A geometrical proof of Shiota's Theorem on a Conjecture of S.P. Novikov

A geometrical proof of Shiota’s theorem on a

conjecture of S. P. Novikov

GIAMBATTISTA MARINI

Dipartimento di Matematica, II Universit `a degli Studi di Roma ‘Tor Vergata’, Via della Ricerca Scientifica – 00133 Roma, Italy; e-mail: marini@axp.mat.utovrm.it

Received 1 March 1995; accepted in final form 6 February 1997

Abstract. We give a new proof of Shiota’s theorem on Novikov’s conjecture, which states that the K.P.

equation characterizes Jacobians among all indecomposable principally polarized abelian varieties.

Mathematics Subject Classifications (1991): 14K25, 14H40.

Key words: characterization of Jacobians, K.P. equation, K.P. hierarchy.

The Kummer variety of a Jacobian has a 4-parameter family of trisecants. Using Riemann’s relations, Fay’s identity and limit considerations, this property has been translated in a hierarchy of non-linear partial differential equations which is satisfied by the theta function of a Jacobian (see [F], [Mu], [Du], [Kr], [AD3]).

Novikov’s conjecture stated that if a theta function associated with an indecom-posable principally polarized abelian variety(

X;

[])satisfies the K.P. equation,

the first equation of the hierarchy, then (

X;

[]) is the Jacobian of a complete

irreducible smooth curve. Shiota originally proved the conjecture in [S] by the use of hard techniques from the theory of non-linear partial differential equations. His proof was later simplified by Arbarello and De Concini (see [AD2]). We give a proof of the theorem which is more geometrical in character; in particular we avoid a technical point, namely Shiota’s Lemma 7, which is instrumental in both Shiota’s and Arbarello–De Concini’s proofs. For our proof, we follow Arbarello and De Concini algebro–geometrical attempt to solve the problem (see [A] and [AD3]) and we go further. First, let us recall that in order to prove Novikov’s conjecture, it suffices to recover the whole K.P. hierarchy from its first equation (because of Wel-ters’ version of Gunning’s criterion). The key point in Arbarello and De Concini geometrical approach is that, no matter what are the parameters in the equations in the K.P. hierarchy, it turns out that the terms to be equated to zero form a sequence of sections of the line bundle O(2)

:

One needs to find parameters that make

this sequence into the identically zero sequence. The difficulty comes from the fact that the theta divisor may have, a priori, a difficult geometry. The key object in the approach in [A] and [AD3] is the subscheme

D

1ofdefined by the zeroes of

the section ofO 

() associated with

D

1

;

where

D

1is an invariant vector field

theta function associated withO

X

()

:

As it was pointed out in [A] and [AD3],

the reduced components of

D

1do not create much trouble. They provide a

geo-metrical proof of the conjecture under the additional hypotheses that the singular locus of the theta divisor has codimension at least 2, and that the scheme

D

1

does not contain components which are invariant under the

D

1-flow. We remove Arbarello–De Concini’s additional hypotheses by proving the following. If the K.P. equation holds, the components of codimension one of the singular locus of the theta divisor are invariant under the

D

1-flow (for this we make use of a result of Koll´ar about the singularities of the theta divisor). Therefore every component of

D

1which creates trouble is

D

1-invariant and, in particular, it contains a translate

of an abelian subvariety of

X:

We then prove that the theta function of an abelian variety which contain an abelian subvariety as above, is not a solution of the K.P. equation. For this we combine an algebraic computation which was discovered by Shiota (namely his Lemmas A and B, which we restate and reprove for the convenience of the reader), and a technical lemma on the obstructions to recover the K.P. hierarchy (Lemma 3.11).

For the discrete analogue to Novikov’s conjecture see [De]. For further discus-sions see [AD3], [Do], [GG], [Ma].

1. Introduction

Let

C

be a smooth complex curve,

J

(

C

)its Jacobian, Pic

d

(

C

)the Picard group of

line bundles of degree

d

on

C

and,the image of

C

via the Abel–Jacobi embedding

associated with an element of Pic,1 (

C

).

Let (

X;

[]) be an i.p.p.a.v. (indecomposable, principally polarized, abelian

variety) of dimension

n

, and letbe a symmetric representative of the polarization.

We shall denote by



a theta function associated with O

X

(); in particular,



is

naturally a nonzero section ofO

X

().

The image of the morphism

K

:

X

!j2j 

associated with the base-point-free linear systemj2jis a projective variety which

is called the Kummer variety of(

X;

[])

:

The Kummer variety of

J

(

C

) has a rich geometry in terms of trisecants and

flexes which is a consequence of the equality

W

0

g

,1 \(

W

0

g

,1 +

p

,

q

)=(

W

1

g

,

q

)[(

W

0

g

,2 +

p

) 8

p;q

2

C; p

6=

q;

where

W

rd

=fj

D

j2Pic

d

(

C

)jdimj

D

j>

r

g

:

Indeed, the inclusion 



\





[



+

,



;

8

;; ;

2,

; 

6=

;

(where

p

:=+

p

)

;

the linear dependence of the sections



(

z

,

)



(

z

,



,



+



)

;

and the collinearity in the projective spacej2j  of the points

K

(



+



)

;

K

(



+



)

;

K

(



+

)

;

8

;; ;

2,

;

8



2 1 2(



,



,



,

)

;

are all different translations (via Abel and Riemann’s theorems) of the previous equality. In particular, once distinct points

;;

are fixed, one has a family of trisecants parametrized by 1₂,

:

Considering the limit situation where



and

tend

to



one obtain a family of flexes parametrized by 1₂,

:

This property has been used to characterize Jacobians among all principally polarized abelian varieties (see [G], [W]). Welters’ improvement of Gunning’s theorem states that an i.p.p.a.v. (

X;

) is a Jacobian if and only if there exists

an Artinian subscheme

Y

of

X

of length 3, such that the algebraic subset

V

= f2



j



+

Y



K

,1

(

l

) for some line

l

 j2j 

ghas positive dimension at some

point (if this is the case it turns out that

V

is isomorphic to the curve

C

)

:

In

particular one has:

PROPOSITION 1.1 [AD1]. Let(

X;

[])be an i.p.p.a.v.. The following statements are equivalent:

(a) the i.p.p.a.v.(

X;

[])is isomorphic to the Jacobian of a curve;

(b) there exist invariant vector fields

D

16=0

;D

2

;:::;

on

X

such that

dimf



2

X

j

K

(



)^

D

1

K

(



)^(

D

2

1+

D

2)

K

(



)=0g>1;

(b0

) there exist invariant vector fields

D

1 6= 0

;D

2

;:::;

on

X

and constants

d

4

;d

5

;:::;

such that

P

m



(

z

):= "

m

D

1,

m

,1 (

D

2+

D

2 1)+

m

X

i

=3

d

i

+1

m

,

i

# [



(

z

+



)



(

z

,



)]j



=0 =0

;

for all

m

>3

;

where the

D

i

operate on the variable

;

and the

j

are defined by

j

= X

i

1+2

i

2+


+

ji

j =

j

1

i

1!

i

2!

i

j

!

