A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
D O D UC T HAI
On the D
∗-extension and the Hartogs extension
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 18, n
o1 (1991), p. 13-38
<http://www.numdam.org/item?id=ASNSP_1991_4_18_1_13_0>
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DO DUC THAI
Introduction
The extension of
holomorphic
maps is the fundamentalproblem
ofcomplex analytic
geometry andcomplex analysis
of several variables.Many
mathematicians were interested in that
problem
and obtainedbig
results. In the extension ofholomorphic
maps, thestudy
ofholomorphic
maps that can be extendedholomorphically
to a hole or to anenvelope
ofholomorphy
was alsofrequently
considered(see Kobayashi [12],
Kwack[13],...).
In this paper, we establish some results
concerning
the extension ofholomorphic
maps to a hole and to anenvelope
ofholomorphy.
These resultsgive
relations between thehyperbolicity
ofcomplex
spaces and the extension ofholomorphic
maps.In the first
section,
westudy
characterizations of D*-extension for compactcomplex
spaces. We also prove ageneralization
of aBrody’s
theorem for compactcomplex
spaces(see [2]).
In the second
section,
we prove a characterization ofHartogs
extension forholomorphically
convex Kahler spaces. That characterization is ageneralization
of an Ivaskowicz’s theorem for
holomorphically
convex Kahler spaces(see
Ivaskowicz[10]).
In the third
section,
westudy
a relation between the D*-extension and theHartogs
extension.In the fourth
section,
we prove a theorem on the inverse invariance of D*-extension andHartogs
extension under somespecial holomorphic
maps.In the fifth
section,
we shallinvestigate
the invariance of the D*-extension and theHartogs
extension under someholomorphic
maps, inparticular
for finiteproper
surjective
maps.In the sixth
section,
weinvestigate
the D*-extension and theHartogs
extension of 1-convex spaces.
Finally,
this paper was written with thesuggestion
of Dr.Nguyen
VanKhue. The author wishes to thank Professor Doan
Quynh
and Dr.Nguyen
VanKhue for their assistance
during
the time this research was in progress.Pervenuto alla Redazione il 6 Marzo 1989.
0. - Definitions
0.1. A
complex space X
is called to have the D*-extension property(shortly D*-EP)
if everyholomorphic
mapf :
D* = X can beextended to a
holomorphic
map F : D --> X.0.2. A
complex
space X is called to have theHartogs
extension property(shortly HEP)
if everyholomorphic
map, from a Riemann domain over a Stein manifold into X, can be extended to anenvelope
ofholomorphy
of that map.In this paper, we shall make use of
properties
ofcomplex
spaces as inGunning-Rossi [7],
andproperties
of theKobayashi pseudodistance
oncomplex
spaces as in
Kobayashi [12]
or Kwack[13].
We alsousually
assume thatcomplex
spaces are connected and have a countable basic of open subsets.For put
Izl r}, D,
= D.1. - Characterizations of D*-extension for
compact complex
spaces First we have thefollowing
definitions.DEFINITION 1.1.
(i)
LetM, N
becomplex
spaces.A map
f :
M - N is called aC°°-map if,
for every local map(U, ~0)
ofM, (Y, ~)
of N, themap 1/;
of
o can be extended to aC°°-map
froman open
neighbourhood
of in Cn into(ii)
LetM,
N becomplex
spaces; A c M.A map
f :
A - N is called aC°°-map
if it can be extended to aC°°-map
from an open
neighbourhood
of A in M into N.(iii)
Let M be acomplex
space.A map
f :
M ---+ R is called aC°°-map if,
for every local map(U, ~p)
of M, the map
f
o can be extended to aC°°-map
from an openneighbourhood
of in Cn into R.(iv)
Let M be acomplex
space,[a, b]
c R.A map
f : [a, b]
-~ M is called aholomorphic
map(or holomorphic curve)
if it can be extended to a
holomorphic
map from an openneighbourhood
of
[a, b]
in C into M.(v)
Let M be acomplex
space,[0, 1]
c R.A map
f : [0, 1
2013~ M is called apiecewise holomorphic
map(or piecewise holomorphic curve)
if there exist numbers0 = ao al ... a,~ = 1
such that] is
holomorphic
forevery j,
1j
k.PROPOSITION 1.2. Let be any open cover
of a complex
space M.Then there exists a
C°°-partition of unity
which is subordinate to thePROOF. As in Narasimhan
[14]. Q.E.D.
