arXiv:1509.02283v2 [math.CV] 30 Aug 2016 · arXiv:1509.02283v2 [math.CV] 30 Aug 2016 A construction of complete complex hypersurfaces in the ball with control on the topology A. Alarcon,

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A construction of complete complex hypersurfaces inthe ball with control on the topology

A. Alarc on, J. Globevnik, and F. J. Lopez

Abstract Given a closed complex hypersurfaceZ ⊂ CN+1 (N ∈ N) and acompact subsetK ⊂ Z, we prove the existence of a pseudoconvex Runge domainD in Z such thatK ⊂ D and there is a complete proper holomorphic embeddingfrom D into the unit ball ofCN+1. For N = 1, we derive the existence ofcomplete properly embedded complex curves in the unit ball of C2, with arbitrarilyprescribed finite topology. In particular, there exist complete proper holomorphicembeddings of the unit discD ⊂ C into the unit ball ofC2.

These are the first known examples of complete bounded embedded complexhypersurfaces inCN+1 with any control on the topology.

1. Introduction and main results

LetD denote the unit disc inC and, forN ∈ N, denote byBN+1 the unit ball inCN+1.

In 1977 P. Yang asked whether there exist complete immersed complex submanifoldsϕ : Mk → C

N+1 (k ≤ N) with bounded image [14, 15]. Here,completemeans that theRiemannian manifold(M,ϕ∗ds) is complete, whereds is the Euclidean metric inCN+1;equivalently, the image byϕ of every divergent path inM has infinite Euclidean length.

P. Jones [13] was the first to construct a bounded complete holomorphic immersionD → C

2, a bounded complete holomorphicembeddingD → C3, and a proper complete

holomorphic embeddingD → B4. Jones’ pioneering results have been extended tothe existence of proper complete holomorphic immersionsR → B2 and embeddingsR → B3, whereR is either an open Riemann surface of arbitrary topology (seeAlarconand Lopez [4] and Alarcon and Forstneric [2]), or a given bordered Riemann surface (seeAlarcon and Forstneric [1, 3]). Further, hereB2 andB3 may be replaced by any convexdomain inC2 andC3, respectively. Moreover, givenk ∈ N, an easy application of these

results furnishes bounded complete holomorphic immersions Rk := R× k· · · ×R → C2k

and embeddingsRk → C3k [1]. In the same direction, B. Drinovec Drnovsek [9]

recently proved that every bounded strictly pseudoconvex domain D ⊂ Ck with C 2-

boundary admits a complete proper holomorphic embeddingD → BN+1 provided thatthe codimensionN + 1− k is large enough.

To find complete bounded holomorphicembeddingsϕ : MN → CN+1 is considerably

more difficult. For instance, the self-intersection pointsof hypersurfaces inCN+1 aregeneric and thus cannot be removed by small perturbations, which is possible when thecodimension is large enough (to be precise, when2k ≤ N ). So, any induction method forconstructing such hypersurfaces will have to take this intoaccount and at no step createself-intersection points.

http://arxiv.org/abs/1509.02283v2

2 A. Alarcon, J. Globevnik, and F. J. Lopez

The embedding problem was settled in the lowest dimensionalcase by Alarcon andLopez, who proved that every convex domain inC

2 contains a complete properly embeddedcomplex curve [6]. Their examples come after a recursive construction process whichapplies, at each step, a self-intersection removal procedure consisting of replacing everynormal crossing in a complex curve by an embedded annulus. This process, which mustbe done while ensuring the completeness of the limit curve, is very delicate and does notprovide any control on the topology of the curve. In principle, the examples could be ofvery complicated topology.

After this, a different approach was used by Globevnik [11].He found an embeddedcomplete holomorphic curve as a level set of some wildly oscillating holomorphic functionon B2. This construction worked also in higher dimensions which lead to the completesolution of the Yang problem in all dimensions by proving that, for anyN ∈ N, there is acomplete, closed complex hypersurface inBN+1 ⊂ C

N+1. The same holds when replacingBN+1 by any pseudoconvex domain inCN+1 [12]. Again, this procedure does not supplyany information about the topology of the hypersurface, which could be very involved.

So at the moment there are two different methods to prove the existence of complete,closed hypersurfaces inBN+1 whenN = 1, and one whenN ≥ 2. Neither of thesemethods provides any information about the topology of sucha hypersurface. In thispaper we develop a conceptually new technique for constructing complete closed complexhypersurfaces in the unit ballBN+1 ⊂ C

N+1, N ∈ N, which in addition admits controlon the topology of the examples. In particular, we show that there is a complete properholomorphic embeddingD → B2 (see Corollary 1.2 below) and thus answer a questionleft open in [6, 11].

Before stating our results we need some background. Given a Stein manifoldX,we denote byO(X) the algebra of holomorphic functionsX → C. A domain (openand connected subset)Ω ⊂ X is called apseudoconvex domainif it has a stronglyplurisubharmonic exhaustion function. In particular,Ω endowed with the induced complexstructure is a Stein manifold as well. A domainΩ ⊂ X is said to be aRunge domain inXif every holomorphic functionΩ → C can be uniformly approximated on compact subsetsof Ω by functions inO(X).

Our main result may be stated as follows.

Theorem 1.1. Let Z ⊂ CN+1 (N ∈ N) be a closed complex hypersurface such that

Z ∩BN+1 6= ∅, letK ⊂ Z ∩BN+1 be a connected compact subset, and letǫ > 0. Thereexists a pseudoconvex domainD ⊂ Z with the following properties:

(i) K ⊂ D.(ii) D is Runge inZ.

(iii) There exists a complete proper holomorphic embeddingψ : D → BN+1 such that|ψ(ζ)− ζ| < ǫ for all ζ ∈ K.

In particular, BN+1 contains complete closed complex hypersurfaces which arebiholomorphic to a pseudoconvex Runge domain inC

N .

These are the first known examples of complete bounded complex hypersurfaces inCN+1,N ∈ N, for which one has any topological information. InC2 we have the following

more precise result:

Corollary 1.2. LetZ ⊂ C2 be a properly embedded complex curve and letK ⊂ Z ∩B2

be a compact connected subset. Givenǫ > 0 there exist a Runge domainD ⊂ Z

Complete complex hypersurfaces in the ball 3

and a complete proper holomorphic embeddingψ : D → B2 such thatK ⊂ D and|ψ(x) − x| < ǫ for all x ∈ K.

As a consequence, the unit ballB2 of C2 carries complete properly embedded complexcurves with any finite topology. In particular, there are proper complete holomorphicembeddingsD → B2.

The second part of the corollary follows from the well-knownfact that all surfaces offinite topology can be realized as Runge domains of properly embedded complex curves inC2 (seeCerne and Forstneric [8] and Section 4).

Our technique is different from the ones in [6, 11]. We begin with a closed complexhypersurfaceZ ⊂ C

N+1, intersectingBN+1, a compact subsetK ⊂ Z ∩ BN+1, andthe natural embeddingZ → C

N+1 given by the inclusion map. In a recursive process,we compose this initial embedding with a sequence of holomorphic automorphisms ofCN+1 which converges uniformly in compact subsets ofBN+1. In this way we obtain

a sequence of proper holomorphic embeddingsZ → CN+1 whose images converge

uniformly on compact subsets ofBN+1 to a closed embedded complex hypersurfaceof BN+1. Moreover, carrying out this process in the right way, we mayensure that aconnected componentD of the resulting hypersurface is biholomorphic to a pseudoconvexRunge domain inZ containingK, and is closed inBN+1 and complete.

