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

ON TAUTNESS MODULO AN ANALYTIC SUBSET OF 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 subset 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 problem 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 conditions on the tautness modulo an analytic subset of complex spaces. First of all, we recall the definitions of the tautness modulo an analytic subset of complex spaces . Let ∆ be the open unit disk in the complex plane and ρ(a, b) := |a − b| tanh−1 be the Poincare distance on ∆. |1 − ab| 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 modulo an analytic subset. The research of the authors is supported by an NAFOSTED grant of Vietnam (Grant No. 101.04-2014.48). 1 2 PHAM VIET DUC*, MAI ANH DUC** AND PHAM NGUYEN THU TRANG** 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 ⊂ S ⊂ X and X is taut modulo S, then X is taut modulo S . 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 analytic 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 = Xi , where {Xi } are irreducible components of i∈I X. Then X is taut modulo S if and only if Xi is 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) = Hol(∆, Xi ). i∈I (⇒) Assume that X is taut modulo S. Fix i ∈ I and assume that {fn } ⊂ Hol(∆, Xi ) is not compactly divergent modulo Si in 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 modulo 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 K ⊂ X \ S, and a subsequence {fnk } of {fn } such that fnk (K) ∩ K = ∅ for all k ≥ 1. Without lost of generality, we may assume that fn (K) ∩ K = ∅ 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 K ⊂ Xi and K ∩ Xi = ∅ for all i ∈ I i∈I and K ∩Xi = ∅ for all i ∈ I. Since fn (K)∩K = ∅ and fn (K) ⊂ fn (∆) for all n ≥ 1, fn (∆) ⊂ Xi for all i ∈ I. Since fn (∆) is a subset of some irreducible component of X, it follows that fn (∆) ⊂ Xi for i∈I all n ≥ 1, and hence, there exist an index i0 ∈ I and a subsequence {fnk } of {fn } such that fnk (∆) ⊂ Xi0 for all k ≥ 1. Without lost of generality, we may assume that fn (∆) ⊂ Xi0 for all n ≥ 1. This yields that fn (K) ∩ (K ∩ Xi0 ) = ∅ for all n ≥ 1, and hence, the sequence ON TAUTNESS MODULO AN ANALYTIC SUBSET 3 {fn } is not compactly divergent modulo Si0 in Hol(∆, Xi0 ). Since Xi0 is taut modulo Si0 , the sequence {fn } contains a subsequence which is convergent uniformly in Hol(∆, Xi0 ) ⊂ Hol(∆, X). Thus, X is taut modulo S. 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 = ∅ for all k ≥ 1. Without lost of generality, we may assume that fn (K) ∩ L = ∅ 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 = ∅. Let A = f −1 (D). It is easy to see that A = ∅ 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 relatively 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) ∈ 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 f ∈ Hol(V, W2 \ D). It implies that f (z0 ) = f (z0 ) ∈ D. This contradics to the hypothesis that f ∈ 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 S \ D. We now give a quantitative criterion on the tautness modulo an analytic subset of complex spaces which is similar to the Royden criterion on the tautness of complex spaces (see [2]). (2) First of all, we define the following function kX . 