Introduction to Several Complex Variables: Lectures by Lipman Bers 1962-1963

iii

PRiFACE These notes reproduce almost verbatim a course taught during the academic year 1962/63. The original notes, prepared by Joan Landman and Marion Weiner, were distributed to the class during the year. The present edition differs from the original only in that many mistakes have been corrected. I am indebted to Miss Weiner who prepared this edition and to several colleagues who supplied lists of errata. I intended the course as an introduction to the modern theory of several complex variables, for people with background mainly in classical analysis. The choice of material and the mode of presentation were determined by this aim. Limitations of time necessitated omitting several important topics. Every account of the theory of several complex variables is largely a report on the ideas of Oka. This one is no exception.

L.B.

Zurich, July 8, 1964.

iv

CONTENTS Preface, • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • ii Chapter 1. Basic Facts about Holamorphic FUnctions ~1. Preliminaries ••••••••••••••••••••••••••••••••••••• · 1 82. An inequality...................................... 5 13. Proof of Hartogs' Theorem 1....................... 7 §4. Holomorphic mappings.: •••••••••••••••••••••••••••• 10 Chapter 2. Domains of Holomorphy 1. Examples and definitions •••••••••••••••••••••••••• 2. Convexity with respect to a family of functions... 3. Domains of convergence of power series ••••• ~ •••••• 4. Bergman domains ••••••••.•••••••••••••••••••••••••• 5. Analytic polyhedra •••••••.•••••••••••••••••.•••••• Chapter 3. Pseudoconvexity 1. P1urisubharmonic and pseudoconvex functions .•••••• 2. Pseudoconvex domains •••••••••••••••••••••••••••••• 3. Solution of the Levi Problem for tube domains •••••

1

12 14

18 22 24

26

30 35

Chapter 4.

1 1. 2.

3. 4.

~5.

Zeros of Holomorphic Functions. Meromorphic Functions. \'leierstrass Preparation Theor. em................... Rings of power series .•••••••••••••••••••••••••••• Meromorphic functions •••••••••••••••••••••••••••.• Removable singularities ••••••••.•.•••••••••••••••• Complex manifolds ••••••••••••••••••••••••• ·.. • • • • • • •

38

44 48

50 53

Chapter 5· The Additive Cousin Problem 1. The additive Problem formulated •••••••••••••• • •• • • ~ 19 2. Reformulation of the Cousin Problem ••••••••••••••• o 3. Reduction of the Cousin Problem to non-homogeneous Cauchy-Riemann equations •••••••••• 63

1

Chapter 6. Cohomology §1. Cohomology of a complex manifold with holomorphic functions as coefficients ••••••••••••• §2. Applications •••••••• ~ •••••••••• ~.................. §3. Other cohomologies ••••••••••••••••••••••••••••••••

69 74

76

Chapter 7. Differential Forms 1. Ring of differential forms in a domain •••••••.•••• 80 2. Differential forms on manifolds ••••••••••••••.•••• 84 3. Poincare Lemmas. • • • • • . • • • • • • • • • • • • • • • • • • • • • • • • • • • • 85

1

Chapter 8. Canonical Isomorphisms 1. De Rham 1 s Theorem. • • • • • • • • . • • • • • • • • • . • • • • . • • • • • • . • 89 2. Dolbeault 1 s Theorem •••••••••••••••••.••••••••••••• 93 3. Complex de Rham Theorem ••••••••••••••.•••••••••••• 96

1

v

Chapter 9. The Multiplicative Cousin Problem §1. The Multiplicative Problem, formulated •••••••••••• 98 ~2. The Multiplicative Cousin Problem is not always solvable ••••••••••••••••••••••• • ••••••••••• 100 §;. The solution of the Multiplicative Cousin Problem for polydiscs •••••••••••••••• -••••••••••••• 10'2: §4. Characteristic classes (From C.II to C.I) ••••••••• 106 ~

Chapter 10. Runge Regions 1. Preliminaries..................................... 110 2. Polynomial polyhedra •••••••••••••••••••••••••••••• 112 ;. Runge domains •••••••••••••••••••••••••••.••••••••• 113

1

Chapter 11. Cohomology of Domains of Holomorphy_ 1. Fundamental Lemma, stated ••••••••••••••• ~ ••••••••• 2. Applications of the Fundamental Lenuna ••••••••••••• .,. Preparation for the proof of the Fundamental Lemma ••••••••••••••••••••••••••••••••• §4. Proof of the Fundamental Lemma ••••••••••••••••••••

1

115 115

'%

117 120

Chapter 12. Some Consequ~nces of the Approximation Theorem 1. Relative convexitY···~···························· 128 2. Unbounded regions of holomorphy .•••••••••••••••••• 129 3. The Behnke... stein Theorem •.••••••••••.••.••••••.••• 130 4. Applications to the Levi Problem ••.•••••.•••••.•.• 132

1

Chapter 13. Solution of the Levi Problem 1. Reduction to a finiteness statement ..•..••.••.••.. 134 2. Reduction to an extension property .••••.•..••••••. 137 3. Proof of Propos! tion 2 •••••...••..••••.••.•.•••••• 140

1

Chapter 14. Sheaves 1. Exact sequences •.•..••••.•••••••.•.•.•..•..••••••• 2. Differential operators ..•••••••..••...•••••••••••• 3. Grcided groups .•... ....•.......•.............•..... 4. Sheaves and pre-sheaves ••••..•••••.•••.••••.•••••• Exact sequences of sheaves and cohomology •••.•••.• §~: Applications of the exact cohomology sequence theorem ••.•.••••..•.••.•.•.•..•.•.•••••.• §7. Proof of the exact cohomology sequence theorem .••• Chapter 15.

142 144 147

149 150

155 158

Coherent Analytic Sheaves

1. Definitions ....................................... 162

2. Oka's coherence theorem ...••••.••••••.•..••..•.•.• 163 3. Heierstrass Preparation Theorem, revisited •••••••• 165

4. The third step ••.••••••••.••••.•••.•.••••••••••••• 168 5. Consequences of Oka 1 s theorem ••.•.•.•••••••• , ••.•. 171 6. The sheaf of ideals of a variety ..••.•••.•.••••••• 173

vi

Chapter 16. FUndamental Theorems (semi-local form) Ill. Statement of the fundamental theorems for a box (semi-local form) .......................... .-.First step of the proof.. • • • • • • • • • • • • • • • • •.• • • • • • • • Reduction of (3) to Cartan 1 s theorem on holomorphic matrices ••••••••••••••••••••••••••• 4. Proof of Cartan's theorem on holomorphic matrices. 5· Ne\'l proof of the Oka-t~eil Approximation Theorem ••• 5. Fundamental theorems for regions of holomorphy (semi-local form) ••••••••••••••••••••••

1

Chapter 17. Coherent Sheaves in Regions of Holomorphy 1. Statement of the fundamental theorems • • • • • • • • • • . • 2. Prepara tiona for the proof •••••••••••••.•••••••••• 3. Proof of Theorem A•••••••••••••••••••••••••••••••• 4. Proof of Theorem· B••••••.•••••••••••••••••••••••••• 5· Applications of the fundamental theorems ••••••••••

175 175

•

17T 180 184 185 187 187 191 194 196

Chapter 18.

Stein Manifolds (Holomorphically Complete Manifolds) §1. Definition and examples ••••••••••••••••••••••••••• §2. An approximation theorem ••••••••••••••••••••••••.• §;. The fundamental theorems for Stein manifolds •••••• §4. Characterization of Stein manifolds .••••••••.•.•••

201 202

203 203

Appendix . •..........•..•.•..••.••.•.•....•••.....••.•... 205

1

Chapter 1.

Basic Facts about Holomorphic Functions § 1.

Preliminaries

We introduce the following notation:

a denotes the field of real numbers. ct denotes the field of complex numbers or the complex plane. ~n . "" denotes the space of n -tuples of complex numbers (z1, ...,zn)

= z.


space over
By function, we will mean a complex -valued function f unless

otherwise stated, for instance f:
lhl-o

f(z 1, ... , z. +h, ... ,z ) - f(z 1., ... , z ) J n n

= ar

h

exists and is finite. We remark that f is holomorphic (in D) if it is holomorphic in each variable separately. Note that f is not a$sumed to be continuous. Hence we obtain immediately: ..

Property l. If r1 and f 2 are holomorphic G.n D) then r1 + r2• r1 - f 2, f 1f 2, and, if r2 4 0, f/ r2 are holomorphic (in D). Property 2. If

(f.} is

a uniformly convergent sequence of holoJ rnorphic functions in D converging to f, then f is holomorphic in D. That is, functions holomorphic on an open set form a ring which is closed under uniform convergence.

2

Property 3. Maximum Principle. Iff is holomorphic in D and has a local maximum at a point p E D then f is identically constant in the component of D containing p. Definition 2. Let Kc
open set De
and there exists a function F holomorphic in D such that F = f on D n K. B.

We remark that for functions f: JB.n-m.~ the existence of all partial

derivatives af I ax .• i = 1, ... In does not imply f is continuous. However, l

-

the corresponding_ theorem for functions of several complex variables is true. Theorem 1 (Hartogs). Every holomorphic function is continuous (in all variables simultaneously). Proof. Given in § 3. For the remainder of this section and Section 2, assume Theorem 1. Definitions 3. A closed polydisc about z 0 in
(
J

n

?I < r - J

·I

J

= 11 ..• , n;

j

0

< r.J
I J J I<- r JJ.1..

denoted f z. -z?

I

t

An ~en polydisc in <en is the interior of a closed polydisc, i.e. the set (
(lzj -zj

I<

rj).

I

z )I n

Iz.J - z J~I < r J.•

.

j

=1, ... , n;

0

< r.J <«>),

.

denoted

.

The boundary of a polydisc is the set

(
J

J-J

••. , nl

IJ

I

and z. - z? = r. for some J

J

j}.

The distinguished boundary of a polydisc is the set

I -

tf.( z1, . .. z ) z . - z 0. + r . e iG j n J J J I

I

0

< -

f'

l,

•

J

< 2rr ) .

Note that the dimension (over 1£.) of the boundar.)' of a polydisc is 2n-l, while the dimension of the distinguished boundary is n.

vr e remark that proofs will be exhibited for the case n

= 2.

The

proofs in the general case are similar and may be completed via an induction argument.

3

Theorem 2 (Cauchy Formula). Let f(z 1•...• z ) be holomorphic no in the closed polydisc {.lz. -z~l·< r.1. Then for{lz. -z.l < r.J. J J - J) . J J J f(zl' ... , zn) = (2 1. )n 1Tl

J

0

I~(zll

f

...

n

TT1

. 1 t f(tl •...• ~ )dt ... dtl . 0 n n t n -z n = r n J= j -z~ · .

I

= rl

I

0

= 2).

We may assume z = 0. f(zr z 2) is, for fixed z 2, Iz 2 1~ r 2• a holomorphic function of z1 in Iz11~ r ':l:'hus by Cauchy's Formula Proof. (for n

r

Similarly, f(tl' z 2) is a holomorphic function of z 2 for each Cauchy's Formula to f(t 1, z 2) gives f(zl. z2) = 2; i

J --

1~ 1 1

r1

~1 \

[ 2; i

J ~ ~z2 lt 1 2

2

f(tl'

~l

. Applying

~2)d~ a)dtl .

r2

By Theorem 1 we may write this as a double integral

Corollary 1. A holomorphic function has derivatives w. r. t. x. and y., i 1

1

=1, ... , n,

of all orders, i. e. holomorphic implies

dXJ.

Proof. Differentiate under integrals in Cauchy Formula as many times as desired. Corollary 2. f holomorphic implies

-

ar I az. holomorphic. J

Corollary 3. Let f be h0lomorphic in the polydisc P -

and let lrl < M on P, then

=[lz.-z~l< r.l, J

J -

J}

4

where z0 denotes the center pf the polydisc. In Corollary 3, f is assumed to be holomorphic in the open polydisc and continuous on its closure. Without assuming continuity on the boundary the inequality still holds with M = sup If

1.

Corollary 4. If f are holomorphic functions in DC
p

uniformly in D thenar I az.- M/ az. normally, i.e. uniformly on comP J J pact subsets of D. · Corollary 5. Power Series Expansion. Let f be holomorphic in int P, then = I: a

0 11 1

0 11 n

(z 1-z1 ) ... (z -z )

vr .. vn

n

n

>

Jl. 0, 1-

which converges absolutely and normally in the open polydisc. Proof. Same as in one variable. Expand 1/(t .-z.) etc. J

Note.

J

Determining the domain of convergence of a given power

series is not trivial. V.Je will do it later. Corollary S. (Continuation of 5)

Proof. By Corollaries 4 and 5. Corollary

~,.

The power series expansion of a holo1norphic func-

tion is unique. Proof. Corollary G gives the coefficients. Corollary

a.

Iff is holomorphic in a domain D and f

neighborhood of a point of D, then f

= 0 in D.

This is the basis of analytic continuation.

= 0 in a

5 § 2.

An inequality

Definition 4. A continuous real-valued function u, of two real variables, in a plane domain

JJ is said to be subharmonic if, for any

closed disc A in ,/) , and for the corresponding function

+such that

; =u on the boundary of A,

in A..

and ; harmonic in 6., u

< .;

Properties of subharmonic functions. 1.

If u and v are subharmonic then max (u, v) is subharmonic.

2.

Being subharmonic is a local property, i.e., if a function is sub-

harmonic in a neighborhood of every point of a domain then it is subharmonic in the domain. 3.

A harmonic function is subharmonic. Remark. If f(z) is

a floloiD'C!JI1)hic function and E > 0 then

(I f(z) I. E) is subharmonic. Proof. At a point where lrl < E, and thus in a neighborhood of this point, log max (I f(z) I, €) = log E = log (constant) is subharmonic. At a point where lrl > E, log max
f ::f 0 and therefore subharmonic. At a point wher~ ffl = E, and in a disc

I

such that ff > e/ 2, max log ( If I. E) is also subharmonic and log max =max log .. Theorem 3. Under the hypothesis of Theorem 2 and for

r. -- 1, z. -- p .eif.J J

log

J

~flz1.J ,znll~12~lJ ... .[1 _ l-p~ <; . eJ + •••

0 0 (n-fold)

_

J-l

.

18

1 2p . cos J .,. 1!1

J

J

2 P. J

I

log If(e 1, ... , e n) dBr .. d8 n . Proof. Take

E

> 0. Let g =log max I. E). E n

If we fix all variables but one, g is a subharmonic function. E

prove the theorem for n = 2. Thus

We will

•

6

g£ =log max clrczl' z2)f, £) . . ;.ix z2' t~en g€(zl' z2) is a subharmonic func-

tion of z 1. Thus for z 1 = p 1e gf(zl' z 2) < 271'

For fixed 61' g£(e

s 271'

1

1., 1

1

·e

2

1

-p 1

. 0 1-2p 1 cos

2 g£(e

<~(e 1 )+p 1

.1

• z 2)de1 .

iOl

• z 2) is a subharmonic function of z 2

= p 2e

if/» 2

•

Thus

Using the continuity of f, write this as a double integral. But loglfCz1, z 2 >1~ gf(zl' z 2), obtaining

I

log f(z 1, z 2)

I

Jn 271'

0

Let €-0. then g +log €

2

1-p

~

il\

ifJ 2

2 g (e ' 'e j=l 1·2p.cos(e.-;J+p. € J J J J

Irl.

)df)ldf) 2 •

Therefore by the Lebesgue Monotone Conver-

gence Theorem

271' 271'

2

c.!..>2JJ n

-< h

0

0

1-p 2

.

2

j=l l-2p. cos (fJ. -f; .)+p. J J J J

i6 1 log f( e , e 2 ) dB 1dB 2 ·

I

i6

I

7 § 3. ~roof of Hartogs' Theorem 1

Use induct1un on n. The theorem is trivially true for n = 1. Assume the theorem, and therefore all corollaries and following theorems for n-i variables. Osgood' s Lemma. Let f(z 1, ... , z ) be holomorphic for n . < R .• 0 < R . < oo. Then there exists t such that t < R1, and p > 0

IzJ.I -

J

II

J

-I t I. and a number M, Iz1-t I <- P and IzJ.I -< R.J for j >- 2.

such that p < ~

I

I

such that f(zl' .•. , zn) < M for

Hartogs' Lemma. Let f(z 1, ... , zn) be holomorphic in lzjl < Rj and bounded for lz1 < p and lz.l < R., j > 2. Then f is continuous in

-tl -

J -

lz.l < R .• J J Proof of Osgood' s Lemma. m(z)

I

=

max f(z, z 2, ••• , z ) lz.l
1.

J

-

Define for Iz I ~ Rl'

This maximum exists by induction

J - J "> 2 l_

hypothesis. Denote

by~ N = { z II z I < R1 and m(z) < N}.

00

l I }I.N. Now A.N is closed, for

{zllzl < R1} =

1.\t(}

if ar
Then

~Nand ar-a,

m(a) =max Irca, z 2 , .•. , z )I =max lim Irca , z 2, .•. , z )j <max N = N. n r n r-oo For instance, by the Baire Category Theorem, one of the ~ N contains a disc. For z1 in this disc f is bounded.

=1, j = l, ... , n, fz 21 <1, ... , lznl9}.

Proof of Hartogs' Lemma. VIe may assume R.

t

=0,

andM•l. LetD1

Let z 0 = (z~~ ...

I

={1zr····znlllz11 ~p< 1,

z~) and z1 = (z~~ ... z~ be interior points of Dr I

Then

8

11

< Izl -zl 0

-

1

{p-lz1 j)(p-fz11h + 0

+

·· ·

lz 0 -z 11 0 1 1 n n n-lz hu-lz 1> n n

by the Cauchy integral representation on 1 variable and the existence of

Irl

the maximum of in D~ Since all the denominators are bounded away from zero as z1-z 0 , lr
>1-o

f at z 0• The arbitrariness of the choice of z 0 implies the continuity of f in the interior of n1• Now fix z 2, ...• zn; zi I < 1. Then

I

=

11

Q()

~

v=O

for

a (z 2••••• z )z 1

n

II

where

..!..( avr ~ 11

a (z·2, •••• z ) = 1 v n 11.

,

az1 / z 1=o

by the holomorphicity of f in z1• Near z1 = 0. (zl' ...• z n) lies in D1 and

£

here f is bounded and continuous. Thus a (z 2, ••• , z ) 11 n

=· _!_ 21T i

It

f{t, z 2••..• zn)

=p 1< P

..JI+l

d~

~

and therefore the a are holomorphic. By our induction hypothesis, the 11

a11 are continuous, and the proof will be complete once we show that the . power series converges normally for

lz11< 1.

Now, 1

lim sup I a (z 2, ...• z ) 111 < 1 11 n

v-ao

9

and Ia (z 2.... , z )I < _!_, II n II p

since the series converges for I z 11 < 1 + TJ1 for some TJ > 0: Set fZ

/\ = t zj = e

w.JI0 ~ ~ j < 211'

1

j

1

I

1\11 = ~rcz 2, ••• zn ) fa11 Cz 2, .•• 1

2~ ••• In

1:

1

zn )

1\II is open in 1\ ~ and • lim m( 1\.II ) = 0, v- co the measure of 1\ . For, if we let II Then

jv >· 1,.

} 1

(z 2, ••• , z n )~

where m ( 1\

II

1\].

) denotes

00

QJ. =

UA.

• II=J

II

and note that

and m(nQ.) = lim m(Q.) J

j-

00

J

then

() ~. = q, (the empty set). J

Thus m( () Q.) = 0 = lim m(c; .). J

J

Since Q. =:J 1\ ., m(/\ .)-0.

J i~ . Now let z. = r .e J,. r. < r J J J-

J

< -

< -

J

1 v(21r )0

1

1 11(21r )0 -

1

j = 2, ...• n.

By Theorem 3,

2 i& 2 i6n (1-r .) log (e •.•.• e II _ _ _ _2=---- dO 2" •• den _ __.,__ ___;_ j=2 l-2r. cos(cb . -6 .)+r. J J J J i8 i8n 2 2 n (1- r .) log a ( e •.•. , e ) _....J_ ___;_II_ _ _--=2-- d8 2" •. cWn. j=2 l-2r. cos (' . -e .)+r. J J J J

f Jnn

Ia

>I

J; Jn _

I

I

..• (\

-

< 1,

...

1\

v

10

Furthermore

1-2r. cos land Ia

II

I <- 1/p11. 1 11

1-r~

2 1-r,

log

<

(~ . -e J+r ~ J J

J

(1-r.)

l+r.

=

2

J

J

J

< l+r

1-r. J

- 1-r

Hence,=

Ia11 I <- (

11

1 27T

=(

1 log-

1

)n -

v

P

1+ n-1

1 271' )n-1

C1 r) -r

1

log- m(;\), p

11

Therefore forE> 0 there exists a number N(e) > 0 such that if v > N(e) then

I I<

-1 log a

v

v-

1E , e log-1 =log(-)

P

P

i.e.

IaII (z 2,. . .,zn>I<

_!_for Ell

p

lz.lJ < r < 1, -

j = 2, ••• ,n,

I J.I <-

Hence, the series converges normally for z

r < 1, j

11

> N(E).

=2, ..• , n and

Iz11 < P E • § 4.

Holomorphic Mappings

In this section we make several remarks about the mappings determined by holomorphic functions. Note that: f is holom~rphic if fE. C00 and

ar az.

=

0

=

-arax.

j

•

J

where 2 a~

az.

J

J

+

. ar

1-

ay.

J

=1•... , n ,

11

We remark that, for n > 1, these criteria for holomorphicity form an overdetermined system of equations. Many of the phenomena associated with functions of several complex variables arise from just this fact. Observe that as an easy consequence of the above remark, a holomorphic function of holomorphic functions is holomorphic. Definition 5. Given f.: D -·a:, DC«!n, j J

=1, .•• , n;

denote

.

f.(z 1, ••. , z ) = t ., f ~ (f.} is a holomorphic mapping if each f. is holoJ . n J lJ _ ·. J morphic in D. Note also that if. ~

. = ~ . + iTJ. ; Z . = X. + iy.

J

J

J

J

J

J

,_

then

=

J

In particular, J orientation. If J

'f o.

~

a<~r

... , ~n; Tlr •.. , TJn)

8(~ •... • xn; Yr ..• , Yn)

a(tl' ... ' \n)

=

a(zl'

o o o 1

2

zn)

0, i. e. , a holomorphic mapping preserves

the map f is locally 1-l, and has an inverse, ( 1,

which is holomorphic. Hence. a holomorphic mapping such that J /= 0 carries holomorphic functions into holomorphic functions. However, under a holomorphic mapping it is possible to map a bounded domain onto all of «!n. There is no simple geometric characterization of a holomorphic map (e. g. angle-preserving), and no Riemann's mapping theorem. Hence, there is no canonical domain, like the disc, and we are forced to consider arbitrary domains.

12

Chapter 2. Domains of Holomorphy § 1.

Examples and definitions

Definition 6. Let D~penc

CCn. Iff holo~orphic in D1 implies that there exists an F holomorphic in o2 such that F o1 = f, then

Dienc:

I

D1 and D 2 are said to exhibit Hartogs' phenomenon.

Note that this does not occur for n = 1. A.

Example 1 (n = 2). Let 1

D•{lz.wlllzl < 1, lwl < 1 and

and not

I

Iz I ~ 2

and

IwI ~

1 "'I

2

j

,

* \
D =

We claim that D and D* exhibit Hartogs' phenomenon. Let f(z, w) be holomorphic in D. Let

D'C D. Hence

f(w, z)

=

00

•

I: a.(w)zJ for fixed w, lwl --oo J 1

and 2

<

1

< Iz I < 1.

Now the a.(w) are holomorphic in w, for J 1

a.(w) :: 27r i J z

J

I I::

Let

f(w,z) "+1 dz • .! + .!.._ ZJ 2 10

D" = [(z,w) ~< lwl < 1, lzl < 1}.

_

D" c D, here:

f(w,z) For w, z such that Therefore

=

00

•

~ b.(w)zJ. j=O J

i < IwI <1 and i < Iz I<1 , the series must agree.

13

,

a.= b. J J

a co

Thus f(w, z)

.

= E a.(w)zl,

*

j

=0

for

.>0 J_ j

<0 .

for every fixed w, lwl < 1 and for every

0 J z, z < 1, i. e. in D •

II

A similar proof gives the holomorphicity in w.

Corollary. (Hartogs 1 Theorem). A holomorphic function of at least two variables cannot have isolated non -removable singularities.

I 1+••. +I 1

Exercise. Consider the domain 0 < r < z1 2 z n 2 < 1. Show that, if f(z 1, •.• , z ) is holomorphic in such a spherical shell, then it n 2 · 2 is holomorphic for 0 ~ z 1 +..• zn < 1.

I 1

+I I

Theorem 4. (Hartogs 1 2nd Theorem). Let DC «!n be any domain homeomorphic to a ball, and bounded by a sufficiently smooth surface E. Then if F is holomorphic in a neighborhood of :E, F can be continued analytically over D. We remark that this is a theorem in overdetermined systems, and shall not be proven here.

{ I

1 'I -1 "" Example 2. Let D = tl2< lzl
I I<

(Proof as above.) Note that D is a cell. B.

.Definition 7. Let Dopen c <en and ~ 6- boundary of D. E is said to be

an essential boundary point of D if and only if there exists an f, holomorphic in D and singular at E (i. e. , f is not the restriction of a function holomorphic in a_ domain

n1 ,::, D such that t • Dr

Definition

a.

DopenC:
if there exists a function f, holomorphic in D, and singular at every boundary

point. We shall show that, if every boundary point of Dis essential, then Dis a region of holomorphy.

14 § 2.

A.

Convexity with respect to a family of functions

Let X denote a topological space and

1- a family of real or complex

valued continuous functions defined in X. Let KC:X.

Kg. , of K is the set of pOints p of X, such that for eac~ function f E 5- which is bounded by 1 on K, If(p) I ~ 1• Definition 9. The f -hull, .....

VIe. remark that K 1 is closed in X. Note: If g.l and

" ~ :::K "" K

g._ •

1

2

1-2 are families

such that

3i

C:

J 2,

then

Definition 10. X is called J.-convex if K compact implies that K5is compact. Example. Suppose X = Dopen c 1ln , and :f is the family of functions which are linear on every component of D. Then D is 1-convex if and only if every component of D is convex, in the ordinary sense. If :f consists of linear functions, L'len D is ~-convex if it is convex.

Lemma 1. Let K c <en . Then K bounded implies

J

Kg.. bounded,

where

is the family of monomials in z 1, .•.,z n. Proof. Let M. =sup 1~.1. which exists and is finite asK is bounded. l

· Zt K z.

1

The functions gi(z) = fu~. f.

J , i = 1, ••• , n. hence Ig.(;) I < 1 for every ~E K:·.~ . Hence, J

I

l

A

l

-

i. e. KJ is bounded.

B.

I

But gi(z) < 1 for every zt K, if

z'~ K..z ' oJ

11;11 <- max IVI. < co, l

·

~· ~n is a normed vector space under the norm,

II z I J = max (I z1I, ... , Jzn I) (and under many norms too,

of course).

Definition 11. Let Dopencd'!n . The distance of a point z from the boundary of D. denoted by 6D(z) or simply 6(z), is 6 (z) = inf llz·t II. . tEbndry D 6(z) satisfies a Lipschitz condition with constant 1.

15

- -Proof. A(z 1 )

=inf II z 1 -t II ~ inf ( fl z 1 -z "II + If z "-t I h t

t

< llz•-z"ll +inf llz"-~11 = fJz•-z"ll +A(z") t·

< liz• -z"lf .

Therefore ja(z' )-A(z">l

Definition 12. For any set KC D A0 (K) C.

= inf

zE K

open

:h

C(C

•

A0 (z) .

Note. Kc c D, read "K is relatively compact in 0 11 , is defined to

mean that the closure of K is compact and contained in D; hence A(cl K)> 0. Theorem 5. (Cartan-Thullen). If Dopenc (Cn , then the following conditions are equivalent: (i) D is a region of holomorphy. (ii) All boundary points of D are essential. " =A(K),

(iii) If KC CD, then A(K)

,.,

where K

"' '1:. =K.J and J is the family of

all functions holomorphic in D. (iv) D is holomorphically convex.

,..

i.e.

K·~-=

c: D implies Kc: ~=D.

Corollary 1. If n

=1,

then every open set is a region of holomorphy.

Corollary 2. Dis a region of holornorphy if and only if every compo• nent of D is a domain of holomorphy.

D1X

Cforollary 3. If o1c a;P and n 2c ([q are domains of holomorphy, then D2 c«:p+q is a domain of holomorphy. Corollary 4. If Da is a region of holomorphy for every a in some set

A, and if

nDa

is open, then

noa

is a region ofllolomorphy.

Proof. Assume Kc c ( f1 - Da ); then KC c Da for every a. Let Ka denote the hull of K in D with respect to all functions holomorphic in D . a a K is compact by assumption and Kc K . Thus K c K , and hence K a ,.. "' ,a a a is compact. Since K c KC ::-( :1 D ). a a

,.

no

I

" n

,.

16 ~·

In general, a union of regions of holomorphy is not a region

of holomorphy. However, if the regions are nested, i.e.

o1co2c ... ,

then their union is a region of holomorphy. This, however, we will prove much later. Cormlary 5. (Exercise) If D is geome.trically convex then Dis a region of holomorphy. Proof of Theorem 5. If D

D.

=
Therefore, assume that D is a proper subset of
~

f

~(K).

Let

~(K)

~

A

= lVI and D.(K) = m.

Then m < M as K ::>K. Choose A

A

~

r, R such that m < r < R < lVl. Thus, there exists a zE K such that A(z)< r •.

K = z ~ K{z Jll z -z II < R}. Then Kis compact. Let f be a holomorphic function in D. rna~ If(z) I = 1J exists and is finite as f is continuous

Let

1

1

z~ K

and K compact. If z E. K then

vr .. .v n"'"' I

<

I

vl+•.. +v

R .

n

A

Since this inequality holds for all z £ K it holds at every point of K, and in A

particular, at z. Thus iff is expanded in a power series about

~-

z~.

the serieE..

A

will converite inside a poly disc of radius R about z. However, this polydisc 0

.

contains, in its interior, points on the boundary of D. Therefore, these boundary points are not essential. (iii) implies (iv). Assume that KC c D. Then K is bounded, and A

A

hence K is bounded, by Lemma 1 above. K is closed relative to D, but since' .a.(K)

1\

=A(K) >

A

0, K is compact.

(iv) implies (i). Let D be any component of D. sequence of sets Dj' j = 1, 2, ... such that o1c c

D = ynj'

(e. g. Dj

=[z

zE

0,

liz

o 2c c

II < j, S 0(z) > j1}

Construct a

0 3... c

c D and

). Let (Pj} be dense

17

D. ([Pj}

D

in could be chosen ;_s the set of points o:_ whose coordinates in :m. 2n are rational). For each j we find a point z .t D such that z. ·p .6.0 (p.) 1 "' " J J J J and z.~ D.. ·we can find such a point since D. doesn't come arbitrarily Jl J J close to the boundary of D. Now, for each j, there exists a function r. boloJ morphic in D, such that < 1 in D. and lr.(zJI =A.> 1. Next choose JJ JJ J ~ integers N. > 0 such that J

II

.II<

lr.l

oo

~ 'l 'A· "

1

-N.

J

Let _

g(z) -

c:~ n j=l

J

< oo.

Nj]j j

~1 - ~fj(z) f ( )

)

zj .

.

Each tern1 of g(z) is holomorphic. In fact, the infinite product converges

I w

w

normally to a function not indentically zero. For f.(z) f.(z)

< 1 for each j, in Dj, implies that

< 1 for all j, in Dr

Furthermore, recall that a necessary and sufficient condition for the aboo

solute and uniform convergence of the product co

TiT (l+r/1.)J is the uniform con-

vergence of the tail end of the series ~ Icp n 1. Now for zED.,. j > j', f.(z) N. ·N. 1 -:t-T. J J < A . J, and by construction E jJ.I. . J < co • Therefore g is j zj J ..... j> j' J a holon1orphic function in D.

Ifn

· "-T! e claim that g cannot be continued holomorphically to a domain D1 ~D.

For, at z .• g has a zero of at least the j th order. Thus J g(z) = u(llz-z.llj> as z- z .. J J

So, the derivatives of g up to order j vanish at z .. On the other hand, if q is a boundary point of

D,

then there exists a

s~bsequence

fp.} of Jp.1 l 1 lv LJ j which converges to q. This means that the p. come arbitrarily close to lv

II -

the boundary, i.e . .6.(p. ) - ~. This implies that liz. -p. 0 and hence ~ ~ ~ that z. - q. Therefore if g is holomorphic at q, all the derivatives of g

lv

18 variish at q. This can only happen if g = 0, a contradiction. Hence g is not holomorphic at any boundary point. Note. Under a holomorphic mapping, !1 region of holomorphy is mapped into a region of holomorphy.

.-

§ 3.

A.

Domains of Convergence of Power Series

In this section we consider power series of the form ~ kn k (z 1-t 1) ..• (z -~ ) kn =0 1: · · n n n

co ao E ••• E

I) =0 for

~

=(~1' •.• , ~n) fixed.

~

(1)

V.le say that the series converges at a point

z =(zl' ... 'zn) if there is some arrangement of terms for which the series converges. Note. In the sequel, we take t = 0 ; i. e. we deal with series of the form co ~ k1 kn :L: • • • I: a. k zl . . . z k =0 k =0 Kl. · · n n

1

(1 I

)

n

Abel's Lemma. If the series (1') converges at some point

"z = (z" 1, z" 2, ..• , zn)' " then the series converges uniformly and absolutely in every compact subset of the polydisc z <

{I J.I IzJ.I} .

Definition 13. (i) A point t is said to be a point of normal convergence if there is a neighborhood of t in which the series (P )converges absolutely and normally (ii) The convergence domain of the series (1 1 ) is the set of points of

normal convergence. From the definition, it is clear that a convergence domain is an open., connected set, star-shaped about the origin.

19

Corollary. If Dis a domain of convergence of the series (1') and (z1, z 2, ••• , zn) € D then (a 1zl' a 2z 2, .•. , anzn)E D where lail ~ 1, i

=1, ••• ,

n.

Definition li. A domain with the above property is called a complete circular domain.

I .I

If a

1

=1,

=1, •.. , n, . the

i

domain is called a circular domain.

Intuitively, a domain is circular if it is rotation-invariant, and com· plete circular if it has no "holes". To any complete circular domain De «:n we associate a set D*CI.n as follows: D* is the image of D under the mapping (zl' z 2, •.• , zn)(I z1I, Iz 2 l, ... , Iz n I). To D* we associate the set log D*, whi~ is the image of D* under the map

Iz.ll f

0, i

(I z11, .. -., Izn I)- (log Iz11, ... ,logj zn I),

defined for

=1, .•. , n.

Example. If Dc«: 2 is the complete circular domaiiJ {I z11< 1, Iz 21< 1}• then D* is the unit square and log D* the third quadrant, i. e.

D* is

B.

0

I and log D* is

zl

Theorem 6. Let D be a complete circular domain. Then the follow-

ing conditions are equivalent: (i) Dis a domain of convergence. (ii) log D* is convex. (iii) Dis a domain of holomorphy. Example. Let Dc«: 2 be the complete circular domain


1, lz 2 1 <~or lz 1 1 <

~·

lz 2 1 < 1 I>. D* is then

lz 2 1

log lz 2 1 1

l

1

and log C* is

2

0

1

fz11

log 2

0

1 lo~"02

log I z11

20

But, by Theorem 6, if there is a power series which converges normally in D, log D* must be convex. Ther.efore the power series must actually converge in

0 1 log Iz11, and thus in log (2)

Hence D exhibit:: Hartogs' Phenomenon. Proof of Theorem 6.

. .,.

------.-

(i) implies (ii).

i;n

1., 1

Let (r1e

) and (R1e

, ...• r n e

i81 i6 •... , Rne n) be

arbitrary points of D. Then log D* is convex if and only if a 1-a it/Jl a 1-a it/J n i¢1 icf>n (r1R1 e , ... ,rnRn e )€Dforeacha, O
I~r .. kn I< ---=~~-A----=k-n where (r1+€1) · · · (rn+€1)

£1 >

Ia

Therefore for 0
0 and A is some constant.

kr .. kn

Ia <

a A

• Similarly, since (R1e ak1 akn (r1+€1) · · · (rn+€1)

i~l

i~

, •.. , Rne n)

is a point of normal convergence, we have for €2 > 0, B some constant, and Bl-a I 11-a 0 < a < 1 , a~ .•. k < . ~ k l-a . Let n (
p.

1

= r a_l-a . .1:t. , 1· = 1, ••• ,n. T hen, Iak_ 1

0

0

k lala

-l.. . . n

1

, where "p . > p . , 1

1

. = 1, ... , n.

1

,_ 11-a=l ~

~... ~n

J.• . .

k 1 n

21 (ii) implies (iii).

By Theorem 5, a sufficient condit1on for (iii) to

hold is that Dis convex with respect to the family 1-of all the functions holomorphic in

D.

If D is convex with respect to a subfamily of ]then D

is convex with respect to 1-. But, (ii) implies that D is convex with respect to monomials, as follows: A

Let KC CD, and

denote~

A

monomial~.

K the hull of K w. r. t.

K is

~nand by Lemma 1, bounded. It remains to prove KC D or equiv-

closed in

A

alently that log (K)* C log D*. Assume D

-=/=

n

n

A

CC , as we already know K C «: .

Let p(z) = az 1Ill, ... z mn be an arbitrary monomial, m. > 0 integers. n 1Then log fp(z)l =log laf +~log lz 1f + ... + mn log fznl' and

I

log (hull of K w. r. t. log p(z) f )* is simply the closed half space.::> log K* defined by the hyperplane P with coefficients Pr,log K* I=

(~,

... , mn) such that

Therefore the intersection of all such half spaces is

cJ> •

log{z e: «:nflogfp(z)j <sup logfp
ao,

then logS* stays away from

that (1) z

E

an implies Cl z

a log 0*.

To prove this it suffices to show

= (alzl, ... ' Cl z ) E: aD for all

n n

Cl E:

ctn with

I 1.I Cl

=1

for all i and ( 2) every z e: ctn with (log Iz11, .... log Iz n I) e: a log D* belongs to ao. For then if log S* comes arbitrarily close to a log D*, given e: > 0 there

an such that (using sup norm) for every < e:, which implies lls.l-ld.ff < E'. e:'- 0 as e:- 0; 1 1

are points s e: Sand d e: i, llogfs.l-logld.lf 1

and by


1

there is a point d' e: an with 1s. - d'.l< e:' for ani: a contra-

1

1

diction. (l) is easily established once we note that z Cl

with

I J = 1, Cl

Cl

z ;. D; because then if z 0

points "A e: D and v -

f.

I

E:

f.

D implies for all

ao and E: >

0 is given, there are

D such that A . -z? I< e: and f11. -z ~I < e: for all i. Hence 1

1

1

1

foralla, la.f =l.la.v. -a.z?l<e:and la.:\.-a.z?l< e:,av 1

a A e: D so that a z 0 e:

11

ao,

11

11

11

f.

Dand

as claimed. (2) is proved similarly by considering

the preimage in ct11 of a neighborhood of a boundary point of log D*. Since elK is compact, there is a ball Bin «:n,
a 6 > 0 such that dist (log K*, a(log B*nlog D*) ) > 6: for Bt'lo is a complete circular domain and KC

c

(BnD).

22

Let T = log B* () log D*. T is convex and c log ·D*

,..

V.l e will show

that log (K)*C T, by considering the convex hull T 1 ofT, i.e. the intersection of all closed half-spaces in .l.n containing T. Now, if~ ~ Tl' then there is a hyperplane P in lln separating ~ from T 1 and such that Pf'l(T~ ( ~ }>=~ Because Bfl Dis a complete circular domain,. if ~o E T, then Thus t(~ ~ m.nl~.1<~?}cT. - 1

expressible as {~

s

T can only be contained in a closed half space

m.s1 1 + •..

+ m ~ < c; c E ll., all m. > o}. Suppose nn1r/ T1' if log I~ i I > log r for at least one i, say i = 1, then we may take P to E

m.nl

be the hyperplane : log Iz11 =log r + E, where e > o is such that logl~ 1 1 > log r +e. If logle.l
-

plane P can already be given with rational (and hence integral) coefficients; or since Pf\ log B* is then compact in 1ln and the distances of~ and T 1 toP depend continuously on the coefficients of P, we can find a P with rational coefficients effecting the separation. Therefore, the convex hull of T, which is cl T, is the intersection of those closed half spaces ~T defined by hyperplanes P with non -negative, integral coefficients and P

n cl T =fb •

Thus, if P is any hyperplane to be considered in (the intersection giving cl T) then if we translate P parallel to itself into the half space :>T until it intersects log K* , P becomes a hyperplane to be considered in (the

"" intersection giving log (K)*), and we have translated P a distance> 6. A

~

Therefore log (K)* c cl T, and the distance of any point of log (K)* to such a hyperplane P is > 6 , s·o that the closed ball of radius 6 about any point of A

log (K)* c the closed half space::> cl T defined by every such hyperplane P and I'

thereforeccl T. By the convexity ofT, it follows that log(K)*C T, as claimt (iii) implies (i). (iii) implies that there exists a function f holomorphi in D and in no larger domain. At any point z e D, we may expand fin a power series which converges normally in a neighborhood of z. Thus there is a power series which converges in D and in no larger domain. § 4.

Bergman Domains

Definition 15. Let f(z 1, ... , z ,X) e «:, z. e «:, i = 1, ... , nand X e lB., n 1 such that f is continuous in all variables simultaneously and holomorphic in

A.

23 some domain DC «:0 for each fixed ). z

=(zl' ... , zn)

€

D, ~

E

E

I

I c J,. Then £z f(zl' ... , z n'). ) • 0.

rJ is called a Bergman Surface.

This is a surface of (real) dimension 2n-l. Definition 16. A Bergman domain is a domain bounded by a finite number of Bergman surfaces.

Example. Let n

= 2.

:

Let B

={ (zl' z 2l 11 z11< 1,

z 2 £ B('"l) }where

the B(z1) a~e domains parametrized by z1 and bounded by Jordan curves: z 2 = g(zl' e~ ), where g is continuous in all variables and holomorphic in z1 for Iz11 < 1 + € • Then B is a Bergman domain. For example, we might choose g(zl' e

iA

1 M ) = 2+zl e

For n arbitrary, D = {cz1, ... , zn)

Iz1l < 1,

z2

€

B(z 1),

z 3 £ B(zl' z 2), ... } . These domains will be called quasi-product domains. They reduce to product domains if B(z 1) =B1 independent of z 1, B(zl' z 2) = B 2, etc. We list the following properties, stated for any Bergman domain and proven for quasi -product domains in ct 2. Property 1. Every Bergman domain is a domain of holomorphy.

or

" ""z 2) is a boundary point then either Proof. If (zl' (a) 1~1 1 =1, and then ~... is singular at (~1 • i 2> and regular inside iA zl zl 1 ~ . s1ngular . ""' (b) "" z 2 = g (zz 1, e 0 ) and then 1s at (zl' z 2) and z 2 -g ( z1, e 0)

regular elsewhere. Property

2~

-

A Bergman domain has a distinguished boundary surface,

defined to be the set of those points of the boundary at which at least n Bergman surfaces intersect. (The distinguished boundary in our example is {< ei.P , g( ei4> , eiA )) fP , ~ E [ .0, 21r ,J ""'

J) .

24 Property 3. Iff is holomorphic in a neighborhood of a Bergman·

I

domain D then the maximum of If in D is achieved on the distinguished boundary. Proof. For each fixed zl' lf(d. z 2>1 has its maximum on the boundary

I

of B(z1); hence, consider f(zl' g(zl' e ) ) tion has its maximum on I z11 =1.

1.

But for each fixed A. this func-

Corollary. Iff is known on the distinguished boundary of a Bergman domain B, it is known throughout B. In fact, we have the following: Bergman Generalization of Cauchy' s Formula.

= (.l.) 2 211'.

211' 211' ('

r

if/> L\ _!& ( i~ iA )dqiUI\. . ~.,. aA e· , e

JJ

e e

00

§ 5.

Analytic Polyhedra

In this section we shall define analytic polyhedra. Vl e shall show that every holomorphy region is a limit of an increasing sequence of analytic polyhedra. A.

Definition 17. Let Dopenc ([n; f1, ... , fk holomorphic in D, and A = {z z E D and fj(z) < 1; j = 1, ...• kJ. If AC C D, A is called an

I

I

I

analytic polyhedron (of dimension n) . Corollary 1. Every analytic polyhedron is a region of holomorphy. Proof. Let B be an analytic polyhedron,

lr.(t)l =1 for J

some j, say j = 1. But then

~

e bndry B. Then

25

is holomorphic at every point of B and singular at

t.

Corollary 2. Every connected analytic polyhedron is a Bergman domain.

:

Proof. Let B be a connected analytic polyhedron, t

€

bndry B. Then

lr.l =1 for some j; i.e. every boundary point satisfies an equation of the J iA form f.(z1•... , z ) -e = 0 . J n B.

Theorem 7. Let D be a region of holomorphy. and K CC D. Choose

D such that 0

KC C Dopenc c D. 0

Then there exists an analytic polyhedron

A.

