an explicit formula for the webster torsion of a pseudo
TRANSCRIPT
An explicit formula for the Webster torsion ofa pseudo-hermitian manifold and its
application to torsion-free hypersurfaces
Song-Ying Li and Hing-Sun Luk
Revised by June 16, 2006
To Sheng Gong, on his seventy-fifth birthday
1 Introduction
Let M be a (2n + 1)-dimensional CR manifold with CR dimension n. LetH(M) be the holomorphic tangent bundle on M . We say that M is a strictlypseudoconvex pseudo-hermitian manifold in the sense of Webster [21] if thereis a real one-form θ so that θ(X) = 0 for all X ∈ H(M), and local complexone-forms {θ1, · · · , θn} which form a basis for H∗(M) with
(1.1) dθ = in∑
α,β=1
hαβ θα ∧ θβ
and (hα β) positive definite. (hαβ) is determined by the Levi-form Lθ withrespect to θ. Then
(1.2) dθα =n∑
γ=1
θγ ∧ ωαγ + θ ∧ τα
where τα are in the linear span of θ1, · · · , θn, and ωβα are 1-forms satisfying
(1.3) 0 = dhαβ − hγβωγα − hαγω
γ
β= dhαβ − ωαβ − ωβα.
1
In particular, if hαβ = δα β then
(1.4) 0 = dhαβ − ωαβ − ωβα = −ωβα − ωα
β, or ωβ
α = −ωαβ.
Let
(1.5) τα = hαγτγ =
n∑β=1
Aαβθβ
Then the torsion on M is defined (see (2.14) below) as
(1.6) Tor(u,v)(z) = i(Aαβuαvβ − Aαβuαvβ)
where u =∑n+1
j=1 uj ∂∂zj
, v =∑n+1
j=1 vj ∂∂zj
∈ Hz(M) and z ∈ M .
Analysis and geometry on a strictly pseudoconvex pseudo-hermitianmanifolds have been studied by many authors (see, for examples, [1], [2],[3], [4], [5], [7], [8, 9, 10], [11, 12], [13], [15], [17], [19], [20, 21] and refer-ences therein). From geometric point of view, it is important to understandcurvatures and torsion. In [16], the authors give an explicit formula for theWebster pseudo Ricci curvature for hypersurfaces and provide an applicationfor the characterization of the ball using the boundary pseudo scalar curva-ture. On many occasions, it is important to find an easier way to computethe Webster torsion. Various formulations through local coordinates weregiven in [20], [11, 12]. The first purpose of this paper is to give an explicitformula for the torsion of a real hypersurface in Cn+1, which can be computedglobally in terms of the defining function ρ. In other words, we shall provethe following theorem.
THEOREM 1.1 Let M be a C4 strictly pseudoconvex hypersurface inCn+1. Let ρ be a defining function for M which is C3 in a neighborhoodof M . Consider the pseudo-hermitian structure defined by θ = 1
2i(∂ρ − ∂ρ)
on M . Then for any u =∑n+1
j=1 uj ∂∂zj
and v =∑n+1
j=1 vj ∂∂zj
∈ Hz(M) we have
(1.7) Tor(u,v)(z) = −det H(ρ)
J(ρ)2 Re
n+1∑q=1
u(ρq)v(ρq)(z), z ∈ M.
Moreover,
(1.8) Tor(u,v) =2
J(ρ)Re
n+1∑p,q=1
upvq(N − det H(ρ)
)(ρpq).
2
where
(1.9) N = det H(ρ)n+1∑`=1
ρ` ∂
∂z`
, and J(ρ) = − det[
ρ ∂ρ(∂ρ)∗ H(ρ)
].
Note: In the definition of ρp appeared in (1.7), we require that H(ρ) isinvertiable at z. However, the operator N in (1.8) is well-defined whether
H(ρ) = [ ∂2ρ∂zj∂zk
] is invertiable on M or not since [det H(ρ)ρpq] is well defined
as the transpose of the adjoint matrix of H(ρ). Moreover, N(ρ) = J(ρ).(Some computations can be found in [13, 14]).
In [11], Lee gave the general formula for the transformation of the torsionunder a conformal change of the contact form in terms of covariant deriva-tives of the conformal factor. We will compute the covariant derivatives inthe hypersurface case, isolating the role of the defining function, and give atransformation formula for the torsion under a conformal change. The secondpurpose of this paper is to apply our formula and obtain the explicit iden-tification of the conformal class of the torsion-free hypersurface (∂Bn+1, θ),where Bn+1 is the unit ball in Cn+1. Let
(1.10) θ0 =1
2i(n+1∑j=1
zjdzj −n+1∑j=1
zjdzj)
be the standard pseudo-hermitian structure for ∂Bn+1, which is torsion-free.Then we will prove the following theorem.
THEOREM 1.2 Let h ∈ C3(∂Bn+1) be positive. Then (∂Bn+1, hθ0) istorsion-free if and only if the harmonic extension of 1/h is a quadratic poly-nomial.
Examples of torsion-free (normal or Sasakian) pseudo-hermitian mani-folds were given by N. Tanaka in [19]; some sufficient conditions in terms ofG−homogeneous structures were given in Musso [18]. We will provide somemore examples here.
A real ellipsoid
(1.11) E(a, b) = {z ∈ Cn+1 : ρ(z) =n+1∑j=1
ajx2j + bjy
2j − 1 = 0}, aj, bj > 0
3
is a very typical example of strictly pseudoconvex hypersurfaces which wasstudied by S. Webster [21], who proves that (E(a, b), θ) with θ = (∂ρ −∂ρ)/(2i) has vanishing Chern-Moser pseudo conformal curvature if and onlyif E(a, b) is complex linearly equivalent to the boundary of unit ball Bn+1.The third purpose of this paper is to show that torsion-free real ellipsoid asabove must be CR equivalent to the unit sphere of the same dimension. Inparticular, we will prove the following theorem.
THEOREM 1.3 Any real ellipsoid (E(a, b), θ) in Cn+1 is torsion-free if andonly if it is the unit sphere after a complex linear change of variables.
