A note on the Kobayashi pseudodistance and the tautness of holomorphic fiber bundles

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POLONICI MATHEMATICI LVIII.1 (1993)

A note on the Kobayashi pseudodistance and the tautness of holomorphic fiber bundles

by Do Duc Thai and Nguyen Le Huong (Ha noi)

Abstract. We show a relation between the Kobayashi pseudodistance of a holomor- phic fiber bundle and the Kobayashi pseudodistance of its base. Moreover, we prove that a holomorphic fiber bundle is taut iff both the fiber and the base are taut.

I. Introduction. In this article we shall show a relation between the Kobayashi pseudodistance of a holomorphic fiber bundle and the Kobayashi pseudodistance of its base. Moreover, we shall give another proof (not using universal coverings as in Nag [5]) that a holomorphic fiber bundle is taut iff both the fiber and the base are taut. An analogous result for hyperbolicity was proved by Nag [5].

In this section we recall some definitions and relevant properties.

We denote the Kobayashi pseudodistance on a complex space M by d M

(see Kobayashi [2], [3] and Lang [4]). M is called (complete) hyperbolic if d M is a (complete) distance. M is called taut [3] if whenever N is a complex space and f i : N → M is a sequence of holomorphic maps, then either there exists a subsequence which is compactly divergent or a subsequence which converges uniformly on compact subsets to a holomorphic map f : N → M . It suffices that this condition should hold when N = D [3, p. 378], where D is the unit disk in C.

Also, a complete hyperbolic space is taut, and a taut complex space is hyperbolic [3, p. 378].

Acknowledgement. The authors are grateful to the referee for many valuable comments.

2. Theorem A. Let E be a holomorphic fiber bundle over M with hy- perbolic fiber F and projection π : E → M , where E, F , M are complex

1991 Mathematics Subject Classification: 32H15, 32L05.

Key words and phrases: holomorphic fiber bundle, Kobayashi pseudodistance, hyper-

bolic space, taut space.

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manifolds. Then d M (x, y) = d E ( e x, π ⁻¹ (y)) = inf _y∈π _˜

⁻¹

_(y) d E ( e x, y) for all e x, y ∈ M and all x ∈ π e ⁻¹ (x).

P r o o f. We have d M (x, y) ≤ d E ( x, π e ⁻¹ (y)) for all x ∈ π e ⁻¹ (x). We must prove the reverse inequality. Take arbitrary points x, y ∈ M and x ∈ π e ⁻¹ (x).

Consider a 1 , . . . , a k ∈ D and f ₁ , . . . , f k ∈ Hol(D, M ) such that f 1 (0) = x, f i (a i ) = f i+1 (0), f k (a k ) = y . Consider the pull-back diagram

D × M E −→ ^σ

¹

E

Θ

1

↓ ↓ π

D −→ ^f

¹

M

By a result of Royden [6], there is an equivalence Φ 1 : D × F → D × M E of holomorphic fiber bundles over M . Thus there exists c 1 ∈ F such that σ 1 ◦ Φ ₁ (0, c 1 ) = x. We define a holomorphic map ϕ e 1 : D → E by

ϕ 1 (z) = σ 1 ◦ Φ ₁ (z, c 1 ) for all z ∈ D . Consider the pull-back diagram

D × M E −→ ^σ

²

E

Θ

2

↓ ↓ π

D −→ ^f

²

M

Reasoning as above, there exists an equivalence Φ 2 : D × F → D × M E of holomorphic fiber bundles over M and a point c 2 ∈ F such that σ 2 ◦ Φ 2 (0, c 2 ) = ϕ 1 (a 1 ). Define a holomorphic map ϕ 2 : D → E by

ϕ 2 (z) = σ 2 ◦ Φ ₂ (z, c 2 ) for all z ∈ D .

Continuing this process we find maps ϕ 1 , . . . , ϕ k ∈ Hol(D, E) such that ϕ 1 (0)= x, ϕ e i (a i )=ϕ i+1 (0), ϕ k (a k )∈π ⁻¹ (y). Thus d M (x, y)≥d E ( x, π e ⁻¹ (y)) for all x ∈ π e ⁻¹ (x).

