p. 1/23 · Longyearbyen (Spitsbergen), July 16, 2014 Joint work with Sébastien Boucksom...

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Augmented base loci andrestricted volumes in normalvarieties

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Frontiers of rationalityLongyearbyen (Spitsbergen), July 16, 2014

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Joint work with

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Joint work with

Sébastien Boucksom (CNRS-Université Pierre etMarie Curie)

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Joint work with

Sébastien Boucksom (CNRS-Université Pierre etMarie Curie)and Salvatore Cacciola (Roma Tre University)

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Overview

• Introduction

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Overview

• Introduction• Results

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Overview

• Introduction• Results• Proofs

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Overview

• Introduction• Results• Proofs• News for real divisors

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Introduction

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Introduction

Stable base loci play a central role in algebraicgeometry.

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Introduction

Stable base loci play a central role in algebraicgeometry.LetX be a normal projective variety defined over analgebraically closed field of any characteristic.

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Introduction

Stable base loci play a central role in algebraicgeometry.LetX be a normal projective variety defined over analgebraically closed field of any characteristic.LetL be a line bundle onX. ToL we can associatethe maps

ϕmL : X 99K PH0(X,mL).

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Introduction



As we all know, knowledge of the behavior of thesemaps often says a lot about the geometry ofX itself.

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Introduction



As we all know, knowledge of the behavior of thesemaps often says a lot about the geometry ofX itself.In particular, there are some closed subsets, associatedtoL, that govern, asymptotically, this behavior.

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Stable base loci

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Stable base lociThe stable base locus ofL is

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B(L) =⋂

m∈N

Bs(|mL|).

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B(L) =⋂

m∈N

Bs(|mL|).

This closed subset is clearly important, but oftendifficult to compute.

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B(L) =⋂

m∈N

Bs(|mL|).


On the other hand, suppose that we want to know if,outside of some closed subset,ϕmL is anembedding.

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B(L) =⋂

m∈N

Bs(|mL|).



To this goal we better assume thatL is big ad wedefine

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B(L) =⋂

m∈N

Bs(|mL|).



To this goal we better assume thatL is big ad wedefine(introduced in 2000 by Nakamaye, and in 2006 byEin, Lazarsfeld, Mustaţă, Nakamaye and Popa)

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Augmented base loci

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Augmented base lociTheaugmented base locusof L is

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B+(L) =⋂

m∈N

B(mL− A)

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B+(L) =⋂

m∈N

B(mL− A)

whereA is ample (the definition does not depend onA). If L is not big, defineB+(L) = X.

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B+(L) =⋂

m∈N

B(mL− A)

whereA is ample (the definition does not depend onA). If L is not big, defineB+(L) = X.It is easy to see that, form ≫ 0, ϕmL is an embeddingoutsideB+(L) (we will come back to this).

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B+(L) =⋂

m∈N

B(mL− A)


Augmented base loci have been recently important inbirational geometry.

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B+(L) =⋂

m∈N

B(mL− A)


Augmented base loci have been recently important inbirational geometry.

Just to mention a few instances, we recall thefundamental papers of Takayama, Hacon andMcKernan on the birationality of pluricanonical maps

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Augmented base loci

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Augmented base locior the finite generation of the canonical ring byBirkar, Cascini, Hacon and McKernan.

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Great, buthow do we compute them?

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Great, buthow do we compute them?To this goal we use another notion, present in thementioned papers, the one of restricted volume.

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For every subvarietyZ ⊆ X not contained inB+(L),it is easily seen that the restriction ofL toZ is big.

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For every subvarietyZ ⊆ X not contained inB+(L),it is easily seen that the restriction ofL toZ is big.But there is more: the space of sections of|mL|Z | thatextend toX has maximal growth. In fact

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For every subvarietyZ ⊆ X not contained inB+(L),it is easily seen that the restriction ofL toZ is big.But there is more: the space of sections of|mL|Z | thatextend toX has maximal growth. In factif we denote byH0(X|Z,mL) the image of therestriction mapH0(X,mL) → H0(Z,mL|Z) and setd = dimZ, then we claim that

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Restricted volumes

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Restricted volumestherestricted volumeof L toZ,

volX |Z(L) = lim supm→+∞

d!

mddimH0(X|Z,mL) > 0 :

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d!


Let sL ∼ A+ E with A ample,E effective and suchthatZ 6⊆ Supp(E).

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d!


