Proof. According to the Oseledec theorem, for every $x\in \Gamma $ , we have that

(*) $$ \begin{align} \lim_{t\rightarrow+\infty}\dfrac{1}{t}\log m(\psi^*_t|_{E_j(x)})=\lim_{t\rightarrow+\infty}\dfrac{1}{t}\log\lVert\psi^*_t|_{E_j(x)}\rVert=\unicode{x3bb}_j,\quad j=1,2. \end{align} $$

Now, for every $0<\epsilon \ll \min \{\lvert \unicode{x3bb} _1\rvert ,\lvert \unicode{x3bb} _2\rvert \}$ and every $\delta \in (0,1)$ , we complete the proof of the lemma. By equation $(*)$ , for each $j=1,2$ , we have that

$$ \begin{align*} \lim_{t\rightarrow+\infty}\bigg\lvert\dfrac{1}{t}\log\lVert\psi^*_t|_{E_j(x)}\rVert-\unicode{x3bb}_j\bigg\rvert=0,~\lim_{t\rightarrow+\infty}\bigg\lvert\dfrac{1}{t}\log m(\psi^*_t|_{E_j(x)})-\unicode{x3bb}_j\bigg\rvert=0\quad\text{for every }x\in\Gamma. \end{align*} $$

Therefore,

$$ \begin{align*} \lim_{t\rightarrow+\infty}\int_{M^3}\bigg\lvert\dfrac{1}{t}\log\lVert\psi^*_t|_{E_j(x)}\rVert-\unicode{x3bb}_j\bigg\rvert \,d\mu(x)=0,\\ \lim_{t\rightarrow+\infty}\int_{M^3}\bigg\lvert\dfrac{1}{t}\log m(\psi^*_t|_{E_j(x)})-\unicode{x3bb}_j\bigg\rvert \,d\mu(x)=0. \end{align*} $$

Thus, there exists a positive number $L^+_j$ such that for every $T\geq L^+_j$ , we have that

$$ \begin{align*} \int_{M^3}\bigg\lvert\dfrac{1}{T}\log\lVert\psi^*_T|_{E_j(x)}\rVert-\unicode{x3bb}_j\bigg\rvert \,d\mu(x)\leq\dfrac{\delta\epsilon}{36}. \end{align*} $$

In addition, there also exists a positive number $L^-_j$ such that for every $T\geq L^-_j$ , we have that

$$ \begin{align*} \int_{M^3}\bigg\lvert\dfrac{1}{T}\log m(\psi^*_T|_{E_j(x)})-\unicode{x3bb}_j\bigg\rvert \,d\mu(x)\leq\dfrac{\delta\epsilon}{36}. \end{align*} $$

Since Oseledec splitting

$$ \begin{align*} \mathcal{N}_\Gamma=E_1\oplus E_2 \end{align*} $$

is a dominated splitting, Oseledec splitting is a continuous splitting [Reference Bonatti, Díaz and Viana3]. Therefore, for fixed $j\in \{1,2\}$ and every natural number $T\geq \max \{L^+_j,~L^-_j\} $ , the two functions

$$ \begin{align*} f_{j,T}(x):=\bigg\lvert\dfrac{1}{T}\log\lVert\psi^*_T|_{E_j(x)}\rVert-\unicode{x3bb}_j\bigg\rvert, \quad \widetilde{f}_{j,T}(x):=\bigg\lvert\dfrac{1}{T}\log m(\psi^*_T|_{E_j(x)})-\unicode{x3bb}_j\bigg\rvert \end{align*} $$

are continuous on $\Gamma $ .

However, $\mu $ is an invariant measure of $\varphi _T$ . By Birkhoff ergodic theorem, there exist two measurable functions $g_{j,T}$ and $\widetilde {g}_{j,T}$ such that for $\mu $ -almost every $x\in M^3$ , we have that

$$ \begin{align*} \lim_{n\rightarrow+\infty}\dfrac{1}{n}\sum_{i=0}^{n-1}f_{j,T}(\varphi_{iT})=\lim_{n\rightarrow+\infty} \dfrac{1}{nT}\sum_{i=0}^{n-1}\lvert\log\lVert\psi^*_T|_{E_j(\varphi_{iT}(x))}\rVert-T\unicode{x3bb}_j\rvert=g_{j,T}(x), \end{align*} $$

$$ \begin{align*} \lim_{n\rightarrow+\infty}\dfrac{1}{n}\sum_{i=0}^{n-1}\widetilde{f}_{j,T}(\varphi_{iT})=\lim_{n\rightarrow+\infty}\dfrac{1}{nT}\sum_{i=0}^{n-1}\lvert\log m(\psi^*_T|_{E_j(\varphi_{iT}(x))})-T\unicode{x3bb}_j\rvert=\widetilde{g}_{j,T}(x), \end{align*} $$

and

(**) $$ \begin{align} \int_{M^3}f_{j,T}(x)\,d\mu(x)=\int_{M^3}\bigg\lvert\dfrac{1}{T}\log\lVert\psi^*_T|_{E_j(x)}\rVert-\unicode{x3bb}_j\bigg\rvert d\mu(x)=\int_{M^3}g_{j,T}(x)\,d\mu(x)\leq\dfrac{\delta\epsilon}{36}. \end{align} $$

(***) $$ \begin{align} \int_{M^3}\widetilde{f}_{j,T}(x)\,d\mu(x)=\int_{M^3}\bigg\lvert\dfrac{1}{T}\log m(\psi^*_T|_{E_j(x)})-\unicode{x3bb}_j\bigg\rvert \,d\mu(x)=\int_{M^3}\widetilde{g}_{j,T}(x)\,d\mu(x)\leq\dfrac{\delta\epsilon}{36}. \end{align} $$

Set

$$ \begin{align*} \Gamma^+_{j,T}=\bigg\{x\in\Gamma:~g_{j,T}(x)\leq\dfrac{\epsilon}{2}\bigg\},\quad\Gamma^-_{j,T}=\bigg\{x\in\Gamma:~\widetilde{g}_{j,T}(x)\leq\dfrac{\epsilon}{2}\bigg\}. \end{align*} $$

By inequalities $(**)$ and $(***)$ , we have that

$$ \begin{align*} \mu(\Gamma^+_{j,T}\cap\Gamma^-_{j,T})\geq 1-\dfrac{\delta}{9} \end{align*} $$

and

$$ \begin{align*} \lim_{n\rightarrow+\infty}\dfrac{1}{nT}\sum_{i=0}^{n-1}\lvert\log\lVert\psi^*_T|_{E_j(\varphi_{iT}(x))}\rVert-T\unicode{x3bb}_j\rvert\leq \dfrac{\epsilon}{2},\\ \lim_{n\rightarrow+\infty}\dfrac{1}{nT}\sum_{i=0}^{n-1}\lvert\log m(\psi^*_T|_{E_j(\varphi_{iT}(x))})-T\unicode{x3bb}_j\rvert\leq\dfrac{\epsilon}{2} \end{align*} $$

for every $x\in \Gamma ^-_{j,T}\cap \Gamma ^-_{j,T}$ . Namely, for every $x\in \Gamma ^+_{j,T}\cap \Gamma ^-_{j,T}$ , there exists $N_j(x)$ such that

$$ \begin{align*} \prod_{i=0}^{n-1}\lVert\psi^*_T|_{E_j(\varphi_{iT}(x))}\rVert&\leq e^{(\unicode{x3bb}_j+\epsilon)nT}, \\ \prod_{i=0}^{n-1}m(\psi^*_T|_{E_j(\varphi_{iT}(x))})&\geq e^{(\unicode{x3bb}_j-\epsilon)nT} \quad \text{for every }n\geq N_j(x). \end{align*} $$

Set

$$ \begin{align*} L:=\max\{L^+_1,~L^-_1,~L^+_2,~L^-_2\}. \end{align*} $$

For every $T\geq L$ , define

$$ \begin{align*} \Lambda^T:=\bigcap_{j=1}^2(\Gamma^+_{j,T}\cap\Gamma^-_{j,T}). \end{align*} $$

Then, $\Lambda ^T\subset \Gamma $ and

$$ \begin{align*} \mu(\Lambda^T)>1-\dfrac{\delta}{2}. \end{align*} $$

Therefore, for every $x\in \Lambda ^T$ , there exists a positive integer $N(x)=\max \{N_1(x),~N_2(x)\}$ such that for every $j\in \{1,2\}$ , one has that

(†) $$ \begin{align} e^{(\unicode{x3bb}_j-\epsilon)nT}&\leq\prod_{i=0}^{n-1}m(\psi^*_T|_{E_j(\varphi_{iT}(x))}) \nonumber\\ & \leq\prod_{i=0}^{n-1}\lVert\psi^*_T|_{E_j(\varphi_{iT}(x))}\rVert\leq e^{(\unicode{x3bb}_j+\epsilon)nT}\quad\text{for every }n\geq N(x). \end{align} $$

For $x\in \Lambda ^T$ , let

$$ \begin{align*} \widetilde{N}(x):=\min\{N(x):\text{ the inequality } (\dagger) \text{ holds for all } n\geq N(x)\}. \end{align*} $$

Denote by

$$ \begin{align*} \Lambda^T_n=\{x\in\Lambda^T:~\widetilde{N}(x)\leq n\}. \end{align*} $$

Then, $\Lambda ^T_n\subset \Lambda ^T_{n+1}$ and

$$ \begin{align*} \Lambda^T=\bigcup_{n=1}^{+\infty}\Lambda^T_n. \end{align*} $$

Consequently, there is $n_0$ satisfying $\mu (\Lambda ^T_{n_0})>1-\delta $ . For proving this lemma, we set

$$ \begin{align*}N=N(T)=n_0,\quad\Gamma^T=\Lambda^T_{n_0}. \\[-37pt] \end{align*} $$

Proof of Theorem A.

