large was! IE, we couldn’t go through the calculation and make explicit a number N_{0} such that if n is odd and greater than N_{0}, then n is the sum of three primes. (Note: this has been removed, and we know have another proof giving an explicit N_{0}).

Theorem 1.1.1 (Agrawal, Kayal, Saxena). There is a procedure which runs in at most O( log^{12} n) steps determines whether n is prime. (Might be a little more than log^{12} n, ie, might be something like log^{12} n(log log n)^{A}).

There were algorithms that were known and faster, but only known to work all the time assuming certain well believed hypotheses (Riemann Hypothesis, RH).

(Go to http://www.math.nyu.edu/∼millerj/problemlist/problems.htm for a copy of their paper).

Need a feel for numbers. When n is big, log n (to any power) is much less than n. For practical applications, the size of the constant is important, as a constant of size 10^{100} would make an algorithm useless for our real world applications (ie, for the ranges we can reach). In AKS, the constants are tractable.

Technical Point: In AKS, they quote a theorem from number theory (they treat this as a black box: someone from this class will hopefully investigate this result further). There are many primes p for which p−1 has a large prime factor q, q > p^{\frac{2}{3}} . Related to Sophie Germain primes: primes p where p − 1 = 2q, q prime. (She showed that for primes like this, x^{p} + y^{p} = z^{p}, you can solve Fermat’s Last Theorem for such primes. It is not known if there are infinitely many primes like this). AKS does not need to know there are infinitely many Sophie Germain primes; fortunately all they need is that there are sufficiently many primes p with p − 1 with large prime factors.

Similar to Twin Primes: primes p_{1}, p_{2} with p_{1}−p_{2} = 2. We don’t know if there are infinitely many twin primes, but we do have heuristics (Hardy-Littlewood) predicting how many twin primes there are (and we observe exactly that many). Sophie Germain primes are more subtle, but should be able to get heuristics. For twin primes and related quantities, see David Schmidt’s report on Prime Investigations (Princeton Undergraduate Math Lab, 2000 − 2001). Anyway, this would be a good project.

1.2 Arithmetic Modulo p

Number Theory: the study of whole numbers. ℤ is the integers, look at ℤ/nℤ = {0, 1, 2, …, n − 1}. This is a finite group (under addition); in fact, it is a finite ring (can also multiply, have inverses for the non-zero elements only if n is prime).

Notation: x ≡ y mod n means x − y is a multiple of n.

Try and solve in ℤ the equation 2x+1 = 2y. The left hand side is odd, the right hand side is even. Thus, there are no solutions. Really, just did arithmetic mod 2 or in ℤ/2ℤ. Harder: x^{2}+y^{2}+z^{2} = 8n+7. This never has a solution. Look modulo 8. The RHS is 7 modulo 8. What are the squares mod 8? 1^{1} = 1, 2^{2} = 4, 3^{2} = 1, 4^{2} = 0, repeats. See there is no way to add three squares and get 7.

Idea: First, try and solve the equation modulo different primes. If you cannot solve it for some prime, then you cannot solve it over the integers.

1.2.1 Algebra Review

ℤ/nℤ: do arithmetic over this ring. (ℤ/nℤ)^{∗} are the invertible (multiplicatively)

elements in the ring ℤ/nℤ, ie, x is in (ℤ/nℤ)^{∗} if there is a y such that xy ≡ 1 mod n. Note: if gcd(x, n) > 1 (ie, x and n have a common prime divisor p), then you cannot invert x (there is no y with xy ≡ 1 mod n). Why? xy ≡ 1 mod n means xy = 1 + λn for some integer λ. But if p|x and p|n, then p|1 which is absurd. Exercise: if gcd(x, n) = 1, there is an inverse (Euclidean Algorithm).

The cardinality (number of elements in the set) of (ℤ/nℤ)^{∗} is the number of x ∈ {0, 1, 2, …, n − 1} such that gcd(x, n) = 1. We denote the number of such x by ϕ(x), the Euler totient function. (Good Reference: H. Davenport: The Higher Arithmetic). Note that if p is prime, ϕ(p) = p − 1. This implies that |(ℤ/pℤ)^{∗}| = p − 1 = ℤ/pℤ − {0}. IE, we have a field if n is a prime, as every non-zero element is invertible.

(ℤ/nℤ)^{∗}, for any n, is a finite Abelian group (we have inverses under multiplication, and order of multiplication doesn’t matter). Finite Abelian Groups is a trivial subject: Structure Theorem for Finite Abelian Groups: product of cyclic groups.

For n = p a prime, (ℤ/nℤ)^{∗} is a cyclic group of order p − 1.

If G is a group (have identity, closed under some binary operation, have inverses with respect to the binary operation, operation is associative), we say the order of x ∈ G, ord(x), is the least positive power m such that x^{m} = e, where e ∈ G is the identity of the group. In a finite group, every element has finite order (proof: use the pidgeonhole principle).

Theorem 1.2.1 (Lagrange). ord(x) | ord(G).

Corollary 1.2.2 (Fermat’s Little Theorem). For any prime p, if gcd(a, p) = 1, then a^{p−1} ≡ 1 mod p.

1.2.2 Using Fermat’s Little Theorem

To check and see if a number n is prime, why not check if a^{n−1} ≡ 1 mod n if gcd(a, n) = 1. It is very easy to quickly compute gcd(a, n) (use Euclid’s Algorithm). Sketch: without loss of generality, let a < n. Write n = b_{0}a + b_{1}, and now the gcd of a and n is the same as that of a and b_{1} (note 0 ≤ b_{1} < a).

How long does it take to raise a to the n − 1 power? Use repeated squares and base 2 expansion. Example 100 = 64 + 32 + 4, or 1 · 2^{6} + 1 · 2^{5} + 1 · 2^{2}. Thus, do a^{2}, a^{2} · a^{2}, a^{4} · a^{4}, a^{8} · a^{8}, a^{16} · a^{16}, a^{32} · a^{32}. Then a^{100} = a^{64} · a^{32} · a^{4}.

If, by choosing an a, you find a^{n−1} ≢ 1 mod n, you have a certificate for compositeness, but you have no idea what the factors of n are!

When a machine uses Fermat’s Little Theorem, it randomly chooses a fixed number of a’s between 1 and n − 1. Problems: how many times should you run these tests to be very confident of the result; are there any numbers which always pass this test, yet are composite?

About eight years ago it was proved there are infinitely many Carmichael numbers (numbers n such that a^{n−1} ≡ 1 mod n for all a, but n is composite).

1.2.3 Quadratic Reciprocity

p, q odd primes. We define \binom{\underline{a}}{p} to be 1 if a is a non-zero square mod p, 0 if a = 0, and −1 otherwise (ie, if a is not a square mod p). Note a is a square mod p if there exists an x ∈ {0, 1, …, p − 1} such that a ≡ x^{2} mod p. For p an odd prime, half the non-zero numbers are squares, half are not.

Exercise: \binom{\underline{a}}{p} = a^{\frac{p−1}{2}} mod p for odd p. Note the above squared is a^{p−1} ≡ 1.

Theorem 1.2.3 (Quadratic Reciprocity). \binom{\underline{q}}{p} = \binom{\underline{p}}{q} · (−1)^{\frac{p−1}{2} \frac{q−1}{2}} , p, q odd primes.

Gauss gave at least four proofs of this deep result. If either p or q is equivalent to 1 mod 4, then one has \binom{\underline{q}}{p} = \binom{\underline{p}}{q}, ie, I’m a square root modulo you if you are a square root modulo me.

Carmichael numbers were behaving like primes. We want to get rid of them. Instead of testing a^{n−1} ≡ 1 mod n, test and see if a^{\frac{n−1}{2}} ≡ \binom{\underline{a}}{n} mod n. Similar to the Euclidean Algorithm, can computer \binom{\underline{a}}{n} in logn steps by constant applications of Quadratic Reciprocity.

Key Test: Will test that a^{\frac{n−1}{2}} ≡ \binom{\underline{a}}{n} mod n for 1 ≤ a ≤ C log^{2} n. If fails for some a, the number is composite. If it passes, by the Riemann Hypothesis (RH), then it is true for all a. Will then show this is a valid test for primality (ie, unlike Carmichael numbers, if this is satisfied for all a up to C log^{2} n, then n is prime. This will take about log^{4} n steps).

Algebra Books: Herstein (Topics in Algebra); Birkhoff-Mclean (Algebra), Lang (Undergraduate Algebra).

1.3 Lecture Next Week

Steve will talk about reciprocity, finite fields.

1.4 WWW Resources

http://mathworld.wolfram.com/ is a good place to look up unfamiliar terms.

Notation, Euclid’s Algorithm, Lagrange’s Theorem, Riemann Zeta Function

We will review notation, Euclid’s Algorithm, Lagrange’s Theorem, and prove there are infinitely many primes (three ways: following Euclid, by studying the Riemann Zeta Function ζ(s) as s → 1, and by analyzing ζ(2). Lecture by Steven J. Miller; notes by Alex Barnett and Steven J. Miller.

2.1 Notation

• a|b : a divides b, i.e. the remainder after integer division \frac{b}{a} is 0.

• (a, b) : Greatest Common Divisor (GCD) of a and b. Sometimes written gcd(a, b).

• x ≡ y( mod n) : An equality once both sides of the equation have been taken modulo n. Equivalently, there exists an integer a such that x = y+an.

• wlog : ‘without loss of generality’. For example, if we have two numbers x and y, it is often convenient to know which is larger and which is smaller. Without loss of generality, we can say x ≤ y, as the case x ≥ y is handled identically (after permuting the variables).

• s.t.: such that.

• ∀ : for all.

• ∃ : there exists.

• big O notation : A(x) = O(B(x)), read “A(x) is of order B(x)”, is shorthand for, there is a C > 0 (which we can explicitly calculate), s.t.|A(x)| ≤ C B(x), ∀x.

• |S| or #S : number of elements in the set S.

• #{condition} : number of objects satisfying the condition.

• p : unless otherwise stated, a prime number.

• ℤ : the set of integers.

• ℤ/nℤ : the additive group of integers mod n.

• (ℤ/nℤ)^{∗} : the multiplicative group of invertible elements mod n.

• ℚ: the set of rational numbers. ℚ = \{x : x = \frac{p}{q}, p, q ∈ ℤ, q ≠ 0\}.

• ℝ: the set of real numbers.

• ℂ: the set of complex numbers.

• (\frac{a}{p}) : Legendre symbol of a and p, defined as

(\frac{a}{p}) = \{ \begin{matrix}0, & if p|a, that is, a = 0( mod p) \\ 1, & if ∃x s.t.x^{2} = a( mod p) \\ −1, & if the above does not exist. \end{matrix} (2.1)

The symbol tests the question, “Does a have a square root in the field of arithmetic modulo p?”

• ‘weak bound’ : an inequality constraining some quantity which does a very poor job of getting close to the true size of the quantity. That is, a not very useful bound.

2.2 Euclid’s algorithm for GCD

Tells you if two positive integers x and y have a GCD greater than 1 by finding it. Therefore it’s a ‘constructive proof’. It is also ‘deterministic’ (involves no random choices). A fast procedure, i.e. takes only O(log y) steps. Remember that the number of digits ∝ log y. wlog we take y > x.

Each step is the ‘black box’ integer division routine we’ll call D, which given the pair x, y returns the pair of integers b, r s.t.r < x and satisfying

y = bx + r.  (2.2)

Note that this step is polynomial in the number of digits (probably ∼ (log y)^{2} — anyone?).

ALGORITHM:

Start with the pair y, x and perform D to get b_{1}, r_{1}.

Perform D on x, r_{1} to get b_{2}, r_{2}.

Perform D on r_{1}, r_{2} to get b_{3}, r_{3}.

⋮

Perform D on r_{n−2}, r_{n−1} to get b_{n}, r_{n}.

Stop when r_{n} is either

• 0, in which case r_{n−1} is the GCD, or

• 1, in which case the GCD is 1, that is, x and y are relatively prime.

The procedure works because the D step gives an r which inherits all common divisors of y and x. This is easy to see by writing D as r = y − bx. Therefore all the adjacent pairs in the sequence y, x, r_{1}, r_{2} ⋯ r_{n−1} share the same GCDs. The sequence is also descending y > x > r_{1} > r_{2} > ⋯ > r_{n−1}, so must reach the case that r_{n−1}|r_{n−2}, in which case r_{n−1} is the GCD and r_{n} = 0, or that a remainder of 1 is reached, which implies no common divisors. We have a worstcase scenario that each remainder is smaller than the previous by a constant factor c < 1 (I believe this is the inverse of the Golden Ratio (\sqrt{5} − 1)/2 ≈ 0.618 ⋯ — anyone?), giving geometric (exponential) shrinkage of the r’s. Therefore the worst case is that the answer is reached in n = O(log y) steps, and the whole algorithm is therefore polynomial in log y.

2.3 Lagrange’s Theorem

2.3.1 Basic group theory

Group G is a set of elements g_{i} satisfying the four conditions below, relative to some binary operation. We often use multiplicative notation (g_{1}g_{2}) or additive notation (g_{1} + g_{2}) to represent the binary operation. For definiteness, we use multiplicative notation below; however, one could replace xy with b(x, y) below.

If the elements of G satisfy the following four properties, then G is a group.

1. ∃e ∈ G s.t.∀g ∈ G : eg = ge = g. (Identity.) We often write e = 1 for multiplicative groups, and e = 0 for additive groups.

2. ∀x, y, z ∈ G : (xy)z = x(yz). (Associativity.)

3. ∀x ∈ G, ∃y ∈ G s.t.xy = yx = e. (Inverse.) We write y = x^{−1} for multiplication, y = −x for addition.

4. ∀x, y ∈ G : xy ∈ G. (Closure.)

If commutation holds (∀x, y ∈ G, xy = yx), we say the group is Abelian. Non-abelian groups exist and are important. For example, consider the group of N × N matrices with real entries and non-zero determinant. Prove this is a group under matrix multiplication, and show this group is not commutative.

H is a subgroup of G if it is a group and its elements form a subset of those of G. The identity of H is the same as the identity of G. Once you’ve shown the elements of H are closed (ie, under the binary operation, b(x, y) ∈ H if x, y ∈ H), then associativity in H follows from closure in H and associativity in G.

For the application to Fermat’s Little Theorem you will need to know that the set {1, x, x^{2}, ⋯ x^{n−1}} where n is the lowest positive integer s.t.x^{n} = 1, called the cyclic group, is indeed a subgroup of any group G containing x, as well as n divides the order of G.

For a nice introduction to group theory see: M. Tinkham, Group Theory and Quantum Mechanics, (McGraw-Hill, 1964) or S. Lang, Undergraduate Algebra.

2.3.2 Lagrange’s Theorem

The theorem states that if H is a subgroup of G then |H| divides |G|.

First show that the set hH, i.e. all the elements of H premultiplied by one element, is just H rearranged (Cayley’s theorem). By closure hH falls within H.

We only need to show that hh_{i} can never equal hh_{j} for two different elements i ≠ j. If it were true, since a unique h^{−1} exists we could premultiply the equation hh_{i} = hh_{j} by h^{−1} to give h_{i} = h_{j}, which is false. Therefore hh_{i} ≠ hh_{j}, and we have guaranteed a 1-to-1 mapping from H to hH, so hH = H.

Next we show that the sets g_{i}H and g_{j}H must either be completely disjoint, or identical. Assume there is some element in both. Then g_{i}h_{1} = g_{j}h_{2}. Multiplying on the right by h_{i}^{−1} ∈ H (since H is a subgroup) gives g_{i} = g_{j}h_{2}h_{1}^{−1}. As H is a subgroup, ∃h_{3} ∈ H such that h = h_{2}h_{1}^{−1}. Thus g_{i} = g_{j}h_{3}. Therefore, as h_{3}H = H, g_{i}H = g_{j}h_{3}H = g_{j}H, and we see if the two sets have one element in common, they are identical. We call a set gH a coset (actually, a left coset) of H.

Clearly

G = ⋃_{g∈G} gH  (2.3)

Why do we have an equality? As g ∈ G and H ⊂ G, every set on the right is contained in G. Further, as e ∈ H, given g ∈ G, g ∈ gH. Thus, G is a subset of the right side, proving equality.

There are only finitely many elements in G. As we go through all g in G, we see if the set gH equals one of the sets already in our list (recall we’ve shown two cosets are either identical or disjoint). If the set equals something already on our list, we do not include it; if it is new, we do. Continuing this process, we obtain

G = ⋃_{i=1}^{k} g_{i}H  (2.4)

for some finite k. If H = {e}, k is the number of elements of G; in general, however, k will be smaller.

Each set g_{i}H has |H| elements. Thus, |G| = k|H|, proving |H| divides |G|.

2.4 Introduction to Riemann zeta function

2.4.1 Prelude: Euclid’s proof of infinity of primes

Given the set of primes p_{1} ⋯ p_{n} you can always construct the number ∏_{i=1}^{n} p_{i} +1 which is indivisible by any of the given p_{1} ⋯ p_{n}. Therefore this number must be divisible only by primes greater than p_{n}, or must be prime itself. Therefore there exists a prime greater than p_{n}. An analysis of this proof gives a very weak lower bound on the number of primes less than x. The worst case scenario is that ∏_{i=1}^{n} p_{i} +1 is the next prime. Thus, if we had n−1 primes up to x = ∏_{i=1}^{n−1} p_{i} +1, we would have n primes up to

∏_{i=1}^{n−1} p_{i} · ( ∏_{j=1}^{n} p_{j} + 1) + 1  (2.5)

Thus, having at least n − 1 primes less than x, we have at least n primes less than (basically) x^{2}. One can quantify this further. One should get something like there are at least n primes less than 2^{n} or 4^{n}.

The zeta function will give us other ways to prove this, and to get a better estimate on the prime counting function,

π(x) ≡ #{p < x},  (2.6)

giving the number of primes below any number x.

2.4.2 Definition, two forms

The Riemann zeta function ζ(s) is defined, for Re(s) > 1, by

ζ(s) = ∑_{n=1}^{∞} \frac{1}{n^{s}} .  (2.7)

We prove the useful fact that, for Re(s) > 1,

∑_{n=1}^{∞} \frac{1}{n^{s}} = ζ(s) = ∏_{p^{primes}} (1 − \frac{1}{p^{s}} )^{−1} ,  (2.8)

which we call LHS and RHS. We call the product over primes an Euler Product.

To show equivalence, we use the Fundamental Theorem of Algebra (FTA) that all positive integers can be expressed as a single, unique, product of prime factors. Expanding all reciprocals in the RHS using the geometric series sum formula (1 − x)^{−1} = 1 + x + x^{2} + x^{3} + ⋯, gives for the RHS,

(1 + 2^{−s} + 2^{−2s} + 2^{−3s} + ⋯)(1 + 3^{−s} + ⋯)(1 + 5^{−s} + ⋯) · · · .

Remarkably, due to the FTA, we can associate 1-to-1 each term (choice of prime factors) on the RHS with each n on the LHS. For instance, n = 12 from LHS is accounted for by the RHS term,

2^{−2s} · 3^{−s} · 1 · 1 · 1 ⋯ = \frac{1}{(2^{2} · 3)^{s}} = \frac{1}{12}.

Each combination of RHS terms corresponds uniquely to a single n.

2.4.3 ζ(s)’s Behaviour and the Infinitude of Primes

We take the limit of s going to 1 and compare sides. LHS gives

lim_{s→1} ζ(s) = lim_{s→1} ∑_{n=1}^{∞} \frac{1}{n^{s}} = ∑_{n=1}^{∞} \frac{1}{n}.  (2.9)

This sum diverges. Why? Crudely, ∑_{n=1}^{N} n^{−1} is close to ∫_{1}^{N} \frac{dy}{y} which equals log N. The definition of ‘close’ can be tightened up. For instance, you can create upper and lower bounds by approximating the integral by rectangular strips, getting

∑_{n=2}^{N} \frac{1}{n} ≤ ∫_{1}^{N} \frac{dy}{y} ≤ ∑_{n=1}^{N−1} \frac{1}{n}.  (2.10)

As the two sums differ by a bounded amount, we see the sum grows like log N. As s goes to 1, if there are only finitely many primes than the product over

primes is well behaved (ie, finite). Therefore, there must be infinitely many primes!

Further study of the zeta function will lead us to a good estimate for π(x). A second proof follows from the fact that ζ(2m) = rational ·π^{2m}, for integer

m. This is known as a Special Values proof, as we are using the value of ζ(s) at a special value. We need the fact that π^{2} is irrational. ζ(2) = ∑_{n=1}^{∞} \frac{1}{n^{2}} = \frac{π^{2}}{6} , which is irrational. Thus, the right hand side (the product over primes) must also be irrational; however, if there are only finitely many primes, when s = 2 the right hand side is rational! Thus, there must be infinitely many primes.

Please see Steve’s notes, and URLs for more information on all of the above.

Legendre Symbols, Gary Miller’s Primality Test and GRH

We review the Legendre Symbol. We discuss Gary Miller’s primality test, and show that if the General Riemann Hypothesis (for Dirichlet Characters) is true, then Miller’s test correctly determines if a number n is prime or composite, and runs in time O(log^{4} n). Lecture by Peter Sarnak; notes by Steven J. Miller.

3.1 Review of the Legendre Symbol

Recall Fermat’s Little Theorem: a^{p−1} ≡ 1 mod p if p is prime.

Given n, check a^{n−1} ≡ 1 mod n for many a’s relatively prime to n. Exercise: If ever this is not satisfied, then n must be composite.

There are composite numbers (called Carmichael numbers) which satisfy a^{n−1} ≡ 1 mod n for all a, yet are not prime. The first Carmichael number is 561 = 3 · 11 · 17; the third is 1729 = 7 · 13 · 19. Exercise: Prove all Carmichael numbers must be square-free.

Aside: 1729 has an interesting history (Ramanujan, Hardy and taxicabs). Hardy visited Ramanujan in the hospital, Hardy remarked that his taxicab’s number was particularly uninteresting; Ramanujan remarks it (1729) is the smallest number which can be written in two different ways as the sum of two cubes. 1729 = 1^{3} + 12^{3} = 9^{3} + 10^{3}.

Recall the Legendre symbol \binom{\underline{a}}{p} is 0 if p|a, 1 if there is an x ≠ 0 with x^{2} ≡ a mod p, and −1 if there is no solution to x^{2} ≡ a mod p. Euler’s condition is \binom{\underline{a}}{p} ≡ a^{\frac{p−1}{2}} mod p.

Really, the Legendre symbol is a function on 𝔽_{p} = ℤ/pℤ. We can extend the Legendre symbol to all integers. We only need to know a mod p, and we define \binom{\underline{a}}{p} = \binom{\underline{a mod p}}{p}.

Initially the Legendre symbol is define only when the bottom is prime. We now extend the definition to all n as follows: let n = p_{1} · p_{2} ⋯ p_{t} be the product of t distinct primes. Then (\frac{a}{n}) = (\frac{a}{p_{1}} ) (\frac{a}{p_{2}} ) ⋯ (\frac{a}{p_{t}} ). Note this is not the same as saying that if a is a square (a quadratic residue) mod n, then a is a square mod p_{i} for each prime divisor.

The main result (which allows us to calculate the Legendre symbol quickly and efficiently) is the celebrated

Theorem 3.1.1 (Quadratic Reciprocity). For m, n odd and relatively prime, (\frac{m}{n}) (\frac{n}{m}) = (−1)^{\frac{m−1}{2} \frac{n−1}{2}} .

3.2 Gary Miller’s Primality Test, 1976

Miller Test: Given n as input (n must be odd), test whether for 2 ≤ a ≤ 70 log^{2} n, (\frac{a}{n}) ≡ a^{\frac{n−1}{2}} mod n (where, of course, a and n are relatively prime). If this test fails for some a in this range, output composite; if it passes the test for all such a, output Prime.

Note that we can very quickly determine if two number are relatively prime (use the Euclidean Algorithm, which takes O(log n) steps).

Theorem 3.2.1 (Miller Test Results). The Miller Test runs in O(log^{4} n) steps. If the output is composite, then the number n is composite (ie, the algorithm’s result is correct). If we assume GRH (the General Riemann Hypothesis, the most important unsolved problem in mathematics), then the output prime is also correct.

Running time: we can compute (\frac{a}{n}) in O(log n) steps. By Quadratic Reciprocity, to compute (\frac{a}{n}) (may assume −\frac{n}{2} ≤ a ≤ \frac{n}{2}), it is enough to compute (\frac{n}{a}) (with a factor of −1).

Why? We only need to know a mod n, so we may reduce a until −\frac{n}{2} ≤ a ≤ \frac{n}{2}. Note the top (in absolute value) is at most half the size of the bottom (n). We then use quadratic reciprocity to evaluate (\frac{a}{n}). Up to a factor of −1, (\frac{a}{n}) = (\frac{n}{a}). Thus, we may reduce n mod a, so that n mod a lies between −\frac{a}{2} and \frac{a}{2}. Again, the top is half the bottom, and the bottom is at most one-quarter of what we started with (n).

We continue this process; we need to do such flippings at most log n times. Why? Each time the size of the denominator is at most half what it was before. If 2^{r} = n, then r = log_{2} n < log n. Thus, after at most log n passes, the denominator would be about 1. So, a stage or two before would give a denominator around 2 or 4. The point is, in log n steps, we can reduce to evaluating the Legendre symbol of something where the bottom is of bounded size. Thus, we can evaluate (\frac{a}{n}) in O(log n) steps. (Have a lookup table for n < 10, et cetera).

We need to evaluate (\frac{a}{n}) for C log^{2} n choices of a, and for each choice we need to evaluate a^{\frac{n−1}{2}} mod n (so we can compare it to (\frac{a}{n})), which takes O(log n) steps. Thus, the number of steps is O(log^{4} n).

The Riemann Hypothesis plays a very important catalystic role. It leads us to statements we feel should be true, statements which can often be proved without the full force of Riemann.

Suppose we pass the test for all a with 2 ≤ a ≤ 70 log^{2} n and a relatively prime to n. Gary Miller proved that, if GRH is true, then knowing that n passes the test for all a in this little segment allows us to conclude that n will pass the test for all a. Clearly, we know if n passes the test for all a up to n − 1, then we know n will pass the test for all a relatively prime to n. The power of GRH is that we need only check log^{2} n values of a.

3.2.1 Aside: Finite Abelian Groups

Let A be a finite Abelian Group, then A is (ie, is isomorphic to) a product of cyclic groups.

Look at (ℤ/nℤ, +), the group of integers mod n under clock addition. This is a cyclic group.

The general statement is:

Theorem 3.2.2 (Structure Theorem for Finite Abelian Groups). Let A be a finite Abelian Group. Then there are integers n_{1} through n_{t} such that A ≊ ℤ/n_{1}ℤ× ⋯ × ℤ/n_{r}ℤ, ie, A is isomorphic to the Cartesian product of groups of the form ℤ/mℤ.

Note ≊ means is isomorphic to; we say two groups are isomorphic if there is a group homomorphism between them which is one-to-one and onto. A group homomorphism ϕ is a map which preserves the group structure. Consider two groups G_{1} and G_{2} and a map ϕ : G_{1} → G_{2}. If ϕ is a group homomorphism, then for x, y ∈ G_{1}, ϕ(x + y) = ϕ(x) ⨁ ϕ(y), where + is addition in the first group and ⨁ is addition in the second group.

We now define the Cartesian Product of two groups. X × Y is the set of pairs (x, y) where x ∈ X and y ∈ Y. If we are writing the group action of X and Y additively (say by ⨁_{X} for addition in X and ⨁_{Y} for addition in Y), then (x_{1}, y_{1}) ⨁_{X+Y} (x_{2}, y_{2}) = (x_{1} ⨁_{X} x_{2}, y_{1} ⨁_{Y} y_{2}).

Let us consider a group written in additive notation. Recall the order of an element is the number of times you must add the element to itself to get the identity. We define the exponent of a group as the least common multiple of the orders of the elements of the group.

Consider ℤ/2ℤ×ℤ/2ℤ, the product of two groups. In ℤ/2ℤ×ℤ/2ℤ, we have the pairs (0, 0), (0, 1), (1, 0) and (1, 1). (0, 0) is the identity under addition; all other elements (check) have order 2. Thus, the exponent of ℤ/2ℤ × ℤ/2ℤ is 2.

Now consider ℤ/4Z = {0, 1, 2, 3}. 0 is the identity, and under addition 1 and 3 have order 4, and 2 has order 2. Thus, the exponent of this group is 4. Exercise: using the exponent, observe that ℤ/2ℤ × ℤ/2ℤ and ℤ/4Z cannot be isomorphic.

Fact: if p is prime, then ℤ/pℤ = 𝔽_{p} is a field. The non-zero elements have multiplicative inverses. (ℤ/pℤ)^{∗} = 𝔽_{p}^{∗} = 𝔽_{p} − {0} (the non-zero elements under multiplication) is a cyclic abelian group with p − 1 elements! IE, there is an element g whose powers generate the group. Exercise: Prove that if n is composite, then ℤ/nℤ is not a field. Hint: show some non-zero element has no multiplicative inverse.

Sketch of Proof of Fact: Suppose 𝔽_{p}^{∗} is not cyclic. Let d be the exponent (the least common multiple of the orders of the elements) of 𝔽_{p}^{∗}. Then d < p − 1. As the order of every element divides p − 1 (Lagrange’s Theorem), we have d|p − 1.

For each x ∈ F pf, x^{d} = 1 (in the field 𝔽_{p}^{∗}). This is because d is the least common multiple of the orders of the elements of the group. Thus, if ord(x) is the order of x, x^{ord(x)} = 1. As ord(x)|d, we have ord(x) = kd for some k ∈ ℤ. Thus, x^{d} = x^{kord(x)} = (x^{ord(x)})^{k} = 1^{k} = 1.

Now use the fact that 𝔽_{p}^{∗} is a field. Consider x^{d} − 1 = 0 over 𝔽_{p} (clearly 0 is not a root). Over 𝔽_{p} means look for solutions to this equation with x ∈ 𝔽_{p}). A polynomial of degree d has at most d roots (another theorem of Lagrange). But every x ∈ 𝔽_{p}^{∗} = 𝔽_{p} − {0} is a root, because d (the exponent of the group) is the least common multiple of the orders of the elements. This is a contradiction: we have p − 1 roots (every x ∈ 𝔽_{p}^{∗} solves x^{d} − 1 ≡ 0 mod p), but by Lagrange there are at most d roots. As d < p − 1, contradiction.

Steve Miller will discuss this needed theorem of Lagrange. See also the handout from Davenport’s The Higher Arithmetic.

3.2.2 Lehmer’s Proposition

Proposition 3.2.3 (Lehmer). If (\frac{a}{n}) ≡ a^{\frac{n−1}{2}} mod n for all a up to C log^{2} n, then n is prime.

We assume n = p_{1} ⋯ p_{t}, t > 1, and all primes are distinct and odd. (When there are repeated primes, the proof is easier, and is left as an exercise to the reader). Then a^{n−1} ≡ 1 mod n for (a, n) = 1 (ie, a and n relatively prime), implies that a^{n−1} ≡ 1 mod p_{j} (p_{j} one of the prime factors of n).

Thus, n − 1 ≡ 0 mod p_{j} − 1. If there were a remainder (ie, if n − 1 wasn’t equivalent to 0 mod p_{j} − 1 but was equivalent to r_{j} ≢ 0 mod p_{j} − 1), if we raised a to this power (r_{j}), we wouldn’t get 1 for all a. Just take a a generator of 𝔽_{p_{j}}^{∗}, ie, an element of maximal order p_{j} − 1. Then a^{n−1} ≡ a^{r_{j}} ≠ 1 mod p_{j}. Note that a^{p_{j}−1} ≡ 1 mod p_{j}, so a^{n−1} ≡ a^{r_{j}} mod p_{j}.

We say p_{j} (a factor of n) is of type 1 if \frac{n−1}{2} ≡ 0 mod p_{j} − 1; we say p_{j} is of type 2 if \frac{n−1}{2} ≡ \frac{p_{j}−1}{2} mod p_{j} − 1.

If at least one of the p_{j}’s (without loss of generality, say p_{1}) is of type 1, take a a quadratic non-residue mod p_{1} and a quadratic residue for p_{2}, p_{3}, …, p_{t}. One can find such an a by the Chinese Remainder Theorem as the primes are distinct. For a statement of the Chinese Remainder Theorem, see the Appendix at the end of the notes.

We have (\frac{a}{p_{1}} ) = −1 and (\frac{a}{p_{j}} ) = 1 for j > 1. As we are assuming n is composite, there are at least two primes.

As (\frac{a}{n}) = (\frac{a}{p_{1}} ) (\frac{a}{p_{2}} ) ⋯ (\frac{a}{p_{t}} ), we obtain (\frac{a}{n}) = −1 · 1 ⋯ 1.

As we are assuming p_{1} is of type 1, we have \frac{n−1}{2} ≡ 0 mod p_{1} − 1. Mod p_{1} − 1, each non-zero element in (𝔽/p_{1}𝔽)^{∗} has order dividing p_{1} − 1. Thus, a^{\frac{n−1}{2}} ≡ 1 mod p_{1}.

We are assuming that the Miller Test is satisfied for all a. Thus, (\frac{a}{n}) = a^{\frac{n−1}{2}} mod n. Mod p_{1}, we have shown the left hand side is −1 and the right hand side is 1, contradiction!

We are left with the case where all the primes are of type 2. We leave this as an exercise for the reader.

3.3 GRH Implies Just Need To Check Up To O(log^{2} n) in Miller

Why does GRH imply that (\frac{a}{n}) ≡ a^{\frac{n−1}{2}} mod n for all a up to 70 log^{2} n (and relatively prime to n) implies that (\frac{a}{n}) ≡ a^{\frac{n−1}{2}} mod n for all a relatively prime to n?

3.3.1 Fourier Analysis

Representation Theory (especially characters of Abelian Groups) is the most important things in Mathematics. Let A be a finite Abelian Group. To each such group, we associate a dual group, denoted Â, where Â is the set of all homomorphisms from A into ℂ^{∗}, ℂ^{∗} are the complex numbers invertible under multiplication.

Recall ψ is a group homomorphism if ψ(a + b) = ψ(a)ψ(b). Note here ψ : G_{1} → G_{2}, and G_{1} has been written additively and G_{2} has been written multiplicatively. We call such a ψ a character.

3.3.2 Examples

Let A = ℤ/nℤ, and ν ∈ ℤ/nℤ. Let e(z) = e^{2πiz} for z ∈ ℂ. Define ψ_{ν}(x) = e(\frac{νx}{n}). Exercise: Show by direct calculation that ψ_{ν}(a+b) = ψ_{ν}(a)ψ_{ν}(b). In fact, for each ν ∈ ℤ/nℤ we get a character, and the characters are distinct (exercise).

We only need to know how ψ_{ν} acts on 1, as ψ_{ν}(k) = ψ_{ν}(1+⋯+1) = (ψ_{ν}(1))^{k}.

Â is canonically isomorphic to A. What this means is, to each ν ∈ A there corresponds a character ψ_{ν} in Â, and to each character ψ ∈ Â there corresponds a number ν_{ψ} ∈ A = ℤ/nℤ.

We can multiply two characters: (ψ_{1}ψ_{2})(x) = ψ_{1}(x)ψ_{2}(x). It is easy to see that this is a character. The trivial character (which sends everything to 1) is the identity of the group Â.

Consider two groups A and B. We can use the characters of A and B to get the characters of the cartesian product A × B. If we want the characters of A × B, take a character ψ_{a} of A and ϕ_{b} of B and form the character ψ_{a}ϕ_{b}, defined by (ψ_{a}ϕ_{b})(x, y) = ψ_{a}(x)ϕ_{b}(y).

3.3.3 Characters of 𝔽_{p}^{∗}: Dirichlet Characters

We denote characters of 𝔽_{p}^{∗} by χ. Dirichlet proved the best theorem of mathematics, introducing a lot of new math to solve the following:

Theorem 3.3.1 (Primes in Arithmetic Progressions, 1836, 1839). Let a and n be relatively prime. There are infinitely many primes which give a when divided by n; moreover, to first order all residue classes a mod n (a, n relatively prime) have the same number of primes!

In other words, there are infinitely many x such that xn + a is prime if a and n are relatively prime.

Hard Question: without using Dirichlet’s Theorem, can you prove that there must be one prime congruent to a mod n if a and n are relatively prime?

Clearly, if a and n are not relatively prime, there cannot be infinitely many primes congruent to a mod n. Dirichlet shows this is also a sufficient condition.

Dirichlet introduced characters without introducing the concept of a group! They didn’t have group notation until later.

Look at (ℤ/nℤ)^{∗} = {x : (x, n) = 1, 0 < x ≤ n − 1}. This is a finite Abelian group. #(ℤ/nℤ)^{∗} = ϕ(n), where ϕ(n) is the Euler totient function, and ϕ(n) is the number of integers (between 0 and n) which are relatively prime to n.

Dirichlet used q instead of n, so we change notation and look at (ℤ/qℤ)^{∗}. A Dirichlet Character is a character χ of (ℤ/qℤ)^{∗}; ie, χ : (ℤ/qℤ)^{∗} → ℂ^{∗}

and χ(ab) = χ(a)χ(b). Thus, χ lies in the dual group of (ℤ/qℤ)^{∗}. Recall that ℂ^{∗} is the set of complex numbers with multiplicative inverses, ie, ℂ^{∗} = ℂ − {0}.

The principal or trivial character takes every x ∈ (ℤ/qℤ)^{∗} to 1; we denote the trivial character by χ_{0}.

If χ is a Dirichlet Character of (ℤ/qℤ)^{∗}, we say χ has modulus (also called conductor) q. We extend χ to be defined on all integers (all of ℤ) by χ(m) = 0 if m and q are not relatively prime, and χ(m) = χ(m mod q) otherwise. Clearly χ is periodic, as χ(x+λq) = χ(x) for any λ ∈ ℤ. Thus, we have the map χ : ℤ → ℂ^{∗}.

If q is prime, we have previously seen the character χ(a) = (\frac{a}{q}). This is an extremely important character.

Another example: Let q = 4. We define χ(n) to be 0 if n is even, 1 if n ≡ 1 mod 4, and −1 if n ≡ −1 mod 4. Exercise: Show this is a character.

3.3.4 General Riemann Hypothesis (GRH)

Below, p will always denote a prime. χ will be a Dirichlet character modulo q (q need not be prime).

Conjecture: GRH: For any q ≥ 1, x ≥ 3, if χ ≠ χ_{0} (ie, if χ is not the principal character) then

| ∑_{p≤x} (1 − \frac{p}{x}) · log p · χ(p)| ≤ C log q · \sqrt{x}.  (3.1)

If χ = χ_{0}, then

∑_{p≤x} (1 − \frac{p}{x}) log p ≤ \frac{x}{2} + O(\sqrt{x}).  (3.2)

C is a universal constant, independent of q.

We are summing primes up to x. The 1 − \frac{p}{x} factors are just weight factors. If q = 4 and χ is the character above, χ(p) gives a positive sign for primes congruent to 1 mod 4 and a minus sign for primes congruent to −1 mod 4.

Analysis means cancellation; you want to see cancellation in a sum. The numbers in these sums are flipping (plus one, minus one); we have about \frac{x}{log x} primes less than x (this is the Prime Number Theorem, first proved in 1896).

Random Drunk: each moment he flips a coin. If it is heads he staggers one unit left, tails he staggers one unit right. After n steps, where do you think he’ll be? Turns out he’ll be about \sqrt{n} units from the origin (with high probability). Steve Miller will prove this in a later lecture or handout.

Motto: random numbers (of size around 1) cancel like the square-root of the number of terms.

What GRH is telling us is that the remainder term behaves like random noise.

3.3.5 Proof of the Miller Test

Theorem 3.3.2. If GRH is true, then if (\frac{a}{n}) ≡ a^{\frac{n−1}{2}} mod n for all a ≤ C log^{2} n for some fixed constant C, then it is true for all a (relatively prime to n, of course).

Remarkable observation: S = \{a ∈ (ℤ/nℤ)^{∗} : (\frac{a}{n}) ≡ a^{\frac{n−1}{2}} mod n\} is a subgroup of (ℤ/nℤ)^{∗}. Why? Exercise (Closure: show if a, b in this set, so is ab. Identity: show 1 is in this set. Inverses: show if a ∈ S, a^{−1} mod n is in S. Note associativity is inherited from (ℤ/nℤ)^{∗}).

By assumption, S contains (at least) the first C log^{2} n elements. We claim this implies S contains all a relatively prime to n.

Suppose S is not all of (ℤ/nℤ)^{∗}. We talked about cosets (of Abelian Groups) last time. (For more information about cosets and quotient groups, see the following lecture by Steven J. Miller, notes by Alex Barnett). As everything is abelian, we can form a quotient group by dividing our group (ℤ/nℤ)^{∗} by the abelian subgroup S.

Thus, A = \frac{(ℤ/nℤ)^{∗}}{S} is a group (a quotient group) and A is a non-trivial group (ie, there is more than one element in this group). Why must A have more than one element? A is the group of representative cosets of (ℤ/nℤ)^{∗} by the abelian subgroup S. As S is not all of (ℤ/nℤ)^{∗}, there must be at least two cosets. Hence A is not just the identity coset (which is 1 · S or just S).

But this quotient (A) is a finite abelian group. We know for a finite abelian group that its dual is isomorphic to itself. This means to each element in A we have a character in Â, and vice-versa. Further, the identity of A is mapped to the trivial character of Â.

Thus, there is a non-trivial χ : A → ℂ^{∗}. Now (ℤ/nℤ)^{∗} → \frac{(ℤ/nℤ)^{∗}}{S} → ℂ^{∗}, and thus we have a non-trivial Dirichlet character of (ℤ/nℤ)^{∗} which is trivial on S (ie, χ(s) = 1 for all s ∈ S).

By construction, χ is a Dirichlet character of (ℤ/nℤ)^{∗}, χ ≠ χ_{0}, and χ|_{S} = 1 (the last means that χ, restricted to s ∈ S is the identity map). We know S is all elements up to at least C log^{2} n. So, χ(a) = 1 for all a ≤ C log^{2} n (from the givens).

So, look at

Sum(p, q, x) = ∑_{p≤x} (1 − \frac{p}{x}) log p · χ(p).  (3.3)

By GRH, Sum(p, q, x) is less than a constant multiple of log q · \sqrt{x}. Calling the constant C_{2}, we have Sum(p, q, x) ≤ C_{2} log q · \sqrt{x}.

If x ≤ C log^{2} n, then there is a contradiction. Why? In Equation 3.3, Sum(p, q, x) is going to be ∑_{p≤x} (1 − \frac{p}{x}) log p, because all the χ(p) = 1 in this range. The Prime Number Theorem states that the number of primes less than x is \frac{x}{log x} plus an error term which is smaller than \frac{x}{log x} (ie, in the limit, the size of the error term divided by the number of primes less than x tends to 0 as x goes to infinity).

Thus,Sum(p, q, x) is going to look like a multiple of x. Why a multiple of x and not a multiple of \frac{x}{log x}? Remember we have the factor log p in the sum; this follows from Partial Summation. Partial Summation is a discrete version of Integration by Parts; see the Appendices at the end for a further statement.

By GRH, Sum(p, q, x) is bounded by C_{2} log q · \sqrt{x}. Therefore, x ≤ C_{2} log q · \sqrt{x}. This implies \sqrt{x} ≤ C_{2} log q, or x ≤ C_{2}^{2} log^{2} q.

Remember, we’ve changed notation from n to q. We are assuming that (\frac{a}{q}) ≡ a^{\frac{q−1}{2}} mod q for all a ≤ C log^{2} q for some fixed constant C. Take C greater than C_{2}^{2}. Then we have a contradiction! If GRH is true, Sum(p, q, x) ≤ C_{2} log q · \sqrt{x} only for x ≤ C_{2}^{2} log^{2} q. But we are assuming that q passes the Miller Test for all a up to C log^{2} q. Thus, we can take x larger than C_{2}^{2} log^{2} q.

3.3.6 Review of Proof

Intuition: For random sequences, expect square-root cancellation in sums.

If we have a non-trivial character, if we look at these weighted sums of χ(p), there is no extra structure; we expect cancellation like \sqrt{x}. There has to be some q-dependence, but GRH says it is like a universal constant times log q.

In the Miller Test, we test something C log^{2} n times. If the condition is always true for these C log^{2} n elements, we have a subgroup S of (ℤ/nℤ)^{∗}. If S isn’t all of (ℤ/nℤ)^{∗}, we can find a character, and we have sums with this character (any sum with characters is called Harmonic Analysis). Further, this is a non-trivial character which is the identity on the original group. The GRH cannot accomodate a character of this type.

Why does the GRH lead to a contradiction? Basically, GRH says a certain weighted sum of χ(p) over the primes less than x (where χ is a Dirichlet character with modulus q) cannot be too large. Specifically, it is at most C_{2} log q · \sqrt{x}.

This implies that there is a lot of noise in the χ(p); basically, we need to have a good mixing of primes which give χ(p) = +1 with primes giving χ(p) = −1.

However, if n satisfies the Miller Test for a up to C log^{2} q (with C > C_{2}^{2}), then we can find a modulus q and a Dirichlet character χ where we have a very long string of primes giving χ(p) = +1. This forces the weighted sum of χ(p) over primes less than x (taking x = C log^{2} q) to be larger than C_{2} log q\sqrt{x}, a contradiction.

3.4 Appendices

3.4.1 Aside: For p Odd, Half the Non-Zero Numbers are Quadratic Residues

Note, for an odd prime p_{j}, half of the non-zero numbers are quadratic residues, and half are quadratic non-residues. The Legendre symbol takes each element a ∈ 𝔽_{p}^{∗} to an element in the group {−1, 1}. This is a homomorphism; not every element has a square. The image is {−1, 1}; the kernel is all the elements of 𝔽_{p}^{∗} which are sent to 1. Thus, half the numbers are residues, half are non-residues.

a → (\frac{a}{p}) = a^{\frac{p−1}{2}} mod p. Thus, we have a homomorphism (given by the Legendre symbol) from 𝔽_{p}^{∗} → {−1, 1}. We claim the map is onto. The kernel is all elements in 𝔽_{p}^{∗} which are mapped by the Legendre symbol to 1, ie, the quadratic residues. (One needs to show the Legendre symbol is a group homomorphism: (\frac{xy}{p}) = (\frac{x}{p}) · (\frac{y}{p})).

Standard Group Theory Arguments: \frac{#𝔽_{p}^{∗}}{kernel} ≊ \{−1, 1\}. Thus, half the numbers in 𝔽_{p}^{∗} are quadratic residues.

3.4.2 Chinese Remainder Theorem

Theorem 3.4.1 (Chinese Remainder Theorem). Let m = m_{1}m_{2}, m_{1} and m_{2} relatively prime. Then ℤ/mℤ ≊ ℤ/m_{1}ℤ × ℤ/m_{2}ℤ.

