Distribution of the outliers: proof of Theorem 2.14

Thus we get

|λ_α+k+−µα(t)| 6 2kN^−2/3ψe⁻¹ (6.36)

for allt∈[0,1]. Choosing

C₂ ..= Ce₂+ 1, C₃ ..= Ce₃+ 1 completes the proof of Theorem 2.7 (recall the definition (6.10)).

7. Distribution of the outliers: proof of Theorem 2.14

7.1. Reduction to the law ofG_v(i)v⁽ⁱ⁾(θ(di)). The following proposition reduces the problem to analysing a single explicit random variable.

Proposition 7.1. There is a constant C₂, depending on ζ, such that the following holds. Suppose that

|di| 6 Σ−1,

|di| −1

> ϕ^C2N^−1/3

for all i = 1, . . . , k. Suppose moreover that for all i ∈ O (2.24) holds. Recall the definitions (2.16) and

Before proving Proposition 7.1, we record the following auxiliary result.

Lemma 7.2. Let C1 denote the constant from Theorem 2.3. For any x ∈

withζ-high probability. More generally, we have, for any normalized v,w∈C^N, ∂_xG_vw(x)−∂_xm(x)hv,wi

6 ϕ^CζN^−1/3κ⁻¹_x (7.2)

withζ-high probability.

Proof. By symmetry, we may assume thatx>0. Moreover, (7.2) follows from (7.1) and polarization.

We therefore prove (7.1) for x>0. We have

∂_xG_vv(x) = X

with ζ-high probability, where in the last step we used Theorem 3.7. (In the proof of Theorem 2.3, the constant C1 was chosen large enough for this application of Theorem 3.7; see (5.6).) A similar calculation using the definition (2.4) yields

Therefore we get, using Theorem 2.3 and Lemma 3.2,

∂xGvv(x)−∂xm(x) withζ-high probability. Choosingη..=N^−1/6κ^3/4yields the claim.

Proof of Proposition 7.1. We only prove the claim for the case di > 1; the case di <−1 is handled similarly.

For 2 +ϕ^C1N^−2/36x6Σ, whereC₁ is the constant from Theorem 2.3, we define thek×kHermitian matricesA(x) andA(x) throughe

Aij(x) ..= G_v(i)v^(j)(x)−m(x)δij+d⁻¹_i δij, Aeij(x) ..= δij

G_v(i)v⁽ⁱ⁾(x)−m(x) +d⁻¹_i .

(Here we subtractm(x)1 so as to ensure that∂xA(x) is well-behaved; see below.) We denote the ordered eigenvalues ofA(x) andA(x) bye a₁(x)6· · ·6a_k(x) andea₁(x)6· · ·6ea_k(x) respectively.

For the rest of the proof we fixi∈Osatisfyingdi >1. We abbreviateθi..=θ(di). We begin by comparing the eigenvalues ofA(θe _i) andD⁻¹. Define the eigenvalue indexr≡r(i) = 1, . . . , kthrough

ear(x) = 1 di

+G_v(i)v⁽ⁱ⁾(x)−m(x). (7.3)

In particular,

ea_r(θ_i) = G_v(i)v⁽ⁱ⁾(θ_i) + 2 d_i. Theorem 2.3 implies that

G_v(j)v^(j)(θi)−m(θi)

6 ϕ^CζN^−1/2(di−1)^−1/2. (7.4) withζ-high probability forj= 1, . . . , k. In particular,

ea_r(θ_i)− 1 di

6 ϕ^CζN^−1/2(d_i−1)^−1/2

withζ-high probability. Moreover, (7.4) and the condition (2.24) yield, forj6=i,

G_v(j)v^(j)(θ_i)−m(θ_i)

|di−d_j| (7.5)

withζ-high probability, providedC₂ is chosen large enough. We therefore conclude that minj6=r

eaj(θi)−ear(θi)

> ϕ^C2−1N^−1/2(di−1)^−1/2 (7.6) withζ-high probability, providedC₂ is large enough.

Next, we compare the eigenvalues of A(θ_i) and A(θe _i) using second-order perturbation theory (the first-order correction vanishes by definition ofAeandA). Theorem 2.3 yields

kA(θ_i)−A(θe _i)k 6 ϕ^CζN^−1/2(d_i−1)^−1/2

withζ-high probability. Therefore (7.6) and nondegenerate second-order perturbation theory yield, for large enoughC₂,

ar(θi) = ear(θi) +O

ϕ^CζN⁻¹(di−1)⁻¹ min_j6=r|ea_j(θ_i)−ea_r(θ_i)|

= ear(θi) +O

ϕ^Cζ−C2N^−1/2(di−1)^−1/2

(7.7)

withζ-high probability.

Next, we analyseA(x) and make the link to µ_α(i). From Lemma 7.2 we find ∂xA(x)

6 ϕ^CζN^−1/3κ⁻¹_x withζ-high probability. In particular, we have for allj= 1, . . . , k that

aj(x)−aj(y)

6 ϕ^CζN^−1/3(κ⁻¹_x +κ⁻¹_y )|x−y| (7.8) withζ-high probability, provided that 2 +ϕ^C1N^−2/36x, y6Σ.

