exeter, january 22nd 2016 - julio c....

Post on 23-Sep-2020

0 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

TRANSCRIPT

Algebraic geometry over finite fields via counting

Tim Browning

University of Bristol

Exeter, January 22nd 2016

What are the simplest varieties?

What are the simplest algebraic varieties X ⊂ PN defined over afield k?

In dimension 1: rational curves (curves of genus 0)

In higher dimension: lots of rational curves P1 → X .

Definition

A rational curve on X is a non-constant morphism P1 → X .

A morphism f : P1 → X of degree e is given by

f = (f0(u, v), . . . , fN(u, v)),

with f0, . . . , fN ∈ k[u, v ] forms of degree e, with no non-constantcommon factor in k[u, v ], such that

Fi (f0(u, v), . . . , fN(u, v)) ≡ 0, (1 6 i 6 R),

where F1, . . . ,FR ∈ k[x0, . . . , xN ] are the polynomials defining X .

What are the simplest varieties?

What are the simplest algebraic varieties X ⊂ PN defined over afield k?

In dimension 1: rational curves (curves of genus 0)

In higher dimension: lots of rational curves P1 → X .

Definition

A rational curve on X is a non-constant morphism P1 → X .

A morphism f : P1 → X of degree e is given by

f = (f0(u, v), . . . , fN(u, v)),

with f0, . . . , fN ∈ k[u, v ] forms of degree e, with no non-constantcommon factor in k[u, v ], such that

Fi (f0(u, v), . . . , fN(u, v)) ≡ 0, (1 6 i 6 R),

where F1, . . . ,FR ∈ k[x0, . . . , xN ] are the polynomials defining X .

What are the simplest varieties?

What are the simplest algebraic varieties X ⊂ PN defined over afield k?

In dimension 1: rational curves (curves of genus 0)

In higher dimension: lots of rational curves P1 → X .

Definition

A rational curve on X is a non-constant morphism P1 → X .

A morphism f : P1 → X of degree e is given by

f = (f0(u, v), . . . , fN(u, v)),

with f0, . . . , fN ∈ k[u, v ] forms of degree e, with no non-constantcommon factor in k[u, v ], such that

Fi (f0(u, v), . . . , fN(u, v)) ≡ 0, (1 6 i 6 R),

where F1, . . . ,FR ∈ k[x0, . . . , xN ] are the polynomials defining X .

Rationally Connected Unirational Rational

X is RC: ∀x1, x2 ∈ X there exists a P1 → X through themX is unirational: ∃ onto rational map Pdim X → XX is rational: if there is also a rational map X → Pdim X

Example

Take X = {x2 + y2 = z2} ⊂ P2. Here P1 → X is given by[s : t] 7→ [s2 − t2 : 2st : s2 + t2] and X → P1 is given by[x : y : z ] 7→ [x/z : y/z ]

Rationally Connected Unirational Rational

X is RC: ∀x1, x2 ∈ X there exists a P1 → X through themX is unirational: ∃ onto rational map Pdim X → XX is rational: if there is also a rational map X → Pdim X

Example

Take X = {x2 + y2 = z2} ⊂ P2. Here P1 → X is given by[s : t] 7→ [s2 − t2 : 2st : s2 + t2] and X → P1 is given by[x : y : z ] 7→ [x/z : y/z ]

{Rationally Connected}??) {Unirational}

Clemens−Griffiths) {Rational}

Suppose X ⊂ Pn is a non-singular hypersurface of degree d .

X is RC if n + 1 > d . When is it unirational?

Example (Harris–Mazur-Pandaripande 1998)

Assume k = k and char(k) = 0. X is unirational if n > d ↑↑ d

What about X in a “reasonable” number of variables?! Can weeven show X contains a rational surface?

Study rational curves on X ...

{Rationally Connected}??) {Unirational}

Clemens−Griffiths) {Rational}

Suppose X ⊂ Pn is a non-singular hypersurface of degree d .

X is RC if n + 1 > d . When is it unirational?

Example (Harris–Mazur-Pandaripande 1998)

Assume k = k and char(k) = 0. X is unirational if n > d ↑↑ d

What about X in a “reasonable” number of variables?! Can weeven show X contains a rational surface?

Study rational curves on X ...

{Rationally Connected}??) {Unirational}

Clemens−Griffiths) {Rational}

Suppose X ⊂ Pn is a non-singular hypersurface of degree d .

X is RC if n + 1 > d . When is it unirational?

