N A L E S E D L ’IN IT ST T U
F O U R IE
ANNALES
DE
L’INSTITUT FOURIER
Lucia ALESSANDRINI & Giovanni BASSANELLI
On the embedding of 1-convex manifolds with 1-dimensional exceptional set Tome 51, no1 (2001), p. 99-108.
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ON THE EMBEDDING OF 1-CONVEX MANIFOLDS WITH 1-DIMENSIONAL EXCEPTIONAL SET
by
L. ALESSANDRINI & G. BASSANELLI *Introduction.
Any
1-convex(i.e. strongly pseudoconvex) space X
is a proper modification atfinitely
manypoints
of a Stein spaceY; X
is calledembeddable if it is
biholomorphic
to a closedanalytic subspace
of xfor suitable m and n. C. Banica
[2]
and Vo Van Tan[15]
had shown thatevery 1-convex surface is
embeddable;
the case when dim X > 2 has been studied in some papers of M. Coltoiu and Vo Van Tan(see [5]
and[16]
for the
references), mainly
when X is smooth and itsexceptional
set S’ is1-dimensional.
Up
to now, the best result is thefollowing:
THEOREM. - Let X be a 1-convex manifold with 1-dimensional
exceptional
set S. Then X isembeddable,
with apossible exception given by
dim X = 3 and S contains a rational curve of
type (-1, -1)
or(0, - 2)
In these three
exceptional
cases, there are someexamples
of non-embeddable
manifolds, only
one of which iscompletely
clarified(see [5]);
anyway, the above result
gives
no criterion in order to check theembedding
* Partially supported by MURST research funds.
Keywords: 1-convex manifolds - Kahler manifolds.
Math. classification: 32FI0 - 53B35.
property
in these situations. In this paper, wegive
aprecise criterion,
whichturns out to be
topological, namely:
THEOREM I. - Let X be a 1-convex manifold with 1-dimensional
exceptional
set S. Thefollowing
statements areequivalent:
(i)
X isKdhler,
(ii)
S does not contain any effective curve C which is aboundary,
i.e.such that
[C]
= 0 inH2 (X, Z).
THEOREM II. - Let X be a 1-convex manifold with 1-dimensional
exceptional
set S.If H2 (X, Z) is finitely generated,
then X is embeddable if andonly
if it is Kahler.The
proofs
do notdepend
on the results of M. Coltoiu and Vo VanTan,
nevertheless
using
them one can show that condition(ii)
in Theorem I isequivalent
to(ii’) dim X =1=
3 or ,S’ does not contain any effective curve which is aboundary
and such that its irreduciblecomponents
are rational curves oftype (-1, -1)
or(0, -2)
or(1, -3).
Theorems I and II
give
apartial
answer to severalquestions
of[16]
(in particular
toQuestion 2.6,
Problem 2.7 and Problem3.9),
andsupport
the idea of a
parallelism
between Moisezon and 1-convex manifolds(see
theIntroduction of
[15]);
moreprecisely,
one should compare Theorem II with the fact that a Moisezon manifold isprojective
if andonly
if it is Kahler.Finally
we remark that all the knownexamples
of non-embeddable 1-convex manifolds with 1-dimensionalexceptional
setsatisfy
ourhypothesis,
that isH2 (X, Z)
isfinitely generated.
We alsohope
that our result willgive
somesuggestion
in order tostudy
thegeneral
case.Let us conclude with a remark on
techniques;
as to the author’sknowledge,
theduality
between forms and currents on anon-compact
manifold is used here for the first time(in
thecompact
case, see[9], [10], [1]
etal.),
and this new toolprovides quite elementary proofs.
Results.
First of
all,
we fix some notations and recall the characterization of 1-convex and embeddable manifolds which we need in what follows.DEFINITION 1. - Let X be a
complex
manifold.A real cp E
~~’1 (X )
is calledpositive (semi-positive)
atx E X if the hermitian form induced
by
cpx on theholomorphic tangent
space
T~ X
ispositive
definite(semi-definite) ;
we shall write ’Px > 00).If(>0 (cp > 0)
for every xE X ,
p is calledpositive (semi- posi ti ve),
and we shall write cp > 0(p > 0).
