Central Limit Theorem for linear eigenvalue statistics of the Wigner and sample covariance random matrices

0.18 MB·English

by  Mariya Shcherbina

Central limit theorem for linear eigenvalue statistics of
1 the Wigner and sample covariance random matrices
1
0
2
M. Shcherbina

n
Institute for Low Temperature Physics, Ukr. Ac. Sci
a
J 47 Lenin ave, 61135 Kharkov Ukraine, e-mail: [email protected]

7
1

Abstract
]
h
We consider two classical ensembles of the random matrix theory: the
p
- Wigner matrices and sample covariance matrices, and prove Central Limit
h Theoremfor linear eigenvalue statistics under rather weak (comparing with
t
a results known before) conditions on the number of derivatives of the test
m functions and also on the number of the entries moments. Moreover, we
[ develop a universal method which allows one to obtain automatically the

bounds for the variance of differentiable test functions, if there is a bound
1
forthe varianceofthe traceoftheresolventofrandommatrix. The method
v
9 isapplicable notonly tothe Wignerandsample covariancematrices,butto
4 any ensemble of random matrices.
2
3
. 1 Introduction
1
0
1 The Wigner Ensembles for real symmetric matrices is a family of n n real
1 ×
symmetric matrices M of the form
:
v
Xi M = n−1/2W, (1.1)
r
a where W = w(n) n with w(n) = w(n) R, 1 j k n, and w(n),
jk j,k=1 jk kj ∈ ≤ ≤ ≤ jk
1 j k n are independent random variables such that
≤ ≤ ≤ (cid:8) (cid:9)
E w(n) = 0, E (w(n))2 =1, j = k, E (w(n))2 = w . (1.2)
jk jk 6 jj 2
Here and(cid:8)below(cid:9)we denot(cid:8)e E . th(cid:9)e averaging with re(cid:8)spect to(cid:9)all random param-
{}
eters of the problem. Let λ(n) n – be eigenvalues of M. Since the pioneer
{ j }i=1
work of Wigner [15] it is known that if we consider the linear eigenvalue statistic
corresponding to any continuous test function ϕ:
n
(n)
[ϕ] = ϕ(λ ), (1.3)
Nn j
j=1
X
1
then n 1 [ϕ] converges in probability to the limit
− n
N
lim n 1 [ϕ] = ϕ(λ)ρ (λ)dλ. (1.4)
− n sc
n N
→∞ Z
where ρ (λ) is the famous semicircle density
sc
1
ρ (λ) = 4 λ21 .
sc [ 2,2]
2π − −
p
The result of this type, which is the analog of the Law of Large Numbers of the
classical probability theory, normally is the first step in studies of the eigenvalue
distribution for any ensemble of random matrices. For the Wigner ensemble this
(n) n
result, obtained initially in [15] for Gaussian W = w , was improved in
jk j,k=1
[11], where the convergence of N (λ) to the semicircle law was shown under the
n (cid:8) (cid:9)
(n) n
minimal conditions on the distribution of W = w (the Lindeberg type
jk j,k=1
conditions).
(cid:8) (cid:9)
The second classical ensemble which we consider in the paper is a sample
covariance matrix of the form
M = n 1XX , (1.5)
− ∗
(n)
where X is a n m matrix whose entries X are independent
× jk j=1,.,n,k=1,.,m
random variables, satisfying the conditions
(cid:8) (cid:9)
E X(n) = 0, E (X(n))2 = 1. (1.6)
jk jk
Corresponding results on(cid:8)the co(cid:9)nvergence(cid:8)of norma(cid:9)lized linear eigenvalue statis-
tics to integrals with the Marchenko-Pastur distribution were obtained in [10].
Central Limit Theorem (CLT) for fluctuations of linear eigenvalue statistics
is a natural second step in studies of the eigenvalue distribution of any ensemble
of random matrices. That is why there are a lot of papers, devoted to the proofs
of CLT for different ensembles of random matrices (see [1, 2, 6, 7, 9, 12, 13, 14]).
