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Let $$A = \begin{bmatrix} 1 & 1 \\ 1 & 0\\ \end{bmatrix}.$$ Then the eigenvalues of $A$ are $1/2(1+\sqrt{5})$ and $1/2(1-\sqrt{5})$. The eigenvector corresponding to the unstable eigenvalue is the line $y = 1/2(-1+\sqrt 5)x$, whereas the stable eigenvalue has eigenvector $y = -1/2 (1+\sqrt 5)x$.

Let $L$ be a hyperbolic linear automorphism of the torus induced by $A$.

Could someone explain how we get the following graphs? To simplify the question, let's jsut discuss how $L$ affects $R_1$. The upper left picture is a Markov partition of the torus. The dark regions represent $R_1$, $L(R_1)$ and $L^{-1}(R_1)$. enter image description here

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The closely spaced line segments in $R_1$ get spaced out more widely because $L$ expands distances along the unstable eigenline. These line segments get shorter because they are parallel to the stable eigenline, along which $L$ contracts distances. Finally, $R_1$ turns into three pieces rather than two because it gets wrapped around the torus. Specifically, the small, top-left piece of $R_1$ turns (mostly) into the small, bottom-right piece of $L(R_1)$, while the big strip of $R_1$ turns into the middle-right piece of $L(R_1)$ plus most of the long strip of $L(R_1)$.

By computing the coordinates of the corners of the various rectangles and doing the linear algebra explicitly, you can make this qualitative explanation quantitative. I'm happy to add a sample calculation to this answer if that would be helpful.

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  • $\begingroup$ Thank you, it is a bit clearer to me now. It would be great if you can add a sample calculation because I find it difficult to see directly how after stretching and contracting we will get those divided regions of $L(R_1)$. $\endgroup$ – user398843 May 14 '20 at 19:11
  • $\begingroup$ Sophie, could you take the action of $L$ on a line segment as an example to explain how to do the calculation? For example, the line segment of $y=0$ in the unit square. $\endgroup$ – user398843 May 19 '20 at 8:43

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