This program will be performed in two different steps, whichwe now describe. Fora while we shall be working in real Euclidean spaceR

n+1, n ∈ N. We let Bn+1 andSn = bBn+1 denote the unit ball and the unit sphere inR

n+1 of center0 and radius1; withthis notation,BN+1 = B2N+2 for all N ∈ N. We denote by〈·, ·〉 and | · | the Euclideanscalar product and norm inRn+1. We writeq + rC for the setq + rp : p ∈ C ⊂ R

n+1

for r > 0, q ∈ Rn+1, and C ⊂ R

n+1. Finally, given p ∈ Rn+1 \ 0, we set

〈p〉⊥ := q ∈ Rn+1 : 〈q, p〉 = 0. Observe thatp + 〈p〉⊥ is the affine tangent hyperplane

to the sphere|p|Sn at the pointp. The following objects play a fundamental role in ourconstruction.

Definition 1.3. Givenp ∈ Rn+1 \ 0 andr > 0, the set

T (p, r) := p+ (rBn+1 ∩ 〈p〉⊥)will be calledthe (closed) tangent ball inRn+1 of centerp and radiusr.

Thus, forp ∈ Rn+1 \ 0 andr > 0, the tangent ballT (p, r) is the closed ball of center

p and radiusr in the affine hyperplanep + 〈p〉⊥. Given a collectionF of tangent balls inRn+1, we denote

|F| :=⋃

T∈FT ⊂ R

n+1.

We will gather tangent balls inRn+1 into collections which we calltidy according to thefollowing

Definition 1.4. A collectionF = T (pj , rj)j∈J of tangent balls inBn+1 ⊂ Rn+1 will be

calleda tidy collectionif the following conditions are satisfied:

• tBn+1 intersects finitely many balls inF for all 0 < t < 1.• If T (p1, r1),T (p2, r2) ∈ F and |p1| = |p2|, then r1 = r2 and T (p1, r1) ∩

T (p2, r2) = ∅.• If T (p1, r1),T (p2, r2) ∈ F and |p1| < |p2|, thenT (p1, r1) ⊂ |p2|Bn+1.


In particular, F consists of at most countably many pairwise disjoint tangent balls and|F|is a proper subset ofBn+1.

Thus, all the balls in a tidy collectionF which are tangent to the same sphere have thesame radius and are pairwise disjoint; ifF contains balls tangent to two spheres then allthe balls tangent to the smaller sphere are contained in the open ball whose boundary is thelarger sphere. IfF′ ⊂ F consists of those balls inF which are tangent to a given sphereλSn,0 < λ < 1, thenF′ is also tidy and there isµ ∈]λ, 1[ such that the boundarybT lies inµSn

for anyT ∈ F′. In particular, for eachT ∈ F′ the affine hyperplane ofRn+1 containingTis disjoint from the convex hull of|F′| \ T . This simple property of tidy collections will becrucial in our construction.

Tangent balls inBn+1 may be viewed as obstacles on the way towards the boundarySn

when we want to reachSn along a path inBn+1 that misses all the balls in a collection.The first step in the proof of Theorem 1.1 is to construct a tidycollectionF of tangent ballsin Bn+1 in such a way that every pathγ : [0, 1[→ Bn+1 satisfyinglimt→1 |γ(t)| = 1 andmissing all the balls ofF, has infinite length.

Theorem 1.5 (Building obstacles). Let n ∈ N and 0 < λ0 < 1. There exists a tidycollectionF of tangent balls inBn+1 ⊂ R

n+1 satisfying|F| ∩ λ0Bn+1 = ∅ and havingthe following property: Every pathγ : [0, 1[→ Bn+1 such thatlimt→1 |γ(t)| = 1 andγ([0, 1[) ∩ |F| = ∅, has infinite length.

In particular, every closed submanifold ofBn+1 missing|F| is complete.

Theorem 1.5 will be proved in Sec. 2 (see the more general Theorem 2.5). It is clear thata collectionF as in Theorem 1.5 will consists of infinitely many tangent balls.

The second step in the proof of Theorem 1.1 is to construct a proper holomorphicembedding from a domainD in any given closed complex hypersurfaceZ ⊂ C

N+1 toBN+1, whose image misses all the balls in a given tidy collectionF of tangent balls inBN+1.

Theorem 1.6(Avoiding obstacles). LetN ∈ N. LetF be a tidy collection of tangent ballsin the open unit ballBN+1 ⊂ C

N+1, and chooseλ0 > 0 such that|F|∩λ0BN+1 = ∅. LetZ ⊂ C

N+1 be a closed complex hypersurface such thatZ ∩ λ0BN+1 6= ∅ and letǫ > 0.

There exist a Runge domainΩ ⊂ CN+1 containingλ0BN+1 and a biholomorphic map

Ψ: Ω → BN+1 such that:

(i) |Ψ(ζ)− ζ| < ǫ for all ζ ∈ λ0BN+1.(ii) Ψ(Ω ∩ Z) ∩ |F| = ∅.

In particular, every connected componentD of Ω ∩ Z 6= ∅ is a pseudoconvex Rungedomain inZ and Ψ|D : D → BN+1 is a proper holomorphic embedding whose imagemisses|F|.

The proof of Theorem 1.6 is included in Sec. 3 and roughly goesas follows. Observe firstthat, given an infinite tidy collectionF = T (pj, rj) : j ∈ N of tangent balls inBN+1,we may assume that the convex hull of

⋃i≤j T (pi, ri) is disjoint from

⋃i>j T (pi, ri) for

eachj ∈ N. Indeed, it suffices to orderF so that|pi| ≤ |pj | if i < j. Thus, thereexists an exhaustionλ0BN+1 ⋐ E1 ⋐ E2 ⋐ · · · ⋐ ⋃

j∈Z+Ej = BN+1 of BN+1 by

smoothly bounded, strictly convex domains, such thatE1 ∩ |F| = ∅, and for eachj ∈ N,


⋃i≤j T (pi, ri) ⊂ Ej+1 andEj+1∩(

⋃i>j T (pi, ri)) = ∅. The key in the proof of Theorem

1.6 is to show that, given a closed complex hypersurfaceX ⊂ CN+1 which does not

intersect⋃i≤j−1 T (pi, ri), there exists a holomorphic automorphismΨj : C

N+1 → CN+1

such thatΨj is close to the identity in the compact convex setEj and such thatΨj(X),which is again a closed complex hypersurface, does not intersect

⋃i≤j T (pi, ri); cf. Lemma

3.1. We obtain the biholomorphic mapΨ: Ω → BN+1 in Theorem 1.6 as the limit of asequence of automorphismsΨjj∈N generated in a recursive way by application of thisresult.

Theorems 1.5 and 1.6 easily imply Theorem 1.1; see Sec. 4. Indeed, Theorem 1.6 appliedto a tidy collection of tangent balls inBN+1 given by Theorem 1.5, provides a complete,closed complex hypersurface inBN+1, which is biholomorphic to a pseudoconvex Rungedomain in a given closed complex hypersurfaceZ ⊂ C

N+1.

2. The Building Obstacles Theorem

In this section we prove Theorem 1.5 in a more general form; see Theorem 2.5 below.

Throughout the section we fixn ∈ N, write B for the open unit ballBn+1 ⊂ Rn+1 and

denote byp : Rn+1 \ 0 → Sn = bB the radial projection

p(x) =x

|x| , x ∈ Rn+1 \ 0.

We denote bydist(·, ·) and diam(·) the Euclidean distance and diameter inRn+1, andı :=

√−1.

We begin with the following result, which is the key in the proof of Theorem 1.5.

