Versal Normal Form for Nonsemisimple Singularities

The theory of versal normal form has been playing a role in normal form since the introduction of the concept by V.I. Arnol'd. But there has been no systematic use of it that is in line with the semidirect character of the group of formal transformations on formal vector fields, that is, the linear part should be done completely first, before one computes the nonlinear terms. In this paper we address this issue by giving a complete description of a first order calculation in the case of the two- and three-dimensional irreducible nilpotent cases, which is then followed up by an explicit almost symplectic calculation to find the transformation to versal normal form in a particular fluid dynamics problem and in the celestial mechanics \(L_4\) problem.


Introduction
In normal form theory for general differential equations or symplectic systems around equilibria, not much attention is usually given to the linear part of the problem. A typical approach in bifurcation theory is to compute the normal form of a general system with respect to a given organizing center and add versal deformation terms (as first considered in [1,2]). One can then analyze all possible bifurcations in a neighborhood of the organizing center. While there is nothing wrong with this approach, it does not answer the question where a given system fits in the analysis. In other words, how does one compute where the given system is in this neighborhood of the organizing center?
It is this question that we attempt to answer for a number of examples. Some of these examples will be very concrete, with only one or two parameters to give us a possibility to actually, see the bifurcations, others are completely general systems where one can use the computation by just filling in the parameter values of a given system with the same type of organizing center.
Ideally, before starting the nonlinear computation, the linear system should be brought in versal normal form in a finite number of steps, as is attempted in [17]. In practice what one does is to put the linear part in normal form in the same way as one does the nonlinear part of the equation, but this may involve infinitely many steps. Since the linear terms influence the computation in every step, this is not very desirable (contrary to the nonlinear computations, which cannot influence the linear part unless there is also a constant term to take into account).
In this paper, we address this problem for a very particular system that has been the subject of several papers already from the versal deformation point of view, namely the L 4problem as described in [6]. This paper contains a very clear discussion of the arguments involved in the versal deformation computation and we will not repeat these here. The issue we want to address here is to change the infinite series approach into a finite explicit computation. Apart from the L 4 -problem, we have added several examples to illustrate the method and to show that it is indeed a method, not a computation that happens to work in the one example. We treat the 2-and 3-dimensional irreducible nilpotent case in section 3 and 4, respectively. We started this research by computing exponential maps using the generators of the Chevalley normal form of the Lie algebra. In the specific L 4 -problem this leads to quartic equations in the flow parameters and even if one is able to explicitly solve these equations the result is a map full of radical expressions which will be very hard to use if one applies the result to the full nonlinear problem as is our goal. We should mention that in the general linear case this does not occur and one can expect that for simply laced simple Lie algebras this approach will work without problems.
In order to simplify the resulting map that puts the linear system in versal normal form, we then decided to drop the requirement that the symplectic form be preserved. As remarked in [12] there is a strong belief that the symplectic form should be preserved, which is a bit strange if one considers the fact that in order to put the symplectic form in its Darboux normal form, one has to use (by definition) transformations that are not symplectic.
Dropping this requirement, which has anyway no consequence for the further analysis since we work with the symplectic vector fields, not with the Hamiltonians, we then proceed as follows. We first determine a theoretical form of the versal normal form, depending on a finite number of versal deformation parameters. Since we want to reach the versal normal form by conjugation, the characteristic polynomial of the original linear vector field and the versal deformation should be equal. From this equality, we determine the versal deformation parameters (this is in the symplectic case the only nonlinear part of the procedure, in the general linear case this part is completely straightforward).
Once we have, given a linear vector field X ε 0 , which consists of an organizing center X 0 0 plus terms in a neighborhood of the organizing center, in order to compute its versal deformation X ε 0 , we need to solve the linear problem X ε 0 T ε = T εX ε 0 in such a way that T 0 reduces to the identity andX ε 0 is in versal normal form. We then can obtain reasonable expressions for the transformation, which can then be put to good use in the nonlinear normal form analysis.

