Exponential mixing for subshifts I asked this question on Math.StackExchange some time ago and got no responses.
Let $G=(V,E)$ be a finite graph with adjacency matrix $A$. Let us consider the associated subshift of finite type
$$
\Sigma_A=\{(v_i)_{i\in\mathbb{Z}} : A_{v_iv_{i+1}}>0, i\in\mathbb{Z}\}\subset V^{\mathbb{Z}}.
$$
For a stochastic matrix $P$ agreed with $A$ (that is $p_{ij}>0$ iff $a_{ij}>0$), there is a shift-invariant probability measure $\mu$ on $\Sigma_A$. Is is well-known that the dynamical system $(\Sigma_A,\sigma,\mu)$ is strongly mixing, i.e., for all measurable $B,C\subset\Sigma_A$,
$$
\mu(\sigma^nB\cap C)\rightarrow\mu(B)\mu(C), \ n\rightarrow\infty,
$$
if and only if the graph $G$ is strongly connected and aperiodic.
I am wondering if it is possible to estimate the speed of convergence. The standard proof given in textbooks uses iterations of $P$ to prove mixing for cylindrical sets. Hence one can estimate the speed of convergence for cylindrical sets in terms of the eigenvalues of $P$. Does it hold for any measurable sets? Is it always exponential, where the exponent comes from the eigenvalues of $P$?
 A: I don't think that it's ever possible to mandate any rate of mixing if you are looking at all measurable sets. Here's a silly example using the full $2$-shift $\{0,1\}^{\mathbb{Z}}$ and the i.i.d. measure $\mu$ with $\mu([0]) = \mu([1]) = 1/2$ (in some sense the "most mixing" SFT and measure).
Define $B$ to be $[0]$, the set of sequences with $x(1) = 0$, i.e. with $0$ in the first coordinate. Clearly $\mu(B) = 0.5$.
Define $C$ to be the set of sequences $x$ satisfying all of the following conditions: $x(1)$ is not $0$, $x(2)$ and $x(3)$ are not both $0$, $x(4), x(5), x(6)$ are not all $0$, $x(7), x(8), x(9), x(10)$ are not all $0$, etc. Then $\mu(C) = \prod_{n=1}^{\infty} (1 - 2^{-n}) > 0$.
But for any $k$, $\mu(\sigma^k B \ | \ C)$ is slightly less than $\mu(B) = 0.5$, in fact it's $0.5 - \frac{1}{2^{n+1} - 2}$, where $n$ is the length of the "condition" that $x(k)$ is part of. (For instance, when $k = 8$, $n = 4$, since $x(8)$ is part of the condition "$x(7), x(8), x(9), x(10)$ are not all $0$."
This already has subexponential mixing rate, but you can make the rate as slow as desired by just spacing your conditions out. For instance, you could wait until $x(n!!)$ to start the $n$th condition, and then $\mu(\sigma^{n!!} B \ | \ C)$ is still roughly $2^{-n}$ away from $\mu(B) = 0.5$.
