Computing N´
eron-Severi groups
Ronald van Luijk
February 27, 2013
Setting
I k a finitely generated field.
I X a nicek-variety (smooth, projective, geometrically integral).
I ks a separable closure ofk with Galois group Γ = Gal(ks/k).
I Xs = X ×k ks.
I Picard group Pic X ⊂ (Pic Xs)Γ.
I Pic0(X ) subgroup of classes algebraically equivalent to0.
I N´eron-Severi groupNS(X ) = Pic X / Pic0(X ) ⊂ NS(Xs)Γ. Goal: ComputeNS(Xs) (or its rank).
Special cases
I Elliptic fibrations: mapNS to the Mordell-Weil group.
I Fibrations into abelian varieties.
I If a finite group G acts on Y andX = Y /G, then
NS(Xs) ⊗ Q → (NS(Ys) ⊗ Q)G
is an isomorphism.
Application (Shioda):
Delsarte surfaces (given by tetranomials inP3) are
Special cases
I Elliptic fibrations: mapNS to the Mordell-Weil group.
I Fibrations into abelian varieties.
I If a finite group G acts on Y andX = Y /G, then
NS(Xs) ⊗ Q → (NS(Ys) ⊗ Q)G
is an isomorphism. Application (Shioda):
Delsarte surfaces (given by tetranomials inP3) are
K3 surfaces of degree 2
Theorem (Hassett, Kresch, Tschinkel)
There is an algorithm that takes as input a K3 surfaceX of degree2 over a number field, and returnsPic Xs= NS Xs. Method: Kuga-Satake correspondence.
Tate conjecture(s)
I Fix0 ≤ p ≤ dim X and prime` 6= char k.
I Zp(X )is group of codimension-p cycles onX.
I V2p= H2pet(Xs, Q`(p)).
I Vtate⊂ V2p is set of Tate classes
(each fixed by some finite-index open subgroupG ⊂ Γ).
Conjecture (Tp(X , `))
The cycle class mapZp(Xs) ⊗ Q
`→ Vtate is surjective.
Theorem (Nijgaard, Ogus, Maulik, Charles, Madapusi Pera)
Supposechar k 6= 2. If X is a K3 surface, thenT1(X , `) holds.
Conjecture (Ep(X , `))
An element inZp(Xs, `) is numerically equivalent to0 if and only if it maps to0in V2p.
Tate conjecture(s)
I Fix0 ≤ p ≤ dim X and prime` 6= char k.
I Zp(X )is group of codimension-p cycles onX.
I V2p= H2pet(Xs, Q`(p)).
I Vtate⊂ V2p is set of Tate classes
(each fixed by some finite-index open subgroupG ⊂ Γ).
Conjecture (Tp(X , `))
The cycle class mapZp(Xs) ⊗ Q
`→ Vtate is surjective.
Theorem (Nijgaard, Ogus, Maulik, Charles, Madapusi Pera)
Supposechar k 6= 2. If X is a K3 surface, thenT1(X , `) holds.
Conjecture (Ep(X , `))
An element inZp(Xs, `) is numerically equivalent to0 if and only if it maps to0in V2p.
Tate conjecture(s)
I Fix0 ≤ p ≤ dim X and prime` 6= char k.
I Zp(X )is group of codimension-p cycles onX.
I V2p= H2pet(Xs, Q`(p)).
I Vtate⊂ V2p is set of Tate classes
(each fixed by some finite-index open subgroupG ⊂ Γ).
Conjecture (Tp(X , `))
The cycle class mapZp(Xs) ⊗ Q
`→ Vtate is surjective.
Theorem (Nijgaard, Ogus, Maulik, Charles, Madapusi Pera)
Supposechar k 6= 2. If X is a K3 surface, thenT1(X , `) holds.
Conjecture (Ep(X , `))
An element inZp(Xs, `) is numerically equivalent to0 if and only if it maps to0in V2p.
Algorithms for general
p
I Nump(X )is group of codimension-p cycle classes up to numerical equivalence.
I Assuming Ep(X , `), the mapNump(X , `) ⊗ Q`,→ Vtate is an injection that is an isomorphism if and only if Tp(X , `) holds.
I For p = 1we haveNum1(X ) ∼= NS(X )/ NS(X )tors and an injectionNS(X ) ⊗ Q` → Vtate.
