Proof. Since $(x,y)\rightarrow W^s_{\mathit {loc}}(\sigma )$ as $x\rightarrow 0$ , the orbit of $(x,y)$ approaches the singularity $\sigma $ . As we assumed $\lambda _1<\lambda _2<0$ and $\lambda _2+\lambda _3>0$ ; thus, and $\lim \limits _{x\rightarrow 0}\big |\partial H(x,y)/\partial y\big |=0$ . Moreover, for every $\alpha \geq 1/(\sqrt {2}-1)$ and , if we denote $DP(v)=a\cdot \partial /\partial x+b\cdot \partial /\partial y$ , then we have $|a|>\sqrt {2}$ and $|b|\leq 1+\alpha \leq \alpha \cdot \sqrt {2}$ . This implies .

Proof. The definition of $\mathcal {F}^s$ implies it is a 2-dimensional $\phi _t^X$ -invariant foliation in $U\setminus \mathcal {F}^{ss}(\sigma )$ . The classical invariant manifold theorem [Reference Hirsch, Pugh and Shub39] shows each leaf of $\mathcal {F}^{ss}$ is $C^r$ -smooth, and so is each leaf of $\mathcal {F}^s$ . The fact that holonomy maps of $\mathcal {F}^s$ are $C^{1}$ -smoothFootnote ¹ has been proved in [Reference Araújo and Melbourne3, Lemma 7.1].

Proof. Since $\Sigma $ is a $C^r$ -smooth cross section of $\phi _t^X$ , we can extend $\Sigma $ to a larger surface $\Sigma _1$ transverse to $\phi _t^X$ . Moreover, Lemma 3.1 shows that the stable foliation $\mathcal {F}^s$ intersects $\Sigma _1$ inducing a $C^{1}$ -smooth foliation in $\Sigma _1$ , where we already have . So we can $C^1$ -smoothly extend the coordinate of $\Sigma $ to $\Sigma _1=[-1-\epsilon ,1+\epsilon ]^2$ for some $\epsilon>0$ , such that

and $l_1=0\times [-1-\epsilon ,1+\epsilon ]=W^s_{\mathit {loc}}(\sigma ,\phi _t^X)\cap \Sigma _1$ .

Let $P_1:\Sigma _1\setminus l_1\rightarrow \Sigma _1$ be the Poincaré map associated to $\Sigma _1$ . Then it has the form $P_1(x,y)=(f_1(x), H_1(x,y))$ and $P_1|_{\Sigma \setminus l}\equiv P$ . In particular, $f_1$ is $C^1$ -smooth and $f_1|_{[-1,1]\setminus \{0\}}\equiv f$ . From the definition of geometric Lorenz map, the one-dimensional quotient map $f:[-1,1]\setminus \{0\}\rightarrow [-1,1]$ satisfies $-1<f(x)<1$ and $f'(x)>\sqrt {2}$ everywhere. This implies $-1<f(-1)<0<f(1)<1$ . Thus, if we choose $\epsilon $ small enough, we also have $-1<f_1(1-\epsilon )<0<f(1+\epsilon )<1$ . Shrinking $\epsilon $ if necessary, $P_1$ is well defined on $\Sigma _1\setminus l_1$ , and $H_1,f_1$ satisfies all estimations of $H,f$ . In particular, we have

$$ \begin{align*} \overline{P_1(\Sigma_1\setminus l_1)}= P_1(\Sigma_1\setminus l_1)\cup\{z^+,z^-\} \subseteq[-1,1]\times(-1,1).\\[-34pt] \end{align*} $$

Notation 3.4. From now on, we use $\Sigma ,P,f,H$ to denote $\Sigma _1,P_1,f_1,H_1$ , respectively, for the simplicity of notations. We denote $\pi _x:\Sigma \rightarrow [-1-\epsilon ,1+\epsilon ]$ the projection to x-coordinate on $\Sigma $ . Since , there exists $\lambda _0>\sqrt {2}$ , such that $f'(x)>\lambda _0$ for every . However, since $f(-1)>-1$ , we must have $\lambda _0<2$ .

We fix $\alpha =1/(\sqrt {2}-1)$ . The proof of Lemma 2.6 shows that the cone field $\mathcal {C}_{\alpha }$ satisfies

New proof of Lemma 3.6

For every $n\geq 1$ , we consider the map $f^n:J\rightarrow [-1,1]$ . Let

We call a discontinuity point of $f^n$ . Then $D_n$ is a finite subset of J with $D_n\subseteq D_{n+1}$ , and $f^n$ is well defined and $C^1$ -smooth on each connected component of $J\setminus D_n$ . Moreover, if $(a,b)\subseteq J$ is a connected component of $J\setminus D_n$ , then $f^n|_{(a,b)}$ is a diffeomorphism from $(a,b)$ to its image

Proof of the claim.

Recall that $f^n$ is well defined and $C^1$ -smooth on I. Let $(c,d)=f^n(I)$ . Since $I\cap D_n=\emptyset $ , one has that $f^m(t)\neq 0$ for any and $1\leq m\leq n-1$ by definition of $D_n$ . If $0\notin (c,d)$ , which means that $f^n(t)\neq 0$ for any , then one has $I\cap D_{n+1}=\emptyset $ by definition of $D_{n+1}$ . As a consequence, $f^{n+1}(I)=f(c,d)$ is an open interval in $(-1,1)$ . Thus, there exists at most one point , such that $f^{n+1}(t)=0$ . This implies $\#\big (I\cap D_{n+2}\big )\leq 1$ .

If, however, , then we denote satisfying $f^n(t_0)=0$ . Then the two open intervals $(a,t_0)$ and $(t_0,b)$ are two connected components of $J\setminus D_{n+1}$ . Moreover, their images by $f^{n+1}$ satisfy

$$ \begin{align*} f^{n+1}(a,t_0)=\big(\lim_{x\rightarrow a^+}f(x),~1\big), \qquad \mathrm{and } \qquad f^{n+1}(t_0,b)=\big(-1,~\lim_{x\rightarrow a^+}f(x)\big). \end{align*} $$

If neither of them contains $0$ , then we have $\#\big (I\cap D_{n+2}\big )=\#\big (I\cap D_{n+1}\big )=1$ .

Otherwise, we have either or . Thus, either

$$ \begin{align*} [0,1)\subset f^{n+1}(a,t_0)\subset f^{n+1}(I), \qquad \mathrm{or} \qquad (-1,0]\subset f^{n+1}(t_0,b)\subset f^{n+1}(I). \end{align*} $$

In both cases, iterating by f again, we have $(-1,1)\subseteq f^{n+1}(I)\cup f^{n+2}(I)$ .

Now we show that there exists some $n>0$ and a connected component I of $J\setminus D_n$ , such that $(-1,1)\subseteq f^{n+1}(I)\cup f^{n+2}(I)$ . Otherwise, Claim 3.7 implies

$$ \begin{align*} \# D_2\leq1, \!\!\!\qquad \# D_4\leq\# D_2+\# D_2+1\leq 3, \!\!\qquad \cdots \!\!\qquad \# D_{2n}\leq \# D_{2n-2}+\# D_{2n-2}+1\leq 2^n-1. \end{align*} $$

This implies $J\setminus D_{2n}$ has at most $2^n$ connected components. So there exists a connected component $I_{2n}\subset J\setminus D_{2n}$ , such that its length satisfies $|I_{2n}|\geq 2^{-n}|J|$ .

