Proof. As $q\ge 1$ , $\partial _x\in \text {ker}(d\xi (0))$ , so any local center manifold of $\xi $ is transverse to $\{x=0\}$ . Let $\gamma :(-\infty ,s)\ni t\mapsto (x(t),{\boldsymbol {\gamma }}(x(t)))\in \mathbb R^{1+n}$ be as in the statement, where t is the natural time associated with $\xi $ . Since $\alpha (\gamma )=0$ , $\gamma $ is included in any local center-unstable manifold (that is, invariant and tangent to the sum of the eigenspaces associated with eigenvalues in $\mathbb R_{\le 0}\oplus i\mathbb R$ ; see [Reference Kelley16] again). We fix a $C^k$ center-unstable manifold $W^{cu}$ , and then a $C^k$ center manifold $W^c\subset W^{cu}$ of the restricted vector field $\xi |_{W^{cu}}$ . Note that $W^c$ is a local center manifold of $\xi $ , so proving the proposition inside $W^{cu}$ suffices. We therefore assume that $W^{cu}$ and $\mathbb R^{1+n}$ coincide locally, so Theorem 3.1 applies. Shrinking our neighborhood of zero if needed, we get the accompanying trajectory $\delta :(-\infty ,s')\ni t \mapsto (\delta _x(t),\delta _{{\mathbf {y}}}(t))\in \mathbb R^{1+n}$ of $\gamma $ in $W^c$ , parameterized by the natural time (that is, satisfying (3.1)). From $\xi _x=x^{q+1}$ , we get that the two functions $x(t)$ , $\delta _x(t)$ satisfy the same differential equation

(E) $$ \begin{align}\frac{dx}{dt}=x^{q+1}. \end{align} $$

By integration, the difference of two different solutions of $(E)$ is always larger than a certain power of $|t|$ when t goes to $-\infty $ . But

$$ \begin{align*} |\delta_x(t)-x(t)|\le \|\delta(t)-\gamma(t)\|=O(\exp(\beta t))\quad \text{ as }t\to-\infty, \end{align*} $$

so $\delta _x(t)$ and $x(t)$ cannot be different solutions: $\delta _x(t) = x(t)$ . Writing $t(x)$ for the inverse function of $x(t)$ , and ${\boldsymbol {\delta }}(x) = \delta _{{\mathbf {y}}}(t(x))$ , we see that $\delta $ is parameterized by $(x,{\boldsymbol {\delta }}(x)), x>0$ , and we have

$$ \begin{align*}\|{\boldsymbol{\delta}}(x)-{\boldsymbol{\gamma}}(x)\|=\|\delta(t(x))-\gamma(t(x))\|= O(\exp(\beta t(x)).\end{align*} $$

From $(E)$ , we also get $t(x)\sim -({ x^{-q}}/{q})$ as ${x\to 0}$ , which, replaced in the previous equation, implies point (i).

Following the notation introduced in Theorem 3.1, notice that $\pi \circ \Psi $ maps a point $(x,{\boldsymbol {\gamma }}(x))$ to $\pi ((\delta _x(t(x)), \delta _{\boldsymbol {y}}(t(x)))$ . We have shown that $\delta _x(t(x))=x$ , which is point (ii).

Proof. In the definition of a (TRS)-form, there appears a factor made of a power $x^e$ and a unit u. Since the proposition only concerns the foliation induced by $\xi $ in the half-space $x>0$ , we might suppose that $x^eu(x,{\mathbf {y}})=1$ . So we assume that $\xi $ is given by

$$ \begin{align*} \xi= x^{q+1}\partial_x+( (D(x)+x^qC )\cdot{\mathbf{y}}+x^{q+N+2}V(x,{\mathbf{y}}))\cdot\partial_{{\mathbf{y}}} ,\end{align*} $$

where D, C and V satisfy the hypothesis.

We prove the proposition by induction on the couple $(n,q)$ , where $1+n$ is the dimension of the ambient space and q is the Poincaré rank of $\xi $ . We initialize the induction when $n=0$ or $q=0$ and prove that the case $(n,q)$ follows from a case $(n',q')$ with $n'<n$ and $q'<q$ .