This allows us to, given a_{1} mod m_{1} and a_{2} mod m_{2}, find an a mod m such that a ≡ a_{1} mod m_{1} and a ≡ a_{2} mod m_{2}. For example, try and solve x ≡ 3 mod 5 and x ≡ 4 mod 7.

See any book on Algebra.

3.4.3 Partial Summation

Lemma 3.4.2 (Partial Summation: Discrete Version).

∑_{M}^{N} a_{n}b_{n} = A_{N}b_{N} − A_{M−1}b_{M} + ∑_{M}^{N−1} A_{n}(b_{n} − b_{n+1})  (3.4)

Lemma 3.4.3 (Abel’s Summation Formula - Integral Version). Let h(x) be a continuously differentiable function. Let A(x) = ∑_{n≤x} a_{n}. Then

∑_{n≤x} a_{n}h(n) = A(x)h(x) − ∫_{1}^{x} A(u)h′(u)du  (3.5)

See, for example, Walter Rudin, Principles of Mathematical Analysis (also known as The Blue Book), page 70.

Cosets, Quotient Groups, and an Introduction to Probability

Quotient groups. Basic probability theory for random walk. Lecture by Steven J. Miller; notes by Alex Barnett and Steven J. Miller.

4.1 Quotient groups

Say we have a finite Abelian group G (this means for all x, y ∈ G, xy = yx) of order m which has a subgroup H of order r. We will use multiplication as our group operation. Recall the coset of an element g ∈ G is defined as the set of elements gH = g{h_{1}, h_{2}, ⋯, h_{r}}. Since G is Abelian (commutative) then gH = Hg and we will make no distinction between left and right cosets here.

The quotient group (or factor group), symbolized by G/H, is the group formed from the cosets of all elements g ∈ G. We treat each coset g_{i}H as an element, and define the multiplication operation as usual as g_{i}Hg_{j}H. Why do we need G to be Abelian? The reason is we can then analyze g_{i}Hg_{j}H, seeing that it equals g_{i}g_{j}HH. We will analyze this further when we prove that the set of cosets is a group.

There are several important facts to note. First, if G is not Abelian, then the set of cosets might not be a group. Second, recall we proved the coset decomposition rule: given a finite group G (with n elements) and a subgroup H (with r elements) then there exist elements g_{1} through g_{k} such that

G = ⋃_{i=1}^{k} g_{i}H.  (4.1)

The choices for the g_{i}’s is clearly not unique. If g_{1} through g_{k} work, so do g_{1}h_{1} through g_{k}h_{k}, where h_{i} is any element of H. Recall this was proved by showing any two cosets are either distinct or identical.

We will show below that, for G Abelian, the set of cosets is a group. Note, however, that while it might at first appear that there are many different ways to write the coset group, they really are the same. For example, the cosets gH and gh_{1}h_{2}^{4}h_{3}H are equal. This is similar to looking at integers mod n; mod 12, the integers 5, −7 and 19 are all equal, even though they look different.

We now prove that the set of cosets is a group (for G Abelian).

Closure. By commutivity g_{i}Hg_{j}H = g_{i}g_{j}HH. What is “HH”? Just the set of all r^{2} possible combinations of elements of H. By closure, and the existence of the identity, this just gives H again (recall no element in a group can appear more than once—duplicates are removed). Therefore g_{i}Hg_{j}H = g_{i}g_{j}H. Now, as G is a group and is closed, g_{i}g_{j} ∈ G. Thus, there is a α such that g_{i}g_{j} ∈ g_{α}H (as G = ⋃_{β=1}^{k} g_{β}H. Therefore, there is an h ∈ H such that g_{i}g_{j} = g_{α}h, which implies g_{i}g_{j}H = g_{α}hH = g_{α}H. Thus, the set of cosets is closed under coset multiplication. Note, however, that while the coset g_{i}g_{j}H is in our set of cosets, it may be written differently.

Identity. If e is identity of G, then eHg_{i}H = g_{i}H and g_{i}HeH = g_{i}H, so eH is the identity of this quotient group.

Associativity. Since as you may have noticed, the quotient group elements behave just like those of G, associativity follows from that of G.

Inverse. It is easy to guess g^{−1}H is the inverse of gH. Check it: g^{−1}HgH = g^{−1}gH = eH = identity, also true the other way round of course by commutativity. Unfortunately, g^{−1}H might not be listed as one of our cosets! Thus, we must be a little more careful. Fortunately, as g^{−1} ∈ G = ⋃_{β=1}^{k} g_{β}H, there is an α such that g^{−1} ∈ g_{α}H. Then, there is an h ∈ H with g^{−1} = g_{α}h. Thus, g^{−1} = g_{α}hH = g_{α}H, and direct calculation will show that the coset g_{α}H is the inverse (under coset multiplication) of gH.

4.2 Random walk and discrete probability

Each step in a random walk is a random event. We first study a single random event, then the combination of two random events, then multiply repeated events.

For other introductory probability theory see

http://engineering.uow.edu.au/Courses/Stats/File24.html

4.2.1 Probability distribution for a single event, mean, variance

Our single event consists of one of a set of choices happening. The choices are labelled by i = 1 ⋯ N, which are exclusive (no more than one can happen), and complete (no less than one can happen). For instance, for a single coin toss,

i = 1  H, heads,
i = 2  T, tails.

We take the choice i as a random variable, meaning all we know is a probability p_{i} ≥ 0 that each choice can happen. p = 0 means it never happens, p = 1 means it always happens, and most things are somewhere in between (the unbiased coin has p_{i} = \frac{1}{2}, ∀i, ignoring the small probabilities of the coin landing on its edge or quantum tunneling through the table.)

The set of {p_{i}} we call the probability distribution. Completeness implies

∑_{i} p_{i} = 1.  (4.2)

Note the abbreviation ∑_{i} for ∑_{i=1}^{N}.

We have some quantity f which has a value f_{i} for each choice i. For instance, f could be the number of dots on each face i of a die, in which case f_{i} = i. Mean and variance are ways to characterize (summarize) the distribution over f.

Mean. The mean or expectation value (expected value) of f over the distribution is

f̄ ≡ E[f] = ∑_{i} p_{i}f_{i}.  (4.3)

This is just a weighted average of f. Check it for the (unweighted) dice. p_{i} = \frac{1}{6}, ∀i. You should get f̄ = \frac{7}{2}.

Variance. The variance is the square of the standard deviation σ_{f}:

σ_{f}^{2} ≡ Var[f] = ∑_{i} p_{i}(f_{i} − f̄)^{2}.  (4.4)

Crudely σ_{f} gives the width of the distribution in f. From the definition you can see σ_{f} is the root mean square (rms) deviation from the mean. Expanding out the square, and using earlier results,

Var[f] = ∑_{i} p_{i}f_{i}^{2} − 2f̄· ∑_{i} p_{i}f_{i} + f̄^{2} · ∑_{i} p_{i}
= ∑_{i} p_{i}f_{i}^{2} − f̄^{2}
= E[f^{2}] − E[f]^{2}.  (4.5)

This is a very useful formula. The first term is often known as the second moment of f. The first moment is just the mean. The m^{th} moment is defined as ∑_{i} p_{i}f_{i}^{m}.

A quick comment about units. If f_{i} and f are measured in feet (for example), the variance (which gives information on how spread out f_{i} is) has units feetsquared. If someone asks how tall the people in our class are, one would answer about 5\frac{1}{2} feet. If one is further pressed to give a range for the heights of our class, one might say 5\frac{1}{2} feet, plus or minus \frac{1}{4} of a foot. One would not give the error range in feet-squared! Thus, in measuring error it is the square-root of the variance that comes into play. Note that if f_{i} is in feet, the variance is in feet-squared, and the square-root of the variance is in feet.

4.2.2 Multiple events

Consider two random events. Let the first have choices i = 1 ⋯ N (with probabilities p_{1} through p_{N}); let the second event have choices j = 1 ⋯ M (with probabilities q_{1} through q_{M}). Then there are NM choices (possibilities) for the combined event, which we could label by k ≡ ij. Since k is also a random variable, it has probability r_{k} = r_{ij} (you could think of this as a matrix in i, j). If the two events are independent (also called uncorrelated), then

r_{ij} = p_{i}q_{j}, ∀ij  independence.  (4.6)

(In other words, the matrix is separable in the i and j directions). Many events we study will be independent. For example, if you flip a fair coin twice, the result of the first flip has no effect on the result of the second flip. Or if you roll a fair die, et cetera.

As before we have a quantity f_{i} associated with each choice i for the first event. For the second event we have a (in general different) quantity g_{j} for its choice j. We want to learn about the sum of these two quantities,

s ≡ f + g.  (4.7)

Note that for each combined choice ij, this quantity s has value s_{ij} = f_{i}+g_{j}. How is s is distributed? We will show that independence of the two events implies a simple law giving the mean and variance of s in term of those of f and g.

We compute the mean of s, following Eq. 4.3, except now we are summing over all combined possibilities ij,

E[s] = ∑_{ij} r_{ij}s_{ij} = ∑_{ij} p_{i}q_{j}(f_{i} + g_{j}) = ∑_{i} p_{i}f_{i} · ∑_{j} q_{j} + ∑_{i} p_{i} · ∑_{j} q_{j}g_{j}
= f̄· 1 + 1 · ḡ = f̄+ ḡ.  (4.8)

So, the means add.

We isolate this important fact:

Lemma 4.2.1. For independent events, the mean of a sum is the sum of the means. Equivalently, the sum of the expected values is the expected value of the sums. Thus, for any independent events A and B, E[A + B] = E[A] + E[B].

What if we multiply A by a constant c? For example, consider outcomes A_{i} with probabilities p_{i}. The mean Ā = E[A]. What is the mean of the new event with outcomes cA_{i} occurring with probabilities p_{i}?

Well,

E[cA] = ∑_{i} p_{i}cA_{i}
= c ∑_{i} p_{i}A_{i}
= cE[A].  (4.9)

We have therefore shown

Lemma 4.2.2. The mean of a multiple is the multiple of the mean. Equivalently, the expected value of a multiple is the multiple of the expected value. Thus, E[cA] = cE[A].

We now calculate the variance of a sum, using the above results. For the variance of s = f + g we can use Eq. 4.5 to get

Var[s] = E[s^{2}] − E[s]^{2}
= E[(f + g)^{2}] − (E[f + g])^{2}
= E[(f + g)^{2}] − (E[f] + E[g])^{2} by Lemma 4.2.1
= E[(f + g)^{2}] − (f̄+ ḡ)^{2}
= E[f^{2}] + 2E[fg] + E[g^{2}] − (f̄+ ḡ)^{2}.  (4.10)

We justify the last step as follows:

E[(f + g)^{2}] = E[f^{2} + 2fg + g^{2}]
= E[(f^{2} + 2fg) + g^{2}]
= E[(f^{2} + 2fg)] + E[g^{2}] by Lemma 4.2.1
= E[f^{2}] + E[2fg] + E[g^{2}] by Lemma 4.2.1
= E[f^{2}] + 2E[fg] + E[g^{2}] by Lemma 4.2.2. (4.11)

Using independence again we can factorize E[fg] = ∑_{ij} r_{ij}f_{i}g_{j} = ∑_{i} p_{i}f_{i} · ∑_{j} q_{j}g_{j} = E[f]E[g] = f̄ḡ. Elegantly, this term is responsible for cancelling the cross-term in (f̄+ ḡ)^{2}, and collecting the remaining terms leaves

Var[s] = Var[f] + Var[g]  (4.12)

So, the variances also add.

We now take the special case when the second event is identical to the first. That is, M = N, q_{i} = p_{i}, g_{i} = f_{i}, ∀i. In this case the above shows that s̄ = 2f̄ and Var[s] = 2Var[f].

We can repeat the above combination law for successively repeated events. Such events are called identical independently distributed (iid) events. Suppose we have K repetitions of iid events, with total s ≡ f + g + ⋯ + z, we can just repeatedly apply the above rules to get

s̄ = K f̄  (4.13)
Var[s] = K Var[f].  (4.14)

The mean and variance are not the only characteristics of the distribution that add like in this way. Amazingly, there is an infinite sequence of special combinations of the higher moments, called cumulants, which add just like this. The mean and variance are just the first two cumulants.

4.2.3 Simplest random walk

We now have the tools to characterize a random walk. We choose N = 2 and p_{1} = p_{2} = \frac{1}{2}, just as with the coin toss, and define the “step” displacements f_{1} = +1 and f_{2} = −1. This corresponds to a drunkard taking (uncorrelated) steps of unit length along the integer line.

We use Eqs. 4.3 and 4.5 to evaluate the mean and variance of a single step event.

f̄ = +\frac{1}{2} − \frac{1}{2} = 0  (4.15)
Var[f] = \frac{1}{2}.(1)^{2} + \frac{1}{2}.(−1)^{2} − (0)^{2} = 1.  (4.16)

For K steps, starting from the origin, we have the final displacement of s, the sum of all the steps, using the formulae above,

s̄ = K · 0 = 0  (4.17)
Var[s] = K · 1 = K.  (4.18)

So the standard deviation, i.e. the width of the distribution of s, has value

σ_{s} ≡ \sqrt{Var[s]} = \sqrt{K}.  (4.19)

This is our first vital fact about all but the most pathological ^{1} random walks: the distribution has width which scales like K^{1/2}. This means that a typical distance from the origin is \sqrt{K}. This is called a diffusion process and is very common in the real world.

Again, remember that if the person walks in feet, the variance (which is a measure of how much the distribution spreads out) will be in feet-squared. By taking the square-root we again have units of feet.

4.2.4 Central Limit Theorem

The Central Limit Theorem (CLT) states that the distribution on s tends to a Gaussian distribution,

p(s) ≈ N (μ, σ^{2}) ≡ \frac{1}{\sqrt{2π}σ} e^{− \frac{(s−μ)^{2}}{2σ^{2}}}  (4.20)

1Random walks which do not exhibit this power law have infinite variance and exhibit anomalous diffusion. For more on this, see M. Bazant’s excellent course at http://www-math.mit.edu/∼bazant/teach/18.325

with mean μ = s̄ and variance σ^{2} = Var[s] as given above, as K → ∞. This is a very common “bell curve” with width σ centered about μ. We have not defined p(s) very rigorously—it is simply the probability of being displaced s from the origin at the end of the K-step random walk. An exact formula for p(s) involves counting all the ways that ±1 can be added K times to get exactly s. We will postpone this, and the proof of CLT, for next time.

Remarkably the CLT applies to any N with any discrete step distribution {p_{i}} and any step displacements {f_{i}}. It also applies to the case of continous-valued steps with distribution p(f) along the real line. However a criterion for validity is always the finiteness of the second moment of p(f).

Quadratic Reciprocity, Central Limit Theorem and Graph Problems

We give Eisenstein’s proof of Quadratic Reciprocity, and then introduce the Graph Theory problems. Lecture by Steven J. Miller and Peter Sarnak; notes by Steven J. Miller.

5.1 Eisenstein’s Proof of Quadratic Reciprocity

5.1.1 Preliminaries

Theorem 5.1.1 (Quadratic Reciprocity). Let p and q be distinct odd primes. Then

\binom{\underline{q}}{p} (\frac{p}{q}) = (−1)^{\frac{p−1}{2} \frac{q−1}{2}} .  (5.1)

As p and q are distinct, odd primes, both \binom{\underline{q}}{p} and (\frac{p}{q}) are ±1. The difficulty is figuring out which signs are correct, and how the two signs are related.

We use Euler’s Criterion, proved in a previous lecture:

Lemma 5.1.2 (Euler’s Criterion).

\binom{\underline{q}}{p} ≡ q^{\frac{p−1}{2}} modp.  (5.2)

The idea behind Eisenstein’s proof is as follows: \binom{\underline{q}}{p} (\frac{p}{q}) is −1 to a power. Further, we only need to determine the power mod 2. Eisenstein shows many expressions are equivalent, mod 2, to this power. Eventually, we arrive at an expression which is trivial to calculate (mod 2).

5.1.2 First Stage

Consider all even multiples of q by an a ≤ p − 1: {2q, 4q, 6q, …, (p − 1)q}. Denote a generic multiple by aq. Recall [x] is the greatest integer less than or equal to x. By integer division,

aq = [\frac{aq}{p} ]p + r_{a}, 0 ≤ r_{a} < p − 1.  (5.3)

Thus, r_{a} is the least non-negative number equivalent to qa mod p.

The numbers (−1)^{r_{a}}r_{a} are equivalent to even numbers in {0, …, p − 1}. If r_{a} is even this is clear; if r_{a} is odd, then (−1)^{r_{a}}r_{a} ≡ p − r_{a} mod p, and as p and r_{a} are odd, this is even.

Lemma 5.1.3. If (−1)^{r_{a}}r_{a} ≡ (−1)^{r_{b}}r_{b}, then a = b.

Proof: We quickly get ±r_{a} ≡ r_{b} mod p. If the plus sign holds, then r_{a} ≡ r_{b} mod p implies qa ≡ qb mod p. As q is invertible mod p, we get a ≡ b mod p, which yields a = b (as a and b are even integers between 0 and p − 1).

If the minus sign holds, then r_{a} + r_{b} ≡ 0 mod p, or qa + qb ≡ 0 mod p. Multiplying by q^{−1} mod p now gives a + b ≡ 0 mod p. As a and b are even integers between 0 and p − 1, 0 < a + b ≤ 2(p − 1). The only integer strictly between 0 and 2p which is equivalent to 0 mod p is p; however, p is odd and a + b is even. Thus, the minus sign cannot hold, and the elements are all distinct. □.

Lemma 5.1.4.

\binom{\underline{q}}{p} = (−1)^{∑_{aeven}r_{a}}.  (5.4)

Proof: For each even a, qa ≡ r_{a} mod p. Thus, mod p:

∏_{a even}qa ≡ ∏_{a even}r_{a}
q^{\frac{p−1}{2}} ∏_{a even}a ≡ ∏_{a even}r_{a}

\binom{\underline{q}}{p} ∏_{a even}a ≡ ∏_{a even}r_{a},  (5.5)

where the above follows from the fact that we have \frac{p−1}{2} choices for an even a (to get q^{\frac{p−1}{2}} and Euler’s Criterion (to replace q^{\frac{p−1}{2}} with \binom{\underline{q}}{p}).

As a ranges over all even number from 0 to p − 1, so too do the numbers (−1)^{r_{a}}r_{a} mod p. Thus, mod p,

∏_{a even}a ≡ ∏_{a even}(−1)^{r_{a}}r_{a}
∏_{a even}a = (−1)^{∑_{a even} r_{a}} ∏_{a even}r_{a}.  (5.6)

Combining gives

\binom{\underline{q}}{p}(−1)^{∑_{a even} r_{a}} ∏_{a even}r_{a} ≡ ∏_{a even}r_{a}.  (5.7)

As each r_{a} is invertible mod p, so is the product. Thus,

\binom{\underline{q}}{p}(−1)^{∑_{a even} r_{a}} ≡ 1 mod p.  (5.8)

As \binom{\underline{q}}{p} is its own inverse, the Lemma now follows by multiplying both sides by \binom{\underline{q}}{p}. □.

Therefore, it is sufficient to determine ∑_{a even} r_{a} mod 2. We make one last simplification. By integer division, we have

∑_{a even}qa = ∑_{a even}([\frac{qa}{p}]p + r_{a})

= ∑_{a even}[\frac{qa}{p}]p + ∑_{a even}r_{a}.  (5.9)

As we are summing over even a, the Left Hand Side above is even. Thus, the Right Hand Side is even, so

∑_{a even}[\frac{qa}{p}]p ≡ ∑_{a even}r_{a} mod 2

p ∑_{a even}[\frac{qa}{p}] ≡ ∑_{a even}r_{a} mod 2

∑_{a even}[\frac{qa}{p}] ≡ ∑_{a even}r_{a} mod 2,  (5.10)

where the last line follows from the fact that p is odd, so mod 2, dropping the factor of p from the Left Hand Side doesn’t change the parity.

We have shown

Lemma 5.1.5. It is sufficient to calculate ∑_{a even}[\frac{qa}{p}]

5.1.3 Second Stage

Consider the rectangle with vertices at A = (0, 0), B = (p, 0), C = (p, q) and D = (0, q). The upward slopping vertical is given by the equation y = \frac{q}{p}x. As p and q are distinct odd primes, there are no pairs of integers (x, y) on the line AC.

We now interpret ∑_{a even} [\frac{qa}{p}]. Consider the vertical line with x coordinate

a. Then [\frac{qa}{p}] gives the number of pairs (x, y) with x-coordinate equal to a and

y-coordinate an integer at most [\frac{qa}{p}]. Thus, ∑_{a even} [\frac{qa}{p}] is the number of integer pairs (in the rectangle ABCD) with even x-coordinate that are below the line AC.

We add some additional points: E = (\frac{p}{2}, 0), F = (\frac{p}{2}, \frac{q}{2}), G = (0, \frac{q}{2}) and H = (\frac{p}{2}, q). We prove

Lemma 5.1.6. The number of integer pairs under the line AC (inside the rectangle) with even x-coordinate is congruent mod 2 to the number of integer pairs under the line AF.

Let a > \frac{p}{2} be an even integer. The integer pairs on the line x = a are (a, 0), (a, 1), …, (a, q). There are q + 1 pairs. As q is odd, there are an even number of integer pairs on the line x = a. As there are no integer pairs on the line AC, for a fixed a > \frac{p}{2}, mod 2 there are the same number of integer pairs above AC as there are below AC.

Further, the number of integer pairs above AC is equivalent mod 2 to the number of integer pairs below AF on the line x = p − a. To see this, consider the map which takes (x, y) to (p − x, q − y). As a > \frac{p}{2} and is even, p − a < \frac{p}{2} and is odd. Further, every odd a < \frac{p}{2} is hit (given a_{odd} < \frac{p}{2}, start with the even number p − a_{odd} > \frac{p}{2}).

Let #F CH_{even} be the number of integer pairs (x, y) in triangle F CH with x even.

Let #EBCH be the number of integer pairs in the rectangle EBCH; #EBCH ≡ 0 mod 2 (we’ve shown each vertical line has an even number of pairs).

Let #AF E_{even} be the number of integer pairs (x, y) in the triangle AF E with x even, and let #AF E be the number of integer pairs in the triangle AF E.

We need to calculate ∑_{a even} [\frac{qa}{p}] mod 2:

∑_{a even}[\frac{qa}{p}] = #AF E_{even} + #EBCH − #F CH
≡ #AF E_{even} + #EBCH + #F CH
= #AF E_{even} + #F CH + #EBCH
= #AF E + #EBCH
= #AF E.  (5.11)

Therefore, μ = ∑_{a even} [\frac{qa}{p}] ≡ #AF E mod 2, and we have

\binom{\underline{q}}{p} = (−1)^{μ}.  (5.12)

Reversing the rolls of p and q, we see that

(\frac{p}{q}) = (−1)^{ν},  (5.13)

where ν ≡ #AF G mod 2, with #AF G equal to the number of integer pairs in the triangle AF G.

Now, μ + ν = #AF E + #AF G, which is the number of integer pairs in the rectangle AEF G. There are \frac{p−1}{2} choices for x and \frac{q−1}{2} choices for y, giving \frac{p−1}{2} \frac{q−1}{2} pairs of integers in the rectangle AEF G.

Thus,

\binom{\underline{q}}{p} (\frac{p}{q}) = (−1)^{μ+ν}
= (−1)^{#AF E+#AF G}
= (−1)^{\frac{p−1}{2} \frac{q−1}{2}} ,  (5.14)

which completes the proof of Quadratic Reciprocity. □.

5.2 Central Limit Theorem

X_{1}, X_{2}, X_{3}, … an infinite sequence of random variables such that the X_{j} are independent identically distributed random variables (abbreviated i.i.d.r.v.) with E[X_{j}] = X̄_{j} = 0 (can always renormalize by shifting) and variance E[X_{j}^{2}] = 1. Let S_{N} = ∑_{j=1}^{N} X_{j}.

Theorem 5.2.1. Fix −∞ < a ≤ b < ∞. Then as N → ∞,

Prob(\frac{S_{N}}{\sqrt{N}} ∈ [a, b]) → \frac{1}{\sqrt{2π}} ∫_{a}^{b} e^{− \frac{t^{2}}{2}} dt.  (5.15)

The probability function is called the Gaussian or the Normal distribution. This is the universal curve of probability. Note how robust the Central Limit Theorem is: it doesn’t depend on fine properties of the X_{j}.

5.3 Possible Problems

5.3.1 Combinatorics and Probability

Combinatorics is the number of ways of doing something. A graph is a set of vertices (V) and edges (E) between them. Thus, edges are unordered pairs of vertices.

Four Color Theorem (proved by an exhaustive search by the computer). Say you have a graph in the plane (thus, if you draw the vertices and edges in the plane, no two edges cross). Call such a graph a planar graph. Can you color the vertices such that if two vertices are joined by an edge, they have different colors? Sure, by using |V | colors! What is the least number of colors needed?

Theorem 5.3.1 (Four Coloring Theorem). You can color the vertices of a planar graph using at most four colors such that no two joined vertices have the same color.

A k-regular graph is a graph such that there are k-edges out of each vertex. A 2-regular graph has no freedom: you get a closed cycle.

Consider 3-regular graphs. To each graph associate the adjacency matrix A. (First to study may be Kirchhoff). Say G = (V, E) has |V | = n vertices. For now, assume there are no multiple edges (ie, between any two vertices is at most one edge, and there are no edges connecting a vertex to itself). A is an n × n matrix, rows and columns indexed by the vertices, and A_{ij} = 1 if there is an edge from v_{i} to v_{j} and 0 otherwise.

Thus, the adjacency matrix is a matrix with 0s and 1s, and is symmetric.

Problem: What is the second largest eigenvalue of A? How does it vary? What do we expect?

In Linear Algebra, we learn we can diagonalize a real symmetric matrix. The eigenvalues are real, and satisfy p_{A}(λ) = det(λI − A), the characteristic polynomial. This is a polynomial in λ of degree n with integer coefficients. Thus, the eigenvalues are algebraic numbers. The leading coefficient is λ^{n}, the constant term is det(A).

Thus, p_{A}(λ) = λ^{n}+⋯+det(A), and by the Fundamental Theorem of Algebra, there are n complex roots. If the leading coefficient of the defining polynomial is 1, we say the roots are algebraic integers. These roots are the eigenvalues of A.

Why must the eigenvalues be real? Want Av = λv, v a non-zero vector.

Fact: if v = (1, 1, …, 1)^{T}, v is an eigenvector of A with eigenvalue k. Why? As each vertex is connected to k distinct vertices, each row has exactly k entries that are 1 and n − k entries that are 0. Thus, k is an eigenvalue, denote by λ_{0}.

Exercise 5.3.2. Show, for such adjacency matrices A, that all eigenvalues satisfy −k ≤ λ ≤ k.

Consider connected graphs G. How big is λ_{1}(G) for the random 3-regular graph?

Theorem 5.3.3 (Kirchhoff’s Theorem). Let det^{∗}(kI − A) be the product of the non-zero eigenvalues of A. Then det^{∗}(kI −A) equals the number of vertices times the number of spanning trees.

A tree is a connected graph with no cycles. A spanning tree is a connected sub-graph containing all the vertices. Sometimes called complexity of the graph.

Efficient Algorithms, Probability, Alg+Transcendental, Pidgeon Hole, Chebychev

We review many basic number theory results. We give efficient algorithms for polynomial evaluation, calculating x^{n}, and finding the greatest common divisor. We briefly review probability theory. After an introduction to Algebraic and Transcendental Numbers, we review Dirichlet’s Box Principle (aka the Pidgeonhole Principle), and give an application. We prove a weak version of Chebyshev’s Theorem on the approximate number of primes. Lecture by Steven J. Miller. Notes by Steven J. Miller (and Florin Spinu, who helped write up the notes on Dirichlet’s Box Principle and Chebyshev).

6.1 Notation

1. 𝕎 : the set of whole numbers: {1, 2, 3, 4, … }.

2. ℕ : the set of natural numbers: {0, 1, 2, 3, … }.

3. ℤ : the set of integers: {…, −2, −1, 0, 1, 2, … }.

4. ℚ: the set of rational numbers: \{x : x = \frac{p}{q}, p, q ∈ ℤ, q ≠ 0\}.

5. ℝ: the set of real numbers.

6. ℂ: the set of complex numbers: {z : z = x + iy, x, y ∈ ℝ}.

7. ℤ/nℤ : the additive group of integers mod n.

8. (ℤ/nℤ)^{∗} : the multiplicative group of invertible elements mod n.

9. a|b : a divides b, i.e. the remainder after integer division \frac{b}{a} is 0.

10. (a, b) : greatest common divisor (gcd) of a and b, often written gcd(a, b).

11. x ≡ y( mod n) : there exists an integer a such that x = y + an.

12. wlog : without loss of generality.

13. s.t.: such that.

14. ∀ : for all.

15. ∃ : there exists.

16. big O notation : A(x) = O(B(x)), read “A(x) is of order B(x)”, means ∃C > 0 such that ∀x, |A(x)| ≤ C B(x).

17. |S| or #S : number of elements in the set S.

18. p : usually a prime number.

19. n : usually an integer.

6.2 Efficient Algorithms

For computational purposes, often having an algorithm to compute a quantity is not enough; we need an algorithm which will compute quickly. Below we study three standard problems, and show how to either rearrange the operations more efficiently, or give a more efficient algorithm than the obvious candidate.

6.2.1 Polynomial Evaluation

Let f(x) = a_{n}x^{n} + a_{n−1}x^{n−1} + ⋯ + a_{1}x + a_{0}. The obvious way to evaluate is to calculate x^{n} and multiply by a_{n} (n multiplications), calculate x^{n−1} and multiply by a_{n−1} (n − 1 multiplications) and add, et cetera. There are n additions and ∑_{k=0}^{n} k multiplications, for a total of n + \frac{n(n+1)}{2} operations. Thus, the standard method leads to O(n^{2}) computations.

Instead, consider the following:

(((a_{n}x + a_{n−1})x + a_{n−2})x + ⋯ + a_{1})x + a_{0}.  (6.1)

For example,

7x^{4} + 4x^{3} − 3x^{2} − 11x + 2 = (((7x + 4)x − 3)x − 11)x + 2.  (6.2)

Evaluating the long way takes 14 steps; cleverly rearranging takes 8 steps.

Exercise 6.2.1. Prove that the second method takes at most 2n steps to evaluate a_{n}x^{n} + ⋯ a_{0}.

6.2.2 Exponentiation

Consider x^{n}. The obvious way to evaluate involves n − 1 multiplications. By writing n in base two, we can evaluate x^{n} in at most 2 log_{2} n steps.

Let k be the largest integer such that 2^{k} ≤ n. Then ∃a_{i} ∈ {0, 1} such that

n = a_{k}2^{k} + a_{k−1}2^{k−1} + ⋯ + a_{1}2 + a_{0}.  (6.3)

It costs k multiplications to evaluate x^{2^{i}}, i ≤ k. How? Consider y_{0} = x^{2^{0}}, y_{1} = y_{0} · y_{0} = x^{2^{0}} · x^{2^{0}} = x^{2^{1}}, y_{2} = y_{1} · y_{1} = x^{2^{2}}, …, y_{k} = y^{k−1} · y^{k−1} = x^{2^{k}}.

Then

x^{n} = x^{a_{k}2^{k}+a_{k−1}2^{k−1}+···+a_{1}2+a_{0}}
= x^{a_{k}2^{k}} · x^{a_{k−1}2^{k−1}} ⋯ x^{a_{1}2} · x^{a_{0}}
= (x^{2^{k}})^{a_{k}} · (x^{2^{k−1}})^{a_{k−1}} ⋯ (x^{2})^{a_{1}} · (x^{1})^{a_{0}}
= y_{k}^{a_{k}} · y_{k−1}^{a_{k−1}} ⋯ y_{1}^{a_{1}} · y_{0}^{a_{0}}.  (6.4)

As each a_{i} ∈ {0, 1}, we have at most k + 1 multiplications above (if a_{i} = 1 we have the term y_{i} in the product, if a_{i} = 0 we don’t).

Thus, it costs k multiplications to evaluate the x^{2^{i}} (i ≤ k), and at most another k multiplications to finish calculating x^{n}. As k ≤ log_{2} n, we see that x^{n} can be determined in at most 2 log_{2} n steps.

Note, however, that we do need more storage space for this method, as we need to store the values y_{i} = x^{2^{i}}, i ≤ log_{2} n.

Exercise 6.2.2. Instead of expanding n in base two, expand n in base three. How many calculations are needed to evaluate x^{n} this way? Why is it preferable to expand in base two rather than any other base?

6.2.3 Euclidean Algorithm

The Euclidean Algorithm is an efficient way to determine the greatest common divisor of x and y, denoted gcd(x, y) or (x, y). Without loss of generality, assume 1 < x < y.

The obvious way to determine gcd(x, y) is to divide x and y by all positive integers up to x. This takes at most 2x steps.

Let [z] denote the greatest integer less than or equal to z. We write

y = \frac{y}{x} · x + r_{1}, 0 ≤ r_{1} < x.  (6.5)

Exercise 6.2.3. Prove that r_{1} ∈ {0, 1, …, x − 1}.

Exercise 6.2.4. Prove gcd(x, y) = gcd(r_{1}, x). Hint: r_{1} = y − \frac{y}{x} · x.

We proceed in this manner until r_{k} equals zero or one. As each execution results in r_{i} < r_{i−1}, we proceed at most x times (although later we prove we need to apply these steps at most 2 log_{2} x times).

x = \frac{x}{r_{1}} · r_{1} + r_{2}, 0 ≤ r_{2} < r_{1}
r_{1} = \frac{r_{1}}{r_{2}} · r_{2} + r_{3}, 0 ≤ r_{3} < r_{2}
r_{2} = \frac{r_{2}}{r_{3}} · r_{3} + r_{4}, 0 ≤ r_{4} < r_{3}
⋮
r_{k−2} = \frac{r_{k−2}}{r_{k−1}} · r_{k−1} + r_{k}, 0 ≤ r_{k} < r_{k−1}.  (6.6)

Exercise 6.2.5. Prove that if r_{k} = 0, then gcd(x, y) = r_{k−1}, while if r_{k} = 1, then gcd(x, y) = 1.

We now analyze how large k can be. The key observation is the following:

Lemma 6.2.6. Consider three adjacent remainders in the expansion: r_{i−1}, r_{i} and r_{i+1} (where y = r_{−1} and x = r_{0}). Then gcd(r_{i}, r_{i−1}) = gcd(r_{i+1}, r_{i}), and r_{i+1} < \frac{r_{i−1}}{2} .

Proof: We have the following relation:

r_{i−1} = \frac{r_{i−1}}{r_{i}} · r_{i} + r_{i+1}, 0 ≤ r_{i+1} < r_{i}.  (6.7)

If r_{i} ≤ \frac{r_{i−1}}{2} , then as r_{i+1} < r_{i}, we immediately conclude that r_{i+1} < r_{i}. If r_{i} > \frac{r_{i−1}}{2} , then we note that

r_{i+1} = r_{i−1} − \frac{r_{i−1}}{r_{i}} · r_{i}.  (6.8)

But \frac{r_{i−1}}{r_{i}} = 1 (easy exercise). Thus r_{i−1} < \frac{r_{i−1}}{2} . □

We count how often we apply Euclid’s Algorithm. Going from (x, y) = (r_{0}, r_{−1}) to (r_{1}, r_{0}) costs one application. Every two applications leads to the first entry in the last pair being at most half of the second entry of the first pair.

Thus, if k is the largest integer such that 2^{k} ≤ x, we see we apply Euclid’s Algorithm at most 1 + 2k ≤ 1 + 2 log_{2} x times. Each application requires one integer division, where the remainder is the input for the next step.

We have proven

Lemma 6.2.7. Euclid’s Algorithm requires at most 1 + 2 log_{2} x divisions to find the greatest common denominator of x and y.

Let us assume that r_{i} = gcd(x, y). Thus, the last equation before Euclid’s Algorithm terminated was

r_{i−2} = \frac{r_{i−2}}{r_{i−1}} · r_{i−1} + r_{i}, 0 ≤ r_{i} < r_{i−1}.  (6.9)

Therefore, we can find integers a_{i−1} and b_{i−2} such that

r_{i} = a_{i−1}r_{i−1} + b_{i−2}r_{i−2}.  (6.10)

Looking at the second to last application of Euclid’s algorithm, we find that there are integers a′_{i−2} and b′_{i−3} such that

r_{i−1} = a′_{i−2}r_{i−2} + b′_{i−3}r_{i−3}.  (6.11)

Substituting for r_{i−1} = r_{i−1}(r_{i−2}, r_{i−3}) in the expansion of r_{i} yields that there are integers a_{i−2} and b_{i−3} such that

r_{i} = a_{i−2}r_{i−2} + b_{i−3}r_{i−3}.  (6.12)

Continuing by induction, and recalling r_{i} = gcd(x, y) yields Lemma 6.2.8. There exist integers a and b such that gcd(x, y) = ax + by. Moreover, Euclid’s Algorithm gives a constructive procedure to find a and b.

Exercise 6.2.9. Find a and b such that a · 244 + b · 313 = gcd(244, 313).

Exercise 6.2.10. Add details to complete an alternate proof of the existence of a and b with ax + by = gcd(x, y):

1. Let d be the smallest positive value attained by ax + by as we vary a, b ∈ ℤ. Such a d exists: consider (a, b) = (1, 0) or (0, 1). Thus, d = ax + by. We now show d = gcd(x, y).

2. gcd(x, y)|d.

3. Let e = Ax + By > 0. Then d|e. Therefore, for any choice of A, B ∈ ℤ, d|(Ax + By).

4. d|x and d|y (consider clever choices of A and B; one choice gives d|x, one gives d|y). Therefore d| gcd(x, y). As we’ve shown gcd(x, y)|d, this completes the proof.

Note this is a non-constructive proof. By minimizing ax + by, we obtain gcd(x, y), but we have no idea how many steps is required. Prove that a solution will be found either among pairs (a, b) with a ∈ {1, …, y − 1} and −b ∈ {1, …, x − 1}, or −a ∈ {1, …, y − 1} and b ∈ {1, …, x − 1}.

6.3 Probabilities of Discrete Events

6.3.1 Introduction

Let Ω = {ω_{1}, ω_{2}, ω_{3}, … } be an at most countable set of events. We call Ω the sample (or outcome) space. We call the elements ω ∈ Ω the events. Let x : Ω → ℝ. That is, for each event ω ∈ Ω, we attach a real number x(ω). We call x a random variable.

Example 6.3.1. Flip a fair coin 3 times. The possible outcomes are Ω = {HHH, HHT, HT H, T HH, HT T, T HT, T T H, T T T }. One possible random variable is x(ω) equals the number of heads in ω. Thus, x(HHT) = 2 and x(T T T) = 0.

Example 6.3.2. Let Ω be the space of all flips of a fair coin where all but the last flip are tails, and the last is a head. Thus, Ω = {H, T H, T T H, T T T H, … }. One possible random variable is x(ω) is the number of tails; another is x(ω) equals the number of the flip which is a head.

We say p(ω) is a probability function on Ω if

1. 0 ≤ p(ω_{i}) ≤ 1 for all ω_{i} ∈ Ω.

2. p(ω) = 0 if ω ∉ Ω.

3. ∑_{i} p(ω_{i}) = 1.

We call p(ω) the probability of event ω.

Often, we have a random variables where x(ω) = ω. In a convenient abuse of notation, we write X for Ω and x for x(ω) and ω. For example, consider two rolls of a fair die. Let X be the result of the first roll, and Y of the second. Then the sample space is X = Y = {1, 2, 3, 4, 5, 6}.

In general, consider X and Y with x_{i} occurring with probability p(x_{i}) and y_{j} occurring with probability q(y_{j}). We analyze the joint probability r(x, y) of observing x and y.

X and Y are independent if ∀x, y, r(x, y) = p(x)q(y). In the example of rolling a fair die twice, r(x, y) = p(x)q(y) = \frac{1}{6} · \frac{1}{6} if x, y ∈ X = Y, and 0 otherwise.

Exercise 6.3.3. Consider again two rolls of a fair die. Now, let X represent the first roll, and Y the sum of the first two rolls. Prove X and Y are not independent.

Events X_{1} through X_{N} are independent if p(x_{1}, …, x_{N}) = p_{1}(x_{1}) ⋯ p(x_{N}).

Exercise 6.3.4. Construct three events such that any two are independent, but all three are not independent. Hint: roll a fair die twice.

6.3.2 Means

If x(ω) = ω, the mean (or expected value) of an event x is defined by

x̄ = ∑_{i} x_{i}p(x_{i}).  (6.13)

More generally, for a sample space Ω with events ω and a random variable x(ω), we have

x̄(ω) = ∑_{i} x(ω_{i})p(ω_{i}).  (6.14)

For example, the mean of one roll of a fair die is 3.5.

Exercise 6.3.5. Let X be the number of tosses of a fair coin needed before getting the first head. Thus, X = {1, 2, … }. Calculate p(x_{i}) and x̄. We could let Ω be the space of all tosses of a fair coin where all but the last toss are tails, and the last toss is a head. Then x(ω) is the number of tosses of ω.

Instead of writing x̄, we often write E[x] or E[X], read as the expected value of x or X. More generally, we would have x̄(ω) and E[x(ω)].

The k^{th} moment of X is the expected value of x^{k}:

E[x^{k}] = ∑_{i} x_{i}^{k}p(x_{i})  (6.15)

or

E[x^{k}(ω)] = ∑_{i} x^{k}(ω_{i})p(ω_{i}).  (6.16)

Lemma 6.3.6 (Additivity of the Means). Let X and Y be two independent events with joint probability r(x, y) = p(x)q(y). Let z = x+y. Then E[z] = E[x+y] = E[x] + E[y].

Proof:

E[x + y] = ∑_{(i,j)}(x_{i} + y_{j})r(x_{i}, y_{j})
= ∑_{i} ∑_{j}(x_{i} + y_{j})p(x_{i})q(y_{j})
= ∑_{i} ∑_{j} x_{i}p(x_{i})q(y_{j}) + ∑_{i} ∑_{j} y_{j}p(x_{i})q(y_{j})
= ∑_{i} x_{i}p(x_{i}) ∑_{j} q(y_{j}) + ∑_{i} p(x_{i}) ∑_{j} y_{j}q(y_{j})
= E[x] · 1 + 1 · E[y] = E[x] + E[y].  (6.17)

The astute reader may notice that some care is needed to interchange the order of summations. If ∑_{i} ∑_{j} |x_{i}y_{j}|r(x_{i}, y_{j}) < ∞, then Fubini’s Theorem is applicable, and we may interchange the summations at will.

We used the two events were independent to go from ∑_{(i,j)} x_{i}r(x_{i}, y_{j}) to ∑_{i} x_{i}p(x_{i}) ∑_{j} q(y_{j}) = E[x]. Lemma 6.3.6 is true even if the two events are not independent.

If the events are not independent, we encounter sums like ∑_{i} ∑_{j} x_{i}r(x_{i}, y_{j}); however, ∑_{j} r(x_{i}, y_{j}) = p(x_{i}). Why? By summing over all possible y, we are asking what is the probability that x = x_{i}; we do not care what y is. Thus, ∑_{i} ∑_{j} x_{i}r(x_{i}, y_{j}) = ∑_{i} x_{i}p(x_{i}) = E[x], and similarly for the other piece.

Exercise 6.3.7. Write out the proof of the generalization of Lemma 6.3.6, where X and Y are not assumed independent.

Given an outcome space X = {x_{1}, x_{2}, … } with probabilities p(x_{i}), let aX be shorthand for the event a times X with outcome space {ax_{1}, ax_{2}, … } and probabilities p_{a}(ax_{i}) = p(x_{i}).

Lemma 6.3.8. Let X_{1} through X_{N} be a finite collection of independent events. Let a_{1} through a_{N} be real constants. Then

E[a_{1}x_{1} + ⋯ + a_{N}x_{N}] = a_{1}E[x_{1}] + ⋯ + a_{N}E[x_{N}].  (6.18)

Lemma 6.3.9. Let X and Y be independent events. Then E[xy] = E[x]E[y].

Exercise 6.3.10. Prove Lemmas 6.3.8 and 6.3.9.

6.3.3 Variances

The variance σ_{x}^{2} (and its square-root, the standard deviation σ_{x}) measure how spread out a probability distribution is. Assume x(ω) = ω. Given an event X with mean x̄, we define the standard deviation σ_{x}^{2} by

σ_{x}^{2} = ∑_{i}(x_{i} − x̄)p(x_{i}).  (6.19)

More generally, given a sample space Ω, events ω, and a random variable x : Ω → ℝ,

σ_{x(ω)}^{2} = ∑_{i} (x(ω_{i}) − x̄(ω))p(ω_{i}).  (6.20)

Exercise 6.3.11. Let X = {0, 25, 50, 75, 100} with probabilities {.2, .2, .2, .2, .2}. Let Y be the same outcome space, but with probabilities {.1, .25, .3, .25, .1}. Calculate the means and the variances of X and Y.

For computing variances, instead of equation 6.19 one often uses Lemma 6.3.12. σ_{x}^{2} = E[x^{2}] − E[x]^{2}.

Proof: Recall x̄ = E[x]. Then

σ_{x}^{2} = ∑_{i} (x_{i} − E[x])^{2}p(x_{i})
= ∑_{i}(x_{i}^{2} − 2x_{i}E[x] + E[x]^{2})p(x_{i})
= ∑_{i} x_{i}^{2}p(x_{i}) − 2E[x] ∑_{i} x_{i}p(x_{i}) + E[x]^{2} ∑_{i} p(x_{i})
= E[x^{2}] − 2E[x]^{2} + E[x]^{2} = E[x^{2}] − E[x]^{2}.  (6.21)

The main result on variances is Lemma 6.3.13 (Variance of a Sum). Let X and Y be two independent events. Then σ_{x+y}^{2} = σ_{x}^{2} + σ_{y}^{2}.

Proof: We constantly use the expected value of a sum of independent events is the sum of expected values (Lemma 6.3.6 and Lemma 6.3.8).

σ_{x+y}^{2} = E[(x + y)^{2}] − E[(x + y)]^{2}
= E[x^{2} + 2xy + y^{2}] − (E[x] + E[y])^{2}
= (E[x^{2}] + 2E[xy] + E[y^{2}]) − (E[x]^{2} + 2E[x]E[y] + E[y]^{2})
= (E[x^{2}] − E[x]^{2}) + (E[y^{2}] − E[y]^{2}) + 2(E[xy] − E[x]E[y])
= σ_{x}^{2} + σ_{y}^{2} + 2(E[xy] − E[x]E[y]).  (6.22)

By Lemma 6.3.9, E[xy] = E[x]E[y], completing the proof. Lemma 6.3.14. Consider n independent copies of the same event (for example, n flips of a coin or n rolls of a die). Then σ_{nx} = \sqrt{n}σ_{x}.

