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Group actions on curves over arbitrary fields Garcia Armas, Mario 2015

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Group actions on curves over arbitraryfieldsbyMario Garcia ArmasLic., Universidad de La Habana, 2008M.Sc., Universidad de La Habana, 2010A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)March 2015c© Mario Garcia Armas 2015AbstractThis thesis consists of three parts. The common theme is finite group actionson algebraic curves defined over an arbitrary field k.In Part I we classify finite group actions on irreducible conic curvesdefined over k. Equivalently, we classify finite (constant) subgroups of SO(q)up to conjugacy, where q is a nondegenerate quadratic form of rank 3 definedover k. In the case where k is the field of complex numbers, these groupswere classified by F. Klein at the end of the 19th century. In recent papers ofA. Beauville and X. Faber, this classification is extended to the case wherek is arbitrary, but q is split. We further extend their results by classifyingfinite subgroups of SO(q) for any base field k of characteristic 6= 2 and anynondegenerate ternary quadratic form q.In Part II we address the Hyperelliptic Lifting Problem (or HLP): Givena faithful G-action on P1 defined over k and an exact sequence 1 → µ2 →G′ → G → 1, determine the conditions for the existence of a hyperellipticcurve C/k endowed with a faithful G′-action that lifts the prescribed G-action on the projective line. Alternatively, this problem may be regardedas the Galois embedding problem given by the surjection G′  G and theG-Galois extension k(P1)/k(P1)G. In this thesis, we find a complete solutionto the HLP in characteristic 0 for every faithful group action on P1 and everyexact sequence as above.In Part III we determine whether, given a finite group G and a base fieldk of characteristic 0, there exists a strongly incompressible G-curve definedover k. Recall that a G-curve is an algebraic curve endowed with the actionof a finite group G. A faithful G-curve C is called strongly incompressibleif every dominant G-equivariant rational map of C onto a faithful G-varietyis birational. We prove that strongly incompressible G-curves exist if Gcannot act faithfully on the projective line over k. On the other hand, ifG does embed into PGL2 over k, we show that the existence of stronglyincompressible G-curves depends on finer arithmetic properties of k.iiPrefaceThe material in this thesis is the result of my own work under the supervisionof Zinovy Reichstein, and it has been published or submitted for publication.• Most of Chapter 3 appears in:Mario Garcia-Armas. Finite group actions on curves of genus zero. J.Algebra, 394:173–181, 2013.• The material in Chapter 4 appears in:Mario Garcia-Armas. Finite group actions on hyperelliptic curves.Submitted for publication.• The results in Chapter 5 appears in:Mario Garcia-Armas. Strongly incompressible curves. Accepted inCanad. J. Math.iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . vi1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Finite subgroups of PGL1(A) . . . . . . . . . . . . . . . . . . 11.2 Finite group actions on hyperelliptic curves . . . . . . . . . . 31.3 Strongly incompressible curves . . . . . . . . . . . . . . . . . 42 Notation and preliminaries . . . . . . . . . . . . . . . . . . . . 72.1 Quaternion algebras . . . . . . . . . . . . . . . . . . . . . . . 72.2 Roots of unity . . . . . . . . . . . . . . . . . . . . . . . . . . 83 Finite group actions on curves of genus zero . . . . . . . . . 93.1 Existence of finite subgroups . . . . . . . . . . . . . . . . . . 93.2 Conjugacy classes of subgroups . . . . . . . . . . . . . . . . . 114 Finite group actions on hyperelliptic curves . . . . . . . . . 214.1 Equivalent characterizations of the HLP . . . . . . . . . . . . 214.2 HLP for cyclic groups . . . . . . . . . . . . . . . . . . . . . . 244.3 HLP for dihedral groups . . . . . . . . . . . . . . . . . . . . 264.4 HLP for polyhedral groups . . . . . . . . . . . . . . . . . . . 304.4.1 Alternating groups . . . . . . . . . . . . . . . . . . . 304.4.2 Symmetric group on 4 letters . . . . . . . . . . . . . . 314.5 Explicit solutions to the HLP . . . . . . . . . . . . . . . . . . 334.5.1 Extensions of cyclic groups . . . . . . . . . . . . . . . 364.5.2 Extensions of the Klein group . . . . . . . . . . . . . 374.5.3 Extensions of dihedral groups . . . . . . . . . . . . . 38ivTable of Contents4.5.4 Extensions of polyhedral groups . . . . . . . . . . . . 405 Strongly incompressible curves . . . . . . . . . . . . . . . . . 435.1 Rational quotients . . . . . . . . . . . . . . . . . . . . . . . . 435.2 Strong incompressibility of curves . . . . . . . . . . . . . . . 445.3 Equivariant maps to projective spaces . . . . . . . . . . . . . 465.4 Some explicit computations . . . . . . . . . . . . . . . . . . . 495.5 Cyclic and dihedral groups: Compressibility of P1 . . . . . . 515.6 Strongly incompressible curves for even cyclic groups . . . . 535.7 Strongly incompressible curves for even dihedral groups . . . 555.7.1 The Klein 4-group . . . . . . . . . . . . . . . . . . . . 555.7.2 Even dihedral groups of order ≥ 8 . . . . . . . . . . . 585.8 Polyhedral groups . . . . . . . . . . . . . . . . . . . . . . . . 615.8.1 Serre’s cohomological invariant . . . . . . . . . . . . . 615.8.2 Computation of the invariant for curves of genus ≤ 1 635.8.3 Strong incompressibility . . . . . . . . . . . . . . . . 676 Conclusions and open problems . . . . . . . . . . . . . . . . . 726.1 Finite group actions on conics . . . . . . . . . . . . . . . . . 726.2 Geometric Galois embedding problems . . . . . . . . . . . . 736.3 Strongly incompressible varieties . . . . . . . . . . . . . . . . 75Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77AppendixA Proof of Theorem 5.2 . . . . . . . . . . . . . . . . . . . . . . . 81vAcknowledgementsI would like to thank my advisor Zinovy Reichstein for introducing me tomany fascinating mathematical problems and ideas. I really appreciate hisextraordinary guidance and support during my time at UBC.I also want to thank my professors in the math department, especiallyJulia Gordon, Kalle Karu and Lior Silberman, for everything I learnt fromthem. Last but not least, thanks to Shane Cernele, Alex Duncan, JeromeLefebvre and Athena Nguyen for useful mathematical conversations.viChapter 1Introduction1.1 Finite subgroups of PGL1(A)The finite subgroups of PGL2(C) have been known for over a century: theseare cyclic, dihedral and the polyhedral groups A4, S4 and A5 (see, e.g.,[15]). Any two isomorphic finite subgroups of PGL2(C) are conjugate. Afinite group is said to be p-regular (resp. p-irregular) if its order is primeto (resp. divides) char(k), where k is a base field. If k is algebraicallyclosed, the classification of finite p-regular subgroups of PGL2(k) and theirconjugacy classes is identical to the complex case (see [28, §2.5]). Using thisresult as a starting point, A. Beauville [2] classified, up to conjugacy, thefinite p-regular subgroups of PGL2(k) over an arbitrary field k. X. Faber [11]completed this picture by classifying the p-irregular subgroups of PGL2(k),and describing their conjugacy classes.We describe the finite (constant) subgroups of (possibly non-split) ad-joint absolutely simple algebraic groups of type A1 over a field k, as wellas their conjugacy classes. To do so, we restrict our attention to arbitraryfields k of characteristic different from 2.It is well known that an adjoint absolutely simple algebraic group of typeA1 over k is of the form PGL1(A) for some quaternion algebra A = (a, b)2,where a, b ∈ k× (see, e.g., [16, §26.A and 26.B]). Alternatively, we may re-gard PGL1(A) as the automorphism group of the conic associated to thequadratic form q = 〈−a,−b, ab〉, which implies the existence of an isomor-phism PGL1(A) ∼= SO(q) (see, e.g., [8, Cor. 69.6]). Any smooth projectivecurve of genus 0 is isomorphic to one such conic, so we completely classifyfaithful actions of finite groups on curves of genus 0, up to equivalence.We will thus be interested in finite constant subgroups of SO(q) for anyternary nondegenerate quadratic form q. Replacing q by a scalar multiple ofitself does not alter its isometry group, so we may assume throughout thatq has discriminant 1.We first deal with the p-irregular subgroups of SO(q). Interestingly, weshow that this case reduces entirely to the classification in [11].11.1. Finite subgroups of PGL1(A)Theorem 1.1. Let k be a field of characteristic p > 2 and suppose thatSO(q) contains a p-irregular subgroup. Then q is isotropic, i.e., there existsan isomorphism SO(q) ∼= PGL2.It remains to classify the p-regular subgroups G of SO(q), so we mayassume henceforth that char(k) is prime to |G|. Over an algebraic closurek of k, we have that SO(q)(k) ∼= PGL2(k). Thus any finite subgroup G ofSO(q) embeds into PGL2(k), so it must be isomorphic to Z/nZ, D2n (thedihedral group of 2n elements), A4, S4 or A5. Theorem 1.2 below classi-fies these subgroups up to isomorphism and Theorem 1.3 up to conjugacy.Taking q ' 〈−1,−1, 1〉 in these theorems, we recover the results in [2].We will prove Theorem 1.2 in Section 3.1. Note that the classificationof polyhedral groups in parts (b) and (c) is hinted at in [29]; here we makeit explicit for completeness. In the next two theorems, we let ωn denote aprimitive n-th root of 1.Theorem 1.2. Let q be a nondegenerate quadratic form of discriminant 1.(a) The group D4 ∼= (Z/2Z)2 is always contained in SO(q). For n ≥ 3, thefollowing are equivalent:(i) The group SO(q) contains Z/nZ.(ii) The group SO(q) contains D2n.(iii) ωn + ω−1n ∈ k, and q represents 4− (ωn + ω−1n )2.(b) The group SO(q) contains A4 if and only if it contains S4, which happensif and only if q ' 〈1, 1, 1〉.(c) The group SO(q) contains A5 if and only if√5 ∈ k and q ' 〈1, 1, 1〉.We will prove Theorem 1.3 in Section 3.2. Our argument relies on Galoiscohomology techniques, building on the approach taken in [Beau10].Theorem 1.3. Let q = 〈−a,−b, ab〉 be a nondegenerate quadratic form.(a) The conjugacy classes of Z/2Z inside SO(q) are in natural bijective cor-respondence with the set D(q) ⊂ k×/k×2 consisting of nonzero squareclasses represented by q.(b) Let Qa,b = {(x, y) ∈ (k×/k×2)2| (ax, by)2∼= (a, b)2}. The symmet-ric group S3 = {s, t|s3 = t2 = (st)2 = 1} acts on Qa,b by settings · (x, y) = (−bxy, abx) and t · (x, y) = (x,−axy) for all (x, y) ∈ Qa,b.Then the conjugacy classes of (Z/2Z)2 inside SO(q) are in natural bi-jective correspondence with Qa,b/S3.21.2. Finite group actions on hyperelliptic curves(c) There is at most one conjugacy class of subgroups isomorphic to Z/nZ(n ≥ 3) inside SO(q).(d) Suppose that D2n is contained in SO(q) (n ≥ 3). We write βn =(ωn+ω−1n )2−4 and let D(〈1,−βn〉) consist of the nonzero square classesrepresented by 〈1,−βn〉, which forms a subgroup of k×/k×2. The squareclass ωn + ω−1n + 2 is contained in D(〈1,−βn〉); let C be the 2-elementsubgroup generated by this class. Then the conjugacy classes of D2n in-side SO(q) are in natural bijective correspondence with D(〈1,−βn〉)/C.(e) There is at most one conjugacy class of subgroups isomorphic to A4, S4or A5 inside SO(q).1.2 Finite group actions on hyperelliptic curvesLet k be an arbitrary base field. A hyperelliptic curve C/k is a smoothprojective geometrically irreducible curve of genus at least 2 endowed witha finite k-morphism C → P1 of degree 2. Equivalently, the function fieldk(C) is a regular quadratic field extension of the rational function field k(x)of genus at least 2.Problem 1.4 (Hyperelliptic Lifting Problem or HLP). Let G be a finitegroup and consider a faithfulG-action on P1 via some embeddingG ↪→ PGL2defined over k. Suppose that we have a central exact sequence1→ µ2 → G′ → G→ 1.Determine the conditions for the existence of a hyperelliptic curve C/kendowed with a faithful G′-action defined over k such that C/µ2 ∼= P1 asG-varieties.Over the complex numbers, finite group actions on hyperelliptic curveshave been studied extensively (see, e.g., [5, 33]). In that case, Problem 1.4 isalways solvable. Indeed, if G is a finite subgroup of PGL2, the correspond-ing G-action on P1 over C is unique up to G-equivariant isomorphism andthe results in [5] imply that any double cover of G acts on some hyperellip-tic curve in the desired way. Over a non-algebraically closed field, severalcomplications may arise in the study of Problem 1.4. First, the embeddingsof a finite group G into PGL2 are not necessarily conjugate, giving rise tonon-equivalent actions on P1 (cf. Theorem 1.3). On the other hand, thereare usually arithmetic constraints on the base field for the existence of ahyperelliptic curve with the required group action.31.3. Strongly incompressible curvesIn Chapter 4, we find a complete solution to the HLP in characteristic 0.More explicitly, for every faithful action on P1 by a finite group G definedover a field k of characteristic 0 and every exact sequence 1 → µ2 → G′ →G → 1, we determine necessary and sufficient conditions on k for the solv-ability of the HLP in Sections 4.2 to 4.4, and we describe the correspondingfamilies of solutions in Section Strongly incompressible curvesLet G be an algebraic group. A G-compression of a generically free G-variety X is a dominant G-equivariant rational map X 99K Y , where Yis also generically free. We say that X is strongly incompressible if everyG-compression of X is birational. This concept was introduced by Z. Re-ichstein in [24, §2], where the author asks for a classification of stronglyincompressible G-varieties (see also [25, §7.1]).A related problem arises when we only consider self rational maps.More precisely, given a generically free G-variety X, is every dominant G-equivariant rational map X 99K X a birational isomorphism? Even whenG is trivial, this appears to be an interesting problem in many contexts. In[6], X. Chen proves that every dominant self rational map of a very generalprojective K3 surface of genus g ≥ 2 is birational (see [7] for generalizations).If a finite group G does not act faithfully on any curve of genus ≤ 1,then there exist strongly incompressible complexG-curves (see [24, Example6]). N. Fakhruddin and R. Pardini have independently found examples ofstrongly incompressible complex G-surfaces for certain finite groups G. Toour best knowledge, no examples of strongly incompressible varieties areknown in higher dimensions.If the base field k has characteristic p > 0, there exist no strongly in-compressible G-varieties for any finite group G. We sketch a proof of thisfact. Let X be a faithful G-variety and let FX/Spec(k) : X → X(p) be the rel-ative Frobenius morphism associated to X (see [21, §3.2.4] for details). Byfunctoriality, we may endow X(p) with an action of the group G(p), which iscanonically isomorphic to G (recall that G is a finite constant group). Thisaction makes FX/Spec(k) into a dominant G-equivariant morphism, which hasdegree pdim(X) by [21, Cor. 2.27]. To complete the proof, we must show thatthe G-action on X(p) is faithful. If N is the kernel of such action, we musthave k(X(p)) ⊂ k(X)N ⊂ k(X), where the inclusion k(X(p)) ⊂ k(X) is apurely inseparable extension induced by FX/Spec(k). Thus k(X)/k(X)N isboth Galois and purely inseparable, which implies that N is trivial.41.3. Strongly incompressible curvesIn Chapter 5, we study the question of existence of strongly incompress-ible G-curves for every finite group G. We may assume that the base field khas characteristic 0. We settle the classification problem for G-curves raisedin [24], by considering finite groups G that can act on a curve of genus ≤ 1.In Section 5.2, we show that strongly incompressible G-curves do exist if Gdoes not act faithfully on any curve of genus 0.Theorem 1.5 (see Theorem 5.4). Suppose that G cannot act faithfully on acurve of genus 0 via k-morphisms. Then there exists a strongly incompress-ible G-curve defined over k.For finite groups G that can act faithfully on a curve of genus 0 overk (recall that these are always cyclic, dihedral, or polyhedral groups), thesituation is more delicate. In particular, it is important to decide whethera faithful G-curve X can be G-compressed to P1, provided that there existsa faithful G-action on the projective line. To this end, we make a smalldetour in Section 5.3 and, given a projective representation G → PGL(V ),we construct a cohomological invariant associated to any faithful G-varietyX, which allows us to determine whether X can be mapped G-equivariantlyto P(V ). In Section 5.4, we compute the invariant for certain group actionson the projective line.We study the existence of strongly incompressible curves for groups thatcan act faithfully on a curve of genus 0 in Sections 5.5 to 5.8. Our resultsare summarized in the following theorem. For a definition of cohomological2-dimension of a field k, denoted by cd2(k), we refer the reader to [31, I.§3].We remark that k has cohomological 2-dimension 0 if and only if everyalgebraic extension of k is quadratically closed (see [9, Lemma 2]).Theorem 1.6. Let ωn be a primitive n-th root of 1 (n ≥ 2).(a) (Thm. 5.4, Prop. 5.14) Let G be either Z/nZ or D2n, where n isodd. Then there exist strongly incompressible G-curves if and only ifωn + ω−1n 6∈ k.(b) (Thm. 5.4, Prop. 5.13 and 5.16) Suppose that n is even. Then thereexist strongly incompressible Z/nZ-curves if and only if ωn 6∈ k.(c) (Prop. 5.17) There exist strongly incompressible (Z/2Z)2-curves if andonly if cd2(k) > 0.(d) (Thm. 5.4, Prop. 5.23) Suppose that n ≥ 4 is even. Then there existstrongly incompressible D2n-curves if and only if either ωn + ω−1n 6∈ k,or cd2(k) > 0.51.3. Strongly incompressible curves(e) (Prop. 5.29) Let G be a polyhedral group, i.e., G = A4, S4, or A5. Thenthere exist strongly incompressible G-curves if and only if cd2(k) > 0.In particular, we note the following corollary of the above results, whichanswers the strong incompressibility problem for curves over an algebraicallyclosed field, as posed in [24].Corollary 1.7. Let G be a finite group and let the base field k be algebraicallyclosed. Then there exists a strongly incompressible G-curve if and only if Gdoes not act faithfully on P1, i.e., G is not cyclic, dihedral, A4, S4 or A5.6Chapter 2Notation and preliminariesIn this thesis, we let k denote a base field and we write ks (resp. k) forits separable (resp. algebraic) closure. A k-variety X is a geometricallyreduced scheme of finite type over k (not necessarily irreducible). A point of avariety means a geometric point, unless stated otherwise. The word “curve”is reserved for a geometrically irreducible smooth projective 1-dimensionalvariety. A curve C/k is said to be hyperelliptic if its genus is at least 2 andthere exists a finite k-morphism C → P1 of degree 2.2.1 Quaternion algebrasGiven a central simple algebra A, we will denote its Brauer class by [A]. Asusual, the symbol (a, b)2 denotes the quaternion algebra with basis 1, i, j, ij,subject to the relations i2 = a, j2 = b and ij + ji = 0. The following simpleobservation will be used repeatedly in the sequel.Lemma 2.1. Let k(x) be a rational function field over k, and suppose thatthe quaternion algebra (f(x), g(x))2 is split over k(x), where f, g ∈ k[x] areseparable. Then f(α) is a square in k(α) for any root α ∈ k of g.Proof. Since the quaternion algebra (f(x), g(x))2 is split, there exist coprimepolynomials p, q, r ∈ k[x] such that the polynomial identityf(x)p(x)2 + g(x)q(x)2 = r(x)2holds. Substituting α in the above identity implies that f(α)p(α)2 = r(α)2.Note that p(α) = 0 implies r(α) = 0. Conversely, suppose that r(α) = 0.Then α is a root of f(x)p(x)2 of multiplicity at least 2, which implies thatp(α) = 0 since f is separable. It follows that r(α) = 0 if and only if p(α) = 0.Assume for the sake of contradiction that p(α) = r(α) = 0. Then itfollows that α is a root of g(x)q(x)2 of multiplicity at least 2. Since g isseparable, we obtain that q(α) = 0. Hence α is a common root of p, q, r,which is impossible since they are relatively prime. This contradiction showsthat p(α)r(α) 6= 0 and therefore f(α) = r(α)2p(α)−2 ∈ k(α)×2.72.2. Roots of unity2.2 Roots of unityFor each positive integer n, let ωn denote a primitive n-th root of unity. Wemay select the system {ωn} satisfying the following compatibility condition:if m/n is a power of 2, then ωn = ωm/nm . For convenience, we also setαn = (ωn + ω−1n )/2 and βn = α2n − 1.The following lemma provides some useful computations.Lemma 2.2. Suppose that k contains αn.