DSpace at VNU: Extensions of holomorphic maps through hypersurfaces and relations to the Hartogs extensions in infinite dimension

Moreover, it is proved that an increasing union of pseudoconvex domains containing no complex lines has the Hartogs extension property.. The above-mentioned theorem of Kwack plays an e

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Extensions in Infinite Dimension

Author(s): Do Duc Thai and Nguyen Thai Son

Source: Proceedings of the American Mathematical Society, Vol 128, No 3 (Mar., 2000), pp 745-754

Published by: American Mathematical Society

Stable URL: http://www.jstor.org/stable/119736

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AMERICAN MATHEMATICAL SOCIETY

Volume 128, Number 3, Pages 745 754

S 0002-9939(99)05033-9

Article electronically published on July 27, 1999

TO THE HARTOGS EXTENSIONS IN INFINITE DIMENSION

DO DUC THAI AND NGUYEN THAI SON (Communicated by Steven R Bell)

ABSTRACT A generalization of Kwack's theorem to the infinite dimensional

case is obtained We consider a holomorphic map f from Z \ H into Y, where

H is a hypersurface in a complex Banach manifold Z and Y is a hyperbolic

Banach space Under various assumptions on Z, H and Y we show that f can

be extended to a holomorphic map from Z into Y Moreover, it is proved that

an increasing union of pseudoconvex domains containing no complex lines has

the Hartogs extension property

INTRODUCTION

We discuss the problems of extending holomorphic maps through hypersurfaces and extending to the envelope of holomorphy, i.e the Hartogs holomorphic exten- sion These two problems are related

In [10], Kwack showed that if f is a holomorphic map from the punctured unit disc A* into a hyperbolic manifold X such that, for a suitable sequence of points

Zk E A* converging to the origin, f(zk) converges to a point po E X, then f extends

to a holomorphic map from the unit disc A into X

The above-mentioned theorem of Kwack plays an essential role and has strongly motivated the study of the extension problem of holomorphic maps Thus this was generalized in many directions In particular, it also generalized to the problem of extending holomorphic maps through hypersurfaces; see, for instance, the survey [11]

Therefore proving Kwack type theorems in the infinite dimensional case is really necessary and motivates the study of the extension problem of holomorphic maps

in infinite dimension

In Sect 1 we discuss the problem of determining when a holomorphic map can

be extended through hypersurfaces in complex Banach manifolds We prove the following theorems

Theorem 1.1 Let X be a hyperbolic Banach analytic space and f: Z \ H -* X a holomorphic map, where H is a hypersurface in a complex Banach manifold Z As- sume that for every z E H there exists a sequence {zn fE Z\H converging to z such

Received by the editors May 27, 1997 and, in revised form, April 20, 1998

1991 Mathematics Subject Classification Primary 32E05, 32H20; Secondary 32F05, 58B12 Supported by the State Program for Fundamental Research in Natural Science

(?1999 American Mathematical Society

745

that the sequence {f(zk)} converges to xz E X Then f extends holomorphically to

Z

Theorem 1.2 Let X be a hyperbolic Banach analytic space which is complete in the Cauchy mean and f: Z\H -* X a holomorphic map, where H is a hypersurface

in a complex Banach manifold Z Assume that for every branch Hog of H there exist z,, E Reg(HO) and a sequence {zc}?l C Z \ H converging to za, such that the sequence {f (Zn)} converges to pc, E X Then f extends holomorphically to Z

We would like to emphasize here that the local compactness of (finite dimen- sional) complex manifolds plays an essential role in proving the finite dimensional Kwack theorems Since complex Banach manifolds do not have the local com- pactness property, the technique for proving Kwack type theorems in the infinite dimensional case required substantial changes The proofs of the above-mentioned theorems are based on the maximum principle for plurisubharmonic functions

In Sect 2 we go further We would like to investigate deep interactions between the Brody hyperbolicity of Banach analytic spaces and the Hartogs extension prop- erty

First we recall the definition of a (finite-dimensional) complex space having the Hartogs extension property (HEP for short) which was introduced by Ivashkovicz [7] A complex space X is said to have HEP if every holomorphic map f from Q into X, where Q is a Riemann domain over Cn, can be extended holomorphically to the envelope of holomorphy Q of Q This definition can be generalized naturally to the case of Banach analytic spaces by replacing Cn by an arbitrary Banach space

