Manin Conjectures for K3 surfaces
Ronald van Luijk CRM, Montreal MSRI, Berkeley PIMS, UBC/SFU
May 17, 2006 Banff
A K3 surface
with 2 singular curves of genus 0 and
Question 1. Is there a K3 surface, defined over a number field, such that its set of rational points is neither empty, nor dense?
Question 1. Is there a K3 surface, defined over a number field, such that its set of rational points is neither empty, nor dense?
Examples of K3 surfaces are smooth quartic surfaces in P3. Question 2 (Swinnerton-Dyer).
Does X ⊂ P3 given by
x4 + 2y4 = z4 + 4w4 have more than 2 rational points?
Question 1. Is there a K3 surface, defined over a number field, such that its set of rational points is neither empty, nor dense?
Examples of K3 surfaces are smooth quartic surfaces in P3. Question 2 (Swinnerton-Dyer).
Does X ⊂ P3 given by
x4 + 2y4 = z4 + 4w4 have more than 2 rational points?
Answer (Elsenhans, Jahnel, 2004):
We will look at the growth of the number of rational points of bounded height. Consider a surface X/K, choose a height H, and set
NU(B) = #{x ∈ U(K) : H(x) ≤ B}.
Conjecture 1 (Batyrev, Manin). Let X be a smooth, geometrically integral, projective variety over a number field K, and let D be a hyper-plane section. Assume that the canonical sheaf KX satisfies −KX = aD for some a > 0. Then there exists a finite field extension L, a constant C, and an open subset U ⊂ X, such that with b = rk Pic(XL) we have
We will look at the growth of the number of rational points of bounded height. Consider a surface X/K, choose a height H, and set
NU(B) = #{x ∈ U(K) : H(x) ≤ B}.
Conjecture 1 (Batyrev, Manin). Let X be a smooth, geometrically integral, projective variety over a number field K, and let D be a hyper-plane section. Assume that the canonical sheaf KX satisfies −KX = aD for some a > 0. Then there exists a finite field extension L, a constant C, and an open subset U ⊂ X, such that with b = rk Pic(XL) we have
NUL(B) ≈ CBa(log B)b−1.
Define the height-zeta function Z(U, s) = X x∈U(K) H(x)−s. From 1 2πi Z c+i∞ c−i∞ x s ds s = 1 if x > 1 1 2 if x = 1 0 if x < 1 (c > 0) we get N (U, x) = 1 2πi Z c+i∞ c−i∞ Z(U, s) x s ds s (c >> 0)
Assuming Z(U, s) is analytic on <(s) > a − ², except for a pole of order b at a, we can write
N (U, x) = 1 2πi Z c+i∞ c−i∞ Z(U, s) x s ds s (c >> 0) = ress=a hZ(U, s)s−1 exp(s log x)i + 1
2πi
Z a−²+i∞
a−²−i∞ Z(U, s) x
s ds
Assuming Z(U, s) is analytic on <(s) > a − ², except for a pole of order b at a, we can write
N (U, x) = 1 2πi Z c+i∞ c−i∞ Z(U, s) x s ds s (c >> 0) = ress=a hZ(U, s)s−1 exp(s log x)i + 1
2πi
Z a−²+i∞
a−²−i∞ Z(U, s) x
s ds
s .
The main term is
xap(log x)
for some polynomial p of degree (
b − 1 if a 6= 0, b if a = 0.
Assuming Z(U, s) is analytic on <(s) > a − ², except for a pole of order b at a, we can write
N (U, x) = 1 2πi Z c+i∞ c−i∞ Z(U, s) x s ds s (c >> 0) = ress=a hZ(U, s)s−1 exp(s log x)i + 1
2πi
Z a−²+i∞
a−²−i∞ Z(U, s) x
s ds
s .
The main term is
xap(log x)
for some polynomial p of degree (
b − 1 if a 6= 0, b if a = 0.
For X a K3 surface: N (U, B) ∼ C(log B)rk Pic X ?
For X a K3 surface: N (U, B) ∼ C(log B)rk Pic X ?
Not if X admits an elliptic fibration (in particular, if rk Pic X ≥ 5). More problems:
• infinitely many curves of genus 0 or 1,
• infinitely many automorphisms (cf. A. Baragar), • Swinnerton-Dyer’s surface has very slow growth, • which height to use?
For X a K3 surface: N (U, B) ∼ C(log B)rk Pic X ?
Not if X admits an elliptic fibration (in particular, if rk Pic X ≥ 5). More problems:
• infinitely many curves of genus 0 or 1,
• infinitely many automorphisms (cf. A. Baragar), • Swinnerton-Dyer’s surface has very slow growth, • which height to use?
We will run experiments on quartic surfaces, comparing exponents, and the main coefficient C to a relatively naive heuristic coefficient ˜C (part of Peyre’s constant for del Pezzo).
