Some effectivity questions for plane Cremona transformations
aa r X i v : . [ m a t h . AG ] A p r SOME EFFECTIVITY QUESTIONS FOR PLANE CREMONATRANSFORMATIONS IN THE CONTEXT OFSYMMETRIC-KEY CRYPTOGRAPHY
N. I. SHEPHERD-BARRON
The 2–dimensional Cremona group Cr ( k ) is the group of k -birational automor-phisms of the projective plane P k over a field k . As such, it is an object of algebraicgeometry, but it is also of interest from the viewpoints of dynamics [M1] and grouptheory, including geometric group theory [CL]. This latter paper introduces anduses a certain infinite-dimensional hyperbolic space H = H k = H ( P k ) on which Cr ( k ) acts as a group of isometries and makes it clear that this action is animportant tool for studying both Cr ( k ) and its individual elements. However,given an algebraic description of a Cremona transformation g in terms of explicitrational functions, it is not always clear how to calculate anything about g thatis relevant to any of these frameworks. For example, there is no known effectiveprocedure for determining the translation length L ( g ) = L ( g ∗ ) of the isometry g ∗ of H that is associated to g .One of our goals here is to make matters more nearly effective in the veryparticular case where g is a special quadratic transformation . That is, g = σα ,where σ is the standard quadratic transformation given in terms of homogeneousco-ordinates x, y, z by σ : ( x, y, z ) ( yz, xz, xy ) and α is a linear involution. Forthis special class of Cremona transformations we give (Theorem 1.1 2 below) asimple and explicit condition, in terms of a constant number of field operations(addition and multiplication in the ground field) to ensure that g is hyperbolic:it is enough to take the three vertices P, Q, R of the triangle ∆ given by xyz = 0,compute the 12 points g i ( P ) etc. for 1 ≤ i ≤ L ( g ) when α is any linear transformation that relieson an unbounded quantity of information about the map g . The upper bound L ( h ) ≤ log D for a Cremona transformation h of degree D is well known [CL];these bounds together give an arbitrarily fine estimate for L ( g ). These bounds willalso determine the type of g ; that is, whether g is elliptic, parabolic or hyperbolic(the word loxodromic is also used for this last class). This is motivated by thefact, explained in Section 2, that certain contemporary encryption algorithms areCremona transformations and that estimates such as those proved here can beseen as a speedy check that the key being used is not obviously weak.Say that points x , x , ..., x n in P are in general position if they are alldistinct and none of them, except for those that happen to equal one of P, Q, R , N. I. SHEPHERD-BARRON lies on ∆. For any natural number n , let w n denote the reduced word in σ, α that has length n and begins with α when reading from right to left. So, forexample, w = 1 , w = α, w = σα . Set P n = w n ( P ), Q n = w n ( Q ) and R n = w n ( R ). So P = P , etc. We say that α , or g , is in ( p, q, r ) -general po-sition if P , ..., P p − , Q , ..., Q q − , R , ..., R r − are in general position. We abbre-viate ( r, r, r )-general position to r -general position. Of course, r -general positionimplies s -general position for any s ≤ r . Theorem 1.1 (1) (= Theorem 4.21) Suppose that g is in ( p, q, r ) -general po-sition and that /p + 1 /q + 1 /r < . Then g is hyperbolic. If also p ≤ q ≤ r then L ( g ) ≥ log(2 − . − p/ ) .(2) (= Theorem 4.22) Assume that < ǫ < / . Then, if p ≥ /ǫ andthe p points P , ..., P p − , ..., R p − are in general position, g is hyperbolic and log 2 − ǫ < L ( g ) ≤ log 2 . We can also give some analogous sufficient conditions that permit the pre-cise determination of the type of a special quadratic transformation g and itslength. Here is an example.Assume that g is in ( p, q, r )-general position and that P p = P p − , Q q = Q q − and R r = R r − . Let λ Lehmer denote
Lehmer’s number , the smallest knownalgebraic integer λ such that | λ | > λ ′ of λ has | λ ′ | ≤ | λ ′ | = 1 for at least one λ ′ . (These are the Salem numbers .) Theorem 1.2 (= Theorem 4.26) (1) The transformation g is biregular on theblow-up of P at these p + q + r points and is a Coxeter element in a Weyl groupof type T p,q,r .(2) If /p +1 /q +1 /r > then g is elliptic and its order is the correspondingCoxeter number. In particular, if p = 2 , q = 3 and r = 5 then g has order .(3) If /p + 1 /q + 1 /r = 1 then g is parabolic.(4) If /p + 1 /q + 1 /r < then g is hyperbolic and L ( g ) is the logarithmof a Salem number of norm . In particular, if p = 2 , q = 3 and r = 7 then L ( g ) = log λ Lehmer . These results, and their proofs, can be summarized by saying that there isa Coxeter–Dynkin diagram associated to the problem and if P , ..., R r − are ingeneral position then the diagram contains the standard tree T p,q,r . Since T , , = E the number 30 appears as the Coxeter number of E . This is a particularinstantiation of the very old idea of relating groups of Cremona transformations toCoxeter–Dynkin diagrams and Weyl groups, although usually the diagrams thathave arisen are of type T , ,r , and also of Steinberg’s idea [S] of describing Coxeterelements as a product of two involutions via a description of the Coxeter diagramas a bipartite graph. This viewpoint has also been exploited by McMullen [M1],whose calculations have inspired some of those that appear here, and Blanc andCantat [BC], who use an infinite group W ∞ that is something like a Coxetergroup of type E ∞ to prove that, for any hyperbolic Cremona transformation h , FFECTIVITY QUESTIONS IN THE CREMONA GROUP spectral radius or dynamical degree λ ( h ) = exp L ( h ) lies in the closure of theset T of Salem numbers and λ ( h ) ≥ λ Lehmer .The other main point of the paper is to extend the results of [CL] thatconcern the rigidity and tightness of elements, or conjugacy classes, in the Cre-mona group. (The definitions of these properties are recalled later, at the start ofSection 5.) Some of these results also complement a recent paper [L] by Lonjou,who exhibits, over any field k , an explicit Cremona transformation some power ofwhich generates a proper normal subgroup of Cr ( k ). She also points out an errorin an earlier version of this paper; the mistake lay in overlooking the possibilitythat a rigid element of Cr ( k ) might normalize a 2-dimensional additive subgroupof Cr ( k ). In consequence, the results of Section 5 below in characteristic p thatrefer to normal subgroups of Cr ( k ) require the assumption that k be algebraic(that is, algebraic over its prime subfield).Fix a plane Cremona transformation g over a field k . Theorem 1.3
Assume that g is hyperbolic.(1) (= Theorem 5.8) g is rigid.(2) (= Theorem 5.13) Suppose that L ( g ) is not the logarithm of a quadraticunit; if char k = p > assume also that k is algebraic and that L ( g ) is not anintegral multiple of log p . Then some power of g is tight. If also n is sufficientlydivisible, then the normal closure hh g n ii does not contain g , so is a non-trivialnormal subgroup of Cr ( k ) . In [CL] it was shown that if g is a very general Cremona transformation ofany degree ≥
2, then some power of g is tight and that g / ∈ hh g n ii for sufficientlydivisible n . (If g is very general then L ( g ) = log deg( g ).)In particular, this applies to special quadratic transformations g if k isalgebraic and char k = 2 (if char k = 2 we must assume also that L ( g ) = log 2),since it is an easy consequence of the other results that we prove about them thatthey satisfy the hypotheses of Theorem 1.3 2.In particular, these hypotheses can be realized over a finite field, since theyimpose three conditions on the 4-dimensional variety of involutions in P GL . Infact, over a finite field more is true. Theorem 1.4 (= Theorems 5.17 and 5.18) Suppose that k is a finite field andthat g is a hyperbolic element of Cr ( k ) . Then g is tight and g
6∈ hh g N ii for allsufficiently divisible N . H´enon introduced certain complex quadratic plane Cremona transformations,now called H´enon maps, as models of (sections of) dynamical systems such as
