Counting fixed points of a finitely generated subgroup of Aff[C]

Publ. Mat. 48 (2004), 127–137 COUNTING FIXED POINTS OF A FINITELY GENERATED SUBGROUP OF Aff[C] F. Loray, M. van der Put and F. Recher Abstract Given a finitely generated subgroup Gof the group of affine transformations acting on the complex line C, we are interested in the quotient Fix(G)/G. The purpose of this note is to establish when this quotient is finite and in this case its cardinality. We give an application to the qualitative study of polynomial planar vector fields at a neighborhood of a nilpotent singular point. Introduction Consider the group of affine transformations acting on the complex line C Aff[C] = {aX +b;a, b ∈C, a 6= 0}. For a finitely generated subgroup G<Aff[C], we denote by Fix(G)⊂C the set of all the points which are fixed by a non-trivial element of G (i.e. whose isotropy group G{c}={g∈G;g(c) = c}is not reduced to the identity {X}) Fix(G) = {c∈C;∃g∈G, g(c) = cand g6=X}. The group Gacts on this set and we denote by Fix(G)/G the set of G-orbits. The purpose of this note is to answer to the following two questions: •When is Fix(G)/G finite? •Suppose that Fix(G)/G is finite, what is its cardinality? Motivations and application More generally, these questions arise naturally for a finitely generated subgroup Gof the group Diff(N) of diffeomorphisms of a smooth manifold Nwhen one wants to study the topology of the leaves of a foliated manifold (M, F) given by suspension of a representation ρ:π1(B, b)−→ Diff(N) 2000 Mathematics Subject Classification. Primary: 34C; Secondary: 11A. Key words. Limit cycles, singularities of vector fields, Riccati equation. 128 F. Loray, M. van der Put, F. Recher where π1(B, b) denotes the fundamental group of another manifold B (see also [G, Chapter I, 2.8]). Let us briefly recall the construction of the suspension (M, F). Consider the canonical representation of the fundamental group σ:π1(B, b)→Diff( e B) in the deck transformations of the universal covering e B. The image of the representation eρ:π1(B, b)−→ Diff( e B×N) γ7−→ (σ(γ), ρ(γ)) acts properly and discontinuously on the product e B×N. The quotient manifold M=e B×N/eρis the total space of a locally trivial bundle Π: M→Bwhose fibers are isomorphic to N. The horizontal foliation on e B×Nwhose leaves are e B× {p},p∈N, is invariant under the action of eρand thus induces a regular foliation Fon M. The leaves of Fare transversal to the fibers and have the same dimension as the basis B. The projection Π induces by restriction a covering of each leaf ΠL:L→Bonto the basis. More precisely, if Lpdenotes the leaf passing through p∈Π−1(b)≃N, any loop γ∈π1(B, b) lifts-up at pas a path ˜γ: [0,1] →Lpjoining ˜γ(0) = pto ˜γ(1) = ρ(γ)(p). The fundamental group of Lpis given by the exact sequence 1−→ ker(ρ)−→ π1(Lp, p)−→ G{p}−→ 1 where G < Diff(N) denotes the image of ρand G{p}={g∈G;g(p) = p}< G, the isotropy subgroup associated to p. In other words, the fundamental group π1(Lp, p) of Lpis isomorphic to ρ−1(G{p}) = {γ∈ π1(B, b); ρ(γ)(p) = p}. Therefore, the leaves Lppassing through generic points p(having trivial isotropy group) are pairwise diffeomorphic (universal property of coverings) with same fundamental group π1(Lp, p)≃ ker(ρ). Finally, the two questions above are related to counting the non-generic leaves, i.e. those leaves having more topology (or having non-trivial holonomy). The number of them is then given by Fix(G)/G and their fundamental group is increased by G{p}. Such foliations naturally appear for instance in the phase portrait of a differential equation of Riccati type y0=a(x)y2+b(x)y+c(x) where a,band cdenote rational functions of x. Denote by Ω ⊂C the