Eigenvalues of Euclidean Random Matrices

Home , Random matrix

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Zee and Mézard, Parisi by n arcsweetecoeffi- the where matrices rns n 3 n opizmatrices Toeplitz and [3] ang n sr stenme fpoints of number the as asure points h slst h adjacency the to esults h nre fthese of entries The . . nso h limiting the of ents n c ftelimiting the of nce − 1 X rbto sthe as tribution h n ( n X admpit in points random = i ta point atial X , { X j )) 1 , 1 · · · ≤ i,j n X , ≤ n n (1) to } , Another field of application is graph theory. Indeed, if F (x) = 11(0 x r), then A ≤ k k ≤ is the adjacency matrix of the proximity (or geometric) graph (refer to Penrose [15]). More generally if F (X) = F ( X) 0, 1 then A is the adjacency matrix of a random graph. The − ∈ { } spectral properties of the adjacency matrix or related matrices are of prime interest in graph theory. For example the probability of hitting times of random walks on graphs is governed by the spectrum of the transition matrix (for a survey on this subject, see e.g. Section 3 in Lovász [12]). Or, in network epidemics, the time evolution of the infected population is also closely related to the spectral radius and the spectral gap of the adjacency matrix, see Draief, Ganesh and Massoulié[8]. For Erdós-Renyi random graphs, some properties of the spectrum can been computed thanks to the seminal work Wigner of [18] and Füredi and Komlós [9]. For power law graphs and related graphs, see Chung, Lu and Vu [4], [5]. Various generalizations of (1) would be worth to consider. Some extra randomness in the model could be added, and the entry of the matrix i, j could be equal to F (X X ), where ij i − j (F ) are i.i.d. mappings independent of the point set . Falls into this framework ij 1≤i,j≤n Xn the adjacency matrix of a random graph where there is an edge between two points with a probability which is deterministic function of their distance, such as the small world graphs (see for example Ganesh and Draief [7]). Another generalization is the original model of Mézard, Parisi and Zee [13] where the entry i, j is equal to F (X X ) uδ F (X X ), i − j − ij i − k Xk where δ is the Kronecker symbol and u R. The case u = 1 is of particular interest, the ij ∈ matrix is then a Markov matrix. In order to obtain the adjacency matrix of more sophisticated geometric graphs, such as the Delaunay triangulation, it would be necessary to consider an entry i, j which depends on the whole point set and not only on X X . Xn i − j We will consider two models in this note. In the first model, Ω = [ 1/2, 1/2]d and F is − 1-periodic function: if x,y Rd and x y Zd then F (x) = F (y). Equivalently, the point ∈ − ∈ set = X , , X could be on the unit torus Td = Rd Zd. We choose a periodic function Xn { 1 · · · n} \ in order to avoid all boundary effects with the hypercube Ω. The matrix A is defined by (1), where F is a measurable function from Rd to C. The discrete Fourier transform of F is defined for all k Zd by Fˆ(k) = F (x)e−2iπk.xdx. Throughout the paper, we assume that a.e. and ∈ Ω at 0, the Fourier series of RF is equal to F: F (x)= Fˆ(k)e2iπk.x. kX∈Zd

A suﬃcient condition is Zd Fˆ(k) < and F continuous at 0. This Fourier transform k∈ | | ∞ plays an important role inP the spectrum of A. As an example, consider U = (Ui)1≤i≤n a vector in Cd and assume F hermitian (F ( x)= F¯(x)), then a.s. − n 2 ∗ 2iπk.(Xi−Xj ) 2iπk.Xi U AU = F (Xi Xj)UiU¯j = Fˆ(k)e UiU¯j = Fˆ(k) e Ui . − Xi,j Xi,j Xk Xk Xi=1

Therefore A is positive if and only if for all k Zd, Fˆ(k) 0. ∈ ≥ We will compute explicitly the spectral measure of the matrix A = A/n as n tends to , n ∞ n

µn = δλi(n)/n, Xi=1

2 where λ (n) is the set of eigenvalues of A. Notice that λ (n)/n is the set of { i }1≤i≤n { i }1≤i≤n eigenvalues of An. We deﬁne the measure:

µ = δFˆ(k). kX∈Zd ˆ Since limkkk→∞ F (k) = 0, µ is a counting measure with an accumulation point at 0. Theorem 1 For all Borel sets K with µ(∂K) = 0 and 0 / K¯ , a.s. ∈ lim µn(K)= µ(K). (2) n

The convergence of the spectral measure µn follows also from Theorem 3.1 in [11]. As an immediate corollary, we obtain the convergence of the spectral radius of An, almost surely, |λi(n)| lim max = max Zd Fˆ(k) . For example if F (x) = 11(max x r) then n→∞ 1≤i≤n n k∈ | | 1≤i≤d | i| ≤ ˆ d d Zd F (k)= r i=1 sinc(2πkj r), where sinc(x) = sin(x)/x and k = (k1, , kd) . The spectral d · · · d ∈ radius of AQn converges a.s. to r and the second largest eigenvalue to r sinc(2πr) if r is small enough, thus the spectral gap is equivalent to rd+2(2π)2/3! as r goes to 0. Our second model is more challenging. Now, = X , , X is the set of n independent Xn { 1 · · · n} points uniformly distributed on the hypercube δ−1Ω = [ δ−1/2, δ−1/2]d where δ goes to 0. In n − n n n this second model, we scale jointly the number of points and the space. We assume that for some γ > 0, lim δdn = γ. (3) n n γ is the asymptotic density of the point set . Let f be a measurable function from RD to C Xn with support included in Ω, the matrix A is defined by (1) (with F replaced by f). Considering the change of variable x δx, the matrix A is equal to the matrix B defined 7→ n by B = (f (X X )) , n δn i − j 1≤i≤j≤n where f : x f(x/δ) and the point set = X , , X is a set of n independent points δ 7→ Xn { 1 · · · n} uniformly distributed on Ω. The spectrum of B is denoted by (λ′ (n), , λ′ (n)), we define n 1 · · · n the empirical measure of its eigenvalues: n 1 νn = δ ′ , n λi(n) Xi=1 We will prove the following:

Theorem 2 For all γ > 0, there exists a measure νγ such that for the topology of the weak convergence, a.s.: lim νn = νγ. n→∞ Moreover γ ν is continuous (for the topology of the weak convergence). 7→ γ The exact computation of νγ is a difficult problem, we will compute the value νγ(Pm), where P is the polynomial t tm (Equation (22)). However, the behavior of ν as γ goes m 7→ γ to infinity is simpler. Indeed, we define the Fourier transform of f by, for all ξ Rd, fˆ(ξ) = ∞ −2iπξ.x ˆ ∈ 0 e f(x)dx. Since f has a bounded support, f is infinitely differentiable. We assume Rthat the following inversion formula holds

a.e. and at 0, f(x)= fˆ(ξ)e2iπξ.xdξ. (4) ZRd

3 Note that if f is hermitian (f( x) = f(x)) then fˆ(ξ) R and for ǫ > 0, 11( fˆ(ξ) ǫ)dξ − ∈ Rd | | ≥ is ﬁnite. Hence by the change of variable formula, there exists a function Rψ such that for all continuous functions h with 0 / supp(h): ∈

h(t)ψ(t)dt = h(fˆ(x))dx. ZR ZRd

ψ is the level sets function of fˆ, if ℓ denotes the d-dimensional Lebesgue measure, for all t> 0, ψ(t) = lim ℓ( x : fˆ(x) t ǫ )/ǫ. If d = 1 and fˆ is a diﬀeomorphism from R to K then ǫ→0 { | − | ≤ } ψ has support on K and is equal to ψ(t) = (fˆ−1)′(t).

Theorem 3 If f is hermitian and (4) holds true, then as γ goes to inﬁnity, for all analytic m m functions h(t)= N h t with h = 0 and N h t ﬁnite for all t: m∈ m 0 m∈ | m| P P −2 t −1 h(t)νγ (dt) h(t)γ ψ( )dt = γ h(γt)ψ(t)dt. ZR ∼ ZR γ ZR

This result states that the measure νγ(dt) is in a weak sense equivalent (not in the measure theory sense) to the measure γ−2ψ(t/γ)dt in the high density asymptotic. The measure ψ(t)dt is the continuous analog of the counting measure µ in Theorem 1. As an example, if d = 1 and f(x) = 11(0 x r) then fˆ(ξ)= rsinc(2πξr) and ψ(t) is plotted in Figure 1. ≤ | |≤

1 50

45 0.8

0.6 35

30 0.4

0.2 20

15 0

−0.2 5

−0.4 0 0 2 4 6 8 10 −0.2 0 0.2 0.4 0.6 0.8

Figure 1: Left: fˆ(ξ) for f(x) = 11(0 x r). Right: the level set function ψ. ≤ | |≤

d Remark. Let γn = nδn, if γn tends to inﬁnity and δn goes to 0, with the material of this note, n we may also prove the convergence of δd δ ′ to the measure ψ(t)dt on all continuous n i=1 λi(n)/γn function h with compact support and 0 /Psupp(h). ∈ The spectral radius of the matrix Bn is not computed explicitly in this paper. However, the following upper bound is available:

Proposition 4 If d 2 and P o(γ) denotes a random variable with Poisson distribution of ≥ intensity γ, then with a probability tending to 1 as n goes to inﬁnity,

′ max λi(n) j(n)sup f(x) , 1≤i≤n | |≤ x∈Ω | | where j(n) is solution of: nP(P o(γ) j(n) + 1) 1

4 For n large enough, using the inequality P(P o(γ) k) exp( k ln( k )), for k e2γ, we deduce ≥ ≤ − 2 γ ≥ that j(n) 3 ln n/ ln ln n for n large enough. ≤ The remainder of this paper is organized as follows. In Section 2, we prove Theorem 1, In Section 3, we prove Theorems 2, 3 and Proposition 4. Finally, in Section 4, we state some simple results on the eigenvectors of A and on the correlation of the eigenvalues. By convention C will denote a constant which does not depend on n. Its exact value may change throughout the paper. Also we deﬁne: F = sup Rd F (x) and B(x, r) will denote k k∞ x∈ | | the open ball of radius r and center x on the torus Td.

2 Proof of Theorem 1

The proof of Theorem 1 relies on the classical Wigner’s method [18] to compute the empirical mean distribution measure of eigenvalues. We will compute for all m N: ∈ n 1 1 EtrAm = EtrAm = λm = µ (P ). n nm nm i n m Xi=1 m We will then use a Talagrand’s concentration inequality to prove that trAn is not far from its mean and conclude. About the rate of convergence of µn to µ, we will state (in the forthcoming Lemma 6) that, if P (t)= tm, m 1, m ≥ m−1 m(m 1) lim n Eµn(Pm) µ(Pm) = qµ(Pq)µ(Pm−q) − µ(Pm). (5) n − − 2 Xq=1 We begin with a technical lemma. Lemma 5 For 0 p m, let Σ be the set of surjective mappings from 1, ,m to ≤ ≤ m,p { · · · } 1, ,p . We have: { · · · } m m E m n trA = F (xφ(j) xφ(j+1))dx1 dxp, (6) p ZΩp − · · · Xp=1 φ∈XΣm,p jY=1

