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By [40], Tate’s limiting argument gives a nef adelicQ-line bundleLf ∈ Pic(X) Q extending L and satisfying fLf = q Lf. We want to generalize the definition to any line bundleM ∈Pic(X)Q. The main theorem of this subsection is the following result.

Theorem 4.9 Assume that X is normal. The projection Pic(X )Q−→Pic(X)Q

has a unique section

M −→Mf

as f-modules. Moreover, the section satisfies the following properties:

(1) If M∈Pic0(X)Q, then Mf is flat.

(2) If M∈Picf(X)Qis ample, then Mf is nef.

Note that this theorem furnishes the construction for the canonical height functions for every line bundleM ∈Pic(X)Q, not only for the polarizationL.

Definition 4.10 An elementMofPic(X )Qis called f -admissibleif it is of the form Mf. In this case, the adelic metric ofM is also called f -admissible.

For any integrable adelicQ-line bundle N on X with a model(NK,XK)over a number fieldK, recall that the height functionhN :X(Q)→Ris defined by

hN(x)= 1

[K(x):Q]N· ˜x, xX(Q).

Herex˜ denotes the image ofxinXK, which is a closed point ofXK. The following simple consequences assert that the definition extends that of [40], and that preperiodic points have height zero under f-admissible adelic line bundles.

Corollary 4.11 Let M ∈Pic(X)Q. The following are true:

(1) If fM=λM for someλ∈Q, then fMf =λMf inPic(X )Q.

(2) For any x ∈Prep(f), one has Mf|x˜ =0inPic( x)˜ Q. Hence, the height function hM

f is zero onPrep(f).

Proof Part (1) follows from the first statement of the theorem. For (2), assume that fm(x)= fn(x)withm >n ≥0. Choose a number fieldK such that(X, f,L)has a model(XK, fK,LK)overKas before, and such thatxandM are defined overK. Start with the identity

Mf|fm(x)=Mf|fn(x). We have

(f)mMf|x =(f)nMf|x.

It follows thatNf|x =0 for anyNin the image of the linear map (fK)m(fK)n:Pic(XK)Q−→Pic(XK)Q.

It suffices to prove that the map is surjective. By Lemma 4.6, Pic(XK)Q is finite-dimensional. Thus we need only prove that the map(fK)m(fK)nis injective. This is true because the eigenvalues of fK have absolute valuesqorq12.

Now we prove Theorem 4.9 by the following steps, which takes up the rest of this subsection. It suffices to prove the corresponding statement over a model (XK,fK,LK)over a sufficiently large number field K, with some extra care about the compatibility between differentK. In the following discussion, we assume that f has a fixed pointx0X(K), and omit the subscriptKin the following discussion by abuse of notation.

Step 1

Denote byPic(X )the group of adelic line bundles onX(with continuous metrics), and denotePic(X )Q=Pic(X )ZQ. In this step, we prove that the canonical projection

Pic(X )Q−→Pic(X)Q has a unique section

M −→Mf

as f-modules. Note thatPic(X )contains the groupPic(X )of integrable adelic line bundles.

DefineD(X)by the exact sequence

0−→D(X)−→Pic(X )Q−→Pic(X)Q−→0.

It is an exact sequence of f-modules. Recall that fhas the minimal polynomial R=P Qon Pic(X)Q.

Lemma 4.12 The operator R(f)is invertible on D(X).

Assuming the lemma, then we have a unique f-equivariant splitting of the exact sequence. In fact, consider the composition

Pic(X)Q R−→(f) D(X)R(f)|

1 D(X)

−→ D(X).

It becomes the identity map when restricted toD(X), and thus gives a splitting of the exact sequence. More precisely, if we denote

E(X)=ker(R(f):Pic(X) Q−→Pic(X) Q),

which is also the kernel of the above composition, then we have an f-equivariant decomposition

Pic(X)Q=D(X)E(X).

The projectionPic(X) Q→Pic(X)Qinduces an isomorphismE(X)→Pic(X)Q. The inverse of this isomorphism gives our desired f-equivariant splitting Pic(X)Q → Pic(X ) . This is the existence of the splitting.

For the uniqueness, note that Pic(X)Qis killed by R(f), so any f-equivariant splitting j :Pic(X)Q →Pic(X )Qhas an image killed by R(f). In other words, the image of jshould lie inE(X). ButE(X)→Pic(X)Qis already an isomorphism, so

jis unique.

The splitting Pic(X)Q→Pic(X)Qcan be written down explicitly. In fact, for any M ∈Pic(X)Q, letMbe any lifting ofMinPic(X )Q. Then the f-equivariant lifting is given by

Mf =MR(f)|D1(X)R(f)M.

One checks that this expression lies inE(X)and maps toM in Pic(X)Q.

Note that our triple(X, f,L)is an abbreviation of a model(XK, fK,LK)over a number fieldK. The compatibility between different K is as follows. If we replace K by a finite extension, then the minimal polynomial Rof fon Pic(X)Qbecomes a multiple of it, andE(X)becomes a space naturally including it.