Let sL ∼ A+ E with A ample,E effective and suchthatZ 6⊆ Supp(E).Form ≫ 0 we haveH1(IZ/X(mA)) = 0 and thediagram

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d!


Let sL ∼ A+ E with A ample,E effective and suchthatZ 6⊆ Supp(E).Form ≫ 0 we haveH1(IZ/X(mA)) = 0 and thediagram

H0(mA) _

��

// //H0(mA|Z) _

��

H0(msL) //H0(msL|Z)– p. 7/23

Restricted volumes andB+

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Restricted volumes andB+shows thath0(X|Z,msL) ≥ h0(mA|Z) ≥ Cmd,whencevolX |Z(L) > 0.

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Restricted volumes andB+shows thath0(X|Z,msL) ≥ h0(mA|Z) ≥ Cmd,whencevolX |Z(L) > 0.So we deduce that

B+(L) ⊇⋃

Z⊆X :volX|Z(L)=0

Z.

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B+(L) ⊇⋃

Z⊆X :volX|Z(L)=0

Z.

The surprising (?) fact is that equality holds!

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B+(L) ⊇⋃

Z⊆X :volX|Z(L)=0

Z.

The surprising (?) fact is that equality holds!Theorem (Ein, Lazarsfeld, Mustaţă, Nakamayeand Popa, 2009)

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B+(L) ⊇⋃

Z⊆X :volX|Z(L)=0

Z.

The surprising (?) fact is that equality holds!Theorem (Ein, Lazarsfeld, Mustaţă, Nakamayeand Popa, 2009)LetX be asmooth complexprojective variety.

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B+(L) ⊇⋃

Z⊆X :volX|Z(L)=0

Z.

The surprising (?) fact is that equality holds!Theorem (Ein, Lazarsfeld, Mustaţă, Nakamayeand Popa, 2009)LetX be asmooth complexprojective variety.LetL be a line bundle onX. Then

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B+(L) ⊇⋃

Z⊆X :volX|Z(L)=0

Z.

The surprising (?) fact is that equality holds!Theorem (Ein, Lazarsfeld, Mustaţă, Nakamayeand Popa, 2009)LetX be asmooth complexprojective variety.LetL be a line bundle onX. Then

B+(L) =⋃

Z⊆X :volX|Z(L)=0

Z.

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Restricted volumes andB+

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Restricted volumes andB+This theorem has a long and complicated proof withdifficult cohomological estimates.

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Restricted volumes andB+This theorem has a long and complicated proof withdifficult cohomological estimates.Our goal is to give a very short and simple proof thatworks forall normal varieties and in arbitrarycharacteristic.

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Restricted volumes andB+This theorem has a long and complicated proof withdifficult cohomological estimates.Our goal is to give a very short and simple proof thatworks forall normal varieties and in arbitrarycharacteristic.These two generalizations are important in birationalgeometry because, as we know, we need to work withnormal varieties. Also recently the MMP in positivecharacteristic is developing.

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Restricted volumes andB+This theorem has a long and complicated proof withdifficult cohomological estimates.Our goal is to give a very short and simple proof thatworks forall normal varieties and in arbitrarycharacteristic.These two generalizations are important in birationalgeometry because, as we know, we need to work withnormal varieties. Also recently the MMP in positivecharacteristic is developing.To be honest, we need to say that ELMNP prove more(and we cannot), namely a sort of continuity statement

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limm→+∞ volX |Z(L+1mA) = 0

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limm→+∞ volX |Z(L+1mA) = 0

for every ampleA and for every irreduciblecomponentZ of B+(L).

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Results

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ResultsIn any case, here is our result:

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ResultsIn any case, here is our result:Theorem (Boucksom, Cacciola, -)

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ResultsIn any case, here is our result:Theorem (Boucksom, Cacciola, -)LetX be a normal projective variety over analgebraically closed field of arbitrary characteristic.

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ResultsIn any case, here is our result:Theorem (Boucksom, Cacciola, -)LetX be a normal projective variety over analgebraically closed field of arbitrary characteristic.LetL be a line bundle onX.

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ResultsIn any case, here is our result:Theorem (Boucksom, Cacciola, -)LetX be a normal projective variety over analgebraically closed field of arbitrary characteristic.LetL be a line bundle onX.Then

B+(L) =⋃

Z⊆X :volX|Z(L)=0

Z.

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B+(L) =⋃

Z⊆X :volX|Z(L)=0

Z.

