OOKA PROPERTIES OF SOME HYPERSURFACECOMPLEMENTS
ALEXANDER HANYSZ
Abstract.
Oka manifolds can be viewed as the “opposite” ofKobayashi hyperbolic manifolds. Kobayashi asked whether thecomplement in projective space of a generic hypersurface of suffi-ciently high degree is hyperbolic. Therefore it is natural to inves-tigate Oka properties of complements of low degree hypersurfaces.We determine which complements of hyperplane arrangements inprojective space are Oka. A related question is which hypersur-faces in affine space have Oka complements. We give some resultsfor graphs of meromorphic functions. Introduction
A complex manifold X is hyperbolic (in the sense of Kobayashi) if,informally speaking, there are “few” maps C → X , and Oka if thereare “many” maps C → X , in a sense to be made precise in Section 2below. Hyperbolic manifolds have been extensively studied since thelate 1960s. Oka theory is a more recent development, motivated byGromov’s paper [5] of 1989; the definition of an Oka manifold was onlypublished in 2009, by Forstneriˇc [2].Many interesting examples of hyperbolic manifolds arise from com-plements of projective hypersurfaces. In particular, Kobayashi asked[8, problem 3 on page 132] whether the complement in P n of a generichypersurface of sufficiently high degree should be hyperbolic. This hasbeen proved for n = 2 by Siu and Yeung [11], but is still an open prob-lem in higher dimensions. The degenerate case of the complement ofa finite collection of hyperplanes is well understood. In particular, thecomplement in P n of at least 2 n + 1 hyperplanes in general positionis hyperbolic, and the complement of a collection of 2 n or fewer hy-perplanes is never hyperbolic. For hyperplanes not in general position,some necessary conditions for hyperbolicity are known. See Kobaya-shi’s monograph [9, Section 3.10] for details. Date : 28th November 2011; revised 19th April 2012.2010
Mathematics Subject Classification.
Primary 32Q28. Secondary 14J70,32H02, 32H04, 32Q45, 32Q55, 52C35.
Key words and phrases.
Stein manifold, Oka manifold, Oka map, subellipticmanifold, spray, hyperbolic manifold, hypersurface complement, hyperplane ar-rangement, meromorphic function. a r X i v : . [ m a t h . C V ] A p r ALEXANDER HANYSZ
Since the Oka property can be viewed as a sort of anti-hyperbolicity,it makes sense to ask which hypersurfaces have Oka complements. InSection 3 of this paper we give a complete answer to this questionfor complements of hyperplane arrangements in projective space. Themain result of this section, Theorem 3.1, states that the complementof N hyperplanes in P n is Oka if and only if the hyperplanes are ingeneral position and N ≤ n + 1. We also investigate the weaker Oka-type properties of dominability by C n and C -connectedness: in thiscontext we find that a non-Oka complement also fails to possess theseweaker properties.In Section 4 we give a sufficient condition for the complement of thegraph of a meromorphic function to be Oka. Our Theorem 4.6 statesthat if m : X → P can be written in the form m = f + 1 /g forholomorphic functions f and g , then the graph complement is Oka ifand only if X is Oka. This is motivated by the open problem of whetherthe complement in P of a smooth cubic curve is Oka: given a cubiccurve, there is an associated meromorphic function and a branchedcovering map from the graph complement of that function to the cubiccomplement. For details, see Buzzard and Lu [1, page 644–645]. Wealso explore the question of when the decomposition m = f +1 /g exists(Lemma 4.2). For meromorphic functions that cannot be written in thisform, further work is required to understand the Oka properties of thegraph complements. Acknowledgements.
I thank Finnur L´arusson for many helpful discus-sions during the preparation of this paper, and Franc Forstneriˇc formaking available preliminary drafts of his book [3]. I am also gratefulto the anonymous referee for constructive suggestions, in particular theproof of the Oka property in Example 4.5 for the case ν = 1.2. Oka manifolds and hyperbolic manifolds
In this section we recall the definitions of Oka manifolds and hyper-bolic manifolds, and collect some results that will be used later. Forbackground, motivation and further details of Oka theory, see the sur-vey article [4] of Forstneriˇc and L´arusson and the recently publishedbook [3] of Forstneriˇc. For more on hyperbolicity, see the monograph[9] of Kobayashi.
Definition 2.1.
A complex manifold X is an Oka manifold if everyholomorphic map K → X from (a neighbourhood of) a convex compactsubset K of C n can be approximated uniformly on K by holomorphicmaps C n → X .This defining property is also referred to as the convex approximationproperty (CAP) . Definition 2.2.
