Infinitesimal Center Problem on zero cycles and the composition conjecture
IINFINITESIMAL CENTER PROBLEM ON ZERO CYCLESANDTHE COMPOSITION CONJECTURE
A. ´ALVAREZ, J.L. BRAVO, C. CHRISTOPHER, P. MARDEˇSI´C
Abstract.
We study the analogue of the classical infinitesimal center problemin the plane, but for zero cycles. We define the displacement function in thiscontext and prove that it is identically zero if and only if the deformation hasa composition factor. That is, we prove that here the composition conjectureis true, in contrast with the tangential center problem on zero cycles. Finally,we give examples of applications of our results. Motivation and the main result
The aim of this paper is to solve the zero dimensional version of the infinitesimalcenter problem. Zero dimensional means that the problem is not in terms of familiesof planar vector fields and one dimensional closed curves (one cycles), but in termsof zero cycles, as we will explain in the sequel. As the problem can be statedin purely algebraic terms, we first introduce the problem and then, we detail themotivation and some applications.1.1.
Infinitesimal center and Hilbert 16th problems.
Given a polynomial f ∈ C [ z ], of degree m ≥
1, a zero cycle C of f is a multivalued function C ( t ) = m (cid:88) i =1 n i z i ( t ) , where the z i ( t ) are roots of f ( z i ( t )) = t , and (cid:80) mi =1 n i = 0. Consider a smallpolynomial deformation f ( z )+ (cid:15)g ( z ) = t of f = t of any degree, and the deformationinduced zero cycle, C (cid:15) ( t ) = m (cid:88) i =1 n i z i ( t, (cid:15) ) , where z i ( t, (cid:15) ), i = 1 , . . . , m , are the multivalued functions determined by ( f + (cid:15)g )( z i ( t, (cid:15) )) = t , z i ( t,
0) = z i ( t ). We shall always assume that the variable t isrestricted to non-critical values of the polynomial f ( z ) so that the deformation iswell-defined. Mathematics Subject Classification.
Key words and phrases. infinitesimal center; tangential center; Abelian integral; compositionconjecture; monodromy.The first two authors are supported by Ministerio de Econom´ıa y Competitividad throughthe project MTM2017-83568-P (AEI/ERDF, EU) and also partially supported by Junta de Ex-tremadura/FEDER Grant Number IB18023. The first and second authors are also partially sup-ported by Junta de Extremadura/FEDER Grants Numbers GR18001 and GR18023, respectively.The last author was supported by Croatian Science Foundation (HRZZ) grant PZS-2019-02-3055from Research Cooperability funded by the European Social Fund and by EIPHI Graduate School(contract ANR-17-EURE-0002). a r X i v : . [ m a t h . D S ] J un A. ´ALVAREZ, J.L. BRAVO, C. CHRISTOPHER, P. MARDEˇSI´C
Definition 1.1.
The displacement function ∆ of the deformation f ( z ) + (cid:15)g ( z ) = t along the zero cycle C is defined by∆( t, (cid:15) ) = (cid:90) C (cid:15) ( t ) f, where by definition (cid:90) C (cid:15) ( t ) f = m (cid:88) i =1 n i f ( z i ( (cid:15), t )) . Note that, since (cid:80) mi =1 n i = 0,∆( t, (cid:15) ) = (cid:90) C (cid:15) ( t ) f = (cid:90) C (cid:15) ( t ) ( f + (cid:15)g − (cid:15)g ) = − (cid:15) (cid:90) C (cid:15) ( t ) g. We formulate two problems:
Problem 1.2.
The infinitesimal center problem for zero cycles:
Caracterize those polynomials f , with their cycles C and deformations g , such thatthe displacement function ∆ of the deformation along C is identically zero. Problem 1.3.
The infinitesimal Hilbert 16 th problem for zero cycles: Bound the number of zeros of the displacement function ∆ of the deformation f + (cid:15)g along cycles C of f as a function of the degree of f ( z ) + (cid:15)g ( z ) as a polynomial in z .The two problems are toy examples of the corresponding two problems for smallpolynomial deformations of Hamiltonian systems in C along one cycles, which canbe traced back to Poincar´e and Arnold [6] respectively. We are considering onlythe polynomial case, but the problems and some of the results can be extended toa more general setting.1.2. Solution of the infinitesimal center problem.
