Nilpotence of orbits under monodromy and the length of Melnikov functions

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Highlights

  • Orbit depth, nilpotence class, derivative length of deformations are studied.

  • Strategy is given for proving that the orbit depth is unbounded, 1 or 2.

  • This would give that if all deformations are of finite length, then it is 1 or 2.

  • This strategy is realized for all Hamiltonians given by a product of four lines.

  • We relate these results to the length of Godbillon–Vey sequences.

Abstract

Let F[x,y] be a polynomial, γ(z)π1(F1(z)) a non-trivial cycle in a generic fiber of F and let ω be a polynomial 1-form, thus defining a polynomial deformation dF+εω=0 of the integrable foliation given by F.

We study different invariants: the orbit depth k, the nilpotence class n, the derivative length d associated with the couple (F,γ). These invariants bind the length of the first nonzero Melnikov function of the deformation dF+εω along γ. We analyze the variation of the aforementioned invariants in a simple but informative example, in which the polynomial F is defined by a product of four lines. We study as well the relation of this behavior with the length of the corresponding Godbillon–Vey sequence. We formulate a conjecture motivated by the study of this example.

Section snippets

Introduction, main results and conjectures

This work is motivated by the 16th Hilbert’s problem or rather its infinitesimal version. As it is known, the second part of Hilbert’s 16-th problem asks for an upper bound in terms of the degree for the number of real limit cycles, i.e. isolated periodic orbits of polynomial vector fields in the plane. The problem is far from been solved and the existence of such a number is open even for quadratic vector fields.

Arnold formulated the infinitesimal Hilbert’s problem, which asks for a bound on

Nilpotence class and derivative length

Definition 2.1

  • (i)

    Given a group G, let Gi be its lower central sequence: G=G1G2,Gj+1=[Gj,G].If there exists jN such that Gj={e}, we say that the group is nilpotent and define its nilpotence class n=n(G) as n=min{j1|Gj+1={e}},where e is the identity element in G.

  • (ii)

    Similarly, the upper central sequence Gj is G=G0G1,Gj+1=[Gj,Gj].If there exists j such that Gj={e}, we say that the group is solvable and define its derived length d as d=min{j1|Gj={e}}.

Note that Gj+1Gj.

This gives dn and in particular any

Germs of diffeomorphisms

Given a germ of diffeomorphism fDiff(,0), we say that it is parabolic if it is of the form f(z)=z+o(z). If f is not the identity, then f=z+azp+1+o(zp+1), with a0. We call p the level of f. Let Diff1(,0)Diff(,0) denote the subgroup of parabolic germs.

The general approach given in 1.4 is based on the following well-known facts about the solvability of the group of parabolic germs Diff1(,0).

Lemma 3.1

Proposition 6.11, [10]

Let f=z+azp+1+ and g=z+bzq+1+. Then [f,g](z)=z+ab(pq)zp+q+1+o(zp+q+1).

Proposition 3.2

Lemma 6.13 [10]

Let G be a finitely

Proof of Theorem 1.5

Proof

In the case (1), of product of lines in generic position, it has been proved in [8] that [π1,π1]O. Therefore, π1Oab is abelian, and k2. Hence, the orbit depth of the real cycle, as well as the length of the first non-zero Melnikov function for any deformation, is bounded by 2, by [4].

To prove that π1Oab is non-solvable for cases (2) and (3) we follow the same strategy. In both cases, by an affine change of coordinates we can assume that the Hamiltonian is given by F=(x1)(x+1)f3f4, where f3

Types of integrability of the deformations

We will study the Godbillon–Vey sequence for the foliation dF+εω=0,with F=f1f2f3f4 and ω=p1(F)df1f1+p2(F)df2f2, where f1=x1, f2=x+1, p1,p2 are polynomials in F, and f3, f4 are linear factors different from f1 and f2. For the parallelogram, f3 and f4 define parallel lines, while for the trapezoid they do not. By explicit computation of Godbillon–Vey sequences we will show that the foliation dF+εdf1f1+Fdf2f2=0 is Liouville integrable, while dF+εFdf1f1+F2df2f2=0 is Riccati integrable. Moreover,

Proof of Theorem 1.7

Proof

We consider the deformation dF+ε(Fdf1f1+F2df2f2), where f1=x1, f2=x+1, and define η0dF+εFx1+F2x+1dx.

Denote φ(x,F)=Fx1+F2x+1. Then η0=dF+εφdx, and since dφ=φFdF+φxdx, where φF is the partial derivative of φ with respect to F: φF=1x1+2Fx+1, we have dη0=εφFdFdx.Thus, we define η1εφFdx. It satisfies the equation dη0=η0η1. Now we consider dη1=εdφFdx. Then, again, writing dφF=φFFdF+φFxdx, we have dη1=εφFFdFdx. So, we can take η2εφFFdx. It satisfies the equation dη1=η0η2. Continuing with

Declaration of Competing Interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

Acknowledgment

We would like to thank Frank Loray for usefull discussions which helped clarify some points in this article.

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This work was supported by Israel Science Foundation grant 1167/17, Papiit (Dgapa UNAM), Mexico IN110520, Croatian Science Foundation (HRZZ) Grant Nos. 2285, UIP 2017-05-1020, PZS-2019-02-3055 from Research Cooperability funded by the European Social Fund .

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