Recall the definition (2.17) ofα(i). From Lemma 6.1 and Theorem 3.7, we know thatµ_α(i)is characterized by the property that there is aq≡q(i)∈ {1, . . . , k} such that

aq(µ_α(i)) = −m(µ_α(i)). By Theorem 2.7 we have

|µ_α(i)−θ_i| 6 ϕ^C3N^−1/2(d_i−1)^1/2 (7.9) withζ-high probability. ProvidedC₂ is large enough (depending onC₃), it is easy to see from (7.9) that

µ_α(i)−2 θ_i−2 (d_i−1)² (7.10)

withζ-high probability. Thus we find, using (7.8), (7.9), and (7.10), that for large enoughC2 we have m(µ_α(i)) = −a_q(θ_i) +O

ϕ^CζN^−5/6(d_i−1)^−3/2

(7.11) withζ-high probability. (Here we absorbed the constantC₃ intoC_ζ.)

We now prove that q=rwithζ-high probability providedC2 is large enough. Assume by contradiction thatq6=r. Then we get, using Theorem 2.3 and the condition (2.24), that

a_q(θ_i)− 1 di

> ϕ^C2−1N^−1/2(d_i−1)^−1/2 (7.12) withζ-high probability. Moreover, (7.8), (7.9), and (7.10) yield

a_q(θ_i) = a_q(µ_α(i)) +O

ϕ^CζN^−5/6(d_i−1)^−3/2

= −m(µα(i)) +O

ϕ^CζN^−5/6(d_i−1)^−3/2

= 1 di

+O

ϕ^CζN^−1/2(di−1)^−1/2+ϕ^CζN^−5/6(di−1)^−3/2

with ζ-high probability, where in the last step we used (6.5). Together with (7.12), this yields the desired contradiction providedC2 is large enough. Henceq=r.

Putting (7.3), (7.11), and (7.7) together, we get m(µ_α(i)) = −G_v(i)v⁽ⁱ⁾(θi)− 2

d_i +O

ϕ^CζN^−5/6(di−1)^−3/2+ϕ^Cζ−C2N^−1/2(di−1)^−1/2

withζ-high probability. Thus we find that, for allxbetweenθi andµ_α(i), we have m⁰(x) = m⁰(θi) +O(ϕ^C3N^−1/2(di−1)^−5/2) = m⁰(θi)(1 +O(ϕ⁻¹))

withζ-high probability, where we used (6.5) and (7.9). Using (6.2), (7.10), and (6.5), we conclude that µ_α(i)−θi = −(1 +O(ϕ⁻¹))G_v(i)v⁽ⁱ⁾(θi) +d⁻¹_i

m⁰(θ_i) +O

ϕ^CζN^−5/6(di−1)^−1/2+ϕ^Cζ−C2N^−1/2(di−1)^1/2 withζ-high probability. The claim now follows for large enoughC₂, using the identity (6.2).

7.2. The GOE/GUE case. By Proposition 7.1, it is enough to analyse the random variable X ..= N^1/2(|d|+ 1)(|d| −1)^1/2

Gvv(θ) +1 d

, (7.13)

wherev∈C^N is normalized,dsatisfies

1 +ϕ^C2N^−1/3 6 |d| 6 Σ−1, (7.14)

and we abbreviatedθ ≡ θ(d). For definiteness, we choosed >1 in the following.

The following notion of convergence of random variables is convenient for our needs.

Definition 7.3. Two sequences of random variables, {AN} and {BN}, are asymptotically equal in distri-bution, denoted AN

∼d BN, if they are tight and satisfy

N→∞lim Ef(AN)−Ef(BN)

= 0 (7.15)

for all bounded and continuousf.

Remark 7.4. Definition 7.3 extends the notion of convergence in distribution, in the sense that Ef(AN) need not have a limit asN → ∞.

Remark7.5. In order to show thatAN

∼d BN, it suffices to establish the tightness of either{AN}or{BN} and to verify (7.15) for allf ∈C_c^∞(R). Indeed, if{AN}is tight then so is{BN}, by (7.15). By tightness of AN andBN, we may replace in (7.15) the bounded and continuousf with a compactly supported continuous functiong. Next, we can approximateg uniformly withC_c^∞-functions.

Remark 7.6. Clearly,AN

∼d BN ifAN

=d BN for allN.

Lemma 7.7. Let A_N ∼^d B_N andR_N satisfy lim_NP(|RN|6ε_N) = 1, where{εN} is a positive null sequence.

ThenAN

∼d BN+RN.

Proof. By Remark 7.5, it suffices to prove (7.15) forf ∈C¹(R) such thatf andf⁰ are bounded. Then Ef(AN)−Ef(BN+RN) = Ef(AN)−Ef(BN)

+ Ef(BN)−Ef(BN+RN)

= o(1) +E

h1(|R_N|6ε_N) f(B_N)−f(B_N +R_N)i

= o(1) where in the last step we used the boundedness off⁰.

Lemma7.8. Let{AN},{A⁰_N},{BN}, and{B_N⁰ }be sequences of random variables. Suppose thatA_N ∼^d A⁰_N, B_N ∼^d B_N⁰ ,A_N andB_N are independent, andA⁰_N andB_N⁰ are independent. Then

AN +BN

∼d A⁰_N +B_N⁰ .

Proof. Without loss of generality, we may assume thatA_N, B_N, A⁰_N, B_N⁰ are independent (after replacing A⁰_N andB_N⁰ with new random variables without changing their laws.) Then for anyλ∈Rwe have

Ee^iλ(AN+BN)−Ee^iλ(A0N+B0N) = E h

e^iλAN(e^iλBN −e^iλBN0 ) + (e^iλAN −e^iλA0N)e^iλBN0 i

= Ee^iλANE(e^iλBN −e^iλBN0 ) +E(e^iλAN −e^iλA0N)Ee^iλBN0

−→ 0 asN → ∞.

