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Smooth normal forms; Equivariant vector fields; Hamiltonian vector fields
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\author {G.R.Belitskii${}^{\dagger}$, A.Ya.Kopanskii${}^{\ddagger}$}
\title
{Sternberg Theorem for Equivariant Hamiltonian Vector Fields}


\maketitle

\begin{center}
 {\footnotesize ${}^{\dagger}$
 Department of Mathematics, Ben-Gurion University of the Negev

 Beer Sheva 84105, Israel ${\tt genrich@cs.bgu.ac.il}$
}
\end{center}

\begin{center}
 {\footnotesize ${}^{\ddagger}$
 Institute of Mathematics, Academy of Sciences of Moldova

 5 Academiei str., Kishinev 2028, Moldova ${\tt alex@revel.moldova.su}$
}
\end{center}


\begin{center}
\noindent
{\sl Keywords:} Smooth normal forms; Equivariant vector fields; Hamiltonian
vector fields
\end{center}

\section*{Introduction}

The well known Sternberg Theorem (see \cite{St}) asserts that if two local
smooth
vector fields are formally conjugate at a hyperbolic singularity
then they are smoothly conjugate. This result reduces local
classification and normalization problems to the formal ones.

In the last years there is a splash of interest to systems with symmetries
and
anti-symmetries ({\em reversible systems}) (for references see
\cite{LR1}). This activity
faces, in particular, a similar ``equivariant`` problem:
could one provide a conjugacy (smooth,
formal) of two symmetric or anti-symmetric vector fields via a
transformation keeping the property?
The linear and the formal aspects of the problem were considered in
\cite{LR2, Ga}.
In \cite{BK} the authors proved a related version of Smooth Conjugacy Sternberg
Theorem.

A similar question arised for hamiltonian systems: is it possible to conjugate
two hamiltonian vector fields via a {\em canonical} coordinate change?
The affirmative answer was given in \cite{Ly, BaLlW, BrK2}.

These two results are in the same streamline and lead to the next very natural
setting: given two hamiltonian (anti-)symmetric vector fields can one choose a
conjugacy which preserves both symmetric and symplectic structures?
The aim of the present paper is to prove a hamiltonian equivariant version of
Sternberg Theorem (Theorem 2.1 below).

We use the so-called {\em deformation method} or {\em homotopy trick}
going back to
J.Mather (see \cite{M, AVG, BrK2}). This method reduces a nonlinear
problem on equivalence of two vector fields
(or more generally, of two local mappings
near a common singularity) to that on solving
a linear functional equation ({\em cohomology equation}).

\section{Definitions}

By a {\em local diffeomorphism} ({\em local vector field}) we mean in what
follows either a germ of diffeomorphisms (vector fields) at a point or a
representative of this germ defined in a neighborhood of the point.
Sometimes, we omit for briefness the word ``local``.

Let $\:\frG\:$ be a group of local diffeomorphisms
$\:\Phi\::\:(\R^{2d},0)\:\to\:(\R^{2d},0)\:$ and
$\:\gs\::\:\frG\:\to\:\R\:$ be a {\em multiplicative character} of
$\:\frG,\:$ {\em i.e.}, a homomorphism into the multiplicative group
$\:\R^*.\:$
\\

{\bf Definition 1.1.}
{\em A local vector field $\:\xi\:$ is
said to be $\:(\frG,\gs)$-equivariant if}
$$
U_*\xi\:=\:\gs(U)\xi\ \ \ (U\:\in\:\frG).
$$

In this definition, $\:U_*\xi\:$ denotes a usual action
of the local diffeomorphism $U$ on the vector field $\xi$:
$$
(U_*\xi)(x)\:=\:DU(U^{-1}x)\xi(U^{-1}x),
$$
where $\:DU(x)\:$ is the Jacobi matrix.

A local transformation $\:G\:$ keeps the property of $\:\xi\:$ to be
$\:(\frG,\gs)$-equivariant if it commutes with every element
$\:U\:\in\:\frG,\:$ {\em i.e.}, $\:UG\:=\:GU.\:$ Such a local
diffeomorphism is said to be $\:\frG$-{\em equivariant}.