Exercise 6.3.15. Prove Lemma 6.3.14.

Note that, if the event X has units of meters, then the variance σ_{x}^{2} has units meters-squared, and the standard deviation σ_{x} and the mean x̄ have units meters. Thus, it is the standard deviation that gives a good measure of the deviations of an event around the mean.

There are, of course, alternate measures one can use. For example, one could consider

∑_{i}(x_{i} − x̄)p(x_{i}).  (6.23)

Unfortunately, this is a signed quantity, and large positive deviations can cancel with large negatives. This leads us to consider

∑_{i} |x_{i} − x̄|p(x_{i}).  (6.24)

While this has the advantage of avoiding cancellation of errors (as well as having the same units as the events), the absolute value function is not a good function analytically. For example, it is not differentiable. This is primarily why we consider the standard deviation (the square-root of the variance).

Exercise 6.3.16. Consider the following set of data: for i ∈ {1, …, n}, given x_{i} one observes y_{i}. Believing that X and Y are linearly related, find the best fit straight line. Namely, determine constants a and b that minimize the error (calculated via the variance)

∑_{i=1}^{n} (y_{i} − (ax_{i} + b))^{2} = ∑_{i=1}^{n} (Observed_{i} − Predicted_{i})^{2}.  (6.25)

Hint: use Multi-variable Calculus to find linear equations for a and b, and then solve with Linear Algebra.

If instead of measuring total error by the squares of the individual error (for example, using the absolute value), closed form expressions for a and b become significantly harder.

If one requires that a = 0, show that the b leading to least error is b = ȳ = \frac{1}{n} ∑_{i} y_{i}.

6.3.4 Random Walks

Consider the classical problem of a drunk staggering home from a lamp post late at night. We flip a fair coin N times. With probability \frac{1}{2} we get heads (tails). For each head (tail) the drunk staggers one unit to the right (left). How far do we expect the drunk to be?

It is very unlikely the drunk will be very far to the left or right.

Exercise 6.3.17. Let x be +1 if we flip a head, −1 for a tail. For a fair coin, prove E[x] = 0, σ_{x}^{2} = 1, σ_{x} = 1.

Exercise 6.3.18. Let p_{N}(y) be the probability that after N flips of a fair coin, the drunk is y units to the right of the origin (lamp post).

1. Prove p_{N}(y) = p_{N}(−y).

2. Consider N = 2M. Prove p_{2M}(2k) = \binom{2M}{M+k} \frac{1}{2^{2M}}, where \binom{n}{r} = \frac{n!}{r!(n−r)!}

3. Use Stirling’s formula (n! ≈ n^{n}e^{−n}\sqrt{2πn} = \sqrt{2π}n^{n+ \frac{1}{2}} e^{−n}) to approximate p_{N}(y).

Label the coin tosses X_{1} through X_{N}. Let X denote a generic toss of the coin, and Y_{N} be the distance of the drunkard after N tosses. By Lemma 6.3.8, E[y_{N}] = E[x_{1} + ⋯ + x_{N}] = E[x_{1}] + ⋯ + E[x_{N}]. As each E[x_{i}] = E[x] = 0, E[y_{N}] = 0.

Thus, we expect the drunkard to be at the lamp post. How spread out is his expected position? By Lemma 6.3.14,

σ_{y_{N}} = σ_{Nx} = \sqrt{N}σ_{x} = \sqrt{N}.  (6.26)

This means that a typical distance from the origin is \sqrt{N}. This is called a diffusion process and is very common in the real world.

6.3.5 Bernoulli Process

Recall \binom{N}{r} = \frac{N!}{r!(N−r)!} is the number of ways to choose r objects from N objects when order does not matter. Consider n independent repetitions of an event with only two possible outcomes. We typically call one outcome success and the other failure, the event a Bernoulli Trial, and a collection of independent Bernoulli Trials a Bernoulli Process.

In each Bernoulli Trial, let there be probability p of success and q = 1 − p of failure. Often, we represent a success with 1 and a failure with 0.

Exercise 6.3.19. For a Bernoulli Trial, show x̄ = p, σ_{x}^{2} = pq, and σ_{x} = \sqrt{pq}.

Let Y_{N} be the number of successes in N trials. Clearly, the possible values are Y_{N} = {0, 1, ⋯, N}. We analyze p_{N}(k). Rigorously, the sample space Ω is all possible sequences of N trials, and the random variable y_{N} : Ω → ℝ is given by y_{N}(ω) equals the number of successes in ω.

If k ∈ Y_{N}, we need k successes and N − k failures. We don’t care what order we have them (ie, if k = 4 and N = 6 then SSF SSF and F SSSSF both contribute). Each such string of k successes and N − k failures has probability of p^{k} · (1 − p)^{N−k}. There are \binom{N}{k} such strings.

Thus, p_{N}(k) = \binom{N}{k}p^{k} · (1 − p)^{N−k} if k ∈ \{0, 1, ⋯, N\} and 0 otherwise. By clever algebraic manipulations, one can directly evaluate the mean \overline{y_{N}} and

the variance σ_{y_{N}}^{2} ; however, Lemmas 6.3.8 and 6.3.14 allow one to calculate both quantities immediately, once one knows the mean and variance for one occurrence.

Lemma 6.3.20. For a Bernouilli Process with N trials, each having probability p of success, the expected number of successes is \overline{y_{N}} = Np, and the variance is σ_{y_{N}}^{2} = Npq.

Exercise 6.3.21. Prove Lemma 6.3.20.

Consider the following problem: Let Z = {0, 1, 2, … } be the number of trials before the first success. What is z̄ and σ_{z}^{2}?

First, we determine p(k), the probability that the first success occurs after k trials. Clearly this probability is non-zero only for k a positive integer, in which case the string of results must be k − 1 failures followed by 1 success. Therefore,

p(k) = p · (1 − p)^{k−1} if k ∈ {1, 2, … }, and 0 otherwise.  (6.27)

To determine the mean z̄ we must evaluate

z̄ = ∑_{k=1}^{∞} k · p · (1 − p)^{k−1}

= p ∑_{k=1}^{∞} kq^{k−1}, 0 < q = 1 − p < 1.  (6.28)

Consider the geometric series

f(q) = ∑_{k=0}^{∞} q^{k} = \frac{1}{1 − q} .  (6.29)

A careful analysis shows we can differentiate term by term if 0 ≤ q < 1. Then

f′(q) = ∑_{k=0}^{∞} kq^{k−1} = \frac{1}{(1 − q)^{2}} .  (6.30)

Recalling q = 1 − p and substituting yields

z̄ = p ∑_{k=1}^{∞} kq^{k−1}
= \frac{p}{(1 − (1 − p))^{2}} = \frac{1}{p}.  (6.31)

Differentiating under the summation sign is a powerful tool in Probability Theory.

Exercise 6.3.22. Calculate σ_{z}^{2}. Hint: differentiate f(q) twice.

6.3.6 Poisson Distribution

Divide the unit interval into N equal pieces. Consider N independent Bernoulli Trials, one for each sub-interval. If the probability of a success is \frac{λ}{N}, then by Lemma 6.3.20 the expected number of successes is N · \frac{λ}{N} = λ.

We consider the limit as N → ∞. Obviously, we still expect λ successes in each interval, but what is the probability of 3λ successes? How long do we expect to wait between successes?

We call this a Poisson process with parameter λ. For example, look at the midpoints of the N intervals. At each midpoint we have a Bernoulli Trial with probability of success \frac{λ}{N} and failure 1 − \frac{λ}{N}.

We determine the N → ∞ limits. For fixed N, the probability of k successes in a unit interval is

p_{N}(k) = \binom{N}{k}(\frac{λ}{N} )^{k}(1 − \frac{λ}{N} )^{N−k}
= \frac{N!}{k!(N − k)!} \frac{λ^{k}}{N^{k}} (1 − \frac{λ}{N} )^{N−k}
= \frac{N · (N − 1) ⋯ (N − k + 1)}{N · N ⋯ N} \frac{λ^{k}}{k!} (1 − \frac{λ}{N} )^{N}(1 − \frac{λ}{N} )^{−k}
= 1 · (1 − \frac{1}{N} ) ⋯ (1 − \frac{k − 1}{N})\frac{λ^{k}}{k!} (1 − \frac{λ}{N} )^{N}(1 − \frac{λ}{N} )^{−k}.(6.32)

For fixed, finite k, as N → ∞, the first k factors in p_{N}(k) tend to 1, (1 −

\frac{λ}{N} )^{N} → e^{−λ}, and (1 − \frac{λ}{N} )^{−k} → 1.

Thus, we are led to the Poisson Distribution: Given a parameter λ (interpreted as the expected number of occurrences per unit interval), the probability of k occurrences in a unit interval is p(k) = \frac{λ^{k}}{k!} e^{−λ} for k ∈ \{0, 1, 2, … \}.

Exercise 6.3.23. Check that p(k) given above is a probability distribution. Namely, show ∑_{k≥0} p(k) = 1.

Exercise 6.3.24. Show, for the Poisson Distribution, that the mean x̄ = λ and the variance σ_{x}^{2} = λ. Hint: let

f(λ) = ∑_{k=0}^{∞} \frac{λ^{k}}{k!} = e^{λ}.  (6.33)

Differentiate once to determine the mean, twice to determine the variance.

6.3.7 Continuous Poisson Distribution

We calculate a very important quantity related to the Poisson Distribution (with parameter λ), namely, how long does one expect to wait between successes?

We’ve discussed that we expect λ successes per unit interval, and we’ve calculated the probability of k successes per unit interval.

Start counting at 0, and assume the first success is at x. What is p_{S}(x)? As before, we divide each unit interval into N equal pieces, and consider a Bernoulli Trial at the midpoint of each sub-interval, with probability \frac{λ}{N} of success.

We have approximately \frac{x−0}{1/N} = Nx midpoints from 0 to x (with N midpoints per unit interval). Let ⌈y⌉ be the smallest integer greater than or equal to y. Then we have ⌈Nx⌉ midpoints, where the results of the Bernoulli Trials of the first ⌈Nx⌉ − 1 midpoints are all failures and the last is a success.

Thus, the probability of the first success occuring in an interval of length \frac{1}{N} containing x (with N divisions per unit interval) is

p_{N,S}(x) = (1 − \frac{λ}{N} )^{⌈Nx⌉−1} · (\frac{λ}{N} )^{1}.  (6.34)

For N large, the above converges to e^{−λx} \frac{λ}{N}.

We say p(x) is a continuous probability distribution on ℝ if

1. p(x) ≥ 0 for all x ∈ ℝ.

2. ∫_{ℝ} p(x)dx = 1.

3. Probability(a ≤ x ≤ b) = ∫_{a}^{b} p(x)dx.

We call p(x) the probability density function.

Thus, as N → ∞, we see the probability density function p_{S}(x) = λe^{−λx}. In the special case of λ = 1, we get the standard exponential decay, e^{−x}.

For instance, let π(M) be the number of primes that are at most M. The Prime Number Theorem states π(M) = \frac{M}{log M} plus lower order terms.

Thus, the average spacing between primes around M is about log M. We can model the distribution of primes as a Poisson Process, with parameter λ = λ_{M} = \frac{1}{log M}. While possible locations of primes (obviously) is discrete (it must be an integer, and in fact the location of primes aren’t independent), a Poisson model often gives very good heuristics.

We can often renormalize so that λ = 1. This is denoted unit mean spacing. For example, one can show the M^{th} prime p_{M} is about M log M, and spacings between primes around p_{M} is about log M. Then the normalized primes, q_{M} ≈ \frac{p_{M}}{log M} will have unit mean spacing and λ = 1.

6.3.8 Central Limit Theorem

X_{1}, X_{2}, X_{3}, … are an infinite sequence of random variables such that the X_{j} are independent identically distributed random variables (abbreviated i.i.d.r.v.) with E[X_{j}] = X̄_{j} = 0 (can always renormalize by shifting) and variance E[X_{j}^{2}] = 1. Let S_{N} = ∑_{j=1}^{N} X_{j}.

Theorem 6.3.25. Fix −∞ < a ≤ b < ∞. Then as N → ∞,

Prob(\frac{S_{N}}{\sqrt{N}} ∈ [a, b]) → \frac{1}{\sqrt{2π}} ∫_{a}^{b} e^{− \frac{t^{2}}{2}} dt.  (6.35)

The probability function is called the Gaussian or the Normal distribution. This is the universal curve of probability. Note how robust the Central Limit Theorem is: it doesn’t depend on fine properties of the X_{j}.

6.4 Algebraic and Transcendental Numbers

6.4.1 Definitions

A function f : A → B is one-to-one if f(x) = f(y) implies x = y; f is onto if given any b ∈ B, ∃a ∈ A with f(a) = b. f is a bijection if f is a one-to-one and onto function.

We say two sets A and B have the same cardinality (ie, are the same size) if there is a bijection f : A → B. We denote this by |A| = |B|. If A has finitely many elements (say n elements), A is finite and |A| = n < ∞.

Exercise 6.4.1. Show two finite sets have the same cardinality if and only if they have the same number of elements.

Exercise 6.4.2. If f is a bijection from A to B, prove there is a bijection g = f^{−1} from B to A.

A is countable if there is a bijection between A and the integers ℤ. A is at most countable if A is either finite or countable.

Recall a binary relation R is an equivalence relation if

1. Reflexive: R(x, x) is true (x is equivalent to x).

2. Symmetric: R(x, y) true implies R(y, x) is true (if x is equivalent to y then y is equivalent to x).

3. Transitive: R(x, y) and R(y, z) true imply R(x, z) is true (if x is equivalent to y and y is equivalent to z, then x is equivalent to z).

We often denote equivalence by ≡ or =.

Exercise 6.4.3. Let x, y, z ∈ ℤ, and let n ∈ ℤ be given. Define R(x, y) to be true if n|(x − y) and false otherwise. Prove R is an equivalence relation. We denote it by x ≡ y.

Exercise 6.4.4. Let x, y, z be subsets of X (for example, X = ℚ, ℝ, ℂ, ℝ^{n}, et cetera). Define R(x, y) to be true if |x| = |y| (the two sets have the same cardinality), and false otherwise. Prove R is an equivalence relation.

6.4.2 Countable Sets

We show several sets are countable. Consider the set of non-negative integers ℕ. Define f : ℕ → ℤ by f(2n) = n, f(2n + 1) = −n − 1. By inspection, we see f gives the desired bijection.

Consider 𝕎 = {1, 2, 3, … } (the positive integers). Then f : 𝕎 → ℤ defined by f(2n) = n, f(2n + 1) = −n gives the desired bijection.

Thus, we have proved

Lemma 6.4.5. To show a set S is countable, it is sufficient to find a bijection from S to either ℤ, ℕ or 𝕎.

We need the intuitively plausible

Lemma 6.4.6. If A ⊂ B, then |A| ≤ |B|.

We can then prove

Lemma 6.4.7. If f : A → C is a one-to-one function (not necessarily onto), then |A| ≤ |C|. Further, if C ⊂ A, then |A| = |C|.

Exercise 6.4.8. Prove Lemmas 6.4.6 and 6.4.7.

If A and B are sets, the cartesian product A × B is {(a, b) : a ∈ A, b ∈ B}.

Theorem 6.4.9. If A and B are countable, so is A ∪ B and A × B.

Proof: we have bijections f : ℕ → A and g : ℕ → B. Thus, we can label the elements of A and B by

A = {a_{0}, a_{1}, a_{2}, a_{3}, … }
B = {b_{0}, b_{1}, b_{2}, b_{3}, … }.  (6.36)

Assume A∩B is empty. Define h : ℕ → A∪B by h(2n) = a_{n} and h(2n+1) = b_{n−1}. We leave to the reader the case when A ∩ B is not empty.

To prove the second claim, consider the following function h : W → A × B:

h(1) = (a_{0}, b_{0})

h(2) = (a_{1}, b_{0}), h(3) = (a_{1}, b_{1}), h(4) = (a_{0}, b_{1})

h(5) = (a_{2}, b_{0}), h(6) = (a_{2}, b_{1}), h(7) = (a_{2}, b_{2}), h(8) = (a_{1}, b_{2}), h(9) = (a_{0}, b_{2})

⋮

h(n^{2} + 1) = (a_{n}, b_{0}), h(n^{2} + 2) = (a_{n}, b_{n−1}), …,

h(n^{2} + n + 1) = (a_{n}, b_{n}), h(n^{2} + n + 2) = (a_{n−1}, b_{n}), …,
h((n + 1)^{2}) = (a_{0}, b_{n})
⋮  (6.37)

Basically, look at all pairs of integers in the first quadrant (including those on the axes). Thus, we have pairs (a_{x}, b_{y}). The above function h starts at (0, 0), and then moves through the first quadrant, hitting each pair once and only once, by going up and over. Draw the picture! □

Corollary 6.4.10. Let A_{i} be countable ∀i ∈ ℕ. Then for any n, A_{1} ∪ ⋯ ∪ A_{n} and A_{1} × ⋯ × A_{n} are countable, where the last set is all n-tuples (a_{1}, …, a_{n}), a_{i} ∈ A_{i}. Further, ∪_{i=0}^{∞}A_{i} is countable. If each A_{i} is at most countable, then ∪_{i=0}^{∞}A_{i} is at most countable.

Exercise 6.4.11. Prove Corollary 6.4.10. Hint: for ∪_{i=0}^{∞}A_{i}, mimic the proof used to show A × B is countable.

As the natural numbers, integers and rationals are countable, by taking each A_{i} = ℕ, ℤ or ℚ we immediately obtain

Corollary 6.4.12. ℕ^{n}, ℤ^{n} and ℚ^{n} are countable. Hint: proceed by induction. For example write ℚ^{n+1} as ℚ^{n} × ℚ.

Exercise 6.4.13. Prove there are countably many rationals in the interval [0, 1].

6.4.3 Algebraic Numbers

Consider a polynomial f(x) = 0 with rational coefficients. By multiplying by the least common multiple of the denominators, we can clear the fractions. Thus, without loss of generality it is sufficient to consider polynomials with integer coefficients.

The algebraic numbers, A, are the set of all x ∈ ℂ such that there is a polynomial of finite degree and integer coefficients (depending on x, of course!) such that f(x) = 0. The remaining complex numbers are the transcendentals.

The algebraic numbers of degree n, A_{n}, are the set of all x ∈ A such that

1. there exists a polynomial with integer coefficients of degree n such that f(x) = 0

2. there is no polynomial g with integer coefficients and degree less than n with g(x) = 0.

Thus, A_{n} is the subset of algebraic numbers x where for each x ∈ A_{n}, the degree of the smallest polynomial f with integer coefficients and f(x) = 0 is n.

Exercise 6.4.14. Show the following are algebraic: any rational, the square-root of any rational, the cube-root of any rational, r^{\frac{p}{q}} where r, p, q ∈ ℚ, i = \sqrt{−1}, \sqrt{3\sqrt{2} − 5}.

Theorem 6.4.15. The Algebraic Numbers are countable.

Proof: If we show each A_{n} is at most countable, then as A = ∪_{n=1}^{∞}A_{n}, by Corollary 6.4.10 A is at most countable.

Recall the Fundamental Theorem of Algebra (FTA): Let f(x) be a polynomial of degree n with complex coefficients. Then f(x) has n (not necessarily distinct) roots. Of course, we will only need a weaker version, namely that the Fundamental Theorem of Algebra holds for polynomials with integer coefficients.

Fix an n ∈ ℕ. We now show A_{n} is at most countable. We can represent every integral polynomial f(x) = a_{n}x^{n} + ⋯ + a_{0} by an (n + 1)-tuple (a_{0}, …, a_{n}). By Corollary 6.4.12, the set of all (n + 1)-tuples with integer coefficients (ℤ^{n+1}) is countable. Thus, there is a bijection from ℕ to ℤ^{n+1}, and we can index each (n + 1)-tuple a ∈ ℤ^{n+1}:

{a : a ∈ ℤ^{n+1}} = ⋃_{i=1}^{∞}{α_{i}},  (6.38)

where each α_{i} ∈ ℤ^{n+1}.

For each tuple α_{i} (or a ∈ ℤ^{n+1}), there are n roots. Let R_{α_{i}} be the roots of the integer polynomial associated to α_{i}. The roots in R_{α_{i}} need not be distinct, and the roots may solve an integer polynomial of smaller degree. For example, f(x) = (x^{2} − 1)^{4} is a degree 8 polynomial. It has two roots, x = 1 with multiplicity 4 and x = −1 with multiplicity 4, and each root is a root of a degree 1 polynomial.

Let R_{n} = {x ∈ ℂ : x is a root of a degree n polynomial}. One can show that

R_{n} = ⋃_{i=1}^{∞} R_{α_{i}} ⊃ A_{n}.  (6.39)

By Lemma 6.4.10, R_{n} is countable. Thus, by Lemma 6.4.6, as R_{n} is at most countable, A_{n} is at most countable.

Therefore, each A_{n} is at most countable, so by Corollary 6.4.10 A is at most countable. As A_{1} ⊃ ℚ (given \frac{p}{q} ∈ ℚ, consider qx − p = 0), A_{1} is at least countable. As we’ve shown A_{1} is at most countable, this implies A_{1} is countable. Thus, A is countable. □

6.4.4 Transcendental Numbers

A set is uncountable if there is no bijection between it and the rationals (or the integers, or any countable set).

Theorem 6.4.16. The set of irrationals in [0, 1] is uncountable.

Proof: Let I = [0, 1] − ℚ = {x : 0 ≤ x ≤ 1 and x ∉ ℚ}. Assume that I is countable (the case where I is finite is even easier).

We can write every number in I in a base two expansion, say y = .y_{1}y_{2}y_{3}y_{4} ⋯, y_{i} ∈ {0, 1}, y = ∑_{i} y_{i}2^{−1}. Certain numbers can be written two different ways. For example, 0.010011111111111 ⋯ = .0101. As we are assuming I is countable, including both representations of these numbers is equivalent to taking the union of two countable sets, which by Theorem 6.4.9 is countable.

Further, we can add back all the rationals in [0, 1], as there are countably many rationals in [0, 1]. Call this set S (the union of the irrationals, the alternate representation of some of the irrationals, and the rationals). As X is contained in the union of three at most countable sets (and two are countable), X is countable by Theorem 6.4.9.

There is therefore a bijection between ℕ and X. We can enumerate the elements by {x_{1}, x_{2}, x_{3}, … }.

For each x_{i}, let .x_{i1}x_{i2}x_{i3} ⋯ x_{ii} ⋯ be its binary expansion. We list the countable members of X:

x_{1} = x_{11}x_{12}x_{13}x_{14} ⋯
x_{2} = x_{21}x_{22}x_{23}x_{24} ⋯
x_{3} = x_{31}x_{32}x_{33}x_{34} ⋯
⋮
x_{n} = x_{n1}x_{n2}x_{n3}x_{n4} ⋯ x_{nn} ⋯
⋮  (6.40)

We construct a real number x ∈ [0, 1] not in X. As this was supposed to be (more than a) complete list of all reals in [0, 1], this will contradict the assumption that I is countable.

Consider the number z = .z_{1}z_{2}z_{3} ⋯ z_{n} ⋯ defined by z_{n} = 1 − x_{nn}. Can z be one of the numbers in our list? For example, could z = x_{m}?

No, as they differ in the m^{th} digit. Thus, z is not on our list, violating the assumption that we had a complete enumeration. Note we had to be careful and make sure we included all equivalent ways of writing the same number. Thus, while z disagrees with the base two expansion of x_{m}, it cannot be an equivalent way of representing x_{m}, as all equivalent ways of representing x_{m} are in our list. This is merely an annoying technical detail.

Thus, the set of irrationals in [0, 1] is not countable. □.

The above proof is due to Cantor (1873 − 1874), and is known as Cantor’s Diagonalization Argument. Note Cantor’s proof shows that most numbers are transcendental, though it doesn’t tell us which numbers are transcendental. We can easily show many numbers (such as \sqrt{3 + 2^{\frac{3}{5}} \sqrt{7}}) are algebraic. What of other numbers, such as π and e?

Lambert (1761), Legendre (1794), Hermite (1873) and others proved π irrational; Legendre (1794) also proved π irrational. In 1882 Lindemann proved π transcental.

What about e? Euler (1737) proved that e and e^{2} are irrational, Liouville (1844) proved e is not an algebraic number of degree 2, and Hermite (1873) proved e is transcendental.

Liouville (1851) showed transcendental numbers exist; we will discuss his construction later.

6.5 Introduction to Number Theory

6.5.1 Dirichlet’s Box Principle

Definition 6.5.1 (Dirichlet’s Box Principle / Pidgeon Hole Principle). Consider n boxes, and place n + 1 objects in the n boxes. Then some box contains at least two objects.

We will use Dirichlet’s Box Principle to find good rational approximations to irrational numbers.

Approximation by Rationals

Let α ∈ ℝ − ℚ be an irrational number. We are looking for a rational number \frac{p}{q}

such that |α − \frac{p}{q} | is small, so that \frac{p}{q} is a good rational approximation to α.

Lemma 6.5.2. Let α ∈ ℝ − ℚ. Then there exist p, q ∈ ℤ, q ≠ 0 such that:

|α − \frac{p}{q}| ≤ \frac{1}{q}  (6.1)

Proof. It is enough to prove this for α ∈ (0, 1). Let q ≥ 1 and divide the interval [0, 1) into q intervals [\frac{p}{q}, \frac{p+1}{q}) of length \frac{1}{q}. Then α belongs to one of these intervals. For some 0 < p < q we then have:

α ∈ [\frac{p}{q}, \frac{p + 1}{q}) ⇒ |α − \frac{p}{q}| ≤ \frac{1}{q}.  (6.2)

To obtain a better approximation, we start with an irrational number α ∈ (0, 1) and an integer parameter Q > 1. As before, divide the interval (0, 1) into Q equal pieces (\frac{a}{Q}, \frac{a+1}{Q}). Consider the Q + 1 numbers inside the interval (0, 1):

{α}, {2α}, ..., {(Q + 1)α},  (6.3)

where {x} denotes the fractional part of x. Letting [x] denote the greatest integer less than or equal to x, we have x = [x] + {x}.

By Dirichlet’s Box Principle, at least two of these numbers, say {q_{1}α} and \{q_{2}α\}, belong to a common interval of length \frac{1}{Q}. Without loss of generality, we may take 1 ≤ q_{1} < q_{2} ≤ Q + 1.

Hence

|\{q_{2}α\} − \{q_{1}α\}| ≤ \frac{1}{Q}  (6.4)

and

|(q_{2}α − n_{2}) − (q_{1}α − n_{1})| ≤ \frac{1}{Q}, n_{i} = [q_{i}α].  (6.5)

Now let q = q_{1} − q_{2}, 1 ≤ q ≤ Q and p = n_{1} − n_{2} ∈ ℤ. Then

|qα − p| ≤ \frac{1}{Q}  (6.6)

and hence

|α − \frac{p}{q}| ≤ \frac{1}{qQ} ≤ \frac{1}{q^{2}} .  (6.7)

We have proven

Theorem 6.5.3. Given α ∈ ℝ, there exist p, q ∈ ℤ, q ≠ 0, such that

|α − \frac{p}{q}| < \frac{1}{q^{2}} .  (6.8)

6.5.2 Counting the Number of Primes

Euclid

Lemma 6.5.4 (Euclid). There are infinitely many primes.

Proof by contradiction: Assume there are only finitely many primes, say p_{1}, p_{2}, …, p_{n}. Consider

x = p_{1}p_{2}..p_{n} + 1.  (6.9)

x cannot be prime, as we are assuming p_{1} through p_{n} is a complete list of primes. Thus, x is composite, and divisible by a prime. However, p_{i} cannot divide x, as it gives a remainder of 1. Thus, x would have to be divisible by some prime not in our list, again contradicting the assumption that p_{1} through p_{n} is a complete enumeration of the primes. □

Exercise 6.5.5. Try, using Euclid’s argument, to find an explicit lower bound (as weak as you like) to the function:

π(X) = #{p : p is prime and p ≤ X}.  (6.10)

Dirichlet’s Theorem

Theorem 6.5.6 (Primes in Arithmetic Progressions). Let a and b be relatively prime integers. Then there are infinitely many primes in the progression an + b. Further, for a fixed a, to first order all relatively prime b give progressions having the same number of primes.

Notice that the condition (a, b) = 1 is necessary. If gcd(a, b) > 1, an + b can never be prime. Dirichlet’s remarkable result is that this condition is also sufficient.

Exercise 6.5.7. Dirichlet’s theorem is not easy to prove, but try to prove it in the particular case a = 4, b = −1, i.e. for the arithmetic progression 4n − 1, using an argument similar to Euclid’s. Proving there are infinitely many primes of the form 4n + 1 is a lot harder.

Prime Number Theorem

Theorem 6.5.8. (Prime Number Theorem or PNT) As X → ∞,

π(X) ∼ \frac{X}{log X}  (6.11)

The Prime Number Theorem was proved in 1896 by Jacques Hadamard and Charles Jean Gustave Nicolas Baron de la Vallee Poussin. Of course, we need to quantify what π(X) ∼ \frac{X}{log X} means. Basically, there is an error function E(X) such that |π(X) − \frac{X}{log X} | ≤ E(X), and E(X) grows slower than \frac{X}{log X}.

A weaker version was proved by Pafnuty Chebyshev (around 1850).

Theorem 6.5.9 (Chebyshev). There exist explicit positive constants A and B such that, for n > 30:

\frac{AX}{log X} ≤ \frac{π(X)}{X} ≤ \frac{BX}{log X} .  (6.12)

Chebyshev showed one can take A = log (\frac{2^{\frac{1}{2}} 3^{\frac{1}{3}} 4^{\frac{1}{4}}}{30^{\frac{1}{30}}} ) ≈ .921 and B = \frac{6A}{5} ≈ 1.105, which are indeed very close to 1. To highlight the method, we will use cruder arguments and prove the theorem for a smaller A and a larger B.

Chebyshev’s argument uses an identity using von Mangoldt’s Lambda function Λ(n), where Λ(n) = log p if m = p^{k} for some prime p, and 0 otherwise.

Define the function

T (X) = ∑_{1≤n≤X} Λ(n)[\frac{X}{n} ] = ∑_{n≥1} Λ(n)[\frac{X}{n} ].  (6.13)

Exercise 6.5.10. Show that T (X) = ∑_{n≤X} log n.

Now, it is easy to see (compare upper and lower sums) that

∑_{n≤X} log n = ∫_{1}^{X} log t dt + O(log X) = X log X − X + O(log X), (6.14)

giving a good approximation to the function T (X). The trick is to look at

T (X) − 2T (\frac{X}{2} ) = ∑_{n} Λ(n)([\frac{X}{n} ] − 2[\frac{X}{2n}])  (6.15)

By the previous remarks, the LHS = X log 2 + O(log X). Also,

RHS ≤ ∑_{p≤X}(log p)\frac{log X}{log p} = π(X) log X.  (6.16)

Hence we immediately obtain the lower bound:

π(X) ≥ \frac{X log 2}{log X} + O(log X)  (6.17)

Exercise 6.5.11. Prove the bound in Equation 6.16.

To obtain an upper bound for π(X), we notice that, since [2α] ≥ 2[α], the sum in equation (6.15) has only positive terms. By dropping terms we get a lower bound.

T (X) − 2T (\frac{X}{2} ) ≥ ∑_{X/2<n≤X} Λ(n)([\frac{X}{n} ] − 2[\frac{X}{2n}])
≥ ∑_{X/2<p≤X} log p

≥ log (\frac{X}{2} ) ∑_{X/2<p≤X} 1

= log (\frac{X}{2} )(π(X) − π(\frac{X}{2} ))  (6.18)

Hence we obtain an upper bound for the number of primes between \frac{X}{2} and X:

π(X) − π(X/2) ≤ \frac{X log 2}{log(\frac{X}{2} )} + O(1)  (6.19)

Now, if we write inequality (6.19) for X, \frac{X}{2} , \frac{X}{2^{2}} , … we get

π(X) − π(X/2) ≤ 2\frac{X/2}{log(X/2)}
π(X/2) − π(X/2^{2}) ≤ 2\frac{X/2^{2}}{log(X/2^{2})}
⋮
π(X/2^{k−1}) − π(X/2^{k}) ≤ 2\frac{X/2^{k}}{log(X/2^{k})}  (6.20)

as long as \frac{X}{2^{k}} ≥ 1, i.e. k ≤ [log_{2} X] = k_{0}. Summing the above inequalities we get on the left hand side a telescoping sum. All the terms cancel, except for the leading term π(X) and π(X/2^{k_{0}}) = 0.

Thus

π(X) ≤ 2 ∑_{k=1}^{k_{0}} \frac{X/2^{k}}{log(X/2^{k})}  (6.21)

To evaluate the sum in the above inequality we split it into two parts, k "small" and k "large". More precisely, let n_{0} = log_{2}(X^{1/10}) so that 2^{n_{0}} = X^{1/10} and note that:

2 ∑_{k>n_{0}} \frac{X/2^{k}}{log(X/2^{k})} ≤ 2 ∑_{k>n_{0}} \frac{X}{2^{k}} ≤ \frac{2X}{2^{n_{0}}} = \frac{2X}{X^{1/10}} = 2X^{9/10}.  (6.22)

Hence the contribution from k "large" is very small compared to what we expect (i.e. order of magnitude \frac{X}{log X}), or we can say that the main term comes from the sum over k small.

We now evaluate the contribution from small k.

2 ∑_{k=1}^{n_{0}} \frac{X}{2^{k}} \frac{1}{log(X/2^{k})} ≤ \frac{2X}{log(X/2^{n_{0}})} ∑_{k=1}^{n_{0}} \frac{1}{2^{n_{0}}} ≤ \frac{2X}{log(X^{9/10})} = \frac{20}{9} \frac{X}{log X} (6.23)

Hence the right hand side of the equation (6.21) is made up of two parts, a main term of size \frac{BX}{log X} coming from equation (6.23), and a lower order term coming from equation (6.22).

For X sufficiently large,

π(X) ≤ \frac{BX}{log X}  (6.24)

where B can be any constant strictly bigger than \frac{20}{9}.

To obtain Chebyshev’s better constant we would have to work a little harder along these lines, but it is the same method.

Gathering equations (6.17) and (6.24) we see we have proven

\frac{AX}{log X} ≤ π(X) ≤ \frac{BX}{log X} .  (6.25)

While this is not an asymptotic for π(X), it does give the right order of magnitude for π(X), namely \frac{X}{log X}.

Exercise 6.5.12. Using Chebyshev’s Theorem, Prove Bertrand’s Postulate: for any integer n ≥ 1, there is always a prime number between n and 2n.

More of an Introduction to Graph Theory

We review some basic definitions of Graph Theory, and prove a simple result about the size of the eigenvalues of adjacency graphs. Lecture by Peter Sarnak; notes by Steven J. Miller.

7.1 Definitions

Definition 7.1.1. For a graph G, let V (G) be the set of vertices and E(G) the set of edges (an edge is a pair (v, w) with v, w ∈ V). We often just write V and E.

Definition 7.1.2 (Connected Graph). A graph G is connected if for any two vertices v, w ∈ G, there is a path of edges in E starting at v and ending at w.

Definition 7.1.3 (Boundary). ∂A = {v ∈ V − A : there is a w ∈ A with (v, w) ∈ E}.

Definition 7.1.4 (k-regular). A graph G is k-regular if there are k edges coming out from each vertex.

Definition 7.1.5 (Expander Graph). Let G be a connected graph with n vertices. Let A ⊂ V (G) be any subset of vertices with |A| ≤ \frac{n}{2} (ie, at most half of the vertices). We say G is a (n, c, k) expander if for any such A, |∂A| ≥ c|A|.

Example 7.1.6. Let G be a 2-regular graph with n vertices. For definateness, label the vertices {1, 2, …, n}. Let the edges be (1, 2), (2, 3), (3, 4), …, (n, 1).

Let A be the first half of the vertices: A = \{1, 2, …, [\frac{n}{2}]\}. Then |∂A| = 2, and G is not an (n, c, 2) expander.

7.2 Size of Eigenvalues

Consider a 3-regular graph with n vertices. We want the graph to have certain desirable connectivity properties.

Let A = (a_{vw}) is the adjacency matrix attached to the graph G, and a_{vw} = 1 if there is an edge from v to w and 0 otherwise. This is a very sparse matrix (only three non-zero entries in each row or column). Compute its eigenvalues (real numbers).

Lemma 7.2.1. The eigenvalues λ_{i} ∈ [−3, 3].

Proof: Let f : V → ℝ, define the action of the adjacency matrix A on f by

Af(v) = ∑_{(v,w)∈E} f(w).  (7.1)

As there are finitely many vertices (n, in fact), we can regard the function f(v) as living in ℝ^{n}, with coordinates (f(v_{1}), …, f(v_{n})).

Suppose ∀w ∈ V, Af(w) = λf(w), f ≠ 0. Then λ is an eigenvalue (which must be real as A is real symmetric).

We use the Maximum Modulus Principle. As f ≠ 0, let w_{0} be such that f(w_{0}) is the maximum value of f(w) (not zero, and exists as we have finitely many vertices). Then

f(w_{0}) = \frac{1}{λ} ∑_{(w,w_{0})} f(w).  (7.2)

If λ > 3, this cannot happen (we’re assuming the graph is k-regular and connected). If λ = 3, then f is constant. Working with absolute values, we similarly obtain λ > −3.

Exercise 7.2.2. Prove a k-regular graph G is connected if and only if λ = k is a simple eigenvalue.

Let λ_{1} be the second largest eigenvalue; λ = k is always an eigenvalue for a k-regular connected graph. How big can the gap be between λ_{1} and k?

Linear Algebra Review, especially Spectral Theorem for Real Symmetric Matrice

We review some basic facts about Linear Algebra and Matrix Groups, and give an introduction to Random Matrix Theory. Lecture by Steven J. Miller; notes by Steven J. Miller and Alex Barnett.

8.1 Linear Algebra Review

Matrices can either be thought of as rectangular (often square) arrays of numbers, or as linear transformations from one space to another (or possible to the same space). The former picture is the simplest starting point, but as Professor Sarnak emphasized, it is the latter, geometric view that gives a deeper understanding.

To connect with the simpler vector case, a vector can be thought of as a list of real numbers which change in a certain way when the coordinate system changes, or as a geometric object with length and direction. The latter object is coordinateindependent, and has different representations in different choices of coordinate axes. Try to keep the geometric picture in mind for matrices.

8.1.1 Definitions

Given an n × m matrix A (where n is the number of rows and m is the number of columns), the transpose of A, denoted A^{T}, is the m × n matrix where the rows of A^{T} are the columns of A (or, equivalently, the columns of A^{T} are the rows of A).

Lemma 8.1.1. (AB)^{T} = B^{T} A^{T} and (A^{T})^{T} = A.

We leave the proof to the reader.

If an n × n matrix (also called a square matrix) A satisfies A^{T} = A, then we say A is symmetric

Example 8.1.2. Let A be the matrix

( \begin{matrix}2 & 2 & 4 & 2 \\ 1 & 1 & −2 & 2 \\ −2 & 0 & 0 & 1 \\ 1 & 1 & 2 & 1 \end{matrix} )  (8.1)

Then A^{T} is

( \begin{matrix}2 & 1 & −2 & 1 \\ 2 & 1 & 0 & 1 \\ 4 & −2 & 0 & 2 \\ 2 & 2 & 1 & 1 \end{matrix} )  (8.2)

Note the above matrix is not symmetric.

The number of degrees of freedom in a symmetric matrix (i.e. independent real numbers needed to completely specify the matrix) is n(n + 1)/2. Why? There are n^{2} entries, n on the diagonal. If you specify all entries above the diagonal and all entries on the diagonal, then you know the symmetric matrix.

There are n^{2} − n non-diagonal entries (half above the diagonal, half below). Thus, one needs to specify \frac{n^{2}−n}{2} + n = \frac{n^{2}+n}{2} entries.

Exercise 8.1.3. If A and B are symmetric, show AB is symmetric.

Matrix multiplication. We call the element in the i^{th} row and j^{th} column a_{ij}. Think of i = 1 ⋯ n going down the left side, and j = 1 ⋯ M going across the top. A vector v we represent as a column of elements with the i^{th} being v_{i}. A nice way to see matrix-vector multiplication is that the v_{i} give the coefficients by which the columns of A are linearly mixed together. For the product w = Av to make sense, the length (dimension) of v must equal m, and the dimension of w will be n. A is therefore a linear transform from m-dim space to n-dim space.

Multiple transformations appear written backwards: if we apply A then B then C to a vector, we write

w = CBAv.  (8.3)

Note that taking the product of two n × n matrices requires O(n^{3}) effort.

Exercise 8.1.4. Show that there are two ways to evaluate triple matrix products of the type CBA. The slow way involves O(n^{4}) effort. How about the fast way? How do these results scale for the case of a product of k matrices?

Definition 8.1.5 (Invertible Martices). A is invertible if a matrix B can be found such that BA = AB = I. The inverse is then written B = A^{−1}. Invertibility requires A to be square.

Transformations of a matrix. Just as with vectors, we can find out how the components of a square matrix A change under transformation. Say we have a scalar quantity x = w^{T} Av. We transform our coordinate system linearly such that the vector v has components v′ = Mv, where M is some invertible matrix representing the transformation. Therefore also w′ = Mw. The only way that x can remain unchanged by the transformation (as any scalar must), for all choices of v and w, is if the transformed matrix is written A′ = M^{−T} AM^{−1}. Check this via

x′ = w^{′T} A′v′ = (Mw)^{T} (M^{−T} AM^{−1})(Mv) = w^{T} IAIv = w^{T} Av = x. (8.4)

This is called a similarity transformation, or a conjugation. Really we have one object, the transformation A, but it may have different representations by a matrix of numbers, depending on the choice of basis.

Definition 8.1.6 (Orthogonal Matrices). Q is an orthogonal n × n matrix if it has real entries and Q^{T} Q = QQ^{T} = I.

Q is invertible, with inverse Q^{T}. The geometric meaning of Q is a rotation: the vector w = Qv is just v rotated (about the origin).

The number of degrees of freedom in an orthogonal matrix is n(n − 1)/2.

Exercise 8.1.7. In 3 dimensions a general rotation involves 3 angles (for example, azimuth, elevation, and ‘roll’). How many angles are needed in 4 dimensions? In 3d you rotate about a line-like axis (the set of points which do not move under rotation); what object do you rotate about in 4d?

Exercise 8.1.8. Show that the identity matrix I, always has representation I_{ij} = δ_{ij} regardless of the choice of basis. Hint: perform orthogonal tranformation on the matrix δ_{ij}.

The set of orthogonal matrices of order n forms a continuous (or topological) group, which we call O(n). (Not to be confused with “of order N”). Group properties:

• Associativity follows from that of matrix multiplication.

• The identity matrix acts as an identity element, since it is in the group.

• Inverse is the transpose (see above): Q^{−1} = Q^{T}.

• Closure is satisfied because any product QR of orthogonal matrices is itself orthogonal.

Exercise 8.1.9. Prove the last assertion.

However, not all the elements of O(n) can ‘talk’ to each other, i.e. you cannot reach all the elements by continuous transformation from the identity I.

Example for n = 2: a general order-2 orthogonal matrix can be written

( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) or ( \begin{matrix}cos θ & − sin θ \\ − sin θ & − cos θ \end{matrix} ) ,  (8.5)

where 0 ≤ θ < 2π is a real angle. The first has determinant +1 and defines the ‘special’ (i.e. unit determinant) group SO(2) which is a subgroup of O(2) with identity I. The second has determinant −1 and corresponds to rotations with a reflection; this subgroup is disjoint from SO(2), and has the weird (reflecting) identity can be written in some basis as

( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) .  (8.6)

Note that SO(2), alternatively written as the family of planar rotations R(θ), is isomorphic to the unit length complex numbers under the multiplication operation:

R(θ) ⟷ e^{iθ}.  (8.7)

Therefore we have R(θ_{1})R(θ_{2}) = R(θ_{1} + θ_{2}). This commutativity relation does not hold in higher n > 2.

Orthogonal transformations. If an orthogonal matrix Q is used for conjugation of a general square matrix A, then the rule Eq. 8.4 for transformation becomes,

A′ = QAQ^{T} .  (8.8)

This tells you how to ‘rotate’ a (square) matrix.

Definition 8.1.10 (Complex Conjugate Transpose). Let A be an n × m matrix. Then the complex conjugate transpose of A, denoted A^{∗}, is obtained by the following: (1) take the complex conjugate of A; ie, replace every entry a_{jk} = x_{jk} + iy_{jk} with \overline{a_{jk}} = x_{jk} − iy_{jk}, and call this matrix A_{1}; (2) take the transpose of A_{1}.

Exercise 8.1.11. Prove that (AB)^{∗} = B^{∗}A^{∗}.

Definition 8.1.12 (Dot or Inner Product). The dot (or inner) product of two real vectors v and w is defined as v^{T} w; if the vectors are complex, we instead use v^{∗}w.

Exercise 8.1.13. Show that the dot product is invariant under orthogonal transformation. That is, show that given two vectors, transforming them using the same orthogonal matrix leaves their dot product unchanged.

Definition 8.1.14 (Length of a vector). The length of a real vector v is |v|^{2} = v^{T} v; for a complex vector, we have |v|^{2} = v^{∗}v.

Definition 8.1.15 (Orthogonality). Two real vectors are orthogonal (also called perpendicular) if v^{T} w = 0; for two complex vectors, the equivalent condition is v^{∗}w = 0.

Definition 8.1.16 (Eigenvalue, Eigenvector). Recall λ is an eigenvalue and v is an eigenvector if Av = λv and v is not the zero vector.

Exercise 8.1.17. If v is an eigenvector of A with eigenvalue λ, show w = av, a ∈ ℂ, is also an eigenvector of A with eigenvalue λ.

Exercise 8.1.18. Show that given an eigenvalue λ and an eigenvector v, you can always find an eigenvector w with the same eigenvalue, but |w| = 1.

To find the eigenvalues, we solve the equation det(λI − A) = 0. This gives a polynomial p(λ) = 0. We call p(λ) the characteristic polynomial.

The trace of a matrix A, denote Tr(A) is the sum of the diagonal entries of A:

Tr(A) = ∑_{i=1}^{n} a_{ii}.  (8.9)

Lemma 8.1.19. Tr(A) = ∑_{i=1}^{n} λ_{i}.

The proof relies on writing out the characteristic equation and comparing powers of λ with the factorized version. By the fact that the polynomial has roots λ_{i} we can write

det(λI − A) = p(λ) = ∏_{i=1}^{n}(λ − λ_{i}).  (8.10)

Note the coefficient of λ^{n} is 1, thus we have ∏_{i}(λ − λ_{i}) and not c ∏_{i}(λ − λ_{i}) for some constant c.

