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Asymptotic expansion of the Bergman Kernel viaPerturbation of the Bargmann-Fock Model

S. Setojoint work with H. Hezari, C. Kelleher, H. Xu

Department of MathematicsUniversity of California, Irvine

[email protected]

Analysis and geometry in several complex variablesTexas A & M University at Qatar

January 8, 2015

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Introduction

In the study of several complex variables, the Bergman kernel is theholomorphic reproducing kernel for the Hilbert space of all squareintegrable holomorphic functions on a domain D in Cn. To be precise,consider the orthogonal projection P : L2 → H0 and let K (x , y) be theholomorphic integral kernel, i.e.

P(f )(x) =

∫D

f (y)K (y , x)dVE .

We call K (x , y) the Bergman kernel.

2 / 49

Introduction

Let (M, ω) be a polarized Kahler manifold of dimension n, withpolarization (L, h). By that, we mean L is a positive line bundle withhermitian metric h such that

ω = −√−1∂∂ log(h)

On a compact Kahler manifold, (M, ω) we consider the Hilbert space ofholomorphic sections of tensor powers of a positive line bundle L,denoted H0(M, Lk) with the L2 inner product: for ϕ, η ∈ H0(M, Lk)

〈ϕ, η〉 :=

∫M

(ϕ, η)hk dVg

where we naturally extend the Hermitian metric h to hk

3 / 49

Introduction

We can define the Bergman kernel as the integral reproducing kernelfor the L2 space of holomorphic sections H0(M, Lk).

An equivalent definition of the Bergman kernel is as follows: LetSα be an orthonormal basis for H0(M, Lk) with L2 inner product.Then the Bergman kernel is defined as

Kk(x , y) =∑α

Sα(x)⊗ Sα(y)

We will be using the integral reproducing definition.

4 / 49

Introduction

The Bergman kernel has the following geometric interpretation

Lemma

Suppose that the map

ϕ :M → CPN

x 7→ [S0(x) : . . . : SN(x)]

is defined on all of M, where Sα is an orthonormal basis of H0(M, L).Then

ϕ∗ωFS = ω +√−1∂∂ log(K (x))

where ωFS is the Fubini-Study metric on CPN

5 / 49

Introduction

Through the works of Zelditch and Catlin, it was shown that theBergman kernel admits an asymptotic expansion:As k →∞

Kk(x) = kn(1 +ρ

2k−1 + O(k−2))

where ρ is the scalar curvature, and the first four coefficients werecomputed by Lu

6 / 49

Introduction

As an application, we immediately get two results

Corollary (Tian, Zelditch)

For large k, an orthonormal basis of H0(M, Lk) gives a mapϕk : M → CPNk , where Nk + 1 = dim H0(M, Lk), and

1

kϕ∗kωFS − ω = O(k−2), in C∞

In other words, any polarized Kahler metric can be approximated byBergman metrics.

Corollary

As k →∞, we have

dim H0(M, Lk) = kn Vol(M) +kn−1

2

∫M

ρωn

n!+ O(kn−2)

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Introduction

We reprove the existence of the asymptotic expansion on the neardiagonal (in a O( 1√

k) neighborhood) by first constructing a “local”

reproducing kernel and using Hormander’s ∂ estimate to show that,asymptotically, is very close to the global kernel.

The localization technique was motivated by Berman, Berndtssonand Sjostrand [1], however, where they use the calculus of Fourierintegral operators and micro-local analysis to construct the “local”reproducing kernel, we use elementary techniques on theBargmann-Fock model and Taylor remainder estimates.

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Introduction

Theorem (Main Result)

Let (L, h)→ (M, ω) be a positive line bundle over a compact Kahler manifoldwith dimC(M) = n. Consider H0(M, L⊗k) the space of holomorphic sections ofk tensor powers of L. Then the scaled off-diagonal Bergman kernel admits anasymptotic expansion. In Bochner coordinates the expansion takes the form

Kk

(u√k,

v√k

)∼ eu·ve

kR(

v√k, v√

k

)(∑j

cj(u, v)√k j

),

where the term R is given by

kϕ(

v√k, v√

k

)= |v |2 + kR

(v√k, v√

k

).