Let $\R^{2d}$ be provided with the
{\it symplectic structure}, {\it i.e.}, with a {\it closed non-degenerate}
2-{\it form} $\omega$. The vector field $\xi$ is said to be {\it
symplectic} if $L_{\xi}\omega =0$. The {\it hamiltonian} $F$,
$i_{\xi}(\omega)=dF$, is a function determined up to a constant. Each
function may serve as a hamiltonian of some symplectic vector field.

Let $\omega$ be a symplectic form on
$\R^{2d}$. According to a theorem due to Darboux,
there exist coordinates $(u,v)=(u_1,\ldots,u_d,v_1,\ldots,v_d)$ on
       $\R^{2d}$ such that
       $$
        \omega(u,v)=du_1\wedge dv_1 +\ldots + du_d\wedge dv_d.
       $$
This form is called {\it standard}. The
changes of variables which respect this form, {\it i.e.},
$$
G^*\omega=\omega
$$
are called {\it canonical}.
By $\:G^*\omega\:$ we denote a usual action
of the local diffeomorphism $G$ on the symplectic form $\omega$.


{\bf Definition 1.2.} {\em A symplectic form $\omega$ is said to be
$\:(\frG,\gs)$-equivariant if}
$$
U^*\omega\:=\:\gs(U)\omega\ \ \ (U\:\in\:\frG).
$$

In what follows, we assume that $\:\frG\:$ is {\em a compact group of
linear operators in $\:\R^{2d}\:$} and $\:\gs\:$ {\em is continuous}.
Without loss of generality
$\:\frG\:$ is supposed to be a closed subgroup of the group $\:O(2d,\R)\:$ of all
orthogonal transformaions. Thus, every element $\:U\:\in\:\frG\:$ keeps
the standard inner product in $\:\R^{2d}.\:$
And besides, the continuity of $\:\gs\:$ implies
$$
\gs(U)\:=\:\pm 1\ \ \ (U\:\in\:\frG).
$$


\section{Main result}

Recall that two local vector fields $\:\xi\:$ and
$\:\eta\:$ are said to be $\:C^k$ conjugate if there exists
a $\:C^k$ diffeomorphism $\:G\:$ such that
$$
G_*\xi\:=\:\eta.
$$
\\

Take a $\:(\frG,\gs)$-equivariant symplectic form $\:\omega\:$ and
a local $\:(\frG,\gs)$-equivariant symplectic $\:C^{\infty}$ vector
field $\:\xi,\ \xi(0)\:=\:0,\ D\xi(0)\:=\:\gL\:$. Choose canonical coordinates
bringing $\:\omega\:$ to the standard form. Let the subspaces
$\:E_u\:=\:\{v\:=\:0\}\:$ and $\:E_v\:=\:\{u\:=\:0\}\:$ be invariant
with respect to the phase flow of $\:\xi$.

Suppose $\:\gL\:$ has no pure imaginary eigenvalues, {\em i.e.}, $\:\xi\:$
has a hyperbolic singularity at the origin.
It is known that $\:\gL\:=\:IS$, where $S$ is a symplectic
linear operator and
$$
 I=\left(\begin{array}{lr}
           0 & -E \\
           E & 0
         \end{array}
         \right) .
$$
According to Williamson's theorem (see \cite{A}),
there exists a canonical linear
transformation which brings $\:\gL\:=\:IS\:$ to the form
$$
 \left( \begin{array}{lr}
          R & 0 \\
          0 & -R
        \end{array}
        \right)
$$
where all the
eigenvalues of the matrix $\:R\:$ have positive real parts.