Lemma 2.1. There exist numbersmn ∈ N, mn ≥ 2, andcn ∈ R, 0 < cn < 1/2, such thatthe following assertion holds.

For every real numberr > 0 there exist finite subsetsF1, . . . , Fmn of Sn such that:

(i) |p− q| ≥ r for all p, q ∈ Fj , p 6= q, j = 1, . . . ,mn.(ii) If F :=

⋃mn

j=1 Fj thenF 6= ∅ anddist(p, F ) ≤ cnr for all p ∈ Sn.

We emphasize that the numbersmn andcn in the lemma only depend onn, the dimensionof the sphere. Possibly some of theFj ’s are empty. Ifr > 2 = diam(Sn), condition (i)implies thatFj consists of at most one point for allj ∈ 1, . . . ,mn.

Let us outline how Theorem 1.5 will follow from Lemma 2.1. We will pick anysequence of numbers0 < s0 = λ0 < s1 < · · · < limj→+∞ sj = 1 such that∑

j∈N√sj − sj−1 = +∞; hereλ0 is given in the statement of the theorem. For each

j ∈ N, we will consider the numberssj,k := sj−1 + ksj−sj−1

mn+1 , k = 1, . . . ,mn + 1 (hence

sj−1 < sj,k ≤ sj), and will takerj > 0 such that the tangent ballT (p, rj) is containedin sj,k+1B for all p ∈ sj,kS

n, k = 1, . . . ,mn. Basic trigonometry gives thatrj can betaken to be larger thana

√sj − sj−1 for some constanta > 0 which does not depend on

j. We will then apply Lemma 2.1 forr = 2rj and obtain subsetsFj,1, . . . , Fj,mn ⊂ Sn,

j ∈ N. Lemma 2.1 (i) and the choice ofrj will ensure that the collection of tangentballsF =

⋃j∈N

(⋃mn

k=1 Fj,k), whereFj,k = sj,kT (p, rj) : p ∈ Fj,k for all j ∈ N and

k ∈ 1, . . . ,mn, is tidy. On the other hand, Lemma 2.1 (ii) will guarantee theexistence ofa constantc > 0, which does not depend onj, such that for anyp ∈ S

n the spherical ball


(p + crjB) ∩ Sn is contained inp(T ) for someT ∈ Fj :=

⋃mn

k=1 Fj,k. This implies thatthe length of every pathγj : [0, 1] → sjB \ sj−1B such that|γj(0)| = sj−1, |γj(1)| = sj ,andγj([0, 1])∩ |Fj| = ∅, is at leastsj−1crj > as0c

√sj − sj−1. (See Lemma 2.3.) Finally,

sincea, s0, andc do not depend onj ∈ N, the choice of thesj ’s will ensure that every pathγ : [0, 1[→ B such thatlimt→1 |γ(t)| = 1 andγ([0, 1[) ∩ |F| = ∅, has infinite length. Thus,the collectionF of tangent balls will prove Theorem 1.5.

Before starting with the proof of Lemma 2.1 observe the following

Claim 2.2. Assume that Lemma 2.1 holds. Then the same statement is validwith the samenumbersmn ∈ N and0 < cn < 1/2 if we replaceSn by tSn = tp : p ∈ S

n for anyt > 0.

Proof. Pick t, r > 0. Since we are assuming that the lemma holds forSn, there existmn

subsetsC1, . . . , Cmn of Sn satisfying (i) and (ii) for the real numberr/t > 0. Therefore, thesubsetstC1, . . . , tCmn of tSn meet (i) and (ii) for the real numberr, proving the claim.

Proof of Lemma 2.1.We proceed by induction onn, the dimension of the sphere.

The basis of the induction(n = 1) admits several proofs. The one we include here willhelp the reader to understand the main point of the inductionstep. We letm1 = 10 andc1 = 1/3 (these numbers are not sharp but fit well with the argument in the inductive step).Chooser > 0. It suffices to find ten subsets ofS1 satisfying conditions (i) and (ii). Wedistinguish cases.

Assumer ≥ 2 = diam(S1). SetFj := eı(j−1)π5 ⊂ S

1 ⊂ C, 1 ≤ j ≤ 10. Condition(i) trivially holds since all theFj ’s are unitary, whereas (ii) follows from the equation

(2.1) |x− xeıσ| = |1− eıσ | = 2 sin(σ2

)for all x ∈ S

1 ⊂ C, σ ∈ [0, 2π].

Indeed, givenp = eıt ∈ S1, t ∈ [0, 2π], there existsj0 ∈ 1, . . . , 10 such that

|t− (j0 − 1)π5 | < π5 , and so

dist(p,

10⋃

j=1

Fj

)≤ dist(p, Fj0) ≤

∣∣p− peı((j0−1)π5−t)∣∣ ≤ 2 sin

( π10

)<

2

3≤ c1r;

take into account that the sinus in increasing in[0, π2 ].

Assume now0 < r < 2. In this case (2.1) ensures that

(2.2) |x− xeıσ| ≥ r, x ∈ S1, for all σ ∈ [α, π],

whereα := 2 arcsin(r/2) ∈]0, π[. Setβ := 2 arcsin(r/6) and observe that

(2.3) 0 < β < α < 5β.

Moreover, since the sinus is increasing in[0, π2 ], (2.1) ensures that

(2.4) |x− xeıσ | ≤ r

3= c1r, x ∈ S

1, for all σ ∈ [0, β].

For eachj ∈ 1, . . . , 5, consider the subset

(2.5) Fj := −ıeı(j−1)βeı5(k−1)β : k ∈ N, ((j − 1) + 5(k − 1))β ≤ π ⊂ S1

generated by rotating the point−ıeı(j−1)β by angles which are multiple of5β. (See Fig.2.1.) Combining (2.2) and (2.3) we obtain thatFj has the property that|p − q| ≥ r if


p, q ∈ Fj , p 6= q, j ∈ 1, . . . , 5. This simply means thatFj satisfies condition (i) in thelemma,j = 1, . . . , 5. The same holds forF−j := −z : z ∈ Fj, j = 1, . . . , 5, that is,

(2.6) F−j = −ıe−ı(j−1)βe−ı5(k−1)β : k ∈ N, ((j − 1) + 5(k − 1))β ≤ π.

Figure 2.1. Basis of the induction forr = 3/4.

To finish, it suffices to show thatF1, . . . , F5, F−1, . . . , F−5 also satisfy (ii). For thatpick p = −ıeıt ∈ S

1, t ∈ [−π, π]. Without loss of generality we assume thatt ∈ [0, π];otherwise we reason in a symmetric way. It is clear that thereexist j0 ∈ 1, . . . , 5 andk0 ∈ N such that((j0−1)+5(k0−1))β ≤ π and|t− ((j0−1)+5(k0−1))β| ≤ β. Thus,sinceeı((j0−1)+5(k0−1))β ∈ Fj0 , (2.4) implies that

dist(p,

5⋃

j=1

(Fj ∪ F−j))≤ dist(p, Fj0) ≤

∣∣∣p− peı(((j0−1)+5(k0−1))β−t

)∣∣∣ ≤ c1r,

which proves (ii) and concludes the basis of the induction.

For the inductive step fixn ∈ N, n ≥ 2, assume that the lemma holds forn− 1, and letus prove it forn.

We begin with some preparations. Givent ∈ [−π2 ,

π2 ] we denote byΠt the affine

hyperplane ofRn+1 given by

Πt := (x1, . . . , xn+1) ∈ Rn+1 : xn+1 = sin t.

Notice that

(2.7) Sn ∩Πt = cos(t)Sn−1 × sin t ∀t ∈

[− π

2,π

2

].

