DIVISION THEOREMS AND TWISTED COMPLEXES
DROR VAROLIN
Abstract.We prove new Skodatype division,or ideal membership,theorems.We work
in a geometric setting of line bundles over Kahler manifolds that are Stein away from an
analytic subvariety.(This includes complex projective manifolds.) Our approach is to
combine the twisted BochnerKodaira Identity,used in the OhsawaTakegoshi Theorem,
with Skoda's basic estimate for the division problem.Techniques developed by McNeal
and the author are then used to provide many examples of new division theorems.Among
other applications,we give a modication of a recent result of Siu regarding eective nite
generation of certain section rings.
Contents
1.Introduction 1
2.Corollaries of Theorem 1 4
3.Hilbert Space Theory 10
4.Classical L
2
identities and estimates 13
5.Twisted versions of L
2
identities and estimates 15
6.Proof of Theorem 1 17
References 19
1.Introduction
A classical problem both in commutative algebra and in several complex variables is the
ideal membership,or division problem.In the setting of commutative algebra,the decisive
result is Hilbert's Nullstellensatz.On the other hand,in several complex variables the
most basic division problem for bounded holomorphic functions is open:given a collection
of bounded holomorphic functions g
1
;:::;g
p
on the unit ball B C
n
(n 2) such that
P
jg
i
j
2
= 1,are there bounded holomorphic functions h
1
;:::;h
p
such that
P
h
i
g
i
= 1?The
case n = 1 is the famous Corona Theorem of Carleson.In higher dimensions,the closest
result thus far is an L
2
version,also known as the celebrated Division Theorem of Skoda.
In this paper,we use the method of the twisted BochnerKodaira Identity together with
Skoda's Basic Estimate to establish a generalization of Skoda's Division Theorem.
Skoda's Theorem has many applications.Some recent applications in algebraic geometry
appear in the work of Ein and Lazarsfeld [1],where Skoda's Theorem plays a key role in
giving an eective version of the Nullstellensatz.Siu has used Skoda's Theorem to prove
results about eective global generation of multiplier ideals,which was a key tool in his
approach to establishing the deformation invariance of plurigenera [3].Siu has also used
Partially supported by NSF grant DMS0400909.
1
Skoda's Theorem in his approach to the problem of nite generation of the canonical ring
[4].
At the same time,several authors have been establishing results that showthe fundamental
role of the OhsawaTakegoshi extension theorem and its variants in the areas of analytic
methods in algebraic geometry and of several complex variables.The applications of the
OhsawaTakegoshi Theorem are too numerous to mention in this introduction.
The results of Skoda and OhsawaTakegoshi are similar in nature and proof.Both re
sults use the BochnerKodaira Identity.However,in the OhsawaTakegoshi technique,the
BochnerKodaira Identity is"twisted".
In Skoda's Theorem,a functional analysis argument is used that is similar to the well
known LaxMilgramLemma.We recall,perhaps with slight modication,Skoda's functional
analysis in Section 3.As usual,the functional analysis requires us to establish an a priori
estimate.This estimate is obtained from the BochnerKodaira Identity together with a non
trivial and very sharp inequality due to Skoda,referred to in this paper as Skoda's Inequality.
The resulting a priori estimate is referred to here as Skoda's Basic Estimate.
The main idea of this paper is to introduce twisting into Skoda's Basic estimate.More
precisely,we twist the BochnerKodaira Identity before applying Skoda's Inequality.The
result is a series of divison theorems whose estimates are dierent from the original result of
Skoda.
Because many recent applications of L
2
theorems have been to complex and algebraic ge
ometry,we will state our results in the more general language of singular metrics and sections
of holomorphic line bundles on socalled essentially Stein manifolds:a Kahler manifold X is
said to be essentially Stein if there exists a complex subvariety V X such that XV is a
Stein manifold.For example,X could be a Stein manifold,in which case we can take V =;,
or X could be a smooth projective variety,in which case V could be the intersection of X
with a hyperplane in some projective space in which X is by hypothesis embedded.A third
interesting class of examples is a holomorphic family of algebraic manifolds bered over the
unit disk or over a more general Stein manifold.
We x on X two holomorphic line bundles E!X and F!X,with singular metrics
e
'
E
and e
'
F
respectively,and suppose given a collection of sections
g
1
;:::;g
p
2 H
0
(X;E)
where p 1 is some integer.
Taking cue from Skoda [5],we seek to determine which sections f 2 H
0
(X;F +K
X
) can
be divided by g = (g
1
;:::;g
p
),in the sense that there exist sections
h
1
;:::;h
p
2 H
0
(X;F E +K
X
)
satisfying the equality
f =
p
X
i=1
h
i
g
i
:
Moreover,if f satises some sort of L
2
estimate,what can we say about estimates for
h
1
;:::;h
p
?We shall refer to this question as the division problem.
To state our main result,it is useful to introduce the following denition.
2
Definition 1.1.For a triple (;F;q) where :[1;1)!R,F:[1;1)![1;1) are C
2
functions and q > 0 is an integer,dene the auxiliary function
(x) = x +F(x):
We call (;F;q) a Skoda Triple if
(1) (x)
0
(x) +1 +F
0
(x) 0 and (x)
00
(x) +F
00
(x) 0:
Given a Skoda Triple (;F;q),we dene
B(x) = 1 +
((x)
0
(x) +1 +F
0
(x))
q(x)
and A(x):=
(
(1+F
0
(x))
2
(F
00
(x)+(x)
00
(x))
1 +F
0
6 0
0 1 +F
0
0
:
Remark.Note that
B
(B 1)
=
q +
0
+1 +F
0
(
0
+1 +F
0
)
=
1
+
q
0
+1 +F
0
Our main result is the following.