CLT for the traces of resolvents for the classical Wigner and sample covari-
ance matrices was proved by Girko in 1975 (see [5] and references therein), but
the expression for the variance found by him was rather complicated. A sim-
ple expression for the covariance of the resolvent traces for the Wigner matrix
in the case E (w(n))2 = 2 was found in [8]. CLT for polynomial test func-
{ ii }
tions for some generalizations of the Wigner and sample covariance matrices was
proved in [1] by using moment methods. CLT for real analytic test functions
for the Wigner and sample covariance matrices was established in [2] under ad-
ditional assumptions that E (w(n))2 = 2, E (w(n))4 = 3E2 (w(n))2 = 3 (or
{ ii } { jk } { jk }
E (X(n))4 = 3E2 (X(n))2 for the model (1.5)). In the recent paper [9] CLT
{ jk } { jk }
2
for the linear eigenvalue statistics of the Wigner and sample covariance matrix
ensemble was proved under assumptions that E (w(n))2 = 2, the third and the
{ ii }
forth moments of all entries are the same, but E (w(n))4 is not necessary 3.
{ jk }
Moreover, the test functions, studied in [9], are not supposed to be real analytic.
It was assumed that the Fourier transform ϕ of the test function ϕ satisfies the
inequality
(1+ k 5)ϕ(k)bdk < , (1.7)
| | | | ∞
Z
which means that ϕ has more than 5 bounded derivatives.
b
In the present paper we prove CLT for the Wigner ensemble (1.1) under the
following assumptions on the matrix entries
E (w(n))4 = w , sup sup E w(n) 4+ε1 = w < , ε > 0. (1.8)
jk 4 | jk | 4+ε1 ∞ 1
n 1 j<k n
≤ ≤
(cid:8) (cid:9) (cid:8) (cid:9)
We consider the test functions from the space , possessing the norm (cf (1.7))
s
H
1
ϕ 2 = (1+2k )2s ϕ(k)2dk, s> 3/2, ϕ(k) = eikxϕ(x)dx. (1.9)
|| ||s | | | | 2π
Z Z
b b
Theorem 1 Consider the Wigner model with entries satisfying condition (1.8).
Let the real valued test function ϕ satisfy condition ϕ < (ε > 0). Then
3/2+ε
|| || ∞
[ϕ] converges in distribution to the Gaussian random variable with zero mean
Nn◦
and the variance
1 2 2 ϕ(λ ) ϕ(λ ) 2 4 λ λ
1 2 1 2
V[ϕ] = − − dλ dλ (1.10)
2π2 Z−2Z−2(cid:18) λ1−λ2 (cid:19) 4−λ21 4−λ22 1 2
2 2
κ 2 2 µ2 p w p2 2 ϕ(µ)µ
4 2
+ ϕ(µ) − dµ + − dµ ,
2π2 Z−2 4−µ2 ! 4π2 Z−2 4−µ2 !
p p
where κ = w 3.
4 4
−
Let us note that similarly to the result of [9] it is easy to check that Theorem 1
remains valid if the second condition of (1.8) is replaced by the Lindeberg type
condition for the fourth moments of entries of W
lim L(4)(τ) = 0, τ > 0, (1.11)
n n ∀
→∞
where
n
1
L(4)(τ) = E (w(n))41 . (1.12)
n n2 { jk w(n)>τ√n}
j,k=1 | jk |
X
3
The proof will be the same as for Theorem 1, but everywhere below n ε1/2 will
−
be replaced by L (τ)/τγ, with some positive γ.
n
The proof of Theorem 1 is based on some combination of the resolvent ap-
proachwithmartingalboundsforthevarianceoftheresolventtraces, usedbefore
by many authors, in particularly, by Girko (see [5] and references therein). An
importantadvantageofourapproachisthatitisshownbythemarginaldifference
method that (see Proposition 2 below)
Var TrG(z) C/ z 4, G(z) = (M z) 1, (1.13)
−
{ } ≤ |ℑ | −
while in the previous papers the martingal method was used only to obtain the
boundsofthetypeVar TrG(z) 1 nC(z). Thebound(1.13)willbecombined
−
{ } ≤
with the inequality
Proposition 1 For any s > 0 and any M
Var [ϕ] C ϕ 2 ∞dye yy2s 1 ∞ Var TrG(x+iy) dx. (1.14)
{Nn } ≤ s|| ||s − − { }
Z0 Z−∞
The proposition allows one to transform the bounds for the variances of the
resolvent traces into the bounds for the variances of linear eigenvalue statistics
of ϕ , where the value of s depends on the exponent of z in the r.h.s. of
s
∈ H |ℑ |
(1.13). Itisimportant,thatProposition1hasarathergeneralformandtherefore
it is applicable to any ensemble of random matrices for which the bounds of the
type (1.13) (may be with a different exponent of z ) are found. This makes
|ℑ |
Proposition1animportanttooloftheproofofCLTforlineareigenvaluestatistics
fordifferentrandommatrices. TheideaofProposition1wastakenfromthepaper
[7], where a similar argument was used to study the first order correction terms
of n 1E [ϕ] for the matrix models. Having in mind Proposition 1, one can
− n
{N }
prove CLT for any dense in set of the test functions, and then extend this
s
H
result to the whole by the standard procedure (see Proposition 3). In the
s
H
present paper for this aim we use a set of convolutions of integrable functions
with the Poisson kernel (see (2.32) and (2.3)). This choice simplifies considerably
the argument in the proof of CLT and makes the proof more short than that in
the previous papers [1, 2, 9].
