On hyperbolicity and tautness modulo an analytic subset of Hartogs domains

Let $X$ be a complex space and $H$ a positive homogeneous plurisubharmonic function $H$ on $X\times\C^m$. Consider the Hartogs-type domain $\Omega_{H}(X):=\{(z,w)\in X\times \C^m:H(z,w)<1 \}$. Let $S$ be an analytic subset of $X$. We give necessary and sufficient conditions for hyperbolicity and tautness modulo $S\times \C^m$ of $\Omega_{H}(X)$, with the obvious corollaries for the special case of Hartogs domains.


Introduction
Let X be a complex space and ϕ : X → [−∞, ∞) be an uppersemicontinuous function on X. The Hartogs domain Ω ϕ (X) := {(x, z) ∈ X × C : |z| < e −ϕ(x) } is a classical object in Several Complex Variables. In particular, in the past ten years, much attention has been given to the properties of Hartogs domains from the viewpoint of hyperbolic complex analysis. For instance, in [11], the authors obtained necessary and sufficient conditions for the hyperbolicity and the tautness of Ω ϕ (X). We refer readers to the articles [10], [2], and references therein for the development of related subjects.
Motivated by studying hyperbolicity and tautness modulo an analytic subset of complex spaces, the main goal of this article is to give necessary and sufficient conditions on hyperbolicity or tautness modulo a "vertical" analytic subset of the Hartogs domains Ω ϕ (X). The results are given in Section 2, but first we recall some basic notions. Definition 1.1. (see [7, p. 68]) Let X be a complex space and S be an analytic subset of X. We say that X is hyperbolic modulo S if for every pair of distinct points p, q of X we have d X (p, q) > 0 unless both are contained in S, where d X is the Kobayashi pseudodistance of X.
If S = ∅, then X is said to be hyperbolic. [7, 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 {f n } in Hol(D, X) one of the following holds: i. There exists a subsequence of {f n } which converges uniformly to f ∈ Hol(D, X) in Hol(D, X); ii. The sequence {f n } is compactly divergent modulo S in Hol(D, X), i.e., for each compact set K ⊂ D and each compact set L ⊂ X \ S, there exists an integer N such that f n (K) ∩ L = ∅ for all n ≥ N.
If S = ∅, then X is said to be taut. It is immediate from the definition that if S ⊂ S ′ ⊂ X and X is taut modulo S, then it is taut modulo S ′ , so in particular if X is taut, it is taut modulo S for any analytic subset S.
Of course, the converse does not hold.
We could have X \ S being taut without X being taut modulo S: On the other hand, there are examples of domains taut modulo S such that X \ S is not taut: just take X a taut domain and S such that the codimension of S is at least 2. Then X \ S is not pseudoconvex, therefore not taut.
Proof of Example 1.3. Since the complex line S is contained in X, it cannot be hyperbolic, thus isn't taut either. On the other hand, X \ S is biholomorphic to (D \ {0}) × D under the map (z, w) → (z, zw), and the latter is clearly taut, which proves (iii). Now suppose (z 0 , w 0 ) = (z 1 , w 1 ) ∈ X, with at least one of them not in S. If z 0 = z 1 , then d X ((z 0 , w 0 ), (z 1 , w 1 )) ≥ d D (z 0 , z 1 ) > 0; if z 0 = z 1 , then z 0 = 0. Given any finite set of points of X, (ζ k , η k ) which connect (z 0 , w 0 ) to (z 1 , w 1 ) via consecutive analytic disks, either there is some k such that |ζ k | ≤ |z 0 |/2 and then the corresponding sum will contribute at least d D (z 0 , z 0 /2), or there is not, and then all points are in X ∩ {|z| > |z 0 |/2} ⊂ {|z 0 |/2 < |z| < 1, |w| < 2/|z 0 |} := P and the sum will be bounded below by d P ((z 0 , w 0 ), (z 1 , Proof of (ii). Assume that {f n } ⊂ Hol(D, X) is not compactly divergent modulo S. Write f n = (g n , h n ) with g n , h n ∈ Hol(D, C) satisfying |g n (z)| < 1, |g n (z)h n (z)| < 1 for all z ∈ D and for n = 1, 2, · · · Since {f n } is not compactly divergent modulo S, {g n } is not compactly divergent modulo {0}. Therefore, by Montel's theorem we may assume, without loss of generality, that {g n } converges uniformly on every compact subset of D to a holomorphic function g ∈ Hol(D, D), not identically zero. Since {g n h n } is not compactly divergent on D and since D is taut, taking a subsequence we may assume that {g n h n } also converges uniformly on every compact subset of D to a holomorphic function γ ∈ Hol(D, C). Hence {h n } converges uniformly on every compact subset of D to a meromorphic function h := γ/g on D. Moreover, by Hurwitz's theorem h is actually holomorphic on D and thus {f n } converges uniformly on every compact subset of D to a holomorphic map f := (g, h) ∈ Hol(D, X).
