Content-Type: multipart/mixed; boundary="-------------0011120614215" This is a multi-part message in MIME format. ---------------0011120614215 Content-Type: text/plain; name="00-449.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="00-449.keywords" Smooth normal forms; Equivariant vector fields; Hamiltonian vector fields ---------------0011120614215 Content-Type: application/x-tex; name="sicilia2.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="sicilia2.tex" \documentstyle[12pt]{article} \textwidth 16.5cm \textheight 23cm \parindent 1.5cm \topmargin -0.5cm \oddsidemargin -0.2cm %%%%%%\pagestyle{empty} %%\input gmac.tex %%\input amac.tex \baselineskip 18pt \begin{document} %gmac.tex \newcommand{\C} {\hbox{${\rm C} \kern -7.5pt \raise 2pt \hbox{\tiny$|$}\kern 7.5pt$}} \newcommand{\G} {\hbox{${\rm G} \kern -7.5pt \raise 2pt \hbox{\tiny$|$}\kern 7.5pt$}} \newcommand{\Q} {\hbox{${\rm Q} \kern -7.5pt \raise 2pt \hbox{\tiny$|$}\kern 7.5pt$}} \newcommand{\ct}{\centerline} \newcommand{\ga}{\alpha} \newcommand{\gga}{\gamma} \newcommand{\gG}{\Gamma} \newcommand{\gb}{\beta} \newcommand{\gd}{\delta} \newcommand{\gD}{\Delta} \newcommand{\gk}{\kappa} \newcommand{\get}{\eta} \newcommand{\gep}{\varepsilon} \newcommand{\gvp}{\varphi} \newcommand{\gf}{\varphi} \newcommand{\gl}{\lambda} \newcommand{\gL}{\Lambda} \newcommand{\Gl}{\Lambda} \newcommand{\gch}{\chi} \newcommand{\gp}{\pi} \newcommand{\gps}{\psi} \newcommand{\gs}{\sigma} \newcommand{\gS}{\Sigma} \newcommand{\gt}{\theta} \newcommand{\Gt}{\Theta} \newcommand{\gT}{\Theta} \newcommand{\gw}{\omega} \newcommand{\gW}{\Omega} \newcommand{\gx}{\xi} \newcommand{\gz}{\zeta} \newcommand{\I}{{\bf I}} \newcommand{\II}{{\bf I\kern -1pt I}} %%\newcommand{\A}{\forall} \newcommand{\E}{\exists} \newcommand{\R}{\hbox{I\kern-.1500em \hbox{\sf R}}} \newcommand{\NN}{\hbox{I\kern-.2500em \hbox{\sf N}}} \newcommand{\Z}{\hbox{\sf Z\kern-0.720em\hbox{ Z}}} \newcommand{\proof}{\par\noindent{\bf Proof:}\quad} \newcommand{\proofo}{\par\noindent{\bf Proof}\quad} \newcommand{\remarks}{\par\noindent{\bf Remark:}\quad} \newcommand{\remarkss}{\par\noindent{\bf Remarks:}\quad} \newcommand{\titles}[1]{\bigskip\ct {\Large\bf #1}\bigskip \bigskip} \newcommand{\titless}[1]{\bigskip\ct {\Large\bf #1}\bigskip } \newcommand{\sections}[1]{\addtocounter{section}{1}\setcounter{theorem}{0} \bigskip \nin {\large\bf #1}\medskip\medskip} \newcommand{\authors}[3]{\bigskip\ct {#1} \bigskip\ct{#2} \bigskip\ct{#3} \bigskip} \newcommand{\authorss}[4]{\bigskip\ct {#1} \ct{#2} \ct{#3} \ct{#4} } %%%%%%%%%%%%%%%%%%%%% %%\documentstyle[11pt]{article} %%\voffset -.8in %%\hoffset -.75in %%\setlength{\textwidth}{6.28 in} %%\setlength{\textheight}{8.7 in} %%\tolerance 10000 \newtheorem{theorem}{Theorem} \font\Bbb=msbm10 scaled 1095 \font\bbb=msbm8 \def\R{\mbox{\Bbb R}} \def\Z{\mbox{\Bbb Z}} \def\r{\mbox{\bbb R}} \def\ls{\mathop{\lim\sup}\limits} \newcommand{\be}{\begin{equation}} \newcommand{\ee}{\end{equation}} \newcommand{\ve}{\varepsilon} \newcommand{\vf}{\varphi} \newcommand{\dok}{\smallskip\indent {\bf Proof}} \newcommand{\dokd}{\smallskip\indent {\bf Proof.\ }} \newcommand{\ti}{\tilde} \newcommand{\Si}{\hbox{\bf S}} \newcommand{\N}{{\cal N}} \newcommand{\B}{{\cal B}} \newcommand{\T}{{\cal T}} \newcommand{\A}{{\cal A}} \newcommand{\cH}{{\cal H}} \newcommand{\Lc}{{\cal L}} \newcommand{\M}{{\cal M}} \newcommand{\id}{\hbox{\rm id}} \newcommand{\Exp}{\hbox{\rm Exp}} \newcommand{\Lip}{\hbox{\rm Lip}} \newcommand{\lip}{\scriptsize{\Lip}} %\newcommand{\r}{\scriptsize{\R}} \newcommand{\pro}{\scriptsize{\pr}} \newcommand{\gr}{\hbox{\rm gr}} \newcommand{\pr}{\hbox{\rm pr}} \newcommand{\ctg}{\/\hbox{\rm ctg}} \font\frak=eufm10 scaled \magstep1 %\font\frak=cmbx10 scaled \magstep1 \def\frM{\mbox{\frak M}} \def\frN{\mbox{\frak N}} \def\frG{\mbox{\frak G}} \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 $ \begin{thebibliography}{99} \bibitem[1]{St} STERNBERG S., On the structure of local homeomorphisms of Euclidean $n$-space. 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