arXiv:2001.01856v2 [math.CV] 7 Jul 2020 steuiu ucino Ω on function unique the is ai of basis ne the under O Introduction 1 e od n hae.BrmnKre,MnmlDmi,SiaConj Suita Domain, Minimal Kernel, Bergman Seconda phrases. 30C40; and words Primary Key Classification. Subject Mathematics 2010 Ω stesto ooopi ucin nΩ h ega pc sas a is Bergman The Ω. on functions holomorphic of set the is (Ω) .Frall For 1. 2. .Frall For 3. e eadmi in domain a be Ω Let iiiytermb h iia on fthe of point minimal the by theorem Rigidity K hc ssrnl ovxadntbhlmrhct h ntba unit the to biholomorphic not and convex strongly is which for ega kernel Bergman sete ikmnsa(osbyepy lsdplrstor set polar closed empty) (possibly a minus disk a either is sn h eosto h zgokre.Fnly eso that show we Finally, Szeg¨o kernel. the of set zero the using lsdplrst hnΩi one with bounded is Ω When set. polar closed ( A ,w z, eueteSiacnetr nwaterm opoeta o a for that prove to theorem) a (now conjecture Suita the use We > n 2 L Ω,te h ega enlfunction kernel Bergman the then (Ω), 2 = ) inrpoutwt eegevlm esr.If measure. volume Lebesgue with product -inner w f ycntutn one opeeRihrtdmi (with domain Reinhardt complete bounded a constructing by 1 ∈ ∈ K A Ω, ( 2 ,z w, Ω and (Ω) K K ( ( ) · · w , , C · oetXnDn n onTreuer John and Dong Xin Robert satisfies ) n ) × n eoeteBrmnsaeo by Ω of space Bergman the denote and ∈ z hc aifisteproperties the satisfies which Ω ∈ A 2 Ω, ega kernel Bergman (Ω) K K f ( ( ( z ,w z, z 0 = ) z , 0 = ) Volume(Ω) = ) Abstract Z Ω C X j f ∞ =0 1 ∞ ( w budr,w rvd ipepofo this of proof simple a provide we -boundary, K φ ) j K ( Ω : z ( ) ,w z, φ × j y3C5 08,30C20 30C85, 30C35, ry − ( w ) Ω 1 dv (1.1) ) o some for → cue Szeg¨o Kernel ecture, ( hstermfist odin hold to fails theorem this { w φ ll C C ) j . B } endby defined iu psil empty) (possibly a minus j ∞ n A =0 ⊂ z 2 Ω = (Ω) 0 ydmi Ω domain ny sacmlt orthonormal complete a is C ∈ leri boundary) algebraic n . fadol fΩ if only and if Ω prbeHletspace Hilbert eparable L 2 (Ω) O ∩ ⊂ C Ω where (Ω) C its n (1.2) We note that the definition of K(z,w) is independent of the particular orthonormal basis chosen. We shall use the notation KΩ for the Bergman kernel of Ω instead of K when we wish to emphasize that Ω is the domain under consideration. If f : Ω Ω is a biholomorphic map, then the 1 → 2 Bergman kernels for the respective domains are related by the transformation law of the Bergman kernel: ′ ′ KΩ1 (z,w)= f (z)KΩ2 (f(z), f(w))f (w). For further background on the Bergman kernel, we refer the readers to Krantz’s book [14]. If Ω is a bounded domain in Cn and v is the Lebesgue R2n-measure, then v(Ω)−1/2 A2(Ω) −1/2 −1/2 ∞ ∈ 2 and v(Ω) 2 = 1. Hence, if v(Ω) φ is a complete orthonormal basis for A (Ω), k kL { }∪{ j}j=1 then by (1.1), 1 K(z, z) , z Ω. (1.3) ≥ v(Ω) ∈ n n Equality in (1.3) is achieved for the unit ball B C , n 1 with z = 0 because KBn (z,w) = ⊂ ≥ v(Bn)−1(1 zw)−n−1. − In this paper, we completely classify the domains in C for which equality in (1.3) holds at some (minimal) point in the domain.

Theorem 1. Let Ω C be a domain. Suppose there exists a z Ω such that ⊂ 0 ∈ 1 K(z , z )= , (1.4) 0 0 v(Ω) where we use the convention v(Ω)−1 =0 if v(Ω) = . ∞ (i) If v(Ω) = , then Ω= C P where P is a possibly empty, closed polar set. ∞ \

(ii) If v(Ω) < , then Ω = D(z0,r) P where P is a possibly empty, polar set closed in the ∞ \ −1 relative topology of D(z0,r) with r = v(Ω)π . p We remark that

1. A set P is said to be polar if there is a subharmonic function u on C such that P 6≡ −∞ ⊂ z C : u(z)= . If P is a closed polar subset of a domain Ω, then A2(Ω P )= A2(Ω) {by∈ [19]. It follows−∞} that if Ω = D(z ,r) P and Ω = C P where P , i =1, 2,\ are relatively 1 0 \ 1 2 \ 2 i closed polar sets, then K extends to Ω Ω and Ωi i × i 1 KΩ1 (z0, z0)= , KΩ2 (0, 0)=0. v(Ω1)

2. Compact subsets of polar sets are totally disconnected. So the polar set P will be empty if, for instance, the boundary of Ω is parametrized by non-trivial simple closed curves.

