We know that there exists a polynomial the Fatou set $F(P)$ is connected, which is just an attracting basin for infinity. 


I have a question: Given a rational function $R$ 
 such that $F(R)$ is connected, is this always true that $F(R)$ is just a attracting basin? 


If the question is not, I wondered whether there exists a rational function $R$, such that $F(R)$ is connected and $F(R)$ is a completely invariant parabolic petal. 


Further question: Is it possible to classify all rational maps $R$ with exactly one one Fatou component. 


Any comments and remarks will be appreciated.



classification of rational map with exactly only one Fatou component
Clearly if $F(R)$ is connected, then $F(R)$ consists of a single attracting or parabolic basin. 


Both cases can occur. Indeed, if you consider the slice $\operatorname{Per}\_1(1)$ of quadratic rational maps having a fixed point with multiplier one, then this slice has an analog of the Mandelbrot set inside it, which is homemorphic to the Mandelbrot set, with corresponding maps being topologically conjugate on their Julia sets, by work of Petersen and Roesch. ("The parabolic Mandelbrot set", see the slides at <http://www.math.univ-toulouse.fr/~roesch/Banff.pdf> .) So there are many maps of this type, having a completely invariant parabolic basin. It won't be hard work to find an explicit example. 


In general, it might be plausible that any example is topologically conjugate to a polynomial (on the corresponding Julia sets). But I am not aware that this has been proven. 