complement of the polar set of a,band c. The regular foliation induced in variables (x, y) on Ω ×Cextends as a regular foliation F on Ω ×P1(C) transversal to the vertical fibration and is actually the suspension of the monodromy representation ρ:π1(Ω, x0)→P GL(2,C) (see [H]). The case N=Cand Diff(N) = Aff[C] considered in the present note occurs when the Riccati equation has a rational solution. Counting Fixed Points 129 The monodromy group Gis therefore affine in a convenient projective coordinate. It would be interesting to answer to the same questions in the more general case of P GL(2,C) acting on the projective sphere. This note underlines the surprising complexity of this simple problem even in the affine case and suggests that the projective case will be difficult. Already, the affine case naturally arises when we study the topology of complex trajectories of a singular analytic vector field in the plane having a Liouvillian first integral. For instance, the trajectories of the nilpotent Hamiltonian vector field V0:= y∂x+nx2n−1∂y,n≥2, are completely understood by means of the first integral f0(x, y) = y2−x2n: all trajectories apart from the singular fiber {y2−x2n= 0}have the same topology at a neighborhood of (x, y) = (0,0) ∈C2. Now, consider a perturbation Vof V0having the form: V= (y∂x+nx2n−1∂y) + (α+x)xn−1(x∂x+ny∂y) for a complex number α∈C. As a direct application of the results of this note, we obtain the: Corollary 1. When α∈C−Qand n≥3, the polynomial vector field V above has infinitely many complex trajectories having non-periodic holonomy in any neighborhood of the singular point (x, y) = (0,0). When α∈Q, then Vhas finitely many trajectories having non-trivial holonomy, the number of which depends on the arithmetic of nand α. For instance, when α= 0 and n=pis an odd prime number, there are 1 + p−1 2+2p−1−1 psuch trajectories. In particular, this corollary provides explicit examples of polynomial planar vector fields having infinitely many complex limit cycles at the neighborhood of a nilpotent singular point. Proof: The phase portrait of the vector field Vis defined by the equation ω= 0 where ωis the holomorphic 1-form ω= (y dy −nx2n−1dx) + (α+x)xn−1(x dy −ny dx). In coordinates t=y/xnand z= 1/x, we obtain the Fuchsian Riccati equation dz dt =(t+α)z+ 1 n(t2−1) =(1 + α)z+ 1 2n(t−1) +(1 −α)z−1 2n(t+ 1) . The monodromy group of this equation is affine, generated by the monodromy maps around 1 and −1 g1(z) = e2iπ (1+α) 2nz+c1and g−1(z) = e2iπ (1−α) 2nz+c−1 for some constants c1, c−1∈C. 130 F. Loray, M. van der Put, F. Recher We claim that g1and g−1commute if, and only if, (1+α) 2nor (1−α) 2n belongs to Z− {0}. Indeed, when α6=−1, the Riccati equation has two singular points over the singular point t= 1, namely z=−1 1+αand z=∞. Therefore, the corresponding monodromy map is conjugated to its linear part (which is the identity) if, and only if, (1+α) 2n∈Z. If α=−1, the Riccati equation has a double singular point at z=∞and g1is a translation which obviously does not commute with g−1. Finally, when g1and g−1both have a non-trivial linear part, then they commute, if, and only if, they share a common fixed point in Ccorresponding to a rational solution z(t) of the Riccati equation. One can see from the phase portrait of this equation in (t, z)∈P1×P1that the graph of z(t) cannot intersect the invariant line z=∞, but must intersect the lines t= 1,−1,∞respectively at the singular points z=−1 1+α,1 1−α,0. In other words, the rational function z(t) has no pole on the Riemann sphere and satisfies z(1) 6=z(−1). This contradicts Liouville Theorem and proves the claim. When αis irrational, g1and g−1have non-periodic linear parts