\magnification = 1200

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\font\sixrm = cmr6
\font\small = cmr9

\def\titlea#1{ \centerline{ \tafont {  {#1}}}
 \vskip 3truecm }
\def\titleb#1{ \bigskip \bigskip {\tbfont {#1}  } \bigskip}

\parindent=0pt
\parskip=10pt
 
\def\G{{\cal G}}
\def\L{{\cal L}}
\def\Lk{{\L_{(k)} }}
\def\S{{\cal S}}
\def\M{{\cal M}}
\def\C{{\cal C}}
\def\F{{\cal F}}
 
\def\phi{\varphi}
\def\eps{\varepsilon}
\def\d{\delta}
\def\D{\Delta}
\def\ga{\gamma}
\def\Ga{\Gamma}
\def\sig{\sigma}
\def\al{\alpha}
\def\la{\lambda}
 
\def\pa{\partial}
\def\x{\times}
\def\o+{\oplus}
\def\Ker{{\rm Ker }}
\def\Ran{{\rm Ran }}
\def\ad{{\rm ad}}
\def\Lra{\Longrightarrow}
\def\LLRA{\Longleftrightarrow}
\def\sse{\subseteq}
\def\ss{\subset}
 
\def\R{\bf R}
\def\Poi{Poincar\'e }
 
\def\~#1{{\widetilde #1}}
\def\^#1{{\widehat #1}}
\def\dsum{{ \sum_{k=0}^\infty }^\o+ }
 
{\nopagenumbers
\titlea{Poincar\'e normal forms and Lie-point symmetries}
 
\centerline{Giampaolo Cicogna}
\centerline{\it Dipartimento di Fisica, Universit\`a di Pisa}
\centerline{\it P.zza Torricelli 2, I-56126 Pisa (Italy)}
\centerline{cicogna@ipifidpt.difi.unipi.it}
\vskip 1 truecm
\centerline{Giuseppe Gaeta}
\centerline{\it C.P.Th., Ecole Polytechnique}
\centerline{\it F-91128 Palaiseau (France)}
\centerline{gaeta@orphee.polytechnique.fr}
 
\vskip 2 truecm
\titleb{Abstract}

{\sl We study \Poi  normal forms of vector fields in the presence of 
symmetry under general - i.e. not necessarily linear - diffeomorphisms. We 
show that it is possible to reduce both the vector field and the symmetry 
diffeomorphism to normal form by mean of an algorithmic procedure similar 
to the usual one for \Poi normal forms without symmetry; this double 
normal form can be given a simple geometric characterization.}

\vfill\eject}
\pageno=1
 
\titleb{Introduction}
 
The Poincar\'e-Dulac theory [1-3] of analytic normal forms (NF) of 
an analytic Ordinary Differential Equation \big(equivalently, vector field 
(VF)\big) in the
vicinity of an isolated fixed point is a venerable topic, but also a
powerful tool in the study of dynamical systems. 
Here we limit ourselves to the (properly speaking, Poincar\'e) case 
in which the linear operator $A$, giving the linearization of the ODE at
the fixed point (see below), commutes with its adjoint, i.e. 
$[ A , A^+ ] = 0$;  in other words, we treat the case $A$ does not
contain Jordan blocks,  or still the algebraic and geometric
multeplicities of its eigenvalues are equal  (we denote this condition
as "Assumption A").
  
Some of the results we will obtain remain valid (in some cases,
if suitably modified)  also if this assumption is not verified, as we
shall occasionally indicate: the  extension goes along the lines of  the
extension of the \Poi to the Poincar\'e-Dulac theory, see e.g. [1].

In the case of generic \Poi NF, these are nicely characterized [2,4,5] 
by the fact that the linear part of, and the full evolution operator, do
commute; in other words, the resonant vectors are those which commute
with the linear operator $A$. In the case of linearly equivariant \Poi
NF, i.e. if the equations admit a linear symmetry, the general form of
the \Poi NF unfolding is restricted by another commutation relation,
i.e. only the equivariant resonant terms can appear [4]. 

Here we generalize this result to the case of nonlinear symmetries. 
We also obtain some result concerning the properties of nonlinear
symmetries admitted by  dynamical systems in NF, and  the connections
existing between these symmetries and the \Poi procedure for
transforming the system into NF. 

We would like to warmly thank prof. H. Duistermaat for convincing us
normal forms are not only useful, but also a very nice subject in
itself, and for introducing us to the particularly convenient
geometrical approach to NF  we use here.

 
\titleb{1. Geometrical setting and notation}
 
Let us consider the space $M \sse \R^N$, and let $\M $ be
the space of analytical vector fields in $M$.
Elements of $\M$ are in correspondence with elements of $V$, the
space of analytical functions $f:M \to \R^N$ such that $f(x) \in T_x
M$; with $x \in M$ we write in component expansion (we will use greek
letters for elements of $\M$, latin ones for elements of $V$)
$$ \phi = f(x) \pa_x \equiv f^i (x) {\pa \over \pa x^i} \eqno(1.1) $$
 
Let us define $V_k \ss V$ as the space of homogeneous polynomial
functions of order $k$ in $V$, and let $\M_k$ be the corresponding
subset of $\M$.
 
In the set $\M$ is naturally defined a bilinear antisymmetric
operation $[.,.]$, the Lie commutator of vector fields, with which 
$\M$ becomes a Lie algebra. This induces
a corresponding Lie-Poisson bracket $\{ .,. \} : V \x V \to V$.
Indeed, if $\phi = f(x) \pa_x$, $\psi = g(x) \pa_x$, then
$$ [\phi , \psi ] = \{ f,g \} \pa_x \eqno(1.2) $$
$$ \{ f , g \}^i = f^j \pa_j g^i - g^j \pa_j f^i \eqno(1.3) $$
 
By means of these we define the (linear) {\it adjoint action} of
$\phi \in \M$ on $\M$ itself, respectively of $f \in V$ on $V$, by
$$ \eqalign{ \ad_\phi (.) =& [ \phi , . ] \equiv L_\phi (.) \cr
\ad_f (.) =& \{ f , . \} \equiv L_f (.) \cr } \eqno(1.4) $$
 
It is clear that $\M$, $V$ can be decomposed as
$$ \M = { \sum_{k=0}^\infty }^\o+ \M_k ~~;~~ V = { \sum_{k=0}^\infty
}^\o+ V_k \eqno(1.5) $$
and it is equally clear that
$$ \eqalign{
\phi \in \M_m \Lra & L_\phi : \M_k \to \M_{k+m-1} \cr
f \in V_m \Lra & L_f : V_k \to V_{k+m-1} \cr } \eqno(1.6) $$
In particular, for $m=1$ this shows that the decomposition (1.5) is
a decomposition in invariant spaces under $\ad_\phi $, $\ad_f$ for
$\phi \in \M_1$, $f \in V_1$.
 