(a) We must have ωrn + ω−rn ∈ k for every natural number r. In particular,the condition k 3 αn is independent of the choice of ωn.(b) If n is odd, then α2n ∈ k.(c) If n is even, then (1 +ω−1n )n = (1 +ωn)n = −(2αn + 2)n/2 = −(2α2n)n.(d) Suppose that n ≡ 0 (4) and −1 ∈ k×2. Then it follows that ωn ∈ k.Proof. (a) This follows from the well known fact that (xr + x−r)/2 is apolynomial in (x+ x−1)/2 with integer coefficients.(b) Note that ζ2n = ωn+22n = −ωn is a primitive 2n-th root of unity, whichsatisfies (ζ2n + ζ−12n )/2 = −αn ∈ k. By part (a), we conclude that α2n ∈ k.(c) The first and last equalities are straightforward. To prove the middleequality, note that(1 + ωn)n = (1 + 2ωn + ω2n)n/2 = ωn/2n (2αn + 2)n/2 = −(2αn + 2)n/2.(d) Note that ω4ωn is an n-th root of unity, so ω4ωn = ωrn for some naturalnumber r. Thus, ωn = 2αn + (ωn − ω−1n ) = 2αn + ω−14 (ωrn + ω−rn ) ∈ k.8Chapter 3Finite group actions oncurves of genus zero3.1 Existence of finite subgroupsLet k denote a base field of characteristic different from 2.Lemma 3.1. Let q be a nondegenerate ternary quadratic form over k andlet M be an element of SO(q)(k). Then the following results hold.(a) Suppose that M is diagonalizable over k. Then its eigenvalues are 1, λ,and λ−1 for some λ ∈ k×. If λ 6= ±1, then q becomes isotropic overk(λ), which is an extension of k of order dividing 2.(b) Suppose that M is a nontrivial unipotent matrix. Then q must beisotropic.Proof. Let Q be the matrix associated to q. Note that M−1 = Q−1 TM Q,whence the characteristic polynomial P of M satisfies P (x) = −x3P (1/x).It follows easily that the eigenvalues of M must be 1, λ, λ−1 for some λ ∈ k.Suppose first that M is diagonalizable. Taking the trace of M , we obtainthat λ + λ−1 ∈ k, whence [k(λ) : k] is 1 or 2. We show that q becomesisotropic over k(λ), provided that λ 6= ±1. Indeed, over this field we canselect an eigenvector v of M associated to the eigenvalue λ. Then, wecompute q(v) = q(Mv) = λ2q(v), whence q(v) = 0. This completes theproof of part (a).Suppose now that M is a nontrivial unipotent matrix. Then we can findnonzero vectors v1, v2 such that Mv1 = v1 and Mv2 = v1 + v2 (this followseasily after conjugating M into Jordan canonical form). Let bq be the sym-metric bilinear form associated to q. Note that bq(v1, v2) = bq(Mv1,Mv2) =q(v1) + bq(v1, v2), whence q(v1) = 0. This finishes the proof.Proof of Theorem 1.1. Let G be a p-irregular subgroup inside SO(q) and letM ∈ G be any element of order p. Note that 0 = Mp−I = (M−I)p, whence93.1. Existence of finite subgroupsM is a unipotent matrix. By Lemma 3.1(b), it follows that q is isotropic, sowe obtain that SO(q) ∼= PGL2.Proof of Theorem 1.2. (a) The first statement is trivial: the diagonal sub-group D0 ⊂ SO(q) is isomorphic to (Z/2Z)2. Assume henceforth that n ≥ 3.It suffices to prove that Z/nZ ⊂ SO(q) ⇒ αn ∈ k and q represents −βn ⇒D2n ⊂ SO(q).To prove the first implication, letM be an element of order n in SO(q)(k).By Lemma 3.1(a), we may assume that the eigenvalues of M are 1, ωn andω−1n , after replacing M by a power of itself if necessary. Using Lemma 3.1(a)again, we see that αn ∈ k and q becomes isotropic over k(ωn) = k(√βn). Itsuffices to prove that q is isotropic over k(√βn) if and only if q represents−βn. If q is isotropic over k, then q is universal, so there is nothing to prove.Suppose that q is anisotropic over k. It follows from [8, Prop. 34.8] thatq ' q0⊗Nk(√βn)/k ⊥ q1 for some nondegenerate quadratic forms q0, q1, whereq1 is anisotropic over k(√βn). If q is isotropic over k(√βn), we conclude thatq 6= q1 and thus dim(q0) = dim(q1) = 1 (since Nk(√βn)/k ' 〈1,−βn〉). Itfollows that q1 ∼= 〈−βn〉 by taking discriminants, whence q represents −βn.Conversely, suppose that the latter holds. Then q ' 〈−βn,−γ, βnγ〉 forsome γ ∈ k×, so it follows that q is isotropic over k(√βn).Suppose next that αn ∈ k and q represents −βn. We may thus assumethat q = 〈−βn,−γ, βnγ〉 for some γ ∈ k×. The matricess =1 0 00 αn βn0 1 αn , t =−1 0 00 1 00 0 −1 ,are contained in SO(q)(k) and satisfy sn = t2 = (st)2 = 1, whence theygenerate a subgroup isomorphic to D2n. The proof of this part is complete.(b) Let q0 = 〈1, 1, 1〉. It clearly suffices to prove that A4 ⊂ SO(q)⇒ q 'q0 ⇒ S4 ⊂ SO(q). We first prove the rightmost implication. Note that S4acts linearly on k3 by rotations on the cube with vertices (±1,±1,±1) andthe corresponding linear representation has determinant 1. The form q0 isinvariant under this action and therefore S4 embeds into SO(q0). If q ' q0,then SO(q) ∼= SO(q0) and the result follows.We now prove that A4 ⊂ SO(q)⇒ q ' q0. Recall that A4 acts linearly onk3 by rotations on the tetrahedron with vertices (1, 2, 12), where i = ±1.This representation is absolutely irreducible and leaves q0 invariant, i.e., wehave a linear representation ρ : A4 ↪→ SO(q0) defined over k. We claim thatany absolutely irreducible representation of a finite group admits at mostone nontrivial invariant quadratic form, up to a scalar. Modulo this claim, it103.2. Conjugacy classes of subgroupsis easy to finish the proof. Indeed, it follows that q ' c · q0 for some c ∈ k×;taking discriminants, we conclude that c = 1.It remains to give a proof of the claim. Let G be a finite group, letV be an absolutely irreducible G-module, and let q1, q2 be a nontrivial G-invariant quadratic form. Since Ker(qi) (i = 1, 2) is G-invariant, it mustbe trivial. It follows that q1, q2 must be nondegenerate, so we may viewthem as G-equivariant isomorphisms V → V ∗. Hence (q2)−1 ◦ q1 is a G-equivariant automorphism of V . Passing to the algebraic closure and usingSchur’s lemma, we conclude that (q2)−1 ◦ q1 must be a scalar.(c) Note that A5 ⊂ SO(q) ⇒ A4 ⊂ SO(q) ⇒ q ' q0 by part (b). SinceA5 contains elements of order 5, it is necessary that ω5 + ω−15 ∈ k, whichhappens if and only if√5 ∈ k.Conversely, if√5 ∈ k, the group A5 acts linearly on k3 by rotationson the icosahedron with vertices (±φ,±1, 0), (0,±φ,±1), (±1, 0,±φ), whereφ = (1 +√5)/2, and this action preserves q0. The result readily follows.Example 3.2. Let k = Q. A primitive n-th root of 1 is given by ωn =exp(2pii/n). Recall that ωn + ω−1n ∈ Q if and only if n = 1, 2, 3, 4 or 6. Thegroup SO(q) contains Z/4Z and D8 if and only if q represents 1. Moreover,SO(q) contains Z/3Z and D6 if and only if it contains Z/6Z and D12 if andonly if q represents 3.3.2 Conjugacy classes of subgroupsThe purpose of this section is to prove Theorem 1.3. We recall the followingconstruction for convenience.Construction 3.3. ([2, §2]) Let G be an algebraic group defined over kand let H ⊂ G(k) be a subgroup. Fix a separable closure ks of k andset Γ = Gal(ks/k). Define the pointed set Embi(H,G(k)) as the set ofembeddings H ↪→ G(k) which are conjugate by an element of G(ks) to thenatural inclusion i : H ↪→ G(k), modulo conjugacy by an element of G(k).Also, define the pointed set Conj(H,G(k)) consisting of subgroups of G(k)which are conjugate to H over G(ks), modulo conjugacy over G(k).The centralizer of H in G, denoted by Z, will be a closed subgroup of Gdefined over k (cf. [4, Ch. 1, §1.7]). The kernel H1(k, Z)0 of the natural mapH1(k, Z)→ H1(k,G) is isomorphic to Embi(H,G(k)) as pointed sets. Thenormalizer N of H in G(ks) acts on 1-cocycles Γ → Z(ks) in the followingway: an element n ∈ N sends σ 7→ aσ to σ 7→ n−1aσσ(n). This (right)113.2. Conjugacy classes of subgroupsaction descends to H1(k, Z) and preserves H1(k, Z)0. Then there is anisomorphism of pointed sets between H1(k, Z)0/N and Conj(H,G(k)).Now recall that any two isomorphic finite subgroups (of order prime tochar(k)) of SO(q)(ks) ∼= PGL2(ks) are conjugate. Therefore, the conjugacyclasses of finite subgroups of SO(q) of the same isomorphism type as someparticular subgroup H ⊂ SO(q) are in natural bijective correspondence withConj(H,SO(q)(k)), independently of the choice of H.We now state some basic facts about the structure of SO(q). The proofsare easy and are left to the reader. In the sequel, we write diag(a1, . . . , an)for the diagonal matrix with entries a1, . . . , an along the diagonal.Lemma 3.4. Let q = 〈−a,−b, ab〉 be a nondegenerate quadratic form. If His a finite subgroup of SO(q), let Z be the centralizer of H in SO(q) and letN be the normalizer of H in SO(q)(ks).(a) Let H ∼= Z/2Z be generated by the diagonal matrix diag(1,−1,−1).Then we have thatZ ={((detM)−1 00 M): M ∈ O(〈−b, ab〉)}∼= O(〈−b, ab〉)and N = Z(ks).(b) Let H ∼= (Z/2Z)2 be the diagonal subgroup inside SO(q). Then we havethat Z = H and N is isomorphic to S4. Explicitly, if we set u =√−aand v =√−b, the matrices0 vu−1 00 0 uv−1 0 0 ,−1 0 00 0 −u0 −u−1 0 ,generate a subgroup N ′ ⊂ SO(q)(ks) isomorphic to S3 and N = HoN ′.(c) Let n ≥ 3 and suppose that H ∼= Z/nZ is contained in SO(q). Us-ing the same notation from Theorem 1.2, we may assume that q =〈−βn,−γ, βnγ〉 and H is generated by the matrixs =1 0 00 αn βn0 1 αn .Then we have thatZ ={(1 00 M): M ∈ SO(〈−γ, βnγ〉)}∼= SO(〈−γ, βnγ〉).123.2. Conjugacy classes of subgroups(d) Let n ≥ 3 and suppose that H ∼= D2n is contained in SO(q). As before,assume that q = 〈−βn,−γ, βnγ〉 and H is generated by the matricess =1 0 00 αn βn0 1 αn , t =−1 0 00 1 00 0 −1 .Then we have that Z ∼= Z/2Z is generated by diag(1,−1,−1) and thematricess′ =1 0 00 α2n 2α2nβ2n0 (2α2n)−1 α2n , t =−1 0 00 1 00 0 −1 ,generate N ∼= D4n inside SO(q)(ks).(e) Let H = A4, S4 or A5 and suppose that H is contained in SO(q). Thenthe centralizer Z is trivial.Remark 3.5. Since any two finite isomorphic subgroups H1 and H2 ofSO(q) are conjugate over ks, their centralizers will also be conjugate over ks(in particular, they must be isomorphic). However, they are not necessarilyisomorphic over k. For a concrete example, take k = R, q = 〈−1,−1, 1〉, H1generated by diag(1,−1,−1) and H2 generated by diag(−1,−1, 1). A sim-ple computation shows that Z(H1) ∼= O(〈−1, 1〉) and Z(H2) ∼= O(〈−1,−1〉).These groups are not isomorphic; the identity component Z(H1)◦ is isomor-phic to Gm while Z(H2)◦ ∼= SO2, which is a non-split torus over R.Remark 3.6. Suppose we are in the situation of Lemma 3.4(c) with γ = 1.Then, q = 〈−βn,−1, βn〉 ' 〈−1,−1, 1〉 and SO(q) ∼= PGL2, so we are dealingwith the case studied in [2]. Any two cyclic subgroups of order n inside SO(q)are conjugate over k (see Theorem 1.3), so the centralizer of such a subgroupis unique up to conjugacy. By Lemma 3.4(c), it must be isomorphic toSO(〈−1, βn〉), which is a split torus if and only if βn = 14(ωn − ω−1n )2 ∈ k×2if and only if ωn ∈ k (since ωn + ω−1n ∈ k). So in general the centralizer isnot isomorphic to the split torus Gm, contrary to an assertion made in theproof of [2, Thm. 4.2] and it might have nontrivial cohomology. However,the final result in [2] is unaffected since the map H1(k, Z)→ H1(k,G) stillhas trivial kernel (see Theorem 1.3).133.2. Conjugacy classes of subgroupsWe now recall some facts about the Galois cohomology of orthogonalgroups of quadratic spaces (see [16, §29.E] for details). Let q be any non-degenerate quadratic form of dimension n defined over k. The cohomologyset H1(k,O(q)) classifies isometry classes of n-dimensional nondegeneratequadratic forms over k, while H1(k, SO(q)) classifies isometry classes of n-dimensional quadratic forms q′ over k such that disc(q′) = disc(q). Thenatural map H1(k, SO(q))→ H1(k,O(q)) is injective. Let D0 ∼= (Z/2Z)n−1and D ∼= (Z/2Z)n be the subgroups of diagonal matrices inside SO(q) andO(q), respectively. We have a commutative diagram1 // D0i// Djdet // Z/2Z // 1SO(q) // O(q)where the top row is exact. This induces a diagram on cohomology1 // H1(k,D0)i∗// (k×/k×2)nj∗p// k×/k×2 // 1H1(k, SO(q)) // H1(k,O(q))where p : (c1, . . . , cn) 7→ c1 . . . cn is the product map. We have thus identifiedH1(k,D0) with the elements (c1, . . . , cn) ∈ (k×/k×2)n such that c1 . . . cn = 1.In what follows, we will abuse notation and refer to quadratic forms aselements of the cohomology sets H1(k,SO(q)) and H1(k,O(q)). The readershould bear in mind that we are tacitly referring to their isometry classes.Lemma 3.7. Suppose that q ' 〈b1, . . . , bn〉 is a nondegenerate quadraticform.(a) The map j∗ takes (c1, . . . , cn) to 〈c1b1, . . . , cnbn〉 and consequently, i∗takes (c1, . . . , cn) with c1 . . . cn = 1, to 〈c1b1, . . . , cnbn〉.(b) Let q′ ' 〈d〉 ⊥ q and define f : SO(q)→ SO(q′) by sendingM 7→(1 00 M).The induced map f∗ : H1(k, SO(q))→ H1(k, SO(q′)) sendsq′′ 7→ 〈d〉 ⊥ q′′.143.2. Conjugacy classes of subgroups(c) Let q′ ' 〈d〉 ⊥ q and define f : O(q)→ SO(q′) by sendingM 7→((detM)−1 00 M).The induced map f∗ : H1(k,O(q))→ H1(k,SO(q′)) sendsq′′ 7→ 〈disc(q′′)disc(q′)〉 ⊥ q′′.Proof. (a) This is well known; see, e.g., the proof of [26, Lemma 4.3].(b) Let iq : D0,q ↪→ SO(q) and iq′ : D0,q′ ↪→ SO(q′) be the embeddingscorresponding to the subgroups of diagonal matrices. It is easy to seethat the restriction f |D0,q : D0,q → D0,q′ induces a map F : H1(k,D0,q) →H1(k,D0,q′) sending (c1, . . . , cn) to (1, c1, . . . , cn). Hence, if q′′ = 〈x1, . . . , xn〉is any quadratic form such that disc(q′′) = disc(q), it follows thatf∗(q′′) = f∗(iq∗(x1/b1, . . . , xn/bn)) = iq′∗(F (x1/b1, . . . , xn/bn)) = 〈d〉 ⊥ q′′.(c) Let jq : Dq ↪→ O(q) and iq′ : D0,q′ ↪→ SO(q′) be as before. Notethat the restriction f |Dq : Dq → D0,q′ induces a map F : H1(k,Dq) →H1(k,D0,q′) sending (c1, . . . , cn) to (c1 . . . cn, c1, . . . , cn). The result followsusing a similar argument to the one in part (b).We are ready to prove the main result of this section.Theorem 3.8. Let q = 〈−a,−b, ab〉 be a nondegenerate quadratic form.(a) The conjugacy classes of embeddings Z/2Z ↪→ SO(q) are in natural bi-jective correspondence with the set D(q) ⊂ k×/k×2 consisting of nonzerosquare classes represented by q.(b) The conjugacy classes of embeddings (Z/2Z)2 ↪→ SO(q) are in naturalbijective correspondence with Qa,b, as defined in Theorem 1.3(b).(c) There is at most one conjugacy class of embeddings Z/nZ ↪→ SO(q)(n ≥ 3).(d) Suppose that D2n is contained in SO(q) (n ≥ 3). Then the conjugacyclasses of embeddings D2n ↪→ SO(q) are in natural bijective correspon-dence with D(〈1,−βn〉).(e) There is at most one conjugacy class of embeddings of A4, S4 or A5 intoSO(q).153.2. Conjugacy classes of subgroupsProof. (a) Let H ∼= Z/2Z be generated by diag(1,−1,−1) inside SO(q). ByLemma 3.4(a), its centralizer Z is isomorphic to O(〈−b, ab〉). By Lemma3.7(c), the inclusion Z ↪→ SO(q) induces a map H1(k, Z) → H1(k, SO(q))sending a binary quadratic form q′ to 〈disc(q′)〉 ⊥ q′. Hence, the kernelH1(k, Z)0 consists of the binary quadratic forms q′ such that 〈disc(q′)〉 ⊥q′ ' q (in particular, disc(q′) ∈ D(q)). Define a map Ψ: H1(k, Z)0 → D(q)sending q′ 7→ disc(q′). If q′, q′′ ∈ H1(k, Z)0 satisfy Ψ(q′) = Ψ(q′′), then〈Ψ(q′)〉 ⊥ q′ ' 〈Ψ(q′′)〉 ⊥ q′′ ' q implies q′ ' q′′ by Witt’s CancellationTheorem, so Ψ is injective. To prove that Ψ is surjective, let d ∈ D(q) bearbitrary. Then q = 〈d〉 ⊥ q′ for some quadratic form q′. Taking discrimi-nants yields disc(q′) = d. This implies that q′ ∈ H1(k, Z)0 and Ψ(q′) = d.This proves that H1(k, Z)0 is in natural bijection with D(q).(b) Let H ∼= (Z/2Z)2 be the subgroup D0 of diagonal matrices insideSO(q). By Lemma 3.4(b), we have that Z = H and the map H1(k, Z) →H1(k, SO(q)) sends (x, y, z), with xyz = 1, to 〈−ax,−by, abz〉. Therefore,(x, y, z) ∈ H1(k, Z)0 if and only if 〈−ax,−by, abz〉 ' 〈−a,−b, ab〉, which isequivalent to (ax, by)2 ∼= (a, b)2. It follows easily that H1(k, Z)0 ∼= Qa,b.(c) We may assume that we are in the situation of Lemma 3.4(c), i.e.,q = 〈−βn,−γ, βnγ〉 and the centralizer Z is isomorphic to SO(〈−γ, βnγ〉).By Lemma 3.7(b), the natural map H1(k, Z) → H1(k, SO(q)) sends a bi-nary quadratic form q′ (with disc(q′) = −βn) to 〈−βn〉 ⊥ q′. By Witt’sCancellation Theorem, the kernel H1(k, Z)0 is trivial and the claim follows.(d) We may assume that q = 〈−βn,−γ, βnγ〉 and H ∼= D2n is as inLemma 3.4(d). The centralizer Z ∼= Z/2Z is generated by diag(1,−1,−1).Let D0 ⊂ SO(q) be the subgroup of diagonal matrices; the natural inclu-sion Z ↪→ D0 induces a map H1(k, Z) ∼= k×/k×2 → H1(k,D0) sendingc ∈ k×/k×2 to (1, c, c). Therefore the natural map H1(k, Z)→ H1(k, SO(q))sends c ∈ k×/k×2 to 〈−βn,−cγ, cβnγ〉. By Witt’s Cancellation Theo-rem, the kernel H1(k, Z)0 is given by those square classes c such that〈−cγ, cβnγ〉 ' 〈−γ, βnγ〉. It follows easily that c ∈ H1(k, Z)0 if and only if〈1,−βn〉 represents c, i.e., H1(k, Z)0 ∼= D(〈1,−βn〉).(e) This is immediate from Lemma 3.4(e).Proof of Theorem 1.3. In view of Theorem 3.8, it suffices to analyze thedifferent actions of the normalizers N described in Lemma 3.4 on H1(k, Z)0.Parts (c) and (e) are immediate because H1(k, Z)0 itself is trivial, so we mayfocus our attention on parts (a), (b) and (d).(a) Note that N = Z(ks), whence the action of N on H1(k, Z) (and afortiori H1(k, Z)0) is trivial. This finishes the proof.(b) Recall that N = H oN ′, where N ′ ∼= S3. Since H acts trivially on163.2. Conjugacy classes of subgroupsH1(k, Z), we only need to determine how N ′ acts on H1(k, Z)0. As we sawbefore, the subgroup N ′ is generated by the matricess =0 vu−1 00 0 uv−1 0 0 , t =−1 0 00 0 −u0 −u−1 0 ,where u =√−a and v =√−b. Note that a 1-cocycle l : Gal(ks/k)→ Z(ks)representing (x, y) ∈ Qa,b is given byσ 7→ lσ = diag(x−11 σ(x1), y−11 σ(y1), x−11 σ(x1)y−11 σ(y1)),where x21 = x and y21 = y. We computes−1lσσ(s) = diag((vx1y1)−1σ(vx1y1), (uvx1)−1σ(uvx1), (uy1)−1σ(uy1)),andt−1lσσ(t) = diag(x−11 σ(x1), (ux1y1)−1σ(ux1y1), (uy1)−1σ(uy1)).Thus, the 1-cocycles σ 7→ s−1lσσ(s) and σ 7→ t−1lσσ(t) correspond to theelements (−bxy, abx) and (x,−axy) in Qa,b, respectively. Hence N ′ ∼= S3acts on Qa,b as claimed and the result follows easily.(d) Recall that N ∼= D4n is generated bys′ =1 0 00 α2n 2α2nβ2n0 (2α2n)−1 α2n , t =−1 0 00 1 00 0 −1 .Note that H is the subgroup of N generated by s = (s′)2 and t. Clearly theaction of H on H1(k, Z)0 is trivial, so we only need to compute the actionof s′ on H1(k, Z)0. Unraveling the identification H1(k, Z)0 ∼= D(〈1,−βn〉),we see that a 1-cocycle l : Gal(ks/k) → Z(ks) representing c ∈ H1(k, Z)0is given by σ → lσ = diag(1, c−11 σ(c1), c−11 σ(c1)), where c1 ∈ k×s satisfiesc21 = c. Then we compute the 1-cocycleσ 7→ (s′)−1lσσ(s′) = diag(1, (α2nc1)−1σ(α2nc1), (α2nc1)−1σ(α2nc1)).It corresponds to the square class of (α2nc1)2, which is precisely 2(αn + 1)c.This completes the proof.Remark 3.9. Suppose we are in the situation of Theorem 1.3(d) and n isodd. Then note that 2(αn+1) = (2α2n)2 ∈ k×2 by Lemma 2.2(b). Therefore,the conjugacy classes of D2n are in natural bijective correspondence withD(〈1,−βn〉) for n odd. This is not necessarily true for even n.173.2. Conjugacy classes of subgroupsWe now make the correspondences in parts (a), (b) and (d) of Theorem3.8 more explicit, by exhibiting representatives for each conjugacy class.• Let q = 〈−a,−b, ab〉, let d ∈ D(q) and let q′ = 〈d, x, y〉 be a quadraticform isometric to q. Select P ∈ GL3(k) such that q = q′ ◦P . Then theelement d ∈ D(q) corresponds to the conjugacy class of the embeddingZ/2Z ∼= P−1HP ↪→ SO(q), where H is generated by diag(1,−1,−1).• Let q = 〈−a,−b, ab〉, let (x, y) ∈ Qa,b and let q′ = 〈−ax,−by, abxy〉.Select P ∈ GL3(k) such that q = q′◦P . The element (x, y) correspondsto the conjugacy class of the embedding (Z/2Z)2 ∼= P−1D0P ↪→ SO(q),where D0 is the subgroup of diagonal matrices in SO(q).