Algebraic Geometry
Birational permutations
Serge Cantat
Département de mathématiques, Université de Rennes, 35042 Rennes, France Received 26 August 2009; accepted after revision 21 September 2009
Available online 9 October 2009 Presented by Étienne Ghys
Abstract
We prove that every permutation ofPn(K), whereKis a finite field with odd characteristic, is induced by a birational transfor- mation with no rational indeterminacy point.To cite this article: S. Cantat, C. R. Acad. Sci. Paris, Ser. I 347 (2009).
©2009 Published by Elsevier Masson SAS on behalf of Académie des sciences.
Résumé
Bijections birationnelles.Nous montrons que toute bijection dePn(K), pourKun corps fini de caractéristique impaire, est induite par une transformation birationnelle sans point d’indétermination rationnel.Pour citer cet article : S. Cantat, C. R. Acad.
Sci. Paris, Ser. I 347 (2009).
©2009 Published by Elsevier Masson SAS on behalf of Académie des sciences.
Version française abrégée
Soitp un nombre premier. SoitK un corps fini de caractéristique p,etml’entier positif tel que le cardinalq deKsoit égal àpm.Le groupe de CremonaCrn(K)est le groupe desK-automorphismes du corpsK(x1, . . . , xn).
Ce groupe coïncide avec le groupe des transformations birationnelles de l’espace projectifPnK.Sif:PnKPnK est une transformation birationnelle, il existe alorsn+1 polynômes homogènesPi(x0, . . . , xn)de même degrédet sans facteur commun tels quef[x0, . . . , xn] = [P0: · · · :Pn].Le lieu d’indétermination Ind(f )def est l’ensemble fini des points satisfaisant le système d’équations Pi(x0, . . . , xn)=0 pour tout i=0, . . . , n.Si Ind(f )et Ind(f−1)ne contiennent pas de point à coordonnées dansK,alorsf induit une bijection de l’ensemble finiPn(K).Notons
BCrn(K)=
f ∈Crn(K)|Ind(f )(K)=Ind f−1
(K)= ∅
le groupe formé de ces transformations rationnelles. Le but de cette Note est de montrer le théorème suivant : Théorème 0.1.SoitKun corps fini de caractéristique impaire, et soitnun entier supérieur ou égal à2. Toute bijection de l’ensemble finiPn(K)est réalisée par un élément deBCrn(K).
E-mail address:[email protected].
1631-073X/$ – see front matter ©2009 Published by Elsevier Masson SAS on behalf of Académie des sciences.
doi:10.1016/j.crma.2009.09.019
Lorsque la caractéristique de K est égale à 2, nous montrerons quetoute bijection alternée est réalisée par un élément deBCrn(K). Si le cardinal deK vaut 2,toute bijection est en fait réalisable, mais lorsqu’il vaut 2m avec m >1,je ne sais pas si l’on peut réaliser les transpositions.
Remarque 0.2.(1) Ce résultat et la preuve qui en est donnée sont analogues à plusieurs énoncés obtenus par Biswas, Huisman, Kollár, Lukacki¯ı et Mangolte à propos des transformations birationnelles deP2(R)qui n’ont pas de point d’indétermination réel (voir [2,4,5] et [7]).
(2) Il doit également être comparé à l’énoncé de Maubach affirmant que les automorphismes polynomiaux du plan affine induisent toutes les permutations de K2 si la caractéristique deK est impaire, et les permutations paires si
|K| =2mavecm >1 (voir [8]).
(3) La proportion d’entiers nN qui sont cardinal d’un espace projectif Pk(K)avec K corps fini etk entier 1 tend vers 0 comme 1/log(N )lorsqueN tend vers l’infini. In fine, peu de bijections sont donc réalisées par des transformations birationnelles.
1. Introduction
Letpbe a prime number. LetKbe a finite field of characteristicp,letq be the cardinal ofKandmthe positive integer such thatq=pm.The Cremona groupCrn(K)is the group ofK-automorphisms of the fieldK(x1, . . . , xn).
It coincides with the group of birational transformations of the projective spacePnK.
Letf:PnKPnKbe a birational transformation. There existsn+1 homogeneous polynomialPi∈K[x0, . . . , xn], 0in,with the same degreed and without common factor, such that
f
[x0: · · · :xn]
= [P0: · · · :Pn].
