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Publ. Mat. 45 (2001), 421–429 ON D∗-EXTENSION PROPERTY OF THE HARTOGS DOMAINS Do Duc Thai and Pascal J. Thomas Abstract A complex analytic space is said to have the D∗-extension property if and only if any holomorphic map from the punctured disk to the given space extends to a holomorphic map from the whole disk to the same space. A Hartogs domain Hover the base X(a complex space) is a subset of X×Cwhere all the fibers over Xare disks centered at the origin, possibly of infinite radius. Denote by φ the function giving the logarithm of the reciprocal of the radius of the fibers, so that, when Xis pseudoconvex, His pseudoconvex if and only if φis plurisubharmonic. We prove that Hhas the D∗-extension property if and only if (i) Xitself has the D∗-extension property, (ii) φtakes only finite values and (iii) φis plurisubharmonic. This implies the existence of domains which have the D∗-extension property without being (Kobayashi) hyperbolic, and simplifies and generalizes the authors’ previous such example. 1. Introduction The “big Picard” theorem states that any holomorphic map ffrom the punctured unit disc D∗into the Riemann sphere P1(C) which omits three points can be extended to a holomorphic map f:D−→ P1(C). Kwack [Kw] extended this theorem to a higher dimensional context. If fis a holomorphic map from D∗into a hyperbolic space Xsuch that, for a suitable sequence of points zk∈D∗converging to the origin, f(zk) converges to a point p0∈X, then fextends to a holomorphic map from Dinto X. 2000 Mathematics Subject Classification. Primary 32H02, 32D20; Secondary 32H20, 32F30. Key words. Kobayashi hyperbolicity, removable singularities, Kontinuit¨atsatz, extension through pluripolar sets.
422 D. D. Thai, P. J. Thomas The above-mentioned theorem of Kwack has strongly motivated the study of the extension problem of holomorphic maps through isolated singularities. At the same time, this result has suggested the study of the class of complex spaces having the following property. Definition. Let Xbe a complex space. We say that Xhas the D∗-extension property (D∗-EP) iff for any holomorphic map ffrom D∗= {z∈C:0<|z|<1}to X, there exists a map f∈Hol(D,X) (where D={z∈C:|z|<1}) such that f|D∗=f. Much attention has been devoted to the D∗-EP and various theorems have been obtained by Kwack [Kw], Thai [Th], Thai and Thomas [Th-Tho], and others, see the monograph [Ko]. The first aim of this note is to prove the following Theorem 1. Let Xϕ:= {(z,w)∈X×C:|w|<e −ϕ(z)}where ϕ:X→ [−∞,+∞)is upper semi continuous. Then Xϕhas the D∗-EP iff: Xhas the D∗-EP, ϕ∈PSH(X)and ϕ(z)>−∞,∀z∈X. Notice that this result admits as a corollary, when X=Dand ϕ is not locally bounded, the existence of a domain in C2which has the D∗-EP without being Kobayashi hyperbolic. We thus simplify the proof of that result given in [Th-Tho], and generalize somewhat the class of counter-examples available. The proof of this result can be carried out by elementary means. However, use of more powerful theorems makes for shorter proofs and more general results. We denote by Λdthe Hausdorff measure in (real) dimension d. Definition. We say that Xhas the n-PEP (resp. n-PPEP, (n, d)-EP) iff for any closed set A⊂Dnwhich is polar (resp. pluripolar, resp. of locally finite Λdmeasure), for any holomorphic map ffrom Dn\Ato X, there exists a map f∈Hol(Dn,X) (where D={z∈C:|z|<1}) such that f|Dn\A=f. Re-using the notations of Theorem 1, we have, for any n≥1, 2n−2< d<2n−1: Theorem 2. Xϕhas the n-PEP (resp. n-PPEP, (n, d)-EP) iff: Xhas the n-PEP (resp. n-PPEP, (n, d)-EP), ϕ∈PSH(X)and ϕ(z)>−∞,∀z∈X. We would like to thank Ahmed Zeriahi and Nguyen Van Khue for their helpful suggestions.
