1. Notation and preliminaries [024W]
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1. Notation and preliminaries
1.1. Notation
Throughout this paper, we fix the following notation.
1.1.1.
Fix a field with a complete and non-archimedean absolute value . The valuation ring of and the maximal ideal of the valuation ring are denoted by and , respectively, that is,
In the case where is discrete, we fix a uniformizing parameter of , that is, .
1.1.2.
A norm of a finite-dimensional vector space over is always assumed to be ultrametric, that is, . A pair is called a normed finite-dimensional vector space over .
1.1.3.
Fix an algebraic scheme over , that is, is a scheme of finite type over . Let be the analytification of in the sense of Berkovich [1]. For , the residue field of the associated scheme point of is denoted by . Note that the seminorm at yields an absolute value of . By abuse of notation, it is denoted by . Let be the completion of with respect to . The extension of to is also denoted by the same symbol . The valuation ring of and the maximal ideal of the valuation ring are denoted by and , respectively. Let be an invertible sheaf on . For , is denoted by .
1.1.4.
By continuous metric on , we refer to a family , where is a norm on over for each , such that for any local basis of over a Zariski open subset , is a continuous function on . We assume that is projective. Given a continuous metric on , we define a norm on such that
Similarly, if is a closed subscheme of , we define a norm on such that
Clearly one has
| (1) |
for any .
1.1.5.
Given a continuous metric on , the metric induces for each integer a continuous metric on which we denote by : for any point and any local basis of over a Zariski open neighborhood of one has
Note that for any section one has . By convention, denotes the trivial metric on , namely for any , where denotes the section of unity of .
Conversely, given a continuous metric on , there is a unique continuous metric on such that . We denote by this metric. This observation allows to define continuous metrics on an element in as follows. Given , we denote by the subsemigroup of of all positive integers such that . We call continuous metric on any family with being a continuous metric on , such that for any and any . Note that the family is uniquely determined by any of its elements. In fact, given an element , one has for any . In particular, for any positive rational number , the family is a continuous metric on , where is a positive integer such that , and the metric does not depend on the choice of the positive integer .
Let be an element in equipped with a continuous metric . By abuse of notation, for we also use the expression to denote the continuous metric on .
1.1.6.
We call model of any projective and flat -scheme such that the generic fiber of is . We denote by the central fiber of . By the valuative criterion of properness, for any point , the canonical -morphism extends in a unique way to an -morphism of schemes . We denote by the image of by the map . Thus we obtain a map from to , called the reduction map of .
Let be an element of such that in . The -invertible sheaf yields a continuous metric as follows.
First we assume that and in . For any , let be a local basis of around and the class of in . For , if we set (), then . Here we set . Note that is continuous because, for a local basis of over an open set of , for all . Moreover,
| (2) |
for all and . Indeed, if we set for , then . Thus
In general, there are and a positive integer such that in and in . Then
Note that the above definition does not depend on the choice of and . Indeed, let and be another choice. As in , there is a positive integer such that in , so that, by using (2),
as desired.
1.1.7.
Let be a model of . As is flat over , the natural homomorphism is injective. Let be a closed subscheme of and the defining ideal sheaf of . Let be the kernel of , that is, . Obviously , so that if we set , then . Moreover, is flat over because is injective. Therefore, is a model of . We say that is the Zariski closure of in .
1.2. Extension obstruction index
In this subsection, we introduce an invariant to describe the obstruction to the extension property. Let be a projective scheme over , be an invertible sheaf on equipped with a continuous metric , and be a closed subscheme of . For any non-zero element of , we denote by the following number (if there does not exist any section extending , then the infimum in the formula is defined to be by convention)
| (3) |
This invariant allows to describe in a numerically way the obstruction to the metric extendability of the section . In fact, the following assertions are equivalent:
- (a)
,
- (b)
for any , there exists such that, for any integer , the element extends to a section such that .
The following proposition shows that, if extends to a global section of for sufficiently positive (it is the case notably when the line bundle is ample), then the limsup defining is actually a limit.