and do not commute. Lemma 3 provides infinitely many trajectories having non-trivial holonomy. Moreover, the holonomy groups of those special trajectories are infinite, and contain contractions as soon as α6∈ R. When αis rational, g1and g−1have periodic linear parts and the Riccati equation has finitely many trajectories having non-trivial holonomy by Theorem 2. For instance, when α= 0, we have after conjugacy g1=ζ2nzand g−1=ζ2nz+ 1. Therefore, the monodromy group Gis also generated by g1=ζ2nXand g−1◦(g1)−1=X+ 1 and the number of exceptional trajectories is given by Theorem 4 with m= 2nand r= 1. Finally, assume that Gis not abelian, and its linear part is not real. For instance, this is the case when n≥3 and αis zero or irrational. Then we claim that any complex trajectory of the Riccati equation having at least one non-trivial holonomy map actually contains loops arbitrary close to z=∞with tbounded providing this holonomy. Therefore, this holonomy will occur in any neighborhood of (x, y) = (0,0) as holonomy of a trajectory of the vector field V. In order to prove the claim, it suffices to show that given a point z∈Cfixed by a non trivial element g∈G, and given arbitrary large constant T0, one can find a conjugate g0= ˜g−1◦g◦˜g, ˜g∈G, with a word decomposition g0= (gε1)k1◦(gε2)k2◦ · · · ◦ (gεM)kM∈G, Counting Fixed Points 131 where εm=±1 and km∈Zfor m= 1,2,...,M, having the following property: the new fixed point z0= ˜g−1(z) and all intermediate iterates (gεm)l◦(gεm+1 )km+1 ◦ · · · ◦ (gεM)kM(z0),m= 0,...,M, l= 0,...,km remain T-far from 0. Obviously, the corresponding loop is in the same leaf (˜g∈G) and may be thought, in the fundamental group of the leaf, as a product of the initial one with another loop having trivial holonomy; they are even not cohomologous in general. The new word g0may be obtained as follows. Start with a word decomposition of glike above. Denote by h, h0∈G two translations that are R-independant in G(the linear part of Gis not real) together with a word decomposition. Consider also ˜ h=g◦h◦g−1. Since z=∞is fixed by the generators g1and g−1, there is a sequence ··· > Tn>··· > T1> T0=Tsuch that any point z0outside the Tn+1-ball remains Tn-far from 0 after one iteration of g1,g−1,g−1 1or g−1 −1. Choose T0:= Tnfor a nbounding the length of the words g,h,h0and ˜ h. Therefore, if z0is T0-far from 0, then its orbit under iteration or g,h,h0 and ˜ h(or one of the inverses) does not intersect the T-ball. Now, we choose a large translation hNsuch that z0=hN(z) and its image z00 =g(z0) are T0-far from 0. Notice that the new transformation g0=h−N◦g◦hNfixes z0and admits the word decomposition g0= h−N◦˜ hN◦g. By construction, the iteration of word gon z0remains T-far from 0. Also, if the sequence z00,˜ h(z00),...,˜ hN(z00), h−1◦˜ hN(z00),...,h−N◦˜ hN(z00) = z0 does not intersect the T0-ball, then the full orbit of z00 under h−N◦˜ hN viewed as a word in g1and g−1will stay T-far from 0. If not, then we can use the commutativity of h,h0or ˜ hto re-arrange the word h−N◦˜ hN. For instance, when hand ˜ hare R-independant, this follows from the fact that the T0-ball cannot disconnect the “lattice” L= (z00 +R·h+Z·˜ h)∪(z00 +Z·h+R·˜ h)∈C. The word h−N◦˜ hNcorresponds to a path in Ljoining z00 to its image z0=h−N◦˜ hN(z00). If this path crosses the T0-ball, then it can be replaced by another path in Lavoiding this ball. When hand ˜ hare R-dependant, then we introduce h0and use the lattice generated by h and h0. This ends the proof of the last claim, as well as the corollary. 