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Kyungpook Mathematical Journal 2022; 62(2): 323-331

Published online June 30, 2022

Copyright © Kyungpook Mathematical Journal.

Pseudohermitian Curvatures on Bounded Strictly Pseudoconvex Domains in 2

Aeryeong Seo

Department of Mathematics, Kyungpook National University, Daegu 41566, Republic of Korea
e-mail : aeryeong.seo@knu.ac.kr

Received: February 25, 2022; Revised: June 16, 2022; Accepted: June 17, 2022

In this paper, we present a formula for pseudohermitian curvatures on bounded strictly pseudoconvex domains in 2 with respect to the coefficients of adapted frames given by Graham and Lee in [3] and their structure equations. As an application, we will show that the pseudohermitian curvatures on strictly plurisubharmonic exhaustions of Thullen domains diverges when the points converge to a weakly pseudoconvex boundary point of the domain.

Keywords: Pseudohermitian manifolds, Thullen domains

Let (M,𝜃) be a pseudohermitian manifold of real dimension 2n+1, i.e. M is an integrable, nondegenerate, real hypersurface in n+1 with CR structure HMTM so that 𝜃 is a 1-form satisfying θ(X)=0 for any XHM and d𝜃 can be expressed by

dθ=ihαβ¯θαθβ¯

where (hαβ¯) is a positive definite n×n matrix with respect to a local basis θ1,,θn for HM. The equivalence problem for such pseudohermitian structures was studied first by Chern using Cartan's method and later it turned out that it was related to the pseudoconformal invariants analyzed by Chern-Moser [1] and Tanaka [4]. In [5], Webster showed that there exists a natural connection in the bundle H1,0M adapted to a pseudohermitian structure, where H1,0M denotes the eigenspace of the endomorphism J:HMHM satisfying JJ=I, which defines the CR structure of M, with an eigenvalue 1. Moreover he gave an expression of pseudoconformal curvature tensor in terms of pseudohermitian curvature tensor.

In this paper, we present an expression of pseudohermitian curvatures on bounded strictly pseudoconvex domains in 2 with respect to the coefficients of adapted frames given by Graham-Lee in [3] and their structure equations. As an application, we will show that the pseudohermitian curvatures on strictly plurisubharmonic exhaustions of Thullen domains diverge when the point converges to the weakly pseudoconvex boundary point of the domain. Our main result is

Main Theorem. For a Thullen domain

Ωm:={(z1,z2)2:|z1|2m+|z2|2<1}

with m>1, the pseudohermitian curvatures of the exhaustions

Ωm,ϵ:={(z1,z2)2:ϕ(z1,z2)=ϵ}

of Ωm with ϕ=K(z,z)m/(2m+1) diverge as the point tends to weakly pseudoconvex boundary points along the complex line {z1=0} in Ωm.

In this section, we calculate the pseudohermitian curvature explicitly in terms of the coefficient of an adapted coframe for a strictly pseudoconvex bounded domain Ω:={(z1,z2)2:ϕ(z1,z2)<0} where ϕ is a defining function of Ω. Let {θ0,θ1} be the adapted coframe and {X0,X1} its dual frame given by Graham-Lee ([3]) satisfying the following.

dθ0=θ1θ1¯rθ0θ0¯dθ1=θ1ωiAθ0θ1¯+r2θ0θ1+r2θ0¯θ1X1rθ0θ0¯dω=Rθ1θ 1¯ +iX 1¯A2ω(X 1¯A ¯)32(X1r)θ1θ 0¯   iX1A ¯2ω(X1A)+32(X 1¯ r)θ0θ 1¯ +32(X1r)θ0θ1  32(X 1¯ r)θ 0¯ θ 1¯ +12Δbr|A|2θ0θ 0¯

Here R denotes the pseudohermitian curvature and Δb:=rααrαα denotes the sublaplacian of the pseudohermitian structure. In coordinates of 2, let

θ0:=A1dz1+A2dz2,θ1:=B1dz1+B2dz2,ω:=C1dz1+C2dz2C1¯d z¯ 1C2¯d z¯ 2,

and

X0=1DB2z1B1z2,X1=1DA2z1+A1z2

where D:=A1B2A2B1 and Aj:=ϕzj for each j=1,2.