Theorem 1.Let X be an essentially Stein manifold of complex dimension n,L;E!X
holomorphic line bundles with singular Hermitian metrics e
and e
respectively,and
g
1
;:::;g
p
2 H
0
(X;E).Let
q = min(p 1;n) and = 1 log(jgj
2
e
):
Let (;F;q) be a Skoda triple.Set = (),A = A() and B = B(),with (x),A(x) and
B(x) as in Denition 1.1.Assume that
p
1@
@ Bq
p
1@
@:
If nonconstant,or if cannot be dened on R and still satisfy (1) of Denition 1.1,
assume further that
jgj
2
e
:=
p
X
j=1
jg
j
j
2
e
< 1:
Then for every section f 2 H
0
(X;L+K
X
) such that
Z
X
B
(B 1)
jfj
2
e
()
e
(jgj
2
e
)
q+1
< +1
there exist sections h
1
;:::;h
p
2 H
0
(X;LE +K
X
) such that
p
X
j=1
h
j
g
j
= f
and
Z
X
jhj
2
e
()
e
( )
( +A)(jgj
2
e
)
q
Z
X
B
(B 1)
jfj
2
e
()
e
(jgj
2
e
)
q+1
:
Theorem 1 implies a large number of division theorems.However,as stated,Theorem
1 does not give a solution to the division problem unless we input a Skoda triple (always
with q = min(n;p 1)).By choosing dierent Skoda triples (;F;q) (with the same q =
min(n;p 1) from Theorem 1),one can obtain numerous division theorems as corollaries of
Theorem 1.In the next section we establish some of these corollaries,and hope the reader
3
is convinced that such corollaries are easy to come by,or equivalently,that Skoda triples are
easy to nd.
2.Corollaries of Theorem 1
Example.Fix > 0.Let q = min(n;p 1),F(x) = 1 x and (x) = ( 1)q(x 1).
Then = 1 so by denition A = 0,and B = .Thus we obtain the following geometric
reformulation of the famous theorem of Skoda.
Theorem 2.1.Let X be an essentially Stein manifold of complex dimension n,F;E!X
holomorphic line bundles with singular Hermitian metrics e
and e
respectively,and
g
1
;:::;g
p
2 H
0
(X;E).Assume that
p
1@
@ q
p
1@
@:
Then for any f 2 H
0
(X;K
X
+F) such that
Z
X
jfj
2
e
(jgj
2
e
)
q+1
< +1
there are p sections h
1
;:::;h
p
2 H
0
(X;K
X
+F E) such that
X
k
h
k
g
k
= f and
Z
X
jhj
2
e
( )
(jgj
2
e
)
q
1
Z
X
jfj
2
e
(jgj
2
e
)
q+1
:
Remark.The proof of Theorem2.1 that we give essentially reduces to Skoda's original proof
since,as we shall see,when is constant the twisting of the
@complex becomes trivial.
Example.In his paper [4],Siu derived from Skoda's Theorem the following result in the
case of algebraic manifolds.We give a slightly dierent proof of Siu's result,in the more
general setting of essentially Stein manifolds.
Theorem 2.2.Let X be an essentially Stein manifold of complex dimension n,L!X
a holomorphic line bundle,and H!X a holomorphic line bundle with nonnegatively
curved singular Hermitian metric e
'
.Let k 1 be an integer and x sections G
1
;:::;G
p
2
H
0
(X;L).Dene the multiplier ideals
J
k+1
= I(e
'
jGj
2(n+k+1)
) and J
k
= I(e
'
jGj
2(n+k)
):
Then
H
0
(X;((n +k +1)L +H +K
X
)
J
k+1
) =
p
M
j=1
G
j
H
0
(X;((n +k)L +H +K
X
)
J
k
):
Proof.By taking G
p
=:::G
n
= 0,we may assume that q = n.Take = (n +k)=n,so that
q = n + k.We are going to use Theorem 2.1 with F = (n + k + 1)L + H,E = L and
g
i
= G
i
.Fix a metric e
for L having nonnegative curvature current,(for example,one
could take = log jG
1
j
2
) and let ='+(n +k +1).Then
p
1@
@ q
p
1@
@ =
p
1@
@'+
p
1@
@ 0:
4
Suppose f 2 H
0
(X;O
X
((n + k + 1)L + H + K
X
)
J
k+1
).By Theorem 2.1 there exist
h
1
;:::;h
p
2 H
0
(X;O
X
((n +k)L +H +K
X
)) such that
P
j
h
j
G
j
= f.Moreover,
Z
X
jhj
2
e
'
jGj
2(n+k)
=
Z
X
jhj
2
e
( )
(jgj
2
e
)
n
1
Z
X
jfj
2
e
(jgj
2
e
)
n+1
=
n +k
n +k 1
Z
X
jfj
2
e
'
jGj
2(n+k+1)
< +1:
Thus h
i
2 J
k
locally (and much more).The proof is complete.
The rest of the results we present here were motivated in part by our desire to say some
thing about the case k = 0 in Theorem 2.2.Of course,Theorem 2.2 as stated is not true
in general if k = 0.But at the end of this section,we will state and prove a result similar
to Theorem 2.2 (namely Theorem 2.8) that does address the question of when sections of
(n +1)L +H +K
X
are expressible in terms of sections of nL +H +K
X
.
The next example again does not make use of twisting.