132 F. Loray, M. van der Put, F. Recher Acknowledgement. We would like to thank the referee who carefully read our paper and motivated us to provide Corollary 1 as a concrete application (with a full proof). The answer to the first question Consider a subgroup G < Aff(C) given by generators G=haiX+bi;i= 1,...,si. Let λ: Aff(C)→C∗;aX +b7→ abe the group homomorphism giving the linear part. Denote by Λ := λ(G) the linear part of Gand by T:= ker(λ:G→Λ) its translation part. We have Λ = hai;i= 1,...,si ⊂ C∗. One can identify Tto a subgroup of C(still denoted by T): T={b;X+b∈G} ⊂ C. From the action of Gby conjugacy on its normal subgroup T, we see that Tis stable under multiplication by elements of Λ and therefore inherits a structure of module over the ring Z[Λ]. In the sequel we will suppose that both Λ and Tare non-trivial, otherwise the two questions we are concerned with become trivial. We take care that, in general, Tis not a finitely generated subgroup of G, but we note that it is finitely generated as a normal subgroup of G. In other words, we claim that Tis a finitely generated module over Z[Λ] (having rank ≤s(s+1) 2). Indeed, the commutator subgroup G0= [G, G] of Gis the subgroup of Tgenerated by all conjugates of the elementary commutators [aiX+bi, ajX+bj], i, j = 1,...,s,i6=jin G. The group G0is thus generated as normal subgroup of G(or, equivalently, as a module over Z[Λ]) by those s(s−1) 2elements. In order to generate T, it suffices to add a set of generators for the quotient T/G0. Since G/G0is a commutative group of rank ≤s, its subgroup T/G0has also rank ≤s. We observe that the map c∈C→G{c}={g∈G;g(c) = c}induces a bijection between Fix(G) and the set of the maximal commutative subgroups Hof Gwith H6=T. Since G{g(c)}=gG{c}g−1, one obtain a bijective correspondance Fix(G)/G ←→ {H < G maximal commutative subgroup, H6=T}/G with the G-conjugacy classes of such subgroups H. The image λ(H)⊂Λ depends only on the G-conjugacy class of H. Counting Fixed Points 133 Theorem 2. With notations above, the set Fix(G)/G is finite if, and only if, the linear part Λof Gis a finite subgroup of C∗. Proof: Suppose that Λ is finite. Then Λ = hζniwhere ζndenotes a primitive nth root of unity and Z[Λ] is equal to the ring of integers Z[ζn] in C. Given a generator g∈Gfor Λ, g=ζnX+t, one may linearize it by a translation and assume without loss of generality that Λ is contained as a linear subgroup in G. Therefore, any element of Gwrites g=ζi nX+t, t∈Tand we have Fix(G) = n−1 [ i=1 1 1−ζi n T. The set of T-orbits of 1 1−ζi nTis equal to the finitely generated module T/(1 −ζi n)Tover the finite ring Z[ζn]/(1 −ζi n). Thus Fix(G)/G is finite. The other implication of Theorem 2 follows from the next lemma. Lemma 3. Suppose that Λis infinite. Then there are infinitely many subgroups of Λof the form λ(H), where H6=Tis a maximal commutative subgroup of G. Proof: Let a∈Λ be an element of infinite order. After conjugation of G by an element of Aff(C), we may suppose that aX lies in G. Choose an integer m > 1. In fact, we prove that, for any integer m > 1, there exists a maximal commutative subgroup H6=Tof Gsuch that λ(H) contains some power of abut ai6∈ λ(H) for i= 1,...,m. The lemma will follow from this. The ring Z[Λ,1 (a−1)(a2−1)···(am−1) ] is finitely generated over Z. Therefore there exists a surjective homomorphism φto a finite field Fq. Set α:= φ(a). Then αi6= 1 for i= 1,...,m and αq−1= 1. Let I⊂Z[Λ] denote the ideal generated by the elements aq−1−1 ad−1for d|q−1 and 1≤d≤m. This is a proper ideal of Z[Λ]: ad−1/∈Isince φ(ad−1) 6= 0 and φsends Ito 0 (φ(aq−1−1) = 0). There exists an element t∈T\IT. Indeed, suppose that IT =T. Let mbe a maximal ideal of R:= Z[Λ] containing I, so mT =T. After localization with respect to S:= R\m, one finds that B:= S−1Tis a finitely generated module over the noetherian