First Steps in Several Complex Variables: Reinhardt Domains (EMS Textbooks in Mathematics)

6MS Textbooks in Mathematics Marek Jarnicki Peter Pflug First Steps in Several Complex Variables: Reinhardt Domains B...

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6MS Textbooks in Mathematics Marek Jarnicki Peter Pflug

First Steps in Several Complex Variables: Reinhardt Domains

BAS Textbooks In Mathematics

EMS Textbooks in Mathematics Is a book series aimed at students or professional mathematicians seeking an introduction into a particular field. The individual volumes are intended to provide not only relevant techniques, results and their applications, but afford insight into the motivations and ideas behind the theory. Suitably designed exercises help to master the subject and prepare the reader for the study of more advanced and specialized literature. Jorn Justesen and Tom Hoholdt, A Course In Error-Correcting Codes Markus Stroppel, Locally Compact Groups Peter Kunkel and Volker Mehrmann, Differential-Algebraic Equations Dorothea D. Haroske and Hans Triebel, Distributions, Sobolev Spaces, Elliptic Equations Thomas Timmermann, An Invitation to Quantum Groups and Duality

Marek Jarnicki Peter Pflug

First Steps in Several Complex Variables: Reinhardt Domains

Luropean Mathematical Society

Authors: Marek Jarnicki Uniwersytet Jagiellofiski Instytut Matematyki ul. Reymonta 4 30-059 Krakow, Poland

Peter Pflug Institut fur Mathematik Carl von Ossietzky Universitat Postfach 2503 26111 Oldenburg, Germany

E-mail: [email protected]

E-mail: [email protected]

2000 Mathematical Subject Classification (primary; secondary): 32-01; 32A05, 32A07, 32A10, 32005, 32E05, 32F12, 32F45 Key words: Reinhardt domains, domains of holomorphy, pseudoconvex domains, invariant metrics and distances The Swiss National Library lists this publication in The Swiss Book, the Swiss national bibliography, and the detailed bibliographic data are available on the Internet at http://www.hetveticat.ch.

ISBN 978-3-03719-049-4 This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must be obtained.

® 2008 European Mathematical Society

Contact address: European Mathematical Society Publishing House Seminar for Applied Mathematics ETH-Zentrum FU C4 CH-8092 Zurich Switzerland

Phone: +41 (0)44 632 34 36 Email: [email protected] Homepage: www.ems-ph.org

Typeset using the author's TEX files: I. Zimmermann. Freiburg Printed on acid-free paper produced from chlorine-free pulp. TCF Printed in Germany

987654321

Preface There are many excellent books for students introducing them to classical complex analysis of one variable, but only a few that cover several complex variables. Thus

we were motivated to write such a book, intended as a textbook for beginning graduate students and as a source book for lectures and seminars. We have developed the main ideas of several complex variables in the context of, but without entering

into too many technical details of, a very simple geometry, known as Reinhardt domains. Though many students may know little about this topic, we think it is a good start for beginners in several complex variables. Using this as a base, we add to all topics a selection of remarks and hints relating the discussion to the general theory. Some of the chapters or sections, those marked with a star (*), are more developed than others and can be skipped in a first reading. Moreover, we present some topics that have never appeared in a textbook or are new findings. We hope that these new ideas will motivate the student studying this book to become more deeply involved in the use of several complex variables. Further toward that end, we include in the Bibliography both direct references and a list of monographs and textbooks in complex analysis, thus providing a source for expansion on topics in our book and extensions to new studies. The book contains many exercises that the reader is asked to work on when encountered, before proceeding with further topics. There are also many points in the proofs that we have marked EXERCISE. By this we mean that the reader should write out the argument in more detail than we have done, to assure mastery of those details in preparation for what is to come. We believe that the study and understanding of mathematics requires continuous interaction between the reader and the text, and this cannot be achieved by passive reading. From time to time we pose open problems (marked by E] ...Q) that to the best of our knowledge have not yet been solved. We encourage the reader to try to solve them and would be most grateful to hear about such attempts, both successes and interesting failures. Note that at many places, in order to simplify formulations, some obvious assumptions that guarantee that the considered objects are non-empty are not stated. For example, if we write "Let D C C" be a Reinhardt domain...", then we always automatically assume that D # 0. We think that the reader will easily be able to complete the missing assumptions. In the interest of consistency of form and notation, we sometimes send the reader to [Jar-Pfl 1993] or [Jar-Pfl 2000] instead of quoting the original research paper. We nevertheless encourage the reader to seek out those original works in their further studies. During the process of proofreading we detected some gaps and misprints. Our thanks go especially to Dr. P. Zapalowski who helped us during that process. Nevertheless, according to our experience with former books, we are sure that a number of errors remain about which we would be happy to be informed.

vi

Preface

We would be pleased if the reader would send any comments or remarks to one of the following e-mail addresses [email protected] [email protected]

We thank the following institutions: - Committee of Scientific Research (KBN), Warsaw (No. I PO3A 005 28) and Deutsche Forschungsgemeinschaft (No. 227/8-1,2) for their financial support without that this project would have been not possible, - the RiP-programm at Oberwolfach for the excellent working atmosphere there, - our universities for their continuous support. We deeply thank Dr. M. Karbe for having encouraged us to write this book and for all his support during the recent years. Krakow and Oldenburg, February 2008


Contents

Preface

I

v

Reinhardt domains 1.1

1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17

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Introduction . . . . . . . . Summable families Domains of convergence of power series Maximal affine subspace of a convex set I .

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Domains of holomorphy . Envelopes of holomorphy

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Holomorphic convexity ......... .... ...... ...

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Reinhardt domains .... ... .... ... ......... . .. Domains of convergence of Laurent series Holomorphic functions Balanced domains .. Extension of holomorphic functions ... . ......... ... Natural Fr6chet spaces .. . .

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Hyperconvexity 1.18* Smooth pseudoconvex domains

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Cartan theory ............. .. ............. 177 Biholomorphisms of bounded complete Reinhardt domains 180 in CZ . .... .

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Biholomorphisms of complete elementary Reinhardt domains

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Miscellanea

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Reinhardt domains of existence of special classes of holomorphic functions 3.1

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General theory .. ... ...... . ..

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220 220 225 230

viii

Contents

3.4 3.5 3.6

4

M'-domains of holomorphy ....... .... ........ 236 Ak-domains of holomorphy ........ ... ........ . 239 Li -domains of holomorphy ........ ... ........ . 241

Holomorphically contractible families on Reinhardt domains 4.1

4.2 4.3*

4.4 4.5 4.6 4.7 4.8 4.9*

251

Introduction ......... ...... ... ....... . .. 251 Holomorphically contractible families of functions ........ 253 Hahn function ........................... 269 Examples I - elementary Reinhardt domains ........... 277 Holomorphically contractible families of pseudometrics .. .. 293 Examples II - elementary Reinhardt domains ........... 310 Hyperbolic Reinhardt domains ... . .. .. 313 .

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Carathdodory (resp. Kobayashi) complete Reinhardt domains The Bergman completeness of Reinhardt domains . . . .

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317 325

Bibliography

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Symbols

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List of symbols

349

Subject index

355

Chapter 1

Reinhardt domains 1.1 Introduction The notion of a holomorphic function of one complex variable can be based on the notion of a power series - a function f : S2 -+ C (where S2 C C is open)

is holomorphic (f E O(S2)) if for every a E S2 there exist a power series FO O

ck (z - a)k centered at a and a neighborhood Ua C S2 of a such that 00

f(Z) = E Ck(z - a)k, k=o

Z E Ua.

It is well known that the domain of convergence of an arbitrary power series 00

bk(Z - a)k

k=0

is the disc K(a, R) := (z E C : I z -a 1 < R) with the radius (radius of convergence)

R=

E [0, +oo]

1

lim sup k Ibk

(where K(a, 0) = 0, K(a, +oo) = C). Moreover, if R > 0, then the function 00

1: bk(Z

P z)

- a)k,

z e K(a, R),

k=0

is holomorphic.

If f E O(0) and f(z) = Ek=OCk(z - a)k, Z E Ua c 12, then the radius of convergence of the series Fk o ck (z - a)k is not smaller than the Euclidean

distance du(a)ofthepoint atoU2(dc(a) Z E K(a.dn(a)).

+oo)and f(z) =

Ek=Ock(z-a)k,

The most elementary is the case where S2 = K(a. r), which, of course, may be reduced to the case 0 = ( = the unit disc. Recall the following well-known 'Equivalently: f is differentiable in the complex sense at every point a e 92. i.e. the function f has at every a E dl the complex derivative

f'(a)

C\t

urn

-+of(a+h)-f(a)

Chapter 1. Reinhardt domains

2

issues, whose analogues will be considered in the sequel in a much more general context:

The structure of the group Aut(D) of holomorphic automorphisms of D. It is well known that

Aut(ID)={ID

I -az

EIDET,aEID,} 111

where T := M. In particular (cf. Exercise 2.1.5 (b)), Aut(ID) acts transitively on ID.

The holomorphic geometry of D. In particular, the theory of holomorphically invariant distances, i.e. those distances d : ID x ID - IR+, for which

d(f(z), f(w)) < d(z,w).

z,w E ID, f E O(ID,ID).

(1.1.1)

where O(D, ID) denotes the set of all holomorphic functions f : ID D. The above condition means in particular that any f E Aut(ID) is an isometry of the metric space (ID, d). examples are:

m(z, w) :_ I z -

w

(Miibius distance),

I

1 - _ iv

P(z. w) :=

I

+

log

2 cf. [Jar-Pfl 1993], Chapter 1.

(Poincare distance).

w) 1

m(Z,

Exercise 1.1.1. (a) Check (1.1.1) ford E {m, p}. (b) Prove that m and p are distances on D.

(c) Prove that p(0, b) = p(0, a) + p(a, b), 0 < a < b < 1. In the next step we substitute power series by Laurent series 00

bk(z -a)k k =-00

and, consequently, discs K(a, r) by annuli

A(a, r

,

r+) :_ {z E C : r- < Iz-al < r+},

-oo < r- < r+ < +oo, r+ > 0.

Note that if r- < 0, then

A(a, r-, r+) = K(a, r+) and A(a, 0, r+) = K(a, r+) \ (a) =: K.(a, r+). A Laurent series with b_k = 0, k = 1.2.... , will be always identified with the power series Ek o bk (z - a)k. The domain of convergence of a Laurent series is

1.1. Introduction

3

an annulus A(a, R-, R+) with R+

R-

I

lim sup k

lim sup

x

Ib-k I

-00

I bk I

if 3kEN : b-k 36 0,

if dkEN : b_k = 0, (1.1.2)

provided that R- < R+. The function 00

f (Z) :_ E bk (z - a)k,

Z E A(a, R-, R+).

k=-oo

is holomorphic. Moreover, for every compact K C A(a. R-, R+) there exist C > 0, 0 E (0,1) such that Ibk(Z - a)k! < Celkl, Z E K, k E Z. Conversely, every function f holomorphic in an annulus A(a. r-, r+) has a unique representation by a Laurent series. We may always assume that a = 0. Notice that for A := A(0. 1/R, R) (R > 1) we have

Aut(A)=(A In particular, the group Aut(A) does not act transitively: cf. Exercise 2.1.5 (c). The holomorphic geometry of an annulus is much more complicated than the one of ID; cf. [Jar-Pfl 1993], Chapter 5.

Notice that for domains (a subset D of a topological space X is said to be a domain if D is open and connected) D C C the following three notions coincide: D is a domain of convergence of a Laurent series centered at 0; D is a domain invariant under rotations, i.e. for any z E D and .l E T the point Az also belongs to D; D is a disc or an annulus centered at 0. The notion of a power series generalizes in a natural way to the case of several complex variables. By an (n fold) power series (centered at 0 E C") we mean any series of the form E aaza (z E C"). aEZ+

where (aa)aEz+ C C. Z := {a E 71" : a > 0}, za := zi ` ... zn" (0° := 1); see § 1.3. The domain of convergence D of a power series (Definition 1.3.3) has the following important properties:

For any a = (a 1. .

.

. , an)

E D, the closed polydisc

{(Zt... ,Zn) E C" : IzzI 5 1ai1. j = I.....n) is contained in D, i.e. D is a complete Reinhardt (n-circled) domain (Definition 1.3.8).

4


The set

logD:= ((loglZll,....loglznl): (z l,.. ,Zn) ED, zt,..zn i4 0) is convex in the geometric sense, i.e. D is logarithmically convex (Definition 1.5.5, Proposition 1.5.16). The series is locally geometrically summable in D, i.e. for any compact K C D there exist C > 0, 9 E (0, 1) such that l aaza 1 < C91al , z E K, a E Z+, where (aI := at + + an (Remark 1.3.5 (f)). In the case n = I the only complete Reinhardt domains are discs K(r) and they are always logarithmically convex. In the case n > 2 the situation is more complicated. There are infinitely many types of complete Reinhardt domains which are not biholomorphically equivalent (e.g. Euclidean balls IB(r) and polydiscs IP(r); cf. Theorem 2.1.17). Moreover, there are complete Reinhardt domains D C C" (n > 2) which are not logarithmically convex, e.g. D := ((Zt,z2) E ID2 : min{Izt(, IZ21) < r}

(r E (0,1)).

The function f (z) := EaEZ+ aaza, Z E D, is holomorphic. Conversely, every function f holomorphic in a complete Reinhardt domain D C C" has a "global" expansion into a power series f (z) = Laezn aaza, z E D (cf. Proposition 1.7.15 (c), (d)). The notion of a Laurent series extends to the notion of an (n fold) Laurent series (centered at 0)

Eaaza; aEZ"

see § 1.6. The domain of convergence V of a Laurent series (Definition 1.6.1) has the following properties:

For any a = (a t .... , a") E D, the torus {(z1..

zn) E C" : 1zj1= 1aj1 j = 1,.. ,n}

is contained in D, i.e. D is a Reinhardt (n-circled) domain (Definition 1.5.2). D is logarithmically convex (Proposition 1.6.5 (d)). Foreveryj E { l , .... n }, if D fl Vj 0,2 where

then for every a = (a 1, ... , a") E 2), the disc

{(a,,....aj_l,zj,aj+t,...,a") : Izjl < Iail} is contained in D (Proposition 1.6.5 (c)). 2Observe that in the case of a power series we have 0 E V and, consequently. D fl vi o 0 for any j.

1.1. Introduction

5

The Laurent series is locally geometrically summable in 1). i.e. for any compact

set K C D there exist C > 0, 0 E (0, 1) such that laazal < COIaI, z E K, a E Z", where lal := Ia1 I + - - - + la" I (Proposition 1.6.5 (a), Lemma 1.6.3).

In the case n = 1 the only Reinhardt domains are discs K(r) and annuli A(r-, r+);3 they are always logarithmically convex. Every function given by a Laurent series is holomorphic. Conversely, every function f holomorphic in a Reinhardt domain D C C" has a "global" expansion into a Laurent series f (z) = EaEZ aaza, z E D (Proposition 1.7.15 (c)). As always, from the point of view of the theory of holomorphic functions, most important are domains of holomorphy, i.e. those domains D which are "maximal" in the sense that all holomorphic functions in D cannot be simultaneously extended through a boundary point of D (Definition 1.11.1); let us mention that for n > 2 there are even pairs of domains D D C C" such that every function f E L9(D) extends holomorphically to D. It turns out that in the category of Reinhardt domains the following conditions are equivalent (Theorem 1.11.13): D is a domain of holomorphy;

D is logarithmically convex and relatively complete, that is, for every j E { 1, ... , n), if D fl V. 54 0, then for every a = (a,_.. a") E D, the disc

((a1,....aj-1.zj.aj+1.....an) : Izjl < lajl) is contained in D;

D=D*\M, where D*:=int

n

D.,c,

M:= U V,

(a,c)ER"xR: DCDa.c

jE(1,...,n): DnVj =o

Da,c := ((z1..... Zn) : Iz1I" ... IZnIa" < ec}: Da,c is called an elementary Reinhardt domain. In particular, the domain of convergence of a Laurent series is always a domain of holomorphy. Such a simple geometric characterization of domains of holomorphy does not occur in any other category of domains. The notion of a domain of holomorphy extends in a natural way to an F-domain of holomorphy, when we are only interested in the extendibility of functions from a family F C O(D) (Definition 1.11.1). If D is not an F-domain of holomorphy, then one can ask whether there exists the maximal domain D C C" (the F-envelope

of holomorphy of D) such that every function f E 37 extends holomorphically to D. The answer is negative in general, even for F = O(D) - the F-envelope of holomorphy of D may be non-univalent, i.e. it is a non-univalent Riemann domain 3A(r-, r+) := A(O, r-, r+)

6


spread over C". In the category of Reinhardt domains the situation is simpler, namely: For an arbitrary Reinhardt domain D C C" and an arbitrary rotationinvariant family of functions F C O(D), the F-envelope of holomorphy of D is again a Reinhardt domain (Theorem 1.12.4). The above results permit us to reduce many problems concerning Reinhardt domains of holomorphy to the case of elementary Reinhardt domains. We will see that many holomorphic properties of D are encoded in geometric properties of log D. In particular, we will discuss the following problem. Given a Reinhardt domain of holomorphy D C C" and a family F O (D), find geometric conditions

under which D is a domain of holomorphy with respect to the family F. For example, we consider as F the following spaces: .7£°'D(D) = the space of bounded holomorphic functions, LP (D) = the space of p-integrable holomorphic functions,

4k (D) = the space of all functions f E O(D) whose derivatives Daf extend continuously to D for all IaI < k. Various geometric characterizations of domains of holomorphy with respect to special families of functions will be presented in Chapter 3. Chapter 2 is devoted to a presentation of different aspects of the problem of biholomorphic equivalence of Reinhardt domains. Finally, Chapter 4 presents a thorough study of the theory of holomorphically invariant functions and pseudometrics on Reinhardt domains.

1.2 Summable families The aim of this auxiliary section is to recall some basic notions related to summable families (cf. for instance [Sch 19671 or [Hdr 1982]).

Let us fix an arbitrary set 0 36 Z C C" and let 1 76 0 be an arbitrary set of indices. Let ap(I) be the set of all non-empty finite subsets of 1. Consider a family

f = (fi )i El of functions f, : Z -+ C. For example (cf. §§ 1.3, 1.6): 1 C Z", fa(z) := aaza, Z E Z C C", a E I, where

C C and the set Z is such that all the powers za, a e 1, are defined on Z. In the case where Z = (a ), instead of a family of functions, we rather should think of a family of complex numbers (f1(a)); E j

.

For A E ap(I) put fA := LEA f . Let, moreover, fo := 0. Definition 1.2.1. We say that the family f is uniformly summable on Z (equivalently: the series EEl j is uniformly summable on Z) if there exists a function

fl : Z -+ C such that Ve>o 3S(e)=S(l,e)eg-(l) dAEg(l): S(e)CA VZEZ I fA(Z) - fl (z)I < e.

(1.2.1)

1.2. Summable families

7

Notice that the case where I is finite is trivial (we take S(1, E) := I for any

e>0). In the case where #Z = 1 we simply say that the family f (considered as a family of complex numbers) is summable or that the series LIE/ f is summable. It is clear (EXERCISE) that the function f, is uniquely determined. We write

f j = LEI f and we say that f j is the sum of the family f . Let $(I, C z) be the set of all families f = (f )i r: f that are uniformly summable

on Z.

More generally, for T C C, let 8(1. Tz) be the set of all uniformly summable families f = with Z -* T, i E I. Exercise 1.2.2. Let (fk)kEN E S(W, CZ). Prove that the series

k fo(k) Is

uniformly convergent in the classical sense for every bijection a : W -+ W.

Exercise 1.2.3. Let I := W. Find a convergent (in the classical sense) series Ek I A of real numbers such that the family (fk)kEw is not summable in the sense of Definition 1.2.1 (cf. Theorem 1.2.12).

Remark 1.2.4. (a) (EXERCISE) If f = (f )ief, g = (gi )aE, E 8(1, CZ), a, /3 E

C.then af + fig := (af +Ygi)iE/ ES(1,Cz)and(af +fig)i =afr+/igj. In particular, 8(1, CZ) is a complex vector space and the mapping

S(1.CZ)9fHfIECZ is C-linear. (b) (EXERCISE) A family f is uniformly summable if the families Re f (Im f, )i E j are uniformly summable. Moreover, Re(fl) (Re f )i ej and lm f

(Ref )I and Im(fl)

(Im f )f. (c) If f E 8(1, CZ) and all the mappings

Z -+ C are bounded, then the

family of functions (fA : A E a-} is uniformly bounded. Indeed, let S := S(1, 1) E '(1) be associated to E = 1 according to (1.2.1). It suffices to prove that the set (fA : A E FV(1 \ S)} is uniformly bounded. Fix an

A E ap(I \S). ThenwehavelfAl = IfAus-fsl < IfAus-f1I+Ifs-fII < 2. (d) If f E $(I, Cz), then the set {i E I : f,, $ 0} is at most countable. Consequently, the most important is the case where I is countable. Indeed, it suffices to show that for every e > 0,

{i E I : 3.-Ez : If (z)I > 2e} C S(E), where S(e) is chosen according to (1.2.1). Fix e > 0 and i E I \ S(e). Then

III = I f(i)us(e) - fs(e) I 5 If(i)us(e) - f, I + I fs(e) - fi I

- 2e.

Proposition 1.2.5 (Cauchy criterion).

f E 8(I, Cz) b V,.>0 3C(e)Ea-(I)VAEW-(I\C(e)) : IfAI < e.

(1.2.2)

8


Notice that the Cauchy condition (1.2.2) permits us to verify the summability

of f without determining fl. Proof. (=): Let f E 8(1. CZ). Take an e > 0 and let S(1, s/2) be associated to s/2 according to (1.2.1). Put C(s) := S(I, E/2). Then for any A E a-(I \ C(e)) we have

IIAI = I fAUC(d) - fC(a)I < 1 fAUC(c) - fl I + Ifc(s) - fj 1 < E.

(=): Suppose that (1.2.2) is fulfilled. Let C := C(1/v), F := fc,,, v E W. Then we have

Fv+k - G1 = I

fc,+k \cv -

1

1< v +

fC

1

v, k E W.

v+k'

satisfies the uniform Cauchy condition on Z and, therefore,

Consequently,

there exists a function Fo : Z - C such that F - Fo uniformly on Z. If k -* +oo, the above inequality implies that

FoI< 1, v

vEW.

Now, let A E R(I ), C, C A. Then we get

IIA - Fol -IIA - fc»I+IFn - FoI < IfA\Cl+

n

0, let S := S (I , v/2) be taken as in (1.2.1), and let

T:=(j EJ:I(j)nSoO). Observe that T E a-(J). We are going to show that T = S(J, e) (with respect to

the family (fl(j)) jEJ ). Take a B E a(J) with T C B. Put N := #B. For any j E J let Sj := S(1(j ), 2N ). We may assume that S n I(j) C Sj. j E J.

1.2. Summable families

9

Let A := UJEB Sj E ap(I). Observe that S C A. Hence, I fA - fr 1 < e/2 and, finally, we get

I fl

- E II(j)l -< I fj - E Isf I + E l fs; -fl(j)1:5 I fr - fAl + e/2 <s. jEB

jEB

jEB

0 Definition 1.2.8. (a) We say that f is absolutely uniformly summable on Z if the family If I := (If I)iEj is uniformly summable on Z, i.e. If I E S(I, R+). In the case where #Z = 1, then we simply say that f is an absolutely summable family. (b) We say that f is normally summable on Z if all the functions f, are bounded on Z and the family of numbers (supz I f I)iEJ is summable. (c) We say that f is locally uniformly summable (resp. locally normally summable) on Z if every point a E Z has an open neighborhood U such that the family (f izr u)iEi is uniformly summable (resp. normally summable) on z fl U. In any of the above cases, instead of the family f , we can say that the series E; EI f is absolutely uniformly summable, normally summable, etc.

Remark 1.2.9 (EXERCISE). (a) Observe that if I f I < gi, i E 1, and (g,),, E S(I, R+), then, by the Cauchy criterion, f E 8(1, CZ). In particular, if If I E 8(1, It+),then f E 8(1, CZ). Moreover, Ifr1 < If 11,i.e. I EiEt f I < LiEI if I. We will see in Theorem 1.2.12 that the converse implication is also true, i.e. if

f E 8(1, CZ), then IfI E 8(1, R+). (b) Using the Cauchy criterion, we conclude that every normally summable family is absolutely uniformly summable. The converse implication is not true as the following standard example shows.

Let Z := [0,1],1:= N, gk : [0,1] -> It,

11gk(x)

if0<xo. Then for every 0 < 8 < 1 we have

«IZ E IP(8r), a E Z Iaazal < CO',

(EXERCISE).

In particular, the series EaEZ+ aaza is locally normally summable in IP(r). Definition 1.3.3. Given a power series S, put

B = Bs

{Z

E C" : sup laaza) < +oo aEZ+

es

{z E C" : the series E aaza is summable}. aEZi.

D = Ds := int C. Clearly D C C C B. The set D is traditionally called the domain of convergence of the power series S.


14

Exercise 1.3.4. Determine BS, CS, and DS for the following power series: (a) E/LEZ+ µ!Z1 Z2;

(b) E

EZ+ Z1 Z2;

(c) Eµ,VEZ+

Z{i1

Z2v'

(d) F µ,VEZ+ µ!Z1 Z2. (e) E,,Ez+(zlz2Y'; 4,v-EN

(g)

µ,vEDJ

czl`_v µ!v! 1 °2'

Remark 13.5. (a) If n = 1 and 0 # D 0 C, then B = C = D = K(R), where R is the radius of convergence of S. (b) If S := E EZ+ µ!z'Iz2, then e = (C x {0}) U ({0} x C) = Vo and D = 0.

In particular, for n > 2 we may have e ¢ D. (c) For every point a = (a 1..... an) E B (resp. e) the closed polydisc

IP((IaiI...., Ianl)) = {(Z1,...,Zn) E Cn : IZjI < IajI, j = 1....,n} is contained in B (resp. e). (d) D = int B = int B. In particular, D is fat. (An open set S2 C IRk is said _ to be fat if S2 = int f2.) Indeed, fix an a = (a 1..... an) E int B. Observe that for small e > 0 the point

b = (b1, .... bn), with

bj.

_ aj(1+e) ifaj 96 0, E

ifaj=0,

_ j-1,.. .n.

also belongs to int B. Let c = (c1.... , cn) E B be such that

Icj _bjI < Eajle

ifaj #0.

j = 1.....n.

Consequently, ajl((I+e)-lajle

rj

Icjl> Ibil - Icj - bjI >

0

if a!j

-IajI, #0 -

j=1, ...n. Thus a E IP(r). Now, by Abel's lemma, we conclude that a E IP(r) C D.

(e) In view of (d), IP((lal I, ... , Ian I)) C D for every a = (a,-.. an) E D. Observe that any closed polydisc IP((Ia 11, . . . , Ianl)) is obviously connected. In particular, D is connected and so, D is really a domain.

13. Domains of convergence of power series

15

(f) For every compact K C D there exist C > 0 and 0 < 0 < I such that (aaza I< C01a1,

Z E K, a E Z.

Consequently, the series S is locally normally summable in D. In particular, the function f (z) EaEz+ aaza, z E V. is continuous (Corollary 1.2.13 (a)). Indeed, take a point a E D and let r E Qt;o n B and 0 < 0 < I be such that a E IP(Or). Next use Abel's lemma (EXERCISE).

Exercise 1.3.6. Let

P z) := F, aaza,

z E DS.

aEZ+,

Prove that for every IP(a, r) C DS there exists a power series F_yEz+ by(z - a)y centered at a such that f(z) _

by(z - a)y,

z E IP(a,r)

yEz+

(cf. Step 3 of the proof of Proposition 1.3.12).

Exercise 1.3.7. Check whether there exists a power series S such that

DS = ((z1, z2) E D2 : either Iz1I < r1 or (z21 < r2} with 0 < rl, r2 < I (cf. Fig. 1.5.2). We are led to the very important notion of a complete Reinhardt set.

Definition 1.3.8. We say that a set A C C" is complete Reinhardt (n-circled) if for every point a = (a 1, ... , an) E A and for every ). = (.Z 1, .... An) E 6n, the point X a = (A 1 a 1..... Ana,,) belongs to A; equivalently,

A=

U

IP(((a11...

,

(an I))

a =(a l ,...,an) E A

Exercise 1.3.9. (a) The domain of convergence of a power series is a complete Reinhardt domain. (b) If A C C" is complete Reinhardt, then A is arcwise connected. (c) If A C C" is complete Reinhardt, then A and int A are complete Reinhardt.

Exercise 1.3.10. Let S = EaEZ+ aaza, T = L,OEZ+ bpzfi be arbitrary power series. Using Theorem 1.2.7, prove that

E aabpza+P = 1` cyzy, a, f EZ+

yEZ+

Z E DS n VT,


16

where

aby-a,

cy :=

y E 7L

n

aeZ+:a(a)isak-linearsymmetric

mapping (C" )k - C, which is, as always, identified with the homogeneous polynomial C" -> C of degree k.


18

If g(k) (a) exists for every k E N, then we define the Taylor series of g at a as the power series Tag(-')

1

aEZ .

a.

Dag(a)(z - a)a.

The number

d (Tag) := sup{r > 0: Tag is uniformly summable in IP(a, r)) E [0, +00] is called the radius of convergence of Tag. Observe that

klg(k)(a)(z - a).

Tag(z) _ k=0

Proposition 1.3.12. Assume that Ds # 0 and let

f (z) .= E aaza, z E Ds. aEZl,}

For fi E Z" let DOS denote the power series as za-B 12 aEZ+: a>p

l

Then f has all complex Frcchet differentials in Ds,t3 Ds C DDns, and as za-p,

Dsf(z)

z E Ds, P E Z n.

(1.3.1)

aEZ+: a>$

In particular, f (z) = To f (z), z c- Ds. Notice the following difference between one and several variables. For n = 1 the radius of convergence of S is equal to the radius of convergence of the series of derivatives. This is no longer true for n > 2, for instance if S is the power series 00

V=0

then Ds=iDx I,but D

00

zt + F, z2' v=0 =0)xC.14

iz(

a.

131n fact. f is holomorphic - cf. Theorem 1.7.19. oo 14,U vz,"

a.,=- V-

-'

1.3. Domains of convergence of power series

19

Proof. Step 1. First observe that, for every j E (1, ... , n), the series as

ajaa za-ej

aZj

aEZn:a>ej

is locally normally summable in DS. It is sufficient to prove that if R E IRn,a fl BS, then the series gaTs is locally normally summable in IP(R). Let C > 0 be such that

I as I RU < C, a E Z". Then for any 0 < 0 < 1 we have sup Iajaa za-ej I < aEZn}: a>ej P(BR)

C 9Rj

ajOlal, aEZ+:a>ej

which gives the normal summability in IP(OR).

In particular, the function Fj defined by the series - is continuous on DS, j = 1, ... , n (Corollary 1.2.13 (a)). Step 2. We have n

n

f(h)=f(0)+Eaejhj+Efj(h)hj, h=(hi.....hn)EDS. j=1

j=1 where

E

ft (h)

aaha-e1

,

F,

aaha-e2

IaI-2, a>e2 a1=0

IaI>_2, a>>e1

fn-1(h) =

1,

f2(h) :=

aaha-en-1

E

fn(h)

IaI>2, a>-en-1 al= ... =an-2 =0

aaha-en .

IaI>2, a>en a 1=...=an-1=0

Observe that all the above series are normally summable in a neighborhood U of 0 (EXERCISE).

that f1(0)

-

In particular. the functions fl,..., fn are continuous in U. Note fn (0) = 0. Thus f'(0) exists and (0) = aej = Fj (0),

j = 1,.. ,n. Step 3. If IP(a. r) C= DS, then the series as (Y)(z -

a)yaa-y

a,yEZ+ a>-y

is normally summable in IP(a, r).

Indeed, let R E BS fl IRn>o and 0 E (0, 1) be such that JayI + rj < ORj,

20


j = 1,.-n, and let Iaa I P < C, a E 7+. Then n

Ia., I(") a,yEZ .

sup zeF(a,r)

f (Izi -aiI +lail)'J

-a)yaa-Yl < E Iaal sup zEP(a.r)j=1 aEZ"

I(z

a>y 01al

laaI(8R)a < C

< +oo.

aEZ+

aEZ"}

Step 4. Fix IP(a, r) C= DS. By Step 3 and Theorem 1.2.7, we have

f(z) = E aa(z + a -a)" = E as E (Y)(z aeZ+

aEZ .

=

YY

a)yaa-Y

/

by(z - a)y.

z E IP(a,r).

yeZ+

Hence, by Step 2, the function F(r) a z H f(a + z) is Frdchet differentiable at 0 and i (0) = be; , j = 1, ... , n. Consequently, f is differentiable at a and ZLj

fzL

Step 5. Iterating the above procedure shows that f has all complex Frdchet differentials and (1.3.1) holds for arbitrary fi (EXERCISE).

Exercise* 1.3.13. Assume that DS # 0,

aaza.

f(Z)

Z E DS.

aEZ"

and f (0) = ao # 0. Find a power series EpEZ+ bpz,6 such that 1

f(z)

= E bpz' 6EZ .

for z in a neighborhood of 0.

1.4 Maximal affine subspace of a convex set I As we have already mentioned in the Introduction, the logarithmic image X := log D C IR" of a Reinhardt domain D C C" will play an important role in various characterizations of the structure of holomorphic functions on D. In all essential cases the domain X will be convex. For the convenience of the reader we collect

1.4. Maximal affine subspace of a convex set I

21

below some basic properties of convex domains in R" which will be used in the sequel. Recall that a set X C R" is said to be convex if for every a, b E X, the segment

[a, b] := {(1- t)a + tb : t E [0, 1]) is contained in X. Remark 1.4.1 (Properties of convex sets; the reader is asked to complete details).

(a) For any family (X1);Ej C R" of convex sets, the set njEf X; is convex. In particular, for any set A C R", there exists the smallest convex set cony A containing A. (b) If A, B C R" are convex, then

conv(AUB)=((1-t)a+tb:aEA. b E B, t E[0,1])=:X. (c) If X C R" is convex, then X is convex. (d) If X C R" is convex, then int X is convex. In particular, for any family (Xi )iE/ C R" of convex sets, the set int ni E! X; is a convex domain. (e) For every a E (R")., c E R, the open halfspace

It,c:={XER":(x.a) 0. X E X)). Observe that E(C) = (0). Indeed, suppose that C + L C C, where L C O" is a real line. Consider any

two-dimensional real space P C R" with L C P and X' := X n P 9& 0. We have Ep (X') = {0}. Let C' be the open cone in P generated by X'. Obviously, C' = C n P. Hence C' + L C C' and the proof is reduced to the case n = 2. In the case n = 2 we only need to observe that if E (X) = (0), then there exist two different half-planes I41,0, 142,0 with X C I ',0 n k2,0; a contradiction. Consequently, there exists a fl° E (R"), such that

C n (X E IR" : (x, f°) = 0) = (0). Indeed, we use induction on n. The case n = 2 is obvious. In the general case,

take any aECR",11x11=1, with CC(xEIR":(x,a) 0. Suppose that for each s > 0 there exists an x8 E C, 11x811 = 1, with (x8, a + su) = 0. Write x8 = v8 + t8u + rya. We have

0 = (x8, a - su) = r8 - et8. Hence r8 = st,. Moreover, t8 = (x', a) < 0 and 0. We may assume that vek , v° to < 0. We have x8k -_). v° + tou E C' and to = - 1 - llv°112. Since

1 = 11x8112 = 11v8112 + t2 (l + s2). Take ek and t8k

v° = (v° + tou) + (-to)u E V n C', we conclude that to = 0 and v° = 0 contradiction. _ It follows that C n {x E 6t" : (x, $) = 0) = (0) for i6 from an open neighborhood of $0, which directly implies (iii). Indeed, suppose that j6' -+ $0 is such that

(yv, fiv) = 0 for some y° E C, y° # 0, v = 1, 2,.... Since C is a cone, we may assume that jly"11 = 1, v = 1, 2, ... , and next that y° -> y° E C, y° # 0. Then

(yo $0) = 0 - contradiction.

0

Lemma 1.4.12. Let X C C" be a convex domain. Then the following conditions are equivalent:

(i) E(X) = (0); (ii) there exist a non-singular matrix A :=

r"' l

E M(n x n, Z) and a vector J c =(c1 , .... c") E Qt" such that X C Ha & ,c, n n II,n,,. ; :

L a^

1.4. Maximal affine subspace of a convex set 1

(iii) there exist a matrix A E GL(n, Z) and a vector e E R" such that X C

27

{A E M(n x n; Z) : I det Al = 1} ,c,

n Ha",c,,.

Proof. By Lemma 1.4.11, we only need to prove that (ii) (iii). Let A and c be as in (ii), X C H&,c1 n n H"",c" _: H(A, c).

Suppose that det A I > 1. Put

S(A,c) := {fJ E Z" : 3d=d0ER : H(A.c) C Hp,d}.

Then S(A,c) = 71" n (Q+al +... + Q+,"). Indeed, obviously the set on the right-hand side is contained on the left-hand one.

Now take a 5 E S(A, c). Then there exists a d E lR such that H(A, c) C Hfi,d.

Write fi = Jj=1 tjaJ = ((tA)1..... (IA)"), where t := (h, ... , t"). Then t = #A-1, i.e. all the ti's are rational numbers. It remains to show that all of them are non-negative. Observe that the linear map L: Qt" -+ IR",

L(x) := ((x,al

(X, an)),

gives an isomorphism satisfying

{y E IR" : y 1 < cj, j = 1,.. ,n} = L(H(A.c)) Hence, t E IR+ (EXERCISE). Note that the set

Q (A. c) = 1.i=1

Pk, a1 : k = 1.... , N }. qk.i

where Pi,k E Z+, qj,k E N and the pairs Pj,k, qj,k are relatively prime. Then we denote by s = s(A, c) the least common multiple of all denominators q1,.

Let x E On with xA E Z". Write xj = uj + vj, where uj := xj - vj E [0, 1) n Cl and vj :_ lxi J E Z. Here AxA := max{k c- Z : k < x} = the integer

part of x E R. Then vA E Z" and (x - v) A = > .1(xj - vj)aJ E Q(A. c). Thus, s(x - v) E Z". Hence, sx E Z". Let r = r(A, c) be the smallest number in N such that if xA E Z" for an on, then rx E z". Comparing with the former paragraph it follows that xE

r < s. Let V denote the j -th row of the inverse matrix A-1 of A. Note that V E Q and &' A E Z". Therefore, r&j E Z" and so rA-' E IM(n x n; Z). Consequently, r" = det(rA-1) det A, i.e. I det Al divides r.

28


Observe that I < I det A 1 < r < Q (A, c) \ {a t , ... , a" }; in particular,

s.

Therefore there exists a vector a E

E Z' . So we may assume that there

exists a r E (RI, rt E (0, 1), such that & _ Ej=t rj ai E S(A, c). Moreover, if A denotes the matrix with rows & 1, a2, .... a", then I det A I = r,"I det A l < I det Al.

If I det AI = 1, then we are done. If not, repeating the above procedure the

0

proof will be finished after a finite number of steps.

Lemma 1.4.13. Let X C IR" be a convex domain. Then the following conditions are equivalent:

(i) there exists a sequence (xk)k t C X such that the sequences (xk,j)k=1, j = 1, ... , n - 1, are bounded and xk," -+ -oo;

(ii) X + IR- e" = X. Proof. The implication (ii) (i) is trivial. Conversely, take an arbitrary xo E X and t < 0. Put Ek : = t/Xk,", k >> 1. We may assume that 0 < sk < 1. Obviously,

sk - 0. Since X is convex, we get yk := (1 - sk)xo + Ekxk E [xo, xkI C X. Moreover, yk -> xo + ten. Hence xo + IR- e" C X. Consequently, X + IR- e" C

0

intX=X.

Definition 1.4.14. Let X C IR" be a domain which is starlike with respect to 0. i.e. [0, x] C X for every x E X. Then the function hX : IR" -, R+ defined by the formula

hX(x) := inf{t > 0 : x/t E X},

x E IR".

is called the Minkowski function of X.

Remark 1.4.15. Before we continue, let us recall the following important notion of semicontinuity. _

Let X be a topological space. We say that a function u : X -+ 1R is upper semicontinuous (u E (!t (X)) if for every t E k the set (x E X : u(x) < 1) is open. We say that u is lower semicontinuous (u E C4 (X)) if -U E et (X). Directly from the definition we get the following properties (EXERCISE):

U E Cl (X) iff for every t E IR the set (x E X : u(x) > t) is open. C(X, N) = et (X) n Cl(X).

uECt(X),fEC(Y,X)= uo f eet(Y). ER>o et(X) = Ct(X). If u, v E et (X) and u (x) + v (x) is well defined for every x E X, then u + v E

Ct(X).

u,v E et(X) max{u, v} E Ct(X). (Ua)aEA C et(X) = inf(u.: a E A) E Ct(X). In particular, if et(X) 3 uv , u pointwise on X, then u E Ct (X ). If et (X, IR) 3 u - u locally uniformly in X, then u E et(X).

1.5. Reinhardt domains

29

If (X, p) is a metric space, then u E et (X) * V aEX : lira supx,Q u(x) _ u(a). (Weierstrass theorem) If (X, p) is a compact space and u E et(X, [R_,,), then there exists a point xo E X such that u(xo) = sup u(X). (Baire theorem; cf. [Loj 1988]) If (X, p) is a metric space, then for every u r= et (X), there exists a sequence 1 C C(X, Ot) such that u,, \ u pointwise on X. Moreover, if u E et (X, R), then the sequence 1 may be chosen in C(X,1R).

Exercise 1.4.16. Let X C iz" be a domain which is starlike with respect to 0. Prove the following properties of the Minkowski function:

(a) hx(tx) = thx(x), t > 0, x E 1k".

(b)X={xEX:hx(x) 0) and (a t = 0, 012 < 0).

In particular, Da,c is complete if Of E QZ+.

(b) Suppose that 0 0 D = int n(a,c)EA Da,c, where A C (R x R. Let

B:= (aEQt":3cER: (a, C)EA). Then, f o r every j E { 1,

... , n). we have:

D(j)=DDf1V #0

Vac-B: aj>0.

Indeed, in view of (a), we only need to observe that if aj > 0 for every a E B.

then DU) = D. In fact, f )(j) C

int n b.7-,c U) = int n Da,c = D. (a,c)EA

(a,c)EA

(c) log Da,c = Ha,c (cf. Remark 1.4.1(e)). (d) If a E Quo x with 0 < s < n. Then

Da,c = ((z1.... , Zn) E Cn : IZI

lal

... Izslas
1, and a (u) 0 Da,c for 0 < u < 1.

s>Iandbt...bs96 0.Puta(t):=(tbt,.. ,tbq,bq+t,.. ,b"),t>0.Then limt.t a(t) = b, a(t) E C"(a), a(t) E Da,c for 0 < t < 1, and a(t) 0 Da,c fort > 1. 1 < s < n - l and bt bs = bs+t . b, = 0. We may assume that bt .. bk 96 0, bk+l = ... = bs = 0(0 < k < s-1), bs+t ... be # 0, be+ t = ... = bn = 0

(s+1£ 0.

Then limt,,,.o a(t, u) = b, a(t, u) E C"(a), a(t, u) E Da,c if t 0 be such that l aaxa 1 < C, l as ya 1 < C.

aEZ

.

Then for every t E [0, 11, we have

laa(xlyj-')" ...

(x'Y1-t)'"1 < Iaaxalnlaayal i-t < C.

a E Z+.

Example 1.5.17. There exists a power series S such that DS = lBn C C". We will see later in Proposition 1.11.11 that, using some Baire category argument, one can prove that there exist many power series S with Vs = 8,,. Here the problem is to find a concrete one.

Indeed, let It,, t2....) C aB,, be an arbitrary countable set which is dense in aBn (EXERCISE: find such a set). Define

M!

S

z".

V! V!

vE(Z"},)

Notice that S is obtained from the series n

00

(Z+tk)k =Mf k=I

k,j)k

_

zj

k=I j=1

k!

00

vkZ"

k=1 vEZ+: Jvl=k

To prove that !B C DS, observe that Ivl! l", z" < (R(z). R( J"I)>J"1 v!

IIzII1"1.

Z E IBn, v E (Z )..

Since Vs is fat (Remark 1.3.5 (d)), we only need to show that VS C i,,. Suppose that Vs \ lB 0. Then there exist ko and t > 1 such that a := ttko E DS. Let C > 0 be such that

v!

"

V!

av I< C.

vE

Put N(k) :_ #{v E Z. : lv1 = k} _ (k+"). Then i

I( k,tko)k1 < vEZ+: JvJ=k

R(tk)"R(tko)" _< N(k)t k, k E N.

Hence

1 = limsupI( k,tko)1 < lim (N(k)C)Ilk1 = 1 < 1; k--+oo

a contradiction.

t

t

1.6. Domains of convergence of Laurent series

41

Exercise* 1.5.18. Given a complex norm N : C" R+ such that N(z) = N(R(z)), z E C", decide whether there exists a power series S such that DS =

{zEC":N(z)o 30E(o,l) : IIaazaIIU < Cola[, a E E(S)}, aEU -open

where IIwIIA

Sup( Irp(z)I

: z E A}. It is clear that int B D B1 = B2 D B3 C

into = D C intB. 2371W is,

sup { FaEA Iaazal : A C E(S), #A < +oo} < +oo. Observe that. by Proposi-

tion 1.2.10, es := (z E C"(E) : the series FaEE(s) aaza is absolutely summable).

'if E(S) = 0, then we put S =e:=C". 25 Proposition 1.6.5 (d) will show that Ds is connected and, therefore. DS is really a domain.

42


Lemma 1.6.2. Let K C C" be a Reinhardt compact set and let r > 0. Put

K(r) := U F(a. r) aEK

and observe that K(r) is also a Reinhardt compact (EXERCISE). Then there exists a 9 E (0, 1) such that for every a E IR" with Kt"l C C"(a) we have

max lzal
0

rj

(Izjle)aj

j:aj 0 be such that pr) C U. Now, we apply Lemma 1.6.2 with K := Or). It remains to show that int B C B 1. Fix an a = (a 1, ... , a") E int B and small

EE(0,1).For a=o,,)E{-1, 1)" define

b(a) = (b1(a)..... b" (a)), where

(1 +s)aj

bj(a)

ifaj # 0andaj = 1,

(I -e)aj ifaj 96 Oand aj = -1.

le

j = 1,...,n.

ifaj = 0,

Taking sufficiently small e E (0, 1), we may assume that b(a) E B for any a. Let

C > 0 be such that laa(b(a))al < C, a E E(S), a E {-1, 1}". Put U(a) Ul (a) x . . . x U. (a), where

ifaj # 0andaj = 1, Uj(a):_ C\K(1-s)lajl) ifaj 96 0andaj =-1, K((1 +s)laj1) K(s)

if a j = 0,

j = 1,...,n.

43


Observe that U := n,,E(-1,1}" U(a) is a neighborhood of a. We will show that

IaazaI < C, z E U, a E E(S). Take such z and a = (a1....,a^) and let if aj < 0 a^) with al := I -11 if aj < or 0 ' Then (aazuI Iaa(b(a))aI C. 0 Remark 1.6.4. Notice that in contrast to the case of power series, the domain of convergence of a Laurent series need not be fat. For example, if S = E= I Z

then Vs = C#. Proposition 1.6.5. Assume Ds 0 0. Then: (a) The series S is locally normally summable in DS. In particular, the function

f(z)

Z E DS,

aaza, aEZ"

is well defined and continuous. (b) as =

1

(27ri)^

I

or(r)

f (o a+1 d , a E 71", r E DS n

where ao 1P(a, r) := 8K(a 1, rl) x ... x dK(an , r"), 1 := (1.... ,1) E QJ", and

I

in

J[O.2")"

0P(r)

p(r - e'9) rie'Bl ... rne'e"dA^(9).

Hence,

laa l
0 and 0 < 0 < 1 such that Ilaazalluauub < C81al,

a E Z"

(EXERCISE).

Define {(e'B'IziI1-'Iwll......

U:= (Z1.

.

Zn) E Ua, (w1.

e`B"Iz"I1-'Iwni1)

,

w") E Ub, (91,

,

9n) E QZ", t E [O, 1]} C C",.

One can easily check that U is open and

IlaazaIlu < C9"°',

a E Z".

Consequently. U C Ds. Since [x, y] C log U. we conclude that DS is log-convex.

0 Proposition 1.6.6. Let a E (Qt")., C E Qt, r E Quo be such that ra = e`. Then the elementary Reinhardt domain Da,c is the domain of convergence of the Laurent series

S= VEZn

N( r )Z

where

N(v) :_ #{k E Z+ : [kaJ = v},

Lkai :_ (Lka1J,..., [ka"J) E Z".

Observe that:

S is obtained by grouping terms in the series

k

Itkoj =o_rL_«


45

Cn(Lka j) = C"(a), k E Z+; in particular, E(S) = C"(a). If a E ((R

then S is a power series.

If a E Z", then S = k o

r

zk"

Proof. We may assume that a E IRn. Moreover, using the biholomorphism C" a (z t , ... , z") H (z1 / r t , ... , z" / r") we may reduce the proof to the case where

rt =

= r" = 1 (c = 0). Notice that hm

kaj = a. k

Hence the classical Cauchy criterion implies that the series E000 z lkal is absolute-

ly convergent in Da. Using Theorem 1.2.7, we conclude that D. C int Cs = Vs. Conversely, let U C= DS be an arbitrary Reinhardt domain. By Lemma 1.6.3, there exist C > 0, 0 E (0, 1) such that N(v)IzvI < CONvl,

Z E U, V E Z".

Therefore, IZLka1/kI

(N(LkaJ)IzLkaJl)llk < (C9IlkalI)1'k,

ZEU,kEW.

Letting k --* +oo we get Iz"I < 01"I < 1, z E U, and, consequently, U C D.

0

Proposition 1.6.7. Let Sj = Eaez" aaz" be a Laurent series, j = 1, ... , m,

n vs.

such that Ds, fl

all For A = (A t...... Xm) E Cm, define

(),,a' +...+Ama'")z".

S(X) = AlS1 +...+ilmSm aeZ"

Then there exists a set C C C" such that (*) C is the union of a countable family of complex (m - 1)-dimensional vector subspaces of Cm and

Ds, n . n Dsm

(L)

Ds(x)

(R)

int Ds, n . n int Dsm , A ECM \ C.

In particular, if Dsj is fat (e.g. Sj is a power series - cf. Remark 1.3.5 (d)),

j = 1, ... , m, then DS(,k) = DS, n ... n Ds,",

A E cm \ C.