The
Kobayashi pseudo-distance on a complex mani-fold X is the largest pseudo-distance such that every holomorphic map KA PROPERTIES OF SOME HYPERSURFACE COMPLEMENTS 3 D → X is distance-decreasing, where D denotes the complex unit discwith the Poincar´e metric. We say that X is hyperbolic if the Kobayashipseudo-distance is a distance.If X is Oka then the Kobayashi pseudo-distance on X is identicallyzero; thus Oka manifolds can be viewed as “anti-hyperbolic”. The mostfundamental examples of Oka manifolds are complex Lie groups andtheir homogeneous spaces; in particular, P n and C n are Oka. Boundeddomains in C n are always hyperbolic. If X is a Riemann surface, then X is Oka if and only if it is one of C , C ∗ (the punctured plane), P ora torus; otherwise it is hyperbolic.Every Oka manifold X of dimension n is dominable by C n , in thesense that there exists a holomorphic map C n → X that has rank n atsome point of C n .Oka manifolds are also C -connected: every pair of points can bejoined by an entire curve, i.e. for any pair of points there exists a holo-morphic map from C into the manifold whose image contains bothpoints. This property is mentioned by Gromov [5, 3.4(B)], and fol-lows easily from the “basic Oka property” described in [4, page 16].(The definition of C -connected is not standardised: the term can alsorefer to the weaker property that every pair of points can be joinedby a finite chain of entire curves, by analogy with the case of rationalconnectedness.)In general it is difficult to verify the condition of Definition 2.1 di-rectly. Instead, sprays (in the sense of Gromov: see below) and fibrebundles are of fundamental importance. If π : X → Y is a holomorphicfibre bundle with Oka fibres, then X is Oka if and only if Y is Oka. (Infact there is a far more general notion of an Oka map which preservesthe Oka property, but this will not be needed here.) In particular,products of Oka manifolds are Oka, and a manifold is Oka if it has acovering space that is Oka.
Definition 2.3. A spray over a complex manifold X consists of aholomorphic vector bundle π : E → X and a holomorphic map s : E → X such that s (0 x ) = x for all x ∈ X . We say that s is dominating at the point x ∈ X if the differential ds x maps the vertical subspace E x of T x E surjectively onto T x X . A family of sprays ( E j , π j , s j ), j =1 , . . . , m , is dominating at x if( ds ) x ( E ,x ) + · · · + ( ds m ) x ( E m,x ) = T x X. The manifold X is elliptic if there exists a spray that is dominating atevery point of X , and weakly subelliptic if for every compact set K ⊂ X there exists a finite family of sprays over X that is dominating at everypoint of K . ALEXANDER HANYSZ
The concept of a spray can be viewed as a generalisation of the expo-nential map for a complex Lie group: for example, see [4, Examples 5.3]or [3, Proposition 5.5.1].Every elliptic or weakly subelliptic manifold is Oka.The following property is equivalent to the CAP.
Definition 2.4.
A complex manifold X satisfies the convex interpola-tion property (CIP) if whenever T is a contractible subvariety of C m for some m , every holomorphic map T → X extends to a holomorphicmap C m → X .(Equivalently, we could take T to be any subvariety of C m that isbiholomorphic to a convex domain in C n ; hence the use of the word convex .)A useful tool for proving hyperbolicity is Borel’s generalisation ofPicard’s little theorem. Kobayashi gives three equivalent formulations(see [9, Theorem 3.10.2 on page 134]), of which we only need the fol-lowing. Theorem 2.5 (Picard–Borel) . Let g , . . . , g N be nowhere vanishingholomorphic functions on C , and suppose g + · · · + g N = 0 . Partition the index set { , , . . . , N } into subsets, putting two indices j and k into the same subset if and only if g j /g k is constant. Then foreach subset J , (cid:88) j ∈ J g j = 0 . Remark . Since the g j are nowhere vanishing, it follows that eachsubset must have size greater than 1. In particular, if N = 2 then thepartition has only one part, hence g , g and g are constant multiplesof each other. 3. Hyperplane complements
Let F , . . . , F N be nonzero homogeneous linear forms in n + 1 vari-ables. We say that the hyperplanes in P n defined by the equations F j =0, j = 1 , . . . , N , are in general position if every subset of { F , . . . , F N } of size at most n + 1 is linearly independent. If N ≤ n + 1, then aset of N hyperplanes is in general position if and only if coordinatescan be chosen on P n so that the given hyperplanes are the coordinatehyperplanes x j = 0, j = 0 , . . . , N − Theorem 3.1.
Let H , . . . , H N be distinct hyperplanes in P n . Thenthe complement X = P n \ (cid:83) Nj =1 H j is Oka if and only if the hyperplanesare in general position and N ≤ n + 1 . Furthermore, if X is not Okathen it is not dominable by C n and not C -connected. KA PROPERTIES OF SOME HYPERSURFACE COMPLEMENTS 5
Before proving this, we state and prove a sharper form of Theo-rem 3.10.15 of Kobayashi’s book [9, page 142]. To state the theorem,it is convenient to introduce the following terminology.
Definition 3.2.
