Berezin–Toeplitz quantization on Kähler manifolds

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BEREZIN-TOEPLITZ QUANTIZATION ON KAHLER MANIFOLDS

XIAONAN MA AND GEORGE MARINESCU

ABSTRACT. We study the Berezin-Toeplitz quantization on Kahler manifolds. We explain

first how to compute various associated asymptotic expansions, then we compute explic-

itly the first terms of the expansion of the kernel of the Berezin-Toeplitz operators, and of

the composition of two Berezin-Toeplitz operators. As application we estimate the norm

of Donaldson’s Q-operator.

0. INTRODUCTION

The Berezin-Toeplitz operators are important in geometric quantization and the prop-

erties of their kernels turned out to be deeply related to various problems in Kahler

geometry (see e.g. [11, 12]). In this paper, we will study the precise asymptotic expan-

sion of these kernels. We refer the reader to the book [18] for a comprehensive study of

the Bergman kernel, the Berezin-Toeplitz quantization and its applications. See also the

survey [17].

The setting of the Berezin-Toeplitz quantization on Kahler manifolds is the following.

Let (X;!; J) be a compact Kahler manifold of dimCX = n with Kahler form ! and

complex structure J . Let (L; hL) be a holomorphic Hermitian line bundle on X, and let(E; hE) be a holomorphic Hermitian vector bundle on X. Let rL;rE be the holomorphic

Hermitian connections on (L; hL), (E; hE) with curvatures RL = (rL)2, RE = (rE)2,respectively. We assume that (L; hL;rL) is a prequantum line bundle, i.e.,! = p�12� RL:(0.1)

Let gTX(�; �) := !(�; J �) be the Riemannian metric on TX induced by ! and J . The

Riemannian volume form dvX of (X; gTX) has the form dvX = !n=n!. The L2–Hermitian

product on the space C1(X;Lp E) of smooth sections of Lp E on X is given by

(0.2) hs1; s2i = ZXhs1; s2i(x) dvX(x) :We denote the corresponding norm with k�kL2 and with L2(X;LpE) the completion of

C1(X;Lp E) with respect to this norm.

Given a continuous linear operatorK : L2(X;LpE) �! L2(X;LpE), the Schwartz

kernel theorem [18, Th. B.2.7] guarantees the existence of an integral kernel with respect

to dvX , denoted by K(x; x0) 2 (Lp E)x (Lp E)�x0, for x; x0 2 X, i.e.,

(0.3) (KS)(x) = ZX K(x; x0)S(x0) dvX(x0) ; S 2 L2(X;Lp E) :Consider now the space H0(X;Lp E) of the holomorphic sections of Lp E on Xand let Pp : L2(X;Lp E) ! H0(X;Lp E) be the orthogonal (Bergman) projection.

Date: September 28, 2010.

2000 Mathematics Subject Classification. 53D50, 53C21, 32Q15.

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http://arxiv.org/abs/1009.4405v2

BEREZIN-TOEPLITZ QUANTIZATION ON KAHLER MANIFOLDS 2

Its kernel Pp(x; x0) with respect to dvX(x0) is smooth; it is called the Bergman kernel.

The Berezin-Toeplitz quantization of a section f 2 C1(X;End(E)) is the Berezin-Toeplitz

operator fTf;pgp2N which is a sequence of linear operators Tf; p defined by

(0.4) Tf; p : L2(X;Lp E) �! L2(X;Lp E) ; Tf; p = Pp f Pp :The kernel Tf; p(x; x0) of Tf; p with respect to dvX(x0) is also smooth. Since End(L) = C,

we have Tf; p(x; x) 2 End(E)x for x 2 X.

We introduce now the relevant geometric objects used in Theorems 0.1, 0.2 and 0.3.

Let T (1;0)X be the holomorphic tangent bundle on X, and T �(1;0)X its dual bundle. LetrTX be the Levi-Civita connection on (X; gTX). We denote by RTX = (rTX)2 the curva-

ture, by Ri the Ricci curvature and by r the scalar curvature of rTX (cf. (3.3)).

We still denote by rE the connection on End(E) induced by rE. Consider the (pos-

itive) Laplacian � acting on the functions on (X; gTX) and the Bochner Laplacian �Eon C1(X;E) and on C1(X;End(E)). Let fekg be a (local) orthonormal frame of(TX; gTX). Then �E = �Xk (rEekrEek �rErTXek ek):(0.5)

Let q; r(X;End(E)) be the space of (q; r)-forms on X with values in End(E), and letr1;0 : q;�(X;End(E))! q+1;�(X;End(E))(0.6)

be the (1; 0)-component of the connection rE. Let (rE)�, r1;0�; �E� be the adjoints

of rE, r1;0; �E, respectively. Let D1;0; D0;1 be the (1; 0) and (0; 1) components of the

connection rT �X : C1(X; T �X)! C1(X; T �X T �X) induced by rTX .

In the sequel, we denote byh� ; �i! : �; �(X;End(E))� �; �(X;End(E))! C1(X;End(E))

the C-bilinear pairing h� f; � gi! = h�; �if � g, for forms �; � 2 �; �(X) and sectionsf; g 2 C1(X;End(E)) (cf. (0.14), (0.16), (0.17)). PutRE� = DRE; !E! :(0.7)

Let Ri ! = Ri (J �; �) be the (1; 1)-form associated to Ri . SetjRi ! j2 =Xi<j Ri !(ei; ej)2 ; jRTX j2 =Xi<jXk<lhRTX(ei; ej)ek; eli2;and let

b2C =� �r48 + 196 jRTX j2 � 124 jRi ! j2 + 1128r2;b2E =p�132 �2rRE� � 4hRi !; REi! +�ERE�� 18(RE� )2 + 18hRE; REi! + 316�E�r1;0�RE;b1 = r8� + p�12� RE� ; b2 = 1�2 (b2C + b2E):(0.8)

We use now the notation from (3.6). By our convention (cf. (3.5)), we have at x0 2 X,D�`m dz`^dzm; �kq dzk^dzqE = �4�`m �m` ; D�mq dzmdzq; �k` dzkdzÈ = 4�mq �mq


(note that jdzqj2 = 2). Then by Lemma 3.1, (5.3) and (5.4), we have at x0 2 X,Ri `k = 2R`kqq = 2R`qqk ; r = 8R``qq; Ri ! = p�1Ri `k dz` ^ dzk ;p�1RE� = 2REkk; b1 = 1� �Rkkmm +REmm� ;b2C = ��r48 + 16Rk`mqR`kqm � 23R``mqRkkqm + 12R``qqRkkmm ;b2E = REqqRkkmm �REmqRkkqm + 12 �REqqREmm �REmqREqm�+ 14 ��REkk ;mm + 3REmk ; km� :(0.9)

We say that a sequence �p 2 C1(X;End(E)) has an asymptotic expansion of the form

(0.10) �p(x) = 1Xr=0Ar(x)pn�r + O(p�1) ; Ar 2 C1(X;End(E)) ;

if for any k; l 2 N, there exists Ck;l > 0 such that for any p 2 N�, we have

(0.11)

��p(x)� kXr=0Ar(x)pn�r��C l(X) 6 Ck;l pn�k�1;

where j � jC l(X) is the C l-norm on X.

Theorem 0.1. For any f 2 C1(X;End(E)), we have

(0.12) Tf; p(x; x) = 1Xr=0 br;f(x)pn�r + O(p�1) ; br;f 2 C1(X;End(E)) :

Moreover,

b0;f =f; b1;f = r8�f + p�14� �RE�f + fRE�� 14��Ef:(0.13)

If f 2 C1(X), then�2b2;f = �2

b2f + 132�2f � 132r�f � p�18 DRi !; ��fE+ p�124 Ddf;rERE�E!+ 124D�f;r1;0�REE! � 124D�f; �E�REE! � p�18 (�f)RE� + 14D��f;REE!:(0.14)

Theorem 0.2. For any f; g 2 C1(X;End(E)), the kernel of the composition Tf; p ÆTg; p has

an asymptotic expansion on the diagonal

(0.15) (Tf; p Æ Tg; p)(x; x) = 1Xr=0 br; f; g(x)pn�r + O(p�1) ; br; f; g 2 C1(X;End(E)) ;

in the sense of (0.11). Moreover, b0; f; g = fg and

b1; f; g = 18�rfg + p�14� �RE�fg + fgRE�� 14��f�Eg + (�Ef)g�+ 12�D �Ef;r1;0gE! :(0.16)


If f; g 2 C1(X), then

b2; f; g = f b2;g + g b2;f � fg b2 + 1�2�� 18D �f; ��gE� 18D ��f; �gE+ 12D �f; �gE�b1 � 14D �f ^ �g;REE! + 116�f ��g + 18DD0;1�f;D1;0�gE�:(0.17)

The existence of the expansions (0.12) and (0.15) and the formulas for the leading

terms hold in fact in the symplectic setting and are consequences of [21, Lemma 4.6 and

(4.79)] or [18, Lemma 7.2.4 and (7.4.6)] (cf. Lemma 2.2). The novel point of Theorems

0.1, 0.2 is the calculation of the coefficients b1;f , b2;f , b1;f;g and b2;f;g . Note that the

precise formula b1;f for a function f 2 C1(X) was already given in [18, Problem 7.2].

In Theorem 5.1, we find a general formula of b2;f for any f 2 C1(X;End(E)).If f = 1, then Tf; p = Pp, and the existence of the expansion (0.12) and the form

of the leading term was proved by [24], [5], [27]. The terms b1, b2 were computed

by Lu [16] (for E = C, the trivial line bundle with trivial metric), X. Wang [26], L.

Wang [25], in various degree of generality. The method of these authors is to construct

appropriate peak sections as in [24], using Hormander’s L2 �-method. In [7, §5.1], Dai-

Liu-Ma computed b1 by using the heat kernel, and in [19, §2], [20, §2] (cf. also [18,

§4.1.8, §8.3.4]), we computed b1 in the symplectic case.

The expansion of the Bergman kernel Pp(x; x) on the diagonal, for E = C, was red-

erived by Douglas and Klevtzov [9] by using path integral and perturbation theory. They

give physics interpretations in terms of supersymmetric quantum mechanics, quantum

Hall effect and black holes (cf. also [10]).

An interesting consequence of Theorem 0.2 is the following precise computation of the

expansion of the composition of two Berezin-Toeplitz operators.

Theorem 0.3. Let f; g 2 C1(X;End(E)). The product of the Toeplitz operators Tf; p andTg;p is a Toeplitz operator, more precisely, it admits the asymptotic expansion

(0.18) Tf; p Æ Tg;p = 1Xr=0 p�rTCr(f;g);p +O(p�1);where Cr are bidifferential operators, in the sense that for any k > 0, there exists k > 0with

(0.19)

Tf; p Æ Tg; p � kXl=0 p�lTCr(f;g);p 6 k p�k�1;where k�k denotes the operator norm on the space of bounded operators. We haveC0(f; g) =fg;C1(f; g) =� 12� hr1;0f; �Egi! 2 C

1(X;End(E));C2(f; g) = b2;f;g � b2;fg � b1;C1(f;g):(0.20)

If f; g 2 C1(X), thenC2(f; g) = 18�2 hD1;0�f;D0;1�gi+ p�14�2 hRi !; �f ^ �gi � 14�2 h�f ^ �g;REi! :(0.21)


The existence of the expansion (0.18) is a special case of [21, Th. 1.1] (cf. also [18,

Th. 7.4.1, 8.1.10]) where we found a symplectic version in which the Toeplitz operators

(0.4) are constructed by using the projection on the kernel of the Dirac operator. Note

that the precise values of b2 are not used to derive (0.21) (cf. Section 5.3).

The existence of the expansion (0.18) for E = C was first established by Bordemann,

Meinrenken and Schlichenmaier [2], Schlichenmaier [23] (cf. also [13]) using the theory

of Toeplitz structures by Boutet de Monvel and Guillemin [3]. Charles [6] calculatedC1(f; g) for E = C.

Note that we work throughout the paper with a non-trivial twisting bundle E. More-

over, we have shown in [21, §5-6] (cf. also [18, §7.5]) that the Berezin-Toeplitz quantiza-

tion holds for complete Kahler manifolds and orbifolds endowed with a prequantum line

bundle. The calculations of the coefficients in the present paper being local in nature,

they hold also for the above cases.

For some applications of the results of this paper to Kahler geometry see the paper

[12] by Fine.

This paper is organized as follows. In Section 1, we recall the formal calculus on Cn for

the model operator L , which is the main ingredient of our approach. In Section 2, we

review the asymptotic expansion of kernel of the Berezin-Toeplitz operators and explain

the strategy of our computation. In Section 3, we obtain explicitly the first terms of

the Taylor expansion of our rescaled operator Lt . In Section 4, we study in detail the

contribution of O2;O4 to the term F4 from (2.20). In Section 5, by applying the formal

calculi on Cn and the results from Section 4, we establish Theorems 0.1, 0.2 and 0.3. We

also verify our calculations are compatible with Riemann-Roch-Hirzebruch Theorem. In

Section 6, we estimate the Cm-norm of Donaldson’sQ-operator, thus continuing [14, 15].

We shall use the following notations. For � = (�1; : : : ; �n) 2 Nm, Z 2 Cm, we setj�j := Pmj=1 �j and Z� := Z�11 � � �Z�mm . Moreover, when an index variable appears twice

in a single term, means that we are summing over all its possible values.

Acknowledgments. We would like to thank Joel Fine for motivating and helpful discussions

which led to the writing of this paper. X. M. thanks Institut Universitaire de France for support.

G. M. was partially supported by SFB/TR 12 and Fondation Sciences Mathematiques de Paris. We

were also supported by the DAAD Procope Program.

1. KERNEL CALCULUS ON CnIn this section we recall the formal calculus on Cn for the model operator L introduced

in [21, §2], [18, §7.1] (with aj = 2� therein), and we derive the properties of the

calculus of the kernels (FP)(Z;Z 0), where F 2 C[Z;Z 0℄ and P(Z;Z 0) is the kernel of

the projection on the null space of the model operator L . This calculus is the main

ingredient of our approach.

Let us consider the canonical coordinates (Z1; : : : ; Z2n) on the real vector space R2n.

On the complex vector space Cn we consider the complex coordinates (z1; : : : ; zn). The

two sets of coordinates are linked by the relation zj = Z2j�1 +p�1Z2j, j = 1; : : : ; n.

