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Hyperbolic SRB Measures and the Leaf Condition

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Abstract

Let M be a Riemannian, boundaryless, and compact manifold, with \(\dim M\ge 2\) and let f be a \(C^{1+}\) diffeomorphism. We show that there is a hyperbolic SRB measure if and only if there exists an unstable leaf with a subset of positive leaf volume of hyperbolic points which return to some Pesin set with positive frequency. This answers a question of Pesin.

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Notes

  1. Admissible means that for any two consecutive letters which represent partition elements a and b resp., \(a\cap f^{-1}[b]\ne \varnothing \).

  2. I.e. its definition does not rely on a specific construction of symbolic dynamics, but only on the quality of hyperbolicity of the orbit of the point.

  3. I.e. finite number of in-going and out-going edges at each vertex.

  4. [Sar13] applies for the case \(d=2\), while [BO18] extends its results to \(d\ge 2\). However, [BO18][Proposition 4.4(1)] refers to [Sar13][Proposition 6.3(1)] for proof, as the proofs for these claims are almost identical. Therefore we refer the reader to the proof in the a-priori 2 dimensional case for the required bound.

  5. Write \(u=dim V^u(\underline{R})\), and let AB be \(u\times u\) matrices s.t. \(|a_{ij}-b_{ij}|\le \delta \) for all \(i,j\le u\). So \(\det A=\sum _{\sigma \in S_{u}}\mathrm {sgn}(\sigma )\prod _{i=1}^{u}a_{i\sigma (i)}\). Then \(|\det A-\det B|\le \sum _{\sigma \in S_{u}}|\prod _{i=1}^{u}a_{i\sigma (i)}-\prod _{i=1}^{u}b_{i\sigma (i)}|\le |S_{u}|\cdot u\Vert B\Vert _{Fr}^{u-1}\delta \). Take maximum over \(u\le d-1\).

  6. If \(a_n\xrightarrow [n\rightarrow \infty ]{} a\), and \(\exists c_n\downarrow 0\) s.t. \(\forall m\ge n\), \(|a_n-a_m|\le c_n\), then \(|a_n-a|\le |a_n-a_m|+|a_m-a|\), \(\forall m\ge 0\). Since the second term tends to 0 as \(m\rightarrow \infty \), we get \(|a_n-a|\le c_n\) for all n. We apply this to the bounds from equation (9).

  7. To see this, show double inclusion. Given \(\underline{S}^+\in [S]\cap \widehat{\Sigma }_R^\circ \), let \(\underline{S}^\pm :=\underline{S}\cdot \underline{S}^+\) be an admissible concatenation, and so \(\underline{S}^+=\tau (\underline{S}^\pm )\) while \(\widehat{\Sigma }^\circ \ni \underline{S}^\pm =\underline{R}({\widehat{\pi }}(\underline{S}^\pm ))\) and \({\widehat{\pi }}(\underline{S}^\pm )\in W^u(\underline{S})\). For the other inclusion, take \(x\in W^s(\underline{S})\), then \(\tau \circ \underline{R}(x)\in [S]\cap \widehat{\Sigma }_R^\circ \).

  8. It implies that \(P_G(\phi )\ge 0\), therefore \(P_G(\phi )=0\) and so \(\sum _{n\ge 1}Z_n(\phi ,R)e^{-nP_G(\phi )}=\sum _{n\ge 1}Z_n(\phi ,R)=\infty \).

  9. By Eq. (36), \(\phi \) is bounded on cylinders of length 2, and there are only finitely many such cylinders contained in \([R_0]\) by the local-compactness of \(\widehat{\Sigma }_L\).

  10. The condition of topological mixing can be assumed w.l.o.g. because of the Spectral Decomposition theorem for topological Markov shifts (see [Sar15, Theorem 2.5]) .

  11. If \(x={\widehat{\pi }}(\underline{R}\cdot \underline{S})\), where the dot means an admissible concatenation, and \(\underline{S}\in \widehat{\Sigma }_R^\circ \), then \(f^{-n}(x)={\widehat{\pi }}(\sigma _R^n\underline{R}\cdot ((R_{-n},...,R_0)\cdot \underline{S}))\) and \((R_{-n},...,R_0)\cdot \underline{S}\in \widehat{\Sigma }_R^\circ \) (by the Markov property); in addition \(\underline{R}\in \widehat{\Sigma }_L^\#\Rightarrow \sigma _R^n\underline{R}\in \widehat{\Sigma }_L^\#\).