The coefficients of cj(u, v) =∑

p,q cp,qj up vq further satisfy

cp,qj = 0 for |p|+ |q| > 2j ,

cp,qj = 0 for |p|+ |q| 6≡2 j .

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Introduction

A key construction in proving the above result is the following proposition

Proposition (Local Reproducing Kernel)

Let k ∈ N and f ∈ H0(B), the set of holomorphic functions on the unitball centered at the origin, χk ∈ C∞c (Cn) such that

χk(x) =

1 if |x | ≤ 1

2 k−14−ε,

0 if |x | ≥ k−14−ε.

Then there exists polynomials cj(u, v) such that it satisfies the scaledlocal reproducing property:

f(

u√k

)=

∫Bχk

(v√k

)f(

v√k

)euv−|v|

2

∑j

cj (u,v)√k j

e−kR

(v√k

) (k∂∂ϕ

(v√k

))n+ O

(kn− N+1

2

)‖f ‖L2(B,kϕ( v√

k)).

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Bargmann-Fock Space

To construct the cj(u, v) polynomials, we use the properties of theBargmann-Fock model.

The Bargmann-Fock Space, F , is the space of entire functions on Cn

that satisfy the weighted square integrability condition:∫Cn

|f (z)|2e−|z|2

dV <∞.

The space F is a closed linear subspace of the space L2(Cn, e−|z|2

) withinner product given by

〈f , g〉 :=

∫Cn

f (z)g(z)e−|z|2

dV .

and thus is a Hilbert space.

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Bargmann-Fock Space

In fact, it is a reproducing kernel Hilbert space on Cn, with reproducingkernel

RCn(v , u) := e u·v .

i.e.f (u) = 〈f (v),RCn(v , u)〉

Lemma (Reproducing Identity)

The following equality holds, for p, q ∈ N with p ≤ q,∫Cn

vpvqeu·v−|v |2 dV =

0 if p > q,

q!(q−p)! uq−p if p ≤ q.

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Formal Construction

For future considerations we shall group together the remainder terms ofthe exponential part of the weight with the Jacobian of the volume formin the following formal expansion:

e−kR

(v√k, v√

k

)det(∂∂ϕ

(v√k

)):=∑m

∑p,q

ar ,sm v rv s

√km

,

To gain some intuition, we give the expansion up to 1k for∑

m

∑p,q

ar,sm v r v s

√km

e−kR

(v√k

)det(∂∂ϕ

(v√k

))= 1− 1

k

(Rickl vkv l − 1

4 (Rm)i jkl(0)v ivkv jv l),

13 / 49

Formal Construction

We will formally construct coefficients cp,qj such that for any polynomials

F , we have

F

(u√k

)=

∫Cn

F

(v√k

)eu·v−|v|2

N∑t=0

∑p,q,r,s

j+m=t

cp,qj ar,s

m√k t

up vqv r v s

dVE

Note that when N = 0, it reduces to the Bargmann-Fock kernel.

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Formal Construction

Note it is sufficient to consider arbitrary degree l monomials in u, andthus due to the homogeneity and the Bargmann-Fock reproducingidentity

ul =

∫Cn

v leu·v−|v |2N∑t=0

∑j+m=t

( ∑p,q,r ,s

cp,qj ar ,sm√

kup vqv rv s

)dV

=N∑t=0

∑m+j=t

∑q+s≤l+r

p

cp,qj ar ,sm√

k tup+l+r−q−s (l + r)!

(l + r − q − s)!.

(1)

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Formal Construction

We can immediately determine the c0 coefficients, as seen in thefollowing lemma.

Proposition

For multiindices p, q ∈ Zn+ the following property holds

cp,q0 =

1 if pi = qi = 0 for all i ,

0 otherwise.

