ON TAUTNESS MODULO AN ANALYTIC SUBSET OF COMPLEX SPACES

Abstract. The main goal of this article is to give necessary and sufficient conditions on the tautness modulo an analytic subset of complex spacesAbstract. The main goal of this article is to give necessary and sufficient conditions on the tautness modulo an analytic subset of complex spaces

COMPLEX SPACES

PHAM VIET DUC*, MAI ANH DUC** AND PHAM NGUYEN THU TRANG**

Abstract The main goal of this article is to give necessary and

sufficient conditions on the tautness modulo an analytic subset of

complex spaces.

1 Introduction The notions of hyperbolicity and tautness modulo an analytic sub-set of complex spaces are due to S Kobayashi (see [4, p.68]) Much attention has been given to these notions, and the results on this prob-lem can be applied to many areas of mathematics, in particular to the extensions of holomorphic mappings For details see [4] and [5]

The main goal of this article is to give necessary and sufficient con-ditions on the tautness modulo an analytic subset of complex spaces First of all, we recall the definitions of the tautness modulo an ana-lytic subset of complex spaces

Let ∆ be the open unit disk in the complex plane and ρ(a, b) := tanh−1 |a − b|

|1 − ab| be the Poincare distance on ∆.

Definition 1.1 (see [5, p.240]) Let X be a complex space and S be

an analytic subset in X We say that X is taut modulo S if it is normal modulo S, i.e., for every sequence {fn} in Hol(∆, X) one of the following holds:

i There exists a subsequence of {fn} which converges uniformly

to f ∈ Hol(∆, X) in Hol(∆, X);

ii The sequence {fn} is compactly divergent modulo S in Hol(∆, X), i.e., for each compact set K ⊂ ∆ and each compact set L ⊂

2000 Mathematics Subject Classification Primary 32M05; Secondary 32H02, 32H15, 32H50.

Key words and phrases Hyperbolicity modulo an analytic subset, tautness mod-ulo an analytic subset.

The research of the authors is supported by an NAFOSTED grant of Vietnam (Grant No 101.04-2014.48).

1

X \ S, there exists an integer N such that fn(K) ∩ L = ∅ for all n ≥ N

If S = ∅, then X is said to be taut By the definition, it is easy to see that if S ⊂ S0 ⊂ X and X is taut modulo S, then X is taut modulo

S0 In particular, if X is taut, then X is taut modulo S for any analytic subset S of X

2 Criteria on the tautness modulo an analytic subset of

complex spaces First of all, we show the inheritance of the tautness modulo an ana-lytic subset of complex spaces under certain assumptions

Theorem 2.1 Let X be a complex space and S be an analytic subset in

X Assume that X = S

i∈I

Xi, where {Xi} are irreducible components of

X Then X is taut modulo S if and only if Xiis taut modulo Si := Xi∩S for all i ∈ I

Proof Assume that {Xi}i∈I are irreducible components of X Then {Xi}i∈I is a locally finite family and Hol(∆, X) = S

i∈I

Hol(∆, Xi) (⇒) Assume that X is taut modulo S Fix i ∈ I and assume that {fn} ⊂ Hol(∆, Xi) is not compactly divergent modulo Siin Hol(∆, Xi) Then there exists {zn} ⊂ ∆ such that {zn} converges to z ∈ ∆ and {fn(zn)} converges to a point p ∈ Xi\ Si Since X is taut modulo S, it implies that {fn} ⊂ Hol(∆, X) converges uniformly to f in Hol(∆, X) Since Xi is closed in X, we have f ∈ Hol(∆, Xi) Thus Xi is taut mod-ulo Xi∩ S for all i

(⇐) Assume that Xi is taut modulo Si for all i and {fn} ⊂ Hol(∆, X)

is not compactly divergent modulo S in Hol(∆, X) Then there exist

a compact subset K ⊂ ∆, a compact subset K0 ⊂ X \ S, and a subse-quence {fnk} of {fn} such that fnk(K) ∩ K0 6= ∅ for all k ≥ 1 Without lost of generality, we may assume that fn(K) ∩ K0 6= ∅ for all n ≥ 1