We adopt the convention0Sn−1 = 0 ⊂ Rn. Obviously,Sn =

⋃t∈[−π

2,π2](S

n ∩Πt).

The idea for the proof is similar to the one in the basis of the induction. In this casethe role of the points−ıe±ı(j−1)β will be played by finite subsetsAtj , j = 1, . . . ,mn−1,of the (n − 1)-dimensional sphereSn ∩ Πt, t ∈] − π

2 ,π2 [, which will be provided by the


inductive hypothesis; take into account Claim 2.2. (See Properties (P1), (P2), and (2.11)below.) Likewise, the role of the setFj will be played by the union of a finite family of setsAtj wherej is fixed andt moves in sufficiently far heights (compare (2.15) below with(2.5)and (2.6), and Properties (P3) and (P4) below with (2.2) and (2.4)).

So, pick any numbercn with

(2.8) 0 < cn−1 < cn <1

2.

Consider the functionf : ]0,+∞[→ R given by

(2.9) f(x) =arcsin

(min

x2, 1)

arcsin(min

(cn − cn−1)x

2, 1) , x > 0.

Observe thatf is continuous, positive,limx→0 f(x) = 1/(cn − cn−1) > 2, andlimx→+∞ f(x) = 1, hencef is a bounded function. Set

(2.10) µ := E(supx>0

f(x))≥ 2,

whereE(·) means integer part. Obviouslyµ only depends oncn−1 andcn. Set

mn := (µ + 1)mn−1,

and let us check that the numbersmn ∈ N andcn ∈]0, 1/2[ satisfy the conclusion of thelemma. For that, chooser > 0 and let us furnish subsetsF1, . . . , Fmn of Sn meetingconditions (i) and (ii) in the statement of the lemma.

For t in the open interval] − π2 ,

π2 [ and in view of (2.7), Claim 2.2 provides subsets

At1, . . . , Atmn−1

of Sn ∩ Πt satisfying the thesis of the lemma for the real numberr > 0.That is:

(P1) Atj is finite and|p− q| ≥ r for all p, q ∈ Atj , p 6= q, j = 1, . . . ,mn−1.(P2) At :=

⋃mn−1

j=1 Atj 6= ∅ anddist(p,At) ≤ cn−1r for all p ∈ Sn ∩Πt.

We also set

(2.11) A±π

2

j := (0, . . . , 0,±1) ⊂ Sn, j ∈ 1, . . . ,mn−1.

To finish the proof, we will distribute a suitable subset of⋃

(t,j)∈[−π2,π2]×1,...,mn−1A

tj ⊂

Sn into mn subsets ofSn satisfying conditions (i) and (ii).

Observe first that, by basic trigonometry,

(2.12) dist(p,Sn ∩Πt) = 2 sin( |t− s|

2

)∀p ∈ S

n ∩Πs, t, s ∈[− π

2,π

2

].

Set

(2.13) α := 2 arcsin(min

r2, 1), β := 2 arcsin

(min

(cn − cn−1)r

2, 1).

In view of (2.8), we have0 < β ≤ α ≤ π. Moreover, since the sinus is increasing in[0, π2 ],(2.12) ensures that:

(P3) dist(Sn ∩Πs,Sn ∩Πt) ≥ minr, 2 for all t, s ∈ [−π

2 ,π2 ] with |t− s| ≥ α.

(P4) dist(p,Sn ∩Πt) ≤ min(cn− cn−1)r, 2 for all p ∈ Sn ∩Πs, for all t, s ∈ [−π

2 ,π2 ]

with |t− s| ≤ β.


Further,α/β = f(r) wheref is the function (2.9), and so (2.10) gives that

(2.14) (µ+ 1)β > α.

Denote byI := 1, . . . ,mn−1 × 0, . . . , µ. For all k ∈ 0, . . . , µ call Ik := l ∈Z : 0 ≤ k + l(µ+ 1) ≤ π/β, and notice thatIk is a (possibly empty) finite set; see (2.10).For (j, k) ∈ I, set

(2.15) Fj,k :=⋃

l∈IkA

−π2+(k+l(µ+1))β

j .

(See Figure 2.2.) To conclude the proof it suffices to check that themn = (µ + 1)mn−1

Figure 2.2. The subsetFj,k ⊂ Sn.

subsets ofSn, Fj,k, (j, k) ∈ I, satisfy Lemma 2.1-(i),(ii).

Pick (j, k) ∈ I. SinceIk is finite, (P1) and (2.11) ensure that the (possibly empty)setFj,k is finite. Suppose thatFj,k 6= ∅ and choose a pair of pointsp, q ∈ Fj,k. If

p, q ∈ A−π

2+(k+l(µ+1))β

j for somel ∈ Ik, (P1) and (2.11) ensure that eitherp = q or

|p − q| ≥ r. Assume now thatp ∈ A−π

2+(k+l1(µ+1))β

j andq ∈ A−π

2+(k+l2(µ+1))β

j withl1, l2 ∈ Ik, l1 6= l2. It follows that

π ≥∣∣− π

2+ (k + l1(µ+ 1))β −

(− π

2+ (k + l2(µ+ 1))β

)∣∣

= |l1 − l2|(µ+ 1)β ≥ (µ+ 1)β > α;

take into account (2.14). This and (2.13) implyr < 2, hence (P3) guarantees that|p− q| ≥ minr, 2 = r. Condition (i) follows.

To check (ii) setF :=⋃

(j,k)∈I Fj,k. Since∅ 6= A−π

2

j ⊂ Fj,0 ⊂ F , F 6= ∅. Pick apoint p ∈ S

n and lets ∈ [−π2 ,

π2 ] be the unique number satisfyingp ∈ S

n ∩ Πs. Since0 < β ≤ π, there existk0 ∈ 0, . . . , µ and l0 ∈ Ik0 such that|t0 − s| ≤ β, wheret0 := −π

2 + (k0 + l0(µ + 1))β ∈ [−π2 ,

π2 ]. Property (P4) providesq ∈ S

n ∩Πt0 such that


|p− q| ≤ min(cn − cn−1)r, 2. Together with (P2) and (2.11), we obtain that

dist(p,

mn−1⋃

j=1

At0j)

≤ |p− q|+ dist(q,

mn−1⋃

j=1

At0j)

≤ min(cn − cn−1)r, 2 + cn−1r ≤ cnr.

Since⋃mn−1

j=1 At0j ⊂ ⋃mn−1

j=1 Fj,k0 ⊂ F , the above inequality proves (ii). This completes theproof.

With Lemma 2.1 in hand, we may find finite tidy collections of tangent balls in a sphericalshell, which are suitable for our purposes.

Lemma 2.3. Givenn ∈ N, there exists a numberan > 0 such that the following holds:

For any 0 < r1 < r2 < 1 there is a finite tidy collectionF of tangent balls inr2B,contained inr2B \ r1B, such that ifγ : [0, 1] → r2B \ r1B is a path such that|γ(0)| = r1,

|γ(1)| = r2, andγ([0, 1]) ∩ |F| = ∅, then the length ofp γ is at leastan

√r2 − r1√r2

. In

particular, the length of suchγ is at leastanr1√r2 − r1√r2

.

Proof. By Lemma 2.1 there aremn ∈ N, mn ≥ 2, andcn ∈ R, 0 < cn < 1/2, such thatgivenr > 0 there are finite setsF1, . . . , Fmn ⊂ S

n satisfying:

(i) |p− q| ≥ r for all p, q ∈ Fj , p 6= q, 1 ≤ j ≤ mn.(ii) If F =

⋃mn

j=1 Fj then for everyp ∈ Sn we havedist(p, F ) ≤ cnr.