D

i

1 1

D

i

j

j

:

In this case, the image curve,is, up to translation, the curve whose parametric

expression is

"

7! 1 X

i

=1

"

i

2

D

i

;

where

"

2C

;

and each

D

i

is viewed as a point of the universal cover of

X

via its

natural identification with

T

0(

X

)

:

2. Shiota’s theorem

First, we observe that

P

3(

D

1

;D

2

;D

3;

d

)



= [, 1 3

D

4 1,

D

2 2+

D

1

D

3+

d

][



(

z

+



)



(

z

,



)]j



=0 = , 2 3

D

4 1





+ 8 3

D

3 1



D

1



,2

D

2 1



D

2 1



+2

D

2



D

2



,2

D

2 2





+2

D

1

D

3





,2

D

3



D

1



+

d

:

(2.0)

THEOREM 2.1 (Shiota [S], conjectured by Novikov). The first non-trivial equation

of the K.P. hierarchy characterizes Jacobians: an i.p.p.a.v.(

X;

[])is a Jacobian if and only if there exist invariant vector fields

D

1 6=0

;D

2

;D

3and a constant

d

such that

P

3(

D

1

;D

2

;D

3;

d

)



=0

:

As we already mentioned, our proof consists in recovering the vanishing of the whole K.P. hierarchy from the equation

P

3



=0

;

i.e. in recovering the curve,from

its third order approximation. We observe that

P

i

(

:::

)



is a section ofO

X

(2)

;

for

all

D

1

;:::;D

i

and

d

4

;:::;d

i

+1

:

Indeed, if

Dis any differential operator, because

of Riemann’s quadratic identity, we have that

D[



(

z

+



)



(

z

,



)]j



=0 = X



2

Z

g

=

2

Z

g D





(0)





(

z

)2

H

0 (

X;

2)

;

where f





g is the basis of

H

0

(

X;

O(2)) having the property that Riemann’s

identity



(

z

+



)



(

z

,



)=







(

z

)





(



)holds. Assuming by induction that

there exist invariant vector fields

D

1

;:::;D

m

,1 and constants

d

4

;:::;d

m

such

that

P

i

(

D

1

;:::;D

i

;

d

4

;:::;d

i

+1

)



=0

;

8

i

6

m

,1

;

one needs to find

D

m

and

d

m

+1such that

P

m

(

:::

)



=0

:

We recall that the vector space

H

0(

;

O()j 

)is the vector space of derivatives

T;

with

T

2

T

0(

X

)

:

We denote by

D

the scheme associated with the section

D

2

H

0

(

;

O()j 

)

;

i.e.

D

 = \f

D

= 0g

:

We shall use the following

remark.

REMARK 2.2 [AD3] (private communication from G. Welters to E. Arbarel-lo). Whenever a section

S

2

H

0

(

X;

O(2)) vanishes on

D



;

there exists an

invariant vector field

E

and a constant

d

such that

S

+

ED



,

E

D

+

d



=02

H

0

As a consequence of this remark , Shiota’s theorem can be stated as follows. THEOREM 2.3. An i.p.p.a.v. (

X;

[]) is a Jacobian if and only if there exist invariant vector fields

D

1 6=0 and

D

2such that

P

3(

D

1

;D

2

;

0; 0)



j

D

1 =0

:

REMARK 2.4. We work with the K.P. differential equation for a theta function, which is an automorphic form



associated with the polarization. If



(

z

)and

~



(

z

)

are automorphic forms associated with the same polarization, there exists a point

z

0in

V;

where

V

is the universal cover of the abelian variety

X;

and a nowhere-vanishing holomorphic function

g

(

z

)on

V;

such that

~



(

z

+

z

0)=

g

(

z

)



(

z

)

:

One

might have

P

3



=0 and

P

3 ~



6=0 but, since

P

3(

g



)=

g

2

P

3



+



2

P

3

g

,

d

g

2



2 ,8(

D

2 1

g

g

,

D

1

g

D

1

g

)
(

D

2 1





,

D

1



D

1



)

;

one has

P

3 ~



j

D

1 =

g

2

P

3



j

D

1

(so that formulation 2.3 of Shiota’s theorem is independent of the theta function representing the polarization). In view of Remark 2.2, there exist

D

1

;D

2

;D

3

;d

such that

P

3(

D

1

;D

2

;D

3;

d

)



=0 if and only if there exist

D

1

;D

2

;

~

D

3

;

d

~such that

P

3(

D

1

;D

2

;

~

D

3;

d

~ ) ~



=0

:

TWO FORMULAS 2.5. We have the general formulas (they can be proved by a direct computation)

P

s

+

s

,3 X

i

=1

i

P

s

,

i

!



=(

D

2 1



,

D

2



)
(, ~

s

,1



) +



"

D

1~

s

,(

D

2 1+

D

2) ~

s

,1 +

s

X

i

=3

d

i

+1 ~

s

,

i

#



+

D

1



(, ~

s

+2

D

1 ~

s

,1 )

;

P

s

+

s

,3 X

i

=1
,

i

P

s

,

i

!



=(

D

2 1



+

D

2



)
(, ~
,

s

,1



) +



" ,

D

1 ~
,

s

,(

D

2 1,

D

2) ~
,

s

,1 +

s

X

i

=3

d

i

+1 ~
,

s

,

i

#



,

D

1



(, ~
,

s

,2

D

1 ~
,

s

,1 )

;

where
,

i

(

D

1

;:::;D

i

) =

i

(,

D

1

;:::;

,

D

i

)

;

~

i

(

D

1

;:::;D

i

) =

i

(2

D

1

;:::;

2

D

i

)

;

~
,

i

(

D

1

;:::;D

i

)=

i

(,2

D

1

;:::;

,2

D

i

)

:

REMARK 2.6 [AD3]. The restriction

P

m



j

D

1does not depend on

D

m

;d

m

+1

:

In

fact

P

m

(

D

1

;:::;D

m

;

d

4

;:::;d

m

+1 )



=

P

m

(

D

1

;:::;D

m

,1

;

0;

d

4

;:::;d

m

;

0 )



+2

D

m

D

1





,2

D

m



D

1



+

d

m

+1



2

:

This equality leads to a crucial point of Arbarello–De Concini’s argument: by Remark 2.2, there exist a

D

m

and a

d

m

+1 which make

P

m



equal to zero if and

only if

P

m



vanishes on

D

1

:

From the formulas in 2.5 and the previous remark, assuming by induction that

P

i



=0 for

i < m;

it follows that the only obstruction to find a

D

m

and a

d

m

+1

which make

P

m



equal to zero is given by those components of

D

1where neither (

D

2

1+

D

2)



nor(

D

2

1,

D

2)



vanish. Since

P

3



equals(

D

2

1+

D

2)



(

D

2

1,

D

2)

;

mod(

;D

1



)

;

and since, by hypothesis,

P

3



=0

;

we have that(

D

2

1+

D

2)



(

D

2 1,

D

2)



vanishes on

D

1

:

Therefore a component of

D

1where neither(

D

2

1+

D

2)



nor(

D

2

1,

D

2)



vanish must be non-reduced.