By the locally finite property of the family {Xi}i≥1, there exists a finite subset I of indices such that K0 ⊂ S

i∈I

Xi and K0∩ Xi 6= ∅ for all i ∈ I and K0∩Xi = ∅ for all i 6∈ I Since fn(K) ∩ K0 6= ∅ and fn(K) ⊂ fn(∆) for all n ≥ 1, fn(∆) 6⊂ Xi for all i 6∈ I Since fn(∆) is a subset of some irreducible component of X, it follows that fn(∆) ⊂ S

i∈I

Xi for all n ≥ 1, and hence, there exist an index i0 ∈ I and a subsequence {fnk} of {fn} such that fnk(∆) ⊂ Xi 0 for all k ≥ 1 Without lost of generality, we may assume that fn(∆) ⊂ Xi0 for all n ≥ 1 This yields that fn(K) ∩ (K0 ∩ Xi ) 6= ∅ for all n ≥ 1, and hence, the sequence

{fn} is not compactly divergent modulo Si 0 in Hol(∆, Xi 0) Since Xi 0

is taut modulo Si0, the sequence {fn} contains a subsequence which is convergent uniformly in Hol(∆, Xi0) ⊂ Hol(∆, X) Thus, X is taut

Theorem 2.2 Let X be a complex space and S be an analytic subset

in X such that X is taut modulo S Let D be a Cartier divisor in X Then X \ D is taut modulo S \ D

Proof Assume that {fn} ⊂ Hol(∆, X \ D) is not compactly divergent modulo S \ D in Hol(∆, X \ D) Then there exist a compact subset

K ⊂ ∆, a compact subset L ⊂ (X \ D) \ (S \ D) = X \ (S ∪ D) and

a subsequence {fnk} of {fn} such that fnk(K) ∩ L 6= ∅ for all k ≥ 1 Without lost of generality, we may assume that fn(K) ∩ L 6= ∅ for all

n ≥ 1 Thus {fn} is not compactly divergent modulo S in Hol(∆, X) Hence {fn} converges uniformly on compact subsets of ∆ to a mapping

f in Hol(∆, X) We now show that f (∆) ∩ D = ∅ Suppose on the contrary that f (∆) ∩ D 6= ∅ Let A = f−1(D) It is easy to see that

A 6= ∅ and A is closed in ∆ We have to prove that A is open in ∆ Take z0 ∈ A and U is a neighbourhood of f (z0) ∈ D in X such that

U ∩ D = {x ∈ X : ϕ(x) = 0}, where ϕ ∈ Hol(∆, C) Choose rel-atively compact complete hyperbolic neighbourhoods W1, W2 of f (z0) such that W1 ⊂ W2 ⊂ W2 ⊂ U Since {fn} converges uniformly to f

in Hol(∆, X), there exists an open neighbourhood V of z0 in ∆ such that fn(V ) ⊂ W1\ D for all n ≥ 1 Since W2 is complete hyperbolic,

W2 \ D is complete hyperbolic too and thus W2 \ D is taut i.e., the family Hol(V, W2\ D) is normal

Assume that there exists z ∈ V such that f (z) 6∈ D Since {fn(z)} converges uniformly to f (z) ∈ W2 \ D, the sequence {fn

V} is not compactly divergent in Hol(V, W2\D) Since Hol(V, W2\D) is normal,

we can assume that {fn

V} converges uniformly to ef ∈ Hol(V, W2\ D)

It implies that ef (z0) = f (z0) ∈ D This contradics to the hypothesis that ef ∈ Hol(V, W2 \ D) Thus f (V ) ⊂ D i.e.,V ⊂ f−1(D) = A Hence A is an open subset ∆ Thus A = ∆ i.e., f (∆) ⊂ D This is a contradition We get f (∆) ∩ D = ∅ and hence X \ D is taut modulo

We now give a quantitative criterion on the tautness modulo an ana-lytic subset of complex spaces which is similar to the Royden criterion

on the tautness of complex spaces (see [2])