Example (Harris–Mazur-Pandaripande 1998)

Assume k = k and char(k) = 0. X is unirational if n > d ↑↑ d

What about X in a “reasonable” number of variables?! Can weeven show X contains a rational surface?

Study rational curves on X ...

Rational curves

k is a field and X = {F = 0} ⊂ Pn, with

F ∈ k[x0, . . . , xn]

a non-singular form of degree d .

Let More(X ) be space of rational curves1 on X :

{f ∈ More(X )} ←→{

open set in P(n+1)(e+1)−1

de + 1 equations of degree d

}

1morphism f : P1 → X of degree e is given by f = (f0(u, v), . . . , fn(u, v)),with f0, . . . , fn ∈ k[u, v ] forms of degree e, with no non-constant commonfactor, such that F (f0(u, v), . . . , fn(u, v)) ≡ 0.

{f ∈ More(X )} ←→{

open set in P(n+1)(e+1)−1

de + 1 equations of degree d

}

Questions

dim More(X )??= (n+1)(e+1)−1−(de+1) = e(n+1−d)+n−1.

Is More(X ) irreducible?

Is More(X )(k) non-empty?

Coskun and Starr (2009): Yes! (if d = 3, n > 5 and k = C).

Goal

Use analytic number theory to answer these questions when k = Fq

{f ∈ More(X )} ←→{

open set in P(n+1)(e+1)−1

de + 1 equations of degree d

}

Questions

dim More(X )??= (n+1)(e+1)−1−(de+1) = e(n+1−d)+n−1.

Is More(X ) irreducible?

Is More(X )(k) non-empty?

Coskun and Starr (2009): Yes! (if d = 3, n > 5 and k = C).

Goal

Use analytic number theory to answer these questions when k = Fq

The players

We are interested in the following data:

k = Fq a finite field

F ∈ k[x0, . . . , xn] a non-singular form of2 degree 3

X ⊂ Pn the hypersurface defined by F = 0

More(X ) the space of rational curves of degree e that arecontained in X

Example

For the case n = 3 of cubic surfaces, Mor1(X ) has 27 componentswhich are permuted by the action of Gal(k/k).

Describing More(X ) for dim(X ) > 2 is difficult!

2For degree > 3 (or anything to do with x) see Alan Lee (2011), a PhDstudent of Trevor Wooley

The players

We are interested in the following data:

k = Fq a finite field

F ∈ k[x0, . . . , xn] a non-singular form of2 degree 3

X ⊂ Pn the hypersurface defined by F = 0

More(X ) the space of rational curves of degree e that arecontained in X

Example

For the case n = 3 of cubic surfaces, Mor1(X ) has 27 componentswhich are permuted by the action of Gal(k/k).

Describing More(X ) for dim(X ) > 2 is difficult!

2For degree > 3 (or anything to do with x) see Alan Lee (2011), a PhDstudent of Trevor Wooley

k-points

Question: When is X (k) 6= ∅?

Theorem (Chevalley–Warning, 1936)

f1, . . . , fr ⊂ k[x0, . . . , xn] homogenous polynomials such that

n >r∑

j=1

deg fj .

Then ∃ x ∈ kn+1 \ {0} such that fj (x) = 0 for 1 6 j 6 r .

Answer: If n > 3

k-points

Question: When is X (k) 6= ∅?

Theorem (Chevalley–Warning, 1936)

f1, . . . , fr ⊂ k[x0, . . . , xn] homogenous polynomials such that

n >r∑

j=1

deg fj .

Then ∃ x ∈ kn+1 \ {0} such that fj (x) = 0 for 1 6 j 6 r .

Answer: If n > 3

Question: When is #X (k) > 2?

Example (Swinnerton-Dyer)

The cubic surface

x30 + x3

1 + x32 + x2

0x1 + x21x2 + x2

2x0 + x0x1x2 + x2x23 + x2

2x3 = 0

has exactly one point over F2.

Answer: This is the only exception!

Question: When is #X (k) > 2?

Example (Swinnerton-Dyer)

The cubic surface

x30 + x3

1 + x32 + x2

0x1 + x21x2 + x2

2x0 + x0x1x2 + x2x23 + x2

2x3 = 0

has exactly one point over F2.

Answer: This is the only exception!

k-rational curves

Question: When is More(X )(k) 6= ∅?