The system of equationsPi(x0, . . . , xn)=0, 0in,determines a (finite) algebraic subvariety of the projective space PnK. This algebraic variety is the indeterminacy locus Ind(f )of f. When Ind(f )and Ind(f−1) contain no rational point (i.e. no point inPn(K)), the birational transformationf induces a bijectionf of the finite setPn(K).
We shall denote by BCrn(K)=
f ∈Crn(K)|Ind(f )(K)=Ind f−1
(K)= ∅
the group made of these birational transformations. Restriction toPn(K)defines a morphism f →f
from the groupBCrn(K)to the groupBij(Pn(K))of all permutations of the finite projective planePn(K).Since this plane containsqn+· · ·+q2+q+1 elements, we get a subgroup of the finite permutation groupBij({1,2, . . . , (qn+1− 1)/(q−1)}).
Theorem 1.1.LetKbe a finite field and letnbe an integer withn2. If the characteristic ofKis odd, the morphism f →f fromBCrn(K)toBij(Pn(K))is onto.
In other words, every permutation ofPn(K)is induced by a birational transformation without rational indetermi- nacy point (if the characteristic ofK is odd). If the characteristic ofK is equal to 2,we shall prove thatthe image off →f contains all even permutations.IfK has two elements, then every permutation is indeed realizable as a birational permutation, but I don’t know whether transpositions of the finite set Pn(K)are in the image off →f when|K| =2m,withm >1.
Remark 1.2.This result and its proof are analogous to several statements due to Biswas, Huisman, Kollár, Lukacki¯ı, and Mangolte. They proved that the group of birational transformations of the real projective plane P2(R)without real indeterminacy points embeds as a dense subgroup into the group of diffeomorphisms of the plane, and acts n-transitively onP2(R)for alln0.The interested reader may consult [2,4,5] and [7].
Remark 1.3.This result should also be compared to a similar statement due to Maubach for automorphisms of the affine plane (see [8]). In Maubach’s case, a set of generators forAut(K2)is known and may be used to prove that odd permutations are not realized by automorphisms when|K| =2r, r >1.
Remark 1.4.The proportion of positive integersnN that are equal to|Pk(K)|for at least one pair(k, K)wherek is an integer andKis a finite field goes to 0 whenN goes to+∞.This proportion is 43/100 forN=100.
The proof follows easily from the prime number theorem. IfnNis an integer of typen= |P1(K)|,whereKis a finite field, thennis equal topm+1 for some pair(m, p)withm1 andpa prime such thatpm< N.The num- ber of primespN is approximately N/log(N ).Ifpm< N andm2, thenp√
N andmlog(N )/log(2) because p2. In particular, for each prime p, the sequence pk +1, k =1, . . . counts at most log(N )/log(2) terms before it reachesN.This implies that the number of integersnN of type|P1(K)| is bounded from above by
N
log(N )+log(N ) log(2)
√N .
For integers n N of type |Pk(K)| with |K| =pm and k2 we have p √
N , mlog(N )/log(2) and klog(N )/log(2).This gives at most
log(N ) log(2)
2√ N
terms. Hence, the proportion of integersnNof type|Pk(K)|for some finite fieldKand integerk1 goes to 0 as 1/log(N )whenNgoes to infinity.
2. From projective transformations to arbitrary permutations
The group of automorphisms of the projective spacePn(K)coincides with the group of projective transformations PGLn(K).This group is a strict subgroup ofBij(Pn(K)).The group of permutations ofPn(K)preserving collinearity is an intermediate subgroup, that is usually denotedPLn(K):
PGLn(K)⊂PLn(K)⊂Bij Pn(K)
.
The groupPLn(K)is generated byPGLn(K)and automorphisms of the fieldK.Depending onK,PLn(K)may be contained, or not, in the groupAlt(Pn(K))of alternating permutations (i.e. with signature+1).
Theorem 2.1.(List [6]; Bhattacharya [1].) LetK be a finite field andn >1be an integer. LetGbe a subgroup of Bij(Pn(K))which contains the automorphism groupPGLn(K).EitherGis contained inPLn(K),orGcontains the alternating subgroupAlt(Pn(K)).
The groupBCrn(K)contains the group of automorphisms ofPnK.Hence, all what we have to do in order to prove Theorem 1.1, is to construct a birational transformationf ofPn(K)such that
(1) f and its inversef−1do not have any rational indeterminacy point;
(2) f does not preserve collinearity;
(3) the signature of the permutationf:Pn(K)→Pn(K)is−1.
