ON THE COMPLETE HYPERBOLICITY AND THE TAUTNESS OF THE HARTOGS DOMAINS

ON THE COMPLETE HYPERBOLICITY AND THE

TAUTNESS OF THE HARTOGS DOMAINS

DO DUC THAI

Department of MathematicsPedagogical Institute of Hanoi, Caugiay, Hanoi

VietnamE-mail: [email protected]

PHAM VIET DUC

Department of MathematicsPedagogical Institute of Thainguyen, Thainguyen

Vietnam

Received 16 March 1998

Let X be a complex space and ϕ a plurisubharmonic function on X. Consider the Hartogsdomain Ωϕ(X) given by Ωϕ(X) = (x, λ) ∈ X × C; |λ| < e−ϕ(x). In this article, thenecessary and sufficient conditions on the complete hyperbolicity and the tautness of theHartogs domain Ωϕ(X) are given.

0. Introduction

Let X be a complex space and ϕ a plurisubharmonic function on X . The Hartogs

domain Ωϕ(X) is defined by

Ωϕ(X) = (x, λ) ∈ X × C; |λ| < e−ϕ(x) .

This domain is the classical subject of complex analysis in several variables. Thus

one was investigated since the beginning of the 20th century, in particular, much at-

tention has been given to the properties of the Hartogs domains from the viewpoint

of hyperbolic analysis since S. Kobayashi introduced the notion of the Kobayashi

pseudodistance and used it to study Geometric function theory in several complex

variables (see [6, 7]). For instance, in 1981 Kerzman and Rosay [5] showed the hy-

perbolic Hartogs domain Ωϕ(∆) such that Ωϕ(∆) is not complete hyperbolic, where

∆ is the open unit disc in C. At the same time, Sibony [11] constructed a partic-

ular plurisubharmonic function ϕ on ∆ such that Ωϕ(∆) is complete hyperbolic.

In 1998, Thai and Huong [13] showed the weakly disc-convex hyperbolic Hartogs

domain Ωϕ(∆), but Ωϕ(∆) is not taut.

Later, Thai and Thomas [14] showed that Ωϕ(∆) is hyperbolic if and only if ϕ

is locally bounded below, i.e. for any K ⊂ ∆, infK ϕ > −∞.

Our aim in this article is to investigate the complete hyperbolicity and the

tautness of the Hartogs domain Ωϕ(X) in general case, where X is any complex

space.

103

International Journal of Mathematics Volume 11, No. 1 (2000) 103–111c©World Scientific Publishing Company

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104 D. D. THAI and P. V. DUC

We now describe more precisely the content of the paper.

In Sec. 2 we show a class of plurisubharmonic functions ϕ for which we can give

the necessary and sufficient conditions for the complete hyperbolicity of Ωϕ(X). We

would like emphasize here that the above-mentioned functions ϕ are quite different

to the functions of Sibony in [11]. It would be nice to have a characterization for the

complete hyperbolicity of Ωϕ(X), but as far as we know, this is an open question.

In Sec. 3 we prove that Ωϕ(X) is taut if and only if X is taut and ϕ is continuous

on X . Our proof is based on Sibony’s idea that a nonsingular complex manifold X

possesses a plurisubharmonic bounded exhaustion function then X is taut (see [11,

Corollary 5]). We would also like to emphasize that the nonsingularity of complex

manifolds plays an essential role in proving Sibony’s theorem. Since complex spaces

do not have this property, the technique for proving the above-mentioned theorem

required technical improvements.

Namely, we prove the following:

Theorem A. Let X be a complex space and ϕ a plurisubharmonic function on

X. Assume that for every x ∈ X, there exists an open neighbourhood U of x in

X and a sequence hj of holomorphic functions on U and a sequence cj of real

numbers in the interval (0, 1) such that the sequence cj log |hj | converges uniformly

on compact subsets of U to the function ϕ. Then the Hartogs domain Ωϕ(X) is

complete hyperbolic if and only if X is complete hyperbolic and ϕ is continuous

on X.

Theorem B. Let X be a complex space and ϕ a plurisubharmonic function on X.

Then the Hartogs domain Ωϕ(X) is taut if and only if X is taut and ϕ is continuous

on X.

1. Basic Notions

In this article, we shall make use of properties of complex spaces as in [4], and

properties of the Kobayashi pseudodistance on complex spaces as in [8] or [9].

