Frobenius trace distributions for K3 surfaces
aa r X i v : . [ m a t h . AG ] F e b FROBENIUS TRACE DISTRIBUTIONS FOR K ANDREAS-STEPHAN ELSENHANS AND J ¨ORG JAHNEL
Abstract.
We study the distribution of the Frobenius traces on K K Introduction
Given a smooth, projective variety X over Q , one may choose a model X of X that is projective over Spec Z . The point counts X p p F p q , at least for the primes p of good reduction, then form a highly interesting set of quantities, related to the va-riety X . For example, for X an elliptic curve, Hasse’s bound states that a p P r´ , s ,for a p : “ p X p p F p q ´ p ´ q{ a p , and it seems natural to ask for the distribution ofthe sequence p a p q p P P in that interval. -2 -1 0 1 2 -2 -1 0 1 2 Figure 1.
Distribution for elliptic curves, general (left) and CM (right)When X does not have CM, p a p q p P P is equidistributed with respect to the mea-sure with density π ? ´ t . This was first observed experimentally by M. Sato.J. Tate gave a partial explanation based on what is now called the Tate conjec-ture [Tat, § § I.A.2, Example 3] provided a proof shortly afterwards.Based on new developments originating with A. Wiles [Wi], the result was finallyestablished unconditionally by R. Taylor et al. [CHT, HST, Tay]. The CM case issubstantially easier and was well understood already in the sixties. Here, the densityfunction is π ? ´ t . A proof may be found in [Su19, Proposition 2.16]. Date : February 14, 2021.2010
Mathematics Subject Classification.
Key words and phrases.
Sato–Tate conjecture, K ANDREAS-STEPHAN ELSENHANS AND J ¨ORG JAHNEL
For curves of higher genus, extensive experiments have been carried out by A.Sutherland. Numerical data are available from his website [Su20]. Theoreticalinvestigations concerning the genus-2 case are made in [FKRS]. An equidistributionstatement is proven in certain cases in which the Jacobian is geometrically isogenousto a direct product of elliptic curves, cf. [FS]. For an arbitrary smooth, projectivevariety X , there is the Sato–Tate conjecture, which we describe in section 2. It is notjust concerned with the traces, but predicts equidistribution of certain elements x p derived from the Frobenii within a compact Lie group, the Sato–Tate group ST i p X q .In the situation of a K p ST p X qq of the Sato–Tate group is known, due to the work of Yu. G. Zarhin [Za] and S. G. Tankeev[Tan90, Tan95]. It depends only on the geometric Picard rank, the degree of theendomorphism field E , and the bit of information whether E is totally real or aCM field. We recall the description of p ST p X qq in Corollary 4.7. Our results. i) The main goal of this article is to report on our experiments con-cerning the Sato–Tate conjecture for certain K K p X q{p ST p X qq is naturally contained in the di-hedral group of order eight. We show that the order is, in fact, at most four andindicate that two non-isomorphic subgroups of order four are possible.2. The Sato–Tate conjecture
The algebraic monodromy group.
Let X be a smooth, projective variety over Q , and S a finite set of primes, outside of which X has good reduction. Then the Lefschetztrace formula in ´etale cohomology [SGA5, Expos´e III, Th´eor`eme 6.13.3], togetherwith the smooth specialisation theorem [SGA4, Expos´e XVI, Corollaire 2.2], show X p p F p q “ X ÿ i “ p´ q i Tr ` Frob p : H i ´et p X Q , Q l q ý ˘ , (1)for p P P z S and any prime l ‰ p . Here, Frob p P Gal p Q { Q q denotes a Frobenius lift.Since Frob p is unique up to conjugation, the trace is independent of that choice.Suppose now that i is even, which is the slightly easier case and the one we studyin this article. ThenTr ` Frob p : H i ´et p X Q , Q l q ý ˘ “ p i ¨ Tr ` Frob p : H i ´et p X Q , Q l p i { qq ý ˘ . (2)The trace on the right hand side is known to be a rational number independentof l . According to the Weil conjectures, proven by P. Deligne [De74, Th´eor`eme 1.6], ROBENIUS TRACE DISTRIBUTIONS FOR K every eigenvalue of Frob p is an algebraic number, all complex (and real) embeddingsof which are of absolute value 1.Moreover, the operation of the Galois group, ̺ iX,l : Gal p Q { Q q ÝÑ GL p H i ´et p X Q , Q l p i { qqq , (3)is continuous, cf. [SGA4, Expos´e VIII, Th´eor`eme 5.2]. Its image is hence an l -adicLie group. The Zariski closure G i, Zar
X,l : “ im p ̺ iX,l q is called the algebraic monodromygroup of X (in degree i ). It is a linear algebraic group over Q l . Inclusion in the orthogonal group.
Fix a hyperplane section H Ă X . Then, byPoincar´e duality and the hard Lefschetz theorem [De80, Th´eor`eme 4.1.1], the coho-mology vector space H i ´et p X Q , Q l p i { qq is equipped with a non-degenerate, symmet-ric, bilinear pairing. For i ď dim X , this is given as follows, H i ´et p X Q , Q l p i { qq ˆ H i ´et p X Q , Q l p i { qq ÝÑ H X ´et p X Q , Q l p dim X qq – Q l , p α, β q ÞÑ x α, β y : “ α Y β Y r H s dim X ´ i . The operation of Gal p Q { Q q respects this pairing, so one actually has an inclusion G i, Zar
X,l Ď O p H i ´et p X Q , Q l p i { qqq . The Sato–Tate group.
Let us fix an embedding Q l ã Ñ C . Then G i, Zar
X,l p C q is acomplex Lie group, equipped with an inclusion ι : G i, Zar
X,l ã Ñ G i, Zar
X,l p C q . Moreover, G i, Zar
X,l p C q is contained in the matrix group O p H i ´et p X Q , Q l p i { qq b Q l C q . In particular,the elements of G i, Zar
X,l p C q have eigenvalues being complex numbers and there is thetrace map tr : G i, Zar
X,l p C q Ñ C . The maximal compact subgroup ST i p X q of G i, Zar
X,l p C q is called the Sato–Tate group of X in degree i . The Sato–Tate group is a compact Lie group, in general discon-nected. For the component group, one clearly hasST i p X q{p ST i p X qq – G i, Zar
X,l p C q{p G i, Zar
X,l p C qq – G i, Zar
X,l {p G i, Zar
X,l q . Remarks . i) The maximal compact subgroups of a Lie group with finitely manyconnected components are mutually conjugate [OV, Theorem IV.3.5]. Thus, theSato–Tate group is well-defined, up to conjugation.ii) According to the Mumford–Tate conjecture, the neutral component p G i, Zar
X,l q of the algebraic monodromy group G i, Zar
X,l coincides with Hg i p X q ˆ Spec Q Spec Q l ,for Hg i p X q the i -the Hodge group of X [Su19, Definition 3.8 and Conjecture 3.10]. Remark . One might want to work without Tate twist, as one is forced to doin the case when i is odd. The algebraic monodromy group is then only containedin GO p H i ´et p X Q , Q l qq and one would impose an orthogonal constraint, i.e. intersectwith the orthogonal group, afterwards. Cf. [Su19], in particular [Su19, Remark 3.3].Such an approach is, however, inferior to the one with Tate twist in the caseof even i , at least as far as the component groups are considered. For example,the algebraic monodromy group might be r SO p Q l qs . Then, working without Tate ANDREAS-STEPHAN ELSENHANS AND J ¨ORG JAHNEL twist, one would find, at first, G r SO p Q l qs . In the next step, however, this leadsto G r SO p Q l qs X O p Q l q “ r SO p Q l qs Y r O ´ p Q l qs , in which, all of a sudden, asecond component appears. In other words, some of the information has been lost. The Sato–Tate conjecture.