Since f is a diffeomorphism on each connected component, and satisfies $(f^{2n})'>\lambda _0^{2n}$ , the length of the interval $f^{2n}(I_{2n})$ satisfies

This is absurd since $f^{2n}(I_{2n})\subseteq (-1,1)$ . We take $N=n+2$ finishing the proof of this lemma.

Proof of Theorem 3.10.

By Proposition 3.8, we can shrink the $C^r$ -neighborhood $\mathcal {V}$ of X if necessary, such that for every , the maximal invariant set $\Lambda _Y=\bigcap _{t>0}\phi _t^Y(U)$ is a geometric Lorenz attractor.

There are two points satisfying $\phi _t^X(z^\pm )\notin \Sigma $ for every $t<0$ . We assume . The proof for $z=z^-$ is the same.

For the critical element , there are two possibilities: either $p=\sigma $ or p is a periodic point.

• • If $p=\sigma $ , there exists , such that $B(z,\delta )\cap W^s_{\mathit {loc}}(\sigma )=\emptyset $ . We only need to show there exists a vector field which satisfies $Y|_{M^3\setminus B(z,\delta )}~\equiv ~ X|_{M^3\setminus B(z,\delta )}$ , such that for some $T>0$ ,

  

• • If p is a periodic point, then we can assume with

  
  There exists , such that $B(z,\delta )\cap \left (\bigcup _{t\geq 0}\phi _t^X(l_p)\cup \{\sigma \}\right ) =\emptyset $ . We need to show there exists a vector field which satisfies $Y|_{M^3\setminus B(z,\delta )}~\equiv ~ X|_{M^3\setminus B(z,\delta )}$ , such that for some $T>0$ ,
  

From now on, we fix which satisfies the above property. Recall that the cross section $\Sigma =[-1-\epsilon ,1+\epsilon ]^2$ for some $\epsilon>0$ and $P:\Sigma \setminus l\rightarrow \Sigma $ is the Poincaré map.

Proof. The fact that $f(-1)>-1$ implies if $\eta $ is small enough, then $f([-1-\eta ,-1+\eta ])\cap [-1-\eta ,-1+\eta ]=\emptyset $ . Thus, $P(\Sigma _z)\cap \Sigma _z=\emptyset $ .

Since $P(\Sigma )\subset [-1,1]\times (-1,1)$ , if a point satisfies , then . The one-dimensional map f satisfies $f(-1-\epsilon )>-1$ , so if $\eta <f(-1-\epsilon )+1$ , then there exists some $\eta '>0$ , such that $P^{-1}(\Sigma _z)\subset (0,\eta ']\times [-1-\epsilon ,1+\epsilon ]$ . If $\eta \rightarrow 0^+$ , then $\eta '\rightarrow 0^+$ , and as . So we only need to shrink $\eta $ to take $\eta '$ small enough, such that $f'(x)>10$ for every . This shows item 1.

To verify item 2, note that $\Sigma _z=[-1-\eta ,-1+\eta ]\times [y^+-\eta ,y^++\eta ]$ is an $\eta $ -square neighborhood of $z=(-1,y^+)$ in the cross section $\Sigma $ and there exists $t_0>0$ such that the orbit segment $\phi _{[-t_0,4t_0]}^X(z)\subset B(z,\delta )$ . Thus, shrinking $\eta $ again if necessary, the flow box would be contained in the open ball $B(z,\delta )$ . Moreover, recall that satisfies $\phi ^X_t(z)\cap \Sigma =\emptyset $ for each $t<0$ and $\Sigma _z\subset \Sigma $ ; thus, $\phi _t^X(z)\notin \phi _{[0,3t_0]}^X(\Sigma _z)$ for each $t<0$ . This proves item 2.

Finally, note that $f'$ is continuous since the quotient map f is $C^1$ -smooth. Thus, by shrinking $\eta $ a last time if necessary, the following satisfies

As a consequence, for every , one has

$$\begin{align*}1-\beta_0<\frac{1}{1+\beta_0}<\frac{f'(x_2)}{f'(x_1)}<1+\beta_0,\end{align*}$$

and thus, $(1-\beta _0)f'(x_1)<f'(x_2)<(1+\beta _0)f'(x_1)$ which is item 3.

The subset $\phi _{[0,3t_0]}^X(\Sigma _z)$ admits a natural coordinate: for every , there exist a unique point and , such that $w=\phi _{t_w}^X(w_0)$ . Then we denote

We point out here that this coordinate is only $C^1$ -smooth.

Let be a smooth vector field satisfying $\operatorname {Supp}(Z)\subset \phi _{[t_0,2t_0]}^X(\Sigma _z)\subset B(z,\delta )$ such that if we represent Z in the coordinate $\{(x,y,t)\}$ as

$$ \begin{align*} Z(x,y,t)= a(x,y,t)\cdot\frac{\partial}{\partial x}+ b(x,y,t)\cdot\frac{\partial}{\partial y}+ c(x,y,t)\cdot\frac{\partial}{\partial t}, \end{align*} $$

then for every , it satisfies the following properties

1. 1. $\max \big \{ |b(x,y,t)|,|c(x,y,t)| \big \}\leq \beta |a(x,y,t)|$ and $|a(x,y,t)|<\beta $ , where $0<\beta <10^{-3}$ is a small constant to be determined in Claim 3.13 below.

2. 2. $a(x,y,t)\geq 0$ and $a(-1,y^+,3t_0/2)>0$ .

Here, since the coordinate $\{(x,y,t)\}$ is $C^1$ -smooth, the three functions $a,b,c$ are only uniformly continuous.

Proof of the claim.

The definitions of Z and $X_s$ imply the first three items directly. The vector field X on $\phi _{[0,3t_0]}^X(\Sigma _z)$ is equal to $\partial /\partial t$ everywhere. For every w in the boundary of $\Sigma _z$ , $Z|_{\phi _{[0,3t_0]}^X(w)}\equiv 0$ , this implies $\phi _t^{X_s}(w)=\phi _t^X(w)$ for every . Moreover, since

hence taking $0<\beta <10^{-3}$ small, for every and every , there exists some $t=t(w,s)$ which is close to $3t_0$ , such that . Thus, the Poincaré map from $\Sigma _z$ to $\Sigma $ associated to $\phi _t^{X_s}$ is well defined for every .

For every , its positive orbit does not intersect $\phi _{[0,3t_0]}^X(\Sigma _z)$ before it enters $\Sigma $ again. This implies for every , the first return map $P_s:\Sigma \setminus l\rightarrow \Sigma $ is well defined for the flow $\phi _t^{X_s}$ and satisfies $P_s|_{\Sigma \setminus \Sigma _z}~\equiv ~ P|_{\Sigma \setminus \Sigma _z}$ . This proves item 4.