First of all, we define the following function ek(2)X

Definition 2.3 Let X be a complex space and S be an analytic subset

of X For each z0 ∈ X \ S, z00∈ X, we set

e

kX(2)(z0, z00) = inf{ρ(0, a) + ρ(0, b)| a, b ∈ ∆ such that ∃ϕ1 ∈ Hol(∆, X \ S),

∃ϕ2 ∈ Hol(∆, X) such that ϕ1(0) = z0, ϕ1(a) = ϕ2(0), ϕ2(b) = z00}

Obviously, ekX(2) ≤ dX, where dX is the Kobayashi pseudodistance of X

We now prove the following criterion for the tautness modulo an analytic hypersurface

Theorem 2.4 Let X be a complex space and S be an analytic hyper-surface in X Then X is taut modulo S if and only if B

e

k(2)X (a, R) := {z ∈ X : ek(2)X (a, z) < R} b X for any R > 0 and a ∈ X \ S

In oder to prove the above theorem, we need the following

Lemma 2.5 (see [1, Lemma 2.5]) Let Z be a complex manifold Let S

be a hypersurface of a complex spaces X If {ϕn}n≥1 ⊂ Hol(Z, X \ S) converging uniformly on every compact subsets of Z to a mapping ϕ ∈ Hol(Z, X), then either ϕ(Z) ⊂ X \ S or ϕ(Z) ⊂ S

Proof of Theorem 2.4

(⇒) Suppose on the contrary Then, there exist a point z0 ∈ X \ S, a number R > 0 and a sequence {zn}n≥1 ⊂ X such that ek(2)X (z0, zn) < R and the sequence {zn} does not contain any convergent subsequence

By the definition, choose sequences {ϕ1

n}n≥1⊂ Hol(∆, X\S), {ϕ2

n}n≥1 ⊂ Hol(∆, X), {a1

n}n≥1 ⊂ ∆, {a2

n}n≥1 ⊂ ∆ such that ϕ1

n(0) = z0, ϕ1

n(a1

n) =

ϕ2n(0), ϕ2n(a2n) = zn and ρ(0, a1n) + ρ(0, a2n) < R ∀n ≥ 1 Since ϕ1n(0) =

z0 ∈ X \ S and X is taut modulo S, there exists a subsequence {ϕ1

n k}

of {ϕ1

n} converging uniformly on compact subsets of ∆ to a mapping

ϕ1 in Hol(∆, X) By Lemma 2.5 then ϕ1(∆) ⊂ X \ S or ϕ1(∆) ⊂ S Since z0 ∈ X \ S and lim

k→∞ϕ1n

k(0) = ϕ1(0) = z0 ∈ X \ S, it implies that

ϕ1 ∈ Hol(∆, X \ S).We can suppose that {a1

n k} converges to a1 ∈ ∆

We have ϕ1(0) = z0 and ϕ1(a1) = lim

k→∞ϕ1n

k(a1n

k) = lim

k→∞ϕ2n

k(0) So, the tautness modulo S of X implies that there is a subsequence of {ϕ2

nk} converging uniformly on compact subsets of ∆ to a mapping

ϕ2 in Hol(∆, X) Without loss of generality, we may assume that the subsequence {ϕ2

n k} converges uniformly on compact subsets of ∆ to a mapping ϕ2 and {a2

n k} converges to a point a2 ∈ ∆ Then lim

k→∞ϕ2n

k(a2n

k) = ϕ2(a2) ∈ X

This yields that lim

k→∞znk = ϕ2(a2) ∈ X This is a contradiction

(⇐) Assume that {fn} is a sequence in Hol(∆, X) such that {fn} is not compactly divergent modulo S in Hol(∆, X) Then there exists a compact subset K ⊂ ∆, a compact subset L ⊂ X \S and a subsequence {fnk} of sequence {fn} such that fnk(K) ∩ L 6= ∅, ∀k ≥ 1 Let ak ∈ K with fnk(ak) = zk ∈ L By the compactness of K and L, we may assume that lim

k→∞ak = a ∈ K and lim

k→∞zk = z0 ∈ L This implies that lim

k→∞fnk(ak) = z0 Without loss of generality, we may assume that ρ(ak, a) < 1 for each k ≥ 1 Since lim