Example.For any"> 0 and any positive integer q the triple (;F;q) where (x) ="log x
and F(x) = 1 x is a Skoda Triple.Indeed,
0
(x) ="=x and
00
(x) = "=x
2
< 0,while
1 +F
0
(x) = 0 and F
00
(x) 0 0,and thus (1) holds.Furthermore,
= 1;A = 0;B = 1 +
"
q
and
B
(B 1)
= 1 +
q
"
;
and thus
Bq q +";while
B
(B 1)
q +"
"
:
Applying Theorem 1 to the Skoda Triple ("log x;0;q) where q = min(n;p 1) gives us the
following result.
Theorem 2.3.Let the notation be as in Theorem 2.1.Fix"> 0 and assume that jgj
2
e
< 1
on X.Set q = min(n;p 1) and = 1 log(jgj
2
e
).Suppose that
p
1@
@ (q +")
p
1@
@:
Then for every f 2 H
0
(X;F +K
X
) such that
Z
X
jfj
2
e
1+"
(jgj
2
e
)
q+1
< +1;
there exist sections h
1
;:::;h
p
2 (H
0
(X;F E +K
X
) such that
p
X
j=1
h
j
g
j
= f and
Z
X
jhj
2
e
( )
"
(jgj
2
e
)
q
q +"
"
Z
X
jfj
2
e
1+"
(jgj
2
e
)
q+1
:
By allowing twisting,we can improve the estimates of Theorem 2.3,at the cost of a little
more curvature from e
.
5
Example.For any"> 0 and any positive integer q the triple (;F;q) where (x) ="log x
and F(x) 0 is a Skoda Triple.Indeed,x
0
(x) ="and
00
(x) = "=x
2
< 0,while
1 +F
0
(x) = 1 > 0 and F
00
(x) 0 0,and thus (1) holds.Moreover
A(x) =
(x)
"
;B(x) =
q(x) +1 +"
q(x)
and
(B 1)
B
=
(1 +")
q +
1+"
;
and thus since 1,
Bq q +1 +";while
B
(B 1)
q +1 +"
1 +"
:
Applying Theorem 1 to the Skoda Triple ("log x;0;q) where q = min(n;p 1) gives us the
following result.
Theorem 2.4.Let the notation be as in Theorem 2.1.Fix"> 0 and assume that jgj
2
e
< 1
on X.Set q = min(n;p 1) and = 1 log(jgj
2
e
).Suppose that
p
1@
@ (q +1 +")
p
1@
@:
Then for every f 2 H
0
(X;F +K
X
) such that
Z
X
jfj
2
e
"
(jgj
2
e
)
q+1
< +1;
there exist sections h
1
;:::;h
p
2 (H
0
(X;F E +K
X
) such that
p
X
j=1
h
j
g
j
= f and
Z
X
jhj
2
e
( )
"
(jgj
2
e
)
q
q +1 +"
"
Z
X
jfj
2
e
"
(jgj
2
e
)
q+1
:
Remark.Note that if,for example,X is a bounded pseudoconvex domain in C
n
and we
take E = O and 0,then Theorem 2.4 is a strict improvement over Theorem 2.3.Thus
twisting can sometimes get us stronger results.
Though less general,Theorems 2.1,2.3 and 2.4 are aesthetically more pleasing than The
orem 1,because the integrands in the conclusions of the former are more natural.Another
method for obtaining such natural integrands is through the use of the notion of denomina
tors introduced in [2] by McNeal and the author.
Definition 2.5.Let D denote the class of functions R:[1;1)![1;1) with the following
properties.
(D1) Each R 2 D is continuous and increasing.
(D2) For each R 2 D the improper integral
C(R):=
Z
1
1
dt
R(t)
is nite.
For > 0,set
G
(x) =
1
1 +
1 +
C(R)
Z
x
1
dt
R(t)
;
6
and note that this function takes values in (0;1].Let
F
(x):=
Z
x
1
1 G
(y)
G
(y)
dy:
(D3) For each R 2 D there exists a constant > 0 such that
K
(R):= sup
x1
x +F
(x)
R(x)
is nite.
A function R 2 D is called a denominator.
The following key lemma about denominators was proved in [2].
Lemma 2.6.If R 2 D and
R
1
1
dt
R(t)
= 1,then the function F = F
given in Denition 2.5
satises
x +F(x) 1;(2a)
1 +F
0
(x) 1;and(2b)
F
00
(x) < 0:(2c)
Moreover,F satises the ODE
(3) F
00
(x) +
(1 +)R(x)
(1 +F
0
(x))
2
= 0;x 1;
where is a positive number guaranteed by Condition (D3) of Denition 2.5.
Using Lemma 2.6 we can prove the following theorem.