Denote by $\:\gd\:$ and $\:\gD\:$
the minimum and the maximum of the real parts of the
eigenvalues of the matrix $\:R$.
Given an integer $\:k\:$ put
$$
\rho(k,\gL)\:=\:k\:{\gD\over\gd}\:+\:k\:+\:1.
$$
\\

{\bf Theorem 2.1.} {\em Let $\:\eta\:$ be a local $\:(\frG,\gs)$-equivariant
symplectic $\:C^K$
vector field, $\:\eta(0)\:=\:0,\:$
$\:K\:\geq\:\rho(k,\gL)$, and all derivatives of the difference
$\:\xi\:-\:\eta\:$ up to the order $\:K\:$ vanish at the origin. Then there
is a local $\:\frG$-equivariant canonical $\:C^k$ diffeomorphism $\:G\:$ which
conjugates $\:\xi\:$ and $\:\eta,\:$ i.e. $\:G_*\xi\:=\:\eta.\:$}

In particular, if $\:K\:=\:\infty\:$ then $\:k\:=\:\infty\:$ as well.

This theorem is the main result of the paper. Its proof is given in
Sections 3 -- 4.


\section{Deformation method.}
$\ \ \ \ \ \ \ \ $
\\
As we have mentioned above, the deformation method reduces a conjugacy
problem to
that of solving a multidimensional linear equation.

Given vector fields $\:\xi\:$ and $\:\eta\:$ consider the {\em
deformation}
$\:h(x,\gep)\:\equiv\:\gep\eta(x)\:+\:(1\:-\:\gep)\xi(x)\:$
and the vector field
$$
\Xi\:\equiv\:h(x,\gep)\frac{\partial}{\partial x}\:
+\:0\cdot\frac{\partial}{\partial\gep}
$$
on $\:\R^{n+1}.\:$ Take another vector field
$$
\Phi(x,\gep)\:\equiv\:\gvp(x,\gep)\frac{\partial}{\partial
x}\:+\:1\cdot\frac{\partial}{\partial\gep}
$$
Let
$$
Q^t(x,\gep)\:=\:(q^t(x,\gep),\gep),\ \ \
P^t(x,\get)\:=\:(p^t(x,\gep),\gep\:+\:t)
$$
be the pase flows of $\:\Xi\:$ and $\:\Phi\:$ respectively.
Put $\:G(x)\:=\:p^1(x,0).\:$
\\

{\bf Lemma 3.1.} {\em Let the vector fields $\:\Phi\:$ and $\:\Xi\:$
commute, i.e.}
$$
[\:\Phi,\Xi\:]\:=\:0.
\eqno (1)
$$
{\em Then $\:G\:$ conjugates $\:\xi\:$ and $\:\eta,\:$ i.e.,
$\:G_*\xi\:=\:\eta.\:$}
\vspace*{.1cm}

{\bf Proof.} Note that the phase flows of $\:\xi\:$ and $\:\eta\:$ are,
respectively, $\:q_0^t(x)\:\equiv\:q^t(x,0)\:$ and
$\:q_1^t(x)\:\equiv\:q^t(x,1).\:$
Since the phase flows $P^t$ and $Q^t$ commute then
$$
Q^{t_2}(P^{t_1}(x,\ve))=(q^{t_2}(p^{t_1}(x,\ve),\ve+t_1),\ve+t_1)=
(p^{t_1}(q^{t_2}(x,\ve),\ve),\ve+t_1)=P^{t_1}(Q^{t_2}(x,\ve)),
$$
or
$$
q^{t_2}(p^{t_1}(x,\ve),\ve+t_1)=p^{t_1}(q^{t_2}(x,\ve),\ve).
$$
For $\ve=0$, $t_1=1$, $t_2=t$ this gives
$$
q^{t}(p^{1}(x,0),1)=p^{1}(q^{t}(x,0),0).
$$
For $G(x)=p^1(x,0)$ we get
$$
q^t(G(x),1)=G(q^t(x,0)),
$$
or
$$
q^t_1(G(x))=G(q^t_0(x)),
$$
{\it i.e.}, $G$ is a conjugacy between $\xi$ and $\eta$.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \Box $
\\

Condition (1) is equivalent to the equation
$$
D_1\gvp(x,\gep)\cdot
h(x,\gep)\:=\:D_1h(x,\gep)\cdot\gvp(x,\gep)\:+\:\eta(x)\:-\:\xi(x),
\eqno (2)
$$
where
$$
D_1\:\equiv\:\frac{\partial}{\partial x}.
$$
This equation is called the {\em cohomology equation}.
\\

Now let us turn to the case where the vector fields $\xi$ and $\eta$
are symplectic.
In order to find a canonical conjugacy we make use of the relationship between
vector fields and differential forms as suggested in \cite{BaLlW}.