4 PHAM VIET DUC*, MAI ANH DUC** AND PHAM NGUYEN THU TRANG** Definition 2.3. Let X be a complex space and S be an analytic subset of X. For each z ∈ X \ S, z ∈ X, we set (2) kX (z , z ) = inf{ρ(0, a) + ρ(0, b)| a, b ∈ ∆ such that ∃ϕ1 ∈ Hol(∆, X \ S), ∃ϕ2 ∈ Hol(∆, X) such that ϕ1 (0) = z , ϕ1 (a) = ϕ2 (0), ϕ2 (b) = z }. (2) Obviously, kX ≤ 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 hypersurface in X. Then X is taut modulo S if and only if Bk(2) (a, R) := X (2) {z ∈ X : kX (a, z) < R} 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 (2) number R > 0 and a sequence {zn }n≥1 ⊂ X such that kX (z0 , zn ) < R and the sequence {zn } does not contain any convergent subsequence. By the definition, choose sequences {ϕ1n }n≥1 ⊂ Hol(∆, X\S), {ϕ2n }n≥1 ⊂ Hol(∆, X), {a1n }n≥1 ⊂ ∆, {a2n }n≥1 ⊂ ∆ such that ϕ1n (0) = z0 , ϕ1n (a1n ) = 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 {ϕ1nk } of {ϕ1n } 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 ϕ1nk (0) = ϕ1 (0) = z0 ∈ X \ S, it implies that k→∞ ϕ1 ∈ Hol(∆, X \ S).We can suppose that {a1nk } converges to a1 ∈ ∆. We have ϕ1 (0) = z0 and ϕ1 (a1 ) = lim ϕ1nk (a1nk ) = lim ϕ2nk (0). So, k→∞ k→∞ the tautness modulo S of X implies that there is a subsequence of {ϕ2nk } converging uniformly on compact subsets of ∆ to a mapping ϕ2 in Hol(∆, X). Without loss of generality, we may assume that the subsequence {ϕ2nk } converges uniformly on compact subsets of ∆ to a mapping ϕ2 and {a2nk } converges to a point a2 ∈ ∆. Then lim ϕ2nk (a2nk ) = ϕ2 (a2 ) ∈ X. k→∞ ON TAUTNESS MODULO AN ANALYTIC SUBSET 5 This yields that lim znk = ϕ2 (a2 ) ∈ X. This is a contradiction. k→∞ (⇐) 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 = ∅, ∀k ≥ 1. Let ak ∈ K with fnk (ak ) = zk ∈ L. By the compactness of K and L, we may assume that lim ak = a ∈ K and lim zk = z0 ∈ L. This implies k→∞ k→∞ that lim fnk (ak ) = z0 . Without loss of generality, we may assume that k→∞ ρ(ak , a) < 1 for each k ≥ 1. Since lim zk = z0 ∈ L, without loss of k→∞ generality, we may assume that for each k ≥ 1, there exists a mapping gk ∈ Hol(∆, X \ S) such that gk (0) = z0 , gk ( 31 ) = 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 f˜nk = fnk ◦ ϕk −1 ∈ Hol(∆, X). Then 1 gk (0) = z0 , gk ( ) = zk = f˜nk (0), f˜nk (˜b) = fnk (b). 3 This implies that 1 (2) kX (z0 , fnk (b)) ≤ ρ(0, ) + ρ(0, ˜b) 3 1 = ρ(0, ) + ρ(ak , b) 3 1 ≤ ρ(0, ) + ρ(ak , a) + ρ(a, b) 3 1 ≤ ρ(0, ) + 1 + R0 = R1 . 3 (2) Thus, if b ∈ ∆ with ρ(a, b) < R0 , then kX (z0 , fnk (b)) < R1 . This means (2) that fnk (∆(a, R0 )) ⊂ kX (z0 , R1 ) X, where ∆(a, R0 ) = {z ∈ C| |z − (2) a| < R0 }. Since kX (z0 , R1 ) is compact, by Arzela-Ascoli’s theorem, the sequence {fnk |∆(a,R0 ) } 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 hypersurface of G. If G is not taut modulo S, then there exist a number 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 (2) i. kG (z0 , zn ) < R; ii. fn (0) = z0 ∈ G \ S; 6 PHAM VIET DUC*, MAI ANH DUC** AND PHAM NGUYEN THU TRANG** 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 S of X such that S ⊂ S and X is