The space of polynomial differential equations of a fixed degree with a center singularity has many irreducible components. We prove that pull-back differential equations form an irreducible component of such a space. The method used in this article is inspired by Ilyashenko and Movasati’s method. The main concepts are the Picard-Lefschetz theory of a polynomial in two variables with complex coefficients, the Dynkin diagram of the polynomial and the iterated integral.
Center conditions: pull-back of differential equations 111 Keywords: Holomorphic foliations, Picard-Lefschetz theory
Let be the set of polynomials in the two variables , and coefficients in of degree less than or equal to . The space of algebraic foliations
is the projectivization of the vector space , and it is denoted by . The maximum degree of the polynomials and is known as the (affine) degree of . The space is a rational variety by taking the coefficients of polynomials as the coordinates of affine variety for some . The set of singularities of the foliation is . If , for an isolated singularity of , then is called reduced singularity. If there is a holomorphic coordinate system in a neighborhood of a reduced singularity with , such that in this coordinate system
then the point is called a center singularity. The closure of the set of algebraic foliations of fixed degree with at least one center in , which is denoted by , is an algebraic subset of (see for instance,  and ). Identifying irreducible components of is the center condition problem in the context of polynomial differential equations on the real plane. The complete classification of irreducible components of is done by H. Dulac in  (see also  p.601). This classification gives applications on the number of limit cycles in the context of polynomial differential equations on the real plane. Ilyashenko in , by computing tangent space at some smooth points of the space of Hamiltonian foliations , , proved the following:
The space of Hamiltonian foliations of degree forms an irreducible component of .
H. Movasati in , by computing the tangent cone at a special point proved the following:
The space of of logarithmic foliations
is an irreducible component of .
Let be the set of foliation
For a generic morphism and foliation , there exist a leaf of such that it has an intersection with at some points with multiplicity 2, where is the curve . Therefore, has a center singularity.
The space of pull-back differential equations
forms an irreducible component of .
This paper is inspired by Ilyashenko’s paper  and H. Movasati’s paper  and a sketch of our proof is the following:
Consider a generic and a generic polynomial of degree . It is clear that the point is in the intersection of and of the algebraic set . It is needed to show that the tangent cone of at the point is equal to , in order to prove Theorem 0.3. The proof will be explained in sections 1,2 and 4.
In §1, by taking the deformation where of , and using Petrov module concept, we show that there is a polynomial 1-form with degree and a polynomial such that is of the form . This paper is organized as follows:
In §2, we are going to calculate the explicit form , by using the iterated integral and Melnikov function . This gives us the proof of Theorem 0.3
In §3, we see some applications of theorem 0.3. We found a maximum lower bound for the cyclicity of a tangency vanishing cycle in a deformation inside which is dependent on a factorization of to two natural numbers.
In §4, we study the action of the monodromy group on a tangency vanishing
cycle in a regular fiber .
1 Pull-back of differential equations
Inspired by H. Movasati’s method (see ), we will calculate the tangent cone of at the point in the intersection of Hamiltonian and pull-back algebraic differential equations. Similar to  and  our methods are based on Picard-Lefschetz theory for the foliations with a first integral.
Let be a foliation of degree , and be a morphism, where and . If a point is the tangent point of and a leaf of the foliation then a point in is called a tangency critical point of the foliation .
Consider the deformation
of the foliation . Let be one of the tangency critical points of foliation . For a generic 222By generic we mean always a non-empty Zariski open subset of the ambient space. choice of and , if the deformed foliation for all small has center singularity near , then is also a pull-back foliation. More precisely, there is a foliation and a polynomial map such that
1.1 Tangent space
The set is an irreducible algebraic subset of (by taking the coefficient of the polynomials as coordinates of the map from the space of polynomials with degree to the projective space). We are going to show that is also a component of . Let us take a point of , then make a deformation and calculate the tangent vector space of at :
For a smooth point of , the tangent space of at is just the set of all vectors , which is contained in the tangent space of at , and in order to prove our main theorem it is enough to prove that the equality happens. Now, we are going to compute the tangent cone of at the point in the intersection of Hamiltonian component and the set .
1.2 A foliation in the intersection of two algebraic sets
Let be defined by
where , and are Morse functions. Let be two polynomials of degree defined by
and meet the following conditions:
All are positive real numbers,
Both equations and have real roots ,
The functions and are Morse, which is a holomorphic function with no degenerate critical points.
If is a critical point of (resp. ) and (resp. ) where is a critical point of (resp. ), then (resp. ). In fact, by moving the roots of and on the real line this is the assumable definition.
Let be defined by
We can suppose that the intersection of the set of the critical values of and is empty. The foliation has three kinds of singularities :
Pull-back of centers of ,
Tangency critical points of the foliation ,
The points in .
Let be the irreducible component of containing .
Consider the deformation
Assume that belongs to . This implies that always has a center singularity near a fixed tangency center of . The set of all differential forms is the tangent cone of at . Note that taking is not sufficient for calculating the tangent cone.
Let be a continuous family of the vanishing cycles around a tangency critical and be a transverse section to at some point of . We are able to write the Taylor expansion of the deformed holonomy
Here is the i-th Melnikov function of the deformation. Since for , then
If is parametrized by the image of ,i,e. , then
See for instance .