In the next section we shall deal with such components. We show that ifW is

a component of

D

1then, assuming by induction that

P

3



=


=

P

m

,1



=0

;

only two cases may occur: either

P

m



vanishes on W

;

or the reduced scheme

underlyingW

;

denoted byWred

;

is invariant under theh

D

1

;D

2i-flow. Moreover,

ifis singular alongWredthen the second case occur (Theorems 3.1 and 3.2).

3. Theh

D

1

;D

2i-invariance

To begin, we observe that we can always assume

D

2 6= 0

;

as well as

D

3 6= 0

:

Indeed, for all complex numbers

b;

we have

P

3(

D

1

;D

2

;D

3;

d

4)=

P

3(

D

1

;D

2+

bD

1

;D

3+2

bD

2+

b

2

D

1;

d

4)

:

(3.0)

LetW be a component of

D

1

:

We assume first thatis smooth at a generic

point ofWred

:

We prove the following.

THEOREM 3.1. Let (

X;

[]) be an i.p.p.a.v. of dimension

n

and assume that

P

3



=


=

P

m

,1



=0

;

where

m

> 4

:

LetW be a component of the scheme

D

1and assume that is non-singular at a generic point ofWred

:

Then either

P

m



vanishes onW

;

orWredis invariant under theh

D

1

;D

2i-flow.

Proof. Let

p

be a generic point of Wred

:

If W is reduced,

P

m



vanishes on W

:

Assume thatWis non-reduced. Since

p

is a smooth point of

;

there exist an

irreducible element

h

2O

X;p

;

an integer

a

>2

;

integers

b;c;

invertible elements

"

2

;"

3 2O

X;p

and elements

g

1

;g

2

;g

3 2O

X;p

such that the ideal ofW inO

X;p

is

of the form(

h

a

;

)

;

and moreover

D

1



=

h

a

+

g

1



D

2



=

"

2

h

b

+

g

2



D

3



=

"

3

h

c

+

g

3



D

2 1



=

a

h

a

,1

D

1

h

+

g

1

h

a

+[

g

2 1+

D

1

g

1]

:

(3.1.1)

We have

b

> 1

;

because

P

3

;

hence(

D

2

1



+

D

2



)
(

D

2

1



,

D

2



)

;

vanishes on Wred

:

If

h

does not divide

D

1

h;

we prove as in [A] that

P

m



vanishes onW: by

substituting the formulas above in the expression of

P

3



and

D

1

P

3

;

one sees that

a

has to equal 2 and by substituting in the expression of

P

m

,1



(which is zero by

inductive hypothesis), one sees that

m

,1



belongs to (

h;

); hence

P

m



2(

h

2

;

)

;

that is

P

m



j W =0

:

If

h

divides

D

1

h;

the varietyWred is invariant under the

D

1-flow. Under this

assumption, theh

D

1

;D

2i-invariance ofWredis a consequence of Lemma 3.5 and

Lemma 3.8 below. 2

Let us now turn to the case

dimsing=

n

,2

;

is singular alongWred

:

(During the revision of the manuscript the preprint by Ein and Lazrsfeld [EL] appeared proving that the casesing =

n

,2 does not actually occur. Therefore,

Theorem 3.2, Lemma 3.6 and Lemma 3.7 below are no longer strictly necessary for the present proof). We want to prove the following.

THEOREM 3.2. Let(

X;

[])be an i.p.p.a.v. of dimension

n:

Suppose the divisor is singular along a reduced subvariety

Z

of codimension 1, and assume that the K.P. equation

P

3



=0 holds. Then

Z

is invariant under theh

D

1

;D

2i-flow.

This theorem is consequence of Lemma 3.5, Lemma 3.7 and Lemma 3.8 below; it will be proved later.

REMARK 3.3. We will make a strong use of the fact that

Z

has codimension 2 in

X:

It is clearly in general false that, if the K.P. equation holds,is

D

1-invariant in

its singular points.

In view of the following general fact proved by J. Koll´ar in [Ko] the theta divisor cannot be ‘too singular’ along

Z:

irreducible hypersurface

Z;

it has a local normal crossing singularity at a generic point of

Z:

LEMMA 3.5. Let (

X;

[]) be an i.p.p.a.v. of dimension

n;

let

Z

be a reduced subvariety ofof dimension

n

,2

;

and let

D

be an invariant vector field on

X:

If is

D

-invariant along

Z;

then

Z

is

D

-invariant.

Proof. If

Z

were not

D

-invariant, the

D

-span of

Z

would be contained in

:

This span would have dimension

n

,1

;

therefore it would be a

D

-invariant

com-ponent of 

:

This is impossible because of the ampleness and the irreducibility

of

:

2

LEMMA 3.6. Suppose the divisor  is singular along

Z;

and assume that the K.P. equation

P

3



=0 holds. Let

p

be a smooth point of

Z

and

T

p

(

Z

)the tangent space to

Z

at

p:

Then

D

1

;D

2

;T

p

(

Z

)are not in general position, i.e.

dim(h

D

1

;D

2

;T

p

(

Z

)i)6

n

,1

:

Proof. Since is singular along

Z;

we have that



j

Z

=

D

1



j

Z

=

D

2



j

Z

=

D

3



j

Z

=0

:

It follows that

P

3



j

Z

=(

D

2 1



) 2 j

Z

;

therefore

D

2 1



j

Z

=0

:

By 2.0 we get

D

2

P

3



j

Z

=[ 8 3

D

1

D

2



D

3 1



]j

Z

and

D

2 1

P

3



j

Z

=[ ,4 3 (

D

3 1



) 2 +4(

D

1

D

2



) 2 ]j

Z

:

Since

P

3



is zero,

D

2

P

3



and

D

12

P

3



are also zero, and therefore we obtain

D

1

D

2



j

Z

=

D

3

1



j

Z

=0

:

(3.6.1)

We now proceed by contradiction. Suppose there exists

p

02

Z

smoothsuch that

h

D

1

;D

2

;T

p

0(

Z

)i=

T

p

0(

X

)

;

then the same equality must hold for every

p

in a neighborhood

U

of

p

0in

Z:

Let

p

2

U:

For every

E

2

T

p

(

X

)

;

there exist

;

such that

E

=

S

+

D

1 +

D

2

;

where

S

2

T

p

(

Z

)

:

As

D

1



j

Z

= 0 and

S

2

T

p

(

Z

) we have

ED

1



(

p

) = (

S

+

D

1+

D

2)

D

1



(

p

) = 0

:

Therefore

ED

1



j

Z

= 0

;

for every

E

2

T

0(

X

)

:

The assumption thath

D

1

;D

2

;T

p

(

Z

)i=

T

p

(

X

)implies that

D

162

T

p

(

Z

)

:

By

The-orem 3.4, the tangent cone toat

p

is a pair of distinct hyperplanes whose

inter-section is

T

p

(

Z

)

:

Therefore, for a generic

E

2

T

p

(

X

)

;

we have that

ED

1



j

Z

6=0

:

This is a contradiction. 2

LEMMA 3.7. Suppose the divisor  is singular along

Z;

and assume that the K.P. equation

P

3



=0 holds. The divisoris

D

1-invariant at each point of

Z:

Proof. From the previous lemma, there exist functions



and



on

Z

smooth not simultaneously vanishing and such that

for all

p

in

Z

smooth

:

If



0 then

Z

is

D

1-invariant. Assume



60; by induction

on



+



, we prove that

D



1

D



2



j

Z

=0

;

for all integers

;:

Let us assume that

D



1

D



2



j

Z

=0

;

for all



+



6



0

:

We need only to show that

D



0+1

1



vanishes on

Z:

In fact, since



(

p

)

D

1+



(

p

)

D

2 is in

T

p

(

Z

)

;

and since



is not identically

zero, the vector

D

2 is a combination of

D

1 and a vector in

T

p

(

Z

)

;

for

p

generic

in

Z

; as

D

₁





vanishes on

Z

for all



6



0+1

;

we have that

D



1

D



2



j

Z

=0

;

for

all



+



6



0+1

:

By 3.6.1,

D

3

1



j

Z

=0; hence we are done if



0 62

:

Assume



0 >3

:

We distinguish two cases:

(a)

D

32h

D

1

;D

2

;T

p

(

Z

)i

;

for all

p

in

Z

;

(b)

D

362h

D

1

;D

2

;T

p

(

Z

)i

;

for

p

generic in

Z:

Let us start with (a). Since

D

3is a combination of

D

1

;D

2and a vector in

T

p

(

Z

)

;

it follows that

D



₁

D

₂



D

₃



j

Z

=0

;

for



+



+

6



0

:

Therefore, the only nonzero

terms in the restriction to

Z

of a derivative of

P

3



are products of derivatives of



of order at least



0+1; as

P

3



=, 2 3

D

14





+ 8 3

D

13



D

1



,2

D

2 1



D

2 1



+‘lower order

terms’ we obtain that the only nonzero term of

D

2



0,2

1

P

3



j

Z

is

D



0+1 1



D



0+1 1

;

with coefficient ,2 , 2



0,2



0,1
+ 8 3 , 2



0,2



0,2
, 2 3 , 2



0,2



0,3

(which is easily seen to be nonzero). Therefore, as

D

2



0,2

1

P

3



j

Z

=0

;

we must have

D



0+1

1



j

Z

=0

:

Let us deal with case (b). Since

D



₁

D



₂



j

Z

=0 for all



+



636



0

;

we have

0=

D

4 1

P

3



j

Z

=(,2

D

4 1



D

4 1



,6

D

4 1



D

1

D

3



)j

Z

;

0=

D

1

D

3

P

3



j

Z

=(2

D

4 1



D

1

D

3



)j

Z

:

It follows that

D

4

1



j

Z

= 0

;

and we may assume



0 > 4

:

We want to compute

D



0+1

1

P

3



j

Z

:

Since any term of

P

3



is a product of derivatives of



of order

i

and

j;

where

i

+

j

64

;

any term of

D



0+1

1

P

3



j

Z

is a product of derivatives of



of order

i

and

j;

where

i

+

j

6



0+5

<

2



0+2

:

Thus, since

D



1

D

2





j

Z

=0 for all



+



6



0

;

any contribution to the restriction to

Z

of

D



0+1

1

P

3



must involve a

D

3; therefore, by 2.0,

D



0+1 1

P

3



j

Z

=

D



0+1 1 [2

D

1

D

3





,2

D

3



D

1



]j

Z

=[,2(



0+1)+2]

D



0+1 1



D

1

D

3



j

Z

;

where the last equality follows because

D

3



j

Z

=0

;D



1



j

Z

=0 for all



6



0

:

Hence, if

D

1

D

3



j

Z

6 0

;

then

D



0+1

1



j

Z

= 0 and we are done. It only

remains to consider the case where

D

1

D

3



j

Z

=0

:

If

D

1is in

T

p

(

Z

)for

p

generic in

Z;

the variety

Z

is

D

1-invariant,is

D

1-invariant along

Z;

and we are done; so we

assume that, for

p

generic in

Z;

the vector

D

1is not in

T

p

(

Z

)

:

Then, for dimensional

reasons,

T

p

(

X

)=h

D

1

;D

2

;D

3

;T

p

(

Z

)i

:

Since

D

2

1

;D

1

D

2



and

D

1

D

3



all vanish on

Z;

we have

D

1

E

j

Z

=0 for all

E

2

T

0(

X

)

:

By Theorem 3.4, the tangent cone

toat

p

is a pair of distinct hyperplanes. Therefore, for a generic

E

2

T

p

(

X

)

;

we

LEMMA 3.8 (Shiota [S], Lemma A, p. 359). Let



be a solution of the equa-tion

P

3



= 0 in a neighborhood of a point

p

0 in C

n

:

If

D



1



(

p

0) = 0 for all integers

;

then

D

₁



D

₂





(

p

0)=0 for all integers



and

:

Proof. Let us denote by

L

1the (local)

D

1-integral complex line through

p

0

:

By hypothesis,

D



₁



(

p

0) =0

;

for all

;

thus



j

L

1 =0

:

We proceed by contradiction,

i.e. we assume that there exists

b >

0 such that

D

₂

b



j

L

160

:

Let



= minf



j

D



2

D

3



j

L

1 60g

;

c

= minf

j



=0g

;

w

= minf



+2

g

;



= maxf

j



+2

=

w

g

;

(3.8.1)

where



and

c

are allowed to be infinite. Note that

w

6



0 6

b <

1

; 

6

1 2

w <

1

; w

=





+2



and

w

6



+2

for all

:

As



j

L

1  0 we have



0

>

0 and

c

>1

:

Moreover

D



1

D



2

D

3



j

L

1=0

;

for all

; < 

:

It follows that

if



+2

< w;

then

D



1

D

2



D

3



j

L

1 =0

;

if

> 

and



+2

6

w;

then

D



1

D



2

D

3



j

L

1=0

:

(3.8.2)

First, we prove that



=

c

(in particular

c <

1)

:

It is clear that



6

c:

If

 < c

then





> 1

;

thus 2





, 2 > 0

:

Let us set

A

0 =

d

4



; A

1 = , 2 3

D

4 1





+ 8 3

D

3 1



D

1



,2

D

2 1



D

2 1

; A

2 = ,2

D

2 2





+2

D

2



D

2

;

A

3=2

D

1

D

3





,2

D

3



D

1



; so that

P

3



=

A

0+

A

1+

A

2+

A

3

:

By 3.8.2 we have

D

2



 ,2 2

D

23



[

A

0]j

L

1 =

D

2



 ,2 2

D

23



[

A

1]j

L

1 =

D

2



,2 2

D

32



[

A

3]j

L

1 = 0

:

Therefore 0 =

D

2



 ,2 2

D

23



P

3



j

L

1 =

D

2



 ,2 2

D

23



[

A

2]j

L

1 = , 2





h 2 , 2



 ,2



 ,1
,2 , 2



 ,2



 ,2
i
(

D



 2

D

3





) 2 j

L

1

;

where the last equality follows by 3.8.2; this is a contradiction

because

D





2

D

3





j

L

1 60

:

Note that



=

c

implies

w

=2

c

and



>

w

,2

=

2

c

,2

>2

;

for all

6

c

,1

:

Let

~

w

= minf



+2

j

< c

g

;

0 = maxf

j

< c;

+2

=

w

~g

:

(3.8.3)

Note that, as

c

> 1 we have

w

~ 6



0

<

1

:

Thus,

w

~ =



0 +2

0

:

Moreover, as

0

< c;

we have



0 >2

:

We want to compute

D



0,2 2

D

c

+

0 3

P

3



j

L

1

:

By 3.8.3 we have if

< c

and



+2

<

w;

~ then

D



1

D



2

D

3



j

L

1 =0

;

if

0

< < c

and



+2

6

w;

~ then

D



1

D



2

D

3



j

L

1=0

:

(3.8.4)

By 3.8.2 and 3.8.4 we get

D



0,2 2

D

c

+

0 3 [

A

0+

A

1]j

L

1 =0

; D



0,2 2

D

c

+

0 3 [

A

2]j

L

1 = ,2 ,

c

+

0

c

(

D



0 2

D

0 3



)
(

D

c

3



)j

L

1; if

0

< c

,1

;

then

D



0,2 2

D

c

+

0 3 [

A

3]j

L

1 =0; if

0 =

c

,1

;

then

D



0,2 2

D

c

+

0 3 [

A

3]j

L

1 =

D



0,2 2 [(2 ,

c

+

0

c

,2 ,

c

+

0

0
)

D

1

D

c

3



D

c

3



]j

L

1 +

D



0,2 2 [

i

+

j

=2

c;i

6=

c

(

:::

)

D

1

D

i

3



D

j

3



]j

L

1 = 0 (in fact, the

coeffi-cient 2 ,

c

+

0

c

,2 ,

c

+

0

0
is zero). Therefore, 0=

D



0,2 2

D

c

+

0 3

P

3



j

L

1 =,2 ,

c

+

0

c

(

D



0 2

D

0 3



)
(

D

c

3



)j

L

1

:

On the other hand, by 3.8.1 and 3.8.3, (

D



0 2

D

0 3



)
(

D

c

3



)j

L

1 60

;

thus a contradiction. 2

Proof. (of Theorem 3.2) By Lemma 3.7,is

D

1-invariant along

Z

; then, by

Lemma 3.5,

Z

is

D

1-invariant. Hence, by Lemma 3.8,  is h

D

1

;D

2i-invariant

along

Z

; so, by Lemma 3.5,

Z

ish

D

1

;D

2i-invariant. 2

We shall use the following algebraic computation about the possible series expan-sion of a solution of the K.P. equation. The following lemma is Lemma B from Shiota, restated in a way that is more convenient to our purpose.

LEMMA 3.9 (Shiota [S], Lemma B, p. 359). Let(

S;

L) be a polarized abelian variety,

D

1 6= 0

;D

2

;

~

D

3 2

T

0(

S

)

:

Assume that

S

is generated byh

D

1

;D

2i

:

Let

Y

be a 2-dimensional disk with analytic coordinates

t

and

:

Let



be a nonzero section ofO

Y

H

0

(

S;

L)and assume that (i)

P

3(

D

1

;D

2

;

~

D

3+

@

t

;

d

)



=0

;

(ii)



(

t;;x

)= X

i;j

>0



i;j

(

x

)

t

i



j

;

where

x

2

S

(observe that



i;j

2

H

0

(

S;

L)for all

i

and

j

)

:

Also assume



0

;

=0

;

where



:=minf

j

j9

i

:



i;j

(
)60g

:

Then there exist local sections at zero ofO

Y

andO

Y

H

0

(

S;

L)

; f

and

;

such that



(

t;;x

)=





f

(

t;

)
(

t;;x

)

;

where (0

;

0

;

)60

;f

(0

;

0)=0 and

f

(

;

0)60

:

Proof. Step I (Shiota), we look for formal power series in

t

and

;f

and as in the lemma. Since

P

3(





[

:::

]) =



2



P

3(

:::

)

;

we can assume



= 0

:

Let



= maxf

i

j



i;

0(
)  0g

;f

0 =

t



and



0(

t;x

) = 

i

>





i;

0

(

x

)

t

i

,



;

so that



=

f

0



0mod(



)

:

Note that

P

3(



0)=0

;

in fact 0=

P

3(



)=

t

2



P

3(



0)mod(



)

:

Note also that



0(0

;x

)=



;

0(

x

)60

:

It suffices to find constants and sections

c

i;j

;

06

i

6



,1

;

16

j;

g

i;j

(

x

)2

H

0 (

S;

L)

;i

>

;j

>1

;

such that



(

t;;x

)= 0 @

f

0+ X

j

>1

f

j

(

t

)



j

1 A
0 @ 



0(

t;x

)+ X

j

>1 



j

(

t;x

)



j

1 A

;

(3.9.1)

where, for

j

>1

;

we define

f

j

(

t

)=



,1 X

i

=0

c

i;j

t

i

;





j

(

t;x

)= X

i

>



g

i;j

(

x

)

t

i

,



:

(3.9.2)

We now proceed by induction: let

l

be a positive integer, and assume that we found constants

c

i;j

;

for all 16

j

6

l

,1

;i

6



,1

;

and sections

g

i;j

(

x

)

;

for all

16

j

6

l

,1

;i

>

;

such that 3.9.1 holds modulo(



l

)

:

Define



0 (

t;x

)by



(

t;;x

) = 0 @

f

0+

l

,1 X

j

=1

f

j

(

t

)



j

1 A
0 @ 



0(

t;x

)+

l

,1 X

j

=1 



j

(

t;x

)



j

1 A +



l



0 (

t;x

) mod(



l

+1 )

:

(3.9.3)

We need to prove that there exist constants

c

i;l

;i

6



,1

;

and sections

g

i;l

;i

>

;

such that



0 (

t;x

)=



,1 X

i

=0

c

i;l

t

i



0(

t;x

)+ X

i

>



g

i;l

(

x

)

t

i

:

In fact, defining

f

l

;





l

as 3.9.2 requires, it is clear that 3.9.1 holds modulo(



l

+1 )

:

We define

P

~ 3(

r;s

)= 1 2[

P

3(

r

+

s

),

P

3(

r

),

P

3(

s

)]

:

By substitution in 2.0 we get ~

P

3(

D

1

;D

2

;

~

D

3+

@

t

;

d

)(

r;s

) =, 1 3(

D

4 1

r

s

+

D

4 1

s

r

)+ 4 3(

D

3 1

r

D

1

s

+

D

3 1

s

D

1

r

) ,2

D

2 1

r

D

2 1

s

,(

D

2 2

s

r

+

D

2 2

r

s

)+2

D

2

r

D

2

s

+

d

r

s

+(

D

1 ~

D

3

r

s

+

D

1 ~

D

3

s

r

),( ~

D

3

r

D

1

s

+ ~

D

3

s

D

1

r

) +(

D

1

@

t

r

s

+

D

1

@

t

s

r

),(

@

t

r

D

1

s

+

@

t

s

D

1

r

)

:

(3.9.4) Note that

P

~

3 is a symmetricC[



]-bilinear operator and that

P

3(

r

) = ~

P

3(

r;r

)

:

If

g

=

g

(

t;

)does not depend on

x;

by a straightforward computation we obtain

~

P

3(

g

r;g

s

)=

g

2
~

P

3(

r;s

) ~

P

3(

t

i

r;t

j

s

)=

t

i

+

j

~

P

3(

r;s

)+(

i

,

j

)

t

i

+

j

,1
(

D

1

r

s

,

D

1

s

r

) (3.9.5) We define

g

=

g

(

t;

) = 

l

,1

j

=0

f

j

(

t

)



j

and



(

t;;x

) = 

l

,1

j

=0 



j

(

t;x

)



j

;

so that



=

g



+



l



0 mod(



l

+1

)

:

Thus, by 3.9.5 the following equalities hold

modulo(



l

+1 ): 0=

P

3(



)=

P

3(

g



+



l



0 )=

P

3(

g



)+2 ~

P

3(

g

;

l



0 )=

g

2

P

3(



)+2



l

~

P

3(

g

;

0 ) =

g

2

P

3(



)+2



l

~

P

3(

t





0

;

0 )

:

In particular we

get

g

2

P

3(



) = 0 mod(



l

)

:

Since

g

2 (

t;

) =

t

2



_mod (



) is nonzero, we get

P

3(



) = 0 mod(



l

)

:

Since

g

2

P

3(



)+2



l

~

P

3(

t





0

;

0 ) = 0 mod(



l

+1 ) and (again)

g

2(

t;

)=

t

2



_mod (



)we get ~

P

3(

t





0

;

0 )=0 mod(

t

2



)

:

(3.9.6)

We now proceed by induction on

i

: assume that



0

(

t;x

)=

i

0,1

i

=0

c

i;l

t

i





0(

t;x

)+



(

x

)

t

i

0

;

_mod (

t

i

0+1 )

;

where 06

i

0 6



,1

:

Since ~

P

3(



0

;



0)=

P

3(



0) =0

;

by 3.9.5 we get

P

~

3(

t







0

;t

i





0)=0

:

Thus, by substitution in 3.9.6 and (again) by 3.9.5

we get that the following equalities hold modulo(

t



+

i

0 ): 0= ~

P

3(

t





0

;



i

0,1

i

=0

c

i;l





0

t

i

+



(

x

)

t

i

0 ) = ~

P

3(

t





0

;

(

x

)

t

i

0 ) = ~

P

3(

t





0(0

;x

)

;

(

x

)

t

i

0 ) = (



,

i

0)

t



+

i

0,1
[

D

1



0(0

;x

)



(

x

),

D

1



(

x

)



0(0

;x

)]=,(



,

i

0)

t



+

i

0,1
[



0(0

;x

)] 2

D

1(



(

x

)

=



0(0

;x

))

:

It follows that



(

x

)

=



0(0

;x

)is

D

1-invariant; on the

other hand, the zeroes of



0(0

;

)do not contain

D

1-integral curves, otherwise, by 3.8

(applied to



0)

;

we would have



;

0(

x

)=



0(0

;x

)=0

:

Thus



(

x

)=

c

j

0

;l



0(0

;x

)

:

It follows that



0 (

t;x

)=

i

0

i

=0

c

i;l

t

i



0(

t;x

)mod(

t

i

0+1

)

;

and we are done.

Step II, we prove that both

f

and can be assumed to be regular functions.

As (0

;

0

;

) 60 we are allowed to fix an

x

0such that (0

;

0

;x

0) 6=0 and

con-sider the formal power series

q

(

t;

) such that (

t;;x

0)

q

(

t;

) =1

:

Consider ~

f

(

t;

):=

f

(

t;

)
(

t;;x

0)and ~ (

t;;x

):= (

t;;x

)

q

(

t;

)

:

It is clear that



(

t;;x

) =





~

f

(

t;

)
~ (

t;;x

)

:

As ~

(

t;;x

0) =1 and



(

t;;x

0) are both

convergent,

f

~

(

t;

)is also convergent. Since



(

t;;x

)and ~

f

(

t;

)are convergent, ~

(

t;;x

) is also convergent. Note that

t



divides ~

f

(

t;

0) = 0

;

~

f

(

t;

0) 6 0 and ~

(0

;

0

;

)60

;

i.e. the properties of

f

and we need still hold for ~

f

and

:

~

2

LEMMA 3.10. As usual, assume that

P

i



= 0

;

for all

i

6

m

,1

:

Let W be a component of the scheme

D

1and let

p

be a generic point ofWred

:

Either

P

m



vanishes onW

;

or there exist irreducible elements

h;k

ofO

X;p

such that

(i) the ideal ofWredat

p

is(

h;k

);

(ii) the hypersurfacesf

h

=0g

;

f

k

=0gare smooth at

p

;

(iii) there exists an integer

l

such that

D

3



62(

k;h

l

)and

D



1

D

2





2(

k;h

l

)

;

for all

;

>0

:

Proof. If  is not singular alongWred we take

k

=



and we define

h

as in

3.1.1. We proved that either

P

m



j W

= 0

;

or

h

divides

D

1

h

inO



;p

:

If

h

divides

D

1

h

inO



;p

;

by substitution in the expression of

P

3

;

we get 2

b

=

a

+

c;

where

the notations are the ones of the formulas 3.1.1. If

b > a

then (

D

2

1 ,

D

2)

;

thus

P

m

;

vanishes onW

:

It follows that either

P

m



j

W

=0

;

or

c < b < a:

Therefore

the lemma holds with

l

=

b:

Let us turn to the case whereis singular alongWred

:

By Theorem 3.4, we can write

where

h

and

k

satisfy (i), (ii) and belong to the analytic completion ofO

X;p

:

We

prove that

h;k

satisfy (iii). Then, taking ~

h

and ~

k

approximating

h

and

k

to the

order

j

(

j

 0)

;

one has that (i), (ii) and (iii) hold. Thus, we can assume that

h;k

2O

X;p

:

As there are no

D

1-invariant components of

;

the element

h

does

not divide

D

1

h

inO

X;p

;

likewise

k

does not divide

D

1

k

inO

X;p

;

(and similarly

for

D

2)and we can write

D

1

h

=

"

1

k

a

+

g

1

h;

D

1

k

=

"

~1

h

~

a

+~

g

1

k;

D

2

h

=

"

2

k

b

+

g

2

h;

D

2

k

=

"

~2

h

~

b

+

g

~2

k;

where

"

1

;

"

~1

;"

2

;

"

~2 are invertible,

a;

a;b;

~ ~

b

>1

:

Note that, by 3.0, we are allowed

to assume

a

>

b;

~

a

> ~

b:

Note that

D



₁

D



₂

h;D



₁

D



₂

k

2 (

h;k

)

;

for all

;;

since,

by Theorem 3.2,Wredish

D

1

;D

2i-invariant. It follows that

D



1



=

D



1(

h

k

)2(

h

k;k

a

+1

;h

~

a

+1 )

;

8



>0;

D



1

D

2





=

D



1

D



2(

h

k

)2(

h

k;k

b

+1

;h

~

b

+1 )

;

8

;

>0

:

We now claim that either

D

3



62 (

h

k;k

b

+1

;h

~

b

+1 )

;

or

P

m



j W = 0

:

Since(

h

k;k

b

+1

;h

~

b

+1 ) = (

h;k

b

+1 )\(

k;h

~

b

+1

) we have that the previous claim (up to

interchanging the roles played by

h

and

k

) implies the lemma. So, let us prove our claim. First, observe that if

D

3



2 (

h;k

b

+1 )\(

k;h

~

b

+1 ) then by substitution in 2.0 we get 0 =

P

3



=2

~ 2

h

2~

b

+2

;

mod(

k;h

~

a

+ ~

b

+2 )

:

Thus 2 ~

b

+2>

a

~+ ~

b

+2

:

Since ~

a >

~

b

we get ~

b

=

a:

~ Similarly, computing

P

3



modulo (

h;k

a

+

b

+2

) we

get

a

=

b:

It follows that(

D

2

1 ,

D

2)



2 (

h

k;k

a

+1

;h

~

a

+1

)

:

Note that the ideal

I

D

1 is (

h

k;"

1

k

a

+1 +

"

~1

h

~

a

+1

)

;

where

"

1

;

"

~1 are invertible. Thus

I

D

1

 (

h

k;k

a

+2

;h

a

~+2

)

:

Since

P

3



=


=

P

m

,1



= 0 by inductive hypothesis, by the

first one of formulas 2.5 (with

s

=

m

)we get

P

m



j W =,(

D

2 1,

D

2)



~

m

,1

;

where we keep the notation of the formulas 2.5. We claim that it suffices to prove that ~

m

,1



2(

h;k

)

:

Indeed, since(

D

2 1,

D

2)



is in(

h

k;k

a

+1

;h

~

a

+1 )

;

if ~

m

,1



is in(

h;k

)

;

then

P

m



j W =,(

D

2 1,

D

2)



~

m

,1



2(

h

k;k

a

+2

;h

a

~+2 )

I

D

1

;

and we are done. By inductive hypothesis, the left-hand side of the first formula 2.5 (with

s

=

m

,1)is zero; it follows that the right hand side must be zero, in

particular we get ,(

D

2 1,

D

2)



~

m

,2



,

D

1



( ~

m

,1 +2

D

1 ~

m

,2 )



=0 mod(



)

:

(3.10.1) It follows that ~

m

,2



2(

h;k

)

;

otherwise we would have,(

D

2

1,

D

2)



2

I

D

1

and we would be done. We now compute the left-hand side of 3.10.1 modulo the ideal(

h

k;k

a

+2

;h

~

a

+2

)(note that this ideal contains(



)=(

h

k

))

:

Since ~

m

,2



2

(

h;k

)and(

D

2 1,

D

2)



2 (

h

k;k

a

+1

;h

~

a

+1 )we have that(

D

2 1 ,

D

2)



~

m

,2



is in (

h

k;k

a

+2

;h

~

a

+2 )

:

Since

D

1



is in (

h

k;k

a

+1

;h

~

a

+1 ) and ~

m

,2

;

hence

D

1~

m

,2

;

is in (

h;k

)

;

also

D

1



D

1 ~

m

,2



is in (

h

k;k

a

+2

;h

a

~+2 )

:

Therefore, by 3.10.1 ,

D

1



~

m

,1



2(

h

k;k

a

+2

;h

~

a

+2 )

:

(3.10.2) Since

D

1



=

D

1(

h

k

)=

"

k

a

+1 +

"

~

h

~

a

+1

mod(

h

k

)it follows that

D

1



is not

in(

h

k;k

a

+2

;h

~

a

+2 )

:

Therefore, by 3.10.2, ~

m

,1



is in

(

h;k

)and we are done.2

4. End of the proof

Let us go back to the K.P. hierarchy. We assume, by induction, that we found invariant vector fields

D

1

;:::;D

m

,1

;

and constants

d

4

;:::;d

m

such that

P

i

(

D

1

;:::;D

i

;

d

4

;:::;d

i

+1

)



=0

;

8

i

6

m

,1

:

We need to find an invariant vector field

D

m

and a constant

d

m

+1 such that

P

m

(

D

1

;:::;D

m

;

d

4

;:::;d

m

+1 )



=0

:

Let

P

m



:=

P

m

(

D

1

;:::;D

m

,1

;

0;

d

4

;:::;d

m

;

0 )

:

Recall that if

P

m



j

D

1 =0

we are done by Remark 2.6. We proved that then the only components of the scheme

D

1 where

P

m



might not vanish are, set-theoretically, h

D

1

;D

2i-invariant. In

order to conclude our proof of Shiota’s Theorem we proceed by contradiction. Let

Wbe a component of

D

1such that

P

m



j W

6=0

:

Thus,Wredish

D

1

;D

2i-invariant.

We denote by

X

0

theh

D

1

;D

2i-invariant minimal abelian subvariety of

X:

Since

D

16=0 we have

X

0

6=0

;

on the other handWcontains a translate of

X

0

;

therefore

X

0

6=

X:

Note thatWredis

T

0(

X

0

)-invariant. Let

X

00

be the complement of

X

0

in

X;

relative to the polarization

:

This means that

X

00

is the connected component containing zero of the kernel of the composite map

X

! Pic

0 (

X

) ! Pic 0 (

X

0 )

:

Here the first map sends

x

to the class of

x

,

;

and the second map is the natural

restriction.

LetR:=(W\

X

00

)red

:

Note thatWred is the

T

0(

X

0

)-span ofR

;

i.e.Wred = R+

X

0

;

and that R has codimension 2 in

X

00

:

In the sequel we shall work on

X

00 

X

0

:

Observe that



is naturally a theta function also for



?

O

X

() via the

sum map



:

X

00 

X

0 !