The result for sample covariance matrices is very similar. We assume that
the moments of the entries of X from (1.5) satisfy the bounds
E (X(n))4 =X , sup sup E X(n) 4+ε1 = X < , ε > 0. (1.15)
jk 4 | jk | 4+ε1 ∞ 1
n 1 j<k n
≤ ≤
(cid:8) (cid:9) (cid:8) (cid:9)
Theorem 2 Consider a random matrix (1.5) – (1.6) with entries of X, satis-
fying the condition (1.15). Let the real valued test function ϕ satisfy condition
4
ϕ < (ε > 0). Then [ϕ] in the limit m,n , m/n c 1
|| ||3/2+ε ∞ Nn◦ → ∞ → ≥
converges in distribution to the Gaussian random variable with zero mean and
the variance
1 a+ a+ ∆ϕ 2 4c (λ1 am)(λ2 am) dλ1dλ2
− − −
V [ϕ] =
SC 2π2 ∆λ (cid:16)4c (λ a )2 4c (λ(cid:17) a )2
Za− Za− (cid:18) (cid:19) − 1− m − 2− m
2
κ a+ pµ a p
4 m
+ ϕ(µ) − dµ , (1.16)
4cπ2 Za− 4c−(µ−am)2 !
p
∆ϕ ϕ(λ ) ϕ(λ )
1 2
where = − , κ = X 3 is the fourth cumulant of entries of X,
4 4
∆λ λ λ −
1 2
a = (1 √c)2, and−a = 1(a +a ).
± ± m 2 + −
2 Proofs
Proof of Proposition 1. Consider the operator
s
D
f(k) = (1+2k )sf(k) (2.1)
s
D | |
It is easy to see that for fixed n Var [ϕ] is a bounded quadratic form in the
d n b
{N }
Hilbert space of the functions with the inner product (u,v) = ( u, v),
s s s
H D D
where the symbol (.,.) means the standard inner product of L (R). Hence there
2
exists a positive self adjoint operator such that
V
Var [ϕ] = ( ϕ,ϕ) = Tr(Π Π )
n ϕ ϕ
{N } V V
where Π is the projection on the vector ϕ
ϕ
(Π f)(x)= ϕ(x)(f,ϕ) ϕ 1
ϕ || ||−0
where . means the norm (1.9) with s = 0. We can write
0
||||
Tr(Π Π ) = Tr(Π 1 1 Π )
ϕV ϕ ϕDsDs− VDs− Ds ϕ
But it is easy to see that
( Π f)(x)= ( ϕ)(x)(f,ϕ) ϕ 1,
Ds ϕ Ds || ||−0
hence,
Π = ϕ = ϕ .
s ϕ s 0 s
||D || ||D || || ||
Therefore we can write
Var [ϕ] = Tr(Π 1 1 Π ) Π 2Tr( 1 1) (2.2)
{Nn } ϕDsDs− VDs− Ds ϕ ≤ ||Ds ϕ|| Ds− VDs−
5
But since for any u,v L (R) we have
2
∈
Γ(2s)( 2u,v) = Γ(2s) (1+2k ) 2su(k)v(k)dk
Ds− | | −
Z
= ∞dye−yy2s−1 e−2|k|yu(k)v(k)dkb= b∞dye−yy2s−1(Py u,Py v)
∗ ∗
Z0 Z Z0
= ∞dye−yy2s−1 dx bPyb(x λ)Py(x µ)u(λ)v(µ)dλdµ,
− −
Z0 Z Z Z
where the symbol means the convolution of functions, and P is the Poisson
y
∗
kernel
y
P (x) = . (2.3)
y π(x2+y2)
This implies
Γ(2s) 2(λ,µ) = ∞dye yy2s 1 dxP (x λ)P (x µ), (2.4)
Ds− − − y − y −
Z0 Z
and so
Γ(2s)Tr( 1 1) = ∞dye yy2s 1 dx P (x .),P (x .)
