Effective computations of HasseWeil zeta functions Edgar Costa - - PowerPoint PPT Presentation

Effective computations of Hasse–Weil zeta functions

Edgar Costa

ICERM/Dartmouth College

20th October 2015 ICERM

1 / 24 Edgar Costa Variation of N´ eron-Severi ranks of K3 surfaces

Variation of N´ eron-Severi ranks of K3 surfaces

Edgar Costa

ICERM/Dartmouth College

20th October 2015 ICERM

2 / 24 Edgar Costa Variation of N´ eron-Severi ranks of K3 surfaces

Setup

k a number field; X a K3 surface over k; p a prime of k where X has good reduction Xp; NS(•) := Pic(•)/ Pic0(•), the N´ eron-Severi group; a Z lattice geometrically associated to the surface • ρ(•) := rk(NS(•)), the arithmetic/geometric Picard number of •. ρ(•) the rank of lattice mentioned above. Question How are the geometric Picard numbers, ρ(X) and ρ(X p), related? How does the geometric Picard number behaves under reduction modulo p?

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Some observations

Question How are the geometric Picard numbers, ρ(X) and ρ(X p), related? How does the geometric Picard number behaves under reduction modulo p? There is a natural specialization homomorphism sp : NS(X) ֒ → NS(X p). Thus, ρ(X) ≤ ρ(X p). 1 ≤ ρ(X) ≤ 20 2 ≤ ρ(X p) ≤ 22, and ρ(X p) is always even. Computing ρ(X) is a hard problem. We can compute ρ(X p) by counting points on Xp.

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Computing ρ(X p)

ZXp(T) := exp ∞

• i=1

#Xp

• Fqi

i T i

• =

1 (1 − T)P2(X, T)(1 − q2T) where P2(X, T) := det

• 1 − T Frobp |H2

∈ Z[t] which have reciprocal roots of absolute value q. Theorem (Tate Conjecture) [Tate], [Charles], [Pera] and [Maulik] X an abelian surface or a K3 surface then: ρ(Xp) = ordT=1/q P2(T) ρ(X p) =

ζ ordT=ζ/q P2(T), where ζ runs over all roots of unity.

(Artin-Tate Conjecture) P2(T) disc(NS(Xp)) mod Q×2

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Partial answer

Question How are the geometric Picard numbers, ρ(X) and ρ(X p), related? Theorem [Charles] Charles constructed a function η(X) ≥ 0a and proved that for all p of good reduction we have min

q ρ(X q) = ρ(X) + η(X) ≤ ρ(X p)

Equality occurs infinitely often. Furthermore, over some finite extension of k, the set of such primes has density one.

aDepends on the Hodge structure underlying the transcendental lattice and its

endomorphism algebra.

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Picard Jumps

Let Πjump(X) :=

• p : ρ(X) + η(X) < ρ(X p)
• =
• p : min

q ρ(X q) < ρ(X p)

• Question

What can we say about Πjump(X)? What about γ(X, B) := # {p ≤ B : p ∈ Πjump(X)} # {p ≤ B} as B → ∞ ?

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Why?

Information about Πjump(X) Geometric statements Theorem [Bogomolov-Hassett-Tschinkel] and [Li-Liedtke] If #Πjump(X) = +∞ or η(X) > 0 then X has infinitely many rational curves.

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Product of Elliptic Curves

Let X ≃ Kummer(E1 × E2) and Ei elliptic curve over Q. ρ(X) = 18 + rk(Hom(E1, E2)) = 18 +

• E1 ∼ E2;

rk(End(E1)) E1 ∼ E2 X ρ

• X
• γ(X, B)

square of CM 20 1/2 + o(1) CM theory square of non-CM 19

log log B log B B

< • < C log B

B1/4

[Elkies] CM times CM 18 1/4 + o(1) CM theory CM times non-CM 18 ? non-CM times non-CM 18 ≫ 0 [Charles] Remark p ∈ Πjump(X) depends uniquely on the pair (aE1(p), aE2(p))