N. I. SHEPHERD-BARRON the Lorenz equations. They are of the form f ( x, y ) = ( ay + q ( x ) , x ) , where q is a quadratic polynomial and a is a non-zero scalar.Then f might have sensitive dependence on initial conditions in this sense:even if initial points x and y are very close, their images f n ( x ) and f n ( y ) canbe far apart for large values of n .In the context where f is a smooth self-map of a compact manifold X this,when stated precisely in terms of Lyapunov exponents, turns out to be equivalentto the topological entropy h ( f ) being strictly positive. (Recall that the topologicalentropy h ( g ) of a self-map g of a compact metric space X is defined by h ( g ) = lim ǫ → lim sup n →∞ log (cid:18) n N ( n, ǫ ) (cid:19) , where N ( n, ǫ ) is the number of g –orbit segments of length n that are at least adistance ǫ apart. Gromov [G] extended this definition to cover correspondences,which include Cremona transformations, as well.) However, this definition in-volves two limits, so the questions arise of finding how large n must be taken,and how small ǫ , in order to estimate it in to a given accuracy in a bounded time.On the other hand, over a finite field, especially one of characteristic 2,Feistel introduced the same kind of Cremona transformations f ( x, y ) = ( y + q ( x ) , x ) , except that he took the parameter a to be a = 1 always and he did not demandthat q be quadratic. Note that, in characteristic 2, this map f is the compositeof two involutions: f = α ◦ σ, where α is the linear map ( x, y ) ( y, x ) and σ ( x, y ) = ( x, y + q ( x )). These mapsare also known as round functions and they are an essential element of Feistelciphers such as DES, the Data Encryption Standard. The Advanced EncryptionStandard, AES, uses different Cremona transformations, but otherwise both DESand AES have a similar structure. In fact, as explained below, AES is a Cremonatransformation that is an element of a Galois twist of the group of standard
Cremona transformations of P F . This (the untwisted group, that is) is thegroup generated by P GL ( F ) and the standard non-linear birational involution x i x − i .Here is a toy model of DES ; it is a toy because it omits the key schedule .Taking the key schedule into account, as is done below, gives something like anoisy dynamical system, but where the noise is wholly determined in advance, aspart of the infrastructure. FFECTIVITY QUESTIONS IN THE CREMONA GROUP K (for example, by using someversion of public key cryptography based, say, on elliptic curves), the key deter-mines, according to a fixed public procedure that is part of the infrastructure ofthe algorithm, a round function f K = α ◦ σ K that is a Cremona transformationof some projective space P nk over a finite field k .Once the key has been established, encryption of a message M is this: break M into blocks M i , each of size n (that is, M i is a k -point of A n ) and then, for afixed integer N that is also part of the infrastructure of the algorithm, apply thetransformation f NK to the plaintext block M i and transmit f NK ( M i ). Decryptionis: apply ( f − K ) N to each block f NK ( M i ) that is received. It is a basic requirementthat, given possession of the key K , decryption should be as fast as encryption;this is achieved by constructing f K as the product of two involutions, and thendecryption is merely the process of applying the same two involutions but in theopposite order.The key schedule amounts to fixing k -linear involutions L , . . . , L N as partof the infrastructure (so that, in particular, the L i are independent of the key),defining f i = f K,i = f K ◦ L i and then taking encryption to be the iterate f N ◦ · · · ◦ f . Since each iterand f i is a product of three involutions, decryption is merely amatter of reversing the order of these involutions, and so is as cheap, in terms oftime and memory, as encryption.AES can be described in similar terms. First, some Galois twist σ of thestandard non-linear birational involution( x , ..., x n ) ( x − , ..., x − n )of P nk is given in advance and is public. Then, after the key K has been estab-lished, as above, it is used to construct a linear transformation of P nk and weset f K = σ ◦ L K . Thereafter the algorithm runs as for DES. (If the process everencounters a base point, meaning that it is trying to invert 0, then it maps 0 to 0.So the iterated Cremona transformation is garbled. However, the security of thescheme does not reside in this garbling.) Since f − K = σ ◦ L − K , decrypting is then,as with DES, as fast and cheap as encrypting (especially if L K is an involution).However, in higher dimensions, the inverse of a general Cremona transformation φ is of higher degree than φ , so that inversion of φ is slower and more expensivethan the execution of φ .As with DES, so AES has a key schedule, and the structure is similar.Encryption should also mix up the points of projective space thoroughly andquickly; in other words, it is desirable that if x and y are distinct basepoints thatare close, then the points f NK ( x ) and f NK ( y ) should be far apart for some large,but fixed, value of N . In other words, the round function should be sensitiveto the parameters that define it, in the sense of (1) above. That is, in highlysimplified terms, over the complex numbers certain Cremona transformationsserve as simple models of a process that is known to be chaotic, while over a N. I. SHEPHERD-BARRON finite field the same Cremona transformations are used to create a process thatmerely has a convincing appearance of chaos.Moreover [M1], positive entropy does not exclude the existence of Siegeldiscs; Siegel discs (whatever their analogues might be over finite fields) are unde-sirable in a cryptographic context because they are regions consisting of plaintextsthat are close and that remain close after encryption. On the other hand, thetheorem of Gromov and Yomdin, that the entropy of an endomorphism g of asmooth projective variety X is the logarithm of the spectral radius of the actionof g on the cohomology of X , shows that, as a consequence of the Lefschetz fixedpoint formula, h ( g ) can be computed by counting fixed points of a certain numberof iterates of g ; how many iterates are required depends on the Betti numbers of X . Despite the fact that the contemporary algorithm AES is a Cremona trans-formation, in this paper we consider only plane Cremona transformations. Thereason is simple: we don’t know any analogous results in higher dimensions, evenfor the group of standard transformations. Here we review the construction and basic properties of the infinite hyperbolicspace H = H k = H ( P k ) and the action of Cr ( k ) on it. This is taken from [CL];we repeat it only in order to establish notation.Let V be any smooth projective surface over the field k . Set Z ( V ) Z =lim −→ Y → V NS( Y ), where the direct limit is taken over all blow-ups Y → V , and Z ( V ) = Z ( V ) Z ⊗ Z R . There is a hyperbolic completion e Z ( V ) of Z ( V ), given by e Z ( V ) = { λ + X e P ∈E n P e P | λ ∈ NS( V ) R , n P ∈ R , X deg( P ) n P < ∞} , where e P is the exceptional curve associated to the closed point P on some blow-up Y , E is the set of such curves, deg( P ) is the degree of the field extension k ( P ) /k and P deg( P ) n P < ∞ means that, for all ǫ > F of E such that, for all finite subsets G of E − F , we have P e P ∈G deg( P ) n P <ǫ . (By definition, P deg( P ) n P is the number l such that for all ǫ > S of closed points P such that, whenever S ⊂ T and T isa finite set of closed points, (cid:12)(cid:12) P P ∈ T deg( P ) n P − l (cid:12)(cid:12) < ǫ. ) On e Z ( V ) there is ahyperbolic inner product denoted by ( x.y ). The hyperbolic space H ( V ) is one ofthe two connected components of the locus { x ∈ e Z ( V ) | ( x.x ) = 1 } . Note that,if x = λ + P n P e P , then ( x.x ) = ( λ.λ ) − P P deg( P ) n P . The distance on H ( V )is denoted by d , so that cosh d ( x, y ) = ( x.y ). The isometries of H ( V ) are thecontinuous linear transformations of e Z ( V ) that preserve the inner product andthe connected component above. FFECTIVITY QUESTIONS IN THE CREMONA GROUP Y → V , there are natural isomorphisms Z ( V ) Z →Z ( Y ) Z , H ( V ) → H ( Y ), etc., so that the group Bir( V ) of birational automor-phisms of V acts as a group of isometries of H ( V ) via g g ∗ .From now on, we take V = P k and write H ( V ) = H k and Bir( V ) = Cr ( k ). Lemma 3.1
The formation of H k is functorial in k and, if k is a subfield of K , then H k is naturally a closed geodesic subspace of H K and, as a subgroup of Cr ( K ) , Cr ( k ) acts on H K so as to preserve H k . PROOF : Immediate from the construction of H k .We shall usually drop the subscript k from H k .Isometries φ of H are of three types: φ is elliptic if it has a fixed pointin (the interior of) H ; φ is parabolic if it has a unique fixed point on the idealboundary ∂ H of H and no fixed point in H ; φ is hyperbolic if the lower bound L ( φ ) = lim inf d ( x, φ ( x )) is strictly positive and is attained in H .In this last case the set { x ∈ H | d ( x, φ ( x )) = L ( φ ) } is a geodesic in H . Itis the axis of φ and is denoted by Ax( φ ). It is the unique geodesic preserved by φ and its endpoints on ∂ H are the unique fixed point on the closure H = H ∪ ∂ H .The quantity L ( φ ) is the translation length of φ . On the other hand, a parabolicisometry does not preserve any geodesic. Lemma 3.2