The paper is organized as follows. In section 2, we will prove Theorem1.1. The transformation formula for the torsion under a conformal changewill be given in Section 3. In Section 4, we will provide some examples oftorsion-free or non-torsion-free real hypersurfaces. The proofs of Theorems1.2 and 1.3 will be given in section 5.
2 Proof of Theorem 1.1
Let u be a real C2 function on an open neighborhood D of M in Cn+1 andH(u) = [ ∂2u
∂zi∂zj ](n+1)×(n+1) be the complex hessian matrix of u. If H(u) is
invertible then we let [ukj] be the inverse of H(u)t so that
(2.1)n+1∑j=1
ukjupj =n+1∑k=1
u`ku`p = δkp.
Let uk = ∂u∂zk , uj = ∂u
∂zj and
(2.2) |∂u|2 = ukjukuj, uj =n+1∑k=1
ujkuk and uj =n+1∑k=1
ukjuk.
Using this notation for ρ (a defining function for M), we write on M
(2.3) θ = −i∂ρ = i∂ρ, dθ = in+1∑k,`=1
ρk`θk ∧ θ`.
4
where
(2.4) θj = dzj − ihjθ, hj = ρj/|∂ρ|2.
We have
(2.5)n+1∑j=1
ρjθj = 0.
Let M1 = {z ∈ M : ρn+1(z) 6= 0}. It was proved in [21] as well as in [16] thaton M1,
(2.6) dθ = in∑
α,β=1
hαβ θα ∧ θβ
where
(2.7) hαβ = ραβ −ρn+1 β ρα
ρn+1
−ραn+1 ρβ
ρn+1
+ ρn+1n+1
ραρβ
|ρn+1|2.
Let
(2.8) Yα =∂
∂zα− ρα
ρn+1
∂
∂zn+1, Y = i
n+1∑j=1
(hj∂j − hj∂j).
Then for any f ∈ C1(M1) we have
(2.9) df =n+1∑j=1
fjdzj + fjdzj =n∑
α=1
(Yαf θα + Yαf θα) + Y (f)θ.
Notice that since θα = dzα − ihαθ, we have
(2.10) dθα = −idhα ∧ θ − ihαdθ =n∑
γ=1
θγ ∧ ωαγ + θ ∧ τα
It was proved in [21] and in (2.16)–(2.21) of [16] with the current notation,that ωα
β can be chosen uniquely so that
(2.11) dhαβ − hγβωγα − hαγω
γ
β= 0
5
and
τα = in∑
β=1
Yβhαθβ.
Then
(2.12) τα = hαγτγ = −i
n∑β=1
n∑γ=1
hαγYβhγθβ =n∑
β=1
Aαβθβ
with
(2.13) Aαβ = (−i)n∑
γ=1
hαγYβ hγ.
The torsion of M with respect to θ is defined as follows:
(2.14) Torθ(zαYα, wβYβ) = i(Aαβzαwβ − Aαβzαwβ).
For any u =∑n+1
j=1 uj ∂∂zj
∈ Hz(M), we have
(2.15) u =n∑
α=1
uαYα +n+1∑j=1
uj ρj
ρn+1
∂n+1 =n∑
α=1
uαYα.
Thus for any u,v ∈ Hz(M), we have
(2.16) Torθ(u,v)(z) = in∑
α,β=1
(Aαβuαvβ − Aαβuαvβ).
The main point of this section is to give an explicit formula for computingthe torsion; in other words, to give a proof of Theorem 1.1.
It was proved in Section 2 in [16] that
(2.17)n∑
γ=1
hβγhγ = −Yβ log ρn+1.
Then
(2.18) Aαβ
6
= −in∑
γ=1
hαγYβ(hγ)
= −in∑
γ=1
hαγYβ(ργ
|∂ρ|2)
= −in∑
γ=1
hαγYβ(ργ)
|∂ρ|2+ i
Yβ(|∂ρ|2)|∂ρ|2
n∑γ=1
hαγhγ
= − i
|∂ρ|2n∑
γ=1
hαγ(ρkYβ(ρkγ) + ρkγYβ(ρk))− iYβ(log |∂ρ|2)Yα log ρn+1.
Since
n∑γ=1
hαγρkγ =
n∑γ=1
(ραγ −ραρn+1γ
ρn+1
−ραn+1ργ
ρn+1
+ρn+1n+1
|ρn+1|2ραργ)ρ
kγ
= δαk − ραn+1ρkn+1 − ρα
ρn+1
δkn+1 +ραρn+1n+1
ρn+1
ρkn+1
−ραn+1ργ
ρn+1
ρkγ +ρn+1n+1
|ρn+1|2ραργρ
kγ
= δαk −ραn+1
ρn+1
ρk − ρα
ρn+1
δkn+1 +ραρn+1n+1
|ρn+1|2ρk
= δαk − ρkYα(log ρn+1)−ρα
ρn+1
δkn+1,
(2.19)n∑
γ=1
hαγρkγYβ(ρk)
=n+1∑k=1
(δαk − ρkYα(log ρn+1)−ρα
ρn+1
δkn+1)Yβ(ρk)
= Yβ(ρα)− Yα(log ρn+1)n+1∑k=1
ρkYβ(ρk)− ραYβ(log ρn+1).
Notice that
(2.20) Yβ(ρkγ) = −n+1∑
p,q=1
ρkqρpγYβ(ρpq),
7
hence
(2.21)n+1∑k=1
n∑γ=1
hαγρkYβ(ρkγ)
= −n+1∑k=1
ρk
n∑γ=1
hαγ
( n+1∑p,q=1
ρkqρpγYβ(ρpq))
= −n+1∑
p,q,k=1
ρkρkqYβ(ρpq)
n∑γ=1
hαγρpγ
= −n+1∑
p,q=1
ρqYβ(ρpq)(δαp − ρpYα(log ρn+1)−ρα
ρn+1
δpn+1)
= −n+1∑q=1
ρqYβ(ραq) +n+1∑
p,q=1
ρpρqYβ(ρpq)Yα log ρn+1 +n+1∑q=1
ρqYβ(ρn+1q)ρα
ρn+1
.