Proposition 1.1.
For any integer , let
Then the sequence is sub-additive, namely one has for any . In particular, if for sufficiently positive integer , the section lies in the image of the restriction map , then ββ in (3) is actually ββ.
Proof.
By (1), one has for any integer . Moreover, if and only if lies in the image of the restriction map . To verify the inequality , it suffices to consider the case where both and are finite. Let and be respectively sections in and such that and , then the section verifies the relation . Moreover, one has
Since and are arbitrary, one has . Finally, by Feketeβs lemma, if for sufficiently positive integer , then the sequence actually converges in . The proposition is thus proved. β
Corollary 1.2.
Assume that the invertible sheaf is ample, then the following conditions are equivalent.
- (a)
,
- (b)
for any , there exists and a section such that and that .
1.3. Normed vector space over a non-archimedean field
In this subsection, we recall several facts on (ultrametric) norms over a non-archimedean field. Throughout this paper, a norm is always assumed to be ultrametric. Let be a finite-dimensional vector space over and a norm of over .
1.3.1. Orthogonality of norms
For , a basis of is called an -orthogonal basis of with respect to if
If (resp. and ), then the above basis is called an orthogonal basis of (resp. an orthonormal basis of ). Let be another basis of . We say that is compatible with if for .
Proposition 1.3.
Fix a basis of . For any , there exists an -orthogonal basis of with respect to such that is compatible with . Moreover, if the absolute value is discrete, then there exists an orthogonal basis of compatible with .
Proof.
We prove it by induction on . If , then the assertion is obvious. By the hypothesis of induction, there is a -orthogonal basis of with respect to such that
for . Choose . As
there is such that . We set . Clearly forms a basis of . It is sufficient to see that
for all . Indeed, as , we have
If , then
Otherwise,
as required.
For the second assertion, it is sufficient to show the following lemma because it implies that the set has the minimal value. β
Lemma 1.4.
If is discrete, then the set is discrete in .
Proof.
Let us consider a map given by
It is sufficient to see that is finite. Let be distinct elements of . We choose with for . If , then for all . Therefore, we obtain
for all . In particular, are linearly independent. Therefore, we have . β
1.3.2. Scalar extension of norms
Let be a vector space over and a norm of .
Lemma 1.5.
For , the set is bounded from above.
Proof.
By the above lemma, we define to be
Note that yields a norm on . We denote by (i.e. the case where and ).
Lemma 1.6.
Let be a subspace of and . For any , there is such that and
Proof.
Let be an -orthogonal basis of such that (cf. PropositionΒ 1.3). We define to be
for . Then . Moreover, note that
so that
for all with . Thus the assertion follows. β
Corollary 1.7.
The natural homomorphism is an isometry.
Proof.
We denote the norm of by , that is,
Note that for all and . In particular, . For , we set and choose with . Then . For any , by LemmaΒ 1.6, there is such that and . As , we have . Thus we obtain by taking . β
Definition 1.8.
Let be an extension field of , and let be a complete absolute value of which is an extension of . We set . Identifying with
we can give a norm of , that is,
The norm is called the scalar extension of . Note that for . Indeed, by CorollaryΒ 1.7,
Proposition 1.9.
For , let be an -orthogonal basis of with respect to . Then also yields an -orthogonal basis of with respect to .
Proof.
Let be the dual basis of . For with ,
and hence . Therefore, for ,
Thus we have the assertion. β
Lemma 1.10.
Let be an extension field of , and let be a complete absolute value of as an extension of . We set . Note that . Let (resp. ) be a norm of obtained by the scalar extension of on (resp. the scalar extension of on ). Then .
Proof.
For , let be an -orthogonal basis of with respect to . Then, by PropositionΒ 1.9, forms an -orthogonal basis of and with respect to and , respectively, so that is also an -orthogonal basis of with respect to . Note that for all . Thus, for ,
and
Thus, we have the assertion by taking . β
Lemma 1.11.