The algorithm
We start with polynomials R[x 1 , · · · , x n ]. We then add to these commuting derivations ∂ 1 , · · · , ∂ n and consider these as a left R[x 1 , · · · , x n ] module, such that [∂ i , x j ] = δ i j . (One could write ∂ i as ∂ ∂x i ). We write ∂P ∂x i ∈ R[x 1 , · · · , x n ] for [∂ i , P ]. We then define a multi- This defines a non-associative algebra with an associator α(x, y, z) = (x ⋆ y) ⋆ z − x ⋆ (y ⋆ z) which is symmetric in its first two variables (this ensures that the Jacobi identity holds, [9]) and from it we can define a Lie algebra, the Polynomial Lie algebra by defining the Lie bracket as [x, y] = x ⋆ y − y ⋆ x. Apart possibly from the notation, this is the usual way of defining polynomial vector fields. We can put a grading on the polynomial vector field by assigning degree 1 to the x i 's and degree −1 to the ∂ j 's. We remark that the ⋆-product is a graded product, that is, the degree of U ⋆ V is the sum of the degree of U and the degree of V, and this makes the Lie algebra into a graded Lie algebra g = ∞ k=0 g k . Among the elements in this Lie algebra, as special position is reserved for those of degree zero. They form a Lie subalgebra gl(n, R).
We start with a given linear vector field which we consider as an element in a reductive Lie algebra g 0 . In our examples, g 0 will be gl(n, R) or sp(n, R). We chose an organizing center (in all our example this will be characterized by the fact that the real part of all its eigenvalues is zero, since this is where bifurcations happen) and introduce for organizational reasons a deformation parameter ε, which at the end of the computation can be set back to 1. In our first two examples we assume that the organizing center X 0 0 is in real Jordan normal form as is usually done in normal form theory. This is not really necessary and might need the knowledge of the spectrum of X 0 0 , something we try to avoid in this paper, so we stress the fact that the whole construction works well without this choice. Alternatively one might want to put X 0 0 in rational normal form before starting the computation, or not all, as in Section 5. All this is a matter of taste and convenience.
Remark 2.1. This construction determines the style of the normal form, since we will choose ker ad(m 0 ) as the complement to im ad(n 0 ) and costyle of the normal form transformation, since we will choose im ad(m 0 ) as the complement to ker ad(n 0 ). The costyle of normal form transformation is the way we choose the free parameters in transformations. As suggested by the terminology, other choices of style are also possible and may in specific problems be preferable.
The versal normal form should be equivalent to the rational (or Frobenius) normal form of the matrix of X ε 0 , although for that normal form one usually chooses a different style. The computation ofX ε 0 from X ε 0 has already been described. The T ε can be computed by linear elimination. If for some ε 0 , T ε fails to be invertible, then we should take |Y ε 0 | < |Y ε 0 0 |. Definition 2.2. We say that X ε 0 = X 0 0 +Ȳ ε 0 is in normal form (in sl 2 -style) with respect to X 0 0 ifȲ ε 0 ∈ ker s 0 ∩ ker m 0 . We say thatX ε 0 = X 0 0 +Ȳ ε 0 is a versal normal form with respect to X 0 0 ifȲ ε 0 is in normal form with respect to X 0 0 and there exists a T ε ∈ GL(n, R) such that X ε 0 T ε = T εX ε 0 and T 0 0 = I.
If the Lie algebra is defined by an invariant bilinear form Ω 0 (for instance, a symplectic form), one has to compute the induced formΩ ε 0 = (T ε ) t Ω 0 T ε . In this case we write g Ω and gΩ. Similar remarks apply to invariant trilinear forms in the less popular (in dynamics) case g 2 , the Lie algebra of G 2 , cf. [3], not to be confused with an element of grade two in g. This ensures that the versal deformation vector field behaves correctly with respect toΩ ε 0 , that is,X ε 0 ∈ gΩ 0 . Here we trade symplecticness of the maps involved against computational convenience. Definition 2.3. Let T ε ∈ GL(2n, R). Then this induces a new symplectic formΩ ε 0 and a new vector fieldX ε 0 as followsΩ Lemma 2.4. The vector fieldX ε 0 isΩ ε 0 -symplectic iff X ε 0 is an Ω 0 -symplectic vector field. The claim is thatX ε 0 is aΩ ε 0 -symplectic vector field, that is, we have to prove that proving the statement of the Lemma.
The next order step is to compute Then we solve in order to obtainX ε 0 +X ε 1 + · · · in gΩ, or, in the general linear case, in g, where X ε 1 is the first order nonlinear term and, with t ε 1 a general vector field of order 1 andX ε 1 is in normal form with respect to X 0 0 in the sl 2 -style. This procedure can then be repeated until the full system is in normal form up to the fixed degree. The ad(s 0 + m 0 ) ensures that the normal form will automatically have the sl 2 -style with respect to X 0 0 . We should remark here that if we start with a general t ε 1 , there may be free parameters in the normal form corresponding to elements in ker ad(s 0 ) ∩ ker ad(n 0 ) in t ε 1 . This is analogous to the way unique normal forms are computed [4,18]. The free parameters may be used to simplify the normal form by removing (typically) higher order ε-terms. There is no style known to us that would be preferable to this simple free-costyle). In most of our examples the transformation turns out to be in sl 2 -costyle. Remark 2.5. In some problems, when one wants to do the calculations by hand, it pays to view the g k , the polynomial vector fields as representation spaces of g 0 , and more specifically of representation spaces of s 0 , n 0 , h 0 , m 0 . For instance, in [4] the g k is shown to be a direct sum (as vector spaces, not as Lie algebras) of two irreducible representations of sl 2 , a k and b k and this gives rise to a basis that is completely natural with respect to the action of the given sl 2 and such that [z k , z l ] ⊂ z k+l for z = a, b.
As formulated, the algorithm follows what might be called the rational approach: no eigenvalues need to be computed, only characteristic polynomials, cf [5]. This makes it suitable not only for Computer Algebra Systems, but also for Symbolic Formula Manipulation Systems like FORM [13] or FERMAT [14], which is nice if the problems get big.
An alternative method, which might also work when the vector fields are not finitely generated at any given order and might be called the spectral approach, is to use the spectrum of s 0 and h 0 , as is done in the averaging method; we refer for this method to [16].

