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Pencils of quadrics and the arithmetic ofhyperelliptic curves

Manjul Bhargava, Benedict H. Gross, and Xiaoheng Wang(with an appendix by Tim and Vladimir Dokchitser)

October 30, 2013

1 Introduction

In this article, for any fixed genusg ≥ 1, we prove that a positive proportion of hyperelliptic curvesoverQ of genusg have points overR and overQp for all p, but have no points globally overanyextension ofQ of odd degree.

By a hyperelliptic curve overQ, we mean a smooth, geometrically irreducible, complete curveC overQ equipped with a fixed map of degree 2 toP1 defined overQ. Thus any hyperelliptic curveCoverQ of genusg can be embedded in weighted projective spaceP(1, 1, g + 1) and expressed by anequation of the form

C : z2 = f(x, y) = f0xn + f1x

n−1y + · · ·+ fnyn (1)

wheren = 2g + 2, the coefficientsfi lie in Z, andf factors into distinct linear factors overQ. Definethe heightH(C) of C by

H(C) := H(f) := max{|fi|}. (2)

Then there are clearly only finitely many integral equations(1) of height less thanX, and we use theheight to enumerate the hyperelliptic curves of a fixed genusg overQ.

We say that a variety overQ is locally solubleif it has a point overQν for every placeν of Q,and issolubleif it has a point overQ. It is known that most hyperelliptic curves overQ of any fixedgenusg ≥ 1 when ordered by height are locally soluble (cf. [26], where it is shown that more than 75%of hyperelliptic curves have this property).

We prove:

Theorem 1. Fix anyg ≥ 1. Then a positive proportion of locally soluble hyperelliptic curves overQof genusg have no points over any odd degree extension ofQ.

Let J = Pic0C/Q denote the Jacobian ofC overQ, which is an abelian variety of dimensiong.The points ofJ over a finite extensionK of Q are the divisor classes of degree zero onC that arerational overK. (WhenC is locally soluble, we will see that everyK-rational divisor class onC isrepresented by aK-rational divisor.) LetJ1 = Pic1C/Q denote the principal homogeneous space forJwhose points correspond to the divisor classes of degree oneonC. A point P onC defined over anextension fieldK/Q of odd degreek gives a rational point onJ1, by taking the class of the divisor ofdegree one that is the sum of the distinct conjugates ofP minus(k−1)/2 times the hyperelliptic classdobtained by pulling backO(1) from P1. Thus Theorem1 is equivalent to the following:

1

Theorem 2. Fix anyg ≥ 1. For a positive proportion of locally soluble hyperelliptic curvesC overQof genusg, the varietyJ1 has no rational points.

We prove Theorems1 and2 by determining the average size of what we call the2-Selmer setSel2(J

1) of J1. Recall that the2-Selmer groupSel2(J) of the JacobianJ of C is a finite subgroup ofthe Galois cohomology groupH1(Q, J [2]), which is defined by certain local conditions. The2-Selmergroup fits into an exact sequence

0 → J(Q)/2J(Q) → Sel2(J) → XJ [2] → 0,

whereXJ denotes the Tate-Shafarevich group ofJ overQ. Geometrically, we may interpret elementsin the 2-Selmer groupSel2(J) of the JacobianJ of C as isomorphism classes of locally soluble two-covers ofJ . A two-coverof a principal homogeneous spaceI of J is an unramified coverπ : Y → I bya principal homogeneous spaceY such that for ally ∈ Y anda ∈ J , we haveπ(y+a) = π(y)+2a ∈ I.The degree of such a covering is22g. The simplest example of a two-cover ofJ is given by the isogeny[2] : J → J. This two-cover corresponds to the identity element ofSel2(J).

Similarly, we define the2-Selmer setSel2(J1) of J1 as the set of isomorphism classes of locallysoluble two-covers ofJ1. The2-Selmer set can be empty, but whenSel2(J1) is nonempty, it forms aprincipal homogeneous space for the finite abelian groupSel2(J). Note that a rational pointP onC(Q) determines a rational pointeP (= the class of the divisor(P )) on J1(Q). Furthermore, anyrational pointe onJ1(Q) gives rise to a locally soluble (in fact soluble) two-cover of J1; namely, afteridentifyingJ1 with J by subtractinge, we pull back the multiplication-by-2map onJ . We say then thatthis two-cover ofJ1 comes frome (or fromP whene = eP ). Theorems1 and2 therefore immediatelyfollow from:

Theorem 3. Fix anyg ≥ 1. Then for a positive proportion of locally soluble hyperelliptic curvesCoverQ, the2-Selmer setSel2(J1) is empty.

To prove this theorem, we compute the average size ofSel2(J1) over all locally soluble hyper-

elliptic curvesC overQ of any fixed genusg ≥ 1:

Theorem 4. Fix any g ≥ 1. Then the average size of the2-Selmer setSel2(J1), taken over all lo-cally soluble hyperelliptic curvesC overQ of genusg ordered by height, is equal to2.

Note that the average size of the 2-Selmer setSel2(J1) over all locally soluble hyperelliptic curvesC

overQ of genusg is independent ofg.Our methods also allow us to count elements, on average, in more general 2-Selmer sets. ForC

a hyperelliptic curve overQ having hyperelliptic classd, andk > 0 any odd integer, define the2-Selmer set of orderk for C to be the subset of elements ofSel2(J

1) that locally come fromQν-rationalpoints onJ1 of the formeν − k−1

2d, whereeν is an effective divisor of odd degreek onC overQν , for

all placesν. Then we show:

Theorem 5. Fix any odd integerk > 0. Then the average size of the2-Selmer set of orderk, over alllocally soluble hyperelliptic curves of genusg overQ, is strictly less than2 provided thatk < g, andtends to0 asg → ∞.

Theorem5 implies that most hyperelliptic curves of large genus have noK-rational points overall extensionsK of Q having small odd degrees:

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Corollary 6. Fix any m > 0. Then asg → ∞, a proportion approaching100% of hyperellipticcurvesC of genusg overQ contain no points over all extensions ofQ of odd degree≤ m.

Corollary 6 allows us to construct many smooth surfaces and varieties ofhigher degree, assymmetric powers of hyperelliptic curves, that fail the Hasse principle:

Corollary 7. Fix any odd integerk > 0. Then asg → ∞, the varietySymk(C) fails the Hasse princi-ple for a proportion approaching100% of locally soluble hyperelliptic curvesC overQ of genusg.

One may ask what is the obstruction to the Hasse principle forthe varietiesJ1 andSymk(C)occurring in Theorem2 and Corollary7, respectively. In both cases, the obstruction arises from thenon-existence of a locally soluble two-cover ofJ1. As shown by Skorobogotov [32, Theorem 6.1.1](see also Stoll [33, Remark 6.5 & Theorem 7.1]), using the descent theory of Colliot-Thelene andSansuc [14], this obstruction yields a case of the Brauer-Manin obstruction for bothJ1 andSymk(C).Therefore, we obtain:

Theorem 8. Fix anyg ≥ 1. For a positive proportion of locally soluble hyperelliptic curvesC overQof genusg, the varietyJ1 of dimensiong has a Brauer–Manin obstruction to having a rational point.

Theorem 9. Fix any odd integerk > 0. Asg → ∞, for a density approaching100% of locally solublehyperelliptic curvesC overQ of genusg, the varietySymk(C) of dimensionk has a Brauer–Maninobstruction to having a rational point.

Recall that the indexI(C) of a curveC/Q is the least positive degree of aQ-rational divisorDonC. Equivalently, it is the greatest common divisor of all degrees[K : Q] of finite field extensionsK/Q such thatC has aK-rational point. Then Theorems1 and2 are also equivalent to:

Theorem 10. For anyg ≥ 1, a positive proportion of locally soluble hyperelliptic curvesC of genusgoverQ have index2.

We will actually prove more general versions of all of these results, where for eachg ≥ 1 werange overany“admissible” congruence family of hyperelliptic curvesC overQ of genusg for whichDiv1(C) (but not necessarilyC) is locally soluble; see Definition42for the definition of “admissible”.

We obtain Theorem3 from Theorem4 by combining it with a result of Dokchitser and Dok-chitser (see Appendix A), which states that a positive proportion of locally soluble hyperelliptic curvesoverQ of genusg ≥ 1 have even (or odd)2-Selmer rank. Indeed, suppose thatC is a locally solublehyperelliptic curve whose2-Selmer setSel2(J1) is nonempty. Then the cardinality ofSel2(J1) is equalto the order of the finite elementary abelian2-groupSel2(J). LetW [2] denote the torsor forJ [2] whoseQ-points consist of elementsP ∈ J1(Q) such that2P is the hyperelliptic classd ∈ J2(Q). SinceCis locally soluble, the class of the torsorW [2] in H1(Q, J [2]) lies in the subgroupSel2(J). Indeed, itsimage inH1(Q, J)[2] is the class of the principal homogeneous spaceJ1 of J .

The class ofW [2] is nontrivial for 100% of hyperelliptic curves; indeed, a rational point onW [2] corresponds to an odd factorization off(x, y) that is rational overQ. An odd(resp.even) factor-izationof f(x, y) is a factorization of the formf(x, y) = g(x, y)h(x, y) whereg, h are odd (resp. even)degree binary forms that are eitherK-rational or are conjugate over some quadratic extension ofK.Since such factorizations rarely exist, we see that, for 100% of locally soluble curves, the2-Selmergroup ofJ containsZ/2Z, and the2-Selmer set ofJ1 (when nonempty) has cardinality at least2.Moreover, if the 2-Selmer rank of the Jacobian is even, then the setSel2(J1) (when nonempty) willhave size at least 4. Therefore, Theorem4 (and Appendix A) implies that for a positive proportion oflocally soluble hyperelliptic curves, the Selmer setSel2(J

1) is empty. This proves Theorem3.

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Remark 11. Another consequence of the fact that odd and even factorizations of a binary formf(x, y)overQ rarely exist is that for 100% of all locally soluble hyperelliptic curvesC overQ, the setJ1(Q)is either empty or infinite. Indeed, ifJ1 has a rational point, then the class ofW [2] in H1(Q, J [2])lies in the image of the groupJ(Q)/2J(Q). If f(x, y) has no odd or even factorization overQ, thenW [2] is nontrivial andJ(Q)[2] = 0. Therefore,J(Q) has positive rank and hence is infinite, and as aconsequenceJ1(Q) is infinite.

We prove Theorem4 by relating the problem to a purely algebraic one involving pencils ofquadrics. LetA andB be two symmetric bilinear forms overQ in n = 2g+2 variables, and assume thatthe corresponding pencil of quadrics inPn−1 is generic. Over the complex numbers, the Fano varietyF = F (A,B) of common maximal isotropic subspaces ofA andB is isomorphic to the JacobianJof the hyperelliptic curve given byC : z2 = disc(Ax − By) := (−1)g+1 det(Ax − By) (cf. [28],[19], [16]); furthermore, all such pairs(A,B) with the same discriminant binary form areSLn(C)-equivalent.

However, as shown in [35], overQ the situation is much different. GivenA andB, the FanovarietyF = F (A,B) might not have any rational points. In general,F is a principal homogeneousspace forJ whose class[F ] in H1(Q, J) has order dividing4 and satisfies2[F ] = [J1]; henceF givesa two-cover ofJ1 (see [35] or §4 for more details on the properties of the Fano variety). Moreover,given a hyperelliptic curveC : z2 = f(x, y) overQ of genusg (equivalently, a binary form of degreen = 2g + 2 overQ with nonzero discriminant), there might not existany pair (A,B) of symmetricbilinear forms overQ such thatf(x, y) = disc(Ax− By)! This raises the natural question: for whichbinary formsf(x, y) of degreen = 2g + 2 and nonzero discriminant overQ does there exist a pair(A,B) of symmetric bilinear forms inn variables overQ such thatf(x, y) = disc(Ax− By)?

In this paper, we give a geometric answer to this question in terms of the generalized Jaco-bianJm of the hyperelliptic curveC : z2 = f(x, y). Assume for simplicity thatf(x, y) = f0x

n +f1x

n−1y + · · · + fnyn has first coefficientf0 6= 0, so that the curveC has two distinct pointsP and

P ′ above the point∞ = (1, 0) onP1. These points are rational and conjugate over the fieldQ(√f0).

Letm = P +P ′ be the corresponding modulus overQ and letCm denote the singular curve associatedto this modulus as in [29, Ch. IV, §4]. ThenCm is given by the equationz2 = f(x, y)y2, and has anordinary double point at infinity. Thegeneralized Jacobianof C associated to the modulusm, denotedby Jm = Jm(C), is the connected component of the identity ofPicCm/Q /Z · d, while J1

m = J1m(C) de-

notes the nonidentity component; hered denotes the hyperelliptic class ofCm in Pic2Cm/Q(Q) obtainedby pulling backO(1) from P1. We prove:

Theorem 12. Let f(x, y) denote a binary form of even degreen = 2g + 2 overQ, with nonzero dis-criminant and nonzero first coefficient. Then there exists a pair (A,B) of symmetric bilinear formsoverQ in n variables satisfyingf(x, y) = disc(Ax−By) if and only if there exists a two-cover of ho-mogeneous spacesFm → J1

m for Jm overQ, or equivalently, if and only if the class of the homogeneousspaceJ1

m is divisible by2 in the groupH1(Q, Jm).

See Theorem23 for a number of other equivalent conditions for the existence of A andBsatisfyingf(x, y) = disc(Ax−By). It is of significance that the singular curveCm and the generalizedJacobianJm appear in Theorem12. As noted in [27, Footnote 2], for the purpose of doing 2-descenton the Jacobians of hyperelliptic curves with no rational Weierstrass point, it is not always enough tostudy only unramified covers ofC; one needs also covers ofC unramified away from the points abovesome fixed point onP1.

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The groupSLn(Q) acts on the spaceQ2 ⊗ Sym2Qn of pairs (A,B) of symmetric bilinear

forms on ann-dimensional vector space, andµ2 ⊂ SLn acts trivially sincen = 2g + 2 is even. Theconnection with Theorem4 arises from the fact that we may parametrize elements ofSel2(J

1) bycertain orbits for the action of the group(SLn /µ2)(Q) on the spaceQ2 ⊗ Sym2Q

n. We say that anelement(A,B) ∈ Q2 ⊗ Sym2Q

n, or its (SLn /µ2)(Q)-orbit, is locally solubleif the associated FanovarietyF (A,B) has a point locally over every place ofQ. Then we prove:

Theorem 13. Let f(x, y) denote a binary form of even degreen = 2g + 2 over Q such that thehyperelliptic curveC : z2 = f(x, y) is locally soluble. Then the(SLn /µ2)(Q)-orbits of locally solublepairs(A,B) of symmetric bilinear forms inn variables overQ such thatf(x, y) = disc(Ax−By) arein bijection with the elements of the2-Selmer setSel2(J1).

To obtain Theorem4, we require a version of Theorem13for integral orbits. LetZ2⊗Sym2 Zn

denote the space of pairs(A,B) of n× n symmetric bilinear forms overZ. Then we show:

Theorem 14. There exists a positive integerκ depending only onn such that, for any integral bi-nary formf(x, y) of even degreen = 2g + 2 with C : z2 = f(x, y) locally soluble overQ, every(SLn /µ2)(Q)-orbit of locally soluble pairs(A,B) ∈ Q2 ⊗ Sym2Q

n such thatdisc(Ax − By) =κ2f(x, y) contains an element inZ2⊗Sym2 Z

n. In other words, the(SLn /µ2)(Q)-equivalence classesof locally soluble pairs(A,B) ∈ Z2 ⊗ Sym2 Z

n such thatdisc(Ax−By) = κ2f(x, y) are in bijectionwith the elements ofSel2(J1).

We will prove Theorem14for κ = 4n but we expect this can be improved. We use Theorem14,together with the results of [1] giving the number ofSLn(Z)-orbits onZ2 ⊗ Sym2 Z

n having boundedheight, and a sieve, to deduce Theorem4.