Proof:

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B+(L) =⋃

Z⊆X :volX|Z(L)=0

Z.

Proof:We can assume thatL is big. LetZ be an irreduciblecomponent ofB+(L). Recall thatdimZ ≥ 1 by awell known result of Zariski.

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Proof

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ProofIf Z ⊆ B(L), thenH0(X|Z,mL) = 0 for everym ≥ 1, whencevolX |Z(L) = 0.

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ProofIf Z ⊆ B(L), thenH0(X|Z,mL) = 0 for everym ≥ 1, whencevolX |Z(L) = 0.Suppose now thatZ 6⊆ B(L).

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ProofIf Z ⊆ B(L), thenH0(X|Z,mL) = 0 for everym ≥ 1, whencevolX |Z(L) = 0.Suppose now thatZ 6⊆ B(L).Letm ≫ 0 be such thatBs(mL) = B(L) andϕmL isbirational onto its image.

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ProofIf Z ⊆ B(L), thenH0(X|Z,mL) = 0 for everym ≥ 1, whencevolX |Z(L) = 0.Suppose now thatZ 6⊆ B(L).Letm ≫ 0 be such thatBs(mL) = B(L) andϕmL isbirational onto its image.Consider the commutative diagram

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Xmµm

��

fm// Ymνm

��

X ϕmL//❴❴❴❴ ϕmL(X)

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Xmµm

��

fm// Ymνm

��


whereµm is the normalized blow-up ofX along thebase ideal of|mL|,

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Proof

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Proof

Xmµm

��

fm// Ymνm

��


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Proof

Xmµm

��

fm// Ymνm

��


νm is the normalization ofϕmL(X) andfm : Xm → Ym is the induced birational morphism.

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Proof

Xmµm

��

fm// Ymνm

��


νm is the normalization ofϕmL(X) andfm : Xm → Ym is the induced birational morphism.By construction we have that

µ∗m(mL) = f∗mAm + Fm

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Proof

Xmµm

��

fm// Ymνm

��


νm is the normalization ofϕmL(X) andfm : Xm → Ym is the induced birational morphism.By construction we have that

µ∗m(mL) = f∗mAm + Fm

with Am ample onYm, Fm effective and such thatSupp(Fm) = µ

−1m (B(L)) ⊆ µ

−1m (B+(L)) .

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Proof

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ProofThe key point is now to write, in two different ways,B+(µ

∗mL).

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∗mL).

By a result of Boucksom-Broustet-Pacienza we have

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∗mL).


B+(µ∗mL) = µ

−1m (B+(L)) ∪ Exc(µm).

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∗mL).


B+(µ∗mL) = µ

−1m (B+(L)) ∪ Exc(µm).

Moreover

B+(µ∗mL) = B+(f

∗mAm+Fm) ⊆ B+(f

∗mAm)∪Supp(Fm)

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∗mL).


B+(µ∗mL) = µ

−1m (B+(L)) ∪ Exc(µm).

Moreover

B+(µ∗mL) = B+(f

∗mAm+Fm) ⊆ B+(f

∗mAm)∪Supp(Fm)

and another application of the result of BBP gives

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∗mL).


B+(µ∗mL) = µ

−1m (B+(L)) ∪ Exc(µm).

Moreover

B+(µ∗mL) = B+(f

∗mAm+Fm) ⊆ B+(f

∗mAm)∪Supp(Fm)


µ−1m (B+(L)) ⊆ B+(µ∗mL) ⊆ Exc(fm) ∪ Supp(Fm).

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∗mL).


B+(µ∗mL) = µ

−1m (B+(L)) ∪ Exc(µm).

Moreover

B+(µ∗mL) = B+(f

∗mAm+Fm) ⊆ B+(f

∗mAm)∪Supp(Fm)



But we also have

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Proof

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ProofExc(fm) ⊆ µ

−1m (B+(L)) .

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ProofExc(fm) ⊆ µ

−1m (B+(L)) .

In fact let me recall the diagram

Xmµm

��

fm// Ymνm

��


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ProofExc(fm) ⊆ µ

−1m (B+(L)) .


Xmµm

��

fm// Ymνm

��


Now, if x ∈ Exc(fm), thenfm is not an isomorphismin a neighborhood ofx.

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ProofExc(fm) ⊆ µ

−1m (B+(L)) .