Strategy for boundingrk Nump(Xs).
1. List cycles to find lower bound (also cycles to intersect with).
2. Bound dimQ`Vtate from above for upper bound. Trivial upper bound: Betti number b2p.
Problem with computingVtate⊂ V2p= H2p
et(Xs, Q`(p)) is that
Algorithms for general
p
I Nump(X )is group of codimension-p cycle classes up to numerical equivalence.
I Assuming Ep(X , `), the mapNump(X , `) ⊗ Q`,→ Vtate is an injection that is an isomorphism if and only if Tp(X , `) holds.
I For p = 1we haveNum1(X ) ∼= NS(X )/ NS(X )tors and an injectionNS(X ) ⊗ Q` → Vtate.
Strategy for boundingrk Nump(Xs).
1. List cycles to find lower bound (also cycles to intersect with).
2. Bound dimQ`Vtate from above for upper bound. Trivial upper bound: Betti number b2p.
Problem with computingVtate⊂ V2p= H2p
et(Xs, Q`(p)) is that
Algorithms for general
p
I Nump(X )is group of codimension-p cycle classes up to numerical equivalence.
I Assuming Ep(X , `), the mapNump(X , `) ⊗ Q`,→ Vtate is an injection that is an isomorphism if and only if Tp(X , `) holds.
I For p = 1we haveNum1(X ) ∼= NS(X )/ NS(X )tors and an injectionNS(X ) ⊗ Q` → Vtate.
Strategy for boundingrk Nump(Xs).
1. List cycles to find lower bound (also cycles to intersect with).
2. Bound dimQ`Vtate from above for upper bound. Trivial upper bound: Betti number b2p.
Problem with computingVtate⊂ V2p= H2p
et(Xs, Q`(p)) is that
Hypothesis. Can computeTi
`n = Hiet(Xs, Z/`nZ)as Γ-module.
Remark. This holds in characteristic0 and for liftable X.
Theorem (Poonen, Testa, vL)
Assume the hypothesis. Then there is an algorithm that takes as input(k, p, X , `) as before, such that, assumingEp(X , `), the algorithm terminates if and only ifTp(X , `)holds, and if the algorithm terminates, it returnsrk Nump(Xs).
Sketch of proof of upper bound for r = dim Vtate.
1. Extend k so thatΓ acts trivially onT`2p0 (p)with
`0 = `for ` > 2and `0 = 4 for ` = 2.
2. Γ acts trivially onM/Mtors with M = H2pet(Xs, Z`(p))tate (Minkovski’s Lemma on finite-order elements in GLn(Z`)).
3. Computet such that `t killsH2pet(Xs, Z`(p))tors (Wittenberg).
4. `r (n−t)≤ #`tM/`nM ≤ #(M/`nM)Γ≤ #T2p `n (p)Γ= O(`rn). 5. Sequence log #T`n2p(p)Γ log `n−t
Hypothesis. Can computeTi
`n = Hiet(Xs, Z/`nZ)as Γ-module.
Remark. This holds in characteristic0 and for liftable X.
Theorem (Poonen, Testa, vL)
Assume the hypothesis. Then there is an algorithm that takes as input(k, p, X , `) as before, such that, assumingEp(X , `), the algorithm terminates if and only ifTp(X , `)holds, and if the algorithm terminates, it returnsrk Nump(Xs).
Sketch of proof of upper bound for r = dim Vtate.
1. Extend k so thatΓ acts trivially onT`2p0 (p)with
`0 = `for ` > 2and `0 = 4 for ` = 2.
2. Γ acts trivially onM/Mtors with M = H2pet(Xs, Z`(p))tate (Minkovski’s Lemma on finite-order elements in GLn(Z`)).
3. Computet such that `t killsH2pet(Xs, Z`(p))tors (Wittenberg).
4. `r (n−t)≤ #`tM/`nM ≤ #(M/`nM)Γ≤ #T2p `n (p)Γ= O(`rn). 5. Sequence log #T`n2p(p)Γ log `n−t
Hypothesis. Can computeTi
`n = Hiet(Xs, Z/`nZ)as Γ-module.
Remark. This holds in characteristic0 and for liftable X.
