3rd orderCanonical form of nonlinear dynamical system I am have been solving this problem since a month. I solved a more difficult ones but I do not know why I stuck at this one. There is a clue I can not understand. I solved the first point and stuck with the second one.
The problem statement mentions **Theorem 1.2 ** in Chapter 3, Section 3.1 in the book Local Bifurcations, Center Manifolds, and Normal Forms in Infinite-Dimensional Dynamical Systems
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 $f(Tx) = Tf(x)$.

*

*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.


*Show that if $f$ is commutative with $T$, then you can choose polynomial $\Phi$ and $N$ in Theorem 1.2 such that $\Phi$ and $N$ are commutative with $T$.


*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?
[(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)-] According to Theorem 1.2, there is a polynomial $\phi: \mathbb{R}^n  \to \mathbb{R}^n $ of degree $2 \geq p$ with
$$\phi(0)=0,\,\,\, \text{ and }\,\,\, D\phi(0)=0.$$
 A: This is my solution of point two, I hope it is right.
Consider the system
\begin{equation}
    \tag{1}
    \dot{x} = Ax + f(x).
\end{equation}
where $f(x)$ commutes with $T$. Having (1) satisfies the hypotheses of Theorem 1.2 then for any $2 \leq p \leq k$, there is a polynomial $h_p:\mathbb{R}^N \to \mathbb{R}^N$ of degree $p$ with
$$h_p(0)=0,\,\,\,\, \text{ and } \,\,\,\, Dh_p(0)=0.$$
and such that the change of variable
$$x = y + h_p(y), \,\,\,\, \text{ where }\, y=Tx$$
defined on a neighbourhood of the origin $U_p \subset \mathbb{R}^N$ transforms (1) into the normal form
\begin{equation}
    \label{nform1}
    \dot{y} = Ay + g_p(y) + r_p(y)
\end{equation}
Now, we apply the theorem again on the system
$$\dot{y}= Ay + f(y).$$
where $y- Tx$. For any $2 \leq q \leq k$, there is a polynomial $h_q:\mathbb{R}^N \to \mathbb{R}^N$ of degree $p$ with
$$h_q(0)=0,\,\,\,\, \text{ and } \,\,\,\, Dh_q(0)=0.$$
and such that the change of variable
$$T^{-1}x = x + h_q(x), \,\,\,\, \text{ where }\, y=Tx$$
defined on a neighbourhood of the origin $U_p \subset \mathbb{R}^N$ transforms (2) into the normal form
\begin{equation}
    \tag{2}
    \dot{x} = Ax + g_q(x) + r_q(x).
\end{equation}
Choosing the neighbourhoods small enough such that  $p=q$ and denote $h=h_p=h_q$ and $g_p = g_q=g$. From $x= y+ h(y)$  and $ x =T^{-1}(Tx) = Tx + T h(x)$ we have
$$Th(x) = h(Tx).$$
\begin{align*}
    \dot{y} & = Ay + g(y) +r(y)\\
    T \dot{x}&= T(Ax) + T g(x) + T r(x).
\end{align*}
Then we get
$$Tg(x) = g(Tx).$$
A: First, let us define the operator $T^*$ on $H_k$;
$$T^* h_k (x) = T^{-1} h_k (Tx),$$
Now, knowing that $L_{A^{(k)}(h_k (y))} = - (Dh_k (y)) A_y -A h_k (y)$. We show that $T^*$ commutes with $L^{(k)}_A$ as the following;
\begin{align}
    L_A^{(k)} (T^* h_k (x))
    & = L_A (T^{-1} h_k (Tx))\\
    &= A T^{-1} h_k (Tx) - D(T^{-1} h_k (Tx)) A x\\
    & =A T^{-1} h_k (Tx) - T^{-1} Dh_k (Tx) T Ax\\
    &= T^{-1} (Ah_k (Tx)) - T^{-1} D h_k (Tx) A Tx\\
    &= T^{-1} (Ah_k (Tx) - Dh_k (Tx)) A Tx\\
    &= T^{-1} L^{(k)} (h_k(Tx))\\