We note that the emptiness ofJ1(Q) for hyperelliptic curvesC overQ has been demonstratedpreviously for certain special algebraic families. In [13], Colliot-Thelene and Poonen constructed one-parameter algebraic families of curvesC = Ct of genus1 and genus2 for which the varietiesJ1 havea Brauer-Manin obstruction to having a rational point for all t ∈ Q. (We note that the family of genus2curves considered in [13] consists of hyperelliptic curvesC overQ with locally solubleJ1(C) but notlocally solubleDiv1(C).) For arbitrary genusg > 5 satisfying4 ∤ g, Dong Quan [20] constructed suchone-parameter algebraic families of locally soluble hyperelliptic curvesC = Ct with emptyJ1(Q) forall t ∈ Q.

This paper is organized as follows. In Section2, we introduce the key representation2 ⊗Sym2(n) of SLn on pairs of symmetric bilinear forms that we will use to studythe arithmetic ofhyperelliptic curves. We adapt the results of Wood [38] to study the orbits of this representation overa general Dedekind domainD whose characteristic is not equal to2. In Section3, we introducehyperelliptic curves and some of the relevant properties oftheir generalized Jacobians. In Section4, wethen relate hyperelliptic curves to generic pencils of quadrics over a fieldK of characteristic not equalto 2, and we review the results that we will need from [35]. In Section5, we then studyregularpencilsof quadrics, which allows us to determine which binaryn-ic forms overK arise as the discriminant ofa pencil of quadrics overK; in particular, we prove Theorem12.

In Section6, we describe how theK-solubleorbits (i.e., orbits of those(A,B) overK such thatF (A,B) has aK-rational point), having associated hyperelliptic curveC overK, are parametrized byelements of the setJ1(K)/2J(K). We study the orbits over some arithmetic fields in more detail inSection7 and then we focus on global fields and discusslocally solubleorbits in Section8. We showthat the locally soluble orbits overQ, having associated hyperelliptic curveC overQ are parametrized

5

by the elements of the finite setSel2(J1), proving Theorem13. The existence of integral orbits (The-orem14) is demonstrated in Section9. We then discuss the counting results from [1] that we need inSection10, and discuss the details of the required sieve in Section11. Finally, we complete the proofsof Theorems4 and5 in the final Section12.

2 Orbits of pairs of symmetric bilinear forms over a Dedekinddomain

In this section, we study the orbits of our key representation 2 ⊗ Sym2(n) over a Dedekind domain.In later sections, we will relate these orbits to the arithmetic of hyperelliptic curves.

Let D be a Dedekind domain, with quotient fieldK. We assume throughout this paper that thecharacteristic ofD is not equal to2. Letn ≥ 2 be an integer. The groupSLn(D) acts on theD-moduleof pairs(A,B) of symmetric bilinear forms on a freeD-moduleW of rankn. After a choice of basisfor W , this is the representationD2 ⊗ Sym2D

n = Sym2Dn ⊕ Sym2D

n.The coefficients of the binaryn-ic form

f(x, y) = disc(xA− yB) := (−1)n(n−1)/2 det(xA− yB) = f0xn + f1x

n−1y + · · ·+ fnyn,

which we call theinvariant binaryn-ic form of the element(A,B) ∈ D2 ⊗ Sym2Dn, give n + 1

polynomial invariants of degreen which freely generate the ring of invariant polynomials over D. Wealso have the invariantdiscriminantpolynomial∆(f) = ∆(f0, f1, . . . , fn) given by the discriminantof the binary formf , which has degree2n(n− 1) in the entries ofA andB.

In Wood’s work [38], the orbits ofSL±2 (T ) = {g ∈ GL2(T ) : det(g) = ±1} onT 2 ⊗ Sym2 T

n

were classified for general rings (and in fact even for general base schemes)T in terms of ideal classesof rings of rankn overT . In this section, we translate these results into a form thatwe will requirelater on, in the important special case whereT = D is a Dedekind domain with quotient fieldK.In particular, we will need to use the actions by the groupsSLn(D) and (SLn /µ2)(D) rather thanSL±

n (D), which causes some key changes in the parametrization data and will indeed be important forus in our later discussion and connection with hyperelliptic curves.

Let us assume thatf0 6= 0 and writef(x, 1) = f0g(x), whereg(x) has coefficients in thequotient fieldK and hasn distinct roots in a separable closureKs of K. LetL = Lf := K[x]/g(x) bethe corresponding etale algebra of rankn overK, and letθ be the image ofx in the algebraL. Theng(θ) = 0 in L. Let g′(x) be the derivative ofg(x) in K[x]; sinceg(x) is separable, the valueg′(θ) mustbe an invertible element ofL. We definef ′(θ) = f0g

′(θ) in L×.Fork = 1, 2, . . . , n− 1, define the integral elements

ζk = f0θk + f1θ

k−1 + . . .+ fk−1θ

in L, and letR = Rf be the freeD-submodule ofL havingD-basis{1, ζ1, ζ2, . . . , ζn−1}. For k =0, 1, . . . , n − 1, let I(k) be the freeD-submodule ofL with basis{1, θ, θ2, . . . , θk, ζk+1, . . . , ζn−1}.ThenI(k) = I(1)k, andI(0) = R ⊂ I(1) ⊂ · · · ⊂ I(n− 1). Note thatI(n− 1) has the power basis{1, θ, θ2, . . . , θn−1}, but that the elements ofI(n− 1) need not be integral whenf0 is not a unit inD.

A remarkable fact (cf. [8], [24, Proposition 1.1], [37, §2.1]) is thatR is a D-order inL ofdiscriminant∆(f), and the freeD-modulesI(k) are all fractional ideals ofR. The fractional ideal(1/f ′(θ))I(n− 2) is the dual ofR under the trace pairing onL, and the fractional idealI(n− 3) willplay a crucial role in the parametrization of orbits in our representation.

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We then have the following translation of [38, Theorem 1.3] in the case of the action ofSLn(D)onD2 ⊗ Sym2D

n, whereD is a Dedekind domain:

Theorem 15. Assume thatf(x, y) is a binary form of degreen over D with ∆(f) 6= 0 and f0 6=0. Then there is a bijection(to be described below) between the set of orbits forSLn(D) on D2 ⊗Sym2D

n with invariant formf and the set of equivalence classes of triples(I, α, s), whereI is afractional ideal forR, α ∈ L×, and s ∈ K×, that satisfy the relationsI2 ⊂ αI(n − 3), N(I) isthe principal fractional idealsD in K, andN(α) = s2fn−3

0 in K×. The triple(I, α, s) is equivalentto the triple(cI, c2α,N(c)s) for any c ∈ L×. The stabilizer of a triple(I, α, s) is S×[2]N=1 whereS = EndR(I) ⊂ L.

From a triple(I, α, s), we construct an orbit as follows. SinceN(I) is the principalD-idealsD, the projectiveD-moduleI of rankn is free. SinceI2 ⊂ αI(n − 3), we obtain two symmetricbilinear forms on the free moduleI by defining〈λ, µ〉A and〈λ, µ〉B as the respective coefficients ofζn−1 andζn−2 in the basis expansion of the productλµ/α in I(n − 3). We obtain anSLn(D)-orbit oftwo symmetricn× n matrices(A,B) overD by taking the Gram matrices of these forms with respectto any ordered basis ofI that gives rise to the basis elements(1∧ζ1∧ζ2∧· · ·∧ζn−1) of the top exteriorpower ofI overD. This normalization deals with the difference betweenSLn(D)- andGLn(D)-orbits.The stabilizer statement follows because elements inS×[2]N=1 are precisely the elements ofL×

N=1 thatpreserve the map1

α× : I × I → I(n− 3).

Conversely, given an element(A,B) ∈ D2 ⊗ Sym2Dn, we construct the ringR = Rf from f

as described above, wheref(x, y) = disc(xA − yB). TheR-moduleI is then constructed by lettingθ ∈ L act onKn by the matrixA−1B. Thenζ1 = f0θ ∈ R preserves the latticeDn. Similarly, formulasfor the action of eachζi ∈ R onDn, in terms of integral polynomials in the entries ofA andB, can beworked out whenA is assumed to be invertible; these same formulas can then be used to show thatDn

is anR-module, even whenA is not invertible. See [38, §3.1] for the details.

Whenn = 2m is even, the larger group(SLn /µ2)(D) acts on the representationD2⊗Sym2Dn,

and distinct orbits for the subgroupSLn(D)/µ2(D) may become identified as a single orbit for thelarger group. SinceH1(D, SLn) = 1, we have an exact sequence of groups

1 → SLn(D)/µ2(D) → (SLn /µ2)(D) → H1(D, µ2) → 1.

By Kummer theory, the quotient groupH1(D, µ2) lies in an exact sequence

1 → D×/D×2 → H1(D, µ2) → Pic(D)[2] → 1.

The image of the groupH1(D, µ2) in H1(K,µ2) = K×/K×2 is the subgroupK×(2)/K×2 of elementst such that the principal idealtD = M2 is a square, and the map toPic(D)[2] is given by mappingsuch an elementt to the class ofM .

With this identification of the quotient, the action of elements t in the groupK×(2)/K×2 onthe equivalence class of triples(I, α, s) giving the orbits forSLn(D) with invariant formf in theTheorem15 is given by

t · (I, α, s) = (MI, tα, tms)

wherem = n/2 andtD = M2. The equivalence classes of triples for this additional action give theorbits of (SLn /µ2)(D) with invariant formf . The stabilizer of the triple(I, α, s) contains the finitegroupS×[2]N=1/D

×[2] whereS = EndR(I) ⊂ L, as that is the image of the stabilizer fromSLn(D).

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Theorem 16.Assume thatf(x, y) is a binary form of degreen overD with∆(f) 6= 0 andf0 6= 0. Thenthere is a bijection between the set of orbits for(SLn /µ2)(D) onD2 ⊗ Sym2D

n with invariant formf and the set of equivalence classes of triples(I, α, s), whereI is a fractional ideal forR, α ∈ L×,ands ∈ K×, that satisfy the relationsI2 ⊂ αI(n− 3), N(I) is the principal fractional idealsD in K,andN(α) = s2fn−3

0 in K×. The triple(I, α, s) is equivalent to the triple(cMI, c2tα,N(c)tms) foranyc ∈ L× andt ∈ K×(2), wherem = n/2 andtD = M2. The stabilizer of the triple(I, α, s) is anelementary abelian2-group which containsS×[2]N=1/D

×[2] whereS = EndR(I) ⊂ L.

Remark 17. We can simplify the statement of Theorem16 when the domainD is a PID and everyfractional ideal for the orderR is principal, generated by an element ofL. In that case, the fractionalidealI of R is completely determined by the pair(α, s) and the identitiesI2 ⊂ (α)I(n− 3), N(I) =(s), andN(α) = s2fn−3

0 . Indeed, together these forceI2 = (α)I(n− 3). There is a bijection from theset of equivalence classes ofα to the set(R×/R×2D×)N=f0. Moreover, we haveS = EndR(I) = RandK×(2) = D×K×2. The stabilizer in(SLn /µ2)(D) of a triple (I, α, s) then fits into the exactsequence

1 → (R×[2])N=1/D×[2] → Stab(SLn /µ2)(D)(I, α, s) → (R×2 ∩D×)/D×2 → 1. (3)

WhenL is not an algebra over a quadratic extension ofK, the quotient(R×2 ∩D×)/D×2 is trivial.

In particular, whenD = K is a field, we recover [5, Theorems 7 and 8]. These versions ofTheorems15 and16 over a fieldK will also be important in the sequel. For convenience, we restatethem below.

Corollary 18. Assume thatf(x, y) is a binary form of degreen overK with ∆(f) 6= 0 andf0 6= 0.Then there is a bijection between the set of orbits forSLn(K) onK2 ⊗ Sym2K

n with invariant formf and the set of equivalence classes of pairs(α, s) with N(α) = s2fn−3

0 in K×. The pair(α, s) isequivalent to the pair(c2α,N(c)s) for anyc ∈ L×. The stabilizer of the orbit corresponding to a pair(α, s) is the finite commutative group scheme(ResL/K µ2)N=1 overK.

It follows from Corollary18 that the set ofSLn(K)-orbits is either in bijection with or has a2-to-1 map to(L×/L×2)N=f0, depending on whetherf(x, y) has an odd degree factor overK or not,respectively. Indeed, the pair(α, s) is equivalent to the pair(α,−s) if and only if there is an elementc ∈ L× with c2 = 1 andN(c) = −1.

Corollary 19. Assume thatf(x, y) is a binary form of degreen overK with ∆(f) 6= 0 andf0 6= 0.Then there is a bijection between the set of orbits for(SLn /µ2)(K) onK2 ⊗ Sym2K

n with invariantform f and the set of equivalence classes of pairs(α, s) with N(α) = s2fn−3

0 in K×. The pair(α, s)is equivalent to the pair(c2tα,N(c)tn/2s) for any c ∈ L× and t ∈ K×(2) = K×. The stabilizer ofthe orbit corresponding to a pair(α, s) is the finite commutative group scheme(ResL/K µ2)N=1/µ2

overK.

It follows from Corollary 19 that the set of(SLn /µ2)(K)-orbits is either in bijection withor has a 2-to-1 map to(L×/L×2K×)N=f0, depending on whetherf(x, y) has an odd factorizationor not, respectively. Here anodd factorizationof f(x, y) is a factorization of the formf(x, y) =g(x, y)h(x, y), whereg andh are odd degree binary forms that are eitherK-rational or are conjugateover some quadratic extension ofK.

8

3 Hyperelliptic curves, divisor classes, and generalized Jacobians

Assume from now on thatn ≥ 2 is even and writen = 2g + 2. Fix a fieldK of characteristic not 2.In order to interpret the orbits forSLn(K) and(SLn /µ2)(K) having a fixed invariant binary form, wefirst review some of the arithmetic and geometry of hyperelliptic curves of genusg overK. As in [21],we define a hyperelliptic curve overK as a smooth, projective curve overK with a 2-to-1 map to theprojective line overK, although now we treat the general case (without any fixedK-rational points atinfinity.)

Let f(x, y) = f0x2g+2+ · · ·+f2g+2y

2g+2 be a binary form of degree2g+2 overK, with∆ 6= 0andf0 6= 0. We associate tof(x, y) the hyperelliptic curveC overK with equation

z2 = f(x, y).

This defines a smooth curve of genusg, as a hypersurface of degree2g + 2 in the weighted projectiveplaneP(1, 1, g+1). The weighted projective plane embeds as a surface inPg+2 via the map(x, y, z) →(xg+1, xgy, . . . , yg+1, z). The image is a cone over the rational normal curve inPg+1, which has asingularity at the vertex(0, 0, . . . , 1) wheng ≥ 1. The curveC is the intersection of this surface witha quadric hypersurface that does not pass through the vertexof the cone. Finally, the linear series onC of projective dimensiong + 2 and degree2g + 2 that gives this embedding is the sum of the all theWeierstrass points (i.e., points withz = 0).

There are two pointsP = (1, 0, z0) andP ′ = (1, 0,−z0) at infinity, wherez20 = f0. If f0 is asquare inK×, then these points are rational overK. If not, then they are rational over the quadraticextensionK ′ = K(

√f0). Letw be the rational functionz/yg+1 onC, and lett be the rational function

x/y onC. Both are regular outside of the two pointsP andP ′ with y = 0, where they have poles oforderg + 1 and1 respectively. The field of rational functions onC is given byK(C) = K(t, w), withw2 = f(t, 1) = f0t

2g+2 + · · ·+ f2g+2, and the subring of functions that are regular outside ofP andP ′

isK[t, w] = K[t,√

f(t, 1)] [21].Let m be the modulusm = P + P ′ onC and letCm be the singular curve constructed fromC

and this modulus in [29, Ch. IV, no. 4]. ThenCm has equation

z2 = f(x, y)y2

of degree2g + 4 in P(1, 1, g + 2). This defines a singular, projective curve of arithmetic genusg + 1whose normalization isC. There is now a single pointQ = (1, 0, 0) at infinity, which is an ordinarydouble point whose tangents are rational over the quadraticextension fieldK.

Let PicC/K andPicCm/K denote the Picard functors of the projective curvesC andCm respec-tively. These are represented by commutative group schemesoverK, whose component groups areboth isomorphic toZ. Let Ks be a fixed separable closure ofK and letE be any extension ofKcontained inKs. TheE-rational points ofPicC/K correspond bijectively to the divisor classes onCover the separable closureKs that are fixed by the Galois groupGal(Ks/E). When the curveC hasnoE-rational points, anE-rational divisor class onC may not be represented by anE-rational divisor.The subgroup of classes inPicC/K(E) that are represented byE-rational divisors is just the image ofPic(C/E) = H1(C/E,Gm) in H0(E,H1(C/Ks,Gm)), under the map induced by the spectral se-quence for the morphismC/E → SpecE. From this spectral sequence, we also obtain an injectionfrom the quotient group to the Brauer group ofK (cf. [32, §2.3], [9, Ch. 8]):

PicC/K(K)/Pic(C/K) → H2(K,Gm) = Br(K).

