Geometry dictates arithmetic

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Ronald van Luijk

February 21, 2013 Utrecht

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Curves

Example. Circle given byx2+ y2 = 1(or projective closure in P2).

Definition.

Genusof a smooth projective curveC overQis the genus ofC (C).

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Example. Circle given byx2+ y2 = 1(or projective closure in P2). Definition.

Genusof a smooth projective curveC overQis the genus ofC (C).

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Genus 0

P2 ⊃ C : x2+ y2= 1 (0, 1) (1, 0) 4 5, 3 5 (−1, 0) y =t(x +1) 1−t2 1+t2, 2t 1+t2

Theorem. If a conic over Qhas

a rational point, then it has in-finitely many.

Theorem. If a conic D overQ

has a rational point, then there is an isomorphismP1(C) → D(C),

so the genus of D is 0.

Theorem. Any curve of genus 0

overQis isomorphic to a conic. Theorem. If a curve of genus 0

overQhas a rational point, then

it is isomorphic to P1 and it has

infinitely many rational points.

B 0 20 40 60 80 100 0 25 50 75 100 125 150 number of a c, b c with |a|, |b|, |c| ≤ B NC(B) = 11 61, 60 61 NC(B) ∼ 4π · B

Theorem. The numberND(B)

of rational points on a conic D

grows linearly with the height B

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Genus 0

P2 ⊃ C : x2+ y2= 1 (0, 1) (1, 0) 4 5, 3 5 (−1, 0) y =t(x +1) 1−t2 1+t2, 2t 1+t2 _{finitely many.}

B 0 20 40 60 80 100 0 25 50 75 100 125 150 number of _c,_c with |a|, |b|, |c| ≤ B 11 61, 60 61 NC(B) ∼ 4π · B

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Genus 0

P2 ⊃ C : x2+ y2= 1 (0, 1) (1, 0) 4 5, 3 5 (−1, 0) y =t(x +1) 1−t2 1+t2, 2t 1+t2

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Genus 0

P2 ⊃ C : x2+ y2= 1 (0, 1) (1, 0) 4 5, 3 5 (−1, 0) y =t(x +1) 1−t2 1+t2, 2t 1+t2 finitely many.

B 0 20 40 60 80 100 0 25 50 75 100 125 150 number of _c,_c with |a|, |b|, |c| ≤ B 11 61, 60 61 NC(B) ∼ 4π · B

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Genus 0

P2 ⊃ C : x2+ y2= 1 (0, 1) (1, 0) 4 5, 3 5 (−1, 0) y =t(x +1) 1−t2 1+t2, 2t 1+t2

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Genus 0

P2 ⊃ C : x2+ y2= 1 (0, 1) (1, 0) 4 5, 3 5 (−1, 0) y =t(x +1) 1−t2 1+t2, 2t 1+t2

B 0 20 40 60 80 100 0 25 50 75 100 125 c c with |a|, |b|, |c| ≤ B 61,61 NC(B) ∼ 4π · B

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Genus 0

P2 ⊃ C : x2+ y2= 1 (0, 1) (1, 0) 4 5, 3 5 (−1, 0) y =t(x +1) 1−t2 1+t2, 2t 1+t2

overQis isomorphic to a conic.

Theorem. If a curve of genus 0

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Genus 0

P2 ⊃ C : x2+ y2= 1 (0, 1) (1, 0) 4 5, 3 5 (−1, 0) y =t(x +1) 1−t2 1+t2, 2t 1+t2

B 0 20 40 60 80 100 0 25 50 75 100 125 c c with |a|, |b|, |c| ≤ B 61,61 NC(B) ∼ 4π · B

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Genus 0

P2 ⊃ C : x2+ y2= 1 (0, 1) (1, 0) 4 5, 3 5 (−1, 0) y =t(x +1) 1−t2 1+t2, 2t 1+t2

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Genus 0

P2 ⊃ C : x2+ y2= 1 (0, 1) (1, 0) 4 5, 3 5 (−1, 0) y =t(x +1) 1−t2 1+t2, 2t 1+t2 _{finitely many.}

B 0 20 40 60 80 100 0 25 50 75 100 125 150 number of a c, b c with |a|, |b|, |c| ≤ B NC(B) = 11 61, 60 61 NC(B) ∼ 4_π · B