The main result we presenthere is the solution of the infinitesimal center problem for zero cycles. In order tostate our result and to relate it with previous results we need to introduce somenotations.Writing the displacement function in power series of (cid:15) , we obtain∆( t, (cid:15) ) = ∞ (cid:88) i =1 (cid:15) i M i ( t ) . We call M i ( t ) the i -th Melikov function. It is easy to check that the first Melnikovfunction is M ( t ) = − (cid:90) C ( t ) g = m (cid:88) i =1 n i g ( z i ( t )) . That is, a zero dimensional abelian integral for the polynomial g along a zero cycle C ( t ) = (cid:80) mi =1 n i z i ( t ) of f , in the terminology of Gavrilov and Movasati [20].Our problem is analogous to the problem studied by Gavrilov and Movasati in[20], but we study the displacement function ∆ whereas they studied the first Mel-nikov function M . They formulated problems corresponding to Problems 1.2 and1.3 in terms of the first Melnikov function M instead of the displacement function∆. We call their first order problems the tangential problems and reserve the adjec-tive infinitesimal for the study of the displacement function ∆ of a deformation ofthe identically vanishing function. Example 3.1 in the final section illustrates thetwo problems.Gavrilov and Movasati [20] also provided a solution for the tangential Hilbert16th problem and a solution of a special case of the tangential center problem. The NFINITESIMAL CENTER PROBLEM 3 complete solution of the tangential center problem is given in [2] (under a genericcondition) and in [21]. See also [3].A key tool for the solution of the tangential center problem is the compositioncondition. Next, we will define it in our context.Assume that there exist h, ˜ f , ˜ g ∈ C [ z ] such that f ( z ) = ˜ f ( h ( z )), g ( z ) = ˜ g ( h ( z ).Then, for each cycle C (cid:15) ( t ) = m (cid:80) i =1 n i z i ( t, (cid:15) ) of f + (cid:15)g , we define the projected cycle h ( C (cid:15) ( t )) of C (cid:15) by h as the cycle of the perturbation ˜ f + (cid:15) ˜ g = t defined by h ( C (cid:15) ( t )) = (cid:88) h ( z i ( t,(cid:15) )) (cid:88) h ( z j ( t,(cid:15) ))= h ( z i ( t,(cid:15) )) n j h ( z i ( t, (cid:15) )) . We say that this projected cycle is trivial if (cid:88) h ( z j ( t,(cid:15) ))= h ( z i ( t,(cid:15) )) n i = 0 , i = 1 , . . . , m. Definition 1.4.
We say that f, g ∈ C [ z ] and a cycle C of f satisfy the compositioncondition if there exist polynomials ˜ f , ˜ g, h ∈ C [ z ] such that f ( z ) = ˜ f ( h ( z )) , g ( z ) = ˜ g ( h ( z )) , and for every (cid:15) , the perturbed cycle C (cid:15) projected by h is trivial. That is,(1.1) (cid:88) h ( z i ( t,(cid:15) ))= h ( z j ( t,(cid:15) )) n i = 0 , i = 1 , . . . , m. Now, we can state our main result.
Theorem A.
A deformation, f + (cid:15)g = t , has an infinitesimal center for a cycle C of f , i.e., ∆( t, (cid:15) ) ≡ t and all (cid:15) small enough, if and only if f , g , C satisfythe composition condition.The sufficiency of the composition condition is easy. Indeed, by (1.1) we get that∆( t, (cid:15) ) = m (cid:88) j =1 n j f ( z j ( t, (cid:15) )) = m (cid:88) j =1 n j ˜ f ( h ( z j ( t, (cid:15) ))) = (cid:90) h ( C (cid:15) ) ˜ f ≡ . The converse is more difficult and will be proved in the next section.
Remark 1.5.
We note that the composition conjecture is not true for the tangentialcenter problem, as showed by Pakovich’s example [23]. (See Examples 3.1 and 3.2.)It is true, however, for the tangential center problem on simple cycles (Theorem1.7 [16].)
Remark 1.6.
In Theorem A we assume that f, g are polynomials, but the problemcan also be considered for analytic functions f, g . In particular, the function f couldhave an infinite number of isolated fibers z i ( t ), so the cycle sould be considered withfinite support.1.3. Motivation.
The motivation for this paper comes from the study of poly-nomial vector fields in the plane. Orbits of polynomial vector fields are relativelysimple and most interesting are periodic orbits, which can belong to continuousfamilies of periodic orbits, which we call by abuse a center , or can be isolated,called limit cycles .Two widely open classical problems, the center problem (Poincar´e) and the
Hilbert 16-th problem , are related to these two situations.