B However, for technical reasons we need the statement that every pseudoconvex Riemann domain over a Banach space B is a domain of existence One only was proved in the case that the Banach space B has a Schauder basis (see Mujica [13, Theorem 54.12, p 390]) Thus in the infinite-dimension case we give the following definition

A Banach analytic space X is said to have the Hartogs extension property (HEP)

if every holomorphic map from a Riemann domain Q over a Banach space B with

a Schauder basis into X can be extended holomorphically to Q, the envelope of holomorphy of Q

We prove the following

Theorem 2.1 Let X be a Banach analytic space which is an increasing union of pseudoconvex domains Assume that X contains no complex lines Then X has the HEP

Finally in this note we frequently make use of the definition and properties of the Kobayashi pseudodistance in Banach analytic spaces as in [18], [19]

IN COMPLEX BANACH MANIFOLDS

First we prove the Kwack theorem [10] in infinite dimensions

1.1 Theorem Let X be a hyperbolic Banach space and f: Z\ H -* X a holomor- phic map, where Z is a Banach manifold and H is a hypersurface in Z Assume that for every z E H there exists a sequence {zn} C Z\ H converging to z such that the sequence {f(zn)} converges to xz E X Then f extends holomorphically to Z

Proof (i) First consider the case where Z = A and H = {O} Choose a pseudo- convex coordinate neighbourhood W of xo in X which is isomorphic to an analytic subset of an open ball in a Banach space B Let V = W/2

The problem is to show that, for a suitable positive number a, the small punc- tured disc {z E A: 0 < lzl < a} is mapped into W by f By taking a subsequence of {Zn } if necessary, we may assume that the sequence { IZ} is monotone decreasing Consider the set of integers n such that the image of the annulus IZn+ lI < IzI < IznI

by f is not entirely contained in V If this set of integers is finite, then f maps

a small punctured disc 0 < lzl < a* into V Assuming that this set of integers is infinite, we shall obtain a contradiction By taking a subsequence, we may assume also that, for every n, the image of the annulus 1zn+I1 < IZj < lznl by f is not entirely contained in V

For each n, put

rn = inf{r < lznl f(r < IzI < lzn ) C V},

Sn= sup{r > lznI f(IznI < IZI < r) C V}l

an= {z E A IZI rn},

AYn = {z E A IZ: lznl=

On = {z E A IZI = Sn}l

Since da* (an) +dA* (-Yn) +dA* (/3n) 0 and by the distance decreasing principle,

it follows that

dx f(n) + dx f(yn) + dx f((n) 0 as n x0

00

Put K= U f(yn U yn+i)

n=I

By the maximum principle, we have

KpSH(w) D U f(an U /n)

n>2 Hence U f(an U o3n) is relatively compact in V By the relative compactness of n>2

U f(an U On) and since dx f (an) 0 and dx f(on) -+ 0, without loss of generality

n>2

we may assume that {f(an)} -xi and {f(13n)} I- X2 By the definition of rn and

Sn it follows that xI, x2 E OV and hence xI, x2 7 xo Choose a continuous linear functional u on B such that u(xI), u(x2) 74 u(xo) = 0

Since f(Arnsn) C V C W, there exist in < rn < Sn < ?n such that f C

W, where Arnsn = {Z E C: rn < IZI < Sn} and = {Z z E C: i? n < IzI < s4n Consider the holomorphic function on = u o f I * Since {fun (3n) } - U(X2)

we have VE > 0,3N, Vn > N, VO: I mn(sne90) - U(X2)1 < 6 Applying the maximum

principle to the function z a-4 Un(Z) - U(X2) on the annulus {z E (C: in < IZI < Sn},

in particular the circle {z E C : IZI = lZnl} C r = {z E C : ? n < IZl < Sn},

it implies that oun(1zn1ei0) - u(x2)1 < c for every 0 Thus u(xo) = u(X2) This is impossible Hence f extends holomorphically to A

(ii) Assume H contains no singular points

Without loss of generality we may assume that the manifold Z has the form

U x A, where U is an open subset of a Banach space, and H = U x {0}

For each z E U consider fZ /A* - X given by

fZ(A) = f(z,A) for each A E A*

Since (z, 0) E H there exists {(Zn, An)} C U x A*, {(Zn, An)} - (z, 0) such that {f(zn, An)} I- x0 By applying the distance decreasing principle for the holomor- phic map f: U x A* - X, we have

dx(f (zI An), f (Zn An)) < duxA* ((Zn, An), (z, An)) = du(z, Zn) 0?