For every
complex
space M, we have the tangent space of M denotedby
TM(see
Fischer[4]).
TM is acomplex
space and the canonicalprojection
7r : TM -; M is
holomorphic.
DEFINITION 1.3.
(i)
A Hermitian form on M is aC°°-map h :
TM EÐ TM - C such thatthe restriction of h on any fibre
TpM
xTpM
is a Hermitianform,
whereT M si T M is the
Whitney
sum of two fibres.(ii)
AHermitian
form h on M is called a Hermitian structure on M ifh(u,
u) > 0 for every0 ¥ u
ETpM,
for any p E M.(iii)
Let M be acomplex
space with a Hermitian structure h.For every v E TM, we
put II vii = h(v, v).
For every
holomorphic
mapf :
X - M(X
is openin C),
we putfor every zo E X.
Clearly Ilf’(zo)11 I
is the normI I Tzo f I I
of the linear mapWe have the
following proposition.
PROPOSITION 1.4. For any
complex
space there exists a Hermitian structure, and when the structure isgiven
that space is called a Hermitiancomplex
space.PROOF. Let U be a atlas of M: U = Choose a
C°°-partition
ofunity
subordinate to the cover{ Ui }iEl.
We have a canonicalinjection
Put a
map 1/;i :
TM - as follows:where 7ri : Ui x C’, -~ C n, is the canonical
projection. Let qi :
X --~ C~be the canonical Hermitian form on Cn, . The function h : TM ED TM - C is defined as follows:
Obviously h
is a Hermitian structure on M.Q.E.D.
DEFINITION 1.5. A Hermitian structure h, on a
complex
space M, thatwas constructed as in
Proposition 1.4.,
is called a canonical Hermitian structure on M.In this paper, we
only
consider canonical Hermitian structures on acomplex
space. For a
complex
space M, with a canonical Hermitian structure h, a distance between twopoints
is defined as follows.Let 1 : [a, b]
- M be aholomorphic
curve.Let y
U -* M be aholomorphic
extension of 1, from an openneighbourhood
U of[a, b]
in C, intoM. Put ,
Let -y : [a, b]
~ M be apiecewise holomorphic
curve, i.e. there existnumbers
a = ao a1 ... a1 - b
suchthat yi =y
] is aholomorphic
curve, for
all i,
1 i l~. Put" " "
Clearly L,~
does notdepend
on the selection of numbersDEFINITION 1.6. Let p, q E M. We say is the set of all
piecewise holomorphic
curves 1 :[0, 1
] ~ M, with7 (0)
= p,~(1)
= q.Put
hM (p, q)
= EClearly
if M is acomplex manifold,
then is a Hermitian metric on M. Hence
hM
is a metric of M and inducesthe
given topology
of M. But this is correct for anycomplex
space with a metric constructed as above.We have the
following proposition.
PROPOSITION 1.7.
hl,,l is a
metric on M and is called the Hermitian metricof M.
PROOF.
(i)
From the definition ofhl,,I,
we have(ii)
If 11 E 12 EQr,q,
then a curve 7 :[0, 1
- M, which is definedby
is in
Qp,q. By L,~
=L~,, + L,~2,
we havehM(p, r)
+hM(r, q)
for everyp,q,r E M.
(iii) Obviously hM(p, p)
= 0, for every p C M. We must prove that > 0, if q E M. Consider the atlas U = that induces h. We can_
0
_
assume that p e V := open c V := compact c
C Ui and q V V.
Assumethat > 0. Consider any
piecewise holomorphic
curve in V, which pevconnects p with a
point
of 8V:Let 0 = ao ai 1 ... ak = 1 such that lj =
-11[aj-,,aj]
i is aholomorphic
curve,for
every j,
1j
k.Let 1j
be aholomorphic
extension of -1j, from an openneighbourhood
of ] in C, into M. We havefor
every t
E[aj- 1, aj].
HenceSince
ai(p) f/. ai(8V)
andhcn,
is a metric we haveHence
L,
>¡e
6. Consider anypiecewise holomorphic
curve a :[0,1]
] --~ M that connects p and q.Then there
exists a minimal number r C[0,1]
such thatE BV, c V.
Put 61 1 = 6|[0.r]
then Lj >L,, .
