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

iii

PREFACE

These notes reproduce almost verbatim a course taught during the academic year 1962/634.

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. exception.

L.B.

Zurich, July 8, 196+.

This one is no

iv

CONTENTS Preface ................................................. Chapter 1.

91. §2. ®3. $4.

Preliminaries ..................................... An inequality.... ........ ...................... Proof of Hartogs' Theorem1 ....................... Holomorphic mappings ..............................

Chapter 2.

3. Meromorphic functions .......................... 4. Removable singularities ............................ @5. Complex manifolds .................................

...

.

.

2. Differential forms on manifolds.. 3. Poincar6 Lemmas

59 61

63

6a 71

76

80 84 85

Canonical Isomorphisms

................................. De Rham's 2. Dolbeault's Theorem . Complex de Rham Theorem. 3 2

38 44 48 50 53

Differential Forms

1. Ring of differential forms inadomain............ .

Chapter B.

26 30 35

Cohomology

§1. Cohomology of a complex manifold with holomorphic functions as coefficients ........ ..... 2. Applications ......... ;. Other cohomologies ... ............................. Chapter 7.

12 14 18 22 24

The Additive Cousin Problem

1. The additive Problem formulated ................... 2. Reformulation of the Cousin Problem ............... 23. Reduction of the Cousin Problem to non-homogeneous Cauchy-Riemann equations .......... Chapter 6.

7 10

Zeros of Holomorphic Functions. Meromorphic Functions.

1. Weierstrass Preparation Theorem ................... 2. Rings of power series .............................

Chapter 5.

5

Pseudoconvexity

1. Plurisubharmonic and pseudoconvex functions ....... 2. Pseudoconvex domains .............................. 3. Solution of the Levi Problem for tube domains ..... Chapter It.

. 1

Domains of Holomorphy

1. Examples and definitions .......................... 2. Convexity with respect to a family of functions... 3. Domains of convergence of power series............ It. Bergman domains ................................... 5. Analytic polyhedra ................................ Chapter 3.

ii

Basic Facts about Holomorphic Functions

89 93 96

V

Chapter 9.

The Multiplicative Cousin Problem

1. The Multiplicative Problem, formulated ............ 98 2. The Multiplicative Cousin Problem is not ......... 100 always solvable ................... 3. The solution of the Multiplicative Cousin 103 loX Problem for polydiscs ......................... §4. Characteristic classes (From C.II to C.I)......... 106 .

Chapter 10.

Runge Regions

1. Preliminaries ..................................... 110 2. Polynomial polyhedra ..........................:... 112 113 3. Runge domains .

Chapter 11.

Cohomology of Domains of Holomorphy

1. Fundamental Lemma, stated ............... :......... 2. Applications of the Fundamental Lemma ............. 3. Preparation for the proof of the Fundamental Lemma ........................... 04. Proof of the Fundamental Lemma ....................

Chapter 12. 1. 2. 3. 4.

117 120

Some Consequences of the Approximation Theorem

.................... Relative convexity... ... Unbounded regions ofholomorphy............. ...... The Behnke-Stein Theorem .......................... Applications to the Levi Problem ..................

Chapter 13.

115 115

128 129 130 132

Solution of the Levi Problem

1. Reduction to a finiteness statement ............... 134 2. Reduction to an extension property ................ 137 3. Proof of Proposition 2 ............................ 140

Chapter 14.

Sheaves

1. Exact sequences ............................... 142 2. 3. 4. 5. 6.

Differential operators ............................ Graded groups ..................................... 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.

1. 2. 3. 4.

144 147 149 150 155 158

Coherent Analytic Sheaves

Definitions .................................... 162 Oka's coherence theorem.. ....... 163 Ueierstrass Preparation Theorem, revisited........ 165 The third step .... .......................... 168

5. Consequences of Oka's theorem ..................... 171 6. The sheaf of ideals of a variety .................. 173

vi

Chapter 16.

Fundamental Theorems (semi-local form)

§1. Statement of the fundamental theorems for a box (semi-local form) ........................... 175 2. First step of the proof ............................ 175 3. Reduction of (3) to on holomorphic matrices ........................... 177: 4. Proof of Cartants theorem on holomorphic matrices. 180 5. New proof of the Oka-Weil Approximation Theorem... 184 6. Fundamental theorems for regions of holomorphy (semi-local form) ...................... 185 Chapter 17. 1. 2. 3. 4. 5.

Coherent Sheaves in Regions of Holomorphy

Statement of the fundamental theorems........... Preparations for the proof ........................ Proof of Theorem A ................................ Proof of Theorem B ............................ Applications of the fundamental theorems..........

Chapter 18.

.

187 187 191 194 196

Stein Manifolds (Holomorphically Complete Manifolds)

31. Definition and examples ........................... 201 2. An approximation theorem..... ..................... 202

3. The fundamental theorems forStein manifolds ..203 94. Characterization of Stein manifolds ............... 203 Appendix ................................................ 205

1

Chapter 1.

Basic Facts about Holomorphic Functions S 1.

Preliminaries

We introduce the following notation:

I denotes the field of real numbers. C denotes the field of complex numbers or the complex plane. Cn denotes the space of n-tuples of complex numbers (zl, ...,zn) = Z. Cn may be considered as an n-dimensional vector space over C or a 2n-dimensional vector space over It. Cn may therefore be identified with Ig2n, which induces a topology in Cn. A. By function, we will mean a complex-valued function f unless

otherwise stated, for instance f: Cn-C. Definition 1. Let DCCn be open and f(zl, ... , z ) a function den

fined in D. f is said to be holomorphic in D if, for every (zl, ... , zn) ED

and each j

=

1, 2, ... , n,

Mil .... z.+h, lira

Ihl-o

8f

h

8z.

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 assumed to be continuous. Hence we obtain immediately: Property 1. If f1 and f2 are holomorphic On D) then f1 + f2, f1- f2, f1f2,

and, if f2 40, fl/f2 are holomorphic(in D).

Property 2. If (f . } is a uniformly convergent sequence of holomorphic 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. If f is holomorphic in D and has a local maximum at a point pE D then f is identically constant in the component of D containing p. Definition 2. Let KC ,n be any set. Let f be defined in K. Then

f is holomorphic in K if and only if to every point p of K there exists an open set DC d;n such that p E D, D fl K is closed in D(i. e. D-K is open in Cn and there exists a function F holomorphic in D such that F = f on DI) K.

We remark that for functions f: In»R the existence of all partial derivatives 8f/ 8xi, i = 1, ... , n does not imply f is continuous. However, the corresponding theorem for functions of several complex variables is true. B.

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 z0 in n is the set

Izj-z°I 0 such that p < Ri - I K 1. and a number M, such that I f(zl, ... , zn) I < M for I zl-K I < p and I zj I 2. i

Hartogs' Lemma. Let f(zl, ... , z n) be holomorphic in I z . I :SR

i and bounded for I zl-r I < p and I z. I < R,, j >_ 2. Then f is continuous in Iz.I < R

Proof of Osgood' s Lemma.

Define for I z I < R1'

max I f(z, z2, ... , zn) This maximum exists by induction Iz.I 2 hypothesis. Denote by A N ={ z I I z I :SR and m(z) < N}. Then m(z) =

1

{z I I z I< R1}

00

tN

Now A

N

is closed, for if a

A

N

and ar-a,

m(a) = max I f(a, z2, ... , zn) = max lim I f(ar, z 2, ... , z) I < max N = N. r-oo For instance, by the Baire Category Theorem, one of the A N contains I

a disc. For zl in this disc f is bounded.

Proof of Hartogs' Lemma. We may assume R.

= 1,

j = 1, ... , n,

J

=0, and M = 1. Let D1=f(zi,...,zn) Iz1I 0 and A is some constant. Therefore for 0 < a < 1,

I ak

1.

.

k I a
0 integers.

Then log I p(z) I = log j a i + ml log (z11 + ... + m n log I zn I , and

log (hull of K w. r. t. log I p(z) I )* is simply the closed half space Dlog K*

defined by the hyperplane P with coefficients (ml, ... , m ) such that n P('ilog K* 0 . Therefore the intersection of all such half spaces is log {z E

cn I logI p(z) I < sup log I p(K ))*, which is log (K)*, by the monotonicity K

of the log.

Now, if a subset S of a complete circular domain D stays away from 8D, then log S* stays away from 8 log D*. To prove this it suffices to show that (1) z e 8D implies a z = (a 1zl, ... , a nzn) e a D for all a e Cn with I a i I= 1 n

for all i and (2) every z E IE with (log I

il

l , ... , 109 1 zn I) E 8 log D* belongs to

D. For then if log S* comes arbitrarily close to 0 log D*, given E > 0 there are points s e S and d c 8D such that (using sup norm) for every i, l logl s. I - logl d. I I < E, which implies I I s I - I d. I I < E' , e' - 0 as e - 0; and by (1) then, there is a point d' E 8D with I si - d'i l< E' for all is a contradiction. (1) is easily established once we note that z j D implies for all a with Iail = 1, az V D; because then if zo c 8D and E > 0 is given, there are points E D and v D such that IX .-zol< a and Iv. -z°I < e for all i. Hence

for all a, Ia.I =1,la.v. -a.z°I D, D is pseudoconvex (by Cor. 4

of Thm. 3).

Establishing the converse of Theorem 9 is the Levi Problem.

Solution of the Levi Problem for tube domains

33.

Definition 27.

D = (zl,...,zn)

Let

zj = xj + iyj.

(x1,...,xn) e B 4Rn

The set where B is some

open subset of Rn, is called a tube domain.

B is called

the base of the tube domain. Example. domain.

D =

(z1,z2)

1

Ix1

< 1,

Ix2I


G.

k

J

Thus every diagonal element is positive, i. e.

xk

This means that w is convex on every straight line segment in B

J

parallel to a coordinate axis, and since ci is pseudoconvex this property is invariant under linear transformations. Hence ;P is convex on every straight line segment in B, and hence inevery component of B. We claim that this implies that every component Bo of B is convex: (a) Firstly, we show that if f X v1 is a sequence of straight line segments in Bo such that Um )L = A and lim aa,, = µ then if µC Bo X C Bo. v v-w

V-w

Indeed, since Vi is convex on each straight line segment in Bo, on each X v max fi(x) = max fi(x), i. e. max (-log zi (x)) = max (-logo (x) ), or equivaB B xEX xEM V xeA xEax v V V lently min AB(x) = min AB(x). Since AB(x) is a continuous function, the XEak

XEA v

v

equality holds in the limit min 6B(x) = min AB(x) But u is a closed set XEµ

XE1,

and µC Bo, therefore min AB(x) > 0; hence L C Bo. XEµ

(b) Now, let x, y c Bo. We must show that the line segment joining them belongs to Bo. Since Bo is connected, there exists a curve W (t), 0 < t < 1, lying in B , joining x and y; 4(0)=x,+(1)=y. Fort sufficiently o

As t - 1 there cannot exist a t 0 such that for all t < to the line segment (x,0 (t) ) C B small, the line segment (x, O (t) )CB

0

.

0 but the segment (x,4(t0))ti B0, because this would violate (a).

Hence B is convex.

0-

37

In the case j C2, it suffices to snow that if Bo is any component of B and if X is a straight line segment in Bo, then max O(x) = max 4i(x), for xeOX xEX then the proof follows as above. So, let X be a straight line segment in Bo Then «C BdomainC CB and the tube domain D over B satisfies DCC D. ^

o

Hence for every c > 0 there is a pseudoconvex function ikE in D such that depending only on the x' s; as we can define smoothing c in D, 10 functions KE depending only on the x' s in the proof of Proposition 2. As in the previous case, each is convex on B. and therefore max cli (x)= max W. XEA

But, max y(x) < max XEA

XEA

max O(X) > max XEX

XEA E

XE aA

(x) + E = max y (x) + E < maxo(x) + 2E, similarly xe aX

E

XEOX

(X) - E = max P (X) - E > max O(x) - 2E. XE aA E XEaX

Letting c \/ 0

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.

Weierstrass Preparation Theorem

91.

A.

Definition 24.

Let

zll... znn be a formal

a v

(This;;;Vines lis not assumed to be

series about the origin. convergent).

Meromorphic Functions..

Then the order of the series is the least

integer K such that

av

10,

v2

?;'normalized v

The series is said to

vl+v2+ ... + vn = K.

with respect to zl

(at the origin) if ord (series) = K and zi occurs with non-zero coefficient

aKO...0. Note that the order of a series is invariant under

holomorphic changes of variables, including non-singular linear changes, which leave the origin fixed

n ajs zs det (ads) / 0 ; Corollary.

If

av

convergent series of order 1) f

2) 3)

..v l,K,

is a

zli ... znn = f(z)

nthen the following are equivalento

is normalized with respect to z1 at the origin

(.4 l #

0

fK(z1,0,...,0)

sum T-vl+...+vn =

K

where

0,

avvn

denotes the partial

fK

zll... znn

of the K th order

homogeneous polynomials in the series f. Note that 1) and 3) are also equivalent for formal power v

series.

We shall write f(z) _

av ., v

zli...znn

sake of brevity, for all formal poweriseries.

for the

When convergence

is assumed it will be mentioned explicitly. Property 1.

If

av

., v

zli...znn = f

is any power

series, it may be normalized witR respect to zl (at the origin) by a linear change of variables: det (a3s) # 0.

zj _

c,sCs, j = 1,...,n; s-

39

Proof. convergent.

Let f be of order K.

Then

Assume first that f is

n

n

f(O = f(

a lscs,..., Z a nsCs)

,

and ord f(C) = ord f(z). llCl,...,anlC,); hence fK(C1,0,...,0)

f(C1,0,...,O) = f(a

is non-zero for some choice of the aj1 complete the matrix

3 = 1,...,n.

We may

(aik) so that det (aik) # 0.

If f(z) is nonconvergent

f= fK+fK+1+fK+2+... where the f3 are homogeneous polynomials in the zi of order J. Now, consider fK as above. Property 2.

If

ff(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 3,

0 =

f(`))

f K

0

f K )

+

+ ...

;

ord

f(,))

= K

+ 1

Consider the spherical hull =

( f (a11,a21,...,anl

jaJl1 2

l

= 11.

(3) Now fK

(zl,...,zn)

is a polynomial, and hence vanishes on

a closed nowhere dense subset of 7;-.

But the union of

countably many nowhere dense sets is noyhgre dense, so there

exists

(all,...,anl) e 2- such that fK ))(allanl)

for every J.

But we may now complete

`)(a11,

...,anl)

0 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 f + ord g

=

ord fg

ord (f+g)

ord f

min(ord f, ord g)

>

=

if and only if

0

if and only if there exists a g

g

f

is a unit; i.e.

such thar

where

fg = 1,

is a formal power series.

We remark here that the set of 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 This ring is an integral domain, with units

rings with unit.

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 B.

c Cn

a

Let

(Weierstrass Preparation Theorem).

Theorem 12.

be a formal power series, normalized w.r.t.

z1,

f

of order

K.

Then

where

h

alzi-l

h(zi +

=

f

is a unit,

+ ... + aK)

al,...,aK power series in

z2,...,zn

and are non units; and this representation is unique. If

is a convergent power series, then h and ai are also.

f

zK + alzi-1 + ...+aK

Note.

as above is called a

Weierstrass polynomial. Proof.

We first make a series of remarks: the theorem is trivial.

For

K = 0,

For

z = z1,

the theorem is also clear.

Furthermore, if

f = h(zi + a1zK-1 + ... + aK), 1

the

must have no constant term, for ord f =

ord h + ord (zi + alzi-1 + ... + aK)

Therefore

K

=

ord (zi + alzK-1 + ... + aK)

If ord ai = 0 for some

a1,

a

41

ord (z1 + alzi-l + ... + 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 ai

is convergent, the convergence of

f

h

and the

is more easily shown by a second proof. f = fK + fK+1 + ...

where

ord

homogeneous polynomials of

f.

;

fK = zl + ...

.

We wish to construct a power series =

XO + X1 + ...

(fK+fK+l+...) (XO+Xl+...)

(1) where

X0 # 0 ,

and all other

YK + YK+l + ...

=

zi + ...

=

yK

such that

K-1

are of order at most

y3

in

zl.

From (1), we obtain fKXO

But

fK = zi + ...

Similarly,

yK = zi + ...

fKX1 + fK+lX0

=

hence

X0 = 1.

YK+1

fK+l

i. e.

We choose zl,

YK

= ,

= - fKXl + YK+l so that YK+1 has order at most

X1

as follows:

Let

fK+l

=

vl+...+vri K+1

K-1

zvl a vl " 'vn 1

µ

fK

=

zl +

z 1 .. .z

b

µl+...+µn K

µ1...µn

1

in

zvn n

µn n

µl# K =

X1

Take

cl = - aK+1

0

c1z1 + c 2 z 2 + ... + cnzn

0

and

cj = - a K,O,...,1,0,...'0 - bK-1,0,...,1,0,...,0 aK+l,Q..,O for j > 2;

Choose

the subscripts 1 appearing in the j th places. X2, etc., similarly.

Note that we have proven uniqueness for both convergent and nonconvergent series.

We now proceed to the proof for

42

the case:

convergent.

f

It is due to Siegel:

As.before, write

f = fK+fK+l+ fK

For

IziI < Izll

zl + ...

=

3 = 2,...,n

,

cllzllK+l + c2Iz1IK+2


K,

and we have obtained a convergent representation f

ev(fKew)

=

as claimed, unique by the above. Corollary 1.

assume

of

f

be a convergent power aeries, and

vanishes at the origin.

f

Then the set of zeroes

in a neighborhood of the origin is of dimension

f

Proof. f

Let

is normalized with respect to f

For

=

z1,

h(zi + alz1-1 + ... + aKK

z21...0zn

small and fixed arbitrarily,

zi + alzi-1 + ... + aK

are small, and

with K

n-1.

By a linear change of variables, we may assume

roots (counting multiplicity).

al,...,aK

is a complex polynomial,

Furthermore, these

zeroes are located in a neighborhood of the origin as they depend continuously on the Let

Definition 25.

a1.

Sclosed C. Dopen

C

Cn.

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 in

D.

44

Corollary 2.

S

Corollary 3.

If n > 1

then the zeroes of

is locally arcwise connected.

f

f- is holomorphic in D C_ 0n

and

are not isolated.

Rings of power series

92.

As we have remarked previously, the set of convergent

A.

power series in

at a point forms a ring, as does

zl,...,zn

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 R

be a commutative unitary ring.

is said to be an integral domain if:

ab = 0

implies

a = 0

called a unit if

b = 0.

or

has an inverse in R.

a

will be called equivalent, written e

is a unit.

a = be,

An element

where

b

a E

and

a e

R

called irreducible or prime.

R

R,

R

a e

Elements

is

a,b e R

where

if a = eb,

a a b,

is called reducible if

are non-units;

c

b e

R,

An element

is otherwise

a

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.

of R rt E

is an ideal if

R

implies

art

A subset

a,b E I

implies

E I.

is a proper ideal if

I

and

a-b E I,

I

e I,

a

I # R,

and maximal if it is proper and such that if t is any ideal

I e 4 e R,

satisfying

R

or

I =

= R.

A ring

is called a local ring if there exists a unique maximal

in R

denotes the ring of polynomials with coefficients

R[t]

ideal.

and

R[t]

coefficients in indeterminate. aj

then

of

f a R[t]

the ring &f power series in

t

with

Let 7- at3c R[t],

a

R,

a

implies

primitive.)

an

aJtj A called primitive if the coefficients

have r.-?, common factor except units.

t

t

f =

ag

with

a E

R

and

(Note that

g e

R[t]

and

Two polynomials will be called relatively prime

or coprime if they have no common polynomial factor.

We use

45

the following results. Lemma a.

(Gauss' Lemma)

Lemma b.

Let

p,q P1,P2

Let

tK + a1tK-1

P2 =

tL + b1tL-1

Then there exists a polynomial called the resultant of

is UFD.

and

q

+ ... + aK + ... + bL

r

in the coefficients

and

have a common factor; R[t],

ai,b

which is zero if and

P2,

deg q > 0,

i.e. if and only such that

Furthermore, there exist polynomials

such that

Lemma d.

P1

p,q,s 6

P2 = sq.

P1 = pq,

and B

P2

and

P1

if there exist

Then p

R an integral domain.

a R[t],

=

Pi

only if

R[t]

is UFD,

('R[t]).

pq primitive.

primitive implies Lemma c.

If R

a R[t]

A

AP1 + BP2 = r.

No proper ideal

I

of a unitary ring R

contains a unit. Lemma e.

R

Is a local ring if the nonunits in

R

form

an ideal. B.

Let do denote the ring of formal

Definition 27.

power series at the origin in

hence

n

complex variables.

Property 1.

pn is an integral domain with unit.

Property 2.

The nonunits of

e9n

Property 3. 0 n Proof.

(9n

form an ideal;

is a local ring. is UFD.

We use induction on

elements of order

1

n.

are irreducible.

For n = 1,

units and

All elements of

order > 1 are reducible, for if ord g = K > 1 g(z)

where

gl(z)

=

zKg1(z)

is a unit, and the decomposition is unique.

Hence, assume 0 n-1

is UFD.

normalized at the origin. f

=

We may assume

f

e On is

Then:

h(zi +

alzi-1

+ ... + aK)

a hp

46

where

p

is a Weierstrass polynomial; i.e.

Now

and is monic.

is prime in Assume

is prime in

f

(9_

p e n-11zi,

if and only if p

for:

p

'9n-1[z1J, is reducible; i.e..p = p1p2 .

to be Weierstrass polynomials.

P1'p2 Conversely, assume

f

Hence

We may take f = (hpl)p2.

is reducible:

f

flf2

=

(hlpl)(h2p2) (h1h2)(p1P2)

But

p1p2 = p But

Now let

by uniqueness, and

C n-1[zlJ f e

n

.

f= But

=

P

where the

pi

h1h2 = h.

is UFD by assumption and Gauss' lemma.

hp . Pl ... Pr

are irreducible Weierstrass polynomials.

f =

Hence

hp1 ... Pr ,

and this decomposition is unique up to order and equivalence, for if

f

=

ti =

ti ... ti hi ri

,

,

by the Weierstrass Theorem and then rl .., ri

by uniqueness.

=

PI ... Pr

Hence

is UFD. 0n-l[ zlI Definition 28.

as

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. 0

Let

e D C 0, n n > 1

f,g

be holomorphic functions in

coprime at the origin, such that

D,

47

f(O) = g(O) = 0.

Then in any neighborhood of the origin,

there exist points at which and points at which

g

We may assume

Proof.

vanishes and

f

vanishes and f

and

g does not,

f does not.

are Weierstrass

g

polynomials

f =

blzL-1

zL +

=

g

-1

Z

+ ... + aK + ... + bL

Suppose there exist no such points. resultant of

and

f

r(ai,bi) For each exist

z1

if

and

f

z2,...,zn

is zero for each

On-1

c

and

f

vanish simultaneously.

g

have a common factor.

f,g

z2,...,zn.

But

near the origin, implies

as polynomials in Lemma g. the origin.

be the

But

have a common zero viewed as polynomials in

one variable, then

r = 0

r(ai,b

in a neighborhood of the origin, there

such that g

Let

g.

Let

r(ai,b

have a common factor

f,g

0

Hence

is analytic, and hence

r

be holomorphic functions coprime at

f,g

They they are also coprime in some neighborhood

of the origin. Proof.

where

p,q

Let

f

=

up

g

=

vq

are Weierstrass polynomials, and let r

be the resultant of

Ap + Bq

=

and

p

the origin so small that Let

a = (al.....,an)

it suffices to show

persists where

N.

E

p,q r

q.

To show

Ci,

f,g

are coprime.

=

are convergent.

are coprime at

a,

The equation

Ap + Bq

r,A,p,B and q

i.e. as series in

be a neighborhood of

N

Let

u,p,v,q,A and B

where

are viewed as series about

a,

48

(i

=

zi - ai .

p

=

C1 + :..

q

=

CL + ...

Then

.

i Now assume

p

e c

h

q

and

have a common factor

p,q are primitive in

But

h

'tat all,

hence h

Cl;

is also.

h where

=

ho + hlC1 + ...

6 nl. But

hi

h r

A

e n-1 and

k

divides =

Therefore

r.

hkA

a primitive power series in

Cl,

k e 0n-1[ c1]

k =

ko + klCl + ...

By comparing coefficients r

Hence

divides

A

a relation in

hokoA ,

=

r

But

h

h

hk.

hk

is primitive,

must be a unit, i.e. hk is a unit, implying that

is a unit.

Thus

p

§ 3.

Let

=

k are primitive, hence

and

hence hoo

-1

and

r,

x

e

and

q are coprime at

a.

Meromorphic functions

Dopen C

fin,

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 is called the germ of If

f

f

at

f

at

x,

denoted by

[f]x,

x.

is a holomorphic function, then

[f] x

amounts to

a convergent power series. Germs at

x

form a ring with the obvious definition of

49

addition and multiplication.

0x of germs of holomorphic functions is a

The ring

commutative integral domain with identity and unique factori-

Topologize the space, of germs Q = (J

zation.

by

0x

defining the following basis for the open BAR Let k e

then

and so

in an e-neighborhood take that class in continuation of

k = [f]X0

where

,

of xo

k e

is defined

f

y e Nxo, D. Nxo Oy containing the direct analytic i.e.

f,

At each

in

Then define

take If] Y'

U [f]

to be an open set and the collection of such

y yeNxo sets to be the basis of open sets. A holomorphic function continuous mapping,

point in D

in D amounts to a

f

f : D ->

0

which assigns to each

a holomorphic germ over that point.

Now form the quotient field Mx of 0 for each x

Topologize M = (J M.

e D.

2 E

then

where

fl

Mxo,

and is represented by 14-11x wand f2

and

are Riolomorphic functions at

because of unique factorization we may take to be coprime at in which At each

f1

y e

[fl]y/[f2]y.

Let

as follows:

Let

xo.

and

xo,

and

fl

f2

Nxo be a neighborhood of x

0

take that class in

Nx ,

[fl] o/[f2]xo

are defined and are still coprime.

f2

and

2 c M,

The union over

Nx

My represented by

of these classes we

define as an open set and the collection of all such sets we take as the basis for the topology.