not taut modulo S 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 = ∅, ∀k ≥ 1. Let ak ∈ K with fnk (ak ) = zk ∈ L. By the compactness of K and L, we may assume that lim ak = a ∈ K and lim zk = z0 ∈ L. This implies k→∞ k→∞ that lim fnk (ak ) = z0 . Obviously, z0 ∈ S. We now show an analytic k→∞ hypersurface S 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 neighborhood Ux of x and holomorphic functions g1x , · · · , glx on Ux such that S ∩ Ux = {z ∈ Ux : g1x (z) = · · · = glx (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 k Uλij . Vx := j=1 gix If z0 ∈ Ux , then there exists ∈ {g1x , · · · , glx } such that gix (z0 ) = 0, and set Sx := {z ∈ Vx : gix (z) = 0}. If z0 ∈ Ux , then set Sx := {z ∈ Vx : g1x (z) = 0}. We now take S is an union all Sx . Then S is the required analytic hypersurface of X. 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}, ON TAUTNESS MODULO AN ANALYTIC SUBSET 7 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 S 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 subset) Let X, X be complex spaces and S be an analytic subset in X. Assume that X is weakly disc-convex and π : X → X is a holomorphic mapping such that for each x ∈ X \ S, there exists an open neighbourhood U of x in X \ S such that π −1 (U ) is taut. Then X is taut modulo S := π −1 (S) if X is taut modulo S. Proof. Suppose that X is taut modulo S and {fn } ⊂ Hol(∆, X) 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 ⊂ X \ S such that fn (K) ∩ L = ∅ for n ≥ 1. For each n ≥ 1, there exists zn ∈ K ⊂ ∆ such that fn (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 fn (zn ) → p ∈ L ⊂ X \ S. Since π : X → X is holomorphic, π(L) is a compact set in X \ S. Therefore, {fn := πo fn } ⊂ 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(fn (zn )) → π(p) and π(fn (zn )) = πo fn (zn ) = fn (zn ) → F (z) as n → ∞. Therefore, we can let p = π(p) = F (z). Since p ∈ X \ S, p = π(p) ∈ S. According to the hypothesis of 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 fn (V ) ⊂ π −1 (U ) for every n ≥ 1. Note that fn (zn ) converges to p, so put K = {zn } ∪ {z} and L = {fn (zn )}∪{p}. Then fn (K)∩L = ∅ for all n. Thus, the sequence {fn V } 8 PHAM VIET DUC*, MAI ANH DUC** AND PHAM NGUYEN THU TRANG** is not compactly divergent modulo S in Hol(∆, X). On the other hand, since π −1 (U ) is taut and fn (V ) ⊂ π −1 (U ), {fn V } converges uniformly to mapping F 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, X \ S) such that there exists a subsequence {fnk W } of {fn W } which converges uniformly to mapping Φ in Hol(W, X \ S). According to the proof above, we have Γ = ∅. In the family Γ, consider the following order relation: (W1 , Φ1 ) ≤ (W2 , Φ2 ) if i) W1 ⊂ W2 and ii) for any subsequence {fnk W1 } of {fn W1 } that converges uniformly to mapping Φ1 in Hol(W1 , X \ S), there exists a subsequence of {fnk } that converges uniformly to mapping Φ2 in Hol(W2 , X \ S). Assume that {(Wα , Φα )}α∈Λ is a complete order subset of Γ. Let W0 = Wα and define a mapping Φ0 ∈ Hol(W0 , X \ S) such that α∈Λ Φ0 Wα = Φα for α ∈ Λ. Take a sequence {(Wi , Φi )}∞ i=1 ⊂ {(Wα , Φα )}α∈Λ such that ∞ (W1 , Φ1 ) ≤ (W2 , Φ2 ) ≤ · · · , W0 = Wi . i=1 As the definition of Γ, the subsequence {fn1 W1 } of {fn W1 } converges uniformly to the mapping Φ1 in Hol(W1 , X \ S). As the definition of the order relation on Γ, there exists a subsequence {fn2 } of {fn1 } which converges uniformly to Φ2 in Hol(W2 , X \ S). By continuing this process, we get the sequence {fnk } such that {fnk } ⊂ {fnk−1 } for all k ≥ 2 and {fnk W } converges uniformly to Φk k in Hol(Wk , X \ S). Thus, a diagonal sequence {fkk } converges uniformly to Φ0 in Hol(W0 , X \ S). 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 {fnk W } is a subsequence of {fn W } such that {fnk W } converges uniformly to Φ in Hol(W, X \ S). 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 ON TAUTNESS MODULO AN ANALYTIC SUBSET 9 z0 in ∆ \ F −1 (S) such that π ◦ fn (W0 ) ⊂ U . Hence fn (W0 ) ⊂ π −1 (U ), for all n ≥ 1. Fix z1 ∈ W0 ∩ W . Then the sequence {fnk (z1 )} is convergent. Since Hol(W0 , π −1 (U )) is a normal family, {fnk W0 } 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 {fnk ∆∗ } is subsequence of {fn ∆∗ }, where {fnk ∆∗ } converges uniformly in Hol(∆∗ , X). Since X is weakly disc-convex, {fnk } converges uniformly in Hol(∆, X). Thus X is taut modulo S. 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 : X → X be a covering space of a complex space X and S be an analytic subset in X. Then X is taut modulo π −1 (S) if and only if X is taut modulo S. Proof. (⇒) Assume that X is taut modulo S := 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 ∈ π −1 (p). Since dX (yn , p) = inf dX (yn , p), for each −1 yn ∈π (yn ) n ≥ 1 there exists yn ∈ π −1 (yn ) such that dX (yn , p) < dX (yn , p) + n1 . Since {yn } converges to p ∈ X \ S as n → ∞, dX (yn , p) → 0 as n → ∞. Moreover, since p ∈ X \ S, p ∈ X \ S and since X is taut modulo S, X is hyperbolic modulo S. Thus yn converges to p as n → ∞. Lifting fn to fn such that πo fn = fn and fn (zn ) = yn . Since yn converges to p, {fn } ⊂ Hol(∆, X) is not compactly divergent modulo S in Hol(∆, X). Since X is taut modulo S, {fn } converges uniformly to a mapping f in Hol(∆, X). Hence {fn } converges uniformly to a mapping f := π ◦ f 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 : X → X is covering holomorphic, π −1 (U ) = Ui , where Ui i∈I are open subsets in X such that Ui ∩ Uj = ∅ (i, j ∈ I, i = j) and 10 PHAM VIET DUC*, MAI ANH DUC** AND PHAM NGUYEN THU TRANG** π Ui : Ui → U is biholomorphic for all i. This implies that Ui is taut. So π −1 (U ) = Ui is taut. i∈I Assume that {fn } ⊂ Hol(∆, X) is not compactly divergent modulo S in Hol(∆, X). Repeating the proof of Theorem 3.2, it follows that {fnk ∆∗ } is a subsequence of {fn ∆∗ } such that {fnk ∆∗ } converges uniformly to Φ in Hol(∆∗ , X) and π ◦ Φ = F ∆∗ . Fix z0 ∈ ∆∗ . Since F ∈ Hol(∆, X) and π : X → X is a holomorphic covering mapping, we can lift F to a mapping Φ ∈ Hol(∆, X) such that π ◦ Φ = F and Φ(z0 ) = Φ(z0 ). Since Φ, Φ ∆∗ are lifting mappings of F ∆∗ with Φ(z0 ) = Φ(z0 ) and by the unique property of lifting mapping, it implies that Φ = Φ ∆∗ . Thus Φ is a holomorphic extension of Φ from ∆∗ to ∆. Hence X is taut modulo S. 