Consider a mapwe have
then
therefore
PROPOSITION 1.8. For a Hermitian
complex
space M, the metrictopology
Th, which was induced
by hM,
coincides with thegiven topology
TMof
M.PROOF. Let e : M be the resolution of
singularities
of M. HenceM
is acomplex
manifold and 0 is properholomorphic surjective.
Consider thediagram:
First we prove that the
map 0 :
:k - (M,
Th) is continuous.Indeed,
let xo E
M,
1 CM,
xo . Take K, a compactneigh-
bourhood of xo in
M.
Choose ê > 0 such thatBM(xo,
ê) c Int K.1 -> xo, we may assume that 1 c
B,.
Consider a mapT M -~
TM ; we haveConsider any
piecewise holomorphic
curve 1 inM,
which connects xn with xo and~([O, 1])
c K. Let ao = 0 al ... ak = 1 ]be a
holomorphic
map, for everyj,
1j
k.Let 1j
be aholomorphic
extension
of lj
from an openneighbourhood
of ] in C intoM,
forevery
j,
1j
k. We haveTherefore
Thus
~8xn }~ l -;
Hence we have theproof.
Now we prove that the map Id :
(M,
r~) 2013~(M,
Th) is continuous.Indeed,
xo for the
topology
and assume 174
XO for thetopology
Th.Then,
there exists an openneighbourhood
U of xo(for
thetopology Th)
and there existsfxn,l-l C fxnl-l
1 such thatXnk ft
U forevery k
> 1. In(M,
TM) we take a compactneighbourhood K
of xo and we may assumeFor k > 1, choose Ynk E c
8-1 (K),
thereforeThe 1 has a
point
of accumulation yoE M.
Without loss ofgenerality,
we can suppose that the sequenceThus,
for the
topology
TM,for the
topology
Th. Hence0(yo)
= xo andfor the
topology
Th. This is a contradiction.Finally
we prove that the map Id :(M,
Th) -(M,
TM) is continuous.Indeed,
1 CM, { xn } ~ 1 -~
xo for thetopology Th
and assume that1
7~
xo for thetopology
There exists a compactneighbourhood
U ofxo
(for
thetopology
there Cfxnl-l
1 such that U, forevery k
> 1. Since is a metric in M, there exists a number c > 0 such thatfor every 1~ > 1.
Hence -
174
Xo for thetopology
Th. This is a contradiction.Q.E.D.
PROPOSITION 1.9. Let
f :
X - Y be afinite
properholomorphic
mapbetween two
complex
spaces. Then,if
Y has the D*-extension property, X also has the D*-extension property.PROOF. Let g be any
holomorphic
map of D* into X. Consider thecomposition
of maps D* Y; this map can be extended to aholomorphic
map G : D - Y. PutG(O)
= p E Y. LetFor every
i,
1 i k, choose arelatively
compact openneighbourhood Ui
and a
hyperbolic
openneighbourhood
of pi such thatWe prove that there exists ê > 0 such that
Assume that there exists 1 c
D*, ~zn}~ 1 ->
0, such thatSince G is
continuous,
we have{G(zn)}~ 1 -~ p,
i.e. On theother
hand,
choose W,compact neighbourhood
of p; we haveg(zn)
EBy
the compactness of the sequence 1 has apoint
ofaccumulation x. Without loss of
generality,
we can suppose thatHence
If
of (x), f (x)
= p and x = pi. Thus pi andg(zn)
EUi,
for every n > N. This is a contradiction. Thus there exists c > 0k
such that
y(De*)
9 6 cU Ui.
Sinceg(De* )
isconnected,
we havei=l
6
Vi
ishyperbolic. Take
any sequence 1 inD~
that converges to 0.By the
compactness
ofUj,
the sequence 1 has apoint
of accumulation inUi.
Without loss of
generality,
we can suppose that{~(~)}~=i
1 converges to apoint
ofVi .
By
a Kwack’s theorem(see Kobayashi [12]),
the restriction of g fromD;
intoVi
can be extended to aholomorphic
map ofDe
intoVi.
Hence the map g can be extended to aholomorphic
map of D into X.Q.E.D.
COROLLARY 1.10.
If
acomplex
space X has the D*-extension property, then X contains nocomplex
lines.PROOF. Assume that there exists a
holomorphic
mapQ
=/constant.