The elements of Mx are called 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

z

0

a

D,

a meromorphic function

efined by the quotient of two functions

fl,f2

g

is

coprime

50

and holomorphic at a)

If

z0.

f2(zo) / 0,

is a unit, then

f2

i.e.

is holomorphic there and hence

is holomorphic at

g

and therefore in a neighborhood of regular point of b)

If

f2(zo) = 0

If

z0.

0

0,

then

g

fl(zo) = 0,

then

zo

fl(zo)

and

f2(z0) = 0

and

points has topological dimension

2n-4.

gIregular pts is holomorphic. Corollary 2. If z is a pole of

exists a neighborhood

N

poles

NM

Corollary 3.

of

zo

g

has no isolated there exists

IgI > M.

A point of indeterminacy is a limit g.

A point of indeterminacy of

point of zeroes of

is open,

in which every point is a

in which

point of zeroes and poles of Exercise.

0

Furthermore,

and, for each number M > 0

(n > 1),

a neighborhood

of z

g

then there

g,

0

pole or a regular point.

is

The set of such

g.

The set of regular points of

Corollary 1.

is

z0.

called a point of indeterminacy of

and

zo

is called a

z

g.

said to have a pole at c)

fl/f2

g - a,

where

a

g

is a limit

is any complex number.

Poincare's Problem 1)

Given a domain

Weak form:

meromorphic in

D

is every function

D,

a quotient of two functions holomorphic

in D? 2)

in

D

D,

Strong. form: is

g

Given a domain

D,

g

meromorphic

the quotient of two functions holomorphic in

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 in one complex variable.

51

A second theorem on removable singularities will be stated but not proven.

Theorem 13 (RadS). Let f(z1,...,z ) be continuous in n Dopen cCn and holomorphic in (D- jz1f(z) = 03 ). Then f

is holomorphic in Proof (Heinz).

D.

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 C 0n,

is any open disc whose closure is contained in

D'

we must prove that

is holomorphic in

f(z)

Without loss of generality, assume that

D'.

is the unit disc

D'

(IzI zo e (a(D'-A) /1 r)

and then

[f(z)-g(z)]-> 0

(ii)

z -> zo E (a(D'-A) /1 A )

and then

[f(z)-g(z)]

remains bounded and

1(z)

and

log If(z)I -> - oo

(D'-A),

.

43(z) -> negative numbers while

(z) -> positive numbers. in

and

=

z s (D'-A),

g(z)

Then, consider the

e (D'-A)

Note that for

or

r.

In either event, 42(z) and

But, since the i are harmonic

they assume both their maximum and their minimum

52

on

6(D'-A).

Yz) and

44(z)

Let a -> 0,

and

41(z)

Hence,

are positive for all

e el(Di-A).

z

then

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

which implies

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

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

which implies

Re [f(z)-g(z)] > 0

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

which implies

Im [f(z)-g(z)] 4 0

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

which implies

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

for all

z e (cl(D'-A) or)

Since

-

cl(D'-A),

equations in

and thus

g s 0.

which means that

has continuous first order

f

hence by continuity, on f = 0

A,

In any

A.

satisfies the Cauchy-Riemann

f

D-A,

In the interior of

z e

f a 0

(D' t) r),

Thus

D'.

partial derivatives.

for

Q,

in

f(z) = g(z)

is harmonic in

consists of points of r and

aA

f(z) = g(z)

component of the interior of Therefore

f(z) = g(z) for all

Therefore

z e cl(D'-A).

points of

f

are negative, and

43(z)

a0

and on r.

and therefore satisfies the

equations in

A.

Hence

equations in

D'

and is therefore holomorphic in

Theorem 14.

holomorphic in

Let D.

D - zjg(z)= 0J . Proof.

h

g

and let

g

D1.

0

be

Let

f

be holomorphie and bounded in

f

is holomorphic in

D.

Consider the function

gf

if g/0

0

if g=0

=

is continuous as

where it is not zero. But

Dopen r_ 0n,

Then

h

Then

f satisfies the Cauchy-Riemann

f

is bounded, and holomorphic

By Radb's theorem,

h

is holomorphic.

is holomorphic by assumption; hence f

is meromorphic. thus holomorphic.

But

f

=

h g

is bounded,

thus without poles,

0

53 Theorem 15.

Let

Dopen C an;

g,h

holomorphic in

D,

not identically zero and relatively prime at every point of Let

D.

Then

f

f be holomorphic in the set is holomorphic in D. Ddomain C 0n Let

Theorem 16. holomorphic in Proof.

D.

Let

Then

(D - {zlg(z) = 03) zlg(z) = 03).

holomorphic in h

D.

g # 0,

h

g $ 0

be

is connected.

Suppose that

S = U U V where U

Define a function

holomorphic where

and let

S = (D -

is not connected, then disjoint sets.

D - [zJg(z) = h(z) = 01.

='1 in U

ZO in V

S

and V are open, h

is

and is bounded; therefore

h

is

This is impossible since it implies that

is identically 1 in D.

5.

Remark.

Complex manifolds

From now on, "differentiable" means "Cco" is a (differentiable) manifold of complex if the following conditions are

Definition 30. X real complex) dimension r, satisfied: (1)

X

(2)

Given an open set in X

is a Hausdorff space. and a function defined in

it, it is possible to say whether or not this function is ,differentiable `

holomorphic every point in X has (3) There exist coordinates: (real, differentiable) functions a neighborhood where r complex, holomorphic are defined such that they give a homeomorphism of this

neighborhood onto a domain in

( Rr),

and every function

C (differentiable)

if holomorphic and only if it I. (differentiable as a function of xl,...,xr holomorphic in each variable of zl,...,zr The coordinates are called local coordinates and such a

defined in this neighborhood is

neighborhood is called a coordinate patch. (4) i.e.

X

There is a countable basis for the open sets of is second countable.

Remarks.

On a complex manifold we may talk about

holomorphic and meromorphic functions; on a differentiable

X,

54

manifold, about differentiable functions. A connected 1-dimensional complex manifold is called a Riemann surface. Every complex manifold is a differentiable manifold. Therefore it is natural to ask on which differentiable manifolds 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,

If

n = 1,

r = 2n.

these conditions are also sufficient;

if n > 1,

however,

In fact, in the latter case,

they are not.

necessary and sufficient conditions are not known. There are other differences between the cases and

n = 1

n > l: (i)

axiom (4) in definition 30 is

When n = 1,

unnecessary as it follows from axioms (1), (2), and (3).

When n > 1, (ii)

axiom (4) is essential.

If n = 1

X

and

is compact, there exist non-

constant meromorphic functions on X

(i.e. on every closed

Riemann surface there exist non-constant meromorphic functions). However, when

there are compact complex manifolds

n > 1,

having no non-constant meromorphic functions. (iii)

If

n = 1

is not compact, there exist

X

and

non-constant holomorphic functions (i.e. on every open Riemann surface), while if

n > l

this is not necessarily so.

Y be a compact connected complex

For example, let

manifold of dimension n > 1,

then

X = (Y-ZpS)

is not

compact and if there existed a non-constant holomorphic function on

it would be holomorphic also at

X,

p

(by

Hartogs' Theorem), and it would be constant (by the maximum modulus theorem).

Examples of Complex Manifolds. 1)

0n

2)

Any open subset of a complex manifold.

3)

If X

and

Y

are complex manifolds then X x Y

is

55

a complex manifold, where a function

f

is holomorphic in

if it is holomorphic in X and holomorphic in

X k Y

Y.

4) The complex projective space Pn(M)

[(zo,zl,...,zn)]Izi

_

[(z0 ...,z )]

where

(zQ,...,z)

Cn+

denotes the equivalence class of points where two points

,

such that

j = 0,1,...,n, with local

Ci = tCj,

coordinates in a neighborhood of for some

and

(CO,...,Cn)

are equivalent if and only if there is a

(CO' " ''Cn)

t # 0

are not all zero}-,

i,

(z0,...,zn)

being

0 < I < n,

zi

0

z0/zi'

.. -'zi-1/zi' zi+1/Zi, On this manifold

is a complex manifold.

..., zn/zi,

where

there exist meromorphic functions; the ratio of two homogeneous polynomials of the same degree is such. Pi(10) = Riemann sphere =

5)

The special case of 4,

6)

Starting with a complex manifold of dimension n > 1,

fC omitting a single point, and imbedding a new complex manifold.

we will obtain

This procedure is known as the starting with

We do this for n = 2,

e~- process.

P1(C),

First we define the space

C2

to consist of two types

X

of points: I

II X

=

L(zl'z2)

I

= i ((Cl,C2)]

C2

(zl,z2) E I

C2

(Cl' 2)E

and

(z1,z2) /

(0,O)}

and

(Cl1C2) /

(0,O)j.

I VII .

=

Secondly, we make X

Into a Hausdorff space by defining

the following basis of open sets:

A neighborhood of a point in the ordinary topology- of not contain

Cl / 0,

satisfying

C2

E I

shall be a neighborhood

such that its closure does

(0,0).

A neighborhood of a point and say

p

it

p c II, p= ((C1'C2)]'

shall be the set of points;

- C2/Cll < E,

and

(zl,z2) E I

[(1,C)]

s II

satisfying

56

Izl) < E

zl / 0,

for some

and

1z21 < E

,

Iz2/zl - C2/Clj < E,

E > 0.

Thirdly, we define local coordinates.. Near a point of

I, take of

[(C11C2)]

and

z1

where say

II,

say, take as local

zl

=

(l+t)T Co

z2

=

T

C1

=

(1+t) to

C2

=

1

On this new manifold

is holomorphic on

Therefore

(0,0).

f

on every point of a

on

For if

f

X-P1(C) = C2-10J.

is also holomorphic at the origin

the value of

P1(0),

II.

approaches some complex number,

as its argument goes to the origin. of

every

the following holds:

is constant on

it is holomorphic on

X,

f

.

X,

holomorphic function on X

and hence

C. # 0

[(Cl,C2)] = [(C1/C2,l)] _ [(c0,1)] coordinates t and T

By Hartogs' Theorem,

Near a point

as coordinates.

z2

but then

P1(C),

Hence near every point

is close to

f

a,

f

a.

Thus

f = a

must be identically

II.

A globally presented, regularly imbedded analytic

7)

subvariety

Y,

complex manifold Let

of codimension r

X

fl,...,fr

such that, for every

in an n-dimensional

is defined as follows:

be holomorphic functions defined on

x e X

at which

X,

fl(x) _ ... = fr(x)= 0,

the rank of the Jacobian matrix

J

= CZ7

is r,

i.e.

J

is of maximal rank.

The derivatives

are to be understood in the following way:

Let

afi/azj

N be a

57

coordinate patch containing

and let 3 : N -> 9"

x,

define the local coordinates (zl,...,zn)

_

of1

Then Then

Y,

ofi 1-1 (z1,...,z n) z

.3

as a subset of X,

is defined as:

Y = j x I f1(x) = ... = fr(x) = 0 Note that Y

is closed in

are clearly satisfied. in

and that axioms

X,

1, 2 and 4

We now define the local coordinates

Y.

Let

with local coordinates I(y) =(zl(y),...,zn(y)

y e Y,

defined in a neighborhood that, at

y,

relabeling the

N

of

where

y,

det (6fi/azj) = det A / 0, if necessary.

zi

=

fl(zl,...,zn)

r =

fr(zl,...,zn)

Cl

Cr+l

Cn

=

=

N G X,

Assume

j = 1,...,r by

Define new coordinates

zr+l Zn

Then the transformation taking

z

to

1

is given by the

square matrix;

1

0

0

which is nonsingular as

det A / 0.

Then the local coordinates

58

in Y are

(C r+lP...ACn).

Note.

Cl = ... = Cr = 0

as

on

Y.

This example exhibits the technique by which

statements in en

that are local, i.e. refer to some neighbor-

hood of a point, are transformed into statements about analytic manifolds

X

of dimension

Hence, when proving results

n.

about manifolds, we shall sometimes assume

X C_ In.

The

modification of notation needed for arbitrary manifolds will be left to the reader. Note.

A regularly imbedded globally presented analytic

subvariety of codimension 1 is called a hypersurface.

8)_ A regularly imbedded analytic subvariety Y

is a closed subset of X

codimension

r

point of Y

has a neighborhood

Y /1 N

N

in X

of

such that every such that

is a globally presented regularly imbedded analytic

subvariety of codimension r

in

N.

59 Chapter 5.

The Additive Problem, formulated

91. A.

The Additive Cousin Problem

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,

and polynomials

a function PV

at

a discrete set of points

Pv(z a ),

av a D,

without constant term, find in

f,

with singular part

D,

av.

As we know, this problem is solvable for all domains

D C. C.

The Additive Cousin Problem is as follows (we shall denote it by "Cousin I", or simply C.I in the sequel): C.I

Let

X be a complex manifold, and U = ui, i e I

be a given open covering of meromorphic functions that

Fi-F3

Pi

is holomorphic in

meromorphic and defined on in

X,

X,

Let

some index set.

I

defined in

ui,

be given, such

Find a function

ui/)uJ.

such that

F-Fi

F,

is holomorphic

u1.

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 presented regularly imbedded Then extension from

analytic hypersurface of codimension 1. Y

is always possible, i.e. given

there exists 1,

holomorphic on

holomorphic on

4,

X,

such that

Y.

$ = I on.Y.

Assuming this theorem for the moment, we exhibit the following application:

Theorem 18 (Cartan). Let X C 92, Cousin problem is always solvable in

X.

open, such that the

Then X

is a

region of holomorphy. Proof.

and that X

We may assume that

X

is a domain, that

C2, as we already know that every

domain of holomorphy.

Hence, assume

b c bdry X;

Cn

0 e X,

is a

b = (bl,b2)

60

where

b1,b2

f(z) = b2z1-b1z2. 0

and b.

f(z) = 0

is an analytic plane through

Let Y = fzjf(z) = 0} .

open set in

hence

C;

and singular at

is an

holomorphic in Y

4>,

But by the extension theorem, there

b.

holomorphic in X

exists a 1,

Y1 = Y/IX

Now

is a region of holomorphy.

Y1

Therefore there exists a function

and I is singular at B.

Then if

are fixed complex numbers.

such that

on

_ 4)

Y1,

b.

Proof of the Extension Theorem

Let Y = {f = 01 ; There exists a covering

ui

holomorphic in

U = luije X of Y

and equal to

on Y /)ui,

4>

Y.

such that Y G .ui;

there exists ai such that Ii

ui

and in each

4 be holomorphic in

let

and

is

by

definition of holomorphicity on closed sets (see proof of Lemma p. 74).

an open set in

u0 = X - Y,

Let

Fi = J1/f ,

Define:

FO

1

=

X.

i / 0

.

This covering and set of associated functions defines a

Cousin problem, for, on u

() ui,

Indeed, on ui 0 UP

is holomorphic except possibly

0

for points on

Y.

Fi-Fj

Hence, assume

Introduce local coordinates

FO F1 is holomorphic. f(z0) = 0,

z0 a ui /) u3.

f = C1.

such that

(C1,..,,Cn)

But now, =

1i(O$C21...PCn) - '$ J(O)'C20 ...,Cn)

and and

as

jj

agree on Y,

and

i(Cl,...,Cn) Fi-Fi

0 ,

.X (Cl,." ,Cn)

_

But in the power series expansion ofi ]V only terms containing powers of

Cl

appear; hence

F1-F5

is holomorphic

in ui (\ u3. By hypothesis, there exists such that

I = fF

gi = F-F1

F,

meromgrphic in

is holomorphic in

is holomorphic in X and equal to

X,

We claim that

u1. 4)

on

Y.

61

Clearly, I is holomorphic on 0 = X - Y. on

u1,

F

Consider

i # 0;

F =

fgi + Ji

=

Hence,

gi +

y1 and

,

in

is holomorphic.

gi

and on Y n ui,

ui,

Reformulation of the Cousin Problem

2.

(Cousin problem belonging to the covering U).

C.I'.

an analytic manifold of dimension

Given X,

U = {u15 and holomorphic functions

covering

n,

defined in

ui /f u,

fi3

satisfying

fA = - fij + f jk + fki f ij find holomorphic

(antisymmetry)

0

=

defined in

fi,

(compatibility),

u1,

such that

fib = fi f,.

C.I' implies C.I.

Claim.

Assume C.I is given, and let

Proof.

where the

F1

fi3 = F1-F3, are meromorphic functions defined In u1.

are holomorphic and satisfy the symmetry

Then the fij

and compatibility conditions. functions of F

C.I',

and define

=

F1 - fi

hence

F

V

i E I

u3),

Then

F3 - fi

Let

U = Iuj3,

(i.e. every

i e i

X,

and assume that

vi

is contained

and let a Cousin problem belonging to the

covering U be given. to

=

ui.

fi - fi

be two coverings of

is a refinement of U

in some

in

ui /I u3

solves C.I.

Induced Cousin Problem. V =

=

fiJ

be the solution

F = Fi ft

is globally well-defined, for in Fi - F3

and

fi

Let

V as follows:

We induce a Cousin problem belonging

Let

be an "affinity" function from

C-

that viC uIn vi nvj,

I

to

J

such

assign fii = fa (i)01- (f)

are antisymmetric and satisfy the

Clearly, the

f1 compatibility condition.

We now reformulate the Cousin problem a secoi}d time:

C.V.

Let a Cousin problem belonging to the covering

Find a refinement V of U

U be given.

such that the

induced problem is solvable with respect to

V.

(We shall

show later that the choice of the affinity function is immaterial.)

C.I" implies C.I.

Claim.

Let U = tuii be a given covering, with associated meromorphic functions

Let V = {vij,

As before, define

Fi.

fij = Fi-Fi. be the solvable induced

with associated

gij Cousin problem, with affinity function

o":

viC- U.(i) , gi,j Let

gi

be the solution functions.

F

in

vi

is globally well defined and solves C.I, for

Fcr Furthermore,

(i) -

F-F1

Then there exists a

F-F1

Define

F.(1) - gi

F

Then

fa (i)0-(j)

=

For,

0)

is holomorphic in vj

containing

gi-g j.

fcraw(j) = gii = u1,

x c u1.

and consider:

x,

F-F6 (,)) + F6 (j) - Fi = - gj + f6. (3) i

for, let

,

defined and holomorphic in v j n u_(3) Cl ui = v j iI ui, and

x e vj /1u1. Remark.

We may now consider the Cousin problem for

locally finite coverings only (i.e., for coverings such that

63

every point of X

has a neighborhood which intersects only

a finite subcollection of the cover),'by virtue of the following observations:

(Paracompactness).

i)

In any manifold every covering

has a locally finite refinement by open sets each relatively compadt in some open set of the origin covering and some coordinate patch (see de Rham, Varietes Differentiables). C.I" implies C.I.

ii)

Reduction of the Cousin Problem to non-homogeneous

03.

Cauchy-Riemann equations A.

Intermediate Problem.

Given a complex manifold

locally finite open covering U = ui functions

defined in

fij

,

X,

and holomorphic

u1 /1 UP antisymmetric and

satisfying the compatibility condition, find functions defined in

such that

ui,

Proposition 1. solvable.

fii = gi - gj

gi,

where the gi a COD

The Intermediate Problem is always

The proof of this proposition will be presented

subsequently (p. 64).

Let the functions

v = 1,...,n be defined on X

aV,

as follows: av

=

in

-

ui

;

v = 1,...,n.

azv We claim the

aV

are globally well defined, i.e.,

agi V

But this is clear, for there:

Furthermore, the

0

in ui /I uJ.

a2V g3-gi = fji aV

and

fji

is holomorphic

satisfy the following

compatibility- ;ondition:

and this is clear.

aaV

aaµ

azIi

azv

Now we can state the final form of C.I:

64

such that

defined on

Find a function A e Cm ,

Final Problem. X,

=

aA/a'zv

where the

v = 1,...,n;

av,

av

are as above. of dimension

More precisely, given a complex manifold X

n and

functions

COD

defined on X and

v = 1,...,n;

av,

satisfying the compatibility conditions 4

aav

µ,v = 1,...,n

3

aZu

find a function

-

aiv

defined on

A,

aA/azv = av,

such that

X,

v = 1,...,n. Note.

These differential equations are known as

non-homogeneous Cauchy-Riemann equations. Proposition 2. Proof.

above.

The final problem implies C.I.

Assume there exists a function

Set

defined in

fi = gi-A,

are given by Proposition 1.

i.e. the fi

where the

ui,

agi

azv

azv

aA

But

are holomorphic.

fi-fi = gi-gj = fib.

X be a differentiable manifold,

Let

defined on

w

U,

1

of U

function, as before

and locally finite

U.

Note that, if V

if there exists a partition

partition wi

at each point of X

there exists a partition of unity

subordinated to the covering Proof.

such that

COD,

on X - u

Given any manifold X

Proposition 3.

open covering

=

Then a partition is a system of

positive and

X,

0

wj

C00

0

of unity subordinated to the covering U wj,

gi a

a"zv

U = i ui a locally finite open covering. functions

defined as

Note that

afi

Definition 31.

A,

is a refinement of wi

for

as follows:

(vi G ud(i)).

V,

Let

U,

and

we may define a a,* be an affinity

Now set

65

For those

on vi

0

otherwise

not as yet included, define wj = 0.

ui

U be given.

Let

wi

U

By Remark i), p. 63,

locally finite refinement V = (v1 some coordinate patch Pi.

such that

has a

vi cc

is paracompact and

Since :X

Hausdorff it is normal, so that there is a locally finite

open covering V'

vi C-Cvi.

such that

For each

i, I

and

be a diffeomorphism of Pi into IRn. Let s = f(vi) siopence sopen. si = f(vi); then In jRn there are

Coo

functions

let

fi

Cl si

satisfying *1 = 1

IRn -> [0,1]

outside ..

and = 0

Let

iii- f = 91,

on

and set

(DI

B.

Proof of Proposition 1.

Let w to

be a fixed partition of unity subordinate

U = Lui

gj =

Set

.

wi fi,

sum is understood as foll unless

x e ui 11uj.

When 0),(x) # 0,

When

fij

for

in

cai(x) = 0,

is defined.

where this

wi(x) = 0

set wifi

Note that

for all but a finite number of indices The

uj,

x e uj,

= 0.

witx) = 0

i.

solve the intermediate problem, for

gj

gi - gj

_

wk f j k

wk fik -

wk (fik+ fkj) _

wk (fij)

=

fii

Hence, we have reduced the Additive Cousin problem to an existence theorem for the nonhomogeneous Cauchy-Riemann We shall exhibit a solution for a polydise

equations. shortly.

Example. manifold,

Let

X

be a simply connected differentiable

U = L ui a locally finite open covering.

We pose

a "Cousin Problem" as follows:

To each

ui /1

ui

let there be assigned a complex

66

number

such that fi3

Then, find constants Let

+ fjk + fki = 0.

and

fii = - fji fi such that

fi3

fi3 = fi-f3.

gi be the solutions of the intermediate problem;

where gi

fi, - gi gj;

is defined in ui

and

gi a C°D .

Define:

Then the

c

in

V = 1,...,n.

ui;

v

are defined globally, as before, and

av

Find a

The final problem becomes:

aaµ/axv.

function A

agi

=

a av

such that =

T

av

v = 1,...,n.

,

V

Such a function exists, by Stokes' theorem.

The

gi - A

=

fi

are constants, as

fi

Set

.

v = 1,...,n.

afi/axv = 0,

Prove the converse of the above example,

Exercise.

in the following form:

X be a domain in [Rn.

Let

Then

solvable implies

C.I

that every curl-free vector field is a gradient.

If

D = (z1,...,zn)

Let

Theorem 19.

are defined and

aj(z1,...,zn)

and satisfy

aaJ/az-k = aak/azJ

function

in D

$

Proof.

Then the

DO = L(zl,...,zn)

We claim that there exists a a smaller neighborhood of

assume

DO

D,

4/azi = ai

are defined and

3 = 1,...,n.

in

Cco

then there is a

such that

Let

1.

a3

Iz3I < R3 < IODJ,

I

I

for

for

3 > k.

Cco

3 = 1,...,n.

IziI < r3 < oo}ce-D.

Ca) in a neighborhood of satisfying

For

k = 0,

DO.

C°D in perhaps

defined and

4

a+/6E

The proof is by induction on

a3 M. 0

3 = 1,...,n,

ail

where we

k,

the problem is

reduced to solving the system of homogeneous Cauchy-Riemann equations,

a4/az3 = 0,

of which

4 = 0

Assume that the problem can be solved for consider the case

and for

C = e + ii1

of

N(e1,DO),

DO,

Choose

k = 1.

and

e > 0

(z1,...,zn)

el < e,

define

is a solution. 1-1,

k =

and

sufficiently small,

in an

e1

neighborhood

67

w(zl,...,zn)

a2(zl,...,zg

l

_

- n

dt dq

z

JJ

Id-ree w(zl,...,zn) N(e1,D0)

is then defined and

and satisfies

in all variables in

C0D

aw/a-zI = of

and

by the compatibility conditions

j > 1,

cb/azj = 0

for

a2/z3 = aaj/azk.

The system

=a,,

(1)

0,

j

aZj

aZj

j = 1+1,...,n;

is then equivalent to the system

j

azj

aw/azA = a2,

but since

a( -W)

(21)

aZ

Now

j

=

aZ j

aZj (2) is actually

aj

aZj

aZ in

(aj- aw/azj) a Coo

N(e1,D0)

compatibility conditions since

j

and satisfies the

a/azk(aI- aw/azj) =

aaj/azk - a/azk(aw/a"zj) = aak/azj - a/azj(&U/azk) =

a/azj(ak aw/azk). has a

Hence, by our induction hypothesis, (21)

2.

is

D.

D0.

Construct open polydiscs

on

Dj

and

c Dj.

If the

satisfied

lim

$

a+j/azk = ak

Choose

(zl,...Izn

Ej > 0

+1'

for

2,...

and

such that 5- Ej < eo

I4j+i-+jl < Ej

in

Dj

then

would exist uniformly on compact subsets of

D,

00

j+p < Ej + Ej+l ... + Ej+p (f)

is a homomorphism of

into Hr(X,0). = [holomorphic functions on Xi.

H°(X,(o)

-If HI(X,D) = 0

then every Cousin Problem is solvable

(of. 2nd reformulation of Cousin Problem).

Holomorphically equivalent manifolds have the same cohomology groups.

2.

Theorem 20.

Let

Applications

X be a complex manifold and Y a

globally presented, regularly imbedded hypersurface,

Y = (z1,,..,zn) 13(zl,...,zn) = o3 where function on

X.

then IF (Y'

) = 0.