A in D such that KCC ACC D . 0

Proof. Lett

€

bndry D . There exists a holomorphic function 0

"'

gt in D such that jg~ l, and lg~(z)l < 1 inK. Hence. there exists a neighborhood Nt of each ~

E

I

I

bndry D0 such that g t (N t) > 1. But bndry D0 ,

being compact, is covered by a finite numrer of such neighborhoods. say Nt{" •.• Ntk .

Let A ={z Ig\ /zll< 1; j =1, ... , k}

Corollary. Let D be a region of holomorphy. Then there exists a sequence A .• j = l, 2, ... of analytic polyhedra in D such that J

A .C..C A "+lC CD, and D = J

J

co

U.J= l AJ..

uco

-

Proof. Choose D1~~C. c o ~~ i=l Di - D. 2 CC: ... c C D such that ~{~1 , Consider the sequence { Dj} ; there exists a subsequence. say DjJ • such that

Then by theorem

o" 1cc

rl.

there exist A. analytic polyhedra, such that

A 1Cc

1

" cc:. A cc ... CC D. o 2 2

26 Chapter 3. Pseudoconvexity § 1.

A,

Plurisubharmonic and pseudoconvex functions V>/ e have already introduced the notion of a continuous sub harmonic

function (I,§ 2). We now extend -this definition as follows: :

Definition 18. Let D be open in«:, 4>: D- B.. Then ,P is said to be subharmonic if: i) -

00

~ cP

< co

' 4>

f . . 00

ii) r,l is upper semi -continuous; i. e. lim sup 9 (p' ) < 9 (p) pr- p

-

-

iii) for any domain DCC D; if h is harmonic in D and continuous on 0

0

ii00 , then h > ~ on SD implies h > t/> in D , -

0

-

0

Property 1 (Mean Value Property, 1). Let

{I z-z0 I < r} C

D, ~ sub-

harmonic in D. Then 211'

f/1 (z ) < -21 0

-

'11

S (z + reiB) df/ • tP

0

0

Property 2 (Mean Value Property, II). Let

{lz-z0 l < r}c D, tP

sub-

harmonic in D. Then r 211'

t/>(z ) 0

1

< -2 7l'r

Jr Jr 0

~ (z 0 + re

ie ) r

drdf) .

0

Property 3 (Strong Maximum Principle). Let 4> be subharmonic in D. Let lVI = sup 0 9 . Then, in each component of D either 9 (z) < lVI or 9 is constant. _Property 4. If ; satisfies i), ii) and the integral condition of Property 1 or 2, then 4> is subharmonic in D. Property 5. If ; and r/1 are subharmonic, then max (9 , 1/J) is subharmonic. Proeerty 6. If ;

€

c 2,

; is subharmonic if and only if A 9 ~ 0 •

27 DC~

Definition 19. Let

n 6

;

:

D - B. • Then t; is said to be pluri-

subharmonic in D if: i) -oo

< ~ < o-J

•

ii) ; is upper semicontinuous iii) if (z16 .•• z ) £ D and a. £ ~ arbitrary6 i =16 ... n, then n 1 . t (t) =~ (z1 +tar •.. , ~n + ~a0) is subharmonic for small J t 6

I.

~

is said to be pseudoconvex if it is plurisubharmonic and continuous.

Remark. Statement iii) above is equivalent to the following: 4> o

T(zl' ... , zn) is subharmonic in each variable separately, for all

linear transformations T . Corollary 1. Let

DC~

n

• f:D- ct. Iff is holomorphic in D, then

log If I is plurisubharmonic. Furthermore, if f :f 0 on D, log

If I is pseudo-

convex. Corollary 2. Plurisubharmonic functions satisfy the strong maximum principle. Corollary 3. If Dc:«:n and cp, 1/J:D- It are plurisubharmonic, then so are

max(~, 1/J), 4>

+1/J, and AcP , A > 0.

Definition 20. Let q;

E

C 2. The the Hessian of 1/) is defined to be the

following matrix:

Note that His Hermitian, if ,P is real valued. Proposition 1. Let DC «:n.

f/J

:D- .IR,

fjJ €

c 2 . Then~

is pseudoconvex

if and only if the Hessian of 4> is positive semidefinite.

Proof. Consider i

.P(~)

= ,P(z 1 + ta1, ... , zn + ~an ) , a.1 £ «:

'

=1, ... , n . Now 0(~) is subharmonic if and only if At~ 0 . But

28

a2ri = 4 atat

6~

n~

a

=

.J=E1 az.J

4ar

n

a2! '\aj azka:Z. . J

4 E: j, k=l

=

a.J

Proposition 2. Let DC <en , D C C D , ~ :D - lB. such that fP is

--~-~---.

0

pseudoconvex in D. Then there exists a sequence [ ~ .) of pseudoconvex, L J. C«) functions in D such that ~ . - ~ uniformly in D . . 0 J 0 Proof. Define the "smoothing functions" K , £

K K

f

f

:«:n (~)

f

> 0--as follows:

-11.

>0

-

Support KfC

{IJ t II <

f };

i.e. Kf(t) = 0 for

lltll>f

Define:

4>

£(~)

=

l

Kf (z-t)

~ (~) dE 1 •••

di'Jn' where we

take~ =0

([!n

where it is undefined.

Then~

f

(z)

f

C

co

• Furthermore, f f

in D , as follows: 0

4> (z)

=

l <en

~ f (z)=

5' fl:,n

Therefore

K f (t)

~ (z)d~ 1 •••

d 1}n

K(t)f(z-t)dE 1 ••• d1}. £ n

~

uniformly

29

I; (z)-; f(z) I ~

S

Kf(t)

a:n

I; (z)-; (z-t) I d~ 1 • • • cbJ0

< max

.-

l;(z)-; (z-t>l- 0 as

II t II < €, z

€

€-

0

D

0

since; is uniformly continuous on D • 0

Now ;

€

is plurisubharmonic since ;

€

(z)

=

S ctn

K (t) ; (z-0 d~ · ••• dTJ , € n

1

is essentially a linear combination of plurisubharmonic functions with positive coefficients. Proposition 3. Let A

domain

C

. ct, g:A - DC ct n; g holomorph1c. Then

; is pseudoconvex in D if and only if cb (g(t) ) is subharmonic and continuous for all such g. Proof. Assume; is pseudoconvex. By proposition 2, we may assume cb

€

00

C • Then 9g

and let

~(t)

€

C

00

• Now,

= fPg(t) .

Then

= as the Hessian of ; is positive definite. The converse is trivial, as the class of holomorphic g:A-D contains all linear transformations. Corollary. The image of a pseudoconvex function under a holomorphic mapping is pseudoconvex.

30 § 2. Pseudoconvex domains

~.

Definition 21. Let

{I t I < 1} C: C.

Then an analytic disc in D is a

mapping g: {It I < 1}- DC ftn, continuous on

{It I ~ 1} and holomorphic

in the interior.

The boundary of the analytic disc is the mapping g restricted to It J=1. z Set g(

{I t I ~ 1)) =:E. Abusing terminology by suppressing mention of g, we shall refer to

:E itself as the analytic disc and to a:E as the boundary. (Note that a:E in

general is not the set-theoretic boundary of the point set :E.) Theorem 8. Let DopenC: tr,n. Then the following are equivalent:

Jbe a sequence" of analytic discs in D. J

i) Let {:E.

then

~

.U :E .c. C D. ("Kontinuita•ssatz").

J=1

00

If

.v1 a :EJ.c c J-

D,

J

ii) -log A(z) is plurisubharmonic in D, where A(z) is taken in any norm. iii) For any analytic disc E in D, A(:E) = A(aE). iv) There exists a pseudoconvex function.; in D such that, for every N > 0 there exists a KC CD for

which~~ Non

D-K. (Informally, f =+co on

the boundary of DJ Exercises. a) In Euclidean space, i) has the following analog: Let Dopenc C Jln. Let { Ej} be a sequence of segments in D. If

Ua :E j c C p,

then U :E j r- c D. Show that this property holds if and only if

every component of D is convex. b) Find the analog of ii) in B. n. Definition 22. A region with any and hence all of the above properties is said to be pseudoconvex.

then

Corollary 1. Let o?penc cr::n, D. pseudoconvex. If n. 001D. is open, J J= J 00 J 0. is pseudoconvex. J=1 J

n.

Corollary 2. Let Dopenc a::n. Then D is pseudoconvex if and only if each component is pseudoconvex.

31 Corollary 3. The holomorphic image of a

pseud~onvex

region is

pseudoconvex. Corollary 4. Let

£A j} be a sequence of pseudoconvex domains such

that A .C A '+l' then UA . is pseudoconvex. J J J B. Proof of Theorem 8.

·

iii) implies i). U8I:.CC D and A(E.) = A(a:t.) imply cl( UE.)CD. J J J J That UI:. is bounded follows from the fact that g assumes its maximum J

over E on

II

{I t I =1}.

and Ua,; j is bounqed.

ii) implies iv). If Dis bounded, we may choose

If Dis unbounded, choose '(z) =max ~(z)

(-log~(z),


(z)

= -logA(z).

lz112 + ... + lzn( 2); where

is taken in the Euclidean norm. ii) implies iii). Let I: be given by:

{It I < 1} - D.

g: Then

-log~(g(t)

) is subharmonic and continuous; hence it has a maximum on

{I~ I < 1], which is assumed on

{J t f = 1}.

Therefore,

-log A(a:t) > -log A (I:) i.e.

£:\(8:£)<

A(l:)

A(8:E) > A(E) .

But clearly,

iv) implies i). Let :E. be given by J

Consider the subharmonic functions -

~

(g.(O ) , j J

iv). Note that max

f/1 (g.(O)

It I<1

i.e.

max

=1, 2, ... ; where q, =max

¢ (z)

Z£3.1;.

J

(g.(t)). =1 3

It I

J

=max

~

is the function given by

4>

Z£1:.

J

(z) .

32 U a 1;.CC 0;

Now.

J

co

hence

sup ~ ( U j =1

But therefore

sup fjl ( U

a~ j) < 1\£ < co

co

E .) l

j=l

< M < co

• •

implying i) implies ii). (Proof due to Hartogs). Since i.\(z) is continuous, -log 6(z) is continuous. In fact -log b.(z) satisfies a Lipschitz· condition on compact subsets of D. To prove that -log

~(z)

is plurisubharmonic in D, it

is sufficient to show plurisubharmonicity at a point z

0

E

D.

Thus we must show

that, for the set of points z = z 0 + ~z 1 where z 0 and z1 are arbitrary points of D and ctn respectively, and t E ct is sufficiently small, -log 6(z 0 + t z 1) .=C/ltt) is subharmonic as a function of measure ~(z), z

E

D for

t.

II z111

For

It I< l.

small enough in the norm used to

Furthermore, 1/1(0 is subharmonic if it sat-

isfies the mean value property of subharmonic functions, and it is enough to show this for

~

= 0; i.e. to show that

s 211'

1/1(0)~ 2~

(! )

1/J(eiB)dB.

0

Let g(t) = h(~) + ih*(~). with h a real-valued function such that

d ~ 1.

h(eW) = f/;(eiB ), g holomorphic for

It I < 1,

claim that such a function exists.

Firstly, the continuity of 1/J implies the

and continuous for J

I

existence of a harmonic function h, in I~ < 1, equal to r/1 on I~

I = 1.

Vie Thus

his defined and is continuous on the closed unit disc. Secondly, take h* to be some conjugate function to h. Since g is now defined and holomorphic, and its

I I = l, its imaginary part satisfies a H~lder condition. Hence h* is continuous on f ~ I = 1. Next, let b be any vector in ctn with II b II = l, and let A satisfy 0 real part satisfies a Lipschitz condition on t

0 < "- < 1 . Consider the analytic disc in Cn 0

33

(1) E(O)C D. This is obvious. (2) If p £ E'(~) and~:- ~ .. then there exist p. £ Z(A.) such that p.-p.

(Namely, p. J (3)

3-g(to)

=z 0 U

O<~<~

-

-

(For if z

£ a:t(~

(4) S

i9

~

e

log~(z +e z)

{AI 0
-

0

1 · = ~ ~(z

o

-0

=

J

) then

-

(3) holds.

J

0

. iC 16 >II -h(e ) ( II z- z + e z1 < ~ e < 0

J

+ t z1 + ~.e b where t is the preimage of p.) 0 J 0 : a:t(~ )C CD.

0

i6 + e z ) and hence 0 1

and E(A. )C:: o1 is open in the space A

J

=[O,A. 0 ].

(5) S is closed in iL (This follows from (2), (3), and the Kontinuit~tssatz.) (6) S is the set [0, A. J. 0

(From (1), (4), and (a), S = A.·) Hence z +A. e 0

-g(O)

!; (A

) C:: D for 0 < 1-.. < 1. Consequently,

I~ I < l; z +~ e-ia e-g(O)b£D

-

b e D for 0 < ~ < 1, in fact for A. complex, -

.

0

for 0 S. ~ < 1 and a real, since we can incorporate e -la into b, as lle1abll = btl ::: 1. This means that a ball about z of radius 1~ e -g(O)I is

II

0

I

contained in D, and therefore A(z ) > I e -g(O) = e -h(O). Hence 0

-

-log A(z ) < h(O), but -log A(z ) = t/J(O), and since his harmonic 0 -

0

2~

(! ) C.

t/J(O)

~ h(O)

1 = 2,(

2~

s

s

0

0

h(e i6 ) dC = 21~

1/J(ei6 ) dC •

Lemma. Let ~ be an analytic disc in D, DC:: Cl:n. Then ~C: bdry ~. the

hull of the boundary of

~

with respect to holomorphic functions.

Proof. Let E be given by the holomorphic mapping: g:

{I t I ~ 1}- D.

For every f holomorphic in D, f(g(t)) is holomorphic and therefore ff(g(t))J assumes its maximum M on

{I d = 1J·

I

I

Therefore f(g(t)) ~ M for g(t)£ aI:

implies !f(g(~))j ~ M for g(~) e 2:. Theorem 9. If D is a region of holomorphy, then D is pseudoconvex. Proof. Jet L2:j~ 'be a sequence of analytic discs in D; such that lJj=l oL:jcc~~'~ But u;l ()~jc 1)~ 1 o i:jc.c D, as D is a region of holomorphy. A . CX) 00 co But by the above lennna, uj=l LjC L)j=l ~ 2:j' hence U j=l ~j c. c D. Now apply theorem 8. Theorem 10. Let DopenC~n. If, for every ~ e oD, there exists a ball N about C such that N ~ D is pseudoconvex, then D is pseudoconvex. Proof. Assume that D is bounded. Then oD is compact. By hypothesis, for ~ e cD, there is a ball N~,about C, such I\ ( that NC1'D is pseudoconvex. The settNc~ is an open covering of oD. The compactness of oD implies that a finite number of the N~ cover oD; call them N1 , ••• ,Np. If N1 fl NJI\ cD ~ ~~ set Qij = Ni~ Nj, for each i 1 j; i,j = 1,2, ••• ,p. In each Qij choose any ball Bij centered at any point C E (Qijnon) such that Bi[~ Qij • Since there are only a finite number of sets Qij' there are only finitely many Bij. Let r = min (radius of B1 j). At each point~ E cD the ball S(C,r) of radius r centered at ~ is contained in some N1, and hence S(~,r)AD is pseudoconvex; because S(~ 1 r)l\ D = S(~,r) ('\ (N1 (\D) and S(~,r) is pseudoconvex by Cor. 5, p. 16 and Thm. 9. Now, consider the fut:lction ~(z) = max(-log ~~ -log 6(z)). We claim that +(z) is pseudoconvex. Clearly +Cz) is continuous. If A= tz I 6(z) ~ r/2~ then for z E A, ~{z) = -log~= constant and therefore is plurisubharmonic. If B = {z I 6(z) < r/2), then for z E B, +Cz) = -log6{z). But fo~ z e B, z E S(~,r) and 6(z) = ~(z) = ~(z) where K = S(C,r) 1'\ D. Since K is pseudoconvex, -log6(z) is pseudoconvex. Thus t(z) is pseudoconvex in D. As z approaches the oD, ~(z) becomes infinite. By part (4) of Theorem 8, D is pseudoconvex. Now, consider the case \'lhen D is unbounded. Set Dj = D f\ lz I Uz IJ<j}, Each Dj is a bounded set. If C E .dD then ;>

35

1

either (i) c e em and ~ e sj = z I liz~ < J}or (ii) c- e OSJ and C i oD, or (iii) C c:_ dD n.nd C E dSj. For case (1), by hypothesis, there exists an NC about C such that NcnD is pseudoconvex. For case (ii), since Sj is convex, any ball about C, NC lying in D, satisfies NC~Dj is pseudoconvex. For case (iii), there exists a ball N ( C, r) such that N/) D is pseudo·convex. But Nl)s j is also pseudoconvex. Therefore N1\Dj is pseudoconvex. Therefore each Dj satisfies the hypothesis of this theorem and is bounded. life have already shotm that therefore D. is pseudoconvex, j = 1,2, •••• Since Djc Dj+l and \ D~j -> D, D is pseudoconvex (by Cor. 4 or Thm. 8). Establishing the converse of Theorem 9 is the Levi Problem.

§3.

Solution of the Levi Problem for tube domains

Definition 23. Let z. = xj + iYj• The set J n D = tJhere B is some open subset ofRn, is called a tube domain. B is called the base of the tube domain. Example. D = t
J

Suppose ;/J

€

a2r/J is positive definite. But azjazk

2

C , then the matrix

( aa

. ~

-i

a~ ) r/J; and since"' is independent of the k

.

Thus every diagonal element is positive# i. e. 2 a r/1 ~ ~ 0.

ax.

This means that r/1 is convex on every straight line segment in B

J

parallel to a coordinate axis, and since r/J is pseudoconvex this property is invariant under linear transformations. Hence r/J is convex on every straight line segment in B# and hence in- every component of B. We claim that this implies that every component B of B is convex: 0

1is a sequence of straight line seg-

(a) Firstly# we show that if { ~ 11

=~

ments in B such that lim ~ 0 v v-co

and lim a~ ll v-oo

=1-' then if pC B0 # ~ C B0 .

Indeed, since 1/J is convex on each straight line segment in B0 , on each ~ v . max 1/J(x) = max 1/J(x), i. e. max (-log 6B(x) ) = max (-log 6B (x) ), or equivaX€~

X€ a~~

XE~

v v v lently min 6B(x) = min AB(x). Since

X€ a~

~B(x)

v is a continuous function# the

X€ a~

X€X

v v equality holds in the limit min AB (x) = min AB(x). But IJ is a closed set X€~..

X€1J

and ~-&C B • therefore min AB(x) > 0; hence il C B . 0

0

X€1J

(b) Now, let x, y

€

B . We must show that the lirie segment joining 0

them belongs to B . Since B is connected# there exists a curve.,; (t), 0

0

0 < t < 1 , lying in B • joining x and y; -

-

0

~ (0 )=x,

+(1 )=y. For t sufficiently

small# the line segment (x# (/) (t) )CB . As t- 1 there cannot exist at such 0

that for all t < t the line segment 0

but the segment

(x,~{t

0

Hence B is convex. o-

))

0

(x.~ (t)

'~B, Y- 0

)CB

0

because this would violate (a).

37

In the case r/J

~ c 2,

it suffices to show that if B0 is any component of .

B and if I\ is a straight line segment in B , then max r/J(x) = max f/J(x), for O

XE/\

XE8/\

then the proof follows as above. So, let. I\ be a straight line segment in B0 • "'domain -'\ "' " Then A c B ~ CB and the tube domain D over B satisfies DC c D. ~-

0

Hence for every

E

> 0 there is a pseudoconvex d>O function f/1 E in D such that•

II/J -t/1E I < E in D, 1/JE depending only on the x' s; as we can define smoothing

functions K depending only on the x' sin the proof of Proposition 2. As in the E

,..

previous case, each 1/J is convex on B, and therefore max 1/J (x)= max f/J (x). € XEA E xea/\ E But, max l/l(x) < max r/J (x) + E X€/\

X€/\

€

max 1/J(x) > max r/J (x) - E X € I\

X €

A

€

=max 1/1 € (x) + E < maxi/J(x) + X€a/\

= max

;/J (x) - E

X € a/\. €

2E, similarly

X€8/\

> max r/J(x) - 2€. Letting

E

\t 0

X € aiL

gives the desired equality. (2) implies (3). Since every component of B is convex, every component of D is convex. Hence D is a region of holomorphy. (3) implies (1) has already been proved.,

38

Chapter 4.

Zeroes of Holomorphic Functions. ~1.

Meromorphic Functions.f

Weierstrass Preparation Theorem

Definition 24. Let .2::: a z1"1 • • • z"n be a formal "1 • • ·"n n • series about the origin. (This series is not assumed to be convergent). Then the order of the series is the least integer K such that av v v ! 0, v1 +v 2+ .• ; + v = K. 1 2· • • n n The aeries is said to oe normalized with respect to z1 (at the origin) if ord (series) = K and z~ occurs with non-zero coefficient aKO~ •• o· ~ that the order of a series is invaria~t under holomorphic changes of variables, including non-singular linear changes, which leave the origin fixed

:A.

n

Cj

=

~=l ajs zs ,

det (ajs)

!

(4) , ozl

1, ••• ,n,

0 ;

Corollary. If L a z1"1 .•. "l···"n convergent series of order K, then the 1) f is normalized with respect to 2)

j =

"n

z = f(z) is a n following are equivalent' z1 at the origin

0

fK(z 1 ,o, ... ,O), O, whe~e fK vdenotes the partial sum Lv + +v = K a, z 1 ••• z n of the K th order 1 •.• n vl···"n 1 n homogeneous polynomials in the series f. Note that 1) and 3) are also equivalent ~or formal power series. We shall write f(z) = 2:: a z11 ••• zn"n for the · ·"n \Vhen convergence sake of brevity, for all formal power"1·series. is assumed it will be mentioned explicitly. ~ av ••• v z"1 ••• zn = f is any power Property 1. If~ 1 series, it may be normalizea witR respec~ to z1 (at the origin) by a linear change or variables: zj = ~=~ aj 8 C8 , j = l, ••• ,n; det (aj 8 ) I= 0.

3)

1·

•

"n

i9 Proof. convergent.

Let f be of order K. Assume first that f is Then n n f(.~) = f( ~=l a ls~s' •• • , ~=l ans~s) ,

and ord r ( ~) = ord f (z ) • f(.~ 1 ,o, ... ,o) = f(a 11 1 , .•. ,anl~l=); hence fK(~ 1 ,o, ... ,o) is non-zero for some choice of the ajl' j = l, .•• ,n. We may complete the matrix (ajk) so that det (ajk) ~ 0. If f(z) is nonconvergent

c

where the fj are homogeneous polynomials in the zi of order j. Now, consider fK as above. Property 2. If {f(j)(z)J is a countable sequence of power series, they may be simultaneously normalized with respect to z1 {at the origin) by one non-singular linear change of variables. Proof. For each j,

(j) + fK +1 + · • • ; j

Consider the spherical hull

;;> {j)

I

= [
1}

Now fK. (z 1 , ••• ,zn) is a polynomial, and hence vanishes on a clos~d nowhere dense subset of 2 . But the union of countably many nowhere dense sets is nof~Tre dense, so there exists (a11 , ••• ,an1 } E ~such that fKj (a11 , ••. ,an1 ) ~ 0 for every j. But we may now complete (a11 , ••• ,an1 ) ~ 0 to a nonsingular matrix. We have shown that, if we consider countable collections of power series, and properties invariant under linear transformations, we may assume these series to be normalized with respect to z1 •

40 Note the following properties of order: (f, g denote formal power series) ord fg = ord f + ord g ord (f+g) ! min(ord f, ord g) ord f = 0 if and on~y if r is a unit; i.e. if and only if there exists a g such thar fg = 1, where g is a formal power series. We remark here that the set or formal power series at a point, as well as the subset of those power series which converge in some neighborhood of the origin, form commutative rings with unit. This ring is an integral domain, with units the series of order zero. We remark also that the definitions and consequences stated above may easily be extended to series whose centers are any point a E en. B. Theorem 12. (Weierstrass Preparation Theorem). Let f be a formal power series, normalized w.r.t. z1 , of order K. Then

where h is a unit, a1 , ••• ,~ power series in z2, ••• ,zn and are non units; and this representation is unique. If f is a convergent power series, then h and ai are also. K-1 Note. zK 1 + a 1 z1 + •.• +aK as above is called a Weierstrass polynomial. Proof. We first make a series of remarks: For K = 0, the theorem is trivial. For z = z1 , the theorem is also clear. K-1 ) Furthermore, if f = h ( zK 1 + a1 z1 + • • • + aK , the ai must have no constant term, for ord f = ord h + ord (z~ + a 1 z~-l + ..• + aK) • Therefore + ••• + ~.) • K = ord (zK1 +a . 1 zK-l 1 ~~.l\ If ord a 1

=

0 for some ai,

41 K-1 ) ord ( zK 1 + a 1 z1 + ... + aK = K- i < K • We now present a proof for the case of formal power series; this (constructive) proof will give uniqueness in both cases. However, if f is convergent, the convergence of h and the a1 is more easily shown by a second proof. f = fK + fK+l + • • • ; f. homogeneous polynomials of K J ·ord j where fK = z1 + •••• We wish to construct a power series 1

11

(1)

( f K + f K+1 +

= xo + xl + · · ·

such that

• ' . ) (XO + Xl + • ' • ) =

where x0 I= 0 , yK = z~ + ••• and all other yj are of order at most K-1 From (1), we obtain

1n

z1 •

K

But fK = z 1 Similarly, i.e.

Take

c1 = -

ap ~ ~.!\.

+1 I 0 I

o o • I

0 and

~ 1 0, ••• , 1 , 0, ••• , 0 - bK -1 , 0, .•• , 1 , 0, ••• , 0 aK+1 , Q•• .,0 for j ~ 2; the subscripts 1 appearing in the j th places. Choose x2, etc., similarly. Note that we have proven uniqueness for both convergent and nonconvergent series. We now proceed to the proof for cj

= -

42 the case: r convergent. As.before, write r

fK + rK+l + • • •

=

lzjl

<

'

K

zl + • • • •

fK = For

It is due to Siegel:

lz 1 1 , · j = 2, ••• ,n

lrK+l + rK+2 + ••• I ~ cllzliK+l + c2lzllK+2 + which is convergent for Therefore as

I rK+l

lz 1 1

<

p small.

+ rK+2 + ••• 1 ~ ~ I z1 1K ;

lfK+l+f~+2+ ••• 1 <~

... ,

I z j I < I z1 1 < p <

under the above conditions.

zl Consider rxfz~; define tj = zj/z 1 , j = 2, ••. ,n. Then fxfz~ is a polynomial in the tj ; rK K = 1 +r , zl where r is a homogeneous polynomial in the t without a constant term. Therefore, for lt 3 1 < p1 , lrt < 1/2. Under these conditions: f

r:K

rK + (rK+l + ••• ) fK rK+l+ ••• /z~ = 1 + K fK/zl rK+l+ • • • 1 = 1 + K l+r zl

=

1

=

+q

I

where q is a power series in z1 and the ourselves to the above inequalities; hence <

1

tj.

We restrict

•

Choosing some determination of log, we obtain: log

~ K

= log (l+q)

= q - q2/2 + q3/3 - q4/4 +

... '

p~

a convergent power series. Substituting, we obtain a convergent power series !n z1 , t 2 , ••• , tn ; replacing t j by z j/z1 we obtain a power series in z2, ••• ,zn and Laurent series in z1 • Now assume also that lz1 1 > E > 0. Hence CD

V

= ?=o avzl

=log+.

K

CD

1\o

co=

zl

+~(j)= v+w

where a v , ~(1) are power series in z2, ••• ,zn.

i.e.

Therefore

log f - log fK = v + w · fe -v = fK ew .

But re-v is an analytic function of z1 , and e·v is a unit. Hence re·v converges in a neighborhood or the origin, by Abel's theorem. Therefore the series fKew cannot contain any negative powers of z1 ; and also no power of z1 > K, and we have obtained a convergent representation f

=

ev(fKew)

as claimed, unique by the above. Corollary 1. Let f be a convergent power series, and assume r vanishes at the origin. Then the set of zeroes of f in a neighborhood of the origin is of dimension n-1. Proof. By a linear change of variables, we may assume f is normalized with respect to z1 , K-1 ) f = h ( zK 1 + a1z1 + . + aK o o

For z2 , •.• ,zn small and fixed arbitrarily, a 1 , ••• ,aK are small, and z~ + a 1 z~-l + •.. + aK is a complex polynomial, with K roots (counting multiplicity). Furthermore, these zeroes are located 1n a neighborhood of the origin as they depend continuously on the ai. Definition 25. Let sclosed C. Dopen C G:n. Then S is said to be a (globally defined) analytic hyPersurface or analytic variety of codimension 1, if S is the set of zeroes of a function f ¢ 0, analytic 1n D.

44 Corollary 2. S is locally arcwise connected. Corollary 3. If n > 1 and f· is holomorphic in D d: en, then the zeroes of f are not isolated.

§ 2. Rings of power series A. As we have remarked previously, the set of convergent power series in z1 , ••• ,zn at a point forms a ring, as does the set of formal power series. We shall now state some algebraic results which will prove useful in the sequel. Refer to any standard algebra text, e.g. van der Waerden, Moderne Algebra, for proofs and details. Definition 26.- Let R be a commutative unitary ring. R is said to be an integral domain if: a E R, b E R, ab = 0 implies a = 0 or b = o. An element a e R is called a unit if a has an inverse in R. Elements a,b e R will be called equivalent, written a = b, if a = eb, where e is a unit. An element a e R is called reducible if a = be, where b and c are non-units; a is otherwise called irreoucible or prime. R is a unique factorization domain (U.F.D.) if every element may be written as a product of primes, unique up to order and equivalence. A subset I of R is an ideal if a, b E I implies a-b E I, and a E I, rt E R implies art E I. I is a proper ideal if I ~ R, and maximal if it is pr•oper and such that if ~ is any ideal satisfying I c. -.:P c.. R, · then I = 'I or ~ = R. A ring R is called a local ring if there exists a unique maximal ideal. R[t] denotes the ring of polynomials with coefficients in R and R[ t] the ring Ill power series in t with coefficients in &,· Let L ajtjE R[t], aj e R, t an indeterminate. L ajtj i~ called primitive if the coefficients aj of t have n8 common factor except units. (Note that r E iU t] implies r = ag with a e R and g e lt[ t] and primitive.} Two polynomials will be called relatively prime or coprime if they have no common polynomial factor. We use

the following results. Lemma a. (Gauss' Lemma) If R is UF.D, R(t] is UFD. Lemma b. Let p,q E R(t] (R[t]). Then p and q primitive implies pq primitive. Lemma c. Let P1 ,P 2 E R[t], R an integral domain. p1

=

p2 =

tK + al tK-1 + . . . +

~

tL + bl tL-1 + • . • + bL

•

Then there exists a polynomial r in.the coefficients a 1 ,bj called the resultant of P1 and P2, which is zero if and only if P1 and P2 have a common factor; i.e. if and only if there exist p,q,s E R[t], deg q > 0, such that P1 = pq, P2 = sq. Furthermore, there exist polynomials A and B such that AP 1 + BP 2 = r. Lemma d. No proper ideal I of a unitary ring R contains a unit. Lemma e. R is a local ring if the nonuni ts in R form an ideal. B. Definition 27. Let C9n denote the ring of formal power series at the origin in n complex variables. Property 1. c?n is an integral dom~in with unit. Property 2. The nonunits of ~n form an ideal; hence ~n is a local ring. Property 3. CJ n is UFD. P~oof. We use induction on n. For n = 1, units and elements of order 1 are irreducible. All elements of order > 1 are reducible, for if ord g = K > 1 , g(z)

=

zKg1 (z)

,

where g1 (z) 1s a unit, and the decomposition is unique. Hence, assume r9 n-l is UFD. We may assume f E (9 n is normalized at the origin. Then: K K-1 ) _ f = h ( z1 + a 1z1 + . • . + ~ = hp ,

46 where p is a Weierstrass polynomial; i.e. p E ~- 1 [z 1 ] and _is monic. Now f is prime in On if and only if p is prime in On-l [ z111 for: Assume p is reducible; i.e •. p = p1p2 • We may take p 1 ,p 2 to be Weierstrass polynomials. Hence f ~ (hp1 )p 2 . : Conversely, assume f is reducible: f =

fl f2

=

(hlpl )(h~2)

=

(hl h2){plp2) •

But p1p2 = p by uniqueness, and h1h2 = h. But (Qn_1[z 1] is UFO by assumption and Gauss' lemma. Now let f EOn· f

=

hp •

But

P = P1 ••• Pr where the pi are irreducible Weierstrass polynomials. f = hp 1 ••• Pr ,

Hence

and this decomposition is unique up to order and equivalence, for if f = tl ••• tt , ti = hi ri

,

by the Weierstrass Theorem and then rl ••• rt

=

P1 ••• Pr

=[

by uniqueness. Hence [r1 , •• • ,r 9 ~ p1 , ••• ,pr\ as C?n_1[z1 ] is UFD. Definition 28. Two holomorphic functions are said to be relatively prime or coprime at a point if their power series expansions at that point have no common irreducible factor other than a unit. Lemma f. Let f,g be holomorphic functions in D, 0 E D c ~n, n > 1 coprime at the origin, such that

47 f(O) = g(O) = 0. Then in any neighborhood of the origin, there exist points at which r vanishes and g does not, and points at which g vanishes and r does not. Proof. We may assume f and g are Weierstrass polynomials f

=

g

=

Suppose there exist no such points. resultant of f and g.

Let

r(a 1 ,bj)

be the

For each z 2, ••• ,zn in a neighborhood of the origin, there exist z1 such that f and g vanish simultaneously. But if f and g have a common zero viewed as polynomials in one variable, then f,g have a common factor. Hence r(ai,bj) is zero for each z 2 , ••. ,zn. But r is analytic, and hence r 0 near the origin, implies f,g have a common factor as polynomials in on. Lemma g. Let f,g be holomorphic functions coprime at the origin. They they are also coprime in some neighborhood of the origin. Proof. Let f = up g = vq ,

=

where p,q are Weierstrass polynomials, and let r

=

Ap + Bq

be the resultant of p and q. Let N be a neighborhood of the origin so small that u,p,v,q,A and B are convergent. Let a= (a 1 , ••• ,an) E N. To show f,g are coprime at a, it suffices to show p,q are coprime. The equation r

= Ap + Bq

persists where r,A,p,B and q are viewed as series about a, i.e. as series in ~i' where

48 ci p

Then

q

= zi - ai • = c~ + • • • = cr + ••• •

Now assume p and q have a common factor h "at a", h E ~-1 ( c1 ) • ~Ut p,q are primitive in t 1 j hence h is also.

where

hi

&n-1"

E

But

h divides r

A

E

k

E

~- 1 and 0 n-1 [ ~1] •

r.

Therefore

= hkA ,

a primitive power series in

k

k

=

ko

t 1,

+ kl ~1 + • • •

By comparing coefficients r

=

a relation in

h0 k0 A ,

Hence A divides

r,

(f)n- 1 •

and r

"X = hk.

But h and k are primitive, hence hk is primitive, hence h0 k0 must be a unit, i.e. hk is a unit, implying that h is a unit. Thus p and q are coprime at a.

§ 3.

Meromorphic . functions

Let x E Dopen C ~n, and consider functions which are each defined in some neighborhood of x in D. Call two such functions equivalent if they coincide on a neighborhood of x. This defines an equivalence relation, and the equivalence class of a function f at x, denoted by [f]x, is called the germ of f at x. If f is a holomorphic function, then [f]x amounts to a convergent power series. Germs at x form a ring with the obvious definition of

addition and multiplication. The ring ~x of germs of holomorphic functions is a commutative integral domain with identity and unique factorization. Topologize the · space. of germs 0 = &x by defining the following basis for the open se~~~ Let k E C), then k E (!)x and so k = [ f] x , where f is defined 0 0 1n an E-neighoorhood Nx of x0 in D. At each y E Mx ·, 0 take that class in c?Y 0 containing the direct analytic continuation of f, i.e. take [f] • Then define f]Y to be an open set and th~ collection of such JENxo sets to be the basis ot open sets. A holomorphic function f in D amounts to a continuous mapping, r : D -> which assigns to each point in D a holomorphic germ over that point. Now form the quotient field Mx of ~ for each x E D. Topologize M = U Mx as follows: Let l E M, then l E Mx , l=[~lx xe:Rnd is represented by [f1 lx /[f2lx 0 .... 0 0 0 where r 1 and f 2 a~ nolomorphic functions at x 0 , and because of unique factorization we may take f 1 and f 2 to be coprime at x0 • Let Nx 0 be a neighborhood of x0 in which f 1 and f 2 are defined and are still coprime. At each y E Nx , take that class in My represented by . [f1lyi[f 2ly· Thg union over Nx of these classes we define as an open set and the co~lection of all such sets we take as the basis for the topology. The elements of Mx are ~alled germs of meromorphic functions over x. Definition 29. A meromorphic function in D is a .continuous mapping which assigns to each point of D a meromorphic germ over that point. Meromorphic functions form a field. At a point z0 E D, a meromorphic function g is efined by the quotient of two functions f 1 ,f 2 coprime

U

U [

dJ

50

and holomorphic at z0 • a) If f 2 (z 0 ) ~ 0, i.e. f 2 is a unit, then f 1/f2 is holomorphic there and hence g is holomorphic at z0 and therefore in a neighborhood of z 0 • z0 i~ called a regular point of g. b) If f 2 (z 0 ) = 0 and f 1 (z 0 ) ~ 0, then g is said to have a pole at z0 • c) If f 2 (z 0 ) = 0 and f 1 (z 0 ) = 0, then z0 is called a point of indeterminacy of g. The set of such points has topological dimension 2n-4. Corollary 1. The set of regular points of g is open, and glregular pts is holomorphic. Corollary 2. If z0 is a pole of g, then there exists a neighborhood N of z 0 in which every point is a pole or a regular point. Furthermore, g has no isolated poles (n > 1), and, for each number M > 0 there exists a neighborhood NM of z0 in which jgj > M. Corollary 3. A point of indeterminacy is a limit point of zeroes and poles of g. Exercise. A point of indeterminacy of g is a limit point of zeroes of g - a, where a is any complex number. Poincar~'s Problem 1) Weak form: Given a domain D, is every function meromorphic in D a quotient of two functions holomorphic in D? 2) Strong. form: Given a domain D, g meromorphic in D, is g the quotient of two functions holomorphic in D and coprime at every point?

§4.

Removable singularities

In this section we shall state three theorems. The / first, Rado's theorem, facilitates the proof of the first theorem on removable singularities which is a direct generalization of the Riemann theorem 1n one complex variable.

51 A second theorem on removable singularities will be stated but not proven. Theorem 13 (Rad6) • Let f( zi, ••• , z .) be continuous in Dopen C. Cn and holomorphic in (D- {zlf(z) = 0""$). Then t is holomorphic in D. Proof (Heinz). A function of n variables is holomorphic if and only if it is holomorphic in each variable separately. It is sufficient to prove the theorem for fUnctions of one variable. If D 1 is any open disc whose closure is contained in D c: ~n, we must prove that f(z) is holomorphio in D1 • Without loss of generality, assume that D' is the unit disc (lzl, it is bounded there, and we may assume lf(z) I < 1 in (D' v ~ By hypothesis, f(z) is holomorphic in (D'-6). Construct a complex-valued harmonic function g(z) in D1 such that g(z) = f(z) on Then, consider the following functions for z E (D 1-6) and a > 0

r

r )

r-.

~l(z)

=

~2(z)

= Re [f(z)-g(z)] - a log I f(z) I

~7\(z)

=

Im [f(z)-g(z)] +a log If(z) I

=

Im [f(z)-g(z)] - a log If( z) I

Re [ f(z)-g(z)] +a log If(z) I

~

~4(z)

•

Note that for Z E (D 1 -6), a log .I f( z) I < 0. Now, as and points where z -> o(D'-6), which consists of r = o, either (i) Z -> Z 0 E (o(D 1-6) () and then [f(z)-g(z)]-> 0 or (ii} z -> z0 E (o(D 1 -6) ~ 6 ) and then [f(z)-g(z)] remains bounded and log lr(z)l ->- oo • In either event, ~ 1 (z} and ~ 3 (z) -> negative numbers while ~ 2 (z) and ~ 4 (z) -> positive numbers. But, since the ~ 1 are harmonic in (D 1 -6), they assume both their maximum and their minimum

r

r}

52 on

~(D 1 -6).

~ 2 (z)

Let

a

Hence, ~ 1 (z) and , 3 (z) and ~ 4 (z) are positive for all -> 0, then

are negative, and z E cl(Dt-6).

Re [ f ( z ) -g ( z ) ]

which implies Re [ ·f ( z) -g ( z) 1 ~ 0

~2(z) -> Re [f(z)-g(z)]

which implies Re [f(z)-g(z)]! 0

~ 3 (z) ->

Im [f(z)-g(z)]

which implies Im [f(z)-g(z)]

~4(z) ->

Im [f(z)-g(z)]

which implies

~l(z)

->

~

0 Im [f(z)-g(z)]! 0

for all z E cl(D'-6). Therefore f(z) = g(z) for all z E (cl(D'-6) Since d6 consists or points of and points of cl(D 1-6), f(z) = g(z) for z E ~6. In any component of the interior of A, f e 0 and thus g e 0. Therefore f(z) = g{z) in (D' U which means that f is harmonic in D'. Thus f has continuous first order partial derivatives. f satisfies the Cauchy-Riemann equations in D-6, hence by continuity, on o6 and on In the interior of 6, f = 0 and therefore satisfies the equations in 6. Hence f satisfies the Cauchy-Riemann equations in D1 and is therefore holomorphic in D1 • Theorem 14. Let Dopen C: ~n, and let g ~ 0 be holomorphic in D. Let f be holomorphio and bQunded in D - fzlg(z)= Then f is holomorphic in D. Proof. Consider the function if g .,. 0 gf

ur>.

r

r->,

r-.

oJ .

h

=

i

if g

0

=

0

Then h is continuous as f is bounded, and holomorphic where it is not zero. By Rad6's theorem, h is holomorphic. But g is holomorphic by assumption; hence f

is meromorphic. But f thus holomorphic.

=

~

is bounded,

thus without poles,

53 Theorem 15. Let Dopen C On; g,h holomorphic in D, not identically zero and relatively prime at every point or D. Let r be holamorphic in the set D- fzjg(z) = h(z) = Then f is holomorphic in D. Theorem 16. Let Ddoma1n c on and 1et g F 0 be holomorphic in D. Then (D - {zlg(z) = oj) is connected. Proof. Let S = (D zlg(z) = Oj). Suppose that S is not connected, then S = U VV where U and V are open, disjoint sets. Define a function h =l5 i~ ~ h is holomorphic where g ~ 0, and is bounded; therefore h is holomorphic in D. This is impossible since it implies that h is identically 1 in D.

oJ.

-l

.

§s.