The set Cl p ST i p X qq of the conjugacy classes of elementsof ST i p X q naturally carries the quotient topology with respect the canonical map π : ST i p X q Ñ Cl p ST i p X qq . As tr | ST i p X q : ST i p X q Ñ C is a continuous class func-tion, it induces a continuous map tr : Cl p ST i p X qq Ñ C satisfying tr ˝ π “ tr | ST i p X q .One equips Cl p ST i p X qq with the measure π ˚ µ Haar , for µ Haar the normalised Haarmeasure on ST i p X q .Moreover, for an arbitrary p P P z S , one puts ξ p : “ ι p ̺ iX,l p Frob p qq P G i, Zar
X,l p C q . This element is uniquely determined, up to conjugation. Write ξ ss p for the semisimplepart of ξ p , according to the Jordan decomposition [Bo, Theorem I.4.4]. Then alleigenvalues of ξ ss p are of absolute value 1. Thus, the group x ξ ss p y Ă G i, Zar
X,l p C q has acompact closure. By [OV, Theorem IV.3.5], x ξ ss p y is, up to conjugation, containedin ST i p X q . Let, finally, x p P Cl p ST i p X qq be the conjugacy class of ξ ss p . Lemma 2.7 below shows that x p is well-defined. Conjecture 2.3 (The Sato–Tate conjecture) . Let X and i be as above. Then thesequence p x p q p P P z S , for p running through the good primes in their usual or-der, is equidistributed with respect to π ˚ µ Haar . In other words, the sequence ` t q P P z S | q ď p u ř q P P z Sq ď p δ x q ˘ p P P of measures on Cl p ST i p X qq converges weakly versus π ˚ µ Haar .Remarks . i) (Equidistribution on the component group.) In particular, theSato–Tate conjecture claims equidistribution among the components of ST i p X q .More precisely, let µ un be the uniform probability measure on the component groupST i p X q{p ST i p X qq . Moreover, let π c : ST i p X q{p ST i p X qq Ñ Cl p ST i p X q{p ST i p X qq q be the canonical map and κ c : Cl p ST i p X qq Ñ Cl p ST i p X q{p ST i p X qq q the map be-tween conjugacy classes induced by the projection κ : ST i p X q Ñ ST i p X q{p ST i p X qq .Then p κ c p x p qq p P P z S is asserted to be equidistributed with respect to p π c q ˚ µ un . Indeed, p κ c q ˚ π ˚ µ Haar “ p κ c ˝ π q ˚ µ Haar “ p π c ˝ κ q ˚ µ Haar “ p π c q ˚ κ ˚ µ Haar “ p π c q ˚ µ un . This part of the Sato–Tate conjecture is known to be true and can be shown as fol-lows. The image of ̺ iX,l : Gal p Q { Q q Ñ G i, Zar
X,l is Zariski dense, hence the inducedhomomorphism Gal p Q { Q q ÝÑ G i, Zar
X,l {p G i, Zar
X,l q – ST i p X q{p ST i p X qq is surjective. The kernel U Ď Gal p Q { Q q is an open subgroup, so corresponding underthe Galois correspondence there is a finite extension field L Ě Q . I.e., ̺ iX,l yieldsan isomorphism Gal p L { Q q – ST i p X q{p ST i p X qq . Consequently, the Chebotarevdensity theorem implies exactly what was claimed. ROBENIUS TRACE DISTRIBUTIONS FOR K ii) (The 0-dimensional case.) In particular, the Sato–Tate conjecture is trivially truewhen p ST i p X qq is the trivial group. For example, this holds for X of dimension 0and i “
0. Indeed, then H p X Q , Q l q – Q π p X Q q l and Gal p Q { Q q acts simply by per-muting the direct summands. Consequently, the algebraic monodromy group G , Zar
X,l must be finite.iii) (Modularity.) For a representation ̺ : ST i p X q Ñ GL d p C q , consider the Artintype L -function L p ̺, s q : “ ź p P P z S p ´ ̺ p x p q p ´ s q , which is clearly holomorphic for Re s ą
1. Assume that, for every irreducible, contin-uous representation ̺ ‰ i p X q , the function L p ̺, s q extends to the closed halfplane Re s ě X and i [Se68, § I.A.2, Theorem 2].iv) (Cohomology.) The group Gal p Q { Q q is compact and hence carries a normalisedHaar measure itself. The conjugacy classes of Frob p , for p running through theprimes in their usual order, are equidistributed with respect to this Haar mea-sure, according to the Chebotarev density theorem. As the representation ̺ iX,l iscontinuous, the image im ̺ iX,l is compact, and the conjugacy classes of ̺ iX,l p Frob p q are equidistributed with respect to the normalised Haar measure on that l -adicLie group. This l -adic kind of equidistribution is certainly of interest. For example,it was studied in detail for elliptic curves, by J.-P. Serre in [Se72]. However, as theembedding Q l ã Ñ C chosen is discontinuous, it does not seem to have any implica-tions towards the Sato–Tate conjecture. A cohomology theory with coefficients in C that provides a continuous Gal p Q { Q q -action would certainly help. But, of course,we have nothing of this kind at our disposal. Remark . When X is a K ξ p is known to be semisimple, for every good prime p ‰ l [De81,Corollaire 1.10]. The step of taking the semisimple part is then superfluous.The Sato–Tate conjecture immediately yields the following prediction for the dis-tribution of the Frobenius traces. Conjecture 2.6 (The Frobenius trace distribution) . Let X and i be as above.Then the sequence Tr p Frob p q p P P z S , for p running through the good primes in theirusual order, is equidistributed with respect to p tr | ST i p X q q ˚ µ Haar . In other words, thesequence ` t q P P z S | q ď p u ř q P P z Sq ď p δ Tr p Frob q q ˘ p P P of measures on R is convergent in the weak sense versus p tr | ST i p X q q ˚ µ Haar . Proof (assuming the Sato–Tate conjecture) . Taking the image measure under acontinuous map commutes with weak convergence, cf. [Di, section 13.4, probl`eme 8].
ANDREAS-STEPHAN ELSENHANS AND J ¨ORG JAHNEL
Hence, the Sato–Tate conjecture implies that t q P P z S | q ď p u ř q P P z Sq ď p δ tr p x q q “ tr ` t q P P z S | q ď p u ř q P P z Sq ď p δ x q ˘ Ñ tr p π ˚ µ Haar q“ p tr ˝ π q ˚ µ Haar “ p tr | ST i p X q q ˚ µ Haar . But tr p x q q “ tr p π p ξ q qq “ tr p ξ q q “ tr p ι p ̺ iX,l p Frob p qqq , for every prime number q . Andtr p ι p ̺ iX,l p Frob p qqq “ Tr p ̺ iX,l p Frob p qq , which is usually denoted shortly as Tr p Frob p q . (cid:3) Due to the lack of a suitable reference, we include the following purely Lie-theo-retic lemma.
Lemma 2.7.
Let G be a faithfully representable complex Lie group and K Ă G amaximal compact subgroup. Then the natural homomorphism Cl K Ñ Cl G betweenconjugacy classes of elements is injective. Proof.