k→∞zk = z0 ∈ L, without loss of generality, we may assume that for each k ≥ 1, there exists a mapping

gk ∈ Hol(∆, X \ S) such that gk(0) = z0, gk(13) = zk

Fix R0 > 0 Consider any point b ∈ ∆ with ρ(a, b) < R0 Take

a biholomorphic mapping ϕk : ∆ → ∆ such that ϕk(ak) = 0 Set

ϕk(b) = ˜b and ˜fnk = fnk ◦ ϕk−1 ∈ Hol(∆, X) Then

gk(0) = z0, gk(1

3) = zk = ˜fnk(0), ˜fnk(˜b) = fnk(b)

This implies that

e

kX(2)(z0, fnk(b)) ≤ ρ(0,1

3) + ρ(0, ˜b)

= ρ(0,1

3) + ρ(ak, b)

≤ ρ(0,1

3) + ρ(ak, a) + ρ(a, b)

≤ ρ(0,1

3) + 1 + R0 = R1. Thus, if b ∈ ∆ with ρ(a, b) < R0, then ekX(2)(z0, fnk(b)) < R1 This means that fnk(∆(a, R0)) ⊂ ekX(2)(z0, R1) b X, where ∆(a, R0) = {z ∈ C| |z − a| < R0} Since ekX(2)(z0, R1) is compact, by Arzela-Ascoli’s theorem, the sequence {fnk|∆(a,R

0 )} contains a subsequence which is convergent

in Hol(∆(a, R0), X) A further refinement gives a subsequence of the sequence {fnk} converging on ∆ Therefore, X is taut modulo S  Corollary 2.6 Let G be a domain in Cn and S be an analytic hy-persurface of G If G is not taut modulo S, then there exist a num-ber R > 0 and sequences {zn}n≥0 ⊂ G, {fn}n≥1 ⊂ Hol(∆, G \ S), {gn}n≥1⊂ Hol(∆, G), {αn}n≥0 ⊂ ∆, {βn}n≥0 ⊂ ∆ such that for n ≥ 1, the following are satisfied

i ek(2)G (z0, zn) < R;

ii fn(0) = z0 ∈ G \ S;

iii fn(αn) = gn(0);

iv gn(βn) = zn, zn→ w ∈ ∂G or |zn| → ∞;

v αn→ α0, βn→ β0

Proposition 2.7 Let X be an irreducible complex space and S be an analytic subset of X Assume that X is not taut modulo S Then there exists an analytic hypersurface S0 of X such that S ⊂ S0 and X is not taut modulo S0 also

Proof If S is hypersurface analytic of X, we end the proof Assume that S is not an analytic hypersurface of X By the assumption, there exists a sequence {fn}n≥1 ⊂ Hol(∆, X) such that the sequence {fn}n≥1

is not convergent uniformly in Hol(∆, X)and the one is not compactly divergent modulo S in Hol(∆, X) either Then there exist a compact subset K ⊂ ∆, a compact subset L ⊂ X \ S and a subsequence {fnk}

of sequence {fn} such that fnk(K) ∩ L 6= ∅, ∀k ≥ 1 Let ak ∈ K with fnk(ak) = zk ∈ L By the compactness of K and L, we may assume that lim

k→∞ak = a ∈ K and lim

k→∞zk = z0 ∈ L This implies that lim

k→∞fnk(ak) = z0 Obviously, z0 6∈ S We now show an analytic hypersurface S0 of X containing S and no meeting z0

Indeed, assume that {Uλ}λ∈Λ is a locally finite open covering of X Since S is an analytic subset of X, for each x ∈ S, there is a neigh-borhood Ux of x and holomorphic functions g1x, · · · , gxl on Ux such that

S ∩ Ux = {z ∈ Ux : g1x(z) = · · · = gxl(z) = 0} On the other hand,

by the local finiteness of the family {Uλ}λ∈Λ, there exists finitely many

Uλi1, · · · , Uλik containing x We put

Vx:=

k

\

j=1

Uλij

If z0 ∈ Ux, then there exists gx

i ∈ {gx

1, · · · , gx

l} such that gx

i(z0) 6= 0, and set

Sx := {z ∈ Vx : gxi(z) = 0}

If z0 6∈ Ux, then set

Sx := {z ∈ Vx : gx1(z) = 0}

We now take S0 is an union all Sx Then S0 is the required analytic

Remark 2.8 The condition on the irreducibility of X is necessary Indeed, consider the following example

