The Zeros of the Bergman Kernel for Some Reinhardt Domains

We consider the Reinhardt domain Dn = {(ζ, z) ∈ C × C : |ζ|2 < (1 − |z1|2) ⋅ ⋅ ⋅ (1 − |zn|2)}. We express the explicit closed form of the Bergman kernel forDn 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.

Over the last decade the Hartogs domain Ω fl {(, ) ∈ C  × Ω :          2 <  ()} was investigated, where  is a suitably chosen continuous function on a bounded domain Ω.The Bergman kernel for Ω 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 kernel   (, ) for a bounded domain  is zero-free for all (, ) ∈  × , then  is called the Lu Qi-Keng domain.One can see many examples of Lu Qi-Keng domains and non-Lu Qi-Keng domains in [8][9][10][11][12][13].
If Ω is symmetric [8] or homogeneous [9], then the main part of the Bergman kernel for Ω is the polynomial 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 is a Lu Qi-Keng domain or not.The existence of zeros of all Hartogs domains Ω 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 for all dimensions, since Routh-Hurwitz theorem involves too many terms when the order of the polynomial is large.

Journal of Function Spaces
In this paper we consider the Reinhardt domain   ⊂ C +1 defined by From Theorem 2.5 in [8], the Bergman kernel for   can be obtained explicitly as the following.
Theorem 1.The Bergman kernel   for   is written as where where (, ) is the Stirling number of the second kind.
In Section 2, we prove Theorem 1 using the result in [8] and express (, ℓ) in terms of the coefficients of a certain generating function (see Theorem 8).We use the well-known formal series for exponential generating function, where (, ) is the Stirling number of the second kind.
For the study of the existence of zeros of   ((, ), (, )), we need to define F () and   () by The zero set of the Bergman kernel   ((, ), (, )) with (, ), (, ) ∈   reduces to the zero set of the polynomial   () with Re  < 0. Now we write In Section 3, we introduce the Routh-Hurwitz theorem that is efficient on checking whether the real polynomial   () has zeros in the right half plane.Using the generating form of coefficients of   () (see Proposition 12), we will show that if  ≥ 3, then   () does not satisfy Routh-Hurwitz conditions, so we obtain the following main result of this paper.
Theorem 2. The Bergman kernel for   is zero-free if and only if  ≤ 2.
For the proof of Theorem 2, we will show that if  ≥ 3, then at least one of (, ), (,  − 1), or (,  − 2) is negative (see Theorem 13).In Section 4, we discuss the properties of (, 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  2 has no zeros and  3 has zeros.The main contribution, Theorem 2 in this paper, is the complete classification of the answer to the Lu Qi-Keng problem for   for all dimensions .

Explicit Form of the Bergman Kernel for 𝐷 𝑛
In [8], we know the explicit form of the Bergman kernel for Cartan-Hartogs domain Ω in the case when where  Ω  is the generic norm with respect to the bounded symmetric domain Ω  .Using the numerical invariants , ,  with respect to the bounded symmetric domain, we define the Hua polynomial where ()  = (+1)(+2) ⋅ ⋅ ⋅ (+−1) for  ≥ 1 and () 0 = 1.
Then the polynomial   () can be written as ) where  (, ℓ) ) Note that Thus we have  (, ℓ) ) Note that (, ) = 0 for  >  and ) Lemma 6.For any nonnegative integer , one has Proof.Note that for any nonnegative integer , which completes the proof.
By Lemma 6, we have ℓ−+ Let [  ] be the operator which gives the th coefficient in the series expansion of a generating function.It is wellknown that the exponential generating function of (, ) is the formal power series Using the above generating function, we prove the following.
Proof.By ( 27) and (32), the coefficient (, ℓ) of   () can be expressed as Note that, by (34), we have which follows that It completes the proof.

Lu Qi-Keng Domains
In  The Routh-Hurwitz criterion is the most efficient method for determining whether the polynomial   () has zeros in the open left half plane.Let with real coefficients and  0 > 0, and define Δ   for  = 1, .
Lemma 11 (Routh-Hurwitz/Liénard-Chipart [15]).All zeros of given polynomial () lie in the open left half plane { ∈ C : Re  < 0} if and only if This condition is also equivalent to any one of the following four forms:

Proposition 12. Let [𝑡 𝑛
] be the operator which gives the th coefficient in the series expansion of a generating function Proof.Note that ℓ− By Theorem 8, By Proposition 12, we have Proof.We will prove it in Section 4.
has two negative real zeros.Thus,  2 is a Lu Qi-Keng domain.
(ii) In fact, we see that has one positive real zero −1/8 + √ 33/24 > 0. Thus, we conclude that the Bergman kernel for  3 has zeros, so  3 is not a Lu Qi-Keng domain.
By (i), (ii), and (iii),   () does not satisfy any condition of Routh-Hurwitz theorem, so the Bergman kernel for   has zeros for all  ≥ 3.
this section we investigate the explicit form of (, ℓ) and prove that the Bergman kernel for   has zeros for any positive integer .