Let us now consider the flow induced in $M$ by the vector field 
$\phi$ given by (1.1); this is described by the equation
$$ {\dot x} = f(x) ~= \phi \cdot x ~~~;~~~x \in M ~,~ f: M \to TM
\eqno(1.7) $$
A VF $\sig \in \M$ will be called a (time independent) Lie-point (LP) 
symmetry of $\phi$ if and only if the flows of $\sig$ and $\phi$ 
commute; that is,
$$ \eqalign{
[ \sig , \phi ] = 0  ~~;~~ & L_\phi (\sig ) = 0 = L_\sig (\phi ) \cr
\{ f , s \} = 0 ~~;~~ & L_f (s ) = 0 = L_s (f) \cr } \eqno(1.8) $$
where the second line is in component notation and we write
$\sig = s(x) \pa_x $, here and in the following.
 
The Lie algebra of LP symmetries of $\phi$ (respectively of $f$) will
be denoted by $\G_\phi \sse \M$ (respectively $\G_f \sse V$); notice
that
$$ \G_\phi = \Ker ( \ad_\phi ) ~~;~~\G_g = \Ker ( \ad_f ) \eqno(1.9) $$
$$ \sig \in \G_\phi \LLRA \phi \in \G_\sig ~~;~~ s \in \G_f \LLRA f \in
\G_s \eqno(1.10) $$
 
{\bf Remark 1.} Notice that if $x = x_0$ is an isolated fixed point for
$\phi$ (an isolated zero for $f$), then it must also be a fixed
point for $\sig$ (a zero for $s$), on the account of (1.8). From now
on we will assume this to be the case, and set $x_0 = 0$.
 
It is therefore natural to consider the linearization of (1.7) at $x =
x_0$; this is given by
$$ {\dot x} = A x = f_0 (x) ~~~~~~~~~~~A = (Df)(x_0 ) \eqno(1.7') $$
The linear operator $A$ will play a central role in the following;
we will make a fundamental assumption on it to simplify our work:
 
{\bf Assumption A.} {\it The linear operator $A = (Df)(x_0)$ commutes 
with its adjoint. }
  
In order to avoid unnecessary duplication of equations, from now on
we will use only the setting in $V$, and leave to the reader the
translation of our statements to the setting in $\M$.
 
Let us now expand $f,s$ in terms of the decomposition (1.5); we write
$$ \eqalign{
f(x) = \sum_{k=0}^\infty f_k (x) ~~~;&~~~~ f_k \in V_{k+1} \cr
s(x) = \sum_{k=0}^\infty s_k (x) ~~~;&~~~~ s_k \in V_{k+1} \cr }
\eqno(1.11) $$
so that by (1.6)
$$ \ad_{f_k} : V_m \to V_{m+k} ~~;~~ \ad_{f_k} : V_m \to V_{m+k}
\eqno(1.12) $$
 
We will consider in particular the linear operator 
$$\ad_{f_0} \equiv L_{f_0} \equiv \L \eqno(1.13)$$ 
which is now decomposed as
$$ \L = \dsum \Lk ~~;~~ \Lk : V_k \to V_k ~~;~~ \Lk = \L
\vert_{V_k} \eqno(1.14) $$
so that in particular
$$ \Ker ( \L ) = \dsum \Ker ( \Lk ) ~~;~~\Ker (\Lk ) = \Ker (\L )
\cap V_k \eqno(1.14') $$
 
{\bf Definition 1.} A function $w \in V_k$ is called a {\it k-resonant
vector} if and only if $w \in \Ker ( \Lk ) \sse V_k$.

{\bf Remark 2.} Written explicitly, the condition $w\in \Ker(\L)$ 
becomes
$$0 = \{Ax,w\}^i =  A^{jk}x_k\pa_jw^i-A^{ij}w_j= 
                            (Ax)^j\pa_jw^i-(Aw)^i\equiv D_Aw^i $$
where $D_A\equiv(Ax)\cdot\pa-A$ is the well known homological operator 
associated to $A$. If $A$ is diagonalized (thanks to Assumption A) with 
eigenvalues $\al_1,\ldots,\al_n$, and $w^i$ is a monomial
$$w^i=x_1^{m_1}x_2^{m_2}\cdot\ldots\cdot x_n^{m_n}$$
the above condition acquires the familiar form [1]
$D_Aw^i=m_j\al_j-\al_i=0$\ .

Notice that under Assumption A, one has \footnote{$^1$}{These 
decompositions can be  easily verified on the basis of 
the monomials  $w^i$ introduced in Remark 2; one could 
also introduce a scalar product in the space $V$, see [4], but this 
is not necessary for our present purposes.}
$$ \eqalign{
V =& \Ker ( \L ) \o+ \Ran ( \L ) \cr
V_k =& \Ker ( \Lk ) \o+ \Ran ( \Lk ) \cr } \eqno(1.15) $$
Let us now consider $h \in V$, and $f,s \in V$ such that (1.8) is
satisfied; by the Jacobi identity, we then have
$$ \{ s , \{ f,h \} \} = \{ f , \{ s,h \} \} \eqno(1.16) $$
which also reads
$$ L_s \cdot L_f = L_f \cdot L_s \eqno(1.16') $$
so that (1.8) implies in particular
$$ \ad_s : \Ker ( \ad_f ) \to \Ker ( \ad_f ) \eqno(1.17) $$
 
Notice also that
$$ \{ f , s \} = 0 \Lra \{ f_0 , s_0 \} = 0 \eqno(1.18) $$
so that when (1.8) is satisfied,
$$ \S \equiv \ad_{s_0} : \Ker (\Lk ) \to \Ker ( \Lk ) \eqno(1.19) $$
 
In terms of the expansion (1.11), the condition (1.8) reads
$$ \sum_{j=0}^\infty \{ f_j , s_{k-j} \} = 0 ~~~~ \forall k \ge 0
\eqno(1.20) $$
 
\vfill \eject

\titleb{2. Poincar\'e normal forms}
 
In the (Poincar\'e) theory of normal forms, one considers
dynamical systems of the form (1.7) and proves that, if Assumption A
is satisfied, by means of formal changes of coordinates they can be
taken to the form
$$ {\dot x} = g(x) ~~= \sum_{k=0}^\infty g_k (x) \eqno(2.1) $$
where $g_k \in V_{k+1}$ and
$$ g_0 = f_0 ~~;~~g_k \in \Ker (\L_{(k+1)} ) ~~~~~  k < k^*
\eqno(2.2) $$ 
for $k^*$ arbitrarily large.
 