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Genus 0

P2 ⊃ C : x2+ y2= 1 (0, 1) (1, 0) 4 5, 3 5 (−1, 0) y =t(x +1) 1−t2 1+t2, 2t 1+t2

B 0 20 40 60 80 100 0 25 50 75 100 125 150 number of a c, b c with |a|, |b|, |c| ≤ B NC(B) = 11 61, 60 61 NC(B) ∼ 4_π · B

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−20 −16 −12 −8 −4 4 8 12 16 20 −4 −3 −2 −1 1 2 3 4 5 6 7 8

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Genus 1 (elliptic)

−20 −16 −12 −8 −4 4 8 12 16 20 −4 −3 −2 −1 1 2 3 4 5 6 7 8 E : y2 = x3− 15x + 19 P Q

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−20 −16 −12 −8 −4 4 8 12 16 20 −4 −3 −2 −1 1 2 3 4 5 6 7 8 E : y2 = x3− 15x + 19 P Q

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Genus 1 (elliptic)

−20 −16 −12 −8 −4 4 8 12 16 20 −4 −3 −2 −1 1 2 3 4 5 6 7 8 E : y2 = x3− 15x + 19 P Q P + Q

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−20 −16 −12 −8 −4 4 8 12 16 20 −4 −3 −2 −1 1 2 3 4 5 6 7 8 E : y2 = x3− 15x + 19 P Q P + Q

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Genus 1 (elliptic)

−20 −16 −12 −8 −4 4 8 12 16 20 −4 −3 −2 −1 1 2 3 4 5 6 7 8 E : y2 = x3− 15x + 19

Fact. E (k)is an abelian group!

Theorem (Mordell-Weil). For any elliptic curve E

overQ, the groupE (Q) is

finitely generated.

Here: rank= 1, and

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−20 −16 −12 −8 −4 4 8 12 16 20 −4 −3 −2 −1 1 2 3 4 5 6 7 8 E : y2 = x3− 15x + 19 2622397863 362178961, 117375339855079 6892627806791 number of _ca,b_c with |a|, |b|, |c| ≤ B log B 0 300 600 900 1200 0 20 40 60 80 100 NE(B) ∼ γ √ log B γ = 2.6768125...

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Genus 1 (elliptic)

−20 −16 −12 −8 −4 4 8 12 16 20 −4 −3 −2 −1 1 2 3 4 5 6 7 8 2622397863 362178961, 117375339855079 6892627806791 number of _ca,b_c with |a|, |b|, |c| ≤ B NE(B) = log B 0 300 600 900 1200 0 20 40 60 80 100 NE(B) ∼ γ √ log B γ = 2.6768125...

Theorem. For any elliptic curveE

overQ withr = rank E (Q), we have NE(B) ∼ c(log B)r /2.

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Genus g ≥ 2

Examples.

I y2= f (x )with f separable of degree2g + 2.

I smooth projective plane curve of degree d ≥ 4 with

g = 1₂(d − 1)(d − 2).

Conclusion.

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Genus g ≥ 2

Examples.

g = 1₂(d − 1)(d − 2).

Theorem(“Mordell Conjecture” by Faltings, 1983).

Any curve overQwithg ≥ 2has only finitely many rational points.

Conclusion.

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Examples.

g = 1₂(d − 1)(d − 2).

Theorem(“Mordell Conjecture” by Faltings, 1983).

Any curve overQwithg ≥ 2has only finitely many rational points. Conclusion.

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Differentials

Definition.

LetX be a smooth projective variety with function fieldk(X ). ThenΩk(X )/k is the k(X )-vectorspace of differential 1-forms,

generated by{df : f ∈ k(X )} and satisfying

I d (f + g ) = df + dg,

I d (fg ) = fdg + gdf,

I da = 0 for a ∈ k.

Proposition. We have dim_{k(X )}Ω_{k(X )/k} = dim X.

Example.

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Definition.

LetX be a smooth projective variety with function fieldk(X ). ThenΩk(X )/k is the k(X )-vectorspace of differential 1-forms,

generated by{df : f ∈ k(X )} and satisfying

I d (f + g ) = df + dg,

I d (fg ) = fdg + gdf,

I da = 0 for a ∈ k.