Let $\omega$ be a symplectic form and $\xi$ and $\eta$
be vector fields which preserve $\omega$.
Fix $\:\gep\:$ and put $\:h_{\gep}(x)\:=\:h(x,\gep)$,
$\:\gvp_{\gep}(x)\:=\:\gvp(x,\gep)$.
It is not hard to verify that
$$
 i_{[h_{\varepsilon},\varphi_{\varepsilon}]}(\omega)=
 d(i_{h_{\varepsilon}}(i_{\varphi_{\varepsilon}}(\omega))),
$$
therefore (2) implies that
$$
 d(i_{h_{\varepsilon}}(i_{\varphi_{\varepsilon}}(\omega)))=
 i_{\eta-\xi}(\omega).
$$
Let $X$ and $Y$ be the hamiltonians of the vector fields $\xi$ and $\eta$.
Denote by $H_{\varepsilon}$ the hamiltonian of the vector field
$\varphi_{\varepsilon}$ we are seeking. Then
$$
 d(i_{h_{\varepsilon}}(dH_{\varepsilon}))=dY-dX.
$$
Taking into account the equality
$$
 L_{h_{\varepsilon}}H_{\varepsilon}=i_{\xi_{\varepsilon}}(dH_{\varepsilon})+
 di_{h_{\varepsilon}}(H_{\varepsilon})
$$
we get
$$
 dL_{h_{\varepsilon}}H_{\varepsilon}=
 di_{h_{\varepsilon}}(dH_{\varepsilon}),
$$
whence
$$
 d(L_{h_{\varepsilon}}H_{\varepsilon}-Y+X)=0.
$$
It suffices to find the hamiltonian $H_{\varepsilon}$ from the equation
$$
 L_{h_{\varepsilon}}H_{\varepsilon}=Y-X
$$
or
$$
D_1H(x,\varepsilon)\cdot h(x,\varepsilon)=Y-X \eqno (3)
$$
which is called the {\it cohomology equation for hamiltonians}.

Thus, in order to prove that two vector fields which preserve the
symplectic form are conjugate via a canonical diffeomorphism we have
to solve the cohomology equation (3), then find the
corresponding vector field $\varphi_{\varepsilon}$ and integrate the system
${\dot x}=\varphi_{\varepsilon}(x),\ {\dot \varepsilon} =1$.

\section{$\frG$-equivariant conjugacy}
$\ \ \ \ \ \ \ \ $
\\
First of all, note that solvability of equation (3) under the assumption
of Theorem 2.1 was proved in \cite{BrK2} (for $\:k\:<\:\infty\:$) and \cite{BK}
(for $\:k\:=\:\infty\:$). Prove that the solution can be choosed
$\:(\frG,\gs)$-equivariant.

Let $\omega$ be a $\:(\frG,\gs)$-equivariant symplectic form.
Take a matrix $U\in\frG$,
$$
 U=\left(\begin{array}{lr}
          A & B \\
           C & D
         \end{array}
         \right) ,
$$
where $\:A,\:B,\:C\:$ and $\:D\:$ are $\:d\times d\:$ matrices. Since
$\omega$ is in the standard form then (see, for example, \cite{KaH}
Proposition 5.5.6) $A^tC$ and $B^tD$ are symmetric and $A^tD-C^tB=\gs(U)E$.
This gives, in particular, that
$$
 U^{-1}=\gs(U)\left(\begin{array}{lr}
          \ D^t & -B^t \\
          -C^t & A^t
         \end{array}
         \right) .
$$


\vspace*{.1cm}
{\bf Lemma 4.1.} {\it
Let $\xi$ be a symplectic vector field and its hamiltonian $F$ is
$(\frG,\gs)$-equivariant. Then $\xi$ is $\frG$-equivariant.}