{\bf Remark 3.} Notice that $f_0 \in \Ker (\L_{(1)} )$ by definition.
 
{\bf Remark 4.} If Assumption A is not satisfied, $\L$ should be 
substituted by its adjoint, and the above Remark would fail; this is
actually the main reason to consider Assumption A.
 
We will take formally $k^* = \infty$, so that the \Poi- Dulac
theorem will read
 
{\bf Theorem 1.}  (Poincar\'e - Dulac) {\it By means of formal changes 
of coordinates, it is possible to take the system {\rm (1.7)} to the
form {\rm (2.1)}, where $g \in \Ker ( \L ) = \G_{f_0}$.}
 
In this sense, the \Poi- Dulac procedures makes explicit the
symmetry of the dynamical system.
 
{\bf Remark 5.} Since $g_0 = f_0$, the operator $\L$ is well defined 
and independent of the form, (2.1) or (1.7), of the system.
 
{\bf Remark 6.} If the system (2.1) satisfies (2.2) with $k^* = n$, we 
say that it is in Poincar\'e normal form up to order $n$; when taking
the formal limit $n \to \infty$, we speak of Poincar\'e normal form,
tout court.
 
The changes of coordinates needed to transform the system to NF are
of the form
$$ x = y + h_k (y ) + R_{k+1} (y) \eqno(2.3) $$
where $h_k \in V_{k+1}$ and $R_k \in {\sum^\o+}_{m=k+2}^\infty V_m$;
notice that this can be seen as corresponding to the (time-one) flow
under the VF $\chi = h_k (x) \pa_x$.
 
Under (2.3), the system
$$ {\dot x} = \sum_m f_m (x) ~~;~~ f_m \in V_{m+1} \eqno(2.4) $$
is changed into
$$ {\dot y} = \sum_m \~f_m (y) ~~;~~ \~f_m = f_m ~~~~ m < k
\eqno(2.4') $$
where
$$ \~f_k = f_k - \{ f_0 , h_k \} \equiv f_k - \L (h_k ) \eqno(2.5) $$
so that if $\pi$ is the projection $\pi : V \to \Ran (\L )$, $\pi
f_k$ can be eliminated by judicious choice of $h_k$. The appropriate
$h_k$ for this, can be determined by solving the equation (also called 
"homological equation") 

$$ \L h_k = \pi f_k \eqno(2.6) $$
 
{\bf Remark 7.} Notice that $h_k$ is only defined up to elements of 
$\Ker (\L_{(k+1)})$; in other words, the changes of coordinates
determined by $h_k$ and by
$$ {h'}_k = h_k + \d h_k ~~;~~ \d h_k \in \Ker ( \L_{(k+1)} )\eqno(2.7) $$
lead to the same $\~f_k$.
 
By this Remark and by (1.15) we can, under Assumption A, decide to
choose
$$ h_k \in \Ran ( \L_{(k+1)} ) \eqno(2.8) $$
Once $h_k$ has been fixed, any $s(x) = \sum_m s_m (x)$ will be
changed according to the same (2.4),(2.5) above; i.e.
$$ \~s_m = s_m ~~~~ m < k ~~~;~~ \~s_k = s_k - \{ s_0 , h_k \} 
\eqno(2.9)$$
It is maybe worth stressing that geometrical objects, such as $\phi ,
\sig \in \M$, are not changed by (2.3), which affects only their
coordinate representation. In particular, $\G_\phi$ remains
unchanged, so that since $\phi = f(x) \pa_x = g(y) \pa_y$ and $f_0
\in \G_g$, then there must be a VF $\sig = f_0 (y) \pa_y \in
\G_\phi$, which will be represented as $\sig = s(x) \pa_x$ in the $x$
coordinates; if $g(y) \not= f_0 (y) \equiv g_0 (y)$, then $\sig \not=
\phi$, and the VF $\phi$ has at least a nontrivial symmetry.
 
Given a linear VF $\phi_0$, one can ask to classify (the local flow
of) all the VF which admit $\phi_0$ as linear part; in terms of the
dynamical system (1.7), this amount to classify (the local behaviour
of solutions of) all the systems $f$ which have the same
linearization $(Df)(x_0) = A$ at the fixed point $x_0$, $f_0 (x) = Ax$.
 
The problem of classifying all the $f$ as above up to formal analytic
transformations, reduces to the problem of classifying the most
general $f(x)$ with linear part $f_0 (x)$, upon reduction to
\Poi NF.
 
If the above classification is meant up to equivalence by formal
analytic transformations, the Poincar\'e NF is a convenient tool; it
should be stressed that if one is satisfied with a classification up
to transformation in a different class, e.g. up to topological or
$C^k$ equivalence, this would lead to different kind of NF and
NF reduction [1]. In the present paper, by NF we will always mean the
Poincar\'e NF.
 

\titleb{3. Symmetries and normal forms}
 
We want now to consider the relations between symmetry properties of
eq.(1.7) and its (reduction to) NF (2.1). The symmetry properties of
systems in NF have already been considered by some authors, see e.g.
[4,6]; in particular Elphick et al.\footnote{$^2$}
{In [2], p.67, this theorem is quoted from [5]; unfortunately this book is 
not
available (to our knowledge) in the western literature.} [4]  have
characterized the NF by means of the commutation properties between
the full VF describing time evolution and its linear part at the fixed
point $x_0$. Indeed, with Assumption A we have from [4]:
 
{{\bf Theorem 2.}}
{\it
Let the VF $\Phi \in \M$ be written in the $x$ coordinates as $\Phi =
f(x) \pa_x = \Phi_0 + \Phi_1$, where $\Phi_0 = f_0 (x) \pa_x$, $\Phi_1
(x) = [f(x) - f_0 (x)] \pa_x$, and $f_m \in V_{m+1}$. Let Assumption 
{\rm A} be satisfied. Then  $\Phi$ is in Poincar\'e NF if and only if 
$[\Phi ,  \Phi_0 ] =  0$}
 
{\bf Corollary 1.} {\it  $\Phi$ is in Poincar\'e NF if and only if 
the following conditions, all equivalent, are verified: }

\parskip 0pt
$i)$ $\{ f,f_0 \} =0$; $ii)$
$\Phi \in \Ker(\ad_{\Phi_0}) = \G_{\Phi_0}$; 
$iii)$ $ \Phi_0 \in \Ker(\ad_{\Phi})=\G_\Phi$.

