Volumes of Arithmetic Okounkov Bodies

0.12 MB·English

by  Xinyi Yuan

Volumes of Arithmetic Okounkov Bodies
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Xinyi Yuan
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January 12, 2012
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Contents

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T
N 1 Introduction 1
.
h
t 2 Lattice points in a filtration 3
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3 The arithmetic Okounkov body 7
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1
v 1 Introduction
1
2
2
This paper is an improvement and simplification of Yuan [Yu]. In particular,
2
. by a simpler method, it proves the identity in [Yu, Theorem A] without
1
0 taking the limit p → ∞. The method also simplifies the proofs of Moriwaki
2 [Mo], where more general arithmetic linear series were treated.
1
Recall that [Yu] explored a way to construct arithmetic Okounkov bodies
:
v
from an arithmetic line bundle, inspired by the idea of Okounkov [Ok1, Ok2]
i
X
and Lazarsfeld–Musta¸tˇa [LM] in the geometric case. Theorem A of [Yu]
r
a asserts that the volumes of the Okounkov bodies approximate the volume of
the arithmetic line bundle. The main result of this paper asserts an exact
identity before taking the limit.
On the other hand, Boucksom–Chen [BC] initiated a different way to
construct Okounkov bodies in the arithmetic setting. From the proof of
the main result in this paper, it is easy to recognize the similarity of the
constructions in [BC] and in [Yu]. We will make a comparison at the end of
this paper.
In the following, we recall the construction of [Yu] and state our main
result. We use exactly the same notations as in [Yu] throughout this paper.
1
Volume of an arithmetic line bundle
Let X be an arithmetic variety of dimension d, i.e., a d-dimensional integral
scheme, projective and flat over Spec(Z). For any hermitian line bundle
L = (L,k·k) over X, denote
H0(X,L) = {s ∈ H0(X,L) : ksk ≤ 1}
sup
b
and
hˆ0(X,L) = log#H0(X,L).
b
Define the volume to be
hˆ0(X,mL)
vol(L) = limsup .
md/d!
m→∞
Recall that a line bundle L is said to be big if vol(L) > 0.
NotethatChen[Ch]provedthat“limsup=lim” inthedefinitionofvol(L).
The result was also obtained in [Yu] by the construction of the Okounkov
body.
Arithmetic Okounkov Body
Assume that X is normal with smooth generic fiber X . Denote by X →
Q
Spec(O ) the Stein factorization of X → Spec(Z). Then K is the algebraic
K
closure of Q in the function field of X. For any prime ideal ℘ of O , denote
K
by F = O /℘ the residue field, and by N the cardinality of F .
℘ K ℘ ℘
Let
X ⊃ Y ⊃ ··· ⊃ Y
1 d
be a flag on X, where each Y is a regular irreducible closed subscheme of
i
codimension i in X. Assume that Y is the fiber X of X above some prime
1 F℘
ideal ℘ of O , and Y ∈ Y (F ) is a rational point.
K d 1 ℘
Define a valuation map
ν = (ν ,··· ,ν ) : H0(X,L)−{0} → Zd
Y. 1 d
with respect to the flag Y. as follows. For any nonzero s ∈ H0(X,L), we first
set ν (s) = ord (s). Let s be a section of the line bundle O(Y ) with zero
1 Y1 Y1 1
locus Y . Then s⊗(−ν1(s))s is nonzero on Y , and let s = s⊗(−ν1(s))s be
1 Y1 1 1 (cid:16) Y1 (cid:17)(cid:12)
(cid:12)Y1
(cid:12)
2
the restriction. Set ν (s) = ord (s ). Continue this process on the section
2 Y2 1
s on Y , we can define ν (s) and thus ν (s),··· ,ν (s).
1 2 3 4 d
For any hermitian line bundle L on X, denote
v (L) = ν (H0(X,L)−{0})
Y. Y.
b
to be the image in Zd. The arithmetic Okounkov body ∆ (L) of L is defined
Y.
1
to be the closure of Λ (L) = v (mL) in Rd. It is a bounded convex
Y. [ md Y.
m≥1
subset of Rd if non-empty. The following is the main result of this paper.
Theorem 1.1. If L is big, then
1
vol(∆ (L))logN = vol(L).
Y. ℘
d!
In [Yu], the author restricted to the case F = F , and could only prove
℘ p
that the left-hand side converges to the right-hand side as p → ∞. The
problem was caused by a large accumulated error term in the estimation.
The solution of this problem is found by the author in the preparation of
[YZ], a recent joint work ofthe author withTong Zhang. The key isTheorem
2.1, which puts the filtration together and makes a more accurate control of
the error term.