Let

$$ \begin{align*} P_{s,1}:\Sigma_z\rightarrow\phi_{3t_0}^X(\Sigma_z), \qquad \mathrm{and} \qquad P_{s,2}:\phi_{3t_0}^X(\Sigma_z)\rightarrow\Sigma \end{align*} $$

be Poincaré maps of $\phi _t^{X_s}$ from $\Sigma _z$ to $\phi _{3t_0}^X(\Sigma _z)$ and from $\phi _{3t_0}^X(\Sigma _z)$ to $\Sigma $ , respectively. Then we have $P_s=P_{s,2}\circ P_{s,1}$ . Let the coordinate on $\phi _{3t_0}^X(\Sigma _z)$ be induced by pushing the coordinate of $\Sigma _z\subset \Sigma $ by $\phi _{3t_0}^X$ . Then $P_{s,2}\equiv P|_{\Sigma _z}$ in this coordinate.

Since depends continuously with respect to s in $\mathscr {X}^r(M^3)$ and the cross section $\Sigma $ is a $C^r$ smooth surface, the curve $P_s(w):[0,1]\rightarrow \Sigma $ is $C^r$ -smooth with respect to s for every .

The property $a(x,y,t)\geq 0$ everywhere implies for every , $\mathrm {d}\pi _x\circ P_{s,1}(w)/\mathrm {d}s\geq 0$ everywhere. Here, $\pi _x$ is the projection to x-coordinate on $\phi _{3t_0}^X(\Sigma _z)$ . Since $f'(x)\geq \lambda _0$ everywhere, we have

$$ \begin{align*} \frac{\mathrm{d}\pi_x\circ P_s(w)}{\mathrm{d}s}~=~ \frac{\mathrm{d}f}{\mathrm{d}x}\cdot \frac{\mathrm{d}\pi_x\circ P_{s,1}(w)}{\mathrm{d}s} ~\geq~0. \end{align*} $$

Moreover, since $|b(x,y,t)|\leq \beta \cdot |a(x,y,t)|\leq 10^{-3}\cdot |a(x,y,t)|$ for every $(x,y,t)$ , hence for every and every $0\leq s_1<s_2\leq 1$ , we have

$$ \begin{align*} \left|\pi_y\circ P_{s_2,1}(w)-\pi_y\circ P_{s_1,1}(w)\right| ~\leq~ 10^{-3}\cdot \left|\pi_x\circ P_{s_2,1}(w)-\pi_x\circ P_{s_1,1}(w)\right|. \end{align*} $$

This implies $\mathrm {d}P_{s,1}/\mathrm {d}s$ is tangent to the cone field $D\phi _{3t_0}^X\left (\mathcal {C}_{\alpha }|_{\Sigma _z}\right )$ everywhere on $\phi _{3t_0}^X(\Sigma _z)$ . Since $P_{s,2}\equiv P|_{\Sigma _z}$ , and P preserves the cone field $\mathcal {C}_{\alpha }$ (Lemma 2.6 and Notation 3.4), the curve $P_s(w):[0,1]\rightarrow \Sigma $ is tangent to $\mathcal {C}_{\alpha }$ everywhere.

For the point , since $a(-1,y^+,3t_0/2)>0$ , there exists a neighborhood of $(-1,y^+,3t_0/2)$ , such that for every $(x,y,t)$ in this neighborhood, we have $a(x,y,t)>0$ . Thus, there exists some $s'>0$ and $\kappa>0$ , such that for every $0<s<s'$ , we have $\pi _x\circ P_{s,1}(z)\geq -1+\kappa \cdot s$ . Thus, we have

$$ \begin{align*} \frac{\mathrm{d}\pi_x\circ P_s(z)}{\mathrm{d}s}|_{s=0}~=~ \frac{\mathrm{d}f}{\mathrm{d}x}|_{x=-1}\cdot \frac{\mathrm{d}\pi_x\circ P_{s,1}(w)}{\mathrm{d}s}|_{s=0} ~\geq~\lambda_0\cdot\kappa>0. \end{align*} $$

Then item 5 is proved, and the proof of the claim is completed.

Recall in Notation 3.4 that we have fixed $\alpha =1/(\sqrt {2}-1)$ and there exists , such that $f'(x)>\lambda _0$ for every . We fix the -vector field Z to satisfy Claim 3.13 and a constant . Also, we take $\beta _0>0$ in Claim 3.12 small enough such that

$$\begin{align*}\lambda(1-\beta_0(1+\alpha)(\lambda-\sqrt{2}))>\sqrt{2}.\end{align*}$$

Proof of the claim.

Since $X_s\rightarrow X$ in $\mathscr {X}^r(M^3)$ and $P_s|_{\Sigma \setminus \Sigma _z}\equiv P|_{\Sigma \setminus \Sigma _z}$ , we have $P_s\rightarrow P$ in $C^1$ -topology under the coordinate $\Sigma =[-1-\epsilon ,1+\epsilon ]^2$ . However,

So there exists $\tau _1>0$ , such that for every , whose Poincaré map $P_s$ satisfies $\partial \pi _x\circ P_s(x,y)/\partial x>\lambda $ and $\partial \pi _x\circ P_s(x,y)/\partial y<\beta _0(\lambda -\sqrt {2})$ for every . The proof for $|\partial \pi _y\circ P_s(x,y)/\partial x|<1$ and $|\partial \pi _y\circ P_s(x,y)/\partial y|<1$ is the same.

Since $P_s(z)$ is $C^r$ -smooth with respect to , the function $\pi _x\circ P_s(z)$ is $C^1$ -smooth with respect to s. There exists $\tau _2>0$ , such that for every , it satisfies

$$ \begin{align*} \frac{\mathrm{d}\pi_x\circ \gamma_0(s)}{\mathrm{d}s}|_{s=s'} ~\geq~\frac{1}{2}\cdot \frac{\mathrm{d}\pi_x\circ \gamma_0(s)}{\mathrm{d}s}|_{s=0}. \end{align*} $$

We take $\tau =\min \{\tau _1,\tau _2\}$ finishing the proof of the claim.

Fix from Claim 3.14. For every $n\geq 1$ , let

From Item 5 of Claim 3.13 and Item 2 of Claim 3.14, both $\pi _x\circ \gamma _0(s)$ and $\pi _x\circ P_s$ are monotonous increasing with respect to the parameter s. If some satisfies

then there exists $\epsilon _s>0$ , such that

This implies $D_n$ is a finite subset of $[0,\tau ]$ and satisfies $D_n\subseteq D_{n+1}$ .

Claim 3.15. For every $n\geq 0$ , the map

$$ \begin{align*} \gamma_n(s)=\left(x_n(s),~y_n(x)\right)=P_s^n\circ\gamma_0(s): ~[0,\tau]\setminus D_n\longrightarrow\Sigma \end{align*} $$

is well defined. For each connected component J of $[0,\tau ]\setminus D_n$ , $\gamma _n$ is a diffeomorphism from J to its image and satisfies

Proof of the claim.