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Formal Construction

The proof proceeds by induction on the length of the multiindex q. First,we consider |q| = 0. Then using the defining formula (1) forl = (0, . . . , 0) and comparing the coefficient of k coefficient yields

1 =∑p

q+s≤r

cp,q0 ar ,s0 up+r−q−s r !

(r − q − s)!.

Since a0,00 = 1 of ar ,s0 = 0 for |r |+ |s| > 0, we have

1 =∑p

cp,00 up.

Immediately we compare the coefficients of u and conclude the result forthis case.

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Formal Construction

Now we assume the induction hypothesis holds for |q| ≤ λ− 1 andconsider the case |q| = λ, and take |l | = λ. Then applying the inductionhypothesis to (1) and parsing apart the right hand summation yields

ul =∑p

∑q≤l

upcp,q0

l!

(l − q)!ul−q

=∑p

∑|q|=λq≤l

upcp,q0

l!

(l − q)!ul−q +

∑p

∑q≤l|q|≤λ−1

upcp,q0

l!

(l − q)!ul−q.

Note in particular that the requirements in the first right side term thatq ≤ l and |q| = λ immediately imply that q = l . Additionally, by theinduction hypothesis, the second right side term reduces to simply ul .Subtracting this from both sides yields

0 =∑p

upcp,l0 .

The coefficients vanish accordingly and we conclude . The desired resultfollows.

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Formal Construction

Determining the coefficientsNow we determine the coefficients cp,q

j for j > 0. By grouping togetherthe terms in (1), we obtain the formula

∑m+j=t

∑s+q≤l+r

∑p

cp,qj ar ,sm up+r+l−q−s (l + r)!

(l + r − q − s)!=

ul t = 0,

0 t ≥ 1.

(2)We induct on j . The base case j = 0 was already determined. Nowassume the induction hypothesis is satisfied for j ≤ τ − 1, which impliesthat the coefficients cp,q

j have been determined for all multiindices p andq and for all such values of j .Take (2) for t = τ ,

τ∑j=0

∑s+q≤l+r

∑p

cp,qj ar ,sτ−ju

p+r+l−q−s (l + r)!

(l + r − q − s)!= 0.

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Formal Construction

By moving all the terms with j ≤ τ − 1 to the other side, we have∑s+q≤l+r

∑p

cp,qτ ar ,s0 up+r+l−q−s (l + r)!

(l + r − q − s)!

= −τ−1∑j=0

∑s+q≤l+r

∑p

cp,qj ar ,sτ−ju

p+r+l−q−s (l + r)!

(l + r − q − s)!.

Since we know ar ,s0 all vanish except at a0,00 = 1, we have∑

q≤l

∑p

cp,qτ up+l−q l!

(l − q)!

= −τ−1∑j=0

∑s+q≤l+r

∑p

cp,qj ar ,sτ−ju

p+r+l−q−s (l + r)!

(l + r − q − s)!.

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Formal Construction

We consider various lengths of l to determine the values of thecoefficients. When |l | = 0, reduces to

∑p

cp,0τ up = −

τ−1∑j=0

∑s+q≤r

∑p

cp,qj ar ,sτ−ju

p+r−q−s r !

(r − q − s)!.

Since, the values of ar ,sm are determined and cp,qj are known due to the

inductive hypothesis for j ≤ τ − 1, we obtain all cp,0τ by comparing the

coefficients of u.

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Formal Construction

We begin a subinduction argument on the values of |q| such that cp,qτ is

known for any |q| ≤ λ− 1. The case λ = 0 was just shown. Considermultiindices l such that |l | = λ. We decompose the summation andrearrange the equality to obtain

∑p

cp,lτ up = −

τ−1∑j=0

∑s+q≤l+r

∑p

cp,qj ar ,sτ−ju

p+r+l−q−s (l + r)!

(l + r − q − s)!

−∑p

∑q≤l|q|≤λ−1

cp,qτ up+l−q l!

(l − q)!.

Due to the induction hypothesis on τ , the first quantity of the right handside is completely determined. Furthermore, by comparing thecoefficients on up we can solve for cp,l

τ . This concludes the induction on|q| which implies that cp,q

τ are completely determined, and thus theinduction on τ is also completed. The result follows.