Case: $n=0$ . Here $\xi = x^{q+1}\partial _x$ . The curve $x, x>0$ is the only trajectory of $\xi $ included in $x>0$ and conditions (i)–(iii) are trivial.

Case: $q=0$ . Here $\xi = x\partial _x + [C\cdot {\mathbf {y}}+x^{2+N}V(x, {\mathbf {y}})]\cdot \partial _{{\mathbf {y}}}$ . The lift $\eta $ of $\xi $ by the composition ${\mathbf {y}}=x^{N} {\mathbf {z}}$ of N full diagonal monomial transformations is given by

$$ \begin{align*} \eta= x\partial_x + [(C-N\text{I}_n)\cdot{\mathbf{z}}+x^2V(x, x^{N} {\mathbf{z}})]\cdot \partial_{{\mathbf{z}}}. \end{align*} $$

The origin is a hyperbolic singularity of $\eta $ , with one positive eigenvalue $1$ associated with $\partial _x$ and n eigenvalues with negative real part (the eigenvalues of $C-N\text {Id}$ ; recall that $\text {Spec}(C)\subset \mathbb R^*_-\oplus i\mathbb R$ ). So the unstable manifold of $\eta $ has dimension one: it is a trajectory issued from the origin and tangent to $\partial _x$ at $x=0$ . In particular, it is not included in $x=0$ , and since $\eta (x)\neq 0$ if $x>0$ , it can be parameterized by x. Let $(x,{\boldsymbol {\delta }}(x))$ be this trajectory. Being tangent to $\partial _x$ , we get ${\boldsymbol {\delta }}(x)=o(x)$ . Then $(x,{\boldsymbol {\gamma }}(x)) :=(x,x^{N}{\boldsymbol {\delta }}(x))$ is a trajectory of $\xi $ which satisfies ${\boldsymbol {\gamma }}(x)=o(x^{N+1})$ . This proves (i).

Now let $\zeta =(x,{\boldsymbol {\zeta }}(x))$ be any trajectory of $\xi $ with $\lim _{x\to 0}{\boldsymbol {\zeta }}(x)=0$ . Such a trajectory is not contained in the stable manifold (which coincides with $x=0$ ). Thus, since the singularity is hyperbolic and $\zeta $ accumulates to it, $\zeta $ is included in the unstable manifold of $\xi $ . This unstable manifold is of dimension one and is made of only one trajectory in $x>0$ , which means that ${\boldsymbol {\zeta }} = {\boldsymbol {\gamma }}$ . So conditions (ii) and (iii) are automatically satisfied.

Induction: $q>0$ and $n>0$ . We reorder the coordinate functions ${\mathbf {y}}$ , in such a way that $1,\ldots , m$ are the indices such that $\text {ker}(D(0))=\text {Span}(\partial _{{\mathbf {y}}_1},\ldots ,\partial _{{\mathbf {y}}_m})$ . Write ${\mathbf {z}}=(y_1,\ldots ,y_m)$ , ${\mathbf {w}} = (y_{m+1},\ldots ,y_n)$ . Recall that ord $_x(D(x)) = 0$ , so $m < n$ . Since $\xi $ has no dominant rotation, the center space (associated with eigenvalues with real part $0$ ) coincide with the kernel of $d\xi (0)$ ; that is, $\text {Span}(\partial _x, \partial _{{\mathbf {z}}})$ . So the vectors $\partial _{{\mathbf {w}}}$ are associated with non-diagonal elements of $D(0)$ . Up to permutation, ${\mathbf {w}} = ({\mathbf {r}},{\mathbf {s}})$ , with ${{\mathbf {r}} = (y_{m+1},\ldots ,y_{m+1+d})}$ , ${\mathbf {s}}=(y_{m+2+d}, \ldots ,y_n)$ , and $\partial _{{\mathbf {r}}}$ (respectively, $\partial _{{\mathbf {s}}}$ ) are associated with positive (respectively, negative) diagonal coefficients of $D(0)$ . So $\xi $ admits a $C^{n(q+1+N)+1}$ center-unstable manifold $W^{cu}$ and a $C^{n(q+1+N)+1}$ center manifold ${W^c\subset W^{cu}}$ , which are graphs over $(x,{\mathbf {y}},{\mathbf {r}})$ and $(x,{\mathbf {y}})$ , respectively. The manifold $W^{cu}$ intervenes only in the proof of point (iii) below, so we focus on $W^c$ . Call ${\boldsymbol {h}}$ the $C^{n(q+1+N)+1}$ map that gives $W^c$ ; that is,