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 (x, π −1 (y)) = inf y∈π −1 (y) dE (x, y) for x, y ∈ M , and x ∈ π −1 (x). Proof. Obviously, dM (x, y) ≤ dE (x, π −1 (y)) for x ∈ π −1 (x). Now we proof dM (x, y) ≥ dE (x, π −1 (y)) for x ∈ π −1 (x). Indeed, take two arbitrary points x, y ∈ M and x ∈ π −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 σ 1 ∆ ×M E −−− → E    π θ 1 ∆ −−−→ M. f1 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 ϕ1 : ∆ → E be a holomorphic map defined by ϕ1 (z) = σ1 ◦ Φ1 (z, c1 ) for z ∈ ∆. ON TAUTNESS MODULO AN ANALYTIC SUBSET 11 Consider a pullback bundle σ 2 ∆ ×M E −−− → E    π θ 2 ∆ −−−→ M. f2 Then there exists an isomorphims Φ2 : ∆ × F → ∆ ×M E between two holomorphic bundles on ∆. Thus, there exists a point c2 ∈ F such that σ2 ◦ Φ2 (0, c2 ) = ϕ1 (a1 ). Let ϕ2 : ∆ → E be a holomorphic map defined by ϕ2 (z) = σ2 ◦ Φ2 (z, c2 ) for z ∈ ∆. Continuing this process, it implies that ϕ1 , ϕ2 , ..., ϕk ∈ Hol(∆, E) satisfying ϕ1 (0) = x, ϕi (ai ) = ϕi+1 (0) for 1 ≤ i ≤ k − 1 and ϕk (ak ) ∈ π −1 (y). Therefore, dM (x, y) ≥ dE (x, π −1 (y)) for x ∈ π −1 (x). Theorem 3.5. Let E be a holomorphic bundle on M with a taut bundle F and the projection π : E → M , where E, F, M are complex manifolds. Let S be an analytic subset in M . Then E is taut modulo S := π −1 (S) if and only if M is taut modulo S. Proof. (⇒) Assume that E is taut modulo S and {fn } ⊂ Hol(∆, M ) is not compactly divergent modulo S in Hol(∆, M ). Then there exists {zn } ⊂ ∆ such that {zn } → z ∈ ∆ and {yn := fn (zn )} → p ∈ M \ S. Take p ∈ π −1 (p) ⊂ E \ S. By Lemma 3.4, it implies thatdM (p, yn ) = inf dE (p, yn ). Hence, for each n ≥ 1, there exists yn ∈ E such −1 yn ∈π (yn ) that 1 . n Since dM (p, yn ) + n1 → 0 as n → ∞, dE (p, yn ) = 0. On the other hand, E is hyperbolic modulo S since E is taut modulo S. Hence yn → p as n → ∞. For each n ≥ 1, consider the pullback bundle θn : ∆ ×M E → ∆ of the bundle π : E → M . We have the following diagram. dE (p, yn ) < dM (p, yn ) + σ n ∆ ×M E −−− → E    π θ n ∆ −−−→ M. fn 12 PHAM VIET DUC*, MAI ANH DUC** AND PHAM NGUYEN THU TRANG** By the proof of Lemma 3.4, there exist an isomorphism Φn : ∆×F → ∆ ×M E between two holomorphic bundles on ∆ and cn ∈ F such that σn ◦ Φn (0, cn ) = yn . Define the holomorphic map ϕn : ∆ → E by ϕn (z) = σn ◦ Φn (z, cn ) for all z ∈ ∆. It is easy to see that ϕn (zn ) → p ∈ E. Since E is taut modulo S, the sequence ϕn converges uniformly to a ϕ in Hol(∆, E) as n → ∞. Thus, {fn } cenverges uniformly to a mapping f := π ◦ ϕ in Hol(∆, M ). Hence, M is taut modulo S. (⇐) Assume that M is taut modulo S. For each x ∈ M \ S, take an open neighbourhood U of x in M \ S such that U is taut. Since U and F are taut, U × F is taut. Since π −1 (U ) is biholomorphic to U × F , π −1 (U ) is taut too. Assume that {fn } ⊂ Hol(∆, E) is not compactly divergent modulo S in Hol(∆, E). Without loss of generality, we may assume that the sequence {fn ∆∗ } converges uniformly to a holomorphic map f in Hol(∆∗ , E). Since {π ◦ fn } converges uniformly on compact subsets of ∆ to a holomorphic map h in Hol(∆, M ), we get F ∆∗ = π ◦ f . Consider the pullback bundle θ : ∆ ×M E → ∆ of the bundle π : E → M . We have the following commutative diagram. σ ∆ ×M E −−−→ E    π θ ∆ −−−→ M. h According to the proof of Lemma 3.4, for a fixed point z0 ∈ ∆∗ , there exists an isomorphism Φ : ∆ × F → ∆ ×M E between two holomorphic bundles on ∆ and a point c ∈ F such that σ ◦ Φ(z0 , c) = y0 , where y0 = f (z0 ) ∈ π −1 (h(z0 )). Define the holomorphic map ϕ : ∆ → E is by ϕ(z) = σ ◦ Φ(z, c) for all z ∈ ∆. Then ϕ is a lifting map of h. By the proof of Theorem 3.3, ϕ is a holomorphic extension of f over ∆. Hence E is taut modulo S. Acknowledgements. This work was done during a stay of the third author at the Vietnam Institute for Advanced Study in Mathematics (VIASM). She would like to thank the staff there, in particular the partially support of VIASM. References [1] DucThai Do, AnhDuc Mai, ThuVan Ninh On limit Brody curves in Cn and (C∗ )2 , Kyushu Journal of Math. 69 (2015). ON TAUTNESS MODULO AN ANALYTIC SUBSET 13 [2] R. Remmert, Classical topics in complex function theory, Springer-Verlag, 1998. [3] Fornaess, J., Narasimhan, R., The Levi problem on complex spaces with singularities, Math. Ann. 248 (1980), 47-72. [4] S. Kobayashi, Hyperbolic manifolds and Holomorphic mappings, N. Y. Dekker, 1970. [5] S. Kobayashi, Hyperbolic complex spaces, Springer-Verlag, Berlin, 1998. [6] Do Duc Thai and Nguyen Le Huong, On the disc-convexity of complex Banach manifolds, Ann. Polon. Math. LXIX. 1 (1998), 1-11. *Department of Mathematics Thai Nguyen University of Education Luong Ngoc Quyen str., Thai Nguyen, Vietnam *Department of Mathematics Hanoi National University of Education 136 XuanThuy str., Hanoi, Vietnam E-mail address: pvduc2002@yahoo.com E-mail address: pnttrangsp@gmail.com E-mail address: ducphuongma@gmail.com [...]... in complex function theory, Springer-Verlag, 1998 [3] Fornaess, J., Narasimhan, R., The Levi problem on complex spaces with singularities, Math Ann 248 (1980), 47-72 [4] S Kobayashi, Hyperbolic manifolds and Holomorphic mappings, N Y Dekker, 1970 [5] S Kobayashi, Hyperbolic complex spaces, Springer-Verlag, Berlin, 1998 [6] Do Duc Thai and Nguyen Le Huong, On the disc-convexity of complex Banach manifolds,... is taut modulo S Acknowledgements This work was done during a stay of the third author at the Vietnam Institute for Advanced Study in Mathematics (VIASM) She would like to thank the staff there, in particular the partially support of VIASM References [1] DucThai Do, AnhDuc Mai, ThuVan Ninh On limit Brody curves in Cn and (C∗ )2 , Kyushu Journal of Math 69 (2015) ON TAUTNESS MODULO AN ANALYTIC SUBSET. .. (0) for 1 ≤ i ≤ k − 1 and ϕk (ak ) ∈ π −1 (y) Therefore, dM (x, y) ≥ dE (x, π −1 (y)) for x ∈ π −1 (x) Theorem 3.5 Let E be a holomorphic bundle on M with a taut bundle F and the projection π : E → M , where E, F, M are complex manifolds Let S be an analytic subset in M Then E is taut modulo S := π −1 (S) if and only if M is taut modulo S Proof (⇒) Assume that E is taut modulo S and {fn } ⊂ Hol(∆, M... disc-convexity of complex Banach manifolds, Ann Polon Math LXIX 1 (1998), 1-11 *Department of Mathematics Thai Nguyen University of Education Luong Ngoc Quyen str., Thai Nguyen, Vietnam *Department of Mathematics