n with φ(m +1) = φ(1) and with the convention that p = 0 for p>n. Proof. By deﬁnition: m trAm = F (X X ), (7) ij − ij+1 i1X,···im jY=1 with i = i and the sum is over all n-tuples of integers i = (i , , i ) in 1,n m. Let p(i) m+1 1 1 · · · m { } be the set of distinct indices in i. We can deﬁne a surjective mapping φi in Σm,p(i) such that

ij = iφi(j). Taking the expectation in Equation (7), we get m E m trA = F (xφi(j) xφi(j+1))dx1 dxp(i), ZΩm − · · · i=(iX1,···im) jY=1 We then reorder the terms. We consider the equivalence relation in Σ , φ φ′ if there m,p ∼ exists a permutation σ of 1, ,p such that σ φ = φ′. The value of m F (x { · · · } ◦ Ωm j=1 φ(j) − x )dx1 dxp is constant on each equivalence class. Let φ Σm,p, theR numbersQ of indices i φ(j+1) · · · ∈ such that φi φ is equal to n!/(n p)! (if n p and 0 otherwise). Since there are p! surjective ∼ − ≥ mappings in the class of equivalence of φ, we deduce Equation (6). ✷

5 Lemma 6 For each m,

m−1 1 m(m 1) 1 Eµ (P )= µ(P )+ qµ(P )µ(P ) − µ(P ) +o( ). n m m n q m−q − 2 m n Xq=1

Proof. We apply Lemma 5 and identify the coeﬃcients in nm and nm−1 in Equation (6). We ﬁrst consider the term in nm, such a term comes from p = m:

m n! F (xj xj+1)dx1 dxm, (n m)! Z m − · · · − Ω jY=1 By induction, we easily obtain that

m ∗m ∗m F (xj xj+1)dx1 dxm = F (0)dx1 = F (0), Z m − · · · Z Ω jY=1 Ω

∗m where denotes the convolution operator: F G(y)= Ω F (y x)G(x)dy and F is F F F ∗ \∗ − ∗ ···∗ (m times). We recall the two properties: F G(k) =R Fˆ(k)Gˆ(k) and F (0) = Zd Fˆ(k), in ∗ k∈ order to get: P m m F (xj xj+1)dx1 dxm = Fˆ(k) = µ(Pm). ZΩm − · · · jY=1 kX∈Zd We thus deduce that: lim Eµn(Pm)= µ(Pm). n It remains to identify the terms in nm−1 in Equation (6). This term comes from two contribu- tions p = m and p = m 1. Since n!/(n m)! = nm nm−1 m−1 i + o(nm−1), the term in − − − i=0 nm−1 in p = m is equal to: P m(m 1) − µ(P ). (8) − 2 m The leading term for p = m 1 is − m n! F (xφ(j) xφ(j+1))dx1 dxm−1, (9) (n m + 1)!(m 1)! ZΩm−1 − · · · − − φ∈ΣXm,m−1 jY=1

−1 Now if φ Σm,m−1, φ (i) is not reduced to a single point for a unique index iφ. Since ∈ m the value of −1 F (x x )dx dx is invariant under permutations of the Ωm j=1 φ(j) − φ(j+1) 1 · · · m−1 indices, withoutR lossQ of generality, we may assume that i = 1, φ−1(1) = 1,q + 1 with φ { } q 1, ,m 1 and φ(j)= j if j q and φ(j)= j +1 if j>q + 1. For such φ, integrating ∈ { · · · − } ≤ over x , ,x ,x , ,x , 2 · · · q q+1 · · · m−1 m ∗(q) ∗(m−q) F (xφ(j) xφ(j+1))dx1 dxm−1 = F (0)F (0)dx1 Z m − · · · Z Ω jY=1 Ω

= µ(Pq)µ(Pm−q).

Finally, for each q, there are (m q) (m 1)! surjective mappings such that, up to a permutation − × − of the indices, φ−1(1) = 1,q + 1 and φ(j)= j if j q and φ(j)= j +1 if j>q + 1. Indeed for { } ≤

6 such φ, there are (m q) possible pairs (i , i + q), 1 i m q such that φ(i )= φ(i + q). − 1 1 ≤ 1 ≤ − 1 1 Therefore Equation (9) can be written as:

m−1 m−1 n!(m 1)! − (m q)µ(P )µ(P )= nm−1 qµ(P )µ(P )+ o(nm−1). (10) (n m + 1)!(m 1)! − q m−q q m−q − − Xq=1 Xq=1 Adding this last term with the term (8), we get the stated formula. ✷ We may now prove Theorem 1. Proof of Theorem 1. We fix n and for each m 1, we define the functional: ≥ 1 Q ( )= trAm = nµ (P ). m Xn nm−1 n m n n If x, y Ω , let d(x, y) = i=1 11(xi = yi) denote the Hamming distance. The functional Qm ∈ 6 l l l is Lipschitz for the HammingP distance d. Indeed, define x = (x )1≤j≤n by x = xj for j = l j j 6 and xl = x , we have: l 6 l m m l 1 l l Qm(x) Qm(x ) = F (xij xij+1 ) F (xij xij+1 ) − nm−1 − − − i1,X··· ,im jY=1 jY=1 2m f m , ≤ k k∞ indeed m F (x x ) m F (xl xl ) is at most 2 f m and it is non zero only if | j=1 ij − ij+1 − j=1 ij − ij+1 | k k∞ there existsQ a index i such thatQi = l. It follows easily that Q is 2m f m -Lipschitz for the j j m k k∞ Hamming distance d. Let Mm denote the median of Qm. We may apply a Talagrand’s Concentration Inequality (see for example Proposition 2.1 of Talagrand [16]),

t2 P( Q M >t) 4 exp( ), | m − m| ≤ −4m2 f 2mn k k∞ integrating over all t we deduce:

nEµ(P ) M E µ(P ) M C √n, | m − m|≤ | m − m|≤ m for some constant Cm and it follows, that for all s>Cm/√n:

(s C /√n)2 P( µ (P ) Eµ(P ) >s) 4 exp( n − m ), | n m − m | ≤ − 4m2 f 2m k k∞ Using the Borel Cantelli Lemma and Lemma 6, a.s. limn µn(Pm)= µ(Pm). ✷

3 Limit Spectral Measure of Scaled ERM

3.1 Proof of Theorem 2 The study of the first model was simplified by the absence of boundary effects with Ω. So in order to prove Theorem 2, we will first discard them in the second model. We define Fδ as the 1-periodic extension of f : for all x Rd, there exists a unique couple (y, u) such that x = y +u, δ ∈ with u Zd and y Ω, and we set F (x)= f (y). ∈ ∈ δ δ

7 We now introduce a matrix and its spectral empirical measure:

n 1 Bñ = (Fδ (Xi Xj))1≤i≤j≤n andν ñ = δ˜ , n − n λi(n) Xi=1 where (λ˜ (n), , λ˜ (n)) is the spectrum of B˜ . The next lemma states that the limiting 1 · · · n n spectral measures ofν ñ and νn are equal. Lemma 7 For the topology of the weak convergence of (signed) measures, a.s. ν ν˜ converges n− n as n goes to infinity to the null measure.

Proof. It is sufficient to prove that for all m 1, a.s. lim ν (P ) ν˜ (P ) = 0. To this end, ≥ n n m − n m we notice that if x,y Ω, f (x y) = F (x y) unless x Ω (1 δ)Ω and y B(x, δ). We ∈ δ − δ − ∈ \ − ∈ write: m m 1 ν (P ) ν˜ (P ) f m(X X ) F (X X ) n m − n m ≤ n δn ij − ij+1 − δn ij − ij+1 i1,X··· ,im jY=1 jY=1 m 1 2 f m 11(X Ω (1 δ )Ω) 11(X B(X , mδ )) ≤ n k k∞ i1 ∈ \ − n ij ∈ i1 n i1,X··· ,im jY=2 2 f m N (Ω (1 mδ )Ω)m, ≤ nk k∞ n \ − n where N is the counting measure N ( )=# i 1, ,n : X . Note that P(X n n · { ∈ { · · · } i ∈ ·} 1 ∈ Ω (1 mδ )Ω) Cδ . By the strong law of large numbers, it follows easily that that N (Ω (1 \ − n ≤ n n \ − mδn)Ω)/n converges almost surely to 0. ✷ By Lemma 7, we may focus on Bñ andν ñ. In order to keep the notations as light as possible we drop the ” ˜ ” in B˜ andν ˜ . · n n We first prove that,

νn converges in probability to a measure νγ for the weak convergence. (11)

By Lemma 5, if m 1, ≥ m m E 1 n νn(Pm)= Fδn (xφ(j) xφ(j+1))dx1 dxp. (12) n p ZΩp − · · · Xp=1 φ∈XΣm,p jY=1

We begin with an elementary lemma.

Lemma 8 If φ Σ , p> 1 the value of ∈ m,p m F (xφ(j) xφ(j+1))dx2 dxp Z p−1 − · · · Ω jY=1

does not depend on x1.

′ Proof. We consider the change of variable, for j > 1, xj = xj x1. The Jacobian of this ′ m − change of variable is 1. If we set x1 = 0, we obtain Ωp−1 j=1 F (xφ(j) xφ(j+1))dx2 dxp = m ′ ′ ′ ′ − · · · p−1 F (x x )dx dx . R Q ✷ Ω j=1 φ(j) − φ(j+1) 2 · · · p R Q

8 Assume m 2, by Lemma 8, we have: ≥ m m E m 1 n νn(Pm) = Fδn (0) + Fδn (xφ(j) xφ(j+1))dx2 dxp n p ZΩp−1 − · · · Xp=2 φ∈XΣm,p jY=1 m m m 1 n = f(0) + ∆(φ)+ fδn (xφ(j) xφ(j+1))dx2 dxp(13) n p ZΩp−1 − · · · Xp=2 φ∈XΣm,p jY=1

m m where ∆(φ)= −1 F (x x ) f (x x )dx dx . Since the Ωp j=1 δn φ(j) − φ(j+1) − j=1 δn φ(j) − φ(j+1) 2 · · · p support of fδ isR includedQ in δΩ, if fδ(x x Q ) = Fδ(x x ) then x ,x φ(j) − φ(j+1) 6 φ(j) − φ(j+1) φ(j) φ(j+1) ∈ Ω (1 δ)Ω. Moreover notice that if m F (x x ) = 0 then x , x B(x , (m 1)δ). \ − j=1 φ(j)− φ(j+1) 6 2 · · · p ∈ 1 − By Lemma 8, from now on, we canQ assume without loss of generality:

x1 = 0, and then ∆(φ)=0 for δ < 1/(2m). Considering the change of variable yi = xi/δn in the integrands of Equation (13), we obtain, for δ < 1/(2m), with y1 = 0,

m d(p−1) m E m δn n νn(Pm)= f(0) + f(yφ(j) yφ(j+1))dy2 dyp. (14) −1 p−1 n p Z(δn Ω) − · · · Xp=2 φ∈XΣm,p jY=1