3. Lemma. Let π : e X → X be a holomorphic map between two com- plex spaces satisfying the following: For every x ∈ X there exists an open neighbourhood U x of x such that π ⁻¹ (U x ) is taut. Then e X is taut if so is X.

P r o o f. Assume that X is taut and a sequence { e f n } ⊂ Hol(D, e X) is not compactly divergent. Without loss of generality we may suppose that there exists a sequence {z n } ⊂ D converging to a point z ∈ D such that { e f n (z n )}

converges to a point p ∈ e e X.

Obviously {f n = π ◦ e f n } ⊂ Hol(D, X) is not compactly divergent and

we can assume that {f n } converges uniformly to a map F ∈ Hol(D, X).

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Put p = π( p) = F (z). Take an open neighbourhood U e p of p such that π ⁻¹ (U p ) is taut. Since {f n } converges to F , there exists an open subset V of D such that f n (V ) ⊂ U p for all n ≥ N .

On the other hand, since { e f n (z n )} converges to p and since π e ⁻¹ (U p ) is taut, we can assume that { e f n |V } ⊂ Hol(V, π ⁻¹ (U p )) converges uniformly to a map e F ∈ Hol(V, e X).

Consider the family V of all pairs (W, Φ), where W is an open subset of D and Φ ∈ Hol(W, e X) such that there exists a subsequence { e f n

k

|W } of { e f n |W } which converges uniformly to Φ in Hol(W, e X).

We define an order relation in V by setting (W 1 , Φ 1 ) ≤ (W 2 , Φ 2 ) if W 1 ⊂ W 2 and for every subsequence { e f n

k

|W 1 } of { e f n |W 1 } converging to Φ 1 in Hol(W 1 , e X) uniformly on compact sets, the sequence { e f n

k

|W 2 } contains a subsequence converging to Φ 2 in Hol(W 2 , e X). Assume that {(W α , Φ α )} α∈Λ is a well-ordered subset of V. Put W 0 = S

α∈Λ W α . Define Φ 0 ∈ Hol(W ₀ , e X) by Φ 0 |W α = Φ α for all α ∈ Λ. Take a sequence {(W i , Φ i )} ^∞ _i=1 ⊂ {(W α , Φ α )} α∈Λ

such that (W 1 , Φ 1 ) ≤ (W 2 , Φ 2 ) ≤ . . . and W 0 = S ∞

i=1 W i . Then there exists a subsequence { e f ¹ _n |W ₁ } of { e f n |W ₁ } converging to Φ ₁ in Hol(W 1 , e X).

Consider { e f ¹ _n |W ₂ }. As above { e f ¹ _n } contains a subsequence { e f ² _n } such that { e f ² _n |W 2 } converges to Φ 2 in Hol(W 2 , e X).

Continuing this process we get sequences { e f ^k _n } such that { e f ^k _n } ∈ { e f ^k−1 _n } for every k ≥ 2 and { e f ^k _n |W _k } converges uniformly to Φ _k in Hol(W k , e X).

Therefore { e f ^k _k } converges to Φ ₀ in Hol(W 0 , e X).

Thus (W 0 , Φ 0 ) ∈ V and hence the subset {(W α , Φ α )} α∈Λ has an upper bound. By Zorn’s Lemma, the family V has a maximal element (W, Φ).

Let { e f n

k

|W } be a subsequence of { e f n |W } converging uniformly to Φ in Hol(W, e X). Take z ∈ W and an open neighbourhood U of F (z 0 ) in X such that π ⁻¹ (U ) is taut. Since {f n } converges uniformly to F ∈ Hol(D, X) in Hol(D, X), there exists an open neighbourhood W 0 of z 0 in D such that (π ◦ e f n )(W 0 ) ⊂ U for all n ≥ 1. Hence e f n (W 0 ) ⊂ π ⁻¹ (U ) for all n ≥ 1.

Take z 1 ∈W ₀ ∩W . Then { e f n

k

(z 1 )} is convergent. By the normality of the family Hol(W 0 , π ⁻¹ (U )) and by the maximality of (W, Φ) we have W 0 ⊂ W and W = D.