Theorem 2.7.Let X be an essentially Stein manifold of complex dimension n,L;E!X
holomorphic line bundles with singular Hermitian metrics e
and e
respectively,and
g
1
;:::;g
p
2 H
0
(X;E).Let
q = min(p 1;n) and = 1 log(jgj
2
e
):
Let R 2 D with constant > 0 and function F = F
determined by Denition 2.5,and set
B = 1 +
1+F
0
()
q(+F())
.Assume that
p
1@
@ qB
p
1@
@ and jgj
2
e
:=
p
X
j=1
jg
j
j
2
e
< 1:
Then for every section f 2 H
0
(X;L+K
X
) such that
Z
X
jfj
2
e
(jgj
2
e
)
q+1
< +1
there exist sections h
1
;:::;h
p
2 H
0
(X;LE +K
X
) such that
p
X
j=1
h
j
g
j
= f
and
Z
X
jhj
2
e
( )
(jgj
2
e
)
q
R()
(1 +q)
(1 +)
C(R) +K
(R)
Z
X
jfj
2
e
(jgj
2
e
)
q+1
:
7
Remark.For the reader that does not like the appearance of the function B in the statement
of Theorem 2.7,we note that qB is always bounded above by q +2 +.Indeed,note that
qB = q +
1+F
0
.Now,in the notation of Denition 2.5,
(1 +F
0
(x)) =
Z
x
1
dy
G
(y)
1
G
(x)
(1 +)
since,from the denition of G
,
1
G
(1 +) while
1
G
0.Thus
(4) qB q +1 +
1 +
q +2 +:
The reason we did not hypothesize that
p
1@
@ (q +2 +)
p
1@
@ is that often can do
better than the bound (4).It is often easy to estimate
1+F
0
in specic examples.
Proof of Theorem 2.7.Let F be the function associated to the denominator R via Lemma
2.6.Observe that in view of Lemma 2.6,(0;F;q) is a Skoda triple.Let ,A and B be the
functions associated to (0;F;q) in Denition 1.1.We claim that
(5)
+A
R
(1 +)
C(R) +K
(R):
Indeed,=R K
(R) by property (D3),while A=R
(1+)
C(R) by the denition of A and
the ODE of Lemma 2.6.(Note that if R 2 D,then C(R)R 2 D and
R
1
1
(C(R)R(t))
1
dt = 1.)
Finally,observe that
(6)
B
(B 1)
= 1 +
q
(1 +F
0
)
1 +q:
Thus
Z
X
jhj
2
e
( )
(jgj
2
e
)
q
R()
Z
X
+A
R()
jhj
2
e
()
e
( )
( +A)(jgj
2
e
)
q
(1 +)
C(R) +K
(R)
Z
X
B
(B 1)
jfj
2
e
()
e
(jgj
2
e
)
q+1
(1 +q)
(1 +)
C(R) +K
(R)
Z
X
jfj
2
e
(jgj
2
e
)
q+1
:
In going from the second to the third line,we used Theorem 1 and The inequality (5),and
in going from the third line to the last we used (6).The proof is complete.
8
The following is a table of denominators and estimates for their corresponding constants.
(7)
R(x) =
(1+)
C(R) +K
(R)
(i) e
s(x1)
(1+)
2
s
+1
(ii) x
2
(2+)
2
(1+)
2
4
(iii) x
1+s
1+
s
+
s
(1+)s
(1+s)
s+1
(iv) R
N
(x)
(1+)(1+s)
s
In entry (iv),
R
N
(x) = x
N2
Y
j=1
L
j
(x)
!
(L
N1
(x))
1+s
;
where
E
j
= exp
(j)
(1) and L
j
(x) = log
(j)
(E
j
x):
Remark.To dene denominators yielding Skoda triples (;F;q) with both and F non
trivial seems a little more complicated.The corresponding ODE that determines the asso
ciated function does have solutions,but since this (rst order) ODE is not autonomous,it
is harder to get explicit properties of the associated function .
We can now state and prove our Siutype division theorem for the case k = 0,under an
additional assumption on the line bundle L.
Theorem 2.8.Let X be an almost Stein manifold of complex dimension n,L!X a
holomorphic line bundle,H!X a holomorphic line bundle with nonnegatively curved
singular Hermitian metric e
'
.Fix sections G
1
;:::;G
p
2 H
0
(X;L) and a singular Hermitian
metric e
for L having nonnegative curvature,and such that
jGj
2
e
< 1 on X:
Fix a denominator R 2 D such that the associated function B satises
p
1@
@'+(n +1 nB)
p
1@
@ 0:
Dene the multiplier ideals
I
1
= I
e
'
jGj
2(n+1)
and I
0
= I
e
'
jGj
2n
R(1 log jGj
2
+)
:
9
Then
H
0
(X;((n +1)L +H +K
X
)
I
1
) =
p
M
j=1
G
j
H
0
(X;(nL +H +K
X
)
I
0
):
Proof.By taking G
p
=:::G
n
= 0,we may assume that q = n.We are going to use Theorem
2.7 with F = (n +1)L +H,E = L and g
i
= G
i
.Let ='+(n +1).Then
p
1@
@ qB
p
1@
@ =
p
1@
@'+(n +1 qB)
p
1@
@ 0:
Suppose f 2 H
0
(X;O
X
((n +1)L +H +K
X
)
I
1
).Then
Z
X
jfj
2
e
(jgj
2
e
)
n+1
=
Z
X
jfj
2
e
'
jGj
2(n+1)
< +1:
By Theorem 2.7 there exist h
1
;:::;h
p
2 H
0
(X;O
X
(nL +H +K
X
)) such that
P
h
i
G
i
= f.
Moreover,
Z
X
jhj
2
e
'
jGj
2n
R(1 log jGj
2
+)
=
Z
X
jhj
2
e
( )
(jgj
2
e
)
2n
R()
.
Z
X
jfj
2
e
(jgj
2
e
)
n+1
< +1:
Thus h
i
2 I
0
locally.The proof is complete.