Let $\mu $ be an ergodic hyperbolic invariant measure of the flow $\varphi _t$ generated by the star vector field $X\in \mathfrak {X}^*(M^3)$ on three-dimensional manifolds and $\unicode{x3bb} _1\leq \unicode{x3bb} _2$ be the Lyapunov exponents of $\psi ^*_t$ with respect to the ergodic hyperbolic invariant measure $\mu $ . Then, there is a measurable $\psi ^*_t$ -invariant splitting

$$ \begin{align*} \mathcal{N}_\Gamma=E_1\oplus E_2 \end{align*} $$

of normal bundle on $\Gamma $ (Oseledec’s basin of $\mu $ ) satisfying that

$$ \begin{align*} \lim_{t\to\pm\infty}\dfrac{1}{t}\log\parallel\psi^*_t(v) \parallel=\unicode{x3bb}_j\quad\text{for all }v\in E_j,~v\neq 0,~j=1,2. \end{align*} $$

By Lemma 2.9, there is a dominated splitting $\mathcal {N}_\Gamma =\widetilde {E_1}\oplus \widetilde {E_2}$ with respect to the scaled linear Poincaré flow $\psi ^*_t$ such that

$$ \begin{align*} \mathrm{dim}(\widetilde{E_1})=\mathrm{Ind}(\mu). \end{align*} $$

Regarding the Lyapunov exponents, we discuss it in two cases.

Case (1). If $\unicode{x3bb} _1$ and $\unicode{x3bb} _2$ are both negative (or positive), then $\mathrm {Ind}(\mu )=2$ (or $0$ ). Therefore,

$$ \begin{align*}\mathrm{dim}(\widetilde{E_1})=2\ (\text{or }0).\end{align*} $$

Thus, $\mu $ is an ergodic measure supported on an attracting (a repelling) periodic orbit.

Case (2). If $\unicode{x3bb} _1<0<\unicode{x3bb} _2$ , then we claim that $\mathcal {N}_\Gamma =E_1\oplus E_2$ is a dominated splitting.

Proof of the Claim.

Since $\mathcal {N}_\Gamma =\widetilde {E_1}\oplus \widetilde {E_2}$ is a dominated splitting with respect to the scaled linear Poincaré flow $\psi ^*_t$ such that $\mathrm {dim}(\widetilde {E_1})=\mathrm {Ind}(\mu )=1$ , we only need to show that the two splittings $\mathcal {N}_\Gamma =E_1\oplus E_2$ and $\mathcal {N}_\Gamma =\widetilde {E_1}\oplus \widetilde {E_2}$ coincide.

Given $x\in \Gamma $ , for the splitting $\mathcal {N}_x=E_1(x)\oplus E_2(x)$ , fix $\epsilon \in (0,{(\min \{\lvert \unicode{x3bb} _1\rvert ,~\unicode{x3bb} _2\})}/{8})$ small. There is a natural number $T_0>0$ such that for every $t\geq T_0$ , one has that

$$ \begin{align*} e^{(\unicode{x3bb}_1-\epsilon)t}\lvert v\rvert\leq\parallel\psi^*_t(v)\parallel\leq e^{(\unicode{x3bb}_1+\epsilon)t}\lvert v\rvert\quad\text{for every non-zero vector }v\in E_1(x), \end{align*} $$

and

$$ \begin{align*} e^{(\unicode{x3bb}_2-\epsilon)t}\lvert v\rvert\leq\parallel\psi^*_t(v)\parallel\leq e^{(\unicode{x3bb}_2+\epsilon)t}\lvert v\rvert\quad\text{for every non-zero vector }v\in E_2(x). \end{align*} $$

For the splitting $\mathcal {N}_x=\widetilde {E_1}(x)\oplus \widetilde {E_2}(x)$ , by the definition of dominated splitting, there are two constants $C\geq 1$ , $\unicode{x3bb}>0$ such that for every $t\geq 0$ , one has that

$$ \begin{align*} \dfrac{\lVert\psi^*_t|_{\widetilde{E_1}(x)} \rVert}{\lVert\psi^*_t|_{\widetilde{E_2}(x)}\rVert}\leq Ce^{-\unicode{x3bb} t}. \end{align*} $$

Conversely, suppose that $E_1(x)\nsubseteq \widetilde {E_1}(x)$ . Then, there is a non-zero vector ${u\in \widetilde {E_1}(x)\backslash E_1(x)}$ . However, since $\mathrm {dim}(E_1(x))=1\leq \mathrm {dim}(\widetilde {E_1}(x))$ , $\widetilde {E_1}(x)\nsubseteq E_1(x)$ . Thus, there is a non-zero vector $v\in E_1(x)\backslash \widetilde {E_1}(x)$ . Therefore, there is a vector decomposition

$$ \begin{align*} u=u_1+u_2,\quad \text{where } u_1\in E_1(x), u_2\in E_2(x), \text{ and }u_2\neq 0, \end{align*} $$

of u under the splitting $\mathcal {N}_x=E_1(x)\oplus E_2(x)$ . Similarly, there is a vector decomposition

$$ \begin{align*}v=v_1+v_2,\quad \text{where }v_1\in\widetilde{E_1}(x), v_2\in\widetilde{E_2}(x), \text{ and }v_2\neq 0,\end{align*} $$

of v under the splitting $\mathcal {N}_x=\widetilde {E_1}(x)\oplus \widetilde {E_2}(x)$ . Thus,

$$ \begin{align*} \begin{aligned} \dfrac{\lvert\psi^*_t(u)\rvert}{\lvert\psi^*_t(v)\rvert} & =\dfrac{\lvert\psi^*_t(u_1)+\psi^*_t(u_2)\rvert}{\lvert\psi^*_t(v)\rvert}\geq\dfrac{\lvert\psi^*_t(u_2)\lvert-\rvert\psi^*_t(u_1)\rvert}{\lvert\psi^*_t(v)\rvert}\geq\dfrac{e^{(\unicode{x3bb}_2-\epsilon)t}\lvert u_2\rvert-e^{(\unicode{x3bb}_1+\epsilon)t}\lvert u_1\rvert}{e^{(\unicode{x3bb}_1+\epsilon)t}\lvert v\rvert}\\ & =\dfrac{e^{(\unicode{x3bb}_2-\unicode{x3bb}_1-2\epsilon)t}\lvert u_2\rvert-\lvert u_1\rvert}{\lvert v\rvert}\rightarrow+\infty\quad(\text{as }t\rightarrow+\infty), \end{aligned} \end{align*} $$

and

$$ \begin{align*} \begin{aligned} \dfrac{\lvert\psi^*_t(v)\rvert}{\lvert\psi^*_t(u)\rvert} & =\dfrac{\lvert\psi^*_t(v_1)+\psi^*_t(v_2)\rvert}{\lvert\psi^*_t(u)\rvert}\geq\dfrac{\lvert\psi^*_t(v_2)\lvert-\rvert\psi^*_t(v_1)\rvert}{\lvert\psi^*_t(u)\rvert}\geq\dfrac{\lvert\psi^*_t(v_2)\lvert-Ce^{-\unicode{x3bb} t}\lvert\psi^*_t(v_2)\lvert}{\lvert\psi^*_t(u)\rvert}\\ & = (1-Ce^{-\unicode{x3bb} t})\dfrac{\lvert\psi^*_t(v_2)\lvert}{\lvert\psi^*_t(u)\rvert}\geq(1-Ce^{-\unicode{x3bb} t})C^{-1}e^{\unicode{x3bb} t}\dfrac{\lvert v_2\rvert}{\lvert u\rvert}\rightarrow+\infty\quad(\text{as } t\rightarrow+\infty). \end{aligned} \end{align*} $$

This leads to a contradiction. Therefore, our assumption that $E_1(x)\nsubseteq \widetilde {E_1}(x)$ is not valid. It means that $E_1(x)=\widetilde {E_1}(x)$ . Thus, the two splittings $\mathcal {N}_\Gamma =E_1\oplus E_2$ and $\mathcal {N}_\Gamma =\widetilde {E_1}\oplus \widetilde {E_2}$ coincide. This completes the proof of the claim.