Ds− VDs− − − V y − y −
Z0 Z (cid:16) (cid:17)
= ∞dye yy2s 1 dxVar [P (x .)]
− − n y
{N − }
Z0 Z
= ∞dye yy2s 1 dxVar TrG(x+iy) .
− −
{ℑ }
Z0 Z
This relation combined with (2.2) proves (1.14).(cid:3)
In what wallows we need to estimate E w(n) 8 (see the proof of Proposition
| jk |
2). Hence, if ε < 4, then it is convenient to consider the truncated matrix
1
(cid:8) (cid:9)
M(τ) = M(τ) n , M(τ) = M 1 , M(τ) E M(τ) . (2.5)
{ ij }i,j=1 ij ij |Mij|≤τ ◦ − { }
f f f f f
Lemma 1 Let [ϕ] = Tr ϕ(M(τ) ) be the linear eigenvalue statistic of the ma-
n ◦
N
trix M(τ) , corresponding to the test function ϕ with bounded first derivative.
◦
Then e f
f eixNn◦[ϕ] eixNen[ϕ] o(1)+C x ϕ′ Ln(τ)/τ3.
| − |≤ | ||| ||∞
6
Proof. Consider the matrix M(t) = M +t(M M). Let λ (t) be eigenvalues
i
− { }
of M(t) and ψ (t) be corresponding eigenvectors. Then
i
{ }
f f
1
E [ϕ] [ϕ] = dtE ϕ(λ (t))λ (t)
Nn −Nn ′ i ′i
n(cid:12) 1 (cid:12)o Z0 nX(cid:12) (cid:12)o
(cid:12) e (cid:12) (cid:12) (cid:12)
ϕ dtE (M (t)ψ (t),ψ (t)) ϕ E Tr M M
′ ′ i i ′
≤ || ||∞Z0 nX(cid:12) (cid:12)o ≤ || ||∞ n | − |o
= ||ϕ′||∞ E u(cid:12)∗kj(M −M)ijujk (cid:12) f
Xk n(cid:12)Xij (cid:12)o
(cid:12) (cid:12)
f
ϕ E (cid:12)(M M) sup ϕ(cid:12) L (τ)/τ3,
′ ij ′ n
≤ || ||∞ − ≤ | |
nXij (cid:12) (cid:12)o
(cid:12) f (cid:12)
where M (t) = dM(t) = (M M), U = u is the unitary matrix such that
′ dt − { ik}
M M = U ΛU, where Λ is a diagonal matrix and M M = U ΛU. Hence,
∗ ∗
− | − | | |
f
feixNn◦[ϕ] eixNen[ϕ]◦ 2Pr M(τ) = M + x E n[ϕf] E n[ϕ]
| − | ≤ { 6 } | | {N }− {N }
o(1)+C x ϕ L (τ(cid:16))/τ3. (cid:17)
≤ f | ||| ′||∞ n e
(cid:3)
It follows from Lemma 1 that for our purposes it suffices to prove CLT for
[ϕ]. Hence, starting from this point we will assume that M is replaced by
Nn◦
M(τ)o, but to simplify notations we will write M instead of M(τ)o just assuming
beelow that the matrix entries of W satisfy conditions
f f
E w = 0, E w2 = 1+o(1),(j = k), E w2 =w +o(1), (2.6)
{ jk} { jk} 6 { jj} 2
E w4 = w +o(1),
{ jk} 4
E w 6 w n1 ε1/2, E w 8 w n2 ε1/2. (2.7)
{| jk| } ≤ 4+ε1 − {| jk| }≤ 4+ε1 −
(n) (n)
Here and below we omit also the superindex (n) of matrix entries w and X .
