On Weil polynomials of K3 surfaces

Andreas-Stephan Elsenhans^{1} and Jörg Jahnel^{2}

1Universität Bayreuth, Mathematisches Institut, Universitätsstraße 30, D-95447 Bayreuth, Germany^{⋆} 2Fachbereich 6, Mathematik, Universität Siegen, Walter-Flex-Straße 3, D-57072 Siegen, Germany

   1stephan.elsenhans@uni-bayreuth.de, ^{2}~jjahnel@mathematik.uni-siegen.de

Abstract. For K3 surfaces, we derive some conditions the characteristic polynomial of the Frobenius on the étale cohomology must satisfy. These conditions may be used to speed up the computation of Picard numbers and the decision of the sign in the functional equation^{∗∗}. Our investigations are based on the Artin-Tate formula.

1 Introduction

An algebraic integer such that all its conjugates have absolute value \sqrt{r} is called an r-Weil number. Correspondingly, a possibly reducible monic polynomial Φ ∈ ℤ[T] such that all roots have absolute value \sqrt{r} is called an r-Weil polynomial.

Let q be a prime power and r = q^{k}. Then, for every smooth projective variety V over 𝔽_{q}, the eigenvalues of the Frobenius endomorphism Frob on the étale cohomology H_{ét}^{k}(V_{\overline{𝔽}_{q}}, ℚ_{l}) are r-Weil numbers [3, Lemme 1.7]. Conversely, every q^{k}-Weil number is an eigenvalue of Frob on H_{ét}^{k}(V_{\overline{𝔽}_{q}}, ℚ_{l}) for a suitable smooth projective variety V over 𝔽_{q}. Actually, this fact is a direct consequence of the results of T. Honda [9].

In this note, we will study the Weil numbers of K3 surfaces. As the second Betti number of a K3 surface is b_{2}(V) = 22 and q is always a root of the characteristic polynomial, the possible Weil numbers are of degree at most 20.

We will show that not all q^{2}-Weil polynomials Φ ∈ ℤ[T] satisfying deg Φ = 22 and Φ(q) = 0 occur as characteristic polynomials of Frob on the étale cohomology of K3 surfaces. Concerning K3 surfaces of fixed degree, even more restrictions result. Our investigations are based on the Artin-Tate formula which we will recall in section 3.

⋆ The first author was partially supported by the Deutsche Forschungsgemeinschaft

(DFG) through a funded research project. ⋆⋆ The computer part of this work was executed on the Sun Fire V20z Servers of the

Gauß Laboratory for Scientific Computing at the Göttingen Mathematisches Institut. Both authors are grateful to Prof. Y. Tschinkel for the permission to use these machines as well as to the system administrators for their support.

An application. The characteristic polynomial of Frob may be computed by counting points over extensions of the ground field. Indeed, for V a K3 surface over 𝔽_{q}, the Lefschetz trace formula [13, Ch. VI, §12] yields tr(Frob^{e}) = #V (𝔽_{q^{e}}) − q^{2e} − 1.

When we denote the eigenvalues of Frob by r_{1}, …, r_{22}, we have tr(Frob^{e}) = r_{1}^{e} + ⋯ + r_{22}^{e} =: σ_{e}(r_{1}, …, r_{22}). Newton’s identity [20]

s_{k}(r_{1}, …, r_{22}) = \frac{1}{k} ∑_{j=0}^{k−1}(−1)^{k+j+1}σ_{k−j}(r_{1}, …, r_{22})s_{j}(r_{1}, …, r_{22})

shows that the knowledge of σ_{e}(r_{1}, …, r_{22}), for e = 1, …, k, is sufficient in order to determine the coefficient (−1)^{k}s_{k} of T^{22−k} of the characteristic polynomial Φ of Frob. Further, there is the functional equation

q^{deg Φ}Φ(T) = ±T^{deg Φ}Φ(q^{2}/T)  (1)

which, as deg Φ = 22, relates the coefficient of T^{k} with that of T^{22−k}.

Nevertheless, this method is time-consuming. The size of the fields to be considered grows exponentially. One would like to avoid point counting over large fields and, nevertheless, determine Φ sufficiently well in order to decide things such as the sign in (1). Algorithms of this type were presented in [6]. For example, Algorithm 22 of [6] verifies that the geometric Picard rank is 2, having counted points over 𝔽_{p}, …, 𝔽_{p^{9}} for p a prime number.

The main result of the present article leads to a more substantial approach to this problem. In fact, we will show that certain hypothetical characteristic polynomials are impossible, in general. This leads to an improvement of [6, Algorithm 22]. Sections 7 and 8 will be devoted to examples showing how this improvement works in practice.

Remark 1. A continuation of this application, which we have in mind, is the computation of the geometric Picard rank for K3 surfaces over ℚ. Here, the general strategy is to use reduction modulo p. One applies the inequality

rk Pic(V\overline{ℚ} ) ≤ rk Pic(V_{\overline{𝔽}_{p}})

which is true for every smooth variety V over ℚ and every prime p of good reduction. Then, the number of eigenvalues of Frob which are roots of unity is an upper bound for the Picard number. More details are given in [6] and [7].

2 The Galois group of a Weil polynomial

For a randomly chosen irreducible polynomial over ℚ, one expects the Galois group to be the full symmetric group. In this sense, the irreducible factors of a Weil polynomial are not very random.

When we consider the operation of Frob on a cohomology group of even degree, cyclotomic factors do arise. They correspond to the algebraic part of the cohomology, i.e., to the image of the Picard group and its analogues in higher codimension. The corresponding Galois group is always abelian.

Concerning the remaining factors, still, there are restrictions on the Galois group. Note that, for each root of an irreducible r-Weil polynomial not of degree 1, the complex conjugate is a root, too. This means, the roots come in pairs. The product of each pair is equal to r. The Galois group therefore acts on the pairs. For a suitable integer n, it is a subgroup of the semi-direct product (ℤ/2ℤ)^{n} ⋊S_{n} ⊂ S_{2n}. Here, each factor (ℤ/2ℤ) acts on one pair by complex conjugation. The complex conjugation itself belongs to the center of the group.