Let $C(M^3)$ be the set of all continuous functions on the manifold $M^3$ and $\{f_i\}_{i=1}^{+\infty }$ be a countable dense subset of $C(M^3)$ . For any two measures $\mu ,~\nu \in \mathcal {M}(X)$ , define

$$ \begin{align*} d_{\mathcal{M}}(\mu,\nu)=\sum_{i=1}^{+\infty}\dfrac{\lvert\int_{M^3}f_i\,d\mu-\int_{M^3}f_i\,d\nu\rvert}{2^i\lVert f_i\rVert}.\end{align*} $$

Thus, $d_{\mathcal {M}}(\cdot ,\cdot )$ is a metric which induces the weak $^*$ topology in the set $\mathcal {M}(X)$ . Given an ergodic hyperbolic invariant regular measure $\mu $ , for every $\varepsilon>0$ , we prove that there exists a periodic measure $\mu _p$ such that

$$ \begin{align*}d_{\mathcal{M}}(\mu,\mu_p)<\varepsilon.\end{align*} $$

First, choose n large enough such that for every invariant measure $\nu $ , one has that

(‡) $$ \begin{align} \sum_{i=n+1}^{+\infty}\dfrac{\lvert\int_{M^3}f_i\,d\mu-\int_{M^3}f_i\,d\nu\rvert}{2^i\lVert f_i\rVert}\leq\sum_{i=n+1}^{+\infty}\dfrac{1}{2^{i-1}}<\dfrac{\varepsilon}{2}. \end{align} $$

According to the Birkhoff ergodic theorem, there is a full $\mu $ -measure $\varphi $ -invariant set $\Lambda _B$ such that for every $x\in \Lambda _B$ and every $f\in C(M^3)$ , one has that

$$ \begin{align*}\lim_{T\rightarrow+\infty}\dfrac{1}{T}\int^{T}_{0}f(\varphi_t(x))\,dt=\int_{M^3}f\,d\mu.\end{align*} $$

Thus, there exists $T_1>1$ such that for every $i=1,2,\ldots ,n$ , every $x\in \Lambda _B$ , and every $T>T_1$ , one has that

(‡‡) $$ \begin{align} \bigg\lvert\dfrac{1}{T}\int_{0}^{T}f_i(\varphi_t(x))\,dt-\int_{M^3}f_i\,d\mu\bigg\rvert<\dfrac{\varepsilon}{4n}\min_{1\leq i\leq n}\{2^i\lVert f_i\rVert\}. \end{align} $$

By the claim, the hyperbolic Oseledec splitting $\mathcal {N}_\Gamma =E_1\oplus E_2$ is a dominated splitting. Let

$$ \begin{align*}\chi=\min\{\lvert\unicode{x3bb}_1\rvert,~\lvert\unicode{x3bb}_2\rvert\},\end{align*} $$

then by Lemma 4.1, for every $0<\epsilon \ll \chi $ and every $\delta \in (0,1)$ , there is a positive number $L=L(\epsilon /2,\delta )$ such that:

• • for every $T_0\geq L$ , there exists a measurable set $\Gamma ^{T_0}=\Gamma ^{T_0}(\epsilon /2,\delta )\subset \Gamma $ with $\mu (\Gamma ^{T_0})\geq 1-\delta $ ;

• • there is a natural number $N=N(T_0)$ such that for every $J\geq N$ and every $x\in \Gamma ^{T_0}$ , we have that

  $$ \begin{align*} \prod_{i=0}^{J-1}\lVert\psi^*_{T_0}|_{E_1(\varphi_{iT_0}(x))}\rVert\leq e^{-(\chi-\epsilon/2)JT_0},~\prod_{i=0}^{J-1}m(\psi^*_{T_0}|_{E_2(\varphi_{iT_0}(x))})\geq e^{(\chi-\epsilon/2)JT_0}, \end{align*} $$
  $$ \begin{align*} \dfrac{\lVert\psi^*_{T_0}|_{E_1(x)}\rVert}{m(\psi^*_{T_0}|_{E_2(x)})}\leq e^{-T_0(\chi-\epsilon/2)}. \end{align*} $$

Fix an integer $T_0\geq \max \{T_1,~ L(\epsilon /2,\delta )\}$ and let $\eta _0=(\chi -\epsilon /2)T_0$ . For $k\in \mathbb {N}^+$ , we can define the Pesin block $\Lambda ^{T_0}_{\eta _0}(k)$ as

$$ \begin{align*} \begin{aligned} \Lambda^{T_0}_{\eta_0}(k) & =\bigg\{x\in\Gamma: \prod_{i=0}^{J-1}\lVert\psi^*_{T_0}|_{E_1(\varphi_{iT_0}(x))}\rVert\leq ke^{-J\eta_0}\text{ for all } J\geq 1,\\ & \quad\quad \prod_{i=0}^{J-1}m(\psi^*_{T_0}|_{E_2(\varphi_{iT_0}(x))})\geq k^{-1}e^{J\eta_0}\text{ for all } J\geq 1,\,d(x,\text{Sing}(X))\geq\dfrac{1}{k}\bigg\}. \end{aligned} \end{align*} $$

According to [Reference Wu, Yang and Zhang37, Proposition 5.3], $\Lambda ^{T_0}_{\eta _0}(k)$ is a compact set and

$$ \begin{align*} \mu(\Lambda^{T_0}_{\eta_0}(k))\rightarrow\mu(\Gamma^{T_0})\quad \text{as } k\rightarrow+\infty. \end{align*} $$

Denote

$$ \begin{align*} K=\max_{1\leq i\leq n}\{\lVert f_i\rVert\}. \end{align*} $$

Take $\gamma>0$ small enough. We may assume that

$$ \begin{align*} (2 K+1)\gamma<\dfrac{\varepsilon}{4n}\cdot\min_{1\leq i\leq n}\{2^i\lVert f_i\rVert\}. \end{align*} $$

Recall that $\lvert X(x)\rvert \leq K_0$ for every $x\in M^3$ . Since $f_i$ ( $1\leq i\leq n$ ) is uniformly continuous on $M^3$ , there exists $\xi \in [0,\gamma ]$ small enough such that for every integer $i\in [1,n]$ and any two points $x,y\in M^3$ with $d(x,y)\leq \xi K_0$ , we have that

$$ \begin{align*} \lvert f_i(x)-f_i(y)\rvert<\gamma. \end{align*} $$

Fix k large enough such that $\mu (\Lambda ^{T_0}_{\eta _0}(k))$ is sufficiently close to $1$ . For this fixed k, there is $j=j(k)\in \mathbb {N}^+$ such that $k<e^{{jT_0\epsilon }/{2}}$ . Thus, for every $x\in \Lambda ^{T_0}_{\eta _0}(k)$ , one has that

$$ \begin{align*} \prod_{i=0}^{J-1}\lVert\psi^*_{jT_0}\lvert_{E_1(\varphi_{ijT_0}(x))}\rVert&\leq e^{-(\chi-\epsilon)JjT_0},\\ \prod_{i=0}^{J-1}m(\psi^*_{jT_0}\lvert_{E_2(\varphi_{ijT_0}(x))})&\geq e^{(\chi-\epsilon)JjT_0}\quad\text{for all } J\geq 1. \end{align*} $$

Set $\eta =(\chi -\epsilon )jT_0$ , $T=jT_0$ . We consider the set

$$ \begin{align*} \begin{aligned} \Lambda^{T}_{\eta}(k) &=\bigg\{x\in\Gamma:\prod_{i=0}^{J-1}\lVert\psi^*_{T}\lvert_{E_1(\varphi_{iT}(x))}\rVert\leq e^{-J\eta}\text{ for all } J\geq 1,\\& \qquad \prod_{i=0}^{J-1}m(\psi^*_{T}\lvert_{E_2(\varphi_{iT}(x))})\geq e^{J\eta}\text{ for all } J\geq 1,\,d(x,\text{Sing}(X))\geq\dfrac{1}{k}\bigg\}. \end{aligned} \end{align*} $$

Take a point $x_0\in \Lambda ^T_\eta (k)\cap \text {supp}(\mu )$ . By the continuity of $X(x)$ , there is $r>0$ small enough such that for every $x\in B(x_0,r)$ , we have that

$$ \begin{align*} B(x_0,r)\subset B(x,\xi\lvert X(x)\rvert). \end{align*} $$

For $\varepsilon '=\min \{\varepsilon ,~\xi \}>0$ and every $\alpha \in (0,1/k)$ , we have a number $\delta _0>0$ by Theorem 3.5. Since $\mu (B(x_0,r)\cap \Lambda ^T_\eta (k)\cap \Lambda _B)>0$ , by the Poincaré recurrence theorem, there is a point $y\in B(x_0,r)\cap \Lambda ^T_\eta (k)\cap \Lambda _B$ and an integer l such that $\varphi _{lT}(y)\in B(x_0,r)\cap \Lambda ^T_\eta (k)\cap \Lambda _B$ with $d(y,\varphi _{lT}(y))<\delta _0$ . Thus, we get an $(\eta ,T)$ - $\psi ^*_t$ -quasi-hyperbolic orbit segment $\varphi _{[0,lT]}(y)$ satisfying:

• • $d(y,\text {Sing}(X))>\alpha $ and $d(\varphi _{lT}(y),\text {Sing}(X))>\alpha $ ;

• • $y\in \Lambda ^T_\eta (k)$ , $\varphi _{lT}(y)\in \Lambda ^T_\eta (k)$ and $d(y,\varphi _{lT}(y))<\delta _0$ .