Let k , k P K be two elements that are conjugate as elements of G . We haveto show that k and k are conjugate in K .According to the decomposition theorem [Le, Theorem 4.43], G – G ⋉ R is isomor-phic to a semidirect product of two closed subgroups, R being simply connected andsolvable and G being reductive. A simply connected solvable group has the trivialgroup as its maximal compact subgroup [Kn, Corollary 1.126]. Thus, the quotienthomomorphism π : G ։ G { R maps K isomorphically onto the maximal compactsubgroup of G { R . Moreover, the elements π p k q and π p k q are clearly conjugatein G { R – G .It therefore suffices to assume G as being reductive. Then G coincides with thecomplexification of K [Le, Theorem 4.31] and there is the Cartan decomposition G “ K ¨ exp p i Lie K q , cf. [Kn, Theorem 6.31.c)]. By assumption, there exist some k P K and X P Lie K such that exp p´ iX q k ´ ¨ k ¨ k exp p iX q “ k P K . As K is fixedunder the Cartan involution Θ, this yieldsexp p iX q k exp p´ iX q “ Θ p exp p´ iX q k exp p iX qq “ exp p´ iX q k exp p iX q , for k : “ k ´ k k . I.e., exp p iX q P C G p k q . Consequently, exp p niX q P C G p k q forevery n P Z , which implies exp p tiX q P C G p k q for every t P R [Kn, Lemma 1.142].In particular, exp p iX q P C G p k q showing k ´ k k “ k , as required. (cid:3) Trace distributions for compact Lie groups
Moment sequences.
Given a connected compact Lie group S Ă GL d p C q , the tracetr | S : S Ñ C is a continuous class function. Let us assume that tr | S is real-valued.Then, for the n -th momentE S r tr n s “ ż R x n d p tr | S q ˚ µ Haar p x q “ ż S tr n p s q dµ Haar p s q , for n P N , one has the Weyl integration formula [Kn, Theorem 8.60],E S r tr n s “ W ż T tr n p t q det p id ´ Ad S { T p t ´ qq dt “ W ż T tr n p t q ź α P ∆ ` | ´ t α | dt . ROBENIUS TRACE DISTRIBUTIONS FOR K Here, T denotes the maximal torus of S , W the Weyl group, ∆ ` a system of pos-itive roots, and Ad : T Ñ GL p S { T q the adjoint representation. The integrand is atrigonometric polynomial, so, for each n , the integral may be computed exactly.The particular compact Lie groups mentioned below are related to the examplesof K Examples A. S Root dim p T q dim p S { T q Moment sequence Label insystem [OEIS]SO p R q – U H p R q A p R q B p R q A A Table 1.
Moments of the trace distributions for some connected com-pact Lie groups
Remarks . i) U is reductive, but not semisimple. Thus, the root system A occurs together with a 3-dimensional torus.ii) In the SO N p R q -cases, we consider the naive trace function, tr : A ÞÑ Tr p A q . In theU N -cases, however, we put tr p A q : “ Tr ` A A ˘ “ p A q . This is compatible withthe embedding GL N p C q ã Ñ SO N p C q , A ÞÑ ` E EiE ´ iE ˘ ´ ` A p A t q ´ ˘` E EiE ´ iE ˘ , of complexLie groups, which is relevant here. Cf. Theorem 4.6.ii), below. Examples B.
For S “ S ˆ S and tr p s , s q “ tr S p s q` tr S p s q , according to Fubini,one has E S r tr n s “ n ÿ i “ ˆ ni ˙ E S r tr iS s E S r tr n ´ iS s . For two Lie groups of this kind, the moment sequences are of interest for us. S Root dim p T q dim p S { T q Moment sequencesystem r SO p R qs r A s r SO p R qs – r U s H Table 2.
Moments of the trace distributions in direct product cases
ANDREAS-STEPHAN ELSENHANS AND J ¨ORG JAHNEL
Example C.
There is a version of the Weyl integration formula for disconnectedcompact Lie groups [We, Proposition 2.3]. We made use of it in the case of O p R q .Component dim p T q dim p S { T q Moment sequence Label in [OEIS]O p R qz SO p R q Table 3.
Moments of the trace distribution for a non-neutral component
Plotting the density of the trace distribution–The moments based approach.
In orderto plot the density of a trace distribution, we used a moments based approach.We split the support interval of the density into subintervals of equal lengths. Thenwe compute a cubic spline function, using the subdivision chosen, that has the samemoments as the distribution. We work with the first 20 to 35 moments.This approach, however, does not work properly for r SO p R qs . As the density inthis case is not a C function, an approximation by cubic splines is not appropriate.The plot shows oscillations that increase when refining the subdivision. In this case,a more naive method has to be applied. The numerical integration approach.
For every continuous function g : R Ñ R , onehas p tr | S q ˚ µ Haar p g q “ ż R g p x q d p tr | S q ˚ µ Haar p x q “ ż S g p tr p s qq dµ Haar p s q ““ W ż T g p tr p t qq ź α P ∆ ` | ´ t α | dt . Thus, the density function den S of the distribution p tr | S q ˚ µ Haar can be described asfollows.For x P R , put T x : “ t t P T | tr p t q “ x u . At least for x P R z Z , this is asubmanifold of T of codimension 1. Thenden S p x q “ W ż T x ś α P ∆ ` | ´ t α | | B tr B ~ n p t q| dt , for ~ n a normal vector of length 1, as usual. Using numerical integration, we evaluatethese p dim p T q ´ q -dimensional integrals at a sufficiently high precision. Remark . Plotting the density with either approach, the results visually coincide.The only exception is the case of r SO p R qs , mentioned above.4. K surfaces Decomposition of cohomology.
Let X be a K Q . One then calls H tr : “ p H alg q K Ă H p X C , Q q the transcendental part of cohomology. Here, H alg : “ im p c : Pic X C Ñ H p X C , Q qq . ROBENIUS TRACE DISTRIBUTIONS FOR K As a quadratic space, H alg is non-degenerate. Indeed, if L P Pic X C , L ≇ O X C hadintersection number 0 with every element of Pic X C then either L or L _ wouldhave a non-trivial section [BHPV, Proposition VIII.3.7.i)]. And hence L ¨r H s ‰ H the hyperplane section, a contradiction. Consequently, H tr “ p H alg q K is non-degenerate, too. Notation . Write r : “ rk Pic X C . Then dim Q H tr “ ´ r .In l -adic cohomology, one puts H l, alg : “ im p c : Pic X Q b Q Q l Ñ H p X Q , Q l qq and H l, tr : “ p H l, alg q K . Then, under the standard comparison isomorphism [SGA4, Ex-pos´e XVI, Th´eor`eme 4.1], H l, tr – H tr b QQ l and H l, alg – H alg b QQ l . The representa-tion ̺ X,l maps H l, alg to itself and therefore H l, tr to itself, too. Thus, ̺ X,l splits intothe direct sum of the two sub-representations ̺ X,l, alg : Gal p Q { Q q Ñ O p H l, alg q and ̺ X,l, tr : Gal p Q { Q q Ñ O p H l, tr q . The image of ̺ X,l, alg is a finite group G Pic – Gal p K Pic { Q q , for K Pic the splitting fieldof Pic X Q . Definition 4.2. i) We call G , Zar
X,l, tr : “ im p ̺ X,l, tr q Ď O p T l q the transcendental part ofthe algebraic monodromy group of X .ii) Moreover, the maximal compact subgroup of G , Zar
X,l, tr p C q is called the transcenden-tal part of the Sato–Tate group of X and denoted by ST p X q . Lemma 4.3. a) For the neutral components, one has p G , Zar
X,l, tr q “ p G , Zar
X,l q , p G , Zar
X,l, tr p C qq “ p G , Zar
X,l p C qq and p ST p X qq “ p ST p X qq . b) Concerning the component groups, G , Zar
X,l, tr {p G , Zar
X,l q “ G , Zar
X,l, tr p C q{p G , Zar
X,l p C qq – ST p X q{p ST p X qq . Proof. a) The first equality is a direct consequence of the fact that G Pic is finite.The second one follows immediately from the first, and, finally, the third is obtainedtaking the maximal compact subgroup on either side.b) follows from the standard facts that the maximal compact subgroup of a Liegroup meets every connected component, and that the maximal compact subgroupof a connected Lie group is connected. (cid:3)
Lemma 4.4.