We know n > 3⇒ ∃ p ∈ X (k)

∃ Fq-lines in X through p ⇐⇒ ∃ q ∈ X (k) such that

0 ≡ F (λp + µq) = λ2µ q.∇F (p)︸ ︷︷ ︸linear

+λµ2 p.∇F (q)︸ ︷︷ ︸quadratic

+µ3 F (q)︸︷︷︸cubic

Answer: Chevalley–Warning ⇒ Mor1(X )(k) 6= ∅ if

n > 1 + 2 + 3 = 6

k-rational curves

Question: When is More(X )(k) 6= ∅?

We know n > 3⇒ ∃ p ∈ X (k)

∃ Fq-lines in X through p ⇐⇒ ∃ q ∈ X (k) such that

0 ≡ F (λp + µq) = λ2µ q.∇F (p)︸ ︷︷ ︸linear

+λµ2 p.∇F (q)︸ ︷︷ ︸quadratic

+µ3 F (q)︸︷︷︸cubic

Answer: Chevalley–Warning ⇒ Mor1(X )(k) 6= ∅ if

n > 1 + 2 + 3 = 6

k-rational curves

Question: When is More(X )(k) 6= ∅?

We know n > 3⇒ ∃ p ∈ X (k)

∃ Fq-lines in X through p ⇐⇒ ∃ q ∈ X (k) such that

0 ≡ F (λp + µq) = λ2µ q.∇F (p)︸ ︷︷ ︸linear

+λµ2 p.∇F (q)︸ ︷︷ ︸quadratic

+µ3 F (q)︸︷︷︸cubic

Answer: Chevalley–Warning ⇒ Mor1(X )(k) 6= ∅ if

n > 1 + 2 + 3 = 6

Main result

Theorem (B. & Pankaj Vishe, 2015)

Assume char(k) > 3 and let e > 1. Let X ⊂ Pn be a non-singularcubic hypersurface over k , with n > 12. Then More(X ) isirreducible and has the expected dimension e(n − 2) + n − 1.

Remarks:

This + Lang–Weil ⇒ More(X )(k) 6= ∅ if #k large enough

Pugin (2011): Special case of diagonal cubic hypersurfacesa0x

30 + · · ·+ anx

3n = 0, for a0, . . . , an ∈ k∗.

What about space More(x ;X ) of rational curves of degree eon X passing through x ∈ X?

Smaller n? Larger degree hypersurfaces?

Main result

Theorem (B. & Pankaj Vishe, 2015)

Assume char(k) > 3 and let e > 1. Let X ⊂ Pn be a non-singularcubic hypersurface over k , with n > 12. Then More(X ) isirreducible and has the expected dimension e(n − 2) + n − 1.

Remarks:

This + Lang–Weil ⇒ More(X )(k) 6= ∅ if #k large enough

Pugin (2011): Special case of diagonal cubic hypersurfacesa0x

30 + · · ·+ anx

3n = 0, for a0, . . . , an ∈ k∗.

What about space More(x ;X ) of rational curves of degree eon X passing through x ∈ X?

Smaller n? Larger degree hypersurfaces?

From geometry to counting

Assume that #k = q and put E = e(n − 2) + n − 1.

Theorem (Lang–Weil, 1953)

Let V ⊂ PN be a variety defined over k . Then

lim supq→∞

q− dim V #V (k) = # irred components of V

It therefore suffices to prove:

lim sup`→∞

q−`E # More(X )(Fq`) 6 1.

Key idea{Fq`-points on More(X )

}←→

{Fq`(t)-points on X of degree e

}

From geometry to counting

Assume that #k = q and put E = e(n − 2) + n − 1.

Theorem (Lang–Weil, 1953)

Let V ⊂ PN be a variety defined over k . Then

lim supq→∞

q− dim V #V (k) = # irred components of V

It therefore suffices to prove:

lim sup`→∞

q−`E # More(X )(Fq`) 6 1.

Key idea{Fq`-points on More(X )

}←→

{Fq`(t)-points on X of degree e

}

Function field dictionary

Idea: Count points in X (Fq(t)) of degree e using Fq(t)-version ofthe Hardy–Littlewood circle method

Q K = Fq(t)

Z O = Fq[t]p prime π ∈ O monic and irreducible

residue field Z/pZ = Fp Fπ = Fqdeg π

∞ t prime at infinityabs. values | · |, | · |p | · | = | · |∞, | · |πcompletions R,Qp K∞ = Fq((t−1)),KπFourier analysis [0, 1] T = {x ∈ K∞ : |x | < 1}add. character exp(2πi ·) ψ(x) = exp(

2πiTrFq/Fp a−1

p ) if

x =∑

i6N ai ti ∈ K∞

Here: |a/b| = qdeg a−deg b if a, b ∈ O and b 6= 0.