By Theorem 3.5, there exists a strictly increasing $C^1$ function $\theta :[0,lT]\rightarrow \mathbb {R}$ and a periodic point $p\in M^3$ such that:

1. (1) $\theta (0)=0$ and $1-\gamma \leq 1-\xi <1-\varepsilon '<\theta '(t)<1+\varepsilon '<1+\xi \leq 1+\gamma $ for every $t\in [0,lT]$ ;

2. (2) p is periodic with period $\theta (lT)$ : $\varphi _{\theta (lT)}(p)=p$ ;

3. (3) $d(\varphi _t(y),\varphi _{\theta (t)}(p))<\varepsilon '\lvert X(\varphi _t(x))\rvert <\xi \lvert X(\varphi _t(x))\rvert $ for every $t\in [0,lT]$ .

Therefore, for every $i=1,2,\ldots ,n$ , we have that

$$ \begin{align*} \begin{aligned} & \bigg\lvert\dfrac{1}{\theta(lT)}\int_{0}^{\theta(lT)}f_i(\varphi_t(p))\,dt-\dfrac{1}{lT}\int_{0}^{\theta(lT)}f_i(\varphi_t(p))\,dt\bigg\rvert\\ &\quad \leq\bigg\lvert\dfrac{\theta(lT)}{lT}-1\bigg\rvert\cdot\dfrac{1}{\theta(lT)}\int_{0}^{\theta(lT)}f_i(\varphi_t(p))\,dt\\ &\quad \leq\gamma K \end{aligned} \end{align*} $$

and

$$ \begin{align*} \begin{aligned} & \bigg\lvert\dfrac{1}{lT}\int_{0}^{\theta(lT)}f_i(\varphi_s(p))\,ds-\dfrac{1}{lT}\int_{0}^{lT}f_i(\varphi_t(y))\,dt\bigg\rvert\\ & \quad=\dfrac{1}{lT} \bigg\lvert\int_{0}^{lT}f_i(\varphi_{\theta(t)}(p))\,d\theta(t)-\int_{0}^{lT}f_i(\varphi_t(y))\,dt\bigg\rvert\\ & \quad\leq\dfrac{1}{lT}\bigg\lvert\int_{0}^{lT}f_i(\varphi_{\theta(t)}(p))(\theta'(t)-1)\,dt\bigg\rvert+\dfrac{1}{lT}\bigg\lvert\int_{0}^{lT} [f_i(\varphi_{\theta(t)}(p))-f_i(\varphi_t(y))]\,dt\bigg\rvert\\ & \quad\leq\gamma K+\gamma. \end{aligned} \end{align*} $$

Thus,

$$ \begin{align*} \bigg\lvert\dfrac{1}{\theta(lT)}\int_{0}^{\theta(lT)}f_i(\varphi_t(p))\,dt-\dfrac{1}{lT}\int_{0}^{lT}f_i(\varphi_t(y))\,dt\bigg\rvert <\dfrac{\varepsilon}{4n}\cdot\min_{1\leq i\leq n}\{2^i\lVert f_i\rVert\}. \end{align*} $$

Denote by $\mu _p$ the invariant measure on the periodic orbit $\mathrm {Orb}(p)$ . From equations $(\ddagger )$ and $(\ddagger \ddagger )$ , we have that

$$ \begin{align*} \begin{aligned} &d_{\mathcal{M}}(\mu,\mu_p)\\&\quad =\sum_{i=1}^{+\infty}\dfrac{\lvert\int_{M^3}f_i\,d\mu-\int_{M^3}f_i\,d\mu_p\rvert}{2^i\lVert f_i\rVert}\\&\quad =\sum_{i=1}^{n}\dfrac{\lvert\int_{M^3}f_i\,d\mu-\int_{M^3}f_i\,d\mu_p\rvert}{2^i\lVert f_i\rVert}+\sum_{i=n+1}^{+\infty}\dfrac{\lvert\int_{M^3}f_i\,d\mu-\int_{M^3}f_i\,d\mu_p\rvert}{2^i\lVert f_i\rVert}\\&\quad \leq\sum_{i=1}^{n}\dfrac{\lvert\int_{M^3}f_i\,d\mu-(1/lT)\int_{0}^{lT}f_i(\varphi_t(y))\,dt+(1/lT) \int_{0}^{lT}f_i(\varphi_t(y))\,dt-\int_{M^3}f_i\,d\mu_p\rvert}{2^i\lVert f_i\rVert}+\sum_{i=n+1}^{+\infty}\dfrac{1}{2^{i-1}}\\&\quad \leq\sum_{i=1}^{n}\dfrac{\lvert\int_{M^3}f_i\,d\mu-(1/lT)\int_{0}^{lT}f_i(\varphi_t(y))\,dt\rvert+ \lvert(1/lT)\int_{0}^{lT}f_i(\varphi_t(y))\,dt-(1/\theta(lT))\int_{0}^{\theta(lT)}f_i(\varphi_t(p))\,dt\rvert}{2^i\lVert f_i\rVert}+\dfrac{\varepsilon}{2}\\&\quad <\sum_{i=1}^{n}\dfrac{(\varepsilon/4n)\cdot\min\limits_{1\leq i\leq n}\{2^i\lVert f_i\rVert\}+(\varepsilon/4n)\cdot\min\limits_{1\leq i\leq n}\{2^i\lVert f_i\rVert\}}{2^i\lVert f_i\rVert}+\dfrac{\varepsilon}{2}\leq\sum_{i=1}^{n}\dfrac{\varepsilon}{2n}+\dfrac{\varepsilon}{2}=\varepsilon. \end{aligned} \end{align*} $$

This completes the proof of Theorem A.

Proof. Let

$$ \begin{align*}\chi=\min\{\lvert\unicode{x3bb}_1\rvert,~\unicode{x3bb}_2\}.\end{align*} $$

Since the Oseledec splitting $\mathcal {N}_\Gamma =E_1\oplus E_2$ is a dominated splitting, by Lemma 4.1, for every $0<\epsilon /4\ll \chi $ and every $\delta _1\in (0,1/9)$ , there is a positive number $L=L(\epsilon /4,\delta _1)$ such that:

• • for every $T\geq L$ , there exists a measurable set $\Gamma ^T=\Gamma ^T(\epsilon /4,\delta _1)\subset \Gamma $ with $\mu (\Gamma ^T)\geq 1-\delta _1$ ;

• • there is a natural number $N=N(T)$ such that for every $n\geq N$ and every $x\in \Gamma ^T$ , we have that

  $$ \begin{align*} \prod_{i=0}^{n-1}\lVert\psi^*_T|_{E_1(\varphi_{iT}(x))}\rVert\leq e^{-(\chi-\epsilon/4)nT}, \quad \prod_{i=0}^{n-1}m(\psi^*_T|_{E_2(\varphi_{iT}(x))})\geq e^{(\chi-\epsilon/4)nT}, \end{align*} $$
  $$ \begin{align*} \dfrac{\lVert\psi^*_T|_{E_1}\rVert}{m(\psi^*_T\|_{E_2})}\leq e^{-T(\chi-\epsilon/4)}. \end{align*} $$

Fix an integer $T_0\geq L(\epsilon /4,\delta _1)$ and let $\eta _0=(\chi -\epsilon /4)T_0$ . For $k\in \mathbb {N}^+$ , we define the Pesin block $\Lambda ^{T_0}_{\eta _0}(k)$ as

$$ \begin{align*} \begin{aligned} \Lambda^{T_0}_{\eta_0}(k) & =\bigg\{x\in\Gamma:\prod_{i=0}^{n-1}\lVert\psi^*_{T_0}|_{E_1(\varphi_{iT_0}(x))}\rVert\leq ke^{-n\eta_0}\text{ for all } n\geq 1,\\ & \qquad\prod_{i=0}^{n-1}m(\psi^*_{T_0}|_{E_2(\varphi_{iT_0}(x))})\geq k^{-1}e^{n\eta_0}\text{ for all }n\geq 1,\,d(x,\text{Sing}(X))\geq\dfrac{1}{k}\bigg\}. \end{aligned} \end{align*} $$