An experimental result. One could ask for further restrictions on the Galois group. For that, we computed the characteristic polynomial of Frob for a few thousand randomly chosen K3 surfaces. In each case, the factorization of that polynomial had precisely one irreducible factor which was not cyclotomic. This coincides with Zarhin’s results [18] for ordinary K3 surfaces.

Furthermore, in the vast majority of the examples, the Galois group of the last factor was actually equal to the semi-direct product (ℤ/2ℤ)^{n} ⋊ S_{n} ⊂ S_{2n}. For example, this was true for 875 out of 1 000 K3 surfaces of degree 2 over 𝔽_{3} and 923 out of 1 000 K3 surfaces of degree 2 over 𝔽_{7}.

The resolvent algebra. Let Φ ∈ ℚ[T] be a polynomial such that its set of roots is of the particular form {r_{1}, r′_{1}, …, r_{n}, r′_{n}} such that r_{1}r′_{1} = … = r_{n}r′_{n} =: r ∈ ℚ. Then, the sums r_{1} + r′_{1}, …, r_{n} + r′_{n} are the roots of a polynomial R ∈ ℚ[T] of half the degree. We will call R the resolvent polynomial and A := ℚ[T]/R the resolvent algebra of Φ.

Remarks 2. a) When Φ is an r-Weil polynomial of even degree, the assumption is satisfied if and only if \sqrt{r} is a root of even multiplicity (or no root) of Φ. In this case, (−\sqrt{r}) has even multiplicity, too. In fact, this means exactly that Φ fulfills the functional equation (1) with the plus sign.

b) On the other hand, when one wants to verify that a given polynomial satisfying the functional equation is, in fact, a Weil polynomial, the resolvent is helpful. Observe that the roots of the initial polynomial are all of absolute value \sqrt{r} if and only if the roots of the resolvent are all real and in the interval [−2\sqrt{r}, 2\sqrt{r}]. That property may easily be checked using Sturm’s chain theorem. This is a fast and exact replacement of [6, Algorithm 23].

3 The Artin-Tate formula

Let us recall the Artin-Tate conjecture in the special case of a K3 surface.

Conjecture 3 (Artin-Tate). Let V be a K3 surface over a finite field 𝔽_{q}. Denote by ρ the rank and by Δ the discriminant of the Picard group of V, defined over 𝔽_{q}. Then,

|Δ| = \frac{lim_{T →q} \frac{Φ(T)}{(T −q)^{ρ}}}{q^{21−ρ}#Br(V)} .

Here, Φ denotes the characteristic polynomial of Frob on H_{ét}^{2}(V_{\overline{𝔽}_{q}}, ℚ_{l}). Finally, Br(V) is the Brauer group of V.

Remarks 4. i) The characteristic polynomial Φ is independent of the choice of the auxiliary prime l as long as l ≠ p for q = p^{e} [3, Théorème 1.6]. ii) For a general non-singular, projective surface, the exponent of q in the numerator is b_{2}(V) − h_{02}(V) − ρ. Here, h_{02}(V) denotes the Hodge number. iii) The Artin-Tate conjecture is proven for most K3 surfaces. Most notably, the Tate conjecture implies the Artin-Tate conjecture [11, Theorem 6.1]. iv) The Tate conjecture claims that all zeroes of Φ of the form qζ for ζ a root of unity belong to the algebraic part of H_{ét}^{2}(V_{\overline{𝔽}_{q}}, ℚ_{l}). I.e., it asserts that the transcendental part never generates a zero of this form. The evidence for this is overwhelming as far as K3 surfaces are concerned. The Tate conjecture is proven for elliptic K3 surfaces [1] and ordinary K3 surfaces [15]. In characteristic different from 2 and 3, even more particular cases were successfully treated [16].

v) It is expected that Br(V) is always a finite group. This is actually equivalent to the Tate conjecture. In this case, #Br(V) is automatically a perfect square. We may therefore compute the square class of Δ making use of the Artin-Tate conjecture.

An unconditional version of the Artin-Tate formula.

Notation 5. i) For n a positive integer, we will denote by μ_{n} the sheaf of n-th roots of unity with respect to the fppf topology. When l is a prime number, we put H_{fppf}^{d}(V_{\overline{𝔽}_{q}}, T_{l}μ) := \underleftarrow{lim}_{e} H_{fppf}^{d}(V_{\overline{𝔽}_{q}}, μ_{l^{e}}). ii) For l a prime number and M an abelian group, the notation M_{l-pow} shall be used for the l-power torsion subgroup of M. Similarly, we will write M_{l-div} ⊆ M_{l-pow} for the subgroup of infinitely l-divisible elements. iii) We will denote by M^{Frob} and M_{Frob} the invariants, respectively coinvariants, under the operation of Frob on the abelian group M. The coinvariants may have torsion even when M is torsion-free. Write M′_{Frob} for the torsion-free quotient.

Proposition 6. Let V be a K3 surface over a finite field 𝔽_{q} and l be any prime. Write Φ for the characteristic polynomial of Frob on the étale cohomology of V_{\overline{𝔽}_{q}} and ρ for the multiplicity of q as a zero of Φ.

i) Then, the Brauer group Br(V) is a torsion group. The quotient

Br_{0}(V, l) := Br(V)_{l-pow}/ Br(V)_{l-div}

is a finite group of square order. ii) Further, H_{fppf}^{2}(V_{\overline{𝔽}_{q}}, T_{l}μ)^{Frob} is a free ℤ_{l}-module of rank ρ. iii) Denote by Δ_{l} the discriminant of the bilinear form

H_{fppf}^{2}(V_{\overline{𝔽}_{q}}, T_{l}μ)^{Frob} × H_{fppf}^{2}(V_{\overline{𝔽}_{q}}, T_{l}μ)^{Frob} ⟶ ℤ_{l}

defined by Poincaré duality. Then,

ν_{l}(Δ_{l}) = ν_{l}(^{\frac{lim_{T →q} \frac{Φ(T)}{(T −q)^{ρ}}}{q^{21−ρ}#Br_{0}(V,l)}}) .