2?Note that:

By Proposition 1.6.5 (d), Ds, n ... n Ds,, is log-convex and, consequently, it is a domain.

IfSj is a powerseries with Vs, 96 0, j = 1,...,m,then obviously 0 E

o 0.

46


Proof. First observe that the inclusion (L) holds for every A E C'".

To prove that there exists a set C C Cm with (*) such that (R) is true for A E Cm \ C, it suffices to show that there exists a set C with (*) such that Ds(x)nQ2."CDs, n...nDsm.

AEC"'\C.

or equivalently,

Q!"\(Ds, n...nDsm)CC"\DS(A), AEC'"\C. n Ds,,,) the vector space

We only need to show that for every b E C; \ (Ds, n

V(b):={AECm:bEDS(x)} has dimension < m - 1. To prove that dim V(b) < m - 1, suppose that for a b E C; \ (Ds, n ... n Dsm) there exist A 1, ... , A' E V (b) such that the matrix P := [At] is non-singular. Let U C= C; n Ds(A,) n ... n DS(Am) be a Reinhardt neighborhood of b. By Remark 1.3.5 (d), there exist C > 0, 8 E (0, 1) such that JAi(z)I < C91«I, where

zEU, aEZ", j =1,...,m. Hence, by the Cramer formulas, we have

saza = qj AI (z) + ... + q/Am(z),

j = 1...

, m,

where Q = [q j ] := P-1. Consequently, there exists a C' > 0 such that Ia,J,,zaI < C'91a1.

which implies that b E Ds, n

Z E U. a E Z". j = l.... , m. n Dsm: a contradiction.

From Propositions 1.6.6 and 1.6.7 one immediately obtains the following

Corollary 1.6.8. For any aj E (Qt"). (resp. (R+),), cj E It, j = L.- M, m, there exists a Laurent (resp. power) series whose domain of convergence coincides with Da, c, n ... n Dam,cm.

Exercise 1.6.9. Find (effectively) a power series whose domain of convergence equals ((zt,z2) E l)2 : 21z1z21 < 1}.

47

1.7. Holomorphic functions

1.7 Holomorphic functions Definition 1.7.1. Let S2 C C" be open. A continuous mapping f : S2 --> C' is holomorphic on S2 (f E O(S2, C')) if f is separately holomorphic, i.e. for any point a = (a 1, ... , a") E S2 and for any k E { l , ... , n), the mapping

X H f(al....,ak-l,A.ak+1,....an) is holomorphic near ak; equivalently, the complex partial derivatives

az (z). j =1,.. .m.k=1. ... n, Zk exist at any point z E S2. Notice that in fact the continuity off follows from the separate holomorphy - cf. Theorem 1.7.13. Put O(S2) := 0(92, C) = the space of all holomorphic functions on S2. Functions holomorphic on C" are called entire holomorphic functions.

Exercise 1.7.2. (a) 0 (92) is a complex algebra.

(b) Let a = (a'. a") E S2 C Ck X C"-k, S2' := {Z' E Ck : (Z', a") E 42}. If

f r= 0(Q),then f(.,a") E 0(S2'). (c) Every polynomial of n complex variables is an entire function, i.e. P(C") C O (C").

Proposition 1.7.3 (Cauchy integral formula). If f E O(IP(a, r)) fl e(i'(a, r)) with

a=(al.....an)EC"and r=(r1....,rn)EQt%,then

f( Z) =

1

(2iri)n iK(ai,rj)( 1

(21ri)"

f

UK(an,/n) (S1 - Z1) ...

yl (L) d .

Zn)

d

d6

y,

Z = (z1..... zn) E IP(a, r).

(1.7.1)

0p(a,r) S - Z

Notice that f o r z = (z1, ... , zn) E IP(a. r), the function

aK(a1, r1) x ... x 8K(an. rn) 9 k; 1, ... , W F+

f(Sl.... , W (6 - ZI)...(Sn - Zn)

is continuous and, therefore, by the Fubini theorem, the above integral is independent of the order of integration.

Proof. We apply induction on n. For n = I the result reduces to the classical Cauchy integral formula (cf. [Con 1973], Chapter IV, Theorem 5.4).

n - 1 M> n: We may assume that a = 0. Fix a z = (z', z") E lP(r') x K(rn) (r = (r', r")). We have

f(z) = f(z'.Z.) =

I (21ri)"-1

Iop(r')M", -Z.) Z'

dr'.

(1.7.2)

48


Observe that fl(, ) E O(K(rn)) fl e(K(rn)) for any ' E BoIP(r'). Then

Indeed, fix a t;' E BoIP(r') and let IP(r')

0(K(rn)) -3 uniformly on K(rn). Hence, by the Weierstrass theorem (cf. [Con 1973], Ch. VII,

Theorem 2.1), f(;', - ) E O(K(rn)). Consequently, by the classical Cauchy formula,

MI. Zn) =

r .f}}( W 27ri JaK(rn) yn - Z. 1

dn.

which together with (1.7.2) gives (1.7.1). Exercise 1.7.4 (Cauchy integral formula). Observe that the following slightly generalized Cauchy integral formula is true (with the same proof). Let Dj C C be a bounded domain whose boundary is a finite union of piecewise et Jordan curves with positive orientation with respect to Dj, j = 1, ... , n. Put

let f E0(D)f1C(D).Then

f(z) =

l

f

.. JBD, (S1 - zt)...(Sn - Zn)

n,

Z = (zt,...,Zn) E D. Exercise 1.7.5. Let T be the Hartogs triangle (Remark 1.5.11(c)) and let f E (9(T) fl e(T ), f (z, w) := z2/w. Prove that f is not a uniform limit of a sequence of functions fk E O(Dk), where Dk is a neighborhood of T, k = 1, 2.... . Compare this result with the theorem of Mergelyan in classical one-variable complex analysis (cf. [Rud 1974], Chapter 20). For more information see [Bed-For 1978].

Theorem 1.7.6. Let Sl C C" be open and let f E O(Q). Then: f has all complex derivatives in Q. For any point a E Sl and a polydisc DP(a, r) C= 12 (r = (r1..... rn)), we have

a! D«f(z) _ (2

ooP(a,r) ( - z)«

zEIP

r). a E Z+,

the Taylor series Ta f is locally uniformly summable in P(a, r), and

f(z) = Taf(z),

Z E IP (a. r),

d(Taf)>ds2(a):=sup{r>O:IP(a.r)CSl}, aEQ. For a function g : Sl - C the following conditions are equivalent:

49


(i) g E 0(Q); (ii) for every point a E S2 there exist a power series F_aEZ+ aa(z - a)a and a polydisc IP(a, r) C S2 such that the power series is locally uniformly summable in IP(a, r) and

a. (z -a)",

g(z) _

z E IP(a, r).

aEZ+

Proof. We may assume that a = 0 and U>(r) C= S2. Observe that for (C, Z) E aoIP(r) x IP(r),

-

1

-z

Za

" a+1 ' aEZ+

and the series is locally normally summable. Hence, by the Cauchy integral formula (Proposition 1.7.3), we get

f(z) _ (27ri)"

f

op(r)

d

((2tri)" fap(r) la+1

a

z E U(r). It remains to apply Proposition 1.3.12.

Lemma 1.7.7. Let f E O(S2) with IP(a', r') x aoIP(a". r") C S2 (r = (r', r") E IR;o X IR;o , a = (a',a") E Ck x C"-k). Define g(z)

1

(2;ri)"-k

f

f(Z'z) d%,

z = (z', z") E P(a, r).

Then g E 0((P(a, r)).

Proof It is obvious that g is continuous. Let

F(z,

f

(z,z0

z = (z', z")

E IP (a, r). C E a0UP(a",r").

Observe that

ah"

ifj = 1,...,k,

a-z")°J +

ifj =k+1,.. ,n.

In particular, the function IP(a, r) x aolP(a", r") 9 (z. C) N

az zj

(z. C)

(1.7.3)

50


is continuous, j = 1, ... , n. Consequently, ag

azj

(z) =

1

(2lri)"-k

J

aF a0p(a",r") azj

Z E IP(a,r), j =

1,....n.

0

exist.

Exercise 1.7.8. Try to generalize Lemma 1.7.7 and find "optimal" assumptions for a continuous function f : IP(a', r') x ao IP(a", r") C under which the function g given by (1.7.3) is holomorphic on IP(a, r).

Exercise 1.7.9. (a) Holomorphic functions are infinitely differentiable in the complex sense.

(b) If f E (9(Q), then D"f E O(12) for arbitrary a E

Z+.

Proposition 1.7.10 (Identity principle). Let f, g e O(D), where D C C" is a domain. Then the following conditions are equivalent:

(i) f = g; (ii) there exists an a E D such that Ta f = Tag; (iii) int{z r= D : f (z) = g(z)} 0 0. Proof. Clearly (i) (ii) * (iii). Since D is connected, to prove the implication (ii) = (i) it is sufficient to note that the set Do :_ {z E D : T, ,f = Tg} is non-empty open and closed in D. 0

Exercise 1.7.11. (a) Let D C C" be a domain such that D fl IR" 0 0. Show that if f E 0(D) is such that f = 0 in D fl IR", then f =_ 0.

(b) Let D C C" be a domain and let G := (2

:

z E D). Assume that

f E O(D x G) is such that f (z, 2) = 0 for z in a neighborhood of a point a E D. Prove that f 0. Proposition 1.7.12. Let f : S2 -> C. The following conditions are equivalent:

(t) f E t9(S2); (ii) f is differentiable in the complex sense at any point of S2; (iii) (Osgood theorem) f is locally bounded and separately holomorphic in S2 (cf. Theorem 1.7.13). Proof. It is clear that (i) 4 (ii) = (iii) (cf. Proposition 1.3.12 and Theorem 1.7.6).

(iii) * (ii): Suppose that If I < C in IP(a, r) C= S2. Then, by the Schwarz lemma (cf. [Con 1973], Chapter VI, Lemma 2.1), we obtain

I f(z) - f(a)I

If(zt, z2.... , zn) - f(at, z2, ... , zn)l + ...

...+If(at,....an-1.zn)-.f(at.....an-t.an)1

r 2C

which shows that f is continuous.

(Izt-all+...+lz"-anl),

zEIP(a.r),

(1.7.4)

0


51

The following result illustrates the essential difference between real and complex analysis.

Theorem* 1.7.13 (Hartogs' theorem on separate holomorphy, cf. [Kra 1992]). Let

S2 C C" be open and let f : S2 -+ C be separately holomorphic, i.e. the partial (z) e x i s t s f o r all z E S 2 and j = 1, ... , n. Then f E 0(Q).

c o m p l e x derivative

Proposition 1.7.12 (ii) implies also

Proposition 1.7.14. The composition of holomorphic mappings is holomorphic.

Proposition 1.7.15. Let D C C" be a Reinhardt domain and let f E O(D). Define

act

(21ri)"

I

p(r)

(+1

aE

", r E D f10t.o.

Then:

(a) For any a E Z", the number as (f, r) is independent of r E D fl Quo. In

particular, we define as = of = aa(f) := aa(f, r). (b) Consequently, D C D1, where D f denotes the domain of convergence of the Laurent series EaEZn aaza; cf. Proposition 1.6.5 (b). (c)

aaza.

P z) =

Z E D.

ac-Z"

(d) If D fl Vj # f, j = 1, ... , n (in particular, if 0 E D, e.g. D is complete), then as = 0 for all a E Z" \ Z+ (cf. Proposition 1.6.5 (c)). Consequently, if 0 E D, then f (z) = To f (z), z E D. Proof. We apply induction on n. For n = 1 the result is well known (cf. [Con 19731, Chapter V). Assume that it is true for n - 1. (a) Since D is connected, it suffices to show that any point a E D has a Reinhardt neighborhood U such that aa(f, r) is independent of r E U fl Qt" >o.

Let U = A"(r-. r+) C D be an arbitrary annulus centered at 028 and let r = (r', r"), s = (s', sn) E U fl (IR ' x QZ>o), a E Z". Write z = (z'. zn) E C"-' X C. 28An (r-, r+) := A(rj, ri) x ... x A(r;, 00.5

r- = (r, , .... r;). r+ = (r1 ..... 0,A(r;-,rj+):=(zEC:r; r , + ) . ).j=1,...,n.

52


Then, using the inductive assumption, we get

as (f s) _

(+1

(21r i )n foPn((s )

d MI. I. w

27ri 8K(sn) \(27ri)"-1 L,fl_I(s')

e

'a'+1

y,l d Sn

d S /1 an+l n

an+1

27r 1 i JaK(sn)

1 fK(sn)aa'(f(

an+1 = aa(f, (r', sn)).

2ni

The same argument with respect to the last variable shows that

aa(f r).

aa(f,

(c) Fix U := A"(r-. r+) C D. By the inductive assumption, using Theorem 1.2.7, for every z = (z', zn) E U C Cn-1 x C. we get:

f(z) _ _

aan(f(Z','))Znn

E ( I Jf 27tt

anEZ

_

aEZn(

1

27ri

=aEZ E ( 217ti JtK(r)

yan1 + 1

aK(rn) Sn

f

(

n

a'EZn-I

aa' (f( , In))

]"+

K(rn)

+dn)zn

Snn

11

L aa(f)za.

aEZn

Corollary 1.7.16 (Cauchy inequalities). If f E O(IP(a, r)) fl e(o'(a. r)), then 0.a

I Daf(a)I

IIf IIaoP(a.r)

a E Z.

Similarly as in the case of one complex variable, the following results are easy consequences of the Cauchy inequalities (ExERclsE).

Proposition 1.7.17 (Liouville theorem). Let f E O(C"), k E Z.t.. Then the following conditions are equivalent:

(i) f is a polynomial of degree < k;

(ii) 3c.Ro>o: If(z)I 2 (with a much more difficult proof).

Theorem* 1.7.27 (Cf. [Nar 1971], p. 86). Let S2 C C" be open and let f = (fl

,..., f.): S2 -+ C" be holomorphic. Then the following conditions are equiv-

alent:

(i) f (S2) is open and f : S2 -+ f (D) is biholomorphic;

(ii) f is injective and Jf(z) 0 0, z E S2; (iii) f is injective. Theorem 1.7.28 (Hurwitz-type theorem). Let S2 C C" be open, a E S2, and let f, fk : S2 -+ C", k E N, be holomorphic mappings with fk -+ f uniformly on S2. Assume that f (a) = 0 and det f'(a) # 0. Then there exist an open neighborhood U C S2 of a and a ko E W such that 0 E fk (U), k> ko. Proof. (The reader is asked to complete details.) First observe that the proof of the inverse mapping theorem (in the real case) implies the following:

Let g : S2 -> C" be a holomorphic mapping with det g(a) 96 0 and let r > 0 be such that

det g'(z) # 0,

Ilg'(z) - g'(a)II < 2 II(g'(a))-t p'

z E B(a, r) C S2.

Then IB(g(a), p) C g(IB(a, r)) with

r

1

2llg'(a)II Using the above remark and the Weierstrass Theorem 1.7.19, we find r, s > 0, and

ko E N such that !B(g(a), s) C g(B(a, r)) for g E { f, fko+t, fko+2, } Since 0 = II f(a) 11 < s, we may assume that II fk (a) II < s fork > ko, which shows that 0 E B(fk (a), s) C f (B(a, r)) fork > ko. -

1.8 Balanced domains Sometimes it is convenient to consider a wider class of domains than complete Reinhardt ones.

56


Definition 1.8.1. We say that a domain D C C" is balanced (complete circular) if Az E D for every z E D and A E ID. Observe that every balanced domain is starlike. Let hD denote the Minkowski function of D (cf. Definition 1.4.14).

Exercise 1.8.2. (a) (Cf. Exercise 1.4.16.) Let D C C" be a balanced domain and let h : C" IR+. Then the following conditions are equivalent: (i) h = hD; (ii) h is upper semicontinuous, D = {z E C" : h(z) < 1), and

h(Az) = IAIh(z),

Z E C", A E C;

(b) Let q : C" -+ IR+ be a C-seminorm (cf. § 1.10) and let B := (z E C" q(z) < 1). Then hB = q. Lemma 1.8.3. Let D C C" be a complete Reinhardt domain.29 Then

hD(A. z) < hD(z),

Z E V. A E D"

(1.8.1)

(in particular, hD(A z) = hD(z), z E C", A E T") and hD is continuous. Consequently, if h : C" -+ IR+ is an upper semicontinuous function such that

h(Az)=IAIh(z),zEC",AEC. h must be continuous.

Proof. The proof of (1.8.1) is left as an EXERCISE. To prove that hD is continuous it suffices to show that hD is lower semicontinuous at any point a E C" such that

. as # 0, . = a" = 0 for some 1 < s < n. Fix a z = (Z 1, ... , z") E C", put m := min{ Izi /ail : j = 1.....s}, and let Ai E ID be such that A1zi /a1 = m, h D (a) > 0. Fix such an a = (a 1, . . . , a.). We may assume that a t

as+l =

j

Then

mhD(a) = hD(ma1,...,mas,0,....0) = hD(Atzl,...,Aszs.Ozs+l,....Ozn) < hD(Z). Consequently,

min{Izi/ail: j = which implies that lim infz..+a hD(z) = hD(a). "In particular. D is balanced.

zEC", 0

1.8. Balanced domains

57

Proposition 1.8.4. Let D C C" be a balanced domain and let f E O(D). Then 00

f(z) = F, Qk(z),

Z E D.

(1.8.2)

k-0 where

Qk(z) :=

I f (k)(0)(z) _

a Daf (0)za,

z E C":

aEZ"}: lal=k

observe that Qk : C" -* C is a homogeneous polynomial of degree k. Moreover, for any compact K C D there exist C > 0 and 0 E (0,1) such that II Qk IIK 5 CBk,

k (=- Z+.

In particular, the series converges locally normally in D.

Proof Take an a E D \ {O}. The function

K(1/hD(a)) 3 A H f(Aa) 30 is holomorphic. Hence 00

f(a) = wa(1) = 00 F, k=0

1gak)(0)

= E Qk(a) k=0

Thus the formula (1.8.2) is true (and the series is pointwise convergent in D). It remains to prove the estimate. Take a compact K C D. Let 8 E (0, 1) be such that

L:={Az:IAI 2. This is one of the main reasons why the theory of several complex variables is not a straightforward generalization of the one-dimensional case. Definition 1.9.3. A set M C C" is called thin if for every point a E M there exist a polydisc (P(a, r) and a function Sp E O(IP(a, r)), V 0- 0, such that M n P(a, r) C

V-10). Remark 1.9.4. (a) If M is thin, then int M = 0. (b) If M is thin and N C M. then N is thin. (c) If MI, M2 are thin, then Mt U M2 is thin. (d) If V E O(D), 9P $ 0, where D C C" is a domain, then 0-1 (0) is thin. In

..., V" are thin. (e) If M is thin, then M x C' is thin.

particular, Yo, Y I,

Lemma 1.9.5. Let 9P E O(i'(r)), Q(0) = 0, 9p $ 0. Then, after a suitable linear change of coordinates, we have cp(O'. - )

0.

Proof. By Theorem 1.7.6, the function 9p may be expanded into a series of homogeneous polynomials 9,(z) = To9,(z) =

E .i=o

D",p(0)za) =

Ec'o

Q, (z),

z E D(r),

i=k

lal=i

with Qk $ 0 (see also Proposition 1.8.4). In particular, the set V := Qk t (0) is thin. Observe that for every X 0 V, JJX 11 = 1, the function

K(r) a A

H

9p(AX )

is not identically zero. Consequently, after a linear change of coordinates L : C"

C" such that L(e") = X, we have (9p o L)(O',z") = 9p(L(z"e")) = 9p(z"X) _ 0 Ox(zn) Exercise 1.9.6. Let co E 0 ((P (r) ), Vk (0) = 0, Spk 0, k E W. Then, after a suitable linear change of coordinates, we have pk (0', ) 14 0, k E W. Hint. Use Baire's theorem.

Proposition 1.9.7. Let D C C" be a domain and let M C D be a thin set. Then the set D \ M is connected. Proof. First observe that it suffices to prove that every point a E D has a convex neighborhood U. C D such that Ua \ M is arcwise connected (cf. Remark 1.5.6 (c)). Indeed, suppose for a moment that this is true and take arbitrary two different points a, b E D \ M. Let y : [0, 11 -+ D be an arbitrary curve with y(0) = a,

60


y(l) = b. For every t E [0, 1] the point y(t) has a convex neighborhood Uy(r) such that UU(t) \ M is connected. One can select a chain of neighborhoods Uy(to) ... , Uy(tt,) 0 = to < ... < tN = 1, Uy(r,-,) n U),(,r) 54 0, i = 1, ... , N. Fix arbitrary points c; E Uy(,,_,) fl Uy(,,) \ M, i = I..... N. Now we connect a with ct in Uy(,0) \ M. Next, we connect cl with c2 in Uy(,,) \ M. etc. Finally, we connect cN with b in Uy(,N) \ M. Fix an a E D. We may assume that a = 0 and that F(r) fl M C rp ' (0), where IP(r) C= D,rp E 0(IP(r)),rp 0- 0. Using Lemma 1.9.5, we easily reduce the situation (EXERCISE) to the case where V (O', ) 0 0, V(0', z") # 0 for 0 < Iz" I < r,,, and

rp(z',z") 0Oforz' E IP(r'),s"
2, be a domain and let M C D be closed in D. Assume that for every a E M there exist an open neighborhood U C D and c1, g°2 E O (U) for which M n U C rp-' (0) n (p2' (0) and rank (

= 2.

(..)

aZk

Z E U. 32

Then everyfunction f r= O(D \ M) extends holomorphically to D.

Proof As in the Riemann theorem, it suffices to extend f locally. Fix an a E M and let U, gyp,, (p2 be as above. We may assume that a = 0 and det 1. aZk

# 0.

(0)Jj.k = 1.2

I

Consider the mapping U

z

H

Then JP(0)

0 and, consequently, by the inverse mapping theorem (Theorem 1.7.26), we may assume (shrinking U if necessary) that 0: U -+ O(U) =: V is biholomorphic. Put

N:={w(=- V: w1 =w2=0), g:= f o(-'Iv\N. 32For example, M C V; n Vk with j 96 k.

62


We only need to extend g to V. The case n = 2 follows directly from the Hartogs extension theorem (Theorem 1.9.1). Thus we may assume that n > 3, U = IP(r) C D. Spy (z) = zj, z E U. j = 1, 2.

Write z = (z', z") E C2 X

Cn-2. For each z" E IP(r"), the function f(.. z") is

holomorphic in IP(r') \ {0'}. By the Hartogs extension theorem, z") f ( , z")(z'), z = (z', z") E IP(r). It remains to prove that f E O(IP(r)). Observe that

f

f

f(z) =

(27ri)2 JaolP(r,)

Z')

(

.

di;.

z = (z',z") E F(r).

0

Hence, by Lemma 1.7.7, f E 0(IP(r)). Now, Corollary 1.9.9 may be extended to the following more general result.

Proposition 1.9.12. Let D C C" be a log-convex Reinhardt domain. Put M D * \ D C Yo. Define Mr to be the set of all a = (a 1, ... , an) E M such that t h e r e e x i s t s e x a c t l y one j E { 1, ... , n } with aj = 0. Let f E 0 (D) be a function of slow growth near M', i.e. every point a E M' has an open neighborhood Ua

such that U0\Y0C Dand (dist(z, 8D))N I f (z)I < C,

Z E Ua \ Yo 33

for some constants 0 < N < 1 and C > 0, which may depend on f and a. Then f extends holomorphically to D* (cf. Corollary 1.11.4).

Proof. Fix a E M', Ua, N, and C as above. We may assume that an = 0, Ua = U' X Un C C;'' x C, and dist(z, 8D) Izn1, z = (z', zn) E Ua, zn 0. Write "0

f(z',zn)=

1k(Z')zn

Z=(Z',Z.)EUa=U'XUn,Z.#0.

k=-o0

By the Cauchy inequalities we get I fk (z')I < C Izn I-N-k. (z', zn) E Ua, Z. # 0. Letting zn -> 0, we conclude that fk - 0 for k < 0. Thus, for every z' E U', the function f (z', - ) extends holomorphically to Un = K(rn ). By the Cauchy integral formula, the extension is given by the formula

f(z) =27ri 1J

8K(s")

Z=(z',zn)EU'xK(s"),0<sn0 dkEN 3voE&1 Vµ,v>vo : gk(fu - fv) < e,

In particular, (fv)v° 1 C F remains a Cauchy sequence in (Y. p'), where p' is the

distance corresponding to a sequence Q' = {qk : k = 1, 2, ...) with T (Q) T(Q').

_

(m)* The topology T (Q) is metrizable if there exists an equivalent countable family of seminorms Q0 such that ngEQO q-1 (0) = (0). Definition 1.10.2. Let F be a complex vector space endowed with the topology gen-

erated by a countable family of seminorms Q = {q1, q2.... } with nk=1qk 1(0) _

(0). We say that F is a Frlchet space if the metric space (F, pQ) is complete (cf. Exercise 1.10.1(k,t)).

Definition 1.10.3. Let F' be a Frtschet space with the topology T = T (Q). A set A C F is said to be bounded if the set q(A) C (R+ is bounded for any q E Q ss The following property of Frtschet spaces will play an important role in the sequel.

Theorem 1.10.4 (Banach theorem, cf. [Gof-Ped 1965], § 5.8). Let F1, F2 be Frechet spaces and let L : T1 --t F2 be an injective continuous linear mapping. Then either L is surjective (and then L-1 is also continuous) or the image L(F1) is of the first Baire category in T2.36 We will be only interested in special Frechet spaces F C 0 (S2), where S2 C C" is open (cf. Chapter 3). "Notice that this property is independent of the family Q with 7 = 7(Q). 36A subset A of a topological space X is said to be of the first Bake category if A = Uk°° I Ak. where each set Ak is nowhere dense, i.e. int Ak = 0.

66


Definition 1.10.5. Let F C O(Q) be a vector subspace endowed with a Frdchet space topology 7 = T (Q). We say that F is a natural Frechet space if for any

sequence (fk ). i C F and fo E F, if fk -> fo in the sense of T, then fk - fo locally uniformly in S2

(1.10.2)

(see also (1.10.4)). In the case where F is a Banach (resp. Hilbert) space, we say that F is a natural Banach (resp. Hilbert) space.

Remark 1.10.6. (a) Let F C 0(0) be a vector subspace endowed with a Frechet

space topology 7 = T(Q). Then F is a natural Frechet space if if for any sequence (fk) 1 C F and fo E F, if fk -> fo in the sense of T, then fk - fo pointwise in Q.

(1.10.3)

Indeed, suppose that (1.10.3) is satisfied. Let T' = T (Q') denote the topology

generated by the family Q' := Q U Q", where Q" stands for the family of all seminorms of the form

T 31H If Ilk := supf. K cf. Example 1. 10.7 (a).

Kc=S2:

In other words, fk -> fo in T' if fk -- fo in T and

fk --> fo locally uniformly in S2. Condition (1.10.3) guarantees that (F, T') is a Frdchet space. The identity operator id: (F, T') -+ (Y, 7) is obviously a continuous bijection. Now, the Banach Theorem 1.10.4 implies that its inverse is continuous, which gives (1.10.2).

(b) E Surprisingly, we do not know an example of a Frdchet space ( , 7) with F C O(Q) such that F is not natural.'] Many classical spaces of holomorphic functions have structures of natural Frdchet spaces.

Example 1.10.7 (Natural Frdchet spaces). The reader is asked to complete all details.

(a) The whole space 0(0) endowed with the topology rn of locally uniform convergence is a natural Frdchet space (cf. Theorem 1.7.19). More precisely, rn 7'(Q). where Q is the following family of seminorms

(9(Q)9I ->II!IIK:=supIf1,

KCcS2.

K

Observe that Q is equivalent to every family (II IIK; )°O_1. where (Kf))__1 is an arbitrary sequence of compact subsets of S2 with KK C int K1+,, Ujt1 Kj = S2. Notice that condition (1.10.2) means that the inclusion operator

(F,T) -> (O (S2). rn) is continuous.

(1.10.4)

1.10. Natural Frechet spaces

67

(b) The space 3l O0 (Q) of all bounded holomorphicfunctions on S2 endowed with

the topology of uniform convergence (i.e. the topology induced by the supremum norm II IIn) is a natural Banach space. Notice that in fact 3eO°(D) is a Banach algebra.

(c) The space LP (S2) := O(S2) fl LP(S2) of all p-integrable holomorphic functions on S2 endowed with the LP-topology (i.e. the topology induced by the LP-norm II IILP(n)) is a natural Banach space, where LP(S2) is taken w.r.t. the Lebesgue measure A2n in C" (1 < p < +oo). Obviously. L° (Q) = 3f°O(Q). To prove that LP(S2) is a natural Banach space we only need to show that the topology induced by LP(S2) on Lh (S2) is stronger than the topology of locally uniform convergence. By Lemma 1.7.22, we get IIf IIK

(nr2)"

f(r)

IfI dA2n,

f E 0(.Q), K Cc S2. 0 < r < dn(K). (1.10.5)

Hence, by the Holder inequality,

Al/9(K(r)) IIf IIK
0. Observe that t4(Q) is a closed subalgebra of Moreover, if S2 is bounded, then n, (Q) = 3Q00(52). (f) The space

If E 0(Q) : I18kf Il a < +oo)

0(k)(S2, 8)

(k > 0)

of all S-tempered holomorphic functions on 92 of degree < k with the norm 0(k)(S2,

8) 9 f H 118kf Ila

is a natural Banach space, where the weight 8: S2 -+ (0, 1] is an arbitrary continuous

function. Note that 0(0)(S2.8) = 3[°O(S2) and 0(k)(S2, 8) C O(k')(S2, 8), k < V. From a certain point of view, the most important is the weight function 8 = 89 given by the formula

8a(z) := min P9(z).

I

} ,

Z E 12,

i + IIz112

where pa(a) := sup{r > 0 : 8(a, r) C Q}, a E $2, denotes the Euclidean distance function to 852; pc" _ +oo, Scn =

=: 8o. Functions from the 1+II 112

space 0(k)(S2) := 0(k) (92, 89) are called holomorphic functions with polynomial growth of degree < k. By the Liouville theorem, Proposition 1.7.17, the space (9(k) (C") coincides with the space ,i lk j (C") of all complex polynomials of degree

< [kJ. (g) Let S: S2 -- (0, 1] be a function such that:

8_ 1, f E O(n).

In particular, Lh (S2) C O(")(S2, 8).

Indeed, fix k, p, f, and a E S2. By (1.10.5) with K := (a) and r da (a), we get Av(1P (aI r)) 1f(a)1
8(a) - Ilz -all ?

28(a), z E IP(a, r).

Consequently, 8(k+2n)lp(a)I

f(a) I

f

O

0(0+)(S2, Sn)

Note that:

Lh(S2) = 3f°O(S2) iff A2i(S2) < +oo.

jeO0(Q) C 0(0+)(52.8) and the inclusion MOO(S2) , 0(0+)(S2,8) is continuous.

(i) Let A C Zn , 0 E A, and let

be a family of natural Fr6chet spaces

in O(R). Let T (Qa) denote the topology of F. generated by a family Q. of seminorms, a E A. Define

F=FA:={f EFo:Daf EFa, aEA}. "Observe that Lh (SJ) = L)(D) n it-(n). In fact, if f e LL (a) n 3[°°(S?), then for every

1 dD (a) there exists an f E 8 such that d (Ta f) < r. Note that the whole space C" is anS-domain of holomorphy for any 0 $C o (C'.). If 8 = (f }, then we say that D is a domain of existence off . If 8 = O(D), then we say that D is a domain of holomorphy. Suppose that we have assigned to each domain D a family F(D) C O(D) (e.g. D -* X°°(D), D Lh (D)). Then, instead of saying that D is an F(D)domain of holomorphy, we shortly say that D is an F-domain of holomorphy (e.g. MI-domain of holomorphy, Lh -domain of holomorphy).

Obviously, if D is an 8-domain of holomorphy, then D is a 7-domain of holomorphy for any family T with 8 C T C O(D). In particular, any $-domain of holomorphy is a domain of holomorphy.

Proposition 1.11.2. Let D C C" be a domain and let 0 0 8 C O(D). Then D is an 8-domain of holomorphy i¶

(*) there are no domains Do, D C C" with 0 t Do C D fl D, D Q D, such that for each f E 8 there exists an f E O(D) with f = f on Do.39 Proof. Suppose that (*) is satisfied, but D is not an 8-domain of holomorphy. Then

there exist a E D and r > dD (a) =: ro such that d (Ta f) > r for any f E 8. Put Do := P(a. ro), D := IP(a, r), and f (z) := Ta f(z), z E D; a contradiction. Conversely, suppose that D is an 8-domain of holomorphy, but (*) is not fulfilled. Let Do, D be as in (*). By the identity principle (Proposition 1.7.10) we may assume that Do is a connected component of D fl D. Then there exists an a E Do such that dD (a) < dD (a) (EXERCISE). Consequently, for any f E 8 we get d (Ta f) = d (Ta f) ? dD (a) > dD (a); a contradiction. 0 38Recall that d(TQ f) a do (a), a E D, f e O(D) (Theorem 1.7.6). 39Notice that if D is fat, then D V! D.

1.11. Domains of holomorphy

73

Figure 1.11.1. For each f E 8 there exists an f E O(D) with f = f on Do.

Remark 1.11.3. (a) Let Do, D be as in Proposition 1.11.2 M. First observe that f is uniquely determined by f . Put S := (f : f E 8}. Then the extension operator

$-3 fHfEB is bijective. Observe that:

(µf) = µ f , provided that f, µf E 8 (µ E C), f + g = f + g, provided that f, g, f + g E 8, D"f = DQ1 f , provided that f, Do f E 8 (a E Z J. In particular, if 8 is a vector space (resp. an algebra), then so is 8 and the above extension operator is an algebraic isomorphism,

if S is stable under differentiation (i.e. f E 8 ..... E 8), then so is 8. (b) Let Do. D be as in Proposition 1. 11.2 (*). Observe that we do not require

f = f on D fl D but only on Do. It may happen that f $ f on the whole of D fl D. Take for example D := C \ (-oo, 0J and 8 := (Log), where Log stands for the principal branch of the logarithm (Log 1 = 0). Put D := {z E C : Re z < 0),

Do := Lz E C :_Re z < 0, Im z > 0). Then the function Log extends to an f E O(D) with f = Log on Do but not on D n D (EXERCISE), which leads to a non-univalent extension.

It is natural to ask whether such an example is possible in the case where 8 contains more functions, in particular, 8 = O(D). Below we will see that for n = 1 such an example with 8 = O(D) is impossible. However, for n > 2 there are such situations (cf. [Jar-Pfl 2000], p. 1).


74

Theorem 1.12.4 will show that if D is a Reinhardt domain and 8 is invariant

under rotations of variables, then f = f on the whole of D n D. Thus, in the category of Reinhardt domains the above phenomena do not occur. (c) Any domain D C t is an 8-domain of holomorphy, where

8:={DE)z" z-a :a¢D}. (d) Any fat domain D

C 1 is an 8-domain of holomorphy, where

8:={D3zH

a

:a

D

In particular, any fat domain D C C' is an 3eO0(D) n 0(b)-domain of holomorphy.40

(e) Let Di be an 8i -domain of holomorphy, i E 1, and let D be a connected li E/ Di. Then D is an 8-domain of holomorphy with

component of int I

8:=U8:ID iEl

In particular, if Di is a domain of holomorphy for every i E I. then D is a domain of holomorphy. Indeed.

dD(a)=inf(dD,(a):i E1)=inf(inf{d(Taf): f E8i):i EI} = inf{d(T0 f) : f E 8},

a E D.

(f) Let Dj C C"i be an 8j -domain of holomorphy, j = 1.... , N. Then D := Dt x x DN is an 8-domain of holomorphy with

8 := {f oprD.: f E 8j. j = 1... .N). where prD. : DI x x DN -+ Dj is the standard projection, j = 1..... N. In particular, if Dj C C J is a domain of holomorphy, j = 1..... N, then D t x .. x DN is a domain of holomorphy. Indeed,

dD(a) = min{dD, (aj) : j = 1,

.... N)

=nun{inf{d(Ta;f): f E8j): j =1,....N} = inf{d(T(a1....,av)f) :.f E 8), 40Recall that O(D) := Unt(IcC O(U)I0. s open

a = (at...,aN) E D.


75

(g) Let D be a domain of holomorphy, let f = (f1, ... , fN) E O(D, CN), and let G be a connected component of the set

f-I(IDN)={ZED:Ifi(z)I inf{d(Ta f) : f E O(G)) > dG(a). If r := dG (a) < dD (a), then there exists a point b E aG fl ao'(a, r). Consequently, there exists a j E { 1, ... , N) with I f j (b) I = 1. Hence the function g : = 1 /(f j fj (b)) is holomorphic in G and d (Tag) = r.

(h) Let D be a domain of holomorphy and let fo E O(D), fo $ 0. Then G := D \ fo 1(0) is a domain of holomorphy.41 In particular, if D C C" is a domain of holomorphy, then D \ (V, U . . . U Y,-k) is a domain of holomorphy for

anyI J in the sense of T (Q) iff locally uniformly on D and f -+ f in the sense of T (Q). Observe that 8 is a Frdchet space. Indeed, if in 8 and

v_ I is a Cauchy sequence in 8, then is a Cauchy sequence is a Cauchy sequence in O(D) in the topology of locally uniform

convergence. Hence there exist functions fo E 8 and go E O(D) such that fv - fo in 8 and f, -> go locally uniformly in D. Since 8 is a natural Frdchet space, we conclude that fv -p fo locally uniformly on D. In particular, Jo = go on Do. Thus go = Jo and, finally, f. -> fo in 8. _ The mapping $ 3 f -> J E 8 is obviously continuous. Since 8 is a Frdchet space, the Banach theorem (Theorem 1.10.4) implies that the above operator is a topological isomorphism, i.e. for each compact K C D there exist a finite set I C Q and c > 0 such that

IIfIIK

cmaxl(f), f E 8.

In particular, if 8 is a natural Banach space with a norm II I18, then for every compact k c D there exists a constant c > 0 such that

IIfIIK.5 cIIf118.

f E8.

In the special case where 8 is a natural Banach algebra, we get more. Namely, 11 f 11 n_ 11 f 118, f E 8 (EXERCISE - cf. Example 1. 10.7 (k)).

(o) Let 8, Do, D be as above. By virtue of (n), if 8 is a closed subspace of O_(D) (in the topology of locally uniform convergence in D), then for each compact

K C D there exist a compact K C D and a constant c > 0 such that IIf IIK < C11f lltc,

f E 8.


77

(p) Let 8, Do, D be as above. In the special case, if 8 is a closed subalgebra of 0(D), then for each compact k C D there exists a compact K C D such that IIf IIg < AA K-

gE8

(EXERCISE).

Proposition 1.9.12 implies the following result.

Corollary 1.11.4. (a) If a Reinhardt domain D C C" is an XZ(D)-domain of holomorphy, then D is fat.

(b) If a Reinhardt domain D C C" is an O(N)-domain of holomorphy with 0 < N < 1, then D is fat. Remark 1.11.5. Let Ta :_ {(21, 22) E ID x [D: IzI 1° < Iz2I}, a = P19 E Q>o. (a) First observe that Ta is an ,e°°-domain of holomorphy. Although it follows from the general results (Theorem 3.4.1), here we give a direct elementary proof. Suppose that Do, D are as in Proposition 1.11.2 (*) with D = Ta and 8 = X°O(T0). Since 0) x I) is obviously an X°0-domain of holomorphy, we conclude

that D C ID x [D. Let f(z) := zi /z2, z = (21,22) E Ta. Then f E 3e°D(Ta) and, therefore, there exists an f E O (D) such that f = f on Do and 11f 11r) < 11f 11 T,

I (Remark 1. 11.3 (k)). Consequently, 4f(z) = zi , z = (z1, z2) E D.

Let b = (b1, b2) E aTa fl D. If b2 # 0, then f (z) = zf /zZ for z = (zl, z2) in an open neighborhood U C D of b. Then Izi /z2I < 1 in U and, by the maximum principle, U C Ta; a contradiction. If b = 0, then f is holomorphic in a small polydisc P(r) C D, f (z) = E k=o aj.kzi z2, z = (21, z2) E IP(r). Consequently, =o aj kz1 z2+y = z1, (zi, z2) E IP(r), which is impossible. (b) The mapping C x C* D (21,22) H (21,21/22) E C x C maps biholomorphically the Hartogs triangle T onto ID x 1D*. Observe that O) x 0)* is not an Y'*-domain of holomorphy. In particular, the notion of an MI-domain of holomorphy is not invariant under biholomorphic mappings.

Proposition 1.11.6. Let D = Ds 0 0 be the domain of convergence of a Laurent series S =

aaza.

aEZ"

Then D is a domain of holomorphy.

Proof. Suppose that D, Do are as in Proposition 1.11.2(*) with 8 = 0(D).

Put fa(z) := aaza, z E D, with a r= E(S).. Observe that b C C"(E(S)) and fa(z) = aaza, Z E D. Indeed, by Remark 1.11.3 (c), (f), C" (E (S)) is a domain of holomorphy. Ob-

viously, O(C"(E(S)))ID C O(D). Hence D C C"(E(S)). To get a contradiction we are going to show that D C DS = D. Suppose that there exists an a E 5 \ D and let K := I3(a, r) C D. By Remark 1. 11.3 (p)


78

with 8 = 0 (D), there exists a compact K C D such that II! 11,k < III 11K for any f E 0(D). By Lemma 1.6.3, there exist C > 0 and 0 E (0, 1) such that

IlaazallK < CO'. a E E(S). Consequently,

Ilaazall,k < C8

,

a E E(S).

Thus int k C Ds = D; a contradiction.

Proposition 1.11.7. For any a E (lR"). and c r= IR. the elementary Reinhardt domain

Da,, = {z E C"(a) : Izal < e`} is a domain of holomorphy.

Proof. Use Propositions 1.6.6 and 1.11.6.

0

Remark 1.11.8. (a) Observe that it is much easier to prove that Da,c is locally a domain of holomorphy, i.e. every a = (at,..., an) E Mc,,, has an open neighborhood U such that each connected component of U n Da,c is a domain of holomorphy.

Indeed, if a E c" n 8D.,,, then let U := P(a, r) C C and let f (z) := ft(zt)... fn(zn),z = (zt,...,zn) E U,wherefj E O(K(aj. r)) isaholomorphic branch of the aj-power, j = 1. ... , n. Then U n Da,c = {z E U : I f(Z) I < e`} and we may apply Remark 1.11.3 (g).

If a E Vo n aDa,c, then let U := P(a, r) C C" be arbitrary. Suppose that a connected component D of u n Da,c is not a domain of holomorphy. Let Do, D be as in Proposition 1.11.2 (*) with 8 = 0(D). Since U is a domain of holomorphy,

we have D C U. We may assume that Do is a connected component of D n D. The first part of the proof shows that 8Do n D C Yo. Thus, it suffices to show that for any point b E Yo n aDa,c there exists a function f E 0 (Da,,) which cannot be continued through b. We may assume that

at,....as >0,as+,.....an o and let I' be the hyperbolic .. A), q = (q 1, , qk)

Take p = (PI .

segment between p and q, {(Pi_tgi'

..' Pk-'q') : t E [0, 1]}.

We want to show that F x {b} C D. Let

U, =U1 x...xUP Xu"_s(s),

Uq=Uq x...xUq xFn_s(E)CD

be Reinhardt neighborhoods of (p, b) and (q, b), respectively. Then

Consequently, the relative completeness of D implies that 1' x {b} c D.

0

Remark 1.11.17. Notice that for a general domain of holomorphy D C C", the projection prck (D) need not be a domain of holomorphy - cf. e.g. [Pfl1978], [Kas 1980], [Shc 1982], [Joi 2000].

Proposition 1.11.18. Let D C C" be a Reinhardt domain. Then the following conditions are equivalent:

(i) D is a domain of holomorphy; (ii)

for any a = (at....,a") E lR;o, a = (al...., of,) E (IR")., and fl _ (01..... On) E {el..... en) C Z" , D..a,B

the set

{(A, s) E (C(a1) n ... n C(an)) X C : (aIIAI°" I

i.I '

.anJXI°`"IlAl'O") E D}

is a Reinhardt domain of holomorphy (provided that Da,a,# # 0). Notice the special role played by two-dimensional Reinhardt domains (cf. Proposition 1.15.9).

Proof. Define fia,Q,p : (C(al) n ... n C(a")) X C - C", 'a,a,.B(A,A) :=

I-1 IanIAI"").


83

First observe that Da,a,p _ Oa- 1,0 (D) is a Reinhardt open set and a

log Da,a,j = {(t, u) E

R2

: loga + to + up E log D).

It is clear that D is log-convex if each Da,a,p is log-convex. It remains to discuss the relative completeness. First assume that each Dc,a,p is relatively complete. We will prove that D is weakly relatively complete. Suppose that D n Vj # 0. We may assume that j = n. Fix a b = (b', 0) E D n ([t;o1 x {0}).

Take a point c = (c'. cn) E D n IR+, c' $ b', cn > 0. We want to prove that (c', 0) E D. We may assume that c' = (0, ... , 0, cs+1.... , cn_t) with c,+,,.... ci_1 > 0 for some 0 < s < n - 1.

S

First, consider the case s = 0. Define

a

(b'.cn),

a:=(a',0)EIRn,

aj

.=

log(cj/bj) log

j = 1,.. n -

#:=en.

= b and 40.,.,p (2, 1) = c, we conclude that (1, 0), (2, 1) E Da,a,p. Thus (2, 0) E Da,a,p and, consequently, (c', 0) _ ba,a,p (2, 0) E D. Now, let s > 0 and suppose that (c', 0) > D. Define Since

a:= ( 1 , .

. .

, 1,cs+l,.. ,cn),

S

a :_ (1. ..,1,0, ..,0),

P := en.

S

Then Pa,a,p (0, 1) = C E D, (0, 0) = (c', 0) ¢ D. Thus (0. 1) E Da,a,p and (0, 0) $ Da,a,O. By the first part of the proof we know that Pa,a,p (e, 0) = (s , s, cs+ 1, ... , cn_ 1, 0) E D for 0 < s o . 8. If D is an 8-domain of holomorphy, then

n

D=int

(zEC"(a):Iafz'I 1 such that 1

K1 C int L12. Put K2 := (L j2)g. Now take a j3 > j2 such that K2 C int L13 etc.

Exercise 1.13.4. Let (K j

1 be an arbitrary sequence of compact subsets of a

domain D C C" such that Kj C int Kj+1 and D = Ujt1 K1. Let A C D be an infinite set without accumulation points in D. Prove that there exist sequences

(ak)r

1

C A and (jk)k 1 C N1, Ik < Ik+1. such thatak E Kjk+, \ Kjk, k E N.

Theorem 1.13.5 (Holomorphic convexity). Let D C C". Then the following conditions are equivalent:

(i) D is a domain of holomorphy; (ii) D is holomorphically convex;

(iii) dD(KO(D)) = dD(K) for every compact set K C D, where dD(A) inf(dD(z) : z E A), A C D; (iv) dD (KO(D)) > O for every compact set K C D; (v) For every infinite subset A C D without accumulation points in D, there exists a function f E 0 (D) such that supA I f I= +oo.

Proof The implications (ii) a (iv), (iii) = (iv) are elementary (EXERCISE). The implication (v) = (i) follows from Remark 1. 11.3 (p) (EXERCISE).

1.13. Holomorphic convexity

(ii)

91

(v): By Remark 1. 13.3 (b) there exists a sequence (K,)',,= t of compact

subsets of D such that (Ky)O(D) = K,, C int K,,+t and U,=t K = D. Using Exercise 1.13.4, we may assume that there is a sequence (a,)',,=, C A such that

aE

t \K,, v > 1. Since a l 0 Kt and Kt = (Kt )O(DI, there exists a function

ft E 8 such that If, (a l) I > 11A II K, . Replacing ft by (aft )N with suitable a > 0 and N E N, we may assume that I ft (a )I > 1, and II ft II K, < 1/2. Repeating the a sequence above argument for the remaining C L9 (D) such that I v+rµ=tt lfu.(a,,)I and IIfvIIK,. < 1/2 Now put f ff. The series is locally normally convergent in D. Hence f E O(D). Moreover,

If (a,,)I > v for every v (EXERCISE).

(i) = (iii): Suppose that for some a E KO(D) we have dD(a) < dD(K) Let 0 < s < r. By the Cauchy inequalities we obtain

-

t

IIDaf IIK r, f E O(D). Finally, since D is

In particular, d(Ta f) a domain of holomorphy, we conclude that D (a, r) C D; a contradiction.

0

Exercise* 1.13.6. Let D C C" be holomorphically convex and let A C D be an infinite set without accumulation points in D. Prove that there exists an f E O(D), f 0- 0, such that sup{orda f : a E A) = +oo, where orda f denotes the order of

zero off at a. Hint. Try to find f as an infinite product. Proposition 1.13.7. Let D C C" be a Reinhardt domain. Then for every Reinhardt compact set K C D we have KO(D) = Kg, where

8 := {zalD : a E Z" is such that D C C"(a)}. Observe that if D is a complete Reinhardt domain, then 8 = {za I D : a E Z }. Proof. We already know (Remark 1. 13.3 (a)) that KO(D) = Kgo, where

80 := { E aazaiD : N E N, as # 0 = D C C"(a)}. QEZ"

Ial -2/p, p(a1 + a2) > -2(1 + r). In particular, l /z2 E Lh (Ta) * p < 2(1 + r). Observe that the function 1 /Z2 explodes at zero. Hence TQ is L' (TQ)-convex for all I < p < 2(1 + r). Suppose that T. is Lh20+0-convex. To get a contradiction it suffices to prove that for every function f E L2(1+r) (T.), the function f (O, - ) extends holomorphically to ID. Write f (Z) = LaEZ2 aaz°t, z E T0. By Example 1.10.7 (c) and the above

criterion (with p = 2(1 + r)), we know that

E(f)C{aE712:a1 >-1/(1+ r), al+a2>-1} ={aE71+x71:a1+a2?0}. Hence a042 ,Z2a2 ,

f (0, Z2) =

Z2 E (pr.

a2EZ+

which implies that the function f (O, - ) extends holomorphically to ID.

Remark 1.13.17. (a) Proposition 1.13.13 and Theorem 3.6.4 will show that every Lh-convex Reinhardt domain D C C" is an Lh-domain of holomorphy. (b) Notice that the following general result is true (cf. [Irg 2002], Theorem IV. 1):

Any bounded Lh-convex domain D C C" is an Lh-domain of holomorphy.