We now prove that f ∈ Hol(D, X). Since g ∈ Hol(D, D), it suffices to show that |g(z)h(z)| < 1 for all z ∈ D. Indeed, suppose not. Then there is z 0 ∈ D such that |g(z 0 )h(z 0 )| = 1. By the maximum principle, gh is a constant function. Therefore, |g(z)h(z)| = 1 for every z ∈ D. This is not possible because {g n h n } ⊂ Hol(D, D) is not compactly divergent. Thus, the proof is complete. ✷

Main Results
We denote S : Theorem 2.1. Let X be a complex space and S be an analytic subset in X.
Then Ω H (X) is hyperbolic modulo S if and only if X is hyperbolic modulo S and the function H satisfies the following condition: The proof is given in Section 3.
The situation for tautness is a bit more complicated, at least in the case of complex spaces.
If furthermore X is a complex manifold and S a (proper) analytic subset, then in addition log H is plurisubharmonic on The proof is given in Section 4. As above, an immediate corollary is obtained for Hartogs domains by observing that H(z, w) := |w|e ϕ(z) is continuous if and only if ϕ is, and log H is plurisubharmonic if and only if ϕ is. Implication (ii) cannot hold as stated for the general case of a complex space, as the following shows.
Note that in this case X is not irreducible and S is a whole component. We don't know what happens if we rule out this degenerate situation.
Using the same argument as in the proof of Remark 3.1.7 and Proposition 3.1.10 in [5], it is easy to see that Lemma 3.1. Let Ω = Ω H (X). Then ℓ Ω ((z, 0), (z, w)) ≤ p(0, H(z, w)) for any (z, w) ∈ Ω, where p is the Poincaré distance. Equality holds if H ∈ P SH(X × C m ).
Proof of Theorem 2.1 (=⇒) Suppose Ω H (X) is hyperbolic. Since X is isomorphic to a closed complex subspace of Ω H (X), we deduce that X is hyperbolic. Next, we will show that H verifies the property (1). Otherwise, there would with lim k→∞ w k = w 0 = 0 such that lim k→∞ H(z k , w k ) = 0. Without loss of generality, we may assume that (z k , w k ) ∈ Ω H (X). Then by Lemma 3.1, we have 0 ≤ k Ω ((z k , 0), (z k , w k )) ≤ p(0, H(z k , w k )), ∀k ≥ 1.

Tautness
Proof of Theorem 2.3 Proof of (i). Since X is isomorphic to a closed complex subspace of Ω H (X), we deduce that X is taut modulo S. We now show that H is continuous on For each k ≥ 1, we define the holomorphic mapping f k : D → Ω H (X) Since Ω H (X) is taut moduloS, by passing to a subsequence if necessary, we may assume that f k converges locally uniformly on D to a holomorphic mapping f ∈ Hol(D, Ω H (X)). It is easy to see that This implies that H(z 0 , w 0 ) < r |λ| , ∀λ ∈ D, and hence H(z 0 , w 0 ) ≤ r.