$$ \begin{align*}W^c = \{{\mathbf{w}} ={\boldsymbol{h}}(x,{\mathbf{z}})\}.\end{align*} $$

We claim that ${\boldsymbol {h}}(x,{\mathbf {z}})=x^{q+1+N}{\boldsymbol {g}}(x,{\mathbf {z}})$ for some $C^{(n-1)(q+1+N)+1}$ map ${\boldsymbol {g}}$ . For this, first remark that $W^c \cap \{x=0\}$ is a center manifold of the linear vector field

$$ \begin{align*}\xi(0,{\mathbf{y}}) = (D(0)\cdot {\mathbf{y}})\cdot\partial_{{\mathbf{y}}} = (D_{{\mathbf{w}}}(0)\cdot{\mathbf{w}})\cdot\partial_{{\mathbf{w}}},\end{align*} $$

where $D_{{\mathbf {w}}}$ is defined by $(D(x)\cdot ({\mathbf {z}},{\mathbf {w}}))_{{\mathbf {w}}} = D_{{\mathbf {w}}}(x)\cdot {\mathbf {w}}$ (we define $D_{\mathbf {z}}$ accordingly, so ${D=D_{{\mathbf {z}}}\oplus D_{{\mathbf {w}}}}$ ). This center manifold is clearly unique and given by ${\mathbf {w}}=0$ , so $W^c \cap \{x=0\}=\{x=0,\; {\mathbf {w}} =0\}$ . This means that ${\boldsymbol {h}}(0,{\mathbf {z}})=0$ , so ${\boldsymbol {h}}(x,{\mathbf {z}})$ is divisible by x and ${\boldsymbol {h}}(x,{\mathbf {z}})/x$ is a $C^{n(q+1+N)}$ map. Let

$$ \begin{align*}s= \sup \{r\le q+1+N;\; {\boldsymbol{h}}(x,{\mathbf{z}})/x^r \text{ is }C^1\},\end{align*} $$

and write ${\boldsymbol {h}}(x,{\mathbf {z}})=x^s {\boldsymbol {g}}(x,{\mathbf {z}})$ . If $s=q+1+N$ , as ${\boldsymbol {h}}$ is $C^{n(q+1+N)+1}$ and divisible by $x^{q+1+N}$ , ${\boldsymbol {g}}$ is $C^{(n-1)(q+1+N)+1}$ and the claim is proved, so showing that $s=q+1+N$ suffices. Suppose $s<q+1+N$ , so ${\boldsymbol {g}}(x,{\mathbf {z}})$ is $C^{n(q+1+N)+1-s}$ and ${\boldsymbol {g}}(0,{\mathbf {z}})\neq 0$ , which gives a contradiction. Since $W^c$ is invariant by $\xi $ and given by ${\mathbf {w}}-{\boldsymbol {h}}(x,{\mathbf {z}})=0$ , we have ${\xi ({\mathbf {w}}-{\boldsymbol {h}})|_{{\mathbf {w}}={\boldsymbol {h}}}} = 0$ , which gives