Fix r > 0 and letFj , 1 ≤ j ≤ mn, be as above.

By (i), given p ∈ Sn there is aq ∈ F such that|q − p| ≤ cnr. So, if y ∈ S

n,|y − p| ≤ (1/2 − cn)r, then |y − q| ≤ |y − p| + |p − q| ≤ (1/2 − cn)r + cnr = r

2 .Thus, for everyp ∈ S

n there arej, 1 ≤ j ≤ mn, andq ∈ Fj such that

(2.16)(p+ (1/2 − cn)rB

)∩ S

n ⊂(q +

r

2B)∩ S

n.

For eachp ∈ Sn denote byT (p) the tangent ballT ((1 − δ)p, η), where0 < δ < 1 and

η > 0 are chosen so that this ball is attached toSn along(p+ r

2Sn) ∩ S

n, that is to say, theboundary

b(T (p)

)=

(p+

r

2Sn)∩ S

n;

see Figure 2.3. This simply means that(1 − δ)2 + η2 = 1 andη2 + δ2 = ( r2 )2, which

implies thatr2 =√2δ.

Note thatT (p) ∩ T (q) = ∅ wheneverp, q ∈ S

n, |p− q| ≥ r,

and that the projection toSn

p(T (p)) =(p+

r

2B)∩ S

n.

Now, let0 < r1 < r2 < 1. Divide the interval[r1, r2] into mn + 1 equal pieces of length

ω =r2 − r1mn + 1


and letsj = r1 + jω, 0 ≤ j ≤ mn + 1, so that

r1 = s0 < s1 < · · · < smn < smn+1 = r2.

We now describe how to get our tidy collection of tangent balls.

Given j, 1 ≤ j ≤ mn, let Gj be the collectionT (p) : p ∈ Fj. This is a collectionof pairwise disjoint tangent balls whose boundaries are contained inSn and whose centersare on(1 − δ)Sn. Now form sjGj by multiplying each ball inGj by sj and thus pullingit insideB; sosjGj = sjT (p) : p ∈ Fj. Observe thatsjGj is a collection of pairwisedisjoint tangent balls with centers onsj(1− δ)Sn and with boundaries contained insjSn.

Consider now the collectionF of all tangent balls obtained in this way (see Figure 2.3),that is:

F =

mn⋃

j=1

sjGj .

The familly F will be a tidy collection of tangent balls inr2B contained inr2B \ r1B

Figure 2.3. The collectionF =⋃mn

j=1 sjGj .

provided that for eachj, 2 ≤ j ≤ mn, the sphere containing the centers of the balls insjGj ,that is,(1− δ)sjS

n, is outside the ballsj−1B which contains the balls ofsj−1Gj−1, that is,provided that

(2.17) sj(1− δ) > sj−1 for all 2 ≤ j ≤ mn,


and since we do not want the balls to meetr1B = s0B, (2.17) will have to hold forj = 1 aswell. Now, (2.17) for1 ≤ j ≤ mn means that

(2.18) δ <sj − sj−1

sj=ω

sj, 1 ≤ j ≤ mn.

Sincesj ≤ r2, 1 ≤ j ≤ mn, (2.18) certainly holds if we choose

δ :=ω

r2=

r2 − r1(mn + 1)r2

.

Recall thatr2 =√2δ. Set

(2.19) r := 2

√2

r2 − r1(mn + 1)r2

at the outset and obtainF as above. ThenF is a tidy collection of balls contained inr2B\r1B.

Let us show thatF has the properties stated in the lemma.

Let q ∈ Sn and setΩ := (q + r

2B) ∩ Sn. Observe thatΩ is an open subset ofSn

whose closure equalsp(T (q)) = p(sT (q)) for anys > 0. Note also that for everys > 0the tangent ballsT (q) cuts the open cone into two components, that is,p−1(Ω) \ sT (q)consists of two components.

Suppose thatγ : [0, 1] → r2B \ r1B is a path with|γ(0)| = r1, |γ(1)| = r2, andγ([0, 1]) ∩ |F| = ∅. Write p = p(γ(0)). By (2.16), there arej ∈ 1, . . . ,mn andq ∈ Fjsuch that

(p+ (12 − cn)rB

)∩ S

n ⊂ (q + r2B) ∩ S

n, and so,

(2.20)(p+

1

2

(12− cn

)rB

)∩ S

n ⊂ Ω,

whereΩ is as above.

Assume for a moment that

(2.21) diam(p(γ([0, 1]))

)<

1

2

(12− cn

)r.

By (2.20), this implies thatγ([0, 1]) ⊂ Ω. On the other hand,sjT (q) is the tangent ballin F contained insjB \ sj−1B ⊂ r2B \ r1B. Since|γ(0)| = r1 and|γ(1)| = r2 it followsthatγ(0) andγ(1) are in different components ofΩ \ sjT (q) and thereforeγ([0, 1]) meetssjT (q), a contradiction. So (2.21) does not hold, which implies that

Length(p γ) ≥ diam(p(π([0, 1])) ≥ 1

2

(12− cn

)r.

By (2.19) it follows that

Length(p γ) ≥ an

√r2 − r1√r2

,

wherean = (12 − cn)√2√

mn+1. This completes the proof.

Finally, a finite recursive application of Lemma 2.3 gives the following extension.

Lemma 2.4. Suppose0 < r1 < r2 < 1 and letρ > 0. There exists a finite tidy collectionF of tangent balls inr2B enjoying the following properties:

(i) |F| ∩ r1B = ∅.(ii) If γ : [0, 1] → r2B\r1B is a path with|γ(0)| = r1, |γ(1)| = r2, andγ([0, 1])∩|F| = ∅,

then the length ofp γ is at leastρ. In particular, the length of suchγ is at leastr1ρ.


Proof. Let an > 0 be the number given by Lemma 2.3. PickN ∈ N,

(2.22) N >r2ρ

2

a2n(r2 − r1).

Setsj := r1 + j

r2 − r1N

, j = 1, . . . , N.

Obviously,r1 = s0 < s1 < · · · < sN = r2. For eachj, 1 ≤ j ≤ N , apply Lemma 2.3 withsj−1, sj in place ofr1, r2 to get a finite tidy collectionFj of tangent balls insjB having thefollowing properties:

(ij) |Fj| ∩ sj−1B = ∅.(ii j) If γ : [0, 1] → sjB \ sj−1B is a path with|γ(0)| = sj−1, |γ(1)| = sj, and

γ([0, 1]) ∩ |Fj | = ∅, then the length ofp γ is at leastan

√sj − sj−1√

sj.

SetF :=⋃Nj=1 Fj. It follows thatF is a tidy finite collection of tangent balls inr2B

satisfying (i). Furthermore, given a pathγ as in (ii), γ contains subpathsγ1, . . . , γN withγj ⊂ sjB \ sj−1B connectingsj−1S

n andsjSn for all j. Therefore,

Length(p γ) ≥N∑

j=1

Length(p γj)

(iij )≥ an

N∑

j=1

√sj − sj−1√

sj

=

N∑

j=1

an√r2 − r1√

r1 +jN(r2 − r1)

1√N

≥N∑

j=1

an√r2 − r1√r2

1√N

=an

√r2 − r1√r2

√N

(2.22)> ρ.

This proves (ii) and completes the proof of the lemma.

We now state and prove the main result in this section, which is a more general versionof Theorem 1.5.