We prove this claim by induction. When $n=0$ , the item 5 of Claim 3.13 shows that for every . And $x_0'(s)\geq \lambda ^0 x_0'(s)$ automatically holds for every . Now we assume for every $0\leq i\leq n-1$ and every ,

For every , we have

$$ \begin{align*} x_n(s)=\pi_x\circ P_s\left(x_{n-1}(s),y_{n-1}(s)\right), \qquad \mathrm{and} \qquad y_n(s)=\pi_y\circ P_s\left(x_{n-1}(s),y_{n-1}(s)\right). \end{align*} $$

Now we fix . There are two possibilities: either $(x_{n-1}(s_0),y_{n-1}(s_0))\notin \Sigma _z$ or . If $(x_{n-1}(s_0),y_{n-1}(s_0))\notin \Sigma _z$ , then there exists a neighborhood of $(x_{n-1}(s_0),y_{n-1}(s_0))$ , such that $P_s\equiv P$ in this neighborhood. Lemma 2.6 shows that

If , item 1 of Claim 3.12 implies $(x_{n-2}(s_0),y_{n-2}(s_0))\notin \Sigma _z$ , $0<x_{n-2}(s_0)<\eta '$ and $f'(x_{n-2}(s_0))>10$ . If we denote

then $\gamma _{n-1}'(s_0)= \left (DP|_{(x_{n-2}(s_0),y_{n-2}(s_0))}\right ) \big (\gamma _{n-2}'(s_0)\big )$ . Thus, we have

$$ \begin{align*} \left|y_{n-2}'(s_0)\right|\leq \alpha\cdot\left|x_{n-2}'(s_0)\right|, \qquad |x_{n-1}'(s_0)|=f'(x_{n-2}(s_0))\cdot|x_{n-2}'(s_0)| \geq 10|x_{n-2}'(s_0)|, \end{align*} $$

and

$$ \begin{align*} |y_{n-1}'(s_0)|~&\leq~ \frac{\partial H}{\partial x}|_{\gamma_{n-2}(s_0)} \cdot|x_{n-2}'(s_0)|~+~ \frac{\partial H}{\partial y}|_{\gamma_{n-2}(s_0)} \cdot|y_{n-2}'(s_0)| \\ &\leq~ |x_{n-2}'(s_0)|~+~|y_{n-2}'(s_0)| \\ &\leq~(1+\alpha)|x_{n-2}'(s_0)|. \end{align*} $$

Now we calculate $\gamma _n'(s_0)$ . Since $x_n(s)=\pi _x\circ P_s\left (x_{n-1}(s),y_{n-1}(s)\right )$ and $y_n(s)=\pi _y\circ P_s\left (x_{n-1}(s),y_{n-1}(s)\right )$ , if we denote $w_{n-1}=\left (x_{n-1}(s_0),y_{n-1}(s_0)\right )$ , then

$$ \begin{align*} x_n'(s_0)= \frac{\partial \pi_x\circ P_{s_0}}{\partial x} \cdot x_{n-1}'(s_0) + \frac{\partial \pi_x\circ P_{s_0}}{\partial y} \cdot y_{n-1}'(s_0) + \frac{\partial \pi_x\circ P_s(w_{n-1})}{\partial s} |_{s=s_0}. \end{align*} $$

From item 1 of Claim 3.14, we have

$$ \begin{align*} \frac{\partial \pi_x\circ P_{s_0}}{\partial x} \cdot x_{n-1}'(s_0) \geq \lambda\cdot x_{n-1}'(s_0) \quad \mathrm{and} \quad \left|\frac{\partial \pi_x\circ P_{s_0}}{\partial y} \cdot y_{n-1}'(s_0)\right| \leq \beta_0(\lambda-\sqrt{2})\cdot|y_{n-1}'(s_0)|. \end{align*} $$

Since $|y_{n-1}'(s_0)|\leq \alpha \cdot x_{n-1}'(s_0)$ and $\lambda (1-\beta _0(1+\alpha )(\lambda -\sqrt {2}))>\sqrt {2}$ , we have

(3.1) $$ \begin{align} x_n'(s_0)\geq \sqrt{2}\cdot x_{n-1}'(s_0) + \frac{\partial \pi_x\circ P_s(w_{n-1})}{\partial s} |_{s=s_0}. \end{align} $$

However, we have

$$ \begin{align*} y_n'(s_0)=\frac{\partial \pi_y\circ P_{s_0}}{\partial x} \cdot x_{n-1}'(s_0) + \frac{\partial \pi_y\circ P_{s_0}}{\partial y} \cdot y_{n-1}'(s_0) + \frac{\partial \pi_y\circ P_s(w_{n-1})}{\partial s} |_{s=s_0}. \end{align*} $$

Again, item 1 of Claim 3.14 shows that $|\partial \pi _y\circ P_{s_0}/\partial x|<1$ and $|\partial \pi _y\circ P_{s_0}/\partial y|<1$ . Thus, we have

(3.2) $$ \begin{align} |y_n'(s)|<(1+\alpha)\cdot x_{n-1}'(s_0)+ \left|\frac{\partial \pi_y\circ P_s(w_{n-1})}{\partial s} |_{s=s_0}\right|. \end{align} $$

Item 5 of Claim 3.13 shows that

which implies

(3.3) $$ \begin{align} \left|\frac{\partial \pi_y\circ P_s(w_{n-1})}{\partial s}|_{s=s_0}\right| ~\leq~\alpha\cdot \frac{\partial \pi_x\circ P_s(w_{n-1})}{\partial s}|_{s=s_0}. \end{align} $$

Since $\alpha =1/(\sqrt {2}-1)$ which is equivalent to $(1+\alpha )/\sqrt {2}=\alpha $ , Equation 3.1 and Equation 3.2 imply

Finally, Equation 3.1 and Equation 3.3 show that

$$ \begin{align*} x_n'(s_0)~\geq~\sqrt{2}\cdot x_{n-1}'(s_0) ~\geq~ 10\sqrt{2}\cdot x_{n-2}'(s_0) ~>~ \lambda^2\cdot\lambda^{n-2} ~=~ \lambda^n. \end{align*} $$

This finishes the proof of the claim.

We can finish the proof of this theorem by using the idea of Lemma 3.6.

Claim 3.16. There exists some integer $N>0$ , such that

$$ \begin{align*} (-1,1)~\subseteq~ \bigcup_{i=0}^N\pi_x\circ\gamma_n\big([0,\tau]\big). \end{align*} $$

Proof of the claim.

If $(a,b)\subseteq [0,\tau ]$ is a connected component of $[0,\tau ]\setminus D_n$ , then

$$ \begin{align*} \pi_x\circ\gamma_n(s)=\pi_x\circ P_s^n\circ\gamma_0(s): ~[0,\tau]\rightarrow[-1,1] \end{align*} $$

is a diffeomorphism from $(a,b)$ to its image.

Similar to Claim 3.7, we show that for every $n\geq 1$ , let $I=(a,b)$ be a connected component of $[0,\tau ]\setminus D_n$ . Then either $\#\big (I\cap D_{n+2}\big )\leq 1$ or $(-1,1)\subseteq \pi _x\circ \gamma _{n+1}(I)\cup \pi _x\circ \gamma _{n+2}(I)$ .

Actually, let $(c,d)=\pi _x\circ \gamma _n(I)$ . If $0\notin (c,d)$ , then $I\cap D_{n+1}=\emptyset $ and

$$ \begin{align*} \pi_x\circ\gamma_{n+1}(I)= \pi_x\circ P_s^{n+1}\circ\gamma_0(I) \end{align*} $$

is an open interval in $(-1,1)$ . Thus, there exists at most one point satisfying $\pi _x\circ \gamma _{n+1}(s_0)=0$ . This implies $\#\big (I\cap D_{n+2}\big )\leq 1$ .

