Fatou-Bieberbach domain in $\Bbb C^*\times\Bbb C^*$ According to Forstneric's book, pag 123, it is a long standing problem whether it exists a Fatou-Bieberbach domain in $\Bbb C^*\times\Bbb C^*$.
The idea is to search for an $F:\Bbb C^2\to\Bbb C^2$ biholomorphic fixing coordinate axis (calling $(x,y)$ and $(x',y')$ the coordinate in the domain/codomain resp. we have $F(\{x=0\})\subseteq\{x'=0\}$ and $F(\{y=0\})\subseteq\{y'=0\}$) having $(1,1)$ as an attractive fixed point.
In this case, the basin of attraction of such $F$ would do the job.
Now, if $F$ is a polynomial automorphism then the jacobian has to be a constant $c\neq0$, and it follows from this paper by Nishimura, that is this case $F$ needs to be of the following form:
$$
x'=cxe^{-\alpha(xy)}\\
y'=ye^{\alpha(xy)}\;\;\;
$$
for some $\alpha\in\mathcal O(\Bbb C)$. Being $F$ polynomial, $\alpha$ needs to be constant and wlog we take it $\alpha\equiv0$, from which $F(x,y)=(cx,y)$; since $(1,1)$ has to be fixed point, it follows that $c=1$ hence $F=\operatorname{id}$ so $(1,1)$ cannot be attractive.
Does exist any development in this direction? Any paper talking about it suggesting some possible way to attack it?
 A: This is a difficult problem and I don't think that we are getting any closer to its solution. I think there is no hope of finding an explicit automorphism that would map both axis to themselves and which would have an attracting fixed point somewhere outside the axis.
There are other ways people have tried to tackle this problem, for example by trying to find an automorphism with a parabolic, semiparabolic, quasiparabolic or resonant fixed point at the origin whose attracting domain (fatou component) avoids both axis. Now most of these domains are known to be biholomorphic to $\mathbb C^2$,
but in the resonant case you could also have   $\mathbb C \times \mathbb C^*$ [Bracci, F., J. Raissy, and B. Stensønes. “Automorphisms of $\mathbb C^k$ with an invariant non-recurrentattracting Fatou component biholomorphic to  $\mathbb C \times (\mathbb C^*)^{k-1}$.”J. Eur. Math. Soc.(2020)]. If I recall correctly in their case the domain almost avoided both axis. Note that if you manage to find a domain that is biholomorphic to  $\mathbb C \times \mathbb C^*$ and which avoids both axis then you have solved the problem. This follows from the fact that $\mathbb C \times \mathbb C^*$ has the density property therefore if in there you can find a copy of  $\mathbb C^2$.   Finally another keyword for you would be a parabolic cylinder.