  12. Hölder continuous on compact sets, while the Hölder exponent is uniform on the space.

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Acknowledgements

This work constitutes a part of a doctoral thesis, conducted in the Weizmann Institute of Science under the guidance of O. Sarig. I would like to thank Professor Pesin for introducing this question to me, for reading this manuscript with care, and for his useful remarks. In addition, I would like to thank the referees for their useful, detailed, and insightful remarks. I would also like to thank the Weizmann Institute of Science for excellent working conditions. This research was partially supported by ISF grant 1149/18 and BSF grant 2016105.

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Correspondence to Snir Ben Ovadia.

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Appendix A. Additional Properties of \(\phi \)

Appendix A. Additional Properties of \(\phi \)

Recall the definition of the potential \(\phi :\widehat{\Sigma }_L\rightarrow \mathbb {R}^-\) from Theorem 4.10: \(\phi (\underline{R})=\log (m_{V^u(\sigma _R\underline{R})}(f^{-1}[V^u(\underline{R})]))\). The questions of the boundedness of \(\phi \) and its cohomology relation with the geometric potential are not necessary for the proof of our main results, but are nonetheless interesting. In this appendix we discuss these questions.

1.1 A.1. Boundness of \(\phi \)

Since the range of \(\phi \) is \(\mathbb {R}^-\), it is clear that \(\phi \) is bounded from above. In this section we show that \(\phi \) is also bounded from below.

Lemma A.1

\(\exists \widehat{\gamma }\ge 1\), a constant depending only on \(\chi ,M\) and f, s.t. \(\mathrm {Vol}\left( V^u(\sigma _R\underline{u})\right) \le \widehat{\gamma } \cdot \mathrm {Vol}\left( V^u(\underline{u})\right) \), \(\forall \underline{u}\in \Sigma _L\).

Proof

Write \(u_{-1}=\psi _{x}^{p^u,p^s}\), \(u_0=\psi _y^{q^u,q^s}\). Let F be the representing function of \(V^u(\underline{u})\) in \(u_{0}\), and G the representing function of \(V^u(\sigma _R\underline{u})\) in \(u_{-1}\).

Step 1: Comparing \(p^u\) and \(q^u\): Since \((u_{-1},u_0)\in {\mathcal {E}}\) (i.e. \(u_{-1}\rightarrow u_0\), see [BO18, Definition 2.23]), \(q^u=\min \{e^\epsilon p^u,Q_\epsilon (y)\}\). Thus

$$\begin{aligned} \frac{p^u}{q^u}=\frac{p^u}{\min \{e^\epsilon p^u,Q_\epsilon (y)\}}\le \max \{e^{-\epsilon }, \frac{Q_\epsilon (x)}{Q_\epsilon (y)}\} \text { }(\because p^u\le Q_\epsilon (x)). \end{aligned}$$
(42)

The fact that \((u_{-1},u_0)\in {\mathcal {E}}\) also implies that \(\psi _{f(x)}^{q^s\wedge q^u}\) \(\epsilon \)-overlaps \(\psi _{y}^{q^s\wedge q^u}\) (see [BO18, Definition 2.18]); which in particular implies (by [BO18, Lemma 2.20]) that \(\frac{\Vert C_\chi ^{-1}(f(x))\Vert }{\Vert C_\chi ^{-1}(y)\Vert }=e^{\pm \epsilon ^3}\). Thus, by definition,

$$\begin{aligned} Q_\epsilon (y)\le \frac{\epsilon ^\frac{90}{\beta }}{3^\frac{6}{\beta }} \Vert C_\chi ^{-1}(y)\Vert ^\frac{-48}{\beta }\le & {} \frac{\epsilon ^\frac{90}{\beta }}{3^\frac{6}{\beta }} \Vert C_\chi ^{-1}(f(x))\Vert ^\frac{-48}{\beta }\cdot (e^{\epsilon ^3})^\frac{48}{\beta }\\\le & {} e^{\frac{\epsilon }{3}}\cdot Q_\epsilon (f(x))\cdot (e^{\epsilon ^3})^\frac{48}{\beta }\le Q_\epsilon (f(x))\cdot e^\frac{\epsilon }{2}, \end{aligned}$$