Suppose that G is a closed hyperbolic subspace of H and φ anisometry of H that preserves G .(1) φ is elliptic if and only if φ | G is elliptic.(2) φ is hyperbolic if and only if φ | G is hyperbolic, and in this case Ax( φ | G ) =Ax( φ ) .(3) φ is parabolic if and only if φ | G is parabolic, and in this case the uniqueideal boundary point of H that is fixed by φ equals the ideal boundary point of G that is fixed by φ | G .(4) L ( φ | G ) = L ( φ ) . PROOF : (1): If P ∈ H − G and φ ( P ) = P , then φ also fixes the unique point Q on G that is closest to P .(2): Assume φ to be hyperbolic, with axis Γ. There is a map Γ → G : x y if y is the closest point to x that lies on G . The image is a geodesic ∆ which ispreserved by φ . Since φ does not preserve two geodesics, Γ = ∆.Conversely, if φ | G is hyperbolic, it preserves a geodesic ∆ in G , so that,again by (1), φ is hyperbolic and its axis is ∆.(3): Obvious, from (1) and (2).(4): Obvious. Corollary 3.3 If g ∈ Cr ( k ) , then whether it is elliptic, parabolic or hyperbolicand its translation length can be calculated after any making any extension of k . Lemma 3.4
Suppose that Γ is a geodesic in H and that φ is an isometry of H that preserves Γ . Then the following statements hold. N. I. SHEPHERD-BARRON (1) Either φ is elliptic and fixes a point on Γ or φ is hyperbolic.(2) If φ is hyperbolic then Γ is the unique geodesic preserved by φ andequals the axis of φ .(3) The translation length of φ acting on on H equals the translation lengthof its restriction φ | Γ to Γ . PROOF : Take G = Γ in Lemma 3.2. Lemma 3.5
Suppose that σ, α are involutions of H .(1) Fix( σ ) and Fix( α ) are hyperbolic subspaces of H .(2) σ preserves each geodesic that is perpendicular to Fix( σ ) , and the samefor α . (3) Fix( σ ) and Fix( α ) meet in H if and only if sα = g , say, is elliptic.(4) Fix( σ ∗ ) and Fix( α ∗ ) are parallel if and only if they meet in a single idealboundary point P if and only if g is parabolic.(5) If Fix( σ ∗ ) and Fix( α ∗ ) are ultraparallel then there is a unique geodesic Γ perpendicular to both and g is hyperbolic. This geodesic is preserved by both α and σ and is the axis of g . PROOF : (1) is clear: both fixed loci are projectivizations of linear spaces.(2) is a simple observation.(3): if Fix( σ ) and Fix( α ) meet in an interior point x of H then g ( x ) = x and g is elliptic.Conversely, suppose x is an interior point and g ( x ) = x . If x ∈ Fix( σ ) ∪ Fix( α ) then it is easy to see that x ∈ Fix( σ ) ∩ Fix( α ), so suppose that this isnot the case. Then there is a unique geodesic segment l from x to α ( x ): this isperpendicular to Fix( α ) and Fix( α ) cuts l at its midpoint. Similarly, there is aunique geodesic segment m from α ( x ) to σα ( x ): this is perpendicular to Fix( σ )and Fix( σ ) cuts m at its midpoint. But σα ( x ) = x , so l = m and the midpointlies in Fix( α ) ∩ Fix( σ ) and (3) is proved.So we can assume that Fix( σ ) ∩ Fix( α ) contains no interior point and that g is not elliptic.Case (a): Fix( σ ) ∩ Fix( α ) contains a boundary point P . Then g ( P ) = P . Supposethat g is hyperbolic, with axis Ax( g ) = l , and suppose that the boundary point Q is the other endpoint of l . Since σgσ = g − and Ax( g − ) = Ax( g ), σ preservesAx( g ) and so fixes Q . Similarly α ( Q ) = Q . Then Fix( σ ) ∩ Fix( α ) contains l ,which is absurd. So g is parabolic.Case (b): Fix( σ ) ∩ Fix( α ) contains no interior nor boundary point. That is, theyare ultraparallel. Then there is a unique geodesic l perpendicular to both of them.Because α, σ are involutions, each of then preserves l , and so g is not elliptic butpreserves a geodesic l . Then g is hyperbolic and l is its axis. This proves (4) and(5). Suppose that δ is a Cremona transformation, of degree D (in that δ isdefined by a net of homogeneous polynomials of degree D ). Then [CL] L ( δ ∗ ) FFECTIVITY QUESTIONS IN THE CREMONA GROUP δ , defined as lim n →∞ (deg( δ n )) /n . Say that δ iselliptic, etc., if δ ∗ is so and that δ is biregular if there is some rational surface X on which δ is a biregular automorphism.Over an algebraically closed field δ is elliptic if and only if δ is biregular,and in this case there is a rational surface X on which δ is biregular and aninteger n > δ n lies in the connected component Aut X of the identityelement in Aut X . The map δ is parabolic if and only if it preserves a pencil ofcurves of genus at most 1.Over any field, if δ is hyperbolic then there is no pencil of curves that ispreserved by δ .Moreover, over C , the entropy h ( δ ) of δ satisfies h ( δ ) ≤ L ( δ ∗ ) and equalityholds if δ is biregular on some smooth projective rational surface. Fix homogeneous co-ordinates x, y, z on V = P k . We denote by σ the standardquadratic involution σ : ( x, y, z ) ( yz, xz, xy ) and by α a linear involution. Weput g = σα .Say that P, Q, R are the base points of σ and Y = Bl P,Q,R V , with excep-tional curves e P , e Q , e R . Note that σ is biregular on Y and α is biregular on V .Let l denote the class of a line in V .Let E denote the set of all exceptional curves e x as x runs over all closedpoints of all blow-ups of Y .Since Z ( Y ) Z ∼ = Z ( V ) Z , the lattice Z ( V ) Z has a Z -basis { l } ∪ { e P , e Q , e R } ∪ E . From now on we shall not always be careful to distinguish between α and α ∗ , nor between σ and σ ∗ . Lemma 4.1 α permutes the set { e P , e Q , e R } ∪ E and σ permutes E . PROOF : Immediate, from the facts that α is biregular on V and that σ is bireg-ular on Y .Let v denote the root v = l − e P − e Q − e R . Lemma 4.2 α preserves l and σ acts on the lattice Z . { l, e P , e Q , e R } as the re-flection s v in v . Our aim is to construct a bipartite graph H that depends upon α . Then g = σα will act as something close to a Coxeter element in the correspondingCoxeter group.We begin by constructing subsets e Γ α and e Γ σ of the Z -lattice spanned by { l, e P , e Q , e R } ∪ E , as follows. These subsets are not necessarily disjoint.The elements (or vertices) of e Γ σ are v and one representative e x − σ ( e x )of each non-zero pair ± ( e x − σ ( e x )) as e x runs over E . The vertices of e Γ α areone representative e y − α ( e y ) of each non-zero pair ± ( e y − α ( e y )) as e y runs over0 N. I. SHEPHERD-BARRON E ∪ { e P , e Q , e R } . Finally, if v lies in e Γ α and ± v lies in e Γ σ , then choose v ratherthan − v in e Γ σ .Put e Γ = e Γ α ∪ e Γ σ . Because α and σ are involutions, two different points in E ∪ { e P , e Q , e R } , (resp., in E ), cannot give the same vertex in e Γ α (resp., in e Γ σ ).We join two distinct vertices in e Γ by an edge of multiplicity equal to theirintersection number, if that number is non-zero. If the intersection number is zero,then the corresponding vertices remain disjoint. So every edge has multiplicity ± − v to be the sum of the absolutevalues of the multiplicities of the edges meeting v . Lemma 4.3 If v lies in the intersection e Γ α ∩ e Γ σ then v is disjoint from all othervertices in e Γ . PROOF : Suppose that v = e x − σ ( e x ) and v = e y − α ( e y ). Note that e x , σ ( e x ) , e y and α ( e y are all classes of irreducible curves. We proceed to consider three casesseparately.(1) v meets e z − σ ( e z ). Since e z and σ ( e z are also classes of irreducible curves,either e x = e z , and then v = e z − σ ( e z ), or e x = σ ( e z ). This latter possibilitycontradicts the construction of e Γ σ .(2) v meets e t − α ( e t ). We reach a similar conclusion.(3) v meets v = l − e P − e Q − e R . Then e x ∈ { e P , e Q , e R } but α ( e P ) = σ ( e P ).So this cannot happen.Now define G α = e Γ α − ( e Γ α ∩ e Γ σ ) and define G σ similarly. Note that v liesin G σ . Lemma 4.4 (1) There are no edges within either G α or G σ .(2) G α ∪ G σ is a bipartite graph G .(3) The vertex v is of valency at most and every other vertex of G is ofvalency at most . PROOF : It is enough to notice that in e Γ the vertex v has valency at most 3and that every other vertex has valency at most 2, so that deleting e Γ α ∩ e Γ σ amounts to deleting those vertices that are joined to themselves and to no othervertex. Equivalently, deleting e Γ α ∩ e Γ σ amounts to deleting all double bonds andthe corresponding vertices.Now define e H to be the connected component of G that contains v . FFECTIVITY QUESTIONS IN THE CREMONA GROUP Lemma 4.5 e H is bipartite and is either a tree T p,q,r consisting of v and threearms of lengths p, q, r ≤ ∞ attached to v or the union ∆ m,r of a cycle of finitelength m together with an arm v , w r − , ..., w of length r ≤ ∞ attached to thecycle at v . PROOF : Immediate.