{\bf Proof.} Let $F$ be a $(\frG,\gs)$-equivariant hamiltonian.
Denote
$$
\xi(u,v)=\sum_{i=1}^d\left( \eta_i(u,v)\frac{\partial}{\partial u_i}
+\zeta_i(u,v)\frac{\partial}{\partial v_i}\right).
$$
By tansition formulas,
$$
\eta_i(u,v)=\frac{\partial F(u,v)}{\partial v_i},\qquad
\zeta_i(u,v)=-\frac{\partial F(u,v)}{\partial u_i}\qquad
(i=1,\dots,d).
$$
Fix a matrix $U\in\frG$.
Since $F(u,v)=\gs(U)F(Au+Bv,Cu+Dv)$ then
$$
\eta(u,v)=\gs(U)(D^t\eta(Au+Bv,Cu+Dv)-B^t\zeta(Au+Bv,Cu+Dv)),
$$
$$
\zeta(u,v)=\gs(U)(-C^t\eta(Au+Bv,Cu+Dv)+A^t\zeta(Au+Bv,Cu+Dv)).
$$
Thus we conclude that
$$
\xi\:=\:U^{-1}\:\xi\circ U,
$$
{\it i.e.}, $\xi$ is $\frG$-equivariant.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \Box $





\vspace*{.1cm}
{\bf Lemma 4.2.} {\it
If a symplectic vector field $\xi$ with the hamiltonian
$F$ is $\:(\frG,\gs)$-equivariant then $F$ is $\:\frG$-equivariant.}

{\bf Proof.}
Let $\xi$ be a $\:(\frG,\gs)$-equivariant symplectic vector field.
Fix $U\in\frG$.  Since $\omega$ is $\:\frG$-equivariant then
$$
\gs(U)U^{-1}=\left(\begin{array}{lr}
          \ D^t & -B^t \\
          -C^t & A^t
         \end{array}
         \right) .
$$
Therefore,
$$
\xi(x)=\left(\begin{array}{lr}
          \ D^t & -B^t \\
          -C^t & A^t
         \end{array}
         \right)\xi(Ux).
$$
Put $\bar u=Au+Bv$, $\bar v=Cu+Dv$. Then
\begin{eqnarray}
F(u,v)&=&\sum_{i=1}^d\int_0^1[\eta^i(ut,vt)v_i-\zeta^i(ut,vt)u_i]dt=
\sum_{i=1}^d\int_0^1[\eta^i(\bar ut,\bar vt)\bar v_i
-\zeta^i(\bar ut,\bar vt)\bar u_i]dt\nonumber\\
&=&F(Au+Bv,Cu+Dv).\nonumber
\end{eqnarray}
Thus, $F$ is $\frG$-equivariant.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \Box $

\vspace*{.1cm}
{\bf Lemma 4.3.} {\it If $\:H\:$ is a $C^k$ solution of equation} (3)
{\it then for every $U\in \frG$ the function $\tilde H$
$$
\tilde H(x,\ve)\equiv \gs(U)H(Ux,\ve),
$$
is a solution of equation} (3) {\it as well.}

{\bf Proof.} Put $\tilde H(x,\ve)\equiv\gs(U)H(Ux,\ve)$. By virtue of
$(\frG,\sigma)$-equivariance of $\xi$ and $\eta$, the following equalities
are true:
\begin{eqnarray}
Uh(x,\ve)&=&\sigma(U)h(Ux,\ve);\nonumber\\
(Y-X)(x,\ve)&=&(Y-X)(Ux,\ve).\nonumber
\end{eqnarray}
Hence,
\begin{eqnarray}
D_1\ti H(x,\ve)h(x,\ve)&=&\gs(U)D_1H(Ux,\ve)Uh(x,\ve)\nonumber\\
&=&\sigma(U)D_1H(Ux,\ve)\gs(U)h(Ux,\ve)\nonumber\\
&=&D_1H(Ux,\ve)h(Ux,\ve)\nonumber\\
&=&(Y-X)(Ux,\ve)\nonumber\\
&=&(Y-X)(x,\ve).\nonumber
\end{eqnarray}
Therefore, $\ti H$ is a solution of the cohomology equation.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \Box $