• Let k contain αn, let q = 〈−βn,−γ, βnγ〉 and let c ∈ D(〈1,−βn〉).Then the quadratic form q′ = 〈−βn,−cγ, cβnγ〉 is isometric to q, so wemay select P ∈ GL3(k) such that q = q′◦P . The element c correspondsto the conjugacy class of the embedding D2n ∼= P−1HP ↪→ SO(q),where H ∼= D2n is as in Lemma 3.4(d).Recall that if q is isotropic, then SO(q) ∼= PGL2, so we recover thecase studied in [2]. Using the above correspondences, we can fully describethe embeddings of finite groups into PGL2, up to conjugacy. We collecttheir descriptions here for convenience, as these results will be needed in theremainder of the thesis. The details of the proof are left to the reader.Proposition 3.10. (a) The embeddings Z/2Z ↪→ PGL2 are parametrizedby k×/k×2, up to conjugacy, and for every a ∈ k×, the embeddingρa : −1 7→(0 a1 0)(3.1)corresponds to a ∈ k×/k×2.(b) The conjugacy classes of embeddings (Z/2Z)2 ↪→ PGL2 are parametrizedby the pairs (a, b) ∈ (k×/k×2)2 such that the quaternion algebra (a, b)2is split. Denote the corresponding embedding by ρ(a,b) and fix generatorse1, e2 of (Z/2Z)2. We have the following three cases:• If both a and b are non-squares, we haveρ(a,b) : e1 7→(λ −a1 −λ), e2 7→(0 a1 0), (3.2)where λ2 − a = b (we can find such λ ∈ k because (a, b)2 is split).183.2. Conjugacy classes of subgroups• If a ∈ k×2, we haveρ(a,b) : e1 7→(0 b1 0), e2 7→(−1 00 1). (3.3)• If b ∈ k×2, we haveρ(a,b) : e1 7→(−1 00 1), e2 7→(0 a1 0). (3.4)(If both a and b are squares, the last two embeddings are conjugate, andwe will assume ρ(a,b) is given by (3.4).)(c) Let k contain αn, where n ≥ 3. Then there exists a unique conjugacyclass of embeddings Z/nZ ↪→ PGL2, which is represented byρ : σ 7→(αn + 1 βn1 αn + 1), (3.5)where σ is a generator of Z/nZ.(d) Let k contain αn, where n ≥ 3. The conjugacy classes of embeddingsD2n ↪→ PGL2(k) are parametrized by the set D(〈1,−βn〉) of nonzerosquare classes represented by the binary quadratic form x2 − βny2. Thecorrespondence is as follows: to the class a of the element a = x2−βny2(x, y ∈ k), we assignρa : σ 7→(αn + 1 βn1 αn + 1), τ 7→(x −yβny −x), (3.6)where σ, τ ∈ D2n satisfy σn = τ2 = (στ)2 = 1.(e) Let k contain elements a, b such that a2 +b2 = −1. Then the embeddingsof A4 and S4 into PGL2 are all conjugate. The matricesR :=(a− b− 1 a+ b+ 1a+ b− 1 b− a− 1), S :=(1 1−1 1), (3.7)satisfy R3 = S4 = (R−1S)2 = 1 in PGL2, and thus generate a sub-group isomorphic to S4. Moreover, R and S′ := S2 satisfy R3 = S′2 =(R−1S′)3 = 1 in PGL2 and generate a subgroup isomorphic to A4.193.2. Conjugacy classes of subgroups(f) Let k contain elements a, b such that a2 + b2 = −1 and suppose that5 ∈ k×2. Then the group A5 embeds into PGL2 in a unique way, up toconjugacy. Write φ = (1 +√5)/2 and consider the matrixU :=(φa+ φ−1 φb− 1φb+ 1 −φa+ φ−1). (3.8)Then R3 = U5 = (R−1U)2 = 1 in PGL2, where R is as in (3.7), andthese two matrices generate a subgroup isomorphic to A5.20Chapter 4Finite group actions onhyperelliptic curves4.1 Equivalent characterizations of the HLPLet k denote a base field of characteristic 0 and suppose that we have acentral exact sequence 1→ µ2 → G′ → G→ 1 of finite groups. A projectiverepresentation ρ : G ↪→ PGL2 defined over k induces a G-variety structureon the projective line, which we denote by P1ρ . We set k(x) = k( P1ρ ) andk(t) = k( P1ρ )G, where k(x)/k(t) is G-Galois.Proposition 4.1. The following are equivalent.(a) There exists a hyperelliptic curve C/k endowed with a faithful G′-actionsuch that C/µ2 ∼= P1ρ as G-varieties.(b) Let c ∈ H1(k(t), G) be the class corresponding to the G-Galois fieldextension k(x)/k(t) and let δ : H1(k(t), G) → H2(k(t), µ2) be the con-necting homomorphism induced by 1 → µ2 → G′ → G → 1. Then δ(c)is trivial.Proof. Suppose first that (a) holds. It follows that the function field ex-tension k(C)/k(t) is G′-Galois and the natural morphism H1(k(t), G′) →H1(k(t), G) maps its class to c. The long exact sequence in cohomologyH1(k(t), µ2)→ H1(k(t), G′)→ H1(k(t), G)→ H2(k(t), µ2)then implies that δ(c) is trivial.Suppose next that (b) holds. From the long exact sequence in coho-mology, it follows that c is the image of a class c′ ∈ H1(k(t), G′), whichis represented by some G′-Galois algebra M/k(t) such that Mµ2 = k(x).We claim that we may assume that M is the function field k(C) of a hy-perelliptic curve C. We may complete the proof modulo this statement, asthe G′-Galois action on k(C) induces a faithful G′-variety structure on C,satisfying the conditions required in (a).214.1. Equivalent characterizations of the HLPIt remains to verify the claim. We prove a stronger statement; namelythat we may assume that M is the function field of a hyperelliptic curve ofarbitrarily large genus. We may write M = k(x)[Y ]/(Y 2−ω(x)), where ω(x)is a separable polynomial. Recall that H1(k(t), µ2) ∼= k(t)×/k(t)×2 acts onthe fiber above c in the following way: the element r · k(t)×2 sends M tothe G′-Galois algebra k(x)[Y ]/(Y 2 − rω(x)). Write t as a rational functiont = p(x)/q(x), where p, q are coprime polynomials in k[x]. Then we canselect α1, . . . , αl ∈ k (l ≥ 6) such that ω(αi)q(αi) 6= 0 for all i, p(αi)/q(αi) 6=p(αj)/q(αj) for all i 6= j, and αi is a simple root of q(αi)p(x)−p(αi)q(x) forall i. If we set r =∏i(t− p(αi)/q(αi)), the Galois algebra k(x)(√rω(x)) isthen the function field of the hyperelliptic curvey2 = ω(x)∏i(p(x)q(x)−p(αi)q(αi)),which ramifies over α1, . . . , αl by construction. The genus of such curve isat least l/2− 1, which can be made arbitrarily large.Corollary 4.2. If G′ = µ2 ×G, then the HLP is solvable.Proof. Left to the reader.The main result of this section relates the solvability of the HLP to theexistence of certain linear representations of G′ that are nontrivial on µ2.Proposition 4.3. The HLP in Problem 1.4 is solvable if and only if at leastone of the following conditions holds:(a) There exists a homomorphism G′ → µ(k) that is nontrivial on µ2, whereµ(k) is the group of roots of unity in k.(b) There exists an embedding G′ ↪→ GL2 lifting G ↪→ PGL2.Proof. By [1, Lemma 6.12], the solvability of the HLP implies that either(a) or (b) must hold. It remains to prove the converse.In the notation of Proposition 4.1(b), it is enough to prove that δ(c) istrivial in each case. Via the natural map H1(k(t), µ2) ↪→ H1(k(t),Gm), wemay regard δ(c) as a Brauer class in Br(k(t)).Suppose first that (a) holds. By a theorem of Karpenko and Merkurjev[14, Thm. 4.4, Rem. 4.5], the index of δ(c) divides dim(V ), for any linearrepresentation V of G′ that is nontrivial on µ2. By assumption, there existssuch linear representation of dimension 1, whence δ(c) must be trivial.224.1. Equivalent characterizations of the HLPSuppose next that (b) holds. Consider the following commutative dia-gram whose rows are central exact sequences1 // Gm // GL2 // PGL2 // 11 // Gm // G′′?ρOO// G?ρOO// 11 // µ2 //?OOG′?OO// G // 1where G′′ is the full preimage of G in GL2. We obtain the correspondingdiagram in cohomologyH1(k(t),PGL2) // H2(k(t),Gm)H1(k(t), G′′) // H1(k(t), G)ρ∗OO// H2(k(t),Gm)H1(k(t), G′)OO// H1(k(t), G) δ // H2(k(t), µ2)?OOWe may endow A2 with a generically free G′′-action via ρ. Then, note thatthe class c ∈ H1(k(t), G) of the G-torsor k(x)/k(t) comes from the class ofa G′′-torsor in H1(k(t), G′′), namely k(A2)/k(A2)G′′. The commutativity ofthe above diagram then implies that δ(c) is trivial.Remark 4.4. The condition in Proposition 4.3(a) above is independent ofthe choice of the embedding G ↪→ PGL2. In other words, the existenceof such homomorphism implies the solvability of the HLP, regardless of thefaithfulG-action on P1. On the other hand, the condition in part (b) dependsheavily on the particular embedding G ↪→ PGL2.Remark 4.5. In [1, Lemma 6.12], it is proven that if the G-action on P1 canbe lifted to a G′-action on a hyperelliptic curve of odd (resp. even) genus,then the condition in Proposition 4.3(a) (resp. (b)) holds. However, theconverse of this statement is not true. In particular, in the following examplethere exists a homomorphism G′ → µ(k) which is nontrivial on µ2 but theG-action on P1 cannot be lifted to a G′-action on any hyperelliptic curve ofodd genus. Consider a field k that contains a primitive 8-th root of 1, andconsider the HLP given by the exact sequence 1→ µ2 → Z/8Z→ Z/4Z→ 1234.2. HLP for cyclic groupsand the Z/4Z-action on P1 defined by z 7→ ω28z. The mapping σ 7→ ω8,where σ is a generator of Z/8Z, defines a homomorphism Z/8Z → µ(k)that is nontrivial on µ2. However, note that k(t) = k(x4) and k(√x)/k(x4)is an element in the fiber above k(x)/k(x4) under the natural morphismH1(k(x4),Z/8Z)→ H1(k(x4),Z/4Z). Hence, any element in such fiber is ofthe form L = k(x)(√xQ(x4)), where we may assume that Q is a separablepolynomial. Then, observe that L is the function field of the hyperellipticcurve Y 2 = xQ(x4), which has even genus 2 deg(Q). Of course, the existenceof such solution to the HLP implies the existence of an embedding G′ ↪→ GL2lifting G ↪→ PGL2, which is given byσ 7→(ω8 00 ω−18).Our goal for the remainder of the chapter is to solve the HLP for everypossible embedding G ↪→ PGL2 and every non-split extension of G by µ2.4.2 HLP for cyclic groupsThroughout this section, we assume that G = Z/nZ. If n is odd, there areno non-split extensions of Z/nZ by µ2. On the other hand, if n is even,there exists a unique non-split extension given by1→ µ2 → Z/2nZ→ Z/nZ→ 1. (4.1)In what follows, we assume that n is even and write n = 2hq, whereh ≥ 1 and q is odd.Proposition 4.6. There exists a homomorphism Z/2nZ → µ(k) that isnontrivial on µ2 if and only if ω2h+1 ∈ k.Proof. Note that Z/2h+1Z is a direct factor of Z/2nZ, so we may assumewithout loss of generality that q = 1. Any homomorphism Z/2h+1Z→ µ(k)that is nontrivial on µ2 must be faithful, which implies the existence of aprimitive 2h+1-root of unity in k. The converse is left to the reader.We now deal with the case G = Z/2Z and we assume that ρa is as in(3.1) for some a ∈ k×2.Proposition 4.7. The HLP given by ρa and 1→ µ2 → Z/4Z→ Z/2Z→ 1is solvable if and only if either −1 ∈ k×2 or −a ∈ k×2.244.2. HLP for cyclic groupsProof. In view of Propositions 4.3 and 4.6, it suffices to prove that we canlift ρa to Z/4Z ↪→ GL2 if and only if −a ∈ k×2. Indeed, such a lift isequivalent to the existence of λ ∈ k such that(0 λaλ 0)2= −I,which can be found if and only if −a is a square in k.For the remainder of the section, we assume that n ≥ 4 and αn ∈ k, andwe define ρ : Z/nZ ↪→ PGL2 as in (3.5).Proposition 4.8. Suppose that k contains αn, where n is even. Then, ρlifts to an embedding Z/2nZ ↪→ GL2 if and only if one of the followingconditions holds:(a) α2n ∈ k.(b) ω4α2n ∈ k and n ≡ 0 (4).Proof. There exists such a lift if and only if we can find λ ∈ k such thatσ˜ = λ(αn + 1 βn1 αn + 1)satisfies σ˜n = −I. Note that the eigenvalues of σ˜ are λ(1+ω±1n ), so it followsfrom Lemma 2.2(c) that σ˜n = −(2λα2n)nI, whence we can lift ρ if and onlyif (2λα2n)n = 1 for some λ ∈ k.If (a) (resp. (b)) holds, we choose λ = (2α2n)−1 (resp. λ = (2ω4α2n)−1)to satisfy the above relation. Conversely, suppose that (2λα2n)n = 1 forsome λ ∈ k. Then we can write 2λα2n = ζn for some n-th root of unity ζn,not necessarily primitive. If ζn is a primitive 4-th root of unity (which is onlypossible if 4 divides n), then we obtain condition (b). Assume henceforththat this is not the case (in particular, ζn + ζ−1n 6= 0). Then, we computeζn + ζ−1n = (4λ2α22n + 1)/(2λα2n) and therefore, we haveα2n =4λ2α22n + 12λ(ζn + ζ−1n )=λ2(2αn + 2) + 12λ(ζn + ζ−1n ).By Lemma 2.2(a), we have ζn + ζ−1n ∈ k, whence α2n ∈ k as desired.Proposition 4.9. Suppose that k contains αn, where n ≡ 2 (4) and n ≥ 6.Then the HLP given by ρ and 1→ µ2 → Z/2nZ→ Z/nZ→ 1 is solvable ifand only if either α2n ∈ k or −1 ∈ k×2.254.3. HLP for dihedral groupsProof. This follows immediately from Propositions 4.3, 4.6 and 4.8.Proposition 4.10. Suppose that k contains αn, where n ≡ 0 (4). Then theHLP given by ρ and 1→ µ2 → Z/2nZ→ Z/nZ→ 1 is solvable if and onlyif either α2n ∈ k or ω4α2n ∈ k.Proof. In view of Propositions 4.3, 4.6 and 4.8, it suffices to show thatω2h+1 ∈ k implies that α2n ∈ k. We actually prove a stronger statement,namely that ω2h+1 ∈ k ⇒ ω2n ∈ k. By Lemma 2.2(d), the conditions αn ∈ kand ω2h+1 ∈ k together imply that ωn ∈ k. Finally, note that ω2h+1ωn is aprimitive 2n-th root of unity contained in k, which implies the result.4.3 HLP for dihedral groupsIn this section, we assume throughout that G = D2n, with the usual presen-tation D2n = 〈σ, τ |σn = 1, τ2 = 1, (στ)2 = 1〉. Let G′ be any double coverof D2n. We denote the nontrivial element of µ2 by −1, which is central inG′, and let s, t by lifts of σ, τ to G′, respectively. We refer the reader to [5,Lemma 4.1] for the cohomology calculations needed below.If n is odd, we have the equality H2(G,µ2) = µ2. The unique non-splitextension (up to equivalence) of G by µ2 is given by1→ µ2 → Dic4n → D2n → 1, (4.2)where Dic4n is the dicyclic group〈s, t | sn = −1, t2 = −1, (st)2 = −1〉.The element t generates a Sylow 2-subgroup of Dic4n, which is isomor-phic to Z/4Z. The corresponding exact sequence of Sylow 2-subgroups isthe unique non-split extension of Z/2Z by µ2.If n is even, it is well known that H2(G,µ2) = µ32 and the extensionsare given by groups of the form G′ = 〈s, t | sn = 1, t2 = 2, (st)2 = 3〉,where (1, 2, 3) ∈ µ32. We introduce some notation for the non-split groupextensions of G by µ2 (cf. [19, §3.7]). For simplicity, we will label theresulting exact sequences by their corresponding element in H2(G,µ2).Element of µ32 Notation for G′(−1, 1, 1) D4n(1,∓1,±1) V4n(1,−1,−1) Z/nZ o Z/4Z(−1,∓1,±1) SD4n(−1,−1,−1) Dic4n264.3. HLP for dihedral groupsRemark 4.11. (a) The exact sequence (1,−1, 1) (resp. (−1,−1, 1)) is notequivalent to (1, 1,−1) (resp. (−1, 1,−1)), as they represent differentelements in H2(D2n, µ2).(b) In the case G′ = Z/nZ o Z/4Z, a generator of Z/4Z acts on Z/nZ byinversion.(c) The groups SD4n and Dic4n are sometimes called semidihedral groupand generalized quaternion group respectively, especially when n is apower of 2.(d) If n ≡ 0 (4), it is not hard to prove that V4n is isomorphic to thepull-back D2n f Z/4Z = D2n ×(f,g) Z/4Z, where f : D2n → Z/2Z andg : Z/4Z→ Z/2Z have kernels 〈σ2, τ〉 and Z/2Z respectively.(e) SD8 ∼= Z/2Z× Z/4Z, V8 ∼= D8, and Dic8 ∼= Q8. In this case, (−1, 1, 1),(1,−1, 1) and (1, 1,−1) are three inequivalent extensions yielding thesame group, and so are (1,−1,−1), (−1,−1, 1) and (−1, 1,−1).Suppose that n = 2hq, with h ≥ 1 and q odd. A Sylow 2-subgroup G′2of G′ is generated by sq and t, and it fits into an exact sequence 1→ µ2 →G′2 → D2h+1 → 1, having the same label as 1→ µ2 → G′ → D2n → 1.Proposition 4.12. Let 1 → µ2 → G′ → G → 1 be either (4.2) if n is oddor the exact sequence (1, 2, 3) if n is even, where i = ±1, not all equalto 1. Then there exists a homomorphism G′ → µ(k) nontrivial on µ2 if andonly if −1 ∈ k×2 and one of the following conditions holds:(a) n is odd.(b) n is even and 1 = (23)n/2.Proof. Write n = 2hq, where h ≥ 0 and q is odd, and let G′2 be a Sylow 2-subgroup of G′. If n is even (resp. odd), we may assume that G′2 is generatedby the elements s′ = sq and t satisfying s′2h= 1, t2 = 2, (s′t)2 = 3 (resp.G′2 ∼= Z/4Z is generated by t). In either case, there exists a surjectionG′ → G′2 sending −1 7→ −1, s 7→ s′ = sq and t 7→ t, with kernel Z/qZ. Itis then clear that the existence of the desired homomorphism G′ → µ(k) isequivalent to the existence of such homomorphism for G′2, so we may assumehenceforth that q = 1 (and therefore G′ = G′2). If n is odd, the result followsimmediately from Proposition 4.6 after this reduction, so we only need todeal with the case where n is even.274.3. HLP for dihedral groupsSuppose that there exists a homomorphism G′ → µ(k) nontrivial on µ2.Then, since s2h+1= t4 = 1, the homomorphism sends s→ ζ2h+1 and t→ ζ4,where ζl is a (not necessarily primitive) l-th root of unity in k. As thehomomorphism is nontrivial on µ2, we must have ζ2h2h+1 = 1, ζ24 = 2 and(ζ2h+1ζ4)2 = 3. Since at least one i is equal to −1, it follows that at leastone of ζ2h−12h+1 , ζ4 or ζ2h+1ζ4 must be a primitive 4-th root of unity, whence−1 ∈ k×2. We also compute the condition1 = (ζ22h+1)2h−1 = (ζ22h+1ζ24 )2h−1(ζ−24 )2h−1 = (23)2h−1 ,which is equivalent to (b).Conversely, suppose that ω4 ∈ k and (b) holds. The existence of thedesired homomorphism is equivalent to the solvability of the system of equa-tionsγ2h1 = 1, γ22 = 2, γ21γ22 = 3for some γ1, γ2 ∈ k. Note that the last equation is equivalent to γ21 =23. Then the first equation becomes superfluous, since (23)2h−1= 1by condition (b). Finally, we can always find γ1, γ2 satisfying γ22 = 2 andγ21 = 23 because ω4 ∈ k by assumption. The proof is complete.We now concentrate on the case n = 2, i.e., G ∼= (Z/2Z)2, and we defineρ(a,b) as in Proposition 3.10.Proposition 4.13. Let 1 → µ2 → G′ → G → 1 be the exact sequence(1, 2, 3), where i = ±1, not all equal to 1 (cf. Remark 4.11(e)). Theembedding ρ(a,b) lifts to an embedding G′ ↪→ GL2 if and only if 123 = −1,1b ∈ k×2, and 2a ∈ k×2.Proof. Let e˜1, e˜2 be lifts of ρ(a,b)(e1), ρ(a,b)(e2) to GL2, respectively. In any ofthe three cases (3.2), (3.3) and (3.4), a computation shows that e˜12 = µ21bI,e˜22 = µ22aI and (e˜1e˜2)2 = −µ21µ22abI, where µ1, µ2 ∈ k. Therefore, theseelements define an embedding G′ ↪→ GL2 lifting ρ(a,b) if and only if thesystem of equationsµ21b = 1, µ22a = 2, −µ21µ22ab = 3is solvable for some µ1, µ2 ∈ k. It is then easy to see this is the case if andonly if 123 = −1, 1b ∈ k×2, and 2a ∈ k×2.We can now solve the HLP for actions of (Z/2Z)2 on the projective line.284.3. HLP for dihedral groupsProposition 4.14. The HLP given by ρ(a,b) and the extension (1, 2, 3) issolvable if and only if one of the following conditions holds:(a) 123 = −1 and 1b, 2a ∈ k×2. (Cases: G′ ∼= D8, Q8.)(b) 123 = 1 and −1 ∈ k×2. (Case: G′ ∼= Z/2× Z/4Z.)Proof. This follows easily from Propositions 4.3, 4.12 and 4.13.For the remainder of the section, we assume that n ≥ 3 and αn ∈ k, andwe define ρa : D2n ↪→ PGL2 as in (3.6) for some a ∈ D(〈1,−βn〉).Proposition 4.15. Let k contain αn and let 1 → µ2 → G′ → G → 1be either (4.2) if n is odd or the exact sequence (1, 2, 3) if n is even,where i = ±1, not all equal to 1. The embedding ρa lifts to an embeddingG′ ↪→ GL2 if and only if one of the following conditions holds:(a) n is odd and −a ∈ k×2.(b) n is even, 1 = −(23)n/2, 2a ∈ k×2, and (23)1/2α2n ∈ k.Proof. Consider respective liftsσ˜ = λ(αn + 1 βn1 αn + 1), τ˜ = µ(u −βnvv −u)of ρa(σ), ρa(τ) to GL2. We then compute σ˜n = −(2λα2n)nI (see the proofof Proposition 4.8), τ˜2 = µ2aI, and (σ˜τ˜)2 = 4λ2µ2α22naI.If n is odd, note that we can have τ˜2 = −I if and only if −a ∈ k×2, so thisis a necessary condition for the existence of the lift G′ ↪→ GL2. Conversely,we prove that the condition −a ∈ k×2 is also sufficient. Indeed, note thatα2n ∈ k by Lemma 2.2 (b), so we can take λ = (2α2n)−1 and µ = (−a)−1/2to produce the desired lift.Assume henceforth that n is even (n ≥ 4). By the computation wecarried out at the beginning of the proof, we see that a lift G′ ↪→ GL2 ispossible if and only if we can find λ, µ ∈ k such that−(2λα2n)n = 1, µ2a = 2, 4λ2µ2α22na = 3.The last equation is equivalent to (2λα2n)2 = 23, which is possible if andonly if (23)1/2α2n ∈ k. On the other hand, the second relation above ispossible if and only if 2a ∈ k×2, and the first relation holds if and only if1 = −(2λα2n)n = −((2λα2n)2)n/2 = −(23)n/2.This completes the proof of the proposition.294.4. HLP for polyhedral groupsWe can now solve the HLP for dihedral actions on the projective line.We separate the odd and even cases for the sake of clarity.Proposition 4.16. Suppose that n is odd and k contains αn. The HLPgiven by ρa and the exact sequence (4.2) is solvable if and only if either−1 ∈ k×2 or −a ∈ k×2.Proof. The proof follows from Propositions 4.3, 4.12 and 4.15.Proposition 4.17. Suppose that n is even and k contains αn. The HLPgiven by ρa and the exact sequence (1, 2, 3) is solvable if and only if oneof the following conditions holds:(a) 1 = (23)n/2 and −1 ∈ k×2.