Finally, since n np/p! as n goes to inﬁnity, we deduce that, for m 2, p ∼ ≥ m p−1 m E m γ lim νn(Pm)= f(0) + f(yφ(j) yφ(j+1))dy2 dyp. (15) n→∞ p! Z(Rd)p−1 − · · · Xp=2 φ∈XΣm,p jY=1

(For m 1, we have ν (P )=1 and ν (P )= f(0)). ≤ n 0 n 1 We check easily that the right hand side of Equation (15) is bounded by (Cm)m for some constant C not depending on m. Therefore, by Carleman’s Condition, there is exists a unique measure νγ such that limn→∞ Eνn(Pm)= νγ (Pm). In particular, the sequence (νn)n∈N is tight and we have proved (11). The continuity of γ ν follows from the comtinuity of γ ν (P ). Indeed, let (γ ) N 7→ γ 7→ γ m n n∈ be a sequence converging to γ < . Since sup ν (P ) < , the sequence (ν ) N is tight. ∞ n γn 2 ∞ γn n∈ Hence for all ǫ > 0, there exists a compact set K such that for all n ν (Kc) ǫ. Now, let h γn ≤ be a continuous function with compact support, we need to prove that limn→∞ νγn (h)= νγ (h). Fix ǫ, there exists a polynomial P such that sup h(x) P (x) ǫ, we deduce that νγ (h) x∈K | − |≤ | n − ν (h) ν (h) ν (P ) + ν (P ) ν (P ) + ν (P ) ν (h) 2ǫ(1+ h )+ ν (P ) ν (P ) . γ | ≤ | γn − γn | | γn − γ | | γ − γ |≤ k k∞ | γn − γ | Letting n tends to inﬁnity, since ǫ is arbitrary small and γ ν (P ) is continuous, we obtain: 7→ γ limn→∞ νγn (h)= νγ(h). It remains to prove the almost sure convergence of ν . We will prove that for each m 1, n ≥ there exists a constant C and

E trBm EtrBm 4 Cn2. (16) n − n ≤ This last equation implies E ν (P ) Eν (P ) 4 C/n2 and by Borel Cantelli Lemma, we n m − n m ≤ deduce that νn(Pm) converges almost surely toward νγ(Pm).

9 It remains to prove Inequality (16). A circuit in 1, ,n of length m is a mapping { · · · } π : Z 1, n such that for all integer r, π(m + r) = π(r). Following Bryc, Dembo and → { · · · } Jiang [3], we introduce the new notation:

m F = F (X X ). π δn π(i) − π(i+1) Yi=1 We then write:

4 E trBm EtrBm 4= E F EF 4= E F EF , (17) n − n π − π πl − πl Xπ π1X,··· ,π4 Yl=1 where the sums are over all circuits in 1, ,n of length m. { · · · } Notice that E 4 F EF = 0 if there exists a circuit π , 1 k 4 such that the l=1 πl − πl k ≤ ≤ image of πk has an emptyQ intersection with the union of the images of πl, l = k. Indeed, due to 6 the independence of the variables (X ) , F EF is then independent of F EF . i 1≤i≤n πk − πk l6=k πl − πl Two circuits π1 and π2 in 1, ,n of length m1 and m2 with a non emptyQ intersection { · · · } of their images may be concatenated into a circuit in 1, ,n of length m + m as follows. { · · · } 1 2 Assume that π (i )= π (j ), we deﬁne the circuit π of length m +m by for i 1, ,m + 1 0 2 0 1,2 1 2 ∈ { · · · 1 m 2} π (i) if 1 i i 1 ≤ ≤ 0 π .π (i)=  π (i i + j ) if i + 1 i i + m 1 2 2 − 0 0 0 ≤ ≤ 0 2  π (i m i ) if i + m + 1 i m + m 1 − − 0 0 ≤ ≤ 1 2 We have: 

Fπ1 Fπ2 = Fπ1·π2 . Using the same reasoning as for Equation (14), we get

2m E d(q−1) Fπ1 Fπ2 = δn f(yπ1.π2(j) yπ1.π2(j+1))dyi2 dyiq , Z −1 q−1 − · · · (δn Ω) jY=1 where is q = q(π ,π ) is the cardinal of the union of the images of π and π and (y , ,y ) 1 2 1 2 i1 · · · iq is the image of π π and y = 0. 1 · 2 i1 If N(π ,π ) is the cardinal of the intersection of the images of π and π , if N(π ,π ) 1, 1 2 1 2 1 2 ≥ we obtain E F F Cn−q(π1,π2)+1. (18) | π1 π2 |≤ Otherwise, N(π1,π2) = 0, if q(πi) is the cardinal of of the image of πi,

E F F = E F E F Cn−q(π1)−q(π2)+2 = Cn−q(π1,π2)+2. (19) | π1 π2 | | π1 | | π2 |≤ Similarly assume that N(π ,π ) 1, N(π .π ,π ) 1, if q(π ,π ,π ) is the cardinal of the 1 2 ≥ 1 2 3 ≥ 1 2 3 union if the images of π1,π2,π3 then Fπ1 Fπ2 Fπ3 = F(π1.π2).π3 and we deduce similarly

E F F F Cn−q(π1,π2,π3)+1. | π1 π2 π3 |≤ Finally assume that N(π ,π ) 1, N(π .π ,π ) 1, N(π .π .π ,π ) 1, if q(π ,π ,π ) is the 1 2 ≥ 1 2 3 ≥ 1 2 3 4 ≥ 1 2 3 cardinal of the union if the images of π1,π2,π3,π4, we obtain:

E F F F F Cn−q(π1,π2,π3,π4)+1. (20) | π1 π2 π3 π4 |≤

10 By (17), it remains to decompose:

4 E E 4! Fπl Fπl ′ − (π1,··· ,πX4)∈S∪S Yl=1

where S is the set of quadruples of circuits such that N(π ,π ) 1, N(π .π ,π ) 1, 1 2 ≥ 1 2 3 ≥ N(π .π .π ,π ) 1 and S′ is the set of quadruples of circuits such that N(π ,π ) 1 and 1 2 3 4 ≥ 1 2 ≥ N(π ,π ) 1 and otherwise for i < j, N(π ,π ) = 0. 3 4 ≥ i j The decomposition of the E 4 F EF gives rise to four types of terms: l=1 πl − πl 4 Q ′ E 1. (π1,··· ,π4)∈S∪S l=1 Fπl , P Q 4 ′ E 2. (π1,··· ,π4)∈S∪S l=1 Fπl , P Q 3. ′ EF F EF F , (π1,··· ,π4)∈S∪S πl1 πl2 πl3 πl4 P 4. ′ EF EF F F , (π1,··· ,π4)∈S∪S πl1 πl2 πl3 πl4 P where (l1, l2, l3, l4) is a permutation of (1, 2, 3, 4). We will apply successively the same method to bound these terms. 4 4 E − Pl=1 q(πl)+4 We begin with the terms of type 1, we have: l=1 Fπl Cn . Since (π1, ,π4) ′ 4 ≤ · · · ∈ S S , q(π1,π2,π3,π4) q(πl) 2, hence:Q ∪ ≤ l=1 − P 4 EF Cn−q(π1,π2,π3,π4)+2. πl ≤ Yl=1

q There are at most Cn quadruples of circuits such that q(π1,π2,π3,π4)= q, therefore the terms of type 1 may be bounded as by

4 E 2 Fπl Cn . ′ | |≤ (π1,··· ,πX4)∈S∪S Yl=1

E 4 We now deal with the terms of type 2. By (20), if (π1, ,π4) S, l=1 Fπl −q(π1,π2,π3,π4)+1 ′ E ·4 · · ∈ −q(π1,π2,π|3,π4|)+2 ≤ Cn otherwise (π1, ,π4) S and, by (19), l=1 Fπl Cn Q . q · · · ∈ | |≤ There are at most Cn quadruples of mappings such that q(π1Q,π2,π3,π4)= q. Hence

4 E 2 Fπl Cn . ′ | |≤ (π1,··· ,πX4)∈S∪S Yl=1

We turn to the terms of type 3: ′ EF F EF F . Assume ﬁrst that the (π1,··· ,π4)S∪S πl1 πl2 πl3 πl4 ′ quadruple (π1, ,π4) S . If l1 =P 1, l2 = 3, l3 = 2, l4 = 4, then EFπ Fπ EFπ Fπ = · · · ∈ l1 l2 l3 l4 4 E l=1 Fπl and we obtain the same bound that the terms of type 1. The other cases reduce E −q(π1,π2)+1 toQ the case l1 = 1, l2 = 2, l3 = 3, l4 = 4 and by (18), Fπ1 Fπ2 Cn . There q+q′ ≤ ′ are at most Cn quadruples such that q(π1,π2) = q and q(π3,π4) = q . We deduce that 2 ′ EF F EF F Cn . (π1,··· ,π4)∈S π1 π2 π3 π4 ≤ P Assume now that that (π1, ,π4) S. We have: · · · ∈

−q(πl ,πl )−q(πl ,πl )+2+11(N(πl ,πl )=0)+11(N(πl ,πl )=0) EFπ Fπ EFπ Fπ Cn 1 2 3 4 1 2 3 4 . l1 l2 l3 l4 ≤

11 ′ If N(π ,π ) = 0, then N(π ,π ) 1 and there are at most Cnq+q −2 quadruples such l1 l2 l3 l4 ≥ that q(π ,π ) = q and q(π ,π ) = q′. Indeed, since (π , ,π ) S, the cardinal of the l1 l2 l3 l4 1 · · · 4 ∈ intersection of the images of (πl1 ,πl2 ) and (πl3 ,πl4 ) is at least 2. The other cases reduce to the case, N(πl1 ,πl2 ) 1 and N(πl3 ,πl4 ) 1, for such cases, we notice that there are at most q+q′ ≥ ≥ ′ Cn quadruples such that q(πl1 ,πl2 )= q and q(πl3 ,πl4 )= q . In all cases, we conclude that: EF F EF F Cn2. Hence, (π1,··· ,π4)∈S π1 π2 π3 π4 ≤ P EF F EF F Cn2. πl1 πl2 πl3 πl4 ′ ≤ (π1,··· ,πX4)∈S∪S