X = {(z1, z2, z3) ∈ C3 : z3 = 0} ∪ {(z1, z2, z3) ∈ C3 : z1 = 0, z2 = 0},

and S = {(0, 0, 1)} Then, it is easy to see that X is not taut modulo

S, but X is taut modulo any analytic hypersurface S0 containing S

3 Invariance of the tautness modulo an analytic subset

of complex spaces under certain holomorphic mappings First of all, we recall the following

Definition 3.1 (see [6]) Let X be a complex space X is said to be weakly disc-convex if every sequence {fn} ⊂ Hol(∆, X) converges in Hol(∆, X) whenever the sequence {fn

∆ ∗} ⊂ Hol(∆∗

, X) converges in Hol(∆∗, X), where ∆∗ = ∆ \ {0}

Theorem 3.2 (Eastwood theorem for tautness modulo an analytic sub-set)

Let X, eX be complex spaces and S be an analytic subset in X As-sume that eX is weakly disc-convex and π : eX → X is a holomorphic mapping such that for each x ∈ X \ S, there exists an open neighbour-hood U of x in X \ S such that π−1(U ) is taut Then eX is taut modulo e

S := π−1(S) if X is taut modulo S

Proof Suppose that X is taut modulo S and { efn} ⊂ Hol(∆, eX) is not compactly divergent modulo S in Hol(∆, X) Hence, without loss

of generality there exist a compact set K ⊂ ∆ and a compact set

L ⊂ eX \ eS such that efn(K) ∩ L 6= ∅ for n ≥ 1 For each n ≥ 1, there exists zn ∈ K ⊂ ∆ such that efn(zn) ∈ L Since K and L are compact sets, by taking subsequences if necessary, we may assume that {zn} ⊂ K ⊂ ∆ such that zn → z ∈ K ⊂ ∆ and efn(zn) → p ∈e

L ⊂ eX \ eS Since π : eX → X is holomorphic, π(L) is a compact set in X \ S Therefore, {fn := πofen} ⊂ Hol(∆, X) is not compactly divergent modulo S in Hol(∆, X) Since X is taut modulo S, we may assume that {fn} converges uniformly to a mapping F ∈ Hol(∆, X) Obviously, π( efn(zn)) → π(p) and π( ee fn(zn)) = πofen(zn) = fn(zn) →

F (z) as n → ∞ Therefore, we can let p = π(p) = F (z).e

Since p ∈ ee X \ eS, p = π(p) 6∈ S According to the hypothesis ofe this theorem, there exists an open neighbourhood U of p in X \ S such that π−1(U ) is taut Taking an open neighbourhood V ⊂ F−1(U ) of

z in ∆ \ F−1(S) and since the sequence {fn} converges uniformly to

a mapping F , we may assume that fn(V ) ⊂ U This implies that e

fn(V ) ⊂ π−1(U ) for every n ≥ 1

Note that efn(zn) converges to p, so put K = {ze n} ∪ {z} and L = { efn(zn)}∪{p} Then efn(K)∩L 6= ∅ for all n Thus, the sequence { efnV}

is not compactly divergent modulo eS in Hol(∆, eX) On the other hand, since π−1(U ) is taut and efn(V ) ⊂ π−1(U ), { efn

V} converges uniformly

to mapping eF in Hol(V, π−1(U ))

Consider a family Γ consist of pairs (W, Φ), where W is an open in

∆ \ F−1(S) and Φ ∈ Hol(W, eX \ eS) such that there exists a subse-quence { efnk

W} of { efn

W} which converges uniformly to mapping Φ in Hol(W, eX \ eS)