Complex manifolds

Remark. From now on, "differentiable" means "C 00 11 • Definition 30. X is a (differentiable) manifold of real complex (com~lex ) dimension r, if the following conditions are satisfied: (1) X is a Hausdorff space. (2) Given an open set in X and a function defined in it, it is possible to say whether or not this function is f1fferentiable) holcmorphic • (3) There exist coordinates: every point in X has a neighborhood where r (real, differentiable) functions complex, holomorphic are defined such that they give a homeomorphism of this neighborhood onto a domain in ( ~~), and every function defined in this neighhorhood is (differentiable) if holomorphic and only if it is (differentiable as a function of x1 , •.• ,~) holomorphic in each variable of z1 , ••• ,zr • The coordinates are called local coordinates and such a neighborhood is called a coordinate patch. (4) There is a countable basis for the open sets of X, i.e. X is second countable. Remarks. On a complex manifold we may talk about holomorphic and meromorphic functions; on a differentiable

54

manifold, about differentiable functions. A connected 1-dimensional complex manifold is called a Riemann surface. Every c6mplex manifold is a differentiable manifold. Therefore it is natural to ask on which differentiable ~hifolds we can introduce the concept of holomorphic functions: that is, which differentiable manifolds can be given a complex structure. Necessary conditions are that the differentiable manifold be orientable and of even dimension, r = 2n. If n = 1, these conditions are also sufficient; however, if n > 1, they are not. In fact, in the latter case, necessary and sufficient conditions are not known. There are other differences between the cases n = 1 and n > 1: (i) When n = 1, axiom (4) in definition 30 is unnecessary as it follows from axioms (1), (2), and (3). When n > 1, axiom (4) is essential. (ii) If n = 1 and X is compact, there exist nonconstant meromorphic functions on X (i.e. on every closed Riemann surface there exist non-constant meromorphic functions). However, when n > 1, there are compact complex manifolds having no non-constant meromorphic functions. (iii) If n = 1 and X is not compact, there exist non-constant holomorphic functions (i.e. on every open Riemann surface), while if n > 1 this is not necessarily so. For example, let Y be a compact connected complex manifold of dimension n > 1, then X= (Y-fpj) is not compact and if there existed a non-constant holomorphic function on X, it would be holomorphic also at p (by Hartogs 1 Theorem), and it would be constant (by the maximum modulus theorem). Examples of Complex Manifolds. 1) vn 2) Any open subset of a complex manifold. 3) If X and Y are complex manifolds then X K Y is

55 a complex manifold, where a function r is holomorphic in · X~ Y if it is holomorphic in X and holomorphic in Y. 4) The complex projective space Pn(IJ)

=

t [(z

0

,z1 , ••• ,zn)] lzi are not all zeroJ·,

where [(z 0 , ••• ,z )] denotes the equivalence class of points (z 9, ••• ,z9l E tn+~, where two points (t0 , ••• ,Cnl and CC0 , ••• ,Cn) are equivale~t if and only if there is a t ~ o such that Cj = tt 3, j = O,l, ••• ,n, with local coordinates in a neighborhood of (z0 , ••• ,zn) where zi ~ 0 for some i, 0 ~ i ~ n, being z0/zi' ••• ,zi_1/zi, zi+l/zi, ••• , zn/zi, is a complex manifold. On this manifold there exist meromorphic functions; the ratio of two homogeneous polynomials of the same degree is such. 5) The special case of 4, P1 (t) =Riemann sphere=

[~e

U{oo3) •

6) Starting with a complex manifold of dimension n

1, omitting a single point, and imbedding P 1 (~), we will obtain a new complex manifold. This procedure is known as the cr-process. We do this for n = 2, starting with t 2 • First we define the space X to consist or two types of points: (z 1 ,z 2) ~ (O,o)j I = (Czl,z2) I (zl,z2) E c2 and (C 1 ,C 2 l -1 (O,O)J. II = { [( cl' C2)] I (cl' C2)E o: 2 and

X

>

I V II • Secondly, we make X into a Hausdorff space by defining the following basis of open sets: A neighborhood of a point p E I shall be a neighborhood in the ordinary topology_ of ~2 such that its closure does not contain (0,0). A neighborhood of a point p E II, p = [(C 1 ,C 2 )], and say cl ~ o, shall be the set of points; [(l,C)] E II satisfying IC - c~cll < E, and (zl,z2) E I satisfying =

z1 # O, lz 1 1 < c , lz 2 1 < c and lz~z 1 - ~~~ 1 1 < E, for some c > 0. Thirdly, we define local coordinates •. Near a point of I, take z1 and z 2 as coordinates. Near a point [(~ 1 ,~ 2 )] ~f II, where say C2 ~ 0 and hence [(~l'C2)] = [(~ 1/~ 2 ,1)] = [(~ 0 ,1)] say, take as local coordinates t and T I

(l+t)T

zl =

~O

z2 = T ~1

(l+t)~ 0

=

~2 = 1

•

On this new manifold X, the following holds: every holomorphic function on X is constant on II. For if f is holomorphic on X, it is holomorphic on X-P 1 (~) = c2-{o]. By Hartogs 1 Theorem, f is also holomorphic at the origin {0,0}. Therefore f approaches some complex number, a, as its argument goes to the origin. Hence near every point of P 1 (~), the value of f is close to a. Thus f =a on every point of P 1 (~), but then f must be identically a on II. 7) A globally presented, regularly imbedded analytic subvariety Y, of codimension r in an n-dimensional complex manifold X is defined as follows: Let f 1 , ••• ,fr be holomorphic functions defined on X, such that, for every x c X at which r1 (x) = ••• = fr(x}= 0, the rank of the Jacobian matrix J

=

ori "\ ( OZ"j)

is r, i.e. J is of maximal rank. The derivatives ofi/ozj are to be understood in the following way: Let N be a

57 coordinate patch containing x, define the local coordinates

and let

l

N -> ,n

Then Then Y,

as a subset or X,

is defined as:

Note that Y is closed in X, and that axioms 1, 2 and 4 are clearly satisfied. We now define the local coordinates in y. Let y E 1, with local coordinates j(y) =(z 1 (y), .•• ,zJy) defined in a neighborhood N of y, where N c.. X, Assume that, at y, det (ofi/ozj) = det A I o, J = l, ••• ,r by relabeling the zj if necessary. Define new coordinates ~1

==

fl(zl, ••• ,zn)

~r ==

fr(zl, ••• ,zn)

• • •

~n

zn • Then the transformation taking square rna trix; ==

z

to

A

is given by the

B

1

0 •

0 0

which is nonsingular as

~

det A I 0.

1

Then the local coordinates

58

are
y

-

59 Chapter 5· §1.

The Additive Cousin Problem

The Additive Problem, formulated

A. This "first" Cousin problem (there is a second one) is a direct generalization of the Mittag-Leffler problem in"one complex variable: : Given a domain D, a discrete set of points a e D, and polynomials Pv(z~ ), without constant te~, find a fUnction f, meromorphic in D, with singular part Pv at av. As we know, this problem is solvable for all domains D C: ~. The Additive Cousin Problem is as follows (we shall denote it by "Cousin ! 11 , or simply C.I in the sequel): C.I Let X be a complex manifold, and U = fui3, i e I be a given open covering of X, I some index set. Let meromorphic functions Fi defined in ui, be given, such that F1 -Fj is holomorphic in uinu 3• Find a function F, meromorphic and defined on X, such that F-Fi is holomorphic 1n ui. This problem is not always solvable, as seen from the following theorems: Theorem 17. Extension Theorem (Oka). Let X be a complex manifold such that the Cousin problem is always solvable. Let Y be a globally present~d regularly tmbedded analytic hypersurface of codimension 1. Then extension from Y is always possible, i.e. given ~~ holomorphic on Y, there exists j, holomorphic on X, such that 1 on.Y. Assuming this theorem for the moment, we exhibit the following application: Theorem 18 (Cartan). _Let X c e2, open, such that the Cousin problem is always solvable 1n X. Then X is a region of holomorphy. Proof. We may assume that X is a domain, that 0 e X, and that X~ e2 , as we already know that every en is a domain of holomorphy. Hence, assume b e bdry X; b = (b 1 ,b 2 )

'=

60 where b1 ,b 2 are fixed complex numbers. Then if f(z) ~ b 2z1-b 1 z 2 o f(z) = 0 is an analytic plane through 0 and b o Let Y = [z If( z) = 0j o Now Y1 = Y fl_X is an open set in C; hence Y1 is a region of holomo~hyo Therefore there exists a function ~~ holomorphic in Y1 and singular at b. But by the extension theorem, there exists a }, holomorphic in X such that l = ~ on Y1 , and l is singular at b. B. Proof of the Extension Theorem Let Y = tf = 0 and let ~ be holomorphic in Yo There exists a covering U = Luii E X of Y such that Y c. and in each u1 there exists a li such that li is holomorphic in ui and equal to ~ on Y 11 ui, by definition of holomorphicity on closed sets (see proof of Lemma p. 74). Let u0 =X- Y, an open set in X. Define: Fi = } 1/f , i .f 0

J;

'fl1 ;

F0

=

1

This covering and set of associated functions defines a Cousin problem, for, on u0 ~ ui, F0 -F1 is holom~rphic. Indeed, on ui n u 3, Fi-Fj is holomorphic except possibly tor points on Y. Hence, assume f(z 0 ) = O, z0 E ui ll uj. Introduce local coordinates (~ 1 , .. ,,~n) such that f = c1 • But now, li(o,c2, •• o,~n)- 1Jco,c2, ••• ,~n) = 0 ' as li and

lj

agree on Y,

and . . 1i(~l, ••• ,Cn) - ]Jcc1, ... ,~n)

Fi-Fj =

(1

.

But in the power series expansion of }i~J' only terms containing powers of C1 appear; hence Fi-Fj is holomorphic in ui (\ u 3• By hypothesis, there exists F, meromQrphic in X, such that gi = F-Fi is holomorphic in ui. We clatm that l = fF is holomorphic in X and equal to ~ on Y.

61 Clearly, 1 is holomorphic on u0 = X - Y. Consider F o~ u1 , 1 J. O· r ' li F = gi + T , and gi is holomorphic. -

Hence,

l =

fgi + li in ui,

and on Y. I) u1 ,

!

= '

•

Reformulation of the Cousin Problem C.I I. (Cousin problem belonging to the covering U). Given X, an analytic manifold of dimension n, covering u = {ui) and holomorphic fUnctions fij defined in ui ~ uj satisfying (antisymmetry) fJi = .. riJ riJ + rJk + fki = find holomorphic

fi,

o

defined in ui,

(compatibility), such that

fiJ = fi-fJ. Claim. C.I' implies C.I. Proof. Assume C.I is given, and let fij = Fi-Fj, where the Fi are meromorphic functions defined in ui. Then the fij are holomorphic and satisfy the symmetry and compatibility conditions. Let fi be the solution functions of C.I 1 , and defi~e F = Fi-fi in ui. Then F is globally well-defined, for in ui ~ uj Fi - FJ =

fiJ = fi - fj ,

hence and F ·solves C.I. Induced Cousin Problem. Let U = ~ u 3), j £ J ; V ={vi~' i £ I be two coverings of X, and assume that V is a refinement of U (i.e. every vi is contained in some u3), and let a Cousin problem belonging to the covering U be given. We induce a Cousin problem belonging to V as follows:

62 Let rr be an "affinity" function fr_om I to J such that vi C. UO'""~ i) • In Vi /) vj, assign fij = f
-

viC- ucr (i) " gij = fo-(i)o- (j) Let gi be the solution functions. F

Define

Fa-( i) - gi

=

•

in vi •

Then F is globally well defined and solves C.I, for

Furthermore, F-Fi is holomorphic 1n ui, for, let x e u1 . Then there exists a v3 containing x, and consider: F-F1 = F-F0"' ( j) + F<J ( j) - Fi =

-

gj + fo-(j)i

..

defined and holomorphic in v J (\ uo-( j) (\ u 1 = v j l) ui, and x E vj ll u i . Remark. We may now consider the Cousin problem for locally finite coverings only (i.e., for coverings such that

every point of X has a neighborhood which intersects only a finite subcollection of the cover),'by virtue of the following observations: i) (Paracompactness). In any ~nifold every covering has a locally finite refinement by open sets each relatively compa~ in some open set of the origin covering and some coordinate patch (see de Rham, Varietes Differentiables). 11) C.I" implies C.I.

§3.

Reduction of the Cousin Problem to non-homogeneous Cauchy-Riemann equations

Intermediate Problem. Given a complex manifold X, locally finite open covering U = {ui1 , and holomorphic functions fij defined in ui ~ uj, antisymmetric and satisfying the compatibility condition, find functions gi, defined in ui, such that fij = g 1 - gj where the gi e CCD • Propos~tion 1. The Intermediate Problem is always solvable. The proof of this proposition will be presented subsequently (p. 64). Let the functions av, v = l, ••• ,n be defined on X as follows: osJ v = 1, ••. ,n. av = in uj ;

A.

()zv

We claim the

av

are globally well defined, i.e.,

But this is clear, for gj-gi = fji and fji is holomorphic there.- Furthermore, the av satisfy the following compatibility ~ondition: ()av

ozJ.L and this is clear.

=

oaJ.l

ozv

Now \'le can state the final form of C.I:

64 Final Problem. Find a function A e C
Oav

-

~aiJ.

=-

oziJ.

;

IJ,,v

= 1, •. • ,n

ozv

find a function A, defined on X, such that oA/ozv = av' v = l,, .. ,n. Note. These differential equations are known as non-homogeneous Cauchy-Riemann equations. Proposition 2. The final problem implies C.I. Proof. Assume there exists a function A, defined as above. Set ri = gi-A" defined 1n ui, where the gi e c00 are given by Proposition 1. Note that

-

afi

ogi

oA

ozv

azv

~zv

----- = .....- ~ ---- =

0

;

i.e. the ri are holomorph1c. But fi-fj = g1-gj = fij. Definition 31. Let X be a differentiable manifold, U = ~ui\ a locally finite open covering. Then a partition of unity subordinated to the coverir,lg U is a system of functions w3, defined on X, positive and C00 , such that ~

wj

=0

wj

= 1 at each point of X •

on X - uj

Proposition 2· Given any manifold X and locally finite open covering u, there exists a partition of unity subordinated to the covering u. Proof. Note that, if V is a refinement of U, and if there exists a partition wj for V, we may define a partition wj of U as follows: Let ~ be an affinity function, as before ( v1 c. ucr" ( i) ) • Now set

5wi

wcr(i)

=(

0

on v1 otherwise

•

For those uj not as yet included, define mj = o. Let U be given. By Remark i), p. 63, U has a locally finite refinement V = ivi) such that vi c:c: some coordinate patch pi" Since =x is paracompact and Hausdorff it 1s normal, so that there is a locally finite '{ open covering V' = .}• z_vis such that vi c.cvi. For eac~ i, 1 let fi be a diffeomorphism of Pi into lRn. Let si = f(v 1 ) and si = f(vi); then s~opencc s~pen. In I'Rn there are C00 t functions t/Ji : IRn -> [ 0, 1] satisfying t/Ji 1 on cl si and= 0 outside s 1 • Let t 1 • f = 91, and set

'

=

IDi = Qi/

B.

2::

Qi,

i

Proof of Proposition 1. Let mj be a fixed partition of unity subordinate to U = [u~ • Set gj = L wi fij in uj, where this sum is understood a~ foll~ for x e uj, wi(x) = 0 unless x e ui (\ u J. When wi (x) = 0, set wifiJ = 0. When wi(x) -1 o, fiJ is defined. Note that wi(x) = 0 for all but a finite number of indices i. The gj solve the intermediate problem, for gi - gj

= ~ (l)k fik- ~ ~ fjk = ~ (l)k (fik + fkj)

= ~ (l)k (fij) = fij • Hence, we have reduced the Additive Cousin problem to an existence theorem for the nonhomogeneous Cauchy-Riemann equations. We shall exhibit a solu~ion for a polydisc shortly. Example. Let X be a simply connected differentiable manifold, U =lui~ a locally finite open covering. We pose a "Cousin Problem" as follows: To each ui n uJ let there be assigned a complex

66

number fiJ such that fiJ = - rJi and fij + rJk + fki = o. Then, find constants r 1 such that r 1J = ri-rJ. Let g 1 be the solutions of the intermediate problem; . m r ij = g1 -g J; where g 1 is defined 1n u1 and g.1 e C • Define: v = 1, ••• ,n. Then the av are defined globally, as before, and ~av/~ = ~a~/~xv. The final problem becomes: Find a function A such that ~A - a I v = 1 I • • •J n " ~ oxv v Such a fUnction exists, by Stokes• theorem. Set fi

=

gi - A •

The ri are constants, as ~r1/~xv = O, v = l, ••• ,n. Exercise. Prove the converse of the above example, in the following form: Let X be a domain 1n tan. Then C. I solvable implies that every curl-tree vector field is a gradient. Theorem 19. Let D = {Cz 1 , ••• ,zn) I lzjl < RJ ~ ooj. If aj(z1, ••• ,zn) are defined and c00 in D, j = l, ••• ,n, and satisfy ~j/ozk = oakf~zj then there is a c00 function ~ in D such that ~~/~zj = aj for j = l, ••• ,n. Proof. 1. Let n0 = z1 , ••• , zn) I IzJ I < r J < oo}~ D. Then the aj are defined and c00 in a neighborhood_.of n0 • We claim that there exists a ' defined and C00 in 'perhaps a· smaller neighborhood of 0 satisfying ~'/ozJ = aJ, j = l, ••• ,n. The proof is by induction on k, where we assume aj e 0 for j > k. For k = o, the problem is reduced to solving the system of homogeneous Cauchy-Riemann equations, o'/ozj = o, of which += 0 is a solution. Assume that the problem can be solved for k = t-1, and consider the case k = t. Choose e > 0 sufficiently small, and for C= ~ + i~ and (z1 , ••• ,zn) in an e1 neighborhood of fi0 , N(e 1 0 ), e1 < e, define

t(

n

,n

--1 .JJ 1f

m(z1 , ••• ,zn)

is then defined and C00

in all variables 1n

,n

N(e 1 0 ) and satisfies an/~zl = a1 and d»/~z 3 = o ror j > l, by the compatibility conditions ~al/~zj = ~a 3/~zl. The system (1)

..2!... =a 3, ~zJ

j = 1, ••• ,l;

~ = ai 3

0,

j

= l+l, ••• ,n;

is then equivalent to the system ( 2)

~ ( t-ro) = aj ~i 3

but since (2 1 )

an/~zl =

an ,

ai 3 al,

j =

1, ••• 'l;

O(!-ru) :

az 3

Q1

j

1

II :XI+ Ito o 1 nj

(2) is actually

~(f~) = aj - d» , j=l, ••• ,l-1; o(f~) =O,j=l, ••• ,n. az 3 ~z 3 ai 3

,o

Now (aj- an/ozj) e c00 in N(e 1 0 ) and satisfies the compatibility conditions since o/ozk(aJ- an/ozj) = oaj/ozk - o/ozk(an/ozj) = oakfozj - o/ozj(an/ozk) = o/ozj(ak- an/ozk). Hence, by our induction hypothesis, (2 1 ) has a ca:> solution and therefore ( 1) has a CCI)) solution in a neighborhood or D0 • 2. Construct open polydiscs n3 c... c. Dj+l whose union is D. 00 By l.,_there exist ~nctions +1 ,+ 2, ••• such that ~j e c on ~j and o+ 3/ozk = ak for k = l, ••• ,n and (z 1 , ••• ,zn) e n3 • Choose Ej > 0 such that ~ ej oo . oo since l~j+p-+JI < Ej + ej+l t, ... + ej+p < ~ Ek in Dj and for j sufficiently large ~ Ek is the ~!1 end of a convergent series. Moreover,~or k fixed, <+J-~k) is holomorphic on Dk for j > k since o/ozi(+j-~k) = 0 on Dk, i = l, ••• ,n. Thsrefore since the sequence {(+J-+k>j ,

-

68 j > k would converge uniformly to (~-~k), all derivatives would also converge, so that + would be c00 and would satisfy the differential equations . ~~/~zk = ak Ain D. So the next step is to construction new pj's, +j' which are also solutions or the differential equations and satisfy A "" A A l+j+l-~jl < Ej on Dj. Let ~l = 1 • Let ~ 2 = ~ 2 - h1 , where h1 is a polynomial. h1 must satisfy 1<+2-+1 >-h1 1<s 1 on n1 • But on fi1 , + 2 -~ 1 is holomorphic and hence has a power series representation which ~an be approximated as closely as desired by a polynomial. Let 3 = ~ 3 -h 2 , etc. Since the polynomials are holomorphic, ~fj/~zk = ~~j/~zk = ak, '-" A k = 1, ••• ,n, on Dj, and ....J+j+l-+JI < sj on Dj by construction. Hence lim = ·~· is a coo solution or o~/ozk = ak in D. j ->CD Note. If X and Y are homeomorphic manifolds and the Cousin Problem is solvable in X then it is solvable in Y. Thus by the previous theorem, the Cousin Problem is solvable in any domain which is the product or simply connected domains in 1.

+

+

+j

Chapter 6. §1.

Cohomology

Cohomology of a complex·manifold with holomorphic fUnctions as coefficients

Let X be a complex manifold and U = {ui)' i e I, be a fixed open covering of X. (We always assume that ui ~ u 3 for i ~ J.) · · · A. Definition 32. An r-cochain ~ on U is a rule which assigns to every ordered intersection of (r+l) sets, Ui Ui , a holomorphiC fUnction fi i (z) o -r o ••• r defined 1n this intersection such that 1. (a) when the ij are distinct and nul ~ ~, fi 0 ••• ir(z) is a function holomorphic in lluiJ• j (b) when the iJ are either nondistinct or fluij = ~, fi o••• i r (z) = o. 2. f (odd permutation of (i0 .,.ir))=- ~(i 0 ••• ir) f (even permutation of (i0 ••• ir) )= f {i0 • • .ir) • Examples. 1. A 0-cochain is a rule f which assigns to every ui e U a holomorphic function f 1 (z) defined in ui. 2. A 1-cochain is a rule f which assigns to every ordered intersection ui ~u 3 , a holomorphic function rij{z) defined in ui A u 3 such that fij{z) = 0 if i = j or u1 f\ u j = ~, f ij = - f j i. Cochains of the same dimension form an Abelian group under addition, cr = Cr{x,u,<9), where 0 refers to holomorphic functions. This group is also a vector space over e and a module over holomorphic functions on X. Definition 33. The coboundary operator, 51 is a linear mapping of cr into cr+l (and therefore a homomorphis• of the group Cr) given by r+l j .-.. r ( 5 10 ••• ir+l) = ~ { -1) .f (i 0 ••• i J ••• ir+l) for f e C , ~ "".1='0 where ij denotes the deletion or ij.

n••• {)

fH

70 8f is called an (r+l)-coboundary. There are no 0-dimensional coboundaries. The coboundaries form an Abel 1m group under addition, Br = Br (X,U, (9). This group is also a vector space and a module, as above. Definition 34. A cochain is qalled a cocycle when its coboundary is o, i.e. of= .0. Since 5 is a ltnear map, the sum and difference of two cocycles is a cocycle, and the cocycles form an Abelian group zr = zr(x,u,(9), a vector space, and a module. ~amples.

i}

For 0 = ( 6/) ( ij} = §(J) - f(i) implies that fi (z) = fj(z) on uil) uj, and hencethat the rk are restrictions or holomorphic functions on X. 2. z1 ={cousin dataJ. For (5§)(1Jk) = f(Jk)-f{ik)+tfij) = 0 implies, since J(ij) =- }(Ji), that }(ij)+ }(Jk)+5(k1) = 0. Therefore we have f 1 j{z) defined in u1 A uj satisfying f 1 j(z) + rjk(~) + fki(z) = 0 and fij(z) = - fj 1 (z). Corollary. 6 = o. Proof. A typical term of {655-) (1 0 ••• ir+l) is A A a= j-(1 0 ••• ij ••• ik ••• ir+l). We must show that this term appears with zero coefficient. If we first delete k and then j, the coefficient of a is (-l)k(-l)J, But this term is also obtained by first deleting J and then k, in which case the coefficient of a is (-l)j(-l)k-l (since J < k). Hence the coefficient of a in 55J is (-l)k(-l)j + (-l)j(-l)k-l = o. By the corollary, every coboundary is a cocycle. Hence Br c. zr c cr. zr/Br is called the r th cohomology group = Hr(x,u,c0). Hr is also a vector space over ~ and a module over holomorphic functions on X. H0 = Z0 = {holomorphic functions on Xj. If H1 = 0, then every cocycle is a coboundary which means that for every r 1 j(z) holomorphic in ui ~uj, 1.

Z0 = tholomorphic functions on

71

=-

rij(z) fji(z) and fij(z) + fjk(z) + fki(z) = 0, there exist fUnctions fi(z) and fj(z), holomorphic in ui a~d uj respectively, such that fij(z) = fj(z) - fi(z). In other words, . H1 = 0 implies that every Cousin Problem is solvable (for the covering U). B. Let V = {vj~ , j E J, be a refinement of the covering U. Then ever! vj is contained in some ui E U, i e I. Let Ol(j) = j be an affinity function which assigns to each v . one ui which ue call u such that v . c u • J J~ J J We assign to every r-cochain ~- in U an r-cochaln r . * r in V as follows: (Call this mapping o- ). Corresponding to every non-empty ordered intersection o~ (r+l) sets, vji' there is an orde:ed intersection of sets, £5o uji' a~d hence an r-cochain ~, defined on

1:b

f)

u:ri .

Assign to

f

its restriction on /}vji'

and call

it f. Then the holomorphic function which__ _-i- assigns to (l vj is the holomorphic function which .f assigns to (I uJ i restricted to nvj • i

i

/) o- * : Cr (X,U,v)

Properties. · 1. c:J is a homomorphism of the group Cr (X, U, D) . 2. ocr * = cr * o. For if J- e cr (X, u,J ) then

*

*~

.

.

r+l k _ "" (ocr })(Jo···Jr+l) = (o~)(Jo···Jr+l) = ~=O (-l) s<Jo··Jk ... Jr+l) and . * ~ . * r+l k _ ~ (a- 0t ) ( J o• ••j r+1 ) = <J ( -l ) J (j 0 • • .j lc • • • Jr+1 )

t; (

=

r+l tho r+l

=

fe

'=O

k (-1)

* _ "' cr f (Jo···Jk···Jr+l) (by

1.)

k ,. "' (-l) .J-(Jo···Jk···Jr+l) •

:;. (a) a-*: zr(x,u,{)) -> zrcx,v,O). For if zr(X,U,t)) then of= o implies that o = o-~~ a f=ocr*f=6~.

72 Br (X,V, D). For if - e Br(x,u,C!>) then f= - 5F implies that rr * f= - r5' * _oF= - 6F• Note. ~ * depends on the choice of the affinity function ~ -Lemma. If r5' and 1' are two affinity functions and a- * rand 1' are their corresponding mappings of Cr (X;U, 0) -> Cr(X,V,c?), respectively, then there exists a linear mapping, r-1 J1 · r r Q: C (x,u,c?) -> c (X,V,tl), such that for f e C (X,U,~),

,,., (b ) cr * : Br (X,U,LI)

f

f-

->

Q5f+ o~f. Proof. Define Q as follows: r-1 (Q t) (io •• ·~-1) = ~l) (-l)k 5-<<1' (io) ••• d' (ik-l)cf (ik)-r(ik)

-r*

cr-*

.f=

••• -r(ir-1)) Now

r

5(Qf) (i 0 ••• ir) = ~ (-1)

r k-1 =~ (-1) k{ f=o r-l

+f=k =

r

~

(-1)

1=O ( k-1

r --s>r l(;U

l

<-1)

t(6 (1 0 )

-1)

l'=k+l

k+l -

(-1)

f

k

~

(of) (i0

2 -

~

••• ik ••• ir)

f< cr' ( i 0 ) •• • <J

1\

••• cr(ik)

A

(i.e )-r( i.e> .. -r( ;tk) •• -r( 1rH

~l ••• <1'(i2+1 )-r(i.e+l) ••• -r(ir))J A

(if ( i 0 ) ••• if' (i.e ) -r (i.e ) •• ,. ( ik) •• -r ( ir))

k+l' A f(o'(1 0 ) •• 6'(ik) •• 6(i21 )-r(i2 ,) •• -r(ir))

73

+A

r

-

- ~ .f{
-

= j-(-r(10 )

••• -r(ir))

-

- j(cr (1 0 ) •• • <1" (ir))

and it is clear that the remaining terms cancel, when one r R:-1 r writes the sums ~ > , for example, as L_ •

*

* Y=0

li, E=O; .t
J;;<5

Hence, -r - C1' = Qo + 5Q, as claimed. _Corollary. If o} = o, E Cr(x,u,~), then ** - is a cocycle on U then 1' J - C1' f = 50]. Hence if f -r*J- and r:r *-~ are cohomologous cocycles on V. Definition 35. If u1 and u 2 are open coverings of X, and if .J- 1 E If(x,u1,&), (i.e • .}.1 is a cohomology class ftf r-cocycles) and 52 E Hr(x,u 2,t9) then :f 1 and :f 2 are equivale~t if they induce the same cohomology class of r-cocycles on some common refinement of u1 and u 2 • Let [f 1 ] denote the equivalence class of :f 1 , ;f 1 is a representative of [~ 1 ]. Define addition of equivalence classes as follows:

J

rtlJ + £J2J

=

fgl + g2J

where t 1 indu_ces g1 and .f 2 induces g2 in some common refinement of u1 and u2 • It must be checked that this definition is independent of the choice of representatives chosen. (Exercise for the reader). Then these equivalence classes of cohomology classes of r-cocycles from an Abelian group under addition, ~(X,£9), called the r th cohomology

74 group of the manifold X. (Hr(X,~) is independent of the choice or· covering or X). ~(X,~)= lim ~(x,u,c9)~ An element or lfCx,{9) can be repres~nt~d by an r-cocycle on a covering of x. If ex,(}) = 0 means that any r-cocycle on a covering U of X induces a coboundary on same refinement or U. Note that f -> [ fl is a homomorphism of arcx,u,~) into arcx.~). H0 (X,O) = [ holomot'phic functions on

Xl•

·If H1 (x,C') = 0 then every Cousin Problem is solvable (cr. 2nd reformulation of Cousin Problem). Holomorphically equivalent manifolds have the same cohomology groups.

§ 2. Applica tiona Theorem 20. Let X be a complex manifold and Y a globally presented, regularly imbedded hypersurface, Y = z1,... • , zn) I cl> ( z1, ••• , zn) = where c1> is a holomorphic function on x. If for some r > o, Hr+l(x, 0) = Hr(x,l!J) = 0, then If (Y, t9 ) = 0 • Proof. We say that a covering U' = ~ ui' l of X is sufficiently fine if tor every (fi~te ui)() Y a fUnction holomorphic in this intersection can be continued to

f(

oJ

,~..

the fiQte u~ • Lemma. There exists a covering u0 of X such that every refinement or u0 is sufficiently fine. Proof. Let y e Y, then in some neighborhood of y, N(y), there are local coordinates t 1 , ••• ,Cn such that Y is given -by (n = 0. Any function holomorphic in N(y) 11 Y is a function of cl, ••• ,(n-1 and hence is holomorphic as a function of C1 , ••• ,(n' i.e. in N(y). Clearly if N1 is any open set contained in N(y) any function holamorph1c in N1 f\ Y can be holomorphically continued to N1 • Take this neighborhood system and add open sets not intersecting Y

75 to form an open covering of X. This covering has the desi~ed property, i.e. is u0 • From now on we consider only such coverings. Now take such a covering U = {ui) of X. We must show that every r-cocycle on Y induces an r-coboundary on some refinement of U. Let .f Y be an r-cocycle on Y, then fY assigns to r every ( j':b ui j) f\ Y a holomorphic function f io ... 1z, (z) de tined in this intersection. Since U is sufficiently fine, fi i ( z) can be continued to ui • Define :f = 0 in o•• r j=O j 1\ u1 when ( (\ u1 ) ('\ Y = ~. There fore, there is an r-cochain ;.x 3on X such tgat Jx = JY on Y. Then ofX = 0 on Y which means that o&x = fg where g is an (r+1)-cochain on X, But 0 = oo+x = +og and hence 6g = 0; g is a cocycle on X. Since If+1 (x,t9) = o, g must induce an (r+l)-coboundary on some refinement ul of u. So consider everything above in this refinement u1 • Then g = 5h where his an r-cochain on X. Therefore 6(tx-+h) = 5~x-+5h = 6~x-+g = 0, and hence tx-+h is an r-cocycle. Since ~(X,6)) = 0, ~x~~h induces an r-coboundary on some refinement u2 of u1 • Considering the above in this refinement u2 , fx-~h = 6F where F is an (r-1)-cochain on x. Hence on Y, 5-x-+h = S..x = ~Y = oF, and since F is an (r-1)-cochain on X, it is an (r-1)-cochain on Y. Note. The following theorem gives sufficient conditions for a domain to be a domain of ho1omorphy. Later on we will prove that these conditions are also necessary. Theorem 21. Let Dopenc<:n, n > 2. If I{(n,CJ) = 0 for 1 ~ r ~ n-1, then D is a region of holomorphy. Proof. Use induction on n. The case n =_2 follows from Theorem 18. Assume that every open set in ~n-l for which If = 0, 1 ~ r ~ n-2, is a region of holomorphy. Rather than working with a component of D, assume that D is a domain. Then let Ddom<:~n and let b E boundary of D. Pass a hyperplane P through b and an interior point of D and set PnD = Y. By Theorem 20, all the

f\

cohomology groups of Y from H1 to Hn- 2 are 0. By our induction hypothesis then, Y is a domain of holomorphy. Hence there is a function g holomorphic on Y and singular at b. Since H1 (n,t/) =.0, the Cousin Problem is solvable in D, and hence g can be continued hoiomorphically to D, and the extended function will be singular at b.

§ 3. Other Cohomoloe;ies A. 1. Had we defined an r-cochain to assign a C00 function, instead of a holomorphic function, to intersections, then we would have gotten Hr(X,C 00 ), where X could be a real manifold. 2. Had we defined an r-cochain to assign a constant function, i.e. complex number, to intersections, then we would have gotten ~(X,C). Here X is any topological space. 3· Had we defined an r-cochain to assign an integer to· intersections, we would have gotten ~(X,~), the Integral Cohomology group of the topological space X. 4. In fact, given any Abelian group we could have in which defined an r-cochain to assign an element of case we would get Hr (X, 1-). B. Definition 36. Let S be a topological space, X a Hausdorff space, and p a mapping of S onto X (called the projection mapping). Denote p- 1 (x) by Sx• called a stalk, and note that SX <: S , · XE l)X Sx = s. The triple (p,S,X) is called a sheaf of Abelian groups over X if a) p is continuous, and for each x e: X and each s e: Sx, p is a homeomorphism of a neighb9rhood of s in S onto a neighborhood of x. b) Each stalk Sx is an Abelian group such that: s -> -s is a continuous mapping of S into S; and (s,t) -> s +t , defined on the set R of pairs (s,t} such that s,t belong to the same stalk, is a continuous mapping of the subset R of S Y S into S.

r-,

r-,

77 For simplicity, we call s the shear. Definition 37. Let Uopen C. X. A section (or cross section) of S over U is a continuous map of u into S such that p•f is the.identity mapping. Denote by r-cu> the additive group of all sections of S over U. Examples. 1. Let X be arbitrary, and let Sx = ~ for all x E X. Let S = l)sx have the discrete topology. Then a section of S over uopenC: X assigns some integer to u. 2. Let X be arbitrary, and let' Sx = G, any Abelian group, for all x E X. Let S = LISx have the discrete topology. A section of UopenC: X is an assignment of an element of G to u. As in example 1, all stalks are isomorphic. Such sheaves are called constant sheaves. 3. Let X be a complex manifold, and Sx = ~~ the set of germs of holomorphic functions. Definition 38. Let x E X, a complex manifold, and consider holomorphic functions at x, each defined in some neighborhood of x. We say that two such functions are equivalent if they coincide on some sufficiently small neighborhood of x. This is clearly an equivalence relation. The set of all holomorphic functions as above, modulo this equivalence relation, is called the set of germs of holomorphic functions at x; which we have denoted by c?x. The set of all germs form a group over each point x. Introduce a topology in the set of all germs S = Usx as follows: Take any element f E S; then f E Sx0 , and is an equivalence class of functions defined in a neighborhood of x0 E X. Take a representative g E Sx ; g is defined in Nx . , a neighborhood of x0 E X. ¥hen for 0 each y E Nx , assign that class in S containing the 0 y , direct analytic continuation of g, say {. gy). Then l) [g ~ is, by definition, an open set; and these are

f

yeN Xo

y

78 to form a basis for the topology of S. Each section over U ls a holomorphic function defined in U. This sheaf is called the sheaf of germs of holomorphic functions. 4. Let X be a differentiable manifold and. Sx = C~, where C~ denotes the set of germs of C00 functions at x, · defined similarly to r.9 x· Then S = Uc~ is made into a sheaf as in example 3 above. With the aid of the concept of a sheaf, we may now define the cohomology groups in a more general setting. Let X be a paracompact space with a sheaf S over it, and U = ~ui) , i e I, an open covering of x. Define the cochains cr(x,u,s) on X associated with the covering u, with coefficients in S, as follows: f e Cr(X,U,S) assigns to each ordered intersection, ui 1'\ • • • 11ui , a section of S over this intersection so t~t S is aKtisymmetric 1n the indices. Note that we can add cochains, and talk of antis~~etry conditions, for we can add their values using the group structure of l-(u 0 fl u1 1"\ ••• llur). Continuing the construction as before, we obtain Hr{X,U,S), and then form the projective (direct) limit, Hr(X,S). Theorem 22. Hr(X,C 00 ) = 0 for all r > 0 and any differentiable manifold X. In fact ~{X,U,C 00 ), r > 0 is already trivial, for every locally finite covering U. Proof. We define a homomorphism Q : cr(x,u,c 00 ) -> cr-l(x,u,c 00 ) r > o, so that f = Qof + oQf, and this is sufficient. Let i, wi} be a partition of unity subordinate to U (see Definition 31 and Proposition 3, 3, Chapter V). Define

= ~ mif(i1 0 ••• ir_1 ) , where this sum is understood as follows: wi f = 0 if f = 0 Q5(i 0 ••• ir-l)

or if wi = 0. Note that the local finiteness of U insures that almost all terms of this sum vanish at any point of X. Now

Note that this proof hinges upon the fact that Cr(X,U,C 00 ) is a module over globally defined C00 functions.

80 Chapter 7. Differential Forms §1. Rins of differential forms in a domain A. Definition 39. Let D <: an, form in D is a formula sum ~

· 'F-a

A differential L ai i dxi 1\ ••• i 1 <1 2<...
open.

1\ dxi

(the symbol 11 /\" is read as "wedge"), where the ai1 ••. rir are functions defined in D, possibly zero (in which case the term dxi I\ .•• .1\dxi may be omitted from the sum). 1 r Identify l•dxi A ••• 1\ dxi and dxi 1\ ••• A. dxi • Note that ihe collectioft of differ~ntial forms ln D constitute a module (Abelian group written additively), closed under multiplication by functions in D, of dimension 2n. The addition is the natural one. We denote this module by R0 • We shall say that a monomial ai •• •i dxi A ••• 1\ dxi 1 r 1 r is of degree r of r-dimensional if it is the sum of monomials or degree at most r, and ~ r-dimensional if it is a sum or monomials or degree precisely r. Observe that any form in D may be written uniquely as a sum of pure r-dimensional forms, 0 ~ r ~ n. We now introduce a multiplication "*" in RD' giving R0 a ring structure, as follows: . j dxj 1\· ••1\dxj)· Define (a 1 i dx1 1\ ..• 1\ dx1 )*(bj 1··· r .,..

_ ~0

1

r

1··· k

1

if ~ = JL for some ·K,L

- J eai .i bj .. j dxq '1 •• r 1 • k l

A .••

Ad~qk+r

otherwise

where each qk is an iK or a jL; qk < qk+l and E is the sign of the permutation ~: (i1 ••• 1r J1 ••• jk)->(ql ••• qk+r). Define "*" on all R0 by postulating that the distributive law holds. The multiplication "*" is clearly associative; hence RD is a ring. We now denote this multiplication by " A", for obvious reasons. (R0 is also called the ring of exterior differential forms).

.

81 We note that RD may also be defined as follows: is a ring ~'ith operators ¢D = ~functions defined in ·generated by the symbols dx1 , ••• ,dxn, where l•dxi

nJ ,

dxi

=

dxi A dxj = - dxj 1\dxi We remark that, if

RD

•

=

are pure dimensional

a,~

aAf3= • Remark. It is to be understood in the sequel that all functions are c00 • B. We now introduce a cohomology structure to R0 , called the d-cohomology, or de Rham cohomology. Define the ring homomorphism d:RD -> Rn as follows: On zero forms, i.e., on functions defined in D, set

df

n

df = ~ axj dxj

.,

and on monomials: d(adxj 1

A. ••• Adxjr) =

( da) 1\ (dx j 1 1\ • • • 1\ dx j r) •

Extend d to RD linearly. Note that d is not an operator-homomorphism, i.e. d(fa) ~ fda , where f is a function on D and a E RD. Example (1). Let D c JR . . . , and observe that da on a zero form acts like a "gradient," on a pure 1-form like "curl," and on a pure 2-form like "divergence." Lemma 1. i) d2a = 0 for every a E RD ii) d (a 1\ ~ ) = da A B+ ( -1) r at\ df3, for every purer-dimensional form a. Proof of this lemma is left as an exercise. Hint: it suffices to consider only monomials Q 1 B. Definition 40. A form a E RD is said to be closed if da = 0. ~

82 A form a

f3

E

RD

is said to be exact if a

= df3 for some

RD. It is clear that ~he closed forms form a subgroup ~ of Rnl' and that the exact forms are a subgroup R~ of RD. Hence, we form the d-cohomology group R~ I R~ = closed forms I exact forma. Lemma 2. The closed forma form a subring of RD · in which the exact forms are a two-sided ideal. ( 1. e. 1) closed A closed = closed ii) closed A exact = exact iii) exact 1\ closed = exact) Proof. We may assume that a,f3 are monomials. 1) If da = df3 = 0, then d(a/\(3) = da/\(3 + (-l)degaaf\d~ E

ii)

If f3 = dy, da

= o.

= 0,

d(a Ar) = da A-y+ (-l)deg aaAdr = [ (-l)dega a] 1\ (3, i.e. a/\ f3 = d ( -1 ) dega a 1\ J' ) ,

and similarly for iii), ( f3 A a) = d(-y A a). Hence, the de Rham group is a ring, (the de Rham cohomology ring). C. NO\IJ assume D C. ron. vie identify an with lR 2n in the usual way, and observe that

= xj + iyj) j z j = xj - iyj zj

-

=

l, ••. ,n

are functions of xj and y j; hence we may apply "d", obtaining dzj = dxj + idyj j = 1, ••• ,n • dzJ = dxj - idyj

r

I

But the dxj,dyj generate the ring of differential forms on D; and the above equ:tions are solvable for dxJ,dyj in terms of dzj and dzj. Hence the dzj and dzj also generate the ring of forms, so that for any form a E R0 :

2n

a =

> L r;Q p+q==r

ai

i -j

-j dzi

1••• p 1•· • q

1

1\ • . •

11 < •••
jl < .... <Jq

Now any element may be written uniquely as a sum of pure dimensional forms; and any pure dimensional form of degree r may be written uniquely as·a sum of monomials of "hi-degree" (p,r-p) , p = o, ... ,r where Def!nition 41. We say the monomial ai i -j -j dzi 1\ m 1\ dzi 1\ dzj 1\ ... 1\dzj is of type 1. • • 1••• q 1 p 1 q· or bideil;'ee (p,q); we denote dz 1 /\ ••• /V:izi I\ dzj 1\ ••• 1\dzj 1 p 1 q by apq. Hence any form in RD may be written uniquely as a sum of monomials of distinct bidegrees (p,q): 2n a=~2_a

Ja. r;o p+q=r il ••• ipJ1···_q pq

Hence, the apq form a basis for the module RD. Let Apq denote the subset of forms of bidegree (p,q) and note that Apq A AP 'q 1 = Ap+p 1 ,q+q 1 • We now introduce homomorphisms o : Apq -> Ap+l,q and Apq -> Ap,q+l as follows:

a:

()a

=

-oa

=

t j

t J

da dzj

dz

-da

d-z

ozJ

j

. J o, -o I

where a is a zero form, and extend to R0 as before. Now d = o + ~~ as can easily be verified. Hence d2 = 0 = (o+~) 2 = o2 + ~ 2 + (o~+~o). Note that (o+~) 2 Apq

= o2 Apq + ~ 2 Apq + (~~+~o) Apq

= Ap+2,q + Ap,q+2 + 2Ap+l,q+l -2

-

-

,.

so that o2 = o, o = 0 and oo = - oo. We nm.z define o-closed, o-closed, o-exa.ct, and o-exact forms, and form the associated cohomology groups:

-

84 0.-closed forms a-exact forms . and

--o-closed forms . a-exact forms

We may again verify that these groups are in fact cohomology rings. Observe that, if we restrict the coefficients to be holomorphic functions in D, becomes trivial and d = o. Hence, we can also form the cohomology ring:

o-

closed holomorRhic forms exact holomorphic forms where we define a holomorphic form as follows: Definition 42. The form L a1 i -j 7 dzi A ••• 1""' p l•••Jq 1 1\ dzip 1\ dz jl 1\ ••• A dz jq is said to be holomorphic if q = 0 and the

a

il ••• ip

are holomorphic functions.

E\2. Differential forms on ma!'lifolds Let D and 6 be domains in Rn and Rm respectively, and let f be a diffeomorphi3m from D onto 6 • Denote a point of D by x = (x1 , ... ,xn) and a point of 6 by ~ = ( ~ 1 , .. ., ~ m) • Then f ( x) = ~ , or ~ j = f j (x1 , •.. , xn), j = l, ••• ,m. There is an induced mapping f * associated with f which maps diffe~(.\ential for•ms in 6 into differential forms in D; if = :;:-- aj j ( ~ ) d~ j A . . . J\ d~ J' r , then ... _._ 1" • • r 1 f*a =

>

1. 2.

3. 1~.

aj 1 .•. j (f(x}) dfj A •.• Adfj r

1

r

and

f * preserves degrees f * ( a+f3) = f *a + f *f3 f * (a 1\f3) = f * aA f * f3 * 1( df a = f da •

Let D,6 be domains in ~n, and let f be a holomorphic mapping from D to 6. Let (z 1 , ••• ,zn) denote a point of D

85

:

and (~1, ..... ~n) denote a point or 6. The induced mapping f* (as above) has the properties 1. f* preserves bidegrees 2. f * (a + ~) = f * a + f.* t3 3· -r* (a A. t3) = f * a 1\ f *t3 4. ~r*a::;: f*~a and ()f*a= f *()a • For the case of holomorphic forms, since f is holomorphic and a holomorphic function of a holomorphic function is holomorphic, f * takes holomorphic forms into holomorphic forms. Let X be a differentiable manifold of dimension n. We define a differential form on X as follows: Definition 43. A differential form on X is a rule which defines a differential form in every ooordinate patch. Each coordinate patch ~a is diffeomorphic to a domain D in mn (by definition). A differential form on X associates with every coordinate patch Pa a differential form a in Da such that if Pa () Pa i ~ then the images of this 1 2 intersection in Da and ~ are diffeomorphic and the 1 2 induced map on forms takes a 1 into a 2 •

,

Poincare Lenunas The Poincare Lemmas state that in sufficiently "nice" domains any closed form not containing a 0-form is exact. More precisely, Theorem 23. ( Ta) Let D = x1 , ••• , xn) I Ixi I < Ri ~ oo j C Rn, and let a be a pure r-dimensional form, r > O, in D. - If da = 0 then a = dt3 for some t3 • (Tb} Let D = z1 , ••• , zn} I Iz j I < Rj ~ ooJC C:n, and let a be a pure r-dimensional holomorphic differential form, r > 0, in D. If da = 0 then a = dt3 for some holomorphic form t3 •

t(

1(

86 {Tc) Let D be the domain in (Tb), and let a be a (p,q)-form in D, q ~ 1. If oa. = 0 then a.= o~ for some (p,q-1)-form f3 • Proof. Lemma. Let i,j be fixed numbers, 1 ~ j, i,j = 1,2, ••• ,n. (La) Let D be the domain in {Ta). If ~ is a C00 • function in D, then there is a C00 function ~ in D such that ~ = o~/oxi and if o~/oxj = 0 then a~/oxj = o. xi Proof. Define 1[J(x1 , ••• ,~) = ~(x 1 , ••• ,xi-l, t ,xi+l, ••• ,xn) dt.