Proof.

If for some

$

Hr(X,9) = 0,

r > 0,

U' = 3ui of X

We say that a covering

sufficiently fine if for every

is a holomorphic

/

ui) () Y ( finite

is

a function

holomorphic in this intersection can be continued to the

finite ui' Lemma. There exists a covering U0

every refinement of U0 Proof. N(y),

Let

y e Y,

C = 0. n is a function of

a function of

N1 A Y

C11 ...'Cn

y,

such that

Y

Any function holomorphic in N(y)1j Y and hence is holomorphic as

C11 ...'Cn-1 C11 ...,1n, i.e. in

is any open set contained in in

such that

then in some neighborhood of

there are local coordinates

is given-by

of X

is sufficiently fine.

N(y)

N(y).

Clearly if N1

any function holomorphic

can be holomorphically continued to

N1.

Take

this neighborhood system and add open sets not intersecting Y

75

to form an open covering of property, i.e. is

This covering has the desired

X.

UO.

Now take

From now on we consider only such coverings.

such a covering U = Jui3 of

r-cocycle on Y of

Let f'y

U.

every

induces an r-coboundary on some refinement

be an r-cocycle on

ui )tlY

(

Since

A

U

when

ui

which means that

= 4g

Since

Hr+1(X,(9) = 0,

g

some refinement U1 of U. refinement Therefore

U1.

where

and hence

0 = 56i-X = 4,6g

= 0

Define

.

Y.

g

in

Then 5S'` = 0

on Y

is an (r+l)-cochain on X,

5g = 0;

g

is a cocycle on

X.

must induce an (r+l)-coboundary on So consider everything above in this

Then g = 5h

h is an r-cochain on X.

where

6(SX-4,h) =

is an r-cocycle.

(z) defined

j

;Xon X such that 4X = Sy on But

-.

Therefore, there is an r-cochain

(n ui )(" Y = 4,

5

fi

is sufficiently fine,

can be continued to J_

uij

y assignd to

then

Y,

a holomorphic function

in this intersection. fio,.ir(z)

We must show that every

X.

0,

and hence

C_X-4,h

Since Hr(X,O) = 0,

TU1.

induces an

r-coboundary on some refinement U2 of Considering the rX-4h above in this refinement U2, SY, = 6F where F is an cX SX-4h Hence on (r-l)-cochain on X. = = S y = 6F,

F

and since on

is an (r-1)-cochain on

X,

it is an (r-l)-cochain

Y.

Note.

The following theorem gives sufficient conditions

for a domain to be a domain of holomorphy.

Later on we will

prove that these conditions are also necessary. Dopen C Cn, n > 2. If Theorem 21. Let for

1 < r < n-1, Proof.

then D

Use induction on

from Theorem 18.

0

is a region of holomorphy. n.

The case

n =-2

Assume that every open set in

IF = 0, 1 < r < n-2, is a region of holomorphy.

follows 0n-1

for which

Rather than

working with a component of D, assume that D is a domain. Then let

DdomC n

and let b c boundary of D. Pass a hyperplane P through b

and an interior point of D and set POD = Y. By Theorem 20, all the

76

cohoinology groups of Y from

Y

induction hypothesis then,

Hence there is a function at

b.

Since

in

D,

and hence

H1

g

H1(D,(9) =.0,

Hn-2

to

are

holomorphic on Y and singular the Cousin Problem is solvable

can be continued ho!omorphically to

g

D,

b.

Other Cohomologies

§3. 1.

By our

is a domain of holomorphy.

and the extended function will be singular at

A.

0.

Had we defined an r-cochain to assign a

C0D function,

instead of a holomorphic function, to intersections, then we would have gotten

Hr(X,COO),

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 3.

X

Here

Hr(X,C).

is any topological space.

Had we defined an r-cochain to assign an integer to

intersections, we would have gotten

Cohomology group of the topological space 4.

X.

r,

In fact, given any Abelian group

defined an r-cochain to assign an element of case we would get B.

we could have

r,

in which

Hr(X, r).

Definition 36.

Let

be a topological space,

S

a mapping of

Hausdorff space, and

p

projection mapping).

Denote

Sx G S

and note that

the Integral

Hr(X,a),

onto X

S

by

p-1(x)

XeX Sx = S.

called a stalk,

Sx,

The triple (p,S,X)

is called a sheaf of Abelian groups over X

if

x e X and each

a)

p

is continuous, and for each

s c Sx,

p

is a homeomorphism of a neighborhood of

onto a neighborhood of b)

s -> -s

S

into

defined on the set

R

of pairs

,

in

is an Abelian group such that:

is a continuous mapping of

s,t

s

x.

Each stalk Sx

(s,t) -> s +t

such that

X a

(called the

S;

and (s,t)

belong to the same stalk, is a continuous

mapping of the subset

R

of

S

r. S

into

S.

S

77

For simplicity, we call

the sheaf.

S

Let UopenC X. A section (or cross over U is a continuous map 5 of U into

Definition 37. section) of S

S

such that

Denote by r(U)

is the. identity mapping.

the additive group of all sections of

over

S

U.

Examples.

X be arbitrary, and let

Let

1.

Let

x e X.

S =

section of 2.

Sx = ' for all

have the discrete topology.

Then a

over UopenC X assigns some integer to

S

Let

USX

X be arbitrary, and let

U.

any Abelian

Sx = a,

group, for all x e X.

Let S = USX have the discrete A section of UopenC X is an assignment of an

topology.

element of a

U.

to

As in example 1, all stalks are isomorphic.

Such sheaves

are called constant sheaves. 3.

Sx = 6x,

X be a complex manifold, and

Let

the

set of germs of holomorphic functions. Let

Definition 38.

x e X,

a complex manifold, and

consider holomorphic functions at neighborhood of

each defined in some

x,

We say that two such functions are

x.

equivalent if they coincide on some sufficiently small neighborhood of

This is clearly an equivalence relation.

x.

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 d -

The set of all germs form a group over each point

x.

S = VSx

Introduce a topology in the set of all germs Take any element

as follows:

then f e Sxo,

5 e S;

and

is an equivalence class of functions defined in a neighbor-

hood of xo a X. defined in each

Nx0 ,

y e Nx

,

Take a representative a neighborhood of

assign that class in Sy

direct analy?ic continuation of

U Yexo

g,

g e Sx

xo e X.

;

g

is

hen for

containing the

say jgyj.

Then

tgyJ is, by definition, an open set; and these are

78

to form a basis for the topology of

Each section over U

S.

is 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°x°

where CX denotes the set of germs of Then

defined similarly to (9x.

C°D

functions at

x,

S = L/Coxo 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

and U = uii , cochains

an open covering of

I e I,

Cr(X,U,S)

with coefficients in

as follows:

S,

U,

assigns

t e Cr(X,U,S)

a section

ui /1 ... (ui ,

S over this intersection so dat J

in the indices.

Define the

on X associated with the covering

to each ordered intersection, of

over it,

S

X.

is antisymmetric

Note that we can add cochains, and talk of

antisymmetry conditions, for we can add their values using the group structure of

1-(u0 0 u1 1) ... /1ur).

Continuing the construction as before, we obtain and then form the projective (direct) limit,

Hr(X,U,S), Hr(X,S).

Theorem 22.

Hr(X,Coo) = 0

differentiable manifold

X.

for all

In fact

and any

r > 0

Hr(X,U,Ceo),

r > 0

is already trivial, for every locally finite covering

U.

We define a homomorphism 9 : Cr(X,U,C,) ->

Proof.

Cr-l(X,U,Co)

r > 0,

so that

f = 96f + 69f,

and this

is sufficient. Let

wi;

be a partition of unity subordinate to

(see Definition 31 and Proposition 3, 9S(io...ir-1)

=

3, Chapter V).

wif(iio...Ir-1)

where this sum is understood as follows:

or if wi = 0.

U Define

,

wif = 0

= 0

if

Note that the local finiteness of U

insures

that almost all terms of this sum vanish at any point of Now

X.

79

(A6 +69j)(io...ir) _ T- wi 4wo...ir)+

(-1)k+l ;(iio... k...ir) + (io...ir)

(0i

(io...ir)

Note

wi -

r

(-1)k wif( io... k...ir

(-1)k wif(iio...ik...1r)

_ (io...ir)

Cr(X,U,CO°)

k...ir)

wi 5(iio..

(-1)k

+

o...ik...ir)

.

that this proof hinges upon the fact that is a module over globally defined

Coo

functions.

80

Differential Forms

Chapter 7.

91. A.

Ring of differential forms in a domain

Definition 39.

form in D

Let

/dxi

(the symbol " A"

case the term

is read as "wedge"), D,

dxi A ... /dxi

A dxi

Note that

a

it 0,

in

holomorphic form

D. D

.

If

da = 0

I

then

a = d(3

for some

86

D be the domain in (Tb), and let a be a

Let

(Tc)

(p,q)-form in some

If

q > 1.

D,

for

then a = aR

as = 0

(p,q-l)-form 0 Proof.

Let

Lemma.

such that

be fixed numbers,

4) = ay/axi

and if

If

is a Cm

4)

function W in D a$/axi = 0 then 4/axi = 0.

then there is a

D,

i,j = 1,2,...,n.

i # J,

D be the domain in (Ta).

Let

(La)

function in

i,j

C°D

Xi

*(x1, " xn) _ f

Define

Proof.

dt.

0

D be the domain in (Tb).

Let

(Lb)

holomorphic function in function

/

D

in

If

is a

$

then there is a holomorphic

D,

such that

and if

4 = axi/azi

a4/az3 = 0

then ail//aZj = 0. Since

Proof.

$

is holomorphic in

D,

OD

ak(zl,

_

n)zi+l

*(zl,...,zn)

If

is a

$

there is a of

D1

k+

Let

(Lc)

C

00

such that

zi then so is

and if

4) =

with

then

$

is holomorphic in

Define

1

n

fJr

r

-z i

Use induction on

(Proof of Ta)).

1 "' Jr

dxj 1'..dxj

dxk+l,dxk+2,...,dxn means that

D1,

in perhaps a smaller neighborhood

ICI k.

dx3,

which means that the

J > k,

whose derivative with respect to on

dxk and

do not depend on

p

Then

j > k.

dxl,...,dxk-l'dxk+11"" dxn'

contains no terms with

coefficients of

respectively

r

0- contains no terms

Since

has no terms with both

do

a' are pure

do' contains only terms with either

J < k or involving

dxk,

and

with

dx3

" ''dxn.

and

p

(r-1)

0 = d a = (-1)r(dxk A dp) + d6 .

with dxk,dxk+11-1 dxn;

dxk+1'

a -(_,)r_1 do = dy

which means that

a = d((-1)r-1 p + y). (b)

The proof is identical with that

(Proof of (Tb)).

of (Ta) with the x's replaced by z's and a a holomorphic

(Lb) gives the existence of a holomorphic form p and

form.

its independence of (c)

Do c c D.

zk+l .. "'zn. (Proof of (Tc)). 1. Let

a

Then

and satisfies

D0

be an open polydisc,

in

N.

a - a p.

of

D0

We claim that there is a form

in perhaps a smaller neighborhood of

p,

N

is defined in a neighborhood

as = 0

D0,

For the proof, use induction in

such that anu an

dzk

argument analogous to that in (a). 2. is

By 1., for each

D.

that

Construct open polydiscs

a = api (i)

bidegree

Di

Di CC D3+l

there exists a form

in a neighborhood of

Assume that (p,0);

i.e.

a

whose union such

P

D3.

has bidegree (p,l),

pj = = bi 1

.,' i

has

then

dzi1 ...dzi p

,

p

and

88

app = a in DJ. For fixed k and 3 > k, Therefore the coefficients of

is a

(p J-PIC)

since (CJ-% ) = 0

holomorphic form on Dk

Dk.

on

can be approximated

(f3J-Bk)

as closely as desired by polynomials, and hence the form

can be approximatedby a form Pjk whose coefficients

(03-Pk)

are these polynomials.

Choose

A

as follows: 102 pil =

a3. lim

131'

01 =

=^a = a on DJ. = 0

g

on on

D2 and Q32 = Q21 + P32, etc.

'2 =

11)-P211 < el I(P3-Y-Q,2I < e2 Since

- bk

Construct 1. T1...Tp where 2 - P21 Dl, 5.A = P1 - P32 where

T1...T

3

such that 7- ej < oo,

> 0

e

-t3}I = = Ib'

and define

on Di

I(33-03+1I < ei

exists uniformly on compact subsets of

the

D,

>03

coefficients of

are

f3

53 = a in

functions and

Coo

D.

(Cf. Part 2 of the proof of Theorem 19.)

If

(ii)

a

has bidegree

has bidegree of

D3.

(p,q-1).

Therefore

in a neighborhood

a((iJ+l-3j) = 0

3+1-

then each

q > 2

(p,q),

for some form

= ayi

y3,

in a

3

neighborhood real valued

N3

by

of Di,

Mi of D3, Then yj is defined in DJ+1

neighborhood D3.

g _

Let

;_l^ al,

a(yl+... n

Set y1 = wi yJ ayj on

=

J+1 - 01

and in general Then

J+1

(RJ+i-pj) -

+ a(y1+...+yj-1) _ (sj+l-fY - Oy 1

at"i = ac3,j = a on Dj.

desired R.

and

a2 = 12

i - a(yl+-2+...1-yJ-1

Let wi be a and ^ 0 outside a

1. of (c).

on D M3 cc N J.

C°D function = 1

Letting 3 -> eo

0,

and

we obtain the

89

Chapter 8.

Canonical Isomorphisms

§1.

De Rham's Theorem

Definition 44a.

A.

with covering

U.

Let

X be a differentiable manifold

We say U

is simple with respect to

differential forms, or d-sim lie, if it is open, locally finite,

and, for the intersection of any finite number of sets of the u0 ll... (fur,

cover

Theorem 24a.

the Poincare lemma for "d" holds.

X be a differentiable manifold.

Let

Then there exist arbitrarily fine d-simple coverings (i.e. every covering of

X open.

X

has a d-simple refinement).

We shall first prove this theorem assuming X GIRn,

Proof.

The case of an arbitrary manifold

X

is treated at

the end of this section. Assume 1 = 1,...,n

Note that in any box jixi aiI

X C Rn.

` ri'

Poincare's lemma holds by Theorem 23.

5 ,

Furthermore, the intersection of any finite number of boxes is again a box; so it suffices to refine any covering

U

to a locally finite covering by boxes and this is easily done. Hr(X,U,CP) = 0;

Lemma la.

r > 0, p > 0,

where C?

denotes the sheaf of germs of p-forms, and U is locally finite.

The sections of 00

Proof. so

over

Hr(X,U,C)0) = Hr(X,U,Coo) = 0,

Since any element of

remains in

that if

A

:

when multiplied by a

function,

Coo

r > 0,

so

then f = Of + 69f precisely

Hence every cocycle is a coboundary.

Corollary.

Theorem 25a.

Hr(X,r?) = 0;

Let

a d-simple covering.

isomorphic, where (P p-forms in

Coo functions,

by Theorem 22.

Cr(X,U,Cp) -> Cr-l(X,U,(p),

f s Cr(X,U,CP),

as before.

are

we may establish this lemma by constructing a

OP,

homomorphism

Up,

U

r > 0,

X:

r > 0,

p > 0,

Cp

as above.

X be a differentiable manifold,

U

Then the following groups are canonically denotes the sheaf of germs of closed

9o

1

M H (X,U,

r?

closed p+l-forms s ,

Hr+l(x,U,cp)

Hr(X,U,CP+1) , r > 0,

=

(ii)

(iii)

p ' 0

exact p+1 form

closed r-forms

Hr(X,Um

ti

exact r- forms

p > 0

r > 0.. '

Before proving this theorem (following A. Weil), we introduce the notion of coelements. B.

Definition 45.

A coelement

f

(r,p)

of bidegree

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 nonempty intersection, then

frp(JO...Jr)

assigns to

this intersection a pure differential form of degree p defined there.

The coelements form a vector space over C. grip+l, Define dfrp = where g assigns to each intersection "d"

Define

Clearly

of the form which f assigns; d2f = 0. 5frp = hr+l,p in the usual way; 52 = 0.

d5 = 5d

(for

d

"adjusts" the domain, and

5

the range, of the coelements). Coelements

f

for which

are cochains with closed

df = 0

forms as coefficients, and if

the coelements

df = Sf = 0,

are cocycles with closed forms as coefficients.

If U then then

is a d-simple covering,

dfrp = 0 f = 6g

implies

f = dg.

f

p > 0,

a coelement,

and 5fr = 0,

If r > 0

by Lemma la.

We say that coelements

fr+l,p

associated if there is a coelement and

f = dg.

C.

Proof of Theorem 25a.

and grP

fr'p+1

such that

are f = 5g

Note first that (i) and (ii) imply (iii), for closed (p+l)-forms exact p+ - o rms ti- H1(Y ,U,C1o

ti Hl+p (x,U,n°) ,

and

°

_

H1+s(XJIUVnp-s)

ti

-

-

p?0

91

Note also that S

Hr(X,U,(

=

)

frp (df'=5f =O 6hr-1,pidhr- ,p=0f

;

L

r>0

We associate a closed (p+l)-form on X (or, flop+1 of bidegree (O,p+l))

(1)

equivalently, a coelement

to each cocycle class in =H1(X,U,CP) f1p

g p such that N

note that

Using Lemma la, there fo,P+l = dg, and

df = 5f = 0.

be any cocycle;

exists a

5g = f.

Let

as follows:

Set

hence, we have assigned a closed

df = 0;

ti

(p+l)-form.

5f = 5dg = d6g = df = 0.) Denote by closed (p+l)-forms containing f. We

(Also,

Pj the class in

exact p+ -orms

make the following assertions:

t f) does not depend on the choice of

a)

f13=

if tfl

f2>

The class mapping

y)

g.

fl-f2 = 5h, dh=0.

i.e. if

is an isomorphism. ti

Proof of a)

Assume

and set

f = 5g1 = 6g ,

a) asserts that fi f2 = dhap

f2 = dg2.

globally defined on

X;

i.e.

where .b P

fl-f2

But

6h = 0.

f1 = dgl; is

= d(g1-g2)

and 6(g1-g2) = 0. Proof of P)

Now

and dh = 0.

= dfl = df2 = 0

z

f1P - f2P - 6h0

Suppose

f2 = 5g2,

Hence fl = dg1 = dg2 + dh,

5(92+h).

,

where

6f1 = 6f2

fl = 6g1 = 6g2+6h

f2 = dg2,

and

fl-f2 = dh = 0. Proof of y)

It is clear that the association map

If.c -> f I is a homomorphism. i.e.

0;

f = 6g, dh0'p-1

means

g OP =

It is one-to-one for, assume

and dpP

so that

= 0.

Furthermore, it is onto, for, assume

ti

Then

form.

Define

df = 0,

f = 6g.

Then

globally defined, and

exists a

df

f

dg OP = 0

hence

f} = 0.

is any closed (p+l)-

f = dg by d-simplicity of ,,U. d6g = 5dg = 61 = 0,

as

f is

6f = 62g = 0.

The proof in this case is essentially the same;

(ii)

let

so

Now,

f = 5g = 6dh:

fr+1'p

grp

satisfy

dfr+l'p = 6fr+l,p = 0.

such that

6g = f.

Set

Then there

r f,p+1 = dg,

and

92

N

N observe

With a similar notation, we prove

df = 6f = 0.

a), 13), and y).

Assume

a)

5grp = 0

fr+l,p

=

6hrp

(dhrpJ = 0,

implies

H"(X,U,CP+l).

dhrp = 0.

as

We again have a homomorphism fr+l,p ->{fr,p+lf

y)

f = §grp

It is one-to-one for, if g = dhr'p-1,

f = 5g = §db ,

and

for assume fr'p+l by simplicity of = 6f = 0 D.

= 6db.

dg =

and

is closed, so: d4 _ {5(dh)j = 0 in

dh

Now

d5hr-l'p

5hr-1'p,

grp =

for

then { dgj = 0,

f = 0 = 6g ;

implies

and

satisfies Set

U.

5f = 0.

Lemma 2a.

Let

X

and so

dgrp = 0,

tf} = 0.

Then

Itluis onto,

Then

df = 6f = 0.

f = 6grp

then

f = dgrp

df = d6g = 6dg

Clearly f} be a differentiable manifold;

d-simple coverings and V a refinement of following diagrams commute, where CP

U.

U,V

Then the

denotes the sheaf

of germs of d-closed p-forms

i) Hr+l(X,U,)

->

Hr(X,U,f?+l)

Fir+l(X,V,Fp)

_>

Hr(X,V,cp+1)

ii) d-closed (p+l)-forms

d-exa ctp i-TJTorms H1(X,Vs CP)

iii)

f;r(X,U,h, d-closed r-forms

d-exact reforms Hr(X,V,C)

where

r > 0 Proof.

and

p > P.

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. commutativity claims are disposed of as easily.

The other

93

Let X be a differentiable manifold for

Theorem 26a.

which there exist arbitrarily fine d-simple coverings-.

the following groups are canonically isomorphic: d-closed r-forms (I) (de Rham): Hr(X,C)

r > 0

d-exact r- forms

Hr(X,U,C) _ Hr(R,C)

(II) (Leray):

for any d-simple covering Proof.

U

r > 0

U.

is a direct limit of groups

Hr(X,C)

any covering of

,

Then

Hr(X,U,C);

where the class of all coverings is

X,

directed by "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 d-closed r-forma for any d-simple Theorem 25a, H"(X,U,C) . 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.

a locally finite refinement of U

Let V be

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 Whitney's imbedding theorem.)

Since every finite intersection of sets of V

is diffeomorphic to an open set in IRS, applies. in

Q3n

the de Rham theorem

r th cohomology group of such an open set

If the

with complex coefficients is trivial, then every

closed r-form is exact:

that this cohomology is trivial is

a known result.

J2.

Dolbeault's Theorem

This section is Section 1 applied to complex manifolds and the operator A. U,

a.

Definition 44b. we say that

U

differential forms,

Given a complex manifold is simple with respect to q > 1,

or

X

and covering

(p,q)

5-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 a holds. Theorem 24b.

X be a complex manifold.

Let

Then there

exist arbitrarily fine a-simple coverings. Proof.

As before, assume X C.9 n,

Once again

open.

use Theorem 23, which establishes the Poincare lemma for

for coverings by polydiscs IzJ- a3I < RJT . The completion of this theorem for a manifold is remarked on at the end of this section.

Lemma lb.

H.r(X,U,CP)

= 0, r > 0;

the sheaf of germs of forms of type

where

(O,p),

now denotes

C?p

and

U

is locally

finite.

Proof.

We may again use a partition of unity argument

as in Lemma la. Theorem 25b.

Let

a-simple covering.

of

X

U

be a complex manifold,

denotes the sheaf of germs

Then, if (P

a-closed forms of type

a

(O,p),

there exist canonical

isomorphisms between the following groups: i)

H1(X,U,C1p)

8-closed forms of type (O,p+l)

ti

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

iii)

Hr+l(X,U,&)

Hr(X,U,5p+l)

ti

Hr(X,U,(9)

a-closed forms of type (O,r) ti

a-exact forms of type (O,r) where

p > 0.

r > 0, Proof.

i) 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 and conversely; i.e. Lemma 2b.

Let

no =

(0,0)

is a holomorphic function,

c9.

-

U,V

X be a complex manifold;

a-simple coverings where V

is a refinement of

following diagrams commute, where C? germs of 5-closed forms of type

(O,p),

U.

Then the

denotes the sheaf of and

r > 0, p > 0:

95 i)

Hr+1(X,U,F')

H,(X,U,np+l)

--->

1'

Hr+l (X,V,fP)

ii)

Hr (X,V,(P+l )

H1(X,U,(P) a-closed forms of type (O,p+l) 3-exact forms of type (O,p+l) H1(X,V,c Hr(X,U,(9)

iii)

6-closed forms of type (O,r) exac t f orms o f type

(

A ,r

)

Hr(X,V,&) Theorem 26b.

Let X be a complex manifold for which

there exist arbitrarily fine 5-simple coverings.

Then-there

exist Canonical isomorphisms between -closed forms of type (0,r) I. (Dolbeault) Hr (X s-exact forms of type (O,r)

Hr (X,

II. (Leray) Hr(X,U,!)) = Corollary. for all

If D r.. &n,

D

r>0 (9)

,

r>0

a polydisc, then

Hr(D, 0) = 0

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:

Hr(X,sheaf of germs of holomorphic forms of degree

s) ti

5-closed forms of type (s,r) a-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

a for holomorphy domains,

so that any locally finite covering by domains of holomorphy is 5-simple.

Hence itssuffices to establish that arbitrarily

fine coverings by domains of holomorphy exist.

96

Complex de Rham theorem

93.

Once again, we attempt to repeat X,

we say that U

U,

d.

Given a complex manifold X and cover-

Definition 44c. ing

1 for complex manifolds

holomorphic forms, and the operator

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

a

holds.

(Recall that

a = d on

holomorphic forms.) Theorem 24c.

X be a complex manifold.

Let

exist arbitrarily fine

For X

Proof.

Then there

a-simple coverings.

Cn,

open, the proof is again immediate

and proceeds as before.

At this point, however, we find that no Lemma lc exists. Hence, we must modify Theorem 250 as follows: Theorem 25c.

X be a complex manifold,

Let

a-simple covering, for which

C? Let

U

a

Hr(X,U,('p) = 0, r > 0, where

denotes the sheaf of germs of holomorphic p-forms. C?

p-forms.

denote the sheaf of germs of closed holomorphic Then the following groups are canonically isomorphic: closed holomo

1:1(X,U,Op)

i)

ti

exact o omorp

hic (p+l)-forms

c

p+ - orms

ii)hr+l(X,U,,np)_ Hr(X,U,(p+l ) iii)

where

Hr(X,U,C)

p > 0,

Proof.

ti

closed holomorphic r-forms exact holomorphic r-forms

r > 0.

As before, under the remark that

Lemma 2c.

00 = C.