Proof. i) Finiteness of Br_{0}(V, l) follows immediately from [8, (8.9)]. Further, there is a non-degenerate alternating pairing Br_{0}(V, l) × Br_{0}(V, l) → ℚ_{l}/ℤ_{l} constructed in [19, Lemma 3.4.1]. This ensures that the group order is a perfect square. ii) and iii) We denote the zeroes of Φ by r_{1}, …, r_{22}. First case. l ≠ p. Here, H := H_{fppf}^{2}(V_{\overline{𝔽}_{q}}, T_{l}μ) = H_{ét}^{2}(V_{\overline{𝔽}_{q}}, ℤ_{l}(1)) is the same as l-adic étale cohomology. It is a free ℤ_{l}-module of rank 22. In the present case, the operation of Frob on H is known to be semi-simple [4, Corollary 1.10]. The eigenvalues are r_{1}/q, …, r_{22}/q. Assertion ii) follows immediately from this.

Further, we have ν_{l}(Δ_{l}) = ν_{l}(#coker(H^{Frob} → Hom(H^{Frob}, ℤ_{l}))), the map being induced by Poincaré duality. Identifying Hom(H, ℤ_{l}) with H, the module Hom(H^{Frob}, ℤ_{l}) goes over into H′_{Frob}. Here, as shown in [19, Proposition 1.4.2], (H_{Frob})_{tors} ≅ Br_{0}(V, l). Further, the order of the cokernel of the canonical homomorphism H^{Frob} → H_{Frob} is equal to the l-primary part of ∏_{r_{j}≠q}(1 − r_{j}/q). Altogether, this implies the claim. Second case. l = p. Here, some modifications are necessary which are described in [11]. More concretely, the short exact sequence

0 → Pic(V_{\overline{𝔽}_{q}})⊗_{ℤ} ℤ_{p} → H_{fppf}^{2}(V_{\overline{𝔽}_{q}}, T_{p}μ) → \underleftarrow{lim} Br(V_{\overline{𝔽}_{q}})_{p^{n}} → 0

immediately shows that H := H_{fppf}^{2}(V_{\overline{𝔽}_{q}}, T_{p}μ) is a torsion-free ℤ_{p}-module. Otherwise, its structure is rather different from the previous case. The rank of H is, in general, less than 22. Eigenvalues of Frob are only those r_{j}/q which are units in \overline{ℚ}_{p} [11, 1.4]. But this is enough to show ii).

Generally, there are unipotent connected quasi-algebraic groups U^{d} and étale group schemes D_{n}^{d} for d = 2, 3 and n ≫ 0 which provide short exact sequences 0 → U^{d}(\overline{𝔽}_{q}) → H_{fppf}^{d}(V_{\overline{𝔽}_{q}}, μ_{p^{n}}) → D_{n}^{d}(\overline{𝔽}_{q}) → 0. For varying n, the vector groups U^{3}(\overline{𝔽}_{q}) are connected by identities. Further, D_{n}^{3} = 0. Hence, if dim U^{3} = s then #H_{fppf}^{3}(V_{\overline{𝔽}_{q}}, T_{p}μ)^{Frob} = q^{s} the operation of Frob being semi-simple. Actually, one has s = 0 except when V is supersingular.

Poincaré duality is available [12, Theorem 5.2 and Corollary 2.7.c)] only at the level of torsion coefficients. Thereby, U^{2}(\overline{𝔽}_{q}) and U^{3}(\overline{𝔽}_{q}) are dual to each other. One has \underleftarrow{lim} U^{2}(\overline{𝔽}_{q}) = 0 and R^{1}\underleftarrow{lim} U^{2}(\overline{𝔽}_{q}) = 0 as the connecting homomorphisms are zero. Hence, H_{fppf}^{2}(V_{\overline{𝔽}_{q}}, T_{p}μ) ≅ \underleftarrow{lim} D_{n}^{2}(\overline{𝔽}_{q}). Further, it turns out that the homomorphism H_{Frob} → Hom(H^{Frob}, ℤ_{p}) does not need to be bijective. It has a cokernel exactly of order q^{s} (cf. [11, Lemma 5.2]).

Summarizing, we find that Δ_{p} has the same p-adic valuation as

q^{s}· ∏_{{νp(r_{j} /q)=0 \atop r_{j} ≠q}} (1 − r_{j}/q) .

For iii), it remains to show the following. Up to p-adic units, the product of the remaining factors, i.e. ∏_{ν_{p}(r_{j}/q)≠0} (1 − r_{j}/q), equals q^{s−1}. This is worked out in [11, sec. 7]. □

Remark 7. The Tate conjecture implies H_{fppf}^{2}(V_{\overline{𝔽}_{q}}, T_{l}μ)^{Frob} ≅ Pic(V_{\overline{𝔽}_{q}}) ⊗_{ℤ} ℤ_{l}. Further, it is equivalent to Br(V)_{l-div} = 0. Thus, Proposition 6 goes over into the Artin-Tate formula in its usual form. However, the Tate conjecture is unknown in general, even for K3 surfaces. For this reason, we prefer to apply the version of the Artin-Tate formula which holds unconditionally.

4 The rank-1 condition

Let V be a K3 surface of degree d over a finite field 𝔽_{q}. Assume that q is a simple zero of the characteristic polynomial of Frob. Then, the Tate conjecture is true for V and the arithmetic Picard rank is equal to 1. The discriminant of Pic(V) is equal to d. A comparison with the analytic discriminant computed via the Artin-Tate formula leads to a non-trivial condition for hypothetical Weil polynomials.