By expanding out the RHS, the coefficient of λ^{n−1} is − ∑_{i=1}^{n} λ_{i}, which we will show is −Tr(A). Expanding the LHS, we want to find the corresponding coefficient in

| \begin{matrix}λ − a_{11} & −a_{12} & ⋯ & −a_{1n} \\ −a_{21} & λ − a_{22} \\ ⋮ & & ... \\ −a_{n1} & & & λ − a_{nn} \end{matrix} | .

We have to remember the expansion of the determinant. Taking the top-left-most 2 × 2 block, we see its determinant is (λ − a_{11})(λ − a_{22}) − a_{12}a_{21} = λ^{2} − (a_{11} + a_{22})λ+(a_{11}a_{22} −a_{12}a_{21}). The determinant of the top-left-most 3×3 block is then formed by (λ − a_{33}) times the above 2 × 2 determinant, plus two other multiples of determinants which can give only a highest power of λ of λ^{1}. Thus we see that the coefficient in λ^{2} is −(a_{11} +a_{22} +a_{33}). Repeating this argument for 4×4 block up to n × n gives us the coefficient of λ^{n−1} in the full determinant is − ∑_{i=1}^{n} a_{ii}. Since the LHS and RHS must be equal ∀λ, the LHS and RHS coefficients in λ^{n−1} are equal. □

Corollary 8.1.20. Tr(A) is invariant under rotation of basis.

The proof follows immediately from the invariance of the eigenvalues under rotation of basis. We need the following:

Lemma 8.1.21. det(AB) = det(A) det(B). Further, by induction one can show det(AB ⋯ Z) = det(A) det(B) ⋯ det(Z). Further, det(I) = 1.

Proof of Corollary: Let A = Q^{T} BQ. We show A and B have the same trace by showing A and B have the same eigenvalues. To find the eigenvalues of A we must solve:

det(λI − A) = det(λI − Q^{T} BQ)
= det(λQ^{T} Q − Q^{T} BQ)
= det(Q^{T} λIQ − Q^{T} BQ)
= det (Q^{T} (λI − B)Q)
= det(Q^{T}) det(λI − B) det(Q)
= det(Q^{T}) det(Q) det(λI − B)
= det(Q^{T} Q) det(λI − B) = det(I) det(λI − B) = det(λI − B).

(8.11)

As the eigenvalues of A and B satisfy the same equation, they are equal. □

8.1.2 Spectral Theorem for Real Symmetric Matrices

The main theorem we will prove is

Theorem 8.1.22 (Spectral Theorem). Let A be a real symmetric n × n matrix. Then there exists an orthogonal n×n matrix Q and a diagonal matrix Λ such that Q^{T} AQ = Λ. Moreover, the n eigenvalues of A are the diagonal entries of Λ.

This result is remarkable: it tells you that any real, symmetric matrix is diagonal when rotated into an appropriate basis (recall the rotation effect of conjugation using Q). In other words, the operation of matrix A on a vector v can be broken down into three steps:

Av = QΛQ^{T} v = (undo the rotation)(stretch along coord axes)(rotation)v.

(8.12)

Recall the ordering of transformations is read like Hebrew, right to left. The rotation is just the rotation into the basis of eigenvectors.

Furthermore, the eigenvalues λ_{i} (= diag els of Λ) are a set of numbers invariant under rotations of A. In other words, if A′ = P AP^{T} is an orthogonallyconjugated (i.e. P is orthogonal) version of A, then A′ has the same {λ_{i}} as A. Of course the ordering of the λ_{i} has to be chosen the same.

For the Spectral Theorem we prove a sequence of needed lemmas:

Lemma 8.1.23. The eigenvalues of a real symmetric matrix are real.

Let A be a real symmetric matrix with eigenvalue λ and eigenvector v. Note that we do not yet know that v has only real coordinates!

Therefore, Av = λv. Take the dot (or inner) product of both sides with the vector v^{∗}, the complex conjugate transpose of v:

v^{∗}Av = λv^{∗}v.  (8.13)

But the left hand side is real. The two sides are clearly complex numbers (ie, 1-dimensional matrices). Taking the complex conjugate transpose of the LHS gives

(v^{∗}(Av))^{∗} = (Av)^{∗}(v^{∗})^{∗} = v^{∗}Av.  (8.14)

Therefore, the LHS is real, implying the RHS is real. But clearly v^{∗}v is real (similar calculation). Thus, λ is real. □

We will only prove the Spectral Theorem when all the eigenvalues are distinct. Henceforth, we shall always assume A is a real symmetric matrix.

Lemma 8.1.24. The eigenvectors of a real symmetric matrix are real.

The eigenvalues solve the equation (λI − A)v = 0. Let λ be an eigenvalue. Then det(λI − A) = 0. Therefore the matrix B = λI − A is not invertible. Therefore it send a vector to 0 (standard linear algebra calculation).

Lemma 8.1.25. If λ_{1} and λ_{2} are two distinct eigenvalues of a real symmetric matrix A, then their corresponding eigenvectors are perpendicular.

We study v_{1}^{T} Av_{2}. Now

v_{1}^{T} Av_{2} = v_{1}^{T} (Av_{2}) = v_{1}^{T} (λ_{2}v_{2}) = λ_{2}v_{1}^{T} v_{2}.  (8.15)

Also,

v_{1}^{T} Av_{2} = v_{1}^{T} A^{T} v_{2} = (v_{1}^{T} A^{T})v_{2} = (Av_{1})^{T} v_{2} = (λ_{1}v_{1})^{T} v_{2} = λ_{1}v_{1}^{T} v_{2}. (8.16)

Therefore

λ_{2}v_{1}^{T} v_{2} = λ_{1}v_{1}^{T} v_{2} or (λ_{1} − λ_{2})v_{1}^{T} v_{2} = 0.  (8.17)

As λ_{1} ≠ λ_{2}, v_{1}^{T} v_{2} = 0. Thus, the eigenvectors v_{1} and v_{2} are perpendicular. □

We can now prove the Spectral Theorem for real symmetric matrices if there are n distinct eigenvectors.

Let λ_{1} to λ_{n} be the n distinct eigenvectors, and let v_{1} to v_{n} be the corresponding eigenvectors chosen so that each v_{i} has length 1.

Consider the matrix Q, where the first column of Q is v_{1}, the second column of Q is v_{2}, all the way to the last column of Q which is v_{n}:

Q = ( \begin{matrix}↑ & ↑ & & ↑ \\ v_{1} & v_{2} & ⋯ & v_{n} \\ ↓ & ↓ & & ↓ \end{matrix} )  (8.18)

The transpose of Q is

Q^{T} = ( ← v_{1} → ; ⋮ ; ← v_{n} → )  (8.19)

Exercise 8.1.26. Show that Q is an orthogonal matrix. Use the fact that the v_{i} all have length one, and are orthogonal (perpendicular) to each other.

Consider Q^{T} AQ. This is a matrix, call it B. To find its entry in the i^{th} row and j^{th} column, we look at

e_{i}^{T} Be_{j}  (8.20)

where the e_{k} are column vectors which are 1 in the k^{th} position and 0 elsewhere:

e_{k} = ( 0 ⋮ 0 1 0 ⋮ 0 )  (8.21)

Thus, we need only show that e_{i}^{T} Be_{j} = 0 if i ≠ j and equals λ_{j} if i = j.

Exercise 8.1.27. Show Qe_{i} = v_{i} and Q^{T} v_{i} = e_{i}.

We calculate

e_{i}^{T} Be_{j} = e_{i}^{T} Q^{T} AQe_{j}
= (e_{i}^{T} Q^{T})A(Qe_{j})
= (Qe_{i})^{T} A(Qe_{j})
= v_{i}^{T} Av_{j}
= v_{i}^{T} (Av_{j})
= v_{i}^{T} λ_{j}v_{j} = λ_{j}v_{i}^{T} v_{j}.  (8.22)

As v_{i}^{T} v_{j} equals 0 if i ≠ j and 1 if i = j, this proves the claim. Thus, the off-diagonal entries of Q^{T} AQ are zero, and the diagonal entries are

the eigenvalues λ_{j}. This shows that Q^{T} AQ is a diagonal matrix whose entries are the n eigenvalues of A. □

Note that, in the case of n distinct eigenvalues, not only can we write down the diagonal matrix, we can easily write down what Q should be. Further, by reordering the columns of Q, we see we reorder the positioning of the eigenvalues on the diagonal.

Central Limit Theorem, Spectral Theorem for Real Symmetric, Spectral Gaps

Proof of the Central Limit Theorem (via the Fourier Transform). Proof of the Spectral Theorem for Real Symmetric Matrices (via maximization). Spectral Gap and Families of Expanders. Lecture by Peter Sarnak; notes by Steven J. Miller.

9.1 Central Limit Theorem

Let X : Ω → ℝ be a random variable (x(ω) ∈ ℝ). Define the density function μ on ℝ by

μ[a, b] = Prob(ω : x(ω) ∈ [a, b]).  (9.1)

Thus, X gives rise to the probability measure μ on ℝ.

Assume

1. X_{1}, X_{2}, … are independent identically distributed random variables (iidrv) with density μ.

2. E(X) = ∫_{Ω} x(ω)dp(ω) = ∫_{ℝ} xdμ(x).

3. Var(X) = ∫_{Ω} x^{2}(ω)dp(ω) = ∫_{R} x^{2}dμ(x) = 1.

Define S_{N} = ∑ X_{j}, and let N(0, 1) denote the standard Gaussian or normal distribution (mean zero, variance one).

Theorem 9.1.1.

\frac{S_{N}}{\sqrt{N}} \overrightarrow{in probability} N(0, 1),  (9.2)

ie,

Prob(\frac{S_{N}}{\sqrt{N}} ∈ [a, b]) → \frac{1}{\sqrt{2π}} ∫_{a}^{b} e^{− \frac{x^{2}}{2}} dx.  (9.3)

Proof: We have a random variable X, induces a probability measure μ on the real line ℝ. Thus, dμ = f(x)dx, f(x) is a nice function. As μ is a probability measure on ℝ, we must have f(x) ≥ 0 and ∫_{R} f(x)dx = 1.

The given assumptions about the X_{i}s imply

1. ∫_{R} xf(x)dx = 0.

2. ∫_{R} x^{2}f(x)dx = 1.

Definition 9.1.2 (Fourier Transform).

f̂(ξ) = ∫_{R} f(x)e^{−2πixξ}dx.  (9.4)

Clearly, |f̂(ξ) ≤ ∫_{R} f(x)dx ≤ 1. Further, f̂(0) = ∫_{R} f(x)dx = 1. Now

f̂′(ξ) = ∫_{R}(2πix)f(x)e^{−2πixξ}dx.  (9.5)

Thus, f̂′(0) = 0 (from E(x) = 0). We will assume f̂is continuous (although this is implied by our assumptions). Further,

f̂′′(ξ) = −4π^{2} ∫_{R} x^{2}f(x)e^{−2πixξ}dx.  (9.6)

Therefore, f̂′′(0) = −4π^{2} (by our assumption on the variance). Using Taylor to expand f̂ we obtain

f̂(ξ) = 1 + \frac{f′′(0)}{2}ξ^{2} + ⋯
= 1 − 2π^{2}ξ^{2} + ⋯  (9.7)

Near the origin, the above shows f̂ looks like a concave down parabola.

Theorem 9.1.3 (Fourier Inversion). If f is a nice function (so all the quantities below make sense, ie, f decays fast enough):

f(x) = ∫_{R} f̂(ξ)e^{2πixξ}dξ.  (9.8)

Definition 9.1.4. e(y) = e^{2πiy}.

Exercise 9.1.5. If ϕ_{ξ}(x + y) = e((x + y)ξ), prove ϕ_{ξ}(x + y) = ϕ_{ξ}(x)ϕ_{ξ}(y). ϕ_{ξ} is a character of (ℝ, +).

Exercise 9.1.6. Is there a ψ : ℝ → ℝ such that ψ(x + y) = ψ(x) + ψ(y)? IE, can you find a homomorphism that takes addition to addition? If yes (of course: see above!) what can you say about ψ? If we assume ψ is continuous, it must be of the form ϕ_{ξ}.

Think of ℝ as a vector space over ℚ (ie, the scalars are ℚ). What is the dimension of ℝ over ℚ, and what is a basis? As the reals are uncountable and the rationals are countable, there are uncountably many basis vectors.

Any linear transformation will satisfy the desired condition! For example, choose any basis (called a Hamel Basis) of ℝ over ℚ (very hard! need the Axiom of Choice).

To show every character of the reals is of the form ϕ_{ξ}, you need more. If you assume the character is continous, then it must be of the form ϕ_{ξ}.

Suppose we have random variables X and Y with measures μ and ν, with induced functions f(x) and g(x). If we choose X and Y independently, what is the distribution of X + Y?

Lemma 9.1.7. The distribution of X +Y clearly cannot be the sum (as that won’t be a probability measure). It is f ∗ g, and f ∗ g is a probability measure.

Definition 9.1.8 (Convolution).

(f ∗ g)(x) = ∫_{R} f(x − y)g(y)dy  (9.9)

We show the convolution is a probability measure. We assume our functions f and g are nice (ie, that we may use Fubini to interchange the order of integration).

∫_{R}(f ∗ g)(x)dx = ∫_{R} ∫_{R} f(x − y)g(y)dydx
= ∫_{R} ∫_{R} f(x − y)g(y)dxdy
= ∫_{R} g(y)( ∫_{R} f(x − y)dx)dy

= ∫_{R} g(y)( ∫_{R} f(t)dt)dy
= ∫_{R} g(y) · 1dy = 1.  (9.10)

Exercise 9.1.9. f ∗ g^{}(ξ) = f̂(ξ) · ĝ(ξ). Thus, Fourier Transform converts convolution to multiplication.

Let Z = X + Y. What is

Prob(Z ∈ [z, z + dz]) = h(z)dz.  (9.11)

x can be anywhere; given x, y must lie between z − x and z − x + dz.

Prob(X + Y ∈ [z, z + dz]) = ∫_{x∈ℝ} Prob(X ∈ [x, x + dx] and

Y ∈ [z − x, z − x + dz])f(x)dx

= ∫_{x∈ℝ} f(x)g(z − x)dx = h(z),  (9.12)

where the last step follows from the definition of the density functions. By Induction, we see X_{1} + ⋯ X_{N} has distribution f ∗ ⋯ ∗ f (as the random

variables are iidrv).

However, we want to study S_{N} = \frac{X_{1}+···X_{N}}{\sqrt{N}}.

Definition 9.1.10 (FT). Let F T (f) = f̂; F T denotes the Fourier Transform.

Lemma 9.1.11.

F T (the distribution of \frac{X_{1} + ⋯ + X_{N}}{\sqrt{N}} ) = [f̂(\frac{ξ}{\sqrt{N}} )]^{N}.  (9.13)

We take the limit as N → ∞ for fixed ξ. Recall we showed that f̂(ξ) = 1 − 2π^{2}ξ^{2} + · · ·. Thus, we have to study

[1 − \frac{2π^{2}ξ^{2}}{N} + O(\frac{|ξ|^{3}}{N^{\frac{3}{2}}} )]^{N}.  (9.14)

But as N → ∞, the above goes to

e^{−2πξ^{2}}.  (9.15)

The universality arises because only terms up to the second order in the Taylor Series contribute. Exercise 9.1.12. Show the Fourier Transform of the Gaussian is the Gaussian.

Key point:

• Used Fourier Analysis to study the sum of independent identically distributed random variables, as it converts convolution to multiplication.

Exercise 9.1.13. Fix g a nice, smooth, rapidly decreasing function. Consider the linear transformation A:

(Af)(x) = ∫_{R} g(x − y)f(y)dy.  (9.16)

Any convolution operator is diagonalized by these characters (exponentials). At a formal level,

(Aϕ_{ξ})(x) = ∫_{R} g(x − y)ϕ_{ξ}(y)dy
= ∫_{R} g(x − y)e^{−2πiξy}dy
= ∫_{R} g(t)e^{−2πiξ(x−t)}dt
= e^{−2πixξ} ∫_{R} g(t)e^{2πiξt}dt = ĝ(ξ)ϕ_{ξ}(x).  (9.17)

Thus, ϕ_{ξ} is an eigenvector of A with eigenvalue ĝ(ξ).

9.2 Spectral Theorem for Real Symmetric Matrices

Let A be a real symmetric matrix acting on ℝ^{n}. Then A has an orthonormal basis v_{1}, …, v_{n} such that Av_{j} = λ_{j}v_{j}.

A simpler proof, assuming all eigenvalues are distinct, is available in the September 25^{th} lecture notes.

Write the inner or dot product ⟨v, w⟩ = v^{t}w. As A is symmetric, ⟨Av, w⟩ = ⟨v, Aw⟩.

Definition 9.2.1. V^{⊥} = {w : ∀v ∈ V, ⟨w, v⟩ = 0}.

Lemma 9.2.2. Suppose V ⊂ ℝ^{n} is an invariant vector subspace under A (ifv ∈ V, then Av ∈ V). Then V^{⊥} is also A-invariant: A(V^{⊥}) ⊂ V^{⊥}.

This proves the spectral theorem. Suppose we find a v_{0} ≠ 0 such that Av_{0} = λ_{0}v_{0}. Take V = {μv_{0} : μ ∈ ℝ} for the invariant subspace.

By Lemma 9.2.2, V^{⊥} is left invariant under A, and is one dimension less. Thus, by whatever method we used to find an eigenvector, we apply the same method on V^{⊥}.

Thus, all we must show is given an A-invariant subspace, there is an eigenvector.

Consider

max_{v with ⟨v,v⟩=1} {⟨Av, v⟩}.  (9.18)

Standard fact: every continuous function on a compact set attains its maximum (not necessarily uniquely). See, for example, W. Rudin, Principles of Mathematical Analysis.

Let v_{0} be a vector giving the maximum value, and denote this maximum value by λ_{0}. As ⟨v_{0}, v_{0}⟩ = 1, v_{0} is not the zero vector.

Lemma 9.2.3. Av_{0} = λ_{0}v_{0}.

Clearly, if Av_{0} is a multiple of v_{0} it has to be λ_{0} (from the definition of v_{0} and λ_{0}).

Thus, it is sufficient to show

Lemma 9.2.4. {μv_{0} : μ ∈ ℝ} is an A-invariant subspace.

Proof: let w be an arbitrary vector perpendicular to v_{0}, and ϵ be an arbitrary small real number. Consider

⟨A(v_{0} + ϵw), v_{0} + ϵw⟩  (9.19)

We need to renormalize, as v_{0}+ϵw is not unit length; it has length 1+ϵ^{2}⟨w, w⟩. As v_{0} was chosen to maximize ⟨Av, v⟩ subject to ⟨v, v⟩ = 1, after normalizing the above cannot be larger. Thus,

⟨A(v_{0} + ϵw), v_{0} + ϵw⟩ = ⟨Av_{0}, v_{0}⟩ + 2ϵ⟨Av_{0}, w⟩ + ϵ^{2}⟨w, w⟩. (9.20)

Normalizing the vector v_{0} + ϵw by its length, we see that in Equation 9.20, the order ϵ terms must be zero. Thus,

⟨Av_{0}, w⟩ = 0;  (9.21)

however, this implies Av_{0} is in the space spanned by v_{0} (as w was an arbitrary vector perpendicular to v_{0}), completing our proof. □

Corollary 9.2.5. Any local maximum will lead to an eigenvalue-eigenvector pair.

The second largest eigenvector (denoted λ_{1}) is

λ_{1} = max_{⟨v,v_{0}⟩=0} \frac{⟨Av, v⟩}{⟨v, v⟩}.  (9.22)

We can either divide by ⟨v, v⟩, or restrict to unit length vectors.

9.3 Applications to Graph Theory

Let G be a k-regular graph, f : V → ℝ. Recall v ∼ w if there is an edge connecting v and w. Let A be the adjacency matrix of G, and define

Af(v) = ∑_{v∼w} f(w)
⟨f, g⟩ = ∑_{v∈V} f(v)\overline{g}(v).  (9.23)

Consider the function f_{0}(v) = 1 for all v ∈ V. Then

Af_{0}(v) = ∑_{v∼w} f_{0}(w) = kf_{0}(v).  (9.24)

Thus, f_{0} is an eigenfunction with eigenvalue k.

Theorem 9.3.1 (Expander Families). Fix k. Suppose we have a sequence of kregular graphs G_{j} with |V_{j}| → ∞ and suppose k − λ_{1}(G) ≥ δ > 0, δ fixed. Then G_{j} is an expander family.

Remark 9.3.2. If you have an algorithm that has a random element, then one can show there is another algorithm which does what this algorithm does without having a random component. (More or less, some slight of hand).

Suppose we have a bipartite graph: there are two sets of vertices I (inputs) and O (outputs). Edges run only between I and O; there are no edges between two vertices in I or between two vertices in O. Let there be n inputs and n outputs, and join each input with k outputs.

Fix δ_{0} > 0. We want, for any B ⊂ I, |∂B| ≥ δ_{0}|B|.

We give a sketch of the proof. We will show that knowledge of a spectral gap ensures that the boundary of any subset of I will be big. We do bipartite for simplicity.

Let B ⊂ I. Define

f(v) = \begin{cases}2n − |B| & for v ∈ B \\ −|B| & otherwise. \end{cases}  (9.25)

Then

∑_{v∈V} f(v) = |B| · (2n − |B|) + (2n − |B|) · (−|B|) = 0.  (9.26)

Then

λ_{1} = max_{⟨f̃,f_{0}⟩=0} = \frac{⟨Af̃, f̃⟩}{⟨f̃, f̃⟩}.  (9.27)

In particular,

\frac{⟨Af_{1}, f_{1}⟩}{⟨f_{1}, f_{1}⟩} ≤ λ_{1}.  (9.28)

Definition 9.3.3 (Laplacian). Δ = kI − A.

Thus, Δf_{0} = kf_{0} − Af_{0} = 0 · f_{0}. The eigenvalues of Δ are trivially related to the eigenvalues of A.

Remark 9.3.4 (Motivation for Laplacian). On the line we have \frac{d^{2}}{dx^{2}} . A discrete version is (exercise)

\frac{f(x + h) − 2f(x) + f(x − h)}{h^{2}}.  (9.29)

In the plane, we would have \frac{d^{2}}{dx_{1}^{2}} + \frac{d^{2}}{dx_{2}^{2}} . Integrating by parts we have

∫_{Ω}(Δf) · gdx_{1}dx_{2} = − ∫_{Ω} ∇f · ∇gdx_{1}dx_{2}
= ∫_{Ω} f · (Δg)dx_{1}dx_{2}.  (9.30)

We want to integrate by parts on a graph!

(ΔF)(x) = kF (x) − ∑_{x∼y} F (y).  (9.31)

Therefore

⟨ΔF, F ⟩ = ∑_{x∈V} (kF (x) − ∑_{y∼x} F (y))F (x)
= k ∑_{x∈V} F (x)^{2} − ∑_{x∈V} ∑_{x∼y} F (x)F (y).  (9.32)

For each edge e, orient it by e^{+} and e^{−}. The analogue of the Laplacian becomes

∑_{e} (F (e^{+}) − F (e^{−}))^{2} = ∑_{e} F^{2}(e^{+}) − 2F (e^{+})F (e^{−}) + F^{2}(e^{−})
= 2⟨ΔF, F ⟩,  (9.33)

where the last line follows by thinking about what it means for vertices to be connected, and how often each vertex is hit.

Recall

k − λ_{1} = min_{⟨F,f_{0}⟩=0} \frac{⟨ΔF, F ⟩}{⟨F, F ⟩} = min_{⟨F,f_{0}⟩=0} \frac{1}{2} \frac{||dF ||^{2}}{⟨F, F ⟩} .  (9.34)

Plug in the function f defined above, namely,

f(v) = \begin{cases}2n − |B| & for v ∈ B \\ −|B| & otherwise. \end{cases}  (9.35)

If the edge runs from input to input or output to output, we get zero. The only way we get non-zero contribution is from an input to an output (or vice-versa). For our f,

\frac{1}{2} ∑_{e∈E} |f(e^{+}) − f(e^{−})|^{2} = \frac{1}{2}(2n)^{2} · #\{edges e running from B to B^{c}\}, (9.36)

where B^{c} is the complement of B. We want 0 < δ_{0} = k − λ_{1}. Divide the previous equation by ⟨F, F ⟩, where ⟨F, F ⟩ = (2n − |B|)^{2} · |B| + |B|^{2} · (2n − |B|) = |B| · (2n − |B|) · 2n.

Thus,

δ_{0} ≤ \frac{\frac{1}{2}(2n)#\{edges\}}{|B| · (2n − |B|) · 2n}.  (9.37)

Therefore,

#\{edges\} ≥ \frac{δ_{0}|B| · (2n − |B|)}{n}
≥ δ_{0}|B|,  (9.38)

as |B| ≤ n. Thus, the total number of edges is at least δ_{0}|B|. But each vertex gets k edges.

Thus,

|∂B| ≥ \frac{δ_{0}|B|}{k}.  (9.39)

9.4 2\sqrt{k − 1}

Let G be a k-regular connected graph. A is the adjacency matrix. Biggest eigenvalue is λ_{0} = k. Eigenvalues cannot be smaller than −k. How big is the gap between k and λ_{1}?

Theorem 9.4.1 (Alon-Boppana). Fix k. Take any sequencce of graphs where the number of vertices |G| → ∞. Then

\overline{lim}_{|G|→∞}λ_{1}(A_{G}) ≥ 2\sqrt{k − 1}.  (9.40)

Remark 9.4.2. In 1-dimension, with probability one the drunk returns home; same in 2-dimensions. He escapes with finite probability in 3 and higher dimensions!

Consider a 3 regular graph with many vertices. Go to v, look locally. If there are no short circuits, know what it looks like locally: it will look like a tree. (This is p-adic hyperbolic geometry).

Let T be the infinite tree where each vertex is connected to three other vertices (and a vertex cannot be connected to itself). Suppose a drunk is walking on a tree. The only way he can get back is to exactly undo what he’s done.

Consider the following operator: consider an infinite dimensional Hilbert space l_{2}(V), the set of all f : V → ℝ such that ∑_{v} |f(v)|^{2} < ∞. This space is infinite dimensional (for each v, take the function f_{v}(w) = 1 if w = v and 0 otherwise).

Using |ab| ≤ \frac{a^{2}+b^{2}}{2} ,

⟨f, g⟩ = ∑_{v} f(v)g(v)  (9.41)

exists (and is our inner product). We define

Af(v) = ∑_{w∼v} f(w).  (9.42)

It is not obvious that there are any eigenvectors (and, in fact, there are no eigenvectors!). There is still a notion of spectrum. We will show the spectrum of this operator (k = 3) is [−2\sqrt{2}, 2\sqrt{2}].

Note the constant function is horrendously not in this space (not even close to being square-integrable.

Properties of Eigenvalues of Adjacency Matrices of Random Graphs

We discuss properties of eigenvalues of adjacency matrices arising from Random Graphs. Lecture by Peter Sarnak; notes by Steven J. Miller.

10.1 Definitions

Let G be a connected, simple (no multiple bonds or edges) k-regular graph with adjacency matrix A = (a_{v,w}). Here

a_{v,w} = \begin{cases}1 & if v ∼ w \\ 0 & otherwise \end{cases}  (10.1)

Thus, a_{v,w} is the number of paths of length one from v to w.

Let A^{2} = (a_{v,w}^{(2)}), and here

a_{v,w}^{(2)} = number of paths of length 2 from v to w.  (10.2)

Thus,

a_{v,w}^{(2)} = ∑_{v′} a_{v,v′}a_{v′,w}.  (10.3)

Similarly let A^{n} = (a_{v,w}^{(n)}), and here

a_{v,w}^{(n)} = number of paths of length n from v to w.  (10.4)

Recall

Trace(B) = T r(B) = ∑_{v} b_{v,v}.  (10.5)

Given an adjacency matrix A, let D be the diagonal matrix of eigenvalues.

D = ( \begin{matrix}λ_{0} \\ & λ_{1} \\ & & . & . & . \\ & & & & & λ_{N−1} \end{matrix} )  (10.6)

A = Q^{−1}DQ for some orthogonal matrix Q. Thus, A^{n} = Q^{−1}DQ, and we find:

Lemma 10.1.1. T r(A^{n}) = T r(D^{n}).

Lemma 10.1.2 (Trace Formula). For any n ≥ 0,

∑_{j=0}^{N−1} λ_{j}^{n} = ∑_{v∈V} a_{v,v}^{(n)}.  (10.7)

10.2 ρ_{k}(2n) and λ_{max}

To count walks of length n from v to v, it is clearly at least as many walks as there are on a k-regular infinite tree.

A tree is a homogeneous object: any vertex looks exactly the same as any other. There is no special vertex on a tree, though we will often name a vertex the root.

Definition 10.2.1. ρ_{k}(n) is the number of paths of length n from v to v, where v is any vertex of the k-regular tree.

Remark 10.2.2. ρ_{k}(n) = 0 for n odd.

Remark 10.2.3. The number of paths of length n from v to v on our graph G is at least the number of paths of length n from any vertex to itself on the infinite tree. Thus,

∀v ∈ V, a_{v,v}^{(n)} ≥ ρ_{k}(n).  (10.8)

Remember that we’ve labeled the N eigenvalues by λ_{0} = k, …, λ_{N−1}. In the trace formula, we find

∑_{j=0}^{N−1} λ_{j}^{2n} ≥ Nρ_{k}(2n).  (10.9)

Therefore

\frac{1}{N} ∑_{j=0}^{N−1} λ_{j}^{2n} ≥ ρ_{k}(2n).  (10.10)

and as λ_{0} = k

\frac{k^{2n}}{N} + \frac{1}{N} ∑_{j=1}^{N−1} λ_{j}^{2n} ≥ ρ_{k}(2n).  (10.11)

Fix n and let N → ∞. Let λ_{max} = max (|λ_{1}|, |λ_{N−1}|). Thus, substituting into Equation 10.11 we find

\frac{k^{2n}}{N} + λ_{max}^{2n} ≥ ρ_{k}(2n)  (10.12)

in the limit as N → ∞ (as we have λ_{max} a total of N − 1 times, and we divide by N; in the limit, \frac{N−1}{N} → 1.

As N → ∞, we find

λ_{max}^{2n} ≥ ρ_{k}(2n)
or λ_{max} ≥ (ρ_{k}(2n))^{\frac{1}{2n}} .  (10.13)

Exercise 10.2.4. Show

1. ρ_{k}(2n) ≥ \frac{1}{m} \binom{2m−2}{m−1}k(k − 1)^{m−1}.

2. (ρ_{k}(2n))^{\frac{1}{2n}} → 2\sqrt{k − 1}.

Using the above exercise, we now find that

λ_{max} ≥ 2\sqrt{k − 1}.  (10.14)

10.3 Measure from the Eigenvalues

The trace formula told us that

∑_{j=0}^{N−1} λ_{j}^{n} = ∑_{v∈V} a_{v,v}^{(n)}.  (10.15)

Definition 10.3.1 (girth). The girth of a graph is the length of the shortest closed cycle that returns to the starting vertex without any backtracking.

Assume that the girth of G_{N} tends to ∞ as N → ∞. Fix n, let N be very large. Then by assumption the girth is greater than say

2n+1. Thus, a_{v,v}^{(n)} cannot have any contribution from cycles without backtracking. Thus, locally, to calculate a_{v,v}^{(n)}, we look like a tree, and we find a_{v,v}^{(n)} = ρ_{k}(n) for every v ∈ V. It is essential that we have fixed n.

Therefore, we now have (for fixed n under our assumption) that

∑_{j=0}^{N−1} λ_{j}^{n} = Nρ_{k}(n)

\frac{1}{N} ∑_{j=0}^{N−1} λ_{j}^{n} = ρ_{k}(n).  (10.16)

The left hand side looks like a Riemann sum. Suppose the density of the eigenvalues of the 3-regular graph is dμ = f(x)dx. We have just shown, for polynomials p_{n}(x) = x^{n}, that

\frac{1}{N} ∑_{j=0}^{N−1} p_{n}(λ_{j}) → ∫_{−3}^{3} p_{n}(x)dμ(x),  (10.17)

where the above converges to ρ_{k}(n) (here k = 3).

Thus, we are looking for a density function f(x) such that

∫_{−3}^{3} x^{n}f(x)dx = ρ_{3}(n), n ≥ 0.  (10.18)

Is there such a function? Is it unique? What does it look like? This is the Inverse Moment Problem.

If there were two such functions, they would have to be equal by the Weierstrass Approximation Theorem, as their difference integrates to zero against any polynomial.

Exercise 10.3.2. Compute the generating function

F (z) = ∑_{n=0}^{∞} ρ_{3}(n)z^{n},  (10.19)

which is something like

\frac{1}{\sqrt{4(k − 1)^{2} − z^{2}}}  (10.20)

if |z| is small (or maybe complex and outside [−k, k]).

∑_{n=0}^{∞} ρ_{3}(n)z^{n} = ∑_{n=0}^{∞} ( ∫_{−3}^{3} x^{n}f(x)dx)z^{n}

= ∫_{−3}^{3} f(x) ∑_{n=0}^{∞}(zx)^{n}dx

= ∫_{−3}^{3} \frac{f(x)dx}{1 − zx} = F (z).  (10.21)

If we let z = \frac{1}{w} we find

F (w) = ∑_{n=0}^{∞} ρ_{3}(n)\frac{1}{w^{n}}

= w ∫_{−3}^{3} \frac{f(x)dx}{w − x}.  (10.22)

Let w ∈ C be such that w ∉ [−3, 3]. Letting w → a ∈ [−3, 3], one gets a different value (a jump) if w approaches from above or below, and the jump is basically f(a).

Look at a + ib and a − ib, b → 0.

10.4 Summary

The above is all based on the assumption that the girth was big. For the random graph, there are very few short closed cycles. Thus, when we use the trace formula, we now have

\frac{1}{N} ∑_{j=0}^{N−1} λ_{j}^{n} = ρ_{k}(n) + O(\frac{1}{N} ).  (10.23)

Spacings of Eigenvalues of Real Symmetric Matrices; Semi-Circle Law

Joint Density Function for eigenvalues of real symmetric matrices; spacing of eigenvalues for 2 × 2 real symmetric matrices; Semi-Circle Rule. Lecture by Steven J. Miller; notes by Steven J. Miller and Alex Barnett.

11.1 Joint density function of eigenvalues of real symmetric matrices (‘GOE’)

11.1.1 Dirac Notation

The derivation handed out in lecture used physics notation which should be explained. The matrix is called the ‘Hamiltonian’ (meaning that it happened to arise in a quantum physics problem). Vectors are often called states (referring to quantum states), however they can be thought of as your usual vectors. (Quantum mechanics is just linear algebra, amazingly). A general vector in 2D is written

|u⟩ equivalent to u = ( \begin{matrix}u_{1} \\ u_{2} \end{matrix} ) ,  (11.1)

the latter being its coordinate representation in some basis. The unit vectors are

|1⟩ , |2⟩ equivalent to ( \begin{matrix}1 \\ 0 \end{matrix} ) , ( \begin{matrix}0 \\ 1 \end{matrix} ) .  (11.2)

The |u⟩ is a column vector, and ⟨u| ≡ |u⟩^{T} is a row vector. Inner product can be written as ⟨v |u⟩ = v^{T} · u. General bilinear product can be written ⟨v |M| u⟩ = v^{T} · M · u.

11.1.2 2 × 2 Gaussian Orthogonal Ensemble (GOE)

We consider 2 × 2 real symmetric matrices,

A ≡ ( \begin{matrix}x & y \\ y & z \end{matrix} ) .  (11.3)

Understanding this case is vital to building intuition about Random Matrix Theory for N × N matrices.

A can always be diagonalized by an orthogonal matrix Q as follows,

Q^{T} ( \begin{matrix}x & y \\ y & z \end{matrix} ) Q = ( \begin{matrix}λ_{1} & 0 \\ 0 & λ_{2} \end{matrix} ) ≡ D.  (11.4)

In 2 × 2 case, the characteristic equation det(A − λI) = 0 is quadratic:

λ^{2} − Tr(A)λ + det(A) = 0,  (11.5)

where

Tr(A) = x + z, det(A) = xz − y^{2}.  (11.6)

Solutions are

λ_{1,2} = \frac{x + z}{2} ± \sqrt{(\frac{x − z}{2} )^{2} + y^{2}},  (11.7)

where 1 is the + case, 2 the −.

If the two eigenvalues are equal, we say the matrix is degenerate. Initially we are in a three-dimensional space (as x, y and z are arbitrary). Degeneracy requires that x, y and z satisfy

(\frac{x − z}{2} )^{2} + y^{2} = 0,  (11.8)

or, equivalently,

x − z = 0, y = 0.  (11.9)

Thus, we lose two degrees of freedom, because there are two equations which must be satisfied. The set of solutions is {(x, y, z) = (x, 0, x)}.

Exercise 11.1.1. Show that λ_{1} − λ_{2} is twice the distance from the origin in this 2D subspace.

Corresponding eigenvectors are,

v_{1} = ( \begin{matrix}c \\ s \end{matrix} ) ,  v_{2} = ( \begin{matrix}−s \\ c \end{matrix} ) .  (11.10)

We use abbreviations c ≡ cos θ and s ≡ sin θ.

Why can we write the eigenvectors as above? We can always normalize the eigenvector attached to a given eigenvalue to have length 1. We have previously shown that, if the eigenvalues are distinct, then the eigenvectors of a real symmetric matrix are perpendicular. This forces the above form for the two eigenvectors, at least when λ_{1} ≠ λ_{2}.

One rotation angle θ defines the orthogonal matrix,

Q = Q(θ) = (v_{1} v_{2}) = ( \begin{matrix}c & −s \\ s & c \end{matrix} ) .  (11.11)

The structure of the eigenvectors is actually quite rich.

Exercise 11.1.2. Find θ in terms of x, y, z. Hint: use trigonometric identities to simplify the resulting form. Hint: solve (A − λ_{1}v_{1}) = 0.

Exercise 11.1.3. Show that a general A can be written

A = α ( \begin{matrix}cos β & sin β \\ sin β & − cos β \end{matrix} ) + γ ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )  (11.12)

Exercise 11.1.4. Find λ_{1,2} in terms of α, β, γ. Show that the eigenvector angle is given by θ = β/2. This result is quite deep; for instance notice that taking a complete 2π cycle in β reverses the signs of the eigenvectors! This isn’t that relevant for the rest of this lecture.

We adopt two assumptions about the joint distribution over A, called p(A) ≡ p(x, y, z):

1. Invariance of p under orthogonal transformations (aka ‘basis-invariance’), p(M^{T} AM) = p(A) for all orthogonal M.

2. Independence of distributions of individual matrix elements, p(x, y, z) = p_{x}(x)p_{y}(y)p_{z}(z).

Section 2C-1 of the handout reminds us that these two assumptions taken together demand a unique form of distribution,

p(x, y, z) ∝ e^{−CTr(A^{2})},  (11.13)

depending on only one parameter C; we choose C = 1. Note ∝ means proportional to; the constant of proportionality is what is needed to make P 9x, y, z) a probability distribution (ie, the integral of ∫ ∫ ∫ p(x, y, z)dxdydz = 1.

This corresponds to Gaussian distributions of matrix elements,

p_{y}(y) = \sqrt{\frac{2}{π}} e^{−2y^{2}} off-diag

p_{x}(x) = p_{z}(x) = \frac{1}{\sqrt{π}} e^{−x^{2}}  diag.  (11.14)

Note that the diag elements have variance \frac{1}{2}, the off-diag variance \frac{1}{4}. We show how to compute the normalization prefactors later on. This form (for general C) is the so-called GOE. The n × n case is derived in Miller’s handout of 9/25/02.

11.1.3 Transformation to diagonal representation

The operation of diagonalizing A can be viewed as the transformation from one 3D space to another 3D space,

r ≡ \underbrace{(x, y, z)}_{A} ⟷ r′ ≡ \underbrace{(λ_{1}, λ_{2}, θ)}_{D,Q} .  (11.15)

This is 1-to-1 apart from the set of measure zero (ie, a lower dimensional subspace) corresponding to degenerate eigenvalues. Looking at Eq. 11.4 we can see the transformation is linear in the eigenvalues, nonlinear in θ. We are interested in the marginal distribution of the eigenvalues,

p′(λ_{1}, λ_{2}) ≡ ∫ dθ p′(λ_{1}, λ_{2}, θ),  (11.16)

in other words we don’t care what θ is. We use primes to signify distributions over final (Q, D) variables.

We want to know how to transform probability density from r space to r′ space. In general this must follow the law,

p(r)dr = p′(r′)dr′,  (11.17)

giving

p′(r′) = det(J)p(r).  (11.18)

The ratio of the volume elements is | det J| where J is the 3 × 3 Jacobean matrix of the transformation. J has elements J_{ij} = ∂r_{j}/∂r′_{i}.

Inverting Eq. 11.4 we can write A(r′) as

( \begin{matrix}x & y \\ y & z \end{matrix} ) = QDQ^{T} = ( \begin{matrix}c & −s \\ s & c \end{matrix} ) ( \begin{matrix}λ_{1} & 0 \\ 0 & λ_{2} \end{matrix} ) ( \begin{matrix}c & s \\ −s & c \end{matrix} )
= ( \begin{matrix}λ_{1}c^{2} + λ_{2}s^{2} & (λ_{1} − λ_{2})sc \\ (λ_{1} − λ_{2})sc & λ_{1}s^{2} + λ_{2}c^{2} \end{matrix} )  (11.19)

We evaluate J for this case,

J ≡ ( \frac{∂x}{∂λ_{2}} ..._{\frac{∂x}{∂θ}}^{\frac{∂x}{∂λ_{1}}  \frac{∂y}{∂λ_{1}}  \frac{∂z}{∂λ_{1}}} ) .  (11.20)

We see λ’s only appear in the bottom three entries, and furthermore they only appear as factors (λ_{1} − λ_{2}) in each entry.

Exercise 11.1.5. Evaluate the bottom row of J to prove the above.

Therefore this factor of a row of J can be brought out in evaluating the determinant:

det(J) = |messy θ-dep 3 × 3 matrix| · (λ_{1} − λ_{2}) = g(θ)(λ_{1} − λ_{2}). (11.21)

Warning! The Jacobian is the absolute value of the determinant. Thus, we need |λ_{1} − λ_{2}| above, or we need to adopt the convention that we label the eigenvalues so that λ_{1} ≥ λ_{2}.

The only dependence on the λ’s is given by the second factor. Plugging into Eq. 11.18 and marginalizing over θ gives,

p′(λ_{1}, λ_{2}) = ∫ dθ g(θ) (λ_{1} − λ_{2})e^{−(λ_{1}^{2}+λ_{2}^{2})}
∝ (λ_{1} − λ_{2})e^{−(λ_{1}^{2}+λ_{2}^{2})}.  (11.22)

Note that we do not need the absolute value sign around (λ_{1} − λ_{2}) because we chose λ_{1} > λ_{2}. This is the joint density of the eigenvalues in 2 × 2 GOE.

11.1.4 Generalization to n × n case

The above generalizes quite easily, with the dimension of the two spaces being N = \frac{1}{2}n(n + 1). The \frac{1}{2}n(n + 1) degrees of freedom in A equal n degrees of freedom in D (namely the eigenvalues \{λ_{i}\}) plus \frac{1}{2}n(n − 1) degrees of freedom in Q (namely the generalized angles Ω). We sort the eigenvalues such that λ_{1} ≥ λ_{2} ≥ ⋯ ≥ λ_{n}.

Theorem 11.1.6. If λ_{i} = λ_{j} for some 1 ≤ i < j ≤ n, then the Jacobean of the transformation A ↔ (D, Q) vanishes, that is det(J) = 0.

The proof relies on realising that the two eigenvectors v_{i} and v_{j} span a 2D subspace, invariant under A. (Recall here we refer to a subspace of the n-dim vector space upon which A operates by multiplication). The invariance means that the choices of directions of the eigenvectors is arbitrary in this 2D plane. Therefore there is one angle degree of freedom in Ω which in not constrained by A, that is, it is independent of the elements of A. Now think of the inverse transformation from (D, Q) → A. An infinitesimal volume element is transformed as

dr = det(J)dr′.  (11.23)

Changes of eigenvector angle within the 2D subspace have no effect on A, so the volume element dr is collapsed to zero. (Another way of putting this is that J acquires a null-space of dimension 1). Therefore det(J) = 0. □.

This vanishing of the Jacobean at degeneracies renders the non-uniqueness of the forward map A → (D, Q) at these points harmless in the following.

The upper n rows of J are messy functions of angles Ω, and the bottom \frac{1}{2}n(n− 1) rows contain entries each which is linear in the eigenvalues. Therefore det(J) is a polynomial of degree \frac{1}{2}n(n − 1) in the eigenvalues λ_{i}. Further, det(J) = 0 if any two eigenvalues are equal.

Consider the polynomial ∏_{1≤i<j≤n}(λ_{i} − λ_{j}). First, note that this polynomial vanishes whenever two eigenvalues are the same. We claim it is a polynomial of degree \frac{1}{2}n(n − 1) in the eigenvalues. For each j, there are j − 1 choices for i. Thus, the degree is

∑_{j=2}^{n} j − 1 = ∑_{k=1}^{n−1} k = \frac{(n − 1)(n − 1 + 1)}{2} = \frac{n(n − 1)}{2}.  (11.24)

Thus, det(J) and ∏_{1≤i<j≤n}(λ_{i} − λ_{j}) both vanish whenever two eigenvalues are equal, and they have the same degree. Therefore, they must be scalar multiples of each other.

So,

det(J) ∝ ∏_{1≤i<j≤n}(λ_{i} − λ_{j}).  (11.25)

Combining with the GOE form of p(A) gives, after marginalizing over Ω as before,

p({λ_{i}}) = ∏_{1≤i<j≤n}(λ_{i} − λ_{j}) · e^{− ∑_{i=1}^{n} λ_{i}^{2}}.  (11.26)

The vanishing of this probability density as any two eigenvalues come close is called level repulsion.

11.2 Eigenvalue spacing distribution in 2 × 2 real symmetric matrices

11.2.1 Reminder: Integral of the Gaussian

We want

I = ∫_{−∞}^{∞}e^{−x^{2}}dx.  (11.27)

Square it and rearrange the summation over area by using polar coordinates:

I^{2} = ∫_{−∞}^{∞}e^{−x^{2}}dx · ∫_{−∞}^{∞}e^{−y^{2}}dy = ∫_{−∞}^{∞}∫_{−∞}^{∞}e^{−x^{2}−y^{2}}dx dy
= ∫_{0}^{2π} dθ ∫_{0}^{∞} r dr e^{−r^{2}} = 2π · [−\frac{1}{2}e^{−r^{2}}]_{0}^{∞}
= π.  (11.28)

Introduction of the radius factor r produced re^{−r^{2}}, a known differential. So,

I = \sqrt{π}.  (11.29)

Changing the variable in the above, and rearranging, gives

\frac{1}{\sqrt{2πσ^{2}}} ∫_{−∞}^{∞}e^{− \frac{x^{2}}{2σ^{2}}} dx = 1.  (11.30)

This is therefore the correct normalization for a 1D Gaussian probability density, of variance σ^{2}.