Next, we observe thatA_N+B_N andA⁰_N+B⁰_N are tight. Therefore, recalling Remark 7.5, we find that it suffices to prove

Ef(AN+BN)−Ef(A⁰_N +B_N⁰ ) −→ 0 f ∈C_c^∞. Denoting by ˆf the Fourier transform off, we find

Ef(AN +BN)−Ef(A⁰_N+B⁰_N) = Z

dλfˆ(λ)h

Ee^iλ(AN+BN)−Ee^{iλ(A0N+BN0 )}i

−→ 0 by dominated convergence.

Proposition 7.9. Let H be a GOE/GUE matrix. Assume that dsatisfies (7.14). Then for large enough C2 we have

X ∼ N^d

0,2(d+ 1) βd²

.

Proof. By unitary invariance, we have G_vv =^d G₁₁, where = denotes equality in distribution.^d In or-der to handle the exceptional low-probability events, we add a small imaginary part to the spectral pa-rameter z ..= θ+ iN⁻⁴. Throughout the following we abbreviate G ≡ G(z) and m ≡ m(z). Writing a^∗..= (h12, h13, . . . , h1N), we get from Schur’s formula and (2.5) that

G11 = 1

h11−z−a^∗G⁽¹⁾a = 1

−m−z+h₁₁− a^∗G⁽¹⁾a−m

= m−m²h11+m² a^∗G⁽¹⁾a−m

+O(|h11|²) +O

a^∗G⁽¹⁾a−m

2

(7.16) with ζ-high probability. Again by unitary invariance, we have a^∗G⁽¹⁾a =^d kak²G⁽¹⁾₂₂. Moreover, both sides are independent ofh11, so that

−m²h11+m² a^∗G⁽¹⁾a−m d

= −m²h11+m² kak²G⁽¹⁾₂₂ −m

. (7.17)

In order to estimate the error term in (7.16), we write kak²G⁽¹⁾₂₂ −m = kak²−1

G⁽¹⁾₂₂ + (G⁽¹⁾₂₂ −m). (7.18)

Using (3.6) to estimateG⁽¹⁾₂₂ −G22, as well as Theorem 2.3, Lemma 3.5, and Lemma 3.2, we therefore find that

kak²G⁽¹⁾₂₂ −m

6 ϕ^CζN^−1/2(d−1)^−1/2 (7.19)

withζ-high probability. Moreover, we have the trivial boundE

kak²G⁽¹⁾₂₂ −m

k6(kN)^Ck fork∈N. From (7.16), (7.17), (7.18), and (7.19), we conclude that there exist random variablesRe1andRe2satisfying

|Re1|+|Re2| 6 ϕ^CζN⁻¹(d−1)⁻¹ (7.20) withζ-high probability, the rough bound

E(|Re₁|+|Re₂|)^k 6 (kN)^Ck, (7.21) and

G⁽²⁾₁₁ −m

+Re1 = −m²h11+m² a^∗G⁽¹⁾a−m

=d −m²h₁₁+m² kak²G⁽¹⁾₂₂ −m

= −m²h₁₁+m³ kak²−1

+m²(G⁽¹⁾₂₂ −m) +Re₂. Defining

Y1 ..= N^1/2(d+ 1)(d−1)^1/2Re G⁽²⁾₁₁ −m

, Y2 ..= N^1/2(d+ 1)(d−1)^1/2Re G⁽¹⁾₂₂ −m , W ..= N^1/2Re

−m²h₁₁+m³ kak²−1

, R_i ..= N^1/2(d+ 1)(d−1)^1/2ReRe_i (i= 1,2), we therefore get

Y₁+R₁ = (d^d + 1)(d−1)^1/2W +m²Y₂+R₂. (7.22) In order to infer the distribution of Y1 from (7.22), we observe that the random variablesY2 andW are independent. Also,Y1

=d Y2. Recalling Theorem 2.3 and (3.6), we find the bounds

|Y_i| 6 ϕ^Cζ, |R_i| 6 ϕ^CζN^−1/2(d−1)^−1/2 (i= 1,2) (7.23) withζ-high probability, and the rough bounds

|Yi| 6 N², E|Ri|^k 6 (kN)^Ck (i= 1,2). (7.24) Moreover, by the Central Limit Theorem

2(d²+ 1) βd⁶

−1

W ∼ N^d (0,1), (7.25)

where we used (6.2).

Next, let B andZ₂ be independent random variables whose laws are given by B =^d N

0,2(d²+ 1) βd⁶

, Z₂ =^d N(0, ξ²),

where we introduced

ξ² ≡ ξ²_N ..= d⁴(d²−1)(d+ 1) d⁴−1

2(d²+ 1)

βd⁶ = 2(d+ 1) βd² . Defining

Z₁ ..= (d+ 1)(d−1)^1/2B+d⁻²Z₂, (7.26) we find thatZ1

=d Z2. Moreover, a standard moment calculation and the definition ofW yield lim

N→∞(EW^k−EB^k) = 0 ; (7.27)

as usual, only the pairings in the moment expansion of EW^k survive the limit N → ∞. (See also (7.25), which however cannot be used to deduce (7.27) directly.)