Nonlinear nilpotent versal normal form
Lemma 2.6. For given X k ∈ g k , k > 0, and parametric vector fieldX ε 0 = s 0 + n 0 +v ε 0 in whichv ε 0 ∈ ker ad(m 0 ) ∩ ker ad(s 0 ) there exists a transformation t ε k ∈ g k to the following problem whereX ε k ∈ ker ad(m 0 ) ∩ g k . The transformation t ε k and the normal formX ε k can be found explicitly from equations (2.7) and (2.8), respectively.
Proof. It should be noted that this proof follows (but with some minor corrections and clarifications) the proof given in [17,Section 2.3].
Our problem is that to find the admissible transformation t ε k and the obstruction term X ε 0 ∈ ker ad(m 0 ) ∩ g k such that the following hold From [16, the procedure is given to solve the following linear problem Denote the transformation t 0 k in equation (2.6) byN X k . Hence from the fact that V = ker ad(m 0 ) ⊕ im ad(n 0 ) one has ad(n 0 )N = π im ad(n 0 ) = 1 − π ker ad(m 0 ) .
Note that the notationNX k shows that the operatorN acts on X k . Let now Q = ad(s 0 +v ε 0 )N andQ =N ad(s 0 +v ε 0 ). We will show that Q andQ are nilpotent operators, so that (1 + Q) −1 and (1 +Q) −1 are both well defined. Observe thatN Q =QN . Proof. We computē and the Lemma is proved.
We claim that t ε k is given by Therefore we have to first show that Q andQ are nilpotent and X k − ad(X ε 0 )t ε k ∈ ker ad(m 0 ) ∩ g k . Assume that the X k has ad(h 0 )-eigenvalue λ; then theNX k has ad(h 0 )-eigenvalue λ + 2 since ad(h 0 )n 0 = −2n 0 . By The proof forQ is the almost the same. It follows that 1 + Q and 1 +Q are invertible. What remains to be done is to show We rewrite this as This concludes the proof of Lemma 2.6.

Nonsemisimple versal normal form
We now extend the versal normal form computation problem from the nilpotent to the nonsemisimple case. We follow [17, Section 2.4]. We consider the problem We observe that the right hand side is by definition in ker ad(m 0 ) ∩ im ad(s 0 ) and ad(s 0 + n 0 ) is invertible on this subspace. We define operators K k : ker ad(m 0 )|g k → ker ad(m 0 )|g k such that K k = I ker ad(m 0 )|g k for ε = 0. Let The projection on ker ad(n 0 ) is necessary, in order not to interfere with the previous normal form calculation in Section 2. We now show that K k : ker ad(m 0 )|g k → ker ad(m 0 )|g k : when the perturbation is zero and this reduces to 1 on ker ad(m 0 ). This in turn implies thatK k is invertible in a neighborhood of ε = 0, which means we can find a transformation generator to bringX ε k into the normal formX ε k . The values of ε for whichK k fails to be invertible are called resonances; they play a role in the bifurcation analysis of the L 4 -problem, cf. Section 6.
The method we describe here does prove that it is possible to compute the transformation explicitly and if the dimension of g k is a bit higher, it may help to reduce the dimension of the linear algebra problem, since one can restrict to ker ad(m 0 ).