According to [Reference Wu, Yang and Zhang37, Proposition 5.3], $\Lambda ^{T_0}_{\eta _0}(k)$ is a compact set and

$$ \begin{align*} \mu(\Lambda^{T_0}_{\eta_0}(k))\rightarrow\mu(\Gamma^{T_0})\quad \text{as } k\rightarrow+\infty. \end{align*} $$

Therefore, one can fix k large enough such that $\mu (\Lambda ^{T_0}_{\eta _0}(k))>1-2\delta _1>0$ . For this fixed k, there is $j=j(k)\in \mathbb {N}^+$ such that $k<e^{{jT_0\epsilon }/{4}}$ . Thus, for every $x\in \Lambda ^{T_0}_{\eta _0}(k)$ , one has that

$$ \begin{align*} \prod_{i=0}^{n-1}\lVert\psi^*_{jT_0}|_{E_1(\varphi_{ijT_0}(x))}\rVert&\leq e^{-(\chi-\epsilon/2)njT_0},\\ \prod_{i=0}^{n-1}m(\psi^*_{jT_0}|_{E_2(\varphi_{ijT_0}(x))})&\geq e^{(\chi-\epsilon/2)njT_0}\quad\text{for all } n\geq 1. \end{align*} $$

Set $\eta =(\chi -\epsilon /2)jT_0$ , $T=jT_0$ . We consider the set

$$ \begin{align*} \begin{aligned} \Lambda^{T}_{\eta}(k) & =\bigg\{x\in\Gamma:\prod_{i=0}^{n-1}\lVert\psi^*_{T}\lvert_{E_1(\varphi_{iT}(x))}\rVert\leq e^{-n\eta}\text{ for all } n\geq 1,\\ & \qquad \prod_{i=0}^{n-1}m(\psi^*_{T}\lvert_{E_2(\varphi_{iT}(x))})\geq e^{n\eta}\text{ for all } n\geq 1,\,d(x,\text{Sing}(X))\geq\dfrac{1}{k}\bigg\}. \end{aligned} \end{align*} $$

For $\epsilon /4>0$ , there is a number $\zeta>0$ such that

$$ \begin{align*} \lvert\log(1+\zeta)\rvert<\dfrac{\epsilon}{4}\quad\text{and}\quad\lvert\log(1-\zeta)\rvert<\dfrac{\epsilon}{4}. \end{align*} $$

Since $\widetilde {\psi }^*_T$ and $(\widetilde {\psi }^*_T)^{-1}$ are both continuous on

$$ \begin{align*} \widetilde{M}=\text{Closure}\bigg(\bigcup_{x\in M^3\backslash \text{Sing}(X)}\dfrac{X(x)}{\lvert X(x)\rvert}\bigg), \end{align*} $$

for every $x_1,~x_2\in M^3\backslash \text {Sing}(X)$ , denote by

$$ \begin{align*} \mathrm{I}(x_1,x_2)=(\psi^*_T|_{\mathcal{N}_{x_1}})^{-1}\circ(\psi^*_T|_{\mathcal{N}_{x_2}}), \end{align*} $$

the map

$$ \begin{align*} (x_1,x_2)\mapsto\mathrm{I}(x_1,x_2)=(\psi^*_T|_{\mathcal{N}_{x_1}})^{-1}\circ(\psi^*_T|_{\mathcal{N}_{x_2}}) \end{align*} $$

is also continuous. Therefore, $\mathrm {I}(x_1,x_2)=(\psi ^*_T|_{\mathcal {N}_{x_1}})^{-1}\circ (\psi ^*_T|_{\mathcal {N}_{x_2}})$ approach to

$$ \begin{align*} (\psi^*_T|_{\mathcal{N}_{\varphi_T(x_2)}})^{-1}\circ(\psi^*_T|_{\mathcal{N}_{x_2}})=I \end{align*} $$

when both the distance between $x_1$ and $\varphi _T(x_2)$ , and the distance between $\varphi _{-T}(x_1)$ and $x_2$ are small enough. Let $\sigma _1>0$ satisfy that

$$ \begin{align*} & \max\{d(x_1,\varphi_T(x_2)),d(\varphi_{-T}(x_1),x_2)\}\\ & \quad\leq\sigma_1 \Longrightarrow 1-\zeta\leq\lVert \mathrm{I}(x_1,x_2)\rVert\leq1+\zeta\Longrightarrow\lvert\log\lVert \mathrm{I}(x_1,x_2)\rVert\rvert<\dfrac{\epsilon}{4}. \end{align*} $$

Since $\widetilde {M}$ is compact, $\sigma _1$ always exists. For $T>0$ , the norm

$$ \begin{align*} \lVert\psi^*_T\rVert=\sup\{\lvert\psi^*_T(v)\rvert:~v\in\mathcal{N},~\lvert v\rvert=1\} \end{align*} $$

is uniformly upper bounded on $\mathcal {N}$ and the norm

$$ \begin{align*} m(\psi^*_T)=\inf\{\lvert\psi^*_T(v)\rvert:~v\in\mathcal{N},~\lvert v\rvert=1\} \end{align*} $$

is uniformly bounded from $0$ on $\mathcal {N}$ . Since $\varphi _t$ is a $C^1$ flow, there is $\sigma ^{\prime }_2>0$ such that for any $t_1,~t_2\in \mathbb {R}$ , we have that

$$ \begin{align*} \lvert t_1-t_2\rvert<\sigma^{\prime}_2\Longrightarrow d(\varphi_{t_1}(x),\varphi_{t_2}(x))<\dfrac{\sigma_1}{2}\quad\text{for every }x\in M^3. \end{align*} $$

For $\zeta>0$ , there is $\sigma ^{\prime }_2>0$ such that for every $s\in (T-\sigma ^{\prime \prime }_2,T+\sigma ^{\prime \prime }_2)$ and every $x\in M^3\backslash \text {Sing}(X)$ , we have that

$$ \begin{align*} \lvert\lVert(\psi^*_T|_{\mathcal{N}_{\varphi_s(x)}})^{-1}\circ\psi^*_s|_{\mathcal{N}_x}\rVert- \lVert(\psi^*_s|_{\mathcal{N}_{\varphi_s(x)}})^{-1}\circ\psi^*_s|_{\mathcal{N}_x}\rVert\rvert\leq\zeta. \end{align*} $$

Let $\sigma _2=\min \{\sigma ^{\prime }_2,\sigma ^{\prime \prime }_2\}$ . Then,

$$ \begin{align*} \lvert\log\lVert(\psi^*_T|_{\mathcal{N}_{\varphi_s(x)}})^{-1}\circ\psi^*_s|_{\mathcal{N}_x}\rVert\rvert<\dfrac{\epsilon}{4}\quad\text{for every }s\in(T-\sigma_2,T+\sigma_2). \end{align*} $$

Now, we give a claim on invertible linear maps.

Claim. Given two invertible linear maps $A,~B:\mathbb {R}^d\longrightarrow \mathbb {R}^d$ . For $\epsilon _1>0$ , there is $\sigma _3>0$ such that if $\lVert B-I\rVert \leq \sigma _3$ , then

(§) $$ \begin{align} \lVert ABv\rVert\geq e^{-\epsilon_1}\lVert Av\rVert\quad \text{for all }v\in\mathbb{R}^d. \end{align} $$

Proof of the Claim.