-

=

-

J 0

(Lb) Let D be the domain in (Tb). If ~ is a holomorphic function in D, then there is a holomorphic function 1[1 in D such that ~ = o1[1/ozi and if o~/ozj = 0 then o1[1/ozj = o. Proof. Since ~ is holomorphic in D, CD A. k ~ = ~ a 1c(z 1 , ••• ,z 1 , ••• ,zn)zi~~ Define k+l "ft;"" CD ak(z 1 , •.• ,zi, ••• ,zn)zi

1[J(zl,. • .,zn)

=~

k+l • k=O (Lc) Let D1 = ~(z 1 , ••• ,zn) I lzjl < rj < ro}C mn. If ~ is a C00 function in a neighborhood of D1 , then there is a C00 function t in perhaps a smaller neighborhood of 1 such that ~ = a,~/az 1 and if ~ is holomorphic in zj then so is 1[1. Proof. Define "' 1 f" ~(z1, ••• ,zi, ••• ,zn) 1[1( zl" ... , zn) = - ;r jj t-zi d~ d11 •

n

ICI
(Proof of Ta)). Use induction on k where a.=~ aj j dxj ••• dxJ. contains no terms with 1••• r 1 r dxk+l'dxk+2, ••• ,dxn. For k = 1, a.= a 1dx1 and da. = 0 means that oa1/oxi = 0, i ~ 2. By (La) there is a bl with ob 1/ox1 = 0, 1 ~ 2 and a. = ob 1/ox1 dx1 • Take = b1 • Assume the theorem to be true for some k-1, and let a be

87 a purer-form with no terms involving dxk+1 , ••• ,dxn. Then a = ( -1 )r-l ( dxk Ap) + lith both dxk and dxj, j > k. Thus dp contains no terms with dxj, j > k, which means that the ~oefficients of p do not depend on xk+1 , ••• ,xn. Now, define p as follows: Replace each coefficient of p by a function whose derivative with respect to xk is this coefficient; call this new form ,1\ p. The coefficients of p"" do not depend A on xk+1 , ••• ,xn. (The existence of p and its independence of xk+l' ••• ,xn' is given by (La).) Hence dp = (dxk A p) + terms involving dxj, j < k. Therefore a - (-l)r-l dp does not contain dxk 1 dxk+1 , ••• ,dxn. But d(a -(-l)r-l d~) = da= 0 (and (a -( -1 f- 1dp) is a pure r-form. By the induction hypothesis, then a -(-l)r-l dp= dy which means that a= d((-l)r-l ~ + y). (b) (Proof of (Tb)). The proof is identical with that of (Ta) \•11 th the x 1 s replaced by z 's and a a holomorphic form. (Lb) gives the existence or a holomorphic form· ~ and its independence of zk+1 •••• ,zn. (c) (Proof of (Tc)). 1. Let D0 be an open poly~isc, D0 c c. D. Then _a is defined in a neighborhood N of D0 and satisfies oa = 0 in N. 'l'le claim that there is a form ~~ :n perhaps a smaller neighborhood of D0 , such that a= o ~. For the proof, use induction Jn dzk anu an argument analogous to that in (a}. 2. Construct open polydiscs Dj c C Dj+l whose union is D. By_l., for each Dj th~re ex~sts a form ~j such that a= o~j in a neighborhood of Dj. (i) Assume that a has bidegree (p,l), then ~j has bidegree (p,O); i.e. ~j = 2:: bij i dz 1 ••• dzi_' and

-

1' •• p

1

-p

88

-()13 j = a

in Dj. For fixed k ~nd j > k, (!3j-f3k) is a holomorphic form on Dk since c(Bj-13k) = 0 on Dk. . Therefore the coefficients of (~j-!3k) can be approximated as closely as desired by polynomials, and hence the form · (~j-13k) can be approximate~ by a form Pjk whose coefficients are these polynomials. Choose ej > 0 such that ~ ej < co, and define 113.-~~1 = ~ lb~ ~ - b~ 1' 1. Construct A J ~ •l••••:p ·1··· p 13 ~ as follO\'lS: 131 = 131 , ~ 2 = f:3 2 - 21 w~ere I f3 2-f31 1 = I ( 13 2 -~ 1 ) -P 21 1 < e1 on D1 , 13~ = !:'-,: - P7, 2 where ,.. "' "' 113 3-13 2 1 = I(!3 3-13 2 )-Q32 1 < c2 on D2 and Q32 = ~ + P32 , etc.

.!

/

-

_.._

-

"'\

/

/

"'

()pj ="'a!3j =a on Dj. Since lf3j-!3j+ll < ej on Dj, llm Bj = 13 exists uniformly on compact subsets of D, oo

j->oo

-

the

coefficients of ~ are C functions and of3 = a in D. (Cf. Part 2 of the proof of Theorem 19.) (ii) If a has bidegr~e (p,q), q > 2 then each Bj ~as bidegree (p,q-1). a(~j+l-~j) = 0 in a neighborhood of Dj. Therefore Bj+l:f3j = ayj for some form yj, in a Nj of Dj, by l._of (c). Let roj be a neighborhood real valued c00 function ~ 1 on Dj and 0 outside a ~ neighb~rhood Mj of Dj, Mj cc Nj. Set y j = rujy j:."'. Then yj ;s defined in Dj+l -~d 13j+l- f3j = ayj on Dj. Let P1 = ~ 1 , 13 2 = ~ 2 - dy 1 , ••• , and in general A. Bj = !3j- a(y 1 ~; 2 + ••• +yj_ 1 ). Then f3j+l-~j = (13j+l-f3j)-

=

1\

-

-

1\

"

.1\

"

-

-"'

'\

.

"

"

;::"'

d(r 1+••• +yj) + d(y 1+ ••• +yj_ 1 ) = (!3j+l-Bj) - orj -A

-

():] j = dP j = a desired f3.

on Dj.

Letting

j

->

oo

= o,

and

we obtain the

Chapter 8.

§ 1.

Canonical Isomorphisms De Rham 1 s Theorem

A. Definition 44a. Let X be a differentiable manifold with covering u. We say U is simple with respect to differential forms, or d-simple, if it is ?pen, locally finite, and, for the intersection of any finite number of sets of the cover u0 II • • • f\ ur, the Poincare lemma for 11 d" holds. Theorem 24a. Let X be a differentiable manifold. Then there exist arbitrarily fine d-simple coverings (i.e. every covering of X has ad-simple refinement). Proof. \'le shall first prove this theorem assuming X C IRn, X open. The case of an arbitrary manifold X is treated at the end of this section. Assume XC. iRn. Note that in any box { lx1-ail < ri, i = 1, ••• ,n ~ , Poincar~ 1 s lemma holds by Theorem 23. Furthermore, the intersection or any finite number of boxes is again a box; so it suffices to refine any covering U to a locally finite cover!~~ by boxes and this is easily done. Lemma la. If (X,U ,oP) = 0; r > 0, p! 0, \'lhere rf denotes the sheaf of germs of p-forms, and U is locally finite. Proof. The sections of 0° over U are C00 functions, so ~(X,U,o0 ) = Hr(X,U,C 00 ) = O, r > 0, by Theorem 22. Since any element of r.P, when multiplied by a C00 function, remains in r.P, we may establish this lemma by constructing a homomorphism Q : cr(x,u,oP) -> cr-l(x,u,rf), r > o, so that if f ~ cr(x,u,rf), then f = Qof + oQf precisely as before. Hence every cocycle is a coboundary. Corollary. Hr(x,oP) = 0; r > o, p! o, rf as above. Theorem 25a. Let X be a differentiable manifold, U a d-simple covering. Then the following groups are canonically isomorphic, where Cf denotes the sheaf of germs of closed p-forms in X:

90 (i) (ii)

( iii)

1( U ~ H X, ,cr ~

closed p+l-forms exact p+l-forms '

P!

0 •

r

>

o, p!

closed r-forms exact r-rorms-, r

>

o .•

Hr+lrx,u,oP) ~ Hr(x,u,QP+l)

r H (X,U,~) ~

I

0 •

Before proving this theorem (following A. Ueil) , we : introduce the notion of coelements. B. Definition 45. A coelement ~ of bidegree (r,p) is an r-cochain on a (fixed) covering with coefficients which are pure dimensional differential forms of degree p, i.e. if uj , ••• ,uj are distinct sets of the covering with nonempt~ inters~ction, then rrp(jo···jr) assigns to this intersection a pure difrerential form of degree p defined there. The coelements form a vector space over €. Define drrp = gr,p+l, v1here g assigns to each intersection "d" of the form which f assigns; d2f = 0. Define 5~P = hr+l,p in the usual way; 52 = 0. Clearly d5 = 5d (for 5 "adjusts" the domain, and d the range, of the coelements). Coelements f for \'lhich df = 0 are cochains with closed forms as coefficients, and if df = 5f = 0, the coelements are cocycles with closed forms as coefficients. If U is a d-simple covering, rrp a coelement, p > O, then a~P = 0 implies f = dg. If r > 0 and 5frp = o, then f = 5g by Lemma la. fr',p+l are l'le say that coelements rr+l,p and associated if there is a coelement grp such that f = 5g and f = dg. C. Proof of Theorem 25a. Note first that (1) and (ii) imply (iii), for closed (p+l)-forms Hl(X U =n) l+s( u ~n-s) exact (p+"l) -forms ~ , ,l r- ~ • • • ~ H X, ,1 r "' ••• ~ Hl+p(X,U,0°) , p ~0 1 -o and o = ~ •

91 Note also that r =n H (X, u ,( r) =

l rP I1df'P=or'P=O l1

~ohr-

.

,

r

> 0 •

,pldhr- ,p=O} (1) Ne associate a closed (p+l)-form on X (or, equivalently, a coelement fo,p+l or bidegree (O,p+l)) to each cocycle class in :H1 (x,u,oP) as follows: Let r1P be any cocycle; df = of = o. Using Lemma la, there exists a g 0 P such that og = r. Set fo,p+l = dg, and "' = 0; hence, we have assigned a closed note that df "' 5dg = dog= df = 0.) Denote by (p+l)-form. (Also, of= Sf~ the class in closed (p+l)-forms containing We ~ J exact (p+l)-forms make the following assertions: ~ "' ? a) ~ f) does not depend on the choice of g. f3) r{>= {r 2~ if (r1\ ,= tf~, i.e. if f 1 -r 2 = oh, dh=O. r) The class mapping tr1->if) is an isomorphism. "' Proof of a) Assume f ~ 5~ 1 = 5g8, and set 0"'r 1 = dg 1 ; r 2 = dg 2 • a) asserts that r 1-f2 = dh P whe~e "'h P is globally defined on X; i.e. 5h = 0. But r 1-r2 = d(g 1-g 2 ) and 5(g1 -g 2 ) = o. Proof of~) Suppose riP - f~P = oh0P, where 5f1 = 5f2 = df1 = df 2 = 0 and dh = o. Now f 2 = og 2, f 1 = og1 = og 2+5h = 5(g 2+h). Hence "'f 1 = dg1 = dg 2 + dh, "'r 2 = dg 2 , and "'r1 -r "' 2 = dh = o. Proof of r) It is clear that the association map fj -> ~ f is a homomor•phism. It is one-to-one for, assume fJ -> o; i.e • r = og, and dp 0P = 0. N0\'1 1 dg0P = 0 means g0P = dhO,p-l so that f = og = 5dh: hence Furthermore, it is onto, for, assume "'f is any closed (p+l)"' "' form. Then df = o, so f = dg by d-simplicity of u. Define f = 5g. Then df-= d5g = 5dg = 61 = 0, as "'f is globally defined, and 5f = 52g = 0. (ii) The proof in this case is essentially the same; let ~+l,p satisfy dtr+l,p = otr+l,p = o. Then there exists a grp such that 5g = f. Set "'r f ' p+l = dg, and

f.

i

i l

J

!rJ·= o.

.

92 'V

'V

observe df = 6f = o. With a similar notation, we prove a), ~), andy). a) Assume f = 0 = 6g ; then { dg~ = 0, for 6grp = o implies grp = 5hr-l,p, and dg = d6hr-l,V = 6dh. Now dh is closed, so:fdr} = f6(dhl).= o in Ffcx,u,oP+l). ~) r+l,p = 6hrp implies idhrp} = o, as dhrp =. o. y) t-le again have a homomorphism fr+l,pj ->trr,p+lj • It is one-to-one for, if f = 6grp and dgrp = o, then g = dhr,p-l, and r = 6g = 5dh , so if}= o. It is onto, 'V +1 'V 'V 'V for assume tr,P satisfies df = Bf = 0. Then f = dgrp by simplicity or u. Set r = 6grP. Then df = d6g = 6dg = 5f = 0 and 6f = o. Clearly {f) -> 't-f • D. Lemma 2a. Let X be a differentiable manifold; U,V d-simple coverings and V a refinement of u. Then the following diagrams commute, where r.P denotes the shear or germs of d-closed p-forms

l

i)

Hr+l(x,u,oP) - >

~

I-Ir+l(x,v,oP) - > ii)

E1 (X,U, (f ) 1

-

H (X,V,cf) iii)

Hr(x,u,~:;)

~ /"~'

~~

W(X,V,iC) ~

d-closed (p+l)-forms d-exacr-{P~fr:lorms

d-closed r-forms d-exact r-forms

where r > 0 and p > o. -Proof. For i), this lemma states that one obtains the same result by first mapping a coelement to any associate and then restricting the domain of definition; or by first restricting the domain and then associating it. The other commutativity claims are disposed or as easily.

93 Theorem 26a. Let X be a differentiable manifold for which there exist arbitrarily fine d-simple covering~. Then the follott1ing groups are canonically isomorphic: ( I) (de Rham): Hr(X ~) ~ d-closed r-rorms r > 0 ' - d-exact r-forms ' Hr(x,u,e) ~ Hr(t,e) , r > o , (II) (Leray):

::

for any d-simple covering u. Proof. Hr(X,I) is a direct limit of groups ~(X,U,~); U any covering of X, where the class of all coverings is directed by 11 is a refinement of." By .Theorem 24a, the d-simple coverings are cofinal, hence it suffices to consider only d-simple coverings in the direct limit. By (iii) of Theorem 25a Hr(X U e) ~ d-closed r-forms for ~ny d-simple ' ' ' - d-exact r-forms U; hence I and II. We now complete Theorem 24a. Note that Theorem 25a does not require the existence of arbitrarily fine d-simple coverings. Now let U be any covering of X. Let V be a locally finite refinement of U such that each v e V lies entirely in a coordinate patch and the intersection of any finite number of v•s is contractible to a point. (Such a covering can be constructed using l4hitney•s imbedding theorem.) Since every finite intersection of sets of V is diffeomorphic to an open set in mn, the de Rham theorem applies. If the r th cohomology group of such an open set in IRn with complex coefficients is trivial, then every clo.sed r-form is exact: that this cohomology is trivial is a lmown result. §2. Dolbeault's Theorem This section is Section 1 applied to complex manifolds and the operator ~. A. Definition 44b. Given a complex manifold X and covering U, we say that U is simple with respect to (p,q) differential forms, q > 1, or • a-simple if it is open,

-

94

-

locally finite, and, for the intersection of any finite number , of sets of the covering, the Poincare lemma for o holds. Theorem 24b. Let X be a complex manifold. Then there exist arbitrarily fine ~-simple coverings. Proof. As before, assume X c. ~n, open. Once again / use Theorem 23, which establishes the Poincare lemma for o, for coverings by polydiacs {lzj- ajl < RjJ • The completion of this theorem for a manifold is remarked on at the end of this section. Lemma lb. Hr(x,u,r.P) = o, r > 0; where r.P now denotes the sheaf of germs or forms of type (O,p), and U is locally finite. Proof. We may again use a partition of unity argument as in Lemma la. Theorem 25b. Let X be a complex manifold, U a ~-simple covering. Then, if f.P denotes the sheaf of germs of ~-closed forms of type (0 1 p), there exist canonical isomorphisms between the following groups: i) Hl(x,u,r.P) ~ a~closed forms or type (O,p+l) I - ~-exact forms of type (O,p+l) ii) Hr+l(x,u,oP) ~ Hr(x,u,oP+1 )

-

~

-

-

~-closed

forms of type (O,r) ~-exact forms of type (O,r)

where r > 0, p ~ 0. Proof. 1) and ii) proceed precisely as in Theorem 25a. iii) is implied by i} and ii), also as before, when one notes that a a-closed form of type (0,0) is a holomorphic fUnction, and conversely; i.e. r.P = (!). Lemma 2b. Let X be a complex manifold; U, V a-simple coverings where V is a refinement of u. Then the following diagrams commute, where f.P denotes the sheaf or germs of a-closed forms of type (O,p), and r > 0, p! 0:

-

- -·----

95 i)

-

ii)

~-closed

forms of type (O,p+l) ~-exact forms of type (O,p+l)

iii)

~ ~-closed forms of type (O,r) ~ o-exact forms of type (O,r)

-

Theorem 26b. Let X be a complex manifold for which there exist arbitrarily fine a-simple coverings. Then -there exist canonical isomorphisms between I. (Dolbeault) Hr(X,(0) "' ~-closed forms of type (O,r) , · - ~-exact forms of type (O,r) r > 0 •

II. (Leray) Hr{x,u,(Q) ~ Hr{X, 0) , r > 0 • Corollary. If D (_ (In, D a polydisc, then W(D, U) = 0 for all r > 0. We shall eventually use this corollary to establish the result for any region of holomorphy. Remark. The general Dolbeault theorem reads as follows: ~r(X,sheaf of germs of holomorphic forms of degree s) ~ ~-closed forms of type (s,r) • ~-exact forms of type (s,r) However, we shall require only the restricted result of 26b, I. B. In order to complete Theorem 24b, we require the result that in domains of holomorphy all the above cohomology groups are trivial (proof in Chap.ll). Assuming this result, we have the Poincare lemma with respect to ~ for holomorphy domains, so that any locally finite covering by domains of holomorphy is o-simple. Hence it~suffices to establish that arbitrarily fine coverings by domains of holomorphy exist. /

-

96

§3. Complex de Rham theorem Once again, we attempt to repeat 1 for complex manifolds X, holomorphic forms, and the operator d. Definition 44c. Given a complex manifold X ·and covering u, we say that U is a-simple or simple with respect to holomorphic forms, if it is open, locally finite, and, for the intersection of any finite number of sets of the covering, / the Poincare lemma for d holds. (Recall that o = d on holomorphic forms.) Theorem 24c. Let X be a complex manifold. Then there exist arbitrarily fine d-simple coverings. Proof. For XC ~n, open, the proof is again immediate and proceeds as before. At this point, hO\Ilever, we find that no Lemma lc exists. Hence, we must modify Theorem 25c as follm'ls: Theorem 2~c. Let X be a complex manifold, U a a-simple covering, for which Hr(x,u,rP) = o, r > 0, where r.P denotes the sheaf of germs of holomorphic p-forms. Let oP denote the shear of germs of closed holomorphic p-forms. TPen the following groups are canonically isomorphic: closed holomorphic (p+l)-forms ) Fl(X u Cf) i ~ exact holomorPEic (p+l)-forms 1

'

'

ii)Rr+lrx,u,OP)~ ~(x,u,r.P+1 ) Hr(X u r,) closed holomorphic r-forms , ' ' ~ exact· holomorphic r-forms where p ~ 0, r > 0. . _ Proof. As before, under the remark that 0° = e. Lemma 2c. We state here merely that the analogous commutativity lemma is valid, assuming the missing Lemma lc for all manifolds and coverings used. Theorem 26c. (Complex de Rham). Let X be a complex manifold for which there exist arbitrarily fine o-simple coverings and such that Hr(x,u,rru) = 0, 'r > o, for all d-simple coverings U. Then, there exists a canonical iii)

97 isomorohism between: Hr[X ~) ~ ~osed holomoQEhic r-forms r>O · · ' - exact holom_f)rp 1c r-forms ' • (We shall not need the complex Leray theorem.) Let us assume that we have already proven that in a domain of holomorphy the cohomolozy with holomorphic coefficients is trivial (not closed forms). Then the hypothesized Lemma lc holds, so that the theorems of this section hold for domains of holomorphy. closed holomorphic r-forms We remark that the group eXac£ holomorphic r-forms is clearly trivial for r > n, where n is the dimension of the manifold. Hence Theorem 27. Let X be a complex manifold of dimension n, such that ~~,u,nP) = o, r > 0 for all o-simple Qoverings u. Then ~(X,C) = O, r > n. This theorem gives a topologically necessary condition for a differentiable manifold X of real dimension 2n to be a complex manifold.

98 Chapter 9· §1.

The Multiplicative Cousin Problem

The Multiplicative Problem, formulated

A. This second Cousin problem is a generalization-of the Weierstrass problem in one complex variable: . Given a domain D ~ 0, a discrete set or points, av, bv and positive integers nv, mv, find a function r, meromorphic in D, with zeroes at av of order nv, and poles at bv or order mv. We.now formulate the multiplicative problem, referred to as C.II in the sequel: Multiplicative Cousin Problem. Let X be a complex manifold U =lui$' 1 e I, an open covering, and let functions Fi be defined and meromorphic in ui' such that Fi/F j is holomorphic in ui /) u 3_. Does there exist a function F; o, defined and meromorphic in X, such that F/Fi is holomorphic in u1? ~· C.II is precisely C.I, written multiplicatively. As in the Weierstrass problem, where it is sufficient to find a fUnction with given zeroes of given order, we shall find we need only consider holomorphic functions Fi. We shall also show that c.II is not always solvable. As before, we shall formulate C.II using sheaves and cohomology groups. B. Definition 46. Let X be a complex manifold, and let ;?t.denote the sheaf, over X, or germs or meromorphic functions unde~ ~1 tiplica tion, where fll = ~EX ~~ and ~ consists or the germs or meromorphic functions at x, and is a multiplicative group. We topologize as \'le did (!) I by defining a subbasis for the topology utilizing the topology or X, as follows: Let m E ·~n i then m E and g E m is defined in a X ~ neighborhood N of x. For each y e N, let ~ g ~ e m. Y be the equivalent class or meromorphic functions in //1. Y .... containing g. Then the sets Ng = {.[ g ~ e MY I y e N for

m

m

J

99 each choice or g e m, and ror each m e ,J?t. , form the subbasis of lfl • Let denote the subsheaf of invertible holomorph1c elements of )ll, ; such that f c. }11 and 1-. C. /rl ror each x E X. Form the quotient groups }ll an~ set ''7'1 -%. X X ~~~ = xeX /'~~x with the quotient topology. The sheaf of germs of divisors of X is the quotient shear A/.:/- . Note that elements of 71!. / j are equivalence classes of germs of meromorphic functions, where germs represented by two functions F1 , F2 are equivalent (at x} if F1/F2 is a local unit, i.e. if F1/F2 is holomorphic and nonvanishing in a neighborhood of x. We note also that a set of Cousin data associated with the covering U may be regarded as a section of l!L;..jover X. Definition 47. A divisor on X is a section over X of AI -f. 1 i.e. is an equivalence class of sets of Cousin data, in the sense that two sets of Cousin data are equivalent if their "mesh" is a set of Cousin data. A divisor on X is integral (positive) if all germs are germs of holomorphic functions. Definition 48. A divisor a on X is principal if there ex1.sts a meromorphic function F defined in X such thatthe divisor it defines (F)= a. Hence, we may state C.II in the following equivalent way: given a divtsor on X, is it princ:ipal? Lemr;a 1. If every integral divisor of the complex manifold X is principal, then every C.II is solvable. Proof. It is enough to show that every divisor is a quotient of integral divisors. Let p E X, and let FP be a meromorphic function defined in a neighborhood NP of p such that (FP) is the restriction of a to NP. FP = ~p/tp' where $P and tp are holomorphic functions defined in Np

j-- ·

-

u

;3':,,

lCC

and coprime at p, and hence in a neighborhood of p, say Np. Let q c NP, and Fq = ~ql'q the corresponding function at q, defined in Nq. He shall be done if ~p "' ~q and ?/lp "' ?/1q' where defined, \'/here ''"'" means equivale~ce modulo 2J-. Not'l Fp"' Fq, hence ~p/'P"' ~q/'q' so ~p?/lq"' ?f!p~q' at q. But ~p?/lq and ~p~q are holomorphic, and are equivalent in a neighborhood of q. We may choose this neighborhood so that ~q' ?/1 are coprime. But ~p divides ?/lp~q; hence ~P divides ~q• Similarly, ~q divides ~p' as ~p' ?/IP were coprime in Np. Therefore ~P/~q is a unit, i.e. ~p "' ~q' and similarly 1frq• Thus (~p I p E X) and (?/IP I p E X) are integral divisors. C. Theorem 28. Let X be a complex manifold such that C.II is allr:ays solvable. Then so is the strong Poincare problem, i.e. every globally defined meromorphic function is the ratio of two holomorphic functions, coprime at every point of X. Proof. Let F be the global meromorphic ~~ction. Then (F) = a/~ , where a and ~ are principal integral divisors; for as was shown in the proof of Lemma 1, on any complex manifold X every divisor is a quotient of integral divisors, and since C.II is solvable every integral divisor is principal; hence F/f/g = rn1/rn 2 is a local unit. Therefore F/f/g is holomorphic and equivalent to 1 at each point of X; so F/f/g is a global unit, say G. Thus F = fG/g, where fG and g are coprime at each point of X. Theorem 29. Let X be a complex manifold such that C.II is always solvable, and Y a regularly imbedded analytic hypersurface~ Then Y may be globally presented. Proof. Exercise for the reader.

'P "'

§2.

The Multiplicative Cousin Problem is not always solvable The following example is due to Oka. Let XC ~ 2 be defined as follO\'lS:

101 X= [Cz 1,z 2 ) I 3/4 < lzjl < 5/4, j = 1,2}. Note that X, as a product domain, is a domain of holomorphy. This shows, incidentally, that ''C.I implies C.II" is false. Set ! = f z1-z 2-1 = I') X, and note that Y is a closed subset of X consisting of two distinct components, for (z1 ,z 2) = (x1 +iy1 , x2 +iy2 ) e Y implies 0 < x1 < 1, where x1 - 1 = x2 • Hence y1 = y2 -# 0; but Y is nonempty and (z 1,z 2 ) e Y implies 1 2 ) e Y. Hence, we see also that the components of Y lie in (Im z1 > O, Im z 2 > 0) and (Im z1 < 0, Im z 2 < 0). Now define the divisor 1 of X as follows: . r(z 1-z 2-l), for (Im z1 > 0, Im z 2 > 0)

OJ

(z ,z

1

-

t

1

outside the upper component of Y

This clearly defines a set of Cousin data, for which, we claim, the Cousin problem has no solution. For, assume there exists a solution F(z 1 ,z 2 )~ and consider its restriction to

[1z1 1 =

1,

lz 2 1 =

If

g(a,p)

=

F( eia, eit3) •

Now g(a,p) is a continuous, periodic fUnction or both a and t3 • Furthermore, g has precisely one zero, for the upper component of y n(lzll=l,lz21=1J ={<ei1T/3,ei 21T/3)J, and g "' 1 elsewhere. Nou consider the edge curve l in the a,B plane as indicated in the figure, and the edge curve r-2 about (v/3,21T/3) within rl and oriented B similarly. 21T · - ( ---: ( 21T 1 21T) Since g is periodic in a and t3 it is clear that arg g(a,p) can be defined as a single-valued function along 1 • Furthermore, by connecting the tHo curves as a in the second figure, we obtain the following:

r

i

r-

102

-~~· _J

rv

J

d log g

=

r2

rl

III

J d log g ;

arg g(a,~) is also single-valued along r- 2 • Now F(z1 ,z 2 ) = h(z 1,z 2 )•(z1-z 2-l), where h(z 1,z 2 ) is a unit. In the region enclosed by 2 , we may define a single-valued branch of the log; so i.e.

r

J d log g = J d log [ ... ei~ + eia r2

\Je

-1] •

r2

may calculate this latter integral explicitly.

S'f3

=

f3'

j'

r2

+-¥

la a +); d log (-eif3 + e = J d log =

obtain

Set

1

1 a -1)

r'2

r2 I

where now encircles the origin in the (a• ,f31 )plane, and -eif3'+2vi/3 + eia•+vi/3_1 has a zero at (0,0). Set -eifj '+2'1fi/3 + eia r +'ITi/3 - 1 = u + iv • Now

= - sin (f3•-a•+w/3) Hence, if ,-.2 was chosen 1 small enough, the region enclosed by 2 • So

J

r-

d log (u+iv)

r;

=

J

I o(u,v)

d(at,pt)

d log (a'+

r;

and this contradition establishes our claim.

•

'< 0

in

i~t) f. 0 ,

103 ~3.

The solution of the Multiplicative Cousin Problem for polydiscs

Theorem 30 (Cousin). Let DC
(

I I

I

I

-.

6 2=l. (z 1 , ... , zn) a~x 1_, 3 ,b 1_, 1~b 2 ; aj-<x~aj't3j-
Let a1 , a2 be open sets (in ~n) containing 61, 62 respectively, and F1 , F2 meromorphic (holomorphic) functions in a1 ,a 2, respectively, such that F1/F 2 ~ 1 in a1 ~G 2 • Then there exists a domain G such that ~ \.) 62 c G C.. ( a1ll G2 ) , and a function F, meromorphic (holomorphic) and defined in G such that Fi/F"' 1 in a1 I) G, i = 1,2. In other words, if C.II is solvable for neighborhoods of 61 and 6 2, it is solvable for a neighborhood of ~ \) 6 2 • ~f. It is easily seen that a1 n a2 contains a neighborhood a0 of 61 n6 2, and that we may choose this neighborhood to be simply connected, for 61 il 62 is. Now, in a simply connected domain every nonvanishing holomorphic funotion F may be written as er, where f is holomorphic, for r is the solution of th~ following set of equations: 1 ()F

Fd"Zj 0

Hence,

.

104

f2(z,z2, • •• ,zn)

1

= - NI

J

f ( C, z 2' • •• , zn)

. t-z

dC '

12

where 1 1 and 1 2 are curves in the projection or G0 on the z1 -plane defined as follows: Let C be a s.imple closed curve in the z1 -plane lying in tbe projection of G0 on that plane, positively oriented, and containing the line {a 2 = ~~ b 1 ~ y1 ~ b2 in its interior. Set

t

11

12

fz = fz

=

I

z

I

z c: C,

E

C,

Re z 1

! a2

T

Re z 1 ~ a 2 )

•

The situation is indicated in the figure. y Now r1 is holomorphic in a neighborhood G1 of ~~ and r2 1s b2 61-+----:.,.._,"'1-11.......,..-~ 6 2 holomorphic in a neighborhood a2 of 6 2 • Take G to be a domain satisfying ( '1, U 62 ) ~ G C a 1 fl 2 xl 1 ) U (G f'\ 2 >] • For· "'--+r--......~~~::;...___. z1 c: interior (1 1 -1?' 2 } , f = f 1 -r2 ; and hence f = f 1 -r2 wherever everything is defined so that F1I F2 = ef = e fl-f2 there. Now define in G1 fl G1

ja3

F

=

l
a

a

in G2 () G2 •

F is meromorphic (holomorphic) in each neighborhood and has the same values in (G1 f\ a1 ) l"t (G 2 1) a2 ) by construction. r I Lennna <;. Let 6 = (Cz 1 , ••• ,_zn) I aj ~ xj ~ aj' Bj ~ yj ~ f3j; j = l, ••• ,nJ be a bounded box, 6c.G, a domain in en. Let a be a divisor in G. Then there exists a neighborhood G1 of 6 in G, and a function F defined and meromorphic in G1 such that (F) = a on G1 ; i.e. given C.II for G ~ 6 1 C.II is solvable in a neighborhood of 6. Proof. t'le cover G by boxes whose edges are parallel to

105 the axes as follows: For each point z e o, there exists a closed box containing z and contained in some set associated with the divisor a (i.e. contained in same set u 1n which the restriction o~ a to u is. principal·) • Let 6~ be an open box containing z and such that :Now 6 is compact; hence 6 is contained in the union of a finite number of open boxe~ ~, ••• ,~. Hence 6C 1 1 6i C G, and a1 N 1'-.:'1 is a closed neighborhood of 6 as it contains 0 D.f, which is open. l~e claim that i=l the restriction or a to a1 is principal; we prove this ' a.. )/2 indirectly. Let 61 = i... (z1 , ••• ,z ) I x1 < ~2 ' n i .1.1 1:::. = ~(z 1 , ••• ,zn) I ('1_·~)/2 ~ x1 ~. Then 6 ~ 6j is a box, for i = 1,2; j = l, ••• ,N. Set .a11 = (6 fl o1 2 2 si 1 G12 = [ 6 fl G1 )• Then Gli is covered by t6 1\ 6JIJ=l, ••• ,N}, i = 1, 2 and we have ''halved" the problem with respect to x1 • Note that a11 V a12 = G1 • Now, if a is principal in a11 and in a12 , it is principal in a1 ; since \CF1 ) on a11 (F) = iLCF ) on G is a divisor in a1 by lemma 2. Hence, assume c.~ is not 2solvable in a1 • Then it is not solvable in either a11 or a12 ; choose one such, and call it a2 • Proceed by halving G2 with respect to y1 , obtaining a3 c: G2 , and so on, each time halving Gn with respect to the "next" coordinate, where "next" is with respect to the following sequence: x1 ,y1 ,x2, ••• ,xn,yn,x1 ,y1 , •••• Now the diameters or the closed nested sets G form a null sequence, and so m 1 contain a point which is contained in some set 6i. Hence, for some integer M, GM C But C.II is therefore solvable in GM as is contained in a set associated -with a ; and this contradiction establishes the lemma. Remark. We could also prove this lemma directly by forming a neighborhood of 6 with a mesh fine enough so that a restricted to each box of the mesh is principal. Construct F by 11 pasting" the boxes of the mesh following Lemma 2.

6!,

6;c6!.

§

t,l

J,

J,

6i

b.i.

106.

B. Proof of Theorem 30. It is clear that it suffices to prove ~his theorem for box domains a. Let Ai be a sequence of closed boxes such that ui fli = 6 and 6i c c ai+l c. c 6. By the above, there exists an Fi meromorphic in a neighborhood ~f Ai such that (Fi) = a there. Hence Fi+l/Fi is a unit in Ai for each i. Clearly, we would be done if the infinite product F1 (F~F1)(F,IF 2 ) ••• converged normally in A to a fUnction F; for at each point z e A, z e An for some n and there exists an N such that (FN+l/FN)(FN+~FN+l) ••• converges uniformly to a unit in An· Hence (F1/F) (FiF1 ) ••• (FN/FN-l) = FN/F "' 1 in an. Generally, the product w.ill not converge. However, our functions are only defined up to units, "' ui and so we replace Fi by Fi =Fie 'A ui a polynomial, so that the product formed from these Fi does converge normally, precisely as in the proof of the Weierstrass theorem in one complex variable. Corollary. C.II is solvable in any domain which is the product or simply connected domains in 1. Exercise. Use the method of the above theorem to solve C.I for a polydisc. ~4.

Characteristic classes

(From C.II to C.I)

A. In the following, we consider only coverings U such that ui and ui ~ uj are simply connected; call such coverings distinguished. We now associate a cohomology group to C.II, as we did for C.I. However, our coefficients are now j, so that our cocycles will be multiplicative. C.II 1 • Let X be a complex manifold, and U any covering of X with meromorphic functions Fi defined in ui such that Fi/F j is a unit in uil\ uj. Set Fij = Fi/Fj • Now Fij is a 1-cochain, for FijFji = 1 in ui ~ uj. It is also a cocycle, for FijFjl!ki = 1 in ui(l uj ll uk.

107

-The Cousin problem is now: Given a cocycle Fij' is it a coboundary; i.e. is the class of Fij the unit in the (multiplicative) cohomology group Hl(x,u,jf)? It is clear that C.II' implies C.II. Now assume that the covering is distinguished. We may thus define 21Tifij : Fij = e , i < j and 21Ti(-fiJ) Fji = e , with fii = o .

\

Now observe that fij = - fji; i.e. { fij)- is an additive 1-cochain, with holomorphic coefficients. However, this is not necessarily a cocycle, for 1

f ij + f jk + fki = ~ log 1

= mijk ,

an integer. But !mijkJ is a 2-cocycle (with integral coefficients); the antisymmetry condition is easily verified, and 6mijk = 0 for mijk is a coboundary of a 1-cochain. Note that mijk is not uniquely determined by the Cousin problem, for the fij are only determined up to an additive integer. However, let fij be replaced by fij + nij" Then mijk is replaced b~ mijk + (nij +njk + nki), and so the class of mijk in H (X,U ,Z) is unchanged. \lie are thus led to: 2 Definition 49. The cohomology class of mijk in H (X,U,~), for distinguished covers U, is called the characteristic class of the Cousin problem. B. The above procedure gives us a mapping of the group of divisors into the group of characteristic classes, for which a commutative lemma holds. This map is clearly homomorphic·- \11e may regard it as a map of H 1 (x,uJ~) -> H 2 (x,u,~, for distinguished covers U, for,suppose the cocycle Fij of C.II 1 is a coboundary; i.e. Fij = Fi/Fj. Then fij = fi-fj and mijk = o, i.e. the map Fij -> fm 13 k)- is a map of classes lFiJl -> mijkr·

l

108

There is a commutativity lemma in this case which we also shall not state explicitly. Furthermore, distinguished coverings are cofinal in the set ot all coverings (easilJ seen tor domains in spac-e) so that we swmnarize in the following theorem, which makes no reference to.particular z coverings: Theorem 31. (Oka~Serre). There exists a canonical homomorphism of the group of divisors of X into H2 (X,Z), with the following properties, where a denotes a divisor and c(a) its class: i) If a is principal then c(a) = o. ii) If H1 (x,c?) = 0 and c (a) = 0, then a is principal. Proof. Note that i) states that C.II solvable implies H2 (x,u,ZJ = 0, and ii) states that if C.I is solvable, then this condition is sufficient. We remark also that i) implies that if two divisors differ by a principal divisor, they have the same class. It is clear that the map is homomorphic, for a multiplication of divisors (on a fixed U) induces an addition of the associated 1-eochains and hence also of the mijk• 1, for

the A f ij now

i) Now let a = (F). Then Fi ~ F on ui for each so that Fij = Fi/Fj ~ 1. Fence we may choose fij = 0 every i,j. ii) c(a) = 0 implies mijk = nij + njk + nki' where n1 j are integral. N01r1 redefi{le f 1 j bY: setting rij2ui rij~i . = f ij A- nij • Note that Fij=e = e • But the fij is a cocycle, for 1'\

"""

A

fij + fjk + fki = mijk- (nij+njk+nki) =

o.

Hence the ~j are also coboundaries as Hl(xLd?> = 0~ so ~ ~fi -~1fj r 1J = r 1-rj, and now we are done, for Fij = e e -2v1fj -2vifi F1/Fj, so Fje = Fie • Hence the function -21Tifi F = Fie in ui is globally defined, and clearly solves C.II.

=

109

Corollarr 1. Let X be a oomplex manifold. It then C.II is solvable. Corollarr 2. Cousin II is solvable in any polydisc X. Proof. We have shown C.I is solvable in any polydisc, i.e. that H1 (x,~) = 0; H 2 (x,~ = 0 is a well-known topological result.

. .n = O, H2 (X,aJ = H1 (X,c;)

110

Chapter 10. Runge Regions Runge regions are regions in which Runge 1s theorem can be generalized: the theorem states that in a simply connected domain in the finite plane a holomorph1c function can be .• expressed as a normally convergent series of polynomials. In other words, given an e > 0, a compact subset K or a simply connected domain D, and a holomorphic fUnction r in D, there is a polynomial p such that lr-pf < e on K. Note that in a multiply connected domain it may be impossible to represent a holomorphic fUnction by an infinite sum or polynomials. ~ 1.

Preliminaries

Let X and Y be complex manifolds of the same dimension with yopen~ x. Definition 50. Y has the Runge property relative to X if every holomorphic function 1n Y can be represented as a series or functions holamorph1c in X converging normally in Y, i.e. given Kcompactc: Y, and e > O, for every holomorphic function r in Y there is a holomorphic function g in X such that lr-gl < e on K. Lemma a. Let X be a complex manifold of dimension n and ~(X 1 ~) = o for all q > o. Let f 1 , ••• ,rr be holomorphic functions in X such that if, at p e X, f 1 = f 2 = ••• = f k = 0 for k fixed 1 k = 1, • • • 1 r 1 then the rank of the·matrix (~fi/ozJ) is min (k,n), where 1 = l, ••• ,k; l = l, ••• ,n; and zJ are suitable local parameters. Let yj = tP eX I fj(p} = Then (1) Z = Y1 1\ Y2 1\ ••• f\ Yr is a regularly imbedded manifold of dimension n-r and Ifl(Z, <9} = 0 for all q > 0. (2) Every holomorphic function on Z is the restriction of a function holomorphic in X. Proof. (1) Define z0 = X; z1 = Y1 ; z2 == Yl " Y2; ••• ; zr = z. zj+l is a globally presented, regularly

oJ.

111 imbedded, analytic hypersurrace in z 3• By Theorem 20 (p. 73), and since z0 has trivial cohomology groups or all positive dimensions, so has zl. But then 80 has z2, etc. Hence ~(z,c0) = o tor q > o. . (2) Since H1 (zr- 1 ,~) = o, the first Cousin problem is solvable in z:r-l. Since z = zr is a globally presented regularly imbedded analytic hypersurface in Zr-l' by the extension theorem, (Theorem 17, p. 59), every holamorphic function in zr can be continued holomorphically to Zr-t• Similarly, every holomorphio function 1n zr·l can be continued holamorphically to zr_ 2, and so on, down to z0 = X. Lemma f3. Let xopenc en and let Xjpen~ X, XjCc.Xj+l' X = U x3, Hq(Xj, r!J) = 0 for all q > O, and let Xj have the Runge property with respect to Xj+t• Then (1) Every x3 has the Runge property with respect to X. (2) Hq(X, (!)) = 0 for q > 0. Proof. (1) Fix j. Let Kjompactc; Xj' e > 0 be given, and let r be any fUnction holomorphic in x3• Choose l e 1~ with L ei = e. Let tKi) i_?: j be a sequence of compact sets K1 cc Ki+l' Ki C Xi+l' U Ki = X, and K1 is the given set Kj for i = j. By hypothesis, there is a function gl holomorphic in xj+l with lf-gll<el compact on Kj. Similarly, for Kj+l C: Xj+l' since Xj+l has the Runge property with respect to x3+2 ' there is a g 2 holomorphic 1n x3+2 with lg 2-g1 1 < e2 on Kj+l' etc. Since 2: ei < oo, lim gk = g exists uniformly on k~oo . compact subsets K or X, because every KcK1 for some l. g is therefore holomorphic in X and satisfies Ir-g I ~ I r-g.t I + Ig .e -gt+ 1 1 + • • • < e.e +a l +l + • • • ~ e on Kj • (2) By Dolbeault 1s theorem, Wl(X, <0) = 0 for q > 0 if and only if ~ closed (O,J) fo~~ = 0 tor q > 01 i e. ~ exact (o,q ror.ms ' • if and only if for a differential form a in X or type (O,q) with ~a = 0 there is a f3 or type (O,q-1) such that a = af3 •

-

112

(a) Let q = 1, and let a, defined in X, be a (O,l) form with oa = O. By Dolbeault 1 s. theorem for Xj, there are differentiable functions_ ~1'~2···· defined in xl,x2, ••• reapectively,~uch that o~j =a in x3• Consider ~he sum: ~1 + (~2-~1) + (~3-~2) + ••• • In Xl and X2' o(~2-~3) = 0 and hence ~ 2 -~ 3 b~ing holomorphic can be approximated on · any compact s~bset o.t ~ 2 by a holomorphic function P23 on X, Set ~ 2 = ~2 , ~ 3 = ~ 3 - P23 , etc., using the Runge trick as before (cr. proof of Theorem 23, p. 88). (b) Let q > 1, and let a , defined in X, be of type (O,q)_ with ~a= o. Again, there are ~j in x3 such t~t opj =a in x3, where ~j ~~s type (O,q~l). ~ince o(~ 3 ~ 2 ) = 0 in x2, (~ 3 -~2 ) = ay 2 in x2 • Let ~3 = f33 2' etc. (of. p. 88).