The morphism is surjective and is a group generated by the action monodromy group on a vanishing cycle around a tangency point.
We will prove this theorem at the end of §4, see Theorem 4.10.
1.3 Brieskorn lattice/Petrov Modules
Consider the Brieskorn lattice/Petrov module
where and are -module and -module respectively, (here and , and also , are the set of polynomial differential forms in .).
A polynomial of degree with homogeneous leading part is called transversal to infinity, if factors out as the product of d pairwise different linear forms.
Consider the Milnor module
with the basis (see e.g. ( chapter 10)). We define
Let be a polynomial transversal to infinity of degree . The -module is free and , where , forms a basis of . Furthermore, every can be written
Pull-backs of for all and are independent in under the map and can be extended to a basis for .
The map is injective, and are linear independent. We have , and the coefficients of are in . By Theorem 1.3 we can write
We are going to show that the functions are constant. We know that and in . By changing the coordinate assume that , therefore
since then the functions ’s are constant. It means that if we write , then all of the coefficients are in . In other words, for all can extend to a basis of . ∎
1.4 Relatively Exact 1-form
Let be a foliation in . If the restriction of a meromorphic 1-form on to each leaf of is exact, then it is called relatively exact modulo , i.e. there is a meromorphic function on L so that .
Note that a meromorphic 1-form is relatively exact modulo if and only if
for all closed cycles in the leaves of .
Every relatively exact polynomial 1-form in of degree modulo a Hamiltonian foliation has the form
where and are polynomials so that divides , and .
See e.g. ( Theorem 4.1.).
1.5 Computing the Tangent Cone
It is not necessary to start from one. Then from the equality (6) we have the following :
There is a polynomial differential 1-form with and a polynomial such that
where is defined by as in (2).
For a regular value of the function , it is clear that the liner map
is surjective. Then
is injective. According to Theorem 1.2, for all , this implies that the linear map
for an element , is well defined. By duality of de Rham cohomology and singular homology there is a differential form in such that
By using Atiyah-Hodge theorem (see e.g. ) the form can be taken algebraically. All these ’s give us a holomorphic global section of cohomology bundle of outside the critical values of ;
We are going to show that it is a holomorphic global section in the whole .
By the Theorem 1.3 we can write
The Periodic matrix
is invertible, where is the rank of , (see e.g.  Proposition 26.44). Therefore, the ’s coefficients are meromorphic functions on , because
and by Theorem 10.7 in Chapter 10 of  each integral for a natural number and close to singular value . Thus, all the elements of the matrices on the right side of the equality have finite growth at critical values. This implies that, there is a polynomial such that is a holomorphic form. We can write , then in . According to Proposition 1.1 the set of for all can be extended to a basis of . Therefore, we have
Since each element of can be written uniquely as a linear combination of the elements on this basis, then for all . In other words, , hence . This implies that is a holomorphic 1-form. By Theorem 4.5, the degree of in the equation (8) is less than or equal to , hence are constant for all . To find the form of we use the Proposition 1.2 and we conclude that for all cycles in the fibers of . This implies that is relatively exact modulo , then by Proposition (1.2) there are polynomials and such that . The fact that implies , so we get our desired equality. ∎
2 Higher order Milnikov function
L.Gavrilov in  has shown that the higher order Melnikov functions can be expressed in terms of iterated integrals. Basic properties of iterated integrals are established by A. N. Parsin in 1969 and a systematic approach for de Rham cohomology type theorems for iterated integrals was made by K. T. Chen around 1977. H. Movasati and I. Nakai in  used the concept of higher order Melnikov functions by iterated integrals.
Let be a piecewise smooth path on . Let be smooth 1-forms on , for the pulled-back of the forms to the interval [0,1]. Recall that the ordinary line integral given by
does not depend on the choice of parametrization of .
Iterated integral of along the path is defined by
Let us consider the deformation
of . The deformed holonomy along the path in is
Since where , then . By using Theorem 3.2 in  (Higher order approximation), we conclude that where , and also
Note that the vector in the case is of the form
The polynomial in the Theorem 1.4 is of the form
Here is a point in the cycle . Now the equality and a similar argument as in the last lemma implies that
Since is a 2-form like , then
This implies that is in the ideal
Now consider the radical ideal and , then it is clear that and , so
We can assume that the curve does not pass any critical point of . By our hypothesis and are generic so we have . This means that thus we have
which states that therefore,we get the result
The point is in , so tangent cone of at the point is
2.1 Proof of Main theorem
Consider a germ of an analytic variety (,0) in (,0).
The analytic path has the Taylor expansion , . Let be the set of all . The tangent cone of at 0 is .
The tangent cone is an algebraic set with pure dimension , i.e. each irreducible component of is of dimension . If 0 is a smooth point of then is the usual tangent space of at 0.
The variety is parametrized by
and so it is irreducible.
Let be an irreducible component of such that . If , where is the irreducible component of , then it must be a subset of , because the equality (10) is union decomposition of to irreducible component. This implies that . Dimension of is equal to dimension , so . Therefore, because and are irreducible algebraic sets and they have the same dimension. ∎
3 Limit cycles
Consider a real planer 1-form where and are polynomials of degree less than or equal to . Let the foliation induced by the 1-form