Density of Positive Eigenvalues of the Generalized Gaussian Unitary Ensemble

# Density of Positive Eigenvalues of the Generalized Gaussian Unitary Ensemble

## Abstract

We compute exact asymptotic of the statistical density of random matrices belonging to the Generalized Gaussian orthogonal, unitary and symplectic ensembles such that there no eigenvalues in the interval . In particular, we show that the probability that all the eigenvalues of an random matrix are positive (negative) decreases for large as where the Dyson index characterizes the ensemble, is some extra parameter and the exponent is a function of which will be given explicitly. For , is universal. We compute the probability that the eigenvalues lie in the interval with and . This generalizes the celebrated Wigner semicircle law to these restricted ensembles. It is found that the density of eigenvalues generically exhibits an inverse square-root singularity at the location of the barriers. These results generalized the case of Gaussian random matrices ensemble studied in [4], [7].
Math Subject classification: 15B52, 15B57, 60B10.
Key-words: Random matrices, Probability measures, Logarithmic potential.

## 1 Introduction

Random matrix theory has been successfully applied in various branches of physics and mathematics, including in subjects ranging from nuclear physics, quantum chaos, disordered systems, and number theory. Of particular importance are Generalized Gaussian random matrices with density is a Gaussian function times a power of the determinant. there are three classes of matrices distributed with such density: real symmetric (Generalized Gaussian Orthogonal Ensemble (GGOE)), complex Hermitian (Generalized Gaussian Unitary Ensemble (GGUE)) and self-dual Hermitian matrices (Generalized Gaussian Symplectic Ensemble (GGSE)). In these models the probability distribution for a matrix in the ensemble is given by

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where is the inner product on the space of matrices invariant, under orthogonal, unitary and symplectic transformations respectively and the parameter is the Dyson index. In these three cases the inner products is defined as follow:

 ⟨X,X⟩=tr(X2),β=1 ⟨X,X⟩=tr(X∗X),β=2 ⟨X,X⟩=tr(X⋆X),β=4.

where denotes the hermitian conjugate of complex valued matrices and denotes the symplectic conjugate on quaternion valued matrices. A crucial result in the theory of random matrices is the celebrated Wigner semi-circle law. It states that for large and on an average, the eigenvalues rescaled by the factor , lie within a finite interval , often referred to as the Wigner ’sea’. Within this sea, the statistical density of eigenvalues has a semi-circular form that vanishes at the two edges , .

 ρ(λ)=1π√2−λ2.

The above result means that, if one looks at the statistical density of eigenvalues of a typical system described by one of the three ensembles above, for a large enough , it will resemble closely to the Wigner semi-circle law. From the semi-circle law, we know that on an average half the eigenvalues are positive and half of them are negative.

One of the main questions posed in [1] and [7] is: for a Gaussian random matrix, what is the probability that all its eigenvalues are positive (or negative).

 Pn=Prob[λ1≥0,...,λn≥0].

This probability has also been studied in the mathematics literature [5] and one can easily compute for smaller values,

 P1=12,P2=12−√24,P3=14−√22π.

The interesting question is how behaves for large ? It was argued in [1] that for large , decays as where the decay constant was estimated to be numerically and via a heuristic argument. In [7] the authors shown that for all the three Gaussian ensembles, to leading order in large

 Pn∼exp(−βθ(0)n2),whereθ(0)=log34=0.2764....

In our case we consider the same question in more general setting, where the ensemble of matrices is the Generalized Gaussian unitary ensemble equipped with the density as in the previous. We will prove that, to leading order in large , and for ,

 Pn∼exp(−βθ(α)n2),

where is a function of which will be given explicitly in proposition 3.5. In the simplest case for ,

 θ(α)=β(log34−Cα)+o(α),

where and is a small terms in .

Another important question will be studied in this work, namely what is the statistical density of the negative (or positive) eigenvalues.

In this paper we will calculate the asymptotic density of eigenvalues in this conditioned ensemble and we will see that it is quite different to the Wigner semi-circle law. We will prove the following result. For a positive real sequence , such that , and after scaling the statistical density of eigenvalues by , it converge as goes to to the density with support , such that: for

 fα,a(x)=12π√b−xx−a(2x+b−a−2α√ab1x),

where and are the unique solutions of the following equations

 b+a−2α√ab=0,34(b−a)2+a(b−a)+2α√a√b−2α−2=0.

