# Topological entropy of non-hyperbolic toral automorphism

I have a matrix $$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$ and its associated toral automorphism $$f_A$$. That is;

$$\begin{equation} {f}^{n}_A := ( x + ny \text{ mod } 1 , y ) \end{equation}$$

I am trying to show that its topological entropy, $$h_{\text{top}}(f_A) = 0$$. My thought process would be to construct its $$(n,\epsilon)$$ - spanning sets and then show that their exponential growth rate is 0, and thus the topological entropy is equal to 0. However I am struggling to construct them, so any tips would be appreciated.

I cannot use that if an toral automorphism has a repeated eigenvalue of 1 then $$h_{\text{top}}(f_A) = \log(1) = 0$$.

• I don't think your description of the iterates is correct: $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x+y \\ y \end{pmatrix}$, so $f^n(x,y) = (x + ny \mod 1, y)$
Since $$f$$ is a continuous map on the compact metric space $$\mathbb{T}^2$$, one form of the variational principle says that $$h_{top}(f) = \sup_{\mu \in \mathscr{M}^f_e(\mathbb{T}^2)} h_\mu(f)$$ where $$\mathscr{M}^f_e(\mathbb{T}^2)$$ is the set of all ergodic $$f$$-invariant probability measures on $$\mathbb{T}^2$$. So it is enough to show that $$h_\mu(f) = 0$$ for any ergodic $$f$$-invariant measure $$\mu$$.
A key observation is the following. If $$L_y = \{(x,y): x \in \mathbb{T}\}$$ is the horizontal line in $$\mathbb{T}^2$$ at height $$y$$, then each $$L_y$$ is an $$f$$-invariant set. Therefore if $$\mu$$ is any ergodic measure, there is a unique $$y$$ such that $$\mu$$ is supported on $$L_y$$. Now when restricted to the line $$L_y$$, the action of $$f$$ is clearly isomorphic to the circle rotation $$x \mapsto x+y$$ on $$\mathbb{T}$$. Since $$\mu$$ is supported on a line, it is naturally identified with a measure on $$\mathbb{T}$$, and so our system $$(\mathbb{T}^2, f, \mu)$$ is naturally isomorphic to the circle rotation $$(\mathbb{T}, x \mapsto x+y, \mu)$$.
It's well known that 1-dimensional circle rotations have zero topological entropy, so in particular any invariant measure for a circle rotation has zero measure entropy as well. Therefore $$h_\mu(f) = 0$$. Since this holds for any ergodic $$f$$-invariant measure $$\mu$$ we are done.