Remark : When we speak of the length of an arm, we count the vertex to which itis joined. So, for example, T , , is the E diagram, T p,q,r has a total of p + q + r − m, is a cycleof length m . Lemma 4.6 If e H = ∆ m,r then m is even. PROOF : Any subgraph of a bipartite graph is bipartite, so the cycle in e H isbipartite, so even.Write m = 2 n . From e H , construct a graph H with the same vertices as e H , but where every edge except at most one has multiplicity +1, by starting at v and proceeding either outwards along one arm at a time (in the case of T p,q,r )or around the cycle and then along the arm (in the case of ∆ n,r ) as follows: ateach step, change an edge of multiplicity − v by its negative, − v . If H is of type T p,q,r then all its edgeshave multiplicity +1, and we write H = T p,q,r ; in the other case all edges, exceptpossibly one edge that meets v and lies in the cycle, are of multiplicity +1, andwe write H = ∆ ± n,r accordingly. Lemma 4.7 (1) H is either of type T p,q,r with p, q, r ≤ ∞ or of type ∆ − n,r with ≤ n < ∞ and ≤ r ≤ ∞ .(2) If p, q, r are finite and the points P , ..., P p − , Q , ..., Q q − , R , ..., R r − are in general position, then H contains the diagram T p,q,r . PROOF : (1): We use the notation of the preceding proof, and in addition let s ′ denote the element of { α, σ } distinct from s . Then there are consecutive nodesin the cycle that are of the form e sw ( P ) − e w ( P ) , e s ′ sw ( P ) − e sw ( P ) and e sw ( Q ) − e w ( Q ) ,and also e s ′ sw ( P ) − e sw ( P ) = e s ′ sw ( Q ) − e sw ( Q ) . However, this is absurd.From the definition of ∆ − n,r it is clear that n, r ≥ H ) the lattice on the vertices v, w of H , with pairing given bythe usual intersection pairing of curves.Put H α = H ∩ G α and H σ = H ∩ G σ , so that v ∈ H σ . For v ∈ H , let s v denote the reflection in v . Lemma 4.8 σ acts on Λ( H ) via the product S = Q v ∈ H σ s v and α acts on Λ( H ) via the product A = Q w ∈ H α s w . PROOF : Immediate observation. Notice that because all the reflections in eachproduct commute with each other, the order in which they are taken is immaterial.2
N. I. SHEPHERD-BARRON
The fact that each product contains infinitely many factors is also immaterial,since there are only finitely many terms in either product that act non-triviallyon any given element of Λ( H ).The next lemma is well known but for lack of a convenient reference wegive a proof. Lemma 4.9
Suppose that ≤ p ≤ q ≤ r < ∞ and H = T p,q,r . Then Λ( H ) isnegative definite if /p + 1 /q + 1 /r > , degenerate if /p + 1 /q + 1 /r = 1 andhyperbolic if /p + 1 /q + 1 /r < . PROOF : If 1 /p + 1 /q + 1 /r > H is one of the Dynkin diagrams classifiedin Bourbaki [GrLie4-6] and Λ( H ) is the corresponding root lattice, twisted by( − /p + 1 /q + 1 /r = 1 then H is an affine Dynkin diagram of type e E n − ,where n = p + q + r −
2, and is degenerate. The radical R is of rank 1 and Λ( H ) /R is isomorphic to the root lattice E n − .If 1 /p + 1 /q + 1 /r < H contains H ′ = T p,q,r − . By induction, Λ( H ′ )is either degenerate or hyperbolic, and then Λ( H ) is hyperbolic. Lemma 4.10 Λ( H ) is degenerate only when H is of finite type T p,q,r with /p + 1 /q + 1 /r = 1 . PROOF : Suppose that H = ∆ − n,r , that r ≥ • v n − − ●●●●●●●●●● • v • w r − • w • v ✇✇✇✇✇✇✇✇✇ (the vertices v , . . . , v n − are arranged in a cycle of length 2 n ). Suppose that η = P n − a i v i + P r − b j w j is in the radical, so that η.v i = η.w j = 0 for all i, j .Put w r = v .By letting i run from 0 to 2 n −
1, we see that a i is a linear function of i ;say a i = λi + a for i = 0 , ..., n −
1. By letting j run from 0 to r , we see that b j = µj + b for j = 0 , ..., r .Since w r = v , we get a = µr + rb . Also, b = 2 b , so that µ = b , a = µ ( r + 1) . From η.v n − = 0 we get a n − − a n − − a = 0 , so that 2 nλ + a = 0.From η.v = 0 we get a − a n − − a + b r − = 0, so that0 = − n − λ − a + µr. FFECTIVITY QUESTIONS IN THE CREMONA GROUP µ = − λ ( n + 1) / r , so that µ = − nλ/ ( r + 1) . Hence ( n + 1)( r + 1) = nr , which is absurd.Finally, suppose that r = 1 (that is, H is a cycle) and that η = P a i v i isin the radical. Then the equations v i .η = 0 for i = 1 , ..., n − a = a + a , ..., a n − = a n − + a n − , so that there is some λ such that a j = a + λj for every j = 0 , ..., n −
1. But v .η = 0 gives a = 2 a + a n − , so that 0 = 2 a + (2 n − λ , while v n − .η = 0gives 0 = 2 a + 2 nλ . Then a = λ = 0 and η = 0.There is an obvious natural homomorphism β : Λ( H ) → Z ( V ) Z of lattices. Lemma 4.11 β is injective. PROOF : Inspection.Let Λ( T ( λ ) p,q,r ) and Λ(∆ − ( λ )2 n,r ) denote the lattices corresponding to the diagrams T p,q,r and ∆ − n,r , but where each vertex v has v = − λ and the other intersectionnumbers are unchanged. So, for example, Λ( H (2) ) = Λ( H ). Lemma 4.12
Λ(∆ − (2)2 n,r ) is hyperbolic if /n + 1 /r < . PROOF : Λ(∆ − (2)2 n,r ) is non-degenerate, by Lemma 4.10. Deleting v leaves a neg-ative definite lattice, and so Λ(∆ − (2)2 n,r ) is either negative definite or hyperbolic.However, deleting the vertex opposite v in the cycle leaves a T n,n,r diagram,which is hyperbolic.Say that a lattice Λ is affine if Λ is degenerate, its radical R (Λ) is of rank1 and Λ /R (Λ) is negative definite. For example, Λ( T (2) p,q,r ) is affine if and only if1 /p + 1 /q + 1 /r = 1.As λ varies, so Λ( T ( λ ) p,q,r ) sweeps out a line in the space of real quadraticforms in n = p + q + r − T ( λ ) p,q,r ) is negative definite if λ ≫ λ ≪ T ( λ ) p,q,r ) is a non-increasing function of λ . (We have adopted the convention that the signature ofa positive definite form of rank n is n and the signature of a negative definiteform of rank n is − n .) For critical values of λ the from will be degenerate.In particular therefore, there exists µ = µ ( p, q, r ) such that Λ( T ( λ ) p,q,r ) isnegative definite if λ > µ while Λ( T ( µ ) p,q,r ) is negative semi-definite and degenerate.For example, µ (2 , ,
6) = µ (2 , ,
4) = µ (3 , ,
3) = 2. Also, define m = m ( p, q, r )by µ = m + m − + 2and m ≥ Lemma 4.13 (1) µ ( p, q, r ) is a strictly increasing function of each of p, q, r .That is, µ ( p, q, r ) > µ ( p, q, r − and the same for p and q . N. I. SHEPHERD-BARRON (2) If /p + 1 /q + 1 /r < then µ ( p, q, r ) > . PROOF : (1): Say s = µ ( p, q, r − T ( s ) p,q,r − ) is affine. Then there is anon-zero vector P n i v i in the radical of Λ( T ( s ) p,q,r − ), and it is easy to see that wecan take every n i to be positive.Say that w is the vertex adjoined in passing from T p,q,r − to T p,q,r and that w meets v r − in T p,q,r − . Then (cid:16)X n i v i + ǫw (cid:17) = 2 n r − ǫ − ǫ s > < ǫ ≪
1, so that Λ( T ( s ) p,q,r ) is hyperbolic By the discussion above, the act ofincreasing s will lead to the lattice becoming negative definite; there is thereforea critical point t = µ ( p, q, r ) at which the lattice becomes degenerate, and t > s .(2) is a consequence of (1) and the the equalities µ (2 , ,
6) = µ (2 , ,
4) = µ (3 , ,
3) = 2.Fix p ≤ q ≤ r < ∞ , with 1 /p + 1 /q + 1 /r <
1. Say µ ( p, q, r ) = µ .Write µ ( p, p, p ) = µ p and m ( p, p, p ) = m p . Lemma 4.14 lim p,q,r →∞ µ ( p, q, r ) = 3 / √ and µ p ≥ √ (1 − − p/ ) . PROOF : Say that f , ..., f p − ; g , ..., g q − ; h , ..., h r − are the vertices of T p,q,r ,reading outwards along the arms from the central vertex v . We can supposethat p = q = r and then, by symmetry, that ξ = b p (0) v + p − X i =1 b p ( i )( f i + g i + h i )is in the radical of Λ( T ( µ p ) p,p,p ).Set b p (0) = 3 and b p ( p ) = 0. Then 0 = ξ.v = − µ p + 3 b p (1) and we havea recurrence relation0 = ξ.f n = b p ( n −
1) + b p ( n + 1) − µ p .b p ( n ) ∀ n ≥ . Therefore, by the well known formula for the solution of such recurrence relations,there are constants C p , D p such that b p ( n ) = C p θ np + D p δ np , where θ p = ( µ p + p µ p − / δ p = ( µ p − p µ p − / θ − p ≤ θ p . Since b p ( p ) = 0, this can be written as b p ( n ) = E p ( θ n − pp − δ n − pp )for some E p that is independent of n . Since b p (0) = 3 and b p (1) = µ p we get µ p b p (1) b p (0) = θ p − p − θ p θ pp − . FFECTIVITY QUESTIONS IN THE CREMONA GROUP µ = lim p →∞ µ p and θ = lim p →∞ θ p ; then µ θ . This leads to µ = √ / . Since µ p is an increasing function of p , we get µ p → ( √ / − . For the speed of convergence, note that µ p = m / p + m − / p and µ p = θ p + θ − p .So m p = θ p . The formula for µ p / m pp = 2 m p − − m p . The same equation is satisfied by m − p . Lemma 4.15 m p → − and m p > − . − p/ , while m − p → (1 / + and m − p < / − p . PROOF : Since µ p → (3 / √ − , m p → − .Solving the above equation for m p when p = 4 gives m ≈ . , so m > √
2. Since m p is an increasing function of p , m p > √ p ≥ − m p = η . Then the equation above gives η = (3 − η )(2 − η ) − p . Since m p > √
2, this leads to η < (3 − η )2 − p/ < . − p/ . The proof for m − p is similar.Since µ p = m p + m − p + 2, we get µ p > − . − p/ + 12 + 2 = 3 (cid:18) − − p/ (cid:19) . So µ p > √ (1 − − p/ ) and Lemma 4.14 is proved.Fix p ≤ q ≤ r < ∞ , with 1 /p + 1 /q + 1 /r <
1. Say µ ( p, q, r ) = µ .Then there is a unique totally isotropic vector ξ = ξ ( p, q, r ) ∈ Λ( T ( µ ) p,q,r )where the coefficient of the branch vertex is 1.Fix a subset E of { p, q, r } and let the members of { p, q, r } − E tend to ∞ .Let H E denote the corresponding infinite graph. Lemma 4.16 µ → µ E for some µ E ≤ / √ and in the completed hyperbolicvector space generated by the vector space Λ( H E ) ⊗ R there is an eigenvector ξ ,with eigenvalue µ E , of the adjacency matrix of H E . PROOF : The existence of µ E follows from Lemmas 4.13 and 4.14.6 N. I. SHEPHERD-BARRON
To prove that ξ exists, we write it down. Define θ = ( µ E + p µ E − / E = { p, q } . Then ξ = p X a i e i + q X b j f j + ∞ X c k g k , where e = f = g = v , the branch vertex, a p = b q = 0 and a = b = c = 1.and we require that ξ be totally isotropic in Λ( T ( µ E ) p,q, ∞ ). This is achieved by taking a i = A ( θ i − θ − i ) + θ − i , b j = B ( θ j − θ − j ) + θ − j , c k = θ − k where A = 1 / (1 − θ p ) and B = 1 / (1 − θ q ). Corollary 4.17 If p, q are fixed then lim r →∞ m ( p, q, r ) exists and ≤ . There isa similar limit if p is fixed and q, r → ∞ . Now suppose that H = ∆ − n,r and that 2 /n + 1 /r <
1. Define µ ∆ (2 n, r ) inthe same way that µ ( p, q, r ) is defined for T p,q,r . Lemma 4.18 (1) µ ( n − , n − , r ) ≤ µ ∆ (2 n, r ) .(2) µ ∆ (2 n, r ) is a strictly increasing function of both n and r .(3) lim r →∞ µ ∆ (2 n, r ) exists and ≤ / √ . PROOF : (1): delete the three vertices u , u , u consisting of the vertex opposite v and the two vertices adjacent to it. This shows that Λ( T n − ,n − ,r ) ⊂ Λ(∆ − n,r ).Say µ ( n − , n − , r ) = λ , and pick ξ in the radical of Λ( T ( λ ) n − ,n − ,r ). Then there isa vector η = ξ + P αu i in Λ(∆ − ( λ )2 n,r ) with η >
0. So µ ( n − , n − , r ) ≤ µ ∆ (2 n, r ).(2) and (3) are proved by the same kind of calculation used in the proof ofLemma 4.14. Corollary 4.19 µ ∆ (2 p, p ) > − . − p +1 . PROOF : Apply Lemma 4.15.