A current T E
(X)’
ispositive (in
the classical sense ofLelong)
ifT (cp) > 0,
for every 0.An effective curve is a finite combination
N 1 nj Cj
of irreduciblecurves
Cj
withintegral
coefficients nj > 0.Let L be a
holomorphic
line bundle on X endowed with a hermitian metich ;
if 6 is the associated curvatureform,
we denoteby cl (L, h) = 2~ O
its Chern form. L is
positive
if andonly
if there is a metric h such that> 0.
Let us denote
by HP(X, Z)
the range of themap i* :
Hp (X, R)
inducedby
the inclusion i : Z - R. So areal,
closedp-form 0
iscalled
integral
if its classM
Ebelongs
toHP(X, Z).
Consider the map
arising
from the exact sequence- - iw iw J. -
It is well known that if L is a
holomorphic
linebundle,
for every hermitian metric h onL, cl (L, h)
Ei* (b(L)).
DEFINITION 2. - A
complex
space X is said 1-convex(or strongly pseudoconvex)
if it satisfies one of thefollowing equivalent
conditions:(i)
there exists a smooth exhaustion function F : X - R which isstrictly plurisubharmonic
outside acompact
subset ofX,
(ii)
there exists a propersurjective holomorphic
mapf :
X -~ Y ontoa Stein space
Y,
and a finite subset B C Y suchthat,
if S := theinduced map
X B
,S’ -~ YB
B isbiholomorphic
andf*Ox.
Remark 3. - The
equivalence
of the above conditions wasproved
in
[11].
The mapf
is called the Remmertreduction,
andS,
which is themaximal
compact analytic
set ofpositive
dimension inX,
is called theexceptional
set of X.DEFINITION 4. - A I-convex space X is called embeddable if there is a
holomorphic embedding j :
X --+ cm x for some m and n.Roughly speaking,
a 1-convex space X "differs" from a Stein space Y(its
Remmertreduction)
in itsexceptional
setS ;
Y is embeddable insome
C~,
and ,S’ is lvloisezon and sometimesprojective.
Therefore it seemsnatural to ask when X is
embeddabje
in C~ xCPn,
for some m and n(this
is notalways
the case, see[5]).
As for thecompact/pro jective
case, toprove that a 1-convex space is
embeddable,
it is useful to search"positive"
holomorphic
line bundles on it:see [14] and [13].
Since we shall assumethat X is a
manifold,
we use asimpler result,
due toEto,
Kazama and Watanabe:PROPOSITION 5
([8],
TheoremIII).
- Let X be a 1-convex manifoldcarrying
apositive holomorphic
line bundle L. Then X is embeddable.To prove Theorems I and
II,
we need acouple
ofLemmata;
the firstone is an
elementary
result in grouptheory.
LEMMA 6. - are elements of an abelian group G such
that,
for suitable real numbers rl,..., rq >0,
gi 0 rl -~ ... + gq 0 rq = 0 in G 0R,
then there existpositive integers
nl,..., nq such that nlgl + ... -~- nqgq = 0 in G.Proof. Let
G’
be thesubgroup
of Ggenerated by
g1, ... , gq. Since is asubgroup
ofG’
satisfies thehypotheses too;
so there is norestriction in
assuming
that G isfinitely generated.
Therefore G = F EBT,
where F is free and T is the torsion
part, and gj = fj
+tj, j
=1,...,
q,with fj E
Fand tj
E T. Thusby
thehypothesis f10rl
+... +fq 0 r ==0;
identifying
F with Z’ C for a suitable n, weget,
inR’~ ,
This turns out to be a linear
system
withintegral
coefhcients: therefore there arepositive integers
m1, ... , mq, such thatmIll
+ ... +mqfq
= 0.For a suitable
positive integer
m:mt, = ... = mtq
=0;
thusLEMMA 7. - Let cp E
Jl",’(X)
be a closed(1, l)-fbrm
on X whichis
positive
on S’(i.e.
px >0,
for every x ES).
Then there exists a smooth function F : X - R such that ppositive
on X.Proof.