It remains to treat the terms of type 4. Assume that (π , ,π ) S S′, we have: 1 · · · 4 ∈ ∪ −q(πl )+1−q(πl ,πl ,πl )+ǫ(π) EFπ EFπ EFπ Fπ Cn 1 2 3 4 , (21) l1 l2 l3 l4 ≤ where ǫ(π) 1, 2 , ǫ(π) = 2 if there exists j 2, 3, 4 such that N(π ,π ) = 0 for all ∈ { } ∈ { } lj lk k 2, 3, 4 j , otherwise, ǫ(π) = 1. ∈ { }\{ } q+q′ If ǫ(π) = 1 then since there are at most Cn quadruples such that q(πl1 ) = q and ′ 2 q(π ,π ,π )= q , we deduce that ′ 11(ǫ(π) = 1)EFπ Fπ EFπ Fπ Cn . l2 l3 l4 (π1,··· ,π4)∈S∪S l1 l2 l3 l4 ≤ If ǫ(π) = 2, then, without loss ofP generality, we may assume N(πl2 ,πl )=0 for k 3, 4 . k ∈ { } It implies that q(π ,π ,π )= q(π )+ q(π ,π ). Since (π , ,π ) S S′, N(π ,π ) 1, l2 l3 l4 l2 l3 l4 1 · · · 4 ∈ ∪ l2 l1 ≥ therefore q(π )+ q(π ) q(π ,π ) + 1 and by Inequality (21), l1 l2 ≥ l1 l1 −q(πl )+1−q(πl ,πl ,πl )+2 −q(πl ,πl )−q(πl ,πl )+2 EFπ EFπ EFπ Fπ Cn 1 2 3 4 Cn 1 2 3 4 . l1 l2 l3 l4 ≤ ≤ q+q′ Finally, we notice that there are at most Cn quadruples such that q(πl1 ,πl2 ) = q and ′ q(πl3 ,πl4 )= q , it follows that:

EF EF F F Cn2. πl1 πl2 πl3 πl4 ′ ≤ (π1,··· ,πX4)∈S∪S

Inequality (16) is proved.

3.2 Proof of Theorem 3 By Equation (15), for m 2, we have (with y = 0): ≥ 1 m p−1 m m γ νγ(Pm)= f(0) + f(yφ(j) yφ(j+1))dy2 dyp. (22) p! Z(Rd)p−1 − · · · Xp=2 φ∈XΣm,p jY=1

The leading term in γ is of order γm−1. Taking p = m in the above expression gives:

m m−1 νγ (Pm) γ f(yj yj+1)dy2 dym. ∼ Z Rd m−1 − · · · ( ) jY=1 A direct iteration leads to: m ∗m f(yj yj+1)dy2 dym = f (0), Z Rd m−1 − · · · ( ) jY=1 where f g(y)= f(x)g(y x)dx, f ∗1(x)= f(x), and for m 2, f ∗m = f ∗(m−1) f. ∗ Rd − ≥ ∗ R 12 m m m Hence −1 f(y y )dy dy = fˆ (ξ)dξ = t ψ(t)dt and for all m 2, (Rd)m j=1 j − j+1 2 · · · m Rd ≥ R Q R R m−1 m m −2 t νγ (Pm) γ t ψ(t)dt = t γ ψ( )dt. ∼ Z Z γ Since tψ(t)dt = fˆ(ξ)dξ = f(0), this formula is still valid for m = 1. Now, let h(t) = Rh tm with R h tm ﬁnite for all t, then since, ν (P ) mγm−1Cm, using Fubini’s m≥1 m m≥1 | m| | γ m |≤ Theorem,P the conclusionP follows. ✷ Remark. We can easily identify the next term in the asymptotic of νγ(Pm). The second leading term in Equation (22) is of order γm−2. As in (10), it is equal to m−1 m−2 p m−p Im = γ p u ψ(u)du v ψ(v)dv. ZR ZR Xp=1 Since if u = v, m−1 pupvm−p = uv(u v)−2((m 1)um mum−1v + vm), we deduce that: 6 p=1 − − − P m m−1 m m−2 (m 1)u mu v + v Im = γ uv − − 2 ψ(u)ψ(v)dudv. ZR2 (u v) − 3.3 Proof of Proposition 4 Let D denote the n n matrix with entry i, j equal to: 11( X X δ ), if I denotes the n × k i − jk≤ n n n n identity matrix, D I is the adjacency matrix of the random geometric graph ( , δ ) × n − n G Xn n where there is an edge between i = j if X X δ . We have component wise: 6 k i − jk≤ n f D B f D . −k k∞ n ≤ n ≤ k k∞ n Since the spectral radius ρ(B ) of B is upper bounded by max n (B ) , we deduce n n 1≤i≤n | j=1 n ij| that: P ρ(B ) f (1+∆ ), n ≤ k k∞ n where ∆ is the maximal degree of the graph ( , δ ). Then, the proposition follows from n G Xn n Theorem 6.6 of Penrose [15]. ✷

4 Further properties of the Euclidean Random Matrices

4.1 Eigenvectors of Euclidean Random Matrices

2iπk.Xi As it is pointed by M´ezard, Parisi and Zee [13], if Ui = (Φk,n)i = e we have:

(AΦ ) = F (X X )e−2iπk.(Xi−Xj ) (Φ ) , (23) k,n i i − j k,n i Xj 2iπk.x In particular, if F (x) = e , then n is an eigenvalue with Φk,n as eigenvector and the rank of A is 1. Note also by the Strong Law of Large Numbers that for all i, a.s.