If , then we denote satisfying $\pi _x\circ \gamma _n(s_0)=\pi _x\circ P_{s_0}^n\circ \gamma _0(s_0)=0$ . Then two open intervals $(a,s_0)$ and $(s_0,b)$ are two connected components of $I\setminus D_{n+1}$ . Moreover, item 4 of Claim 3.13 shows that $P_s|_{\Sigma \setminus \Sigma _z}\equiv P|_{\Sigma \setminus \Sigma _z}$ . This implies

$$ \begin{align*} \lim_{s\rightarrow s_0^-}\pi_x\circ\gamma_{n+1}(s)=1, \qquad \mathrm{and} \qquad \lim_{s\rightarrow s_0^+}\pi_x\circ\gamma_{n+1}(s)=-1. \end{align*} $$

Thus, there exists , such that

$$ \begin{align*} \pi_x\circ\gamma_{n+1}\big( (a,s_0) \big)=(e_1,1), \qquad \mathrm{and} \qquad \pi_x\circ\gamma_{n+1}\big( (s_0,b) \big)=(-1,e_2). \end{align*} $$

If both of them do not contain $0$ , then we have $\#\big (I\cap D_{n+2}\big )=\#\big (I\cap D_{n+1}\big )=1$ .

Otherwise, we have either or . Thus, either

$$ \begin{align*} [0,1)\subset \pi_x\circ\gamma_{n+1}\big( (a,s_0) \big) \subset \pi_x\circ\gamma_{n+1}(I), \qquad \mathrm{or} \qquad (-1,0]\subset \pi_x\circ\gamma_{n+1}\big( (s_0,b) \big) \subset \pi_x\circ\gamma_{n+1}(I). \end{align*} $$

In both cases, we can apply $P_s|_{\Sigma \setminus \Sigma _z}\equiv P|_{\Sigma \setminus \Sigma _z}$ and show that

$$ \begin{align*} (-1,1)\subseteq \pi_x\circ\gamma_{n+1}(I)\cup \pi_x\circ\gamma_{n+2}(I). \end{align*} $$

Now we claim there exists some $n>0$ and a connected component I of $[0,\tau ]\setminus D_n$ , such that $(-1,1)\subseteq \pi _x\circ \gamma _{n+1}(I)\cup \pi _x\circ \gamma _{n+2}(I)$ . Otherwise, we have

$$ \begin{align*} \# D_2\leq1, \qquad \# D_4\leq\# D_2+\# D_2+1\leq 3, \qquad \cdots \qquad \# D_{2n}\leq \# D_{2n-2}+\# D_{2n-2}+1\leq 2^n-1. \end{align*} $$

This implies $[0,\tau ]\setminus D_{2n}$ has at most $2^n$ connected components. So there exists a connected component $I_{2n}\subset [0,\tau ]\setminus D_{2n}$ , such that its length satisfies $|I_{2n}|\geq 2^{-n}\tau $ .

Since $\pi _x\circ \gamma _{2n}$ is a diffeomorphism on each connected component of $[0,\tau ]\setminus D_{2n}$ , and Claim 3.15 shows that

the length of the interval $\pi _x\circ \gamma _n(I_{2n})$ satisfies $|\pi _x\circ \gamma _n(I_{2n})| \geq \lambda ^{2n}\cdot |\gamma _0(I_{2n})|$ .

Finally, item 2 of Claim 3.14 shows that

$$ \begin{align*} |\gamma_0(I_{2n})|~\geq~\frac{1}{2}\cdot \frac{\mathrm{d}\pi_x\circ \gamma_0(s)}{\mathrm{d}s}|_{s=0} \cdot|I_{2n}| ~\geq\frac{1}{2}\cdot \frac{\mathrm{d}\pi_x\circ \gamma_0(s)}{\mathrm{d}s}|_{s=0}\cdot 2^{-n}\tau. \end{align*} $$

Thus, we have

This is absurd because $\pi _x\circ \gamma _n(I_{2n})\subseteq (-1,1)$ . So there exists $n>0$ and a connected component I of $[0,\tau ]\setminus D_n$ , such that $(-1,1)\subseteq \pi _x\circ \gamma _{n+1}(I)\cup \pi _x\circ \gamma _{n+2}(I)$ . We take $N=n+2$ to complete the proof of this claim.

For the critical point , recall that $l_p=\{x_p\}\times [-1-\epsilon ,1+\epsilon ]\subset \Sigma $ with (if $p=\sigma $ , then $x_p=0$ ). Claim 3.16 shows there exist and $n>0$ , such that

$$ \begin{align*} x_p~=~\pi_x\circ P_s^n\circ\gamma_0(s) ~=~\pi_x\circ P_s^n\circ P_s(z). \end{align*} $$

This implies for the vector field , there exists $T>0$ , such that .

Since $X_s|_{M^3\setminus B(z,\delta )}\equiv X|_{M^3\setminus B(z,\delta )}$ and $B(z,\delta )\cap \left (\bigcup _{t\geq 0}\phi _t^X(l_p)\cup \{\sigma \}\right ) =\emptyset $ , we have $l_p\subset W^s_{\mathit {loc}}(\operatorname {Orb}(p),\phi _t^Y)$ . The fact implies

$$ \begin{align*} \operatorname{Orb}(z,\phi_t^Y)~\subset~ W^u(\sigma,\phi_t^Y) \cap W^s(\operatorname{Orb}(p),\phi_t^Y). \end{align*} $$

This completes the proof of this theorem.

Proof. Since the two hyperbolic periodic orbits $\operatorname {Orb}(p)$ and $\operatorname {Orb}(q)$ are homoclinically related, there exists a hyperbolic horseshoe $K\subset \Lambda $ of X containing p and q. Note here that the hypberolic horseshoe K of X is the suspension of a hyperbolic horseshoe $\tilde {K}$ for a diffeomorphism (the time one map of the flow) with a continuous roof function. By [Reference Sigmund66, Theorem B], the space $\mathcal {M}_{erg}(\tilde {K})$ is path connected. However, there is a one-to-one correspondence between $\mathcal {M}_{erg}(\tilde {K})$ and $\mathcal {M}_{erg}(K)$ ; see, for, instance [Reference Parry and Pollicott54, Chapter 6]. Hence, $\mathcal {M}_{erg}(K)$ is path connected. Moreover, $\mathcal {M}_{per}(K)$ is dense in $\mathcal {M}_{erg}(K)$ since K is a hyperbolic horseshoe. Thus, $\delta _{\operatorname {Orb}(p)}$ and $\delta _{\operatorname {Orb}(q)}$ can be connected by a path contained in $\mathcal {M}_{erg}(K)\subset \overline {\mathcal {M}_{per}(\Lambda )}\cap \mathcal {M}_{erg}(\Lambda )$ .