\vspace*{.1cm}
It is well known that since $\frG$ is compact then one can find a
Haar measure $\mu$ on $\frG$ which is right invariant with respect to
$\frG$. We can suppose without loss of generality that $\mu(\frG)=1$.
Let $H$ be a $C^k$ solution of equation (3).
Define a $C^k$ function $\cH:\R^{2d+1}\to\R$ by
$$
\cH(x,\ve)=\int\limits_{\frG}\gs(U)H(Ux,\ve)d\mu(U)\qquad(x\in\R^{2d},\
\ve\in\R).
\eqno (4)
$$
\\

{\bf Lemma 4.4.} {\em The function $\:\cH\:$ is a $\:(\frG,\gs)$-equivariant
solution of the cohomology equation} (3).

\vspace*{.1cm}
{\bf Proof.} By linearity of the cohomology equation,
it follows from Lemma 4.3 that
$\:\cH\:$ is a solution. Since $\:\mu\:$ is right invariant,
for any $\:V\:\in\:\frG\:$ we have
$$
\cH(Vx,\ve)\:=\:\int\limits_{\frG}\gs(U)H(UVx,\gep)d\mu(U)
\:=
$$
$$
=\:\gs(V)\int\limits_{\frG}\gs(V)\gs(U)H(UVx,\gep)d\mu(U)\:=\:
\gs(V)\int\limits_{\frG}\gs(W)H(Wx,\gep)d\mu(W)\:=\:\gs(V)\cH(x,\gep).
$$
Here we put $\:W\:=\:UV.\:$ As a result we get
$$
\cH(Vx,\ve)\:=\:\gs(V)\cH(x,\gep),
$$
{\em i.e.}, $\:\cH\:$ is $\:(\frG,\gs)$-equivariant.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Box $
\\

{\bf Proof of Theorem 2.1.} Take a local $\:C^k\:$ solution $\:H\:$ of
equation (3). Then the function $\:\cH\:$ given by formula (4) is a local
$\:(\frG,\gs)$-equivariant $\:C^k\:$ solution of (3). Let
$\:F^t(x,\gep)\:\equiv\:(f^t(x,\gep),\gep\:+\:t)\:$ be the phase flow
of the vector field
$$
\Psi(x,\gep)\:=\:\psi(x,\gep)\frac{\partial}{\partial x}\:+\:
1\cdot\frac{\partial}{\partial\gep},
$$
where $\:\psi(\cdot,\ve)\:$ is a symplectic vector field generated by
the hamiltonian $\:\cH(\cdot,\ve)$.
Put $\:G(x)\:=\:f^1(x,0).\:$ Since $\:[\Psi,\Xi]\:=\:0\:$
then, by Lemma 3.1, the local
$\:C^k$ diffeomorphism $\:G\:$ conjugates $\:\xi\:$ and $\:\eta.\:$
Show that $\:G\:$ is $\:\frG$-equivariant and canonical.
In fact, since $\:G\:$ is a shift along the trajectories of the symplectic
phase flow it presrves the symplectic structure.
Note that the vector field $\:\psi(\cdot,\ve)\:$ is $\:\frG$-equivariant.
Choose
$\:U\:\in\:\frG\:$ and denote
$\:\tilde{f}^t(x,\gep)\:\equiv\:U^{-1}f^t(Ux,\gep).\:$ Then
$$
\left.\frac{d\tilde{f}^t(x,\gep)}{dt}\right|_{t=0}\:=
\:\left.U^{-1}\frac{df^t(Ux,\gep)}{dt}\right|_{t=0}\:=\:
U^{-1}\psi(Ux,\gep)\:=\:\psi(x,\gep).
$$
Hence $\:\tilde{F}^t(x,\gep)\:\equiv\:(\tilde{f}^t(x,\gep),\gep\:+\:t)\:$
is also a phase flow of the vector field $\:\Psi.\:$ It follows from
Uniqueness Theorem that $\:\tilde{F}^t\:=\:F^t\:$ and
$\:G(x)\:=\:U^{-1}G(Ux).\:$ We conclude that $\:G\:$ is
$\:\frG$-equivariant
completing the proof of Theorem 2.1.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Box $


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