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Properties of the coefficients

We establish the following property of bivariate power series which willallow us to eliminate the amount of coefficients to consider andultimately yield finiteness and bounds of the degrees of cj .

Definition (Parity property of power series)

We say the coefficients of the bivariate power series

B(x , y) :=∑m,p,q

Bp,qm√km

xpyq,

has the parity property if given p, q ∈ Zn+ with |p|+ |q| 6≡2 m, then

Bp,qm = 0.

It can be shown by direct computation that the parity property is closedunder addition and multiplication.

23 / 49


Lemma

The quantity∑

m

∑p,q ap,qm vpvq has the parity property.

Proof.Consider the expansion

e−kR

(v√k

)=∑n

((−1)n(

√k)2∑ 1√

kmRp,qm vpvq

)n

= (−1)n(√

k)2n∑n

(∑ 1√km

Rp,qm vpvq

)n

.

The factors of k−1/2 come from evaluating the expansion of theexponential at ( v√

k), the Rp,q

m terms have the parity property. Since the

parity property is closed under addition and multiplication, andmultiplication by k2n preserves the parity property, the entire power seriesadmits the property. Furthermore since Ω( v√

k) also has the parity

property, the product also has the parity property. The result follows.

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It can be proved using the same strategy that the coefficients cp,qm also

have the parity property

Lemma

Given p, q ∈ Zn+ such that |p|+ |q| 6≡2 m, we have cp,q

m = 0.

Furthermore, the coefficients cp,qm also have an upper bound for the

degree.

Lemma

Given m ∈ Z+ and multi-indices p, q ∈ Zn+ such that |p + q| > 2m, then

cp,qm = 0.

The upper bound in the degree aids us our estimate of the localreproducing kernel.

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Remainder Estimates

We now begin our analysis on the remainder term of our localreproducing kernel.

Proposition (Local reproducing property)

Let f ∈ H0(B), and cj quantities as constructed above. Then for u ∈ B,the following equality holds

f

(u√k

)=

⟨χk

(v√k

)f

(v√k

), e u·v

(N∑j

cj (u, v)√k j

)⟩L2

(B(√k),kϕ

(·√k

))

+ O

(1

√kN+1−n

)‖f ‖L2(B,kϕ).

26 / 49

Remainder Estimates

We need to prove a series of estimates on the truncations and Taylorseries remainders of volume form and the holomorphic functionbeing reproduced.

For the following section we use the following notation: given adomain U ⊂ Cn and f ∈ H0(U) we expand f via the Taylor seriesexpansion

f (x) =∞∑j=0

((D j f )(0)

j!x j

).

Then we set

fN(x) :=N∑j=0

((D j f )(0)

j!x j

).

27 / 49

Remainder Estimate

SetΩ(z) := det

(∂∂ϕ(z)

),

where the determinant is taken over each matrix coefficient of the matrixvalued two-form ∂∂ϕ(z)

Lemma (Remainder of exponential term)

For N ≥ 0 and any f ∈ H0(B),

∫B(√k)χk

(v√k

)f(

v√k

)euv−|v|

2

N∑j=0

cj (u,v)√k j

(e−kR

(u√k

)−(e−kR2N+5

(u√k

))2N+1

)

× Ω(

v√k

)dV

= ‖f ‖L2(B,kϕ)O

(kn−N+1

2

).

28 / 49

Remainder Estimate

First note that since |v | ≤ k14−ε we have

kR

(v√k

)= O(k−ε).

We decompose

e−kR − e−kR2N+5 = e−kR(

1− ek(R−R2N+5)).

By Taylor expansion

k

∣∣∣∣(R − R2N+5)

(v√k

)∣∣∣∣ ≤ k sup|α|=2N+6

|ξ|≤ |v|√k

∣∣∣∣DαR(ξ)

(α)!