(b) 1 = −(23)n/2, 2a ∈ k×2, and (23)1/2α2n ∈ k.Proof. Again, the proof follows from Propositions 4.3, 4.12 and HLP for polyhedral groupsRecall that a polyhedral group G (= A4, S4, or A5) embeds into PGL2 ifand only if −1 is a sum of two squares over k, with the additional conditionthat√5 ∈ k if G = A5. Under these conditions, the embeddings intoPGL2 are all conjugate, so the solvability of HLP’s for polyhedral groupsare independent of the chosen embedding. For any such group G, we fix theembedding G ↪→ PGL2 to be the one described in Proposition (3.10).4.4.1 Alternating groupsIn this subsection, let G = An for n = 4, 5. Recall that H2(G,µ2) = µ2,whence G has a unique non-split central extension1→ µ2 → G˜→ G→ 1. (4.3)Actually, it is well known that A˜4 ∼= SL2(F3) and A˜5 ∼= SL2(F5) (see,e.g., [32, §9.1.3, Ex. 1]). Since G is generated by squares, it follows that G˜embeds into GL2 as a lift of G ↪→ PGL2. Alternatively, the lifts 12R and12S′generate A˜4 inside GL2, while 12R and12U generate A˜5 inside GL2. (Here,R, S′ and U are as defined in 3.10.) This fact implies the following result.Proposition 4.18. Suppose that −1 = a2 + b2 for a, b ∈ k, and√5 ∈ k ifG = A5. Then the HLP given by G ↪→ PGL2 and (4.3) is solvable.304.4. HLP for polyhedral groupsProof. The proof follows immediately from Proposition 4.3.Remark 4.19. It is not hard to see that any homomorphism A˜n → µ(k)(n = 4, 5) must be trivial on µ2. Indeed, this follows from the well knownfact that µ2 is contained in the commutator subgroup [A˜n, A˜n].4.4.2 Symmetric group on 4 lettersIn this subsection, we assume that G = S4. Recall that H2(S4, µ2) = µ22(see [5, Lemma 4.1]), which gives rise to three non-split central extensions ofS4 by µ2. As we did in the case of even dihedral groups, we label the exactsequences by their corresponding element of H2(S4, µ2). Select generatorsr, s of S4 satisfying r3 = s4 = (r−1s)2 = 1, and pick respective lifts r′, s′of r, s to G′ (changing r′ by −r′ if necessary, we assume without loss ofgenerality that r′3 = 1). Then, the extensions are parametrized by thenontrivial pairs (1, 2) ∈ µ22 via the relations s′4 = 1, (r′−1s′)2 = 2. Thedifferent cases are outlined below.(i) The pair (1,−1) corresponds to the group S4, which is the only non-trivial extension of S4 by µ2 that restricts to the trivial extensionof A4 by µ2. (This group is denoted by S′4 in [32, §9.1.3].) Thetranspositions (resp. products of disjoint transpositions) lift to ele-ments of order 4 (resp. 2) in S4. It is isomorphic to the pull-backS4fZ/4Z = S4×(χ,g)Z/4Z, where χ : S4 → Z/2Z is the sign characterand g : Z/4Z→ Z/2Z has kernel Z/2Z. Its Sylow 2-subgroups are thusisomorphic to V16.(ii) The pair (−1, 1) case corresponds to the group S˜4, which is isomorphicto GL2(F3) (see [32, §9.1.3, Ex. 2(a)]). The transpositions (resp.products of disjoint transpositions) lift to elements of order 2 (resp. 4)in S˜4. Its Sylow 2-subgroups are isomorphic to SD16.(iii) The pair (−1,−1) corresponds to the group Ŝ4, which is also knownas the binary octahedral group. Both transpositions and the productsof disjoint transpositions lift to elements of order 4 in Ŝ4. Its Sylow2-subgroups are isomorphic to Dic16.Proposition 4.20. Let 1 → µ2 → G′ → G → 1 be the exact sequence(1, 2), i = ±1, (1, 2) 6= (1, 1). Then there exists a homomorphism G′ →µ(k) nontrivial on µ2 if and only if (1, 2) = (1,−1) and −1 ∈ k×2.314.4. HLP for polyhedral groupsProof. We claim that if (1, 2) = (−1,±1), such a homomorphism does notexist. Indeed, in that case G′ contains A˜4 as a subgroup of index 2, so theclaim follows by Remark 4.19.If (1, 2) = (1,−1), G′ is generated by r′, s′ such that r′3 = 1, s′4 = 1and (r′−1s′)2 = −1. The latter relation implies that −1 ∈ k×2 is a necessarycondition for the desired homomorphism to exist. Conversely, r′ 7→ 1, s′ 7→ω4 satisfies the required properties if −1 ∈ k×2.Proposition 4.21. Let 1 → µ2 → G′ → G → 1 be the exact sequence(1, 2), i = ±1, (1, 2) 6= (1, 1). Then there exists G′ ↪→ GL2 lifting theprescribed embedding G ↪→ PGL2 if and only if 1 = −1 and −22 ∈ k×2.Proof. LetR˜ = λ(a− b− 1 a+ b+ 1a+ b− 1 b− a− 1), S˜ = µ(1 1−1 1)be lifts of R and S to GL2, respectively. A computation shows that R˜3 =8λ3I, S˜4 = −4µ4I and (R˜−1S˜)2 = −(2λ2)−1µ2I. Hence, a lift G′ ↪→ GL2 ispossible if and only if we can find λ, µ ∈ k such that8λ3 = 1, −4µ4 = 1, −(2λ2)−1µ2 = 2.Assume first that λ and µ exist. From the first equation, λ = ζ3/2 forsome third root of unity ζ3. Squaring the last equation, we obtain 1 = 22 =(ζ3/4)−1µ4 = (ζ3/4)−1(−1/4) = −1/ζ3. It follows easily that 1 = −1 andζ3 = 1. The last equation then becomes µ2 = −2/2, which implies that−22 ∈ k×2.Conversely, if we assume 1 = −1 and −22 ∈ k×2, it suffices to takeλ = 1/2 and µ = (−22)1/2/2. The proof is complete.Proposition 4.22. Suppose that −1 = a2 + b2 for a, b ∈ k. Then the HLPgiven by G ↪→ PGL2 and the exact sequence (1, 2) is solvable if and only ifone of the following conditions holds:(a) (1, 2) = (1,−1) and −1 ∈ k×2.(b) (1, 2) = (−1,±1) and −22 ∈ k×2.Proof. This follows immediately from Propositions 4.3, 4.20 and 4.21.324.5. Explicit solutions to the HLP4.5 Explicit solutions to the HLPSuppose that the HLP given by an embedding G ↪→ PGL2 and a non-split extension 1→ µ2 → G′ → G→ 1 is solvable. Recall that the completefamily of solutions to this HLP is the set of hyperelliptic curves with functionfield of the form k(x)(√rω(x)), for some fixed separable polynomial ω(x) ∈k[x] and r ∈ k(t)× arbitrary (see the proof of Proposition 4.3).In this section, we compile the solutions to the HLP’s that we studiedearlier. Note that it suffices to indicate ω(x), so that is exactly what we willdo.With the aid of the following lemma, we can compute the group actionson the explicit solutions to the HLP’s.Lemma 4.23. Let σ be an element of order n ≥ 2 in PGL2(k), fix a liftσ˜ =(a bc d)∈ GL2(k) of σ, and let µ1, µ2 ∈ k be its eigenvalues. Considerthe polynomial Q(x, y) =∏n−1i=0 (yix− xiy) ∈ k(x0, y0)[x, y], where x0, y0 areindeterminates and (xi : yi) = σi(x0 : y0) for i = 0, . . . , n−1. Then we haveQ(ax+ by, cx+ dy) = µn1Q(x, y) = µn2Q(x, y).Proof. Note first that µn1 = µn2 , since σ has order n in PGL2(k). This provesthe second equality.The homogeneous forms Q(ax+ by, cx+ dy) and Q(x, y) have the sameroots, namely (xi : yi) for 0 ≤ i ≤ n − 1. Hence they differ by a constantfactor, which a priori could depend on x0, y0. Let (u, v)T be an eigenvectorof σ˜, say corresponding to the eigenvalue µ1. Then Q(au + bv, cu + dv) =Q(µ1u, µ1v) = µn1Q(u, v). It follows that the constant factor equals µn1 = µn2 ,independently of x0, y0.Definition 4.24. Given z ∈ P1(k), we denote its orbit under G by Oz.Then we define the orbit polynomial Pz ∈ k[x] as follows:Pz(x) =∏z′∈Oz−{∞}(x− z′),and we write Qz = P|G|/|Oz |z . (Note that degQz = |G|(1 − 1/|O∞|) ifOz = O∞, and degQz = |G| otherwise.) If Pz (or equivalently Qz) hascoefficients in k, we say that Oz is a k-rational orbit. This is the case if andonly if Gal(k/k)Oz = Oz.334.5. Explicit solutions to the HLPProposition 4.25. Let σ ∈ G be an element of order n ≥ 2 and select alift σ˜ ∈ GL2(k) as in Lemma 4.23. Define U = {l ∈ Oz : σl = l}. If l ∈ U ,we denote the eigenvalue of σ˜ corresponding to any lift of l by µl, and wedenote the other eigenvalue by µ′l. Then we have the equalityPz(σ(x)) =∏l∈Uµ′lµ|Oz |−|U |1(cx+ d)|Oz |Pz(x).Proof. Let S be the stabilizer of z in G and for every gS ∈ G/S, writegz = (xg : yg) ∈ P1(k). We may assume that yg = 1 if gz 6= ∞ andxg = −1, yg = 0 when gz =∞. Define the following homogenization of Pz:Pz(x, y) =∏gS∈G/S(ygx− xgy) ∈ k[x, y].Note that we can recover Pz(x) = Pz(x, 1).Let g be any element of G. We claim that either σgz = gz, or σigz 6=σjgz for every 0 ≤ i < j < n. Assume to the contrary that σgz 6= gz, butσi0gz = σj0gz for some 0 ≤ i0 < j0 < n. If we write m = j0 − i0, it followsthat σmgz = gz, where 0 < m < n. However, note that σ has exactly twodistinct fixed points in P1(k) (corresponding to eigenvectors of σ˜), which aretherefore fixed points of σm as well. Since σm has at most two fixed points,we conclude that σgz = gz, in contradiction with our assumption.Let F ⊂ G be a set of representatives of classes gS ∈ G/S such thatgz ∈ U (0 ≤ |F | ≤ 2). By the claim above, we can expressG/S =∐g∈FgS unionsqs∐i=1n−1∐j=0σjgiS,for some g1, . . . , gs in G. A convenient way to think about these elementsis by noting that F unionsq {g1, . . . , gs} is a set of representatives for 〈σ〉\G/S. Ifwe define Qj(x, y) =∏n−1i=0 (yσigjx− xσigjy), we can then writePz(x, y) =∏g∈F(ygx− xgy)s∏j=1Qj(x, y).By Lemma 4.23, we have that Qj(ax + by, cx + dy) = µn1Qj(x, y). On theother hand, if g ∈ F , then gz = (xg : yg) is fixed by σ by assumption andthe lift (xg, yg)T of gz is an eigenvector of σ˜ with eigenvalue µgz. It follows344.5. Explicit solutions to the HLPthat axg + byg = µgzxg and cxg + dyg = µgzyg. We now computeyg(ax+ by)− xg(cx+ dy) = (ayg − cxg)x− (dxg − byg)y= (a+ d− µgz)(ygx− xgy)= µ′gz(ygx− xgy).We thus obtain the equalityPz(ax+ by, cx+ dy) =∏g∈Fµ′gz µns1 Pz(x, y) =∏l∈Uµ′l µ|Oz |−|U |1 Pz(x, y).It follows thatPz(σ(x)) =1(cx+ d)|Oz |Pz(ax+ b, cx+ d)=∏l∈Uµ′lµ|Oz |−|U |1(cx+ d)|Oz |Pz(x, 1)=∏l∈Uµ′lµ|Oz |−|U |1(cx+ d)|Oz |Pz(x).Corollary 4.26. Let σ ∈ G and select a lift σ˜ ∈ GL2(k) as in Lemma 4.23.ThenQz(σx) =µ|G|1(cx+ d)|G|Qz(x).Proof. The result is obvious if σ is the identity in G, so we may assumehenceforth that σ has order n ≥ 2. Using the notation from Proposition4.25, we computeQz(σx) =∏l∈Uµ′ |G||Oz |lµ|G|−|U | |G||Oz |1(cx+ d)|G|Qz(x).Hence it suffices to prove that∏l∈Uµ′ |G||Oz |l = µ|U | |G||Oz |1 .If |U | = 0, there is nothing to prove. If |U | = 1 or 2, it means that Ozcontains an element l fixed by σ. Consequently, 〈σ〉 is a subgroup of Stab(l)and therefore, n divides |Stab(x)| = |G|/|Oz|. Thus µ|G||Oz |1 = µ|G||Oz |2 and theresult follows.354.5. Explicit solutions to the HLPCorollary 4.27. The fixed field k(x)G is equal to the rational field k(t),where t = Qz(x)/Q∞(x) for any k-rational orbit Oz 6= O∞.Proof. Since x is a root of the polynomial Qz(x)− tQ∞(x), which is a monicpolynomial of degree |G|, it follows that [k(x) : k(t)] ≤ |G|. Since [k(x) :k(x)G] = |G|, it suffices to prove that k(t) ⊂ k(x)G. To see this, consider anarbitrary element σ in G. By Corollary 4.26, it follows thatσt = Qz(σx)/Q∞(σx) = Qz(x)/Q∞(x) = t.Since σ is arbitrary, this completes the proof of the corollary.We use the notation we have introduced in this section to produce explicitsolutions to the HLP’s we already studied. In every case, we describe theaction of G′ on k(x)[y]/(y2 − ω(x)).4.5.1 Extensions of cyclic groupsIn this subsection, we exhibit the solutions for the HLP given by an extensionof cyclic groups.Extension (4.1): 1→ µ2 → Z/4Z→ Z/2Z→ 1.Embedding: ρa : Z/2Z ↪→ PGL2, where a ∈ k× (see (3.1)).Conditions for solvability of the HLP: Either −a ∈ k×2 or −1 ∈ k×2 (seeProposition 4.7).Solutions:Condition ω Generator of Z/4Z−a ∈ k×2 x2 − a (x, y) 7→(ax ,√−ax y)−1 ∈ k×2 x(x2 − a) (x, y) 7→(ax ,ω4ax2 y)Example 4.28. We illustrate here how to compute different solutions tothe HLP above. Assume we are dealing with the case where −1 ∈ k×2.Note that the equation y2 = x(x2 − a) defines a curve of genus 1, whichis therefore not a solution to the HLP. However, we are free to modify theright hand side by an element r ∈ k(t)×. Using Corollary 4.27, we may taket = (x2 + a)/x. Choosing, say, r = t(t+ 1), we obtain a hyperelliptic curvewith equationy2 = x(x4 − a2)(x2 + x+ a),which is a solution to the above HLP. The Z/4Z-action is given byσ′ : (x, y) 7→(ax,ω4a2x4y).364.5. Explicit solutions to the HLPWe now switch our attention to the case n ≥ 4 (n even) and let k containαn. Assume that ρ : Z/nZ ↪→ PGL2 is defined as in (3.5). The orbits of theelements ±√βn are singletons and hence P±√βn(x) = x∓√βn. On the otherhand, |O∞| = n. By Proposition 4.25, we may computeP√βn(σ(x))P−√βn(σ(x)) =2αn + 2(x+ αn + 1)2P√βn(x)P−√βn(x)andP∞(σ(x)) = −(2αn + 2)n2(x+ αn + 1)nP∞(x).Extension (4.1): 1→ µ2 → Z/2nZ→ Z/nZ→ 1.Conditions for solvability of the HLP: See Propositions 4.9 and 4.10.Solutions:Condition ω Generator of Z/2nZα2n ∈ k x2 − βn (x, y) 7→(σ(x), 2α2nx+αn+1y)ω4α2n ∈ k (x2 − βn)P∞(x) (x, y) 7→(σ(x), 2ω4α2n(2αn+2)n4(x+αn+1)n+22y)−1 ∈ k×2 (x2 − βn)P∞(x) (x, y) 7→(σ(x), ω4(2αn+2)n+24(x+αn+1)n+22y)In the second (resp. third) row above, we assume that n ≡ 0 (4) (resp.n ≡ 2 (4)). In each case, note that the Z/2nZ-action is defined over k.4.5.2 Extensions of the Klein groupWe turn our attention to the HLP for extensions of (Z/2Z)2 by µ2. In thissubsection, we always consider one of the embeddings ρ(a,b) : (Z/2Z)2 ↪→PGL2 that we specified in (3.2), (3.3) and (3.4). In every case, it is not hardto see that we can choose respective liftse˜1 =(a1 b1c1 d1), e˜2 =(a2 b2c2 d2)of ρ(a,b)(e1), ρ(a,b)(e2), such that e˜1 has eigenvalues ±√b, e˜2 has eigenvalues±√a, and e˜1e˜2 has eigenvalues ±√−ab.Let P1 (resp. P2, resp. P3) be the orbit polynomial of the solutionsof e1z = z (resp. e2z = z, resp. e1e2z = z), which lie in the same orbit.For example, if both a and b are non-squares (cf. (3.2)), we have P1(x) =374.5. Explicit solutions to the HLPx2 − 2λx + a, P2(x) = x2 − a, and P3(x) = x2 − 2 aλx + a if λ 6= 0 (resp.P3(x) = x if λ = 0). By Proposition 4.25, we can computeP1(e1(x)) =−bP1(x)(c1x+ d1)2, P1(e2(x)) =aP1(x)(c2x+ d2)2,P2(e1(x)) =bP2(x)(c1x+ d1)2, P2(e2(x)) =−aP2(x)(c2x+ d2)2,P3(e1(x)) =bP3(x)(c1x+ d1)2, P3(e2(x)) =aP3(x)(c2x+ d2)2.Extension (1, 2, 3): 1→ µ2 → G′ → (Z/2Z)2 → 1.Conditions for solvability of the HLP: Either 123 = 1 and −1 ∈ k×2,or 123 = −1 and 1b, 2a ∈ k×2 (see Proposition 4.14).Solution: Let ηi =1−i2 for i = 1, 2, 3; i.e., ηi = 1 if i = −1, and ηi = 0if i = 1. Then we can takeω(x) = P1(x)η1P2(x)η2P3(x)η3 .The G′-action on k(x)[y]/(y2 − ω(x)) is generated byσ′ : (x, y) 7→(e1(x),(−1)η12 bη1+η2+η32(c1x+ d1)η1+η2+η3y)andτ ′ : (x, y) 7→(e2(x),(−1)η22 aη1+η2+η32(c2x+ d2)η1+η2+η3y).An easy computation shows that σ′2 = 1, τ ′2 = 2 and (σ′τ ′)2 = 3.To prove that σ′ and τ ′ are defined over k, it suffices to prove that thenumerators of the coefficients of y in each map belong to k. We considerboth sets of conditions for solvability separately:Case 123 = 1 and −1 ∈ k×2: In this case, we haveη1+η2+η32 = 1.Thus, the numerators are (−1)η12 b and (−1)η22 a, which belong to k.Case 123 = −1 and 1b, 2a ∈ k×2: In this situation, a compu-tation shows that (−1)η12 bη1+η2+η32 = (1b)12 bη2η3 and (−1)η22 aη1+η2+η32 =(2a)12aη1η3 .4.5.3 Extensions of dihedral groupsIn this subsection, we let k be a base field containing αn, where n ≥ 3, anddefine ρa : D2n ↪→ PGL2 as in (3.6).384.5. Explicit solutions to the HLPThe solutions ±√βn to σz = z form a two element orbit O1 permutedby τ , with orbit polynomial P1(x) = x2 − βn. If n is odd, the orbits of thesolutions of τz = z are disjoint and each one contains a solution of στz = z;we denote them by O2 and O3. (Note that these orbits are not necessarilyk-rational, but O2 ∪ O3 is Gal(k/k)-stable.) If n is even, both solutionsof τz = z (resp. στz = z) lie in the same orbit, which we denote by O2(resp. O3). (In particular, this implies that O2 and O3 are k-rational for neven.) We denote the corresponding orbit polynomials by P2 and P3. Forconvenience, we define P4 to be the orbit polynomial of any fixed k-rationalorbit of cardinality |G|. (Note that O∞ may not satisfy this condition).Using Proposition 4.25, we computeP1(σ(x)) =(2αn + 2)P1(x)(x+ αn + 1)2, P1(τ(x)) =aP1(x)(vx− u)2,P2(σ(x)) =−(2αn + 2)n2 P2(x)(x+ αn + 1)n, P2(τ(x)) =−an2 P2(x)(vx− u)n,P3(σ(x)) =−(2αn + 2)n2 P3(x)(x+ αn + 1)n, P3(τ(x)) =an2 P3(x)(vx− u)n,P4(σ(x)) =(2αn + 2)nP4(x)(x+ αn + 1)2n, P4(τ(x)) =anP4(x)(vx− u)2n.We first deal with the unique nontrivial extension of odd dihedral groups.Extension (4.2): 1→ µ2 → Dic4n → D2n → 1, where n is odd.Conditions for solvability of the HLP: Either −1 ∈ k×2, or −a ∈ k×2(see Proposition 4.16).Solutions:Condition ω Generators of Dic4n−1 ∈ k×2∏3i=1 Pi(x) σ′ : (x, y) 7→(σ(x), (2α2n)n+1(x+αn+1)n+1y)τ ′ : (x, y) 7→(τ(x), ω4an+12(vx−u)n+1 y)−a ∈ k×2∏4i=1 Pi(x) σ′ : (x, y) 7→(σ(x), (2α2n)2n+1(x+αn+1)2n+1y)τ ′ : (x, y) 7→(τ(x), (√−a)2n+1(vx−u)2n+1 y)Recall that if n is odd, then αn ∈ k implies that α2n ∈ k by Lemma 2.2(b), so σ′ and τ ′ are defined over k. On the other hand, a simple computationshows that σ′n = τ ′2 = (σ′τ ′)2 = −1.Extension (1, 2, 3): 1→ µ2 → G′ → D2n → 1, where n is even.394.5. Explicit solutions to the HLPConditions for solvability of the HLP: Either 1 = (23)n/2 and −1 ∈k×2, or 1 = −(23)n/2, 2a ∈ k×2, and (23)1/2α2n ∈ k. (see Proposition4.17).Solution: As before, define ηi =1−i2 for i = 1, 2, 3, and chooseω(x) = P1(x)η1P2(x)η2P3(x)η3 .The G′-action on k(x)[y]/(y2 − ω(x)) is then generated byσ′ : (x, y) 7→(σ(x),(−1)η2+η32 (2αn + 2)η12 +n(η2+η3)4(x+ αn + 1)η1+n(η2+η3)2y)andτ ′ : (x, y) 7→(τ(x),(−1)η22 aη12 +n(η2+η3)4(vx− u)η1+n(η2+η3)2y).A computation shows that σ′n = 1, τ ′2 = 2 and (σ′τ ′)2 = 3. We needto show that σ′ and τ ′ are defined over k. Again, we separate the proof intotwo cases:Case 1 = (23)n/2 and −1 ∈ k×2: In this case, it is not hard to see thatN = η12 +n(η2+η3)4 is an integer. Therefore, the numerators (−1)η2+η32 (2αn+2)N and (−1)η22 aN belong to k by our assumption −1 ∈ k×2.Case 1 = −(23)n/2, 2a ∈ k×2, and (23)1/2α2n ∈ k: In this situation,it is not hard to verify that we can write(−1)η2+η32 (2αn + 2)η12 +n(η2+η3)4 = 2(−1)η2η3(23)12α2n(2αn + 2)η1−12 +n(η2+η3)4and(−1)η22 aη12 +n(η2+η3)4 = (2a)12aη1−12 +n(η2+η3)4 .Moreover, note that 1 = −(23)n/2 implies thatη1−12 +n(η2+η3)4 is an integer.4.5.4 Extensions of polyhedral groupsRecall that polyhedral groups have three orbits O1,O2,O3 of order < |G|,which correspond to the vertices, faces and edges of the corresponding Pla-tonic solids. The sizes of the orbits are detailed below.G |O1| |O2| |O3|A4 4 4 6S4 6 8 12A5 12 20 30404.5. Explicit solutions to the HLPWe denote the corresponding orbit polynomials by P1, P2, P3. ApplyingGal(k/k) to an orbit yields another orbit of the same size, so it follows thatthe above orbits are all k-rational, except possibly for O1 and O2 if G = A4.The computations of Pi(σ(x))/Pi(x), where σ is a generator of G, can becarried out with the help of Proposition 4.25 as usual; we omit the details forthe sake of brevity, but still describe the group actions derived from thesecomputations.In what follows, we will use the notation for the embeddings of polyhedralgroups introduced in Proposition 3.10.Extension (4.3): 1→ µ2 → A˜4 → A4 → 1.Conditions for solvability of the HLP: Always solvable (see Proposition4.18).Solution:ω Generators of A˜4P3(x) r : (x, y) 7→(R(x), 8((a+b−1)x+b−a−1)3 y)s′ : (x, y) 7→(− 1x ,yx3)Extension (4.3): 1→ µ2 → A˜5 → A5 → 1.Conditions for solvability of the HLP: Always solvable (see Proposition4.18).Solution:ω Generators of A˜5P3(x) r′ : (x, y) 7→(R(x), 32768((a+b−1)x+b−a−1)15 y)u : (x, y) 7→(U(x), 32768((φb+1)x−φa+φ−1)15 y)Remark 4.29. If G = A4, note that P3(x) = (x2 + 1)(x2 + 2 bax − 1)(x2 −2abx − 1) if a, b 6= 0, while P3(x) = x(x4 − 1) if ab = 0 (i.e., if −1 ∈ k×2).The general expression of P3 for the group A5 is very cumbersome, so wedecided to omit it.Extension (1,−1): 1→ µ2 → S4 → S4 → 1.Conditions for solvability of the HLP: −1 ∈ k×2 (see Proposition 4.22).Solution:ω Generators of S4P3(x) r′ : (x, y) 7→(R(x), 64((a+b−1)x+b−a−1)6 y)s′ : (x, y) 7→(S(x), 8ω4(−x+1)6 y)414.5. Explicit solutions to the HLPExtension (−1, 1): 1→ µ2 → S˜4 → S4 → 1.Conditions for solvability of the HLP: −2 ∈ k×2 (see Proposition 4.22).Solution:ω Generators of S˜4P1(x) r′ : (x, y) 7→(R(x), 8((a+b−1)x+b−a−1)3 y)s′ : (x, y) 7→(S(x), 2√−2(−x+1)3 y)Extension (−1,−1): 1→ µ2 → Ŝ4 → S4 → 1.Conditions for solvability of the HLP: 2 ∈ k×2 (see Proposition 4.22).Solution:ω Generators of Ŝ4P1(x)P3(x) r′ : (x, y) 7→(R(x), 