Remarks 8. a) This is a condition for rank-1 surfaces of a given degree d. It is not a condition for K3 surfaces, in general.

b) The degree of a K3 surface may be any even integer greater than zero. On the other hand, when the arithmetic Picard rank is 1, the number (−q) is necessarily among the Frobenius eigenvalues. Hence, the Artin-Tate formula can generate only even numbers.

c) The Artin-Tate conjecture implies the inequality #Br(V)|Δ| ≤ 2^{22−ρ}q. Thus, the left hand side is O(q). Observe the following striking consequence. Over the field 𝔽_{q}, there is no K3 surface of a square-free degree d > 2^{21}q and arithmetic Picard rank 1.

Remark 9. The rank-1 condition may be extended to other situations where a subgroup of the Picard group is known. For this, one has to compare the predicted ranks and discriminants with the known ones.

5 The field extension condition

Notation 10. For q a positive integer, let Φ be a q^{2}-Weil polynomial. Then, we will write

E_{Φ}^{(c)} := ∏_{r_{j}≠q}\frac{q^{c} − r_{j}^{c}}{q − r_{j}} /q^{(c−1)(21−ρ)} .

Here, r_{j} runs over all the zeroes of Φ. Further, ρ is the multiplicity of the zero q.

Observation 11 (Field extension for the characteristic polynomial). Let V be any smooth, projective variety over 𝔽_{q} and ∏_{j}(T − r_{j}) the characteristic polynomial of Frob on H_{ét}^{2}(V_{\overline{𝔽}_{q}}, ℚ_{l}). Then, the corresponding polynomial for V_{𝔽_{q^{d}}} is ∏_{j}(T − r_{j}^{d}).

Theorem 12. Let V be a K3 surface over 𝔽_{q}. Further, let c be a positive integer. Then, for Φ the characteristic polynomial of Frob, the expression E_{Φ}^{(c)} is a perfect square in ℚ. Proof. If there is an r_{j} ≠ q such that r_{j}^{c} = q^{c} then E_{Φ}^{(c)} = 0. Otherwise, for every prime l, H_{fppf}^{2}(V_{\overline{𝔽}_{q}}, T_{l}μ)^{Frob^{q^{c}}} is a sublattice of finite index in H_{fppf}^{2}(V_{\overline{𝔽}_{q}}, T_{l}μ)^{Frob^{q}}. In particular, the discriminants differ by a factor being a perfect square. Dividing the Artin-Tate formulas for V_{𝔽_{q^{c}}} and V_{𝔽_{q}} through each other yields that ν_{l}(E_{Φ}^{(c)}) is even for every l. Finally, it is easy to see that E_{Φ}^{(c)} > 0. □

Remark 13. Assume the Tate conjecture. Then, E_{Φ}^{(c)} is non-zero if and only if rk Pic(V_{𝔽_{q}}) = rk Pic(V_{𝔽_{q^{c}}}).

Definition 14. We will call the condition on E_{Φ}^{(c)} to be a perfect square, the field extension condition for the field extension 𝔽_{q^{c}}/𝔽_{q}.

Explicit computation of the expression E_{Φ}^{(c)}. Our goal is now to describe the square class of E_{Φ}^{(c)} more explicitly. It will turn out that, for an arbitrary Weil polynomial, E_{Φ}^{(c)} may be a non-square. In other words, Theorem 12 provides a non-trivial condition.

Remark 15. A priori, there are infinitely many conditions, one for each value of c. The main result of this section is that there is in fact only one condition. Further, this condition may be checked easily.

Lemma 16. Let f ∈ ℚ[T] be a q^{2}-Weil polynomial. Suppose f(q) ≠ 0 and f(−q) ≠ 0. Then, for r_{1}, …, r_{2l} the zeroes of f,

∏_{j=1}^{2l} \frac{q^{c} − r_{j}^{c}}{q − r_{j}} ∈ \{ \begin{matrix}(ℚ^{∗})^{2} ∪ \{0\} & for c odd, \\ f(−q)(ℚ^{∗})^{2} ∪ \{0\} & for c even. \end{matrix}

Further, the left hand side is actually in f(−q)(ℚ^{∗})^{2} for c = 2. Proof. First observe that, for c = 2, the numerators q^{2} − r_{j}^{2} are all non-zero according to the assumption. Hence, the additional assertion is clear once we showed the main one.

For that, let us start with the contribution of one pair of complex conjugate roots. Put r_{j} = q(u + iv). Then, the corresponding factor is

\frac{(q^{c} − r_{j}^{c})(q^{c} − \overline{r}_{j}^{c})}{(q − r_{j})(q − \overline{r}_{j})} = \frac{(q^{c} − q^{c}(u + iv)^{c})(q^{c} − q^{c}(u − iv)^{c})}{(q − q(u + iv))(q − q(u − iv)}

= q^{2(c−1)} ∏_{k=1}^{c−1}(1 − ζ_{c}^{k}(u + iv))(1 − ζ_{c}^{k}(u − iv)) .

Using (u + iv)(u − iv) = 1, we get

q^{2(c−1)} ∏_{k=1}^{c−1}(1 − 2ζ_{c}^{k}u + ζ_{c}^{2k}) .

Next, for k ≠ c/2, let us multiply the factors for k and c − k. This yields

(1 − 2ζ_{c}^{k}u + ζ_{c}^{2k})(1 − 2ζ_{c}^{c−k}u + ζ_{c}^{2c−2k}) = 2 + 4u^{2} − 8u Re(ζ_{c}^{k}) + 2 Re(ζ_{c}^{2k}) .

As Re(ζ_{c}^{2k}) = 2 Re(ζ_{c}^{k})^{2} − 1, the latter term is the same as

4u^{2} − 8u Re(ζ_{c}^{k}) + 4 Re(ζ_{c}^{k})^{2} = (2u − 2 Re(ζ_{c}^{k}))^{2} .

Multiplying over all k such that 1 ≤ k < c/2, we find a square in ℚ(u). Consequently, up to the factor for k = c/2, if present, the contribution of the pair \{r_{j}, \overline{r}_{j}\} is a square in the resolvent algebra A of f.

Multiplying over all l pairs means to form a norm for the extension A/ℚ. As the norm of a square is a square, the result is a perfect square in ℚ. For c odd, this completes the argument.

