# The Zeros of the Bergman Kernel for Some Reinhardt Domains

Authors: Jong-Do Park
Journal: Journal of Function Spaces
 (2016)

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## Abstract

We consider the Reinhardt domainDn={(ζ,z)∈C×Cn:|ζ|2<(1-|z1|2)⋯(1-|zn|2)}. We express the explicit closed form of the Bergman kernel for Dn using the exponential generating function for the Stirling number of the second kind. As an application, we show that the Bergman kernel Kn for Dn has zeros if and only if n≥3. The study of the zeros of Kn is reduced to some real polynomial with coefficients which are related to Bernoulli numbers. This result is a complete characterization of the existence of zeros of the Bergman kernel for Dn for all positive integers n.


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## Body

## 1. Introduction

LetD be a bounded domain in Cn. The space La2(D) is denoted by the set of all holomorphic functions f on D satisfying(1)∫Dfz2dVz<∞,where dV is the volume measure on D. For any z∈D,  Φz:La2(D)→C defined by Φz(f)=f(z) is a bounded linear functional on La2(D). By Riesz representation theorem, there exists the unique element Kz(·)∈La2(D) such that Φz(f)=f·,Kz·; namely, (2)fz=∫DfwKzw¯dVw,for all f∈La2(D). Define the Bergman kernel function KD(z,w)≔Kz(w)¯ for D. It is defined for arbitrary bounded domains, but it is difficult to obtain the explicit form of the Bergman kernel for general bounded domains. Recently, the Bergman kernels for various domains have been computed explicitly in [1–6].Over the last decade the Hartogs domain(3)Ω^m≔ζ,z∈Cm×Ω:ζ2<pzwas investigated, where p is a suitably chosen continuous function on a bounded domain Ω. The Bergman kernel for Ω^m was obtained explicitly in [7] when Ω is an irreducible bounded symmetric domain. This result was generalized to the cases when Ω is the product of bounded symmetric domains in [8] and when Ω is a bounded homogeneous domain in [9].Also the problem of determining whether the Bergman kernels are zero-free has been a well-known open problem in several complex variables ever since Lu Qi-Keng raised the question related to the existence of Bergman representative coordinates. If the Bergman kernelKD(z,w) for a bounded domain D iszero-free for all (z,w)∈D×D, then D is called theLu Qi-Keng domain. One can see many examples of Lu Qi-Keng domains and non-Lu Qi-Keng domains in [8–13].IfΩ is symmetric [8] or homogeneous [9], then the main part of the Bergman kernel for Ω^m is thepolynomial whose coefficients are written as the forms containing the Stirling number of the second kind. The Routh-Hurwitz theorem (see Lemma 11) gives the condition that a real polynomial has no zeros in the closed right half-plane, and using this criterion we have the algorithmic method of determining whether the Hartogs domain Ω^m is a Lu Qi-Keng domain or not. The existence of zeros of all Hartogs domains Ω^m is classified in [8, 9] only when the dimension of the base domain Ω is low (less than 4). However it looks hard to study Lu Qi-Keng problem forall dimensions, since Routh-Hurwitz theorem involves too many terms when the order of the polynomial islarge.In this paper we consider the Reinhardt domainDn⊂Cn+1 defined by (4)Dn≔ζ,z=ζ,z1,z2,…,zn∈C×Cn:ζ2<∏j=1n1-zj2.From Theorem  2.5 in [8], the Bergman kernel for Dn can be obtained explicitly as the following.Theorem 1.
The Bergman kernelKn for Dn is written as (5)Knζ,z,η,w=1πn∏l=1n11-zlw¯l3Fnζ,η∏l=1n1-zlw¯l,where (6)Fnt=∑j=1n-1n-jSn,jj+1!1-tj+2,where S(n,j) is the Stirling number of the second kind.In Section2, we prove Theorem 1 using the result in [8] and express a(n,l) in terms of the coefficients of a certain generating function (see Theorem 8). We use the well-known formal series (7)et-1jj!=∑n=0∞Sn,jtnn!,for exponential generating function, where S(n,j) is the Stirling number of the second kind.For the study of the existence of zeros ofKn((ζ,z),(η,w)), we need to define F~n(t) and Gn(τ) by(8)F~nt=1-t2Fnt,Gnτ=F~n1-1τ+1/2=∑j=1n-1n-jSn,jj+1!τ+12j.The zero set of the Bergman kernel Kn((ζ,z),(η,w)) with (ζ,z),(η,w)∈Dn reduces to the zero set of the polynomial Gn(τ) with Re⁡τ<0. Now we write (9)Gnτ=∑k=0nan,n-kτk=an,n+an,n-1τ+⋯+an,1τn-1+an,0τn.In Section3, we introduce the Routh-Hurwitz theorem that is efficient on checking whether the real polynomial Gn(τ) has zeros in the right half plane. Using the generating form of coefficients of Gn(τ) (see Proposition 12), we will show that if n≥3, then Gn(τ) does not satisfy Routh-Hurwitz conditions, so we obtain the following main result of this paper.Theorem 2.
The Bergman kernel forDn is zero-free if and only if n≤2.For the proof of Theorem2, we will show that if n≥3, then at least one of a(n,n), a(n,n-1), or a(n,n-2) is negative (see Theorem 13). In Section 4, we discuss the properties of a(n,l) and prove Theorem 13 using properties of the Bernoulli numbers and Genocchi numbers.Remark 3.
In Theorem 5.2(ii) and Theorem 5.3 of [8], one can see that K2 has no zeros and K3 has zeros. The main contribution, Theorem 2 in this paper, is the complete classification of the answer to the Lu Qi-Keng problem for Dn for all dimensions n.