jk jk
Proposition 2 If the conditions (2.6) are satisfied, then for any 1 > δ > 0
n
Var γ Cn 1 E G (z)1+δ / z 3+δ, Var γ C/ z 4. (2.8)
n − ii n
{ } ≤ {| | } |ℑ | { } ≤ |ℑ |
i=1
X
If the conditions of (2.7) are also satisfied, then
E γ 4 Cn 1 ε1/2/ z 12. (2.9)
{| n◦| } ≤ − − |ℑ |
7
Proof. Denote E the averaging with respect to w . Then, according
k ij 1 i j k
≤ { } ≤≤ ≤
to the standard martingal method (see [4]), we have
n
Var γ = E E γ E γ 2 . (2.10)
n k 1 n k n
{ } {| ≤ − { }− ≤ { }| }
k=1
X
Denote E the averaging with respect to w . Then, using the Schwarz
k ki 1 i n
{ } ≤≤
inequality, we obtain that
E γ E γ 2 = E γ E γ E γ E γ 2 .
k 1 n k n k 1 n k n k 1 n k n
| ≤ − { }− ≤ { }| | ≤ − { − { }| ≤ ≤ − {| − { }| }
Hence
n
Var γ E γ E γ 2 . (2.11)
n n k n
{ } ≤ {| − { }| }
k=1
X
Let us estimate the first summand (with k = 1) of the above sum. The other
ones can be estimated similarly. Denote M(1) the (n 1) (n 1) matrix which
− × −
is the main bottom (n 1) (n 1) minor of M
− × −
G(1) = (M(1) z) 1, m(1) = n 1/2(w ,...,w ) Rn 1. (2.12)
− − 12 1n −
− ∈
We will use the identities
1+(G(1)G(1)m(1),m(1)) 1+B(z)
Tr G Tr G(1) = =: . (2.13)
− −z+n 1/2w +(G(1)m(1),m(1)) − A(z)
− 11
G = A 1, G G(1) = (G(1)m(1))2/A,
11 − − ii− ii − i
where (.,.) means the standard inner product in Cn 1.
−
The first identity of (2.13) yields that it suffices to estimate E BA 1
−
{| −
E BA 1 2 and E A 1 E A 1 2 . We will estimate the first expression.
1 − − 1 −
{ }| } {| − { }| }
Thesecondonecanbeestimatedsimilarly. Denoteξ = ξ E ξ foranyrandom
1◦ − 1{ }
variable ξ and note that for any a independent of w we have
1i
{
E ξ 2 E ξ a 2 .
1{| 1◦| } ≤ 1{| − | }
Hence it suffices to estimate
B E B B A B B A
1{ } = 1◦ ◦1 1◦ + ◦1
A − E A E A − E A A ≤ E A zE A
(cid:12) 1{ }(cid:12) (cid:12) 1{ } 1{ } (cid:12) (cid:12) 1{ }(cid:12) (cid:12)ℑ 1{ }(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
Let us u(cid:12)se also the id(cid:12)enti(cid:12)ties that follow from(cid:12)the(cid:12)spectra(cid:12)l th(cid:12)eorem (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(G(1)m(1),m(1)) = z(G(1) 2m(1),m(1)), Tr G(1) = zTr G(1) 2. (2.14)
ℑ ℑ | | ℑ ℑ | |
8
where G(1) = (G(1)G(1) )1/2. Thefirstrelationyields,inparticular,that B/A
∗
| | | | ≤
z 1. Moreover, using the second identity of (2.14), we have
−
|ℑ |
n 1Tr G(1) 2 (n 1Tr G(1) 2)δ(n 1Tr G(1) 2)1 δ z 1 δ
− − − − − −
| | = | | | | C |ℑ | .
z+n 1Tr G(1) 2 z+n 1Tr G(1) 1+δ z+n 1Tr G(1) 1 δ ≤ E A 1+δ
| − | | − | | − | − | 1{ }|
(2.15)
Since
A = n 1/2w +n 1 G(1)w w +n 1 G(1)(w2 ) , (2.16)
◦1 − 11 − ij 1i 1j − ii 1i ◦
i=j i
X6 X
E A 2 Cn 2Tr G(1) 2+Cn 1,
1{| ◦1| }≤ − | | −
we get by (2.15) and the second identity of (2.14):
A 2
E1 ◦1 C z E1 A −1−δ. (2.17)
E A ≤ |ℑ || { }|
1
n(cid:12) { }(cid:12) o
(cid:12) (cid:12) (cid:0) (cid:1)
Similarly (cid:12) (cid:12)
B 2 Cn 2Tr G(1) 4 C z 2n 2Tr G(1) 2 n 1 z 3 δ
E 1◦ − | | |ℑ |− − | | C − |ℑ |− − .