Since
(2.22)n+1∑
p,q=1
ρpρqYβ(ρpq)−n+1∑k=1
ρkYβ(ρk) + Yβ(|∂ρ|2)
=n+1∑
p,q=1
ρpρqYβ(ρpq)−n+1∑
p,q=1
ρpYβ(ρpqρq) + Yβ(|∂ρ|2)
= −n+1∑q=1
ρqYβ(ρq) + Yβ(|∂ρ|2)
=n+1∑q=1
ρqYβ(ρq)
=n+1∑q=1
ρq(ρβq −ρβ
ρn+1
ρn+1q)
= ρβ −ρβ
ρn+1
ρn+1
= 0,
combining (2.18), (2.19), (2.21) and (2.22), we have
Aαβ
=i
|∂ρ|2[ n+1∑
q=1
ρqYβ(ραq)−n+1∑
p,q=1
ρpρqYβ(ρpq)Yα log ρn+1 −n+1∑q=1
ρqYβ(ρn+1q)ρα
ρn+1
]
8
− i
|∂ρ|2[Yβ(ρα)− Yα(log ρn+1)
n+1∑k=1
ρkYβ(ρk)− ραYβ(log ρn+1)]
−iYβ(log |∂ρ|2)Yα(log ρn+1)
=i
|∂ρ|2[ n+1∑
q=1
ρqYβ(ραq)−n+1∑q=1
ρqYβ(ρn+1q)ρα
ρn+1
− Yβ(ρα) +ρα
ρn+1
Yβ(ρn+1)]
− i
|∂ρ|2( n+1∑
p,q=1
ρpρqYβ(ρpq)−n+1∑k=1
ρkYβ(ρk) + Yβ(|∂ρ|2))Yα log ρn+1
=i
|∂ρ|2[Yβ(ρα)−
n+1∑q=1
ραqYβ(ρq)− Yβ(ρn+1)ρα
ρn+1
+n+1∑q=1
ρn+1qYβ(ρq)ρα
ρn+1
− Yβ(ρα) +ρα
ρn+1
Yβ(ρn+1)]
=i
|∂ρ|2[−
n+1∑q=1
ραqYβ(ρq) +n+1∑q=1
ρα
ρn+1
∂ρq
∂zn+1
Yβ(ρq)]
= − i
|∂ρ|2n+1∑q=1
Yα(ρq)Yβ(ρq)
Notice that for any w =∑n+1
j=1 wj ∂∂zj
∈ Hz(M), we have w =∑n
α=1 wαYα.
Therefore, for any u,v ∈ Hz(M), we have
(2.23) Torθ(u,v) = −2Re[i
n∑α,β=1
Aαβuαvβ]
= − 2
|∂ρ|2Re (
n+1∑q=1
u(ρq)v(ρq)).
Notice that on M
(2.24) J(ρ) = |∂ρ|2 det H(ρ).
Hence (2.23) implies that (1.7) holds.
To prove (1.8), we notice that
(2.25)n+1∑q=1
u(ρq)ρpq =
n+1∑j=1
ujδpj = up.
and
(2.26)n+1∑q=1
u(ρq)v(ρq) =n+1∑p=1
n+1∑q=1
u(ρq)v(ρpqρp)
9
=n+1∑p=1
n+1∑q=1
u(ρq)ρpqv(ρp) +
n+1∑p=1
n+1∑q=1
u(ρq)ρpv(ρpq)
=n+1∑p=1
upv(ρp)−n+1∑p=1
n+1∑q=1
u(ρq)ρp
n+1∑k,`=1
ρkqρp`v(ρk`)
=n+1∑
p,q=1
ρpqupvq −
n+1∑k,`=1
ukρ`v(ρk`)
=n+1∑
p,q=1
ρpqupvq −
n+1∑k,`,p=1
ukρ`vpρpk`
=n+1∑
p,q=1
ρpqupvq − 1
det H(ρ)
n+1∑k,p=1
ukvpN(ρpk)
where N is defined by (1.9). Therefore,
(2.27) Torθ(u,v) =2
J(ρ)Re
n+1∑p,q=1
upvq(N − det H(ρ)
)(ρpq).
This proves (1.8). Therefore, the proof of Theorem 1.1 is complete.
3 Torsion under conformal change of θ
The purpose of this section is to consider the transformation formula of thetorsion of a hypersurface when its pseudo-hermitian structure is changed interms of the defining function. A general transformation formula for torsionin terms of covariant derivatives was obtained by J. Lee in [11]. Here we letr(z) = h(z)ρ(z) with h(z) > 0, and consider
(3.1) θρ =1
2i(∂ρ− ∂ρ), θr =
1
2i(∂r − ∂r).
Then θr = hθρ. In order to derive a formula for the torsion transformation,we first try to understand the relation between Nr and Nρ.
Lemma 3.1 If r(z) = h(z)ρ(z) with J(r) > 0, then
(3.2) T ρ,h = h−n−1N r −Nρ
is a tangent vector field in H(M).
10
Proof. Let [Bij(r)] be the co-factor matrix of H(r) so that
(3.3)n+1∑j=1
Bij(r)rpj = det H(r) δip.
Since, on M , we have
(3.4) rpq = ρphq + ρqhp + hρpq,
(3.5) N r =n+1∑
p,q=1
Bpq(r)rp∂
∂zq
= hn+1∑
p,q=1
Bpq(r)ρp∂
∂zq
.
For z0 ∈ M , by a complex rotation, we may assume that zn+1 is complexnormal at z0. Thus ρα(z0) = 0 for all 1 ≤ α ≤ n. Moreover, at z = z0 wehave
(3.6) Bn+1n+1(r) = hnBn+1n+1(ρ)
and
(3.7) Bn+1β(r)(z0) = hnBn+1β(ρ)− hn−1n∑
α=1
hαρn+1(−1)α+β det Hn(ρ)αβ.
Thus
T ρ,h = h−n−1N r −Nρ = −|ρn+1|2h−1n∑
α,β=1
(−1)α+β det Hn(ρ)αβ hα∂
∂zβ
is a tangent vector in Hz0(M). Therefore, the proof of the lemma is complete.
For convenience, we introduce the following notation:
(3.8) TORρ(u,v) =2
J(ρ)
n+1∑p,q=1
upvq(Nρ − det H(ρ))ρpq(z).