Acknowledgments. This paper is based on a result of lattice points in a joint work of the
author with Tong Zhang. The author would like to thank Tong Zhang for many inspiring
discussions. The author is very grateful for Huayi Chen who provided a way to extend
the result of lattice points from the field of rational numbers to general number fields.
2 Lattice points in a filtration
In this section, we state and prove the result on lattice points in Theorem
2.1. It is the key for Theorem 1.1.
Fix a number field K. By a normed O -module, we mean a pair (M,{k·
K
k } )consisting ofalocallyfreeO -moduleM offiniterank, andacollection
σ σ K
{k·k } of C-norms k·k on M ⊗ C, indexed by σ : K ֒→ C and invariant
σ σ σ σ
under the complex conjugation.
Let M = (M,{k·k } ) be normed O -module. Define
σ σ K
H0(M) = {m ∈ M : kmk ≤ 1, ∀σ},
σ
b
3
and
hˆ0(M) = log#H0(M).
b
Denote by
O hH0(M)i, ZhH0(M)i
K
respectively the O -submoduble and the Zb-submodule of M generated by
K
H0(M). For any α ∈ R, denote
b M(α) = (M,{e−αk·k } ).
σ σ
For a normed O -module L of rank one, define
K
deg(L) = log#(L/sO )− logksk .
K X σ
d σ
Here s ∈ L is any non-zero element, and the definition is independent of the
choice of s. It is just the usual arithmetic degree of a hermitian line bundle
over Spec(O ).
K
∨
It is clear that the dual L gives a natural normed O -module of rank
K
one, and that the tensor product M ⊗ L is a natural normed O -module.
K
The identity of the tensor is given by the trivial normed O -module O =
K K
(O ,{|·| } ), where |·| is just the usual absolute value.
K σ σ σ
In some literatures, a normed O -module is also called a metrized vector
K
bundle over Spec(O ). A normed O -module of rank one is also called a
K K
metrized line bundle over Spec(O ), or just a hermitian line bundle over
K
Spec(O ).
K
The innovation of this paper is the application of the following result.
Theorem 2.1. Let K be a number field. Let M be a normed O -module,
K
and L ,L ,··· ,L be a sequence of normed O -modules of rank one. Here
0 1 n K
L is the trivial normed O -module of rank one. Denote
0 K
∨
α = deg(L ), r = rank O hH0(M ⊗L )i, i = 0,··· ,n.
i i i OK K i
d b
Assume that
0 = α ≤ α ≤ ··· ≤ α .
0 1 n
Then there is a constant C > 0 depending only on K such that
n
h0(M) ≥ r (α −α )−C(r logr +r ),
X i i i−1 0 0 0
b i=1
n
∨
h0(M) ≤ h0(M ⊗L )+ r (α −α )+C(r logr +r ).
n X i−1 i i−1 0 0 0
b b i=1
4
The case K = Q isequivalent to [YZ, Proposition2.3], which follows from
the successive minima of Gillet–Soul´e [GS]. The extension to the general K
is provided by Huayi Chen in a private communication to the author. We
start with a simple lemma in the following.
Lemma 2.2. Denote κ = [K : Q] and denote by D the discriminant of K.
K
(1) For any hermitian normed O -module L of rank one, we have
K
1
hˆ0(L) > deg(L)−κlog2− log|D |.
K
2
d
(2) There is a constant δ ≥ 0 depend only on K such that
1
rank O hH0(M(−δ))i ≤ rank ZhH0(M)i ≤ rank O hH0(M)i
OK K κ Z OK K
b b b
for any normed O -module M.
K
Proof. The result in (1) follows from Minkowski’s theorem. The second in-
equality of (2) follows from the inclusion
ZhH0(M)i ⊂ O hH0(M)i.
K
b b
For the first inequality of (2), fix a basis (a ,··· ,a ) of O over Z. Set
1 κ K
δ = max{log|a | : i = 1,··· ,κ, σ : K ֒→ C}.
i σ
Then
O hH0(M(−δ))i
K
⊂ Zha Hb 0(M(−δ)),··· ,a H0(M(−c))i
1 κ
⊂ ZhH0b(M)i. b
b
The inequality also follows.
Proof of Theorem 2.1. If K = Q, then L is isometric to Z(α ), and thus
i i
∨
M ⊗L = M(−α ).
i i
5
The theorem is equivalent to the inequalities for hˆ0 in [YZ, Proposition 2.3].
Next we consider the general case.
1
Denote κ = [K : Q] and c = log2 + log|D |. Denote β = α /κ for
K i i
2κ
convenience. View M as a normed Z-module, and apply [YZ, Proposition
2.3] to the sequence
0 = β ≤ β +c ≤ ··· ≤ β +c.