We consider the L2-norm k � kL2 = � RR2n j � j2 dZ�1=2 on R2n, where dZ = dZ1 � � �dZ2n is

the standard Euclidean volume form. We define the differential operators:bi = �2 ��zi + �zi ; b+i = 2 ��zi + �zi ; b = (b1; : : : ; bn) ; L =Xi bi b+i ;(1.1)


which extend to closed densely defined operators on (L2(R2n); k � kL2). As such, b+i is the

adjoint of bi and L defines as a densely defined self-adjoint operator on (L2(R2n); k � kL2).The following result was established in [19, Th. 1.15] (cf. also [18, Th. 4.1.20]).

Theorem 1.1. The spectrum of L on L2(R2n) is given by

(1.2) Spe (L ) = �4�j�j : � 2 Nn� :Each � 2 Spe (L ) is an eigenvalue of infinite multiplicity and an orthogonal basis of the

corresponding eigenspace is given by

(1.3) B� = �b��z�e��Pi jzij2=2� : � 2 Nn with 4�j�j = �, � 2 Nn�and

SfB� : � 2 Spe (L )g forms a complete orthogonal basis of L2(R2n). In particular, an

orthonormal basis of Ker(L ) is

(1.4)

�'�(z) = ��j�j�! �1=2z�e��Pi jzij2=2 : � 2 Nn� :Let P(Z;Z 0) denote the kernel of the orthogonal projection P : L2(R2n) �! Ker(L )

with respect to dZ 0. Let P? = Id�P. We call P(�; �) the Bergman kernel of L .

Obviously P(Z;Z 0) = P� '�(z)'�(z0) so we infer from (1.4) that

(1.5) P(Z;Z 0) = exp �� 2 Pni=1 �jzij2 + jz0ij2 � 2ziz0i�� :In the calculations involving the kernel P(�; �), we prefer however to use the orthogo-

nal decomposition of L2(R2n) given in Theorem 1.1 and the fact that P is an orthogonal

projection, rather than integrating against the expression (1.5) of P(�; �). This point of

view helps simplify a lot the computations and understand better the operations. As

an example, if '(Z) = b��z�e��Pi jzij2=2� with �; � 2 Nn, then Theorem 1.1 implies

immediately that (P')(Z) = ( z�e��Pi jzij2=2 ; if j�j = 0;0 ; if j�j > 0:(1.6)

The following commutation relations are very useful in the computations. Namely, for

any polynomial g(z; z) in z and z, we have[bi; b+j ℄ = bib+j � b+j bi = �4�Æi j;[bi; bj℄ = [b+i ; b+j ℄ = 0 ;[g(z; z); bj℄ = 2 ��zj g(z; z);[g(z; z); b+j ℄ = �2 ��zj g(z; z) :(1.7)

For a polynomial F in Z;Z 0, we denote by FP the operator on L2(R2n) defined by the

kernel F (Z;Z 0)P(Z;Z 0) and the volume form dZ according to (0.3).

The following very useful Lemma [18, Lemma 7.1.1] describes the calculus of the

kernels (FP)(Z;Z 0) := F (Z;Z 0)P(Z;Z 0).


Lemma 1.2. For any F;G 2 C[Z;Z 0℄ there exists a polynomial K [F;G℄ 2 C[Z;Z 0℄ with

degree degK [F;G℄ of the same parity as degF + degG, such that

(1.8) ((FP) Æ (GP))(Z;Z 0) = K [F;G℄(Z;Z 0)P(Z;Z 0):2. EXPANSION OF THE KERNEL OF BEREZIN-TOEPLITZ OPERATORS

In this section, we review some results from [19], [21] (cf. also [18, § 7.2]). We explain

then how to compute the coefficients of various expansions considered in this paper. We

keep the notations and assumptions from the Introduction.

Kodaira-Laplace operator. Let �LpE;� be the adjoint of the Dolbeault operator �LpE.

Let �p = �LpE;��LpE be the restriction of the Kodaira Laplacian to C1(X;LpE). Let�LpE be the Bochner Laplacian on C1(X;Lp E) associated to rL, rE , gTX , defined

as in (0.5). Then we have (cf. [18, Remark 1.4.8])2�p = �LpE � p�12 RE(ei; Jei)� 2np:(2.1)

Moreover, by Hodge theory (cf. [18, Th. 1.4.1]) we haveKer�p = H0(X;Lp E):(2.2)

This identification is important since the computations are performed by rescaling �pand expanding the rescaled operator.

Normal coordinates. Let aX be the injectivity radius of (X; gTX). We denote by BX(x; ")andBTxX(0; ") the open balls inX and TxX with center x and radius ", respectively. Then

the exponential map TxX 3 Z ! expXx (Z) 2 X is a diffeomorphism from BTxX(0; ")onto BX(x; ") for " 6 aX . From now on, we identify BTxX(0; ") with BX(x; ") via the

exponential map for " 6 aX . Throughout what follows, " runs in the fixed interval℄0; aX=4[.Basic trivialization. We fix x0 2 X. For Z 2 BTx0X(0; ") we identify (LZ ; hLZ), (EZ; hEZ)and (Lp E)Z to (Lx0 ; hLx0), (Ex0 ; hEx0) and (Lp E)x0 by parallel transport with respect

to the connections rL, rE and rLpE along the curve Z : [0; 1℄ 3 u! expXx0(uZ). This

is the basic trivialization we use in this paper.

Using this trivialization we identify f 2 C1(X;End(E)) to a family ffx0gx02X wherefx0 is the function f in normal coordinates near x0, i.e. fx0 : BTx0X(0; ") ! End(Ex0),fx0(Z) = f Æ expXx0(Z). In general, for functions in the normal coordinates, we will

add a subscript x0 to indicate the base point x0 2 X. Similarly, Pp(x; x0) induces in

terms of the basic trivialization a smooth section Z;Z 0 7! Pp; x0(Z;Z 0) of �� End(E) overf(Z;Z 0) 2 TX �X TX : jZj; jZ 0j < "g, which depends smoothly on x0. Here we identify

a section S 2 C1�TX �X TX; ��End(E)� with the family (Sx)x2X , where Sx = Sj��1(x).Coordinates on Tx0X. Let us choose an orthonormal basis fwigni=1 of T (1;0)x0 X. Thene2j�1 = 1p2(wj + wj) and e2j = p�1p2 (wj � wj), j = 1; : : : ; n form an orthonormal basis ofTx0X. We use the coordinates on Tx0X ' R2n where the identification is given by

(2.3) (Z1; : : : ; Z2n) 2 R2n �!Xi Ziei 2 Tx0X:In what follows we also introduce the complex coordinates z = (z1; : : : ; zn) on Cn ' R2n.

Volume form on Tx0X. If dvTX is the Riemannian volume form on (Tx0X; gTx0X), there

exists a smooth positive function �x0 : Tx0X ! R, Z 7! �x0(Z) defined by

(2.4) dvX(Z) = �x0(Z)dvTX(Z); �x0(0) = 1;


where the subscript x0 of �x0(Z) indicates the base point x0 2 X.

Sequences of operators. Let �p : L2(X;Lp E) �! L2(X;Lp E) be a sequence of

continuous linear operators with smooth kernel �p(�; �) with respect to dvX (e.g.�p =Tf;p). Let � : TX �X TX ! X be the natural projection from the fiberwise product

of TX on X. In terms of our basic trivialization, �p(x; y) induces a family of smooth

sections Z;Z 0 7! �p; x0(Z;Z 0) of ��End(E) over f(Z;Z 0) 2 TX �X TX : jZj; jZ 0j < "g,which depends smoothly on x0.

We denote by j�p; x0(Z;Z 0)jC l(X) the C l norm with respect to the parameter x0 2 X.

We say that �p; x0(Z;Z 0) = O(p�1) if for any l;m 2 N, there exists Cl;m > 0 such thatj�p; x0(Z;Z 0)jCm(X) 6 Cl;m p�l.Notation 2.1. Recall that Px0 = P was defined in (1.5). Fix k 2 N and "0 2 ℄0; aX[ . LetfQr; x0g06r6k;x02X be a family of polynomials Qr; x0 2 End(E)x0[Z;Z 0℄ in Z;Z 0, which is

smooth with respect to the parameter x0 2 X. We say that

(2.5) p�n�p;x0(Z;Z 0) �= kXr=0(Qr; x0Px0)(ppZ;ppZ 0)p�r=2 +O(p�(k+1)=2) ;on f(Z;Z 0) 2 TX �X TX : jZj; jZ 0j < "0g if there exist C0 > 0 and a decomposition

(2.6) p�n�p;x0(Z;Z 0)� kXr=0(Qr; x0Px0)(ppZ;ppZ 0)p�r=2 = p;k;x0(Z;Z 0) +O(p�1) ;where p;k;x0 satisfies the following estimate on f(Z;Z 0) 2 TX �X TX : jZj; jZ 0j < "0g:for every l 2 N there exist Ck; l > 0, M > 0 such that for all p 2 N

(2.7) jp;k;x0(Z;Z 0)jC l(X) 6 Ck; l p�(k+1)=2(1 +pp jZj+pp jZ 0j)M e�C0pp jZ�Z0j :The sequence Pp . By [7, Proposition 4.1] we know that the Bergman kernel decays very

fast outside the diagonal ofX�X. Namely, for any l;m 2 N, " > 0, there exists Cl;m;" > 0such that for all p > 1 we havejPp(x; x0)jCm 6 Cl;m;" p�l on f(x; x0) 2 X �X : d(x; x0) > "g :(2.8)

Here the Cm-norm is induced by rL, rE, rTX and hL; hE; gTX .

By [7, Th. 4.180], there exist polynomials Jr; x0(Z;Z 0) 2 End(E)x0 in Z;Z 0 with the

same parity as r, such that for any k 2 N, " 2℄0; aX=4[ , we have

(2.9) p�nPp; x0(Z;Z 0) �= kXr=0(Jr; x0Px0)(ppZ;ppZ 0)p� r2 +O(p� k+12 ) ;on the set f(Z;Z 0) 2 TX �X TX : jZj; jZ 0j < 2"g, in the sense of Notation 2.1.

The sequence Tf; p . From (2.9), we get the following result (cf. [21, Lemma 4.6], [18,

Lemma 7.2.4]).

Lemma 2.2. Let f 2 C1(X;End(E)). There exists a family fQr; x0(f)gr2N; x02X , depending

smoothly on the parameter x0 2 X, whereQr; x0(f) 2 End(E)x0[Z;Z 0℄ are polynomials with

the same parity as r and such that for every k 2 N, " 2℄0; aX=4[ ,(2.10) p�nTf; p; x0(Z;Z 0) �= kXr=0(Qr; x0(f)Px0)(ppZ;ppZ 0)p�r=2 +O(p�(k+1)=2) ;


on the set f(Z;Z 0) 2 TX �X TX : jZj; jZ 0j < 2"g, in the sense of Notation 2.1. Moreover,Qr; x0(f) are expressed by

(2.11) Qr; x0(f) = Xr1+r2+j�j=rK �Jr1; x0 ; ��fx0�Z� (0)Z��! Jr2; x0� :Especially, Q0; x0(f) = f(x0):(2.12)

Our goal is of course to compute the coefficientsQr; x0(f). For this we need Jr; x0 , which

are obtained by computing the operators Fr; x0 defined by the smooth kernels

Fr; x0(Z;Z 0) = Jr; x0(Z;Z 0)P(Z;Z 0)(2.13)

with respect to dZ. Our strategy (already used in [18, 19]) is to rescale the Kodaira-

Laplace operator, take the Taylor expansion of the rescaled operator and apply resolvent

analysis. In the remaining of this section we outline the main steps and continue the

calculation in Section 3.

Rescaling �p and Taylor expansion. For s 2 C1(R2n; Ex0), Z 2 R2n, jZj � 2", and fort = 1pp , set (Sts)(Z) := s(Z=t);r0t := S�1t trLpESt;rt := S�1t t �1=2rLpE��1=2St = �1=2(tZ)r0t ��1=2(tZ);Lt := S�1t �1=2 t2(2�p)��1=2St:(2.14)

Then by [18, Th. 4.1.7], there exist second order differential operators Or such that we

have an asymptotic expansion in t when t! 0,

Lt = L0 + mXr=1 trOr + O(tm+1):(2.15)

From [18, Th. 4.1.21, 4.1.25] (cf. also Theorem 3.2), we obtain

L0 =Xj bjb+j = L ; O1 = 0:(2.16)

Resolvent analysis. We define by recurrence fr(�) 2 End(L2(R2n; Ex0)) byf0(�) = (��L0)�1; fr(�) = (��L0)�1 rXj=1Ojfr�j(�):(2.17)

Let Æ be the counterclockwise oriented circle in C of center 0 and radius �=2. Then by

[19, (1.110)] (cf. also [18, (4.1.91)])

(2.18) Fr; x0 = 12�i ZÆ fr(�)d�:Since the spectrum of L is well understood we can calculate the coefficients Fr; x0 .

Recall than P? = Id�P. From Theorem 1.1, (2.16) and (2.18), we get

F0; x0 =P; F1; x0 = 0;F2; x0 =�L

�1P

?O2P �PO2L �1P

?;F3; x0 =�L

�1P

?O3P �PO3L �1P

?;(2.19)


and

F4; x0 = 12�i ZÆ �(��L )�1P?(O2f2 +O4f0)(�) + 1�P(O2f2 +O4f0)(�)�d�= L�1

P?O2L �1

P?O2P �L

�1P

?O4P+ PO2L �1P

?O2L �1P

? �PO4L �1P

?+ L�1

P?O2PO2L �1

P? �PO2L �2

P?O2P�PO2PO2L �2

P? �P

?L

�2O2PO2P:(2.20)

In particular, the first two identities of (2.19) implyJ0; x0 = 1; J1; x0 = 0:(2.21)

Remark 2.3. Lt is a formally self-adjoint elliptic operator on C1(R2n; Ex0) with respect

to the norm k � kL2 induced by hEx0 ; dZ (cf. Section 1). Thus L0 and Or are also formally

self-adjoint with respect to k � kL2. Therefore the third and fourth terms in (2.20) are

the adjoints of the first and second term, respectively. In Lemma 4.1, we will show that

PO2P = 0, hence the last two terms in (2.20) vanish. Set

F41 = L�1

P?O2L �1

P?O2P �L

�1P

?O4P:(2.22)

3. TAYLOR EXPANSION OF THE RESCALED OPERATOR LtIn this section we compute the operators L0 and Oi (for 1 6 i 6 4) from (2.15) (see

Theorem 3.2), which will be used in Sections 4, 5 for the evaluation of the coefficients

of the expansion of the kernels of the Berezin-Toeplitz operators.