Let H , . . . , H k be distinct hyperplanes in P n definedby linear forms F , . . . , F k , and suppose the forms satisfy a minimallinear relation of the form c F + · · · + c k F k = 0where c j (cid:54) = 0 for all j . (By “minimal” we mean that (cid:80) j ∈ J c j F j (cid:54) = 0for every proper nonempty subset J of { , . . . , k } .) Then the diagonalhyperplanes of the linear relation are the hyperplanes defined by thelinear forms (cid:80) j ∈ J c j F j where J is a subset of { , . . . , k } with 2 ≤ | J | ≤ k −
2. (If k ≤
3, there are no diagonal hyperplanes.) The associatedsubspaces of { H , . . . , H k } are the linear subspaces of P n which contain (cid:84) kj =1 H j with codimension 1. (If (cid:84) H j = ∅ , the associated subspacesare exactly the points of P n .) Remark . If p ∈ P n \ (cid:84) H j , then p is contained in exactly oneassociated subspace for each minimal linear relation. Example 3.4. On P consider the linear forms F = x ,F = x ,F = x − x ,F = x − x . If we consider x = 0 to be the line at infinity, then the lines F j = 0, j = 1 , , ,
4, are the sides of a “unit square” in the affine plane. Thelinear relation F − F − F + F = 0 has three diagonal lines (notingthat J = { , } and J = { , } give the same line, and so on). Theyare the two diagonals of the square ( x = x and x + x = x ), andthe line at infinity ( x = 0). Example 3.5.
Let P be any point of P and let F , F and F belinear forms defining three distinct lines through P . Then there existsa linear relation among F , F and F , and the associated subspacesare the lines through P . Theorem 3.6.
Let H , . . . , H N be distinct hyperplanes in P n definedby linear forms F , . . . , F N , and let f : C → P n \ (cid:83) H j be a holomorphicmap. Suppose that F , . . . , F N are linearly dependent. Then for eachsubset of F , . . . , F N satisfying a minimal linear relation, there is adiagonal hyperplane or an associated subspace containing the imageof f .Remark . In the case where (cid:84) H j = ∅ , the associated subspacesare points, so the conclusion is that either f is constant or the imageis contained in a diagonal hyperplane. ALEXANDER HANYSZ
Proof of Theorem 3.6.
First we note that f can be lifted to a holomor-phic map ˜ f : C → C n +1 \ { } . To see this, observe that the quotientmap C n +1 \ { } → P n can be regarded as the universal line bundleover P n with the zero section removed. Thus lifting f is equivalent tofinding a nowhere vanishing section of the pullback by f of the uni-versal bundle. But the vanishing of the cohomology group H ( C , O ∗ )guarantees that line bundles over C are trivial, and therefore a nowherevanishing section always exists.By reordering and rescaling the defining forms, we can put a minimallinear relation in the form F + · · · + F k = 0 . Define entire functions h , . . . , h k by h j = F j ◦ ˜ f . Then the h j satisfythe hypotheses of the Picard–Borel theorem (Theorem 2.5): each h j vanishes nowhere (because the image of f misses all the hyperplanes),and the h j sum to the zero function (because of the linear relationbetween the F j ). Theorem 2.5 tells us that there is a subset J ⊂{ , . . . , k } with (cid:88) j ∈ J h j = 0and such that all the ratios h µ /h ν are constant for µ, ν ∈ J . There aretwo possibilities.First, if J is a proper subset of { , . . . , k } then J must have size atleast 2 and at most k −
2. (If J either consisted of or omitted only asingleton j , then the corresponding h j would be identically zero.) Inthis case the linear form F = (cid:88) j ∈ J F j defines a diagonal hyperplane in P n . (The minimality of the linear rela-tion implies that F is nonzero.) The image of f lies in this hyperplane.The second case is that J = { , . . . , k } . Then there exist nonzeroconstants c . . . , c k − such that h j = c j h k for j = 1 , . . . , k −
1. This means that the image of ˜ f lies in each of thehyperplanes F j = c j F k . Let A and B be the linear subspaces of C n +1 given by A = k − (cid:92) j =1 { F j − c j F k = 0 } ,B = k (cid:92) j =1 { F j = 0 } . Clearly B ⊂ A . It remains to show that B has codimension at most1 in A . Equivalently, we wish to show that given x, y ∈ A \ B , some KA PROPERTIES OF SOME HYPERSURFACE COMPLEMENTS 7 nontrivial linear combination of x and y lies in B . The numbers α = F k ( x ) and β = F k ( y ) are both nonzero. Then F k ( βx − αy ) = 0, so F j ( βx − αy ) = 0 for all j , hence βx − αy ∈ B . (cid:3) Remark . In the above proof, the subspaces A and B can neverbe equal (because the image of f misses all of the H j ). A naive di-mension argument might suggest that A and B have the same dimen-sion. However, the fact that the h j sum to zero implies the relation c + · · · + c k − = −
1, so the forms F j − c j F k are linearly dependent:their sum is zero. Remark . In the case J = { , . . . , k } , Kobayashi states that f isconstant ([9, page 142, proof of Theorem 3.10.15, last paragraph]).In fact there exist nonconstant maps whose images lie in associatedsubspaces, but this does not invalidate the conclusion of Kobayashi’sTheorem 3.10.15. For an example of such a map, take n = 4 with linearforms F = x − x , F = x − x , F = x − x , F = x − x , F = x so that F + F + F + F = 0. Let f : C → X be a function that liftsto ˜ f ( t ) = ( t, t + 1 , t + 2 , t + 3 , . Then h , . . . , h are the constant functions − , , − , f is not constant. Corollary 3.10.