11.2.2 Spacing distribution

Figure 11.1: Change of coordinates to get spacing distribution in E. Light shading suggests form of density across the 2D plane. Dark shading shows a graph of its projection onto the E axis.

Here for convenience we present a slightly simpler derivation than in lecture. Given the 2D density p′(λ_{1}, λ_{2}) we want the 1D density of the difference E ≡ λ_{1} − λ_{2}. This will require marginalizing again, since there is a reduction in dimensionality. We define S ≡ λ_{1} + λ_{2}. The linear transformation (λ_{1}, λ_{2}) → (E, S) has fixed Jacobean (it is a rotation by −45^{o} and a compression by \sqrt{2} in each axis). See Fig. 11.1.

Therefore, substituting in λ_{1} = (S +E)/2 and λ_{1} = (S −E)/2 into Eq. 11.22 gives

p′(E, S) ∝ p′(λ_{1}(E, S), λ_{2}(E, S)) = Ee^{− \frac{1}{4} [(S+E)^{2}+(S−E)^{2}]}
= Ee^{−E^{2}/2} · e^{−S^{2}/2},  (11.31)

which is separable. Therefore integrating over S gives an E-independent number, and

p′(E) ∝ Ee^{−E^{2}/2}.  (11.32)

This is the so-called ‘Wigner Surmise’ for the eigenvalue spacing density. Remarkably, in the n × n case, even for large n, this density is very close to the true spacing distribution of adjacent eigenvalues. The limiting powerlaw lim_{E→0} p′(E) ∝ E^{β} with β = 1 is intimately related to the matrix symmetry class GOE that we close. It is also possible to achieve β = 2 and β = 4 by choosing different symmetry classes.

Finally, let’s say you couldn’t be bothered to construct your second variable S. Instead you could derive the above using the Dirac delta-function (see below) to marginalize:

p′(E) = ∫_{−∞}^{∞}dλ_{1} ∫_{−∞}^{λ_{1}} dλ_{2} p′(λ_{1}, λ_{2}) δ(E − (λ_{1} − λ_{2})).  (11.33)

Apart from the unusual limits (due to ordering of eigenvalues), this is the standard procedure to extract a marginal density.

Exercise 11.2.1. Simplify the above to arrive at p′(E).

11.3 Delta Function(al)

Let f(x) be a nice function; for example, let f(x) be an infinitely differentiable function whose Taylor Series converges to f(x):

f(x) = f(0) + \frac{f′(0)}{1!}x + \frac{f′′(0)}{2!}x^{2} + ⋯  (11.34)

Let

δ_{n}(x) = \begin{cases}n & if x ∈ [ − \frac{1}{2n}, \frac{1}{2n}] \\ 0 & otherwise \end{cases}  (11.35)

Exercise 11.3.1. Show that

∫_{−∞}^{∞} f(x)δ_{n}(x)dx = f(0) + O(\frac{1}{n}).  (11.36)

Let δ be the limit as n → ∞ of δ_{n}. We find / define

lim_{n→∞} ∫_{−∞}^{∞} f(x)δ_{n}(x)dx = ∫_{−∞}^{∞} f(x)δ(x)dx = f(0).  (11.37)

Exercise 11.3.2. Show that

∫_{−∞}^{∞} f(x)δ(x − a)dx = f(a).  (11.38)

A good analogy for the δ functional is a point mass. A point mass has no extension (no length, width or height) but finite mass. Therefore, a point mass has infinite density.

A probability density must integrate to one. This corresponds to ∫ 1·δ(x)dx = 1. We often refer to δ(x) as a point mass at the origin, and δ(x − a) as a point mass at a.

11.4 Definition of the Semi-Circle Density

Consider

P (x) = \begin{cases}\sqrt[\frac{2}{π}]{1 − x^{2}} & if |x| ≤ 1 \\ 0 & otherwise \end{cases}  (11.39)

Exercise 11.4.1. Show that P (x) is a probability density. IE, show that it is nonnegative and integrates to 1. Graph P (x).

We call P (x) the semi-circle density.

11.5 Semi-Circle Rule: Preliminaries

Let λ_{j} be the eigenvalues of a real, symmetric N × N matrix A. We normalize the eigenvalues of A by dividing by 2\sqrt{N}.

Define

μ_{A,N}(x) = \frac{1}{N} ∑_{j=1}^{N} δ(x − \frac{λ_{j}(A)}{2\sqrt{N}}).  (11.40)

δ(x − \frac{λ_{j}(A)}{2\sqrt{N}}) is a point mass at \frac{λ_{j}(A)}{2\sqrt{N}}. By summing these point masses and dividing by N, we have a probability distribution. For example,

∫_{−∞}^{∞} f(x)μ_{A,N}(x)dx = ∑_{j=1}^{N} f(\frac{λ_{j}(A)}{2\sqrt{N}}).  (11.41)

We will show that, as N → ∞, the above converges to the integral of f against the semi-circle density:

∫_{−∞}^{∞} f(x)P (x)dx.  (11.42)

What does this mean?

∑_{j=1}^{N} f(\frac{λ_{j}(A)}{2\sqrt{N}})  (11.43)

looks like a Riemann Sum. The statement that, for nice f(x),

∑_{j=1}^{N} f(\frac{λ_{j}(A)}{2\sqrt{N}}) → ∫_{−∞}^{∞} f(x)P (x)dx  (11.44)

means that as N → ∞, the number of eigenvalues of a random A in [a, b] equals

∫_{a}^{b} P (x)dx.  (11.45)

Theorem 11.5.1. Choose the entries a_{ij} of a real, symmetric matrix independently from a fixed probability distribution p with mean zero, variance one, and finite higher moments. For each A, form the probability measure μ_{A,N}. As N → ∞, with probability one the measures μ_{A,n}(x)dx converge to the semi-circle probability P (x)dx.

This is not the most general version; however, it is rich enough for our purposes.

11.6 Sketch of Proof of the Semi-Circle Law

11.6.1 Calculation of Moments via Trace Formula

We will show that the expected value of the moments of the μ_{A,N}(x) equal the moments of the semi-circle.

Definition 11.6.1. M_{A,N}(k) is the k^{th} moment of the probability measure attached to μ_{A,N}(x)dx:

M_{A,N}(k) = ∫ x^{k}μ_{A,N}(x)dx = \frac{1}{N} ∑_{j=1}^{N} (\frac{λ_{j}(A)}{2\sqrt{N}})^{k}.  (11.46)

Note that ∑ λ_{j}(A)^{k} = Trace(A^{k}). Thus, we have

M_{A,N}(k) = \frac{1}{2^{k}N^{1+ \frac{k}{2}}} Trace(A^{k}).  (11.47)

We now calculate the expected values of the first few moments (k = 0, 1, 2 and 3). Lemma 11.6.2. The expected value of M_{A,N}(0) = 1.

Proof:

E[M_{A,N}(0)] = \frac{1}{N} E[Trace(I)] = 1.  (11.48)

Note that summing the eigenvalues to the zeroth power is the same as taking the trace of the identity matrix. □ Lemma 11.6.3. The expected value of M_{A,N}(1) = 0.

Proof:

E[M_{A,N}(1)] = \frac{1}{2N^{3/2}} E[Trace(A)]
= \frac{1}{2N^{3/2}} E[ ∑_{i} a_{ii}]
= \frac{1}{2N^{3/2}} ∑_{i} E[a_{ii}] = 0,  (11.49)

because we have assumed that each a_{ij} is drawn from a probability distribution with mean zero. □ Lemma 11.6.4. The expected value of M_{A,N}(2) = \frac{1}{4}.

Proof: Note that

Trace(A^{2}) = ∑_{i} ∑_{j} a_{ij}a_{ji}.  (11.50)

As our matrix is symmetric, a_{ij} = a_{ji}. Thus, the trace is ∑_{i} ∑_{j} a_{ij}^{2}. Now

E[M_{A,N}(2)] = \frac{1}{4N^{2}} E[Trace(A^{2})]
= \frac{1}{4N^{2}} E[ ∑_{i} ∑_{j} a_{ij}^{2}]

= \frac{1}{4N^{2}} ∑_{i} ∑_{j} E[a_{ij}^{2}] = \frac{1}{4},  (11.51)

where the last line follows from each a_{ij} has variance 1. As their means are zero, the variance E[a_{ij}^{2}] − E[a_{ij}]^{2} = 1 implies E[a_{ij}] = 1. There are N^{2} pairs (i, j). Thus, we have \frac{1}{4N^{2}} · (N · 1) = \frac{1}{4}. □

Lemma 11.6.5. The expected value of M_{A,N}(3) = 0 as N → ∞.

We need

Trace(A^{3}) = ∑_{i} ∑_{j} ∑_{k} a_{ij}a_{jk}a_{ki}.  (11.52)

We find

E[M_{A,N}(3)] = \frac{1}{8N^{2.5}} E[Trace(A^{3})]
= \frac{1}{8N^{2.5}} E[ ∑_{i} ∑_{j} ∑_{k} a_{ij}a_{jk}a_{ki}]

= \frac{1}{8N^{2.5}} ∑_{i} ∑_{j} ∑_{k} E[a_{ij}a_{jk}a_{ki}].  (11.53)

There are three cases. If the subscripts i, j and k are all distinct, then a_{ij}, a_{jk}, and a_{ki} are three independent variables. Hence

E[a_{ij}a_{jk}a_{ki}] = E[a_{ij}] · E[a_{jk}] · E[a_{ki}] = 0.  (11.54)

If two of the subscripts are the same (say i = j) and the third is distinct, we have

E[a_{ii}a_{ik}a_{ki}] = E[a_{ii}] · E[a_{ik}^{2}] = 0 · 1 = 0.  (11.55)

If all three subscripts are the same, we have

E[a_{ii}^{3}]  (11.56)

This is the third moment of a_{ii}. It is the same for all variables a_{ii}, and is finite by assumption. There are N triples where i = j = k.

Thus,

E[M_{A,N}(3)] = \frac{1}{8N^{2.5}} · NE[a_{11}^{3}] = \frac{E[a_{11}^{3}]}{8} · \frac{1}{N^{1.5}} .  (11.57)

Thus, as N → ∞, the expected value of the third moment is zero. □

To calculate the higher moments requires significantly more delicate combinatorial arguments.

11.6.2 Calculation of Moments from the Semi-Circle

We now calculate the moments of the semi-circle. For k ≤ 3, the k^{th} moment of the semi-circle C(k) equals the expected k^{th} moment of μ_{A,N}(x) as N → ∞.

C(k) = ∫_{−∞}^{∞} x^{k}P (x)dx = \frac{2}{π} ∫_{−1}^{1} x^{k}\sqrt{1 − x^{2}}dx.  (11.58)

We note that, by symmetry, C(k) = 0 for k odd, and C(0) = 1 as P (x) is a probability density.

For k = 2m even, we change variables x = sin θ.

C(2m) = \frac{2}{π} ∫− \frac{π}{2}^{\frac{π}{2}} sin^{2m} θ · cos^{2} θdθ.  (11.59)

Using sin^{2} θ = 1 − cos^{2} θ gives

C(2m) = \frac{2}{π} ∫− \frac{π}{2}^{\frac{π}{2}} sin^{2m} θdθ − \frac{2}{π} ∫− \frac{π}{2}^{\frac{π}{2}} sin^{2m+2} θdθ.  (11.60)

The above integrals can be evaluated exactly. We constantly use

cos^{2}(ϕ) = \frac{1}{2} + \frac{1}{2} cos(2ϕ)
sin^{2}(ϕ) = \frac{1}{2} − \frac{1}{2} cos(2ϕ).  (11.61)

Repeated applications of the above allow us to write sin^{2m}(θ) as a linear combination of 1, cos(2θ), …, cos(2mθ).

Let

n!! = \begin{cases}n · (n − 2) ⋯ 2 & if n is even \\ n · (n − 2) ⋯ 1 & if n is odd \end{cases}  (11.62)

We find (either prove directly or by induction) that

\frac{2}{π} ∫− \frac{π}{2}^{\frac{π}{2}} sin^{2m} θdθ = 2\frac{(2m − 1)!!}{(2m)!!}.  (11.63)

Exercise 11.6.6. Show the above gives

C(2m) = 2\frac{(k − 1)!!}{(k + 2)!!}.  (11.64)

Also, show C(2) agrees with our earlier calculation.

More Graphs, Maps mod p, Fourier Series and n alpha

More on Graphs; Arithmetic maps mod p. Lecture by Peter Sarnak; notes by Steven J. Miller. Appendix: Introduction to Fourier Series and {nα}, by Steven J. Miller.

12.1 Kesten’s Measure

For a k-regular graph, define

dμ_{k}(t) = \begin{cases}^{\frac{c_{k}\sqrt{(k−1)− \frac{t^{2}}{4}}}{1−( \frac{t}{k})^{2}}} dt & if |t| ≤ 2\sqrt{k − 1} \\ 0 & otherwise \end{cases}  (12.1)

Fix a and b, and consider the N eigenvalues λ_{j}. Count

\frac{#\{j : λ_{j} ∈ [a, b]\}}{N}.  (12.2)

Then Claim 12.1.1.

lim_{N→∞} \frac{#\{j : λ_{j} ∈ [a, b]\}}{N} = μ_{k}([a, b]).  (12.3)

We have

(ρ_{k}(2n))^{\frac{1}{2n}} → 2\sqrt{k − 1}.  (12.4)

12.2 Generating Functions on k-Regular Trees

12.2.1 R(z)

Fix k. The generating function R(z) is

R(z) = ∑_{n=0}^{∞} r_{n}z^{n},  (12.5)

where

r_{n} = \frac{ρ(n)}{k^{n}} = Probability that we return to v in n steps.  (12.6)

For |z| < 1, the series expansion for R(z) converges.

Let q_{n} be the probability of starting at v and ending at v for the first time (after n steps).

12.2.2 Q(z)

Define

Q(z) = ∑_{n=0}^{∞} q_{n}z^{n}.  (12.7)

Exercise 12.2.1. Prove

R(z) = \frac{1}{1 − Q(z)}.  (12.8)

12.2.3 T (z)

Define

T (z) = ∑_{n=0}^{∞} t_{n}z^{n},  (12.9)

where for w adjacent to v, t_{n} is the probability of going from w to v in n-steps for the first time.

Further, let t_{w,v}(n) be the probability of going from w to v in n-steps first time and d(w, v) = m ≥ 1. Remember that d(w, v) is the distance from w to v.

Exercise 12.2.2. Prove

1. Q(z) = zT (z).

2. ∑_{n=0}^{∞} t_{w,v}(n)z^{n} = (T (z))^{n}.

Exercise 12.2.3. Prove

T (z) = \frac{z}{k} + \frac{k − 1}{k}zT^{2}(z).  (12.10)

Note that this explicitly gives us T (z) by application of the quadratic formula:

T (z) = \frac{1 ± \sqrt{1 − 4(\frac{k−1}{k}z)\frac{z}{k}}}{2\frac{k−1}{k}z} .  (12.11)

Now that we have T (z) we have Q(z), from which we get R(z). T (z) will have a square-root – it will be an algebraic function of z.

12.3 Recovering the Measure f(x) from R(z)

We have

R(z) = ∑_{n=0}^{∞} r_{n}z^{n}.  (12.12)

As we know T (z), we know R(z), hence we know the numbers r_{n}.

Now,

r_{n} = ∫_{∞}^{∞} x^{n}f(x)dx = ∫_{−k}^{k} x^{n}f(x)dx.  (12.13)

How do we recover f(x) given the numbers r_{n}? We’ve now normalized the eigenvalues to lie in [−1, 1].

R(z) = ∑_{n=0}^{∞} ( ∫_{−1}^{1} x^{n}f(x)dx)

= ∫_{−1}^{1} ( ∑_{n=0}^{∞}(xz)^{n})f(x)dx

= ∫_{−1}^{1} \frac{f(x)}{1 − xz} dx
B(z) = \frac{1}{z}R(\frac{1}{z}) = ∫_{−1}^{1} \frac{f(x)}{z − x}dx.  (12.14)

Suppose we know the LHS. Can we recover f(x)? If z ∈ [−1, 1], the function will have a singularity. Thus, if f(x) is a nice function, we do not expect to be able to make sense of the above relation if z ∈ [−1, 1]. We will, however, consider z close to the interval [−1, 1].

Let z = ξ + iy, ξ ∈ [−1, 1], y > 0. Later we will take z = ξ − iy. Look at

B(ξ + iy) − B(ξ − iy) = ∫_{−1}^{1} f(x)[\frac{1}{ξ + iy − x} − \frac{1}{ξ − iy − x}]dx

= 2i ∫_{−1}^{1} \frac{yf(x)}{(ξ − x)^{2} + y^{2}} dx.  (12.15)

We will study the above as y → 0.

12.3.1 Poisson Kernel

Recall ξ ∈ [−1, 1], f(x) fixed, we are integrating f(x) against the Poisson Kernel

\frac{y}{(ξ − x)^{2} + y^{2}} .  (12.16)

As y → 0, the above looks singular at x = ξ. At x = ξ, the kernel has height \frac{1}{y}, which is quite large. If x = ξ + ϵ, then as y → 0, the kernel goes to 0 very rapidly. Basically, as y → 0, the kernel becomes a higher, thinner spike centered at ξ. Now

∫_{∞}^{∞} \frac{y}{(x − ξ)^{2} + y^{2}} dx = ∫_{−∞}^{∞} \frac{y}{t^{2} + y^{2}} dt
= ∫_{−∞}^{∞} \frac{y^{2}}{y^{2}η^{2} + y^{2}} dη, from \frac{t}{y} = η
= ∫_{−∞}^{∞} \frac{1}{1 + η^{2}} dη
= π.  (12.17)

This is an approximation to the identity. Thus,

B(ξ + iy) − B(ξ − iy) → 2πif(ξ).  (12.18)

12.3.2 Cauchy Integral Formula

If you have an analytic function f(z) and γ is a curve enclosing z then

\frac{1}{2πi} ∫_{γ} \frac{f(ζ)}{z − ζ} dζ  (12.19)

In our case above, we cannot apply Cauchy’s Integral Formula, as our function f(x) is not analytic. It is compactly supported, and no non-zero analytic function is compactly supported.

Call this permutation ϕ:

ϕ : 𝔽_{p}^{∗} → 𝔽_{p}^{∗},  (12.20)

where ϕ^{2} is the identity.

Question 12.3.1. Does ϕ behave like a random permutation?

12.4 Third Problem

12.4.1 Introduction

Let p be a large prime, and consider the map

x ↦ x^{−1} mod p, x ≠ 0.  (12.21)

This is a map from 𝔽_{p} → 𝔽_{p}. The map is not completely random, as

1 ↦ 1
2 ↦ \frac{p + 1}{2}
p − 1 ↦ p − 1.  (12.22)

The map which sends x → x^{−1} is a permutation of 𝔽_{p}^{∗}. It is not a completely arbitrary permutation, as it pairs x with x^{−1} (away from a few very special x’s, such as p − 1 and 1).

Thus, this permutation is a product of transpositions.

12.4.2 Character Sums

Let

1 ≤ A ≤ B ≤ p, B − A large.  (12.23)

Let

\overline{m} be the inverse of m mod p  (12.24)

IE, m\overline{m} ≡ 1 mod p. Let

e(z) = 2^{2πiz}.  (12.25)

For ν ∈ ℤ/pℤ, consider

S = ∑_{A≤m≤B} e(\frac{\overline{m}ν}{p} ).  (12.26)

These sums will measure how equidistributed or random the map m → \overline{m} is.

Exercise 12.4.1. Prove the trivial bound for |S|:

|S| ≤ B − A.  (12.27)

Let N = B − A + 1. By the Central Limit Theorem, with high probability if we add N random numbers of modulus one we expect square-root cancellation. Thus, we expect (if the inverse map is random) that |S| ≈ \sqrt{N}.

12.4.3 Completing the Square

Let a, b ∈ ℤ, and consider the Kloosterman Sum

Kl(a, b, p) = ∑_{{x mod p \atop x≠0}} e(\frac{ax + b\overline{x}}{p}).  (12.28)

How large can the Kloosterman Sum be? If a = b = 0, then trivially Kl(0, 0, p) = p − 1.

If b = 0 and a ≠ 0 (or, by symmetry, the other way around) then

Kl(a, b, p) = ∑_{{x mod p \atop x≠0}} e(\frac{ax}{p})

= ∑_{{y mod p \atop y≠0}} e(\frac{y}{p})

= ∑_{y=0}^{p−1} e(\frac{y}{p}) − 1 = −1.  (12.29)

Exercise 12.4.2. Prove ∑_{y=0}^{p−1} e(\frac{y}{p}) = 0. Hint: Let T be this sum. Then show

e(\frac{1}{p})T = T; thus T = 0.

12.4.4 Weil’s Bound

Let a ≢ 0 mod p. Then

|Kl(a, b, p)| ≤ 2\sqrt{p}.  (12.30)

This is a very deep result.

How big is

∑_{a mod p} |Kl(a, 1, p)|^{2}  (12.31)

If we believe Weil’s bound, each term is of size at most 2\sqrt{p}, we square, then sum p terms. Thus, we expect a size of at most 4p^{2}. We will show on average that Weil’s bound is correct.

∑_{a mod p} |Kl(a, 1, p)|^{2} = ∑_{a(p)} | ∑_{{x mod p \atop x≠0}} e(\frac{\overline{x} + ax}{p})|

= ∑_{a(p)} ∑_{{x_{1},x_{2} mod p \atop x_{1},x_{2}≠0}} e(\frac{\overline{x}_{1} − \overline{x}_{2} + a(x_{1} − x_{2})}{p})

= ∑_{{x_{1},x_{2} mod p \atop x_{1},x_{2}≠0}} e(\frac{\overline{x}_{1} − \overline{x}_{2}}{p} ) ∑_{a(p)} e(\frac{a(x_{1} − x_{2})}{p})
= (p − 1)p,  (12.32)

where the last line follows from the a-sum vanishes unless x_{1} = x_{2}, which then collapses the sums. There are p − 1 ways x_{1} = x_{2}, and when this occurs, the a-sum gives p.

Exercise 12.4.3. Consider

∑_{a(p)} |Kl(a, 1, p)|^{4}.  (12.33)

Above there are p-terms, each term of size (2\sqrt{p})^{4} = 16p^{2}. Thus, show the sum is at most 16p^{3}. You will find cp^{3} for some c independent of p.

By looking at one term, as every summand is positive, we find

|Kl(a, 1, p)|^{4} ≤ cp^{3}.  (12.34)

Thus, taking the fourth-root yields

|Kl(a, 1, p)| ≤ c^{\frac{1}{4}} p^{\frac{3}{4}} .  (12.35)

12.4.5 Fourier Expansion of Sums

Define the indicator function

I(y) = \begin{cases}1 & A ≤ y ≤ A + N \\ 0 & otherwise \end{cases}  (12.36)

Consider

S = ∑_{A≤x≤A+N} e(\frac{ν\overline{x}}{p})
= ∑_{x(p)} e(\frac{ν\overline{x}}{p})I(x).  (12.37)

We want to write I(x) in terms of its Fourier Coefficients

Î(m) = ∫_{0}^{1} e(−mt)I(t)dt.  (12.38)

Then

I(y) = ∑_{m=−∞}^{∞} Î(m)e(\frac{my}{p} ).  (12.39)

12.4.6 Brief Review of Fourier Series

Consider the unit interval [0, 1]. Define

ϕ_{m}(x) = e(mx).  (12.40)

Then (if our function is sufficiently nice)

f(x) = ∑_{m∈ℤ} f̂(m)e(mx),  (12.41)

where

f̂(m) = ∫_{0}^{1} f(x)e(−mx)dx.  (12.42)

12.5 Fourier Analysis and the Equi-Distribution of {nα}

12.5.1 Inner Product of Functions

We define the exponential function by means of the series

e^{x} = ∑_{n=0}^{∞} \frac{x^{n}}{n!} ,  (12.43)

which converges everywhere. Given the Taylor series expansion of sin x and cos x, we can verify the identity

e^{ix} = cos x + i sin x.  (12.44)

Exercise 12.5.1. Prove e^{x} converges for all x ∈ ℝ (even better, for all x ∈ ℂ. Show the series for e^{x} also equals

lim_{n→∞} (1 + \frac{x}{n})^{n},  (12.45)

which you may remember from compound interest problems.

Exercise 12.5.2. Prove, using the series definition, that e^{x+y} = e^{x}e^{y}. Use this fact to calculate the derivative of e^{x}. If instead you try to differentiate the series directly, you must justify the derivative of the infinite sum is the infinite sum of the derivatives.

Remember the definition of inner or dot product: for two vectors v⃗ = (v_{1}, ⋯, v_{n}), w⃗ = (w_{1}, ⋯, w_{n}), we take the inner product v⃗ · w⃗ (also denoted ⟨v, w⟩) to mean

v⃗ · w⃗ = ⟨v, w⟩ = ∑_{i} v_{i}w̄_{i}.  (12.46)

Further, the length of a vector v is

|v| = ⟨v, v⟩.  (12.47)

We generalize this for functions. For definiteness, assume f and g are functions from [0, 1] to ℂ. Divide the interval [0, 1] into n equal pieces. Then we can represent the functions by

f(x) ⟷ (f(0), f(\frac{1}{n}), …, f(\frac{n − 1}{n})),  (12.48)

and similarly for g. Call these vectors f_{n} and g_{n}. As before, we consider

⟨f_{n}, g_{n}⟩ = ∑_{i=0}^{n−1} f(\frac{i}{n}) · ḡ(\frac{i}{n}).  (12.49)

In general, as we continue to divide the interval (n → ∞), the above sum diverges. For example, if f and g are identically 1, the above sum is n.

There is a natural rescaling: we multiply each term in the sum by \frac{1}{n}, the size of the sub-interval. Note for the constant function, the sum is now independent of n.

Thus, for good f and g we are led to

⟨f, g⟩ = lim_{n→∞} ∑_{i=0}^{n−1} f(\frac{i}{n}) · ḡ(\frac{i}{n})\frac{1}{n} = ∫_{0}^{1} f(x)\overline{g(x)}dx.  (12.50)

The last result follows by Riemann Integration.

Definition 12.5.3. We say two continuous functions on [0, 1] are orthogonal (or perpendicular) if their dot product equals zero.

Exercise 12.5.4. Prove x^{n} and x^{m} are not perpendicular on [0, 1] for n ≠ m.

We will see that the exponential function behaves very nicely under the inner product. Define

e_{n}(x) = e^{2πinx} for n ∈ ℤ.  (12.51)

Then a straightforward calculation shows

⟨e_{n}(x), e_{m}(x)⟩ = \begin{cases}1 & if n = m \\ 0 & otherwise. \end{cases}  (12.52)

Thus e_{0}(x), e_{1}(x), e_{2}(x), ⋯ are an orthogonal set of functions, which means they are pairwise perpendicular. As each function has length 1, we say the functions e_{n}(x) are an orthonormal set of functions.

Exercise 12.5.5. Prove ⟨e_{n}(x), e_{m}(x)⟩ is 1 if n = m and 0 otherwise.

12.5.2 Fourier Series and {nα}

Fourier Series

Let f be continuous and periodic on ℝ with period one. Define the nth Fourier coefficient f̂(n) of f to be

f̂(n) = a_{n} = ⟨f(x), e_{n}(x)⟩ = ∫_{0}^{1} f(x)e^{−2πinx}dx.  (12.53)

Returning to the intuition of ℝ^{m}, we can think of the e_{n}(x)’s as an infinite set of perpendicular directions. The above is simply the projection of f in the direction of e_{n}(x). Exercise 12.5.6. Show

⟨f(x) − f̂(n)e_{n}(x), e_{n}(x)⟩ = 0.  (12.54)

This agrees with our intuition, namely, that if you remove the projection in a certain direction, what is left is perpendicular to that direction.

The N^{th} partial Fourier series of f is

s_{N}(x) = ∑_{n=−N}^{N} f̂(n)e_{n}(x).  (12.55)

Exercise 12.5.7. Prove 1. ⟨f(x) − s_{N}(x), e_{n}(x)⟩ = 0 if |n| ≤ N.

2. |f̂(n)| ≤ ∫_{0}^{1} |f(x)|dx.

3. If ⟨f, f⟩ < ∞, then ∑_{n=−∞}^{∞} |f̂(n)|^{2} ≤ ⟨f, f⟩.

4. If ⟨f, f⟩ < ∞, then lim_{|n|→∞} f̂(n) = 0.

As ⟨f(x) − s_{N}(x), e_{n}(x)⟩ = 0 if |n| ≤ N, we might think that we just have to let N go to infinity to obtain a series s_{∞} such that

⟨f(x) − s_{∞}(x), e_{n}(x)⟩ = 0.  (12.56)

Assume that for a periodic function g(x) to be orthogonal to e_{n}(x) for every n it must be zero for every x. Then f(x) − s_{∞}(x) = 0, and hence f = s_{∞}. Voilá – an expression for f as a sum of exponentials! Be careful, however. We have just glossed over the two central issues – completeness and, even worse, convergence. We will now see a way of avoiding some of our problems.

Weighted partial sums

Define

D_{N}(x) = ∑_{n=−N}^{N} e_{n}(x) = \frac{sin((2N + 1)πx)}{sin πx},

F_{N}(x) = \frac{sin^{2}(Nπx)}{N sin^{2} πx} = \frac{1}{N} ∑_{n=0}^{N−1} D_{n}(x).

(12.57)

Here F stands for Féjer, D for Dirichlet. In general, functions which we are interested in taking their inner product against f are called kernels; thus, the Dirichlet kernel, the Féjer kernel, etc.

Note that, no matter what N is, F_{N}(x) is positive for all x.

We say that a sequence f_{1}(x), f_{2}(x), f_{3}(x), … of functions is an approximation to the identity if

1. f_{N}(x) ≥ 0 for all x and every N;

2. ∫_{0}^{1} f_{N}(x)dx = 1;

3. lim_{N→∞} ∫_{δ}^{1−δ} f_{N}(x)dx = 0 if 0 < δ < \frac{1}{2}.

Theorem 12.5.8. The Féjer kernels F_{1}(x), F_{2}(x), F_{3}(x), … are an approximation to the identity.

Proof: The first property is immediate. The second follows from the observation that F_{N}(x) can be written as

F_{N}(x) = e_{0}(x) + \frac{N − 1}{N}(e_{−1}(x) + e_{1}(x)) + ⋯,  (12.58)

and all integrals are zero but the first, which is 1.

To prove the third property, note that F_{N}(x) ≤ \frac{1}{N sin^{2} πδ} for δ ≤ x ≤ 1 − δ. □

Let f be a continuous, periodic function on ℝ with period one. Thus, we can consider f as a function on just [0, 1], with f(0) = f(1). Define

T_{N}(x) = ∫_{0}^{1} f(y)F_{N}(x − y)dy.  (12.59)

Recall the following definition and theorem:

Definition 12.5.9 (Uniform Continuity). A continuous function is uniformly continuous if given an ϵ > 0, there exists a δ > 0 such that |x − y| < δ implies |f(x) − f(y)| < ϵ. Note that the same δ works for all points.

Theorem 12.5.10. Any continuous function on a closed, compact interval is uniformly continuous.

Exercise 12.5.11. Show x^{n} is uniformly continuous on [a, b] for −∞ < a < b < ∞.

Theorem 12.5.12. Given ϵ > 0, there is an N such that

|f(x) − T_{N}(x)| ≤ ϵ  (12.60)

for every x ∈ [0, 1].

Proof. For any positive N,

T_{N}(x) − f(x) = ∫_{0}^{1} f(x − y)F_{N}(y)dy − f(x) · 1

= ∫_{0}^{1} f(x − y)F_{N}(y)dy − ∫_{0}^{1} f(x)F_{n}(y)dy (property 2 of F_{N})

= ∫_{0}^{δ} (f(x − y) − f(x))F_{N}(y)dy
+ ∫_{δ}^{1−δ} (f(x − y) − f(x))F_{N}(y)dy
+ ∫_{1−δ}^{1} (f(x − y) − f(x))F_{N}(y)dy.

(12.61)

Let δ ∈ (0, 1/2). Then, using the fact that the F_{N}(x)’s are an approximation to the identity, we find

|∫_{δ}^{1−δ} (f(x − y) − f(x))F_{N}(y)dy| ≤ 2 max |f(x)| · ∫_{δ}^{1−δ} F_{N}(y)dy. (12.62)

Since

lim_{N→∞} ∫_{δ}^{1−δ} F_{N}(y)dy = 0,  (12.63)

we obtain

lim_{N→∞} ∫_{δ}^{1−δ}(f(x − y) − f(x))F_{N}(y)dy = 0.  (12.64)

Thus, by choosing N large enough (where large depends on δ), we can insure that this piece is at most \frac{ϵ}{3}.

It remains to estimate what happens near zero. Since f is continuous and [0, 1] is compact, f is uniformly continuous. Thus, we can choose δ small enough that |f(x − y) − f(x)| < \frac{ϵ}{3} for any x and any positive y < δ. Then

|∫_{0}^{δ} (f(x − y) − f(x))F_{N}(y)dy| ≤ ∫_{0}^{δ} \frac{ϵ}{3}F_{N}(y)dy ≤ \frac{ϵ}{3} ∫_{0}^{1} F_{N}(y)dy ≤ \frac{ϵ}{3}.

(12.65)

Similarly

|∫_{1−δ}^{1} (f(x − y) − f(x))F_{N}(y)dy| ≤ \frac{ϵ}{3}.  (12.66)

Therefore

|T_{N}(x) − f(x)| ≤ ϵ  (12.67)

for all N sufficiently large.

Definition 12.5.13 (Trigonometric Polynomials). Any finite linear combination of the functions e_{n}(x) is called a trigonometric polynomial.

From Theorem 12.5.12 we immediately get the Stone-Weierstrass theorem:

Theorem 12.5.14 (Stone-Weierstrass). Any continuous period function can be uniformly approximated by trigonometric polynomials.

12.5.3 Equidistribution

We say that a sequence {x_{n}}, x_{n} ∈ [0, 1] is equidistributed if

lim_{N→∞} \frac{1}{2N + 1}#\{n : |n| ≤ N, x_{n} ∈ (a, b)\} = b − a  (12.68)

for all (a, b) ⊂ [0, 1].

Theorem 12.5.15 (Weyl). Let α be an irrational number in [0, 1]. Let x_{n} = {nα}, where {y} denotes the fractional part of y. Then the sequence {x_{n}} is equidistributed.

Proof. We will estimate \frac{1}{2N+1} ∑_{−N}^{N} χ_{(a,b)}(x_{n}) as N → ∞, where χ_{(a,b)} is the function taking the value 0 outside (a, b) and 1 inside (a, b). We call χ_{(a,b)} the characteristic function of the interval (a, b).

Thus, we must show

lim_{N→∞} \frac{1}{2N + 1} ∑_{n=−N}^{N} χ_{(a,b)}(x_{n}) = b − a.  (12.69)

Consider e_{k}(x) = e^{2πikx}. Since x_{n} = {nα} = nα − [nα] and e_{k}(x) = e_{k}(x + m) for every integer m,

e_{k}(x_{n}) = e^{2πiknα}.  (12.70)

Hence

\begin{aligned}\frac{1}{2N + 1} ∑_{n=−N}^{N} e_{k}(x_{n}) & = \frac{1}{2N + 1} ∑_{n=−N}^{N} e_{k}(nα) \\ & = \frac{1}{2N + 1} ∑_{n=−N}^{N}(e^{2πikα})^{n} \\ & = \begin{cases}1 & if k = 0 \\ \frac{1}{2N+1} \frac{e_{k}(−Nα)−e_{k}((N+1)α)}{1−e_{k}(α)} & if k > 0. \end{cases} \end{aligned}  (12.71)

Now for a fixed irrational α, |1 − e_{k}(α)| > 0. Therefore if k ≠ 0:

lim_{N→∞} \frac{1}{2N + 1} \frac{e_{k}(−Nα) − e_{k}((N + 1)α)}{1 − e_{k}(α)} = 0.  (12.72)

Let P (x) = ∑_{k} a_{k}e_{k}(x) be a finite sum (ie, P (x) is a trigonometric polynomial). By possibly adding some zero coefficients, we can write P (x) as a sum over a symmetric range: P (x) = ∑_{k=−K}^{K} a_{k}a_{k}(x).

Exercise 12.5.16. Show ∫_{0}^{1} P (x)dx = a_{0}.

By the above arguments, we have shown that for any (finite) trigonometric polynomial P (x):

lim_{N→∞} \frac{1}{2N + 1} ∑_{n=−N}^{N} P (x_{n}) → a_{0} = ∫_{0}^{1} P (x)dx.  (12.73)

Consider two approximations to the characteristic function χ_{(a,b)}:

1. f_{1m}: f_{1m}(x) = 1 if a + \frac{1}{m} ≤ x ≤ b − \frac{1}{m}, drops linearly to 0 at a and b, and is zero elsewhere.

2. f_{2m}: f_{1m}(x) = 1 if a ≤ x ≤ b, drops linearly to 0 at a − \frac{1}{m} and b + \frac{1}{m}, and is zero elsewhere.

Note there are trivial modifications if a = 0 or b = 1. Clearly

f_{1m}(x) ≤ χ_{(a,b)}(x) ≤ f_{2m}(x).  (12.74)

Therefore

\frac{1}{2N + 1} ∑_{n=−N}^{N} f_{1m}(x_{n}) ≤ \frac{1}{2N + 1} ∑_{n=−N}^{N} χ_{(a,b)}(x_{n}) ≤ \frac{1}{2N + 1} ∑_{n=−N}^{N} f_{2m}(x_{n}).

(12.75)

By Theorem 12.5.12, for each m, given ϵ > 0 we can find trigonometric polynomials P_{1m}(x) and P_{2m}(x) such that |P_{1m}(x) − f_{1m}(x)| < ϵ and |P_{2m}(x) − f_{2m}(x)| < ϵ.

As f_{1m} and f_{2m} are continuous functions, we can replace

\frac{1}{2N + 1} ∑_{n=−N}^{N} f_{im}(x_{n}) with \frac{1}{2N + 1} ∑_{n=−N}^{N} P_{im}(x_{n})  (12.76)

at a cost of at most ϵ. As N → ∞,

\frac{1}{2N + 1} ∑_{n=−N}^{N} P_{im}(x_{n}) → ∫_{0}^{1} P_{im}(x)dx.  (12.77)

But ∫_{0}^{1} P_{1m}(x)dx = (b − a) − \frac{1}{m} and ∫_{0}^{1} P_{2m}(x)dx = (b − a) + \frac{1}{m}. Therefore, given m and ϵ, we can choose N large enough so that

(b − a) − \frac{1}{m} − ϵ ≤ \frac{1}{2N + 1} ∑_{n=−N}^{N} χ_{(a,b)}(x_{n}) ≤ (b − a) + \frac{1}{m} + ϵ. (12.78)

Letting m tend to ∞ and ϵ tend to 0, we see \frac{1}{2N+1} ∑_{n=−N}^{N} χ_{(a,b)}(x_{n}) → b − a. Exercise 12.5.17. Rigorously do the necessary book-keeping to prove the previous theorem. Exercise 12.5.18. Prove

1. If α ∈ ℚ, then {nα} is periodic.

2. If α ∉ ℚ, then no two {nα} are equal.

Liouville’s Theorem Constructing Transcendentals

1. We prove Liouville’s Theorem for the order of approximation by rationals of real algebraic numbers.

2. We construct several transcendental numbers.

3. We define Poissonian Behaviour, and study the spacings between the ordered fractional parts of {n^{k}α}.

Lecture by Steven J. Miller; notes for the first two by Steven J. Miller and Florin Spinu; notes for the third by Steven J. Miller.

13.1 Review of Approximating by Rationals

Definition 13.1.1 (Approximated by rationals to order n). A real number x is approximated by rationals to order n if there exist a constant k(x) (possibly depending on x) such that there are infinitely many rational \frac{p}{q} with

|x − \frac{p}{q}| < \frac{k(x)}{q^{n}}.  (13.1)

Recall that Dirichlet’s Box Principle gaves us:

|x − \frac{p}{q}| < \frac{1}{q^{2}}  (13.2)

for infintely many fractions \frac{p}{q}. This was proved by choosing a large parameter Q, and considering the Q + 1 fractionary parts {qx} ∈ [0, 1) for q ∈ {0, …, Q}. The box principle ensures us that there must be two different q’s, say:

0 ≤ q_{1} < q_{2} ≤ Q  (13.3)

such that both \{q_{1}x\} and \{q_{2}x\} belong to the same interval [\frac{a}{Q}, \frac{a+1}{Q}), for some 0 ≤ a ≤ Q − 1. Note that there are exactly Q such intervals partitioning [0, 1), and Q + 1 fractionary parts! Now, the length of such an interval is \frac{1}{Q} so we get

|\{q_{2}x\} − \{q_{1}x\}| < \frac{1}{Q}.  (13.4)

There exist integers p_{1} and p_{2} such that

{q_{1}x} = q_{1}x − p, {q_{2}x} = q_{2}x − p.  (13.5)

Letting p = p_{2} − p_{1} we find

|(q_{2} − q_{1})x − p| ≤ \frac{1}{Q}  (13.6)

Let q = q_{2} − q_{1}, so 1 ≤ q ≤ Q, and the previous equation can be rewriten as

|x − \frac{p}{q}| < \frac{1}{qQ} ≤ \frac{1}{q^{2}}  (13.7)

Now, letting Q → ∞, we get an infinite collection of rational fractions \frac{p}{q} satisfying the above equation. If this collection contains only finitely many distinct fractions, then one of these fractions, say \frac{p_{0}}{q_{0}} , would occur for infintely many choices Q_{k} of Q, thus giving us:

|x − \frac{p_{0}}{q_{0}} | < \frac{1}{qQ_{k}} → 0,  (13.8)

as k → ∞. This implies that x = \frac{p_{0}}{q_{0}} ∈ ℚ. So, unles x is a rational number, we can find infinitely many distinct rational numbers \frac{p}{q} satisfying Equation 13.7. This means that any real, irrational number can be approximated to order n = 2 by rational numbers.

13.2 Liouville’s Theorem

Theorem 13.2.1 (Liouville’s Theorem). Let x be a real algebraic number of degree n. Then x is approximated by rationals to order at most n.

Proof. Let

f(X) = a_{n}X^{n} + ⋯ a_{1}X + a_{0}  (13.9)

be the polynomial with integer coefficients of smallest degree (minimal polynomial) such that x satisfies

f(x) = 0.  (13.10)

Note that deg x = deg f and the condition of minimality implies that f(X) is irreducible over ℤ. Further, a well known result from algebra states that a polynomial irreducible over ℤ is also irreducible over ℚ.

In particular, as f(X) is irreducible over ℚ, f(X) does not have any rational roots. If it did, then f(X) would be divisible by a linear polynomial (X − \frac{a}{b}). Let G(X) = \frac{f(X)}{X− \frac{a}{b}}. Clear denominators (multiply throughout by b), and let g(X) = bG(X). Then deg g = deg f − 1, and g(x) = 0. This contradicts the minimality of f (we choose f to be a polynomial of smallest degree such that f(x) = 0). Therefore, f is non-zero at every rational.

Let

M = sup_{|z−x|<1} |f′(z)|.  (13.11)

Let now \frac{p}{q} be a rational such that |x − \frac{p}{q} | < 1. The Mean Value Theorem gives us that

|f (\frac{p}{q}) − f(x)| = |f′(c) (x − \frac{p}{q})| ≤ M |x − \frac{p}{q}|  (13.12)

where c is some real number between x and \frac{p}{q}; |c − x| < 1 for \frac{p}{q} moderately close to x.

Now we use the fact that f(X) does not have any rational roots:

0 ≠ f (\frac{p}{q}) = a_{n} (\frac{p}{q})^{n} + ⋯ + a_{0} = \frac{a_{n}p^{n} + ⋯ a_{1}p^{n−1}q + a_{0}q^{n}}{q^{n}} (13.13)

The numerator of the last term is a nonzero integer, hence it has absolute value at least 1. Since we also know that f(x) = 0 it follows that

|f (\frac{p}{q}) − f(x)| = |f (\frac{p}{q})| = \frac{|a_{n}p^{n} + ⋯ a_{1}p^{n−1}q + a_{0}q^{n}|}{q^{n}} ≥ \frac{1}{q^{n}} . (13.14)

Combining the equations 13.12 and 13.14, we get:

\frac{1}{q^{n}} ≤ M |x − \frac{p}{q}| ⇒ \frac{1}{Mq^{n}} ≤ |x − \frac{p}{q}|  (13.15)

whenever |x − \frac{p}{q}| < 1. This last equation shows us that x can be approximated by rationals to order at most n. For assume it was otherwise, namely that x can be approximated to order n + ϵ. Then we would have an infinite sequence of distinct rational numbers \{\frac{p_{i}}{q_{i}} \}_{i≥1} and a constant k(x) depending only on x such that

|x − \frac{p_{i}}{q_{i}} | < \frac{k(x)}{q_{i}^{n+ϵ}}.  (13.16)

Since the numbers \frac{p_{i}}{q_{i}} converge to x we can assume that they already are in the interval (x − 1, x + 1). Hence they also satisfy Equation 13.15:

\frac{1}{q_{i}^{n}} ≤ M |x − \frac{p_{i}}{q_{i}} | .  (13.17)

Combining the last two equations we get

\frac{1}{Mq_{i}^{n}} ≤ |x − \frac{p_{i}}{q_{i}} | < \frac{k(x)}{q_{i}^{n+ϵ}},  (13.18)

hence

q_{i}^{ϵ} < M  (13.19)

and this is clearly impossible for arbitrarily large q since ϵ > 0 and q_{i} → ∞.

Exercise 13.2.2. Justify the fact that if \{\frac{p_{i}}{q_{i}} \}_{i≥1} is a rational approximation to order n ≥ 1 of x, then q_{i} → ∞.

Remark 13.2.3. So far we have seen that the order to which an algebraic number can be approximated by rationals is bounded by its degree. Hence if a real, irrational number α ∉ ℚ can be approximated by rationals to an arbitrary large order, then α must be transcendental! This provides us with a recipe for constructing transcendental numbers.

13.3 Constructing Transcendental Numbers

13.3.1 ∑_{m} 10^{−m!}

The following construction of transcendental numbers is due to Liouville.

Theorem 13.3.1. The number

x = ∑_{m=1}^{∞} \frac{1}{10^{m!}}  (13.20)

is transcendental.

Proof. The series defining x is convergent, since it is dominated by the geometric series ∑ \frac{1}{10^{m}}. In fact, the series converges very rapidly and it is this high rate of convergence that will yield x is transcendental.