We only need to prove the inequality $(\S )$ for $v\in \mathbb {R}^d$ with $\lVert v\rVert =1$ . Now, we always assume that $\lVert v\rVert =1$ . For any $\epsilon _1>0$ , let $\epsilon _2>0$ satisfying that $1-\epsilon _2>e^{-\epsilon _1}$ and

$$ \begin{align*} \sigma_3=\dfrac{m(A)}{\lVert A\rVert}\epsilon_2. \end{align*} $$

Therefore,

$$ \begin{align*} \lVert A\rVert\cdot\lVert B-I\rVert\leq\epsilon_2m(A). \end{align*} $$

Consequently, if $\lVert B-I\rVert \leq \sigma _3$ , we have that

$$ \begin{align*} \lVert ABv\rVert & =\lVert Av+A(B-I)v\rVert\geq\lVert Av\rVert-\lVert A\rVert\cdot\lVert B-I\rVert\cdot\lVert v\rVert=\lVert Av\rVert-\lVert A\rVert\cdot\lVert B-I\rVert\\ & \geq\lVert Av\rVert-\epsilon_2m(A)\geq(1-\epsilon_2)\lVert Av\rVert\geq e^{-\epsilon_1}\lVert Av\rVert. \\[-35pt] \end{align*} $$

For a given $x\in M^3$ and a sufficiently small neighborhood $U(x)$ of x, similarly to the technique used in [Reference Pollicott27], by parallel transporting vector along the unique geodesic in the Sasaki metric, we can locally trivialize the tangent bundle $T_{U(x)}M^3=U(x)\times \mathbb {R}^3$ of $U(x)$ . Thus, for every $y\in U(x)$ , we can translate the vectors of $T_xM^3$ to $T_yM^3$ . Thus, we can translate the vectors of $\mathcal {N}_x$ to $\mathcal {N}_y$ . According to Lemma 3.1, there is a $\beta ^*_0>0$ such that for every $x\in M^3\backslash \text {Sing}(X)$ , every point $y\in U(x)=B(x,\beta ^*_0\lvert X(x)\rvert )$ is not a singularity. Thus, for every $v\in \mathcal {N}_y$ and $\rho>1$ , we can define the $\rho $ -cone in $\mathcal {N}_y$ :

$$ \begin{align*} \mathcal{C}_\rho(y)=\{v\in\mathcal{N}_y:~\lVert v_2\rVert\geq\rho\lVert v_1\rVert\}, \end{align*} $$

where $v=v_1+v_2$ , $v_j\in E_j(x)$ , and $E_j(x)$ is the Oseledec subspace with respect to the Lyapunov exponent $\unicode{x3bb} _j$ for $j=1,2$ . Namely, translating the splitting $E_1(x)\oplus E_2(x)$ of $\mathcal {N}_x$ to $\mathcal {N}_y$ , and for $v\in \mathcal {N}_y$ , $v=v_1+v_2$ is the expression corresponding to the splitting. For computation, we can define an equivalent norm $\lVert \cdot \rVert '$ on $\mathcal {N}_y$ :

$$ \begin{align*} \lVert v\rVert'=\max\{\lVert v_1\rVert,\lVert v_2\rVert\}. \end{align*} $$

Since the Oseledec splitting

$$ \begin{align*} \mathcal{N}_\Gamma=E_1\oplus E_2 \end{align*} $$

is a dominated splitting for $\psi ^*_T$ , there exist two positive numbers $\gamma \in (0,1)$ and $\rho>1$ such that $\gamma \rho>1$ and

$$ \begin{align*} \psi^*_T\mathcal{C}_{\gamma\rho}(x)\subset\mathcal{C}_{\rho}(\varphi_{T}(x))\quad\text{for all }x\in\Lambda^{T}_{\eta}(k). \end{align*} $$

Furthermore, we can choose $0<\sigma _4=\sigma _4(\gamma ,\rho )$ such that for every homeomorphism $\mathcal {T}$ , we have that

$$ \begin{align*} \lVert\mathcal{T}-I\rVert\leq\sigma_4\Longrightarrow \mathcal{T}\mathcal{C}_{\rho}\subset\mathcal{C}_{\gamma\rho}. \end{align*} $$

Take $y\in \Lambda ^{T}_{\eta }(k)$ . By Poincaré recurrence theorem, there is an increasing integer sequence $\{l_n\}$ such that

$$ \begin{align*} d(y,\varphi_{l_nT}(y))\longrightarrow 0. \end{align*} $$

Let

$$ \begin{align*} \sigma_3=\dfrac{m(\psi^*_T)}{\lVert\psi^*_T\rVert}(1-e^{-\epsilon/4}). \end{align*} $$

For $0<\varepsilon <\min \{\beta ^*_0,~\xi ,~{r_0}/{K_0},~{\epsilon }/{4},~{\sigma _1}/{2K_0},~{\sigma _2}/{K_0},~\sigma _3,~\log (1+\sigma _4)\}$ and ${\alpha \in (0,1/k)}$ , we get a constant $\delta =\delta (\varepsilon ,\alpha )$ by Theorem 3.5. For $l_n$ large enough, we obtain a $(\eta ,T)$ - $\psi ^*_t$ -quasi-hyperbolic orbit segment $\varphi _{[0,l_nT]}(y)$ satisfying that:

• • $d(y,\text {Sing}(X))>\alpha $ and $d(\varphi _{l_nT}(y),\text {Sing}(X))>\alpha $ ;

• • $y\in \Lambda ^{T}_{\eta }(k)$ , $\varphi _{l_nT}(y)\in \Lambda ^{T}_{\eta }(k)$ , and $d(y,\varphi _{l_nT}(y))<\delta $ .

According to Theorem 3.5, there is a strictly increasing $C^1$ function $\theta :[0,l_nT]\to \mathbb {R}$ and a periodic point $p\in M^3$ such that:

1. (1) $\theta (0)=0$ and $1-\varepsilon <\theta '(t)<1+\varepsilon $ for every $t\in [0,l_nT]$ ;

2. (2) p is periodic, $\varphi _{\theta (l_nT)}(p)=p$ ;

3. (3) $d(\varphi _t(y),\varphi _{\theta (t)}(p))<\varepsilon \lvert X(\varphi _t(y))\rvert $ for every $t\in [0,l_nT]$ .

Moreover, by [Reference Wu, Yang and Zhang37, Proposition 4.4], there is constant $N=N(\eta ,T)$ such that

$$ \begin{align*}\lvert\theta(iT)-iT\rvert\leq Nd(y,\varphi_{l_nT}(y))\quad\text{for }i=1,2,\ldots,l_n.\end{align*} $$

Take $l_n$ large enough such that

$$ \begin{align*}d(y,\varphi_{l_nT}(y))<\dfrac{\sigma_2}{2N}.\end{align*} $$

We write $l_n$ simply as l when no confusion can arise. Thus, we get a periodic orbit of p with period $\theta (lT)$ . Then, we have an invariant measure

$$ \begin{align*}\mu_p=\dfrac{1}{\theta(lT)}\int_{0}^{\theta(lT)}\delta_{\varphi_t(p)}\,dt\end{align*} $$

supported on the periodic orbit of p. Let

$$ \begin{align*}\unicode{x3bb}_1(p)\leq\unicode{x3bb}_2(p)\end{align*} $$

be the Lyapunov exponents of $\mu _p$ . To estimate the largest exponent and the smallest exponent, we give the the following claim.

Claim. For every $i=0,1,\ldots ,l-1$ , let

$$ \begin{align*} T_i(y,p)=(\psi^*_T|_{\mathcal{N}_{\varphi_{(i+1)T}(y)}})^{-1}\circ\psi^*_{\theta((i+1)T)-\theta(iT)}|_{\mathcal{N}_{\varphi_{\theta(iT)}(p)}}. \end{align*} $$

We have that

$$ \begin{align*} \lvert\log\lVert T_i(y,p)\rVert\rvert<\dfrac{\epsilon}{2}. \end{align*} $$

Proof of the Claim.

For every $i=0,1,\ldots ,l-1$ , we have that

$$ \begin{align*} \psi^*_T|_{\mathcal{N}_{\varphi_{\theta((i+1)T)-T}(p)}}\circ(\psi^*_T|_{\mathcal{N}_{\varphi_{\theta((i+1)T)}(p)}})^{-1} \end{align*} $$

is identity. Thus, we consider

$$ \begin{align*} (\psi^*_T|_{\mathcal{N}_{\varphi_{(i+1)T}(y)}})^{-1}\circ\psi^*_T|_{\mathcal{N}_{\varphi_{\theta((i+1)T)-T}(p)}}\circ (\psi^*_T|_{\mathcal{N}_{\varphi_{\theta((i+1)T)}(p)}})^{-1}\circ\psi^*_{\theta((i+1)T)-\theta(iT)}|_{\mathcal{N}_{\varphi_{\theta(iT)}(p)}}.\end{align*} $$

Since

$$ \begin{align*} \lvert\theta(iT)-iT\rvert\leq Nd(y,\varphi_{lT}(y))\quad\text{for }i=0,1,2,\ldots,l, \end{align*} $$

we have that

$$ \begin{align*} \begin{aligned} \lvert\theta((i+1)T)-T-\theta(iT)\rvert&= \lvert\theta((i+1)T)-\theta(iT)-((i+1)T-iT)\rvert\\ & = \lvert\theta((i+1)T)-(i+1)T-(\theta(iT)-iT)\rvert\\&\leq\lvert\theta((i+1)T)-(i+1)T\rvert+\lvert(\theta(iT)-iT)\rvert\\ & \leq 2N\,d(y,\varphi_{lT}(y))<\sigma_2. \end{aligned} \end{align*} $$