local ring S−1Rsuch that mB =B. This implies that B= 0 (Nakayama’s lemma) [La]. Since the elements of Sare not zero divisors on T, one obtains the contradiction that T= 0. Finally, let Hbe a maximal commutative subgroup containing h:= aq−1X+t. Let d≥1 be minimal such that ad∈λ(H). Since we have also aq−1∈λ(H), then d|q−1. It follows that Hcontains an element 134 F. Loray, M. van der Put, F. Recher of the form k:= adX+bwith b∈Tsince aX ∈G. Now kq−1 d=h(if not, there exists a non-trivial translation in Hwhich is a contradiction since His commutative) and thus t=aq−1−1 ad−1b. This implies that d > m and Hhas the required properties. The answer to the second question We have to consider a group Ggenerated by a finite linear subgroup hζnXiof order nand by a translation subgroup {X+t;t∈T} where T⊂Cis a finitely generated Z[ζn]-module of rank r. We note that the module Thas no torsion and thus Tis a projective Z[ζn]-module. For the computation of # Fix(G)/G, we start with two examples. Example 1. Assume that n=pk+1 with pprime and k≥0.As in the beginning of the proof of Theorem 2, one may assume without loss of generality that any element of Gwrites g=ζi nX+t,t∈T. If g∈Ghas a fixed point (g6∈ T), then a convenient iterate g◦g◦ · · · ◦ ghas linear part ζpwith the same fixed point. Therefore, we have Fix(G) = 1 1−ζp ·T and we may consider Fix(G)/T ≃T/(1 −ζp)T as a module over the ring Z[ζpk+1 ]/(1−ζp). We have to count the number of Λ-orbits on T/(1 −ζp)T, where Λ = hζpk+1 i. First, we have Z[ζpk+1 ]/(1 −ζp)≃Z[X]/(Φpk+1 (X), Xpk−1), where Φn=Xpk+1 −1 Xpk−1=X(p−1)pk+···+Xp+ 1 denotes the nth cyclotomic polynomial. The ideal is also generated by pand Xpk−1. After substitution X= 1 + Y, one finds that Z[ζpk+1 ]/(1 −ζp)≃Fp[Y]/(Ypk). In particular, the ideal (1 −ζp) is contained in only one maximal ideal, say m. Let Sbe the multiplicative system S:= Z[ζpk+1 ]\m. Then S−1Tis a free module of rank rover the local ring S−1Z[ζpk+1 ]. Since T/(1−ζp)Tis isomorphic to S−1T/(1−ζp)S−1T, one has that T/(1−ζp)T is a free module of rank rover Z[ζpk+1 ]/(1 −ζp) and we can identify T/(1 −ζp)T≃Fp[Y]/(Ypk)r . Counting Fixed Points 135 We have to count the number of Λ-orbits on Fp[Y]/(Ypk)r where Λ is the cyclic group of order pkgenerated by (1 + Y) modulo (Ypk). The set Fix(G)/T ≃Fp[Y]/(Ypk)r splits into the disjoint union of C(ps) = nc∈Fp[Y]/(Ypk)r ; the Λ-orbit of chas length pso for s= 0,...,k. On the other hand, one observes that the set of orbits having length at most psis given by ts i=0C(pi) = Ypk−sFp[Y]/(Ypk)r and thus s X i=0 #C(pi) = # Ypk−sFp[Y]/(Ypk)r =prps. One concludes that C(ps) contains exactly prorbits if s= 0 and prps−prps−1 psorbits for s= 1,...,k. This leads to the formula # Fix(G)/G =pr+ k X s=1 prps−prps−1 ps. Example 2. Assume that n=pk+1ql+1 with distinct primes p,qand k, l ≥1.Let T⊂Cbe a finitely generated Z[ζpk+1ql+1 ]-module of rank rand let G⊂Aff(C) be the group generated by ζpk+1ql+1 Xand {X+t;t∈T}. As in Example 1, the fixed point of a non-trivial element g∈G\Tis also the fixed point of a convenient iterate of ghaving ζp or ζqas linear part. Therefore, Fix(G) = 1 1−ζp T[1 1−ζq T. The intersection of these two sets is T. Indeed, a 1−ζp=b 1−ζqimplies (1 −ζp)b= (1 −ζq)a. Since the image of (1 −ζq) in Z[ζpk+1ql+1 ]/(1 −ζp) is invertible, one has that ais divisible in Tby (1 −ζp). Let Np,Nq denote the respective numbers of Λ-orbits on the set of the non-zero elements of 1 1−ζpT/T and 1 1−ζqT/T. Then # Fix(G)/G = 1 + Np+Nq. Now we concentrate on the counting of Np. The natural homomorphism Z[ζpk+1 ]⊗Z[ζql+1 ]−→ Z[ζpk+1ql+1 ] is an isomorphism. The same arguments as in Example 1 yield Z[ζpk+1ql+1 ]/(1 −ζp)≃Fp[Y]/(Ypk)⊗Fp[X]/(Φql+1 ).