512((a+b−1)x+b−a−1)9 y)s′ : (x, y) 7→(S(x), 16√2(−x+1)9 y)Remark 4.30. Note that P1(x) = (x2 + 1)(x2 + 2 bax− 1)(x2 − 2abx− 1) ifab 6= 0, while P1(x) = x(x4 − 1) if ab = 0. We can also computeP3(x) =(x4 +4aba2 − 1x3 +6b2a2 − 1x2 −4aba2 − 1x+ 1)×(x4 −4abb2 − 1x3 +6a2b2 − 1x2 +4abb2 − 1x+ 1)×(x4 +8aba2 − b2x3 − 6x2 −8aba2 − b2x+ 1),if (a2−1)(b2−1)(a2− b2) 6= 0. Note that the latter product can only vanishif −2 ∈ k×2; in this case, we can take a = b = 1/√−2, which impliesP3(x) = x(x10 +119x8 +223x6 −223x4 −119x2 − 1).42Chapter 5Strongly incompressiblecurves5.1 Rational quotientsAs usual, a rational map X 99K Y of k-varieties is an equivalence class ofk-morphisms U → Y , where U is a dense open subset of X. We denote thealgebra of rational functions of X by k(X). In general, k(X) is the directsum of the function fields of the irreducible components of X.An algebraic group G over k is a smooth affine group scheme of finitetype over k. We say that X is a G-variety if G acts morphically on X. Theinclusion k(X)G ↪→ k(X) induces a rational quotient map piX : X 99K W ,where k(W ) = k(X)G (see [23, §2.3]). The variety W is denoted by X/G andis unique up to birational isomorphism. IfN is a normal subgroup ofG, thereexists a model of X/N with a regular action of G/N (see [23, Remark 2.6]).It is uniquely defined up to G/N -equivariant birational isomorphism. Arational map X 99K Y of G-varieties gives rise to a G/N -equivariant rationalmap f : X/N 99K Y/N such that f ◦ pi′X = pi′Y ◦ f , where pi′X : X 99K X/Nand pi′Y : Y 99K Y/N are the rational quotient maps.A G-action on X is said to be generically free if there exists a dense G-invariant open subset of X with trivial scheme-theoretic stabilizers. (Inparticular, a faithful action of a finite group is generically free.) A G-compression is a G-equivariant dominant rational map X 99K Y , whereX and Y are generically free G-varieties. A generically free G-variety Xcontains a dense G-invariant open subset U which is the total space of aG-torsor piU : U → U/G (see [3, Thm. 4.7]). We say that X is primitive if Gtransitively permutes the irreducible components of X (equivalently, if X/Gis irreducible). Under this condition, the fiber at the generic point of U/Gis a G-torsor T → Spec(K), where K ∼= k(X)G. The class of this torsorin H1(K,G) will be denoted by [X]. Conversely, given a finitely generatedfield extension K of k, any class in H1(K,G) determines a generically freeprimitive G-variety X endowed with a k-isomorphism k(X)G ∼= K, uniquely435.2. Strong incompressibility of curvesup to G-equivariant birational isomorphism. In what follows, we assume allG-varieties to be primitive, unless stated otherwise.5.2 Strong incompressibility of curvesIn the sequel, the word “curve” will be reserved for smooth projective geo-metrically irreducible varieties of dimension 1, unless stated otherwise.Let G be a finite group. Recall that a faithful G-variety X is said to bestrongly incompressible if any G-compression X 99K Y onto a faithful G-variety Y is birational. We are interested in the study of strong incompress-ibility of G-curves. We remark that the existence of strongly incompressibleG-curves depends not only on the group G, but also on the base field k.Note also that G-compressions of curves extend naturally to surjectivefinite G-equivariant morphisms, so we will regard G-compressions of curvesas morphisms in the sequel. The following simple lemma is extremely usefulin our analysis.Lemma 5.1 (cf. [24, Example 6]). Suppose that there exists a faithful G-curve X that cannot be G-compressed to any G-curve of genus ≤ 1. Thenthere exists a strongly incompressible G-curve.Proof. Consider the set S consisting of faithful G-curves Y such that thereexists a G-compression X → Y . By assumption, the genus g(Y ) ≥ 2 forall Y ∈ S. Select a curve Y0 ∈ S having minimal genus. We claim that Y0is strongly incompressible. Indeed, suppose that we have a G-compressionf : Y0 → Y ′, which implies that Y ′ ∈ S. In particular, we must have g(Y ′) ≥g(Y0) ≥ 2. However, by Hurwitz Formula (see [21, Thm 7.4.16]) it alsofollows that g(Y0) ≥ g(Y ′), whence equality must hold. This implies that fis birational.The following result will be instrumental in the sequel. It is a specialcase of [27, Prop. 8.6] (see also [27, Rem. 9.9]), whose proof dependson resolution of singularities. We include an alternative proof because itworks over any base field of characteristic 0, it is more elementary and, inparticular, does not rely on resolution of singularities.Theorem 5.2. There exists a faithful G-curve X defined over k such thatevery element of G fixes some geometric point of X.Proof. See Appendix A.445.2. Strong incompressibility of curvesWe now recall some facts about the automorphism group of an ellipticcurve.Lemma 5.3. Let E be an elliptic curve defined over a field k.(a) There exists a split exact sequence1 // E i // Aut(E) pi // Aut0(E) // 1 , (5.1)where E acts on itself by translations and Aut0(E) denotes the group ofautomorphisms of E that preserve the origin.(b) There exists a natural isomorphism Aut0(E) ∼= µn, wheren =2, if j(E) 6= 0, 1728;4, if j(E) = 1728;6, if j(E) = 0.(c) If j(E) = 1728 (resp. 0), we have Aut0(E)(k) = Z/4Z (resp. Z/6Z) ifand only if ω4 ∈ k (resp. ω3 ∈ k).(d) The translation by P0 ∈ E and the automorphism α ∈ Aut0(E) commuteif and only if α(P0) = P0.Proof. (a) See, e.g., [34, §X.5]. Note that in [34], Aut(E) and Aut0(E) aredenoted by Isom(E) and Aut(E), respectively.(b) See [34, Cor. III.10.2].(c) This follows directly from part (c).(d) Let τP0 denote the translation by P0. Then note that τP0 and αcommute if and only if α(P ) +α(P0) = α◦ τP0(P ) = τP0 ◦α(P ) = α(P ) +P0for all P ∈ E, which implies the desired result.Theorem 5.4. Suppose that G cannot act faithfully on a curve of genus 0via k-morphisms. Then there exists a strongly incompressible G-curve.Proof. By Lemma 5.1, it suffices to prove that there exists a faithful G-curveX that cannot be G-compressed to any curve of genus 1.Note that G is not isomorphic to Z/nZ for n = 1, 2, 3, 4, 6, because thesegroups act faithfully on P1 over k (see, e.g., [2]). By Theorem 5.2, thereexists a faithful G-curve X such that every g ∈ G fixes a geometric point ofX. For the sake of contradiction, suppose that there exists a G-compressionX → E, where E is a curve of genus 1 endowed with a faithful G-action.In the algebraic closure, we obtain a G-compression Xk → Ek. Regard G455.3. Equivariant maps to projective spacesas a subgroup of Aut(Ek). By the exact sequence (5.1) and the fact thatG 6∼= Z/nZ for n = 1, 2, 3, 4, 6; we conclude that G ∩ i(Ek) 6= ∅. Since i(Ek)acts on Ek by translations, G ∩ i(Ek) acts freely on Ek. However, everyg ∈ G must fix a point on Ek by our assumption on Xk. This contradictionshows that X cannot be G-compressed to any G-curve of genus 1.In view of the above theorem, it remains to study the existence ofstrongly incompressible G-curves when G can act faithfully on a curve ofgenus 0. We will devote Section 5.3 to the study of equivariant rationalmaps to projective spaces, and we will use these results to understand com-pressions onto curves of genus 0.5.3 Equivariant maps to projective spacesLet G be an algebraic group defined over a field k. A projective representa-tion ρ : G ↪→ PGL(V ) gives rise to a G-action on P(V ). We will denote theresulting G-variety by P(V )ρ . If ρ and σ are projective G-representations, itis clear that P(V )ρ and P(V )σ are G-equivariantly isomorphic if and only ifρ and σ are conjugate. In what follows, we always assume that the G-actionon P(V )ρ is generically free.Consider the commutative diagram whose rows are central exact se-quences1 // Gm // GL(V ) // PGL(V ) // 11 // Gm // G′?ρOO// G?ρOO// 1(5.2)where G′ is the full preimage of G in GL(V ). Given a field extension K/k,we obtain the corresponding diagram in cohomology1 // H1(K,PGL(V )) // H2(K,Gm)1 // H1(K,G′)ρ∗OOϕ// H1(K,G)ρ∗OO∆ρ// H2(K,Gm)(5.3)(Note that H1(K,Gm) and H1(K,GL(V )) are trivial by Hilbert’s Theorem90.) This construction defines a cohomological invariant ∆ρ : H1(K,G) →H2(K,Gm) = Br(K). If X is a generically free primitive G-variety andL = k(X)G, we denote the Brauer class associated to [X] ∈ H1(L,G) by∆ρ(X). Note that ∆ρ(X) is trivial if and only if [X] lifts to a G′-torsor[X ′] ∈ H1(L,G′).465.3. Equivariant maps to projective spacesConstruction 5.5. Let Y be a primitive closed G-subvariety of P(V )ρ .Endow V with a linear G′-action via ρ and define Y˜ ⊂ V to be the affinecone over Y with the origin removed. It is not hard to see that Y˜ is aprimitive G′-variety. Moreover, it is well known that Y˜ is a Gm-torsorand Y is isomorphic to the geometric quotient Y˜ /Gm. Note also that thegroup G′/Gm ∼= G acts naturally on Y˜ /Gm, in such a way that the aboveisomorphism is G-equivariant.Lemma 5.6. Let Y be a generically free primitive closed G-subvariety ofP(V )ρ . Then ∆ρ(Y ) is trivial.Proof. We need to show that [Y ] is in the image of the map ϕ : H1(K,G′)→H1(K,G), where K = k(Y )G. Let Y˜ be as in Construction 5.5. If x ∈ Yhas trivial stabilizer in G, then any lift x˜ ∈ Y˜ of x has trivial stabilizer inG′. It follows that Y˜ is a generically free primitive G′-variety and clearlyϕ([Y˜ ]) = [Y˜ /Gm] = [Y ].Proposition 5.7. Let G be a finite group, let ρ : G ↪→ PGL(V ) be a projec-tive representation and let X be a faithful primitive G-variety.(a) Suppose that there exists a G-equivariant rational map f : X 99K P(V )ρ .Then ∆ρ(X) is trivial.(b) Conversely, suppose that ∆ρ(X) is trivial. Then, given any G-invariantopen subset U ⊂ P(V )ρ , there exists a G-equivariant rational map X 99KU .Proof. (a) We write Y = f(X), K = k(Y )G and L = k(X)G. We separatethe proof into two cases.Case 1: Suppose that Y is a faithful G-variety. This case follows fromthe fact that ∆ρ is a cohomological invariant. The G-compression f : X 99KY naturally induces a k-field homomorphism i : K ↪→ L and we have acommutative diagramH1(K,G)∆Kρ//i∗H2(K,Gm)H1(L,G)∆Lρ// H2(L,Gm)(5.4)It is well known that in the above situation, we must have i∗([Y ]) = [X]. ByLemma 5.6, we have ∆Kρ (Y ) = 1. The commutativity of the above diagramthen implies that ∆Lρ (X) = 1.475.3. Equivariant maps to projective spacesCase 2: Suppose that the G-action on Y has a kernel H. Let Y˜ be as inConstruction 5.5, and let H ′ be the kernel of the G′-action on Y˜ . We claimthat pi−1(H) splits as Gm ×H ′, where pi : G′ → G is the natural projection.Since G/H is finite and acts faithfully on Y , it also acts generically freely.Hence, we can select a geometric point y ∈ Y such that StabG(y) = H. Fixany lift y˜ ∈ Y˜ of y; by construction, it follows that StabG′(y˜) = H ′.Let h ∈ H and let h∗ ∈ pi−1(h) be any lift. Since h acts trivially on y,there exists λh∗ ∈ Gm such that h∗ ·y˜ = λh∗ y˜. It follows that λ−1h∗ h∗ stabilizesy˜, whence it must be contained in H ′. Since Y˜ is a Gm-torsor, it is easy to seethat λ−1h∗ h∗ is the unique element in pi−1(h) contained in H ′; in particular,it is independent of the lift h∗. It follows that the section s : H → pi−1(H)given by h→ λ−1h∗ h∗ is a well-defined homomorphism satisfying s(H) = H ′.Hence the exact sequence 1→ Gm → pi−1(H)→ H → 1 splits in the desiredway. This finishes the proof of the claim.We thus have a commutative diagram with exact rows1 // Gm // G′pi //G //11 // Gm // G′/H ′ // G/H // 1Since H acts trivially on Y , the dominant G-equivariant rational map X 99KY induces a G/H-compression X/H 99K Y , which gives rise to a k-fieldhomomorphism i : K ↪→ L. Using the bottom sequence above, we obtain acommutative diagram in cohomologyH1(K,G′/H ′) //H1(K,G/H) //i∗H2(K,Gm)H1(L,G′/H ′) // H1(L,G/H) // H2(L,Gm)The G/H-variety Y represents a class [Y ] ∈ H1(K,G/H), which maps to[X/H] ∈ H1(L,G/H) under i∗. It is easy to see that the G′/H ′-action on Y˜is generically free, so it follows that [Y ] comes from a class in H1(K,G′/H ′)and therefore its image in H2(K,Gm) is trivial. By the commutativity ofthe above diagram, the image of [X/H] in H2(L,Gm) is also trivial.To complete the proof of Case 2, note that we have a commutative485.4. Some explicit computationsdiagramH1(L,G′) //H1(L,G)∆ρ//H2(L,Gm)H1(L,G′/H ′) // H1(L,G/H) // H2(L,Gm)The image of [X] ∈ H1(L,G) under the middle vertical map is precisely[X/H]. It thus follows that ∆ρ(X) is trivial.(b) By assumption, [X] can be lifted to a class in H1(L,G′), i.e., thereexists a generically free primitive G′-variety X ′ such that X ′/Gm is bira-tionally isomorphic to X as a G-variety. Without loss of generality, we mayidentify X ′/Gm with X.We may view V as a generically free linear G′-variety and the naturalprojection piV : V 99K P(V )ρ as a rational quotient map. Let U′ = pi−1V (U),which is clearly a G′-invariant open subset of V . Note that V is a versalG′-variety (see, e.g., [13, Example 5.4]), whence there exists a G′-equivariantrational map X ′ 99K U ′. Taking quotients by Gm, we obtain a G-equivariantrational map X = X ′/Gm 99K U ′/Gm = U .We record the following corollary for future reference.Corollary 5.8. Let ρ : G ↪→ PGL2 be a projective representation of a non-trivial finite group G and let X be a faithful irreducible G-variety. Thenthere exists a G-compression X 99K P1ρ if and only if ∆ρ(X) = 1.Proof. The “only if” part follows directly from Proposition 5.7(a). On theother hand, suppose that ∆ρ(X) = 1. Since G is nontrivial, P1ρ has afinite number of G-fixed points. Therefore, we can find a G-invariant openU ⊂ P1ρ not containing any G-fixed points. By Proposition 5.7(b), thereexists a G-equivariant rational map f : X 99K P1ρ such that f(X) ⊂ U .The closure f(X) is a G-invariant closed irreducible subset of P1ρ . Byconstruction, it cannot be a fixed point, so it coincides with P1ρ itself. Thisproves that f is dominant.5.4 Some explicit computationsIn this section, we explicitly compute the invariant introduced in Section5.3 for certain actions on the projective line. We will use these results laterto study the strong incompressibility of G-curves in the case where G actsfaithfully on P1.495.4. Some explicit computationsRecall that the conjugacy classes of embeddings of (Z/2Z)2 ↪→ PGL2(k)are parametrized by the pairs (a, b) ∈ (k×/k×2)2 such that the quaternionalgebra (a, b)2 is split. The corresponding embedding is denoted by ρ(a,b),which is defined in Proposition 3.10. For simplicity, denote the (Z/2Z)2-variety P1ρ(a,b) by P1(a,b) . Clearly, P1(a,b) and P1(a′,b′) are isomorphic as(Z/2Z)2-varieties if and only if a = a′ and b = b′.Lemma 5.9. Let ρ(a,b) be as above, let K/k be a field extension and identifyH1(K, (Z/2Z)2) with (K×/K×2)2. Then ∆ρ(a,b)(c, d) = [(ac, bd)2] for allc, d ∈ K×.Proof. It suffices to prove that ρ(a,b)∗ : (K×/K×2)2 → H1(K,PGL2) maps(c, d) to (ac, bd)2. Let U, V ∈ GL2 be lifts of ρ(a,b)(e1), ρ(a,b)(e2), respectively.Note that U2 = b′I, V 2 = a′I and UV + V U = 0, where a′ = a and b′ = b.Rescaling the lifts if necessary, we may assume that a′ = a and b′ = b. LetA be the split quaternion algebra (b, a)2. Note that there is a k-algebraisomorphism A ∼= M2 given by i 7→ U , j 7→ V , which induces isomorphismsGL1(A) ∼= GL2 and PGL1(A) ∼= PGL2. By construction, ρ(a,b) factors as(Z/2Z)2  ϕ// PGL1(A)∼= // PGL2,where the embedding ϕ is given by e1 7→ [i], e2 7→ [j]. We have thereforereduced the problem to showing that ϕ∗ : (K×/K×2)2 → H1(K,PGL1(A))sends (c, d) to (ac, bd)2 for all c, d ∈ K×.We give a proof of this fact by Galois descent. Let L = K(√c,√d);then we may view ϕ∗(c, d) as an element of H1(Gal(L/K),PGL1(A)(L)).For simplicity, assume that c, d and cd are non-squares; the remaining casesare easier and left to the reader. Define generators σ1, σ2 ∈ Gal(L/K) suchthat σ1 fixes√d and sends√c to −√c, while σ2 fixes√c and sends√d to−√d. Note that the 1-cocycle v : Gal(L/K) → PGL1(A)(L) representingϕ∗(c, d) is given by σ1 7→ [i], σ2 7→ [j]. Then we twist the Galois action onγ = x+ yi+ zj + tij ∈ A⊗K L by settingσ1 ∗ γ = vσ1(σ1(γ)) = i−1σ1(γ)i = σ1(x) + σ1(y)i− σ1(z)j − σ1(t)ij;σ2 ∗ γ = vσ2(σ2(γ)) = j−1σ(γ)j = σ2(x)− σ2(y)i+ σ2(z)j − σ2(t)ij.It follows that γ is invariant under the twisted Galois action if and onlyif γ = x + y1√d i + z1√c j + t1√cd ij for x, y1, z1, t1 ∈ K. This impliesthat ϕ∗(c, d) is generated as a K-algebra by i′ =√d i and j′ =√c j, whichsatisfy i′2 = bd, j′2 = ac and i′j′ + j′i′ = 0. Consequently, we obtain thatϕ∗(c, d) = (bd, ac)2 ∼= (ac, bd)2.505.5. Cyclic and dihedral groups: Compressibility of P1Let ρa : Z/2Z ↪→ PGL2 be as in (3.1) and denote P1ρa simply by P1a .Note that P1a and P1a′ are isomorphic as Z/2Z-varieties if and only if a = a′.By [20, Example 6], it follows that P1a is versal if and only if a ∈ k×2.Corollary 5.10. Let ρa be as above, let K/k be a field extension and identifyH1(K,Z/2Z) with K×/K×2. Then ∆ρa(c) = [(c, a)2] for all c ∈ K×.Proof. We need to show that ρa∗ : K×/K×2 → H1(K,PGL2) maps c to(c, a)2 for all c ∈ K×. Note that we may write ρa = ρ(1,a) ◦ φ, whereφ : Z/2Z→ (Z/2Z)2 sends −1 7→ e1. Therefore we must haveρa∗(c) = ρ(1,a)∗ ◦ φ∗(c) = ρ(1,a)∗(c, 1) = (c, a)2,where the last equality follows from Lemma Cyclic and dihedral groups: Compressibilityof P1Recall that the groups Z/nZ and D2n act faithfully on some curve of genus0 if and only if they act faithfully on P1, which happens if and only if αn ∈ k(see Theorem 1.2). If the latter condition does not hold, the existence ofstrongly incompressible curves for Z/nZ and D2n follows from Theorem 5.4.Lemma 5.11. Let n ≥ 3 be any integer, let k be a field containing αn, anddefine the embedding ρ1 : D2n ↪→ PGL2 by sendingσ 7→(αn + 1 βn1 αn + 1), τ 7→(1 00 −1), (5.5)where σ, τ are the usual generators of D2n. (This results from choosinga = x = 1, y = 0 in (3.6).) Then P1ρ1 is not strongly incompressible.Proof. We need to exhibit a G-equivariant map P1ρ1 → P1ρ1 that is notinjective. Select a square root of βn (possibly in a quadratic extension of k)and defineQ =(1 1−β−1/2n β−1/2n),in such a way thatQ−1ρ1(σ)Q =(1 + ωn 00 1 + ω−1n), Q−1ρ1(τ)Q =(0 11 0).515.5. Cyclic and dihedral groups: Compressibility of P1Let F : P1 → P1 be given by F (x : y) = (xn+1 : yn+1). A calculation showsthatF ◦ (Q−1ρ1(σ)Q) = (Q−1ρ1(σ)Q) ◦ FandF ◦ (Q−1ρ1(τ)Q) = (Q−1ρ1(τ)Q) ◦ F.It follows that Q ◦ F ◦Q−1 is a G-equivariant map P1ρ1 → P1ρ1 definedover k(β1/2n ). Explicitly, note that Q◦F ◦Q−1 sends (x : y) to (u : v), whereu = (x+ β1/2n y)n+1 + (x− β1/2n y)n+1;v = β−1/2n((x+ β1/2n y)n+1 − (x− β1/2n y)n+1).In particular, it follows that Q ◦F ◦Q−1 is actually defined over k. Since ithas degree n+ 1, it is not injective and we are done.Remark 5.12. Restricting the embedding (5.5) to Z/nZ, the above lemmaproves a fortiori that the projective line is not strongly incompressible as aZ/nZ-variety.Proposition 5.13. Let n ≥ 2 be any integer and let k be a field containingωn. Then there are no strongly incompressible Z/nZ-varieties.Proof. This is proved in [24, Example 5]; we supply a short alternativeproof. Recall that the embedding ρ : Z/nZ ↪→ PGL2 sending a generator ofZ/nZ to the diagonal matrix diag(ωn, 1), is generic, i.e., P1ρ is versal. Anyfaithful Z/nZ-variety can thus be Z/nZ-compressed to P1ρ . Moreover, P1ρis not strongly incompressible, as shown by the nontrivial Z/nZ-compression(x : y) 7→ (xn+1 : yn+1).The techniques introduced above can be used to show that there are nostrongly incompressible varieties for odd cyclic and odd dihedral groups ifthey act faithfully on the projective line.Proposition 5.14. Let n ≥ 3 be an odd integer, let k be a field containingαn, and let G be either Z/nZ or D2n. Then there are no strongly incom-pressible G-varieties.Proof. We focus on the case G = D2n; the cyclic case follows along the samelines. Note that the embedding ρ1 defined in (5.5) is generic for odd n, i.e.,the G-variety P1ρ1 is versal (see [20, Thm. 8]). It follows that any faithfulG-variety can be G-compressed to P1ρ1 . It thus suffices to prove that P1ρ1is not strongly incompressible, which follows directly from Lemma 5.11.525.6. Strongly incompressible curves for even cyclic groups5.6 Strongly incompressible curves for evencyclic groupsLet G = Z/nZ, where n ≥ 4 is even, and let k be a field containing αn.Define ρ : G ↪→ PGL2 as in (3.5), which is unique up to conjugacy. By theresults in Proposition 5.13, it remains to analyze the case where ωn 6∈ k. (Inthis situation, P1ρ is not versal.) Interestingly, we will prove that there existstrongly incompressible G-curves under this assumption. With this goal inmind, we first prove the following technical lemma.Lemma 5.15. Let k be a field such that αn ∈ k. Then there exists s ∈ k[x]square-free satisfying the following properties:(a) s(a) = 0 for some a ∈ k.