Theorem (Poonen, Testa, vL)
Assume the hypothesis. Then there is an algorithm that takes as input(k, p, X , `) as before, such that, assumingEp(X , `), the algorithm terminates if and only ifTp(X , `)holds, and if the algorithm terminates, it returnsrk Nump(Xs).
Sketch of proof of upper bound for r = dim Vtate.
1. Extend k so thatΓ acts trivially onT`2p0 (p)with
`0 = `for ` > 2and `0 = 4 for ` = 2.
2. Γ acts trivially onM/Mtors withM = H2pet(Xs, Z`(p))tate (Minkovski’s Lemma on finite-order elements in GLn(Z`)).
3. Computet such that`t killsH2pet(Xs, Z`(p))tors (Wittenberg).
4. `r (n−t)≤ #`tM/`nM ≤ #(M/`nM)Γ≤ #T2p `n (p)Γ= O(`rn). 5. Sequence log #T`n2p(p)Γ log `n−t
Finite fields
Supposek is finite. LetVµ⊂ V2p = H2pet(Xs, Q`(p))be the largest subspace on which all eigenvalues of Frobenius are roots of unity.
Nump(Xs) ⊗ Q` → Vtate⊂ Vµ
Theorem
AssumingEp(X , `), the following are equivalent.
1. rk Nump(Xs) = dim Vtate.
2. Conjecture Tp(X , `)holds.
3. rk Nump(Xs) = dim Vtate= dim Vµ.
Proof. 1 ⇔ 2. Under Ep(X , `), the first map is injective, so it is surjective if and only if 1 holds.
2 ⇒ 3. Vtate= Vµ follows asEp(X , `)andTp(X , `) together imply that Frobenius acts semi-simple onVµ. 3 ⇒ 1. Obvious.
Finite fields
Supposek is finite. LetVµ⊂ V2p = H2pet(Xs, Q`(p))be the largest subspace on which all eigenvalues of Frobenius are roots of unity.
Nump(Xs) ⊗ Q` → Vtate⊂ Vµ
Theorem
AssumingEp(X , `), the following are equivalent.
1. rk Nump(Xs) = dim Vtate.
2. Conjecture Tp(X , `)holds.
3. rk Nump(Xs) = dim Vtate= dim Vµ.
Proof. 1 ⇔ 2. Under Ep(X , `), the first map is injective, so it is surjective if and only if 1 holds.
2 ⇒ 3. Vtate= Vµ follows asEp(X , `)andTp(X , `) together imply that Frobenius acts semi-simple onVµ. 3 ⇒ 1. Obvious.
Finite fields
Supposek is finite. LetVµ⊂ V2p = H2pet(Xs, Q`(p))be the largest subspace on which all eigenvalues of Frobenius are roots of unity.
Nump(Xs) ⊗ Q` → Vtate⊂ Vµ
Theorem
AssumingEp(X , `), the following are equivalent.
1. rk Nump(Xs) = dim Vtate.
2. Conjecture Tp(X , `)holds.
3. rk Nump(Xs) = dim Vtate= dim Vµ.
Proof. 1 ⇔ 2. Under Ep(X , `), the first map is injective, so it is surjective if and only if 1 holds.
2 ⇒ 3. Vtate= V
µ follows asEp(X , `) andTp(X , `) together imply that Frobenius acts semi-simple onVµ. 3 ⇒ 1. Obvious.
Finite fields
Theorem
There is an algorithm that takes as input(k, p, X , `), with k a finite field, and that, assumingEp(X , `), terminates if and only if
Tp(X , `)holds, and if it terminates, it returns rk NumpXs.
Proof. By searching for cycles, we get a lower bound for
rk NumpXs that eventually is sharp. To verify that it is, it suffices to computedim Vµ. Say k = Fq. The degree of the zeta-function
ZX(T ) = 2 dim X Y i =0 det 1 − T · Frob∗| Hiet(Xs, Q`) (−1)i +1
is bounded by the sum of Betti numbers: computable boundB. Computing#X (Fqn) forn = 1, . . . , 2B gives enough information
to determineZX(T ). Thendim Vµ equals the number of poles of
Finite fields
Theorem
There is an algorithm that takes as input(k, p, X , `), with k a finite field, and that, assumingEp(X , `), terminates if and only if
Tp(X , `)holds, and if it terminates, it returns rk NumpXs.