3. We can see that in Theorem 1, Statement (i) still holds in the Riemann surface setting, since on any non-planar open Riemann surface the Bergman kernel does not vanish. However, Statement (ii) cannot be generalized to an arbitrary open Riemann surface.

2 For example, let X := X u be an open Riemann surface obtained by removing one τ,u τ \{ } single point u from a compact complex torus X := C/ (Z + τZ), for τ C and Im τ > 0. τ ∈ By the removable singularity theorem, the Bergman kernel on Xτ,u is the 2-form Kτ,u = (Im τ)−1dz dz,¯ where z is the local coordinate induced from the complex plane C (see [7,8]). ∧ Here v(Xτ,u) is precisely Im τ, the area of the fundamental parallelogram. So (1.4) holds true for X , which is not biholomorphic to D(z ,r) P . τ,u 0 \ n ′ 4. A bounded domain Ω C is called a minimal domain with a center z0 Ω if v(Ω) v(Ω ) for any biholomorphism⊂ ϕ : Ω Ω′ such that det(Jϕ(z )) = 1, where∈ Jϕ denotes≤ the → 0 Jacobian matrix of ϕ. It is known [15] that equivalently a domain Ω is minimal with the center at z0 if and only if 1 K(z, z ) , z Ω. 0 ≡ v(Ω) ∈ For more information about minimal and representative domains, see [12,13,15,22]. Theorem 1 classifies all minimal domains in C. More precisely, we have the following corollary. Corollary 2. Let Ω C be a domain. Suppose there exists a z Ω such that ⊂ 0 ∈ K(z, z ) C, for any z Ω. 0 ≡ ∈ Then C = v(Ω)−1 and (i) If C =0, then Ω= C P where P is a closed polar set. \ (ii) If C > 0, then Ω= D(z0,r) P where P is a possibly empty, polar set closed in the relative \ −1 topology of D(z0,r) with r = v(Ω)π . p Consequently, all minimal domains with the center z0 in C are disks centered at z0 possibly minus closed polar sets. In Cn, n 1, Equality (1.4) is achieved for the unit ball and more generally for complete Reinhardt domains.≥ A domain Ω in Cn is said to be complete Reinhardt if for all z =(z , ..., z ) Ω 1 n ∈ (λ z , ..., λ z ) Ω, λ 1. 1 1 n n ∈ | i| ≤ α 2 For a bounded complete Reinhardt domain, z α∈Nn is a complete orthogonal system of A (Ω). It −1 { } follows from (1.1) that KΩ(0, 0) = v(Ω) . Equality (1.4) also holds for complete circular domains (cf. [6]). When n> 1, the unit polydisk D(0, 1)n is an example of a complete Reinhardt domain which is not biholomorphic to Bn. Thus, to generalize Theorem 1 to Cn, n 1, D(z ,r) cannot simply ≥ 0 be replaced by a translation and rescaling of Bn. However, the polydisk does not have smooth boundary whereas the unit ball is strongly convex with algebraic boundary. So, we also consider whether Theorem 1 generalizes to Cn if Ω is required to be complete Reinhardt, strongly convex with algebraic boundary. The answer is no as the next theorem shows. 2 4 2 2 Theorem 3. Let Ω = z C : z1 + z1 + z2 < 1 be a domain with algebraic boundary. Then Ω is complete Reinhardt,{ ∈ strongly| | convex| | and| | not biholomorph} ic to B2. The paper is organized as follows. In Section 2, we give a proof of Theorem 1 using only complex analysis of one variable for the case where Ω has C∞-boundary. In Section 3, Theorem 1 is proved in full generality using the Suita conjecture and Corollary 2 is proved. In Section 4, Theorem 3 is proved.