We will prove the following theorem.
11
THEOREM 3.2 If ρ is a defining function for M and r = hρ with h > 0and J(r) > 0, then for any u,v ∈ Hz(M),
(3.9) TORr(u,v)(z) = TORρ(u,v)(z)− 2hn+1∑
p,q=1
upvq ∂2
∂zp∂zq
(1
h
)
+2
J(ρ)
n+1∑p,q=1
upvq(−Nρ log h + T ρ,h
)ρpq
Note: The relation between Torθr(u, v)(z) = Re TORr(u, v)(z) andTorθρ(u, v) follows by taking real parts on both sides of (3.9).
Proof. Notice that
(3.10) Nr(r) = J(r) = hn+2J(ρ) = hn+2Nρ(ρ).
Also
hN r(ρp) = N r(hρp)− ρpN r(h)
= N rrp −N r(ρhp)− ρpN r(h)
= (det H(r)− N r(h)
h)rp − hpN r(ρ)
and
N rrpq
= N r(hpqρ + hpρq + hqρp + hρpq)
=hpq
hN rr + N r(hp)ρq + hpN r(ρq) + N r(hq)ρp + N r(ρp)hq + N r(h)ρpq + hN r(ρpq)
Thus for any u,v ∈ Hz(M) we have
n+1∑p,q=1
upvq(N r − det H(r))rpq(z)
=n+1∑
p,q=1
upvq (− det H(r)rpq + N rrpq)
12
=n+1∑
p,q=1
upvq(− det H(r)hρpq + N r(r)hpq
h+ hpN r(ρq) + hqN r(ρp) + N r(h)ρpq + hN r(ρpq))
= hn+1∑
p,q=1
upvq(
(− det H(r) + N r(log h))ρpq + N rρpq − 2hn−1Nρ(ρ)hphq + hnNρ(ρ)hpq
)
= hn+1∑
p,q=1
upvq(− det H(r) + N r(log h) + N r
)ρpq
+hnNρ(ρ)n+1∑
p,q=1
upvq(− 2hphq + hhpq
)
At z0, by a complex rotation, we may assume that ρα(z0) = 0 for all1 ≤ α ≤ n. Then
det H(r)−N r(log h)(z0)
=n+1∑β=1
rn+1βBn+1β −Bn+1βrn+1∂β log h
=n+1∑β=1
Bn+1β(hρn+1β + ρn+1hβ + hn+1ρβ − ρn+1hβ)
=n+1∑β=1
Bn+1β(r)(hρn+1β + ρβhn+1)
= hBn+1n+1(r)ρn+1n+1 + hn∑
β=1
Bn+1β(r)ρn+1β + hnBn+1n+1(ρ)ρn+1hn+1
= hn+1 det H(ρ)− hnn∑
α,β=1
(−1)α+β det Hn(ρ)αβhαρn+1ρn+1β + hnBn+1n+1(ρ)ρn+1hn+1
= hn+1 det H(ρ) + hnρn+1
n∑α=1
Bαn+1(ρ)hα + hnBn+1n+1(ρ)ρn+1hn+1
= hn+1 det H(ρ) + hnn+1∑
α,β=1
ρβBαβ(ρ)∂
∂zα
h
= hn+1 det H(ρ) + hnNρh
Therefore,
TORr(u, v)(z)
13
=2hn+2
J(r)
n+1∑p,q=1
upvq(− det H(ρ)ρpq − h−1Nρhρpq + Nρρpq − (Nρ − h−n−1N r)ρpq
)
+2hnJ(ρ)
J(r)
n+1∑p,q=1
upvq(− 2hphq + hhpq
)
=2
J(ρ)
n+1∑p,q=1
upvq(Nρ − det H(ρ))ρpq(z)
+2
J(ρ)
n+1∑p,q=1
upvq(−Nρ log h + T ρ,h
)ρpq − 2h
n+1∑p,q=1
upvq ∂2
∂zp∂zq
(1
h
)
= TORρ(u, v)(z) +2
J(ρ)
n+1∑p,q=1
upvq(T ρ,h −Nρ log h
)ρpq − 2h
n+1∑p,q=1
upvq∂pq
(1
h
).
Therefore, the proof of the theorem is complete.
4 Examples of torsion-free real hypersurfaces
in Cn+1
In this section, we want to understand what kind of pseudo-hermitian man-ifolds (M, θ) of hypersurface type with zero torsion. We first point out asimple perturbation of the unit sphere in Cn+1 which is not torsion-free.
By Theoreon 1.1, Tor(u,v)(z) = 0 for all u,v ∈ Hz(M) and z ∈ M ifand only if
(4.1) Ren+1∑
p,q=1
upvq(N − det H(ρ)
)(ρpq) = 0 for all u,v ∈ Hz(M), z ∈ M.
If we replace u by iu then (4.1) implies that
(4.2) Re in+1∑
p,q=1
upvq(N−det H(ρ)
)(ρpq) = 0 for all u,v ∈ Hz(M), z ∈ M.
Therefore, we have
(4.3)n+1∑
p,q=1
upvq(N − det H(ρ)
)(ρpq) = 0 for all u,v ∈ Hz(M), z ∈ M.
14
Thus
(4.5) Torθ = 0 ⇐⇒ TORρ = 0, θ = i∂ρ.
By formula (1.8), it is obvious that (S2n+1, θ0) with θ0 = 12i
∑n+1j=1 (zjdzj −
zjdzj) is a torsion-free pseudo-hermitian manifold. However, if one considers
r(z) = |z|2 + εn+1∑j=1
|zj|4 − 1,
(4.6) D = {z ∈ Cn+1 : r(z) < 0}, M = ∂D, θr =1
2i(∂r − ∂r),
then we have the following proposition.
Proposition 4.1 With the notation above, (∂D, θr) is not a torsion-freepseudohermitian manifold for any ε > 0.