Theorem 2.5. Let 0 < s0 < s1 < · · · < limj→+∞ sj = 1 and 0 = ρ0 < ρ1 < · · · <limj→+∞ ρj = +∞ be sequences of real numbers. There exists a tidy collectionF oftangent balls inB satisfying the following properties:

(i) |F| ∩ s1B = ∅.(ii) If γ : [0, 1] → B is a path with|γ(0)| = sj−1, |γ(1)| = sj, andγ([0, 1]) ∩ |F| = ∅,

then the length ofp γ is at leastρj . In particular, the length of suchγ is at leastsj−1ρj.

(iii) If γ : [0, 1[→ B is a path such thatlimt→1 |γ(t)| = 1 and whose range intersects atmost finitely many balls inF, thenp γ andγ have infinite length.

Thus, every closed submanifold ofB missing|F| is complete.


Proof. For eachj ∈ N, Lemma 2.4 applied tosj−1, sj in place ofr1, r2, furnishes a tidyfinite collectionFj of tangent balls insjB satisfying that:

(ij) |Fj| ∩ sj−1B = ∅.(ii j) If γ : [0, 1] → sjB \ sj−1B is a path with|γ(0)| = sj−1, |γ(1)| = sj, and

γ([0, 1]) ∩ |Fj | = ∅, then the length ofp γ is at leastρj . In particular, the length ofsuchγ is at leastsj−1ρj .

SetF :=⋃j∈N Fj. By (ij), j ∈ N, F is a tidy collection of tangent balls inB satisfying

(i), whereas (ii) clearly follows from (iij), j ∈ N.

To finish, assume thatγ : [0, 1[→ B is a proper path intersecting at most finitely manyballs inF. SinceF is tidy in B, there existsj0 ∈ N such thatγ([0, 1[) ∩ |Fj| = ∅ for allj ≥ j0. Up to enlargingj0 if necessary, we infer thatγ contains a subpathγj with range insjB \ (|Fj | ∪ sj−1B) connectingsj−1S

n andsjSn, j ≥ j0. Item (ii) ensures that

Length(p γ) ≥∑

j≥j0Length(p γj) ≥

∑

j≥j0ρj = +∞

and so

Length(γ) ≥∑

j≥j0Length(γj) ≥

∑

j≥j0sj−1ρj ≥ sj0−1

∑

j≥j0ρj = +∞.

This proves (iii) and completes the proof.

3. The Avoiding Obstacles Theorem

This section is devoted to the proof of Theorem 1.6. Throughout the section we fixN ∈ N and writeB for the unit ballBN+1 ⊂ C

N+1. Numbers inC will be denoted byroman letters, typicallyz andw, whereas elements ofCN orCN+1 will be denoted by greekletters such asζ andξ.

As a preliminary step to the proof of Theorem 1.6 we shall prove the following.

Lemma 3.1. Let D andE be compact sets inCN+1 and assume thatE is convex andD is contained in an affine real hyperplaneL ⊂ C

N+1 which does not intersectE. LetZ ⊂ C

N+1 be a closed complex hypersurface. Givenǫ > 0 there exists a holomorphicautomorphismΦ: CN+1 → C

N+1 such that:

(i) Φ(Z) ∩D = ∅.(ii) |Φ(ζ)− ζ| < ǫ for all ζ ∈ E.

The authors wish to thank the referee for pointing out how to prove this lemma whichreplaces a similar one proved in a previous version of the paper under the extra assumptionthat the hypersurfaceZ is algebraic.

Proof. Let π1 : CN+1 → C denote the orthogonal projection into the first component givenby π1(z1, . . . , zN+1) = z1. By an affine complex change of coordinates we may assumethat

L = (z1, . . . , zN+1) ∈ CN+1 : ℜ(z1) = 0,

whereℜ(·) means real part, andE is contained in the half spaceℜ(z1) < 0. SinceD ⊂ Lis compact, there exist a compact segmentD′ in the real linez ∈ C : ℜ(z) = 0 = π1(L)


and a real numberλ > 0 such that

(3.1) D ⊂ D′ × (λBN ) ⊂ C× CN .

Further, by dimension reasons we may assume thatZ ∩ (D′ × 0) = ∅ (here0 ∈ CN ;

this can be achieved, for example, by an arbitrarily small translation of the hypersurfaceZ).Thus, sinceZ is closed andD′ is compact, there isη > 0 such that

(3.2) Z ∩ (D′ × (ηBN )) = ∅.On the other hand,E′ := π1(E) ⊂ C is a compact convex subset ofz ∈ C : ℜ(z) < 0.

Pick a smallτ > 0, which will be specified later, and choose a holomorphic functionψ : C → C such that

(3.3) |ψ(z)| < τ for all z ∈ E′ = π1(E)

and

(3.4) ℜ(ψ(z)) > 1/τ for all z ∈ D′ ⊃ π1(D).

Such exists by the classical Runge approximation theorem; observe thatD′ andE′ areconnected, simply-connected, disjoint compact subsets ofC, and henceD′ ∪ E′ is Rungein C. We claim that, ifτ > 0 is chosen small enough, then the holomorphic automorphismΦ: CN+1 = C× C

N → CN+1 = C× C

N given by

Φ(z, ξ) = (z, eψ(z)ξ), z ∈ C, ξ ∈ CN ,

satisfies the conclusion of the lemma. Indeed, for(z, ξ) ∈ E we have

|Φ(z, ξ) − (z, ξ)| = |(1− eψ(z))ξ| ≤ |1− eψ(z)|max|ζ| : ζ ∈ E < ǫ,

where the last inequality is ensured by (3.3) provided thatτ > 0 is sufficiently small; recallthatE is compact. This implies (ii). On the other hand, forz ∈ D′ we have

Φ(z × (ηBN )) = z × (eψ(z)ηBN ) ⊃ z × (λBN ),

where the last inclusion is guaranteed by (3.4) provided that τ > 0 is sufficiently small.Thus, in view of (3.1),

D ⊂ D′ × (λBN ) ⊂ Φ(D′ × (ηBN )).

Since (3.2) ensures thatΦ(Z) ∩ Φ(D′ × (ηBN )) = ∅, the above inclusion guarantees (i).This concludes the proof.

We now prove Theorem 1.6 by a recursive application of Lemma 3.1.

Proof of Theorem 1.6.LetF be a tidy collection of tangent balls in the unit ballB ⊂ CN+1

(see Def. 1.3 and 1.4) and fix numbersǫ > 0 andλ0 > 0 such that|F| ∩ λ0B = ∅. LetZ ⊂ C

N+1 be a closed complex hypersurface such thatZ ∩ λ0B 6= ∅.

Pickλ0 < λ1 < 1 such that|F| ∩ λ1B = ∅. With no loss of generality we may assumethat the collectionF is infinite; otherwise we simply replaceF by any infinite tidy collectionin B containing it and disjoint fromλ1B. SetT0 = ∅. SinceF is tidy, it is clear that wemay writeF = Tj : j ∈ N so that there exists an exhaustion ofB by smoothly bounded,strictly convex domains

E0 := λ0B ⋐ E1 := λ1B ⋐ E2 ⋐ . . . ⋐⋃

j∈Z+

Ej = B,


such that

(3.5)j−1⋃

i=0

Ti ⊂ Ej and Ej ∩ (⋃

i>j−1

Ti) = ∅, j ∈ N.

Recall the short discussion which follows Theorem 1.6 in theintroduction.