-.

or

§2. Polynomial Polyhedra Definition 51. Let p1, ••• ,pr be polynomials in I IPj(z)l < 1 for j = l, ••• ,ri. 1 , ... ,zn. Let A= A is an open set. If Acc en, then A is called a polynomial polyhedra~ (of dimension n). Note that a polynomial polyhedron is a region of ho1omorphy. Theorem 32. (Oka-Weil). Let X be a polynomial poly~ hadron of dimension n, then (1) Hq(x,c?> = o for all q > o (2) If f(z) is holomorphic -·in X then f = ~ qj where the qj are polynomials in z1 , ••• ,zn and the pk used to define X, and the sum converges normally in X. Proof. (1) X is a bounded set and therefore lies in a polydisc; assume X lies in the unit po1ydisc, i.e. XC: (lzjl < 1, j = l, ••• ,n). Consider cn+r where r is the number of polynomials defining X. In cn+r consider L: = (z 1, ••• ,zn, t 1, ••• ,Cr) I lz 3 1.! 1, J=l,, •• ,n, ci = Pi
z

(z

,rJ. >

113 polydisc. Consider the analytic hypersurraces 0 = fi(z 1 , ••• 1 zn~C 11 ••• ,Cr) = Ci- pi(z1 , ••• 1zn). The fi are defined everywhere and are clearly holomorphic fUnctions in cn+r. With (lzjl < 1 1 ICil < 1) as the complex manifold X in lemma a and noting that the Jacobian or the fi has maximal rank everywhere because ari/~Ck = 5ik~= the hypothesis of lemma a is satisfied and since Z = ~ 0 , Hq( 2: 0 ~ D> = 0 1 tor all q > 0 and every holomorphic function on ~ 0 is a restriction of a holomorphic fUnction in the open polydisc (lzjl < 1, ICil < 1). But a holomorphic function in the open polydisc can be written as a power series. Hence every holomorphic function on ~0 is the restriction of a series, 1 ~ J1 Jn 11 r ~ aj ••• j i ••• 1 zl ••• zn cl ···Cr I converging normally inn 1~ 0 : We claim that ~ 0 is holomorphically equivalent to X. For, define the mapping (z 1 , .•• ,zn) -> (z 1 , ••• ,zn, p1 (z), ••• ,pn(z)). It is of rank n and one-to-one. The preimage of Lo is X. Hence Hq(X, f9) = Hq( ~ 0 , g) = 0 for all q > 0. (2) We have already obtained that every holomorphic function on ~ 0 is a restriction of a series jl i 2: aj 1 ••• ir z1 ••• err converging normally in ~ 0 • But~ ~ 0 , Ck = pk(z) and thus the above series is a series in only the zj, converging normally in x. §3.

Runge domains

Definition 52. Let K C.c r,n. The polynomial hull of K, = 0 I for every polynomial p with lp(z)l ~-1 on K, IP cz0 >I ~ l} . Definition 53. Xopenc: en is polynomially convex if K c ~X implies that K* c::: c. X. Definition 54. A Runge region is a region of holomorphy in which every holomorphic function can be expanded in a

* K

,-tz

114

normally convergent series of polynomials. Theorem 33. Let Xopen C: en. The following statements are equivalent: (1) X is polynom1ally convex. (2) X = L)Xj, where the Xj are polynomial polyhedra, xj c c xj+l. (3) X is a Runge region. Proof. (1) implies (2). SL~ce X is polynomially convex, X is holomorphically convex, By the Cartan-Thullen theorem, X is a region of holomorphy. Continue the proof by adapting the proof for analytic polyhedra, Theorem 7 and its corollary, (p. 25).

(2) implies (3). By the Oka-Weil Theorem, Theorem 32, each polynomial polyhedron is Runge in the next one, and Hq(Xj,£9) = 0 for all q < o. Apply lemma~. Hq(X,(?) = 0 for q > 0. Hence by Theorem 21 (p. 75), X is a region of holomorphy. Let f be holomorphic in X and Kcompac\: x. Then Kc Xj, for some j, and by the Oka-Weil Theorem, f can be approximated as closely as desired by a polynomial in K. Hence X is a Runge region. (3) implies (1). Since X is a Runge region it is a region of holomorphy and therefore is holamorphically convex, A A i.e. if Kccx, then KccX where K is the hull of K with respect to holomorphic functions on X. We claim that "K = K* and hence K* ~c X. Indeed, K "' c: K* since the family of all holomorphic functions on X is larger than the A family of polynomials. It r~ains to show that K* c: K. Let z0 e K* and let f be a holomorphic function in X such that jf(z)l ~ 1 for z e K. For every e > O, there is a polynomial p(z) satisfying jp(z}-f(z)l ~ £ on K V f z0 ~ , since X is a Runge region. But then Ip (z ) I ~ 1 + e on K, and since z0 e K* , Ip ( z0 ) I ~ 1 + e • Hence lf(z 0 ) I~ 1 + 2e, "and because e is arbitrary, If(z 0 ) I ~ 1, i.e. z0 e K.

115 Chapter 11. §1.

Cohomology of Domains of Ho1omorphy Fundamental Lemma, stated

( 1) Let Kcompact c G:n. Let x* denote the polyn·omial hull or K; x* = {z I IP(z)l <max IP(C)I for every polynomial 7 * K PJ. Note_that K is bounded and compact and that K*

=llf D

K c Dj

I D is a polynomial polyhedron,

(Proof easy) • ( 2) Recall that an analytic polyhedron De ~n was . defined as follows: there exists an open set G c mn and functions rl' ••• I r holomorphio in G such that D c. c. G v ' and D-c~z I z e G, lrj(z)l < 1, j = l, ••• ,v.f. ~ince 7 Dec G, hence bounded, He shall assume that G c. II z II < lJ, the unit polydisc. (3) Let D be an analytic polyhedron. Then ~' the Oka image of the closure of D, is defined as follows

i

L- = [Cz,C) I lz 1 1, ICJI i

L

~

1, z

E

= l, ••• ,n;

G, CJ=fJ(z 1 , ••• ,zn); j

l = 1, ••• 1 VJ•

c a:n+v, and is closed. We now stcte the fundamental lemma. Fundamental Len11a ( Oka) • Let ~ be as above.

---·-

--.;;.

>=~·· L.--

'--

Then

"'---

•

We shall prove the lemma in this chapter; the proof of a more general form of the lemma will come later (p. 196).

§2. !l2Plications of the Fundamental Lemma Let D ,_._ c Gopen (: en be an analytic polyhedron. ~ Observe that the Oka mapping of c1 D -> 2._ c: 111~ n+v 1

A.

(zl, ••• lzn)

->

(zl, ••• ,znlcl~···,Cv>

I

given by: zi = zi, cj = fj(zl, ••• ,zn); is defined on all of G: call the image of G under this mapping 11 2:: " . Note 1

116

c.z:=

that 2:: is closed and 1 • Hence there exists an e > 0 such that if dist [(z1 , ••• ,zn,C 1 , •.• ,cv>' L J < e:, then (z1 , ••• ,zn) e: G; where the distance of a point p e: ¢m from a set S c. a:m is defined as infse:siiP-sll = infse:s·lmax IP1-s 1 1} ir=l, •••,m Let > e: denote th~ e:-neighborhood of)! 1 defined by E = { w I dis t ( w, <; ) <E Applying the fundamental lemma, there exists a polynomial ~ ~e: polyhedron 61 c. Cn+v , such that L -c.c 61 <- '--- • Let 6 = 61 :'l1:_lzil< 1, ICjl < 1$, and note that 6 is al~o a polynomial polyhedron. \·le now claim that ~ 0 = ~l)i Izil <1, ICjl
>

r.

og,

c

aek

)

is of maximal rank. Hence lemma a applies. Since ~ 0 is holomorphically equivalent to D, we have established: Theorem 34. If D is an analytic polyhedron, then Hq(D,J}) = O, q > O. B. Using the Oka mapping, we may consider any function f, holomorphic in D, as a holomorphic function on Lo• By the above argument and lemma a, f is the restriction of a function g holomorphic on 6. Now by Theorem 32 (p. 112), g = '5 l/1 j ( z1 , ••• , zn' C1 , ••• , Cv }, normally convergent on 6, where the o/j are polynomials. This series therefore converges normally on 2:: 0 , hence on D: f(z 1 , ••• ,zn) = l/lj(z 1 , ••• ,zn,f1 , ••• ,fv)' and we have also established the following: Theorem 35 (Oka-Weil Approximation Theorem). Let D be an analytic polyhedron in G defined by r 1 , ••• ,rv; and let f be holomorphic in D, Then f = 2:: *J(z1 , ••• ,zn,r1 , ••• ,rv) where the lf!j are polynomials and the series converges normally.

L

117 Corollary. If D is an analytic polyhedron in G, then any fUnction holomorphic in D may be represented by a sequence of holomorphic functions defined 1n G and normally convergent 1n .D, i.e. D has the Runge property relative to G. Theorem 36 (Oka). If D is a region of holomorphy, then Hq(D,l9) = 0 1

q > 0.

Corollary. C.I is solvable in every region of holomorphy. Remark. Note that this theorem completes the proof of the de Rham theorems of Chapter 8. Proof (of Theorem 36). Note that every analytic polyhedron is a region of holomorphy. Exhaust D by a sequence of analytic polyhedra Dj cc Dj~lenc:::... D, 1 Dj = D; compare corollary to Theorem 7, p. 25. Now, by Theorem 34, Hq(Dj,(?) = O, q > o, j = 1,2, ••• ; and by the Corollary to Theorem 35, Dj has the Runge property relative to Dj+l• Hence we may apply lemma ~.

P-

§3.

Preparation for the proof of the fundamental lemma

A. Proposition. Let Kcornpactc: t.n, and let K* denote its polynomial hull. Let Gopenc. en, and [a~t~b]CTopencc. «. Let F(z,t) be holomorphic in G x T, continuous in G and such that {z I z e bdry G, F(z, t) = o, t e "T}/\ K* = ~~ and F(z,t) -:f 0 on (K "G)>< T. Then either, for each t e [a,b] there exists z e K* ~ G such that F(z,t) = 0; or F(z,t) -:f 0 on (K * A G) x [a,b] • . Proof. We may assume T to be a polynomial polyhedron. Let U be an open neighborhood of K* , which we may also take to be a polynomial polyhedron, for (K*) * = K* • Further, since F(z,t) -:f 0 in (K* ()bdry G)x T, we may choose U such that F(z,t) f. 0 in (U llbdry G).x T. Therefore there is an open set H c.c G such that F( z, t) -:f 0 in U fl (G-el H) x T. Now Ux T is a polynomial polyhedron, hence C.I is solvable ( cf. Theorem 32). l·le are able to find a meromorphic function G(z,t) in UxT with poles at the zeroes of F(z,t).

x,,

118

To this end, we pose the following problem: In (U II G) x T, set n r 1 = 1/F(z,t); in (U t1 (C -cl H))xT, set r 2 = 1. This is a well-posed problem, for F(z,t) ~ 0 in (UI\(G-cl H))xT. Let G(z,t) be the solution of this problem. Set yt = tz I z E u llG; F(z,t) = oj. Assume the P~oposition * . is false; i.e. that Yt I) K {: + tor some t 0 e [a, b); Yt 1\ K* = + for some 0t 1 e [a, b). Now 1.5 tiYt I)K* {: 4> 1z is clSsed; henee there exist numbers b0 ,b1 e [a,b] such that 0 1 A* * .L for b < t < b , Yt '' K = ~ and Y I) K f: "'. Choose t e (b 0 ,b 1 ). Then there exis~~ an open neighborhood V or K* such that K*c. V c.U, and Yt 1\ V = We may assume V is a polynomial polyhedron, as any smaller neighborhood has the same property or non-intersection. Consider G(z,t) for this fixed t; it is holomorphic in V, and can therefore be expanded in a normally convergent series or polynomials in V (Theorem 32). Now G(z,t), for a~ t ~ b, is continuous in K; so IGI has a maximum here: IG(z,t)l 0 be given, and let N be sufficiently large so that j > N implies IG-Pjl < E on K. Hence IPj(z)l N, so IPjl ~A+ Eon K*, implying IG(z,t)l_! A+ 2e for z e K* • But there exists a C e Y 0 ll K* , and b 0 l1mt~b 0 IG(t,t)l ~A+ 2e, although G has a pole at (t,b ); and this contradiction establishes the proposition. B. We now establish some properties of subharmonic functions; their definition may be restated as follows: Let Dopen~ c. Then +Cz) is subharmonic in D if: (i) - oo < +Cz) < oo (ii) + is upper semicontinuous; i.e. zl~ +
+•

-

-

~(z) ~

-Jp

J

lt-zl=p

+

dC

•

119

+

Note. (1) and (11)_ imply + is measurable, for may be-approximated from above by continuous functions. Pro2osition 1. Assume to be subharmonic in D; then + possesses the following properties: (1) + is bounded from above on compact sets ( 2) If 2 - oo in a neighborhood of some point z or D, then + oo on the component or D containing z. (3) (Maximum principle). Let Kcompactc D. Then +Cz) _! A on bdry K implies +Cz) _:: A in K. (4) Let D(z0 ,r) < D, and assume + = constant in D(z0 ,r). Then + • constant on D(z0 ,r). Proof. (1) follows from (ii) and (2), (3) from (iii). We prove (4): We may assume z0 = 0, r = 1, and + 0 on D(O,l) = [lzl < 1}. Then + ~ 0 on {lzl = 1} , using (ii). Assume, e.g. +Cl) > 0; and let 1 > e > 0, R > 1 such that D(O,l-e) ~ D(O,l) D(O,R) CD. Then + ~ A in D(O,R).

+

+ =-

=

7

Define u(z)

=

log lzl -log A

log R -log

11-e(

11-e:

•

Then u is harmonic in D(O,R) - D(O,l-e), + < u and u = 0 for lzl = 1- e. Hence, +Cl) < A log ~/log~ ; ~-E ~-E letting e ~ O, we obtain +Cl) = 0. Proposition 2. If + ~ 0 and log + is subharmonic, then + is subharmonic. Proof. (i) and (ii) follow from the monotonicity of "log", and (iii) from the inequality:

Proposition 3. Let +Cz) ~ 0 be defined and upper is not sub harmonic semicontinuous in Dc. a such that log at z0 e D, and +
+

120

and continuous in D(z0 ,r)., such that +11/11 < 1 on llz-z0 1=rJ and +1 = 1. Proof. We may assume z0 = o. Let r ~ 0 be chosen such that D(O,r) c. D and

log:~(O)

>

~

j"'

log 4>(z) dz

•

lzl=r Using upper semicontinuity, let +n be continuously differentiable in D, +n .J. cf>. Let hn(z) be the harmo~ic function for which log +n(z) = hn(z) on lzl=r. Let hn be the conj~gate harmonic function, and set -h -ih 7/1

n

=e n

n.

Now on tlzl=rJ, log ~17/lnl = (log 4>> - hn =log+ • log +n < o, i.e. +11fln1 < 1. If 4>(0) 11/Jn(o) I = K;: 1, then 4> 11/lnJ ~ ~11/lnJ, so 11/ln(z)l is the required function. If +(0) + ln lfn{O)I < o, i.e. ln +Co) < hn(O) = 2~ hn(z) dz = 2.~} ln 4>n(z) dz, Iz I=r Iz I=r therefore ln 4>(0) ~ n!~ l/21rr. ln cf>n(z) dz, so lzl=r ln ~(0) ~ 2~ ln +(z) dz, contradicting choice of disc. lzl=r

i

i

J

f

J

§4.

Proof of the fundamental lemma

Statement. Let Gopencc ,n and assum~ (for the sake of s1mplici1ty) that G cc(lzj(1 ~ 1 for 1 = 1, ... ,v}: l5 is compact. Let L: =f
121 (z 1,z 2, ••• ,zn,~l'''''~v) = (z1 ,Z,C). Let ~(z) mean ~/\tz 1=z] and ~* (z) mean ~* I) [ z1=z]. Denote by S = { z I there is a (Z,C) for which (z,Z,~) e ~*f: S is a closed bounded set. Let = [z I there is a (Z,C) tor which (z,Z,C) e ~* and (z,Z) I. GJ: n0 is a closed set C:: S. Call the complement of C0 1 0 : 0 is an open unbounded set. 1) For z e 0 and j = l, ••• ,v, set if z I. s

CO

ro

Rj(z)

=

l

max

I~ 3-r j ( z,Z) I*over all ( Z 1 Z 1 C) E ~

~

if

Z E

S • A

If, for fixed z and all j = l, ••• ,v, Rj(z) = O, ~ A L(z) = Z:::*"" (z). Proof: If Az I. S, then L *""\ (zJ = 'I'.1. h A and, since 2: c.Z:::* and hence L(z) c 2:* (z), " L(z) = ~. If A. z E s, then for all j = l, ••• ,v, cj = J\ A * "' )\ fj(z,Z) for every (z,Z,C) E ~ • Since z E o, z l o0 , "' " and so (z,Z) E G for every (z,Z,C) E 2:: * • Hence for J\ * .1\, A. every {z,Z,~) E ~ (z), ~ cj = fj{z,Z) and (z,Z) E G. Therefore {z,Z,C) E 2::{z). 2) log Rj(z) is subharmonic for each j = l, ••• ,v. Proof: It is enough to consider R1 • R1 ~ 0 and is upper semicontinuous, i.e. lim R1 (zj) ~ R1 (z). For, if z I. S zj->z then R1 (z) = o. If z E S and zj -> z, then for large j, either zj is and then limR 1 (zj) = 0 ~ R1 (z), or zj E S zJ•>z and then R1 (zj) =max lc 1-r1 1 = lci-f1 (zJ,zJ)I for some (zJ,zJ,Cj) e 2:*· By the compactness of L*, there is a subsequence [(zJ,zJ,Cj)7 converging to a point (z,Z,C)e 2:* j A j 1\ and z !zR1 ( z ) = IC1- f 1 ( z, Z) I ~ max IC1- f 1 ( z, Z) i = R1 (z) •

J

Hence R1 is upper semicontinuous. Now, assume that log R1 is not subharmonic at some z0 Eo, then there is a closed disc in n, [lz-z0 1 ~ p}, and a function t(z) holomorphic in the open disc and

122

continuous on its closure, such that R1{z) lt(z)l < 1 on lz-z0 1 = p, ,and therefore < 1 - 2E for some 1/4 > E > o, and R1 (z0 ) lt(z0 )1 = 1. Define F(z,Z,~,t) = (~ 1 -f1 (z,Z))t(z) - (l+t)eia for all ~~ (z,Z) E G, lz-z0 f < p, ~nd t E T = [t E e I lt-t0 1 < E and t 0 E [O,al] , wher~ a and a are fixed real numbers to be determined later. We claim that this F satisfies the hypothesis of the proposition with K = L c en+v , K* = 2:* , T = T and G in the propos! tion, call it GP, = G f\(lz-z 0 1 < p) and all~. Indeed: a) F :l.s holomorphic in Gp X T and continuous in cl GP J( cl T. b) If r- = ~z I z E bdry {G A(lz-z0 l" = ~. Proof: Bdry (G ll (I z-z0 I(max 1~ 1 -f1 (~,z)l) lt(z)l ~ 1~ 1 -f 1 (~,Z)I 17/J(z)l while ll+tl leial ~ 1-E and thus F(~ 1 Z,~,t) ~ o, i.e. i a contradiction. Hence

r

l

r

r (\

cr

"'

r (\ L

"'

* = ~.

"'

1\1

"'

z r- ;

>

c) F(z,Z,~,t) ~ 0 for (z,Z,~) E and t E cl T because on > , ~l = f 1 and therefore IFI = (l+t) > O. However, the conclusion of the proposition is violated. Fo~, by choosing ~ sufficiently large, l(l+a)l can be made greater than 1(~ 1 -f 1 )tl, since the latter is bounded, so that F ~ 0 at t = a. But for t = 0, F = (~ 1 -f 1 )t - eia = 0 at a point of 2::* II G: for at z0 , R1 17f11 = 1 implies that R1 (z 0 ) ~ 0 andAhence zQ E S and R1 (z 0 ) ~ * max 1~ 1 -f 1 (z0 ,z) I = I~ 1 -f1 (z0 ,z) I for some (z 0 ,z,~) E L .

12:5 A

1\

J\

-

18

A

lcl-tl.(zo,z>t I~Czo>l = 1 or ccl-tl(zo,Z)) t{l(zo> = e • Choose a= Q.. Hence log R1 (z) is subharmonic, and R1 is then subharmonic by Proposition 2. Let o1 be an unbounded component or o. (o1 ~ exists because 0 contains the complement of s, a compact set 1n c.) :5) ·L(z) = :L:* (z) tor z £ o1 • Proof: Firstly, there is a z £ o1 , for which a3(z) = 0 for all j = l, ••• ,v. For, since o1 is unbounded and S is bounded, (o1 1l {I-S)) ~ Therefore there is a z £ ~ and z i S and for this z, R3(z) = o for all j. Secondly, since (o1 (\ (C-S)) is an open set, there is a neighborhood of this z contained in ('1_ I) (c-s)) 1n which RJ = o tor all J. By 2), since Rj vanishes 1n a neighborhood of z, it vanishes 1n the largest component containing z, i.e. 1n o1 • By 1), then, ~(z) = L * (z) tor z e·~· If we can prove that ~ = e, then we are done. So, assume o1 rf. e, then o1 has a boundary. point ~ • ~ is also a boundary point of f.b 1 and since ~ is closed, Thus

+

+•

~

£

Oa· 4>

L<~> ~ ~.

Proof: Since E e o0 C s, either ~ e int S or ~ £ bdry s. a) If ~ e 1nt s, then there is a sequence of points ~ j -> ~ , ~ j e iS"'\). Since ~ j e S, there are points (~3,z.1,cJ> e L, and ·by 3), c~3~z3,c3> e L:• By the compactness of ,.-1 there is a convergent subsequence j j j jj .. j (~ , Z 1 C ) , ~ -> ~ 1 Z -> Z, C -> C and (~ 1 Z, C) E L • b) If ~ e bdry S and L<~) = +1 then for every .1\ point (~ 1 Z,C) with first coordinate sufficiently close A.. to ~~ £.-.. ~(~) = ,, by compactness, i.e. there is an£> 0 h 1\ such that if ~~-~I < E then L:<~) a ' · Since ~ £ bdry S, there is a point ~ 1 with I~ -~ I < e and ~ 1 I. S. Consider the function F(z,Z,C,t) = z- ((1-t)~+ t~ 1 ) for

124

t e T ={t e C I lt-t0 1 < e1, and t 0 e [0,1]}, where e1 is cho~en such that (1-t)~+ t~ 1 for t e T is contained in (1~-~l<e). F is holomorphic for all Cz,Z,C,t). F ~ 0 for z e L and t e T because z = (1-t)~ + .t~ 1 lies in the disc or radius e about ~ and by assumption, then, L:<·z) = 4>. Hence, F satisfies the hypothesis or the proposition. However, for t = 0, F = z - ~~ and since ~ e S, [F=O] (lL * = [z=~] () L * = L::* (~) p ~; while for t = 1 ~ = z -*~ 1 and since ~liS, [F=OJ ()2::* = [z=~ 1 l IlL: = L: (~ 1 ) = ~~ contradicting the conclusion of the proposition. Thus ~(~) ~ ~. * 5) L:(z) = L:Cz). Proof~ Use induction on n. Either n = 1, or n = k+l and the fundamental lemma holds for n~ k. If n = 1, then L(z) is a point, namely the point (z,r1 (z), ••• ,rv(z)), and a point is its own polynomial hull. If n = k+l, then :L:(z) = f(z 1 , ••• ,zk+l'c 1 , ••• ,cv) I lzjl ~ 1, lcil ~ 1~ (zl,Z) E G f\[zl=z] and ci = ri(z,Z) for i = l, ••• ,vj and therefore is the Oka image or an analytic polyhedron in ek. By the induction hypothesis, then, L:Cz) = Z::Cz) * • 6) There is an e > 0, a number b, 0 < b < 1, and N polynomials PJ(Z,C) such that a) if lz-~1 < e and for all j, IPj(Z,C)I < 1, then (z,Z) e G and b) if lz-~ I < e and (z,Z,~) e L: then for all j, IP j(Z,C) I < b. Proof. a) If (~ 1 Z,C) e 2: then (~ 1 Z) e G, and there is a point (~ 1 Z,C) e ~ by 4). Since G is open, there is a neighborhood of (~,Z,C) such that every point in this neighborhood has its first n-coordinates in G; i.e. there is a 6 > 0 such that if the distance from (z,Z,C) to ~(~} is less than 6, then (z,Z) e G. But since

125 = Z:<~) * , any neighborhood of 2:(~) contains a smaller neighborhood which is a polynomial polyhedron. Hence we can find an E1 , and polynomials Pj with the required property a). b) At every point (~,Z,~) E 2:, IP j(Z,~) I < 1, j = 1, ••• ,N. VIe claim that f-br z sufficiently close to ~, and (z,Z,C) E 2:, IPj(Z,~)l < 1 for all j. Suppose not, then there is a sequence ~ k -> ~, and Zk, ~lc such that (~k,zk,Ck) E ~ and max IPj(~k,zk,~k)l > 1 for each k. Since ;;- i~ompact ~here is a subsequ~nce (~k,zk,Ck) ."'\ /\ which converges to a point (~ ,z, C) E > • renee in the limit we obtain a point E 2:<~) \'lhere at least one of the polynomials Pj has absolute value> 1; a contradiction. Therefore there is an E2 neighborhood N of ~- sue~ that IP j{Z,C) I < 1 for all j and (z,Z,~) E {Li1[z E Choose E =min (E 1 ,E 2 ). Then if lz-~1 < E and {z,Z,C) E ~~ IPj(Z,C)I < 1 for all j, and since IPjl is continuous, there is a number o < b < 1 for which IPj(Z,C)I * and j = l, ••• ,N if z c S. log ~(z) is subharmonic. Proof: ~(z) ~ b > 0 and is upper semicontinuous. (The proof is similar to that of part 2).) Assume that log ~(z) is A. not subharmonic. Then there is a disc, C = (lz-~l o, ~~d ~(~) lt(~)l = 1. By definition, ~{z) ~b. Therefore on lz-~1 = p, l~(x>l < 1/b. Since ~ is holomorphic in A int C, I1.H~) J. < 1/b and hence II, ~(~) >b. This means that ~ E S and ~(~)=max IPjl. Pick out one polynomial assuming the maximum value, say P1 : ,, All. .A.AA * ~(~) = IP 1 (Z,C)I, {~,Z,~) sZ:. Let F(z,Z,C,t) = P1 (Z,C) *(z) - eia{l+t) for all z, C, lz-gl < 9 1 and

2:<~)

Nk).

126

t e T = lte C I lt-t0 1 < e and t 0 e [O,al] where· a and a will be determined later. F satisfies the-hypothesis or the proposition. A * a) If z e (lz~~l=p) and (z,Z,C) e ~ t~en F ~ 0 since then z e S and ~=max (b,IP f) and satisfies +1~1 < 1 or +1~1 < 1-2~. But all .jpjf < + implies that IP 1 1 1~1 < 1- 2e on lz-~1 = p. BuX leial ll+tl ~ 1-e. b) If ( z, Z, C) e ( 2: ()( I z-~ I 1-e. However, F contradicts the conclusion of the proposition •. For 2:** C ( lz 3 1~ 1, ICil.~ 1) means that P1 is bounded on 2: , and for lz-~1 "' < p, ~is bounded; hence for a large enough choice or a, ll+al can be made > IP 1 1 ltl so that F(z,Z,C,a) # 0 * "' for "'<.~, Z, C) ~ L:. . For t = 0, 1 = i )"' 11/1 ( ~ ) I = IP1 (Z,C)I 1~(~)1. Hence at the point (~,Z,C), there is a (!" such that P1t = eiO'"'. Take a == cr. Then F = 0 at AJ\1\ ""i\.'\ * ( ~ , Z, C, 0 ) and (~ , Z, C) e 2: . 8) Consider only z e Clz-el<e). Since ~ £ o0 , there is a point with first coordinate ~ belonging to (~*-~), since there is a point £ ~* (~) with (z,Z) i G. At this point ~=max (b,IPjl) and at least one polynomial is ~ 1, i.e. ~(~) ~ 1, for if not, then (z,Z) e G by 6a). Now ~ e boundary o1 means that 1n ( I z-~ I <e) , there is ··lin ~. e o1 . with ~(~) =b. For a) If ~ i S, ~ ( ~) = b. b) If ~ e S, consider any (~,Z,C) £ ~ * • Since ~ e o1 , ( 11 , Z, C) e ~ by 3) • By 6b ) , IP j ( Z1 C) I < b for all j at (~,Z,C) , and hence ~(~) = b. Next, consider the set of points (in (lz-~1 <e)) where ~ ~ 1. This set is closed because ~ is upper semicontinuous, and at ~~ ~ < 1. This means that there is a closed set with c1 boundary 6 C:(lz-~l<e), containing ~, such that ~ < 1 1n int 6 and cl> ~ 1 at some boundary point of 6. Now in

!(

A.

127 1. Therefore either z i S and hence z i o0 , a subset of s, or z E s. If z E S, then-since 4> ~ b. Let S be a closed disc c 6 and tangent to o6 at a point p where 4> ~ 1. Then 4> • b in int S and 4> (p) ! 1. But a sub harmonic function which is b in an open disc is E b on the boundary of the disc (by property{4)of subharmonic functions, p. 119). But ~! 1 at a boundary point of 6; a contradiction. Hence o1 = e and by 3), then, L_{z) = ~* {z) for all z E ~~ i.e. * int 6,

~(z) <

=

L:=Z:·

128

Chapter 12.

Some Consequences of the Approximation Theorem

§1. Relative con~exity Definition 55. Let D0 c. D, both open in an. We say that- D0 _!!! D-convex if, for every K
= [ z0

e D0

I

If( z0 ) I ~ sup If( z) !

for every f

holomorphic in

,

z e K,

D} .

" the hull of K with respect to Note that K1 JK, functions holomorphic in D0 • Theorem 37. Let n0 C. D, both open in ¢n. Then the following statements are equivalent: 1) D0 is D-convex. 2) There exists a sequence P jc GP j+l' ljP j = D0 , where each Pj is an analytic polyhedron in D.J 3) n0 is a region of holomorphy and (D0 ,D) is a Runge pair. Note. This is in some sense an extension of Theorem 33 of Chapter 10, g3; \'lhere D = Cn. Proof. 1) implies 2) : Let Kc. c D0 • There exists an open set D1 c..cn0 such that K1 c....c.D1 • Now, for every ~ E bdry n1 there exists a function f£ such that If~ I < 1 on K1 and lr~(~)l > 1, where f~ is holomorphic in D. Hence,there exists a neighborhood N~ D of ~, such that If~ I > 1 on N~. But n1c c. D0 ; hence bdry n1 is compact and a finite number of neighborhoods N1 , ••• ,Nr suffice. Let ~ = [z I zED, lfj(z)l < 1, j = l, ••• ,rJ. Now P is an analytic polyhedron in D, with P cc n1 , and K c. K1 c <:. P ; hence every compact subset of D0 may be ~nclosed in an analytic polyhedron in D. So, let K c. K c ...c D0 be a sequence of compact al. a2 sets such that ~ Ka = D0 • As above, there exists an analytic polyhedron ini D, P1 ~ K • Now, cl P1 UK is al a2

129 compact; so there exists an analytic polyhedron in n; P2 :;:;J cl P1 UK~, and so on. . 2) implies 3). Each Pj is Runge in Pj+l• Hence lemma ~ (p. 111) is applicable. Th~refore D0 is a region of holomorphy (by Theorem 21, p. 75) and (D0 ,D) is a Runge pair. ...... 3) implies 1). · Let K c<. n0 , and let K, K1 be as ~ A ~ in definition 551 K C. K1 • ~le c~aim K = K1 • Assume q i K; we must shovl q i K1 • Now q i K implies that there exists a function f(z), holomorphic in n0 , such that lf(z)l < 1 on K and lf(q)j > 1; in fact, there exists an E > 0 such that lf(z)l < 1- 2E on K, and lf(q)l > 1 + 2e. Since (D0 ,D) is a Runge pair, there exists a function g, holomorphic in D, such that lg-fl < e on the compact set K U fqf • But lg(z) I = lg(z)-f(z)+f(z) I ~lg(z)-f(z) I+ lf(z) I < 1-E on K; and lg(q) I~ jf(q) 1-lg(q)-f(q) I > l+e; so q f. K1 •

S2. Unbounded regions of holomorphl

r. Let Xopenc en; let xj c

xj-~l

be such that Xj =X, where Xj is a region of holomorphy and (Xj,Xj+l) ts a Runge pair. Then X is a region of holomorphy. Proof. As indicated by the proof of lemma ~ (Chapter 10, §1), we may assume Xj is Run~e in X. Recall that, for every Kc..c..Xc~n, 6(K) = 6(K) if and only if X is a region of holomorphy (Theorem 5, Chapter 2, S2}. Hence, A,. assume there exists a K c.c. X for which . 6(K) < 6(K), where 6 is taken in the maximum norm; i.e. there exists a q e "K - K such that 6(q) < 6(K). For j sufficiently large, Xj ~K u tq~. Now for every f holomorphic on Xj, lrl < 1 on K implies lf{q)l < 1, i.e. q e [Hull of K with respect to functions holomorphic in Xj]. But Xj is a region of holomorphy, so dist (q, bdry Xj) ~ dist (K, bdry Xj); i.e. letting j -> co, 6(q) ~ 6(K), which is a contradiction.

V

Lemma

130

Note. One obtains another proof by constructing a I I I sequence Xj(,:'c.Xj+l such that V Xj = X, using analytic polyhedra as in 11 1 implies 2" of' ~heorem 37. Remark. (Oka) Let XC (Jn; we assume 0 e X for simplicity. Let Xr = X ntll z II < by assumi>tion. Then X is a region of' holomorphy if and only if' ~ is a region of holomorphy for every r > 0. !!~· Assume Xr is a region of' holomorphy for every r > o. If Xr is Runge in XR for r < R, we are done by lemma 1• Using the preceding theorem, it suffices to show Xr is XR-convex. Let K c.c.Xr; let K1 be the XR hull of K. K1 c c XR as ~ is a domain of' holomorp_hy. But, f'or every z e K, II z II < r - e. Hence, in K1 II z II ~ r - e ; where the norm is the maximum norm. Hence, K1 cc.. xr. The reverse implication is clear. -

:J, f '

§3.

The Behnka-Stein Theorem

Theorem 38. (Behnke-Stein). If Xj C.. Xj+l C Oln, and Xj is a region of' holomorphy, then X = U Xj is a j region of holomorphy. We first require a lemma: Lemma 5. Let n1 c.. c n2cc. n3 c c. ¢n be regions, n3 a region of holomorphy, such that dist (z 2-bdry n3 ). z1e bdry n1 z2e bdry D2 z3e bdry n3 . (This means that any subset or n3 whose distance to oD3 is~ the distance of cl D1 to oD 3 -lies in int D2 .) Then there exists an analytic polyhedron P in n3 such that D1 cc.PccD2 • Proof. Let K = cl n1 , and let K1 denote the n3-hull of K. Then t;,(K)=~(K 1 ), so K1 cc.D 2 • Butnowthere exists an analytfc polyhedron P such that K1 c. P cc D2 , as in "1 implies 2" or Theorem 37. min

II z1-z 3 11

>

max

131

Proof or Theorem 38. Utilizing Oka 1 s remark, we may assume X<:.C ff. We may also assume Xj ccxj+l as every region or holamorphy can be written as a strictly increasing (i.e. ~elatively co!DPact) sequence or analytic polyhedra (see corollary to theorem 7, Chapter 2, § 5). Define: M j

= max

z e bdry X j

M

jl, j2; jl <j2

min

= max

II z-C II

C e bdry X

min II z-C II z e bdry xjl e bdry xj2

c

minz e bdry X mine e bdry ~I z-C II j

mjl,J2;.1l<j2

=

minz e bdry xjl mine ebdry x,;lz-CII.

We select a sequence v1 ,v 2, ••• or subscripts as follows: Set v1 = 1. Choose v2 such that Mv < mv , possible 2 1 as r4v 2 -> 0 as v2 -> oo • Select v3 > v2 such that Mv 2,v < m · 3 vl,v3' Mv < mv , possible as lim Mv v = M < m = lim mv v 3 2 v3->oo 2' 3 v2 vl v3->m 1' 3 and Mv -> 0 as v3 -> en • Continuing, select vj > v j-l such t~at M < m v and M < m ; possible vj-l'vj vj-2• j vj vj-1 as before. Consider now the subsequence Xv ; any three successive terms Xv c c X c c Xv satisfy1 the conditions of q Vq+l q+2 1 emma u for II:

minz c bdry. X II z-C II = m > M v Vq vq,vq+2 vq+l'vq+2 Ce bdry Xv q+2 maxz E bdry X mine E bdry X II z-C vq+l vq+2 Hence, there exists a sequence fPjJ of analytic polyhedra such that X c c Pjc c x.. , where Pj is an analytic vj v.1+1 polyhedron in Xv and nence in Pj+l<: Xv • j+2 j+2 =

n.

132

Appealing to the Oka-Weil theorem (theorem 35), (Pj,Pj+l) is a Runge pair. Now use lemma r (or 13) to obtain the desired result.

§4.

ARElications to the Levi problem

We recall the Levi probl~: Is every pseudoconvex domain a domain of holomorphy? We shall reduce this problem to the consideration of "strictly pseudoconvex" domains, defined as follows: Definition 56. Let Gopenc: ~n, and let '(z) be a real valued, c00 function on G. Let ' be strongly plurisubharmonic i.e., (d 2 ~; dZidzj ) > 0 for every z E G; assume also that ~ > 0 near the boundary of G and ~ < 0 somewhere in G; more precisely, there exists an e > 0 such that for every z e ~z I diet {z, bdry G) < E, z E aj, ~(z) > 0 and there exists a z0 E G such that ~(z 0 ) < 0. Now consider the open (nonempty) set G_= I z EG, ~(z) < 0; it is called strictly pseudoconvex; any region D for which there exists a Gopen<: ~n and strongly plurisubharmonic (real-valued, c00 ) function ~ for which D = G_ is called a strictly pseudoconvex region. We remark that a strictly pseudoconvex region is pseudoconvex; consider log (-1/~). The strict pseudoconvexity is essentially a "smooth boundary" condition. Example. The unit ball is strictly pseudoconvex: set n ~ = ~ zjzj ~ 1. The unit polydisc is not, and analytic polyhedra are generally not, strictly pseudoconvex. Theorem 39. Every pseudoconvex region is the limit of an increasing sequence of strictly pseudoconvex regions. Proof. Let D be pseudoconvex; i.e. there exists a real-valued, continuous, plurisubharmonic function t in D such that t -> + CD on the boundary of D. We claim that for every compact subset K of D there is a strictly

tz

133 pseudoconvex region D0 satisfying K c.. D0 c c.. D. To prove this, notice first that since '1/1 -> + 00 on an, there is a positive integer N so large that K c s c T Lcompactc D, where S = fz e D I '1/J(z) < N_), T = 1z e D I 1/J(z). < N+Y and L is any compact subset or D containing T. Give e > 0 small. By Proposition 2, p. 28, there exists a c00 plurisubharmonic function '1/Jo on L such that lt(z)-'1/10 (z)l< e on L. Define ):_(z) = 1fJ0 (z) + e/M zizi;

c

n

f.:.

~1

M = maxL ~ zizi. Then X(z) is coo and strongly plurize _ Y subharmo l on L. For z e S, ,_(z) < N + 2e; for z e (L-cl T), X(z) > N+ 1-E. Therefore if we set D0 = ''fz e int L I~ (z) < N +1/2], then- n0 is strictly pseudoconvex and K c D0 c L, as claimed. Let Kj be an increasing sequence of compact subsets of D such that 'iKj = D. For each j, let Dj be a strictly pseudoconvex region, Kj c. Dj c:: c. D. We choose a sequence v j of subscripts as follows: Set v1 = 1; D1 C Kj 1 for some .1_ ; hence D1 c. c Kj • Set v 2 = jl; and so on. Note that Kjc.D c D ••• 1 vj vj+l so that D c C D and U D = D. vj Vj+l j vj Proposition. The Levi problem is reduced to the following: Is every strictly pseudoconvex region with compact closure a region or holomorphy? Proof. Use the preceding theorem together with the Behnke-Stein theorem. ----.;;.~,_;_..

Chapter 13. Solution of the Levi Problem The object of this chapter is to prove the following theorem. Theorem 40. If Dopen c¢n is ~seudoconvex, it is a region of holomorphy. Since every pseudoconvex region is the union or an increasing sequence or strictly pseudoconvex regions (Theorem 39) 1 and the union or an increasing sequence or regions of holomorphy is a region or holomorphy (Theorem 38), it suffices to prove Theorem 40 I • If Dopenc c en is strictly pseudoconvex, it is a region of holomorphy. The Levi problem has been solved first by Olm (for n = 2), then by Bremermann, Norguet and Oka for any n. There exist today many proofs of this theorem using either Oka's original method, or functional analysis.methods (Ehrenpreis, Grauert, Narasimhan, Andreotti-Gr~uert, etc.), or partial differential equations (J. J. Kahn). The proof given here follows mainly Grauert and Narasimhan. The idea of using an "Extension Lenuna" plus a result by L. Schwartz to establish finiteness of cohomology groups is due to Cartanand Serre. Reduction to a finiteness statement Proposition 1. If Dope~.c mn is strictly pseudoconvex, th en dim Hl(D , ..0) < oo. , . Let U = f ui<~ be a fixed, locally finite covering. The 1 "' ~ dim H (D, ) ) = m < oo means that given (m+l) cocycles on U, a nontrivial linear combination or them is a coboundary on a refinement of u. But a cocycle on U is a set of Cousin I data, and a cocycle induces a coboundary when the induced Cousin I is solvable. Since \-Je can add Cousin data and multiply by constants, dim H1 (D,0 ) = m < oo means that a nontrivial linear combination or (m+l) CI data is solvable. Proposition 1 implies Theorem 40 1 • Proof~ Recall that Dope~c.¢n is strictly pseudoconvex if there is an open set Gc".:. 0 such that if m is a real-valued §1.

135 000 function 1n 0 and if lrot I Iem/ozj II Io~/ozjozkl are all~ e 1 n1 =- [~-Ku
n1c..c e •

since ro is small and ~-ftu is a real-valued C~ function 1n G. ~-kl> is strongly plurisubhar:monic since the sum of a small quadratic form and positive definite quadratic form is a positive definite quadratic form. Lemma 2. Let Dope~c en be a strictly pseudoconvex region. There is an e > 0 such that if q e bdry D and B is the polydisc or radius e about q, there is a quadratic polynomial Q, in z1 , ••• ,zn, with Q(q) = 0 and Q(z) ~ 0 in D f\ B. Proof. Since ~ e 000 1n G and is real-valued, by Taylor's theorem, n ~(z) = ~ [Aj(zj-qj) +Aj(zj-qj)] + ~or.

D

c

e.G

where H = (Hij) = H* and R involves third order terms. So, ~(z) = 2 Re Q(z) + (z-q) * H (z-q) + R where

Q(z)

~ (~

AJ(zFqJ) +

~

A1 J(z1-q1 )(zJ-qJ) •

i~l No~-since the hessian of is positive definite, H is a positive definite hermitian matrix of second order derivatives of at q. Since H varies continuously with q, and at each boundary point q of D it has a positive smallest eigenvalue, the minimum eigenvalue of H over oD, call it a, is positive. Hence (z-q)*H (z-q) ~ allz-qll2 • Similarly by estimating third order derivatives of ~ over oD, we can find an M > 0, independent of q, such that - Mllz-q 11 3 < R < Ml~-q113 • Hence (z-q)*H (z-q) + R > (a-MIIz-qlllllz-qll2: - E = a/M; then for II z-qll < E, (z-q) - *H(z-q)-1-R > 0 Take j-

+

+

except at q where it vanishes. But in D" B ~ < 0; therefore Q f 0. Lemma 3. Let Dope~c. _,n be a strictly pseudoconvex region. There is an E > 0 such that if q E bd~y D and B is the polydisc of radius~ E about q, then B tj D is : a region or holomorphy. Proof: We will show that every boundary point is essential. Take E = 1/2 E2 , the E of lemma 2. Bdry (B flD) = ( (bdry B)() D) V(B /)bdry D) U(bdry Bnbdry D). If y E bdry B, then since B is a domain of holomorphy, there is a function holomorphic in B and singular at y. If y E (bdry D)A B, then the polynomial Q, of lemma 2, corresonding to y, is holomorphic, Q(y) = 0, and Q(z) F 0 in D A B. Hence 1/Q is holomorphic in DAB and singular at y. B. Assuming proposition 1, we will show that every boundary point or a strictly pseudoconvex D is essential. Let q E bdry D. Let B be the polydisc of lemma 2 and Q(z) the quadratic polynomial. Let ru > 0 be a c00 function whose support lies in the interior of B and ru(q) > o. Consider ~ - tn, where t > 0 is small. By lemma 1, if t is small enough, n1 = [~ - tm < 0] is strictly pseudoconvex. DC Dl cc.
-

137 holomorphic in u1 and F(z) - 0 = F(z) is holomorphic 1n D. This function F(z) is holomorphic in D and has a_ pole at q. ~2.

Reduction to an extension property

Propositio~.