## 2. Explicit Form of the Bergman Kernel forDn

In [8], we know the explicit form of the Bergman kernel for Cartan-Hartogs domain Ω^m in the case when(10)Ω=Ω1×⋯×Ωn,pz1,…,zn=∏l=1nNΩlzl,zlμl,μl>0,where NΩl is the generic norm with respect to the bounded symmetric domain Ωl. Using the numerical invariants a,  b,  r with respect to the bounded symmetric domain, we define the Hua polynomial (11)χs=∏j=1rs+1+j-1a21+b+r-ja,where (s)k=s(s+1)(s+2)⋯(s+k-1) for k≥1 and (s)0=1. Then for any μ>0, we define c(μ,j) by(12)χkμχ0=∑j=0dcμ,jk+1j,where d≔r+rb+r(r-1)a/2. We also define djj1,…,jn satisfying(13)∏l=1nk+1jl=∑j=maxj1,…,jnj1+⋯+jndjj1,…,jnk+1j.Theorem 4 (see [8]).
LetΩ^m≔{(ζ,z)∈Cm×Ω:ζ2<p(z)}, where Ω and p are defined as in (10). Then the Bergman kernel for Ω^m is given by (14)K^mζ,z,η,w=1m!∏l=1nKΩlzl,wlNΩlzl,wlmμlFnζ,z∏l=1nNΩlzl,wlμl,where(15)Fnt=∑j1=0dimΩ1⋯∑jn=0dimΩn∏l=1ncμl,jl∑j=maxj1,…,jnj1+⋯+jndjj1,…,jnj+m!1-tj+m+1.For anyn≥k, the number of the partitions of {1,2,…,n} into k blocks is denoted by S(n,k) and called theStirling number of the second kind.Lemma 5 (see [14]).
For any positive integerp, it holds that (16)xp=∑j=1pSp,jx-j+1j.Proof of Theorem1.
For any1≤l≤n, let Ωl={z∈C:z<1} be the unit disk and m=μ1=⋯=μn=1. Then Ω^m=Dn and (17)r=1,a=2,b=0,g=2,NΩlz,w=1-zw¯,KΩlz,w=1-zw¯-2.Since the Hua polynomial is χ(s)=s+1, by (12) we have (18)k+1=c1,0+c1,1k+1,so that c(μ,0)=0 and c(μ,1)=1.
Thus the functionFn(t) in Theorem 4 is reduced to(19)Fnt=∑j=1ndj1,…,1j+1!1-tj+2.Now we claim that dj1,…,1 is equal to (-1)n-jS(n,j). If j1=⋯=jn=1, then, by (13), we have(20)k+1n=∑j=1ndj1,…,1k+1j.By Lemma 5, we have (21)-xn=∑j=1nSn,j-x-j+1j=∑j=1nSn,j-1jxj,since (-x-j+1)j=(-x-j+1)(-x-j+2)⋯(-x)=(-1)j(x)j. Thus we obtain(22)xn=∑j=1n-1n-jSn,jxj.If we compare (20) and (22), then dj1,…,1=(-1)n-jS(n,j) for all 1≤j≤n. Then from (19), it completes the proof of Theorem 1.Then the polynomialGn(τ) can be written as(23)Gnτ=∑j=1n-1n-jSn,jj+1!