Consider aholomorphic map # : CB{0} -->
X definedby
By
the mentionedcondition,
the map(3
can be extended to aholomorphic
map~3 :
C - X. Hence a can be extended to aholomorphic
X.Obviously a-
is proper and finite.By
theProposition 1.9., CP~
1 has the D*-extension property. This is a contradiction. Thus X contains nocomplex
lines.
Q.E.D.
PROPOSITION 1.11. Let M be a Hermitian
complex
space. Let Y be acomplex subspace of
M and e : Y - M be the canonicalembedding.
Then,
Y ishyperbolic if
In
particular,
acomplex
space M ishyperbolic if
PROOF. Put
We prove that
dy(p, q)
> 0, for everyp =/
q E Y. On D we consider thePoincare-Bergman
metric w that is definedby
Hence
For every a E D, choose a map J : D 2013~ D defined
by
We have
Assume that there exist
p fl
q E Y such thatdy(p, q)
= 0. Choose 0 A 1such that A
1}
ifAssume that
Then
there exist
a sequence 1 cHol(D, Y)
and a sequence ofpoints
such that
For every n > 1, we choose a
holomorphic
map(3n :
: D 2013~ D such that,8n(0) = =
1 -IZ"12
> 1 -a 2
> 0, for every n > 1. We havewhen n - oo. This is a contradiction. Thus
By
the definition of theKobayashi pseudodistance,
there exist aI, ... , ak ED1j2,
Hol(D, Y)
such thatFor every
i,
1 i k we define the maps andClearly
e ofi
oJi(0)
= e ofi
oQi(1)
= pi, e ofi
o ui EThus
Hence
Therefore
dy(p, q)
> 0. This is a contradiction.Q.E.D.
THEOREM 1.12.
(Theorem
ofBrody
for compactcomplex spaces).
Let Mbe a compact Hermitian
complex
space. Then thefollowing
areequivalent.
(i)
M ishyperbolic.
(ii)
M has the D*-extension property(iii)
M contains no(nontrivial) complex
lines.’
(iv) f
EHol(D, M) }
+oo.PROOF.
i
ii)
Theproposition
is deduced from a Kwack’s theorem(see
Kwack[13]).
ii -
iii)
Theproof
followsimmediately
fromCorollary
1.10.iii -
iv)
Assume thatThen,
there exists a CHol(D, M)
such thatConsider a
holomorphic
map gn :Drn -i
M definedby
We
= 1,
for every n > 1. Forevery t
E[o, 1 ),
putClearly s(t)
oo, forevery t
E[o, 1 )
and > iscontinuous;
Thus there exists a number to E
(o, 1]
such thats(to)
= 1. If to = 1, we chooserpn = If to 1, the supremum is
actually
attained at some interiorpoint
zo.Let L be an
automorphism
ofDrn
withL(O)
= zo. DefineHence rpn E
Hol(Drn’ M)
andTherefore
Hence,
for every 0 r E R, thecontains
asubsequence
whichis
equicontinuous
onDr.
Since M is compact,by
Arzela-Ascoli’stheorem,
there exists asubsequence
of the 1converging
on D to a limit map y~. A further refinementgives
asubsequence converging
onD2; continuing
inthis way allows us to extend ~o
analitically
to allof c. rp
cannot be a constantmap, = lim = 1. The
implication
isproved. (Remark:
onn--+ 00
the above
proof,
seeBrody [2],
Lemma2.1).
iv -
i)
Theproof
followsimmediately
fromProposition
1.11.Q.E.D.
2. - A characterization of
Hartogs
extension forholomorphically
convexKahler spaces
The definition of Kahler forms on
complex
spaces can be found in[5].
Acomplex
space X is called a Kahler space if X has a Kahler form.In
[10]
Ivaskowicz hasgiven
a characterization of theHartogs
extension forholomorphically
convex Kahler manifolds. He hasproved
that aholomorphically
convex Kahler manifold has the
Hartogs
extensionproperty
if andonly
if everyholomorphic
map J : X is constant.The aim of this section is to
generalize
the result of Ivaskowicz toholomorphically
convex Kahler spaces.THEOREM 2.1. Let X be a
holomorphically
convex Kahler space.Then,
X has the
Hartogs
extension propertyif
andonly if
everyholomorphic
mapo~ : ~ P I -~
X is constant.’