Setǫ0 = ǫ andΨ0 = Id: CN+1 → CN+1. We shall inductively use Lemma 3.1 to find a

sequenceΦjj∈N of holomorphic automorphisms ofCN+1 and a sequenceǫjj∈N suchthat if

Ψj := Φj · · · Φ1, j ∈ N,

then for eachj ∈ N the following conditions hold:

(aj) ǫj < ǫj−1/2.(bj) ǫj < dist(Ej−1,C

N+1 \ Ej).(cj) ǫj < dist(Ej ,C

N+1 \ Ej+1).(dj) |Φj(ζ)− ζ| < ǫj for all ζ ∈ Ej.(ej) Ψj(Z) ∩ (

⋃ji=0 Ti) = ∅, and there is an open neighborhoodUj ⊂ Ej of

⋃j−1i=0 Ti

(see (3.5)) such that ifΘ is an automorphism ofCN+1 such that|Θ(ζ)− ζ| < 2ǫjfor all ζ ∈ Ej , thenΘ(Ψj−1(Z)) ∩ Uj = ∅.

Assume for a moment that we have sequencesΦjj∈N andǫjj∈N satisfying (aj)–(dj )above,j ∈ N. By (bj) and (dj ), we have that

(3.6) Φj(Ej−1) ⊂ Ej, j ∈ N.

If we writeLj = Ψ−1j (Ej), j ∈ N, we infer from (3.6) that

(3.7) E0 = λ0B ⋐ Lj ⋐ Lj+1 for all j ∈ N;

for the former inclusion take into account thatΨ1 = Φ1. Thus, since (aj) ensures that∑j∈N ǫj < +∞, (3.7) and properties (bj) imply that limj→∞Ψj = Ψ exists uniformly on

compacta inΩ :=

⋃

j∈NLj ⊂ C

N+1

andΨ is a biholomorphic map fromΩ onto⋃j∈NEj = B; see [10, Proposition 4.4.1].

On the other hand, (cj) and (dj ) ensure that

(3.8) Φj(Ej) ⊂ Ej+1, j ∈ N,

and soΨj(E1) ⊂ Ej+1 for all j ∈ N. Thus, forζ ∈ E1, (dk) and (ak), k ≤ j, give

|Ψj(ζ)− ζ| ≤ |Ψ1(ζ)− ζ|+j−1∑

k=1

|Ψk+1(ζ)−Ψk(ζ)|

= |Φ1(ζ)− ζ|+j−1∑

k=1

|Φk+1(Ψk(ζ))−Ψk(ζ)| ≤j∑

k=1

ǫj < 2ǫ1.

Sinceλ0B = E0 ⋐ Ω ∩ E1 (see (3.7) and the definition ofΩ), passing to the limit andtaking into account (a1) we get that|Ψ(ζ) − ζ| ≤ 2ǫ1 < ǫ0 = ǫ for all ζ ∈ λ0B. Thisproves (i) in the statement of the theorem.

Assume now that, in addition, the sequencesΦjj∈N andǫjj∈N satisfy (ej ), j ∈ N.We show that this implies thatΨ(Z ∩ Ω) misses|F|, which proves (ii). Indeed, for each


j ∈ N andk ≥ j call Θj,k := Φk · · · Φj. ThenΘj,k is an automorphism ofCN+1. By(3.8),Θj,k(Ej) ⊂ Ek+1 for all k ≥ j. Thus, for eachζ ∈ Ej, (di) and (ai), j ≤ i ≤ k, give

|Θj,k(ζ)− ζ| ≤ |Θj,j(ζ)− ζ|+k∑

i=j+1

|Θj,i(ζ)−Θj,i−1(ζ)|

= |Φj(ζ)− ζ|+k∑

i=j+1

|Φi(Θj,i−1(ζ))−Θj,i−1(ζ)| ≤k∑

i=j

ǫi < 2ǫj .

Therefore, (ej) guarantees thatΘj,k(Ψj−1(Z)) ∩ Uj = Ψk(Z) ∩ Uj = ∅ for everyk ≥ j,and soΨ(Z ∩Ω)∩Uj = ∅. Since this holds for everyj ∈ N and|F| ⊂ ⋃

j∈NUj , it followsthatΨ(Z ∩ Ω) ∩ |F| = ∅ as claimed.

It remains to show that there are sequencesΦjj∈N andǫjj∈N, and for eachj ∈ N anopen neighborhoodUj ⊂ Ej of

⋃j−1i=0 Ti (see (3.5)), such that ifΨj = Φj Φj−1 · · · Φ1

then (aj )–(ej ) hold for all j ∈ N. We proceed by induction.

For the basis of the induction we choose any numberǫ1 > 0 satisfying (a1), (b1), and (c1).By Lemma 3.1 there is a holomorphic automorphismΦ1 of CN+1 such that|Φ1(ζ)−ζ| < ǫ1onE1 andΦ1(Z) ∩ T1 = ∅. SetU1 = ∅. Conditions (d1) and (e1) are clear; recall thatΨ1 = Φ1 andT0 = ∅.

For the induction step, letj ≥ 2, assume that we haveΦj−1, ǫj−1, andUj−1 satisfying(aj−1)–(ej−1), and let us provideΦj, ǫj, andUj meeting (aj)–(ej ). Fix a numberǫj > 0which will be specified later. Assume thatǫj satisfies (aj), (bj), and (cj). Thus, Lemma 3.1furnishes a holomorphic automorphismΦj of CN+1 satisfying (dj) and such thatΨj(Z) =Φj(Ψj−1(Z)) missesTj . Furthermore, since (ej−1) and (3.5) ensure thatΨj−1(Z) misses⋃j−1i=0 Ti = Ej ∩ |F|, we may guarantee thatΨj(Z) misses

⋃ji=0 Ti provided thatǫj is

chosen small enough. This proves the former assertion in (ej ). For the latter one, recallthatΨj−1(Z) ∩ (

⋃j−1i=0 Ti) = ∅ and choose an open neighborhoodUj of

⋃j−1i=0 Ti whose

closure is a compact set contained inEj disjoint fromΨj−1(Z)∩Ej. Takeη > 0 such thatUj ⊂ ζ ∈ Ej : dist(ζ,C

N+1 \Ej) ≥ η and such that(Ψj−1(Z) ∩ Ej) + ηB is disjointfrom Uj . It then follows that ifΘ is an automorphism ofCN+1 such that|Θ(ζ) − ζ| < η

for all ζ ∈ Ej thenΘ(Ψj−1(Z)) ∩ Uj = ∅. Thus, (ej) is fully satisfied provided that wechooseǫj < η/2. This closes the induction and completes the proof of the existence ofΩandΨ satisfying (i) and (ii) in Theorem 1.6.

The domainΩ is biholomorphically equivalent toB and hence it is pseudoconvex. SinceEj is a compact convex set inCN+1 it is polynomially convex and hence every holomorphicfunction in a neighborhood ofEj can be, uniformly onEj , approximated by polynomials.SinceΨj is an automorphism ofCN+1 it follows that every holomorphic function in aneighborhood ofLj = Ψ−1

j (Ej) can be, uniformly onLj , approximated by entire functionsfor all j ∈ N. It follows thatΩ =

⋃j∈N Lj ⊂ C

N+1 is a Runge domain. (See [10].)

Notice thatΩ∩Z ⊃ Z∩λ0B 6= ∅ and pick a connected componentD ⊂ Ω∩Z. ThenDis a closed submanifold ofΩ. SinceΩ is pseudoconvex it follows thatD is a pseudoconvexdomain inZ, and, moreover, by Cartan’s extension theorem every holomorphic functionϕonD extends holomorphically to a holomorphic functionϕ onΩ and, sinceΩ is Runge inCN+1, ϕ can be, uniformly on compacta inΩ, approximated by entire functions onCN+1.