For c even, the factors for k = c/2 are still missing. These are the ones for ζ_{c}^{k} = −1. We find the product

∏_{j=1}^{l}(1 + r_{j}/q)(1 + \overline{r}_{j}/q) = q^{−2l}f(−q) .

The assertion follows. □

Proposition 17. Let Φ be a q^{2}-Weil polynomial of even degree. Then,

E_{Φ}^{(c)} ∈ \{ \begin{matrix}(ℚ^{∗})^{2} ∪ \{0\} & for c odd, \\ qΦ(−q)(ℚ^{∗})^{2} ∪ \{0\} & for c even. \end{matrix}

For c = 2, we actually have E_{Φ}^{(c)} ∈ qΦ(−q)(ℚ^{∗})^{2}. Proof. First case: c is odd. Then, the denominator q^{(c−1)(21−ρ)} is a perfect square. The zeroes (−q) contribute factors q^{c−1} which are squares, too. Finally, the contribution to E_{Φ}^{(c)} of the zeroes not being real is a perfect square according to Lemma 16. Second case: c is even. If (−q) is a zero of Φ then E_{Φ}^{(c)} = 0. This coincides with the claim as Φ(−q) = 0. Otherwise, write Φ(T) = (T − q)^{ρ}f(T) where f(q) ≠ 0 and f(−q) ≠ 0. By assumption, ρ is even. Hence, q^{(c−1)(21−ρ)} is in the square class of q. Further, the zeroes of Φ differing from q are exactly the zeroes of f. Their contribution is in f(−q)(ℚ^{∗})^{2} for c = 2 and in f(−q)(ℚ^{∗})^{2} ∪ {0}, in general. As ρ is even, f(−q)(ℚ^{∗})^{2} is the same class as Φ(−q)(ℚ^{∗})^{2}. The assertion follows. □

Corollary 18. Let f ∈ ℤ[T] be a q^{2}-Weil polynomial.

i) Then, all field extension conditions for 𝔽_{q^{c}}/𝔽_{q} are satisfied if only if the condition for the quadratic extension 𝔽_{q^{2}}/𝔽_{q} does hold. ii) For extensions of odd degree, the field extension condition is always satisfied. iii) If 𝔽_{q} and 𝔽_{q^{2}} lead to different Picard ranks then all the field extension conditions are satisfied.

Remark 19. One might want to study the field extension conditions for 𝔽_{q^{ac}}/𝔽_{q^{a}}, i.e., for an extended ground field. Our calculations show that this does not lead to new conditions.

Simplification of the field extension test. Denote by ϕ_{n} the n-th cyclotomic polynomial. Correspondingly, there is the monic polynomial ψ_{n} given by ψ_{n}(T) := q^{φ(n)}ϕ_{n}(T/q). This is a q^{2}-Weil polynomial.

Lemma 20. Let n > 1 be an integer. Then,

ψ_{n}(−q) ∈ \{ \begin{matrix}(ℚ^{∗})^{2} & if n is not a power of 2 , \\ 2(ℚ^{∗})^{2} & for n = 2^{m}, m ≥ 2 , \\ \{0\} & for n = 2 . \end{matrix}

Proof. It is well known (see, e.g., [14, sec. 3]) that ϕ_{n}(−1) = 1 unless n is a power of 2. Further, the formula ϕ_{2^{e}}(t) = t^{2^{e−1}} + 1 shows ϕ_{2}(−1) = 0 and ϕ_{2^{e}}(−1) = 2 for e > 1. Observe, finally, that φ(n) is always even for n > 2. □

Remark 21. The result used here is a very special case of the value of a cyclotomic polynomial at a root of unity.

Theorem 22. Let Φ ∈ ℤ[T] be a q^{2}-Weil polynomial of even degree. Factorize Φ as

Φ(T) = (T − q)^{r}(T + q)^{s}ψ_{n_{1}}(T) · . . . · ψ_{n_{k}}(T)Φ_{1}(T)

such that Φ_{1} has no root being a root of unity multiplied by q. Denote by M the number of the powers of 2 among the n_{1}, …, n_{k}. Then,

i) if c is odd then E_{Φ}^{(c)} ∈ (ℚ^{∗})^{2} ∪ {0}. ii) If c is even and s > 0 then E_{Φ}^{(c)} = 0 for every c. iii) Finally, if c is even and s = 0 then E_{Φ}^{(c)} ∈ 2^{M}qΦ_{1}(−q)(ℚ^{∗})^{2} ∪ {0}. Furthermore, for c = 2, one actually has

E_{Φ}^{(2)} ∈ 2^{M}qΦ_{1}(−q)(ℚ^{∗})^{2} .

Proof. i) and ii) are immediate consequences from Proposition 17. For iii), observe the assumption implies that r is even. In particular, (−2q)^{r} is a perfect square. The assertion now follows from Proposition 17 together with Corollary 20. □

Remark 23. Suppose Φ ∈ ℤ[T] is a q^{2}-Weil polynomial of degree 22. In order to show that Φ may not be the characteristic polynomial of the Frobenius for a K3 surface over 𝔽_{q}, it suffices to verify that s = 0 and 2^{M}qΦ_{1}(−q) is a non-square.

Example 24. As an example, we look at K3 surfaces of Picard rank 18 such that the Picard group is defined over an extension of odd degree. Then, (−q) is not an eigenvalue of the Frobenius. The transcendental part of the characteristic polynomial is given by (T^{4} + aT^{3} + bT^{2} + aq^{2}T + q^{4}). Hence, the field extension condition usually requires that (2q^{2} − 2aq + b)q is a perfect square. If, however, the cyclotomic factors contain an odd number of type ψ_{2^{n}} then 2(2q^{2} −2aq +b)q is required to be a square.

6 The special case of a degree-2 surface – Twisting

When a K3 surface has a non-trivial automorphism, one can hope to get more conditions by inspecting the corresponding twist. This is the case for degree-2 surfaces.