It follows that { f (z, An)} I- xO

By (i), fZ extends to a holomorphic map fz A - X Define the mapf

U x A X by f(z,A) = fz(A) for (z,A) E U x A Since X is hyperbolic, f is continuous

Indeed, let (z, 0) E H and { (zn, An) } C U x A be such that { (zn A)} An (z ?0)

Choose {A0} C A* such that {An} 0 We have

dx(f(znl An)i f(zi 0)) < dx(f(znl An)i f(zni An)) + dx(f(zni An)i f(zi An))

+ dx(f (z, An), f (z, 0))

= dx(fzn(An), fzn(An)) + dx(f(zn,An), f(z,A\n)) + dx(fz(An), fz(o))

<dix(An,An)+du(zn,z)+dA(An,O) > O as n +oo

Thus f is a holomorphic extension of f

(iii) Let ae E H be an arbitrary point of H By Ramis [14, p.14, Corollaire] there exists a neighbourhood U which is isomorphic to a neighbourhood V x Ae of 0 E B and a Weierstrass polynomial

P(x, A) = AP + ap-i(x)AP-1 + * + ao(x) such that Zero (P) = Hn u, for some decomposition B = E ED Ce of B

We have U n H = Zero (P) = (Zero(,9P) n Zero(P)) U (Zero(P) \ Zero( ))-

By (ii) f extends to the holomorphic map fi: U \ Zero (,9P) -+ X We now prove, for every zo E Zero( (9P), there exists {zn} C U \ Zero( (9P) converging to zo such

that {f1(zn)} converges to xzo E X

Indeed, by hypothesis there exists {Zn} C U \ H converging to zo such that

{f(zn)} converges to xzo E X

Choose {zn } C U \ (Zero(,9P) U H) such that du\H (zn, Z$) -O 0 Then the

sequence {zn} satisfies the above requirement

Repeating this process f can be extended holomorphically to U \ Zero (N& ),

1.2 Theorem Let X be a hyperbolic Banach analytic space which is complete

in the Cauchy mean, and f: Z \ H X a holomorphic map, where H is a hypersurface in a Banach analytic manifold Z Assume that for every branch H,, of

H there exist za, E Reg(H, ) and a sequence {z' }I'1 C Z\ H converging to zc such

that the sequence {f (Zc4)} converges to pc, E X Then f extends holomorphically to

Z

Proof (i) First of all we prove that X satisfies the weakly disk-convex condition, i.e if every sequence {S n} c Hol(A, X) converges in Hol(A, X) whenever the sequence {SWn I* } converges in Hol(A*, X), where Hol(X, Y) denotes the space of all holomorphic maps from X into Y with the open-compact topology

Indeed, let

{(Pn} C Hol(/\,X) be such that (On * fo in Hol(A*,X) Since

dx ((Qn (O), (Pm (O)) < dx (sOn (O), ,n (Z)) + dx (On (Z) , (Om (Z)) + dx (Pm (Z) , (Pm (?))

< 2dA (O, z) + dx (Pn (Z), (Pm (Z))

and (On fo in Hol(A*, X), it follows that {(On(?)} converges to xo E X as n 00 Define the map 7: A X by 71A* = fo and 7(0) = xo Since X is hyperbolic, 7

is continuous This yields 7 E Hol(A, X) and hence SRn ( in Hol(A, X)

(ii) By (i) X has the HEP (see [16] or [17]) Thus, it suffices to prove that f

extends holomorphically over (Z \ H) U W, where W is an open subset of Z such that W n Hc, is not empty for every branch Hc, of H

(iii) Assume that Hc, is a branch of H and zO, E Reg(H,) as in the assumption of Theorem 1.2 We prove that f extends holomorphically over an open neighbourhood

Wc, of z, in Z

Since the problem is local, we may assume that Z = B x A with Hc, = B x {O}, where B is the open unit ball in a Banach space E, and let zn Zo = (x, 0) Let F be an arbitrary finite dimensional subspace of E containing x We write

Za = (x",A") E B x A* Let P: E F be a continuous linear projection of E

onto F We may assume that Pxl e B n F for every n> 1

Since B is hyperbolic and Px" - Px = x, we have

dx (f (xn, A'), f (Pxn, A")) = dx (fA, (x?), fv, (Pxn))