Notice that for the hyperbolicity the above lemma was proved by East- wood [1].

4. Corollary. Let π : e X → X be a holomorphic covering map between two complex spaces. Then e X is taut if and only if so is X.

P r o o f. ⇒ Assume that e X is taut and a sequence {f n } ⊂ Hol(D, X) is

not compactly divergent. Then there exists a sequence {z n } ⊂ D converging

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to z 0 ∈ D such that {f _n (z n )} converges to p ∈ X. Put f n (z n ) = y n .

Since e X is taut, X is hyperbolic. Take p ∈ π e ⁻¹ (p). Since d X (y n , p) = inf y ˜

n

∈π

⁻¹

(y

n

) d X ˜ ( e y n , p) (see [2]), there exists e e y n ∈ π ⁻¹ (y n ) such that d X ˜ ( e y n , p) < d e X (y n , p) + 1/n. Lift f n to a map e f n such that π ◦ e f n = f n

and e f n (z n ) = y e n . Since { e f n (z n )} converges to p and e e X is taut, we can as- sume that { e f n } converges uniformly to a map e f ∈ Hol(D, e X) in Hol(D, e X).

Then {f n } converges to π ◦ e f in Hol(D, X).

⇐ The proof is deduced immediately from Lemma 3.

5. Theorem B. Let E be a holomorphic fiber bundle over M with fiber F and projection π : E → M , where E, F , M are complex manifolds. Then E is taut if and only if both F and M are.

P r o o f. ⇒ Assume that E is taut. Since F embeds as a closed complex subspace in E, F is taut. Assume that a sequence {f n } ⊂ Hol(D, M ) is not compactly divergent. Without loss of generality we may suppose that there exists a sequence {z n } ⊂ D converging to z ∈ D such that {y n = f n (z n )} converges to p ∈ M . Take p ∈ π e ⁻¹ (p). By Theorem A, d M (y n , p) = inf x

n

∈π

⁻¹

(y

n

) d E ( p, x e n ). Hence there exists x n ∈ E such that d _E ( p, x e n ) <

d M (p, y n ) + 1/n.

Pull back the fiber bundle π : E → M to a fiber bundle Θ n : D × M E → D getting a commutative diagram

D × M E −→ ^σ

ⁿ

E

Θ

n

↓ ↓ π

D −→ ^f

ⁿ

M

As in the proof of Theorem A there exists an equivalence Φ n : D × F → D × M E of holomorphic fiber bundles over D and c n ∈ F such that σ n ◦ Φ n (z n , c n ) = x n . Put ϕ n (z) = σ n ◦ Φ _n (z, c n ) for all z ∈ D. It is easy to see that {ϕ n (z n )} converges to p ∈ E. Without loss of generality we may e assume that {ϕ n } converges to a map ϕ ∈ Hol(D, E) in Hol(D, E). Hence {f _n } converges to π ◦ ϕ in Hol(D, M ).

⇐ The converse is deduced immediately from Lemma 3.

References

[1] A. E a s t w o o d, A propos des vari´ et´ es hyperboliques compl` etes, C. R. Acad. Sci. Paris 280 (1975), 1071–1075.

[2] S. K o b a y a s h i, Hyperbolic Manifolds and Holomorphic Mappings, Dekker, New York 1970.

[3] —, Intrinsic distance, measures and geometric function theory , Bull. Amer. Math.

Soc. 82 (1976), 357–416.

[4] S. L a n g, Introduction to Hyperbolic Complex Spaces, Springer, 1987.

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A note on the Kobayashi pseudodistance and the tautness of holomorphic fiber bundles

POLONICI MATHEMATICI LVIII.1 (1993)

A note on the Kobayashi pseudodistance and the tautness of holomorphic fiber bundles

by Do Duc Thai and Nguyen Le Huong (Ha noi)

Abstract. We show a relation between the Kobayashi pseudodistance of a holomor- phic fiber bundle and the Kobayashi pseudodistance of its base. Moreover, we prove that a holomorphic fiber bundle is taut iff both the fiber and the base are taut.