The Twist. Let the K3 surface V be given by the equation

w^{2} = f_{6}(x, y, z) .

Then, for n a non-square in 𝔽_{q}, consider the twist Ṽ of V given by

nw^{2} = f_{6}(x, y, z) .

Fact 25. Assume that q, r_{2}, …, r_{22} are the eigenvalues of Frob for V. Then, the eigenvalues for Ṽ are q, −r_{2}, …, −r_{22}. Proof. For e even, V_{𝔽_{q^{e}}} and Ṽ_{𝔽_{q^{e}}} are isomorphic. When e is odd, we have

#V (𝔽_{q^{e}}) + #Ṽ(𝔽_{q^{e}}) = 2·#P^{2}(𝔽_{q^{e}}) = 2q^{2e} + 2q^{e} + 2 .

It is easy to check that the Lefschetz trace formula, applied to the eigenvalues q, −r_{2}, …, −r_{22}, implies exactly this relation. □

Proposition 26. Let V be a K3 surface of degree 2 over 𝔽_{q}. Denote by Φ the characteristic polynomial of Frob for V and by Φ̃ the corresponding polynomial for the twist Ṽ.

i) Then, Φ has a simple zero at q if and only if Φ̃ does not have a zero at (−q). I.e., the rank-1 condition can be applied to the one precisely when the field extension condition is non-empty for the other one. ii) The two conditions are equivalent to each other.

Proof. i) immediately follows from Fact 25. ii) By assumption, we can write Φ(T) = (T − q)(T + q)^{2n−1}f(T). Here both, f(q) and f(−q) are non-zero. Fact 25 shows, the corresponding polynomial for the twist is Φ̃(T) = (T − q)^{2n}f(−T). Using these two formulas, one can make the conditions explicit. The rank-1 condition for Φ simply means (2q)^{2n−1}f(q) = 2 in ℚ^{∗}/(ℚ^{∗})^{2} which is equivalent to saying that qf(q) is a perfect square. This is precisely the field extension condition for Φ̃. □

7 Examples

Let us show in detail the data for a few examples. Our goal is to illustrate how the Artin-Tate conditions work in practice.

Example 27 (A K3 surface of degree 2 over 𝔽_{7}). Consider the surface V over 𝔽_{7}, given by

w^{2} = y^{6} +3z^{6} +5xz^{5} +5x^{2}y^{4} +x^{2}z^{4} +3x^{3}y^{3} +x^{3}z^{3} +5x^{4}y^{2} +x^{4}z^{2} +5x^{5}y +2x^{6} .

Over 𝔽_{7}, …, 𝔽_{7^{9}}, there are exactly 66, 2 378, 118 113, 5 768 710, 282 535 041, 13 841 275 877, 678 223 852 225, 33 232 944 372 654, and 1 628 413 551 007 224 points. We claim that rk Pic(V_{\overline{𝔽}_{7}}) = 2.

Assuming the characteristic polynomial of the Frobenius has more than two zeroes of the form 7 times a root of unity, [6, Algorithm 22] leaves us with three candidates Φ_{1}, Φ_{2}, Φ_{3}.

Φ_{i}(t) = t^{22} − 16 t^{21} + 140 t^{20} − 1 029 t^{19} + 5 831 t^{18} − 36 015 t^{17} + 268 912 t^{16}
− 1 882 384 t^{15} + 11 529 602 t^{14} − 46 118 408 t^{13} + a_{i}t^{12} + b_{i}t^{11} + c_{i}t^{10}
+ (−1)^{j_{i}}[−110 730 297 608 t^{9} + 1 356 446 145 698 t^{8} − 10 851 569 165 584 t^{7}
+ 75 960 984 159 088 t^{6} − 498 493 958 544 015 t^{5} + 3 954 718 737 782 519 t^{4}
− 34 196 685 556 119 429 t^{3} + 227 977 903 707 462 860 t^{2}
− 1 276 676 260 761 792 016 t + 3 909 821 048 582 988 049]

for

j_{1} = 0,  (a_{1}, b_{1}, c_{1}) = (161 414 428, −1 129 900 996, 7 909 306 972) ,
j_{2} = 1,  (a_{2}, b_{2}, c_{2}) = ( 80 707 214,  0, −3 954 653 486) ,
j_{3} = 1,  (a_{3}, b_{3}, c_{3}) = (121 060 821,  0, −5 931 980 229) .

Each of the three polynomials leads to an upper bound of 4 for the rank of the geometric Picard group. All three have roots of absolute value 7, only. Applying the Artin-Tate formula, we find the following.

Table 1. Hypothetical ranks and discriminants

polynomial field arithmetic #Br(V)|Δ|

Picard rank

Φ_{1}  𝔽_{7}  2  58
𝔽_{49}  2  4524
Φ_{2}  𝔽_{7}  1  4
𝔽_{49}  2  1996
Φ_{3}  𝔽_{7}  1  6
𝔽_{49}  2  2997

The polynomial Φ_{1} is excluded by the field extension condition as the two values in the rightmost column define different square classes. On the other hand, the rank-1 condition excludes Φ_{2} and Φ_{3} since we have a degree-2 example. Thus, relative to the Tate conjecture, geometric Picard rank 2 is proven.

Example 28 (continuation). On the same surface, point counting over 𝔽_{7^{10}} leads to a number of 79 792 267 067 823 523. For the characteristic polynomial of the Frobenius, we find the two candidates Φ_{4}, Φ_{5},

Φ_{i}(t) = t^{22} − 16 t^{21} + 140 t^{20} − 1 029 t^{19} + 5 831 t^{18} − 36 015 t^{17} + 268 912 t^{16}
− 1 882 384 t^{15} + 11 529 602 t^{14} − 46 118 408 t^{13} + 40 353 607 t^{12} + a_{i}t^{11}

+ (−1)^{j_{i}}[ −1 977 326 743 t^{10} + 110 730 297 608 t^{9} − 1 356 446 145 698 t^{8} + 10 851 569 165 584 t^{7} − 75 960 984 159 088 t^{6} + 498 493 958 544 015 t^{5}

− 3 954 718 737 782 51 9t^{4} + 34 196 685 556 119 429 t^{3} − 227 977 903 707 462 860 t^{2} + 1 276 676 260 761 792 016 t

− 3 909 821 048 582 988 049]

for j_{4} = 0, a_{4} = 0, j_{5} = 1, and a_{5} = 564 950 498. Φ_{4} corresponds to the minus sign in the functional equation, Φ_{5} to the case of the plus sign. Both candidates, according to the Tate conjecture, imply geometric Picard rank 2.