(b) The hyperelliptic curve C with equation y2 = s(x) can be endowed witha faithful G-action, in such a way that the unique element of order 2 inG acts as the hyperelliptic involution on C.Proof. Suppose first that n = 4. In that case, we can take s(x) = x5 − xand a = 0. The Z/4Z-action on the hyperelliptic curve y2 = s(x) is givenby σ · (x, y) 7→ (−1/x, y/x3), where σ is a generator of Z/4Z.Assume henceforth that n ≥ 6, and consider the exact sequence 1 →Z/2Z → G → Z/mZ → 1, where m = n/2. Using Proposition 4.9 (resp.4.10) if m is odd (resp. even) and the fact that α2m ∈ k, it follows that thereexists a hyperelliptic curve C/k with a faithful G-action satisfying (b). Weclaim that we can select C such that (a) holds as well.Let L = k(x) and K = k(t) be rational function fields such that L =k(C)Z/2Z and K = LZ/mZ. By the results from Section 4.5, it follows thatk(C) = L(√(x2 − βm)r), where we are free to choose r ∈ k(t)×. Recallthat t is a rational function of x, say t = p(x)/q(x). Choose a ∈ k, which isnot a zero of q or qp′−pq′, and let r = q(a)t−p(a). A hyperelliptic equationfor C is then given by y2 = s(x), where s(x) is the square-free part of(x2 − βm)(q(a)p(x)− p(a)q(x))q(x).By construction, s(a) = 0, which finishes the proof.Proposition 5.16. Let k be a field such that αn ∈ k and ωn 6∈ k. Thenthere exists a strongly incompressible G-curve.535.6. Strongly incompressible curves for even cyclic groupsProof. Let C be the hyperelliptic curve constructed in Lemma 5.15. Itsuffices to prove that C cannot be G-compressed to any curve of genus ≤ 1,as the result then follows from Lemma 5.1.We claim that C cannot be G-compressed to any curve of genus 0. Firstof all, such a curve would be forced to be P1ρ since C has k-rational points.For the sake of contradiction, suppose that there exists a G-compressionC → P1ρ . Regard this map as a Z/2Z-compression. Note thatρ(σm) =(0 βn1 0)and therefore P1ρ is isomorphic to P1βn as a Z/2Z-variety. On the otherhand, if we regard C as a Z/2Z-variety, its class [C] ∈ H1(L,Z/2Z) =L×/L×2 is given by s(x). (Here, we have L = k(C)Z/2Z = k(x) as in theproof of Lemma 5.15.) If we denote the restriction of ρ to Z/2Z by ρ′, weconclude using Corollary 5.10 that ∆ρ′(C) = [(s(x), βn)2]. This class mustbe trivial over L = k(x) by Corollary 5.8. If we apply Lemma 2.1 to theroot a of s, we obtain that βn ∈ k×2, i.e., ωn − ω−1n ∈ k. However, thiscontradicts the fact that ωn 6∈ k.It remains to show that C cannot be G-compressed to any G-curve ofgenus 1. Since C has k-rational points, it suffices to prove that there isno G-compression C → E, where E is an elliptic curve endowed with afaithful G-action. Suppose there is such a G-compression and regard G asa subgroup of Aut(E). By Lemma 5.3(a), we can write G ∼= G0 × pi(G),where G0 = G ∩ E and pi(G) ⊂ Aut0(E). Since G is cyclic, we concludethat G0 and pi(G) are cyclic groups of relatively prime order. We claim thatσm (the unique element of order 2 inside G) belongs to G0, or equivalentlythat G0 has even order. Suppose on the contrary that the order of pi(G)is even, i.e., pi(G) ∼= Z/dZ for d = 2, 4, or 6. By Lemma 5.3(d), thetranslation by P0 ∈ E and the automorphism α ∈ Aut0(E) commute if andonly if α(P0) = P0. Since pi(G) has even order, it contains the inversionmap P 7→ −P . Therefore any point of E fixed by pi(G) has order dividing 2.Since we are assuming that G0 is a cyclic group of odd order that commuteswith pi(G), it must be trivial. It follows that G = pi(G) ∼= Aut0(E) ∼= Z/nZfor n = 4 or 6. By Lemma 5.3(c), this contradicts the assumption thatk does not contain the appropriate roots of unity. We have proved thatσm ∈ G0, and hence acts as a translation on E. On the other hand, notethat σm fixes a k-rational point in C, namely (a, 0). Hence, it must also fixa point in E. This contradiction completes the proof.545.7. Strongly incompressible curves for even dihedral groups5.7 Strongly incompressible curves for evendihedral groups5.7.1 The Klein 4-groupThroughout this subsection, let G denote the Klein 4-group with generatorse1, e2. Recall that G acts faithfully on P1 over any field k, but such an actionis never versal. Our goal is to prove the following proposition.Proposition 5.17. The following are equivalent:(i) There are no strongly incompressible G-curves over k.(ii) cd2(k) = 0.Proof of (ii) ⇒ (i). Assume that k has cohomological 2-dimension 0 and letX be any faithful G-curve. The field K = k(X)G is a transcendence degree1 extension of k, so cd2(K) ≤ 1 by [31, Prop. II.4.2.11]. Then, by [31, Prop.II.2.3.4], it follows that Br2(K) is trivial. Let ρ : (Z/2Z)2 ↪→ PGL2 be anyembedding. We claim that X can be G-compressed to P1ρ . Indeed, notethat ∆ρ(X) is the class of a quaternion algebra defined over K and thereforetrivial. The result then follows from Corollary 5.8. To conclude the proofof the sufficiency, we need to prove that P1ρ is not strongly incompressible.We are free to select ρ conveniently, so we may assume that ρ is as in (3.4)with a = 1, i.e., ρ = ρ(1,1). Then, it is obvious that (x : y) 7→ (x3 : y3) is aG-compression of P1ρ to itself that is not birational.It remains to prove that (i)⇒ (ii). To achieve this, we need the followingresult.Proposition 5.18. Let P,Q ∈ k[x] be separable polynomials of degree ≥ 1satisfying the following conditions.(i) P and Q have no common roots.(ii) P (0) 6= 0, Q(0) 6= 0.(iii) There exists a root x1 ∈ k of P (resp. x2 ∈ k of Q) such that x1Q(x1) ∈k×2 (resp. x2P (x2) ∈ k×2).Then the curve X with function field L = K(√xP (x),√xQ(x)), whereK = k(x) is rational, can be endowed with a faithful G-action such thatevery element of G fixes at least one geometric point of X.555.7. Strongly incompressible curves for even dihedral groupsProof. Let A3 be the affine 3-space over k and let Y ⊂ A3 be the affinevariety cut out by the ideal I = 〈y2 − xP (x), z2 − xQ(x)〉. Note that Yis an irreducible affine curve having a unique singularity at (0, 0, 0) and itsfunction field is precisely L. We can endow Y with a faithful G-action bysetting e1 ·(x, y, z) = (x,−y, z) and e2 ·(x, y, z) = (x, y,−z). This action canbe lifted to the unique nonsingular projective curve X which is birationalto Y , in such a way that the natural birational isomorphism X 99K Y is G-equivariant. Note also that X can be seen as a Galois G-cover of P1 inducedby the inclusion K ↪→ L.We claim that every element of G fixes at least one geometric point of X.We first prove the assertion for e1 ∈ G to illustrate the procedure. Note thatA = (x1, 0,√x1Q(x1)) is a nonsingular k-rational point of Y fixed by e1.Therefore, the natural G-equivariant rational map Y 99K X must be definedat the point A and its image in X is fixed by e1 as desired. Analogously, wesee that B = (x2,√x2P (x2), 0) is a nonsingular k-rational point of Y fixedby e2 and the result follows along the same lines.It remains to prove that e1e2 fixes a point in X. Unfortunately, theonly fixed point of e1e2 in Y is O = (0, 0, 0), which is not smooth. Toovercome this difficulty, we consider the blowup of A3 at the origin O withexceptional divisor E and consider the strict transform Y ′ of Y . The G-action lifts naturally to Y ′, in such a way that the birational morphismY ′ → Y is G-equivariant. We claim that Y ′ has a smooth point fixed bye1e2, which has to be contained in Y ′ ∩ E. Recall thatBlOA3 = {((x, y, z), (t0 : t1 : t2)) ∈ A3 × P2 | xt1 = yt0, xt2 = zt0, yt2 = zt1}is covered by three affine charts Ui = {ti 6= 0} isomorphic to A3. We pickcoordinates y, u = t0/t1, v = t2/t1 in U1 (so that x = yu and z = yv)and compute Y ′ in this coordinates. Any point in Y ′ ∩ U1 must satisfy theequations y − uP (yu) = 0 and yv2 − uQ(yu) = 0. Moreover, note that thepolynomial Q(0)(y − uP (yu)) − P (0)(yv2 − uQ(yu)) is divisible by y andconsequently we obtain thatQ(0)− P (0)v2 − u2Q(0)P1(yu) + u2P (0)Q1(yu) = 0,for all points (y, u, v) ∈ Y ′∩U1, where P1(x) = (P (x)−P (0))/x and Q1(x) =(Q(x)−Q(0))/x. Then it is easy to see that the above three equations defineY ′ ∩ U1 and that Y ′ ∩ U1 ∩ E = {(0, 0,±√Q(0)/P (0))}. (Actually one cansee by looking at the other two charts that Y ′∩E consists only of these twopoints.) We now look at the G-action on Y ′∩U1. Note that e1e2 · (y, u, v) =(−y,−u, v), since e1e2 · (x, y, z) = (x,−y,−z) in Y . Therefore, the points565.7. Strongly incompressible curves for even dihedral groups(0, 0,±√P (0)/Q(0)) are fixed by e1e2. Moreover, by applying the Jacobiancriterion to the three polynomials defining Y ′ ∩U1, one can show that bothpoints are smooth; the details are left to the reader. Since the G-equivariantrational map Y ′ → Y 99K X is defined at all smooth points, it follows thate1e2 has a fixed point in X.Lemma 5.19. Let X be any (smooth projective) G-curve obtained fromProposition 5.18. Then X cannot be G-compressed to any curve of genus 1.Proof. Suppose that such a G-compression X → E exists. We may assumethat E is an elliptic curve since X has k-rational points. By parts (a) and(b) of Lemma 5.3, some element of G must act freely on E. This contradictsthe fact that every element of G fixes a point in X.We are ready to prove that (i) ⇒ (ii) in Proposition 5.17. Suppose thatk does not have cohomological 2-dimension 0. We will produce a faithfulG-curve that cannot be G-compressed to any G-curve of genus ≤ 1 by usingProposition 5.18. The following well known lemma provides more manage-able conditions on k.Lemma 5.20. Let k be a field. The following are equivalent:(i) cd2(k) = 0.(ii) k is hereditarily quadratically closed, i.e., every algebraic extension ofk is quadratically closed.(iii) ξ is a square in k(ξ) for every ξ ∈ k.(iv) Br2(k(x)) = 0.Proof. The equivalence (ii) ⇔ (iii) is straightforward and left to the reader,while (i)⇔ (ii) follows directly from [9, Lemma 2]. We now prove that (i)⇒(iv). If cd2(k) = 0, it follows from [31, Prop. II.4.1.11] that cd2(k(x)) ≤ 1.Then we conclude that Br2(k(x)) = 0 by [31, Prop. II.2.3.4]. To completethe proof, it suffices to show that (iv) ⇒ (iii). Suppose that (iv) holds,but there exists ξ ∈ k, which is not a square in k(ξ). Let h ∈ k[x] be theminimal polynomial of ξ over k. The quaternion algebra (x, h(x))2 must besplit over k(x), which implies that ξ is a square over k(ξ) by Lemma 2.1.This contradiction completes the proof.575.7. Strongly incompressible curves for even dihedral groupsConstruction 5.21. In view of Lemma 5.20, given that cd2(k) 6= 0, we canchoose an element ξ algebraic over k, which is not a square in k(ξ). Leth ∈ k[x] be the minimal polynomial of ξ over k and define polynomialsP (x) = (x− α)((x− α− 1)h(x− α) + (α+ 1)h(−α)(α+ 1)h(−α)x),Q(x) = α(α+ 1− x)h(0)h(x− α),where α ∈ k is taken such that P has no multiple roots, P (0) 6= 0 andQ(0) 6= 0. (It is not hard to see that such a selection of α is always possible.)We conclude that P and Q satisfy the conditions of Proposition 5.18; in whatfollows, let X denote the corresponding curve.Lemma 5.22. Let X be the curve obtained in Construction 5.21. Thenthere is no G-compression X → Y , where Y is a curve of genus 0.Proof. Note that X has k-rational points. Hence, such a G-compressioncould only be possible if Y ∼= P1(a,b) for some embedding ρ(a,b) : G ↪→ PGL2.From Proposition 5.18, we observe that k(X)G = K = k(x) and the class[X] ∈ H1(K,G) corresponds to (xP (x), xQ(x)) ∈ (K×/K×2)2. By Lemma5.9, we obtain that ∆ρ(a,b)(X) = [(axP (x), bxQ(x))2] ∈ Br(K).Suppose that there exists a G-compression X → P1ρ . Then, by Corol-lary 5.8, the quaternion algebra (axP (x), bxQ(x))2 must be split over K.Applying Lemma 2.1 to the roots α+ 1 and α+ ξ of bxQ(x), we obtain thata ∈ k×2 and aξ ∈ k(ξ)×2, respectively. This contradicts the assumption thatξ is not a square in k(ξ).To finish the proof of Proposition 5.17, we use Lemmas 5.19 and 5.22 toconclude that X cannot be G-compressed to any curve of genus ≤ 1. Thus,it follows from Lemma 5.1 that there exist strongly incompressible G-curvesif cd2(k) > 0. The proof is now complete.5.7.2 Even dihedral groups of order ≥ 8In this subsection, G will always denote the dihedral group D2n, where n ≥ 4is an even integer. A result similar to Proposition 5.17 holds in this case.Proposition 5.23. Let k be a field such that αn ∈ k. Then there exist nostrongly incompressible G-curves defined over k if and only if cd2(k) = 0.Proof. Suppose first that cd2(k) = 0. Similarly to the proof of Proposition5.17, it follows that any faithful G-curve X can be G-compressed to P1ρ ,585.7. Strongly incompressible curves for even dihedral groupswhere the embedding ρ : G ↪→ PGL2 is as in (5.5). Moreover, it follows fromLemma 5.11 that P1ρ is not strongly incompressible.To prove the converse, assume that cd2(k) > 0. We must show thatthere exists a strongly incompressible G-curve under this assumption. Wefirst study the special case where ωn 6∈ k, i.e., βn is not a square in k. We onlysketch the argument, as it is very similar to the proof of Proposition 5.16.Using the results from Section 4.5, one can construct a hyperelliptic G-curveC such that the central element σn/2 ∈ G acts as the hyperelliptic involutionof C, and there exists a k-rational point of C fixed by σn/2. Suppose that Ccan be G-compressed to P1η , where η is any embedding G ↪→ PGL2. Regardthe G-compression as a Z/2Z-compression with respect to the center of G.As a Z/2Z-variety, P1η is isomorphic to P1βn . As in the proof of Proposition5.16, we must have βn ∈ k×2, contradicting our assumption. Similarly, itfollows that C cannot be G-compressed to any curve of genus 1. By Lemma5.1, there exists a strongly incompressible G-curve in this case.In what follows, assume that ωn ∈ k. By Lemma 5.20, there exists ξ ∈ ksuch that ξ is not a square in k(ξ). Using this information, we construct ahyperelliptic G-curve C that cannot be G-compressed to any curve of genus≤ 1. Let m = n/2, and define C to be the hyperelliptic curve with equationy2 = xf(xm + x−m2),where f ∈ k[t] will be determined later. This curve can be endowed with afaithful G-action given by σ : (x, y) 7→ (ω2nx, ωny), τ : (x, y) 7→ (x−1, yx−1).Note that we can regard C as a (Z/2Z)2-variety under the action of thesubgroup 〈σm, τ〉. We can write the function field of C in the formk(C) = k(x, y)/(y2 − xf(xm + x−m2)).It is not hard to see thatk(C)〈σm〉 = k(x),k(C)〈τ〉 = k(x+ x−12, y(1 + x−1))/(y2 − xf(xm + x−m2)),k(C)〈σm,τ〉 = k(x+ x−12).Recall that there exists a polynomial Tm such that Tm((x + x−1)/2) =(xm + x−m)/2. For simplicity, write s = (x+ x−1)/2. Note thaty2(1 + x−1)2 = xf(xm + x−m2)(1 + 2x−1 + x−2) = (2s+ 2)f(Tm(s)),595.7. Strongly incompressible curves for even dihedral groupswhence k(C)〈τ〉 is obtained from k(s) by adjoining√(2s+ 2)f(Tm(s)). Onthe other hand, note that k(C)〈σm〉 is obtained from k(s) by adjoining√s2 − 1. It follows that the class [C] ∈ H1(k(s), (Z/2Z)2) is equal to(2(s+ 1)f(Tm(s)), s2 − 1).Suppose that C can be G-compressed to a curve of genus 0. Since C hask-rational points, such genus 0 curve must be G-equivariantly isomorphicto P1ρa , where ρa is the embedding (3.6) for some class a. Note that we areassuming that βn is a square in k, so the binary form 〈1,−βn〉 is universaland therefore, a ∈ k×/k×2 is arbitrary. Since ωn ∈ k, the embedding ρa canbe conjugated to the embeddingη : σ 7→(ωn 00 1), τ 7→(0 a1 0),so it follows that P1η ∼= P1ρa as G-varieties. It is then straightforward tosee that P1η ∼= P1(a,1) as (Z/2Z)2-varieties. Using Lemma 5.9, we com-pute ∆ρ(a,1)(C) = [(2a(s + 1)f(Tm(s)), s2 − 1)2]. If we regard the assumedG-compression C → P1η as a (Z/2Z)2-compression, then it follows that∆ρ(a,1)(C) is trivial in Br(k(s)).We will now select f to arrive to a contradiction. Let γ, δ ∈ k be suchthat γ2 = 1 + ξ and δ2 = 1 + ξ−1. Replacing ξ by another element in ξ · k×2if necessary, we can choose f satisfying the following properties:• f(Tm(γ)) = f(Tm(δ)) = 0.• The polynomial (s+ 1)f(Tm(s)) is separable.Since (2a(s+1)f(Tm(s)), s2−1)2 is split over k(s), we can apply Lemma2.1 to γ and obtain that ξ is a square in k(γ). It follows that k(γ) = k(√ξ),since [k(√ξ) : k(ξ)] = 2 by assumption. We can thus write γ = l1 + l2√ξ forsome l1, l2 ∈ k(ξ). Squaring, we obtain that ξ+1 = l21+l22ξ+2l1l2√ξ, whencel1l2 = 0. If l2 = 0, it follows that ξ+ 1 is a square in k(ξ), contradicting thefact that [k(γ) : k(ξ)] = 2. Hence we must have l1 = 0, which implies that1 + ξ−1 is a square in k(ξ), i.e., k(δ) = k(ξ). However, applying Lemma 2.1to δ implies that ξ−1 (and hence ξ) is a square in k(δ), which contradicts ourassumption. This proves that a G-compression C → P1ρa is not possible.It remains to prove that C cannot be G-compressed to any curve ofgenus 1. Suppose there is such a G-compression C → E. By construction,the hyperelliptic involution of C, namely σm, fixes a k-rational point ofC. Hence, σm must fix some k-rational point of E, which we may assumeto be an elliptic curve. We adopt the notation of Lemma 5.3(a), where605.8. Polyhedral groupspi : Aut(E) → Aut0(E) denotes the natural projection. Since Aut0(E) isabelian, the relation (στ)2 = 1 implies that pi(στ)2 = pi(σ2)pi(τ2) = pi(σ2) =1. It follows that σ2 acts as a translation on E. We claim that σ acts as atranslation as well. For the sake of contradiction, assume the contrary. ByLemma 5.3(a), we may write σ = τP0 ◦ α, where τP0 denotes the translationby P0 ∈ E, and α ∈ Aut0(E) is nontrivial. Since σ2 acts as a translation, itfollows that σ2(P )− P = α2(P )− P + α(P0) + P0 must be constant for allP ∈ E. This implies that the isogeny α2 − id is constant, so it is the zeromap. This proves that α has order 2 in Aut0(E), whence α is the inversionmap P 7→ −P . It follows that σ2 = id in Aut(E), which is a contradictionbecause σ has order n ≥ 4. We have proved that σ acts as a translation onE, and therefore so does σm. Hence σm cannot fix any point of E.5.8 Polyhedral groupsIt remains to study the incompressibility of curves endowed with a faithfulaction of a polyhedral group G, i.e., G = A4, S4, or A5.5.8.1 Serre’s cohomological invariantLet Ĝ be the binary polyhedral group associated to G. If G is an alter-nating group, then Ĝ coincides with the unique nontrivial central extensionof G by Z/2Z. If G = S4, then Ĝ is the unique central extension of G byZ/2Z in which transpositions and products of disjoints transpositions lift toelements of order 4. (This is not the double cover studied in [30], in whichtranspositions lift to involutions). We have a central exact sequence1 // Z/2Z // Ĝ // G // 1 ,which yields a corresponding sequence in cohomologyH1(K, Ĝ) // H1(K,G) ∆̂ // Br2(K),for any field extension K/k. Note that ∆̂ : H1(K,G) → H2(K,Z/2Z) =Br2(K) defines a cohomological invariant. If X is a faithful primitive G-variety and L = k(X)G, we denote the Brauer class associated to [X] ∈H1(L,G) by ∆̂(X). Note that ∆̂(X) is trivial if and only if [X] lifts to aĜ-torsor [X̂] ∈ H1(L, Ĝ). The following result follows from the definition ofcohomological invariant.615.8. Polyhedral groupsProposition 5.24. Let X,Y be faithful primitive G-varieties and supposethat there exists a G-compression f : X 99K Y . Let i : k(Y )G ↪→ k(X)G bethe natural inclusion induced by f and define i∗ : Br2(k(Y )G)→ Br2(k(X)G)as the corresponding functorial map. Then i∗(∆̂(Y )) = ∆̂(X).Proof. Left to the reader.J.