Entropy of a Linear Toral Automorphism I'm trying to calculate the entropy of the Linear Toral Automorphism
induced by
$$f(x,y,z)=(x,y+x,y+z)$$
This is an exercise in the Katok book.  
This map has all eigenvalues ​​equal to 1. But I do not want to use that $~~ h_{top}(f)= log (max|\lambda_i|)$. would like to use Katok's suggestion that says that the cardinality separate sets grow quadratically with $ n $  where $ n $ is the size of the orbit. But I can not see it clearly.
 A: We have
$$f^n(x,y,z)=(x,\,y+nx,\,z+ny+\tbinom n 2 x)$$
Taking $\|\cdot\|_\infty$ as a metric on $(\mathbb R/\mathbb Z)^3$, this implies
$$d_n(a,b) \le (1+n+n(n-1)/2) \|a-b\|_\infty$$
where $d_n$ is the maximum distance between the two orbits $(a,f(a),\dots,f^n(a))$.
So that an $(n,\varepsilon)$-separated set must be $(0,\Omega(\varepsilon/n^2))$-separated (in other words, the metric $d_n$ grows at most quadratically) and therefore, since we are in dimension $d=3$, for fixed $\varepsilon$ its cardinality grows as $O(n^{2d})$, which suffices to conclude that $h_{top}(f)=0$.

Note that the cardinality itself does grow faster than quadratically, as can be seen with the following $n^2(n-1)/2$ points $M_{uv}$:
$$\left\{\begin{aligned}
x=&u/\tbinom n 2\\
y=&v/n\\
z=&0
\end{aligned}\right.$$
If $d_n(M_{uv},M_{u'v'})<\varepsilon<1/4$ for some odd $n$, then we have
$$|n(y'-y)+\tbinom n 2 (x'-x)|<\varepsilon\\
|v'-v+u'-u|<1\\
v'-v = -(u'-u)$$
$$|(n+1)/2\cdot(y'-y)+\tbinom{(n+1)/2}{2}(x'-x)|<\varepsilon\\
|u'-u|<\frac{n}{(n+1)/4}\varepsilon<1\\
u=u'\\
v=v'$$
So that we have a set of $\Omega(n^3)$ points that is $(n,\varepsilon)$-separated.