3.Hilbert Space Theory
Summation convention.We use the the complex version of Einstein's convention,where
one sums over (i) repeated indices,one upper and one lower,and (ii) an index and its complex
conjugate,provided they are both either upper or lower indices.In addition,we introduce
into our order of operations the rules
ja
i
b
i
j
2
= ja
1
b
1
+:::j
2
;while ja
i
j
2
jb
i
j
2
= ja
1
j
2
jb
1
j
2
+::::
The functional analysis.Let H
0
,H
1
,H
2
and F
1
be Hilbert spaces with inner products
(;)
0
,(;)
1
,(;)
2
and (;)
respectively.Suppose we have a bounded linear operator
T
2
:H
0
!H
2
and closed,densely dened operators T
1
:H
0
!H
1
and S
1
:H
1
!F
1
satisfying
S
1
T
1
= 0:
Let K = Kernel(T
1
).We consider the following problem.
Problem 3.1.Given 2 H
2
,is there an element 2 K such that T
2
= ?If so,what
can we say about jj
0
?
Problem 3.1 was solved by Skoda in [5].The dierence between Skoda's solution and the
one we present here is that Skoda identied the subspace of all for which the problem can
be solved,whereas we aim to solve the problems one at a time.This is a dierence in
presentation only;the two approaches are equivalent.
Proposition 3.2.Let 2 H
2
.Suppose there exists a constant C > 0 such that for all
u 2 T
2
(K ) and all 2 Domain(T
1
),
(8) j(;u)
2
j
2
C
jT
2
u +T
1
j
2
0
+jS
1
j
2
:
Then there exists 2 K such that T
2
= and jj
2
0
C.
10
Proof.In (8) we may restrict our attention to 2 Domain(T
1
)\Kernel(S
1
).Note that (i)
since S
1
T
1
= 0,the image of T
1
agrees with the image of the restriction of T
1
to Kernel(S
1
),
and (ii) the image of T
1
is dense in K
?
.Thus the estimate (8) may be rewritten
(9) j(;u)j
2
Cj[T
2
u]j
2
;
where we denote by [ ] the projection to the quotient space H
0
=K
?
and the norm on the
right hand side is the norm induced on H
0
=K
?
in the usual way.As is well known,with
this norm H
0
=K
?
is isomorphic to the closed subspace K.(The isomorphism sends any,
and thus every,member of u+K
?
to its orthogonal projection onto K.) We dene a linear
functional`:[Image(T
2
)]!C by
`([T
2
u]) = (;u)
2
:
Then by (9)`is continuous with norm
p
C.By extending`constant in the directions
parallel to [Image(T
2
)]
?
in H
0
=K
?
,we may assume that`is dened on all of H
0
=K
?
with norm still bounded by
p
C.The Riesz Representation Theorem then tells us that`
is represented by inner product with respect to some element of H
0
=K
?
which we can
identify with K at this point.Evidently we have jj
2
0
C and
(T
2
;u)
2
= (;T
2
u +K
?
) =`([T
2
u]) = (;u)
2
:
The proof is complete.
Hilbert spaces of sections.Let Y be a Kahler manifold of complex dimension n and
H!Y a holomorphic line bundle equipped with a singular Hermitian metric e
'
.Given a
smooth section f of H +K
Y
!Y,we can dene its L
2
norm
jjfjj
2
'
:=
Z
Y
jfj
2
e
'
:
This norm does not depend on the Kahler metric for Y.Indeed,we think of a section of
H + K
Y
as an Hvalued (n;0)form.Then the functions jfj
2
e
'
transforms like the local
representatives of a measure on Y,and may thus be integrated.
We dene
L
2
(Y;H +K
Y
;e
'
)
to be the Hilbert space completion of the space of smooth sections f of H +K
Y
!Y such
that jjfjj
2
'
< +1.
More generally,we have Hilbert spaces of (0;q)forms with values in H+K
Y
.Given such
a (0;q)form ,dened locally by
=
J
dz
J
;
where J = (j
1
;:::;j
q
) 2 f1;:::;ng
q
is a multiindex and dz
J
= dz
j
1
^:::^ dz
j
q
and the
J
are
skewsymmetric in J,we set
J
= g
j
1
k
1
:::g
j
q
k
q
K
and jj
2
=
J
J
;
where g
j
k
is the inverse matrix of the matrix g
j
k
of the Kahler metric g of Y.It follows that
the functions
jj
2
e
'
11
transform like the local representatives of a measure,and may thus be integrated.We then
dene
jjjj
2
'
=
Z
Y
jj
2
e
'
:
We now dene the Hilbert space
L
2
0;q
(Y;H +K
Y
;e
'
)
to be the Hilbert space closure of the space of smooth (0;q)forms with values in H +K
Y
such that jjjj
2
'
< +1.Of course,these norms depend on the Kahler metric g as soon as
q 1.
We are only going to be (explicitly) interested in the cases q = 0 and q = 1,although
q = 2 will enter in an auxiliary way.
Choices.In employing Proposition 3.2,we shall consider the following spaces.
H
0
:= (L
2
(
;K
X
+F E;e
'
1
))
p
H
1
:= (L
2
(0;1)
(
;K
X
+F E;e
'
1
))
p
H
2
:= L
2
(
;K
X
+F;e
'
2
)
F
1
:= (L
2
(0;2)
(
;K
X
+F E;e
'
1
))
p
Next we dene our operators T
1
and T
2
.Let
T:L
2
(
;K
X
+F E;e
'
1
)!L
2
0;1
(
;K
X
+F E;e
'
1
)
be the densely dened operator whose action on smooth forms with compact support is
Tu =
@u:
As usual,the domain of T consists of those u 2 L
2
(
;K
X
+ F E;e
'
1
) such that
@u,
dened in the sense of currents,is represented by an of L
2
(0;1)
(
;K
X
+ F E;e
'
1
)form
with values in F +K
X
.We let
T
1
:H
0
!H
1
be dened by
T
1
(
1
;:::;
p
) = (T
1
;:::;T
p
):
We remind the reader that T has formal adjoint T
= T
'
1
given by the formula
T
= e
'
1
@
(e
'
1
):
It follows that
T
1
(
1
;:::;
p
) =
e
'
1
@
e
'
1
1
;:::;e
'
1
@
e
'
1
p
:
We will also use the densely dened operators
S:L
2
0;1
(
;K
X
+F E;e
'
1
)!L
2
0;2
(
;K
X
+F E;e
'
1
)
dened by
@ on smooth forms,and the associated operator
S
1
(
1
;:::;
p
) = (S
1
;:::;S
p
):
However,we will not need the formal adjoint of S.