    &= T^{*} L^{(k)}_A (h_k (x)).
\end{align}
From above we can see that the image and the kernal of the operator $L^{(k)}_A$ are invariant under the influence of $T^*$. Now, we will show that the image and the kernal of the operator $L^{(k)_{A^*}}$ are invariant under $T^*$.
\begin{align}
    L_{A^*}^{(k)} (T^* h_k (x)) 
    &= L^{(k)}_{A^*} (T^{-1} h_k (Tx))\\
    &= A^* T^{-1} h_k (Tx) - D (T^{-1} h_k (Tx)) A^* x\\
    &= A^* T^{-1} h_k (Tx) - T^{-1} Dh_k (Tx) T A^* x\\
    &=  T^{-1} A^* h_k (Tx) - T^{-1} Dh_k (Tx) A^* Tx\\
    &=  T^{-1} L^{(k)}_{A^*} (h_k (Tx))\\
    &= T^* L_{A^*}^{(k)} (h_k (x)).
\end{align}
Using the above facts we have
\begin{align*}
    \dot{x} &= T^{-1} A Tx + T^{-1} F_2(Tx) + T^{-1} F_3(Tx)+\dots\\
    &= Ax + F_2(x) + F_3 (x) + \dots.
\end{align*}
Considering simplifying the second order terms by making the change of variables
$$x \mapsto y + h_2(y),$$
gives
$$\dot{x} = A x + L^{(k)}_A (h_2 (x)) + F_2 (x) + O(3).$$
Now we let $y \mapsto Ty$ we get
$$\dot{y} = T^{-1} A T y + T^{-1} L_A^{(2)} (h_2 (Ty)) + T^{(k)} P_{Im (L^{(2)}_A)} + T^{-1} P_{\ker L^{(2)}_{A^*}} F_2 (Ty)+ O(3),$$
where $P_{Im(L^{(2)}_{A^{*}})}$ and $P_{\ker L^{(2)}_{A^*}}$ are the projections onto $Im L_A^{(k)}$ and $ \ker L_{A^*}^{(2)} $ respectively. Since $T^*$ commutes with the projection operators, the we can write
$$\dot{y} = Ay + T^* \left( L^{(k)_A} (h_2 (x)) + P_{Im (L_A^{(k)})} F_2(y) \right) + P_{\ker L_{A^*}^{(2)}} T^{-1} F_2 (Ty) + O(3).$$
Since $h_2 (y)$ can be chosen such that
$$L_A^{(2)} (h_2(y)) + P_{\ker L_{A^*}^{(2)}}=0,$$
it follows that
$$\dot{y} = A y  + P_{\ker L_{A^*}^{(2)}} T^{-1} F_2 (T x) + O(3).$$
We can do the same steps for higher orders too.
[(3)-]  Considering the system
\begin{align}
        \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}
Let's define the change of variable
$$
\begin{pmatrix}
z_1 \\
z_2
\end{pmatrix} = \begin{pmatrix}
1 &&0 \\
0&&-1
\end{pmatrix} \begin{pmatrix}
z\\
\overline{z}\\
\end{pmatrix}$$
$$
\begin{pmatrix}
z \\
\overline{z}
\end{pmatrix} = \begin{pmatrix}
1 &&0 \\
0&&-1
\end{pmatrix} \begin{pmatrix}
z_1\\
z_2\\
\end{pmatrix}$$
$$
\begin{pmatrix}
\dot{z} \\
\overline{\dot{z}}
\end{pmatrix} = \begin{pmatrix}
1 &&0 \\
0&&-1
\end{pmatrix} \begin{pmatrix}
\dot{z_1}\\
\dot{z_2}\\
\end{pmatrix}$$
$$
\begin{pmatrix}
\dot{z} \\
\overline{\dot{z}}
\end{pmatrix} =  
i \omega \begin{pmatrix}
1 &&0 \\
0&&-1
\end{pmatrix} \begin{pmatrix}
 z_1\\
 z_2
\end{pmatrix} +  \begin{pmatrix}
1 &&0 \\
0&&-1
\end{pmatrix} \begin{pmatrix}
f_1(z_1,\overline{z_1},z_2,\overline{z_2})\\
f_2(z_1,\overline{z_1},z_2,\overline{z_2})
\end{pmatrix}$$
\begin{align*}
    \dot{z}&=i \omega z + F_1(z,\overline{z})\\
\overline{\dot{z}}&=- i \omega \overline{z} + F_2(z, \overline{z}).
\end{align*}
From theorem 5.1, and considering $Q: \mathbb{R} \to \mathbb{C}$ such the standard form
\begin{align*}
    \dot{z}&=i \omega z +zQ (|z|^2) + r(z,\overline{z})\\
\overline{\dot{z}}&=- i \omega \overline{z} + \overline{z} \overline{Q (|z|^2)} + \overline{r(z, \overline{z})}.
\end{align*}