PROOF. The
proof
is as in[10].
Let M ={z
E u CC2 :
=0}
bea
strongly pseudo-convex hypersurface,
where U is a domain inC~ 2 and V
isa
C2 -function
on U. Put{ z
e U : >0}.
Letf : Zl + -~
X be aholomorphic
map. We haveproved
thatf
can be extended to aholomorphic
map on a
neighbourhood
of everypoint belonging
to U+. For each n putu n
=Bn
n U+ and let B = nf ( Zl ~ ) .
We shall prove that B containsonly
apoint.
n
For each z E
U+,
let ={w : (w -
z,grad p(z)) = 0)
nun.
As in[10]
we have
LEMMA 2.2.
If
U+ issufficiently small,
LEMMA 2.3. For every sequence
(n
isfixed) converging
toLlö,
we can
find
asubsequence
such converges to ananalytic
set A in Y
of
dimension 1 with 8A cf(afYô).
Moreover the map
f I o, 0 ~{o}
into X can be extended to aholomorphic
mapon
A’
00 Thus, by
Lemma2.3.,
AU U Bj,
whereBj
are compactanalytic
j=i
00
sets in X. Observe
that U Bj
is contained in a compact set in X. Thisimplies
j=l
that there exist q E X and a
neighbourhood
U of q such thatB j
nU fl 0,
forall j
> 1. We may also assume that there exists ananalytic covering
map 0 of U intoBk,
where k = dim X andBk
={ z
E C1 ~
with8(q)
= 0.Put
Bj
=8(Bj ). By
the propermapping theorem, Bj
is ananalytic
subsetof
Bk
forevery j
> 1.By
a theorem in Alexander[ 1 ],
we have VolB j >
c - .1r,00 -
for
every j
> 1. This isimpossible,
sinceVol Bj)
+oo. Let p be a}=l
metric
defining
thetopology
of X and S’ acompact complex
curve in X. LetL(S) =
X2, ...,xn}
and E > 0 such thatB(xj, c) _ {x
E X :p(x, xj) c I
aredisjoint neighbourhoods
of x j .n --
Put
UE = U
= =YBUê.
Observe thatSe
is a closed}=l
Stein manifold in
Y,.
Hence,by
a result of Siu[17],
there exists a Steinneighbourhood We of Se
inYe.
Let e : anembedding.
As in[10]
we have the
following.
LEMMA
2.4. Assume that,for
every 6 > 0, there exists ananalytic
diskf :
D --~ X such that(i) Sê
is contained in the6-neighbourhood f(D)8 of f (D);
(ii) y) :
x Ef (o9D),
y E > 26;(iii) f (D)
na(Sê)8
c(aSê)8.
PROOF. Let 7r : N -
e(Sê)
denote the normal bundle fore(S,)
in en.Then,
for 6 > 0sufficiently small,
there existneighbourhoods Ns
ofe(S,)
in N andV8
ofe(S,)
and abiholomorphic V5
such that = Id.Put 7T
= 1r . rp-l : V8
->e(S,).
Observe thatif2
= ~r.From the
hypothesis
of thelemma,
there exists ananalytic
diskf :
D - Xsuch that
f(D) D
andThen
f (D) n e-1 (V5)
~S,,
where 7r, =e-l
0 1f o e, is propersurjective.
Thisimplies
that G =f -1 [ f (D)
r1e-I(V8)]
-->S,
is ananalytic covering
map.Hence, as in the
proof
of Lemma 7 in[10],
we haveFrom Lemma
2.4,
as in[10],
we haveLEMMA 2.5.
Bj is a
rational curvefor every j
> 1.PROOF OF THEOREM 2.1. From Lemmas 2.3 and
2.5,
as in[ 10], f :
u+ --> Xcan be extended to a
holomorphic
map on aneighbourhood
ofu+
and thetheorem is
proved. Q.E.D.
3. - D* -extension and
Hartogs
extensionIn this
section,
we shall prove thefollowing
THEOREM 3.1. Let X be a
holomorphically
convex spacehaving
theD* -extension property. Then, X has the
Hartogs
extension property.PROOF. STEP 1. We shall prove that
for every
relatively compact
open subset Q of X, where e : Q - X is a canonicalembedding.