Proposition 1.13.18 ([Pfl 1984]). Let D C C" be an arbitrary domain and let a E D. Put

3 (D) :_ (f E 9(D, ID) : f(a) = 0}. Then the following conditions are equivalent:

96


(i) for any infinite set A C D without accumulation points in D we have

sup{I f(b)I : f E Fa(D), b E A} = 1; (ii) for any infinite set A C D without accumulation points in D, there exists a function fo E Fa(D) such that sup(I fo(b)I : b E A) = 1.

Observe that (ii) implies that D is MOO-convex. Indeed, suppose that A C K,Koo(D) has no accumulation points and let fo be as in (ii). Then sup(Ifo(b)I b E A) -< II fo II K < 1; a contradiction.

Proof. (i) = (ii): Take sequences (bk)k fk(bk) > 1 1/22A, k = 1, 2..... Put

-

1

C A and (fk)k t C FF(D) such that

l+fk gk E O(D, H+) and gk(a) = 1, where H+ := (A E C : Red. > 0). Let 00

g := E 2k1

9k-k=1

Let

MD(Z) := Sup{If(=)I : I E Fa(D)) (cf. Example 4.2.3). Observe that, by Lemma 1.7.23, MD is continuous, and by the Montel theorem (Theorem 1.7.24), MD < 1. Hence, for any compact set K C D

there exists 0 = 9(K) < 1 such that I1.fk 11K < 9, k = 1, 2..... Consequently, Ilgk IIK < 2/(1 - 9), k = 1, 2,.... It follows that the series is convergent locally uniformly in D and so g E O(D, H+). Note that g(a) = 1. We have

Ig(bk)I >- IReg(bk)I > I gk(bk) > 2k _ +00. Now, put

f

g-1 g+1

Then f E Y,,(D) and

If(bk)I >

Ig(bk)I-1 >

2k-I

Ig(bk)I + 1

2k + 1

-+ 1.

Theorem 1.13.19 ([Pfl 1984], [Fu 1994]). Let D C C" be a bounded Reinhardt domain of holomorphy satisfying the Fu condition. Then for any points a E D. b E 3D and for any sequence D a bk -+ b, there exists a sequence (fk )k t C O(D, D) such that fk(a) = 0and I fk(bk)I -+ 1. In particular, by the remark after Proposition 1.13.18, D is M"-convex.

97


Proof. We may assume that D C D". Since D satisfies the Fu condition, we may assume that D fl Y. 0 0, j = l . . . . , s, and D fl V = 0, j = s + 1.... , n, for some 0 < s < n. Thus there exists an q o E (0, 1) such that

DCG:={zE(D":Izjl>qo, j =s+l,...,n}. b. Fix a E D, b E 8D, and a sequence D 9 bk First consider the case where b E C . We may assume that b E Ut;o. Let

:IloglzjII 1 and sup{I(pk(a)I : k E N) < 1/2. 1

Indeed, suppose that we have found such a sequence. For c E ID let

-C

hc(A) := 1

- cA

A E C \ {I/C};

he I p is a MSbius automorphism of 0) with he (c) = 0. Define fk h1,k (a) ° (Pk Then, obviously, fk E )(D, ID) and fk(a) = 0. To show that I fk(bk)I 1, take an arbitrary convergent subsequence I fkt (bkt) I - to E (0,1 ]. We may assume that Sokt (bkt) -+ Co E olD, (pkt (a) -). C1 E D. Then

to = lim Ifkt(bkt)I = t-++oo

lim

(Pke(bke)-(Pkt(a) I 1 - (pkt(a) . (pkt(bkt) I

-

C0-C1 I 1 - elco

1.

Since X := log D is convex and xo := (log Ib1 I, ... , log Ibn I) E aX, there

exists a vector a E (Ot"), such that X C Ha°. Put c :_ (xo,a). Observe that D C Da,c. Since D fl Vj 0 0, j = 1, ... , s, we conclude that a1, ... , as > 0 (ExEltclsE). We may assume that a 1 = = at = 0, at+1, ... , as > 0,0:5 t < s. Take an arbitrary s > 0. By the Kronecker theorem,47 there exist sequences C N such that (Pv,j )v=1 C Z, j = 1, ... , n, I P,j - gvaj I < ts,

sgn p,, j = sgn a j ,

j = l , ... , n, qv

+oo.

47 There are the following two equivalent formulations of the Kronecker theorem (cf. [Har-Wri 19791,

Theorems 442 and 444).

Theorem. Assume that a I , ... , an, 1 are linearly independent over Q. Let µ t

.... 11L" E R, e > 0

and C > 0 bearbitrar. Then there exist pi...., pn, q E Z such that q > C and l qai - pj -µ j I
qv(-c+ E ajloglzj1)-EMb, j=t+1

where Mb := 2 j=t+1 I log bj I (note that Mb depends only on b). In other words, I*e.v(z)I

(e-clzal)gve-aMb,

z E U.

Consequently, letting D fl U z -+ b, we conclude that II *e,v 11D > e-eMb. Let / II *,,v II D. To estimate Ws,v (a) we argue as follows. There are two Ve, v : _ cases.

There exists a jo E {t + 1. ... , s} such that a j = 0: Then, obviously, Pe,v(a) = 0. a1+1 as # 0: Then a E Ct x C;-t and n

logl*e,v(a)l C and

Igaj -Pj - Nil < s. j = I,...,n. In the case /c, =

= !.e" = 0. as a direct consequence one obtains the following approximation

theorem (ExEacisr).

Theorem. Let a, , ... , an E R. e > 0 and C > 0 be arbitrary. Then there exist p, , ... , p" , q E Z

such that q>Cand Igaj -Pj1 <e,sgnpj = sgn aj, j = t,...,n.


99

Recall that q --> +oo. Consequently, we may assume that I cpe,v (a) I < 1/2'V E N. Now, we are going to estimate 11 V/,,, lI D from above. There are two cases:

t < s: Then we have n

log I*e,v(z)I

E e l log Iz1I I

e((s - t)I log qI + (n - s)I log j1oI) _: eMn,

i=1+1

E A, n (Ct x

C;-I).

where

A,,.={zEGnD.., Iz;I>q, j =t+1, ...s} and 0 < n < l is so small that (z E G : Iz1I < rl, j = t + 1,...,s) C= D,,,. The maximum principle implies that I*,,, I 1. (i) If u 1, u2 are log-psh, then u 1 + u2 is log-psh.

Proposition 1.14.4 (Maximum principle). Let D C Cn be a domain and let u E PSJ{(D). If u < u(a) for some a E D, then u =_ u(a). In particular, if D C C" is a bounded domain, u E PSX(D), and u $ const, then

u(z) < sup { lim sup u(w) :

D3w-

E aD},

z E D.

Proof. Let Do := {x E D : u(x) = u(a)}. Observe that the set

D\Do={xED:u(x) R is twice Qt-differentiable at a point a E S2, then we define the Levi

form ofu at a: n

.u(a;X)

l

2

a

u

1a

X = (X1,...,Xn) E Cn.

z;aik(a)X;Xk,

(1.14.1)

Notice that .(II 112)(a; X) = 11X 112 for any a. X E C. Observe that

Zu(a; X) =

a2uO'X (0).

aaaA

Consequently, we have the following result:

Proposition 1.14.5. Let u E e2(12, FR). Then

u E PS}{(S2) b V°Esi, XECn : Y-u(a: X) > 0. Exercise 1.14.6. Assume that I C Qt is an open interval, u E e2(S2, Ut), U(S2) C 1, and V E e2(1, FR). Prove that 12

au (a)Xj

.SE(q o u)(a: X) = w"(u(a))I 1=1

for aES2,XECn.

dz;

+ co'(u(a)).u(a; X)

1.14. Plurisubharmonic functions

103

Notice that the above formula and Proposition 1.14.5 give a direct proof of Remark 1. 14.3 (h) for the case where u and Sp are of class e2.

Exercise 1.14.7. Let F : S2' -> S2 be holomorphic, where S2' C C' is open. Prove that for u E C2(S2, (R) we have

X(u o F)(b; Y) = Xu(F(b): F'(b)(Y)),

b E S2'. Y E C'".

Consequently, if u E PS3{(S2) n 6(0, 6t), then u o F E PS5C(S2'); cf. Proposition 1.14.34. Definition 1.14.8. A function u E C2(S2, 6R) is called strictly plurisubharmonic if

Y-u(a; X) > 0,

a E S2, X E (C")..

The following proposition gives a very useful tool for constructing new psh functions.

Proposition 1.14.9. Let G C S2 C C' be open and let v E PS9{(G), U E PS1C(S2).

Assume that

lim sup v(z)
Qt, is bounded from above and measurable, then we define

A(u, a, r) :

(nr2 1 (nr,2,) P(a,,)

u dA2n =

-1"

f u(a + r w) dA2n(w)

Exercise 1.14.11. Let S2 C C" be open and let u : S2 - Ot_00 be upper semicontinuous. (a) Prove that the functions {(z, r) E S2 X R" 0 : doIP(z, r) c 92} n (z, r) H J(u; z, r),

{(z, r) E Sl x IR'o : IP(z, r) C S2}

(z, r) H A(u: z, r),

CSl}-3 (z,r.X) ff2,1

2n

u(z + re'BX) dO

are upper semicontinuous (in particular, measurable).

Hint. Use Fatou's lemma. (b) Prove that prn 2 2 rrl A(u;a.r) = 2... 2 J ... J J(u;a,(ri.....Tt))r ...T"dT1 ... r, I" 0 0

dTn

t

t

=2"...J J(u:a,(Ttrt,...,Tnrn))r1...T"drt... dTn. fo

0

Pr oposition 1.14.12. Let S2 C Cn be open and let u E TSX(Q), a E S2. Then

J(u;a,r) \..,, u(a) when r \ 0, A(u;a,r) \ u(a) when r \ 0. 49Recall that 8o J(a, r) := 8K(a 1, rt) x

x dK(an, rn ).

"That is, the function [0, 2n)" 9 8 7y u(a + r e' N) is Lebesgue measurable.


105

Proof. By Exercise 1.14.11 (b) it is enough to consider only J (u; a, ). First, we

prove that J (u; a r') _< J (u, ; a, r") for r' = (r' , .. r'n), r" = (r t ",..., rn") ,

0 -oo

for all j. We may assume that vJ < 0 on IP(j) and a (b) ? -2'j, j E (N s2 Define v := F_j=1 vj. Then V E PSX(C") (cf. Remark 1.14.3(d)), v(b)

-1,

and M C v-1(-o0). Proposition 1.14.23. Let S2 C C" be open and let a sequence (u, ,),,GIN C PS7{(S2) be locally bounded from above. (a) Put u := u,,. Then the set {z r= S2 : u(x) < u*(x)) is of zero measure.53 (b) Put u := lim u,,. Then the set {z E S2 : u(z) < u*(z)} is of zero measure. supI(j) v j and an 521t suffices to substitute vi by a function of the form e j (vj - cC) with cj appropriate sj > 0. 53 The result remains true in the case where u := supi E1 u; (with arbitrary I) and (u; ); E f C PS`}C(Q) is locally bounded from above - cf. [Jar-Pf120001, Prop. 2.1.38(a).


108

Proof. (a) Observe that the function u is measurable. To prove that u = u* a.e., it suffices to show that A(u; a, r) = A(u*; a, r) for any a E Sl and 0 < r < do (a). Fix a and r as above. We have

u(z) < P(u; a. r; z), z E iP(a, r), 0 < r < r. Hence

u*(z) < P(u; a, r; z), z E F(a, r), 0 < r < r. Observe that

P(P(u: a, r: ); a. r'; z) = P(u; a, r; z), z E IP(a, r'), 0 < r' < r < r. Thus

P(u*; a, r; z) < P(u; a, r, z),

z E IP (a, r), 0 < r < r.

In particular, J(u*; a, r) < J(u; a, r), 0 < r < r. Consequently, A(u; a. r) _ A(u*;a, r). (b) Let Vk := sup,,,k u,,, k E M. Then Vk \ u and

{Z E S2 : u(z) < u*(z)} C U{Z E S1 : vk(Z) < v (Z)) k=1

0

and we apply (a).

In fact the following more general result is true.

Theorem* 1.14.24 (Bedford-Taylor theorem; cf. [Kli 1991], Theorem 4.7.6). (a) Assume that a family (u ; )i Et C PSl{(Q) is locally bounded from above. Put u := sup;EJ u;. Then the set {z E Sl : u(z) < u*(z)} is pluripolar. (b) Assume that a sequence (u,,)' 1 C PSJ{(SQ) is locally bounded from above. u,,. Then the set (z E Sl : u(z) < u*(z)} is pluripolar. Put u := lim Proposition 1.14.25 (Removable singularities of psh functions). Let M be a closed pluripolar subset of Q. (a) Let U E PS}{(Q \ M) be locally bounded from above in 51.54 Define

5(Z):= lim sup u(w), Z E Sl s2\Mnw-*z

(notice that ii is well defined). Then u E PS9{(Q). (b) For everyfunction u E PSJ{(SQ) we have

u(z) = lint sup u(w). Z E Q. s2\M3w-.z

(c) If Sl is a domain, then the set Sl \ M is connected. MThat is every point a E S2 has a neighborhood Va such that u is bounded from above in V, \ M.


109

Proof. (a) The result has a local character. Thus we may assume 42 = D is

connected, u< 0 in D \ M. and M C u-'(-oo) with v E TS7{(D), v< 0, v 0 -00. Put

u,.._ u+(1/v)v onD\M. -00

on M.

Then U, E PSX(D), V E N (EXERCISE). Put uo = u,,. Observe that uo = u on D \ P and uo = -oo on P, where P := v-I (-oo) (P is pluripolar). By Proposition 1.14.16, (uo)* E PSX(D). By Proposition 1.14.23(a), the set A := {z E D : uo(z) < (uo)*(z)) is of zero measure. Then (uo)* = uo = u on D \ (P U A). Hence, by Proposition 1.14.13, (uo)* = u on D \ M. It remains to prove that (uo)* = u. Obviously, (uo)* = u = u on D \ M. Take an a E M. Then

u(a) = limsup u(z) = limsup (uo)*(z) < hmsup(uo)*(z) = (uo)*(a) z-'a D\M3z-+a D\M3z-+a = 1im sup uo(z) $ lira sup uo(z) = lira sup u(z) D\P3z-+a

z--oa

D\P3z-.a

lim sup u(z) = u(a). (b) Let

u(z) := lim sup u(w),

z E Q.

By (a), u E P&JI(Q). Moreover, u = u on Sl \ M. Now, since AU(M) = 0, we use Proposition 1.14.13.

(c) Suppose that Sl \ M = UI U U2, where Ut and U2 are disjoint and nonempty open sets. Then, in view of (a), the function u(z) := j for z E U; would

0

extend to a psh function on S2, which contradicts the maximum principle.

Definition 1.14.26. Let S2 be an open subset of C". A function u E C2(S2, Ut) is pluriharmonic on S2 (u E PJ{(Q)) if a2u aZ; azk

(1.14.5)

Remark 1.14.27. (a) If n = 1, then RX(Q) = .7l(S2). (b) TX(S2) is a vector space; P}C(Q) C PS1{(Q). (c) For a function u E C2(Q, Qt) the following conditions are equivalent:

(i) u E P }C(Q); (ii) ua,X E 3C(Qa,X) for any a E 12 and X E C";

(iii) Xu(a; X) = 0 for any a E Q and X E C".


110

(d) Condition (1.14.5) is equivalent to the following system of equations a2u a2u (z), axl d yk (`') = axkayl

a2U

a2u

(z)

0.

+ a Yj ayk (_) = z E S2, j,k = 1,...,n.

axljax k

(1.14.6)

In particular, a2u

ax;

a2u

(z) + -- f (z) = 0,

z E S2, j = 1, ... , n,

which shows that TX(S2) C 3e(S2) C CO0(S2).

(e) If f = u+ i v E O(Q), then u E PJf(S2). Proposition 1.14.28. if D C C" is a starlike domain with respect to a point a E D, then for any u E TX(D) there exists an f E O (D) such that u = Re f . In particular, any pluriharmonic function is locally the real part of a holomorphic function.

Proof. (Cf. Remark 1.19.8.) We may assume that a = 0. Define /'1 "

v(z) := -i J

o j=1 \Zj aZ_ J (t-)

- zj u (tz))dt. ZED.

j

Then f := u + i v E CI(D) and using (1.14.5) we get au

Z (z) aZk

aZk

"

1

ak au

a2u

k2a

2

2

+ fo, (" (zj a2kaZj (tz)t - Zj a 'j (tz)ta (tz))dt k j=1 "

-JO (Z(Zjaaa3k(rZ)+yja2 a2i k(tZ)1 + a d

au

= a Zk ( ` z ) - fo, at (t a Zk

1

(tz))dt = 0, k =1, .... n.

0

Corollary 1.14.29. Let S2j C C"J be open, j = 1, 2, and let F E O(521. S22). Then u o F E TX(S21) for any u E P X(S22). Proposition 1.14.25 implies the following important corollary.

Corollary 1.14.30. Let M be a closed pluripolar subset of S2. (a) Let u E P }C(S2 \ M) be locally bounded in S2. Then u extends pluriharmonically to S2.

(b) Let f E 0 (S2 \ M) be locally bounded in S2. Then f extends holomorphically to S2.


111

Proof. Since U E TSx(S2 \ M) and u is locally bounded from above in S2, Proposition 1.14.25 implies that u extends to a function u+ E PSJ{(S2). We can repeat

the same for the function -u. Thus -u extends to a function u_ E PSH(S2). Then u+ + u_ E PSX(S2) and u+ + u_ = u + (-u) = 0 on S2 \ M. Hence, by Proposition 1.14.13, u+ + u_ = 0, which implies that It extends to a function

uEC(Q). By Proposition 1.14.14, for any a E M and 0 < r < da(a), we get it (z) P(ii; a, r; z), z E IP(a, r). In particular, u is of class e°° in 92. Since the interior of M is empty, we see that u must be pluriharmonic in S2. (b) follows from (a) - EXERCISE.

C PSJ{(42) be a sequence

Proposition 1.14.31 (Hartogs lemma). Let

locally bounded from above. Assume that for some m E OZ,

lim sup u < M. v-*+oo

Then for every compact subset K C S2 and for every e > 0, there exists a vo such that max u < m + e, v > vo. K

Notice that the above result gives a tool to prove Theorem 1.7.13.

Proof. Take an s > 0. It is sufficient to show that for every a E S2 there exist 8(a) > 0 and v(a) such that u < m + s in D'(a, 8(a)) for v > v(a). Fix a and 0 < R < dg (a)12. We may assume that u, < 0 in D'(a, 2R) for any v >_ 1, and

m 0. Then u is log-psh i ff f o r any a E C and j E { 1, ... , n } the function Va.j

Sl

z - +

IeaZ)Iu(z)

is psh.

Proof. We only need to prove that if va, j is psh (for any a and j), then log u is psh. By definition, we have to check that for any zo E 92 and X E (C")., the function A H log u(zo +AX) is subharmonic (in the region where it is defined), equivalently (cf. [Vla 1966], Chapter 2, § 15), we have to prove that for any wo E C, the function A

+ le'ozlu(zo + AX)

is subharmonic. Let k be such that Xk 0 0. Put a := wo/Xk. Then V(A) _ le_azo.k

Iva k(zo + AX). Thus g is subharmonic provided va,k is psh.

Proposition 1.14.37. (a) Any C-seminorm q : C" --> IR+ is log-psh. (b) Let h : C" -> R+ be such that

h(,lz)=JAIh(z), AEC. ZEC". Then h is psh iff h is log-psh.

Proof. (a) By Exercise 1. 14.2 (b) we have q E PS l{(C" ). Now we can apply Lemma 1.14.36 because lelzi Iq(z) = q(eazi z) and the right-hand side is psh by Proposition 1.14.34. (b) follows from the proof of (a).

Exercise 1.14.38. Let S2 C C" be open and let u E e2(Q, IR). Prove that

4tu((x + iy): (a + ib)) = Xu((x. y); (a, b)) + 3'fu((x, y); (b. -a)),

X+iyESQ, a+ibEC"=IR"+iIR".


114

where X denotes the real Hessian: if U C RN is open and v E e2(U. IR), then N

2

,7e v(X; o := E j,k=1

ax aXk

()j tk ,

XEU,=(1,....N)E IRN. (1.14.8)

Proposition 1.14.39. Let U be a domain in IR" and let v : U -+ IR_,o. Define

U := U + i IR" C C",

u(x + iy) := v(x), x + iy E U.

Then u E P8J{(U) iff v is convex on U.ss

Proof First consider the case where v is of class e2. By Exercise 1.14.38 we get

41v'(x +iy;a + ib) =.7fv(x;a)+3fv(x:b), which, of course, implies the required result. In the general case, assume that v is psh and let (0)8 denote the e-regularization of v. Observe that (U)8 + i IR" = (U)e. Hence (U)5 = U8 + i IR" for an open set U8 C IR" (EXERCISE). Moreover,

(0)8(z + it) = fDn v( z + it + ew)P(w) dA2n(w)

=

f

u(z + ew)0(w) dA2n(w) _ (v)E(z),

z E (U)s, t E R".

"

R. Note that Hence, (v)8(X + iy) = v8(x), x + iy E (U)8, where v8: U8 v8 \ v. By the first part of the proof, v8 is convex in U8 for any e > 0. Consequently, v is convex (EXERCISE).

Conversely, assume that v is convex and let v8 be the e-regularization of v (in R"):

v8(x) := f

v(x + ey)e(y) dAn(y),

x E U8 :_ {XEU : IB(x, e) C U),

B(1)

where U is a "regularization" function in R".16 Put U8 := U8 + i IR" C C", u8(x +iy) := v8(X), x + iy E U8. Note that v8 \ ». By_the first part of the proof, u8 is psh in Ue for any e > 0. Consequently, v is psh in U. "That is. if [x, y] C U, then v(tx + (I - t)y) < tv(x) + (1 - t)v(y), t e [0, 1]. If v E e2(U, R), then v is convex iff Xv(x; ) > 0 for any x E U and t E R".

That is o E C (U, R+), supp © = R(1), fat u 0 dA" = 1, and 8(x) = 0(Ix.I, ... , Ix,,1) for any x = (x1 , ... , x") E R" . It is known that if v is convex in U, then ve is convex in Ue, ve E e°°(U8), and of \, v.

115


Proposition 1.14.40. Let D C C" be a Reinhardt domain and let u : D -* It-,,o be such that

U(Z1,...,Zn) = u(Iz1I,...,IZnI).

(ZI....,Zn) ED.

Put

u(r1,...,rn):= u(r1,....rn), (TI ,...,rn) E R(D). u(x) := u(ex), x E log D.

For any I =(i1,....ik)with 1 0. One easily checks that Z j Zk (r)--XjXk + rj rk arj ark 82 ii

4Yu(z; X) = j,k=1

n

all

1 (r)-IXj12.

j=1 arj

rj

z=(z1, ...z") ED. r=(r1... ,rn):=(Iz1I... .Iznl)


116

In particular,

4tu(ex; X) = 4Re(3fu(ex; X)) 2,.

n

_ 1,k=1

a I

n

ar (ex) Re(Xi Xk) + E ar. (ex)e-x' I Xi I2 k

2-

it

it

+bbbk)+ L (ex)e-xi(aj +bj). j=t J xElogD, X =a+ibEC".

_ i.k=1

J

J

j=1

k

On the other hand, n

3eu(x: > _ 1,k=1

it

a2-

as/ ar(ex)ex' k

j=1

aft

arJ (ex)exJ

X E log D. i E IV. Finally,

4.u(ex;a+ib) = 3eu(x,a-x a)+3eu(x,a-x b),

xElogD, X =a+ibEC", which implies the required relation. Now, let u be arbitrary and assume that u is psh. Let uE denote the e-regulari-

zation of u (u8 is psh and uE \, u). Observe that DF is Reinhardt and for any z E DE and 0 E 1R" we get ue(e'e

z) =

r JD=

=ff

u(eie

+ ew)P(w) dA2n(w)

u(eie z + e8 w)ds(w) dA2(w) ^

u(z +

dA2n(w) = uF(z).

^

Thus us(z1,...,zn) = us(Iz11.....Izn1) for any (zl,...,zn) E D8. By the first part of the proof, uE is convex in convex.

Since U.

u, we conclude that u is

Conversely, assume that v := u is convex in G := log D. Let v1 denote the standard e-regularization of v:

v8(x) := J

a0)

v(x + sy)U(y) dAn(y).

x E Gf :_ {x E G : Q3(x. e) C G}

1.15. Pseudoconvexity

117

(v, is convex in G. and vs \ v). Define

D' := {(z1,.. . , z") E C" : (log lzl 1, ... , log lzn I) E Gf}, u'(Z) := vs(IogIZl1,...,10gIZ.1). (z1,. ..,Z") E D'. Observe that Ds / D and u8 \ u. By the first part of the proof, U' E PSJ{(DE). Hence U E PS}-C(D).

(b) First assume that u E PS7{(D). Then it is clear that (i) is satisfied. Observe

that log D = log(D n Q. Hence, by (a), condition (ii) is satisfied for k = 0. F o r k = 1, ... , n - 1, the function ul is psh on D1. Consequently, we can repeat the above argument and so (ii) is satisfied for any k. To prove (iii) observe that the function

K(81)aAFiu(r°,...,r° 1,A,

i+1,....rn)

is well defined, radial, and subharmonic. In particular, the function

+1.....r,°°)

I0,8)3riH!(v;O,ri)=u(r1....,rj-,, ri,

is increasing. Now, assume that (i), (ii), and (iii) are satisfied. Then obviously u is upper continuous on D. Moreover, by (a), u is psh on D n C* and, more generally, each function u l is psh in Dj n C"-k, I = (i 1.... , ik).

Takeana = (a1,...,a") E Dwithal an = OandX = (X1,...,X") E C" with 11 X ll,o = 1. We want to prove that n

u(a)
lR+, define

dD,q(a) := sup(r > O : Bq(a, r) C D),

a E D,

where Bq(a, r) := {z E C" : q(z - a) < r}. Obviously, dD,. p". = dD. Notice that the function dD,q is continuous.

Exercise 1.15.4. dD,q = inf{BD,X : X E C", q(X) = 1). For a compact K C D and a family 8 C PSi{(D) let

KB:={zED:VUEB: u(z)<maxu}. K

By Remark 1. 14.3 (c), KTSat(D) C Ko(D). Moreover, the set Kpssc(D)ne(D) is relatively closed in D. A function u : D -> IR is called an exhaustion function if for any t E IR the set

(z E D : u(z) < t) is relatively compact in D. The next result contains various equivalent descriptions of pseudoconvexity.

Theorem 1.15.5. Let D be an open subset of C". Then the following conditions are equivalent: (i) - log 8D,X E PS{-C(D) for every X E C"; (ii) - log dD,q E PS)C(D) for every C-norm q; (iii) D is pseudoconvex; (iv) there exists an exhaustion function u E PS1{(D) fl C(D); (v) there exists an exhaustion function u E PSX{(D); (vi) KPSSt(D)ne(D) is compact in D for every compact K C D; (vii) K'9Sx(D) is relatively compact in D for every compact K C D; (viii) every point a E 8D has an open neighborhood Ua such that Ua fl D is pseudoconvex, i.e. D is locally pseudoconvex; (ix) (Kontinuitatssatz) for every sequence Wk E C(lD, D) fl O(lD, c"), k = 1, 2, ... , if Uk I pk (T) C D, then Uk t (pk ([D) C D; _ (x) (Kontinuitatssatz) for every continuous mapping (p: [0, 1] x lD --* C", if gp(t, - ) E 0 (ID, C"), t E [0,1), and (p(([0,1) x m) U (1} x T)) C D, then

rp({1}xI)CD.

120

Chapter I . Reinhardt domains

Proof. The case D = C" is elementary. Thus we may assume that D C". (i) = (ii) follows from Exercise 1.15.4 and Proposition 1.14.16. The implication (ii) (iii) is trivial. For the proof of (iii) = (iv) take u(z) := max(- logdD(z), lull), z E D. (v) and (vi) = (vii) are trivial. The implications (iv) (v) = (vii): If u is as in (v), then

Kpgx(D)C{zED:u(z)<mK u}cD. (vi). In the same way we check that (iv) (vii) (i): (This is the main part of the proof.) Fix a E D, X. Y E (C").. We want to show that the function

Da,y 9 A -+ -logSD,x(a +AY) is subharmonic.

First consider the case where X and Y are linearly dependent. We may assume that X = Y. Since SD,x (a + AX) = dDa X (A), A E Da,x, we can use Remark 1. 15.2 (a). Now assume that X, Y are linearly independent. It is sufficient to prove (cf. Re-

mark 1. 14.3 (a) (iv)) that if K(r) C Da,Y, and if p E ?(C) is such that

-logSD,x(a +AY) < Rep(A),

A E aK(r),

then the same inequality holds for all A E K(r). In other words, if

SD,x (a + AY) > e-R` p("),

A E aK(r),

then the same is true for all A E K(r). Thus we have to show that if

a + AY + K(Ie-p(-)I) X C D.

A E aK(r),

then the same inclusion holds for all A E K(r).

For O < 8 < I let Ke

Me

{a + AY + K(8je-P(")p X : A E dK(r)}, (a + AY + K(8Ie-p(a)d) X : A E K(r)).

Observe that KB and Me are compact and KB, C KBD., Me, C Me,, for 0 < 8' < 8" < 1. Our problem is to show that if KB C D for all 0 < 8 < 1, then Mg C D for all 0 < 8 < 1. Thus assume that Ke C D for all 0 < 8 < I and let

Io:={9E[0,1):M9CD}. Notice that Mo = a + K(r) Y C D. Hence to 96 0. Suppose that 80 E J. Since Moo is compact, there exists a 0 E (80,1) such that Mg C D. Consequently,


121

Io is open. It remains to prove that 10 is closed in [0, 1), i.e. if Me C D for

0 0 be such that u < -e on K. Then

Kys9((n) C {Z E D : u(z) < -s} n Kysx(D) C S2. Now, let u be arbitrary. Let us denote the e-regularization of u (cf. Proposition 1.14.33). By the first part of the proof (applied to the function - log dD + loge) the open set

D6={ZED:dD(z)>e}={ZED:-logdD(z)+loge 0,

hj(Az) = IAIhj(z), A E C, Z E C",

hjlv,u...uv, = 0.

Moreover, by the maximum principle, there are points z j E 8IB" such that h j (z j) = 1. By the Hartogs Lemma for psh functions (Proposition 1.14.31), it turns out that there "Notice that, by Proposition 1. 14.37 (b), log h e PS }f (C" ).


124

is a point z*, llz* 11 < 2, with Jim supj -.oo h j (z*) > 2/3. So taking an appropriate

subsequence (h) C (h j) j with h j,, (z*) > 1/2 and defining 00

h(z) :=Fl (hj (Z))2 -v = exp (E 2v log Ihj,.(z)I).

ZE

v=1

V=I

we obtain a psh function on C" (EXERCISE) with 1

h(z*) > 2,

h1 v = 0,

h(Az) = IAIh(z). A E C. Z E V.

The following lemma will be used in the proofs of the next propositions.

Lemma 1.15.13. Let 0 # E C (FR")* and let (ca)aEE C IRO be such that sup{ca/Ial : a E E} < +00. 59

(1.15.1)

Define

u(z) := sup(IcaZa11/1al : a E E).

Z E C"(E).

In the case where the set E C Z" is unbounded put

v(z) := limsup Icazajt/I«1,

z E C"(E).

lal-*+oo

Then:

(a) the family 0cazaIt11a1)aeE is locally bounded in C" (E);

(b) U*, V* E PSJ(C"(E)); (c) u*, v* are invariant with respect to n-rotations; 60

(d) U. v E C(C ); (e) D. {z E C"(E) : u*(z) < 1) = intnuEE{z E C"(E) : caIZaj < 11;

(f) D :_ {z E C"(E) : V*(Z) < 1} = U'I(intnaEE {Z E C"(E) lal>v

Ca1za1 < 1});

(g) ifDO #f?I,then :u(z)v

If a E G, then there exist a Reinhardt neighborhood U C C" (E) of a and vo e N such that calzal < 1, z E U, of E E, lal > vo. Consequently, by (e), w*(a) < 1,

where w(z) := sup{lcazall/Ial : a E E, lal > vo}, z E C"(E). Hence v*(a) < w*(a) < 1. Conversely, assume that v*(a) < 8' < 8 < I for some a E V(E) and let V C= U C= C"(E) be neighborhoods of a with v(z) < 8', z E U. Then, by the 61Recall that A(r) := UaEA IP(a, r).

126


Hartogs lemma (Proposition 1.14.31), there exists a vo E N such that lcazaI1/Ial
vo. Consequently, V C G. (g) The only problem is to show that if b E 9t;o is such that u(b) = 1, then b E

Fixana E D nIR;o. Putz(t) := (a1-tb',,...,a,,-'bn).tp(t) := u(z(t)), t E [0, 1]. Then tp is continuous. V(O) = u(a) < 9 < I and tp(1) = u(b) = 1. We only need to prove that tp(t) < 1, t E (0, 1). Suppose that V(to) = 1 for some to E (0, 1). Take 0 < E 1 -E.

l

Since (caa")111al < 0, we conclude that 91-to((caba)1/1"I)t°

> I -e >

01-to.

Hence, u(b) ? (cab")1/11I > 1; a contradiction. (h) Take a b E IR;o such that v(b) = 1. Fix an a E n IR".o. Put z(t) (a'-'bi, ... , a,1,-'bt,), tp(t) := v(z(t)), t E [0, 1]; (o is continuous, tp(0) = v(a) < 9 < 1 and tp(1) = v(b) = 1. We want to prove that V(t) < 1, t E (0, 1). Suppose that V(to) = 1 for some to E (0, 1). Take 0 < E < 1 - 91-'0. Then there exists a sequence (a(k))k 1 C E such that la(k)I -+ +oo and (Ca(k)(z(to))a(k))1/1a(k)1 >

I - e. Thus ((ca(k)aa(k))1/la(k)I)1-10((Ca(k)ba(k))1/Ia(k)Ito

> 1 -e.

k E N.

W,//we

Since lim a ko E N such that 01-to((Ca(k)ba(k))1/1a(k)l)to

9, k E

> 1 -s >

conclude that there exists

91-to,

k > ko.

Hence, v(b) > ((I - e)010-1)1/'0 > 1; a contradiction. (i) We have s = n, i.e. C"(E) = C". It is obvious that u and v satisfy (1.15.2) and (1.15.3). Moreover, v(z) < u(z) < CoI1 z E C". In particular, u and v are continuous at 0 and u * (0) = v * (0) = 0. To prove that u * (resp. v *) satisfies (1.15.2) and (1.15.3) we need the following general observation.

If a function h: C" -+ R+ with h(z) < Coflzflm, z E C", satisfies (1.15.2) and (1.15.3), then so does h*. _ Condition (1.15.2) with A # 0 and condition (1.15.3) with A E FD" \ Yo are elementary. It remains to check (1.15.3) with A E (®» n Vo)*. Fix such a A. We

.. A, # 0, Ar+ 1 = ... = An = 0 with I < r < n - 1. Fix an a E (C")*. We may assume that ar+1 = =at = 0 and a,+1 an 0 with may assume that A 1

r < t < n. Observe that if C" a z -+ A a, then z = µ(z) w(z), where w(z) := (Z1/A1, ... , Zr/Ar, Zr+1, ... , zt, at+1, ... , a") -+ a,


A(Z) _ (A1...... r,1,..

127

,

(A1... ,Ar,1,.. .1,0....,0). In particular, µ(z) E ID" for z near A a. Consequently,

h*(A a) = lim suph(z) < lim sup h(it w) < lim suph(w) = h*(a). w-+a

W

z

IDn 3µ-A0

0

Remark 1.15.14. (a) The functions u, v need not be continuous on the whole C" (E). For example:

n:=2,E :={(t,l-t) :t E(0,1)},ca:=1,aEZ.Then C2(E)=C2 and u(z1, z2)

m0

ax{Izt I, Iz21} if ZIZ2 $ 0

n := 2, E := M2, ca := 1, a E E. Then v coincides with the above u.

(b) If D. = 0 (resp. D. = 0), then the formula in (g) (resp. (h)) need not be true. For example:

n := 1, E := (-1, 1), ca := 1, a E E. Then C(E) = C. and u(z) _ max{IzI, l/IzI}. Hence Dt, = 0, but (z E C. : u*(z) < 1) = T. n := 1, E := Z*, ca := 1, a E E. Then v coincides with the above u. Proposition 1.15.15. Consider a Laurent series f (z) _ EaeZn aaza whose domain of convergence D is non-empty and the set E {a E Z" : as # 0} is unbounded.62 Put

v(z) := limsup laazall/lal,

ZE

C"(E*).63

Then D={zEC"(E*):v*(z) ko. Then U C D, and finally, Do C D. Since v < u, we get D 1 C Do = D. By the Cauchy inequalities, we have Ilaaza l11) < 11f 11v < 1, a E E. Hence, by Lemma 1.6.2, D C D1. 62Notice that E is finite if f (z) = P(z)/zY, where P is a polynomial and y E Z+. 63Note that if n = 1, then v(z) = max{ Iz I/R+, R-/ Iz 11. where R- and R+ are given by (1.1.2).

128


Proposition 1.15.16. Let o 96 D conditions are equivalent:

C" be a Reinhardt domain. Then thefollowing

(i) D is a fat domain of holomwrphy; (ii) there exist E C ((R"), and a family (ca)aEE C Ut>o with (1.15.1) .such that

D = {zEC"(E):u*(z)o such that

D=into{zEC"(E):caIZaI u (zj) > ti (zj) - I - 0; a contradiction.

1.17. Hyperconvexity

133

In the case of a Reinhardt domain an even stronger version of Theorem 1.17.5 is true.

Proposition 1.17.6. Let D C C" be a pseudoconvex Reinhardt domain. Assume that for any a E 8D fl Yobs (including a = oo if D is unbounded) there exists a weak psh barrier function. Then D is hyperconvex.

Proof. Let a E 8D \ Yo. Put l; := log a. Then t is a boundary point of the convex domain log D. Therefore, we may take a real linear functional L : iQ" -+ Qt,

L(x) _

such that L(x) < L(t), X E logD. Put u: C -+ R,

n u(z) aj log 1z11. Then v := u - u(a) E 9183{(C11) n e(c;), v < 0 on D fl C n, such that limD3z--Q v(z) = 0. Observe that v is locally bounded from

above on D \ Yo. In virtue of Theorem 1.14.25, v extends to a psh function on D, which is everywhere negative (use the maximum principle). Hence this extension gives a weakly psh barrier at a. Hence, using the assumption, Theorem 1.17.5 completes the proof.

In the general theory, hyperconvex domains do not contain non-trivial entire holomorphic curves.

Definition 1.17.7. A domain D C C" is called Brody hyperbolic if any rp E O(C. D) is identically constant. Exercise 1.17.8. Observe that any elementary Reinhardt domain Da,c is not Brody hyperbolic.

Proposition 1.17.9. Let D C C" be a hyperconvex domain. Then D is Brody hyperbolic.

Proof Let u E PSJ{(D) denote the negative exhaustion function from the defini-

tion of hyperconvexity. If q E O(C, D), then v := u o (p E SX(C) and v < 0. Hence, in virtue of the Liouville theorem for subharmonic functions (see 1. 14.3 (g)),

v = C E QR. Applying (1.17.1) implies that V(Q is bounded and therefore, (p is a constant function according to the classical Liouville theorem for holomorphic functions.

In the case of Reinhardt domains there are the following results for Brody hyperbolic domains. We begin with a direct consequence of Lemma 1.5.14 (iii).

Lemma 1.17.10. Let D C C" be a Brody hyperbolic Reinhardt domain of holoat 1

morphy. Then there exist a matrix A

such that D C Day c, fl

fl Da",c"

a" J

E GL (n, 71) and a vector c E R"

6'Recall that V0 = ((Z1'...' z") E C" : Z t ... Z. = 0).


134

Proof. We only need to observe that an affine line L = a + Otb, b 0 0, is contained in log D if the entire curve C 9 A H (ea, exbt

,

can ezb" )

.

0

has its image in D.

Theorem 1.17.11. Let D C C" be a Brody hyperbolic Reinhardt domain of holomorphy. Then D is algebraically equivalent to a bounded domain (cf. Def-

inition 1.5.12), i.e. there exists a matrix A :=

I

:t

L an

E GL(n, Z) such that

J D C C"(A) and 'PA maps biholomorphically D onto a bounded Reinhardt do-

main of holomorphy, where (PA: D

C",

(PA(z) := (zat ,.. .za"),

ZED.

Proof. The proof is done by induction. For n = 1 the only unbounded Reinhardt

domains in C are C, C., and A(r, oo) with r > 0. The first two are not Brody hyperbolic. The annulus can be algebraically mapped by z H 1 /z onto a bounded Reinhardt domain. Now let n > 1 and assume that the theorem is true for all lower dimensions. If D C C", then Remark 1.5.13 (b) and Lemma 1.17.10 apply.65 In order to discuss the remaining case let us assume, without loss of generality,

that D f1 V" # 0. Then, in virtue of Corollary 1.11.16, D := prCn-1(D) = prCn-t (D n V") is a Reinhardt domain of holomorphy in C"-1. Moreover, it is easily seen that D is Brody hyperbolic. Applying the induction hypothesis we find a matrix A E GL(n - 1, 7L) with the same properties as above. So'PA is defined on D and maps b biholomorphically onto its image 'PA (D), a bounded Reinhardt domain of holomorphy. Now put A

_ 10A 0

1].

Then A E GL(n, 71) and OA maps D biholomorphically onto a Brody hyperbolic Reinhardt domain of holomorphy OA (D) which is contained in lP"-I (r) x C (ExERCISE).

Therefore, we may assume from the very beginning that D satisfies

DnV"7 0, DChn_1(r)xC for some r > 0. Without loss of generality let

Dnv 00, j = 1

. . . . ,

k.

DnL, =0. j =k+1,...,n-i,

where kE{0....,n-1}. 65 Notice that, in general. if A is as in Lemma 1.17.10. then (OA)ID need not be biholomorphic. For

example, if A is as in Remark 1.5.13 (c). then ®2 C (Iz,z21 < 1, lz,zz < 1), but 0,(0,0) _

(0,0) 0 C"(A-1).

1.17. Hyperconvexity

135

Then we find an a = (a,_., a") = (0, a") E Zk x 7Z"-k, an # 0, and an m e Qt+ such that ZEDn(C"-1xCe).

IZat<m,

Indeed, if k = 0, then Lemma 1.17.10 applies directly (we take a := aj, where j is such that a,,, 34 0). Now let k > 1. Put k

D" := prCn-k (D) = prcn-k (D nn vi), j=1

recall that D is relatively complete. Then D" is a Brody hyperbolic Reinhardt domain of holomorphy. In virtue of Lemma 1.17.10, we find a" = (ak+l,..., an) E Z"-k, an # 0, and m E R+ such that (Z")or I< m,

z" E D", zn 96 0.

I

Then, setting a :_ (0, ... , 0. a") completes the argument. In a last step put

_

if j < k,

0

la. l of

j+1 ifk<j 0, j = k + 1.... , n - 1. Then, for z = (z', z") E D, zn 36 0, we get

Iz'I < IWY"I1""'1Izk+llsk+l

...IZn-IIsn-1

<ml/lanlrn-k-1

By continuity, this estimate remains true on D. Finally, we introduce

A :=

[IIn0-1 ?]Elx?:766

Note that det A = 1, 'PA is defined on D with a non-vanishing Jacobian, and 'PA is injective on D. Hence, 'PA gives an algebraic biholomorphic mapping from D 0 onto its image 'A (D) which is a bounded Reinhardt domain of holomorphy. Based on the former theorem we have the following result for Brody hyperbolic Reinhardt domains of holomorphy.

Proposition 1.17.12. Let D C C" be a Brody hyperbolic Reinhardt domain of holonwrphy. Then the following conditions are equivalent: 6`'IIk denotes the (k x k)-dimensional unit matrix.


136

(i) D is algebraically equivalent to an unbounded Reinhardt domain of holomorphy;

(ii) D is algebraically equivalent to a bounded Reinhardt domain of holomorphy G which does not satisfy the Fu condition. (i): We may assume that D is bounded and does not satisfy the Fu Proof. (ii) condition. Therefore, V n D = 0 and D n Vj 0 0 for a certain j. Take simply the following map

F: D

C",

F(z):=(zt,.. .zj-t, L,z1+r,....z")

Then G := F(D) is an unbounded Reinhardt domain of holomorphy and F is an algebraic biholomorphism from D onto G. (i) = (ii): Without loss of generality, we assume that D is unbounded. In virtue of Theorem 1.17.11 there are a matrix A = [a j,k ] t < j.k D is biholomorphic. Obviously, G is a Reinhardt domain of holomorphy. Suppose that G does satisfy the Fu condition. So we may assume that

VnGOO, j=l,...,k.

and

GnVj=O, j=k+l....,n.

where k is a suitable number in 10..... n). Therefore, a,,j > 0 if j E { 1, ... , k) and I < r < n. Moreover, there is an m > 0 such that I zj I > m if z E G and j = k + I..... n. Denote now by a' the r-th row of A. Then I('PA(z))r I = Izar

I = Izl' ... z:r.n I < m'

if z E G, I < r < n,

where m' is a suitable real number. So, PA(G) = D is bounded; a contradiction.

0

Now we are able to present a complete description of hyperconvex Reinhardt domains. Before presenting the result we need an additional definition which sharpens the notion of hyperconvexity.

Definition 1.17.13. A domain (resp. a Reinhardt domain) D C C" is said to be strictly hyperconvex (resp. strictly R-hyperconvex) if there exist a domain (resp. Rein-

hardt domain) D' and a function u E PS5{(D') n C(D') (resp. U E PS}{(D) n C(D') with u(z) = u(Izr I, ... , Izn I), Z E D') such that

DCD', u jK,L), or a subsequence (fjt)e with fir -+ f E O(D, D) locally uniformly on D. Exercise 1.17.18. When is a planar domain D C C taut ? Example 1.17.19. The Hartogs triangle T is taut. Indeed, let cpk = (Vk,t, (Pk,2) : ID -+ T, cpk --' Vo locally uniformly in ID locally uniformly in lD with with Vo =(Vo,1,(Po,2) E (9 (®, T), ok,t/Wk,2 -+ * E 0(ID, !D). Note that (Qo,1 = Or rpo,2. By the Hurwitz theorem (cf. [Con 1973], Chapter VI, Theorem 2.5), either {po,2 - 0 or VO,2 has no zeroes. In the first case (po(ID) = ((0.0)) C aT. In the second case, either V0.2 (ID) n T 0 0 (and then Spo(D) C aT), or (po,2(D) C ID. In the latter case, either *(D) n T 0 (and then Po(D) C aT), or Vo(D) C T. Exercise 1.17.20. Decide whether the domain T° := ((ZI, z2) E

D2

: IZi I° < IZ21)

(a > 0)

is taut.

Hint. Use the maximum principle for subharmonic functions. Using Theorem 1.15.5(x) it is easy to solve the following exercise.

Exercise 1.17.21. Prove that any taut domain is pseudoconvex.

142


On the other hand it turns out that hyperconvexity implies tautness.

Theorem 1.17.22. Any hyperconvex bounded domain D C C" is taut. Proof. Let us start with a sequence (f) j C O (ID. D). By the Montel theorem we may assume that fj -+ f E O((D, C") locally uniformly. We have to show that f E O(U), D). Otherwise, f (,Io) E 8D for a ko E 0D. Note that all values off are in D. By assumption there is a continuous function u E 5SX(D), u < 0, satisfying

(1.17.1). Puttingu := Oon8D,weextendcontinuouslyutoD. Thenuofj -+ uof locally uniformly on ID. Hence, u o f E PS7{(D) with u < 1. Observe that u o f(,to) = 1. Therefore the maximum principle gives the contradiction. For Reinhardt domains of holomorphy we even have the following characterization of taut domains. Theorem 1.17.23. Let D C C" be a taut Reinhardt domain ofholomorphy. Then D is algebraically equivalent to a bounded domain.

Remark 1.17.24. It has to be pointed out that the converse statement is also true; its proof will be given later in Theorem 4.7.2. The proof of Theorem 1.17.23 will use the following lemma.

Lemma 1.17.25. Any taut domain D C C" is Brody hyperbolic.

Proof Otherwise there exists (p E O(C. D) which is not identically constant. Put ppj : ID -+ D, cpj (.l) := o(j l). Since (pj(O) = (p(0), no subsequence diverges locally uniformly. Assume there is a subsequence (qjk )k with (Pjk -* f E 0(m, D) locally uniformly. Then I cjk (A) - f(A) 1 < 1, ICI < 1/2, if k > ko. Therefore,

Iwjk(A)I < If(A)I+Iwjk(A)-f001 < IIfII(ti2)m+l =: C.

IAI < 1/2, k > ko.

Hence, (p is bounded and so identically constant; a contradiction.

Proof of Theorem 1.17.23. The proof follows directly from Lemma 1.17.25 and Theorem 1.17.11.

1.18* Smooth pseudoconvex domains This section collects terminology and basic results related to the pseudoconvexity of smooth domains (proofs and details may be found e.g. in [Jar-Pfl 20001, § 2.2). The reader may skip this section during the first reading.

1.18'. Smooth pseudoconvex domains

143

Definition 1.18.1. Let D C C" be a bounded domain. We say that aD is smooth of class Ck (or Ck-smooth) in a neighborhood of a point a E 8D if there exist an open neighborhood U of a and a function u c- Ck(U, IR) such that

DnU={zEU:u(z)0},

(1.18.1)

gradu(z) 960 for z E U n aD;

(1.18.3)

(1.18.2)

here k E N U {oo} U {w}, where u E Cw means that u is real analytic. The function u is called a local defining function for D. Observe that if u r Ck (U, IR) satisfies (1.18.1) and grad u (z) # 0 for all z E U, then u satisfies (1.18.2) in a sufficiently small neighborhood of a, i.e. u is a local defining function in a suitable neighborhood of a. We say that D is Ck-smooth or has a Ck-smooth boundary if 8D is & -smooth at any point a E 8D. Put au

"

Tb (dD):= {X EC" : E8z (b)Xj =O}, bE UndD. j=1

!

The complex space T, (SD) is called the complex tangent space to 8D at b; notice that the condition

az (z)Xj = 0 i=1

j

means that X I grad u (b) in the sense of the Hermitian scalar product in C". The definition of Tb (8D) is independent of u (this will follow from Proposition 1. 18.2 (b)). Observe that if n = 1, then Tb (SD) _ {0}. Proposition* 1.18.2 ([Jar-Pfl 2000], Proposition 2.2.3). Let D C C" be a bounded domain, a E 8D, and let U be an open neighborhood of a. (a) Let u1, u2 E Ck(U, IR) be two local defining functions (k E W U {oo}). Then u2 = vu 1 with v E Ck-' (U, IR>o).