\parskip 10 pt
The proof can easily be obtained from the discussion of sects.1 and
2; indeed, in the present notation this amounts to a corollary of
the \Poi- Dulac theorem as given in sect.2.

{\bf Remark 8.} From the point of view of symmetry
properties, an interesting result comes  from $iii)$ of the Corollary 
above, which states that the linear part $f_0(x)=Ax$ determines a linear
LP symmetry $\Phi_0=Ax\pa_x$ for the full problem $\dot x=f(x)$ [4,6]. 

{\bf Remark 9.} We notice that if Assumption A is not satisfied, the 
above theorem would be stated with the commutator condition $[ \Phi_1 ,
\Phi_0^+ ] = 0$ (and correspondingly modified conditions in the
corollary). In this case, the statement of Remark 8, would be
substituted by the weaker  result that the linear operator
$\Phi_0^+=A^+x\pa_x$ is a linear symmetry for the nonlinear part $\dot
x=\Phi_1\cdot x$ (and not for the full problem) .
 
We want to consider here the symmetries of the original system (1.7),
and how these are reflected into the NF coordinates. The motivation
for this comes from the following obvious but interesting fact (see
later discussion).
Let $\phi , \sig \in \M$, expressed in two systems of coordinates
$x$ and $y$ in $M$ as
$$ \eqalign{
\phi = f(x) \pa_x =& \~f (y) \pa_y \equiv g(y) \pa_y \cr
\sig = s(x) \pa_x =& \~s (y) \pa_y \equiv t(y) \pa_y \cr } \eqno(3.1) $$
The relation $[ \sig , \phi ] = 0$, equivalent to $\phi \in \G_\sig$,
$\sig \in \G_\phi$, is independent of the coordinate choice, so that
$$  \{ f,s \} = 0 \LLRA \{ \~f , \~s \} = 0 \eqno(3.2) $$
Therefore, the presence of a symmetry for eq.(1.7) will pose some
restriction to the NF (2.1): while in general the NF satisfies only
$$ \~f \in \Ker ( \ad_{f_0} ) \equiv \Ker ( \L ) \eqno(3.3) $$
in the presence of the symmetry we will also have from (3.2) that
$$ \~f \in \Ker ( \ad_{\~s } ) \eqno(3.4) $$

The combination of (3.3) and (3.4) can lead to a 
simplification of the NF unfolding; see sections 5,6.
 
It could be worth checking explicitely (3.2) in the following way: let
us rewrite (1.20) at order $k$ as

$$\eqalign {\C_k=&\ u_k \qquad (k=0,1,2,\ldots) \qquad {\rm where} \cr
      \C_k\equiv & \{ f_0 , s_k \} + \{ f_k , s_0 \} \quad {\rm and}
      \quad
       u_k\equiv -\sum_{j=1}^{k-1} \{ f_j , s_{k-j} \}
      \quad , \quad {\rm with} \quad u_0=u_1=0 }\eqno (3.5)$$ 

and let us consider a transformation (2.3): under this
$$ \C_k \to \~\C_k = \C_k - \{ f_0 , \{ s_0 , h_k \} \} - \{ \{ f_0 ,
h_k \} , s_0 \} $$
or, using Jacobi identity,
$$ \~\C_k = \C_k + \{ \{ s_0 , f_0 \} , h_k \} \eqno(3.6) $$
so that $\{ s_0 , f_0 \} = 0 \Lra \~\C_k = \C_k$; the r.h.s. of (3.5)
remains unchanged in the change of coordinates, since it contains
only terms of degree smaller than $k$, and so the whole equation (3.5)
is invariant.

{\bf Remark 10.} Notice that if $S$ is a linear operator such that 
$[A,S]=0$ and if $A$ satisfies Assumption A, then also $[A,S^+]=0$;
which implies that  $S+S^+$ and $S-S^+$ commute with $A$. Then, we can
assume that the linear part $S\equiv (Ds)(x_0)$ of the symmetry $\sig$
satisfies Assumption A.  In the following we will use freely the fact
that both $S$ and $A$ satisfy Assumption A, and denote $\ad_{s_0}$ by
$\S$. It should be stressed that if Assumption A is not verified, the
results stated in Theorem 3 below fail to be true, in general.
 
We will find it useful to have the following Lemma, which easily follows 
from Jacobi identity.

{\bf Lemma 1.}  {\it Let $v, w\in V$ such that $v,w\in \Ker(\L)$. Then }
$\{v,w\}\in\Ker(\L).$

We also have the following

{\bf Theorem 3.}
{\it Let $\Phi \in \M$ be expressed in Poincar\'e NF as $\Phi = f(x)
\pa_x$; then any $\sig$ such that $[\sig , \Phi ] =0$ is expressed in
the $x$ coordinates as $\sig = s(x) \pa_x$ where $s \in \Ker (\L )$.
In other words, all LP symmetries $\sig$ of a dynamical system in NF, 
which are obtained as formal series expansion, 
are also LP symmetries of the linearized system} $\dot x=f_0(x)=Ax$.
 
{\it Proof.} We can proceed recursively using the set of equations 
(3.5). For  $k=0$, $\{f_0,s_0\}=0$ is satisfied, see (1.18); for $k=1$ we
have $$\{f_0,s_1\}+\{f_1,s_0\}=0\eqno(3.7)$$
Applying $\L$ to this equation, we obtain
$$\L(\L(s_1))\equiv \{f_0,\{f_0,s_1\}\}=0\eqno(3.8)$$
being $\{f_0,\{f_1,s_0\}\}=0$. Using (1.15) one has
$$\L(s_1)=0 \qquad {\rm or} \qquad s_1\in\Ker(\L)\eqno(3.9)$$
The argument can be repeated recursively for each $k$: indeed, 
$\{f_k,s_0\}\in\Ker(\L)$ for Lemma 1, and similarly
if $s_j\in\Ker(\L) ,\ \forall j<k$, then also $u_k\in\Ker(\L)$. 
Therefore,  using again (1.15), eq. (3.5) can be solved only  with
$$\L(s_k)=0 \qquad {\rm and} \qquad \S(f_k)=u_k \eqno(3.10)$$ 
and this completes the proof.