Let H , . . . , H N be distinct hyperplanes in P n , not ingeneral position. Let p be any point of X = P n \ (cid:83) H j . Then there isa finite collection of proper subspaces of T p X with the property that forevery map f : C → X with f (0) = p , the derivative df (0) lies in oneof those subspaces.Proof. By Theorem 3.6, the image of f is restricted to a proper linearsubspace of X . Thus there is a corresponding subspace of T p X con-taining df (0). We just need to verify that there are only finitely manypossible subspaces. But each point of X is contained in exactly oneassociated subspace for each minimal linear relation among the F j , andthere are only finitely many diagonal hyperplanes. (cid:3) Corollary 3.11.
If the distinct hyperplanes H , . . . , H N in P n are notin general position, then P n \ (cid:83) H j is not dominable by C n .Proof. Let f be a map from C n into P n \ (cid:83) H j , with f (0 , . . . ,
0) = p .The image of df (0) is spanned by the n vectors d ( t (cid:55)→ f ( te j )) | t =0 ( j = 1 , . . . , n )where { e , . . . , e n } is a basis for C n . If df (0) is surjective, those vec-tors are linearly independent, so there will be no finite set of propersubspaces containing d ( t (cid:55)→ f ( tv )) | t =0 ALEXANDER HANYSZ for all v ∈ C n , contradicting the previous corollary. (cid:3) Proof of Theorem 3.1.
Write X for the space P n \ (cid:83) Nj =1 H j . Case 1: hyperplanes in general position and
N > n + 1 . In thiscase, Kobayashi’s Theorem 3.6.10 [9, page 136] tells us that the imageof a nonconstant holomorphic map C → X must lie in one of a finitecollection of hyperplanes. Therefore X is not dominable by C n . Also, X is not C -connected: distinct points outside the finite collection ofhyperplanes cannot be joined by an entire curve. Case 2: hyperplanes in general position and N ≤ n + 1 . If N = 0,then X = P n is Oka. For N >
0, the fact that the hyperplanes arein general position means that we can choose coordinates so that H j is the hyperplane x j − = 0 for j = 1 , . . . , N . Then we see that X ∼ = C ∗ × · · · × C ∗ × C × · · · × C with N − C ∗ and n + 1 − N factors C . This is a product of Oka manifolds, hence Oka. Case 3: hyperplanes not in general position.
The fact that X isnot dominable, and therefore not Oka, is just Corollary 3.11 above.To see that X is not C -connected, choose any point p ∈ X . ThenTheorem 3.6 gives a finite collection of hyperplanes containing everyentire curve through p . If q is a point of X outside that finite collection,then p and q cannot be joined by an entire curve. (cid:3) Remark . In fact if X is not Oka, then it does not satisfy the weakerversion of C -connectedness referred to above: there exist pairs of pointsthat cannot be connected even by a finite chain of entire curves. Incase 1 of the above proof this is immediate. For case 3, some furtherwork is needed, since the finite collection of hyperplanes referred tocan vary with the choice of the point p . The key ingredients are thefact that there are only finitely many diagonal hyperplanes in total,and that given an associated subspace A and a diagonal hyperplane D of the same configuration, either A ⊂ D or A ∩ D ⊂ (cid:83) H j . In otherwords, points inside a diagonal hyperplane cannot be joined to pointsoutside via associated subspaces.4. Complements of graphs of meromorphic functions
Buzzard and Lu [1, Proposition 5.1] showed that the complement in P of a smooth cubic curve is dominable by C . Their method of proofwas to construct a meromorphic function associated with the cubic,and a branched covering map from the complement of the graph of thatfunction to the complement of the cubic, and then show that the graphcomplement is dominable. We will show that the graph complementis in fact Oka; this result can be generalised to meromorphic functionson Oka manifolds other than C , subject to an additional hypothesis.(Note that our result is not enough to settle the question of whether thecubic complement is Oka. We know that the Oka property passes down KA PROPERTIES OF SOME HYPERSURFACE COMPLEMENTS 9 through unbranched covering maps, but no similar result is known forbranched coverings.)For a holomorphic map m : X → P on a complex manifold X , thatis to say either a meromorphic function with no indeterminacy or elsethe constant function ∞ , we will write Γ m for the affine graphΓ m = { ( x, m ( x )) ∈ X × C : m ( x ) (cid:54) = ∞} . This is a closed subset of X × C , so the set ( X × C ) \ Γ m is a complexmanifold. (If m is identically ∞ , then Γ m is the empty set.)Buzzard and Lu’s result relies on the fact that meromorphic functionson C can be written in the following form. Lemma 4.1.