$$ \begin{align*} (D_{{\mathbf{w}}}+x^q C_{{\mathbf{w}}}){\boldsymbol{h}} & = \partial_{{\mathbf{z}}} {\boldsymbol{h}} \cdot(D_{{\mathbf{z}}}+x^q C_{{\mathbf{z}}})\cdot{\mathbf{z}} +x^{q+1}\partial_x {\boldsymbol{h}} \cdots \\ &\quad +\;x^{q+1+N} (\partial_{{\mathbf{z}}} h \cdot V_{{\mathbf{z}}}(x,{\mathbf{z}},{\boldsymbol{h}})-V_{{\mathbf{w}}}(x,{\mathbf{z}},{\boldsymbol{h}})), \end{align*} $$

where $C=C_{{\mathbf {z}}}\oplus C_{{\mathbf {w}}}$ and $V(x,{\mathbf {y}})=(V_{{\mathbf {z}}}(x,{\mathbf {y}}),V_{{\mathbf {w}}}(x,{\mathbf {y}}))\in \mathbb R^{m}\times \mathbb R^{n-m}$ . Replacing ${\boldsymbol {h}}$ by $x^s{\boldsymbol {g}}$ , dividing by $x^s$ and setting $x=0$ in this equation leads to

$$ \begin{align*} D_{{\mathbf{w}}}(0)\cdot {\boldsymbol{g}}(0,{\mathbf{z}}) = 0, \end{align*} $$

which is a contradiction since $D_{{\mathbf {w}}}(0)$ is invertible and ${\boldsymbol {g}}(0,{\mathbf {z}})\neq 0$ .

Now let $\eta $ be the pull-back of $\xi $ by the inclusion $W^c\hookrightarrow \mathbb R^{n+1}$ . In the coordinate system $(x,{\mathbf {z}})$ of $W^c$ , $\eta $ is given by

$$ \begin{align*} x^{q+1}\partial_x + ((D_{{\mathbf{z}}}+x^q C_{{\mathbf{z}}})\cdot{\mathbf{z}} +(D_{{\mathbf{w}}}+x^qC_{{\mathbf{w}}})\cdot {\boldsymbol{h}}(x,{\mathbf{z}}) + x^{q+1+N}V_{{\mathbf{z}}}(x, {\mathbf{z}},{\boldsymbol{h}}(x,{\mathbf{z}})))\cdot\partial_{{\mathbf{z}}}. \end{align*} $$

But

$$ \begin{align*}(D_{{\mathbf{w}}}(x)+x^qC_{{\mathbf{w}}})\cdot {\boldsymbol{h}}(x,{\mathbf{z}})= x^{q+1+N}(D_{{\mathbf{w}}}(x)+x^qC_{{\mathbf{w}}})\cdot {\boldsymbol{g}}(x,{\mathbf{z}})\end{align*} $$

so

$$ \begin{align*} \eta = x^{q+1}\partial_x + ((D_{{\mathbf{z}}}(x)+x^q C_{{\mathbf{z}}})\cdot{\mathbf{z}} + x^{q+1+N}{\boldsymbol{\nu}}(x,{\mathbf{z}}))\cdot \partial_{{\mathbf{z}}} \end{align*} $$

for some $C^{(n-1)(q+1+N)+1}$ germ ${\boldsymbol {\nu }}$ . Let $v=\text {ord}_x(D_{{\mathbf {z}}}(x)+x^qC_{{\mathbf {z}}})$ , so $1\le v\le q$ , since $\text {ord}_x(D_{{\mathbf {z}}})\ge 1$ and $C_{{\mathbf {z}}}\neq 0$ (recall that $0$ is not an eigenvalue of C, or of $C_{{\mathbf {z}}}$ ). The vector field $\eta $ is divisible by $x^v$ , and $\eta /x^{v}$ is in (TRS)-form of type $(q-v,N+1,0)$ in coordinates $(x,{\mathbf {z}})$ and satisfies that:

1. (1) the exponential part $D_{{\mathbf {z}}}(x)$ has no dominant rotation;

2. (2) the residual part $C_{{\mathbf {z}}}$ has spectrum included in $\mathbb R^*_-\oplus i\mathbb R$ ; and

3. (3) the vestigial part ${\boldsymbol {\nu }}(x,{\mathbf {z}})$ is a $C^{(n-1)(q+1+N)+1}$ germ, and

   $$ \begin{align*}(n-1)(q+1+N)+1 \ge m(q-v+1+N)+1.\end{align*} $$

So, since $m<n$ and $q-v<q$ , the induction hypothesis applies to $\eta /x^v$ .