(b) The space Th (SD) is independent of the local defining function uEC'(U,IR),bEUnWD. (c) Let u 1, u2 E Ck (U, IR), k > 2, be two local defining functions with u2 = vu 1, where v E

Ck-t (U, UZ>0) is

as in (a). Then

Xu2(b; X) = v(b)Xu1(b; X),

b E U n SD, X E Tb (al)),

where X denotes the Levi form (cf. (1.14.1)). (d) Let k E N U loo). Then the following conditions are equivalent:

(i) D is Ck-smooth;


144

(ii) there exists a function u E ek (C", [R) satisfying (1.18.1), (1.18.2), (1.18.3)

with U := C". The above function u is called a global defining function for D.

Propositions 1.18.3([Jar-Pfl 2000], Proposition 2.2.23). Let D C C" be a bounded e2-smooth domain. Then D is pseudoconvex i,¢ for any local defining function u E e2 (V, [R) we have:

I u (b; X) > 0.

b E V n aD, X E Tb (dD) (Levi condition).

Notice that by Proposition 1. 18.2 (c), the Levi condition is independent of u.

Definition 1.18.4. Let D C C" be a bounded domain. We say that aD is strongly pseudoconvex in a neighborhood of a point a E aD if there exist an open neighborhood U of a and a local defining function u E e2(U, [R) such that

.u(b; X) > 0,

b E U n aD, X E (Tb (aD))..

(1.18.4)

Observe that by Proposition 1. 18.2 (c), the definition is independent of u. We say that D is strongly pseudoconvex if aD is strongly pseudoconvex at any

point a E W. Remark 1.18.5. (a) Obviously, if n = 1, then any C2-smooth domain D t C is strongly pseudoconvex. (b) The notion of the strong pseudoconvexity is invariant under local biholomorphic mappings (EXERCISE).

(c) We will see (Proposition 1. 18.8 (a)) that any strongly pseudoconvex domain is hyperconvex and, consequently, pseudoconvex.

(d) Recall that a bounded domain D C C" is said to be strongly convex at a point a E aD if there exist an open neighborhood U of a and a local defining function u r C2(U, IR) for D such that .7eu(z;t) > 0,

z E U n 8D, t E (Tf(aD)).,

where a denotes the real Hessian (cf. (1.14.8)) and T R(aD) is the real (2n - 1)dimensional tangent space to aD at z. The definition is independent of u. In particular, any strongly convex domain D C C" is strongly pseudoconvex (EXERCISE).

Proposition* 1.18.6 ([Jar-Pfl 2000], Proposition 2.2.5). Let D C C" be a bounded domain.

(a) Assume that aD is e2-smooth at a E aD. Let U be an open neighborhood ofa and let u E ('.2(U, IR) be strictly psh with (1.18.1) and (1.18.2). Then u satisfies (1.18.3). In particular, u is a local defining function.

1.18*. Smooth pseudoconvex domains

145

(b) Let U be an open neighborhood of aD and let u e ek (U, Qt), k > 2, be a local defining function with (1.18.4). Then there exists a c > 0 such that for the function uc := (eC° - 1) we have: c

SEue(b; X) > 0.

b E aD, X E (C")..

In particular, u,, is strictly psh in a neighborhood of aD (notice that u, is a local ek -defining function).

(c) For k > Z, the following conditions are equivalent:

(i) D is ek-smooth and strongly pseudoconvex; (ii) there exist an open neighborhood U of aD and a strictly psh function u E ek (U, OZ) with (1.18.1) and (1.18.2).

With respect to (b) and (c) compare [For 1979] and [Beh 1985] for the case of general pseudoconvex domains.

Exercise 1.18.7 (Complex ellipsoids; cf. [Jar-Pfl 1993], § 8.4). For n > 2, p = (P1, ... , p") E Quo, define the complex ellipsoid

(:p := {(Z1,...,Z") E Cn : IZzI2pi < 1}. !`

(1.18.5)

j=1

Obviously E 1 = 03" . Prove the following properties of Q=P.

(a) Ep C D' is a complete Reinhardt domain of holomorphy (use Theorem 1.11.13).

(b) Ep is convex iff pl,... , p" > 1/2. (c) Ep is geometrically strictly convex if and only if p1, ... , p" > 1/2 and

#{j:pj=1/2} Qt+ such that d(z, z) = 0, d(z, w) = d (w, z). and d (z, w) < d (z, u) + d (u, w) for arbitrary z, w, u E D. It is adistance if, in addition, d(z, w) > O for z w. 71 Note that any two points in D can be connected even by a e°O-curve in D. Moreover, observe that d, (z I , z2) = inf (LQ (y)), where the infimum is taken over all C°°-curves in D connecting z 1, Z2 (EXERCISE).

1.19*. Complete Kahler metrics

147

If g is a Hermitian metric on D, then dg is a distance (EXERCISE). We say that g is a complete Hermitian metric on D iff g is a Hermitian metric on D and

Bg(a,r):=Bd,(a,r)={zED:dg(a,z)r

0

is in C°O(C) and has a compact support. Therefore,

!P r-+ C,

g(z)

f

I

,

a"`W'X'z")d d

A=

belongs to C°O(P(r)). Moreover, we have a = X(zm)am(z) = am(z), when z E IP(rl),73

(z) = 0, when z E P(r) and j > m (use the assumption and the integrability conditions). Instead of dealing with a j we are going to study the functions

tip := aj - - g E COO(IP(r)), a2j

1 < j < n.

Note that this new system of functions fulfills the conditions of (*)m on IP(rt). Therefore, by the induction hypothesis, there exists an f E C°O(P(r')) such that of azj

_«;, I<j 1, k = 1, ... , k(j ), where fj,k := fzk. Since 1Ifj,k11L1 < 1, we may choose an exponent Xj E W such that II fi,k IIL, < 2, ), where fj,k := j fj k . In a next step we discuss the series

k(j

00

k(j)

j=1

k=1

_

u.=E(Efjkfj,k

In virtue of the above construction it is clear that this series converges locally uniformly on D. Reading this sequence on D x D as k(j) 00 (I:fj,k(Z)fj,k(w)) u(Z, w) = L..: j=1 k=1 shows that the series, in fact, gives a holomorphic function on D x D. In virtue of the Weierstrass theorem (Theorem 1.7.19), we finally obtain (k

00

a2u a-µ

j=1

afj,k(z) afj.k(z))

k=1

°µ

Observe that the g,,,,,'s are real analytic functions on D which define a Hermitian pseudometric g . Put g,,,µ (z) := g,, (z) Then the g,,,,,,'s define a C°'-Kahler metric j on D, i.e. g (z; X) = g (z; X) + II X 112. What remains is to show that this metric is a complete one on D. Note that

dg < dg. Therefore, B1(a, r) C Bg(a. r), a E D, r > 0. Hence, it suffices to show that Bg(a, r) lies relatively compact in D. In fact, fix a E D and r > 0. We may assume that a E int L 1. Assume now that there is a point b E lB(a, r) \ Ls+1. Take an arbitrary C1-curve y: [0, 1] --1- D with y(0) = a, y(1) = b. Then there exists a to E (0, 1) with the following properties:

y(t) E int Ls+1, 0 < t < to, and y(to) E 8L,+1. We find an index jb, 1 < jb < k(s), such that y(to) E Vs, jb. Therefore, I fs, jh(y(to))I > 1. Thus, by definition,

fs,j,,(Y(to))I > s. On the other hand, recall that Ifs, j, (y(0)) I < val =: const. Thus, y := fs, jb o y : [0, to] -> C defines a Cl -curve in the complex plane; in particular,

Jrclddtt)Idt>IY(to)-Y(0)I ?s - const. 0

It remains to estimate Lg(y) from below. In fact, we have

L, (Y) ?

Jg((t);Y'(t))dt > J o

0

Ilfs jb(Y(t))Y (t)II ? f ' I dd I dt. 0


154

Hence, d8 (a. b) > s -const -+ oo when s -+ oo. Therefore. Ba (a, r) is contained in some La; in particular, it is a relatively compact subset of D. 0 Hence the proof is completed. In virtue of Theorem 1.13.5 we have the following consequence.

Corollary 1.19.10. Any Reinhardt domain of holomorphy carries a complete C°1-Kdhler metric.

In a next step we discuss some examples of domains carrying a complete CKahler metric.

Lemma 1.19.11. Let D := (C").. Then D carries a complete C0-Kahler metric.

Proof. Put u: (0,oo) -> IR, u(t) := r'IOg

Then u is a real analytic function

satisfying the following properties (EXERCISE):

u(t) > 0, t E (0, 00), fo tu2(t)dt =: d E IZ+,

fo u(t)dt = oo. Moreover, put v(t) := fo ru2(r)d r, t E (0, oo), and let U be a primitive of r -p t on (0, oo). Finally, set h: D -+ 1t, h(z) := U(IIzI12), and define e2ti

Then the g,,,,, are real analytic functions on D and they give a Kahler pseudometric g = (g,,,N,) I< v,j D be a e I -curve. Then

La(y) = f 1 0

=

g(Y(t): y'(t))dt

.

J01 ((')"f 1 ) f2yr

2n...

2n 2rr

/

f 2n ... o

f22r

f

1/2

f2a g(Ye(r): y' (t))dOl ... d6

dt

fg(ye(1):Y(t))dtd9i...d9n

2n

0

infIdg(z. w) : z E Ty(o) W E Ty(t)} = dg(Ty(o),Ty(I)). where TO := {(1; j a a1 , ... ,

:.1.....

E T} denotes the n-dimensional torus

through a E C" and

Yo (t) := (yo,t....,Ye,")(t)

(e;9'y1(t),...,eiOnyn(t)).76

Using the last inequality we continue proving that g is complete on D. Fix aD (or aj -- oo (if possible)). Then da (a. aj) >

points a. aj E D with aj

76(a) Observe that B H LR(yd) is continuous. (b) The first inequality is a consequence of the Schwarz inequality (EXERCISE).

1.19*. Complete Kahler metrics

157

dg (T0 , Ta j) for any j. Thus, for suitable b j E T. and c j E Ta j , it follows that

da (a, a j) + 1 > dg (b j , c j ). We may assume that c j -> aD (or c j -+ oo) and d1 (a,aj) > dg (a, cj) - dg (bj, a) - 1 > dg (a,cj)-M for a suitable number M (observe that dg (a. - ) is continuous on the compact torus Ta). Therefore, applying that g is complete, we get dg (a, a j) -* oo, i.e. g is a complete Kahler metric on D. Now, take the pullback h of g via the holomorphic mapping

0:T:=logD+ilR"-+D. ds(w):_(e' ....,e'"), i.e. h := 0-1(g) = (h,,,r,,). In particular, h,,u(w) = 8,,,,,(0(w))ew"ew" = h,,,µ (u) when w = u +i v E T, i.e. the functions h,,,,, depend only on the variable u. on T. So we Exploiting the Kahler conditions for h we see that = a al,-

obtain n closed one-forms on log D," namely au := F_v=l h,µdu,,, 1

µ da(0(Yro(0)), 4'(yro(s)))2 s 00, since j is a complete metric on D.

For 0 < t < to, put y, : [0, 1] - log D, y, (s) := u' + s(y(t) - u') and X(t) := y(t) - W. Note that Fu =1 fylio.rol auk < M, M > 2. Now, chooses E [0, so) near so such that F_µ-1 X,, fyfollo.sl a, > 2M + 1. nA one form (or a Pfa,Bian form) a on a domain Q C R" can bethought as an n-tuple (ft , ... , fn ) of continuous functions fj : S2 -+ C written in the form a = E j% _ t fj dxj. Such one-forms can be integrated along e '-curves y : [a, b] - S2 defining

f

v

n "

a :=

Ja

E fj(Y(t))Yy(r)dr. j=1

In case that all f j E e1 (92), a is called closed if- _, I < j, k < n.


158

Then, for a t, t < to, sufficiently near to, it follows that

2M rt=t

Xµ f tp1(os) aµ - 1 < E Xµ(t) u=)

Lh(Yr) = E Xl(t) 14=1

f au

t*1

Yr

f

al = Lh (Yt I [o,s])

Yrl(os]

I E Xu(t)f jL=t

au,

M + 1;

Y1(0.r)

a contradiction. Observe that equality (*) is a consequence of the fact that the curves yt and yl[o,r] are homotopic (EXERCISE) and the one-forms au are closed. Hence, the Stokes theorem applies and gives (*).

Thus Theorem 1.19.14 shows that a Reinhardt domain with a complete Kahler metric is almost holomorphically convex. Hence, by Remark 1.12.7, we get

Corollary 1.19.15. Let D be a Reinhardt domain in C". Assume that there is a complete Kahler metric on D. Then D* \ M(D) is the envelope of holomorphy of D.

In fact a stronger result holds. In order to be able to formulate it we need the following definition. Definition 1.19.16. For a Reinhardt domain D C C". let b be the set of all points a E C" such that there is a neighborhood U = U(a) with

U

U\

I;,

c D.

Yi n...nWk nD=O, 1- 3, which is not an analytic set but, nevertheless, D \ A allows a complete Kdhler metric.

(e) Observe that for a real closed C'-submanifold A in a domain D of real dimension 2n - 2 the following is true: A is analytic if and only if A is nowhere linearly generated (for a point a E A, A is called linearly generated at a if the smallest complex linear subspace of C" containing the real tangent space of A at a coincides with C"). This observation allows the following generalization of (b) (cf. [Die-For 1984]): Let A be a closed real CO°-submanifold of a domain D C C" of

real dimension 2n - d, d > 3, such that D \ A admits a complete Kahler metric. Then A is nowhere linearly generated.

78Compare the notion of weak relative completeness (cf. Theorem 1.11.13).

79That is, there is a neighborhood U = U(a) such that any connected component of D fl U is a domain of holomorphy.

Chapter 2

Biholomorphisms of Reinhardt domains 2.1 Introduction Let G, D C C" be domains. Recall that Bih(G. D) denotes the set of all biholomorphic mappings F : G -+ D. For a E G, b E D, put

Biha,b(G, D) := IF E Bih(G, D) : F(a) = b}. Define

Aut(G) := Bih(G, G), Auta,b(G) := Biha,b(G, G), Auta(G) := Auta,a(G). Recall that Aut(G) with the operation

Aut(G) x Aut(G) 9 (0, W) -> W o 0 E Aut(G)

is a group and Auta(G) is a subgroup of Aut(G). Observe that if F E Bih(G, D), then the mapping

Aut(G) 9 0

F o 0 o F-1 E Aut(D)

is a group isomorphism. Moreover, if F E Biha,b (G, D), then E F (Auto (G)) _ Autb(D). From the point of view of the theory of holomorphic functions, domains which bib

D) may be identified - thus it is important to know when Bih(G. D) 0 0 or (more precisely) when Biha,b(G, D) 0. On the other hand, the group Aut(G) characterizes the holomorphic geometry of G and, therefore, it is important to describe the structures of Aut(G) and are biholomorphic (G

Auta,b (G ).

Definition 2.1.1. We say that a domain G is homogeneous if the group Aut(G) acts transitively on G, which means that Auta,b(G) 36 0 for any a, b E G, i.e. for any

a, b E G there exists a ' E Aut(G) with '(a) = b. We say that a domain G is symmetric at a point a E G if there exists a 0 E Auta(G) such that 02 = id and a is an isolated point of the fixed point set

Fix(O) := {z E G : O(z) = z}. We say that G is symmetric if G is symmetric at every point a E G.

2.1. Introduction

161

Remark 2.1.2. (a) G is homogeneous if there exists a point ao E G such that Autoo,b(G) $ 0 for every b E G. (b) The notion of a homogeneous (resp. symmetric) domain is invariant under biholomorphic mappings. Indeed, let F E Bih(G, D) and assume that D is homogeneous. Fix a, b E G.

Let 0 E AutF(o),F(b)(D). Then F-' o 0 o F E Auta,b(G). In the case where D is symmetric, a E G, and fi E AutF(a)(D) is such that := F-' o 0 o F E 02 = id and the point F(a) is isolated in Fix('P), then Autb(G), W2 = id, and the point a is isolated in F-' (Fix(O)) = Fix(W). (c) If G1 C C"% is homogeneous (resp. symmetric), j = 1, 2, then G, x G2 is homogeneous (resp. symmetric). (d) If G is homogeneous and symmetric at a point ao e G, then G is symmetric.

Indeed, let 0 E Autao(G) be such that 02 = id and ao is an isolated point of the set Fix('P). Take a b E G and let F E Autao,b(G). Put 1 := F o df o f-' E Autb(G). Then W2 = id and Fix(W) = F(Fix('P)). (e) Let G be a homogeneous domain with 0 E G. By (d), if G is symmetric with

respect to 0 in the geometric sense (i.e. Z E G = -z E G), then G is symmetric in the sense of Definition 2.1.1.

(f) If G is homogeneous and Bih(G, D) $ 0, then Biha,b(G, D) : 0 for all points a E G, b E D. Theorem 2.1.3. Let D C C" be a bounded homogeneous domain. Then D is a domain of holomorphy.

Proof. Suppose Do and b to be as in Proposition 1. 11.2 (_*) with 8 = O(D). We may assume that Do is a connected component of D n D. Fix points a E Do, b E (8D0) n D. Let (bk )°k° I C Do be such that bk -> b. Since D is homogeneous, there exists a Ok E Auta,bk (D), k E W. Since D is bounded, we may assume that Ok -+ 0 locally uniformly in D, where df : D -> D is a holomorphic mapping with 'P(a) = b. We are going to show that J(P(a) #_0. Indeed, put Wk := Oj' E Autbk,a(D). Let Yik : D -+ C" be the holomorphic

extension of #Pk with 'ilk = Pk on Do, k E W. Since D is bounded, we may assume that Wk --* W locally uniformly in D, where W : D -* D is a holomorphic mapping. Let W be the holomorphic extensions of 4P. Since the extension operator is continuous (cf. Remark 1.11.3 (p)), we conclude that 171k -+ 17, locally uniformly in D. Let U C= Do be a connected neighborhood of a such that P(U) C= D. We may assume that 'k (U) C D, k E N. Thus Ok (U) is a domain contained in D n D

with bk ='Pk(a) E Ok(U). Hence 'k(U) C Do. Consequently, for Z E U, we obtain

Z = Pk(Ok(z)) _ k('Pk(z)) and, therefore, JO(z) : 0, z E U. Hence, we may assume that 01u: U -- V is biholomorphic, where V := O(U) is an open neighborhood of b.

Chapter 2. Biholomorphisms of Reinhardt domains

162

Recall that V C D. If D is fat, then we have a contradiction. In the general case we argue as follows. Take a Euclidean ball B C= U centered at a. For a we compact K C C" let p(b, K) := inf(Ilb - wil : w E K). Since b get p(b, dik(aB)) = p(b, Ok(B)), k E N (EXERCISE). Then

0 < p(b, a4'(B)) = p(b, 4y(aB)) = lim p(b. bk(3B)) lim p(b. 0k(B)) = p(b. 4s(B)) = 0;

k -*+00

a contradiction.

Exercise 2.1.4. Prove that (C2). is homogeneous, but is not a domain of holomorphy.

Exercise 2.1.5 (Examples of groups of automorphisms of planar domains).

xC},

(a) Aut(C)={C

Auto(C)={C zHazEC:aEC }. In particular: Aut(C) acts transitively on (C; Aut(C) depends on 4 real parameters; Auto(C) depends on 2 real parameters.

(b) Given an a E I), put

h (z) _ a

z - a

1 -az.

z EC\{l/a}DID.

(2.1.1)

Observe that ha E Aut(D), ha(a) = 0, and hat = h_a. Then: Aut(D) =

E T. a E m),

Auto(D)={i;ho:SE T). In particular: Aut(ID) acts transitively on ID; Aut(m) depends on 3 real parameters; Auto(ID) depends on I real parameter.

(c) Let A = A(R) :_ {z E C : 1/R < IzI < R}. Observe that every annulus {z E C : rt < Izi < r2} with 0 < rt < r2 < +oo is biholomorphic to A(R) with

R := Jr2/rt. Then Aut(A)=(A3Zr-+> In particular: Aut(A) does not act transitively on A;

2.1. Introduction

163

Aut(A) depends on I real parameter.'

(d) Aut(C. x C.): The following mappings F : C -+ C are biholomorphic: (zle"2f(zi l Z22), z2e-"t f(z1' Z22)), where nl, n2 E Z and F(z,, z2)

f E 9(C*);

al.2 z1a2., zza2.2 ), where a j k E Z with at,ta2,2 F(zl, zz) (z1al, z2 al,2a2,1 = ±1; F(zl,z2) (ctzt.c2z2), where CI, C2 E C. The full description of Aut(C) seems to be not known; ['] it is conjectured that the mappings above generate Aut(C ). IT (e) Aut(C x C.): Due to [Nis 19861 the following theorem is true. 0) = 0 (in particular, FIC,C. E Aut(C x C.)) Let F E Aut(C2) with

and det F' = c E C. Then F(z,, Z2) = (Czi

a-"(Z, z2.z2) +

$(z t

z2,

z2). z2ea(=, Z2,Z2)).

where a, i3 E O(C2).

0 According to our knowledge the full description of Aut(C x C.) remains still open. 0 (f) Aut(C2): Recall that the set of holomorphic automorphisms of C is quite simple. In contrast, for n > I the situation for Aut(C") is much more complicated. There are, for example, automorphisms F of the following types: F(z1, z2) := (zl, 22 + f (zt )), where f E O(C); mappings of this type are usually called shears; F(z,, z2) := (zI, eh(Z')z2 + fl Z2)), where f, h E O(C); mappings of this type are the so-called overshears; F(zl , z2) := (zte"12, zzeZ1Z2) It is known that mappings of the third type do not belong to the group of automorphisms generated by the overshears. Moreover, the following result shows how complicated Aut(C") may be. Theorem 2.1.6 ([FMV 2006]). Let n. k r= DJ, n ? 2 and let a 1, ... , ak E C" be pairwise distinct points. Then there is a polynomial automorphism F E Aut(C")

such that Fix(F) = {aj : j = 1,...,k}. Proof. (Details are left to the reader as an EXERCISE.) Let a j =

E

C x C"-I. One may assume that the numbers a are all distinct (take a suitable invertible linear transformation). Denote by f : C polation polynomial for a"..., ak and a " j , ... , a",j,

C the Lagrange inter-

= 2,.... n, i.e. f

'The description of Aut(A) in the case where A := (z E C : r, < Jzl < r2) is a degenerated annulus (i.e. ri = 0 or/and r2 = +oo) is left for the reader. In particular.

Aut(C.)=(C.9z-+azEC.:aEC.)U(C.3zHa/zEC.:aEC.).


164

(f2, .... f") is a polynomial mapping with f(a) = aj". Then F : C" -> C", F(z) = F(zr, z') := (zt, z' + f(zj)), is a polynomial automorphism of C" with F (aj', 0) = aj. In a second step, consider the polynomial automorphism G : C" -+ C"

G(z) := (zr +z2+(zt -ai) ... (zi -ak), z2+(zt -a') ... (zi -a'), iz3.... , iz"). Then Fix(G) = ((a'1. 0), ... , (ak, 0)). Finally, put F := F o G o F-r and observe that F has all the required properties. From now on we will be concentrated on bounded domains in C".

Theorem 2.1.7 (Cartan). Let G C C" be a bounded domain, let a E G, and let fi : G -+ G be a holomorphic mapping such that (P(a) = a and 'P'(a) = id. Then

0 = id. Notice that the assumption that G is bounded is essential - take for instance

G .=C,a:=0,(P(z):=z(1-z). Proof We may assume that a = 0. Suppose that 0 $ id. Fix r, R > 0 such that I' (r) C D C IP(R). We have ao

O(z) = E Qk(z),

z E IP(r).

k=0

where Qk: C" -* C" is a homogeneous polynomial mapping of degree k (cf. Pro-

position 1.8.4). We know that Qo = 0 and Q, = id. Let ko > 2 be such that = Qko-r = 0, Qko # 0. Denote by 0" the v-th iterate of the mapping Q2 = 0, i.e. 00 := id, Ov+I ._ Ov o P. Then 00

V(Z) = z + vQko(z) +

Qv,k(z).

z E [P(r),

k=ko+i

where Q,,,k : C" -+ C" is a homogeneous polynomial of degree k. To prove this formula use induction on v. Suppose that the formula is true for a v > 1. Let 8 > 0 be such that '(IP(8)) C IP(r). Then for Z E IP(8) we get (EXERCISE) (Pv+r(z)

00

= V(NZ)) = (P(z) + uQk0((P(z)) + E Qv,k((P(z)) k=ka+r 00

=z+

Qk (z) + vQka (z + higher order terms) k=ko 00

+ E Q,,,k (z + higher order terms) k=ko+i

2.1. Introduction

165

00

= Z + (V + 1)Qk0(Z) +

Qv+l.k(Z)

k=ko+l

It remains to use the identity principle to conclude that the formula holds on the whole of P(r). Hence, by the Cauchy inequalities, for any z E F(r) we get

l; E T) < R.

IV(Qk0)j(Z)l

j = 1,...,n.

Letting v -+ +oo, we obtain Qko = 0; a contradiction.

Proposition 2.1.8 (Cartan). Let G, D C C" be circular domains2 with 0 E G, 0 E D, such that G is bounded, and let F E Biho,o(G, D). Then F is a linear isomorphism.

Notice that the assumption that G is bounded is essential - take for instance

G = D = C2, F(z1, z2) :_ (zI + f (Z2), z2), where f E 0(C) is a nonlinear entire function with f (O) = 0; cf. Example 2.1.5 (f).

Z E G. Then Ht E Auto(G) Proof. For l; E T put H{(z) := and H,'(0) = id. Therefore, by Theorem 2.1.7, HH = id, i.e. F(ez) = F(z), z E G, E T. Expand F into a series of homogeneous polynomials in a polydisc P(r) C G:

F(z) = Eoo Q,(z), z E F(r). k=1

Then

00

F(z) _

k-1Qk(Z),

z E P(r), E T. k=l This means that Qk = 0 in F(r) for k > 2 (EXERCISE), and so, by the identity principle, F = Q 1. Therefore F is a linear mapping. Since F is biholomorphic, it must be a linear isomorphism.

Proposition 2.1.9. Let II 11j be a C-norm in C"J, let

B1 := {zEC"i :Ilzlli C.

Observe that L is a C-linear mapping with L(b) = Lo(b) = nbll and IL(w)l = ILo(P(w))I = UP(w)I12 < IIwII2. W E C"2.

2.1. Introduction

167

the group Aut(ED") depends on d(n) := 3n real parameters; the group Auto (DI) depends on do(n) := n real parameters. Indeed, it is easy to see that e is a subgroup of Aut(U)"), (s'o is a subgroup of Auto(ID"), and & acts transitively on ID". We only need to show that Auto(®") C

(o. By Propositions 2.1.8, 2.1.9, any automorphism di E Auto(DI) is a linear isomorphism with 110(z)II,, = IIzIIoo, Z E C". Let [dsj,k]j,k=t,....n E (&L(n,C) denote the matrix representation of (P. We have n

max

{I-Vj,kzkl : j = ].....n}

Z=(z1....Zn)EC".

= IIzIIoo.

k=1

In particular, max{I(Pl,kl.

,I'n.kl)= 1,

I'j,iI+...+Imj,n1

1,

k

1,...,n,

j =1,...,n.

Thus the matrix ['Pj,k] has in each row, and each column, exactly one nonzero element (which must have absolute value 1). This means that (P E bro. (b) Unit Euclidean ball IBn. For a E ((Bn)., let

ha(Z)

Ila ll

2

l-

I1all2(IIaII2z i (z,a)a)

- IIaII2a + (z.a)a

ZEC"\((z,a)=1)DBn, denotes the standard Hermitian complex scalar product in C". Let, moreover, ho := id. Observe that in the case where n = I the above definition of ha agrees with that from (2.1.1). Denote by U(n) the group of all unitary automorphisms of C".4 Under above notation we have: where

Aut(IBn) = {U o ha : U E UJ(n), a E IB,), Auto(IB") = UJ(n). In particular: the group Aut(IB.) acts transitively on IBn; B. is homogeneous and symmetric; the group Aut(B") depends on b(n) := n2 + 2n real parameters; the group Auto(Bn) depends on bo(n) := n2 real parameters. 4A C-linear mapping L : C" - C" is unitary if one of the following equivalent conditions is satisfied:

(L(z'), L(z")) = (z', z"), z', z" E C"; IIL(z)I1=Wzp,zEC":

_

LL* = L* L = In (L is identified with its matrix representation; L` := L'). The group 11(n) depends on n2 real parameters.


168

Indeed, the fact that Auto(IB") = 4J(n) follows immediately from Propositions 2.1.8 and 2.1.9. Since ha(a) = 0, we only need to prove that ha E Aut(1B"). Direct calculations show that Fix an a E 1 - (ha(z). ha(w)) _

(I - (a . a))(l - !z w)) (I - (z, a))(1 - (a, w))

z. w E IB".

Taking w = z, we conclude that ha(IB") C 18n and_ha(alBn) C aIB". In particular, ha o h_a is well defined in a neighborhood of 8". Using once again direct calculations, we prove that ha o h_a = id. Hence ha E Aut(IB") and hat = h_a. (c) Unit Lie ball L". Let

L,,:={zEC":L"(Z)< 1) = {zEB.:211z112-I(:,2)12< I), where

Ln(z)

= (11z112 +

IIzll4 - I(z.

2)I2)1/2

= (11x112 + 11y112 +2 IIxII21Iy1I2

- (x,y)2)1/'2,

Z=X+iyEIR"+i(R"_C". The Lie norm L" is the maximal C-norm q: C" -+ (R+ with q(x) = Ilxll for all x E IR"

IR" + i 0 3 (cf. Exercise 2.1.14). Observe that: IL = 11).

L2 = {(zt, Z2) E C2: Izl + iz21 < 1, IZt - iZ2l < 1}

b"h ID2.

For n > 2 the ball L. is not Reinhardt. One can prove that for any a E L" there exists an ha E Aut(L,,) such that ha(a) = 0 (cf. [Hua 1963], p. 86-87; for n > 3 the proof is heavily "technical" and, therefore, we skip it); in the case where n = I the above mapping ha agrees with that in (2.1.1). Under the above notation we have:

Auto(Ln) =

1; E T, aELn}, E T, A E (D(n)) _:'so,

where 0(n) := the group of all orthogonal6 isomorphisms A : 1R" - 1R" acting on

C" according to the formula C" a x + iy H A(x) + iA(y) E C". 5Recall that 11 11 stands for the Euclidean norm.

6An IR-linear mapping A : R" -. IR" is orthogonal if one of the following equivalent conditions is satisfied:

(A(x'), A(x")) _ (x', x"), x', x" E Rn;

IIA(x)II=lx!1,xER"; AA' = A'A = In (A is identified with its matrix representation). The group 0(n) depends on (2) real parameters.

2.1. Introduction

169

In particular:

the group Aut(Ln) acts transitively on L"; L. is homogeneous and symmetric;

the group Aut(Ln) depends on £(n) (2) + 2n + I real parameters; the group Auto(L") depends on eo(n) := (2) + 1 real parameters. Indeed, since (s'o is obviously contained in Auto(L"), we only need to prove that any automorphism 0 E Auto(Ln) belongs to 0 0. We already know that 0 is C-linear and Ln o 0 = Ln (Proposition 2.1.9). As always, we identify ds with its matrix representation. Write 0 = A + i B, where A, B E MI (n x n, R). Then the identity Ln o di - Ln implies that

IIAx-Byil2+ IIAy+BxII2

+ 2IIAx-BYII2IIAY+Bxl12-(Ax-By,Ay+Bx)2 = 11X112+11Y112+2 IIxlI2IlYII2-(x,y)2, X+iy EC".

(2.1.2)

Suppose that we already know that A, B are It-linearly dependent. Then we write A + iB = i;P with E T and P E Ml(n x n, It). Putting in (2.1.2) y = 0, we get IIPx1I = Ilxll, X E C", which shows that P E 0(n). Thus, it suffices to show that A, B are It-linearly dependent. We may assume that A 57E 0 and B # 0. Put U := Ker A, V := Ker B. Suppose that U and V are non-zero. Then identity (2.1.2) implies that

IIAy+BxII2 = IIXII2+ llY1l2+2 IIxlI2IlYII2-(x,y)2,

(x,y) E U X V.

In particular, if y = 0, we get IIBxlI = IIxII, x E U. Similarly, IIAYli = Ilyll, y E V. Consequently, (Ay, Bx) =

-III- (x, y)2,

llx1I2

11 y

(x, y) E U x V.

Since the left-hand side is bilinear, we conclude that either U or V is trivial. Suppose, for instance, that A is non-singular. Observe that for x + iy 0 0, the right-hand side of (2.1.2) is differentiable if Ilxll2llyll2 54 (x, y)2, i.e. x and y are It-linearly independent. Consequently, for

x + iy # 0 we get: x and y are 6t-linearly dependent iff Ax - By and Ay + Bx are 1R-linearly dependent.

Take an arbitrary x E (It"). and let y := ax with a E R. Suppose that Ax - By = fl(Ay + Bx) for some fl E 1R. Hence (1- af)Ax = (a + $)Bx, and consequently, Ax and Bx must be Qt-linearly dependent.

In the remaining case we have Bx = -aAx. Thus Ax and Bx are It-linearly dependent for any x E C". Put C := A-' B. We only need to show that there exists a y E It such that C = y IIn . Fix x, y E (R').

and let y(x), y(y) E Il be such that Cx = y(x)x, Cy = y(y)y. We want to prove


170

that y(x) = y(y). If x and y are IR-linearly independent, then x + y # 0 and C(x+y) = y(x +y)(x +y), which directly implies that y(x) = y(y) = y(x+y). If x and y are IR-linearly dependent, then the result is obvious.

(d) Observe that d(1) = f(1) = b(1) = 3.

d(2) = 42) = 6 < 8 = b(2),

d(n) < f(n) < b(n), do(1) = fo(1) = bo(1) = 1,

n > 3,

do(2) = fo(2) = 2 < 4 = bo(2),

do(n) < fo(n) < bo(n),

n > 3.

Consequently, it is intuitively clear that Bih(ID", IB") = 0 for n > 2 and that Bih(m", lL") = Bih(IB", IL") = 0 for n > 3. A precise proof will be presented in Theorem 2.1.17.

Exercise 2.1.13. Let N : C" - IR+ be an arbitrary complex norm, n > 2, and let

B := {z E C" : N(z) < 1), B(r) := {z E B : N(z) < r} and A = A(r)

{zEB:r C is C-linear, II0I1R" ll < 1) = sup(I Ilxsin 0+ycos011. Step 3. By Step 2, IIZ 112,;n =

min{

II2 + (11x 112

- IIY 112) sine 0 + (x, y) sin(20) : 0 E [0, 2n]}

(Ilx112-IIY112)2+4(x,y)2), Z=x+iyEC".

= 2(IIxll2+IIYII2+

Step 4. Using Lagrange's multipliers gives max(I(z,a)I:aElR",

Ila11=1)=IIzIlmin,

Remark 2.1.16. Notice that:

Z=X+iyEC".

-

Ilzll, Z = X + iy E C". max{Ilxil, IIYII} < 11Zilmin 11 Z 11 min = II z II if x, y are linearly dependent. In particular, if n = 1, then IIZllmin = IZ1, Z E C. IIZllmin = max{IIxll, llyll} iff (x,y) = 0. inf{q : q E 11m} is not a norm.

Indeed, let q8(z) := (1 - e)Izt + iz21 + elzt - iz21, z = (z,, z2) E C2. Then q8 E F., but q8(1. i) = 2e -- 0 (cf. [Hah-Pfl 1988]). The minimal ball M,, (z E C" : ll z ll min < 1) is not a Reinhardt domain for n > 2. If n = 2, then 1Z1 2iZ21 + IZi

2iz21'

IIZllmin =

z = (Zi,Z2) E C2.

In particular, M{2 is biholomorphic to the domain {w E C2: Iwt I + Iw21 < 0*

Aut(Mn) =

E T, A E 0(n)) (cf. [Kim 19911, [Zwo 1996]).

In particular, M. is neither homogeneous nor symmetric. The group depends on (2) + 1 real parameters. Consequently, Bih(M,,. D) = 0 for D E {®", B", L"),

n>2.

The following theorem is a generalization of the famous Poincarrs theorem saying that Bih(IB", U)") = 0 for n > 2.

Theorem 2.1.17. Let 2 < n = nt

mt

Bµ E {B",,,

µ = 1, ... , k, B;, E {Mm , Lm,), v = 1, ... , f. Assume that if BA = L.,, (resp. B;, = Lmv ), then nu, > 3 (resp. m > 3) - cf. Example 2.1.12 (c). Then

Bih(Bl x ... x Bk, B' x

x Be) 36 0


174

iff f = k and there exists a permutation or E Gk such that matµ) = nu and

Bat,) = Bµ,µ = 1.... , k. Moreover, every biholomorphic mapping F : Bt x is, up to a permutation of Bi .... . Bk, of the form

x Bk -+ Bi x

x Bk

Z = (Z I, ...Zk) E B1 x ... x Bk,

F(z) = (Fl(Z1)...., Fk(Zk)), where Fu E Aut(B.), µ = 1. ... , k.

Remark 2.1.18. (a) In the case where B. = (B,,, µ = 1.... , k, B;, = B, , v = I-- P, Theorem 2.1.17 shows that Bih(IBn, x X (Bnk , IBm X ... X Bmt) 0 0 if f = k and there exists a permutation a E Cyk with ma(,) = nip, µ = 1.... , k. (b) In particular, in the case where k = 1, Bt = IBn, e = n > 2, Theorem 2.1.17 reduces to the Poincarr theorem. 1

(c) In the case k = f = n, Theorem 2.1.17 reduces to the description of Aut(IDn ) given in Example 2.1.12 (a). (d) In the case k = 1, B1 = IL,,, By = Bm,,, v = 1, .... P, Theorem 2.1.17 shows that Bih(Ln, Bm, x x IBmt) = 0 for n > 3 (cf. Example 2.1.12 (c)).

Proof of Theorem 2.1.17. (The main idea of the proof is due to W. Jamicki.) Since Bµ is homogeneous, µ = 1, .... k, Remark 2.1.2 (c) implies that the domain B1 x x Bk is also homogeneous. Now, by Corollary 2.1.10, Bih(B, x ... x Bk. Bi x

x BL) 36 0 iff there exists a C-linear isomorphism F = (F1, .... Ft): C" -+ Cn = Cm, x . . . x Cn't such that max{II F1(Z)Ifi

IIF,(Z)II'',

Z= (Z1.

.:k) E C"1 X ... X C"k

where 11

II

II

IIu

II' -_

11 = Euclidean norm

Lno = Lie norm 11

if B11,

=

(B",,.

if B,L = Ln,,,

11 = Euclidean norm

Lm,, = Lie norm

µ=1,

if B; = IBm,,. if B;, = Lm,, .

,k.

V = 1,

,1.

First consider the set Ap C C P, p > 3, on which the Lie norm Lp is not real analytic, i.e. Ap := {w E C1': IIw1I4 = I(w, 0) 12)

={w=(w,....,wp)EC':w,i , E IR, i,j = 1,...,p} Observe that Appt is closed and (Ap). = Uf I Mi, where

*i((tl, .Ip-1)

(t1/

... ,ti-I

M,

(EXERCISE).

>/' (C. x [RP-1),

....tpl0;

2.1. Introduction

175

O,j is a real analytic mapping. In particular, , P+l(*;(K)) < +oo for every compact K C C. x IRP-1, where 3e P+' denotes the (p + 1)-Hausdorff measure in IR2P (EXERCISE). Note that p + I < 2p - 1. Consequently, AP is a countable union of compact sets with finite (p + 1)-dimensional Hausdorff measure ([Fed 1969], § 2.10) and therefore CP \ AP is connected (see [Rud 19801, Theorem 14.4.5, [Jar-Pfl 2000], p. 226).

Now, let C :_ {(z, w) E CP X Cq : N1(z) = N2(w)), where Nl (resp. N2) stands for the Euclidean or Lie norm in CP (resp. Cq). If Nl (resp. N2) is the Lie norm, then we assume that p > 3 (resp. q > 3). Then C is nowhere dense. Indeed, define St C CP, S2 C Cq,

S'

{0}

if Nl = 11

AP

if Nl = LP,

11,

S2

{0}

if N2 = 11

Aq

if N2 = Lq.

11,

Then S := (Si X Cq) U (C" x S2) is a closed set being a countable union of compact sets with finite t-dimensional Hausdorff measure where t < 2(p + q) - 1. Hence CP X Cq \ S is connected. Suppose that int C # 0. Then, by the identity principle for real analytic functions, CP X Cq \ S C C. Therefore, by continuity, C = CP X Cq; a contradiction.

Thus, for every A' 0 µ" and v' # v", the sets

{(Zl,....Zk)EC", x...xC"k {(w. ... , wt) E C"'' X ... X C" : Qwv' IIv' = IlW,are nowhere dense. Consequently, since F is homeomorphic, the set

U {ZEC":IIFv'(Z)IIv'=IIFv"(Z)IIv,,}U U v'3Ev"

is nowhere dense. In particular, for every j E { 1, . . open set Sly C C" and an s E 11, ... , e} such that

.,k

} there exist a non-empty

Z E 52j.

IIFs(z)IIS = Ilzilll Let

Tl

{0} A ";

if II IIi = 11 11, if II Ili = L" ; ,

_ T2

{0}

if II III = II II

A..,

if 11

11' = Lm,.

Then

T:=(C"" x...xC"i-1 xTl xC"i+I x...xC"k) U F-t(Cmi x ... X C`-' x T2 x Cms+I x ... X Cmr)

176


is a closed set being a countable union of compact sets with finite t-dimensional

Hausdorff measure where t < 2n - 1. Hence C' \ S is connected and, by the identity principle for real analytic functions, we conclude that IIF..(z)11' = lIzjllj,

z E V.

(2.1.4)

In particular, s =: or(j) is uniquely determined. Moreover, a (j') 0 a (j ") for # j ". Hence k < e. Since the Euclidean norm in CP is real analytic on (C"),, but the Lie norm is real analytic only on CP \ AP, we conclude that both norms II IIj and II IIS must be of the same type (i.e. both must be Euclidean or both must be Lie). Moreover,

n - ms = dim Ker Fs = n - n j, which implies that m3 = n j and Bs = Bj. It is also clear that F, depends only on z, i.e. F, (z) = Uj (z), where Uj : C"' - C"j is a linear isomorphism. Condition (2.1.4) guarantees that Uj E Aut(Bj). Finally, + nk = n. k = e because ma(1) + + m0(k) = n 1 +

Exercise 2.1.19. Let B1, ... , Bk be as in Theorem 2.1.17. Find a generalization of Example 2.1.12 (a) and characterize the group Aut(B1 x . . . x Bk) in terms of Aut(BI),... , Aut(Bk). The phenomenon described in Theorem 2.1.17 appears also under other assumptions. Recall, for example, the following classical general result.

Theorem* 2.1.20 ([Nar 1971 ], p. 77). Let D j, G j be bounded domains in C"J such that there is no non-constant holomorphic curve cp : ID -- aG j, j = 1.... , k. Then any biholomorphic mapping

1L: D1x...xDk-+ Glx...xGk is, up to a permutation of G1..... Gk, of the form

CZ1, Zk) =

71k(zk)),

(Z1,...,Zk) E D1 x ... x Dk,

wherey'j EBih(Dj,Gj), j = 1....,k. Notice that the above theorem applies for instance to complex ellipsoids (in particular, to Euclidean balls), which is a direct consequence of the following lemma.

Lemma 2.1.21. If P E N", then there is no non-constant holomorphic curve (p: ID -+ aIp.

X. is holomorphic. First consider the case p = 1. Then Ilwll = 1. Obviously, IcjI < 1, j = 1, ... , n. Composing Sp with a rotation we may assume that V(0) = (1, 0.... , 0). Then, by the maximum principle, lpl =_ 1. Consequently, cp j - 0, j = 2, ... , n. For arbitrary p we only need to observe that (dpi' , ... , (pp,"): L UnProof. Suppose that (p = (1P1, .... V,,): lD

2.2*. Cartan theory

177

Remark 2.1.22. On the other hand, the Lie ball 1L", n > 2, does not satisfy the condition from Theorem 2.1.20. Indeed,

UE)

In particular, Theorem 2.1.20 does not imply Theorem 2.1.17.

2.2* Cartan theory Summarizing, the previous results show that for n < 3 we have the following bounded homogeneous domains (which are not biholomorphically equivalent). n 1 domain 1

1

ID

2

182, ID2

3

183, V, IDXIB2,L3

Theorem* 2.2.1 ([Car 1935]). If n < 3, then any bounded homogeneous domain G C C" is biholomorphic to one of the above canonical homogeneous domains. In particular, any bounded homogeneous domain G C C" with n < 3 is symmetric (Remark 2.1.2 (c), Example 2.1.12).

The first example of a 4-dimensional homogeneous non-symmetric bounded domain was given by I. Piatetsky-Shapiro in [Pia-Sha 1959]. Theorem* 2.2.2 ([Car 1935]). Every bounded symmetric domain is homogeneous. Moreover, every bounded symmetric domain G C C" is biholomorphic to a Cartesian product of canonical symmetric domains belonging to the following six Cartan types:

n=pq,12, IIp:=(ZED.I(pxp,C):Z'=-Z, IIp-ZZ*>0}; n=(PZl),p2 1,IIIp:=(ZE DvI(pxp,C):Z' = Z, IIp-ZZ*>0); IV" := 1Ln.

The domains of types 1-117 are called classical. They are balanced - cf. Definition 1.4.14. n = 16, an exceptional domain V16; n = 27, an exceptional domain V127. 'For A E M(m x m,C), "A > 0" means that A is positive definite. i.e. X'AX = m E'P t ai.k X; Xk > 0 for all X 6 (C"')*.


178

The above Cartan domains are not biholomorphically equivalent except for the following cases: bih

(a) 12,2 '=' La,

(b) 112

ID, 113 Lh (B3, 114

(c) III1 ^_- I), II12

bib

bib

L6; thus type 11 is essential only for p ? 5,

L3; thus type III is essential only for p > 2,

bih

(d) L t = 9), L2 '_- ID2; thus type IV is essential only for n > 5.

Let 1/i(n) denote the number of biholomorphically non-equivalent canonical bounded symmetric domains G C C" (from the above list) and let W(n) be the number of non-equivalent bounded symmetric domains G C C" (which are biholomorphic to Cartesian products of canonical symmetric domains). The following table describes the situation for I < n < 30.

n

I

*(n) 1 1 1 11)

1

1

2

1

3

2

4

2

5

2

6

4

136,12.3

7

2

8

9

II

(112D) (1111=D)

182 183

N

111

(113.183)

1112

1 104,12.2

Y/VI

SY(n)

(LI=D)

1

(L2-D2)

2

(L3

1112)

4

(L4

12.2)

7

L5

11

L6

21

187

L7

31

3

188,12,4

L8

51

3

189, 13.3

L9

80

10

5

181o, 12,5

Llo

126

11

2

B11

L11

187

12

4

1812, 12.6, 13.4

L12

292

13

2

1813

L 13

427

14

3

1 1 1814,12.7

L14

638

15

5

615,13.5

16

5

816,12.8,14.4

L16

17

2

18j7

L 17

1960

18

4

818,12.9, 13.6

LI8

2843

191 1

2

B19

L19

4024

20

4

1820,12.10,14.5

L20

5724

21

5

1821,13,7

12,

8046

L22

11303

L23

15687

22

3

23

2

185

1

822, 12.11 1

1823

L6)

(114

115

116

117

1113

1114

1115

1116

935

L15 V16

1371

2.2*. Cartan theory

I

n 1 1*(n)J I

IV :'l (n)

Ill

11

24

5

1 1IB24,12.12,13.8,14,6

C24

21840

25

3

1 1821, 11,5

L25

30058

26

3

1 1 ®26,12.13

126

41366

27

4

827, 13.9

28

6

829,

29

2

30

5

56525

V127

127 128

77126

829

L29

1 1104490

830,

_30

1

1117

118

179

11415261

Remark 2.2.3. The biholomorphisms in (a)-(c) are given by the following formulas: (a)

iZ2+Z31 E 12,2

+iZ4

L4 B (Z 1, Z2, Z3, Z4) H IZ1 iZ2-Z3

Z1-iZ4J

(EXERCISE).

(b)

0

63 9 (Z1,Z2,Z3) H -Z1 L-z2

ZI

Z2

0

Z3

-Z3

0

E 113

(EXERCISE),

and (cf. [Mor 1956]) 0

+iZ2) L6 9 (ZI....,Z6) H -(Z1 -(z3 +iZ4) -(Z5 + i Z6)

Z1+Iz2

Z3+iZ4

0

Z5-IZ6

-(Z5 -IZ6) -(-Z3 + iZ4)

-(Z1 - iZ2)

0

Z5+iz6 -Z3+iZ4 EII4. Z 1 -122 0

(c)

13 9 (ZI. Z2, Z3) -*

LZ1

2iZ3

Z1

1.Z3] E 1112.

Exercise 2.2.4. Find a formula for fl n). Remark 2.2.5. Let us mention the following two results related to various characterizations of a bounded domain D C C" by its automorphism group Aut(D). (a) Assume that b E aD is a point such that aD is strongly pseudoconvex at

b (such a point always exists if aD E e2). Moreover, assume that there exist a compact K C D and sequences (ak)k° I C K, ('k)°k° 1 C Aut(D) such that bih

Ok(ak) -+ b. Then D ^- IB" ([Ros 1979]). (b) We say that a bounded domain D C C" has piecewise ek-boundary if aD is a topological (2n -1)-dimensional manifold, and there exist an open neighborhood pm = 0 on aD and for U of aD and p E ek (U, Qt"') (with some m) such that p1 every

allzEUwith p11(z)_

have .

=ple(z)=0.

180


Let D C C" be a bounded homogeneous domain with piecewise e2-boundary. bih

B", x ... x (Bno ([Pin 1982]). In particular, every bounded homogeneous Then D domain D C C" with smooth boundary is biholomorphically equivalent to B". (c) In virtue of Theorem 2.1.17 (and Remark 2.1.18), the above result implies that the boundary of 1" (n > 3) is not piecewise e2.

Exercise* 2.2.6. Prove directly (without using Remark 2.2.5 (c)) that the boundary of 1,, (n > 3) is not piecewise e2.