Proposition. We have dim_{k(X )}Ω_{k(X )/k} = dim X.

Example.

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Holomorphic differentials on curves

Definition. For a point P on a smooth projective curve C with local parametertP ∈ k(C )and a differentialω ∈ Ωk(C )/k, we write

ω = fPdtP; then ω is holomorphicatP iffP has no pole atP.

Example.

CurveC : y2= f (x ) withf separable of degree d ≥ 3. Then

ω = 1 yd (x − c) = 1 ydx = 2 f0_{(x )}dy is holomorphic everywhere.

Definition. Set ΩC /k = {ω ∈ Ωk(C )/k : ω holom. everywhere}.

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Definition. For a point P on a smooth projective curve C with local parametertP ∈ k(C )and a differentialω ∈ Ωk(C )/k, we write

ω = fPdtP; then ω is holomorphicatP iffP has no pole atP.

Example.

CurveC : y2= f (x ) withf separable of degree d ≥ 3. Then

ω = 1 yd (x − c) = 1 ydx = 2 f0_{(x )}dy is holomorphic everywhere.

Definition. Set ΩC /k = {ω ∈ Ωk(C )/k : ω holom. everywhere}.

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Holomorphic differentials in general

Recall. IfX smooth, projective, thendim_{k(X )}Ω_{k(X )/k} = dim X.

Fact. IfV is a vector space withdim V = n, thendimVn

V = 1.

Definition(unconventional notation for(dim X )-forms). SetΩ_{X /k} = {ω ∈Vdim X

Ω_{k(X )/k} : ω holom. everywhere}.

Definition

(Wrong: use tensor powers ofVdim X

Ωk(X )/k.)

For ak-basis (ω0, ω1, . . . , ωN) ofΩX /k, we getfi ∈ k(X )

such thatωi = fiω0. TheKodaira dimension κ(X ) ofX is

−1ifdimkΩX /k = 0, or the dimension of the image of the map

X → AN, P 7→ (f1(P), f2(P), . . . , fN(P)).

Proposition. For a curve C we get

κ(C ) =      −1 g = 0 0 g = 1 1 g ≥ 2

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Holomorphic differentials in general

V = 1.

Definition

X → AN, P 7→ (f1(P), f2(P), . . . , fN(P)).

κ(C ) =      −1 g = 0 0 g = 1 1 g ≥ 2

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Holomorphic differentials in general

V = 1.

Definition

(Wrong: use tensor powers ofVdim X

Ωk(X )/k.)

X → AN, P 7→ (f1(P), f2(P), . . . , fN(P)).

κ(C ) =      −1 g = 0 0 g = 1 1 g ≥ 2

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Holomorphic differentials in general

V = 1.

Definition(Wrong: use tensor powers ofVdim X

Ωk(X )/k.)

X → AN, P 7→ (f1(P), f2(P), . . . , fN(P)). κ(C ) =     −1 g = 0 0 g = 1 1 g ≥ 2

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Holomorphic differentials in general

V = 1.

Definition(Wrong: use tensor powers ofVdim X

Ωk(X )/k.)

X → AN, P 7→ (f1(P), f2(P), . . . , fN(P)).

κ(C ) =      −1 g = 0 0 g = 1 1 g ≥ 2

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Varieties of general type

In general,−1 ≤ κ(X ) ≤ dim X (complexX ⇒ highκ(X )).

Definition. We say thatX is ofgeneral typeif κ(X ) = dim X. (“many” holom. differentials, “canonical bundle ispseudo-ample”)

points lie in aZariski closed subset, i.e., a finite union of proper subvarieties ofX.

Corollary. Let X ⊂ P3 be a smooth, projective surface over Qof

degree≥ 5. Then the rational points are all contained in some finite union of curves.

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Varieties of general type

Conjecture(Lang).

IfX is a variety overQthat is of general type, then the rational

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Conjecture(Lang).

IfX is a variety overQthat is of general type, then the rational

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Fano varieties

Definition. AFano varietyis a smooth, projective variety X with ample anti-canonical bundle.

We haveκ(X ) = −1and X is geometrically “easy”.

Conjecture(Batyrev-Manin).

SupposeX overQis Fano. Set ρ = rk Pic X.

There is an open subset U ⊂ X and a constant c with

NU(B) ∼ cB(log B)ρ−1.