It was remarked in sect.2 that the $h_k$ identifying the
normalizing transformation are identified by solutions of the
homological equation (2.6) only modulo $\Ker (\L )$, see Remark 7.
This means that once the system is in NF we can still apply changes
of coordinates of the form
$$ x = y + \d h_k (y) ~~~~~~~~~ \d h_k \in \Ker (\L_{(k+1)} ) 
\eqno(3.11) $$ 
without modifying $f(x)$. Under this change of coordinates, $s_k$
will be changed to
$$ \~s_k = s_k - \{ s_0 , \d h_k \} \equiv s_k - \S (\d h_k )
\eqno(3.12) $$ 
Notice that if $s_k \in \Ker (\L )$, also $\~s_k$ does,
see (1.19). By opportunely choosing $\d h_k \in \Ker (\L_{(k+1)} )$, we
can therefore eliminate the component of $s_k$ in $\Ker (\L ) \cap \Ran
(\S )$. We have therefore proved the
 
{\bf Proposition 1.} {\it Let $f(x)$ be in NF, and let $s(x)$ be a 
symmetry of $f(x)$. Then it is possible to choose coordinates in which
$f(x)$ is still in NF and such that } $s \in \Ker (\L ) \cap \Ker
(\S)$.
 
{\bf Remark 12.} Just as in the reduction to Poincar\'e NF, this 
change of coordinates will in general be purely formal.
 
{\bf Remark 13.} We always have $f_0 \in \Ker (\L ) \cap \Ker (\S )$; 
notice that for $s=f$ we have $\S = \L$, and indeed $s=f \in \Ker (\L )
\cap \Ker  (\S) \equiv \Ker ( \L )$; the symmetry $f$ will be called
trivial.
 
The above Proposition suggests the
 
{\bf Definition 2.} A VF $\Phi = f(x) \pa_x$ is in NF if $f \in 
\Ker ( \L )$, with $\L = \ad_{f_0}$; its symmetry VF $\sig = s(x) 
\pa_x$ is in NF if $s \in \Ker ( \L ) \cap \Ker (\S ) $, with $\S =
\ad_{s_0 }$.

 
\titleb{4. Determining equations for vector fields in normal forms}
 
Let us consider $f$ given and try to determine its symmetries $s$; in
order to do this we have to consider again the relation (1.20), to be
regarded as the determining equation for $s$. We will assume $f(x)$
is in NF, and look for solutions $s(x)$ which are in NF as well, i.e.
$ s \in \Ker (\L ) \cap \Ker ( \S )$; the Proposition 1 above ensures
that we are legitimate to restrict to such $s$.
 
At order $k$ the determining equations are, see (3.5),
$$ \L ( s_k ) - \S (f_k ) = u_k ~~~~~~~~ ;~~~~~ u_0 = u_1 = 0 
\eqno(4.1) $$
 
{\bf Remark 14.} If for all $j<k$ it happens that $f_j \in \Ker (\S )$, 
and necessarily then $f_j \in \Ker (\L ) \cap \Ker (\S )$ since $f$ is
in NF, then also $u_k \in \Ker (\L ) \cap \Ker (\S )$. Indeed, this just
follows from lemma 1 of the previous section.

{{\bf Lemma 2.}}
{\it If $f(x)$ and its symmetry $s(x)$ are in NF according to the above
definition, then }
$$ f \in \Ker (\L ) \cap \Ker (\S ) \qquad s \in \Ker (\L ) \cap 
\Ker (\S )$$ 
{\it Proof.}
The result for $s$ comes directly from Def. 2. If $f$ is in NF, 
then $f \in \Ker (\L ) $; it follows from (4.1) that if there 
is a $m$ such that $f_j \in \Ker (\S)$ for all $j < m$, then $f \in 
\Ker (\S )$.  Indeed, if $u_k \in \Ker (\S )$, then $\S (f_k ) = 0$. 
It suffices now to check that $f_1 \in \Ker (\S )$, as it follows 
from (4.1)  for $k=1$.
 
{\bf Remark 15.} The above lemma restores the symmetry of roles 
between $f$ and $s$.
 
{\bf Remark 16.} If $f$ and $s$ are in NF, we can still make a change

Date: Tue, 11 May 1993 14:38:27 +0200
From: gaeta@orphee.polytechnique.fr (Giuseppe Gaeta)
To: cicogna@ipifidpt.difi.unipi.it

of coordinates (2.3) generated by $\D h_k \in \Ker (\L ) \cap \Ker
(\S)$; such a transformation will not change $f_k$ neither $s_k$.
 
It is remarkable that equations (4.1) are of the same form as the full
determining equations (1.20) for the fields $(f-f_0 )$ and $(s - s_0 )$;
we could therefore repeat the above discussion and continue
iteratively. With the notation ($r \ge 1$)
$$ f^{[r]} = \sum_{m=r}^\infty f_m \equiv f - \sum_{m=0}^{r-1} f_m
\eqno(4.2) $$
$$ \L^{[r]} = \ad_{f^{[r]}} ~~,~~ \L \equiv \L^{[0]} ~~~;~~~    \S^{[r]}
= \ad_{s^{[r]}} ~~,~~ \S \equiv \S^{[0]} \eqno(4.3) $$
(so that $f^{[0]} = f$, $s^{[0]}=s$) we would arrive to the conclusion 
that if $\Phi , \sig \in \M$ satisfy $[ \Phi , \sig ] = 0$ and are in NF
when expressed as $\Phi = f(x) \pa_x$ , $\sig = s(x) \pa_x$, then for
any $r \ge 0$ both $f^{[r]}$ and $s^{[r]}$ are in $\Ker ( \S^{[r]} ) 
\cap \Ker ( \L^{[r]})$.
 