3 2 Proof of C∞ boundary case of Theorem 1

The main ingredient in the proof of Theorem 1 will be (2.1). The theory that follows can be found in Bell’s book [1]. We begin by recalling the Szeg¨okernel. Let Ω be a bounded domain with C∞-smooth boundary and denote its boundary by bΩ. Then Ω is n-connected with n < and the boundary consists of n simple closed curves parametrized by C∞ functions z : [0, 1] ∞ C. Without loss of generality, let z parametrize the outermost j → n boundary curve; that is, zn parametrizes the boundary component which bounds the unbounded component of the complement of the domain. Additionally, the boundary component parametrized by zj, j =1, ..., n, is denoted by bΩj . Let T (z) denote the unit tangent vector to the boundary and 2 ds denote the arc-length measure of the boundary. Define L (bΩ) = f : bΩ C : f 2 < { → k kL (bΩ) ∞} where the norm 2 is induced by the inner product k·kL (bΩ) f,g = fg¯ ds. h i ZbΩ Let A∞(bΩ) denote the boundary values of functions in (Ω) C∞(Ω). The of bΩ denoted H2(bΩ) is the L2(bΩ) closure of A∞(bΩ). If PO: L2(∩bΩ) H2(bΩ) is the orthogonal → projection, then the Szeg¨okernel for Ω, S(z, a), is defined by 1 T (z) P (C ( ))(z)= S(z, a), a, z Ω, C (z)= a · ∈ a 2πi z a − [1, Section 7]. It can be shown that S(z, a)= S(a, z), and from the proof of the Ahlfor’s Mapping Theorem, for each a, S( , a) has n 1 zeros counting multiplicity [1, Theorem 13.1]. We note that · − the proof of the Ahlfor’s Mapping Theorem just cited requires C∞-boundary regularity. Since we will need the fact about the n 1 zeros of S( , a), we have imposed a C∞ boundary regularity assumption on Ω in this section.− ·

Let ωj be the (unique) solution to the Dirichlet boundary-value problem ∆u(z)=0 z Ω  ∈ u(z)=1 z bΩj  ∈ u(z)=0 z bΩ , k = j ∈ k 6  and define Fj : Ω C by Fj(z)=2∂ωj/∂z. Then the Bergman kernel and Szeg¨okernel are related by → n−1 2 K(z, a)=4πS(z, a) + λjFj(z) (2.1) Xj=1 where λ are constants in z and depend on a [1, Theorem 23.2]. Since ω C∞(Ω) is harmonic, j j ∈ F (Ω) C∞(Ω) A2(Ω). We now prove Theorem 1 when Ω is bounded with C∞ boundary. j ∈ O ∩ ⊂ Proof. After a translation we may assume that z = 0. Let v(Ω)−1/2 φ ∞ be a complete 0 { }∪{ j}j=1 orthonormal basis for A2(Ω). Then 1 1 ∞ = K(0, 0) = + φ (0)φ (0), v(Ω) v(Ω) j j Xj=1

4 which implies that φ (0) = 0, for all j. It follows that K(0, a) = v(Ω)−1, and for any f A2(Ω) j ∈ by the reproducing property (1.2), 1 f(0) = f(w)dv(w). v(Ω) ZΩ In particular for F , j =1, ..., n 1, j − 1 ∂ωj Fj(0) = 2 dw¯ dw 2iv(Ω) ZΩ ∂w ∧ 1 = − ωjdw¯ iv(Ω) ZbΩ 1 = − 1dw¯ iv(Ω) ZbΩj = 0.

Hence setting z = 0 in (2.1), 1 = K(0, a)=4πS2(0, a) v(Ω) Since S(0, )= S( , 0) has n 1 zeros counting multiplicity, n = 1; that is, Ω is simply-connected. · · − Let F : D(0, 1) Ω be the inverse of the Riemann map with F (0) = 0, F ′(0) > 0. By the → transformation law of the Bergman kernel,

1 F ′(z)F ′(0) = K (z, 0) = F ′(z)K (F (z), 0)F ′(0) = . π D(0,1) Ω v(Ω) So F is linear; hence Ω = D(0, F ′(0)).

3 Proof of Theorem 1

The proof of Theorem 1 in the previous section used methods specific to bounded domains with C∞-boundary and cannot be used to prove the general case where Ω C is a domain. So instead, we will use the Suita conjecture to prove the general case. ⊂ Let SH(Ω) be the set of subharmonic functions on Ω. The (negative) Green’s function of a domain Ω C is defined by ⊂ g(z,w) = sup u(z): u< 0,u SH(Ω), lim sup(u(ζ) log ζ w ) < . { ∈ ζ→w − | − | ∞}

A domain admits a Green’s function if and only if there exists a non-constant, negative subhar- monic function on Ω. In particular any bounded domain has a Green’s function. Definition 4. The Robin constant is defined by