Proof. Since
(4.7) rp = zp(1 + 2ε|zp|2), H(r)(z) = Diag(1 + 4ε|z1|2, · · · , 1 + 4ε|zn+1|2)
and
(4.8) rq =n+1∑p=1
rpqrp =1 + 2ε|zq|2
(1 + 4ε|zq|2)zq,
N
det H(r)=
n+1∑q=1
1 + 2ε|zq|2
1 + 4ε|zq|2zq
∂
∂zq
,
we have
(4.9) rpq −N
det H(r)rpq = 2εz2
pδpq − 2εδpq1 + 2ε|zp|2
1 + 4ε|zp|2zp2zp
= 2εz2pδpq(1− 2
1 + 2ε|zp|2
1 + 4ε|zp|2)
= − 2ε
1 + 4ε|zp|2z2
pδpq.
Therefore,
J(r)
4 det H(r)TORr(u,v)(z) = ε
n+1∑p=1
upvp z2p
1 + 4ε|zp|2
15
where u,v ∈ Hz(∂D). It is obvious that TORθr(u,v) = 0 when z = a ei forany 1 ≤ i ≤ n+1. When z ∈ ∂D and z 6= aei for any a ∈ C and 1 ≤ i ≤ n+1,without loss of generality, we may assume z1 6= 0 and zn+1 6= 0. Then we let
u1 =−zn+1
1 + 2ε|z1|2, un+1 =
z1
1 + 2ε|zn+1|2v1 =
−zn+1
1 + 2ε|z1|2, vn+1 =
z1
1 + 2ε|zn+1|2;
and uj = vj = 0 for all 2 ≤ j ≤ n. It is easy to see that u,v ∈ Hz(∂D) and
J(r)
4 det H(r)TORr(u,v)
= −εz21z
2n+1
[ 1
(1 + 4ε|z1|2)(1 + 2ε|z1|2)2+
1
(1 + 4ε|zn+1||2)(1 + 2ε|zn+1|2)2
]6= 0,
and the proof of the proposition is complete.
Next we let f(z) be a holomorphic function with f(0) = 0 and α be apositive integer. In [19], Tanaka shows that if f is a weighted homogeneouspolynomial, then M0 = {z ∈ Cn+1 : |z|2 = 1, f(z) = 0} = (∂Bn+1) ∩ Z(f) isa torsion-free pseudo-hermitian manifold of CR-dimension (n− 1). Now weconsider M = ∂D(f, α) where
D(f, α) = {z ∈ Cn+1 : ρ(z) = |z|2 + |f(z)|2α < 1}, f(0) = 0,
and
θα,f =1
2i(∂(|z|2 + |f(z)|2α)− ∂(|z|2 + |f(z)|2α)).
We try to understand for what kind of f(z) and α is (∂D(f, α), θα,f ) torsion-free. For this purpose, we calculate TOR in terms of f and α as follows.
Proposition 4.2 With notation above, we have
(4.10) −1
2J(ρ)TOR(u,v) = [(1−
n+1∑j=1
zj∂j)f(z)α]n+1∑i,j=1
uivj(∂ijf(z)α)
Proof. Since
ρi = zi + α|f(z)2|α−1f(z)∂if(z), ρij = δij + α2|f(z)|2(α−1)∂if∂jf(z)
16
and
ρij(z) = α|f(z)2|α−1f(z)∂ijf(z) + α(α− 1)|f(z)2|α−2f(z)2∂if∂jf,
H(ρ) = In+1 + α2|f(z)2|α−1(∂f)∗(∂f).
Then
H(ρ)−1 = In+1 −α2|f(z)2|α−1
1 + A(∂f)∗(∂f), det H(ρ) = 1 + A
whereA = A(f)(z) = α2|f(z)2|α−1|∂f |2.
Thus
ρpq = δpq −α2|f(z)2|α−1
1 + A∂qf(z)∂pf(z)
and
N
det H(ρ)=
n+1∑q=1
ρq∂q =n+1∑q=1
ρq∂q −α2|f(z)2|α−1
1 + A
n+1∑p=1
ρp∂pfn+1∑q=1
∂qf∂q
Then
Nf
det H(ρ)=
n+1∑q=1
ρq∂qf(1− A
1 + A) =
1
1 + A(n+1∑q=1
zq∂qf +A
αf)
Hence
(1− α
det H(ρ)N)f =
1
1 + A(1− α
n+1∑j=1
zj∂
∂zj
)f
and
Nρij = αN(|f(z)2|α−1f)∂ijf + α(α− 1)N(|f(z)2|α−2f(z)2)∂if(z)∂jf(z)
= α2|f(z)2|α−1Nf∂ijf + α2(α− 1)|f(z)2|α−2fNf∂if∂jf
= α2|f(z)2|α−1[∂ijf + (α− 1)|f(z)2|−1f∂if∂jf ]Nf
Thus
ρij −1
det H(ρ)Nρij
= α|f(z)2|α−1(f − α
det H(ρ)Nf)∂ijf
+α(α− 1)|f(z)2|α−2f(f − α
det H(ρ)Nf)∂if∂jf
17
Therefore
−J(ρ)
2TOR(u, v)
= det H(ρ)α|f(z)2|α−1(f − αRf)(n+1∑i,j=1
uivj(∂ijf +α− 1
f∂if∂jf))
= det H(ρ)f(z)α−1(f − αRf)n+1∑i,j=1
uivj(∂ijf(z)α)
= [(1−n+1∑j=1
zj∂j)f(z)α]n+1∑i,j=1
uivj(∂ijf(z)α)
Thus, the proof of the proposition is complete.
Proposition 4.3 With the notation above, the following two statementshold:
(i) If fα is linear, then D(f, α) is biholomorphic to Bn+1 and(∂D(f, α), θα,f ) is torsion free;
(ii) If (∂D(f, α), θα,f ) is torsion free, then either f is a linear holomorphicfunction or f satisfies
n+1∑p,q=1
upvq ∂2fα
∂zp∂zq
(z) = 0, fo all z ∈ ∂D,u,v ∈ Hz(∂D).
Proof. We first prove Part (i). Since f(z)α is linear holomorphic function,the vanishing of torsion is clear from Proposition 4.2. Let us write
f(z)α =n+1∑j=1
cjzj
and
D = {z ∈ Cn+1 : |z|2 + |n+1∑j=1
cjzj|2 < 1}.