Next we let
T
2
:H
0
!H
2
12
be dened by
T
2
(h
1
;:::;h
p
) = h
i
g
i
:
We have
(T
2
u;h)
0
= (u;T
2
h)
2
=
Z
u
h
i
g
i
e
'
2
=
Z
e
('
2
'
1
)
g
i
u
h
j
e
'
1
;
And thus
T
2
u =
e
('
2
'
1
)
g
1
u;:::;e
('
2
'
1
)
g
p
u
:
Remark.Let us comment on the meaning of this a priori local formula.The sections g
j
,
1 j p,are sections of E,and e
('
2
'
1
)
is a metric for F (F E) = E.Thus for each
j,jg
j
j
2
e
('
2
'
1
)
is a globally dened function,and so the expressions
e
('
2
'
1
)
g
j
= e
('
2
'
1
)
jg
j
j
2
=g
j
transform like sections of E.Since u takes values in K
X
+F,the expressions
e
('
2
'
1
)
g
j
u
transform like sections of K
X
+F E,which is what we expect.
4.Classical L
2
identities and estimates
In this section we collect some known L
2
identities.
The BochnerKodaira Identity.Let
be a domain in a complex manifold with smooth,
Rcodimension1 boundary @
.Fix a proper smooth function on a neighborhood of
such that
= f < 0g and j@j 1 on @
:
Let H!
be a holomorphic line bundle with singular Hermitian metric e
'
.
The following identity is a basic fact known as the
BochnerKodaira Identity:
For any smooth (0;1)form with values in K
X
+H and lying in the domain of
@
,
Z
j e
'
@
(
e
'
)j
2
e
'
+
Z
j
@j
2
e
'
=
Z
(@
@
')e
'
+
Z
j
rj
2
e
'
+
Z
@
(@
@
)e
'
:
Remark.The formal case,in which the boundary termdisappears is due to Kodaira,and is
a complex version of earlier work of Bochner.With the boundary termincluded,the identity
above is due to C.B.Morrey.(For higher degree forms,it is due to Kohn.)
Skoda's Identity.For u 2 T
2
(Kernel T
1
) and = (
1
;:::;
p
) 2 Domain(T
1
) we have
(T
2
u;T
1
)
0
= (T
1
(T
2
)u;)
1
=
Z
u
n
k
@
(g
k
e
('
2
'
1
)
)
o
e
'
1
:
13
It follows that if is also in Domain(S
1
) then
jjT
1
+T
2
ujj
2
0
+jjS
1
jj
2
= jjT
1
jj
2
0
+jjS
1
jj
2
+jjT
2
ujj
2
0
+2Re (T
2
u;T
1
)
0
=
p
X
k=1
jjT
k
jj
2
'
1
+jjS
k
jj
2
'
1
+
Z
e
2('
2
'
1
)
jgj
2
juj
2
e
'
1
+2Re
Z
u
n
k
@
(g
k
e
('
2
'
1
)
)
o
e
'
1
:
By applying the BochnerKodaira identity,we obtain the identity we have called
Skoda's Identity:
jjT
1
+T
2
ujj
2
0
+jjS
1
jj
2
=
Z
e
('
2
'
1
)
jgj
2
juj
2
e
'
2
(10)
+2Re
Z
u
n
k
@
(g
k
e
('
2
'
1
)
)
o
e
'
1
+
Z
(
k
k
@
@
'
1
)e
'
1
+jj
rjj
2
+
Z
@
(
k
k
@
@
)e
'
1
:
Here jj
rjj
2
= jj
r
1
jj
2
'
1
+:::+jj
r
p
jj
2
'
1
.
Skoda's inequality.To obtain an estimate from Skoda's identity,one makes use of the
following inequality of Skoda.
Theorem 4.1 (Skoda's Inequality).[5] Let g = (g
1
;:::;g
p
) be holomorphic functions on a
domain U C
n
,and let q = min(n;p 1).Then
q(
k
k
@
@
log jgj
2
) jgj
2
k
@
(g
k
jgj
2
)
2
Remark.When passing to a global setting,it is helpful to keep in mind that g
k
jgj
2
trans
forms like a section of the antiholomorphic line bundle
E,and thus @
(g
k
jgj
2
))dz
trans
forms like a (
E)valued (1;0)form.In particular,both sides of Skoda's inequality consist
of globally dened functions.
Skoda's Basic Estimate.From Theorem 4.1 and Skoda's Identity (10),we immediately
obtain the following slight extension of a theorem of Skoda.
Theorem 4.2 (Skoda's Basic Estimate).Let X be an essentially Stein manifold,E;F!X
holomorphic line bundle with singular metrics e
and e
respectively,
X a pseudo
convex domain,B:
!(1;1) a function,and g
1
;:::;g
p
2 H
0
(X;E) holomorphic sections.