when \(\epsilon >0\) is sufficiently small. Similarly, \(Q_\epsilon (y)\ge e^{-\frac{\epsilon }{2}}Q_\epsilon (f(x))\); and so \(\frac{Q_\epsilon (f(x))}{Q_\epsilon (y)}=e^{\pm \frac{\epsilon }{2}}\). In addition, by [BO18, Lemma 2.15], \(\exists \omega _0\ge 1\) depending only on \(\chi , M\) and f, s.t. \(\frac{Q_\epsilon (f(x))}{Q_\epsilon (x)}=\omega _0^{\pm 1}\). Plugging these inequalities in Eq. (42), we obtain

$$\begin{aligned} \frac{p^u}{q^u}\le \max \{e^{-\epsilon }, \frac{Q_\epsilon (x)}{Q_\epsilon (f(x))}\cdot \frac{Q_\epsilon (f(x))}{Q_\epsilon (y)}\} \le \max \{e^{-\epsilon }, \omega _0\cdot e^\frac{\epsilon }{2}\}\le e^\epsilon \cdot \omega _0. \end{aligned}$$
(43)

Step 2: Write \({\widetilde{F}}(t):=(F(t),t)\), \({\widetilde{G}}(t):=(G(t),t)\). Following from Definition 4.15, \(\mathrm {Jac}(d_\cdot {\widetilde{F}}),\mathrm {Jac}(d_\cdot {\widetilde{G}})=e^{\pm \epsilon }\), for \(\epsilon >0\) sufficiently small.

Step 3: Comparing \(\mathrm {Jac}(C_\chi (x))=|\det C_\chi (x) |\) and \(\mathrm {Jac}(C_\chi (y)) =|\det C_\chi (y) | \): By [BO18, Theorem 2.4], \(\forall \xi \in T_xM\),

$$\begin{aligned} |C_\chi ^{-1}(x)\xi |=\sqrt{S^2(x,\xi _s)+U^2(x,\xi _u)}, \end{aligned}$$

where \(\xi =\xi _s+\xi _u\), \(\xi _s\in H^s(x)\), \(\xi _u\in H^u(x)\), and

$$\begin{aligned} S^2(x,\xi _s)=2\sum _{m=0}^\infty |d_xf^m\xi _s|^2 e^{2\chi m}<\infty ,\text { } U^2(x,\xi _u)=2\sum _{m=0}^\infty |d_xf^{-m}\xi _u|^2 e^{2\chi m}<\infty . \end{aligned}$$

The same formula holds with f(x) replacing x. One can then check that

$$\begin{aligned} \frac{U^2(f(x),d_xf\xi _u)}{U^2(x,\xi _u)}=e^{\pm \log (e^{2\chi }+M_f^2)}, \frac{S^2(f(x),d_xf\xi _s)}{S^2(x,\xi _s)}= e^{\pm \log (e^{2\chi }+M_f^2)}. \end{aligned}$$

It follows then, that \(\frac{|C_\chi ^{-1}(f(x))d_xf\xi |}{|C_\chi ^{-1}(x)\xi |}= e^{\pm \log \sqrt{e^{2\chi }+M^2_f}}\). It follows that \(\frac{\mathrm {Jac}(C_\chi (x))}{\mathrm {Jac}(C_\chi (f(x)))}=e^{\pm d\log \left( e^{2\chi }+M^2_f\right) }\). In addition, by [BO18, Lemma 2.20], \(\frac{\mathrm {Jac}(C_\chi (f(x)))}{\mathrm {Jac}(C_\chi (y))}=e^{\pm d\cdot \epsilon ^2}=e^{\pm \epsilon }\) (for \(\epsilon >0\) sufficiently small). Thus, in total,

$$\begin{aligned} \frac{\mathrm {Jac}(C_\chi (x))}{\mathrm {Jac}(C_\chi (y))}= \frac{\mathrm {Jac}(C_\chi (x))}{\mathrm {Jac}(C_\chi (f(x)))}\cdot \frac{\mathrm {Jac}(C_\chi (f(x)))}{\mathrm {Jac}(C_\chi (y))}= e^{\pm \left( \epsilon +d\log \left( e^{2\chi }+M^2_f\right) \right) }. \end{aligned}$$

