But the adjunction formula (Hartshorne V.1.5) states that
' KX D2g0 2 '2: We deduce that
2
' D .2g 2/d
which is negative, because of our assumption ong Dg.C /. Note thatd is bounded: the Hurwitz formula says that
2g 2Dd.2g0 2/C.positive/:
Thus, there is an integerN (independent of') such that N '2< 0:
For each polynomialP there exists a Hilbert scheme, HilbP;classifying the curves onX with Hilbert polynomialP. We know that HilbP is a finite union of varietiesVi (when the ambient space isPn, it is even connected), and that if 2Vi, thendimVi DdimH0. ; N / (by deformation theory) where N is the normal bundle. In our case, N D 0 since
2
' < 0. Thus eachVi is a point. We deduce that
f ' j' 2Homnoncnst.C; C0/g
is finite, and since a map is determined by its graph, this proves the theorem. Alterna-tive approach: Use differential geometry. The condition g.C / > 1implies that C.C/ is
hyperbolic. 2
6 The Faltings Height.
To any abelian variety A over a number field K, Faltings attaches a canonical height H.A/2R.
The Faltings height of an elliptic curve overQ
Consider first an elliptic curve E overC. We want to attach a numberH.E/toE which is a measure of its “size”. The most natural first attempt would be to write E C=, and define H.E/to be the reciprocal of the area of a fundamental domain for , i.e., if DZ!1CZ!2, then
H.E/D j!1^!2j 1:
Unfortunately this doesn’t make sense, because we can scale the isomorphism to make the area of the fundamental domain any positive real number we choose. In order to get a height, we need additional data.
PROPOSITION6.1. LetE be an elliptic curve overC. Then each of the following choices determines the remainder:
(a) an isomorphismC=!E.C/;
(b) the choice of a basis for .E; ˝1/, i.e., the choice of a nonzero holomorphic differ-ential on E;
(c) the choice of an equation
Y2D4X3 g2X g3 (*) forE.
PROOF. (a)!(c). There are associated with a lattice , a Weierstrass function}.z/and numbersg2./,g3./for which there is an isomorphism
E.C/DC=!E0P2; z7!.}.z/W}0.z/W1/
whereE0is the projective curve given by the equation (*).
(c)!(b). Take!D dXY .
(b)!(a). From a differential!onEand an isomorphism˛WC=!E.C/we obtain a differential˛.!/onCinvariant under translation by elements of. For example, if˛is the map given by}and!D dXY , then˛.!/D d }.z/}0.z/ Ddz. Thus we should choose the˛ so that˛.!/Ddz. This we can do as follows: consider the mapP 7!RP
0 !WE.C/!C. This is not well-defined because the integral depends on the choice of the path. However, if 1and2are generators forH1.E;Z/, then (up to homotopy), two paths from0toP will differ by a loopm11Cm22, and because!is holomorphic, the integral depends only on the homotopy class of the path. Therefore, we obtain a well-defined mapE.C/ ! C=, DZ!1CZ!2,!i DR
When the elliptic curve is given overQ(rather thanC), then we choose an equation Y2D4X3 g2X g3; g2; g3 2Q;
and take the differential!to bedX=Y. When we change the choice of the equation,! is only multiplied by a nonzero rational number, and so
H.E/Ddf H.E; !/
When we consider an elliptic curve over a number field K two complications arise.
Firstly,K may have several infinite primes, and so we may have to take the product over their separate contributions. Secondly, and more importantly, OK may not be a principal ideal domain, and so there may not be a global minimal equation. Before describing how to get around this last problem, it is useful to consider a more general construction.
6. THE FALTINGS HEIGHT. 151 The height of a normed module
Anormon a vector spaceM overRorCis a mappingk kWM !R>0such that kxCyk kxk C kyk; kaxk D jajkxk; x; y 2M; ascalar.
Herej jis the usual absolute value.
Now letK be a number field, and let R be the ring of integers in K. Recall that a fractional ideal inK is a projectiveR-module of rank 1; conversely, ifM is a projective R-module of rank1, thenM ˝RK K, and the choice of an isomorphism identifiesM with a fractional ideal in K. LetM be such anR-module. Suppose we are given a norm k kvonM ˝RKvfor eachvj1. We define theheightofM (better, of.M; .k kv/vj1/)
LEMMA6.2. The definition is independent of the choice ofm.
PROOF. Recall that, for a finite primev corresponding to a prime idealp, the normalized absolute value is defined by,
jajvD.R Wp/ ordv.a/; ordvWK Z; and that for any infinite primev,
jajv D jaj"v:
Moreover, for the normalized absolute values, the product formula holds:
YjajvD1:
It is obvious that the expression on the right is unchanged whenmis replaced witham. 2 LEMMA6.3. In the expression (6) forH.M /,mcan be taken to be any element ofM ˝R
K. When we define,
h.M /D 1
ŒK WQ logH.M /;
then, for any finite extensionLofK,
h.RL˝RM /Dh.M /:
PROOF. Exercise in algebraic number theory. 2
The Faltings height of an abelian variety
PROPOSITION6.4. LetV be a smooth algebraic variety of dimensiongover a fieldk.
(a) The sheaf of differentials˝V =k1 onV is a locally free sheaf ofOV-modules of rank g.
(b) IfV is a group variety, then˝1is free.
PROOF. See T. Springer, Linear Algebraic Groups, Birkh¨auser, 1981, 3.2, 3.3. 2 COROLLARY6.5. LetV be a smooth algebraic variety of dimensiongover a fieldk. Then
˝g Ddf g˝1is a locally free sheaf of rank1, and it is free ifV is a group variety.
PROOF. Immediate from (6.4). 2
LetMbe a coherent sheaf on a varietyV. For any pointv2V we obtain a vector space M.v/over the residue fieldk.v/. For example, ifV is affine, sayV DSpecm.R/, thenM corresponds to theR-moduleM D .V;M/, and ifv$m, thenM.v/DM=mM. Note that, for any open subsetU ofV containingv, there is a canonical map .U;M/!M.v/.
PROPOSITION6.6. LetV be a complete geometrically connected variety over a field k, and let Mbe a free sheaf of finite rank onV. For any v 2 V .k/, the map .V;M/ ! M.v/is an isomorphism.
PROOF. For M D OV, .V;M/ D k (the only functions regular on the whole of a complete variety are the constant functions), and the map is the identity mapk ! k. By assumptionM.OV/nfor somen, and so the statement is obvious. 2 PROPOSITION6.7. LetAbe an abelian variety of dimensiongover a fieldk. The canoni-cal maps
.A; ˝1/!˝1.0/; .A; ˝g/!˝g.0/
are isomorphisms.
PROOF. By0we mean the zero element ofA. For the proof, combine the last two results.2 Now letAbe an abelian variety over a number fieldK, and letRbe the ring of integers inK. Recall from I,17, that there is a canonical extension ofAto a smooth group scheme AoverSpecR(the N´eron model). The sheaf˝A=Rg of (relative) differentialg-forms onA is a locally free sheaf ofOA-modules of rank1(it becomes free of rank1when restricted to each fibre, but is not free on the whole ofA). There is a sectionsWSpecR !Awhose image in each fibre is the zero element. DefineM Ds˝A=Rg . It is a locally free sheaf of rank1onSpecR, and it can therefore be regarded as a projectiveR-module of rank1. We have
M ˝RK D˝A=Kg .0/D .A; ˝A=Kg /
—the first equality simply says that˝A=Rg restricted to the zero section ofAand then to the generic fibre, is equal to˝A=Rg restricted to the generic fibre, and then to the zero section;
the second equality is (6.7).