We now compare the distributions ofY₁andZ₁by computing moments. Note that the family{EZ₁^k}_N∈N is bounded for eachk∈N. We claim that

lim

N→∞ EY₁^k−EZ₁^k

= 0 (7.28)

for all k ∈ N. (This will imply that Y1

∼d Z1.) We shall prove (7.28) by induction on k. Taking the expectation of (7.22) yields

EY1 = m²EY1+O

ϕ^CζN^−1/2(d−1)^−1/2 where we used (7.23), (7.24), andEW =O(N^−1/2). Therefore

EY1 6 Cϕ^CζN^−1/2(d−1)^−3/2 = o(1) providedC₂ in (7.14) is large enough. Here we used that

m(z) = d⁻¹+O(N⁻³), (7.29)

as follows from the definition of z = θ+ iN⁻⁴, (5.5), Lemma 3.2, and (6.2). Therefore (7.28) for k = 1 follows usingEZ1= 0.

For the induction step, we assume that (7.28) holds for allk⁰ 6k−1. From (7.22) we find

EY₁^k+

k

X

l=1

k l

E R^l₁Y₁^k−l

= E (d+ 1)(d−1)^1/2W+m²Y₂^k +

k

X

l=1

k l

E

R^l₁ (d+ 1)(d−1)^1/2W +m²Y₂k−l

. (7.30) We estimate the summands on the left-hand side by

E R^l₁Y₁^k−l

6 N^Cexp(−ϕ^C) +

ϕ^CζN^−1/2(d−1)^−1/2l

E|Y1|^k−l 6 C

ϕ^CζN^−1/2(d−1)^−1/2l ϕ^Cζ 6 ϕ^CζN^−1/2(d−1)^−1/2,

where in the first step we used (7.23) and (7.24), in the second step the estimateE|Y1|^k−l6ϕ^Cζ as follows from the induction assumption (7.28) applied to even moments (recall thatY1is real) as well as (7.23) and (7.24), and in the third step the fact thatl >1. Note that the constant Cζ is independent of k. A similar estimate applies to the summands on the right-hand side of (7.30). Thus (7.30) yields

EY₁^k = E (d+ 1)(d−1)^1/2W +m²Y₂^k where in the second step we used the induction assumption and the estimateEW =O(N^−1/2). Therefore we get

In order to conclude the proof of (7.28), we deduce from (7.26) that EZ₁^k = 1

Using the induction assumption (7.28) fork⁰ =k−l, (7.29), and the condition l >2, we get from (7.31), (7.32), and (7.27) that

lim

N→∞ EY₁^k−EZ₁^k

= 0 for large enoughC2. This concludes the proof of (7.28).

Next, by definition we have ξ⁻¹Z₁ =^d N(0,1). Moreover, we have that ξ ∈ [c, C] for some positive

for any continuous bounded functionf. Next, we estimate G11(θ)−G⁽²⁾₁₁(z) withζ-high probability, where in the second step we used Lemma 7.2, (5.5), and Lemma 3.2 to estimate the first term, and Theorem 2.3 and (6.1) to estimate the second term. Therefore

X =^d N^1/2(d+ 1)(d−1)^1/2 G11(θ) +d⁻¹

= Y1+O ϕ^CζN^−1/2(d−1)^−1/2

= Y1+o(1)

with ζ-high probability, where in the second step we used (7.29). Therefore (7.33), the fact that Z =^d Z₁, and dominated convergence yield

lim

N→∞ Ef(ξ⁻¹X)−Ef(ξ⁻¹Z)

= 0. (7.34)

The claim now follows from Lemma 7.10 below.

Lemma 7.10. Let {ξN} be a bounded deterministic sequence. Let A_∞, A1, A2, . . . be random variables such that AN converges weakly to A_∞. Then we have for any bounded continuous functionf

Ef(ξ_NA_N)−Ef(ξ_NA_∞) −→ 0 asN → ∞.

Proof. By Skorokhod’s representation theorem, there exist new random variablesAe_∞,Ae1,Ae2, . . . such that A∞

=d Ae∞, AN

=d AeN for allN ∈N, and AeN →Ae∞ almost surely. Let ω be such that AeN(ω)→Ae∞(ω).

By assumption onξ_N, we find that there exists aC≡C(ω) such thatξ_NAe_N(ω)∈[−C, C] andξ_NAe_∞(ω)∈ [−C, C] for allN ∈N. Sincef is uniformly continuous on [−C, C], we find that

lim

N→∞

f(ξNAeN(ω))−f(ξNAe_∞(ω))

= 0. The claim now follows by dominated convergence.

7.3. The almost-GOE/GUE case. As it turns out, replacing the matrix element hij with a Gaussian in the Green function comparison step below (Section 7.4) is only possible if|vi|6ϕ^−D and |vi| 6ϕ^−D, for some large enough constant D >0. If this assumption is not satisfied, we first have to replace hij with a Gaussian using a different method, which effectively keeps track of the fluctuations of Gvv resulting from large components ofv. Thus we shall proceed in two steps:

(i) We compare the original Wigner matrix H withHb, a Wigner matrix obtained from H by replacing the (i, j)-th entry ofH with a Gaussian whenever |vi|6ϕ^−D and|vj|6ϕ^−D.

(ii) We compare the matrixHb to a Gaussian matrix.

The step (ii) is performed in this section. To simplify notation, we write H instead of Hb throughout this section. The step (i) is performed using Green function comparison in Section 7.4 below.

The following shorthand will prove useful.