Step 4: It follows by the definition of \({\widetilde{G}},{\widetilde{F}}\), that \(V^u(\underline{u})=\psi _y\circ {\widetilde{F}}[R_{q^u}(0)]\) and \(V^u(\sigma _R\underline{u})=\psi _x\circ {\widetilde{G}}[R_{p^u}(0)]\), where \(R_{q^u}(0)=\{u\in \mathbb {R}^{\mathrm {dim}H^u}:|u|_\infty \le q^u\}\), \(R_{p^u}(0)=\{u\in \mathbb {R}^ {\mathrm {dim}H^u}:|u|_\infty \le p^u\}\). Recall, \(\psi _x=\exp _x\circ C_\chi (x), \psi _y=\exp _y\circ C_\chi (y)\). Thus

$$\begin{aligned} \mathrm {Vol}(V^u(\sigma _R\underline{u}))=\,&\int _{R_{p^u}(0)} \mathrm {Jac}(d_{ C_\chi (x)\circ {\widetilde{G}}(t)}\exp _x)\cdot \mathrm {Jac}(C_\chi (x))\cdot \mathrm {Jac}(d_t{\widetilde{G}})d\mathrm {Leb}(t)\\ \le \,&2^d\cdot e^{\epsilon } \cdot (2p^u)^{\mathrm {dim}H^u(x)}\mathrm {Jac}(C_\chi (x))\,\\&\,(\because \Vert d_\cdot \exp _x\Vert \le 2\text { by Definition} 4.14)\\ \le \,&4^d e^{2\epsilon } e^{d(\epsilon +\log \omega _0)}e^{\epsilon +d\log \left( e^{2\chi }+M^2_f\right) }\\&\int _{R_{q^u}(0)} \mathrm {Jac}(d_{ C_\chi (y)\circ {\widetilde{F}}(t)}\exp _y)\cdot \mathrm {Jac}(C_\chi (y))\cdot \mathrm {Jac}(d_t{\widetilde{F}})d\mathrm {Leb}(t)\\ =\,&4^d \cdot e^{2\epsilon }\cdot e^{d(\epsilon +\log \omega _0)}\cdot e^{\epsilon +d\log \left( e^{2\chi }+M^2_f\right) }\cdot \mathrm {Vol}(V^u(\underline{u})), \end{aligned}$$

The lemma follows with \(\widehat{\gamma }:= 4^d \cdot e^{2}\cdot e^{d(1+\log \omega _0)}\cdot e^{1+d\log \left( e^{2\chi }+M^2_f\right) }\). \(\square \)

Lemma A.2

Let \(\underline{u},\underline{v}\in \Sigma ^\#\) s.t. \(\pi (\underline{u})= \pi (\underline{v})=p \). Then \(\exists A\subseteq V^u(\underline{u})\cap V^u(\underline{v})\) an open set in \(V^u(\underline{u})\) which depends only on \(\underline{u}\) s.t. \(\mathrm {Vol}(A) \ge \mathrm {Vol}(V^u(\underline{u}))\cdot \widehat{\alpha } \), where \(\widehat{\alpha }\in (0,1)\) depends only on M.

Proof

Write \(u_0=\psi _x^{p^s,p^u}, v_0=\psi _y^{q^s,q^u}\). By [BO18, Lemma 4.12], \(\frac{p^u}{q^u}=e^{\pm \epsilon ^\frac{1}{3}}\). Let \(B:=R_{\frac{p^u}{100\sqrt{d}}}(0)\) be a ball in \(|\cdot |_\infty \)-norm in \(\mathbb {R}^d\) centered at 0 with a radius of \(\frac{p^u}{100\sqrt{d}} \). By [BO18, Theorem 4.13], \(\psi _y^{-1}\circ \psi _x[B]\subseteq R_{\frac{p^u}{100}+\frac{q^u}{10}+\frac{p^u}{100}\frac{1}{2}\epsilon ^\frac{1}{3}}(0)\subseteq R_{\frac{q^u}{5}}(0)\). Thus \(\psi _x[B]\subseteq \psi _y[R_{\frac{q^u}{5}}(0)] \subseteq \psi _y[R_{q^u}(0)]\), and so \(\psi _x[B]\cap V^u(\underline{u})\subseteq V^u(\underline{v})\) (since \(V^u(\underline{u})\) spans over the window of \(u_0\), and \(V^u(\underline{v})\) spans over the window of \(v_0\), and since both are local unstable leaves of p, they must coincide where the windows intersect).