9

SinceC has a rational point over the quadratic extensionK ′ = K(√f0), the image of this injection

is contained in the subgroupBr(K ′/K) = K×/N(K ′×). Every class inBr(K ′/K) corresponds to aquaternion algebraD overK that is split byK ′, or equivalently, to a curve of genus zero overK withtwo conjugate points rational overK ′.

Proposition 20. If a hyperelliptic curveC overK has a rational divisor of odd degree, or equivalentlya rational point over an extension ofK of odd degree, then everyK-rational divisor class is representedby aK-rational divisor. IfK is a global field andDiv1(C) is locally soluble, then everyK-rationaldivisor class is represented by aK-rational divisor.

Indeed, a quaternion algebra split by an odd degree extension of K is already split overK.Similarly, a quaternion algebra over a global field that splits locally everywhere is split globally.

The distinction betweenK-rational divisor classes andK-rational divisors does not arise forthe curveCm, which always has theK-rational singular pointQ. Hence the points ofPicCm/K overEcorrespond to the classes of divisors that are rational overE and are prime tom, modulo the divisorsof functions withf ≡ 1 modulom. We have an exact sequence of smooth group schemes overK :

0 → T → PicCm/K → PicC/K → 0, (4)

whereT is the one-dimensional torus that is split byK ′. Taking the long exact sequence in Ga-lois cohomology, and noting that the image ofPicCm/K(K) in PicC/K(K) is precisely the subgroupPic(C/K) = H1(C/K,Gm) represented byK-rational divisors, we recover the injection

PicC/K(K)/Pic(C/K) → H1(K, T ) = K×/N(K ′×) = Br(K ′/K).

To see this geometrically, note that the fiber over aK-rational pointP of PicC/K is a principal homoge-neous space forT overK, which is a curve of genus zero with two conjugate points overK ′ removed.This curve of genus zero determines the image ofP in Br(K ′/K).

The connected components of the identity of the Picard schemes J = Pic0C/K and Jm =

Pic0Cm/K are the Jacobian and generalized Jacobian of [29, Ch. V]. They correspond to the divisorclasses of degree zero on these curves. The exact sequence in(4) restricts to the following exact se-quence [29, Ch. V,§3]

0 → T → Jm → J → 0. (5)

There is a line bundle of degree2 onCm (and hence onC) which is the pull-back of the linebundleO(1) from the projective line under the map(x, y, z) → (x, y). This is represented by theK-rational divisord = (R) + (R′) prime tom consisting of the two points above a point(x0, y0)on the projective line, whenevery0 is nonzero. The quotient groupsPicC/K /Z · d = J ⊔ J1 andPicCm/K /Z · d = Jm ⊔ J1

m both have two connected components, represented by the divisor classes ofdegree0 and1. There are morphisms

C −→ J1

C − {P, P ′} = Cm − {Q} −→ J1m

defined overK, which take a point to the corresponding divisor class of degree1 [29, Ch V, §4].

Proposition 21. Letf(x, y) = f0x2g+2 + · · ·+ f2g+2y

2g+2 be a binary form with nonzero discriminantand nonzerof0. LetC : z2 = f(x, y) andCm : z2 = f(x, y)y2 denote the associated hyperellipticcurve and singular curve with JacobianJ and generalized JacobianJm. LetL = K[x]/f(x, 1) denotethe correspondingetale algebra of rank2g + 2. Then:

10

1. The2-torsion subgroupJm[2] of Jm is isomorphic to the group scheme(ResL/Kµ2)N=1. ItsK-rational points correspond to the even degree factors off(x, y) overK.

2. The2-torsion subgroupJ [2] of J is isomorphic to the group scheme(ResL/Kµ2)N=1/µ2. ItsK-rational points correspond to the even factorizations off(x, y) overK.

3. The2-torsionWm[2] in the componentJ1m ofPicCm/K /Z·d = Jm⊔J1

m, is a torsor forJm[2] whoseK-rational points correspond to the odd degree factors off(x, y) overK.

4. The2-torsionW [2] in the componentJ1 of PicC/K /Z · d = J ⊔ J1 is a torsor forJ [2] whoseK-rational points correspond to the odd factorizations off(x, y) overK.

Here an odd (resp. even) factorization off(x, y) overK is a factorization of the formf = gh,whereg andh are either odd (resp. even) degree binary forms overK or are conjugate over somequadratic extension ofK.

Proof. To prove the proposition, we observe that the2-torsion points ofJm over the separable closureKs are represented by the classes of divisors of the form(P1) + (P2) + · · · + (P2m) − md, whereeachPi = (xi, 1, 0) comes from a distinct rootxi of f(x, 1) [22, §4]. Hence the points ofJm[2] overKs correspond bijectively to the factors of even degree off(x, y). Since the Galois group acts bypermutation of the roots, we have a canonical isomorphismJm[2] ≃ (ResL/Kµ2)N=1. On the quotientJ , there is a single relation:(P1)+· · ·+(P2g+2)−(g+1)d = div(y) ≡ 0, soJ [2] ≃ (ResL/Kµ2)N=1/µ2.The last two statements of Proposition21 follow similarly.

Finally, we note that the Weil pairingJ [2]×J [2] → µ2 gives the self-duality of the finite groupscheme(ResL/Kµ2)N=1/µ2, and the connecting homomorphismH1(K, J [2]) → H2(K,µ2) whosekernel is the image ofH1(K, Jm[2]) is cup product with the class ofW [2] (see [27, Proposition 10.3]).

4 Generic pencils of quadrics

In this section, we relate hyperelliptic curves to pencils of quadrics. In particular, we will see howpencils of quadrics yield two-covers ofJ1 for certain hyperelliptic curves.

LetW = Kn be a vector space of dimensionn ≥ 3 overK and letA andB be two symmetricbilinear forms onW . LetQA andQB be the corresponding quadric hypersurfaces inP(W ), soQA isdefined by the equation〈w,w〉A = 0 andQB is defined by the equation〈w,w〉B = 0. LetY be the baselocus of the pencil spanned byA andB, which is defined by the equations〈w,w〉A = 〈w,w〉B = 0in P(W ). ThenY has dimensionn − 3 and is a smooth complete intersection if and only if thediscriminant of the pencildisc(xA− yB) = f(x, y) has∆(f) 6= 0. In this case we say that the pencilspanned byA andB is generic. In this section, we will only consider generic pencils. The Fano schemeF = F (A,B) is the Hilbert scheme of maximal linear subspaces ofP(W ) that are contained inY .

Whenn = 2g+ 1 is odd, the Fano scheme has dimension zero and is a principal homogeneousspace for the finite group schemeResL/Kµ2/µ2 ≃ (ResL/Kµ2)N=1. HereL is the etale algebra of rank2g+1 determined by the separable binary formf(x, y). The22g points ofF over the separable closureof K correspond to the subspacesZ of W of g that are isotropic for all the quadrics in the pencil, andthe schemeF depends only on theSLn(K)-orbit of the pair(A,B).

11

Whenn = 2g + 2 is even, the Fano schemeF is smooth and geometrically connected ofdimensiong, and is a principal homogeneous space for the JacobianJ of the smooth hyperellipticcurveC with equationz2 = f(x, y). A point of F corresponds to a subspaceZ of W of dimensiong that is isotropic for all of the quadrics in the pencil, whereas a point ofC corresponds to a quadricin the pencil plus a choice of one of the two rulings of that quadric. This interpretation can be used todefine a morphismC × F → F overK, which in turn gives a fixed point free action ofJ on F . Inthis case, the Fano varietyF depends only on the(SLn /µ2)(K)-orbit of the pair(A,B). Proofs of allassertions on the Fano scheme can be found in [35].

Theorem 22. ([35, Theorem 2.7])Let F be the Fano variety of maximal linear subspaces containedin the base locus of a generic pencil of quadrics generated bysymmetric bilinear forms(A,B) ∈K2 ⊗ Sym2K

n. Let f(x, y) denote the invariant binary form of(A,B). LetC : z2 = f(x, y) denotethe corresponding hyperelliptic curve with JacobianJ . Then the disconnected variety

X := J ⊔ F ⊔ J1 ⊔ F (6)

has a commutative algebraic group structure overK. In particular, [F ] as a class inH1(K, J) is4-torsion and2[F ] = [J1].

The groupX contains the subgroupPicC/K /Z · d = J ⊔ J1 with index two. LetF [4] bethe principal homogeneous space forJ [4] consisting of the points ofF of (minimal) order4 in thegroupX. Multiplication by2 in X gives finite etale coverings

F → J1

F [4] → W [2]

of degree22g with an action of the group schemeJ [2]. This shows that the class[F ] of the principalhomogeneous spaceF satisfies2[F ] = [J1] in the groupH1(K, J). Similarly, the class ofW [2]in H1(K, J [2]) is the image of the classF [4] in H1(K, J [4]) under the map2 : H1(K, J [4]) →H1(K, J [2]).

Consequently, a necessary condition on the existence of a pencil (A,B) overK with discrim-inant curveC is that the class ofJ1 and the class ofW [2] should be divisible by2 in H1(K, J) andH1(K, J [4]) respectively. However, this condition is not sufficient. Consider the curveC of genus zerowith equationz2 = −x2 − y2 overR. In this case, bothJ andJ [2] reduce to a single point, so anyhomogeneous space forJ or J [2] is trivial, and hence divisible by2. On the other hand, sinceL = Candf0 = −1 is not a norm, by Corollary18 (or 19) there are no pencils overR with discriminantf(x, y) = −x2 − y2. To obtain a geometric condition that is both necessary and sufficient for theexistence of a pencil, we will have to consider non-generic pencils whose invariant binary form definesthe singular curveCm. This is the object of the next section.

5 Regular pencils of quadrics

In this section, we give a list of equivalent conditions for the existence of a pencil overK whosediscriminant is some given binary formf(x, y). In particular, we prove Theorem12.

Let (A,B) generate a generic pencil of bilinear forms on a vector spaceW of even dimensionn = 2g+2 overK, and letf(x, y) = disc(xA−yB) be the associated binary form of degree2g+2 and

12

discriminant∆(f) 6= 0. We continue to assume thatf0 = disc(A) is also nonzero inK. Let (A′, B′)be a pair of bilinear forms on the vector spaceW ′ = W ⊕K2 of dimensionn+ 2 = 2g + 4, whereA′

is the direct sum ofA and the rank one form〈(a, b), (a′, b′)〉 = aa′ onK2 andB′ is the direct sum ofB and the split form〈(a, b), (a′, b′)〉 = ab′ + a′b of rank2. The invariant binary form of this pencil

disc(xA′ − yB′) = f(x, y)y2

then has a double zero at(x, y) = (1, 0), and the pencil is not generic. The base locus defined by theequationsQA′ = QB′ = 0 in P(W ⊕ K2) has an ordinary double point at the unique singular pointR = (0W ; 0, 1) of the quadricQA′. There are exactly2g + 3 singular quadrics in the pencil and allof them are simple cones. TheK-algebraL′ associated to the pencil is not etale, but is isomorphic toL ⊕ K[y]/y2. Even thoughL′ is not etale, the vector spaceW ′ is a freeL′-module of rank 1, so thepencil is regular in the sense of [35, §3]. Since the norms fromK[y]/y2 toK are precisely the squaresin K, we have an equality of quotient groupsK×/K×2N(L×) = K×/K×2N(L′×).

The Fano schemeFm of this pencil consists of the subspacesZ of dimensiong + 1 in W ⊕K2

that are isotropic for all of the quadrics in the pencil and donot contain the unique line that is theradical of the formA′ (so the projective spaceP(Z), which is contained in the base locus, does notmeet the unique double pointR). The Fano scheme is a smooth variety of dimensiong + 1. However,in this caseFm is not projective. It is a principal homogeneous space for the generalized JacobianJm

associated to the singular curveCm of arithmetic genusg+1 and equationz2 = f(x, y)y2 in weightedprojective space.

For example, wheng = 0, the curveC is the non-singular quadricz2 = ax2 + bxy + cy2 in P2,with a = f0 andb2 − 4ac = ∆(f) both nonzero inK. The pencil(A′, B′) has discriminantf ′(x, y) =ax2y2 + bxy3 + cy4. Its base locusD in P3 is isomorphic to a singular curve of arithmetic genus one,with a single nodeR whose tangents are rational over the quadratic extensionK ′ = K(

√f0). The

Fano varietyFm in this case is just the affine curveD − {R}, andJ1m is the affine curveCm − {Q} =

C − {P, P ′}. Both are principal homogeneous spaces for the one-dimensional torusT = Jm which issplit byK ′. We shall see that there is an unramified double coverFm → J1

m that extends to a doublecover of complete curves of genus zeroM → C which is ramified atP andP ′.

Since the pencil is regular and its associated hyperelliptic curve has only nodal singularities,we again obtain a commutative algebraic group

Xm = Jm ⊔ Fm ⊔ J1m ⊔ Fm (7)

overK with connected componentJm and component groupZ/4, which contains the algebraic groupPicCm/K /Z · d = Jm ⊔ J1

m with index two [35, §3.2]. Just as in the generic case, multiplication by2 inthe groupXm gives an unramified cover

Fm → J1m

of degree22g+1 with an action ofJm[2], and shows that2[Fm] = [J1m] in the groupH1(K, Jm) of

principal homogeneous spaces forJm. Hence a necessary condition for the existence of such a pencil(A′, B′) is that the class ofJ1

m is divisible by2. In this case, the necessary condition is also sufficient.

Theorem 23.Letf(x, y) = f0x2g+2+ · · ·+ f2g+2y

2g+2 be a binary form of degree2g+2 overK withf0 and∆(f) both nonzero inK. Writef(x, 1) = f0g(x) with g(x) monic and separable. LetL be theetale algebraK[x]/g(x) of degree n overK and letβ denote the image ofx in L. LetC be the smoothhyperelliptic curve of genusg with equationz2 = f(x, y) and letCm be the singular hyperelliptic

13

curve of arithmetic genusg + 1 with equationz2 = f(x, y)y2. Then the following conditions are allequivalent.

a. There is a generic pencil(A,B) overK with disc(xA− yB) = f(x, y).

b. There is a regular pencil(A′, B′) overK with disc(xA′ − yB′) = f(x, y)y2.

c. The coefficientf0 lies in the subgroupK×2N(L×) of K×.

d. The class of the homogeneous spaceJ1m is divisible by2 in the groupH1(K, Jm).

e. The class of the homogeneous spaceWm[2] is in the image of the multiplication-by-2mapH1(K, Jm[4]) →H1(K, Jm[2]).

f. There is an unramified two-cover of homogeneous spacesFm → J1m for Jm overK.

g. The maximal unramified abelian coverU → C − {P, P ′} of exponent2 overKs descends toK.

h. The maximal abelian coverM → C of exponent2 overKs that is ramified only at the points{P, P ′}descends toK.

Note the maximal abelian covers above all have degree22g+1. The equivalence of conditionsa, d, andf proves Theorem12.

Proof. a ⇔ b ⇔ c. We have already seen that the existence of orbits in both cases is equivalent toconditionc, that the class off0 in K×/K×2 is in the image of the norm map fromL×/L×2.

b ⇒ d ⇔ e ⇒ f. When a regular orbit(A′, B′) exists inb, the Fano varietyFm of the baselocus of the pencil provides a homogenous space forJm whose class is a square-root of the class ofJ1

m

in the groupH1(K, Jm). The equivalence of conditionsd, e andf is clear.f ⇒ g ⇒ h. Assuming that an unramified coveringF → J1

m exists overK, we obtain theunramified covering ofC−{P, P ′} by taking the fiber product with the morphismC−{P, P ′} → J1

m,and the ramified covering ofC by completing the unramified cover ofC − {P, P ′}.

h ⇒ c. Finally, assuming the existence of the ramified coveringM → C overK, we show thatf0 lies in the subgroupK×2N(L×) of K×, which will complete the proof of Theorem23. The coveringM → C corresponds to an inclusion of function fieldsK(C) → K(M). OverKs, the function fieldKs(M) is obtained fromKs(C) by adjoining the square-roots of all rational functions onC whosedivisors have the form2d1 or 2d1 + (P ) + (P ′) for some divisord1 onC. Since the characteristic ofK is not equal to2, these square-roots either give unramified extensions ofC or extensions that areramified only at the two pointsP andP ′, where the ramification is tame. More precisely, there are22g+1 − 1 distinct quadratic extensions ofKs(C) of this form that are contained inKs(M), and theircomposition is equal toKs(M).