{\bf Remark 17.} With this notation, and recalling (1.18), it is immediate 
to check that $\{ f , s \} = 0$ enforces $ \{ f^{[1]} , s^{[1]} \} =0$: 
indeed, $\{ f , s \} = \{ f_0 + f^{[1]} , s_0 + s^{[1]} \}$
and due to $f,s \in \Ker (\L ) \cap \Ker ( \S ) $, we just have $\{ f ,
s \} = \{ f^{[1]} , s^{[1]} \}$.
 
Notice that now we can define
$$ u_k^{[r]} = - \sum_{j=1+r}^{k-r-1} \{ f_j , s_k \} ~~~~~ u_k^{[r]}
\equiv 0 {\rm ~~ for ~} k \le r $$
(so that $u_k^{[0]} \equiv u_k$) and $ \{ f^{[1]} , s^{[1]} \} =0$ reads 
now
$$ \L^{[r]} (s_k ) - \S^{[r]} (f_k ) = u_k^{[r]} ~~~~ k \ge r 
\eqno(4.4) $$
with $r=1$. By repeating the discussion iteratively, we get the
equation for generic $r \ge 0$.
 
We will summarize our discussion by stating the following
 
{\bf Theorem 4.}
{\it Let $\Phi = g(y) \pa_y$, $\sig = t(y) \pa_y$, where
$g(y) , t(y) \in V$ satisfy $\{ s , f \} = 0$. Then by means of formal
changes of coordinates {\rm (1.3)} we can take them to the form $\Phi =
f(x) \pa_x$, $\sig = s(x) \pa_x$, where $g_0 = f_0$, $t_0 = s_0$ and,
with $\L = \ad_{f_0}$, $\S = \ad_{s_0}$,
$$ \eqalign{
f \in & \Ker ( \S ) \cap \Ker ( \L ) \cr
s \in & \Ker ( \S ) \cap \Ker ( \L ) \cr } $$
Moreover, with the notation {\rm (4.3),(4.4)}, in the same coordinates 
we also have
$$ \eqalign{
f^{[r]} \in & \Ker ( \S^{[r]} ) \cap \Ker ( \L^{[r]} ) \cr
s^{[r]} \in & \Ker ( \S^{[r]} ) \cap \Ker ( \L^{[r]} ) \cr } $$ }
 
{\bf Remark 18.} This theorem can be seen as a generalisation of 
theorem 4 of [4] to the case of nonlinear symmetries; see this also for
the generalisation to the case in which $f_0$ does not meet Assumption
A.
 
{\bf Remark 19.} After the completion of the present work, prof. 
Duistermaat pointed out that this result can be obtained in an
alternative way based on the remark that the changes of variables (2.3)
amount to the adjoint action of $h \in V$ on $V$, so that the kernels
considered in the above Theorem 4 are necessarily invariant and provide
a classification of NF equations. This can also be seen as a
consequence of arguments concerning filtration of Lie algebras
applied to NF reduction, which are contained in the thesis by Broer
[7]. Our present discussion and statement of results has
nevertheless the advantage of being completely elementary and
explicit.

 
\titleb{5. Unfoldings of equivariant normal forms}
 
\def\KLS{ \Ker ( \L ) \cap \Ker ( \S ) }
 
We want now to discuss how the above Theorem 4 is of help in the
determination of equivariant normal forms unfoldings, i.e. in the
classification of systems ${\dot x} = f(x)$ and symmetries $s(x)$ of
these with given (necessarily commuting) linear parts $f_0$, $s_0$.
 
In this respect, it is useful to remark that in view of the
discussion in the previous section, we can restate our Theorem 4 as
 
{\bf Proposition 2.}
{\it Let $f(x)$ be in NF; the necessary and sufficient condition for
$s(x)$ to be a symmetry of $f$ in NF is that, with the notation
introduced above, the equation
$$ \L^{[r]} (s_k ) - \S^{[r]} (f_k ) = u_k^{[r]} $$
is satisfied for all $r$ and for all $k \ge r$.}
 
In many (and most of) concrete applications, one has to deal with
symmetry vectors that are linear or quadratic; it is therefore worth
discussing briefly these special cases, which will also make clear
the general procedure.
 
If the symmetry is linear,
$$ s = s_0 \in \Ker ( \L ) \eqno(5.1) $$
so that $s \in \Ker ( \L ) \cap \Ker ( \S )$ is trivially satisfied.
The determining equations, or equivalently the condition $f \in
\KLS$, imply now that
$$ \{ f_k , s_0 \} = 0 ~~~~ \forall k \ge 0 \eqno(5.2) $$
Correspondingly, the equivariant NF unfolding for (5.1) can be determined
order by order.
 
Notice that, writing $s_0 (x) = Sx$, eq.(5.2) is equivalent to
$$ f_k (Sx) = S f_k (x) ~~~~ \forall k \ge 0 \eqno(5.3) $$
In the same vein, writing $f_0 (x) = Ax$, the condition $f \in \Ker (
\L )$ reads
$$ f_k (Ax) = A f_k (x) ~~~~ \forall k \ge 0 \eqno(5.4) $$
Notice that
$$ \{ f_0 , s_0 \} = 0 \LLRA [ A , S ] = 0 \eqno(5.5) $$
so that we fully recover the setting of [4].
 
In the case of given quadratic symmetries (already in NF), $ s = s_0 +
s_1$, we can proceed in a similar way. First of all, $f \in \Ker 
( \L)$ ensures that (5.4) does apply, and $f \in \KLS$ yields again (5.3).
Now we do also have $f^{[1]} \in \Ker ( \S_1 )$, which enforces $$
\{f_k , s_1 \} = 0 ~~~~ \forall k \ge 1 \eqno(5.6) $$
 
We can therefore still determine the equivariant normal form
unfolding order by order.
 