Denote by F the representing function of \(V^u(\underline{u})\) in \(u_0\). Write \({\widetilde{F}}(t):=(F(t),t)\). We now wish to compare the volume of \(A:=\psi _x\circ {\widetilde{F}}[R_{\frac{p^u}{\sqrt{d}100}}(0)]\subseteq \psi _x[B]\). We follow the estimates in steps 2 and 4 of Lemma A.1:

$$\begin{aligned} \mathrm {Vol}(A)=&\int _{R_{\frac{p^u}{\sqrt{d}100}}(0)}\mathrm {Jac}(d_{{\widetilde{F}}(t)}\psi _x)\mathrm {Jac}(d_t{\widetilde{F}})d\mathrm {Leb}(t)\\ =&\int _{R_{\frac{p^u}{\sqrt{d}100}}(0)}\mathrm {Jac}(d_{C_\chi (x){\widetilde{F}}(t)}\exp _x)\mathrm {Jac}(C_\chi (x))\mathrm {Jac}(d_t{\widetilde{F}})d\mathrm {Leb}(t)\\ \ge&2^{-d}\cdot e^{-\epsilon }\cdot (2\frac{p^u}{100\sqrt{d}})^{\mathrm {dim}H^u(x)}\mathrm {Jac}(C_\chi (x))\\ \ge&2^{-2d}\cdot e^{-2\epsilon }\cdot \frac{1}{(100\sqrt{d})^{d}} \int _{R_{p^u}(0)}\mathrm {Jac}(d_{{\widetilde{F}}(t)}\psi _x)\mathrm {Jac}(d_t{\widetilde{F}})d\mathrm {Leb}(t)\\ =&2^{-2d}\cdot e^{-2\epsilon }\cdot \frac{1}{(100\sqrt{d})^{d}}\cdot \mathrm {Vol}(V^u(\underline{u})). \end{aligned}$$

Define \(\widehat{\alpha }:= 2^{-2d}\cdot e^{-2}\cdot \frac{1}{(100\sqrt{d})^{d}} \in (0,1)\). \(\square \)

Corollary A.3

\(\exists \widehat{\omega }\ge 1\) which depends only on \(\chi ,M\) and f, s.t. \(\forall \underline{R}\in \widehat{\Sigma }_L\), \(\mathrm {Vol}(V^u(\sigma _R\underline{R}))\le \widehat{\omega }\mathrm {Vol}(V^u(\underline{R}))\).

Proof

Fix some extension \(\underline{R}^{\pm }\in \widehat{\Sigma }^\#\) s.t. \(R_i=R^{\pm }_i\) \(\forall i\le 0\). By Definition 4.3, \(\mathrm {Vol}(V^u(\sigma _R\underline{R}))=\mathrm {Vol}(\bigcap \limits _{\underline{u}\curvearrowright \sigma _R\underline{R}}V^u(\underline{u}))\). Let \(\underline{u}\in \Sigma _L^\#\) be a chain which covers \(\underline{R}\), then \(\sigma _R\underline{u}\curvearrowright \sigma _R\underline{R}\). Fix a chain \(\underline{v}\in \Sigma _L^\#\) s.t. \(\underline{v}\curvearrowright \underline{R}\). Let \(A=A(\underline{v})\) be the set given by Lemma A.2. It follows that \(A\subseteq \bigcap \limits _{\underline{u}\curvearrowright \underline{R}}V^u(\underline{u})\). Then,