Indeed, by Galois theory, these quadratic extensions correspond to the subgroups of index2in Jm[2](K

s), or equivalently to nontrivialKs-points in the Cartier dualResL/Kµ2/µ2. Let w bethe rational functionz/yg+1 on C, and lett be the rational functionx/y on C, so w2 = f0g(t).The nontrivial points in(ResL/Kµ2/µ2)(K

s) correspond bijectively to the nontrivial monic factoriza-tions g(x) = h(x)j(x) overKs, and the corresponding quadratic extension ofKs(C) is given byKs(C)(

√h(t)) = Ks(C)(

√j(t)). When bothh(x) andj(x) have even degree, the divisors of the

rational functionsh(t) andj(t) are of the form2d1 and the corresponding quadratic cover of the curveC is unramified. When the factors both have odd degree, these divisors are of the form2d1+(P )+(P ′)and the quadratic cover is ramified at the pointsP andP ′.

14

Since there might be no nontrivial factorizations ofg(x) overK, there might be no nontrivialK-rational points ofResL/Kµ2/µ2 and hence no quadratic field extensions ofK(C) contained inK(M).However, overL we have the factorizationg(x) = (x−β)j(x) = h(x)j(x), so the algebraL(M) mustcontain a square-rootu of some constant multiple of the functionh(t) = (t−β). (The need to adjoin asquare-root oft−β whose divisor has the form2d1+(P )+(P ′) is the main reason for the appearanceof the generalized JacobianJm (cf. [27, Footnote 2]).) Writeu2 = α(t− β) with α in L× and take thenorm toK(M) to obtain the equationN(u)2 = N(α)g(t). Then the two rational functionsN(u) andwin K(M)× have the same divisor, so they are equal up to a constant factor in K×. Writing bN(u) = wwith b in K×, we findw2 = b2N(u)2 = b2N(α)g(t). However,w2 = f0g(t), sof0 = b2N(α) is in thesubgroupK×2N(L×) of K×. This completes the proof of Theorem23.

In fact, the obstruction classes for the eight conditions inTheorem23 are all equal. Moreprecisely, the obstruction class for conditionsa, b, c is the class off0 in K×/K×2(NL×). This groupcan be viewed as a subgroup ofH2(K, Jm[2]) via

coker(N : H1(K,ResL/Kµ2) → H1(K,µ2)

)−→ H2(K, (ResL/Kµ2)N=1).

We denote the image off0 in H2(K, Jm[2]) by [f0]. This is the cohomological classdf whose non-vanishing obstructs the existence of rational orbits with invariantf for (all pure inner forms of)SLn;see [5, §2.4 and Theorem 9]).

The obstruction class for conditionsd, e is the classδ[J1m] in H2(K, Jm[2]) whereδ is the con-

necting homomorphismH1(K, Jm) → H2(K, Jm[2]) arising from the exact sequence1 → Jm[2] →Jm

2−→ Jm → 1.The obstruction class for conditionsf, g, h comes from Galois descent. There is an unramified

two-coverπ : J1m → J1

m overKs obtained by identifyingJ1m with Jm using aKs point of J1

m, thentaking the multiplication-by-2 map onJm. The descent obstruction of this cover toK is the image inH2(K, Jm[2]) of the class[π : J1

m → J1m] under the following map from the Hochschild-Serre spectral

sequence:H0

(K,H1(C ×K Ks − {P, P ′}, Jm[2])

)−→ H2(K, Jm[2]).

This obstruction class equalsδ[J1m] for formal reasons (cf. [32, Lemma 2.4.5]). We have the following

strengthening of Theorem23.

Theorem 24. Let f(x, y) = f0x2g+2 + · · · + f2g+2y

2g+2 be a binary form of degree2g + 2 overKwith f0 and∆(f) both nonzero inK. LetC be the smooth hyperelliptic curve of genusg with equationz2 = f(x, y) and letJm denote its generalized Jacobian. Then the obstruction classes for conditionsathroughh in Theorem23 are all equal inH2(K, Jm[2]), i.e.,[f0] = δ[J1

m].

Proof. Consider the following commutative diagram:

1 // Jm[2] //� _

��

Jm

2//

� _

��

Jm//

=

��

1

1 // Jm ⊔ J1m[2]

//

����

Jm ⊔ J1m

2//

����

Jm// 1

µ2=

// µ2 .

15

Here the mapJm ⊔ J1m

2−→ Jm is given by[D] 7→ 2[D] − deg([D]) · d. Theorem24 follows from thefollowing two results.

Proposition 25. For any a ∈ K, there exists a class[J1/2a ] ∈ H1(K, Jm ⊔ J1

m) such that2[J1/2a ] =

[J1m] in H1(K, Jm) and such that the image of[J1/2

a ] in H1(K,µ2) = K×/K×2 equalsf0g(a) =f0NL/K(a− β).

Lemma 26. Let 1 → A1 → B1 → C → 1 and1 → A2 → B2 → C → 1 be central extensions ofalgebraic groups overK such that the following diagram commutes:

1 // A1//

� _

��

B1//

� _

��

C //

=

��

1

1 // A2//

����

B2//

����

C // 1

D=

// D

Then the following diagram commutes up to sign:

H1(K,B2) //

��

H1(K,C)

��

H1(K,D) // H2(K,A1)

Lemma26 follows from a direct cocycle computation. We now prove Proposition 25. Fixa ∈ K, let Pa ∈ C(K(

√α)) be a point withx-coordinatea whereα = f0g(a), and letP ′

a be theconjugate ofPa under the hyperelliptic involution. The class[J1

m] ∈ H1(K, Jm) is given by the 1-cocycleσ 7→ σ(P ′

a)− (P ′a). In other words,

[J1m]σ =

{0 if σ(

√α) =

√α

(Pa)− (P ′a) if σ(

√α) = −√

α.

Let [J1/2a ] denote the following 1-cochain with values in(Jm ⊔ J1

m)(Ks) :

[J1/2a ]σ =

{0 if σ(

√α) =

√α

(Pa) if σ(√α) = −√

α.

Since2(Pa) − d = 2(Pa) − ((Pa) + (P ′a)) = (Pa) − (P ′

a), we see that2[J1/2a ]σ = [J1

m]σ for allσ ∈ Gal(Ks/K). Moreover, a direct computation shows that[J

1/2a ] is a 1-cocycle and its image in

H1(K,µ2) is the 1-cocycleσ 7→ σ√α/

√α. This completes the proof of Proposition25, and thus

Theorem24.

6 Soluble orbits

In the previous section, we gave necessary and sufficient conditions for the existence of pencils ofbilinear forms(A,B) ∈ K2 ⊗ Sym2K

n having a given invariant binary form. In this section, we

16

considersolublepencils of bilinear forms(A,B), i.e., those for which the associated Fano varietyF = F (A,B) has aK-rational point.

Fix a binary formf(x, y) of degreen = 2g + 2 overK with ∆(f) andf0 nonzero inK, andlet C be the smooth hyperelliptic curve with equationz2 = f(x, y). Suppose that(A,B) is a genericpencil of bilinear forms onW overK with invariant binary formf(x, y) = disc(Ax − By) and let(A′, B′) be the regular pencil of bilinear forms onW ⊕ K2 having invariant binary formf(x, y)y2

constructed above. We say that(A,B) lies in asolubleorbit for SLn if the Fano varietyFm of the baselocus of(A′, B′) has aK-rational point. Similarly, we say that the pencil(A,B) lies in asolubleorbitfor SLn /µ2 if the Fano varietyF of the base locus of(A,B) has aK-rational point. In this section,we classify the soluble orbits forSLn andSLn /µ2.

Since we have constructed an unramified two-coverFm → J1m, a necessary condition for the

existence of soluble orbits forSLn is thatJ1m(K) is nonempty. (Recall that everyK-rational point inJ1

m

is represented by an odd degreeK-rational divisor on the curveC that is relatively prime to the divisorm = (P ) + (P ′).) In this case, the groupJm(K) acts simply transitively on the set of pointsJ1

m(K).

Theorem 27. Let f(x, y) be a binary form of degreen = 2g + 2 overK with ∆(f) andf0 nonzeroin K. Then soluble orbits for the action ofSLn(K) onK2 ⊗ Sym2K

n having invariant binary formf(x, y) exist if and only if there is aK-rational divisor of odd degree on the curveC : z2 = f(x, y).In that case, they are in bijection with the elements ofJ1

m(K)/2Jm(K).

Proof. Suppose first that soluble orbits with invariant binary formf(x, y) exist. Let (A,B) be inK2⊗Sym2K

n with invariant binary formf(x, y) such that the Fano varietyF (A,B)m of the associatedregular pencil(A′, B′) in W ⊕K2 has a rational point. The stabilizer of(A,B) in SLn is isomorphicto Jm[2] by Corollary18 and Proposition21. SinceH1(K, SLn) = 1, we see that the rational orbitswith invariant binary formf(x, y) are in bijection with the elements in the Galois cohomology groupH1(K, Jm[2]). This bijection depends on the choice of the initial solubleorbit (A,B) which maps tothe trivial class inH1(K, Jm[2]).

Explicitly, suppose the pair(A1, B1) ∈ K2 ⊗ Sym2Kn has invariant binary formf(x, y) and

corresponds to the classc ∈ H1(K, Jm[2]). Let (A′1, B

′1) be the associated regular pencil with Fano

varietyF (A1, B1)m. Then as elements ofH1(K, Jm)[4], we have, up to sign1, the formula

[F (A1, B1)m] = [F (A,B)m] + c. (8)

Hence we see thatF (A1, B1)m is the trivial torsor ofJm if and only if c is in the Kummer image ofJm(K)/2Jm(K). Therefore, the set of soluble orbits with invariant binary form f(x, y) is in bijectionwith the elements of the quotient groupJm(K)/2Jm(K), once the fixed soluble orbit(A,B) has beenchosen.

On the other hand, ifx ∈ F (A,B)m(K) is any rational point, then the sumx + x = 2x in thealgebraic groupXm in (7) gives a rational point ofJ1

m well-defined up to hyperelliptic conjugation (cf.Footnote1). HenceJ1

m(K) is nonempty. Therefore, the setJ1m(K)/2Jm(K) is also in bijection with

Jm(K)/2Jm(K).To complete the proof of Theorem27, it remains to show that ifJ1

m(K) is nonempty, thensoluble orbits with invariant binary formf(x, y) exist. We show this first in the special case where thecurveCm has a non-singularK-rational pointQ = (x0, 1, z0). Let L = K[x]/f(x, 1) denote as usual

1The ambiguity of sign comes from the fact that we cannot distinguish between[Fm] and−[Fm] in H1(K, Jm). In otherwords, we cannot distinguish the two copies ofFm in the groupXm defined in (7).

17

the etale algebra of rankn associated tof(x, y) and letβ denote the image ofx in L. The rationalorbit corresponding to(Q) is given by the equivalence class of a pair(α, s) (see Corollary18) whereα = (x− T )(Q). Here “x− T ” is the descent map introduced by Cassels [12]:

J1m(K)/2Jm(K) → (L×/L×2)N≡f0.

We note thats is not uniquely determined whenWm[2] is a nontrivial torsor ofJm[2]. In this case, thefibers of the abovex − T map also have size2. From the definition of the bijection between the setof rational orbits and the set of equivalence classes of pairs (α, s) in Section2, we see that if the orbitcorresponding to a pair(α, s) is soluble, then the orbit corresponding to any pair(α′, s′) with α′ = αis also soluble.

Consider the two binary forms(A′, B′) onL⊕K2 given by

〈(λ, a, b), (µ, a′, b′)〉A′ = (coefficient ofβn−1 in αλµ) + aa′,

〈(λ, a, b), (µ, a′, b′)〉B′ = (coefficient ofβn−1 in αβλµ) + ab′ + a′b.

We show that forα = (x − T )(Q), there is a rational(g + 1)-planex′ isotropic with respect to bothbilinar forms.

Whenz0 6= 0, we haveα = (x− T )(Q) = x0 − β. Seth1(t) = (f(t, 1)− f(x0, 1))/(t− x0).Then

x′ = Span{(1, 0, 0), (β, 0, 0), . . . , (βg−1, 0, 0), (βg, 1,−1

2(x0 +

f1f0))}

is isotropic with respect to both bilinear forms. To check this, we note that(x0 − β)β2g+1 has leadingcoefficientx0 + f1/f0.

Whenz0 = 0, we seth0(t) = t − x0 andh1(t) = f(t, 1)/(t− x0). Thenα = (x − T )(Q) =h1(β)− h0(β) and the following(g + 1)-plane is isotropic with respect to both bilinear forms:

x′ = Span{(h1(β)−h1(x0), 0, 0), (β−x0, 0, 0), . . . , ((β−x0)g−1, 0, 0), ((β−x0)

g, 1,−1

2((2g+1)x0+

f1f0))}.

This can be checked by a simple calculation noting thath1(β)h1(x0)(h1(β)− h1(x0)) = h0(β)h1(x0).Before moving on to the general case, we make an important observation. Using this pencil

with α = (x − T )(Q) as the base point, we obtain a bijection between the set of therational orbitswith invariant binary formf(x, y) andH1(K, Jm[2]) as described above. If(A1, B1) is an element ofK2 ⊗ Sym2K

n with invariant binary formf(x, y) and such that its associatedα equals to(x− T )(D)for someD ∈ J1

m(K)/2Jm(K), then the orbit of(A1, B1) corresponds to the classD−(Q) orD−(Q′)in Jm(K)/2Jm(K) whereQ′ denote the hyperelliptic conjugate ofQ. Hence the orbit of(A1, B1) issoluble.

We now treat the general case, assuming only thatJ1m(K) is nonempty. NowCm has a non-

singular pointQ defined over some extensionK ′ of K of odd degreek. Let D ∈ J1m(K) denote the

divisor class of degree1 obtained by taking the sum of the conjugates ofQ and subtractingk−12

timesthe hyperelliptic class. We claim that the orbits correspond to D are soluble thereby completing theproof of Theorem27. Let (A,B) is an element ofK2 ⊗ Sym2K

n with invariant binary formf(x, y)and such that its associatedα equals to(x − T )(D) and letF (A,B)m denote the Fano variety ofthe associated regular pencil. SinceC has a point overK ′, we have seen that theK ′-rational orbits(α, s) with α = (x − T )(Q) and hence withα = (x − T )(D) are soluble overK ′. In other words,F (A,B)m(K

′) is nonempty. Thus, as an element ofH1(K, Jm), the class ofF (A,B)m becomes trivial

18

when restricted toH1(K ′, Jm). A standard argument using the corestriction map shows thatthis classis killed by the degreek of K ′ overK. SinceF (A,B)m is a torsor ofJm of order dividing4 andk isodd, we see thatF (A,B)m must be the trivial torsor.

The same argument also classifies the soluble orbits forSLn /µ2, provided thatC has aK-rational divisor of odd degree. The descent map “x− T ” gives a map of sets

J1(K)/2J(K) → (L×/L×2K×)N≡f0

and is either 2-to-1 or injective (depending on the triviality of the classW [2] in H1(K, J [2])). To seethat there are no soluble orbits whenC has no divisors of odd degree, we use the exact sequence ofcommutative algebraic groups [35, Corollary 3.22]:

1 → T → Xm → X → 1.

If J1m(K) is empty but bothJ1(K) andF (K) are nonempty, then the quotient ofX(K) by the image

of Xm(K) maps onto the component groupZ/4Z of X. On the other hand, this quotient injects intoH1(K, T ), which has exponent2, a contradiction. Hence we have proved the following:

Theorem 28. Letf(x, y) be a binary form of degreen = 2g + 2 overK with ∆(f) andf0 nonzero inK. Then soluble orbits for the action of(SLn /µ2)(K) with invariant binary formf(x, y) exist if andonly if there is aK-rational divisor of odd degree on the curveC : z2 = f(x, y). In that case, they arein bijection with the cosets ofJ1(K)/2J(K) and the groupJ(K)/2J(K) acts simply transitively onthe set of soluble orbits.