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 a-simple coverings and such that

Hr(X,U,(?) = 0, `r > 0,

for all

3-simple coverings U. Then, there exists a canonical

97

closed holomo hic r-forms r>0. - exac o omorp c r- orms (We shall not need the complex Leray theorem.)

isomorphism between:

Hr(x,s) ti

Let us assume that we have already proven that in a domain of holomorphy the cohomology 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.

c --r-forms

is

closed holomor hic r-forms We remark that the group oph c clearly trivial for the manifold.

such that

coverings

U.

where

n

is the dimension of

Hence

Theorem 27. n,

r > n,

Let X be a complex manifold of dimension

Hr(X,U,CP) = 0,

Then

r > 0

H"(X,C) = 0,

for all a-simple

r > n.

This theorem gives a topologically necessary condition for a differentiable manifold be a complex manifold.

X

of real dimension

2n

to

98

Chapter 9.

The Multiplicative Problem, formulated

§1. A.

The Multiplicative Cousin Problem

This second Cousin problem is a generalization of the

Weierstrass problem in one complex variable:

Given a domain D G C, aV, by

a discrete set of points,

and positive integers

function

f,

meromorphic in

of order

nv,

and poles at

find a

nv, mv,

with zeroes at

D,

of order

bV

av

mv.

We,now formulate the multiplicative problem, referred to as C.II in the sequel:

Multiplicative Cousin Problem. manifold

U = j ui ,

functions F1/Fi

i e I,

an open covering, and let

be defined and meromorphic in

F1

is holomorphic in

F 9 0,

ui /I u3.

defined and meromorphic in

holomorphic in Note.

Let X be a complex such that

ui,

Does there exist a function X,

F/F1

such that

is

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 shall

we need only consider holomorphic functions

Fi.

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

denote the sheaf, over X,

of germs of meromorphic

functions under multiplication, where )? = xeX yj

and

x

consists of the germs of meromorphic functions at

x,

and is a multiplicative group.

We topologize n as we did

(9,

by defining a subbasis

for the topology utilizing the topology of

as follows:

X,

then m c Mx and g e m is defined in a

Let m e neighborhood

N of x.

For each

y e N,

let

g

a

In y be

the equivalent class of meromorphic functions in )!Zy containing

g.

Then the sets

Ng = Zg e

My

I

y e Nj for

99

each choice of

g e m,

and for each m e

, form the

subbasis of

Let j denote the subsheaf of invertible

elements of k ;

/

such that

/YC and

c

x

holomorphic

C

for

x

x /` x' and set with the quotient topology.

Form the quotient groups At

each x e X.

- xeX

The sheaf of germs of divisors of X

is the quotient

sheaf Note that elements of

1-4

are equivalence classes

of germs of meromorphic functions, where germs represented

by two functions

F1, F2

are equivalent (at x) if

is a local unit, i.e. if F1/F2 vanishing in a neighborhood of

F1/F2

is holomorphic and nonx.

We note also that a set of Cousin data associated with

U may be regarded as a section of A/.-

the covering over

X.

Definition 47.

of AI-4

A divisor on X

X

is a section over

i.e. is an equivalence class of sets of Cousin

,

data, in the sense that two sets of Cousin data are equivalent

A divisor on X

if their "mesh" is a set of Cousin data.

is integral (positive) if all germs are germs of holomorphic functions.

Definition 48.

on X

A divisor a

there exists a meromorphic function thatthe divisor it defines (F) =

F

is principal if

defined in X

such

a.

Hence, we may state C.II in the following equivalent way:

given a divisor on

Proof.

is it principal?

If every integral divisor of the complex

Lemra 1.

manifold X

X,

is principal, then every C.II is solvable.

It is enough to show that every divisor is a

quotient of integral divisors.

Let

and let

p e X,

a meromorphic function defined in a neighborhood such that

(Fp)

where p and

is the restriction of gyp

a

to

Np.

Np

Fp

be

of p Fp = V*p

are holomorphic functions defined in

Np

1CC

and coprime at

at

defined in

q,

. at

But

$pzyq

equivalent in a neighborhood of neighborhood so that

as

gyp,

(4p

$q, *

were coprime in

*p

unit, i.e.

and

Similarly, Np.

But $q

Therefore

and similarly *p ti ?p q.

sp ti 4q,

p s X)

1

We may choose this

q.

are coprime.

hence p divides q.

*pOq;

(*p

I

$p*q - *p$q,

so

are holomorphic, and are

yp$q

and

and

$p ti $q

"u" means equivalence modulo

$p/dip ti q/*q,

hence

FIB , Fq,

Now q.

We shall be done if

Nq.

where defined, where

dip ti ?Pq,

say

p,

Fq = q/*q the corresponding function

and

q e Np,

Let

Np.

and hence in a neighborhood of

p,

p e X)

$p

divides

divides Tip/4q

Thus

are integral divisors.

X

be a complex manifold such that

C.II is always solvable.

Then so is the strong Poincar6'

C.

Theorem 28.

Let

+p

is a

problem, i.e. every globally defined meromorphic function is the ratio of two holomorphic functions, coprime at every point of

X.

Proof. (F) = a/p

,

Let

F be the global meromorphic function.

where

a

and

p

Then

are principal integral divisors;

for as was shown in the proof of Lemma 1, on any complex manifold X

every

is a quotient of integral divisors, and

divisor

since C.II is solvable every integral divisor is principal; F/f/g = w1/w2

hence

Therefore

is a local unit.

F/f/g

is holomorphic and equivalent to 1 at each point of X; is a global unit, say

F/f/g and

g

Theorem 29.

Let X

hypersurface..

Proof.

F = fG/g,

Thus

are coprime at each point of

where

so fG

X.

be a complex manifold such that

C.II is always solvable, and

§2.

G.

Y a regularly imbedded analytic

Then Y may be globally presented. Exercise for the reader.

The Multiplicative Cousin Problem is not always solvable The following example is due to Oka. M2 be defined as follows: Let X -

101

X = {(zl,z2) 13/4
0,

is nonempty

Im z2 > 0)

(Im z1 < 0, IM z2 < 0).

Now define the divisor r(zl-z2 1), 7

of X as follows:,

y

for

(Im zl )-. 0, IM z2 > 0)

=

outside the upper component of Y

t1

This clearly defines a set of Cousin data, for which, we claim, the Cousin problem has no solution. a solution {IzlI

F(zl,z2), Iz2I = 1S

= 1,

g(a,3) Now a

g(a,3)

and

3

.

F(eia,

=

is a continuous, periodic function of both Furthermore,

has precisely one zero, for

g

the upper component of Y n and

g .

For, assume there exists

and consider its restriction to

1

zll=l,lz

z = (e1ir/,e 21=11

;

Now consider the edge curve

elsewhere.

rl

in the a,3 plane as indicated in the figure, and the edge

curve

r2 about

(Tr/3,2vr/3)

3

within

rl and oriented

similarly.

(2y, 2Y)

Since g and

1

6

is periodic in a

it is clear that arg g(a,p)

can be defined as a single-valued

r2

4-

rl

function along Tl.

Furthermore,

by connecting the two curves as a

in the second figure, we obtain the following:

102

f

d log g

=

rl

i.e.

arg g(a,P)

Now a unit.

f dlog g; r2

is also single-valued along

F(zl,z2) =

1),

r2,

In the region enclosed by

r2.

where

h(zl,z2)

is

we may define a

single valued branch of the log; so

f

= f d log [-eio + eia

d log g

r2

r2

We may calculate this latter integral explicitly.

Set

B'+T

5.

at +

a = obtain

r d log (-e1

+ eia -1)

=

f d

LP I

+

i2T

log (-e

1a + -1),

e

r2'

r2

now encircles the origin in the where r2 _eV'+2rri/3 + eia'+7ri/3_1 has a zero at (0,0). Set and -eij3+2ri/3

+

eiaI +rri/3 I

- 1 = u + iv.

a(u,v)

a1 1N17

I

Now

- sin 0'-a'+ir/3)

=

0 in

Hence, if T was chosen small enough, i

the region enclosed by

fd r2

7'2.

log (u+iv)

.

I

So =

f d log (a++ ifl1) # 0

r2

and this contradition establishes our claim.

,

103 93.

The solution of the Multiplicative Cousin Problem for

polydiscs Theorem 30

Let

(Cousin).

D C

&n

Then

be a polydise.

C.II is always solvable. We prove this theorem twice; the first proof, given in this section, is due to Cousin.

§4,

The second is given in

and gives more. Lemma 2.

an defined as

d1,A2 be box domains in

Let

follows:

agx9'J'

={(z1, ,zn)

ij, J=2, ...,n7J

A2=(z1, «.,zn)Iagxl fmijki is a map of classes

U,

for,suppose the cocycle

i.e.

jFijj

108

There is a commutativity lemma in this case which we also shall not state explicitly.

Furthermore, distinguished

coverings are cofinal in the set of all coverings (easily seen for domains in space) so that we summarize in the following theorem, which makes no reference to particular coverings:

Theorem 31.

There exists a canonical homo-

(Oka-;Serre).

morphism of the group of divisors of X the following properties, where

a

with

H2(X,Z),

into

denotes a divisor and

c(a) its class: i)

If a

is principal then

ii) if H'(X,0) = 0

and

c(a) = 0.

c (a) = 0,

then a

is

principal.

Note that i) states that C.II solvable implies

Proof.

H2(X,U,Z) = 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-cochains and hence also of the mijk. i) i,

for every ii)

a

Now let

= M. Then Fi .

F1j = F1/FJ

so that

F

ui

on

for each

Fence we may choose

= 0

f1

i,j.

c(a) = 0

implies

are integral.

the

1.

A nib fij = fij^ n1j. now the

milk = nij + njk Now redefie ifi1 b

Note that

i

=

Fij=e

+ nki,

e

where

setting i' .

But

is a cocycle, for fij

fi3 + fjk + fki = Hence the

f

fib

mijk - (nij +n + nki) = 0. jk

are also coboundaries as

= f -f , and now we are done, for je-2Trifj

Fi/FJ, is Fi F = Fie solves C.II.

in u i

_

Fie-2Trifi.

F

H1(X (9) = 0 = e

so

fi erif

Hence the function

is globally defined, and clearly

log

Corollary 1. Let X be a complex manifold. H3. (X,(9) = 0, H2(X,ac) = then C.II is solvable. Corollary 2. Proof. i.e. that

If

Cousin II is solvable in any polydisc

X.

We have shown C.I is solvable in any polydisc,

H1(X,0) = 0;

topological result.

H2(X,Z) = 0

is a well-known

110

Runge Regions

Chapter 10.

Runge regions are regions in which Runge's theorem can be generalized:

the theorem states that in a simply connected

domain in the finite plane a holomorphic function can be expressed as a normally convergent series of polynomials. In other words, given an a simply connected domain f

in

and a holomorphic function

D,

there is a polynomial

D,

on K.

a compact subset K of

e > 0,

such that

p

If-pl < e

Note that in a multiply connected domain it may be

impossible to represent a holomorphic function by an infinite sum of polynomials.

91. Let

X

and

Preliminaries

Y be complex manifolds of the same

dimension with YopenG X. has the Runge property relative to

Y

Definition 50. X

if every holomorphic function in Y

can be represented

as a series of functions holomorphic in X converging normally in

i.e. given KcompactG Y,

Y,

for every holomorphic function holomorphic function Lemma a.

for all

holomorphic functions in X f2 = ... = fk = 0

for

rank of the matrix

k

(1)

q > 0.

there is a If-gj < e on K.

fixed, is

and

Yj = T p e X

Z = Y1 A Y2 /i... n Yr

Let

fl,...,fr be

such that if, at

(afi/az2)

i = 1,...,k; A = 1,...,n; Let

in Y

such that

X be a complex manifold of dimension

Let

n and Hq(X,0) = 0

parameters.

f

in X

g

e } 0,

and

z2 I

p e X,

fl =

then the

k =

min (k,n),

where

are suitable local

fj(p) = 01.

Then

is a regularly imbedded manifold

of dimension n-r and Hq(Z, (9) = 0 for all q > 0. (2)

Every holomorphic function on

of a function holomorphic in Proof. ...; Zr = Z.

(1)

Zj+1

Define

Z

is the restriction

X.

Z0 = X;

Z1 = Y1;

Z2 = Yl A Y2;

is a globally presented, regularly

in imbedded, analytic hypersurface in (p. 73), and since

has trivial cohomology groups of

ZO

all positive dimensions, so has etc.

Hence (2)

Hq(Z, D) = 0

Since

Z2,

for q > 0.

Since

Zr_1.

But then so has

Z1.

the first Cousin problem

H1(Zr_1,(9) = 0,

is solvable in

By Theorem 20

ZJ.

Z = Zr

is a globally

presented regularly imbedded analytic hypersurface in

Zr-1'

by the extension theorem, (Theorem 17, P. 59), every holomorphic function in

can be continued holomorphically to

Zr

Similarly, every holomorphio function in holomorphically to Lemma R.

and so on, down to

Zr-21

for all

q > 0,

the Runge property with respect to (1)

Every X

(2)

Hq(X, (9) = 0

Proof.

for J.

gl

Similarly, for

K3.

tKij

Let

Ki c :c Ki+l'

K3

is the given set

there is a function on

for

Kj+lpaetc 3 1g2-g11

k- oo g k

K

of

X,

=g

I f-gJ I + ( gA-g j+l j (2)

be a

By hypothesis,

+1,

< E2

If-gll<el

With

since 3+1 has there is a

on Kj+1,

g2 etc.

exists uniformly on

+ ... < E2 +E.2+l + ... < e

closed (O ,q) forms

for some L. If-gJ
0

0 for q > 0, tv) q if and only if for a differential form a in X

exact

orms'

i.e.,

of type

(0,q)

with

that a =

a13

as - 0

.

there is a

0

of type (0,q-l) such

112

Let

(a)

form with

and let

q = 1,

defined in

a,

differentiable functions- pl,p2,... respectively, such that

app = a in

pl + (p2-pl) + (p3-p2) + ...

X3,

there ale

X1,X2,...

defined in

Consider the sum:

XJ.

In X1

.

be a (0,1)

X,

By Dolbeault's.theorem for

as = 0.

and

X2,

a(p2 p3) = 0

p2-p3 being holomorphic can be approximated on

and hence

any compact subset of X2 by a holomorphic function

on X.

Set

p2 = p2,

P23

etc., using the Runge

03 = 03 - P23,

trick as before (cf. proof of Theorem 23, p. 88). Let

(b)

type

A

p3 =

in

a(p3-02 = 0 3 -

in

Xi,

where

01

has type

p

be of

in

Xj

(0,q-l).

in X2.

ay2

X2,

X.

Let

'2' etc. (cf. p. 88). 92.

Definition 51. zl,...,zn. A

defined in

a ,

Again, there are

as = 0.

api = a

such that

Since

and let

q > 1,

with

(0,q)

Polynomial Polyhedra Let

pl,...,pr be polynomials in

A = z I Ipi(z)I < 1

Let

is an open set.

If

A cc Cn,

then

for j =

A

is called a

polynomial polyhedron (of dimension n). Note that a polynomial polyhedron is a region of holomorphy. Theorem 32.

(Oka-Weil).

hedron of dimension

n,

(1)

Hq(X,(9) = 0

(2)

If

where the

then

q > 0

for all

are polynomials in

(1)

X be a polynomial poly.

is holomorphic in X

f(z)

qj

used to define Proof.

Let

then

zl,...,zn

f =

and the sum converges normally in

X,

X

qj

and the pk X.

is a bounded set and therefore lies in

a polydisc; assume X

lies in the unit polydisc, i.e. n+r where r is X C (IziI < 1, 3 = 1,...,n). Consider the number of polynomials defining X. In Cn+r consider

I- =

(zl,...,zn, Cl1"Cr)

I

IziI' 1, J=l,...,n,

Ci = pi(z), i = 1,...,r,. r' is closed. Define =p = M I Z iI < 1, Itil < 1). =o is closed in the open

113

polydisc.

Consider the analytic hypersurfaces

0 = fi(zl,...,zn,Sl,...,Y = Si - pi(zl,...,zn).

fi

The

are defined everywhere and are clearly holomorphic functions in

X

With

Cn+r,

lemma

in

a

(1ziI < 1,

as the complex manifold

l(il < 1)

and noting that the Jacobian of the

maximal rank everywhere because

fi

has

afilaCk = 6ik,' the

Z = I:o,

hypothesis of lemma a is satisfied and since

Hq( 7-0, d) = 0,

for all

function on ;;-o

is a restriction of a holomorphic function

q > 0

(1zjI < 1,

in the open polydisc

and every holomorphic

But a

JCiI < 1).

holomorphic function in the open polydisc can be written as a power series.

Hence every holomorphic function on

z:o is the restriction of aj

series, i

zll... znn ..j i

normally inn 1

converging

Or

We claim that 7-0 is holomorphically equivalent to X.

For, define the mapping

pl(z),...,pn(z)).

preimage of =p is for all

(zl,...,zn) -> (zl)...,zn,

It is of rank X.

n

and one-to-one.

The

H'(X,(9) = Hq( 2:0, p) = 0

Hence

q > 0.

We have already obtained that every holomorphic

(2)

function on 5-0 is a restriction of a series

converging normally in 1:0 jl,.ir z 1 ...Crr But on moo, Ck = Pk(z) and thus the above series is a a

series in only the

Definition 52.

K

= iz0

I

zj,

converging normally in

§3.

Runge domains

Let

K Cc 0n.

X.

The polynomial hull of K,

for every polynomial p with jp(z)j

0.

and by the Oka-Weil Theorem,

can be approximated as closely as desired by a polynomial K.

Hence

Is a Runge region.

X

(3) implies (1).

Since X

is a Runge region it is a

region of holomorphy and therefore is holomorphically convex, A

then K cc X

i.e. if K C cX,

where

is the hull of

K

with respect to holomorphic functions on

K = K

and hence

K c-c X.

Indeed, K

We claim that

X.

since the

. K

family of all holomorphic functions on X

K

is larger than the

It remains to show that K C K.

family of polynomials. *

Let

z0 e K

such that

and let jf(z)j < 1

is a polynomial

K 0 fzo ,

p(z)

since

Ip(z)I < 1 + s on

Hence

x K,

f

i.e.

z e K.

satisfying

For every

z0 c K*,

and because

z0 a K.

s > 0,

tp(z)-f(z)I < c

is a Runge region.

and since

If(z0)+ < 1 + 2e,

lf(z0)I < 1,

be a holomorphic function in

for

c

X there

on

But then Ip(z0)l < 1 + E.

is arbitrary,

115

Cohomology of Domains of Holomorphy

Chapter 11.

Fundamental Lemma, stated

f1.

K

Let

(1)

hull of

compact

K* _

K;

P. Note that K* = 11J D

z

jP(z)

K*

denote the polynomial

max IP(C)I

for every polynomial

is bounded and compact and that

K* I

I

Let

On.

D is a polynomial polyhedron,

K C D3

(Proof easy).

Recall that an analytic polyhedron Dc

(2)

holomorphic in G

fl,...,fv

functions and

such that D c

D C"G, 'hence bounded, we shall assume that

G

Since

Ifi(z)I < 1, 1 = 1,...,vj.

z s G,

I

was.

G c 0n and

there exists an open set

defined as follows:

n

G r-

I I z II < 1 j ,

the unit polydisc. Let

(3)

D

Oka image of the closure of _

t(z,C)

I

is defined as follows

D,

Izij, IC1

1, z E G, =fn) ;

i = 1,...,n; n+v,

Then -, the

be an analytic polyhedron.

j = 1,...,v 3.

and is closed.

We now state the fundamental lemma. Fundamental Lemma (Oka).

Let ` be as above.

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. A.

Applications of the Fundamental Lemma

D-_r Gopen , Cn be an analytic polyhedron.

Let

Observe that the Oka mapping of (zl,...,zn) given by: of

G:

->

(z11 ...,zn,C11...,Cv)

zi = zi, C = fi (zl,...,zn);

n+v

c

cl D ->

,

is defined on all

call the image of G under this mapping " = ". Note 1

116

that = is. closed and C =l. Hence there exists an e s 0 such that if

(zl,...,zn) s G; a set

e

Om

S C.

= )

then < c, " ''zn'c1'...PCV)' where the distance of a point p s On' from

dist [(zl'

is defined as

infsesIlp-s 11 = infs5S.£max Ipi-siIS

11, ...,m denote thh s-neighborhood of L, defined by

Let

=

w I dist (w, ) < cf.

Applying the fundamental lemma, there exists a polynomial polyhedron

n+v

Al c

Izi < 1,

A = Al

such that =cc Al c5£. Let is also a

We now claim that =o = = m1zi1 c > 0,

D(O,R) C D.

Define

and

r = 1,

on Jzj = 13

$.t 0

and let

4(1) > 0;

D(O,1)

D(0, 1-e)

z.

4 = constant on D z0,r .

Then

Proof.

KcompactC D.

Let

implies

z

containing

Then

4 < A

4 = 0

on

using (ii).

,

R

1

in

D(O,R).

such that

log IzI - log l1-El u(z)

A

log R - log ll-E

Then u for

is harmonic in

(zI = 1 - E.

letting

s y 0,

$

4(1)


4(C) dC)

IC-z0I

IC-z01=r

Proposition 3.

Let

semicontinuous in Dc 0 at

z0 a D,

and

4(z0)

4(z) > 0 such that 0.

log 4(C) dC =r

be defined and upper log 4,

is not subharmonic

Then there exists a disc

D z0,r C D and a function '

holomorphic in

D(z0,r) C -D,

120

and continuous in D z0,r , and

Iz-z0I=rJ

4(z0)

such that

on

'(z0)l = 1.

I

z0 = 0.

We may assume

Proof.

1

be chosen

r > 0

Let

such that D 0,r c.D and

log-%O) >, J, log 4,(z) dz

Izl=r $n be continuously

Using upper semioontinuity, let differentiable in

D,

Let

$n4 4.

hn(z)

log 4n(z) = hn(z)

function for which

on

be the harmonic

Let

jzI=r.

hn

be the conJj}gate harmonic function, and set

*n= e

-hri ihn

IIzI=rS

Now on

i.e. 1*n1 < 1.

log $n < 0,

If '(0) n(z)I

log fl &I = (log 4) - hn = log 4 -

,

1 *n(0) I

= K > 1,

then 4 R I&( < 41'nI,

is the required function.

If CO) 1* (0)1 < 1,

In 4(0) + In

then

In $(0) < hn(0) _

i.e.

hn(z) dz J

=

in 4(0) < nl

>aD

1

7n

(0)I < 0,

in 4n(z) dz,

IzI=r

lzJ=r

therefore

so

1/2rr f In 4n(z) dz,

so

zI=r

r in 4(z) dz,

In 4(0)

contradicting choice of disc.

lzl=r 34.

Statement.

Proof of the fundamental lemma Let

of simplicity) that

Gopen Cc $n

and assume (for the sake

1z and then R1(zJ) = max ICl-fl1 = ICI-fl(zJ,ZJ)I for some zJ j S

(zJ,ZJ,CJ) e

and then

=*. By the compactness of =*, there is

a subsequence

(z3, J,CJ)J, converging to a point

(z,Z,C)e

and im Rl(zJ) = IC? fl(z,Z)I < max ICl-fl(z,Z)1 = Rl(z). Hence

->z Rl

is upper semicontinuous.

Now, assume that z0 a (),

log Rl

is not subharmonic at some

[Iz-z0I < PS, holomorphic in the open disc and

then there is a closed disc in

and a function

bi(z)

0,

122

on

continuous on its closure, such that R1(z) I*(z)I < 1

and R1(z0) I*(z0)I = 1. - (l+t)eia for all It E C

Define

and

1/4 > e > 0,

for some

F(z,Z,C,t) = (C1-f1(z,Z))*(z)

(z,Z) e G,

C,

It-t01 < e

I

< 1 - 2e

and therefore

Iz-zOI = p,

t e T =

and

Iz-z01 < p,

where. a and

t0 a [O,a)J ,

are

a

We claim that this

fixed real numbers to be determined later.

satisfies the hypothesis of the proposition with

F

C Cn, K*

K = call it

= G (1(lz-z&) < p)

Gp,

F

a)

T = T

in the proposition,

G

and

and all K

.

Indeed:

Gp X T and continuous in

is holomorphic in

cl Gp X cl T. b) If r = z z e bdry (G ()(Iz-zOI t

,

compactness of T , (tJ,ZJ,CJ), b)

point to

tJ ->

If

t, = (t) A with 4',

CJ

-> C

and 5-(t) _ 4,

and

(t,Z,C) e 5-

then for every

first coordinate sufficiently close by compactness, i.e. there is an e > 0

It -tI < e

there is a point tl the function

ZJ -> Z,

,

a bdry S

(4,Z,C)

such that if

there is a convergent subsequence

then

with

I tl

t) _ 4. e

and

F(z,Z,C,t) = z - ((l-t)t+ ttl)

Since tl j S. for

t e bdry 3,

Consider

124

t e T = {t e C

is chosen such that

He-eke). F for z e 5 and in

tel

about

a

F

Hence,

proposition.

because

T

where

t0 a [0,1)1, for

el

is contained

t e "T

is holomorphic for all t o

in the disc of radius 4).

and

It-t01 < Cl.,

1

F # 0

(z,Z,C,t).

z = (1-t)e + tel

lies

and by assumption, then,

e

satisfies the hypothesis of the

However, for

F = z - , and since

t = 0,

; while (F=O] n =* = [z=f) () =* _ *( ) F = z - 1 and since S, [F=Oj ()=* _ [z=tll [)=* _ =*(tl) = 4), contradicting the conclusion

t e S, for

t = 1

of the proposition.

Thus

5) =(z)* = =(z). Proof:

Use induction on

n.

Either n = 1,

and the fundamental lemma holds for

=(z)

n < k.

is a point, namely the point

i(zl,...,zk+l)Cl,...,Cv)

(zl,Z) e G (1[zI=z)

and

n = k+l

If n = 1,

then

(z,fl(z),...,fv(z)),

and a point is its own polynomial hull.

=(z) =

or

If n = k+l,

then

IzjI < 1, Icil 1, Ci = fi(z,Z) for i = 1,...,vf and I

therefore is the Oka image of an analytic polyhedron in

Ck.