$$\begin{aligned} \mathrm {Vol}(V^u(\sigma _R\underline{R}))=&\mathrm {Vol}\left( \bigcap \limits _{\underline{u}\curvearrowright \sigma _R\underline{R}}V^u(\underline{u}) \right) = \mathrm {Vol}\left( \bigcap \limits _{\underline{v}'\curvearrowright \underline{R}}V^u(\sigma _R\underline{v}') \right) \le \mathrm {Vol}\left( V^u(\sigma _R\underline{v})\right) \\ \le&\widehat{\gamma } \mathrm {Vol}\left( V^u(\underline{v}) \right) \le \frac{\widehat{\gamma }}{\widehat{\alpha }}\mathrm {Vol}\left( A \right) \le \frac{\widehat{\gamma }}{\widehat{\alpha }} \mathrm {Vol}\left( \bigcap \limits _{\underline{u}\curvearrowright \underline{R}}V^u(\underline{u}) \right) \\ =&\frac{\widehat{\gamma }}{\widehat{\alpha }} \mathrm {Vol}\left( V^u(\underline{R})) \right) \text { (}\because \text {Lemma} A.1,\hbox {Lemma} A.2). \end{aligned}$$

Define \(\widehat{\omega }:=\frac{\widehat{\gamma }}{\widehat{\alpha }}\). \(\square \)

Corollary A.4

\(\phi :\widehat{\Sigma }_L\rightarrow (-\infty ,0]\) is bounded.

Proof

Let \(\underline{R}\in \widehat{\Sigma }_L\). Recall, \(m_{V^u(\sigma _R\underline{R})}=\frac{\rho _{\sigma _R\underline{R}}}{\mathrm {Vol}(V^u(\sigma _R\underline{R}))}\cdot \lambda _{V^u(\sigma _R\underline{R})}\), where \(\rho _{\sigma _R\underline{R}}:V^u(\sigma _R\underline{R})\rightarrow [e^{-\epsilon },e^\epsilon ]\) and \(\lambda _{V^u(\sigma _R\underline{R})}\) is the (not normalized) Riemannian volume measure of \(V^u(\sigma _R\underline{R})\). As in Claim 4.12, \(\phi (\underline{R})=\log m_{V^u(\sigma _R\underline{R})}\left( f^{-1}[V^u(\underline{R})]\right) \). Then,

$$\begin{aligned} \phi (\underline{R})=&\log m_{V^u(\sigma _R\underline{R})}\left( f^{-1}[V^u(\underline{R})]\right) =\pm \epsilon +\log \lambda _{V^u(\sigma _R\underline{R})} \left( f^{-1}[V^u(\underline{R})]\right) \\&+\log \frac{1}{\mathrm {Vol}\left( V^u(\sigma _R\underline{R})\right) }\\ =&\pm \epsilon \pm \log M_f+\log \frac{\mathrm {Vol}(V^u(\underline{R}))}{\mathrm {Vol}(V^u(\sigma _R\underline{R}))}\ge -\epsilon -\log M_f-\log \widehat{\omega }>-\infty . \end{aligned}$$

\(\square \)

Remark

Claim 4.21 together with the boundness of \(\phi \) imply that \(\phi :\widehat{\Sigma }_L\rightarrow \mathbb {R}^-\) is in fact Hölder continuous.

1.2 A.2. Cohomology of \(\phi \) to the geometric potential

Recall the geometric potential from Definition 6.1: \(\varphi :\widehat{\Sigma }\rightarrow [-d\cdot \log M_f,d\cdot \log M_f]\), \(\varphi (\underline{R}):=\log \mathrm {Jac}(d_{{\widehat{\pi }}(\underline{R})}f^{-1}|_{T_{{\widehat{\pi }}(\underline{R})}V^u(\underline{R})})\). In Part 1 of Theorem 6.2 we saw: \(\forall n\ge 1\),

$$\begin{aligned} |\phi _n(\underline{R})-\varphi _n(\underline{R}^\pm )|\le 2\epsilon , \end{aligned}$$
(44)

where \(\underline{R}\) is any n-periodic chain in \(\widehat{\Sigma }_L\), and \(\underline{R}^\pm \) is the unique periodic extension of \(\underline{R}\) to \(\widehat{\Sigma }\).