The versal normal form of the linear system
In this section, we intend to study the versal normal form of two-dimensional nilpotent singularities. We use this example to illustrate the method in great detail. This leads at times to statements that sound a bit simplistic; these are nevertheless stated explicitly so that it is clear what the flow of the argument is in the later examples, where the complexity of the calculation can obscure what is going on.
Consider the following two-dimensional perturbed singular system.
where we regardm i,j for all i, j = 1, 2 as elements of a commutative ring R of functions of certain parameters taking their values in R (since we want to work with real differential equations) andm 2,1 ∈ R * where R * denotes the invertible elements in the ring R. Invertible in this context means that if we use asymptotic estimates, dividing by an invertible element does not produce big numbers, which could ruin the asymptotic estimate. As a consequence one is not allowed to divide by the noninvertible elements in the course of the normal form computation.
Sincem 2,1 is invertible, there exists an invertible linear transformation (where m 1,1 =m 1,1 , m 2,2 =m 2,2 , and m 1,2 = −m 1,2m2,1 ), so that −X 0 0 is in Jordan normal form (the minus sign is there to be consistent with the definitions of the A and B-families to follow shortly).
We now rewrite Equation (3.2) to the operator form and express X ε 0 to the A and B families introduced by [4] (but with A and B interchanged) as We now want (this is the choice of normal form style) (the fact that the ε A are on the diagonal and will stay there if we go to higher dimensions prompted the interchange of A and B with respect to the definitions in [4]) and the differential operatorX We want to find the transformation that is named The necessary condition under which such transformation exists is that the characteristic polynomial of X ε 0 andX ε 0 be the same. In what follows using the characteristic polynomial of X ε 0 and X ε 0 we find the ε A , ε B . The characteristic polynomial of X ε 0 andX ε 0 are given, respectively by We define the invariants of X ε 0 as and we identify ∆ 1 as the trace of X ε 0 and ∆ 2 as the determinant. Since the equivalent matrices have the same characteristic polynomial then we find that We close this part by the following theorem (this is not much of a theorem in this particular problem, but we formulate it as such because it is a basic step in this paper).
Proof. The transformation (3.7) is obtained using equation (1) and the existence of a solution is shown here explicitly.

Some representation theory
Following [4] we describe vector fields of arbitrary order in a bigraded infinite dimensional Lie algebra a ⊕ b, where a and b are bigraded Lie subalgebras and the ⊕ denotes the direct sum of modules, not of Lie algebras, as can be seen from the Lie brackets below, and spanned by elements We can now write an arbitrary order s vector field as A general element of order s in ker ad(B −1 0 ) can be written as

Nonlinear normal form reduction
We now have to solve the equation (in t ε s ) Recall thatX We have to solvē Thus we find, if we look at the B s+1 s -term, that where β s+1 s is a free parameter, to be determined later at our convenience. Similarly, looking at the A s s terms we find where α s s is the free parameter. For 0 ≤ k ≤ s we find, looking at the B k s , Let us now specialize to s = 1. We find Choosing β 2 1 = 0 and α 1 1 = 0 (in accordance with the sl 2 -costyle) we find It follows from the definitions in [4] that where the elementsm 2,1 ,m 3,2 ∈ R * . By applying the following invertible transformation where α = − 1 2 ε 2m 3,1 (εm 2,3m3,1 +m 2,1m2,2 ) + 1 2m 2,1 ε 2m 3,1m3,3 +m 2,1m3,2 , the system (4.1) transforms to the following system in which Remark 4.1. Note that due to the assumptionm 2,1 ,m 3,2 ∈ R * the transformation given by (4.2) when ε = 0 is invertible. Now, we writing down (4.3) in terms of vector fields from A , B, C given in [8] to find the following Due to sl 2 -style normal form, in order to find the versal normal form of X ε 0 we seek the vector fields which belong to ker ad(B −1 0,0 ). Hence the following special structure constants associated to the B −1 0,0 are given Therefore we obtain thatX 4) and the correspondence differential equation ofX ε 0 is (4.5) Now we are ready to find the versal parameters ε A , ε B and ε C . As before by computing the characteristic polynomial ofX ε 0 andX ε 0 we get These define the invariants ∆ i , i = 1, 2, 3 by Hence we find Theorem 4.2. There exists invertible transformation T ε (1) as in which which brings the matrix (4.3) to (4.5).
Proof. The transformation T ε (1) is obtained using equation
Proof. In order to find the transformation the following linear system should be solved or in the different basis it equals to By solving Equation (4.11) one can find the coefficients of transformation t ε 1 as are given in Appendix A, see Equation (A.1). On the other hand, by solving the equation below we find the coefficients of normal form which has four free parameters as α 2,0,0 , α 2,0,0 , α

