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Erratum: Linear projections and successive minima 2 1 0 2 Christophe Soulé∗ 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] 1 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   3 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 4 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. 5 References [1] Morrison,I.Projectivestabilityofruledsurfaces.Invent.Math.56,269-304 (1980). [2] Soulé, C. Linear projections and successive minima. Nagoya Math. J. 197 (2010), 45-57. 6