7 Finite and archimedean local fields

In this section we consider the orbits for the action of(SLn /µ2)(K) onK2 ⊗ Sym2Kn when the base

field K is a finite field or an archimedean local field. In particular, we compute the number of theseorbits with a fixed invariant binary formf(x, y).

7.1 Finite fields

Let K be a finite field of odd cardinalityq. Let f(x, y) be a binary form of even degreen overK withnonzero first coefficientf0 and nonzero discriminant∆ and writef(x, 1) = f0g(x). We factor

g(x) =m∏

i=1

gi(x)

wheregi(x) has degreedi. ThenL is the product ofm finite fieldsLi of cardinalityqdi. Since finitefields have unique extensions of any degree, we see that either one ofLi has odd degree ofK or all oftheLi contain the unique quadratic extension ofK. Therefore,f(x, y) always has either an odd or aneven factorization overK.

Since the norm mapL× → K× is surjective,f0 is always a norm. Therefore, by Corollary19,the number of(SLn /µ2)(K)-orbits with binary formf(x, y) is 2m if all Li have even degree andn iscongruent to0 (mod4); 2m−1 if all Li have even degree andn is congruent to2 (mod4); and2m−2 if

19

someLi has odd degree overK. The size of the stabilizer equals to the number of even factorization off(x, y). Hence each stabilizer has size2m if all Li have even degree andn is congruent to0 (mod4);2m−1 if all Li have even degree andn is congruent to2 (mod4); and2m−2 if someLi has odd degreeoverK. Therefore, the number of pairs(A,B) ∈ K2 ⊗ Sym2K

n with invariant binary formf(x, y) is|(SLn /µ2)(K)| = |SLn(K)|. This agrees with [1, §3.3]. Note for the purpose of application in Section12, the main ingredients that we need are the number of orbits and the fact that all the orbits with thesame invariant binary form have the same number of elements.

By Lang’s theorem, we haveH1(K, J) = H1(K, Jm) = 0. Then the Fano varietiesF andFm

associated to an orbit always have aK-rational point, and every orbit is soluble.

7.2 R and C

We now classify the orbits overK = R andK = C. Let f(x, y) be a binary form of degreen overK with nonzero first coefficientf0 and nonzero discriminant∆, and writef(x, 1) = f0g(x). OverCthere is a single orbit with binary formf(x, y).

In the case whenK = R, we factor

g(x) =

r1∏

i=1

gi(x)

r2∏

j=1

hj(x)

where eachgi(x) has degree one and eachhj(x) has degree two. Then the algebraL is the productof r1 copies ofR andr2 copies ofC, with r1 + 2r2 = n. Note thatr1 has the same parity asn. Thequotient groupR×/R×2N(L×) is trivial unlessr1 = 0, in which case it has order2. Just as in the caseof finite fields,f(x, y) always has either an odd or an even factorization overK.

If the form f is negative definite, then there are no orbits having invariant binary formf(x, y).Indeed, in this caser1 = 0 and the leading coefficientf0 is negative. In this case, the hyperellipticcurveC with equationz2 = f(x, y) has no real points, and the mapPicC/R(R) → Br(C/R) = Z/2Zis surjective. The real divisor classes that are not represented by real divisors have degrees congruentto g − 1 modulo2. Wheng is even, the JacobianJ(R) is connected and every principal homogeneousspace forJ is trivial. In particular,J1 has real points (which are not represented by real divisors of odddegree). Wheng is odd, the real points of the JacobianJ(R) have two connected components, andJ1

is the unique nontrivial principal homogeneous space forJ . The points in the connected component ofJ(R) are the real divisor classes of degree zero that are represented by real divisors.

If f is not negative definite, then the elementf0 is a norm fromL× toR×. Hence rational orbitsexist. Whenr1 = 0, sof is positive definite, there are two orbits ifn is congruent to0 (mod4) andthere is only one orbit ifn is congruent to2 (mod4). In both cases, the real points of the hyperellipticcurveC(R) and its JacobianJ(R) are both connected and the orbits are all soluble. Ifr1 > 0 and theform f takes both signs, then the number of orbits is2r1−2. The hyperelliptic curveC with equationz2 = f(x, y) hasm = r1/2 connected components in its real locus, andJ(R) has2m−1 connectedcomponents. Since the subgroup2J(R) is equal to the connected component ofJ(R), 2m−1 of theseorbits are soluble.

The computation for the sizes of the stabilizers is similar to the finite field case. Ifr1 = 0, thenthe size of the stabilizer is2n/2 if n is congruent to0 (mod4) and is2n/2−1 if n is congruent to2 (mod4). If r1 > 0, then the size of the stabilizer is2n/2+m−2 wherem = r1/2.

20

8 Global fields and locally soluble orbits

In this section, we assume thatK is a global field of characteristic not 2. Letf(x, y) be a binary formof degreen = 2g + 2 overK with nonzero discriminant. LetC : z2 = f(x, y) denote the associatedhyperelliptic curve. Recall that an element(A,B) of K2 ⊗ Sym2K

n (or its (SLn /µ2)(K)-orbit) withinvariant binary formf(x, y) is locally solubleif the associated Fano varietyF (A,B) overK haspoints over every completionKv. We wish to determine when rational orbits and locally soluble orbitsfor the action of(SLn /µ2)(K) on K2 ⊗ Sym2K

n exist. Theorem23 gives a list of necessary andsufficient conditions for the existence of rational orbits over general fields. In this section, we assumethat there exists a locally soluble two-cover ofJ1 overK and thatDiv1(C) is locally soluble. The mainresult is that these two conditions are sufficient for the existence of a global orbit and a locally solubleorbit.

Recall the torsorW [2] of J [2] which consists of pointsP ∈ J1 such that2P = d whered isthe hyperelliptic class ofC. The class ofW [2] in H1(K, J [2]) maps to the class ofJ1 in H1(K, J)[2].SinceJ1(Kv) is nonempty for allv, we see that a prioriW [2] lies in the2-Selmer subgroupSel2(J/K)of H1(K, J [2]). Letπ : F0 → J1 denote a locally soluble two-cover ofJ1 overK. LetF0[4] denote thetorsor ofJ [4] consisting of pointsx ∈ F0 such thatπ(x) ∈ W [2]. Then the class ofW [2] is twice theclass ofF0[4] in H1(K, J [4]). SinceF0(Kv) is nonempty for allv, the class ofF0[4] is in the4-SelmersubgroupSel4(J/K) of H1(K, J [4]).

Conversely ifC is any hyperelliptic curve overK with locally solubleDiv1(C). If W [2] isdivisible by 2 inSel4(J/K), then a locally soluble two-cover ofJ1 overK exists. Indeed, supposeW [2] = 2F [4] for someF [4] ∈ Sel4(J/K). Let F denote the principle homogeneous space ofJwhose class inH1(K, J) is the image ofF [4] in H1(K, J)[4]. Then2F = [J1] and hence there existsa mapF → J1 realizingF as a two-cover ofJ1.

Theorem 29. SupposeC : z2 = f(x, y) is a hyperelliptic curve over a global fieldK of characteristicnot 2 such thatDiv1(C)(Kν) 6= ∅ for all placesν of K and such thatJ1 admits a locally solubletwo-cover overK (equivalently,W [2] is divisible by2 in Sel4(J/K)). Then orbits for the action of(SLn /µ2)(K) onK2 ⊗ Sym2K

n with invariant binary formf(x, y) exist.

Proof. LetT = (ResK ′/K Gm)N=1 be the kernel ofJm → J as in (5) whereK ′ = K(√f0). Letφ, i2, i4

be defined by the following diagram arising as part of the longexact sequence in Galois cohomology:

H1(K, T [4])i4

//

��

H1(K, Jm[4])φ

//

��

H1(K, J [4])

��

H1(K, T [2])i2

// H1(K, Jm[2]) // H1(K, J [2])

Let F [4] be a class inSel4(J/K) such that2F [4] = W [2]. By Theorem23, it suffices to show that theclass of the torsorWm[2] of Jm[2] is in the image of the mapH1(K, Jm[4]) → H1(K, Jm[2]) inducedby multiplication by 2. We break up the proof of Proposition29 into three steps.

Step 1: There exists a classFm[4] in H1(K, Jm[4]) such thatφ(Fm[4]) = F [4].For any placeν of K, let φν : H1(Kν , Jm[4]) → H1(Kν , J [4]) denote theν-adic restriction of

φ, and letF [4]ν denote theν-adic restriction ofF [4]. SinceF [4] is in Sel4(J/K), the restrictionF [4]νcomes from aKν-rational point on the Jacobian for every placeν of K. SinceDiv1 is locally soluble,everyKν-rational divisor class is representable by aKν-rational divisor (Proposition20). Therefore,

21

F [4]ν is in the image ofφν for every placeν of K. To show thatF [4] is in the image of the globalφ,note that the cokernel ofφ maps injectively toH2(K, T [4]). Hence it suffices to show thatH2(K, T [4])satisfies the local-global principle. SinceT [4] = (ResK ′/K µ4)N=1, we have the exact sequence

1 → K×/N(K ′×)K×4 → H2(K, T [4]) → Br(K ′)[4]N=1 → 1.

Now Br(K ′) satisfies the local-global principle with respect to the places ofK, andK×/N(K ′×)K×4 =K×/N(K ′×) also satisfies the local-global principle sinceK ′/K is cyclic.

Step 2: The class2Fm[4] in H1(K, Jm[2]) is in the 2-Selmer group. That is, its image inH1(K, Jm) islocally trivial.

LetFm denote the image ofFm[4] inH1(K, Jm). Fix any placeν ofK and denote byFm,ν andFν

theν-adic restrictions ofFm andF , respectively. SinceFm,ν maps toFν under the mapH1(Kν , Jm) →H1(Kν , J) andF is locally soluble, we see thatFm,ν is in the image ofH1(Kν , T ) → H1(Kν , Jm).Therefore,2Fm,ν is the trivial torsor sinceH1(Kν , T ) has exponent2.

Step 3: There existsc′ ∈ H1(K, T [4]) such that2(Fm[4] + i4(c′)) = Wm[2]. This will complete the

proof of Theorem29.Sinceφ(Fm[4]) = F [4], we see thatWm[2] − 2Fm[4] = i2(c) for somec in H1(K, T [2]). By

Step 2, the image ofc under the composite mapH1(K, T [2]) → H1(K, Jm[2]) → H1(K, Jm) is locallytrivial. SinceDiv1 is locally soluble, the mapH1(k′, T ) → H1(k′, Jm) is injective fork′ = K or Kν

for any placeν of K. Therefore, the image ofc under the mapH1(K, T [2]) → H1(K, T ) is locallytrivial. SinceH1(K, T ) = K×/N(K ′×) has the local-global principle, the image ofc in H1(K, T ) isin fact 0. Therefore,c is in the image of the Kummer mapT (K)/2T (K) → H1(K, T [2]). Hence it isalso in the image ofH1(K, T [4]) → H1(K, T [2]) induced by multiplication by 2. Writec = 2c′ forsomec′ ∈ H1(K, T [4]). Then2(Fm[4] + i4(c

′)) = Wm[2].

If we assume thatDiv1(C) is locally soluble, then the existence of a locally soluble two-coverof J1 is in fact equivalent to the existence of a locally soluble orbit for the action of(SLn /µ2)(K) onK2⊗Sym2K

n. We will see thatSel2(J/K) acts simply transitively on the set of locally soluble orbits.Therefore, we see that every locally soluble two-cover ofJ1 is isomorphic to the Fano varietyF (A,B)associated to the pencil of quadrics determined by some(A,B) ∈ K2 ⊗ Sym2K

n. This also provesTheorem13.

Theorem 30. SupposeC : z2 = f(x, y) is a hyperelliptic curve over a global fieldK of characteristicnot 2 such thatDiv1(C)(Kν) 6= ∅ for all placesν of K. Then locally soluble orbits for the action of(SLn /µ2)(K) onK2⊗Sym2K

n exist with invariant binary formf(x, y) if and only ifW [2] is divisibleby 2 in Sel4(J/K), or equivalently thatJ1 admits a locally soluble two-cover overK. Furthermore,when these conditions are satisfied, the groupSel2(J/K) acts simply transitively on the set of locallysoluble orbits and this set is finite.

Before proving Theorem30, we note that the notion of locally soluble orbit is a tricky one.There could exist an integral binary quartic formf(x, y) that has locally soluble orbits but no solubleorbits overQ. For a specific example (suggested by John Cremona, see also [32, §8.1]), consider theelliptic curveE defined by the equationy2 = x3 − 1221. This curve has trivial Mordell-Weil groupE(Q) = 0 and Tate-Shafarevich group isomorphic to(Z/4Z)2. The binary quartic formf(x, y) =3x4 − 12x3y + 11xy3 − 11y4 of discriminant∆ = −40252707 = −35112372 corresponds to a classb in the Tate-Shafarevich group ofE that is divisible by 2. Any of the elementsc of order4 in the

22

Tate-Shafarevich group with2c = b gives a locally soluble orbit with invariant binary formf(x, y).The hyperelliptic curvez2 = f(x, y) is locally soluble but has no global points; hence, by Theorem 28,there is no soluble orbit having invariant binary formf(x, y).

There are also examples where rational orbits exist but there are no locally soluble orbits. Forexample, consider the binary quartic formf(x, y) = −1x4 + 2x3y + 104x2y2 − 104xy3 − 2764y4 ofdiscriminant∆ = −28571. The associated quartic fieldL has discriminant−24571 = −9136 and ringof integersZ[θ], whereθ is a root of the polynomialF (t) = t4 − 2t2 + 2t − 3. SinceF (1) = −2,F (0) = −3, andF (−1) = −6, the elementθ3− θ in L× has norm−62 ≡ −1 = f0. So there are orbitsoverQ with this invariant quartic form. On the other hand, the hyperelliptic curveC : z2 = f(x, y)of genus one is a principal homogeneous space of order2 for its JacobianE, which is an elliptic curvewith equationy2 + xy = x3 − x2 − 929x − 10595 and prime conductor571. This curve has trivialMordell-Weil groupE(Q) = 0 and Tate-Shafarevich group isomorphic to(Z/2Z)2. Hence Sel2(E/Q)and Sel4(E/Q) are both isomorphic to(Z/2Z)2. The curveC represents one of the nontrivial locallytrivial principal homogeneous spaces forE. Since its class is not in the image of multiplication by2from Sel4(E/Q), there are no locally soluble orbits. (Thanks to John Cremona and Noam Elkies forhelp with computation in this example.)

Proof of Theorem 30: Suppose locally soluble orbits with invariant binary formf(x, y) exist. Weprove first thatSel2(J/K) acts simply transitively on the set of locally soluble orbits with invariantbinary formf(x, y). Indeed, suppose that(A,B) is a rational pencil with Fano varietyF (A,B) andinvariant binary formf(x, y). Any other rational pencil(A1, B1) with the same binary form corre-sponds to a class[c] in H1(K, J [2]) that is in the kernel of the composite mapγ : H1(K, J [2]) →H1(K, SLn /µ2) → H2(K,µ2) ([3, Proposition 1]). The mapγ is cup product with the classW [2] ∈H1(K, J [2]) ([27, Proposition 10.3]). LetF (A1, B1) denote the Fano variety associated to the pencil(A1, B1). Then one has, up to sign (cf. Footnote1),

[F (A1, B1)] = [F (A,B)] + [c] (9)

in H1(K, J)[4]. Since the subgroupJ(Kv)/2J(Kv) of H1(Kv, J [2]) maps to the trivial class inH1(Kv, SLn /µ2) for all placesv, the Hasse principle for the cohomology of the groupSLn /µ2 showsthat the subgroupSel2(J/K) of H1(K, J [2]) also lies inker γ. It is then clear from (9) that if (A,B)is locally soluble, then[c] ∈ Sel2(J/K) if and only if (A′, B′) is locally soluble. HenceSel2(J/K)acts simply transitively on the set of locally soluble orbits with invariant binary formf(x, y). Sincethe2-Selmer group is finite, the set of locally soluble orbits with invariant binary formf(x, y) is alsofinite. Moreover, if(A,B) is locally soluble, thenF (A,B) gives a locally soluble two-cover ofJ1

overK.We now consider the sufficiency of the above two conditions for the existence of locally soluble

orbits. LetF denote the Fano variety corresponding to one rational orbitwith invariant binary formf(x, y). The existence of this rational orbit follows from Theorem29. LetF [4] = F [4]0 denote the liftof F to a torsor ofJ [4] consisting of elementsx ∈ F such thatx+x+x+x = 0 in the groupX of fourcomponents defined in Theorem22. We need to show that there exists a classc ∈ H1(K, J [2]) suchthatc ∪W [2] = 0 andF [4] + c ∈ Sel4(J/K). Let d0 be a class inSel4(J/K) such thatW [2] = 2d0. Itsuffices to show that

(F [4]− d0) ∪W [2] = 0. (10)

For ease of notation, we denote the above cup product bye2(F [4] − d0,W [2]) since the cup productis induced from the Weil pairinge2 onJ [2]. Sinced0 ∈ Sel4(J/K) is isotropic with respect toe4, we

23

havee2(F [4]− d0,W [2]) = e4(F [4]− d0, d0) = e4(F [4], d0).