2.3 Biholomorphisms of bounded complete Reinhardt domains in C2 Let us look more thoroughly into the problem of biholomorphic classification of

bounded Reinhardt domains D C C2 with Vj n D # 0, j = 1.2. The case where DI, D2 C C2 are bounded convex complete Reinhardt domains was first considered by K. Reinhardt in [Rei 19211. The general case was completely solved by P. Thullen in [Thu 1931 ] (see also [Car 1931 ]). Observe that, by Proposition 1.12.8, we may always assume that D 1, D2 are bounded complete Reinhardt domains of holomorphy. By rescaling variables, we may reduce the situation to the case where D j is no)7alized, i.e.

{Z E C : (ZI,O) E Dj} = {z2 E C : (0,Z2) E Dj} = D.

j = 1.2.

(2.3.1)

In particular, D C D2.

Lemma 2.3.1. Let D C C2 be a normalized complete Reinhardt domain of holomorphy. Then 1D2 D = {(z1. z2) E : Izi l < R(Iz21)}. where R = RD: [0. 1) -+ (0, 1] is a continuous function with R(0) = 1. Prof. Since log D is convex, we conclude (EXERCISE) that for every u E (0, 1) there exists exactly one t =: R(u) E (0, 1) such that (t. U) E 8R(D) (recall that R(D) := {(Iz11,1z21) : (zI, z2) E D}). It remains to observe that the function R: [0, 1) -+ (0, 1] is continuous (EXERCISE).

Example 2.3.2. Let IEP = {(Z1 2) E C2 ; 1ZI 12P1 + IZ212P2 < l },

P = (PI P2) E

be a complex ellipsoid; cf. (1.18.5). If (PI = 1, P2 # 1) or (pi

lR>0,

1, P2 = 1),

then EP is traditionally called a Thullen domain. The domain IEP is a normalized complete Reinhardt domain of holomorphy (cf. Exercise 1.18.7). Notice that

RE,,(t) := (I -t2P2)I/(2P1),

t E [0, 1).

2.3. Biholomorphisms of bounded complete Reinhardt domains in C2

181

Exercise 2.3.3. Determine the function RD for the domain D :_ {(z1, z2) E ID 2 : Izr dal Iz2Ia2 < 9},

where 011, a2 > 0, 0 < 9 < 1.

Recall that for i; = i;2) E T2, we have put TT(z) _ z = (z1, z2) E C2. Let S(zr, z2) := (z2. z1), TT := TT O S.

z = (6Z I, 6z2),

The following three results due to P. Thullen [Thu 1931) give the full characterization of biholomorphic equivalence of bounded normalized complete Reinhardt domains of holomorphy in C2.

Theorem 2.3.4. Let a > 0, a # 1. Then the group Aut(l£(a.1)) coincides with the set of all mappings of the form Ic12

iE(a,l) 9 Z H

(.Izl((_)2)

p .S2hc(z2))

E I(a.l)

(2.3.2)

where c E ID, (l; I, 2) E T2 , and the branch of the power () 11(2a) may be arbitrarily chosen.

In particular, IE(a,l), a $ 1, is not homogeneous. Observe that Aut(i£(a,t)) (a 34 1) depends on four real parameters (cf. Example 2.1.12 (d)).

Theorem 2.3.5. Let D C C2 be a normalized bounded complete Reinhardt domain of holomorphy. (a) The following conditions are equivalent:

(i) Aut(D) acts transitively on D;

(ii) either D = ID2 or D = 82 (cf Example 2.1.12 (a), (b), (d), Theorem 2.1.17).

(b) Assume that D

{ID'. 82}. Then the following conditions are equivalent:

(i) there exist b E V and Ob E Aut(D) such that Ob(z) _ (zl fb(22), mb(z2)),

z = (z1, z2) E D,

wherefb E 0* (D) 8 and mb E Aut((D), mb (b) = 0 (note that fib (0, b) _ (0, 0));

(ii) for every c r= lD there exists a 1C E Aut(D) such that

0c(z) _ (z1fc(z2), mc(z2)).

z = (zl , z2) E D,

where ff E 0*(ID) and m, E Aut(D), mc(c) = 0 (45c(0, c) = (0, 0));

80*(G):=If e0(G): f(z)

0,

zeG).

182


(iii) D = IE(a,t) for some a # 1. (c) The following conditions are equivalent:

(i) there exists an a E D such that Aut(D) = Auta(D); (ii) Aut(D) = Auto(D); (iii) any automorphism P E Aut(D) is of the form TT or T, with second case is possible only if S (D) = D.

E

T2 (the

Obviously, (b) may be formulated also with respect to the second variable.

Theorem 2.3.6. Let Dt, D2 C C2 be normalized bounded complete Reinhardt domains of holomorphy and let F E Bih(D1, D2). Then we have the following possibilities:

Dc =D2=1D2;

Dt=D2=1B2; DI = D2 = IE(a,1) with a 34 1 and F = tyc,C for some c E ID and i E T2, where tfic, is as in (2.3.2);

DI = IE(a,t), D2 = lE(t,a) with a # 1 and F = S o tPc,t, for some c E ID and E T2, where +Lc, is as in (2.3.2);

DI = D2 = IE(t,a) with a # 1 and F = S o tPc,t o S for some c E ID and E T2, where 'J' , is as in (2.3.2); in all remaining cases, either D2 = DI and F = TT with i; E T2, or D2 = S(D1) and F = T* with E T2; in particular, F E Biho,o(DI, D2). Our method of proof of the Thullen theorems needs the notion of the so-called Wu ellipsoid introduced by H. Wu in [Wu 19931, see also [Joh 1948] and [Jar-Pfl 2005], § 1.2.6.

Exercise 2.3.7. Let L, Lt, L2 E GL (n, C). Then:

(a) L-t (IBn) = {z E C"

:

zt M2 < 1), where M := Lt L, and therefore,

L-t ((n) is given by the Hermitian scalar product (z. w) H z' Mw.

(b) A2n(L(IBn)) = d (c) L-1(0,) is a Reinhardt domain iff L-t((Bn) = {z E C" : r z E B.) for some r E IR;o.

(d) Li t(lBn) = L2 t(8,) if L2 o L, t E OJ(n). Lemma 2.3.8. For every bounded domain 0 96 D C C", the family

{L-'(ln) : L E GL(n,C), D C L-' (B,,))9 contains exactly one domain (the Wu ellipsoid) IE(D) with minimal volume. 9Observe that the family is not empty because D is bounded.


183

Proof. Let .F(D) :_ {L E GL(n,C) : D C L (ln)). By Exercise 2.3.7(b), we want to maximize the function .F(D) 9 L H I det LI. Let B be the smallest balanced domain containing D, B = int n GDD G. Then B C L-' (S.) for G is balanced

every L E .P(D). Hence IIL(z)II < hB(z), z E Cl, where hB stands for the Minkowski function of B. In particular, the set Y(D) is bounded in D,I(n x n, Q. Consequently, there exists an Lo E F (D) such that I det Lol = C := sup{ I det L I : L E F(D)} (EXERCISE).

Let F0 (D) := J L E F (D) : I det L I = C). We have to show that M LZ o L I E OJ(n) for any Lt, L2 E .Fo(D) (Exercise 2.3.7(d)). We may assume that det L I = det _L2 = C. Suppose that the Hermitian matrix M1 M ,54 IIn. Then we can write MI M = PAP-I , where P E OJ(n) and A is a diagonal matrix with elements dl,..., d" > 0. Since det M = 1, we have d, . d" = 1. Moreover, since M' R # IIn, we conclude that dJ 54 1 for at least one j. Observe that

17'(L;LI + L2' T,2)2 = 2(IILI(Z)II2 + IIL2(=)II2) 2(hB(z) +hB(z)) = hB(z).

Z E C".

Write 1(L i L1 + LZL2) = L' L with L E GL(n, C). The above inequality shows that L E F (D ). Thus I det L I < C. On the other hand we have

IdetLI2 = 2" det(L'LI +L2L2)

= 2" det(Li)det(1, + (Li)-IL2L2(LI)-I)det(LI)

= ZC2 n det(II" + MM) = I C2det(P)det(IIn + P-I M'MP)det(P-I ) = ZnC2det(II"+A)=C21

2dI

...

1

2d"

>CZ dI...dn =C2:

a contradiction.

Remark 2.3.9. (a) If A E GL(n, C), then for every bounded domain D C C" we have A(L(D)) = L(A(D)). In particular, if A(D) = D, then A(I=(D)) = E(D). (b) If D is a bounded Reinhardt domain, then lE(D) = (z E C" : r - z E (Bn} for some r E UZ;o.

Lemma 2.3.10. Let D 1, D2 C C" be bounded Reinhardt domains with 0 E D I fl D2. Let F E Biho,o(DI, D2). Then

F(z) = p-t U(r z). where U E OJ(n), r. p E UZ;o, and p I := (1/pl,..., 1/pn).

184


Proof By Remark 2.3.9 (b),

IE(D1)={z E

GL(n.C). By Remark 2.3.9(a), F(IE(D1)) _

E

IE(D2). Consequently, the linear isomorphism

C"

Z

p F(r-I Z) E C"

0

maps Mn onto 1Bn, and therefore, it belongs to U(n).

Proposition 2.3.11. Let DI, D2 C C2 be bounded complete Reinhardt domains of holomorphy and let F E Biho,o(D1, D2). Then: either Dj = E(DD ), j = 1, 2, and F has the form from Lemma 2.3.10, or

IE(Dj), j = 1, 2, and F(zi, z2) = (rI I za(1), E T2, a E 62 (r1, r2) E R2 >O' Dj .

where

Proof. Using Remark 2.3.9 (b) and Lemma 2.3.10, we may assume that, after rescaling variables, we have DJ IE (DJ) = 82, j = 1, 2, and F E (IJ (n). Next, by permuting and rotating variables, we may also assume that

F(zI,z2) = (z1cosa+z2sina,-zIsina+z2cosa) with a E [0,;r/2) (EXERCISE). We have to prove that a = 0. Suppose that a E (0, 7r/2) and consider the following construction.

T a k e a point x° = (r cos 9, r sin 9) E Q Q. n aDI (r >0,0< 9 0. Thus D = IE(c,I). Step 6. Observe that if D = E(,. 1) with a 96 1, then Ob(ZI,Z2)

_

(IzI(()2) 11 b y

12

.

2hb(z2)).

z = (zl,z2) E IE (a,I),

where (6, %2) E T2. Indeed, since (1 - 1b12)(1 - 1z212)

1 - 1hb(z2)12 =

z2 E (D.

(2.3.9)

11 - bz212

we get (cf. (2.3.7))

Ilb(Z2)I -

R(Ihb(z2)1) R(IZ2I)

(I - Ihb(z2)12 I - 12212

21.

)

I - IbI2 2L. = (11 -bZ212)

Z2 E D.


191

(b) Let +P := Ob I. Observe that W(z) = (z2g(zt),k(zt)), z = (zt,z2) E D, where k = Mb I E Aut(ID), k(0) = b, and g E O `(ID). Fix A E T \ (1 } and put a := mb(,kb) E D.. Then (Pb o T(t,I/,A) o (P(a,0) _ Ob o T(I,/A)(0,k(a))

= 'Pb o T(I.I/x)(0.116) = 45b(0, b) = (0.0). Moreover,

45Q (zt, z2) : = 'Pb o T(I,I/x) 0 111(21, 22) = 45b o T([,1/x)(Z2g(z0 k(zt))

_ Ob(z2g(z1), (l/A)k(zI)) = (mb((1/A)k(z,)),z2g(z,).f((1/A)k(z,))) (mQ(za), z2fa (z1)). (z1, z2) E D.

0

Proof of Theorem 2.3.4. We already know (cf. the proof of Proposition 2.3.14 (a), Steps 4 and 6) that 2

Aut(E(a,I)) J I (E(a,1) 3 Z H

(,zl((l1 - I cZ2)2)

-

.

J

2hc(z2)) E E(a,t)

CE(D.

(ry

111.

2)ET2}

yy

_ (45 E Aut(IE(a,t)) : $ has form (2.3.5)} Moreover, (S' is a subgroup of Aut(E(a,I)) (EXERCISE).

Fix a 45 E Aut(IE(a,t)). If 45(0, 0) = (0, 0), then, by Corollary 2.3.12, either 45 = TT or 45 = T . The second case is impossible because E(., 1) is not symmetric. Hence 45 E 6r.

Now, assume that 0(0,0) = (a, b) 0 (0, 0). Then F := +vb,l 0 45 E Aut(IE(a,I)) and F(0, 0) = (c, 0) for some c E D. If c = 0, then F E (Sf and hence 45 E (SS.

Suppose that c 54 0. Put G := S o

F-I

o S E Aut(I_(t,a)). By Proposition 2.3.13, G is either of the form (2.3.5) or (2.3.6). In the first case Proposition 2.3.14 (a) implies that E(1,«) = E(p,t), which is impossible. In the second case, since G(8E(, )) = 8E(,,,) we get Ihc(z2)I2 + (1-

IZ212a)aIfc(z2)12a

1,

z2 E D.

Hence

1- Ihc(z)I2 (2.3.9) 1- IC12 Ifc(z)I2a = ( I - IZI2a)a I

Consequently, a = 1; a contradiction.

1-1=12 a

ZEID.

192


Indeed, suppose that 0 < a < 1. Then for every

E T we get

I

lim

II-

(1 - ta)a

= 0.

Hence, by the maximum principle, fc - 0; a contradiction. If a > 1, then lim

1

- 11 -

(1 - ta)a

1- Ic12 lim r,l- 1 -t

=mot Iff(z)120

-0

and we have again a contradiction.

Proof of Theorem 2.3.5. (a) Assume that D is homogeneous and D # 182. In

particular, for any b E D there exists a (bb E Aut(D) such that Pb(0, b) = (0, 0). By Proposition 2.3.13, 'b is either of the form (2.3.5) or (2.3.6). Now, by Proposition 2.3.14, either D = ID2 or D = lE(«,t) or D = IE(t,a) with a 96 1. By Theorem 2.3.4, the only homogeneous case is D = ID2. (b) follows from Proposition 2.3.14 (a) and Theorem 2.3.4. (c) follows from Corollary 2.3.12 with DI = D2 = D.

Proposition 2,3.15. Let D C C2 be a normalized bounded complete Reinhardt domain of holomorphy and let ' E Aut(D) be such that 0(0.0) = (a. b) with ab $ 0. Then there exists a W E Aut(D) such that l'(0, c) = (0.0) or P(c, 0) _ (0, 0) for some c E D..

Proof. Put Vo := ((zl,z2) E D : ztz2 = 0) = (D x {0)) U ((0) x ID), V. Vo \ {(0, 0)) = (D. x (0)) U ((0) x DD.). Suppose that the result is not true, i.e. (*) F(0, 0) ll V, for every F E Aut(D) (equivalently, (0, 0) 0 W(V.) for every E Aut(D)). Define Wt

M

0-t o Tt 0 0 E Aut(D), {T,r(P(t)) : q, E T2},

lpt(0,0),

E T2, 1' t (Vo),

E T2.

Vo for all E T2 \ ((1,1)). Indeed, in view of (*), P(1;) E Vo if Note that (1, 1). W (0, 0) = (0, 0), which means that TT o 0(0, 0) = cP(0, 0) and hence

Moreover, m fl Indeed, it is clear that

E M fl Fix q. l; E T2 and suppose that i.e. W I(T0(P())) = p t o T. o i' (0.0) E Vo. By (*) we E get W t (T,(P())) = (0,0). Thus T,7(P()) = P(i;). We are going to show that

(**) there exists a point

NCM.

which lies on a smooth 3-dimensional surface


193

Assume for a moment that (**) is already proved. Then the intersection N fl cannot be a point, which leads to a contradiction. Indeed, N' :_ PP(N) is also a smooth 3-dimensional surface. It suffices to

prove that N' fl Vo 0 ((0, 0)). Assume that

N' = {(xl,YI,x2,q(xt,Yt,x2)): (XI,Yt,x2) E U3), where U C IR is an open neighborhood of 0 and (p is a smooth function in U. Then

N' fl Vo contains the curve {(0, 0, x2, 00, 0,x2)) : x2 E U). All other cases are similar (EXERCISE).

We come back to (**). Consider the mapping

i

(0, 2,r) x (0, 2n) D a i

(I

(P(e;a))1I

I(P(eia))21)

E (>o.

Put W := f((0, 2ir) x (0, 27r)). Observe that if W contains a smooth curve, then M contains a smooth 3-dimensional 2-circled surface. Indeed, suppose that W contains a graph u = ap(t), t E U, where U C IR>o is open, .p is smooth in U and V(t) > 0, t E U. Then M contains the set

M'

(e'# t, e'Y(p(t)) : t E U,

y E IR}.

Consider the mapping Ux!R2

(t,#,y) H (e'Ot,e'YV(t))_(tcosP,tsin$,cp(t)cosy,(p(t)siny) E 1R4

and calculate g'(t, P, y):

g,(t

y) _

cos$

-tsinP

sin fl

t cos 6

Ip'(t) cosy _V (t) sin y

0 0

0 0

-V(I) sin y VQ) Cosy

Then rank g'(t, f, y) = 3 (EXERCISE), which implies that M' locally contains a smooth 3-dimensional surface. Now we prove that W contains a smooth curve. The mapping f is real analytic (EXERCISE). If there exists an a with rank f'(a) = 2, then W contains an open set and, therefore, a curve. Thus we may assume that rank f'(a) < 1, of E (0, 27r) x (0, 2n). Obviously, if rank f'(a) = I on a non-empty open set, then W contains a curve. It remains to exclude the case where rank f' = 0 on (0, 2n) x (0, 27r). Then

f is constant. Thus I P(%) I I = cI > 0, 1 P(02I = C2 > 0. ' E (T \ 11))2. By continuity, (0.0) = (I P((1, 1))1I,IP((l, l))21) = (cl.c2); a contradiction. 0 Proof of Theorem 2.3.6. Since F E Bih(DI, D2), we see that DI is homogeneous if D2 is homogeneous. Thus, by Theorem 2.3.5 (a), DI E {ID2, 'B2) if D2 E


194

{m2, 82}. By the Poincare Theorem 2.1.17, the only possible cases are D1 = D2 = ID2 and D, = D2 = lB2. Assume that Dj is not homogeneous, j = 1, 2. If there exists an a E Dt such that Aut(D1) = Auta(Di), then Aut(D2) = AutF(a)(D2). Consequently,

by Theorem 2.3.5 (c), a = F(a) = 0 and by Corollary 2.3.12, F = TT (and D2 = DI) or F = TT (and D2 = S(D1)) for some E 72. It remains to consider the case where Dj is not homogeneous and Aut(Dj)

Auta (Dj) for any a E Dj, j = 1, 2. Then, by Theorem 2.3.5 and Proposition 2.3.15, Dt = ln, D2 = E. for some

p,q E (11) x (IR>o \ {1))) U ((IR>o \ {1)) x {1)). In view of Theorem 2.3.4, we only need to prove that p = q or p = S (q).

The case F(0, 0) = (0, 0) follows from Corollary 2.3.12, so assume that F(0, 0) # (0, 0). In the case where F(0, b) = (0, 0) for some b E ID we use Proposition 2.3.13 and we conclude that F is either of the form (2.3.3) or (2.3.4). In fact, substituting D2 by S (D2), if necessary, we may assume that

F(z) = (Z1fb(z2),mb(Z2)),

Z = (Z1,Z2) E DI,

where fb E 0*(D) and mb E Aut(tD). Mb (h) = 0. Recall that

Dj _ 1(Zt.z2) E

[D2

. 1211 < Rj(I221)),

j = 1,2,

where

RI(t) := (I - t2P2)1/(2P1)

R2(t) := (1 -f2g2)t/(2qI)

I E [0, 1).

Since F(8Dt fl (D x ID)) = dD2 fl (D x ID), we get

Rl(I:21)Ifb(z2)I = R2(Ihb(z2)I), (1 _ IZI2v2)1/(2vl)Ifb(z)I

Z2 E D,

= (1 -

ZEID,

and, consequently, IA(z)1 =

(I - Ihb(Z)I2g2)t/(2qi) (I - 1z12P2)h/(2Pt)

ZEID.

We have to consider the following three cases: F E Bih(IE(a,,). E(p,t)). Then we have 2

Ifb(2)I = (Il -162112)

(1- IZ12)

-

,

ZEID.


195

Since fb E 0`(D), letting I z I -+ 1, we conclude that a = . F E Bih(IE(a,l), (E(1,p)). Then we have

Ifb(Z)I =

1/2 (

( I - I hb(z)12'0 I - Ihb(z)12

1 -1b 12

1/2

E D.

11 -bzI2

Consequently, a = 1; a contradiction. F E Bih(IE(l,a), IE(l,p)). Then we have

(

I fb(z)1=

1 - Ihb(z)I2 112 I - IzI2a ) I - Ihb(Z)I2p

1 - lhb(z)120 I - Ihb(z)12 I - IZ12 1/2 IzI2a ) 1-1=12

- (1 - Ihb(z)12

I - Ib12

I - IZ12

1/2

t-Ihb(z)12 I1-6:121-1z12a) Letting z -*

E lp.

,

E T, we conclude that I -1b 12

Zm I fb(z)I=(811 -

-

1 a)1/2_

I1const

l

-hi;12

ET.

Hence, by the identity principle, we get

fb(z)=

I

nh`,

ZED,

with rl E C.. We have - Z12* Ig12I11-Ibz12

= 1 - Ihb(:)I20,

Z E D.

Since both sides of the above equality are real analytic functions on C \ { 1/b}, we get 1?1121-1Z12a

1l -hz12

= 1-Ihb(z)I20.

2EC\{I/b}.

Letting z -+ 1/b, we conclude that $ = I (EXERCISE); a contradiction. The case where F(a, 0) = (0, 0) for some a E D. is analogous.

In the case where F(a, b) = (0, 0) for ab # 0 we may assume that D1 = IE(a,1). Then +Irb,I(a,b) = (a*,0) (Wb,t is as in Theorem 2.3.4). Consequently, F o J/ (a*, 0) = (0, 0). Thus the problem is reduced to the previous situation, for some ' E T2. Finally, which implies that D2 = E(a,l) and F o Wbl = F = Wb,1 0 't'a* t E AUt(IE(a,I)).


196

2.4 Biholomorphisms of complete elementary Reinhardt domains in C2 Recall that a Reinhardt domain of the form Da,c = {z E C2 : IZI la' IZ2Ia2 < ec},

a = (a1,a2) E (IR+)., C E IR,

is a so-called elementary Reinhardt domain. Because of the restriction on the exponent a it is complete; moreover, it is a domain of holomorphy. Remark 2.4.1. Observe that Da,c is algebraically equivalent (cf. Definition 1.5.12) to Da = Da,o (EXERCISE). Therefore, we will only study domains of type Da, a E (IR+).. In fact, we will only consider the following three types of normalized elementary Reinhardt domains, namely:

a l a2 = 0: then either D. = D(l,o) = ID x C or D. = D(o,1) = C X ID (obviously, both domains are biholomorphically equivalent);

Of l a2 # 0 and a 1102 = p/q with p, q E 04. p, q relatively prime: then D. = {z E C2: IZII°IZ2Iq < 1}; ala2 36 0 and a1/a2 0 0: then Da = D(r,ll with t := at/a2 E IR,o \ Definition 2.4.2 (Cf. Definition 1.4.8). Let a = (a1,a2) E (IR+).. (a) If a 102 = 0 or a r , a2 E W, a 1, a2 relatively prime, then the domain Da :_ (z E C2 : Izl Ia' Iz2Ia2 < 1) is called an elementary Reinhardt domain of rational type. (b) If ala2 0 and at 0 Q, a2 = 1, then D. = {z E C2 : Izl Ia' Iz2I < 1) is called an elementary Reinhardt domain of irrational type. Remark 2.4.3. Let D. be an elementary Reinhardt domain. Then its logarithmic image contains the straight line L := ((t1, t2) E IR2 : all + a2S2 = 1), 1 < 0, if a 1 a2 0 0, or { (S l . ?; 2) : 52 E R), t j < 0. if a = (1, 0). Conversely, any unbounded

C2 whose logarithmic image contains a straight line is of the form D = Da,c (EXERCISE) and so biholomorphic

complete Reinhardt domain of holomorphy D

to D. Exercise 2.4.4. Let Da, a E W2 (a1, a2 relatively prime). For an f E 0*(ID) Put SI>

Of (z) := (ZI (f (za))-a2. z2(f (Zf))a' ).

z E Da.

Prove that g := {g f : f E 0*(ID)} is a subgroup of Aut(DD). In the following theorem all automorphisms of an elementary Reinhardt domain, normalized as before, are described.

Theorem 2.4.5 ([Shi 1991], [Shi 1992]). Let D. be a complete elementary Reinhardt domain.

2.4. Biholomorphisms of complete elementary Reinhardt domains in C2

197

(a) If Da = D(t,o), then

Aut(DD) = {Da -3 z H (m(zt) f(zt)z2 + g(zl)) m E Aut(D). f E 0*(®), g r= O(D)). (a') If Da = D(o, I), then

Aut(D) = S o Aut(D(t,o)) o S. to (b) If Da is of rational type with at , a2 E W, relatively prime, then

Aut(Da) = {T o gf o a f E O*(D), i; E T2, a E Fs(Da)). where V (Da)

IS. id) if at = a2 = I. {id}

if ala2 54 1.

(c) If D. is of irrational type (i.e. a = (at,1), at f 0), then

Aut(Da)={Da -3 zH

8>0}.

Exercise 2.4.6. Let Da be as in (b). Prove that F E Aut(Da) iff there exist E T, f E O*(D), and A E GL(2, z) n MI(2 x 2, Z+) with aA = a such that F = T(t,t) o of o OA (for the definition of OA := PI,A see Definition 1.5.12). Moreover, the following equivalence result will be discussed.

Theorem 2.4.7 ([Shi 1991], [Shi 1992]). Let D. and Dp be normalized complete elementary Reinhardt domains (in the sense of Remark 2.4.1). (a) If Da is of rational and DD of irrational type, then D. is not biholomorphically equivalent to Ds. (b) D(t,o) and Da, a = (at, a2) E N2, at, a2 relatively prime, are not biholomorphically equivalent.

(c) If D. and Dp are biholomorphically equivalent, then either Da = Dp or D. = S(DO). The proof will be based on the following notion of a Liouville foliation.

Definition 2A.8. Let D C Q;" be a domain. A system (F,t),,EA (A a suitable index set) of sets F,j C D is called a holomorphic (resp. psh) Liouville foliation of D if the following conditions are fulfilled:

F,,1nFee=fifnt0n2, D = UjEA Fn,

10Recall that S (z, w) = (w, 4

198


if u E 3t°°O(D) (resp. U E 98J{(D), bounded from above), then uI F is identically constant, q E A, for q, . R2 E A, q, # q2, there exists a u E MOO (D) (resp. an upper bounded

U E PSX(D)) such that uIF,, 0 uIFn2'

Example 2.4.9 (A holomorphic Liouville foliation). Let D = Da C C2 be an elementary Reinhardt domain of rational type.

If a = (1,0), put FF := {i;} x C,

E D. Then (Ft)tEu is a holomorphic

Liouville foliation of D. In fact, if u E 3e°O(D), then, ford E lD, u(t;, ) E 3e°°(C). Hence, in virtue of the Liouville Theorem, it follows that u IF, is constant. Finally,

observe that the function D a z H z, is a bounded holomorphic function on D which separates the fibers FF.

If a1a2 # 0, put Ft := 1z E D : za = }, E D. Then, again, (FF)tEp is a holomorphic Liouville foliation of D. In fact, we mention that for E Q) \ (0) the map Wt : C. -> D, (pt (A) := (.1-'12, All ), where f 12 = , is holomorphic. Therefore, if u E .lf°O(D), then u o (Pt E 3e°D(C.). Note that q (C.) = Ft. Applying the Riemann removable singularity theorem and then the Liouville theorem, we conclude that u IF. is identically constant. In case of % = 0 the fiber F0

equals (10} x C) U (C x 10}). By the same reasoning as above it is easily seen that if u r= .7e°O (D ), then u I F0 is a constant function. Moreover, the bounded holomorphic function g: D C, g(z) := z", separates the different fibers. In order to be able to present an elementary Reinhardt domain of irrational type as an example of a psh Liouville foliation we will need the following result due to Kronecker (cf. (Har-Wri 1979], see also p. 97).

Lemma 2.4.10. Let C E IR \ 0, b E C. Moreover, put

Lc,b := {z E C2 : Cz, + z2 = b} and P : C2 -+ C;, 4S(z) := (e2"zl, Then 4'(Lc,b) is a dense subset of F := 1z E 2 : izt I`Iz2I = e2'rReb} e2'rz2).

Proof. Take a point z E Lc,b. Then c Re z, + Re Z2 = Re b, hence dy(L',b) C F. On the other hand fix a point z° E F. Then choose an w E C with e2"w = z2. Setting (b - w)/c we have that (l;, w) E L,,b and so ({; + i I. W - i ct) E Lc,b, t E IR. Then ds((, (o) = (z1 e's, z2) for a suitable s E R. Moreover, it is well known (recall that the number c is irrational) that the set

1'P(t; + i t, w - ict) : t E R) =

{(z°ei(s+t),

zee-ict) :

t E IR)

is dense in F.

Example 2.4.11 (A psh Liouville foliation). Let D = D. C 2 (Cl = (a,, 1)) be a normalized complete elementary Reinhardt domain of irrational type. Put


199

Ft := {z E D : Izt Ial IZ21 = t }, t E (0, 1). Then (Fr),Elo,1) is a psh Liouville foliation.

In fact, if t = 0, then Fo is as in Example 2.4.9. Therefore, if u E PSJ{(D) is bounded from above, then u(0, - ) E S3f(C) and 0) E SJ{(C) are upper bounded and so, in virtue of the Liouville theorem for psh functions (see Remark 1. 14.3 (g)), identically constant. Hence, uIF0 is a constant function.

Now let t E (0, 1). Fix a U E PSJ((D) bounded from above. Using the holomorphic mapping 0 : C2 __+ C2, O(z) := (e2rr2, , e2' 2), we see that 0 maps the domain S2 := {z E C2 : at Re zt + Re z2 < 0} holomorphically onto D fl C. Thus u o 45 is a psh function on 92 which is bounded from above. Fixing a point b = (e2irf1, e2i $2) E Ft we define La1 2L105r

E C2 : ate+(0 = 2a log1) C 12.

Then C D A H u o O(A, -L 2jr log t - a t A) is an upper bounded subharmonic function and therefore identically constant. Hence, u is constant on the 0-image of Lat log, that is dense in Ft. Applying that u is upper semicontinuous we conclude that u(b) < u(p) for any p E Ft. Changing the role of b and p we see that UI IF, is identically constant.

Finally, it remains to mention that u : D - IR, u(z) := Iz1 Ia' IIz21, is bounded psh and separates different fibers. In the sequel the following observation will serve as a basic argument.

Lemma 2.4.12. Let W : D -* D' be a biholomorphic mapping between domains

D. D' C V. Assume that (Fa)aEA (resp. (F)pes) is a holomorphic Liouville foliation of D (resp. D'). Then there exists a bijective map r : A -+ B such that W (F,) = F'(a), a E A. The same result is true for psh Liouville foliations. Proof. We restrict ourselves to proving this lemma for holomorphic foliations. The analogous argument in the case of psh foliations is left as an EXERCISE. In a first step we assume D = D' and = ido. Observe that if F. fl F' 34 0, then Fa = F'. Indeed, suppose that both fibers are different. Then we may assume

that F. \ F' ¢ 0. Fix points p E Fa fl F, and q E F. \ F'. In view of the last condition in Definition 2.4.8 there is a bounded holomorphic function h on

D with h(p) # h(q). On the other hand, p.q E Fa, therefore, h(p) = h(q); a contradiction. The remaining properties of Definition 2.4.8 then prove the lemma.

Now let D and D' be arbitrary. We have only to observe that (P(F,)),EA defines a holomorphic Liouville foliation of D' (EXERCISE). Then the first step 0 completes the proof. Proof of Theorem 2.4.5 (a) and (a'). The proof will be based on the holomorphic x C (see Example 2.4.9). Let rp E Aut(D,). foliation (FF)tEg>, where FF :=

200


In virtue of Lemma 2.4.12, there is a bijective mapping r : ID (p(FF) = FFq), E ID. Therefore, (p may be written in the form

D such that

(P(z) = ((Pt(z),(P2(z)) = (r(z0.(P2(z))

Therefore, r - (pt ( , 0) is a biholomorphic map from D to D. What remains is to describe the second component function. Let us fix a Ao E ID;

then (p2(Ao, ) is a biholomorphic map from C to C, i.e. (p2(Ao, w) = y(Ao)w + S(A0), w E C, where y(Ao) E C., S(zo) E C. Now, we write 502 as its Hartogs E O(D)." Then yi(Ao) = 0, j > 2. series (p2(z) = Since A0 was arbitrarily chosen, we get 62(z) = yo(zt) + y1 (zt)z2. Observe that

yt E O`(U) Obviously, any mapping given in the lemma is an automorphism of Da. It remains to mention that S gives a biholomorphic mapping between Da and

0

CxlD.

To be able to continue the proof of Theorem 2.4.5, it is necessary to study another automorphism group.

The automorphism group of D. Let a = (at, a2) E Z2, a

(0, 0), at, a2

relatively prime in the case where a t a2 0 0. We set

D.:_(zEC*:(zlI"Iz2112 Da, any point z° E D, and any point w° E T. with t(w°) _ f(z°) there exists a uniquely determined holomorphic function f : D -> T. with f (z°) = w° such that c o f = f . f is called the lifting off . We advise the reader to look into general books on topology for this result. In particular, for any pair of points w', w" E Ta with O(w) = O(w") there

is an f E Aut(Ta) such that 0 o f= 0 and f (w') = w"; f is uniquely defined. In fact, f is uniquely defined since it is the lifting of 0: T. -> Da. In virtue of the former property of the universal covering we find a holomorphic map f : T. -->

T. with f (w') = w" such that 0 o f = (P. We have to show that f E Aut(Ta). Changing the role of w' and w" we also have a holomorphic g : T. -+ T. with

(w") = w' such that 'P = 0 o g. Then 0 o (f o g) _ 0, 4i o (g o f) = Qi, o f (w') = w', and o g(w") = w". Using again the first property of the

f

universal covering we conclude that f o g = id ITa and g o f = id ITa. Therefore,

f E Aut(Ta). Moreover, it is easily seen that

Aut"(Ta):={* where a,,(z) := z + iq, z E Ta. Now we are going to apply the above lifting properties for a given q. E Aut(Da ). Then there is a lifting rp E Aut(Ta) such that 0 o Sp = (p o 0 (EXERCISE). Moreover, for a fixed q E Z2, 0 o Cr'? 0 0-I E Aut"(Ta). Therefore, we find an q' E 72 such that 0 o an = or,,, o gyp. It is easily seen that the mapping q --> q' leads

to a group isomorphism of Z2. Therefore, there exists a matrix P E GL(2, Z)

such that oa,,=alp00,gE7L2. So we are led to study the group of automorphisms of the domain Ta. We get the following lemma.

Lemma 2.4.17. Let (p E Aut(D.'). Assume that its lifting gyp: T. + Ta is a complex affine transformation, i.e. 0 (i;) = A + f, where A E GL(2, C) and


203

al.1'al.2,a2Za2.1Za2.2

E C 2. Then P is of the form rp(z) = = 0a,n(z). where A E 4L(2, Z), a = (a,, a2) a C2., i.e. rp is an algebraic automorphism ofDa'. Proof. From the discussion before we get

i; ETa,rlEZ2.

0

Using the form of cp it follows that A = P. Finally, the equality gP o 0 = ds o leads to the form of (p claimed in the lemma. After all these preparations we proceed with the proof of Lemma 2.4.16.

Proof of Lemma 2.4.16. Let V E Aut(D,x) and 0 E Aut(Ta) be its lifting. In virtue of Lemma 2.4.17 we have to show that 0 is a complex affine transformation. In

a first step we will show that there are a r E Aut(H-), an f E O*(H-), and an h E O(H) (lw- := {z E C : Re z < 0}) such that

(f (g*)gt +

where _ (6, 6) E T. and (to simplify notation) i * := at Sl + In fact, put tib : C2 - C2, *(%) := (i;1, i;*). Obviously, * is a biholomorphic mapping and 1r IT,, maps T. biholomorphically onto C x H -. Its inverse mapping is given by *-1(z, w) = (z, w - at z). Thus, Aut(Ta) = 1/r-1 o Aut(C x Oil-) o *.13 Moreover, let g: Of- -+ O) be any biholomorphic map. Then

g: C x 0h- -+ C x 0),

8(z, w) := (z, g(w)),

is also biholomorphic. So Aut(Ta) = 0: 1 o k-l o Aut(C x O)) o k o 1/r. Then, in virtue of Theorem 2.4.5 (a), there are f E O*(O)), h E 0(O)), and m E Aut(D) such that for i; E Ta we get

OW = 1/r-1 ° k-1 (f ° W*)6 + h ° and therefore, for

m ° W*)),

E Ta,

0(0 = (f°g(t*)St+hog(t*) g-lomogW)-at (f°g(t*)St+hoga*))), which proves the above claim with f := fog, h := h o g, and z := g-1 o m 0g. In virtue of Lemma 2.4.17 it remains to verify that cp is a complex affine mapping.

Indeed, in virtue of the properties of the covering mappings we have a matrix P = [ n s ] E GL(2, Z) such that for all pairs (k. e) E Z2 the following identities are true:

i k, y2 + it) = 13To be precise {b should be understood as {Gl rQ.

i (k, e) P,

i; E Ta.

204


In particular,

+ik)+h(i;*+i(atk+l)) i(kp + er), r(r* + i (atk + e))

=

+ik)+h(i;*+i(a2k+t)) =

h(i;*)) + i(kq + fs).

From these identities we deduce that f is a constant function. In fact, fix a point wo E H-. Then, applying the first of the above identities for points (fit, wo - a t S t) E Ta, gives

f(wo + i (atk +

=

ik) + h(wo + i (atk + e)) h(wo) + i(kp + tr), t E C.

Hence, f (wo + i (at k + e)) = f(wo) =:Ao 0 0, k, e E Z. Recall that the number at is an irrational one. So the set (wo + i(atk + 8)) : k, j E Z} has an accumulation point in the plane. Then, in virtue of the identity theorem, it follows

that f = Ao. Applying the first of the above identities, we claim that h is a complex linear function. Indeed, the above identity implies that

i.lok+h(Z+i(atk+t)) = h(Z) + i(kp + fr) for all Z = * E H-. Differentiation in direction of Z leads to

h'(Z + i (atk + 8)) = h'(Z). Fixing some Z = Zo E H- we have h'(Zo + i (at k + e)) = h'(Zo) := µt . So h' is constant, i.e. h(Z) = µt Z + µo for a suitable µo. It remains to show that r is a complex affine mapping. In fact, using the second identity, we arrive at the following equality:

r(Z) - at

µt Z + µo) + i (kq + ts),

Z E H-.

Again differentiation gives r'(Z + i (atk + e)) = r'(Z), Z E H-. As above, fixing Z = Zo and using the identity theorem, we arrive at r' - r'(Zo) =: Ti 36 0 (recall that r is a biholomorphic mapping). As a consequence we conclude that r(Z) = rt Z + to for a suitable TO. Finally, rewriting gyp, we see that

OM =

+A I W) + µo, rt W) +ro -of 1 (Ao)t +111 W) + µo)).


205

or

µatk t EA°

at (riTI

t

A0 _ I

LIal)] + (120. To -

Obviously, the matrix is a non-singular one, and therefore, fP is complex affine. Hence, Lemma 2.4.17 gives the end of the proof. Now we return to the proof of Theorem 2.4.5.

Proof of Theorem 2.4.5 (b). Obviously, any of the given mappings belongs to Aut(DD). To prove the converse we will use the holomorphic Liouville foliation

where FF := {z E C2 : za = }. Fix a yP E Aut(DD). Then, applying Lemma 2.4.12, there exists a bijective mapping T : (D -+ ID such that rp(FF) = Fm),

E 0). Observe that the fiber F0 is the only one with a "singularity". So one concludes that r(0) = 0 (EXERCISE) which means that rplD defines an automorphism of D.*. Using Lemma 2.4.15 shows that

z E D.

w(z) = where

(yI, S2) E 72, f E O*(D ), and P = [; s ] E GL(2, Z) with

aP = a. Since aj E Z+, one concludes that P = II2 if ata2 # 1, and (P = II2 or p = s = 0, q = r = 1) if aI = a2 = 1, which gives the description of a. We will only discuss the case when a = id (the case when a = S may be taken as an EXERCISE). Observe that K := 1 L* x (1) C Da. Therefore,

(p2(z1, 1) = i;2(f(zz1 E

zL*, is bounded. Applying the Riernann theorem of removable singularities we see that f extends holomorphically to D. Taking into account that rp is bijective, it even follows that f E O*(U). Finally, a continuity argument leads to the description of rp on the whole of Da.

Finally, we discuss the case of normalized elementary Reinhardt domains of irrational type. Proof of Theorem 2.4.5 (c). Here weusethepshLiouvillefoliation (FF)rElo,I)from Example 2.4.11. Let rp E Aut(Da). Then there is a bijection r: [0. 1) -> [0, 1)

such that (p(FF) = Fy), t E [0, 1). In particular, F0 is homeomorphic to F,(o), which implies that r(0) = 0. So (pIDa E Aut(D,*,). Applying Lemma 2.4.16, cIDa is of the form

IP(z) _ (fit

z°,., Za,.2

t

2

z 2.2), 2 za2., t 2

zE

Da*,

where A = [ a2:; a2 .2 ] E GL(2, 7L). Observing that the coordinate axes belong

to Da, it follows that a'j E Z+ and, therefore A = II2, which implies Theorem 2.4.5 (c) (EXERCISE).

206


Summarizing, Theorem 2.4.5 has been completely proved.

Now we turn to the proof of Theorem 2.4.7. We start with the discussion of bounded holomorphic functions on elementary Reinhardt domains. First, recall that an elementary Reinhardt domain D" of rational type carries at least one bounded holomorphic function that is not identically constant. On the other hand, we have

Lemma 2.4.18. Any bounded holomorphic function on an elementary Reinhardt domain of irrational type is identically constant.

Proof. Take an f E 3e°o(D"). Then If I E PSJ{(D"). Therefore, in virtue of Example 2.4.11, f IF, is identically equal to a constant st, where (Ft)tE[o.t) denotes the psh Liouville foliation from that example. Recall that

Ft= {zED":Iztl"1Iz21=t}. In particular, f(zt. 1) = st whenever Iztl"1 = t. Applying the identity theorem, it follows that st = s, t E [0, 1). Hence, f = s on D. Proof of Theorem 2.4.7. (a). In virtue of Lemma 2.4.18 and the remark before, it is clear that D" and DS are not biholomorphically equivalent. (b) Suppose that there is a biholomorphic map tp : D(o, t) -+ D. Then, using the holomorphic Liouville foliation (FF)tED of D. and (F )tE[) of D(t,o), respectively (see Example 2.4.9), there is a bijection r : ID -+ ID such that tp(FF) = F=(t), E D. In particular, tp I Ft, = F. Using that Fo has a singularity at (0.0) we get, as before, a contradiction. (c) Assume that D" and Dp are biholomorphically equivalent. In virtue of (a) there are two cases. Case 1: Assume that D. and Dp are of rational type. Suppose that a = (1, 0), then f = (1, 0) or P = (0, 1) and a biholomorphic map is given either by the identity

or by S. Therefore we only have to discuss the case where ala2 54 0 0 0102, a t , a2, resp. f t , i62, relatively prime. Take a biholomorphic map tp : D" -+ D'8. As in the proof before using holomorphic Liouville foliations we conclude that svl Da E Aut(D°*). Following Corollary 2.4.14 there is a biholomorphic mapping tJr : D; - D°* of the form t/r ='A, A E GL(2, 71) with det A = 1 and aA = P. Hence, , := tP o >(i E Aut(DS ). In virtue of Lemma 2.4.15, >1i may be written as

Vf(z)=Tt-gf-Op(z), zED*. Therefore,

tpIDa =TtoSforhpoOA-i = TToSfo4PA-1 and PA -I = a. Thus, Tt

' 0 (P(z) = ((f(z$))-112zr zZ, (f(zP))"1 zrz2),

z E D°*,

2.5*. Miscellanea

207

PA-1 where =: [ p s ]. Observe that T o (P defines a biholomorphic mapping from Da onto Do. Recall that the left-hand side is holomorphic on D. In particular, the functions

D. a l i-*

(f(A$1+82))-Q2Ap+q

and D. 3 X H

(f(AP1+52))S1Ar+s extend

holomorphically to D. Therefore, f has a pole at 0, i.e. f(A) = Ak f (A), A E lD*, where f E O*(ID) and k E Z. Hence, TTI

o

(p(Z) = ((f

(-10))-Q2Zp-kQ1fi2Z4-kfi

(f (-S) )S'

-1

+k, 1 `2+kfl I 02 ) .

Z E Da .

Finally, we define an automorphism of DS, namely,

X(z) := g1(z),

: E D.

Then, for : E D,,*,. we get

X o Ti o (P(Z) = (:Ip-kflr#2Z29-k#

r+kp _s+k0192 ).

Z1

..2

Taking into account that the mapping on the left-hand side is holomorphic on Da, it is easily seen that X : = X o TT o (p = id I Da or X = S I Da . Hence, Case I is verified.

Case 2: Assume that D. and DS are of irrational type, i.e. a = (&1, 1), P = (#1, 1), where a,, fl, r= IR+ \ Q. Applying psh Liouville foliations, one gets that cOI D : D." -> D6 is a biholomorphic map. Following the proof of Lemma 2.4.16, one may show (EXERCISE) that (PI Da = Applying now that the left mapping is holomorphic on Da, it follows that either (P(:) = TT(z) or co(:) _ TT o S(z) whenever z E Da. In the first case observe that ICI IS' 1R2I02 = 1. Then of-12 IDq E Aut(DS), 1, X21). Hence 0 where 0 (P = id on Da. A similar argument for the >o \ 11), the set {FH, : H E Aut(ISk). E Tn-k) is a subgroup of Aut(IEp) (ExERctsE); Cf. [Jar-Pfl 1993]. Lemma 8.5.2. Q We do not know whether (2.5.1) remains true.

0

Theorems 2.5.10, 2.1.20 and Lemma 2.1.21 imply the following

n2 + 2, then D = S. up to rescaling of variables. If D IB", then d E [n, n2 + 21 and d is of the same parity as n. d = n2 + 2 if D = ID x IBn_1 up to permutation and rescaling of variables. If d = n2, then D is algebraically equivalent to one of the following domains:

(i) {z E C" : r < IIzil < R}, 0 < r < R < +oo (cf. Exercise 2.1.13); (ii) ID3 (n = 3); (iii) (82 x(62 (n = 4); (iv) E(t ,,. p" 96 1 (cf. Theorem 2.5.10);

212


(v) {(z',zn) E (n-1 x C : r(1 - IIz'112)a < Iznl < R(I - IIz'II2)*), 0 < r
Fq,l is a proper holomorphic mapping;

(ii) F(zl,z2) if P21PI

=

W.

eq2/ql - kP2/PI E Z, (6 zZ9z/4jB(zl z2 D2/Pl ). 2ZD if P2/PI E W. eq2/ql E W.

where S I S2 E T, k, e E N. and B is a finite Blaschke product.

Remark 2.5.22. (a) In the case n > 3 the implication (ii) = (i) is elementary and remains true for arbitrary p. q E IRao. Indeed n-1

n-1

IFn(Z)I-2q,, E IFj(Z)I2qj

j=1 n-1

1:IZn

=IZnI-2tq.

EIZnl2sjqjIZlI2r'qj

j=11

n-1 12(fgn/qj -Pn/9j )qj -2tgn

1Zj 12Pj = IZn I-2P" E IZj I2P' .

j=1

j=1

Observe that F is biholomorphic if e = r1 =

= rn_ I = I iff p j = q j .

j = L.... n - 1, and (pn - qn )/ p j E Z. j = 1, ... , n -

1.

In particular,

there are p,q E N" such that IFP,n_1 0 Fq,n_I but tP,n-1 (Fq,n_I. Take for instance p i = qj, j = 1, ... , n - 1, and pn 0 qn such that (pn - qn )l Pi E Z. n - 1.

0 We do not know whether the implication (i) = (ii) remains true.

(b) In the case n = 2 the implication (ii) for arbitrary p, q E R2>01 Indeed

(i) is elementary and remains true

IF2(z)I-292IF1(Z)I291 = (Iz1IIz2I-P2/PI)2kg1 IB(ZIZ2P21P1)I2gl

if p2/PI 0 N, ifp2/PI E W.

2.5*. Miscellanea

215

Observe that F is biholomorphic if

£ = k = 1, q2/q - P2/P1 E 7L f= 1, B E Aut(D),Q2/q1 E N

if p2/P1 0 W. if p2/p1 E W.

0 We do not know whether the implication (i) = (ii) remains true. 0 Theorem* 2.5.23. Assume that p E W", 2 < s < n - 2. (a) ([Che-Xu 2001 ]) The following conditions are equivalent:

(i) there exists a proper holomorphic mapping F : [Fp,s -* Fp,s:

(ii) there exist permutations or E G5 and 8 E G"_S such that pa(j)/pj E N, j = 1, .. ..s, Ps+8(k)/Ps+k E N, k = n - s. (b) ([Che-Xu 2002]) Any proper holomorphic mapping F : 1Fp,s -+ Op,s is an automorphism (cf Theorem 2.5.16). Let V E e°0([0, 1], 1R+) be such that there exists an h E (0, 1) for which (PI[a,h] = 0, co(t) = 1,

(p'>0and cp">0on [0,1], gyp'>0andlp">0on(h,1). Define D,e,h :_ ((z1, Z2) E D2 : IZ1 I2 + Oz2I2) < 1).

Exercise 2.5.24. (a) Dp,h is a normalized (cf. (2.3.1)) bounded pseudoconvex complete Reinhardt domain with (8D) x (h FD) C dD,e,h.

(b) D,y,h 0 {D2, IE(l,a), E(a,,), a > 0). Consequently, by the Thullen Theorem 2.3.6, every biholomorphic mapping F : D9)1 ,h D,e2,h2 is of the form

,-

bih

F = Ti for some ' E T2. Hence, D(p1.h1 Dm2,h2 iff hl = h2 and cO1 = c02 (c) D9,,h is strongly pseudoconvex at a boundary point a = (a,. a2) E BD,P,h if Ia2l > h (cf. § 1.18*). In particular, the set of weakly pseudoconvex boundary points is not contained in Yo.

