## Formalized Name Existence and Uniqueness of the SRB Measure for a Transitive Anosov Diffeomorphism ## Formalized Statement Let $M$ be a compact connected $C^\infty$ Riemannian manifold, let $m$ denote its Riemannian volume measure, and let $f: M \to M$ be a topologically transitive $C^{1+\alpha}$ Anosov diffeomorphism for some $\alpha \in (0,1]$. Let \begin{align*} TM = E_s \oplus E_u \end{align*} denote the continuous $Df$-invariant stable and unstable splitting, let $W_{u,\mathrm{loc}}(x)$ denote a local unstable manifold through $x \in M$, and let $m_{u,x}$ denote the Riemannian volume measure induced on $W_{u,\mathrm{loc}}(x)$. Then there exists a unique $f$-invariant Borel probability measure $\mu_{\mathrm{SRB}}$ on $M$ whose conditional measures, with respect to every measurable partition subordinate to local unstable manifolds equivalently in every local product box away from partition-boundary ambiguity sets, are absolutely continuous with respect to the corresponding induced measures $m_{u,x}$. This measure is ergodic. Moreover, its basin \begin{align*} B(\mu_{\mathrm{SRB}}) := \left\{x \in M : \frac{1}{N}\sum_{k=0}^{N-1} \varphi(f^k(x)) \to \int_M \varphi \, d\mu_{\mathrm{SRB}} \text{ for every } \varphi \in C^0(M;\mathbb{R})\right\} \end{align*} has full Riemannian volume: \begin{align*} m(B(\mu_{\mathrm{SRB}})) = m(M). \end{align*} ## Proof [proofplan] This proof uses the standard Sinai-Bowen-Ruelle construction. A Markov partition codes the transitive Anosov diffeomorphism by a transitive subshift of finite type, and the unstable Jacobian defines a Hölder geometric potential on the symbolic system. Thermodynamic formalism gives a unique Gibbs equilibrium state for this potential; its pushforward to $M$ is invariant, ergodic, and has absolutely continuous unstable conditional measures by bounded distortion. Finally, uniqueness of the symbolic Gibbs state gives uniqueness of the SRB measure, and absolute continuity of stable holonomy upgrades full conditional measure on unstable plaques to full Riemannian volume in $M$. [/proofplan] [step:Code the Anosov diffeomorphism by a transitive Markov shift] Let $E_u \subset TM$ and $E_s \subset TM$ denote the unstable and stable subbundles of the Anosov splitting for $f$. Define the unstable Jacobian \begin{align*} J_u f: M \to (0,\infty) \end{align*} by \begin{align*} J_u f(x) := \left|\det\left(df_x\big|_{(E_u)_x}: (E_u)_x \to (E_u)_{f(x)}\right)\right|. \end{align*} Since $f$ is $C^{1+\alpha}$ and Anosov, the unstable bundle is Hölder continuous, so $J_u f$ is Hölder continuous and positive. By the Markov partition theorem for transitive Anosov diffeomorphisms (citing a result not yet in the wiki: Markov partitions for Anosov diffeomorphisms), there exist a finite alphabet $\mathcal{A}=\{1,\dots,q\}$, a transitive transition matrix $A \in \{0,1\}^{q \times q}$, a two-sided subshift of finite type \begin{align*} \Sigma_A := \left\{\omega=(\omega_k)_{k \in \mathbb{Z}} \in \mathcal{A}^{\mathbb{Z}} : A_{\omega_k\omega_{k+1}}=1 \text{ for every } k \in \mathbb{Z}\right\}, \end{align*} the left shift map \begin{align*} \sigma: \Sigma_A &\to \Sigma_A \end{align*} \begin{align*} (\omega_k)_{k \in \mathbb{Z}} &\mapsto (\omega_{k+1})_{k \in \mathbb{Z}}, \end{align*} and a continuous surjective coding map \begin{align*} \pi: \Sigma_A &\to M \end{align*} such that \begin{align*} \pi \circ \sigma = f \circ \pi. \end{align*} The map $\pi$ is finite-to-one outside the standard Markov partition boundary ambiguity set, and this exceptional set is negligible for the Gibbs measures constructed below. [guided] We first replace the smooth dynamical system by a symbolic one. The object that makes this possible is a Markov partition. The Markov partition theorem for transitive Anosov diffeomorphisms says that the orbit of every point can be coded by the itinerary of the rectangles it visits, and the allowed transitions are encoded by a finite matrix $A$. Formally, let $\mathcal{A}=\{1,\dots,q\}$ be the finite set of Markov rectangles and let $A \in \{0,1\}^{q \times q}$ record which rectangles may follow which under $f$. The associated symbolic