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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
cπ
⇒ 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
cπ
⇒ 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
cπ
⇒ 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
cπ
⇒ 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?