The homomorphism G , Zar
X,l {p G , Zar
X,l q ÝÑ G Pic ‘ G , Zar
X,l, tr {p G , Zar
X,l q induced by the decomposition, is a subdirect product. I.e., it is an injection, but theprojections to either summand are surjective. Proof.
One has that G , Zar
X,l is the image of ̺ X,l “ ̺ X,l, alg ‘ ̺ X,l, tr , while G Pic and G , Zar
X,l, tr are the images of the direct summands. Therefore, the decomposition inducesa homomorphism G , Zar
X,l Ñ G Pic ‘ G , Zar
X,l, tr that is a subdirect product. The assertionfollows immediately from this. (cid:3) In certain cases, the trace of the Frobenius on ST p X q is related to the pointcount on a singular model of the K X . Lemma 4.5.
Let X be a K surface over Q and pr : X Ñ X a birational mor-phism. Write r : “ rk Pic X Q and let r be the number of p´ q -curves blown downunder pr Q . Suppose that these generate Pic X Q , together with p r ´ r q further lin-early independent classes that are defined over Q .Then, for every prime p of good reduction, X p p F p q “ ` p p r ´ r q ` p ¨ Tr p ̺ X,l, tr p Frob p qq ` p . Proof.
For a K H p X Q , Q l q “ H p X Q , Q l q “
0, while H p X Q , Q l q and H p X Q , Q l q are one-dimensional. Hence, in view of (2), formula (1) specialises to X p p F p q “ ` p ¨ Tr p ̺ X,l p Frob p qq ` p “ ` p ¨ Tr p ̺ X,l, alg p Frob p qq ` p ¨ Tr p ̺ X,l, tr p Frob p qq ` p . Thus, one has to show that X p p F p q ´ X p p F p q ` p p r ´ r q “ p ¨ Tr p ̺ X,l, alg p Frob p qq .For this, let us consider the r points blown up. These are permuted by Frob p and the difference X p p F p q ´ X p p F p q may be written as p times the number offixed points. Which is the same as p ¨ Tr p ̺ bl p Frob p qq , for ̺ bl the correspondingpermutation representation, a sub-representation of ̺ X,l, alg . As the complementof ̺ bl is, by assumption, trivial of rank p r ´ r q , the claim follows. (cid:3) The neutral component of the Sato–Tate group.
The transcendental part of coho-mology is a weight-2 Hodge structure H tr Ă H p X C , Q q , the endomorphism ring E of which is always a field. There are exactly three possibilities [Za, Theorem 1.6.a)].i) One has E “ Q . This is the generic case.ii) E % Q is a totally real field. Then X is said to have real multiplication (RM) .Put δ : “ r E : Q s and let e P E be a primitive element. It is known that e acts on H tr as a self-adjoint linear map [Za, Theorem 1.5.1]. Thus, H tr b Q C “ H ‘ ¨ ¨ ¨ ‘ H δ splits into eigenspaces that are mutually perpendicular. The eigenspaces H i , for i “ , . . . , δ , are not defined over Q , but form a single orbit under conjugationby Gal p Q { Q q . In particular, dim Q H “ ¨ ¨ ¨ “ dim Q H δ “ ´ rδ . Let us note, inaddition, that each H i is a simultaneous eigenspace for all elements of E .iii) E is a CM field. Then X is said to have complex multiplication (CM) .Write E “ E p?´ τ q , for E Ă E the maximal totally real subfield and τ P E atotally positive element. We put δ : “ r E : Q s . Then, as above, the action of E splits H tr b Q C “ H ‘ ¨ ¨ ¨ ‘ H δ into simultaneous eigenspaces, which are mutuallyperpendicular.Under the action I of ?´ τ P E , each H i “ H i, ` ‘ H i, ´ , for i “ , . . . , δ , is split intotwo eigenspaces. For v and w in the same eigenspace, one has [Za, Theorem 1.5.1] ´ τ ¨x v, w y “ x Iv, Iw y “ x v, ´ I w y “ τ ¨x v, w y , which yields that the eigenspaces H i, ` and H i, ´ are both isotropic. ROBENIUS TRACE DISTRIBUTIONS FOR K Theorem 4.6 (Zarhin, Tankeev) . Let X be a K surface over Q , H tr Ă H p X C , Q q the transcendental part of cohomology, and E its endomorphism field. i) If E is totally real of degree δ then p G , Zar
X,l p C qq – SO p H q ˆ ¨ ¨ ¨ ˆ SO p H δ q – r SO ´ rδ p C qs δ . For δ “ , this includes the generic case E “ Q . ii) If E is a CM field of degree δ then p G , Zar
X,l p C qq – O p H q p H , ` ,H , ´ q ˆ ¨ ¨ ¨ ˆ O p H δ q p H δ, ` ,H δ, ´ q – r GL ´ r δ p C qs δ . Proof.
Due to the work of S. G. Tankeev [Tan90, Tan95], together with [Za, Theo-rem 2.2.1], one has p G , Zar
X,l p C qq – p C E p O p H tr b Q C qqq . Since a linear map commutes with the action of E if and only if it maps each ofthe eigenspaces H , . . . , H δ , or H , ` , H , ´ , . . . , H δ, ` , H δ, ´ , respectively, to itself, allassertions follow, except for the final isomorphism claimed in part ii).For this, note that, for every i , the subspaces H i, ` and H i, ´ are both isotropic,while H i “ H i, ` ‘ H i, ´ is non-degenerate. Thus, the cup product pairing identifies H i, ´ with the dual H _ i, ` . But then, for an arbitrary element g P GL p H i, ` q , the map p g, p g _ q ´ q P GL p H i, ` q ˆ GL p H _ i, ` q Ă GL p H i q is orthogonal, and there is no otherchoice for the second component that would lead to this property. (cid:3) Corollary 4.7.
Let X be a K surface over Q , H Ă H p X C , Q q the transcendentalpart of cohomology, and E its endomorphism field. i) If E is totally real of degree δ then p ST p X qq – r SO ´ rδ p R qs δ . For δ “ , thisincludes the generic case E “ Q . ii) If E is a CM field of degree δ then p ST p X qq – r U ´ r δ s δ . Proof.
The maximal compact subgroup of SO n p C q is SO n p R q and that of GL n p C q is U n , cf. [Kn, Table (1.144)]. (cid:3) Upper estimates for the component group.
Lemma 4.8.