Lemma 4.20 If n is fixed and r = ∞ then there is an eigenvector ξ , witheigenvalue lim r →∞ µ ∆ (2 n, r ) , of the infinite adjacency matrix. PROOF : Put µ E = µ ∆ (2 n, ∞ ) and θ = ( µ E + p µ E − /
2. As in the proof ofLemma 4.16 the easiest thing is to write ξ down: ξ = n − X i =0 a i v i + ∞ X j =0 b j w j where b j = θ − j a i = A θ i + A θ − i and A , A are determined by the equations0 = − s + A ( θ − θ n − ) + A ( θ − − θ − (2 n − ) + θ − , A ( θ n − − sθ n − ) + A ( θ − (2 n − − sθ − (2 n − ) − . FFECTIVITY QUESTIONS IN THE CREMONA GROUP /p + 1 /q + 1 /r < /n + 1 /r <
1, as appropriate.Then at this point we have shown, whether H is finite or infinite, the existenceof a totally isotropic vector ξ in Λ( H ( µ ) ) for some µ >
2. Moreover, ξ generatesthe radical R (Λ( H ( µ ) )).Let M denote the adjacency matrix of H ; then the matrix B = − µ. M is the Gram matrix of Λ( H ( µ ) ) and ξ is an eigenvector, with eigenvalue 0, of B .The adjacency matrix is, according to the bipartite decomposition H = H σ ⊔ H α ,of the form M = (cid:20) t CC (cid:21) , where each row and column of C contains at most three non-zero entries, andeach non-zero entry is 1. (In the case of ∆ − n,r some entries will be − ξ generates R (Λ( H ( µ ) )), the other eigenvalues of B are real andstrictly negative. In particular, µ is the maximum eigenvalue of M , and is ofmultiplicity 1.Write ξ = (cid:20) uz (cid:21) . Then t Cz = µu and Cu = µz .Recall that g = σα and notice that, from the bipartite description of H , α acts as the matrix A = (cid:20) C − (cid:21) , while σ acts as the matrix S = (cid:20) − t C (cid:21) .Then S (cid:20) u (cid:21) = (cid:20) − u (cid:21) , S (cid:20) z (cid:21) = (cid:20) µuz (cid:21) ,A (cid:20) u (cid:21) = (cid:20) uµz (cid:21) , A (cid:20) z (cid:21) = (cid:20) − z (cid:21) . Thus S and A preserve the real 2-plane Π based by (cid:26)(cid:20) u (cid:21) , (cid:20) z (cid:21)(cid:27) and act on Πwith respect to this basis as the matrices (cid:20) − µ (cid:21) and (cid:20) µ − (cid:21) , respectively.The matrix of g ∗ is then SA = (cid:20) µ − − µµ − (cid:21) = δ , where δ = (cid:20) µ −
11 0 (cid:21) . Since Tr δ = µ >
2, it follows that δ , and so g ∗ , is hyperbolic.The projectivization of Π is a geodesic ℓ in the hyperbolic space H and ℓ is preserved by σ, α and g ∗ . Therefore, by Lemma 3.4, the translation length of g ∗ acting on H equals the translation length of g ∗ acting on ℓ . We shall calculatethis from the preceding discussion.Write µ = m + m − + 2 , so thatlog m = 2 cosh − ( µ/ . The next result contains the first part of Theorem 1.1.8
N. I. SHEPHERD-BARRON
Theorem 4.21
Suppose that g is in ( p, q, r ) -general position and that /p +1 /q +1 /r < . Then g is hyperbolic. If also ≤ p ≤ q ≤ r then L ( g ) ≥ log(2 − . − p/ ) . PROOF : The hypotheses imply that H is either T p,q,r with 1 /p + 1 /q + 1 /r < − n,r with 2 /n + 1 /r <
1. We now use the notation of the preceding discussion.The vector x = (cid:20) (cid:21) ∈ Π satisfies ( x.x ) = 1 and ( x.δ ( x )) = µ/
2, so that d ( x, δ ( x )) = cosh − ( µ/ L ( δ ) = cosh − ( µ/ g ∗ preserves thegeodesic ℓ . Therefore g ∗ is hyperbolic and L ( g ) = 2 cosh − ( µ/
2) = log m > . For the second part, note that, since µ E is an increasing function of eachargument and, in the notation of Lemma 4.15 and the formula immediately pre-ceding it, µ p = m p + m − p + 2 , we have L ( g ) ≥ log m p > log(2 − . − p/ ), byLemma 4.15.Finally we give the proof of the rest of Theorem 1.1. Theorem 4.22 (= Theorem 1.1 2) Assume that < ǫ < / . Then, if p ≥ /ǫ and the p points P , ..., P p − , ..., R p − are in general position, the Cremonatransformation g = σα is hyperbolic and log 2 − ǫ < L ( g ) ≤ log 2 . PROOF : We know that L ( g ) ≥ log(2 − . − p/ ), so that it is enough to verifythat log(2 − . − p/ ) ≥ log 2 − ǫ if p > /ǫ . This is elementary.These techniques also give Cremona transformations with prescribed prop-erties, as follows.For example, let us calculate µ (2 , , T , , diagram • u • u • u u • • u • u • u • u • u • u we have to find the value of s for which the lattice Λ( T ( s )2 , , ) has a non-trivialtotally isotropic vector ξ = P b i u i . Then µ (2 , ,
7) = s . That is, we musteliminate the variables b , ..., b from the equations ξ.u i = 0for all i = 1 , ...,
10 where u i = − s and the other intersection numbers are givenby the diagram. We can normalize ξ by assuming that b = 1.This process of elimination is mechanical and yields s − s + 27 s − s + 12 s − . FFECTIVITY QUESTIONS IN THE CREMONA GROUP µ (2 , , > m (2 , , > m (2 , ,
7) = λ Lehmer ≈ . . (Recall that µ = m + m − + 2.)Exactly similar calculations show that µ (2 , ,
5) is a solution of s − s − s + 17 s − , so that m (2 , ,
5) is a zero of m − m − m − m + 1; µ (3 , ,
4) is a zero of s − s + 8 s − m (3 , ,
4) is a zero of m − m − m − m + 1; µ (4 , ,
4) = q (5 + √ / µ (5 , ,
5) = q √ . The approximate values are given in this table.( p, q, r ) (2,3,7) (2,4,5) (3,3,4) (4,4,4) (5,5,5) ( ∞ , ∞ , ∞ ) µ / √ m λ Lehmer p, q, r . Then we can force H to be equal to the finitetree T p,q,r by putting stringent conditions on the linear involution α , as follows: P , ..., P p − , ..., R r − must be in general position while P p = P p − , Q q = Q q − , R r = R r − . Note that, since σ has only finitely many fixed points while α has a line offixed points, and (if char k = 2) one isolated fixed point, these are two conditionson α (which moves in a 4-dimensional subvariety of the 8-dimensional group P GL ,k ) for each of p, q, r that is odd, and one condition on α for each that iseven. As before, l is the class of a line in P . Lemma 4.23 H is a finite diagram of type T p,q,r and its vertices are v = l − e P − e Q − e R ; e P − e P , ..., e P p − − e P p − ; e Q − e Q , ..., e Q q − − e Q q − ; e R − e R , ..., e R r − − e R r − . PROOF : Immediate.0
N. I. SHEPHERD-BARRON
Lemma 4.24
Both α and σ are biregular on the blow-up Y of P at the p + q + r points P , ..., R r − . PROOF : Also immediate.Of course, g is then also biregular on Y . Lemma 4.25
The orthogonal complement Λ( H ) ⊥ of Λ( H ) in NS( Y ) has a Z -basis ( x P , x Q , x R ) given by x P = l − P e P i , etc. Each of α, σ acts trivially on Λ( H ) ⊥ . Theorem 4.26 (= Theorem 1.2) The biregular quadratic map g acts on NS( Y ) as a Coxeter element in the Weyl group of the lattice Λ( T p,q,r ) . In particular, if ( p, q, r ) = (2 , , then g has order ; if ( p, q, r ) = (2 , , then g is parabolic;while if ( p, q, r ) = (2 , , r ) with r ≥ then g is hyperbolic and L ( g ) = log λ r +3 .In particular, if r = 7 then L ( g ) = log λ Lehmer . PROOF : This follows from the discussion above.