In this quadratic case, we can write
$$ s_1 = {1 \over 2} \Ga_{\ell m} x^\ell x^m ~~;~~ \ga_{[\ell ]} =
\Ga_{\ell m} x^m \eqno(5.7) $$
with $\Ga$ a symmetric tensor; eq.(5.6) does now read
$$ \{ f_k , \ga_{[ \ell ]} \} = 0 ~~~~\forall \ell ~,~\forall k \ge 1
\eqno(5.8) $$
 
It is clear that the same procedure can be applied for the
determination of the equivariant NF unfolding under given arbitrary $s
\in V$ (provided of course that $s \in \KLS$); at order $k$, we just
have
$$ f_k \in \F_k = \Ker ( \L ) \cap \Ker ( \S ) \cap \Ker ( \S^{[1]} )
\cap \dots \cap \Ker ( \S^{[k]} ) \eqno(5.9) $$
Notice that at any finite order we have to solve a finite number of
algebraic equations of the form
$$ \{ f_j , s_m \} = 0 \eqno(5.10) $$
 
For $k = r+1$, eq.(4.4) reduces to
$$ \L^{[r]} (s_{r+1} ) = \S^{[r]} (f_{r+1 } ) \eqno(5.11) $$
which also means, repeating previous considerations,
$$ \eqalign{f_{r+1} \in & \Ker ( \L^{[r]} ) \cap \Ker ( \S^{[r]} ) \cr
 s_{r+1} \in & \Ker ( \L^{[r]} ) \cap \Ker ( \S^{[r]} ) \cr } 
\eqno(5.12) $$
 
In particular, if $f_j , s_j $ have been determined for $j<k=r+1$,
then (5.11) and $\{ f_k , s_k \} = 0$ and $\{ f_k , s_k \}$ determine $f_k 
, s_k $.
 
This means that we can just limit ourselves to solve
recursively equations of the form
$$ \{ f_k , s_k \} = 0 ~~;~~ \L^{[k]} (s_{k+1} ) = \S^{[k]} (f_{k+1} )
\eqno(5.13) $$
 
This also permits to classify the commuting pairs $(f,s)$ with given
(necessarily commuting) linear parts $(f_0 , s_0 )$, up to formal
analytical equivalence, i.e. of solving the general problem of
equivariant unfoldings of the NF. Indeed, our discussion can be
summarized as follows.
 
{\bf Theorem 5.}
{\it Let $\Phi_0 , \sig_0 \in \M_1$ be linear VFs, $\Phi_0 = f_0 (x)
\pa_x$, $\sig_0 = s_0 (x) \pa_x$, with $f_0 (x) = Ax$, $s_0 (x) =
S(x)$, $A$ and $S$ satisfying Assumption {\rm A}, and $[ \Phi_0 , 
\sig_0 ] = \{ f_0 , s_0 \} = [ A , S ] = 0$. Then for any pair $\Phi ,
\sig \in \M$ of VFs, $\Phi = g (y) \pa_y$, $\sig = t (y) \pa_y$, such
that $[\Phi , \sig ] = \{ g , t \} = 0$ and $(Dg)(0) = A$, $(Dt)(0) =
S$, there is a formal change of coordinates taking them to the form
$\Phi = f(x) \pa_x$, $\sig = s(x) \pa_x$, with $f_0 = g_0 = Ax$, $s_0 =
t_0 = Sx$, $f_m , s_m \in \Ker (\L^{[k]} ) \cap \Ker (\S^{[k]} )$ for any 
$ m
\le k$, and $\L^{[k]} ( s_{k+1} ) = \S^{[k]} (f_{k+1})$.}
 
{\bf Corollary 2. }
{\it For given commuting $A,S$, satisfying Assumption {\rm A}, the
equivariant normal form unfolding can be determined by solving
recursively equations of the form $\L^{[k]} ( s_{k+1} ) = \S^{[k]} 
(f_{k+1})$ and $\L^{[k]} ( s_k ) = 0 = \S^{[k]} (f_k)$.}


\titleb{6. Examples and discussion}

The classical problem of finding the most general form of a dynamical
system in NF once its linear part $f_0(x)=Ax$ is given, has been already
examined from the point of view of the symmetry properties [2,4-8]: for
a generic $A$ (i.e. not necessarily satisfying Assumption A), the result
can be written   
$$\dot x= Ax + K(\kappa(x)) x \eqno(6.1)$$
where $K$ is the most general matrix such that
$$KA^+ = A^+K\eqno(6.2)$$ 
and the entries $K_{ij}$ of $K$ are functions of the constants of 
motions $\kappa=\kappa(x)$ of the {\it linear} system 
$$\dot x=A^+x \ .\eqno(6.3)$$ 
Some special cases are discussed in [8]. The fact that $K$ satisfies 
(6.2) and its elements depend on the system (6.3) involving $A^+$ and
not $A$, clarifies the relevance of Assumption A in the context of
symmetry properties. In particular, it is now clear why, if $[A,A^+]\ne
0$, the linear symmetry $A^+x\pa_x$ is a symmetry for the nonlinear
part, and {\it not} for the full problem (see Remarks 8 and 9). 

{\bf Remark 20.} If the nonlinear terms were resonant with $A^+$
(not with $A$), then $\sig=(Ax)\pa_x$ would be a linear symmetry for the
full problem (for some other considerations on this situation, see
[6]).   

A very well known case of a reduction to NF concerns the
classical 2-dimensional Hopf periodic bifurcation problem: the matrix
$A$ has eigenvalues  $\pm i$, and the problem in NF
exhibits an explicit rotation covariance. As a trivial application of 
our above results, let us notice that this problem in NF cannot possess,
in agreement with Theorem 3, a scaling symmetry along one axis,
i.e.  a symmetry of the form 
$$\sig'=u\pa_u$$
where $u$ is either $x$ or $y$ \big(with $(x,y)\in R^2$\big). 
Instead, the presence of a scaling symmetry in the plane 
$$\sig=x\pa_x+y\pa_y \eqno(6.4)$$
which in fact is admitted by Theorem 3, would imply that the problem is 
trivially a linear problem.

\def \a{\alpha}
\def \b{\beta}
\def \g{\gamma}
A less trivial example is the following. With  $(x,y,z)\in 
R^3$, let us consider the system
$$\eqalign{\dot x =& x - yr^2\cr
           \dot y =& y + xr^2 \qquad \qquad (r^2=x^2+y^2) \cr
           \dot z =& (y + xr^2)z }           \eqno(6.5)$$ 
which corresponds, in the expansion in homogeneous terms $\phi_k\in
V_{k+1}$, to 
$$\eqalign{ \phi_0=&x\pa_x+y\pa_y\cr
            \phi_1=&yz\pa_z\cr
            \phi_2=&r^2(x\pa_y-y\pa_x)\cr
            \phi_3=&r^2xz\pa_z\cr
            \phi_j=&0 \quad j\ge 4 } \eqno(6.6)$$
This is symmetric under [9]
$$\sig =s(x)\pa_x=s_0\pa_x+s_1\pa_x\equiv (x\pa_y-y\pa_x)+xz\pa_z 
\eqno(6.7)$$
The above system is not in NF (actually, all nonlinear terms in 
(6.5) are nonresonant): accordingly, we have that $s_1\notin \Ker(\L)$
and $s\notin \Ker(\L)$ (or, which is the same, $\sig$ is not a LP
symmetry for the linear part of (6.5) - see Theorem 3). 

