Erratum : Linear projections and successive minima

0.09 MB·English

by  Christophe Soule

Erratum: Linear projections and successive
minima
2
1
0
2 Christophe Soul´e∗

l
u July 31, 2012
J

7
2
Let E be a finitely generated projective modules over the ring of integers

] in a number field K, h an hermitian metric on E ⊗C, and X ⊂ P(E∨) a
G Z K K
smooth projective curve. In [2], Prop. 1, we formulated an inequality relating
A
the successive minima of the euclidean lattice (E,h) to the Faltings height of
.
h X .
K
t
a As explainedin§1below,theproofofthis propositionisincorrect. Weshall
m
prove another result instead (see §2.1, Theorem 1).
[ I thank J.-B. Bost, C. Gasbarri and C. Voisin for their help.

1
v
1 Erratum
3
3
6 The proof of Proposition 1, hence Theorem 2, in [2] is incorrect. Indeed, §2.5
6 and §2.7 in op.cit contain a vicious circle: the definition of the filtration V ,
i
.
7 1≤i≤n, in §2.5 depends on the choice of the integers ni, when the definition
0 oftheintegersn in§2.7depends onthe choiceofthefiltration(V ). Thus,only
i i
2
Theorem 1 and Corollary 1 in [2] are proved.
1
:
v
2 An inequality
i
X
r 2.1
a
LetK beanumberfield,O itsringofalgebraicintegersandS =Spec(O )the
K K
associated scheme. Consider an hermitian vector bundle (E,h) over S. Define
the i-th successive minima µ of (E,h) as in [2] §2.1. Let X ⊂ P(E∨) be a
i K K
smooth,geometricallyirreduciblecurveofgenusganddegreed. Weassumethat
X ⊂P(E∨)is definedbyacompletelinearseriesonX ,andthatd≥2g+1.
K K K
The rank of E is thus N =d+1−g. Let h(X ) be the Faltings height of X
K K
([2] §2.2).
∗CNRSetIHE´S,LeBois-Marie,35routedeChartres,91440Bures-sur-Yvette, France
[email protected]
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For any positive integer i≤N we define the integer f by the formulas
i
f =i−1 if i−1≤d−2g,
i
f =i−1+α if i−1=d−2g+α, 0≤α≤g,
i
and f =d.
N
Fix a natural integer t≤N −3. When i≥t+2 we let
f2
A (t)= i ,
i i−1
(i−t−1)f − f
i j
j=Pt+2
and
A(t)= max A (t).
i
t+2≤i≤N
Theorem 1. Let t be an integer such that 0 ≤ t ≤ N −3. There exists a
constant c(N) such that the following inequality holds:
N−t−1
h(X )
K
[K :Q] +(2d−A(t)(N −t−1))µ1+A(t) µα+c(N)≥0.
αX=1
2.2
To prove Theorem 1 we start by the following variant of theorem 4.1 in [1].
Proposition 1. Fix an increasing sequenceof integers e1 ≤e2 ≤...≤eN such
that e =0 if i≤t+1. Set
i
ℓ−1
e +e
T(r1,...,rN)=0=i1<m...i<niℓ=NXj=1 (rij −rij+1) ij 2 ij+1 .
 
N
Subject to the constraints rj = 1, r1 = r2 = ... = rt+1, and r1 ≥ r2 ≥
j=Pt+1
...≥rN =0, the maximal value of T(r1,...,rN) is
1 e2
Tmax = 2 t+m2≤aix≤N i i−1 .
(i−t−1)e − e
i j
j=Pt+2
Proof. As in [1], loc.cit., we may firstassume thatT =1 andseek to minimize
the sum of the r ’s. If we graphthe points (e ,r ), T is the areaof the Newton
j j j
polygontheydetermineinthefirstquadrant. Movingthepointsnotlyingonthe
N
polygondownontoitonlyreduces r ,sowemayassumethatallthepoints
j
j=Pt+1
2
actually lie on the polygon. In particular we may assume that (e ,r )=(0,r )
j j j
lies on this polygon when 1≤j ≤t+1. For such r we have
i
N−1
ei+et+1
T(r1,...,rN)= (ri−ri+1) .
2
Xi=1
Let σi =ri−1−ri, i=2,...,N. The conditionthat the points (ei,ri) lie on
their Newton polygon and that the r decrease becomes, in terms of the σ ,
i i
σt+2 σt+3
≥ ≥...≥0. (1)
et+2−et et+3−et+2
Furthermore
σ2 =...=σt+1 =0.
N
The constraint r =1 becomes
j
j=Pt+1
N
(j−t−1)σ =1. (2)
j
j=Xt+2
Inthe(N−t−2)-dimensionalsubspaceofthepointsσ =(σt+2,...,σN)defined
by (2), the inequality (1) defines an (N −t−2) simplex. The linear function
N
ej−1+ej
T = σ
j
2
j=Xt+2
must achieve its maximum on this simplex at one of the vertices, i.e. a point
where, for some i and α, we have
σt+2 σi σi+1
α= =...= > =...=0.
et+2−et+1 ei−ei−1 ei+1−ei
We get
σ = α(ej −ej−1) if t+2≤j ≤i
j (cid:26)0 else.
Then
N i
1 = (j−t−1)σj =α (j−t−1)(ej −ej−1)
j=Xt+2 j=Xt+2
i−1
= α(i−t−1)ei− ej ,
j=Xt+2
 