Hanoi National University of Education 136 XuanThuy str., Hanoi, Vietnam E-mail address: pvduc2002@yahoo.com E-mail address: pnttrangsp@gmail.com E-mail address: ducphuongma@gmail.com .. .ON TAUTNESS MODULO AN ANALYTIC SUBSET 11 Consider a pullback bundle σ 2 ∆ ×M E −−− → E    π θ 2 ∆ −−−→ M f2 Then there exists an isomorphims Φ2 : ∆ × F → ∆ ×M E between two holomorphic bundles on ∆ Thus, there exists a point c2 ∈ F such that σ2 ◦ Φ2 (0, c2 ) = ϕ1 (a1 ) Let ϕ2 : ∆ → E be a holomorphic map defined by ϕ2 (z) = σ2 ◦ Φ2 (z, c2 ) for z ∈ ∆ Continuing this process,... According to the proof of Lemma 3.4, for a fixed point z0 ∈ ∆∗ , there exists an isomorphism Φ : ∆ × F → ∆ ×M E between two holomorphic bundles on ∆ and a point c ∈ F such that σ ◦ Φ(z0 , c) = y0 , where y0 = f (z0 ) ∈ π −1 (h(z0 )) Define the holomorphic map ϕ : ∆ → E is by ϕ(z) = σ ◦ Φ(z, c) for all z ∈ ∆ Then ϕ is a lifting map of h By the proof of Theorem 3.3, ϕ is a holomorphic extension of f over ∆ Hence... → p as n → ∞ For each n ≥ 1, consider the pullback bundle θn : ∆ ×M E → ∆ of the bundle π : E → M We have the following diagram dE (p, yn ) < dM (p, yn ) + σ n ∆ ×M E −−− → E    π θ n ∆ −−−→ M fn 12 PHAM VIET DUC*, MAI ANH DUC** AND PHAM NGUYEN THU TRANG** By the proof of Lemma 3.4, there exist an isomorphism Φn : ∆×F → ∆ ×M E between two holomorphic bundles on ∆ and cn ∈ F such that σn ◦ Φn (0,... {fn } ⊂ Hol(∆, E) is not compactly divergent modulo S in Hol(∆, E) Without loss of generality, we may assume that the sequence {fn ∆∗ } converges uniformly to a holomorphic map f in Hol(∆∗ , E) Since {π ◦ fn } converges uniformly on compact subsets of ∆ to a holomorphic map h in Hol(∆, M ), we get F ∆∗ = π ◦ f Consider the pullback bundle θ : ∆ ×M E → ∆ of the bundle π : E → M We have the following... is easy to see that ϕn (zn ) → p ∈ E Since E is taut modulo S, the sequence ϕn converges uniformly to a ϕ in Hol(∆, E) as n → ∞ Thus, {fn } cenverges uniformly to a mapping f := π ◦ ϕ in Hol(∆, M ) Hence, M is taut modulo S (⇐) Assume that M is taut modulo S For each x ∈ M \ S, take an open neighbourhood U of x in M \ S such that U is taut Since U and F are taut, U × F is taut Since π −1 (U ) is biholomorphic... compactly divergent modulo S in Hol(∆, M ) Then there exists {zn } ⊂ ∆ such that {zn } → z ∈ ∆ and {yn := fn (zn )} → p ∈ M \ S Take p ∈ π −1 (p) ⊂ E \ S By Lemma 3.4, it implies thatdM (p, yn ) = inf dE (p, yn ) Hence, for each n ≥ 1, there exists yn ∈ E such −1 yn ∈π (yn ) that 1 n Since dM (p, yn ) + n1 → 0 as n → ∞, dE (p, yn ) = 0 On the other hand, E is hyperbolic modulo S since E is taut modulo S Hence ... complex spaces First of all, we show the inheritance of the tautness modulo an analytic subset of complex spaces under certain assumptions Theorem 2.1 Let X be a complex space and S be an analytic subset. .. X and X is taut modulo S, then X is taut modulo S In particular, if X is taut, then X is taut modulo S for any analytic subset S of X Criteria on the tautness modulo an analytic subset of complex. .. 0}, ON TAUTNESS MODULO AN ANALYTIC SUBSET 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 S containing S Invariance of the