For every meromorphic function m : C → P there existholomorphic functions f, g : C → C such that m = f + 1 g . In other words, the projection map from C \ Γ m onto the first co-ordinate has a holomorphic section given by x (cid:55)→ ( x, f ( x )).The result follows from the classical theorems of Mittag-Leffler andWeierstrass; see [1, page 645] for details.The analogous result in higher dimensions is not true. We have thefollowing topological criterion. Lemma 4.2.
Let m be a holomorphic map from C n to P , and write m = h/k for holomorphic functions h, k : C n → C with no commonzeros. Then the following statements are equivalent. (1) There exist holomorphic functions f, g : C n → C such that m = f + 1 g . (2) The function h has a logarithm on the zero set Z ( k ) of k . (3) The function h has a logarithm on a neighbourhood of Z ( k ) .Proof. (1 ⇒ g = m − f = h − kfk has no zeros. Therefore h − kf is a nowhere vanishing entire function,so h − kf = e µ for some holomorphic µ : C n → C . Then µ | Z ( k ) is the desired logarithmof h .(2 ⇒ λ on Z ( k ) suchthat e λ = h | Z ( k ) . We wish to find a neighbourhood U of Z ( k ), smallenough that h vanishes nowhere on U , such that the inclusion map Z ( k ) (cid:44) → U induces an epimorphism of fundamental groups. Givensuch a neighbourhood, we have the following situation. C exp (cid:15) (cid:15) Z ( k ) (cid:31) (cid:127) (cid:47) (cid:47) λ (cid:53) (cid:53) U h (cid:47) (cid:47) λ (cid:48) (cid:59) (cid:59) C ∗ Then the existence of λ , together with the epimorphism π ( Z ( k )) → π ( U ), tells us that h | U satisfies the lifting criterion for the coveringmap exp : C ∗ → C . Therefore there exists a holomorphic function λ (cid:48) such that e λ (cid:48) = h on U .To find a suitable neighbourhood U , we start by realising C n as asimplicial complex with Z ( k ) as a subcomplex. (The existence of sucha simplicial complex is guaranteed by standard results on the topologyof subanalytic varieties: see for example [10, Theorem 1].) Then wecan find a basis of neighbourhoods of Z ( k ) such that Z ( k ) is a strongdeformation retract of each basis set (this is a general fact about CW-complexes: see [6, Prop. A.5, p. 523]). Finally, we choose a basis set U small enough that h does not vanish on U .(3 ⇒ λ : Z ( k ) → C is a logarithm for h . We wish to find a suitable holomorphicfunction µ : C n → C which extends λ . Then we can define f = h − e µ k and g = ke µ . In order for such f to be a well defined holomorphic function, we requirethat h − e µ should vanish on Z ( k ) to order at least the order of vanishingof k , in other words that the divisors should satisfy( h − e µ ) ≥ ( k ) . By hypothesis, we in fact have a neighbourhood U of Z ( k ) and a log-arithm λ (cid:48) : U → C for h on U . Then the above condition is equivalentto requiring that µ agrees with λ (cid:48) on Z ( k ) up to the order of vanishingof k .We write O for the sheaf of germs of holomorphic functions on C n .Consider the exact sequence of sheaves0 → k O → O → O /k O → . The sheaf k O is coherent, so by Cartan’s Theorem B, H ( C n , k O ) = 0,and therefore the map O ( C n ) → ( O /k O )( C n ) is surjective. Notingthat the stalk of O /k O at any point outside Z ( k ) is zero, we see thatthe function λ (cid:48) represents an element of ( O /k O )( C n ). Then we canchoose µ to be any preimage of that element. (cid:3) KA PROPERTIES OF SOME HYPERSURFACE COMPLEMENTS 11
Remark . When n = 1 in the above lemma, the zero set of k is dis-crete, and so a logarithm of h always exists on Z ( k ). Thus Lemma 4.1is a special case of Lemma 4.2. Remark . The only properties of C n used in the above proof are thatit is Stein and simply connected and that all meromorphic functions on C n can be written as a quotient. Thus we can generalise the result: if X is a simply connected Stein manifold, h, k ∈ O ( X ) have no commonzeros and m = h/k , then the three statements given in the lemma areequivalent. Example 4.5.
For positive integers ν , the functions m ν : C → P given by m ν ( x, y ) = xxy ν − f + 1 /g . At present it is not knownwhether any of the spaces C \ Γ m ν for ν ≥ ν = 1, the Oka property for C \ Γ m follows from work of Ivarssonand Kutzschebauch [7, Lemmas 5.2 and 5.3]. Specifically, let p bethe polynomial p ( x, y, z ) = xyz − x − z . Then Γ m is the level set p − (0), and the complement is the union of all the other level sets.The complement is isomorphic to the product of C ∗ with the level set p − (1) via the map C ∗ × p − (1) → C \ Γ m given by( λ, x, y, z ) (cid:55)→ ( λx, λ − y, λz ) . Now p − (1) is smooth, and by the results of Ivarsson and Kutzschebauch,its tangent bundle is spanned by finitely many complete holomorphicvector fields. This implies that the set is Oka (see for example [3,Example 5.5.13(B)]), so C ∗ × p − (1) is Oka.With these considerations in mind, we are ready to state the mainresult of this section. Theorem 4.6.