Definition7.11. Let{σN}be a bounded positive sequence. IfA_N andB_N are independent random variables withBN

∼ Nd (0, σ_N²), and if SN

∼d AN +BN, then we write SN

∼d AN+N(0, σ_N²). For the following we write

X = νN^1/2

Gvv(θ) +1 d

, ν ≡νN ..= (d+ 1)(d−1)^1/2.

Proposition7.12. FixD >0. Letv∈C^N be normalized andH be a Wigner matrix such that if|v_i|6ϕ^−D and|vj|6ϕ^−D thenhij is Gaussian. Then we have

X ∼ −νN^{d 1/2}d⁻²hv, Hvi+N 0, 2(d+ 1)

βd⁴ +4ν²Q(v)

d⁵ +ν²R(v) d⁶

! ,

whereQ(v)andR(v)were defined in (2.22).

Proof. As before, we consistently drop the spectral parameterz=θ from our notation.

Let M ∈ N denote the number of entries of v satisfying|vi| > ϕ^−D. Since v is normalized, we have M 6ϕ^2D. To simplify notation, we assume (after a suitable permutation of the rows and columns of H) that the entries ofvsatisfy|vi|> ϕ^−D fori6M and|vi|6ϕ^−D fori > M. Splitv= _w^u

, whereu∈C^M andw∈C^N−M. (Throughout the following we assume that w6= 0; the casew= 0 may be easily handled by approximation with nonzerow.) We also split

H =

where the second equality defines the right-hand side using self-explanatory notation. Note that, by defini-tion,kxk=kvk= 1.

Next, we claim that

(F^∗F)ij = δij+O ϕ^CζN^−1/2

(7.36) withζ-high probability. In order to prove (7.36), write

F^∗F = B^∗Se^∗SBe B^∗Se^∗a a^∗SBe a^∗a

! .

We consider four cases. First, if 16i6=j6M we find using (3.15) that

withζ-high probability. Second, if 16i6M we find using (3.13) and (3.14) that (F^∗F)ii−1

withζ-high probability. This completes the proof of (7.36).

Next, abbreviateG1(z) ..= (H1−z)⁻¹. SinceN^1/2(N−M−1)^−1/2H1is an (N−M−1)×(N−M−1) GOE/GUE matrix, we find from (7.36), Theorem 2.3, and Lemma 3.2 that

withζ-high probability. Therefore Schur’s formula yields Γ = x^∗ with ζ-high probability, where in the second step we expanded using (2.5), and estimated the error term using (7.37) as well as the boundsM 6ϕ^Cζ and|Eij|6ϕ^CζN^−1/2. Recalling thatkxk= 1, we find

Applying (3.12) tohw, Bui=P

i,jwiujBij (withN in (3.12) replaced byM(N−M)), we find hw, Bui

2 6 ϕ^CζN⁻¹. Similarly, using (3.13) and (3.14) we find that

kBuk² = kuk²+O ϕ^CζN^−1/2

, kSBuke ² = kuk²+O ϕ^CζN^−1/2 withζ-high probability, using (3.15) that

Bu,Se^∗a

6 ϕ^CζN^−1/2 withζ-high probability, and using (3.13) that

kak² = 1 +O ϕ^CζN^−1/2

withζ-high probability. Usingkuk>ϕ^−D (by definition ofu), we therefore conclude that kFxk² = kBuk²+ 2 Re kukkwk

kSBe ukkak

SBu,e a

+kwk²kak²+O ϕ^CζN⁻¹

= 1 +O ϕ^CζN^−1/2

(7.40) with ζ-high probability. Using Theorem 2.3 applied to G1 (recall that F and H1 are independent), we therefore get from (7.39) that

Γ−m = −m²

hu, Aui+kwk²g+ 2 Rehw, Bui

+m² 1

kFxk²hFx, G₁Fxi −m

!

+m³ kBuk²− kuk²+ 2 Re kukkwk kSBukkake

SBu,e a

+kwk² kak²−1

! +O

ϕ^CζN⁻¹(d−1)⁻¹

(7.41)

withζ-high probability. We write this as

Γ−m = Γ1+· · ·+ Γ6+O

ϕ^CζN⁻¹(d−1)⁻¹

(7.42) withζ-high probability, where

Γ1 ..= −m²hu, Aui, Γ2 ..= −m²kwk²g , Γ3 ..= m² 1

kFxk²hFx, G1Fxi −m

! ,

Γ4 ..= −2m²Rehw, Bui+m³ kBuk²− kuk²

, Γ5 ..= 2m³Re kukkwk kSBukkake

SBu,e a , Γ₆ ..= m³kwk² kak²−1

.