We denote by h�; �i the C-bilinear form on TX R C induced by gTX . Let RTX be

the curvature of the Levi-Civita connection rTX . Let Ri and r be the Ricci and scalar

curvature of rTX . Then we have the following well know facts: for U; V;W; Y vector

fields on X, RTX(U; V )W + RTX(V;W )U +RTX(W;U)V = 0;hRTX(U; V )W;Y i = hRTX(W;Y )U; V i = �hRTX(V; U)W;Y i:(3.1)

Now we work on Tx0X ' R2n as in (2.3). Recall that we have trivialized L;E. Let rUdenote the ordinary differentiation operator on Tx0X in the direction U .

We adopt the convention that all tensors will be evaluated at the base point x0 2 Xand most of time, we will omit the subscript x0.

For W 2 Tx0X, Z 2 R2n, let fW (Z) be the parallel transport of W with respect to rTXalong the curve [0; 1℄ 3 u! uZ. Because the complex structure J is parallel with respect

to rTX , we know that JZfW (Z) = Jx0W (Z):(3.2)

Recall that feig is a fixed orthonormal basis of (Tx0X; gTX). Then for U; V 2 Tx0X,Ri x0(U; V ) = �hRTX(U; ej)V; ejix0 ; rx0 = �hRTX(ei; ej)ei; ejix0:(3.3)


We define RTX; � 2 (T �X �2(T �X) End(TX))x0 ;RTX; (�;�) 2 ((T �X)2 �2(T �X) End(TX))x0 ;Ri ; � 2 (T �X (T �X)2)x0 ;RE; � 2 (T �X �2(T �X) End(E))x0 ;RE; (�;�) 2 ((T �X)2 �2(T �X) End(E))x0 ;by DRTX; ek(em; ej)eq; eiE = �rek DRTX(eem; eej)eeq; eeiE �x0 ;DRTX;(ek; e`)(em; ej)eq; eiE = �re`rek DRTX(eem; eej)eeq; eeiE�x0 ;Ri ; ek(ei; ej) = (rek Ri (eei; eej))x0;RE; ek(ei; ej) = (rekRE(eei; eej))x0;RE; (ek ; e`)(ei; ej) = (re`rekRE(eei; eej))x0 :(3.4)

We will also use the complex coordinates z = (z1; : : : ; zn). Note that

(3.5) e2j�1 = ��Z2j�1 = ��zj + ��zj ; e2j = ��Z2j = p�1� ��zj � ��zj�; �� zj ��2 = 12 :Set Rkm`q = *RTX� ��zk ; ��zm� ��z` ; ��zq+x0 ; REk` = REx0� ��zk ; ��z`�;Ri k` = Ri x0 � ��zk ; ��z`� ;Rkm`q; s = *RTX; ��zs � ��zk ; ��zm� ��z` ; ��zq+ ; REkq; s = RE; ��zs � ��zk ; ��zq�;(3.6)

and in the same way, we define Rkm`q; s , Rkm`q; ts , Ri kq; s , REkq; s, REkq; ts .

Since RTX is a (1; 1)-form and rE is the Chern connection on (E; hE), we deduce from

(3.1) and (3.2) the following.

Lemma 3.1.

(1) Rkm`q = R`mkq = Rkq`m = R`qkm , r = 8Rmqqm , (REkq)� = REqk ,

(2) Rkm`q ; s = R`mkq ; s = Rkq`m ; s = R`qkm ; s ,

(3) Ri is a symmetric (1; 1)-tensor and Ri mq = 2Rmkkq , Ri mq ; s = 2Rmkkq ; s ,

(4) RTX; ek , RTX; (ek ; e`) are (1; 1)-forms with values in End(Tx0X) which commute with Jx0 ,

(5) RE; ek , RE; (ek; e`) 2 End(Ex0).Let div(Ri ) be the divergence of Ri . By [22, §2.3.4, Prop. 6],dr = 2div(Ri ) :(3.7)


Lemma 3.1 and (3.7) entailR``mm ;k = R``mk ;m ; R``mm ; k = R``km ;m;�(�r) x0 = 2eqem(Ri (eq; em))x0 = 32Rkmqq ;mk ; �(�r) x0 = 32Rmmqq ;kk :(3.8)

Set

(3.9) R :=Xi Ziei = Z; r0;� := r� + 12RLx0(R; � ):Thus R is the radial vector field on R2n. We also introduce the vector fields z = Pi zi ��ziand z = Pi zi ��zi . By [18, Prop. 1.2.2], (3.2), we haveR =Xi Zieei; z =Xi zig��zi ; z =Xi zig��zi :(3.10)

By (0.1) and (1.1), we get

(3.11) bi = �2r0; ��zi ; b+i = 2r0; ��zi ; RLx0 = �2�p�1 hJ � ; � ix0 :Let A1Z; A2Z 2 (T �X)2 be polynomials in Z with values symmetric tensors, defined byA1Z(ei; ej) = DRTX;(Z;Z)(R; ei)R; ejEx0 ;A2Z(ei; ej) = DRTXx0 (R; ei)R; RTXx0 (R; ej)REx0 :(3.12)

Recall that the operator L was defined in (1.2). SetO 02 =13 DRTXx0 (R; ei)R; ejEr0;eir0;ej � 2REkk+ ��3RTXx0 (z; z)R; ej�+ 23 Ri x0(R; ej)� REx0(R; ej)�r0;ej ;(3.13)

and O41 = 120(A1Z � 43A2Z)(ei; ej)r0;eir0;ej ;O42 =�L ;�� 180A1Z � 1360A2Z�(ej; ej)� 1288 Ri (R;R)2�+ L144 Ri (R;R)2;O43 =� 1144 Ri (R;R)L Ri (R;R);O44 =� �30A1Z(z; ei)� �10A2Z(z; ei) + ��Zj� 120A1Z + 245A2Z�(ei; ej)� ��Zi� 140A1Z + 145A2Z�(ej; ej)�r0;ei;(3.14)


andO45 =�29 DRTXx0 (R; ek)R; RTXx0 (R; ek)eÈx0 � 19 DRTXx0 (R; e`)R; ekEx0 Ri (R; ek)+ 14 DRTXx0 (R; e`)R; emEx0 REx0(R; em)� 14RE;(Z;Z)(R; e`)�r0;e`;O46 =� �236A2Z(z; z) + �30 DRTX;(Z;e`)(z; z)R; eÈx0� �20 DRTXx0 (z; z)R; emEx0 Ri (R; em) + 49 Ri kmRi m` zkz` � 49Rk`mq Ri `m zkzq+ 16 D�RTXx0 (z; z)R; emEREx0(R; em) + 18 Ri (R; em)REx0(R; em)+ 12(REkmREm` + REm`REkm)zkz` � 14RE;(Z;e`)(R; e`)� RE;(Z;Z)( ��z` ; ��z` ):

(3.15)

The following result extends [18, Th. 4.1.25] where L0;O1;O2 were computed.

Theorem 3.2. The following identities hold for the operators Or introduced in (2.15) :

L0 =Xj bjb+j = L = �Xi r0;eir0;ei � 2�n; O1 = 0;(3.16a) O2 =O 02 � 13 Ri x0(R; ej)r0;ej � rx06 ;(3.16b)

and O3 =16 DRTX;Z (R; ei)R; ejEx0 r0;eir0;ej(3.17a) + �2�15 DRTX;Z (z; z)R; eiEx0 + 16 Ri ;Z(R; ei)+ 16 DRTX;ej (R; ej)R; eiEx0 � 23RE;Z(R; ei)�r0;ei+ �15 DRTX;ej (z; z)R; ejEx0 � 16 Ri ;ei(R; ei)� 112 Ri ;Z(ei; ei)� 13RE;ei(R; ei)� p�12 RE;Z(ei; Jei);O4 =O41 +O42 +O43 +O44 +O45 +O46:(3.17b)

Proof. Recall that eei(Z) is the parallel transport of ei with respect to rTX along the curve[0; 1℄ 3 u! uZ. Let e�(Z) = (�ij(Z))2ni;j=1 be the 2n� 2n-matrix such thatei =Xj �ji (Z)eej(Z); eej(Z) = (e�(Z)�1)kjek:(3.18)

Taking into account the Taylor expansion of �ij at 0 we have (cf. [18, (1.2.27)])Xj�j>1(j�j2 + j�j)(��ij)(0)Z��! = DRTX(R; ej)R; eeiEZ :(3.19)

From this equation, we obtain first thatej(Z) = eej(Z) + 16 DRTXx0 (R; ej)R; ekEx0 eek(Z) + O(jZj3):(3.20)


From (3.4), (3.12), (3.19) and (3.20), we get further�ij = Æij + 16 DRTXx0 (R; ei)R; ejEx0 + 112 DRTX;Z (R; ei)R; ejEx0+ 120�12A1Z(ei; ej) + 16A2Z(ei; ej)�+ O(jZj5) :(3.21)

Set gij(Z) = gTX(ei; ej)(Z) = hei; ejiZ and let (gij(Z)) be the inverse of the matrix(gij(Z)). Then by (3.12), (3.18) and (3.21), we havegij(Z) = �ki (Z)�kj (Z) = Æij + 13 DRTXx0 (R; ei)R; ejEx0 + 16 DRTX;Z (R; ei)R; ejEx0+ 120A1Z(ei; ej) + 245A2Z(ei; ej) + O(jZj5):(3.22)

In view of the expansion (1 + a)�1 = 1� a+ a2 + : : : , we obtaingij(Z) = Æij � 13 DRTXx0 (R; ei)R; ejEx0 � 16 DRTX;Z (R; ei)R; ejEx0� 120A1Z(ei; ej) + 115A2Z(ei; ej) + O(jZj5):(3.23)

If �ij are the Christoffel symbols of rTX with respect to the frame feig, then (rTXei ej)(Z)= �ij(Z)e`. By the explicit formula for rTX , we get (cf. [18, (4.1.102)]) with �j := ��Zj�ij(Z) = 12g`k(�igjk + �jgik � �kgij)(Z)= 13� DRTXx0 (R; ej)ei; eÈx0 + DRTXx0 (R; ei)ej; eÈx0 �+ O(jZj2) :(3.24)

For j fixed, �jj(Z) = 12g`k(2�jgjk � �kgjj)(Z), thus by (3.22),�jj(Z) = 23 DRTXx0 (R; ej)ej; eÈx0+ 112�4 DRTX;Z (R; ej)ej; eÈx0 + 2 DRTX;ej (R; ej)R; eÈx0 + DRTX;e` (R; ej)ej;REx0 �� 29 DRTXx0 (R; e`)R; RTXx0 (R; ej)ejEx0 + ��Zj� 120A1Z + 245A2Z�(ej; e`)� 12 ��Zl� 120A1Z + 245A2Z�(ej; ej) + O(jZj4):(3.25)

Note that det(Æij + aij) = 1 +Xi aii +Xi<j(aiiajj � aijaji) + : : :and (1 + a)1=4 = 1 + 14a� 332a2 + : : :


By (3.3) and (3.22), we get�(Z)1=2 = j det(gij(Z))j1=4 = 1 + 112 DRTXx0 (R; ej)R; ejEx0� 124 Ri ;Z(R;R) + 180A1Z(ej; ej) + 190A2Z(ej; ej)+ 136Xi<j �DRTXx0 (R; ei)R; eiEx0 DRTXx0 (R; ej)R; ejEx0 � DRTXx0 (R; ei)R; ejE2x0 �� 196�Xj DRTXx0 (R; ej)R; ejEx0 �2 + O(jZj5)= 1� 112 Ri (R;R)� 124 Ri ;Z(R;R) + � 180A1Z � 1360A2Z�(ej; ej)+ 1288 Ri (R;R)2 + O(jZj5):(3.26)

Thus �(Z)�1=2 = 1 + 112 Ri (R;R) + 124 Ri ;Z(R;R)� � 180A1Z � 1360A2Z�(ej; ej)+ � 1144 � 1288�Ri (R;R)2 + O(jZj5) :(3.27)

Observe that J is parallel with respect to rTX , thus hJ eei; eejiZ = hJei; ejix0 . From

(3.9), (3.10), (3.18) and (3.21), we getp�12� RLZ(R; e`) =�j(Z) hJ eei; eejiZ Zi = �j(Z) hJR; ejix0= hJR; eìx0 + 16 DRTXx0 (R; JR)R; eÈx0+ 112 DRTX;Z (R; JR)R; eÈx0 + 140 DRTX;(Z;Z)(R; JR)R; eÈx0+ 1120 DRTXx0 (R; JR)R; RTXx0 (R; e`)REx0 + O(jZj6):(3.28)

Let �� = �E;�L and R� = RE ; RL, respectively. By [18, Lemma 1.2.4], the Taylor

coefficients of ��(e`)(Z) at x0 up to order r are only determined by those of R� up to

order r � 1, and Xj�j=r(��)x0(el)Z��! = 1r + 1 Xj�j=r�1(��R�)x0(R; e`)Z��! :(3.29)

Thus by (3.28), (3.29) and since RTX is a (1; 1)-form, we obtainRTX(R; JR) = �2p�1RTX(z; z) ; DRTX;(Z;Z)(R; JR)R; eiEx0 = �2p�1A1Z(z; ei)and t�1�L(ei)(tZ) = ��p�1 hJR; eiix0 � t2�6 DRTXx0 (z; z)R; eiEx0� t3 �15 DRTX;Z (z; z)R; eiEx0 � t4 �60A1Z(z; ei)� t4 �180A2Z(z; ei) + O(t5):(3.30)


By (2.14), (3.4), (3.20), (3.29) and (3.30), for t = 1pp , we get�E(ei)(Z) = 12REx0(R; ei) + 13RE;Z(R; ei)+ 18�RE;(Z;Z)(R; ei) + 13 DRTX(R; ei)R; ekEx0 REx0(R; ek)�+ O(jZj4);r0t;ei = rei + 1t�L(ei)(tZ) + t�E(ei)(tZ)= r0;ei � t26 D�RTXx0 (z; z)R; eiEx0 + t22 REx0(R; ei) + O(t3):(3.31)

By (2.1) and (2.14), we get,

Lt = ��(tZ)1=2gij(tZ)�r0t;eir0t;ej � t�lij(t �)r0t;el�(Z)�(tZ)�1=2�t2p�12 RE(eei; J eei)(tZ)� 2�n:(3.32)

We will derive now (3.16a) and (3.16b) (they were already obtained in [18, Th. 4.1.25]).