<dB(xn,Pxn) - 0 as n 00,

where fv (x) = f (x, Ac) Hence f (Px", Ac) pO, as n - oo Choose a pseudocon- vex coordinate neighbourhood W of pO, in X which is isomorphic to an open ball

in a Banach space B Let V = W/2

Consider fF = fI (BnF)x A* Put BF = B n F By taking a subsequence of {Zn}

if necessary, we may assume that the sequence { lAc } is monotone decreasing By making use of the same argument in the proof of Theorem 1.1 (part (i)) suppose that, for every n, the image of the set aBF x {jAce+1J < lzl < lAa} by fF is not entirely contained in V Put

rn= inf{r E (0, IA" 1) fF (BF xA) C V, VA E C, such that r < JAI < IAal}

Sn =sup{r EA (IA" v1) fF (BFXA) C V, VA E C, such that IA" I < JAI <r}

Let an = {A E A :JA = r I}, = IA {A A: : = IA-}l,3n = {A E A JAI = Snl Then

as n 00

Consider the compact subset K in W given by

By the maximum principle we have

KPsH(W)D Uf((BF B )U ( O)) F

and hence U f ((9BF X an) U (9BF X 3n)) is relatively compact in V

n>2

Without loss of generality we may assume that

f (&n X an )

and

f(BF >~ +2

f ( < On x BW 2

By the definition of rn and Sn, it follows that w1, w2 E aV and hence w1, w2 ? p,

Choose a continuous linear functional u such that u(wl), U(W2) # U(pa) = 0

Since f (BF Xrnsn) C V C W, there exist in < rn <Sn < S such that

f (n X >

nn) C W

Take a point xo E &BF Consider the holomorphic function

an (A) = U o f (-, A) on Arnsnn

It is easy to see that {Un(-yn)} -+ U(p,) 0 O and {jn(i3n)} -+ U(W2)

By making use of the same argument as in Theorem 1.1, we get a contradiction Thus fF is extended holomorphically to BF X A

By (iii) the family {fF} defines a Gateaux holomorphic map f: Z\S(H) ->X, where S(H) denotes the singular locus of H

Since [3]

dBxA(U, v) = inf {dBnFxA(U, V): F 3 u, v and dimF < oo},

we have

dx(f(u), f(v)) < inf {dBnFxA(U,vV) F 3 u,v and dimF < oo} = dBxA(u,v)

Hence f is continuous This yields the holomorphicity of f Q.E.D 1.3 Remark Theorem 1.2 was proved by Fujimoto [5] when Z is a finite-dimen- sional complex space and X is a taut complex space Howeover, since tautness

is not defined in the infinite-dimensional case, the assumption on the complete hyperbolicity of X is a natural substitute

?2 HARTOGS HOLOMORPHIC EXTENSION

We now prove the following

2.1 Theorem Let X be a Banach analytic space which is an increasing union of pseudoconvex domains Assume that X contains no complex lines Then X has the HEP

HYPERSURFACES AND HARTOGS EXTENSIONS 751

Proof (i) First we assume that X is pseudoconvex Let f: Q X be a holomor- phic map Consider the commutative diagram

>

r X f

B where Qf is the domain of existence of f with the canonical extension f: Qf X and e, -y, 7r are locally biholomorphic canonical maps

We need the following

2.2 Lemma The map f: Qf -i X is locally pseudoconvex, i.e for every x E

X there exists a pseudoconvex neighbourhood U of x in X such that f (U) is pseudoconvex

Proof Given x E X Choose a neighbourhood V of x in X which is isomorphic to

an analytic set in an open ball of a Banach space

Consider the restriction fIf 1(V) Let g: Af-1(V) -+ V be a holomorphic extension of flf- 1(V) to the envelope of holomorphy Af-1 (V) of f-1 (V) Since Qf

is the domain of existence of f, it follows that Afi-(V)cQf

On the other hand, from the relation

f(Af-1(V)) = g(Afil(V))CV

we have f-1(V) = Af-1(V)

Consequently f: QfJ> X is locally pseudoconvex

In order to show that Qf = AQ it remains to check that Qf satisfies the weakly disc-convex condition