From this we deduce the three points of the lemma. For point (i), $\eta /x^v$ has a trajectory $(x,{\boldsymbol {\delta }}(x)), x>0$ such that ${\boldsymbol {\delta }}(x)=o(x^{N+1})$ . If ${\boldsymbol {\gamma }}(x) := ({\boldsymbol {\delta }}(x),{\boldsymbol {h}}(x,{\boldsymbol {\delta }}(x)))$ , the curve $(x,{\boldsymbol {\gamma }}(x)), x>0$ is a trajectory of $\xi $ , and since ${\boldsymbol {\gamma }}(x)= ({\boldsymbol {\delta }}(x), x^{q+1+N}{\boldsymbol {g}}(x,{\boldsymbol {\delta }}(x)))$ , it also verifies that ${\boldsymbol {\gamma }}(x) = o(x^{N+1})$ .

For point (ii), let $\zeta =(x,{\boldsymbol {\zeta }}(x)), x>0$ be a trajectory of $\xi $ satisfying $\lim _{x\to 0}{\boldsymbol {\zeta }}(x) = 0$ . We are in the conditions of Proposition 3.3, so $\zeta $ admits an accompanying trajectory $(x,{\boldsymbol {c}}(x))$ of $\xi $ contained in $W^c$ (that is, ${\boldsymbol {c}}(x)$ has flat contact with ${\boldsymbol {\zeta }}(x)$ ). As usual, we write ${\boldsymbol {c}} = ({\boldsymbol {c}}_{{\mathbf {z}}}, {\boldsymbol {c}}_{{\mathbf {w}}})$ . Then $(x,{\boldsymbol {c}}_{{\mathbf {z}}}(x))$ is a trajectory of $\eta /x^v$ and $\lim _{x\to 0}{\boldsymbol {c}}_{{\mathbf {z}}}(x)=0$ . From the induction hypothesis, this implies that ${\boldsymbol {c}}_{{\mathbf {z}}}(x)$ and ${\boldsymbol {\delta }}(x)$ have flat contact. Since ${\boldsymbol {h}}$ is differentiable, ${\boldsymbol {h}}(x, {\boldsymbol {c}}_{{\mathbf {z}}}(x))$ and ${\boldsymbol {h}}(x, {\boldsymbol {\delta }}(x))$ have flat contact also, so ${\boldsymbol {c}}(x)=({\boldsymbol {c}}_{{\mathbf {z}}}(x), {\boldsymbol {h}}(x, {\boldsymbol {c}}_{{\mathbf {z}}}(x)))$ and ${\boldsymbol {\gamma }}(x)=({\boldsymbol {\delta }}(x),{\boldsymbol {h}}(x,{\boldsymbol {\delta }}(x)))$ have flat contact. Finally, since ${\boldsymbol {\zeta }}$ has flat contact with ${\boldsymbol {c}}$ and ${\boldsymbol {c}}$ has flat contact with ${\boldsymbol {\gamma }}$ , we have that ${\boldsymbol {\zeta }}$ and ${\boldsymbol {\gamma }}$ have flat contact. This proves point (ii).

