1

We say $\Lambda \subset M$ is a hyperbolic set for $f:M \to M$ ($\Lambda$ is a closed invariant set) if there are constants $C \gneq 0$ , ($C \geq 1$) , $0 \lneq \lambda \lneq 1$ and two subspaces $E_p^s$ and $E_p^u$ of $T_pM$ for each $p \in \Lambda$ s.t:

1- $T_pM=E_p^s \oplus E_p^u$

2- $E_p^s$ and $E_p^u$ are $D_pf$-invariant i.e $D_pf(E_p^s)=E_{f(p)}^s$ , $D_pf(E_p^u)=E_{f(p)}^u$

3- $\forall$ $v \in E_p^s $ : $\| D_pf^n(v)\| \leq C \lambda^n\|v\| , n\geq0$ , $\forall$ $u \in E_p^u $ : $\| D_pf^{-n}(u)\| \leq C \lambda^n\|u\| , n\geq0$

If $\Lambda=M$ then we say $f$ is Anosov.

Now let $A=a_{(ij)} \in GL_n(\mathbb{Z})$ i.e $det(A) \in \{+1,-1\}$

Let $A = \left(\begin{matrix} 2 & 1\\1 & 1\end{matrix}\right)$, infact $A^{-1}$ has integer entrires.

Let $L_A:\mathbb{R}^n \to \mathbb{R}^n$ be the induced automorphism.

Then we have $L_A(\mathbb{Z}^n)= \mathbb{Z}^n$. Define a relation on $\mathbb{R}$ by $x \sim y$ iff $x-y \in \mathbb{Z}^n$, the quotion space $\frac{\mathbb{R}^n}{\sim}$ is the $n$-dimensional torus, $\mathbb{T}^n \equiv S^1 \times \cdots \times S^1$.

Let $\pi: \mathbb{R}^n \to \mathbb{T}^n$ be the projection, $ x \mapsto [x]:=\{y \in \mathbb{R}^n, x \sim y\}$. Then the map $f_A: \mathbb{T}^n \to \mathbb{T}^n$ induced by $L_A$ is a diffeomorphism, which $f_A([x])=(f_A \circ \pi)(x) = ( \pi \circ L_A)(x) = [Ax]$. Suppose $A$ has no eigenvalue of modulo 1 then $\mathbb{T}^n$ is hyperbolic for $f_A$; i.e $f_A$ is Anosov.

Now I want to solve an example in which we have $A = \left(\begin{matrix} 2 & 1\\1 & 1\end{matrix}\right)$, then $L_A: \mathbb{R}^2 \to \mathbb{R}^2$ ,$L_A(x,y)=(2x+y,x+y)$.

We have $sp(A)=\{\lambda_1=\frac{3-\sqrt5}{2},\lambda_2 =\frac{3+\sqrt5}{2} \}$,so $|\lambda_1 \neq 1|$ and $|\lambda_2 \neq 1|$. we have $Av_1=\lambda_1v_1$ so $v_1=(1, \frac{\sqrt 5 -1}{2})$ and $Av_2 = \lambda_2v_2$ so $v_2 = (1 , \frac{-\sqrt5-1}{2})$ w.r.t $\lambda_1$ and $\lambda_2$.

Put $E_0^s=<v_1>=\{tv_1, t\in \mathbb{R} \}$ and $E_0^u=<v_2>=\{tv_2, t\in \mathbb{R} \}$, then we have $T_0\mathbb{R}^2 = E_0^s \oplus E_0^u$. we have $\pi : \mathbb{R}^2 \to \mathbb{T}^2$ we may consider $\pi(0)=p$ by the translation above. then $ D_0\pi: T_0\mathbb{R}^2 \to T_p \mathbb{T}^2$. Put $D_0\pi(E_0^s)=E_p^s$, $D_0\pi(E_0^u)=E_p^u$.

I checked the number 1 and 2 in the definition of hyperbolic set for this example but I could not check the number 3 for this. Any help?

Alp Uzman
  • 10,742
Reza Yaghmaeian
  • 1,055

1 Answers1

1

A minor point: To make the toral automorphism compatible with the general framework for compact Riemannian $M$ and a $C^1$-diffeomorphism of it $f$ one would first need to establish what $\Vert \cdot \Vert$ stands for. There is a natural choice for it (due to translation invariance on $\mathbb{T}^2$), and indeed up to rescaling the natural choice is the only one: of course it is the Euclidean norm on $\mathbb{R}^2$. On that note, in the general definition there is a $\Vert\cdot\Vert'$ such that $C=1$ (this is traditionally attributed to Mather in dynamics literature, starting from Smale's survey "Differentiable Dynamical Systems").

Hint: Observe that the derivative of a linear map is itself (at any point $p$). Also the splitting $T_p\mathbb{T}^2=E^s_p(f_A)\oplus E^u_p(f_A)$ for $f_A$ corresponds to the eigenspace decomposition $\mathbb{R}^2=\langle v_1\rangle \oplus \langle v_2\rangle$ of $L_A$.

Alp Uzman
  • 10,742