Set
q = min(n;p 1);'
1
= + +q log(jgj
2
e
) and'
2
='
1
+log jgj
2
:
14
For any ptuple of F Evalued (0;1)forms = (
1
;:::;
p
) 2 Domain(T
1
)\Domain(S
1
)
and any u 2 T
2
(Kernel(T
1
)) we have the estimate
jjT
1
+T
2
ujj
2
0
+jjS
1
jj
2
(11)
Z
B 1
B
juj
2
e
'
2
+
Z
k
k
(@
@
Bq@
@
) e
'
1
+
Z
k
k
@
@
q(B 1)@
@
log(jgj
2
e
)
e
'
1
:
Proof.We are going to use Skoda's Identity (10).First note that,by the CauchySchwartz
Inequality,for any open set U we have
2Re
Z
U
u
n
k
@
(g
k
e
('
2
'
1
)
)
o
e
'
1
= 2Re
Z
U
ujgj
1
n
jgj
k
@
(g
k
jgj
2
)
o
e
'
1
Z
U
1
B
juj
2
jgj
2
e
'
1
Z
U
Bjgj
2
k
@
(g
k
jgj
2
)
2
e
'
1
Z
U
1
B
juj
2
e
'
2
Z
U
Bq(
k
k
@
@
log jgj
2
)e
'
1
;(12)
where the second inequality follows from Skoda's Inequality (Theorem 4.1).Since the inte
grands are globally dened,we may replace U by
.Substituting'
1
= + q log jgj
2
,
combining the inequality (12) with Skoda's Identity (10) and dropping the positive terms
jj
rjj
2
and
Z
@
k
k
@
@
e
'
1
nishes the proof.
5.Twisted versions of L
2
identities and estimates
The twisted BochnerKodaira Identity.Let'be the weight function in the Bochner
KodairaHormander identity.Suppose given a second weight function ,and set = e
'
.
Then
p
1@
@'=
p
1@
@
p
1@
@
+
p
1@ ^
@
2
;
and from the BochnerKodaira identity we have
Z
j e
@
(
e
)
1
@
j
2
e
+
Z
j
@j
2
e
=
Z
@
@
@
@
+
1
(@
)(@
)
e
+
Z
j
rj
2
e
+
Z
@
@
@
e
:
Expanding the rst term,we obtain the socalled
15
Twisted BochnerKodaira Identity:
jj
p
T
jj
2
+jj
p
Sjj
2
=
Z
(@
@
@
@
)
e
+2Re
Z
(@
)
T
e
+
Z
j
rj
2
e
+
Z
@
@
@
e
:
Twisted version of Skoda's Identity.We shall now twist the weights'
1
and'
2
by the
same factor .Let
= e
1
'
1
= e
2
'
2
:
Then
@
@'
1
= @
@
1
@
@
+
@ ^
@
2
;
T
'
1
= e
'
1
@
(e
'
1
) =
e
1
@
(e
1
)
= e
1
@
(e
1
)
@
= T
1
1
@
:
The operator T
2
remains unchanged.We thus calculate that
jjT
1;'
1
+T
2
ujj
2
'
1
= jj
p
T
1;
1
p
1
@
+
p
T
2
ujj
2
1
= jj
p
T
1;
1
+
p
T
2
ujj
2
1
2Re
Z
(
k
@
)
(T
1;
1
+T
2
u)
k
e
1
+
Z
1
k
k
(@
)(@
)e
1
;
that
Z
u
n
k
@
(g
k
e
('
2
'
1
)
)
o
e
'
1
=
Z
u
n
k
@
(g
k
e
(
2
1
)
)
o
e
1
and that
e
'
1
@
@'
1
= e
1
(@
@
1
@
@
1
@ ^
@):
Substitution of these three calculations into Skoda's Identity (10) yields the
Twisted version of Skoda's Identity:
jj
p
T
1;
1
+
p
T
2
ujj
2
1
+jj
p
S
1
jj
2
= 2Re
Z
(
k
@
)
(T
1;
1
+T
2
u)
k
e
1
(13)
+
Z
e
(
2
1
)
jgj
2
juj
2
e
2
+2Re
Z
u
n
k
@
(g
k
e
(
2
1
)
)
o
e
1
+
Z
k
k
(@
@
1
@
@
)e
1
+
p
r
2
+
Z
@
(
k
k
@
@
)e
1
:
Twisted version of Skoda's Basic Estimate.The following Lemma is trivial.
Lemma 5.1.T
2
(Kernel(T
1
)) = (T
2
p
+A)(Kernel(T
1
p
+A)):
Next we have the following result.
16
Theorem 5.2 (Twisted version of Skoda's Basic Estimate).Let X be an essentially Stein
manifold,E;F!X holomorphic line bundle with singular metrics e
and e
respectively,
X a smoothly bounded pseudoconvex domain,;A:
!(0;1) and B:
!(1;1)
functions,and g
1
;:::;g
p
2 H
0
(X;E) holomorphic sections.Set
q = min(n;p 1);
1
= + +q log(jgj
2
e
) and
2
=
1
+log jgj
2
:
For any ptuple of F Evalued (0;1)forms = (
1
;:::;
p
) 2 Domain(T
1
)\Domain(S
1
)
and any u 2 (T
2
p
+A)(Kernel(T
1
p
+A)) we have the estimate
jj
p
+AT
1
+
p
+AT
2
ujj
2
0
+jj
p
S
1
jj
2
(14)
Z
B 1
B
juj
2
e
2
+
Z
k
k
(@
@
Bq@
@
) e
1
+
Z
k
k
@
@
@
@
(@
)(@
)
A
e
1
:
+
Z
k
k
q(B 1)@
@
log(jgj
2
e
)
e
1
:
Proof.First,by Lemma 5.1 we may make use of (13).