Proof. By searching for cycles, we get a lower bound for
rk NumpXs that eventually is sharp. To verify that it is, it suffices to computedim Vµ. Say k = Fq. The degree of the zeta-function
ZX(T ) = 2 dim X Y i =0 det 1 − T · Frob∗| Hiet(Xs, Q`) (−1)i +1
is bounded by the sum of Betti numbers: computable boundB.
Computing#X (Fqn) forn = 1, . . . , 2B gives enough information
to determineZX(T ). Thendim Vµ equals the number of poles of
Finite fields
Theorem
There is an algorithm that takes as input(k, p, X , `), with k a finite field, and that, assumingEp(X , `), terminates if and only if
Tp(X , `)holds, and if it terminates, it returns rk NumpXs.
Proof. By searching for cycles, we get a lower bound for
rk NumpXs that eventually is sharp. To verify that it is, it suffices to computedim Vµ. Say k = Fq. The degree of the zeta-function
ZX(T ) = 2 dim X Y i =0 det 1 − T · Frob∗| Hiet(Xs, Q`) (−1)i +1
is bounded by the sum of Betti numbers: computable boundB. Computing#X (Fqn)for n = 1, . . . , 2B gives enough information
to determineZX(T ). Thendim Vµ equals the number of poles of
Finite fields
Example. Let X ⊂ P3 overF2 be given bydet M = 0 with M =
x0 x2 x1+ x2 x2+ x3 x1 x2+ x3 x0+ x1+ x2+ x3 x0+ x3 x0+ x2 x0+ x1+ x2+ x3 x0+ x1 x2 x0+ x1+ x3 x0+ x2 x3 x2 . Φ = Frob∗| H2(Xs, Q`(1)) t n= Tr Φn #X (F2n) = 1 + 2ntn+ 22n n 1 2 3 4 5 6 7 8 9 10 #X (F2n) 6 26 90 258 1146 4178 17002 64962 260442 1044786 tn 12 94 258 161 12132 8164 617128 −575256 −1703512 −37911024 fΦ= palindromic or antipalindromic = t22− 1 2t 21− t20− 1 2t 19+ t18−1 2t 15+ t14+1 2t 13− 2t11+ . . . = (t − 1)2(t20+32t19+ t18−1 2t 13+ t11+ 2t10+ . . .). Conclusion. We haverk NS(Xs) = 2.
Finite fields
Example. Let X ⊂ P3 overF2 be given bydet M = 0 with M =
x0 x2 x1+ x2 x2+ x3 x1 x2+ x3 x0+ x1+ x2+ x3 x0+ x3 x0+ x2 x0+ x1+ x2+ x3 x0+ x1 x2 x0+ x1+ x3 x0+ x2 x3 x2 . Φ = Frob∗| H2(Xs, Q`(1)) t n= Tr Φn #X (F2n) = 1 + 2ntn+ 22n n 1 2 3 4 5 6 7 8 9 10 #X (F2n) 6 26 90 258 1146 4178 17002 64962 260442 1044786 tn 12 94 258 161 12132 8164 617128 −575256 −1703512 −37911024 fΦ= palindromic or antipalindromic = t22− 1 2t 21− t20− 1 2t 19+ t18−1 2t 15+ t14+1 2t 13− 2t11+ . . . = (t − 1)2(t20+32t19+ t18−1 2t 13+ t11+ 2t10+ . . .). Conclusion. We haverk NS(Xs) = 2.
Finite fields
Example. Let X ⊂ P3 overF2 be given bydet M = 0 with M =
x0 x2 x1+ x2 x2+ x3 x1 x2+ x3 x0+ x1+ x2+ x3 x0+ x3 x0+ x2 x0+ x1+ x2+ x3 x0+ x1 x2 x0+ x1+ x3 x0+ x2 x3 x2 . Φ = Frob∗| H2(Xs, Q`(1)) t n= Tr Φn #X (F2n) = 1 + 2ntn+ 22n n 1 2 3 4 5 6 7 8 9 10 #X (F2n) 6 26 90 258 1146 4178 17002 64962 260442 1044786 tn 12 94 258 161 12132 8164 617128 −575256 −1703512 −37911024 fΦ= palindromic or antipalindromic = t22− 1 2t 21− t20− 1 2t 19+ t18− 1 2t 15+ t14+ 1 2t 13− 2t11+ . . . = (t − 1)2(t20+32t19+ t18−1 2t 13+ t11+ 2t10+ . . .). Conclusion. We haverk NS(Xs) = 2.