Remark : Blanc and Cantat prove [BC] that L ( g ) ≥ log λ Lehmer for any hy-perbolic element g of Cr ( k ), while in [M1] McMullen proves the existence ofbiregular Cremona transformations g with L ( g ) = log λ r +3 for all r ≥
7. Hisexamples also preserve a cuspidal cubic curve but are less explicit than ours. Forexample, sufficient conditions on the linear involution α for g to be biregular oftype (2 , ,
7) are that α ( P ) should be a fixed point of σ and that σα ( Q ) and( σα ) ( R ) should lie on the line that is the 1-dimensional part of Fix α . Theseare 4 explicit polynomial conditions on the 4-dimensional family of involutions in P GL . Remark : There is a much shorter argument that suffices to prove merely that g is hyperbolic if the diagram H is hyperbolic, without any estimates, as follows.Consider the action of the involutions σ, α on the completed hyperbolicspace H associated to Λ( H ). Lemma 4.27
Fix( α ∗ ) and Fix( σ ∗ ) are ultraparallel. PROOF : Fix( α ∗ ) ∩ Fix( σ ∗ ) is spanned by common non-zero eigenvectors p = (cid:20) xy (cid:21) of α ∗ and σ ∗ such that ( p.p ) ≥
0. Here, x (resp., y ) lies in the Hilbert spacecompletion of the negative definite vector space Λ( H α ) ⊗ R (resp., Λ( H σ ) ⊗ R ),each of which is naturally embedded in the hyperbolic completion of Λ( H ) ⊗ R .We shall check that no such vectors p exist. There are three cases toconsider.(1) α ∗ ( p ) = p and σ ∗ ( p ) = p . Then Cx = 2 y and t Cy = 2 x . So ( − M ) p = 0.Since − M is the Gram matrix of Λ( H ), p is then orthogonal to everyvector in Λ( H ). However, Λ( H ) is hyperbolic, so non-degenerate.(2) α ∗ ( p ) = − p . Then x = 0 and y = 0, so that ( p.p ) = − t yy < FFECTIVITY QUESTIONS IN THE CREMONA GROUP σ ∗ ( p ) = − p . Then y = 0 and x = 0, so that ( p.p ) = − t xx < Recall from [CL] the crucial notions of rigidity and tightness : given ǫ, B >
0, ahyperbolic conjugacy class C in G = Cr ( k ) is ( ǫ, B ) -rigid ifdiam(Tub ǫ Ax( g ) ∩ Tub ǫ Ax( h )) ≤ B whenever g, h ∈ C and Ax( g ) = Ax( h ). If ǫ ′ > ǫ and C is ( ǫ, B )-rigid then itis also ( ǫ ′ , B ′ )-rigid for some explicit B ′ = B ′ ( ǫ, ǫ ′ , B ) ([CL], 2.3.2). So we canspeak of rigidity without reference to the precise values of ǫ and B .A hyperbolic element is rigid if its conjugacy class is rigid. Lemma 5.1 If g is rigid then so is g n , for every n = 0 . PROOF : Taking n th powers of elements in a conjugacy class gives a new conju-gacy class but does not change the set of axes.A rigid hyperbolic conjugacy class C is tight if whenever g, h ∈ C andAx( g ) = Ax( h ), then h = g ± . An element g is tight if its conjugacy class istight. For the rest of this section g will denote a fixed hyperbolic element of Cr ( k ).Suppose that, for all i ∈ N , Σ i is a segment of Ax( g ) of length i and thatall the Σ i have the same midpoint. For ǫ > V Σ i ,ǫ = V i,ǫ = { f ∈ Cr ( k ) | d ( x, f ( x )) < ǫ ∀ x ∈ Σ i } . Remark : Note that, if g is not rigid, then, by Prop. 3.3 of [CL], for allbounded segments Σ of Ax( g ) and for all ǫ >
0, there exists f ∈ Cr ( k ) such that d ( x, f ( x )) < ǫ for all x ∈ Σ while f does not preserve Ax( g ). So, if g is not rigid,then, for every i and every ǫ > V i,ǫ contains elements f of Cr ( k ) thatdo not preserve Ax( g ). In particular, if g is not rigid then every V i,ǫ is infinite.We now isolate the main part of the argument and formulate it as a separateresult. Proposition 5.2
Assume that k is algebraically closed and that V i,ǫ is infinitefor all i and for all ǫ > . There is a positive-dimensional affine algebraic groupvariety S over k acting biregularly and effectively on a k -rational surface Y suchthat, when S ( k ) is identified with a subgroup of Cr ( k ) ,(1) for all sufficiently large i and for all ǫ ≪ V i,ǫ is a Zariski dense subsetof S ( k ) and N. I. SHEPHERD-BARRON (2) g normalizes S ( k ) . PROOF : Fix i ∈ I , x ∈ Σ i and ǫ >
0. Put V i,ǫ, ≤ r = { f ∈ V i,ǫ | deg f ≤ r } . Let ℓ ∈ H be the class of a line in P and suppose that f ∈ V i,ǫ . Then d ( ℓ, f ( ℓ )) ≤ d ( ℓ, x ) + d ( x , f ( x )) + d ( f ( x ) , f ( ℓ )) ≤ d ( ℓ, x ) + ǫ. That is, the degree of f is bounded. In other words there is an integer D suchthat V i,ǫ = V i,ǫ, ≤ D .Now suppose that f, h ∈ V i,ǫ, ≤ D . Then f h ∈ V i, ǫ , so that d ( ℓ, f h ( ℓ )) ≤ d ( ℓ, x ) + 2 ǫ . Since cosh − ( N ) is discrete in R , it follows that deg( f h ) ≤ D if,as we now assume, 0 < ǫ ≪
1. So multiplication gives a map V i,ǫ, ≤ D × V i,ǫ, ≤ D → V i, ǫ, ≤ D . According to [BF], the set of Cremona transformations whose degree is ex-actly d is naturally the set of k -points of a reduced quasi-projective scheme ( Cr ) d ,while ( Cr ( k )) ≤ D = ⊔ d ≤ D ( Cr ) d ( k ) has no natural structure as a scheme or alge-braic space, although ( Cr ( k )) ≤ D is naturally a noetherian topological space. Let Z i,ǫ, ≤ D denote the closure of V i,ǫ, ≤ D in ( Cr ( k )) ≤ D , so that multiplication definesa map m : Z i,ǫ, ≤ D × Z i,ǫ, ≤ D → Z i, ǫ, ≤ D . By the noetherian property, Z i, ǫ, ≤ D = Z i,ǫ, ≤ D for 0 < ǫ ≪
1, so that m is a map m : Z i,ǫ, ≤ D × Z i,ǫ, ≤ D → Z i,ǫ, ≤ D . We can, and do, suppose that D is minimal with respect to the two properties(1) Z i,ǫ, ≤ D is infinite and(2) m ( Z i,ǫ, ≤ E × Z i,ǫ, ≤ E ) is not contained in Z i,ǫ, ≤ E for any E ≤ D − Z i,ǫ, ≤ D = ⊔ d ≤ D Z i,ǫ,d ( k ), where the k -scheme Z i,ǫ,d is the Zariskiclosure of V i,ǫ,d in ( Cr ) d . Lemma 5.3 m ( Z i,ǫ,D ( k ) × Z i,ǫ,D ( k )) meets Z i,ǫ,D ( k ) non-trivially. PROOF : Assume otherwise, so that m ( Z i,ǫ,D ( k ) × Z i,ǫ,D ( k )) lies in Z i,ǫ, ≤ D − . Weuse the notation of [BF]: W d is the projectivized space of triples of homogeneousdegree d polynomials in three indeterminates, H d is the locally closed subschemeof W d consisting of triples that define a Cremona transformation of degree atmost d , and, for any e ≤ d , H e,d is the locally closed subscheme of H d of triplesthat define a Cremona transformation of degree exactly e . There are surjections π e,d : H e,d → ( Cr ) e and π d : H d ( k ) → ( Cr ( k )) ≤ d . The morphism π d,d is anisomorphism.Set Z ′ i,ǫ,D = π − D,D ( Z i,ǫ,D ), so that there is a commutative diagram Z ′ i,ǫ,D ( k ) (cid:31) (cid:127) / / ∼ = (cid:15) (cid:15) H D,D ( k ) (cid:31) (cid:127) open / / ∼ = (cid:15) (cid:15) H D ( k ) π D (cid:15) (cid:15) Z i,ǫ,D ( k ) (cid:31) (cid:127) / / ( Cr ) D ( k ) (cid:31) (cid:127) / / ( Cr ( k )) ≤ D FFECTIVITY QUESTIONS IN THE CREMONA GROUP E ≤ D − U of Z i,ǫ,D × Z i,ǫ,D such that m ( U ( k )) ⊂ Z i,ǫ,E ( k ). Write U ′ = π − D,D ( U ) and let m D : H D × H D → H D denote the multiplication. Then m D ( U ′ ) ⊂ H E,D .Let Z ′ i,ǫ,D be the closure of Z ′ i,ǫ,D in H D . Since Z ′ i,ǫ,D maps onto Z i,ǫ,D under π D , it follows that m ( Z i,ǫ, ≤ D × Z i,ǫ, ≤ D ) ⊂ Z i,ǫ, ≤ E . In particular, m ( Z i,ǫ, ≤ E × Z i,ǫ, ≤ E ) ⊂ Z i,ǫ, ≤ E , so that, by the minimality assumption, Z i,ǫ, ≤ E is finite. Butthen m ( Z i,ǫ, ≤ D × Z i,ǫ, ≤ D ) is finite, which contradicts the fact that Z i,ǫ, ≤ D is infinite.Now define Z i as the closure of Z ′ i,ǫ,D ∪ { } in H D . By what we have provedso far, m then defines a rational map m : Z i × Z i − − → Z i . Since Z i has anidentity element and is preserved under taking inverses and since multiplicationis associative in Cr , m is a group chunk. By the theorem of Weil and Rosenlichtthere is then a rational surface Y over k , a positive-dimensional group variety S over k that is k -birational to Z and an embedding of S into the group schemeAut Y . Since Y is rational, the group variety S is affine (it has no abelian part),and (1) of Proposition 5.2 is proved.For (2), note first that, if we fix i ≫ j, i with j ≥ i + L ( g )and i ≥ i , then V i,ǫ ⊃ V j,ǫ and V j,ǫ is Zariski dense in S ( k ).Let f ∈ V j,ǫ ; then f ∈ V i,ǫ and then gf g − ∈ V i,ǫ , since L ( g ) ≤ j − i .Therefore there is a Zariski dense subset Σ of S ( k ) such that g Σ g − ⊂ S ( k ). Itfollows that g normalizes the subgroup S ( k ) of Cr ( k ). This concludes the proofof Proposition 5.2. Lemma 5.4
