# Action of a toral automorphism on a Markov partition

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 $$L^{-1}(R_1)$$.

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 $$L(R_1)$$.
• 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 $L(R_1)$. – user398843 May 14 '20 at 19:11
• 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. – user398843 May 19 '20 at 8:43