0 1 n
We obtain
n
h0(M) ≤ h0(M(−β −c))+ r′ (β −β )+O(r logr ).
n X i−1 i i−1 0 0
b b i=1
Here
r′ = rank ZhH0(M(−β −c))i, i = 0,··· ,n−1.
i Z i
b
We claim that r′ ≤ κr by our special choice of c. It simply implies the
i i
∨
second equality we need to prove. By Lemma 2.2 (1), H0(L (β + c)) 6= 0.
i i
Since b
∨ ∨
M ⊗L = M(−β −c)⊗L (β +c),
i i i i
we have
r′ = rank ZhH0(M(−β −c))i ≤ rank ZhH0(M ⊗L∨)i ≤ κr .
i Z i Z i i
b b
The last inequality follows from Lemma 2.2 (2). It proves the claim.
As for the first inequality, apply [YZ, Proposition 2.3] to M(δ+c) viewed
as a normed Z-module, and the sequence
0 = β ≤ β ≤ ··· ≤ β .
0 1 n
Here δ is as in Lemma 2.2 (2). We obtain
n
h0(M(δ +c)) ≥ r′′(β −β )+O(r logr +r ).
X i i i−1 0 0 0
b i=1
Here
r′′ = rank ZhH0(M(−β +δ +c))i.
i Z i
b
Lemma 2.2 (1) gives H0(L (−β +c)) 6= 0, and thus
i i
b
r′′ ≥ rank ZhH0(M ⊗L∨(δ))i ≥ κr .
i Z i i
b
6
Here the last inequality follows from Lemma 2.2 (2). The proof is finished
by the basic inequality
h0(M) ≥ h0(M(δ +c))−κr (δ +c+log3).
0
b b
See [YZ, Proposition 2.1] or the original form [GS, Proposition 4].
3 The arithmetic Okounkov body
In this section, we prove Theorem 1.1, and compare our construction with
that of [BC].
Proof of the main theorem
Recall that X is a normal arithmetic variety with smooth generic fiber, geo-
metrically irreducible over O . Recall that we have the flag
K
X ⊃ Y ⊃ Y ⊃ ··· ⊃ Y
1 2 d
of regular vertical subvarieties. We have assume that Y = X for some
1 F℘
prime ideal ℘ of O , and Y ∈ Y (F ) is a rational point.
K d 1 ℘
By [Yu, Proposition 2.5], if L is a big hermitian line bundle on X,
#v (mL)
Y.
lim = vol(∆ ).
m→∞ md Y.
Hence, Theorem 1.1is reduced to thefollowing result, which strengthens [Yu,
Theorem 2.6].
Theorem 3.1. For any L ∈ Pic(X),
c
#v (mL) hˆ0(X,mL)
lim (cid:12) Y. logN − (cid:12) = 0.
m→∞(cid:12)(cid:12) md ℘ md (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
Now we prove the theorem. Recall that
v(L) = v (L) = ν (H0(X,L))
Y. Y.
b
7
is the image in Zd of the valuation map
ν = ν = (ν ,··· ,ν ) : H0(X,L)−{0} → Zd
Y. 1 d
defined by the flag
X ⊃ Y ⊃ Y ⊃ ··· ⊃ Y .
1 2 d
The flag
Y ⊃ Y ⊃ ··· ⊃ Y
1 2 d
on the ambient variety Y induces a valuation map
1
ν◦ = (ν ,··· ,ν ) : H0(X,L )−{0} → Zd−1
2 d F℘
of dimension d − 1 in the geometric case. Similarly, we have a valuation
map ν◦ on the line bundle (L ⊗ ℘i)| for any i ∈ Z. There are natural
Y1
isomorphisms (L⊗℘i)| ∼= L , which are compatible with the valuations.
Y1 F℘
The valuations ν and ν◦ are compatible in the sense that
ν(s) = ν (s),ν◦((s⊗(−ν1(s))s)| ) , s ∈ H0(X,L)−{0}.
(cid:16) 1 Y1 Y1 (cid:17)
b
Here s is the section of O(Y ) defining Y .
Y1 1 1
Decompose v(L) = ν (H0(L)) according to the first component in Zd.
Y.
We have b
v(L) = i, ν◦(H0(L⊗℘i)| ) .
a(cid:16) Y1 (cid:17)
i≥0 b
Here ℘ is naturally a hermitian line bundle on Spec(O ), endowed with the
K
metric induced from the trivial metric of O via the inclusion ℘ ⊂ O . It is
K K
also viewed as a hermitian line bundle on X by pull-back.
By the geometric case in [LM],
#ν◦(H0(L⊗℘i)| ) = dim F hH0(L⊗℘i)| i.