Theorem* 2.5.25 ([Lan-Pat 1993]). The following conditions are equivalent: (1) there exists a proper holomorphic mapping F : D,,, ,h, -> D02,h2; (ii) there exist m E N and ti 1. 6 E T such that: t E [0. 1], h2 = hm, (p,(t) = F(c) =1 (6z1,6z2 ), Z = (zI,:2) E Dwl,hi. p2(tm),

Theorem* 2.5.26 ([Lan 1994]). Let D C C2 be a bounded smooth pseudoconvex complete Reinhardt domain whose weakly pseudoconvex boundary points are contained in Yo. Then any proper holomorphic mapping F : D -> D is an automorphism.

216


Theorem* 2.5.27 ([Lan-Pin 1995]). Let D1, D2 C C2 be bounded pseudoconvex complete Reinhardt domains such that there exist a complex analytic variety W and an open neighborhood U of a point a c- 8D 1 such that W fl U c 8D 1. Then any

proper holomorphic mapping F = (Fl, F2): D1 -- D2 is such that Fl and F2 depend only on one variable. Moreover, if D1 = D2 is not a bidisc, then F has the form

F(zl,z2) = (6Za(1),S2za(2)), where Sl, 6 E T, Cr E 62 Theorem* 25.28 ([Ber-Pin 1995], [Lan-Spi 19961, [Spi 1998]). Let D1, D2 C C2 be bounded complete Reinhardt domains such that at least one of them is neither a bidisc nor a complex ellipsoid. Then any proper holomorphic mapping F : D1 -> D2 has the form F(z1, z2) = (cl a(i), C2 0(2)

where c3 E C, mj E W, j = 1, 2, a E 62. Moreover, if D1 = D2, then F has the form F(z1, z2) = (6Za(1), S2Za(2)),

where 1,1;2ET,QE82. Remark 2.5.29. The full description of proper holomorphic mappings F : D,

D2, where D1, D2 C C2 are bounded Reinhardt domains, may be found in [Isa-Kru 2006].

For the case of proper holomorphic mappings between unbounded Reinhardt domains we mention the following result.

Theorem* 2.5.30 ([Edi-Zwo 1999]). Let a, f E Z and let

C2JDaF ) DpCC2 be a proper holomorphic mapping. Then

F(z) _ (H 11,01 (za)z21, zi2 H-1/fl2(za)). where H E O(D,C.), E T, and kI,k2,E1,E2 E Z+ are such that a2181k1 = aif2k2, compare with Theorems 2.4.5, 2.4.7.

alrllel = a2182e2:

2.5*. Miscellanea

217

Theorem* 2.5.31 ([Din-Pri 1987]). Let D C C" be a Reinhardt domain and let

F: D -+ IEp be a polynomial proper mapping with p E N". Then F(z) _ (zit , ... , zn") with d l , ... , d" E W, up to action of T" on D and an automorphism bih of Ep. 'S Moreover, D IEq with qj := dj pj, j = I.... , n. Theorem* 2.5.32 ([Din-Pri 1988]). Let D C C" be a Reinhardt domain with 0 E D and let p r= N". Then the following conditions are equivalent:

(i) there exists a proper holomorphic mapping F : D -+ lip; (ii) there exists a proper polynomial mapping F : D -+ CEp.

Theorem* 2.5.33 ([Din-Pri 1989]). Let D1 C C" be a Reinhardt domain and let D2 C C" be a bounded simply connected strictly pseudoconvex domain with

C' boundary. Then any proper holomorphic map F : D 1 -+ D2 is, up to an automorphism of D2, of the form F(z) = (zi' , ... , zn") with dl .... , do E N. Remark 2.5.34. For a = (al, a2) E (IR2)* and 0 < r- < r+ < +oo let Da,r-,r+ := {(Z1,z2) E C2(a) : r- < Izal < r+}. Recently L. Kosinski [Kos 2007] gave the full characterization of all proper holo-

morphic mappings F : Da,r-,r+ -+

Pr :='4(1/r, r),

D$,R-,R+. More precisely, let

Dy,r := {(Z1, z2) E C2 : 1/r < IZIIIZ2Iy < r},

yEIR\Q,r>1. One may prove (EXERCISE) that Da,r-,r+ is algebraically equivalent to a domain of one of the following three types:

PrxC,

ifala2=0,

Pr X C,.,

if a2/al E Q,,

Dy,r,

if Y := a2/a1 $ Q.

(2.5.4)

If D 1, D2 are of type (2.5.4), then there are no proper holomorphic mapping F : D 1 -+ D2 except for the following four cases: (1) D, = Pr xC, D2 = Prm XC (m E N), F(z) = (l;zf"'. P(z)), where l; E T.

s E {-I, I}, P(z) = E50P(z1)z, N E N, Pp,... , PN E O(Pr), P(z1, ) const, z1 E P. (2) D1 = Pr x C., D2 = Pr- X C (m E N), F(z) = z2 k P(z)), where E Ir, s E (-1, 1), P(z) _ FN O Pi (zl)z2, k, N E N, 0 < k < N, Po,..., PN EO(Pr),Ek=OIPi(Zl)I >0,FNk+lIPi (Z1)I>0,Z1 E Pr. '5That is, there exist

ZED.

E T" and 0 E Aut(IEP) such that df o F o Tt(z) = (z d' , ... , zd,"),

218


(3) D1 = Pr X C., D2 = Prm X C. (m E W), F(z) =

zig(zl)), where

ET,eE{-1.1),kEZ.,gE(9*(Pr).

(4) D1 = Dr,r, D2 = Da,R with tog R

=

kl + 18.

l + .e28 } tog r = k2

for some k = (k1, k2), t = (e 1, e2) E Z2, F(z) = (azsk, bzse), E E {-1,1}, a,b E C, lallbl8 = 1.

2.5.4 Non-compact automorphism groups Theorem* 2.5.35 ([Bed-Pin 1998] (see also [Bed-Pin 1988])). Let D C C2 be a bounded domain with real analytic boundary such that Aut(D) is non-compact.t6 Then

D 2h (l,m) = {(z1, Z2) E C2 : 12112 +

11,

where m E W. In particular, D admits a proper holomorphic mapping onto 132.

We say that a bounded domain D C C" with smooth boundary is of finite type if there exists an m E N such that for every point a E aD and for every complex one-dimensional manifold V passing through a, the order of contact of aD and V does not exceed m, i.e. for any a E 8D and (P E O(1D, C") with (p(0) = a we have ordo(u o (p) < m for any local defining function u: U -+ Q( defined in a neighborhood U of a with gp(m) C U. Notice that Theorem 2.5.35 remains true if D is a pseudoconvex domain with smooth boundary of finite type.

Theorem* 2.5.36 ([Bed-Pin 1991]). Let D C C"+1 be a bounded pseudoconvex domain with smooth boundary of finite type such that Aut(D) is non-compact. Assume that the Levi form of a defining function of D has rank at least n - 1 at each boundary point. Then X (C

bih

{(Z1..... Zn, W) E C"

D

n

: 1w12m + E iZj 12 < 1 }. J=1

where m E N.

Theorem* 2.5.37 ([Bed-Pin 1994]). Let D C C"+1 be a convex bounded domain with smooth boundary of finite type such that Aut(D) is non-compact. Then there exist m 1, ... , m" E N and aa,p = apq,a E C such that bih

D = {(z. W) E C" X C : 1w12 + I: aa,pza $ < 11. 'That is, there exist a point a E D and a sequence

C Aut(D) such that

8D.

2.5*. Miscellanea

219

Qwhere the sum is taken over all a. # E Z with a 1 / m 1 + ...+a,, /m = 1 and

N1/ml +-+Pn/mn = 1. Theorem* 2.5.38 ([FIK 1996a]). Let D C Cn+1 be a bounded Reinhardt domain with C°O-smooth boundary such that Aut(D) is non-compact. Then D

Nh

{(z, w) E C" X C : Iwi2 + P(IZI I..... IZnI) < 1}.

up to permutation and rescaling of variables, where P is a non-negative polynomial with real coefficients.

The case where 3D is only of class Ck was solved in [Isa-Kra 19971 - in this case P is a non-negative Ck-function.

Theorem* 2.5.39 ([Isa-Kra 1998])). Let D C C2 be a hyperbolic Reinhardt domain with Ck-smooth boundary (k > 1) such that D fl Yo # 0 and Aut(D) is non-compact. Then D is algebraically equivalent to one of the following three types of domains:

{(z1,z2) E C2 : IZ1I2 +

Iz2I2,

< 1), wherem < 0orm > k or In E M;

{(zl,z2) E ID x C : t1-I 17J.- < IZ2I < (I_Ia }. where I < R < +oo, >

a0;

Q Q {(z1.z2) E C2 : exp(NIz1I2) < IZ21 < Rexp(iBIZ1I2)}. where I < R < +00, ,8 E IR. and (R = +oo =" > 0).

Remark 2.5.40. (a) Some of the above results may give the impression that every domain with non-compact automorphism group is biholomorphic to a Reinhardt domain. This is not true - the following bounded pseudoconvex circular domain with real analytic boundary and non-compact automorphism group is not biholomorphic to any Reinhardt domain ([FIK 1996b]): {(zl. Z2, Z3) E C3 : IZ1 I2 + IZ2I4 + IZ3I4 + (52Z3 + Z3Z2)2 < I }.

(b) In [KKS 20051 the reader may find a characterization of those "analytic polyhedra" in C2 whose automorphism groups are not compact. (c) For general domains with non-compact automorphism groups the reader may contact the survey article [Isa-Kra 1999].

Chapter 3

Reinhardt domains of existence of special classes of holomorphic functions 3.1 General theory Let D be a Reinhardt domain of holomorphy and let 8 C 0 (D) be a natural Frrchet space (cf. § 1.10). e.g. 8 = ,7e-A(D), Ak(D), Lh.k(D), O(N) (D), O(o+)(D);

cf. Example 1.10.7. Our aim is to find geometric characterizations of those Reinhardt domains D which are 8-domains of holomorphy. We like to point out that such geometric characterizations are not known for more general classes of domains (e.g. balanced domains of holomorphy). Except for § 3.1, all results presented in this chapter are more elaborated and detailed versions of some results from [Jar-Pfl 2000], § 4.1.

Remark 3.1.1. Consider the case where 8 = O(N)(D) (Example 1.10.7 (f)). (a) First recall some known general results. Let G C C" be a domain of holomorphy (Reinhardt or not). Then: ([Jar-Pfl 2000], Corollary 4.3.9.) G is an Ot2n+6)(G)_domain of holomorphy for any e > 0. ([Jar-Pft 2000], Corollary 4.3.9.) If G is a bounded domain, then G is an 00+e)(G)-domain of holomorphy for any e > 0. ([Jar-Pft 2000], Theorem 4.2.7.) If G is bounded and fat, then G is an L2 (G)-domain of holomorphy; in particular, in this case G is an 0(")(G)-domain of holomorphy (cf. Example 1. 10.7 (c), (f), (g)). We do not know whether the above results are optimal, e.g. whether there exists a µ < n such that every bounded fat domain of holomorphy is an 0 (A) -domain of holomorphy. (b) In contrast to the above general situation, in the case where D is a Reinhardt domain of holomorphy, we are able to show that: D is an 0(') -domain of holomorphy (Theorem 3.4.4).

0

If D is fat, then D is an 04)-domain of holomorphy for any e > 0 (Theorem 3.4.3).

The following notion will be useful in the sequel. Definition 3.1.2. Let D C C" be a Reinhardt domain. We say that a natural Frdchet space 8 C 0(D) is regular if for every function f E 8 with the Laurent expansion

f(z) _

afza, z E D. arez"

3.1. General theory

221

we have:

Z" E $,a E E (f) _ (a E Z" : a f # 0), the set {at za : a E E (f) } is bounded in $ (cf. Definition 1.10.3). Remark 3.1.3. Observe that there are natural Frdchet spaces which are not regular.

For example, 8 := C f , where f E 0(D) is not a "monomial" of the form cz". Example 3.1.4 (Examples of regular natural Frdchet spaces). (a) 8 = O(D). Indeed, by the Cauchy inequalities, for any Reinhardt compact set K C D we

have Ilafz"IIK < IlfIIK,a E E(f). (b) $ = 3e1°o°(D).

Indeed, by the Cauchy inequalities, we have Ilafzal1B(r)nD < III IIB(r)f1D

aEE(f),r>0.

(c) $ _ MI(D). Indeed, by the Cauchy inequalities we have II a f za lID < II f IID, a E E (f ).

(d) $ = O(N)(D) (N > 0). Indeed, the function 8D is invariant under n-rotations. Hence, using once again the Cauchy inequalities, for r E D fl R'O, a E E (f ), we get

sD(r)Iafral +oo. 'That is, sup(c,,,"«' : a e E) < +oo.

3.1. General theory

223

Observe that condition (iii') gives an effective geometric characterization of $-domains of holomorphy. Notice that the result need not be true for non-regular natural Fr6chet spaces (cf. Remark 3.1.3). Proof. The equivalences (ii)

' (ii') and (iii) q (iii') follow from Lemma 1.15.13. (ii): By Proposition 1.11.11, there exists an f E 8 such that D is the domain of existence of f. Let f (z) = LaEZn of za, z E D, be the Laurent (i)

expansion off . By Proposition 1.11.6 the domain of convergence V y of the above series is a domain of holomorphy. Thus D = Df. In the case where E(f) is finite we have D f = C" (E (f) ), which contradicts our assumption that D C" is fat. Now, it remains to use Proposition 1.15.15.

(ii) = (iii): Put E := E (f)*, ca := laa 1. The regularity of 8 implies that the set {af za : a E E} is bounded in 8. _ (iii) (iv): Fix a point a E C; \ D. By Lemma 1. 15.13 (h), v(a) = v* (a) > n > 1. Thus, there exists a sequence (a(k)) t C E such that la(k)l -> +oo and la(k)I -> +oo, k -, +00. (iv) (i): Suppose that D is not an 8-domain of holomorphy and let Do, D be as in Proposition 1.11.2(*). Since D is fat, we may assume that D C C; and that there exists a point a E D \ D. Let Q = (q; : i E N) be a countable family of seminorms generating the topology of 8 with q; _< qi+t, i E W. Since 8 is a natural Frechet space, the extension operator 8 g H g E O(D) is Ca(k)Iaa(k)I

continuous (Remark 1. 11.3 (n)). In particular, there exist C > 0 and io E W such that Ig(a)I < Cq;o(g), g E 8. Since the set {d(k)za(k) : k E W) is bounded in 8, there exists a constant M > 0 such that q;o (d (k)za(k)) < M, k E W. In particular, d(k)Iaa(k)I < CM, k E W; a contradiction.

Proposition 3.1.7. Let 0 0 D C C" be a Reinhardt domain and let 8 C MOO (D) be a natural Banach algebra which is moreover regular (e.g. 8 = with the norm III 118 := 2k max{ 1 1 D f 11D : 1 v 1 < k); cf. Example 1.10.70)). Then the

following conditions are equivalent:

(i) D is an 8-domain of holomorphy; (ii) there exists an f E $, II f 11,8 < 1 II f 11D < 1, f (z) _ such that the set E (f) is unbounded and

aEzn as za, ZED,

D = {ZEC"(E(f)):u*(Z)o with (1.15.1) such that:

D = {z r= C"(E) : u*(z) < 1}, where u(z) := sup{Icazalt/Ial : a E

i }, Z E C"(E),

z" E 8 and IIcaZ" II8

1, a E E;

(iii') there exist E C (Z")* (E C (Z"). if 0 E D) and (ca)aEE C IR>o with (1.15.1) such that.

D = intnaEE{Z E C"(E) : calz,I < 1},

ZaES and llcazall8 Iz21,

(3.2.2)

Indeed, fix an a = (a1,...,an) E (aD) \ C"(a). Note that a1 as = . an = 0. We only need to prove that there exists a neighborhood U of a such that U'(z,1z21) C G for any z E U n D \ Yo (EXERCISE). Let U be a neighborhood of a such that Iz2-ej I < 1, z E U, j = I--, n, and as+1

s

n

fl(1+Iz2-ej1)aj

j=I

(I-1Z2-ejI)aj <e8, fl j=s+1

- EU.

Then n

s

fl (Izj I + IZ21)aj

J1 (IZj I -

IZ21)aj

j=s+1

j=1

S

n

11 (1 = 1=1Ia1 ...IZnlan 11 (1 + 1Z2-ej I)ai j=I j=s+1

Iz2-ej

I)aj < ec+e,

zEUnDu,which shows that We need the following lemma.

Lemma 3.2.2. Let S2 C Cn be open and let S : S2 -+ (0, 1) be a function such that S _< pn,

f

IS(=') - 3(Z" )l < IIZ' - Z"II, Z' E S2, Z" E IB(:', As1(z')) (c. Example 1.10.7 (g)). Then

IISN+ITIDrgll.n < r!(,/n)Ir12N+1rIll8"glls2 N> 0. T E Z", g E (9(Q). Proof of Lemma 3.2.2. Fix N, r, g, and a E S2. Observe that S(z) > 8N+IrI(a)I

Let r := Z(f
0. Fix 0 < N < s and x° E I. Note that x° + tw E III,,, t E IR. Put r(t) := ex°+tw, t E IR. Since r(t) E D, the Cauchy inequalities imply (cf. Example 3.1.4 (d)): l

a

fI
M-2 Consequently, T is bounded. Therefore, 8D(r(t)) = PD(r(t)) fort > to. Now, we estimate PD.

Let d := dist(x° + IRw, 2II.,,) = dist(x°, (8D) \ Vo such that pD(r(t))

Fix a t E Qt and Z E 2IIz - r(t)II. Write Iz11 = rj(t)e"i (u1 E R),

j = 1--n. Note that I I u I I > d. Fix a j = j(1) such that I ujI ? d / /

.

Then

we obtain

pD(r(t)) ? ilzz - rj(t)I ?

2IIzzI -r3(t)I = Zrj(t)Ie"; - lI > dorj(t),

3.2. Elementary Reinhardt domains

229

where d° := i(1 - e-d/1). Finally, choose jo such that there is a sequence (tk)k C [t0, +oo) with j(tk) = jo for all k and limk,,, tk = oo. Then 1

doe"'/p+tkwjp+tk{w.v}/N

M > PD(r(tk))etkslN >

) 00; k-P. oo

a contradiction.

(d) Let f E 3eO0(D), f(z) =

z E D. In view of Proposi-

av z

tion 1.6.5 (b), for v E E (f ), we have

c(v) := log IIIIID

Ia,c C Hv,c(v),

Consequently, E (f) C (1R+ a) n z,. In particular, if a V IR . Z", then f = const. (e) For f = FpEZ" of zfl E 3e°O(D) define 00

ofkaekckk

g(A)

AEID.

k =O

Since D. C (e-cza : z E D) C ID, for every A E D. there exists a z E D such that k = e-cza. We know that E(f) C (IR+ a) f1 Z". Observe that in fact

E (f) C Z+ a (because 00

are relatively prime). Thus 00

laf ekc(e-cza)k I =

la aekcAk I = k=0

k=0

laf -01 < +oo OEZ"

and, therefore, g is well defined, g E 0(D), and g(z) = g(e-`za) = f (z), z E D. Hence, g E 3e°O(D) and fi(g) = f . (f) Recall (Example 1.10.7 (j)) that the inclusion 3e°O(D) -+ O10+1(D) is continuous. Consequently, by the Banach theorem (Theorem 1.10.4), either 3e°° (D) _ O10+i(D) or 3e°°(D) is of first Baire category in the Frdchet space 0(0+)(D). Define

f(z) = Log 1- e-cza I

ZED.

where Log stands for the principal branch of the logarithm. Obviously, f is holomorphic and unbounded. We are going to prove that f E O (°+) (D). Fix an N > 0. Then

If W1
l and p
2f + I with 76 0. Let jo be the smallest of such j's. We may assume that tj Put x := b2 + b4 + b6 + + b2" = Ej=t xjbi, where q := ["21 J. Then x E X C Kt(X). Consider the vector x + a E Kt(X). Using (l)t, write x + a = z + c with z = En=, zz bi E X and c = Ej ti 1 ca b' . Observe that t1

z j = z, + c. = x, + aj, j = 2e + 2, ... , jo. In each of the following three cases we get a contradiction with the definition of X:

jo is odd: Then zjo = ado = -1, zj0_1 = xjo-t = 1; jo < n and jo is even: Then zio = I - 1 = 0; jo = n and jo is even: Then z1o = an = -I. (2)1

(3)t: It suffices to observe that (2)1 implies

E(Kt(X))1non ={a=(at.....a")EOn :(a,bl)=...=(a,b2t+1)=0)

aEQ"

=0

ifkC+1/2

a2kr + a2k+l s + a2k+2 + ... + an = 0 ifk < f _ (a E On : a t = ... = a2t+1 = 0, a2t+2 + a2t+3 + ... + an = 0).

I

3.3. Maximal affine subspace of a convex set II

233

(1)t+l: The inclusion "C" in (I)j+1 follows trivially from (3)t. To (3)t prove the opposite inclusion, take x x j bf E X and t = 21+3 t j bf . Define t+l

z := x1b' + E((Ix2jl + It2j+lI)b2' j=1 n

xjbf E X,

+ (x2f+l + 121+1)b2l+l) + j =2f+4 t+l

i:= x1b' + 1:(x2j - Ix2 j I + 12j - It2j+l Db2j E F (K1(X )) 1=1

Since x + t = i + 1, we conclude that x + t E X + F (Kt (X )) C Kt (X) +

F(Kt(X)) = Kt+l(X) Exercise 3.3.7. Let X be the domain from the above example. Write X in the form

X=hic,n...nHU2p,C2p, p=

n 2

so that

Kt(X)=Halt+I,c2t+I n...nlj2,c2p, a=0, 1,2,...,L Hint. It suffices to choose of, c j , j we have

n-1 2

= 1, ... , 2p, so that for every x =

(x, a2j-1) < C21_1 q t21+1 < t2j,

(x, a2j) < C2j q t2j+1 > -12j, (x , a"-') < Cn -

l q to < 1, (x, a") < Cn to > -1,

F_'=1 ij b1

n j=1,2,....LI-1 2

j

Ln 2 1 J,

n = 2p, n = 2p.

Proposition 3.3.8. Let X = I1, , n

n I.N,CN, where a1... . aN E (Ot"), and let r := rank [a' , .. . , a"]. Further, let

J:={1 =(i1,...,Ir):1 2). We may assume that N > 2 and

n

Xj0Xo, k=1,...,N.5

(3.3.1)

It suffices to prove that r = N. Consider the following two cases:

Case 1. r dD (a) be such that each derivative DOf extends holomorphically to IP(a. r), 13ESk.

240

Chapter 3. Reinhardt domains of existence

Observe that if g E O(D) is such that each derivative a extends to a function g j E 0(UP(a, r)), j = 1 ,n , then the function g itself extends holomorphically to IP(a. r). Indeed, the extension may be given by the formula n

g(z) = g(a)+J(zj -aj)

r

I

gj (a+t(z-a)) dt, z E IP(a,r)

(EXERCISE).

o

j=I

Consequently, using the above remark inductively, we easily conclude that every

function f E 8 extends holomorphically to P(a, r); a contradiction.

O

Lemma 3.5.4. Let D C" be a Reinhardt domain. Assume that D is an domain of holomorphy,9 where S is such that there exists a ko E Z+ such that Sko C S. Then

.7(',S-

D = int

n {z E V(a) : Iaf f fEXoo.s(D)

!(a)za-fi

< IIDfif IID}

aEE(f), flES

a0fl, (s)#o

Moreover, if S = (Z")., then D satisfies the Fu condition.

Proof. By Lemma 3.5.3, D is an .F :_ {Dpf : f E ,7(°O,S(D), 6 E S}-domain of holomorphy. Observe that Y C J°O(D) is invariant under n-rotations (in the sense of (1.12.1)). Thus we may apply Proposition 1. 12.6 (b) (EXERCISE). Now, assume that S = (Z"). and let (SD) fl V j o 0 0 f o r some jo E 11, ... , n). By Remark 1.5.7 (e), to prove that D (jo) = D we only need to show that a jo -fjo

Oforanya E E(f), fi E S witha # )4, (s) 96 0. Suppose that there exists f E J'°O'S(D) and a E E(f) with ajo < 0. Put max{0, av }, v = 1, ... , n. Then iB a, and 0. # = (ft, .... /+n), Therefore

za-p =

T7

1!

z.0v

vE(I..,n) av 0 because of (SD) fl Vo # 0; a contradic-

0

tion.

Now we are able to verify Theorem 3.5.2.

Proof of Theorem 3.5.2. We may assume that D V. The implications (v) (iii) (iv) (ii) are evident. (ii) = (i): Recall that D is fat (Corollary 1.11.4). Suppose that aD fl V0 0 0 for some jo E (1, ... , n ). Observe that for any r > 0 the open set Dr := D fl F(r) 'Recall that

J(oo.s(D)

:_ (f E O(D) : Vaes : DO1f E Je°"(?)}, 0 96 S C Z.

3.6. Lh -domains of holomorphy

241

is fat and log-convex. We know that D, is an JOo.Z+-domain of holomorphy (cf. Remark 1. 11.3 (e)). Hence, by Lemma 3.5.4, if D, fl Vo 0 0, then Drj°) _ Dr. Consequently, D(i0) = D. (i) (iv) (resp. (i) (v) if D is an X'-domain of holomorphy)): Suppose that D is not an O(D)-domain of holomorphy (resp. Jt°(D) fl 0(b)-domain

of holomorphy). Let Do, D be as in Proposition 1.11.2(*) with 8 = O(D) (resp. $ = J'°O(D) fl O(D)). Recall (cf. Theorem 1.1 (b) (resp. 3.4.1 (iv))) that D can be written as

D = int n Da,c, (a,c)EA

where A C (tR"). x R (rest. A C (7L"). x R). Fix (a, c) E A and a> 0 such that D C Da,c C Da,c+s and D 0 Da,c+,. Since Da,c+e is a domain of holomorphy (resp. an J°O(Da,c+e)-domain of holomorphy), we only need to observe that, by Remark 1.5.11 (b), we have D C Da,c C Da,c+e; a contradiction. 0

Example 3.5.5 ([Sib 1975]). Let T = {(z1, z2) E D X D.: Izt I < Iz21) be the Hartogs triangle. Then: (a) T is an Ak-domain of holomorphy for arbitrary k E Z+ (cf. Theorem 3.5.1). (b) T is not an A'-domain of holomorphy (cf. Theorem 3.5.2).

Exercise 35.6. Complete the following direct proof showing that T is an Ak (T)domain of holomorphy for every k E Z +. Fix a k E 71+ and let Do, D be as always with D =_T and 8 = .Ak(T). We may assume that D ¢ T. Let a = (a 1, a2) E (D \ Yo) \ T . We have the following two cases:

tat I > 1: Then the function f(zt, z2) := 1/(z1 - a1) belongs to O(T) C dk (T) and is not extendible to D; a contradiction. 1a21 < lai I < 1: Then the function f(zt, z2) := zi+2/(a1z2 - a2z1) belongs to .,6k (T) and is not extendible to D ; a contradiction.

Exercise 3.5.7. Find a power series f(z) _ Fk oakzk, z E m, such that f E db°O(D) \ 0(i)).

3.6 L h -domains of holomorphy Our aim in this section is to discuss the problem of geometric characterization of Reinhardt $-domains of holomorphy D C C" with 8 C Lh (D). Recall the general Proposition 3.1.6 and Theorem 3.2.1(g) for the elementary Reinhardt domains. The following lemma, which will be used in the proof of Theorem 3.6.4, is of independent interest.

242


Lemma 3.6.1. Let D C C" be a Reinhardt domain. Then Moo,st (D) C db(D). Consequently, 3oo.k(D) C ,bk-I(D) fork E N. In particular, MOO-°°(D) C Exercise 3.6.2. Find a fat domain D C C \ (-oo, O] such that Log E .3e°D'S1(D) \ .4(D), where Log stands for the principal branch of logarithm. Before presenting the proof we recall a few well known facts from the theory of metric spaces. Let (X, p) be an arcwise connected metric space. For a continuous curve y : [a, b] -+ X define its p-length L,p(y) E [0, +oo] by the formula N

Lp(Y) := sup { E i=I

: a = to < ... < tN = b. N E N}.

Obviously, p(y(a), y(b)) < Lp(y). One can easily prove (EXERCISE) that

a < c < b.

Lp(Y) = Lp(YI [Q,c]) + Lo(YI [c,bl),

We define the inner metric p' associated to p by the formula

p'(x, y) := inf{Lp(y) : y E C([a, b], X), y(a) = x. y(b) = y},

x, y r= X.

Clearly, p < p'. One can easily prove (EXERCISE) that p' is a metric. In the case where X is a subdomain of E, where (E, 11 11) is a normed space, with P(x,Y) = IIx -YII, x, y E X, we have 11x - YII = P(x,Y) provided

that the segment [x, y] is contained in X. In particular, if B(x°, r) _ {x E E

IIx-x°IIo, (f J) j __ l C Z" ((fl! 1 C 7L+ provided that D is complete) such that

D C C"(PJ).

IIcizfli. 1IL2(D)

Proof The implications (v) = (ii) and (iv) cation (i)

1. j E W,

ilm cila0' I = +oo.

(i) = (ii) are obvious. The impli-

(iv) follows directly from Lemma 3.6.1.

(cf. Example 1.10.7(h)). "Recall that Loh .k(D) = L,'' (D) ( 12Recall that E (log D) = (0) for any bounded Reinhardt domain.

246


(iii)

(v): Fix a 0 D U Y° and j E W. Put X := log D and let x° := (log jai 1, ... , log Ia"1)

Note that x°

X. Since E (X) = {0}, there exist linearly independent vectors a 1,

an E Z" (a 1..... a" E Z+ provided that D is complete) such that n

x C n Ha° _: Xo; i=1

cf. Lemma 1.4.11 (iii). Let A := [ak]i,k=1,...,n E GL(n, Q. We may assume that

IdetAI > j2n"1aI ...an12. Put pj =: a1 +---+a" -1,

X1 ={WEIR":ti 1, Iz1I < Iz212},

and the unbounded function f(z) := z1/z2, z = (Z1,z2) E Ga, belongs to MOO-S, (Ga) (EXERCISE).

The problem of characterizing circular (in particular, balanced) F -domains of holomorphy is still open for many of the natural Frdchet spaces .' we discussed so far. Q (Cf. [Sib 1975], [Sic 1982], [Sic 1984], [Sic 1985], [Jar-Pfl 1996] for positive results.)

Chapter 4

Holomorphically contractible families on Reinhardt domains

4.1 Introduction Recall from Chapter 2 that Bih(O". U3,,) = Bih(l8", lL") = 0 for n > 2 and Bih(f". L") = 0 for n > 3 (see Theorem 2.1.17). In this chapter we will study other methods which may be useful to decide whether two given domains in C" are not biholomorphically equivalent. The idea here is that two biholomorphically equivalent domains should have the same amount of functions of a special class (e.g. bounded holomorphic functions or psh functions with specific singularities) or geometric data (e.g. analytic discs through corresponding pairs of points). To give a rough idea of what we are going to deal with let us discuss again whether ID" and B" are biholomorphically equivalent domains. We introduce the following function:

inD : D -+ [0, oo).

AD (z) := SuP{I f(z)I

:f

E 0(D,1)). f(O) = 0),

where D is a domain in C" with 0 E D. Let D C C" and G C C"' be domains, both containing the origin. Note that if F E O(G, D), F(0) = 0, then (EXERCISE)

rD(F(z)) < mr,(z),

z E G.

(4.1.1)

In particular, if F is biholomorphic, then inD o F = mG In the case where D E (D', lB") we get

AD(z) = qD(z),

z E D,

(4.1.2)

where

qD(z) :=

Ilzll00

if D = lD",

Ilzll

ifD=(B",

ZEC".

Indeed, let f E 0(D, ID), f (0) = 0. Then, in virtue of Proposition 2.1.9, it follows that I f(z)I < qD(z), z E D. Hence, AD < qD, D E (ID", lB"). On the other hand, in the case of D = lD" and z E lD" \ {0) put

gz:

gz(w):=wj,

252

Chapter 4. Holomorphically contractible families on Reinhardt domains

when qpn (z) = Iz j I. Therefore, mpy (z) > Ig.(z)I = qyn (z). Now let D = I" and Z E ie" \ (0). Choose a rotation AZ such that Azz = (21, 0); in particular, lid = IIA=zII = I4iII. Put gz: 1B. -+ D,

gz(w) :_ (Azw)i.

Obviously, gz E O(C3,,, ID), g,(0) = 0; hence men(z) > Igz(z)I = 1IzII = gen(z) So the above equations are verified. Exercise 4.1.1. Prove formula (4.1.2) for an arbitrary norm ball in C".

Now let F : to -+ ID" be a biholomorphic mapping, n > 2. Using a Mobius

transform (p: Dn -+ IDn, (G(z)

Fn(z) - F.(0)

F1 (z) - F1(0)

I-

71(0)F1(z)'...

1- F.(0)F.(z)

"

Z E O,

we may even assume that F(0) = 0. Then, by (4.1.1),

mmn(F(z)) =

z E 1$n.

Therefore, we get the following equation:

IIF-'(t, 1/2,..., 1/2) 11 = max{t, 1/2),

t E (0, 1).

Note that the left function is differentiable on (0, 1), but, obviously, the right one is not; a contradiction. Observe that in the above argument the number of bounded holomorphic functions on ID" and I" is compared and this strategy finally has led to the result that both domains cannot be biholomorphically equivalent. Instead of dealing with the function AD. i.e. with the family of bounded holomorphic functions, we may take all analytic discs p E 0 (1), D) in D through two given points, where D is a domain in C" with 0 E D. We define

kD(z) := inf{r E [0. 1) : 3q,Eo(n,D) : p(0) = 0, W(r) = z},

z E D.'

Remark 4.1.2. Let G C C'", D C C" be domains both containing the origin. If F E O(G, D), F(0) = 0, then (EXERCISE)

kD o F < kG; see (4.1.1). In particular, if F is biholomorphic, then kD o F = kG. Note that ko (0) = 0.

(4.1.3)

253

4.2. Holomorphically contractible families of functions

We claim that kD = qD, where D E {m", In }. Indeed, fix a Z E D \ {0}. Then V(A) := A9v(s) gives a function rp E O(U). D)

with (p(0) = 0 and rp(gD(z)) = z. Therefore, kD(z) < qD(z). On the other hand, let tp E O(ID, D) with rp(0) = 0 and p(r) = z, r > 0. If D = D", then using the Schwarz lemma for (pj we get that IzjI _ Iq. j(r)I < r, j = 1, ... , n. Therefore, q 1D. (z) < k®n (z). In the case when q E 0(i), is") with

p(0) = 0 and V (r) = z we see that qe" (z) = qe ((p(r)) = rqs" (ip(r)), where p(A) = AO(A), A E ID, and W E 0(U),C"). Observe that qe"(ip(A)) < l/1A1, A E iJ \ {0}. Applying the maximum principle for the subharmonic function qe" o SP we obtain that 0 E 0 (m, [B" ). Hence, qen (z) < r and since r was arbitrarily chosen,

we have qen (z) < ken (z). Observe that we may also use this geometric function to disprove the biholomorphic equivalence of U)" and B,, (EXERCISE).

Let us summarize what we have done so far. We have introduced a family (dD)oEDCCn. nEN

of functions dD : D [0, 00) (dD E (AD, kD)) satisfying the following property: (*) for any domains G C C'", 0 E G, D C C", 0 E D. and for any

F E O(G, D), F(0) = 0, we have dD(F(z)) < dG(z), z E G. In particular, dD(F(z)) = dG(z), z E G, if F E Biho,o(G, D). Moreover, these functions were explicitly described in terms of the geometry of D.

4.2 Holomorphically contractible families of functions Let us begin with the following definition of a holomorphically contractible family which puts the functions of the introduction in a general context. The interested reader is referred to [Jar-Pfl 19931 and [Jar-Pfl 2005] for more information than is given in this chapter.

Definition 4.2.1. A family (dD)D of functions dD : D x D --> R+, where D runs over all domains D C C" (with arbitrary n E N), is said to be holomorphically contractible if the following two conditions are satisfied: (A) dm (a, z) = m (a, z) _ I l aza2 1, a, z E D (m is the Mobius distance),

(B) for arbitrary domains G C C"', D C C", any F E O(G, D) works as a contraction with respect to dG and dD, i.e.

dD(F(a), F(z)) < dG(a.z),

a,z E G.

(4.2.1)

(Compare condition (B) and (*) from Section 4.1.) Notice that there is another version of the definition of a holomorphically contractible family in which the normalization condition (A) is replaced by the condition

254

Chapter 4. Holomorphicaily contractible families on Reinhardt domains

(A') dm = p, where p =

2

log i±' is the Poincare distance on D.

Both definitions are obviously equivalent in the sense that (dD)D fulfills (A) and (B) iff the family (tanh-I dD)D satisfies (A') and (B). In our opinion the normalization condition (A) is more handy in calculations.

Remark 4.2.2. (a) Recall that m and p are distances on m. (b) If F E Bih(G, D), then F-I E O(D, G). Therefore,

dD(F(a), F(z)) = dc(a,z),

a, z E G.

In particular, if F E Aut(D), then dD(a, z) = dD(F(a), F(z)), a, z E D. (c) Moreover, if Di C C"J, j = 1.2, are domains, then

dD1XD2((a,,a2),(bi,b2)) ? max{dD1(a,,bl),dD2(a2,b2)},

(4.2.2)

whenever (a 1,a2), (b1,b2) E DI x D2 (EXERCISE, use (4.2.1) for the projection maps). If in (4.2.2) always equality holds, then we say that (dD )D satisfies the product property. The following holomorphically contractible families of functions seem to be the most interesting ones.

Example 4.2.3 (Mobius pseudodistance). MD (a, z) :

= sup{m(f (a), f(z)) : f e 0(D, m)} = sup{I f(z)I : f E 0 (D,1D), f(a) = 0},

(a, z) E D x D;

the function CD := tanh-I mD is called the Carathesodory pseudodistance. Indeed, to prove (B) it suffices to note that for F E O(G. D) and f E O(D, ID) one has f o F E O (G, ID). Moreover, the fact that

m(a. z) = m(f (a). f (z)),

f E Aut(D),

a, z E m,

gives the second equality in the definition of mD. To obtain condition (A) it suffices to observe that m m (0. ) = m (0. ). And this equation follows immediately by using the Schwarz lemma. Observe that mD (resp. CD) is positive semidefinite, symmetric and it satisfies the triangle inequality (EXERCISE). So mD and CD are, in fact, pseudodistances

on D. For D = C, Liouville's theorem immediately gives that me = 0; thus, in general, mD (resp. CD) is not a distance.

Example 4.2.4 (Mobius function of higher order).

mDk)(a. z) := suP{If(z)II1k : f E O(D, ID), orda f > k},

(a,z)EDxD, kEW,


255

where orda f denotes the order of zero of f at a. Indeed, in order to see (B) it suffices to observe that for a, z E D, F E O (G, D) and f E O(D, iD), ordF(a) f > k, one has f o F E O(G, ID) and orda f o F > k.

In particular, mD)(a, z) = mD)(F(a), F(z)) if F E Aut(D). Therefore, to see (A) it suffices to show m(k)(0, z) = m(0, z) which is a simple consequence of the Schwarz lemma.

Remark 4.2.5. Note that, in virtue of Montel's theorem (see Theorem 1.7.24), there exist extremal functions for mD), i.e. for any domain D C C", any pair

(a, z) E D x D, and any k E N there exists an f E O(D, D) with orda f> k such that mD)(a, z) = I f(z)I'lk Example 4.2.6 (Pluricomplex Green function).

gD(a,z) := sup(u(z) : u: D -+ [0. 1), logu E TSX(D),2 3C=C(u,a)>0 WED : u(w) < CIIw - all),

(a,z) E D x D.3

The point a is called the pole of the pluricomplex Green function.' Indeed, to see (B) let F E O(G. D), where D C C" and G C C'" are arbitrary domains. Fix an a E G. If u : D --> [0,1) is log-psh satisfying u < C II -F(a) II on D, then log u o F E PSX(G) and

(u o F)(z) = u(F(z)) < CIIF(z) - F(a)II < CIlz -a1I,

z E IB(a,r) Cs G,

where e and r are suitably chosen. Therefore, u o F < gC, (a, z). Since u was arbitrarily chosen, it follows that gD (F(a), F( - )) < gG (a, ). In the case where D = ID we fix a u : U) -* [0,1) such that log u is psh and u(A) < C IAI. Observe that then u/I ido I E SJ{(!D,) and that this function is locally bounded in D. Therefore, it extends to a sh function on the whole of D. By the maximum principle it follows that u < I id[) I. Hence we have that u = I idm I, i.e. gp(0, - ) = m(0, ). The situation for a general pole follows immediately using (B) and a MObius transformation.

While (dD)D, dD E {MD, mD), gD}, are based on families of functions we turn now to families defined by geometric conditions, namely by a set of analytic discs. 2Recall that 9S9-C(D) denotes the family of all functions plurisubharmonic on D. ;Note that it suffices to have u(z) < C (I z - a i( for all z E I(a, r) \ {a } when r > 0 is sufficiently small. Moreover. u(a) = 0 (Ext=atcISE). 4For relations between the pluricomplex and the classical Green functions in the unit ball see (Car 1997]. For a different pluricomplex Green function see [Ceg 1995]. [Edi-Zwo 19981.

256


Example 4.2.7 (Lempert function).

ko(a, z) := inf{m(1l, µ) : A, /I E m, 30EO(D.D) : w(),) = a, (p(/L) = z) =inf{µ E [0,1) : 3yvEo(D,D) : V (O) = a. (p(µ) = z) = inf{µ E (0, 1) : W(0) = a, V(µ) = z),

where (a, z) E D x D. Put k'D := tanh-t ko. Indeed, first we have to show that the above definition makes sense. So let us fix

points a, z E D. Connect them by a continuous curve, i.e. take a y E e([0,11, D) with y(0) = a and y(1) = z. In virtue of the Weierstrass approximation theorem, we may approximate y uniformly by a sequence of polynomial mappings (p1 )j. Taking a sufficiently large j we may assume that 411 Pi

- YII[o,l] < dist(y([0,1]), 8D).

Put p := pj and then

P(A) := P(A) + (a - P(0))(1 -A) +A(z - p(l)),

A E C.

Note that p([0,1]) C D, p(0) = a, and p(1) = z. Then p maps even a simply connected domain U, [0, 1] C U, into D (EXERCISE). Applying the Riemann E O (ID, U) with *(0) = 0 and W(p) = 1 for a mapping theorem leads to a suitable A E (D. Hence, 0 := p o tlr gives an analytic disc through the points a and z.5

Observe that if F E O(G, D), a, z E G, and Q E O(D, G), (p(0) = a and V(µ) = z for some it E [0, 1), then F o rp E O(0). D) with F o cp(0) = F(a) and F o V (It) = F(z). Hence (B) is fulfilled. To prove (A) use the Schwarz lemma in order to see that k®(0, z) > Izl = m(0, z), z E m \ {0}. To get the inverse inequality take simply qp c (9(U, ID), p(A) := x 1 1.

Exercise 4.2.8. Prove that

k p(a, z) = inf{lul : µ E m, 3,Eo(m,D) : qp(0) = a, (p(µ) = z),

a. Z E D.

For many purposes it is important to know whether the infimum in the definition of the Lempert function is taken by some analytic disc.

Definition 4.2.9. Let D C C" and a, b r= D. A mapping (p E O(U, D) is called a k'D-geodesic for the pair (a, b) 6 if there are 1t, A2 E 1) such that cp(.AI) = a, co(A2) = b, and kp(a, b) = m(AI, A2). 'Note that the same remains true in the case when D is a connected complex manifold (see [Win 2005)). 6We also say that rp is an extremal disc through a and b.


257

Remark 4.2.10. In general such a geodesic need not exist. For example, take D := iB2 \ ((1/2,0)). Observe that D is a non-taut domain. Fix now the points (1/4, 0) from D. For R E (0, 1) put (PR E 0(U), D), a (0, 0) and b (PR(A) :_ (RA, s(R)A(A -1/(4R))), where s(R) /r(0) = a,

0 (U, U)) with Ort (0) = 0. The Schwarz lemma gives 1 /4 < kD (a, b) and therefore, *1 = idp. Then, taking into account that * maps ID into 1B2 leads to the fact that

0 (use the maximum principle). And therefore, *(1/2) = (1/2,0); a *2 contradiction. In the case of taut domains we always know that such geodesics exist.

Proposition 4.2.11. Let a, b be two points of a taut domain D C V. Then there exists a kD-geodesic for (a, b). Proof. By definition we have a sequence ((p) j C 0(ID, D) such that rpj (0) = a,

kn(a,b). By assumption, D is taut and V(Qj) = b with aj E (0, 1), and aj (p j (0) = a, j E N. Therefore, we may choose a subsequence (V jk) such that rpjk -+ tp E 0(ID, D) locally uniformly. Then (p(0) = a and rp(k,(a, b)) = b (EXERCISE), i.e. (p is a geodesic we were looking for.

0

Exercise 4.2.12. (a) Let D C C" be a domain and let a, b E D. A map rp E O(U, D) is an MD-geodesic for the pair (a, b) if there exist At, A2 E lD such that (p(At) = a, 9P(A2) = b, and mD(a, b) = m(A1, A2) Prove that any MD -geodesic for (a, b) is a k; -geodesic for (a, b). (b) Let cP E (9(1), D) be an MD-geodesic for ((p(A1), O A2)), whereAI 96 A2. Prove that mD(9 (A'), (P(A")) = m(A', A"), A', A" E i7, i.e. (p is an MD-geodesic for all pairs (cp(A'), Sometimes such a q is simply called an MD -geodesic. and use properties of Hint. Study the function m \ (At } 9 A H m subharmonic functions. (c) Prove that any complex mD-geodesic op E O(0), D) is a proper injective mtv(p

mapping.

The next example relies on the normalization (A'). Example 4.2.13 (Kobayashi pseudodistance).

kD(a, z) := sup{d(a. z) : dD a pseudodistance on D, d < kD} _: tanh-t kD(a, z), a. z E D.

258


Indeed, let F E O(G, D) be as in (4.2.1). To any pseudodistance dD < kD on D x D we associate a new pseudodistance dG on G x G by dG (w', w") dD(F(w'), F(w")). Then

dG(w', w") < kD(F(w'), F(w")) < kG(w', w"). Therefore, dD(F(w'), F(w")) < kG(w', w"), w', w" E G. Since dD was arbitrarily chosen we end up with (4.2.1). For the normalization (A') it suffices to mention that p is a distance.

_ (a) kD is the largest pseudodistance on D below of kD; (b) CD kDi+, > k*D,

In particular, we havej -+oo lim kD. > kD. Now fix points a, z E D and choose a jo

'

Jim k )., (a. z) > k,(a. z). Then

such that a, z E Dj, j > jo. Suppose that p

there exist an analytic disc tp E O (U), D) and a number 00 r E (0. p) such that cp(0) = a

and rp(r) = z. Select an e > 0 such that (1 + s)r < p. Put O(A) := VOA/(1 + s)), A E [D. Then 0 E O(ID, D), 0 (0) = a, and 0 (r(1 + s)) = z. Note that 0 (lD) C= D. Therefore, 0 E O(UD, Dj), j >> 1, which implies that kDj (a, z) < (I + s)r < p; a contradiction.

Recall that the pluricomplex Green function gD (a, - ) = h, a E D, need not be continuous (see Remark 4.2.28), where D = Dh denotes a pseudoconvex balanced domain with Minkowski function h. With the help of Lemma 4.2.30 we get the following continuity result for the pluricomplex Green function.

Proposition 4.2.31. Let D C C" be a domain. Then gD is upper semicontinuous

onDxD. Proof. In view of Lemma 4.2.30 we may restrict ourselves to study only a bounded domain D. To be able to continue we need the following lemma.

Lemma 4.2.32. Let D C C" be bounded, assume that IB(a, r) C D, and lets > 0. Then there exists a 8 E (0, r) such that (gD(z, W))' +e < 9D (a. w).

z E B(a. 8), w e D \ 8(a, r).

Proof. Puts := r/3 and R := diam D. Then, by (4.2.1) and Proposition 4.2.21,

gD(z. w) < ge(z,2s)(z, w)
0 and choose a positive 8 E (0, s/3) such that

33)1+6

S

2s 0 such that b 0 IB(a, r) C D. Assume now that 9D (a, b) < a < f < 1. Then fix an s < 0 such that a < 9D (Z, w)

01+E

Taking the corresponding 8 from Lemma 4.2.32 we see that (9D (a,

w))1/(1+E)

z E IB(a, 8). W E D \ IB(a, r).

Recall that gD (a, -) E T85C(D); in particular, gD (a, -) is upper semicontinuous in b. Therefore, gD (a, w) < a, when w E IB(b, >?) C D \ IB(a, r) for a sufficiently small >) > 0. Hence, gD(z, w) < a'/(1+E) < P, z E IB(a, 8), w E IB(b, >)), which proves the upper semicontinuity. Exercise 4.2.33. Prove the following slight generalization of Lemma 4.2.32. (a) Let D, a, r, and s be as in Lemma 4.2.32. Then there is a 8 E (0, r) such that (9D (Z, w))1+E < 9D W, w),

Z. Z' E @(a, 8), WED \ R(a. r).

(b) Show (using (a)) that for a bounded domain D C C" the function gD (-, w) is continuous if w E D is fixed. Moreover, we have the following deep result due to Demailly (cf. [Dent 1987]; see also [Kli 1991]) which we will not prove in this book.

Theorem* 4.2.34. Let D C C" be a bounded hyperconvex domain. Then the function gD is continuous on D x D, where 91DxaD

I-

To get explicit formulas for the "invariant" objects on Cartesian products we discuss the following result which is extremely useful.

Proposition 4.2.35. The family (k p)D satisfies the product property; i.e. for all pairs of domains Dj C C"J, j = 1, 2, and points (a1, a2), (b1, b2) E D1 x D2 one has

kD,xD2((al,a2),(b1,b2)) = max{kp1(a1,b1),ko2(a2,b2)}.

266


Proof. Because of (4.2.2) only the remaining inequality has to be verified. Suppose that

p := kD,XD2((a1,a2) (b1.b2)) - r

> max{i 1(a,,b1),kD2(a2,b2)} = kD1(a,,b1) for some r > 0. We find analytic discs (pj E O(ID, Dj) with (pf (0) = aj and

(pj(p,) =bb,where kD,(a1,b1) kD,,,D2((a,, a2), (b1, b2)); a contradiction.

0

Exercise 4.2.36. Use Proposition 4.2.35 and Corollary 4.2.24 to prove that ID" x IB," is not biholomorphically equivalent to EBm+".