∣∣∣∣ ∣∣∣∣ v√k

∣∣∣∣α

≤ CNk

(|v |√

k

)2N+6

≤ CNk−N+1

2 .

Applying the above to the decomposition, we have∣∣∣∣e−kR(v√k

)− e−kR2N+5

(v√k

)∣∣∣∣ ≤ CNk−N+1

2 .

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Remainder Estimate

Next we consider the difference∣∣∣∣e−kR2N+5

(v√k

)−(e−kR2N+5

(v√k

))M

∣∣∣∣≤ sup|x|≤|−kR2N+5(

v√k

)

∣∣∣∣ |x |M+1

M + 1!

∣∣∣∣ ∣∣∣∣ v√k

∣∣∣∣2N+2

≤ Ck−ε(M+1) ≤ Ck−N+1

2 ,

where M ≤ N+12ε . Combining the two estimates on the truncation of the

exponential, we have∣∣∣∣e−kR(

v√k

)−(

e−kR2N+5

(v√k

))M

∣∣∣∣ ≤ CNk−N+1

2 .

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Remainder Estimate

Lemma (Remainder of determinant)

The following estimate holds: Let M = [N+12ε ] + 1,∣∣∣∣∣

∫B

χk

(v√k

)f(

v√k

)euv−|v|2

(N∑j=0

cj (u,v)√k j

)(e−kR2N+5

(v√k

))M

× (Ω− Ω2N+1)(

v√k

)dV∣∣∣

= ‖f ‖L2(B,kϕ)O(

kn− N+12

).

32 / 49

Remainder Estimate

We first observe the following estimate∣∣∣∣(Ω− Ω2N+1)

(v√k

)∣∣∣∣ ≤ sup|α|=2N+2

|ξ|≤∣∣∣ v√

k

∣∣∣

∣∣∣∣DαΩ(ξ)

α!

∣∣∣∣ ∣∣∣∣ v√k

∣∣∣∣2N+2

≤ CNk−N+1

2 .

Using the above estimate with a similar manipulation as the previouslemma, we conclude the result.

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Remainder Estimate

Lemma (Estimate Outside Disk)

The following estimate holds∫Cn

(1− χk

(v√k

))fN

(v√k

)euv−|v|2

(N∑t=0

∑m+j=t

cj(u, v)am(v , v)√k t

)dV

≤ ‖f ‖L2(B,kϕ)kne−

132

k12−2ε

.

34 / 49

First we observe that

∂

(∑i

e v(u−v) 1

ui − v idV

i

)= −ne v(u−v)dV .

Integrating by parts yields∫Cn

(1− χk

(v√k

))F

(v√k

)e v(u−v)

(N∑t=0

∑m+j=t


)dV

= −1

n

∫Cn

(1− χk

(v√k

))F

(v√k

)( N∑t=0

∑m+j=t


)

× ∂

(∑i

e v(u−v) 1

ui − v idV

i

)

= −1

n

∫Cn

F

(v√k

)∑i

∂ i

((1− χk

(v√k

))( N∑t=0

∑m+j=t


))

× e v(u−v) 1

ui − v idV .

35 / 49

Remainder Estimate

Iterating the above manipulation precisely 2N times gives

=(−1)2N+1

n2N+1

∫Cn

F

(v√k

) ∑I=(i1,...,i2N+1)

|I |=2N+1

∂I

(1− χk

(v√k

)) N∑t=0

∑m+j=t

cj (u, v)am(v , v)√kt

×e v(u−v)

(u − v)IdV

Recalling that the degree of ar ,sm and cp,qj are 2m and 2j respectively, we

will always take the derivative on (1− χk). Therefore, we only consider

the integral on the annulus 12 k

14−ε ≤ |v | ≤ k

14−ε.