By the induction hypothesis, then, :(z) _ =(z)*. 6)

There is an

N polynomials a)

then

if

P (Z,C)

if

a number

b,

0 < b < 1, and

such that

and for all

iz-ef < e

(z,Z) E G b)

e > 0,

j,

IP(Z,C)I < 1,

and

1z-C < e

and

(z,Z,1) e = then for all

j,

IPi(Z,C){ < b. Proof.

a)

there is a point

e = then

If

(e,Z,') e = by 4+).

there is a neighborhood of

(t,Z) e G,

Since

G

such that every point

in this neighborhood has its first n-coordinates in there is a

S > 0

such that if the distance from

is less than

5,

then

and

is open,

(z,Z) a G.

G;

i.e.

(z,Z,C) to

But since

125

any neighborhood of (y) contains a smaller neighborhood which is a polynomial polyhedron.

we can find an

el,

and polynomials

Hence

with the required

Pi

property a).

(S7Z,C) e =, IPj(Z,VI < 1,

At every point

b)

j = 1,...,N.

We claim that fbr

sufficiently close to

z

(z,Z,C) e =, IPj(Z,C)( < 1

and

Ck

then there is a sequence

for all

-> , and

and max IPjR k,ZkItk)I > 1

(Ck,zk,Ck) a

Since 5 is compact

IPj(Z,C)l < 1

Choose

for all

and

j

for all

For

7)

and

IP

for which

(z,Z,C) e

is continuous,

I

TPj(Z,C)I < b.

set

Iz-el < e,

if z j S

(b (z)

e Nt).

Iz-CI < e

and since

j,

such that

N of

(z,Z,C) e

Then if

0 < b < 1

there is a number

a contradiction.

neighborhood

e2

e = min (el,e2).

IPj(Z,C)l < 1

where at least one of the

e

has absolute value > 1;

Therefore there is an

k.

(Sk,Zk,Ck)

E Z. fence in the

which converges to a point

polynomials Pi

such that

for each

here is asubsequence

limit we obtain a point

Suppose not,

J.

Zk, Ck

= max

(z,Z,C) a r>

over

(b,IPj(Z,C)I)

j = 1,...,N if z e S.

and

log 4(z) is subharmonic. Proof:

and is upper semicontinuous.

4(z) > b > 0

not subharmonic.

C(Iz-Cj<e) on

C

and

Then there is a disc, '(z)

satisfying for some

4(z) > b.

Therefore on

ai

W) > b.

1 ^ on

z-C = p, C e S

and continuous

A1.

*(z),

is holomorphic in ^int C,

This means that

C

int C

e > 0, and

is < 1-2e

Since

holomorphic in

4(z) Iy(z)l


1. in an open disc is

then- since

(z,Z,C) a 2-*, Hence all points

z J 00.

is connected and therefore

p

Thus

s 01.

11

Let

= b.

at a point

z s S,

and all

j

But

C-{)0.

and hence

z j S

S

where

? 1.

A;

and by 3), then, =(z) _

Then

+

b

But a subharmonic function which = b

on the boundary of the disc

(by property(4)of subharmonic functions, p. 119).

at a boundary point of

int A C Of.

be a closed disc c A

a contradiction. *(z)

for all

But

4 > 1

Hence nl = C z s d,

i.e.

128

Chapter 12.

Some Consequences of the Approximation Theorem

§l.

Let DOG D,

Definition 55. that

both open in

denotes the hull of K with respect to

K1

K1

Note

jzO E DO

=

!f(zO)! < sup !f(z)!

!

We say

dn.

K1CC DO,

is D-convex if, for every K c G DO,

D o

where

Relative convexity

D,

i.e.

z e K,

,

for every f holomorphic in D} n K1 K, the hull of K with respect to

that

functions holomorphic in_ Do. Let

Theorem 37.

both open in Vn.

D0 C. D,

Then the

following statements are equivalent: D-convex.

1)

Do

2)

There exists a sequence P3C C P3+1,

where each 3)

is

P3

is an analytic polyhedron in

is a region of holomorphy and

Do

U P3 = DO, D.J

is a

(DO,D)

Runge pair.

This is in some sense an extension of Theorem 33

Note.

of Chapter 10,

§3; where D = Cn.

1) implies 2):

Proof.

an open set

D1 e e Do

e bdry D1

Let

K C C Do .

such that

K1 G c.D1.

there exists a function

on K1 and

!fe(e)! > 1,

where

on

Ne.

D

N

But D1C C -DO;

P = {z

z e D ,

!

!f3(z)! < 1,

is an analytic polyhedron in K C K1 C c-P

;

!fI < 1

of , such that is compact

bdry D1

hence

and a finite number of neighborhoods Let

Now, for every such that

ff is holomorphic in D.

Hence,there exists a neighborhood If eI > 1

fe

There exists

N1,...,Nr

suffice.

1 = 1,...,r].

D, with

P r-c D1,

Now

P

and

hence every compact subset of D0 may be

inclosed in an analytic polyhedron in

D.

Ka C Ka2 e ...C Do be a sequence of compact sets such that 1 i Ka = Do. As above, there exists an So, let

analytic polyhedron ini D,

P1

Ka1.

Now,

cl P, L/ Ka

is

2

129

compact;

so there exists an analytic polyhedron in

and so on.

a1 P1 U Ka2,

P2

D,

2) implies 3).

Each

is Runge in Pj+l.

P i

lemma 0 (p. 111) is applicable.

Therefore DO

of holomorphy (by Theorem 21, p. 75) and

Hence

is a region is a

(DO,D)

Runge pair. 3) implies 1). in definition 55;

we must show a function

on K Since

q / K

K1

K,

We claim K = Ki.

such that

D0,

on

q / K;

If(z)I < 1

in fact, there exists an

If(q)I > 1;

be as

Assume

implies that there exists

holomorphic in

If(z)I < 1 - 2e

(DO,D)

and let

K CC DO,

Now

q j K1. f(z),

and

such that

Let K C K1.

e > 0

If(q)I > 1 + 2e.

and

K,

is a Runge pair, there exists a function

holomorphic in

D,

K U q j .

Ig(z)I = Ig(z)-f(z)+f(z)I _ If(q)I-Ig(q)-f(q)I > 1+e;

Unbounded regions of holomorph

= X,

let

Xi C Xj+1

be such that

is a region of holomorphy and

XJ

where

Proof.

XopenC en,

Let

Then X

s a Runge pair.

is a region of holomorphy.

is Runge in X.

X

A(K) = A(K)

every K c c. X C Cn,

Recall that, for

if and only if

region of holomorphy (Theorem 5, Chapter 2, §2).

assume there exists a

X

A(q) < A(K).

Hence,

j

sufficiently large,

Now for every

f

holomorphic on

If(q)I < 1,

i.e.

to functions holomorphic in

Xi].

X D K v q on K implies

For

is a

for which.. A(K) < A(K),

K C r- X

is taken in the maximum norm; i.e. there exists a

such that

(Xj,XJ+1)

As indicated by the proof of lemma P (Chapter 10,

31), we may assume

A

g,

on the compact set

Ig-fI < e

K1.

§2.

X

such that

X,,

where

q e K - K

IfI

But

Xi

is a region of

holomorphy, so dist (q, bdry X

> dist (K, bdry Xi);

letting

which is a contradiction.

j -> eo,

A(q) > A(K),

-C 1

q e [Hull of K with respect

i.e.

130 One obtains another proof by constructing a

Note.

sequence XIc

X

such that

using analytic

= X,

polyhedra as in "1 implies 2" of Yheorem 37. Remark.

(Oka)

Let

simplicity. Let Xr = X Then X

(,n;

X C

we assume

< J,

z II

Assume Xr

Proof.

If Xr

by lemma y.

show Xr XR

4)

is a region of holomorphy if and only if Xr

a region of holomorphy for every

r > 0.

for

0 e X

by assumption.

is a. region of holomorphy for every

is Runge in XR

r < R,

for

we are done

Using the preceding theorem, it suffices to

Let K

XR-convex.

is

hull of

K.

K1 C C XR

But, for every IIzII < r - e

is

r > 0.

let

is a domain of holomorphy.

XR

as

Hence, in K1

114 < r - e.

z e K,

K1 be the

where the norm is the maximum norm.

;

Hence,

K1Cc Xr The reverse implication is clear.

D. The Behnke-Stein Theorem Theorem 38. and

X3

If X3C. Xj+1 C on,

(Behnke-Stein).

U X3

is a region of holomorphy, then X =

is a

region of holomorphy. We first require a lemma:

Lemma 5. Let D1c.c D2CC D3 C c. n be regions, D3 a region of holomorphy, such that min

IIzl-2311

>

max

z1c bdry D1

dist (z2-bdry D3). z2e bdry D2

e bdry D3

z

3

(This means that any subset of is > the distance of

cl D1

to

D3

whose distance to int D2.)

aD3 "lies in

Then there exists an analytic polyhedron

P

aD3

in

D3

such

that D1cc Pcc D2. Proof.

Let

K = cl D1,

and let

K1

denote the D3 hull

K. Then (K) _ tD3(Kl), so K1 cc D2. But now there AD exists an analytic polyhedron P such that K1 C P CC D2,

of

as in "1 implies 2" of Theorem 37.

131

Utilizing OkaIs remark, we may

Proof of Theorem 38.

assume X cc en.

We may also assume

as every

Xi e C XJ+1

region of holomorphy can be written as a strictly increasing (i.e. relatively compact) sequence of analytic polyhedra (see corollary to theorem 7, Chapter 2, 3 5).

Mi = max M

z e bdry Xi

J1'J2'J1<J2

= sax

IIz-C I I

C e bdry X

z e bdry X

min. a bdry Xi

=

m3

min

minC

min

I I z-C II

C e bdry X

i1

a bdry Xj1

}C) z-C II

minC a bdry X

II z-C II J2

of subscripts as

vl,v2,...

We select a sequence

J2

II

mini E bdry

mJl' J2' J1<J2

Define:

follows: Set

Choose v2

vi = 1.

such that

Mv2 < mvl,

possible

as Mv2->0 as v2-> OD. Select

v3 > v2

such that

possible as vu

Mv3 < mv2,

My


ae.

as

such Rat M

< m

Continuing, select and

VJ-2'v

m

MV

vj > vJ-1

possible

;

as before.

Consider now the subsequence terms

CC X

X c c X

lemma 5 for minz

q+1

Xv ;

any three successive

satisfyithe conditions of g+2

Xvgllz-CII

a bdry

mvq,vq+2 >

Mvq+l'vq+2

C e bdry Xv q+2

maxi a bdry X

minC

IIz-C II a bdry XVq+2

vq+l

Hence, there exists a sequence jPjj of analytic polyhedra

such that

Xv c _ C P C C X V

i

and

polyhedron ink XvJ+2

,

where P i

ce in

is an analytic

PJ+1 C Xvj+2 '

132

Appealing to the 0 ka-Weil theorem (theorem 35), is a Runge pair.

(Pj,PJ+1)

to obtain the

Now use lemma y (or p)

desired result.

34.

Applications to the Levi problem

We recall the Levi problem:

Is every pseudoconvex domain a domain of

holomorphy?

We shall reduce this problem to the consideration of "strictly pseudoconvex" domains, defined as follows:

Let

Definition 56. real valued,

Gopen G dn,

function on G.

Coo

plurisubharmonic i.e.,

and let

Let

(a2$/ aziazj

)

4>

> 0

4>(z)

be strongly for every

$ > 0

somewhere in

more precisely, there exists an

such that for every

z e j z

and there exists

4>(z) > 0

D

$ < 0

e > 0

such that

z0 e G

$(z0) < 0.

G_= z I z e G

it is called strictly pseudoconvex; any region

for which there exists a

Gopen C

plurisubharmonic (real-valued, D = G_

z e G;

dist (z, bdry G) < e, z e Gl,

I

a

Now consider the open (nonempty) set 4>(z) < 0;

and

near the boundary of G

assume also that G;

be a

Or' and strongly

C°D) function

$

for which

is called a strictly pseudoconvex region.

We remark that a strictly pseudoconvex region is pseudoconvex; consider

log (-1/4).

The strict pseudo-

convexity is essentially a "smooth boundary" condition. Example. $ =

The unit ball is strictly pseudoconvex:

zizi - 1.

set

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

' in D We claim that

real-valued, continuous, plurisubharmonie function such that

* -> + oD

on the boundary of

for every compact subset

K of D

D.

there is a strictly

133

pseudoconvex region D0

this, notice first that since a positive integer

S = iz e D

where L

and

e > 0

N

4' -> + 00 on

so large that

T = Iz a D ( 4i (z). < N+ j

4'(z) < N3,

on L

zizi. L T_;

Then

subharmofiit on L.

pseudoconvex and

For z e S,

C0O

Therefore if we set

K c D0 e L,

For each J,

as follows : D

l.

let

then- Do

as claimed.

is strictly Let

D3

v2 = jl;

K3

be an

such that

be a strictly pseudoconvex

We choose a sequence

Set vl = 1; Set

I4(z)-*0(z)t< e

zizi;

increasing sequence of compact subsets of D K3 C Die C D.

Coo

and strongly pluriX(z) < N + 2e; for

D0 = z e int LI X(z) < N +1/2j,

region,

Give

1=1

is

X(z)

X(z) > N+ 1-e.

z e (L-ci T),

U KJ = D.

T.

such that

*0 Xz) = 4i0(z) + e/M

Define

n

M = ma

containing

By Proposition 2, p. 28, there exists a

small.

L.

there is

aD,

K c S e T C Lcompact C D,

is any compact subset of D

plurisubharmonic function on

To prove

satisfying K e D0 c c.D.

for some

D1 C Ki and so on.

v3

Jl ;

Note that

of subscripts

hence Kjc

c Dv D.+1

so that Dvic G DvJ+1

and

U Dv

= j

Proposition. following:

The Levi problem is reduced to the

Is every strictly pseudoconvex region

with compact closure a region of holomorphy? Proof.

Use the preceding theorem together with the

Behnke-Stein theorem.

134

Chapter 13.

Solution of the Levi Problem

The object of this chapter is to prove the following theorem. on is pseudoconvex, it is a Theorem 40. If Dopen region of holomorphy.

Since every pseudoconvex region is the union of an increasing sequence of strictly pseudoconvex regions (Theorem 39),

and the union of an increasing sequence of regions of holomorphy is a region of holomorphy (Theorem 38), it suffices to prove Theorem 401.

If

Dope c c 0n

is strictly pseudoconvex,

it is a region of holomorphy.

The Levi problem has been solved first by Oka (for n = 2), then by Bremermann, Norguet and Oka for any n. There exist today many proofs of this theorem using either Okays original method, or functional analysis. methods (Ehrenpreis, Grauert, Narasimban,

Andreotti-Grauert, etc.), or partial differential equations here follows mainly Grauert

The proof given

(J. J. Kohn).

The idea of using an "Extension Lemma" plus

and Narasinhan.

a result by L. Schwartz to establish finiteness of cohomology groups is due to Cartanand Serre.

Reduction to a finiteness statement

61.

Proposition 1.

If

Dopen ` Cn

is strictly pseudoconvex,

then dim H (D, hJ) < co . Let

U = ,u1S be a fixed, locally finite covering.

dim H1(D, ')) = m < co

The

means that given (m+l) cocycles on U,

a nontrivial linear combination of 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 we can add Cousin data and multiply by constants,

dim H1(D,O ) = m < oo

combination

means that a nontrivial linear

of (m+l) CI data is solvable.

Proposition 1 implies Theorem 401. open Recall that On is strictly pseudoconvex if Di_

there is an open set G,::Tn such that D-:zl:G and in G there is a

real-valued C00 function 4 which is strongly plurisubharmonic and D = [$ 0 such that if u) is a real-valued

135

function in

C0D

are all < e,

and if

G

00z1I, 13 w/azJazk)

Iwl,

is strictly pseudoconvex and

D1 = [4iw 0

region.

the polydisc of radius

polynomial

Q,

in

such that if about

e

q e bdry D and B is

there is a quadratic

q,

with Q(q) = 0

z1,...,zn ,

Q(z) # 0

and

in D /1 B. Since

Proof.

and is real-valued, by

G

in

$ e COD

Taylor's theorem,

4,(z) _

5-n [Ai(zj-qj) +AJ(ZJ-qJ)] + 7- [A1j(zi-gi)(zj-q3) +

Aij(zi-gi)(2j-gj)]

+

n Hij(zi-gi)(zfgi) + R ,,T7= where

H = (Hij) = H*

R

and

involves third order terms.

So,

where

=

2 Re Q(z) + (z-q)* H (z-q) + R n A1i(zi-gi)(zj-g3) Aj(zj-qj) Q(z) _ 1 , -1

4,(z)

Now, since the hessian of

4,

is positive definite,

H

is a

positive definite hermitian matrix of second order derivatives of

4,

at

q.

Since

each boundary point

H

varies continuously with

value, the minimum eigenvalue of is positive.

Hence

q,

and at

q of D it has a positive smallest eigenH

over

aD,

(z-q)*H (z-q) > ajlz-g112.

estimating third order derivatives of

4,

over

call it a,

Similarly by aD,

we can

find an M > 0, independent of q, such that - Mjjz-gfl3 _< R< Take

Hence e = a/M;

(z-q)*H (z-q) + R

then for

jjz-qjI < E,

> (a-Mjz-gIJ)(Jz-gjI2. * (z-q) H(z-q)+R > 0

136

except at

q where it vanishes.

therefore

Q # 0.

Lemma 3. region.

B

Let Dope cc Cr

There is an

e > 0

But in D r B

4' < 0;

be a strictly pseudoconvex

q e bdry D and

such that if

is the polydisc of radius

then B (I D

about q,

a

is

a region of holomorphy. Proof:

essential.

We will show that every boundary point is Take e = 1/2 e2,

a of lemma 2.

the

Bdry (BAD) = ((bdry B)()D) V(Bfbdry D) U(bdry B()bdry D). then since B

y e bdry B,

If

is a domain of

holomorphy, there is a function holomorphic in B singular at If

y e (bdry D)( B,

Hence

Q(y) = 0,

is holomorphic in

1/Q

and singular at y.

DAB B.

is holomorphic,

y,

in D A B.

Q(z) / 0

Q, of

then the polynomial

lemma 2, corresonding to and

and

y.

Assuming proposition 1, we will show that every boundary

point of a strictly pseudoconvex D q e bdry D.

Let

B be the polydisc of lemma 2 and

Let w > 0

quadratic polynomial.

be a

support lies in the interior of B 4) - tw,

where t > 0

w(q) > 0.

and

small enough,

D1 = [4) - tw < 0]

q e D1-D,

Q(z)

the

t

Consider is

is strictly pseudoconvex.

and since supp w C B, D1-D C B.

is strictly pseudoconvex, by hypothesis

dim H1(D1,(9) = m < w .

Construct

data as follows.

Take

covering of Dl.

For each

F1 = 1/Qk,

F2 = 0.

(m+l)

U1 = B A D1,

F1, F2

k = 1,2,...,m+l,

consider

in D n B,

(U1 (1U2)C )C(DI)B)

1/Qk is holomorphic in U1AU2.

al, ...,am+1

as an open

are meromorphic functions and

F1-F2 =

of these data is solvable.

sets of Cousin I

U2 = D,

because

constants

Let

function whose

C00

By lemma 1, if

is small.

D G D1 cc Cn, Since D1

is essential.

and

Q

0

A linear combination

Therefore there are complex

not all zero,

m+ a

function F(z) Q(z)-j

meromorphic in

D1

such that

F(z) -

a1

is

137

holomorphic in U1 This function

F(z) - 0 = F(z)

and

is holomorphic in D.

is holomorphic in D and has a pole at q.

F(z)

Reduction to an extension property

§2.

Proposition 2.

DopenC a d

Let

be strictly pseudoconvex.

There is another strictly pseudoconvex region D0

Dc C D0c c

on

aj dzi is a

such that if a

differential form of type

(0,1)

there exists a

P = Z- b3 dz3

In D0 in

and a

form

Coo

D

in

function x in

C0D

and

with Coo

in D0

with

aO = 0

a - j = TX

such that

D,

then

act = 0,

D.

means that every closed

2

cohomologous in

D-to a

(0,1)

form in D

(0,1)

is

form defined and closed in a

larger region. implies 1.

2

By Leray's theorem, the cohomology groups of

Proof:

are isomorphic to the cohomology groups

Hr(D, (9)

of D with respect to any simple covering of

D.

D,

Hr(D,U,(9)

Hence we

need only consider cohomology with respect to a simple covering in order to prove 1.

D

for

Cover

and

Let

el,e2 be the

respectively.

D0,

Take

ets

given by Lemma 3

e = 1/2 min (e1,E2)

by finitely many open polydiscs of radius

aD0

a

.

such

that (*) the closures of the polydiscs are disjoint when their interiors are disjoint.

Complete this covering to a finite

covering V _ Lvj ' of

D0 by adding open polydiscs of

radius a satisfying (*) and whose closures do not intersect aD0.

About each

vj

'

take a slightly larger open polydisc v If

I

It

such that vi cc v j, if I'cl v / aD0 = 4 then cl vi /) aD0 = if

v4i n vk = c

then

v4 /1vk = 4,

and such that for each

j,

v' n DO is still a region of holomorphy. Set ,U = vI n D and U" = Zu j = vj 11 D0 . Then U' and U are a-simple coverings of

D

and

D0,

i

u

respectively, since they

are finite coverings by regions of holomorphy. With each covering we have the groups of cochains,

_

138

Cr(DO,U", 0),

Cr(D,U'9(9),

Zr(D,U', (/')

and

and the groups of cocycles

Zr(DO,U', 0).

As we have already noted,

these groups are also linear vector spaces.

With the

following notion of convergence, it can be shown that and hence

Zr

,

is a topological vector space:

of elements of 0r(D)U,0)

converges if in each

Cr,

a sequence u1 /1 ... 4 ur+l/+

the sequence of holomorphic functions assigned 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: Co (D,U',(9)

Z1(D,U',O)

to be the coboundary operator and

r

:

->

Zl(DO,U",D)

to be the restriction map; i.e. if

Zl(D,U',(9)

z s Z1(D0,U",l9)

U1.

then

is the restriction of

z

to the covering

a continuous linear map and

r

is a completely continuous

r(z)

To show that

(i.e. compact) linear map.

r

6

is

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"

holomorphic functions

with

fib

to it

Consider one intersection functions

fij

assigned there with r,

functions

Ifi3I < e.

fij,

11

11

assign to

if kI < c, ij

If13! < e.

Il

ui/1u3 ? $

cocycles, under

these functions,

ui n uJ

which assign

ui/lu

and the holomorphic IfijI < e.

The image

the holomorphic

Take any sequence i.f13 and a compact set

of

K C D, 0

(ui Au ) 7 , K . D (ui /1 u') . Since the derivatives of the fib

are uniformly bounded on

and uniformly bounded on convergent subsequence.

K,

K,

is equicontinuous

and hence contains a normally

This shows sequential compactness

of the image, but in a metric space sequential compactness implies compactness.

139

Now, consider the following two maps of the direct sum

of C°(D,U', O)

Z1(DO,U", 3) into Z1(D,Ut, I9),

and

6

Co(D,U t, (q) + Zl (DO,U",

r=A

='-- je, z (D,U', -r=B

where 5 @r means that 6 operates on Co and r on Z1 and -r is just the map

operating on

-r

B

continuous and linear. A

Both maps are

is compact, and we claim that

is onto (see below). (L. Schwartz).

Theorem. and

Z1.

E, F be Frechet spaces

Let

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.

(The proof

dim F(A+B)E < o D.

is given in the appendix.)

Hence if A

A + B = 6, this theorem

is onto, since

implies that the space of cocycles on

modulo coboundaries

U'

is finite dimensional. A

It remains to show that Proof.

is onto.

onto means that a cocycle in

A

U'

cohomologous to the restriction of a cocycle in fij

be a cocycle in

assigned to

and

Ut

ui /1 ul # 4.

COD functions on

is

U".

Let

the holomorphic function

fij

fij = gi-gj

where

gi, gj

are

ui, u; respectively; since the intermediate

Cousin I problem is always solvable. closed form independent of

In ui, agi = a ,

because on

i

ui ( u,

By proposition 2, there is a closed (0,1) form p

Do and a COD function x in D

,

a

agi = agi.

defined over

a = 13 + 5f- in

such that

D.

Un

In

ui,

Let

G = ahi3

= hi - hJ*

f

in ui / ui.

covering

hi E^COD, fib

because

is defined on the larger

If we restrict this cocycle to

a cocycle cohomologous to

5-simple.

is defined and holomorphic

Hence a cocycle 3-ij

U".

is

Indeed, in Indeed,

Ut we get ui /) UP

fij - fij =(gi-gi) - (hi-h j) _ (gi hi- X) - (gj-hj- X) and

a(gi-hi X) = a - F - 5X= 0 a coboundary on

U'.

and therefore

fij -

is iJ

140

Proof of Proposition 2

§3.

We will prove a proposition 2' and then show that 2' implies 2.

Let

Take a to be the

q e bdry D.

the polydisc about a

D1 =

Then every closed

w(q) > 0

and

be

B1

and let w > 0

e/2,

form in D

(0,1)

be

so that

is cohomologous in D

form defined and closed in

(0,1)

Let

small is strictly pseudoconvex.

t > 0

for

that

of radius

q

be strictly pseudoconvex. of lemma 3.

a

function with support in B1

COD

to a

DopenCG Cn

Let

Proposition 2'.

(We will say

D1.

satisfies 2' with respect to D.)

D1

B be the polydisc about

q of radius n and Di-DC B1. Let o' be a COD function, Note that D4= D1 CC Proof.

d

and

o'= 0

outside

D.

Since

D/)B

form in

(0,1)

and

B1

in

1

Let

a is closed in

Let

B.

cn

be a closed

is a region of holomorphy C00

there is a

DI'B,

e.

function X in

such that a = &:. The function Y- C' is C°D in D and .X1= x. in B1/)D. Thus 0 = a - a(xQ') is a closed D /1B

form in D

cohomologous to

a

In B1/1D,

.

and therefore can be continued as

p = a - af_= 0

to all of

0

D1 - D.

2' implies 2.

Since

DopenEG,n

Let

Proof.

be strictly pseudoconvex.

by definition, there are open sets

CGopen,

D C

E, F

such that D CC E CC F e n and the distance from D to is greater than

e.

E by a finite number of polydiscs,

e/2

such that every polydisc containing

of radius

B1, ...,BN,

is centered about a boundary point of

a boundary point of D D.