This is a contradiction. It remains to show that log H is plurisubharmonic. According to the theorem of Fornaess and Narasimhan [4], it suffices to show that u(z) := log H • g(z) = log H(g 1 (z), g 2 (z)) is subharmonic for every g = (g 1 , g 2 ) ∈ Hol(D, X \S ×C m )∩C(D, X \S ×C m ). Suppose the contrary. Then ∃z 0 ∈ D, r > 0 such that D(z 0 , r) ⊂ D and a harmonic function h such that h(z) ≥ u(z) for any z = z 0 +re iθ , ∀θ ∈ R, but u(z 0 ) > h(z 0 ). Leth denote a harmonic conjugate to h. We For any n ≥ 1, we set ϕ n (λ) := g 1 (z), e −h(z)−ih(z)−ε 0 − 1 n g 2 (z) , where z ∈ D(z 0 , r). Then ϕ n ∈ Hol (D(z 0 , r), Ω H (X)) ∩ C (D(z 0 , r), Ω H (X)), ∪ ∞ n=1 ϕ n (∂D(z 0 , r)) ⋐ Ω H (X), and ϕ n (0) tend to a boundary point. This contradicts the tautness of Ω H (X). Proof of (ii). By the removable singularity theorem for plurisubharmonic functions [6, Theorem 2.9.22] since log H is locally bounded above in X × C m and plurishubharmonic in X \ S × C m , it can be extended acrossS to a function logĤ ∈ P SH(X × C m ) given, for We claim thatĤ = H. Since H is upper semi continuous on X × C m , this conclusion can only fail if there exists (z 0 , w 0 ) ∈S such that H(z 0 , w 0 ) > lim sup (z,w)→(z 0 ,w 0 ),(z,w) / ∈S H(z, w). Since X is a manifold, we can go to a coordinate patch and find an analytic disk f such that f (0) = (z 0 , w 0 ), f (D) ⊂S. Then f −1 (S) must be a discrete subset of D, and reducing the disk we may assume that f −1 (S) = {0} and sup 0<|ζ| H(f (ζ)) = H(z 0 , w 0 ) − δ, δ > 0. The proof then proceeds essentially as above. We have a contradiction. Proof of (iii). Assume that X is taut modulo S and log H is continuous on X \ S × C m , and plurisubharmonic on X × C m . We now show that Ω H (X) is taut modulo S.
Consider the projection π : Ω H (X) → X defined by π(x, z) = x. We now prove, for each x ∈ X \S, that there exists an open neighbourhood U of x in X \ S such that π −1 (U) is taut. Indeed, choose a hyperconvex neighbourhood U of x in X \ S. It is easy to see that π −1 (U) = {(u, z) ∈ U × C m : H(z, w) < 1} = Ω H (U). Suppose that ρ is a negative plurisubharmonic exhaustion function of U. Then (u, z) → max(ρ(u), log H(u, z)) is also a negative plurisubharmonic exhaustion function of Ω H (U). Thus, Ω H (U) is hyperconvex. By a Theorem of Sibony [9] and [11], Ω H (U) is taut. Thus, π −1 (U) is taut.
Without loss of generality, we may assume that there exist a compact set K ⊂ D and a compact set L ⊂ X such that f n (K) ∩ L = ∅ for n ≥ 1. For each n ≥ 1, there exists z n ∈ K ⊂ D such that f n (z n ) ∈ L. Since K and L are compact sets, by taking subsequences if necessary, we may assume that {z n } ⊂ K ⊂ D such that z n → z ∞ ∈ K ⊂ D and f n (z n ) → p ∈ L ⊂ X. It is easy to see that {f n := π • f n } ⊂ Hol(D, X) is not compactly divergent modulo S in Hol(D, X). Since X is taut modulo S, we may assume that {f n } converges uniformly to a mapping F ∈ Hol(D, X). Obviously, π( f n (z n )) → π( p) and π( f n (z n )) = π • f n (z n ) = f n (z n ) → F (z ∞ ) as n → ∞. Therefore, we can let p = π( p) = F (z ∞ ).
Since p ∈ X, p = π( p) ∈ S. Then 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 D \ F −1 (S) and since the sequence {f n } converges uniformly to a mapping F , we may assume that f n (V ) ⊂ U. This implies that f n (V ) ⊂ π −1 (U) for every n ≥ 1.
Consider the family Γ of all pairs (W, Φ), where W is an open set in D \ F −1 (S) and Φ ∈ Hol(W, X) such that there exists a subsequence { f n k W } of { f n W } which converges uniformly to mapping Φ in Hol(W, X).