space is \begin{align*} \Sigma_A := \left\{\omega=(\omega_k)_{k \in \mathbb{Z}} \in \mathcal{A}^{\mathbb{Z}} : A_{\omega_k\omega_{k+1}}=1 \text{ for every } k \in \mathbb{Z}\right\}. \end{align*} The left shift is the map \begin{align*} \sigma: \Sigma_A &\to \Sigma_A \end{align*} \begin{align*} (\omega_k)_{k \in \mathbb{Z}} &\mapsto (\omega_{k+1})_{k \in \mathbb{Z}}. \end{align*} The coding map \begin{align*} \pi: \Sigma_A &\to M \end{align*} sends an admissible itinerary to the unique point whose orbit follows that itinerary, except for the usual finite boundary ambiguity. The compatibility relation \begin{align*} \pi \circ \sigma = f \circ \pi \end{align*} means that shifting the symbolic itinerary corresponds exactly to applying $f$ on $M$. The transitivity hypothesis on $f$ is used here to choose the symbolic model transitive. This matters later because uniqueness and ergodicity of the Gibbs equilibrium state hold for Hölder potentials on transitive subshifts of finite type, without requiring the stronger mixing assumption. [/guided] [/step] [step:Construct the symbolic Gibbs state for the geometric potential] Define the geometric potential \begin{align*} \phi_u: M &\to \mathbb{R} \end{align*} \begin{align*} x &\mapsto -\log J_u f(x). \end{align*} Define its symbolic lift \begin{align*} \psi: \Sigma_A &\to \mathbb{R} \end{align*} \begin{align*} \omega &\mapsto \phi_u(\pi(\omega)). \end{align*} Because $\phi_u$ is Hölder continuous on $M$ and $\pi$ is Hölder continuous for the standard symbolic metric, $\psi$ is Hölder continuous on $\Sigma_A$. By the Ruelle-Perron-Frobenius theorem and the existence and uniqueness theorem for equilibrium states of Hölder potentials on transitive subshifts of finite type (citing a result not yet in the wiki: existence and uniqueness of equilibrium states for Hölder potentials on transitive subshifts of finite type), there exists a unique $\sigma$-invariant Borel probability measure $\nu$ on $\Sigma_A$ which is the equilibrium state for $\psi$. Equivalently, $\nu$ is the unique Gibbs measure for $\psi$. [/step] [step:Push the Gibbs state forward to obtain an invariant measure on $M$] Define the Borel probability measure \begin{align*} \mu_{\mathrm{SRB}} := \pi_* \nu \end{align*} on $M$ by \begin{align*} \mu_{\mathrm{SRB}}(B) := \nu(\pi^{-1}(B)) \end{align*} for every Borel set $B \subset M$. Since $\nu$ is $\sigma$-invariant and $\pi \circ \sigma = f \circ \pi$, for every Borel set $B \subset M$ we have \begin{align*} \mu_{\mathrm{SRB}}(f^{-1}(B)) = \nu(\pi^{-1}(f^{-1}(B))). \end{align*} Using the semiconjugacy relation, $\pi^{-1}(f^{-1}(B))=\sigma^{-1}(\pi^{-1}(B))$, and therefore \begin{align*} \mu_{\mathrm{SRB}}(f^{-1}(B)) = \nu(\sigma^{-1}(\pi^{-1}(B))) = \nu(\pi^{-1}(B)) = \mu_{\mathrm{SRB}}(B). \end{align*} Thus $\mu_{\mathrm{SRB}}$ is $f$-invariant. [/step] [step:Identify the pushforward measure as an SRB measure] Let $W_{u,\mathrm{loc}}(x)$ denote a local unstable manifold through $x \in M$, and let $m_{u,x}$ denote the Riemannian volume measure induced on $W_{u,\mathrm{loc}}(x)$. In a local product box, disintegrate $\mu_{\mathrm{SRB}}$ along the measurable partition by local unstable plaques. For the plaque through $x$, let $\mu^u_x$ denote the corresponding conditional probability measure, and define $\rho_x: W_{u,\mathrm{loc}}(x) \to [0,\infty)$ to be the Radon-Nikodym density of the absolutely continuous part of $\mu^u_x$ with respect to $m_{u,x}$. The Gibbs property for $\nu$ and the choice $\psi=-\log J_u f \circ \pi$ imply the standard product formula for conditional densities along local unstable leaves: if $y,z \in W_{u,\mathrm{loc}}(x)$ lie outside the Markov partition boundary ambiguity set, then the density ratio is given by the convergent distortion product \begin{align*} \frac{\rho_x(y)}{\rho_x(z)} = \prod_{k=1}^{\infty} \frac{J_u f(f^{-k}(z))}{J_u f(f^{-k}(y))}. \end{align*} The infinite product converges because backward iterates of $y$ and $z$ converge exponentially along the unstable leaf and $\log J_u f$ is Hölder continuous. Bounded distortion gives constants $0<c_x\le C_x<\infty$ such that \begin{align*} c_x \le \rho_x(y) \le C_x \end{align*} for $m_{u,x}$-almost every $y \in W_{u,\mathrm{loc}}(x)$. Hence the conditional measure of $\mu_{\mathrm{SRB}}$ on $W_{u,\mathrm{loc}}(x)$ is absolutely continuous with respect to $m_{u,x}$. This is precisely the SRB property for a $C^{1+\alpha}$ Anosov diffeomorphism, so $\mu_{\mathrm{SRB}}$ is an SRB measure. The bounded distortion and conditional-density assertion is the standard Sinai-Bowen-Ruelle SRB-Gibbs correspondence (citing a result not yet in the wiki: SRB-Gibbs correspondence for Anosov diffeomorphisms). [/step] [step:Deduce ergodicity from transitivity of the symbolic Gibbs state] The transitive subshift $\sigma:\Sigma_A \to \Sigma_A$ has a unique Gibbs equilibrium state $\nu$ for the Hölder potential $\psi$. By the ergodicity part of the thermodynamic formalism theorem for transitive subshifts of finite type (citing a result not yet in the wiki: ergodicity of Hölder equilibrium states on transitive subshifts of finite type), $\nu$ is ergodic. Let $B \subset M$ be an $f$-invariant Borel set. Then $\pi^{-1}(B) \subset \Sigma_A$ is $\sigma$-invariant because $\pi \circ \sigma=f \circ \pi$. Since $\nu$ is ergodic, \begin{align*} \nu(\pi^{-1}(B)) \in \{0,1\}. \end{align*} Therefore \begin{align*} \mu_{\mathrm{SRB}}(B) \in \{0,1\}. \end{align*} Thus $\mu_{\mathrm{SRB}}$ is ergodic. [/step] [step:Use the symbolic uniqueness theorem to prove uniqueness of the SRB measure] Let $\mu$ be any SRB measure for $f$. Since $\pi:\Sigma_A \to M$ is a finite-to-one Markov coding and its ambiguity set is carried by Markov partition boundaries, one may lift $\mu$ to a $\sigma$-invariant Borel probability measure $\tilde{\nu}$ on $\Sigma_A$ with $\pi_*\tilde{\nu}=\mu$. The SRB property of $\mu$, namely absolute continuity of conditional measures on unstable leaves, is equivalent under the Markov coding to the Gibbs property for the lifted potential $\psi=-\log J_u f \circ \pi$; this is again the Sinai-Bowen-Ruelle SRB-Gibbs correspondence (citing a result not yet in the wiki: symbolic characterization of SRB measures for Anosov diffeomorphisms). Therefore $\tilde{\nu}$ is an equilibrium state for $\psi$. By uniqueness of the equilibrium state for Hölder potentials on the transitive subshift $\Sigma_A$, we have \begin{align*} \tilde{\nu}=\nu. \end{align*} Pushing forward by $\pi$ gives \begin{align*} \mu=\pi_*\tilde{\nu}=\pi_*\nu=\mu_{\mathrm{SRB}}. \end{align*} Hence the SRB measure is unique. [/step] [step:Spread the basin from unstable leaves to full Riemannian volume] Since $\mu_{\mathrm{SRB}}$ is ergodic, Birkhoff's ergodic theorem applied to the measure-preserving system $(M,\mu_{\mathrm{SRB}},f)$ gives \begin{align*} \mu_{\mathrm{SRB}}(B(\mu_{\mathrm{SRB}}))=1. \end{align*} Because $\mu_{\mathrm{SRB}}$ has conditional measures equivalent to unstable Riemannian volume on local unstable plaques, it follows that for a full family of local unstable plaques, the basin has full $m_{u,x}$-measure inside each plaque. The stable foliation of a $C^{1+\alpha}$ Anosov diffeomorphism is absolutely continuous (citing a result not yet in the wiki: absolute continuity of stable foliations for $C^{1+\alpha}$ Anosov diffeomorphisms). Moreover, if $y$ lies on the local stable manifold of $x$, then for every $\varphi \in C^0(M;\mathbb{R})$, \begin{align*} \left|\frac{1}{N}\sum_{k=0}^{N-1}\varphi(f^k(y))-\frac{1}{N}\sum_{k=0}^{N-1}\varphi(f^k(x))\right| \to 0 \end{align*} as $N \to \infty$, because $\operatorname{dist}(f^k(y),f^k(x)) \to 0$ exponentially and $\varphi$ is uniformly continuous on compact $M$. Hence the basin $B(\mu_{\mathrm{SRB}})$ is saturated by local stable manifolds, up to endpoints of local product boxes. Applying absolute continuity of stable holonomy in each local product box transfers full conditional measure on unstable plaques to full Riemannian volume in that box. A finite collection of local product boxes covers the compact manifold $M$, so \begin{align*} m(M \setminus B(\mu_{\mathrm{SRB}}))=0. \end{align*} Equivalently, \begin{align*} m(B(\mu_{\mathrm{SRB}}))=m(M). \end{align*} This proves the full-volume basin statement and completes the proof. [/step]