We now claim that Γ1, . . . ,Γ6are independent. In order to prove this, letf1, . . . , f6be indicator functions of Borel sets in R. Write a=aωin polar coordinates, wherea >0 andω∈S^N−M−2. Sincea is Gaussian,

aandωare independent. Denote by ρ1, . . . , ρ6 the laws ofA, B, g, a,ω, H1 respectively. Then we get

where the second equality follows by definition of the Γ’s, and the third from the invariance of the law of ω under rotations (applied to Γ5) and from the invariance of the law of H1 under orthogonal/unitary conjugations (applied to Γ3). This proves the independence of Γ1, . . . ,Γ6. The variance of the term in parentheses is

E

where we abbreviated

Q(w,u) ..= N^−1/2ReX

i,j

w_iM_ij⁽³⁾u_j|uj|², R(u) ..= 1 N

X

i,j

M_ij⁽⁴⁾−4 +β

|uj|⁴,

and used that 3−β = 2β⁻¹ for β = 1,2. Since kuk 6 1 and kwk 6 1, we find that Q(w,u) 6 C and R(u)6C for some positive constantC. Next, using Γ5= 2m³kwkRehSBue ,ai+O(ϕ^CζN⁻¹) withζ-high probability and

E 2 RehSBu,e ai²

= 4

βN²(N−M−1)kuk², we find from the Central Limit Theorem and Lemma 7.10 that

νN^1/2Γ5

∼ Nd 0,4ν²β⁻¹m⁶kuk²kwk²

. (7.46)

Finally, we havekak²−1 =kak²−Ekak²+O(M/N) and E |ai|²−E|ai|²2

= 2β⁻¹N⁻². Thus we conclude from the Central Limit Theorem and Lemma 7.10 that

νN^1/2Γ6

∼ Nd 0,2ν²β⁻¹m⁶kwk⁴

. (7.47)

Next, (7.43) – (7.47) imply that νN^1/2Γ2, . . . , νN^1/2Γ6 are tight (as N-dependent random variables).

Moreover, an easy variance calculation shows that νN^1/2Γ1 is also tight. Therefore we get from (7.35), (7.42), (7.43) – (7.47), Lemma 7.7, and Lemma 7.8 that (recall the notation from Definition 7.11)

X ∼ −νN^{d 1/2}m²hu, Aui+N(0, V1), where

V₁ ..= 2(d+ 1) βd⁶ + 2ν²

βd⁴ kwk⁴+ 2kuk²kwk² +4ν²

d⁵ Q(w,u) +ν²

d⁶R(u) + 2ν² βd⁶

kuk⁴+ 2kuk²kwk²+kwk⁴ . Here we used (6.2).

Next, from

hv, Hvi = hu, Aui+ 2 Rehw, Bui+hw, H₀wi, the Central Limit Theorem, Lemma 7.10, and Lemma 7.8 we find

νN^1/2hv, Hvi ∼^d νN^1/2hu, Aui+N

0,2ν²

β (1− kuk⁴)

. (7.48)

Moreover, using that the dimensionM ofusatisfiesM 6ϕ^2D and the fact that maxi|wi|6ϕ^−D, we find Q(w,u) = Q(v) +O(ϕ^−D), R(u) = R(v) +O(ϕ^−2D).

Therefore we get, using Lemma 7.8 and recalling that 1 =kvk²=kuk²+kwk², X ∼ −νN^{d 1/2}d⁻²hv, Hvi+N

0, 2(d+ 1) βd⁶ +4ν²

d⁵ Q(v) +ν²

d⁶R(v) + 2ν² βd⁶

. This concludes the proof.

7.4. Conclusion of the proof of Theorem 2.14. In this section we compute the distribution ofGvv(θ)−m(θ) for a general Wigner matrixH, and hence complete the proof of Theorem 2.14. We use the Green function comparison method from the proof of Lemma 3.9.

LetH = (hij) = (N^−1/2Wij) be an arbitrary real symmetric / Hermitian Wigner matrix,V = (N^−1/2Vij) a GOE/GUE matrix independent ofH, and v∈C^N be normalized. For D >0 define the subset

ID ..=

i= 1, . . . , N ..|vi|6ϕ^−D . Define a new Wigner matrixHb = (bhij) = (N^−1/2cWij) through

Wcij ..=

(Vij ifi∈ID andj ∈ID

Wij otherwise. Thus,Hb satisfies the assumptions of Proposition 7.12. Let

JD ..=

16i6j6N..i∈ID andj∈ID

be the set of matrix indices to be replaced. Similarly to (3.21), we choose a bijective map φ .. JD → {1, . . . , γmax(D)} and denote byHγ = (h^γ_ij) the matrix defined by

h^γ_ij ..=

(N^−1/2Wij ifφ(i, j)6γ N^−1/2Wcij otherwise.

In particular, H₀ =Hb andH_γmax(D)=H. Let now (a, b)∈J_D satisfy φ(a, b) = γ. Similarly to (3.22), we write

H_γ−1 = Q+N^−1/2V where V ..= V_abE^(ab)+1(a6=b)V_baE^(ba), and

Hγ = Q+N^−1/2W where W ..= WabE^(ab)+1(a6=b)WbaE^(ba).

In order to avoid singular behaviour on exceptional low-probability events, we add a small imaginary part to the spectral parameterθ, and setz..=θ+ iN⁻⁴. Abbreviate

x ..= νN^1/2Re(Gvv(z)−m(z)). (7.49) Thus we have the rough bound|x|6N⁴ which we shall tacitly use in the following. We use the notation (3.23), which gives rise to the quantities xR, xS, xT defined through (7.49) with G replaced by R, S, T respectively. We may now state the main comparison estimate.

Lemma 7.13. Provided D is a large enough constant, the following holds. Letf ∈ C³(R)be bounded with bounded derivatives and q≡qN be an arbitrary deterministic real sequence. Then

Ef(x_T+q) = Ef(x_R+q) +Y_abEf⁰(x_R+q) +A_ab+O ϕ⁻¹Eb_ab

, (7.50)

Ef(xS+q) = Ef(xR+q) +Aab+O ϕ⁻¹Ebab

, (7.51)

whereA_ab satisfies|A_ab|6ϕ⁻¹,

Yab ..= −νN⁻¹Re

m⁴M_ab⁽³⁾vavb+m⁴M_ba⁽³⁾vbva

,

and

Ebab ..=

2

X

σ,τ=0

N−2+σ/2+τ /2|va|^σ|vb|^τ+δab 2

X

σ=0

N^−1+σ/2|va|^σ.