Indeed, let {Jk} C Hol(A, Qf) be such that the sequence {akI*A} converges

to a in Hol (A*,Qf) Since X is pseudoconvex and X contains no complex lines,

X satisfies the weakly disc-convex condition (see [17, Proposition 2.3]) Thus the sequence {f o ok} C Hol(A , X) converges to f o a in Hol(A, X) Choose a pseu- doconvex neighbourhood V of f o a(O) which is isomorphic to an analytic set in

an open ball of a Banach space and fl (V) is pseudoconvex Since f -(V) is

a pseudoconvex Riemann domain over a Banach space B with a Schauder ba- sis, f-1 (V) is a domain of holomorphy It is easy to see that there exist ko and

E > 0 such that (fo(Jk) (A,6) C V for every k > ko and fooa(A,) C V, where /\ = {z E C: izl < E} Hence ak(A6:) C f'-(V) for every k > ko It follows that {oJkIzyj -+ (J in Hol(AL, f-1(V)) (see [6, Theorem 5 and Lemma 6] Thus the sequence {ak } is convergent in Hol(A, Qf)

00

(ii) Assume that X = U Xn where Xn are pseudoconvex domains and X1 C

n=1

X2 C

Put Qn = f'(Xn) for each n > 1

By (i), for each n > 1, the map fn = fl,,, extends to a holomorphic map

fn : Qn -+Xn-

It is easy to see that for each n > 1 there exists a unique locally biholomorphic map en :A Qn _A Qn+1 such that the following diagram is commutative:

AQn AQn+1

B and fn+len = fn for n > 1, where irn :A Qn -* B defines AQn as a Riemann domain over B Thus we can define maps f: Q lim indAQn -* X and 7r Q-+ B

by f|AQn = fn for all n > 1 and 1rF/AQ A r for all n > 1 Since irn is a local homeomorphism for n > 1, it follows that 7r is also a local homeomorphism Moreover, we have dn(z) < dn+1(en(z)) for all z E A Qn and n > 1, where dn denotes the boundary distance with respect to 7rn: AQn+ B for each n > 1 Since AQn is pseudoconvex, -log dn is plurisubharmonic for all n > 1 Hence the function - log d(z) = lim - log dn (Z), for every z E Q, is plurisubharmonic This

n-oo

means that Q is pseudoconvex and hence Q is a domain of holomorphy

This yields Q = AQ The theorem is proved Q.E.D

2.3 Remark There exists a complex manifold X which is not pseudoconvex such

00

that X U Xn, where Xn is Stein

n=1

Indeed, as in [4] for each n we put

Xn= {(Z7 w) C E3 :wz =pn(z),p(z) f j (z - ) }

Obviously, Xn are closed submanifolds of C3 and hence, Xn are Stein For each

n, consider the map n: Xn- Xn+j defined as follows:

AYn(Z) W7 77) = (Z7 W7 77 (Z n+1)

Clearly, an is biholomorphic from Xn onto Xn+j \ {f 1 x C2 } Thus we can define X = lim (Xn7 -yn) We shall prove that X is not pseudoconvex For the converse case, we assume that X is pseudoconvex and hence, in our case X satisfies the weakly disc-convex condition Let {ffn} C Hol(A, X) be a sequense of maps defined by

fn(A) = (p7 A + n (A)) Then fn (A) C Xn+1 We prove that {fn } is uniformly convergent in Hol(A*, X) For each k, consider fn E Hol(\ 1,X) defined by

fn (v n + 1A- 1+ ) \+1 k ,7

where

Akal,1 {z E : k? 1 < Z| <1

Note that

fn eHl(A k+ l,Xk)

For every n > k, {f1} converges in Hol((Al 1,X) to the map fk given by

fk (A) = (A, PkAA)) A On the other hand, since

"YnO Yn-1o *.* o Yk fnk = fn for every n, k > 1,

we have ,q .p P- f q , where p, q are natural numbers with p < q Thus we can define a map f: /\* -* X by setting f(z) = fk(z) for every z E A/ 1 1 Since

k+1 -0,the sequence {fn} converges to f in Hol(A*,X) By hypothesis, {fn}

converges to f in Hol(A,X) Consider A, = {z E C: zl < },e E (0, 1) Since

00

{fn} is uniformly convergent on A., it follows that U fn(A,) is compact Since

n=1

00

X = U Xk, Xk C Xk+j and Xk is open in X for every k > 1, there exists k0 such

n=1

that U fn A() C Xko Hence

n=1

f ko (A) (A, APko (A)) for all AeA AE*

Thus f(ko) (A) can be extended holomorphically to A,, This is impossible, be- cause Pko (0) ? 0 Hence X is not pseudoconvex

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