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Remainder Estimate

The integral outside the disk is then bounded by(∫12k

14−ε≤|v|≤k

14−ε

∣∣∣∣F ( v√k

)∣∣∣∣2 e−|v|2dV) 1

2

×

∫

12k

14−ε≤|v|≤k

14−ε

∣∣∣∣∣∣∣∣∣∑

I=(i1,...,i2N+1)

|I |=2N+1

∂I

(1− χk

(v√k

)) N∑t=0

∑m+j=t


×e|u|2−|u−v|2

2

(u − v)I

∣∣∣∣∣∣2

dV

12

≤ CN‖f ‖L2(B,kϕ)kne−

132

k12−2ε

37 / 49

Remainder Estimate

Proof of local reproducing property By our construction, we have

F

(u√k

)=

∫Cn

F

(v√k

)eu·v−|v|

2

N∑t=0

∑m+j=t


dV

for N ≥ 0. We note that the N is independent of the polynomial F . We then split theabove into two labeled pieces

F

(u√k

)=

∫Cn

(1− χk

(v√k

))F

(v√k

)eu·v−|v|

2

N∑t=0

∑m+j=t


dV

I1

+

∫Cnχk

(v√k

)F

(v√k

)eu·v−|v|

2

N∑t=0

∑m+j=t


dV

I2

.

38 / 49

Remainder Estimate

As a consequence of the estimate outside the disk, we have

I1 ≤ ‖F‖L2(B,kϕ)kne−

132

k12−2ε

.

For the second integral, by applying the estimates for the truncation of the volumeform, we have

I2 ≤

∣∣∣∣∣∣∫Bχk

(v√k

)F(

v√k

)euv−|v|

2

N∑j=0

cj (u,v)√k j

(e−kR2N+5

(v√k

))M

Ω2N+1

(v√k

)dV

∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣⟨χk

(v√k

)F(

v√k

), e u·v

N∑j

cj (u,v)√k j

⟩L2

(B(√k),kϕ

(·√k

))∣∣∣∣∣∣∣∣

+ ‖F‖L2(B,kϕ)O(kn− N+1

2

).

This concludes the proof of the local reproducing property.

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Local to Global

Let K (x , y) be the global Bergman kernel, as a section of L⊗ L. Thenorm as a section of this bundle is the Bergman function B(x), which inthe standard local frame is

B(x) = |K (x , x)|h = |K (x , x)|e−ϕ(x)

Where K (x , x) is the coefficient function of the Bergman kernel withrespect to this local trivialization. We also have an extremalcharacterization of the Bergman function given by

B(x) = sup‖s‖L2≤1

|s(x)|2h

where s ∈ H0(M, L). Henceforth we will denote B(x) to be the Bergmanfunction associated to Lk .

Lemma (Uniform upper bound on Bergman function)

There exists C ∈ R dependent on M which is uniform over all k ∈ Z+

and x ∈ M such thatBk(x) ≤ Ckn.

40 / 49

Local to Global

Let K (x , y) = Ky (x) be the global Bergman kernel of H0(M, L⊗k). It isdefined by the global reproducing property

f (y) = 〈f ,Ky 〉L2

for any element f ∈ H0(M, Lk). Using the local reproducing property onthe global Bergman kernel, we show that the local Bergman kernel isequivalent to the global Bergman kernel up to a small error. The proof isessentially the same as the one given by Berman, Berndtsson, andSjostrand [1] but we include here for completeness.

Theorem (Local to Global)

The following equality relates the truncated local Bergman kernel K locN to

the global Berman kernel Kk .

K

(u√k,

v√k

)= K loc

N

(u√k,

v√k

)+ O

(k2n− N+1

2

).

41 / 49

Local to Global

Fix u, v ∈ B. We apply the local reproducing property to the globalBergman kernel

K

(v√k,

u√k

)=

⟨χk(·)K

(·, u√

k

),K loc

N

(·, v√

k

)⟩L2(B,kϕ)

+ O(

kn− N+12

)‖K‖L2(B,kϕ).

By the reproducing property and uniform upper bound of the Bergmanfunction,

‖K u√k‖2L2(B,kϕ) ≤ ‖K u√

k‖2L2(M,Lk ) = K

(u√k,

u√k

)= B

(u√k

)ekϕ( u√

k) ≤ Ckn

where K u√k(w) means section with respect to w and coefficient function

with respect to u. Thus we have

K

(v√k,

u√k

)=

⟨χk(·)K

(·, u√

k

),K loc

N

(·, v√

k

)⟩L2(B,kϕ)

+O(

k2n− N+12

).