1.1. We denote the Kobayashi pseudodistance on a complex space X by dX ·X is

called to be (complete) hyperbolic if dX is a (complete) distance.

1.2. A complex space X is called to be taut if whenever Y is a complex space and

fi : Y → X is a sequence of holomorphic maps, then either there exists a subse-

quence which is compactly divergent or a subsequence which converges uniformly

on compact subsets to a holomorphic f : Y → X . It suffices that this condition

should hold when Y = ∆ [1], where ∆ is the unit disc in C.

Also, a complete hyperbolic space is taut, and a taut complex space is hyperbolic

[1].

1.3. Let X be a complex space. A plurisubharmonic function on X is a function

ϕ : X → [−∞,∞) having the following property. For every x ∈ X there exist

an open neighbourhood U with a biholomorphic map h : U → V onto a closed

complex subspace V of some domain G ⊂ Cm and a plurisubharmonic function

ϕ : G→ [−∞,∞) such that ϕ|U = ϕ h. (see [10, p. 225]).

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COMPLETE HYPERBOLICITY AND THE TAUTNESS OF THE HARTOGS DOMAINS 105

Some remarks should be made at this point. First, the definition of plurisub-

harmonicity does not depend on the choice of local charts. Second, Fornaess and

Narasimhan proved [3] that an upper semicontinuous function ϕ : X → [−∞,∞) on

a complex space X is plurisubharmonic iff ϕ f is subharmonic for all holomorphic

maps f : ∆→ X , where ∆ is the open unit disc in C.

1.4. A complex space X is called to be hyperconvex if X is Stein and X has a

negative plurisubharmonic exhaustion function.

1.5. If u is a function of class C2 in an open set of Cn and w = (w1, . . . , wn) ∈ Cnwe denote

〈Lu(p)w,w〉 =

n∑i,j=1

∂2u(p)

∂zi∂zjwiwj

the Levi form of u at p.

2. The Complete Hyperbolicity of the Hartogs Domains

Theorem A. Let X be a complex space and ϕ a plurisubharmonic function on X.

Assume that for every x ∈ X, there exist an open neighbourhood U of x in X and a

sequence hj of holomorphic functions on U and a sequence cj of real numbers

in the interval (0, 1) such that the sequence cj log |hj | converges uniformly on

compact subsets of U to the function ϕ. Then the Hartogs domain Ωϕ(X) is complete

hyperbolic if and only if X is complete hyperbolic and ϕ is continuous on X.

Proof. (⇒) Since X is isomorphic to a closed complex subspace of Ωϕ(X), X is

complete hyperbolic.

Assume that ϕ is not continuous at z0 ∈ X . Then we can find a sequence

zj ⊂ X and a number r ∈ R such that zj → z0 and

e−ϕ(z0) < r < e−ϕ(zj) for every j ≥ 1 .

For every j ≥ 1, we define the holomorphic map fj : ∆→ Ωϕ(X) by fj(λ) = (zj , rλ)

for each λ ∈ ∆.

Since fj(0) → (z0, 0) ∈ Ωϕ(X) and by the tautness of Ωϕ(X), without loss

of generality we may assume that the sequence fj converges uniformly to the

holomorphic map f in Hol(∆,Ωϕ(X)), where f is a holomorphic map from ∆ into

Ωϕ(X) given by f(λ) = (z0, rλ) for every λ ∈ ∆.

It implies that r|λ| < e−ϕ(z0) for every λ ∈ ∆. Thus r ≤ e−ϕ(z0). It is

impossible.

(⇐) Consider the canonical projection π : Ωϕ(X)→ X given by π(z, λ) = z.

By a theorem of Eastwood [2], it suffices to show that for every z0 ∈ X

there exists an open neighbourhood U of z0 in X such that π−1(U) is complete

hyperbolic.

Indeed, assume that z0 ∈ X and U is an open relatively compact neighbourhood

of z0 in X such that U is isomorphic to an analytic subset of the open unit ball in

Cn.

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Since ϕ is continuous on X,π−1(U) is relatively compact in X×C. Thus π−1(U)

is hyperbolic.

On the other hand, by the hypothesis, there exist a sequence hj of holomorphic

functions on U and a sequence cj of real numbers in the interval (0, 1) such that

cj log |hj | converges uniformly on compact subsets of U to the function ϕ.