On D∗-Extension Property of the Hartogs Domains 423 2. Proof of Theorem 1 1) Sufficiency: Let H∈Hol(D∗,X ϕ),H =(h1,h 2). Since Xhas the D∗-EP, h1 extends to a map h1, h1:D→X. Let z0= h1(0). We need the following result on the local growth of (pluri-)subharmonic functions. Theorem ([Ho1, Corollary 4.4.6, p. 98] or [Ho2, Corollary 4.2.10, p. 261]).If ϕ∈PSH(Ω) \ {−∞}, where Ωis a connected pseudoconvex open set, then e−ϕis locally integrable in a dense open subset G containing all points zwhere ϕ(z)>−∞. In particular, since ϕ(z0) is finite, e−2ϕ∈L1 loc in a neighborhood of z0. But then h2∈L2 loc, which implies that h2extends holomorphically across 0, as can easily be deduced from the Laurent series expansion. Finally, to see that we actually have log | h2(0)|<−ϕ( h1(0)), we apply the maximum principle to the subharmonic function log | h2|+ϕ( h1), as was done in [Th-Tho]. A direct proof of the extendability of h2may be given without recourse to H¨ormander’s result. Since ϕis u.s.c, there exists r0>0, M∈Rsuch that ϕ(z)≤M, ∀z∈D(z0,r 0). Without loss of generality, suppose M≤0. Therefore we have log |h2(ξ)|≤−ϕ( h1(ξ))(1) and since ϕ≤0, log+|h2(ξ)|≤−ϕ1(ξ), where ϕ1(ξ):=ϕ( h1(ξ)), therefore ϕ1∈SH(D,R−). By the mean value inequality for subharmonic functions, (1) implies: ∀r∈(0,r 0), 1 πr2D(0,r) log+|h2(ξ)|dλ2(ξ)≤−ϕ1(0) <+∞. We want to show that this implies that h2has a removable singularity at the origin. Expand h2as a Laurent series h2(ξ)= n∈Z anξn. Then for rsmall enough, n>0 anξn≤e,
424 D. D. Thai, P. J. Thomas so log+|h0(ξ)|≤1 + log+|h2(ξ)|, where we set h0(ξ):= n≤0 anξn, and we are reduced to ∀r>0 1 πr2D(0,r) log+|h0(ξ)|dλ2(ξ)≤C<+∞. Set f(ξ):=h0(1/ξ). This is now an entire function. Under the change of variable ψ=1 ξ, we get 1 πr2D(0,r) log+|h0(ξ)|dλ2(ξ)= 1 πr2C\¯ D(0,1/r) log+|f(ψ)|1 |ψ|4dλ2(ψ). Now log+|f|∈SH(C), so m(ρ):=2π 0 log+|f(ρeiθ)|dθ 2π is an increasing function of ρ. Passing to polar coodinates, we get that C≥1 πr2∞ 1 r2π 0 log+|f(ρeiθ)|dθ1 ρ3dρ =2 r2∞ 1 r m(ρ) ρ3dρ ≥2 r2m(1/r)∞ 1 r dρ ρ3=m1 r. Therefore m(ρ) is bounded as ρ→∞. But then, since log+|f|∈SH(C), it must be bounded above on C, since by the Poisson formula log+|f(z0)|≤2π 0 1−z0 ρ 2 z0 ρ−eiθ 2log+|f(ρeiθ)|dθ 2π≤ 1+z0 ρ 1−z0 ρ m(ρ)≤3C, for ρ≥2|z0|,sofis constant by Liouville’s theorem. Therefore hhas a removable singularity at 0.