1 −2iπk.(Xi−Xj ) lim F (Xi Xj)e = Fˆ(k). n→∞ n − Xj if An = A/n, by Equation (23), for all i, a.s.:

lim (AnΦk,n)i = Fˆ(k)(Φk,n)i. n→∞ This last equation is consistent with Theorem 1: a.s. for n large enough there exists an eigenvalue of An close to Fˆ(k). It is possible to strengthen this last convergence as follows:

13 Proposition 9 For p 1, let U = U p 1/p and U = sup U . For all ≥ k kp i≥1 | i| k k∞ i≥1 | i| p (2, ], a.s. for all k Zd, P ∈ ∞ ∈

lim AnΦk,n Fˆ(k)Φk,n p = 0. n→∞ k − k Moreover, lim E A Φ Fˆ(k)Φ 2 = f 2 Fˆ(k) 2 = Fˆ(l) 2. n→∞ k n k,n − k,nk2 k k2 − | | l6=k | | P Proof. To simplify notation, we write Φ = Φ and f = F (X X ). k,n ij i − j n n p 1 p P AnΦk,n Fˆ(k)Φk,n p >ǫ = P fijΦj Fˆ(k)Φi >ǫ k − k n − Xi=1 Xj=1 n 1 p ǫp nP f1jΦj Fˆ(k)Φ1 > ≤ n − n Xj=1 n 1−1/p nP f1jΦj nFˆ(k)Φ1 > ǫn . ≤ − Xj=1

From Equation (23), n f Φ nFˆ(k)Φ = n F (X X )e−2iπk.(X1−Xj ) nFˆ(k) . | j=1 1j j − 1| | j=1 1 − j − | Hence: P P n P A Φ Fˆ(k)Φ >ǫ nP F (X X )e−2iπk.(X1−Xj ) nFˆ(k) > ǫn1−1/p k n k,n − k,nkp ≤ | 1 − j − | Xj=1 n nE P F (X X )e−2iπk.(X1−Xj ) (n 1)Fˆ(k) ≤ | 1 − j − − | h Xj=2 1−1/p > ǫn F (0) Fˆ(k) X1 − | | − | | i 1−1/p 2 max(0, (ǫn F (0) Fˆ(k) )) 2n exp( − | | − | | ), ≤ − F (n 1) k k∞ − where the last equation is Hoeﬀding’s Inequality. We then apply Borel Cantelli Lemma. It remains to prove the statement of the proposition for p = 2. Similarly, we obtain: 1 E A Φ Fˆ(k)Φ 2 = E F (X X )e−2iπk.(X1−Xj ) nFˆ(k) 2. k n k,n − k,nk2 n | 1 − j − | Xj

We then write E F (X X )e−2iπk.(X1−Xj ) nFˆ(k) 2 = E[E[ F (X X )e−2iπk.(X1−Xj ) | j 1− j − | | j 1− j − Fˆ(k) 2 X ]] = PF (0) Fˆ(k) 2 + (n 1) F (x)e−2iπk.x Fˆ(kP) 2dx . The statement follows. | | 1 | − | − Ω | − | ✷ R

4.2 Correlation of the Eigenvalues In this paragraph, we state an elementary lemma on the m-correlation of the eigenvalues of A (m n): ≤ m n M = 1/ E (λ F (0)), m m ij − {i1,··· ,imX}⊂{1,···n} jY=1 where the sum is over all subsets of 1, ,n of cardinal m. Note that M = 0 and that { · · · } 1 M is related to the factorial moment measure ρ (dz , , dz ) (also called the joint intensity m m 1 · · · m

14 measure, refer to Daley and Vere-Jones [6]) of the point process λ F (0), , λ F (0) as { 1 − · · · n − } follows: m Mm = zjρm(dz1, , dzm), ZCm · · · jY=1 Heuristically, ρ (dz , , dz ) is the infinitesimal probability of having an eigenvalue at F (0)+ m 1 · · · m z for each i 1, ,m . We define A¯ = A F (0)I, where I is the n n identity matrix (note i ∈ { · · · } − × that “¯“ is not the complex conjugate of the matrix A). A¯(x , ,x ) is the m m matrix · 1 · · · m × where the coefficient i, j is equal to F (x x ) δ F (0). i − j − ij Lemma 10

Mm = det A¯(x1, ,xm)dx1 dxm. ZΩm · · · · · · For m = 2 we get: 2 M2 = F (x) dx, − ZΩ the point process of eigenvalues is thus repulsive. ¯ ¯ n Proof. The characteristic polynomial of A is χA¯(t) = det(A tI) = i=1(λi f(0) t) = n n−m m − − − am( t) , where, am = (λi f(0)). However,Q by Newton formula, we m=0 − {i1,··· ,im} j=1 j − alsoP have, a = detA¯P , whereQ for a set of indices i = i , , i , A¯i is the m {i1,··· ,im} {i1,··· ,im} { 1 · · · m} m m extracted matrixP obtained from A¯ by keeping the raws and columns i , , i (i.e. A¯i is × { 1 · · · m} a principal minor). Taking expectation, we deduce that Ea = n det A¯(x , ,x )dx dx . m m Ωm 1 · · · m 1 · · · m ✷ R In Lemma 10, we have computed the mean value of the symmetric polynomials:

m α (x , ,x )= x m 1 · · · n j {i1,··· ,imX}⊂{1,···n} jY=1 for the vector λ¯ = (λ F (0), , λ F (0)). Actually, it is possible to compute the mean 1 − · · · n − value of the symmetric polynomials:

m α (x , ,x )= xk. m,k 1 · · · n j {i1,··· ,imX}⊂{1,···n} jY=1

k n k for the vector λ¯. To this end simply consider, χ ¯k (t) = det(A¯ tI)= ((λ f(0)) t). A − i=1 i − − We obtain similarly: Q

m n E ¯ k ¯k 1/ αm,k(λ)= zj ρm(dz1, , dzm)= det Am(x1, ,xn)dx1 dxn, m ZCn · · · Z n · · · · · · jY=1 Ω where m = 1, ,m and for a set of indices i = i , , i , A¯i is the m m extracted { · · · } { 1 · · · m} × matrix obtained from A¯ by keeping the raws and columns i , , i . { 1 · · · m} Acknowledgments The author thanks Neil O’Connell for suggesting this problem and Florent Benaych-George for valuable comments.

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