In this section we recall some definitions and relevant properties.

We denote the Kobayashi pseudodistance on a complex space M by d M

Also, a complete hyperbolic space is taut, and a taut complex space is hyperbolic [3, p. 378].

Acknowledgement. The authors are grateful to the referee for many valuable comments.

2. Theorem A. Let E be a holomorphic fiber bundle over M with hy- perbolic fiber F and projection π : E → M , where E, F , M are complex

1991 Mathematics Subject Classification: 32H15, 32L05.

Key words and phrases: holomorphic fiber bundle, Kobayashi pseudodistance, hyper-

bolic space, taut space.

manifolds. Then d M (x, y) = d E ( e x, π −1 (y)) = inf y∈π ˜

(y) d E ( e x, y) for all e x, y ∈ M and all x ∈ π e −1 (x).

P r o o f. We have d M (x, y) ≤ d E ( x, π e −1 (y)) for all x ∈ π e −1 (x). We must prove the reverse inequality. Take arbitrary points x, y ∈ M and x ∈ π e −1 (x).

Consider a 1 , . . . , a k ∈ D and f 1 , . . . , f k ∈ Hol(D, M ) such that f 1 (0) = x, f i (a i ) = f i+1 (0), f k (a k ) = y . Consider the pull-back diagram

D × M E −→ σ

E

Θ

↓ ↓ π

D −→ f

M

By a result of Royden [6], there is an equivalence Φ 1 : D × F → D × M E of holomorphic fiber bundles over M . Thus there exists c 1 ∈ F such that σ 1 ◦ Φ 1 (0, c 1 ) = x. We define a holomorphic map ϕ e 1 : D → E by

ϕ 1 (z) = σ 1 ◦ Φ 1 (z, c 1 ) for all z ∈ D . Consider the pull-back diagram

D × M E −→ σ

E

Θ

↓ ↓ π

D −→ f

M

Reasoning as above, there exists an equivalence Φ 2 : D × F → D × M E of holomorphic fiber bundles over M and a point c 2 ∈ F such that σ 2 ◦ Φ 2 (0, c 2 ) = ϕ 1 (a 1 ). Define a holomorphic map ϕ 2 : D → E by

ϕ 2 (z) = σ 2 ◦ Φ 2 (z, c 2 ) for all z ∈ D .

Continuing this process we find maps ϕ 1 , . . . , ϕ k ∈ Hol(D, E) such that ϕ 1 (0)= x, ϕ e i (a i )=ϕ i+1 (0), ϕ k (a k )∈π −1 (y). Thus d M (x, y)≥d E ( x, π e −1 (y)) for all x ∈ π e −1 (x).

3. Lemma. Let π : e X → X be a holomorphic map between two com- plex spaces satisfying the following: For every x ∈ X there exists an open neighbourhood U x of x such that π −1 (U x ) is taut. Then e X is taut if so is X.

P r o o f. Assume that X is taut and a sequence { e f n } ⊂ Hol(D, e X) is not compactly divergent. Without loss of generality we may suppose that there exists a sequence {z n } ⊂ D converging to a point z ∈ D such that { e f n (z n )}

converges to a point p ∈ e e X.

Obviously {f n = π ◦ e f n } ⊂ Hol(D, X) is not compactly divergent and

we can assume that {f n } converges uniformly to a map F ∈ Hol(D, X).

Put p = π( p) = F (z). Take an open neighbourhood U e p of p such that π −1 (U p ) is taut. Since {f n } converges to F , there exists an open subset V of D such that f n (V ) ⊂ U p for all n ≥ N .

On the other hand, since { e f n (z n )} converges to p and since π e −1 (U p ) is taut, we can assume that { e f n |V } ⊂ Hol(V, π −1 (U p )) converges uniformly to a map e F ∈ Hol(V, e X).

Consider the family V of all pairs (W, Φ), where W is an open subset of D and Φ ∈ Hol(W, e X) such that there exists a subsequence { e f n

|W } of { e f n |W } which converges uniformly to Φ in Hol(W, e X).