-P. Serre has described an effective way to compute ∆̂. An element ofH1(K,G) can be viewed as (the isomorphism class of) an e´tale K-algebraE, which has trivial discriminant if G is alternating. Then we have thefollowing result.Proposition 5.25 (cf. [30, Th. 1]). Let qE is the trace form of E/K. Then∆̂(E) = w2(qE) + [(−2, dE)2],where w2 denotes the second Stiefel-Whitney class and dE is the discriminantof E.Proof. See [30, Th. 1] or [35, §2].Remark 5.26. If the field k satisfies some additional conditions, we mayview ∆̂ as a particular case of the cohomological invariant defined in Section5.3. Suppose that the following assumptions hold.(i) There exists an embedding ρ : G ↪→ PGL2. This is the case if and onlyif −1 is the sum of two squares over k, with the additional requirementthat√5 ∈ k if G = A5 (see [2]).(ii) There exists an embedding ρ : Ĝ ↪→ GL2 that fits in a commutativediagram1 // Gm // GL2 // PGL2 // 11 // Z/2Z //?OOĜ?ρOO// G?ρOO// 1This is automatic if G is alternating. In the case G = S4, it is true ifand only if√2 ∈ k.Passing to cohomology in the above diagram, we conclude that ∆̂ coin-cides with ∆ρ, if we regard both their images to lie in the Brauer group.625.8. Polyhedral groups5.8.2 Computation of the invariant for curves of genus ≤ 1We first compute the cohomological invariant ∆̂ for polyhedral actions oncurves of genus 0. Recall that, up to equivariant birational isomorphism,there is only one action of a polyhedral group G on a curve of genus 0.In what follows, let q0(x, y, z) = x2 + y2 + z2 and denote by X0 ⊂ P2 thecorresponding quadric. Then, G acts on X0 via the standard embeddingρ : G ↪→ SO(q0) as a rotation group. If G = A4 or S4, the action is definedover any field k, while for G = A5 the action is defined over k if and only if√5 ∈ k. Recall also that K := k(X0)G is isomorphic to a rational functionfield, i.e., X0/G ∼= P1.Proposition 5.27. (a) If G is alternating, then ∆̂(X0) = [(−1,−1)2] inBr2(K).(b) If G = S4, then ∆̂(X0) = [(−1,−1)2] + [(2, t)2] in Br2(K), where t issome generator of K/k.Proof. (a) Let k′/k be a field extension, and suppose that q0 is isotropic overk′. We claim that ∆̂(X ′0) is trivial in Br2(k′(X ′0)G), where X ′0 = X0×Spec(k)Spec(k′). Indeed, note that PGL2 ∼= SO(q0) over k′, whence there exists anembedding ρ : G ↪→ PGL2 defined over k′ and a G-equivariant isomorphismX ′0 ∼= P1ρ . It follows from Remark 5.26 that ∆̂(X′0) = ∆ρ(X′0) = ∆ρ( P1ρ ),which is trivial by Lemma 5.6. This completes the proof of the claim.Let E be the e´tale algebra corresponding to [X0] ∈ H1(K,G). ThenqE ∼= 〈1, a, b, ab〉 for some a, b ∈ K if G = A4 (resp. qE ∼= 〈1, a, b, c, abc〉 forsome a, b, c ∈ K if G = A5). It follows that ∆̂(X0) = w2(qE) = [(−a,−b)2]+[(−1,−1)2] if G = A4 (resp. [(−ac,−bc)2] + [(−1,−1)2] if G = A5). In anycase, we can write ∆̂(X0) = [(u, v)2] + [(−1,−1)2] for some u, v ∈ K, soit suffices to prove that (u, v)2 is split over K. Since q0 is isotropic overk′ := k(s, t)/(s2 + t2 + 1) and (−1,−1)2 splits over k′, it follows from theprevious paragraph that (u, v)2 splits over k′(X ′0)G ∼= K(s, t)/(s2 + t2 + 1).Equivalently, the Pfister form 〈1,−u,−v, uv〉 is hyperbolic over K(s, t)/(s2+t2 +1), which is the function field of the quadratic form 〈1, 1, 1〉 defined overK. By [17, Th. X.4.5], either 〈1,−u,−v, uv〉 is isotropic (hence hyperbolic)over K, or 〈1,−u,−v, uv〉 ∼= 〈1, 1, 1, 1〉 over K. Equivalently, either (u, v)2splits, or (u, v)2 ∼= (−1,−1)2. The former case yields the desired result,while the latter implies that ∆̂(X0) is trivial. Hence, it suffices to prove that∆̂(X0) is nontrivial whenever (−1,−1)2 is not split over K (equivalently overk, since K is purely transcendental over k).Assume for the sake of contradiction that (−1,−1)2 is not split over kand ∆̂(X0) is trivial. This implies that [X0] comes from a class in H1(K, Ĝ),635.8. Polyhedral groupsi.e., there exists a faithful primitive Ĝ-variety X̂0 such that X̂0/(Z/2Z) isbirationally isomorphic to X0 as a G-variety. Note that X̂0 must be geomet-rically irreducible since 1→ Z/2Z→ Ĝ→ G→ 1 is not split. Thus, we mayassume without loss of generality that X̂0 is a (smooth projective) Ĝ-curve,endowed with a 2-1 quotient morphism X̂0 → X0. It follows that X̂0 is ahyperelliptic curve (in the sense that its canonical divisor is not very ample).Moreover, note that Aut(X̂0)(k) contains Ĝ, which equals SL2(F3) if G = A4(resp. SL2(F5) if G = A5). By [33, Table 1], it follows that the genus of X̂0 iseven. However, it is well known that this implies that X̂0/(Z/2Z) = X0 hasa k-rational point (see, e.g., [22, §2.1]), which is equivalent to the splittingof (−1,−1)2 over the field k. This contradiction concludes the proof.(b) Note that S4 embeds into SO(q0) as the matrices of the form DP ,where D is diagonal with entries ±1 and P is a permutation matrix. (Thereare 24 such matrices of determinant 1.) The e´tale K-algebra correspondingto [X0] ∈ H1(K,S4) is the field extension k(X0)H/K, where H is any copyof S3 inside S4. For convenience, we choose the subgroup H generated byσ =0 1 00 0 11 0 0 , τ =0 0 −10 −1 0−1 0 0 .Note that S4 = V oH, where V is the subgroup of diagonal matrices in S4.We write k(X0) = k(a, b)/(a2 + b2 + 1), where a = x/z and b = y/zin the usual coordinates of X0. Note that σ(a) = b/a and σ(b) = 1/a,while τ(a) = 1/a and τ(b) = b/a. An easy computation then shows thatk(X0)H = k(α), where α = a+ b/a+ 1/b+ 1/a+ b+a/b. By Galois Theory,the minimal polynomial of α over K is equal toP (Y ) =∏g∈V(Y − g(α)) = Y 4 − 6Y 2 + 8Y + t+ 24,wheret =(a− 1)2(a+ 1)2(2a2 + 1)2(a2 + 2)2a4(a2 + 1)2is a generator of K/k, which proves that k(X0)H = K[Y ]/(p(Y )). By asimple computation, the trace form of K[Y ]/(p(Y )) over K is isomorphicto 〈1, 3,−(t + 27),−3t(t + 27)〉. It follows that its Stiefel-Whitney class isequal to [(−3t, t(t + 27))2] + [(−1,−t)2]. The first quaternion algebra issplit over K because (−3t)32 + t(t + 27) = t2. It follows that ∆̂(X0) =[(−1,−t)2] + [(−2, t)2] = [(−1,−1)2] + [(2, t)2]. The proof is complete.645.8. Polyhedral groupsWe now focus our attention on polyhedral actions on curves of genus 1.In this case, we only need to consider A4-actions, since S4 and A5 cannotact faithfully on curves of genus 1.Proposition 5.28. Let C be a curve of genus 1 endowed with a faithfulA4-action defined over a field k. Then the following properties hold.(a) The Jacobian E ∼= Pic0(C) has j-invariant 0.(b) The elliptic curve E can be endowed with a faithful A4-action definedover k.(c) The curve C is A4-equivariantly isomorphic to E over some extensionk′/k of odd degree.(d) We have the equality ∆̂(C) = [(−1,−1)2] in Br2(k(C)A4).Proof. We will extensively use the results and notation from [34, §X.3] (seealso [18]). Recall that C is a principal homogeneous space under E. A k-automorphism g : C → C induces a group automorphism of Pic0(C), hencealso a k-automorphism gˆ : E → E fixing the origin. Explicitly, it is not hardto see that gˆ(P ) = g(p0 +P )− g(p0), where the definition is independent ofp0 ∈ C(k). Note also that gˆ is the identity if and only if g is a translationby an element of E(k). This proves that we have an exact sequence1 // E(k) // Aut(C)(k) pi // Aut0(E)(k).Regard A4 as a subgroup of Aut(C)(k). It follows that E(k)∩A4 ∼= (Z/2Z)2and pi(A4) = Z/3Z ⊂ Aut0(E)(k). By Lemma 5.3(b), it follows that j(E) =0.We now proceed with the proof of part (b). Note that E(k) contains asubgroup isomorphic to (Z/2Z)2, whence the 2-torsion points of E are k-rational. Using Lemma 5.3(b), we conclude from part (a) that k contains aprimitive third root of unity ω3 and Aut0(E)(k) = Z/6Z. We now explicitlyconstruct the A4-action on E. Since E has j-invariant 0, it has a Weierstrassequation y2 = x3 + b for some b ∈ k×. Let the normal subgroup V =(Z/2Z)2 ⊂ A4 act on E via translation by 2-torsion points (as it does on Cas well). Then we can write A4 = V oH, and let H ∼= Z/3Z act on E byα · (x, y) = (ω3x, y), where α is a generator of H. For convenience, we fixthe above notation for the remainder of the proof.To prove part (c), we first show that C has a k′-rational point over someextension k′/k of odd degree. Fix an element g ∈ A4 ⊂ Aut(C)(k) of order655.8. Polyhedral groups3 and assume without loss of generality that gˆ = α. Note that g(q) =g(p) +α(q−p) for any p, q ∈ C(k). Taking q = pσ for any σ ∈ Gal(k/k) andusing the fact that g is defined over k, we obtain that g(p)σ−g(p) = α(pσ−p),i.e., (1−α)(pσ−p) = P σ−P for P = p−g(p) ∈ E(k). By [34, Thm. X.3.6],it follows that the class {C/k} ∈ H1(k,E) belongs to the kernel of the map(1− α)∗ : H1(k,E)→ H1(k,E) induced by 1− α ∈ End(E). However, notethat (2 + α) ◦ (1 − α) = 3, which implies that the class {C/k} is 3-torsion.It follows that there exists an extension k′/k such that [k′ : k] is a powerof 3 and C has a k′-rational point (see [18, Prop. 5] and the remark thatfollows).We claim that after possibly taking a cubic extension of k′, we can findan A4-equivariant isomorphism C → E. Fix a point p0 ∈ C(k′). We wouldlike to find P0 ∈ E(k) such that (1− α)(P0) = g(p0)− p0 ∈ E(k′). It is nothard to see that such a point P0 can be found over some cubic extensionof k′. (For example, this can be done by noting that the coordinates of P0satisfy cubic polynomials with coefficients in k′.) Without loss of generality,assume that P0 ∈ E(k′) and define q0 = p0 + P0 ∈ C(k′). Note thatg(q0) = g(p0) + α(P0) = p0 + P0 = q0. We claim that the k′-isomorphismϕ : C → E defined by q 7→ q−q0 is A4-equivariant. Since it clearly commuteswith translations, it suffices to show that ϕ(g(q)) = α(ϕ(q)). We computeϕ(g(q)) = g(q)− q0 = g(q)− g(q0) = α(ϕ(q)), which completes the proof ofthe claim.It remains to prove part (d). We reduce the problem to curves of genus1 with k-rational points. Assume the result is true in this case. Then wemust have ∆̂(E) = [(−1,−1)2] in Br2(k(E)A4), where E is the Jacobianof C. By part (c), we can find an odd degree extension k′/k such thatEk′ ∼= Ck′ as A4-varieties. Therefore, we must have ∆̂(Ck′) = [(−1,−1)2]in Br2(k′(C)A4). The natural map Br2(k(C)A4)→ Br2(k′(C)A4) is injectivesince [k′(C)A4 : k(C)A4 ] is odd, so it follows that ∆̂(C) = [(−1,−1)2] inBr2(k(C)A4). This implies that it suffices to prove the statement for E.We explicitly compute ∆̂(E) ∈ Br(k(E)A4). It is easy to check that therational map E 99K P1 given by(x, y) 7→ t =(y4 + 18by2 − 27b2)y3is an A4-invariant map of degree 12, so it coincides with the rational quotientmap E 99K E/A4. We may view the element [E] ∈ H1(k(t), A4) as theA4-Galois field extension k(E)/k(t). Therefore, its corresponding e´tale k(t)-algebra is (isomorphic to) the fixed field k(E)H = k(y) (recall that A4 =665.8. Polyhedral groupsV oH). Note that y is a root ofp(Y ) = Y 4 − tY 3 + 18bY 2 − 27b2,so it follows that k(y) = k(t)[Y ]/(p(Y )). A computation shows that the traceform of this e´tale algebra is isomorphic to 〈1, A,B,AB〉, whereA = 3t2−144band B = (192b− 3t2)(144b− 3t2). It follows that its Stiefel-Whitney class isequal to [(−A,−B)2] + [(−1,−1)2]. By Proposition 5.25, it suffices to showthat (−A,−B)2 is split over k(t). Note that we have an isomorphism(−A,−B)2 ∼= (144b− 3t2, 192b− 3t2)2.Recall that −3 is a square in k because k contains a primitive third root ofunity. Hence the identity(144b− 3t2)22 + (192b− 3t2)(√−3)2 = (√−3 t)2holds over k(t), which proves that the above quaternion algebra is split.5.8.3 Strong incompressibilityProposition 5.29. Let G be a polyhedral group. The following are equiva-lent:(i) There are no strongly incompressible G-curves defined over k.(ii) cd2(k) = 0.Proof of (ii) ⇒ (i). Suppose that cd2(k) = 0. By Lemma 5.20, it followsthat k satisfies the hypotheses of Lemma 5.30 below. In particular, thereexists an embedding ρ : G ↪→ PGL2 defined over k. We claim that anyfaithful G-curve X can be G-compressed to P1ρ . Indeed, the field K =k(X)G satisfies cd2(K) = 1 and therefore, Br2(K) = 1. Hence ∆ρ(X) = 1and the claim follows from Corollary 5.8. To finish the proof, we must showthat P1ρ is not strongly incompressible. This is achieved in Lemma 5.30.Lemma 5.30. Let G be a polyhedral group. Suppose that ω4 ∈ k if G = A4or S4 (resp. ω5 ∈ k if G = A5), and let ρ : G ↪→ PGL2 be an embeddingdefined over k (it is unique up to conjugacy). Then the G-variety P1ρ is notstrongly incompressible.675.8. Polyhedral groupsProof. As the group A4 is contained in S4, it suffices to find non-birationalcompressions for S4 and A5.Case 1: Suppose that G = S4. The matrices(ω4 00 1),(ω4 ω4−1 1),generate a subgroup isomorphic to S4 inside PGL2, whence we may assumethat ρ(G) is this particular subgroup. Then a computation shows that(x : y) 7→ (7x4y3 + y7 : −x7 − 7x3y4)is a G-compression P1ρ → P1ρ , which is clearly not birational.Case 2: Suppose that G = A5. Consider the matrices(ω5 00 1),(ω5 + ω−15 11 −ω5 − ω−15);they generate a subgroup isomorphic to A5 inside PGL2(k). Again, assumethat ρ(G) coincides with this subgroup. Then the morphism(x : y) 7→ (x11 + 66x6y5 − 11xy10 : −11x10y − 66x5y6 + y11)is a non-birational G-compression P1ρ → P1ρ .It remains to prove (i) ⇒ (ii) in Proposition 5.29. The following lemmawill be useful in the sequel.Lemma 5.31. Let k be a field, let (a, b)2 be a quaternion algebra definedover k and let n ≥ 4 be an integer. Then we have the following properties.(a) There exists an n-dimensional e´tale k-algebra E1 with trivial discrimi-nant such that the Stiefel-Whitney class w2(qE1) = [(a, b)2]+[(−1,−1)2],(b) If (a, b)2 6∼= (−1,−1)2, there exists an n-dimensional e´tale k-algebraE2 with nontrivial discriminant dE2 such that w2(qE2) = [(a, b)2] +[(−1,−dE2)2].Proof. It suffices to prove the results for n = 4, as adding copies of the triviale´tale algebra k to E does not change the discriminant of E, or w2(qE). By[13, Lemma 31.19], the k-algebra E[A,B] = k[X]/(X4− 2AX2 +B) is e´talewhen AB(A2 −B) 6= 0, has discriminant 64B(A2 −B)2, and its trace formis isomorphic to 〈1, A,A2 −B,AB(A2 −B)〉. A computation shows thatw2(qE[A,B]) = [(−A,−B(A2 −B))2] + [(−1,−B)2].685.8. Polyhedral groupsTo prove part (a), select c ∈ k× such that b2c4 − 1 6= 0, and put A =−a(bc2−1)2 and B = a2(b2c4−1)2. It is easy to see that −A ≡ a mod k×2and −B(A2−B) ≡ b mod k×2, whence E1 = E[A,B] satisfies the requiredproperties.To prove part (b), we may assume without loss of generality that −b 6∈k×2 and b 6= 1, by changing the presentation of (a, b)2 if necessary. DefineA = −a and B = −4ba2/(b − 1)2; then we obtain that A2 − B ∈ k×2. Thealgebra E2 = E[A,B] has discriminant −b 6∈ k×2 and satisfies w2(qE2) =[(a, b)2] + [(−1, b)2].Remark 5.32. The conclusion in part (b) of the above theorem mightfail if (a, b)2 ∼= (−1,−1)2. Indeed, suppose that k = R. By [13, Thm.31.18], we observe that the trace form of any 4-dimensional e´tale algebra Ehas the form qE = 〈1, A,A2 − B,AB(A2 − B)〉, which has second Stiefel-Whitney invariant w2(qE) = [(−A,−B(A2 − B))2] + [(−1,−B)2]. Since wewant the discriminant to be nontrivial, B must be negative, so w2(qE) =[(−A,A2 − B)2]. This class is obviously trivial because A2 − B > 0, so wecannot obtain [(−1,−1)2].Proof of (i) ⇒ (ii) in Proposition 5.29. Suppose that cd2(k) > 0 and letK = k(x). Note in particular that the field K is Hilbertian (see [12, Prop.13.2.1]).Case 1: Suppose that G = An, where n = 4 or 5. By Lemma 5.20,there exists a nonsplit quaternion algebra A defined over K. Using Lemma5.31(a), we can construct an n-dimensional e´tale K-algebra E with trivialdiscriminant such that w2(qE) = [A] + [(−1,−1)2]. By [10, Thm. 1], thereexists a field extension L/K of degree n whose trace form is isometric toqE ; moreover, we may assume that its Galois closure L˜/K has Galois groupG. Therefore, the class of L (viewed as an e´tale K-algebra) in H1(K,G)corresponds to a faithful G-curve X defined over k with function field L˜. ByProposition 5.25, we must have ∆̂(X) = [A] + [(−1,−1)2].We claim thatX cannot beG-compressed to any curve of genus≤ 1. Anyfaithful G-curve of genus 0 is G-equivariantly isomorphic to X0. Supposethat there exists a G-compression X → X0. By Proposition 5.24, the imageof ∆̂(X0) in Br2(K) under the induced map is equal to ∆̂(X) = [A] +[(−1,−1)2]. By Proposition 5.27(a), it follows that [A] is trivial, which is acontradiction.If G = A5, the claim follows because A5 does not act on any curve ofgenus 1. On the other hand, suppose that there exists a A4-compressionX → C, where C has genus 1. (A word of warning: Here we cannot assure695.8. Polyhedral groupsthat C is an elliptic curve because it might not have k-rational points.) Asbefore, it follows that ∆̂(C) maps to ∆̂(X) ∈ Br2(K) under the map inducedby the compression. However, Lemma 5.28(d) contradicts the fact that A isnot split. This completes the proof of the claim. By Lemma 5.1, there existstrongly incompressible G-curves.Case 2: Suppose that G = S4. We claim that there exists a quaternionalgebra A 6∼= (−1,−1)2 over K which does not split over k′(x), where k′ =k(√2). If 2 is a square in k and (−1,−1)2 is split over K, the result followsimmediately from Lemma 5.20. If 2 is a square but (−1,−1)2 is not split overK, we choose A = (−1, x)2. Note that A ∼= (−1,−1)2 over K if and onlyif (−1,−x)2 is split. By Lemma 2.1, if either A is split or A ∼= (−1,−1)2,it would follow that −1 is a square in k, which contradicts our assumptionthat (−1,−1)2 is not split.Finally, if 2 is not a square over k, we choose A = (x, x2−4x+ 2)2. Sup-pose for the sake of contradiction that A splits over k′(x). By Lemma 2.1,2+√2 is a square over k′, i.e., 2+√2 = (l1 + l2√2)2 for some l1, l2 ∈ k. Tak-ing norms with respect to k′/k yields 2 = (l21 − 2l22)2, which contradicts ourassumption. We now prove that A 6∼= (−1,−1)2, where we may assume that(−1,−1)2 is not split. Indeed, such an isomorphism would imply that thequadratic forms 〈1, 1, 1〉 and 〈−x,−(x2−4x+2), x(x2−4x+2)〉 are isomor-phic over K. It follows that 〈1, 1, 1〉 represents −x, i.e., there exist coprimepolynomials p, q, r, s ∈ k[x] such that p(x)2 +q(x)2 +r(x)2 = −xs(x)2. Mak-ing x = 0 yields p(0) = q(0) = r(0) = 0 since we are assuming that 〈1, 1, 1〉is anisotropic over k. This implies that p(x), q(x), r(x) are divisible by x,whence s(x) is divisible by x as well. This contradicts the fact that p, q, r, sare coprime.By Lemma 5.31(b), we can construct a 4-dimensional e´tale K-algebraE with nontrivial discriminant dE such that w2(qE) = [A] + [(−1,−dE)].By [10, Thm. 1], we can find a field extension L/K of degree 4 whosetrace form is isometric to qE , whose Galois closure L˜/K has Galois groupG. As before, its class in H1(K,G) corresponds to a faithful G-curve Xdefined over k with function field L˜. By Proposition 5.25, it follows that∆̂(X) = [A] + [(−1,−1)2] + [(2, dE)2].As in Case 1, suppose that there is a G-compression f : X → X0 andlet f ′ : X ′ → X ′0 be the base extension of f to k′ = k(√2). There exists a705.8. Polyhedral groupscommutative diagramBr2(k(X0)G)i∗ //j0Br2(K)jBr2(k′(X ′0)G)i′∗ // Br2(k′(x))where the vertical arrows are induced by base extension and the horizon-tal arrows are induced by f and f ′. By Proposition 5.24, we must havei∗(∆̂(X0)) = ∆̂(X) in Br2(K). By Proposition 5.27(b), it follows thatj0(∆̂(X0)) = [(−1,−1)2], since 2 is a square in k′. Consequently, we con-clude that[(−1,−1)2] = i′∗(j0(∆̂(X0))) = j(i∗(∆̂(X0))) = j(∆̂(X)) = [A]+[(−1,−1)2],whence A must be split over k′(x). This contradicts our initial assumption.Since G does not act faithfully on any curve of genus 1, it follows fromLemma 5.1 that there exist strongly incompressible G-curves.71Chapter 6Conclusions and openproblems6.1 Finite group actions on conicsLet q be a nondegenerate quadratic form of rank 3 over k. Recall that qdefines a (smooth projective) k-curve Xq ⊂ P2 of genus 0. Such a curve isusually referred to as a conic. It is well known that any smooth projectivecurve of genus 0 is isomorphic to some conic Xq, where q is unique up toscalar multiplication. Note