To decide which sign is the right one, one would first check the absolute values of the roots. Unfortunately, both polynomials only have roots of absolute value 7. The Artin-Tate formula provides the picture given in the table below.

Table 2. Hypothetical ranks and discriminants

polynomial field arithmetic #Br(V)|Δ|

Picard rank

Φ_{4}  𝔽_{7}  1  2
𝔽_{49}  2  997
Φ_{5}  𝔽_{7}  2  55
𝔽_{49}  2  4125

Thus, Φ_{5} is excluded by the field extension condition. The minus sign in the functional equation is correct.

Example 29 (A K3 surface of degree 8 over 𝔽_{3}). Consider the complete intersection V of the three quadrics in P_{𝔽_{3}}^{5} , given by q_{1}, q_{2}, and q_{3},

q_{1} := −xy + xz + xu + xv + xw − y^{2} − yz − yv + yw

+ z^{2} + zu + zw − u^{2} − uw + v^{2} + w^{2} ,

q_{2} := −x^{2} + xy + xz − xv + xw − y^{2} + yz − yu − yv

+ yw − zu − zw + uw − v^{2} + vw ,

q_{3} := xu − yz .

V is smooth and, therefore, a K3 surface. As q_{3} is of rank 4, V carries an elliptic fibration. There are precisely 14, 98, 794, 6 710, 59 129, 532 460, 4 784 990, 43 049 510, and 387 374 024 points over 𝔽_{3}, …, 𝔽_{3^{9}}. From these data, let us check whether one can prove rk Pic(V_{\overline{𝔽}_{3}}) = 2.

Assume that the characteristic polynomial of the Frobenius has more than two zeroes of the form 3 times a root of unity. Then, [6, Algorithm 22] leaves us with five polynomials Ψ_{1}, …, Ψ_{5},

Ψ_{i}(t) = t^{22} − 4 t^{21} + 27 t^{18} + 81 t^{17} − 243 t^{16} + 6 561 t^{13} + a_{i}t^{12} + b_{i}t^{11} + c_{i}t^{10}
+ (−1)^{j_{i}}[531 441 t^{9} − 14 348 907 t^{6} + 43 046 721 t^{5} + 129 140 163 t^{4}
− 13 947 137 604 t + 31 381 059 609]

for j_{1} = 0,  (a_{1}, b_{1}, c_{1}) = (−59 049, 236 196, −531 441) ,

j_{2} = 0, (a_{2}, b_{2}, c_{2}) = ( 0, −118 098, 0) ,
j_{3} = 0,  (a_{3}, b_{3}, c_{3}) = ( 19 683, −236 196, 177 147) ,
j_{4} = 1,  (a_{4}, b_{4}, c_{4}) = (−59 049,  0, 531 441) ,
j_{5} = 1,  (a_{5}, b_{5}, c_{5}) = (−39 366,  0, 354 294) .

Applying the Artin-Tate formula to these polynomials, we obtain the following data.

Table 3. Hypothetical ranks and discriminants

polynomial field arithmetic #Br(V)|Δ|

Picard rank

Ψ_{1}  𝔽_{3}  2  24
𝔽_{9}  4  1116
Ψ_{2}  𝔽_{3}  2  27
𝔽_{9}  2  81
Ψ_{3}  𝔽_{3}  2  28
𝔽_{9}  2  112
Ψ_{4}  𝔽_{3}  3  144
𝔽_{9}  4  1152
Ψ_{5}  𝔽_{3}  1  2
𝔽_{9}  2  65

Observe that an elliptic surface of Picard rank 2 automatically has a discriminant of the form (−n^{2}) for n an integer. We may therefore exclude everything except for Ψ_{4}. Note that Ψ_{2} is, in addition, incompatible with the field extension condition.

Thus, using the numbers of points over the fields up to 𝔽_{3^{9}}, we only obtain that, either the geometric Picard rank is equal to 2, or Ψ_{4} is the characteristic polynomial of the Frobenius in which case it is 4.

Example 30 (continuation). The number of points over 𝔽_{3^{10}} is 34 871 648 631. This additional information reproduces Ψ_{1} and Ψ_{4} as possible characteristic polynomials of Frob. Consequently, the minus sign holds in the functional equation and the geometric Picard rank of V is equal to 4.

8 Statistics

We tested the Artin-Tate conditions on samples of K3 surfaces of degrees 2, 4, 6, and 8. The possibilities of computing are limited by the fact that point counting over large finite fields is slow. In degree 2, decoupling [6, Algorithm 17] (see also [5]) leads to a substantial speed-up. In higher degrees, one may focus on elliptic K3 surfaces and exploit the fact that point counting on the elliptic fibers is fast. The numbers and particularities of the examples treated are listed in Table 4.

Table 4. Numbers of examples computed

p = 2 p = 3 p = 5 p = 7
d = 2 1000 rand 1000 rand 1000 dec 1000 dec
d = 4 1000 rand 1000 ell
d = 6 1000 rand 1000 ell
d = 8 1000 rand 1000 ell

dec = decoupled, ell = elliptic, rand = random

The remaining parameters of the surfaces were chosen by a random number generator. We stored the equations and the numbers of points over 𝔽_{p}, …, 𝔽_{p^{10}} in a file.