Surfaces over global fields
I Global field K, discrete valuation ring R ⊂ K, residue fieldk.
I X a nice surface over K, integral modelX overR with good reduction.
NS(Xs) ⊗ Q`,→ NS(Xks) ⊗ Q`
rk NS(Xs) ≤ rk NS(Xks)
Problem. IfT1(X
ks, `) holds, then rk NS(Xks) ≡ b2(X ) (mod 2).
Proof.
The roots offΦ that are not roots of unity come in conjugate pairs.
Question.
Surfaces over global fields
I Global field K, discrete valuation ring R ⊂ K, residue fieldk.
I X a nice surface over K, integral modelX overR with good reduction.
NS(Xs) ⊗ Q`,→ NS(Xks) ⊗ Q`
rk NS(Xs) ≤ rk NS(Xks)
Problem. IfT1(X
ks, `) holds, then rk NS(Xks) ≡ b2(X ) (mod 2).
Proof.
The roots offΦ that are not roots of unity come in conjugate pairs.
Question.
The injection
Num1(Xs) ,→ Num1(Xks)
respects the intersection pairing.
Lemma. IfΛ0 is a sublattice of finite index of Λ, then we have
disc Λ0 = [Λ : Λ0]2disc Λ.
Hence,disc Λ = disc Λ0 in Q∗/(Q∗)2.
Corollary(vL). Ifv , w are two places of good reduction with 1)rk Num1(Xk(v )s) = r = rk Num1(Xk(w )s), and
2)disc Num1(Xk(v )s) 6= disc Num1(Xk(w )s) inQ∗/(Q∗)2,
thenrk Num1(Xs) < r.
The injection
Num1(Xs) ,→ Num1(Xks)
respects the intersection pairing.
Lemma. IfΛ0 is a sublattice of finite index of Λ, then we have
disc Λ0 = [Λ : Λ0]2disc Λ.
Hence,disc Λ = disc Λ0 in Q∗/(Q∗)2.
Corollary(vL). Ifv , w are two places of good reduction with 1)rk Num1(Xk(v )s) = r = rk Num1(Xk(w )s), and
2)disc Num1(Xk(v )s) 6= disc Num1(Xk(w )s)in Q∗/(Q∗)2,
thenrk Num1(Xs) < r.
Example. Let X ⊂ P3
Q be given by
wf = 3pq − 2zg
withf ∈ Z[x, y , z, w ]and g , p, q ∈ Z[x, y , z]equal to
f = x3− x2y − x2z + x2w − xy2− xyz + 2xyw + xz2+ 2xzw + y3 + y2z − y2w + yz2+ yzw − yw2+ z2w + zw2+ 2w3, g = xy2+ xyz − xz2− yz2+ z3, p = z2+ xy + yz, q = z2+ xy . Thenrk NS(Xs) = 1.
Proof. Two primes of good reduction: 2and3. For both we obtaindim Vµ= 2 as before.
ReductionXF2 contains conic C given byw = p = 0. ReductionXF3 contains line L given byw = z = 0.
ThenNum1(XF2) enNum
1(X
F3) contain finite-index sublattices
4 2 2 −2 and 4 1 1 −2
with discriminants−12 and−9in Q∗/(Q∗)2.
Example. Let X ⊂ P3
Q be given by
wf = 3pq − 2zg
with . . . Thenrk NS(Xs) = 1.
Proof. Two primes of good reduction: 2and3. For both we obtaindim Vµ= 2 as before.
ReductionXF2 contains conic C given byw = p = 0.
ReductionXF3 contains line Lgiven by w = z = 0.
ThenNum1(XF2) enNum
1(X
F3) contain finite-index sublattices
4 2 2 −2 and 4 1 1 −2
with discriminants−12 and−9in Q∗/(Q∗)2.
Example. Let X ⊂ P3
Q be given by
wf = 3pq − 2zg
with . . . Thenrk NS(Xs) = 1.
Proof. Two primes of good reduction: 2and3. For both we obtaindim Vµ= 2 as before.