Let Dopenc:~ mn be strictly pseudoconvex. There is another stri~tly pseudoconvex region D0 with Dec D0 cc o:n such that if a= L a 3 dzj i~ a c00 differential form of type (0,1) in D and ~a= 0, then there exists a c00 form ~ = ~ bj dzj in D0 with ~~ = o in D0 and a C00 function ;:_ 1n D, such that a - ~ = 'd.X in D. 2 means that every closed (0,1) form in D is cohomologous 1n D-to a (0,1) form defined and closed in a larger region. 2 implies 1. Proof: By Leray's theorem, the cohomology groups of D, Hr(D,~) are isomorphic to the cohomology groups ~(D,U,(9) of D with respect to any simple covering of D. Hence we need only consider cohomology with respect to a simple covering 1n order to prove 1. Let £ 1 ,£ 2 be the e 1 s given by Lemma 3 for D and D0 , respectively. Take £ = l/2.m1n (£ 1 ,£ 2 ) • Cover ~Do by finitely many open_polydiscs of radius £ such that (*) the closures of the polydiscs are disjoint when their interiors are disjoint. Complete this covering to a finite 1 ' . covering V = v j ~ of D0 by adding open polydiscs of radius £ satisfying (*) and whose closures do not intersect " oD0 • About each v3 take a slightly larger open polydisc v 3 I II I . II such that v j c:c:: v3, if "cl vJII ~Do = ~ then cl v j f) ~Do = ~~ if v~ 1\ vk = ~ then v j t) vk = ~~ and such that for each j, v~' A D0 is still a region or holomorphy. Set u' = Su 3' = 1 II ': II fl .., 1 11 l ~j/)D and U = Luj = v 3 t1 DoJ· Then U and U are ~-simple coverings of D and n0 , respectively, since they are finite coverings by regions of holomorphy. With each covering we have the groups of cochains,

f

'

cr(D,U 1 , &>, Cr(D0 ,u" 1 0) 1 and the groups of cocycles zr (D 1 U' 1 (!}) and zr (D0 , u~' 0) • As we have already noted, these groups are also linear vector spaces. With the following notion or convergence, it can be shown .that cr, and hence zr , is a topological vector space: a sequence or elements or Cr (D, U, (9 ) converges if in each u1 ll ••• I) ur+l#+ the sequence of holomorphic functionsassigned there converges normally. In fact, these spaces are Frechet spaces, i.e. Hausdorff, locally convex, metrizable and complete under the metric. Define the mappings 6: 0° (D,U 1 ,(9)

->

z1 (D,U 1 ,(9)

to be the coboundary operator and r:

z1 (n0 ,u",0)

->

z1 (n,u•,<9) z1 (D0 ,u",&)

to be the restriction map; i.e. if z e then r{z) is the restriction of z to the covering u1 • 6 is a continuous linear map and r is a completely continuous (i.e. compact) linear map. To show that r is indeed compact, we must show that there is a neighborhood of the origin which is mapped into a relatively compact set. A neighborhood of the origin is the set of all 1-cocycles on U" which assign II If holomorphic functions fij ~0 ~if\ uj :J ~ with Irij I < E. Consider one intersection uil\ u j {; ~ and the holomorphic functions fij assigned there with ·.1 fijI < e. The 1mage I 1\ I cocycles, under r, assign to ui''uj the holomorphic functions r ij, Ifij I < e. Take any sequence r~j ~ or these functions, lrikjl < E 1 and a compact set KcD0 , II II I (uil\uj) :::>.:>K -)~(ui f\uj). Since t}1e derivatives of the rl1j are uniformly bounded on K, 'i, f i~} is equicontinuous and uniformly bounded on K, and hence contains a normally convergent subsequence. This shows sequential compactness of the image, but in a metric space sequential compactness implies compactness.

l.

139 Now, consider the follO't'ling two maps of th.e direct sum of C0 (D,U 1 ,0) and z1 (o0 ,u",L)) into z1 (n,u•,U),

c

0

(D,U 1

,cJ)

tez1 (o0 ,u",())

5.$ r=A

=-~ z1 (o,u•,0) -r=B

where 6 (±) r means that 5 operates on c0 and r on z1 and -r is just the map -r operating on z1 • Both maps are continuous and linear. B is compact, and we claim that A is onto (see below). Theorem. (L. Schwartz). Let E, F be Frechet spaces and A, B : E -> F be continuous linear maps from E into F. If A is onto and B is compact then the range of A + B has finite codimension, i.e. dim F(A+B)E
t·

14.0

§3. Proof of Proposition 2 We will prove a proposition 2 1 and then show that 2' implies 2. Proposition 2•. Let Dope~~ Cn be strictly pseudoconvex. Let q E bdry D. Take E to be the E of lemma 3. L~t B1 be the· polydisc about q of radius e/2, and let t•) ~ 0 be a C00 function with support in B1 and ru(q) > 0 so that n1 = [~-tm 0 small is strictly pseudoconvex. Then every closed (0,1) form in D is cohomologous in D to a (0,1) form defined and closed in n1 • (We will say that D1 satisfies 2 1 with respect to D.) Proof. Let B be the polydisc about q of radius e. Note that D c n1 c.c ~n and D1-Dc B1 • Let r::r be a C00 function, c1' = 1 in B1 and r:l e 0 outside B. Let C'. be a closed (0,1) form in D. Since DI)B is a region of holomorphy and a. is closed in Drm, there is a C00 function X in DflB such that a. = -a]: • The function ]:. rs- is Cm in D and :L if= J- in B1 fln. Thus t3 = a. - ~(J cr) is a ~losed form in D cohomologous to a. • In B1 (\ D, t3 = a. - a..t = 0 and therefore can be continued as 0 to all of n1 - D. 2 1 implies 2. Proof. L~t Dopenc:c:ren be strictly pseudoconvex. Since DccGopen, by definition, there are open sets E, F such that Dc.cE c..cF c G:n and the distance from D to E is greater than g. Co~er E by a finite number of polydiscs, B1 , ••• ,BN' of rad-ius E/2 such that every polydisc containing a boundary poi~t of D is centered about a boundary point or D. Let ro1 , ••• ,ruN be a partition of unity subordinated to the covering: ru j ~ 0 and C00 in E with support in Bj and ~ mj = l at each point or E. For l = l, ••. ,N

consi~

the regions D£ =

[~- ~

tj mj

<

0],

where the

tj

are positive numbers so small tnat each n2 is strictly pseudoconvex and Dt ":]_en. Since ~-I w j = 1 at every boundary point of D, tj ruj > 0 here. Hence DCcDN. Also,

l=l

141 D1 satisfies 2' with respect to D and D2 satisfies 2 1 with respect to D2 ~ 1 , 2 = 2, .•. ,~. Hence every closed (0,1) form in D is cohomologous in D to a closed (0,1) form in D1 which in turn is cohomologous in n1 (and hence D) to a closed (0,1) form in D2, etc. up to DN. Take for D0 , in 2, the set DN.

142

Chapter 14.

§1.

Sheaves

Exact sequences

In the following, all groups are Abelian and ·all maps are homomorphisms. : A. Definition 57. A sequence is a collection of groups AJ and maps ~J : A;--> AJ+l' written [Aj,~J) or: •••

.

>

j-1

Aj-l

>

~j

Aj

Aj+l

>

> • •• •

The sequence is said to be exact at Aj if 1m ~j"l=ker ~j' where 1m ~j-l = [ala E Aj , there exists b E Aj-l such that ~j-lb = aJ ker ~ j

= [ al a

E

Aj

1

~j a =

0} .

The sequence { Aj~~;I is called Aj' for every j. Definition 58. A collection of to form a commutative diagram if all leading from a group A to a group the same result: e.g. the diagram A

<j>

exact if it is exact at maps and groups is said compositions of maps B in the collection give

> B

~w c commutes if t~(a) = Q(a) for every a E A. Remarks. 1) Clearly, 0 -> A -> 0 is exact if and only if A = 0. J. 2) 0--> ~> B is exact if and only if ~ is one-to-one. In this case, we may regard A as a subgroup of B, for A1 = ~(A)C: B, and ~ : A -> A1 is an isomorphism. Hence the diagram ~ 0->A->B

t~

tid

0 ->Ali> B

143 is commutative (1 will always denote the inclusion map and id the identity map). 3) A !_> B -> 0 is exact if and only if 4> . is onto. Here,.the homomorphism theorem of group theory implies B"' A/ker denote!3 "is isomorphic to"). Hence, we may "factor" ~ as follows:

+ ("'

-

A _j_> B j " ' -/

A/ker

++1

This diagram commutes, where here j (as always) denotes the canonical projection and is onto; and 1 , the map induced is an isomorphism. by 4) Combining 2) and 3), 0 -> A i.> B L> C -> 0 is called a short exact sequence if and only if ' is 1-1, t is onto and 1m = ker ~. Remark. Utilizing the above remarks, the following · diagram commutes and both horizontal sequences are exact: ~

+

+,

+

+

0-> A-> B->

tid

~t

Note that ' : A -> A1 , id : B -> B and ~l-1 : C -> B/A1 are all isomorphisms. Isomorphic groups may be identified; hence short exact sequences should be thought of as being in the rorrri: 0

->

A

..!....>

B

..J.....>

B/A - > 0 •

Proposition 1. Let I=[ a,l3,-y, •••j be directed by " > "; and let there be given sequences {Aj,+j~ exact for each a, such that, for a < 13 the following diagram exists and is commutative:

B.

,

144

:

where the a~ a

< f3 < 1

satisfy the following compatibility condition: implies

a

r

=

aj1 ar

•

Then the limit sequence lAj = 11m Aj , ~j = 11m ~~ exact and the following diagram commutes:

a,!J> Aaj+l ->

• • • -> Aj

t1 ti

J

is

•••

• • • -> A • ~> A -> • • • J 'flj j+l

•

Remark. The Aj are defined as follows: Set Sj =lJ Aaj; define an equivalence relation """" on aer a a2 Sj as follows: s 1 E Aj 1 , s 2 E Aj are equivalent, s 1 ~ s 2 , if there exists a f3 such that ~ < ~~ a 2 < f3 and alf3 a2f3 aj (s~= aj (s 2 ). Then Aj = Sj/~, and the group structure is canonical. Alternatively, define a thread to be a set of elements tga) such that for every a E I, g 0 E Aj; and for every pair a,p E I such that a<~, g 13 = aj~(g 0). The group structure is again the obvious one, and the group formed by the threads is denoted Aj. The homomorphisms ~j are defined in the obvious way, utilizing the ~j and the compatibility condition: The proof of the proposition then follows from the directedness of the set I. §2. Differential operators A. Definition 60. Let A be an abelian group. A homomorphism d : A -> A satisfying d2 = 0 is called a

145 differential operator. The sequence A ~> A~> A is not necessarily exact, but ker d ~1m d as d2 = 0. Hence \'1e may define: H(A)

3

ker d/im d,

the der}ved group of A. H(A) is a measure of the deviation from exactness or the above sequence, in the sense that H(A) = 0 if and only if the sequence is exact. We say x E A is closed if dx = 0 ; x E A is exact if there exists a y E A such that X = dy. Denote the homology class in H(A) of an element x E A by [x]. Definition 61. If A,B are groups with differential opera tors, l'le say that f : A -> B is an allot'lable homomorphism if fd 1 = d2 f; i.e. if the following diagram commutes: dl A

B

> A

d

2

> B

Examples. The group of cochains on a space, with boundary operator; continuous maps are allowable. The additive group of differential forms, d the differential; differential maps are allowable. Chains on a simplex, boundary operator; simplicial maps are allowable. Proposition 2. An allowable map f : A --> B induces a homomorphism f*

such that if g

H(A) - > H(B) ,

B -> C allowable, then: (gf) * = g* f *

and

146

Proof. Define f * [x] = [ fx] mapping of H(A) -> H(B} I for implies dx = 0 and X = dy implies

1

X

r* is

Then

e A.

d(fx)

=

rx =

f(dy)

f(dx)

=

a

= od(fy) •

B. Proposition 3. Let A,B,C be groups with differential operators, and let the following be a short exact sequence of allowable maps: O->A!_>B£_>C->O. Then there exists a canonical homomorphism D such that the follO\dng diagram is exact (viewed as an infinite, repeated sequence) H{A) *

D/ 'J

H(C) ~ HlB) g*

Proof. (Recall the \'Tell proof of de Rham 1 s theorem!} Exactness at H(B): {Does not need D) i) Let a e: H(A); \\le must show g*f * (a) = o. But, let x e: a ; g* f * (a) = [gf{x)] = 0 as gf = 0. 1i) Let y e: B, dy = o, such that g* [y] = 0. We wish to find x e: A, dx = 0, such that f * [x] = [y]. Now g* [y] = [gy] = 0, implies gy = dz, z e: c. But g is onto; hence there exists a y1 e: B such that gy1 = z, which implies gy = dz = dgy1 = g(dy1 ). Hence y-dy1 e: ker g, so there exists an x e: A ·such that fx = y-dy1 , and x is closed for dfx = f{dx) = dy = 0, and f is one-to-one. But f * [x] = [y-dy1 ] = [y], as required. Construction of D: Let z e: C, dz = o. Now g is onto, so there exists a y e: B such that z = gy. But 0 = dz = dgy = g( dy), so dy e: ker g implies that there exists an x e: A such that fx = dy, and x is closed as before. Note that x is unique once y has been chosen. Set D[z] = [x]; D is well defined if [x) is independent of y. Bence, let y e: B such that O=gy.

147 Then y -e ker g, so there exists a unique x e A such that y = fx. Therefore dy = f dx, so D[O] = [dx] = O. D is clearly homomorphic. Exactness at H(C): i) Let y e B, dy = o. Then ng*[y] = D[gy] = [x), where rx = dy. But f is one-to-one, so d7 = 0 implies X=

0.

ii) Let z e C, dz = o such that D[z] = 0; i.e. there exists a y e B such that z = gy; dy = f.x and D[z] = [x] = 0. Hence, x = dx1 • Set y1 = y-rx1 • Then y1 is closed, !or dy1 = dy- dfx1 = fx- f(dx1 ) = f(x-dx1 )=0. Furthermore, g [y1 ] = [gy-gfx1 ] = [gy] = [z], as gf = o. Exactness at H(A): i) Let z e C, dz = 0. Then f *D[z] = [fx), where z = gy and dy = fx. But then f *D[z] = [dy] = O. ii) Let x e A, dx = o such that r* [x] = 0; i.e., fx = dy. Set z = gy. Then z is closed, for dz = dgy = gdy = gfx = 0; and D[z] = [x].

§3. Graded groups Definition 62. A group A is called graded if, for every integral J, there exists a subgroup AJ such that X E A implies X= Xjl+ ••• + Xj ; Xji E Aji' k < m and the representation is unique.k Note that this uniqueness implies Aj nAk = O, j ~ k. An element xJ E Aj is called pure (j-)dimensional. Examples. Chains and cochains on a simplicial complex are graded by their dimension. Differential forms are graded by their degrees. In both these cases, Aj = 0 for j < 0. Definition 63. A differential operator d on a graded group A is said to respect the grading if there exists an integer r, called the shift of d, such that d AjC: Aj+r for every j. (In practice, r is almost always i 1.)

148 A map f : A -> B of graded groups with differential operators is called allowable if the differential operators have the same shift and f preserves dimension. Corollary. If A is a graded group with differential operator which respects the grading, then th! derived group H(A) is graded; and [x e A.o dx;::~ j H (A)

t em j-r j

=

Corollary. An allowable map groups induces homomorphisms

·

f : A -> B of graded

f* : Hj(A) -> Hj(B) •

Proposition 4. Let 0--> A_£> B_s> G--> 0 be a short exact sequence of graded groups and allowable homomorphisms. Then there exist maps d such that the following sequence is exact:

* Hj(B) lL> * Hj(C) .2_> Hj+r(A) -> ••. -> Hj(A) .f._> ••• where r is the shift of the differential operators. Note: There are lrl distinct sequences. Proof. Utilizing proposition 3, there exists a D: H(C) -> H(A). As a map of graded groups, D : Hj(C)->Hj+r(A),. for: suppose z e C is pure j-dimensional, dz = o. Then there exists a y such that gy = z, and dim y = j. There exists an x such that dy = fx and dim (dy) = j+r = dim x. But D[z) = [x], so shift D = r =shift d; renameD "d"; then the exactness result of proposition 3 and the above corollaries conclude the proof. Example. Let X be a topological space; A c. X a subspace. Let C(Z) denote the graded group of chains over Z, with standard boundary operator, a. Then 0 -> C(A)

1 ->

C(X}

j ->

.

C(X)/C(A) :: C(X,A}

->

0

is an allO\-mble short exact sequence of graded groups, and proposition 4 implies the exactness of the sequence ••• -> Hj(A}

2>

Hj(X)

~>

Hj(X,A)

~>

Hj-l(A)

- > •••

149

§4.

Sheaves and pre-sheaves

A. Recall the defini~ion of "sheaf''· (Chapter 6, §;, Defn. 36); we rephrase it as follol·Js: Definition 64. A sheaf of Abelian groups, ~~ is defined . as follous: Let X, the base space, be a paracompact Hausdor~f spa~e. For every x E X, let Sx be an associated Abelian group called the stalk of the sheaf over x; and set S = ~EX Sx, whose topology is smallest such that i) the projection map p : S -> X, defined by p(s) = x if s E Sx' is continuous and a local homeomorphism. ii) the group operations in the stalks are continuous; i.e. s -> -s is a continuous map of S into S; and (s 1 ,s 2 )->s 1+s 2 , defined on the set R of pairs (s1 ,s 2 ) such that s 1 ,s 2 belong to the same stalk, is a continuous map of the subset R of S x S into S. vie remark that the stalks are discrete. Let Y c. X; then S (Y) = UxEY Sx, with the induced topology, is called the induced sheaf of S ~ Y. A section over X is a map t : X -> S, continuous, such that p • t = idx. A section over Y c X is a section of S(Y) over Y. Remarks. Every sheaf has at least one section, the zero section, given by t : x -> 0 E Sx. If two sections coincide at a point, they coincide in a neighborhood of this point. Corollary. Let Yopel~ X. Then t(Y) is open in S. B. Definition 65. Let X. be a paracompact Hausdorff space. Let U = { u1 \ be an open covering of X such that u1 , u2 E U implies u1 ~ u2 E U. Let r-(ui) be an Abelian group associated to each ui E U, such that, if ui c. uj there exists a homomorphism '~'ij : r(uj) -> r(ui) satisfying the compatibility condition: ui c uj c uk

implies

~'ji

?'kj

= Yki •

The collection (x,r-(ui),rij) is called a prehseaf. Proposition 5. To each presheaf there may be associated a sheaf, called the sheaf defined by the presheaf.

150

Proof. Take X as the base space. For each x e X the collection tui I ul e U, x e ui} is directed; set Sx equal to the direct limit of the groups r(ui). Set S = SX, with topology defined as follows: s e S implies . s e Sx. Now x e Ux e U; then [sy I sy e SY; y e is an open set and the collection of all such sets is a basis for the topology of S • Note that the 1-(ui) form sections of the sheaf S defined by the presheaf. Proposition 6. Every sheaf is defined by same presheaf. Proof. Take = I uopen in Let r-(u) be the sections over u; and define the rij by restriction. c. A subset T of a sheaf S is itself a sheaf if and only if T is open and Tx = T llsx is a subgroup of sx. Then T is called a subsheaf of s. Example. S is the sheaf of germs of continuous functions and T is that subset of ~ consisting of all the germs of ceo functions.

4ex

ux'f

u lu

xJ .

§5. Exact sequences of sheaves and cohomology Unless otherwise stated, all sheaves have the same fixed base space X. Definition 66. Let s 1 and s 2 be two sheaves. A continuous map of s1 into .§ 2 such that ~(s 1 ,x> C...8 2,x and ~ I 8l,x is a group homomorphism, is called a homomorphism of the sheaf s 1 . into the sheaf ~ 2 • The subset of ~l mapped into the neutral elements of 8 2, Sxt' is called the kernel of ~; denoted ker ~. The ker ~ is an open set, for the set of neutral elements of ~ 2 is the image of the null section of 82 over X and this is open in .§ 2 (cr. corollary of §4, Chap. 14), and since ~ is continuous, the preimage of the set of neutral elements of 82 is open in 81 . The ker ~ f'\8 1 ,x is the kernel of the group homomorphism ' I 8l,x and thus

foe

151 is a subgroup or s1 ,x. Therefore ker' is a subshear or ~ 1 • The image of ,, ~{~l) c~ 2 ; denoted 1m ,, is a subshear of ~ 2 : that is open follows from the· continuit7 of ~~ the commutivity of ' and the projection map, i.e. p2' = p1 , and the fact that p1 and p2 are local homeomorphisms, and, as before, 1m ' f)s 2,x is a subgroup of s2,x, so that 1m ' is a ·subsheaf of ~ 2 • Hence we can form the quotient sheaves s2 11m ~ and ~ 1 1 ker ~' the cokernel of ' and coimage of ,, respectively. Definition 67 The se~uence or sheaves and shear homomorphisms I!j h..> I!j+l :.1±1> .§j+2 is called~ exact when, for each x e X, the sequence sj,x ~~--~j+l,x .l+l> Sj+2,x is exact. Example. Let T be a subsheaf of S, and let 0 denote the null sheaf, i.e. the sheaf whose stalks are the trivial i groups over each point. The sequence 0 -> T -> S -> S/:£ -> .Q is exact by definition. We have already defined the cohomology groups, Hq{X,S), q > 0, of a paracompact space X with coP.fficients in a sheaf S • For convenience, define Hq(X,~) = 0 for q < 0. Note that H0 (X,S) is the group of global sections of the sheaf. Let S and T be two aheaves and let ' be a homomorphism of ~ into T. We claim that for each q, ' induces a homomorphism ~*of Hq(X,S) into Hq(X,T). Consider any open covering of X, U =lui)-. The group of cochains C{X,U,S) is a graded group with differential operator (the coboundary) which respects grading (the shift is +1). Hence the derived group H(C(X,U,S)) is graded and its pure dimensional parts are the cohomology groups of the covering with coefficients in S. Similarly we have C(X,U,T) and H(C(X,U,T)). Now, an element of C(X,U,S) is an assignment of sections of ~~ and ~ maps S continuously into T, thus ~ maps C(X,U,S) into C(X,U,T). ~ is in fact an allowable homomorphism and hence induces a homomorphism of the derived

im'

152

groups H(C(X,U,S)) and H(C(X,U,T)). Taking the direct * . limit, we obtain the desired homomorphism Theorem 41. (Exact cohomology sequence). Let 0 ->·A!> B i> C -> 0 be a short exact sequence of sheaves.

+•

-

-

-

-

-

~> H2(X,!)

--> •••

Assume the theorem for now. (It is proved in §7, p. 158.) Definition 68. A sheaf S is fine if and only if, for any locally finite open co~ering~X, U = fui!' i E I, there exist homomorphisms ~ 1 of S into S such that 1. ~i(Sx) = 0 for xi ui and 2.

fu

~1 =

identity.

(The sum is finite at each point because U is locally finite and ~ satisfies 1.) Example. Let X be a paracompact differentiable manifold, and let § be the sheaf of germs of differential forms of degree p. Let U be a locally finite covering of X and let wi) be a partition of unity subordinate to U. Defin~ ~i to be multipli'cation by wi' Then 11~· are ·homomorphisms of § into § satisfying 1. and 2. above., so that S is a fine sheaf. Theorem 42. If S is a fine sheaf, then Hq(X,§) = 0 for all q > 0. Proof. The proof is the exact analogue of the C00 case: theorem 22, p. 78. Let q > 0 be fixed and let U = fui~ be a locally finite covering of X. Define the homomorphism Q: Cq{X,U,S) -> Cq-l{X,U,S) by Qf(i ••• i 1 ) = ~~if(ii 0 ••• iq 1 ) •

i

-

f

-

0

q-

ne:

,

-

Verify that . S- = Q5 )- + 5Q j~ exactly as before. Hence if ot =0, then f = 5QJ, Thus Hq{X,U,S) = 0 and then the direct limit Hq(X,S) = O.

153 Definition 69. A resol¥tion of~a shea€ ~ is an exact -1=1 0 1 sequence of sheaves .Q -> ..§ -> ~ - > A1 - > • • • such that Hq(X,._!j) = 0 for all j ~ 0 and q > o. A resolution is called a fine resolution when all the !j are fine sheaves.

=

~amples.

Let X be a connected differentiable manifold and let ! = ~ (Sx = ~ and the topology is the discrete one). Let Aj be the sheaf of germs of .differential forms of i d d degree j. The sequence .Q -> tl! -> !a -> _!1 -> • • • is a fine resolution of e. Proof. Note that if X were not connected we would have to take for ~ the sheaf of germs of functions which are constant on each component of X in order that the sequence 0 -> ~ -> !o -> .!h -> • • • be exact at A0 • We have already established that Hq(X,A j) ;: 0 for all j ~ 0 and q > 0 (cr. corollary p. 89) and that the Aj are fine sheaves. The exactness of the sequence at ~ and at A0 is immediate, and exactness at Aj, j > O, follows from the Poincare lemmas. 2. Let X be a complex manifold and let S be C9 and Aj be the sheaf of germs o! differential forms of type (O,j). The sequence .Q -> (!) .!> !o ...2> A1 d> • • • is a fine resolution of () • . Proof. As in example 1., the Aj are fine sheaves, and the exactness of the sequence at Aj, j > O, follows from the Poincare lemmas. Th~orem 43.~ (Abstract de Rham). Let -1=1 o/Q ~1 0 -> .§. > !Q -> ,!1 -> • , • be a resolution of a sheaf-~· Consider the inguced cohomology sequence * 1.

0

0

-> H

(X,S)

~ ~0 -=s: H (X,A0 }

1m ~;_ 1 C ker ~;

1

0

0

-> H

(X,A1 )

~1

-> , • , •

Then

and HP(x,S) ~ ker ~;/1m ~;-l for all p > o. canon. isom.

154

Applying this theorem to the above examples or resolutions or sheaves, we obtain for 1. HP(x ~) closed p-forms P > 0 1 ' ~ exact P•forms 1 i.e. the de Rham Theorem (Theorem 26a, p. 93) 1 and for 2. = HP(x 1 t/) ~ ~-closed (O,p) forms 1 P > 0 , - ~-exact (0 1 p) forms i.e. the Dolbeault Theorem (Theorem 26b 1 p. 95). Proof of Theorem 43. For j = 0 1 1 1 2 1 • • • 1 set BJ = ker = 1m 'J-l since the resolution sequ~nce is exact. For each j 1 the sequence 0 -> ker ~j !_> .!j ~> 1m ~j - > .Q is exact by cons¥ruction; rewrite it as .Q -> ~j .!_> AJ .!J> ~j+l -> 0. By the exactness theorem (Theorem 41) 1 the sequence ~.

'j

Hq(X,~j)

Hq(X,~j+l)

Hq+1 (X 1 Bj) -> Hq+l(X,Aj) is exact for q ~ 0 and j ;: 0. By hypothesis, ~(X,~j) = 0 for q > o and j ! 0. Hence Ifi(X,~j+l) ~ Hq+ (X,BJ) for q > 0 and j > 0. Then HP(X,S) = HP(X,Bn) ~ HP-l(X,B1 ) 2 -1 --v"' HP- (X,B 2 ) ~ ••• ~ H (X,Bp_1 ). Now BP = ker 'P is a sub sheaf of Ap. t1e claim that X0 (X,BP) ~ ker ~;. Indeed, the ker ~~ is the set of those global sections of AP that ~~ maps into the null section of ~+1, but, by the definition of ~;, this set is precisely the set of global sections of ~p· Consider, next, the exact sequence ->

0

->

0

.

0

0 -> H (X,~j) - > H (X;~·j) -> H (X,~j+l)

E...> H 1 (x,~j)

->

H 1 (x,~J)

-> • • •

-

for j = p-1, p

>

i.e. _

J..* onto H1 (X,BP_ ), Since 5 is a homomorphism from ker ~P 1 1 ) "' ker ~P* I ker 5 1m ~*p- 1 = ker 5 C ker ~*p and H (X,B -p- 1 J..* J..* = ker o/p I 1m ~p-l' by exa;tness. * Hence HP(x,s) ~ ker ~P I 1m ~p-l.

o,

155

66.

Applications of the exact cohomology sequence theorem

I.

Let X be a complex manifolq.

Let

~:

sheaf of germs of homomorphic functions

?It:

sheaf of germs of meromorphic functions

We may view be the inclusion. Al

JtL ;

-dJ

as a subsheaf of let i \'le form the exact sequence i

~

j

.,

: & ->h,.

.()

0->~->I!J:->~1~-> 0.

Recall that a section of .t!:J: I Q. over X is an equivalence class of sets of data for a C.I problem. Using the exact cohomology sequence theorem, there exists an exact sequence o ~ i* o lAo j* 0 .),. /tl 1 AI 0->H (X, C1) - > -R (X, rrl) -> H (X,nt/ V) - > H (X, • h

-

-

--

----

Now j* sends a meromorphic function (a section of~) into the Cousin I problem it solves, hence C.I is always solvable if j* is "onto", i.e. Theorem I. H 1 (X,~) = 0 implies C.I always solvable ( cf. Chapter 6 , §1) • II. Let X be a complex manifold, Y a globally defined hypersurface: Y = [f=O] ; f holomorphic in X. Assume f has no critical points where it vanishes [i.e. maximal Jacobian rank on Y]. Consider cO, the sheaf of germs of holomorphic functions on X. Let c2..Y denote the sheaf-of germs of homomorphic functions on X vanishing on Y. f2Y is clearly a subsheaf of c? ; and we form (as before} the e~act sequence:

Q_y .!...> () L> f2. 1 <2._y -> o • Q I !2_y is the induced sheaf of _!2 over

.Q - >

claim that Y. For points off Y, the stalks are trivial, for any stalk of c?Y over points not in Y is identical with the corresponding stalk of For any point Yo e Y, two functions representing elements of ( r.JIOy>Yo are equivalent if and only if they t'le

c?.

coincide in a neighborhood of y0 • (If they coincide in a neighborhood in Y, they coincide in a slightly larger neighborhood in X, as Y is closed); hence they represent the same germ in the induced sheaf of crJ over Y~ Using the exact cohomology sequence theorem, we have the following exact sequence for q~ 0:

-

•••

->

Hq(X,

!2_)

Hq+l(X,~y) Note that Hq(x,Q

Hq(X, Q/Qy)

-> ->

Hq+l(X, ~)

I 2,y) ~ Hq(Y,

->

->

•••

Q) .

Ue nO\'l claim that: q

->

Theorem II. Hq(X, (9) = 0 = Hq+l(X, {9) for fixed o implies Hq(Y, (0)= o. (cr. Chapt-; 6, §2, Theorem 20.) For, using exactness, tze obtain immediately:

-

q

t)

0.

('.I hT (X, ;.;;; __/~y)

~

pq+l (X I ;.:_y ,/) ) l

,

so that it is enough to show Hq+l (X, :!) ) = 0. -Y Let a be a cochain in Hq+l (X, 0 ) • Then r • a is a cochain with L~y coefficients, a~ f is holomorphic in x. Hance, multiplication by f induces a homomorphism f*: Hq+l (X, {_?) -> Hq+l (X, {9 y). Clearly f * is onto and one-to-one; therefore an isomorphism. III. As a last application, we obtain another old result: Let X be a complex manifold; and let ~* denote the sheaf of germs of invertible holomorphic functions under multiplication. Note that the sections are the nowhere

157 vanishing globally defined holomorphic functions. Let ~* denote the sheaf of germs of meromorphic functions under multiplication. Then {!) * C /It, and \lle obtain the exact sequence: , * . I * * o -> {) *. -i > [1 L> iJJ. I Q. - > .Q • The sections of fit* I~* are divisors in ~' i.e. equivalence classes of sets of data for the C.II problem. Using the exact cohomology sequence theorem, we obtain the exact sequence: * 0 .-}1 * j 0 k * ., * 5 1 ;() * ••• - > H (X, ttl ) - > H (X,~ I C.t ) - > H (X,£ ) -> •••

Here j * takes a meromorphic function into the C.II problem it solves: therefore any C.II problem a can be solved if 5a = 0.

This, however, is not particularly illuminating for we know little about H1 {x,C?*); so, we imbed this group in another exact sequence involving "simpler" coefficient groups. We have an exact sequence: 0 ->

i /-f'\ exn (11 ~-> ~ ~> .;!__

* ->

0 ,

where here ~ is viewed as a sub sheaf of CJ, giving exactness at 2.; 11 exp 11 iS the map: exp (s)-: 21fiS and rll (1 21ris the exactness at ~ is clear since ker exp =[ s , e = and exactness at 0* follovJS from the fact that every nonvanishing holomorphic function is locally an exponential. Hence, \•le obtain th~ exact sequence: 1 . exp 1 If * d 2 .

e

••. - > H

(x,O)

- > B (X,£ ) - > H (X,~) - >

... ,

but this gives rise to a (canonical) map: C

=

d • 5 : H0

- *I (X,ii~

0*) ->

2

H {X,
assigning to each divisor D E H0 (x,f,; *I(.!/*) its Chern class C(D) c. H 2 {x,~. Clearly, if D is principal, i.e. DE im j*, C{D} = 0 for 5(D) = 0 by exactness. Now assume H1 (X,tl) = o. Then d : H1 {x,c9*) -> H 2 (X,~ is one-to-one. Thus C(D) = 0 implies D is principal.

158 -··

So, we have re-established the Oka-Serre Theorem: Theorem III. There exists a map C : H 0 (X,~*/c?*)->H 2 (x,ZJ such that 1) D principal implies C(D) = 0 2) H1 (x,i2) = o and C(D) = o implies n· is principal. • (Cf. Chapter 9, §4, Theorem 31.)

--

.

~7.

Proof of the exact cohomology sequence theorem

nottl restate and then prove the theorem: Theorem 44. Let 0 -> A !> B ~> C -> 0 be an exact sequence of sheaves. Then there exists an exact sequence: \~e

*

f • • • - > Hj ( X, A ) -> Hj ( X,B )

-

Proof. Let U = consider the sequence:

}

-

L*> Hj ( X, C) -> 5 Hj+l (X , A ) -> , •• -

--

-

ui~ be a covering of X;

and

0 -> Cq(U,~) !_> Cq(U,~) 1L> Cq(U,C) where Cq(U,A) denotes the group of q-cochains on the covering Us and f and g denote the induced mappings. We claim this sequence is exact. At Cq(U,A) we must show ker f = 0; therefore assume a e Cq(U,A); f(a) = 0. Now f(a) = 0 means f • ~ 0 ••• iq(x) = 0 e Bx for all x e ui 0 n ... /)uiq. By exactness, f is one-to-one so ai i (x) = 0 e Ax for 0 ••• q every x E ui ll ... /)ui ; i. e. a = 0. At ·_cq{U~B), \rle m8st show im f = ker g. Let a e Cq(U ,A). Then g • f{a)i i (x) = gfai i (x) = 0, since gf = 0, -o• • • q

o· • • q

for every x e ui (l ••• (}ui ; hence g • f( al 0 q im f c. ker g. Now let ~ e ker g; i.e. g~~i 0 ••• 1q (x)

=

= 0 for every

x e ui fl .•. (l u1 • For each x e ui () ••• (', ui 0

q

0

0 so

q

there exists

an aio· •• iq(x) e Ax such that

ra1 ••• i (x) = f3 i ••• 1 (x). 0 q 0 q We claim that the assignment a defined by the ai 1 (x) is a cochain; i.e. that ai i is a section over 0 ·~~ ~ ••• 1• o· · • q

o

nu

q

159 ai ••• i (x) e Ax so p • ai i = i~ ·i 1) ••• () ui • o q o• • • q ""U 0 q To show continuity, let L. e S = ai i (ui ll ... ()ui ) ·1 --u a o• • • q o q = f f3i ••• i (ui () ••• I) ui ) , an open set for Now

-

q

0

ui /) ••• ,1 ui o

continuous.

q

0

is open,

f3i

q

.

i

o• • • q

is a section and

Let Nx= be a neighborhood of XO

f

is

in Sa ; then

0

a-il •• i [Nx] =a-il i • r·l. f(Nx) = (f• ai i )-lf(Nx) o • q o o· • • q o o• • • q o = pi-l i f(Nx ) which is open for f is an open mapping. o· • • q o Hence, 1m f = ker g. We cannot complete this sequence to a short exact sequence, for g may not be onto [e.g. take sequence 0 ...(") 0 l0 .),.; c (X,z) -> c (X,~) -> c (X,d/.Q)] • So, define C~(U,C) = g[Cq(U,B)], a subgroup of Cq(U,_Q) comprised of "liftable" cochains. Hence, we now have the follo't'ling short exact sequence of groups and allol'lable maps: 0 -> Cq(U,A) _1> Cq(U,B) ~> Cq(U,C) -> 0 • -

-

a

-

Hence, we obtain the exact cohomology sequence: ••• -> Hq(U,A) -> Hq(U,~) -> Hi(U,C) -> Hq+l(U,A)->,. For refinements of U, we have the desired commutativity, so that we may appeal to the proposition of Chapter 14, §1, to obtain the exactness of the limit sequence:

*

*

5

••• -> Hq(X,A)f-> Hq(X,~)g-> Hi(X,C) -> Hq+l(X,!) -> 1-'le

• ••

shall therefore be done if we can shot'l Hi(X,C) ~ Hq(X,C)

(canonically!), and this shall be proven by showing that for each cochain ~ in -Cq(U,C) for a locally finite covering U, there is a refinement V in which ~ is liftable; then the limit groups are isomorphic, since we have an injective {one-to-one) map i : ci(U,C) --> Cq(U,_Q) which commutes with the boundary operator and the"refinement" maps of the direct limit procedure.

160

(Since X is paracompact, we may restrict ourselves to the locally finite coverings U.) Let U = [ui~ be a gi~en locally finite covering. Let W= !wii be a new locally finite covering, refining. U and such that wi C c.. ui for each i, possible since paracompactness implies normality •. Let ~! Cq(U,C). To each x eX we assign an open vxc X such that

-

i)

X E VX

ii) x E wi iii) x E uj iv) x i uj v)

implies implies implies

X E Ui () ••• {)ui 0

q

Observe that once we establish the existence of the theorem is proved. Let x E X; using the local finiteness of U, vi there exist integers r,s < oo such that x E wj , ••• ,wj; r

i

uk , • •., uk 1

s

and no other wi' uk.

v~

wj {\ ••• 1l wj t\ uk f) ••• /)uk ; clearly v! is 1 r 1 s an open neighborhood of x and satisfies i), 11), and iii). Note that any smaller neighborhood of x will also satisfy these. For each k F k1 , ••• ,ks; xi uk' which implies Set

=

X t Wk' SO there_exiS~S an Open 2neigibor[tJ~d Vk,X Of~X such that vk,x(\ wk = ..,. Set vx = vx (\ k,x , k

1

k 1 , ••• , k 8

an open neighborhood of x l'Jhich now satisfies i), ••• , iv). To satisfy v), observe that X E u1 /1 ••• /)ui can only 0 2 q happen if ui E { uk 1 • • • , uk 1 and that vx C ui f\ ••• () ui • j 1 s 0 q 2 2 Now ~ 1· 0 ••• i q I vx is a section of C over vx • Hence i (v 2 ) is an open. ~i ... i (v~) is open in C, so g- 1 ~

I

0

q

non-empty subset of B {g is onto).

o/i 0 •·• q

X

There exists a

'

161 bx

E

g- 1

~i•••i (v~) such that p ~bx = x, and neighborhood q B

0

of bX in - such that1 p : NX -> p(NX ) is a homeo. 2 morphism. Let MX = Nx()g- ~i •• • i (vX ), and set 0 v = p o g(M ) C v2 • q X X X 1 Now g I Mx is one-to-one; and g- ~i i : v -> M cr•q X X: is a section of B over vx , mapped by g onto the section

N

x.

~io••iq : vx

->

g(Mx) of C.

162

Chapter 15.

Coherent Analytic Sheaves §1.

Definitions

Definition 71. An analytic sheaf ~ is a ,sheaf whose base space X is a complex manifold (or subspace of one); and such that each element of the sheaf can be multiplied by the germ of a holomorphic function; more precisely, each stalk ~x is an c?x-module, and this multiplication is continuous. vie have the notions of subsheaf, sheaf homomorphism, induced sheaf, factor sheaf, etc., as before. r Examples. t9 , the sheaf of germs of (r-dimensional) vector-valued holomorphic functions is an analytic sheaf. The sheaf of germs of continuous functions is an analytic sheaf. The sheaves Q_ * and ftt~ are not analytic, for there is no distributive 1a-v1 for multiplication by germs of holomorphic functions {recall that the operation in the stalks of these sheaves is multiplication). Definition 72. An analytic sheaf ~ is ~lobally finitely generated if there exist a finite number of global sections ·"' x' s 1 ,s 2, ,sk Sl.,tch that for every x E X, t E ;::r t = ~l ( s 1 )x + . • • + ~k ( sk) x where ~ j E CJ x. Examples. The sheaf 0; section "1". The sheaf CJ r ; r s~tions (0, ••• ,0,1,0,. .,0). Definition 73. An analytic sheaf is locally finitely generated if every point x E X has a neighborhood Nx such that the induced sheaf d- {N ) over Nx is globally X finitely generated. Let be an analytic sheaf, and s 1 , ••• , sk sections of ~(U), uopenc X. Let x E U. If there exists a tuple (~ 1 ,~ .. ,~k) E {9 ~ such that 1 ••

1

;1:

~l(sl)x + ··~ + ~k(sk)x = 0 ' the tuple (~ 1 ,.~.,~k) is called a relation between the sections s 1 , •.• ,sk at x.

+--

Note that the collection of all such relations forms an analytic "sheaf of relations" (between s 1 , ••• ,sk) contained 1n t!) k (over the space U). . Definition 74. The analytic sheaf ~ is called a coherent analytic sheaf if: 1) It is locally finitely generated • . 2) For every open U C X, the sheaf of relations of any finite number of sections over U is also locally finitely generated. Note that the definition is local. Remark. For convenience, we will call a coherent analytic sheaf, a coherent sheaf.

-

§2.

Oka's coherence theorem

The aim of this section is the statement and two steps of the proof of a three-step theorem due to Oka. The last section of the proof will be postponed until two theorems are established. Theorem 45. (Oka) The sheaf of germs of vector-valued holomorphic functions is coherent; i.e. (noting the local character of coherence) let Dopenc..:: ¢n, and let { aij ( z) i=l, ••• sq, j=l, ••• ,p be holomorphic fUnctions defined in D. Let xED; then there exists an open D1 C: D, x E D1 , with the following property: For any ~ E D1 the holomorphic solutions (~ 1 , ••• ,~p) of

J,

p

* :

1=~ aij(z) ~j(z)

defined in some neighborhood of

= 0 ~"

,

i

= 1, ••. ,q

may be written as:

L

~j(z) = (=l ~v(z) ~~(z) ,

j = 1, ..• ,p

where the ~j; j=l, ••• ,p, v=l, ••• ,L<® are a fixed finite set of solutions of*, holomorphic in a fixed neighborhood of ~; and the ~v arc defined and holomorphic in some neighborhood of ~.

164 Note. The ~r are said to form a finite psuedo basis. Proof. We proceed by a doub~e induction on q = number of equations and n = dimension of space. The three steps or the proof are the following: I. The theorem is true for n = 0. II. If the theorem holds for a fixed n and < q equations, then it is true for q equations, where here q > 1. III. If the theorem is true for same n and all q, then it is true for n+l and q = 1. It is clear that these steps complete the theorem; and that I holds, since a holomorphic function of no variables is a constant. We may assume x = 0 e D, with no loss of generality. II. Let us first introduce the following abbreviations: . p

(a)

tj; a1 j (z)

-

-

p

~j(z)

=

0,

?

{f3 ) -

'i J~ 1 ai j ( z )

~j ( z)

=

0,

(y)

-

~

~j(z)

=

0

(~j)

=

tl

(~l,

aqJ(z)

t

i = 1, ••• ,q

i = 1 , ••• , q

-15

J

... ,~p) .

,,

By hypothesis, (y) has a finite pseudobasis (~j), rt=l, ••• ,K
These

~j

=

~ '1t;-r

~,, .lfC 'YfC'V j

j

I

=

1

1 • • • 1

p

•

(e), hence: •c aij 7/1,~~~

must also satisfy 0

=

Lte,j

=

Lrc( L j aij~j)'l/lrc ,

rc

r!

Set birc =L j aij ~j ; But the set of equations

i = l, ••• ,q-1.

these functions are known.

= 1, ••• ,q-1 (~), p

has by hypothesis a finite pseudobasis Hence:

= l, ••• ,R.

So: R

K

~j = ~ ~ and therefore the functions

(

K

IDp

~~ ~j

~~I ¥~ ~~)

1

form a finite

pseudobas1s for (a); proving step II. We interrupt the proof to establish two needed theorems.

§3.

vleiel'•strass preparation theorem, revisitea

A. Theorem 46. be holomorphic at (z 1 , ••. ,z }, Z = co n j f = ~ aj(Z} z ,

(Weierstrass Preparation Theorem) Let f(Z,z) the origin, f(O,z) ~ 0 [where (Z,z) = (z 1 , ••• ,zn_ 1 ) and z = zn]' so that with a 0 (o) = ••• = as_ 1 (o) = 0, while

as(oi=~ o. Then: f(Z,z) where h is a unit, i.e. h(O,O) ~ 0; and b1 , •.• ,b 8 are holomorphic in some neighborhood of the origin toli th b1 (0) = •••

= b s (0) = o.