Let X be a complex manifold, and let m : X → P be aholomorphic map. Suppose m can be written in the form m = f + 1 /g for holomorphic functions f and g . Then ( X × C ) \ Γ m is Oka if andonly if X is Oka.Remark . The existence of the decomposition m = f + 1 /g is a geo-metric condition that is of some independent interest: it is equivalentto the condition that the projection map from ( X × C ) \ Γ m onto thefirst factor has a holomorphic section. The projection map is an ellipticsubmersion in the sense defined in [4, page 24]. (It is easy to see that itis a stratified elliptic submersion, as defined in [4, page 25]. For a sketchof why it is an (unstratified) elliptic submersion, see Remark 4.12 be-low.) However, unless either m has no poles or m = ∞ , the projectionis not an Oka map, because it is not a topological fibration. In the case where X is Stein, it follows from [4, Theorem 5.4 (iii)]that the existence of a continuous section of the elliptic submersionimplies the existence of a holomorphic section. For general X , onemight expect that this ellipticity property could be applied to yield asimpler proof than the one presented below. So far, such a proof hasbeen elusive.We will first prove Theorem 4.6 for the special case X = C n , andthen show how the convex interpolation property (Definition 2.4) forgeneral X reduces to the special case. The proof for X = C n involvesa variation of Gromov’s technique of localisation of algebraic subellip-ticity (see [5, Lemma 3.5B] and [3, Proposition 6.4.2]). This relies onthe following lemma. Lemma 4.8.
Let g : C n → C be a holomorphic function, not identicallyzero, and suppose x ∈ C n satisfies g ( x ) = 0 . Then for all s ∈ C n , lim x → x g ( x ) (cid:54) =0 (cid:18) g ( x ) − g ( x + g ( x ) s ) (cid:19) = g (cid:48) ( x )( s ) . Proof.
First, in order for the limit to make sense, we need to verify that x has a neighbourhood on which g ( x + g ( x ) s ) vanishes only when g ( x )vanishes. We use the approximation g ( x + h ) = g ( x ) + g (cid:48) ( x )( h ) + O ( | h | ) . (1)With h = g ( x ) s , this gives g ( x + g ( x ) s ) = g ( x ) + g (cid:48) ( x )( g ( x ) s ) + O ( | g ( x ) | ) . When x (cid:54) = x and g ( x ) (cid:54) = 0, if x is close to x , then the second andthird terms of the right hand side are much smaller than the first, so g ( x + g ( x ) s ) (cid:54) = 0, as required.Now, using (1) again, we obtain1 g ( x ) − g ( x + h ) = g ( x + h ) − g ( x ) g ( x ) g ( x + h )= g (cid:48) ( x )( h ) + O ( | h | ) g ( x )( g ( x ) + g (cid:48) ( x )( h ) + O ( | h | )) . (In the event that g ( x + h ) vanishes, we interpret the fractions as mero-morphic functions.) Replacing h with g ( x ) s , and using the fact that g (cid:48) ( x ) is a linear map, gives1 g ( x ) − g ( x + g ( x ) s ) = g (cid:48) ( x )( g ( x ) s ) + O ( | g ( x ) | ) g ( x )( g ( x ) + g (cid:48) ( x )( g ( x ) s ) + O ( | g ( x ) | ))= g (cid:48) ( x )( s ) + O ( | g ( x ) | )1 + g ( x ) g (cid:48) ( x )( s ) + O ( | g ( x ) | ) . As x → x this expression tends to g (cid:48) ( x )( s ). (cid:3) KA PROPERTIES OF SOME HYPERSURFACE COMPLEMENTS 13
Remark . The exponent 2 in the lemma corresponds to a doublytwisted line bundle in the proof of Proposition 4.10 below. A singletwist would not be sufficient: for example, if we take n = 1 and g ( x ) = x then for s (cid:54) = 0 the expression 1 g ( x ) − g ( x + g ( x ) s ) does not have afinite limit as x → Proposition 4.10.