Assume that k is algebraically closed and that every V i,ǫ is infinite.(1) Every positive-dimensional subgroup B of S that is normalized by g has an open orbit in Y .(2) Every positive-dimensional characteristic subgroup of S has an openorbit in Y . PROOF : (1): Suppose that B has no open orbit in Y . Then its orbits form apencil of curves. This pencil is then preserved by g , which is impossible since g is hyperbolic.(2) follows at once. Lemma 5.5 (Assume that k is algebraically closed and that every V i,ǫ is infinite.) S is not semi-simple. PROOF : There is an S -equivariant desingularization e Y → Y . Then there isan S -equivariant blowing-down e Y → Y ′ where Y ′ is a minimal rational surface:that is, either P or a Hirzebruch surface Σ n (that is, a P -bundle Σ n → P where n ≥ n = 1 and Σ n has a unique negative section σ with σ = − n ). Thesemi-simple part of the automorphism group of Σ n is isogenous to GL if n ≥ P GL × P GL if n = 0, while if Y ′ = P then S ⊂ P GL and so equals P GL if it stabilizes a conic or equals P GL otherwise.4 N. I. SHEPHERD-BARRON If S = P GL then Y ′ is either P or Σ n . If Y = Σ n then the set of rulingson Y ′ that are preserved by S is finite and non-empty. Some power of g then fixesat least one of them, which is impossible since g is hyperbolic. If Y ′ = P then S = P SO and preserves a conic Γ such that Γ and P − Γ are the only S -orbitson P . Then every h ∈ S preserves the base locus of g , so that g ∈ P GL .This argument also covers the case where S = P GL .If S = P GL × P GL then Y ′ = P × P and then, by a similar argument, g preserves the two rulings on Y ′ . But g is hyperbolic. Lemma 5.6
If a torus T acts effectively on a rational surface Y then dim T ≤ . PROOF : As in the proof of the previous lemma, we reduce to an action of T on P or Σ n . Then T lies in P GL or P GL or P GL × P GL or a group isogenousto GL . Lemma 5.7 (Assume that k is algebraically closed and that every V i,ǫ is infinite.)In Cr ( k ) the element g normalizes a commutative group A which is iso-morphic to either G m or G sa with s ≥ . PROOF : Take H to be the connected component of the centre of the radicalof S ; since H acts effectively on Y and has an open orbit on Y , it follows thatdim H ≥
2. Therefore H = G rm × G sa with r + s ≥ r ≤ G sa is characteristic in S , so s = 1, since each characteristicsubgroup has an open orbit in Y . Theorem 5.8 g is rigid. PROOF : Suppose not, and assume first that k is algebraically closed. So every V i,ǫ is infinite. Note that each set V i,ǫ contains a subset V i,ǫ, that is Zariski densein S ( k ), where S and Y are the objects provided by Proposition 5.2.Regard Y as the rational surface on which Cr ( k ) acts. In the inversesystem of all blow-ups X → Y there is an inverse subsystem of S -equivariantblow-ups; the normalizer N of S ( k ) in Cr ( k ) acts on this subsystem. Note that g lies in N .Taking the direct limit of the N´eron–Severi groups of the surfaces in thissystem and then constructing infinite-dimensional hyperbolic spaces gives a closedhyperbolic subspace H N of H that is preserved by N . Now g acts hyperbolicallyon H and therefore acts hyperbolically on every g -invariant closed hyperbolicsubspace of H . Therefore Ax( g ) ⊂ H N . However, A ( k ) acts trivially on theN´eron–Severi group of every A -surface in the subsystem above, so acts triviallyon H N and so on Ax( g ). Then S ( k ), and so V i,ǫ, , preserves Ax( g ), which, as inthe Remark preceding Proposition 5.2, contradicts our assumption.So we have proved that g is rigid when k is algebraically closed.Now suppose that k is an arbitrary field and that g is not rigid. FFECTIVITY QUESTIONS IN THE CREMONA GROUP K of k . Then the hyperbolic space H k is, from itsdefinition, a closed geodesic subspace of H K that is preserved by g . By assump-tion, g acts hyperbolically on H k , so preserves Ax( g ), which is a geodesic γ in H k and so a geodesic in H K . Suppose now that g is not hyperbolic on H K ; then g isnot parabolic, since it preserves γ , so is elliptic. Suppose that P ∈ H K is a fixedpoint; then g preserves the hyperbolic plane Π spanned by γ and P . However,no non-trivial isometry of Π can both fix a point and preserve a geodesic. So g is hyperbolic on H K and preserves γ . It follows that γ is the axis of g when g isregarded as an isometry of H K .Pick ǫ, B > g is ( ǫ, B )-rigid as an element of Cr ( K ). Supposethat f, h lie in the conjugacy class of g in Cr ( k ). Thendiam (Tub ǫ Ax( f ) ∩ Tub ǫ Ax( h )) < B where the tubular neighbourhoods are taken in H K , and then the same inequalityholds in the subspace H k . That is, g is rigid. Theorem 5.9 If k is algebraically closed and no power of g is tight then g normalizes a copy of either G m or G sa . PROOF : We know that g is rigid. Fix an orientation of Ax( g ) and consider thesubgroup N = { f ∈ Cr ( k ) | f (Ax( g )) = Ax( g ) } of Cr ( k ) and its subgroup N + of index at most 2 consisting of transformationsthat preserve the orientation. Then there is a group homomorphism π : N + → ( R , +), where h ∈ N + acts on Ax( g ) as a shift by π ( h ). Note that an element h of N + is hyperbolic if and only if π ( h ) = 0, and then | π ( h ) | = L ( h ). Set H = ker π ,so that H = { f ∈ Cr ( k ) | f ( x ) = x ∀ x ∈ Ax( g ) } .Since [BC] the spectrum of Cr ( k ) (the set of lengths of its hyperbolicelements) is bounded away from zero, the image im π of π is discrete, so infinitecyclic. Choose g ∈ N + such that π ( g ) generates im π , so that N + is a semi-directproduct N + = H ⋊ h g i .If H is not finite then every V i,ǫ is infinite, so that, by Proposition 5.2 andresults following, there is a k -group variety S normalized by g such that S ( k ) ⊂ H and S contains a subgroup isomorphic to either G m or G a that is also normalizedby g . So we may, and do, assume that H is finite.We can write g = hg m for some h ∈ H , m ∈ Z . Consider the conjugationaction of g on the finite group H ; then g s centralizes H for some s >
0. That is, g s lies in the centre Z ( N + ). Also, g s = c.g ms for some c ∈ H , so that g st = c t g mst for all t ; choosing t to be divisible by the order of H leads to g st ∈ Z ( N + ). Then f g st f − = g st whenever f ∈ N + .If f ∈ N \ N + , put g = f gf − . Then g = h g − m and g s = c g − m for some h , c ∈ H . It follows that g st = c t g − mst ; choosing t as before gives g st = g − mst = g − st and f g st f − = g − st , so that g st is tight.6 N. I. SHEPHERD-BARRON
Proposition 5.10
Suppose that k is algebraically closed and that g normalizesa copy A of G sa in Cr ( k ) . Then s = 2 and A acts biregularly on A via thestandard additive action. PROOF : We have already observed that s ≥ Y on which A acts effectively and biregularly and Y is either P or Σ n with n = 1. If a groupacts with a dense orbit then the stabilizers are conjugate, so that, if the groupis commutative, its dimension is 2. Therefore, if s ≥