Fix a placeν and denote byF [4]ν , d0,ν, e4,ν theν-adic restrictions. Pick anyDν ∈ J1(Kν). SinceFarises from a pencil of quadrics, we define

F [2]Dν = {x ∈ F : x+ x = Dν}.

The image of this torsor ofJ [2] in H1(Kν , J [4]) is the following torsor:

F [4]2Dν−d = {x ∈ F : x+ x+ x+ x = 2Dν − d},

whered denotes the hyperelliptic class as before. Therefore, as elements ofH1(Kν , J [4]), we have

F [4]ν − F [2]Dν = δ4,ν(2Dν − d),

whereδ4,ν is the Kummer mapJ(Kν)/4J(Kν) → H1(Kν , J [4]). Sinced0 ∈ Sel4(K, J), we see thatd0,ν is in the image ofδ4,ν . SinceJ(Kν)/4J(Kν) is isotropic with respect toe4,ν , we have

e4,ν(F [4]ν, d0,ν) = e4,ν(F [2]Dν , d0,ν) = e2,ν(F [2]Dν ,W [2]ν). (11)

Choosing a differentDν ∈ J1(Kν) changesF [2]Dν by an element ofJ(Kν)/2J(Kν). AsJ(Kν)/2J(Kν)is isotropic with respect toe2, the last term in (11) does not depend on the choice ofDν . Theorem30then follows from the following general lemma.

Lemma 31. SupposeK is any local field of characteristic not 2. Letf(x, y) be a binary form of degree2g+2 with nonzero discriminant such that the associated hyperelliptic curveC : z2 = f(x, y) satisfiesDiv1(C)(K) 6= ∅. Suppose there is a rational orbit for the action of(SLn /µ2)(K) onK2⊗ Sym2K

n

with invariant binary formf(x, y) and letF denote the associated Fano variety. Then

e2(F [2],W [2]) = 0, (12)

whereF [2] denotes any lift ofF to a torsor ofJ [2] using a point ofJ1(K).

Proof. The first key point is that if (12) holds for one rational orbit, then it holds for any rationalorbitwith the same invariant binary form. Indeed, ifF ′ denotes the torsor ofJ coming from a different orbit,thenF ′ − F ∈ ker γ, whereγ : H1(K, J [2]) → H2(K,µ2) is cup product withW [2]. In other words,e2(F

′ − F,W [2]) = 0. Hencee2(F [2],W [2]) = e2(F′[2],W [2]).

The second key point is that sinceDiv1(C)(K) 6= ∅, there exists a soluble orbit by Theorem28.LetF denote the corresponding torsor arising from this soluble pencil. ThenF [2] ∈ J(K)/2J(K) andhencee2(F [2],W [2]) = 0.

This completes the proof of Theorem30. �

We conclude by remarking that the natural generalization ofthe fake2-Selmer setSel2, fake(C)of C ([10]), namely thefake2-Selmer setSel2, fake(J1) of J1, is in natural bijection with the set oflocally soluble orbits for the groupSL±

n /µ2, whereSL±n denotes as before the subgroup of elements of

GLn with determinant±1. Using the groupSLn instead ofSL±n allows us to “unfake” this fake Selmer

set (cf. [34]).

24

9 Existence of integral orbits

The purpose of this section is to prove Theorem14. More precisely, we prove:

Theorem 32. Assume thatn ≥ 2 is even. Letf(x, y) be a binary form of degreen = 2g + 2 withcoefficients in16nZ such that the hyperelliptic curveC : z2 = f(x, y) has locally solubleDiv1. Thenevery locally soluble orbit for the action of(SLn /µ2)(Q) onQ2⊗Sym2Q

n with invariant binary formf(x, y) has an integral representative, i.e., a representative inZ2 ⊗ Sym2 Z

n.

By Theorem16 with D = Z andZp, it suffices to find a representative overZp for everysoluble orbit overQp with f(x, y) ∈ Zp[x, y]. We begin by recalling from [1, §2] the construction ofan integral orbit associated to a rational point onC, or ap-adically integral orbit associated to ap-adicpoint onC. For this we recall some of the notations in Section2. Without loss of generality, we mayassumef0 6= 0. (By our convention,C being a hyperelliptic curve is equivalent to∆(f) 6= 0.) Writef(x, 1) = f0g(x) and letL = Qp[x]/g(x) be the corresponding etale algebra of rankn overQp. Fork = 1, 2, . . . , n− 1, there are integral elements

ζk = f0θk + f1θ

k−1 + · · ·+ fk−1θ

inL. LetRf be the freeZp-submodule ofL havingZp-basis{1, ζ1, ζ2, . . . , ζn−1}. Fork = 0, 1, . . . , n− 1,let If(k) be the freeZp-submodule ofL with basis{1, θ, θ2, . . . , θk, ζk+1, . . . , ζn−1}. By Theorem16,an integral orbit is an equivalence class of triples(I, α, s) whereI is an ideal ofRf , α ∈ L×, s ∈ K×

such thatI2 ⊂ αIf(n − 3), N(I) = sZp andN(α) = s2fn−30 . The rational orbit is given by the

equivalence class of the pair(α, s).Via a scaling and change of variable, we may assume that we have an integral pointP = (0, 1, c)

on the curvez2 = f(x, y), so that the coefficientfn = c2 is a square. Then setα = θ, and note that

θIf(n− 3) = SpanZp{c2, θ, θ2, . . . , θn−2, f0θ

n−1}. (13)

Let I = SpanZp{c, θ, θ2, . . . , θ(n−2)/2, ζn/2, . . . , ζn−1}. Then it is easy to check thatI is an ideal ofRf

and thatI2 ⊆ αIf(n− 3) and

N(I)2 = N(θ)N(If (n− 3)) = [c/f(n−2)/20 ]2Zp.

Let s = ±c/f(n−2)/20 be such that(α, s) corresponds to the rational orbit determined byP . The triple

(I, α, s) gives an integral orbit representing the soluble orbit given byP in J1(Qp)/2J(Qp). We notethat this association of an integral orbit to aQ-rational point, and the paucity of integral orbits, is thekey to the arguments of [1] which show that rational points are rare.

Given one suchf(x, y) = f0xn + · · · + fny

n with coefficients in16nZ, we put f(x, y) =

f0xn + · · ·+ fny

n wherefi = fi/16n−i. If (A,B) ∈ Z2 ⊗ Sym2 Z

n has invariant binary formf(x, y),then (4A,B) has invariant binary formf(x, y). Therefore, Theorem32 follows from the followingproposition where the assumption on the coefficients is asymmetrical in contrast to Theorem32.

Proposition 33. Assume thatn ≥ 2 is even. Letf(x, y) = f0xn + · · · + fny

n be a binary form ofdegreen = 2g + 2 satisfyingf0 6= 0 and24i | fi/f0 for i = 1, . . . , n, such that the hyperelliptic curveC : z2 = f(x, y) has locally solubleDiv1. Then every locally soluble rational orbit for the action of(SLn /µ2)(Q) onQ2 ⊗ Sym2Q

n with invariant binary formf(x, y) has an integral representative.

25

Proof. We work overZp and give an explicit construction of the idealI, in a manner similar to the one-point case shown above (cf. [1, §2]) and the corresponding statements in [4, Proposition 8.2] and [30,Proposition 2.9]. We begin by noting that the condition24i | fi/f0, which is nontrivial only whenp = 2,implies that ifa ∈ Qp is non-integral, thena−θ ∈ L× in fact lies inL×2Q×

p . SinceDiv1(C)(Qp) 6= ∅,everyQp-rational divisor class can be represented by a rational divisor by Proposition20. Therefore,by [36, Lemma 3.8], we can represent any class inJ1(Qp)/2J(Qp) by a Galois-invariant divisor ofdegreem ≤ g + 1, say

D = (P1) + (P2) + · · ·+ (Pm)−D∗, (14)

where eachPi = (ai, bi, ci) is defined over some finite Galois extensionK of Qp with ci 6= 0 andD∗ is adivisor supported on points above∞. We may assume thatm is odd, for otherwisef0 is a square inQ×

p

and we are reduced to the case of [30, Proposition 2.9]. By Theorem16, if D1, D2 ∈ J1(Qp)/2J(Qp)have the same image in(L×/L×2Q×

p )N=f0 under the “x − T ” map, then a triple(I, α, s) exist forD1

if and only if it exists forD2. We may assume that the coordinates of all thePi’s lie in the ring ofintegers ofK with bi = 1, and that theai’s are all distinct. Indeed, the sum of all the pointsPi with ainon-integral is rational and maps to1 in L×/L×2Q×

p .Define the monic polynomial

P (x) = (x− a1)(x− a2) · · · (x− am)

of degreem overQp. LetR(x) be the polynomial of degree≤ m− 1 overQp obtained from Lagrangeinterpolation with the property that

R(ai) = ci, 1 ≤ i ≤ m.

ThenR(x)2 − f(x, 1) = h(x)P (x) in Qp[x] for someh(x) ∈ Qp[x].The polynomialP (x) clearly has coefficients in the ringZp. WhenR(x) also has coefficients

in Zp, we define the idealID of Rf as theRf -submodule ofL = Kf given by

〈R(θ), P (θ)If((n− 3−m)/2)〉.

For example, whenm = 1, the divisorD has the form(P ) with P = (a, b, c), and the idealID is givenby

ID = SpanZp{c, θ − a, (θ − a)2, . . . , (θ − a)(n−2)/2, ζn/2, . . . , ζn−1}.

We claim thatI2D is contained in the idealαIf(n − 3) whereα = P (θ), and thatN(ID)2 =

N(P (θ))N(If(n − 3)). Let s be the square root ofN(α) such that(α, s) corresponds to the rationalorbit determined byD. Thus we obtain an integral orbit mapping to the rational orbit corresponding toD. Since the casem = 1 has been dealt with already, we assume thatm ≥ 3.

The fact that the idealI2D = 〈R(θ)2, P (θ)R(θ)If((n − 3 − m)/2)), P (θ)2If (n − 3 − m)〉 iscontained inP (θ)If(n− 3) follows by observing thatf(θ, 1) = 0, soR(θ)2 = h(θ)P (θ). It thereforesuffices to show that〈h(θ), R(θ)If((n − 3 − m)/2), P (θ)If(n − 3 − m)〉 is contained inIf(n − 3),and this follows from the fact thatP has degreem, R(x) has degree less thanm and hence degree atmost(n − 3 +m)/2, andh has degree at mostn −m and thus at mostn − 3. (Note that the leadingcoefficient ofh is f0. Therefore, even whenm = 1, one still hash(θ) ∈ If (n− 3).)

To compute the norm ofID, we use a specialization argument. LetR denote the ring

R = Zp[f0, . . . , fn, a1, . . . , am][√f(a1, 1), . . . ,

√f(am, 1)],

26

wheref(x, 1) = f0xn + · · ·+ fn. Write ci =

√f(ai, 1) for eachi = 1, . . . , m. InsideR[x]/f(x, 1),

we defineζ1, . . . , ζn−1 as before and denote the correspondingRf , If , ID by Rf , If , ID. One has thenotion ofNID as anR-submodule of the fraction field ofR.

We claim thatNID is the principal ideal generated bys = c1 · · · cm/f (n−3+m)/20 . Specializing to

particularf0, . . . , fn, a1, . . . , am then completes the proof. We prove this claim by first invertingf0. Inthis case, the result follows from [3, Proposition 8.5]. Next we localize at(f0) to check that the correctpower off0 is attained. Since every ideal is invertible now, it sufficesto show thatI2

D = P (θ)In−3f

which follows from the statements

(θ − ai)(Rf)(f0) = (If )(f0) (15)

for i = 1, . . . , m. To see that (15) is sufficient, we note thatP (θ)(Rf)(f0) = (If(m))(f0) sinceP (θ) isthe product of allθ − ai. Therefore,

P (θ)(If(n− 3))(f0) = P (θ)2(If(n− 3−m))(f0) ⊂ (ID)2(f0) ⊂ P (θ)(If(n− 3))(f0).

To prove (15), note that the containment⊂ is clear sinceθ − ai ∈ If ; equality follows becausethey have the same norm. We now give another more explicit proof. Note that it remains to showthat 1 ∈ (θ − ai)(Rf )(f0). Consider the polynomialhi(t) = (f(t, 1) − c2i )/(t − ai). By definitionhi(θ)(θ − ai) = −c2i . Moreover, writing outhi(t) explicitly, one sees that

hi(θ) = ζn−1 + aiζn−2 + a2i ζn−3 + · · · an−2i ζ1 + hi(0) ∈ Rf .

This shows thatc2i ∈ (θ − ai)(Rf )(f0), and hence1 ∈ (θ − ai)(Rf)(f0) sinceci is a unit in(Rf )(f0).Suppose now thatR(x) is not integral. Letz2 = f(x, 1) denote an affine model for the given

hyperelliptic curve. A Newton polygon analysis onf(x, 1) − R(x)2 shows thatdiv(z − R(x)) − [D]has the formD∗ + E with D∗ ∈ J1(Qp) andE ∈ J(Qp), whereD∗ can be expressed in (14) with mreplaced bym − 2 and thex-coordinates of the non-infinity points inE have negative valuation. Thecondition of divisibility on the coefficients off(x, y) ensures that(x − θ)(E) ∈ L×2Q×

p . The proofnow concludes by induction on the numberm of points inD. Oncem ≤ 1 the polynomialsP (x) andR(x) are both integral.

The following result will be important to us in Section11.

Proposition 34. Let p be any odd prime, and letf(x, y) ∈ Zp[x, y] be a binary form of even de-green such thatp2 ∤ ∆(f) and f0 6= 0. Let C denote the hyperelliptic curvez2 = f(x, y). Sup-pose thatDiv1(C)(Qp) 6= ∅. Then the(SLn /µ2)(Zp)-orbits onZ2

p ⊗ Sym2 Znp with invariant binary

form f(x, y) are in bijection with soluble(SLn /µ2)(Qp)-orbits onQ2p ⊗ Sym2Q

np with invariant bi-

nary formf(x, y). Furthermore, if(A,B) ∈ Z2p ⊗ Sym2 Z

np with invariant binary formf(x, y), then

Stab(SLn /µ2)(Zp)(A,B) = Stab(SLn /µ2)(Qp)(A,B).

Proof. As noted earlier, we only need to focus on the pair(I, α). The conditionp2 ∤ ∆(f) impliesthat the orderRf is maximal and that the projective closureC of C over Spec(Zp) is regular. ByTheorem28, the assumption thatDiv1(C)(Qp) 6= ∅ implies that solubleQp-orbits with invariantbinary formf(x, y) exist. Sincep is odd, thep-adic version of Proposition33 implies thatZp-orbitswith invariant binary formf(x, y) exist. Therefore, by Remark17, the set of equivalence classes ofpairs(I, α) is nonempty and is in bijection with(R×

f /R×2f Z×

p )N≡1. LetJ denote the Neron model ofJ

27

overSpec(Zp). SinceC is regular,J is connected and we have via flat cohomology ([30, Proposition2.11]) that

J (Zp)/2J (Zp) ≃ (R×f /R

×2f Z×

p )N≡1.