Before proving Lemma 7.13, we show how it implies Theorem 2.14.

Proof of Theorem 2.14. FixD >0 large enough that the conclusion of Lemma 7.13 holds. By Remark 7.5, we may assume thatf ∈C_c^∞(R). Letγ=φ(a, b). Since|v_a|6ϕ^−D and|v_b|6ϕ^−D, we find

Y_ab² 6 ϕ⁻¹Eb_ab. (7.52)

Applying (7.50) and (7.51) withf replaced byf⁰ yields

YabEf⁰(xT +q) = YabEf⁰(xR+q) +YabAab+O ϕ⁻¹Ebab

. Subtracting this from (7.50) and using|Aab|6ϕ⁻¹ yields

Ef(x_T +q) = Ef(x_R+q) +Y_abEf⁰(x_T +q) +A_ab+O ϕ⁻¹Eb_ab+ϕ⁻¹|Y_ab| . Subtracting (7.51) yields

Ef(xγ+q) = Ef(xγ−1+q) +YabEf⁰(xγ+q) +O ϕ⁻¹Ebab+ϕ⁻¹|Yab| , where we introduced the notationxγ ..=νN^1/2Re (Hγ−z)⁻¹_vv−m(z)

. Using (7.52) we therefore get Ef(xγ+q−Yab) = Ef(xγ−1+q) +O ϕ⁻¹Ebab+ϕ⁻¹|Yab|

. (7.53)

We now iterate (7.53), starting atγ = 1 andq= 0. Using that P

a,bEb_ab6C and P

a,b|Y_ab|6C, we find afterγ_max iterations of (7.53)

Ef x_γmax(D)−

γ_max(D)

X

γ=1

Y_ϕ−1(γ)

!

= Ef(x0) +O(ϕ⁻¹).

Moreover, using|va|6ϕ^−D and|vb|6ϕ^−D, we find that

γ_max(D)

X

γ=1

Y_ϕ−1(γ) = −νN⁻¹Re X

a,b∈ID

1(a6b)m(z)⁴

M_ab⁽³⁾v_av_b+M_ba⁽³⁾v_bv_a

= −νN⁻¹Re

N

X

a,b=1

m(z)⁴M_ab⁽³⁾vavb+O(ϕ^−2D). Using Lemma 7.8 we find

νN^1/2

(H−z)⁻¹_vv −m(z) _d

∼ νN^1/2

(Hb −z)⁻¹_vv−m(z)

−νN⁻¹Re

N

X

a,b=1

m(z)⁴M_ab⁽³⁾vavb.

Using Lemma 7.2, it is now easy to remove the imaginary partN⁻⁴ ofz to get νN^1/2

(H−θ)⁻¹_vv+d⁻¹ _d

∼ νN^1/2

(Hb −θ)⁻¹_vv+d⁻¹

−νS(v) d⁴ . SinceHb satisfies the assumptions of Proposition 7.12, we find

νN^1/2

(H−θ)⁻¹_vv+d⁻¹ _d

∼ −νN^1/2hv, Hvi

d² −νS(v)

d⁴ +N 0, 2(d+ 1)

βd⁴ +4ν²Q(v)

d⁵ +ν²R(v) d⁶

! ,

using the notation of Definition 7.11. Now Theorem 2.14 follows from Proposition 7.1 and Lemma 7.7.

Proof of Lemma 7.13. As before, we consistently drop the spectral parameterz=θ+ iN⁻⁴fromGand m. We focus on (7.50). From Theorem 2.3, (3.29), and (3.28) (withS replaced byT), we find

|Tva| 6 ϕ^CζN^−1/2(d−1)^−1/2+C|va|, |Rva| 6 ϕ^CζN^−1/2(d−1)^−1/2+C|va|+ϕ^CζN^−1/2|vb| (7.54) with ζ-high probability, and similar results hold for T_av, T_vb, T_bv, R_av,R_vb, andR_bv. Similarly, from the first inequality of (3.30) (withS replaced byT), we get

Tvv−Rvv

6 ϕ^CζN^−1/2

N⁻¹(d−1)⁻¹+N^−1/2(d−1)^−1/2(|va|+|vb|) +|va||vb|+N^−1/2(|va|²+|vb|²) withζ-high probability. This yields

|xT−xR| 6 ϕ^Ceζh

N⁻¹(d−1)^−1/2+N^−1/2(|va|+|vb|) + (d−1)^1/2|va||vb|+N^−1/2(d−1)^1/2(|va|²+|vb|²)i (7.55) withζ-high probability for some constantCe_ζ. Now chooseD>Ce_ζ+ 1. By definition of J_D, we have that

|va|6ϕ^−D and|vb|6ϕ^−D. Therefore

|x_T−x_R|³ 6 ϕ⁻¹Eb_ab withζ-high probability. This yields

Ef(x_T +q) = Ef(x_R+q) +E f⁰(x_R+q)(x_T−x_R) +1

2E f⁰⁰(x_R+q)(x_T−x_R)²

+O ϕ⁻¹Eb_ab

. (7.56) In order to analysexT −xR=νN^1/2Re(Tvv−Rvv), we write

x_T −x_R = y₁+y₂+y₃+y₄, where

yk ..=

(νN^1/2−k/2Re (−RW)^kR

vv ifk= 1,2,3 νN^−3/2Re (−RW)⁴T

vv ifk= 4.

