I come a cross this problem in my nonlinear analysis course. I know how to find the normal forms of any order. However, the commutative isometry! And in the third point the professor put two transformations! Could you please help me with some enlightening.
Let $f: \mathbb{R}^N \to \mathbb{R}^N$ be a smooth vector field and let $T: \mathbb{R}^N \to \mathbb{R}^N $ be a isometric linear transform. $f$ is called to be commutative with $T$ if $\forall x \in \mathbb{R}^N$ we have $f(Tx) = Tf(x)$. \begin{enumerate} 1)- Show that if $f$ is commutative with $T$ then for any solution for the ODE $\dot{x} = f(x), , x(t) \in \mathbb{R}^N$, we have $y(t) := T x(t)$ also solve the ODE. 2)- Show that if $f$ is commutative with $T$, then you can choose polynomial $h$ and $g$ in Theorem 2.1 (Haragus Loose, attached picture) such that $h$ and $g$ are commutative with $T$.
3)- Let $\omega >0$. Consider the ODE with respect to $z = (z_1,z_2) \in \mathbb{C}^2$
\begin{align} \label{complex-sys-1} \dot{z_1} & = i \omega z_1 + f_1(z_1,\overline{z_1},z_2,\overline{z_2}) \\ \dot{z_2} & = i \omega z_2 + f_2(z_1,\overline{z_1},z_2,\overline{z_2}),\nonumber \end{align} where $f_1,f_2$ are complex-valued functions and the right hand side of (\ref{complex-sys-1}) is commutative with $R$ and $T_\theta \,\, (\forall \theta \in \mathbb{R})$ wiht
\begin{align*} R(z_1,z_2) &= (z_2,z_1)\\ T_\theta (z_1,z_2) &= (e^{i \theta} z_1, e^{-i \theta} z_2), \,\, \theta \in \mathbb{R}. \end{align*}
Then, what is the 3-order canonical form? And what is the higher order canonical form?
\end{enumerate}
My proof
\begin{enumerate}
\item [(1)]- Let $f$ is commutative with $T$, and let $s(t) \in \mathbb{R}^N$ be a solution for $\dot{x}=f(x)$, i.e. $\dot{s}(t)=f(s(t)),, \forall x \in \mathbb{R}^N,,t >0$. Let $t>0$ and let $x \in \mathbb{R}^N$. Differentiating the given $ y(t) = T x(t)$ w.r.t time we get;
\begin{align*} \dot{y}(t) & = T \dot{x}(t)\\ &= Tf(x(t))\\ & = f(Tx(t))\\ & = f(y(t)).\\ & = f(Tx(t)). \end{align*} Thus, $\dot{y} = f(y)$ is a solution for the system $\dot{x}=f(x)$.
[(2)-]