Xmµm

��

fm// Ymνm

��


Now, if x ∈ Exc(fm), thenfm is not an isomorphismin a neighborhood ofx.If x ∈ Exc(µm), thenµm(x) ∈ B(L) ⊆ B+(L),whencex ∈ µ−1m (B+(L)).

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ProofExc(fm) ⊆ µ

−1m (B+(L)) .


Xmµm

��

fm// Ymνm

��


Now, if x ∈ Exc(fm), thenfm is not an isomorphismin a neighborhood ofx.If x ∈ Exc(µm), thenµm(x) ∈ B(L) ⊆ B+(L),whencex ∈ µ−1m (B+(L)).If x 6∈ Exc(µm) andµm(x) 6∈ B+(L), thenµm, ϕmLandνm are isomorphisms in a neighborhood ofx,contradicting the fact thatfm is not an isomorphism ina neighborhood ofx.

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Proof

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ProofFrom the previous inclusions

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we conclude that

µ−1m (B+(L)) = Exc(fm) ∪ Supp(Fm).

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we conclude that


Recall that we have an irreducible componentZ ofB+(L) such thatZ 6⊆ B(L) and that we want to provethatvolX |Z(L) = 0.

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we conclude that


Recall that we have an irreducible componentZ ofB+(L) such thatZ 6⊆ B(L) and that we want to provethatvolX |Z(L) = 0.Now letZm be the strict trasform ofZ onXm, so thatZm ⊆ µ

−1m (B+(L)) is an irreducible component.

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we conclude that




Of courseZm 6⊆ Supp(Fm), sinceZ 6⊆ B(L).

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we conclude that




Of courseZm 6⊆ Supp(Fm), sinceZ 6⊆ B(L).HenceZm is an irreducible component ofExc(fm)

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Proof

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Proof

Xmµm

��

fm// Ymνm

��


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Proof

Xmµm

��

fm// Ymνm

��


and then it is contracted byfm, whence also byνm ◦ fm and we deduce thatZ is contracted byϕmL.

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Proof

Xmµm

��

fm// Ymνm

��


and then it is contracted byfm, whence also byνm ◦ fm and we deduce thatZ is contracted byϕmL.It follows thatZ is contracted by the map

Z 99K PH0(X|Z,mL).

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Proof

Xmµm

��

fm// Ymνm

��



Z 99K PH0(X|Z,mL).To show that this implies thatvolX |Z(L) = 0, we willuse the following lemma.

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Proof

Xmµm

��

fm// Ymνm

��



Z 99K PH0(X|Z,mL).To show that this implies thatvolX |Z(L) = 0, we willuse the following lemma.We recall a definition

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Graded linear series

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Graded linear seriesLetL be a line bundle on a varietyZ.A graded linear seriesW is a sequence of subspaces

Wm ⊆ H0(Z,mL),m ∈ N such that

Wm ⊗Ws ⊆ Wm+s for everym, s ≥ 0.

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Wm ⊗Ws ⊆ Wm+s for everym, s ≥ 0.LetΨm : Z 99K PWm be the associated rational mapand let

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κ(W ) = max {dimΨm(Z) | m ∈ N such thatWm 6= 0} .

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Lemma

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LemmaThere existsC > 0 such that

dimWm ≤ Cmκ(W ).

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LemmaThere existsC > 0 such that

dimWm ≤ Cmκ(W ).

This lemma follows from a deep result ofKaveh-Kovanskii (using Okounkov bodies), but wewill give a simple proof ispired by a paper of DiBiagio-Pacienza. – p. 17/23

Conclusion of proof

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Conclusion of proofThe Lemma implies the theorem:

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Conclusion of proofThe Lemma implies the theorem:SetWm = H0(X|Z,mL). SinceZ is contracted byΨm, it follows thatκ(W ) < dimZ =: d, whence

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d!

mddimH0(X|Z,mL) ≤

≤ lim supm→+∞

d!

mdCmκ(W ) = 0.

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d!

mddimH0(X|Z,mL) ≤

≤ lim supm→+∞

d!

mdCmκ(W ) = 0.

Now the proof of the Lemma, that isdimWm ≤ Cm

κ(W ).

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d!

mddimH0(X|Z,mL) ≤

≤ lim supm→+∞

d!

mdCmκ(W ) = 0.

Now the proof of the Lemma, that isdimWm ≤ Cm

κ(W ).We can assume that the base filed is uncountable,since the estimate is invariant under base fieldextension.

