Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
Hint: start by computing eigenvalues of $Df$. The largest eigenvalue will tell you about expansion rate along that tangent direction. What does this tell you about entropy? –  William Jun 22 '12 at 4:04
1  
@William: This map has all eigenvalues ​​equal to 1. But I do not want to use that $~~ h_{top}(f)= log (max|\lambda_i|)$. –  user27456 Jun 22 '12 at 16:40
    
Could anybody tell me what branch of math is this? And what is that entropy? Wikipedia article or something similar will suffice. –  Yrogirg Jun 25 '12 at 6:13
2  
@ Yrogirg: Look at this page, maths.bristol.ac.uk/~maxcu/DynSysErgTh.html . For something more specific look at the reading titled topological entropy –  user27456 Jun 25 '12 at 15:57

1 Answer 1

up vote 2 down vote accepted
+100

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.