ReductionXF2 contains conic C given byw = p = 0.
ReductionXF3 contains line Lgiven by w = z = 0. ThenNum1(XF2) enNum
1(X
F3) contain finite-index sublattices
4 2 2 −2 and 4 1 1 −2
with discriminants−12 and−9in Q∗/(Q∗)2.
Extension by Kloosterman
I k = Fq. I X /k a nice surface. I Φ = Frob∗| H2(Xs, Q `). I fΦ(T ) = det(1 − T · Φ).I ρ = rk Num1(X ) and∆ = disc Num1(X ).
I b2= b2(X )and α = χ(X , OX) − 1 − dim Pic0(X ).
Conjecture(Artin–Tate). lim T →q−1 fΦ(T ) (1 − qT )ρ = (−1)ρ−1· # Br X · ∆ qα(# NS(X ) tors)2 . Facts. T1(X , `)⇒ Artin–Tate. T1(X , `)⇒ # Br X ∈ (Q∗)2 (Liu–Lorenzini–Raynaud).
Extension by Kloosterman
I k = Fq. I X /k a nice surface. I Φ = Frob∗| H2(Xs, Q `). I fΦ(T ) = det(1 − T · Φ).I ρ = rk Num1(X ) and∆ = disc Num1(X ).
I b2= b2(X )and α = χ(X , OX) − 1 − dim Pic0(X ).
Conjecture(Artin–Tate). lim T →q−1 fΦ(T ) (1 − qT )ρ = (−1)ρ−1· # Br X · ∆ qα(# NS(X ) tors)2 . Facts. T1(X , `)⇒ Artin–Tate. T1(X , `)⇒ # Br X ∈ (Q∗)2 (Liu–Lorenzini–Raynaud).
Application
Theorem (Kloosterman)
The elliptic K3 surfaceπ : X → P1 overQgiven by y2 = x3+ 2(t8+ 14t4+ 1)x + 4t2(t8+ 6t4+ 1)
Extension by Elsenhans-Jahnel, I
Main idea. If you consider Galois action, you may not need
r = rk Num1(Xs) + 1.
Example. (Elsenhans, Jahnel)
LetX : w2 = f (x , y , z)be a K3 surface of degree2 overQwith f ≡ y6+ x4y2− 2x2y4+ 2x5z + 3xz5+ z6 (mod 5) and
f ≡ 2x6+ x4y2+ 2x3y2z + x2y2z2+ x2yz3+ 2x2z4+ xy4z + xy3z2+ xy2z3+ 2xz5+ 2y6+ y4z2+ y3z3 (mod 3).
Extension by Elsenhans-Jahnel, I
Main idea. If you consider Galois action, you may not need
r = rk Num1(Xs) + 1.
Example. (Elsenhans, Jahnel)
LetX : w2 = f (x , y , z)be a K3 surface of degree2 overQwith f ≡ y6+ x4y2− 2x2y4+ 2x5z + 3xz5+ z6 (mod 5) and
f ≡ 2x6+ x4y2+ 2x3y2z + x2y2z2+ x2yz3+ 2x2z4+ xy4z + xy3z2+ xy2z3+ 2xz5+ 2y6+ y4z2+ y3z3 (mod 3).
Extension by Elsenhans-Jahnel, I
LetLdenote the pull-back of a line inP2(x , y , z).
The characteristic polynomial of Frobenius acting on the space
(NS X
F3⊗ Q)/hLi
equals(t − 1)(t2+ t + 1), so only finitely many Galois-invariant subspaces ofNS XF
3⊗ Q containingL; dimensions are1, 2, 3, 4.
The characteristic polynomial of Frobenius acting on the space
(NS X
F5⊗ Q)/hLi
equals(t − 1)Φ5(t)Φ15(t), whereΦn denotes then-th cyclotomic polynomial. So only finitely many Galois-invariant subspaces of
NS X
F5⊗ QcontainingL; dimensions are 1, 2, 5, 6, 9, 10, 13, 14.
Only common dimensions are1and 2. Comparing discriminants up to squares of the subspaces of dimension2 yields rk NS(Xs) = 1.
Extension by Elsenhans-Jahnel, I
LetLdenote the pull-back of a line inP2(x , y , z).