A. ´ALVAREZ, J.L. BRAVO, C. CHRISTOPHER, P. MARDEˇSI´C
The center problem asks for a geometric caracterization of polynomial vectorfields in the plane having a center. The 16-th Hilbert problem asks for a bound, asa function of the degree, for the number of limit cycles.Each of these problems has its infinitesimal version . One starts with a poly-nomial system dF = 0 having a family of cycles C called a center. Consider itsdeformation(1.2) dF + (cid:15)ω = 0of degree n , where ω is a polynomial one-form. Infinitesimal center problem on -cycles: Find all polynomial deformations (1.2)of dF preserving the center i.e. for which the family of periodic solutions is tran-formed to a nearby family of periodic solutions. Infinitesimal 16-th Hilbert problem (Arnold [6]): Give an upper bound for thenumber of limit cycles born in (1.2) as a function of the degree n .In order to deal with these problems, one considers a transversal T parametrizedby the values of F , the (not necessarily closed) trajectory C (cid:15) ( z ) of (1.2) with endpoints on T deforming C ( z ) and one defines the displacement function ∆ of thedeformation (1.2) by (1.3) ∆( z, (cid:15) ) = (cid:90) C (cid:15) dF = − (cid:15) (cid:90) C (cid:15) ω. For the center problem one searches for a characterization when ∆ is identicallyequal to zero. For the infinitesimal 16-th Hilbert problem, one searches for a boundfor the number of isolated zeros of ∆. Our Definition 1.1 of the displacementfunction of a deformation along zero cycles is directly inspired by (1.3).The previous problems also have their tangential versions, obtained by expandingin series of functions the function ∆ with respect to powers of (cid:15) and asking theanalogous questions for the first term M ( t ) of ∆( t, (cid:15) ).A special case is when the planar system can be reduced to a family of Abelequations, x (cid:48) = A ( t ) x + (cid:15)B ( t ) x , where A , B are trigonometric polynomials. The tangential center problem for spe-cial types of cycles first appeared when studying this family (see [9, 10, 11]) where,for simplicity, Briskin, Fran¸coise and Yomdin considered A, B polynomials ratherthan trigonometric polynomials. In that context, the composition conjecture wasformulated [10], conjecturing that a certain sufficent condition (the compositioncondition) for vanishing of the abelian integrals was also necessary. The composi-tion condition defines all the irreducible components of the center variety in manyfamilies of Abel equations [14], in some planar systems (see [27, 28]), and accountsfor most of the irreducible components when studying the tangential centers ofthe Abel equation at infinity [12]. However, not all centers satisfy the compostionconjecture in the trigonometric Abel equation [4], or even the polynomial Abelequation [22].The tangential center problem on zero cycles also has appeared when studyinghyperelliptic planar systems [16]. In the last section, we show that the infinitesimalcenter problem also appers in these contexts, although in a more general versionthan the problem solved in this paper.2.
Monodromy group of perturbations and proof of the main result
In this section we prove Theorem A. The key of the proof will be to definea convenient monodromy for the deformation. To that end, we start recalling
NFINITESIMAL CENTER PROBLEM 5 the monodromy of a polynomial and then extending it to the deformation of apolynomial.Given a nonconstant polynomial f ∈ C [ z ], recall that z ∈ C is a critical point of f if f (cid:48) ( z ) = 0, and its associated critical value is t = f ( z ). If t is not a criticalvalue, we say that t is regular .We will denote by Σ (cid:48) the set of all critical values of f , which is a finite set.Let m > f . Then, for t ∈ C \ Σ (cid:48) the set f − ( t ) consists of m different points z i ( t ), i = 1 , . . . , m . By the implicit function theorem, one can pushlocally each solution z i ( t ) to nearby values of t thus defining multi-valued analyticfunctions z i ( t ), t ∈ C \ Σ (cid:48) .For a polynomial f of degree m with Σ (cid:48) the set of critical values, each loopbased at t ∈ C \ Σ (cid:48) defines a permutation of the roots z ( t ) , . . . , z m ( t ) of f ( z ) = t .Thus, we have a mapping from the fundamental group π ( C \ Σ (cid:48) , t ) to the automor-phism group Aut ( f − ( t )), whose image forms a group G f , called the monodromygroup of f .The monodromy group G f acts transitively on the fibre f − ( t ) (see [16]). More-over, G f ⊆ S m is the Galois group of the Galois extension of C ( t ) by the m pre-images z ( t ) , . . . , z m ( t ) of t ∈ C \ Σ (cid:48) by f (see, for instance, Theorem 8.12 of [19]).That is, G f = Aut C ( t ) C ( z , . . . , z m ) . The monodromy group G f induces an action on cycles. In fact, reordering thepreimages z i ( t ) appearing in a cycle C after the action of the monodromy, we canconsider the monodromy as acting on the coefficients n i of a zero cycle of f bypermuting them: σ ( C ( t )) = m (cid:88) i =1 n i σ ( z i ( t )) = m (cid:88) i =1 n i z σ ( i ) ( t ) = m (cid:88) i =1 n σ − ( i ) z i ( t ) . Monodromy group of a polynomial perturbation.