It means that $\theta ((i+1)T)-\theta (iT)\in (T-\sigma _2,T+\sigma _2)$ for every $i=0,1,\ldots ,l-1$ . Therefore,

$$ \begin{align*} d(\varphi_{\theta((i+1)T)-T}(p),\varphi_{\theta(iT)}(p))<\dfrac{\sigma_1}{2}, \end{align*} $$

$$ \begin{align*} \lvert\log\lVert(\psi^*_T|_{\mathcal{N}_{\varphi_{\theta((i+1)T)}(p)}})^{-1}\circ\psi^*_{\theta((i+1)T)-\theta(iT)}|_{\mathcal{N}_{\varphi_{\theta(iT)}(p)}} \rVert\rvert\!<\dfrac{\epsilon}{4}\quad \text{for every }i=0,1,\ldots,l-1. \end{align*} $$

However, since

$$ \begin{align*} d(\varphi_t(y),\varphi_{\theta(t)}(p))<\varepsilon\lvert X(\varphi_t(y))\rvert\quad\text{for every }t\in[0,lT], \end{align*} $$

we have that

$$ \begin{align*} d(\varphi_{iT}(y),\varphi_{\theta(iT)}(p))<\varepsilon\lvert X(\varphi_{iT}(y))\rvert<\dfrac{\sigma_1}{2}\quad\text{for every }i=0,1,2,\ldots,l. \end{align*} $$

Thus,

$$ \begin{align*} & d(\varphi_{iT}(y),\varphi_{\theta((i+1)T)-T}(p))\leq d(\varphi_{iT}(y),\varphi_{\theta(iT)}(p))\\ & \quad +d(\varphi_{\theta((i+1)T)-T}(p),\varphi_{\theta(iT)}(p))<\sigma_1. \end{align*} $$

Consequently,

$$ \begin{align*} \lvert\log\lVert(\psi^*_T|_{\mathcal{N}_{\varphi_{(i+1)T}(y)}})^{-1}\circ\psi^*_T|_{\mathcal{N}_{\varphi_{\theta((i+1)T)-T}(p)}}\rVert\rvert<\dfrac{\epsilon}{4}. \end{align*} $$

Then,

$$ \begin{align*} \lvert\log\lVert T_i(y,p)\rVert\rvert<\dfrac{\epsilon}{2}. \\[-42pt] \end{align*} $$

Since

$$ \begin{align*} \psi^*_{\theta(lT)}|_{\mathcal{N}_p}=\psi^*_T|_{\mathcal{N}_{\varphi_{(l-1)T}(y)}}\circ T_{l-1}(y,p)\circ\cdots\circ\psi^*_T|_{\mathcal{N}_{\varphi_T(y)}}\circ T_1(y,p)\circ\psi^*_T|_{\mathcal{N}_y}\circ T_0(y,p), \end{align*} $$

we have that

$$ \begin{align*} \begin{aligned} \unicode{x3bb}_2(p) & =\lim_{j\rightarrow+\infty}\dfrac{1}{j\theta(lT)}\log\lVert\psi^*_{j\theta(lT)}|_{\mathcal{N}_p}\rVert =\lim_{j\rightarrow+\infty}\dfrac{1}{j\theta(lT)}\log\lVert (\psi^*_{\theta(lT)}|_{\mathcal{N}_p})^j\rVert\\ & \leq\dfrac{1}{\theta(lT)}\log\lVert\psi^*_{\theta(lT)}|_{\mathcal{N}_p}\rVert\leq\dfrac{1}{\theta(lT)} \bigg(\!\sum_{i=0}^{l-1}\log\lVert\psi^*_T|_{\mathcal{N}_{\varphi_{iT}(y)}}\rVert+\sum_{i=0}^{l-1}\log\lVert T_i(y,p)\rVert\bigg). \end{aligned} \end{align*} $$

Since $1-\varepsilon <\theta '(t)<1+\varepsilon $ , for every $t\in [0,lT]$ , we have that

$$ \begin{align*} (1-\varepsilon)lT<\lvert\theta(lT)\rvert<(1+\varepsilon)lT. \end{align*} $$

Therefore,

$$ \begin{align*} \unicode{x3bb}_2(p)\leq\dfrac{lT(1+\varepsilon)}{\theta(lT)}\cdot\dfrac{1}{lT}\bigg(\!\sum_{i=0}^{l-1}\log\lVert\psi^*_T|_{\mathcal{N}_{\varphi_{iT}(y)}}\rVert+ \sum_{i=0}^{l-1}\log\lVert T_i(y,p)\rVert\bigg)\leq\unicode{x3bb}_2+\epsilon. \end{align*} $$

However, since $\varepsilon <\log (1+\sigma _4)$ , by the similar discussions as the claim, we have that

$$ \begin{align*} \lVert T_i(y,p)-I\rVert\leq\sigma_4\quad\text{for every }i=0,1,\ldots,l-1 \end{align*} $$

and

$$ \begin{align*} \lVert T_i(y,p) v\rVert'\geq e^{-{\epsilon}/{2}}\lVert v\rVert'\quad\text{for all } v\in\mathcal{N}_{\varphi_{\theta(iT)}(p)}\backslash\{0\},\text{ for every }i=0,1,\ldots,l-1. \end{align*} $$

Therefore, for each $i=0,1,\ldots ,l-1$ , we have that

(▵) $$ \begin{align} \psi^*_{T}|_{\mathcal{N}_{\varphi_{iT}(y)}}\circ T_i(y,p)\mathcal{C}_{\rho}(\varphi_{\theta(iT)}(p))\subset\mathcal{C}_{\rho}(\varphi_{\theta((i+1)T)}(p)). \end{align} $$

For every $v=(v_1,v_2)\in \mathcal {C}_\rho (p)$ and $i=0,1,\ldots ,l-1$ , denote by

$$ \begin{align*} \bar{v}^0=v,\quad \bar{v}^i=T_i(y,p)\circ\psi^*_{T}|_{\mathcal{N}_{\varphi_{(i-1)T}(y)}}\circ T_{i-1}(y,p)\circ\cdots\circ\psi^*_{T}|_{\mathcal{N}_y}\circ T_0(y,p)(v) \end{align*} $$

and

$$ \begin{align*} v^{i+1}=\psi^*_{T}|_{\mathcal{N}_{\varphi_{iT}(y)}}\circ T_i(x,p)\circ\psi^*_{T}|_{\mathcal{N}_{\varphi_{(i-1)T}(y)}}\circ T_{i-1}(x,p)\circ\cdots\circ\psi^*_{T}|_{\mathcal{N}_y}\circ T_0(y,p)(v). \end{align*} $$

Corresponding to the Oseledec splitting

$$ \begin{align*} \mathcal{N}_\Gamma=E_1\oplus E_2, \end{align*} $$

denote by

$$ \begin{align*} \bar{v}^i=\bar{v}^i_1+\bar{v}^i_2,\quad v^i=v^i_1+v^i_2. \end{align*} $$

Then, for each $i=0,1,\ldots ,l-1$ , we have that

(1) $$ \begin{align} \lVert v^{i+1}\rVert'&=\lVert v^{i+1}_2\rVert =\lVert\psi^*_T(\bar{v}^i_2)\rVert\geq m(\psi^*_T|_{E_2(\varphi_{iT}(y))})\lVert\bar{v}^i_2\rVert=m(\psi^*_{T}|_{E_2(\varphi_{iT}(y))})\lVert\bar{v}^i\rVert' \nonumber\\& =m(\psi^*_{T}|_{E_2(\varphi_{iT}(y))})\lVert T_i(x,p)v^i\rVert'\geq e^{-{\epsilon}/{2}}m(\psi^*_{T}|_{E_2(\varphi_{iT}(y))})\lVert v^i\rVert'\nonumber\\& \geq e^{-{\epsilon}/{2}(i+1)}\prod^{i}_{j=0}m(\psi^*_{T}|_{E_2(\varphi_{jT}(y))})\lVert v\rVert'. \end{align} $$

Since $\mathcal {C}_\rho (p)$ contains the subspace $E_2(p)$ , formulas $(\Delta )$ , $(1)$ , and Lemma 4.1 indicate that

$$ \begin{align*} \unicode{x3bb}_2(p)\geq\unicode{x3bb}_2-\epsilon. \end{align*} $$

This proves the proposition.