We now show point (iii). Recall that ${\mathbf {y}}=({\mathbf {z}},{\mathbf {r}},{\mathbf {s}})$ where the center manifold $W^c$ is given by $({\mathbf {r}},{\mathbf {s}})={\boldsymbol {h}}(x,{\mathbf {z}})$ , and the center-unstable manifold $W^{cu}$ is a graph over ${\mathbf {s}}=\boldsymbol {0}$ . Let $\widetilde {\Psi }$ be the homeomorphism introduced in Remark 3.2. According to Proposition 3.3, point (2), $\widetilde {\Psi }$ maps a point $(x,{\mathbf {z}},{\mathbf {r}},{\mathbf {s}})\in W^{cu}$ to $((x,{\mathbf {z}}',{\boldsymbol {h}}(x,{\mathbf {z}}')),{\mathbf {r}})\in W^c\times \mathbb R^d$ , where the trajectory issued from $(x,{\mathbf {z}}',{\boldsymbol {h}}(x,{\mathbf {z}}'))$ is the accompanying trajectory of the one issued from $(x,{\mathbf {z}},{\mathbf {r}},{\mathbf {s}})$ . We let $\pi $ be the projection onto the first factor of $W^{cu}\times \mathbb R^d$ .

Given a neighborhood U of a, we fix an open neighborhood $U_0\subset U$ of a in such a way that $\widetilde {\Psi }$ restricts to a homeomorphism $\widetilde {\Psi }_0$ from $U_0\cap W^{cu}$ onto a product $U^c_0\times B$ of two connected open sets of $W^c$ and $\mathbb {R}^d$ , respectively. Let $V^c\subset U^c_0$ be the open neighborhood of a in $W^c$ , and let $S^c$ be the closed connected topological submanifold of $V^c\cap \{x>0\}$ given by point (iii) for the vector field $\eta /x^v$ . Choose an open neighborhood $V\subset U$ of a such that $V\cap W^{cu}=\widetilde {\Psi }_0^{-1}({V^c\times B})$ , and let S be the subset of V given by

$$ \begin{align*} S:=\widetilde{\Psi}_0^{-1}({S^c\times B}). \end{align*} $$

Since $\widetilde {\Psi }_0$ is a homeomorphism and preserves x, S is a closed connected $C^0$ submanifold of $V\cap \{x>0\}$ owing to the corresponding property for $S^c$ . The dimension of S is $\dim (S^c)+d$ , and according to the inductive hypothesis, $\dim (S^c)=1+u(x^{-v}D_{\mathbf {z}}(x))=1+u(D_{\mathbf {z}}(x))$ . Since ${\mathbf {r}}$ is a d-tuple, and corresponds to the positive diagonal elements of $D(0)$ , we conclude that $\dim (S)=1+u(D(x))$ (point (iii,a)).

Let $b\in V\cap \{x>0\}$ and call $\gamma $ the trajectory of $\xi _{|V}$ passing through b and parameterized as $(x,{\boldsymbol {\gamma }}(x)), x\in (\alpha ,\omega )$ . If $b\in S$ , then $\pi (\widetilde {\Psi }(b))\in S^c$ , so the trajectory $\delta $ passing through $\pi (\widetilde {\Psi }(b))$ is included in $S^c$ and has $\alpha $ -limit point $0$ . By definition of $\widetilde {\Psi }_0$ , $\delta $ is the accompanying trajectory of $\gamma $ , so we deduce that $\gamma $ is included in S and $\alpha =0$ . The first property shows that S is locally invariant for $\xi $ (point (iii, b)). The second property means that S is composed by trajectories accumulating to $0$ for negative time (point (iii, c) direct implication). Suppose now that $b\notin S$ . If $\alpha =0$ and $\lim _{x\to 0}{\boldsymbol {\gamma }}(x)=0$ , then $\pi (\widetilde {\Psi }(\gamma ))$ is a trajectory in $W^c\cap \{x>0\}$ accumulating to $0$ , and is thus contained in $S^c$ , which contradicts the fact that $b\not \in S$ . This proves point (iii, c) reciprocal.

Proof.

1. (1) $U_R$ admits $U_{-R}$ as a reciprocal and is clearly smooth; the expression of $U_R$ in restriction to its axis is the identity.

2. (2) For any fixed $x>0$ , the map ${\mathbf {y}}\mapsto U_R(x,{\mathbf {y}})$ is an isometry (with respect to Euclidean distance).

3. (3) From the shape of R, we notice that $\int _x^{+\infty } ({R(t)}/{t^{q+1}}) \; dt$ commutes with its derivative. The result follows classically.