that Xq ∼= P1 if and only if q is isotropic over k.Recall that the automorphism group of Xq is isomorphic to the group ofprojective similitudes PGO(q) (see [8, Cor. 69.6]). Since rank(q) = 3 is odd,there exists an isomorphism of algebraic groups PGO(q) ∼= SO(q) (cf. [16,Prop. 12.4 and 12.6]), which is an absolutely simple adjoint group of typeB1. The accidental isomorphism of Dynkin diagrams A1 = B1 implies thatSO(q) ∼= PGL1(Aq),where Aq is a quaternion algebra with Severi-Brauer conic Xq.In the case where q is isotropic, we have that SO(q) ∼= PGL2. Theclassification, up to conjugacy, of finite subgroups of PGL2 (or equivalently,finite group actions on the projective line) over an arbitrary field can befound in [2, 11]. In Chapter 3 of this thesis, we classify the finite subgroupsof SO(q) up to conjugacy, for every nondegenerate ternary quadratic formq and every base field of characteristic different from 2.To complete the classification of finite subgroups of absolutely simpleadjoint group of type B1 = A1, it remains to consider the non-split case overa base field of characteristic 2. To the author’s best knowledge, this caseremains wide open. To fix ideas, let q be an anisotropic ternary quadraticform over a base field k of characteristic 2. After rescaling, we may assumethat q ' 〈1〉 ⊥ [a, b] for some a, b ∈ k, where [a, b] stands for the binaryquadratic form ax2 + xy + by2 (cf. [8, Cor. 7.32]).726.2. Geometric Galois embedding problemsQuestion 6.1. What are the finite subgroups of SO(q)? What can we sayabout their conjugacy classes?We do not even know whether SO(q) can contain a 2-irregular subgroup(i.e., a subgroup of even order) when q is anisotropic. We showed in Theorem1.1 that this assertion fails if 2 is replaced by an odd prime p, but the proofrelies on Lemma 3.1(b), which breaks down in characteristic 2.6.2 Geometric Galois embedding problemsIn this section, all groups are assumed to be finite. Let M/K be Galoisextension with Galois group G, and let pi : E  G be a group epimorphism.The Galois embedding problem given by M/K and pi asks for the existenceof an E-Galois extension F/K, such that M ⊂ F and pi coincides with thenatural surjection E ∼= Gal(F/K) Gal(M/K) ∼= G (cf. [19, §2.2]).Let X be a geometrically irreducible faithful G-variety defined over abase field k. It is well known that k(X)/k(X)G is a G-Galois extension.Given this extension and an epimorphism pi : E  G, we can then definea Galois embedding problem, which we refer to as the geometric Galoisembedding problem given by X and pi.How can we interpret a solution to the above problem geometrically?Suppose that F/k(X)G is such a solution. Then, there exists a faithfulE-variety Y/k such that k(Y ) ∼= F as k-fields with an E-action. (Suchan E-variety is commonly referred to as a model for F .) In particular, thisimplies that Y/Ker(pi) and X are birationally isomorphic as G-varieties. Wehave thus proven that the geometric Galois embedding problem given by Xand pi is equivalent to the following lifting problem.Question 6.2. Determine necessary and sufficient conditions on X/k andpi : E  G for the existence of a faithful E-variety Y/k such that Y/Ker(pi)is G-equivariantly birationally isomorphic to X.In Chapter 4, we completely answered the above question in the casewhere X = P1 over an arbitrary field of characteristic 0 and pi is any dou-ble cover of a Kleinian group G acting on the projective line. That case isparticularly interesting because the typical solutions to the lifting problemare hyperelliptic curves endowed with a group action. To the author’s bestknowledge, the analogous problem is still open over a base field of positivecharacteristic. Even over an algebraically closed field k, the literature avail-able on the complete classification of automorphism groups of hyperellipticcurves often assumes that char(k) = 0 (cf. [5, 33]).736.2. Geometric Galois embedding problemsWe illustrate an interesting connection between Galois cohomology andthe geometric Galois embedding problem for X and pi considered above.Suppose that N := Ker(pi) is abelian and consider the exact sequence1 // N // E pi // G // 1.This induces to an exact sequence in cohomologyH1(K,E) // H1(K,G) ∆ // H2(K,N),where K := k(X)G. Recall that X gives rise to a cohomology class [X] ∈H1(K,G) (see Section 5.1). We claim that the solvability of the liftingproblem is closely related to the vanishing of the class ∆([X]) ∈ H2(K,N).Indeed, if Y is a solution to the lifting problem, then [X] is the image of [Y ]in the above exact sequence, which implies that ∆([X]) = 0. Conversely, if∆([X]) = 0, it follows that [X] arises from a class in c ∈ H1(K,E). How-ever, a variety Y representing c is not necessarily irreducible. (In general,Y is a primitive E-variety, i.e., E permutes the irreducible components ofY .) In the usual language of Galois embedding problems, such a Y wouldcorrespond to a weak solution to the problem (cf. [19, §2.4]). Under certainmild hypotheses though, one can guarantee that a weak solution is a propersolution as well, e.g., when N ∼= µp for some prime p 6= char(k), and kcontains all p-th roots of unity (see [19, Thm. 2.4.1 and Cor. 2.4.2]).The above interpretation can be used to solve the lifting problem incertain cases. To fix ideas, let us assume that N is cyclic, i.e., N = µm.Then, we can interpret H2(K,µm) as the m-th torsion part of the Brauergroup Br(K), and therefore ∆([X]) represents the Brauer class of somecentral simple algebra of period dividing m.In this thesis, we compute such Brauer class in certain instances. Forexample, suppose that we can find a diagram with exact rows1 // Gm // GL(V ) // PGL(V ) // 11 // N //?OOE?ρOOpi // G?ρOO// 1Then Lemma 5.6 implies the lifting problem for P(V )ρ and pi is (weakly)solvable. (Recall that P(V )ρ is simply the projective space P(V ) endowedwith a G-action via ρ.) In the case where N = µ2 and dim(V ) = 2, werecover some of the results in Chapter 4. Unfortunately, the technique above746.3. Strongly incompressible varietiesdoes not work for all exact sequences 1→ N → E → G→ 1, as they mightnot fit into the required commutative diagram (see, e.g., Remark 5.26).A more systematic method to compute the class ∆([X]) based on geo-metric properties of the G-variety X would be very desirable.6.3 Strongly incompressible varietiesLet G be an algebraic group. Recall that a G-compression of a genericallyfree G-variety X is a dominant G-equivariant rational map onto anothergenerically free G-variety. We say that X is strongly incompressible if everyG-compression of X is birational.It is natural to pose the following question.Question 6.3. Let k be a base field, let G/k be an algebraic group and letn be a positive integer. Does there exist a strongly incompressible G-varietyof dimension n defined over k? If so, can one construct explicit examples ofsuch G-varieties?In dimension 1, Z. Reichstein ([24]) showed the existence of stronglyincompressible G-curves for finite groups G that cannot act faithfully oncurves of genus ≤ 1. In Chapter 5 of this thesis, we answer the existence partof the above question for every field of characteristic 0 and every finite groupG. Roughly, the classification goes as follows: There always exist stronglyincompressible G-curves if G cannot act faithfully on a curve of genus 0.If, on the contrary, G has a faithful action on a curve of genus 0, then theexistence of strongly incompressible G-curves depends on the arithmetic ofthe field k. In particular, after a large enough algebraic extension of k, onecan prove that all G-curves are compressible.In higher dimensions, the situation is largely unexplored. However, someprogress has been made in dimension 2. N. Fakhruddin proved in an un-published note that if G is a finite group that does not act faithfully on arational surface or a surface of Kodaira dimension 0 or 1 with finite funda-mental group, then there exist a strongly incompressible complex G-surface.In particular, his results imply that all but finitely many non-abelian sim-ple groups have strongly incompressible actions on complex surfaces. Also,R. Pardini has shown that any group containing (Z/2Z)6 or (Z/pZ)5 (p anodd prime) acts strongly incompressibly on some complex surface. Perhapsmore importantly, she produces an explicit example of a strongly incom-pressible complex surface: if C is a Hurwitz curve of genus g ≥ 2, then(Aut(C))2 o (Z/2Z) acts strongly incompressibly on C × C.756.3. Strongly incompressible varietiesIn most cases, the problem of constructing examples of strongly incom-pressible G-varieties remains open, even in dimension 1. Our argument inLemma 5.1 is essentially non-constructive and though in Chapter 5 we giveseveral examples of G-curves that compress to some strongly incompressiblecurve, we do not know if they are strongly incompressible themselves.76Bibliography[1] Matthew H. Baker, Enrique Gonza´lez-Jime´nez, Josep Gonza´lez, andBjorn Poonen. Finiteness results for modular curves of genus at least2. Amer. J. Math., 127(6):1325–1387, 2005.[2] Arnaud Beauville. Finite subgroups of PGL2(K). In Vector bundlesand complex geometry, volume 522 of Contemp. Math., pages 23–29.Amer. Math. Soc., Providence, RI, 2010.[3] Gre´gory Berhuy and Giordano Favi. Essential dimension: a functorialpoint of view (after A. Merkurjev). Doc. Math., 8:279–330 (electronic),2003.[4] Armand Borel. Linear algebraic groups, volume 126 of Graduate Textsin Mathematics. Springer-Verlag, New York, second edition, 1991.[5] Rolf Brandt and Henning Stichtenoth. Die Automorphismengruppenhyperelliptischer Kurven. Manuscripta Math., 55(1):83–92, 1986.[6] Xi Chen. Self rational maps of k3 surfaces. arXiv:1008.1619 [math.AG],2010.[7] Xi Chen. Rational self maps of Calabi-Yau manifolds. In A celebrationof algebraic geometry, volume 18 of Clay Math. Proc., pages 171–184.Amer. Math. Soc., Providence, RI, 2013.[8] Richard Elman, Nikita Karpenko, and Alexander Merkurjev. The alge-braic and geometric theory of quadratic forms, volume 56 of AmericanMathematical Society Colloquium Publications. American Mathemati-cal Society, Providence, RI, 2008.[9] Richard Elman and Adrian R. Wadsworth. Hereditarily quadraticallyclosed fields. J. Algebra, 111(2):475–482, 1987.[10] Martin Epkenhans and Martin Kru¨skemper. On trace forms of e´talealgebras and field extensions. Math. Z., 217(3):421–434, 1994.77Bibliography[11] Xander Faber. Finite p-irregular subgroups of PGL2(k).arXiv:1112.1999 [math.NT], 2011.[12] Michael D. Fried and Moshe Jarden. Field arithmetic, volume 11 ofErgebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. A Seriesof Modern Surveys in Mathematics [Results in Mathematics and Re-lated Areas. 3rd Series. A Series of Modern Surveys in Mathematics].Springer-Verlag, Berlin, third edition, 2008. Revised by Jarden.[13] Skip Garibaldi, Alexander Merkurjev, and Jean-Pierre Serre. Cohomo-logical invariants in Galois cohomology, volume 28 of University LectureSeries. American Mathematical Society, Providence, RI, 2003.[14] Nikita A. Karpenko and Alexander S. Merkurjev. Essential dimensionof finite p-groups. Invent. Math., 172(3):491–508, 2008.[15] Felix Klein. Lectures on the icosahedron and the solution of equationsof the fifth degree. Dover Publications, Inc., New York, N.Y., revisededition, 1956. Translated into English by George Gavin Morrice.[16] Max-Albert Knus, Alexander Merkurjev, Markus Rost, and Jean-PierreTignol. The book of involutions, volume 44 of American Mathemati-cal Society Colloquium Publications. American Mathematical Society,Providence, RI, 1998. With a preface in French by J. Tits.[17] Tsit Y. Lam. Introduction to quadratic forms over fields, volume 67of Graduate Studies in Mathematics. American Mathematical Society,Providence, RI, 2005.[18] Serge Lang and John Tate. Principal homogeneous spaces over abelianvarieties. Amer. J. Math., 80:659–684, 1958.[19] Arne Ledet. Brauer type embedding problems, volume 21 of Fields In-stitute Monographs. American Mathematical Society, Providence, RI,2005.[20] Arne Ledet. Finite groups of essential dimension one. J. Algebra,311(1):31–37, 2007.[21] Qing Liu. Algebraic geometry and arithmetic curves, volume 6 of Ox-ford Graduate Texts in Mathematics. Oxford University Press, Oxford,2002. Translated from the French by Reinie Erne´, Oxford Science Pub-lications.78Bibliography[22] Jean-Franc¸ois Mestre. Construction de courbes de genre 2 a` partir deleurs modules. In Effective methods in algebraic geometry (Castiglion-cello, 1990), volume 94 of Progr. Math., pages 313–334. Birkha¨userBoston, Boston, MA, 1991.[23] Zinovy Reichstein. On the notion of essential dimension for algebraicgroups. Transform. Groups, 5(3):265–304, 2000.[24] Zinovy Reichstein. Compressions of group actions. In Invariant theoryin all characteristics, volume 35 of CRM Proc. Lecture Notes, pages199–202. Amer. Math. Soc., Providence, RI, 2004.[25] Zinovy Reichstein. Essential dimension. In Proceedings of the Interna-tional Congress of Mathematicians. Volume II, pages 162–188. Hindus-tan Book Agency, New Delhi, 2010.[26] Zinovy Reichstein and Angelo Vistoli. Birational isomorphisms betweentwisted group actions. J. Lie Theory, 16(4):791–802, 2006.[27] Zinovy Reichstein and Boris Youssin. Splitting fields of G-varieties.Pacific J. Math., 200(1):207–249, 2001.[28] Jean-Pierre Serre. Proprie´te´s galoisiennes des points d’ordre fini descourbes elliptiques. Invent. Math., 15(4):259–331, 1972.[29] Jean-Pierre Serre. Extensions icosae´driques. In Seminar on NumberTheory, 1979–1980 (French), pages Exp. No. 19, 7. Univ. Bordeaux I,Talence, 1980.[30] Jean-Pierre Serre. L’invariant de Witt de la forme Tr(x2). Comment.Math. Helv., 59(4):651–676, 1984.[31] Jean-Pierre Serre. Galois cohomology. Springer Monographs in Mathe-matics. Springer-Verlag, Berlin, english edition, 2002. Translated fromthe French by Patrick Ion and revised by the author.[32] Jean-Pierre Serre. Topics in Galois theory, volume 1 of Research Notesin Mathematics. A K Peters, Ltd., Wellesley, MA, second edition, 2008.With notes by Henri Darmon.[33] Tanush Shaska. Determining the automorphism group of a hyperel-liptic curve. In Proceedings of the 2003 International Symposium onSymbolic and Algebraic Computation, pages 248–254 (electronic). ACM,New York, 2003.79[34] Joseph H. Silverman. The arithmetic of elliptic curves, volume 106 ofGraduate Texts in Mathematics. Springer, Dordrecht, second edition,2009.[35] Nu´ria Vila. On stem extensions of Sn as Galois group over numberfields. J. Algebra, 116(1):251–260, 1988.80Appendix AProof of Theorem 5.2Lemma A.1. Let P,Q be two polynomials in k[x], not both zero, and letA ⊂ k be the set of their common roots. Then for all but finitely many c ∈ k,the polynomial P + cQ has no multiple roots outside of A.Proof. It suffices to show that given two coprime polynomials P,Q ∈ k[x],the polynomial P + cQ has simple roots for all but finitely many c ∈ k. Ifboth polynomials are constant, the result is immediate, so we may assumethat is not the case. Note that ξ ∈ k is a multiple root of P +cQ if and onlyif P (ξ) + cQ(ξ) = P ′(ξ) + cQ′(ξ) = 0, which implies that P (ξ)Q′(ξ) −P ′(ξ)Q(ξ) = 0. The polynomial PQ′ − P ′Q cannot be identically zerobecause P and Q are coprime and not both constant, so it has finitelymany roots. If we take c ∈ k outside the finite set{−P (ξ)/Q(ξ) | ξ ∈ k satisfies P (ξ)Q′(ξ)− P ′(ξ)Q(ξ) = 0, Q(ξ) 6= 0},it follows that P + cQ has simple roots. The proof is complete.Definition A.2. We define a ramification condition to be an l-tuple ofintegers P = (b1, . . . , bl), where l ≥ 1 and bi ≥ 2 for all i. We say thatP ∈ k[x] has a local decomposition of type P at β ∈ k, if there exists afactorizationP (x)− β = a(x− α1)b1 . . . (x− αl)bl(x− αl+1) . . . (x− αr),where a is the leading coefficient of P , and α1, . . . , αr are distinct elementsin k.Proposition A.3. Let Pi = (bi,1, . . . , bi,li) (1 ≤ i ≤ n) be a collectionof ramification conditions (not necessarily distinct), and let β1, . . . , βn bedistinct points in k. Then there exists a polynomial P ∈ k[x] that satisfieslocal decompositions of type Pi at βi for 1 ≤ i ≤ n. Moreover, we can choosedeg(P ) to be any sufficiently large positive integer.81Appendix A. Proof of Theorem 5.2Proof. Choose distinct points aij ∈ k for 1 ≤ i ≤ n, 1 ≤ j ≤ li. By theChinese Remainder Theorem, there exists Q ∈ k[x] such thatQ(x) ≡ βi + (x− aij)bij mod (x− aij)bij+1,for 1 ≤ i ≤ n, 1 ≤ j ≤ li. We defineH(x) =∏i,j(x− aij)bij+1and we let A = {aij}i,j be the set of its roots. Applying Lemma A.1 togi = Q − βi and H for 1 ≤ i ≤ n, we conclude that there exists a finite setS ⊂ k such that if c ∈ k lies outside S, the polynomials gi + cH containno multiple roots outside of A for 1 ≤ i ≤ n. Choose any such c and defineP = Q+ cH. We claim that P satisfies the desired conditions. Indeed, notethat the following properties hold.(i) For 1 ≤ i ≤ n, 1 ≤ j ≤ li, the polynomial P − βi has a root ofmultiplicity bij at the point aij .(ii) If i′ 6= i, we have P (ai′j) = βi′ 6= βi and therefore P − βi cannot haveany root of the form ai′j .(iii) By construction, P − βi does not have multiple roots outside of A.It remains to prove that we can take deg(P ) to be any sufficiently largepositive integer d. To show this, take n = max(deg(Q),deg(H)). We claimthat there exists P satisfying the desired properties such that deg(P ) = d forany d > n. Indeed, if we replace H(x) by (x− a11)d−degHH(x) and ensurethat c 6= 0 in the definition of P , it follows easily that deg(P ) = d.Proof of Theorem 5.2. Without loss of generality, we may assume that G =Sm for some m ≥ 2. Given a partition b1 + . . .+ bs of m, where b1 ≥ . . . ≥bl > 1 = bl+1 = . . . = bs for some l ≥ 1, we can define a ramificationcondition P = (b1, . . . , bl). Let P1, . . . ,Pn be the ramification conditionsobtained as we range over all possible partitions of m, except for 1 + . . .+ 1.By Proposition A.3, we can construct a polynomial P ∈ k[x] satisfying localdecompositions of type Pi at distinct points βi for 1 ≤ i ≤ n. Moreover, wemay assume that deg(P ) is some sufficiently large prime number p.Let the group Sp act on p letters and embed Sm inside Sp as the subgroupthat fixes the last p−m letters. We want to construct X as a ramified Sp-cover of P1. Let Pt(x) = P (x) − t, where t is an indeterminate, and defineL as the splitting field of Pt over k(t). It is clear that Gal(L/k(t)) is a82Appendix A. Proof of Theorem 5.2transitive subgroup of Sp; we claim that equality holds. Since Gal(Lk/k(t))is a subgroup of Gal(L/k(t)), it suffices to prove that the former is isomorphicto Sp. (Note that this also implies that L is regular, i.e., L∩k = k.) We usea technique similar to [32, Thm. 4.4.5]. We may view the polynomial Pt asa ramified cover P1 → P1 of degree p. Note that β1, . . . , βn,∞ are amongthe ramification points. If Pi = (b(i)1 , . . . , b(i)li), the inertia subgroup at βiis generated by an element of Sp of cycle type (b(i)1 , . . . , b(i)li, 1, . . . , 1), whilethe inertia group at∞ is a p-cycle. In particular, Gal(Lk/k(t)) contains thesubgroup generated by a p-cycle and a transposition, which is all of Sp sincep is prime. The claim follows immediately.Let X be the (unique) smooth projective curve defined over k with func-tion field L, which is geometrically irreducible since L/k is regular. Note thatX can be endowed with a natural faithful Sp-action via the Galois action onL. If Q is a closed point in Xk lying above βi, then its stabilizer is a cyclicsubgroup generated by an element of Sp of cycle type (b(i)1 , . . . , b(i)li, 1, . . . , 1).Since any two subgroups of this form are conjugate, they all occur as stabi-lizers of points in the fibre above βi. Clearly, any nontrivial element of Smhas one of the above cycle types inside Sp, so it fixes at least one geometricpoint in X. The proof is complete.83


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