Results I. Point counting until �_{p^{9}}. First, we tried to show that the geometric Picard-rank was equal to 2 only using the numbers of rational points over 𝔽_{p}, …, 𝔽_{p^{9}}. I.e., we applied [6, Algorithm 22]. This algorithm produces a list of hypothetical Weil polynomials for each surface. If one is able to exclude all of them then, relative to the Tate conjecture, rank 2 is proven. To exclude a particular polynomial, we first checked whether the roots are of absolute value p. When a surface was known to be elliptic over 𝔽_{p}, we checked in addition that the predicted Picard rank over 𝔽_{p} was at least equal to 2.

Then, we applied the Artin-Tate conditions to the polynomials. We checked the field extension condition and the rank-1 condition. For surfaces known to be elliptic over 𝔽_{p}, we observed the fact that arithmetic Picard rank 2 forces the discriminant to be minus a perfect square. The results are summarized in Table 5.

Table 5. Distribution of the remaining hypothetical characteristic polynomials

Number of polynomials  0  1  2  3  4  5  6
d = 2, p = 2 without  84 479 312 89 21 12  3
with A-T conditions  149 598 218 28  7  0  0
d = 2, p = 3 without  116 480 285 88 24  4  3
with A-T conditions  214 573 193 20  0  0  0
d = 2, p = 5 without  85 581 209 96 25  4  0
with A-T conditions  158 651 169 20  2  0  0
d = 2, p = 7 without  92 534 232 98 37  7  0
with A-T conditions  214 611 154 21  0  0  0
d = 4, p = 2 without  40 532 303 87 29  8  1
with A-T conditions  81 638 249 27  5  0  0
d = 4, p = 3 without  22 669 242 57  9  1  0
with A-T conditions  53 785 161  1  0  0  0
d = 6, p = 2 without  39 549 312 70 22  6  2
with A-T conditions  83 645 257 14  1  0  0
d = 6, p = 3 without  16 713 217 47  7  0  0
with A-T conditions  50 797 148  5  0  0  0
d = 8, p = 2 without  25 657 268 38  8  4  0
with A-T conditions  29 723 239  5  4  0  0
d = 8, p = 3 without  12 720 236 27  4  1  0
with A-T conditions  20 803 175  2  0  0  0

Results II. Point counting until �_{p^{10}}. Using data up to 𝔽_{p^{10}}, one obtains two hypothetical Weil polynomials for each of the surfaces. The two polynomials correspond to the possible signs in the functional equation (1). One has to exclude one of them. For this, we first checked the absolute values of the roots. For surfaces known to be elliptic over 𝔽_{p}, we then tested whether the predicted arithmetic Picard rank is at least 2. Then, we applied the Artin-Tate conditions. We checked the field extensions and the rank-1 condition. For elliptic surfaces, supposed to be of arithmetic Picard rank 2, we tested, in addition, whether the predicted discriminant was minus a square.

Table 6 shows the number of surfaces with known signs. In the case that the sign is not known, we computed the numbers of points predicted over further extensions of 𝔽_{p}. Comparing these numbers for both hypothetical polynomials indicates whether further point counting would lead to a decision of the sign. We count how often which fields had to be considered in order to decide the sign.

Table 6. Sign decision in the functional equation

p  2  3  5  7  2  3  2  3  2  3
d  2  2  2  2  4  4  6  6  8  8
Known signs without A-T  768 843 864 869 761 876 790 888 822 897
Known signs using A-T  863 940 940 961 863 943 868 933 867 944
Remaining unknown signs  137 60 60 39 137 57 132 67 133 56
Data up to 𝔽_{p11} insufficient  84 23 15 12 69 19 77 25 72 21
Data up to 𝔽_{p12} insufficient  41 11  2  1 39  3 42 11 47  7
Data up to 𝔽_{p13} insufficient  22  5  1  0 24  2 20  2 24  2
Data up to 𝔽_{p14} insufficient  13  2  0  0 12  0 13  1  8  0
Data up to 𝔽_{p15} insufficient  7  0  0  0  8  0  7  0  5  0
Data up to 𝔽_{p16} insufficient  4  0  0  0  3  0  2  0  4  0
Data up to 𝔽_{p17} insufficient  4  0  0  0  2  0  2  0  0  0
Data up to 𝔽_{p18} insufficient  4  0  0  0  0  0  1  0  0  0
Data up to 𝔽_{p19} insufficient  2  0  0  0  0  0  1  0  0  0
Data up to 𝔽_{p20} insufficient  0  0  0  0  0  0  0  0  0  0

Using these data, we repeated our attempt to prove that the geometric Picard rank is equal to 2. More precisely, we checked whether only two roots of the characteristic polynomial are of the form p times a root of unity. The numbers of surfaces for which we succeeded are listed in Table 7.

Table 7. Numbers of rank-2 cases using 𝔽_{p10} -data

rank 2 proven rank 2 possible
p = 2, d = 2 without  271  330
with A-T conditions  278  301
p = 3, d = 2 without  397  460
with A-T conditions  409  428
p = 5, d = 2 without  353  425
with A-T conditions  360  382
p = 7, d = 2 without  460  511
with A-T conditions  464  476
p = 2, d = 4 without  132  197
with A-T conditions  138  163
p = 3, d = 4 without  79  114
with A-T conditions  79  81
p = 2, d = 6 without  145  183
with A-T conditions  152  163
p = 3, d = 6 without  74  101
with A-T conditions  74  81
p = 2, d = 8 without  65  93
with A-T conditions  65  74
p = 3, d = 8 without  23  47
with A-T conditions  23  25

Conclusion. The Artin-Tate conditions usually halve the number of cases with unknown signs. Furthermore, they double the number of cases where geometric Picard rank 2 may be proven only using data up to 𝔽_{p^{9}}. Comparing Table 5 with Table 7, we see, however, that still only about one half of the cases with Picard rank 2 may be detected when counting until 𝔽_{p^{9}}.

Remark 31. Let us finally mention that the Artin-Tate conditions came to us as a big surprise. It is astonishing that the Artin-Tate formula may be incompatible with itself under field extensions. Thus, it seems not entirely unlikely that there are even more constraints and one can still do better.

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