X:

In fact, as

T

0(

X

00 )

T

0(

X

0 )  =

T

0 (

X

)(canonically),

there is a canonical identification of the universal cover of

X

00 

X

0

with the one of

X

which commutes with the isogeny



:

X

00



X

0 !

X;

(

x

00

;x

0 ) 7!

x

00 +

x

0

:

In particular, this property allows us to write



instead of



?



while working on

X

00 

X

0

:

Let us fix general points

b

2R

;x

0 2

X

0

;

so that

p

:=(

b;x

0 )is a general point of



,1

(Wred)

:

Let us decompose

D

3as

D

0 3+

D

00 3

;

where

D

0 3 2

T

0(

X

0 )

;D

00 3 2

T

0(

X

00 )

:

Since

X

0

is generated by theh

D

1

;D

2i-flow,

D

00

3 is nonzero by Lemma 3.10 (iii). Let

L

be the (analytic) germ at zero of the

D

00

3-integral line in

X

00

Cbe the germ at

b

of a smooth curve in

X

00

meeting

L

+

b

transversally only at

b;

and let

Y

be the surfaceC+

L

in

X

00

:

Letbe the subvariety

Y



X

0

of

X

00 

X

0

:

Let



be a parameter on C vanishing at

b

and let

t

be the coordinate on

L

(vanishing at zero) with

@

t

=

D

00

3

:

Thus

;t

are parameters on

Y;

likewise they are naturally parameters on the product=

Y



X

0

:

Note that[

D



1

D



2

D

3

(

:::

)]j =

D



1

D



2

D

3

(

:::

j )

:

Onwe write



(

t;;x

)= X

i;j

>0



i;j

(

x

)

t

i



j

;

(4.1) where

x

is in

X

0

:

We recall that by the definition of the complement of

X

0

there is an isomorphism (

t



x

O())j

X

0  = O()j

X

0 for all

x

2

X

00

;

where

t

x

denotes the translation

y

7!

y

+

x:

Thus the



(

t;;

)’s are sections of the restriction j

X

0

:

Note that



i;j

depends on the point

b

and the curve

C chosen, and that



i;j

= (1

=i

!

j

!)((

@

j

=@

j

)

D

00

i

3



)(0

;

0

;

) is in

H

0 (

X

0

;

j

X

0

)

:

Indeed, since the



(

t;;

)’s are sections of the restrictionj

X

0

;

so are its derivatives with respect to

t

and

:

We use Lemmas 3.9 and 3.10 to reach a contradiction. Our analysis is divided naturally in two cases which correspond to whether the variety R is not

D

00

3 -invariant, or it is

D

00

3-invariant.

Let us first assume that R is not

D

00

3-invariant. Let us choose C in such a

way that it meets R transversally only at

b;@



62 h

T

b

(R)

;D

00

3i

:

This is possible

because R has codimension 2 in

X

00

:

We have

Y

\R = f



=

t

= 0g

;

thus \



,1 (Wred) = f



=

t

= 0g

X

0

:

It follows that



i;

0 6 0 for some

i;

and, moreover,



0

;

0(

x

) =0 (otherwise we would not have



j

b

+

X

0 =0)

:

Because

of Lemma 3.9 we have



=

f

(

t;

)
(

t;;x

)

;

where

f

(0

;

0) = 0

:

We have \



,1

W = \f



= 0g\f

D

1



= 0g  \f

f

= 0g

:

Moreover, since

f

(0

;

0) =0

;

it follows that\



,1

W has codimension 1 in

:

This contradicts \



,1

(Wred)=f



=

t

=0g

X

0

:

Let us now assume thatRis

D

00

3-invariant. ChooseC

;

depending on the point

x

0

;

in such a way that it meetsRtransversally only at

b;

andCf

x

0

gf

k

=0g

;

where

k

is as in Lemma 3.10. Since the locif

h

=0gandf

k

=0gare transverse

by 3.10 (i) , and C meets R transversally at

b;

we may assume that



is the

restriction of

h

toCf

x

0 g  = C

:

We have that\



,1 (Wred) = f



= 0g

:

Let



=minf

j

j9

i

:



i;j

(
)60g

:

Note that, asCdepends on

x

0

;

i;j

depends on

x

0

:

We want to prove that



0

;

=0

:

For this it suffices to prove that

D



1

D

2





0

;

(

x

0

) = 0

;

for all



and

;

since the flow generated by

D

1 and

D

2 is dense in

X

0

:

Since C=f

t

=0g

;

by 4.1 we have

D



1

D

2





j C

X

0 =

D



1

D

2





(0

;;

)=





D



1

D

2





0

;

(
)

;

mod(





+1 )

;

D

3



j C

X

0 =

D

3



(0

;;

)=0

;

mod(





)

:

(4.2) By Lemma 3.10, in the local ringO

f

k

=0g

;p

we have that

D



1

D

2





2 (

h

)

l

;D

3



62 (

h

)

l

;

for some

l;

where

h

is as in 3.10. Since



is the restriction of

h

toCf

x

0 g

 =

in the local ring O Cf

x

0 g

;p

;

we have that

D



1

D

2





2 (



)

l

;D

3



62 (



)

l

:

We have

l > 

by the second formulas in 4.2. On the other hand, since

l > 

we must have

D



₁

D



₂



0

;

(

x

0

)= 0 for all



and

;

by the first formulas in 4.2. Since

X

0

is generated by the h

D

1

;D

2i-flow and

D



1

D

2





0

;

(

x

0

) = 0 for all



and

;

we get



0

;

=0

:

Hence we can apply Lemma 3.9. It follows that the equality 4.1 takes the

form



(

t;;x

) =





f

(

t;

)
(

t;;x

)

;

so that

f

divides both



j

and

D

1



j

:

Therefore, \



,1

(Wred)  \f

f

= 0g

:

By Lemma 3.9,

f

(0

;

0) = 0 and

f

(

;

0)60

:

As\



,1

(Wred)\f

f

=0g

;

the locus\



,1

(Wred)contains

(locally at

p

)a component which is not the componentf



=0g

:

This contradicts

the fact that, locally at

p;

\



,1

(Wred)=f



=0g

:

2

REMARK 4.3. If one could show thatW (not only its underlying reduced scheme Wred) wereh

D

1

;D

2i-invariant it would easily follow by the very expression of

P

m



that

P

m



vanishes onW

:

In fact, in this case,



(

z

+

a

)(where

a

2

X

0

)would

vanish onW

:

Hence,

D

2

1



and

D

2



would vanish onW as well.

Acknowledgements

This paper is the core of my thesis at the University of Rome. I heartily thank my advisor Enrico Arbarello for being very important in my mathematical development and introducing me to this problem. I am grateful to Riccardo Salvati Manni and Corrado De Concini for many stimulating conversations on the subject of this paper. I am grateful to Olivier Debarre for pointing out a mistake in a previous version of this paper, and also to the referee who suggested several improvements.

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