The characteristic polynomial of Frobenius acting on the space
(NS X
F3⊗ Q)/hLi
equals(t − 1)(t2+ t + 1), so only finitely many Galois-invariant subspaces ofNS XF
3⊗ Q containingL; dimensions are1, 2, 3, 4.
The characteristic polynomial of Frobenius acting on the space
(NS X
F5⊗ Q)/hLi
equals(t − 1)Φ5(t)Φ15(t), whereΦn denotes then-th cyclotomic polynomial. So only finitely many Galois-invariant subspaces of
NS X
F5⊗ QcontainingL; dimensions are 1, 2, 5, 6, 9, 10, 13, 14.
Only common dimensions are1and 2. Comparing discriminants up to squares of the subspaces of dimension2 yields rk NS(Xs) = 1.
Extension by Elsenhans-Jahnel, I
LetLdenote the pull-back of a line inP2(x , y , z).
The characteristic polynomial of Frobenius acting on the space
(NS X
F3⊗ Q)/hLi
equals(t − 1)(t2+ t + 1), so only finitely many Galois-invariant subspaces ofNS XF
3⊗ Q containingL; dimensions are1, 2, 3, 4.
The characteristic polynomial of Frobenius acting on the space
(NS X
F5⊗ Q)/hLi
equals(t − 1)Φ5(t)Φ15(t), whereΦn denotes then-th cyclotomic polynomial. So only finitely many Galois-invariant subspaces of
NS X
F5⊗ QcontainingL; dimensions are 1, 2, 5, 6, 9, 10, 13, 14.
Only common dimensions are1and2. Comparing discriminants up to squares of the subspaces of dimension2 yieldsrk NS(Xs) = 1.
Extension by Elsenhans-Jahnel, II
I p 6= 2 prime.
I X a scheme that is proper and flat over Z.
Theorem(Elsenhans-Jahnel). If the special fiber Xp is nonsingular, then the cokernel of the specialization homomorphism
spQ: Pic(XQ) → Pic(Xp) is torsion-free.
Corollary(E-J, Hassett–V´arilly-Alvarado).
LetX be a double cover ofP2, ramified over a smooth plane
sexticC. Letp, p0 denote two odd primes of good reduction. Assume that there is a tritangent line`to the curveCp.
SupposePic(Xps) has rank2 and is generated by the components in the pull-back of`. If there are no tritangent lines toCp0, then
Extension by Elsenhans-Jahnel, II
I p 6= 2 prime.
I X a scheme that is proper and flat over Z.
Theorem(Elsenhans-Jahnel). If the special fiber Xp is nonsingular, then the cokernel of the specialization homomorphism
spQ: Pic(XQ) → Pic(Xp) is torsion-free.
Corollary(E-J, Hassett–V´arilly-Alvarado).
LetX be a double cover ofP2, ramified over a smooth plane
sexticC. Letp, p0 denote two odd primes of good reduction. Assume that there is a tritangent line`to the curveCp.
SupposePic(Xps) has rank2 and is generated by the components in the pull-back of`. If there are no tritangent lines toCp0, then
“It works” by Charles
Question.
1) Given a nice surfaceX over a number field k, is there always a primep of good reduction with rk Num1(Xps) ≤ rk Num1(Xs) + 1? 2) Are there two so that the discriminant trick works?
Answer(Charles).
Not always, but if not, then the minimal jumps are still controllable!
Consequence(Charles).
There is an algorithm with input a K3 surfaceX over a number field that either returnsrk NS(Xs) or does not terminate.
IfX × X satisfies the Hodge conjecture for codimension2 cycles, then the algorithm applied to X terminates.
“It works” by Charles
Question.
1) Given a nice surfaceX over a number field k, is there always a primep of good reduction with rk Num1(Xps) ≤ rk Num1(Xs) + 1? 2) Are there two so that the discriminant trick works?
Answer(Charles).
Not always, but if not, then the minimal jumps are still controllable!
Consequence(Charles).
There is an algorithm with input a K3 surfaceX over a number field that either returnsrk NS(Xs) or does not terminate.
IfX × X satisfies the Hodge conjecture for codimension2 cycles, then the algorithm applied to X terminates.
“It works” by Charles
Question.