Proposition 2.1. Let Ω={(z1,z2)2:ϕ(z1,z2)<0} be a strictly pseudoconvex domain with a defining function ϕ. Then the pseudohermitian curvature is given by

R=1|B1|22Rei(X1A¯2ω(X 1 ¯ A¯))+32(X 1 ¯ r)A1B1¯C1 z ¯ 112Δbr|A|2|A1|2.

Proof. By (2.1), we have

dθ0=θ1θ1¯rθ0θ0¯  =(B1dz1+B2dz2)(B1dz1+B2dz2¯)  r(A1dz1+A2dz2)(A1dz1+A2dz2¯)  =(|B1|2+r|A1|2)dz1d z¯ 1(B1 B¯ 2+rA1 A¯ 2)dz1d z¯ 2  (B2 B¯ 1+rA2 A¯ 1)dz2d z¯ 1(|B2|2+r|A2|2)dz2d z¯ 2.

On the other hand, since

dθ0=d(A1dz1+A2dz2)  =A1 z¯1dz1d z¯1A1 z¯2dz1d z¯2A2 z¯1dz2d z¯1A2 z¯2dz2d z¯2,

by comparing equations (2.3) and (2.4)we have

|B1|2+r|A1|2=A1z¯1,
B1 B¯2+rA1 A¯2=A1z¯2,

and

|B2|2+r|A2|2=A2z¯2.

Since we have

|B1|2|B2|2=A1z¯1r|A1|2A2z¯2r|A2|2    =A1z¯2rA1 A¯2A1 ¯ z2rA2 A¯1

by (2.5), (2.7), (2.6), we obtain

r=|A1z¯2|2A1z¯1A2z¯2A1 A¯2A1¯ z2+A2 A¯1A1z¯2|A2|2A1z¯1|A1|2A2z¯2.

By substituting (2.8) into (2.5) and (2.7), we have

|B1|2=A1A2¯A1 z¯ 1A1 ¯z2+A2A1¯A1 z¯ 1A1 z¯ 2|A2|2A1 z¯1 2|A1|2|A1 z¯ 2|2A1 A¯ 2A1 ¯z2+A2 A¯ 1A1 z¯ 2|A2|2A1 z¯ 1|A1|2A2 z¯ 2,

and

|B2|2=A1 A¯2A1 ¯ z2A2 z¯2+A2 A¯1A1 z¯2A2 z¯2|A1|2A2 z¯2 2|A2|2|A1 z¯2|2A1 A¯2A1 ¯ z2+A2 A¯1A1 z¯2|A2|2A1 z¯1|A1|2A2 z¯2.

By (2.1) and a straightforward calculation we obtain

dθ1=θ1ωiAθ0θ1¯+r2θ0θ1+r2θ0¯θ1X1rθ0θ0¯=(B1dz1+B2dz2)(C1dz1+C2dz2C1¯d z¯ 1C2¯d z¯ 2)iA(A1dz1+A2dz2)(B1dz1+B2dz2)¯+r2(A1dz1+A2dz2)(B1dz1+B2dz2)+r2(A1dz1+A2dz2)¯(B1dz1+B2dz2)X1r(A1dz1+A2dz2)(A1dz1+A2dz2)¯=B1C2B2C1+r2A1B2r2A2B1dz1dz2+B1C1¯iAA1B1¯r2A1¯B1(X1r)|A1|2dz1d z¯ 1+B1C2¯iAA1B2¯r2A2¯B1(X1r)A1A2¯dz1d z¯ 2+B2C1¯iAA2B1¯r2A1¯B2(X1r)A2A1¯dz2d z¯ 1+B2C2¯iAA2B2¯r2A2¯B2(X1r)|A2|2dz2d z¯ 2.