It will be proved that for given , the previous equations has a unique solutions and .

For

 f0,0(x)=12π√b−xx(2x+b),

where and

More general for and . We will prove that the statistical density of eigenvalues such that there are no eigenvalues in the interval scaled by converge to some density with support , where and are the unique solution in the interval of the following equations

 b+a−2α√ab≥0,34(b−a)2+a(b−a)+2α√a√b−2α−2=0. (1.1)

For it is assumed that .

More precisely for , if , then and . If , then and is the unique solution of (1.1).

If , and , then and , if , then and .

For example for , one recover’s the famous Wigner semi-circle law. One can se that for , must be strictly positif. This can be explain because the singularity at and there no single support of the density of eigenvalues when and . This latter case will be studied by the author in forthcoming paper.

The paper is organized as follows. In the first section we begin by recalling some result about potential theory and equilibrium measure and we enunciate our fundamental equilibrium measure which is the key of the work (theorem 2.4). In section 2 we prove the existence of such measure. Section 3 is dedicated to prove that the measure of theorem 2.4 is an equilibrium measure and we calculate explicitly the equilibrium energy of such measure.

In section 4 we defined the ensemble of Generalized Gaussian random matrices ensemble and we prove the convergence of the statistical density of eigenvalues to the equilibrium measure of theorem 2.4. In the same way, we describe the probability of atypical and large fluctuations of around its mean, say over a wider region of width . Since , this requires the computation of the probability of an extremely rare event characterizing a large deviation of to the left of the mean. Such result has been proved in [?] in the case of Gaussian random matrices ensemble and our result follows in the same way.

## 2 Solution on one single interval

Let be a closed interval, and be a lower semi-continuous function on . If is unbounded we assume that

 lim|x|→+∞(Q(x)−log(1+x2))=∞.

For given and , we wish to compute the equilibrium measure. We start by some general results.

For any probability measure on , we defined the potential of by: for

 Uμ(x)=∫Σlog1|x−y|μ(dy),

and the energy by

 EQ,Σ(μ)=∫ΣUμ(x)μ(dx)+∫ΣQ(x)μ(dx).

From the inequality

 |x−y|≤√1+x2√1+y2,

it can be seen that is bounded from below. Let

 E∗Q,Σ=infμ∈M(Σ)EQ,Σ(μ),

where is the set of probability measures on the closed set .

If , where is continuous function with compactly support , the potential is a continuous function, and .

###### Proposition 2.1

— There is a unique measure such that

 E∗Q,Σ=EQ,Σ(μ∗),

moreover the support of is compact.

This measure is called the equilibrium measure.

See theorem II.2.3 [6]. The next proposition is a method to find the equilibrium measure

###### Proposition 2.2

— Let with compact support. Assume the potential of is continuous and, there is a constant such that
(i) on .
(ii) on supp. Then is the equilibrium measure .

The constant C is called the (modified) Robin constant. Observe that

 E∗Q,Σ=C+12∫ΣQ(x)μ∗(dx).
###### Remark 2.3

— Let be a closed interval of , if we consider the restriction of the function initially defined on to the closed interval and if the equilibrium measure associate to satisfies supp. Then the equilibrium measure for the couple is .

We come to our first result. Let , . Consider the closed interval and

 Qα(x)=x2+2αlog1x,

if , it is assumed that .

One can observe that is lower semi-continuous on the closed interval . Moreover , hence the energy is correctly defined. Let be the equilibrium measure associate to .

###### Theorem 2.4

(1)

If and , then there is a unique and a unique such that , and the equilibrium measure on is given by the measure with support and with density

 fα,0(x)=1π√(bc−x)(x−ac)(1+α√acbc1x),

where

 bc=23(√6(α+1)−2a2c−ac2).
(2)

If and , then the equilibrium measure on still the same as in (1) ().