Recall: X = {F = 0} ⊂ Pn defined by a non-singular cubic formF ∈ Fq[x0, . . . , xn].

Goal

Estimate

N(q, e) = #

{x ∈ On+1 : F (x) = 0, max

06i6n|xi | 6 qe

}as q →∞

Can use this to make deductions about # More(X )(Fq)...

Enter the adelic circle method...

N(q, e) = #

{x ∈ On+1 : F (x) = 0, max

06i6n|xi | 6 qe

}=

∫TS(α)dα, where S(α) =

∑x∈On+1: maxi |xi |6qe

ψ(αF (x))

by orthogonality of characters

=∑|r |<qQ

r monic

∑|a|<|r |

gcd(a,r)=1

∫|θ|<|r |−1q−Q

S(a/r + θ)dθ

by Dirichlet + ultrametric inequality (any Q > 1 )

= qe(n+1)∑|r |<qQ

r monic

|r |−(n+1)

∫|θ|<|r |−1q−Q

∑c∈On+1

Sr (c)Ir (c)dθ

by adelic Poisson summation

N(q, e) = qe(n+1)∑|r |<qQ

r monic

|r |−(n+1)

∫|θ|<|r |−1q−Q

∑c∈On+1

Sr (c)Ir (c)dθ

where

Sr (c) =∑|a|<|r |

gcd(a,r)=1

∑y∈On+1

|yi |<|r |

ψ

(aF (y)− c.y

r

)complete exp. sum

Ir (c) =

∫Tn+1

ψ

(θdqdeF (x) +

qec.x

r

)dx oscill. integral

Main contribution: c = 0Error term: c 6= 0

N(q, e) = qe(n+1)∑|r |<qQ

r monic

|r |−(n+1)

∫|θ|<|r |−1q−Q

∑c∈On+1

Sr (c)Ir (c)dθ

where

Sr (c) =∑|a|<|r |

gcd(a,r)=1

∑y∈On+1

|yi |<|r |

ψ

(aF (y)− c.y

r

)complete exp. sum

Ir (c) =

∫Tn+1

ψ

(θdqdeF (x) +

qec.x

r

)dx oscill. integral

Main contribution: c = 0Error term: c 6= 0

Back to varieties over finite fields

Need to analyse

Sr (c) =∑|a|<|r |

gcd(a,r)=1

∑y∈On+1

|yi |<|r |

ψ

(aF (y)− c.y

r

)

r = πν11 . . . πνs

s ⇒ Sr (c) = Sπν11

(c) . . . Sπνss

(c)

ν > 1: Elementary treatment...ν = 1: Then

Sπ(c) = |π| {|π|#Xc(Fπ)−#X (Fπ) + 1}

where Xc = X ∩ {∑n

i=0 cixi = 0} has dim(Xc) = n − 2

Back to varieties over finite fields

Need to analyse

Sr (c) =∑|a|<|r |

gcd(a,r)=1

∑y∈On+1

|yi |<|r |

ψ

(aF (y)− c.y

r

)

r = πν11 . . . πνs

s ⇒ Sr (c) = Sπν11

(c) . . . Sπνss

(c)

ν > 1: Elementary treatment...ν = 1: Then

Sπ(c) = |π| {|π|#Xc(Fπ)−#X (Fπ) + 1}

where Xc = X ∩ {∑n

i=0 cixi = 0} has dim(Xc) = n − 2

Sπ(c) = |π| {|π|#Xc(Fπ)−#X (Fπ) + 1}

For generic3 values of c, Deligne (Weil I) ⇒

Sπ(c) = (−1)n−2|π|2bn−2∑j=1

ωn−2,j + O(|π|n+1

2 ),

where ωn−2,j are eigenvalues of Frobenius acting on Hn−2et (X c,Q`).

These satisfy |ωn−2,j | = |π|n−2

2 .

Hence r square-free ⇒ Sr (c) = O(|r |n+2

2 )

...this is square-root cancellation!

3i.e. [c] 6∈ X dual

Sπ(c) = |π| {|π|#Xc(Fπ)−#X (Fπ) + 1}

For generic3 values of c, Deligne (Weil I) ⇒

Sπ(c) = (−1)n−2|π|2bn−2∑j=1

ωn−2,j + O(|π|n+1

2 ),

where ωn−2,j are eigenvalues of Frobenius acting on Hn−2et (X c,Q`).

These satisfy |ωn−2,j | = |π|n−2

2 .