Remark 4.2.37. Note that also the family of Mobius pseudodistances (resp. of pluricomplex Green functions) fulfills the product property; for details see [Jar-Pfl 2005] (resp. [Jar-Pfl 1995], [Edi 1999], and [Edi 2001]). Obviously, if the product property holds one can get new formulas for the invariant functions for Cartesian products.

Proposition 4.2.38. Let Di C C" be a domain, j = 1, 2, and let F E 49 (D1. D2) be such that daED2 dbEDl, F(b)=a d9)EO(D,D2),Qp(°)=a 3*EO(®,D1) : *(0) = b, F o * = (p. (4.2.4)

Then, for points a1, a2 E D2 and bI E DI with F(b1) = a1, one has

kD2(a1, a2) = inf{kD, (b1, b2) : b2 E D1, F(b2) = a2}:

(4.2.5)

kD2(a1,a2) = inf{kDI (bl, b2) : b2 E D1, F(b2) = a2}.

(4.2.6)

Remark 4.2.39. (a) Recall the following definition: an F E O(DI, D2), DI, D2 domains in C", is said to be a holomorphic covering if for any z E D2 there exists a neighborhood V = V(z) C D2 such that F-I (V) = UJEJ U1, where Uj is an open subset of D 1, such that F I u, : Uj - V is biholomorphic, j E J. Any holomorphic covering F : D I -+ D2 satisfies (4.2.4). The reader is referred

to [Con 1995]. Even more is true: for any b E DI, A E ID, and 1p E O(U, D2) with (p(2.) = F(b) there exists a unique tfi E O(1), DI) such that *(A) = b and F o * = (p; * is called the lifting of (p with respect to F. (b) Recall that for any plane domain D C C there is a simply connected domain

D° E {C, ID) and a mapping F E O(D°, D) with the property (4.2.4). D° is the


267

universal covering domain of D (cf. the uniformization theorem in the classical complex analysis of one complex variable). (c) In Example 4.4.16 (see [Zwo 1998]), an example will be given where the infimum in equation (4.2.5) is not attained. This gives a negative answer to a long standing question asked by S. Kobayashi. Proof of Proposition 4.2.38. In view of (4.2.1) we obviously have

kD2(a,,a2) < inf{kD, (bl, b2) : b2 E D1. F(b2) = a2}. Assume that the above inequality is a strict one. Then there exists an analytic disc cP E O(0). D2) with (p(0) = al and (p(µ) = a2, where

inf{k,,(bl,b2) : b2 E D1, F(b2) = a2} > µ > kD2(a1.a2). Applying property (4.2.4), we find an analytic disc * E 19(0), D1) with Or(0) = b1

and F o /r = V. Therefore, A

kD1(bt, *(A)) > inf{kDl (bl, b2) : b2 E DI, F(b2) = a2} > /A:

a contradiction. Hence (4.2.5) has been verified. The equality (4.2.6) could be proved in a similar way. Its proof is left as an exercise for the reader.

Exercise 4.2.40. Find the formula for

D

where O+ :_ {A E C : Red, > 0).

One of the most important results in the theory of invariant functions is the following one due to Lempert. Its proof is beyond the scope of this book. Therefore, the reader is referred to [Jar-Pft 1993].

Theorem* 4.2.41 (Lempert theorem). (a) Let D C C" be abounded convex domain

and let a, b E D. Then there exists acomplex MD-geodesic W E O(0), D) such that a, b E (p(ID). In particular, CD = kD on D x D. (b) Assume that D C C" is a domain which can be exhausted by an increasing sequence (Dj) j of domains Dj C C", where each of them is biholomorphically equivalent to a convex domain. Then CD = kD on D x D. Note that (b) is a simple consequence of (a) and general properties of the MObius and the Lempert functions (EXERCISE).

As an application for pseudoconvex Reinhardt domains we have

Theorem 4.2.42. Let D C C be a pseudoconvex Reinhardt domain. Then

kD=kDon DxD. In particular, kD is continuous.

268


Proof. Recall that the logarithmic image log D is convex. Hence the tube domain

TD := log D + i IR" is convex. Therefore, by the Lempert theorem, we have kTD = kTo on TD X TD. D, F(z) := eZ, is a On the other hand, observe that the mapping F: TD holomorphic covering. Therefore, Proposition 4.2.38 immediately gives the proof of the equality in the theorem. Recall from Remark 4.2.26 that for a general pseudoconvex Reinhardt domain D the Kobayashi pseudodistance kD and the Lempert function kD are, in general, different. Example 4.2.43. Another application of Theorem 4.2.41 gives the following result (see Exercise 2.1.13). Let N be a complex norm on C", n > 2. Recall that !B = (z E C" : N(z) < 1),

B(r)={zEC":N(z) D.

Example 4.3.1 (Hahn function). HL (a, z) := inf{m(,1, µ) : 3,Eo(1D,D) : tp is injective, 1p(A) = a, (P(µ) = z) = inf{IuI : 34 eO(D.D) : sp is injective, V(0) = a, (p(µ) = z),

where (a, z) E D x D, satisfies (A) and (B').10

Remark 4.3.2. Observe that the infimum in the above definition is taken over a

non-empty set. Indeed, fix points a, b E D. a 0 b. Then there is an injective C1-curve a : [0, 1) -+ D connecting a and b such that a'(t) 0 0, t E [0, 11. By the Weierstrass approximation theorem, we find a sequence (p j) j e 14 of polynomial

mappings p j : C -> C" such that

Pj (0) = a.

P1(l) = b,

11

k)

- a(k) II

Io.11

0,

k=0,1.

and

pj([0, 1]) C D,

j E N.

If j >> 1, then pj I(o,11 is injective. Indeed, suppose the contrary, i.e. there exist 1 ', with ,o1 tj', tj" E [0, 1],1 j j E N. By the compactness of [0, 1] P1 we may assume that ti -. t' and t". Then the uniform convergence of (pj)j

implies that a(t') = a(t"). Applying the fact that a is injective gives t' = t". Therefore, n

0

IIPi(ti)-Pi(tj")II2

= IPj.k(ti.k)I2Itf - tjlz k=1

n

Ilak(rj,k)I - I PJ,k(tj,k) -

a'(ij,k)II2Itjl - tfil2

k=1

2IIa'II[0,1 It'

_,j"12, if j >> 1,

where tj,k is between tj' and r,". Hence tj' = tj for j >> 1; a contradiction. 1°Observe that the definition of HD is similar to the one of k,.

270


Fix a j such that pj is injective on [0, 11. Then there exists a simply connected

domain G C C with [0, 1] C G, pj (G) C D, and pj lG injective (EXERCISE). Arguing as in the case of the Lempert function we end up with an injective analytic disc in D passing through a and b.

0 0 H.

Remark 4.3.3. Obviously, kD < H. But

Indeed, fix two different points a, b E C,. Let V E O(U, C.) be injective with cp(0) = a, rp(µ) = b for a suitable µ E D. Applying the Koebe distortion theorem (see [Pom 1992], Theorem 1.3 and Corollary 1.4) we have lb - a I = I q (µ) - V(0)I < Iw'(0)I

1111

(I - (µl)2

< 4dist(a. af(tD))

1µl

(I -

1µl)2.

Taking into account that f (D) is simply connected we get

lb - al < 4lal

1µ1 -1µl)2.

(I

Hence Hey (a, b) > 0. On the other hand, the following result for n > 3 is due to M. Overholt.

Theorem 4.3.4 ([Ove 1995]). If D C C", n > 3, is a domain, then ko = HD on D x D.

Proof. Fix a, b e D, a # b. Without loss of generality, we may assume that a = 0 E D (EXERCISE). Let e > 0. Then there is an analytic disc (p E 49(0), D) with cp(0) = 0, cp(µ) = b for a suitable µ E m such that 0 < 1µI < kD (0, b) +e/2.

We choose an R E (0. 1) such that µ/R E U) and lµ/RI < k,(0, b) + e. Put cPR(A) := w(RA), IAI < 1/R. Obviously, VRlp E O(m, D) with WR(0) = 0 and cpR(µ./R) = b. Since 'pR is continuous on ID, we have dist('R(lD), aD) =: 2s > 0. Now we take a polynomial mapping p : C -+ C" coming from the power series expansion of'pR such that dist(p(m), aD) < s/2 and (PRCR)

PCR)II D j the universal covering mapping." Then there are two different points q1, q2 E lD and automor-

phisms fj E Aut(ID), j = 1, 2, such that pj (.f (q 1)) = pj (fj (q2)), j = 1, 2, and det

r(pi ° fi)'(gi) (P1 ° fi)'(g2)1 (P2 ° f2)'(g1)

(P2 ° f2)'(g2)J # 0.

Note that, in virtue of the uniformization result, the universal covering of D1 is given by the unit disc.

4.3*. Hahn function

273

Proof. By assumption the map p j is not injective, j = 1, 2. Therefore, there exists tlrj E Aut(ID) \ {idm} such that pj o 1lrj = pj, j = 1, 2; in particular, 1/rj is a lifting of p j . Note that *j has no fixed points in lD (otherwise, applying the uniqueness of the lifting, it would be equal to ide). Therefore, it has one or two fixed points on am (see Exercise 2.1.4 (b)). Fix A' E am with t/rj (A) # A' for j = 1.2. Then m(tA', * ( : A ' ) ) -+ 1 when t / 1, j = 1, 2. Hence we find zI, z2 E m with m(Z1, *I (Z 0) = m(z2, 0'2 (Z2)) E (0,1).

Let d E (0, 1) with m (-d , d) = m (zl , *I (z1)). Then, by Exercise 2.1.4(b), there exist h j E Aut(ID) with

hj(-d) = zj, hj(d) = IGj(zj) j = 1,2. Assume that (p j o h j)'(-d) 0 ±(p j o h j)'(d) for at least one of the j's, say for j = 1. Then one of the following determinants does not vanish: det

det

(PI °hI)'(-d) (pI ohi)'(d) (P2 o h2)'(-d)

(P2 o h2)'(d)

L(PI ohI o (- idr))'(-d) (Pt ohI o (- idr))'(d) JJ (P2 ° h2)'(-d) (P2 ° h2)'(d)

L

(EXERCISE, use that (P2 o h2)'(d)

0.)

So we may put fi = h 1, f2 = h2 (resp. f, = h I o (- id[), f2 = h2) and

qt=-d,q2=d.

Now, for the remaining part of the proof we may assume that

((pj ohj)'(d))2 = ((pj ohj)'(-d))2,

j = 1,2.

(4.3.1)

Put t%rj := hi I o t/ij o hj and Pj := pj o hj, j = 1,2. Then t%rj(-d) = d

and Pj'(-d) = (Pj o j)'(-d) = Pj'(jj (-d ))t/rj'(-d ).

Taking squares on

both sides we get ((-d))2 = I (see (4.3.1)). Therefore, either i/rj (-d) = d, jj(-d) = -1 or I%rj(-d) = d, t%rj(-d) = 1. Applying Exercise 2.1.5 (b) to 'p = - idm (in case of -) or rp = he and the fact that 3/j has no fixed points in ID, it follows that

-2d I =Vr2=hc with c:= 1+d 2' Now fix an a E D and choose cp E Aut(D) such that W(a) = I/r (a) and qp(* (a)) = a (see Exercise 2.1.4 (b)). Note that such a 'p exists.

Suppose that Sp'(a) = *'(a). By Exercise 2.1.4(b), SP = t/' and therefore (r o t/r(a) = a. So t/r o >/i has a fixed point in m and, therefore, none on ao). On the

274


other hand, Or is without fixed points on D. So it has at least one fixed point on 8D, say b E 8U). Then i/i o tf(b) = b; a contradiction. Fix an ao E ID n IR. Let rp E Aut(D) with tp(ao) = }G(ao) and tp(t/r(ao)) = ao. Then gp = h_a0 o (- ide) o hhao(*(ao)) o hao. By a direct calculation it follows that

g'(ao) -Vr'(ao) Summarizing, we know that if tP E Aut(D) is such that g(ao) = t/r(ao) and g(tf (ao)) = ao, then g'(ao) # ±t/r'(ao). Then, by Exercise 2.1.4(b), tp o (p = id® (note that gp o go has two fixed points in U)) and so tp'(t/r(ao)) _5. Finally, we put ql := ao, q2 := *(ao), .ft := ht, and f2 := h2 o ip. Then Pt (fi (g2)) = (P1 o h l)(f(ao)) = (Pt o t/rl)(h1(ao)) = (P1 o h1)(g1) = (PI o ft)(g1), P2(f2(g2)) _ (P2 o h2)((P(tG(ao))) = (P2 o h2)(ao) = (P2 0 *2)(h2(ao))

= (P2 o h2)(tr(ao)) = (P2 o (h2 o (p))(ao) = (P2 0 f2)(gl)Moreover, we have Idet

(P1 0 .fi)'(gt) 0 .f2)'(g0

r

(P1 0 fi)'(g2)(P2

(P2 0

f2)'(q2)]

(Pt oh1)'(ao) o

h2)'(go(ao))go'(ao)

r(P1 ohl)'(tr(ao))tr'(ao)

det lL(P2

0

(P1

(p2° (pt oh1)'(tf(ao)) 1 (P2° h2)'(Vr(ao))/go'(ao)]

_ (P1 ohl)'(*(ao))(P2oh2)'(*(ao))det

*'(ao) Sv (ao)

1

Vr (ao)/go

(ao)]

0.

0

Hence this lemma is proved.

Proof of Theorem 4.3.6. (a) Without loss of generality, we may assume that D1 is simply connected (EXERCISE). Our task is to apply Exercise 4.3.5. So let (p =

(g1, g2) E 0(ID, D1 x D2) and 0 < a < 8 < 1 with tp(0) 0 to(a). Assume that tpl (0) 54 VI (a). Recall that koi = HD1. Hence there exists an injective t%rl E 0 (U), D 1) with t (0) = rpt (0) and it (8) = (Pt (a). Put

* (A) :_ ( l (A), p2('A)). Then

A E ID.

E 0(ID, D1 x D2), t1' is injective, and one obtains *(0) = (p(0) and

tr(8) _ 0a) Now let (pl (0) = VI (a) and tp2(0)

(p2(a). Take a d E (0, dist((pl (0), i)D1))

275

4.3*. Hahn function

and put h(A)

V2(04 -'i(0)

(4.3.2)

W2 40 - 'P2(0)

*I (A) := VIP + M8 + 1(h (A)

A E ID,

(4.3.3)

S

where M := IIhII fi. Observe that I/it E O(L, DI). Finally, define *(A) :=

(i/ri(A). (M'10), A E D. Then Ir E O(D, D1 x D2) with *(0) = V(O) and cp(a). Moreover, one easily sees that I/r is an injective analytic disc. Hence, (a) is proved. C (EXERCISE). Let, as in (a), (b) We may assume that DI = C. and D2 54 '(a). Moreover, ' P = ('PI , V2) E 0 (1D, DI x D2), 0 < a < 8 < 1 , a n d

applying a suitable automorphism of C. we may even assume that 'PI (0) = I (EXERCISE). _ rp2(0) + dist('p2(0), aD2)I). In the case where w2(0) = 'p2(a) define D2 Obviously, D2 is a simply connected domain, 0 _ (VI, 02) E 0(D, DI X D2), where 02(l) := V2(0), A E D. In virtue of (a), there exists an injective analytic disc /r E O(D, D1 X D2) with I/r(0) = (p(0), *(8) = sp(a) = cp(a). Next, we discuss the situation when Ip2 (0) # cp2 (a). For the moment we assume,

in addition, that (pt (a) = 1 + 8. Put

I/i(A) :_ (1 + A, (2(gA)),

A E D.

Then ly E O(D, C* x D2) is injective and satisfies VV (0) = p(0) and I/r(8) = Ip(a). Now we turn to the remaining case IpI (a) # 1 + S. Then, fork E N, we choose and Arg(dk) 0 when k -> oo. numbers dk E C \ (1) such that dk = Note that dk -+ 1. Put (P2(00 - 'P2(0) Ck

Since Ick I

1 - dk

k E W.

oo we choose a ko such that Ickol > M := sup{IIp1(A)I : IAI
/r E 0(iD, DI x D2) with IJr(0) = (1, W2 (0)) = V(0). Moreover, a short calculation leads to'r(8) = rp(a). If *(A') = *(A"), then h(A') = h(A"), and therefore, A' = A", i.e. 1/r is also injective. Hence, the proof of (b) is complete.


276

(c) Recall that the universal covering of D j is ID and that the covering mapping pj : ID -+ Dj is locally biholomorphic and surjective, but both are not injective, j = 1, 2. Applying Lemma 4.3.7 we find a point q = (q1, q2) E 0)2, qI q2, and

automorphisms fj E Aut(ID), j = 1, 2, such that with fj := pj o fj, j = 1.2, the following is true:

Pj(gt) = Pj(g2), j = 1,2,

and

det p; (qt)

P2(qt)

P;(42) # O. P2(g2)

Moreover, choose an r E (0, 1) such that both mappings fj are injective on K(r) and

put a := (a,, a2) = (P1(0), h(0)), b := (bt. b2) = (fit (r) P2(r)) E DI x D2. Note thataj 36 bj, j = 1,2. Then, in virtue of Proposition 4.2.35, Proposition 4.2.38, and the choice of r, we have

kptxD2(a,b) = max{k,1(a,,bl),kD2(a2,b2)} = r. Assume now that k,1 xD2 = Hot xD2 in particular, r = kpl D (a, b) _ HDIXD2(a.b). Then there exist a sequence of analytic discs ((pj)j C ( (ID, DI x D2) and a sequence of numbers (aj )j C (1, 1 / 4) with a j \,, I such that (p j (0) _ a and (p j (a j r) = b for all j. Then, applying Exercise 4.3.5, we find /,, = (j(lj,I, *j,2) E O(ID. DI x D2) injective such that *j (0) = a and t/rj (a4 r) = b, j E W. Recall that P j are covering mappings. Therefore, we can lift the functions *j,k, k = 1, 2, i.e. there are holomorphic mappings 1Jij,k E 19(0), ID) such that Pk ° %j,k = *j,k and tyj,k(0) = 0. Note that (fk ° t%ij.k)(c r) = fk(r). Recall that fk is injective on K(r) and therefore injective on K(r +s), where e E (0,1-r) is sufficiently small (EXERCISE). Then, for large j, we have that 1/ij,k (a?r) = r,

k=1,2. By the Montel theorem we may assume that tifrj,k - t%rk E O(ID, ID) locally

uniformly, k = 1, 2. Since >%i(0) = 0 it follows that, in fact, j E 0(ID, ID2). Moreover, because of the previous remark, ik (r) = r, k = 1, 2. Then, by the Schwarz lemma, we have 1/rk = idp, k = 1, 2.

Put g = (gI.g2): 0)2 C2, gk(A./.t) := Pk(A) - Pk(µ). Note that g(q) = 0 with q := (q1, q2) and det g'(q) 9& 0. Hence we find neighborhoods U = K(ql, s) x K(q2, s) C D2 of q and V = V(O) C C2 such that g maps U biholomorphically to V and K(qt, s) fl K(q2, s) = 0.12 Letnowgj: 0)2 C2, j E W, g j (A, A) := (*j',1(A) - *j,1(l4 Vuj,2(A) - *j,2(IL))

(A. A) E D2.

By the result before we conclude that gj -). g uniformly on U. Then, in virtue of Theorem 1.7.28, there exists a large index jo such that g1° vanishes in at least one 12Recall that q1 96 q2.

4.4. Examples I - elementary Reinhardt domains

277

point (ti, t2) E U, i.e. >lijo(tt) = t/'jo(t2), which contradicts the injectivity of 'f

.0

4.4 Examples I - elementary Reinhardt domains In this section we will establish effective formulas for dq,,, a E R", where k

dDa E and

Da={ZEC"(a):IZIa 0, then D. = ID and all the invariant functions coincide with m on ID x D. If n = I and a < 0, then D. = C \ ID. Thus D. is biholomorphically equivalent to D.. Then it is easy to prove that m = mo. = gp. on D. X lD* (EXERCISE). Moreover, applying Proposition 4.2.38, we are able, at least in principle, to calculate k*.. Firstly, we give the formula for the Lempert function even for an arbitrary annulus.

Theorem 4.4.1. For R > 1 put A = A(1 /R, R). If a E (11R. R), then

x2 + 1- 2x cos(n(s - t))11/2 l X2 + 1- 2x cos(rr(s + t))1 where a =

Rt-2s

IzI =

RI-2r, and x

= exp (

z = IzIe'B(Z) E A. -7r < 9(z) < s,

Ilo' gR).

Proof. Note that

A3A-*A/aEQ:={wEC:rl m(0, b") < (m(0,ba))UI"1 = gDa(0,b)


281

Therefore, in general, gD ( , b) need not be continuous. Recall that also gD (a, is not necessarily continuous.

)

Using the formulas for the pluricomplex Green function for elementary Reinhardt domains, we get the following result (see Lemma 2.4.12).

Lemma 4.4.6. Let Da, Do be elementary Reinhardt domains in C", n >_ 2, of rational type with s(a) = s(fl) = 0, i.e. 0 E Da n Dp, and let F E O(Da, Dp). Then there exists a V E 9 (D, D) with F(V (Da, A)) C V (DD, (o(A)),

A E D.

where V(Da,A):={zEDa:za=A). Proof. Fix a A E D. Put

a = a(X) :=

1,A11a"),

where At/an is a certain root of A. Obviously, a E V (Da, A) and k = k(a) = n. Set

(p(A) := Fj"'(a)...Fhn(a) = F" (a) E D. Then F(a) E V (Do. rp(A)). Now take another point z in V (Da, A). Then

0

I

as - za - aaza

11/rcr(a)

= gDa (a, z)

gD8(F(a), F(z)) =

_F# (a) - Ffi(z)

I1/rs(F(a))

1 - F$(a)FP(z)

Therefore, F(z) E V (DO, V (X)). Note that by taking locally a holomorphic root A1/a^ in D. it follows directly from its definition that V is holomorphic in D.. Hence it extends holomorphically

to the whole of ID and this extension coincides with (p(0) = lim FO(a(t)) = lim to(t). So rp E O(D, ID).

r-o+

Corollary 4.4.7. Let Da and DD be as in the previous lemma. Assume now that

F E 9(Do , DO) is even biholomorphic. Then there is a V E Aut(ID) with F(V (Da, A)) = V (DS, (p(A)),

A E ID.

To show how to use invariant functions like the pluricomplex Green function we are going to prove the following result [Edi-Zwo 1999] (see also Section 2.3).

Theorem 4.4.8. Two elementary Reinhardt domains Da, Do C C", s(a) = s(p) = 0, are biholomorphically equivalent if and only if there exist a permutation a of { l .... , n) and a t > 0 such that aj = t f, (J ), j = 1, ... , n.


282

Proof. From the very beginning we may assume that n > 2 and that both domains are either of rational or of irrational type. In the first case we know by Lemma 4.4.6 that F(Vo) = V (Do, A) " for a certain it E D. Observe that if µ E D*, then the latter set is an analytic set with only regular points, but Vo has 0 as an irregular point. Therefore, µ = 0, or F(Vo) = Vo. Now let both domains be of irrational type. Suppose that F(V0) ¢ Vo, i.e. there is an a E Vo with F(a) 0 Vo. By Theorem 4.4.4 we conclude that

0 # gDa (a, -) = gDe (F(a), -) = 0: a contradiction. So, F(Vo) = Vo also in this case. 1,

Hence, there is a permutation a of { 1.... , n) such that F(Vj) = VQ(j), j = ... , n. In particular, F(0) = 0. Applying the formulas for the pluricomplex

Green function it follows that

(IZllat

...IZ.1,,)1/(a1+...+.an)

= gDa(0.z) = gDS(0, F(z))

1

IFn(z)1'in)11(B1+...+Bn)

= (IFI(z)Ist ...

ZEDa.

Moreover, for the points dd = (1, ... ,1, 0, 1, ... , 1) E V1 we have that F(dj) is contained in Va(t) \ Ucn=1, t &a(j) V . Therefore,

Izalt/aj = gDD(dj, z) = gDS(F(dj). F(z)) = I FB(z)11 /Ba(i).

Z E Da.

Combining the last equalities gives

Oil + ... + en

O!,

fit +-'-+On

A, (j)

j=

n,

0

which finishes the proof

Now we come back to prove almost all of the formulas stated in Theorem 4.4.4.

Proof of Theorem 4.4.4. The proof will be given in several steps.

Proof for m Da - the rational case. Define

f(W)

/ wa - a°`\ f r l :

1\

1 - aallJa

wEDa.

Then f E O(Da, ID) with orda f = r [ l > e. Hence m() (a, z) > If(z)I 1l I fr . which implies that mDa (a. z) > (m(aa, za))7 1-'Note that Vo C D. fl Dg.

283


Now let f E O(Da, U)) with ordr f > P. Put 45(w) := wa, w E Da. Applying Theorem 3.2.1 (e), we get f = tp o 4>, where (P E O (ID, D) and order f = r ord,a V.

Hence, ordaa V > r i and therefore, If f(z)I" t < (m (aa za)) l f;1, which proves the remaining inequality. Proof for m( .?f°O(Da)

- the irrational case. According to Theorem 3.2.1(d) we know that C, which immediately proves the claimed formula.

ProofforgDa -therationalcase. Let # :_ (la1I

Ob-

serve that Da and DO := Dp fl C; are biholomorphically equivalent via the following mapping

F: Cn(a) -* Cn(a),

z H (zi I,...,zy.

zs+l,...,zn)

Hence gga(a, z) = gDUo(F(a), F(z)), Z E D, In virtue of Proposition 1.14.25, we even know that gD0o(F(a), ) = gDS(F(a), - ) (EXERCISE).14 Therefore, from now on we assume that s = 0. Using the equation for m D. from above, we know that gDa (a, z) > m D. (a, z) = (m (a', za)) tl'. To get the converse inequality let u : D.

[0, 1) be a log-psh function satisfying

u(z) < C 11z -all, Z E Da; in particular, u(a) = 0. Fix a A E ID. with µan = A. Then C n-1 -r Da,

1/r(wl, ... , wn-1)

-nn

(w1 .... , wn-1' WIat

It an_t wn-t

is a holomorphic mapping onto V (Da, A) (EXERCISE15). So u o t/r E 3'8J{(C;-1) is

bounded. By Proposition 1.14.25 we conclude that this function is, in fact, psh on the whole of Cn-1. Moreover, it is bounded from above. Therefore, the Liouville type theorem for psh functions gives that u o v(A) E D. So we have constructed a function v : ID. -+ ID. such that uIV(Da,x)

vW,

A E D*-

v is sh on iD,. Indeed, fix Ao E D, and p > 0 with K(Ao, p) C D.. Choose a holomorphic an-th root of idx(xo,p), i.e. a g E 0(K(Ao. p), ID) with gan (A) = A. A E K(Ao, p). Then

v(A) = u(1, ... , 1, g(A)),

A E K(Ao, p).

14Note that Do \ Vo = D. " Use the assumption that or t , ... , an are relatively prime and therefore Z = a 1 Z +

+ an Z.

284


Hence, v is locally log-sh on D. and therefore log-sh on DD,.. As above, v extends to a log-sh function v on D.

First we discuss the case k < n, i.e. a01 = 0. Recall that r = ak+I + For A E D. we have

+ an.

v(A) = v(A) = u(al....,ak,Allr....,A'/r,Al/r(a11 ...akk)-'Ian) (a71...akk)-'/an)II D. be defined as

pA:_ {p1

2knaA (A),....wn-2(A),a-:pn-r(A),a

aj 2knirA anT

1

E O(H,D.),hp(0) =a,andhp(iv) _ (zr,....z.-I.e an 2knn i

T h e n

(pn(A))

Now we continue modifying the other coordinates in the same way as above 0 which finishes the proof. In the same spirit is the following lemma.

Lemma 4.4.11. Let L 1, L2 C= I), L C= C., and a E IR;. Assume that

m(L1, L2) = inf{m(Ai,A2) : )Li E Li, j = 1,2} > 8 > 0. Then there exists a set L C C., L C L, such that for any 71- z2 E L and Ai E Li,

j = 1, 2, thereisa V1 EO(ID,4.)suchthat*(A3)=zi, j = 1.2,andi/'(D)C L. Moreover, there is a set K C= C. such that for any z r , ... , :n E L with I za l =

1

and any wr.... , wk E L, k < n, there are fimetions h/ri E 0(m. C.) such that V/' (ID) C K, V*1(A1) = zi, j = I..... n, and V' (A2) = wi, j = I..... k, and ji r (A) ... *an (A) = e!6 A E 9).

286


Prof. Without loss of generality we may assume that Lt = (A1 = 0) and L2 =

0-2 = 5).16 Put L := exp 1(L) n (IR + i [0.2n)). Then L C (log El, log E2) + i [0.27r), where 0 < e1 < e2 and the e1's depend only on L. Set

L :_ (eaz+h : A E D, a, b E C such that aA1 + b E L, j = 1, 2). Then L C= C.. What remains is to choose 11i := eh, where h is an appropriate function of the form as it appears in the definition of L. wn_1 E L To prove the last part of the lemma we take, in addition, in an arbitrary way. For the pairs z1, w1 we fix functions 1Jrj E O(D, C.) with

f1(l)CL,

Y'17.1)=z1,

1Gi(A2)=wf.

Put

eiB (,I,i 1 (A) ...

*I,

1Yn

jnar^

t

I

AEID,

where the branches of the pow ers are chosen arbitrarily and 0 is taken such that *n (An) = zn. Exercise 4.4.12. Prove the following statement using the ideas from the proof of Lemma 4.4.11. (a) Let a E C., X E C, and A E ID be given. Then there exists a t/' E O(D, C.) such that V'(A) = a and *'(A) = X.

(b) Moreover, if A E ID, a E C', X E Ck, where k < n, and a E IR; with Iaa I = 1, then there is a 1/r =

t//n) E O(ID, C;) such that

4/(A) = a, (t/r1(A).... ,1//k(A)) = X, V,01' ... *n- = eiB idm

.

Applying the previous lemmas in the case of elementary Reinhardt domains leads to the following results.

Lemma 4.4.13. Let Da be of irrational type and a, z E Da n CZ. Then

k

(a,z)=k,(a',z'). a'ETa, z'ET.

Proof. Note that it is enough to prove that k, (a. z) = k% (a, z') (use the symmetry), whenever z' E T,. Suppose the contrary, i.e. there are points z', z" E T with

k , (a z') < k, (a, z") =: e. '6To get the general case. recall that if m(A 1, )2) > 8 = m(0, 8), then there is a 4p c- 490D, ID) with (p(.A 1) = 0 and rp(.l2) = 8.

287


In virtue of Lemma 4.4.10, we have k ,(a, z) = C, z E TZ""(a). Observe that z' E T = Ti" = Therefore, z' is an accumulation point of Ti""(a), D

which contradicts the upper semicontinuity of k a (a, ).

Corollary 4.4.14. Let a E Da fl C. where D. is of irrational type. Then k

(a,z) = 0,

z E Ta.

Proof for k Da - the case k < n. In virtue of Lemma 4.4.9, we know that

kDa(a,z) > gD (a.z) > 1z.111'. In order to prove the converse inequality for a z E Da \ Vo, put r Then all the points z1, ... , zk, z* ...... Aa belong to C. with n

k

ITIj)=i.

(1I,Ia1)( j=1

IZ XI

j=k+1

Moreover, a 1, ... , ak E C.. Then, in virtue of Lemma 4.4.11, we find functions

*j E O(D, C.) such that

ft *7'(A) = e'6,

AEID,

j=1

*j (r) = zj, 1 , , ( r ) =

- .

*j (0) = aj,

j = 1.....k, j = k + 1,...,n,

j = 1....,k.

Put

(p(A) := (*1(A)..... IVk(A), A1//k+1(A)......X*n(A)), A E D. Then (P E 0(Q), Da) with tp(0) = a and (p(r) = z. Hence, k%(a, z) < r, i.e. the proof is complete. Now we discuss the remaining case when z E D. f1 V0. Note that necessarily

we have -j # 0, j < s. First suppose that there is a coordinate zj = 0 with j > k + 1. Then the holomorphic mapping

CsX C n-s-1 3w = (wt .... , wj-l, wj+l.... , wn) (Wl,...,wj-1.0,wj+l,..., Wn) E Da leads to the following inequality

k%(a, z) < kCs xC^-S-t (a. i) = 0, "Recall that one can map C onto C via the exponential map.

17

288


where a :_ (at,...,aj-l,aj+t,...,an)and1 :_ (zt,...,zj-t,zj+t,...,zn) What remains is the case where I z j I + 1a11 0 for all j 's. For fi E (0, 1) we are going to define a SP = (apt .... , 0.) E O(m, D.), where

i = *j (A)

if a j = 0,

*j (A)

if Zj = 0,

Vj

.X E D.

ifajzj 54 0,

*j (A)

The *j E O (D, C,) have to be chosen in a correct way.

We need that the *j's satisfy rj;=t e'9 on m and that tp(f) = a, and (p(- fi) = z. Note that then we would get k, (a. z) < m(fl, -P) -+ 0 if P -+ 0. Fix some jt such that a j, = 0. Then we would like the functions tlrj to attain the following values in C.:

'Vi(i) =

ifajzj # 0,

aj a(

ifzj =0,0aj,

20

if aj = 0 0 zj, j 0 ji,

1

*h(8)

f I*j(0)lai rj ItVj(0)lai zi =0

aj zj #0

Moreover, at -P we only need to have

tVi (-f) =

zj

ifajzj 76 0,

z, (t-0 -20 2)

if aj = 0

zj

Note that there are fewer than n values we want to specify at -P. Therefore, Lemma 4.4.1 1 works and gives such a mapping t/r E O (D, Cl.) which completes the proof.

The remaining case, i.e. s = n or k = n, for elementary Reinhardt domains will be discussed later in this section. First we establish the formulas for the pluricomplex Green function in the irrational case.

Proof for gam, - the irrational case. As in the proof of Theorem 4.4.4 we may as-

sume that s = s(a) = 0. In the case where k = n we have gD (a. z) = 0, z E Ta (see Corollary 4.4.14). Then the maximum principle for psh functions implies that g D,,, (a, - ) = 0 on (P(0, (Ia i E, .... lan p). Recall that log g j (a, ) E P83{(Da) and that either log g

(a, ) _ -oo or the level set {z E D.

pluripolar set. Thus, gI (a, ) = 0 on D..

:

log g,Q (a, z) = -oo) is a


If k < n, then, by the formula for k

289

in that case, we know that

gj (a, z) < kva(a, z) =

Iza1l/r.

To conclude the proof apply Lemma 4.4.9.

The last part in this proof is devoted to prove some of the remaining formulas for the Lempert function.

Lemma4.4.15.Ifa,zED=DtnYo (1=(1....,1)),then

(a,z)=0. If aEDtnC;,zEDi,then k,, (a,z) _

(m(aI...an,zr...zn))r/`

where r := max{#{ j : zj = 0}, 1). Proof. The first formula is a direct consequence of the one for k* which has been proved before. Now we will discuss the case where a, z E C n.. In a first step suppose that µ := a r an = zr zn. Then the holomorphic mapping "-I

BO (w1....,wn-t) H (e'I....,eWn-I,µewI-..._wn_I) E DI

is onto V (Dt, µ), i.e. F(w) = a and F(w") = z for certain w', w" E C"-r. Hence,

kvt (a. z) < kin-I (w , w") = 0. zn. Put So we may assume that, from now on, a t . an zr A l :=a, ... an E I).

A2 := Zr ... Zn E D.

Applying Lemma 4.4.11 we find a i/rj E O(I), C) such that

fj(A1)=a1,'fj(A2)=zj, j =1,...,n-1, and *1

*n(Ai)=(at...a"-t)-'

;fin - ere on D1. Put

(A)

(A), ... , n-t (A)

(A)),

A E ID.

Then V E O(U), DI) such that (p(A1) = a, t/r(A2) = z. Hence,

m(ar...an,zr...Zn)?kD,(a,z)?m(ar...an,zr...zn), where the last inequality is a consequence of the property of holomorphic contractibility.

It remains the case that a E C and Z E Vo. Then again Lemma 4.4.11 gives the desired formula.

290


What we have just discussed is the simplest case of an elementary Reinhardt domain of rational type. The proof of the formula in case s < n needs deep results on geodesics which are beyond the scope of this book. Therefore we skip its proof. Details may be found in [Pfl-Zwo 19981 and [Zwo 2000]. Note that the case k < n is contained in Theorem 4.4.4. So the difficult case is the one with k = n. Now we turn to the irrational case for s < n.

Proof for k* - the irrational case with s < n. The case k < n was already verified. So we assume that k = n and z 0 V0. Recall that

kpa(a,z) = kpa,((IaiI...., lanl),(Iz1I,....Iznl)) (see Lemma 4.4.13). Now we approximate the ay's by rational vectors. We choose

a sequence (a('))i C 0' such that

a(i)>a, t(a(i))=1,

a1

j E N.

Applying Theorem 4.4.4 for points x, y E Da(i) fl IR+ we get a(i) , y1 U)

*

ynn

kD (.) (x. y) = m(xl

a(i) ... yn

jEN.18

(4.4.1)

By employing a biholomorphic reordering of the coordinates we may assume that

t(a)=a,,. Now suppose that

k,(a,z) < m(Iaal, Izal). Then there exist an analytic disc (p E O(1, Du) and A 1. A2 E ID such that cp(A 1) _

(Jail,..., IanI),W(A2) =

IznI),and

m(A1, A2) < m(laaI, Izal). Since tp(ID) is a compact subset of D. we can choose a large jo such that VIm E O(ID, Da(io)). Therefore, (io)

m(A1, A2) < m(I ai Ia(

(io

(io)

... Ian la"

,

Izi Ia,

(io)

... Izn Ian

):

a contradiction to (4.4.1).

On the other hand put Al := Iaal,

A2 := IzaI. If At # A2, then we find

analytic discs 1/!j = ehi E O (D, C.) with h j (,11) = log l a j l and hi (A2) = log Izi I,

j = 1, ... , n - 1 (use Lemma 4.4.11). Moreover, define eXP(-alh1

A E ID.

"Note that m(x,e'By) > m(x, y), t9 E IR, when x, y E [0, 1) (ExERcisE).


291

Then A 1 i/r" (A 1) = Ian 1. Put

(P(A) := (* (A), ....1//n-1(A), A*n(A))

A E 0).

Then (pn (A 1) = Ian l and r,n (A2) = Izn 1. Hence,

k, (a, z) <m(laa1.1za1).

IfA1 =A2,putAE:=A2+EEQ),e>0small. Put hs(A)=

A-A1 - As I -A1A I -AsA'

AE[.

Then h, E O (D. D) with hE (A 1) = hs (As) = 0. Then, using an appropriate Mobius transformation, we find an hs E O(D, I)) such that hr(A1) = hv(AE) = Al.

As in the case before there are 1/ri E O (ID, C.) with Or, (A 1) = l ai I and *j (AE) = Izi 1 , j = 1.... , n - 1. Put now W(A) := (Or1(A).... , *.-I (,X), hE(A)(

1

A E D.

()L) ...1/in"1

Then V E O(D, Da), (P(A 1) = (la l l.... , Ian l), and (p(AE) = (lz1 I, ... , Izn l). Taking into account that s may be taken arbitrarily small, we end up with k % (a, z) = 0, which finishes the proof. 0

In a last step we discuss the case when s = n, i.e. ai < 0 for all j.

Proof for k, - the case s = n. First observe that the map F : C"-1 x D. FA

Da,

A

is a holomorphic covering. Note that F(A) = w iff

Ai =

an

(log lw1I+i(Argwi+21in)), n-1

1

An

where 11,

j = 1....,n-1.

n Ej=isi(Argaj +21i rr) .

= 1=1

... , In_1 E Z. Applying Propositions 4.2.38 and 4.2.35 we are led to

k% (a. z) = inf ji*.

i (la'l-Vane_«n

Za -1/aye-an

Ejnn=,

aj Area,

Ej-;aj(Argz;+2rjn)

.1

In the rational case we apply Corollary 4.4.3 and get k*,(a, z) = k*. (aa. za).

292


In the irrational case we conclude via the Kronecker theorem that

k

(a z) = ko+(laal-an, IzaI-a").

It just remains to mention that kpy(x, y) = W;. (x'. y') fort > 0. Applying this remark with t = -a gives the desired formula. In a final step we discuss the Kobayashi pseudodistance in the case of elementary Reinhardt domains with s < n, which so far we have not mentioned.

Proof for k

- the case s < n. In the rational case it is easy to see that

Da x Da 9 (z, w) H

y2) : yt, y2 E D, as

za = 2})

=: d(z. w) satisfies the triangle inequality and is majorized by i,%. Hence it follows that kDo, > d and both are equal outside of the axes. Then the continuity of the Kobayashi pseudodistance gives the desired result. In the irrational case the reasoning is analogous to the one before and therefore left to the reader as an EXERCISE.

Proof for k I - the case s = n. This step is left as an EXERCISE. In particular, the effective formulas from above make it possible to give a negative answer to the following old question asked by S. Kobayashi (see [Kob 19701, p. 48), namely: is the infimum in Proposition 4.2.38 taken by a certain point in the holomorphic covering.

Example 4.4.16 ([Zwo 1998]). Let D = DI_,i _tl and >(r : C x D. -* D,

(e1,_'_e'11). 112

be a holomorphic covering. Take a := (r, r) E D with r > 0 and z := (r, ir). Then kD (a, z) = 0. Fix the following preimage (- log r, r-t -") of the point a. Then

0 = inf kE,o, ((- log r, keZ

(-logr + 27rik, ri exp(/(-logr + 2nik)))) = inf k p, (r-t kEz

r

exp(f(-log r+21rik))).

293

4.5. Holomorphically contractible families of pseudometrics

Suppose that there is a k' E Z such that

kD.

exp (v(- log r + 2nik'))) = 0.

Then 1/2 + k'J E 71;19 a contradiction.

4.5 Holomorphically contractible families of pseudometrics Recall from classical analysis that the Euclidean distance between two points x. Y E

G = lR" can be also given by the "minimal" length of all piecewise el-curves in G connecting these points. Let y : [0,1 ] --)- G be such a curve. Then the length of y is given by L(y) = fo II y'(t)Ildt, i.e. along the curve the lengths of its tangent vectors II y'(t) II, t E [0,1 ], are summed up. Hence we have an assignment

G x R" 9 (x, X) H ac(x;X) :=

him

IIx-(x+tX)II

hR.3t-+o

ItI

with the following property: a(x; sX) = IsIa(x; X), X E G, S E QR, and X E IR". This procedure will be transformed into the context of families of holomorphically contractible families of functions (dD)D. Let us start with a general definition.

Definition 4.5.1. A family (8D)D of pseudometrics SD : D x C" domain in C", i.e.

Z E D, A E C, X E C",

SD(Z; AX) = IAISD(Z; X),

9t+, D a (4.5.1)

is said to be holomorphically contractible if the following two conditions are satisfied:

(A) 81D(a; X) = y(a; X) := 1

I

IQI2,

a E D, X E C.

(4.5.2)

(B) for arbitrary domains G C C, D C C", any F E O(G, D) works as a contraction with respect to SG and SD, i.e.

SD(F(a); F'(a)X) < & (a; X), a E G, X E C"'.

(4.5.3)

Note that the Hermitian pseudometrics discussed in Section 1.19 are pseudometrics in the sense of Definition 4.5.1. In the following we will discuss the most important holomorphically contractible families of pseudometrics. 19Note that m(A, µ) = kp(A, µ.) [0, 1), log u is psh, and u (z) < C 11z - all, when z E IB(a, r) C D for certain C, r > 0.

If D =

I

Dj C C", DJ C Dj +t then y..) (a; X)

Montel's theorem). Moreover, we have the following results.

yD)(a; X) (use

297


Lemma 4.5.7. Let a E D C C", D a domain. Then:

EC".

(a)yDk)(a;X)=C lira

(b) yD) (a; ) is continuous and yo) is upper semicontinuous. (c) If we additionally assume that D is bounded and II X II = 1, then yo) is continuous on D X C" and (k) (z I 9z "

IID'-z"i

, _z" IF,-ell

Y( )(a;X), when z',

X.

Proof. It is obvious that the left-hand side is majorized by the right-hand side. Now let C. 9 A, -+ 0. Choose extremal functions f , E O (D, m), orda f,, > k, such that

MD)(a,a+At,X) =

Ilk I

IaI=k

a.

f locally uniformly on D. Note that

By the Montel argument we get fv,U

orda f > k and f E C9 (D, D). Then 1

Yp)(a:X)> I EatDa.f(a)Xal

1/k

lal=k

= µm IAv

=µmIA

lfvu(a+AvuX)I1/k vu

IMD)(a,a +A,, X).

The proof of the remaining points are left to the reader as an EXERCISE.

What we saw up to now is that the Mdbius functions have led via (4.5.5) or (4.5.6) to holomorphically contractible families of pseudometrics. In the case of the pluricomplex Green function we put

XD (a) := {u: D -> [0,1) : logu E PSK(D), 3M,r>o : u(z) < M II z - all, z E IB(a, r) C D},

where D is a domain in C" and a E D.21 Recall from Corollary 4.2.27 that gD (a, - ) E BCD (a). Moreover, for any X E C" the limit I

lim sup u (a + AX) C.aA-0 IAI

always exists. Thus we can define 21 Notethat M '(a) C Xo(a).

298


Example 4.5.8 (Azukawa pseudometric).

AD(a;X ) : =sup {limsup -u(a +AX) : u E 9Cp(u)} 1

(4.5.7)

IAI

= lim sup

1

I

gD(a, a+ AX),

a E D, X E V.

(4.5.8)

C.3A-+o IAI

Indeed, for D = ID we have lim sup

1

go)(a, a + AX) = lim sup

C.3A- 0 JAI

I

m(a, a + AX) = y(a; X).

c.91-0-o JAI

Note that u o F E XG(a) whenever F E O(G, D) and U E XD(F(a)). Hence, property (4.5.3) follows.

Lemma 4.5.9. Let D be as before. Then AD is upper semicontinuous.

Proof Fix a E 8(a, 2r) C D and Xo E C". Suppose that AD(a; X0) < M' < M. Then 19D(a.a + AX0) < M' when 0 < IAI < 2e for a certain positive E < 2(xo +1 In particular, fgD (a, a + AX0) < eM', IAI = E. Then, applying the upper semicontinuity of gD, there is a positive 8 < r such that

gD(b, b + AX) < eM'.

b E IB(a, 8). IIX

- XoII 1. Then 0. Therefore, h(z) < 1 /5 for large j and so h (l. z2 /z i) there is a sequence (a) j C C, a j -> 0, such that e"i = zZ /z J . Using a similar argument we find another zero sequence (fl, ) j C C satisfying h (1, A) I (EXERCISE22).

Let F E Aut(C2), F(z) := (zI,z2exp(-zl)). By Corollary 1.15.6, D' F' (D) is a bounded pseudoconvex domain in C2. In virtue of Proposition 4.2.21 it follows that

-9D'(0, (ak, ak)) =

k

k gD (0, (ak. ak exp(ak))) h(ak,ak exp(ak)) = h(1,exp(ak)) < 1/4, 1 ak

when k oo. In a similar way, we get gD'(0, (bk, bk)) -> 1, when k -+ oo. Hence, this different behavior of (0, A0, 1)), when A --* 0, verifies that we cannot take the limit in (4.5.8). Exercise 4.5.12. Try to construct a simpler example of a bounded pseudoconvex balanced domain in C2 with the same phenomenon as above. Nevertheless, in the case when D is bounded and hyperconvex the "limsup" in the definition of the Azukawa pseudometric can be substituted by just taking the limit ([Zwo 2000a)).

Proposition 4.5.13. If D is a bounded hyperconvex domain in C", then: (a) AD is continuous,

(b) AD(a;X) =

lim Cs A- o

I gD(a,a+AX),

(a, X) E D X C".

We should mention that there are similar results even under weaker hypotheses; for more details see [Zwo 2000a]. The proof of the above proposition needs some preparation which will be discussed first. Let D be as in the proposition and a E D. Put

DE := D,(a) :_ (z E D : gD(a. z) < e-E), where e E IR,o. Obviously, D. is open, a E DE, and, by Theorem 4.2.34, DE C D. Even more is true. 22Use the mean value inequality for psh functions.

300


Lemma 4.5.14. Under the above conditions, Ds is a domain. Proof. Suppose the contrary, i.e. that D. has a connected component U with a 0 U. e_8 We know that gD < on U and gD(a, z) > e-8, z E aU fl D. Consequently, the function if z E U, e-8 v(z)

gD(a,z) ifz E D \ U.

is psh on D (EXERCISE). Noting that a > U we have v < 9D (a. - ) on D. In particular, gD (a, - ) I U> e-8; a contradiction. 11

Lemma 4.5.15. Let a, D, D8 be as before. Then gDf(a, z) = 9D (a, z) ee,

z E D8;

ZED8, X EC".

(4.5.9) (4.5.10)

Proof. Note that gD (a, )e8 < 1 on D8. Thus, gD (a, z)ee < gDE (a, z), z E D8. On the other hand, gD (a, z) > e'e when z E aD8. Therefore, the function gDE(a,

v(Z)

z) e-8 if z E D8,

9D (a, z)

if z E D \ D8,

is psh on D. Hence, v < gD (a, - ) on D, which finally gives (4.5.9). The remaining equation is a simple consequence of the definition of the Azukawa pseudometric.

0 Proof of Proposition 4.5.13. (a) Fix (a, X) E D x C" with AD (a; X) # 0;23 in particular, X 54 0. Suppose that there is a number M > AD (a; X) and a sequence ((aj. Xj))j E D x (C" \ {0}) converging to (a, X) such that AD(aj; Xj) > M, j E W. Fix then e E R>0 such that M > AD(a; X) el. Put e' := 2E. Then D8'(a) C= D8(a) (note that gD is continuous on D x D, where gD = 0 on D x aD). Now we choose affine isomorphisms Fj E Aut(C") (j >> 1) such that

Fj(aj) = a,

Fj(aj)Xj = X,

D6'(a) C Fj(D6(aj)),

j E N large enough.

Then, by (4.5.10),

AD(aj;Xj)-e8 = AD,(aj)(aj;Xj) = AF,(De(aj))(Fj(aj):FJ(aj)Xj) = AFj(D.(a,))(a: X) < AD, (a) (a; X) = AD(a; X) e8', i.e. AD (a j ; X j) < AD (a; X) e8 < M ; a contradiction. Hence, A D is continuous. 23Note that AD is upper semicontinuous.


301

(b) Without loss of generality we may assume that a = 0 and AD (0; X) > 0 (EXERCISE); in particular, X # 0. Suppose now that there is a sequence (A j) j C C., A ,j -+ 0, such that 1

gD(O,O+AjX) < AD(0:X)e- 28,

I

j E N.