Since ϕ(z) 6= −∞ for every z ∈ U , there exists a complete hyperbolic open

neighbourhood U1 of z0 in U such that hj(z) 6= 0 for every z ∈ U1 and for every

j ≥ j0. Thus we can write hj(z)cj = egj(z) for every z ∈ U1 and every j ≥ j0. It

implies that |egj(z)| converges uniformly to eϕ(z) on U1.

Since π−1(U1) ⊂ π−1(U), π−1(U1) is also hyperbolic. We now prove that

π−1(U1) is complete hyperbolic.

Assume that (zj, λj) is a Cauchy sequence in π−1(U1). By the distance de-

creasing principle, zj is the Cauchy sequence in U1 and hence, zj converges to

z0 ∈ U1.

Since π−1(U1) is relatively compact in X×C, without loss of generality we may

assume that (zj , λj) converges to a point (z0, λ0) of X × C.

Suppose that (z0, λ0) ∈ ∂π−1(U1). It implies that |λ0| · eϕ(z0) = 1. Choose

εj ↓ 0 such that e−ϕ(z) < |e−gj(z)+εj | for every z ∈ U1. For each j ≥ 1, we define

the holomorphic function fj : π−1(U1)→ C given by fj(z, λ) = λegj(z)−εj for every

(z, λ) ∈ π−1(U1). It is easy to see that |fj(z, λ)| < 1 for every (z, λ) ∈ π−1(U1),

fj(z, 0) = 0 for every z ∈ U1 and |fj(zj , λj)| = |λj | · |egj(zj)−εj | → |λ0|.eϕ(z0) = 1.

On the other hand, since the sequence (zj , λj) is a Cauchy sequence and the

sequence (zj, 0) converges to (z0, 0), there exists a positive integer N such that

dπ−1(U1)((zj , λj), (zN , λN )) < 1 and dπ−1(U1)((zj , 0), (z0, 0)) < 1 for every j ≥ N .

It implies that

dπ−1(U1)((zj , λj), (zj , 0)) ≤ dπ−1(U1)((zj , λj), (zN , λN ))

+ dπ−1(U1)((zN , λN ), (z0, 0))

+ dπ−1(U1)((z0, 0), (zj, 0)) < C <∞

for every j ≥ N .

Denote the Caratheodory distance of π−1(U1) by cπ−1(U1) (see [6, p. 49]).

We have

∞ > lim supj→∞

dπ−1(U1)((zj , λj), (zj , 0))

≥ lim supj→∞

cπ−1(U1)((zj , λj), (zj , 0))

≥ limj→∞

log1 + |fj(zj , λj)|1− |fj(zj , λj)|

=∞ .

It is impossible.

Thus π−1(U1) is complete hyperbolic and hence, Ωϕ(X) is complete hyperbolic.

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3. The Tautness of the Hartogs Domains

Definition 3.1. Let X be a complex space and TX be the Zariski tangent space

of X .

Put ∆r = |z| < r ⊂ C, ∆1 = ∆, e0 = ddz|z=0 ∈ T0∆r is the unit tangent

vector.

The Royden–Kobayashi cone of X is defined by

ConX := v ∈ TX ; ∃ϕ ∈ Hol(∆r, X), ∃u ∈ T0∆r such that ϕ′(u) = v .

Moreover, we define the Royden–Kobayashi (differential) pseudometric on TX

as follows:

FX(v) =

inf

1

r; ∃ϕ ∈ Hol(∆r , X), such that v = ϕ′(e0)

, if v ∈ ConX

∞ , if v /∈ ConX

Lemma 3.2. Let X be a complex space and H a Finsler metric on X. Then X

is hyperbolic if and only if for each p ∈ X, there are a neighbourhood U of p and a

constant C > 0 such that FX(ξx) ≥ CH(ξx) for all ξx ∈ TxX with x ∈ U.

Proof. (⇒) Let D be a coordinate polydisc about a point p. Since X is hyperbolic,

(X, dX) is tight (see [1]) and hence the family Hol(∆, X) is an even family. Thus

there is a disc ∆δ about 0 and a neighbourhood U of p such that if Φ(0) = x ∈ U ,

then Φ(∆δ) ⊂ D. Hence for x ∈ U , we have δFD(ξx) ≤ FX(ξx). We may suppose U

is a compact subset of D. Then for x ∈ U , ξx ∈ TxX , we have FX(ξx) ≥ δFD(ξx) ≥CH(ξx) for some positive constant C.