On D∗-Extension Property of the Hartogs Domains 425 2) Necessity: To prove that Xhas the D∗-EP, if h∈Hol(D∗,X), then the map H given by H(ξ):=(h(ξ),0) is holomorphic from D∗to Xϕand must therefore admit an extension Hsuch that (by continuity) H(D)⊂X× {0}⊂Xϕ. Writing H=( h, 0), we obtain the required extension of h. If there exists z0∈Xsuch that ϕ(z0)=−∞ then the complex line {(z0,w):w∈C}⊂Xϕ,soXϕdoes not have the D∗-EP (take h(ξ)=(z0,1/ξ), see [Th]). There remains to show that ϕ∈PSH(X). We first do this for the case where Xis an open set in Cn. Lemma. Let Ω⊂Cnbe a domain with the D∗-EP. Then Ωis pseudoconvex. This lemma (which we alluded to in [Th-Tho]) is a consequence of a theorem of Shiffman [Si] (see also [So-Th]): if for any sequence {fn}⊂ Hol(D,X), convergence of {fn|D∗}in Hol(D∗,X) implies convergence of {fn}(“weak disk condition”), then Xhas the Hartogs extension condition (which implies pseudoconvexity for open sets in Cn). But a domain in Cnwith the D∗-EP verifies the weak disk condition (simply extend the limit mapping and then apply the maximum principle on all coordinates). However, there is a direct and elementary proof which avoids the use of Shiffman’s theorem. For the reader’s convenience, and since some colleagues of ours seemed to find it nice, we include it here. Direct Proof of the Lemma: Let Φ be a holomorphic embedding of the closed unit bidisk D2into Cn. Call Hartogs figure the image under Φ of the set H0:= {|z1|≤1,z 2=0}∪{|z1|=1,|z2|≤1}. Recall that Ω is pseudoconvex if and only if for every Hartogs figure contained in Ω, Φ(D2) is also contained in Ω [Ra]. Therefore, assuming Ω is not pseudoconvex, we obtain the following situation: there exists a holomorphic embedding Φ such that Ω1:= Φ−1(Ω) ∩D2is open, D2\Ω1=∅, and D2\Ω1∩H0=∅. Let r2:= inf{|z2|:(z1,z 2)∈D2\Ω1}; our hypotheses mean that 0 <r 2<1, and they also imply that the set K:= D2\Ω1∩|z2|≤1+r2 2
426 D. D. Thai, P. J. Thomas is compact in D2, and therefore r1:= max(z1,z2)∈K|z1|<1. Now set δε(z):=ε|z1|2−|z2|2. There exists a point z0=(z0 1,z0 2) so that δε(z0) = maxKδε.Forεsmall enough (|ε|r2 1−(1+r2 2)2<−r2 2), z0∈K\{|z2|=1+r2 2}, so there exists a neighborhood Vof z0such that V∩Ω1=V\K. It is now enough to find an analytic disk fwith center f(0) ∈Kand f(D(0,r)\{0})⊂V\K, for r>0 small enough. We may in fact pick an affine disk, namely f(ξ):=(z0 1+¯z0 2ξ,z0 2+ε¯z0 1ξ). Observe that f(C) is a line tangent to the level hypersurface of δεcorresponding to the value δε(z0), and in fact an elementary calculation shows that δε(f(ξ)) = δε(z0)+ε|ξ|2(|z0 2|2−ε|z0 1|2)>δ ε(z0) for ε>0 and small enough (z0 1and z0 2do depend on ε, but we have the condition as soon as r2 2−εr2 1>0), which completes the proof of the lemma. We may remark that setting ft(ξ):=f(ξ)+t∇δε(z0), the map Φ ◦ fgives a disk violating the D∗-EP for the Ω which we had assumed non-pseudoconvex, and the maps Φ ◦ftprovide a refined failure of the Kontinuit¨atsatz for Ω (contact with the boundary occurs at one point exactly). Together with the above lemma, the following will complete the proof of necessity. Claim. If Xϕhas the D∗-EP, and ϕ/∈PSH(X)then there exists Ω⊂ C2, having the D∗-EP and Ωnot pseudoconvex. We need the following characterization. Denote by Ha(D) the space of harmonic functions on the disk. Fact. ϕ∈PSH(X)iff∀f∈Hol(D,X), ∀u∈Ha(D)∩C 0(D), such that ϕ◦f(eiθ)≤u(eiθ)∀θ∈R, then ϕ◦f(0) ≤u(0). This follows immediately from the theorem of Fornaess and Narasimhan [Fo-Na] which characterizes plurisubharmonic functions on complex spaces as those whose pullback under any analytic disk is subharmonic, and the characterization of subharmonicity by the mean value inequality (see e.g. [Ho1, Theorem 1.6.3, p. 16]). Now suppose ϕ/∈PSH(X). Then ∃f∈Hol(D,X),u∈Ha(D)∩C0(D), such that ϕ(f(0)) >u(0), ϕ(f(eiθ)) ≤u(eiθ), ∀θ∈R.