Now we define themonodromy group G for a family of polynomials f ( z ) + (cid:15)g ( z ), considered as poly-nomials in z . Let us denote F : CP × C → CP ( z, (cid:15) ) (cid:55)→ F ( z, (cid:15) ) = f ( z ) + (cid:15)g ( z ) , z ∈ C , and F ( ∞ , (cid:15) ) = ∞ and H : CP × C → CP × C ( z, (cid:15) ) (cid:55)→ H ( z, (cid:15) ) = ( F ( z, (cid:15) ) , (cid:15) ) . Let D ( t, (cid:15) ) be the discriminant of f ( z ) + (cid:15)g ( z ) − t as a polynomial in z , and c ( (cid:15) )the coefficient of the monomial of the highest degree. DefineΣ := { ( t, (cid:15) ) ∈ C : c ( (cid:15) ) D ( t, (cid:15) ) = 0 } ∪ {∞} × C . The fiber of every ( t, (cid:15) ) ∈ CP × C \ Σ, consists of n different points ( z i ( t, (cid:15) ) , (cid:15) ),where n is the degree of F as a polynomial in z . By Ehresmann’s fibrationlemma [18], the map H : CP × C \ H − (Σ) → C \ Σ defines a fibration whosefibers are zero dimensional, given by the union of n distinct points z i ( (cid:15), t ).The fact that H defines a fibration guarantees that any closed path γ in thebasis CP × C \ Σ at a point ( t , (cid:15) ) lifts to a path joining two points, z i ( t, (cid:15) ) and z j ( t, (cid:15) ), of the fiber. Moreover, closed paths homotopic to γ lift to homotopic pathsbetween the points of the fiber z i ( t, (cid:15) ) and z j ( t, (cid:15) ). As CP × C \ Σ is path-connected,its fundamental groups with different base poins are conjugated and one can con-sider the basepoint free homotopy group. We obtain a well-defined applicationfrom π ( CP × C \ Σ , ( t, (cid:15) )) to Sym ( H − ( t, (cid:15) )) (cid:39) S n . We call the subgroup G of Sym ( H − ( t, (cid:15) )) thus obtained as the image of π ( CP × C \ Σ , ( t, (cid:15) )) the monodromygroup G of f + (cid:15)g . A. ´ALVAREZ, J.L. BRAVO, C. CHRISTOPHER, P. MARDEˇSI´C
Consider also the projection p : CP × C → C , p ( t, (cid:15) ) = (cid:15) . DefineΣ = { ( t, (cid:15) ) ∈ Σ : dp | Σ=0 . } Note that since Σ is an algebraic variety, the projection of the points where thevariety is orthogonal to the projection is just a finite set. (cid:15)(cid:15)(cid:15)zt H − (Σ)ΣΣ p (Σ ) CP × C (cid:121) H CP × C (cid:121) p C Figure 1.
Diagram of the fibrations
Proposition 2.1.
Given a closed path γ in the basis CP × C \ Σ and (cid:15) ∈ C \ p (Σ ) ,the path γ is homotopic within CP × C \ Σ to a closed path in ( C \ Σ (cid:15) ) × { (cid:15) } ,where Σ (cid:15) is the set of critical values of f ( z ) + (cid:15) g ( z ) .Proof. The proof relies on Thom’s first isotopy lemma (see e.g. [7, p.180]), whichwe apply to the mapping p : W → C \ p (Σ ) , where W = ( CP × C ) \ Σ , is stratified asΣ \ Σ < W \ Σ . The map H restricted to W is proper and its restriction to Σ is a submersion.Therefore, p : W → C \ p (Σ )is a topologicaly locally trivial fibration compatible with the stratification. Weconsider a connection associated with this stratification.Now, consider a closed path in W \ Σ. Then, we can deform the path by theconection without crossing Σ until we obtain a closed path in CP × { (cid:15) } , with (cid:15) (cid:54)∈ Σ . (cid:3) Note that the monodromy induced by the original closed path and the deformedpath are conjugated.For fixed (cid:15) ∈ C , denote by G (cid:15) the monodromy group of f ( z ) + (cid:15)g ( z ). The nextresult shows the relation between G and G (cid:15) . NFINITESIMAL CENTER PROBLEM 7
Corollary 2.2.
For every (cid:15) ∈ C such that c ( (cid:15) ) (cid:54) = 0 , G (cid:15) is a subgroup of G , upto conjugation, and for every (cid:15) (cid:54)∈ p (Σ ) , G (cid:15) is G , up to conjugation. More-over, G is the Galois group of the Galois extension of C ( t, (cid:15) ) by the n preimages z ( t, (cid:15) ) , . . . , z n ( t, (cid:15) ) .Proof. Let us consider a base point ( t , (cid:15) ) ∈ C \ Σ and let G be the monodromygroup induced by closed paths with that base point. The monodromy group G (cid:15) isgenerated by the closed paths based at t , which are in the hyperplane ( C \ Σ (cid:15) ) ×{ (cid:15) } ⊂ C \ Σ. Hence, they generate the same element in G and obviously G (cid:15) ⊂ G .On the other hand, by the previous proposition, closed paths in CP × C \ Σ basedat ( t , (cid:15) ) with (cid:15) (cid:54)∈ p (Σ ) can be continuously deformed into closed paths in thehyperplane { (cid:15) = (cid:15) } . Therefore, permutations corresponding to closed paths in G also belong to G (cid:15) , so G ⊂ G (cid:15) .Let ¯ G denote the Galois group of the Galois extension of C ( t, (cid:15) ) by the n preim-ages z ( t, (cid:15) ) , . . . , z n ( t, (cid:15) ), that is¯ G = Aut C ( t,(cid:15) ) ( C ( z ( t, (cid:15) ) , . . . , z n ( t, (cid:15) ))) . Since the elements of G are automorphisms of the fibers, then G is a subgroup of¯ G . On the other hand, we have the projection ¯ G → G (cid:15) , as any automorphism canbe restricted to a hyperplane. Take (cid:15), (cid:15) (cid:48) (cid:54)∈ p (Σ ), then the following commutativediagram holds G (cid:15) ←→ G (cid:15) (cid:48) (cid:45) (cid:37) ¯ G If the image of an element of ¯ G is the identity element of G (cid:15) , then the diagramabove proves that it is the identity element of G (cid:15) (cid:48) for every (cid:15) (cid:48) (cid:54)∈ p (Σ ) and, inconsequence, the element of ¯ G is the identity element. Therefore, the morphism¯ G → G (cid:15) = G is injective, and ¯ G and G are conjugated. (cid:3) Proof of the main result.