Hence r square-free ⇒ Sr (c) = O(|r |n+2

2 )

...this is square-root cancellation!

3i.e. [c] 6∈ X dual

e-aspect?

Since we are interested in q →∞ limit, need to make all theimplied constants uniform in q in above strategy.

The circle method can also be applied to deal with e-aspect. Infact, for fixed q, we have

n > 7⇒ lime→∞

q−e(n−2)N(q, d) = c∞∏π

⇒ lime→∞

q−e(n−2)N(q, d) > 0 if X (Kπ) 6= ∅ ∀ primes π

Theorem (B. & Pankaj Vishe, 2015)

Let K = Fq(t) and assume that char(Fq) > 3. Let X ⊂ Pn be anon-singular cubic hypersurface over K , with n > 7. Then Xsatisfies the Manin conjecture and the Hasse principle over K .

Newsflash! (Zhiyu Tian, 2015)

Hasse Principle for n > 5 and char(Fq) > 5.

e-aspect?

Since we are interested in q →∞ limit, need to make all theimplied constants uniform in q in above strategy.

The circle method can also be applied to deal with e-aspect. Infact, for fixed q, we have

n > 7⇒ lime→∞

q−e(n−2)N(q, d) = c∞∏π

⇒ lime→∞

q−e(n−2)N(q, d) > 0 if X (Kπ) 6= ∅ ∀ primes π

Theorem (B. & Pankaj Vishe, 2015)

Let K = Fq(t) and assume that char(Fq) > 3. Let X ⊂ Pn be anon-singular cubic hypersurface over K , with n > 7. Then Xsatisfies the Manin conjecture and the Hasse principle over K .

Newsflash! (Zhiyu Tian, 2015)

Hasse Principle for n > 5 and char(Fq) > 5.

e-aspect?

Since we are interested in q →∞ limit, need to make all theimplied constants uniform in q in above strategy.

The circle method can also be applied to deal with e-aspect. Infact, for fixed q, we have

n > 7⇒ lime→∞

q−e(n−2)N(q, d) = c∞∏π

⇒ lime→∞

q−e(n−2)N(q, d) > 0 if X (Kπ) 6= ∅ ∀ primes π

Theorem (B. & Pankaj Vishe, 2015)

Let K = Fq(t) and assume that char(Fq) > 3. Let X ⊂ Pn be anon-singular cubic hypersurface over K , with n > 7. Then Xsatisfies the Manin conjecture and the Hasse principle over K .

Newsflash! (Zhiyu Tian, 2015)

Hasse Principle for n > 5 and char(Fq) > 5.

e-aspect?

Since we are interested in q →∞ limit, need to make all theimplied constants uniform in q in above strategy.

The circle method can also be applied to deal with e-aspect. Infact, for fixed q, we have

n > 7⇒ lime→∞

q−e(n−2)N(q, d) = c∞∏π

⇒ lime→∞

q−e(n−2)N(q, d) > 0 if X (Kπ) 6= ∅ ∀ primes π

Theorem (B. & Pankaj Vishe, 2015)

Let K = Fq(t) and assume that char(Fq) > 3. Let X ⊂ Pn be anon-singular cubic hypersurface over K , with n > 7. Then Xsatisfies the Manin conjecture and the Hasse principle over K .

Newsflash! (Zhiyu Tian, 2015)

Hasse Principle for n > 5 and char(Fq) > 5.

What next?

Higher degree.Extend investigation of More(X ) to hypersurfaces of degree > 3over finite fields (a la Alan Lee)

A Manin conjecture for Campana orbifolds.The Manin conjecture makes predictions about equidistribution ofK -rational points on projective varieties over global fields K .

There is a version emerging for integral points on Campanaorbifolds (P1,∆) with χ(∆) > 0.

Example

e-aspect: is it true that

#

{(x , y , z) ∈ Fq[t] :

x , y , z square-full & degree ex + y = z

}??= O(qe/2).

q-aspect: is this any easier?

What next?

Higher degree.Extend investigation of More(X ) to hypersurfaces of degree > 3over finite fields (a la Alan Lee)

A Manin conjecture for Campana orbifolds.The Manin conjecture makes predictions about equidistribution ofK -rational points on projective varieties over global fields K .

There is a version emerging for integral points on Campanaorbifolds (P1,∆) with χ(∆) > 0.

Example

e-aspect: is it true that

#

{(x , y , z) ∈ Fq[t] :

x , y , z square-full & degree ex + y = z

}??= O(qe/2).

q-aspect: is this any easier?

top related