Let

and

be a partition of unity subordinated to

w,,---,,a)N

w

the covering:

wi = 1

> 0

and

COD

in

at each point of

consi er the regions

D2 = [4-

E.

point of

Since

D Cc. Cn. D,

'

t

w

> 0

2 = 1,...,N

For

tj wj < 0],

wi = 1 there.

B

with support in

E

are positive numbers so small tat each convex and

E

Cover

D2

where the

tj

is strictly pseudo-

at every boundary

Hence D C C DN.

Also,

141

D1

satisfies

with respect to form in

D1

D2_1,

(0,1)

2, the set

D and

2 = 2,...,1t.

is cohomologous in D

which in turn

a closed in

D

2' with respect to

form in

D2,

satisfies 21

to a closed (0,1) form in

is cohomologous in

DN.

D2

Hence every closed (0,1)

D1

etc. up to

(and hence DN.

D)

Take for

to D0,

142

Chapter 14.

Sheaves

Exact sequences

61.

In the following, all groups are Abelian and all maps are homomorphisms. A.

A sequence is a collection of groups

Definition 57.

and maps

written

Ai -> A +l,

...

{AJ 4J3

The sequence is said to be exact at where

or:

> A .> A +1

> A -1

im $J-l = {ala s Ai

> ... if

A i

.

im $J-1=ker P

b e A,-1

there exists

,

A

such that

$ J-lb = as

ker 4i

=

[al ae Aj,

is called exact if it is exact at

The sequence f

AP for every

4.a= 01 .

J.

A collection of maps and groups is said

Definition 58.

to form a commutative diagram if all compositions of maps A

leading from a group

to a group

in the collection give

B

e.g. the diagram

the same result:

A

B

4>

\9 C

commutes if

for every

?P4>(a) = 8(a)

Remarks.

1)

Clearly,

a e A.

0 -> A -> 0

is exact if and only

if A=O. 2)

0--> A-> B

is exact if and only if

In this case, we may regard Al = +(A) C B,

and

4>

Hence the diagram

:

A -> Al

is one-to-one.

is an isomorphism.

0-> I

4>

as a subgroup of

A

tid

0`>A1i>B

B,

for

143

is commutative

will always denote the inclusion map and

(i

id the identity map). A AL> B -> 0

3)

is exact if and only if

4)

is onto.

Here,-the homomorphism theorem of group theory implies B .. A/ker

"factor"

Hence, we may

(ti denotes "is isomorphic to").

as follows:

4)

A >B J \ / 4)1

A/ker 4)

This diagram commutes, where here

j

canonical projection and is onto; and

by

$,

(as always) denotes the

the map induced

4l,

is an isomorphism.

0 -> A A-> B ± > C -> 0

Combining 2) and 3),

4)

called a short exact sequence if and only if onto and

4,

is

is 1-1,

im $ = ker ?i.

Utilizing the above remarks, the following

Remark.

diagram commutes and both horizontal sequences are exact:

0->A->B->C>0 1 4

$i

lid

t?l

0 -> A1--1> B J>B/A1-> 0 Note that

4)

:

A -> Al,

id

:

B -> B

and

C -> B/A1

are all isomorphisms.

Isomorphic groups may be identified; hence short exact sequences should be thought of as being in the form:

0-> A l> B L> B/A-> 0 B. It

Proposition 1. > ";

for each

Let

I =

a,p,y,...

be directed by

and let there be given sequences LAa,$aj , a,

such that, for

exists and is commutative:

a < 0

exact

the following diagram

is

144

... -> A

Aa J+l

ja

a aO

... -> where the

a

A --> AO +1

-' ... ,

satisfy the following compatibility condition:

a043

a'

< y implies


A

d2 = 0.

= ker d/im d,

H(A)

the derived group of A.

is not necessarily exact, Hence we may define:

H(A)

is a measure of the deviation

from exactness of 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

x e A

is exact if there exists a

dx = 0 ;

y e A

such that

x = dy.

Denote the homology class in

of an element

H(A)

x e A

by [x]. If A,B are groups with differential operators, we say that f : A -> B is an allowable homomorphism Definition 61.

if

i.e. if the following diagram commutes:

fd1 = d2 f;

d1

> A

A

ft

I

B

Examples.

d

f

>B 2

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 homomorphism

f* such that if

g

:

:

H(A) -> H(B) ,

B -> C

allowable, then: * *

(gf)*

g f

=

and (idA)*

=

idH(A)

A -> B

induces

146

Define

Proof.

mapping

of

f*[x] = [fx]

H(A) -> H(B)

dx

=

,

Then

x e A.

is a

f*

for :

,

0

implies

d(fx)

dy

implies

fx

=

_

f(dx)

and

x = B.

Proposition 3.

=

f(dy)

d(fy)

.

be groups with differential

A,B,C

Let

=

operators, and let the following be a short exact sequence of allowable maps:

0 ---> A f> B 21-> C --> 0 Then there exists a canonical homomorphism D

such that the

following diagram is exact (viewed as an infinite, repeated sequence)

.;

D

H(B)

H(C) g

(Recall the Weil proof of de Rham's theorem!)

Proof.

Exactness at i)

x E a ;

Let

(Does not need D) g*f*(a)

H(B):

we must show

a e H(A);

g*f*(a) = [gf(x)] = 0 y s B,

Let

ii)

wish to find

x e A,

dx = 0,

g*(y] = [gy] = 0, implies

hence there exists a implies

closed for

Let

a y E B

z e C,

fx = dy,

y [x]

x

But

is onto;

g

which

gy1 = z,

y-dy1 E ker g,

fx = y-dy1,

x

and

so is

is one-to-one.

f

as required.

Now But

g

is onto, so there exists 0 = dz = dgy = g(dy),

is closed as before.

has been chosen.

is independent of

Note that

Set

D[z] = [x];

y.

Eence, let

D

so

such that

x e A

implies that there exists an and

Now

f*[x] _ [y].

Hence

and

We

D:

dz = 0.

such that z = gy.

dy e ker g

if

such that

f*[x] = [y-dy1] = [y],

g*[y] = 0.

z e C.

such that

dfx = f(dx) = dy = 0,

Construction of

once

y1 E B

x e A

gf = 0.

such that

gy = dz,

gy = dz = dgy1 = g(dy1).

there exists an

But

as

such that

dy = 0,

But, let

= 0.

x

is unique

is well defined

y e B

such that 0=gy.

147

Then y e ker g, y = fx.

so there exists a unique dy = f dx, so

Therefore

x e A

such that

D

D[0] _ [dx] = 0.

is

clearly homomorphic.

Exactness at Let

i)

where

H(C):

But

fx = dy.

Then Dg*[y] = D[gy] _ [x],

dy = 0.

y e B,

dy = 0

is one-to-one, so

f

implies

x=0. Let

ii)

Hence,

is closed, for

Furthermore,

Let

z = gy and

Set

x = dxl.

dz = 0.

x E A,

fx = dy.

Set

z = gy.

dx = 0

Definition 62. J,

Then

f*D[z] = [dy] = 0.

z

f*[x] = 0;

is closed, for

i.e.,

dz = dgy =

Graded groups

A group

A

is called graded if, for

there exists a subgroup

x

;

and the representationlis unique.

i

xj e Ai

e A

Ai

i

,

such that k < co

A 11Ak = 0, j # k.

Note that this uniqueness implies

Examples.

where

f*D[z] = [fx],

such that

implies x= x + ... + x

An element

gf = 0.

as

D[z] = [x].

§3.

every integral

Then

But then

dy = fx.

Let

and

Then

yl = y-fxl.

H(A):

z e C,

gdy = gfx = 0;

i.e.

dy = fx and

dyl = dy - dfxl = fx - f(dxl) = f(x-dxl)=0.

ii)

x e A

D[z] = 0;

z = gy;

g*[y1] = [gy-gfxl] = [gy] = [z],

Exactness at i)

such that

such that

y e B

D[z] = [x] = 0. yl

dz = 0

z e C,

there exists a

is called pure

(J-)dimensional.

Chains and cochains on a simplicial complex

are graded by their dimension.

Differential forms are graded by their degrees. In both these cases, Definition 63. group integer

Ai = 0

for

j < 0.

A differential operator

d

on a graded

is said to respect the grading if there exists an

A

called the shift of

r,

for every

J.

(In practice,

r

d,

such that

is almost always

d Ai C Aj+r + 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

If A

Corollary.

preserves dimension.

is a graded group with differential

operator which respects the grading, then the derived group is graded; and

H(A)

x c 1, Adx=0}

j

H

(A)

j-r

Corollary.

An allowable map

f

:

of graded

A -> B

groups induces homomorphisms f*

Proposition 4.

:

Hj(A) -> H3(B)

be a short

Let 0-> A -S> B-E> C--> 0

exact sequence of graded groups and allowable homomorphisms. Then there exist maps

such that the following sequence

d

is exact:

... -> Hj(A) f > HO(B) L-> Hj(C) d -> Hj+r(A) -> where

r

is the shift of the differential operators.

Note:

There are

for:

H(C) -> H(A). suppose

But

distinct sequences.

x

y

gy = z,

dy - fx

so shift

and

and

HO(C)->H3 +r(A),

dz = 0.

is pure j-dimensional,

such that

such that

D[z] = [x],

D

As a map of graded groups,

z c C

there exists a

exists an

Iri

Utilizing proposition 3, there exists a

Proof. D :

...

Then There

dim y = j.

dim (dy) = j+r = dim x.

D = r = shift d;

rename D "d";

then the exactness result of proposition 3 and the above corollaries conclude the proof. Example. subspace. Z,

Let

X

Let C(Z)

be a topological space,

A C X

a

denote the graded group of chains over

with standard boundary operator, . Then i

0 -> C(A)

> C(X)

L > C(X)/C(A) a C(X,A) -> 0

is an allowable short exact sequence of graded groups, and proposition 4 implies the exactness of the sequence 3**>

... -> Hi (A) i**> Hi(X)

HJ(X,A) L> Hj-1(A) ->

...

149

Sheaves and pre-sheaves

§4. A.

Recall the definition of "sheaf" (Chapter 6, §3, Defn. 36);

we rephrase it as follows: A sheaf of Abelian groups,

Definition 64. as follows:

Let

the base space, be a paracompact

X,

For every x E X,

Hausdorff space.

Sx be an associated

let

Abelian group called the stalk of the sheaf over S = L)XEX i)

if

the projection map

and set

p

defined by p(s) = x

S -> X,

:

is continuous and a local homeomorphism.

s e Sx,

the group operations in the stalks are continuous;

s -> -s

is a continuous map of S into S; and

(sips 2)->s1+s2

that

x;

whose topology is smallest such that

Sx,

ii)

i.e.

is defined

S,

the subset

defined on the set R of pairs (sl,s2) such

,

belong to the same stalk, is a continuous map of

s1,s2

R

of

S x S

into

S.

t Je remark that the stalks are discrete.

S(Y) = XEY Sx,

then

Let Y e-- X;

with the induced

topology, is called the induced sheaf of

A section over such that of

S(Y)

is a map

X

Remarks.

: X -> S,

A section over

p -t = idX. over

t

over

S

Yr:: X

Y.

continuous,

is a section

Y.

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.

Definition 65.

Let

U =

implies

1 u1

Y°pe11C X.

Let

B.

Let

Then

be an open covering of X

associated to each

r(ui)

Let

u1 /1 u2 E U.

is open in

t(Y)

S.

X. be a paracompact Hausdorff space.

such that, if

ui E U,

such that

ul,u2 E U

be an Abelian group ui c. u

there J

exists a homomorphism yij

:

r(uj) -> r'(ui)

satisfying

the compatibility condition:

uic uj C. uk

implies

yji ykj = 'ki is called a presseaf.

The collection

(X, r(ui),Yij)

Proposition 5.

To each presheaf there may be associated

a sheaf, called the sheaf defined by the presheaf.

150

Take

Proof.

collection

tut

I

X

For each x e X

as the base space.

ui a U, x e uij

is directed; set

to the direct limit of the groups

r(ui).

with topology defined as follows:

s e S

Now

then Isy

x e Ux a U;

I

Set implies

Sx

the equal

S = VxeX sX' s e Sx.

y e Uxf is an open

sy a Sy;

set and the collection of all such sets is a basis for the topology of S.

Note that the

form sections of the sheaf

1-(u1)

S

defined by the presheaf. Every sheaf is defined by some presheaf. open u r(u) be the Take U = 'u in X, . Let

Proposition 6. Proof.

sections over

u; and define the

A subset

C.

T

of a sheaf

S

only if

T

Then

is called a subsheaf of

T

is open and

Example.

C0D

functions.

is a subgroup of

Sx.

S.

is the sheaf of germs of continuous functions

S

T

by restriction.

is itself a sheaf if and

Tx = T ('Sx

is that subset of

and

yij

S

consisting of all the germs of

Exact sequences of sheaves and cohomology

95.

Unless otherwise stated, all sheaves have the same fixed base space

X.

Definition 66. continuous map of and

$

Let

1

The subset of 10 a Sxj,

S2,

ker 4.

The

ker $

elements of

X

and

S2

S2

be two sheaves.

such that

A

$(Sl,x) - S2 x

is a group homomorphism, is called a homo-

Sl,x morphism of the sheaf

of

S1

into

S1

S2

S1

$

S2.

mapped into the neutral elements

is called the kernel of

4 ;

denoted

is an open set, for the set of neutral

is the image of the null section of

and this is open in

and since

into the sheaf

S1

S2

S2

over

(cf. corollary of 34, Chap. 14),

is continuous, the preimage of the set of

neutral elements of

S2

is open in

The ker OS 1,x

S1.

is the kernel of the group homomorphism

4'

I

S1,x and thus

151

is a subgroup of The image of

subsheaf of S2 continuity of map, i.e.

Therefore

S1,x.

denoted

4)(S1)C S2;

4),

that

:

p2$ = pl,

im $

so that

S2,x,

and the projection

4)

and the fact that

2,x is a"subsheaf of

im $

Hence we can form the quotient sheaves the cokernel of

211 ker $,

for each

S

Ei+i the sequence

x e X,

S2.

and

S211m 4)

and coimage of

$

are is a

respectively.

4),

The sequence of sheaves and sheaf

Definition 67 homomorphisms

p2

and

pl

im 4 OS

local homeomorphisms, and, as before, subgroup of

is a

im 4),

is open follows from the

the commutivity of

4),

is a subsheaf of S1.

ker 4)

> S S

is called exact when,

J+2

> S

j,x

j+l,x

4)>

S

J+2,x

is exact. Example.

T be a subsheaf of

Let

and let

S,

0

denote

the null sheaf, i.e. the sheaf whose stalks are the trivial groups over each point.

The sequence

0 -> T i> S -> S/T -> 0

is exact by definition.

We have already defined the cohomology groups, of a paracompact space

q > 0,

sheaf S

For convenience, define

.

Note that Let of

S

and T.

homomorphism covering of

Hq(X,S) = 0

T

be two sheaves and let

We claim that for each

4)* of

X,

with coefficients in a q < 0.

for

is the group of global sections of the sheaf.

H°(X,S)

into

S

X

Hq(X,S),

U =

Hq(X,S) I_u1

.

into

q,

Hq(X,T).

$

be a homomorphism

$

induces a

Consider any open

The group of cochains

C(X,U,S)

is a graded group with differential operator (the coboundary) which respects grading (the shift is +1). group

H(C(X,U,S))

Hence the derived

is graded and its pure dimensional parts

are the cohomology groups of the covering with coefficients in

S.

Similarly we have

an element of and

4)

C(X,U,S)

maps into

C(X,U,S) S

C(X,U,T)

continuously into C(X,U,T).

and

H(C(X,U,T)).

Now,

is an assignment of sections of

$

T,

thus

4)

S,

maps

is in fact an allowable

homomorphism and hence induces a homomorphism of the derived

152

groups

H(C(X,U,S))

and

H(C(X,U,T)).

Taking the direct

limit, we obtain the desired homomorphism 4. .Theorem 41.

0 ->.A

Let

(Exact cohomology sequence).

> B 2> C -> 0 be a short exact sequence of sheaves.

Then there exists, canonically, an exact sequence

H1(x,c) -

0 -> H°(X,A) E> H°(x,B) ±. H°(x,C) > H1(x,B)

s> H1(X,A)

s> H2(X,A) -> ... Assume the theorem for now.

A sheaf

Definition 68.

(It is proved in §7, p. 158.) is fine if and only if,

S

U = uib

for any locally finite open covering of X, i e I,

there exist homomorphisms

rli

of

S

into

S

such that 1.

ii(Sx) = 0

for

x i ui

and

Tji = identity.

2. is

(The sum is finite at each point because finite and

satisfies

q

Example.

Let

manifold, and let forms of degree

X

and let

p.

X

U

is locally

1.)

be a paracompact differentiable

be the sheaf of germs of differential

S

Let

U be a locally finite covering of

wi3 be a partition of unity subordinate to U.

are ni to be multiplication by WI. Then { r homomorphisms of S into S satisfying 1. and 2. above,

Define

so that

S

is a fine sheaf.

Theorem 42. for all

If

S

is a fine sheaf, then O(XS) = 0

q > 0.

Proof.

The proof is the exact analogue of the

theorem 22, p. 78.

Let

q > 0

a locally finite covering of

X.

be fixed and let

U = {ui be

Define the homomorphism

9: Cq(X,U,S) -> Cq-1(X,U,S) by then

- = 963- + 5G S- exactly = 69.F. Thus Hq(X,U,S) = 0

Hq(X,S)

= 0.

Verify that

Coo case:

2-Tli'1(iio...iq-1). iEI

as before. Hence if 6f =0, and then the direct limit

153

Definition 69.

A resolution of,a sheaf

is an exact

S

90 l 0 --> S -> 10 -> Al -> ... l=.l

sequence of sheaves Hq(X,Aj) = 0

for all

j > 0

such that

q > 0.

and

A resolution is called a fine resolution when all the Aj

are fine sheaves. Examples.

S = 0

Let

Aj

degree

X be a connected differentiable manifold and

Let

1.

let

(Sx = 0

and the topology is the discrete one).

be the sheaf of germs ofidifferrentiall forms of

C.

Note that if

Proof.

S

X

were not connected we would

the sheaf of germs of functions which

are constant on each component of X sequence

q > 0

be exact at

Aj

The exactness of the sequence at

is immediate, and exactness at

C

j > 0

are and at

follows

j > 0,

Aj,

We

A0.

for all

Hq(X,AJ) = 0

(cf. corollary p. 49) and that the

fine sheaves. _AO

in order that the

0 -> S -> _AO -> A1 -> ...

have already established that and

-> ...

1

is a fine resolution of

have to take for

> AO -> A

0 -> 9

The sequence

J.

from the Poincare lemmas. 2.

Aj

Let

X

be a complex manifold and let

S

be

be the sheaf of germs of differential forms of type

The sequence

AO a> Al a> ...

0

C9 and (0,j).

is a fine resolution

of 0 . As in example 1., the

Proof.

the exactness of the sequence at

are fine sheaves, and

Aj

Aj,

j > 0,

follows from

the Poincare lemmas. 43., (Abstiract de Rham).

ThTh

0 -> S

> AO o> Al 1>

...

Let

be a resolution of a sheaf S.

Consider the in!uced cohomology sequence

Ho(X,AO) -> Ho(X,A1) 1>

0 ---> H0(X,S)

C im 4* P-1

ker

and p

...

.

Then

HP(X,S) for all p > 0. ker 4*/im +* p-1 p canon. isom.

154

Applying this theorem to the above examples of

Note.

resolutions of sheaves, we obtain for 1. Hp(X,(C)

p > 0

closed p-forms

,

exact p- forms

J.

i.e. the de Rham Theorem (Theorem 26a, p. 93), and for 2. p > 0

a-closed (O,p) forms

HP (X, Q) ti

J,

5-exact (0,p) forms

i.e. the Dolbeault Theorem (Theorem 26b, p. 95). Proof of Theorem 43.

For j = 0,1,2,... ,

set

B = ker +i = im J-l since the resolution sequ;nce is exact. For each

0 -> ker +i i > A -.i. im

the sequence

j,

0

is exact by construction; rewrite it as 0 -> B'j i > Ai U> BJ+l

-> 0.

By the exactness theorem

(Theorem 41), the sequence

Hq(X,Ai) -> Hq(X,BJ+1) -> q > 0

exact for for

q > 0

and

for

q > 0

and

= Hp-2M B2)

ti

and

Hence

3 > 0.

Then

Now

Bp = ker ker 4p.

global sections of

of Bp+l,

By hypothesis, Hq(X,BJ+1)

Hq(X,A.) = 0 Hq+1(X,Bj)

HP(X,S) = Hp(X,B0) '

Hp-1(X,B1)

... _ H1(X,Bp-1).

X (X,Bp)

o

J.> 0.

J.> 0.

is

Hq+l(X,A

p

is a subsheaf of

Indeed, the Ap

Ap.

We claim that

is the set of those

ker $*

that p maps into the null section

but, by the definition of

gyp,

precisely the set of global sections of

this set is Bp.

Consider, next, the exact sequence

0 -> H°(X,B3) -> H°(X,A3) -> H°(X,B3+l) 5

H1(X,A3)

>

i.e.

-> ...

for 3 = p-1, p > 0,

_

0 -> H°(X,Bp-l) -> H°(X,Ap-1) -> ker 6

Since in

-

> Hl(X,B

0 ... p-1 ) ->

.

is a homomorphism from

6

= ker S C ker

A-1* = ker 4 / im Hence

*

1,

and

p -

Hl(X,B

p by exactness.

onto

ker

p) ti ker

-p-l

HP(X,S) , ker $* / im p-1.

H1(X,B p

p-1 ker S

155 §6.

Applications of the exact cohomology sequence theorem

I.

Let X be a complex manifold.

Let

69:

sheaf of germs of homomorphic functions

A:

sheaf of germs of meromorphic functions

We may view U as a subsheaf of X ; be the inclusion.

let i

:

©

-> //lam

We form the exact sequence

0->0 i>LIp L> 1,/D->0 Recall that a section of

over

class of sets of data for a C.I problem.

X

.

is an equivalence

Using the exact

cohomology sequence theorem, there exists an exact sequence

I*

0->Ho(X,D) Now

y,

Ho(X,)) --> H0(X,4/(9)

>

-> Hl(X,Q)--> ...

sends a meromorphic function (a section of A) into

J*

the Cousin I problem it solves, hence C.I is always solvable if J, is "onto", i.e. Hl(X,(9) = 0

Theorem I.

implies C.I always solvable

(cf. Chapter 6, §1). II.

X

Let

Y

be a complex manifold,

a globally defined

hypersur f ace:

Y = Assume

[f=0]

f

;

holomorphic in

X.

has no critical points where it vanishes [i.e.

f

maximal Jacobian rank on Y]. Consider (9,

on X.

functions on of

(9

the sheaf of germs of holomorphic functions

Let (2Y ;

X

denote the sheaf of germs of homomorphic vanishing on

We claim that

For points off

0Y

Y.

VY

is clearly a subsheaf

and we form (as before) the exact sequence:

Y,

(2 / O y

elements of

£ over

Y.

the stalks are trivial, for any stalk of

over points not in Y

stalk of 0.

is the induced sheaf of

For any point

is identical with the corresponding y0 e Y,

two functions representing

( 0/0Y)y0 are equivalent if and only if they

156

coincide in a neighborhood

a neighborhood in

Y,

neighborhood in X,

as

of

y0.

(If they coincide in

they coincide in a slightly larger

Y

is closed); hence they represent (9

the same germ in the induced sheaf of

over Y.

Using the exact cohomology sequence theorem, we have the following exact sequence for ...

->

Hq(X, (?)

jY)

:

Hq(X,

->

_>

q > 0

Hq+l(X, _9

...

)

Note that Hq(X,_' / 2 ) N Hq(Y, 2/! ?;re now claim that:

n

Hq(X, :9) = 0 = Hq+l(X, .%)

Theorem II. q > 0

implies

Hq(Y,

for fixed

(Cf. Chapter 6, §2, Theorem 20.)

0.

For, using exactness, we obtain immediately:

Hq(X,

)

so that it is enough to show

Let

a be a cochain in

is a cochain with in

*.

CJ

_ Hq+l(X, ;'2Y) Hq+1(X, : Y) = 0.

coefficients, as

Hence, multiplication by

X.

vq+l,,. Clearly

Then

Hq+1(X, f

f

f a

is holomorphic

induces a homomorphism

T.q+1,,, inY1. 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 (i

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

Then 9 * C A*, and we obtain the exact

multiplication. sequence:

The sections of

9

)'47 /

are divisors in

X,

i.e.

equivalence classes of sets of data for the C.II problem. Using the exact cohomology sequence theorem, we obtain the exact sequence:

*

... -> Here

j

*) s H1(X,0*) ->

H°(X,nt*/

...

takes a meromorphic function into the C.II problem therefore any C.II problem

it solves:

can be solved if

a

Sa = 0. This, however, is not particularly illuminating for we (X,('*);

know little about

so, we imbed this group in

H

another exact sequence involving "simpler" coefficient groups. We have an exact sequence:

!_O exp> (9

0

0

X. is viewed as a subsheaf of

where here

giving exp (s) = e27ris and e2vis = s is clear since ker exp

a; "exp" is the map:

exactness at

the exactness at

C9

and exactness at

FIT

I

follows from the fact that every

nonvanishing holomorphic function is locally an exponential. Hence, we obtain thi exact sequence: exp d

... ->

-> El(x,6)

Hl(x

)

> H2(X,Ta) -> ...

;

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

=

d-6

Ho(x,0*) -> H2(X,a)

:

assigning to each divisor C(D) s H2 (X,73.

C(D) = 0

for

Now assume is one-to-one.

D s H°(X,)1 / &) its Chern class

Clearly, if 5(D) = 0

D

is principal, i.e.

D s im j

by exactness.

H1(X,O) = 0. Thus

;

C(D) = 0

Then d implies

:

H1(X,O*) -> D

H2(X,Tc)

is principal.

158 So, we have re-established the Oka-Serre Theorem: There exists a map

Theorem III.

2 )->H (X,Z)

Ho(a,Ih

C

such that

D principal implies

1)

0

2)

C(D) = 0

C(D) = 0

and

D' is principal.

implies

(Cf. Chapter 9, §4, Theorem 31.) Proof of the exact cohomology sequence theorem

87.

We now restate and then prove the theorem: Theorem 44.

