Measures of rectangles in Markov partitions for the cat map Let $\mathcal{R}$ be a Markov partition for the cat map. (How) can it be shown that the Lebesgue measure of a rectangle $R_j \in \mathcal{R}$ satisfies $\mu(R_j) = \phi^{-n}$ for some $n$, where $\phi = \frac{1+\sqrt{5}}{2}$? 
A "physicist's proof" would be based on the Ansatz that $\mathcal{R}$ can be constructed extending local stable and unstable manifolds around the origin à la Gallavotti, but it's not clear to me how to build a rigorous argument along these lines.
Any references to work informing an answer would be particularly appreciated. Best of all would be a pointer accounting for the relative measures and multiplicities of all rectangles.
 A: Lebesgue measure is both the unique SRB measure and the unique measure of maximal entropy for the cat map.  Any Markov partition into $p$ rectangles gives a topological (semi-)conjugacy between the cat map and a subshift of finite type on $p$ symbols with some 0-1 transition matrix $A$.  This conjugacy carries the measure of maximal entropy for the cat map (Lebesgue measure) into the measure of maximal entropy for the subshift, which is a Markov measure (namely, the Parry measure).  In particular, the measures of the rectangles in the partition are the entries in the probability vector that defines the Parry measure.  But these are given very explicitly in terms of the transition matrix $A$:  if $u$ and $v$ are the left and right eigenvectors for $A$ corresponding to the maximal eigenvalue (which is $e^{h_\mathrm{top}(f)} = \phi$), then the entries in the probability vector are proportional to $u_i v_i$ (one needs to normalise so that they sum to 1).
At this point it becomes linear algebra; given a 0-1 matrix whose maximal eigenvalue is $\phi$, show that the corresponding eigenvectors $u$ and $v$ have the property that $u_i v_i / (\sum_j u_j v_j)$ is of the form $\phi^n$.  I'm not sure how difficult this is, but that's the direction I'd attack the problem from.