In order to prove Theorem A, we distinguishdifferent cases, given by the following proposition. We will follow [16] (see also [15]or [2] for more details).A permutation group G acting on X is said to be imprimitive if there exists aproper subset B ⊂ X , card( B ) >
1, such that for any σ ∈ G , σ ( B ) ∩ B = ∅ or σ ( B ) = B . If G is not imprimitive, then, it is called primitive . Proposition 2.3.
Let f, g ∈ C [ z ] and let G be the monodromy group of f + (cid:15)g .Then, one of the following cases holds:(i) G is two transitive.(ii) G is isomorphic to the monodromy group of z p with p prime.(iii) G is isomorphic to the monodromy group of T p ( z ) with p prime.(iv) G is imprimitive.Proof. Let c k ( (cid:15) ), k = 0 , . . . , n := max(deg f, deg g ), be the coefficients of the poly-nomial f ( z ) + (cid:15)g ( z ) as a polynomial in z . That is, f ( z ) + (cid:15)g ( z ) = c n ( (cid:15) ) z n + c n − ( (cid:15) ) z n − + . . . + c ( (cid:15) ) , where c k are affine in (cid:15) .Choose (cid:15) such that c n ( (cid:15) ) (cid:54) = 0. If t is large, then z k ( t, (cid:15) ) can be expanded as z k ( t, (cid:15) ) = ω k c ( (cid:15) ) /n t /n + O ( t /n − ) , where ω is a primitive nth-root of unity. Hence, if we consider a cycle with t largeenough, it produces a cyclic permutation of the z k . Therefore, there exists a cycle(1 , . . . , n ) ∈ G . A. ´ALVAREZ, J.L. BRAVO, C. CHRISTOPHER, P. MARDEˇSI´C
Now, applying Burnside-Schur theorem (see [17]), either G is imprimitive or, twotransitive, or permutationally isomorphic to the affine group Aff( p ), where p is aprime. If G is two-transitive, then we are in case (i). Assume G is permutationallyisomorphic to the affine group Aff( p ), where p is a prime.Arguing as in [16], we obtain that for any fixed (cid:15) , the monodromy group of f + (cid:15)g is either isomorphic to the monodromy group of z p or to the monodromy group of T p , for p prime. By Corollary 2.2, the same holds for G for any (cid:15) (cid:54)∈ p (Σ ). (cid:3) The case G two-transitive is generic. In particular, it contains the case when allcritical points are of Morse type with different critical values. We first solve theinfinitesimal center problem in this case. Proposition 2.4.
Assume that for some polynomials f, g , the group G is two-transitive. Let C be a zero cycle of f . Then, f + (cid:15)g has an infinitesimal center forthe cycle C if and only if f, g, C satisfy the composition condition.Proof. Fix (cid:15) (cid:54)∈ Σ . For any fiber z i ( t, (cid:15) ), denote G i the stabilizer of the fiber.Acting on the zero cycle by G i , since G is two-transitive, we get0 = (cid:88) σ ∈ G i (cid:88) j n σ ( j ) f ( z j ) = | G i | n i f ( z i ) + | G i | (cid:80) j (cid:54) = i n j n − (cid:88) j (cid:54) = i f ( z j ) , where n is the degree of f + (cid:15)g as a polynomial in z . Since C is a cycle then n i + (cid:80) j (cid:54) = i n j = 0. Assume that n i (cid:54) = 0, dividing by n i | G i | in the previous equation,we obtain 0 = f ( z i ) − n − (cid:88) j (cid:54) = i f ( z j ) . Assume that n , n (cid:54) = 0 (reordering the roots if necessary). Then,0 = f ( z ) − n − (cid:88) j (cid:54) =1 f ( z j ) − f ( z ) − n − (cid:88) j (cid:54) =2 f ( z j ) = nn − f ( z ) − nn − f ( z ) . Arguing as in Proposition 2.3, by L¨uroth’s theorem, L = C ( h ( z )), so we obtain f + (cid:15)g = F ( h ( z ) , (cid:15) ) for a certain F ∈ C [ z, (cid:15) ], affine in (cid:15) , and the compositioncondition holds. (cid:3) In the proof of Propostion 2.4, we have not used essentially that f and g arepolynomials, so the proposition could be generalized to the class of rational func-tions. Proof of Theorem A.