Proof. We consider the flow $\overline {\varphi _t}$ generated by the $C^1$ star vector field $-X\in \mathfrak {X}^1(M)$ on a compact three-dimensional smooth Riemannian manifold. Then, $\mu $ is an ergodic hyperbolic invariant regular measure of the flow $\overline {\varphi _t}$ . By Lemma 2.9, every star vector field admits a dominated splitting $\mathcal {N}_{\text {supp}(\mu )\backslash \text {Sing}(X)}=E\oplus F$ with respect to the scaled linear Poincaré flow $\overline {\psi ^*_t}$ such that $\mathrm {dim}(E)=\mathrm {Ind}(\mu )$ . According to the similar discussions on Lyapunov exponents as the two different cases in the proof of Theorem A, we only need to consider the case where one Lyapunov exponent is positive and the other is negative. Namely, the Oseledec splitting $\mathcal {N}_\Gamma =\overline {E_1}\oplus \overline {E_2}$ of the ergodic hyperbolic invariant measure $\mu $ is a dominated splitting and we have that

$$ \begin{align*} \lim_{t\to\pm\infty}\dfrac{1}{t}\log\parallel\overline{\psi^*_t}(v) \parallel=\overline{\unicode{x3bb}_j}\quad\text{for all } v\in\overline{E_j},~v\neq 0,~j=1,2, \end{align*} $$

where $\overline {\unicode{x3bb} _1}<0<\overline {\unicode{x3bb} _2}$ .

Let

$$ \begin{align*} \overline{\chi}=\min\{\lvert\overline{\unicode{x3bb}_1}\rvert,~\overline{\unicode{x3bb}_2}\}. \end{align*} $$

Since the Oseledec splitting $\mathcal {N}_\Gamma =\overline {E_1}\oplus \overline {E_2}$ is a dominated splitting, by Lemma 4.1, for every $0<\epsilon /4\ll \overline {\chi }$ and every $\delta _2\in (0,1/9)$ , there is a positive number $\overline {L}=\overline {L}(\epsilon /4,\delta _2)$ such that:

• • for every $T\geq \overline {L}$ , there exists a measurable set $\overline {\Gamma ^T}=\overline {\Gamma ^T}(\epsilon /4,\delta _2)\subset \Gamma $ with $\mu (\overline {\Gamma ^T})\geq 1-\delta _2$ ;

• • there is a natural number $N=N(T)$ such that for every $n\geq N$ and every $x\in \overline {\Gamma ^T}$ , we have that

  $$ \begin{align*} \prod_{i=0}^{n-1}\lVert\overline{\psi^*_T}\lvert_{\overline{E_1}(\overline{\varphi_{iT}}(x))}\rVert\leq e^{-(\overline{\chi}-\epsilon/4)nT},\quad \prod_{i=0}^{n-1}m(\overline{\psi^*_T}|_{\overline{E_2}(\overline{\varphi_{iT}}(x))})\geq e^{(\overline{\chi}-\epsilon/4)nT}, \end{align*} $$
  $$ \begin{align*} \dfrac{\lVert\overline{\psi^*_T}|_{\overline{E_1}(x)}\rVert}{m(\overline{\psi^*_T}|_{\overline{E_2}(x)})}\leq e^{-T(\overline{\chi}-\epsilon/4)}. \end{align*} $$

Fix an integer $T_0\geq \overline {L}(\epsilon /4,\delta _2)$ and let $\eta _0=(\overline {\chi }-\epsilon /4)T_0$ . For $k\in \mathbb {N}^+$ , we define the Pesin block $\Lambda ^{T_0}_{\eta _0}(k)$ with respect to the scaled linear Poincaré flow $\overline {\psi ^*_t}$ as

$$ \begin{align*} \begin{aligned} \Lambda^{T_0}_{\eta_0}(k) & =\bigg\{x\in\Gamma:\prod_{i=0}^{n-1}\lVert\overline{\psi^*_{T_0}}|_{\overline{E_1}(\overline{\varphi_{iT_0}}(x))}\rVert\leq ke^{-n\eta_0}\text{ for all } n\geq 1,\\ & \qquad \prod_{i=0}^{n-1}m(\overline{\psi^*_{T_0}}|_{\overline{E_2}(\overline{\varphi_{iT_0}}(x))})\geq k^{-1}e^{n\eta_0}\text{ for all } n\geq 1,~d(x,\text{Sing}(X))\geq\dfrac{1}{k}\bigg\}. \end{aligned} \end{align*} $$

According to [Reference Wu, Yang and Zhang37, Proposition 5.3], $\Lambda ^{T_0}_{\eta _0}(k)$ is a compact set and

$$ \begin{align*} \mu(\Lambda^{T_0}_{\eta_0}(k))\rightarrow\mu(\overline{\Gamma^{T_0}})\quad\text{as } k\rightarrow+\infty. \end{align*} $$

Therefore, one can fix k large enough such that $\mu (\Lambda ^{T_0}_{\eta _0}(k))>1-2\delta _2>0$ . For this fixed k, there is $j=j(k)\in \mathbb {N}^+$ such that $k<e^{{jT_0\epsilon }/{4}}$ . Thus, for every $x\in \Lambda ^{T_0}_{\eta _0}(k)$ , one has that

$$ \begin{align*} \prod_{i=0}^{n-1}\lVert\overline{\psi^*_{jT_0}}|_{\overline{E_1}(\overline{\varphi_{ijT_0}}(x))}\rVert\leq e^{-(\overline{\chi}-\epsilon/2)njT_0},\\ \prod_{i=0}^{n-1}m(\overline{\psi^*_{jT_0}}|_{\overline{E_2}(\overline{\varphi_{ijT_0}}(x))})\geq e^{(\overline{\chi}-\epsilon/2)njT_0}\quad\text{for all } n\geq 1. \end{align*} $$

Set $\eta =(\overline {\chi }-\epsilon /2)jT_0$ , $T=jT_0$ , we consider the set

$$ \begin{align*} \begin{aligned} \Lambda^{T}_{\eta}(k)& =\bigg\{x\in\Gamma: \prod_{i=0}^{n-1}\lVert\overline{\psi^*_{T}}\lvert_{\overline{E_1}(\overline{\varphi_{iT}}(x))}\rVert\leq e^{-n\eta}\text{ for all } n\geq 1,\\&\qquad \prod_{i=0}^{n-1}m(\overline{\psi^*_{T}}|_{\overline{E_2}(\overline{\varphi_{iT}}(x))})\geq e^{n\eta}\text{ for all } n\geq 1,\,d(x,\text{Sing}(X))\geq\dfrac{1}{k}\bigg\}. \end{aligned} \end{align*} $$

Take $y\in \Lambda ^{T}_{\eta }(k)$ . By Poincaré recurrence theorem, there is an increasing integer sequence $\{l_n\}$ such that

$$ \begin{align*} d(y,\overline{\varphi}_{l_nT}(y))\longrightarrow 0. \end{align*} $$

By a discussion similar to Proposition 5.1, we conclude that there exists a periodic point p of $\overline {\varphi }_t$ such that

$$ \begin{align*} \lvert\overline{\unicode{x3bb}_2}(p)-\overline{\unicode{x3bb}_2}\rvert<\epsilon, \end{align*} $$

where $\overline {\unicode{x3bb} _1}(p)\leq \overline {\unicode{x3bb} _2}(p)$ are the Lyapunov exponents of $\mu _p$ with respect to the scaled linear Poincaré flow $\overline {\psi ^*_t}$ . According to the relationship between the Lyapunov exponents of the scaled linear Poincaré flow $\overline {\psi ^*_t}$ and those of the scaled linear Poincaré flow $\psi ^*_t$ , we have that

$$ \begin{align*} \overline{\unicode{x3bb}_1}(p)=-\unicode{x3bb}_2(p),\quad\overline{\unicode{x3bb}_2}(p)=-\unicode{x3bb}_1(p),\quad \overline{\unicode{x3bb}_1}=-\unicode{x3bb}_2,\quad\overline{\unicode{x3bb}_2}=-\unicode{x3bb}_1.\end{align*} $$

Thus,

$$ \begin{align*} \lvert\unicode{x3bb}_1(p)-\unicode{x3bb}_1\rvert<\epsilon. \end{align*} $$

This completes the proof.

Proof of Theorem B.

Given $\varepsilon>0$ small enough, let

$$ \begin{align*} L=\max\bigg\{L\bigg(\frac{\varepsilon}{4},\delta_1\bigg),\overline{L}\bigg(\frac{\varepsilon}{4},\delta_2\bigg)\bigg\}, \end{align*} $$

where $L({\varepsilon }/{4},\delta _1)$ given in Proposition 5.1 and $\overline {L}({\varepsilon }/{4},\delta _2)$ given in Proposition 5.2. For $T\geq L$ , the set

$$ \begin{align*} \Lambda:=\Gamma^T\cap\overline{\Gamma^T} \end{align*} $$

is a $\mu $ -positive set. For the $\varepsilon>0$ , by Propositions 5.1 and 5.2, there is a periodic point p such that the Lyapunov exponents of $\mu _p$ satisfy that

$$ \begin{align*} \lvert\unicode{x3bb}_i-\unicode{x3bb}_i(p)\rvert<\varepsilon\quad\text{for every }i=1,2, \end{align*} $$

where $\unicode{x3bb} _1(p)\leq \unicode{x3bb} _2(p)$ are the Lyapunov exponents of $\mu _p$ .