Let g :
X -CN
be the properholomorphic
mapgiven by
the
composition
of the Remmert reductionf :
X - Y and anembedding j :
Y - Denoteby
z =(z~ , ... , zN )
coordinates in(CN
and leta := then 0 on X and
g*a
> 0 on XB
S. Since ~p > 0 in aneighborhood
ofS ,
for everycompact
subset K of X there exists apositive
constantMK
such that p +MKg*a
> 0 on K.Let
Kl
~ and for every n >1, Therefore,
for suitableMan
> 0:Choose a smooth function u :
[0, +oo) -
R such thatu’ (t)
>0,
for everyt >, 0,
and >Mn
onKn,
for ?~ ~ 1.Let
v(s) fo u(t)dt, s >
0. Weget
The first term in the
right
hand sideis > 0,
hence onKn
we havetherefore we
get
on X:Let us go to the
proofs
of Theorem I and Theorem II.THEOREM I. - Let X be a 1-convex manifold with 1-dimensional
exceptional
set S. Thefollowing
statements areequivalent:
(i)
X isKiihler,
(ii)
S does not contain any effective curve C which is aboundary,
i.e.such that
~C~ =
0 inH2 (X, Z).
Proof.
(i)
~(ii):
In a Kahler manifold nopositive
combination ofcompact analytic
subsets can be aboundary.
(ii)
~(i):
Letand
Let us argue
by
contradiction and assume that A n B -0.
Since A isa
non-empty
open cone in the naturaltopology
of and B is alinear
subspace, by
the Hahn-Banach Theorem([12], 11.3.1),
weget
a closedhyperplane
in thetopological
vector space.61"(X), containing
B and notintersecting
A. Thatis,
there exists a current T E61"(X)’
such that> 0 for every cp E
A,
and = 0 for every cp E B.(i)
> 0 for every EA,
then T ispositive
andsupp(T)
CS;
(ii)
= 0 for every cp EB,
then188T
= 0.Proof of the claim. -
Let 0
EE 1’ 1 ( X ) ~ ~ > 0, and
let p E A: thenfor every E >
0, ~
EA,
henceNow let
so that
this
implies T( 1/J)
= 0.To prove the second
assertion,
notice that for everyz99/
E B so that iThe claim allows to
apply
Theorem 4.10 in[3]
to the current Tarising
from the Hahn-Banach Theorem:
therefore,
there are some irreduciblecomponents
ofS,
sayCl,..., Cq,
and somepositive
real numbers rl , ... , rq such thatn
(in particular,
T isclosed).
Let cp -4’02
+ + be a closed real 2-form onX ;
define7r(p)
:= Then 7r induces a linear map II :H2(X, JR) ---+ H~°’2 (X ) - ~(X,0) .
But S is1-dimensional,
thus([4],
Lemma
1) H2 (X, C7) - 0;
since II is the null map, every classbelonging
to
H2(X, JR)
can berepresented by
means of an element of B. Hence"T vanishes on B" means that vanishes on every class of
H2(X, JR).
ButH2(X, II8) = Hom(
thus2:J=l ]
- 0 inH2 (X, R) = H2 (X, Z) 0
R.Applying
Lemma 6 weget
a contradiction with(ii).
Therefore there exists cp E A nB ; by
means of Lemma 7 the form pcan be modified in order to obtain a closed Kahler form for X.
Q.E.D.
Remark 8. - When ,5’ is
irreducible,
condition(ii)
in Theorem Imeans that the class of S’ in
H2 (X, Z)
is not a torsion class: therefore the mapH2 (,S’, Il~) ~ H2 (X, R)
turns out to beinjective and, : H2 (X, H2(S, R)
issurjective.
In this case there is anotherproof1
of Theorem I.Denote
by H
the sheaf of germs of realpluriharmonic functions;
as in[9],
we
get easily
that exact and that isthe space of real closed
(1, l)-forms,
modulo the space of -exact forms.Now for a suitable embeddable
neighborhood
U of ,S’(see [6],
p.563)
weget
that in thefollowing
commutativediagram
with exact lines:a is an
isomorphism, H2 (U, R)
=H 2(S, R), -~
issurjective.
Hence/3
issurjective,
thus we can extend a Kahler form on U to a closed(1, I)-form
on
X; by
Lemma7,
X is Kahler.THEOREM II. - Let X be a 1-convex manifold with 1-dimensional
exceptional
set S. IfH2 (X, íl)
isfinitely generated,
thefollowing
statementsare
equivalent:
(i)
X isKalller,
(ii)
S does not contain any effective curve C which is aboundary,
i.e.such that
[C]
= 0 inH2 (X, Z), (iii)
X is embeddable.Proof.