Let X be a K surface over Q , H tr Ă H p X C , Q q the transcendentalpart of cohomology, and E its endomorphism field. i) If E is totally real then N O p H tr b QC q pp G , Zar
X,l p C qq q “ r O p H q ˆ ¨ ¨ ¨ ˆ O p H δ qs ⋊ S δ ,the group S δ permuting the δ direct factors. ii) If E is a CM field then N O p H tr b QC q pp G , Zar
X,l p C qq q “ r O p H q p H , ` ,H , ´ q ˆ ¨ ¨ ¨ ˆ O p H δ q p H δ, ` ,H δ, ´ q s ⋊ p Z { Z q δ ⋊ S δ . Here, e i P p Z { Z q δ interchanges H i, ` with H i, ´ , while S δ permutes the δ direct fac-tors. Proof. i) “ Ě ” is clear.“ Ď ”: The natural action of the group SO p H q ˆ ¨ ¨ ¨ ˆ SO p H δ q on H tr b Q C setwisestabilises the subvector spaces H , . . . , H δ and no others of dimension ´ rδ . Hence, alinear map normalising SO p H q ˆ ¨ ¨ ¨ ˆ SO p H δ q must permute H , . . . , H δ . As everyorthogonal map sending H , . . . , H δ to themselves lies in O p H q ˆ ¨ ¨ ¨ ˆ O p H δ q , theassertion is proven.ii) Again, “ Ě ” is clear.“ Ď ”: Here, the group O p H q p H , ` ,H , ´ q ˆ¨ ¨ ¨ˆ O p H δ q p H δ, ` ,H δ, ´ q stabilises the subvectorspaces H , ` , H , ´ , H , ` , . . . , H δ, ´ and no others of dimension ´ r δ . Thus, a linearmap normalising O p H q p H , ` ,H , ´ q ˆ ¨ ¨ ¨ ˆ O p H δ q p H δ, ` ,H δ, ´ q must permute the spaces H , ` , H , ´ , H , ` , . . . , H δ, ´ . Furthermore, for i “ , . . . , δ , the space H i, ` is perpen-dicular to both, H j, ` and H j, ´ , for j ‰ i , but it is not perpendicular to H i, ´ . Thus,the sets t H i, ` , H i, ´ u form a block system. The proof is therefore complete. (cid:3) By construction, one has G , Zar
X,l, tr p C q Ď O p H tr b QC q . Moreover, in every Lie group,the neutral component is a normal subgroup. Thus, in the RM as well as the CMcases, for the component group, one finds an inclusion i X,l : C tr : “ ST p X q{p ST p X qq – G , Zar
X,l, tr p C q{p G , Zar
X,l p C qq “ G , Zar
X,l, tr {p G , Zar
X,l q (4) ã Ñ N O p H tr b QC q pp G , Zar
X,l p C qq q{p G , Zar
X,l p C qq – p Z { Z q δ ⋊ S δ . For δ “ i X,l p C tr q Ă p Z { Z q δ ⋊ S δ is always a proper subgroup, as the nextTheorem shows. Remark . We do not discuss in this article the question whether the homomor-phism i X,l is independent of l . The component group C tr itself certainly is, as followsfrom [Se81, p. 16, Th´eor`eme], cf. [Se12, § Theorem 4.10.
Let X be a K surface over Q . Write E for the endomorphismfield of H tr Ă H p X C , Q q and let E Ď E be the maximal totally real subfield.Suppose that E { Q is Galois and that its Galois group is cyclic. Then, for anyprime l that is totally inert in E , the following statements hold. a) The image of π X,l : C tr Ñ S δ is a permutation group that is regular on any ofits orbits. In other words, only the identity element has a fixed point. b) Moreover, the kernel of π X,l is either trivial or of order , generated by the centralelement p´ , . . . , ´ q P p Z { Z q δ . Proof. a) Suppose, to the contrary, that there is an element A P G , Zar
X,l, tr that fixes aneigenspace H i , but does not fix another, H j . We know that H i and H j are conjugateunder Gal p E { Q q . As l is totally inert, this means Frob kl p H i q “ H j , for a certain k P N . Since A is Q l -linear, this yields H j “ Frob kl p H i q “ Frob kl p A p H i qq “ A p Frob kl p H i qq “ A p H j q , a contradiction. ROBENIUS TRACE DISTRIBUTIONS FOR K b) The kernel of π consists of the elements stabilising each of the H i , for i “ , . . . , δ .In the CM case, the counter assumption is that some element in G , Zar
X,l, tr fixes theeigenspaces H i, ` and H i, ´ , for some i , but interchanges H j, ` and H j, ´ , for a cer-tain j ‰ i . This is contradictory for exactly the same reason as in the proof of a).In the RM case, the counter assumption is that there is some element A P G , Zar
X,l, tr being contained in O p H q ˆ ¨ ¨ ¨ ˆ O p H δ q and having determinant 1 on some H i , butdeterminant p´ q on another, H j . Again, this is contradictory, as there is some k P N of the kind that Frob kl p H i q “ H j .Indeed, one has the Q l -linear map Ź ´ rδ A : Ź ´ rδ p H tr b Q Q l q Ñ Ź ´ rδ p H tr b Q Q l q , induced by A : H tr b Q Q l Ñ H tr b Q Q l . The base extension to Q l contains theone-dimensional subspaces Ź ´ rδ H i , on which Ź ´ rδ A acts as the identity, and Ź ´ rδ H j , on which it acts as the multiplication by p´ q . The eigenspaces for theeigenvalues p` q and p´ q are, however, Q l -subvector spaces of Ź ´ rδ p H tr b Q Q l q ,so that Frob kl p Ź ´ rδ H i q “ Ź ´ rδ H j is impossible. (cid:3) Examples . i) For δ “ p Z { Z q ⋊ S is the dihedral group of order eight.Part ii) of the Theorem forbids exactly two of its elements. In a somewhat symbolicnotation, these are ` ´ ` ˘ and ` ` ´ ˘ . Thus, two of the three conjugacy classes ofsubgroups of order four are still allowed, the cyclic subgroup being one of them.ii) For δ an odd prime, Theorem 4.10 implies that the component group is alwayscyclic of an order dividing 2 δ .5. Experimental results
The approach in general.
According to the Sato–Tate conjecture, one can use thetheory of Lie groups in order to make a prediction on the distribution of the Frobe-nius traces. We tested this in the situation of K p -adic point count-ing algorithm [Ha] in order to determine the number of F p -rational points on thereduction mod p , for all primes p up to 10 . We implemented the moving simplexidea [Ha, § r´ , s for the trace into300 classes of equal widths and counted the number of hits for each class. Represent-ing the numbers of hits as columns, we then plotted the corresponding histogram. Running times.
It took around eight hours per surface on one core of an Intel(R)Core(TM)i7-7700 CPU processor running at 3 . w “ xyzf p x, y, z q . For Example 5.8, which is ofthe slightly more general shape w “ xyf p x, y, z q , it took 58 hours.Note that the main step in the algorithm is to compute a small number of co-efficients in huge powers of f p x, y, z q . When working with a form of a particular shape as above, only the powers of a cubic, respectively quartic, form have to beconsidered, which leads to a massive reduction of the resulting computation. Remark . For two of the seven K Constraints concerning the endomorphism field.
Lemma 5.2.
Let X be a K surface over C . a) Suppose that rk Pic X “ . Then the endomorphism field is E “ Q . b) Suppose that rk Pic X “ . Then the endomorphism field is either Q , or aquadratic number field, or a CM field of degree six. Proof.
One has dim H tr “ ´ rk Pic X . Furthermore, r E : Q s | dim H tr , as H tr carries the structure of an E -vector space.a) Then r E : Q s “ r E : Q s “ E is not a CM field, since 5 is odd.Moreover, in the RM case, one has dim H tr r E : Q s ě E “ Q .b) Then r E : Q s “
1, 2, 3, or 6. The assumption r E : Q s “ r E : Q s “ r E : Q s “ E cannot be totally real, since dim H tr r E : Q s “ ă (cid:3) Lemma 5.3.