10 20 30 40 # pt’s ˜ C ∼ 6.24 line : C(log B) − 9˜ ρ(XQ) = 1 # Aut XQ < ∞ no elliptic fibration f = x3w − 3x2y2 − x2yw − x2zw + x2w2
−xy2z − xy2w − 4xyz2 − xyzw + 2xyw2 − 2xz3 +xz2w + 2xzw2 + y3w + y2zw − y2w2 − 5yz3 +yz2w + yzw2 − yw3 − z4 + z2w2 + zw3 + 2w4
2 4 6 8 10 # pt’s ˜ C ∼ 0.57 line : C(log B)˜ ρ(X) = 1 ρ(XQ) = 20 # Aut XQ = ∞
no elliptic fibration over Q
10 20 30 40 # pt’s ˜ C ∼ 7.8 line : 34C(log B) − 10˜ ρ(XQ) = 1 # Aut XQ < ∞ no elliptic fibration f = x3w − 3x2y2 + 4x2yz − x2z2 −x2z2 + x2zw − xy2z − xyz2 +xw3 + y3w + y2z2 + z3w 17
Consider X given by
f + 6m(w4 + 2x4 − (y + w)4 − 4z4) for m ∈ {±1, ±2, ±3} with f as on the previous page.
Then X has geometric Picard number 1.
m C˜ C · log(12000) #{x : H(x) ≤ 12000}˜ 0 7.8 73 46 1 0.30 2.8 5 −1 0.58 5.4 9 2 0.39 3.7 2 −2 0.35 3.3 6 3 0.18 1.7 4
5 10 15 20 25 30 # pt’s ˜ C ∼ 6.66 line : 12C(log B)˜ ρ(X) = 1 ρ(XQ) = 2 (over Q(√3)) Aut XQ = {1} no elliptic fibration f = x3w − x2y2 − 2x2yz −x2z2 + x2zw + xy2z + xyz2 xw3 + y3w − y2z2 + z3w 19
20 40 60 80 100 # pt’s ˜ C ∼ 0.96 parabola : 1.1(log B)2 ρ(X) = 2 ρ(XQ) = 2 Aut XQ = {1} no elliptic fibration f = 3x3w − x2y2 − 2x2yz
−x2z2 − 4xy2z + 3xy2w − 4xyz2 +4xyw2 + 3xzw2 + 4xw3 + y3w −3y2z2 + 4y2w2 + 2z3w + w4 20
1 2 3 4 5 log(# pt’s) ˜ C ∼ 0.96
line : 2(log log B) + log 1.1 ρ(X) = 2
ρ(XQ) = 2 Aut XQ = {1}
no elliptic fibration
f = 3x3w − x2y2 − 2x2yz
−x2z2 − 4xy2z + 3xy2w − 4xyz2 +4xyw2 + 3xzw2 + 4xw3 + y3w −3y2z2 + 4y2w2 + 2z3w + w4 21
10 20 30 40 # pt’s ˜ C ∼ 0.53 parabola : 32C((log B) − 2)˜ 2 + 2 ρ(X) = 2 ρ(XQ) = 2 Aut XQ = {1} no elliptic fibration f = −6x3y − 6x3z − 3x3w − x2y2 + 16x2yz +35x2z2 + 2xy2z + 3xy2w + 8xyz2 − 2xyw2 +6xz3 + 3xzw2 − 2xw3 + y3w + 3y2z2
1 2 3 4 y log(# pt’s) ˜ C ∼ 0.53
line : 2(log log B) − 0.75 ρ(X) = 2
ρ(XQ) = 2 Aut XQ = {1}
no elliptic fibration
50 100 150 200 250 # pt’s ˜ C ∼ 0.012 cubic : 0.9((log B) − 3)3 + 10 ρ(X) = 3 ρ(XQ) = 3 # Aut XQ = ∞ (A.Baragar) no elliptic fibration f = −7x3z + 11x3w + 7x2y2 − 11x2yz − 11x2yw − x2z2 +16x2zw − 9x2w2 + 10xy2z + 5xy2w − 44xyz2 + 10xyzw
−14xyw2 − 10xz3 + 18xz2w − 15xzw2 + 5xw3 + 2y4 + 23y3z − 7y3w + 16y2z2 −3y2zw + 12y2w2 − yz3 − 15yz2w − 21yzw2 − 3yw3 + 18z4 + 16z3w + 3z2w2
1 2 3 4 5 y log(# pt’s) ˜ C ∼ 0.012
line : 3(log log B) − 1.2 ρ(X) = 3
ρ(XQ) = 3
# Aut XQ = ∞
no elliptic fibration
10 20 30 40 50 60 # pt’s ˜ C ∼ 0.00061 parabola : 0.074(log B)3 ρ(X) = 3 ρ(XQ) = 3 # Aut XQ = ∞ (A.Baragar) no elliptic fibration f = · · · 26
1 2 3 4 log(# pt’s) ˜ C ∼ 0.00061
line : 3(log log B) − 2.6 ρ(X) = 3
ρ(XQ) = 3
# Aut XQ = ∞
no elliptic fibration