λ(z0) = lim g(z, z0) ln z z0 . z→z0 − | − |

5 In 1972, Suita [20] made the following conjecture.

Suita Conjecture. If Ω is an open Riemann surface admitting a Green’s function, then πK(z, z) e2λ(z) ≥ and if equality holds at one point, then Ω is biholomorphic to D(0, 1) P , where P is a possibly \ empty, closed (in the relative topology of D(0, 1)) polar set. The inequality part of the Suita conjecture for planar domains was proved by Blocki [3]. See also [2]. The equality part of the Suita conjecture was proved by Guan and Zhou [10]. See also [9] for related work. We will use the following result in Helm’s book [11, Theorem 5.6.1], which is attributed to Myrberg [16]. Theorem 5. Let Ω be a non-empty open subset of R2. Then R2 Ω is not a polar set if and only \ if Ω admits a Green’s function. Proof of Theorem 1. First suppose, v(Ω) = . Since e2λ(z0) > 0, the inequality part of the Suita ∞ conjecture implies that Ω does not have a Green’s function. By Theorem 5,Ω= C P where P is a polar set. P is closed because it is the complement in C of an open set. \ Now suppose v(Ω) < . Since polar sets have two-dimensional Lebesgue measure 0, by Theorem 5, Ω admits a Green’s∞ function. As in the proof of the C∞ boundary case, after a translation, z = 0 and K ( , 0) v(Ω)−1. Let Ω = z Ω: g(z, 0) < τ . Let 0 Ω · ≡ τ { ∈ } τ−λ(0)−ǫ τ−λ(0)+ǫ r0 := e , r1 := e . Then for τ < 0 sufficiently negative, D(0,r ) Ω D(0,r ) 0 ⊂ τ ⊂ 1 (cf. [2, 4].) Hence, e−2ǫe2λ(0) e2τ e2ǫe2λ(0) . π ≤ v(Ωτ ) ≤ π Letting ǫ 0+, → e2τ e2λ(0) , as τ . v(Ωτ ) ≈ π → −∞ 2τ By Theorem 3 of [5], e is a decreasing function on ( , 0]; hence, v(Ωτ ) −∞ 1 e2λ(0) K(0, 0) = K(0, 0). v(Ω) ≤ π ≤ By the equality part of the Suita conjecture, there exists a biholomorphic map f : D(0, 1) P Ω where P is a closed polar set. After a M¨obius transformation of the unit disk, we may assume\ 0→ ′ 2 6∈ P , f(0) = 0 and f (0) > 0. Since P is removable for functions in A (D(0, 1) P ), KD(0,1)\P ( , )= \∞ · · KD(0,1)( , ) when both sides are well-defined. As in the case where Ω has C -boundary, by the transformation· · law of the Bergman kernel, f is linear. Hence Ω = D(0, f ′(0)) f(P ) and f(P ) is a closed polar set. \

6 −1 Proof of Corollary 2. Since 1 = Ω K(w, z0)dv(w), it follows that C = v(Ω) and K(z0, z0) = v(Ω)−1. The result now follows fromR Theorem 1.

4 Proof of Theorem 3

In this section, we provide a short proof of Theorem 3. Historically, biholomorphic mappings between Reinhardt domains have been an active area of research since the earliest days of several complex variables (see [17, 21]). Proof of Theorem 3. It is easy to see that Ω is complete Reinhardt with algebraic boundary. To verify that Ω is strongly convex, one lets ρ(z) = z 4 + z 2 + z 2 1 and verifies that the | 1| | 1| | 2| − real-Hessian of (ρ(z)) satisfies H wτ (ρ(z ))w> 0, z ∂Ω, w R4 0 . H 0 0 ∈ ∈ \{ } Suppose towards a contradiction that there exists an F : B2 Ω which is biholomorphic. Since the holomorphic automorphism group of B2 is transitive, we→ may suppose that 0 0. By Henri 7→ Cartan’s theorem, [18, Theorem 2.1.3.], F is linear; that is F (z)=(a1z1 + a2z2, a3z1 + a4z2). Consequently, F : bB2 bΩ. After composing with a holomorphic rotation of B2, we may also suppose F ((0, 1)) = (0,→1). Then, a =0, a = 1. Since for all θ [0, 2π], 2 4 ∈ 1 1 a a eiθ bΩ F ( , eiθ)=( 1 , 3 + ), ∋ √2 √2 √2 √2 √2 we see that a 4 a 2 a 2 1 | 1| + | 1| + | 3| + Re a , eiθ + =1, 4 2 2 h 3 i 2 which implies that a3 = 0. Thus, a 4 +2 a 2 =2. (4.1) | 1| | 1| Since (a1, 0) = F ((1, 0)) bΩ, ∈ a 4 + a 2 =1. (4.2) | 1| | 1| Equations (4.1) and (4.2) do not have a simultaneous solution. Thus, F does not exist.

Acknowledgements The authors sincerely thank L´aszl´oLempert, Song-Ying Li, and the anonymous referee for the valuable comments on this paper.

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Department of Mathematics, University of California, Irvine, CA 92697-3875, USA

Emails: [email protected] [email protected]

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