∑k=0jjk12j-kτk=∑k=0n∑j=kn-1n-jSn,jj+1!jk12j-kτk=∑k=0nan,n-kτk,where (24)an,l≔∑j=n-ln-1n-jSn,jj+1!jn-l12j-n+l.Note that (25)jn-l=jj-1⋯j-n+l+1n-l!=j!n-l!j-n+l!=n-l+1n-l+2⋯jj-n+l!=-1j-n+ll-n-1l-n-2⋯-jj-n+l!=-1j-n+ll-n-1l-n+j.Thus we have (26)an,l=-1l∑j=n-lnSn,jj+1!l-n-1l-n+j12j-n+l.Note that S(n,j)=0 for j>n and l-n-1l-n+j=0 for j<n-l; we have(27)an,l=-1l∑j=0∞Sn,jj!j+1l-n-1l-n+j12j-n+l.Lemma 6.
For any nonnegative integert, one has (28)∑j=0∞j+1-t-1-t+jxj=ddxxt+11+xt+1.Proof.
Note that for any nonnegative integert, (29)∑j=0∞j+1-t-1-t+jx-t+j=∑j=0∞j+t+1-t-1jxj.
Note that(30)ddx∑j=0∞-t-1jxj+t+1=∑j=0∞-t-1jj+t+1xj+t=∑j=0∞j+t+1-t-1jxj+t.
Thus(31)∑j=0∞j+t+1-t-1jxj=1xtddxxt+11+xt+1,which completes the proof.By Lemma6, we have(32)∑j=0∞j+1l-n-1l-n+j12l-n+juj=12l-n∑j=0∞j+1l-n-1l-n+ju2j=12l-nddxxn-l+11+xn-l+1x=u/2.Definition 7.
Define(33)Hn,lu≔12l-nddxxn-l+11+xn-l+1x=u/2.Let[tn] be the operator which gives the nth coefficient in the series expansion of a generating function. It is well-known that the exponential generating function of S(n,j) is the formal power series(34)et-1jj!=∑n=0∞Sn,jtnn!.Using the above generating function, we prove the following.Theorem 8.
LetGn(τ) be the polynomial defined as in (8) with (35)Gnτ=∑k=0nann-kτk.Then the coefficient a(n,l) is written as (36)an,l=-1ltnn!Hn,let-1,where 0≤l≤n.Proof.
By (27) and (32), the coefficient a(n,l) of Gn(τ) can be expressed as(37)an,l=-1l∑j=0∞Sn,jj!ujHn,lu.
Note that, by (34), we have (38)-1l∑j=0∞et-1jujHn,lu=-1l∑j=0∞∑n=0∞Sn,jj!ujHn,lutnn!=∑n=0∞-1l∑j=0∞Sn,jj!ujHn,lutnn!=∑n=0∞an,ltnn!,which follows that (39)an,l=-1ltnn!∑j=0∞et-1jujHn,lu=-1ltnn!Hn,let-1.It completes the proof.