Remarlc. This representation is unique, but we shall not prove or use this. In the Weierstrass preparation theorem proved earlier ( 4 , §1, Thm 12) we had also assumed that ord f = s. Proof. For s = 0, this theorem is a triviality; buvo no hence, tako s > 1. Note that constant term. N Let s < N, an integer. Set zj = ~j' j = l, ••• ,n-1; and define: G(~l'"""'~n-l'z)

=

Then G is a holomorphic function of its variables in a

166 neighborhood of the origin and of order s, so that Theorem 12 applies~ Hence: G((l, ••• ,(n-l'z) = H((l, ••• ,(n-l'z)[zs+Bl(Cl, ••• ,(n-l)zs-1+ 0

••• + Bs(Cl~ ••• ,cn-1>1 and the B1 vanish at the origin. It is now enough to show that the Ci occur only as C~n. Take Q to be a primitive N th root of unity; ~hen: G(C1, ••• ,Q(i, ••• ,cn-1'z) = H(C1, ••• ,9(i ••• ,cn-1z[zs + Bl(Cl, ••• ,Q(i, ••• ,cn-1)zs-1~ •• ] =

G(Cl, ••• ,ci, ••• ,tn-1'z)

But the expansion of the ~leierstrass Theorem 12 is unique. Corollary. Let P(Z,z) = zs + a1 (z)zs-l + ••• + as(Z), where the aj are holomorphic in a neighborhood D of the origin. Let (C,c) e ~n, C c D. Then P(Z,z) = PI (Z,z) PII (Z,z) where

P·t(Z,z)

=

PII( Z, z )

zr + a1 (Z)zr-l + ••• + ar(Z)

= zt + 13 1 ( Z ) zt-1 + ••• + f:3t ( Z )

with Bi' a. holomorphic in a neighborhood of C, that pi"(c .. z)J= (z-c)r and PII(C,c) f. o. Proof. Changing variables, set:

z' = z - c,

z'

=

such

z - c

I

and define: Q(Z ,z•)· = P(Z 1 +C,z'+c). Then Q(Z 1 ,z') = QI(z 1 ,z')QII(z 1 ,z'); where QII is a unit and QI(z',z•) = z'r + a1 (Z 1 )z 1r-l + ••• + ar(Z 1 ). Set: PI{z,z) = QI(z-c,z-c) pii(z,z)

=

QII(z-c,z-c) •

Then PII(C,c) I 0 and PI(C,z) = QI(O,z-c) = (z-c)r. It is clear that pi is a monic polynomial.

167 B. Theorem· 47. (Division Theorem) _ I.et there be given a polynomial P = zs + a 1 (Z)zs-l + •.• + as(Z) with the aj holomorphic in a-neighborhood of the origin, and aj(O) = 0. Let f(Z,z) be holomorphic near the origin. Then: f(Z,z}

=

q(Z,z) P(Z,z) + R(Z,z)

where R is a polynomial in z of degree < s, with coefficients holomorphic in a neighborhood of the origin, and q is a holomorphic function in a neighborhood of the origin. Furthermore, this representation is unique. Proof. We first establish uniqueness. Suppose 0 =

qP + R 00

=

Let

v ~ 1;

( ~ qj(Z)zj)P + R •

we equate the coefficients of zs+v, 0

=

qv + qv+lal + ••• + qv+sas

obtaining •

But ai(O) = O, hence q (0) = 0. But then ~q./~zj m_o, v M ml v n-1 j = l, ••• ,n-1. Similarly, one finds ~ qv/~ z1 ••• ~ zn_ 1=0, hence q 0. But then R = 0. We now assume f = ~ fj(Z)zJ. This is no loss of generality, since the te -s of order < s in z may be included in R. Assume aj(Z) =NO (llzllj); this may be achieved by replacing zi by ~i' N > j, i = l, ••• ,n-1 in f and P, since aj(O) = o.·. The transformation back to the zi is achieved as in Theorem 46 since we have already established uniqueness.

=

NOl'l

zs = AsP + asl (Z)za-1 + • •• +ass (Z) where A = 1 and a . = -aj' Hence, multiplying successively S SJ by zm, m = 1,2, ••• and substituting appropriately gives: z s+m = As+m ( Z,z }P + as+m,l ( Z) za-1 + .•• +as+m,s(Z)

168 where in fact:

As+m+l =

zAs+m +.Asas+m,l

as+m+l,p = as+m,l asp + as~,p+l;

0

<

P

<

and

a s+m+l,s = a s+.m,l a ss • m+l Hence Astm+1 1=~ as+m-j, 1 zj; a 8 _ 1 , 1 = 1 is bounded in a neighborhood of the origin since la8 PI < [qlzii]P implies

=

las+m,pl < (CIIZII>m+p, (l'lhich is easily established by an induction over m). But this means that all series in the following expression for f(Z,z), obtained by substitution for zs+m, converge in a neighborhood of the origin. 00

f(Z,z)

=

P(Z,z) • ( ~ fj Aj) + zs~l ( J=S (I)

+ • . ' + ( ¥=s

fj

aj s ) •

This is the required representation of f.

§4.

The third step

Recall the statement of the missing step in the proof of Theorem 45: III. If Theorem 45 is true for some n and all q, then it is true for n+l and q = 1. Proof. Consider the equation:

~ f=l ai(zl, ••• ,zn+l) fi(zl, ••• ,zn+l)

=

0

•

We may assume that not all ai e 0. Write ai = h1Pi" valid in some neighborhood of the origin, using a linear change of variables if necessary ( 4, Sl, Property 2), where hi is a unit and Pi is a polynomial in zn+l" It suffices to consider the equation P

f=l Pi ci

= 0 ,

i'lhere, by renumbering if necessary, we may assume a·= deg PP ~ deg Pj.

s

We first show that there exists a neighborhood of the origin in which every solution may be represented as a linear combination, with holomorphic coefficients, of polynomials in zn+l of bounded degreesf Let (C,c) be any point at which the Pi are analytic. Write: p = pi pii p

.

where PII(C,c) ~ 0 and PI(C,z) = (z-c)r, where PI, pii are polynomials in z = zn+l' of degree ~ a. Let (Ci) be a holomorphic solution of: ~ Pici = O, in a neighborhood of (C,c). Using the division ~orem: ci

=

~i

"' PI + ci,

i =

l, ••• ,p-1;

1'\

where ci is a polynomial in z with coefficients holomorphic in a neighborhood of c and deg ci < deg pi Consider: cl

~1

•• •

cP

0

pp

I 0

1\

•

• •

pi I I

I

\-PJ

I

0

~2

•

I

~-1,

0

••

i= prr

I

pp\

0 I

\-P \

+ ••• +

~ 1 :tr p

81pii 62pii

t\ •

+1

••

pn

0

\

a •

\ I

•

••

cp-1pii cppii

pp -P p-

<

I

CP is chosen so that this equation holds identically. Each of the column vectors, except perhaps the last, is a solution, so the last is also. Furthermore, we will have expressed {Ci) as a sum of polynomial solutions if we can A show that C pii is a polynomial in z. Note also that all p entries of these polynomial solutions (except perhaps "C PII ) A. p are bounded in degree by 2a. Now Cp is defined by the equation: wh~re

170

+ so

PII C~p

and is holomorphic.

IJ.p-1 pp-1 p!I

P + r~ = IJ.1 p 1

=

pii C

I

--

But

piC

=

p-1

A

(

I

1

( 1=I > (ci -c i)

p

= -

1 ( r~ prr =

Using the division theorem, p-1

-"'

Cp =

q

'\

0 i Pi)

=

(Pi ci)

where q, R are polynomials and p II

JJ.i P Pi

1

P cP + :n

-t=t

~ 1=!

I

P = P cP + prr

I

p.L

Pi)

•

I q P

+R ,

deg R < deg PI.

+ ..!1,

)

Hence

•

But PII ~ is holomorphic, so R vanishes of order I p (deg P ) > deg R; hence R o. Thus:

=

PII C = q p

'

i = l, •.. ,p,

i = l, •.. ,p;

But so

Hence

~

i

Pi

c1 = 0

if and only if

171

~ .J'- gij(Z)

j+.:~=S f;r

1TB(Z)

=

0 ;

s = 0, ••• ,3a ,

whel"e the 1ri.t are knol'm functions; and this sys tern of equations involves n variables. Hence the solutions have a finite pseudobasis by the induction bypothesis, thus completing the proof.

.-

§s.

~sequences of Oka's theorem A. Remarks on coherent sheaves. 1. Coherence is a local property. A sheaf is coherent if and only if every x E X has a neighborhood ia uhich the induced sheaf is coherent. 2. A subsheaf Q of a coherent sheaf vf , is coherent if and only if it is locally finitely generated. For, any section of G is a section of ·d, since Gx is a subgroup of ~ • Hence, the sheaf of relations R of any finite X number of sections of G is the sheaf of relations between these sections, conside;ed as sections of .~. Since ~?L is coherent, R is locally finitely generated. B. Corollary. If~4 is coherent and s 1 , ••• ,sk are sections of 2:, then the sheaf of relations R(s 1 , ••• ,sk) is coherent. ,..<) k , a co h eren t snea , f, Proo f • R 1 s a sub s h ea f 01.n ··~ and is locally finitely generated, by the definition of coherence of J:.L• Theorem Let .__f· and G be coherent sheaves, G a subsheaf of ~. The~the qu~tient sheaf vL;a is coherent. Proof. For every x EX, ·there is a neighborhood Nx in which a finite number of sections s 1 , ••• , sk of ~ (Nx) generate the stalk at every y E Nx. The images of s 1 , ••• ,sk under the natural homomorphism of ~ into ~/G are sections of ::r {Nx)/G(Nx) generating the factor stalk at every y e Nx. Hence ~~/G is globally finitely generated. To show that the sheaf of relations of any finite number of sections of ~G is locally finitely generated,

48.

f

172

consider an open set vex, sections t 1 , ••• ,tk of ~V)/G(V), a point x_ E V, and a neighborhood Nx of x in V~ Construct sections s 1 , ••• ,sk of .J- as follows: E f.. /G ia the image of an For each fixed j, ( tj) 1 X XX ;:::1_ fj E ~x under the natural homomorphism h of ~ · into J-· /G. is coherent; thus in perhaps a smaller neighborhood of x, 1 there exists a section sj of with 1 (sj)x = fj. Let tj be the ~mage of sj under h. Then (tj)x = (tj)x' implying that tj = tj near x. Therefore, there is a neighborhood U of x in which we may assume that tj is the image of sj, j = l, ••• ,k, and since G is coherent, that r·1 , ••• ,r are sections of G(U) generating the stalk p at every y E u. Now, consider an~ element of (R(t1 , ••• ,tk))x: it is a relation (~l' ... ,~k) e Ox with ~j(tj)x = 0.

?/:-

:!::

'f.,\

But L ~j(tj)x = 0 me~ns that L ~j(sj)x E Gx which means ~ ~. ( s.} = - 2._ '1/Ji(ri) , so that j:I J J X i=l X ( ~l, • • •, ~k' 7./11 ,. ·., '1/JP) E l?J~+p is a re1a tion bet\•Ieen the sections s 1 , ••• ,sk,r1 .•• ,rp of~, a coherent sheaf. Thus in some neighborhood N C: U of x there are sections ,~~~\fr, ,'¥~, v = 1,~ • • ,N of R(s 1 , ••• ,rp)(Nx) over Nx such that

...

qir,. ..

t\"'1 ?=1

~:

N

'=

Jp)

~

.

•

•

·v

'l'p

v

= ~=l mv (jj)x, j = 1, ••• ,k. Note that the vectors (1)v, ••• ,lkv), v = 1, •.• ,N are sections of R(t 1 , ••• , -1 -

Hence

~j

'fr

(j)v

,

tk)(Nx) over Nx since they are continuous maps of Nx into (~ 1 , ••• ,~k) E (~ ~ such that ~ ~j sj + 2:: 7/Jiri = 0, i.e. L ~ .s. E G , i.e. ~. tj = 0. J J

X

>

J

173 Theorem 49. Suppose A and ~ are coherent sheaves and f : A -> B is a homomorphism of ~ into ~. Then 1m f, ker f, coker f = B /im f, and coim f = A I ker f are coherent sheaves. Proof. It is sufficient to prove that 1m f and ker f are coherent, as the coherenc~of coker f and coim f is then given by Theorem 48. To establish the coherence of 1m f = f(~) and ker f, we need only show that they are locally finitely generated since we already know that 1m f is a sub sheaf of B and l{er f is a sub sheaf of A ( cf. p. 151). 1. Since A is locally finitely generated, every x E X has a neighborhood Nx in \'lhich a fin! te number of sections s 1 , ••• ,sk of A(Nx) over Nx generate (A(Nx))Y, y E Nx. Their images under f generate ((1m f)(Nx))y' y E Nx. Thus 1m f is locally finitely generated. 2. For kerf, take x, Nx and s 1 , ••• ,sk as above. Since f(s 1 ), ••• ,f(sk) are sections of B(Nx) over Nx' R(f(s 1 ), ••• ,f(sk)) is a coherent sheaf. Consider the following mapping g : Ry -> Ay of Ry into Ay' y E Nx

01

k

->

~ ~i(si)y •

Since f( ~ ~ 1 (si)y) = ~ ~ 1 f(s 1 )Y = 0, g is a homomorphism of R into (ker f) I N • But every element X of {{ker f) I Nx)y is of the form L ~i ( s 1 )Y with ~ 1 E CJx and ~ ~ 1 f(si)y = 0. Hence g is onto. Therefore ker f I Nx is the image under a homomorphism of a coherent sheaf and thus is coherent, by part 1.

§6.

The sheaf of ideals of a variety

Definition 75. Let X be a complex manifold. An analytic set V in X is a closed subset of X such that

174 for every v e: V there is a neighborhood Nv of v and functions ~ 1 , ••• , ~k holomorphic in Nv such that V/) Nv = fx e: X I ~ 1 (x) = ~ 2 (x) = ••• = ~k(x) = ~· Definition 76. For x e: X and V an analytic set in X, let e/~ be the subset of C?x of all the germs of holomorphic functions vanishing on V. (If xi V then &x = (? ~ ; if V E X then 0~ is trivial.) C)~ is a subgroup of l9x and an ideal. ~, (9 ~ \'lith /'("'\ XEA induced topology is a subsheaf of ~ called the sheaf of germs of the ideals of the analytic set V, denoted ~V(X). Theorem 50. (Cartan) If V is an analytic set in a complex manifold X then ~v(X) is coherent. He \'lill only prove a wealcer form of this theorem. Theorem 51. If V is a regularly imbedded, analytic subvariety of codimension lc in an n-dimensional complex manifold X, then ~v(X) is coherent. Proof. ~ v(X) is a subsheaf of a coherent sheaf. Hence it suffices to ShOH that .;) V(X) is locally finitely generated. Let x e: X. If x i V then there is a neighborhood Nx of x \·lith Nx 11 V = ~. In Nx, .J V(X) = Q, and l'Ie are done. If x e: V, then by the definition of V we can introduce local coordinates z1 , ••• ,z such that n ~ V = (Cz 1 , ••• ,zn) EX I z1 = z 2 = ••• = zk = Oj, in some neighborhood Nx of X in X. Now, if f e (d V(:NX) )X, then f is the germ of u functi~n holomorphic in Nx and vanishing on V, so that f = O.)i zi, where the -w1 are holomorphic in Nx. Since z1 , ••• , zk ~ sections of v-iv(Nx) over Nx and the mi e &x, the proof is complete. Remark:. CJ (X)/~Jv(X) is a coherent sheaf, by Theorem 48. Its stalk over every point off of V is trivial. On V, since (cJ v(V) )x is trivial, its stalk is the stalk of 0 (V). Hence, on V this sheaf can be ictentitied vJith the sheaf C) (V) of germs of holomorphic functions on V. c0 (X) /,J~(X) is the trivial extension of i2(v) to X.

c?,

f=I

175 Chapter 16.

§I.

Fundamental Theorems (semi-local form)

Statement of the fundamental theorems for a box {semi-local form)

Notation. By an open box in 0n, we mean a [(z1 , ••• ,zn) ea.n I ai<xi o. Note that these theorems hold for polydiscs as \~Jell as for boxes. Remarlc:. Theorem 52". implies the Fundamental Lemma (proof later, p. 196). §2.

First step of the proof

By a degenerate closed box we mean a closed box given by, say ai ~ x1 ~ aj_ and b1 ~ y1 ~ bl, 1 = 1, ••• ,n Hhere some I · I of the a 1 = a 1 an.d/or b1 = b 1 , i.e. some of the intervals degenerate into points. The number of non-degenerate intervals is the (real) dimension of the box. For r = 0,1, ••• ,2n we formulate :1.. a Theorem 52Ar• Let X be an open box in ~ n , ~ coherent sheaf ofer X, and K LX a degenerate closed box of dimension r. Then th~re exists an open box x0 such that K cc.x0 c.c..X and . .f (X0 ) is globally finitely generated. and Theorem 52Br. Under the same hypothesis as in Theorem 52Ar, there exists an open box x0 such that Kcc x0 cc. X and

Hq(x0 ,~)

= o for all q

>

o.

Theorems 52A 2n and B 2n~ Theorems 52A and B. Proof of Theorems 52Ar and Br for r = 0,1, ••• ,2n. The proof consists of verifying the following three st~tements. (1) Theorem 52A0 is true. {2) If Theorem 52Aj is true for all j ~ r, then Theorem 52Br is true. (3) If Theorems 52Ar and Br are true, then Theorem 52Ar+l is true. First i'le prove that if X is an open box in an and is any sheaf, then Hq(X,~) = 0 for q > 2n. It is enough to consider Hq(x,u,Jt) for U a locally finite covering of X. Since if X c E2n we can refine U to a covering in which more than 2n+l sets always have an empty intersection, we are done. That (1) is true follovJs from the definition of a coherent sheaf as being locally finitely generated. To prove (2), assume Theorem 52Aj for j ~ ~· Let K be .··/ the given degenerate box of dimension r and ~ the coherent sheaf over X. Then there is a box x0 with K c c. x0 c. c X and jl(X0 ) globally finitely generated; call the generating sections s 1 , ••• ,sm. This means that there exists a homomorp~ism f 1 of1 .9ml(x0 } onto :.£.(x0 }. For, define

.J:

f

1

:

(9 ·~1 -> 1:.X .1\.

1

X E

X0 by(f !1: )-> >

~i ( si) X.

~mr generate ( :f(x0 ) )x at every

It is

onto because the s 1 x E x0 • Therefore, denoting ker r1 by a1 , we have the following i ,/'lml f, .-1 exact sequence 0 -> a1 -> £. -""> it:_ -> 0 • By Theorem 49, a1 is coherent. Apply Theorem 52Ar to a1 over x0 . We get a new x0 , call it X~ and G 1 (X~) globally finitely generated. Then there exist m2 generating sections •••• Denoting ker r 2 by G2, we get the exact sequence 11m2 -> G -> 0. Again, G is coherent and .Q -> G2 -> L./ 2 1 we can continue this process, as far as we want, up to 4n,

177

§3. Reduction of (3) to Cartan's theorem on holomorphic matrices Lemma 1. Let K be an r+l-dimensional degenerate ,.. closed box as in Theorem 52 A +l' given by: K = l a:l:<x-t < ai, "' 7 r .~... - J..B1~Y 1~ f3 1 5 • Assume, e.g. that a 1 < a 1 (or similarly, .,.-

/'

that

f3i

<

0

f3i ).,

0

for some

0

a 2n-l-dimensional hyperplane o

t( r

A

~-

0

A

-.

= (a 1 +~ )/2j( is ° 0 . 0 '\0 Set K1 = K A. (f (a1 0 -kx1 )/2<x1 l,

1

o

H =ic'x 1

0

o)

a 10-Kl:i 0 ) /2 ~ x10S . Then, if Theorem Ar+l holds for both K1 and K2 , it holds for K. Note. In view of the following claim, it is enough to prove Lemma 1 using only Theorems 52Ar and Br. Claim. Lemma 1 implies (3), Proof. Assume {3} is false. Order all the nondegenerate dimensions cyclically. Cut K along a first nondegenerate dimension as in Lemma 1. Then (3) is false for at least one of the two resulting boxes; choose one and call it K1 o Now cut K1 in the second nondegenerate dimension; (3) is then false for a still smaller box. Call it K2 •

K2 = K f)

178 Proceed, halving each nondegenerate dimension successively, obtaining a sequence Kj :> Kj+l of closed, nest~d boxes such that in no neighborhood or any box is the sheaf induced by_~i globally finitely generated. But the Kj intersect in a point, and a point surely has such ~ neighborhood; and this contradiction establishes the claim. . . A )\ Lemma 2. Let K1 , K2 be two closed boxes, given as follO\'JS: ( "' "" 1\ . Pl. ~ xl ~ ~' 131 ~ Y1 ~ f31

~1

=

K2 =

~j ~ j ~ 7xj (;;:1-E ~ X1 ~ ; (

X

~ Yj ~ ~ j (11 ~ Y1 ~ ;1

f3 j

1

1

1

where a1 ~ ~1 -E~ Le~ ~ be an analytic sheaf over a neighborhood or K1 U K2 • Let there be givxn sections a 1 , ••• ,ar of ~over a neighborhood of K1 generating the stalks of ~(K 1 ) at every point; and, similarly, sections b1 , ••• , b s of d- over a neighborhood o~ 2;._ Furthermore, assume that in some neighborhood of K1 f'. K2 ,

J:..

K

a 1 = ~ ~ij bj and bj = ~ tjk ak ; ~ij' tjk functions holomorphic in this neighborhood. Then there exist sections c1 , ••• ,cN over a neighborhood A A of K1 U K2 such that, in ~ neighborhood of K1, ai= L:~ijcj; and, in a neighborhood of K2 , bi = ~ tijcj. Claim. Lemma 2 implies Lenuna 1. Let K1 , K2 be as in Lemma 1, where we tal<:e. 10 = 1, We ~irst show that, in some neighborhood x0 or ·K1 n K2, a1 = ~ij bj and bj = ~ o/jiai; where ~ij' tj 1 .. are holomorphic in x0 ; and ai, bj are the generating sections over K1 and K2, respectively, given by the hypothesized Theorem 52 A +l' But then the sheaf over "' <, "r 1\ ( K2 = ([(a1+C)_)/2 - E)_! x1 ~ C)_ 1 • • • J will also b~ generated by b1 , ••• ,bs for E > 0 so small that Kl f1 K2 ()X0 • By the induction hypothesis, there exists a neighborhood X0 of K1 K2 such that Theorems 52Ar and Br apply; now ~

>

n

179 consider the map rJ._s -> i_(x0 ) given by . (~ 1 , ••• ,~s) -> ~ ~ibi. This is a sheaf homomorphism, and the bi generate the stalks of ~(X0 ) at every point, so that the sequence d) s _.., ( ) ~ -> ~- x0 -> ~ is exact. We comple~e to a short exact sequence:

,.

and G is coherent.

Hence, we have an exact cohomology sequence:

Ho(Xo,.Q. s) - > Ho(Xo,i·(xo)) - > Hl(xo,g) •

If x0 is small enough, Theorem 52Br applies, so is onto; but H1 (x0 ,G) = o. Hence H0 (X0 ,f2_s) -> E0 (x0 this means that every section of over x0 is a linear combination of the bi ; in particular, the aj are. In a similar manner, the bi are a linear combination of the aj. Proof of Lemma 2. Under the hypothesis of Lemma 2,

,J::)

.1:

ai = ~ ~ijbj and bj = ~ ~jka~ wh~re holomorphic in a neighborhood of K1 /! K2 • vle adopt the following notation: . a

=

~

=

and b = (

{~ij)

and

~

= 1\

l: );

~ij'

tjk are

column vectors

{tij);

matrices holo-

..,

morphic in a neighborhood of K1 /.) K2 • Then the hypothesis takes the form: ~b

= a

~a

=

b.

Now consider the {r+s) )l..(r+s) matrices defined below, which satisfy the follO\'Jing relations: . ·' 0 t a ' \ I a.l I • 1

o\ -r ~

\f ~

cr

\'

li or

1

8

'

I

/U J

I ;1~ ; ( -Ir = { ar / I

\

~1 J bS/

~

I

I

I

\ 0

I \

~rb

.•

:

~: j

=

0

~1 •

•

bs

,.

180

where Ij

denotes the

j

j

;w.

Define

unit matrix.

,/

'

!

(_-r_r.---· ( Ir M

=

\

:

\

0

0

\ "''.

~

Then M( ~) = (~), and M is a" nons~lar matrix holomorphic in a neighborhood of K1 K2 • -1M , where M is a holomorphic If we can write M = M1 1 2 A nonsingular matrix defined in a neighborhood of K2 , and M2 a holomorphic no)nsingular matrix defined in a neighborhood of Kl. Then M2\~ = Ml ( Set

n

n

/' cl c

\

= (I :•

)

\.. cr+s Then the cj are global sections generating Sf in a /1 ""\ ; ) neighborhood of K1 t) K2 , since the rows of M ( ~ are sections generating .J- in a n;._ighborhood ~~, ~ , M1 ( ~ )

1

j'

sections in a neighborhood of K2 ; and M2 { /\

"'

\0

=

M1 ( 0 )

b

in a neighborhood of K1 tl K2; and both M1 , M2 are invertible. Hence, the proof of Theorems A and B is reduced to: Lerruna 3. (Cartan 1 s Theorem on Holomorphic Matrices) Let M be a holomorphic nonsingular matrix defined in a neighborhood " 1\ of K 1 ~ K2 • Then there exist hol~morphic nonsingular matrices A, defined in a ~eighborhood 'of K1 , and B, defined in a neighborhood of K2 , such that M = ~A.

§4. A.

Proof of Cartan•s Theorem on Holomorphic Matrices

Let A be an

II z II

= max

Iz j j=l, •• ,n N )< N rna trix; set

Recall that

II AII = Note that

I aij I ~ II AII ~

sup

I

where

i ~x II

N max I a 1 j

I •

.

z = ( z1 , ••• , zn)

E

en.

181

Let A( z) denote an N >< N matrix whose entries are fUnctions holomorphic in a domain De ¢n. Set:

Ill A Ill

max

=

Z E

and

Ill Alila where 0 < a < 1 and

Ia1 j(z)-aij(z)l J\

~

H A. ct.

Hlz-zl

=

D

II A( z) n

Ill AIll + H

,

is the smallest constant such that

•

Note that if D1 c c D, domains, then IliA lila ,D ~ kii(A lin· 1 It is lmown that Ill Ill is a norm; and if n a DC ct , the N)(N matrices whose entries are fUnctions holomorphic in D form a Banach space under this norm. Furthermore-: IIIABIII

~

IIIAIII ·IIIBIII

Ill AB lila

~

c

I

Ill A II Ia nIB lila •

(The proof of the above statements is left as an exercise.) Note that, for any matrix A, eA e ~ ~~ is dominated by

)

111~111n.

A3/3- A4/4

If

IliA Ill

< 1,

then

log

(I~A) = A -

A2/2 +

+ ••• is dominated by JIIAIII + IIIAIII 2/2 + ••••

Furthermore, eA is always nonsingular, and eloe(I+A) = I+A. B. The following propositions establish Lemma ;: Proposition 1. Let D be a polydisc, D1 c c D and M a holomorphic nonsingular matrix defined in D. Given e > 0, there exists a nonsingular·_ entire matrix P such that M = PM1 in n1 and II II -M1 ~ ID < e • Proposition 2. There exists an 1 e· ·> 0 such that, if Ill I-r-i lla < E then Lemma 3 holds for M. Claim. Propositions 1 and 2 imply Lemma 3. Proof. By Propos! tion 1, M = PM1 , Ill I-r-i1 lll D < e • 1 Then in a smaller domain, IIII-M1 111a < e • Eence, M1 = BA by Proposition 2. But P is holomorphic everywhere, so M = (PB)A.

182

of Proposition 1.

~roQf

o.

about

M(z)

Assume that D is a polydisc

~·lrite:

=

M(O)lM(0)-1M(.fz)JtM-1 (tz) M(.fz>} •••

[M-1(¥z> L

an integer.

M(z~ ,

:

Now

llli-M- 1 (~z)M(K~lz) Ill ~ IIIM- 1 (~z) Ill IIIM(!z)-M(¥!z> Ill for L sufficiently large,

0 < K < L.

z) K+l M-1 (K L M(L z)

as the log series converges. M(z)

=

< e

Hence: =

NK(z) e

,

So:

N0 (z)

M(O) e

••• e

NL_1 (z)

•

h

By going to a smaller polydisc, D, if necessary, for each NK(z) there exists a polynomial sequence PKj(z) such that Poj(z) PlJ(z) PL :j,f) IIPKj(z)-NK(z) 11 0 -> o. Set Mj(z) = M(O)e e ••• e - •.•

o,

and det M I 0 for j sufficien~ly large. DHence M(z) = Mj(z~ [M~ 1 (z) M(z)I , -1 J and Mj (z) M(z) converges uniformly to I with j; and M. is a matrix of entire functions, for all j. J "" .. "' Proof of Proposition 2. Let K1 , K2 be given as follows: t 1\ ·" .-.. ) 0 1 ~ xl ~ al ' 131 ~ Y1 ~ t31 Then

UIM.(z)-r~(z)

Kl

=

l\

1{2

=

111.'\

->

1

1\

(laj~xj~aj, J

1\

cal-£"" xl

.-.

A

13j~Yj~f3j;

j=2, ••• ,n

A

~ al_. tll ~ yl ~ tll

a 1 < d1-E, as in lemmas 2 and 3. "\ A. M is defined in a neighborhood N of K1 /) K2; we write M = M(z), where z ~ z1 and the dependence on z 2, ••• ,zn is suppressed. Now M = I + X, and lllxllla < E where g is to be chosen later. In the z1-plane, let r be

where

""'

1\

a smooth analytic Jordan curve· containing K1 () K2 and .lying in N (more precisely, their projections on the z1-plane). Let 1 1 and r 2 d~~ote disjoint closed arcs, segments of 1, such that 1 I) K1 C.. int 1 1 . and 1 I) K2 < 1nt 1 2 , where "int 1i" denotes the segment 1i without its endpoints, as indicated in the diagram. Let G denote the bounded component of the complement of 1 in the z1-plane, and G its closure. Take o- to be a real-valued C0 function defined on 1, 0 < ~< 1, -t-~-H-+---H-+11-----+---> such that cr= 0 on 1 1 and xl cr· = 1 on 1 2 • For any matrix Y holomorphic in G and continuous inn, define:

-

T (Y) 1

=

1

2~i

f

Y( ~) CT ( ~) d'"

t-z

~

1 r

T2(Y)

=

y -

T (Y) 1

,

= __!_ fY(~>l_l-cJ(C)j dC 2~i.

~-z

1 I\

Then T1 (Y) is holomorphic in a neighborhood of Ki; and the linear operator Y -> T.(Y) is bounded in G; ~ IIIT1 (Y) !Ia ~ cii!YIIIa· tie ltdsh to solve the equation: .

I

+X

+ T1 (Y))

(I

+ T2 (Y))

=

(I

=

(I + T1 (Y)) (I + Y - T1 (Y)) ,

for some matrix Y holomorphic in G and continuous in for Ill XIll a sufficiently small; i.e. we wish to solve: X

=

Y- T1 (Y) T1 (Y) + Y T1 (Y)

=

Y +

F(Y) • Therefore, define T(Y) =X- F{Y), for Y E S = tY I Y holomorphic in G and continuous in G, IIIY-XIIIa < E, lllxlla < Note that T(S) c. s. l'le claim that, if

EJ·

E

G;

184 is small enough, T is contracting; so that the contracting mapping_ principle applies and there exists a unique Y0 c: S such that Y0 =X- F(Y0 ), as desired. IIIT1 (Y) Ill a~ c lilY Ill a lc( c,C) depending on

Now constant

implies that for some c and C (of p. ·181),

IIIT(Y)-XIII a= IIIF(Y) Ill a-~ kii!YIII~~ k[ IIlY-XIII a+ IIIXIII al 2 < 4ke 2 ;: so we require E < l/4k. Furthermore:

Ill T(Y) -T ( Z) I!.Ia

=

Ill F (Y) - F {Z) II Ia

=

Ill (Y-Z)Tl (Y)+ZTl (Y-Z)+Tl (Z)Tl (Z-Y)

+ T1 (Y) ~

T1 (Z-Y)111a

IIIY-Zllla K(K+l) (IIIYII6 + lllzllla)

IIIa

for some constant K(c,C). Now IIlY-X < e:, henc~ 0 ~ IIIYIIIa ~ 2c:, so IIIT(Y)-T(Z) Ill a~ 4c: K(IC+l) IIIY-ZIIIa, and we require also E < l/[4K(K+l)]. §5.

New proof of the Oka-Weil Approximation Theorem

Theorem 53. Let X be an analytic polyhedron, xc:::_Gopen c ~n, X= c: G I lrj(z) 1<1, j=l 1 • • • ,r; fj holomorphic in~· Then, given ~(z} ho1omorphic in X, ~ can be approximated on any Kc c X by functions holomorphic in G, in fact by polynomials in z1 , •.. ~zn,f 1 , ••• ,fr. Proof. Assume that G G ( Iz j I
fZ

=!

("

=1

1

1

q

J

E

vx(De:) = vxe:(DE}~ H (De:,vXe:) = 0 for all q > 0. Now XE and X are regularly imbedded analytic subvarieties of codimension r in D and D, respectively. E vle claim that the function ~ ( z} can be extended holomorphically into D • Consider the sheaf vfX (D). It is a subsheaf of E

E

E

185

f2. (DE).

Q

Since .f2 ~X {D ) !.> (..?{D )r-> 0(x ) -> 0 -

gE-E

-E-

is exact: i = inclusion, r = restriction. Therefore, the cohomology sequence H0 (DE ,0) !:.> H0 (XE , 0) -> H1 (DE is exact. But Hence ~ holomorphic in ~a. i Jl• •• k

~

E

E

<:.1

XE • Then 1/1 = .? p j ( z1 , ••• , zn, ~l, ••• 1 ~r) tfie p. are polynomials. Therefore on K,

D~t;.l ; ~

H1 (D '~X ) = O,

implying that r is onto.E is the restr•iction of a function 7fJ D • Write t as a power series jl E ik zl ••• ck converging uniformly in some D~ I :

cc

K cc XE

,JX )

on

.J

=~

pj(z 1 , •.• ,zn,f1 , ••• ,fr). Since z1 , .,.,z11 ,f1 (z), ••• , are holomorphic on G, so are the pj.

fr{z)

S6.

Fundamental Theorems for regions of holomorphy (semi-local form)

Theorem 54A. Let X be a region of holomorphy, . .-/ a coherent sheaf over X and Kc c: X. Then there exists an analytic polyhedron x1 such that Kcc x1 ,·.:..c X and a finite nwnber of sections of ·~(X 1 ) generate 0fx at every x s x1 • Theorem 54B. Under the same hypothesis as in Theorem 54A, there exists an analytic polyhedron x1 such that K cc. x1 c.c X q and H (X 1 ,~) = 0 for all q > 0. Proof of A. Exhause X by analytic polyhedra, Xjcc Xj+l cc X, U Xj = X. Pick one of the Xj, _call it x0 , satisfying Kc.c x0 ccx. Let f 1 , ••• ,fr be the functions defining x0 , and let D be its Oka image. x0 is a regularly imbedded analytic subvariety of codimension r in D. v Define the sheaf ~- over D as follows: let -~

~

~x

-

f j-

if X E XO I 0 other\dse

1

"

X

j

,

and let

Ar -

_.._v_,

~

be open if

186 and only if pA is open in D and A I) .'f is open in :1 (p is the projection map in the sheaf Since ;l_ (x0) is coherent 1 in particular ., analytic~ ~ is an analytic sheaf. Ue claim that is coherent. Introduce local coordinates ~ 1 , ••• ,~n+r in a neighborhood N in· D of a~point of x0 such th~t x0 ~ N is given by ~ 1=···=~r = o. ~ is locally finitely generated. Consider its sheaf of rela~ons _!!(s 1 , •• • ,s.t), i.e. let s 1 , •• • ,s.e be sections of :1:. over some open set U c:. {D /l N). A relation is a set of functions ~j(TI 1 , ••• ,T\ ..J..,) eC2, x e U, j=l, ••• ,.e, o nTr x

:.,F.

iJ:

N

satisfying ~ ~j(sj)x so that the

-

= ~· But {sj)x = 0 for xi x0 ,

~j satis-fy ~ ~j{O, ••• ,0 1 T)r+l' ••• ,,~n+r)(sj)x = O,

x e u/)x0 • Thu~ the f~J(~, •.. ,O,T)r+l~···'~n+r>r are relations of ,7 over U 11x0 and hence are locally finitely generated, say by {'1/!r, ... , ,~~), v = 1, ••• ,N in a neighborhood Nx of x. Therefore as generators of the sheaf R in Nx take the !\!li, ..• ,lf!~) and add the .e-tuples {11i,o, ••• ,o}, v {0, T\ i, 0, ••• , 0), •.• , { 0, ••• , 0, ~ i) for i = 1, ••• , r. Hence ..dis a coherent sheaf over D.~ Apply Theorem 52A to ..f (D). Then there exists an Olea image n1 of an x1 , x1 = { zve G I lfj(z)l 0. For, consider any covering . U of x1 , think of it as a covering of n1 • A cochain on U I x1 is an assig~~ent of a section of ~~ and hence by the trivial extension, it can be considered on U and is an assignment A of a section of -!:.._: hence a cochain on U. A cocycle on U I x1 is a cochain satisfying a certain relation, By the trivial extension, it is a cocycle on U. Hence if v q ·1 a j H (D1 ,z:_) = 0 then H ·{x 1 ,_) = 0 for q > o. By Theorem 52B, q v, H (D 1 ,~} = 0 for all q > o.

Chapter 17.

Uoherent Sheaves in Regions of Holomorphy

§1. Statement of the Fundamental Theorems Theorem 55A. Let X be a region of holomorphy and da coherent sheaf over X. Then global sections of .::J generate ~x at every x E X. Theorem 55B. Under the same hypothesis as in Theorem 55A,, Hq(X,~) = 0 for all q > 0. These theorems have numerous applications which will be given later on. .. A

:

~2.

Preparations for the proof

Notation. (z 1 , ••• ,zn_1 ) = Z, zn = z so that (z 1 , ••• ,zn) = (Z,z). Theorem 56. (Cartan) Let P(Z,z) be a Ueierstrass polynomial or degree s, P(Z,z) = z8 + a1 (Z)zs-l + ••• + as(Z); the aj are holomorphic in a neighborhood of the origin and aj(O) = o. Let r 1 , ••• ,rn-l'rn = r be> 0 and such that each aj is holomorphic for lzjl ~ rj and let P{~,z) 1 0 for L(Z,z) I lzjl~ rj, j = 1, ••• ,n~l and lzl = r; • Let f(Z,z) be holomorphic in D = L{Z,z) I lzjl~ rj for all ~J; and let I rl ~ 1 on D. Then (1) f = QP + R where Q is holomorphic in 1nt D and R is a polynomial of degree s-1 in z with holomorphic S-1 j coefficients in int D, R = bj(Z)z , and

f-o

IQ(Z,z) I ~ K and bj(Z) I ~ K \'lhere IC does not depend on r. Proof. (This proof is independent of the Division Theorem, Theorem 47, and hence gives a new proof of 1t.) · 1 r {z, C) dC Let Q(Z,z) = 2-iii. ICI=r P(Z,~)(~-z) ; Q is (2)

j'

holomorphic in int D.

