# cat map re-transformation

Hi,

Is there any way of moving from one cat map transformation to the other without resetting parameters? For example, suppose you have two matrices '$A$'and '$B$' each permuted with different cat map parameters namely '$p_1$','$q_1$' and '$p_2$','$q_2$'. The goal is by having the permuted matrix of '$A$' under '$p_1$', '$q_1$' which we call '$A_p$' and parameters of '$B$' in this case '$p_2$' and '$q_2$', we get a matrix called '$A_bp$' which is equal to permuting the original '$A$' with '$p_2$' and '$q_2$'? obviously, we are not allowed to re-permute '$A_p$' to get '$A$' and apply '$p_2$', and '$q_2$' on '$A$'.

Thx

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Can you clarify what your definitions are? By "cat map" I assume you mean "toral automorphism", that is a map $\mathbb{R}^d/\mathbb{Z}^d$ given by a matrix $A\in SL(d,\mathbb{Z})$, and probably you want $A$ to have no eigenvalues on the unit circle. But I don't know what you mean by "cat map parameters" or "permuted matrix", so right now I don't understand your question. –  Vaughn Climenhaga Oct 10 '12 at 21:12
@Vaughn: Thanks for your comments and sorry for the confusion. Yes I meant "toral automorphism" with the application of something like Arnold Cat Map where we have a map $\begin{bmatrix} x_n\\y_n \end{bmatrix} =\begin{bmatrix} 1&p\\ q&1+pq \end{bmatrix} \begin{bmatrix} x_{n-1}\\y_{n-1} \end{bmatrix} mod~ n$. Since $det\begin{pmatrix} \begin{bmatrix} 1 & p\\ q & pq+1 \end{bmatrix} \end{pmatrix}=1$ then the map is area preserving. So we permute $A$ and $B$ using the map with two different parameter $p_1$,$q_1$ and $p_2$ and "q_2" which gives us $A_p$ and $B_p$. –  Shanti Oct 11 '12 at 3:02
Then, we want to use $B_p$ and $p_2$, $q_2$ ,$p_1$, and $q_1$ to permute $B_p$ further such that the result is like permuting $B$ with $p_1$ and $q_1$ using the above mapping. –  Shanti Oct 11 '12 at 3:05
Still don't understand the question: what do you mean by "moving from one transformation to another"? –  Anthony Quas Oct 11 '12 at 10:57
@Quas: I mean instead of permuting $B_p$ back to $B$ with its parameters and then permuting the resulted $B$ with $A$'s parameter to get the same permuted matrix as $A_p$, directly go from $B_p$ to $A_p$ without doing back and forth. –  Shanti Oct 11 '12 at 14:45