As we mention is the introduction, the composition conditionis sufficient for f + (cid:15)g to have an infinitesimal center for the cycle C .Now, assume that the deformation f + (cid:15)g has an infinitesimal center for the cycle C . By Proposition 2.3, the monodromy group G of f + (cid:15)g is either two-transitive,equivalent to a monomial z p or a Chebyshev polynomial T p with p prime, or isimprimitive. If G is two-transitive, we conclude by Proposition 2.4. We now dealwith each of the remaining cases.Assume that G is isomorphic to the monodromy group of z p or T p , with p prime.In the first case, f + (cid:15)g has a unique critical value for every (cid:15) (as G (cid:15) is a subgroupof G ), and therefore, f ( z ) + (cid:15)g ( z ) = K ( (cid:15) )( z − a ( (cid:15) )) p for some functions K, a . Butequaling the coefficients, a must be constant, and K ( (cid:15) ) affine, so f ( z ) + (cid:15)g ( z ) = K ( (cid:15) )( z − a ) p . In the second case, arguing analogously, the two critical valuesmust remain constant and therefore, f and g are multiple of the same Chebyshev NFINITESIMAL CENTER PROBLEM 9 polynomial. Note that in all cases, the cycle projected by the composition factoris trivial. To resume, we have obtained that if the group G is primitive, then thecycle can by projected to a trivial one by f .Assume that G is imprimitive. Then there exists a subset of fibers B such thatfor every σ ∈ G , we have either σ ( B ) ∩ B = ∅ or σ ( B ) = B . Assume B containsthe fiber z . Denote G and G B the stabilizers of z and B , respectively. We havethe groups inclusions G ⊂ G B ⊂ G. Since G is the Galois group of f + (cid:15)g , by the fundamental theorem of Galois theory,we have the inclussions of fields C ( (cid:15) )( t ) ⊂ L := C ( (cid:15) )( z , . . . , z n ) G B ⊂ C ( (cid:15) )( z ) . Now, applying L¨uroth’s theorem (see e.g. [26]), we obtain f + (cid:15)g = F ◦ h for certain F, h ∈ C ( z, (cid:15) ), where F, h have degree > z . As F ◦ h is a polynomial, then forevery (cid:15) the preimage of ∞ by h is a point, α ( (cid:15) ), and the preimage of α ( (cid:15) ) by F isagain ∞ . Then, composing with a convenient M¨oebius function (see, for instance,Lemma 3.5 of [16]), we can assume that α ( (cid:15) ) ≡ ∞ , so F, h ∈ C [ z, (cid:15) ]. Since thecomposition F ( h ( z, (cid:15) ) , (cid:15) ) has degree one in (cid:15) , then either F or h does not deppendon (cid:15) . Moreover, as deg F > F must be affine in (cid:15) , that is, F ( z, (cid:15) ) = F ( z )+ (cid:15)F ( z ),for certain F , F ∈ C [ z ].That is, f ( z ) + (cid:15)g ( z ) = F ( h ( z ) , (cid:15) ), were F, h are polynomials and F is affine in (cid:15) , and consider the projected cycle h ( C (cid:15) ). We consider the projected problem, thatis, 0 ≡ (cid:90) C (cid:15) ( t ) f = (cid:90) h ( C (cid:15) ( t )) F ( · , . where the degree of F ( · ,
0) is strictly lower than the degree of f .If the projection h ( C (cid:15) ) is trivial we conclude.If not, we continue projecting until the monodromy group of the projected prob-lem is primitive or until we get a trivial projection. In the first case, as proved above,the cycle can be proyected to a trivial one, so we assume we are in the second case.Assume that h, h , h , . . . , h k are the succesive projections, that F, F , F , . . . , F k are the succesive composition factors, and that h k ( h k − ( . . . h ( h ( C (cid:15) )))) ≡
0. Then, f ( z ) + (cid:15)g ( z ) = F k (( h k ◦ h k − ◦ . . . ◦ h ◦ h )( z ) , (cid:15) ) , and the cycle projected by h k ◦ h k − ◦ . . . ◦ h ◦ h is trivial, so it satisfies thecomposition condition. (cid:3) Remark 2.5.
Recall that the composition condition is not necessary for g to be asolution of the tangential center problem. This is due to the fact that monodromyin the tangential center problem is just the monodromy group of f , and the problemis linear in g , so it is only required that the summands in g satisfy the compositioncondition, while in the infinitesimal center problem, we are considering a groupthat contains both the monodromy groups of f and g , and therefore, compositionfactors considered must be common composition factors of f and g . (See Example3.1.) 3. Examples and applications
In this section, we show some examples and applications to illustrate the problemand its relation with dynamical systems. The first two examples show that theproblem is not trivial, even in simple cases. The third example introduces theinfinitesimal problem for planar vector fields and shows how, for certain planar fields, the one dimensional problem reduces to a zero dimensional problem. Thelast two examples shows the same but for the Abel equation. Moreover, in the lastexample we shall apply the results of this paper to obtain a new proof of a recentresult.
Example 3.1. Tangential problem.