In what follows, we focus on the first non-trivial case, namelyn=2.The general case is obtained along similar lines.
In order to construct the desired transformationf:P2KP2K,we shall first construct a birational transformation gof a smooth quadricQ,and then transport the construction onto the plane by stereographic projection.
3. A smooth quadric
Lemma 3.1.There exists a field extensionK ofK of degree2,a smooth quadric Q⊂P3K,a lineL⊂P3K,and a rational pointN∈Q(K)(all defined overK)such that:
• the lineLdoes not intersectQ(K);
• the tangent planeTNQintersectsQ(K)in two conjugate linesDandD,but does not intersectQ(K)\ {N};
• the planeTNQcontains the lineL;
• there is a rational pointM∈Q(K)such that the planeHspanned byLandMcutsQon a smooth conic.
Proof. LetK be an extension ofKof degree 2,and letabe an element ofK \K.Since the extension has degree 2, there are elementsu,andvinKsuch thataand its conjugatea are the two roots ofX2+uX+v=0.Let us first assume that the characteristic ofKis different from 2.In this case, we can chooseuto be 0; we then chooseQ⊂P3K to be the quadric
x2+vy2+z2=t2,
where[x:y:z:t]are homogeneous coordinates for the projective spaceP3K.The pointN= [0:0:1:1]is on the conicQ,and the tangent plane toQatN intersectsQon the pair of lines(x−ay)(x−ay)=0.
The line Ldefined byz=t=0 does not intersect Q(K).ForM,we choose the point [1:0:0:1].The plane H which containsLandMis the planez=0.It intersectsQalong the conicx2+vy2=t2.This conic is smooth because the equationX2+uX+vdoes not have any rational root.
When the characteristic of K is equal to 2,the previous formula does not define a smooth quadric, but one can chooseQto be given by the following equationx2+uxy+vy2+z2+x(z+t )+y(z+t )+zt=0. 2
Let us now use this lemma. By a projective change of coordinates, we can and shall assume that N is the point [0:0:1:1], M is the point[1:0:0:1], Lis the linez=t=0,andH is the planez=0,where[x:y:z:t]are the (new) chosen homogeneous coordinates of P3K. The planeH is isomorphic toP2K,with projective coordinates [x:y:t].
LetΦN be the stereographic projection from the poleN to the planeH.By definition, ΦN:Q→H=P2K is a birational map which blows down the two lines ofQthroughN on two distinct points ofP2(K),namely[a:1:0] and[a :1:0],and blows upNinto the line through these two points, i.e. to the lineL.
Letgbe a birational transformation ofQ,which is defined overK.Let us assume thatgdoes not have any inde- terminacy point onQ(K),and thatgfixes both lines tangent toQthroughN pointwise (in particular, the differential ofgatNis the identity). ThenΦN◦g◦ΦN−1does not have any rational indeterminacy point either. We shall use this fact to construct our desired birational transformation.
4. Construction and conclusion
Letπ:P3KP1K be the projection fromP3K to the pencil of planes containing the lineL.In coordinates, π
[x:y:z:t]
= [z:t].
This rational map has indeterminacies along the lineL(i.e. alongz=t=0). This line does not contain any rational point of the quadricQ.If[x:y:z:t]is a point ofQ,we shall denote byH[z:t]the fiber ofπ:P3KP1K through [x:y:z:t]; this is a plane which intersectsQon a conic containing[x:y:z:t].For example, starting with the point M= [1:0:0:1],the planeH[0:1]coincides with the planeHin Lemma 3.1, and intersectsQalong a smooth conic.
LetGbe the group of birational transformations ofP3Kwhich preserveQand the fibers ofπ,acting in each fiber H[z:t]ofπ by a projective automorphism which letsH[z:t]∩Qinvariant. In affine coordinates(x, y, z),witht=1, elements ofGare of the form
g(x, y, z)=
Az(x, y), z
whereA:z→Aztakes its values in the groupAut(P2K)=PGL3(K).Let us now construct an elementgofGwithout indeterminacy point inQ(K).
Being a smooth conic with a rational point, H[0:1]∩Qis isomorphic toP1K (the stereographic projection from a rational point is defined overK). Moreover, if his a homographic transformation ofP1(K)there is a projective transformationh of the planeH[0:1]such thath preserves the conic and the restriction ofhto the conic is conjugate tohby the stereographic projection. We callh aliftofh.