We define an order relation in V by setting (W 1 , Φ 1 ) ≤ (W 2 , Φ 2 ) if W 1 ⊂ W 2 and for every subsequence { e f n

|W 1 } of { e f n |W 1 } converging to Φ 1 in Hol(W 1 , e X) uniformly on compact sets, the sequence { e f n

|W 2 } contains a subsequence converging to Φ 2 in Hol(W 2 , e X). Assume that {(W α , Φ α )} α∈Λ is a well-ordered subset of V. Put W 0 = S

α∈Λ W α . Define Φ 0 ∈ Hol(W 0 , e X) by Φ 0 |W α = Φ α for all α ∈ Λ. Take a sequence {(W i , Φ i )} ∞ i=1 ⊂ {(W α , Φ α )} α∈Λ

such that (W 1 , Φ 1 ) ≤ (W 2 , Φ 2 ) ≤ . . . and W 0 = S ∞

i=1 W i . Then there exists a subsequence { e f 1 n |W 1 } of { e f n |W 1 } converging to Φ 1 in Hol(W 1 , e X).

Consider { e f 1 n |W 2 }. As above { e f 1 n } contains a subsequence { e f 2 n } such that { e f 2 n |W 2 } converges to Φ 2 in Hol(W 2 , e X).

Continuing this process we get sequences { e f k n } such that { e f k n } ∈ { e f k−1 n } for every k ≥ 2 and { e f k n |W k } converges uniformly to Φ k in Hol(W k , e X).

Therefore { e f k k } converges to Φ 0 in Hol(W 0 , e X).

Thus (W 0 , Φ 0 ) ∈ V and hence the subset {(W α , Φ α )} α∈Λ has an upper bound. By Zorn’s Lemma, the family V has a maximal element (W, Φ).

Let { e f n

Take z 1 ∈W 0 ∩W . Then { e f n

(z 1 )} is convergent. By the normality of the family Hol(W 0 , π −1 (U )) and by the maximality of (W, Φ) we have W 0 ⊂ W and W = D.

Notice that for the hyperbolicity the above lemma was proved by East- wood [1].

4. Corollary. Let π : e X → X be a holomorphic covering map between two complex spaces. Then e X is taut if and only if so is X.

P r o o f. ⇒ Assume that e X is taut and a sequence {f n } ⊂ Hol(D, X) is

not compactly divergent. Then there exists a sequence {z n } ⊂ D converging

to z 0 ∈ D such that {f n (z n )} converges to p ∈ X. Put f n (z n ) = y n .

Since e X is taut, X is hyperbolic. Take p ∈ π e −1 (p). Since d X (y n , p) = inf y ˜

∈π

(y

) d X ˜ ( e y n , p) (see [2]), there exists e e y n ∈ π −1 (y n ) such that d X ˜ ( e y n , p) < d e X (y n , p) + 1/n. Lift f n to a map e f n such that π ◦ e f n = f n

and e f n (z n ) = y e n . Since { e f n (z n )} converges to p and e e X is taut, we can as- sume that { e f n } converges uniformly to a map e f ∈ Hol(D, e X) in Hol(D, e X).

Then {f n } converges to π ◦ e f in Hol(D, X).

⇐ The proof is deduced immediately from Lemma 3.

5. Theorem B. Let E be a holomorphic fiber bundle over M with fiber F and projection π : E → M , where E, F , M are complex manifolds. Then E is taut if and only if both F and M are.

∈π

(y

) d E ( p, x e n ). Hence there exists x n ∈ E such that d E ( p, x e n ) <

d M (p, y n ) + 1/n.

Pull back the fiber bundle π : E → M to a fiber bundle Θ n : D × M E → D getting a commutative diagram

D × M E −→ σ

E

Θ

↓ ↓ π

D −→ f

M

manifolds. Then d M (x, y) = d E ( e x, π ⁻¹ (y)) = inf _y∈π _˜

_(y) d E ( e x, y) for all e x, y ∈ M and all x ∈ π e ⁻¹ (x).