Let m : C n → P be a holomorphic map, andsuppose m can be written in the form m = f + 1 /g for holomorphicfunctions f and g . Then C n +1 \ Γ m is Oka.Proof. If g = 0, so that m = ∞ , then Γ m = ∅ , so C n +1 \ Γ m = C n +1 isOka. For the rest of the proof, assume that g (cid:54) = 0.We will write points of C n +1 as ( x, y ) or ( s, t ) where x, s ∈ C n and y, t ∈ C . Let X denote the complement of the graph of 1 /g ; i.e. X = { ( x, y ) ∈ C n +1 : g ( x ) y (cid:54) = 1 } . The map X → C n +1 \ Γ m given by ( x, y ) (cid:55)→ ( x, y + f ( x )) is a biholo-morphism. Hence it suffices to prove that X is Oka.We begin by describing a covering space Y of X . Then we shallexhibit sprays on trivial bundles over certain subsets of the coveringspace. Finally, these sprays will be extended to sprays on twistedbundles over Y , using the above lemma. (This is the localisation stepreferred to above.) This will be sufficient to establish that Y is weaklysubelliptic, hence Oka. Therefore X is Oka.The covering space Y is constructed as follows. Define an equivalencerelation ∼ on C n +1 × Z by(2) ( x, y, k ) ∼ ( x (cid:48) , y (cid:48) , k (cid:48) ) if x = x (cid:48) , g ( x ) (cid:54) = 0 and g ( x )( y − y (cid:48) ) = ( k − k (cid:48) )2 πi. Then Y is the quotient space ( C n +1 × Z ) / ∼ . From now on we willwrite [ x, y, k ] as shorthand for the equivalence class in Y of ( x, y, k ),and Y k for the k th “layer” { [ x, y, k ] : ( x, y ) ∈ C n +1 } .Note that Y can be described in concrete terms as a hypersurface in C n +2 : see Remark 4.11. The description of Y used here is chosen toemphasise the simple form of the sprays σ k described below.It is straightforward to verify that Y is a Hausdorff space. We canmap each Y k bijectively to C n +1 by sending [ x, y, k ] to ( x, y ). Thus Y has the structure of an ( n + 1)-dimensional complex manifold.By way of motivation for this construction, observe that if x ∈ C n with g ( x ) (cid:54) = 0, then the set { [ x, y, k ] ∈ Y : x = x } is a copy of C ,whereas if g ( x ) = 0, then { [ x, y, k ] ∈ Y : x = x } is a countable unionof disjoint copies of C . The covering map described below looks like anexponential map when g (cid:54) = 0, but the identity map when g = 0. Theconstruction involves a holomorphically varying family of holomorphicmaps which include both exponentials and the identity. We follow Buzzard and Lu [1, page 645] in defining a function φ on C by φ ( x, y ) = (cid:40) e xy − x if x (cid:54) = 0 y if x = 0(3) = y + xy x y
3! + · · · . From the series expansion we see that φ is holomorphic. Then we define π : Y → X by π [ x, y, k ] = ( x, − φ ( g ( x ) , y )) . If ( x, y ) is a point of X with g ( x ) = 0, then the fibre over ( x, y ) is theset π − ( x, y ) = { [ x, − y, k ] : k ∈ Z } . If g ( x ) (cid:54) = 0, then a set of unique representatives for π − ( x, y ) is givenby { [ x, log(1 − g ( x ) y ) g ( x ) , } for all possible branches of the logarithm. It follows that all fibres of π are isomorphic to Z . It can be verified that every point of X has aneighbourhood that is evenly covered by π , and so π is a covering map.For each layer Y k of Y there is a dominating spray σ k on the trivialbundle Y k × C n +1 , given by σ k ([ x, y, k ]; s, t ) = [ x + s, y + t, k ] . We wish to construct a bundle E k over Y and a spray ˜ σ k : E k → Y such that ˜ σ k agrees with σ k with respect to a trivialisation of E k | Y k .Since every compact subset of Y is covered by finitely many Y k , thiswill establish that Y is weakly subelliptic (Definition 2.3).To simplify the notation, we will only describe E and ˜ σ ; the con-struction for k (cid:54) = 0 is similar. Define open subsets U and U of Y by U = { [ x, y, k ] : k (cid:54) = 0 } ,U = { [ x, y, k ] : k = 0 } = Y . As each [ x, y, k ] is an equivalence class, these sets are not in fact disjoint.(This is the only part of the proof where the assumption g (cid:54) = 0 isrequired.) The intersection U ∩ U is the set of points [ x, y,
0] with g ( x ) (cid:54) = 0. The bundle E is described by local trivialisations E | U α → U α × C n +1 , α = 1 ,
2, with transition map θ : ( U ∩ U ) × C n +1 → ( U ∩ U ) × C n +1 given by(4) θ ([ x, y, s, t ) = ([ x, y, g ( x ) s, t ) . KA PROPERTIES OF SOME HYPERSURFACE COMPLEMENTS 15
Define ˜ σ by˜ σ | U ([ x, y, k ]; s, t ) = (cid:26) [ x + g ( x ) s, y − k πi/g ( x ) + t,
0] if g ( x ) (cid:54) = 0 , [ x, y − k πig (cid:48) ( x )( s ) + t, k ] if g ( x ) = 0 , ˜ σ | U ([ x, y, s, t ) = σ ([ x, y, s, t ) = [ x + s, y + t, . The fact that ˜ σ | U is continuous follows from Lemma 4.8 together withequation (2). It is easy to verify from equations (2) and (4) that ˜ σ | U and ˜ σ | U agree on U ∩ U . Thus ˜ σ is a well defined holomorphic mapfrom E to Y extending σ . Finally, ˜ σ ([ x, y, k ]; 0 ,
0) = [ x, y, k ], so ˜ σ is a spray. This completes the proof. (cid:3) Remark . The covering space Y from the above proof can be em-bedded into C n +2 by the map [ x, y, k ] (cid:55)→ ( x, − φ ( g ( x ) , y ) , g ( x ) y + 2 πik ).The image of this map is the set Z = { ( x, y, z ) ∈ C n × C × C : 1 − g ( x ) y = e z } and a covering map Z → X is given by ( x, y, z ) (cid:55)→ ( x, y ). Proof of Theorem 4.6.