3, then the orbits of A are1-dimensional and preserved by g . However, g is hyperbolic. So s = 2.It remains to show that, whether Y = Σ n or P , there is an open subvariety,homogeneous under A , which is isomorphic to A on which A acts additively.If Y = P then A ⊂ P GL ; conjugating A into the subgroup representedby strictly upper triangular matrices shows that A acts additively on A .If Y = Σ = P × P then conjugating A into a suitable subgroup of P GL × P GL shows the same thing.Recall that, if n ≥
1, then the automorphism group of Σ n is isogenous toa semi-direct product L n ⋊ GL where L n = H ( P , O ( n )) and A is conjugate toa subgroup of L n ⋊ U where U ⊂ GL is the group of strictly upper triangularmatrices. In terms of suitable inhomogeneous co-ordinates ( x, y ) on Σ n , L n is thespace of polynomials in x of degree ≤ n , and the element ( v,
0) of L n × { } actsvia ( v, x, y ) = ( x, y + v ( x )) while the element (0 , α ) of { } × U ∼ = G a acts via(0 , α )( x, y ) = ( x + α, y ). Since A has a dense orbit in Y , A is not contained in L n . It is clear that ( v,
0) and (0 , α ) commute for all α ∈ G a if and only if v is aconstant polynomial, and the result is proved. Proposition 5.11
Suppose that k = F p and that g normalizes a copy A of G a in Cr ( k ) . Then L ( g ) is an integral multiple of log p . PROOF : We know that A acts additively on A . That is, a ( x ) = a + x . Since g normalizes this action, g is biregular on A and we can write gag − = h ( a )for a ∈ A . The equation h ( a )( g ( x )) = g ( a + x ) can be re-written as g ( a + x ) = h ( a ) + g ( x ). Pick an origin 0 ∈ A ; this provides an isomorphism A → A .Setting x = 0 gives g ( a ) = h ( a ) + g (0). Set g n (0) = r n ; then g n ( a ) = h n ( a ) + r n . Since r n is constant, it follows that g n and h n are defined by polynomials ofthe same degrees, so that λ ( g ) = λ ( h ). (Recall that L ( g ) = log λ ( g ) and λ ( g ) = lim n →∞ (cid:0) deg( g n ) /n (cid:1) .) Therefore it is enough to prove the propositionfor automorphisms g of the group variety G a .Choose a finite subfield F = F q = F p n of k over which A and g are defined.Take the non-commutative polynomial ring F [ V ], the quotient of W ( F )[ F, V ](the Dieudonn´e ring of F ) by the ideal ( F ). Then, after fixing an identification A = G a, F , every F -endomorphism of A is a F [ V ]-linear endomorphism Φ of a FFECTIVITY QUESTIONS IN THE CREMONA GROUP F [ V ]-module M of rank 2. Moreover, deg( g ) = p deg(Φ ) , where deg(Φ ) isthe maximum of the degrees, with respect to V , of the entries of a matrix thatrepresents Φ ; this is independent of any choice of basis.Let R denote the restriction of scalars R = R F / F p A and let r : R → R be the automorphism induced by g . Then deg r m = deg g m for all m and, afterfixing a F p -isomorphism R → G na, F p , we can identify r with a 2 n × n matrix Φover the commutative polynomial ring F p [ V ].Taking degrees of polynomials, with respect to V , defines a discrete valu-ation deg : F p ( V ) → Z . We define a norm v on F p ( V ) by v ( f ) = p − deg( f ) . Let K = F p (( V )) be the v -adic completion of F p ( V ), K an algebraic closure of K and C K the v -completion of K ; then C K is algebraically closed, and v extends to anorm on it.We can regard matrices over F p [ V ] as having entries lying in C K . Then v defines a norm on such matrices Ψ , Ω with the properties that v (ΨΩ) ≤ v (Ψ) v (Ω)and v (Ψ + Ω) ≤ max { v (Ψ) , v (Ω) } .Then, by definition and by Gel’fand’s theorem on the spectral radius, λ ( g ) = λ ( r ) = lim m →∞ ( v (Φ m ) /m ) = sup v ( α ) , where α ∈ C K runs over the spectrum of Φ. Since Φ has finite F p [ V ]-rank, thissupremum is achieved by an eigenvalue α that is algebraic over F p ( V ). Thendeg( α ) ∈ Q , so that λ ( g ) = p deg( α ) is a rational power p a/b ≥ p . On the otherhand, by [DF] (see also [BC], Th. 1.2) λ ( g ) is either a Salem number or a Pisotnumber, so is a positive integral power of p .The case where char k = 0, k is algebraically closed and g normalizes acopy of G a does not arise. For then g would be linear, so not hyperbolic. Proposition 5.12 If g normalizes a -torus, then its length L ( g ) equals its lengthas an isometry of the upper half-plane and so is the logarithm of a real quadraticunit. PROOF : The same argument as in the additive case above shows that it is enoughto prove this for the action of g on the torus itself. This appears in [BC], p.4.The next result is now immediate. As we have already observed, given anyoverfield K of k , L ( g ) is the same, whether taken in Cr ( k ) or in Cr ( K ). Theorem 5.13
Assume that char k = 0 or that k is algebraic.Suppose that L ( g ) is not the logarithm of a quadratic unit; if char k = p suppose also that L ( g ) is not an integral multiple of log p .(1) Some power of g is tight.(2) For all sufficiently divisible n , the normal subgroup hh g n ii of Cr ( k ) isproper. PROOF : (1): The previous results can be applied to show tightness over k . Butenlarging the ground field is irrelevant, and the result follows.8 N. I. SHEPHERD-BARRON (2) now follows from Theorem 2.10 of [CL]. (Note the typo in loc. cit. : thephrase “either h is a conjugate of g , or...” should read “either h is a conjugateof g ± , or...”.) Corollary 5.14 If g is a hyperbolic quadratic Cremona transformation and L ( g ) = log( √ ) (if char k = 2 then assume also that L ( g ) = log 2 ) then somepower of g is tight and, for all sufficiently divisible n , the normal closure hh g n ii does not contain g and so is a non-trivial normal subgroup of Cr ( k ) . PROOF : It is only necessary to check that √ is the unique quadratic unitbetween 1 and 2. Now suppose that the ground field k is finite.Proposition 5.15 For every point z ∈ H and every r > , the set of f ∈ Cr ( k ) such that d ( z, f ( z )) < r is finite. In particular, Cr ( k ) acts properlydiscontinuously on H . PROOF : Fix z, r and suppose that d ( z, f ( z )) < r . Let ℓ be the class of a line in P . Then d ( ℓ, f ( ℓ )) ≤ d ( ℓ, z ) + d ( z, f ( z )) + d ( f ( z ) , f ( ℓ )) , so that d ( ℓ, f ( ℓ )) < d ( ℓ, z ) + r. Since cosh d ( ℓ, f ( ℓ )) = ( ℓ.f ( ℓ )) = deg( f ), itfollows that deg( f ) is bounded. Because k is finite, we are done.We know that, over any field, there exist hyperbolic elements g of Cr ( k );for example, a hyperbolic element of SL ( Z ) acting on G m will do. Fix a hyper-bolic element g .Let M = Fix(Ax( g )) denote the set of elements f of Cr ( k ) that fix everypoint on Ax( g ). Lemma 5.16 M is finite. PROOF : The argument involving the triangle inequality that was used in theproof of Proposition 5.2 shows that ( ℓ.f ( ℓ )) is bounded independently of f ∈ M ,and now finiteness follows again. Theorem 5.17
Some power of g is tight. PROOF : We know that all powers of g are rigid. Note that g acts on M byconjugation; choose n > g n is trivial.Put N = Stab(Ax( g )); then N is a semi-direct product N = M ⋊ h γ i ,where γ is a hyperbolic element of N whose length is minimal, and g n is a centralelement of N . So, for every h ∈ N , we have hg n h − = g n , so that g n is tight. Theorem 5.18 If k is finite then each sufficiently divisible power of g generatesa normal subgroup of Cr ( k ) that does not contain g . PROOF : Apply Theorem 2.10 from [CL] once more.I am grateful to Anthony Manning, Geoff Robinson, Caroline Series andColin Sparrow for several valuable conversations and emails, and particularly toSerge Cantat for pointing out and correcting a significant error.
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