This is proved in many cases for surfaces.

False in higher dimension, but no counterexamples to lower bound.

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Fano varieties

There is an open subsetU ⊂ X and a constant c with

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Fano varieties

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K3 surfaces

Definition. AK3 surface overQis a smooth, projective surfaceX

withX (C) simply connected and with trivial canonical bundle. There is a unique holomorphic differential and we haveκ(X ) = 0.

Examples

I Smooth quartic surfaces in P3.

I Double cover of P2 ramified over a smooth sextic.

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Definition. AK3 surface overQis a smooth, projective surfaceX

withX (C) simply connected and with trivial canonical bundle. There is a unique holomorphic differential and we haveκ(X ) = 0.

Examples

I Smooth quartic surfaces in P3.

I Double cover of P2 ramified over a smooth sextic.

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Theorem(Tschinkel-Bogomolov).

Ifrk Pic X ≥ 5, then there is a finite extension K ofQ such that

theK-rational points are Zariski dense onX, i.e., rational points arepotentially dense onX.

Question. Is there a K3 surface X over a number field with

rk Pic X = 1and rational points potentially dense?

rk Pic X = 1and rational points not potentially dense?

Question. Is there a K3 surface X over a number field K with

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Question. Is there a K3 surface X over a number fieldK with

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K3 surfaces

Theorem(Logan, McKinnon, vL).

Takea, b, c, d ∈ Q∗ with abcd ∈ (Q∗)2. LetX ⊂ P3 be given by

ax4+ by4+ cz4+ dw4.

IfP ∈ X (Q)has no zero coordinates and P does not lie on one of the48lines (no two terms sum to 0), then X (Q)is Zariski dense.

Question. Are the conditions on P necessary?

Conjecture(vL) Every t ∈ Qcan be written as

t = x

4_{− y}4

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Theorem(Logan, McKinnon, vL).

Takea, b, c, d ∈ Q∗ with abcd ∈ (Q∗)2. LetX ⊂ P3 be given by

ax4+ by4+ cz4+ dw4.

IfP ∈ X (Q)has no zero coordinates and P does not lie on one of the48lines (no two terms sum to 0), then X (Q)is Zariski dense.

Question. Are the conditions on P necessary?

Conjecture(vL) Every t ∈ Qcan be written as

t = x

4_{− y}4

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S : x3− 3x2_y2_{+ 4x}2_{yz − x}2_z2 + x2z − xy2z − xyz2+ x + y3+ y2z2+ z3 = 0 log B NU(B) 0 1 2 3 4 5 0 8 16 24 32 40 48 56 N ∼ 13.5 · log B

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Conjecture(vL).

SupposeX is a K3 surface overQ withrk Pic XC= 1.

There is an open subsetU ⊂ X and a constant c such that

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Hasse principle

Theorem(Hasse).

LetQ ⊂ Pn be a smooth quadric overQ. Suppose that Q has

points overRand overQp for every p. ThenQ(Q) 6= ∅.

Proposition(Selmer).

The curveC ⊂ P2 given by3x3+ 4y3+ 5z3= 0 has points over R

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Theorem(Hasse).

LetQ ⊂ Pn be a smooth quadric overQ. Suppose that Q has

points overRand overQp for every p. ThenQ(Q) 6= ∅.

Proposition(Selmer).

The curveC ⊂ P2 given by3x3+ 4y3+ 5z3= 0 has points over R

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Brauer-Manin obstruction

To every varietyX we can assign theBrauer group Br X.

Every morphismX → Y induces a homomorphism Br Y → Br X. For every pointP over a fieldk we haveBr(P) = Br(k).

LetX be smooth and projective. X (Q) // Q vX (Qv) φ %%K K K K K K K K K K Br(Q) //L vBr(Qv) //Q/Z Corollary. If Q vX (Qv) Br := φ−1(0) is empty, thenX (Q) = ∅.

Conjecture(Colliot-Th´el`ene).

ThisBrauer-Manin obstructionis the only obstruction to the existence of rational points forrationally connected varieties.

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Brauer-Manin obstruction

LetX be smooth and projective. X (Q) // Q vX (Qv) φ %%K K K K K K K K K K Br(Q) //L vBr(Qv) //Q/Z

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Brauer-Manin obstruction

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