Indeed,
assume that supf)’(O)11 : f
EHol(D, SZ) }
= oo, where Q is arelatively
compact open subset of X and e : Q - X is a canonicalembedding.
Then,
there exists a CHol(D, Q)
such thatAs in Theorem
1.12,
there exists a sequence ofholomorphic
maps rpn :Drn ---+
)Q,
rn - oo, such that the sequence 1uniformly
converges, on every fixeddisk,
to aholomorphic
map -. C - Xand (e
oI
= 1, forall n > 1.
Clearly V
is not constant, since - lim))(e
o = 1.n->00
Hence X contains a nontrivial
complex
line. This is a contradiction. Thus +oo, for everyrelatively
compact open subset Q of X.By Proposition
1.11, Q ishyperbolic.
STEP 2.
Finally
we prove that X has theHartogs
extensionproperty.
By
a result of Shiffman[15],
it suffices to check that X satisfies the disk condition. Now assume that 1 CHol(D,X)
and 1 converges to a inH(D*, X).
Since X isholomorphically
convex, the set K is compact, where K = U X - Z be the Remmert reduction of X,i.e.,
8 is a proper
holomorphic surjection
and Z is a Stein space. Sinceok
is aholomorphically
convex compact set, there exists aneighbourhood S2
of8K
such that
S2
is acomplete C-space.
HenceQ
is acomplete hyperbolic
space.This
implies
that themap 0:
Q =0- 1 (C2)
-S2
can be extended to a continuousmap ->
!C2,
whereQ
is acompletion
of Q for theKobayashi
metricdo
ofQ.
We prove that
Q
= Q.Indeed,
let z ESZ
and 1 C Q such that z.By
the compactness of the subsetdZI,
it follows that the subset(zn)
isrelatively
compact in Q. Thus z E Q. Therefore Q is a
complete hyperbolic
space and the sequence 1 converges to j inHol(D, Q) (see [11]). Q.E.D.
4. - Inverse invariance of D* -extension and
Hartogs
extensionWe have the
following
theorems.THEOREM 4.1. Let 7r : X --~ Y be a
holomorphic
map between twocomplex
spaces
satisfying
thefollowing
condition.For every y E Y, there exists a
neighbourhood
Uof
y such that1r-I(U)
has the D*-extension property
(the Hartogs
extensionproperty).
Then, if
Y has the D*-extension property(resp.
theHartogs
extensionproperty),
so does X.PROOF.
(i)
Assume that Y has the D*-extensionproperty.
Let a : D* - Xbe a
holomorphic
map. Consider theholomorphic
mapfl
= 7rJ : D* ---~ Y.By hypothesis, fl
can be extended to aholomorphic
mapPut
,Q(0)
= yo.By
the mentionedcondition,
we find aneighbourhood
U of yo such that~-1 (U)
has the D*-extension property.Since ~
iscontinuous,
thereexists 6 > 0 such that
Ø(Dc)
c U. Thisimplies that J(D§)
c7r"~(!7).
Thugs acan be extended to a
holomorphic map 8 :
D --~ X.(ii)
Assume that Y has theHartogs
extension property.To
prove X
has theHartogs
extension property, it suffices to show that theenvelope
toholomorphy Of
of everyholomorphic
mapf :
Q - X, where S2 is a Riemann domain over a Steinmanifold,
ispseudoconvex.
Assume that there exists zo E
9Q/
suchthat,
for everyneighbourhood
U ofzo in Q, there exists a
neighbourhood
V of zo in U such thatBy hypothesis, g = 7rf :
K2 , Y can be extended to aholomorphic
S2 - Y.Put yo =
g(zo).
Take aneighbourhood W
of yo in Y such that has theHartogs
extension property. Letf :
X denote the canonical extension off.
Take aneighbourhood
V of zoin f2
such that V and/(V)
C W.This
yields f (Y n SZ f)
c~r-1 (W).
Thus can be extended to aholomorphic
map of V into X. This is a contradiction.
thus
the theorem isproved.
Q.E.D.
THEOREM 4.2. Let 0 : X - Y be a proper
holomorphic surjection
betweencomplex
spaces, such that8-1 (y)
has the D*-extension property(resp.
has theHartogs
extensionproperty) for
all y E Y.Then
(i) If
Y has the D*-extension property, so does X.(ii) If
Y has theHartogs
extension property and X is a Kahler space, then X has theHartogs
extension propertyPROOF.