Let X be a K surface over C . Suppose that X has CM by a quadraticfield Q p?´ δ q , for δ P N . a) If dim H tr ” p mod 4 q then, for the discriminant [Se70, Chapitre IV, § , onehas disc p H tr q “ δ P Q ˚ { Q ˚ . b) If dim H tr ” p mod 4 q then disc p H tr q “ P Q ˚ { Q ˚ . Proof.
Take an anisotopic vector v P H tr and let I : H tr Ñ H tr be the endomor-phism corresponding to ?´ δ . Then p Iv, v q “ p v, ´ Iv q “ ´p Iv, v q , i.e. p Iv, v q “ p Iv, Iv q “ p v, ´ I v q “ p v, δv q “ δ p v, v q .Thus, the two-dimensional I -invariant quadratic subspace x v, Iv y is of discrim-inant ` det ` p v,v q δ p v,v q ˘ mod Q ˚ ˘ “ δ P Q ˚ { Q ˚ . The assertion follows inductivelyfrom this. (cid:3) The surfaces inspected.
Each of the seven surfaces inspected is represented by asingular degree 2 model of the shape X i : w “ f i p x, y, z q , where f i , for i “ , . . . ,
7, is a ternary sextic form over Q . In all cases, the ramifica-tion curve V p f i q Ă P Q has only ordinary double points. Thus, blowing up each ofthem once yields a K X i , to which Lemma 4.5 applies.Moreover, if there are N singular points then the exceptional curves E , . . . , E N to-gether with the pull-back of a general line in P Q generate a subgroup of rank p N ` q ROBENIUS TRACE DISTRIBUTIONS FOR K in Pic X i, Q . If, in particular, V p f i q geometrically splits into a union of six lines thenrk Pic X i, Q ě
16. If rk Pic X i, Q “
16 holds exactly thendisc Pic X i, Q “ p det diag p , ´ , . . . , ´ q mod Q ˚ q “ ´ P Q ˚ { Q ˚ and hence disc H tr “ P Q ˚ { Q ˚ . Thus, Q p?´ q is the only imaginary quadraticfield that is possible for CM.We list the bad primes as well as the jump character for each of the seven samplesurfaces in a table at the very end of this article. By bad primes, those of the obviousmodel over Z are meant, which is constructed from the double cover of P Z , definedby the equation f i “
0, by the blow-ups centred in the finitely many singular pointsof the generic fibre. The jump characters are obtained using [CEJ, Algorithm 2.6.1].Note that, in each case, not only the geometric Picard rank is known, but the geo-metric Picard group as a Gal p Q { Q q -module (in Example 5.10, at least conjectural). A generic example of Picard rank 16.Example . Let X be the double cover of P Q , given by w “ xyz p x ` y ` z qp x ` y ` z qp´ x ` y ´ z q and X the K X .a) Then the geometric Picard rank of X is 16.b) The endomorphism field of X is E “ Q . Proof. a) One has a lower bound of 16, as the ramification locus has 15 singu-lar points. An upper bound of 16 is provided by the reduction modulo 31, which isof geometric Picard rank 16.b) The reduction modulo 17 is of geometric Picard rank 18, which, by [EJ20a,Lemma 6.2] implies that r E : Q s ď
2. Furthermore, RM is excluded, since there isa reduction of geometric Picard rank 16 [EJ14, Corollary 4.12]. Finally, if E were aCM field then, by Lemma 5.3, the only option would be E “ Q p?´ q .In that case, one would have a decomposition H tr b Q C “ H ` ‘ H ´ , the sum-mands being eigenspaces for the eigenvalues ˘?´
1, and hence defined over Q p?´ q .By Lemma 4.8, the algebraic monodromy group G , Zar
X,l, tr has at most two components.The non-neutral component, if present, interchanges the eigenspaces and hence allelements are of trace zero. The neutral component stabilises H ` and H ´ and hence,the characteristic polynomial of every element factors over Q l p?´ q into two cu-bic polynomials. However, the characteristic polynomial of ̺ X, , tr p Frob q P G , Zar X, , tr is t ´ t ` t ` t ` t ´ t `
1, which splits over Q p?´ q “ Q into ir-reducible factors of degrees two and four. Moreover, the trace of ̺ X, , tr p Frob q isnonzero, a contradiction. (cid:3) In view of the results above, Corollary 4.7 shows that p ST p X qq – SO p R q .Moreover, ST p X q{p ST p X qq “ Z { Z , i.e. ST p X q – O p R q . Indeed, for thecomponent group, we have an upper bound of Z { Z by (4), and the trivial group isexcluded, due to the nontrivial jump character, cf. Table 4. An example of Picard rank 16 with trivial jump character.Example . Let X be the double cover of P Q , given by w “ xyz p x ` y ´ z qp x ´ y ´ z qp x ` y ` z q and X the K X .a) Then the geometric Picard rank of X is 16.b) The endomorphism field of X is E “ Q . Proof. a) One has a lower bound of 16, as the ramification locus has 15 singu-lar points. An upper bound of 16 is provided by the reduction modulo 19, which isof geometric Picard rank 16.b) The reduction modulo 13 is of geometric Picard rank 18, which, as in Exam-ple 5.4, leaves E “ Q p?´ q as the only nontrivial option. Moreover, this is ex-cluded by observing that the characteristic polynomial of ̺ X , , tr p Frob q P G , Zar X , , tr is t ` t ` t ´ t ` t ` t `
1, which splits over Q p?´ q “ Q intoirreducible factors of degrees two and four. Note that the trace is nonzero. (cid:3) Here, one has ST p X q – SO p R q . Indeed, as above, p ST p X qq – SO p R q , andthe component group is trivial, due to the trivial jump character, cf. Table 4. An example of Picard rank 17 with trivial jump character.Example . Let X be the double cover of P Q , given by w “ xyz p x ` y ` z qp´ x ´ y ´ z qp x ` y ` z q and X the K X .a) Then the geometric Picard rank of X is 17.b) The endomorphism field of X is E “ Q . Proof. a) The 16 elements π ˚ r l s , r E s , . . . , r E s P Pic X , Q are linearly independent,as before. Moreover, the inverse image of the conic C : “ V p xy ` yz ` zw q Ă P in X splits into two curves, C and C , as a Gr¨obner base calculation shows.We claim that r C s is independent of the 16 elements above. Indeed, otherwise r C s would be invariant under the involution of the double cover π . Since one has r C s`r C s “ π ˚ r l s and C is interchanged with C under the involution, this implies r C s “ π ˚ r l s P Pic p X q Q . But C is rational, and hence a p´ q -curve, while π ˚ r l s hasself-intersection number p` q , a contradiction. Thus, there is a lower bound of 17.Concerning the upper bound, the reductions modulo 13 and 23 are both of geo-metric Picard rank 18. The characteristic polynomials of the Frobenii are p t ´ q p t ` t ` t ` t ` q and p t ´ q p t ` t ` t ` t ` q , so that the Artin–Tate formula [Mi, Theorem 6.1] determines the discriminants ofthe four-dimensional lattices to 6 and 10 P Q ˚ { Q ˚ , respectively. I.e., the latticesare incompatible and van Luijk’s method [vL] lets the upper bound drop to 17.b) follows immediately from a), in view of Lemma 5.2.a). (cid:3) ROBENIUS TRACE DISTRIBUTIONS FOR K Corollary 4.7 shows that p ST p X qq – SO p R q . This is the only component, asthe jump character is trivial, cf. Table 4. The generic trace distributions. -6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6
Figure 2.