Proof. Since $\operatorname {Orb}(p)$ and $\operatorname {Orb}(q)$ are homoclinically related, there exist hyperbolic horseshoes such that every point in which spends arbitrarily large fractions of its orbit shadowing $\operatorname {Orb}(p)$ and $\operatorname {Orb}(q)$ as closely as we want. Thus, we can choose a hyperbolic horseshoe K containing p and q such that any ergodic measure supported on K is $\eta $ -close to $\delta _{\operatorname {Orb}(p)}$ or $\delta _{\operatorname {Orb}(q)}$ . As a consequence, the horseshoe K is $2\eta $ -close to $\mu $ in the measure sense: any ergodic measure supported on K is contained in the $2\eta $ -neighborhood of $\mu $ . By [Reference Sigmund66, Theorem B], there is a continuous path connecting $\delta _{\operatorname {Orb}(p)}$ and $\delta _{\operatorname {Orb}(q)}$ , satisfying $d(\mu ,\nu _t)<2\eta $ for any . Then Lemma 4.3 concludes if we take $\eta =\varepsilon /2$ .

Proof of Proposition 4.1.

Take two ergodic measures . We only need to show there exists a path such that $\mu _0=\mu $ and $\mu _1=\nu $ . We follow the idea of [Reference Gorodetski and Pesin31, Theorem 1.1]. By assumption, there exist two sequences of periodic points $\{p_n\}$ and $\{q_n\}$ in $\Lambda $ such that $\delta _{\operatorname {Orb}(p_n)}\rightarrow \mu $ and $\delta _{\operatorname {Orb}(q_n)}\rightarrow \nu $ in the weak*-topology. Taking subsequences if necessary, we assume

$$\begin{align*}d(\delta_{\operatorname{Orb}(p_n)},\mu)<\frac{1}{2\cdot3^n} \text{~~and~~} d(\delta_{\operatorname{Orb}(q_n)},\nu)<\frac{1}{2\cdot3^n}, \text{ for all }n\geq 1. \end{align*}$$

By Lemma 4.2, there exists a continuous path such that $\nu _{\frac {1}{3}}=\delta _{\operatorname {Orb}(p_1)}$ and $\nu _{\frac {2}{3}}=\delta _{\operatorname {Orb}(q_1)}$ . By Lemma 4.3, for any $n\geq 1$ , there exist two paths

in $\overline {\mathcal {M}_{per}(\Lambda )}\cap \mathcal {M}_{erg}(\Lambda )$ connecting $\delta _{\operatorname {Orb}(p_n)},\delta _{\operatorname {Orb}(p_{n+1})}$ and $\delta _{\operatorname {Orb}(q_n)},\delta _{\operatorname {Orb}(q_{n+1})}$ , respectively, and the length of the paths decreases to 0 exponentially when n goes to . Therefore, is a continuous path in $\overline {\mathcal {M}_{per}(\Lambda )}\cap \mathcal {M}_{erg}(\Lambda )$ satisfying that $\nu _0=\mu $ and $\nu _1=\nu $ . This completes the proof.

Proof. Let $d=\dim M$ and $s=\dim E^s(\sigma )$ be the index of $\sigma $ . Then $\dim E^u(\sigma )=d-s$ . Since $\Gamma =\operatorname {Orb}(z,\phi _t^X)$ is a homoclinic loop of X, we have

By Grobman-Hartman Theorem [Reference de Melo and Palis27, 4.10 Theorem], there exists a small neighborhood $U(\sigma )$ of $\sigma $ , such that the flow $\phi _t^X$ is orbit equivalent to the linear system generated by $DX(\sigma )$ . Moreover, shrinking $U(\sigma )$ if necessary, we can assume there exist $T_1<0<T_2$ satisfying

where and .

There exists a $(d-1)$ -dimensional -smooth cross section $\Sigma _1$ which is transverse to X everywhere and satisfies the following properties:

• • , and $\phi _{[0,1]}^X(\Sigma _1)\cap \Gamma =\phi _{[T_1-1,T_1]}^X(z)$ .

• • There exists a s-dimensional -smooth disk $\Sigma _1^s\subset \Sigma _1$ such that $\Sigma _1^s$ is transverse to $W^u_{\mathit {loc}}(\sigma )$ at $\phi _{T_1-1}^X(z)$ .

Symmetrically, there exists a $(d-1)$ -dimensional -smooth cross section $\Sigma _2$ which is transverse to X everywhere and satisfies the following properties:

• • , and $\phi _{[-1,0]}^X(\Sigma _2)\cap \Gamma =\phi _{[T_2,T_2+1]}^X(z)$ ;

• • there exists a $(d-s)$ -dimensional -smooth disk $\Sigma _2^u\subset \Sigma _2$ such that $\Sigma _2^u$ is transverse to $W^s_{\mathit {loc}}(\sigma )$ at $\phi _{T_2+1}^X(z)$ .

By the $\lambda $ -lemma (Inclination Lemma) (see [Reference de Melo and Palis27, 7.2 Lemma]), there exists a sequence of points and , such that

1. 1. The point $x_n\rightarrow \phi _{T_2+1}^X(z)$ in $\Sigma _2^u$ as .

2. 2. The point , and $y_n\rightarrow \phi _{T_1-1}^X(z)$ in $\Sigma _1^s$ as .

3. 3. The orbit segment $\phi _{[0,S_n]}^X(x_n)\subset U(\sigma )$ for every n.

Now we consider two flow boxes $V_1=\phi _{[T_1-1,T_1]}^X(\Sigma _1)$ and $V_2=\phi _{[T_2,T_2+1]}^X(\Sigma _2)$ . Since $x_n\rightarrow \phi _{T_2+1}^X(z)$ in $\Sigma _2^u$ and $y_n\rightarrow \phi _{T_1-1}^X(z)$ in $\Sigma _1^s$ , there exists a sequence of vector fields , such that

1. 1. The vector field $X_n$ converges to X in $C^r$ -topology.

2. 2. $X_n|_{M\setminus (V_1\cup V_2)}\equiv X|_{M\setminus (V_1\cup V_2)}$ .

3. 3. There exist $a_n,b_n\rightarrow 1$ as , such that

   $$ \begin{align*} \phi_{a_n}^{X_n}(y_n)=\phi_{T_1}^X(z), \qquad \mathrm{and} \qquad \phi_{b^n}^{X_n}\big( \phi_{T_2}^X(z) \big)=x_n. \end{align*} $$

This implies the vector field $X_n$ satisfies $\phi _{b_n+S_n+a_n}^{X_n}\big (\phi _{T_2}^X(z)\big )=\phi _{T_1}^X(z)$ . So $\Gamma _n=\operatorname {Orb}(z,\phi _t^{X_n})$ is a periodic orbit of $X_n$ with period $\pi _n=a_n+b_n+(T_2-T_1)+S_n$ . Moreover, we have

Recall that and $y_n\rightarrow \phi _{T_1-1}^X(z)$ . Since $(\pi _n-S_n)$ is uniformly bounded above, the fact that

$$ \begin{align*} \phi_{[0,S_n]}^{X_n}(x_n)=\phi_{[0,S_n]}^X(x_n)\subseteq U(\sigma) \end{align*} $$

and $\phi _t^X$ is orbit equivalent to the linear system generated by $DX(\sigma )$ in $U(\sigma )$ implies $\Gamma _n$ converges to $\Gamma \cup \{\sigma \}$ in the Hausdorff topology and the periodic measure $\delta _{\Gamma _n}$ converges to $\delta _{\sigma }$ .

Proof. Assume there exists a sequence of ergodic measures $\mu _n$ approximating the Dirac measure $\delta _{\sigma }$ . We only need to find a continuous function $\varphi :M\rightarrow {\mathbb R}$ , such that does not converge to . Then we get a contradiction.