According to the proof above, we have Γ 6= ∅ In the family Γ, consider the following order relation: (W1, Φ1) ≤ (W2, Φ2) if

i) W1 ⊂ W2 and

ii) for any subsequence { efnk

W 1} of { efn

W 1} that converges uniformly

to mapping Φ1 in Hol(W1, eX \ eS), there exists a subsequence of { efnk} that converges uniformly to mapping Φ2 in Hol(W2, eX \ eS)

Assume that {(Wα, Φα)}α∈Λ is a complete order subset of Γ Let

W0 = S

α∈Λ

Wα and define a mapping Φ0 ∈ Hol(W0, eX \ eS) such that

Φ0

W α = Φαfor α ∈ Λ Take a sequence {(Wi, Φi)}∞i=1⊂ {(Wα, Φα)}α∈Λ

such that

(W1, Φ1) ≤ (W2, Φ2) ≤ · · · , W0 =

∞

[

i=1

Wi

As the definition of Γ, the subsequence { ef1

n

W 1} of { efn

W 1} converges uniformly to the mapping Φ1 in Hol(W1, eX \ eS) As the definition of the order relation on Γ, there exists a subsequence { efn2} of { efn1} which converges uniformly to Φ2 in Hol(W2, eX \ eS)

By continuing this process, we get the sequence { efnk} such that { efk

n} ⊂ { efk−1

n } for all k ≥ 2 and { efk

n

Wk} converges uniformly to Φk

in Hol(Wk, eX \ eS) Thus, a diagonal sequence { efkk} converges uni-formly to Φ0 in Hol(W0, eX \ eS) Hence (W0, Φ0) ∈ Γ and the subset {(Wα, Φα)}α∈Λ of Γ has a supremum

By Zorn lemma, there exists a maximum element (W, Φ) of the family

Γ Assume that { efnk

W} is a subsequence of { efn

W} such that { efnk

W} converges uniformly to Φ in Hol(W, eX \ eS)

We now show that W = ∆ \ F−1(S) Suppose that there exists

z0 ∈ W ∩ (∆ \ F−1(S)) Take an open neighbourhood U of F (z0) in

X \ S such that π−1(U ) is taut Since {fn} converges uniformly to mapping F in Hol(∆, X), there exists an open neighbourhood W0 of

z0 in ∆ \ F−1(S) such that π ◦ efn(W0) ⊂ U Hence efn(W0) ⊂ π−1(U ), for all n ≥ 1

Fix z1 ∈ W0∩ W Then the sequence { efnk(z1)} is convergent Since Hol(W0, π−1(U )) is a normal family, { efnk

W 0} converges uniformly to

Φ0 in Hol(W0, π−1(U )) Thus (W0, Φ0) ∈ Γ It implies that W0 ⊂ W Hence W = ∆ \ F−1(S)

Since F−1(S) is an analytic subset in the open unit disc ∆, F−1(S)

is a discrete set and F−1(S) does not have any accumulation point Therefore, for each z ∈ F−1(S), there exists a number 0 < r < 1 such that B(z, r) ∩ F−1(S) = {z} Thus, ∆ \ F−1(S) is ∆∗ Hence the sequence { efnk

∆ ∗} is subsequence of { efn

∆ ∗}, where { efnk

∆ ∗} con-verges uniformly in Hol(∆∗, eX) Since eX is weakly disc-convex, { efnk} converges uniformly in Hol(∆, eX) Thus eX is taut modulo eS 

We now prove the invariance of the tautness modulo an analytic subset under holomorphic covering mappings and holomorphic fiber mappings

Theorem 3.3 Let π : eX → X be a covering space of a complex space

X and S be an analytic subset in X Then eX is taut modulo π−1(S) if and only if X is taut modulo S

Proof (⇒) Assume that eX is taut modulo eS := π−1(S) and {fn} ⊂ Hol(∆, X) is not compactly divergent modulo S in Hol(∆, X) Then there exists {zn} ⊂ ∆ such that {zn} converges to z ∈ ∆ and {yn :=

fn(zn)} converges to p ∈ X \ S

Take p ∈ πe −1(p) Since dX(yn, p) = inf

e

y n ∈π −1 (y n )dXe(yen,p), for eache

n ≥ 1 there exists yen ∈ π−1(yn) such that d

e

X(yen,p) < de X(yn, p) + 1n Since {yn} converges to p ∈ X \ S as n → ∞, d

e

X(yen,p) → 0 as n → ∞.e Moreover, since p ∈ X \ S, p ∈ ee X \ eS and since eX is taut modulo eS, e