Since DE = D8(0) C= D, we find a Oo E (0, rr) such that e'0De 9 D, 101 < Oo. Moreover, we may assume that A j X E D8, j E N. Then l gD(0,e'BAjX) < 1 IAjI IAj1

gei6Dr(0,ei8),jX)

=

141

a; X e-8.

e8 < A

O. A X

9DE(O+AjX)

0< 00,

E N.

Fix r > 0 such that 8(r11 X 11) C D. Put TffgD(0, M

AD (0; X)

if E K.(r), if l; = 0.

Then log u E S}((K(r)). By the upper semicontinuity of u there is a jo such that A j E K(r), j > jo, and

u(e'B)j) < u(0)e

2! e35

u(0)es,

0 E [-7r, 7r].

On the other hand we already know that

u(ef°Aj)
0 9 t < A is suitably chosen. Then (p(lD) is a compact subset of D. In particular, a full s-neighborhood of V(FD) is contained in D. Now take z E IB(a. E14) and Y E C" with (1/t)11Y - X 11 < s/4. Then

*(A) :_ (p(A) + (z - a) + (Alt)(Y - X)

leads to a t/i E O(ID, D), ;fr'(0) = z and tt/r'(0) = tgo'(0) + Y - X = Y, i.e. MD (Z; Y) < t < A. (c) is left as an EXERCISE.

As a direct application we get

Corollary 4.5.20. Let Dj = Dhj C C" be balanced pseudoconvex domains, j = 1,2. If F E Biho,o(D1, D2), then F'(0): DI --* D2 is a linear isomorphism. Proof. Let X E D1. Then

h2(F'(0)X) = xD2(0; F'(O)X) = xDt (0; X) = ht(X),

0

which immediately gives the proof.

Remark 4.5.21. Let Dj be bounded balanced pseudoconvex domains in C". By a deep result of Kaup-Upmeier (see [Kau-Upm 1976], [Kau-Vig 1990]) it is known

that if Bih(D1, D2) 36 0, then Biho,o(D1, D2) j4 0. In particular, if Dt is biholomorphically equivalent to D2, then DI is linearly equivalent to D2. Exercise 4.5.22. Applying Lemma 4.5.19. prove:

x

2

2

1/2

42)2) . 25 (a) Yen (a, X) = xe, a: X) = 2 + (b) IB" is not biholomorphically equivalent to ID", n > 2. (c) Decide whether 033 and thedomainD := {z E C3 : IziI4+Iz214+Jz3I4 < 1) are biholomorphically equivalent (use Remark 4.5.21). t_1t

There is also another way to define the Kobayashi-Royden pseudometric which will be important in the proof of Proposition 4.5.25. 25To get the equation for y use the fact that l6 is convex.

304


Proposition 4.5.23. Let (a, X) E D x C", X # 0, where D is a domain in C". Then

xD (a; X) = inf {t E 1R,0 : 3FEo(Bn,D) : F(0) = a, t

- ( 0 ) = X. det F'(0) # 0}. 1

Proof. Only during this proof we will denote the right-hand side by WD (a; X). Obviously, any mapping F in the formula for ID (a; X) induces an analytic disc !p E O(D, D) by (p(A) = F(A, 0, ... , 0). Hence, xD(a; X) < iD(a; X).

Now suppose that xD (a; X) < m < x`D (a; X). Then there exist a t E (xD(a; X). m) and a V E 0(D, D) such that qp(0) = a and t(p'(0) = X. Put zE(DxCn-1,

FE(Z):=(SPI(Zt).

s>0.

Obviously, det F;(0) # 0, F&( , 0..... 0) = (p on D, t (0) = X. Now fix r < 1, near 1. Then F1(rID. 0,.... 0) is a compact subset in D. Thus we can choose a 8 small enough such that F1(rD x 80) x ... x SD) C= D. Hence, if a is sufficiently small, Fs E O(In(r), D). Finally, define F(z) := FE(z/r), Z E Bn, which gives the desired contradiction for r very near to 1.

Lemma 4.5.24. Let a. Z E 8n and let (P E O(O, B,), (p(A') = a, (p(Ag) = z, and Ao # Ao such that k® (a, z) = k1Bn (cp(A0), (p(A0)) = m(A0, Aa). Then A E D. In particular, (p is a xen -geodesic for (rp(A). V'(A)), A E D.

xBn (co(A); (p'(A))

Proof Observe that kBn = men and xen = Yen (EXERCISE). Put

u(A) .- mm'A' A ))

A E ID \ (A0').

Recall that men (a, - ) is continuous; so, by its definition, it is log-psh. Therefore, u is sh, u < 1, and u(A") = 1. Then, by the maximum principle, it follows that

u=Ion ID \{Ao}. So men (a, rp(A)) = m(Ao, A),

A E ID.

Now we can repeat the same argument w.r.t. the first variable to get finally m(A', A"),

men

A'. A" E ID.

Fix Ao E D. Then, by Lemma 4.5.3, we have

Y(Ao; l) =

lim

lim-

m()Lo' A)

IAo - Aj men ((POLO), V(A)) PLO -'XI

= Yen

i

co (Ao)).


305

Note that the definition of the Kobayashi-Royden pseudometric is of different type than the ones of the previous pseudometrics. Nevertheless, if the domain D is taut, then we have the following result (see [Pan 1994]).

Proposition 4.5.25. Let D C C" be a taut domain. Then

xD(a;X)_C.liim lim

oI

IkD(a,a+AX) 1

RI

k(a,a + AX), (a, X) E D x V.

0 We do not know any example of a non-taut domain for which this result becomes false although such an example seems very probable. 0 Proof. Note that it suffices to prove only the second formula. Suppose it is not true.

Then there are a point (a, X) E D x C", X 96 0, and a sequence C. a Aj - 0 such that

' kp(a,a+AjX)-xD(a:X)I ?so>0 1

(4.5.12)

IAiI

for some so. Applying Proposition 4.2.11, we may choose k;-geodesics (pj for

(a,a + AjX), i.e. cpj E (9(D, D), cpj(0) = a, cpj(tj) = a + AjX, and tj = kv (a, a + A j X) > 0 (recall that D is bounded). Moreover, D is taut, so we may, without loss of generality, assume that rpj -> .p E O(U), D), op(0) = a, locally unifonnly. Fix IB(a, r) C D. Then, if j is sufficiently large, 1 _ k* (a, a + Ai X) < "IB(a,r)((Pi (0), Vj (tj))

tj

tj < 111 Vj (o) - cci (ti) II r tj

.- 1 Ilty to)II r

Hence, wp'(0) 96 0.

Fix an s E IR,o. Then, by Proposition 4.5.23, we find an F E 0(8, D) and a t > 0 such that F(0) = a, det F'(0) , 0, t (0) = V(0), and 0 < ND (a: c'(0)) < t < XD (a: V(0)) + s.

Now we choose open neighborhoods U = U(0) C IB" and V = V(a) C D such that Flu: U -* V is a biholomorphic mapping. Moreover, fix Jo such that

a+AjX E Vand define gj :=(Flu)-t(a+AjX), j > jo. Note that l3" is taut. Hence we have &

-geodesics for all pairs (0, qj), i.e. there exist >lij E (9 (ID, (B"),

f(rj(0)=0,*j(rj)=gj,andkBn(0,gj)=rj,J ?Jo.

306


In virtue of Lemma 4.5.24, we conclude that men (0: tlr (0)) = 1, j ? jo. Applying again that @n is taut we may assume (without loss of generality) that *j -> * E O(DD, llln) locally uniformly. Obviously, *(0) = 0. Then, because of Lemma 4.5.19, we obtain

I = xBn (0: V'j (0)) - men (0 Vr'(0)) = I, i.e. >/i is a xBn-geodesic for the pair (0. Vr'(0)). Note that

9j -0

(Flu)-'(a+,XjX)-(FIu)-'(a)

tj

tj _ (Flu)-'((pj(tj)) - (Flu)-'((pj(0)) -+ tj

(F-' o(p)'(0)

In particular, this limit exists and it is different from zero. Observe that F o *j (O) = F(0) = a = SDj (0),

Fot/rj(Tj) = F(9j) =a+A1X =cpj(tj). Therefore,

tj

for j > jo. In particular, 1 < Recall that Fo*j(ij)-Fo*j(0)

lim

j*oo

=(Fov/r)'(0)#0

Tj

and Jim'pi(ts)-Vj (0)

0 54 (R' 0

= lim

a+AjX - a

= Jim F kj(Tt)-Fo%rf(0)Ti j-+00

Ti

tj

Hence, lim 1 =: A > 1 exists. Moreover, we have

i

V'(0) = A(F o *)'(0) = AF'(0)*'(0)

Taking into account that t-(0) = cp'(0) finally leads to Ai/r'(0) = tet = t (1, 0, ... , 0). Then I = xmn (0; Vr'(0)) = AxBn (0: et) =

A.


307

i.e. A = t and, therefore, I < A = t < XD (a; cp'(0)) + e. Then, letting E \ 0, gives 1 < XD (a; q'(0)) < 1. Hence, cp is a XD-geodesic for the pair (a, (p'(0)). Note that

fpj(tj!pj(0) _

V (O) = lim

tj

1-400

a+AjX -a = X lim .lj j- 00 j-.oo tj lim

tj

So we conclude

k,(a,a+AjX)

II

ju +00 rn

t

= XD (a; (p'(0)) jl m

I = xD (0; X):

a contradiction to (4.5.12).

11

Working with the Kobayashi pseudometric we always have

Proposition 4.5.26. Let D C C" be a domain and (a, X) E D x C". Then 1

kD(a,a+AX) lim sup C.3z-.o IAI

0 is given, then we find an analytic disc Sp E O(ID, D) with the following properties:

tcp'(0) = X,

(p(0) = a,

0 < t < XD (a; X) + E.

Then rp can be written as

rp(A) = a+

i X+ A20(A),

A E ID,

where 0 E 0(1D, C"). Fix IB(a, r) C D and then take only such A E C. with IAIt < 1 and a + AX E 1B (a. r). Hence,

IkD(a,a +.1X) < I,XIkD(a,v(tA)) + I IkD(w(tA),a + AX) I

I1 kD(co(0).rp(ll)) + rill IIt2A2w(1A)II ICI p(0,tA.) +

121AI II0(1A)II

r

Letting A -+ 0 leads to lim

I IAI

kD(a,a+AX) C", F, (z1,...,zn-1) := (eanz`,

..,ea"zn-1,µe-a,z,-..._an-izn-I

belongs to O(Cn-1, Da). Then there are a µo E I* and a 1 = (z1, ... , Zn_1) E

C"-1 such that FN,o(z) = a and F 0(2)Y = X, where Y = (X1, ... , Xn_1). Hence,

0 = xCn-1 (2; Y) ? xDa (a; X), i.e. the proof is finished.

4.7. Hyperbolic Reinhardt domains

313

Proof for xq, - the case s < n = k. First we recall that (Proposition 4.5.26)

xDa,(a;X)> lim sup

k°r°'(a.a+AX) ICI

Evaluating the right-hand side leads (by a trivial calculation) to the claimed formula for xD, (a; X). Hence the estimate from below is verified. We continue with the estimate from above. In virtue of Lemma 4.6.3, we may assume 54 0 and t = an (recall that t = 1 in the irrational case). Ejn= t °`

By the symmetry of Da we also assume that aj > 0, j = 1, ... , n. Now put AO := (aa)t/«n E ID n (0, 1) and r := k0 Ej_t ° Exercise 4.4.12, we find a V E 0(ID, D") such that (p(Ao) = a,

.

In virtue of

rfp'(ko) = X.

Hence, r > xD,, which gives the desired estimate from above.

Proof for xD. - the case s = n. We leave this last case as an EXERCISE for the reader. Use the ideas of the corresponding case for the Lempert function and Exercise 4.5.18 (e).

4.7 Hyperbolic Reinhardt domains We know that, in general, CD (resp. kD) need not be distances. We define

Definition 4.7.1. Let D C C" be a domain and let dD E {CD, kD, k;). D is said to bed -hyperbolic if dD (a, b) = 0, a, b E D, implies that a = b.

In particular, D is c- (resp. k-) hyperbolic if and only if CD (resp. kD) is a distance on D (in the sense of metric spaces). Note that (EXERCISE)

any bounded domain is c-hyperbolic; any c-hyperbolic domain is k-hyperbolic; any k-hyperbolic domain is k-hyperbolic. In the class of pseudoconvex Reinhardt domains we have a complete description of those domains which are hyperbolic.

Theorem 4.7.2 ([Zwo 19991). Let D C C" be a pseudoconvex Reinhardt domain. Then the following properties are equivalent:

(i) D is c-hyperbolic; (ii) D is k-hyperbolic; (iii) D is k-hyperbolic; (iv) D is Brody hyperbolic;

314


(v) there exists A = [a j,k ] t < f,k 0 such that S(a, 2r) C U.

Then kD(a, z) > 0 whenever z E al6(a, r). By assumption, kD(a,.) > c > O on 81B(a, r) (recall that kD(a, - ) is continuous).

Now take a point z E D \ IB(a, r) and a piecewise et-curve a: [0, 1] - D. which connects a and z. Then

f

I.

LXD (a) >

XD (a(t ); a'(t ))dt > kD (a. co(to)) > c,

wherea(t) E 8(a, r), t E [0, to), and a(to) E 31B(a, r). Therefore, kD(a. z) > c. Hence we have shown that the kD-ball BkD(a,c) C IB(a.r) C U.6 Put U _ Bk0 (a, c/2) and V := Bkm(0. c/2) C D. Now let (p E O(0). D) with tp(0) E U. If A E V, then kD(a.co(A))

kD(a,to(0))+kD(V(0),V(A))

26Recall that Bd (a, r) := (x E X : d (x, a) < r).

c/2+ko(0,A) $c.

4.7. Hyperbolic Reinhardt domains

315

Hence, rp(V) C U. (ii) (iii): Fix an a E D and put U := B(a, r) C= D. Then choose V = V(O) C m and Cl = CI (a) C U according to the assumption in (ii). If now

rp E O(lD, D) with V(0) E U and tcp'(0) = X, then rp(V) C U. Fix an s > 0 such that K(s) C V. Then we conclude that VI K(,) E O(K(s), U). Put O(A) := rp(sA). Then cp E O(1D, U) satisfying 0(0) = rp(0) and sip'(0) = X. Hence

xD(Z;X)>SxU(z:X)>CIIXII,

zEU, X EC",

where c is a certain positive number. (iii) = (i): Fix two different points a, b E D. By assumption, we may choose a neighborhood U = U(a) C D such that XD (z; X) > c II X II for a certain c > 0,

z E U. Fix now an r > 0 such that b 0 B(a, 2r) C U. Let a : [0, 1] -+ D be a piecewise e t -curve in D which connects a and b. Let to be that time such that a(t) E (B (a, r) for all t E [0, to) and a(to) E 8B (a, r). Then to

LXD(a) > Lxo(aI[o401) > cJ

IIa'(t)Ildt > cr > 0. 0

Hence, kD (a, b) > Cr; in particular, D is k-hyperbolic.

0

Exercise 4.7.4. (a) Prove that a domain D is k-hyperbolic if top D = topkD. Here top D means the standard topology on D, where top kD is the topology on D that is induced by the Kobayashi pseudodistance. (b) Prove the following generalization of Cartan's theorem (see Theorem 2.1.7):

Let D C C" be a k-hyperbolic domain, let a E D. and let 0: D -+ D be a holomorphic mapping such that 45(a) = a and df'(a) = id. Then ds = id. Use (a) to get the bounded situation. Moreover, we have

Proposition 4.7.5. Any taut domain D C C" is k-hyperbolic. Proof. Suppose that D is not k-hyperbolic. Then, in virtue of Lemma 4.7.3 (ii), we

find a z' E D, a neighborhood U = U(z') C D, a sequence Aj -+ 0 in D, and a sequence (rpj)j C O (m, D) such that tpj (0) -+ z', but rpj (,kj) 0 U, j E W. Since tp, (0) -+ z', there is no subsequence which is locally uniformly divergent. And because of rpj (Aj) 0 U, j E W, there is no subsequence tending locally uniformly to a rp E 0(lD, D); a contradiction. To get another large class of k -hyperbolic domains we prove the following result which is due to [DDT Tho 1998].

316


Proposition 4.7.6. Let u : lD --> [-oo, oo) be upper semicontinuous and locally bounded from below. Then

D := {zEIDxC:lz21e"(z1) 0 such that

xD(Z;X)>ClIX11,ZEU,XEC2. By assumption we have

g(r) := inf{u(;L) : RI < r} > -oo,

r E (0, 1).

Fix S E (0, 1), z' E (sD x C) fl D, and X E C2, X # 0. Now choose an analytic disc (p = ((pt, V2) E O(Q), D) such that V(O) = z' and tV'(0) = X for a certain t E (0, 1). In virtue of the Schwarz lemma we have ico (0)1 < I - Izi 12 < 1. Put so := 1+2s s0 for a A0 E D. Then )I _ 2+s Note that so < 1. Suppose hvt (Ao> the Schwarz lemma implies that

p1(Ao) - z'1 > Iwt (Ao)I -Izi 1 > so -Izi 1 > so - s IAoI > ll-z,V,(,Xo)l- 1-Iwt(Ao)IIz'l - I-so1z'l - 1-sos

Put S2 := (A E ID

2

IVI(A)l < so). Then I(p2(A)I < e-g(so), A E S2, and

:

K(1/2) C S2. Thus, I(pl (0)l < 2e-g4O) (Schwarz lemma). Hence

t > max I IX1I,

se-94lo)

>

I

I

min

{

1,

2e-s(so) IIXII =: t(s)II X II

Since c was arbitrarily chosen we have

XD(z; X) 2 t(s)IIX II,

Hence, D is k-hyperbolic.

(z, X) E ((sm x C) fl D) x C2.

0

The result allows us to present a k-hyperbolic pseudoconvex domain which is not c-hyperbolic. Thus the general situation is more complicated than the one inside the class of pseudoconvex Reinhardt domains.

Example 4.7.7 ([Sib 1981]). Choose a sequence (a1)j C ID of pairwise different points aj such that any boundary point E 8D is the nontangential limit of a subsequence of (aj)j. The reader is asked (EXERCISE) to construct such a sequence. Moreover, we choose natural numbers m j and nj, j E W, such that

nj <mj, j E N, 2 < -00, E1=, nt log -lug

4.8. Caratheodory (resp. Kobayashi) complete Reinhardt domains

K(a j . 3e-!'j)

317

fl K(ak, a-kmk) = 0, j # k,

K(aj, 3e-jmi) C D. Finally, we define

u(,.):=En maxjmj,logJA j=I i

)LE(.

2 ajI

Then U E e(ID) fl 83{(ID) (EXERCISE). In particular, u is locally bounded from below. Therefore, by Proposition 4.7.6, we see that

D:={Z E IDxC:I:2Ieu(=t) < 1} is a k-hyperbolic domain. Observe that D is even pseudoconvex (EXERCISE).

On the other hand, let f E X1 (D). Then f(z) =

r

hj(zI)z2, where the

h j's are holomorphic on D. Then, using the Cauchy inequalities and the maximum principle for holomorphic functions leads to h j = 0, j ? 1. So f depends only on

the variable z1. In particular, CD((zt,z2),(zt,0)) =0whenever z = (zi,z2) E D. Hence, D is not c-hyperbolic.

4.8 Caratheodory (resp. Kobayashi) complete Reinhardt domains Let D C C" be a bounded domain. Then we know that (D, CD) (resp. (D, kD)) are metric spaces in the usual sense. In general, let dD E {CD, kD }. We define

Definition 4.8.1. Let D C C" be a given domain.

(a) D is said to be d -complete if it is d -hyperbolic and any dD-Cauchy sequence (zj)JEN C D converges in the standard topology to a point z* E D, i.e. Ilzj - z*II -> 0. (b) D is said to be d -finitely compact27 if it is d -hyperbolic and if for any a E D

and any r > 0 the dD-ball Bdo (a, r) is relatively compact in D in sense of the standard topology of D.

Remark 4.8.2. (a) Observe that if D is d-finitely compact, then D is d-complete (use that dD is continuous). (b) Any c-complete domain is d-complete. (c) Any c-complete domain is a domain of holomorphy (EXERCISE).

(d) Recall that if D is k-hyperbolic, then top D = topkD. Therefore one may formulate k-complete (resp. k-finitely compact) by using top kD instead of top D. Note that in case of c there are examples showing that the topologies top CD and top D are different (see [Jar-Pfl 19931). 27This notion is taken from standard differential geometry. see the theorem of Hopf-Rinow.

318


(e) For a c-hyperbolic plane domain D we have: D is c-complete if D is c-finitely compact. This result is due to [Sel 1974] and [Sib 1975] (see also [Jar-Pfl 1993], Theorem 7.4.7). (f) Q In the case of a domain D in C', n > 2, it is still an open question, whether c-completeness implies c-finitely compactness. Q On the other hand, in the class of general complex spaces there are counterexamples; see [Jar-Pfl-Vig 1993]. (g) If F E Bih(D1, D2) and if DI is d-complete, then D2 is d-complete, where

d E {c, k). Dealing with the Kobayashi distance we have that both notions are the same.

Proposition* 4.8.3. Let D be a k-hyperbolic domain in C". Then the following properties are equivalent: (i) D is k-complete; (ii) D is k-finitely compact. Here we will not present a proof of this result. But we mention that the former result is a particular case of a result where one deals with continuous inner distances

(note that kD satisfies these properties). The main idea is taken from differential geometry. Details may be found in [Jar-Pfl 1993], Theorem 7.3.2. Next we mention the following necessary conditions for a domain to be k-complete.

Lemma 4.8.4. Any k-complete domain is taut. In particular, it is a domain of holomorphy (use Exercise 1.17.21 and the solution of the Levi problem).

Proof Take a sequence (Vj)jew C (9(ID, D). Suppose that this sequence is not locally uniformly divergent. So we find compact sets K C ID and L C D such that, without loss of generality, qp j (A. j) E L, where k j E K. Fix a point a E L and an r E (0, 1) such that K C K(r). Then, for any A E K(r), we have

kD((oj(A),a) < kD(cof(A),coJ(Af)) +kD(j(AJ),a) < p(,1Jj) + sup{kD(z,a) :: E L} $ C(r). Hence,

U cpj(K(r)) C BkD(a,C(r)) C= D. jeN What remains is to apply Montel's theorem.

0

In case of Reinhardt domains even the following converse statement is true.

Theorem 4.8.5. Any hyperbolic pseudoconvex Reinhardt domains D C C" is k -complete.

4.8. Carathdodory (resp. Kobayashi) complete Reinhardt domains

319

In order to be able to prove Theorem 4.8.5 we need the following localization result due to Eastwood (see [Jar-Pfl 1993], Theorem 7.7.5).

Lemma 4.8.6. Let D C C" be a bounded domain. Assume that for any b E aD there exists a bounded neighborhood U = U(b) of b such that any connected component U' of D fl U satisfies the following condition: if a E U' and U' 3 bk -+ b, then kU-(a, bk) _+ oo. The,, D is k-complete. Proof Suppose the contrary. Then we find a b E aD and a kD-Cauchy sequence

(zJ)jEN C D such that z1 -+ b E aD. Let U = U(b) be the neighborhood whose existence is known from the assumption. Choose R > 0 such that U U D C 1B" (R) =: V. By Exercise 4.7.4, U is open in topkv, i.e. Bkt, (b, 2s) C U for a certain positive s. Then we find a jo E D 1 such that

kD(zj, zt) < s/3

Z1 E Bk1, (b, s/3).

and

j, e > jo.

Fix a j > jo. Then there exist k E N, V,, E O(D, D), and A E 1D, v = 1.... , k, such that (P1(0) = Z

Wv+1 (0),

v = 1.... , k - 1.

k

p(0. Av) < s/3.

cPk(Ak) = Z', v=1

LetAEBp(0,s)CIDandI jo; a contradiction.

320


Proof of Theorem 4.8.5. By the hyperbolicity condition we may assume that D is bounded and D C D". The proof is done by induction. The one-dimensional case is obvious (use the explicit formulas from Section 4.4). Now let D C C", n > 2, and assume that the theorem is true for all smaller dimensions. We will show that D fulfills the condition in Lemma 4.8.6. So let us fix a b E 8D. There are different cases to discuss. Case b E C;: Let us choose a polycylinder I'(b, r) C= C. Put

V := {zEC":Ilzkl-IbkII b (if it exists). In virtue of Theorem 1.13.19, there is a function f E O(V', D) such that f (a) = 0 and I f (bk) I -> 1. Therefore,

kv,(a.bk) > tanh-t mv,(a,bk) > tanh-t If(bk)I -> oo, i.e. b fulfills the condition in Lemma 4.8.6.

Case b E Yo \ {0}: We may assume that b = (bt , ... , bk, 0,.... 0) = (b', 0) E

C; XC"-k,where I oo, i.e. b satisfies the condition in Lemma 4.8.6 with U = C".

Assume that there is a jo E {k + 1.....n} with D n Vjo = 0; without loss of generality, let jo = n. Then D 3o z -t z,, defines a holomorphic map

F E O(D,C,) and F(D) is bounded. Therefore, F(D) C K, (r). Note that K,(r) is k-complete. Hence, if a E D and D E) bk -r b, then

kD(a,bk) > kx.(r)(F(a), F(bk)) -> oo, i.e. b fulfills the condition in Lemma 4.8.6 with U = U(b) = C". Case b = 0: If D n Vo = 0 for a jo, then one argues as just before. On the other hand, D n Vj # 0 for all j = 1, .... n is impossible, since D is a Reinhardt domain of holomorphy. In case of Carath6odory finitely compactness we have the following reformulation in terms of bounded holomorphic functions.

Lemma 4.8.7. Let D be a c-hyperbolic domain in C". Then the following properties are equivalent:

4.8. Carathdodory (resp. Kobayashi) complete Reinhardt domains

321

(i) D is c-finitely compact; (ii) for any a E D and any sequence (zj)j EN C D without accumulation points in D there exists an f E 0 (D, D) with f (a) = 0 and sup{ I f (zz) I : j E

N} = I. In particular, any c-finitely compact domain is X'-convex. Proof. Obviously, (ii) implies (i). For (i)

(ii) just apply Proposition 1.13.18.

Remark 4.8.8. According to this result one can conclude that a lot of smooth pseudoconvex domains whose boundary points are general peak points are c-finitely compact. For example, any strongly pseudoconvex domain is c-finitely compact. Recall that a boundary point a of a domain D is said to be a general peak point if

for any sequence (zj)jEN C D, zj -- a, there exists an f E 0(D, ID) such that sup{I.f(zi)I : j E N} = 1. In case of Reinhardt domains the following geometric characterization is true (see [Fu 1994] and [Zwo 2000b]).

Theorem 4.8.9. Let D be a pseudoconvex Reinhardt domain in C. (a) If D is hyperbolic, then D is Kobayashi complete. (b) The following properties are equivalent:

(i) D is c-finitely compact; (ii) D is c-complete;

(iii) there is no sequence (zj)j C D having no accumulation points in D such that E,'?°_t gD(zJ,z1+1) < oo; (iv) D is bounded and satisfies the Fu condition. Proof. (a) Note that D is biholomorphically equivalent to a bounded pseudoconvex

Reinhardt domain. Hence, from the very beginning we may assume that D is bounded. Suppose that D is not k-finitely compact, i.e. there exist a point a E D fl C;, an R > 0, and a sequence (zj)jE N C D such that

kD(a,zj) oo; a contradiction.

In the case (ii), if k = 0, then z* = 0 E D; a contradiction. So we only have to discuss the case when k > 0. Then kD(a, zi) > kprck (D)(prck (a), prck ("i ))

Note that prck (zi) -+ prCk (z*) E 8(prck (D)) and the limit has no vanishing coordinates. This means we are back in the situation we started with; a contradiction. (b) The implication (i) = (ii) is trivial, (ii) (iii) is a consequence Of MD < gD,

the triangle inequality for CD, and the growth of tanh-t at 0. Finally (iv) =(i) follows directly from Theorem 1.13.19 and Lemma 4.8.7. So it remains to prove (iii) = (iv). Suppose (iv) is not true. In case that D is not bounded we know that D is an unbounded hyperbolic pseudoconvex Reinhardt domain. Then, in virtue of Proposition 1.1 7.12, we conclude that D is algebraically equivalent to a bounded pseudoconvex Reinhardt domain that does not satisfy the Fu condition. Hence for the rest of the proof we may assume that D is bounded and does not fulfill the Fu condition. So, without loss of generality, it suffices to deal with the following situation:

DnV 34 0, DnV. =0, 1 < j Ce,

F(A, µ)

µ2exa,

(eA.

.... p. ezaf ).

Note that F is a locally biholomorphic mapping and DR := F(HR x A) is a pseudoconvex Reinhardt domain (EXERcisE). Moreover, we have DR / D :_ F(C x A) C C;, when R -+ oo, and D,,. is a pseudoconvex Reinhardt domain. Hence, we get

kD,,.(F(-1. 1...., 1). F(A. 1,.... 1)) < kC(-1.A) = 0.

A E C.

In virtue of Theorem 4.2.42 we know that kD,o is continuous. Therefore,

1).z)=0, zED,, nF(Cx{(1..... 1)})=:M. Note that Do x (1) C D but (0, ... , 0, 1) 0 D, where 1 E C"-e t rj < oo. To get a contradiction to Choose positive numbers rj such that (iii) it remains to find points Z j E Do, j E IN, such that -j -> 0 and

j

9D((zj,1), (zj+t, 1))
N L = YD2 (0; X)

(EXERCISE).

We mention that there is also a different way to describe PD which is often very useful.

Lemma* 4.9.12. Let D C C" with kD > 0 on D. Then PD (Z; X) =

kD(Z) suP{I

f'(z)X I : f E L2 (D), 'If IIL2(D) = 11 .f(Z) = 0),

ZED,XEC". With fD we can measure the length of all tangential vector X E C" at any point z E D. Therefore, we introduce a new pseudodistance on D, the Bergman pseudodistance bD, defining

bD(z, w) := inf $ fo f D(a(t):a'(t))dt : a E el([0,11, D), 1

a(0)=z,a(l)=w

Z,WED.

It is easy to check that bD is a pseudodistance on D and that bG(F(z). F(w)) _ bD(z. W), z, w E D, whenever F E Bih(D, G), i.e. the Bergman pseudodistance is invariant under biholomorphic mappings.

4.9*. The Bergman completeness of Reinhardt domains

329

Exercise 4.9.13. Prove that the sequence ((0, ))JEt.J is a bT-Cauchy sequence, where T is the Hartogs triangle. In particular, T is not Bergman complete (see Definition 4.9.15). In comparison to the objects we have discussed before we have the following result (see [Jar-Pfl 1993]).

Lemma* 4.9.14. Let D C C" be such that kD(z) > 0, z E D. Then (a) YD (b) CD

8D:

CD < bD

We should mention that there is no comparison between the Bergman pseudometric and the Kobayashi pseudometric. Now we start to discuss the notion of Bergman complete domains. To be more precise we set

Definition 4.9.15. Let D C C" such that kD > 0. D is said to be Bergman complete if bD is a distance, any bD-Cauchy sequence (z j) j C D (i.e. bD (z f , zk)

0 if j, k

oo)

converges to a point a in D (i.e. lim zJ = a in the topology of D). J-+00

There is a long history of studying Bergman complete domains. An old result by Bremermann shows that any Bergman complete bounded domain is necessarily pseudoconvex (see [Jar-Pfl 1993]).

Proposition 4.9.16. If D C C" is a bounded Bergman complete domain, then it is a domain of holomorphy.

Proof. Suppose the contrary. Then there are a point a E D and positive numbers r < R such that

i'"(a,r) C D, i'"(a, R) ¢ D, df EO(D) IP"(a,r) _ .f I P (a,.). In particular, by the Hartogs theorem, the Bergman kernel function KD extends to IP" (a, R) x IP" (a, R), or more precisely, there exists an f : IP"(a, R) x Fn (a. R) -+ C such that

.f [R. (a, R) x Fn (a, R) 9 (z, w) H f (z, w) is holomorphic. By construction we find a point b r= IP" (a, R) f13D such that [a, b) C D. Apply-

ing that log kD (z) = log f (z, z), z near [a, b), leads to the fact that PD (a + t(b - a); b - a) is bounded on (0, 1). Hence, (a + (1 - 1/j)(b - a))J is a bD-Cauchy sequence tending to the boundary point b; a contradiction. Therefore, in order to characterize the Bergman complete domains it suffices to restrict on domains of holomorphy.

330


The most useful sufficient criteria for being Bergman complete is that of Kobayashi (see [Jar-Pfl 1993], [Blo 2003), and [Blo 2005]).

Proposition* 4.9.17. Let D C= C" be a bounded domain. (a) If D3zmaD If(z)1

< Ilf IILh(D)

f E Lh(D) \ {0},

then D is Bergman complete. (b) Assume for a dense subspace H C Lh (D) that for any sequence (z j ) C D, zj -+ zo E aD, and any f E H there exists a subsequence (zjk )k such that lim k-'oo

If(zik)I

= 0.

kD (zik )

Then D is Bergman complete.

After a long development the following result was found (see [Blo-Pfl 19981, [Her 1999]). Theorem* 4.9.18. Any bounded hyperconvex domain is Bergman complete.

But conversely, there are a lot of Bergman complete domains which are not hyperconvex; examples will be given later. In the class of bounded pseudoconvex Reinhardt domains there is a complete characterization of Bergman complete domains in geometric terms due to Zwonek (see [Zwo 1999a], [Zwo 2000]). To do so we have to introduce the set

E(D) := (v E (K(D) : exp(a + R+v) C D}, where a E D fl C'. Then we have the following result. Theorem* 4.9.19. Let D be a bounded pseudoconvex Reinhardt domain. Then the following conditions are equivalent:

(i) D is Bergman complete;

(ii) ti:'(D) fl on = 0, where V(D) :='(D) \ 5(D). We conclude this discussion presenting two examples.

Example 4.9.20. (a) Put Dt := {Z E C2 : IZ112/2 < Iz2I < 21zt I2. Izt I < 1). Then D is a bounded pseudoconvex Reinhardt domain. Note that

IS'(D1) = IR >o - (-l, -2).

4.9*. The Bergman completeness of Reinhardt domains

331

In particular, W(D) contains the rational vector (-1. -2). Hence, DI is not Bergman complete. We add also a direct proof which may give some idea of how to prove (i) = (ii)

in Theorem 4.9.19. Recallthatza E L2(Dt)iff(a+1,(-1,-1)) o (-1, -/), i.e. V(D) does not contain any rational vector. Hence, D2 is Bergman complete. It is easy to see that D2 is not hyperconvex. Iz2I2e-t1IZi 12 < 1). Calculate log D3, Exercise 4.9.21. Put D3 := {z E m. x C : ((D3), and decide whether D3 is Bergman complete. Is D3 hyperconvex?

Note that D1, j = 1, 2, 3, does not fulfil the Fu condition, hence it is not c-complete.

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Moscow 1976 (in Russian); see also Introduction a l'analyse complexe/Introduction to complex analysis. Tome 1. Fonctions d'une variable, Mir, Moscow 1990; Part II: Functions of several variables. Transl. Math. Monogr. 110, Amer. Math. Soc., Providence, RI, 1992. [Sch 2005]

V. Scheidemann, Introduction to complex analysis in several variables. Birkhauser Verlag, Basel 2005.

[Via 1966]

V. S. Vladimirov, Methods of the theory of functions of many complex variables. The MIT Press, Cambridge, Mass., 1966.

Symbols

General symbols N := the set of natural numbers, 0 0 N; No Z := the ring of integer numbers; 0 := the field of rational numbers; I2 := the field of real numbers; It_,p := {-oo} U IR, R+".:= It U {+oo}; C := the field of complex numbers;

In E N n > k};

W U {0}; Wk

LxJ := max(k E 7 : k < x) = the integer part of x E It;

(xl:=min(kEZ:k>x),xEIR; Re z := the real partof z E C, lm z :=the imaginary partof z E C; 2:=x - t the conjugate of z = x + iy; IzI := x2 + y2 = the modulus of a complex number z = x + iy; the Cartesian product of n copies of the set A, e.g. C"; M(m x n; A) = the set of all (m x n)-dimensional matrices with entries from a set

A"

ACC; II" := the (n x n)-dimensional unit matrix; (L E IM(n x n; I") : det L 0 0), I' E {0, R, C}; GL(n, F) (L E Dal (n x n; Z) : I det LI = 1); GIL(n, Z)

x_y:sxj 0), e.g. Z+, IR+; A+ := (A+)", e.g. Z ., It+;

A_:={aEA:ao

(a E A : a > 0), e.g. It>o;

A,o(aEAao :_ (A>o)", e.g. It>o;

A+B:={a+b:aeA, bEB},a+B:={a}+B, A,BCX,aEX,Xis a vector space;

8, k :=

,

bEB), ACC,BCC"; if j 96 =k

j

= the Kronecker symbol;

=k e = (el,...,en) := the canonical basis inC",ee := 1 = In = (1, ... ,1) E W"; 2:= 2.1 = (2, ... , 2) E W"; I.

if

81,n), j = 1,...,n;

zj wj = the Hermitian scalar product in C"; (z, w) :_ w := (01, ... ,1,On), w = (W1,..., wn) E C";

z=(z1,...,zn),w=(w1....,w")EC";

e=:=(es',...,ez^), z=(z1 .,zn) C"; IiZll

(z z)1"2 = (Ej=1 IZi I2)1/2 =the Euclidean norm in C";

Ilzllao := max{Iz1 I

Iz"1} = the maximum norm in C";

346

Symbols

Ilz1I1 :=

=theft-norm in C";

idA,x : A -> X, idA,X(x) := x, x E A; idA := idA,x if A = X or it is clear what the outer space X is, #A := the number of elements of A; diam A := the diameter of the set A C C" with respect to the Euclidean distance; cony A := the convex hull of the set A; A C= X :4==> A is relatively compact in X ;

prx : X x Y X, prx(x, v) := x, or prx : X ®Y -)-X, prx(x + y) := x; Bd (a, r) := {x E X : d (a, x) < r), a E X, r > 0 ((X, d) is a pseudometric space; d:XxX IR+, d(x, x) = 0, d(x, y) = d(y. x), d(x. y) < d(x, z) + d(z, y)); Bq(a, r) :_ {x E E : q(x - a) < r}, a E E, r > 0 ((E, q) is seminormed space; q : E --> IR+ q(0) = 0, q(Ax) = IAIq(x), q(x + y) 5 q(x) + q(y)); Euclidean balls (B(a, r) = M,, (a, r) := {z E C" : 11z - a11 < r} = the open Euclidean ball in C" with center_a E C" and radius r > 0; IBn(a.0) := O; IB(a, +oo) := C"; l3(a, r) = lin (a, r) := LB (a. r) = {z E C" : 11z -aII < r} = the closed Euclidean (a); ball in C" with center a E C" and radius r > 0; lBn(a.0)

B(r) _ tn(r) :_ tn(0,r); IB(r) _ IB (r) :_ n(O,r); : = ,, (1) = the unit Euclidean ball in C'; K(a, r) := IB, (a, r); K(r) := K(0. r); K(a, r) K(r) := K(0, r); 1Bl (a, r); IB = iB

K. (a. r) := K(a, r) \ (a); K. (r) := K.(0, r);

lD:=K(1)=(AEC:I.lI 0,

r- = (ri .....

j=1,

A" (r-, r+) := A" (0. r-. r+); Laurent series

za:=ZI

...zn". Z=(Z1,. ,Zn)EC",a=(a,.....a")E7" (00:= 1);

a! ali...a"!, a = (a1,....00 E Z+; IaI := Iai I + ... + Ian I , a = (a1..... an) E IR"; () := a s-1 a- +1), a E 7, N E 7+;

(0) '-(#I)...(fin), a=(al,...,an)EZ" Functions

11fIIA :=sup( If(a)I :a E A}, f: A -+ C; supp f (x : f (x) 0 0) = the support off ; .P (C") := the space of all polynomials f : C" -* C;

Rd(C"):_ {F E.P(C"):degF k) ............. XD(a) := {u: D -+ [0, 1 ) : logu E TSx(D), 3M,r,o u(z) < MIIz --

294 294 296

296

.. . .... .. ... .. ...................... XD = the Kobayashi-Royden pseudometric ................. 302 ................ ...................... 309 aII, z E B(a,r) C D}

... .

.

.

.

.

297

.

AD =the Azukawa pseudometric

298

xD = the Kobayashi-Busemann pseudometric f ND = the integrated form of xD

308

top ....................................... 315 KD = the Bergman function

...... .. .. . .... . ... .. .. .

.

(£(D)_E(D,a):_{vEIR":log a+IR+vElog D},aEDnC,

.

.

.

.

.

.

.. ... ... . .. .. . ... . . ... .. .. .. .. . ... .. ... ... . . ... .. . ... ... . {v E E(D) : exp(a + I+v) C D) . .... . . ... ... .

kD = the Bergman kernel OD = the Bergman pseudometric bD = the Bergman pseudodistance

a (D) W(D)

.. ..

.

.

.

.

.

.

.

.

.

.

.

.

325 326 327

327 328 330

.(D) \ .(D) ........................... 330

Subject index

Abel's lemma, 1.3 absolute convergence, Ll

image, 30 absolutely

summable family, 9 series, 9 uniformly summable family, 9 series, 9 algebraic automorphism, 200 equivalence, 34 mapping, 38 analytic set, 63 automorphism, 2 group

of I,,, 162 of ID", L66 of Ln , L6.8

Azukawa pseudometric, 298 balanced domain, 56 Banach theorem, 65 Bedford-Taylor theorem, 1118 Bergman exhaustive, 32.2 function, 325 kernel, 32.2 pseudodistance, 328.

pseudometric, 322 biholomorphic mapping, 54 boundary of class Ck, 143 bounded holomorphic function, 62 set, 65 Brody hyperbolicity, L33

Carathdodory-Reiffen pseudometric, 294 Carath6odory pseudodistance, 254 Cartan domains, 177 theorem, 164, 165, 177, 315

Cauchy condition, 8 criterion, 2 inequalities, 43, 52 integral formula, 47 48 product, 16 c-complete, 317 c-finitely compact, 317 c-hyperbolic, 313 circular domain, 85 Ck-smooth boundary, L43 domain, 143 CW-Kahler metric, 141 complete circular domain, 55 Hermitian metric, L42 Kahler metric, L42 a-circled domain, 3 15 set, 15 Reinhardt

domain, 3 15 set, 15 complex derivative, 1 ellipsoid, 145 Frdchet differential, 16 manifold, 53 partial derivative, 16 tangent space, 143 contraction, 253, 293

356

Subject index

convex function, L14 set, 21

C-seminorm, 63 curve rectifiable, 258

function for YD, 294 function for m(k), 255 fat

hull, 35 set, 14

finite type, 218

dD -length, 25$ defining function, 143, L44 3-tempered holomorphic function, 69 dimension of an analytic set, 63 domain, 3 n-circled, 30 balanced, 56 circular, 85 complete n-circled, 3 circular, 56 Reinhardt, 3 homogeneous, L60 of convergence of a Laurent series, 4 41 of a power series, 1, 3, 13

of existence of f, 22 of holomorphy, 5 22 Reinhardt, 30 relative complete, 5, 8Q starlike, 85 symmetric, 161 weakly relative complete, 84 D-point, 15$ elementary Reinhardt domain, 5, 33 normalized form, 196 of irrational type, 196. 222 of rational type, 126, 222 entire holomorphic function, 42 envelope of holomorphy, 84 equivalent families of seminorms, 64 exhaustion function, 119 extension operator, 73 extremal disc, 256

first Baire category, 65 formal partial derivatives, 17 Frdchet differentiability, 16 differential, 16 space, 65 Fu condition, 32 function of slow growth. 62

generalized complex ellipsoid, 208 Hartogs triangle, 212 geometric series, 12 g-length, L46 global defining function, L44 g-pseudodistance, L46 g-pullback, 142 Green function, 255 pole, 255.

group of automorphisms of IB", 162

ofd",166 of IL" , L68

Hahn function, 269 halfspace, 21 harmonic function, 1D1 Hartogs extension theorem, 5a lemma for psh functions, LLl theorem on separate holomorphy, 51 triangle, 32 Hermitian metric, L46 pseudometric, 146

Subject index

Hessian, L14 higher order complex Frdchet differentials, 12 partial derivatives, 12 holomorphic covering, 266 automorphism, 2 convexity, 89 function, 42 with polynomial growth, 69 Liouville foliation, 197 mapping, 42 holomorphically contractible family of functions, 253 pseudometrics, 293 homogeneous domain, 160 Hurwitz-type theorem, 55 hyperbolicity, 313 hyperconvexity, 130

identity principle, 50 inner metric, 242 integrated form of XD, 309 inverse mapping theorem, 54 irrational type, 22 25, 230 Josefson theorem, L07

Kahler metric, 147 KAhler pseudometric. 147 XD-geodesic, 302 k-complete, 317 k-finitely compact, 317 k-hyperbolic, 313 k_-hyperbolic, 313 kD-geodesic, 256 Kobayashi pseudodistance, 252 Kobayashi-Busemann pseudometric, 308 Kobayashi-Royden pseudometric, 302 Kronecker theorem, 97 k-th M6bius function, 255 k-th Reiffen pseudometric, 296

Laurent series, 2 441 Lempert function, 256 Levi condition, 144 form, 102 problem, 122 Lie ball, 168 norm, 168 lifting, 202 266 linearly generated, 152 Liouville theorem, 52 for psh functions, L01 local defining function, L43 potential, L48 pseudoconvexity, 119 locally complete pluripolar set, 106 normal convergence, 43 normally summable family, 9 series, 9 uniformly summable family, 9 series, 2 logarithmic convexity, 4 31 image, 352

plurisubharmonicity, 191 lower semicontinuity, 28 matrix orthogonal, 168 positive definite, 177 unitary, 162 maximal norm. 110 maximum principle, 53 for psh functions, 102 mD-geodesic, 252 minimal ball, 123

357

358

Subject index

norm, 172 representation of a convex set, 21 of a Reinhardt domain. 36 Minkowski function, 28 5L MSbius distance, 2 253 function of higher order, 255 pseudodistance, 254 Montel theorem, 54 natural Banach space, 66 Frdchet space, 66 Hilbert space, 65 a-circled

domain, 4 30 set, 30 n-fold Laurent series, 4 power series, 3 13 normalized elementary Reinhardt domains, 196 Reinhardt domain, 180 normally summable family, 9 series, 9 a-rotation, 29

open halfspace, 21 order of zero, 91, 103, 255 orthogonal complement, 22 matrix, 1.68 operator, L68 Osgood theorem, 511

overshears, 163 partial derivative, 116

Pfaffian form, 151 piecewise ek-boundary, 179 p-integrable holomorphic function, 57 pluricomplex Green function, 255

pluriharmonic function, 149 pluripolar set, L06 plurisubharmonic function, 1514

measure, 131 Poincard

distance, 2 254 theorem, L24 positive definite, 177

power series, 1.3 13 product property, 254 proper mapping, 207 pseudoconvexity, 118 pseudodistance, 254 pseudometric, 293 psh Liouville foliation, L92 radius of convergence of a power series, I of a Taylor series, 1$ rational type, 22. 25, 234 real Frdchet differential, 16 partial derivative, 16 tangent space, L44 regular Frdchet space, 220 point- 63 regularization of a function, 112 regularized relative extremal function, 131 Reinhardt

domain, 4 30 normalized, 180 set, 30 relative completeness, 5 80 extremal function, 13.1 removable singularities of psh functions, 108 p-length, 242

Subject index

Riemann removable singularities theorem, 60

8-convexity, 89 .8-domain of holomorphy, 72 segment, 21 seminorm, 29 63 $-envelope of holomorphy. 84 separately holomorphic function, 42 mapping, 42 shears, 153 Sierpinski theorem, 10 singular point, b3 smooth boundary of class ek, 143 starlike domain, 28, 85 Stein neighborhood basis, 13.1 strict hyperconvexity, 135 R-hyperconvexity, L3.6 strictly plurisubharmonic function, 103 strong convexity, L44 pseudoconvexity, L44 sum of a summable family, 2 series, Z summable family, 6 series, 6 symmetric domain, 150

tautness, 141 Taylor series, 18 theorem Banach theorem, 65 Bedford-Taylor theorem, 108 Cartan theorem, 164, 165, 177, 315 Hartogs extension theorem, 58 Hartogs' theorem on separate holomorphy, 51

359

Hurwitz-type theorem, 55 inverse mapping theorem, 54 Josefson theorem, 102 Kronecker theorem, 92 Liouville theorem, 52 for psh functions, 191 Montel theorem, 54 Osgood's theorem, 50 Poincard theorem, 174 Riemann removable singularities theorem, 60 Sierpitiski theorem, 10 Thullen theorem, 181. 182 Vitali theorem, 54 Weierstrass theorem, 53 thin set, 59 Thullen domain, 180 theorem, 181. 182 topology generated by seminorms, 64 transitivity, 160 triangle inequality, 29 unconditional convergence, 11 uniformly summable family, 6 series, 6 unit disc, 2 unitary matrix, 162 operator, 152 upper semicontinuity, 28 semicontinuous regularization, 105 Vitali theorem, 54 weak Fu condition, 92 psh barrier function, 132 relative completeness, 80 Weierstrass theorem, 53 Wu ellipsoid, 182


First Steps in Several Complex Variables: Reinhardt Domains This book provides a comprehensive introduction to the field of several complex variables in the setting of a very special but basic class of domains, the so-called Reinhardt domains. In this way the readers may learn much about this area without encountering too many technical difficulties. Chapter 1 describes the fundamental notions and the phenomenon of simultaneous holomorphic extension. Chapter 2 presents a fairly complete discussion of biholomorphisms of bounded (complete) Reinhardt domains in the two-dimensional case. The third chapter gives a classification of Reinhardt domains of existence for the most important classes of holomorphic functions. The last chapter deals with invariant functions and gives explicit calculations of many of them on certain Reinhardt domains. Numerous exercises are included to help the readers with their understanding of the material. Further results and open problems are added which may be useful as seminar topics. The primary aim of this book is to introduce students or non-experts to some of the main research areas in several complex variables. The book provides a friendly invitation to this field as the only prerequisite is a basic knowledge of analysis.

ISBN 978-3-03719-049-4

www.ems-ph

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First Steps in Several Complex Variables: Reinhardt Domains (EMS Textbooks in Mathematics)

First Steps in Several Complex Variables: Reinhardt Domains (EMS Textbooks in Mathematics)

First Steps in Several Complex Variables: Reinhardt Domains