Step 2

Proof of Lemma4.12 Recall that we have assumed that the base field is a number fieldK, andx0X(K)is a rational point. Denote by Pic(X,x0)andPic(X, x0)the isomorphism classes of bundles with a rigidification atx0. More precisely, an element of Pic(X,x0)consists of an element M ∈ Pic(X)and an isomorphism M(x0)K; an element ofPic(X, x0) consists of an element M ∈ Pic(X) and an isometry M(x0)K, whereK is endowed with the trivial adelic metric at every place. Then we have identifications:

Pic(X)Q=Pic(X,x0)Q, Pic(X)Q=Pic(X,x0)Q⊕Pic(K)Q,

wherePic(K)Q is embedded intoPic(X)Qby pull-back via the structure morphism XSpecK. The kernel of the projectionPic(X, x0)Q→Pic(X)Qcan be identified with metrics onOXrigidified atx0. By taking−log1, it is identified with

C(X,x0):= ⊕vC(Xanv ,x0),

where the sum is over all placesvof K, andC(Xanv ,x0)is the space of continuous functions onXvanvanishing atx0.

Thus we have shown the equality

D(X)=C(X,x0)⊕Pic(K )Q.

This decomposition is stable under fsincex0is a fixed point of f. OnPic(K )Q, the operator facts as the identity map. ThusR(f)acts asR(1)on Pic(K)Q, which is non-zero since all eigenvalues of fhave absolute valuesq1/2orq.

It remains to show that R(f)is invertible overC(X,x0). By the action of the complex conjugation, it suffices to show thatR(f)is invertible overC(X,x0)C= C(X,x0)RC. DecomposeR(T)over the complex numbers as follows:

R(T)=a

is absolutely convergent on each Banach spaceC(Xvan,x0)Cdefined by the norm φvsup,φvC(Xvan,x0)C.

The convergence is clear. This finishes the proof of the lemma.

Step 3

The previous steps have constructed a unique lifting Mf ∈Pic(X)Q for any M ∈ Pic(X)Q. To prove that Mf actually lies inPic(X) Q, which means thatMf is inte-grable, it suffices to prove statements (1) and (2) of the theorem since Pic0(X)Q and the ample elements of Picf(X)Qgenerate Pic(X)Q.

In this step, we prove (1); namely, ifM ∈Pic0(X)Q, thenMf is flat. We refer the construction of flat adelic line bundles to “Appendix (Flat metrics)”.

We first reduceX to abelian varieties. As above, by extendingK and replacing f by an iteration, we can assume that f has a fixed pointx0X(K). As in “Appendix the expression of Mf at the end of Step 1, this isomorphism carries f-admissible adelic metrics on elements of Pic0(A)Qto f-admissible adelic metrics on elements of Pic0(X)Q. Hence, it suffices to prove that any f-admissible adelic metrics on Pic0(A)Qis flat. In other word, we only need to treat abelian varieties.

Now assume that X is an abelian variety, and f : XX is an endomorphism (fixing the originx0). LetM ∈Pic0(X)Qbe a general element. Note that f commutes with[2] :XX, and thus[2]carriesf-admissible adelic metrics to f-admissible adelic metrics. It follows that

[2]M = [2]M =2M .

Then the adelic metric ofMf is invariant under the dynamical system[2] : XX.

It is flat by Theorem5.16 Step 4

It remains to prove (2) of the theorem. LetM ∈Picf(X)Qbe ample (and f-pure of weight 2). We need to prove thatMf is nef.

Let M0be any nef adelicQ-line bundle extendingM and consider the sequence Mm = (q1f)mM0of nef bundles. We will pick a subsequence “convergent” to Mf.

First, denote byMm,f the admissible adelic line bundle extendingMm, and denote Cm:=MmMm,f.

Then Cm lies in C(X) := ⊕C(Xanv ), the direct sum of the spaces of continuous functions onXvan. The relationMm =(q1f)mM givesMm,f =(q1f)mMf.It follows that

Cm =(q1f)m(M0Mf).

HereM0Mf also lies inC(X). Since fdoes not change the norm ofC(X), the sequenceCmconverges to 0.

Second, take a basisNi (i =1, . . . ,r) of Picf(X)Q. Write

MmM = r

i=1

ai,mNi, ai,m ∈Q.

Then

Mm,fMf = r

i=1

ai,mNi,f.

It follows that

MmMf =Cm+ r

i=1

ai,mNi,f.

This is a relation inPic(X )Q.

Third, notice that the orthogonal transformationq1f has eigenvaluesμi with absolute values 1 on Picf(X)Q. We can find an infinite sequence{mk}k of integers such thatμmi k →1 for everyi. Thus the operator(q1f)mkconverges to the identity operator on Picf(X)Q. In other words,ai,mk →0 for alli. Then the theorem follows from the nefness of the limits of nef bundles in a finite-dimensional space in Proposition 5.8.

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