On the other hand, since

dθ1=d(B1dz1+B2dz2)  =B1 z¯1dz1d z¯1B1z2dz1dz2B1 z¯2dz1d z¯2  B2z1dz2vdz1B2 z¯1dz2d z¯1B2 z¯2dz2d z¯2,

by comparing equations (2.11) and (2.13), we obtain

B1C2B2C1+r2A1B2r2A2B1=B1z2+B2z1,
B1C1¯+iAA1B1¯+r2A1¯B1+(X1r)|A1|2=B1 z¯ 1,B1C2¯+iAA1B2¯+r2A2¯B1+(X1r)A1A2¯=B1 z¯ 2,B2C1¯+iAA2B1¯+r2A1¯B2+(X1r)A2A1¯=B2 z¯ 1,

and

B2C2¯+iAA2B2¯+r2A2¯B2+(X1r)|A2|2=B2z¯2.

By considering B2×(2.14)B1×(2.15), we have

A=1iB1¯DB1z¯ 1B2B2z¯ 1B1(X1r)A1¯D.

By substituting (2.16) into (2.14), we obtain

C1¯=1B1B1z¯11DA1B1B1z¯1B2B2z¯1B1r2A1¯.

Similarly by considering B2×(2.15)B1× (2.15) we have

A=1iB2¯DB1z¯ 2B2B2z¯ 2B1(X1r)A2¯D

and by substituting (2.17) into (2.15), we have

C2¯=1B1B1z¯21DA1B1B1z¯2B2B2z¯2B1r2A2¯.

Now using the expression (2.1) and (2.2), we obtain

dω=R(B1dz1+B2dz2)(B1dz1+B2dz2)¯  +iX 1¯A2ω(X 1¯A¯)32(X1r)(B1dz1+B2dz2)(A1dz1+A2dz2)¯  iX1A¯2ω(X1A)+32(X 1¯ r)(A1dz1+A2dz2)(B1dz1+B2dz2)¯  +32(X1r)(A1dz1+A2dz2)(B1dz1+B2dz2)  32(X 1¯ r)(A1dz1+A2dz2)¯(B1dz1+B2dz2)¯  +(Δbr|A|2)(A1dz1+A2dz2)(A1dz1+A2dz2)¯  =d(C1dz1+C2dz2C1¯d z¯ 1C2¯d z¯ 2).

By comparing the coefficient of dz1d z¯1 in the equation (2.18), one has

C1z¯1C1¯z1=R|B1|2+iX 1¯A2ω(X 1¯ A ¯ )32(X1r)B1A1¯      iX1 A ¯ 2ω(X1A)+32(X 1¯r)A1B1¯+12Δbr|A|2|A1|2.

As a result, we complete the proof.

For m, let

Ωm:={(z1,z2)2:|z1|2m+|z2|2<1}

be the Thullen domain in 2 with m1. The diagonal of the Bergman kernel of Ωm is given by

K(z,z)=1mπ2(m+1)1|z2|21/m(1m)|z1|21|z2|221/m1|z2|2 1/m|z1|23.

For the detail, see [2]. Then ϕ(z,z¯):=K(z,z)m/(2m+1) gives a defining function of Ωm.

For a point aΔ:={z:|z|<1}, denote by ψa an automorphism of Ωm obtained by

ψa:(z1,z2)(1|a|2)1/2mz1 (1a¯z2)1/m,z2 a1a¯z2 .

Note that ψa is the inverse of ψa.

Then we have

Aut(Ωm)={(z1,z2)eiθ1 (1|a|2)12mz1 (1a¯z2)1m,eiθ2 z2 a1a¯z2 :θ1,θ2,aΔ}

and W:={(0,w):|w|=1}Ωm is the set of weakly pseudoconvex boundary points of Ωm for m>1. Note that ΩmW is the set of strictly pseudoconvex boundary points.