The Neron mapping property implies thatJ (Zp)/2J (Zp) = J(Qp)/2J(Qp). For the stabilizer state-ment, consider the exact sequence (3). SinceL×[2] = R×

f [2], it remains to compare(L×2 ∩ Q×p )/Q

×2p

and(R×2f ∩ Z×

p )/Z×2p . These two sets are only nonempty whenL contains a quadratic extensionK ′

of Qp. The conditionp2 ∤ ∆(f) implies thatK ′ = Qp(√u) can only be the unramified quadratic

extension ofQp. In other words,u ∈ Z×p . Hence in this case(L×2 ∩ Q×

p )/Q×2p and(R×2

f ∩ Z×p )/Z

×2p

both are equal to the group of order 2 generated by the class ofu.

10 The number of irreducible integral orbits of bounded height

Let V = Sym2(W∗) ⊕ Sym2(W

∗) be the scheme of pairs of symmetric bilinear forms onW . Definethe height of an elementv ∈ VZ by H(v) := H(fv). In [1, §4], the asymptotic number of irreducibleSL±1

n (Z)-orbits onV (Z) having height less thanX was determined, and also the number of such orbitswhose invariant binary formsf(x, y) satisfy any finite set of congruences. The same computationapplies also withG = SLn /µ2 in place ofSL±1

n . We assume henceforth thatn is even.To state this counting result precisely, recall ([1, §4.1.1]) that we may naturally partition the set

of elements inV (R) with ∆ 6= 0 and nonnegative definite invariant binaryn-ic form into∑n/2

m=0 τ(m)components, which we denote byV (m,τ) for m = 0, . . . , n/2 andτ = 1, . . . , τ(m) whereτ(m) =22m−2 if m > 1; τ(0) = 2 if n is congruent to0 (mod4); andτ(0) = 1 if n is congruent to2 (mod4).In [1, §4.1.1], whenm > 0, there are22m−1 choices for the parameterτ as it runs over all assignmentsof signs± to the2m real linear factors off(x, y) subject to the condition that the product of the signsmatches the sign of leading coefficient off(x, y). When using the group(SLn /µ2)(R), two suchτ1, τ2are equivalent if and only if the sign assignments associated to τ1 are exactly the negatives of thoseassociated toτ2.

For a given value ofm, the componentV (m,τ) in V (R) maps to the componentI(m) of non-negative definite binaryn-ic forms inRn+1 having nonzero discriminant and2m real linear factors.LetF (m,τ) denote a fundamental domain for the action ofG(Z) onV (m,τ), and set

cm,τ = Vol(F (m,τ) ∩ {v ∈ V (R) : H(v) < 1}).

The number ofτ ’s that correspond to orbits soluble atR is #(J1(R)/2J(R)) whereJ denotes theJacobian of a hyperelliptic curvez2 = f(x, y) with f(x, y) ∈ I(m). The size of this quotient does notdepend on the choice off(x, y) ∈ I(m). Then from [1, Theorems 9 and 17], we obtain:

Theorem 35. Fix m, τ . For anyG(Z)-invariant setS ⊂ V (Z)(m,τ) := V (Z) ∩ V (m,τ) defined byfinitely many congruence conditions, letN(S;X) denote the number ofG(Z)-equivalence classes ofelementsv ∈ S satisfyingH(v) < X. Then

N(S;X) = cm,τ ·∏

p

µp(S) ·Xn+1 + o(Xn+1),

whereµp(S) denotes thep-adic density ofS in V (Z).

28

11 Sieving to locally soluble orbits

Since local solubility is defined by infinitely many congruence conditions, we need a weighted versionof Theorem35 in which we allow weights to be defined by certain infinite setsof congruence condi-tions. The technique for proving such a result involves using Theorem35 to impose more and morecongruence conditions. While doing so, we need to uniformlybound the error term. To this end, wehave the following proposition proven in [2].

Proposition 36. For each primep, let Wp denote the set of elementsv = (A,B) ∈ V (Z) such thatp2 | ∆(disc(Ax− By)). Then there existsδ > 0 such that, for anyM > 0, we have

∑

p>M

N(Wp;X) = O(Xn+1/M δ),

where the implied constant is independent ofX andM .

To describe which weight functions onV (Z) are allowed, we need the following definition motivatedby the above proposition:

Definition 37. SupposeW = AM is some affine space. A functionφ : W (Z) → [0, 1] is said tobedefined by congruence conditionsif there exist local functionsφp : W (Zp) → [0, 1] satisfying thefollowing conditions:

1. For allv ∈ W (Z), the product∏

p φp(v) converges toφ(v).

2. For each primep, the functionφp is locally constant outside some (p-adically) closed subset ofW (Zp) of measure0.

3. Thep-adic integral∫

W (Zp)

φp(v)dv is nonzero.

A subsetW ′ of W (Z) is said to bedefined by congruence conditionsif its characteristic function isdefined by congruence conditions.

A function φ : V (Z) → [0, 1] is said to beacceptableif, for all sufficiently largep, we haveφp(v) = 1 wheneverp2 ∤ ∆(disc(Ax− By)).

Then we have the following theorem.

Theorem 38. Let φ : V (Z) → [0, 1] be a function that is defined by congruence conditions via localfunctionsφp : V (Zp) → [0, 1]. Fix m, τ . LetS be anG(Z)-invariant subset ofV (Z)(m,τ) defined bycongruence conditions. LetNφ(S;X) denote the number ofG(Z)-equivalence classes of irreducibleelementsv ∈ S having height bounded byX, where each equivalence classG(Z)v is counted withweightφ(v). If φ is acceptable, then

Nφ(S;X) = cm,τ

∏

p

∫

v∈V (Zp)

φp(v)dv + o(Xn+1).

Theorem38 follows from Theorem35 and Proposition36 just as [6, Theorem 2.20] followedfrom [6, Theorems 2.12 and 2.13].

LetF0 denote the set of all integral binary formsf(x, y) of degreen. Identify the scheme of allbinaryn-ic forms overZ with An+1

Z . If F is a subset ofF0, denote byF (Fp) the reduction modulop ofthep-adic closure ofF in An+1

Z (Zp).

29

Definition 39. A subsetF of F0 is large if the following conditions are satisfied:

1. It is defined by congruence conditions,

2. There exists a subschemeS0 of An+1Z of codimension at least 2 such that for all but finitely

manyp, we haveF0(Fp)\F (Fp) ⊂ S0(Fp).

We identify hyperelliptic curves with their associated binary forms. Thus, we say that a family ofhyperelliptic curvesz2 = f(x, y) is large if the set of binary formsf(x, y) appearing is large.

As an example, the subsetF1 of F0 consisting of binaryn-ic formsf(x, y) such that the cor-responding hyperelliptic curvesC given byz2 = f(x, y) have locally solubleDiv1 is large. The setF2 ⊂ F1 of integral binaryn-ic forms such that the corresponding hyperelliptic curveC is locally sol-uble is also large. These two statements follow from [25, Lemma 15]. Our aim is to prove Theorem4for large families of hyperelliptic curves whose associated binary forms are contained inF1.

LetF be a large subset ofF0 contained inF1. Since the curvesz2 = f(x, y) andz2 = κ2f(x, y)are isomorphic overQ whereκ is the constant in Theorem14, we assume without loss of generalitythat the coefficients of everyf(x, y) in Inv(F ) lie in κ2Z. To prove Theorem4, we need to weight eachlocally soluble elementv ∈ V (Z) having invariants inInv(F ) by the reciprocal of the number ofG(Z)-orbits in theG(Q)-equivalence class ofv in V (Z). However, in order for our weight function to bedefined by congruence conditions, we instead define the following weight functionw : V (Z) → [0, 1]:

w(v) :=

(∑

v′

#StabG(Q)(v′)

#StabG(Z)(v′)

)−1

if v is locally soluble with invariant inInv(F ),

0 otherwise,

(16)

where the sum is over a complete set of representatives for the action ofG(Z) on theG(Q)-equivalenceclass ofv in V (Z). We then have the following theorem:

Theorem 40. LetF be a large subset ofF0 contained inF1. Moreover, suppose that the coefficients ofeveryf(x, y) in Inv(F ) lie in 16nZ. Then

∑

C∈FH(C)<X

#Sel2(J1) =

n/2∑

m=0

∑

τ soluble

Nw(V (Z)(m,τ)F ;X) + o(Xn+1). (17)

whereV (Z)(m,τ)F is the set of all elements inV (Z)(m,τ) with invariant binary forms inInv(F ).

Proof. By Theorems30and32, the left hand side is equal to the number ofG(Q)-equivalence classesof elements inV (Z) that are locally soluble, have invariants inInv(F ), and have height bounded byX.Given a locally soluble elementv ∈ V (Z) with invariant inInv(F ), let v1, . . . , vk denote a completeset of representatives for the action ofG(Z) on theG(Q)-equivalence class ofv in V (Z). Then

k∑

i=1

w(vi)

#StabG(Z)(vi)=

( k∑

i=1

#StabG(Q)(v)

#StabG(Z)(vi)

)−1k∑

i=1

1

#StabG(Z)(vi)=

1

#StabG(Q)(v). (18)

WhenStabG(Q)(v) is trivial, which happens for all but negligibly manyv ∈ V (Z) by [1, Proposi-tion 14], (18) simplifies to

k∑

i=1

w(vi) = 1. (19)

30

Since the size ofStabG(Q)(v) is bounded above by22g, (19) always holds up to an absolutely boundedfactor. Therefore, the right hand side of (17) counts the number ofG(Q)-equivalence classes of ele-ments inV (Z) that are locally soluble, have invariants inF , and have height bounded byX.

In order to apply Theorem38 to computeNw(V (Z)(m,τ);X), we need to know thatw is anacceptable function defined by congruence conditions. The proof thatw is a product

∏pwp of local

weight functions is identical to the proof of [6, Proposition 3.6]. To prove thatw is acceptable, itremains to show that forp sufficiently large,w(v) = 1 wheneverp2 ∤ ∆(disc(Ax− By). This followsfrom Proposition34. Therefore, to computeNw(V (Z)(m,τ);X), it remains to computecm,τ and thep-adic integrals

∫wp(v)dv. For these, we have the following results:

cm,τ =|J |Vol(G(Z)\G(R))

#J (m)[2](R)Vol({f ∈ I(m)|H(f) < X});

∫

v∈V (Zp)

wp(v)dv = |J |pVol(G(Zp))Vol(Invp(F ))#(J1(Qp)/2J(Qp))

#J [2](Qp).

HereJ is a nonzero rational constant;J (m) denotes the Jacobian of any hyperelliptic curve defined byz2 = f(x, y) wheref(x, y) ∈ Z[x, y] ∩ I(m); andJ denotes the Jacobian of any hyperelliptic curvedefined byz2 = f(x, y) wheref(x, y) ∈ Inv(F ); andInvp(F ) is thep-adic closure ofInv(F ). Thefirst equation is proved in [1, §4.4]. The second equation follows from the identical computation as in[30, §4.5].

For every placeν of Q, we letaν denote the following quotient:

aν =#(J1(Qν)/2J(Qν))

#J [2](Qν).

Because of the assumption thatJ1(Qν) 6= ∅, this quotient depends only onν, g. Indeed, it is equal to2−g for ν = ∞, 2g for ν = 2, and1 for all other primes (see, e.g., [3, Lemma 12.3]). Theaν ’s satisfythe product formula

∏ν aν = 1.

We now combine Theorem38, Theorem40, and the product formula∏

ν |J |ν = 1 to obtain:

Theorem 41. LetF be a large subset ofF0 contained inF1. Moreover, suppose that the coefficients ofeveryf(x, y) in Inv(F ) lie in 16nZ. Then

∑

C∈FH(C)<X

#Sel2(J1) = τ(G)Vol({f ∈ I(m)|H(f) < X})

∏

p

Vol(Invp(F )) + o(Xn+1), (20)

whereτ(G) = 2 denotes the Tamagawa number ofG.

12 Proofs of main theorems

All the results stated in the introduction, with the exception of Theorems1–3, hold even if for eachg ≥ 1 we range over any large congruence family of hyperelliptic curvesC overQ of genusg forwhichDiv1(C) is locally soluble. (See Definition39 for the definition of “large”.)

We prove Theorems4 and5 in this generality.

31

Proof of Theorem 4: Let F be a large family of hyperelliptic curves with locally solubleDiv1. SinceCondition 2 in Definition39 is a modp condition, a similar uniformity estimate as in Proposition36counting the number of hyperelliptic curves inF with bounded height is given by [26]. Theorem38then gives

∑

C∈FH(C)<X

1 = Vol({f ∈ I(m)|H(f) < X})∏

p

Vol(Invp(F )) + o(Xn+1), (21)

and dividing (20) by (21) gives Theorem4. �

Proof of Theorem 5: Let F be a large family of hyperelliptic curvesC with locally solubleDiv1. Letk > 0 be an odd integer. Recall that the 2-Selmer set of orderk is defined to be the subset of elementsof Sel2(J1) that locally everywhere come from points inJ1(Qν) of the formd1 − k−1

2d whered1 is an

effective divisor of degreek andd is the hyperelliptic class. To obtain the average size of this 2-Selmerset of orderk, we need to perform a further sieve from the whole 2-Selmer set to this subset. Letϕν ≤ 1 denote the local sieving factor at a placeν of Q. Therefore, to prove that the average size ofthe 2-Selmer set of orderk is less than 2, it suffices to show thatϕν < 1 for someν.

We use the archimedean place. Supposef(x, y) is a degree2g + 2 binary form having2mreal linear factors and letC be its associated hyperelliptic curve withm > 0. ThenC(R) hasmconnected components andJ(R)/2J(R) has size2m−1. Let σ denote complex conjugation. Then foranyP ∈ C(C) with x-coordinatet ∈ C×, we have that(t− β)(σt− β) = NC/R(t− β) ∈ R×2 for anyβ ∈ R. Hence the descent “x− T ” map sends the class of(P ) + (σP )− d to 1 inL×/L×2R whereLdenotes the etale algebra of rankn associated tof(x, y). Thus(P )+(σP )−d ∈ 2J(R). Therefore, theimage of(Symk(C))(R) in J1(R)/2J(R) is equal to the image ofSymk(C(R)) in J1(R)/2J(R).Sincem is positive,C has a rational Weierstrass point overR. Hence ifP ∈ C(R), then2(P )− d ∈ 2J(R).SinceC(R) hasm connected components, we see that the image ofSymk(C(R)) in J1(R)/2J(R) hassize at most

Sm(k) =

(m

1

)+

(m

3

)+ · · ·+

(m

k

).

There is a positive proportion of hyperelliptic curvesC : z2 = f(x, y) in F such thatf(x, y) splitscompletely overR. For any odd integerk < g, we haveSg+1(k) < 2g = |J1(R)/2J(R)|. Therefore,ϕ∞ < 1.

Consider now the second statement that the average size of the 2-Selmer set of orderk goes to 0asg approaches∞. We may use the archimedean place again. Supposef(x, y) is a degreen = 2g+ 2binary form having2m real linear factors and letC be its associated hyperelliptic curve. For a fixedodd integerk > 0, we have

limm→∞

Sm(k)

2m−1= 0. (22)

On the other hand, [15, Theorem 1.2] states that the density of real polynomials ofdegreen havingfewer thanlogn/ log logn real roots isO(n−b+o(1)) for someb > 0. Therefore, the result now followsfrom this and (22) since2m−1 = |J1(R)/2J(R)|. �

Our approach to Theorem3 (which in turn implies Theorems1 and2) using a result of Dok-chitser and Dokchitser (Appendix A) does not work in the generality of large families, but does workfor “admissible” families as defined below.

32

Definition 42. A subsetF of the setF0 of all integral binary forms of degreen is admissibleif thefollowing conditions are satisfied:

1. It is defined by congruence conditions;

2. For large enough primesp, thep-adic closure ofF contains all binary formsf(x, y) of degreenoverZp such that the hyperelliptic curvez2 = f(x, y) has aQp-rational point.

We say that a family of hyperelliptic curvesz2 = f(x, y) is admissibleif the set of binary formsf(x, y)appearing is admissible.