## 3. Lu Qi-Keng Domains

In this section we investigate the explicit form ofa(n,l) and prove that the Bergman kernel for Dn has zeros for any positive integer n.Note that if(ζ,z),(η,w)∈Dn, then (40)ζ,η∏l=1n1-zlw¯l≤ζη∏l=1n1-zl21-wl2=ζ2∏l=1n1-zl2η2∏l=1n1-wl2<1.Thus by Theorem 1 and (8), we obtain the following.Lemma 9.
The zero set(41)ζ,z,η,w∈Dn×Dn:Knζ,z,η,w=0is equal to the zero set (42)ζ,z,η,w∈Dn×Dn:F~nt=0,t<1,where t≔〈ζ,η〉/∏l=1n(1-zlw¯l).Since the holomorphic functiont↦1/(1-t)-1/2 maps the unit disk onto the right half plane, we obtain the following consequence of Theorem 1 and Lemma 9.Lemma 10.
For any positive integersn, the domain Dn is a Lu Qi-Keng domain if and only if all zeros of the polynomial Gn(τ) lie in the closed left half plane {z∈C:Re⁡z≤0}.The Routh-Hurwitz criterion is the most efficient method for determining whether the polynomialGn(τ) has zeros in the open left half plane. Let (43)fτ=a0τn+a1τ-1+⋯+an-1τ+an,with real coefficients and a0>0, and define Δjn for j=1,…,n by (44)Δjn≔a1a3a5⋯a2j-1a0a2a4⋯a2j-20a1a3⋯a2j-3⋮⋮⋮⋱⋮000⋯aj,where aj=0 if j<0 or j>n.Lemma 11 (Routh-Hurwitz/Liénard-Chipart [15]).
All zeros of given polynomialf(τ) lie in the open left half plane {τ∈C:Re⁡τ<0} if and only if (45)Δ1n>0,…,Δnn>0.This condition is also equivalent to any one of the following four forms: (i)
an>0,  an-2>0,  an-4>0,…;  Δ1n>0,  Δ3n>0,…,(ii)
an>0,  an-2>0,  an-4>0,…;  Δ2n>0,  Δ4n>0,…,(iii)
an>0;  an-1>0,  an-3>0,…;  Δ1n>0,  Δ3n>0,…,(iv)
an>0;  an-1>0,  an-3>0,…;  Δ2n>0,  Δ4n>0,….Proposition 12.
Let[tn] be the operator which gives the nth coefficient in the series expansion of a generating function(46)an,l=-1ln-l+12n-l+2tnn!et-1n-let+1n-l+2.Proof.
Note that(47)Hn,lu=12l-nddxxn-l+11+xn-l+1x=u/2=12l-nn-l+1xn-l1+xn-l+2x=u/2=n-l+12n-l+2un-lu+2n-l+2.
By Theorem8, (48)an,l=-1ln-l+12n-l+2tnn!et-1n-let+1n-l+2.By Proposition12, we have (49)an,n=4-1ntnn!1et+12,an,n-1=16-1n-1tnn!et-1et+13,an,n-2=48-1n-2tnn!et-12et+14.Theorem 13.