Since $\sigma $ is a hyperbolic singularity, each $\mu _n$ does not support on a singularity. Let be a generic point of $\mu _n$ (i.e., the empirical measure equidistributed on the orbit segment $\phi _{[0,T]}(x_n)$ converges to $\mu _n$ as ). We must have $\operatorname {Ind}(p)<d-1$ . Otherwise, $\operatorname {Orb}(p)$ is a periodic sink. Hence, as the positive orbit of $x_n$ approaches $\sigma $ , it also approaches $W^u(\sigma )$ and enters the attracting region of $\operatorname {Orb}(p)$ . This contradicts , which implies that $x_n$ is a recurrent point.

As we have assumed $\dim E^u(\sigma )=1$ , let $\lambda _1,\cdots ,\lambda _d$ be eigenvalues of $DX(\sigma )$ which satisfy $|\mathrm {Re}\lambda _1 |\leq \cdots \leq |\mathrm {Re}\lambda _{d-1}| < 1< |\mathrm {Re}\lambda _d|=|\lambda _d |$ . We fix a constant $\epsilon _1$ satisfying $0<\epsilon _1<\min \big \{-\log |\mathrm {Re}\lambda _{d-1}|,\log |\lambda _d|\big \}$ , and let

(4.1) $$ \begin{align} u_1=\log|\mathrm{Re}\lambda_{d-1}|+\epsilon_1<0. \end{align} $$

Claim 4.7. There exists a neighborhood $U(\sigma )$ of $\sigma $ admitting a $C^1$ -coordinate

such that

1. 1. The singularity , and the local stable and unstable manifolds of $\sigma $ in $U(\sigma )$ are

   $$ \begin{align*} W^s_{\mathit{loc}}(\sigma)=\big\{(x,0):~\|x\|\leq1\big\}, \qquad \mathit{and} \qquad W^u_{\mathit{loc}}(\sigma)=\big\{(0,y):~|y|\leq1\big\}. \end{align*} $$

2. 2. For every , let $\phi _t(z)=\big (x(t),y(t)\big )$ be the orbit of z under $\phi _t$ where $\big (x(0),y(0)\big )=(x,y)$ . Assume the orbit segment $\phi _{[T_1,T_2]}(z)\subset U(\sigma )$ , where $T_1<0<T_2$ . Then it satisfies $|y'(t)-\lambda _d\cdot y(t)|\leq \epsilon _1\cdot |y(t)|,~~\forall T_1\leq t \leq T_2$ ,

   $$ \begin{align*} \|x(t)\|\leq\exp(u_1 t)\|x(0)\|, ~~\forall 0\leq t \leq T_2; \quad\!\! \mathit{and}\!\! \quad \|x(t)\|\geq\exp(u_1 t)\|x(0)\|, ~~\forall T_1\leq t\leq0. \end{align*} $$

3. 3. $U(\sigma )\cap \operatorname {Orb}(p,\phi _t)=\emptyset $ , and for , they satisfy $\phi _t(z^\pm )\notin U(\sigma )$ for every $t>0$ .

For every $0<\delta <1$ , we denote . Let

Then $\Sigma ^{\pm }\subseteq \partial U(\sigma )$ . Moreover, the second item of Claim 4.7 shows that the vector field X is transverse to $\Sigma ^{\pm }$ and points outside $U(\sigma )$ . For every $\delta>0$ , $\Sigma ^\pm $ are local cross sections of X centered at $z^\pm $ , respectively.

Proof of Claim 4.8.

From the second item of Claim 4.7, for $z=(x,y)$ satisfying $x\neq 0$ and $y>0$ , the flow $\big (x(t),y(t)\big )$ satisfies

• • $y(t)>0$ and monotonous increasing with respect to t, which implies $\exists T_2>0$ such that ;

• • $\|x(t)\|\geq \exp (u_1 t)\|x(0)\|$ for every $t\leq 0$ , which implies $\exists T_1<0$ , such that .

Moreover, we have for every . This proves the first item. The proof of second item is the same.

As $\|x(T_1)\|=1$ and $\|x(T_2)\|=\|x_w\|\leq \exp (u_1 T_2)\|x\|<\delta $ , we apply the fact that $\|x(t)\|\geq \exp (u_1 t)\|x(0)\|$ for every $t\leq 0$ , which shows

$$ \begin{align*} T_2'-T_1'~\leq~T_2-T_1~\leq~ \frac{1}{u_1}\cdot\big(\log\|x_w\|-\log1\big) ~=~\frac{1}{u_1}\cdot\log\|x_w\|.\\[-34pt] \end{align*} $$

Recall p is a hyperbolic periodic point of X with index $0<\operatorname {Ind}(p)<d-1$ . We denote $k=\operatorname {Ind}(p)$ . Then we have the following claim.

Claim 4.9. There exist a cross section $\Sigma _p$ transverse to X everywhere admitting a $C^1$ coordinate

and two constants $\kappa>1,~T_0>0$ , such that

1. 1. The orbit $\operatorname {Orb}(p,\phi _t)\cap \Sigma _p=\{p\}$ and .

2. 2. The local stable and unstable manifolds of the orbit of p intersect $\Sigma _p$ as

   $$ \begin{align*} W^s_{\mathit{loc}}(\operatorname{Orb}(p))\cap\Sigma_p= \big\{ (r,0):\|r\|\leq1 \big\}, \qquad W^u_{\mathit{loc}}(\operatorname{Orb}(p))\cap\Sigma_p= \big\{ (0,s):\|s\|\leq1 \big\}. \end{align*} $$

3. 3. The Poincaré return map

   
   is well defined and $C^1$ -smooth on $\Sigma _p'$ . For every , if we denote $(r_1,s_1)=R(r,s)$ , then
   $$ \begin{align*} \|r_1\|<\|r_0\|, \qquad \mathrm{and} \qquad \|s_1\|\leq\kappa\cdot\|s_0\|. \end{align*} $$

4. 4. For every , let $T_z>0$ be the first return time of z to $\Sigma _p$ (i.e. ). Then $T_z\geq T_0$ and

   

Proof of Claim 4.9.

Let $\pi (p)$ be the period of p and $u_2>0$ be the largest Lyapunov exponent of $\operatorname {Orb}(p)$ . There exists a $C^1$ cross section $\Sigma _1$ containing p and transversing to X everywhere. By shrinking $\Sigma _1$ if necessary, we can assume $\Sigma _1\cap \operatorname {Orb}(p,\phi _t)=\{p\}$ .

Since p is a hyperbolic periodic orbit of X, there exists a small neighborhood $\Sigma _2\subseteq \Sigma _1$ , such that the Poincaré map $R:\Sigma _2\rightarrow \Sigma _1$ is well defined and $C^1$ -smooth. Moreover, p is a hyperbolic fixed point of R with largest Lyapunov exponent equal to $\big (\pi (p)u_2\big )$ . For every , let $T_z$ be the first return time of z in $\Sigma _1$ (i.e., $R(z)=\phi _{T_z}(z)$ ). By shrinking $\Sigma _2$ if necessary, there exists $T_0>0$ which is close to $\pi (p)$ , such that $T_z\geq T_0$ for every . Moreover, since $U(\sigma )\cap \operatorname {Orb}(p)=\emptyset $ , we can assume

Finally, we fix a small constant $\epsilon _2>0$ and take

(4.2) $$ \begin{align} \kappa=\exp\big[\pi(p)\cdot(u_2+\epsilon_2)\big]. \end{align} $$

Similar to Claim 4.7, we can take an adapted $C^1$ -coordinate on $\Sigma _2$ , such that in this coordinate and the second and third items of the claim are satisfied on

See [Reference Shilnikov, Shilnikov, Turaev and Chua63, Theorem 2.4 & 2.5]. This finishes the proof of the claim.