Fix N large, and let n > N. Write

\frac{p_{n}}{q_{n}} = ∑_{m=1}^{n} \frac{1}{10^{m!}}  (13.21)

with p_{n}, q_{n} > 0 and (p_{n}, q_{n}) = 1. Then \{\frac{p_{n}}{q_{n}} \}_{n≥1} is a monotone increasing sequence converging to x. In particular, all these rational numbers are distinct. Not also that q_{n} must divide 10^{n!}, which implies

q_{n} ≤ 10^{n!}.  (13.22)

Using this, we get

0 < x − \frac{p_{n}}{q_{n}} = ∑_{m>n} \frac{1}{10^{m!}} = \frac{1}{10^{(n+1)!}} (1 + \frac{1}{10^{n+2}} + \frac{1}{10^{(n+2)(n+3)}} + ⋯)
< \frac{2}{10^{(n+1)!}} = \frac{2}{(10^{n!})^{n+1}}
< \frac{2}{q_{n}^{n+1}} ≤ \frac{2}{q_{n}^{N}} .  (13.23)

This gives an approximation by rationals of order N of x. Since N can be chosen arbitrarily large, this implies that x can be approximated by rationals to arbitrary order. We can conclude, in view of our precious remark 13.2.3 that x is transcendental.

13.3.2 [10^{1!}, 10^{2!}, …]

Theorem 13.3.2. The number

y = [10^{1!}, 10^{2!}, …]  (13.24)

is transcendental.

Proof. Let \frac{p_{n}}{q_{n}} be the continued fraction of [10^{1!} ⋯ 10^{n!}]. Then

|y − \frac{p_{n}}{q_{n}} | = \frac{1}{q_{n}q′_{n+1}} = \frac{1}{q_{n}(a′_{n+1}q_{n} + q_{n−1})}
< \frac{1}{a_{n+1}} = \frac{1}{10^{(n+1)!}} .  (13.25)

Since q_{k} = a_{n}q_{k−1} + q_{n−2}, it implies that q_{k} > q_{k−1} Also, q_{k+1} = a_{k+1}q_{n} + q_{k−1}, so we get

\frac{q_{k+1}}{q_{k}} = a_{k+1} + \frac{q_{k−1}}{q_{k}} < a_{k+1} + 1.  (13.26)

Hence writing this inequality for k = 1, ⋯, n − 1 we obtain

q_{n} = q_{1} \frac{q_{2}}{q_{1}} \frac{q_{3}}{q_{2}} ⋯ \frac{q_{n}}{q_{n−1}} < (a_{1} + 1)(a_{2} + 1) ⋯ (a_{n} + 1)
= (1 + \frac{1}{a_{1}} ) ⋯ (1 + \frac{1}{a_{n}})a_{1} ⋯ a_{n}
< 2^{n}a_{1} ⋯ a_{n} = 2^{n}10^{1!+···+n!}
< 10^{2n!} = a_{n}^{2}  (13.27)

Combining equations 13.25 and 13.27 we get:

|y − \frac{p_{n}}{q_{n}} | < \frac{1}{a_{n+1}} = \frac{1}{a_{n}^{n+1}}
< (\frac{1}{a_{n}^{2}} )^{\frac{n}{2}} < (\frac{1}{q_{n}^{2}} )^{\frac{n}{2}}
= \frac{1}{q_{n}^{n/2}} .  (13.28)

In this way we get, just as in the previous theorem, an approximation of y by rationals to arbitrary order. This proves that y is transcendental.

13.3.3 Buffon’s Needle and π

Consider a collection of infinitely long parallel lines in the plane, where the spacing between any two adjacent lines is d. Let the lines be located at x = 0, ±d, ±2d, . . .. Consider a rod of length l, where for convenience we assume l < d.

If we were to randomly throw the rod on the plane, what is the probability it hits a line? This question was first asked by Buffon in 1733.

Because of the vertical symmetry, we may assume the center of the rod lies on the line x = 0, as shifting the rod (without rotating it) up or down will not alter the number of intersections. By the horizontal symmetry, we may assume −\frac{d}{2} ≤ x < \frac{d}{2}. We posit that all values of x are equally likely. As x is continuous distributed, we may add in x = \frac{d}{2} without changing the probability. The probability density function of x is \frac{dx}{d}.

Let θ be the angle the rod makes with the x-axis. As each angle is equally likely, the probability density function of θ is \frac{dθ}{2π}.

We assume that x and θ are chosen independently. Thus, the probability density for (x, θ) is \frac{dxdθ}{d·2π}.

The projection of the rod (making an angle of θ with the x-axis) along the x-axis is l · | cos θ|. If |x| ≤ l · | cos θ|, then the rod hits exactly one vertical line exactly once; if x > l · | cos θ|, the rod does not hit a vertical line. Note that if l > d, a rod could hit multiple lines, making the arguments more involved.

Thus, the probability a rod hits a line is

p = ∫_{θ=0}^{2π} ∫_{x=−l·| cos θ|}^{l·| cos θ|} \frac{dxdθ}{d · 2π}
= ∫_{θ=0}^{2π} \frac{l · | cos θ|}{d} \frac{dθ}{2π}
= \frac{2l}{πd}.  (13.29)

Exercise 13.3.3. Show

\frac{1}{2π} ∫_{0}^{2π} | cos θ|dθ = \frac{2}{π} .  (13.30)

Let A be the random variable which is the number of intersections of a rod of length l thrown against parallel vertical lines separated by d > l units. Then

A = \begin{cases}1 & with probability \frac{2l}{πd} \\ 0 & with probability 1 − \frac{2l}{πd} \end{cases} .  (13.31)

If we were to throw N rods independently, since the expected value of a sum is the sum of the expected values (Lemma 6.3.8), we expect to observe

N · \frac{2l}{πd}  (13.32)

intersections. Turning this around, let us throw N rods, and let I be the number of observed

intersections of the rods with the vertical lines. Then

I ≈ N · \frac{2l}{πd} → π ≈ \frac{N}{I} · \frac{2l}{d}.  (13.33)

The above is an experimental formula for π!

Poissonian Behavior and {n^{k}α}

We now define Poissonian Bevahior, and investigate the normalized spacings of the fractional parts of n^{2}α. Lecture and notes by Steven J. Miller.

14.1 Equidistribution

We say a sequence of number x_{n} ∈ [0, 1) is equidistributed if

lim_{N→∞} \frac{#\{n : 1 ≤ n ≤ N and x_{n} ∈ [a, b]\}}{N} = b − a  (14.1)

for any subinterval [a, b] of [0, 1].

Recall Weyl’s Result: If α ∉ ℚ, then the fractional parts {nα} are equidistributed. Equivalently, nα mod 1 is equidistributed.

Similarly, one can show that for any integer k, {n^{k}α} is equidistributed. See Robert Lipshitz’s paper for more details.

14.2 Point Masses and Induced Probability Measures

Recall from physics the concept of a unit point mass located at x = a. Such a point mass has no length (or, in higher dimensions, width or height), but finite mass. As mass is the integral of the density over space, a finite mass in zero volume (or zero length on the line) implies an infinite density.

We can make this more precise by the notion of an Approximation to the Identity.

Definition 14.2.1 (Approximation to the Identity). A sequence of functions g_{n}(x) is an approximation to the identity (at the origin) if

1. g_{n}(x) ≥ 0.

2. ∫ g_{n}(x)dx = 1.

3. Given ϵ, δ > 0 there exists N > 0 such that for all n > N, ∫_{|x|>δ} g_{n}(x)dx < ϵ.

We represent the limit of any such family of g_{n}(x)s by δ(x).

If f(x) is a nice function (say near the origin its Taylor Series converges) then

∫ f(x)δ(x)dx = lim_{n→∞} ∫ f(x)g_{n}(x) = f(0).  (14.2)

Exercise 14.2.2. Prove Equation 14.2.

Thus, in the limit the functions g_{n} are acting like point masses. We can consider the probability densities g_{n}(x)dx and δ(x)dx. For g_{n}(x)dx, as n → ∞, almost all the probability is concentrated in a narrower and narrower band about the origin; δ(x)dx is the limit with all the mass at one point. It is a discrete (as opposed to continuous) probability measure.

Note that δ(x − a) acts like a point mass; however, instead of having its mass concentrated at the origin, it is now concentrated at a.

Exercise 14.2.3. Let

g_{n}(x) = \begin{cases}n & if |x| ≤ \frac{1}{2n} \\ 0 & otherwise \end{cases}  (14.3)

Prove g_{n}(x) is an approximation to the identity at the origin.

Exercise 14.2.4. Let

g_{n}(x) = c \frac{\frac{1}{n}}{\frac{1}{n^{2}} + x^{2}} .  (14.4)

Find c such that the above is an approximation to the identity at the origin.

Given N point masses located at x_{1}, x_{2}, …, x_{N}, we can form a probability measure

μ_{N}(x)dx = \frac{1}{N} ∑_{n=1}^{N} δ(x − x_{n})dx.  (14.5)

Note ∫ μ_{N}(x)dx = 1, and if f(x) is a nice function,

∫ f(x)μ_{N}(x)dx = \frac{1}{N} ∑_{n=1}^{N} f(x_{n}).  (14.6)

Exercise 14.2.5. Prove Equation 14.6 for nice f(x).

Note the right hand side of Equation 14.6 looks like a Riemann sum. Or it would look like a Riemann sum if the x_{n}s were equidistributed. In general the x_{n}s will not be equidistributed, but assume for any interval [a, b] that as N → ∞, the fraction of x_{n}s (1 ≤ n ≤ N) in [a, b] goes to ∫_{a}^{b} p(x)dx for some nice function p(x):

lim_{N→∞} \frac{#\{n : 1 ≤ n ≤ N and x_{n} ∈ [a, b]\}}{N} → ∫_{a}^{b} p(x)dx.  (14.7)

In this case, if f(x) is nice (say twice differentiable, with first derivative uniformly bounded), then

∫ f(x)μ_{N}(x)dx = \frac{1}{N} ∑_{n=1}^{N} f(x_{n})

≈ ∑_{k=−∞}^{∞} f(\frac{k}{N} )\frac{#\{n : 1 ≤ n ≤ N and x_{n} ∈ [\frac{k}{N} , \frac{k+1}{N}]\}}{N}

→ ∫ f(x)p(x)dx.  (14.8)

Definition 14.2.6 (Convergence to p(x)). If the sequence of points x_{n} satisfies Equation 14.7 for some nice function p(x), we say the probability measures μ_{N}(x)dx converge to p(x)dx.

14.3 Neighbor Spacings

We now consider finer questions. Let α_{n} be a collection of points in [0, 1). We order them by size:

0 ≤ α_{σ(1)} ≤ α_{σ(2)} ≤ ⋯ ≤ α_{σ(N)},  (14.9)

where σ is a permutation of 123 ⋯ N. Note the ordering depends crucially on N. Let β_{j} = α_{σ(j)}.

We consider how the differences β_{j+1} − β_{j} are distributed. We will use a slightly different definition of distance, however.

Recall [0, 1) is equivalent to the unit circle under the map x → e^{2πix}. Thus, the numbers .999 and .001 are actually very close; however, if we used the standard definition of distance, then |.999 − .001| = .998, which is quite large. Wrapping [0, 1) on itself (identifying 0 and 1), we see that .999 and .001 are separated by .002.

Definition 14.3.1 (mod 1 distance). Let x, y ∈ [0, 1). We define the mod 1 distance from x to y, ||x − y||, by

||x − y|| = min {|x − y|, 1 − |x − y|}.  (14.10)

Exercise 14.3.2. Show that the mod 1 distance between any two numbers in [0, 1) is at most \frac{1}{2}.

In looking at spacings between the β_{j}s, we have N − 1 pairs of neighbors:

(β_{2}, β_{1}), (β_{3}, β_{2}), …, (β_{N}, β_{N−1}).  (14.11)

These pairs give rise to spacings β_{j+1} − β_{j} ∈ [0, 1).

We can also consider the pair (β_{1}, β_{N}). This gives rise to the spacing β_{1}−β_{N} ∈ [−1, 0); however, as we are studying this sequence mod 1, this is equivalent to β_{1} − β_{N} + 1 ∈ [0, 1).

Henceforth, whenever we perform any arithmetic operation, we always mean mod 1; thus, our answers always live in [0, 1)

Definition 14.3.3 (Neighbor Spacings). Given a sequence of numbers α_{n} in [0, 1), fix an N and arrange the numbers α_{n} (n ≤ N) in increasing order. Label the new sequence β_{j}; note the ordering will depend on N. Let β_{−j} = β_{N−j} and β_{N+j} = β_{j}.

1. The nearest neighbor spacings are the numbers β_{j+1} − β_{j}, j = 1 to N.

2. The k^{th}-neighbor spacings are the numbers β_{j+k} − β_{j}, j = 1 to N.

Remember to take the differences β_{j+k} − β_{j} mod 1.

Exercise 14.3.4. Let α = \sqrt{2}, and let α_{n} = \{nα\} or \{n^{2}α\}. Calculate the nearest neighbor and the next-nearest neighbor spacings in each case for N = 10.

Definition 14.3.5 (wrapped unit interval). We call [0, 1), when all arithmetic operations are done mod 1, the wrapped unit interval.

14.4 Poissonian Behavior

Let α ∉ ℚ. Fix a positive integer k, and let α_{n} = {n^{k}α}. As N → ∞, look at the ordered α_{n}s, denoted by β_{n}. How are the nearest neighbor spacings of β_{n} distributed? How does this depend on k? On α? On N?

Before discussing this problem, we consider a simpler case. Fix N, and consider N independent random variables x_{n}. Each random variable is chosen from the uniform distribution on [0, 1); thus, the probability that x_{n} ∈ [a, b) is b − a.

Let y_{n} be the x_{n}s arranged in increasing order. How do the neighbor spacings behave?

First, we need to decide what is the correct scale to use for our investigations. As we have N objects on the wrapped unit interval, we have N nearest neighbor spacings. Thus, we expect the average spacing to be \frac{1}{N}.

Definition 14.4.1 (Unfolding). Let z_{n} = Ny_{n}. The numbers z_{n} = Ny_{n} have unit mean spacing. Thus, while we expect the average spacing between adjacent y_{n}s to be \frac{1}{N} units, we expect the average spacing between adjacent z_{n}s to be 1 unit.

So, the probability of observing a spacing as large as \frac{1}{2} between adjacent y_{n}s becomes negligible as N → ∞. What we should ask is what is the probability of observing a nearest neighbor spacing of adjacent y_{n}s that is half the average spacing. In terms of the z_{n}s, this will correspond to a spacing between adjacent z_{n}s of \frac{1}{2} a unit.

14.4.1 Nearest Neighbor Spacings

By symmetry, on the wrapped unit interval the expected nearest neighbor spacing is independent of j. Explicitly, we expect β_{j+1} − β_{j} to have the same distribution as β_{i+1} − β_{i}.

What is the probability that, when we order the x_{n}s in increasing order, the next x_{n} after x_{1} is located between \frac{t}{N} and \frac{t+Δt}{N}? Let the x_{n}s in increasing order be labeled y_{1} ≤ y_{2} ≤ ⋯ ≤ y_{N}, y_{n} = x_{σ(n)}.

As we are choosing the x_{n}s independently, there are \binom{N−1}{1} choices of subscript n such that x_{n} is nearest to x_{1}. This can also be seen by symmetry, as each x_{n} is equally likely to be the first to the right of x_{1} (where, of course, .001 is just a little to the right of .999), and we have N − 1 choices left for x_{n}.

The probability that x_{n} ∈ [\frac{t}{N} , \frac{t+Δt}{N}] is \frac{Δt}{N}.

For the remaining N − 2 of the x_{n}s, each must be further than \frac{t+Δt}{N} from x_{n}. Thus, they must all lie in an interval (or possibly two intervals if we wrap around)

of length 1− \frac{t+Δt}{N}. The probability that they all lie in this region is (1− \frac{t+Δt}{N})^{N−2}.

Thus, if x_{1} = y_{l}, we want to calculate the probability that ||y_{l+1} − y_{l}|| ∈ [\frac{t}{N} , \frac{t+Δt}{N}]. This is

Prob(||y_{l+1} − y_{l}|| ∈ [\frac{t}{N} , \frac{t + Δt}{N}]) = \binom{N − 1}{1} · \frac{Δt}{N} · (1 − \frac{t + Δt}{N})^{N−2}
= (1 − \frac{1}{N} ) · (1 − \frac{t + Δt}{N})^{N−2}Δt.

(14.12)

For N enormous and Δt small,

(1 − \frac{1}{N} )  ≈ 1
(1 − \frac{t + Δt}{N})^{N−2} ≈ e^{−(t+Δt)} ≈ e^{−t}.  (14.13)

Thus

Prob(||y_{l+1} − y_{l}|| ∈ [\frac{t}{N} , \frac{t + Δt}{N}]) → e^{−t}Δt.  (14.14)

Remark 14.4.2. The above argument is infinitesimally wrong. Once we’ve located y_{l+1}, the remaining x_{n}s do not need to be more than \frac{t+Δt}{N} units to the right of x_{1} = y_{l}; they only need to be further to the right than y_{l+1}. As the incremental gain in probabilities for the locations of the remaining x_{n}s is of order Δt, these contributions will not influence the large N, small Δt limits. Thus, we ignore these effects.

To rigorously derive the limiting behavior of the nearest neighbor spacings using the above arguments, one would integrate over x_{m} ranging from \frac{t}{N} to \frac{t+Δt}{N}, and the remaining events x_{n} would be in the a segment of length 1 − x_{m}. As

|(1 − x_{m}) − (1 − \frac{t + Δt}{N})| ≤ \frac{Δt}{N},  (14.15)

this will lead to corrections of higher order in Δt, hence negligible.

We can rigorously avoid this by instead considering the following:

1. Calculate the probability that all the other x_{n}s are at least \frac{t}{N} units to the right of x_{1}. This is

p_{t} = (1 − \frac{t}{N} )^{N−1} → e^{−t}.  (14.16)

2. Calculate the probability that all the other x_{n}s are at least \frac{t+Δt}{N} units to the right of x_{1}. This is

p_{t+Δt} = (1 − \frac{t + Δt}{N})^{N−1} → e^{−(t+Δt)}.  (14.17)

3. The probability that no x_{n}s are within \frac{t}{N} units to the right of x_{1} but at least one x_{n} is between \frac{t}{N} and \frac{t+Δt}{N} units to the right is p_{t+Δt} − p_{t}:

p_{t} − p_{t+Δt} → e^{−t} − e^{−(t+Δt)}
= e^{−t}(1 − e^{−Δt})
= e^{−t}(1 − 1 + Δt + O((Δt)^{2})
→ e^{−t}Δt.  (14.18)

Definition 14.4.3 (Unfolding Spacings). If y_{l+1} −y_{l} ∈ [\frac{t}{N} , \frac{t+Δt}{N}], then N(y_{l+1} − y_{l}) ∈ [t, t + Δt]. The new spacings z_{l+1} − z_{l} have unit mean spacing. Thus, while we expect the average spacing between adjacent y_{n}s to be \frac{1}{N} units, we expect the average spacing between adjacent z_{n}s to be 1 unit.

14.4.2 k^{th} Neighbor Spacings

Similarly, one can easily analyze the distribution of the k^{th} neighbor spacings when each x_{n} is chosen independently from the uniform distribution on [0, 1).

Again, consider x_{1} = y_{l}. Now we want to calculate the probability that y_{l+k} is between \frac{t}{N} and \frac{t+Δt}{N} units to the right of y_{l}.

Therefore, we need exactly k −1 of the x_{n}s to lie between 0 and \frac{t}{N} units to the right of x_{1}, exactly one x_{n} (which will be y_{l+k}) to lie between \frac{t}{N} and \frac{t+Δt}{N} units to the right of x_{1}, and the remaining x_{n}s to lie at least \frac{t+Δt}{N} units to the right of y_{l+k}.

Remark 14.4.4. We face the same problem discussed in Remark 14.4.2; a similar argument will show that ignoring these affects will not alter the limiting behavior. Therefore, we will make these simplifications.

There are \binom{N−1}{k−1} ways to choose the x_{n}s that are at most \frac{t}{N} units to the right of x_{1}; there is then \binom{(N−1)−(k−1)}{1} ways to choose the x_{n} between \frac{t}{N} and \frac{t+Δt}{N} units to the right of x_{1}.

Thus,

Prob(||y_{l+k} − y_{l}|| ∈ [\frac{t}{N} , \frac{t + Δt}{N}]) =

= \binom{N − 1}{k − 1}(\frac{t}{N} )^{k−1} · \binom{(N − 1) − (k − 1)}{1}\frac{Δt}{N} · (1 − \frac{t + Δt}{N})^{N−(k+1)}
= \frac{(N − 1) ⋯ (N − 1 − (k − 2))}{N^{k−1}} \frac{(N − 1) − (k − 1)}{N} \frac{t^{k−1}}{(k − 1)!}(1 − \frac{t + Δt}{N})^{N−(k+1)}Δt
→ \frac{t^{k−1}}{(k − 1)!}e^{−t}Δt.  (14.19)

Again, one way to avoid the complications is to integrate over x_{m} ranging from \frac{t}{N} to \frac{t+Δt}{N}.

Or, similar to before, we can proceed more rigorously as follows:

1. Calculate the probability that exactly k − 1 of the other x_{n}s are at most \frac{t}{N} units to the right of x_{1}, and the remaining (N − 1) − (k − 1) of the x_{n}s are at least \frac{t}{N} units to the right of x_{1}. As there are \binom{N−1}{k−1} ways to choose k − 1 of the x_{n}s to be at most \frac{t}{N} units to the right of x_{1}, this probability is

p_{t} = \binom{N − 1}{k − 1}(\frac{t}{N} )^{k−1}(1 − \frac{t}{N} )^{(N−1)−(k−1)}
→ \frac{N^{k−1}}{(k − 1)!} \frac{t^{k−1}}{N^{k−1}} e^{−t}
→ \frac{t^{k−1}}{k − 1!}e^{−t}.  (14.20)

2. Calculate the probability that exactly k − 1 of the other x_{n}s are at most \frac{t}{N} units to the right of x_{1}, and the remaining (N − 1) − (k − 1) of the x_{n}s are at least \frac{t+Δt}{N} units to the right of x_{1}. Similar to the above, this gives

p_{t} = \binom{N − 1}{k − 1}(\frac{t}{N} )^{k−1}(1 − \frac{t + Δt}{N})^{(N−1)−(k−1)}
→ \frac{N^{k−1}}{(k − 1)!} \frac{t^{k−1}}{N^{k−1}} e^{−(t+Δt)}
→ \frac{t^{k−1}}{(k − 1)!}e^{−(t+Δt)}.  (14.21)

3. The probability that exactly k−1 of the x_{n}s are within \frac{t}{N} units to the right of x_{1} and at least one x_{n} is between \frac{t}{N} and \frac{t+Δt}{N} units to the right is p_{t+Δt} − p_{t}:

p_{t} − p_{t+Δt} → \frac{t^{k−1}}{(k − 1)!}e^{−t} − \frac{t^{k−1}}{(k − 1)!}e^{−(t+Δt)} → \frac{t^{k−1}}{(k − 1)!}e^{−t}Δt.

(14.22)

Note that when k = 1, we recover the nearest neighbor spacings.

14.5 Induced Probability Measures

We have proven the following:

Theorem 14.5.1. Consider N independent random variables x_{n} chosen from the uniform distribution on the wrapped unit interval [0, 1). For fixed N, arrange the x_{n}s in increase order, labeled y_{1} ≤ y_{2} ≤ ⋯ ≤ y_{N}.

Form the induced probability measure μ_{N,1} from the nearest neighbor spacings. Then as N → ∞ we have

μ_{N,1}(t)dt = \frac{1}{N} ∑_{n=1}^{N} δ(t − N(y_{n} − y_{n−1}))dt → e^{−t}dt.  (14.23)

Equivalently, using z_{n} = Ny_{n}:

μ_{N,1}(t)dt = \frac{1}{N} ∑_{n=1}^{N} δ(t − (z_{n} − z_{n−1}))dt → e^{−t}dt.  (14.24)

More generally, form the probability measure from the k^{th} nearest neighbor spacings. Then as N → ∞ we have

μ_{N,k}(t)dt = \frac{1}{N} ∑_{n=1}^{N} δ(t − N(y_{n} − y_{n−k}))dt → \frac{t^{k−1}}{(k − 1)!}e^{−t}dt. (14.25)

Equivalently, using z_{n} = Ny_{n}:

μ_{N,k}(t)dt = \frac{1}{N} ∑_{n=1}^{N} δ(t − (z_{n} − z_{n−k}))dt → \frac{t^{k−1}}{(k − 1)!}e^{−t}dt.  (14.26)

Definition 14.5.2 (Poissonian Behavior). We say a sequence of points x_{n} has Poissonian Behavior if in the limit as N → ∞ the induced probability measures μ_{N,k}(t)dt converge to \frac{t^{k−1}}{(k−1)!}e^{−t}dt.

Exercise 14.5.3. Let α ∈ ℚ, and define α_{n} = {n^{m}α} for some positive integer m. Show the sequence of points α_{n} does not have Poissonian Behavior.

Exercise 14.5.4. Let α ∉ ℚ, and define α_{n} = {nα}. Show the sequence of points α_{n} does not have Poissonian Behavior. Hint: for each N, show the nearest neighbor spacings take on at most three distinct values (the three values depend on N). As only three values are ever assumed for a fixed N, μ_{N,1}(t)dt cannot converge to e^{−t}dt.

14.6 Non-Poissonian Behavior

Conjecture 14.6.1. With probability one (with respect to Lebesgue Measure), if α ∉ ℚ, if α_{n} = {n^{2}α} then the sequence of points α_{n} is Poissonian.

There are constructions which show certain irrationals give rise to non-Poissonian behavior.

Theorem 14.6.2. Let α ∈ ℚ such that |α − \frac{p_{n}}{q_{n}} | < \frac{a_{n}}{q_{n}^{3}} holds infinitely often, with a_{n} → 0. Then there exist integers N_{j} → ∞ such that μ_{N_{j},1}(t) does not converge to e^{−t}dt.

As a_{n} → 0, eventually a_{n} < \frac{1}{10} for all n large. Let N_{n} = q_{n}, where \frac{p_{n}}{q_{n}} is a good rational approximation to α:

|α − \frac{p_{n}}{q_{n}} | < \frac{a_{n}}{q_{n}^{3}} .  (14.27)

Remember that all subtractions are performed on the wrapped unit interval. Thus, ||.999 − .001|| = .002.

We look at α_{k} = {k^{2}α}, 1 ≤ k ≤ N_{n} = q_{n}. Let the β_{k}s be the α_{k}s arranged in increasing order, and let the γ_{k}s be the numbers \{k^{2} \frac{p_{n}}{q_{n}} \} arranged in increasing order:

β_{1} ≤ β_{2} ≤ ⋯ ≤ β_{N}
γ_{1} ≤ γ_{2} ≤ ⋯ ≤ γ_{N}.  (14.28)

14.6.1 Preliminaries

Lemma 14.6.3. If β_{l} = α_{k} = \{k^{2}α\}, then γ_{l} = \{k^{2} \frac{p_{n}}{q_{n}} \}. Thus, the same permutation orders both the α_{k}s and the γ_{k}s.

Proof. Multiplying both sides of Equation 14.27 by k^{2} ≤ q_{n}^{2} yields

|k^{2}α − k^{2} \frac{p_{n}}{q_{n}} | < k^{2} \frac{a_{n}}{q_{n}^{2}} ≤ \frac{a_{n}}{q_{n}} < \frac{1}{2q_{n}} .  (14.29)

Thus, k^{2}α and k^{2} \frac{p_{n}}{q_{n}} differ by at most \frac{1}{2q_{n}}. Therefore

||\{k^{2}α\} − \{k^{2} \frac{p_{n}}{q_{n}} \}|| < \frac{1}{2q_{n}} .  (14.30)

As the numbers \{m^{2} \frac{p_{n}}{q_{n}} \} all have denominators of size at most \frac{1}{q_{n}}, we see that \{k^{2} \frac{p_{n}}{q_{n}} \} is the closest of the \{m^{2} \frac{p_{n}}{q_{n}} \} to \{k^{2}α\}.

This implies that if β_{l} = \{k^{2}α\}, then γ_{l} = \{k^{2} \frac{p_{n}}{q_{n}} \}, completing the proof.

Exercise 14.6.4. Prove the ordering is as claimed. Hint: about each β_{l} = {k^{2}α}, the closest number of the form \{c^{2} \frac{p_{n}}{q_{n}} \} is \{k^{2} \frac{p_{n}}{q_{n}} \}.

14.6.2 Proof of Theorem 14.6.2

Exercise 14.6.5. Assume ||a − b||, ||c − d|| < \frac{1}{10}. Show

||(a − b) − (c − d)|| < ||a − b|| + ||c − d||.  (14.31)

Proof of Theorem 14.6.2: We have shown

||β_{l} − γ_{l}|| < \frac{a_{n}}{q_{n}} .  (14.32)

Thus, as N_{n} = q_{n}:

||N_{n}(β_{l} − γ_{l})|| < a_{n},  (14.33)

and the same result holds with l replaced by l − 1. By Exercise 14.6.5,

||N_{n}(β_{l} − γ_{l}) − N_{n}(β_{l−1} − γ_{l−1})|| < 2a_{n}.  (14.34)

Rearranging gives

||N_{n}(β_{l} − β_{l−1}) − N_{n}(γ_{l} − γ_{l−1})|| < 2a_{n}.  (14.35)

As a_{n} → 0, this implies the difference between ||N_{n}(β_{l}−β_{l−1})|| and ||N_{n}(γ_{l}−

γ_{l−1})|| goes to zero.

The above distance calculations were done mod 1. The actual differences will differ by an integer. Thus,

μ_{N_{n},1}^{α}(t)dt = \frac{1}{N_{n}} ∑_{l=1}^{N_{n}} δ(t − N_{n}(β_{l} − β_{l−1}))  (14.36)

and

μN_{n},1^{\frac{pn}{qn}}(t)dt = \frac{1}{N_{n}} ∑_{l=1}^{N_{n}} δ(t − N_{n}(γ_{l} − γ_{l−1}))  (14.37)

are extremely close to one another; each point mass from the difference between adjacent β_{l}s is either within a_{n} units of a point mass from the difference between adjacent γ_{l}s, or is within a_{n} units of a point mass an integer number of units from a point mass from the difference between adjacent γ_{l}s. Further, a_{n} → 0.

Note, however, that if γ_{l} = \{k^{2} \frac{p_{n}}{q_{n}} \}, then

N_{n}γ_{l} = q_{n}\{k^{2} \frac{p_{n}}{q_{n}} \} ∈ ℕ.  (14.38)

Thus, the induced probability measure μN_{n},1^{\frac{pn}{qn}}(t)dt formed from the γ_{l}s is sup-

ported on the integers! Thus, it is impossible for μN_{n},1^{\frac{pn}{qn}}(t)dt to converge to e^{−t}dt. As μ_{N_{n},1}^{α}(t)dt, modulo some possible integer shifts, is arbitrarily close to

μN_{n},1^{\frac{pn}{qn}}(t)dt, the sequence \{k^{2}α\} is not Poissonian along the subsequence of Ns given by N_{n}, where N_{n} = q_{n}, q_{n} is a denominator in a good rational approximation to α. □

14.6.3 Measure of α ∉ ℚ with Non-Poissonian Behavior along a sequence N_{n}

What is the (Lebesgue) measure of α ∉ ℚ such that there are infinitely many n with

|α − \frac{p_{n}}{q_{n}} | < \frac{a_{n}}{q_{n}} , a_{n} → 0.  (14.39)

If the above holds, then for any constant k(α), for n large (large depends on both α and k(α)) we have

|α − \frac{p_{n}}{q_{n}} | < \frac{k(α)}{q_{n}^{2+ϵ}}.  (14.40)

Exercise 14.6.6. Show this set has (Lebesgue) measure or size 0.

Thus, almost no irrational numbers satisfy the conditions of Theorem 14.6.2, where almost no is relative to the (Lebesgue) measure.

Exercise 14.6.7. In a topological sense, how many algebraic numbers satisfy the conditions of Theorem 14.6.2? How many transcendental numbers satisfy the conditions?

Exercise 14.6.8. Let α satisfy the conditions of Theorem 14.6.2. Consider the sequence N_{n}, where N_{n} = q_{n}, q_{n} the denominator of a good approximation to α. We know the induced probability measures μN_{n},1^{\frac{pn}{qn}}(t)dt and μ_{N_{n},1}^{α}(t)dt do not converge to e^{−t}dt. Do these measures converge to anything?

Remark 14.6.9. In The Distribution of Spacings Between the Fractional Parts of {n^{2}α} (Z. Rudnick, P. Sarnak, A. Zaharescu), it is shown that for most α satisfying the conditions of Theorem 14.6.2, there is a sequence N_{j} along which μ_{N_{n},1}^{α}(t)dt does converge to e^{−t}dt.

More Graphs, Kloosterman, Randomness of x → \overline{x} mod p

More on Graphs, Kloosterman, and the Third Problem on the Randomness of x ↦ \overline{x} mod p. Review of projective geometry and fractional linear transformations. Lecture by Peter Sarnak; notes by Steven J. Miller.

15.1 Kloosterman Sums

Recall

S(a, b, p) = ∑_{x mod p}^{∗} e(\frac{ax + b\overline{x}}{p}),  (15.1)

where ∑^{∗} means sum over all x relatively prime to p, and \overline{x} ≡ x^{−1} mod p.

Theorem 15.1.1 (Weil). For p an odd prime and a, b ∈ ℤ,

|S(a, b, p)| ≤ 2\sqrt{p}.  (15.2)

The above captures the randomness. We add p − 1 numbers of modulus 1, and we see square-root cancellation. Weil’s Theorem says the cancellation is "like" random numbers. Recall when we added ±1, we expected to observe a sum around \sqrt{N} if we had N summands.

15.2 Projective Geometry

Consider the map

ℂ → ℂ ∪ {∞}
z ↦ \frac{az + b}{cz + d}.  (15.3)

We define the above as the action of the matrix

( \begin{matrix}a & b \\ c & d \end{matrix} )  (15.4)

on z. ℙ^{1}(ℝ) is identified with the perimeter of a circle, with antipodal points (points

on a diagonal, separated by π radians) identified.

15.3 Example

Definition 15.3.1 (ℙ^{1}(𝔽_{p})). ℙ^{1}(𝔽_{p}) is the projective line,

ℙ^{1}(𝔽_{p}) = {0, 1, …, ∞}.  (15.5)

We construct a 3-regular graph G_{p} on p + 1 vertices as follows:

1. Join x to x + 1 mod p.

2. Join x to x − 1 mod p.

3. Join x to −\overline{x} mod p,

where \frac{1}{∞} = 0. Form the adjacency matrix of the above graph. Is there a spectral graph?

Theorem 15.3.2. There is a spectral gap!

λ_{1}(G_{p}) ≤ 2.99.  (15.6)

These graphs are not Ramanujan in general.

We can look at the three maps as matrices

T = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) , S = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) , T^{−1} = ( \begin{matrix}1 & −1 \\ 0 & 1 \end{matrix} )  (15.7)

Exercise 15.3.3. (ST)^{3} = ±I.

Exercise 15.3.4. Show the three maps we used to create G_{p} can be given by

1. x → T x (corresponding to x → x + 1),

2. x → T^{−1}x (corresponding to x → x − 1),

3. x → Sx (corresponding to x → −\overline{x}.

15.4 Stereographic Projections and Fractional Linear Transformations

What are the analytic, 1 − 1 invertible maps from ℂ → ℂ? What if we include ∞. First, one might ask what is ∞? Take a sphere, call the north pole N. Consider the infinite plane z = 0. To each point P on the sphere, draw the line from N to P, and write down the

point of intersection on the plane z = 0. Call this map S (stereographic projection; preserves angles), and call the sphere S^{2} = ℙ^{1}.

Thus,

S(P) ∈ ℂ, S(∞) ↔ N, S^{2} ≅ ℂ ∪ {∞}.  (15.8)

Recall GL_{2} are the 2 × 2 matrices with non-zero determinant. SL_{2}(ℂ) is the group of 2 × 2 matrices with determinant 1.

Fractional Linear Transformation: z ↦ \frac{az+b}{cz+d}. If we have two linear transformations

γ = ( \begin{matrix}a & b \\ c & d \end{matrix} ) , δ = ( \begin{matrix}a_{1} & b_{1} \\ c_{1} & d_{1} \end{matrix} )  (15.9)

then

γ(δz) = (γδ)z,  (15.10)

where (γδ) is usual matrix multiplication.

15.5 More Kesten

Let Γ be a group generated by

A_{1}, A_{1}^{−1}, A_{2}, A_{2}^{−1}, …, A_{k}, A_{k}^{−1}.  (15.11)

We make a 2k-regular graph by joining x and y with an edge (x ∼ y) if and only if x = A_{j}^{±1}y for some j.

Define, for f : Γ → ℂ,

Bf(x) = ∑_{x∼y} f(y).  (15.12)

To make the sum make sense if we have infinitely many vertices we require f ∈ l^{2}(Γ), the space of functions g where ∑ |g(γ)|^{2} < ∞.

One could ask about the spectrum of B on this space. The HW problem was on a k-regular tree. Make words using the generators. Have the notion of a free group: no relation (ie, no word is the identity word) except the trivial ones (A_{i}A_{i}^{−1} = I).

Γ = ⟨A_{1}, A_{2}, …, A_{k}⟩.  (15.13)

Definition 15.5.1 (Free Group). Γ is a free group if the only relations in Γ giving the identity are the obvious ones (ie, the only word in A_{1}, …, A_{k} that is the identity word is words of the form A_{k}^{−1}A_{i}A_{i}^{−1}A_{j}^{−1}A_{j}A_{k}, and so on).

The graph we just spoke about, G(Γ) on A_{1}, A_{1}^{−1} up to A_{k}, A_{k}^{−1} is a 2k-regular tree if and only if Γ is a free group on A_{1}, …, A_{k}. This is called a Cayley Graph (relative to the given generators).

Theorem 15.5.2 (Kesten). The spectrum of B when Γ is free on k generators is

spectrum(B) = [ − 2\sqrt{2k − 1}, 2\sqrt{2k − 1}].  (15.14)

Further, Γ is free on A_{1}, …, A_{k} if and only if

spectrum(B) ⊂ [ − 2\sqrt{2k − 1}, 2\sqrt{2k − 1}].  (15.15)

Thus, the spectrum contains a point outside this interval if and only if Γ is not free.

Finally, 2k is in the spectrum if and only if Γ is amenable (for example, abelian).

Corollary 15.5.3. Our graphs G_{p} cannot be Ramanujan, as there is a relation among the generators, namely (ST)^{3} = ±I.

15.6 SL_{2}(ℤ)

Definition 15.6.1 (SL_{2}(ℤ)). SL_{2}(ℤ) is the group of 2 × 2 matrices with unit determinant and integer coefficients.

Exercise 15.6.2. Consider the matrices

T = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) , S = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) , T^{−1} = ( \begin{matrix}1 & −1 \\ 0 & 1 \end{matrix} )  (15.16)

Show these three matrices generate SL_{2}(ℤ); as (ST)^{3} = ±I, this shows SL_{2}(ℤ) is not a free group.

Where should SL_{2}(ℤ) act? Lubotsky: you don’t understand a group until it acts on something you know. Galois had groups acting on roots of polynomials (permuting roots).

What does SL_{2}(ℤ) act on? It does act on the sphere, but it is too big a space. We want to study the smallest space where it acts reasonably.

Look at SL_{2}(ℝ), the group of 2 × 2 matrices with determinant 1 and real entries. Let z be in the upper half plane, so z = x + iy, y > 0. Then

z ↦ \frac{az + b}{cz + d}, Im(\frac{az + b}{cz + d}) > 0.  (15.17)

A similar statement holds for z in the lower half plane. Thus, SL_{2}(ℝ) maps the upper (lower) half plane to itself.

By shifting by an integer, we can bring any x ∈ ℝ to x′ ∈ [0, 1).

Gauss was the first to draw the fundamental domain for SL_{2}(ℤ) acting on the upper half plane:

Draw a circle of radius 1 with center 0; draw vertical lines at x = ±\frac{1}{2}, going from the point on the circle to infinity. The region formed is called the fundamental domain for SL_{2}(ℤ) on the upper half plane. This means that any z in the upper half plane can be brought into this region.

15.7 Is x → \overline{x} mod p Random?

15.7.1 First Test

Question 15.7.1. To what extent is / how random is x → \overline{x} mod p?

There are p numbers.

If look at all numbers 1 ≤ x ≤ p − 1, we get all the numbers back (in some new order). We can look at a long segment (on what scale?), say

\sqrt{p} ≤ x ≤ 2\sqrt{p}.  (15.18)

Now \overline{x} will be all over [1, p − 1]. This generates \sqrt{p} numbers between 1 and p − 1. The average spacing is is approximately \frac{p−1}{\sqrt{p}≈\sqrt{p}}. How are numbers spaced?

First, we need to write them in increasing order

1 ≤ a_{1} < a_{2} < ⋯ < a_{l} ≤ p − 1.  (15.19)

Let us denote the nearest neighbor spacings by Δ_{1} = a_{2} − a_{1}, Δ_{2} = a_{3} − a_{2} and so on. Let

δ_{j} = \frac{Δ_{j}}{\sqrt{p}}.  (15.20)

Note the δ_{j}s have unit mean spacing.

Conjecture 15.7.2 (Naive Conjecture). We expect the spacings to follow Poissonian Statistics. Explicitly, the distribution of spacings we see here should be the same as that from choosing \sqrt{p} numbers independently from the uniform distribution on [0, 1).

15.7.2 Second Test

Question 15.7.3 (Jim Propp). Does x → \overline{x} behave like a random permutation of order 2?

What is the distribution of the longest increasing sub-sequence?

Given a permutation of 1, 2, …, N, we have i ↦ σ(i). Permutations are often denoted by

({1 \atop σ(1)} {2 \atop σ(2)} ⋯ {N \atop σ(N)})  (15.21)

Look at the distribution of the longest increasing sub-sequence about the mean. Normalized appropriately, what does it look like?

15.7.3 Third Test: Hooley’s R^{∗}

∑_{x=A}^{A+N} e(\frac{\overline{x}}{p}),  (15.22)

where 1 ≤ A ≤ A + N ≤ p − 1. When we add numbers of modulus one, we expect square-root cancellation. Then

Conjecture 15.7.4 (Hooley’s R^{∗}). For every ϵ > 0,

| ∑_{x=A}^{A+N} e(\frac{\overline{x}}{p})| ≪_{ϵ} N^{\frac{1}{2}} p^{ϵ}.  (15.23)

Note ≪_{ϵ} means the left hand side is less than c_{ϵ} times the right hand side (for some c_{ϵ} > 0).

Note there is no dependence on A – the only dependence is on the size of summation N and the prime p.

If N > \sqrt{p}, the above can be proven. by Weil’s bound.

15.8 Note on Non-trivial Bound of Fourth Powers of Kloosterman Sums

Note on conditions arising in non-trivial bound on sum of fourth powers of Kloosterman sums (Heath-Brown review). Supplemental notes by Alex Barnett.

Please refer to Professor Sarnak’s lecture of 10/16/02, and Heath-Brown’s review article on Kloosterman Sums.

There are six summations inherent in the desired sum

∑_{a=0}^{p−1} ∑_{b=0}^{p−1} |Kl(a, b, p)|^{4} ,  (15.24)

namely the two sums shown and one internal sum in each of the Kl’s. Using the result

∑_{a=0}^{p−1} e (\frac{an}{p}) = \{ \begin{matrix}p − 1, & n = 0 (modp) \\ 0, & otherwise \end{matrix}  (15.25)

twice, turns the two e(·) functions into counting conditions on the variables internal to the four Kl’s. Calling these variables w, x, y, z, we want the total number of ways that, with 1 ≤ w, x, y, z ≤ p − 1 (the four surviving sums), we satisfy both

w + x − y − z = 0 (p)  (15.26)

and

w̄ + x̄ − ȳ − z̄ = 0 (p).  (15.27)

Multiplying Eq. 15.27 by wxyz gives

(w + x)yz − (y + z)wx = 0 (p).  (15.28)

Substituting Eq. 15.26 gives

(w + x)(yz − wx) = 0 (p).  (15.29)

So either w + x = 0 (p) or yz − wx = 0 (p), or both.

The first set of cases has x = −w from (15.26) giving z = −y, so there are (p−1)^{2} choices of w and y. For each choice x and z are fixed uniquely. Therefore these cases contribute (p − 1)^{2} ways.

For the second set, we have two equations

y + z = w + x (p)  (15.30)
yz = wx (p)  (15.31)

for two unknowns y, z, for any of the arbitrary choices of w, x. You could combine these equations into the single quadratic

y^{2} − y(w + x) + wx = 0 (p).  (15.32)

Two solutions for y are y = w and y = x (check by substitution). Since it is a quadratic, these are the only two solutions. Therefore the number of ways contributed is at most 2 times the (p − 1)^{2} ways of choosing w, x.

Over-counting due to the and/or is at least of order p or smaller, but also can only reduce the number of ways. Therefore the total number of ways ≤ 3(p − 1)^{2}, which is O(p^{2}). From this follows the bound on the sum given in the article and lecture.

Introduction to the Hardy-Littlewood Circle Method

Introduction to the Hardy-Littlewood Circle Method. Lecture and notes by Steven J. Miller.

16.1 Problems where the Circle Method is Useful

For each N, let A_{N} be a set of non-negative integers such that

1. A_{N} ⊂ A_{N+1},

2. |A_{N}| → ∞ as N → ∞.

Let A = lim_{N→∞} A_{N}.

Question 16.1.1. Let s be a fixed positive integer. What can one say about a_{1} + ⋯ + a_{s}? Ie, what numbers n are representable as a sum of s summands from A?

We consider three problems; we will mention later why we are considering sets A_{N}.

16.1.1 Waring’s Problem

Let A be the set of k^{th} powers of non-negative numbers, and let

A_{N} = {0^{k}, 1^{k}, …, N^{k}}.  (16.1)

Question 16.1.2. Fix a positive integer k. For what positive integers s can every integer be written as a sum of s numbers, each number a k^{th} power?

Thus, in this case, we are trying to solve

n = a_{1}^{k} + ⋯ + a_{N}^{k}.  (16.2)

16.1.2 Goldbach’s Problem

Let A be the set of all prime numbers, and let A_{N} be the set of all primes at most N. Question 16.1.3. Can every even number be written as the sum of two primes?

In this example, we are trying to solve

2n = a_{1} + a_{2},  (16.3)

or, in more suggestive notation,

2n = p_{1} + p_{2}.  (16.4)

16.1.3 Sum of Three Primes

Again, let A be the set of all primes, and A_{N} all primes up to N. Question 16.1.4. Can every odd number be written as the sum of three primes?

Again, we are studying

2n + 1 = p_{1} + p_{2} + p_{3}.  (16.5)

16.2 Idea of the Circle Method

16.2.1 Introduction

Definition 16.2.1 (e(z)). We define e(z) = e^{2πiz}.