For eachn≥3, at least one of a(n,n),  a(n,n-1), or a(n,n-2) is negative. More precisely, one has (i)
a(n,n)<0 if n=3,4,7,8,11,12,15,16,…,(ii)
a(n,n-1)<0 if n=5,9,13,…,(iii)
a(n,n-2)<0 if n=6,10,14,….Proof.
We will prove it in Section4.Remark 14.
(i) Note that(50)G2τ=12+6τ+4τ2has two negative real zeros. Thus, D2 is a Lu Qi-Keng domain.
(ii) In fact, we see that(51)G3τ=-12+2τ+18τ2+24τ3has one positive real zero -1/8+33/24>0. Thus, we conclude that the Bergman kernel for D3 has zeros, so D3 is not a Lu Qi-Keng domain.
(iii) By using a computer program (Maple or Mathematica), we computed the explicit values ofa(n,n),  a(n,n-1), and a(n,n-2) for 2≤n≤14. One can check Theorem 13 holds for 3≤n≤14 in Tables 1 and 2.Table 1
2 ≤ n ≤ 8.
G 2 G 3 G 4 G 5 G 6 G 7 G 8 a ( n , n ) 1/2 −1/2 −1 1 17/4 −17/4 −31 a ( n , n - 1 ) 6 2 −8 −13 34 107 −248 a ( n , n - 2 ) 4 18 6 −90 −129 693 1896Table 2
9 ≤ n ≤ 14.
G 9 G 10 G 11 G 12 G 13 G 14 131 691/2 −691/2 −5461 5461 929569/8 −1258 2764 20462 −43688 −885881/2 929569 −7920 −34014 126918 776661 −2723175 −44729673/2Now we can prove the main theorem of this paper using Theorem13.Theorem 2 (again).
The Bergman kernel for(52)Dn=ζ,z1,z2,…,zn∈Cn+1:ζ2<∏j=1n1-zj2is zero-free if and only if n≤2.Proof.
(i) Ifn=4k-1 or n=4k for k∈N, then a(n,n)<0 by Theorem 13. So, the polynomial Gn(τ) does not satisfy any condition in Lemma 11. It follows that Dn is not a Lu Qi-Keng domain.
(ii) Ifn=4k+1 for k∈N, then a(n,n-1)<0 by Theorem 13. So, the polynomial Gn(τ) does not satisfy conditions (iii) and (iv) in Lemma 11. It follows that Dn is not a Lu Qi-Keng domain.
(iii) Ifn=4k+1 for k∈N, then a(n,n-2)<0 by Theorem 13. So, the polynomial Gn(τ) does not satisfy conditions (i) and (ii) in Lemma 11. It follows that Dn is not a Lu Qi-Keng domain.
By (i), (ii), and (iii),Gn(τ) does not satisfy any condition of Routh-Hurwitz theorem, so the Bergman kernel for Dn has zeros for all n≥3.


