is related to $ \theta  $ and $ B
$ by the relation $ y=B-\alpha \theta -[B-\alpha \theta ] $. So the point $(\theta ,B)$ maps to $$(\theta',B')=
(\theta +AB/\alpha -A[\theta ]-A/\alpha [B-\alpha \theta ],B)$$ since $$\begin{aligned}
{\left(\begin{array}{c} {\theta'} \\ {v} 
\end{array}\right)} &=&
\left( \begin{array}{cc}
1+A & A/\alpha \\
-A\alpha  & 1-A
\end{array}\right) \left( \begin{array}{c}
\theta -[\theta ]\\
B-\alpha \theta -[B-\alpha \theta ]
\end{array}\right) +\left( \begin{array}{c}
[\theta ]\\
\left[ B-\alpha \theta \right] 
\end{array}\right) \\
&=& \left( \begin{array}{c}
\theta +AB/\alpha -A[\theta ]-A/\alpha [B-\alpha \theta ]\\
-\alpha \theta +B-AB+A\alpha [\theta ]+A[B-\alpha \theta ]
\end{array}\right). \end{aligned}$$ We have added integers $ [\theta ] $ and $ [B-\alpha \theta ] $ so that $\theta'\in[0,s)$ and $v=B-\alpha\theta$ (see eg [@Zyc98]).

Thus the action of the map (\[eqmap\]) on the coordinate $\theta$ is simply $$\theta'=T_B(\theta)=\left(\theta+\frac{s}{r}AB-A[\theta]-A\frac{s}{r}
[B-\frac{r}{s}\theta] \right) ~ (\bmod s).
\label{map1d}$$ Because the variable $B\in [\frac{1-s}{s},1)$ is invariant during the dynamics it is treated as a parameter. Again, the slope of these 1D maps is equal to one implying that all Liapunov exponents are zero.

In the case of a rational map, ie where both $A=p/q$ and $\alpha=r/s$ are rational with $(r,s)=(p,q)=1$ we can understand a lot about the mapping $T$ using the following factor map. 