(3)

If and , in this case the equilibrium measure on is supported by , and density , where is the unique solution of the following equations

 σ+b−2α√σb≥0,and34(b−σ)2+σ(b−σ)+2α√σb−2α−2=0,

and

 fα,σ(x)=12π√b−xx−σ(2x+b−σ−2α√σb1x).
(4)

If and , then two cases are present:

(a)

If , the equilibrium measure on has support and density , where

 b=b(0,σ)=23(√σ2+6+σ2).

and

 f0,σ(x)=12π√b−xx−σ(2x+b−σ).
(b)

If , the equilibrium measure on is the semicircle law with density , and support .

Before proving the theorem let gives same remarks.

One can remark that, if and , the equilibrium measure is supported by and have as density

 f0,0(x)=12π√b0−xx(2x+b0),

where

 b0=b(0,0)=23√6.

Such density appear in the work ([4], [7]) where the authors studies the density of positive eigenvalues of the Gaussian random matrices ensemble. It can be deduced from the first step. In fact and,

 bc=b(ac,α)=23(√6(α+1)−2a2c−ac2),

hence as , we obtain and
Moreover one can obtained such result from step (4).

The density in Step (4) appear in ([4], [7]) where the authors studies the density of eigenvalues bigger then in the case of the Gaussian random matrices ensemble. Such density can be obtained from step (3). Letting one gets,
and,

 34(b−σ)2+σ(b−σ)−2=0.

The last equation has a unique solution with . Such solution can be find explicitly, it is given for all by

Moreover in this case, it will be seen that the measure still an equilibrium measure for all , and when which is the limit case, we obtain , , and became the semi-circle law.
Observe that in (4),(b) the equilibrium measure is independent of the support . In fact for all , the equilibrium measure relatively to the set , has semicircle law density.

To prove the theorem we need some preliminary results.

## 3 Existence of the probability measure

We will prove in the next proposition that, in each cases of the previous theorem, defined a probability measure. In section 3 we show that such measure is an equilibrium measure.

For , let defined on ,

 φα(x,a)=x+a−2α√ax,
 ψα(x,a)=34(x−a)2+a(x−a)+2α√a√x−2α−2,

and

 Eα={a>0∣∃x>a,φα(x,a)≥0andψα(x,a)=0}.

For , and , will be defined on and

 E0={a∈R∣∃x>a,φ0(x,a)≥0andψ0(x,a)=0}.
###### Proposition 3.1

— Let , then

(1)

is a closed set. Moreover is correctly defined.

(2)

If , then

(a)

for all , there is a unique such that . Furthermore defined an increasing function and

(b)

For the unique element , satisfies and it is given explicitly by

 bc=23(√6(α+1)−2a2c−ac2).

Moreover .

(3)

If , then , moreover,

(a)

if , then the unique solution of the equation in is

 b=23(√a2+6+a2).
(b)

If , then is the unique solution of the equation in .

Using mathematica we obtain, for ,

 ac=√53+5α3−γ3−23√2+4α−4α2+2(1+α)γ,

where

To prove the proposition we need a technical lemma

###### Lemma 3.2

—Let be a real number, and be two times continuous differentiable function on such that , , and for all . Then there exist a unique such that .

Proof.— The function increase in the interval .
First case. If , then , for all . It follows that increase in . Since , and . Then the equation admit a unique solution in the interval .
Second case. If . Since , and is an increasing function hence there exist a unique , such that , and by monotony of the function , in , and, in . Hence decrease in and increase in . It follows that for all . Furthermore . Then there exist a unique such that . Which complete the proof.

Proof of proposition.
Step 1): Let , and be a positive real sequence convergent to some . Then by definition of there is a sequence , such that

 φα(xn,an)≥0,andψα(xn,an)=0. (3.2)

Since hence

 an

The sequence converge, it follows that is a bounded sequence, and there is some subsequence convergent to , and . From the inequality valuable for all ,

 φα(xn,an)=xn+an−2α√anxn≥0,

it follows that . Using equation(3.2), then when , there is , such that

 φα(x0,a)≥0,andψα(x0,a)=0.

Since , hence . which prove that , and is closed.

Existence of : Let . We will prove that . Since

 φα(√α,√α)=0.