We will write π and π for the projections ofthe complement ( X × C ) \ Γ m onto X and C respectively. The map σ : X → ( X × C ) \ Γ m given by σ ( x ) = ( x, f ( x )) is a holomorphicsection of π .First suppose ( X × C ) \ Γ m is Oka. The convex interpolation prop-erty for X can easily be verified as follows. Let φ : T → X be a holo-morphic map from a contractible subvariety T of some C n . Then bythe CIP of ( X × C ) \ Γ m , the composite map σ ◦ φ : T → ( X × C ) \ Γ m has a holomorphic extension ψ : C n → ( X × C ) \ Γ m . The composition π ◦ ψ is a map C n → X extending φ . Therefore X is Oka.Conversely, suppose X is an Oka manifold, and m : X → P aholomorphic map with m = f + 1 /g as in the statement of the theorem.We will verify the CIP for ( X × C ) \ Γ m .Suppose T is a contractible subvariety of C n for some n , and let φ : T → ( X × C ) \ Γ m be a holomorphic map. We want to find aholomorphic map µ : C n → ( X × C ) \ Γ m which extends φ .First of all we can use the CIP for X to extend the composite map π ◦ φ to a holomorphic map ψ : C n → X . This is indicated in thefollowing diagram. T (cid:127) (cid:95) ι (cid:15) (cid:15) φ (cid:47) (cid:47) ( X × C ) \ Γ mπ (cid:15) (cid:15) π (cid:47) (cid:47) CC n µ (cid:63) (cid:63) ψ (cid:47) (cid:47) X m (cid:47) (cid:47) P Now we have a holomorphic map m ◦ ψ : C n → P . In fact m ◦ ψ = f ◦ ψ + 1 / ( g ◦ ψ ), so we know by Proposition 4.10 that C n +1 \ Γ m ◦ ψ is Oka. We want to map T into C n +1 \ Γ m ◦ ψ , then use the CIP to extendthis map.Define α : T → C n +1 by α ( x ) = ( ι ( x ) , π ( φ ( x ))) . Since φ ( x ) is an element of ( X × C ) \ Γ m , it follows that π ( φ ( x )) isnever equal to m ( π ( φ ( x ))). By the definition of ψ , this means that π ( φ ( x )) (cid:54) = m ( ψ ( ι ( x ))) for all x ∈ T . Therefore the image of α iscontained in C n +1 \ Γ m ◦ ψ .The CIP for C n +1 \ Γ m ◦ ψ tells us that α extends to a map β : C n → C n +1 \ Γ m ◦ ψ , as in the following diagram. (The map π : C n +1 \ Γ m ◦ ψ → C is the restriction to C n +1 \ Γ m ◦ ψ of the projectionof C n × C onto the last coordinate. The use of π for two different pro-jection maps should not cause any confusion, as the domain is alwaysclear from the context.) C C n +1 \ Γ m ◦ ψπ (cid:111) (cid:111) T α (cid:111) (cid:111) (cid:127) (cid:95) ι (cid:15) (cid:15) φ (cid:47) (cid:47) ( X × C ) \ Γ mπ (cid:15) (cid:15) π (cid:47) (cid:47) CC nβ (cid:94) (cid:94) µ (cid:63) (cid:63) ψ (cid:47) (cid:47) X m (cid:47) (cid:47) P Finally, we can define µ : C n → ( X × C ) \ Γ m by µ ( x ) = ( ψ ( x ) , π ( β ( x ))) . Since π ( β ( x )) can never equal m ( ψ ( x )), we see that the image of µ isindeed contained in ( X × C ) \ Γ m . And from the definitions, the factthat β is an extension of α implies that µ is an extension of φ . (cid:3) Remark . As mentioned in Remark 4.7 above, the projection mapfrom ( X × C ) \ Γ m onto the first factor is an elliptic submersion, inthe case where m can be written as f + 1 /g . To prove this, it isnecessary to construct a dominating fibre spray ([4, page 24]). Thiscan be done using the function φ defined by equation (3) in the proofof Proposition 4.10 above. As previously, there is no loss of generalityin assuming f = 0. Denoting points of X × C by ( x, y ) with x ∈ X and y ∈ C , define a map s : (( X × C ) \ Γ m ) × C → ( X × C ) \ Γ m by s ( x, y, t ) = ( x, ye tg ( x ) − φ ( g ( x ) , t )) . The verification that the image of s is indeed contained in ( X × C ) \ Γ m ,and that s is a dominating fibre spray, is routine. References [1]
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