(iii)
~(i)
isobvious,
soby
theprevious
theorem it isenough
to show(ii) ~ (iii) .
Letfor some
holomorphic
line bundle L on X and some hermitian metric h onL}.
Let us In order to argue
exactly
as in TheoremI,
with B’ instead ofB, only
two remarks are necessary.First,
from A n B’ -0
itfollows An
(B’
0If8) = 0
where B’ 0 R is the linearsubspace generated by B’;
so A and B’ can beseparated
as above.Second,
letf
be a real smooth1 Private communication of M. Coltoiu.
function on X and L a
holomorphic
line bundle with a hermitian metrich;
then
Thus Claim
(ii)
in theproof
of Theorem I holds with B’ instead of B.Therefore there is a current ’-’ , with rj >
0,
which vanishesAs we have
just noted, H2 (X, 0)
=0,
thus themap 6 : Hl (X, 0*)
-~H2 (X, Z)
issurjective.
Therefore eachintegral
class can berepresented by
an element of
B’,
thus vanishes onH’(X, Z).
SinceH2 (X, Z)
isfinitely generated,
=H2 (X, Z)
thus rj vanishes onH2 (X, JR).
As we havejust
seen in theprevious proof,
thisgives
acontradiction,
so we conclude that there are L and h such thatE A n B’.
Finally, applying
Lemma7,
it follows thatis
positive
on X.Q.E.D.
COROLLARY 9. - In the
hypotheses
of TheoremI,
the statements(i), (ii)
are alsoeq ui valent
to(ii’) dim X =I
3 or S does not contain any effective curve which is aboundary
and such that its irreduciblecomponents
are rational curves oftype (-l, -1)
or(0, -2)
or(1, -3).
Proof.
Obviously (ii)
~(ii’).
Assume that(ii)
isfalse,
thatis,
there are
positive integers
nul, ... , nq and irreduciblecomponents Cl , ... , Cq
of ,S’ such that T =
r-’=l njcj
is aboundary.
Then the canonical bundleKx
vanishes onT,
thatis,
0njKx.Cj,
soKx.Cj
= 0 forevery j
=1, ... ,
q.By
theStep
1 of Theorem 1.5 in[16],
everyCj
is theexceptional
set of a 1-convex
neighborhood Uj
of it. Now 0 =Kx.Cj = KUJ.Cj;
soKu,
is notample and, by
means of Theorem 3 in[5],
thisimplies
thatCj
is
rational,
dim X = 3 and(see [5] ,
Remark p.463) Cj
must be oftype
(-1, -1), (0, -2) or (1, -3). Q.E.D.
Remark 10. - All the
examples given
in the literature in connection to theembedding problem satisfy
theproperty "H2 (X, Z)
isfinitely
gen-erated" ; nevertheless, using
theexample given
in[7],
it is easy to build a1-convex threefold whose second
homology
group is notfinitely generated.
Now assume that the
exceptional
curve ,S’ is irreducible. One canuse the
proof
of Theorem II orapply
theargument
of Remark 8 to the sequence 0 - Z - O - 0* ---+0,
to see that(without
thehypothesis
"H2(X, Z)
isfinitely generated")
X is embeddable if andonly
if thereexists a
homomorphism
inHom(H2 (X, Z), Z)
which does not vanish onS,
that
is,
the mapH2 (X, 7 ) -~ H2 (S, Z)
is not the zero map.Therefore a
comparison
beetwen this last condition and Theorem I maysuggest
the way to acounterexample (if
itexists!)
of a 1-convex Kahlermanifold,
with 1-dimensionalexceptional set,
which is non-embeddable.BIBLIOGRAPHY
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Manuscrit regu le 24 mars 2000, accepté le 7 septembre 2000.
Lucia ALESSANDRINI & Giovanni BASSANELLI, Università di Parma
Dipartimento di Matematica via Massimo D’Azeglio,
85/A
43100 Parma
(Italy).
giovanni.bassanelli@unipr.it lucia.alessandrini@unipr.it