Furthermore is an increasing function hence , for all . So it is enough to prove that admit a solution . Take the derivative with respect to , we obtain

 ∂ψα∂x(x,√α)=32(x−√α)+√αx−α√√αx√x.

and

 ∂2ψα∂2x(x,√α)=32+√α+32α√√αx2√x.

Hence , for all . Furthermore , . Hence By the previous lemma, with , it follows that for fixed , the equation admit a unique solution , in , which satisfies .

This prove that .

For , is bounded below by , and by closeness exist. It is obvious to see that because is not defined for .
For ,

 φ0(x,a)=x+a,andψ0(x,a)=34(x−a)2+a(x−a)−2.

Hence for every , the solutions of the equation are

 x1=23(√a2+6+a2),andx2=−23(√a2+6−a2).

There is a unique solution , such that . Such solution exist if , and is given by

 x=23(√a2+6+a2).

It follows that .
Step(2): Let . We begin by proving the existence, uniquness and the growth of .
Case (a): We saw that , hence there exist some , such that

 φα(bc,ac)≥0,andψα(bc,ac)=0. (3.3)

For all , the function satisfies all the hypotheses of the previous lemma in the interval , in fact

 ∂ψα∂x(x,a)=32(x−a)+ax−α√ax√x,

and

 ∂2ψα∂2x(x,a)=32+a+32α√ax2√x>0,

, . it follows that,

 ∃!x∈]a,+∞[;ψα(x(a),a)=0. (3.4)

By equations (3.3), (3.4) the set where is a the unique solution in of the problem

 φα(x,a)≥0,andψα(x,a)=0

is not empty. By unicity, it follows that defined a function on .

Growth of the function . Let , and be a solution of the problem

 ∃x>a1>0,φα(x,a1)≥0,ψα(x,a1)=0,

and the unique element in the interval , such that

 ψα(x(a2),a2)=0.

We want to prove . Assume the contraire. Then

 a1

Take the derivative of the function , it yields

 ∂ψα∂a(x,a)=−12φα(x,a),

and

 ∂2ψα∂2a(x,a)=−12(1+αa√ax)<0.

Hence for all , the function decreases on . Thus for all ,

 ∂ψα∂a(x(a1),a)≤∂ψα∂a(x(a1),a1)=−12φα(x(a1),a1)≤0,

which mean that the function decreases on .

Since it follows that

 ψα(x(a1),a2)≤ψα(x(a1),a1)=0.

By the assumption and the unicity of the solution of the equation in , on gets

 ψα(x(a1),a2)<0. (3.5)

Now, consider the function . Such function satisfies for all ,

 ∂2ψα∂2x(x,a2)>0,

then increases in .
If . Then , for all , and the function increases in . Using the assumption and equation(3.5), one gets

 0=ψα(x(a2),a2)≤ψα(x(a1),a2)<0.

Assume . The function is strictly increasing on , and , hence there is a unique , such that It follows that decreases on , and increases on . Moreover , hence . From the assumption , and equation(3.5) we obtains

 0=ψα(x(a2),a2)≤ψα(x(a1),a2)<0.

This gives contradiction. Which is the desired result.

Let . From the previous, there is a unique , such that . Moreover and , and the function increase, hence . Furthermore the two functions , and increases. Thus

 φα(b,σ)≥φα(bc,σ)≥φα(bc,ac)≥0.

Which means that is the unique solution in of the problem

 φα(x,σ)≥0,ψα(x,σ)=0.

(b) Assume . We saw that , hence there exist , such that for all , . By the definition of , for all , , or . Furthermore the equation has a unique solution in , for each . It follows that for all , there is some sequence such that , and .

Since

 φα(xn,an)=xn+an−2α√anxn<0,

Hence

 an

The sequence converge to , it follows that is bounded, and there is some subsequence convergent to , and Since

 ψα(xnk,ank)=0,andφα(xnk,ank)<0.

letting to infinity, we obtain by continuity

 ψα(x0,ac)=0,andφα(x0,ac)≤0.

By unicity of the solution of the equation in the interval , it follows that . Hence and by the definition of , it follows that and . Which means that

 bc+ac−2α√acbc=0,

and

 34(bc−ac)2+ac(bc−ac)+