By using the Taylor expansion of the expressions from (3.32) (see (3.23), (3.24), (3.26),

(3.27), (3.31)) we obtain immediately the formulas for L0 and O1 given in (3.16a).

In order to computeO2, observe first that by (3.1) and the fact that RTX is a (1,1)-form

with values in End(TX), we get

(3.33) rej DRTXx0 (z; z)R; ejE = 2 � ��zj DRTXx0 (z; z)z; ��zj E+ ��zj DRTXx0 (z; z)z; ��zj E� = 0:Thus from (3.13), (3.23), (3.26), (3.31) and (3.32), we haveO2 = O 02 + �

L0; 112 Ri (R;R)�:(3.34)

By the formula of L0 (see (3.16a)) and since[L0;Ri (R;R)℄ = �4Ri (R; ej)r0;ej � 2Ri (ej; ej) ;we get from (3.34) the formula for O2 given in (3.16b).

From (3.32), we have alsoO3 = 16 DRTX;Z (R; ei)R; ejEx0 r0;eir0;ej� �� 2�15 DRTX;Z (z; z)R; eiEx0 + 23RE;Z(R; ei)�r0;ei� ��Zi �� 15 DRTX;Z (z; z)R; eiEx0 + 13RE;Z(R; ei)�+ 112�4Ri ;Z(R; el) + 2 DRTX;ej (R; ej)R; elEx0 + Ri ;el(R;R)�r0;el+ �L0; 124 Ri ;Z(R;R)�� p�12 RE;Z(ei; Jei):

(3.35)

In (3.35), the first (resp. second and third, resp. fourth, resp. fifth) term is the contribu-

tion of the coefficient of t3 in gij(tZ) (resp. r0t;ei, resp. t�lii(t�), resp. �1=2(tZ)). By the


same argument in (3.33) and the formula of L0 given in (3.16a), we get��Zi DRTX;Z (z; z)R; eiEx0 = DRTX;ej (z; z)R; ejEx0 ;[L0;Ri ;Z(R;R)℄ =� 2(Ri ;ei(R;R) + 2Ri ;Z(R; ei))r0;ei� 4Ri ;ei(R; ei)� 2Ri ;Z(ei; ei):(3.36)

From (3.35) and (3.36) we get the formula for O3 asserted in (3.17a). Moreover,O4 = O42 + �O 02; 112 Ri (R;R)�+O43 +O41+ 13 DRTXx0 (R; ei)R; ejEx0 �� 13 D�RTXx0 (z; z)R; ejEx0 +REx0(R; ej)�r0;ei� 16 ��Zi D�RTXx0 (z; z)R; ejEx0 � 23 DRTXx0 (R; ei)ej; elEx0 r0;el�� 16 D�RTXx0 (z; z)R; eiEx0 + 12REx0(R; ei)� 23 DRTXx0 (R; ej)ej; eiEx0 �� 16 D�RTXx0 (z; z)R; eiEx0 + 12REx0(R; ei)�� O44 + �� 9A2Z(z; ei) + 112 DRTX(R; ei)R; emEx0 REx0(R; em)+ 29 DRTXx0 (R; ei)R; ekEx0 Ri (R; ek) + 14RE;(Z;Z)(R; ei)�r0;ei�� Zi�� 60A1Z(z; ei)� �180A2Z(z; ei) + 18RE;(Z;Z)(R; ei)+ 124 DRTXx0 (R; ei)R; ekEx0 REx0(R; ek)�� p�14 RE;(Z;Z)(ei; Jei) :(3.37)

Here � O42 is the contribution of the coefficients of t4 in �1=2(tZ),� the second term is the contribution of the coefficients of t2 in �1=2(tZ), ��1=2(tZ)and �gij(tZ)(r0t;eir0t;ej � t�ij(tZ)r0t;e`),� O43 is the contribution of the coefficients of t2 in �1=2(tZ) and ��1=2(tZ),� O41 is the contribution of the coefficients of t4 in gij(tZ),� the fifth term is the contribution of the coefficients of t2 in �gij(tZ) and in(r0t;eir0t;ej � t�ij(tZ)r0t;e`),� the sixth, seventh and eight terms are the contributions of the coefficients of t4 in�(r0t;eir0t;ei � t�ii(tZ)r0t;e`): the sixth term is the contribution of the coefficients

of t2 in r0t;ei, �t�ii(tZ) and t2 in r0t;ei; the seventh and eighth terms are the

contributions of the coefficients of t4 in r0t;ei and �t�ii(tZ).Now by (3.13),112[O 02;Ri (R;R)℄ = 136 DRTXx0 (R; ei)R; ejEx0 (4Ri (R; ej)r0;ei + 2Ri (ei; ej))+16 ��3RTXx0 (z; z)R; ej�+ 23 Ri (R; ej)�REx0(R; ej)�Ri (R; ej):(3.38)


and the same argument used to obtain (3.33) shows thatDRTXx0 (R; ei)R; ejEx0 ��Zi DRTXx0 (z; z)R; ejE = DRTXx0 (R; ei)R; RTXx0 (z; z)eiE ;��Zi (A1Z(z; ei)) = ��Zi DRTX;(Z;Z)(z; z)R; eiEx0 = 2 DRTX;(Z;ei)(z; z)R; eiEx0 ;��Zi (A2Z(z; ei)) = ��Zi DRTXx0 (z; z)R; RTXx0 (R; ei)REx0= DRTXx0 (z; z)ei; RTXx0 (R; ei)REx0 + DRTXx0 (z; z)R; ejEx0 Ri (R; ej):(3.39)

Finally, by (3.1) and since RTX is a (1; 1)-form, we obtainDRTXx0 (z; z)e`; RTXx0 (R; e`)RE = DRTXx0 (z; z)e`; emE DRTXx0 (R; e`)R; emE = 0;DRTXx0 (R; e`)R; emEx0 Ri (e`; em) = �8Rk`mq Ri `m zkzq:(3.40)

Thus O4 = 4X�=1O4� � �236 DRTXx0 (z; z)R; RTXx0 (z; z)RE� 14REx0(R; ej)2+ 16 D�RTXx0 (z; z)R; ejEREx0(R; ej) + �30 DRTX;(Z;ei)(z; z)R; eiEx0� �20 DRTXx0 (z; z)R; ejEx0 Ri (R; ej) + 19 Ri (R; ej) Ri (R; ej)+ 118 DRTXx0 (R; ei)R; ejEx0 Ri (ei; ej) + 18 Ri (R; ej)REx0(R; ej)� 14RE;(Z;ei)(R; ei)�RE;(Z;Z)( ��zi ; ��zi ):(3.41)

Putting together Lemma 3.1, (3.12), (3.14), (3.40), (3.41) and the fact that RTX is a(1; 1)-form, we infer (3.17b). The proof of Theorem 3.2 is completed. �

4. EVALUATION OF F4 FROM (2.20)

We calculate in this section an explicit formula for the operator F4, defined by (2.13)

and appearing in the Bergman kernel expansion (2.9). This is necessary in Section 5 in

order to evaluate the expansion of the kernel of the Berezin-Toeplitz operators. We use

formula (2.20) to achieve our aim. Recall that explicit formulas for the operators L , O2,O4 appearing in (2.20) were given in Theorem 3.2.

This section is organized as follows. In Section 4.1, we determine the terms in (2.20)

which involve O2. In Section 4.2, we calculate the terms in (2.20) which involve O4. In

Section 4.3, we obtain the formula for F4(0; 0) (cf. Theorem 4.5).

We adopt the convention that all tensors will be evaluated at the base point x0 2 Xand most of time, we will omit the subscript x0.4.1. Contribution of O2 to F4(0; 0).


Lemma 4.1. The following identities hold:

PO2P = 0;(4.1a) (L �1O2PO2L �1)(0; 0) = 14�2�Xkm Rmmkk +Xk REkk�2;(4.1b) (PO2L �2O2P)(0; 0)(4.1c) = 136�2Rmkq`Rkm`q + 14�2�43Rqmm` +REq`��43R`kkq + REq�:Proof. Note that by (1.1) and (1.5),

(4.2) (b+i P)(Z;Z 0) = 0 ; (biP)(Z;Z 0) = 2�(zi � z 0i)P(Z;Z 0):For � 2 T �X, by (3.5), (3.11) and (4.2), we have�(ei)ei = 2�( ��zj ) ��zj + 2�( ��zj ) ��zj ; �(ei)r0;ei = �( ��zj )b+j � �( ��zj )bj;�(ei)r0;eiP(Z; 0) = �2��(z)P(Z; 0):(4.3)

By Lemma 3.1, (3.1), (3.6), (3.16b), (4.3) and the fact that RTX is a (1; 1)-form, we getO2 = 13Rkm`qzkz`bmbq + 13Rkq`mzkzm(bqb+ + b+bq) + 13Rkm`qzmzqb+k b+ � 43Rmmqq+ �23Rkm`kzm � �3Rkm`qzkzmzq�b+l � �23R`kkqz` + �3Rkm`qzkzmz`�bq� 2REqq � RE(z; ��z` )b+ +RE(z; ��zq )bq :(4.4)

By Lemma 3.1, (1.7) and (4.4), we get [Rkm`qzkz`; bmbq℄ = 8bqRkk`qz` + 8Rmmqq, andO2 = 13bmbqRkm`qzkz` + bq�� 3Rkm`qzkz`zm + 2R`kkqz` + REqz`�+ �2bq3 Rkm`qzkzm � �3Rkm`qzkzmzq + 2Rkk`mzm +REmzm�b++ 13Rkm`qzmzqb+k b+:(4.5)

Thus Lemma 3.1, (4.2) and (4.5) yieldO2P = �13bmbqRkm`qzkz` + bq�� 3Rkm`qzkz`(bm2� + z0m) + 2R`kkqz` + REqz`��P= �16bmbqRkm`qzkz` + 43bqR`kkqz` � �3 bqRkm`qzkz`z0m + bqREqz`�P:(4.6)

Now, (4.1a) follows from Theorem 1.1, (1.6) and (4.6). These imply also

L�1O2P = �bmbq48� Rkm`q zk zl + bq3�R`kkq z` � bq12Rkm`q zk z` z 0m + bq4�REq z`�P:(4.7)

Due to (1.7) and (4.7) we have(L �1O2P)(Z; 0) = �bmbq48� Rkm`qzkz` + bq4��43R`kkq +REq�z`�P(Z; 0);(L �1O2P)(0; Z) = � 12� (Rmmqq +REqq)P(0; Z):(4.8)


Since O2, L are symmetric (as explained in Remark 2.3) and (REq)� = REq` , we get by

(1.1), (3.6) and (4.8),(PO2L �1)(0; Z) = ((L �1O2P)(Z; 0))�= �P

�Ruvqszvzs b+u b+q48� + �43Rqvvs +REqs�zs b+q4��(0; Z);(PO2L �1)(Z; 0) = ((L �1O2P)(0; Z))� = � 12� (Rkkqq + REqq)P(Z; 0):(4.9)

Note that P(0; 0) = 1 by (1.5). From (1.7), (4.2), (4.8) and (4.9), we get (4.1b), and(PO2L �2O2P)(0; 0) = P

� 32�2(48�)2RmsqtzsztRkm`qzkz`+ 14��43Rqsst + REqt�zt�43R`kkq +REq�z`�P(0; 0):(4.10)

Let � 2 C[b; z℄ be a polynomial in b; z. By (1.7), (4.2), we have(zk�(b; z)P)(Z; 0) = �(b; z) bk2�P(Z; 0) = � bk2��+ 1� ��zk�P(Z; 0);(zkzl�(b; z)P)(Z; 0) = (�(b; z)zkzlP)(Z; 0)= �bkbl4�2�+ bk2�2 ��zl + bl2�2 ��zk + 1�2 �2��zk�zl�P(Z; 0):(4.11)

Let F (Z) be a homogeneous degree 2 polynomial in Z. By (4.2),(F (Z)P) (Z; 0) = �12 �2F�zi�zj zizj + �2F�zi�zj zi bj2� + 12 �2F�zi�zj bibj4�2�P(Z; 0):(4.12)

Thus from (1.6), (4.11) and (4.12), we have(PFP)(Z; 0) = � Xj�j=2 �2F�z� z��! + 1� �2F�zi�zi�P(Z; 0);(PFzizjP)(0; 0) = 1�2 �2F�zi�zj :(4.13)

By (4.10) and (4.13), we get (4.1c). The proof of Lemma 4.1 is completed. �

Lemma 4.2. The following identity holds�2(L �1P

?O2L �1O2P)(0; 0) = � 2523 � 33Rmkq`Rkm`q � 4754Rkkq`Rmm`q+18Rkk``Rmmqq + 14RERmmqq � 76REq`Rmm`q + 18(REREqq � 3REq`REq):(4.14)

Proof. Set

I1 =�13bkb`Rskt`zszt + b`�� 3Rskt`zsztzk + �2Rtss` +REt`)zt�� bmbq48� Rkm`qzkz` + bm4��43R`kkq + REq�z`�;I2 =�2b`3 Rskq`zszk � �3Rskj`zszkz` + �2Rssqk +REqk�zk�� bm6 Rkm`qzkz` + 43R`kkqz` + REqz`�:(4.15)


Then by Lemma 3.1, (1.7), (4.2), (4.5) and (4.7), we get as in (4.10) that(O2L �1O2P)(Z; 0) = �I1 + I2 + 2�9 Rmsqt zsztRkm`qzkz`�P(Z; 0):(4.16)

Let h(z) = Pi hizi, h0(z) = Pi h0izi with hi; h0i 2 C, and let F (Z) be a homogeneous

degree 2 polynomial in Z. By Theorem 1.1, (1.6), (1.7), (4.2), (4.11) and (4.12), we

have (P?FP)(0; 0) = � 1� �2F�zi�zi ;(4.17a) �L

�1P

?hbiP� (0; 0) = �L

�1bihP

� (0; 0) = � 12�hi ;(4.17b) �L

�1P

?FP

� (0; 0) = � 14�2 �2F�zi�zi ;(4.17c) �L

�1bjFbiP� (0; 0) = � �L �1bibjFP

� (0; 0) = � 12� �2F�zi�zj ;(4.17d) �L

�1P

?FbibjP� (0; 0) = L�1

P?�bjFbi + 2�F�zj bi�P(0; 0) = � 32� �2F�zi�zj ;(4.17e) �

L�1

P?FzizjP� (0; 0) = L

�1P

?F bibj4�2P(0; 0) = � 38�3 �2F�zi�zj ;(4.17f) �L

�1bkFzlP� (0; 0) = L�1bkF bl2�P(0; 0) = � 14�2 �2F�zk�zl :(4.17g)

In (4.17c) and (4.17d), we have used Fbi = biF + 2 �F�zi . Observe that (1.5), (1.7) imply

that for every homogeneous degree k polynomial G in Z, and every � 2 Nn, we have(b�GP)(0; 0) = 8><>:0 ; if j�j 6= k;(�2)k��G�z� ; if j�j = k :(4.18)

By Theorem 1.1, (1.6) and (1.7), we also have�L

�1bihbjh0P� (0; 0) = bjbi8� hh0 + bi2�hjh0!P(0; 0) = 12� (hih0j � hjh0i);(4.19a) �L

�1P

?hbih0bjP� (0; 0) = � 12�hjh0i � 32�hih0j:(4.19b)

where we use in the last equation that hbih0bj = bihh0bj + 2hih0bj and further (4.17b),

(4.17d).