To prove that D is biholomorphic to the unit ball Bn+1 in Cn+1, it sufficesto prove that there are
(4.11) ϕj(z) =n+1∑k=1
cjkzk
18
so that
(4.12)n+1∑j=1
|ϕj|2 = |z|2 + |n+1∑j=1
cjzj|2
orn+1∑j,k=1
n+1∑j=1
cjkcj`zkz` =n+1∑k,`=1
(δk` + ckc`)zkz`
Let
(4.13) B = [cjk]n+1×n+1, Ct = (c1, · · · , cn+1).
Since In+1 +C∗C is positive definite, there is a non-singular matrix B so that
(4.14) B∗B = In+1 + C∗C.
Then ϕ : D → Bn+1 is a biholomorphism, and Part (i) is proved.To prove Part (ii). Since TOR(u,v) = 0. If (1 − ∑n+1
j=1 zj∂j)fα = 0
in an open set on ∂D, then it equals 0 in an open set in Cn+1 by uniquecontinuation of holomorphic functions. This implies that fα must be a linearholomorphic function and α = 1. Otherwise,
(4.15)n+1∑i,j=1
uivj(∂ijf(z)α) = 0 for all u, v ∈ Hz(∂D) and z ∈ ∂D.
The proof of the proposition is complete.
Note: It seems that (4.15) implies that f(z)α is a linear function withα = 1, but we cannot prove it at this moment.
5 Real Ellipsoids in Cn+1
The real ellipsoid E(a, b) = ∂D(a, b) where
(5.1) D(a, b) = {z :n+1∑j=1
ajx2j +
n+1∑j=1
bjy2j < 1}, aj, bj > 0,
19
is a very important example in pseudo-hermitian geometry and Chern-Mosertheory. Many interesting questions on it were studied by S. Webster [20, 21],Huang and Ji [6], and many others.
It is easy to see that E(a, b) can be written as
(5.2) E(A, B) = ∂D(A, B), D(A, B) = {z : ρ(z) < 1},
where
(5.3) ρ(z) =n+1∑j=1
Aj|zj|2 + Bj(z2j + z2
j), Aj =aj + bj
2, Bj =
aj − bj
4.
Thus
(5.4) ρpq = Apδpq, ρpq = 2Bpδpq, Nρ(ρ) = det H(ρ)n+1∑j=1
A−1j |ρj|2
Since ρpq are constant, by (3.8) and (3.9),
TORρ(u,v) = −2det H(ρ)
J(ρ)
n+1∑p,q=1
upvqρpq
and
TORhρ(u,v) = TORρ(u,v)− 2Nρ log h
J(ρ)
n+1∑p,q=1
upvq ρpq − 2hn+1∑
p,q=1
upvq ∂2
∂zp∂zq
(1
h
)
Therefore TORhρ(u,v) = 0 if and only if
(5.5) 2(− det H(ρ)−Nρ log h)n+1∑p=1
Bpupvp − J(ρ)h
n+1∑p,q=1
upvq∂pq(1
h) = 0.
Without loss of generality, we may assume that Aj = 1 for all 1 ≤ j ≤ n+1,
otherwise, we can make a complex linear change of variables z′j =√
Ajzj for
1 ≤ j ≤ n + 1. Then |Bj| < 1/2 and
(5.6) det H(ρ) = 1, J(ρ) = |∂ρ|2 = Nρ(ρ).
20
Thus(5.7)
TORhρ(u,v) = 0 ⇐⇒ 2(1+Nρ log h)n+1∑p=1
Bpupvp+J(ρ)h
n+1∑p,q=1
upvq∂pq(1
h) = 0.
Then we get the following theorem:
THEOREM 5.1 Let E(B) = {z ∈ Cn+1 : ρ(z) = |z|2 +∑n+1
j=1 Bj(z2j + z2
j)−1 = 0} with |Bj| < 1/2 and let h(z) ∈ C3(E(B)) be positive on E(B) withan extension in a neighborhood of E(B) so that Nρ log h 6= −1 on E(B).Then the following statements are equivalent
(i)E(B) = ∂Bn+1, the sphere in Cn+1, and∑n+1
p,q=1 upvq ∂2
∂zp∂zq(1/h) = 0
for all u,v ∈ Hz(E(B)) and z ∈ E(B);(ii) Torθhρ
= 0 on E(B) and∑n+1
p,q=1 upvq ∂2
∂zp∂zq(1/h) = 0 for all u,v ∈
Hz(E(B)) for two points z0 and z1 in E(B) so that z0j = 0 and z1
j 6= 0 forsome 1 ≤ j ≤ n + 1.
In particular, (E(B), θρ) is torsion-free if and only if E(B) = ∂Bn+1.
Proof. First, we prove (i) implies (ii). When E(B) = ∂Bn+1, Bj = 0 forall 1 ≤ j ≤ n + 1. Moreover,
∑n+1p,q=1 upvq∂pq(1/h) = 0 for all u,v ∈ Hz(∂Bn)
and all z ∈ ∂Bn+1 implies (5.7), and therefore, Torθhρ(u,v)(z) = 0 for all
u,v ∈ Hz(∂Bn+1) for all z ∈ ∂Bn. Thus (ii) holds.
Next we prove (ii) implies (i). By (4.5), Torθhρ= 0 on E(B) if and only
if TORhρ = 0 on E(B). By (5.7), TORhρ = 0 on E(B) implies that
(5.8) 2(1 + Nρ log h)n+1∑p=1
Bpupvp + J(ρ)h
n+1∑p,q=1
upvq∂pq(1
h) = 0
for all u,v ∈ Hz(E(B)) and z ∈ E(B).Since
∑n+1p,q=1 upvq∂pq(1/h) = 0 for all u,v ∈ Hz(E(B)) at two points
z = z0 and z1 ∈ E(B), and since Nρ log h 6= −1 on E(B), one has that
(5.9)n+1∑p=1
Bpupvp = 0
for all u,v ∈ Hz(E(B) with z = z0 and z = z1.
21
Note that
(5.10) ρp = zp + 2Bpzp.