Then

188

r

(

~ P(Z,~)-P(Z,z) ] d'" • But PTZ;0 lc l=r s-j s-j ( ) ( ) s-1 P Z,~ -P Z,z = ~ a (Z) C - z , a = 1. Carrying t-z j=O j ~ - z _1 0 out the division gives R(Z,z) ~ 0 bj(Z) zj, where

R(Z z) = 1

211-I J

I

e-z

';,

jt

bJ(Z)

=

2-;Ij

~~~~~~ (Cs-j-1 + alcs-J-2 + a2cs.-J-3+ •••

ICI=r + as-j-1) dC. Hence R is of the required form. Now to estimate Q and the coefficients of R. For ICI = r and lzjl ~ rj, j = l, ••• ,n-1, lrl ~ 1 and !PI has a lO't'ler bound b > 0 and laj I < c, a constant, for all j. Hence lbj{Z)I ~ K1, a constant independent of r. To estimate Q write Q =(f-R)/P. Then l+K 1 (l+r+ ••• +rs-l) IQI ~ = K2, a constant independent of r. Take K =max (K1 ,K 2 ). /()q. Theorem 57. Let Mj be a submodule of L/ 0 J (vectorvalued holomorphic functions in ¢nnear the origin) of dimensi~n q., j = l, ••• ,L; i.e. each Mj is a set of columns

rh\ ~ J ;( : )C ~i l\ ~qji)

holomorphic near zero, such that this set is

closed under addition, and under multiplication by holomorphic functions near the origin. Then (1) Each M: has a finite basis B., i.e. for every J J Mj there exist a finite number Nj of elements of Mj such that any other element of r~j is a linear combination of these with holomorphic coefficients. (2) After a linear change of variables, \~e can find a SEquence of polydiscs D about the origin D1 =' D2 ·:;) ••• , l v 1\ D = 1OJ , and the finite bases Bj are defined v -> in D1, such that if ~ E ~1 . and is holomorphicNin some Dv, and Iff (z) II~ s~p l~i(z~l < 1 then r~ ?~1 j1{1£ ~ I ZEDV

189 ->

->

where (~ 1 , ••• ,+N) = B; and the tg are holomorphi~ in Dv and II 'If nil < J K v ~a constant depending on v). Proof. We use a double induction. First we show that if for a fixed n this theorem holds for ideals (1-dimensional modules), it holds in general. Then we use induction on n. ;J

-

Consider Mj, · j

Let ( tl )

fixed.

£

Mj, Consider the

9qJ/ ideal I 1 of the ideal r2 for 1 = 0; = ~q _1 = O.

+

all those ~ which can occur as a ~ 1 ; then of all those 9 which ·can occur as a ~ 2 etc. till Iqj of +q 1 s for 91 = 2 = ••• By hypothesis each Ik j has a finite basis

+

" jfk k .. ~k = j , •. • ,jr ~.

Then the basis Bj

1

(1~ \ I o \ k I o . I 0 1) I 12 ., (: I I i2 i 0 . \ Jj '\ ~ j ''"'\ j~J 0

I

for -~J

is

t

l? ,

13

= l, ... ,r3)

\

qj Part (2) follows similarly. Now, for n = 0 the l-1j are the finite (qj) dimensional vector spaces of all qj-tuples of constants, and the statements are obvious • Assume the theorem for n-1 and every finite number of modules. We must prove it for n and any finite number of ideals. Assume that none of the ideals Ij is identically zero, so that \'le may pick a non-ideritic~lly zero element from each. Make a linear transformation such that these elements are normalized with respect to the variable zn. Consider any one of the ideals I with element 9 ~ 0. Then 9 = ~ 0p; where 0 is a unit and p is a lleierstrass polynomial, say of degree s, p = ~ s + a 1 {Z}zs-1 + .•. + a 8 {Z), a.j(O} = 0. Assume, without loss of generality, that 0 doesn't appear, then p e I. Let t be any element of I. t is holomorphic in some closed neighborhood N of the \

I

\

+j

+

+

190 origin. In N, 1~1 ~ c, a constant. Hence ~/c = JC e I, is holomorphic 1n N, and IJI ~ 1 . there. In perhaps a smaller closed neighborhood D or the origin we may apply Theorem 56. Then J( = qp + r; q is bolomorphic in int D, lql < K, and r = b0 (z) + b1 (Z)z + ••• + b 8 _ 1 (z)zs-l 1 where·tb; bj are holomorphic in 1nt D and lb j I ~ K. Since and qp e I so does r. Consider all a-tuples

.J.

(b 0 (z)

)

/ : such that b0 + b1z + ••• + b 8 1zs-l e I. They \ bs-l(Z) form a module in the n-1 variables z. By hypothesis, this . ~~(Z) ( •v

module has a finite basis

)

v = 1 1 • • • ,m;

1

Bs-l(Z)

so that j

m bj(Z) = ~ av(Z) Bj(Z);(*) av holomorphic in int D, v~ ~ j ~-1 _E!.,_ v.

= 0,1, ••• ,s-1. Then r

=

>

j=O

bj(Z)z = /

3=0

(

> _av(Z)B]Z)) z j

V=I

and therefore

~

Hence

_.

iP

=

.r· and

=

qp

+L

=

qp

+ "v

>s-1

j;n

v j

Bjz

,

(2: avBj)zj "'v(

~ B~zj

) '

(**)

.

v = l, ••• ,mf is a finite basis

\

I. Do this for all the ideals.·. Ij. f ~g(z) \ Now for each module L. consisting of all s-tuplesl :j ) \ b s-1 J (Z 1 such that > b~ zk e I., l·le find polydiscs DvCO:n- \ s-l for

1{;0

J

with the required property {2). Take a sequence or polydiscs I n I D in ~ such that every hyperplane z = constant /)Dv is V I I .,1. ( contained in DV 1 DV +l C DV , /I , DV = { 0J 1 and the bases for the Ij are defined in D1 • Consider any ~ e Ij, say given by(**}. Then by our induction hypothesis, in(*) the II aj II~ Kv and we already have Iql ~ K. Therefore q and aj,

191 the coefficients 1n the basis expansion of t, do have norms bounded by a constant depending on v. -> Theorem 58. Let f.i be a submodule of Let ~vi be a sequence of elements of M which are all ho!~morphic in a fixed neighborhood ~~ of the ~~igin. If the ~v converge uniformly in N to ~ , then ~ e M. Proof. The ~: converge componentwise; ~~ ->·~j uniformly 1n N, j = l, ••• ,q. In any compact subset of N the ~~ are uniformly bounded; assume ~~~~ ~ 1. By Theorem 57, perhaps a smaller polydisc D, ~~ = t~

t

06 .

i:t

:e

iJe i

}j

where the are the basis vectors and the 'if/~ are holomorphic in D with ll¢,e II~ KD. H~nce) for j fixed and for each fixed .e we have a sequence (1/lp,'s of uniformly bounded holomorphic functions in D. Thus ~ 1/J~ ~ contains a subsequence which converges normally, say to '¥_e1, . Therefore for each j there isNa subs:quence of .<~~J which c~nverge~ normally on D to 'f,e J.~ Then ~J = -> N

i:t 'f.e li

f:t

on D,

l'Jhere

'1' £

t •

.•

.

is holomorphic in D.

Hence ~ e M.

§3. Proof of Theorem A Now we are ready to prove Theorem A. Exhaust X by analytic polyhedra, Xj cc Xj+l' Jxj = X. Let Kj = cl Xj. Apply Theorem 54A to each Kj C Xj+l• Then for each j there exists an analytic polyhedron Nj with Kj c_ Nj c. c Xj+l and a finite number of sections of .;l (N j) generate -5-x at every x E Nj ; call them sj 1 ,sj 2, ••• ,sjtj• ConsidEr,

for fixed

j,

all sections of the following form,

where t~e

~

are holomorphic and bounded in Xj.

->

~j

a = ~=l~vsjv' The

( .1 \ v

=\ : \

\ form a Banach space Aj of vector-valued

~Lj I

holomorpl:.ic functions under the norm,

192

Aj is complete by Theorem 5~; Let Aj denote the subspace of Aj of relations, i.e. + e Aj if and only if s = ~ 'v sjv = 0 • Aj is a closed linear subspace of Aj, by The~rem 58. Hence we may form -~J/A~ = BJ.-> r+J • Denote an element of Bj, represen~~d by ~I· by Bj is a Banach space with norm Jl[+llj =-~nf_> max l~vl •

+ e [ cV]

v ,xj

Proposition. (Oka-Weil Theorem for sections). Let be a s·:;tion of .£.: over x3 such that fS= 2:v ~v sjv

~

and [~] e: Bj. Let e: > 0 be given. Then there exists a se~tion -r -~f over X such that II rr- -r 113 < e:, i.e. lfcr--rllj = II[}J IIJ < e , where (o-·- -r) I x3 = >--v lv sjv'

.1:

Proof. sjk:

We claim that every section

~;l~~k)

sj+l,l ,

where the JC1k) are holomorphic in

a neighborhood of K.; k = l, ••• ,L .• For, consider the J ( ,f\Lj+l J(Nj))x -> (jt(Nj))x, ma~pin~ f: ~; x e Nj, given by ( f;l

\

Ij \

Lj+l

)'->

f=IJ._e(sj+l,.e>x.

Lj+l

~Lj+l(Nj)

into ..

generate

.,

~(Nj)

r isahomomorphismof

and is onto because the sj+l,.t

( .;T (N j)) x at every x

E

Nj.

Hence

where Q : ker r, is an exact sequence of coherent sheaves. By Theorem 54B there I\ A is an an~lytic p_olyhcdron Nj such that Kj C. N 3 ~c Nj L and Hq(N3,a) = 0 for all q > o. Therefore HO(Nj, J2. j+l)->

0 -> G -> QLj+l(Nj) -> _i(Nj) -> O,

0

1\

./

H (Nj 1~) -> 0 is an exact sequence, implying that the mapping j_ ,.. is onto. Hence every section of ·L~j) is a LJ:.esj+l,.t where the Ap, are holomorphic in Nj; therefore so is each sjk' h Nou, ()= ~ sj and sjv = ~·r(v) ~~·! sj+l,.e' w ere ~v and Jl~v) are h~lom~rphic in a neighborhood of Kj. Thus

>

193 ; ?/I.e = 2:v ~v .X.lv) is ho1omorphic in a neighborhood of Kj. On Xj, the ?/I.e can be approximated by functions ?/J~l) ho1omorphic 1n a neighborhood of Kj+1 ' rf'=

2:~ 1fr.esj+l,£

I?/J2-?/11 1 ) I

<

by Theorem 53.

c./2 ;

is a section of

Hence

cr1 = ~.e ?/IJ 1 ) ·s j+1 , .e

~

over=a neighborhood of KJ+1 ' and (1) II 0" - <J1 11J < c./2. Similarly "i = 2:.e 1/1.e s J+1 , .e = ~ (1) .!1±2 ·v (.e) fl) (.t) -'-.t *.e ..Am sj+2,m) = 2:m ( L.e *l X m 8 j+2,m' 71

--

Cj

~-e) are ho1omorphic in a neighborhood where the L.e 7/1~ 1 of Kj+l' and therefore can be approximated on Xj+1 by 2 ) in a neighborhood of Kj+2 ' holomorphic functions

>j.:

tj

< c./4 Then a: - ~ ?'' ( 2 ) s J-m • 2 - L- y t j+2,t is ~ section over a neighborhood of Kj+2 and II 0""2-0]_11 j+l < e/4, etc. obtaining ~~ o4, ••• such that II Oj_ -Oj_+1 11J+i < c./21+1. Since the BJ are complete, on each Xk' k = j, j+l, j+2, ••• , there is a section Tk of such that II Tk-Oj_llp < c. for i large enough and p ~ k. But then for each k, Tk must be the restriction to Xk of Tk+l' Hence there exists a section T of over X such that II T- max X I~ I for [ ~1 v' J v y- ~v sJ.V = s and II s llj+l = inf -> maxv X ltv I. for [1/1] , j+1

1,,, ( 2 ) '~',t

~ ?'' (1 ) ·;r ( .e ) I

- L--

T"

£

J-

Take any representation of the in terms of the

-~ J-(v)

s j -1:1, v - 2-I.L

->

->

over all representatives

'if!

I.L

E

[1/1 ] ,

s =

s jJ..L •

sj+l,v

Then

2: ~(!.,sj+l,v'

194 Since v depends only on

j,

say v

II s I j ~

= l, ••• ,Lj+l; X j, IJ.(~ ) I

Lj +1 inf max I~v II X.J v ) I • On !J.,v,Xj bounded by a constant bj depending only on

j.

is

~herefore

lslj ~ bj Lj+l

1~! max I~vI ~ b jL j+lll s llj+l. This [~] :V,Xj proves the proposition. To complete the proof or Theorem A, we must show that -·l every stalk ~x' x e X, can be generated by global sections. So, let ~ e X. Then x lies in some Xj. The sections sj 1 , ••• ,sjL or Xj generate the stalk at every point of Xj. By t~e proposition, there are global sections tj 1 , ••• ,tjL such that on Xj j L t jk =

( *)

f=~

(

Bkt

+ 1/tkt ) s jt

'

where the tke are holomorphic in Xj and l~ktl < e there. For e sufficiently small, the transformation (*) is nonsingular, so that we may solve for the sjt in terms of the tjk" Hence the tjk generate ~·

§4.

Proof of Theorem B

First we define the tensor product sheaf -j- ® 0°' q over X, where 0°'q is the sheaf of germs of differential forms of type (O,q). This sheaf is called the sheaf of germs of differential forms of type (O,q) with values in the sheaf j-. Let x e X. Both .J:X and 0°'q. are r0x modules. X · Consider al~ finite sum~ (~jfj~tjllj), for ~j' 1/Jj e c.9x and f j E j-x , ID j E Ox' q. Define addition of t\110 sums in the natural way, -

>

:;= aj + N

M

I

~ aj =

I

f

a 1 + ••• +aN +a 1 + ••• + aM.

AllOt'/

interchanges of the order of terms of a sum and drop any term w:tth ~j' fj, "if/j or IDj equal to zero. Identify the terms (~jf,fJ\ljjiDj) and (fj®~i/JJIDj). Then these finite

195 sums modulo the identification form an Abelian group Gx = J'$-o~,q· Under the follO\'ling topology, we obtain X ..., 0 q the sheaf :r~ 0 ' : Take a representative of any element _g e Gx , ~(~Jfj~V!Jcoj). In a sufficiently small neighborhood Nx of x, the ~j and V!j are holomorphic functions, the f j are sections of .;i: over Nx, the co j are differential forms, and the projection maps of the sheaves ::1- and 0°'q are homeomorphisms. Then, for each y E Nx, .

a~sign tha~ class in

(~

l.t

I~

I'))

Gy" ~ ('f1Jfjtt9V!fj 1 for which +j and ~J are the 1direct a~alytic continuation~ or +j and "J' and the fj and coj are sections of ~ and Oo,q through rj and coj, respectively. We define the collection of all these classes to be an open set; and these open sets are to form a basis for the topology. _

N~l" ~efine ~ ( 2:_ (~Jf l~~ Jcoj)) = 2: (+Jr J®~ imJ). Then ~: 4 ® 0° 1 q -> .:; ® oo, q+1 is a homomorphism of the

sheaves; ~d ~ 2 = 0. Since Hq(x,Q) = 0 for all q > O, (cf. Theorem 36, p. 117), by Dolbeault's theorem (Theorem 26B, / p. 95) we have the Poincare lemma with respect to in X. i ·tr_ Ao "d · ~J Cl o 1 o ·-:1..r.:\ o 2 j Hence the sequence .2 ->_->~iS 0 - > L 0' 0 ' ->.£_r,b0 ' -> • • • is exact. For, at J-@ 0°' 0 ker ~ is :Jr~J CO, and else\'lhere exactness follo~IS from the Poincare lermna. This sequence is a resolution of '<-t-. Indeed, J 0 0°'P, p ~ 0 CD are fine sheaves. For, define multiplication by a C function;

-o

~

if a E c00 then o:( '> (~jfj~ 1/lrj)) = ~ (~jfj~) V'j(aiDj)),; and then proceed as in the example on p. 152. - Then by the Abstract de Rham Theorem (p. 153), Hq(X~J.:!: (~ closed (O,q) forms with values in:/-)/(~ exact (O,q) forms with values in :f). Now exhaust X by analytic polyhedra Xj. Tben for every j, Hq(X ., )-) = 0 for all q > 0 by Theorem 54B. As in the proof J -that Hq(X, 0) = 0 for all q > 0 (Theorem ;6), we obtain a lemma ~ for-;he sheaf ~~ and hence Hq(X,~) = 0 for all q>O.

-

196 §5.

Applications of the Fundamental Theorems

The following results are all obtained relatively easily from the fundamental theorems A and B. Some of these results have been obtained previously, with more effort. Note. In the following, X is a region of Molomorphy and ~ a coherent analytic sheaf over X. A. Theorem 59a. Let V be an analytic set in X; i.e. V C X, and every .point p e V has a neighborhood Np such that X ~NP is the set of common zeroes of a finite number of functions defined and holomorphic in Np. Then V = fx eX I fi(x) = 0 for every i E where the· fi are fUnctions holomorphic in X and I is some index set. We cannot prove this theorem, since it relies on the fact that the sheaf ~V(X) is coherent (Theorem 50). However, we can establish: Theorem 59. If V is a regularly imbedded analytic subvariety in X, then there are ~nctions fi' i E I, holomorphic in X, such that V = 1 x EX I fi(x) = 0 for 7 every i E IJ. Proof. ~V(X) is coherent by Theorem 51. Hence, by Theorem 55A, the global sections of ~V(X) generate the stalk at every point. For any point p E X - V, the stalk (;j V(X) )P contains the germ "1" j hence "1" is a linear combination of functions holomorphic on X and vanishing on V (with appropriate coefficients). Hence at least one function does not va~ish at pj hence there is a function holomorphic in X, = 0 on V and ~ 0 at p. Theorem 60. Oka 1 s Fundamental Lemma; a general form. Let V be a regularly imbedded subvariety in X; P the closure of a polynomial polyhedron in X. Then cl (V(\ P) = (cl (VAP)) * , its polynomial hull. Pro~f. cl (V f) P) C P which is defined by polynomial inequalities. Hence there exists a polynomial polyhedron P' such that PC(. P •c:·c X and JV(P 1 ) is globally finitely

rJ

197

generated (applying Theorem A to the coherent sheaf ~v{X)). At any point p e P - V, {Jv

= o. But C) is coherent, so Theorem B applies. Theorem 62. In X, every ~-closed form is ~-exact. Proof. Appeal to the Dolbeault isomorphism theorem and Theorem B. Theorem 6;. Suppose there exist finitely many local -1 sections s 1 , ••• ,sr generating all the ~~x' x eX. Then every global section s is of the form s = ~ ~jsj where the are holomorphic in X. Proof. Consider the sheaf homomorphism (Q r(X) -> ~

'j

defined by: hypothesis;

.+1 (:

~r x

-

->

2::

~i ( si) x

-

• This map is onto by .

hence we may form the exact sequence:

o

->

Q

->

~r

->

.J-

->

.Q

•

G is also coherent; hence we obtain the exact cohomology sequence: H0 (x,<9 r) -> H0 (x,E-) -> H1 (X,G) • Now H1 (X,G) = 0 by Theorem B; and since H0 (X,~r), H0 (x,j_) -;:re the global sections in ;J_r, respectively; Theorem 63 is complete. Corollary 1. Let Ucc X, U open. Then there exist finitely many global sections s 1 , ••• , sr of such that every section s of' l_(u) is of the form s = L ~jsj, where the ~j are holomorphic in u.

J:.

,-1

Proof. By Theorem 6:;, it is enough to shou that there exist a finite number of global sections of ~ generating the stalks of ~(U) at. every point. ~t this is Theorem A. Corollary 2. Let D c· ~n, D open. Then the following are equivalent: i) D is a region of holomorphy ii) lfuenever ~l, ••• , ~r are holomorphic functions in D without common zeroes, there exist holomorphic functions t 1 , •• ,'/fr in D such that 2: ~ j'/1 j ;; 1. Proof. i) implies ii). View the ~i as global sections of the sheaf ~(D). It is thus enough to show that they generate tbe stalks {J.... at every point x E D, as "1" is a global section and Theorem 63 applies. But this is just the hypothesis of ii). ii) implies i). If D has no boundary points, D = ~n and so is a region of holomorphy. Hence, assume D has boundary points; we shall show that every such point is essential. Let "a" E bdry D; a = (a1 , ••• ,an). Consider the n holomorphic functions ~. = zj-a .• They have a common zero at the point a, J . J only; hence they have no common zero in D. Hence there exist '/fj, holomorphic in D, such that 2: V!j(z) (zj-a .) 1 in D. If the ?{lj are all holomorphic in a neighborhood Y.a", ( 2:: V!j(z)(zj-aj))a = 1. But this is clearly a contradiction, so at least one of the ?frj is singular at "a". B. Recall that, in a region of holomorphy, vJe have an extension theorem for functions defined on regularly imbedded, globally presented hypersurfaces. This theorem extends as follows (X still denotes a region of holomorphy): Theorem 64. Let Yc X be a regularly imbedded subvariety. Then every function holomorphic on Y is the restriction of a function holomorphic on X. Proof. Consider the sheaf Jy(X). lrJe may form the exact sequence: A

=

->

~{X)

->

SJ - ;J.y

->

0 '

199

J

where Q; y = Q (Y) ( cf. Remark p. 174). have the exact cohomology sequence:

H0 (X,~) and

->

H0 (Y,i2)

->

t:le therefore

H 1 (X,~y)

H1 (x,~yl = 0 by Theorem B. Theorem 65. Let points z j e

-, x, f z j) discrete, be given together with numbers aj. Then there exists a function ~~ holomorphic ~n X, such that ~(zj) = aj. Proof. z j ~ is a regularly imbedded subvariety, of dimension zero. Theorem 66. lvith the zj as above, let polynomials Pj(z), or degree Nj, be given. Then there exists a function ~~ holomorphic in X, such that in some neighborhood N.+l or zj, ~(z) = Pj(z) + O(llz~ J ); i.e. has any given Taylor expansion up to any given order. Proof. Consider the sheaf ~ 1 defined by its stalks as follows. If for x e X, xI= zj set = (2 X • If ::1_ r X for x e X, x = z j set :;-x = germs of functions l'lhose Taylor expansions about zj have no terms of order< N.; . J i~e. which vanish at zj of order at least Nj+lj( • .:}- is an open subsheaf of (X). For coherence, l'le must ---~ show that ~ is locally finitely generated. But, for points x I= zj, this is clear; and at zj the stalks are generated by the polynomials in z-zj of degree Nj+l. Hence, we may form the exact sequence:

!

+

J

i

CJ-

-

o

->

J-

->

(9

->

C) I

f._

->

o

and therefore the exact cohomology sequence: H0 (x,t..l)

by Theorem B.

->

H0 (X,

;9/.f.)

->

But (o

o

(fJ/j)x = J l germs of polynomials of degree
c.

/

if x I= zj if x

= zj

Recall our attempt to solve the Poincare problem in the strong sense. This is not possible in an arbitrary region of

•

200

holomorphy; but is possible in its weak sense: Theorem 67. Given a function g, meromorph1c in X, then g = h/f where h and f are holomorphic in X• . Proof. Locally, we can find holomorphic functions f --I such that fg is holomorphic. Hence, define the sheaf ~ 1 ~ as follO'l'IS: J.x those germs fx _e t~,~ such that fxgx is holomorphic. is a sub sheaf of J:9 • For coherence, it suffices to show that is locally finitely generated. ,., Recall that the Poincare problem is solvable in the strong sense in any polydisc: g = h1/f1 ; h1 , r 1 are coprime and the representation is unique up to units. lle claim r 1 generates the stalks at every point (in the disc): If (r1 )x ~ 0, this is clear. If {r1 )x = 0, the only functions regularizing g are then multiples of r 1 • Hence, ~ is a coherent nontrivial sheaf. By Theorem A, there exists some global section r. But fg is then holomorphic at every point; set fg = h.

= ::J:

Jl

201 Chapter 18.

Stein Manifolds (Holomorphically Complete Manifolds)

Stein manifolds were designed to generalize the more characteristic properties of regions or holomorphy. §1.

Definition and example~

Definition 77. A complex manifold X is called a Stein manifold if: Condition 0: It is the union or countably many compact sets. Condition 1:A X is holomorP.hically convex, i.e. for every K c c X, Kc c X, l'lhere K denotes the hull with respect to functions holomorphic on the manifold X. Condition 2: The holomorphic functions separate points; i.e. for every distinct p, q e X there exists a function g holomorphic on X such that g(p) ~ g(q). Condition 3: The collection or functions holomorphic on X contain for each point a set of local coordinates at that point. Examples. i) Any region or holomorphy. ii) Regularly imbedded n-dimensional subvariety X of eN. l'Je note first that X, being regularly imbedded, is closed. Condition 1 is established by observing that if K c..c.X but ~ _/ A K )le X, then there is a dis crete sequence ( zn) e K. Hence by Theo~em 65, there is a holomorphic function ~ on en with I~( zn) I -> oo; contradicting ( zn) e "K. Condition 2 is trivial, as is Condition 3 once it is observed that every point of X has local coordinates such that zn+l = • •• = zN = 0 describe X. iii) .If X is a Stein manifold, and r is holomorphic on X, then the set x I f{x) -1 0~ is also a Stein manifold. iv) The product of two Stein manifolds is also one. He leave it to the reader to verify that iii) and iv) are Stein manifolds, while stating the following theorems (without pl''Oof). ~

t

202

Theorem 68. (Stein) Every universal covering space or a Stein manifold is again a Stein manifold. Theorem 69. (Behnke-Stein) Every open Riemann surface is a Stein manifold. vie remark that there exist manifolds which are not Stein man1folds; that conditions 2 and 3 can be replaced by a ''K-completeness" condition. (A complex manifold X is K··complete if for every x E X there exist finitely many functions f 1 , ••• ,rK holom~rphic on X such that x is a~ isolated point of the set (y EX I f 1 y = f 1x, ••• ,fKy = fKxj.), and that: Theorem 70. (Grauert) Conditions 1, 2, and 3 imply condition 0.

§2.

An approximation theorem

Definition 78. An analxt~c polyhedron Y in a complex manifold X is defined as follows: Y<...c. X, such that there exist a set x0 and functions r1 , ••• ,fr holomorphic in X such that: Y c.. c x0 c. c. x and Y = { z I z E x0 , If j ( z ) I < lJ • Theorem 71. Let Y be an analytic polyhedron in the complex manifold X, as above, and let g be a function defined and holomorphic in Y. Then g can be expanded in and the a normally convergent series of functions of the coordinate functions fi' holomorphic in X• .::·Proof. By adding functions fj, we;;-may assume that: Y ='l_z I z E x0 ; If (zJI < 1, j = l, ••• ,NJ and that the Oka map (z 1 , ••• ,znY -> (f ~z), ••• ,fN(z)) is one to one, of maximal ranlc, of Y into ll~jl < 1, j = 1, ••• ,~IJ.. The image of Y in the disc is a regularly imbedded analytic subvariety of the disc. Therefore by Theorem 64, g can be extended to a function G holomorpbic in the disc. Hence in the disc G = ci i ~ 1 1 ••• ~NN and this series l' • • N converges normally. But, setting ~i = fi(z 1 , ••• ,zn), vle

1

2::

203

obtain the desired normally convergent expansion.

§;.

The fundamental theorems· for Stein manifolds

Theorem 72. Theorems A and B hold for Stein manifolds. Corollary. All consequences of these theorems, except Corollary 2, hold also. In particular, we have the complex de Rham theorem: I

~) closed holomorphic q-forms • ,. ~ exact fiolomorphic q-rorms Note that this result shows also that the cohomology of differential forms on any Stein manifold is trivial. These statements need no proof!

Hq(X

§4.

Characterization of Stein manifolds

Theorem 73. Let X be a manifold satisfying condition 0. Then the follo\·ling are equivalent: i) X is Stein. ii) H1 (x, :f:.) = 0 for every coherent sheaf of ideals ~i i.e. for every coherent subsheaf of (9 . Proof. i) implies ii): Theorem B. ii.) implies i) : Recall the corollaries of theorems A and B: Given a discrete sequence of points, there exists a function taking prescribed values. This implies holomorphic convexity and separation of points. At every point there exists a function with a prescribed expansion in terms of local coordinates. This ii~lplies the cxio"Cencc of' local coordin:J:~c::.: ~1l1ich

are holomorphic functions. Now recall that the proof of these corollaries required only Theorem Bin the form of 11). Theorem 74. (Grauert-Harasimhan) Let X be a complex manifold satisfying condition 0. Then the follol'ring are equivalent:

204

i) X is Stein. ii) There exists a strongly plurisubharmonic realvalued function ' on X such that {' < aJccx, for every a • Proof of this theorem is essentially that of the solution • to the Levi problem, and will not be given here. Theorem 75· (Bishop; Narasimhan) Let X be a complex manifold of dimension n. Then the following are equivalent: i) X is holomorphically equivalent to a regularly imbedded subvariety or e2n+l. ii) X is Stein. Note that this gives an imbedding theorem for regions or hol--omorphy. Proof. i) implies ii). Clear by the examples. ii) llnplies i) will not be proved here. One must find 2n+l functions such that the mapping defined by them is one to one, of maximal rank, and "proper'' in that the inverse image of a compact set is compact. Ue do not establish this, but make the following remarks: This mapping is not unique. However, in the space of all holomorphic maps X -> c2n+l , under the topology of normal convergence, the fUnctions of 1) are dense.

205

appendix This appendix is concerned with proving the theorem of L. Schwartz

ap~earing

in chapter 13 on

139. We actually

pa~e

prove a weaker theorem than is stated there, but one which nonetheless suffices for our purposes. We assume that E and F are vector spaces, each having a nested sequence

!"I '

!'In ' .

of norms defined on it;

!!

II II n +1

, n = 1, 2, • • • •

Furthermore, E and F are metric spaces with metric

• Under the topology induced by each metric, we assume that E and F are complete and separable, property that fer each n, {x f E bounded (relatively compact) The theorem He are Theorem (L.

l.Ji

~oing

Scbt-rart.~).

alld

that E has the r.dded

r II X lln+l ~

th respect

1} is totally

to the norm

II II n.

to prove is the follovdng: If A and B are continuous linear

mappine-s of E into F and A is onto and B is compact, then

F/(A+B)E is finite dimensional.

206

Before proceeding with the proof,

= C0 (D·, U @')~Z 1 (D0 , U", {f)

E

1,

and F

note that if

~e

= z1 (D, U1 , {j), then

E and F

have the properties assumeJ. above, for: The nestej sequence of norms on each space is defined as follows. Kij1C~

First, in each

1 c Ki,j+l /~ u1 1 .

u 1 1 ~U 1

take a sequence of subsets

I n eac h uj, ""1 uk " tr

It

¢ or s e ts o.~.r-- u" take a

1 F

...,

sequence of subsets Kjkl c c KJk,J+l / (u/'{)uk"); and in each u. 1 nu. 1 /:¢of sets of U1 ta:.e a sequence of subsets J

l

I

I

KiJkc C.Kij,k+l

/'

"\

(ui 1 / !UJ 1 ) .

, Then for J C: E; i.e. .f = g + h,gE; C0

aSSiO.:nS the hOlOmOri)hiC fUnCtion g.l tO U.l ~

I

and hE_ z1 assigns

the holomorphic function hjk to uj"/\uk" /: ¢; define

,•

=max i·zG K ' in

Clearly these are norms and

II

lin ~

II II

The separability of E and F is obvious.

n+l for all n

= l,

2, ••••

Finally, the totally boundedness of {xE::E in the norm

II II n

I II xll n+l ~ lJ

follows from the fact that a uniformly

bounde1 sequence o ~. . holomorphic functions contains a normally . convergent subsequenqe. I.

Preliminaries. Hencerorth x and y shall 1enote elements of E and F,

respectively.

(e and f shall 1enote elements of the dual spaces).

Let sn dS n

so n

= LX t: E = {xE E = ~X €: E

Note.

'-

llxlln

I II xlln I II x lin

~ 1},

Ln

= ·[yt:

F ,. 11 Yltn

~ ~~

I II Ylln I II Ylln

= ~},

.,

= lJ, )

<

l.f,

aCn Con

= tYt: F = iyt: . ., F

<

1':) .

1.€rcE (or F) is open if and only if for every point

~€ ~there is an n and a 2.

k > 0

such that

2 + kS n c: (l.

xi~ x in E (or F) means llxi- xlln

1

0

for all n.

E* denotes the space of continuous linear functionals on E. The roll owing remarlcs, although stated .f'or E* , are true as l't'ell for F* • For e EE* define

II e II n*

= sup

le(x) 1.

These are n.ot really

xESn norms because for some n, llelln* may be infinite. each fixed n, the elements of E* -with finite norm a Banach space.

However for

II II n*

form

208

Proof.

e is

Since

o,

set containing

continuous, {x t E

I (S(x) 1 < 1}

~

oo •

is an

~pen

and therefore contains a set kSn for some

Then le(x) I <=1 for all x ~ kS , and by linearity

n and k >

o.

le (x)J ~ ~

for all

-

X f

sn, implying

II~

n: ~ l· n

2.

II 11~+ 1 ~ II 11; •

).

Given e € E*, if there is an n for which

II e II:

= 0,

= o.

then e

If e ~ 0 then there is an x E E and e(x)

Proof. Since

11"11* e n

For ·each ''· e E E* there is an n such that

1.

II xlln

= N
contradicting

4•

II e 11:

=

j

= a F 0.

~Sn and e(j) =

N'I o,

o.

Since E is separable, the usual argument based on the

Cantor diagonal process shows that from any sequence

ei~

E*

with lle 1 11~ ~ C for some fixed n, we may extract a subsequence e1

converging at every xE E.

and

II e II~

j

.,

s:~

~ C.

= {e <:: E~~

Set e(x) = l r eiJ(x), then e E-E{~

Let

I II e It: ~ lJ

and

Cn = {f E: F* I II r n: :: lJ.

Define mappings A*, B~~: F~~ -+E~~ by (A*f)x

(B*r )~

= r(Bx)

=f(Ax) -an]

for f t p* and X~ E• ....

1. Since A is onto, AA is 1-1.

2. B compact means that there is an n such that B(Sn) is z totally bounded in F.

), A* and B* are continuous. linear maps.

II. Easy Results. Propos! tion A. every f ~ F*,

Given n, there exi.s t mn and

~ c II A*ru:. mn n oo Proof. Since A is onto, F = UA(kSn)

II fll*_

1=1

=

en such that

for

oo (the bar

lJ.... A(,..RJ.,...S....),

t=l

n

denoting closure). By the BaireCategory Theorem, one of the closed sets A(j.S ) contains an open set, therefore a set of the n

form y + Cm , y E F and k > 0. If to~e make l.. even la.rger, we n

can get ~A(-~_~sn-) J Cm • For, y f A(iSn) implies y = Ax, xi n x j x1t t Sn' so that A(x+LSn) :> ki::m , implying A{k~n) ~

n

u

Let f ~ F~r. For yEA(fSn)' y =Ax, 1t(y)l

= lf(Ax)J

=

l(A{:·r)xl

llxlln

~ ,iand

....;>

x,

Lmn.

~ llxllnliA*rll~ ~ iiiA~,}fll~· By the

continuity of f, this inequality is valid for y::_ AL.Sn) and hence for y~·Cm

.

•

n

Take

Proposition B.

en= t. There is a fixed positive integer p such

that for every m, B~~(r:::) is totally bounded in the norm

II

u;.

210

Proof.

Choose n so that B(Sn) is totally bounded in F.

(p will be n + 1).

B(S n )CF::

.P..r-o ~

Then B(Sn) is compact in F.

~ (~. '--ro

Given any ro,

By compactness, a finite number of _ the

.

K.~

cover B(Sn)' and since they are nested, choosing the largest

{gives B(Sn) c B(Sn)c.[C~·

Let fE c~ and X€ sn, then

I
Therefore, B~J-Q=·~) is a

set of functionals which are uniformly bounded on Sn' and since the functionals are linear, this set is equicontinuous with respect to the norm II

lin· Furthermore, Sn+i.::sn is totally bounded in Hence by the Arzel~-Ascoli theorem, from any

the norm II lin.

sequence in B~~ (r=::) we may extract a subsequence which converges uniformly on sn+l' i.e. B~~(c:> is totally bounded t-Tith respect Jt.

to the norm II

II~ +l. If f t F* , then IIB.;~fll* < co. p

Corollary 1.

~(.

(For, fEF"

implies fE: ~ for some m, K < oo.) If {f ~ E F~~ are uniformly bounded in some norm,

Corollary 2.

,

..,

then there is a subsequence ~.f i :; such that Theorem 1. Proof.

N

=

. "I

-~f t-F;" \..

j_,,_

Jt.

(A"+B,..)f

= Oj )

u;1k .> 0. .,

.,

II B;~ (f1. -f i

)

J k is finite dimensional.

Since A~~ is 1-1, if suffices to show that A*(N) is

finite dimensional.

Let p be as in Proposition B.

Then A~:-(N) is·

211

II 11;,

a Banach space under the norm

II e II*p < norm II

00

n;

because for every e~A-~(N),

since e = A*f, f E: N, implies

e=

-B~~-f ·which has finite

by Corollary 1. •c.

Suppose A"''(N) is infinite dimensional.

Then there is a

*

sequence ~~ a 2 , ••• of linearly independent elements of A (N). ~

*

~

We claim that there exists -e 1..{ tF:A .. (N) satisfying llekllp = 1 . and II ek - e. II .!J- > -12 for k 'f j. The proof is based on the followini. J

Lemma.

p -

If E is a Banach space and G is a closed, linear,

proper subset of a linear set DeE, then there is an x 0 ED with

II x 0 II

-

II x 0

= 1 and

1

GII ~ 2•

I

Proof. I

II

there is a y E: G with d ~

and let Di 2 =

..

~

1_a1 , a 2~

•

= diatance

I

to G. Then I I I I X -y x - y II ~ 2d. Set X • 0 tlx 1 -y' II the linear space spanned by ~~

Take x ' D - G and let d

of x

Since every finite dimensional linear

subspace of a Banach space is closed, we may apply the above len,ma to Gl and Dl2 as subsets of A-:~ (N) with norm exists e 1 ~ D12 with

II e1 11; = 1

Next, apply the lemma to G12 get e 2 ( D123 with

II e 2 11;

and

II e 1

= {a1 , a2}

= 1 and

II e 2

u;.

II

~~for all e E G1 •

-ell;

= {~, a 2 , a3f

and n123 - e·H;

"

spanned by

and

II ek

~

- e II~

,::

1

and

~ ~ for all e f. a12 •

Continue this process, obtaining a sequence {ek·~

II ek 11; = 1

Hence, there

t

A~~'(N) with

..

2 for all e belonging to the space

~, ••• , ~; but then II ek - e j ~~~- ~ ~ for k 'f

On the other hand, since -[ek~~A-:t-(N), ek = AJ.~ofk' fk E: N.

j.

212

By Proposition A, there are

{eiJ

mp and CP such that

Corollary 2, (B*fk) contains a subsequence

II fk II"~ ~ CP. Then by which ~ is Cauchy in the and

consta~ts

norm

II

*

*

However, B*fk = .~A fk = -ek,

liP •

can have no Cauchy subsequence in the norm

II I~

by

construction. Theorem 2.

Given n, there exist mn and Kn < oo such that if

e '"(A~~+B.;~)F.;~ and and II f

II: ~ n

Proof.

II e II~

Kn II e

< oo, then there is an f€ F* ·t-rith e = (A:~+B*)r

II~ •

Let p be as in Proposition B, and let mn be given by

Proposition A.

It suffices to consider only n

have established the theorem for n

=p

of {A.;~ +B.;:-), as before.

ll.;~+l n

< ·II -

ll.;t- • n

p;

since once we

we have it at once for all

n < p by choosing· for such n, the constants mn and recalling that II

~

=mp

and Kn

= Kp ,

Let N denote the null space

~ie claim that it is sufficient to prove ~

,,

.,

.,

that there is a Kn < oosuch that for all fE F"'' 1rlith II(A"'"'+B'.. )fll~ <en, * Kn .. .. .~. .. .. It f - Nlllrb ~ 2ll (A"''"+B-''")fll~. Indeed, if (A"'"'+B.,..)f = e, then there will exist a ~t-- f - N satisfying II~~~~!- < K II (A:~+B:~)fll{:- = K llell* m - n n n n ... ,,. " •t. n and (A·.r+B''" )~ = (A·.r+B"'' )f = e.

213 By Proposition B, a subsequence of the Vi' call them again

o,

satisfies IIB*(wi-fj>ll; i,,

and therefore

with .,

II ' i

"'u: r

-

v

=

((A'~+a·,;-)'IJ )x

II 'ti o,

1

I:t o.

O, implying

11: r,1 o. Henge there is a 'It~ F* II"' - nNII* ! ~· Then t t N, lvhile

and

~

~

t 1 ( (A+B)x) =::

,, ... lim ( (A''"+s''" )\lfi )x = 0

1-+ro N, a contradietion.

i~oo

III.

1j'

"l

- vj

n t( (A+B)x) = lim

for all x E E means that "'

,·

IIB*('lti-wj)ll~

Thenbythe triangle inequality, IIA.~~(\Ifi'•tj)ll~ by Proposition A that

iVi1

Main Results. Let YoE (A+B)E, and let M ={r

fF~~

I f(yo)

=

o}.

Call

(A~t-+B* )M = L.

Lemma 1.

Suppose e 1 € L,

and sup:pose ei (x)

y+ e(x) for all

x

~ C for some n and C

EE.

< oo,

Then e €- L.

By Theorem 2, we can find f 1 E F~~ with e 1 = (A-:~ +B"'.t-) f i

Proof. and

II ei II~

II f.ll~~ l. m

< K. n _

~le claim that fit H. ~

for l-lhich e 1 = (A-""+B'

)t 1 •

Set

w1 =

Indeed, there are ~-£ N

.. " fi - ~i' then (A''"+B'~)\jti

l.

= O.

Since y o E (A+B)E, there is a sequence 4x ~ ~ E such that .. a (A+B)x ~ y • Hence ti (y ) =lim vi( (A+B)x ~ =lim (L\~~+B-::}~.. ~=0, a a o o ~ 00 a a-;.oo J. implying that \It i E lJf, but then f i = \If i + 'if M. Now, because the

tf i}

is uniformly b0unded in the norm

it follot-rs that a subsequence of the

{r~,

call them again

II II~ ,

{ri~,

converges at every point y ofF, fi (y)--;:. f(y), f =:-F~~ and f(y 0

)

so that f € 11. But then, ( (A-:~ +B·::- )f )x = f ( (A+B )x} = lim f i ( (A +B )x) .. ., i~oo lim ( (A';-+B-·")fi )x lim ei (x) = e (x) for all xE E. Thus i~oo.. .. i-+oo e = (A"+B"';-)f! L.

=

=

n

=0

214

Before proceeding, He recall some notions from functional The Hahn-Banach Theorem states that if S is a linear

analysis.

subspace of T, a Hausdorff locally convex topological vector space, and if x{ (j, an open convex subset of T such thatcf.!ls = ~~ then there is a closed hyperplane H:JS with HncJ= ~.

= H +' ~x~,

theorem it follows that since T

(,.

~

From this

i.e. H + the linear

space spanned by x, if we define fer t f T, ~( t) = 'A where t = h + 'Ax, then ~ is a continuous linear functional on T whose nullspace is H. Hence there exists a continuous linear functional ~ on T satisfying ~(x)

=1

and ~(S) = 0.

Theorem

=1

that e(x ) 0

Proof. If n

= 1,

3.

If e E- E.:~ but e

t L,

= 0 for

while g(x ) 0

then there is an x 0

,

every g tL.

(1) Since e E E.::·, there is an n for Hhich

Therefore !!ell~:.< oo for some n > 2. we will assume

-

II e II~

II e II~

< oo.

II II~~ II l!~,llell~

i.e. llell{!- < oo, then, since n

E E such

< ro.

In order to simplify notation

< oo : the proof in the general case is

(essentially) the same. (2) then

There is an

r\

>

0 such that if gE::L and llg- ell~<

1,

llg- ell~ >tt_. such~

Proof. ·Assume that no (gj )fL satisfying llgj-

every xE E, Jgj(x) - e(x)l llgj II~ ~ 1 +

II e II~

exists.

Then there is a sequence·

ell~~"~ land llgj- ell~ j

j

O. But

II

gj -ell; ~

O, i.e. for 1 implies

< oo so that by Lemma l, e f: L; a contradiction.

215 Take t1_ smaller so that 2 Yl_ ~

(3)

II g- ell~ ~ 2rt•·

llg-

1, then if g{ L and

ell~ >'1.·

as1 •

(4) Let {x1 (l)} be a dense sequence in exists since

as1 C

E a separable metric space.

II g

an. integer N1 > 0 such that if gEL and

Such a sequence

Then there exists

- e II~ ~ 2 yt the follow•

\

ing inequalities cannot hold simultaneously

Proof.

Assume the contradiction, then there is a sequence

(gN)t L trJith 1

= 1,

II~- ell~~ 2yt

~rLfor

ell~~ 2f1. implies that II gNII~ ~ 2rt_+ llell; < oo

llgN-

••• , N.

and lgN(x 1 (l))- e(xi(l))l

so that there is a subsequence, call it again (p.N), such that at every xE E, gH(x) 1r> g(x), g t L by Lemma 1, and lg(xi (l)) - e(xi (l))

I·~ 'l for

i

= 1,

2, ••••

llg-

e II~ ~ 2 rt.'

Hence }g(x) -e(x)j

~ tl,

for xt ~s 1 , which implies that llg- e II~ ~rv contradicting (3). (5)

.

Let {xi

(2)7

J

be a dense sequence in

exists an N2 such that if g' L and llg- e

os2 •

11; ~ 3y

Then there

the following

inequalities cannot hold simultaneously

i = 1 i

Proof:

= 1,

=1,

Assume the contradiction, then there is a sequence

(gN) E L with

for i

,

II gN

••• , N1

- e II; j

~ 3rL

and lgN(x1 (l)) - e(x1 ( 1 )) I

lgN(x 1 ( 2 )) - e(x 1 ( 2 )) I

~ 2fl

for 1

~ rL

= 1,

••• , N2 •

216

II gl'J - e II;~

3rt_ implies that llgN

11; ~ 3f'L + lie II;~ 3yt + lie II; < oo

so that there is a subsequence of the (gN) conver.ging at every point of E to a gtL satisfying lg(x.(l))- e(x_(l))l
1

and Ig (xi ( 2 ) ) - e (;i ( 2 ) ) I

~ 2 rt

1

-

= 1,

for i

Hence

••• , N2 •

* ~ 2'l,; but jg(x) - e(x) I ~ 2 rt, for x(: os2' so that llg - ell2 and

Ig{xi {1) ) ( 6)

- e(xi (1)

>I~

ytfor i

= 1,

g€ L,

. (4) ••• , N1 ,• contradict1ng •

* ~k Continue this process: hence if g ~ L and llg - e Ilk

rt

then the following inequalities cannot hold simultaneously

Ig {x.1 (s) )

- e(x. (s) ) I < s 1

-

rt,,

i

.

= 1,

•. . , Ns , s = 1, ... , k- 1.

x(l) x(l) { (7) Let {an} denote the sequence ~ , ~ , x(l) (2) (2) x( 2 ) (3) .., Nl xl x2 N2 xl ] 27l' 2V(_' ••• , 2rt' 3yt' .•. EE. Since

. x(k)

\I ·

and

II

lin~

II

lln+l' for any fixed k,

II

anllk

7

Jl

1

~'l,ik = k~

... , tl'

0, i.e. an ~0

in E. (8)

II

* +l ~ Ilk

* IfgE:Lthenforsomek, llg- ellk
II II *k

, for k sufficiently large llg - e

Since

II *k

~ k yt.

By ( 6),

at least one of the inequalities lg(x. (s)) - e(x. (s)) I < sY(' 1

for i = 1, ... , Ns; s

= 1,

1

•.. , k - 1; is invalid..

-

This means

that there is ann, depending on g, such that lg{an) - e(an) Then for all gE:-L, sup lg(a ) - e(.· n)

-n

(9 ) q

n

--n->

>

1.

> 1.

Set P = { ( g ( a 1 ), g {a 2 ), ••• , g (an), ••• ) Ig

= (e(a1 ),e(a2 ), ••• ,e(an), ••• ).

g(an)

I

I

c L}

and

Since g and e are continuous onE,

0 and e(an) ~0, by (7).

P, qc

P., 0 ,

the Banach space

of sequences of complex numbers tending to zero under the norm,

"1""' ~ f

II v II = s ~P Ivi I

(

v 1 , v 2 , ••• ) •

p is a linear

- q II = inf sup Ig ( a. ) - e ( a1 ) I > 1 by ( 8 ) • 1 gEP i -: there is a continuous linear functional on 'f._ 0 \-Tbich is 0

subspace of Th~refore

,. 0 where vE: )L , v --

II P

I! 0 L •

on P and 1 at q.

This means that there is a sequence of complex

numbers ak l'lith [:lakl < oo such-that if gf: L then [::akg('1c) = 0

= 1.

and Cake(~)

From (7) and Clakl < oo, we see that the sequence of

(10)

partial sums an x 0

EE.

of r=:ak'1c is Cauchy in E and therefore converges to

Then e (x 0

)

= 1, lvhile g(x 0 )

=0

for every gt Lj

completing the proof. Lemma 2.

(A+B)E is closed. Let y 0 f:. (A+B)E.

Proof.

so we may as well assume y 0 f (y ) o o

= 1.,

so that f

f

M.

o·

If y 0

F 0.

=0

then we know y 0 E (A+B)E,

.

Then there is an f 0 f p'H' such that

Set e

= (A*+B-:~)f o'

o

e

J L:

o.,..

for if

= {A~:-+B~~ )' 0 , ~ 0 E· M, so that if w0 = f 0 - ~ 0 , then = o, but wo (y o ) = lim w ( {A+B)x ) = lim ( (A;:-+B*)w )x = o a o a o a

e0 E L. then e , 0

(A~:-+B.;~)w

o

~~

implies ''''~'o f L and hence f o E L. -

there exists an x f E such that e (x ) 0

gEL.

0

Let fE F~:-, then f - f(y 0 )f 0

then e - f(y )e 0

0

E L,

0

0

= (A+B)x

0

E

t

M.

=1

=0

and g{x ) 0

Therefcre. if e 0

&

e{x ) = f(y ), i.e. { (A"+3" )f)x Hence y

0

for every

= (A.;;.+B.;~)f,

which means t~at e(x ) - f(y )e (x ) = 0, or &

0

¢.

a

Since e o t E" but e o - L, by '11heorern 3

(A+B)E.

0

0

0

= f( (A+B)x 0 ) = f{y 0 )

0

A

fer all fE: F". .

218

Theorem

4.

Proof.

Let K = (A+B)E. By Lemma 2, K is a closed subspace of F,

F/(A+B)E is finite dimensional.

If F/K is infinite dimensional, then there ~s a sequence of linearly independent y i ~ F, so that K0

=K, K1 =K0 + {y1~, K2 =K1

+ {12}, •• , are

closed subspaces ofF satisfying K1 properly contained in Ki+l' For each i, then, there is a pi such that IIKi- Yi+liiPi > 0. Hence we can find continuous linear ftL."'lctionals fiEF{~ vlith fi (Ki) =0 and fi(yi+l) = 1. i, f i (K)

= o.

The fi a~e linearly independent and for every

Therefore for every x EE, ( (A-:~o +B-:1- )f i )x =r1 {{A+B)x)

so that f.c N, and N therefore is infinite dimensional. I

l

contradicts Theorem 1 and completes the proof.

This

=0

1