Y1 F℘ ℘ Y1
b b
Here F hH0(L ⊗ ℘i)| i denotes the F -vector subspace of H0(L ⊗ ℘i| )
℘ Y1 ℘ Y1
generatedbby H0(L⊗℘i)| . The reduction map gives an injection
Y1
b
H0(L⊗℘i)/℘H0(L⊗℘i) −→ H0(L⊗℘i| ).
Y1
It follows that
dim F hH0(L⊗℘i)| i = rank O hH0(L⊗℘i)i.
F℘ ℘ Y1 OK K
b b
8
Here O hH0(L ⊗℘i)i denotes the O -submodule of H0(L ⊗ ℘i) generated
K K
by H0(L⊗b℘i). Therefore,
b
#v(L) = rank O hH0(L⊗℘i)i. (1)
X OK K
i≥0 b
Note that it is essentially a finite sum.
For any i ≥ 0, denote L = ℘−i. Then
i
deg(L ) = ilogN .
i ℘
d
Apply Theorem 2.1 to the normed O -module
K
M = (H0(L),{k·k } )
σ,sup σ
and the sequence {L } . It is easy to obtain
i i≥0
h0(L) = rank O hH0(L⊗℘i)i·logN +O(h0(L )logh0(L )). (2)
X OK K ℘ Q Q
b i≥0 b
Compare (1) and (2). We have
#v(L)·logN = h0(L)+O(h0(L )logh0(L )).
℘ Q Q
b
Replace L by mL. We obtain Theorem 3.1.
Construction of Boucksom–Chen
In [BC], Boucksom and Chen constructed Okounkov bodies in a very gen-
eral arithmetic setting. Here we briefly compare their construction with our
construction in the setting of this paper.
We first recall the construction of [BC]. Let (X,L) be as above. Denote
by Z = X the generic fiber. Fix a flag
1 K
Z ⊃ Z ⊃ ··· ⊃ Z
1 2 d
on the generic fiber Z . It gives an Okounkov body ∆ (L ) ⊂ Rd−1 of the
1 Z. K
generic fiber L .
K
Restricted to the generic fiber, we have a valuation
ν◦ = (ν′,··· ,ν′) : H0(X,L)−{0} → Zd−1.
Z. 2 d
b
9
It gives a convex body ∆◦ (L) contained in ∆ (L ).
Z. Z. K
More generally, for any t ∈ R, denote L(−t) = (L,etk · k). Then we
have a convex body ∆◦ (L(−t)) of L(−t) contained in ∆ (L ). The se-
Z. Z. K
quence {∆◦ (L(−t))} is decreasing in t. Define the incidence function
Z. t∈R
G : ∆ (L ) → R by
Z. Z. K
G (x) = sup{t ∈ R : x ∈ ∆◦ (L(−t))}.
Z. Z.
Its graph gives a (d+1)-dimensional convex body
∆ (L) = {(x,t) ∈ ∆ (L )×R : 0 ≤ t ≤ G (x)}.
Z. Z. K Z.
This is the Okounkov body constructed in [BC]. One main result of [BC]
asserts that
1
vol(∆ (L)) = vol(L).
Z.
d!
Go back to our construction. Recall that we have a flag
X ⊃ Y ⊃ ··· ⊃ Y
1 d
on X. On the special fiber Y = X , we have a flag
1 F℘
Y ⊃ Y ⊃ ··· ⊃ Y .
1 2 d
It gives a valuation of the line bundle L on Y . Thus we get an Okounkov
F℘ 1
body ∆ (L ) in Rd−1.
Y. F℘
Restricted to the special fiber, we have a valuation
ν◦ = (ν ,··· ,ν ) : H0(X,L)−{0} → Zd−1.
Y. 2 d
b
It gives a convex body ∆◦ (L) contained in ∆ (L ).
Y. Y. F℘
The hermitian line bundle L ⊗ ℘t gives a convex body ∆◦ (L ⊗ ℘t) in
Y.
Rd−1 for any t ∈ Z. By passing to tensor powers, extend the definition of
∆◦ (L⊗℘t) to any t ∈ Q. The sequence {∆◦ (L⊗℘t)} is decreasing in t.
Y. Y. t∈Q
Define the incidence function G : ∆ (L ) → R by
Y. Y. F℘
G (x) = sup{t ∈ Q : x ∈ ∆◦ (L⊗℘t)}.
Y. Y.
Its graph gives a (d+1)-dimensional convex body
∆ (L) = {(x,t) ∈ ∆ (L )×R : 0 ≤ t ≤ G (x)}.
Y. Y. F℘ Y.
It recovers our previous construction.
Hence, we see that the similarity of these two constructions. The con-
struction of [BC] is archimedean in nature, while the construction of [Yu] is
non-archimedean in nature.
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