Let cbe a generator of the cyclic group K∗. Multiplication bycdetermines a linear transformation ofK, and therefore a homographic transformationhcofP1(K),namelyhc([a:b])= [ca:b].The multiplicative groupK∗has q−1 elements, and multiplication bycis a cyclic transitive permutation of this set. As a consequence, the signature
of the permutationhcis equal to−1 if the characteristic ofKis odd, and+1 if the characteristic is 2.Lethcbe a lift ofhctoH[0:1].
Remark 4.1.If we replacehcby the translationgc([a:b])= [a+cb:b],wherecis an element ofK,thengcfixes [1:0],and the cardinal of all other orbits is equal to the order ofcin the abelian group (K,+).In particular, gc
is a transposition ifK is the field with 2 elements andc=1. In all other cases, the signature of the permutation gc:P1(K)→P1(K)is equal to+1.
Interpolation in finite fields shows that there is a rational mapA:z→Az,with values inPGL3(K),such thatA0 coincides withhcandAzis the identity for all other values ofz, z= ∞included (see [9], Chapter I.6 for interpolation in finite fields). The rational transformationg∈Gwhich is defined by such a choice forA
(1) is an element of the groupG;
(2) does not have any indeterminacy point inQ(K);
(3) is the identity at the north poleN (fixing the two lines contained inQ(K)throughN pointwise).
As a consequence, the stereographic projectionΦNconjugatesgto an elementf ofBCr2(K),such that (1) f fixes the whole lineLpointwise (the “image” ofN byΦN);
(2) f preserves the imageC of the conicQ∩H[0:1]: this conicC is smooth (isomorphic to P1(K)) and does not intersect the lineL;
(3) f fixes two points onC,and permutes theq−1 remaining points ofCcyclically;
(4) all other points ofP2(K)are fixed byf;
(5) if the characteristic ofKis different from 2,the permutationf ofP2(K)has signature−1,as forhc.
Letmbe one of the fixed points off alongC.Letm be a point ofC which is not fixed byf (such a point exists as soon as|K| =2). The line containingmandm intersectsLon a third pointm .Bothmandm are fixed points off, m is not fixed, andm is the unique point ofC\ {m}on the line(mm ).This shows that
(6) ifKhas more than 2 elements,f does not preserve collinearity.
Hence, if|K|>2, f is an element ofBCr2(K)that does not preserve collinearity. The existence off and Theorem 2.1 show that the image of the morphism
BCr2(K) → Bij(P2(K))
f → f
contains the alternating group. Property (5) concludes the proof of the theorem.
Remark 4.2. Let us assume that K has two elements. With the help of Remark 4.1, one can change hc into the translationg1[a:b] = [a+b:b].Doing that, the bijectionf ofP2(K)becomes a transposition. SincePGL3(K)is 2-transitive, all transpositions are in the image of the morphismf →f ,and the image coincides withBij(P2(K)).
5. Complement
Higher dimension.The proof is the same in higher dimension. In order to construct the required elementf:PnK PnK,we start with a smooth quadricQinPnK+1and a lineLwhich does not intersectQ(K).The family of planes of dimension 2 that containLcutsQalong a family of conics. We may assume that one of this planesH intersectsQ along a smooth conic. The birational transformationgofQis defined as in the previous section: It preserves the family of planes, coincides withhcalongH,and is the identity for the remaining planes. We choose a pointN inQ(K)such that the plane throughN containingLis tangent toQatN.We then conjugategto an elementf ofBCrn(K)by a stereographic projectionΦNfrom the poleN.The mapf does not preserve collinearity, and its signature is−1 ifp is odd or|K| =2.
Question.Let us fix the dimensionnand the fieldK.LetBCrn,d(K)be the finite set of elementsf ∈BCrn(K)which are defined by homogeneous polynomials of degree at mostd.Letσ be an element ofBij(Pn(K)),andN (d, σ )be the number of elementsf inBCrn,d(K)such thatf =σ.What is the asymptotic behaviour ofN (d, σ )/|BCrn,d(K)|?
Ifσ is an element ofBij(Pn(K)),letL(σ )be the length of its longest orbit. What is the expectation ofL(f )forf in BCrn,d(K)? Similar questions have been studied for projective transformations instead of birational transformations (see [3]).
References
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