P r o o f. We have d M (x, y) ≤ d E ( x, π e ⁻¹ (y)) for all x ∈ π e ⁻¹ (x). We must prove the reverse inequality. Take arbitrary points x, y ∈ M and x ∈ π e ⁻¹ (x).

Consider a 1 , . . . , a k ∈ D and f ₁ , . . . , f k ∈ Hol(D, M ) such that f 1 (0) = x, f i (a i ) = f i+1 (0), f k (a k ) = y . Consider the pull-back diagram

D × M E −→ ^σ

D −→ ^f

By a result of Royden [6], there is an equivalence Φ 1 : D × F → D × M E of holomorphic fiber bundles over M . Thus there exists c 1 ∈ F such that σ 1 ◦ Φ ₁ (0, c 1 ) = x. We define a holomorphic map ϕ e 1 : D → E by

ϕ 1 (z) = σ 1 ◦ Φ ₁ (z, c 1 ) for all z ∈ D . Consider the pull-back diagram

D × M E −→ ^σ

D −→ ^f

ϕ 2 (z) = σ 2 ◦ Φ ₂ (z, c 2 ) for all z ∈ D .

Continuing this process we find maps ϕ 1 , . . . , ϕ k ∈ Hol(D, E) such that ϕ 1 (0)= x, ϕ e i (a i )=ϕ i+1 (0), ϕ k (a k )∈π ⁻¹ (y). Thus d M (x, y)≥d E ( x, π e ⁻¹ (y)) for all x ∈ π e ⁻¹ (x).

3. Lemma. Let π : e X → X be a holomorphic map between two com- plex spaces satisfying the following: For every x ∈ X there exists an open neighbourhood U x of x such that π ⁻¹ (U x ) is taut. Then e X is taut if so is X.

Put p = π( p) = F (z). Take an open neighbourhood U e p of p such that π ⁻¹ (U p ) is taut. Since {f n } converges to F , there exists an open subset V of D such that f n (V ) ⊂ U p for all n ≥ N .

On the other hand, since { e f n (z n )} converges to p and since π e ⁻¹ (U p ) is taut, we can assume that { e f n |V } ⊂ Hol(V, π ⁻¹ (U p )) converges uniformly to a map e F ∈ Hol(V, e X).

α∈Λ W α . Define Φ 0 ∈ Hol(W ₀ , e X) by Φ 0 |W α = Φ α for all α ∈ Λ. Take a sequence {(W i , Φ i )} ^∞ _i=1 ⊂ {(W α , Φ α )} α∈Λ

i=1 W i . Then there exists a subsequence { e f ¹ _n |W ₁ } of { e f n |W ₁ } converging to Φ ₁ in Hol(W 1 , e X).

Consider { e f ¹ _n |W ₂ }. As above { e f ¹ _n } contains a subsequence { e f ² _n } such that { e f ² _n |W 2 } converges to Φ 2 in Hol(W 2 , e X).

Continuing this process we get sequences { e f ^k _n } such that { e f ^k _n } ∈ { e f ^k−1 _n } for every k ≥ 2 and { e f ^k _n |W _k } converges uniformly to Φ _k in Hol(W k , e X).

Therefore { e f ^k _k } converges to Φ ₀ in Hol(W 0 , e X).

Take z 1 ∈W ₀ ∩W . Then { e f n

(z 1 )} is convergent. By the normality of the family Hol(W 0 , π ⁻¹ (U )) and by the maximality of (W, Φ) we have W 0 ⊂ W and W = D.

to z 0 ∈ D such that {f _n (z n )} converges to p ∈ X. Put f n (z n ) = y n .

Since e X is taut, X is hyperbolic. Take p ∈ π e ⁻¹ (p). Since d X (y n , p) = inf y ˜

) d X ˜ ( e y n , p) (see [2]), there exists e e y n ∈ π ⁻¹ (y n ) such that d X ˜ ( e y n , p) < d e X (y n , p) + 1/n. Lift f n to a map e f n such that π ◦ e f n = f n

) d E ( p, x e n ). Hence there exists x n ∈ E such that d _E ( p, x e n ) <

D × M E −→ ^σ

D −→ ^f