Let �(z) = �ijzizj; = ijzizj be degree 2 polynomials in z with symmetric matrices(�ij), ( ij). Then by Theorem 1.1, (1.7), (4.11), (4.17d), (4.17e) and (4.19a), we obtain�L

�1b`bk�bjhP

� (0; 0) = b`bkbj12� �h+ b`bk4� ��zj h!P(0; 0)(4.20a) = �23� �3(�h)�z`�zk�zj + 2� (�`jhk + �kjh`);�L

�1bk�b`bjhP

� (0; 0) = L�1 b`bk�+ 2bk ��z`! bjhP(0; 0)(4.20b) = �23� �3(�h)�z`�zk�zj + 2� (�kjh` + �`khj);�� L �1bk�z`bjhP

� (0; 0) = �12L�1bk�bj (b`h+ 2h`)P(0; 0)(4.20c) = 13� �3(�h)�z`�zk�zj � 1��`khj;

and �L

�1bkhbibj P

� (0; 0) = bkbibj12� h + bk4� (bihj + bjhi) !P(0; 0)(4.21a) = �23� �3(h )�zk�zi�zj + 2� ( kihj + kjhi);(L �1bkhz`bj P)(0; 0) = L�1bkhbj b`2� + 1� � �z`!P(0; 0)(4.21b) = �13�2 �3(h )�z`�zk�zj + 1�2 ( kjh` + `jhk);�� L �1hzkz`bj P

� (0; 0) = 16�2 �3(h )�z`�zk�zj + 32�2 `khj ;(4.21c)

In fact, by (4.2), (4.11) implies�(hzkz`bj P)(Z; 0) = 12�bkhbj z` + �2hkbj + 2hbj � �zk� b`2��P(Z; 0) ;so from (4.17e), (4.19b) and (4.21b) we get (4.21c).

Set Rmq(z) = Rkmlq zkzl. By (1.7) and Lemma 3.1,Rsk(z)bmbqRmq(z) = �bmbqRsk(z) + 8bmRqs`k z` + 8Rmsqk�Rmq(z) ;


thus Theorem 1.1, (4.11), (4.20b) and (4.21a) show that�L

�1bsbkRskbmbqRmqP� (0; 0)= bsbk�bmbq16� Rsk + 2bm3� Rqs`kz` + 1�Rmsqk�RmqP(0; 0)= 1� �4(RskRmq)�zs�zk�zm�zq � 163� �3(Rqs`kz`Rmq)�zs�zk�zm + 8�RmsqkRsmkq ;� ��L�1bkzsRskbmbqRmqP� (0; 0)= �L

�1bk�12bsRskbmbqRmq + �Rsk�zs bmbqRmq +Rskbmbq �Rmq�zs �P(0; 0)= 16� �4(RskRmq)�zs�zk�zm�zq + 83� �3(Rqs`kz`Rmq)�zs�zk�zm � 4�RmsqkRsmkq � 16� RssqkRmmkq :

(4.22)

Due to (4.15), (4.17a)–(4.22), we get

(4.23) �2(L �1I1P)(0; 0) = 1144�76 �4(RstRmq)�zs�zt�zm�zq � 83 �3(Rqs`tzvRmq)�zs�zt�zm + 4RmsqtRsmtq� 16RssqtRmmtq � 2�3((2Rvuut +REvt)zvRmq)�zt�zm�zq + 12(2Rquut + REqt)Rtmmq�+ 112�� 13 �3(Rst(43R`kkq +REq)z`)�zs�zt�zq + 4Rssqt�43Rtkkq + REtq��Rsstt�43Rqkkq +REqq��+ 18��2Rtuut + REtt��43Rqkkq +REqq�� 2Rquut +REqt��43Rtkkq + REtq��:

But from Lemma 3.1, we have�4(RstRmq)�zs�zt�zm�zq = 4RmsmtRsmtq + 4RssttRmmqq + 16RssmtRmmtq;�3(RssvtzvRmq)�zt�zm�zq = 2RssttRmmqq + 4RssqtRmmtq;�3(RmsvtzvRmq)�zs�zt�zq = 2RmsqtRsmtq + 4RmsstRtmqq:(4.24)

Plugging (4.24) in (4.23) we see that the coefficient of RssqtRmmtq in the term 1144 [� � � ℄ of

(4.23) is 1144(563 � 323 � 16� 16 + 24) = 0 and�2(L �1I1P)(0; 0) = 1144 103 RmsqtRsmtq � 127RssqtRmmtq+ � 1144 � �103 + 112 � �209 + 13�RssttRmmqq + ��4144 + 112 � �53 + 512�REttRmmqq+ � 4144 + 112 � 83 � 103 � 8�REqtRmmtq + 18(REttREqq � REqtREtq)= 523 � 33RmsqtRsmtq � 127RssqtRmmtq + 18RssttRmmqq+ 14REttRmmqq � 16REqtRmmtq + 18(REttREqq � REqtREtq):

(4.25)


From (4.15), (4.17c), (4.17f), (4.17g), (4.21b), (4.21c) and (4.24), we get�2(L �1I2P)(0; 0) = 118�� 12 �3(RusqtzuRmq)�zs�zt�zm + 4RssqtRtmmq+ 32RmsqtRsmtq + 3(2Ruuqs + REqs)(�24)Rmmsq�+ �� 112Rtsqt � 14(2Ruuqs + REqs)��43Rskkq + REsq�= 136RmsqtRsmtq � 56RssqtRmmtq � REsqRtsqt � 14REqtREtq:(4.26)

By (4.17f), we get�L

�12�9 RmsqtzsztRkm`qzkz`P�(0; 0) = � 16�2RmsqtRsmq:(4.27)

Relations (4.16), (4.25), (4.26) and (4.27) imply the desired formula (4.14). �

4.2. Contribution of O4 to F4(0; 0). We will use the following remark repeatedly in our

computation.

Remark 4.3. Let � be a polynomial in b+; z; b; z. Due to (1.7) and (4.2), the value of the

kernels of P�P, P?�P, L �1P?�P at (0; 0) consists of the terms of � whose total

degree in b and z is the same as the total degree in b+ and z.

Lemma 4.4. We have the following identity :��2(L �1O4P)(0; 0) = � �r96 + 23108RmsqtRsmtq + 4154RssqtRmmtq+ RmmqkREkq + 18(�REmm;qq + 3REqm;mq) + 14REkqREqk :(4.28)

Proof. By (1.7), (3.14), (4.2) and (4.3), as in (4.4), we have�(O41P)(Z; 0) = �� 120�A1Z � 43A2Z�( ��zi ; ��zj )bibj� 120�A1Z � 43A2Z�( ��zi ; ��zj )b+i bj�P(Z; 0)= �� 25 �A1Z � 43A2Z�(z; z) + �5�A1Z � 43A2Z�( ��zj ; ��zj )�P(Z; 0):(4.29)

and �L�1O42P =P

?�4� 180A1Z � 1360A2Z�( ��zj ; ��zj )� 172 Ri (z; z)2�P:(4.30)


From (3.12), (3.14), (4.2) and (4.3), and since RTX is (1; 1)-form, we have�(O44P)(Z;0) = 2�� 30A1Z(z; z) + � ��Zj� 120A1Z + 245A2Z��(z; ej)� �10A2Z(z; z)� zi ��zi� 110A1Z + 445A2Z�( ��zj ; ��zj )�P(Z; 0);�(O45P)(Z;0) = 2��29 DRTX(R; ek)z; RTXx0 (R; ek)zE� 19 DRTX(z; z)R; ekERi (R; ek)+ 14 DRTX(z; z)R; ejERE(R; ej)� 14RE;(Z;Z)(z; z)�P(Z; 0):(4.31)

Let ijk be degree 3 polynomials in z which are symmetric in i; j. By (1.7), (4.2), we get( ijkzizjzkP)(Z; 0) = 18�3 ( ijkbibjbkP)(Z; 0) = 18�3�bibjbk ijk+ 2bibj � ijk�zk + 4bibk� ijk�zj + 8bi �2 ijk�zj�zk + 4bk �2 ijk�zi�zj + 8 �3 ijk�zi�zj�zk�P(Z; 0):(4.32)

Thus by Theorem 1.1, (1.1) and (4.32), we obtain�2(L �1P

? ijkzizjzkP)(0; 0) = 18�2�bibjbk12 ijk + 14bibj � ijk�zk+ 12bibk � ijk�zj + 2bi �2 ijk�zj�zk + bk �2 ijk�zi�zj�P(0; 0) = � 1124�2 �3 ijk�zi�zj�zk :(4.33)

Since j ��zj j2 = 12 , Lemma 3.1, (3.6), (3.12) and the fact that RTX is a (1; 1)-form entailA1Z(z; z) = DRTX;(Z;Z)(z; z)z; zE ;A2Z(z; z) = 2 DRTX(z; z)z; RTX(z; z)zE = �4RusvtRkmtqzuzszvzkzmzq;A1Z( ��zq ; ��zq ) = DRTX;(Z;Z)(z; ��zq )z; ��zq E ;A2Z( ��zq ; ��zq ) = DRTX(z; ��zq )R; RTX(z; ��zq )RE= 2(RqsktR`qtu +RqstuR`qkt)zszuzkzl:(4.34)

By Remark 4.3, we can replace A1Z(z; z) by 2Rkm`q;stzkzmz`zqzszt in our computation.

We deduce from (4.33) and (4.34) that�2(L �1P

?A1Z(z; z)P)(0; 0) =� 116�2 (Rmmqq;tt + 2Rsmqq;mt);�2(L �1P

?A2Z(z; z)P)(0; 0) = 113�2 (RmsqtRsmtq + 2RssqtRmmtq):(4.35)

Let Fij be homogeneous degree 2 polynomials in Z. Then by (1.1), (4.11) and (4.17f),

we obtain (P?FijzizjP)(0; 0) = � 1�2 �2Fij�zi�zj ;(4.36a) (L �1P

?zk�(Fijzizj)�zk P)(0; 0) = � 34�3 �2Fij�zi�zj :(4.36b)


(Note that by Remark 4.3 the contributions of zk �(Fijzizj)�zk and 2Fijzizj to (4.36b) are the

same, so (4.36b) follows from (4.17f).)

By Remark 4.3 and (4.34), only the term �2Rqmkq;`tzmzkztz` from A1Z( ��zq ; ��zq ) has

a nontrivial contribution in our computation at (0; 0), and from (4.17f), (4.36a) and

(4.36b), we get�P

?A1Z( ��zq ; ��zq )P� (0; 0) = 2�2 (Rqmmq;tt + Rqmtq;mt);12 L�1

P?zm ��zmA1Z( ��zq ; ��zq )P! (0; 0) = �

L�1

P?A1Z( ��zq ; ��zq )P� (0; 0)= 34�3 (Rqmmq;tt +Rqmtq;mt);�

P?A2Z( ��zq ; ��zq )P� (0; 0) = � 2�2 (RqsstRuqtu + 3RqsutRsqtu);12 L�1

P?zm ��zmA2Z( ��zq ; ��zq )P! (0; 0) = �

L�1

P?A2Z( ��zq ; ��zq )P� (0; 0)= � 34�3 (RqsstRuqtu + 3RqsutRsqtu):

(4.37)

Remark 4.3, (4.3), (4.17c) and (4.17f) yield�L

�1P

? DRTX;(Z;em)(z; z)R; emEx0 P

� (0; 0)= 2 �L �1P

?(Rkq`m;ms �Rkqms ; `m)zkzqz`zsP� (0; 0) = 0;�L

�1P

? DRTXx0 (z; z)R; ekEx0 Ri (R; ek)P� (0; 0)= 2 �L �1P

?(�R`skq Ri mk+R`smkRi kq)z`zszqzmP

� (0; 0) = 0;�L

�1P

?Ri (R; eq)REx0(R; eq)P� (0; 0)= 2 �L �1P

?(�Ri kq REqs + Ri qsREkq)zkzsP� (0; 0) = 0;�L

�1P

?RE;(Z;es)(R; es)P� (0; 0)= 2 �L �1P

?(REks;sq �REsq;ks)zkzqP� (0; 0) = 0:(4.38)

By (3.39), (3.40), (4.3) and (4.38), we know that for � = 1; 2,�L

�1P

? ��Zq�A�Z(z; eq)�P

�(0; 0) = 0;�L

�1P

?� ��ZqA�Z�(z; eq)P�(0; 0) = �2�L�1

P?A�Z( ��zq ; ��zq )P�(0; 0):(4.39)


By (3.5) and (4.17f), we have�L

�1P

? DRTXx0 (R; ek)z; RTXx0 (R; ek)zEx0 P

� (0; 0)= 4 �L �1P

?(RksùRmkuq + RmkùRkiuq)zsz`zmzqP� (0; 0)= � 32�3 (RkssuRqkuq + 3RskquRksuq);�L

�1P

? D�RTXx0 (z; z)R; ekEREx0(R; ek)P� (0; 0)= 2� �L �1P

?(�R`skqREmk � R`smkREkq)z`zszqzmP

� (0; 0) = 3�2RssqkREkq;�L

�1P

?RE;(Z;Z)(z; z)P� (0; 0) = � 34�3 (REss ; qq +REqs ; sq):(4.40)

Note that by Lemma 3.1, Ri (R;R) = 2Ri kq zkzq, and by (1.2), (1.7) and (4.2),

L Ri (R;R)P = 2bib+i Ri kq zkzqP = 4b`Ri k` zkP:(4.41)

Thus by (3.14), (4.19b) and (4.41), we obtain�(L �1P

?O43P)(0; 0) = 118(L �1P

?Ri lq zlbmRi km zk bq2�P)(0; 0)= � 172�2 (Ri mmRi qq+3Ri mq Ri qm):(4.42)

From (4.29)-(4.39) and (4.42), we get��2(L �1(O41 +O42 +O43 +O44)P)(0; 0) = � 215 � �116 (R``qq ;uu + 2Ru`qq ; ù)+ 115 � 113 (R`squRsùq + 2RssquR`ùq)+ �� 25 � 34 + 110�(Rqmmq;uu + Rqmuq;mu)� ��1215 � 34 � 290�(RqssuRvquv + 3RqsvuRsquv)+ 172(Ri ``Ri qq+Ri `q Ri q`)� 172(Ri ``Ri qq+3Ri `q Ri q`)= 245R``qq ;uu + 1345Ru`qq ; ù + 199 R`squRsùq + RssquR`ùq :(4.43)

Moreover, (4.31), (4.38) and (4.40) yield��2(L �1O45P)(0; 0) =� 23(Rk`ùRqkuq + 3R`kquRkùq)+ 32R``qkREkq + 38(RE;qq + REq` ; `q):(4.44)

Further, (3.14), (4.17c), (4.35), (4.38) and (4.40) imply� �2(L �1O46P)(0; 0) = 11108(R`squRsùq + 2RssquR`ùq) + 19 Ri kq Ri qk� 19Rk`qkRi `q�12R``qkREkq + 14REkqREqk � 12RE ;qq= 11108R`squRsùq + 2354RssquR`ùq � 12R``qkREkq + 14REkqREqk � 12RE ; qq:(4.45)


By (3.8), (3.17b), (4.43)– (4.45), we get (4.28). The proof of Lemma 4.4 is completed.