X is hyperbolic modulo eS Thus yen converges to p as n → ∞.e

Lifting fn to efn such that πofen = fn and efn(zn) = yen Since yen converges to p, { ee fn} ⊂ Hol(∆, eX) is not compactly divergent modulo e

S in Hol(∆, eX) Since eX is taut modulo eS, { efn} converges uniformly

to a mapping ef in Hol(∆, eX) Hence {fn} converges uniformly to a mapping f := π ◦ ef in Hol(∆, X) So X is taut modulo S

(⇐) Assume that X is taut modulo S For each x ∈ X \ S, take an open neighbourhood U of x in X \ S such that U ∩ S = ∅ and U is taut Since π : eX → X is covering holomorphic, π−1(U ) = S

i∈I

Ui, where Ui are open subsets in eX such that Ui ∩ Uj = ∅ (i, j ∈ I, i 6= j) and

U i : Ui → U is biholomorphic for all i This implies that Ui is taut

So π−1(U ) = S

i∈I

Ui is taut

Assume that { efn} ⊂ Hol(∆, eX) is not compactly divergent mod-ulo eS in Hol(∆, eX) Repeating the proof of Theorem 3.2, it follows that { efnk

∆ ∗} is a subsequence of { efn

∆ ∗} such that { efnk

∆ ∗} converges uniformly to eΦ in Hol(∆∗, eX) and π ◦ eΦ = F

∆ ∗ Fix z0 ∈ ∆∗ Since F ∈ Hol(∆, X) and π : eX → X is a holomorphic covering mapping, we can lift F to a mapping Φ ∈ Hol(∆, eX) such that π ◦ Φ = F and eΦ(z0) = Φ(z0)

Since eΦ, Φ

∆ ∗ are lifting mappings of F

∆ ∗ with eΦ(z0) = Φ(z0) and

by the unique property of lifting mapping, it implies that eΦ = Φ

∆ ∗ Thus Φ is a holomorphic extension of eΦ from ∆∗ to ∆ Hence eX is

Lemma 3.4 Let E be a holomorphic bundle on M with a hyperbolic bundle F and the projection π : E → M , where E, F, M are complex manifold Then

dM(x, y) = dE(ex, π−1(y)) = inf

e y∈π −1 (y)dE(ex,y)e for x, y ∈ M , and ex ∈ π−1(x)

Proof Obviously, dM(x, y) ≤ dE(ex, π−1(y)) for ex ∈ π−1(x) Now we proof dM(x, y) ≥ dE(x, πe −1(y)) forx ∈ πe −1(x)

Indeed, take two arbitrary points x, y ∈ M andx ∈ πe −1(x) Assume that a1, a2, , ak ∈ ∆ and f1, f2, , fk ∈ Hol(∆, M ) such that f1(0) =

a1, fi(ai) = fi+1(0) for 1 ≤ i ≤ k − 1, and fk(ak) = y

Consider a pullback bundle

∆ ×M E σ1

−−−→ E

θ 1



 y





yπ

f 1

M

Then there exists an isomorphims Φ1 : ∆ × F → ∆ ×M E between two holomorphic bundles on ∆ Thus, there exists a point c1 ∈ F such that σ1◦ Φ1(0, c1) =x Let ϕe 1 : ∆ → E be a holomorphic map defined

by ϕ1(z) = σ1◦ Φ1(z, c1) for z ∈ ∆

We now give a quantitative criterion on the tautness modulo an ana-lytic subset of complex spaces which is similar to the Royden criterion

on the... modulo any analytic hypersurface S0 containing S

3 Invariance of the tautness modulo an analytic subset

of complex spaces under certain holomorphic mappings First of all,...

Proposition 2.7 Let X be an irreducible complex space and S be an analytic subset of X Assume that X is not taut modulo S Then there exists an analytic hypersurface S0 of X such that