(⇐) Let dCH be the distance function generated on X by CH (see [9]). By the

hypothesis, f∗(CH) ≤ ds2∆ for all f ∈ Hol(∆, X), where ds2

∆ denotes the Bergman–

Poincare metric of ∆. Thus we have

dCH(x, y) ≤ dX(x, y) for x, y ∈ X .

It implies that X is hyperbolic.

Definition 3.3. Let X be a complex space. Let Sp be the family of functions u

defined on X satisfying 0 ≤ u ≤ 1, u(p) = 0, u of class C2 in a neighbourhood of p

and log u plurisubharmonic on X .

We define the Sibony pseudometric PX on the Royden–Kobayashi cone of X as

following:

PX(p, ξ) = sup

(∂2u σ(0)

∂λ∂λ

)1/2

; u ∈ Sp, σ ∈ Hol(∆, X)

such that σ(0) = p, σ′(e0) = ξ

,

where (p, ξ) is an element of the Royden–Kobayashi cone Con X .

It is easy to see that PX(p, ξ) ≤ FX(p, ξ) for every (p, ξ) ∈ ConX .

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Lemma 3.4. Let X be a complex space. Suppose there exists a plurisubharmonic

bounded function u on X which is strictly plurisubharmonic on a neighbourhood of

p ∈ X, then X is hyperbolic at p, i.e. there exist a neighbourhood V of p and a

constant C > 0 such that FX(q, ξ) ≥ C‖ξ‖ for all q ∈ V, ξ ∈ TqX, where ‖ · ‖denotes the fixed Finsler metric on X.

Proof. As in [11, Theorem 3], we may assume that u is of class C2 in a neighbour-

hood of p.

Let Φ be a chart from a neighbourhood U of p to the ball B(0, 2r) in Cn such

that Φ(p) = 0. Choose a relatively compact neighbourhood V of p in X such that

V ⊂ U and Φ(z)− Φ(p) ∈ B(0, r) for all z ∈ V .

Assume that q is any point of V .

Put qΦ(z) = Φ(z)− Φ(q), ∀ z ∈ V . Then qΦ(q) = 0.

We can suppose that the function u is negative and uqΦ−1 extends to a strictly

plurisubharmonic u q Φ−1 on B(0, 2r) such that

〈Lu q Φ−1(z)w,w〉 ≥ C‖w‖2

for w ∈ Cn, where C is a positive constant which is independent of q ∈ V .

Let θ be a smooth nondecreasing function defined on R+ such that θ(x) = x for

x ≤ 12 and θ(x) = 1 for x ≥ 3

4 .

For t > 0 we define

Ψqt (z) = θ

(|qΦ(z)|2r2

)exp(tu(z)) for z ∈ U

Ψqt (z) = exp(tu(z)) for z ∈ X \ U .

It is clear that on a neighbourhood of X \ U and on qΦ−1(B(0, r2 )) the function

log Ψqt is plurisubharmonic. Consider in B(0, 2r) the function

Ψqt (z) = θ

(‖z‖2r2

)exp(t(u q Φ−1)(z)) .

Let h = log θ. An easy computation shows that there is a constant A > 0 such that

for all z ∈ B(0, 1) and w ∈ Cn

〈Lh(‖z‖2)w,w〉 ≥ −A · ‖w‖2 .

But then

〈L log Ψqt (z)w,w〉 ≥

(−Ar2

+ Ct

)‖w‖2 .

If we choose t = Ar2C

we find that Ψqt ∈ Sq.

Assume that σ ∈ Hol(∆, X) such that σ(0) = q, σ′(e0) = ξ. We have

∂Ψqt σ(λ)

∂λ= θ′

(|qΦ(σ(λ))|2

r2

)1

r2

n∑j=1

∂qΦj(σ(λ))

∂λqΦj(σ(λ)) exp(tu(σ(λ)))

+ θ

(|qΦ(σ(λ))|2

r2

)t∂u(σ(λ))

∂λexp(tu(σ(λ)))

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and

∂2Ψqt σ(λ)

∂λ∂λ= θ′′

(|qΦ(σ(λ))|2

∂λ

)1

r4

n∑j=1

qΦj(σ(λ))∂qΦj(σ(λ))