Since z0j = 0, (5.9) with u = v = ej implies that Bj = 0. If z1
k = 0 then (5.9)implies Bk = 0. Since z1
j 6= 0, for any k with z1k 6= 0, (5.9) with u = v implies
thatBkρj(z
1)2 + Bjρk(z1)2 = 0,
Thus Bk = 0 for all 1 ≤ k ≤ n + 1. Therefore, E(B) = ∂Bn+1. Applying(5.8) again, we have
∑np,q=1 upvq∂pq(1/h)(z) = 0 for all u,v ∈ Hz(∂Bn) and
all z ∈ ∂Bn+1. The proof of the theorem is complete.
As a direct consequence, we have proved Theorem 1.3.
Observe that for all u,v ∈ Hz(E(B)) where ρn+1(z) 6= 0,
n+1∑p,q=1
upvq∂pq =n+1∑
p,q=1
upvq(YpYq +ρpq
ρn+1
∂n+1), Yp = ∂p −ρp
ρn+1
∂n+1,
and
Nρ −J(ρ)
ρn+1
∂n+1 = Nρ −Nρ(ρ)
ρn+1
∂n+1 =n+1∑q=1
ρqYq,
hence (5.7) can be written as TORhρ(u,v) = 0 if and only if
2(1 +n+1∑q=1
ρqYq log h)n+1∑p=1
Bpupvp + J(ρ)h
n+1∑p,q=1
upvqYpYq(1
h) = 0.
Similarly, for all u,v ∈ Hz(E(B)) where ρ`(z) 6= 0,
(5.11) 2(1+n+1∑q=1
ρqYq(`) log h)n+1∑p=1
Bpupvp+J(ρ)h
n+1∑p,q=1
upvqYp(`)Yq(`)(1
h) = 0,
with Yq(`) = ∂q − ρq
ρ`∂`.
Let Xp` = z`∂
∂zp− zp
∂∂z`
. Then on ∂Bn+1, since ρp = zp, if
Yp(`)Yq(`)H(z) = 0 then Xp`Xq`H(z) = 0. The following question is nat-ural and very interesting:
Question: Let H ∈ C2(∂Bn+1) satisfy the following equation:
(5.12) Xp`Xq`H = 0, z ∈ ∂Bn+1
for all p, q 6= ` and 1 ≤ ` ≤ n + 1. Then what can one say about H ?
22
THEOREM 5.2 Let H ∈ C2(Bn+1
) be real, harmonic in Bn+1 so thatXp`Xq`H(z) ≡ 0 on ∂Bn+1 for all 1 ≤ p, q, ` ≤ n + 1. Then H is a quadraticpolynomial.
Proof. Since H is harmonic in Bn+1, we claim that Xp`H is also harmonicin Bn+1. In fact,
n+1∑k=1
∂2
∂zk∂zk
Xp`H(z) = Xp`1
4∆H(z)−
n+1∑k=1
(δpk∂2H
∂zk∂z`
− δk`∂2H
∂zk∂zp
) = 0.
Since Xp`H is CR on ∂Bn+1, we have that Xp`H can be extended to beholomorphic in Bn+1. By the uniqueness of harmonic extension, we havethat Xp`H(z) is holomorphic in Bn+1. Now let us write the Taylor series forH near z = 0 as follow:
(5.13) H(z) =∞∑
|α|+|β|=0
aαβzαzβ.
Then
(5.14) −Xp`H(z)
=∑
|α|+|β|≥1
aαβzα(|zp|2β` − |z`|2βp)zβ−ep−e`
=∑
α≥ep
aα−epβ+e`(β` + 1)zαzβ +
∑αp=0
aαβ+e`(β` + 1)zα+epzβ
−∑α≥e`
aα−e`β+ep(βp + 1)zαzβ −∑
α`=0
aαβ+ep(βp + 1)zα+e`zβ
=∑
α≥ep+e`
(aα−epβ+e`(β` + 1)− aα−e`β+ep(βp + 1))zαzβ
+∑
α≥ep,α`=0
aα−epβ+e`(β` + 1)zαzβ +
∑αp=0
aαβ+e`(β` + 1)zα+epzβ
−∑
α≥e`,αp=0
aα−e`β+ep(βp + 1)zαzβ −∑
α`=0
aαβ+ep(βp + 1)zα+e`zβ
=∑
α≥ep+e`
(aα−epβ+e`(β` + 1)− aα−e`β+ep(βp + 1))zαzβ
+∑
αp>1,α`=0
aα−epβ+e`(β` + 1)zαzβ + 2
∑αp=1,α`=0
aα−epβ+e`(β` + 1)zαzβ
23
−∑
α`>1,αp=0
aα−e`β+ep(βp + 1)zαzβ − 2∑
α`=1,αp=0
aα−e`β+ep(βp + 1)zαzβ
+∑
αp=1,α`≥1
aα−epβ+e`(β` + 1)zαzβ −
∑α`=1,αp≥1
aα−e`β+ep(βp + 1)zαzβ
=∑
αp>1,α`>1
(aα−epβ+e`(β` + 1)− aα−e`β+ep(βp + 1))zαzβ
+∑
αp=α`=1
(2aα−epβ+e`(β` + 1)− 2aα−e`β+ep(βp + 1))zαzβ
+∑
αp=1,α`>1
(2aα−epβ+e`(β` + 1)− aα−e`β+ep(βp + 1))zαzβ
+∑
αp>1,α`=1
(aα−epβ+e`(β` + 1)− 2aα−e`β+ep(βp + 1))zαzβ
+∑
αp>1,α`=0
aα−epβ+e`(β` + 1)zαzβ + 2
∑αp=1,α`=0
aα−epβ+e`(β` + 1)zαzβ
−∑
α`>1,αp=0
aα−e`β+ep(βp + 1)zαzβ − 2∑
α`=1,αp=0
aα−e`β+ep(βp + 1)zαzβ
Notice that Xp`H(z) is holomorphic in Bn+1. All terms bαβzαzβ in its powerseries expansion will vanish if |β| > 0. This fact with the above expansionimply that
(i) If α` = 0, αp ≥ 1 and |β| > 0 then
aα−ep β+e`= 0.
(ii) If αp = 0, α` ≥ 1 and |β| > 0 then
aα−e` β+ep = 0.
Thus, (i) and (ii) imply that
(5.15) aαβ+e`= 0 if α` = 0, |β| > 0; aαβ+ep = 0 if αp = 0, |β| > 0.