Trace distributions for Examples 5.4, 5.5, and 5.6The red lines in Figure 2 show the densities of theoretical trace distributions, aspredicted by the Sato–Tate conjecture. For example 5.4, it is the superposition ofthe distributions for the two components, as explained in Section 3.
An example with CM by Q p?´ q .Example . Let X be the double cover of P Q , given by w “ xyz p x ` y ` z qp x ` y ` z qp x ` y ` z q and X the K X .a) Then the geometric Picard rank of X is 16.b) The endomorphism field of X is E “ Q p?´ q . Proof. a) One has a lower bound of 16, as the ramification locus has 15 singu-lar points. An upper bound of 16 is provided by the reduction modulo 13, which isof geometric Picard rank 16.b) The reduction modulo 11 is of geometric Picard rank 18, which, as in Example 5.4,leaves E “ Q p?´ q as the only nontrivial option. The result follows from Theo-rem A.1 below, as the endomorphism does not shrink under specialisation [EJ20a,Corollary 4.6]. (cid:3) Here, Corollary 4.7 yields p ST p X qq – U . Moreover, for the component group,one has ST p X q{p ST p X qq “ Z { Z . Indeed, (4) gives an upper bound of Z { Z ,and there must be a second component due to the nontrivial jump character, cf.Table 4. -6 -4 -2 0 2 4 6 Figure 3.
Trace distribution for Example 5.7
In the figure above, the spike is of mass 1 { An example with RM and a cyclic component group of order four.Example . Let X be the double cover of P Q , given by w “ xy p x ´ x y ´ x z ` x y ` x yz ` x z ´ xy ´ xy z ´ xyz ´ xz ` y ` y z ` y z ` yz ` z q . and X the K X .a) Then the geometric Picard rank of X is 16.b) The endomorphism field of X is E “ Q p? q . Proof. a) One has a lower bound of 16, as the ramification locus has 15 singu-lar points. As far as upper bounds are concerned, the reductions modulo 7 and 19are both of geometric Picard rank 18. The characteristic polynomials of the Frobeniiare p t ´ q p t ` q p t ` q p t ´ t ` q and p t ´ q p t ` q p t ´ t ` t ´ t ` q , so that the Artin–Tate formula [Mi, Theorem 6.1] determines the discriminants ofthe four-dimensional lattices to 1 and 5 P Q ˚ { Q ˚ , respectively. I.e., the lattices areincompatible and van Luijk’s method [vL] lets the upper bound drop to 17.On the other hand, the endomorphism field of X contains Q p? q , which excludesthe option of rank 17. Indeed, X is isomorphic to the specialisation to t “ P Q , given by the matrix ¨˝ ˛‚ . b) In view of a), this follows from [EJ20a, Example 1.5.iv)]. (cid:3) Corollary 4.7 shows that p ST p X qq – r SO p R qs . Theorem 5.9.
The representation ̺ X ,l, tr induces an isomorphism ̺ : Gal p Q p ζ q{ Q q ÝÑ ST p X q{p ST p X qq . Proof.
First step.
Generalities.The jump character is trivial, so ST p X q{p ST p X qq is bound to the cosets ` ` ` ˘ , ` ´ ´ ˘ , ` `´ ˘ , and ` ´` ˘ . Quite generally, there exists a unique number field L , forwhich ̺ X ,l, tr induces an isomorphism ̺ : Gal p L { Q q – ÝÑ ST p X q{p ST p X qq , cf.Remark 2.4.i). In our situation, we find that L is cyclic of a degree dividing four. Second step. L % Q p? q .According to Chebotarev, the elements τ ´ Frob p τ , for τ P Gal p Q { Q q and p ” , p mod 5 q , are dense in the nontrivial coset C : “ Gal p Q { Q qz U of the open subgroup U : “ Gal p Q { Q p? qq Ă Gal p Q { Q q of index two. Moreover, [EJ20a, Lemma 6.7] ROBENIUS TRACE DISTRIBUTIONS FOR K shows together with Lemma 4.5 that Tr p ̺ X ,l, tr p Frob p qq “
0, for every prime p ” , p mod 5 q . Consequently, Tr p ̺ X ,l, tr p σ qq “ , (5)for every σ P C .Since, U Ă Gal p Q { Q q is a subgroup of finite index, ̺ X ,l, tr p U q Ď G , Zar X ,l, tr has thesame neutral component, only the component group may differ. Moreover, dueto (5), ̺ X ,l, tr p C q Ď G , Zar X ,l, tr is certainly a nontrivial coset. In particular, ̺ X ,l, tr p U q must be a proper subgroup of G , Zar X ,l, tr , which yields that L Ě Q p? q .Furthermore, (5) shows that the coset ̺ X ,l, tr p C q consists only of components oftype ` `´ ˘ and ` ´` ˘ . In particular, ST p X q{p ST p X qq is indeed of order four. Third step.
Conclusion.A standard argument involving the smooth specialisation theorem for ´etale cohomol-ogy groups [SGA4, Exp. XVI, Corollaire 2.2] shows that L is unramified at everyprime p ‰
2, 5, and l , cf. [CEJ, Lemma 2.2.3.a)]. The field L is, moreover, knownto be independent of l [Se81, p. 16, Th´eor`eme], cf. [Se12, § l “ L may ramify only at 2 and 5.Besides Q p ζ q , there are only three cyclic number fields of degree four that areunramified outside 2 and 5 and contain Q p? q . These are the quadratic twists of Q p ζ q by Q p? δ q , for δ “ ´
1, 2, and p´ q . I.e., the unique further cyclic subfieldof degree four in Q p ζ , ? δ q . Indeed, let L be such a field. Then, since ? P L and ? P Q p ζ q , L p ζ q has Galois group Z { Z ˆ Z { Z . Thus, L p ζ q “ Q p ζ , ? δ q , forsome δ P Z . The claim follows, as L p ζ q is unramified outside 2 and 5.Suppose that L is the quadratic twist of Q p ζ q by Q p? δ q , for δ “ ´
1, 2, or p´ q .Then Frob p P Gal p L { Q q is not the neutral element for p “
11 in the first two cases,and for p “
31 in the third. However, an experiment shows that ̺ X ,l, tr p Frob q and ̺ X ,l, tr p Frob q are contained in the neutral component, which completes the proof. (cid:3) An example with RM and the Klein four group as the component group.Example . Let X be the double cover of P Q , given by w “ xyz p x ´ x z ` xy ´ xz ` y ´ y z ´ yz ` z q and X the K X .a) Then the geometric Picard rank of X is 16 or 17.b) The endomorphism field of X is at most quadratic. Proof. a) Concerning the upper bound, the reductions modulo 19 and 59 are bothof geometric Picard rank 18. The characteristic polynomials of the Frobenii are p t ´ q p t ` q p t ` t ` q and p t ´ q p t ` q p t ´ t ` t ´ t ` q , so that the Artin–Tate formula [Mi, Theorem 6.1] determines the discriminants ofthe four-dimensional lattices to 1 and 6 P Q ˚ { Q ˚ , respectively. I.e., the lattices areincompatible and van Luijk’s method [vL] lets the upper bound drop to 17. On the other hand, for the lower bound, the situation is analogous to [CEJ,Example 2.7.3]. One immediately has a lower bound of 13, as the ramificationlocus has twelve singular points. Among them, ten are Q -rational, the two othersare defined over Q p?´ q , and conjugate to each other. Moreover, there are a Q -rational line, the inverse image of which splits over Q p? q , and two conics thatare defined over Q p?´ q and conjugate to each other, the inverse images of whichsplit over Q p?´ , a ?´ q . Thus, there is a sublattice P Ď Pic X Q of rank 16,such that P b Z C “ χ ‘ χ Q p?´ q ‘ χ Q p? q ‘ χ Q p? q ‘ χ Q p?´ q (cf. [CEJ] fornotation).b) In the case of geometric Picard rank 17, this assertion follows from Lemma 5.2.a).Otherwise, as there are reductions of rank 18, it is a consequence of [EJ20a, Lemma6.2]. (cid:3) There is strong evidence that the endomorphism field of X is in fact E “ Q p? q .Then the geometric Picard rank is 16. The evidence has been described in [EJ16,Section 5]. Note that X “ V p q , in the notation of [EJ16, Conjectures 5.2]. Thus,conjecturally, p ST p X qq – r SO p R qs .The observation that Tr p ̺ X ,l, tr p Frob p qq “ p “ ˘ p mod 12 q hasmeanwhile been extended to p ă . As these are exactly the primes, at whichthe jump character evaluates to p´ q , the component group ST p X q{p ST p X qq is bound to the elements written symbolically as ` ` ` ˘ , ` ´ ´ ˘ , ` `` ˘ , and ` ´´ ˘ .The component ` ´ ´ ˘ is indeed met, thus the component group is isomorphic tothe Klein four group. According to our experiments, x p P r O ´ p R qs if and only if p ” ˘ p mod 12 q and p ´ ¨ ¨ p q “ ´ The trace distributions in the RM examples. -6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6
Figure 4.