1 E A ≤ z+n 1Tr G(1) 2 ≤ z+n 1Tr G(1) 2 ≤ E A 1+δ
(cid:26)(cid:12) 1{ }(cid:12) (cid:27) | − | | − | | 1{ }|
(cid:12) (cid:12)
Then(cid:12)(cid:12), using(cid:12)(cid:12)the Jensen inequality E1 A −1 E1 A−1 , and the second iden-
| { }| ≤ {| | }
tity of (2.13), we conclude that
C
E (γ (z)) 2 E G (z)1+δ .
{| n ◦1| } ≤ n z 3+δ {| 11 | }
|ℑ |
Then (2.11) implies (2.8).
To prove (2.9) we use the inequality similar to (2.11) (see [4])
n
E γ 4 Cn E γ E γ 4 . (2.18)
{| n◦| }≤ {| n − k{ n}| }
k=1
X
Thus, in view of (2.13), it is enough to check that
E A 4 Cn 1 ε1/2 z 4, E B 4 Cn 1 ε1/2 z 8. (2.19)
1{| ◦1| } ≤ − − |ℑ |− 1{| ◦|1} ≤ − − |ℑ |−
The first relation here evidently follow from (2.16), if we take the forth degree of
the r.h.s., average with respect to w , and take into account (2.7). The second
1i
{ }
relation can be obtained similarly. (cid:3)
Proposition 2 gives the bound for the variance of the linear eigenvalue statis-
tics for the functions ϕ(λ) = (λ z) 1. We are going to extend the bound for a
−
−
wider class of test functions.
9
Lemma 2 If ϕ , with any ǫ > 0, then
3/2+ǫ
|| || ≤ ∞
Var [ϕ] C ϕ 2 . (2.20)
{Nn } ≤ ε|| ||3/2+ε
Proof. In view of Proposition 1 we need to estimate
∞
I(y) = Var γ (x+iy) dx
n
{ }
Z−∞
Take in (2.8) δ = ε/2. Then we need to estimate
∞ E G (x+iy)1+ε/2 dx, j = 1,...,n.
jj
{| | }
Z
−∞
We do this for j = 1. For other j the estimates are the same. The spectral
representation
N (dλ)
11
G =
11
λ x iy
Z − −
and the Jensen inequality yield
N (dλ)
∞ G1+ε/2(x+iy)dx ∞ dx ∞ 11 C y ε/2.
| |11 ≤ (x λ 2+y2)(1+ε/2)/2 ≤ | |−
Z−∞ Z−∞ Z−∞ | − |
Taking s = 3/2+ε in (1.14), we get
Var [ϕ] ϕ 2 C ∞e yy2+2εy 3 εdy C ϕ 2 .
{Nn } ≤ || ||3/2+ε − − − ≤ || ||3/2+ε
Z0
(cid:3)
To simplify formulas we will assume below that w are i.i.d. and
jk 1 j<k n
{ } ≤ ≤
w are i.i.d. Note that this assumption does not change the proof seri-
jj 1 j n
{ } ≤ ≤
ously, it just allows us to write the bounds only for G instead of all G .
11 ii
The next lemma collects relations which we need to prove CLT.
Lemma 3 Using notations of (2.13) we have uniformly in z ,z : z > a with
1 2 1,2
ℑ
any a > 0:
E (A )3 , E A 4 , E (B )3 , E B 4 = O(n 1 ε1/2), (2.21)
◦ ◦ ◦ ◦ − −
{ } {| | } { } {| | }
2
nE A (z )A (z ) = TrG(1)(z )G(1)(z )+w (2.22)
1 ◦ 1 ◦ 2 1 2 2
{ } n
κ4 (1) (1) (1) (1)
+ n Gii (z1)Gii (z2)+γ◦n (z1)γ◦n (z2)/n,
i
X
d
nE A (z )B (z ) = n E A (z )A (z ) , (2.23)
1 ◦ 1 ◦ 2 1 ◦ 1 ◦ 2
{ } dz { }
2
10