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The simplest case

Let X ≃ Kummer(E1 × E2) and Ei elliptic curve over Q. ρ(X) = 18 + rk(Hom(E1, E2)) = 18 +

• E1 ∼ E2;

rk(End(E1)) E1 ∼ E2 X ρ

• X
• P (p ∈ Πjump(X))

γ(X, B) square of CM 20 1/2 1/2 square of non-CM 19 ∼ 1/√p c/ √ B CM times CM 18 1/4 1/4 CM times non-CM 18 ∼ 1/√p c/ √ B non-CM times non-CM 18 ∼ 1/√p c/ √ B Remark p ∈ Πjump(X) depends uniquely on the pair (aE1(p), aE2(p))

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Sato-Tate for Abelian Surfaces

Let A be an abelian surface. Let G ⊂ Sp4 be the “smallest” group such that for each p of good reduction there is an g ∈ G such that det(1 − Tg) = det(1 − T/√q Frobp |H1(A)). Then STA ⊂ USp4 is the maximal compact subgroup of G. Conjecture [Sato-Tate] The conjugacy classes of Frobp /√q|H1 are equidistributed with respect to the Haar measure on STA. Theorem [Fit´ e–Kedlaya–Rotger–Sutherland] The Galois structure on End(A) ⊗ R determines STA. Only 52 groups up to conjugacy can be realized as STA Example, if End(A) = Z, then STA = USp4.

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Sato-Tate for Abelian Surfaces

Conjecture [Sato-Tate] The conjugacy classes of Frobp /√q|H1 are equidistributed with respect to the Haar measure on STA. If det

• 1 − T/√q Frobp |H1

:= 1 + a1T + a2T 2 + a1T 3 + T 4, then (a1, a2) ∼ (Tr(M), Tr(M ∧ M)) with M ∈ STA. Question What about det

• 1 − T/q Frobp |H2

? H2 = H1 ∧ H1, thus det

• 1 − T/q Frobp |H2

= (T − 1)2ψp(T) ψp(T) := 1 + (2 − a2)T + (2 + a2

1 − 2a2)T 2 + (2 − a2)T 3 + T 4

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Abelian Surfaces

Let X ≃ Kummer(A), A an abelian surface over Q. End(A) ρ(A) and STA ρ(X) = 16 + ρ(A) A ρ

• A
• γ(A, B) as B → ∞

the generic case 1 c/ √ B non-CM times non-CM (non Galois) 2 c/ √ B non-CM times non-CM (Galois) or simple RM 2 1/2 CM times non-CM 2 c/ √ B simple CM 2 3/4 CM times CM 2 1/4 simple QM or square of non-CM 3 c/ √ B square of CM 4 1/2 Remark Easy to verify the conjectures numerically!

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What about K3 surfaces?

Let X be a K3 surface. In this case H1(X) is trivial and dim H2(X) = 22. In general, we need to compute det(1 − T/q Frobp |H2(X)) to deduce ρ(X p)! The definition of STX follows closely the one for an abelian surface. We just need to replace det(1 − T/√q Frobp |H1(A)) by det(1 − T/q Frobp |H2(X)). STX ⊂ O22−ρ(X) and ST0

X ⊂ SO22−ρ(X).

Very little is known about what subgroups of O21 can show up as STX!

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Kummer surface

If X ≃ Kummer(A), where A is an abelian surface, then ρ(X) = 16 + ρ(A) and STX = STA/{±1} ⊂ SO5 ∼ = USp4/{±1}.

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Computing ρ(X)

Let TX be the orthogonal complement of NS(X top

C ) in H2(X top C , Q).