Exercise 16.2.2. Let m, n ∈ ℤ. Prove

∫_{0}^{1} e(nx)e(−mx)dx = \begin{cases}1 & if n = m \\ 0 & otherwise \end{cases}  (16.6)

Let A, A_{N} be as in any of the three problems above. Consider

f_{N}(x) = ∑_{a∈A_{N}} e(ax).  (16.7)

We investigate (f_{N}(x))^{s}:

(f_{N}(x))^{s} = ∏_{j=1}^{s} ∑_{a_{j}∈A_{N}} e(a_{j}x)
= ∑_{m} r_{N}(m)e(mx).  (16.8)

The last result follows by collecting terms. When you multiply two exponentials, you add the exponents.

Thus, when we multiply the s products, how can we get a product which gives e(mx)?

We have s products, say e(a_{1}x) through e(a_{N}x). Thus,

e(a_{1}x) ⋯ e(a_{N}x) = e((a_{1} + ⋯ + a_{N})x) = e(mx).  (16.9)

Thus, the coefficient r_{N}(m) in (f_{N}(x))^{s} is the number of ways of writing

m = a_{1} + ⋯ + a_{N},  (16.10)

with each a_{j} ∈ A_{N}. As the elements of A_{N} are non-negative, if N is sufficiently large r_{N}(m) is

equal to the number of ways of writing m as the sum of s elements of A.

The problem is, if m is larger than the largest term in A_{N}, then there may be other ways to write m as a sum of s elements of A.

Lemma 16.2.3.

r_{N}(m) = ∫_{0}^{1} (f_{N}(x))^{s}e(−mx)dx.  (16.11)

Proof: direct calculation.

Note that, just because we have a closed form expression for r_{N}(m), this does not mean we can actually evaluate the above integral. Recall, for example, the inclusion - exclusion formula for the number of primes at most N. This is an exact formula, but very hard to evaluate.

16.2.2 Useful Number Theory Results

We will use the following statements freely:

Theorem 16.2.4 (Prime Number Theorem). Let π(x) denote the number of primes at most x. Then

π(x) = ∑_{p≤x} 1 = \frac{x}{log x} + smaller.  (16.12)

Upon applying Partial Summation, we may rewrite the above as

∑_{p≤x} log p = x + smaller.  (16.13)

Theorem 16.2.5 (Siegel-Walfisz). Let C, B > 0, and let a and q be relatively prime. Then

∑_{{p≤x \atop p≡a(q)}} log p = \frac{x}{ϕ(q)} + O(\frac{x}{log^{C} x})  (16.14)

for q ≤ log^{B} x, and the constant above does not depend on x, q or a (ie, it only depends on C and B).

For completeness, we include a review of partial summation as an appendix to these notes.

16.2.3 Average Sizes of (f_{N}(x))^{s}

Henceforth we will consider f_{N}(x) arising from the three prime case. Thus, s = 3. For analytic reasons, it is more convenient to instead analyze the function

F_{N}(x) = ∑_{p≤N} log p · e(px).  (16.15)

Working analogously as before, we are led to

R_{N}(m) = ∫_{0}^{1} (F_{N}(x))^{3}e(−mx)dx.  (16.16)

By partial summation, it is very easy to go from R_{N}(m) to r_{N}(m).

Exercise 16.2.6. Prove the trivial bound for |F_{N}(x)| is N. Take absolute values and use the Prime Number Theorem.

We can, however, show that the average square of F_{N}(x) is significantly smaller:

Lemma 16.2.7. The average value of |F_{N}(x)|^{2} is N log N.

Proof: The following trivial observation will be extremely useful in our arguments. Let g(x) be a complex-valued function, and let \overline{g}(x) be its complex conjugate. Then |g(x)|^{2} = g(x)\overline{g}(x).

In our case, as \overline{F}_{N}(x) = F_{N}(−x) we have

∫_{0}^{1} |F_{N}(x)|^{2} = ∫_{0}^{1} F_{N}(x)F_{N}(−x)dx
= ∫_{0}^{1} ∑_{p≤N} log p · e(px) ∑_{q≤N} log q · e(−qx)dx

= ∑_{p≤N} ∑_{q≤N} log p log q ∫_{0}^{1} e((p − q)x)dx
= ∑_{p≤N} log^{2} p.  (16.17)

Using ∑_{p≤N} log p = N + small and Partial Summation, we can show

∑_{p≤N} log^{p} = N log N + smaller.  (16.18)

Thus,

∫_{0}^{1} |F_{N}(x)|^{2} = N log N + smaller.  (16.19)

Thus, taking square-roots, we see on average |(F_{N}(x))|^{2} is N log N, significantly smaller than the maximum possible value (N^{2}). Thus, we see we are getting almost square-root cancellation on average. □

16.2.4 Definition of the Major and Minor Arcs

We split the unit interval [0, 1) into two disjoint parts, the Major and the Minor arcs.

Roughly, the Major arcs will be a union of very small intervals centered at rationals with small denominator (relative to N). Near these rationals, we will be able to approximate F_{N}(x) very well, and F_{N}(x) will be of size N.

The minor arcs will be the rest of [0, 1); we will show that F_{N}(x) is significantly smaller than N here.

Major Arcs

Let B > 0, and let Q = (log N)^{B} ≪ N.

For each q ∈ {1, 2, …, Q} and a ∈ {1, 2, …, q} with a and q relatively prime, consider the set

M_{a,q} = \{x ∈ [0, 1) : |x − \frac{a}{q}| < \frac{Q}{N} \}.  (16.20)

We also add in one interval centered at either 0 or 1, ie, the "interval" (or wrapped-around interval)

[0, \frac{Q}{N} ] ∪ [1 − \frac{Q}{N} , 1].  (16.21)

Exercise 16.2.8. Show, if N is large, that the major arcs M_{a,q} are disjoint for q ≤ Q and a ≤ q, a and q relatively prime.

We define the Major Arcs to be the union of each arc M_{a,q}:

M = ⋃_{q=1}^{Q} ⋃_{{a=1 \atop (a,q)=1}} M_{a,q},  (16.22)

where (a, q) is the greatest common divisor of a and q.

Exercise 16.2.9. Show |M| < \frac{2Q^{3}}{N}. As Q = log^{B} N, this implies as N → ∞, the major arcs are zero percent of the unit interval.

Minor Arcs

The Minor Arcs, m, are whatever is not in the Major Arcs. Thus,

m = [0, 1) − M.  (16.23)

Clearly, as N → ∞, almost all of [0, 1) is in the Minor Arcs.

16.3 Contributions from the Major and Minor Arcs

16.3.1 Contribution from the Minor Arcs

We bound the contribution from the minor arcs to r_{N}(m):

| ∫_{m} F_{N}^{3}(x)e(−mx)dx| ≤ ∫_{m} |F_{N}(x)|^{3}dx
≤ ( max_{x∈m} |F_{N}(x)|) ∫_{m} |F_{N}(x)|^{2}dx
≤ ( max_{x∈m} |F_{N}(x)|) ∫_{0}^{1} F_{N}(x)F_{N}(−x)dx
≤ ( max_{x∈m} |F_{N}(x)|)N log N.  (16.24)

As the minor arcs are most of the unit interval, replacing ∫_{m} with ∫_{0}^{1} doesn’t introduce much of an over-estimation.

In order for the Circle Method to succeed, we need a non-trivial, good bound for

max_{x∈m} |F_{N}(x)|  (16.25)

This is where most of the difficulty arises, showing that there is good cancellation in F_{N}(x) if we stay away from rationals with small denominator.

We will show that the contribution to the major arcs is

𝔖(N)\frac{N^{2}}{2} + smaller,  (16.26)

where ∃c_{1}, c_{2} > 0 such that, for all N, c_{1} < 𝔖(N) < c_{2}. Thus, we need the estimate that

max_{x∈m} |F_{N}(x)| ≤ \frac{N}{log^{1+ϵ} N} .  (16.27)

Relative to the average size of |F_{N}(x)|^{2}, this is significantly smaller; however, as we are showing that the maximum value of |F_{N}(x)| is bounded, this is a significantly more delicate question. We know such a bound cannot be true for all x ∈ [0, 1) (see below, and not that F_{N}(0) = N). The hope is that if x is not near a rational with small denominator, we will get moderate cancellation.

While this is very reasonable to expect, it is not easy to prove.

16.3.2 Contribution from the Major Arcs

Fix a q ≤ Q and an a ≤ q with a and q relatively prime. We evaluate F (\frac{a}{q}).

F (\frac{a}{q}) = ∑_{p≤N} log p · e^{2πip \frac{a}{q}}

= ∑_{r=1}^{q} ∑_{{p≡r(q) \atop p≤N}} log p · e^{2πi \frac{ap}{q}}

= ∑_{r=1}^{q} ∑_{{p≡r(q) \atop p≤N}} log p · e^{2πi \frac{ar}{q}}

= ∑_{r=1}^{q} e^{2πi \frac{ar}{q}} ∑_{{p≡r(q) \atop p≤N}} log p  (16.28)

Note the beauty of the above. The dependence on p in the original sums is very weak – there is a log p factor, and there is e(\frac{ap}{q}). In the exponential, we only

need to know p mod q. Now, p runs from 2 to N, and q is at most log^{B} N. Thus, in general p ≫ q.

We use the Siegel-Walfisz Theorem. We first remark that we may assume r and q are relatively prime. Why? If p ≡ r mod q, this means p = αq + r for some α ∈ ℕ. If r and q have a common factor, there can be at most one prime p (namely r) such that p ≡ r mod q, and this can easily be shown to give a negligible contribution.

For any C > 0

∑_{{p≡r(q) \atop p≤N}} log p = \frac{N}{ϕ(q)} + O(\frac{N}{log^{C} N} ).  (16.29)

Now, as ϕ(q) is at most q which is at most log^{B} N, we see that the main term is significantly greater than the error term (choose C much greater than B).

Note the Siegel-Walfisz Theorem would be useless if q ≈ N^{ϵ}. Then the main term would be like N^{1−ϵ}, which would be smaller than the error term.

This is one reason why, in constructing the major arcs, we take the denominators to be small.

Thus, we find

F (\frac{a}{q}) = ∑_{{r=1 \atop (r,q)=1}}^{q} e^{2πi \frac{ar}{q}} \frac{N}{ϕ(q)} + smaller

= \frac{N}{ϕ(q)} ∑_{{r=1 \atop (r,q)=1}}^{q} e^{2πi \frac{ar}{q}} .  (16.30)

We merely sketch what happens now. First, one shows that for x ∈ M_{a,q} that F_{N}(x) is very close to F (\frac{a}{q}). This

is a standard analysis (Taylor Series Expansion – the constant term is a good approximation if you are sufficiently close).

Thus, as the major arcs are distinct,

∫_{M} F_{N}^{3}(x)e(−mx)dx = ∑_{q=1}^{Q} ∑_{{a=1 \atop (a,q)=1}} ∫_{M_{a,q}} F_{N}^{3}(x)e(−mx)dx.  (16.31)

We can approximate F_{N}^{3}(x) by F (\frac{a}{q}); integrating a constant gives the constant

times the length of the interval. Each of the major arcs has length \frac{2Q^{3}}{N}. Thus we find that, up to a smaller correction term, the contribution from the Major Arcs is

∫_{M} F_{N}^{3}(x)e(−mx)dx = \frac{2Q^{3}}{N} ∑_{q=1}^{Q} ∑_{{a=1 \atop (a,q)=1}} (\frac{N}{ϕ(q)} ∑_{{r=1 \atop (r,q)=1}}^{q} e^{2πi \frac{ar}{q}} )^{3}e(\frac{−2πima}{q})

= N^{2} · 2Q^{3} ∑_{q=1}^{Q} \frac{1}{ϕ(q)^{3}} ∑_{{a=1 \atop (a,q)=1}} ( ∑_{{r=1 \atop (r,q)=1}}^{q} e^{2πi \frac{ar}{q}} )^{3}e(\frac{−2πima}{q}).

(16.32)

To complete the proof, we need to show that what is multiplying N^{2} is nonnegative, and not too small.

We will leave this for another day, as it is getting quite late here.

16.4 Why Goldbach is Hard

Using

F_{N}(x) = ∑_{p≤N} log p · e^{2πipx},  (16.33)

we find we must study

∫_{0}^{1} F_{N}^{s}(x)dx,  (16.34)

where s = 3 if we are looking at p_{1} + p_{2} + p_{3} = 2n + 1 and s = 2 if we are looking at p_{1} + p_{2} = 2n. Why does the circle method work for s = 3 but fail for s = 2?

16.4.1 s = 3 Sketch

Let us recall briefly the s = 3 case. Near rationals \frac{a}{q} with small denominator (small means q ≤ log^{B} N), we can evaluate F_{N}(\frac{a}{q}). Using Taylor, if x is very close to \frac{a}{q}, we expect F_{N}(x) to be close to F_{N}(\frac{a}{q}).

The Major Arcs have size \frac{log^{B} N}{N}. As F_{N}(x) is around N near such rationals, we expect the integral of F_{N}^{3}(x)e(−mx) to be N^{2} times a power of log N.

Doing a careful analysis of the singular series shows that the contribution is actually 𝔖(N)N^{2}, where there exist constants independent of N such that 0 < c_{1} < 𝔖(N) < c_{2} < ∞.

A direct calculation shows that

∫_{0}^{1} |F_{N}(x)|^{2}dx = ∫_{0}^{1} F_{N}(x)F_{N}(−x)dx = N.  (16.35)

Thus, if m denotes the minor arcs,

| ∫_{m} F_{N}^{3}(x)e(−mx)dx| ≤ max_{x∈m} |F_{N}(x)| ∫_{0}^{1} |F_{N}(x)|^{2}dx
≤ N max_{x∈m} |F_{N}(x)|.  (16.36)

As the major arcs contribute 𝔖(N)N^{2}, we need to show

max_{x∈m} |F_{N}(x)| ≪ \frac{N}{log^{D} N} .  (16.37)

Actually, we just need to show the above is ≪ o(N). This is the main difficulty – the trivial bound is |F_{N}(x)| ≤ N. As F_{N}(0) = N plus lower order terms, we cannot do better in general.

Exercise 16.4.1. Show F_{N}(\frac{1}{2}) = N − 1 plus lower order terms.

The key observation is that, if we stay away from rationals with small denominator, we can prove there is cancellation in F_{N}(x). While we don’t go into details here (see, for example, Nathanson’s Additive Number Theory: The Classical Bases, Chapter 7), the savings we obtain is small. We show

max_{x∈m} |F_{N}(x)| ≪ \frac{N}{log^{D} N} .  (16.38)

Note that Equation 16.35 gives us significantly better cancellation on average, telling us that |F_{N}(x)|^{2} is usually of size N.

Thus, it is our dream to be so lucky as to see | ∫_{I} |F_{N}(x)|^{2}dx| for any I ⊂ [0, 1), as we can evaluate this extremely well.

16.4.2 s = 2 Sketch

What goes wrong when s = 2? As a first approximation, if s = 3 has the Major Arcs contributing a constant times N^{2} (and F_{N}(x) was of size N on the Major Arcs), one might guess that the Major Arcs for s = 2 will contribute a constant times N.

How should we estimate the contribution from the Minor Arcs? We have F_{N}^{2}(x). If we just throw in absolute values we get

| ∫_{m} F_{N}^{2}(x)e(−mx)dx| ≤ ∫_{0}^{1} |F_{N}(x)|^{2}dx = N.  (16.39)

Note, unfortunately, that this is the same size as the expected contribution from the Major Arcs!

We could try pulling a max_{x∈m} |F_{N}(x)| outside the integral, and hope to get a good savings. The problem is this leaves us with ∫_{m} |F_{N}(x)|dx.

Recall

Lemma 16.4.2.

∫_{0}^{1} |f(x)g(x)|dx ≤ ( ∫_{0}^{1} |f(x)|^{2}dx)^{\frac{1}{2}} · ( ∫_{0}^{1} |g(x)|^{2}dx)^{\frac{1}{2}} .  (16.40)

For a proof, see Lemma 16.5.1.

Thus,

| ∫_{m} F_{N}^{2}(x)e(−mx)dx| ≤ max_{x∈m} |F_{N}(x)| ∫_{0}^{1} |F_{N}(x)|dx
≤ max_{x∈m} |F_{N}(x)|( ∫_{0}^{1} |F_{N}(x)|^{2}dx)^{\frac{1}{2}} · ( ∫_{0}^{1} 1^{2}dx)^{\frac{1}{2}}
≤ max_{x∈m} |F_{N}(x)| · N^{\frac{1}{2}} · 1.  (16.41)

As the Major Arcs contribute something of size N, we would need

max_{x∈m} |F_{N}(x)| ≪ o(\sqrt{N}).  (16.42)

There is almost no chance of such cancellation. We know

∫_{0}^{1} |F_{N}(x)|^{2}dx = N plus lower order terms.  (16.43)

Thus, the average size of |F_{N}(x)| is N, so we expect |F_{N}(x)| to be about \sqrt{N}. To get o(N) would be unbelievably good fortune!

While the above sketch shows the Circle Method is not, at present, powerful enough to handle the Minor Arc contributions, all is not lost. The quantity we need to bound is

| ∫_{m} F_{N}^{2}(x)e(−mx)dx|.  (16.44)

However, we have instead been studying

∫_{m} |F_{N}(x)|^{2}dx  (16.45)

and

max_{x∈m} |F_{N}(x)| ∫_{0}^{1} |F_{N}(x)|dx.  (16.46)

Thus, we are ignoring the probable oscillation / cancellation in the integral ∫ F_{N}(x)e(−mx)dx. It is this cancellation that will lead to the Minor Arcs contributing significantly less than the Major Arcs.

However, showing there is cancellation in the above integral is very difficult. It is a lot easier to work with absolute values.

16.5 Cauchy-Schwartz Inequality

Lemma 16.5.1. [Cauchy-Schwarz]

∫_{0}^{1} |f(x)g(x)|dx ≤ ( ∫_{0}^{1} |f(x)|^{2}dx)^{\frac{1}{2}} · ( ∫_{0}^{1} |g(x)|^{2}dx)^{\frac{1}{2}} .  (16.47)

For notational simplicity, assume f and g are real-valued, positive functions. Working with |f| and |g| we see there is no harm in the above.

Let

h(x) = f(x) + λg(x), λ = −\frac{∫_{0}^{1} f(x)g(x)dx}{∫_{0}^{1} g(x)^{2}dx}  (16.48)

As ∫_{0}^{1} h(x)^{2}dx ≥ 0, we have

0 ≤ ∫_{0}^{1} (f(x) + λg(x))^{2}dx
= ∫_{0}^{1} f(x)^{2}dx + 2λ ∫_{0}^{1} f(x)g(x)dx + λ^{2} ∫_{0}^{1} g(x)^{2}dx

= ∫_{0}^{1} f(x)^{2}dx − 2\frac{( ∫_{0}^{1} f(x)g(x)dx)^{2}}{∫_{0}^{1} g(x)^{2}dx} + \frac{( ∫_{0}^{1} f(x)g(x)dx)^{2}}{∫_{0}^{1} g(x)^{2}dx}

= ∫_{0}^{1} f(x)^{2}dx − \frac{( ∫_{0}^{1} f(x)g(x)dx)^{2}}{∫_{0}^{1} g(x)^{2}dx}

\frac{( ∫_{0}^{1} f(x)g(x)dx)^{2}}{∫_{0}^{1} g(x)^{2}dx} ≤ ∫_{0}^{1} f(x)^{2}dx

( ∫_{0}^{1} f(x)g(x)dx)^{2} ≤ ∫_{0}^{1} f(x)^{2}dx · ∫_{0}^{1} g(x)^{2}dx
∫_{0}^{1} f(x)g(x)dx ≤ ( ∫_{0}^{1} f(x)^{2}dx)^{\frac{1}{2}} · ( ∫_{0}^{1} g(x)^{2}dx)^{\frac{1}{2}} .  (16.49)

Again, for general f and g, replace f(x) with |f(x)| and g(x) with |g(x)| above. Note there is nothing special about ∫_{0}^{1}. □

The Cauchy-Schwarz Inequality is often useful when g(x) = 1. In this special case, it is important that we integrate over a finite interval.

Exercise 16.5.2. For what f and g is the Cauchy-Schwarz Inequality an equality?

16.6 Partial Summation

Lemma 16.6.1 (Partial Summation: Discrete Version). Let A_{N} = ∑_{n=1}^{N} a_{n}. then

∑_{n=M}^{N} a_{n}b_{n} = A_{N}b_{N} − A_{M−1}b_{M} + ∑_{n=M}^{N−1} A_{n}(b_{n} − b_{n+1})  (16.50)

Proof. Since A_{n} − A_{n_{1}} = a_{n},

∑_{n=M}^{N} a_{n}b_{n} = ∑_{n=M}^{N}(A_{n} − A_{n−1})b_{n}
= (A_{N} − A_{N−1})b_{N} + (A_{N−1} − A_{N−2})b_{N−1} + ⋯ + (A_{M} − A_{M−1})b_{M}
= A_{N}b_{N} + (−A_{N−1}b_{N} + A_{N−1}b_{N−1}) + ⋯ + (−A_{M}b_{M+1} + A_{M}b_{M}) − a_{M−1}b_{M}
= A_{N}b_{N} − a_{M_{1}}b_{M} + ∑_{n=M}^{N−1} A_{n}(b_{n} − b_{n+1}).  (16.51)

Lemma 16.6.2 (Abel’s Summation Formula - Integral Version). Let h(x) be a continuously differentiable function. Let A(x) = ∑_{n≤x} a_{n}. Then

∑_{n≤x} a_{n}h(n) = A(x)h(x) − ∫_{1}^{x} A(u)h′(u)du  (16.52)

See, for example, W. Rudin, Principles of Mathematical Analysis, page 70. Partial Summation allows us to take knowledge of one quantity and convert

that to knowledge of another.

For example, suppose we know that

∑_{p≤x} log p = x + O(x^{\frac{1}{2} +ϵ}).  (16.53)

We use this to glean information about ∑_{p≤x} 1.

Define

h(n) = \frac{1}{log n} and a_{n} = \begin{cases}log n & if n is prime \\ 0 & otherwise. \end{cases}  (16.54)

Applying partial summation to ∑_{p≤x} a_{n}h(n) will give us knowledge about ∑_{p≤x} 1. Note as long as h(n) = \frac{1}{log n} for n prime, it doesn’t matter how we define h(n) elsewhere; however, to use the integral version of Partial Summation, we need h to be a differentiable function.

Thus

∑_{p≤x} 1 = ∑_{p≤x} a_{n}h(n)

= (x + O(x^{\frac{1}{2} +ϵ}))\frac{1}{log x} − ∫_{2}^{x} (u + O(u^{\frac{1}{2} +ϵ}))h′(u)du. (16.55)

The main term (A(x)h(x)) equals \frac{x}{log x} plus a significantly smaller error. We now calculate the integral, noting h′(u) = −\frac{1}{u log^{2} u}. The error piece in the

integral gives a constant multiple of

∫_{2}^{x} \frac{u^{\frac{1}{2} +ϵ}}{u log^{2} u}du.  (16.56)

As \frac{1}{log^{2} u} ≤ \frac{1}{log^{2} 2} for 2 ≤ u ≤ x, the integral is bounded by

\frac{1}{log^{2} 2} ∫_{2}^{x} u^{− \frac{1}{2} +ϵ} < \frac{1}{log^{2} 2} \frac{1}{\frac{1}{2} + ϵ}x^{\frac{1}{2} +ϵ},  (16.57)

which is significantly less than A(x)h(x) = \frac{x}{log x}. We now need to handle the other integral:

∫_{2}^{x} \frac{u}{u log^{2} u}du = ∫_{2}^{x} \frac{1}{log^{2} u}du.  (16.58)

The obvious approximation to try is \frac{1}{log^{2} u} ≤ \frac{1}{log^{2} 2}. Unfortunately, plugging this in bounds the integral by \frac{x}{log^{2} 2}. This is larger than the expected main term, A(x)h(x)!

As a rule of thumb, whenever you are trying to bound something, try the simplest, most trivial bounds first. Only if they fail should you try to be clever.

Here, we need to be clever, as we are bounding the integral by something larger than the observed terms.

We split the integral into two pieces:

∫_{2}^{x} = ∫_{2}^{√\overline{x}} + ∫_{√\overline{x}}^{x}  (16.59)

For the first piece, we use the trivial bound for \frac{1}{log^{2} u}. Note the interval has

length \sqrt{x} − 2 < \sqrt{x}. Thus, the first piece contributes at most \frac{x^{\frac{1}{2}}}{log^{2} 2}, significantly less than A(x)h(x).

The reason trivial bounds failed for the entire integral is the length was too large (of size x); there wasn’t enough decay in the function.

The advantage of splitting the integral in two is that in the second piece, even though most of the length of the original interval is here (it is of length x − \sqrt{x} ≈ x), the function \frac{1}{log^{2} u} is small here. Instead of bounding it by a constant, we now bound it by substituting in the smallest value of u on this interval, \sqrt{x}. Thus, the contribution from this integral is at most \frac{x−\sqrt{x}}{log^{2} \sqrt{x}} < \frac{4x}{log^{2} x}. Note that this is significantly less than the main term A(x)h(x) = \frac{x}{log x}.

Multiplicative Functions, Kloosterman, p-adic Numbers, and Review of the Three Problems: Germain Primes, λ_{1}(G) for Random Graphs, Randomness of x → \overline{x} mod p

Multiplicative Functions, Kloosterman Sums and Bounds, p-adic Numbers. Review of the Three Problems (Germain Primes, Randomness of x → \overline{x} mod p, Random Graphs). Lecture by Peter Sarnak; notes by Steven Miller.

17.1 Multiplicative Functions, Kloosterman and p-

adic Numbers

17.1.1 Multiplicative Functions

Definition 17.1.1 (Multiplicative Functions). Let f be defined on the positive integers ℕ. f is multiplicative if

f(mn) = f(m)f(n) if m and n are relatively prime.  (17.1)

This is the same as

∑_{n=1}^{∞} \frac{f(n)}{n^{s}} = ∏_{p} (1 + f(p)p^{−s} + f(p^{2})p^{−2s} + ⋯).  (17.2)

We call the above an Euler Product. The standard example is the Riemann Zeta Function,

ζ(s) = ∑_{n=1}^{∞} \frac{1}{n^{s}} = ∏_{p} \frac{1}{1 − p^{−s}} .  (17.3)

There are many multiplicative functions (some, like Dirichlet Characters, do not even require m and n to be relatively prime).

The Kloosterman sums are not multiplicative.

17.1.2 Kloosterman Sums

If c_{1} and c_{2} are relatively prime,

K(a, b, c_{1}c_{2}) = K(∗, ∗, c_{1}) · K(∗, ∗, c_{2}),  (17.4)

where the ∗s are functions of a, b, c_{1} and c_{2}. Say we show |K(a, b, p)| ≪ p^{ν}, where ν does not depend on a or b or p. Then, if c = ∏_{i} p_{i}^{r_{i}}, we have

K(a, b, ∏_{i} p_{i}^{r_{i}}) = ∏_{i} K(∗, ∗, p_{i}^{r_{i}}).  (17.5)

Thus, we just need to get bounds over prime powers to bound a general Kloosterman sum.

Theorem 17.1.2 (Salie). If α ≥ 2,

K(a, b, p^{α}) ≤ 2p^{\frac{α}{2}} .  (17.6)

Proof: elementary.

∑_{x mod p^{2}} = ∑_{x_{1}=0}^{p−1} ∑_{x_{2}=0}^{p−1},  (17.7)

where x = x_{1} + x_{2}p and x_{1}, x_{2} ∈ {0, 1, …, p − 1}. Thus, when we encounter terms like e(\frac{\overline{x_{1}+px_{2}}}{p^{2}} ), we need the inverse of \overline{x_{1} + px_{2}}.

Let x_{1}^{−1} be the inverse of x_{1} mod p. Then

(x_{1} + px_{2})^{−1} = x_{1}^{−1}(1 + x_{1}^{−1}x_{2}p)^{−1}.  (17.8)

Note that

(1 − pb)^{−1} = 1 + pb + O(p^{2}).  (17.9)

Say we want the inverse mod p^{2} of (1 − bp). Try multiplying by (1 + bp). We get

(1 − pb)(1 + pb) = 1 − b^{2}p^{2} ≡ 1 mod p^{2}.  (17.10)

The above arguments is Hensel’s Lemma.

17.1.3 p-adic numbers

We define the p-adic norm of a rational α = \frac{a}{b}, a and b relatively prime, by

||α||_{p} = p^{−m}, where \frac{a}{b} = p^{m}l, (p, l) = 1.  (17.11)

Note that numbers that are highly divisible by p are small p-adically. We have the rationals ℚ and the p-adic norm || ∗ ||_{p}. Similar to completing

the rationals ℚ with the usual norm to get ℝ, we can complete the rationals with respect to this norm. The resulting field is called the p-adic numbers, ℚ_{p}.

ℚ ⊂ ℝ, and ℚ is dense in ℝ. Similarly ℚ ⊂ ℚ_{p} and ℚ is dense in ℚ_{p}. Let x ∈ ℚ_{p}. Then

x = \frac{a_{−m}}{p^{m}} + \frac{a_{−m+1}}{p^{m−1}} + ⋯ + a_{0} + a_{1}p + a_{2}p^{2} + ⋯,  (17.12)

where 0 ≤ a_{i} ≤ p − 1. Suppose we have a solution f(x_{0}) ≡ 0 mod p. We then try and find x_{1} such

that f(x_{0} + px_{1}) ≡ 0 mod p^{2}. Hensel noted that all we need to find x_{1} is some knowledge of the derivative at the previous stage.

17.2 Germain Primes

p − 1 = 2q, p and q prime. What are the statistics? How many are there up to x? Do they know about each other? What are their correlations? What about p − 3?

The Circle Method is a way of trying to solve additive equations (Waring’s Problem, Goldbach’s Problem p_{1}+p_{2} = 2n, Vinogradov’s Three Primes Theorem p_{1} + p_{2} + p_{3} = 2n + 1, Twin Primes p_{2} − p_{1} = 2).

Definition 17.2.1 (Germain Primes). If p is prime and p−1 = 2q for some prime q, we say p is a Germain Prime.

Definition 17.2.2 (π_{G}(x)). Recall π(x) is the number of primes at most x. Then π(x) ∼ \frac{x}{log x}. Let π_{G}(x) be the number of Germain primes at most x. If the probability of getting a prime is log x, then we might expect that

π_{G}(x) = ∑_{{p≤x \atop p−1=2q}} 1 ∼ Const · \frac{x}{log^{2} x}.  (17.13)

Using the circle method, we will try and estimate the above constant, and hope the minor arcs do not contribute to the main term.

Major McMahan (from the army, friendly with Hardy and Littlewood) made tables of primes to the millions. He checked, and Hardy and Littlewood’s constant (for twin primes) was correct and Sylvester (who made a probabilistic argument) was shown to be slightly off.

See Hardy and Littlewood, Acta Mathematica, v44, 1923, Partitio numerorum. III: On the expression of a number as a sum of primes.

We will then investigate the nearest neighbor and k^{th}-nearest neighbor spacings.

Also look at Robert Vaughn, The Hardy Littlewood Method.

17.3 Randomness of x → \overline{x}

Given a prime p, look at x → \overline{x}. How do we compute \overline{x}?

One can compute \overline{x} by using the Euclidean Algorithm (very fast, log p steps). Recall the Euclidean Algorithm gives a and b such that ax + bp = 1. Thus, mod p, ax ≡ 1, or a = \overline{x} mod p.

We now study the spacings between \overline{x} as x ranges in some interval mod p. If the interval is very small, we don’t expect randomness. What if we take an interval of length \sqrt{p}. Do we see Poissonian Behavior there for a fixed prime?

Now, fix a number a. Look at \frac{a mod p}{p} as you vary p.

Theorem 17.3.1 (Duke-Iwaniec). Suppose p ≡ 1 mod 4. There is a square root of −1 mod p; ie, ∃x such that x^{2} ≡ 1 mod p. Now we can take 1 ≤ x ≤ \frac{p−1}{2}, so

\frac{x}{p} ∈ [0, \frac{1}{2}]. Then \frac{x}{p}, as we vary p, is equi-distributed.

One can also look at the longest increasing sub-sequence.

Knuth, volume 2 of the Art of Computer Programming. Look at the stuff on generating random numbers.

17.4 Random Graphs / Ramanujan Graphs

Bollobas, Random Graphs: he will have a model of the random 3-regular graphs (what it means, how to generate, how many are there). Very hard if you don’t distinguish between isomorphic graphs (which have the same spectrum).

Look at the distribution of the second largest eigenvalue λ_{1} of the random 3-regular graph. Find the mean and the variance, graph.

Professor Sarnak will give a lecture on the Tracy-Widom distribution (which is the distribution of the biggest eigenvalue in some random ensemble – will we see the same distribution here)?

What is the scale for normalizing?

Take an interval, see how many eigenvalues in it, slide the interval down, and see how the number varies.

Random Graphs, Autocorrelations, Random Matrix Theory and the Mehta-Gaudin Theorem

Random Graphs (especially graphs with large girth and chromatic number), Autocorrelation, Random Matrix Theory (Vandermonde determinants, orthonormal polynomials) and the Mehta-Gaudin Theorem (large N limits of quantities related to the joint density function of the eigenvalues). Lecture by Peter Sarnak; notes by Steven Miller.

18.1 Random Graphs

Definition 18.1.1 (Chromatic Number). The chromatic number is the least number of colors such that each vertex has a different color than all of its neighbors.

Example 18.1.2. A bi-partite graph is 2-colorable, as is a tree (alternate colors as you go through the generations).

What can force you to have a lot of colors? If a vertex is joined to n_{v} vertices, you need a lot of colors if the vertices it is joined to are joined to each other.

Definition 18.1.3 (Girth). The girth is the shortest closed cycle.

If the girth is large, make a vertex blue, yellow next level, blue on next level, et cetera until you come back on yourself.

Question 18.1.4. Can you make a graph with large girth and large chromatic number?

The two fight each other. Erdos solved this problem by showing that if you take a Random Graph (with suitable properties), then that graph will have large girth and large chromatic number with high probability.

Take a Random Graph with n vertices and basically n^{1+ϵ} edges placed at random among the \binom{n}{2} = O(n^{2}) possible edges.

The random graph has short cycles, but the number of short cycles is small. Erdos removes certain graphs with small girth, and shows with high probability the graphs left have large girth and large chromatic number.

See Mckay’s paper: he proves Kesten’s measure holds for the random graph as the number of vertices goes to infinity.

18.2 Baire Category

Given α ∉ ℚ and C_{α}, how often can we find \frac{a}{q} ∈ ℚ such that

|α − \frac{a}{q}| ≥ \frac{C_{α}}{q^{2+ϵ}} .  (18.1)

In Lebesgue Measure, almost all α satisfy the above infinitely often.

In the Baire Category, this inequality does not hold infinitely often.

18.3 Autocorrelation

Note: Alex Barnett lectured on this section.

x-axis is number of swaps n, y-axis is number of graphs with given λ_{1}, λ_{1}(n). Say takes 100 swaps to randomize. Then the y-value at 101 swaps should be independent of the y-value at 1 swap.

But we don’t know the number of swaps before we have moved far enough. Let λ′_{1}(n) = λ_{1}(n) − \overline{λ}_{1}, where \overline{λ}_{1} is the average value. Autocorrelation: Say the x-axis runs to m.

A(c) = \frac{1}{m} ∑_{n} λ′_{1}(n)λ′_{1}(n + c).  (18.2)

The above is a function of c, symmetric in c. As c gets large, A(c) dies to zero, and has largest value at c = 0.

Look for c such that say 10% of the area is from c onward.

18.4 Gaudin’s Method

18.4.1 Introduction

From Random Matrix Theory: we have a probability distribution on ℝ^{n}:

p_{β}(x_{1}, …, x_{N}) = c_{N}(β) ∏_{j<k} |x_{j} − x_{k}|^{β}e^{− ∑_{j=1}^{N} x_{j}^{2}}dx_{1} ⋯ dx_{N}.  (18.3)

Start off with a real N ×N matrix, diagonalize with eigenvalues x_{1}, …, x_{N}. If you choose the matrix at random, you get N numbers, and you have a probability distribution on the eigenvalues.

We’ve derived the probability above for N × N matrices. For convenience, we order the eigenvalues.

If β = 1 we call the ensemble GOE (Gaussian Orthogonal Ensembles); if β = 2 we have GUE (Unitary) and if β = 4 we have GSE (Symplectic).

What is the correlation between two eigenvalues? What is the probability of observing a given spacing between two eigenvalues? We’ve done this in the 2 × 2 case.

In the N × N case, we would need to integrate out most of the eigenvalues. The difficulty is ∏ |x_{j} − x_{k}|^{β}.

For β = 1, 2 or 4, we can evaluate these integrals; we are fortunate that these values are the ones that arise in practice.

In fact, even just determining c_{N}(β) is difficult. This is called the Selberg Integral, which A. Selberg solved in high school!

We will only consider β = 2, and will be interested in the limit as N → ∞ (under appropriate re-scaling).

R_{N}(x_{1}, …, x_{N}) = ∫_{ℝ} ⋯ ∫_{ℝ} p_{2}(x_{1}, x_{2}, …, x_{n}, x_{n+1}, …, x_{N})dx_{n+1} ⋯ dx_{N}.

(18.4)

This will be a symmetric function of the first n variables. If we integrate all but 1 variable we get the density of eigenvalues; if we integrate all but two we get information on pairs of eigenvalues.

Remark 18.4.1. β = 0 is Poissonian.

18.4.2 Vandermonde Determinants

Notation: dp means

dp(θ_{1}, …, θ_{N}) = c_{N} ∏_{j<k} |e^{iθ_{j}} − e^{iθ_{k}} |^{2}dθ_{1} ⋯ dθ_{N}.  (18.5)

We are now working on the N-torus [0, 2π] × ⋯ × [0, 2π]. This goes under the name CUE (Circular Unitary Ensemble).

Remember the group

U(N) = {N × N matrices A with AA^{⋆} = I}.  (18.6)

Similar to the diagonalization of symmetric matrices, for any unitary matrix U there is a unitary matrix V such that V^{−1}UV is diagonal; further, the eigenvalues have absolute value 1, and hence can be written as e^{iθ}.

Suppose we have f_{1}, …, f_{N}. We form the Vandermonde of the N-variables

Van(f_{1}, …, f_{N}) = ∏_{i<j}(f_{i} − f_{j}).  (18.7)

Today we will only use the square, so we don’t worry about ordering so that f_{i} < f_{j}.

Exercise 18.4.2.

Van(f_{1}, …, f_{N}) = det (f_{i}^{j−1})_{1≤i,j≤N}.  (18.8)

Thus, we have

| \begin{matrix}1 & 1 & ⋯ & 1 \\ f_{1} & f_{2} & ⋯ & f_{N} \\ ⋮ & ⋮ & ... & ⋮ \\ f_{1}^{N−1} & f_{2}^{N−1} & ⋯ & f_{N}^{N−1} \end{matrix} |  (18.9)

18.4.3 Orthonormal Polynomials

On the unit circle T, we have the measure

dμ(t) = \frac{dt}{2π} .  (18.10)

Let f(t) be a function such that

∫_{T} f(t)dμ(t) = 0, ∫_{T} |f(t)|^{2}dμ(t) = 1.  (18.11)

Define a sequence of monic polynomials P_{n}(x) for n ∈ ℕ and ϕ_{n}(t) with

ϕ_{n}(t) = P_{n}(f(t)), ϕ_{0}(t) = \frac{1}{\sqrt{μ(T)}}, ∫_{T} ϕ_{i}(t)\overline{ϕ}_{j}(t)dμ(t) = δ_{ij}. (18.12)

This is Gramm-Schmidt, where the inner product between two functions f and g is given by

⟨f, g⟩ = ∫_{T} f(t)\overline{g}(t)dμ(t),  (18.13)

and the ‘Kronecker delta’ symbol (the discrete analog of the continuous delta ‘function’ δ(·)) is defined by

δ_{ij} = \begin{cases}1 & if i = j \\ 0 & otherwise \end{cases}  (18.14)

We introduce orthogonal polynomials to handle the integral. The above process (constructing the P_{n}s and the ϕ_{n}s) gives an orthonormal sequence of polynomials.

18.4.4 Kernel K_{N}(x, y)

Define the kernel

K_{N}(x, y) = ∑_{j=0}^{N−1} ϕ_{j}(x)\overline{ϕ}_{j}(y).  (18.15)

Exercise 18.4.3. Prove the following:

1. ∫_{T} K_{N}(x, x)dμ(x) = N.

2. ∫_{T} K_{N}(x, y)K_{N}(y, z)dμ(y) = K_{N}(x, z).

Remark 18.4.4.

∫_{T} K_{N}(x, y)g(y)dμ(y) = ∑_{j=0}^{N−1} [ ∫_{T} \overline{ϕ}_{j}(y)g(y)dμ(y)]ϕ_{j}(x).  (18.16)

Thus, integrating g against K_{N} projects g onto the first N vectors.

Define, for 1 ≤ n ≤ N,

D_{n,N}(t_{1}, …, t_{n}) = det (K_{N}(t_{j}, t_{k}))_{1≤j,k≤n}.  (18.17)

For example,

D_{1,N} = K_{N}(t_{1}, t_{1})  (18.18)

and

D_{2,N} = | \begin{matrix}K_{N}(t_{1}, t_{1}) & K_{N}(t_{1}, t_{2}) \\ K_{N}(t_{2}, t_{1}) & K_{N}(t_{2}, t_{2}) \end{matrix} |  (18.19)

18.4.5 Gaudin-Mehta Theorem

Theorem 18.4.5 (Gaudin-Mehta). We have

1.

\frac{1}{μ(T)}Van(f(t_{1}), …, f(t_{N})) = det_{N×N}(ϕ_{i−1}(t_{j}))_{1≤i,j≤N}. (18.20)

2.

\frac{1}{μ(T)}|Van(f(t_{1}), …, f(t_{N}))|^{2} = D_{N,N}(t_{1}, …, t_{N}).  (18.21)

3. For 1 ≤ n ≤ N,

∫_{T} D_{n,N}(t_{1}, …, t_{n})dμ(t_{n}) = (N + 1 − n)D_{n−1,N}(t_{1}, …, t_{n−1}). (18.22)

The third statement is the beef, allowing us to integrate out one variable at a time by induction.

Remember

D_{n,N}(t_{1}, …, t_{n}) = det_{n×n}(K_{N}(t_{j}, t_{k}))_{1≤j,k≤n}.  (18.23)

Corollary 18.4.6. Let F be a symmetric function of t_{1}, …, t_{n}, and define

F_{n,N}(t_{1}, …, t_{N}) = ∑_{1≤i_{1}<i_{2}···<i_{n}<N} F (t_{i_{1}}, …, t_{i_{n}})

dμ_{n,N}(t_{1}, …, t_{N}) = \frac{1}{n!}D_{n,N}(t_{1}, …, t_{N})dμ(t_{1}) ⋯ dμ(t_{N}). (18.24)

Then

∫_{T^{N}} F_{N,N}(t_{1}, …, t_{N})dμ_{n,N}(t_{1}, …, t_{N}) = ∫_{T^{n}} F (t_{1}, …, t_{n})dμ_{n,N}(t_{1}, …, t_{n}).

(18.25)

How might we use the above? For example, consider for 1 ≤ j, k ≤ N, and consider for f even

∑_{1≤j<k≤N} f(x_{j} − x_{k}).  (18.26)

What is the expectation of the above? In this case, F is a function of two variables, and F (x_{1}, x_{2}) = f(|x_{1} −x_{2}|) and we now just need to integrate f(|x_{1} − x_{2}|) against the determinant of a 2 × 2 matrix, and this is the only place where N will arise.

Suppose we had

dp(x_{1}, …, x_{N}) = e^{− ∑ x_{j}^{2}} ∏_{j<k} |x_{j} − x_{k}|^{2}dx_{1} ⋯ dx_{N}.  (18.27)

Consider the expectation of

∑_{1≤j<k≤N} f(|x_{j} − x_{k}|).  (18.28)

According to the corollary, the answer is just

∫_{ℝ} ∫_{ℝ} f(|x_{1} − x_{2}|)\frac{1}{2!} | \begin{matrix}K_{N}(x_{1}, x_{1}) & K_{N}(x_{1}, x_{2}) \\ K_{N}(x_{2}, x_{1}) & K_{N}(x_{2}, x_{2}) \end{matrix} | e^{−x_{1}^{2}−x_{2}^{2}}dx_{1}dx_{2}. (18.29)

This is enormous progress – we started with N variables; we now have 2 variables. We need to take the N → ∞ limit of the determinant, a much easier question.

18.4.6 Example

T = [0, 2π], dμ(x) = \frac{dx}{2π}, f(x) = e^{ix}, f^{n}(x) = e^{inx}, and P_{n}(x) = x^{n}. These P_{n}s are monic, ϕ_{n}(x) = P_{n}(f(x)) is orthonormal, which gives ϕ_{n}(x) = e^{inx}, and clearly

∫_{0}^{2π} e^{inx}e^{−imx}dx = δ_{ij}.  (18.30)

Finally, we obtain a geometric progression

K_{N}(x, y) = ∑_{n=0}^{N−1} e^{in(x−y)}

= \frac{1 − e^{iN(x−y)}}{1 − e^{i(x−y)}} .  (18.31)

We will symmetrize (and go from −N to N), and when we take the 2 × 2 determinant, we get something like

\begin{matrix}sin & ( & \frac{N(x−y)}{2} & ) \\ sin & ( & \frac{x−y}{2} & ) \end{matrix}.  (18.32)

The most famous pair correlation: we have N eigenvalues so that the mean spacing is 1. The pair correlation is

1 − [\frac{sin (π(x − y))}{π(x − y)}]^{2}.  (18.33)

Increasing Length Subsequences and Tracy-Widom

More on increasing length subsequence and the Tracy-Widom distribution. Lecture by Peter Sarnak, notes by Steven Miller.

19.1 Increasing Length Subsequences

Consider the set of permutations S_{n} on n numbers. Let σ ∈ S_{n} be a random permutation, and let L_{σ} be the length of the longest increasing sub-sequence. What is the expected value of L_{σ}?

Conjecture 19.1.1 (Ulam).

E[L_{σ}] ∼ 2\sqrt{n}.  (19.1)

Proved by several people (Schepp, Vircheck (?), ...).

At Bell Labs, many people (including Odlyzko) investigated. Monte-Carlo simulations for variance (beginning 1993). Dividing variance by n^{\frac{1}{3}} was good. Looking at the expectation of

\frac{L_{σ} − 2\sqrt{n}}{n^{\frac{1}{6}}}  (19.2)

and investigated whether or not it went to a limit. Noticed this distribution is negative (shifted to the left). Prefers to be less than 2\sqrt{n}.