On D∗-Extension Property of the Hartogs Domains 427 Let Ω:={(z,w)∈D×C:(f(z),w)∈Xϕ} ={(z,w)∈D×C:|w|<e ϕ◦f(z)}=: Dϕ◦f. Since ϕ◦fis not subharmonic, the classical result about Hartogs domains implies that Ω is not pseudoconvex. The following claim then completes our proof. Claim. Ωhas the D∗-EP. Proof of the Claim: Let h∈Hol(D∗,Ω), h(ξ)=(h1(ξ),h 2(ξ)). Now h1(ξ)∈Dfor all ξ,soh1(ξ) extends to h1∈Hol(D,D). The map ξ→ (f◦h1(ξ),h 2(ξ)) is holomorphic from D∗to Xϕby construction, so it extends to F∈Hol(D,X ϕ). Let F(ξ)=(F1(ξ),F 2(ξ)). F2provides an extension of h2. It remains to see that h=( h1,F 2)∈ Hol(D,Ω), that is, that |f( h1(0))|<e −ϕ(F2(0)). Since |f( h1(0))|= limξ→0|f(h1(ξ))|= limξ→0|F1(ξ)|=|F1(0)|< e−ϕ(F2(0)) because F∈Hol(D,X ϕ), we are done. 3. Proof of Theorem 2 The direct implication proceeds as in the previous section, noticing that each of the extension properties we have defined implies the D∗-EP. (In the case where n≥2, given a map f∈Hol(D∗,X), simply consider the map F∈Hol(Dn\{z1=0},X) given by F(z1,...,z n):=f(z1).) To prove the converse implication, recall the following result of extension. Theorem ([Ha-Po, Theorem 1, (d)]).Suppose Ais a closed subset of an open set Ω⊂Cnand that f∈Hol(Ω \A).Let2≤p<∞and pbe the conjugate exponent (1 p+1 p=1).Iff∈Lp loc(Ω) and Λ2n−p(A)is locally finite, then f∈Hol(Ω). Now given das in the theorem, set p:= 2n−dand pits conjugate exponent. Given a map h=(h1,h 2)∈Hol(Dn\A, Xϕ), h1extends to h1∈Hol(Dn,X) by the extension property for Xwhich is included in the hypothesis, and pϕ ◦ h1is a finite-valued plurisubharmonic function, so locally integrable, therefore |h2|≤e−ϕ◦ h1verifies all the hypotheses of the Harvey-Polking theorem.
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On D∗-Extension Property of the Hartogs Domains 429 [Th-Tho] D. D. Thai and P. J. Thomas,D∗-extension property without hyperbolicity, Indiana Univ. Math. J. 47(3) (1998), 1125–1130. Do Duc Thai: Department of Mathematics Institute of Pedagogy no 1 Cau Giay Hanoi Vietnam E-mail address:[email protected] Pascal J. Thomas: Laboratoire de Math´ematiques Emile Picard CNRS UMR 5580 Universit´e Paul Sabatier 118 route de Narbonne 31062 Toulouse Cedex France E-mail address:[email protected] Rebut el 9 d’octubre de 2000.