4. (4) By induction on $\lfloor {M}/{q+1}\rfloor $ , if $q\ge M\ge 1$ , $x^M\Omega _R$ has a limit (zero) as x goes to $0$ , then admits a continuous extension. Otherwise, $ M\ge q+1$ and the differential equation satisfied by $\Omega _R$ gives

   $$ \begin{align*}(x^M\Omega_R)' = (Mx^q \text{Id} + R(x)) \cdot x^{M-(q+1)}\Omega_R.\end{align*} $$

   From the induction hypothesis, $x^{M-(q+1)}\Omega _R$ has a $C^{\lfloor {M}/{q+1}\rfloor -1}$ extension. Then $x^M \Omega _R(x)$ admits a $C^{\lfloor {M}/{q+1}\rfloor }$ extension.

5. (5) $\Omega _R$ and $\Omega _R^{-1}$ have the same block diagonal structure as R.

Proof. As in Lemma 3.4, the factor $x^e u(x,{\mathbf {y}})$ of the (TRS)-form does not intervene, since the result only involves the foliation induced by $\xi $ in the half-space $x>0$ . So we suppose that

$$ \begin{align*} \xi = x^{q+1}\partial_x+ ( (D(x)+x^qC)\cdot{\mathbf{y}}+x^{q+1+N+M}V(x,x^M{\mathbf{y}}))\cdot\partial_{{\mathbf{y}}}, \end{align*} $$

where D, C, V satisfy the hypothesis. Up to reordering the coordinate functions ${\mathbf {y}}$ , we assume that $y_1,\ldots ,y_m$ carry all dominant rotations. More precisely, we suppose that

$$ \begin{align*}D = \Theta(\text{Diag}(c_1,\ldots,c_{m/2}))\oplus\Theta(\text{Diag}(c_{m/2+1},\ldots,c_{n_1}))\oplus\text{Diag}(d_{1},\ldots,d_{n_2}),\end{align*} $$

where:

1. (1) $c_1,\ldots ,c_{m/2}\in \mathbb C_{q-1}[x]$ , and $\text {ord}_x(\text {Re}(c_j))>\text {ord}_x(\text {Im}(c_j))$ for $j=1,\ldots ,m/2$ ;

2. (2) $c_{m/2+1},\ldots ,c_{n_1}\in \mathbb C_{q-1}[x]$ , and $\text {ord}_x(\text {Re}(c_j))\le \text {ord}_x(\text {Im}(c_j))$ for $j=({m}/{2})+1,\ldots ,n_1$ ; and

3. (3) $d_1,\ldots ,d_{n_2}\in \mathbb R_{q-1}[x]$ .

If $m=0$ , then $(\xi ,\Gamma ,(x,{\mathbf {y}}))$ is in (TRS)-form of type $(q,N,0)$ without dominant rotation and Lemma 3.4 applies, which gives the result, taking into account that $\Gamma $ has contact at least $N+1$ with the x-axis.

Otherwise, for $l=1,\ldots ,m/2$ , we let

$$ \begin{align*}v_l=\text{ord}_x \text{Re} (c_l(x)) \quad\text{and}\quad b_l(x)=j_{v_l-1}(c_l(x)),\end{align*} $$

so $b_l(x)$ is the initial pure imaginary part of $c_l(x)$ . Define the rotational matrix

$$ \begin{align*} R(x) := \Theta(\text{Diag}(b_1(x),\ldots, b_{{m}/{2}}(x))\oplus \boldsymbol{0}_{n-m}. \end{align*} $$

We apply the transformation $(x,{\mathbf {y}})=U_R(x,{\mathbf {z}})$ . The trajectories $(x,{\mathbf {y}}(x))$ of $\xi $ included in $\mathbb R_+^*\times \mathbb R^{n}$ are in one-to-one correspondence with the trajectories $(x,{\mathbf {z}}(x))$ of $\eta =U_R^*\xi $ included in $\mathbb R_+^*\times \mathbb R^{n}$ , where

$$ \begin{align*} \eta = x^{q+1}\partial_x + [(D(x)-R(x) + x^q C )\cdot{\mathbf{z}} + x^{q+1+N+M}\Omega_{R}^{-1}(x)\cdot V(x, x^M \Omega_{R}(x) \cdot{\mathbf{z}})]\cdot\partial_{{\mathbf{z}}}. \end{align*} $$

The crucial point to get this expression is that C and $D(x)$ commute with $\Omega _{R}(x)$ and $\Omega _{R}^{-1}(x)$ (Lemma 4.2(5)).