1) Given a nice surfaceX over a number field k, is there always a primep of good reduction with rk Num1(Xps) ≤ rk Num1(Xs) + 1? 2) Are there two so that the discriminant trick works?
Answer(Charles).
Not always, but if not, then the minimal jumps are still controllable!
Consequence(Charles).
There is an algorithm with input a K3 surfaceX over a number field that either returnsrk NS(Xs) or does not terminate.
IfX × X satisfies the Hodge conjecture for codimension2 cycles, then the algorithm applied toX terminates.
Saturation
Theorem (Poonen, Testa, vL)
There is an algorithm that takesk, X, and a finite setD of divisors as input, and computes the saturation insideNS(Xs) of the
Γ-submodule generated by the classes of divisors in D.
Saturation
Goal. Given a surface X over a global fieldK and a sublattice
G ⊂ Num1(Xs), show that G is primitive.
If not primitive, thenG has nontrivial index in its saturation G˜, so there is a primer | [ ˜G : G ]with r2| [ ˜G : G ]2· disc ˜G = disc G. ThenG ⊗ Fr → Num1(Xs) ⊗ Fr is not injective.
For all primesp of good reduction and H ⊂ Num1(Xps)the map
Num1(Xs) → Hom(H, Z)
induces a non-injective composition
G ⊗ Fr → Num1(Xs) ⊗ Fr → Hom(H ⊗ Fr, Fr).
Sufficient for primitivity. Find for each r with r2| disc G a prime p
and a subgroupH ⊂ Num1(Xps)for which the composition is injective (linear algebra overFr).
Saturation
Goal. Given a surface X over a global fieldK and a sublattice
G ⊂ Num1(Xs), show that G is primitive.
If not primitive, thenG has nontrivial index in its saturation G˜, so there is a primer | [ ˜G : G ]with r2| [ ˜G : G ]2· disc ˜G = disc G. ThenG ⊗ Fr → Num1(Xs) ⊗ Fr is not injective.
For all primesp of good reduction and H ⊂ Num1(Xps)the map
Num1(Xs) → Hom(H, Z)
induces a non-injective composition
G ⊗ Fr → Num1(Xs) ⊗ Fr → Hom(H ⊗ Fr, Fr).
Sufficient for primitivity. Find for each r with r2| disc G a prime p
and a subgroupH ⊂ Num1(Xps)for which the composition is injective (linear algebra overFr).
Saturation
Goal. Given a surface X over a global fieldK and a sublattice
G ⊂ Num1(Xs), show that G is primitive.
If not primitive, thenG has nontrivial index in its saturation G˜, so there is a primer | [ ˜G : G ]with r2| [ ˜G : G ]2· disc ˜G = disc G. ThenG ⊗ Fr → Num1(Xs) ⊗ Fr is not injective.
For all primesp of good reduction and H ⊂ Num1(Xps)the map
Num1(Xs) → Hom(H, Z)
induces a non-injective composition
G ⊗ Fr → Num1(Xs) ⊗ Fr → Hom(H ⊗ Fr, Fr).
Sufficient for primitivity. Find for each r with r2| disc G a prime p
and a subgroupH ⊂ Num1(Xps)for which the composition is injective (linear algebra overFr).
Saturation
Goal. Given a surface X over a global fieldK and a sublattice
G ⊂ Num1(Xs), show that G is primitive.
If not primitive, thenG has nontrivial index in its saturation G˜, so there is a primer | [ ˜G : G ]with r2| [ ˜G : G ]2· disc ˜G = disc G. ThenG ⊗ Fr → Num1(Xs) ⊗ Fr is not injective.
For all primesp of good reduction and H ⊂ Num1(Xps)the map
Num1(Xs) → Hom(H, Z)
induces a non-injective composition
G ⊗ Fr → Num1(Xs) ⊗ Fr → Hom(H ⊗ Fr, Fr).
Sufficient for primitivity. Find for each r with r2| disc G a primep
and a subgroupH ⊂ Num1(Xps)for which the composition is injective (linear algebra overFr).
Application
Theorem (Mizukami (m = 4), Sch¨utt–Shioda–vL (m ≤ 100))
For any integer1 ≤ m ≤ 100the N´eron-Severi group of the Fermat surfaceSm⊂ P3 overCgiven by
xm+ ym+ zm+ wm = 0