Trace distributions for Examples 5.8 and 5.10In the figures above, the spike is of mass 1 {
2. As before, the red line shows thedensity of theoretical trace distribution. It is the superposition of the distributionsfor the two components r SO p R qs and r O ´ p R qs . The first is obtained as explainedin Section 3, the second is constructed by mirroring on the y -axis. An example with CM by an endomorphism group of order six.Example . Let X be the double cover of P Q , given by w “ xyz p x ´ x z ´ xy ´ xyz ` y ` y z ` yz ` z q and X the K X . ROBENIUS TRACE DISTRIBUTIONS FOR K a) Then the geometric Picard rank of X is 16.b) The endomorphism field of X contains Q p?´ q . Proof. a) One has a lower bound of 16, as the ramification locus has 15 singu-lar points. An upper bound of 16 is provided by the reduction modulo 5, which isof geometric Picard rank 16.b) The automorphism of P Q , given by the matrix ¨˝ ´ g ` g ` g ´ g ´ g ˛‚ , for g : “ ζ ` ζ ´ ´
1, transforms X into a fibre of the family q : X Ñ B , consideredin Theorem A.1. The assertion follows, as the endomorphism field does not shrinkunder specialisation. (cid:3) There is strong evidence that X has complex multiplication by the endomorphismfield E “ Q p ζ ` ζ ´ , ?´ q , which is abelian of degree six. The evidence has beendescribed in [EJ16, last subsection]. Note that X “ V p´ ,µ q in the notation of[EJ16, Conjectures 5.2]. Thus, conjecturally, p ST p X qq – r U s .The observation that Tr p ̺ X ,l, tr p Frob p qq “ p ı ˘ p mod 36 q hasmeanwhile been extended to p ă . If one knew this unconditionally then Exam-ple 4.11.b) would show that ST p X q{p ST p X qq – Z { Z . Note that the maximaltotally real subfield E “ Q p ζ ` ζ ´ q of the conjectural endomorphism field is cyclicof degree 3. -6 -4 -2 0 2 4 6 Figure 5.
Trace distribution for Example 5.11In the figure above, the spike is of mass 5 { Conclusion.
For each of the seven K Appendix A. A family that is acted upon by Q p?´ q Theorem A.1.
Let B Ă P Q ˆ P Q be the closed subscheme given by the equations a b ` a b ´ a b “ and a b ` a b ´ a b “ , and let, moreover, q : X Ñ B be the family of double covers of P given by w “ l ¨ ¨ ¨ l , for l , . . . , l the linear forms l : “ x , l : “ y , l : “ z , l : “ x ` y ` z , l : “ a x ` a y ` a z , and l : “ b x ` b y ` b z . a) Then the generic fibre X η is normal surface, the minimal desingularisation ofwhich is a K surface X η of geometric Picard rank . b) The endomorphism field of X η is Q p?´ q . Proof. a) The singularities of X η are caused by those of the ramification curve,and are therefore isolated and of type A . The surface X η is a K X , cf. Example 5.7.c).b) The specialisation to X is known to have an endomorphism field of degree ď X η contains Q p?´ q .For this, we blow up P k p B q in the seven points V p l i , l j q , for t i, j u “ t , u , t , u , t , u , t , u , t , u , t , u , and t , u . Since no four of these points are collinear, theresult is a weak del Pezzo surface S of degree 2 [Do, Corollary 8.1.24]. The linearsystem of the cubic forms vanishing in the seven blown-up points defines a birationalmorphism S Ñ S to a singular model S : W “ Q p X, Y, Z q . Here, Q defines a planequartic having only simple singularities [Do, Theorem 8.3.2.(iv)]. A calculationshows that, in our particular situation, the quartic V p Q q splits over k p B q into theunion of two conics, Q “ Q Q .The double cover X η of P k p B q goes over, under blowing up, into a double coverof S , and therefore also into one of S . The special choice of B makes sure that theramification locus V p l ¨ ¨ ¨ l q is mapped to V p Q q . In other words, a linear algebracalculation over the function field k p B q shows that three linearly independent cubicforms vanishing in the seven blown-up points, together with the coordinate w , fulfilexactly one quartic relation, which is of the kind W “ Q p X, Y, Z q . I.e., X η has asingular model X η of degree four, which is given by the equation W “ Q p X, Y, Z q Q p X, Y, Z q . There is an automorphism of X η , given by I : p W : X : Y : Z q ÞÑ p iW : X : Y : Z q ,cf. [EJ20b, Example 6.17]. We claim that the operation of I on H tr gives rise tocomplex multiplication. For this, we have to show that J : “ I ˝ I acts on H tr asthe multiplication by p´ q . Let us note that H tr is a simple Q -Hodge structure [Za,Theorem 1.6.a)], so it suffices to exclude multiplication by p` q .For this, let us observe that X η has four singularities of type A . Hence,dim H p X η p C q , Q q “
10. The fixed point set of J is the union of two conics, whichhas topological Euler characteristic 0. Therefore, the Lefschetz trace formula [Ed,Theorem 8.5] shows that Tr p J | H p X η p C q , Q q q “ ´
2. In other words, J | H p X η p C q , Q q has ROBENIUS TRACE DISTRIBUTIONS FOR K the eigenvalue p` q with multiplicity 4, while the eigenvalue p´ q occurs with mul-tiplicity 6. In particular, H tr , which is of dimension six, cannot be contained in the p` q -eigenspace, which completes the proof. (cid:3) Appendix B. A table i X i , , , , , , , , , , , , , , , , , , , ,
29 7 ,
11 11 , , , tr P Q ˚ { Q ˚ of X i ´ ´ p q ´ Table 4.
Bad primes and jump character for the surfaces in the sample
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