Using (7.54), it is easy to check thaty1is bounded by the right-hand side of (7.55), and that

|yk| 6 ϕ^CζN^1/2−k/2

N⁻¹(d−1)^−1/2+ (d−1)^1/2 |va|²+|vb|²

(k= 2,3,4) (7.57)

withζ-high probability. In particular,

xT −xR = y1+y2+y3+O ϕ⁻¹Ebab

withζ-high probability. Moreover, using,|va|6ϕ^−D, |vb|6ϕ^−D, (7.57) fork= 2, and the fact that y₁ is bounded by the right-hand side of (7.55), we find that

|y1||y2| 6 ϕ⁻¹Ebab

withζ-high probability, providedD is chosen large enough. Similarly, using (7.57) we find that|yk||yk⁰|6 ϕ⁻¹Ebabfork, k⁰ >2 for large enough D. Thus we conclude from (7.56) that

Ef(xT +q) = Ef(xR+q) +E f⁰(xR+q)y3

+Aab+O ϕ⁻¹Ebab , where

Aab ..= E f⁰(xR+q)(y1+y2) +1

2E f⁰⁰(xR+q)y²₁

depends on the randomness only throughR and the first two moments ofWab. Moreover, from (7.57) and the fact thaty1 is bounded by the right-hand side of (7.55), we conclude that|Aab|6ϕ⁻¹.

What remains is the analysis of the term E f⁰(xR+q)y3

. We shall prove that

E f⁰(x_R+q)y₃

−Y_abEf⁰(x_R+q)

6 Cϕ⁻¹Eb_ab. (7.58)

Ifa=b, it is easy to see from (7.57) and the definition ofY_abthat

|y3|+|Yab| 6 ϕ⁻¹Ebab, from which (7.58) follows.

Let us therefore assume thata6=b. We multiply out the matrix product in (−RW)³R

vv and regroup the resulting eight terms according to the number, r, of off-diagonal matrix elements (Rab or Rba) of R.

(By convention, the endpoint matrix elements Rv· andR·v are not counted as off-diagonal.) This gives, in self-explanatory notation,y3=P2

r=0y3,r. Using Theorem 2.3 and (7.54), we find

|y3,1|+|y3,2| 6 ϕ^CζN^−3/2

N^−1/2(d−1)^−1/2+|va|+|vb|2

6 ϕ⁻¹Ebab

withζ-high probability . Therefore it suffices to prove that

E f⁰(xR+q)y3,0

−YabEf⁰(xR+q)

6 Cϕ⁻¹Ebab (7.59)

fora6=b. By definition,

y_3,0 = −νN⁻¹Re

R_vaW_abR_bbW_baR_aaW_abR_bv+R_vbW_baR_aaW_abR_bbW_baR_av . Using (7.54) and Theorem 2.3 we find

y3,0+νN⁻¹Re

m²|Wab|² RvaWabRbv+RvbWbaRav

6 ϕ⁻¹Ebab

withζ-high probability. We only deal with the first term of y3,0; the second one is dealt with analogously.

Recalling the definition ofYab, we conclude that, in order to establish (7.59), it suffices to prove withζ-high probability; here we used thatR is independent ofWab.

Setting u= (ui) withui..=1(i /∈ {a, b})vi and recalling (3.9) and (3.10), we get see (4.20). The second and third terms are estimated using (7.54) and Theorem 2.3:

|Raa−m|+|Rba| 6 ϕ^CζN^−1/2(d−1)^−1/2 (7.62) with ζ-high probability. Moreover, sinceR^(a)=T^(a), we find from Lemma (3.12), Theorem 2.3, and (3.8) that

withζ-high probability. A similar estimate holds forRbu. Using (7.61), (7.62), (7.63), and (7.54) we get νN⁻¹

What remains is to estimate the right-hand side of (7.64). Defining x^(a)_R ..= νN^1/2Re(R^(a)_vv −m), we find from (3.8) and (7.54) that

x_R−x^(a)_R

6 ϕ^Cζ

N^−1/2(d−1)^−1/2+N^1/2(d−1)^1/2|v_a|²+N^−1/2(d−1)^1/2|v_b|²

with ζ-high probability. Using (7.63) and using that the derivative of f is bounded, we may estimate the first term of (7.64) as

νN⁻¹

with ζ-high probability. In the second step we used that x^(a)_R is independent of the the a-th column of Q and thatEaRua= 0. The second term of (7.64) is similar. In order to estimate the third, we have to make Rbuindependent of thea-th column ofQ. (See the definition (3.22).) We estimate, using (3.8),R^(b)=T^(b), (3.12), and (7.54) withζ-high probability. Thus we may estimate the third term of (7.64) by

νN⁻¹ where in the second step we again used that E^aRua = 0. This concludes the proof of (7.60), and hence of (7.50).

The proof of (7.51) is almost identical to the proof of (7.50), except that E|Vab|²Vab = 0, so that the left-hand side of the analogue of (7.60) vanishes. Note that, by definition, Aab depends only on R and on the first two moments ofWab, which coincide with those ofVab. HenceAabis the same in (7.50) and (7.51).

This concludes the proof.

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