To prove Theorem3 where we range over any admissible family of hyperelliptic curves overQof genusg ≥ 1 with locally solubleDiv1, we note that the result of Dokchitser and Dokchitser holds foradmissible families (TheoremA.2). The rest of the proof is identical to that given in the introduction.

We conclude by giving a version of Theorem1 in the most general setting that our methodsallow.

Theorem 43. SupposeF is a large congruence family of integral binary forms of degreen = 2g + 2for which there exist two primesp, q neither of which is a quadratic residue modulo the other suchthatthe following conditions hold for a positive proportion off(x, y) in F :

1. The four integral binary formsf(x, y), pf(x, y), qf(x, y), pqf(x, y) all lie insideF and thecorresponding hyperelliptic curves have points overQp andQq.

2. If J denotes the Jacobian of the hyperelliptic curvez2 = f(x, y), thenJ has split semistablereduction of toric dimension1 at p and good reduction atq.

Then for a positive proportion of binary formsf(x, y) in F , the corresponding hyperelliptic curveC : z2 = f(x, y) has no points over any odd degree extension ofQ (i.e., the varietyJ1 has no rationalpoints), and moreover the2-Selmer setSel2(J1) is empty.

Appendix A: A positive proportion of hyperelliptic curves h aveodd/even 2-Selmer rank

by Tim and Vladimir Dokchitser

In this appendix we show that both odd and even 2-Selmer ranksoccur a positive proportion ofthe time among hyperelliptic curves of a given genus.

For an abelian varietyA defined over a number fieldK, write rk2(A/K) = dimF2 Sel2(A/K)for the 2-Selmer rank, andrk2∞(A/K) for the2∞-Selmer rank2. We will say ‘rank of a curve’ meaning‘rank of its Jacobian’.

Theorem A.1. The proportion of both odd and even2∞-Selmer ranks in the family of hyperellipticcurves overQ,

y2 = anxn + an−1x

n−1 + · · ·+ a0 (n > 3),

ordered by height as in(2) is at least2−4n−4. In particular, assuming finiteness of the2-part ofX, atleast these proportions of curves have Jacobians of odd and of even Mordell–Weil rank.

2 Mordell–Weil rank + number of copies ofQ2/Z2 in XA/K ; if X is finite, this is just the Mordell–Weil rank.

33

Theorem A.2. LetK be a number field with ring of integersO. Fix n > 3. Consider the family of allhyperelliptic curves

y2 = anxn + · · ·+ a0, ai ∈ O,

or any other “admissible” family (see Definition42). Then a positive proportion of the hyperellipticcurves in the family, when ordered by height, have even2-Selmer rank and a positive proportion haveodd2-Selmer rank. The same conclusion holds for the2∞-Selmer rank.

The proofs resemble that of [7, §4.1] for elliptic curves overQ. Recall that the conjecture ofBirch and Swinnerton-Dyer implies, in particular, that theparity of the rank of an elliptic curveE isdetermined by whether its root number — that is, the sign of the functional equation of theL-functionL(E, s) of E — is +1 or −1. The proof in [7] uses that twisting by−1 does not affect the height ofthe curve but often changes the root number, and that the parity of the Selmer rank is (unconjecturally)compatible with the root number.

This compatibility is not known for hyperelliptic curves (but see the forthcoming work ofA. Morgan for 2-Selmer ranks for quadratic twists). Instead, we tweak the argument to use Brauerrelations in biquadratic extensions, where it is known in enough cases. To illustrate the method, con-sider an elliptic curveE/Q with split multiplicative reduction at 2. Then it has root number−1 overF = Q(i,

√2), since the unique place above 2 inF contributes−1, while every other rational place

splits into an even number of places inF and so contributes+1. In other words, the sum of theMordell–Weil ranks for the four quadratic twists

rk(E/F ) = rk(E/Q) + rk(E−1/Q) + rk(E2/Q) + rk(E−2/Q) (∗)

should be odd, and so both odd and even rank should occur amongthe 4 twists. The point is thatfor the 2∞-Selmer rank, the parity in(∗) can be computed unconditionally, using a Brauer relationin Gal(F/Q) ∼= C2 × C2. Moreover, this works for general abelian varieties and over a generalnumber fieldK, replacingQ(i,

√2) by a suitable biquadratic extension ofK. The fact that most of the

decomposition groups are cyclic allows us to avoid all the hard local computations and restrictions onthe reduction types, and varying the curve in the family gives the required positive proportions.

The exact result we will use is:

Theorem A.3. LetF = K(√α,

√β) be a biquadratic extension of number fields. Suppose that some

primep0 of K has a unique prime above it inF . LetC/K be a curve with JacobianJ , such that

1. C(Kp0) 6= ∅ andJ has split semistable reduction of toric dimension1 at p0;

2. C(Kp) 6= ∅ andJ has good reduction atp for everyp 6= p0 that has a unique prime above it inF/K.

Thenrk2∞(J/K) + rk2∞(Jα/K) + rk2∞(Jβ/K) + rk2∞(Jαβ/K) ≡ 1 mod 2.

If, in addition,Cα(Kp), Cβ(Kp) andCαβ(Kp) are non-empty for all primesp of K that have a uniqueprime above them inF , then the same conclusion holds for the2-Selmer rank as well.

Postponing the proof of this theorem, we first explain how it implies TheoremsA.1 andA.2.

34

Proof of TheoremA.1

For TheoremA.1, it suffices to prove the following:

Proposition A.4. Consider a squarefree polynomialf(x) ∈ Q[x],

f(x) = anxn + · · ·+ a0 (n > 3, n = 2g + 1 or 2g + 2),

whose coefficients satisfya2 ≡ 1 mod 8, a2g+1 ≡ 4 mod 8 and all otherai ≡ 0 mod 8. Thenamong the four hyperelliptic curves

y2=f(x), y2=−f(x), y2=2f(x), y2=−2f(x)

at least one has even and at least one has odd2∞-Selmer rank.

Proof. Replacingy 7→ 2y + x in C : y2 = f(x) and dividing the equation by 4, yields a curve withreduction

C/F2 : y2 + xy = x2g+1.

This equation has a split node at(0, 0) and no other singularities, so Jac(C) has split semistable reduc-tion at 2 of toric dimension 1. Hensel lifting the non-singular point at∞ onC we find thatC(Q2) 6= ∅.Now apply TheoremA.3 with K = Q, F = Q(i,

√2) andp0 = 2. (Note that all odd primes split in

F/Q, and that Jac(Cα) = (Jac(C))α.)

Proof of TheoremA.2

Lemma A.5. LetK be a finite extension ofQp (p odd), with residue fieldFq. Take a hyperelliptic curve

C : y2 = anxn + · · ·+ a0, ai ∈ OK ,

and letf(x) ∈ Fq[x] be the reduction of the right-hand side.

1. If f is squarefree of degreen and has anFq-rational root, then Jac(C) has good reduction, andCα(K) 6= ∅ for everyα ∈ K×.

2. If f(x) = (x − a)2h(x) for somea ∈ Fq and some squarefree polynomialh(x) of degreen − 2that possesses anFq-rational root and satisfiesh(a) ∈ F×2

q , then Jac(C) has split semistablereduction of toric dimension 1, andCα(K) 6= ∅ for everyα ∈ K×.

3. If f(x) is not of the formλh(x)2, λ ∈ Fq, andq > 4n2, thenC(K) 6= ∅.

Proof. In the first case,C has good reduction, and therefore so does Jac(C). In the second case,Chas one split node and no other singular points, and so its Jacobian has split semistable reduction oftoric dimension 1. In both cases,f(x) has a simple rootb ∈ Fq, by assumption. Lifting it by Hensel’slemma, we get a point(B, 0) on C/K. This point gives aK-rational point on every quadratic twistof C.

For (3), this is the argument in [25, Lemma 15]: writef(x) = l(x)h(x)2 with l andh coprimeandl non-constant and squarefree. By the Weil conjectures, the curvey2 = l(x) has at leastq + 1 −n√q > n rational points overFq. So there is at least one whosex-coordinate is not a root off . It is

non-singular ony2 = f(x), and by Hensel’s lemma it lifts to a point inC(K).

35

Proof of TheoremA.2. WriteO for the ring of integers ofK, andFp for the residue field atp.Suppose we are given an admissible familyF of hyperelliptic curves. In other words, for every

primep the curves lie in some open setFp of p-adic curvesC/Op, defined by congruence conditionsmodulopmp, and outside a finite set of primesΣ of O these sets include all curves withC(Op) 6= ∅.EnlargeΣ to include all primesp|2, with mp chosen so that units of the form1 + pmp are squares inOp, and all primes of norm6 4n2.

Take a primep0 /∈ Σ. Pickα, β ∈ O with α ≡ β ≡ 1 mod∏

p∈Σ pmp, and such thatα hasvaluation 1 atp0 andβ is a non-square unit modulop0. Thenp0 ramifies inK(

√α) and is inert in

K(√β), soF = K(

√α,

√β) is a biquadratic extension with a unique prime abovep0. There is a finite

set of primesU of K that have a unique prime above them inF , andU ∩ Σ = ∅. (The set is finitesince such primes must ramify inF/K.)

Within our familyF consider those curvesC : y2 = f(x) whose reductions are as in LemmaA.5(2) atp0, as in LemmaA.5(1) at allp ∈ U \ {p0}, and such thatf modp is not a unit times thesquare of a polynomial at anyp /∈ Σ ∪ U . (This is a positive proportion of curves inF by [26].)For each such curveC, TheoremA.3 implies that both odd and even 2-Selmer ranks occur among thetwists of Jac(C) by 1, α, β andαβ, in other words the Jacobians ofC,Cα, Cβ andCαβ . Note that thesetwists are inF , since forp ∈ Σ this twisting does not change the class modulopmp, while for p 6∈ Σthese twists are all locally soluble by LemmaA.5(3).

Because quadratic twists byα, β andαβ only change the height by at mostNK/Q(αβ)n, we get

the asserted positive proportion.

Proof of TheoremA.3

We refer the reader to [18, §2] for the theory of Brauer relations and their regulator constants.

Notation A.6. Let F/K be a Galois extension of number fields with Galois groupG, andA/K anabelian variety. Fix a global invariant exterior formω onA/K. ForK ⊂ L ⊂ F and a primep, wewrite

X[p]A/L p-primary part ofXA/L modulo divisible elements(a finite abelianp-group).

cA/L

∏cv|ω/ωo

v|v, where the product is taken over all primes ofL, cv is the Tamagawa numberof A/L at v, ωo

v the Neron exterior form and| · |v the normalised absolute value atv.In the theorem below we write

S the set of self-dual irreducibleQpG-representations.Θ =

∑niHi a Brauer relation inG (i.e.

∑i ni Ind

GHi

1 = 0).C(Θ, ρ) the regulator constant

∏i det

(1

|Hi|〈, 〉|ρHi

)ni ∈ Q∗p/Q

∗2p ,

where〈, 〉 is some non-degenerateG-invariant pairing onρ.

Finally, as in [17] we let3

SΘ = {ρ ∈ S | ordp C(Θ, ρ) ≡ 1 mod 2}.Theorem A.7. SupposeA/K is a principally polarised abelian variety. Forρ ∈ S write mρ for itsmultiplicity in the dualp∞-Selmer group ofA/F . Then

∑

ρ∈SΘ

mρ ≡ ordp

∏

i

cA/FHiX[p]

A/FHimod 2.

3[17] also includes representations of the formT ⊕ T ∗ for some irreducibleT 6∼= T ∗ (T ∗ is the contragredient ofT ),but these have trivial regulator constants by [18, Cor. 2.25].

36

Proof. This is essentially [17, Thm. 1.6], except for theX[p] term in the right-hand side. For oddp,this term is a square and does not contribute to the formula. For p = 2, this is the formula that comesout of the proof of [17, Thm. 1.6]. There the main step of the proof ([17, Thm. 3.1]) assumes thatA/Khas a principal polarisation induced by aK-rational divisor to get rid of theX[2] term coming from[17, Thm. 2.2].

Corollary A.8. LetF = K(√α,

√β) be a biquadratic extension of number fields. For every princi-

pally polarised abelian varietyA/K,

(†)

rk2∞(A/K) + rk2∞(Aα/K) + rk2∞(Aβ/K) + rk2∞(Aαβ/K) ≡

≡ ord2cA/K(

√α) cA/K(

√β) cA/K(

√αβ)

cA/F (cA/K)2

+ord2|X

[2]

A/K(√

α)||X

[2]

A/K(√β)||X

[2]

A/K(√αβ)

|

|X[2]A/F

||X[2]A/K

|2mod 2.

Proof. Write 1,Ca2,C

b2,C

c2 for the proper subgroups ofG = Gal(F/K), and1, ǫa, ǫb, ǫc for its 1-

dimensional representations (soC[G/C•2]∼=1 ⊕ ǫ• for • = a, b, c). Thus the four2∞-Selmer ranks in

question are the multiplicities of these four representations in the dual2∞-Selmer group ofA/F . Nowapply the theorem to the Brauer relation

Θ = {1} − Ca2 − Cb

2 − Cc2 + 2G. (23)

Its regulator constants are (see [18, 2.3 and 2.14])

CΘ(1) = CΘ(ǫa) = CΘ(ǫb) = CΘ(ǫc) = 2 ∈ Q×/Q×2,

and soSΘ = {1, ǫa, ǫb, ǫc} in this case.

Proof of TheoremA.3. We write the two expressions inord2(...) on the right-hand side of CorollaryA.8 as a product of local terms. The modified Tamagawa numberscJ/K, cJ/K(

√α), . . . are, by definition,

products over primes ofK, K(√α), . . ., and we group all terms by primes ofK. Similarly, as shown

by Poonen and Stoll in [25, §8], the parity oford2X[2] is a sum of local terms that are 1 or 0 depending

on whetherPicg−1(C) is empty or not over the corresponding completion, and againwe group themby primesp of K. This results in an expression

rk2∞(J/K) + rk2∞(Jα/K) + rk2∞(Jβ/K) + rk2∞(Jαβ/K) ≡∑

p

tp mod 2.

There are three cases to consider forp:If there are several primesqi|p in F , then the decomposition groups ofqi are cyclic, and this

forcestp = 0. This is a general fact about Brauer relations and functionsof number fields that areproducts of local terms, see [18, 2.31, 2.33, 2.36(l)].

If there is a unique primeq|p in F , thenC(Kp) 6= ∅ by assumption. SoPicg−1(C) is non-empty in every extension ofKp, and all the local terms forX[2] abovep vanish. AlsoJ has semistablereduction, again by assumption, so its Neron minimal modelstays minimal in all extensions. The term|ω/ωo

v|v always cancels in Brauer relations in this case, see e.g. [18, 2.29]. So the only contribution totp comes from Tamagawa numbers.

37

Whenp 6= p0, the JacobianJ has good reduction and the Tamagawa numbers are trivial, sotp = 0. Finally, if p = p0, thenJ has split semistable reduction atp of toric dimension 1. In this case,the Tamagawa number term atp multiplies to2 ∈ Q×/Q×2, in other wordstp = 1. This follows e.g.from [18, 3.3, 3.23] for the Brauer relation (23). This proves the claim for the2∞-Selmer rank.

It remains to deduce the formula forrk2 from the one forrk2∞. The difference betweenrk2andrk2∞ comes fromX[2] and the 2-torsion in the Mordell–Weil group onJ , Jα, Jβ, andJαβ. Two-torsion is the same for all four twists, and so gives an even contribution. As forX[2], the local termsthat define its parity give an even contribution at every prime ofK that splits inF , as the twists thencome in isomorphic pairs. At the non-split primes, all four twists have local points by assumption, andso the local terms are 0.

Acknowledgments

We thank Jean-Louis Colliot-Thelene, John Cremona, NoamElkies, Tom Fisher, Bjorn Poonen, PeterSarnak, Arul Shankar, and Michael Stoll for many helpful conversations. The first and third authorswere supported by a Simons Investigator Grant and NSF grant DMS-1001828. The second author wassupported by NSF grant DMS-0901102.

References

[1] M. Bhargava, Most hyperelliptic over Q curves have no rational points,http://arxiv.org/abs/1308.0395.

[2] M. Bhargava, The geometric sieve and the density of squarefree values of polynomial discrimi-nants and other invariant polynomials, in progress.

[3] M. Bhargava and B. Gross, Arithmetic invariant theory,http://arxiv.org/abs/1208.1007.

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