�

4.3. Evaluation of F4(0; 0).Theorem 4.5. The following identity holds:

F4; x0(0; 0) = b2:(4.46)

Proof. By Lemmas 4.2, 4.4 and (2.22), we have

(4.47) �2F41(0; 0) = ��r96 + 772RmsquRsmuq � 19RssquRmmuq + 18RssuuRmmqq+ 14REuuRmmqq � 16REquRmmuq + 18(REuuREqq �REquREuq � REmm;qq + 3REqm ;mq) :

Remark 2.3, Lemmas 3.1, 4.1, (0.9), (2.20), (4.47) and formula P(0; 0) = 1 entailJ4;x0(0; 0) =F41(0; 0) + F41(0; 0)� + 14�2�Xmq Rmmqq +Xq REqq�2� 136�2Rmkq`Rkm`q � 14�2�43Rqvv` +REq`��43R`kkq + REq� = b2 :(4.48)

The proof of Theorem 4.5 is completed. �

5. THE FIRST COEFFICIENTS OF THE ASYMPTOTIC EXPANSION

The lay-out of this section is as follows. In Section 5.1, we explain the expansion of

the kernel of the Berezin-Toeplitz operators and verify its compatibility with Riemann-

Roch-Hirzebruch Theorem. In Section 5.2, we establish Theorem 0.1. The results from

Sections 4.1, 4.2 play an important role here. In Section 5.3, we prove Theorems 0.2,

0.3, i.e. the expansion of the composition of two Berezin-Toeplitz operators.

We use the notations and assumptions from Introduction and Section 3.

5.1. Expansion of the kernel of the Berezin-Toeplitz operators. For U 2 TX, we have

(cf. (3.2)) rT �XU gdzj = 2hrTXU g��zj ;g��zm igdzm; rT �XU gdzj = 2hrTXU g��zj ; g��zm igdzm:(5.1)

For � = Pkq �kqgdzk ^ gdzq 2 1;1(X;End(E)), by [18, Lemma 1.4.4], (0.6) and (5.1), we

get r1;0�� =� �2rEg��zm�mq + 4�kq�rTXg��zm g��zk ;g��zm�+ 4�ml�rTXg��zm f��zl ;g��zq��gdzq;�E�� =�2rEg��zm �km + 4�kq�rTXg��zmg��zq ; g��zm�+ 4�lm�rTXg��zmg��zl ;g��zk��gdzk:(5.2)

We evaluate now (5.2) at the point x0 (identified to 0 2 R2n). By using (3.29) applied

for r = 0; 1 associated with the vector bundles E, T (1;0)X, we get(r1;0��)x0 = �2��mq�zm (0) dzq; (�E��)x0 = 2��km�zm (0) dzk;(�E�r1;0��)x0 = 4 ��km�zm�zk (0) + 2 hREmk ; �km(0)i :(5.3)


Note that by (0.9), (3.6) and (5.3), we have at x0,! = p�12 dzq ^ dzq; Tr[RT (1;0)X ℄ = Ri kq dzk ^ dzq = �p�1Ri !;RE = REkqdzk ^ dzq; r1;0�RE = �2REmq;mdzq; �E�RE = 2REkq;qdzk:(5.4)

For f 2 C1(X;End(E)), recall that Tf; p(x; x0) is the smooth kernel of the Berezin-

Toeplitz operator Tf; p defined according to (0.4). Then by (3.29), at x0,�2fx0�zq�z` (0) = (rEg��zqrEg��z` f)(x0)� 12[REq`; f(x0)℄; �Ef = �4 �2f x0�zq�zq (0):(5.5)

In view of Lemma 3.1, (0.7) and (5.5), we introduce the following coefficients:

bCf :=Rmmqq �2f x0�zk�zk (0)� R`kkq �2fx0�zq�z` (0)=� r32�Ef � p�18 DRi !;r1;0�Ef � 12[RE; f ℄E! ;(5.6)

and

bEf1 :=�fx0�zu (0)�16REkk ;u � 512REqu ; q�+ 14REmu ;m�fx0�zu (0)+ �16REkk ;u � 512REuq ; q��fx0�zu (0) + 14 �fx0�zu (0)REum ;m= 148�r1;0f; 2p�1 �ERE� + 5r1;0�RE�! � 116hr1;0�RE;r1;0fi!+ 116h�Ef; �E�REi! + 148�2p�1r1;0RE� � 5�E�RE; �Ef�!;bEf2 :=12 �2fx0�zk�zk (0)REqq � 12 �2fx0�zq�z` (0)REq + 12REqq �2fx0�zk�zk (0)� 12REq �2fx0�zq�z` (0)=� p�116 �RE��Ef + (�Ef)RE��+ 18�r1;0�Ef � 12[RE; f ℄; RE�!+ 18�RE;r1;0�Ef � 12[RE; f ℄�!:

(5.7)

The following result implies Theorem 0.1.

Theorem 5.1. Let f 2 C1(X;End(E)). There exist smooth sections br;f(x) 2 End(E)xsuch that (0.12) and (0.13) hold and�2

b2;f = b2Cf(x0) + 12�b2E + 116(RE� )2�f(x0) + 12f(x0)�b2E + 116(RE� )2�� 116RE�f(x0)RE� + 132(�E)2f + bCf + bEf1 + bEf2:(5.8)

Before giving the proof, we verify that Theorem 0.1 is compatible with the Riemann-

Roch-Hirzebruch Theorem. Note that by (0.1), the first Chern class 1(L) of L is rep-

resented by !. By the Kodaira vanishing Theorem and the Riemann-Roch-Hirzebruch


Theorem, we have for p large enough:dimH0(X;Lp E) = ZX Td(T (1;0)X) h(E) ep!=rk(E) ZX !nn! pn + ZX � 1(E) + rk(E)2 1(X)�!n�1pn�1(n� 1)!+ ZX �rk(E)fTd(T (1;0)X)g(4) + 12 1(X) 1(E) + f h(E)g(4)�!n�2pn�2(n� 2)!+ O(pn�3) :(5.9)

As usual, h(�); 1(�);Td(�) are the Chern character, the first Chern class and the Todd

class of the corresponding complex vector bundles, f�g(4) is the degree 4-part of the

corresponding differential forms. Note thatx1� e�x = 1 + x2 + x212 + : : : ;thus fTd(T (1;0)X)g(4) = 112( 1(X)2 + 2(X)). Let RT (1;0)X be the curvature of the Chern

connection on T (1;0)X which is the restriction of the Levi-Civita connection in our case.

Then by (5.4), we have the following identities at the cohomology level:f h(E)g(4) = � 18�2 Tr[(RE)2℄; 1(X) = 12� Ri !;fTd(T (1;0)X)g(4) = 132�2 (Ri !)2 + 196�2 Tr �(RT (1;0)X)2�:(5.10)

By Lemma 3.1 and (5.4), we have132 D(Ri !)2; !2=2E = 12(RuuvvRkkqq � RuuqvRkkvq);196 �Tr �(RT (1;0)X)2�; !2=2� = 16(RkuqvRukvq � RuuqvRkkvq);hf h(E)g(4); !2=2i = 12�2 Tr �REkkREqq � REkqREqk�;�12 1(X) 1(E); !2=2� = 12�2 Tr(REkk Ri qq�Ri qkREkq):(5.11)

We set now f = 1 in Theorem 0.1, take the pointwise trace of the expansion (0.12)

relative to E and then integrate the result overX with respect to the volume form !n=n! .Taking into account (0.9) and (5.10), (5.11), we recover the expansion up to O(pn�3)given in (5.9) for the Hilbert polynomial. Thus the value of b2 obtained in Theorem 0.1

is compatible with the Riemann-Roch-Hirzebruch Theorem.

5.2. Proof of Theorem 5.1. The first part of Theorem 5.1 follows from Lemma 2.2.

Moreover, by (2.10), we have for any r 2 N,

br;f(x0) = Q2r;x0(f)(0; 0):(5.12)

Thus by (2.12), the formula b0;f = f , and by (2.11) and (2.21), we get

(5.13) Q2; x0(f) = K

h1; f(x0)J2;x0i+ K

hJ2;x0; f(x0)i+ Xj�j=2K

�1 ; ��fx0�Z� (0)Z��! � :


Further, (1.8), (2.13) and (2.19), entail

K

h1; f(x0)J2;x0iP = �f(x0)PO2L �1P

?;K

hJ2;x0; f(x0)iP = �(L �1P

?O2P)f(x0):(5.14)

From (1.8) and (4.13), we deduceXj�j=2K

�1 ; ��f x0�Z� (0)Z��! �P(Z; 0) = Xj�j=2 ��fx0�z� (0)z��! + 1� �2fx0�zi�zi (0)!P(Z; 0):(5.15)

Lemma 3.1, (3.31), (4.8), (4.9) and (5.12)–(5.15) yield the formula for b1;f from (0.12).

It remains to compute b2;f for a self-adjoint section f 2 C1(X;End(E)) in order to

complete the proof of Theorem 5.1. Set

K2f = Xj�j=2K

�1; ��f x0�Z� (0)Z��! J2;x0�:(5.16)

By (2.11) and (2.21), we getQ4; x0(f) = K

�1; f(x0)J4;x0�+ K

�J2;x0; f(x0)J2;x0�+ K

�J4;x0; f(x0)�+ Xj�j=2K

�J2;x0 ; ��fx0�Z� (0)Z��! �+ K2f+ K

�1; �fx0�Zi (0)ZiJ3;x0�+ K

�J3;x0 ; �fx0�Zi (0)Zi�+ Xj�j=4K

�1; ��fx0�Z� (0)Z��! � :(5.17)

Since L0 and Or are formally self-adjoint, (2.17) and (2.18) show that (Fr;x0)� = Fr;x0 .Hence, in the right hand side of (5.17), the first, fourth and sixth terms are adjoints of

the third, fifth and seventh terms, respectively. When we take f = 1 in (5.17), we getJ4;x0 = K

h1; J4;x0i+ K

hJ2;x0 ; J2;x0i+ K

hJ4;x0 ; 1i;(5.18)

which is also a direct consequence of (2.19), (2.20) and (4.1a), as by (1.8),

K

h1; J4;x0iP = PO2L �1P

?O2L �1P

? �PO4L �1P

? �PO2L �2O2P;K

hJ4;x0 ; 1iP = L�1

P?O2L �1

P?O2P �L

�1P

?O4P �PO2L �2O2P;K

hJ2;x0 ; J2;x0iP = L�1

P?O2PO2L �1

P? + PO2L �2O2P:(5.19)

Set K41 := ��r96 + 572RmuqvRumvq � 59RuuqvRmmvq + 18RuuvvRmmqq;K42 := 14REvvRmmqq � 56REqvRmmvq + 18�REvvREqq � 3REqvREvq �REmm;qq + 3REqm;mq�;K2f := 14(Rmmqq + REqq)f(x0)(Ruuvv + REvv) + 136Rmkq`Rkm`qf(x0)+ 14�43Rqss` +REq`�f(x0)�43R`kkq + REq�:(5.20)

By (2.22), (4.1c), (4.47) and (5.19), we have

K

hJ4;x0 ; 1i(0; 0) = 1�2 (K41 +K42):(5.21)


By (1.8) and (2.19), we see as in (5.19) that

K

hJ2;x0 ; f(x0)J2;x0iP = L�1

P?O2f(x0)PO2L �1

P?+PO2L �1f(x0)L �1O2P:(5.22)

Thus by (4.8), (4.9) and (5.22), as in Lemma 4.1, we get

K

hJ2;x0 ; f(x0)J2;x0i(0; 0) = 1�2K2f :(5.23)

We next compute the fifth term in (5.17). From (2.19) and (5.16), we get

K2f = PXj�j=2 ��f x0�Z� (0)Z��! ��L

�1O2P �PO2L �1P

?� :(5.24)

For a degree 2 polynomial F (Z) we have by Remark 4.3, (1.7), (4.2) and (4.8)�(PFL�1O2P)(0; 0)= ��P

�2F�zu�zv zuzv�bmbq48� Rkmlqzkzl + bq4��43Rlkkq + RElq�zl�P

�(0; 0)= ��P�2F�zq�zv zv2��43Rlkkq +RElq�zlP�(0; 0) = � 12�2 �2F�zq�zl�43Rlkkq + RElq�;(5.25)

where we have used (1.6), (1.7) and (4.13) in the last two equalities.