Let $s\in \{0,\ldots ,q\}$ be the order of $D(x)-R(x)+x^qC$ , and let

$$ \begin{align*} {\boldsymbol{g}}(x,{\mathbf{z}})=x^M\Omega_{R}^{-1}(x)\cdot V(x, x^M \Omega_{R}(x)\cdot{\mathbf{z}}). \end{align*} $$

From Lemma 4.2(5), ${\boldsymbol {g}}$ admits a $C^{\lfloor {M}/{q+1}\rfloor }$ extension on $\mathbb R_+\times \mathbb R^n$ . The expression of the vector field $\eta /x^s$ is

$$ \begin{align*} \eta/x^s = x^{q-s+1}\partial_x + \bigg[\bigg(\frac{1}{x^s}(D-R)(x) + x^{q-s} C \bigg)\cdot {\mathbf{z}} + x^{q+1-s+N}{\boldsymbol{g}}(x, {\mathbf{z}})\bigg]\cdot \partial_{{\mathbf{z}}}. \end{align*} $$

We deduce that $(x,{\mathbf {z}})$ is a system of (TRS)-coordinates of type $(q-s,N,0)$ for $\eta /x^s$ , and that:

• • the exponential part $x^{-s}(D-R)(x)$ has no dominant rotation;

• • the residual part C has spectrum included in $\mathbb R^*_-\oplus i\mathbb R$ ; and

• • the vestigial part ${\boldsymbol {g}}(x,{\mathbf {y}})$ is $C^{\lfloor {M}/{q+1}\rfloor }$ , and $\lfloor {M}/{q+1}\rfloor \ge n(q+1+N)+1$ .

Then Lemma 3.4 applies to $\eta /x^s$ , and since $\eta $ and $\eta /x^s$ have the same trajectories, the conclusion of that result applies to $\eta $ . We have that $\eta $ admits a trajectory $(x,{\boldsymbol {\delta }}(x))$ which has contact of order at least $N+1$ with the x-axis. So $(x,{\boldsymbol {\gamma }}(x)):=(x,\Omega _R(x){\boldsymbol {\delta }}(x))$ is a trajectory of $\xi $ which also has contact of order at least $N+1$ with the x-axis, and, consequently, with $\Gamma $ . This proves point (i). Now if $(x,{\boldsymbol {\zeta }}(x))$ is a trajectory of $\xi $ with $\lim {\boldsymbol {\zeta }}(x)=0$ , then $(x,\Omega _{R}(x){\boldsymbol {\zeta }}(x))$ is a trajectory of $\eta $ that also satisfies $\lim \Omega _{R}(x){\boldsymbol {\zeta }}(x))=0$ . By Lemma 3.4, (ii), $\Omega _{R}(x){\boldsymbol {\zeta }}(x)$ and ${\boldsymbol {\delta }}(x)$ have flat contact. Since $\Omega _R$ preserves $\text {ord}_x$ , ${\boldsymbol {\zeta }}(x)$ and ${\boldsymbol {\gamma }}(x)$ have flat contact, which gives point (ii). Finally, point (iii) is obtained from the corresponding item (iii) of Lemma 3.4 applied to $\eta $ , taking into account that $\Omega _R$ provides a diffeomorphism from the half-space $\{x>0\}$ to itself, preserving the accumulation of trajectories to the origin. Concerning the dimension of S, notice that, according to Definition 3.1, we have equality of the instability indices $u(D(x))=u(x^{-s}(D(x)-R(x)))$ .