0 -> A E> B L C -> 0

Let

sequence of sheaves.

be an exact

Then there exists an exact sequence:

... -> H`)(X,A) f > H3(X,B) I.> H')(X,C) 6 > Hj+l(X,A)->... Proof.

U =

Let

ui

be a covering of

and

X;

consider the sequence:

f > Cq(U,B) .-> Cq(U,C)

0 -> Cq(U,A) where

denotes the group of q-cochains on the

Cq(U,A)

covering

U, and f

and

denote the induced mappings.

g

We claim this sequence is exact. At

Cq(U,A)

a E Cq(U,A); f

we must show

f(a) = 0.

ai0...iq(x) = 0 e Bx

exactness,

f

for all

; i.e.

x e ui00 .../)uiq. ai

At

we mat show

Cq(U,B),

0

in f c ker g. Now let

;

.

.

o q aio...iq(x) e Ax

a cochain; i.e. that

since

g,oi ..,i (x) = 0 o q

o

such that a

ai ., i 0 q

gf = 0,

f(a) = 0 so

For each x e ui /} ... () ui

We claim that the assignment

Let a c Cq(U A).

im f = ker g.

hence g

0 e ker g; i.e.

x e ui (1 ... 0 ui an

q

for

0q

a=

g f(a)io...iq(x) = gfaio...iq(x) = 0,

for every x e ui (1 ... nui

By

i (x) = 0 e Ax

0

Then

therefore assume

means

f(a) = 0

is one-to-one so

x e ui f .../)ui

every

Now

ker f = 0;

q

for every

there exists

fai .. i (x) = a 1 ...i (x).

o q o q, defined by the ai ...i (x) is a section over o ui

0

is

.. n ui'

q

159

Now

aio...iq(x) a Ax

so

p- ai0...iq = idui /1,..flui

-

To show continuity, let =

x0 6 Sa = ai ...i (ui /I... n ui o q o q f-1 i ..,i (u1 0 .../ ui ), an open set for

q

o

q

o ui°/).../)ui

is open,

q continuous.

pi

o .. iq

is a section and

Nx= be a neighborhood of x0

Let

f-1- f(Nx

ail.o.1 ail..,i [Nx o o o o q which is open for = 0- ... i f(Nx ) o o q Hence, im f = ker g.

lq

in

f

is

Sa ; then )-lf(Nx

(f. ai

o ---1 q

o)

is an open mapping.

f

We cannot complete this sequence to a short exact sequence, for

may not be onto [e.g. take sequence

g

So, define

-> C°(X, ,t/(9)] .

C°(X,G9) -> C°(X,)l.) Ca(U,C) = g[c (U,B)],

a subgroup of

0 (U,C)

comprised of

Hence, we now have the following short

"liftable" cochains.

exact sequence of groups and allowable maps:

0 -> Cq(U,A) f> Cq(U,B) J&-> Ca(U,C) -> 0 Hence, we obtain the exact cohomology sequence:

... ->

Hq(U,A) -> Hq(UB) -> Ha(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:

f

... -> Hq(X,A) -> 11 (X,B)g > Ha(X,C) S> gq+1(X,A) ->

...

We shall therefore be done if we can show Ha(X,C)

ti

Hq(X,C)

(canonically!), and this shall be proven by showing that for each cochain U,

in -Cq(U,C)

$

there is a refinement V

for a locally finite covering in which

$

is liftable; then

the limit groups are isomorphic, since we have an injective (one-to-one) map I :

Ca(U,C) -> Cq(U,C)

which commutes with the boundary operator and the"refinement" maps of the direct limit procedure.

160

X

(Since

is paracompact, we may restrict ourselves to

the locally finite coverings

{ui be a given locally finite covering.

U =

Let

U.)

W = wij be a new locally finite covering, refining U such that

for each

wi C G ui

ness implies normality. .Let we assign an open

Let and

possible since paracompaet-

i,

4 e Cq(U,C).

vx C X

such that

x e wi

implies

vx C wi

x e u

implies

vx C u

implies

vx 0wj _ 4)

To each

x e X

i) xe vx ii)

iii)

x

iv) v)

uj

x e ui

i ..,i

implies

... (1ui

q

o

vx e Caq(V,C).

q

o

Observe that once we establish the existence of V = £ vxI the theorem is proved.

using the local finiteness of

x e X;

Let

exist integers

r,s < w

such that

and no other

uk ,...,uk

w1, uk.

1

Set

v1 = w (1 ... ()w

il

an open neighborhood of

there

U,4!

x e L1 i, ...,w. ; r

x

it

/) uk /1... luk ; clearly

v1

is

1 s and satisfies i), ii), and iii).

will also satisfy

Note that any smaller neighborhood of x these.

For each x

k,

k / kl,...,ks;

which implies

x / uk,

so there exists an open neighborhood

such that vk,x(1 wk = 4). Set

vk,x

of x

Uvk,x

vx = vx

k / k1,...,ks an open neighborhood of

x

which now satisfies i),...,iv).

To satisfy v), observe that

0

happen if

Now

ui euk ,...,uk s 1

o" q

I

v2

,

2

4...i (v2) is open in C, so g-1 o q non-empty subset of B (g is onto).

c

(v2)

o

q

vx C ui f1...n ui o q over v2. Hence

and that

is a section of

-

can only

x e ui (1... n ui

is an open,

q

There exists a

.

161

bx e g-1

q(v2 )

such that

of bx in B such that morphism. Let Mx = Nx/'?g 1 Nx

p

and neighborhood

p o- bx = x, :

Nx -> p(Nx)

i...i (v2), o q

is a honieo-

and set

vx = p ci g (Mx) C vx Now

g

I

Mx

is a section of B

is one-to-one; and over

io..iq : vx -> g(DIM) of

g-1 4icr,.iq : vx -> Mx

vx , mapped by g onto the section C.

162

Coherent Analytic Sheaves

Chapter 15.

§1.

An analytic sheaf iY is a,sheaf whose

Definition 71.

base space X

Definitions

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

X-module, and this multiplication is continuous.

is an

We have the notions of subsheaf, sheaf homomorphism, Induced sheaf, factor sheaf, etc., as before. r Examples. 0 , 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

and

are not analytic, for there

;7-t

is no distributive law for multiplication by germs of holomorphic functions (recall that the operation in the stalks of these sheaves is multiplication). Definition 72.

is globally finitely

An analytic sheaf .

generated if there exist a finite number of global sections such that for every

sips 21...Isk

x E X,

The sheaf 0;

Definition 73.

section "1".

sections

r

The sheaf (9 r ;

x,

e 0x

t = 4l(sl)x + ... + Ysk)x where Examples.

t e

(0,...,0,1,0,...,0).

An analytic sheaf is locally finitely

generated if every point that the induced sheaf

x e X

has a neighborhood

Z+ (Nx)

over

Nx

such

is globally

Nx

finitely generated.

be an analytic sheaf, and

Let

Uopen C X.

of ,9'(U),

''Y

x

Let

x e U.

sections

(4

,...,4

sl,...,sk

sections

If there exists a tuple

such that

$1(sl)x + ... + 1k(sk)x the tuple

sl,...'sk

)

at

=

0

is called a relation between the x.

163 Note that the collection of all such relations forms an analytic "sheaf of relations" (between

in Q k

(over the space

sl,...,sk)

The analytic sheaf

Definition 74.

contained

U).

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.

Oka's coherence theorem

§2.

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 i=l,...,q, J=1,...,p Let

Dopen1 On,

"'a

ij(z)l,

be holomorphic functions defined in

then there exists an open

x e D;

and let

Dl C D,

D.

x e D1, with

the following property:

For any

C c D1

the holomorphic solutions

aij(z) 4i(z)

*

defined in some neighborhood of

=

0

may be written as:

L (z)

where the

_ 2 Pv(z) 0z) ,

i = 1,...,p

cv; j=l,...,p, v=1,...,L 1.

q,

then

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.

Let us first introduce the following abbreviations:

II. (a)

=

aij(z) j(z)

=

0,

1 = 1,...,q

aij(z)

j(z)

=

0,

1 = 1,...,q-i

agj(z) j(z)

=

03

J- 1

J= CC

(Y) J-

By hypothesis, Since

(y)

has a finite pseudobasis

any solution of (a) satisfies (y), of

K

o

Set

_

J = 1, ...,p.

'Pre$

must also satisfy

$j

), !'=1,...,K 1;

0 a (0) = 0,

But

J = l,...,n-1. q = 0.

hence

qv + qv+lal

=

hence

+ ... +

q (0) = 0. V

Similarly, one finds

obtaining

zs+v

qv+sas

But then

aq./az

= 0,

Jmn-lzn-1=0'

aMgv/amlzi....

But then R = 0.

We now assume

f =

generality, since the to

fJ(Z)zJ.

This is no loss of

s of order < s in z may be included

in R. Assume replacing

aJ(Z) = 0 JIZ11J);

zi

by

a1(0) = 0.

since

Ci,

this may be achieved by

N > J, i = 1,...,n-1

in

The transformation back to the

and

f

zi

P,

is

achieved as in Theorem 46 since we have already established uniqueness.

Now zs

AS = 1

where

by

zm,

=

and

m = 1,2,...

zs+m

=

ASP +a

sl

(Z)Z"-l + ... + ass(Z)

as3 = -a3. Hence, multiplying successively and substituting appropriately gives:

AS+M (Z,z)P + as+m,l(Z)zB-1 + ... +as+M,s(Z)

168 where in fact:

zAs+m + Asas+m 1

=

As+m+l as+m+l,p

and Hence

as+m,l asp + as

as+m+l,s m+.

=

a,+$,1 ass

As+m+l = = as+m-j,1

But this means that all series in the

following expression for zs+m,

f(Z,z)

implies

[cIZII]p

jaspI

(which is easily established by an

las+m,pI ` (CllZtI)m+p,

for

a

z');

neighborhood of the origin since

induction over m).

0 < p

,.p+l;

obtained by substitution

f(Z,z),

converge in a neighborhood of the origin. =

00 ( Z- f, A,) + zs_l

P(Z,z)

+ ... + (

T

OD

fj aj1

fi ajs)

This is the required representation of

f.

The third step

§4.

Recall the statement of the missing step in the proof of Theorem 45: III.

If Theorem 45 is true for some

then it is true for Proof.

n+l

and

n and all

q,

q = 1.

Consider the equation:

ai(zl,...,zn+l) fi(zl,...,zn+l)

=

0

We may assume that not all

ai s 0. Write ai = hips' valid in some neighborhood of the origin, using a linear

change of variables if necessary ( 4, 91, Property 2), where hi

is a unit and

P i

is a polynomial in

zn+1'

to consider the equation p Pi ci

0

where, by renumbering if necessary, we may assume a'= deg Pp > deg Pj.

It suffices

169 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 Let

of bounded degreest (C,c)

Write:

be any point at which the Pp = PI PII

PII(C,c) # 0

where

are polynomials in Let

(Ci)

Pi

PI(C,z) _ (z-c)r,

and

z = zn+1'

of degree


deg R;

11* PII C p

=

vanishes of order

R

Thus:

q

A

t

Pi C a polynomial. 1

deg (

A

Pi C

deg (C PII) = deg

Furthermore,

i

)

- deg (P

interested in solutions

)

i=

p

I

< 2a .

PI

Hence we are only

of the form:

(Ci)

a

But so

C1(Z,z)

=

Pi(Z,z)

=

PiC1

i = 1, ...,P;

grit (Z)zI , ¢+j

=

a

E-0 Hence Z Pi Ci = 0

i = 1,...,p,

a gij(Z)zj

2a j

--0

gij 7ri2

z 2+j

z

if and only if

(

gi j 'iz )

171

gij(Z) iri,(Z) j

s = 0,...,3a ,

0 ;

=

s

where the

are known functions; and this system of

rib

equations involves n

Hence the solutions have a

variables.

finite pseudobasis by the induction hypothesis, thus completing the proof.

Consequences of Oka's theorem

§5. A.

Remarks on coherent sheaves. 1.

Coherence is a local property.

if and only if every x e X

A sheaf is coherent

has a neighborhood in which the

induced sheaf is coherent. 2.

A subsheaf G

of a coherent sheaf J ,

is coherent

if and only if it is locally finitely generated.

section of G of

;

x.

is a section of L, since

Gx

Hence, the sheaf of relations R

number of sections of

G

For, any

is a subgroup

of any finite

is the sheaf of relations between

these sections, considered as sections ofSince coherent, B.

is

is locally finitely generated.

R

Corollary.

sections of

If 7L

is coherent and

sl,...,sk

then the sheaf of relations

,

are

R(sl,...,sk)

is coherent. Proof.

R

is a subsheaf of (9 k,

a coherent sheaf,

and is locally finitely generated, by the definition of

coherence of _L. Theorem 48.

a subsheaf of Proof.

Let .

and

be coherent sheaves,

G

Then the quotient sheaf

For every

x c X,

y e Nx.

under the natural homomorphism of

of i-(Nx)/G(Nx)

Hence ;/G

is coherent.

there is a neighborhood

in which a finite number of sections generate the stalk at every

/G

sl,...,sk

of

The images of into Z-/G

G

Nx (N3,)

sl,...,sk

are sections

generating the factor stalk at every

y c Nx.

is globally finitely generated.

To show that the sheaf of relations of any finite

number of sections of ZG is locally finitely generated,

172

consider an open set V C X, a point

(V)/G(V),

in

x .e V,

Nx

of x

sl'...,sk of JL as follows:

Construct sections

V.

of

tl,...,tk

sections

and a neighborhood

(ti )x e x/Gx is the image of an fj ex under the natural homomorphism h of into

For each fixed

is coherent; thus in perhaps a smaller neighborhood

/G. of

j,

x,' there exists a section

be the image of s under

Let

implying that

the image of

x

of

(sJ)x

C?x+p

sl,...Isk,rl...,rp V

Nx

is

41(t1)x = 0.

j(si)x a Gx which

*(r) x

-

Thus in some neighborhood

over

t i

is coherent,

generating the stalk

G(U)

means that

*p) e

$k''P1'

sections

and since G

...,k) e Ox with

41(t1)x = 0

($1'

(ta)x = (tax

Now, consider any element of

it is a relation

means

Then

h.

fj

CsJ)x =

in which we may assume that

are sections of

y e U.

But

with

Therefore, there is a

x.

J = 1,...,k,

sj,

ri,...,rp

at every

near

ti = t1

neighborhood U that

of -

si

so that

is a relation between the of ..,

a coherent sheaf.

of x there are sections

Nx C U

of R(sl,...,rp)(Nx

V = 1,...,N

such that v

V

k v

v

wv (IV) x

tk)(Nx)

1,...,k.

v = l,...,N

vectors over

wv e

x'

\ip/

,p1 N

Hence

'

Note that the

are sections of R(tl,...,

since they are continuous maps of Nx

Nx

into

e

i.e.

c .s a Gx,

X

such that

i.e. L 4i tj = 0.

s

+

Viri = 0,

173

Theorem k9. and

Suppose

A and B

are coherent sheaves

is a homomorphism of

f : A -> B

into

A

Then

B.

and coim f = A / ker f

im f, ker f, coker f = B /im f,

are

coherent sheaves. Proof.

It is sufficient to prove that

are coherent, as the coherence-of

im f

ker f

and

and coim f

coker f

is

then given by Theorem 48.

To establish the coherence of

im f = f(A)

we need only show that they are

ker

and

f,

locally finitely generated since we already know that

in f

is a subsheaf of

(cf.

B

ker

and

f

is a subsheaf of

A

p. 151).

is locally finitely generated, every x e X

A

Since

1.

s1)...,sk

of

Since

generate

f

y c Nx.

(A(Nx))yI

((im f)(Nx))yI

y e Nx.

is locally finitely generated.

im f For

2.

over Nx generate

A(Nx)

Their images under Thus

in which a finite number of sections

Nx

has a neighborhood

ker f,

take

are sections of

f(s1),...,f(sk)

R(f(sl),...,f(sk))

Nx and

x,

is a coherent sheaf. g : Ry -> Ay

following mapping

of Ry

sl,...,sk

as above.

over

B(Nx)

Nx,

Consider the into

y c Nx

Ay,

k (si)y

1=1 Since

f( 2- 4i(si)y) R

homomorphism of

of

((ker f)

I

$i f(si)y = 0,

into

(ker f)

Nx)y

ker f

I

Nx.

But every element

is of the form 7- i(si)y with

i e X and 7- $1 f(si)y = 0. Therefore

I

is a

g

Hence

g

is onto.

is the image under a homomorphism

Nx

of a coherent sheaf and thus is coherent, by part 1.

§6.

The sheaf of ideals of a variety

Definition 75.

analytic set V

in

Let X

X

be a complex manifold.

is a closed subset of

X

An

such that

174

Nv

there is a neighborhood

for every

v c V

functions

41,...,4k holomorphic in

of

and

v

Nv^ such that V/ Nv =

fx E X I Yx) = Yx) = ... = k(x) = Of .

and V an analytic set

For x e X

Definition 76. in

let oX be the subset of x of all the germs

X,

of holomorphic functions vanishing on then v

DX = (

if V = X

;

V.

(If

X

then

x / V

is trivial.)

X v with is a subgroup of 1 and an ideal. X x XEX x induced topology is a subsheaf of( called the sheaf of

germs of the ideals of the analytic set Theorem 50.

If V

(Cartan)

complex manifold

X

V,

denoted J V(X).

is an analytic set in a

then JV(X)

is coherent.

We will only prove a weaker form of this theorem.

If V is a regularly imbedded, analytic

Theorem 51.

subvariety of codimension manifold

then .j (X)

X,

Proof. J V(X)

in an n-dimensional complex

Ic

is coherent.

is a subsheaf of C9,

is locally finitely

Hence it suffices to show that %/ V(X) generated. Nx

Let

then

f

NX

x

in

so that

V,

holomorphic in NX

zl = z9 _ ... = zk = 01 ,

I

X.

Now, if

Since

NX.

and the

Remark.

f =

a) i E

(/ (X)/

and

where the wi

are

V(NX

the proof is complete. is a coherent sheaf, by Theorem 48.

Its stalk over every point off of V is trivial.

(c,7V(V)

Nx

are sections of

z1,...,zlc

vX,

V(X)

031zi,

in some

f e (d V(`gx))x'

is the germ of a functin holomorphic in

vanishing on

over

of

such that

z1,...,zn

introduce local coordinates

V = L(zl,...,zn) E X

and we

then by the definition of V we can

If x e V,

neighborhood

then there is a neighborhood

Nx tl V = . In NX, iV(X) = 0,

of x with

are done.

If x / V

x e X.

a coherent sheaf.

On V, since

is trivial, its stalk is the stalk of (9 (V).

)X Hence, on V this sheaf can be identified with the sheaf (0(V)

of germs of holomorphic functions on the trivial extension of

O(V)

to

V. O(X) /')V(X) X.

is

175

Fundamental Theorems (semi-local form)

Chapter 16.

1. Statement of the fundamental theorems for a box (semi-local form) Notation.

By an open box in

en

ai < x1

where

zi = xi + iyi

or

ai

0n,

we mean a

bi < yi

xeXO by,

x

l

onto J(X0).

fl oft 9 ml(XO)

homomorphism

f

K be

the coherent

.

Then there is a box X0 with K cc. XOc C X

X.

-A (X0)

sections

and

For, define

''i(si)x. It is

Pmt onto because the

si

generate

by

Therefore, denoting exact sequence G1

ker f1 0 -> G1 i>

is coherent.

0

G1,

1f-1,

Apply Theorem 52Ar to

X0,

call it X and

G1(Xl)

Then there exist

m2

Denoting kerf2 m2 by 0 -> G2 -> v

->

G2,

x c X0.

we have the following

j -> 0

generated.

a new

at every

( 54(XO))x

G

.

By Theorem 49, over

X0.

We get

globally finitely

generating sections ...

.

we get the exact sequence Again,

G2

is coherent and

we can continue this process, as far as we want, up to

4n,

177

obtaining

and

G., G41 ...,G4n

Call

XOn-1,'

the sequence

GO =

Then for mXO and every ,o, = 0,1,...,4n-1 0 -> G2 +1 ->J 1+1 -> GI -> 0 is exact; XO.

Hence the cohomology sequence, Hk+l(XO'o

Hk(X0,%) -> Hk+l(XO'G1+1) -> ml

exact for

But

k > 1.

Hk+1(X0',-

because Hq(XOr9) = 0 Hk+1(X

»X0n-1

X0 J...

X0

0 ,,) - +1 for

Iterating we get O,1,...,2n.

Hk(XO,O+1) ->

I+1) _> ...

is

yk(X0,Lm

4+1) = 0

+1) =

q > 0.

for

...

Thus loom.

k > 1

and

= O,l,...,4n-1. n H2k+"(XO'Gk+2)'

Hk(XO,GO) N I = 2n,

Let

k = 1,...,2n;

Hk(XO,GO) _ H2k+2n(XO'Gk

then

'

k = 1,...,2n. for all

2k+2n )- 2n, _Hk(XO,GO) = Hk(XO,:

Since

) = 0

k > 0.

It remains to prove (3).

§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

Ar+1'

given by: K= Jai<xi< ai'

Assume, e.g. that

ai


xi i . Then, if Theorem Ar+l holds that

p i

00

0

for both K1 Note.

0

0

and

K2,

it holds for

K.

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.

dimensions cyclically. dimension as in Lemma 1.

Cut

K

Order all the nondegenerate

along a first nondegenerate

Then (3) is false for at least one

of the two resulting boxes; choose one and call it Now cut

Kl

K1.

in the second nondegenerate dimension;

(3) is then false for a still smaller box.

Call it

K2.

178

Proceed, halving each nondegenerate dimension successively, obtaining a sequence

K3 , K3+1

of closed, nested boxes

such that in no neighborhood of any box is the sheaf induced by .,/L globally finitely generated.

But the

intersect

K3

in a point, and a point surely has such a neighborhood; and this contradiction establishes the claim. Lemma 2.

Let

follows:

K1, K2 be two closed boxes, given as

< x

A

K1

-

3

A K2

01

Let

of .'

> 3 = 2,...,n

,

03

Let there be given sections

over a neighborhood of K1

IT(K1) "

the stalks of

01

be an analytic sheaf over a

.

neighborhood of K1() K2.

sections


0

then Lemma 3 holds for

such that,

M.

Propositions 1 and 2 imply Lemma 3.

Proof. By Proposition 1,

M=

IIII-M IIID

PM1,

IIII-M1IIIa < e

by Proposition 2.

is holomorphic everywhere,

M = (PB)A.

But

P

.

<e

Pence, 1 Ml = BA

Then in a smaller domain,

so

and

Given

D.

there exists a nonsingular. entire matrix

Proposition 2. if

. elog(I+A) = I+A.

The following propositions establish Lemma 3:

B.

M

IIIAIII

182

about 0.

is a polydisc

D

Assume that

P-ropf of Proposition L.

Write:

M(z)

l`M`1(TIz)

M(0)[M(0)-1M(. z)

=

M( z)

...

sM-l(- LL1-z) M(z)1j ,

L an integer.

Now

IIII-M"l(Kz)M(KLlz) III for

L


is exact:

r = restriction.

i = inclusion,

(XE) -> 0

Therefore,

the cohomology sequence H°(DE,O) r> Ho(XE, ©) -> H1(DE'JX But

is exact.

Hence

implying that

Hl(DE,c;/XE) = 0,

r

is onto.

is the restriction of a function V

4

Write as a power series ik zil... converging uniformly in some Ck ajl...ik

holomorphic in

DE.

DE ;

are polynomials.

pj

t?ie

' = L pj(zl,...,zn,Cl,...,Cr)

Then

K cc XEllC G XE.

DE1

Therefore on

on

K,

1

Since

re are"

fr(z)

holomorphic on

so are the

G,

zl,...,zn,fl(z),..., pj.

Fundamental Theorems for regions of holomorphy

96.

(semi-local form)

Theorem 54A.

be a region of holomorphIT,

Let

coherent sheaf over X and analytic polyhedron X1

Then there exists an

K c c X.

such that

a

K

X

and a

finite number of sections of (X1) generate Tx at every x s X1. Theorem 54B.

Under the same hypothesis as in Theorem 54A,

there exists an analytic polyhedron and

Hq(X1,_1) = 0 Proof of A.

cc X, U Xj = X.

for all

Exhause

XO,

and let

D

such that K cc X1 cc X

X by analytic polyhedra,

Pick one of the

satisfying K c c XO Cc, X. Let defining

X1

q > 0.

X3,

fl,..., fr

call it

C C X XJ+1

X0,

be the functions

be its Oka image.

XO

is a

regularly imbedded analytic subvariety of codimension r in

D.

V

Define the sheaf

over

D

j- ifxEXO and let x

0

1

11

otherwise

as follows:

A c -4

let

be open if

186

A/) Y is open in is the projection map in the sheaf). Since

and only if (p

is open in D

pA

and

J'1

is coherent, in particular analytic, o is an analytic tie claim that

sheaf.

coordinates a.point of

n+r in a neighborhood

nl' XO

Introduce local

is coherent.

in' D of

N

is given by

X0 n N

such that

nl=...=nr = 0.

7" is locally finitely generated.

Consider its sheaf of

relations

s1,...,s

of

:-

R(sl,...,sg),

i.e. let

over some open set

set of functions

satisfying > 0j(sj)x = 0. so that the

x e U/1 X0.

over U /)X0

.j

of

x.

take the

j=1,...,,

e U,

for

(sj)x = 0

But

j(0, ...,O,nr+1'

Thus the

generated, say by Nx

x

(ail,..., N)

0,

'nn+r)r are

and hence are locally finitely in a neighborhood

v = 1,...,N

Therefore as generators of the sheaf (l,..,,arv)

x j X0,

j(0,...,O,nr+1'

satisfy

relations of

A relation is a

U C (D /1 N).

$j(nl' " ''fin+r) e 0x,

be sections

and add the 2-tuples

R

to :t (D).

of an

X1 = Z ze G

D1

such that

X1,

K c -c X1 c c XO

generated, say by to

Hence

D.,,

Apply Theorem 52A image

ATx

(ni,0,...,0),

(01r1i,O,...for I = 1,...,r. is a coherent sheaf over

in

Then there exists an Oka I

and } (D1

If(z)kl -e, Is globally finitely

The restrictions of the

t1,...,t1{.

ti

generate ( (X1))x at every x e X1. It is enough to show that Hq(D1,) = 0

X1

Proof of B. for all

q > 0.