(i)
Assume that Y has the D*-extension property and J : D* -~ X is anyholomorphic
map.By hypothesis,
0J : D* - Y can be extended to aholomorphic map fl :
D - Y. Put yo= ~(0).
Take ahyperbolic neighbourhood
Uof yo in Y.
By hypothesis,
contains nocomplex
lines. Thisimplies
thatf-’(V)
ishyperbolic
for everyneighbourhood
V of yo in U. Take arelatively
compact
neighbourhood W
of yo in V, and r > 0 such that(3(Dr)
C W. Since8 is proper, it follows that there exists a sequence 1 C
Dr, { zn } ~ 1 -~
0,such that xo in
f -1 (Y)
and ishyperbolic. By
a Kwack’stheorem
[13], ~
can be extended to aholomorphic map 8 :
D --> X.(ii)
Assume now that Y has theHartogs
extension property.By
Theorem4.1,
it suffices to showthat,
for every y E Y, there exists aneighbourhood
U of y in Y such that0- 1 (U)
has theHartogs
extension property.Let U be a Stein
neighbourhood
of y in Y. For every compact set K in8-1 (U),
we havewhere 01
l = It follows that9- l (U)
isholomorphically
convex. On the otherhand,
since0-’(y)
contains no rationalcurves, for every y E U, it follows that contains no rational curves.
By
Theorem
2.1, 8-1 (U)
has theHartogs
extensionproperty. Q.E.D.
5. - Invariance of D * -extension and
Hartogs
extensionWe have the
following
theoremsTHEOREM 5.1. Let B : X --+ Y be a
finite
properholomorphic surjection
between two
complex
spaces. Then(i) If H°°(X)
separatespoints of
X, then X has the D*-extension propertyif
andonly if
so does Y.(ii) If H(X)
separatespoints of
X, then X has theHartogs
extensionproperty
if
andonly if
so does Y.PROOF.
(i)
Assume that X has the D*-extensionproperty
andConsider the commutative
diagram
Since D* is normal and
B
is finite propersurjective, 8 : ~ *8
--~ D* is ananalytic covering
map of order n. As in[4],
we infer thatH’(u*O)
is aninteger
extension of of order n such
that,
for eachf
E there existsa
polynomial Pf
EH°°(D*)[a]
of order n f,n, such that
P f ( f )
= 0. Moreoverfor all w E
SHOO(D*),
whereV(p¡,w)
= and R : --~SH’(D*)
is the restriction map. It is easy to see that R is finite propersurjective.
Put Z =Ro
=Rlz :
Z --> D.Observe that D is an open subset of
SHI(D*) (see [9])
andC Z, where e : J*0 -
SH°°(J*0)
denotes the canonical map.By hypothesis,
is open in and e : u*O ~-- Let W be the subset of all w E D such that there existneighbourhoods
G of w in D andUj
ofWj
ERo 1 (w) _ ~wl , ... , wP~ satisfying
thefollowing
conditions:for
every j,
1j
p.Obviously
W is open in D andR° : Ro 1 (W ) --i
W is a finite propertopological covering
map. Takewo
E D such that andf
EH°°(J *0)
such thatÎ(wJ) =I f (w°),
for all ifl j,
where&§, CoO
EBy
(*), we havewhere
D f
denotes the discriminant ofP f .
ObserveDj fl 0.
We shall prove thatDBW
cV(.bf). -
Assume that there exists
w°
EDBW
such that Let~wl , ... , wq},^ and Uj
bedisjoint neighbourhoods
of SinceRo :
Z - D isproper and we find a
neighbourhood
G ofw°
such thatThis
implies
thatHence w E W. This is a contradiction. Thus
Ro :
Z --> D is ananalytic covering
map.
By
a Grauert-Remmerttheorem,
there exists acomplex
structure on Z suchthat c
H(Z).
On the otherhand,
since #(ZBe(~ *e)) _
# 00and every 1-dimensional
complex
normal space isnon-singular,
it follows X can be extended to aholomorphic map a :
Z --> X. PutObviously
7r is finite propersurjective. By
thedirect
image
theorem of Grauert[6],
it follows that is ananalytic
set inD x Y, where
r 8
denotes thegraph of 8 .