Consider f ( z ) = z , and a cycle C ( t ) = (cid:88) j =0 n j z j ( t ) , z j ( t ) = t e j πi , where n = 2, n = 1, n = − n = − n = − n = 1.There are two possible decompositions of f , with factors h ( z ) = z and h ( z ) = z . Consider a perturbation with g ( x ) = z + z = h ( z ) + h ( z ).The projection of C ( t ) by h consists of identifying the roots z j such that h ( z j )has the same value and assign the sum of the weights. That is, h ( C ( t )) = ( n + n + n ) t / + ( n + n + n )( − t / ) = 0 . Analogously, h ( C ( t )) is a trivial cycle.Recall that ∆ (cid:15) ( t ) = − (cid:15) (cid:90) C ( t ) g + O ( (cid:15) ) . It is known (see e.g. [2]) that for this election of f, g and C , the return map at firstorder is identically zero, as (cid:90) C ( t ) g = (cid:90) C ( t ) h + h = (cid:90) C ( t ) h + (cid:90) C ( t ) h = (cid:90) h ( C ( t )) z + (cid:90) h ( C ( t )) z = 0 . On the other hand, when calculating∆( t ) = − (cid:15) (cid:90) C (cid:15) g = − (cid:15) (cid:90) C (cid:15) h + h , but C (cid:15) is a cycle of f + (cid:15)g , which has no non-trivial composition factors, so theabove argument does not work.Indeed, as we will prove, the second Melnikov function is not zero, so the solutionfor the tangential problem is not a solution of the infinitesimal problem. Indeed, ifwe derive in f ( z j ( t, (cid:15) )) + (cid:15)g ( z j ( t, (cid:15) )) = t with respect to (cid:15) , we obtain ∂z j ∂(cid:15) ( t, (cid:15) ) = − g ( z j ( t, (cid:15) )) g (cid:48) ( z j ( t, (cid:15) )) f (cid:48) ( z j ( t, (cid:15) )) . Then, ∆ (cid:15) ( t ) = (cid:15) M ( t ) / O ( (cid:15) ), where M is obtained differentiating (cid:82) C (cid:15) ( t ) g ( z )with respect to (cid:15) , obtaining M ( t ) = (cid:90) C ( t ) − g ( z ) g (cid:48) ( z ) f (cid:48) ( z ) . But, replacing the roots by its value, M ( t ) = − (cid:88) i =0 n j g ( z j ( t )) g (cid:48) ( z j ( t )) f (cid:48) ( z j ( t )) = − t . Therefore, it is not an infinitesimal center.
Example 3.2. A simple monomial perturbation.
Consider f ( z ) = z , aperturbation of the form g ( x ) = az + bz + c , and a generic cycle C ( t ) = (cid:88) j =0 n j z j ( t ) , z j (1) = i j . NFINITESIMAL CENTER PROBLEM 11
By direct computation, it is easy to obtain that the roots of f ( z ) + (cid:15)g ( z ) = t are z j ( t, (cid:15) ) = ( − (cid:98) j/ (cid:99) (cid:115) − (cid:15)b + ( − j (cid:112) (cid:15) b − (cid:15)c − t )(1 + (cid:15)a )2(1 + (cid:15)a ) . Then,∆ (cid:15) ( t ) = (cid:88) j =0 n j f ( z j ( t, (cid:15) )) = (cid:15)b ( − n + n − n + n ) (cid:112) (cid:15) b − (cid:15)c − t )(1 + (cid:15)a )2(1 + (cid:15)a ) . That is, if b = 0 or n + n = n + n , then ∆ (cid:15) ( t ) ≡ t, (cid:15) ∈ C . In thefirst case, g ( z ) = af ( z ) + c , and in the second case, if we denote h ( z ) = z , then f ( z ) = h ( z ), g ( z ) = ah ( z ) + bh ( z ) + c , and, since n + n + n + n = 0, then n + n = 0 and n + n = 0. In particular, the projection of C (cid:15) by h is identicallyzero. Example 3.3. Infinitesimal center problem and hyper-elliptic planar sys-tems.
Let us consider a perturbed center problem in the plane, that is, con-sider F ∈ C [ x, y ], a polynomial one-form ω , and the deformation of the foliation F ( x, y ) = t , dF + (cid:15)ω = 0 . Take a regular value t of F and γ ( t ) ⊂ F − ( t ) a closed path. Let T be atransversal section to the leaves of F parametrized by t . Consider the one cycle γ (cid:15) ( t ) obtained by deformation of γ ( t ) with respect to t and (cid:15) . Note that γ (cid:15) ( t ) isnot necessarily closed, as we are considering it up to the first intersection with T .Let ∆ (cid:15) be the associated displacement map∆ (cid:15) ( t ) = (cid:90) γ (cid:15) ( t ) dF = − (cid:15) (cid:90) γ (cid:15) ( t ) ω. Then, the deformation preserves the center (foliation with closed paths) defined by γ ( t ) if and only if ∆( t, (cid:15) ) ≡
0, for every t, (cid:15) ∈ C .Now, assume that we are in the hyper-elliptic case in the real plane, that is F ( x, y ) = y + f ( x ), f ∈ R [ x ]. Consider a transversal section T in the axis y = 0and x , t = f ( x ) such that f (cid:48) ( x ) (cid:54) = 0 and the curve γ ( t ) is closed. As for (cid:15) = 0the foliation defined by F ( x, y ) = t consists of closed curves, there exists a firsttime t ( t ) > γ ( t ) belongs to the axis y = 0. We may extend thisfunction to t, (cid:15) by the implicit function theorem. Moreover, as γ ( − t ) also belongto the axis y = 0, we also define a function t ( t, (cid:15) ) as the first negative time suchthat γ ( t ) intersects the axis y = 0. Then, the displacement map can be written asthe zero dimensional integral∆ (cid:15) ( t ) = (cid:90) γ (cid:15) ( t ) dF = (cid:90) ( x ( t ,(cid:15) ) , x ( t ,(cid:15) ) , dF = F ( x ( t , (cid:15) ) , − F ( x ( t , (cid:15) ) , f ( x ( t , (cid:15) )) − f ( x ( t , (cid:15) )) . The tangential version of this problem has been solved in [25] for a vanishing cyclewith a Morse point and in [21] for a vanishing cycle in general. Note that in thisproblem the deformation is not polynomial, so we can not directly apply our results.