∂λ

×n∑j=1

∂qΦj(σ(λ))

∂λqΦj(σ(λ)) exp(tu(σ(λ)))

+ θ′(|qΦ(σ(λ))|2

r2

)1

r2

n∑j=1

∂qΦj(σ(λ))

∂λqΦj(σ(λ))

× t∂u(σ(λ))


+ θ′(|qΦ(σ(λ))|2

r2

)1

r2

n∑j=1

qΦj(σ(λ))∂qΦj(σ(λ))

∂λ

× t∂u(σ(λ))


+ θ

(|qΦ(σ(λ))|2

r2

)t∂2u(σ(λ))

∂λ∂λexp(tu(σ(λ)))

+ θ

(|qΦ(σ(λ))|2

r2

)t2∂u(σ(λ))

∂λ

∂u(σ(λ))

∂λexp(tu(σ(λ))) .

It implies that

∂2Ψqt σ(0)

∂λ∂λ≥ t exp(tu(q))

∂2u(σ(0))

∂λ∂λ

≥ Ct exp(tu(q))‖ξ‖2 .

Therefore

PX(q, ξ) ≥√A

rexp

(A

2r2Cu(q)

)‖ξ‖ for all ξ ∈ ConX ∩ TqX .

Since F (q, ξ) ≥ P (q, ξ) and the majorization above is independent of q ∈ V we

see that X is hyperbolic at p.

Lemma 3.5. Let X be a complex space admitting a smooth strictly plurisubhar-

monic function. If there exists on X a continuous bounded plurisubharmonic

exhaution function, then X is hyperbolic.

Proof. By reasoning as in [11, Corollary 4], X is hyperbolic at every point. Hence

by Lemma 3.2, X is hyperbolic.

Also by reasoning as in [11, Corollary 5] we have the following:

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Lemma 3.6 (Sibony’s theorem for complex spaces). Let X be a complex

space admitting a smooth strictly plurisubharmonic function. If there exists on X a

continuous bounded plurisubharmonic exhaustion function, then X is taut.

We now prove the main theorem of this section.

Theorem B. Let X be a complex space and ϕ a plurisubharmonic function on X.

Then the Hartogs domain Ωϕ(X) is taut if and only if X is taut and ϕ is continuous

on X.

Proof. (⇒) Since X is isomorphic to a closed complex subspace of Ωϕ(X), X is

taut.

Assume that ϕ is not continuous at z0 ∈ X . Then we can find a sequence

zj ⊂ X and a real number r > 0 such that zj → z0 and e−ϕ(z0) < r < e−ϕ(zj)

for every j ≥ 1.

For every j ≥ 1, we define the holomorphic map fj : ∆ → Ωϕ(X) by fj(λ) =

(zj , rλ) for each λ ∈ ∆.

Since fj(0) → (z0, 0) ∈ Ωϕ(X) and by the tautness of Ωϕ(X), without loss

of generality we may assume that the sequence fj converges uniformly to the

holomorphic map f in Hol(∆,Ωϕ(X)), where f is a holomorphic map from ∆ into

Ωϕ(X) given by f(λ) = (z0, rλ) for every λ ∈ ∆.

It implies that r|λ| < e−ϕ(z0) for every λ ∈ ∆.

Thus r ≤ e−ϕ(z0). It is impossible.

(⇐) Consider the canonical projection π : Ωϕ(X)→ X given by π(z, λ) = z.

By a result of Thai and Huong [12], it suffices to prove that for every z ∈ Xthere exists an open neighbourhood U of z in X such that π−1(U) is taut.

Indeed, choose a hyperconvex neighbourhood U of z in X . Then

π−1(U) = (z, λ) ∈ U × C : |λ| < e−ϕ(z) = Ωϕ(U) .

Assume that ρ is a negative plurisubharmonic exhaustion function of U . Then

the function (z, λ) 7→ max(ρ(z), log |λ| + ϕ(z)) is also negative plurisubharmonic

exhaustion function of Ωϕ(U). Thus Ωϕ(U) is hyperconvex. By Lemma 3.6, Ωϕ(U)

is taut, i.e. π−1(U) is taut. Hence Ωϕ(X) is taut.

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ON THE COMPLETE HYPERBOLICITY AND THE TAUTNESS OF THE HARTOGS DOMAINS

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