Clearly 1r(r8)
n(D*
xY)
=r~
and the canonicalprojection
p :r8 ---+
D isproper. This
implies
that the map J : D* - Y can be extended to ameromorphic
map 6 :
D ---> Y. Since codimpea)
> 2(see [15]),
wherepea)
denotes theindeterminacy
locusof 6,
it follows that 8 isholomorphic
and it is an extension of J .Conversely,
assume that Y has the D*-extension property. Since 0 is finite for every y EY, e-1 (y)
has the D*-extension property.Hence, by
Theorem4.2,
X has the D*-extension property and
(i)
isproved.
(ii)
Assume now that X has theHartogs
extension property. To prove that Y has also theHartogs
extension property, it suffices to show that everyholomorphic- map g : Hk(r) -
Y, whereHk(r)
is theHartogs diagram given by
can be extended to a
holomorphic map g :
=Dk
---+ Y.Indeed,
consider the commutativediagram
where Z = =
(Hk(r)
X yX)red.
Z -~Hk(r)
is ananalytic covering
map. We may assume that Z is normal.
Then,
the branch locusof 6,
H, has codimension 1.By
a result ofDloussky [3],
there exists ahypersurface H
inDk
such thatHk(r)BH
=Obviously H
nHk(r)
C H. Since H can be written in the form H= H
nHk(r)) U H’,
where H’ is ahypersurface
inHk(r)
such that = without loss ofgenerality,
we may assume thath
nHk(r)
= H. Consider the commutativediagram
where a is induced
by 6
and eby
the canonicalembedding Zo -~
Z. As in(i),
we may prove that a :
SH(Z) - Dk
is ananalytic covering
map.Without loss of
generality,
we may assume thatSH(Z)
is irreducible normal.By hypothesis,
it follows that e and(3
are openembeddings.
Thisimplies
thate(2o) = SH(Z)Ba-I(H). Thus g :
Z - X can be extended to aholomorphic
mapConsider the
holomorphic
mapwhere
~3 :
W ---+SH(Z)
is the resolution ofsingularities
ofSH(Z).
Since Xhas the
Hartogs
extension property and0-’(Z)
meets every irreducible branch of(3g
can be extended to aholomorphic
mapf :
W - X.By hypothesis, H(X)
separatespoints
of X, it follows thatf
can be factorizedthrough fl.
Thus(3g
can be extended to aholomorphic
mapf :
X.As in
(i), f
induces ameromorphic map 0 : D~ 2013~
Y which is an extension of g. Consider the commutativediagram
in which
8, 8
are finite. Thus the canonicalprojection Dk
is finite.By
the
meromorphicity
ofg,
it followsthat g
isholomorphic. Conversely,
assumethat Y has the
Hartogs
extension property. For each y E Y, we find a Steinneighbourhood
V of yo. Then is Stein(see [4]).
Hence8-1(Y)
hasthe
Hartogs
extension property.By
Theorem4.1,
X has theHartogs
extensionproperty. Thus
(ii)
isproved. Q.E.D.
PROPOSITION 5.2.
Every a, ffzne algbraic variety of
the typewhere al,..., an are
holomorphic functions
on C withan fl
0, has not theD*-extension property.
PROOF. Assume that V has the D*-extension property. Take r > 0 and
zt fl
0 such that .Put
VI > r ~ .
Thenis proper.
By (*),
we have7r(VI)
c whereis the canonical
projection.
Since and> r}) separates
thepoints
ofC)(0, z#)
and ofI
>r } respectively,
it follows thatseparates
thepoints
of Vi.Let J : D* ---+
VI
be aholomorphic
map. Then the map Q can be extendedto a
holomorphic map 6 :
D -~ V. Sincewe have
Therefore,
Vi has the D*-extension property. Hence thecomplex subspace {0
ofC~ 2
also has the D*-extensionproperty.
This is acontradiction.
Q.E.D.
THEOREM 5.3.
finite
propersurjective
mapof
a non-compactcomplex
space X into aholomorphically
convex Kahlercomplex
space Y. Let dim X 2. Then, X has theHartogs
extension propertyif
andonly if
so doesY.
PROOF. Assume that X has the
Hartogs
extension property. To prove that Y has theHartogs
extension property,by
Theorem 2.1, it sufficies to show that Y contains no rational curves.Let Q :
C~ P 1 ~
Y be aholomorphic
map andlet J fl
constant. Considerthe commutative