From now on, we will find the pseudohermitian curvature of the exhaustion

Ωm,ϵ:={(z1,z2)2:ϕ(z1,z2)=ϵ}

with ϵ1. In particular, we will observe the behavior of the pseudohermitian curvature when (0,z2)(0,w) with |w|=1, i.e. weakly pseudoconvex points of Ωm. At z=(0,z2), we have the following:

K(z,z)=1π2m+1m1(1|z2|2)2+1m,ϕ= 1 π 2 m+1 m m2m+1(1|z2|2),A1=A2z1=A1z2=A2 z¯1=A1 z¯2=A1z1=2A1( z¯1)2=0A2=ϕ z¯21|z2|2= 1 π 2 m+1 m m2m+1 z¯2,A2 z¯2= 1 π 2 m+1 m m2m+1.

Hence at z=(0,z2), by straightforward calculation we have

r=1|A2|2A2z¯2= 1 π 2 m+1m m2m+11|z2|2,|B1|2=A1z¯1=2ϕz1z¯1=2mm+11(1| z2|2)1/mϕ,B2=0,A=1iA2B1¯B1z¯1,C1¯=1B1B1z¯1,C2¯=1B1B1z¯2r2A2¯,R=2|B1|2ReC1z¯1.

By differentiating the equation (2.5) with z¯1, we obtain

B1z¯1B1¯+B1B1¯z¯1=z¯1A1z¯1r|A1|2      =2A12z¯1rz¯1|A1|2r|A1|2z¯1.

Since A1=0 and A1z1=0 at z=(0,z2), we have

B1z¯1B1¯+B1B1¯z¯1=0,

and hence

1B1B1z¯1=1B1¯B1¯z¯1.

By differentiating the equation (3.2) with z1, we obtain

2B1z1 z¯1B1¯+ B 1 z ¯ 1 2+ B 1 z 1 2+B12B1¯z1 z¯1=3A1z12 z¯12rz1 z¯1|A1|2r z¯1|A1|2z1rz1|A1|2 z¯1r2|A1|2z1 z¯1.

Evaluating the above equation at z=(0,z2) gives

2Re1B12B1z1 z¯1+21B1B1 z¯12=1|B1|23A1z12 z¯1rA1 z¯1 2.

Since

C1¯z1=z11B1B1 z¯ 11DA1B1B1 z¯ 1B2B2 z¯ 1B1r2A1¯  =1B12B1z1B1 z¯ 1+1B12B1z1 z¯ 1r2A1¯z1  = 1 B 1 B 1 z 1 2+1B12B1z1 z¯ 1r2A1¯z1,

by Proposition 2.1 and equations (3.3), (3.1) we obtain

R=2|B1|21B1 B1z¯1 2+Re1B12B1z1 z¯1r2A1¯z1=1|B1|21|B1|23A1z12 z¯1rA1 z¯1 2rA1¯z1=1|B1|43A1z12 z¯1+2r1|B1|2A1 z¯1=1|B1|44ϕ2z12 z¯1+2r,
ϕz1=ϕz1log(ϕ)=pϕz1logK  =pϕz1log(m+1)1|z2|21/m(1m)|z1|2  +3pϕz1log1|z2|21/m|z1|2  =pϕz¯1(1m)(m+1)(1|z2|2)1m (1m)|z1|2+3(1|z2|2)1m |z1|2

with p=m2m+1, and

2ϕ2z1=p2z¯12ϕ (1m) (m+1) (1| z 2 | 2) 1m (1m)| z 1 | 2 +3 (1| z 2 | 2) 1m | z 1 | 2 2pϕz¯12(1m)2 (m+1) (1| z2 |2 )1 m (1m)|z1 |2 2+3 (1| z2 |2 )1 m |z1 |2 2.

Let

M:=(1m)(m+1)(1|z2|2)1m(1m)|z1|2+3(1|z2|2)1m|z1|2

and

N:=(1m)2(m+1)(1|z2 |2 )1m (1m)|z1|22+3(1|z2 |2 )1m |z1|22.

Then

2ϕ2z1=p2z¯12M2ϕpϕz¯12N.

Hence at z=(0,z2), we have

4ϕ2z12 z¯1=2pϕ(pM2N)

and by (3.4)

R=(3m+1)(m1)m(2m+1)1ϕ+2r.

As a result we obtain our main theorem.

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