Honors 1: Undergraduate Math Lab^{∗}

Peter Sarnak^{†}, Steven Miller^{‡}, Alex Barnett^{§}

November 6th, 2002

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

Abstract

Random Graphs, Autocorrelation, Random Matrix Theory and the Mehta-Gaudin Theorem. Lecture by Peter Sarnak; notes by Steven Miller.

1 Random Graphs

Definition 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 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 1.3 (Girth). The girth is the shortest closed cycle.

∗Homepage: http://www.math.nyu.edu/∼millerj/

†E-mail: sarnak@math.princeton.edu

‡E-mail: millerj@cims.nyu.edu or sjmiller@math.princeton.edu

§E-mail: barnett@nmr.mgh.harvard.edu

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 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.

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}} .  (1)

In Lebesgue Measure, almost all α satisfy the above infinitely often. In the Baire Category, this inequality does not hold infinitely often.

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).  (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.

4 Gaudin’s Method

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}.  (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}.

(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 4.1. β = 0 is Poissonian.

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}.  (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}.  (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}).  (7)

Today we will only use the square, so we don’t worry about ordering so that f_{i} < f_{j}.

Exercise 4.2.

Van(f_{1}, …, f_{N}) = det (f_{i}^{j−1})_{1≤i,j≤N}.  (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} |  (9)

4.3 Orthonormal Polynomials

On the unit circle T, we have the measure

dμ(t) = \frac{dt}{2π} .  (10)

Let f(t) be a function such that

∫_{T} f(t)dμ(t) = 0, ∫_{T} |f(t)|^{2}dμ(t) = 1.  (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}. (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),  (13)

and

δ_{ij} = \begin{cases}1 & if i = j \\ 0 & otherwise \end{cases}  (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.

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).  (15)

Exercise 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 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).  (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}.  (17)

For example,

D_{1,N} = K_{N}(t_{1}, t_{1})  (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} |  (19)

4.5 Gaudin-Mehta Theorem

Theorem 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}. (20)

2.

\frac{1}{μ(T)}|Van(f(t_{1}), …, f(t_{N}))|^{2} = D_{N,N}(t_{1}, …, t_{N}).  (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}). (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}.  (23)

Corollary 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}). (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}).

(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}).  (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}.  (27)

Consider the expectation of

∑_{1≤j<k≤N} f(|x_{j} − x_{k}|).  (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}.  (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.

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}.  (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)}} .  (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}.  (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}.  (33)