An example on sp(4, R)
In [15,Equation (48)] the versal deformation problem is studied using formal power series. We refer to this paper for more references to the literature and a general introduction of the importance of the versal deformation in applied mathematics. We mention that to keep things simple, we use an almost symplectic map to obtain the versal normal form, a trade off we have discussed in Section 2.
In this section we shall find the near identity transformation T ε as discussed in the previous section to bring the symplectic matrices given by [15,Equation (48)], describing oscillations of a simply supported elastic pipe conveying fluid, to its versal normal form. Set where p 1 , p 2 are two real parameters. Define The normal form of X ε 0 consists of those elements which are in ker ad(m 0 ); in fact Hence the normal form is given bȳ The characteristic polynomial of X ε 0 given in the (5.1) andX ε 0 are as χ(X ε 0 ) = λ 4 + 10 ε 1 λ 2 + 9 ε 1 2 + 54 ε 2 .
Remark 5.2. In this example, we did not put the X 0 0 into the symplectic normal form.
6 Three body problem 6.1 The versal normal form at L 4 In the theory of the restricted three body problem, the Langrange equilibria play a very practical role, since they are used to park satellites in orbit, as has been the case for L 1 and L 2 . The Trojan points L 4 and L 5 are considered as positions for space colonies, since they are stable, unlike L 3 which only made it into science fiction sofar. Consider (Cf. [6, Equation 1.8]) the four-dimensional L 4 -singularity The bifurcation point of X δ 0 is δ = δ 0 := 4 √ 2 3 . Set δ = δ 0 + ε to find The normal form of (6.1) is given bȳ where ε N , ε S are obtained as follows. The characteristic polynomial of X ε 0 is Comparing the characteristic polynomial of X ε 0 with characteristic polynomial ofX ε 0 as Then equations (6.3) and (6.4) imply that Now, by solving Equation (6.6) respect to ε N we find two solutions. The negative root is the right solution, since for ε = 0 we have hence, Now substitute ε N into (6.5) and solve for ε S to get Here also we choose the negative root, since from ε = 0 we obtain Note that to have a real normal form we should make this restriction: ≈ 0.114381918.
6.2 Finding the transformation generator t ε 0 Let t ε 0 ∈ GL(4, R) We solve the following equation for {t i , i = 1 · · · 16}. The solutions of above equation respect to four free parameters t 1 , t 2 , t 5 , t 6 are given in Appendix B. Now we should find parameters t 1 , t 2 , t 5 , t 6 . By substituting parameters in t ε 0 and Taylor expansion around ε = 0 we find Due to Equation (6.8) transformation T ε should be near identity. Hence, it requires t 2 = −t 5 , t 1 = t 6 + 9 4 . We have two free parameters t 5 , t 6 which can be taken as t 5 = 0, t 6 = − 9 4 . Hence t 1 = t 2 = 0. Thereby we find Theorem 6.1. The following transformation takes X δ 0 to its versal normal fromX ε 0 through of equation Note that t 4 , t 7 are in order ε and t 10 , t 11 , t 13 , t 16 are in order ε 2 .

Concluding remarks
We have shown that the correct implementation of versal normal form in normal form computations is possible. It does give, and this was to be expected, an added level of complexity. In any practical computation, this will have to be balanced against the added level of correctness. It will be interesting to see whether these considerations can also be applied in practice to the theory of unique normal form. This would, after all, be the holy grail of normal form theory: unique versal normal forms!