Let be a surjective homomorphism of finite-dimensional vector spaces over . Let and be norms of and , respectively. We assume that and is the quotient norm of in terms of the surjection . We set and . Let and be the norms of and obtained by the scalar extensions of and , respectively. Then is the quotient norm of in terms of the surjection .
Proof.
Let be the quotient norm of with respect to the surjection . Let be an non-zero element of . As , it is sufficient to show that . Note that
so that we have . Let us consider an inequality . For , let be an -orthogonal basis of such that forms a basis of . Clearly we may assume that . Then
Therefore, we have by taking . β
Lemma 1.12.
We assume that the absolute value of is trivial. Let be a finite-dimensional normed vector space over . Then we have the following:
- (1)
The set is a finite set.
- (2)
Let be a field and a complete and non-trivial absolute value of such that and is an extension of . Let be the valuation ring of and the maximal ideal of . We assume the following:
- (i)
The natural map induces an isomorphism .
- (ii)
If an equation holds for some and , then .
Let be a norm of over such that for all . If is an orthogonal basis of , then forms an orthogonal basis of . In particular, .
- (i)
Proof.
(2) First we assume that
Then, for any ,
Let us see that
for . Clearly we may assume that
We set . We fix with . By the assumption (i), for each , we can find and such that
Note that
Moreover, as , we have
| and | ||||
Therefore,
In general, we take positive numbers and non-empty subsets of such that for and . Note that for . Let us consider
where . Note that forms an orthogonal basis of and for all . Therefore, by the above observation,
so that it is sufficient to see that
Clearly we may assume that . We set
For with , we have . Indeed, we choose and with and . If , then
so that, by the assumption (ii), , which is a contradiction. Therefore,
as required. β
Remark 1.13.
We assume that is discrete and
for . If
then the assumption (ii) holds. Indeed, we suppose that for some and . Then
so that , and hence , as required.
1.3.3. Lattices and norms
From now on and until the end of the subsection , we assume that is non-trivial. Let be an -submodule of . We say that is a lattice of if and
for some norm of . Note that the condition does not depend on the choice of the norm since all norms on are equivalent. For a lattice of , we define to be
Note that forms a norm of . Moreover, for a norm of ,
is a lattice of .
Proposition 1.14.
Let be a lattice of . We assume that, as an -module, admits a free basis . Then is an orthonormal basis of with respect to .
Proof.
For and ,
so that . β
Let us consider the following lemmas.
Lemma 1.15.
A subgroup of is either discrete or dense in .
Proof.
Clearly we may assume that , so that . We set . If , then . Indeed, for , let be an integer such that . Thus , and hence . Therefore, is discrete.
Next we assume that . Then there is a sequence in such that for all and . If we set , then and . For an open interval of (), we choose and an integer such that and . Then we have and
so that . Thus is dense. β
Lemma 1.16.
Let be a norm of and . Then
Moreover, and for all with .
Proof.
The first assertion is obvious because, for , if and only if .
For , let with . Then , that is, , and hence .
Finally we consider the second inequality, that is, for . Clearly we may assume that . As , there is with . By the first assertion, we can choose such that . If , then
Thus . This is a contradiction, so that . Therefore,
as required. β
Proposition 1.17.
We assume that is discrete. Then we have the following:
- (1)
Every lattice of is a finitely generated -module.
- (2)
If we set for a norm of of , then .
Proof.
(1) Let be an orthogonal basis of with respect to (cf. PropositionΒ 1.3). As is discrete, there is with . If we set for , then forms an orthonormal basis of with respect to . Therefore,
Thus we have (1) because is noetherian.
(2) follows from LemmaΒ 1.16. β
Proposition 1.18.
We assume that is not discrete. If we set for a norm of of , then .
Proof.
Proposition 1.19.
We assume that the absolute value is not discrete. Let be a norm of and . For any , there is a sub-lattice of such that is finitely generated over and .