For, consider any covering U

of it as a covering of

D1.

A cochain on U

assignment of a section of -41

I

of

X1

X1,

think

is an

and hence by the trivial

extension, it can be considered on U

and is an assignment

4

of a section of': hence a cochain on

on U

I

X1

the trivial extension, it is a cocycle on

Hq(D1,.

U.

A cocycle

is a cochain satisfying a certain relation.

0

U.

By

Hence if

then Hq(X1,J) = 0 for q > 0. By Theorem 52B,

Hq(D ,) = 0 for all q > 0. 1

187

Chapter 17.

Coherent Sheaves in Regions of Holomorphy

Statement of the Fundamental Theorems

31.

Theorem 55A.

X be a region of holomorphy and

Let

a coherent sheaf over

Then global sections of

X.

generate fx 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.

§2.

Notation.

Preparations for the proof

(zl,...,zn-1) = Z,

zn = z

(z1,...,zn)

so that

_ (Z,z). Theorem 56.

polynomial of degree the

aj

aj

(1)

R

rl,...,rn-l,rn = r

Let

IzjI< rj,

I

f(Z,z)

and let

be a lUleierstrass

P(Z,z) = zs + a1(Z)zs-1 + ... + as(Z);

is holomorphic for

for L(Z,z) Let

s,

P(Z,z)

are holomorphic in a neighborhood of the origin and

aj(0) = 0.

each

Let

(Cartan)

j = 1,...,n-1

be holomorphic in on

IfI < 1

D.

f = QP + R

and such that

be > 0

Izjl ^ rj

and let

IzI = rs

and

D = t(Z,z)

P(Z,z)

I

.

Izl< rj for all j

Then

where

Q

is holomorphic in

int D

is a polynomial of degree

coefficients in (2)

int D,

IQ(Z,z)I < K

not depend on Proof.

s-1 in z with holomorphic s-1 R = bj(Z)zj, and

and

b (Z)I < K

where

K

does

f.

(This proof is independent of the Division

Theorem, Theorem 47, and hence gives a new proof of It.) Let

aZ

Q(Z,z) _ I JItI=r

holomorphic in

int D.

0

Then

Pf(Z,

; )

Q

is

and

188

R(Z,z) _

P(Z,

2R j

----L-- _

dC

Cs-J_ zs-1

s-1

P(Z z)

aJ(Z)

z

_p out the division gives

1 f

ap = 1.

z- ,

bJ(Z) zJ,

R(Z,z) _ (Cs-J-l

b(Z) =

)

But

.

I=r

I

P(Z

( P(Z C)_P(Z,z)

)

Carrying where

+ alts-J-2 + a2Cs-

+...

)

P(Z,

ICI=r

+ as-J-1) dC.

Hence R

is of the required form.

Q and the coefficients of R.

Now to estimate

IzjI < rJ, ICI = r and has a lower bound b > 0

all

J.

Hence

For

and

J = 1,...,n-1,

Ifi < 1

and

a constant, for

Ia3I < c,

IPI

a constant independent of

Ib3(Z)I < K1,

f.

To estimate Q

write Q =(f-R)/P. Then l+K1(l+r+...+rs-1) = K2, a constant independent of IQI


Let

f vi be a sequence of elements of M which are all holomorphic in a fixed neighborhood _

uniformly in N Proof.

to,,.

of the origin. ,

4)

The

then

$ e

If the

4y

the

10

In any compact subset of N

j = 1,...,q.

N,

are uniformly bounded; assume

D,

4v =

D

with

for each fixed

I

we have a sequence

I i2I < KD.

bounded holomorphic functions in

are

Hence for

Therefore for each

of uniformly

4,

Thus

D.

contains

there is a subsequence of

j

converges normally on on

D

where

D,

93.

to

Tj

J

Then

.

is holomorphic in

`Y

Xjcc Xj+1,

Apply Theorem 54A to each

D.

which =v

call them

X by

Let

Kj = cl Xj.

Nj

with

j,

where the _> ( 1

4v

generate

x

Consider,

sjl,aj2,...,sjL

j =

form a Banach space .

j

all sections of the following form, a = are holomorphic and bounded in Aj

Xj.

The

of vector-valued

/

holomorphic functions under the norm,

j

Kj r N3C c Xj+l

J*

for fixed

>

Hence 4)e M.

Then for each

and a finite number of sections of .L(N3) x e NJ.;

$

Exhaust

JXj = X.

Kj C Xj+l.

there exists an analytic polyhedron

at every

pY

Proof of Theorem A

Now we are ready to prove Theorem A. analytic polyhedra,

fixed and

j

a subsequence which converges normally, say to

JJ

*V

are the basis vectors and the

holomorphic in

`Y

By

I4JI < 1.

Theorem 57, in perhaps a smaller polydise

where they

v

M.

converge componentwise; 4

uniformly in

converge

$

max v,X3

192

is complete by Theorem 58.

Aj

of s

Aj

of relations, i.e.

=

of

=

4v sjv

0

is a closed linear subspace ;>

represented by

Denote an element of Bj, Bj

if and only if

e AJ

Hence we may form Aj/Ao = Bj. >

by Theorem 58.

Aj,

41

A0

.

denote the subspace

A0

Let

,. by

inf >

is a Banach space with norm

max I4v1 V,X3

E[0]

Proposition.

over

Xj

e > 0

Let

a Bj.

[41]

Let

(Oka-Weil Theorem for sections).

be a section of and

[41]

or

such that 0= v v sjv Then there exists a

be given.

section T ->f 4- over x such that IId- r I I < e, i.e. j Xj = II6TII1 = 11111111 < e , where (o-- T) v Iv sjv' I

Proof. s jk =

We claim that every section

1 j(k) N

s+1 J, 2

a neighborhood of

l

where the

,

(k)

are holomorphic in

For, consider the

k = 1,...,L j*

mapping f 1

1

:

(LO j+Kj; (Nj))x -> (J (Nj))x,

x a NJ,

given by

Lj+l (sj+l,N)x

f

is a homomorphism of

)LLj+l

n Lj+1(Nj) generate

0 -> G ->

into

(Nj)

and is onto because the

(t (Njx at every x e Nj.

Lj+l(ITj) ->

Hq(Idj,G) = 0

and

0

H°(N3,

for all

Hence

Nj

By Theorem 54B there

such that K

q > 0.

N

Therefore

each

N

L

H°(Nj, 9 j+1)_>

is an exact sequence, implying that the mapping

is onto. Hence every section of 4 (N ) is a are holomorphic in

where the

is

(N3) -> 0, where G = ker f,

an exact sequence of coherent sheaves. is an analytic polyhedron

sj+l,,

NJ;

Naj+l,1

therefore so is

sjk.

Now,

and

(3-=

v sjv

and

sjv =

L1v)310,

are holomorphic in a neighborhood of

where Kj.

Thus

193

` =, *,O s j+1,.2 ' neighborhood of

= 2:v +v LT) is holomorphic in a

*1

On XV

KJ.

by functions

is a section of

Hence

by Theorem 53.

Similarly cj

where the =1 iQl)x m

=

*(2)

and

:2::, *1 sj+1,.2 = s,J+2,m'

rn

cr-(2) s J+2,t 2 t

Then

is a section over a neighborhood of etc. obtaining Since the

Bi

o,, off,

be the restriction to Xk

of J- over

of

Note that if

In particular

i.

where

only on

inf

For

ci is a constant depending i maxv P

+v sjv = s

inf

and II s IIi+1 =

in terms of the sjv;

maxµ,x

j

for

Xi

maxv X

J+l I*v I

Take any representation of the

-v *V sj+l,v = s.

= w ( =v *v

IT-al < 2e. over X,+1'

is any section of -4

s

s1 k1,v =

(0 )sAµ

Lv ?rv

over all representatives

Yi

for

Tk must

k,

Hence there exists a

Tk+1'

then II s III < ciII s II +1 , J.

II Tk ojillp < e

such that II T-olllp < e for all

X

and sufficiently large

< e/4,

k = j, J+1, J+2,...,

But then for each

p < k.

j+l

Ii ai (ri+lllJ+i -< e/21+1.

Xk,

of 4 such that

Tk

large enough and

T

such that

...

and II 62 crlII

K3+2

are complete, on each

there is a section

Vi

aJ+1,

in a neighborhood of

M

< inf[ j

01)

are holomorphic in a neighborhood

o2) _(1)(2)I a3 +

a

$ () c)o' q

is the sheaf of germs of differential

germs of differential forms of type

sheaf

Hence

tJk.

171-.

34.

over

there.

sufficiently small, the transformation (*) is nonsingular,

c

so that we may solve for the the

,

(O,q)

with values in the

x and C)X'q

are for

(4Jfj

modules. +J,

e

Define addition of two sums t

t

a, = al + ... +a,, +a, + ... + a M.

Allow

interchanges of the order of terms of a sum and drop any

term with J, fJ, ,yJ

terms

( 4 Jf J`X);U Jw J)

or wJ and (fJ

equal to zero. 4 Jai Jw

j).

Identify the

Then these finite

x

195

sums modulo the identification form an Abelian group

Gx = 1X 'q. the sheaf g E

Under the following topology, we obtain Take a representative of any element

-,.r-{ O°'q:

*jwj).

Gx ,

neighborhood

of

Nx

In a sufficiently small and

the

x,

functions, the

are sections of

fj

holomorphic

j are

4J

over

the wj

Nx,

are differential forms, and the projection maps of the sheaves ,5- and

are homeomorphisms.

assign that class in and Aj

and

*j,

through

Gy,

y e Nx ,

Then, for each

for which

(4

are the direct anal ytic continuations of

hj and the

4j

fJ

and

fj

coil

and wj

are sections of

and n°'q

-

We define the collection

respectively.

of all these classes to be an open set; and these open sets are to form a basis for the topology. Now define

(y jp,Pjawj)

a ( 2::

5: -4 0 no, q -> 7Z® C)o, q+l is a homomorphism of the

Then

sheaves; and

a2 = 0.

Since

Hq(X,(9) = 0

for all

q > 0,

(cf. Theorem ;6, p. 117), by Dolbeault's theorem (Theorem 26B,

p. 95) we have the Poincare lemma with respect to

Hence the sequence 0 -> ; i> zr; C)c'° - > -> ...

is exact.

For, at 3 (, C)°'°

ker

a

T

'*) 9

is

elsewhere exactness follows from the Poincare lemma.

sequence is a resolution of f .

Indeed, / U C)°'p

are fine sheaves.

For, define multiplication by a

if

a,(

then

a e C00

forms with values in ;

(4)jf3'3'pJwj))

0

that

for all

Hq(X, (9) = 0

,

and

This p > 0 CCD function;

Then by the

Hq(X A) ti (5 closed (0,q)

)/(a exact (0,q) forms with values in -4 ).

X by analytic polyhedra

Now exhaust

X.

(4)jfj`)*ij(awj));

and then proceed as in the example on p. 152. Abstract de Rham Theorem (p. 15 ?),

in

1 >1U°' 2 6) C)°'

q > 0

for all

lemma P for the sheaf . ',

Xj.

Then for every

by Theorem 54B. q > 0

J,

As in the proof

(Theorem 36), we obtain a

and hence

Hq(X, ) = 0 for all q>O.

196

Applications of the Fundamental Theorems

§5.

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,

is a region of 13olomorphy

X

and I a coherent analytic sheaf over

Let V be an analytic set in

Theorem 59a.

A.

V C X, that

such

Np

is the set of common zeroes of a finite number

of functions defined and holomorphic in

V = f x e X

X; i.e.

p e V has a neighborhood

and every.point X n Np

X.

i e I where the

for every

fi(x) = 0

I

are functions holomorphic in

X

Then

Np.

and

I

fi

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:

If V

Theorem 59.

subvariety in

holomorphic in every

is a regularly imbedded analytic

then there are functions

X,

such that V =

X,

x e X

JV(X)

Proof.

fi(x) = 0

I

is coherent by Theorem 51.

stalk at every point.

for

For any point

p e X - V,

combination of functions holomorphic on X (with appropriate coefficients).

function does not vanish at

holomorphic in

X, = 0

Theorem 60. Let

the stalk

p;

and vanishing

Hence at least one

hence there is a function

on V and

# 0

at

p.

Oka's Fundamental Lemma; a general form.

V be a regularly imbedded subvariety in X;

the closure of a polynomial polyhedron in

cl (V(1 P) = (ci (V(IP))*,

cl (V(P)C P

Proof. inequal'_ties.

such that

Hence, by

generate the

contains the germ "1"; hence "1" is a linear

(j v(X))p

P

i e I,

i e I

Theorem 55A, the global sections of 'V(X)

on V

fi,

X.

Then

its polynomial hull.

which is defined by polynomial

Hence there exists a polynomial polyhedron

Pcc P'cc X

and JV(P')

is globally finitely

P'

197

generated (applying Theorem A to the coherent sheaf JV(X)). At any point

p e P - V,

contains the germ "1";

(,/ V(P')) p

which can be expressed as a linear combination, with ! P

coefficients, of global sections of A(P') at :vanishing on

V.

But

p.

these global sections are holomorphic functions on

P'

Hence, using a sufficiently high partial

sum, we obtain a polynomial which is close to 1 at p, and close to 0 on

Theorem 61. Proof.

Hence

cl V/1P.

p / (cl (V()P))*,

as desired.

Cousin I is solvable in X.

It suffices to show that

H1(X,0) = 0.

But

0 is coherent, so Theorem B applies. Theorem 62. Proof.

X, every 5-closed form is c-exact.

In

Appeal to the Dolbeault isomorphism theorem

and Theorem B.

Suppose there exist finitely many local

Theorem 63. sections

generating all the X x e X.

sl,...,sr

every global section the

+3

are holomorphic in

s =

where

s

X.

Consider the sheaf homomorphism (9 r(X) ->

Proof.

+i(si)x

->

defined by:

r hypothesis;

.

This map is onto by

x

hence we may form thenn exact sequence:

->

0 G

is of the form

s

Then

G

-> :J r

->

,L

->

0

is also coherent; hence we obtain the exact cohomology

sequence:

H°(X,(9 r) Now

H1(X,G) = 0

H°(X,4) Theorem 617

->

H°(x,?)

->

H1(X,G)

by Theorem B; and since

are the global sections in

H°(X,(9 r)

respectively;

r

is complete.

Corollary 1.

Let

U cc X,

finitely many global sections every section

s

of J (U)

U

open.

sl,...,sr

Then there exist

is of the form

where the i are holomorphic in

U.

such that

of s =

4jsj,

198

By Theorem 6 it is enough to show that there

Proof.

exist a finite number of.global sections of T generating the stalks of T (U)

Let D C on,

Corollary 2.

But this is Theorem-A.

at every point.

D

Then the following

open.

are equivalent:

D

i)

is a region of holomorphy +1,...,4r

Whenever

ii)

are holomorphic functions in D

without common zeroes, there exist holomorphic functions

r in

D

such that

i) implies ii).

Proof.

of the sheaf

1.

* View the

$i

as global sections

It is thus enough to show that they

't (D).

generate the stalks

at every point

,9,

as "1" is

x e D,

But this is just

a global section and Theorem 63 applies. the hypothesis of ii).

If D

ii) implies i).

D = 0n

has no boundary points,

and so is a region of holomorphy.

Hence, assume D

has

boundary points; we shall show that every such point is essential. a = (al,...,an).

Let "a" e bdry D;

functions 3 = z-a3. holomorphic in

If the

holomorphic

They have a common zero at the point

only; hence they have no common zero in *3,

n

Consider the

D.

such that 7- y3(z)(z3-a1i) ° 1

D,

a,

Hence there exist in D.

are all holomorphic in a neighborhood f'a",

P 3

( 7- *3(z)(z3-a3))a = 1.

But this is clearly a contradiction,

so at least one of the

is singular at "a". V3

B.

Recall that, in a region of holomorphy, we have an

extension theorem for functions defined on regularly imbedded, globally presented hypersurfaces. follows (X

still denotes a region of holomorphy):

Theorem 64. subvariety.

This theorem extends as

Let

Y C X be a regularly imbedded

Then every function holomorphic on Y

restriction of a function holomorphic on

is the

X.

Proof. Consider the sheaf .Y(X) . We may form the exact sequence:

2 --

' (x)

-> iP (x)

->

9 /-/Y

->

o,

199

where J/JY = 0(Y) (cf. Remark p. 174). We therefore have the exact cohomology sequence: H°(X,(9) and Hl(X, J Y) = 0 Theorem 65.

H0(Y,A

->

by Theorem B. points

Let

given together with numbers

holomorphic in Proof.

Hl(X,J )

->

zj a Xi

such that

X,

discrete, be

j ZJ'

Then there exists a function

aj.

4)(zi) = aj.

is a regularly imbedded subvariety, of

I zj?

dimension zero. Theorem 66. Pi(z),

function of

With the

as above, let polynomials

zj

of degree NP be given. 4,,

holomorphic in

X,1 such that in some neighborhood

N

zi,

Then there exists a

+

4)(z) = Pj(z) + OflIzO J

); i.e.

has any given

4,

Taylor expansion up to any given order. Proof.

Consider the sheaf

as follows.

If for

x e X,

for x e X,

x = zj

set

zi

points

x

Nj+iJ

this is clear; and at

generated by the polynomials in

.

For coherence, we must

is locally finitely generated.

x # zj,

= C9x. If

have no terms of order < N of order at least

is an open subsheaf of CJ (X). show that

set

zi

3'x =germs of functions whose

Taylor expansions about zi i.e. which vanish at

x

defined by its stalks

z-zj

But, for

the stalks are

zj

of degree

Ni +1.

Hence, we may form the exact sequence: 0

->

%j

-> ()

->

I//

->

0

and therefore the exact cohomology sequence:

H°(X,C)) by Theorem B.

->

H°(X, ;9/-4)

->

0

But

(0 germs of polynomials of degree co;

contradicting

4

but

(f-c- X

c

(zn) a K.

by Theorem 65, there is a holomorphic function

with

K

on

Hence 0n

(zn) a K.

Condition 2 is trivial, as is Condition 3 once it is observed that every point of 0

X

has local coordinates such that

describe

X. zn+l = ... = Z N = iii) If X is a Stein manifold, and

on

X, iv)

then the set x I f(x) # 01

f

is holomorphic

is also a Stein manifold.

The product of two Stein manifolds is also one.

We leave it to the reader to verify that iii) and iv)

are Stein manifolds, while stating the following theorems (without proof).

202

Theorem 63.

Every universal covering space

(Stein)

of a Stein manifold is again a Stein manifold. Theorem 69.

Every open'Riemann surface

(Behnke-Stein)

is a Stein manifold.

We remark that there exist manifolds which are not Stein manifolds; that conditions 2 and 3 can be replaced by a "K-completeness" condition.

(A complex manifold

if for every x e X functions

holomorphic on X

fl,...,fK

isolated point of the set

X

is

there exist finitely many

iy e X

I

such that

x

is an

fly = flx,...,fKy = fKx3.),

and that:

Conditions 1, 2, and 71 imply

(Grauert)

Theorem 70. condition 0.

An approximation theorem

§2.

Definition 78.

An analytic polyhedron Y

manifold X

is defined as follows:

exist a set

XO

and functions

in a complex such that there

Y c c.X,

holomorphic in X

fl,...,fr

such that:

YccX0cc X Theorem 71.

complex manifold

Let X,

Y=

and Y

z

z E XO , If j(z)f < 1l.

I

be an analytic polyhedron in the

as above, and let

defined and holomorphic in

Y.

Then

g

be a function

can be expanded in

g

a normally convergent series of functions of the coordinate functions Proof. Y = j z

Oka map

I

f1,

holomorphic in

By adding functions

fj,

(zi,...,zn -> (fl(z),...,fN(z))

The image of

Y

into

X.

and that the is one to one, of

j = 1,...,N .

in the disc is a regularly imbedded analytic Therefore by Theorem 64,

subvariety of the disc.

extended to a function in the disc

ItjI < 1,

and the

we may assume that:

z s XO; If (z)I < 1, j = 1,...,Nj

maximal rank, of Y

zj

G

holomorphic in the disc.

G = = ci1 ...iN

converges normally.

But, setting

g

can be

Hence

and this series ti = fi(zl,...,zn), we

203

obtain the desired normally convergent expansion.

§3.

The fundamental theorems for Stein manifolds

Theorem 72. Corollary.

Theorems A and B hold for Stein manifolds. All consequences of these theorems, except`

Corollary 2, hold also.

In particular, we have the

complex de Rham theorem: Hq(X,C)

ti

closed holomorphic g-forms exact holomorp c q-?orms

Note that this result shows also that the cohomology of differential forms on any Stein manifold is trivial. These statements need no proof!

§4.

Characterization of Stein manifolds

Theorem 73.

Let

X be a manifold satisfying condition 0.

Then the following are equivalent: i)

11)

Y-;

X

is Stein.

F-l(X,

f) = 0

for every coherent sheaf of ideals

i.e. for every coherent subsheaf of 1.1 . Proof. ii.)

i) implies ii):

implies i):

Theorem B.

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 terns of local coordinates. This implies the existence of local coordinate:: :rhich

are holomorphic functions.

Now recall that the proof of these corollaries required only Theorem B in the form of ii). Theorem 74.

(Grauert-Idarasimhan)

manifold satisfying condition 0. equivalent:

Let

X be a complex

Then the following are

204 i)

X

is Stein.

There exists a strongly plurisubharmonic real-

ii)

valued function

4 on

X

such that

< a CC X,

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)

manifold of dimension i)

X

n.

X be a complex

is holomorphically equivalent to a regularly

imbedded subvariety of ii)

Let

Then the following are equivalent:

X

02n+1

is Stein.

Note that this gives an imbedding theorem for regions of hol-omorphy.

Proof.

i) implies ii).

Clear by the examples.

ii) implies i) will not be proved here. 2n+l

One must find

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: is not unique. X _. C2n+1 ,

This mapping

However, in the space of all holomorphic maps under the topology of normal convergence, the

functions of i) are dense.

205

Appendix

This appendix is concerned with proving the theorem of L. Schwartz appearing in chapter 13 on pate 139.

We actually

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

it

of norms defined on it;

Iln_II Iln+l ,n= 1,2,....

Furthermore, E and F are metric spaces with metric _

OD

d(xl,x2)

2

n

1Ixl-x21In

1+

xl-x2 11

n

Under the topology induced by each metric, we assume that E and F are complete and separable, and that E has the edded

property that for each n, {x EE

I

II xIIn+1 < lJ is totally

bounded (relatively compact) with respect

to the norm

11

lin'

The theorem we are going to prove is the following: Theorem (L. Schwartz).

If A and B are continuous linear

mappings 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,

!e note that if

E = Co(D,U',O)®Z1(Do,U",6) and F = Z1(D,U',( ), then E and F have the properties assume-1 above, for:

The nested sequence of norms on each space is defined as follows.

First, in each ui'(; U' take a sequence of subsets

Kijc. C Kij+1 /1 u1'.

In each ujVI uk A 0 of sets of U" take a

sequence of subsets Kjkt C C

Kjk,t+l I (uj"Ouk"); and in each

ui'/luj' A 0 of sets of U' tale a sequence of subsets

Kijk`- C_Kij,k+l

E; i.e. f = g + h,geCo

(ui';"!ujI). Then for

assigns the holomorphic function gi to ui' and h6 Z1 assigns the holomorihic function hjk to uj"n uk" A 0; define

Il+ IIn = max

Igi(z)I + max

"

j,k;z EKjkn

i;zE Kin

Ihjk(z)I

and for ktF; i.e. kEZ1(D,U',JJ) assigns kij to ui'r uj'; define 11 kiln = max

i,j;zeKijn

Clearly these are norms and

II

Iki.(z)I J

IIn ` II

The separability of E and F is obvious.

IIn+1 for all n = 1, 2, ...

.

Finally, the totally boundedness of Ix tE I in the norm II

II xtI n+l

IIn follows from the fact that a uniformly

bounded sequence o= holomorphic functions contains a normally convergent subsequence.

I.

Preliminaries.

Henceforth x and y shall ienote elements of E and F, respectively.

(e and f shall lenote elements of the dual spaces).

Let

Sn=LX&EI IIxIInfin=jyeFI

IIYIIn
x in E (or F) means 11x1- xI In - -> 0

i

for all n. E

denotes the space of continuous linear functionals on E. *

The following remarks, although stated for E , are true as well *

for F For e E E* define IIeIIn = sup

xeSn

Ie(x)I.

These are not really

norms because for some n, IIeIIn may be infinite. each fixed n, the elements of E*-with finite norm II a Banach space.

However for IIn form

208

For each a EEC there is an n such that

1. Proof.

Since a is continuous, Zx E E

I

11

e 11*

< oD

lg(x)I < 1 is an open

set containing 0, and therefore contains a set kSn for some n and k > 0.

Then

l e ( x ) I

< 1 for all x E kSn, and by linearity

for all x E Sn, Implying

l e" (x) I
x,

inplyinp A(

E

.k

k'n Let f E F*. For yEA(gSn), y = Ax, IIxIIn < $ and

If(y)I = If(Ax)I = I(A*f)xl < IIxIInIIA"flln

_
2.

In order to simplify notation

we will assume Ile112 < co : the proof in the general case is (essentially) the same.

There is an rl > 0 such that if g t L and g - e l I2 < then IIg - eIIl > ft. Proof. Assume that no such Il exists. Then there is a sequence (g ) E L satisfying II gj - e II2 < 1 and Ilgj - e IIl 0, i.e. for every xE E, Igj(x) - e(x)1 0. But ll gj - eII2 < 1 implies (2)

-

llgjlli' 2 < 1 +

J

e 112

< co so that by y Lemma

1,

e f L; r a contradiction.

215

(3)

Take 0, smaller so that 2q < 1, then if g 4E L and

11g - eli2 < 2h, jig -ell (4)

>Q.

Let {x1(1)J. be a dense sequence in aSl.

exists since aS1 G E a separable metric space.

an integer N1 > 0 such that if g E L and

Such a sequence

Then there exists

g - e 1 12 < 211 the follow-

ing inequalities cannot hold simultaneously lg(xi(1)) - e(xi(1))I < V 1 = 1, ..., N1.

Proof.

Assume the contradiction, then there is a sequence

(gN)t L with llgN - ell2 < 2n and lgN(xi(1)) - e(xi(1))1