Let EX be the endomorphism algebra of TX that respects the Hodge structure EX is a totally real field or a CM-field. Theorem [Charles] ρ(X p) ≥

• ρ(X)

if EX is a CM or dimEX (TX) is even, ρ(X) + [EX : Q] if EX is a totally real or dimEX (TX) is odd. Further, assume that we are in the second case, then exist infinitely many pairs (p, q) such that the equality holds and disc(NS(Xp)) ≡ disc(NS(Xq)) mod Q×2

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Numerical experiments for ρ(X) = 1

X a quartic K3 surface with ρ(X) = 1 and EX = Q. γ(X, B) ∼ cX/ √ B, B → ∞ Prob(p ∈ Πjump(X)) ∼ 1/√p Heuristics for the 1/√p behaviour?!?

k 1 2 3 4 5 E Tr(Frobp /p)k 0.002 0.988

• 0.016

2.866

• 0.218
• 0.008

1.016

• 0.065

3.096

• 0.658

EO[Tr(M)k] 1 3

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Numerical experiments for ρ(X) = 2

Now ρ(X) = ρ(X) = 2 and EX = Q or CM No obvious trend . . . Is it related to the splitting of primes in a quadratic extension over Q? Hint: Q(√DX).

k 1 2 3 4 5 E Tr(Frobp /p)k 0.022 1.009 0.0573 2.959 0.136 0.001 0.990

• 0.039

2.941

• 0.454

0.009 1.013 0.055 3.069 0.015 EO[Tr(M)k] 1 3 ∼9000 CPU hours per example.

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Discriminant of a K3 surface

Let X be a quartic K3 surface over Q and DX be its discriminant. Theorem [Elsenhans-Jahnel] The functional equation of the Frobenius action on H2(X) has the plus sign if and only if DX is square modp. Theorem If ρ(X) = 2r then ρ(X p) ≥ 2r + 2 at all primes p such that DX is not square mod p. Corollary If DX is not a square, ρ(X) = ρ(X) = 2r and η(X) = 0, then

• p : p is inert in Q
• DX
• ⊂ Πjump(X),

lim inf

B→∞ γ(X, B) ≥ 1/2,

and X has infinitely many rational curves.

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Numerical experiments for ρ(X) = ρ(X) = 2

If we “ignore”

• p : p is inert in Q

√DX

• ⊂ Πjump(X)

γ

• XQ(√DX ), B
• ∼ c/

√ B, B → ∞

100 1000 104 105 0.10 1.00 0.50 0.20 0.30 0.15 0.70 γ(XQ(√DX), B) 5.13B−0.448 100 1000 104 105 0.10 1.00 0.50 0.20 0.30 0.15 0.70 γ(XQ(√DX), B) 6.00B−0.439 100 1000 104 105 0.10 1.00 0.50 0.20 0.30 0.15 0.70 γ(XQ(√DX), B) 5.34B−0.428

Prob(p ∈ Πjump(X)) =

• 1

if p is inert in Q √DX

• ,

∼

1 √p

if p splits in Q √DX

• Heuristics for the 1/√p behaviour?!?

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More examples?

Does Q √DX

• play the same role when ρ(X) < ρ(X) = 2r ? No!

But we have a similar trend! lim inf

B→∞ γ(X, B) ≥ 1/2.

Sometimes we can guess the right quadratic extension K such that {p : p is inert in K} ⊂ Πjump(X) What about when ρ(X) is odd? γ(X, B) ∼ cX/ √ B, B → ∞.

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Summary and Questions

Computing zeta functions of K3 surfaces via p-adic cohomology Experimental data for Πjump(X) Results regarding Πjump(X) New examples of K3 surfaces with infinitely many rational curves Questions Heuristics for the 1/√p behaviour? “Equidistribution” of the characteristic polynomials doesn’t seem to be enough. Can we compute (or statistically guess) EX? Can we distinguish CM vs Q? For abelian surfaces A, STA is determined by its Galois type, i.e., ρA : Gal(K/k) ֒ → AutR−alg(End(AK)R). Can we say something similar for K3 surfaces? involving EX?

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References

Edgar Costa and Yuri Tschinkel: Variation of N´ eron-Severi ranks of reductions of K3 surfaces Experimental Mathematics 23 (2014), 475-481. Edgar Costa: Effective computations of Hasse–Weil zeta functions Ph.D. Thesis

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Thank you!

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