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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.

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Cat map is 21/11? If not, then I don't see any reason for the measure to be related to the golden mean. If yes, then there're only two rectangles and you can compute their measures by hand. –  Andrey Gogolev Sep 27 '10 at 22:33
    
@Andrey: I think the point is that one could consider other Markov partitions for the same system (including, but not limited to, refinements of the standard one), and then it makes sense to ask if the measures of the rectangles for these other partitions are also of the form $\phi^n$. –  Vaughn Climenhaga Sep 28 '10 at 0:09
    
Oh I see, anyway kind of surprising. Is it really so you think? –  Andrey Gogolev Sep 28 '10 at 0:22
    
I'm not entirely sure if it's true. It's true for the three examples I've considered, along with various types of refinements (not just the obvious ones). I'm pretty sure however that a refinement always exists for which it's true, though still no idea how to prove this. –  Steve Huntsman Sep 28 '10 at 2:47

1 Answer 1

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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.

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Thanks, this looks promising. –  Steve Huntsman Sep 27 '10 at 17:47
    
...but upon reflection, also difficult. –  Steve Huntsman Sep 27 '10 at 18:18
    
You need to impose extra assumptions on the matrix, or your linear algebra problem will not be equivalent to the original statement. For example, you can always complement a matrix by block which is any square matrix whatsoever, with its own associated eigenvalues. Thus you can always add extra eigenvalues that are (real or complex) Perron numbers that are less than $\phi.$ Irreducibility would take care of this particular issue, but there may be others. –  Victor Protsak Sep 28 '10 at 2:15
    
@Victor: Good point. I suppose then one may need to consider the same problem for only those matrices that can arise as transition matrices for a Markov partition of the system in question, which is a much more restrictive class. However, it also seems to not be particularly well understood... –  Vaughn Climenhaga Sep 29 '10 at 18:10
    
One place to start would be to see if this property is preserved under state splittings, which correspond to elementary conjugacies between the subshifts that model the system. –  Vaughn Climenhaga Sep 29 '10 at 18:11

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