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Proof of Lemma

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Proof of LemmaIf κ(W ) = dim(Z) then

dimWm ≤ h0(mL) ≤ Cmκ(W ).

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If κ(W ) < dim(Z), we choose a completeintersectionT ⊆ Z of very ample divisors general intheir linear systems and such that

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dimT = κ(W ) and Ψm(T ) = Ψm(Z).

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dimT = κ(W ) and Ψm(T ) = Ψm(Z).(here we use that the base field is uncountable, as weimpose countably many conditions).

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dimT = κ(W ) and Ψm(T ) = Ψm(Z).(here we use that the base field is uncountable, as weimpose countably many conditions).We now claim that the restriction map

Wm → H0(T,mL|T ) is injective for m ≫ 0

(and this will prove the Lemma).– p. 19/23

Proof of Lemma

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Proof of LemmaIn fact, if there is a nonzero sections ∈ Wm vanishingonT and with zero divisorD,

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Proof of LemmaIn fact, if there is a nonzero sections ∈ Wm vanishingonT and with zero divisorD,thenΨm(Supp(D)) = Ψm(Z). But this is acontradiction sinceΨm(Supp(D)) is, by definition ofΨm, a hyperplane section ofΨm(Z).

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Proof of LemmaIn fact, if there is a nonzero sections ∈ Wm vanishingonT and with zero divisorD,thenΨm(Supp(D)) = Ψm(Z). But this is acontradiction sinceΨm(Supp(D)) is, by definition ofΨm, a hyperplane section ofΨm(Z).So this concludes the proof of the Lemma and of theTheorem.

– p. 20/23


Now I would like to briefly come back to the fact,mentioned in the beginning, that, given a big linebundleL on a varietyX, then

– p. 20/23


Now I would like to briefly come back to the fact,mentioned in the beginning, that, given a big linebundleL on a varietyX, thenthe mapsϕmL are an isomorphism onX −B+(L) form ≫ 0.

– p. 20/23


Now I would like to briefly come back to the fact,mentioned in the beginning, that, given a big linebundleL on a varietyX, thenthe mapsϕmL are an isomorphism onX −B+(L) form ≫ 0.Using the same method of proof of the previoustheorem, we can prove

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Another result (folklore)

– p. 21/23

Another result (folklore)Theorem (Boucksom, Cacciola, -)

– p. 21/23

Another result (folklore)Theorem (Boucksom, Cacciola, -)LetX be a normal projective variety over analgebraically closed filed of arbitrary characteristic.

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Another result (folklore)Theorem (Boucksom, Cacciola, -)LetX be a normal projective variety over analgebraically closed filed of arbitrary characteristic.LetL be a big line bundle onX.

– p. 21/23

Another result (folklore)Theorem (Boucksom, Cacciola, -)LetX be a normal projective variety over analgebraically closed filed of arbitrary characteristic.LetL be a big line bundle onX.ThenX −B+(L) is the largest Zariski open subsetU ⊆ X −B(L) such that

– p. 21/23

Another result (folklore)Theorem (Boucksom, Cacciola, -)LetX be a normal projective variety over analgebraically closed filed of arbitrary characteristic.LetL be a big line bundle onX.ThenX −B+(L) is the largest Zariski open subsetU ⊆ X −B(L) such thatfor m ≫ 0 and divisible, the restriction toU of themorphism

ϕmL : X −B(L) → PH0(X,mL)

is an isomorphism onto its image.

– p. 21/23

Another result (folklore)Theorem (Boucksom, Cacciola, -)LetX be a normal projective variety over analgebraically closed filed of arbitrary characteristic.LetL be a big line bundle onX.ThenX −B+(L) is the largest Zariski open subsetU ⊆ X −B(L) such thatfor m ≫ 0 and divisible, the restriction toU of themorphism

ϕmL : X −B(L) → PH0(X,mL)

is an isomorphism onto its image.

So the two theorems work forQ-divisors. What aboutreal divisors? (I must say that ELMNP’s theorem, inits "continuity version", also holds for real divisors)

– p. 21/23

Real divisors

– p. 22/23

Real divisorsHere there are several problems to generalize ourapproach in the case of real divisorsD:

– p. 22/23


(a) the mapsϕmD do not exist;

– p. 22/23


(a) the mapsϕmD do not exist;(b) the restricted volumevolX |Z(D) is not defined (itis defined, by ELMNP, only whenZ 6⊆ B+(D)).