## 4. Proof of Theorem13

In this section, we investigate the properties ofa(n,n),  a(n,n-1), and a(n,n-2) and prove Theorem 13. For convenience, we denote the functions A,  B,  C,  D by (53)At≔1et+1=∑k=0∞aktk,Bt≔1et+12=∑k=0∞bktk,Ct≔1et+13=∑k=0∞cktk,Dt≔1et+14=∑k=0∞dktk.It is interesting that the numbersak’s are related to the following Bernoulli numbers. Bernoulli [16] introducedBernoulli numbersB2n for n≥1 satisfying the identity (54)tet-1=1-t2+∑n=1∞-1n+1B2nt2n2n!,where B2=1/6, B4=1/30, and B6=1/42,…. TheGenocchi numbersG2n for n≥1 are defined [17] by (55)G2n≔222n-1B2n.Lemma 15 (see [18]).
Letn be any positive integer. Then (i)
B2n’s are positive for all positive integers n;(ii)
2t/(et+1)=t+∑n=1∞‍(-1)nG2n(t2n/2n!).Lemma 16.
For anyk∈N, one has (56)a2k=0,a4k-1>0,a4k-3<0.Moreover a0=1/2.Proof.
By Lemma15(ii), we have (57)1et+1=12+∑n=1∞-1nG2n2·2n!t2n-1.It follows that (58)a0=12,a2k=0k≥1,a2n-1=-1nG2n2·2n!.Note that G2k>0 for all k≥1 by Lemma 15(i). Thus a4k-1>0 and a4k-3<0 for all k≥1.Lemma 17.
For anyk∈N, one has (i)
b2k+1=a2k+1,(ii)
bk=(k+1)ak+1+ak,(iii)
2c2k=3b2k,(iv)
2ck=(k+1)bk+1+2bk,(v)
2d2k+1=4a2k+1-6b2k+1+4c2k+1,(vi)
3dk=(k+1)ck+1+3ck.Proof.
From the identity(59)B-t=e2tet+12=et+12-2et+1+1et+12,we obtain (60)Bt-B-t=2At-1.It follows that (61)2∑k=0∞b2k+1t2k+1=2∑k=0∞aktk+1.Thus we obtain (i). From the identity (62)A′t=-etet+12=-et+1+1et+12,we obtain (63)At+A′t=Bt.It follows that (64)∑k=0∞aktk+∑k=1∞kaktk-1=∑k=0∞bktk.Thus we obtain (ii). From the identity (65)Ct+C-t=1-3At+3Bt,we have 2c2k=-3a2k+3b2k. Since a2k=0 for k≥1 in Lemma 16, we obtain (iii). From the identity (66)B′t+2Bt=2Ct,we obtain (iv). Similarly as the previous proofs, we can easily see that (v) comes from the identity (67)Dt-D-t=-1+4At-6Bt+4Ct,and (vi) comes from the identity (68)C′t+3Ct=3Dt.Proposition 18.
For anyk∈N, it holds that (i)
a2k=0,  a4k-1>0,  a4k+1<0,(ii)
b4k<0,  b4k+1<0,  b4k+2>0,  b4k+3>0,(iii)
c4k+1>0,  c4k+2>0,(iv)
d4k+2<0 for k≥2.Proof.
(ii) Note that(69)b4k=4k+1a4k+1<0,b4k+1=a4k+1<0,b4k+2=4k+3a4k+3>0,b4k+3=a4k+3>0.
(iii) Note thatc4k+2=(3/2)b4k+2>0. Since C(t)=A(t)·B(t), we have (70)c4k+1=∑l=04k+1alb4k+1-l=a0b4k+1+a1b4k+a4k+1b0+∑l=12k-1a2l+1b4k-2l>a0b4k+1+a1b4k+a4k+1b0=12b4k+1-14b4k+14a4k+1=12a4k+1-144k+1a4k+1+14a4k+1=142-4ka4k+1>0.Here by Proposition 18(i) and (ii), (71)∑l=12k-1a2l+1b4k-2l=∑l=1k-1a4l+1b4k-4l+∑l=1ka2l-1b4k-4l+2>0.
(iv) Similarly as the proof of (iii), we obtaind4k+2<0 for k≥2.Now we prove Theorem13 using the above proposition.Theorem 13 (again).
For any positive integersk, one has (i)
a(n,n)<0 if n=4k-1 or n=4k,(ii)
a(n,n-1)<0 if n=4k+1,(iii)
a(n,n-2)<0 if n=4k+2.Proof.
(i) Note that(72)an,n=4-1ntnn!1et+12=4-1nbn.By Proposition 18(ii), it follows that (73)a4k-1,4k-1=-4b4k-1<0,a4k,4k=4b4k<0.
(ii) Note that(74)an,n-1=16-1n-1tnn!et-1et+13=16-1n-1bn-2cn,since (75)et-1et+13=et+1-2et+13=1et+12-2et+13.By Proposition 18(ii) and (iii), it follows that (76)a4k+1,4k=16b4k+1-2c4k+1<0.
(iii) Note that(77)an,n-2=48-1n-2tnn!et-12et+14=48-1n-2bn-4cn+4dn,since (78)et-12et+14=et+1-22et+14=1et+12-4et+13+4et+14.By Lemma 17(iii), we have (79)a4k+2,4k=48b4k+2-4c4k+2+4d4k+2=48-5b4k+2+4d4k+2<0fork≥2.Moreover a(6,4)=48(-5b6+4d6)=48·(-43/11520)<0.



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