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Local to Global

We next estimate the difference of the local Bergman kernel with the projectionof the local kernel.

gk,v

(w√k

):=χk

(w√k

)K locN

(w√k,

v√k

)−⟨χk (·)K

(·,

w√k

),K loc

N

(·,

v√k

)⟩L2(B,kϕ)

=χk

(w√k

)K locN

(w√k,

v√k

)−⟨χk (·)K loc

N

(·,

v√k

),K

(·,

w√k

)⟩L2

We can regard gk,v as a global section of Lk because of the cut-off function χk

for each fixed v . Since⟨χkK loc

N, v√k,K u√

k

⟩L2

= PH0

(χkK loc

N, v√k

),

where PH0 is the Bergman projection, gk,v is the L2-minimal solution to

∂gk,v = ∂(χkK loc

N, v√k

).

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Local to Global

Now we estimate ∂(χkK(N)v√k

) by using Hormander’s L2-estimate:

∂(χkK loc

N, v√k

)∣∣∣w√k

=

(∂ (χk) K loc

N, v√k

+ χk∂

(K locN,

v√k

))∣∣∣∣w√k

= ∂ (χk) K locN,

v√k

∣∣∣∣w√k

.

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Local to Global

Considering the right hand side of the equality, we have that the secondterm vanishes due to analyticity. The first term of the right hand sideensures that |w − v | ≥ 1

4 k14−ε. Furthermore, since we have

K locN, v√

k

( w√k

) = ewv(

1 + O(

1√k

))then we observe that

|ewv |2e−|w |2

= e2 Re wv−|w |2 = e−|w−v |2+|v |2 ≤ Ce−

116 k

12−2ε

.

Therefore for some constant δ > 0,∥∥∥∂(χk)K locN, v√

k

∥∥∥L2(M,Lk )

≤ e−δk12−2ε

.

So by the Hormander’s L2-estimate, the following inequality holdsuniformly for v ∈ B(1),

‖gk,v‖L2(M,Lk ) ≤ Ce−δk12 .

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Local to Global

By a standard argument using Cauchy integral formula, for the fixedu ∈ B we obtain the pointwise estimate from the L2 estimate∣∣∣gk,v

(u√k

)∣∣∣ ≤ Ce−δk12 .

Note this δ here may be smaller than the above. Finally we conclude theestimate, for any u, v ∈ B,∣∣∣∣K loc

N

(u√k,

v√k

)− K

(u√k,

v√k

)∣∣∣∣ ≤ Ck2n− N+12 .

The result follows.

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Computation of coefficients

We next explicitly compute the coefficient c0,02 of K loc

By applying reproducing property to f (z) = 1 we obtain c002 :

1 =

∫Cn

eu·v−|v |2(

1 +c00

2

k+

c(1),(1)2

kuiv j +

c(2)(2)2

kuiukv jv l

)×(

1− 1k

(Rickl vkv l − 1

4 (Rm)i jkl(0)v ivkv jv l))

dV .

Collecting the 1k terms, we obtain

c002 =

∫Cn

Rici j v iv jeuv−|v |2 dVE −1

4

∫Cn

(Rm)i jklvivkv jv leuv−|v |2 dV

The first integral on the right hand side is nonzero when i = j . The leftside is nonzero when i = j , k = l and i = l , j = k . We therefore obtain

c002 =

ρ

2

47 / 49

References

R. Berman, B. Berndtsson, J. Sjostrand.A direct approach to Bergman kernel asymptotics for positive linebundles.Ark. Mat., 46, (2008), 197-217.

H. Hezari, C. Kelleher, S. Seto, H. Xu.Asymptotic expansion of the Bergman Kernel via PerturbedBargmann-Fock Model.arXiv:1411.7438

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Thank you for your attention!

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