By the pseudoconvexity of
,we may drop the last term on the right hand side of (13).
The second last term is clearly nonnegative and may thus also be dropped.
Since
2
1
= log jgj
2
,we see that
2Re
Z
u
n
k
@
(g
k
e
(
2
1
)
)
o
e
1
Z
B
juj
2
e
2
Z
Bq
k
k
(@
@
log jgj
2
)e
1
:
Moreover,by the CauchySchwarz inequality we have
2Re
Z
(
k
@
)
(T
1
+T
2
u)
k
e
1
Z
k
k
(@
)(@
)
A
e
1
Z
AjT
1
+T
2
uj
2
e
1
:
Applying these inequalities to (13) easily yields (14).
6.Proof of Theorem 1
We now plan to use Proposition 3.2 to prove Theorem 1.
An a priori estimate.Assume that
fjgj
2
e
< 1g:
With q = min(p 1;n),x a Skoda Triple (;F;q),and let
:= 1 log(jgj
2
e
):
Following Denition 1.1,let
:= () = +F() and A:= A() =
(F
00
() +
00
())
(1 +F
0
())
2
1
:
17
Let
= ():
Then if is nonconstant,we have
p
1@
@
p
1@
@
p
1
A
@ ^
@
= (
0
() +1 +F
0
())(
p
1@
@)
F
00
() +
00
() +
(1+F
0
())
2
A
p
1@ ^
@
= (
0
() +1 +F
0
())(
p
1@
@):
The last equality follows from the denition of A.On the other hand,if is constant,
we need not apply the twisted Skoda estimate;we can just appeal to the original Skoda
estimate,which would yield (by convention)
p
1@
@
p
1@
@
p
1
A
@ ^
@ =
0
()(
p
1@
@)
00
()
p
1@ ^
@
(
0
() +1 +F
0
())(
p
1@
@)
Now,@
@ = @
@ log(jgj
2
e
).Thus from (14) we obtain the a priori estimate
jj
p
+AT
1
+
p
+AT
2
ujj
2
1
+jj
p
S
1
jj
2
(15)
Z
B 1
B
juj
2
e
2
+
Z
k
k
(@
@
Bq@
@
)e
1
+
Z
(q(B 1) +(
0
() +1 +F
0
()))
k
k
(@
@
log(jgj
2
e
))e
1
:
From the denition of B (See Denition 1.1) we have
q(B 1) =
0
() +1 +F
0
():
Then
B =
q +
0
() +1 +F
0
())
q
;
and thus
(B 1)
B
=
(
0
() +1 +F
0
())
q +
0
() +1 +F
0
()
:
Thus we obtain from (15) the estimate
jj
p
+AT
1
+
p
+AT
2
ujj
2
1
+jj
p
S
1
jj
2
(16)
Z
(B 1)
B
juj
2
e
2
+
Z
k
k
(@
@
Bq@
@
)e
1
:
Conclusion of the proof of Theorem 1.Now suppose that
p
1@
@ Bq
p
1@
@:
It follows from (16) that
j(f;u)
1
j
2
Z
B
(B 1)
jfj
2
e
()
e
jgj
2q+2
Z
(B 1)
B
juj
2
e
2
18
In view of Proposition 3.2,we nd sections H
1
;:::;H
p
such that
@(
p
+AH
i
) = 0;(g
i
H
i
)
p
+A = f
and
Z
jHj
2
e
()
e
jgj
2q
Z
B
(B 1)
jfj
2
e
()
e
jgj
2q+2
:
Letting h
i
=
p
+AH
i
,we obtain
Z
jhj
2
e
()
e
( +A)jgj
2q
Z
B
(B 1)
jfj
2
e
()
e
jgj
2q+2
:
Since the estimates are uniform,we may let
!XV.Since V has measure zero,we may
replace X V by X.This completes the proof of Theorem 1.
Acknowledgment.I am indebted to Je McNeal,from whom I have learned a lot about
the method of twisted estimates and with whom I developed the theory of denominators.I
am grateful to YumTong Siu for proposing that I study Skoda's Theorem and try to say
something nontrivial about the case = 1,which I hope I have done at least somewhat.
References
[1] Ein,L.,Lazarsfeld,R.,A geometric eective Nullstellensatz.Invent.Math.137 (1999),no.2,427{448.
[2] McNeal,J.,Varolin,D.,Analytic Inversion of Adjunction.L
2
extension theorems with gain.Ann.Inst.
Fourier (Grenoble) 57 (2007),no.3,703{718.
[3] Siu,Y.T.,Extension of Twisted Pluricanonical Sections with Plurisubharmonic Weight and Invariance
of Semipositively Twisted Plurigenera for Manifolds Not Necessarily of General Type.Complex geometry
(Gottingen,2000),223{277,Springer,Berlin,2002.
[4] Siu,Y.T.,Multiplier Ideal Sheaves in Complex and Algebraic Geometry.Sci.China Ser.A 48 (2005),
suppl.,1{31.
[5] Skoda,H.,Application des techniques L
2
la theorie des ideaux d'une algebre de fonctions holomorphes
avec poids.Ann.Sci.
Ecole Norm.Sup.(4) 5 (1972),545{579.
Department of Mathematics
Stony Brook University
Stony Brook,NY 117943651
Email address:dror@math.sunysb.edu
19
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