Mixing but non-exponentially mixing Axiom A flows: Ruelle’s example

Bowen extended Sinai’s construction of Markov partition to Axiom A diffeomorphism, and proved that the mixing rate of every Gibbs measure $\mu$ with respect to a Holder potential over a mixing basic set is always exponential, that is,
for all smooth functions $\phi,\psi$ on $M$ , the correlations $\rho(t)=\int \phi\cdot\psi\circ \sigma^n d\mu-\mu(\phi)\cdot\mu(\psi)\to 0$ exponentially.
However the situation is quite different for Axiom A flows. Ruelle gave the first class of examples: mixing but non-exponentially mixing Axiom A flows.

Ruelle first constructed a symbolic example and then embedded it into an Axiom A flow. Let $\Omega=\{0,1\}^{\mathbb{Z}}$ and $\sigma$ be the shift on $\Omega$ . Pick two positive numbers $\lambda_0<\lambda_1$ such that $\frac{\lambda_0}{\lambda_0}$ is irrational. Define a ceiling function $\tau:\omega\in\Omega\mapsto \lambda(\omega_0)$ . Then let $(\Omega_\tau,\sigma_t)$ be the suspension flow over $(\Omega,\sigma)$ with respect to $\tau$ . It is well known that there is a one to one correspondence between the $\sigma_t$ -invariant measures $\nu$ and $\sigma$ -invariant measures $\mu$ : $d\nu=\frac{dt\times d\mu}{\mu(\tau)}$ . Ruelle examined the measure of maximal entropy, $\mu$ (corresponding the zero potential) and showed that the corresponding measure $\nu$ does not mix exponentially under $\sigma_t$ .

Let $\phi$ be a smooth function supported on $[a,b]\subset (0,\lambda_0)$ . Without loss of generality we assume $\int \phi dx=0$ . Note that $\Phi(\omega,x)=\phi(x)$ is well defined on $\Omega_\tau$ and $\nu(\Phi)=0$ . We will derive a contradiction by assuming the following converges to 0 exponentially: $\rho(t)=\int \Phi\cdot \Phi\circ \sigma_t d\nu$ .

A useful trick here is to represent $\Phi(\sigma_t(\omega,x))=\sum_{n\ge0}\int_0^{\lambda(\omega_n)}\Phi(\sigma^n\omega,y)\delta_{t+x-y-\tau(\omega,n)} dy$ , where $\delta$ is the $\infty$ -pulse at $0$ . So the Fourier dual
$\hat{\rho}(s)=\int_{\mathbb{R}}e^{ist}\rho(t)dt=\int_{\mathbb{R}} e^{ist}\int_{\Omega_\tau}\Phi\cdot \Phi\circ \sigma_t d\nu dt$
$=\frac{2}{\lambda_0+\lambda_1}\int_{\mathbb{R}}e^{ist}\int_{\Omega} \int_{0}^{\lambda(\omega_0)} \phi(x)\cdot \sum_{n\ge0}\int_0^{\lambda(\omega_n)}\phi(y)\delta_{t+x-y-\tau(\omega,n)} dy dx d\mu dt$
$=\frac{2}{\lambda_0+\lambda_1}\int_{\Omega} \int_{0}^{\lambda(\omega_0)} \phi(x)\cdot \sum_{n\ge0}\int_0^{\lambda(\omega_n)}\phi(y)(\int_{\mathbb{R}}e^{ist}\delta_{t+x-y-\tau(\omega,n)} dt) dy dx d\mu$
$=\frac{2}{\lambda_0+\lambda_1}\int_{\Omega} \int_{0}^{\lambda(\omega_0)} \phi(x)\cdot \sum_{n\ge0}\int_0^{\lambda(\omega_n)}\phi(y)e^{is(y+\tau(\omega,n)-x)} dy dx d\mu$
$=\frac{2}{\lambda_0+\lambda_1}\sum_{n\ge0}\int_{\Omega} \int_{0}^{\lambda(\omega_0)} \phi(x)e^{-isx} dx\cdot \int_0^{\lambda(\omega_n)}\phi(y)e^{isy} dy \cdot e^{is\tau(\omega,n)} d\mu$
$=\frac{2\hat{\phi}(-s)\cdot \hat{\phi}(s)}{\lambda_0+\lambda_1}\sum_{n\ge0} \cdot\int_{\Omega} e^{is\tau(\omega,n)} d\mu$
$=\frac{2\hat{\phi}(-s)\cdot \hat{\phi}(s)}{\lambda_0+\lambda_1} \cdot\sum_{n\ge0}\sum_{(\omega)_0^{n-1}}\prod_{0\le j<n} \frac{e^{is\lambda(\omega_j)}}{2}$
$=\frac{2\hat{\phi}(-s)\cdot \hat{\phi}(s)}{\lambda_0+\lambda_1}\cdot\sum_{n\ge0} \left(\frac{e^{is\lambda_0}+e^{is\lambda_1}}{2}\right)^n=\frac{2\hat{\phi}(-s)\cdot \hat{\phi}(s) }{\lambda_0+\lambda_1}\cdot\left(1-\frac{e^{is\lambda_0}+e^{is\lambda_1}}{2}\right)^{-1}$ .

Remark 1. Ruelle’s assumption on $\phi$ simplifies the expression greatly: $\hat{\phi}(s)=\int_{0}^{\lambda(\omega_j)} \phi(x)e^{isx} dx$ since $\phi$ is supported on $[a,b]\subset (0,\lambda_0)$ . Moreover as pointed out by Ruelle, there are poles of $\left(1-\frac{e^{is\lambda_0}+e^{is\lambda_1}}{2}\right)^{-1}$ near the real axis and it may be arranged that these are not zeros of $\hat{\phi}(\pm\bullet)$ . So $\rho(t)$ does not decay exponentially.

Remark 2. Now let’s embed $\Omega$ into a diffeomorphism $(M,f)$ such that $f|_{\Omega}=\sigma$ . Clearly two cylinders $[0]$ and $[1]$ are $3\delta$ -separated for some $\delta>0$ . Assume the suspension function is also extended to $\tau:M\to[\lambda_0,\lambda_1]$ with $B([i],\delta)\subset\tau^{-1}(\lambda_i)$ . Then we see that the local stable manifold $W^s_\delta(x,f)$ gets embedded to the local stable manifold $W^s_\delta((x,t),f_{\bullet})$ , so is the local unstable manifolds . In particular $E^s\oplus E^u$ is locally integrable at $\Omega_\tau$ (although $\Omega_\tau$ may not be open). So integrability may be one of the obstructions for exponential decay. In fact uniform non-integrability condition is the key assumption of Dolgolpyat’s paper here for the exponential decay of SRB measures of Anosov flows with $C^1$ foliations.