By (4.9), (4.13), (5.24) and (5.25), we get

K2f (0; 0) = 12�2 �2fx0�zk�zk (0)(Rmmqq + REqq)� 12�2 �2fx0�zq�zl (0)�43Rlkkq +RElq�:(5.26)

By Lemma 3.1, (5.6), (5.7) and (5.26), we get

K2f(0; 0) + K2f (0; 0)� = 1�2 (bCf + bEf2)� 13�2Rlkkq �2f x0�zq�zl (0):(5.27)

We compute now the last term in (5.17). By Remark 4.3, (1.8) and (4.13), we haveXj�j=4K

�1; ��fx0�Z� (0)Z��! �(0; 0) = 12�2 �4fx0�zi�zq�zi�zq (0):(5.28)

We next turn to the computation of the sixth term in (5.17). SetK3f : = �16Rkkmm;u � 13Rkkmu;m��fx0�zu (0) + �16Rkkmm;u � 13Rkkum;m��fx0�zu (0)+ �fx0�zu (0)�16REkk;u � 12REqu;q�+ 13REmu;m�fx0�zu (0)+ �16REkk;u � 12REuq;q��fx0�zu (0) + 13 �fx0�zu (0)REum;m:(5.29)

Lemma 5.2. The following identity holds with K3f defined in (5.29):

K

�J3; x0 ; �fx0�Zu (0)Zu�(0; 0) + K

�1; �fx0�Zu (0)ZuJ3; x0�(0; 0) = 1�2K3f :(5.30)


Proof of Lemma 5.2. SetB1(b; Z) :=�16Rks`q;mzkz`bsbq + 2�3 Rqskq;mzszk(5.31a) � �2�15Rkq`s;mz`zq + 13R``ks;m � 13Rksqm;q � 23REks;m�zkbs� 2�15Rksqm;qzkzs � 23R``qm;q � 23R``qq;m + 23REqm;q � 2REqq;m�zm;B1(Z) :=2�25 Rks`q;mzkz`zszqzm + 8�15Rksqm;qzkzszm(5.31b) + 4�3 REks;mzkzszm � 23�R``qm;q + R``qq;m � REqm;q + 3REqq;m�zm:Then by Lemma 3.1, (4.2), we haveB1(b; Z)P(Z; 0) = �4�26 Rks`q;mzkz`zszq + 2�3 Rqskq;mzszk� �2�15Rkq`s;mz`zq + 13R``ks;m � 13Rksqm;q � 23REks;m�zk � 2�zs� 2�15Rksqm;qzkzs � 23R``qm;q � 23R``qq;m + 23REqm;q � 2REqq;m�zmP(Z; 0)= B1(Z)P(Z; 0):(5.32)

Observe that the commutation relations (1.7) imply thatRks`q;mzkz`bsbqbm = bsbqbmRks`q;mzkz` + bsbm(8Rqskq;m + 4Rksqm;q)zk+ bm(8Rqssq;m + 16Rssqm;q) :By (1.7), (4.2) and (5.31b), we haveB1(Z)P(Z; 0) = 1�� 120Rkslq;mzkz`bq + 215Rksqm;qzk + 13REks;mzk�bsbm�13�R``qm;q +R``qq;m �REqm;q + 3REqq;m�bm�P(Z; 0)= 1��bsbqbm20 Rks`q ;mzkz` + bsbm�25Rqskq ;m + 13Rksqm;q + 13REks;m�zk+bm� 115Rssqq;m + Rssqm;q � 13REss;m + REqm;q��P(Z; 0):(5.33)

Note that by Theorem 1.1 and (4.2), we have (P?zuP)(Z; 0) = (PzuP)(Z; 0) = 0.

Taking into account that P(0; 0) = 1 and relations (1.8), (2.13) and (2.19), we get

K

�J3;x0 ; �fx0�Zu (0)Zu�(0; 0) = �(L �1P

?O3P �fx0�zu (0)zuP)(0; 0)�(PO3L �1P

? �fx0�zu (0)zuP)(0; 0):(5.34)


By Remark 2.3, (PO3L �1zuP)(0; 0) is the adjoint of (PzuL �1P?O3P)(0; 0), thus we

will compute only the latter. By Lemma 3.1, (1.7), (3.17a) and (4.3), we get as in (4.4),O3 = 16Rks`q;Zzkz`bsbq + 2�3 Rqskq;Zzszk + s(b; b+; Z)b+s� �2�15R`qks;Zz`zq + 13R``ks;Z + 13�R`qks;qz` � Rqmks;qzm�� 23REks;Z�zkbs+ 2�15 (Rks`q;qz` � Rksqm ; qzm)zkzs � 23�R``kq;qzk +R``qm;qzm�� 23R``qq;Z � 23�REkq;qzk �REqm;qzm�� 2REqq;Z;(5.35)

and s(b; b+; Z) are polynomials in b; b+; Z, whose precise formula will not be used, andRkq`s;Z, REks;Z are defined by replacing ��zs by Z in (3.6).

From Lemma 3.1, (1.7), (5.31a) and (5.35) we deduce that the only term in O3 not

containing b+ and having total degree in b; z bigger than its degree in z, is B1(b; Z). Now,

Theorem 1.1, Remark 4.3, (1.6), 1.7, (5.32), (5.33) and (5.35) imply�(PzuL �1P

?O3P)(0; 0) = �(PzuL �1P

?B1(b; Z)P)(0; 0)= ��Pzu bm4�2� 115Rssqq;m +Rssqm;q � 13REss;m + REqm;q�P

�(0; 0)= � 12�2� 115Rssqq;u +Rssqu;q � 13REss;u + REqu;q�:(5.36)

By Remark 4.3, (1.6), (1.7) and (5.35), we also have�(L �1P

?O3PzuP)(0; 0) = �(L �1P

?B1(b; Z)zuP)(0; 0)= ��L�1

P?hzuB1(b; Z)� 2 ��buB1(b; Z)iP�(0; 0):(5.37)

Lemma 3.1, (4.17c), (4.17f), (4.33), (5.31b) and (5.32) yield�(L �1P

?zuB1(b; Z)P)(0; 0) = �(L �1P

?zuB1(Z)P)(0; 0)= 1124�2 � 25�4Rssmu;m + 2Rssqq;u�+ 38�2 � �1615Rssqu;q + 43REss;u + 43REmu;m�� 16�2�R``qu;q + R``qq;u �REqu;q + 3REqq;u�= 1�2�2930Rssmu;m + 15Rssqq;u + 23REmu;m� :(5.38)

Moreover, by (5.31a),��buB1(b; Z) = 13Rksù;mzkz`bszm� �2�15Rkqù;mz`zq + 13R``ku;m � 13Rkuqm;q � 23REku;m�zkzm:(5.39)

Lemma 3.1, (4.2), (4.17c), (4.17f) and (5.39) yield(L �1P

?( ��buB1(b; Z))P)(0; 0)= � 38�2 � 815 � 2Rssmu;m + 14�2�13Rmqqu;m � 13Rmuqm;q � 23REmu;m�= � 25�2Rssmu;m � 16�2REmu;m:(5.40)


Formulas (5.37)–(5.40) entail altogether

(5.41) � �2(L �1P

?O3PzuP)(0; 0) = 16Rssmu;m + 15Rssqq;u + 13REmu;m:Combining (5.34), (5.36) with (5.41), we get�2

K

�J3;x0 ; �fx0�Zu (0)Zu�(0; 0) = �16Rssmu;m + 15Rssqq;u + 13REmu;m��fx0�zu (0)+�� 130Rssqq;u � 12Rssuq;q + 16REss;u � 12REuq;q��fx0�zu (0):(5.42)

Since K

�1; �fx0�Zu (0)ZuJ3;x0� is the adjoint of K

�J3;x0 ; �fx0�Zu (0)Zu�, Lemma 3.1 yield�2K

�1; �fx0�Zu (0)ZuJ3;x0�(0; 0) = �fx0�zu (0)�16Rssum;m + 15Rssqq;u + 13REum;m�+�fx0�zu (0)�� 130Rssqq;u � 12Rssqu;q + 16REss;u � 12REqu;q�:(5.43)

Finally, (5.42) and (5.43) deliver (5.30). The proof of Lemma 5.2 is completed. �

We continue with the proof of Theorem 5.1. We’ll write now the formulas in terms of

connections. For f 2 C1(X;End(E)), we obtain by (2.1) (as in (3.32)) the following

formula in normal coordinates:(�Ef)(Z) = �gij(rEeirEej � �lijrEel)f; rEf = df + [�E(�); f ℄:(5.44)

By (3.31),rEeirEei = �2�Z2i +RE(R; ei) ��Zi + 14RE(R; ei)RE(R; ei) + 23RE;Z(R; ei) ��Zi+ 13RE;ei(R; ei) + 18�2RE;(Z;ei)(R; ei) + 13 DRTX(R; ei)ei; ekERE(R; ek)�+ O(jZj3):(5.45)

Note that by (3.4),�2�Z2mRE;(Z;ei)(R; ei) = 2RE;(em;ei)(em; ei) = 0. Thus from (3.1), (4.3),

(5.45), and taking into account that RE(em; ei) is anti-symmetric in m; i, and Ri (em; ei)is symmetric in m; i, we infer

(5.46)

� �2�Z2mrEeirEeif�(0) = �4fx0�Z2m�Z2i (0) + 2"RE(em; ei); �2fx0�Zi�Zm (0)#+ 12�RE(em; ei); �RE(em; ei); f(x0)��+ 23"RE;em(em; ei); �fx0�Zi (0)#= 16 �4fx0�zi�zq�zi�zq (0)�4�REk`; �REk; f(x0)��+ 83"REmq;m; �fx0�zq (0)#� 83"REqm;m; �fx0�zq (0)#:


By (3.3), (3.25) and (3.31),� �2�Z2m (�iirEe`f)(0)= �43 Ri (em; e`)� �2fx0�Z`�Zm (0) + 12 hRE(em; e`); f(x0)i �� 16�4Ri ;em(em; e`)� 2Ri ;eq(eq; e`) + Ri ;e`(em; em)��fx0�Z` (0)= �323 Ri m` �2fx0�z`�zm (0)� 43��Ri m`;m+Ri mm;` ��fx0�z` (0) + �Ri `m;m+Ri mm;` ��fx0�z` (0)� :(5.47)

Formulas (3.3), (3.23), and (5.44)–(5.47) yield((�E)2f)(x0) = �� 2�Z2m�Ef�(0)= 23 Ri (ei; eq)(rEeirEeqf)(0) + � �2�Z2mrEeirEeif�(0)� �2�Z2m (�liirEelf)(0)= 16 �4fx0�zi�zq�zi�zq (0)� 163 Ri m` �2fx0�z`�zm (0)� 4 hREk` ; [REk ; f(x0)℄i+ 83"REmq;m ; �fx0�zq (0)#� 83"REqm;m ; �fx0�zq (0)#� 43��Ri m`;m+Ri mm;` ��fx0�z` (0) + �Ri `m;m+Ri mm;` ��fx0�z` (0)� :(5.48)

By Lemma 5.2, the discussion after (5.17), formulas (5.17), (5.21), (5.23), (5.27) and

(5.28), we have�2Q4; x0(f)(0; 0) = 2K41f(x0) +K42f(x0) + f(x0)K42 +K2f +K3f+ bCf + bEf2 � 13R`kkq �2fx0�zq�z` (0) + 12 �4f x0�zi�zq�zi�zq (0):(5.49)

Note thathREk` ; [REk ; f(x0)℄i = REk`REkf(x0)� 2REk`f(x0)REk + f(x0)REk`REk , so by (3.8),

(5.12), (5.48) and (5.49), we get (5.8). The proof of Theorem 5.1 is completed.

5.3. Composition of Berezin-Toeplitz operators: proofs of Theorems 0.2, 0.3.

Proof of Theorem 0.2. By Lemma 2.2, we deduce as in the proof of Lemma 2.2, that forZ;Z 0 2 Tx0X, jZj; jZ 0j < "=4, we have (cf. [21, (4.79)], [18, (7.4.6)])p�n(Tf; p Æ Tg; p)x0(Z;Z 0) �= kXr=0(Qr; x0(f; g)Px0)(ppZ;ppZ 0)p� r2 +O(p� k+12 );(5.50)

where Qr; x0(f; g) = Xr1+r2=r K [Qr1; x0(f); Qr2; x0(g)℄ 2 End(E)x0[Z;Z 0℄;(5.51)

is a polynomial in Z;Z 0 with the same parity as r.


The existence of the expansion (0.15) and the expression of b0; f; g follow from (2.12),

(5.50) and (5.51); we get also

br; f; g(x0) = Q2r; x0(f; g)(0; 0):(5.52)

By (5.51),Q2; x0(f; g) = K [f(x0); Q2; x0(g)℄ + K [Q1; x0(f); Q1; x0(g)℄ + K [Q2; x0(f); g(x0)℄:(5.53)

Formulas (1.8), (5.13)–(5.14) yield

K [Q2; x0(f); g(x0)℄P = (Q2; x0(f)P)Pg(x0)= �P

Xj�j=2 ��fx0�Z� (0)Z��! P �L�1O2Pf(x0)�g(x0);

K [f(x0); Q2; x0(g)℄P = f(x0)P(Q2; x0(g)P)= f(x0)�PXj�j=2 ��gx0�Z� (0)Z��! P � g(x0)PO2L �1

P?�:(5.54)

Using by (0.9), (4.8), (4.9), (4.13), (5.44) and (5.54), we obtain

K [Q2; x0(f); g(x0)℄(0; 0) =�� 14� (�Ef)(x0) + 12(b1f)(x0)� g(x0);K [f(x0); Q2; x0(g)℄(0; 0) =f(x0)�� 14� (�Eg)(x0) + 12(gb1)(x0)� :(5.55)

By (1.8), (2.11) and (2.21), we have (cf. [18, (7.4.14)]),Q1; x0(f)(Z;Z 0) = K

�1; �fx�Zq (0)Zq�(Z;Z 0) = �fx0�zi (0)zi + �fx0�zi (0)z 0i :(5.56)

Thus from (4.13), (5.55), we get as in [18, (7.4.15)] (cf. also [18, (7.1.11)]),

K [Q1; x0(f); Q1; x0(g)℄(0; 0) = nXi=1 1� �fx0�zi (0) �gx0�zi (0):(5.57)

Now, (5.52), (5.55) and (5.57), imply the formula for b1; f; g given in (0.16).