hence
−1
i−1
α=(i−t−1)ei− ej ,
j=Xt+2
 
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and, as in [1] loc.cit., we deduce that
1 e2
T(r1,...,rN)= 2 i i−1 .
(i−t−1)e − e
i j
j=Pt+2
Corollary 1. Let t ≥ 0 be an integer less than N − 2. Fix an increasing
sequence of integers e1 ≤ e2 ≤ ... ≤ eN and a decreasing sequence of numbers
r1 ≥r2 ≥...≥rN. Assume that e1 =e2 =...=et+1 =0 and r1 =r2 =...=
rt+1. Let
ℓ−1
S = min (r −r )(e +e ).
0=i0<...<iℓ=N Xj=0 ij ij+1 ij ij+1
Then
  e2  N
S ≤ max i (r −r ).
t+2≤i≤N(i−t−1)ei− i−1 ejj=Xt+2 j N
  j=Pt+2 
2.3
We come back to the situation of Theorem 1. For every complex embedding
σ :K →C, the metric h defines a scalarproducth onE ⊗ C. Ifv ∈E we let
σ
OK
kvk=max h (v,v).
σ
σ
p
ChooseN elementsx1,...,xN inE,linearlyindependentoverK andsuchthat
logkxik=µN−i+1, 1≤i≤N.
Let y1,...,yN ∈EK∨ be the dual basis of x1,...,xN. Let A(d) be the constant
appearing in [2], Theorem 1.
Lemma 1. Assume0≤t≤N−3. Wemay choose integers n , i=1,...,t+1,
i
such that
i) For all i≤t+1, |n |≤A(d)+d
i
ii) Let wi =yi+niyi+1 if i≤t+1 and wi =yi if i≥t+2.
Let hw1,...,wii⊂EK∨ be the subspace spanned by w1,...,wi, and
Wi =EK∨/hw1,...,wii
(W0 =EK∨). Then, when i≤t+1, the linear projection from P(Wi−1) to P(Wi)
does not change the degree of the image of X .
K
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Proof. The image of wi in P(Wi−1) is the center of the linear projection
P(Wi−1)...→P(Wi). When n6=m are two integers, the images of the vectors
yi +nyi+1 and yi +myi+1 in Wi−1 are K-linearly independent, hence their
images in P(Wi−1) are distinct. Lemma 1 follows from this and [2], Theorem 1
and Corollary 1.
2.4
Let (v ) ∈ EN be the dual basis of (w ). Since the matrix expressing (w ) in
i K i i
terms of (y ) has integral coefficients and determinant one, the same is true
i
for the matrix expressing (v ) in terms of (x ). Furthermore these matrices
i i
have coefficients bounded by a function of N. This implies that there exists a
constant c1(N) such that
logkvik≤µN +c1(N)
when 1≤i≤t+1.
On the other hand, when t+2≤i≤N, we get v =x , hence
i i
logkxik=µN+1−i if i≥t+2.
2.5
Let e be the drop in degree of X when projected to W . From Lemma 1 ii)
i K i
we get
e1 =e2 =...=et+1 =0.
Let
r = µN +c1(N) if i≤t+1 (3)
i (cid:26)µN+1−i+c1(N) if i≥t+2.
From Corollary 1 we deduce, as in [2] pp. 52–53, that there exists a constant
c2(N) such that
N
h(X )
K
[K :Q] +2drN ≥−A(t) (rj −rN)−c2(N),
j=Xt+2
where
f2
A(t)= max i .
t+2≤i≤N i−1
(i−t−1)f − f
i j
j=Pt+2
From (3) we deduce that there exists c(N) such that
N−t−1
h(X )
K
[K :Q] +(2d−A(t)(N −t−1))µ1+A(t) µα+c(N)≥0.
αX=1
Theorem 1 is proved.
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References
[1] Morrison,I.Projectivestabilityofruledsurfaces.Invent.Math.56,269-304
(1980).
[2] Soul´e, C. Linear projections and successive minima. Nagoya Math. J. 197
(2010), 45-57.
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