In this section, we show further that in fact \(\phi \) and \(\varphi \) are cohomologous:

Claim A.5

There exists a locally Hölder continuous function \({\mathcal {L}}:\widehat{\Sigma }\rightarrow \mathbb {R}\) s.t.

$$\begin{aligned} \phi =\varphi + {\mathcal {L}}-{\mathcal {L}}\circ \sigma ^{-1} \end{aligned}$$

when restricted to irreducible components of \(\widehat{\Sigma }\).

\({\mathcal {L}}\) is called a Hölder continuous transfer function, and \({\mathcal {L}}-{\mathcal {L}} \circ \sigma ^{-1}\) is called a coboundary.

Proof

Assume w.l.o.g. that \(\widehat{\Sigma }_L\) is irreducible. By Part 2 in Theorem 6.2, there exists a bounded and Hölder continuous function \(A:\widehat{\Sigma }\rightarrow \mathbb {R}\) s.t.

$$\begin{aligned} \varphi ^-:=\varphi +A-A\circ \sigma ^{-1} \end{aligned}$$

is a well-defined and Hölder continuous function \(\varphi ^-:\widehat{\Sigma }_L\rightarrow \mathbb {R}\) (i.e. depends only on the negative coordinates of chains).

We continue to show that \(\varphi ^-\) and \(\phi \) are cohomologous with a Hölder continuous transfer function, which in turn implies that \(\phi \) and \(\varphi \) are cohomologous with a Hölder continuous transfer function.

Let \(n\ge 1\), and let \(\underline{w}=(R,w_{-(n-2)},\ldots ,w_{-1},R)\) be an admissible word of length n. Let \(\underline{W}\) be its periodic extension to \(\widehat{\Sigma }_L\).

Then it follows that for any \(m\in \mathbb {N}\),

$$\begin{aligned} \phi _{n\cdot m}(\underline{W})-\varphi ^-_{n\cdot m}(\underline{W})=m\cdot \left( \phi _n(\underline{W})-\varphi _n^-(\underline{W}) \right) . \end{aligned}$$

At the same time, \(|\phi _{n\cdot m}(\underline{W})-\varphi ^-_{n\cdot m}(\underline{W}) |\le 2(\epsilon +\Vert A\Vert _\infty )\) for all n and m. Therefore it follows that \(\forall n\ge 1\), and for every n-periodic chain \(\underline{R}\in \widehat{\Sigma }_L\),

$$\begin{aligned} \phi _n(\underline{R})-\varphi ^-_n(\underline{R})=0. \end{aligned}$$

By Livsic’s theorem (see [Sar09, Theorem 1.1] for the statement and the proof in the context of countable Markov shifts), there exists a locally Hölder continuous functionFootnote 12\({\mathcal {L}}':\widehat{\Sigma }_L\rightarrow \mathbb {R}\) s.t.

$$\begin{aligned} \phi =\varphi ^-+{\mathcal {L}}'-{\mathcal {L}}' \circ \sigma _R. \end{aligned}$$

Thus, in total,

$$\begin{aligned} \phi = \varphi +A-A\circ \sigma ^{-1}+{\mathcal {L}}'-{\mathcal {L}}' \circ \sigma ^{-1}= \varphi +(A+{\mathcal {L}}')-(A+{\mathcal {L}}') \circ \sigma ^{-1}. \end{aligned}$$

Set \({\mathcal {L}}=A+{\mathcal {L}}'\). \(\square \)

Remark

Carrying out (38) without the assumption \(\underline{R}=\sigma _R^n\underline{R}\) yields the more careful estimate \(e^{\phi _n-\varphi _n}=e^{\pm 2\epsilon }\frac{\mathrm {Vol}(V^u(\underline{R}))}{\mathrm {Vol}(V^u(\sigma _R^n\underline{R}))}\), which in general may not be bounded in n, even on irreducible components. This implies that while both \(\phi \) and \(\varphi \) are Hölder continuous (and bounded), and \({\mathcal {L}}\) is locally Hölder continuous, and \(\phi =\varphi +{\mathcal {L}}-{\mathcal {L}}\circ \sigma ^{-1}\), in general \({\mathcal {L}}\) may still be unbounded.

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Ben Ovadia, S. Hyperbolic SRB Measures and the Leaf Condition. Commun. Math. Phys. 387, 1353–1404 (2021). https://doi.org/10.1007/s00220-021-04208-6

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