Therefore,ϕ = ϕ|D can be, uniformly on compacta inD, approximated by restrictions ofentire functions onCN+1 to Z. Thus, every holomorphic function onD can be, uniformly


on compacta inD, approximated by holomorphic functions onZ, soD is a Runge domainin Z. This completes the proof.

4. Proof of the main results

In this final section we make use of Theorems 1.5 and 1.6 in order to prove Theorem 1.1and Corollary 1.2.

Proof of Theorem 1.1.LetN ∈ N, let Z be a closed complex hypersurface inCN+1 suchthatZ ∩BN+1 6= ∅, letK ⊂ Z ∩BN+1 be a connected compact subset, and letǫ > 0.

Choose0 < λ0 < 1 such thatK ⊂ λ0BN+1, let F be a tidy collection of tangent ballsin BN+1 given by Theorem 1.5, satisfying|F| ∩ λ0BN+1 = ∅.

Apply Theorem 1.6 toF, Z, λ0, andǫ, and consider the arising Runge pseudoconvexdomainΩ of CN+1, which containsλ0BN+1, and biholomorphismΨ: Ω → BN+1. LetD ⊂ Ω∩Z be the connected component containingK, which ensures (i), and consider theproper holomorphic embeddingψ := Ψ|D : D → BN+1. Item (ii) in Theorem 1.1 followsstraightforwardly. To get (iii), observe that the completeness ofψ is a direct consequenceof the choice ofF (see the last sentence in Theorem 1.5), and take into accountTheorem1.6-(i) and thatK ⊂ λ0BN+1.

Before proving Corollary 1.2, recall that every open Riemann surfaceR is Stein (seeBehnke-Stein [7]), whereas a domainD ⊂ R is a Runge domain inR if and only if R \Dcontains no relatively compact connected components; in particular a domainD ⊂ C isRunge if and only if it is simply connected.

Proof of Corollary 1.2.The first part of the corollary trivially follows from Theorem 1.1.

For the second part, recall first that there are properly embedded complex curves inC2

with arbitrary topology (see Alarcon and Lopez [5]; the case of finite topology, which isrequired in our proof, is due toCerne and Forstneric [8]). LetZ be a connected finitetopology properly embedded complex curve inC

2 and assume, up to applying a homothetictransformation toZ if necessary, that all the topology ofZ is contained in12B2. This meansthatZ intersects the boundary12S

3 of 12B2 transversely and so thatZ \ 1

2B2 consists offinitely many open annuliA1, . . . , Am (herem denotes the number of topological ends ofZ) with pairwise disjoint closures, properly embedded inC

2 \ 12B2, such that the boundary

bAi of Ai in Z is a smooth Jordan curve in12S3 for all i = 1, . . . ,m. It follows that

Z ∩ 12B2 is homeomorphic toZ andK := Z ∩ 1

2B2 is a Runge compact connected subsetof Z. Thus, the first part of the corollary provides a Runge domainD ⊂ Z, containingK, and a complete proper holomorphic embeddingψ : D → B2. SinceZ \ D has norelatively compact connected components inZ andD containsK = Z ∩ 1

2B2, it followsthatD \ 1

2B2 consists of finitely many open annuliA′1, . . . , A

′m such thatA′

i ⊂ Ai andbAi = (bA′

i) ∩ 12S

3 for all i = 1, . . . ,m. This guarantees thatD is homeomorphic toZ ∩ 1

2B2, hence toZ, which concludes the proof.

We finish the paper with the following

Question 4.1. We have proved that the unit discD ⊂ C properly embeds into the unit ballB2 ⊂ C

2 as a complete complex curve, so it is natural to ask whether, givenN ≥ 2, thereexists a complete proper holomorphic embeddingBN → BN+1 (cf. [11]). A less difficult


question would be whether there is a complete closed complexhypersurface inBN+1 whichis homeomorphic toBN .

Acknowledgements.The authors wish to thank Jean-Pierre Demailly for suggesting theuse of tangent balls instead of faces of convex polytopes. They are also grateful to thereferee for very useful comments that led to improvement of the original manuscript.

A. Alarcon is supported by the Ramon y Cajal program of the Spanish Ministry ofEconomy and Competitiveness.

A. Alarcon and F. J. Lopez are partially supported by the MINECO/FEDER grant no.MTM2014-52368-P, Spain.

J. Globevnik is partially supported by the research programP1-0291 and the grant J1-5432 from ARRS, Republic of Slovenia.

References

[1] Alarcon, A.; Forstneric, F.: Every bordered Riemann surface is a complete proper curve in a ball. Math.Ann. 357(2013), 1049–1070

[2] Alarcon, A.; Forstneric, F.: Null curves and directedimmersions of open Riemann surfaces. Invent. Math.196(2014), 733–771

[3] Alarcon, A.; Forstneric, F.: The Calabi-Yau problem,null curves, and Bryant surfaces. Math. Ann.363(2015), 913–951

[4] Alarcon, A.; Lopez, F. J.: Null curves inC3 and Calabi-Yau conjectures. Math. Ann.355(2013), 429–455[5] Alarcon, A.; Lopez, F. J.: Proper holomorphic embeddings of Riemann surfaces with arbitrary topology

intoC2. J. Geom. Anal.23 (2013), 1794–1805

[6] Alarcon, A.; Lopez, F. J.: Complete bounded embedded complex curves inC2. J. Eur. Math. Soc. (JEMS)18 (2016), 1675–1705

[7] Behnke, H.; Stein, K.: Entwicklung analytischer Funktionen auf Riemannschen Flchen. Math. Ann.120(1949), 430–461

[8] Cerne, M.; Forstneric, F.: Embedding some bordered Riemann surfaces in the affine plane. Math. Res.Lett. 9 (2002), 683–696

[9] Drinovec Drnovsek, B.: Complete proper holomorphic embeddings of strictly pseudoconvex domains intoballs. J. Math. Anal. Appl.431(2015), 705–713

[10] Forstneric, F.: Stein Manifolds and Holomorphic Mappings (The Homotopy Principle in ComplexAnalysis). Ergebnisse der Mathematik und ihrer Grenzgebiete, 3. Folge, 56. Springer-Verlag, Berlin-Heidelberg (2011)

[11] Globevnik, J.: A complete complex hypersurface in the ball of CN . Ann. of Math. (2)182(2015), 1067–1091

[12] Globevnik, J.: Holomorphic functions unbounded on curves of finite length. Math. Ann.364 (2016),1343–1359

[13] Jones, P.W.: A complete bounded complex submanifold ofC3. Proc. Amer. Math. Soc.76 (1979), 305–

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[15] Yang, P.: Curvatures of complex submanifolds ofCn. J. Differential Geom.12 (1977), 499–511

Antonio Alarcon

Departamento de Geometrıa y Topologıa e Instituto de Matematicas (IEMath-GR),Universidad de Granada, Campus de Fuentenueva s/n, E–18071Granada, Spain.


e-mail:[email protected]

Josip Globevnik

Department of Mathematics, University of Ljubljana, and Institute of Mathematics, Physicsand Mechanics, Jadranska 19, SI–1000 Ljubljana, Slovenia.


Francisco J. Lopez

Departamento de Geometrıa y Topologıa e Instituto de Matematicas (IEMath-GR),Universidad de Granada, Campus de Fuentenueva s/n, E–18071Granada, Spain.


Documents

arXiv:1509.02283v2 [math.CV] 30 Aug 2016 · arXiv:1509.02283v2 [math.CV] 30 Aug 2016 A construction of complete complex hypersurfaces in the ball with control on the topology A. Alarcon,