Writing now a generic polynomial VF as
$$h=\pmatrix{\a(x)\cr\beta(x)\cr\g(x)}\eqno(6.8)$$ 
the action of $\L$ is  given by
$$\L(h)=\pmatrix{x\a_x+y\a_y-\a\cr x\b_x+y\b_y-\b\cr
x\g_x+y\g_y}\eqno(6.9)$$
and it is immediate to check that
$$h\in \Ker(\L) \iff \pmatrix{\a=&a(z)x+b(z)y\cr
\b=&c(z)x+d(z)y\cr
\g=&e(z)}\eqno(6.10)$$
As for $\S={\rm ad}_{s_0}$, its action is given by
$$\S(h)=\pmatrix{x\a_y-y\a_y+\b\cr
x\b_y+y\b_x-\a\cr
x\g_y-y\g_x}\eqno(6.11)$$
and one can check that
$$h\in \Ker(\S) \iff \pmatrix{\a=&\^a(z, r^2)x+\^b(z,r^2)y\cr
\b=&-\^b(z,r^2)x+\^a(z,r^2)y\cr
\g=&\^e(z,r^2)}\eqno(6.12)$$
so that in particular
$$h\in \Ker(\L)\cap\Ker(\S) \iff \pmatrix{\a=&a(z)x+b(z)y\cr
                                           \b=&-b(z)x+a(z)y\cr
                                           \g=&e(z)}\eqno(6.13)$$
Let us now proceed to the normalizing quadratic transformation; 
the homological equation is
$$\L(h_1)=\pmatrix{0\cr 0\cr yz}\eqno(6.14)$$
which gives trivial equations for $\a$ and $\b$, and
$$x\g_x+y\g_y=yz\eqno(6.15)$$
so that
$$\a=\b=0 ; ~~~~~~~\g=yz+cz^2, ~~~~ c\in R\eqno(6.16)$$
i.e. $x,y$ are not changed, and
$$z=(1+y)\~z$$
With the $h_1$ given by (6.16),
$$\~f_1=f_1-\{f_0,h_1\}=0\eqno(6.17)$$
$$\~s_1=s_1-\{s_0,h_1\}=s_1-\S(h_1)\eqno(6.17')$$
Using (6.11) we immediately have
$$\S(h_1)=\pmatrix{0\cr 0\cr xz}\eqno(6.18)$$
which means
$$\~s_1=0\ .\eqno(6.19)$$
The above calculations can be extended to the higher orders, and one 
can see that the results (6.17) and (6.19) are true to all orders:
indeed, as already remarked, all terms in (6.5) are nonresonant;
once reduced to NF, the system becomes a linear system, and 
$$\sig=s_0\pa_x=x\pa_y-y\pa_x\eqno(6.20)$$
is (trivially) a symmetry for it.

We can also give examples containing resonant terms. Consider e.g., 
again with $(x,y,z)\in R^3$,   
$$\eqalign{\dot x =& \la x - y(1+r^2)\cr
           \dot y =& \la y + x(1+r^2) \qquad \qquad \qquad 
                      (r^2=x^2+y^2)\cr
           \dot z =& z^2+2\la y^2+2xy(1+r^2)-2y^2z+y^4 } \eqno(6.21)$$
If $\la=0$, then $yr^2, \ xr^2$ and $z^2$ are resonant terms; if 
$\la\ne 0$, then only $z^2$ is resonant. In both cases a LP symmetry for
the system is [9] 
$$\sig = x\pa_y-y\pa_y+2xy\pa_z\eqno(6.22)$$
Once the system (6.21) is reduced to NF, and all nonresonant terms 
are dropped, its symmetry becomes (both for $\la=0$ and $\ne 0$) the
rotation symmetry (6.20), in agreement with Theorem 3.

Let us point out finally that all our results concern symmetries 
$\sig$ which are obtained as series expansions. Other symmetries are
actually possible, as this example shows. Consider the problem in
$R^2$, in NF  
$$\eqalign{\dot x=&-x^3\cr
           \dot y=&-y}\eqno(6.23)$$
One of the symmetries of this problem is $\sig={\rm e}^{-1/2x^2}\pa_y$, 
which cannot be obtained as a series expansion and which 
is {\it not} a symmetry for the linearized problem (or  equivalently 
$s\notin \Ker(\L)$): this is not in contrast with the conclusion of 
Theorem 3; in fact, the series expansion of the VF defining this
symmetry would be identically zero. Other symmetries of the above
dynamical system,  e.g. $x^3\pa_x$ or $y\pa_y$ do actually satisfy the
hypotheses (and the conclusions as well) of the theorems given above.

\vfill\eject

\titleb{References}
 
[1] V.I. Arnold, "Geometrical methods in the theory of differential 
equations"; Springer, Berlin, 1982
 
[2] V.I. Arnold and Yu.S. Il'yashenko, "Ordinary differential 
equations"; in {\it Encyclopaedia of Mathematical Sciences - vol. I,
Dynamical Systems I}, (D.V. Anosov and V.I. Arnold eds.), p. 1-148
Springer, Berlin, 1988
 
[3] A.D. Bruno. "Local methods in nonlinear differential equations",
Springer,  Berlin, 1989
 
[4] C. Elphick, E. Tirapegui, M.E. Brachet, P. Coullet and G. 
Iooss, Physica D {\bf 29} (1987) 95
 
[5] G.R. Belitsky: {\it Normal forms, invariants, and local mappings}, 
Kiev, Naukova Dumka 1979 (Russian)
 
[6] G. Cicogna and G. Gaeta, J. Phys. A {\bf 23} (1990) L799; and 
{\bf 25} (1992) 1535.

[7] H.W. Broer, "Bifurcations of singularities in volume preserving vector 
fields"; Thesis, Groningen, 1979

[8] G. Cicogna and G. Gaeta, "Symmetry invariance and center manifolds for 
dynamical systems"; preprint 1993 (in preparation)

[9] G. Cicogna and G. Gaeta, Phys. Letters A {\bf 172} (1993), 361

\bye
