$x'=(1+A) x + A/\alpha (B+K-\alpha (x+L))+M$ with $M,N$ integers. Rearranging this we have $$y'= B+(K+M+N)-\alpha(x+L)$$ and so the family of lines $L_B$ is invariant under the map for any given $B$. We define the maximal invariant set within $L_B$ as $$X_B^+= X^+ \cap L_B.$$

Semirational and rational parabolic maps {#secrational}
----------------------------------------

For semirational maps ($\alpha=r/s>0$) and for any given value of $B$ the set $L_B$ consists of $r+s$ (or exceptionally $r+s-1$ if it contains the origin) intervals that are parallel to the eigenvector $v=(1,-\alpha)$ of the matrix $M$. We can parametrise any particular $L_B$ by $\theta\in[0,s)$. We will consider $\alpha>0$ from here on. More precisely we re-parametrize $X$ by $$\begin{array}{l}
x = \theta - [\theta]\\
y = B- \alpha\theta -[B-\alpha\theta]
\end{array}$$ For $\theta\in[0,s)$ and $B\in [\frac{s-1}{s},1)$ the mapping $(x,y) \leftrightarrow (\theta,B)$ is one-to-one and has unit Jacobian everywhere.

Since each $L_B$ is invariant the maximal invariant set must nontrivially intersect $L_B$ and so no single trajectory is dense in $X^+$. In this case we need to approximate $X^+$ using a distribution of trajectories on a dense set of lines.

The relation between $ \theta  $ and $ x $ is $ x=\theta -[\theta ]. $ The variable $ y $ 