Example 3.4. Moment problem.
Around 2000, in a series of papers [8, 9,10, 11], Briskin, Fran¸coise and Yomdim proposed the problem of determing thetrigonometric polynomials a, b such that the following family of Abel differentialequations has a center at the origin for every (cid:15) ∈ R (3.4) x (cid:48) = a ( t ) x + (cid:15)b ( t ) x , where an equation of the family is said to have a center at the origin if everybounded solution is closed.A necessary condition, called composition condition [5], is the existence of func-tions, ˜ A, ˜ B, h , with h (0) = h (2 π ), such that A ( t ) := (cid:90) t a ( s ) ds = ˜ A ( h ( t )) , B ( t ) := (cid:90) t b ( s ) ds = ˜ B ( h ( t )) . Let us denote x ( t, x , (cid:15) ) the solution of (3.4) determined by the initial condition x (0 , x , (cid:15) ) = 0. Assume that for (cid:15) = 0, (3.4) has a center at the origin. That is (cid:82) π a ( t ) dt = 0. Differentiating x − ( t, x , (cid:15) ) with respect to t , evaluating at t = 2 π and denoting t = 1 /x (0), we obtain that (3.4) is a center for every (cid:15) if and only if(3.5) (cid:90) π b ( t ) t − (cid:82) t a ( s ) + (cid:15)b ( s ) x ( s ) ds dt ≡ , for every (cid:15), t ∈ R . By the change of variables t (cid:55)→ z = exp( it ), and complexifying the variables, (cid:73) | z | =1 b ( z ) t − (cid:82) z a ( w ) + (cid:15)b ( w ) x ( w ) dw dz ≡ , for every (cid:15), t ∈ C . By Proposition 3.1 of [1], this is equivalent to (cid:88) n i b ( z i ( t, (cid:15) )) ≡ , for every (cid:15) ∈ R , t ∈ C , where z i are the preimages of t by (cid:82) z a ( w ) + (cid:15)b ( w ) x ( w ) dw , and n i are related tothe branches of A − . Obviously, the functions involved are not polynomials, noteven rational, but we expect some ideas developped here could be used for thatproblem. Example 3.5. Polynomial moment problem.
The same problem can be con-sidered when a, b are polynomials and a closed solution is a solution x ( t ) of (3.4)such that x (0) = x (2 π ). This polynomial version of the moment problem has beenrecently solved by Pakovich [24], proving that if (3.4) has a center for every (cid:15) ∈ R ,then a, b satisfy the composition condition.Pakovich’s solution of the polynomial moment problem is based strongly on thesolution of the tangential version of the problem [25], which ask for centers at firstorder in (cid:15) . More precisely, taking (cid:15) = 0 in (3.5), a necessary condition for (3.4) tohave a center for every (cid:15) is that(3.6) (cid:90) b ( t ) t − (cid:82) t a ( s ) ds dt ≡ , for every t ∈ R . When (3.6) holds, we say that (3.4) has a tangential center.A stronger condition is to consider a ( t ) = a ( t ) + (cid:15)b ( t ), and assume that (3.6)holds for every (cid:15) ∈ R . It was proposed by Cima, Gasull and Ma˜nosas [13]. Theycall it highly persistent center . They proved that if (3.4) has a highly persistentcenter, then (3.4) has a composition center.Now, we show an alternative proof of this result, using Theorem A. Denote f ( z ) = (cid:82) z a ( w ) dw , g ( z ) = (cid:82) z b ( w ) dw , and t = 1 /x (0). By Proposition 8.2 of [2], (cid:90) b ( z ) t − f ( z ) dz = (cid:90) C ( t ) g ( z ) , where C is the cycle C ( t ) = n n (cid:88) i =1 z a i ( t ) − n n (cid:88) i =1 z b i ( t ) , NFINITESIMAL CENTER PROBLEM 13 where z a i ( t ) are all the solutions of f ( z a i ( t )) = t with z a i ( t ) close to 0 for t closeto f (0), and analogously for z b i ( t ) and 1.Now, if a ( z ) = a ( z ) + (cid:15)b ( z ), then f ( z ) = f ( z ) + (cid:15)g ( z ), the cycle C becomes C (cid:15) .So, if (3.4) has a highly persistent center, then (cid:90) C (cid:15) ( t ) g ( z ) ≡ , for every t, (cid:15) ∈ C . By Theorem A, there exists F ∈ C [ (cid:15), z ], affine in (cid:15) , and h ∈ C [ z ], such that f ( z ) + (cid:15)g ( z ) = F ( h ( z ) , (cid:15) ) and h ( C (cid:15) ( t )) ≡
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