(iii) If α` = 1 and |β| > 0 then we have two cases:a) if αp > 1, then
aα−epβ+e`(β` + 1) = 2aα−e` β+ep(βp + 1);
b) if αp = 1, then
aα−epβ+e`(β` + 1) = aα−e` β+ep(βp + 1).
24
a) and b) imply that
(5.16) aα+e`β+e`(β` + 1) = (2− δ0αp)aα+epβ+ep(βp + 1) if α` = 0.
By (5.15) and (5.16),
(5.17) aαβ+2e`= 0, if α` = 1, |β| ≥ 0.
(iv) If αp = 1, α` > 1 and |β| > 0 then
(5.18) 2aα−epβ+e`(β` + 1) = aα−e` β+ep(βp + 1);
(v) If αp > 1, α` > 1 and |β| ≥ 1 then
(5.19) aα−epβ+e`(β` + 1) = aα−e` β+ep(βp + 1)
Consider α` = 2. By (5.17) and (v), if αp > 1 then
aα−ep β+3e`=
βp + 1
β` + 3aα−e` β+2e`+ep = 0,
and by (5.17) and (iv), if αp = 1, we have
aα−epβ+3e`=
(βp + 1)
2(β` + 3)aα−e`β+2e`+ep = 0.
Thus
(5.20) aαβ+3e`= 0 if α` = 2.
A similar computation for α` > 2, together with (5.15) and (5.17), gives
(5.21) aαβ+(α`+1)e`= 0 if α` ≥ 1, |β| ≥ 0; or α` = 0, |β| ≥ 1.
Since H is real-valued, we have
aβα = aαβ
Thus
(5.22) aαβ = 0 if α 6= β, |α| > 0, |β| ≥ 0; or α = 0, |β| ≥ 2.
25
Therefore
H(z) = a00 +∑|α|=1
(aα0zα + a0αzα) +
∑|α|=|β|=1
aαβzαzβ +∑|α|≥2
aαα|zα|2
LetPk(z, z) =
∑|α|=k
aαα|zα|2
where Pk(z, z) is a homogeneous polynomial of degree 2k. Since Xp`H(z) isholomorphic, one can see that
Xp`
∞∑k=2
Pk(z, z) =∞∑
k=2
Xp`Pk
is holomorphic. Since Xp`Pk(z, z) is a homogeneous polynomial of degree 2kor 0, we have that Xp`Pk(z, z) is holomorphic for all k ≥ 2. We want toprove Pk = 0 when k ≥ 2.
If k ≥ 2 then
−Xp`Pk(z, z)
=∑|α|=k
aααzα(zpα`zα−e` − z`αpz
α−ep)
=∑|α|=k
aααzα+epα`zα−e` −
∑|α|=k
aααzα+e`αpzα−ep
=∑
|α|=k−1
aα+e`α+e`zα+ep+e`(α` + 1)zα −
∑|α|=k−1
aα+ep α+epzα+e`+ep(αp + 1)zα
=∑
|α|=k−1
[(α` + 1)aα+e`α+e`− (αp + 1)aα+ep α+ep ]z
α+e`+epzα
is holomorphic. Thus
(α` + 1)aα+e`α+e`− (αp + 1)aα+ep α+ep = 0, if |α| > 0.
That is,
(5.23) aα+e` α+e`=
αp + 1
α` + 1aα+ep α+ep , for all |α| > 0.
26
Since H is harmonic, we have
0 = ∆H(z) = 4∑|α|=1
aαα +∞∑
k=2
∆Pk(z, z)
It is easy to see that
(5.24)∑|α|=1
aαα = 0, ∆Pk(z, z) = 0 for all k ≥ 2.
For convenience, we let aα = aαα. Then
∆Pk(z, z) = 4∑|α|=k
aα
n+1∑j=1
|∂jzα|2
= 4∑|α|=k
aα
n+1∑j=1
α2j |zα−ej |2
= 4∑
|α|=k−1
n+1∑j=1
aα+ej(αj + 1)2|zα|2
Together with (5.23), we have
(5.25)n+1∑j=1
aα+ej(αj + 1)2 = 0, aα+e`
=αp + 1
α` + 1aα+ep for all |α| > 0.
By (5.25), we have
0 = aα+e1(α1 + 1)2 +n+1∑j=2
(αj + 1)(α1 + 1)aα+e1 = (α1 + 1)aα+e1
n+1∑j=1
(αj + 1)
which implies thataα+e1 = 0, for all |α| ≥ 1.
By the same argument, we have
aα+ej= 0, for all |α| ≥ 1, j = 1, 2, · · · , n + 1.
Therefore
(5.26) aα = 0 for all |α| ≥ 2.
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In other words, Pk(z, z) = 0 when k ≥ 2. Hence
(5.27) H(z) = b0 +n+1∑j=1
2Re (bjzj) +n+1∑j,k=1
bjkzjzk
with b0 being real, bjk = bkj and
(5.28)n∑
j=1
bjj = 0.
Thus H(z) is a harmonic quadratic polynomial. The proof of the theorem iscomplete.
To prove Theorem 1.2. We let H(z) = 1/h(z) be the harmonic extensionof 1/h from ∂Bn+1 to Bn+1. By (5.8) with Bp = 0 for 1 ≤ p ≤ n + 1,TORhρ(u, v) = 0 on ∂Bn+1 if and only if
(5.29) Xp`Xq`(1/h) = 0, z ∈ ∂Bn+1
for all 1 ≤ p, q, ` ≤ n + 1. Since h is real-valued, by Theorem 5.2, weknow that 1/h is a harmonic quadratic polynomial. Therefore, the proof ofTheorem 1.2 is complete.
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Mailing Address:
Partial work was done when the first author was visiting Fujian NormalUniversity, China. The address is:
School of Mathematics and Computer Secience, Fujian Normal Univer-sity, Fuzhou, Fujian, China
Current address for Li and Luk are:Department of Mathematics, University of California, Irvine, CA 92697–3875, USA. E-mail: [email protected]
Department of Mathematics, Lady Shaw Building, The Chinese University ofHong Kong, Shatin, N. T., Hong Kong. E-mail: [email protected]
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