19.2 Tracy-Widom

On ℝ^{n}, we have

P_{N}(x)dx = e^{− ∑_{j=1}^{N} x_{j}^{2}} ∏_{j<k} |x_{j} − x_{i}|^{β} dx_{1} ⋯ dx_{n}, β ∈ {1, 2, 4}.  (19.3)

Will use the weight e^{−x^{2}}dx, which will give rise to Hermite polynomials.

Definition 19.2.1 (F_{N,β}(s)). F_{N,β}(s) is the probability that there is no x ∈ [s, ∞).

We can write this as a determinant (see the papers by Mehta and Mehta-Gaudin). We will have again

K_{N}(x, y) = ∑_{0≤j≤N−1} ϕ_{j}(x)\overline{ϕ}_{j}(y).  (19.4)

Remember the semi-circle rule, that (with some normalization) the eigenvalues lie in [−2\sqrt{N}, 2\sqrt{N}].

What is the expected value of the largest eigenvalue? We know most are in [−2\sqrt{N}, 2\sqrt{N}]. We haven’t discussed whether or not there are outliers. With probability one, we can show that there will be no such outliers.

We have N eigenvalues. Near 0 is called the bulk. As there are N numbers, we expect eigenvalues in an interval of size \frac{1}{N} near the origin.

What about eigenvalues near the edges, ±2\sqrt{N}? Let s ∈ (−∞, ∞). Look at

F_{N}(2\sqrt{N} + \frac{s}{N^{\frac{1}{6}}} ).  (19.5)

This is the scaling limit. Say s = 0. this would give us what is happening at the origin.

Theorem 19.2.2 (Tracy-Widom).

lim_{N→∞} F_{N}(2\sqrt{N} + \frac{s}{N^{\frac{1}{6}}} ) = F_{β}(s), F_{β}(s) = \frac{dF_{β}(s)}{ds}.  (19.6)

Here the N^{\frac{1}{6}} arises from the particular problem we’re interested in. Here, we are looking at eigenvalues from random matrices.

Circle Method and Germain Primes

Using the Hardy-Littlewood Circle Method (and assuming no main term contribution from the Minor Arcs), we calculate the expected number of Germain primes. Calculations and notes by Steven Miller.

20.1 Preliminaries

20.1.1 Definitions

Let

e(x) = e^{2πix}  (20.1)

and

λ(n) = \begin{cases}log p & if n = p is prime \\ 0 & otherwise \end{cases}  (20.2)

Finally, define

c_{q}(a) = ∑_{{r=1 \atop (r,q)=1}}^{q} e(r\frac{a}{q}).  (20.3)

20.1.2 Partial Summation

Lemma 20.1.1 (Partial Summation: Discrete Version). Let A_{N} = ∑_{n=1}^{N} a_{n}. then

∑_{n=M}^{N} a_{n}b_{n} = A_{N}b_{N} − A_{M−1}b_{M} + ∑_{n=M}^{N−1} A_{n}(b_{n} − b_{n+1})  (20.4)

Proof. Since A_{n} − A_{n−1} = a_{n},

∑_{n=M}^{N} a_{n}b_{n} = ∑_{n=M}^{N}(A_{n} − A_{n−1})b_{n}
= (A_{N} − A_{N−1})b_{N} + (A_{N−1} − A_{N−2})b_{N−1} + ⋯ + (A_{M} − A_{M−1})b_{M}
= A_{N}b_{N} + (−A_{N−1}b_{N} + A_{N−1}b_{N−1}) + ⋯ + (−A_{M}b_{M+1} + A_{M}b_{M}) − a_{M−1}b_{M}
= A_{N}b_{N} − a_{M_{1}}b_{M} + ∑_{n=M}^{N−1} A_{n}(b_{n} − b_{n+1}).  (20.5)

Lemma 20.1.2 (Abel’s Summation Formula - Integral Version). Let h(x) be a continuously differentiable function. Let A(x) = ∑_{n≤x} a_{n}. Then

∑_{n≤x} a_{n}h(n) = A(x)h(x) − ∫_{1}^{x} A(u)h′(u)du  (20.6)

See, for example, W. Rudin, Principles of Mathematical Analysis, page 70.

20.1.3 Siegel-Walfisz

Theorem 20.1.3. [Siegel-Walfisz] Let C, B > 0, and let a and q be relatively prime. Then

∑_{{p≤x \atop p≡a(q)}} log p = \frac{x}{ϕ(q)} + O(\frac{x}{log^{C} x})  (20.7)

for q ≤ log^{B} x, and the constant above does not depend on x, q or a (ie, it only depends on C and B).

20.1.4 Germain Integral

Define

f_{1N}(x) = ∑_{p_{1}≤N} log p_{1} · e(p_{1}x)
f_{2N}(x) = ∑_{p_{2}≤N} log p_{2} · e(−2p_{2}x)
f_{N}(x) = ∑_{p_{1}≤N} ∑_{p_{2}≤N} log p_{1} log p_{2} · e((p_{1} − 2p_{2})x).  (20.8)

Consider

∫− \frac{1}{2}^{\frac{1}{2}} f_{N}(x)e(−x)dx = ∑_{p_{1}≤N} ∑_{p_{2}≤N} log p_{1} log p_{2} ∫− \frac{1}{2}^{\frac{1}{2}} e((p_{1} − 2p_{2} − 1)x)dx.

(20.9)

Note

∫− \frac{1}{2}^{\frac{1}{2}} e((p_{1} − 2p_{2} − 1)x)dx = \begin{cases}1 & if p_{1} − 2p_{2} − 1 = 0 \\ 0 & if p_{1} − 2p_{2} − 1 ≠ 0 \end{cases}  (20.10)

Thus, we get a contribution of log p_{1} log p_{2} if p_{1} and p_{2} = \frac{p_{1}−1}{2} are both primes. Thus,

∫− \frac{1}{2}^{\frac{1}{2}} f_{N}(x)e(−x)dx = ∑_{{p_{1}≤N \atop p_{2}= \frac{p_{1}−1}{2} prime}} log p_{1} log p_{2}.  (20.11)

The above is a weighted counting of Germain primes.

20.1.5 Major and Minor Arcs

Let B be a positive integer, Q = log^{B} N, and define the Major Arc M_{a,q}

M_{a,q} = \{x ∈ [0, 1) : |x − \frac{a}{q}| < \frac{Q}{N} \}.  (20.12)

We also add in one interval centered at either 0 or 1, ie, the "interval" (or wrapped-around interval)

[0, \frac{Q}{N} ] ∪ [1 − \frac{Q}{N} , 1].  (20.13)

For convenience, we often use the interval [−\frac{1}{2}, \frac{1}{2}] instead of [0, 1], in which case we would have

[ − \frac{1}{2}, −\frac{1}{2} + \frac{Q}{N} ] ⋃ [\frac{1}{2} − \frac{Q}{N} , \frac{1}{2}].  (20.14)

For functions that are periodic of period one, we could instead consider

[\frac{1}{2} − \frac{Q}{N} , \frac{1}{2} + \frac{Q}{N} ].  (20.15)

The Major Arcs are defined by

M = ⋃_{q≤Q} ⋃_{{a=1 \atop (a,q)=1}}^{q} M_{a,q}.  (20.16)

The Minor Arcs, m, are whatever is not in the Major Arcs. Then

∫− \frac{1}{2}^{\frac{1}{2}} f_{N}(x)e(−x)dx = ∫_{M} f_{N}(x)e(−x)dx + ∫_{m} f_{N}(x)e(−x)dx. (20.17)

We will assume that there is no net contribution over the minor arcs. Thus, in the sequel we investigate

∫_{M} f_{N}(x)e(−x)dx.  (20.18)

20.1.6 Reformulation of Germain Integral

f_{1N}(x) = ∑_{m_{1}≤N} λ(m_{1}) · e(m_{1}x)
f_{2N}(x) = ∑_{m_{2}≤N} λ(m_{2}) · e(−2m_{2}x)
f_{N}(x) = ∑_{m_{1}≤N} ∑_{m_{2}≤N} λ(m_{1})λ(m_{2}) · e((m_{1} − 2m_{2})x). (20.19)

We investigate

∫_{M} f_{N}(x)e(−x)dx.  (20.20)

We will show the Major Arcs contribute, up to lower order terms, T_{2}N, where T_{2} is a constant independent of N. The length of the Major Arc M_{a,q} is \frac{Q}{N}. We sum over (a, q) = 1 and q ≤ Q. Thus, the total length is bounded by

∑_{q≤Q} q · \frac{Q}{N} ≪ \frac{Q^{3}}{N} ≪ \frac{log^{B}}{N} .  (20.21)

By choosing B sufficiently large, we will be able to make all the errors from the Major Arc calculations less than the main term from the Major Arcs. Of course, we have absolutely no control over what happens on the minor arcs, and we will simply assume there is no contribution from the minor arcs.

Thus, on the Major Arc M_{a,q}, success will be in finding a function of size N^{2} such that the error from this function to f_{N}(x) on M_{a,q} is much smaller than N^{2}, say N^{2} divided by a large power of log N.

Similarly, when we integrate over the Major Arcs, we will find the main terms will be of size N; again, success will be in showing the errors in the approximations are much smaller than N, say N divided by a large power of log N.

We are able to do this because of the Siegel-Walfisz Theorem (Theorem 20.1.3). Given any B > 0, we can find a C > 0 such that, if q ≤ log^{B} N, then

∑_{{p≤N \atop p≡r(q)}} log p = \frac{N}{ϕ(q)} + O(\frac{N}{log^{C} N} ),  (20.22)

(r, q) = 1. Thus, we can take C enormous, large enough so that even when we multiply by the length of the Major Arcs (of size \frac{log^{3B} N}{N}, we still have something small.

20.2 f_{N}(x) and u(x)

20.2.1 f(\frac{a}{q})

We now calculate f_{N}(\frac{a}{q}) for q ≤ log^{B} N.

Up to lower order terms,

f_{N}(\frac{a}{q}) = ∑_{p_{1}≤N} log p_{1} · e(p_{1} \frac{a}{q}) ∑_{p_{2}≤N} log p_{2} · e( − 2p_{2} \frac{a}{q})

= ∑_{r_{1}=1}^{q} ∑_{{p_{1}≤N \atop p_{1}≡r_{1}(q)}} log p_{1} · e(p_{1} \frac{a}{q}) ∑_{r_{2}=1}^{q} ∑_{{p_{2}≤N \atop p_{2}≡r_{1}(q)}} log p_{2} · e( − 2p_{2} \frac{a}{q})

= ∑_{r_{1}=1}^{q} e(r_{1} \frac{a}{q}) ∑_{r_{2}=1}^{q} e(r_{2} \frac{−2a}{q}) ∑_{{p_{1}≤N \atop p_{1}≡r_{1}(q)}} log p_{1} ∑_{{p_{2}≤N \atop p_{2}≡r_{2}(q)}} log p_{2}

= \frac{N^{2}}{ϕ^{2}(q)} ∑_{{r_{1}=1 \atop (r_{1},q)=1}}^{q} e(r_{1} \frac{a}{q}) ∑_{{r_{2}=1 \atop (r_{2},q)=1}}^{q} e(r_{2} \frac{−2a}{q})

= \frac{N^{2}}{ϕ^{2}(q)}c_{q}(a)c_{q}(−2a),  (20.23)

where the second to last line follows from the Siegel-Walfisz Theorem (Theorem 20.1.3). We restrict to (r_{i}, q) = 1 because if (r_{i}, q) > 1, there is at most one prime p_{i} ≡ r_{i} mod q.

20.2.2 u(x)

Let

u(x) = ∑_{m_{1}≤N} ∑_{m_{2}≤N} e((m_{1} − 2m_{2})x).  (20.24)

We will often look at

\frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)}u(x).  (20.25)

Note

u(0) = N^{2}.  (20.26)

20.3 f_{N}(α) − \frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)}u(α − \frac{a}{q}), α ∈ M_{a,q}

Let

C_{q}(a) = \frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)}.  (20.27)

We write α as β + \frac{a}{q}, β ∈ [ − \frac{Q}{N} , \frac{Q}{N} ], Q = log^{B} N. As always, we ignore lower order terms.

Note f_{N}(x) is approximately C_{q}(a)N^{2} for x near \frac{a}{q}. We now expand and show

f_{N}(α) is C_{q}(a)u(α − \frac{a}{q}) plus errors of size \frac{N^{2}}{log^{C−2B} N} for α ∈ M_{a,q}.

20.3.1 Setup

S_{a,q}(α) = f_{N}(α) − C_{q}(a)u(α − \frac{a}{q})
= ∑_{m_{1},m_{2}≤N} λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})α) − C_{q}(a) ∑_{m_{1},m_{2}≤N} e((m_{1} − 2m_{2})β)

= ∑_{m_{1},m_{2}≤N} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]e((m_{1} − 2m_{2})β)

= ∑_{m_{1}≤N} [ ∑_{m_{2}≤N} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]e(−2m_{2}β)]e(m_{1}β)

(20.28)

We now apply Partial Summation multiple times. First, we apply Partial Summation to the m_{2}-sum:

S_{2;a,q} = ∑_{m_{2}≤N} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]e(−2m_{2}β)
= ∑_{m_{2}≤N} a_{m_{2}}b_{m_{2}}
= A_{2}(N)e(−2Nβ) + 4πiβ ∫_{0}^{N} ∑_{m_{2}≤u} a_{m_{2}}e(−uβ)du.  (20.29)

We hit the above with e(m_{1}β), and sum from m_{1} = 1 to N. We get two pieces:

S_{1 ∑;a,q} = ∑_{m_{1}≤N} A_{2}(N)e(−2Nβ) · e(m_{1}β)
S_{1 ∫;a,q} = ∑_{m_{1}≤N} 4πiβ ∫_{0}^{N} ∑_{m_{2}≤u} a_{m_{2}}e(−uβ)du · e(m_{1}β)
S_{a,q} = S_{1 ∑;a,q} + S_{1 ∫;a,q}.  (20.30)

20.3.2 S_{1 ∑;a,q}

S_{1 ∑;a,q} = ∑_{m_{1}≤N} A_{2}(N)e(−2Nβ) · e(m_{1}β)
= e(−2Nβ) ∑_{m_{1}≤N} A_{2}(N)e(m_{1}β)
= e(−2Nβ) ∑_{m_{1}≤N} ∑_{m_{2}≤N} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]e(m_{1}β)
= e(−2Nβ)[A_{1}(N)e(Nβ)

−2πiβ ∫_{0}^{N} ∑_{m_{1}≤t} ∑_{m_{2}≤N} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]e(tβ)dt.

(20.31)

First Piece

The first piece, the A_{1}(N)e(Nβ) term, is small for q ≤ Q. Why? We have (up to lower order terms)

A_{1}(N)e(Nβ) = ∑_{m_{1},m_{2}≤N} λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − ∑_{m_{1},m_{2}≤N} C_{q}(a)
= C_{q}(a)N^{2} − N^{2}C_{q}(a) = 0.  (20.32)

Thus, because of our choice of functions, the leading terms vanish, and the remaining term is small.

Second Piece

We now study the second piece. Note |β| ≤ \frac{Q}{N} = \frac{log^{2} B}{N}, and C_{q}(a) = \frac{c_{q}(a)}{ϕ^{2}(q)} \frac{c_{q}(−2a)}{ϕ^{2}(q)}. Up to lower order terms, the m_{2}-sum will leave us with

β \frac{c_{q}(−2a)N}{ϕ(q)} ∫_{0}^{N} ∑_{m_{1}≤t} [λ(m_{1})e(m_{1} \frac{a}{q}) − \frac{c_{q}(a)}{ϕ(q)}]e(tβ)dt.  (20.33)

Note f_{N}(x) is a multiple of N^{2} for x near \frac{a}{q}. Thus, we want to make sure the above is well dominated by N^{2}.

For t ≤ \sqrt{N}, this is immediate. For t ≥ \sqrt{N}, using Siegel-Walfisz (Theorem 20.1.3), we can make the bracketed quantity in the integrant dominated by \frac{N}{log^{C} N} for any C when q ≤ log^{B} N. Thus, we integrate a quantity that is at most \frac{N}{log^{C} N} over an interval of length N, we multiply by Nβ ≪ Q = log^{B} N.

Thus, choosing C appropriately, the integral contributes \frac{N^{2}}{log^{C−B} N}, and hence is negligible.

Remark 20.3.1. Note, of course, that the contribution is only negligible while |β| ≤ \frac{Q}{N}.

Lemma 20.3.2. S_{1 ∑;a,q} is a lower order correction.

20.3.3 S_{1 ∫;a,q}

We must evaluate

S_{1 ∫;a,q} = ∑_{m_{1}≤N} 4πiβ ∫_{0}^{N} ∑_{m_{2}≤u} a_{m_{2}}e(−uβ)du · e(m_{1}β), (20.34)

where

a_{m_{2}} = [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)].  (20.35)

We bring the sum over m_{1} inside the integral and again use Partial Summation. We will ignore the integration and β for now, as these will contribute βN ≪

Q = log^{B} N times the maximum value of the integrand. We will leave the e(−uβ)du with this integration.

When u ≤ \sqrt{N}, we can immediately show the above is a lower order correction. Thus, below we always assume u ≥ \sqrt{N}.

First Piece

We have

S_{1 ∫ ∑;a,q} = ∑_{{m_{1}≤N \atop m_{2}≤u}} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]e(Nβ)

= e(Nβ)[ ∑_{{m_{1}≤N \atop m_{2}≤u}} λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a) ∑_{{m_{1}≤N \atop m_{2}≤u}} 1].

= e(Nβ)[C_{q}(a)uN − C_{q}(a)uN + Lower Order Terms], (20.36)

where by the Siegel-Walfisz Theorem (Theorem 20.1.3), the error in the bracketed quantity is of size \frac{uN}{log^{C} N}.

We then integrate from u = \sqrt{N} to N and multiply by β, giving a contribution bounded by

βN · \frac{N^{2}}{log^{C} N} ≪ \frac{log^{B}}{N} \frac{N^{3}}{log^{C} N} ≪ \frac{N^{2}}{log^{C−B} N} ,  (20.37)

again getting a lower order correction to f_{N}(x) for x near \frac{a}{q} (remember f_{N}(x) is of size N^{2}).

Second Piece

Again, u ≥ \sqrt{N}, and we have

2πiβ ∫_{0}^{N} ∑_{m_{1}≤t} [ ∑_{m_{2}≤u} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]]e(tβ)dt.

(20.38)

Again, for t ≤ \sqrt{N}, the contribution will be a lower order correction. For t, u ≥ \sqrt{N},

Again, executing the sum over m_{1} and m_{2} will give us

C_{q}(a)ut − C_{q}(a)ut + Lower Order Terms,  (20.39)

with the lower order terms of size \frac{ut}{log^{C} N}. Integrating over t (from \sqrt{N} to N), then integrating over u (from \sqrt{N} to N)

and then multiplying by β^{2} gives an error bounded by

β^{2}N^{2} · \frac{N^{2}}{log^{C} N} ≪ \frac{log^{2B} N}{N^{2}} \frac{N^{4}}{log^{C} N} ≪ \frac{N^{2}}{log^{C−2B} N} ,  (20.40)

again a lower order correction.

20.4 Integrals of u(x)

20.4.1 Formulations

Remember

u(x) = ∑_{m_{1},m_{2}≤N} e((m_{1} − 2m_{2})x).  (20.41)

We need to study ∫− \frac{1}{2}^{\frac{1}{2}} f_{N}(x)e(−x)dx. We have shown that

f_{N}(α) = C_{q}(a)u(α − \frac{a}{q}) + O(\frac{N^{2}}{log^{C−2B} N} ), α ∈ M_{a,q}.  (20.42)

Thus, we must evaluate

∫_{M_{a,q}} u(α − \frac{a}{q}) · e(−α)dα = ∫_{\frac{a}{q} − \frac{Q}{N}}^{\frac{a}{q} + \frac{Q}{N}} u(α − \frac{a}{q}) · e(−α)dα

= ∫− \frac{Q}{N}^{\frac{Q}{N}} u(β) · e( − \frac{q}{q} − β)dβ

= e( − \frac{a}{q}) ∫− \frac{Q}{N}^{\frac{Q}{N}} u(β)e(−β)dβ. (20.43)

20.4.2 ∫−\frac{1}{2}^{\frac{1}{2}} u(x)e(−x)dx

∫− \frac{1}{2}^{\frac{1}{2}} u(x)e(−x)dx = ∫− \frac{1}{2}^{\frac{1}{2}} ∑_{m_{1},m_{2}≤N} e((m_{1} − 2m_{2})x) · e(−x)dx

= ∑_{m_{1},m_{2}≤N} ∫− \frac{1}{2}^{\frac{1}{2}} e((m_{1} − 2m_{2} − 1)x)dx. (20.44)

If m_{1} − 2m_{2} − 1 = 0, the integral gives 1. There are approximately \frac{N}{2} ways to choose m_{1}, m_{2} ≤ N such that m_{1} − 2m_{2} − 1 = 0.

Assume now m_{1} − 2m_{2} − 1 ≠ 0. Then the integral vanishes. Hence,

Lemma 20.4.1.

∫− \frac{1}{2}^{\frac{1}{2}} u(x)e(−x)dx = \frac{N}{2} + O(1).  (20.45)

20.4.3 ∫_{−\frac{1}{2}}^{−\frac{Q}{N}} + ∫ \frac{1}{2} ; \frac{Q}{N} u(x)e(−x)dx

Define

I_{1} = [ − \frac{1}{2}, −\frac{1}{2} + \frac{Q}{N} ]
I_{2} = [ − \frac{1}{2} + \frac{Q}{N} , −\frac{Q}{N} ]
I_{3} = [\frac{Q}{N} , \frac{1}{2} − \frac{Q}{N} ]
I_{4} = [\frac{1}{2} − \frac{Q}{N} , \frac{1}{2}]
I = I_{1} ∪ I_{2} ∪ I_{3} ∪ I_{4}.  (20.46)

20.4.4 Integral over I_{2}, I_{3}

We have

∫_{I_{i}} u(x)e(−x)dx = ∫_{I_{i}} ∑_{m_{1},m_{2}≤N} e((m_{1} − 2m_{2} − 1)x)dx

= ∫_{I_{i}} ∑_{m_{1}≤N} e(m_{1}x) ∑_{m_{2}≤N} e(−2m_{2}x) · e(−x)dx

= ∫_{I_{i}} \frac{e(x) − e((N + 1)x)}{1 − e(x)} \frac{e(−2x) − e(−2(N + 1)x)}{1 − e(−2x)}e(−x)dx.

(20.47)

On I_{2} and I_{3}, the integral is

≪ ∫_{I_{i}} \frac{2}{x} \frac{2}{x}dx ≪ \frac{N}{Q} = \frac{N}{log^{B} N} ,  (20.48)

see, for example, Nathanson (Additive Number Theory: The Classical Bases, Chapter 8).

20.4.5 Integral over I_{1}, I_{4}

Each of these intervals has length \frac{Q}{N} = \frac{log^{B} N}{N}. There are \frac{N}{2} + O(1) pairs such that m_{1} − 2m_{2} − 1 = 0. Each of these pairs will contribute (bound the integrand by 1) \frac{Q}{N}. As there are at most \frac{N}{2} pairs, these contribute at most \frac{N}{2} \frac{Q}{N} ≪ log^{B} N.

Henceforth we assume m_{1} − 2m_{2} − 1 ≠ 0. We write

I_{1} ∪ I_{4} = [\frac{1}{2} − \frac{Q}{N} , \frac{1}{2} + \frac{Q}{N} ] = I′.  (20.49)

We have

∑_{{m_{1},m_{2}≤N \atop m_{1}−2m_{2}−1≠0}} ∫_{I′} e((m_{1} − 2m_{2} − 1)x)dx

= e( − \frac{1}{2}) ∑_{{m_{1},m_{2}≤N \atop m_{1}−2m_{2}−1≠0}}(−1)^{m_{1}} ∫− \frac{Q}{N}^{\frac{Q}{N}} e((m_{1} − 2m_{2} − 1)x)dx

= e( − \frac{1}{2})\frac{1}{2πi} ∑_{{m_{1},m_{2}≤N \atop m_{1}−2m_{2}−1≠0}}(−1)^{m_{1}} \frac{2 sin ((m_{1} − 2m_{2} − 1)\frac{Q}{N} )}{m_{1} − 2m_{2} − 1},(20.50)

because, changing variables by sending x to (x − \frac{1}{2}) + \frac{1}{2} gives factors of

e((m_{1} − 2m_{2} − 1)\frac{1}{2}) = e(−\frac{1}{2})e(\frac{m_{1}}{2} )e(−m_{2}), and e(\frac{m_{1}}{2} ) = (−1)^{m_{1}}.

0 < |m_{1} − 2m_{2} − 1| ≤ N^{1−ϵ}

Let w = m_{1} − 2m_{2} − 1. We will do the case 0 < w ≤ N^{1−ϵ}, the case with −N^{1−ϵ} > w > 0 being handled similarly.

For each w, there are at most N pairs of m_{1}, m_{2} giving rise to such a w. For such w, \frac{sin(w \frac{Q}{N})}{w} ≪ \frac{Q}{N} (because we are taking the sin of a quantity very close to zero).

Thus, these pairs contribute at most

≪ N · \frac{Q}{N} ≪ Q = log^{B} N.  (20.51)

Inserting absolute values in Equation 20.50 gives a contribution of at most log^{B} N for such w, 0 < w < N^{1−ϵ}.

N^{1−ϵ} < |m_{1} − 2m_{2} − 1| ≤ N

Again, let w = m_{1} − 2m_{2} − 1 and assume N^{1−ϵ} < |w| ≤ N. We will only consider w > 0; w < 0 is handled similarly.

The cancellation is due to the presence of the factor (−1)^{m_{1}}; note that for the pair (m_{1}, m_{2}) we only care about the parity of m_{1}.

Consider w and w − 1. For m_{1} − 2m_{2} − 1 = w, the solutions are

m_{1} = w + 3, m_{2} = 1
m_{1} = w + 5, m_{2} = 2
m_{1} = w + 7, m_{2} = 3  (20.52)

and so on; thus there are about \frac{N−w}{2} pairs, all with parity −(−1)^{w}. For m_{1} −2m_{2} −1 = w −1, we again have about \frac{N−w}{2} pairs, but now the parity

is (−1)^{w}. Thus, each of the \frac{N−w}{2} pairs with m_{1} − 2m_{2} − 1 = w is matched with one of the \frac{N−w}{2} pairs with m_{1} − 2m_{2} − 1 = w − 1, and we are off by at most O(1) pairs, which will contribute

≪ ∑_{w=N^{1−ϵ}}^{N} \frac{1}{w} ≪ log N.  (20.53)

For the remaining terms, we subtract in pairs, using the first order Taylor Expansion of sin(x). We have

∑_{w=N^{1−ϵ}}^{N} [\frac{sin (w\frac{Q}{N} )}{w} − \frac{sin (w\frac{Q}{N} − \frac{Q}{N} )}{w − 1}].  (20.54)

The Main Term of the Taylor Expansion gives ≪ \frac{1}{w^{2}} , which when summed over w gives \frac{1}{N^{1−ϵ}}. As we have about \frac{N−w}{2} ≪ N pairs for each w, this contributes at most N · \frac{1}{N^{1−ϵ}} ≪ N^{ϵ}.

We also have the first order term from the Taylor Expansion:

sin (w\frac{Q}{N} − \frac{Q}{N} ) = sin (w\frac{Q}{N} ) + O(\frac{Q}{N} ).  (20.55)

This error leads to (remembering there are \frac{N−w}{2} ≪ N pairs for each w)

≪ N ∑_{w=N^{1−ϵ}}^{N} \frac{\frac{Q}{N}}{w − 1} ≪ Q log N^{ϵ} ≪ log^{B+1} N.  (20.56)

20.4.6 Collecting the Pieces

We have shown

∫_{[− \frac{1}{2} , \frac{1}{2} ]} u(x)e(−x)dx = \frac{N}{2} + O(1)
∫_{[− \frac{1}{2} , \frac{1}{2} ]−[− \frac{Q}{N} , \frac{Q}{N}]} u(x)e(−x)dx = O(\frac{N}{log^{B} N} ).  (20.57)

Therefore Lemma 20.4.2.

∫− \frac{Q}{N}^{\frac{Q}{N}} u(x)e(−x)dx = \frac{N}{2} + O(\frac{N}{log^{B} N} ).  (20.58)

Remembering that we had

∫_{M_{a,q}} u(α − \frac{a}{q}) · e(−α)dα = ∫_{\frac{a}{q} − \frac{Q}{N}}^{\frac{a}{q} + \frac{Q}{N}} u(α − \frac{a}{q}) · e(−α)dα
= ∫− \frac{Q}{N}^{\frac{Q}{N}} u(β) · e( − \frac{q}{q} − β)dβ
= e( − \frac{a}{q}) ∫− \frac{Q}{N}^{\frac{Q}{N}} u(β)e(−β)dβ, (20.59)

we see that Lemma 20.4.3.

∫_{M_{a,q}} u(α − \frac{a}{q}) · e(−α)dα = e( − \frac{a}{q})\frac{N}{2} .  (20.60)

20.5 Determination of the Main Term

We now calculate the contribution from the Major Arcs. Up to lower order terms,

∫_{M} f_{N}(x)e(−x)dx = ∑_{q≤Q} ∑_{{a=1 \atop (a,q)=1}}^{q} ∫_{\frac{a}{q} − \frac{Q}{N}}^{\frac{a}{q} + \frac{Q}{N}} f_{N}(α)e(−α)dα
= ∑_{q≤Q} ∑_{{a=1 \atop (a,q)=1}}^{q} ∫_{\frac{a}{q} − \frac{Q}{N}}^{\frac{a}{q} + \frac{Q}{N}} C_{q}(a)u(α − \frac{a}{q})e(−α)dα
= ∑_{q≤Q} ∑_{{a=1 \atop (a,q)=1}}^{q} e( − \frac{a}{q}) ∫− \frac{Q}{N}^{\frac{Q}{N}} C_{q}(a)u(β)e(−β)
= ∑_{q≤Q} ∑_{{a=1 \atop (a,q)=1}}^{q} C_{q}(a)e( − \frac{a}{q})\frac{N}{2}
= \frac{N}{2} ∑_{q≤Q} ∑_{{a=1 \atop (a,q)=1}}^{q} \frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)} · e( − \frac{a}{q})
= \frac{N}{2} ∑_{q=1}^{Q} [ ∑_{{a=1 \atop (a,q)=1}}^{q} C_{q}(a)e( − \frac{a}{q})]
= \frac{N}{2} ∑_{q=1}^{Q} ρ_{q}
= 𝔖_{N} \frac{N}{2} ,  (20.61)

where we have defined

c_{q}(a) = ∑_{{r=1 \atop (r,q)=1}}^{q} e(r\frac{a}{q})

C_{q}(a) = \frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)}
ρ_{q} = ∑_{{a=1 \atop (a,q)=1}}^{q} C_{q}(a)e( − \frac{a}{q})

𝔖_{N} = ∑_{{a=1 \atop (a,q)=1}}^{q} ρ_{q}.  (20.62)

20.5.1 Properties of C_{q}(a) and ρ_{q}

We will follow the presentation of Nathanson (Additive Number Theory: The Classical Bases, Chapter 8 and Appendix A).

c_{q}(a) is Multiplicative

We follow Nathanson, Pages 320 − 321, Theorem A.23. Note that we are labeling by r what he labels a, and we are labeling by a what he labels n.

Lemma 20.5.1. c_{q}(a) is multiplicative; ie, if (q, q′) = 1, then c_{qq′}(a) = c_{q}(a)c_{q′}(a).

Proof: We have

∑_{{r̃=1 \atop (r̃,qq′)=1}}^{qq′} e(r̃\frac{a}{qq′} ).  (20.63)

Exercise 20.5.2. Show that we can write the r̃s above as r̃ ≡ rq′ + r′q mod qq′, where 1 ≤ r ≤ q, 1 ≤ r′ ≤ q′, and (r, q) = (r′, q′) = 1.

Thus

c_{q}(a)c_{q′}(a) = ∑_{{r=1 \atop (r,q)=1}}^{q} e(r\frac{a}{q}) ∑_{{r′=1 \atop (r′,q′)=1}}^{q′} e(r′ \frac{a}{q′} )
= ∑_{{r=1 \atop (r,q)=1}}^{q} ∑_{{r′=1 \atop (r′,q′)=1}}^{q′} e(\frac{(rq′ + r′q)a}{qq′})
= ∑_{{r̃=1 \atop (r̃,qq′)=1}}^{qq′} e(r̃\frac{a}{q}) = c_{qq′}(a).  (20.64)

c_{q}(a) for (a, q) = 1 Exercise 20.5.3. Show that

h_{d}(a) = ∑_{r=1}^{d} e(r\frac{a}{d}) = \begin{cases}d & if d|a \\ 0 & otherwise \end{cases}  (20.65)

Recall the moebius function:

μ(d) = \begin{cases}(−1)^{r} & if d is the product of r distinct primes \\ 0 & otherwise \end{cases}  (20.66)

Exercise 20.5.4. Prove

∑_{d|(r,q)} μ(d) = \begin{cases}1 & if (r, q) = 1 \\ 0 & otherwise \end{cases}  (20.67)

Then

c_{q}(a) = ∑_{{r=1 \atop (r,q)=1}}^{q} e(r\frac{a}{q})

= ∑_{r=1}^{q} e(r\frac{a}{q}) ∑_{d|(r,q)} μ(d)

= ∑_{d|q} μ(d) ∑_{{r=1 \atop d|r}}^{q} e(r\frac{a}{q})

= ∑_{d|q} μ(d) ∑_{l=1}^{\frac{q}{d}} e(l \frac{a}{\frac{q}{d}})
= ∑_{d|q} μ(d)h_{\frac{q}{d}} (a)
= ∑_{d|q} μ(\frac{q}{d})h_{d}(a)
= ∑_{{d|q \atop d|a}} μ(\frac{q}{d}) · d

= ∑_{d|(a,q)} μ(\frac{q}{d})d.  (20.68)

Note that if (a, q) = 1, then there is only one term above, namely d = 1, which yields

c_{q}(a) = μ(q) if (a, q) = 1.  (20.69)

Corollary 20.5.5. If q = p^{k}, k ≥ 2 and (a, q) = 1, then c_{q}(a) = 0.

C_{q}(a) is Multiplicative

We have shown c_{qq′}(a) = c_{q}(a)c_{q′}(a) if (q, q′) = 1. Recall the Euler phi-function, ϕ(q), is the number of numbers less than q which are relatively prime to q.

Exercise 20.5.6. Prove that ϕ(q) is multiplicative; ie, if (q, q′) = 1, then ϕ(qq′) = ϕ(q)ϕ(q′).

We now have Lemma 20.5.7. C_{q}(a) is multiplicative.

Proof: Assume (q, q′) = 1. We have

C_{qq′}(a) = \frac{c_{qq′}(a)c_{qq′}(−2a)}{ϕ^{2}(qq′)}
= \frac{c_{q}(a)c_{q′}(a)c_{q}(−2a)c_{q′}(−2a)}{ϕ^{2}(q)ϕ^{2}(q′)}
= \frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)} · \frac{c_{q′}(a)c_{q′}(−2a)}{ϕ^{2}(q′)}
= C_{q}(a)C_{q′}(a).  (20.70)

ρ_{q} is Multiplicative We first prove a needed lemma. Lemma 20.5.8. Consider C_{q_{1}}(a_{1}q_{2}). Then

C_{q_{1}}(a_{1}q_{2}) = C_{q_{1}}(a_{1})  (20.71)

if (q_{1}, q_{2}) = 1.

Proof:

C_{q_{1}}(a_{1}q_{2}) = ∑_{{r_{1}=1 \atop (r_{1},q_{1})=1}}^{q_{1}} e(r_{1} \frac{a_{1}q_{2}}{q_{1}} )
= ∑_{{r_{1}=1 \atop (r_{1},q_{1})=1}}^{q_{1}} e(r_{1}q_{2} \frac{a_{1}}{q_{1}} )
= ∑_{{r=1 \atop (r,q_{1})=1}}^{q_{1}} e(r\frac{a_{1}}{q_{1}} ) = C_{q_{1}}(a),  (20.72)

because (q_{1}, q_{2}) = 1 implies that as r_{1} goes through all residue classes that are relatively prime to q_{1}, so too does r = r_{1}q_{2}. □

Lemma 20.5.9. ρ_{q} is multiplicative.

Recall

ρ_{q} = ∑_{{a=1 \atop (a,q)=1}}^{q} C_{q}(a)e( − \frac{a}{q}).  (20.73)

Assume (q_{1}, q_{2}) = 1. Then we can write the congruence classes mod q_{1}q_{2} as a_{1}q_{2} + a_{2}q_{1}, with 1 ≤ a_{1} ≤ q_{1}, 1 ≤ a_{2} ≤ q_{2} and (a_{1}, q_{1}) = (a_{2}, q_{2}) = 1.

ρ_{q_{1}q_{2}} = ∑_{{a=1 \atop (a,q_{1}q_{2})=1}}^{q_{1}q_{2}} C_{q_{1}q_{2}}(a)e( − \frac{a}{q_{1}q_{2}} )
= ∑_{{a=1 \atop (a,q_{1}q_{2})=1}}^{q_{1}q_{2}} C_{q_{1}}(a)C_{q_{2}}(a)e( − \frac{a}{q_{1}q_{2}} )
= ∑_{{a_{1}=1 \atop (a_{1},q_{1})=1}}^{q_{1}} ∑_{{a_{2}=1 \atop (a_{2},q_{2})=1}}^{q_{2}} C_{q_{1}}(a_{1}q_{2} + a_{2}q_{1})C_{q_{2}}(a_{1}q_{2} + a_{2}q_{1})e( − \frac{a_{1}q_{2} + a_{2}q_{1}}{q_{1}q_{2}} ).

(20.74)

Exercise 20.5.10. With a_{1}, a_{2}, q_{1}, q_{2} as above,

C_{q_{1}}(a_{1}q_{2} + a_{2}q_{1}) = C_{q_{1}}(a_{1}q_{2}) and C_{q_{2}}(a_{1}q_{2} + a_{2}q_{1}) = C_{q_{2}}(a_{2}q_{1}). (20.75)

Thus, we have

ρ_{q_{1}q_{2}} = ∑_{{a_{1}=1 \atop (a_{1},q_{1})=1}}^{q_{1}} ∑_{{a_{2}=1 \atop (a_{2},q_{2})=1}}^{q_{2}} C_{q_{1}}(a_{1}q_{2})C_{q_{2}}(a_{2}q_{1})e( − \frac{a_{1}q_{2} + a_{2}q_{1}}{q_{1}q_{2}} )

= ∑_{{a_{1}=1 \atop (a_{1},q_{1})=1}}^{q_{1}} C_{q_{1}}(a_{1}q_{2})e( − \frac{a_{1}}{q_{1}} ) ∑_{{a_{2}=1 \atop (a_{2},q_{2})=1}}^{q_{2}} C_{q_{2}}(a_{2}q_{1})e( − \frac{a_{2}}{q_{2}} )

= ∑_{{a_{1}=1 \atop (a_{1},q_{1})=1}}^{q_{1}} C_{q_{1}}(a_{1})e( − \frac{a_{1}}{q_{1}} ) ∑_{{a_{2}=1 \atop (a_{2},q_{2})=1}}^{q_{2}} C_{q_{2}}(a_{2})e( − \frac{a_{2}}{q_{2}} )
= ρ_{q_{1}} · ρ_{q_{2}}.  (20.76)

Thus, ρ_{q} is multiplicative. □

Calculation of ρ_{q} Lemma 20.5.11. ρ_{p^{k}} = 0 if k ≥ 2 and p is a prime.

Proof: This follows immediately from C_{p^{k}}(a) = 0. □ Lemma 20.5.12. If p > 2 is prime, ρ_{p} = −\frac{1}{(p−1)^{2}} .

Proof:

ρ_{p} = ∑_{{a=1 \atop (a,p)=1}}^{p} C_{p}(a)e( − \frac{a}{p})
= ∑_{a=1}^{p−1} \frac{c_{p}(a)c_{p}(−2a)}{ϕ^{2}(p)}e( − \frac{a}{p}).  (20.77)

But as p > 2, c_{p}(a) = c_{p}(−2a) = μ(p) as (a, p) = 1. As μ^{2}(p) = 1 and ϕ(p) = p − 1 we have

ρ_{p} = ∑_{a=1}^{p−1} \frac{1}{(p − 1)^{2}} e( − \frac{a}{p})
= \frac{1}{(p − 1)^{2}} [ − e( − \frac{0}{p}) + ∑_{a=0}^{p−1} e( − \frac{a}{p})]
= −\frac{1}{(p − 1)^{2}} .  (20.78)

Lemma 20.5.13. If p = 2, then ρ_{2} = 1.

Proof:

ρ_{2} = ∑_{{a=1 \atop (a,2)=1}}^{2} C_{2}(a)e( − \frac{a}{2})
= C_{2}(1)e( − \frac{1}{2})
= \frac{c_{2}(1)c_{2}(−2)}{ϕ^{2}(2)} · e^{−πi}
= \frac{e^{πi}e^{−2πi}}{1^{2}} · e^{−πi} = 1,  (20.79)

where we have used c_{2}(1) = e^{πi} and c_{2}(−2) = e^{−2πi}.

Exercise 20.5.14. Prove c_{2}(1) = e^{πi} and c_{2}(−2) = e^{−2πi}.

20.5.2 Determination of 𝔖_{N} and 𝔖

Recall

𝔖_{N} = ∑_{q≤Q} ρ_{q}.  (20.80)

We define

𝔖 = ∑_{q} ρ_{q}.  (20.81)

Exercise 20.5.15. Let h_{q} be any multiplicative sequence (with whatever growth conditions are necessary to ensure the convergence of all sums below). Then

∑_{q} h_{q} = ∏_{p prime}(1 + ∑_{k=1}^{∞} h_{p^{k}}).  (20.82)

𝔖

We have

𝔖 = ∑_{q} ρ_{q}

= ∏_{p prime}(1 + ∑_{k=1}^{∞} ρ_{p^{k}})
= ∏_{p} (1 + ρ_{p})  (20.83)

because ρ_{p^{k}} = 0 for k ≥ 2 and p prime by Lemma 20.5.11. We have previously shown (see Lemmas 20.5.12 and 20.5.13) that ρ_{2} = 1 and ρ_{p} = −\frac{1}{(p−1)} for p > 2 prime. Therefore

𝔖 = ∏_{p} (1 + ρ_{p})
= (1 + ρ_{2}) ∏_{p>2}(1 + ρ_{p})
= 2 ∏_{p>2} [1 − \frac{1}{(p − 1)^{2}} ]
= 2T_{2},  (20.84)

where Definition 20.5.16 (Twin Prime Constant).

T_{2} = ∏_{p>2} [1 − \frac{1}{(p − 1)^{2}} ] ≈ .6601618158  (20.85)

is the twin prime constant.

𝔖_{N}

We need to estimate |𝔖 − 𝔖_{N}|. As ρ_{q} is multiplicative and zero if q = p^{k} (k ≥ 2), we see we need only look at sums of ρ_{p}. As ρ_{p} = −\frac{1}{(p−1)^{2}} , one can show that the difference between 𝔖 and 𝔖_{N} tends to zero as N → ∞.

Thus, Lemma 20.5.17.

𝔖 = 2T_{2}.  (20.86)

20.5.3 Number of Germain Primes and Weighted Sums

Combining the above arguments, we have shown that, up to lower order terms,

∑_{{p≤N \atop p, \frac{p−1}{2} prime}} log(p) · log (\frac{p − 1}{2}) = 𝔖\frac{N}{2}
= 2T_{2} \frac{N}{2}
= T_{2}N.  (20.87)

Note that we are counting Germain prime pairs by (\frac{p−1}{2}, p) and not (p, 2p+1). Such a difference in counting will introduce a factor of 2.

We can pass from this weighted sum to a count of the number of Germain prime pairs (\frac{p−1}{2}, p) with p ≤ N.

Again we follow Nathanson, Chapter 8. Define

π_{G}(N) = ∑_{{p≤N \atop p, \frac{p−1}{2} prime}} 1

G(N) = ∑_{{p≤N \atop p, \frac{p−1}{2} prime}} log(p) · log (\frac{p − 1}{2}).  (20.88)

Clearly

G(N) ≤ log^{2} N · π_{G}(N).  (20.89)

Therefore,

Lemma 20.5.18. Up to lower order terms,

π_{G}(N) ≥ \frac{G(N)}{log^{2} N} = \frac{T_{2}N}{log^{2} N} .  (20.90)

We now provide a bound in the opposite direction.

π_{G}(N^{1−δ}) = ∑_{{p≤N^{1−δ} \atop p, \frac{p−1}{2} prime}} 1 ≪ \frac{N^{1−δ}}{log N} .  (20.91)

Then

G(N) ≥ ∑_{{p≥N^{1−δ} \atop p, \frac{p−1}{2} prime}} log p · log (\frac{p − 1}{2})
= (1 − δ)^{2} log^{2} N ∑_{{p≥N^{1−δ} \atop p, \frac{p−1}{2} prime}} 1
= (1 − δ)^{2} log^{2} N(π_{G}(N) − π_{G}(N^{1−δ}))
≥ (1 − δ)^{2} log^{2} Nπ_{G}(N) + O((1 − δ)^{2} log^{2} N · \frac{N^{1−δ}}{log N} ).(20.92)

Therefore

log^{2} N · π_{G}(N) ≤ (1 − δ)^{−2} · G(N) + O( log^{2} N · \frac{N^{1−δ}}{log N} )
0 ≤ log^{2} N · π_{G}(N) − G(N) ≤ [(1 − δ)^{−2} − 1]G(N) + O( log N · N^{1−δ}). (20.93)

If 0 < δ < \frac{1}{2}, then (1 − δ)^{−2} − 1 ≪ δ. We thus have

0 ≤ log^{2} N · π_{G}(N) − G(N) ≪ N[δ + O(\frac{log N}{N^{δ}} )]. (20.94)

Choose δ = \frac{2 log log N}{log N}. Then we get

0 ≤ log^{2} N · π_{G}(N) − G(N) ≤ O(N \frac{log log N}{log N} ).  (20.95)

Recalling G(N) ≈ T_{2}N gives Lemma 20.5.19.

π_{G}(N) ≤ \frac{T_{2}N}{log^{2} N} .  (20.96)

Combining with the other bound we have finally shown Theorem 20.5.20. Assuming there is no contribution to the main term from the Minor Arcs, up to lower order terms we have

π_{G}(N) = \frac{T_{2}N}{log^{2} N} ,  (20.97)

where T_{2} is the twin prime constant

T_{2} = ∏_{p>2} [1 − \frac{1}{(p − 1)^{2}} ] ≈ .6601618158.  (20.98)