Let which is a neighborhood of $\operatorname {Orb}(p)$ . Since $W^u(\sigma )\setminus \{\sigma \}\subseteq W^s(\operatorname {Orb}(p))$ , for , there exist $T^\pm>0$ and such that

$$ \begin{align*} (r^+,0)=\phi_{T^+}(z^+), \qquad \mathrm{and} \qquad (r^-,0)=\phi_{T^-}(z^-). \end{align*} $$

Since $\operatorname {Orb}^+(z^\pm )\cap U(\sigma )=\emptyset $ , there exist two constants $\delta _0>0,K>0$ with $K\cdot \delta _0\ll 1$ , such that

• • The Poincaré map $P^+:\Sigma ^+(\delta _0)\rightarrow \Sigma _p$ is well defined and $C^1$ -smooth, satisfying $P^+(z^+)=P^+(0,1)=(r^+,0)$ , and for every , if we denote $P^+(w)=(r_w,s_w)$ , then it satisfies

  $$ \begin{align*} \|r_w-r^+\|\leq K\cdot\|x_w\|, \qquad \mathrm{and} \qquad \|s_w\|\leq K\cdot\|x_w\|. \end{align*} $$
  Moreover, if we denote $T^+_w>0$ the time satisfying $P^+(w)=\phi _{T^+_w}(w)$ , then the orbit segment $\phi _{(0,T^+_w)}(w)\cap U(\sigma )=\emptyset $ .

• • The Poincaré map $P^-:\Sigma ^-(\delta _0)\rightarrow \Sigma _p$ is well defined and $C^1$ -smooth, satisfying $P^-(z^-)=P^-(0,-1)=(r^-,0)$ , and for every , if we denote $P^-(w)=(r_w,s_w)$ , then it satisfies

  $$ \begin{align*} \|r_w-r^-\|\leq K\cdot\|x_w\|, \qquad \mathrm{and} \qquad \|s_w\|\leq K\cdot\|x_w\|. \end{align*} $$
  Moreover, if we denote $T^-_w>0$ the time satisfying $P^-(w)=\phi _{T^-_w}(w)$ , then the orbit segment $\phi _{(0,T^-_w)}(w)\cap U(\sigma )=\emptyset $ .

Claim 4.10. For every with $x_z\neq 0,y_z>0$ , let $T_1<0<T_2$ be defined as in Claim 4.8 and . Then for the point , let

$$ \begin{align*} T_p=\left(-\frac{\log(K\|x_w\|)}{\log\kappa}-2\right)\cdot T_0. \end{align*} $$

We have $\phi _{[0,T_p]}(P^+(w))\subseteq V(p)$ .

In particular, combined with the third item of Claim 4.8 that $T_2-T_1\leq \log \|x_w\|/u_1$ , we have

(4.3) $$ \begin{align} \frac{T_p}{T_2-T_1}~\geq~ \frac{-u_1\cdot T_0}{\log\kappa}\cdot \left(1+\frac{\log K+2\log\kappa}{\log\|x_w\|}\right). \end{align} $$

The same conclusion holds for with $x_z\neq 0,y_z<0$ .

Proof of Claim 4.10.

We only prove the case that $y_z>0$ . From the property of Poincaré map $P^+:\Sigma ^+(\delta _0)\to \Sigma _p$ , we have $\|s_w\|\leq K\cdot \|x_w\|$ . Moreover, for the Poincaré map $R:\Sigma _p'\to \Sigma _p$ in the neighborhood of $\mathrm {Orb}(p)$ , we can apply the third item of Claim 4.9, which shows that for the point , it will iterate by R at least $([-\log \|s_w\|/\log \kappa ]-1)$ -times hitting $\Sigma _p'$ inside $V(p)$ . So for

$$ \begin{align*} T_p~=~\left(-\frac{\log(K\|x_w\|)}{\log\kappa}-2\right)\cdot T_0 ~\leq~ \left(\left[\frac{\log\|s_w\|}{\log\kappa}\right]-1\right)\cdot T_0, \end{align*} $$

we have $\phi _{[0,T_p]}(P^+(w))\subseteq V(p)$ .

Moreover, by the fact $0<T_2-T_1\leq \log \|x_w\|/u_1$ , we have the following estimates:

$$ \begin{align*} \frac{T_p}{T_2-T_1}&\geq \left(-\frac{\log(K\|x_w\|)}{\log\kappa}-2\right)\cdot T_0\cdot \frac{u_1}{\log\|x_w\|}\\ &= \frac{-u_1\cdot T_0}{\log \kappa}\left(\frac{\log(K\|x_w\|)+2\log\kappa}{\log\|x_w\|} \right)\\ &= \frac{-u_1\cdot T_0}{\log\kappa}\cdot \left(1+\frac{\log K+2\log\kappa}{\log\|x_w\|}\right). \end{align*} $$

This proves the claim.

Now we take small enough, such that $(\log K+2\log \kappa )/\log \delta _1>-1/2$ . Then for every , the corresponding estimation of Equation (4.3) is

(4.4) $$ \begin{align} T_p~\geq~ L\cdot(T_2-T_1), \qquad \mathrm{where} \qquad L=\frac{-u_1\cdot T_0}{2\log\kappa}. \end{align} $$

Now we take a $C^1$ smooth function $\varphi :M\rightarrow {\mathbb R}$ satisfying:

1. 1. $\varphi (z)\geq -1$ for every and $\varphi (\sigma )=-1$ .

2. 2. $\varphi (z)\geq 0$ for every .

3. 3. $\varphi (z)\geq L^{-1}$ for every .

Then for the generic point $x_n$ of $\mu _n$ , let

$$ \begin{align*} 0~\leq~ T_1^1<T_2^1<T_3^1<T_4^1~<~ \cdots\cdots ~<~T_1^m<T_2^m<T_3^m<T_4^m~<~\cdots\cdots \end{align*} $$

be the time sequence where $\{T_1^m\}_m$ are all the times that $\phi _t(x_n)$ enters $U(\sigma ,\delta _1)$ and $T_2^m$ is the time when $\phi _t(x_n)$ enters $U(\sigma ,\delta _1)$ after $T_1^m$ ; $T_3^m$ is the time when $\phi _t(x_n)$ enters $V(p)$ after $T_2^m$ and $T_4^m$ is the time when $\phi _t(x_n)$ leaves $V(p)$ after $T_3^m$ .

For every m, we have

$$ \begin{align*} (T_4^m-T_3^m)~\geq~L\cdot (T_2^m-T_1^m). \end{align*} $$

This implies

However, we have . This contradicts $\mu _n\rightarrow \delta _\sigma $ ; thus, $\delta _\sigma $ is isolated in $\mathcal {M}_{erg}(X)$ .