We prove next (0.17). It suffices to consider f; g 2 C1(X;R), which we henceforth

assume. By (2.9), we have Qr; x0(1) = Jr; x0 . Hence, taking f = 1 in (5.51), we get

(5.58) Q4; x(g) = K [1; Q4; x(g)℄ +K [J2; x; Q2; x(g)℄ +K [J3; x; Q1; x(g)℄ + g(x)K [J4; x; 1℄:Taking g = 1 in (5.51) yields

(5.59) Q4; x(f) = f(x)K [1; J4; x℄+K [Q1; x(f); J3; x℄+K [Q2; x(f); J2; x℄+K [Q4; x(f); 1℄:By (2.12) and (5.51), we getQ4; x0(f; g) = K [f(x0); Q4; x0(g)℄ + K [Q1; x0(f); Q3; x0(g)℄+ K [Q2; x0(f); Q2; x0(g)℄ + K [Q3; x0(f); Q1; x0(g)℄ + K [Q4; x0(f); g(x0)℄:(5.60)


Set eQ3; x0(g) = K

�1; �gx0�Zq (0)ZqJ2; x0�+ K

�J2; x0 ; �gx0�Zq (0)Zq�+ K

�1; Xj�j=3 ��gx0�Z� (0)Z��! �;I4;f;g = K

"K

�1; Xj�j=2 ��gx0�Z� (0)Z��! �; K

�1; Xj�j=2 ��gx0�Z� (0)Z��! �#:(5.61)

Note that by (2.13) and (2.19), we haveJ3; x0 = K [1; J3; x0 ℄ + K [J3; x0 ; 1℄:(5.62)

and (2.11), (2.21) together with (5.62) implyQ3; x0(g) = g(x0)J3; x0 + eQ3; x0(g):(5.63)

Since g 2 C1(X;R), (5.13) entailsQ2; x0(g) = g(x0)J2; x0 + K

�1; Xj�j=2 ��gx0�Z� (0)Z��! �:(5.64)

By (5.18), (5.58)-(5.64), we getQ4; x0(f; g) = f(x0)Q4; x0(g) + g(x0)Q4; x0(f)� f(x0)g(x0)J4; x0+ K [Q1; x0(f); eQ3; x0(g)℄ + K [ eQ3; x0(f); Q1; x0(g)℄ + I4;f;g:(5.65)

By (4.13) and (5.15), we get(I4;f;gP)(0; 0)= �P

Xj�j=2 ��fx0�Z� (0)Z��! � Xj�j=2 ��gx0�z� (0)z��! + 1� �2gx0�zi�zi (0)�P

�(0; 0)= 1�2�12 �2fx0�zi�zq (0) �2gx0�zi�zq (0) + �2fx0�zq�zq (0) �2gx0�zi�zi (0)�:(5.66)

By Remark 2.3, Q1; x0(f), Q3; x0(g) are self-adjoint for f; g 2 C1(X;R), thus by (5.63),

K [ eQ3; x0(f); Q1; x0(g)℄ = K [Q1; x0(g); eQ3; x0(f)℄�:(5.67)

Thus we only need to compute the fourth term in (5.65).

An examination of (4.4) shows that in each term of the sum giving O2, the total degree

in b+; z equals the total degree in b; z. Hence Remark 4.3, (1.8), (2.19), (4.8), (4.9),

(5.25) and (5.56) yield

K

�Q1; x0(f);K �1; �gx0�Zq (0)ZqJ2; x0��P(0; 0)= �P�fx0�zu (0)zu�gx0�zv (0)zvF2; x0�(0; 0)= 12�2 �fx0�zu (0)�gx0�zv (0)�Æuv(Rssqq + REqq)� 43Rukkv � REuv�:(5.68)


Using (2.19), and the formula (P?ziP)(Z; 0) = (PziP)(Z; 0) = 0 (cf. (1.6)), we get(K [J2; x0 ; �gx0�Zq (0)Zq℄P)(Z; 0) = (F2; x0 �gx0�Zq (0)ZqP)(Z; 0)= (�L�1O2P �gx0�zq (0)zqP �PO2L �1

P?�gx0�zq (0)zqP)(Z; 0):(5.69)

By Remark 4.3, (1.7), (1.8), (5.56) and (5.69), we obtain as in (5.68)

K

"Q1; x0(f);K �J2; x0 ; �gx0�Zq (0)Zq�#P(0; 0)= P�fx0�zu (0)zuK �J2; x0 ; �gx0�Zq (0)Zq�P!(0; 0)= ��P�fx0�zu (0)zuL �1O2P �gx0�zq (0)zqP�(0; 0) = 0;(5.70)

since by (1.6) and (4.11) we have (Pzub�z�P)(Z; 0) = 0 for j�j > 1.

Note that for homogeneous degree 3 polynomials H in Z the analogue of formula

(4.12) holds for (HP)(Z; 0). Using this analogue together with (1.6) and (4.11) we

obtain (PHP)(Z; 0) = � Xj�j=3 �3H�z� z��! + zq� �3H�zq�zi�zi�P(Z; 0):(5.71)

Finally, (1.8), (4.13), (5.56), (5.71) and the equality P(0; 0) = 1 imply

K

�Q1; x0(f);K �1; Xj�j=3 ��gx0�Z� (0)Z��! ��(0; 0) = 1�2 �fx0�zu (0) �3gx0�zu�zi�zi (0):(5.72)

By Lemma 3.1, (3.23), (3.25) and (5.44) for E = C, we get

(5.73)��zq (�g)(0) = �4 �3gx0�zq�zi�zi (0) + 43 Ri q` �gx0�z` (0):

Lemma 3.1, (5.61) and (5.68)–(5.73) entail�2K [Q1; x0(f); eQ3; x0(g)℄(0; 0)= �fx0�zu (0) �3gx0�zu�zi�zi (0) + 12 �fx0�zu (0)�gx0�zv (0)�Æuv(Rssqq + REqq)� 43Rukkv � REuv�= �18h�f; ��gi+ 14h�f; �gi(Rssqq + REqq)� 18h�f ^ �g;REi!:(5.74)

From (5.67) and (5.74), we have�2K [ eQ3; x0(f); Q1; x0(g)℄(0; 0)= �18h��f; �gi+ 14h�f; �gi(Rssqq + REqq)� 18h�f ^ �g;REi!:(5.75)

By (0.9), (5.6), (5.65), (5.66), (5.74) and (5.75), we get (0.17). The proof of Theorem

0.2 is completed. �

Proof of Theorem 0.3. The existence of the expansion (0.18) and formula C0(f; g) = fgwere established in [21, Th. 1.1] (cf. also [18, Th. 7.4.1]) in general symplectic settings.


By (0.18), (2.10), (5.12), (5.50) and (5.52), we obtain (cf. also [18, (7.4.9)]),

(5.76) C1(f; g) = (Q2;x(f; g)�Q2;x(fg))(0; 0) = b1; f; g � b1; fg:Hence (0.13), (0.16), (5.44) and (5.76) yield the formula for C1(f; g) given in (0.20).

Moreover, (0.18), (5.12) and (5.52) imply the formula for C2(f; g) from (0.20).

We will prove (0.21) now. Let feig be an orthonormal frame of (TX; gTX), and fwigbe an orthonormal frame of T (1;0)X. Let � = �� + � �� be the Kodaira Laplacian on�(T �X) R C, and let � be the Bochner Laplacian on �(T �X) R C associated with

the connection r�(T �X) on �(T �X) induced by rTX (cf. (0.5)). Let R�(T �(1;0)X) be the

curvature of the holomorphic Hermitian connection on �(T �(1;0)X). By Lichnerowicz

formula [18, Remark 1.4.8], and (5.4), we haveR�(T �(1;0)X) = �hRTXwl; wkiwl ^ iwk ;2� = ��R�(T �(1;0)X)(wq; wq) + (2R�(T �(1;0)X) + Ri )(wl; wk)wk ^ iwl:(5.77)

Since (X;!; J) is Kahler, � commutes with the operators �; �; d (cf. [18, Cor. 1.4.13]),

and (5.77) shows that 2�f = �f for any f 2 C1(X). From Lemma 3.1, (5.4) and

(5.77), we have for any f 2 C1(X):��f = ��f � Ri (�; wk)wk(f);��f = ��f � Ri (�; wk)wk(f);�df = d�f � Ri (�; ej) ej(f):(5.78)

Thus (0.5), (5.78) yield for any f; g 2 C1(X):�(fg) = g�f + f�g � 2hdf; dgi;�h�f; �gi = h��f; �gi+ h�f;��gi � 2hrT �X�f;rT �X�gi= h��f; �gi+ h�f; ��gi � 2hrT �X�f;rT �X�gi � 2Ri (wm; wq)wm(g)wq(f):(5.79)

Using (5.78) and (5.79), we infer�2(fg) =f�2g + g�2f + 2(�f)�g � 4hd�f; dgi � 4hdf; d�gi+ 4hrT �Xdf;rT �Xdgi+ 4Ri (ei; ej) ei(g)ej(f):(5.80)

We examine now closely the expression of �2C2(f; g) given by (0.20). Using (0.14),

(0.17), (0.20), we see that the term of differential order 0 in f; g from the expression of�2C2(f; g) is zero, and the term of total differential order 2 in f; g, disregarding the term

involving RE, in �2C2(f; g) isC22 = p�18 DRi !;�f��g � g��f + ��(fg)E :(5.81)

The term of total differential order 4 in f; g in the expression of �2C2(f; g) isC24 = 132f�2g + 132g�2f � 18h�f; ��gi � 18h��f; �gi+ 116(�f)�g+ 18hD0;1�f;D1;0�gi � 132�2(fg)� 18�h�f; �gi:(5.82)

By (5.80), (5.82) and by the formula hD1;0�f;D0;1�gi = hD0;1�f;D1;0�gi, we getC24 = 18hD1;0�f;D0;1�gi+ p�18 DRi !; �f ^ �g � �g ^ �fE :(5.83)


Finally, by inspecting (0.14), (0.17), (0.20), we see that the term involving RE in the

expression of �2C2(f; g) isp�18 �� f�g � g�f +�(fg)�RE� + 14hg��f + f��g � ��(fg); REi!+p�14 h�f; �giRE� � 14h�f ^ �g;REi! + p�14 h�f; �giRE� :(5.84)

Combining (5.81), (5.83) and (5.84), we get (0.21). The proof of Theorem 0.3 is com-

pleted. �

6. DONALDSON’S Q-OPERATOR

In this section we study the asymptotics of the sequence of operators introduced by

Donaldson [8]. We suppose henceforth that E = C. Set Vol(X; dvX) := RX dvX . Follow-

ing [8, §4], setKp(x; x0) := jPp(x; x0)j2hLpx hLp�x0 ; Rp := (dimH0(X;Lp))=Vol(X; dvX):(6.1)

Let Kp, QKp be the integral operators associated to Kp, defined for f 2 C1(X) by

(6.2) (Kpf)(x) := ZX Kp(x; y)f(y)dvX(y); QKp(f) = 1RpKpf:Recall that, just as the Bergman kernel appears when comparing a Kahler metric ! to its

algebraic approximations !p (i.e. pull-backs of the Fubini-Study metrics by the Kodaira

embeddings), the operators QKp appear when one relates infinitesimal deformations of

the metric ! to the corresponding deformations of the approximations !p . The asymp-

totics of the operator Kp were obtained in [15, Th. 26]1 and used in [11]. The following

result refines [15, Th. 26] and is applied in the recent paper [12].

Theorem 6.1. For every m 2 N, there exists C > 0 such that for any f 2 C1(X), p 2 N�,�� 1pnKpf � f + 18�p(�rf + 2�f)��Cm(X) 6 Cp�3=2 jf j

Cm+3(X) or Cp�2 jf jCm+4(X) :(6.3)

Proof. By (2.9) with Z = 0, (2.19) and (2.21), we get�� 1p2nKp;x0(0; Z 0)�x0(Z 0)� �1 + kXr=2 p�r=2J 0r(0;ppZ 0)�e��pjZ0j2��Cm(X)

6 Cp�(k+1)=2(1 + jppZ 0j)N exp(�C0ppjZ 0j) + O(p�1);(6.4)

with J 02(0; Z 0) = (J2 + J2)(0; Z 0):(6.5)

1Note that in the present context [15, Th. 26] should be modified as follows��QKpf � Vol(X; �)Vol(X; dvX) �f��Cm(X) 6 Cp�1=2 jf j

Cm+1(X) or Cp�1 jf jCm+2(X) ;

since the right hand side of the second equation of [15, (33)] and [15, (34)] should readCp�1=2 jf jCm+1(X)

or Cp�1 jf jCm+2(X).


Now we have the analogue of [15, (32)],��p�nKpf � pn ZjZ0j6" �1 + kXr=2 p�r=2J 0r(0;ppZ 0)�e��pjZ0j2fx0(Z 0)dvTx0X(Z 0)��Cm(X)

6 Cp�(k+1)=2 jf jCm(X) :(6.6)

But as in the proof of [1, Th .2.29 (2)], we get��pn ZjZ0j6" J 0r(0;ppZ 0)e��pjZ0j2fx0(Z 0)dvTx0X(Z 0)��Cm(X) 6 C jf j

Cm(X) :(6.7)

Moreover ��pn ZjZ0j6" e��pjZ0j2fx0(Z 0)dvTx0X(Z 0)� f(x0) + 14�p(�f)(x0)��Cm(X)

6 Cp�3=2 jf jCm+3(X) or Cp�2 jf j

Cm+4(X) :(6.8)

Finally, by (4.9) (cf. also [18, (4.1.110)]), we haveZZ02Cn J2(0; Z 0)jPj2(0; Z 0)dZ 0 = ZZ02Cn P(0; Z 0)J2(Z 0; 0)P(Z 0; 0)dZ 0= (PJ2P)(0; 0) = �(PO2L �1P

?)(0; 0) = 116�r:(6.9)

Thus ��pn ZjZ0j6" J 02(0;ppZ 0)e��pjZ0j2fx0(Z 0)dvTx0X(Z 0)� � r8�f�(x0)��Cm(X)

6 Cp�1=2 jf jCm+1(X) or Cp�1 jf j

Cm+2(X) :(6.10)

The proof of Theorem 6.1 is completed. �

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Berezin–Toeplitz quantization on Kähler manifolds