– p. 22/23


(a) the mapsϕmD do not exist;(b) the restricted volumevolX |Z(D) is not defined (itis defined, by ELMNP, only whenZ 6⊆ B+(D)).(note that it does not seem to work to use⌊mD⌋).

– p. 22/23


(a) the mapsϕmD do not exist;(b) the restricted volumevolX |Z(D) is not defined (itis defined, by ELMNP, only whenZ 6⊆ B+(D)).(note that it does not seem to work to use⌊mD⌋).Nevertheless we are able to generalize toR-divisorsusing a recent idea of Birkar (used to proveNakamaye’s theorem on any scheme).

– p. 22/23

Real divisors

– p. 23/23

Real divisorsGiven anR-divisorD we can write

– p. 23/23


(∗) D ∼R t1H1 + . . . + tsHs

where, for1 ≤ i ≤ s, Hi is a very ample Cartierdivisor onX andti ∈ R.

– p. 23/23


(∗) D ∼R t1H1 + . . . + tsHs

where, for1 ≤ i ≤ s, Hi is a very ample Cartierdivisor onX andti ∈ R.Form ∈ N we set

〈mD〉 = ⌊mt1⌋H1 + . . . + ⌊mts⌋Hs

– p. 23/23


(∗) D ∼R t1H1 + . . . + tsHs


〈mD〉 = ⌊mt1⌋H1 + . . . + ⌊mts⌋HsΦ〈mD〉 : X 99K PH

0(X, 〈mD〉)

– p. 23/23


(∗) D ∼R t1H1 + . . . + tsHs



0(X, 〈mD〉)and

volX |Z(D, (∗)) = lim supm→+∞

h0(X|Z, 〈mD〉)

md/d!.

– p. 23/23


(∗) D ∼R t1H1 + . . . + tsHs



0(X, 〈mD〉)and

volX |Z(D, (∗)) = lim supm→+∞

h0(X|Z, 〈mD〉)

md/d!.

Now the two theorems above go through.– p. 23/23

OverviewOverviewOverviewOverview

IntroductionIntroductionIntroductionIntroductionIntroductionIntroduction

Stable base lociStable base lociStable base lociStable base lociStable base lociStable base lociStable base loci

Augmented base lociAugmented base lociAugmented base lociAugmented base lociAugmented base lociAugmented base lociAugmented base loci

Augmented base lociAugmented base lociAugmented base lociAugmented base lociAugmented base lociAugmented base lociAugmented base loci

Restricted volumesRestricted volumesRestricted volumesRestricted volumesRestricted volumes

Restricted volumes and $B _+$Restricted volumes and $B _+$Restricted volumes and $B _+$Restricted volumes and $B _+$Restricted volumes and $B _+$Restricted volumes and $B _+$Restricted volumes and $B _+$Restricted volumes and $B _+$

Restricted volumes and $B _+$Restricted volumes and $B _+$Restricted volumes and $B _+$Restricted volumes and $B _+$Restricted volumes and $B _+$Restricted volumes and $B _+$Restricted volumes and $B _+$

ResultsResultsResultsResultsResultsResultsResultsResults

ProofProofProofProofProofProofProof

ProofProofProofProofProof

ProofProofProofProofProofProofProofProof

ProofProofProofProofProofProof

ProofProofProofProofProofProofProofProof

ProofProofProofProofProofProof

Graded linear seriesGraded linear seriesGraded linear seriesGraded linear seriesGraded linear seriesGraded linear seriesGraded linear series

Conclusion of proofConclusion of proofConclusion of proofConclusion of proofConclusion of proofConclusion of proof

Proof of LemmaProof of LemmaProof of LemmaProof of LemmaProof of LemmaProof of Lemma

Proof of LemmaProof of LemmaProof of LemmaProof of LemmaProof of LemmaProof of LemmaProof of Lemma

Another result (folklore)Another result (folklore)Another result (folklore)Another result (folklore)Another result (folklore)Another result (folklore)Another result (folklore)

Real divisorsReal divisorsReal divisorsReal divisorsReal divisorsReal divisors

Real divisorsReal divisorsReal divisorsReal divisorsReal divisorsReal divisorsReal divisors

p. 1/23 · Longyearbyen (Spitsbergen), July 16, 2014 Joint work with Sébastien Boucksom...

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Transcript of p. 1/23 · Longyearbyen (Spitsbergen), July 16, 2014 Joint work with Sébastien Boucksom...